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17601 | https://www.nature.com/scitable/topicpage/the-information-in-dna-determines-cellular-function-6523228/ | This page has been archived and is no longer updated
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The Information in DNA Determines Cellular Function via Translation
The ribosome assembles the polypeptide chain
To manufacture protein molecules, a cell must first transfer information from DNA to mRNA through the process of transcription. Then, a process called translation uses this mRNA as a template for protein assembly. In fact, this flow of information from DNA to RNA and finally to protein is considered the central dogma of genetics, and it is the starting point for understanding the function of the genetic information in DNA.
But just how does translation work? In other words, how does the cell read and interpret the information that is stored in DNA and carried in mRNA? The answer to this question lies in a series of complex mechanisms, most of which are associated with the cellular structure known as the ribosome. In order to understand these mechanisms, however, it's first necessary to take a closer look at the concept known as the genetic code.
What is the genetic code?
More on translation
How did scientists discover how ribosomes work?
What are ribosomes made of?
Is prokaryotic translation different from eukaryotic translation?
At its heart, the genetic code is the set of "rules" that a cell uses to interpret the nucleotide sequence within a molecule of mRNA. This sequence is broken into a series of three-nucleotide units known as codons (Figure 1). The three-letter nature of codons means that the four nucleotides found in mRNA — A, U, G, and C — can produce a total of 64 different combinations. Of these 64 codons, 61 represent amino acids, and the remaining three represent stop signals, which trigger the end of protein synthesis. Because there are only 20 different amino acids but 64 possible codons, most amino acids are indicated by more than one codon. (Note, however, that each codon represents only one amino acid or stop codon.) This phenomenon is known as redundancy or degeneracy, and it is important to the genetic code because it minimizes the harmful effects that incorrectly placed nucleotides can have on protein synthesis. Yet another factor that helps mitigate these potentially damaging effects is the fact that there is no overlap in the genetic code. This means that the three nucleotides within a particular codon are a part of that codon only — thus, they are not included in either of the adjacent codons.
Figure 1: In mRNA, three-nucleotide units called codons dictate a particular amino acid. For example, AUG codes for the amino acid methionine (beige).
Figure Detail
The idea of codons was first proposed by Francis Crick and his colleagues in 1961. During that same year, Marshall Nirenberg and Heinrich Matthaei began deciphering the genetic code, and they determined that the codon UUU specifically represented the amino acid phenylalanine. Following this discovery, Nirenberg, Philip Leder, and Har Gobind Khorana eventually identified the rest of the genetic code and fully described which codons corresponded to which amino acids.
Reading the genetic code
Redundancy in the genetic code means that most amino acids are specified by more than one mRNA codon. For example, the amino acid phenylalanine (Phe) is specified by the codons UUU and UUC, and the amino acid leucine (Leu) is specified by the codons CUU, CUC, CUA, and CUG. Methionine is specified by the codon AUG, which is also known as the start codon. Consequently, methionine is the first amino acid to dock in the ribosome during the synthesis of proteins. Tryptophan is unique because it is the only amino acid specified by a single codon. The remaining 19 amino acids are specified by between two and six codons each. The codons UAA, UAG, and UGA are the stop codons that signal the termination of translation. Figure 2 shows the 64 codon combinations and the amino acids or stop signals they specify.
Figure 2: The amino acids specified by each mRNA codon. Multiple codons can code for the same amino acid.
Figure Detail
What role do ribosomes play in translation?
As previously mentioned, ribosomes are the specialized cellular structures in which translation takes place. This means that ribosomes are the sites at which the genetic code is actually read by a cell. Ribosomes are themselves composed of a complex of proteins and specialized RNA molecules called ribosomal RNA (rRNA).
Figure 3: A tRNA molecule combines an anticodon sequence with an amino acid.
Figure Detail
During translation, ribosomes move along an mRNA strand, and with the help of proteins called initiation factors, elongation factors, and release factors, they assemble the sequence of amino acids indicated by the mRNA, thereby forming a protein. In order for this assembly to occur, however, the ribosomes must be surrounded by small but critical molecules called transfer RNA (tRNA). Each tRNA molecule consists of two distinct ends, one of which binds to a specific amino acid, and the other which binds to a specific codon in the mRNA sequence because it carries a series of nucleotides called an anticodon (Figure 3). In this way, tRNA functions as an adapter between the genetic message and the protein product. (The exact role of tRNA is explained in more depth in the following sections.)
What are the steps in translation?
Like transcription, translation can also be broken into three distinct phases: initiation, elongation, and termination. All three phases of translation involve the ribosome, which directs the translation process. Multiple ribosomes can translate a single mRNA molecule at the same time, but all of these ribosomes must begin at the first codon and move along the mRNA strand one codon at a time until reaching the stop codon. This group of ribosomes, also known as a polysome, allows for the simultaneous production of multiple strings of amino acids, called polypeptides, from one mRNA. When released, these polypeptides may be complete or, as is often the case, they may require further processing to become mature proteins.
Initiation
Figure 4: During initiation, the ribosome (grey globe) docks onto the mRNA at a position near the start codon (red).
Figure Detail
At the start of the initiation phase of translation, the ribosome attaches to the mRNA strand and finds the beginning of the genetic message, called the start codon (Figure 4). This codon is almost always AUG, which corresponds to the amino acid methionine. Next, the specific tRNA molecule that carries methionine recognizes this codon and binds to it (Figure 5). At this point, the initiation phase of translation is complete.
Figure 5: To complete the initiation phase, the tRNA molecule that carries methionine recognizes the start codon and binds to it.
Figure Detail
Elongation
Figure 6: Within the ribosome, multiple tRNA molecules bind to the mRNA strand in the appropriate sequence.
Figure Detail
Figure 7: Each successive tRNA leaves behind an amino acid that links in sequence. The resulting chain of amino acids emerges from the top of the ribosome.
Figure Detail
The next step in translation, called elongation, begins when the ribosome shifts to the next codon on the mRNA. At this point, the corresponding tRNA binds to this codon and, for a short time, there are two tRNA molecules on the mRNA strand. The amino acids carried by these tRNA molecules are then bound together. After this binding has occurred, the ribosome shifts again, and the first tRNA, which is no longer connected to its corresponding amino acid, is released (Figure 6). Now, the third codon in the mRNA strand is ready to bind with the appropriate tRNA (Figure 7). Once again, the tRNA binds to the mRNA strand, the third amino acid is added to the series, the ribosome shifts, and the second tRNA (which no longer carries an amino acid) is released. This process is repeated along the entire length of the mRNA, thereby elongating the polypeptide chain that is emerging from the top of the ribosome (Figure 8).
Figure 8: The polypeptide elongates as the process of tRNA docking and amino acid attachment is repeated.
Figure Detail
Termination
Eventually, after elongation has proceeded for some time, the ribosome comes to a stop codon, which signals the end of the genetic message. As a result, the ribosome detaches from the mRNA and releases the amino acid chain. This marks the final phase of translation, which is called termination(Figure 9).
Figure 9: The translation process terminates after a stop codon signals the ribosome to fall off the RNA.
Figure Detail
What happens after translation?
For many proteins, translation is only the first step in their life cycle. Moderate to extensive post-translational modification is sometimes required before a protein is complete. For example, some polypeptide chains require the addition of other molecules before they are considered "finished" proteins. Still other polypeptides must have specific sections removed through a process called proteolysis. Often, this involves the excision of the first amino acid in the chain (usually methionine, as this is the particular amino acid indicated by the start codon).
Once a protein is complete, it has a job to perform. Some proteins are enzymes that catalyze biochemical reactions. Other proteins play roles in DNA replication and transcription. Yet other proteins provide structural support for the cell, create channels through the cell membrane, or carry out one of many other important cellular support functions.
Watch this video for a summary of translation in eukaryotes
Further Exploration
Key Questions
What other functions does RNA have in the cell?
What happens to proteins after they are translated?
Who discovered the relationship between DNA and proteins?
Key Concepts
mRNA | transcription | ribosome
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17602 | https://www.youtube.com/watch?v=-LHN8v2zJ0I | Pythagorean Theorem: Irrational Side length
Euler's Academy
9400 subscribers
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Description
850 views
Posted: 18 Sep 2022
This video continues with the idea of using the Pythagorean Theorem by looking at a specific example from the Khan Academy exercise linked below that deals with an irrational side length in a right triangle.
The Pythagorean Theorem shows the relationship between the side lengths of a right triangle.
A theorem is a statement in mathematics that has been proven true. This particular theorem is named after the Greek mathematician and philosopher Pythagoras who lived around 2500 years ago. It states that the sum two legs in a right triangle when squared are equal to the hypotenuse of the triangle squared (a^2 + b^2 = c^2).
This theorem is useful for finding missing side lengths in a right triangle.
Khan Academy exercise to practice:
1. Use Pythagorean Theorem to Find Right Triangle Side Lengths:
EulersAcademy.org
4 comments
Transcript:
so let's look at another example problem of using the pythagorean theorem and remember the pythagorean theorem relates the side lengths in a right triangle so the triangle needs to have a right angle a 90 degrees angle and this theorem which can be written as a squared plus b squared is c squared where if you had some random right triangle the two legs of the triangle that are next to the right angle are called a and b and the long side of the triangle is called the hypotenuse so this theorem this pythagorean theorem named after the greek mathematician pythagoras will help us figure out a missing side length and a right triangle and this particular example will be a little bit different than the intro video examples since we're going to deal with an irrational number here so this will not be a whole number and that adds a little bit of difficulty to the problem but overall it's pretty much the same process so to deal with these problems we really just need to label a b and c and then plug them into the formula and then solve for whatever our missing side length is and when dealing with the legs of the triangle it doesn't matter what we call a and what we call b so let's just call this a and we can call this one b though if we want to switch it that's fine and the important thing to remember is that the side opposite the right angle that is what we have to call c since that's the hypotenuse of our right triangle so let's just plug everything into the formula now so we have that a squared so that's 4 squared plus b squared so 5 squared is equal to c squared and we called x c so now just simplifying 4 times 4 is 16 5 squared that's 5 times 5 that is 25 and this is equal to x squared and we can add these together so if we add 10 to 25 that brings us to 35 adding 6 more goes to 41. now remember at this point we need to essentially cancel out this square and the way to do that is to take a square root of each side of the equation since squaring and taking square roots those are inverse operations they cancel each other out so on the right hand side the square root and the square cancel and we just get x and usually when we do this we would have to put plus or minus the square root of 41 but we don't care about the negative answer since this is a physical side length it has positive value so we just have the square root of 41 and we can ask ourselves what number when multiplied by itself gives us 41 and that number is not a whole number since we can compare this to perfect squares so if we look at the square root of 36 and let's say we look at the square root of 49 the square root of 36 is just 6 and the square root of 49 is 7. so the square root of 41 is somewhere in the middle of this in fact it's going to be an irrational number and remember that an irrational number is a number that cannot be written as a fraction so rational number notice it has the word ratio in it those are fractional numbers and irrational just means it is not a fraction number so we can't write this as a fraction and in fact these numbers have infinitely long decimal expansions so if you put this into your calculator you get 6.403 then it goes 1 2 4 and it's going to go on forever it has an infinitely long expansion that does not have a pattern now your calculator can't go to infinity so at some point it's going to round the number so even though it looks like it ends it actually does not so these numbers are a little bit more complicated but they are valid numbers and this would just be our final answer we would want to leave it as the square root of 41. since if we write down a decimal answer we would need to be careful about that since this is only approximately equal to about 6.403 since at some point you would have to round that so when you write down a decimal just be careful since that's no longer exactly the right answer that is only fairly close to the right answer |
17603 | https://geographicalimaginations.com/wp-content/uploads/2012/07/gregory-and-castree-human-geography-introduction.pdf | 1 HUMAN GEOGRAPHY: AN INTRODUCTION † Derek Gregory and Noel Castree Introduction When we were invited by Sage to identify published work in human geography that represents what is best and most distinctive about the field it seemed an impossible task (it still does) because there is such a rich volume of material to draw from. We decided to focus on English-language and to a lesser extent other European contributions, although we are acutely aware of the irony, even the imperialism of limiting a field like human geography to knowledges rooted in only a fraction of the world. We discuss below the dangers of delimiting Geography as a European or Euro-American science, and several of our selections return to this issue again and again. If there is a much richer geography of Geography than this, there is also a much longer history than our selections might imply. Our focus on the last thirty years is not an exercise in progressivism or triumphalism which treats the present as the climactic moment in a chain of contributions that reaches back into an ever more distant and ever more imperfect past. Here too our decision was a purely pragmatic way to confine our search. Even within these geographical and historical limits it was difficult to make a judicious selection – and we know how many contributions we had to excise at the eleventh hour for fear of turning five volumes into fifty – † References to selections that appear in these five volumes are shown in bold. 2 because human geography, perhaps more than most disciplines, is so heterodox. Its practitioners set their intellectual compasses according to no one philosophy; no dominant theoretical framework overshadows all others; methodological pluralism is the order of the day; topical diversity is the norm, as is heterogeneity in the research questions asked and data generated; putatively ‘objective’ approaches rub shoulders with avowedly ‘political’ ones; the rigorously cerebral and insistently practical jostle for attention; and human geographers draw intellectual inspiration from every conceivable field, with some favouring the arts and humanities, others the wider social sciences, and still others the environmental and life sciences. This may sound like a discipline in crisis – indeed, something scarcely worthy of the title ‘discipline’. And yet the selections that we include in these five volumes show that – even within our narrow pre-determined limits – the field of human geography is remarkably fecund: it would undoubtedly seem even richer if we extended those geographical and historical horizons. In large part this positive judgement rests on a rethinking of what we mean by a discipline. An older meaning – inculcation into a canonical body of knowledge, a sort of academic holy writ, upon which one slowly builds to become a ‘disciple’ – has given way to a newer meaning: exposure to a variety of knowledges that share a family resemblance and which provide the means for critical, creative inventions not mere additions or supplements to the existing stocks of knowledge. Geographers John Agnew and James Duncan suggest that it is ‘the absence of a disciplining orthodoxy’ and ‘the openness to fresh thinking that now makes the field so interesting.’ This might be qualified in several ways. Human geography has not altogether abandoned a canon, and its working practices are still ‘disciplined’ in senses 3 that owe less to the monastery than Michael Foucault. Its courses and syllabuses, its textbooks and journals, its examinations and dissertations, its refereeing and reviewing: all work to produce ‘disciplinary subjects’ of a sort – students, teachers, researchers – and to normalise particular conceptions of what does and does not count as successful and significant ‘geography’. But that last sentence is full of plurals, and it is the plurality of conceptions, their co-existence but still more their interpenetration – the astonishing proliferation of hybrid geographies that combine different approaches, old sub-disciplines and new inquiries without ever congealing into a single orthodoxy – that Agnew and Duncan celebrate: to them, human geography is ‘amazingly pluralistic’. 1 So it is, but those who are less sanguine worry that human geography has become a house divided: a large building with many rooms and lots of occupants but too few doorways, stairwells and communal spaces – and then there’s the noise from the neighbours and a nagging anxiety about the foundations. This invites another qualification because the neighbours (other disciplines) issue invitations and come to visit, so that the ‘openness to fresh thinking’ is not confined to human geography; it is widely acknowledged that no one field, perspective or approach can ask all the important questions or provide all the interesting answers. What we find particularly encouraging is the reciprocity contained in the metaphor of invitation and visit. Where human geographers once borrowed freely from others in a one-directional 1 John Agnew and James Duncan (eds), ‘Introduction’, The Wiley-Blackwell Companion to Human Geography (Oxford: Wiley-Blackwell, 2011) p. 1. On hybrid geographies in a methodological sense, see Mei-Po Kwan, ‘Beyond Difference: From Canonical Geography to Hybrid Geographies’, Annals of the Association of American Geographers, 94(4) (2004) 756–763. The closing sections of our introduction describe quite other, or rather more radically hybrid geographies. 4 series of appropriations, their work is now taken up – and taken seriously – elsewhere. Today, creative human geography is practised outside Human Geography as well as inside, and much of it in concert with scholars in other fields as well as with artists, photographers, film-makers, playwrights and journalists. 2 This openness is common to all intellectually alive disciplines; the borders between them are no longer policed with the same vigour that obtained thirty years ago. It is perfectly true that all of this busy intellectual commerce and trafficking has put a strain on the foundations but, again, this is not confined to human geography and, as we will see, there are reasons for suspecting that the old foundations provided little more than an illusion of stability and security. Like many other disciplines, human geography is now guided by other, less structural metaphors that provide for a much more fluid and lively sense of inquiry. We have organised these volumes and our introductory essay as four loose but powerful themes that run throughout human geography: ‘Histories, philosophies and politics’; ‘Theories, methods and practices’; ‘Space, place and landscape; and ‘Nature, environment and the non-human’. These are not rigid categories and it is important to notice the resonances, references and the counter-arguments that flow back and forth between them. We hope 2 See, for example, geographer Michael Watts’ award-winning collaboration with photographer Ed Kashi in Curse of the Black Gold: 50 years of oil in the Niger Delta (New York: powerhouse Books, 2008) – and the multimedia and video at – and geographer Geraldine Pratt’s theatricalization (with Caleb Johnston) of her research with Filipina domestic caregivers in Nanay – a testimonial play, which has been performed in Vancouver and Berlin: Caleb Johnston and Geraldine Pratt, ‘Nanay (Mother): a testimonial play’, Cultural geographies 17 (2010) 123-33. These works radically extend the traditional conception of geography as an art: cf. D.W. Meinig, ‘Geography as an Art’, Transactions of the Institute of British Geographers, NS, 8 (1983) 314–328. . 5 that readers will find the result exciting, edifying and even surprising. We hope too that those who, like us, have been involved in the field for many years will find new things to think about, while those unfamiliar with human geography will be able to use these selections as springboards for their own intellectual journeys. Histories, philosophies and politics Re-telling geography’s story The history of geography involves many stories that start at different times in different places; they slowly become intertwined, and their narrative threads can be unpicked and rewoven into many different designs. It’s important to keep this image in mind because there is no one History (with that imperial capital H) of geography, and intellectual historians have chosen different starting-points for their stories: the traditions of chorography and geography in the classical world; their successor projects, the ‘special’ and ‘general’ geographies of early modern Europe; the modern discipline that comes into view in the nineteenth and early twentieth centuries on both sides of the Atlantic; and the ‘new geographies’ that emerged during and after the Second World War. 3 Remarkably, and regrettably, few historians have 3 For detailed accounts of these and other technical terms used in this essay, see Derek Gregory, Ron Johnston, Geraldine Pratt, Michael Watts and Sarah Whatmore (eds) The dictionary of human geography (Oxford: Wiley Blackwell, 5th edition, 2009); see also Rob Kitchin and Nigel Thrift (eds) The international encyclopedia of human geography (London: Elsevier Science, 2009) (12 volumes), which may also be available online at your university library. For more succinct accounts, see Noel Castree, Rob Kitchin and Alisdair Rogers (eds) Dictionary of Human Geography (Oxford: Oxford University Press, forthcoming). 6 been interested in the worlds beyond the Atlantic until Europe reached out to explore, occupy and often devastate them. Those processes of colonialism and imperialism relied on and resulted in various geographical knowledges, of course, and the emergence of a postcolonial critique in human geography has helped recover some of those appropriations and erasures. And yet there has been far too little effort made to recover (for example) older Arab, Chinese or Indian traditions of geography. This reminds us that there is no one Geography (with its own imperial capital G) either. Most orthodox histories of geographical inquiry have been directed towards its institutionalization and formalization, in which moments are clipped together like magnets until the present becomes the climax of the past, a ‘proper’ discipline that erases earlier mistakes and eclipses previous experiments. Geography is made to appear as the logical result of Science vanquishing fantasy and Reason triumphing over superstition, finally taking the place that had been allotted to it within the intellectual landscape. This is an odd sort of history as well as a dull one: courses in geography were taught in universities long before the modern creation of separate ‘disciplines’, and many of the figures usually placed on pedestals as the principal architects of modern geography displayed an intellectual range that was rarely bounded by a single field of expertise or interest. Consider a man like Alexander von Humboldt, who intended to join Napoleon’s military expedition to Egypt in 1798 as a scientific observer but missed the boat and travelled to Spain’s colonies in central and South America instead: his 30-volume account of his travels ranged from botany and zoology to history and political economy, and his magnum opus, Cosmos, promised nothing less than ‘a sketch of the physical description of the universe’. Orthodox histories are inadequate for 7 another reason: they provide ‘internalist’ narratives that focus on the inside of intellectual inquiry and rarely look at its outside, let alone wonder about the boundary between the two. Although they usually pay some attention to biography they are remarkably disinterested in history, in the wider currents in which Humboldt (and all the others) were caught up. There are important debates about the relations between knowledge and society, and while most writers would agree that these are not purely matters of choice, how they are to be theorised and analysed remains an open question. 4 But it is hard to imagine making much sense of – or stimulating much interest in – the work of previous scholars without taking these debates, and the connections that they identify, with all possible seriousness. So it is reassuring to notice that those orthodox, cloistered histories have been unsettled by two moves. The first involves re-territorializing geography, and the second de-territorializing geography, and we need to think about each of them in turn. To re-territorialize geography is to see geographical inquiry not as the progressive realization of disembodied Reason but as the continuing product of shifting networks of social practices. This literally makes geography come alive: it means filling its pages with people who exist beyond the text, flesh and blood characters who animate intellectual inquiry as something 4 A range of contributions has attracted the attention of human geographers. Some were drawn to the early work of Jürgen Habermas, who identified a series of ‘cognitive interests’ embedded in the structures of society that ‘constituted’ or shaped particular forms of knowledge and their evaluation: see Allen J. Scott, ‘The Meaning and Social Origins of Discourse on the Spatial Foundations of Society’, in Peter Gould and Gunner Olsson (eds), A Search for Common Ground (London: Pion, 1982) pp. 141–156. Others preferred Michel Foucault, who traced an intimate connection between power and knowledge: power-knowledge. Still others have turned to the sociology of science and science studies to show how human geography, like any other field of knowledge, is a profoundly social practice and as such caught up in networks of social practice that reach far beyond the academy. 8 more than a purely contemplative pursuit, sometimes competitive, at others collaborative (and usually both at the same time). Universities have never been ivory towers; they have always been caught up in the societies in which they are embedded. Scholars respond to events and situations in different ways, sometimes swept along by the tides of history, sometimes riding the waves (or commanding the tides to retreat), but almost always involved. This doesn’t mean that we can reduce a text to its context, but it does require us to think about the ways in which discoveries, ideas and claims emerge out of particular situations. This isn’t confined to geography, of course, but the realization that all knowledge is situated – that it is produced or reworked by somebody from somewhere – has an obvious special significance for a field that attaches so much importance to place and space. Human geographers have started to examine the different sites at which geographical knowledges have been produced – field sites, laboratories, libraries, archives, museums and a host of others – and the reciprocal relationships between these spaces and the social practices of knowledge production that take place there. 5 None of this need make geographical knowledge parochial; it may be marked in all sorts of ways by context and circumstance, but knowledges do travel – sometimes aggressively, under the banners of colonialism and imperialism, sometimes in a spirit of collaboration or solidarity – and they rarely survive the journey intact. They are examined on arrival, re-made and put to work in different contexts and different circumstances. In short, there is always a restlessness to our knowledge of the world. It may settle for a time in places where it is stored – hence the importance of libraries, archives and what are sometimes called ‘centres of calculation’ – but it is also usually 5 See John Agnew and David Livingstone (eds), The SAGE Handbook of Geographical Knowledge (London: Sage, 2011) Part II: ‘Geography’s venues’. 9 mobile, especially in our liquid world, moving through changing circuits and pathways, sometimes openly, and sometimes furtively. Once we start to think about knowledge like this, it’s really a small step to de-territorialize geography and to release the history of geographical knowledge from the confines of any one discipline. Geographical knowledge has always been produced at multiple sites and circulated through multiple networks. 6 Some of this is more or less formal. The list of organisations that keep an eye on the world as part of their standard operating procedure is endless. The United Nations, the World Bank and NATO; departments of government, militaries and intelligence agencies; major oil companies, banks and transnational corporations; non-governmental organisations like Human Rights Watch, Oxfam and Save the Children; print, TV and online news organisations: they are all producers and consumers of geographical knowledge. But this can be expressed in other, less ‘data-driven’ forms too. Advertisements, art, film, literature, magazines, music, video: all of these are media through which other imaginative geographies are created that shape our sense of places and people around the world. ‘Academic’ geography spirals in and out of all those sites and networks too, so that the production of geography is, by its very nature, all over the place. And yet some knowledges are typically privileged while others are marginalized or even ignored. Sometimes this is a matter of style rather than substance: for example, research in spatial statistics may be seen as central 6 This too has a history; see, for example, Miles Ogborn, ‘Writing Travels: Power, knowledge and ritual on the English East India Company’s early voyages’, Transactions of the Institute of British Geographers, NS, 27 (2002) 155–171. 10 to geographical inquiry by some practitioners, while travel-writing may be rejected as the impressionistic work of the amateur. Sometimes, and rarely unconnected, it is a matter of what is called ‘positionality’: for example, indigenous or subaltern knowledges are often discounted in order to promote particular versions of ‘Science’ or ‘Development’. And sometimes texts are cast as heroes or villains. The standard accounts of human geography in the English-speaking world today usually measure their distance from two milestones (or millstones): Richard Hartshorne’s The nature of geography (1939) and David Harvey’s Explanation in Geography (1969), and we need to consider each in turn. Hartshorne was an American political geographer who had left the United States in July 1938 for Germany, intending to spend his sabbatical leave studying the political geography of the Danube Basin. But his plans were thrown into disarray by Hitler’s geopolitical ambitions. Hartshorne arrived four months after Nazi Germany had annexed Austria as part of the Third Reich, and he retreated to the library at the University of Vienna to complete a draft historical-cum-philosophical essay that he had taken with him. This grew into a major book that offered a rigorous prescription for Geography as a discipline. 7 It is of interest for four reasons. First, Hartshorne’s history of geography was ruthlessly selective: he was determined to establish a continuous intellectual tradition – what he called geography’s ‘continuity of life’ – that would at once identify the legitimate line of intellectual descent (so that professional geographers could ‘keep on the track’) and renounce its bastard children: ‘deviations from the course of 7 Richard Hartshorne, The nature of geography: a critical survey of current thought in light of the past (Lancaster PA: Association of American Geographers, 1939). 11 historical development.’ His was not a disinterested history, then, but history with a purpose, a normative history. In Hartshorne’s telling it was a story that privileged German writers. One of his heroes, Alfred Hettner, had declared that ‘Geography is a German science’, and Hartshorne readily agreed: ‘the foundation of geography as a modern science was primarily the work of German students.’ Other stories are possible, but many of those written after Hartshorne (and often against him) still insisted that modern Geography had its origins as a distinctively European science. 8 Second, Hartshorne’s purpose was to confirm Geography as a distinctive discipline, different from (say) Botany or Geology, Economics or Sociology. He drew on a distinction made by the philosopher Immanuel Kant between ‘logical classifications’, which direct our attention to formal or functional similarities between things, and ‘physical classifications’ that direct our attention to the co-existence of things in time or space. In Hartshorne’s view, logical classifications formed the basis for the systematic sciences, which included Biology and Geology, Economics and Sociology, while physical classifications were the foundation for what he saw as the ‘exceptional sciences’ of History and Geography. The unique object of geographical inquiry was thus the region, the assemblage of things found together in the same space, and the discipline’s task was to account for the differences between one region and another, which Hartshorne called ‘areal differentiation’. This is routine stuff for histories of Geography, but it begs a critical question, and this is the third reason why his work is significant: given the 8 See for example David Stoddart, On geography and its history (Oxford: Blackwell, 1985). 12 circumstances in which Hartshorne set out these ideas, how on earth was it possible for a political geographer (of all people) to withdraw so completely into the world of books – and to reduce Geography to a succession of texts – whose pages were unmarked by the monstrous events taking place all around him? There is little doubt that Hartshorne was deeply affected by what he saw; he wrote of his good fortune in conversing regularly with a professor at the Geographical Institute in Vienna who ‘was permitted to do research but not to teach’ because his wife was Jewish, and recorded their fundamental agreement over ‘basic political and human issues in the irrational world of Nazi Germany.’ And he was certainly not indifferent to the rise of fascism; in 1941 Hartshorne was summoned to Washington to form a Geography section in what would eventually become the Office of Strategic Services, supervising the production of vital intelligence for the war against the Reich and its allies. 9 The clue is in the word ‘irrational’. Hartshorne turned away from the desperately contested, acutely physical borders between states in a Europe on the brink of war to plot ones that were idyllic and imaginary: borders between disciplines that would reveal an enduring rational order in a world rapidly descending into madness. In doing so, Hartshorne not only declared his belief in Geography as a pure, objective Science but also his faith in a radically different German intellectual tradition to the debased fantasies being peddled by the Nazis. If this interpretation can be sustained, then it confirms that texts cannot be reduced their context in any simple or direct fashion. In this case, the context in which Hartshorne sketched out his ideas is present in the text through its absence. 9 See Trevor Barnes and Matthew Farish, ‘Between Regions: Science, Militarism, and American Geography from World War to Cold War’, Annals of the Association of American Geographers, 96(4) (2006) 807–826. 13 There is a fourth reason for returning to Hartshorne’s work. Thirty years later The Nature of Geography was the object of Harvey’s critique in Explanation in Geography. 10 Others had disagreed with Hartshorne before, notably the American cultural geographer Carl Sauer who had objected to the barriers Hartshorne installed between history and geography. To Sauer this was the ‘Great Retreat’, and the hybrid ‘historical geography’ was not a mutant but a vital mode of inquiry. Harvey would not have disagreed since he had been trained as an historical geographer. But like many others of his generation he rejected the view, which was common to both Hartshorne and Sauer (though in different ways), that the distinctiveness of geography was to be found outside the mainstream scientific tradition. Rather than turn to a disciplinary history to provide his warrant – not least because he was part of a movement that sought to break with the discipline’s past: the so-called ‘Quantitative Revolution’ – Harvey turned to Philosophy and specifically the philosophy of science. In his view, a properly scientific geography had to use ‘the standard model of scientific explanation’ derived from the physical sciences. Its distinctive object would be (planetary) space not the region – Harvey insisted that space was ‘the central concept on which Geography as a discipline relies for its coherence’ – and its distinctive method would then be spatial analysis. In fact, Hartshorne had also described Geography as a ‘spatial science’, but his was a remarkably limited conception that treated each region as unique and required it to be analysed through an idiographic approach that promised a full understanding (usually an inventory) of the elements assembled within each distinctive regional constellation. This ruled 10 David Harvey, Explanation in Geography (London: Edward Arnold, 1969). 14 out the prospect of generalizations, whereas Harvey’s contrary view was to insist on the active search for a spatial order – a ‘spatial structure’ – existing beneath all these surface variations that could then be explained through generative processes. Seen like this, geography was to be a consciously theoretical project: in Harvey’s words, ‘By our theories you shall know us.’ The backdrop to the Quantitative Revolution and Harvey’s summation of its philosophical basis was more than an argument about the direction of the discipline. The 1960s were a time of principled social protest on both sides of the Atlantic: civil rights marches in the United States, rallies and demonstrations against the Vietnam War, struggles for political freedoms behind the Iron Curtain, student demonstrations and sit-ins across Europe and the Americas, the anti-apartheid campaign, the rise of the environmental movement, the continuing march of the women’s liberation movement, and the rise of the gay rights movement. As Explanation in Geography appeared in print at the very end of that turbulent decade, Harvey moved from Britain to the United States where he embarked on a determined attempt to bring about what he would later call a (new) ‘revolution in geographical thought’. Dismayed by what he now saw as the poverty of spatial science, he criticised the ‘clear disparity between the sophisticated theoretical and methodological frameworks we are using and our ability to say anything really meaningful about events as they unfold around us.’ This was a bold challenge to both the ‘objectivity’ demanded by Hartshorne and the objectivism of spatial science. Instead, Harvey proposed a radically new, politically engaged human geography. The project was new partly because it was an activist rather than merely an ‘applied’ geography; spatial science had forged all sorts of links between human geography and urban and regional planning 15 but these were largely instrumental, ‘expert knowledges’ that were directed at policy rather than politics. It was new too because Harvey sought its foundation in historical materialism, which he claimed provided not only the analytical depth missing from spatial science but also a spur to political action. The radical geography he advertised in Social Justice and the city (1973) was not a purist philosophical venture: as Karl Marx insisted – the sentiment is engraved on his tombstone – ‘philosophers have only interpreted the world in various ways; the point is to change it.’ Harvey’s early engagements with classical Marxism were exuberant but unformed, and he would devote the rest of his career to a closer reading of Marx and to the construction of what he came to call historico-geographical materialism. 11 As this suggests, space remained one of his central concerns, but it was now conceived in a different way: if, as Marx argued, capitalism should be theorised as a mode of production then it was essential to see that it produced not only commodities but also space. Philosophies and politics We will return to these claims later. For now, Harvey’s trajectory provides a template to gauge two other important developments. First, not all human geographers shared his impatience with philosophical exploration. Many of them endorsed his critique of spatial science but suspected that the root of the problem was the philosophy of science on which the Quantitative 11 David Harvey, Social justice and the city (London: Edward Arnold, 1973; republished Athens GA: University of Georgia Press, 2009). For more detailed views of Harvey’s work, see Noel Castree and Derek Gregory (eds), David Harvey: a critical reader (Oxford: Wiley-Blackwell, 2006), and for more immediate views of Harvey’s work see his blog, Reading Marx’s Capital with David Harvey, at 16 Revolution appeared to rest: positivism. This was perhaps premature; few of those who developed its first- or even second-generation spatial models and dreamed of what Peter Haggett, another British geographer, once called ‘a general theory of locational relativity’ had been much interested in philosophy, and Harvey’s attempt to provide a philosophical warrant for Explanation in Geography had come late in the day. In fact, ‘positivism’ didn’t even appear in the index. But there was more than a family resemblance between spatial science and positivism: the privilege given to empirical observations (‘the facts’); the obsession with hypothesis testing; the search for an order of things that could be enshrined in general laws that in principle could even unite physical and human geography; and the belief in neutral, value-free inquiry. Each of these could be challenged, and – like many social sciences – human geography was soon caught up in a sustained exploration of post-positivist philosophies. There was a dazzling parade of –isms and –ologies through the discipline, each one accompanied by a considerable fanfare. The largest crowds were attracted by phenomenology, (critical) realism, structuralism and post-structuralism. The only one of these that retained an affiliation with something approaching ‘the scientific method’, or at least one that would be recognised by physical scientists, was realism. Most of its architects were philosophers of science, and its emphasis on providing causal explanations (rather than establishing correlations) gave a new impetus to empirical work in many areas of human geography. 12 In some of its forms phenomenology also had a close relation to science, in so 12 See Andrew Sayer, ‘Postmodernist Thought in Geography: A Realist View’, Antipode, 25(4) (1993) 320–344; see also his classic text, Method in social science: a realist approach (London: Hutchinson, 1984; second edition Routledge, 1992). By the 1990s critical realism was also one of the few areas of philosophical overlap between human and physical geographies; several prominent fluvial geomorphologists made substantive use of realist protocols in their field research. 17 far as it sought to criticise science’s ‘natural attitude’ in order to disclose the way in which the objects of its inquiries were brought into view. In doing so, however, it ranged far beyond the natural sciences and the social sciences that aped them; so too, in different ways, did structuralism and post-structuralism. In human geography most of these philosophies (the exception is the cluster of approaches grouped under post-structuralism) were used, as often as not, to replace one foundationalism – positivism – with another. In other words, they sought to establish a secure and certain foundation for claims to knowledge. The cardinal assumption was that Philosophy, what American philosopher Richard Rorty called ‘Philosophy-with-a-capital-P’, occupied a special vantage point from which to lay down the rules and issue instructions for the conduct of substantive inquiries to be carried out by underlabourers in other fields. Rorty was deeply sceptical about this assumption (in fact, he was wonderfully rude about it). Of course, Philosophers are not the only pretenders to the throne, and Rorty also listed the Party, Priests, Physicists and Poets, all surrogates for larger political and cultural formations. Harvey, for all his impatience with philosophizing, retains a strong faith in foundationalism and repeatedly distinguishes the ‘surface appearance’, ‘froth’ and disorder of things from the invariant laws and logics of capital accumulation that drive those transformations. But Rorty was a philosopher and so he took Philosophy as his exemplar. In his view, Philosophy can never provide a single, canonical language into which all questions can be translated and in whose terms all disputes can be resolved. Those who think it can, he said, simply fail to take language seriously. Feminist scholar Donna Haraway – whose first book was on metaphors in twentieth-century 18 biology and who evidently takes science rather more seriously than Rorty – agrees. In one of her most celebrated essays she calls this ‘the God-trick’, the claim to see everything from nowhere in particular. What it conveniently ignores, she argues, is the worldliness of intellectual inquiry: the fact that all knowledge is situated, so that there is no position from which to freely and fully observe the world in all its complex particulars. All our knowledges provide partial perspectives, and acknowledging this is not a barrier to objectivity but the very condition of it because only then, through dialogue with others, can we start to understand how limited our own views are. 13 Rorty and Haraway are helpful guides, because they suggest why the relationship between philosophy and geography slowly changed. At least since Hartshorne the modern terms of exchange had enforced a monologue in which Philosophy dictated and Geography obeyed, but this has given way to something much more like a conversation. Today philosophy is increasingly treated as resource rather than writ, used to inform rather than police inquiry. The parade of –isms and –ologies has been dismissed, to be replaced by an interest in the writings of particular philosophers whose interventions spiral far beyond the philosophy of science to address urgent questions of political and moral philosophy. Indeed, the work of philosophers like Giorgio Agamben, Alain Badiou, Judith Butler, Gilles Deleuze, Jacques Derrida or Michel Foucault spirals far beyond philosophy too, and human geographers have found in their writings inspiration for their 13 Donna Haraway, ‘Situated knowledges: the science question in feminism and the privilege of partial perspective’, Feminist studies 14 (1988) 575-599; see also Gillian Rose, ‘Situating Knowledges: Positionality, Reflexivities and Other Tactics’, Progress in Human Geography 21(3) (1997) 305–320. 19 own investigations of the human and the non-human, subjectivity and spatiality, power and violence, gender and sexuality and a host of other substantive issues. At its best, this is not about ‘applying’ their insights but reading their texts at once closely and creatively. We may seem to have travelled far from Harvey’s corpus, but in fact we have circled back to it because a second development from the baseline of Social justice and the city, which the engagement with political and moral philosophy illustrates, has been to widen the political and ethical address of human geography. 14 Harvey’s project has been animated, above all, by a strong sense of class politics, and although he has addressed other axes of oppression and discrimination these have always been secondary. But other human geographers have insisted that there are multiple forms of injustice that cannot be reduced to class or convened within the plenary discourse of historical or even historico-geographical materialism. Two braiding streams of research are particularly important: feminist geography and postcolonial geography, both influenced by various forms of post-structural thought. We now have a far richer understanding of the ways in which race, gender and sexuality are embedded in and reproduced through places and landscapes, but feminist geography and postcolonial geography have also shown how discriminations are reproduced in – and legitimated through – geographical knowledges. Feminist geographies have posed a major challenge to the assumption that knowledge claims derive their authority from being 14 Not surprisingly this has occurred from a variety of philosophical perspectives: compare David M. Smith, ‘Social Justice Revisited’, Environment and Planning A, 32(7) (2000) 1149–1162; Stuart Corbridge, ‘Development Ethics: Distance, Difference, Plausibility’, Ethics, Place and Environment, 1(1) (1998) 35–53; Jeffrey Popke, ‘Poststructuralist Ethics: Subjectivity, Responsibility and the Space of Community’, Progress in Human Geography, 27(3) (2003) 298–316. 20 universal and somehow gender-free; they have shown to the contrary that conventions and concepts, theories and methods – the working practices of the academy and most other sites of knowledge production at large – have worked to advance particular, gendered ways of knowing (and being in) the world. These have typically privileged a highly restricted model of the masculine and used it to regulate – in fact to authenticate – what counts as reliable, acceptable or professional work. 15 The various geographies written under the sign of colonialism have not been free from masculinism – think of the hideous ideology of ‘the white man’s burden’ – but their special effect has been to privilege the powers and knowledges that accrued to what today would be called the global North. 16 During the long history of European colonialism and imperialism what Felix Driver calls ‘geography militant’ functioned not only in a directly practical sense to advance occupation, dispossession and appropriation – surveying territories, compiling resource inventories and the like – but also in an epistemological sense to situate ‘Europe’ at the centre of the advance of 15 A classic statement was Gillian Rose, Feminism and geography: the limits of geographical knowledge (Cambridge: Polity, 1993); see also Juanita Sundberg, ‘Masculinist Epistemologies and the Politics of Fieldwork in Latin Americanist Geography’, Professional Geographer, 55(2) (2003) 180–190. To de-limit these enclosures requires more than a critique of masculinist epistemology; it also requires theorizing the spaces through which human subjects are constantly constituted as knowledgeable agents: see Geraldine Pratt, ‘Spatialising the Subject of Feminism’, in her Working Feminisms (Philadelphia: Temple University Press/Edinburgh University Press, 2004) pp. 12–37. 16 Postcolonialism is, in part, a political and ethical project directed at uncovering and resisting those privileges but it provides no guarantee of purity or innocence; all knowledges are caught up in the play of power, so that the postcolonial project involves a constant struggle to dis-cover and undo its own complicities: see Tariq Jazeel and Colin McFarlane, ‘The Limits of Responsibility: A Postcolonial Politics of Academic Knowledge Production’, Transactions of the Institute of British Geographers, NS, 35 (2010) 109–124. 21 rational knowledge. If modern Geography was indeed a European science, as many of its historians claim, it was also a profoundly Eurocentric one. 17 Eurocentrism was never a static enterprise, and in the course of the long twentieth century it was transformed into a sort of ‘Euro-Americanism’ that will surely be disrupted though not necessarily displaced by the resurgence of Asia in the twenty-first century. But even before it assumed today’s hyphenated form, Eurocentrism was divided internally (so that British, French and German colonialisms were distinct and rival enterprises, for example) and it assumed different forms in different places. One of its most pervasive and pernicious versions was Orientalism, in which European and later American politicians and generals, writers and artists constructed ‘the Orient’ as at once an exotic and bizarre space, at the limit a monstrous and pathological space – what Edward Said famously called ‘a living tableau of queerness’ – and as a space that had to be domesticated, disciplined and normalized – straightened out – through a forceful (imperial) imposition of the order it was supposed to lack: ‘framed by the classroom, the criminal court, the prison, the illustrated manual.’ 18 Again, this matrix was infinitely divisible: there were multiple ‘Orients’, from the ‘Near East’ and the ‘Middle East’ to the ‘Far East’. Other places and other peoples were the subject of other imaginative geographies, notably primitivism and savagism for indigenous peoples in sub-Saharan Africa, the Americas and the Pacific archipelago. Other natures were also enrolled in the project, and both scientific and non-scientific discourses (including art and travel writing) worked to normalise temperate nature as ‘normal’ nature: ‘all that is modest, 17 See Derek Gregory, ‘Cook’s Tour: Anthropology and Geography’, in his Geographical Imaginations (Oxford: Blackwell, 1994) pp. 16–33. 18 Edward Said, Orientalism (London: Penguin, 1978). 22 civilized, cultivated’. In particular, the discourse of tropicality constructed ‘the tropics’ as a paradise of excess and abundance, a veritable Garden of Eden, or as a rotting, fallen nature: the distance between Gauguin’s Tahiti and Conrad’s Congo. 19 Here, as is so often the case, talking about ‘nature’ was also a way of talking about ‘culture’. None of these constructions are creatures of the past: Orientalism helped to shape British geographer-politician Halford Mackinder’s imperial vision of a ‘heartland’ in the early twentieth century, and it was reactivated in stunningly violent ways in the wars in Afghanistan, Iraq and elsewhere that were launched in the shadows of 9/11. 20 None of them is confined to human geography either, but feminist and postcolonial geographies, in addition to their other, vital contributions, draw our attention to the ways in which assumptions about what is normal – ‘universal’ – have been smuggled into our field to normalise a particular (and particularly limited) conception of the ‘human’ in human geography. The challenge is to recognise and resist the ways in which those assumptions also diminish everyday lives inside and outside the classroom and the lecture theatre. 21 19 David Arnold, The problem of nature: environment, culture and European expansion (Oxford: Blackwell, 1999); Felix Driver and Luciana Martins (eds) Tropical visions in an age of empire (Chicago: University of Chicago Press, 2005); David Livingstone, ‘Tropical Hermeneutics and the Climatic Imagination’, in his Science, Space and Hermeneutics (Heidelberg: Hettner Lectures, Heidelberg, 2001), pp. 43–73. 20 See Gerry Kearns, ‘The Political Pivot of Geography’, Geographical Journal, 170 (4) (2004) 337–346; Derek Gregory, The colonial present: Afghanistan, Palestine, Iraq (Oxford: Blackwell, 2004). 21 See, for example, Michael Haldrup, Lasse Koefoed and Kirsten Simonsen, ‘Practical Orientalism – Bodies, Everyday Life and the Construction of Otherness’, Geografiska Annaler, 88B (2006) 173–184. . 23 In enlarging its sense of the human in these and other ways, human geography has moved into an interdisciplinary space, which is where most disciplines now find themselves. This has perplexed some writers, however, who have returned to the quest for a disciplinary identity. This is, in part, a response to the changes that have taken place in post-secondary education and advanced research under contemporary neo-liberalism. In the not very brave new world of the modern corporate university the commitments of a critical human geography (like those of other disciplines) are put under a microscope whose lenses have been cut to reveal a highly particular vision of knowledge. Accountability contracts to accountancy, politics to policy, and the very idea of critique (except in the ultimately empty and supposedly marketable form of ‘critical thinking’) all but disappears. Once hailed, by geographers at any rate, as Geography’s strength – its multiple allegiances to the sciences, social sciences and humanities – threatens to become a liability. If physical and human geographies look outwards and rarely at each other, physical geography to the earth, ocean and atmospheric sciences and to the biological sciences, and human geography to philosophy, political economy, sociology, history and literature, the administrative-cum-fiscal temptation to ‘rationalize’ and re-brand is not always easy to resist. Geography may not be coming apart at the seams, as Ron Johnston once feared, but there are always willing fingers prepared to unpick the stitches from the outside. 22 But the renewed debate over the nature of Geography – conducted in terms that Hartshorne would surely have enormous difficulty in recognising 22 Ron Johnston, ‘Geography – Coming Apart at the Seams?’, in N. Castree, A. Rogers, and D. Sherman (eds), Questioning Geography: Fundamental Debates (Oxford: Wiley Blackwell, 2005), pp. 9–25. 24 – is also driven by the sheer range of its inquiries and the demands placed upon its students by the theories, methods and practices that these involve. It is to these that we now turn. Theories, methods and practices Revolution and transformations in geography Many writers have argued that the ‘Quantitative Revolution’ of the 1960s is better understood as a theoretical revolution. For them, its most significant and perhaps even lasting contribution was the emphasis it placed on theoretical work in contrast to the fact-grubbing geographies of the past: the regional inventories that were the ever-present corollary of Hartshorne’s problematic of areal differentiation. The contrast is real enough, even if his critics would be surprised to learn that Hartshorne himself acknowledged the significance of select studies in location theory, including J.H. von Thünen’s model of agricultural land use (in fact not so surprising in an exegesis of a German intellectual tradition, since German scholars had been prominent in the development of location theory). But it was never part of his vision of Geography as what he called, only in passing and in relation to astronomy, ‘a spatial science’. After the Second World War the most advanced work in human geography was increasingly concerned with the development and substantiation of formal theory, notably central place theory and general models of the space-economy (Walter Christaller, August Lösch), industrial location theory (Alfred Weber), diffusion theory (Torsten Hägerstrand) and 25 theories of urban residential structure (E.W. Burgess and Homer Hoyt). 23 These avowedly ‘scientific’ investigations were not peculiarly geographical preoccupations; all the social sciences were re-shaped by their service in the war, and the Cold War continued to influence and often to fund academic research. 24 In geography, however, this was a sea change that repudiated Hartshorne’s prospectus and replaced it with a self-consciously ‘new’ and emphatically modern geography. There was a degree of irony in this, because most of the theoretical bases for spatial science could be traced back to the pre-war years; the only exception in the list above is Hägerstrand (and even then much of his data came from the 1920s and 30s). The seductive post-war gloss was applied through new means of computation – including computers themselves, and also new modes of analysis including operations research – that were created or transformed during the war and considerably expanded the scope of mathematical and statistical analysis. 25 For many of those most closely involved in spatial science, it seems that quantitative methods were always a means to an end. They were seen as highly rigorous – a way of putting ‘the’ scientific method into practice – but also wonderfully suggestive. For if human geography was now about the search for spatial order, it was clear that spatial order was not immediately apparent to the casual observer but would have to be ferreted out using 23 These contributions were reviewed and extended in Peter Haggett, Locational analysis in human geography (London: Edward Arnold, 1965). 24 Barnes and Farish, ‘Between regions’, below; see also Matthew Farish, The contours of America’s Cold War (Minneapolis: University of Minnesota Press, 2010) pp. 138-146 and passim. 25 The Second World War and the Cold War also had a direct bearing on the development of behavioural geographies: for a general discussion, see Joel Isaac, ‘The human sciences in Cold War America’, Historical journal 50 (2007) 725-46. 26 spatial statistics in an almost forensic fashion. 26 En route, however, some human geographers undoubtedly became so fascinated that they mistook the means for the end. They successfully alerted their colleagues to a central dilemma of standard statistical inference, which assumes that observations are independent from one another – the search for spatial order is predicated on spatial dependence, which means that this assumption is violated in the domains of most geographical interest – and in doing so made major contributions to the mathematics of spatial autocorrelation. 27 But whether these technical achievements increased the explanatory power of available spatial theories was another matter entirely. And it was theoretical power that was supposed to be decisive: remember Harvey’s injunction, ‘By our theories you shall know us.’ This was at once an argument about the distinctiveness of geography – although what a purely ‘geographical’ theory might be remained unanswered – and about the elevation of theory over method: hence the retrospective re-coding of the Quantitative Revolution as a theoretical revolution. But we need to add two riders to this reading, one about quantitative methods and the other about theory itself. In human geography the excesses of spatial modelling, which were most visible in the various point-process models (Poisson, negative binomial and the rest) that were used to generate 26 A favourite quotation of spatial scientists was Christoph von Sigwart’s claim, ‘That there is more order in the world than appears at first sight is not discovered till the order is looked for’: see Peter Haggett and R.J., Chorley, ‘Models, paradigms and the new geography’, in R.J. Chorley and Peter Haggett (eds) Models in Geography (London: Methuen, 1967) p. 20. The American philosopher William James quoted the remark in his address on ‘The dilemma of determinism’, but Chorley and Haggett, not surprisingly, found it in N.R. Hanson’s philosophy of science, Patterns of discovery (1958). 27 A.D. Cliff and J.K. Ord, Spatial autocorrelation (London: Pion, 1973). 27 spatial patterns and distance-decay curves, resulted in a growing conviction that, as Gunnar Olsson put it, ‘our statements often reveal more about the language we are talking in than the things we are talking about.’ Like Harvey, Olsson had been part of the avant-garde of spatial science, but the two friends were now drawing a distinction (in different ways) between the abstract regularity of spatial form – the isotropic planes and hexagonal grids of spatial science – and the turbulent dialectics of social process. ‘In the realm of intentions, hopes and fears,’ Olsson warned, ‘two times two is not always equal to four.’ 28 In his case, unlike Harvey’s, the argument was driven by a continuing, astonishingly creative engagement with philosophy, but the narrower, more sober-sided critique of positivism licensed a general and no doubt premature withdrawal from quantitative methods altogether by many human geographers. It was also reinforced by Harvey’s own, highly influential transition ‘from models to Marx’ and by the development of a range of other critical geographies that affirmed their distance from spatial analysis. 29 The newfound interest in human agency and the human subject seemed to demand radically other skills and sensibilities. The collective turn to all manner of qualitative methods proved remarkably fruitful (though these could not escape their Cold War shadows either) and explorations of ethnography, textual analysis and other interpretative approaches did much to re-humanise and radicalize a human geography that had become virtually 28 Gunnar Olsson, ‘The dialectics of spatial analysis’, Antipode 6 (3) (1974) 50-62; unpersuaded by Harvey’s turn to Marx, Olsson’s subsequent work involved a dazzling journey into philosophy and modern art: see his Abysmal: a critique of cartographic reason (Chicago: University of Chicago Press, 2007). 29 David Harvey, ‘From Models to Marx: Notes on the Project to ‘Remodel’ Contemporary Geography’, in Bill Macmillan (ed.), Remodelling Geography (Oxford: Blackwell, 1989), pp. 211–216. 28 eviscerated. 30 But the use of quantitative methods is not a diagnostic test for positivism, and the inability to interrogate large, complex datasets and tease out the relationships within them runs the risk of blunting the critical force of human geography. 31 There is a crucial, reciprocal relationship between theory and data – as even hard-core social science approaches like EITM (Empirical Implications of Theoretical Models) recognise – and most human geographers would now probably agree that it is actively unhelpful to oppose quantitative and qualitative methods or to see GIS as the work of the devil. 32 But they would also insist that the sources and media with which they work are not limited to ‘data’. The situation is further complicated by changes in the theory used in human geography. For the most part standard location theory relied on a mix of psychology and economic theory – ideas about economic rationality and price signals, supply and demand schedules, and equilibrating markets – complemented (or confounded) by elementary theses about the ‘friction of distance’. The critique of spatial science had many sources, but some of the earliest and most penetrating arguments took direct aim at its economic base by drawing upon radical political economy. These focused attention on capital accumulation and crisis formation, on the intersections between 30 Gillian Hart, ‘Denaturalizing Dispossession: Critical Ethnography in the Age of Resurgent Imperialism’, Antipode, 38(5) (2006) 977–1004; Steve Herbert, ‘For Ethnography’, Progress in Human Geography, 24(4) (2000) 550–568. For a survey, see Dydia DeLyser, Steve Herbert, Stuart Aitken, Mike Crang and Linda McDowell (eds) The SAGE Handbook of Qualitative Geography (London: Sage, 2010). 31 Eric Sheppard, ‘Quantitative Geography: Representations, Practices, and Possibilities’, Environment and Planning D: Society and Space, 19 (2001) 535–554; Elvin Wyly, ‘Strategic Positivism’, The Professional Geographer, 61(3) (2009) 310– 322. 32 Michael F. Goodchild, ‘Geographical Information Science’, International Journal of Geographical Information Systems, 6 (1) (1992) 31–45. 29 labour markets and housing markets, and on global processes of combined and uneven development that together established a problematic centred on the production of space under capitalism. As the agenda for critical human geography gradually became more extensive, however, other theoretical resources were tapped, and a second wave crested under the impetus of modern social theory. 33 Much of this was directed at explicating modalities of power that reached beyond the economic sphere – its sites included the state, the community and the family – but it too sought to elucidate the intrinsic spatiality of social life. 34 This second wave was not independent from the first – it was, in part, a critical response to it, but there was also a tradition of ‘Western Marxism’ that sought to extend historical materialism beyond the economic preoccupations of Marx and Engels, and human geographers discovered that Walter Benjamin, Guy Debord, Jürgen Habermas, Frederic Jameson, Henri Lefebvre, Nicos Poulantzas and a host of other writers had much to teach them – but the object of inquiry was now not only capitalism and its transformations (capitalism was a moving target: hence the distinctions between industrial capitalism and finance capitalism and, later, between Fordism and post-Fordism) but more insistently capitalist modernity and, eventually, the putative formation of the ‘post-modern’. 33 The term ‘critical human geography’ became widely used from the early 1990s to describe research that was broadly ‘progressive’ in orientation; it virtually eclipsed the term ‘radical geography’ which, from the late 1960s (especially in the United States), was used to describe research informed by historical materialism and anarchism in particular. Critical human geography drew on a wider bank of theoretical and political sources. 34 See Derek Gregory and John Urry (eds), Social relations and spatial structures (London: Macmillan, 1985). 30 Here too there was what Edward Soja called a powerful ‘reassertion of space in social theory’. 35 Both the first and second waves relied on theories that were markedly different from those of spatial science because, unlike the frozen lattices and equilibrium worlds of those early models, they described geographies in constant motion, where (as Marx put it) ‘all that is solid melts into air’ and where the contemporary ‘space of flows’ springs from a ‘liquid’ modernity. And yet, like spatial science, they were all marked by an extraordinary, almost imperial ambition. This was true in the more or less literal sense that these were all still Euro-American theories put to work to make sense of human geographies everywhere – there was still too little interest in what David Slater called ‘learning from other regions’ 36 – but they were also often so many versions of what is sometimes called Grand Theory. This was partly a matter of range, an architectonic impulse to construct a conceptual system that can reveal the central generating mechanisms that produce the (dis)order of things, and partly a matter of style, an epistemological desire to master the world and domesticate its differences. Seen like this, these theoretical preoccupations and privileges became vulnerable to critiques from both feminism and postcolonialism. 37 In response, the advocates of 35 Edward Soja, Postmodern geographies: the reassertion of space in critical social theory (London: Verso, 1989); for a review of these developments, see Derek Gregory, Geographical imaginations (Oxford: Blackwell, 1994). 36 David Slater, ‘On the borders of social theory: learning from other regions’, Environment and Planning D: Society & Space 10 (1992) pp. 307-27; see also David Slater, Geopolitics and the postcolonial: rethinking North-South relations (Oxford: Blackwell, 2004). 37 Linda McDowell, ‘Understanding Diversity: The Problem of/for “Theory” ’, in R.J. Johnston, P. J. Taylor and M. J. Watts (eds), Geographies of Global Change (Second Edition) (Oxford: Blackwell, 2002), pp. 280–294. 31 postmodernism protested that their work displayed an acute sensitivity to difference – hence Michael Dear’s insistence that ‘there can be no grand theory for human geography!’ 38 – and that their research programme could rehabilitate ‘areal differentiation’ in a theoretically informed and politically charged fashion. But to many critics it was just that – fashion (or worse) – and Cindi Katz issued an eloquent plea for theory in a minor key that would refuse the theoretical allegiances demanded by such grandiose schemes and work instead in the awkward spaces-in-between different traditions. There would be no grand synthesis, no totalizing vision, only the constant effort to understand and, where necessary, to un-do. 39 Similarly – but differently – Nigel Thrift argued that a more ‘modest’ form of theorizing was necessary for human geography to avoid a ‘theory-centred’ style of research ‘which continually avoids the taint of particularity’, though several critics evidently regard his non-representational theory as another exorbitation of Theory (and a radical diminution of its political possibility). 40 In the course of these exchanges about method and theory another change came into view. Human geography had been one of the last fields in the English-speaking world to take Marx seriously (in contrast, for example, to Francophone geography). Many human geographers were excited by the prospect of deeper theorizations of the capitalist space-economy, and there was a considerable interest in the structural logics – what Harvey sometimes 38 Michael Dear, ‘The postmodern challenge: reconstructing human geography’, Transactions, Institute of British Geographers 13 (1988) 262-74. 39 Cindi Katz, ‘Towards Minor Theory’, Environment and Planning D: Society and Spac, 14 (1996) 487–499. 40 Nigel Thrift, Non-representational theory: space, politics, affect (London: Routledge, 2007); Ben Anderson and Paul Harrison (eds) Taking-place: non-representational theories and geography (London: Ashgate, 2010). 32 called the ‘laws’ – of the contemporary capitalist mode of production. Other human geographers were attracted by the historical sensibility of what was, after all, historical materialism, and their analysis of processes and dynamics was driven less by formal theory and more by the particularities of archival research. Political-economic theory flourished in a host of different forms, from regulation theory through analytical Marxism and beyond, but so too did an ostensibly more traditional cultural-historical scholarship inspired by the work of historian E.P. Thompson and cultural critic Raymond Williams. This too was theoretically informed and had affinities with some versions of social theory – when Thompson railed against ‘the poverty of Theory’ it was Grand Theory, and strictly speaking structural Marxism, that he had in his sights – but it was much closer to the humanities than to the social sciences. These currents flowed into a more general ‘humanistic geography’ that had many sources. 41 Some of its practitioners were indifferent, even hostile to discussions of theory or method (and much more invested in philosophical speculation); their style was often particularistic but also individualistic, even idiosyncratic, and they relied as much on contemplation and reflection as analysis. 42 Others were more analytical, and while they drew on social theory their work also harvested the resources of the arts and humanities. This had major consequences. These allied fields were as theoretical in their sensibilities as the social sciences, but they were not characterized by 41 Stephen Daniels, ‘Arguments for a Humanistic Geography’, in R. J. Johnston (ed.), The Future of Geography (London: Methuen, 1985) pp. 143–158. 42 See, for example, the stream of books produced by Yi-Fu Tuan, from Space and place: the perspective of experience (London: Edward Arnold, 1977; Minneapolis: University of Minnesota Press, reprinted 2001) through Morality and imagination: paradoxes of progress (Madison WI: University of Wisconsin Press, 1989) to Cosmos and hearth: a cosmopolite’s viewpoint (Minneapolis: University of Minnesota Press, 1996). 33 any theoretical dominant: human geographers developed a keen appreciation of Michel Foucault and Edward Said, Jacques Derrida and Terry Eagleton, Roslayn Deutsche and John Berger. These enthusiastic readings intensified the heterodox nature of human geography. They opened its doors not only to the general admission of post-structuralisms of various kinds but also to the particular contributions of art history, literary theory and psychoanalytic theory to the analysis of core concerns like landscape, place and identity. This in turn has sustained new cross-fertilizations between human geography and the humanities. Artists, historians and literary scholars (among others) have come to recognise the critical potential in the conceptual and technical contributions of human geography – one recent collection even announces the emergence of the ‘geohumanities’ 43 – and the American literary theorist Stanley Fish, reviewing these developments, has argued that ‘the humanities have been the victors in the theory wars; nearly everyone now dances to their tune.’ 44 Not surprisingly, Fish emphasizes the interplay ‘between a literary and a geographical vocabulary’, which has assuredly been important. It would be impossible to make sense of the conversations between human geography and (for example) postcolonial studies without a close reading of the contributions of literary scholars Edward Said, Homi Bhabha and 43 Michael Dear, Jim Ketchum, Sarah Luria and Douglas Richardson (eds) Geohumanities: art, history, text and the edge of place (New York: Routledge, 2011); see also Stephen Daniels, Dydia DeLyser, J. Nicholas Entrikin and Douglas Richardson (eds) Envisioning landscapes, making worlds: Geography and the humanities (New York: Routledge, 2011). Others, with perhaps a keener eye on the ‘digital humanities’, prefer the term ‘spatial humanities’: see David Bodenhamer, John Corrigan and Trevor Harris (eds) The spatial humanities: GIS and the future of humanities scholarship (Bloomington IN: Indiana University Press, 2010). 44 Stanley Fish, ‘The triumph of the humanities’, New York Times, 13 June 2011. 34 Gayatri Chakravorty Spivak. 45 There is also a long and rather less theoretically informed history of human geographers fretting over what they used to call ‘the problem of geographical description’, which they understood as ‘the inherent difficulty of conveying a visual impression in a sequence of words’. 46 But, as that remark makes clear, a key term passed over by Fish is the visual: and it is perhaps here that the exchanges between the (other) humanities disciplines and human geography have been most energising. Envisioning human geography The philosopher Martin Jay described vision as the ‘master sense of the modern era’ – the gendering of the gaze is not incidental – and metaphors of sight constantly surface in our claims to know something: the emphasis on observation, on evidence (from the Latin videre meaning ‘to see), and the common use of ‘I see’ when we mean ‘I understand’. Visualization is also hidden in the word ‘theory’ itself, which combines the Greek thea (‘outward appearance’) and horao (‘to look closely’). These are general features, but many writers have identified a special affinity between visualization and the working practices of modern geography. This intimacy has been scrutinised in projects as outwardly different as Mackinder’s geopolitics – with its purportedly disembodied and detached gaze 47 – and 45 See Bart Moore-Gilbert, Postcolonial theory: contest, practices, politics (London and New York: Verso, 1997). 46 H.C. Darby, ‘The problem of geographical description’, Transactions of the Institute of British Geographers 30 (1962) 1-14; see also Pierce Lewis, ‘Beyond description’, Annals of the Association of American Geographers 75 (1985) 465-78. 47 Kearns, ‘Political pivot’, below; more generally, see Gearóid Ó Tuathail, Critical geopolitics: the politics of writing global space (London: Routledge, 1996). 35 spatial science, whose apprehension of ‘the world-as-exhibition’ separates observer and observed to produce the ‘perspective’ that is supposed to guarantee objectivity and order. 48 Some of the liveliest interventions have focused on a number of human geography’s central concepts – like landscape, which we discuss in the next section 49 – and on perhaps its most basic method: mapping. Mapping is usually represented as a technical process, and the history of cartography as a journey from error (‘here be monsters’) to Truth. Seen like this, the modern map and the atlas become purely technical artefacts, the products of a carefully controlled and recognisably scientific process that combines topographical or geodetic survey with the mathematics of map projection. If the history of cartography was an invitation to measure – and marvel at – the accuracy and fidelity of the modern map, then reading the map was a technical exercise too, involving a knowledge of projections and scale, contours and symbols. The map itself was inert and innocent: you could read it or you couldn’t, and apart from the manipulations of so-called ‘propaganda maps’ it could serve multiple purposes, from bombing cities to rebuilding them. This state of grace was interrupted by two interventions. The first, like the other initial de-stabilising encounters with the humanities, was largely historical. In several seminal essays historical geographer J.B. Harley sought to subvert ‘the apparent naturalness and innocence of the world shown in maps’. He used a vivid series of historical vignettes to 48 Gregory, Geographical imaginations, pp. 34-37, 53-69. 49 For a discussion of the European historical geography through which a particular conception of landscape emerged as a ‘way of seeing’, see Denis Cosgrove, ‘Prospect, Perspective and the Evolution of the Landscape Idea’, Transactions of the Institute of British Geographers, NS, 10(1) (1985) 45–62. 36 demonstrate the multiple ways in which maps were routinely enlisted in the service of political and economic power. 50 Harley’s arguments sparked a firestorm of controversy, but they also licensed a new, critical history of cartography that was much more aware of the ways in which cartographic ‘science’ was a vehicle for the promotion of interests and ideologies. In his quest to ‘deconstruct’ the map Harley invoked both Derrida and Foucault, but his real strengths lay in historical inquiry rather than conceptual acrobatics, and even those who were sympathetic to his project (and there were many) remained sceptical about his theoretical gestures. Partly in consequence, the second intervention was more rigorously theoretical but also, as it happened, directed squarely at the present rather than the past. This involved a searching interrogation of what was called ‘cartographic reason’, which had two entailments. On one side maps fixed a capricious world and represented it as a stable and ordered totality, while on the other side they were ‘performative’ so that, under specified conditions, they had the power to produce the effects they named: mapping, wrote John Pickles, ‘even as it claimed to represent the world, produced it.’ 51 This may seem frustratingly abstract, but in William Boyd’s novel An Ice-Cream War there is a marvellous passage that speaks directly to these propositions. During the First World War a young English officer is posted to East Africa, 50 Harley’s contributions have been collected in J.B. Harley, The new nature of maps: essays in the history of cartography (ed. Paul Laxton) (Baltimore MD: Johns Hopkins University Press, 2001); see also Matthew Edney, ‘Cartography without “Progress”: Reinterpreting the Nature and Historical Development of Mapmaking’, Cartographica, 30 (2&3) (1993) 54–67. 51 John Pickles, A history of spaces: cartographic reason, mapping and the geo-coded world (Londion: Routledge, 2004); see also Jeremy Crampton, Mapping: a critical introduction to cartography and GIS (Oxford: Wiley-Blackwell, 2010). 37 where his regiment is ordered to attack a detachment of German colonial troops. The mission is meticulously planned on a map, but when he and his comrades plunge over the side of the troopship and wade ashore they find themselves in a terrifying, perplexing battle for which they were almost wholly unprepared. ‘Gabriel thought maps should be banned,’ Boyd writes; ‘they gave the world an order and a reasonableness which it didn’t possess.’ This is an instructive example because it also directs our attention beyond the map: ‘mapping’ is not something that lies wholly behind the map, the historical process that culminates in its production, because mapping is also ‘beyond’ the map, what happens every time we interact with and through a map. 52 This shifts the focus from the map as a technical object or a cultural representation to maps as practices. Rob Kitchin and Martin Dodge capture this change when they insist that ‘maps are of-the-moment, brought into being through practices (embodied, social, technical), always remade every time they are engaged with.’ On their reading, then, ‘maps are transitory and fleeting, being contingent, relational and context-dependent.’ In short, ‘maps are practices – they are always mappings...’ 53 This realization turns cartography into something more than a means of representation; it becomes a medium of critical, political intervention. Human geographers, artists and others have collaborated in a range of projects – the Counter Cartographies Project, the Atlas of Radical Cartography and a host of others, many of them online – that continue and rework a tradition that can be traced back to the modernist cartographic experiments of the Situationists in the early twentieth 52 This is most readily grasped through the inherent interactivity of GIS. 53 Rob Kitchin and Martin Dodge, ‘Rethinking Maps’, Progress in Human Geography, 31(3) (2007) 331–344; for a detailed example, see Derek Gregory, ‘Seeing Red: Baghdad and the event-ful city’, Political Geography 29 (5) (2010) 266-79. 38 century. What they have in common is the recognition that the map can be not only the object of critique but a means of critique. 54 These developments have been closely aligned to human geography’s deepening engagements with other visual media – from those that have long been part of the modern geographical repertoire (photography, satellite and remotely sensed imagery) to those that have only more recently attracted the attention of human geographers (television, film, video) 55 – and, in tandem, to the interest in what Gillian Rose calls ‘a critical visual methodology’ that ‘thinks about the visual in terms of the cultural significance, social practices and power relations in which it is embedded.’ 56 Central to all of this has been a vital distinction between vision as a biological-physiological capacity – which naturalizes vision – and visuality as a culturally or techno-culturally mediated way of seeing. This distinction does not reduce visualization to techno-culture; on the contrary, questions of embodiment and corporeality – 54 David Pinder, ‘Subverting Cartography: The Situationists and Maps of the City’, D. Pinder Environment and Planning A, 28 (1996) 405–427; see also Alan Ingram, ‘Art and the geopolitical: remapping security at Green zone/Red zone’, in Alan Ingram and Klaus Dodds (eds) Spaces of security and insecurity: geographies of the war on terror (London: Ashgate, 2010) pp. 257-77. 55 Stuart Aitken and Deborah Dixon, ‘Imagining Geographies of Film’, Erdkunde, 60 (2006) 3265–336; Lisa Parks, ‘Digging into Google Earth: An Analysis of “Crisis in Darfur”’, Geoforum, 40 (2009) 535–545. An important complement to Parks’s essay is David Campbell, ‘Geopolitics and visuality: sighting the Darfur conflict’, Political geography 26 (2007) 357-82; Campbell has provided some of the most consistently revealing analyses of photography and visual imagery; see his blog on Photography, Multimedia, Politics at 56 Gillian Rose, Visual methodologies: an introduction to the interpretation of visual materials (London: Sage, 2007; second edition); although Rose doesn’t discuss it, GIS – like cartography – is also a visual method and can be pressed into critical service: see Mei-Po Kwan, ‘Feminist Visualization: Re-envisioning GIS as a method in feminist geographic research’, Annals of the Association of American Geographers, 92(4) (2002) 645–661. For more of Rose’s work, see her blog, visual/method/culture at 39 the refusal of the disembodied eye and the unmarked gaze – are focal to this way of thinking about human geographies. So too is the sociality of seeing, so that perspective, in its literal or metaphorical senses, is not the individual construction of an isolated observer. 57 If seeing is no longer taken for granted, and human geography is now exploring different theories and different methods that can illuminate what happens through different visual practices, it is also clear that ‘seeing things differently’ is what all our theories and methods claim to do: they promise to dis-close things we hadn’t seen before, to reveal relations and consequences we hadn’t noticed. This was the promise of spatial science and it remains at the heart of critical human geography. The difference is that we now know that seeing is never innocent and, from Haraway and others, that there is no single point of overview – no Archimedean point – from which the world can be ‘objectively’ disclosed as a fully transparent space. If our theories and methods establish spaces of constructed visibility, these are also always spaces of constructed invisibility. The price of seeing this is not to see that. And yet most human geographers would be reluctant to limit their work to contemplation. Teaching and research are also ways of intervening in the world, of ‘making a difference’ or electing not to, and this activates another sense of vision: the image of a future and somehow better world, which requires us to think again about questions of theory, method and practice. Practice here carries a profoundly political and ethical charge. For if we do not care about the world – if we treat it as merely a screen on which to 57 Fraser MacDonald, ‘Visuality’, International Encyclopedia of Human Geography (London: Elsevier, 2009), pp. 151–156; see also Fraser MacDonald, Rachel Hughes and Klaus Dodds (eds), Observant states: Geopolitics and visual culture (London: I.B. Tauris, 2010). 40 display our command of Technique or as a catalogue that furnishes examples of our Theory – we abandon any prospect of a genuinely human geography. We do not want to be misunderstood: of course theories and methods are important, but it is simply wrong to encounter the world and render it in such exorbitantly and exclusively instrumental ways. Just like the extremists of spatial science, this is to mistake the means for the end. Human geographers have made political and ethical interventions in a number of ways. ‘Applied geography’ has a long history, which has been transformed through contracted research for private and public interests and the involvement of human geographers in the formulation and assessment of public policy. These are muddy waters; some practitioners have despaired at the abstract elaboration of Technique or Theory as a studied disengagement from the messiness of the world, while others have challenged the normative claims that are covertly advanced through the ‘application’ of techniques and theories. 58 Here, as elsewhere, human geographers have to negotiate the various interests that shape their claims to ‘expert knowledge’. 59 This has prompted some of them to work outside the privileged worlds of the state, the corporation or the think-tank, and to engage instead in research with non-profit, non-governmental organizations and with disadvantaged or marginalized groups. They seek to lend their voice to those who are often denied a voice, but they also learn from them as well as about them in a 58 Ron Martin, ‘Geography and Public Policy: The Case of the Missing Agenda’, Progress in Human Geography, 25(2) (2001) 189–210; Ann Markusen, ‘Fuzzy Concepts, Scanty Evidence, Policy Distance: The Case for Rigour and Policy Relevance in Critical Regional Studies’, Regional Studies, 37(6&7) (2003) 701–717. 59 Scott, ‘Meaning and social origins of discourse’, below. 41 collaborative, participatory process of making human geography. 60 Others have preferred to engage in, even to provoke public debate about matters of urgent political, economic, social or environmental concern. At the start of the twenty-first century the President of the Association of American Geographers lamented that ‘critical geographical perspectives and idea are largely missing from public discussion of issues and events.’ 61 What a difference a decade makes. Human geographers are now actively involved in the production and circulation of ‘public geographies’ that reach far beyond the academy and, in doing so, they are involved in the simultaneous production of the spaces and publics that compose the public sphere. 62 Space, place and landscape Enlivening space The production of geographical knowledge has always involved claims to know terrestrial space in particular ways. Historically special importance was attached to the power to fix the locations of places, people and physical phenomena on the surface of the Earth and to represent these 60 For discussions, see Paul Chatterton, ‘ “Give up Activism” and Change the World in Unknown Ways: Or, Learning to Walk with Others on Uncommon Ground’, Antipode, 38 (2006) (2006): 259–281; Christine Dunn, ‘Participatory GIS – a People’s GIS?’ Progress in human geography 31 (5) (2007) 615-37. 61 Forum: ‘The role of geography in public debate’, Progress in human geography 29 (2005) 165-193. 62 Don Mitchell, ‘Radical Scholarship: A Polemic on Making a Difference Outside the Academy’, in Duncan Fuller and Rob Kitchin (eds), Radical Theory, Critical Praxis: Making a Difference Beyond the Academy? ACME e-book series (Place: Praxis (e)Press, 2004, pp. 21–31; Ulrich Oslender, ‘The resurfacing of the public intellectual: towards the proliferation of public spaces of critical intervention’, ACME: an international e-journal for critical geographies 6 (1) (2007) 98-123. 42 on maps. But, as we have just seen, the capacity to ‘write’ the earth in this way – the literal meaning of ‘Geo-graphy’ is ‘earth-writing’ – is not a purely technical affair because it is always implicated in the production of particular constellations of power. The relations between power, knowledge and geography animate contemporary discussions of a series of concepts, including place, landscape, region and territory, which have a direct bearing on how we understand the spatiality of life on Earth. Two general debates frame those more particular discussions. First, several writers have treated the nineteenth century as the epoch of time and the twentieth century as the epoch of space. This could mean many things, but it has usually been taken to address imagination (an accent on time in the work of major nineteenth-century philosophers and artists, for example) and substance (an aggressive preoccupation with geopolitics in the twentieth century, for example). The contrast between the two can be traced to a remark Foucault made in a lecture in 1967 – ‘the great obsession of the nineteenth century was, as we know, history’, whereas ‘the present epoch will perhaps be above all the epoch of space’ 63 – and several commentators later claimed, with a confidence that one might have expected Foucault’s conditional ‘may’ to have qualified, that as the modern gave way to the postmodern so critical social theory was compelled to recognize (or, as Edward Soja preferred, to ‘reassert’) the intrinsic spatiality of social life. The distinction is deceptively simple; the high modernism of the 1950s and 63 Michael Foucault, ‘Of other spaces’, Diacritics 16 (1) (1986) 22-27; it’s a tantalizing quotation, endlessly repeated, but Foucault never published the lecture (that only happened after his death) and his published works, especially Discipline and punish, are much better guides to his spatial analytics. See also Jeremy Crampton and Stuart Elden (eds) Space, knowledge and power: Foucault and geography (Aldershot UK: Ashgate, 2007) 43 60s, particularly in the United States, privileged change and transformation, and its functionalism planed away all local particularity, but the modernisms of the late nineteenth and early twentieth centuries, especially in Europe, were by no means silent about space. 64 Be that as it may, other writers have traded on global changes in communications and financial infrastructures to advance the opposite view, announcing the contemporary ‘death of distance’ and the imminent ‘end of geography’ in the same late, liquid or postmodern world: as Thomas Friedman put it in his ‘brief history of the twenty-first century’, ‘the world is flat’. 65 This too is deceptively simple – in fact simply wrong – because it advances and celebrates a model of globalization that ignores the constitutive relations between power and space. 66 This group of writers has noticed the process that David Harvey calls time-space compression – echoing Marx’s account of the ‘annihilation of space by time’ under capitalism – but failed to recognise its inherent variation: combined and uneven development is not an accidental by-product of capitalism but rather inheres in the very structures through which it reproduces exaction and inequality. 67 The ‘flattening’ of the world is about more than the almost 64 Edward Soja, Postmodern geographies, has been the most vigorous advocate of the reassertion of space in postmodern social theory; for a critique that turns in part on the spatiality of early modernism, see Gregory, Geographical imaginations, and on the spatiality of still earlier modernisms see Stephen Kern, The culture of time and space1880-1918 (Cambridge MA: Harvard University Press, 1983). 65 Thomas Friedman, The world is flat: a brief history of the twenty-first century (New York: Farrar, Straus and Giroux, 2005); see also Richard O’Brien, Global financial integration: the end of geography (London: Council on Foreign Relations Press, 1992); Francis Cairncross, The death of distance: how the communication revolution will change our lives (London: Orion, 1997). 66 Deborah Cowen and Neil Smith, ‘After Geopolitics? From the Geopolitical Social to Geoeconomics’, Antipode 41(1) (2009) 22–48. 67 Harvey’s original discussion of time-space compression said remarkably little about its geographical unevenness, but it was nevertheless predicated on the relations between (class) power and space under capitalism and so recognized at least some of the social 44 frictionless mobility of capital, information and commodities, and the no less oiled movement of executives, tourists and even troops from the privileged zones of the global North. Mobility is differentiated, and elsewhere people are violently displaced by flood, famine and war: as Bhabha has it, ‘the globe shrinks for those who own it’, but ‘for the displaced or dispossessed, the migrant or refugee, no distance is greater or more awesome than the few feet across borders or frontiers.’ 68 Like Mark Twain’s, reports of the death of distance have been greatly exaggerated. William Gibson, who devised the term ‘cyberspace’ in the 1980s and who has an acute awareness of the transformations Friedman and others fetishize, has often claimed that ‘the future is here – it’s just not evenly distributed yet.’ The point has been sharpened by accounts of globalization that are closely attuned to its powers of destruction as well as creation, and invested in elucidating its complex and compound geographies. 69 In short, the debate over whether ‘space matters’ depends very much on how space is conceptualized. The second debate follows directly from the first and concerns the ‘nature’ of space. Many writers inside and outside human geography have treated planetary space as either a framework within which social life inequalities intrinsic to the process: see Harvey, The condition of postmodernity: an enquiry into the origins of cultural change (Oxford: Blackwell, 1989). 68 Homi Bhabha, ‘Double visions’, Artforum 305 (1992) pp. 82-90: 88; see also Alison Mountz, ‘Stateless by Geographical Design’, in her Seeking Asylum: Human Smuggling and Bureaucracy at the Border (Minneapolis: University of Minnesota Press, 2010), pp. 121–203. Distance and dispossession are also inflected by class and gender: artist Barbara Kruger entitled one of her most famous artworks, ‘It’s a small world, but not if you have to clean it.’ 69 J.-K. Gibson-Graham, ‘Querying Globalization’, in The End of Capitalism (As We Knew It) (Minneapolis: University of Minnesota Press, 1996/2006), pp. 120–147; Eric Sheppard, ‘The Spaces and Times of Globalization: Place, Scale, Networks, and Positionality’, Economic Geography, 78 (2002) 307–330. . 45 happens or as the terrain on which human history unfolds. One represents space as an empty and unchanging grid of mutually exclusive points within which objects exist and events occur. This ‘absolute’ conception of space provided the basis for the system of areal differentiation that Hartshorne devised in The nature of geography, and its coordinate system is translated directly into the conventional map. The other treats space as the physical stage for the drama of human history, in which geography is assigned the task of painting the scenery without being drawn into the action. When Paul Vidal de la Blache, one of the founders of the French school of human geography, protested that ‘the stage itself is alive’, he had in mind a dynamic ‘nature’ on which ‘culture’ would work over the course of human history to produce a distinctive region. Either way, space was treated as what Foucault called, in that same 1967 lecture, ‘the dead, the fixed, the undialectical, the immobile.’ Against these views is a torrent of research in human geography that proposes a much livelier conception of space. It focuses on space not only as the outcome of social and biophysical processes, a commonplace of human geography, but also as the medium through which they take place. This is a radical reformulation of the geographical prospectus, for if space is involved in both the outcome and the operation of social and biophysical processes, then we can make sense of what some commentators (inside and outside geography) have seen as a ‘spatial turn’ across the spectrum of the humanities and social sciences, and even beyond. 70 The fact that we inhabit 70 See Mike Crang and Nigel Thrift (eds), Thinking space (London: Routledge, 2000). This collection charts a ‘spatial turn’ (p. xi), including a move away from absolute conceptions of space towards what the editors call ‘space as process and space in process’, and registers their claim that ‘Space is the everywhere of modern thought’. Although they were concerned at what they saw as the use of a ‘geographical idiom’ in social theory that was ‘resolutely ignorant of geographers and geography as a discipline’, however, puzzlingly none of the essays addressed the work of geographers, whose 46 a world in which things and events are distributed in time and space is not an elementary observation of no great consequence, something that sometimes makes drawing a map or describing the context useful: it becomes central to our apprehension and explanation of the world. We begin with three basic propositions about the ‘enlivening’ of space, which we’ll then use to examine the reformulation of the attendant concepts of place and landscape, region and territory. First, time and space are now theorized and analysed conjointly. Most human geographers have abandoned the project of an autonomous science of the spatial, rejected conceptions of space as the fixed and frozen ground on which events take place or processes leave their marks, and now work with concepts of time-space. 71 This project has taken many different forms, from a ‘time-geography’ that not only narrates but visibly choreographs the ways in which time and space are woven into the conduct of everyday life, through placing a revived historical geography at what Cole Harris calls ‘the heart of a reconstructed human geography’, to the development of the far wider field of historico-geographical materialism. 72 These and other developments all signal a decisive reversal of the ‘Great Retreat’ that so perturbed Sauer. Like him, historical geographers have long believed that ‘all geography is historical geography’ – the phrase can be traced back to collective role appeared to be reduced to commentator on and critic of social theory. Ten years on, the spatial turn is a multi-disciplinary affair in which human geographers are no longer perched in the bleachers watching the action on the field below. 71 Jon May and Nigel Thrift (eds), TimeSpace: geographies of temporality (London: Routledge, 2001). 72 Cole Harris, ‘Power, Modernity, and Historical Geography’, Annals of the Association of American Geographers, 81(4) (1991) 671–683. 47 Derwent Whittlesey and H.C. Darby 73 – but this is a radically different historical-geographical scholarship, less defensive about its disciplinary identity and with a far more developed sense of theoretical, methodological and political possibility. Second, this has directed attention towards the co-production of time and space. Time-space is not an external grid that enframes and contains life on Earth, but is folded into the flows and forms of the world in which we find ourselves. This is the basis for time-geography, in which time-space is conceived as a ‘resource’ on which individuals must draw in order to realize particular projects. In doing so they reproduce or transform the differential relations of power that enable or constrain their freedom of movement, and they do so by performing a ‘place ballet’, what Hägerstrand – the Swedish originator of time-geography – called a ‘weaving dance in time and space’ that is also a dance of time-space. In its original form this was all rather skeletal – Anne Buttimer described the formal time-geography diagrams as a danse macabre – but in the creative hands of Allan Pred it becomes clear that the folding of time-space into social life can be conveyed through a narrative that owes as much to the arts and humanities as it does to the social sciences. 74 Similar ideas reappear in historico-geographical materialism. 73 Derwent Whittlesey, ‘The horizon of geography’, Annals of the Association of American Geographers 35 (1945) 1-36; H.C. Darby, ‘On the relations of geography and history’, Transactions of the Institute of British Geographers 19 (1953) 1-11. 74 Allan Pred, ‘A Day in the Life, a (Some)body in Motion: A Docker’s Daily Path’, in his Lost Words and Lost Worlds: Modernity and the Language of Everyday Life in (Cambridge: Cambridge University Press, 1990), pp. 229–245 and 291–294; cf. Derek Gregory, ‘Suspended animation: the stasis of diffusion theory’, in Derek Gregory and John Urry (eds) Social relations and spatial structures (London: Macmillan, 1985) pp. 296-336. 48 One of Harvey’s cardinal achievements was to demonstrate that capitalism’s production of space is not incidental to its production of commodities, so that any viable political economy must incorporate the turbulent spatialities of production and circulation as a central dimension of its critique. 75 But, as Noel Castree emphasizes, Harvey’s project is simultaneously an historical and a geographical materialism, and the hyphen joining them is called upon to do a considerable amount of work. Harvey insists that capitalism ‘is not a system whose operation occurs in space and through time, as if these were empty matrices waiting to be filled with the diverse products of human activity’; instead space and time are ‘co-constituted’ and, as Castree puts it, capitalism ‘is spatio-temporal “all the way down”.’ 76 None of these authors can be assimilated to a single project – remember the heterodoxy of human geography – but Hägerstand and Pred, Harvey and Castree bring in to view the stubborn materialism of these ways of thinking about time-space. This is not the airy stuff of philosophical speculation (though it can be); these are all attempts to capture the sheer physicality of human geographies. 77 In a similar vein, Nigel Thrift proposed the idea of ‘spatial formations’ to convey a sensuous ontology of practices and encounters between diverse, distributed bodies and things. This is closer to Hägerstrand than Harvey, because it operates through an analytics of the 75 This marks a considerable distance from neoclassical economics, whose famously dimensionless, frictionless economy is perched on the head of a pin; whatever angels it may encounter there, diabolically Harvey shows that space is integral to Marx’s political economy: see David Harvey, The limits to capital (Oxford: Blackwell, 1982; reprinted London: Verso, 2006). 76 Noel Castree, ‘The Spatio-temporality of Capitalism’, Time & Society, 18(1) (2009) 265–60. 77 Some commentators see this as a renewed interest in ontology, compared to human geography’s obsession with epistemology in the 1980s and 90s: cf. Ulf Strohmayer, ‘The Culture of Epistemology’, in Steve Pile and Nigel Thrift (eds), Handbook of Cultural Geography (London: Sage, 2003) pp. 520–531. 49 surface rather than the depth-models of mainstream Marxism, yet here too time-space is not apart from the world (which would be another version of the God-trick) but emerges as a process of continual co-construction ‘through the agency of things encountering each other in more or less organized circulations’. 78 Third, human geographers are now much more willing to accept the unruliness of time-space. Most of them would probably agree that spatial science and conventional social theory made too much of pattern and systematicity, labouring in different registers to solve what they called ‘the problem of order’, without recognizing the multiple ways in which life on Earth evades and exceeds those orders. They were both attempts to order what is now most often seen as a partially ordered world – to tidy it up. As the philosopher A.N. Whitehead warned, ‘Nature doesn’t come as clean as you can think it’, and it is in this spirit that much of human geography is increasingly exercised by the ways in which the coexistence of different time-spaces perturbs, disrupts and transforms the fields through which social and bio-physical processes operate. To be sure, time-space is not infinitely plastic: ‘certain forms of [time-]space tend to recur,’ Rose reminds us, ‘their repetition a sign of the power that saturates the spatial.’ 79 And yet, while modalities of power often work to condense particular spatio-temporalities as ‘natural’ outcomes through architectures of surveillance and regulation, Doreen Massey insists that time-space is not a coherent system of 78 Nigel Thrift, ‘Space: The Fundamental Stuff of Human Geography’, in Sarah Holloway, Stephen Rice and Gill Valentine (eds), Key Concepts in Geography (London: SAGE, 2003), pp. 95–107; for a fuller account, see Nigel Thrift, Spatial formations (London: Sage, 1996). 79 Gillian Rose, ‘Performing space’, in Doreen Massey, John Allen and Philip Sarre (eds), Human geography today (Cambridge: Polity, 1999) pp. 247-59 50 discriminations and interconnections, a grid of ‘proper places’. She argues that it necessarily entails plurality and multiplicity. Hence spatial formations for her involve (and invite) ‘happenstance juxtapositions’ and ‘accidental separations’, so that time-space becomes a turbulent field of constellations and configurations: a world of structures and solidarities, disruptions and dislocations that provides for the emergence of genuine novelty. 80 ‘Emergence’ is not necessarily progressive or emancipatory, of course, and the argument may also be put in reverse: contemporary spaces of exception trade on paradoxical orderings of space whose very ambiguity is used to foreclose possibilities for political action. Either way, however, far from space being ‘the dead’, it is now theorized as being fully involved in the modulations of tension and transformation. You might think that all of this returns us to Hartshorne’s region, and the ‘physical classifications’ that enclose the co-existence of things in time or space. But if you re-read our last three paragraphs you will see that this is a return with a difference. Geography is no longer ruthlessly partitioned from History; time and space are no longer absolutes but defined in relation to people, events and objects, and these are not located ‘in’ time and space but enter into the co-production of time-space; and ‘physicality’ now carries a much livelier, more sensuous charge. Reclaiming conceptual spaces 80 Doreen Massey, ‘Philosophy and Politics of Spatiality: Some Considerations’, in her Power-Geometries and the Politics of Space-Time (Heidelberg: Hettner- Lecture, Heidelberg, 1998), pp. 27–42; for a fuller account, see her For space (London: Sage, 2005). 51 These developments can be traced in the genealogies of other spatial concepts, but these have their own particularities too. During the heyday of spatial science place and landscape were relegated to the margins of human geography while concepts like region and territory were reduced to abstract geometries. ‘Place’ was marginalized because it was seen as subjective, a jumble of attachments that was not immediately susceptible to scientific analysis; all a human geographer could do was describe a place in all its particularity. Some certainly saw this as a higher calling, and early humanistic geography became deeply invested in the meaning of place and its more or less literary evocation. But this often congealed into a conservative, romanticised sense of place. ‘To be human,’ Edward Relph declared, ‘is to live in a world that is filled with significant places: to be human is to know your place.’ This sense of attachment, of belonging and feeling at home, was supposed to be affirmative, and many writers including Relph endorsed what Yi-Fu Tuan called ‘topophilia’ (love of place) and gave grateful thanks for its distance from ‘placelessness’ and the homogenised ‘non-places’ of the modern world (malls, airports, hotels). 81 On this reading, ‘place’ denoted an older, slower, more authentic world: a still point in the now spinning spaces of modernity. More recent writings have troubled these constructions. Places are inhabited by multiple, conflicting meanings. People may refuse to ‘know their place’; they may transgress the codes that regulate a place, or they may take back places that 81 Edward Relph, Place and placelessness (London: Pion, 1976); Yi-Fu Tuan, Topophilia: a study of environmental perception, attitudes and values (Englewood Cliffs NJ: Prentice Hall, 1974); Marc Augé, Non-places: introduction to an anthropology of supermodernity (London: Verso, 1995). Tim Cresswell, ‘The Genealogy of Place’, in his Place: A Short Introduction (Oxford: Blackwell, 1996) provides an appreciation of all three authors. 52 have been taken from them or to which they have been denied access. 82 For everyone who feels at home, comfortable and safe in a place, there are others who feel lost, frightened and vulnerable. Places can excite pleasure, security and affection, but also pain, fear and revulsion, and all of them can invest places with profound and contradictory meanings. In these ways, place is bound in to both the play of power and the construction of identities – so that there is a doubled politics of place and identity 83 – and the erasure of a place can have traumatic consequences for those whose lives are affected. 84 Places, like identities, are always in the process of becoming, and at the limit many human geographers now treat place as process. 85 In doing so, they have come to see places as almost always impure, not tightly bounded but open and porous. Places, writes Doreen Massey, are meeting places in which various trajectories collide, ‘woven together out of ongoing stories, as a moment within power-geometries, as a particular constellation within wider topographies of space.’ 86 Far from being fixed and still, places are knottings, tied and untied, entangled and disentangled. Seen like this, place is not a refuge from the modern but, like space, is made much livelier: in 82 Tim Cresswell, In place/out of place: geography, ideology and transgression (Minneapolis: University of Minnesota Press, 1996). For more on Cresswell’s work, see his blog, Varve, at 83 Michel Keith and Steve Pile (eds) Place and the politics of identity (London: Routledge, 1993). 84 Gearóid Ó Tuathail (Gerard Toal), ‘The Effacement of Place? Us Foreign Policy and the Spatiality of the Gulf Crisis’, Antipode, 25(1) (1993) 4–31. The conversion of a peopled place into an abstract target is a standard condition for its attack and erasure. 85 Pred, A day in the life’, below; for an elaboration, see Allan Pred, ‘Place as historically contingent process: structuration and the time-geography of everyday life’, Annals of the Association of American geographers 74 (1984) 279-97. 86 Massey, For Space, p. 131; see also Doreen Massey, ‘Power-geometry and a progressive sense of place’, in Jon Bird, Barry Curtis, Tim Putnam, George Robertson and Lisa Tickner (ed), Mapping the futures: local cultures, global change (London: Routledge) pp. 60-70; ‘A global sense of place’, in her Space, place and gender (Cambridge: Polity, 1994) pp. 146-56. 53 Peter Merriman’s words, ‘much more contingent, open, dynamic and heterogeneous’ than the standard usages allowed. 87 ‘Landscape’ too seemed out of joint with spatial science. Unless it could be idealised as an isotropic plane, it was seen as belonging to an older, plodding geography that was limited to reconstructing its morphologies. But there were close links between the mathematical and geometric logics of spatial science and what Denis Cosgrove and others, drawing on art history, called landscape as a ‘way of seeing’. For the staging of the world as a landscape depended on the mathematics of linear perspective, and Cosgrove and his colleagues showed that this visual ideology emerged in Europe in the fifteenth and early sixteenth centuries and, in line with the developing materialism of human geography, that it was saturated in the tonalities of early capitalism. Its trick of representing three-dimensional space on a two-dimensional surface was achieved by rendering space ‘the property of an individual, detached observer, from whose divine location it is a dependent, appropriated object.’ For Cosgrove, this vision asserted a class privilege; it was a bourgeois gaze whose production simultaneously declared possession, delimited property, and depended on patronage. 88 Rose soon added a powerful rider: this was also a profoundly gendered way of seeing, and functioned to naturalize the masculinism of the gaze. 89 These interventions stirred up the complacent view that limited geography’s ‘art of landscape’ to 87 Peter Merriman, ‘Driving Places: Marc Augé, Non-places, and the Geographies of England’s M1 Motorway’, Theory, Culture & Society, 21(4&5) (2004) 145–167. 88 Cosgrove, ‘Prospect, perspective’; it could also be an imperial gaze: for an elaboration, see Judith Kenny, ‘Climate, Race, and Imperial Authority: The symbolic landscape of the British Hill Station in India’, Annals of the Association of American Geographers, 85(4) (1995) 694–714. 89 Rose, Feminism and geography, pp. 86-112. 54 the field sketch, but other human geographers worried about the danger of losing the physicality of the landscape in the chase after its representations. Don Mitchell argued for a re-cognition of its material presence: landscapes, he claimed, in a remark that would not have been out of place in an older school of historical geography, are the products of work, ‘of human labour.’ But he left that tradition in the dust when he also claimed that landscape does work. Cosgrove’s deep interest in visual ideology was relocated to a brilliantly physical register. ‘In many respects,’ he explained, landscapes produced under capitalism – what Harvey calls landscapes of accumulation – are like the commodity: they conceal (‘fetishize’) the labour that goes into their making. When he writes about ‘the lie of the land’, therefore, Mitchell is simultaneously reactivating a traditional concern with the forms and features of the visible landscape and drawing attention to its duplicity. His purpose is to unsettle, disrupt and call into question the outer unity of landscape – its aesthetic harmony and ‘natural’ integrity – and disclose the struggles and conflicts that seethe below its surface. 90 But when John Wylie turned to ‘practices of landscape and, especially, towards the simultaneous and ongoing shaping of self, body and landscape via practice[s] and performance[s]’, he did not have in mind the experiences of the migrant workers whose lives (and deaths) are concealed in the agrarian landscapes of southern California. His project was based on the walker’s or traveller’s corporeal encounter with a landscape. In one way, his project unsettles the visual ideology that was the object of Cosgrove’s critique, 90 Don Mitchell, ‘Dead Labor and the Political Economy of Landscape – California Living, California Dying’, in Kay Anderson et al. (eds), Handbook of Cultural Geography (London: SAGE, 2002), pp. 233–248; see also his The lie of the land: migrant workers and the California landscape (Minneapolis: University of Minnesota Press, 1996). 55 because it ‘turns landscape from a distant object or spectacle to be visually surveyed to an up-close, intimate and proximate material milieu of engagement and practice.’ And yet, when landscape thus becomes ‘the close-at-hand, that which is both touching and touched, an affective handling through which self and world emerge and entwine’, it is surely not difficult to translate this into Mitchell’s migrants feeling the earth of the California valleys trickle through their bruised fingers or the stonemasons running their hands over the blocks of Brunelleschi’s dome that is the focal point of Cosgrove’s account. 91 Either way, landscape, like place, becomes alive. Regions and territories were retained by spatial science but in purely formal terms, as nodal regions or marketing territories that marked ‘spheres of influence’ whose ‘influence’ was strangely purged of any tincture of political or economic power. This too has been dramatically reversed. A revitalized history of geography has shown that regional geography had a strategic dimension from its very beginning. When Strabo developed the classical Greek conception of chorography, of regional description, he wrote as an admirer of the Roman emperor Augustus and his successor, Tiberius, and his Geography was intended to be of direct service to imperial administrators and military commanders. 92 Fast forward to the twentieth century, and a traditional regional geography was pressed into military 91 John Wylie, Landscape (London: Routledge, 2007) pp. 166-7; see Kenneth Olwig, ‘Performing on the Landscape versus Doing Landscape: Perambulatory Practice, Sight and the Sense of Belonging’, in Tim Ingold and Jo Lee Vergunst (eds), Ways of walking: Ethnography and Practice on Foot (Place: Ashgate, 2008), pp. 81–91. This is a suggestion – Wylie’s own approach is resolutely personal, and there is more than a hint of narcissism in some versions of non-representational theory in which the world is made to revolve around the researcher’s own experience. 92 Daniela Dueck, Strabo of Amasia: a Greek man of letters in Augustan Rome (London: Routledge, 2000) 56 service during the two World Wars and the Cold War. 93 In both cases these enlistments also required a technical capacity to fix and locate: in a word, to map. One might say much the same about the concept of territory, which relies on a strategic discourse and a political technology too, except that its conceptual armature is also wrapped in legal formularies. 94 Critical human geography today is more likely to resist these deadly complicities, and its conceptualizations of region and territory share in the general ‘enlivening’ that we have identified for space, place and landscape. Regions are now rarely seen as so many building blocks, a device that is at once partitional (it assumes that the world can be exhaustively divided into bounded spaces) and aggregative (these spaces can be fitted together to form a larger totality). Our present understanding of regions suggest that they have never been closed, cellular spaces, and that much of traditional regional geography – rather like traditional descriptions of ‘place’ – may turn out to have been about inventing a ‘traditional’ world of supposedly immobile, introspective and irredeemably localized cultures. Many anthropologists, geographers and historians now accept that non-capitalist societies have always been actively engaged in other worlds, and that they always been constituted through their involvements in more extensive networks. Those involvements have been intensified by capitalist modernity, even transformed by them, but they are not the creation of the 93 Barnes and Farish, ‘Between regions’, below. 94 Stuart Elden ‘Land, Terrain, Territory’, Progress in Human Geography, 34(6) (2010): 799–817; see also Elden, Terror and territory: the spatial extent of sovereignty (Minneapolis: University of Minnesota Press, 2009) and his The birth of territory (forthcoming), which provides a genealogy of the concept from the classical world through to early modern Europe. For a still wider view of Elden’s interests, his blog Progressive Geographies, at 57 modern. There is a broad consensus that regional formations are more or less impermanent condensations of institutions and objects, people and practices that are intimately involved in the operations and outcome of local, trans-local and trans-regional processes. Once regions are conceptually ‘unbound’, as Ash Amin puts it, then bounding and b/ordering become the precarious historico-geographical achievements of political and military, economic and cultural power. 95 Similarly, territory comes to be seen as what Stuart Elden calls ‘a historical question: produced, mutable and fluid’ but also simultaneously as a geographical question ‘not simply because it is one way of ordering the world, but also because it is profoundly uneven in its development.’ 96 For much the same reason, many human geographers have become much more attentive to the ways in which these scalar distinctions have been produced and to their sedimentations in imaginative geographies and public policies. 97 All of this has produced new ways of writing about regions and borderlands. Soja’s early experiment in ‘taking Los Angeles apart’ has been followed by a host of others, many of them multi-media presentations that draw on film, video and music too. Others make use of the interactivity of new media, including blogs, to present multi-perspectival views of places to wider publics. These are by their very nature usually non-disciplinary or 95 Ash Amin, ‘Regions Unbound: Towards a New Politics of Place’, Geografiska Annaler, 86B(1) (2004) 33–44; Anssi Paasi, ‘Bounded Spaces in the Mobile World: Deconstructing “Regional Identity”’, Tijdschrift voor Economische en Social Geografie, 93(2) (2002) 137–148.. 96 Elden,‘Land, terrain, territory’. 97 Sallie A. Marston, John Paul Jones III and Keith Woodward ,‘Human Geography without Scale’, Transactions of the Institute of British Geographers, NS, 30 (2005) 416– 432. 58 interdisciplinary projects, but the ‘problem’ of geographical description has never seemed less of a problem and more of an opportunity. 98 These experiments with fluidity, mobility and hybridity can seem intoxicating, which makes it all the more important not to lose sight of their other dimensions. By this we mean not only the fixities and immobilities that limit the lives of millions of people – a counter-geography to the ‘liquid world’ celebrated by those with the freedom to take advantage of it: again, we draw attention to the figure of the migrant and the refugee 99 – but also the violence and immiseration that often inheres in these paradoxical spaces. Gloria Anzaldúa famously described the US-Mexico borderlands as ‘an open wound’ where ‘the Third World grates against the First and bleeds.’ Before a scab can form, she continued, ‘it hemorrhages again, the lifeblood of two worlds merging to form a third country – a border culture.’ Matthew Coleman thus sees the US-Mexico border as a trickster figure, at once being opened to the passage of capital and (licit) commodities under the sign of neoliberalism and closed to the movement of ‘undocumented’ migrants who are often the victims of neoliberalism. For him, the border is performed through the countervailing operations of ‘de-bordering’ and ‘re-bordering’, in the course of which Anzaldúa’s sanguinary metaphor has become ever more appropriate as the border is increasingly militarized as part of the ‘war on drugs’ that since 9/11 has morphed into the ‘war on terror’. 100 On those 98 Edward Soja, ‘Taking Los Angeles Apart: Some Fragments of a Critical Human Geography’, in his Postmodern Geographies: The Reassertion of Space in Critical Social Theory (London: Verso, 1989) pp. 190–221. 99 Mountz, ‘Stateless by design’. 100 Gloria Anzaldúa, Borderlands/La Frontera: the new mestizo (San Francisco CA: Aunt Lute Books, 1987); Matthew Coleman, ‘US statecraft and the US-Mexico border as security/economy nexus’, Political Geography 24 (2005) 185-209; see, more generally, 59 borders where military violence is an ever-present reality, where the fixity of the line has yielded to the fluidity of attack and counter-attack, the result is often a ‘space of exception’ in which legal protections are removed from people who are knowingly exposed to death. As Derek Gregory has shown, across the Green Line from Israel into the occupied Palestinian territories on the West Bank of the Jordan, these zones of indistinction have proliferated and Palestinians are trapped in ‘a frenzied cartography of mobile frontiers’ in which time and space are twisted and deformed so that ‘nothing is fixed, nothing is clear’. Or again, across the Durand Line from Afghanistan into Pakistan’s Federally Administered Tribal Areas, a liminal space that has been turned into a war zone, ordinary people are exposed to death from orbiting drones as the United States seeks to find and kill Taliban insurgents and members of al-Qaeda. 101 These may all be ‘lively’ spaces but they are also, and by virtue of their liveliness, spaces of death. Nature, environment and the non-human The Great Divide What David Livingstone once called ‘the geographical experiment’ was founded on a Big Idea: that it was possible to examine the relations between people and their environments, or ‘culture’ and ‘nature’, within the confines of one great scientific enterprise. There has long been John Agnew, ‘Borders on the Mind: Re-framing Border Thinking’, Ethics & Global Politics, 1(4) (2008) 175–191. 101 Derek Gregory, ‘Besieging Cartographies’, in his The Colonial Present: Afghanistan, Palestine, Iraq (Oxford: Blackwell, 2004) pp. 117–138; Derek Gregory, ‘The everywhere war’, Geographical Journal 177 (2011) 238-50. 60 disagreement over the nature of that enterprise – as Livingstone shows, it’s an essentially contested tradition 102 – but for much of the eighteenth and nineteenth centuries it was conducted within what David Stoddart called the ‘great tradition’ of the natural sciences. As we’ve explained, this is a story that can be told in many different ways. Stoddart invoked a radically different lineage from what he dismissed as ‘conventional wisdom’, whose heroes were ‘the Ritters, Ratzels, Hettners’. These were three largely desk-bound German geographers who were admitted to Hartshorne’s canon, not least through their ‘spatial’ predilections, but Stoddart’s oppositional sense of geography as an intellectual enterprise sprang from the exploratory field sciences of Georg Forster, Charles Darwin and T.H. Huxley: ‘and it works,’ he added triumphantly. 103 One would surely have to add Humboldt to this trinity, whose vision of the Cosmos was an exquisite version of the exploratory tradition that Stoddart eulogized and the synoptic project of geography that he too endorsed. But even those who follow different routes, perhaps closer to Hartshorne’s, that by the end of the nineteenth century tracked through the humanities and the emergent social sciences, would concede that these were profoundly affected by the natural sciences. These other routes turned out to vitally important because they led from a pre-disciplinary world – in which the scholar roamed the world and roamed the library in equal measure – to a more cloistered, disciplinary world. By then, 102 David Livingstone, The geographical tradition: episodes in the history of a contested enterprise (Oxford: Blackwell, 1993 ); see especially Ch. 6, ‘The geographical experiment’. 103 Stoddart, On geography. For a more recent (and more rigorous) investigation of field science and its claims to objectivity, a world away from the South Pacific that enthralls Stoddart, see Richard Powell, ‘“The Rigours of an Arctic Experiment”: The Precarious Authority of Field Practices in the Canadian High Arctic, 1958–1970’, Environment and Planning A, 39 (2007) 1794–1811. 61 modern geography had gained a significant but small foothold in English-speaking and European universities, and its practitioners more or less agreed that that geography crossed an emerging divide between what today we would call the earth and environmental sciences (including geology, geomorphology, botany and climatology) on one side and the humanities and social sciences (including history, anthropology, economics and sociology) on the other. As the young Halford Mackinder put it in his 1887 address to the Royal Geographical Society in London, geography would bridge ‘one of the greatest of all intellectual gaps’: it would, he advertised, ‘trace the interaction of man in society and so much of his environment as varies locally’. 104 The challenge was enormous. Given the vast range of topics that geography claimed to cover, what ideas and approaches could it claim as its own, and what signature insights could it provide? Could it equal the intellectual status of the specialist sciences like physics or economics by somehow making a virtue of its totalizing or synthesising perspective on the world? These questions did not apply only to geography, since it was closely allied to an anthropology that also placed the relations between people and their environment at the centre of its concerns. Even so, the early university geographers were in select company: few of their academic peers chose to make the human and biophysical worlds – from the scale of everyday life up to the globe – the combined focus of their inquiries. 104 Halford Mackinder (1887) ‘On the scope and methods of geography’, Proceedings of the Royal Geographical Society, 9: 141-60: 143. 62 Consistent with these grand ambitions, Mackinder and his colleagues refused any clear distinction between ‘human’ and ‘physical’ geographies. Although Mary Somerville had devoted a book to physical geography in 1849 – defining it as ‘a description of the earth, the sea, and the air, with their inhabitants animal and vegetable, of the distribution of these organized beings, and the causes of their distribution’ 105 – few university geographers followed her lead, and most tended to examine human populations in relation to their immediate biophysical surroundings. These studies took several forms. First, there were detailed investigations of different places and regions, the sort of work that Hartshorne would later represent as the core of geographical inquiry. They typically sought to demonstrate the distinctive connections between economy, politics, society and culture and the physical landscape, climate and relative location of an area. 106 Second, there were much grander attempts to describe the world’s geography as a patchwork of biophysical regions that were associated with characteristic patterns of human habitation. These authors – like Ellen Semple in her Influences of geographic environment (1911) – took their readers on a grand tour of the earth’s differentiated surface but, like most tourists, paid selective attention to the places they visited. 107 Third, some geographers, working on an equally large canvas, focused on one element of human practice (such as state-craft) and related it to environmental causes and circumstances. Mackinder’s own Britain and the British Seas (1902) is one example among 105 Mary Somerville, Physical geography (London: John Murray, 1849). 106 See, for example, H.J. Fleure, Wales and her people (Wrexham: Hughes & Son, 1926). 107 Ellen Semple (1911) Influences of Geographic Environment (New York: Henry Holt, 1911). 63 many: it was an attempt to understand the geopolitical manoeuvrings of Britain as a maritime nation.108 All of these studies were highly descriptive; the early university geographers found it virtually impossible to treat causation and process in any detail because they tried to cover so much ground, often on a planetary scale. Where explanations were ventured they were typically speculative, even breezy, and possessed none of the rigour of the spatial science that would eclipse them after 1945. Yet these speculations were offered with supreme confidence, as if they possessed substantial empirical warrant. This can in part be explained by a combination of Darwinian thinking and neo-Lamarkianism that was a major influence on intellectual culture at the turn of the twentieth century (in fact, this is what Livingstone had in mind when he invoked the geographical experiment). Like many other American and European intellectuals, university geographers saw no problem in extending Darwin’s paradigmatic inquiries into species evolution to humans. 109 The ‘races of man’, as they were called, were seen as the products of adaptation to more or less favourable natural environments, with Jean-Baptiste Lamarck’s thesis about ‘fast evolution’ pressed into service (implicitly or otherwise) to argue that Anglo-Europeans had been able to progress more rapidly than other ‘races’ with less advantageous physical geographic 108Halford Mackinder (1902) Britain and the British Seas (Oxford: Clarendon Press); se Kearns, ‘Political pivot’ below. 109Charles Darwin (1859) The Origin of Species (London: John Murray). Darwin was very careful not to generalise his findings to homo sapiens, beyond the key insight that humans are as much products of natural history as any other animal, insect, vegetable or microbiological species. See D.R. Stoddart, ‘Darwin’s impact on geography’, Annals of the Association of American Geographers 56 (1966) 683-9; David Livingstone, ‘The geography of Darwinism’, Interdisciplinary Science Reviews 31 (2006) 32-41. 64 conditions. The result was a curiously asymmetrical application of what became known as environmental determinism. Europeans were held to have domesticated their temperate, ‘normal’, so to speak ‘natural’ natures, obliging them to give up their secrets to Science and their energies to Industry, while other cultures were held to be creatures of their non-temperate, abnormal and even ‘un-natural natures’.110 To describe the asymmetry in these terms is to reveal its abiding racism: thus, for example, Ellsworth Huntington’s reflections on The character of races (1924) claimed that various ‘natives’ in the world’s tropical and ‘frigid’ zones could never rise above the challenging environments that made them what they were. Regrettably these were not exceptional views; anxieties about the dangers of tropical nature for ‘temperate’ cultures bedevilled the colonial and imperial project, and at that time eugenics – the ‘science of racial improvement’ – enjoyed widespread respectability before Hitler’s Third Reich took it to its still more hideous and ultimately genocidal conclusion. By the time Hartshorne visited Nazi-occupied Vienna, the prospects for the unified Geography that had captivated scholars in a pre-disciplinary world were dimming, even though Hartshorne and his critics continued to affirm their faith in human and physical geography as conjoint moments in disciplinary inquiry. Dramatic advances were being made elsewhere in the 110 This is a corollary of the discourse of tropicality (above, p. 00); see also Livingstone, ‘Tropical hermeneutics’. In his prescient contributions, American George Perkins Marsh had reversed the causal arrows and pointed – at early as the 1860s – to the large-scale impacts of humans upon their natural environments. Few others in Geography’s early decades appeared to share his concern, despite clear evidence – in late nineteenth-century North America, for example – of ‘ecocidal behaviour’ (on that continent European immigrants succeeded in virtually exterminating bison and north Pacific fur seals, among other species). 65 academy in the natural sciences and the social sciences – from Einstein’s epoch-making discoveries in physics to J.M. Keynes’ seminal contributions to economics – and these had thrown geography’s shortcomings into stark relief and threatened to consign it to a backwater far from the mainstream of scientific progress. The predicament was heightened during the Second World War, when many geographers on both sides of the Atlantic served in the military and intelligence services and learned an object lesson in the importance of precision and measurement, systematicity and objectivity.111 But there were other pressures from other directions. Many philosophers had argued that the physical sciences were intrinsically different from the humanities and the social sciences by virtue of their subject matter. The geographical experiment was doomed to failure: there could be no unified science of people-and-nature because people were distinctly different from rocks, rivers or ravines. Unlike atoms or molecules, people inhabit a world of socially constructed meanings that are indispensable for its interpretation and transformation; they are able to create their own history and geography in reflexive, conscious and even unconscious ways that are unavailable to stones rolling along riverbeds. Though cusped between biology and culture, people are not reducible to their genes or their physiologies, let alone to forces exerted by the physical environment. This was not only a rebuff to environmental determinism and its derivatives but also a counter-claim that the ‘non-biological’ aspects of humanity could be analysed in their own right 111For instance, Edward Ackerman published an essay entitled ‘Geographic training, wartime research and immediate professional interests’ in America’s premier geographical journal: Annals of the Association of American Geographers 35 (2) (1945) 121-43. This was something of a manifesto written on behalf of a generation of future academic geographers in the US whose entire outlook had been altered by their war-time service. 66 (which is, of course, what sociologists, economists, musicologists, art historians and cultural anthropologists had been arguing for years). After 1945 academic Geography splintered into human geography and physical geography, with each fragmenting into a series of systematic sub-disciplines. The terms ‘human geography’ and ‘physical geography’ began to name two distinct projects within a single disciplinary space. Other disciplines were bi-polar too – physical and social anthropology for example – and, like them, geography gave these distinctions a substantive rather than purely nominal significance. Perhaps more importantly, these were all so many versions of an even greater divide in a minor key. In 1956 C.P. Snow sketched the outlines of a thesis about ‘two cultures’ that, just a few years later, he would delivered as a lecture that would cause a sensation. 112 Snow had trained as a physicist in the 1920s, and during the war served as Chief of Scientific Personnel for the Ministry of Labour in Great Britain; after the war he embarked on a parallel, highly successful career as a novelist. ‘By training I was a scientist,’ he later wrote, ‘but ‘by vocation a writer.’ In his 1959 Rede Lecture at Cambridge, ‘The two cultures and the scientific revolution’, Snow declared that he had felt as though he were ‘moving among two groups’ who ‘had almost ceased to communicate at all’, and he was quick to add that this was not a personal odyssey but a pervasive feature 112 That same year a major collection of essays provided a spirited case for a continued (or renewed) conversation. William L. Thomas (ed), Man’s role in changing the face of the earth (Chicago: University of Chicago Press, 1956) was derived from symposium that drew seventy scholars together from more than 20 disciplines, including human geographers like Andrew H. Clark, H.C. Darby, Estyn Evans, Clarence Glacken, Pierre Gourou and Carl Sauer, earth scientists and climatologists like Luna Leopold, Arthur Strahler and C.W. Thornthwaite, and luminaries like Pierre Teilhard de Chardin and Lewis Mumford. 67 of intellectual life in the West. ‘At one pole we have the literary intellectuals,’ he explained, and at the other ‘scientists, and as the most representative the physical scientists.’ A ‘gulf of incomprehension’ lay between them – ‘sometime hostility and dislike, but most of all lack of understanding. They have a curious distorted image of each other.’ Snow thought this a tragedy, an exceptionally destructive loss to both sides that was particularly damaging at the height of the Cold War. Insisting that ‘scientific culture really is a culture,’ Snow complained that ‘there seems to be no place where the cultures meet.’ 113 And yet, of course, geography had long claimed to be that place. It would be absurd to read the rise of spatial science as a response to Snow, though his arguments were (and remain) influential, but it did seek to capitalise on the second part of Snow’s title – ‘the scientific revolution’ – and to show that his ‘scientific culture’ was capable of addressing both the human and the physical worlds. Its inspiration was not Snow, however, but Thomas Kuhn, whose account of The structure of scientific revolutions just three years later provided a powerful rhetorical model for the ‘Quantitative Revolution’. This was ironic; Kuhn’s account fastened on the physical sciences – in fact, he wondered if it was not limited to the physical sciences – but he drew on a series of methods from the humanities to develop his 113 C.P. Snow, The two cultures and the scientific revolution (Cambridge and New York: Cambridge University Press, 1959). There are intriguing parallels between Snow and Donna Haraway; trained in zoology and biology, her first major study addressed the functions of metaphor in shaping biological research, and her subsequent writings on techno-science and techno-culture reveal a dazzling capacity to traffic in the space between the arts and the sciences. She opens a ‘meeting place’ too, but whether Snow would look through the window let alone go in the door is a more difficult question. 68 concept of a ‘paradigm’ for scientific inquiry. 114 During the Quantitative Revolution, however, geography turned its face – or, more accurately, the face of most of its practitioners – from the stuff of the ‘literary humanities’ to the geometries of the earth’s surface. In fact many of the models of spatial science were derived from the physical sciences: the gravity model calibrating the ‘friction of distance’ on the interaction between two locations is the most obvious, but the neoclassical economics on which most standard location theory depended was closely related to statistical mechanics. If this achieved a precarious unity between human and physical geography, however, it was remarkably short-lived. During the 1960s, two British geographers, one a human geographer (Peter Haggett) and the other a physical geographer (R.J. Chorley), tried to jumpstart the stalled geographical experiment. In a series of publications they proposed to unite human and physical geography through a common object (spatial order), a common method (the ‘scientific method’), and a common conceptual apparatus (systems analysis). 115 But the project was abandoned. On one side, physical geographers retained (and in fact reinforced) their commitment to science, though now usually phrased in different terms 114 T.S. Kuhn, The structure of scientific revolutions (Chicago: University of Chicago Press, 1962); the book had its origins in a course in science that Kuhn was required to give to arts majors, and he subsequently explained that it was the enormity of the task that prompted him to turn to familiar ground for his students – history – and to take the model of a ‘paradigm’ from language-learning. 115 The flagship of the enterprise was R.J. Chorley and Peter Haggett (eds) Models in Geography (London: Methuen, 1967), but see also Peter Haggett and R.J. Chorley, Network analysis in geography (London: Arnold, 1969). The last hurrah was R.J. Bennett and R.J. Chorley, Environmental systems: philosophy analysis and control (London: Methuen, 1978). 69 that turned from positivism to non-positivist philosophies of science and directed attention from form to process. This is not the place to review the increasingly separate history of the subdiscipline, but in general physical geographers directed their research and teaching to fact-based descriptions, explanations and predictions of earth surface phenomena. Specialisation, new databases and remote sensing capabilities, new field techniques and computer technologies, and new physical and mathematical models made this possible, but the price was, at first, further internal division. Physical geographers divided their field into five major areas – geomorphology, biogeography, climatology, hydrology, and Quaternary environmental change – and fostered increasingly close connections and collaborations with scientists in cognate fields like geology, botany and atmospheric science. It is only recently, and in part through these extra-disciplinary conversations, that reintegration has been set in motion through avowedly interdisciplinary projects like Earth Systems Science and the rise of the ‘biogeosciences’ (particularly in North America). 116 On the other side, the critique of spatial science drew many human geographers deeper into the modern social sciences while at the same time prompting a series of calls for the traditional ties with the arts and the humanities to be reaffirmed (though not always in traditional ways). Here 116 For brief introductions to physical geography see Keith Richards, ‘Geography and the physical sciences tradition’, in Nicholas Clifford, Sarah Holloway, Stephen Rice and Gill Valentine (eds) Key Concepts in Geography, 2nd edition (London: Sage, 2009), pp. 21-45; Peter Sims, ‘Previous actors and current influences: trends and fashions on physical geography’ in Stephen Trudgill and Andre Roy (eds) Contemporary Meanings in Physical Geography (London: Arnold, 2003) pp. 3-24; and Nick Clifford’s entry on ‘physical geography’ in Gregory, Johnston, Pratt, Watts and Whatmore, eds, Dictionary of human geography. Significantly, this fifth edition was the first to include an entry on Physical Geography – and the first to include an entry on Geography. 70 too the consequence was a series of divisions. One axis was sub-disciplinary – the formation and consolidation of separate economic, political, social and cultural geographies, for example, with often only historical geography and historico-geographical materialism to muddle things up – and the other was procedural: a divergence between a spatial-analytic geography, markedly less interested in geometric order than its predecessors, and a social-theoretic geography, much more interested in political critique. In the face of such diversity, it was sometimes hard to see the forest for the trees – and the different woodcutters hacking away at them (though, in parallel with physical geography, close connections were developed with economists, political scientists, sociologists, anthropologists, historians and literary scholars). In fact, however, most of us saw neither the forest nor the trees. Human geography typically abstracted economic, political, social and cultural practices from their biophysical circumstances, an abstraction indexed most visibly by the rise of an urban geography that analysed the created, ‘artificial’ environment of the post-war metropolis. Human geography was ‘de-naturalised’, a process that was equally apparent in the humanities and social sciences from which it drew its inspiration, while physical geography – apart from consultancy projects to do with problems of coastal management, soil erosion slope failure and the like – was effectively ‘de-socialised’. Where the human-environment nexus remained a subject of concern – as in Gilbert White’s research into how people perceive and respond to the threat of natural hazards – it was conducted in a way that reflected the commitment to a model of science that much of human geography had abandoned. Even when the environmental movement took its first steps, few geographers trailed along. The era of the 71 first Earth Day, when Greenpeace and Friends of the Earth were founded, did not inspire a revival of human-environment study in geography. Instead, people like White and his students went about their business without generating any sea-change in the topical focus of their colleagues in human and physical geography. These differences were reflected and reproduced in the divergent publication practices of human and physical geographers: they increasingly turned the pages of different journals (or different pages in the same journal), and for every supposedly general journal there seemed to be at least ten specialist journals in the different sub-disciplines of physical and human geography. In the United States many programmes in geography were predominantly and sometimes exclusively programmes in human geography, while in most Scandinavian countries human and physical geographers occupied separate departments within the same university (and still do). Perhaps unsurprisingly, by the mid-1980s, many geographers started to wonder if there was any longer any reason to think that human and physical geography could be sustained as two halves of a unitary disciplinary field.117 The re-naturalisation of human geography Over the last twenty-five years there has been a gathering reaction to the ‘denaturalisation’ of human geography that has gained momentum until now nature, in all its attendant varieties, is one of the central terms of contemporary human geography. The reasons for this are as much external 117 Johnston, ‘Geography – coming apart at the seams?’, below. 72 and internal, and here as elsewhere there is no simple separation between text and context, inside and outside. Environmental incidents – oil spills, species extinctions, landslides, earthquakes, tsunamis and much more besides – have rarely been out of the news this last quarter century. Less dramatically, but no less seriously, the evidence for human-induced climate change is now unequivocal. The growth in global population and levels of consumer demand – within the context of capitalist globalization – have ramped up levels of natural resource use, producing relative scarcity, price fluctuations and vast volumes of waste. The technical prowess of applied science is now such that it can splice genes and clone organisms, challenging ethical norms about how we should regard our own biological ‘nature’ and that of non-human species. Environmental protest movements and pressure groups remain as visible now as they were during the era of the first Earth Day, four decades ago. Research councils and organisations funding the full range of university disciplines have promoted research into the relations into the society-environment nexus, as well as the wider impacts of the life and biomedical sciences. The cultural critic Raymond Williams once described Nature as one of the most complicated words in the English language, so it is not surprising that, in responding to all the predicaments and possibilities sketched in the last paragraph, ‘nature’ should have been given various interpretations and required the supplements we have signposted: ‘environment’ and the ‘non-human’. The latter is not simply a synonym for the other two; it describes all those densely material phenomena – from buildings to domestic gardens to commodities – that are neither strictly ‘natural’ nor part of ‘the environment’ in its conventional sense. These bear the marks of human intentionality in 73 their creation, use and meaning but, as with all materials, they have a specific texture, shape, composition and efficacy of their own. Indeed, this is so important that some commentators have described this as a process of ‘re-materialisation’ too, by which they mean a new focus on the substance of the world, including our own bodies. The re-naturalisation/materialisation of human geography through work conducted under these three banners has deepened and widened our understanding of the field. But it has not completely dug it over: the roots of these changes lie, in part, in research traditions that were, if not exactly fallow, then cultivated at the margins. But they also involve the creation of new hybrids by grafting a concern with the biophysical world onto (and into) theories and perspectives that previously paid remarkably little attention to that world. There are two modern baselines for today’s re-naturalised approaches: research into hazards and studies of land use change. The contributions of Gilbert White and his students to our knowledge of ‘natural hazards’ in the decades following the Second World War were of considerable practical importance; they interrogated the ways in which communities calculated and responded to risks from floods, droughts, earthquakes and other hazards, and this was of obvious and direct interest to international organizations, state agencies and insurance companies. Yet by the end of the twentieth century more and people were being harmed by ‘natural disasters’ despite decades of research into risk minimization and mitigation. In an important critical intervention Kenneth Hewitt and his collaborators argued that the root of the problem was the term itself. ‘Natural’ and ‘disaster’ had to be prised apart 74 because so-called ‘natural’ events like flooding may be caused in part by human action (or inaction), but the epicentre of the ‘disaster’ was almost always political and social. Restricting policy prescriptions to technocratic solutions like flood defences or zoning restrictions on building, so they claimed, deflected attention from the differential vulnerability of populations to hazard events. 118 This critique was of double significance. It reaffirmed the importance of what, for those who knew their Marx, was his materialist dialectic between ‘nature’ and ‘society’, which was already providing the mainspring for projects from political economy through to political ecology. These in their turn would provoke a new round of critical responses from cultural constructions of what counts as ‘nature’ to a revitalized analysis of human populations – and life itself – under the sign of what Foucault called biopolitics. This critique also reinforced the developing political and ethical sensibilities of human geography. It required the category of people ‘at risk’ to be deconstructed by locating the space of vulnerability (and the space of resilience) within a socio-economic matrix of inequality and information. It also demanded an involvement with politics as much as policy or, rather, an awareness that policy, in both its formulation and its implementation, is never a narrowly managerial exercise involving expert knowledges but is also always a profoundly political practice. 119 And yet it is important not to ignore the continued development and even enlargement of the technical base for studying these questions. The 118 Kenneth Hewitt (1983) Interpretations of a Calamity (New York: Allen & Unwin, 1983). 119 Piers Blaikie, Terry Cannon, Ian Davis and Ben Wisner, At risk: natural hazards, people’s vulnerability and disasters (London: Routledge, 1994); Michael Watts and Hans Bohle, ‘The Space of Vulnerability: The Causal Structure of Hunger and Famine’, Progress in Human Geography, 17(1) (1993) 43–67. 75 formation of two avowedly interdisciplinary fields, ‘land change science’ and ‘sustainability science’, has depended on the use of remote sensing imagery and GIS techniques to create, manipulate and analyse macro-scale, multi-dimensional data bases that capture, display and monitor land use and other environmental changes. 120 Other human geographers have been drawn into other large-scale projects to gauge the human involvements (not merely impacts) of global climate change. The ultimate objective of these ‘Big Science’ projects is to record humanity’s ecological footprint on the planet and, like White’s work, they have been of considerable importance to policy makers, state bureaucracies and government officials. They have been a particularly important (re)source for international and state actors seeking to manage, conserve and preserve areas of the biophysical world deemed to be of special ecological, aesthetic or cultural value (the iconic example is the Amazon rainforest). The provision of expert knowledges does not make these scholars politically reactionary – though when their parent sciences turn into Sciences (with that imperial capital again) that discount lay or indigenous knowledges it is not surprising they should attract spirited criticism 121 – and there have been fruitful reciprocal exchanges between them and an overtly political ecology; but neither does it make these scientists ‘neutral’, providers of ‘facts’ free from political judgements of the sort that state actors are charged with making. 122 120 B. L. Turner II, ‘Land change (systems) science’, in Noel Castree, David Demeritt, Diana Liverman and Bruce Rhoads (eds), A Companion to Environmental Geography (Oxford: Wiley-Blackwell) pp. 167-180, and Robert Kates, William Clark, and others, ‘Sustainability science’, Science 292 (2001) 641–642. 121 James Fairhead and Melissa Leach, Reframing deforestation (London: Routledge, 1998). 122 B. L. Turner II and Paul Robbins. ‘Land change science and political ecology, Annual Review of Environment and Resources, 33 (2008) 295-316. 76 Other approaches have addressed the politics of environmental change (and, ultimately, of ‘nature’) much more directly. Two derive directly from human geography’s engagement with historical materialism. Many of its early conversations were strangely silent about the nature-society dialectic, which usually appeared only in preliminary and usually abstract accounts of the ‘material base’ of the mode of production where it served to prepare the ground for the analytics of the capitalist production of space. This state of affairs was disrupted with extraordinary brio when Neil Smith insisted that the production of space could not be understood apart from what he called ‘the production of nature’. 123 The recognition that capitalism produced – not merely dominated, exploited or appropriated – nature was central to the political economy of environment and resources and to a self-consciously political ecology. The first of these typically focused on specific ‘regional capitalisms’ – on the co-production of particular spaces, exemplified by Richard Walker’s study of California’s ‘Golden Road to Riches’ (for some, at any rate) which lay through minerals, forests and water, or by Michael Watts’s compelling studies of petro-capitalism in the Niger Delta 124 – but it also spiralled beyond the circles, cycles and crises of capital accumulation to capture the physical presence of the non-human world: for Watts, ‘the devil’s excrement’ that was oil, or for Scott Prudham the spotted owl that 123 Neil Smith, Uneven Development (Oxford: Blackwell, 1984); for a summary, see Neil Smith, ‘The Production of Nature’, in George Robertson, Melinda Mash, Lisa Tickner, Jon Bird, Barry Curtis and Tim Putnam (eds), FutureNatural: Nature, science, culture (London: Routledge, 1996) pp. 35–54. 124 Richard Walker, ‘California’s Golden Road to Riches: Natural Resources and Regional Capitalism, 1848–1940’, Annals of the Association of American Geographers 911 (2001) 167–199; Michael Watts, ‘Resource Curse? Governmentality, Oil and Power in the Niger Delta, Nigeria’, Geopolitics 9 (2004) 50–80. 77 raised a hue and cry over the exploitation of old-growth forests in the Pacific Northwest. 125 The animating concept for these studies was the commodification of nature. That this was a process pre-existing the present bears emphasis, and as such it overlaps with histories that reach back far beyond the eighteenth and nineteenth centuries. These were meat and drink to a distinguished tradition of research into agrarian-ecological change in cultural geography, but this is now reinforced by research in environmental history that traces the paths through which resources have been turned into commodities. In a seminal study of the ecological transformation of New England between 1600 and 1800, William Cronon described in exquisite detail how ‘changes in the land’ were brought about by labile interactions between indigenous peoples and European settlers that culminated, as Edward Johnson wrote in 1653, in ‘the wilderness turned a mart’: the ‘wilderness’ turned into a market. 126 The process, in part (but only in part) a conjunction of economic and ecological imperialism, intensified with industrialization, including the industrialization of agriculture itself, which later prompted Cronon to describe Chicago, the artificial heart of the Great Plains, as ‘Nature’s 125 Scott Prudham, ‘The political economy of an ecological crisis’, in his Knock on Wood: nature as commodity in Douglas fir country, New York: Routledge, 2003; Michael Watts, ‘Oil as Money: the devil’s excrement and the spectacle of black gold,’ in Stuart Corbridge, Ron Martin and Nigel Thrift (eds) Money, power and space (Oxford: Blackwell, 1994) pp. 406-45. 126 William Cronon, Changes in the land: Indians, colonists and the ecology of New England (New York: Hill and Wang, 1983). Cronon emphasizes the co-production of the land, but draws an important social distinction: ‘Ironically, though colonists perceived fewer resources in New England’s ecosystems than did the Indians, they perceived many more commodities.’ 78 metropolis’. 127 The cross-fertilisations between these fields have been immensely fruitful, spawning investigations into commodity chains like the contemporary ‘agro-food’ networks that connect investors, farms, seed firms, pesticide manufacturers and others with (usually) far distant food consumers.128 By revealing the socio-spatial relations that materially enact the passage from resource to commodity these studies have shown how the biophysical world has become a means not an end in a process that produces not only nature but also social and environmental injustice. These consequences, it is now clear, cannot be attributed simply to ‘corrupt’ politicians or to ‘greedy’ elites – even when corruption and greed are plain to see – but are systemic, rooted in the basal logics of political economy, even in the most war-torn, autocratic and unstable countries.129 The second set of studies, in political ecology, can also be traced back to older traditions of research, especially in cultural geography and cultural ecology, but its modern foundation stones were laid by investigations into peasant cultures and rural economies in the global South. These investigations, led by Piers Blaikie and Harold Brookfield, traced the ways in which rural communities in South Asia and elsewhere were influenced by distant forces like international trade, inward investment and political decision-making at the national level. This implied that local land use decisions had to be explained through a causal cascade that extended up to the global level, and which often involved unequal relations of power that 127 William Cronon, Nature’s metropolis: Chicago and the Great West (New York: Norton, 1992). 128 Michael Watts and David Goodman (eds), Globalizing Food (New York: Routledge, 1997). 129 Philippe Le Billon, ‘The Geopolitical Economy of “Resource Wars”’, Geopolitics 9 (2004) 1–28. 79 allowed land users limited room for manoeuvre. 130 This did not displace analysis, still less concern away from local land users, however, and these wider perspectives required a complementary, closely textured analysis of the local economy, culture and society that was sensitive to the varied roles of land users and unpacked generic categories like ‘peasant’, ‘small holder’ or ‘herder’. 131 As a result of this multi-scalar approach, political ecologists downgraded the causal role of local ecology (‘environment’) in explaining land use patterns in the global South – which was a standard ruse of environmental determinism – and they challenged the no less imperial, no less shop-worn doctrine of neo-Malthusianism, which invoked regional ‘over-population’ to explain (for example) famine. 132 In the wake of these seminal studies, subsequent research has – like many of their original subjects – moved into the cities of the global South; it has also arced back to the global North, not as a distant actor but as the site of other, equally local and trans-local political ecologies. 133 130 Piers Blaikie, The Political Economy of Soil Erosion in Developing Countries (London: Methuen, 1985); Piers Blaikie and Harold Brookfield, Land Degradation and Society London: Methuen, 1987). The radical political implications of this approach were developed in Michael Watts and Richard Peet (eds), Liberation ecology (London and New York: Routledge, 1996; second edition, 1994); the title plays off the ‘liberation theology’ whose revolutionary doctrines excoriated the deepening poverty of the poor and animated peasant movements in South and Central America in the 1970s. 131 See, for instance, Judith Carney, ‘Converting the Wetlands, Engendering the Environment: The Intersection of Gender with Agrarian Change in The Gambia’, Economic Geography 69 (1993) 329–348. 132 Michael Watts Silent violence: food, famine and peasantry in northern Nigeria (Berkeley CA: University of California Press, 1983); David Harvey had already criticized neo-Malthusianism in ‘Population, resources and the ideology of science’, Economic Geography 50 (1974) 256-77. 133 See Matt Gandy, ‘Landscapes of Disaster: water, modernity, and urban fragmentation in Mumbai’, Environment and Planning A, 40 (2008) 108–130; Paul Robbins and Julie Sharp, ‘Producing and consuming chemicals: the moral economy of the American lawn’, Economic Geography 79 (2003) 425–51. 80 There are no hard and fast lines between the approaches of political economy and political ecology, so that the differences are mainly ones of focus or emphasis. The two streams have braided into one another, and many geographers swim in both. But they also find another common, and to their critics, more uncertain ground. Smith’s emphasis on the production of nature placed its explanatory weight on the productive (and destructive) capacities of the social; so too the political economy of environment and resources. Even political ecology in most of its versions privileged political and economic processes over biophysical processes, which prompted some commentators to wonder ‘where the ecology had gone’ and to direct human geographers to developments in biogeography and ecological science. 134 The same might be said of a further round of research that builds on and responds to these approaches. That resonant phrase, ‘the wilderness turned a mart’, conceals a double movement. For it points not only forwards – to the transformation of resources into commodities – but also backwards. It was a commonplace of an older ‘resource geography’ that, as Erich Zimmerman argued, ‘resources are not; they become’. 135 But if resources (and markets) are cultural constructions – matters and materialities of human appraisal, imagination and invention – then so too, as Cronon urged, is 134Karl Zimmerer, ‘Human geography and the new ecology’, Annals of the Association of American Geographers 84 (1994) 108-25; Peter A. Walker, ‘Political ecology: where is the ecology?’ Progress in Human Geography 29 (2005) 73-82; see also Matthew Turner, ‘Ecology: natural and political’, in Noel Castree, David Demeritt and Diana Liverman (eds) A companion to environmental geography (Oxford: Wiley-Blackwell, 20009) pp. 181-98. 135 Erich Zimmerman, World resources and industries: a functional appraisal of the availability of agricultural and industrial resources (New York: Harper, 1933). 81 ‘wilderness’. 136 It’s then a small step to radicalize this insight and to see that every element of ‘nature’ – far from standing outside ‘culture’ – is always already culturally ‘constructed’. Mountains, forests and bears are simply unintelligible without a great deal of work that is typically unrecognised by expert and lay actors alike: hence Braun’s emphasis on the ‘buried epistemologies’ that construct and normalize particular ‘natures’. As he subsequently showed, these are more than epistemologies – theories of what counts as knowledge – and their exhumation requires an analysis of the productive work that is done by discourse more generally. 137 Discourse is to be understood not only as a torrent of words and images but also as a series of techniques and practices that, in certain circumstances, produce the objects that they name: in other words, they are performative. Just as the discourse of tropicality produces ‘the tropics’, so a series of discourses, both inside and outside science, work together to produce the still wider, taken-for-granted, ostensibly ‘natural’ concept of ‘nature’. For, as Nancy Stepan remarks, ‘Nature is not “natural” but is created as natural’, so that what counts as ‘nature’ is not given in nature. 138 These discourses have their own topographies and circulations. Sometimes they are confined to – and in fact help to reproduce – specific discursive communities, so that the promissory note that treats science as a social practice is redeemed in full measure in 136 William Cronon, ‘The trouble with wilderness, or getting back to the wrong nature’, in Cronon (ed) Uncommon ground: rethinking the human place in nature (New York: Norton, 1995) pp. 69-90. 137 Bruce Willems-Braun. ‘Buried epistemologies: the politics of nature in (post)colonial British Columbia’, Annals of the Association of American Geographers 87 (1997) 1: 3-32; see also Bruce Braun, The intemperate rainforest: nature, culture and power on Canada’s west coast (Minneapolis: University of Minnesota Press, 2002). 138 Nancy Leys Stepan, Picturing tropical nature (Ithaca: Cornell University Press, 2010); see more generally Noel Castree and Bruce Braun (eds), Social nature: theory, practice, politics (Oxford: Blackwell, 2001). 82 (for example) studies of the ways in which atmospheric scientists have come to define and understand ‘normal’ climatic behaviour. 139 But, as the heated debate over global climate change shows, these discourses often spill over into other, more public circles – and, similarly, work to produce publics – and enter into the collective, consciously articulated identity of particular sections of society. 140 Bio-political geographies have also inflected these discussions of what counts as nature. Many of them, inspired by Foucault’s luminous writings and lectures on biopolitics, have recognised that ‘population’ is a central category of state power and governmentality, whose production derives from and inheres in the power to make, sustain or remove life. These ideas have opened up new conversations between population geography and medical geography that travel far beyond the conventional, spatial-analytic framing of their work to open up searching interrogations of what is made to count as ‘life’. 141 More proximately, many bio-political geographies circle around the concept of ‘bare life’. According to Giorgio Agamben, classical Greek philosophy made a vital distinction between political life (bios) and merely existent, biological life (zoe). Bare life is poised between the two, as life that is excluded from political participation and which can legitimately be 139 Mike Hulme, Suraje Dessai, Irene Lorenzoni and Donald Nelson, ‘Stable climates: exploring the statistical and social construction of ‘normal’ climate’, Geoforum 40 (2009) 197-206. 140 Wendy Wolford, ‘Agrarian moral economies and neoliberalism in Brazil: competing worldviews and the state in the struggle for land’, Environment & Planning A 37 (2005) 241-61. 141 Stephen Legg, ‘Foucault’s Population Geographies: Classifications, Biopolitics and Governmental Spaces’, Population, Space and Place, 11 (2005) 137–156; Stephen Legg, Spaces of colonialism: Delhi’s urban governmentalities (Oxford: Wiley-Blackwell, 2007). 83 abandoned to violence and death. 142 This bears directly on our discussion of the ‘more-than- human’ in two ways. On one side, those who are abandoned to the space of exception and who embody the spectral figure of what Agamben calls homo sacer become, in their very abjectness, limit cases for what is to count as human: in effect, they are rendered as ‘less-than-human’, produced through what Judith Butler calls ‘exclusionary conceptions of who is normatively human’. These reductions were the stock-in-trade of the wars conducted in the shadows of 9/11 by the Bush administration; their most visible locus was the US war prison at Guantanámo Bay. 143 On the other side, the space of vulnerability to environmental hazards or disasters can be seen as also always a potential space of exception in which marginalized or disadvantaged groups are wilfully exposed to disaster and death. This has prompted inquiries into the politically modulated effects of the Indian Ocean tsunami in 2004 and Hurricane Katrina in 2005, for example, and other equally probing analyses of the political roots of famine in the past and the present. 144 142 Giorgio Agamben, Homo sacer: sovereign power and bare life (Stanford CA: Stanford University Press, 2008); for a geographical discussion, see Claudio Minca, ‘Giorgio Agamben and the new biopolitical nomos’, Geografiska Annaler B 88 (2006) 387-403; Minca, ‘Agamben’s geographies of modernity’, Political geography 26 (2007) 78-97. 143 Judith Butler, Precarious life: the power of mourning and violence (London: Verso, 2004); Derek Gregory, ‘The Black Flag: Guantánamo Bay and the space of exception’, Geografiska Annaler B 88 (2006) 405-27; Simon Reid-Henry, ‘Exceptional Sovereignty? Guantánamo Bay and the Re-Colonial Present’, Antipode 39 (2007) 627– 648. An important limitation of Agamben’s work is its marginalization of struggles against the proliferation of spaces of exception and the (collective) agency of those who are consigned to them. 144 Kris Olds, James Sidaway and Mathew Sparke, ‘White death’, Environment and Planning D: Society and Space, 23 (2005) 475–479; Bruce Braun and James McCarthy, ‘Hurricane Katrina and abandoned being’, Environment and Planning D: Society & Space 23 (2005) 802-809; Jennifer Hyndman, ‘The Securitization of Fear in Post-Tsunami Sri Lanka’, Annals of the Association of American Geographers, 97(2) (2007) 84 As with political economy and political ecology, however, it seems that an emphasis on constructions of nature and on biopolitics places the emphasis squarely on the ‘social’ in multiple forms: on cultural formations, on political and military violence, on the political regulation of the life of human populations. It fails to deal in equal measure with the ‘natural’ and, in particular, still maintains the divide or, marginally better, the dialectic between them. This has generated two responses. The first is to insist on hybridity, the claim that the world is not, and has never been, a tabula rasa waiting to be inscribed as we wish. Instead, as Sarah Whatmore emphasizes, it’s a world that is always already part of us, just as we are a part of it. 145 This implies that the conceptual dualisms organising Western thought – such as reason/instinct and human/animal – are not always easy to maintain. On this reading the world is composed of more or less indissoluble relations between entities; these entities are at once the medium and the outcome of those relations. The research in this vein is both descriptive and explanatory – for example, Morgan Robertson’s research into the creation of markets in ‘wetland ecosystem services’ shows how and why these markets must adapt to the specific biophysical character of wetland environments (even as wetlands are managed and physically recreated according to market logics).146 But much of this work is also ethical in character or else focussed on the somatic dimensions of human engagement with the non-human world. It’s a call not only to pay closer attention to the sheer existence of all 361–372; David Nally, Human encumbrances: political violence and the Great Irish Famine (Notre Dame: University of Notre Dame Press, 2011). 145Sarah Whatmore, Hybrid Geographies: natures, cultures, spaces (London: Sage, 2002). 146Morgan Robertson, ‘The nature that capital can see’, Environment and Planning D: Society and Space 24 (2006) 367-87. 85 the species and materials with whom our own fates are entwined; it’s also a call to explore their moral and affective importance, for ourselves (whoever we happen to be) and for non-human entities. A vivid example is provided by the emergence of an ‘animal geography’, radically different in tone and temper from a far older zoogeography, and part of a wider exploration of the varied relationships different people have with living species – animal, vegetable, insect and even microbiological – that Steve Hinchliffe and Sarah Whatmore describe as a new politics of ‘conviviality’. 147 In these various ways this work has plainly enlarged our sense of the ‘human’ in human geography, to the point that some have identified a ‘more than human geography’ or even a ‘post-human’ geography. To develop this project still further, some of those most closely involved have turned to a form of materialism that owes less to Hegel and Marx than to Spinoza and Deleuze. Its purpose is to confound the distinction between ‘dull stuff’ (things) and ‘vibrant life’ (us) in order to bring into view what Jane Bennett calls a ‘vibrant materiality’. She understands this to mean not only the capacity of things to impede, disrupt or even destroy the designs of humans ‘but also to act as quasi-agents or forces with trajectories, propensities or tendencies of their own.’ 148 When she says ‘things’ she means just that – 147 Jennifer Wolch, Kathleen West and Thomas Gaines, ‘Transspecies urban theory’, Environment and Planning D: Society and Space 13 (1995) 735-60); Steve Hinchliffe and Sarah Whatmore, ‘Living cities: towards a politics of conviviality’, Science as Culture 15 (2) (2006)123-38. See also Jennifer Wolch and Jody Emel (eds), Animal geographies (London: Verso, 1998). 148 Jane Bennett, Vibrant matter: a political ecology of things (Durham: Duke University Press, 2010) p. viii; see also Bruce Braun and Sarah Whatmore (eds) Political Matter: technoscience, democracy and public life (Minneapolis: University of Minnesota 86 including everything from metals to worms 149 – but if this seems unsettling it is not difficult to think in terms of the conventional actants of much of physical geography, biogeography and zoogeography (though they are not usually thought of like this). This work is important for three overlapping reasons that can be aligned directly with other research in human geography. First, where critical human geographies inspired by the humanities and by certain forms of ‘humanist’ social theory privileged human agency – working to people the skeletal geometries and empty landscapes of spatial science – this body of work directs attention to the material agency of non-human or ‘not-quite-human’ things. Second, just as postcolonial geographies attempt to provide critiques of Eurocentrism, so these studies seek to displace anthropocentrism: to challenge a view of the world in which human beings are always the privileged beings at its centre. Third, the waywardness of nature and the non-human touches (literally so) on the liveliness and unruliness of space that has also captivated human geographers. These developments are intended to be profoundly, actively political, but when Bennett describes her project as a ‘political ecology’ she means this in a radically different sense from the previous authors. She directly challenges their sense of the social ‘productions’ and ‘constructions’ of nature by insisting that these still inhabit ‘the image of dead or thoroughly instrumentalized matter [that] feeds human hubris and our earth-destroying Press, 2010). The ghost in the machine here is Bruno Latour, whose actor-network theory and subsequent writings (and exhibitions) have done much to challenge conventional conceptions of action and agency. 149 The example of worms is in fact Darwin’s, but for a brilliant rendering of the agency of other living creatures in what he calls the ‘para-sites of capitalism’ see Timothy Mitchell, ‘Can the mosquito speak?’, in his Rule of experts: Egypt, techno-politics, modernity (Berkeley CA: University of California Press, 2002) Chapter 1. 87 fantasies of conquest and consumption.’ 150 Her enlarged sense of ecology is shared by other human geographers too, particularly those interested in embodiment. The body, at once a natural’ given and yet shaped by the full range of circumstances and relations in which people find themselves, is an important medium through which to better understand not only the cognitive but also the sensibilities, emotions and feelings that must be part of any viable politics. If, at times, some of this research has revelled in the pre-cognitive aspects of human engagement with the non-human, other research reminds us that non-human entities – by virtue of their specific material affordances and affective capacities, as well as their plasticity – can be enrolled into attempts to control nature and people.151 Not all the non-human world is as intransigent as Robertson’s wetlands that offer resistance to the logic of capitalist commodification. At times, even a more symmetrical approach that gives equal weight to the non-human must concede that metaphors of ‘production’ and ‘construction’ are not always entirely inappropriate: metaphors are devices that take us so far, even if they ultimately break down. Indeed, the critical sense of ‘production’ and ‘construction’ – and devastation and destruction – are consistent with the belief that we inhabit the Anthropocene, a period of momentous, potentially calamitous environmental change in which human interventions are so dominant as to constitute a radically new geological epoch. 152 To study the Anthropocene ‘symmetrically’ is not to downplay what Thomas 150 Bennett (2011), Vibrant matter (….) 151See, for example, Jake Kosek (2010) ‘Ecologies of empire: on the new uses of the honeybee’, Cultural Anthropology 25, 4: 650-78. 152The term was proposed by ecologist Eugene Stoermer, and atmospheric scientist Paul Crutzen has argued that it constitutes a new geological era 88 and his collaborators more than half a century ago called humanity’s ‘role in changing the face of the earth’ but to attend equally closely to biophysical changes and responses. 153 Bennett’s point about nature, the environment and the non-human ‘having its own trajectories’ – even when these are in part the product of human activities – has been sharpened by Nigel Clark. He argues that attempts to achieve a more symmetrical understanding of how the social and the natural intertwine in what is at once a life-giving – in fact life- defining – and yet deadly embrace are confounded by two problems. Not only have they directed attention to the non-human ‘close in’ – to phenomena we encounter directly each day or to our own ever-present corporeality – but they have also avoided consideration of those processes and events that greatly exceed our capacity to cope. Where a term like ‘vitality’ is derived from the life sciences, Clark explores the more catastrophic vocabulary of the geological sciences. In his view, we need to take far more seriously the power of nature to overwhelm us – tsunamis, earthquakes, volcanic eruptions – and to consider how this power might oblige us to re-think at a fundamental level our relations with other people and with the non-human world. Clark’s manifesto goes far beyond the concerns of ‘natural hazards’ to explore existential questions of human vulnerability and precariousness on a dynamic planet. While it makes a certain sense to analyse the human-non-human nexus symmetrically, as critics have urged, Clark insists on our 153See also Simon Dalby, ‘Anthropocene geopolitics: globalization, empire, environment and critique’, Geography compass 1 (2007) 103-18; Dalby, ‘Welcome to the Anthropocene! Climate change, biopolitics and the end of the world as we know it’, Paper presented to the Annual Meeting of the Association of American Geographers, Seattle, April 2011. 89 recognising that there are situations where earthly forces display a vastly greater power than even the most technologically advanced societies. The distance between symmetry and asymmetry cannot be calibrated by philosophical or theoretical imprimatur: it requires precisely the careful, substantive analysis of the imbrications of society-nature, of the human and non-human, that has long been geography’s (distant) goal. Clark’s arguments may seem redolent of environmental determinism, but the similarity is superficial. Inhuman nature is a plea to remember how fleeting our presence on the planet is, and to consider how much we can profit from responding to the ‘ethical’ call of the earth when it too inflicts damage and destruction. 154 New horizons What next for human geography? All we know – if the last 50 years are anything to go by – is that change will be the only constant. The subject’s extraordinary intellectual richness and diversity has set it on a productive course: as the wider world changes, and intellectual tides turn, human geography will be well placed to respond. It is pointless to try to legislate on its future directions of travel: there is simply (and we think fortunately) no available mechanism for steering the ship. All we can hope for is that human geographers continue to practice ‘engaged pluralism’ – the habit of paying respectful attention to, borrowing from, and critically appraising the work of others. Out of such engagement – which must, of course, extend beyond as well across the field – human geography will 154 Nigel Clark, Inhuman nature: sociable life on a dynamic planet (London: Sage, 2011). 90 continue to make important contributions to thought and practice. It will do do so on a range of fronts in the belief that a properly human geography has to learn from and speak to the concerns of the humanities, social sciences and environmental sciences at one and the same time. And, above all, that it shares in the wider responsibility to produce and foster a more humane geography. 91 Acknowledgements Selecting contributions that together represent the strength and breadth of human geography today was relatively easy; getting the number down to around eighty was very challenging indeed. We received good advice from Trevor Barnes, Gavin Bridge, Nick Clifford, Paul Cloke, Stuart Corbridge, Martin Dodge, Stewart Fotheringham, Matt Gandy, Rod Neumann, Chris Perkins, Paul Robbins, Eric Sheppard and Billie Lee Turner II. Robert Rojek and Judi Burger remained extraordinarily patient as successive deadlines came and went: though most of our promises were hollow, they continued to believe that one of them would not be. We thank them warmly, so too Bhairav Sharma and the production team in India - they did a marvellous job working to very tight deadlines. Finally, we'd like to thank all the authors whose writings are collected here for giving us their permission to do so. We hope our readers will realize that none of these essays should be taken as emblematic of our authors’ work as a whole; they all have an even richer portfolio than our selections can convey. Even then, having to leave out so much wonderful work from other authors has been immensely difficult: but it is also a testament to human geography's continued intellectual vitality. Derek Gregory and Noel Castree 92 |
17604 | https://www.youtube.com/watch?v=CCX4RiJtecw | Math150/151: Section 5.2 Polar Curves - Plotting with GeoGebra
Jamie Mulholland
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Posted: 31 Mar 2014
An introduction of how to plot curves defined by a polar equation using GeoGebra.
Transcript:
hi everyone in this video we're gonna look at how to plot polar curves using geogebra if you haven't grabbed a copy of geogebra yet you can download your own free copy at wwg org so here's the idea we've got a polar curve R equals f of theta and we'd like to plot it now how do we do this well generally most graphing utilities don't recognize polar coordinates or at least equations of polar curves and plot them accordingly the way you get around that is by converting a polar curve into some parametric equations so a parametric curve and here we're just going to use the conversion between polar and Cartesian coordinates so for polar coordinates R and theta we can convert to their cartesian coordinates by x equals R coast at y equals R sine theta and that's all we've done here we've written down the parametric curves to be x equals our coast theta but R is f of theta and Y is equal to our sine theta but again R is f of theta so now we've got the x and y coordinates defined entirely in terms of a single variable theta the parameter of the parametric curve and so then we take this and we just feed it into geogebra so here we just use the curve command and we use the X function as f of T times cosine of T and the Y function is f of T times sine of T the very the parameter is T notice I've switched from theta 2 T it's just easier to type when I I go to fire the stuff into geogebra and the interval we're interested over is well 0 to 2 pi R really zero to wherever you want so this alone won't plot a curve you have to have a particular function f in mind so for example the line before in geogebra may be to take f of t to be 1 plus cos of T and then that curve command will now plot it so let's go ahead and fire this into geogebra so we start by going down to the input field and I'm going to type in our function R of and we could use theta you get theta by going over to the right hand side opening up the window which has all the Greek letters in it clicking on theta and that's equal to one here we'll do 1 plus coast of theta so I'll use theta again there we go so there's our R function we hit enter notice though it's plotted it as if it was an equation in Cartesian coordinates we wanted to plot this in polar coordinate so I'm just gonna hide it and I'm now going to use parametric curves to construct the plot we can force it to plot the points whose polar coordinates satisfy the equation by getting it to plot the Cartesian coordinates of that corresponding point using the curve command and so we plot the curve in Cartesian coordinates by R of T times coast of T R of T times sine of T parameter T ranges from 0 and we can go around to 2 pi and there's the corresponding plot so what we've managed to do is we took the polar curve and forcefully plotted all of the points whose polar coordinates satisfy this by converting them to their corresponding Cartesian coordinates and allowing the parametric curve to plot the Cartesian coordinates of those points we can do the same thing we've done with other curves we can change the color we can go ahead and change the style the line width and there's our corresponding curve and we can play around with this we can change it to plotting the corresponding curve 1 plus sine theta and it automatically updates so you can go ahead that's really all there is to plotting polar curves in is just converting them to the corresponding parametrized curves and plotting those now one thing we can do is we can sort of just a little bit she's gonna blow it up move it to the center and what I'd what I'd like to do is have the point yet drawn as the angles increase as theta increases here the theta in the original equation is really represented by this parameter T so I'd like it to trace the curve so how can I do that well I'm going to introduce a variable called capital T I'm gonna show the slider that corresponds to that variable this allows me to change that number capital T and maybe will adjust where it ranges over so capital T will range from 0 to we can go around to 2pi or 4 pi I'm gonna I'm just gonna go around to 4 pi so that just says I've got a slider now where those T values could go from 0 all the way up to 4 pi now if I go to the curve and I edit the properties of the curve notice the parametric curve is plotting from little T going from 0 out to what was here was 2 pi instead I'm gonna have it go all the way out to big T now if I click OK see how it's only plotted a little bit of it it's plotted all the points on the curve where little T goes from 0 out to big T so now if I let Big T range it'll now start to trace the curve out so there was we just passed T equals pi there's T equals 3 PI by 2 there's T equals 2 pi and then it keeps going of course I lost track at a point now I could have it plot the point at the end as well that's sort of laying this trail down and it's behind it and I'll do that by saying well I want to plot the point a remember the curve is a of T that's going to plot the point on the curve whose corresponding parameter is capital T and there it is right at the end go object properties I'm going to color of red as well I'll bump up the points now seven point I'm gonna hide the label hide the label on the curve as well and now as we drag the slider looks like that point is laying out the trail behind it and then there's T equals two pi and then it since we graphed a capital T and I going out to four pi it makes these two revolutions so very nice and we can do things like okay change the change the function here maybe we'll go 1 plus and now let's say 3 times sine of and maybe we'll go 2 times stage I just for fun or gonna plot the polar curve 1 plus 3 sine 2 theta everything's updated now we'll just let the slide or drag it out okay I'm gonna zoom out a bit because we losing it and there we go no that's pretty cool so there's our polar curve and it's going around twice there's one shape completely traced and we're now at around T equals 2 pi so since T went up to 4 pi it traces it out twice alright so that's how you can plot polar curves by first converting them to parametric equations so feel free to start playing with these things and you can now sketch all of the curves that are given to you and the exercises in the textbook and create the visualizations that go along with them |
17605 | https://atb.nrel.gov/electricity/2023/hydropower | Hydropower
2023 ATB data for hydropower include data for two broad resource categories: non-powered dams (NPD) and new stream-reach developments (NSD). Cost projections are derived independently for the two resource categories. The data for NPD are based on a reduced-form model estimated using data from (Oladosu et al., 2021), and those for NSD are retained from the 2020 ATB, which is based on projections developed for the Hydropower Vision study (DOE, 2016).
The NPD and NSD data combine calculations of costs based on a baseline and potential future cost reductions using a mix of bottom-up modeling, technological learning, input from a technical team of Oak Ridge National Laboratory researchers, the experience of expert hydropower consultants, and other assumptions. The three scenarios for technology innovation are:
Resource Categorization
Hydropower resources can be broadly categorized depending on analytical needs and technological characteristics. Three hydropower categories are included in the 2023 ATB:
Representative Technology
In the 2023 ATB, CAPEX estimates are provided for eight representative NPD plants. The NPD sites are grouped into lock and lake design categories, each with four cost groups (Low Cost, Medium Cost, High Cost, and Very High Cost). CAPEX estimates for NPD sites are derived from a reduced-form model estimated with data for 19 reference sites selected to capture variations in the U.S. resource base. The reference sites are analyzed in detail to understand the drivers of costs (Oladosu et al., 2021). The detailed cost analysis is extended to other NPD sites using the reduced-form model. The NPD reference plants shown below are plants near the midrange of costs within each design/cost category previously highlighted.
The 2023 ATB retains NSD data for four representative plants from the 2022 ATB. These are based on binning approximately 8,000 potential NSD sites by head and capacity estimates from the Oak Ridge National Laboratory resource assessment (Kao et al., 2014). Analysis of these bins is used to estimate future hydropower deployment. The table below shows resource and technology characteristics for the hydropower categories included in the 2023 ATB; details are provided in the ATB data spreadsheet. The reference plants for NSD are developed using the average characteristics (weighted by capacity) of the resource groups within each set of ranges. For example, NSD 1 is constructed from the capacity-weighted average values of NSD sites with 3–30 feet of head and 0.5–10.0 MW of capacity. The weighted-average values are used as inputs to the cost formulas (O'Connor et al., 2015a) to calculate site CAPEX and O&M costs for NSD sites.
The 2023 ATB does not include cost estimates for hydropower upgrade potential. This is currently modeled in the Regional Energy Deployment System (ReEDS) model as becoming available at the relicensing date or plant age of 50 years or both.
Design Values of Representative Hydropower Plants
| Resource Characteristics | Technology Characteristics
Plants | Resource Detail 1 | Resource Detail 2 | Turbine Type | Flow (cfs) | Head (ft) | Water Conveyance Length (ft) | Distance to Substation (mi) | Capacity (MW) | Capacity Factor
NPD 1 | Lake | Low Cost | Francis | 507 | 240 | 2,400 | 0.87 | 9.85 | 0.335
NPD 2 | Lake | Medium Cost | Kaplan | 872 | 137 | 2,050 | 2.80 | 9.51 | 0.414
NPD 3 | Lake | Medium Cost | Kaplan | 1,095 | 71 | 1,000 | 1.85 | 6.31 | 0.333
NPD 4 | Lake | Very High Cost | Kaplan | 500 | 50 | 900 | 4.87 | 2.03 | 0.375
NPD 5 | Lock | Low Cost | Kaplan | 51,666 | 22 | 226 | 7.19 | 86.19 | 0.442
NPD 6 | Lock | Medium Cost | Kaplan | 51,666 | 17 | 1,125 | 8.08 | 66.52 | 0.441
NPD 7 | Lock | High Cost | Kaplan | 36,000 | 9 | 1,200 | 2.72 | 20.11 | 0.623
NPD 8 | Lock | Very High Cost | Kaplan | 2,895 | 8 | 2,435 | 6.40 | 1.70 | 0.399
NSD 1 | Head: 3-30 ft | Capacity: 1-10 MW | Bulb-Kaplan | 3,000 | 16 | Site-dependent | 5–15 | 3.72 | 0.656
NSD 2 | Head: 3-30 ft | Capacity: 10+ MW | Bulb-Kaplan | 30,000 | 20 | Site-dependent | 5–15 | 44.12 | 0.662
NSD 3 | Head: 30+ ft | Capacity: 1-10 MW | Bulb-Kaplan | 12,000 | 47 | Site-dependent | 5–15 | 4.25 | 0.624
NSD 4 | Head: 30+ ft | Capacity: 10+ MW | Bulb-Kaplan | 27,500 | 45 | Site-dependent | 5–15 | 94.04 | 0.665
Scenario Descriptions
In general, differences among the technology scenarios for hydropower in the ATB reflect different levels of adoption of innovations. Reductions in technology costs, particularly in the Advanced Scenario, reflect opportunities outlined in the Hydropower Vision study (DOE, 2016), which includes road map actions to lower technology costs. The tables below provide an overview of potential hydropower innovations. A recent analysis of NPD sites also identified potential near-term opportunities for reducing the cost of NPD hydropower under the Moderate Scenario (Oladosu et al., 2021).
Summary of Technology Innovation: Moderate Scenario (2030)
| Learning-by-Doing | New Materials | New Turbines
Technology Description | Widespread implementation of value engineering and design/construction best practices | Use of alternative materials in place of steel for water diversion (e.g., penstocks) | Matrix of smaller turbines
Impacts | Facility cost reduction | Material costs reduction | Reduced footprint and civil works cost
Resource Type | NPD and NSD | NPD and potentially NSD | NPD and potentially NSD
References | (DOE, 2016) | (Oladosu et al., 2021) | (Oladosu et al., 2021)
Summary of Technology Innovation: Advanced Scenario (2030)
| Modularity | New Materials | Automation/Digitalization | Eco-Friendly Turbines
Technology Description | Use of drop-in systems that minimize civil works and maximize ease of manufacture to reduce both capital investment and O&M expenditures | Alternative materials for water diversion (e.g., penstocks); advanced manufacturing with new composite materials for electromechanical systems | Implementation of standardized "smart" automation and remote monitoring systems to optimize scheduling of maintenance | Research and development on environmentally enhanced turbines to improve performance of the existing hydropower fleet
Impacts | Civil works cost reduction | Material costs reduction | Reduced maintenance cost | Reduced environmental mitigation costs
References | (DOE, 2016) | (DOE, 2016)(Oladosu et al., 2021) | (DOE, 2016) | (DOE, 2016);(Oladosu et al., 2021)
Scenario Assumptions
The scenarios for hydropower described above are not associated with specific deployment assumptions. Scenarios assume that ongoing and future R&D efforts will gradually overcome technological challenges facing new U.S. hydropower development.
Methodology
This section describes the methodology to develop assumptions for CAPEX, O&M, and capacity factor. For standardized assumptions, see labor cost, regional cost variation, materials cost index, scale of industry, policies and regulations, and inflation.
Capital Expenditures (CAPEX)
Definition: Hydropower capital costs are defined according to the breakdown structure described by O'Conner et al. (O'Connor et al., 2015a). The broad components of CAPEX include initial capital costs (e.g., site preparation, water conveyance system, powerhouse, electromechanical system, and electrical infrastructure) and development costs.
Recent Trends: Data from the literature and other sources are reviewed in the historical and literature review charts below. An attempt is made to identify potential CAPEX reduction for resources of similar characteristics over time (e.g., the estimated cost to develop the same site in 2019, 2030, and 2040 based on different technology, installation, and other technical aspects). Because the sample size is limited, all literature projections are analyzed together without distinguishing the types of technologies. Note that costs for the three 2023 ATB hydropower scenarios are for the lowest-cost category (i.e., NPD 1); however, such sites have dwindled in number as the most productive sites are built first.
For CAPEX, estimates represent actual and proposed projects from 1981 to 2014. Year represents the commercial online date for a past or future plant.
IEA-ETSAP: International Energy Agency (IEA)-Energy Technology Systems Analysis Program (ETSAP); IRENA: International Renewable Energy Agency
Base Year: CAPEX estimates for NPD sites are based on a detailed analysis of 19 reference sites using simulations with the bottom-up smHIDEA (Hydropower Integrated Design and Economic Analysis) model. The bottom-up simulations include estimates for the broad components of capital costs as outlined previously (Oladosu et al., 2021). The results of the simulations are extended to other NPD sites using a reduced-from model estimated with the bottom-simulations results. Estimates for NSD plants are based on a statistical analysis of historical plant data from 1980 to 2015 as a function of key design parameters, plant capacity, and hydraulic head (O'Connor et al., 2015b) as given in the following equation:
$$ \text{NSD CAPEX} = \left(9,605,710 \times P^{0.977} \times H^{-0.126}\right) + \left(610,000 \times P^{0.7}\right) $$
where P is capacity in megawatts, and H is head in feet. Actual and proposed NPD and NSD CAPEX from 1981 to 2014 (O'Connor et al., 2015b) are shown in box-and-whiskers format for comparison to the ATB base year CAPEX estimates and future projections.
Base year estimates of overnight capital cost (OCC) for NPD sites in the 2023 ATB range from $2,820/kW to $18,700/kW. These estimates reflect the variation in resource potentials and site features considered in the NPD reference cost analysis as outlined in the above table (Design Values of Representative Hydropower Plants). In general, the higher-cost sites reflect lower head and smaller flow that may also require significant civil works and transmission connections. These sites have fewer analogs in the historical data because the most productive NPD sites have already been built. Base Year estimates of OCC for NSD, based on the above equation, range from $6,244/kW to $7,973/kW. The estimates reflect potential sites with 3 feet of head to more than 60 feet of head and from 1 MW to more than 30 MW of capacity. The higher-cost ATB sites generally reflect small-capacity, low-head sites that differ from the historical data sample's generally larger-capacity and higher-head facilities. These characteristics lead to higher ATB base year CAPEX estimates than past data suggest. For example, the NSD projects that became commercially operational in this period are dominated by a few high-head projects in the mountains of the Pacific Northwest and Alaska.
Future Years: Projections for NPD sites are based on baseline and near-term innovation cost estimates in a 2021 study of reference sites (Oladosu et al., 2021) and on other assumptions, including technological learning, for the specific years in the 2023 ATB. Projections for NSD are based on the Hydropower Vision study (DOE, 2016) using technological learning assumptions and analysis of process and/or technology improvements to estimate future costs. The three scenarios for hydropower in the 2023 ATB are as follows:
Use the following table to view the components of CAPEX.
Components of CAPEX
Operation and Maintenance (O&M) Costs
Definition: Operation and maintenance (O&M) costs represent average annual fixed expenditures (and depend on rated capacity) required to operate and maintain a hydropower plant over its lifetime, including items noted in the table below (Components of O&M Costs).
Base Year: The core metric chart shows the Base Year estimate and future year projections for fixed O&M (FOM) costs for each technology innovation scenario. The estimate for a given year represents the annual average FOM costs expected over the technical lifetime of a new plant that reaches commercial operation in that year.
A statistical analysis of long-term plant operation costs from Federal Energy Regulatory Commission Form-1 results in the relationships in the following equation between annual FOM costs and plant capacity. Values are updated to 2021$.
$$ \text{Lesser of Annual O&M} = \left(227,000 \times P^{0.547}\right) \text{ or } \big(2.5\% \text{ of CAPEX}\big) $$
This equation is used to estimate O&M costs for both NPD and NSD plant groups.
Future Years: Projections developed for the Hydropower Vision study (DOE, 2016) using technological learning assumptions and bottom-up analysis of process and/or technology improvements provide a range of future cost outcomes. Three different FOM projections are developed for scenario modeling as bounding levels:
Use the following table to view the components of O&M.
Components of O&M Costs
Capacity Factor
Definition: The capacity factor represents the expected annual average energy production divided by the annual energy production, assuming the plant operates at rated capacity for every hour of the year. The capacity factor is intended to represent a long-term average over the plant's lifetime; it does not represent interannual variation in energy production. Future year estimates represent the annual average capacity factor over the technical lifetime of a new plant installed in a given year. The capacity factor is influenced by site hydrology, design factors (e.g., exceedance level), and operation characteristics (e.g., dispatch or run-of-river).
Recent Trends: Actual energy production from about 200 run-of-river plants operating in the United States from 2003 to 2012 (EIA, 2016) is shown in the chart below. This sample includes some old plants that might have lower availability and efficiency. It also includes plants that have been relicensed and may no longer be optimally designed for the current operating regime (e.g., a peaking unit now operating as run-of-river). This contributes to the broad range, particularly on the low end. Interannual variation of hydropower output for run-of-river plants might be significant because of hydrological events such as drought. And this impact might be exacerbated by climate change over the long term.
For the capacity factor, historical data represent energy production from about 200 run-of-river plants operating in the United States from 2003 through 2021, where the year represents the calendar year. Projection data represent the expected annual average capacity factor for plants with the commercial online date specified by year.
Base Year: Base Year capacity factors for NPD are based on the simulation results used to estimate CAPEX values, as discussed previously. Base Year capacity factors for NSD are assumed to be near the 80th percentile of the historical range and to have a small range.
Future Years: The capacity factor remains largely unchanged from the Base Year through 2050, except in cases where technological changes lead to slight differences in capacity, plant efficiency, or flow/head range of operation.
References
The following references are specific to this page; for all references in
this ATB, see References.
Oladosu, Gbadebo, Lindsay George, and Jeremy Wells. “2020 Cost Analysis of Hydropower Options at Non-Powered Dams.” Oak Ridge, TN: Oak Ridge National Laboratory, 2021.
DOE. “Hydropower Vision: A New Chapter for America’s Renewable Electricity Source.” Washington, D.C.: U.S. Department of Energy, 2016.
Hadjerioua, Boualem, Yaxing Wei, and Shih-Chieh Kao. “An Assessment of Energy Potential at Non-Powered Dams in the United States.” Oak Ridge National Laboratory, 2012.
Kao, Shih-Chieh, Ryan A. McManamay, Kevin M. Stewart, Nicole M. Samu, Boualem Hadjerioua, Scott T. DeNeale, Dilruba Yeasmin, M. Fayzul K. Pasha, Abdoul A. Oubeidillah, and Brennan T. Smith. “New Stream-Reach Development: A Comprehensive Assessment of Hydropower Energy Potential in the United States.” Oak Ridge, TN: Oak Ridge National Laboratory, April 2014.
O’Connor, Patrick W., Qin Fen (Katherine) Zhang, Scott T. DeNeale, Dol Raj Chalise, and Emma E. Centurion. “Hydropower Baseline Cost Modeling.” Oak Ridge, TN: Oak Ridge National Laboratory, 2015a.
O’Connor, Patrick W., Scott T. DeNeale, Dol Raj Chalise, Emma Centurion, and Abigail Maloof. “Hydropower Baseline Cost Modeling, Version 2.” Oak Ridge, TN: Oak Ridge National Laboratory, 2015b.
EIA. “Capital Cost Estimates for Utility Scale Electricity Generating Plants.” Washington, D.C.: U.S. Energy Information Administration, 2016.
Developed with funding from the U.S. Department of Energy’s Office of Energy Efficiency and Renewable Energy.
The National Renewable Energy Laboratory is a national laboratory of the U.S. Department of Energy, Office of Energy Efficiency and Renewable Energy, operated by the Alliance for Sustainable Energy LLC. |
17606 | https://web.engr.oregonstate.edu/~webbky/ESC440_files/Section%204%20Curve%20Fitting.pdf | ESC 440 – Numerical Methods for Engineers SECTION 4: CURVE FITTING K. Webb ESC 440 Introduction 2 K. Webb ESC 440 3 Curve Fitting Often, we have data, 𝑦, that is a function of some independent variable, 𝑥, Possibly noisy measurement data Underlying relationship is unknown Know 𝑥’s and 𝑦’s (approximately) But, don’t know 𝑦= 𝑓𝑥 K. Webb ESC 440 4 Curve Fitting May want to determine a function (i.e., a curve) that ‘best’ describes relationship between 𝑥 and 𝑦 An approximation to (the unknown) 𝑦= 𝑓𝑥 This is curve fitting K. Webb ESC 440 5 Regression vs. Interpolation We’ll look at two categories of curve fitting: Least-squares regression Noisy data – uncertainty in 𝑦 value for a given 𝑥 value Want “good” agreement between 𝑓(𝑥) and data points ◼Curve (i.e., 𝑓𝑥) may not pass through any data points Polynomial interpolation Data points are known exactly – noiseless data Resulting curve passes through all data points K. Webb ESC 440 Before moving on to discuss least-squares regression, we’ll first review a few basic concepts from statistics.
Review of Basic Statistics 6 K. Webb ESC 440 7 Basic Statistical Quantities Arithmetic mean – the average or expected value ത 𝑦= σ 𝑦𝑖 𝑛 Standard deviation (unbiased) – a measure of the spread of the data about the mean 𝜎= 𝑆𝑡 𝑛−1 where 𝑆𝑡 is the total sum of the squares of the residuals 𝑆𝑡= 𝑦𝑖−ത 𝑦2 K. Webb ESC 440 8 Basic Statistical Quantities Variance – another measure of spread The square of the standard deviation Useful measure due to relationship with power and power spectral density of a signal or data set 𝜎2 = 𝑆𝑡 𝑛−1 = σ 𝑦𝑖−ത 𝑦2 𝑛−1 or 𝜎2 = σ 𝑦𝑖2 −σ 𝑦𝑖2 𝑛 𝑛−1 K. Webb ESC 440 9 Normal (Gaussian) Distribution Many naturally-occurring random process are normally-distributed Measurement noise Very often assume noise in our data is Gaussian Probability density function (pdf): 𝑓𝑥= 1 2𝜋𝜎2 𝑒−𝑥−𝜇2 2𝜎2 where 𝜎2 is the variance, and 𝜇 is the mean of the random variable, 𝑥 K. Webb ESC 440 10 Random Number Generation – default_rng() Very often useful to generate random numbers Simulating the effect of noise Monte Carlo simulation, etc.
First, construct a random-number generator object using NumPy: rng = np.random.default_rng(seed) seed: optional initialization seed for generator rng: initialized generator object – will run methods on this object to generate random numbers K. Webb ESC 440 11 Normally-Distributed Random Numbers Generate random values from a normal (Gaussian) distribution x = rng.normal(loc=0, scale=1, size=1) rng: generator object created with default_rng() loc: optional mean of distribution – default: 0.0 scale: optional standard deviation – default: 1.0 size: optional dimension of resulting array x: resulting array of random values K. Webb ESC 440 12 Uniformly-Distributed Random Numbers Generate random values from a uniform distribution on the interval [low, high) x = rng.uniform(low=0, high=1, size=1) rng: generator object created with default_rng() low: optional lower bound of interval – default: 0.0 high: optional upper bound of interval – default: 1.0 size: optional dimension of resulting array – default: 1 x: resulting array of random values Half-open interval: Resulting values are ≥ low and < high K. Webb ESC 440 13 NumPy Statistical Functions NumPy includes many statistical functions, including: np.max() np.min() np.mean() np.std() np.median() np.var() np.cov() K. Webb ESC 440 14 Histogram Plots Histogram plots Graphical depiction of the variation of random quantities ◼Plots the frequency of occurrence of ranges (bins) of values Provides insight into the nature of the distribution plt.hist(x, bins=20, edgecolor='k’) x: data to be histogrammed bins: optional number of bins edgecolor: optional color of bin outlines – default: none K. Webb ESC 440 15 Statistics in NumPy, matplotlib K. Webb ESC 440 Linear Least-Squares Regression 16 K. Webb ESC 440 17 Linear Regression Noisy data, 𝑦, values at known 𝑥 values Suspect relationship between 𝑥 and 𝑦 is linear i.e., assume 𝑦= 𝑎0 + 𝑎1𝑥 Determine 𝑎0 and 𝑎1 that define the “best-fit” line for the data How do we define the “best fit”? K. Webb ESC 440 18 Measured Data Assumed a linear relationship between 𝑥 and 𝑦: 𝑦= 𝑎0 + 𝑎1𝑥 Due to noise, can’t measure 𝒚 exactly at each 𝒙 Can only approximate 𝑦 values ො 𝑦= 𝑦+ 𝑒 Measured values are approximations True value of 𝑦 plus some random error or residual ො 𝑦= 𝑎0 + 𝑎1𝑥+ 𝑒 K. Webb ESC 440 19 Best Fit Criteria Noisy data do not all line on a single line – discrepancy between each point and the line fit to the data The error, or residual: 𝑒= ො 𝑦−𝑎0 −𝑎1𝑥 Minimize some measure of this residual: Minimize the sum of the residuals ◼Positive and negative errors can cancel ◼Non-unique fit Minimize the sum of the absolute values of the residuals ◼Effect of sign of error eliminated, but still not a unique fit Minimize the maximum error – minimax criterion ◼Excessive influence given to single outlying points K. Webb ESC 440 20 Least-Squares Criterion Better fitting criterion is to minimize the sum of the squares of the residuals 𝑆𝑟= 𝑒𝑖 2 = ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖2 Yields a unique best-fit line for a given set of data The sum of the squares of the residuals is a function of the two fitting parameters, 𝑎0 and 𝑎1, 𝑆𝑟𝑎0, 𝑎1 Minimize 𝑆𝑟 by setting its partial derivatives to zero and solving for 𝑎0 and 𝑎1 K. Webb ESC 440 21 Least-Squares Criterion At its minimum point, partial derivatives of 𝑆𝑟 with respect to 𝑎0 and 𝑎1 will be zero 𝜕𝑆𝑟 𝜕𝑎0 = −2 ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖= 0 𝜕𝑆𝑟 𝜕𝑎1 = −2 ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖𝑥𝑖= 0 Breaking up the summation: ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖= 0 𝑥𝑖ො 𝑦𝑖−𝑎0𝑥𝑖−𝑎1𝑥𝑖 2 = 0 K. Webb ESC 440 22 Normal Equations 𝜕𝑆𝑟/𝜕𝑎0 = 0 and 𝜕𝑆𝑟/𝜕𝑎1 = 0 form a system of two equations with two unknowns, 𝑎0 and 𝑎1 𝑛 𝑎0 + 𝑥𝑖 𝑎1 = ො 𝑦𝑖 (1) 𝑥𝑖 𝑎0 + 𝑥𝑖 2 𝑎1 = 𝑥𝑖ො 𝑦𝑖 (2) In matrix form: 𝑛 𝑥𝑖 𝑥𝑖 𝑥𝑖 2 𝑎0 𝑎1 = ො 𝑦𝑖 𝑥𝑖ො 𝑦𝑖 (3) These are the normal equations K. Webb ESC 440 23 Normal Equations Normal equations can be solved for 𝑎0 and 𝑎1: 𝑎1 = 𝑛σ 𝑥𝑖ො 𝑦𝑖−σ 𝑥𝑖σ ො 𝑦𝑖 𝑛σ 𝑥𝑖 2 −σ 𝑥𝑖2 𝑎0 = σ ො 𝑦𝑖−𝑎1 σ 𝑥𝑖 𝑛 = ത 𝑦−𝑎1 ҧ 𝑥 Or solve the matrix form of the normal equations, (3), in Python using np.linalg.solve() K. Webb ESC 440 24 Linear Least-Squares - Example Noisy data with suspected linear relationship Calculate summation terms in the normal equations: 𝑛, Σ𝑥𝑖, Σො 𝑦𝑖, Σ𝑥𝑖 2, Σ𝑥𝑖ො 𝑦𝑖 K. Webb ESC 440 25 Linear Least-Squares - Example Assemble normal equation matrices Solve normal equations for vector of coefficients, 𝐚, using np.linalg.solve() K. Webb ESC 440 26 Goodness of Fit How well does a function fit the data?
Is a linear fit best? A quadratic, higher-order polynomial, or other non-linear function?
Want a way to be able to quantify goodness of fit Quantify spread of data about the mean prior to regression: 𝑆𝑡= ො 𝑦𝑖−ത 𝑦2 Following regression, quantify spread of data about the regression line (or curve): 𝑆𝑟= ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖2 K. Webb ESC 440 27 Goodness of Fit 𝑆𝑡 quantifies the spread of the data about the mean 𝑆𝑟 quantifies spread about the best-fit line (curve) The spread that remains after the trend is explained The unexplained sum of the squares 𝑆𝑡−𝑆𝑟 represents the reduction in data spread after regression explains the underlying trend Normalize to 𝑆𝑡 - the coefficient of determination 𝑟2 = 𝑆𝑡−𝑆𝑟 𝑆𝑡 K. Webb ESC 440 28 Coefficient of Determination 𝑟2 = 𝑆𝑡−𝑆𝑟 𝑆𝑡 For a perfect fit: No variation in data about the regression line 𝑆𝑟= 0 → 𝑟2 = 1 If the fit provides no improvement over simply characterizing data by its mean value: 𝑆𝑟= 𝑆𝑡 → 𝑟2 = 0 If the fit is worse at explaining the data than their mean value: 𝑆𝑟> 𝑆𝑡 → 𝑟2 < 0 K. Webb ESC 440 29 Coefficient of Determination Calculate 𝑟2 for previous example: K. Webb ESC 440 30 Coefficient of Determination Don’t rely too heavily on the value of 𝑟2 Anscombe’s famous data sets: Same line fit to all four data sets 𝑟2 = 0.67 in each case Chapra K. Webb ESC 440 Linearization of Nonlinear Relationships 31 K. Webb ESC 440 32 Nonlinear functions Not all data can be explained by a linear relationship to an independent variable, e.g.
Exponential model 𝑦= 𝛼𝑒𝛽𝑥 Power equation 𝑦= 𝛼𝑥𝛽 Saturation-growth-rate equation 𝑦= 𝛼 𝑥 𝛽+ 𝑥 K. Webb ESC 440 33 Nonlinear functions Methods for nonlinear curve fitting: Linearization of the nonlinear relationship Transform the dependent and/or independent data values Apply linear least-squares regression Inverse transform the determined coefficients back to those that define the nonlinear functional relationship Nonlinear regression Treat as an optimization problem – more later… K. Webb ESC 440 34 Linearizing an Exponential Relationship Linearize the fitting equation: ln 𝑦= ln 𝛼+ 𝛽𝑥 or ln 𝑦= 𝑎0 + 𝑎1𝑥 where 𝑎0 = ln 𝛼, 𝑎1 = 𝛽 Have noisy data that is believed to be best described by an exponential relationship 𝑦= 𝛼𝑒𝛽𝑥 K. Webb ESC 440 35 Linearizing an Exponential Relationship Determine 𝑎0 and 𝑎1: ln 𝑦= 𝑎0 + 𝑎1𝑥 Can calculate 𝑟2 for the line fit to the transformed data Note that original data must be positive Fit a line to the transformed data using linear least-squares regression K. Webb ESC 440 36 Linearizing an Exponential Relationship Exponential fit: 𝑦= 𝛼𝑒𝛽𝑥 where 𝛼= 𝑒𝑎0 , 𝛽= 𝑎1 Note that 𝑟2 is different than that for the line fit to the transformed data Transform the linear fitting parameters, 𝑎0 and 𝑎1, back to the parameters defining the exponential relationship K. Webb ESC 440 37 Linearizing an Exponential Relationship K. Webb ESC 440 38 Linearizing a Power Equation Linearize the fitting equation: log 𝑦= log 𝛼+ 𝛽log 𝑥 or log 𝑦= 𝑎0 + 𝑎1 log 𝑥 where 𝑎0 = log 𝛼, 𝑎1 = 𝛽 Have noisy data that is believed to be best described by an power equation 𝑦= 𝛼𝑥𝛽 K. Webb ESC 440 39 Linearizing a Power Equation Determine 𝑎0 and 𝑎1: log 𝑦= 𝑎0 + 𝑎1 log 𝑥 Can calculate 𝑟2 for the line fit to the transformed data Note that original data – both 𝑥 and 𝑦 – must be positive Fit a line to the transformed data using linear least-squares regression K. Webb ESC 440 40 Linearizing a Power Equation Power equation: 𝑦= 𝛼𝑥𝛽 where 𝛼= 10𝑎0 , 𝛽= 𝑎1 Note that 𝑟2 is different than that for the line fit to the transformed data Transform the linear fitting parameters, 𝑎0 and 𝑎1, back to the parameters defining the power equation K. Webb ESC 440 41 Linearizing a Power Equation K. Webb ESC 440 42 Linearizing a Saturation Growth-Rate Equation Linearize the fitting equation: 1 𝑦= 1 𝛼+ 𝛽 𝛼 1 𝑥 or 1 𝑦= 𝑎0 + 𝑎1 1 𝑥 where 𝑎0 = 1 𝛼 , 𝑎1 = 𝛽 𝛼 Have noisy data that is believed to be best described by a saturation growth-rate equation 𝑦= 𝛼 𝑥 𝛽+ 𝑥 K. Webb ESC 440 43 Linearizing a Saturation Growth-Rate Equation Determine 𝑎0 and 𝑎1: 1 𝑦= 𝑎0 + 𝑎1 1 𝑥 Can calculate 𝑟2 for the line fit to the transformed data Fit a line to the transformed data using linear least-squares regression K. Webb ESC 440 44 Linearizing a Saturation Growth-Rate Equation Saturation growth-rate equation: 𝑦= 𝛼 𝑥 𝛽+ 𝑥 where 𝛼= 1 𝑎0, 𝛽= 𝑎1 𝑎0 Note that 𝑟2 is different than that for the line fit to the transformed data Transform the linear fitting parameters, 𝑎0 and 𝑎1, back to the parameters defining the saturation growth-rate equation K. Webb ESC 440 45 Linearizing a Saturation Growth-Rate Equation K. Webb ESC 440 Polynomial Regression 46 K. Webb ESC 440 47 Polynomial Regression So far we’ve looked at fitting straight lines to linear and linearized data sets Can also fit mth-order polynomials directly to data using polynomial regression Same fitting criterion as linear regression: Minimize the sum of the squares of the residuals ◼m+1 fitting parameters for an mth-order polynomial ◼m+1 normal equations K. Webb ESC 440 48 Polynomial Regression Assume, for example, that we have data we believe to be quadratic in nature 2nd-order polynomial regression Fitting equation: ො 𝑦= 𝑎0 + 𝑎1𝑥+ 𝑎2𝑥2 + 𝑒 Best fit will minimize the sum of the squares of the residuals: 𝑆𝑟= ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖−𝑎2𝑥𝑖 2 2 K. Webb ESC 440 49 Polynomial Regression – Normal Equations Best-fit polynomial coefficients will minimize 𝑆𝑟 Differentiate 𝑆𝑟 w.r.t. each coefficient and set to zero 𝜕𝑆𝑟 𝜕𝑎0 = −2 ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖−𝑎2𝑥𝑖 2 = 0 𝜕𝑆𝑟 𝜕𝑎1 = −2 𝑥𝑖ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖−𝑎2𝑥𝑖 2 = 0 𝜕𝑆𝑟 𝜕𝑎2 = −2 𝑥𝑖 2 ො 𝑦𝑖−𝑎0 −𝑎1𝑥𝑖−𝑎2𝑥𝑖 2 = 0 K. Webb ESC 440 50 Polynomial Regression – Normal Equations Rearranging the normal equations yields 𝑛 𝑎0 + Σ𝑥𝑖𝑎1 + Σ𝑥𝑖 2 𝑎2 = Σො 𝑦𝑖 Σ𝑥𝑖𝑎0 + Σ𝑥𝑖 2 𝑎1 + Σ𝑥𝑖 3 𝑎2 = Σ𝑥𝑖ො 𝑦𝑖 Σ𝑥𝑖 2 𝑎0 + Σ𝑥𝑖 3 𝑎1 + Σ𝑥𝑖 4 𝑎2 = Σ𝑥𝑖 2 ො 𝑦𝑖 Which can be put into matrix form: 𝑛 Σ𝑥𝑖 Σ𝑥𝑖 2 Σ𝑥𝑖 Σ𝑥𝑖 2 Σ𝑥𝑖 3 Σ𝑥𝑖 2 Σ𝑥𝑖 3 Σ𝑥𝑖 4 𝑎0 𝑎1 𝑎2 = Σො 𝑦𝑖 Σ𝑥𝑖ො 𝑦𝑖 Σ𝑥𝑖 2 ො 𝑦𝑖 This system of equations can be solved for the vector of unknown coefficients using NumPy’s linalg.solve() K. Webb ESC 440 51 Polynomial Regression – Normal Equations For mth-order polynomial regression the normal equations are: 𝑛 Σ𝑥𝑖 ⋯ Σ𝑥𝑖 𝑚 Σ𝑥𝑖 Σ𝑥𝑖 2 ⋯ Σ𝑥𝑖 𝑚+1 ⋮ ⋮ ⋱ ⋮ Σ𝑥𝑖 𝑚 Σ𝑥𝑖 𝑚+1 ⋯ Σ𝑥𝑖 2𝑚 𝑎0 𝑎1 ⋮ 𝑎𝑚 = Σො 𝑦𝑖 Σ𝑥𝑖ො 𝑦𝑖 ⋮ Σ𝑥𝑖 𝑚ො 𝑦𝑖 Again, this system of 𝑚+ 1 equations can be solved for the vector of 𝑚+ 1 unknown polynomial coefficients using NumPy’s linalg.solve() K. Webb ESC 440 52 Polynomial Regression – Example K. Webb ESC 440 53 Polynomial Regression – np.polyfit() p = np.polyfit(x,y,m) x: 𝑛-vector of independent variable data values y: 𝑛-vector of dependent variable data values m: order of the polynomial to be fit to the data p: (𝑚+ 1)-vector of best-fit polynomial coefficients Least-squares polynomial regression if: 𝑛> 𝑚+ 1 i.e., for over-determined systems Polynomial interpolation if: 𝑛= 𝑚+ 1 Resulting fit passes through all (x,y) points – more later K. Webb ESC 440 54 Polynomial Regression – np.polyfit() Note that the result matches that obtained by solving normal equations K. Webb ESC 440 55 Exercise Determine the 4th-order polynomial with roots at 𝑥= {1, 5, 16, 19} Generate noiseless data points by evaluating this polynomial at integer values of 𝑥 from 0 to 20 Add Gaussian white noise with a standard deviation of 𝜎= 180 to your data points Use np.polyfit() to fit a 4th-order polynomial to the noisy data Calculate the coefficient of determination, 𝑟2 Plot the noisy data points, along with the best-fit polynomial Polynomial Regression Using np.polyfit() K. Webb ESC 440 Multiple Linear Regression 56 K. Webb ESC 440 57 Multiple Linear Regression We have so far fit lines or curves to data described by functions of a single variable For functions of multiple variables, fit planes or surfaces to data Linear function of two independent variables: multiple linear regression ො 𝑦= 𝑎0 + 𝑎1𝑥1 + 𝑎2𝑥2 + 𝑒 Sum of the squares of the residuals is now 𝑆𝑟= ො 𝑦𝑖−𝑎0 −𝑎1𝑥1,𝑖−𝑎2𝑥2,𝑖 2 K. Webb ESC 440 58 Multiple Linear Regression – Normal Equations Differentiate 𝑆𝑟 w.r.t. fitting coefficients and equate to zero The normal equations: 𝑛 Σ𝑥1,𝑖 Σ𝑥2,𝑖 Σ𝑥1,𝑖 Σ𝑥1,𝑖 2 Σ𝑥1,𝑖𝑥2,𝑖 Σ𝑥2,𝑖 Σ𝑥1,𝑖𝑥2,𝑖 Σ𝑥2,𝑖 2 𝑎0 𝑎1 𝑎2 = Σො 𝑦𝑖 Σ𝑥1,𝑖ො 𝑦𝑖 Σ𝑥2,𝑖ො 𝑦𝑖 Solve as before – now fitting coefficients, 𝑎𝑖, define a plane K. Webb ESC 440 General Linear Least-Squares Regression 59 K. Webb ESC 440 60 General Linear Least-Squares We’ve seen three types of least-squares regression Linear regression Polynomial regression Multiple linear regression All are special cases of general linear least-squares regression ො 𝑦= 𝑎0𝑧0 + 𝑎1𝑧1 + ⋯+ 𝑎𝑚𝑧𝑚+ 𝑒 The 𝑧𝑖’s are 𝑚+ 1 basis functions Basis functions may be nonlinear This is linear regression, because dependence on fitting coefficients, 𝑎𝑖, is linear K. Webb ESC 440 61 General Linear Least-Squares ො 𝑦= 𝑎0𝑧0 + 𝑎1𝑧1 + ⋯+ 𝑎𝑚𝑧𝑚+ 𝑒 For linear regression – simple or multiple: 𝑧0 = 1, 𝑧1 = 𝑥1, 𝑧2 = 𝑥2, … 𝑧𝑚= 𝑥𝑚 For polynomial regression: 𝑧0 = 1, 𝑧1 = 𝑥, 𝑧2 = 𝑥2, … 𝑧𝑚= 𝑥𝑚 In all cases, this is a linear combination of basis function, which may, themselves, be nonlinear K. Webb ESC 440 62 General Linear Least-Squares The general linear least-squares model: ො 𝑦= 𝑎0𝑧0 + 𝑎1𝑧1 + ⋯+ 𝑎𝑚𝑧𝑚+ 𝑒 Can be expressed in matrix form: ො 𝐲= 𝐙 𝐚+ 𝐞 where 𝐙 is an 𝑛× 𝑚+ 1 matrix, the design matrix, whose entries are the 𝑚+ 1 basis functions evaluated at the 𝑛 independent variable values corresponding to the 𝑛 measurements: 𝐙= 𝑧01 𝑧11 ⋯ 𝑧𝑚1 𝑧02 𝑧12 ⋯ 𝑧𝑚2 ⋮ ⋮ ⋱ ⋮ 𝑧0𝑛 𝑧1𝑛 ⋯ 𝑧𝑚𝑛 where 𝑧𝑖𝑗 is the 𝑖𝑡ℎ basis function evaluated at the 𝑗𝑡ℎ independent variable value. (Note: 𝑖 is not the row index and 𝑗 is not the column index, here.) K. Webb ESC 440 63 General Linear Least-Squares The least-squares model is: 𝑧01 𝑧11 ⋯ 𝑧𝑚1 𝑧02 𝑧12 ⋯ 𝑧𝑚2 ⋮ ⋮ ⋱ ⋮ 𝑧0𝑛 𝑧1𝑛 ⋯ 𝑧𝑚𝑛 𝑎0 𝑎1 ⋮ 𝑎𝑚 + 𝑒1 𝑒2 ⋮ 𝑒𝑛 = ො 𝑦1 ො 𝑦2 ⋮ ො 𝑦𝑛 More measurements than coefficients 𝑛> 𝑚+ 1 𝐙 is not square – tall and narrow Over-determined system 𝐙−𝟏 does not exist K. Webb ESC 440 64 General Linear Least-Squares – Design Matrix Example For example, consider fitting a quadratic to five measured values, ො 𝐲, at 𝐱= 1, 2, 3, 4, 5 𝑇 Model is: ො 𝑦= 𝑎0 + 𝑎1𝑥+ 𝑎2𝑥2 + 𝑒 Basis functions are 𝑧0 = 1, 𝑧1 = 𝑥, and 𝑧2 = 𝑥2 Least-squares equation is 1 1 1 1 2 4 1 3 9 1 4 16 1 5 25 𝑎0 𝑎1 𝑎2 = ො 𝑦1 ො 𝑦2 ො 𝑦3 ො 𝑦4 ො 𝑦5 − 𝑒1 𝑒2 𝑒3 𝑒4 𝑒5 K. Webb ESC 440 65 General Linear Least-Squares – Residuals Linear least-squares model is: ො 𝐲= 𝐙 𝐚+ 𝐞 (1) Residual: 𝐞= ො 𝐲−𝐲= ො 𝐲−𝐙 𝐚 (2) Sum of the squares or the residuals: 𝑆𝑟= σ 𝑒𝑖 2 = 𝐞𝐓𝐞= ො 𝐲−𝐙 𝐚𝐓ො 𝐲−𝐙 𝐚 (3) Expanding, 𝑆𝑟= ො 𝐲𝐓ො 𝐲−𝐚𝐓𝐙𝐓ො 𝐲−ො 𝐲𝐓𝐙𝐚+ 𝐚𝐓𝐙𝐓𝐙𝐚 (4) K. Webb ESC 440 66 Deriving the Normal Equations Best fit will minimize the sum of the squares of the residuals Differentiate 𝑆𝑟 with respect to the coefficient vector, 𝐚, and set to zero 𝑑𝑆𝑟 𝑑𝐚= 𝑑 𝑑𝐚ො 𝐲𝐓ො 𝐲−𝐚𝐓𝐙𝐓ො 𝐲−ො 𝐲𝐓𝐙𝐚+ 𝐚𝐓𝐙𝐓𝐙𝐚= 𝟎 (5) We’ll need to use some matrix calculus identities: 𝑑 𝑑𝐚𝐚𝐓𝐙𝐓𝐲= 𝐙𝐓𝐲 𝑑 𝑑𝐚𝐲𝐓𝐙𝐚= 𝐙𝐓𝐲 (6) 𝑑 𝑑𝐚𝐚𝐓𝐙𝐓𝐙𝐚= 𝟐𝐙𝐓𝐙𝐚 K. Webb ESC 440 67 Deriving the Normal Equations 𝑑𝑆𝑟 𝑑𝐚= 𝑑 𝑑𝐚ො 𝐲𝐓ො 𝐲−𝐚𝐓𝐙𝐓ො 𝐲−ො 𝐲𝐓𝐙𝐚+ 𝐚𝐓𝐙𝐓𝐙𝐚= 𝟎 Using the matrix derivative relationships, (6), 𝑑𝑆𝑟 𝑑𝐚= −2𝐙𝐓ො 𝐲+ 2𝐙𝐓𝐙𝐚= 𝟎 (7) Equation (7) is the matrix form of the normal equations: 𝐙𝐓𝐙𝐚= 𝐙𝐓ො 𝐲 (8) Solution to (8) is the vector of least-squares fitting coefficients: 𝐚= 𝐙𝐓𝐙 −𝟏 𝐙𝐓ො 𝐲 (9) K. Webb ESC 440 68 Solving the Normal Equations 𝐚= 𝐙𝐓𝐙 −𝟏 𝐙𝐓ො 𝐲 (9) Remember, our starting point was the linear least-squares model: 𝐲= 𝐙 𝐚 (10) Couldn’t we have solved (10) for fitting coefficients as 𝐚= 𝐙−𝟏𝐲 (11) No, must solve using (9), because: Don’t have 𝐲, only noisy approximations, ො 𝐲 We have an over-determined system ◼𝐙 is not square ◼𝐙−𝟏 does not exist K. Webb ESC 440 69 Solving the Normal Equations Solution to the linear least-squares problem is: 𝐚= 𝐙𝐓𝐙 −𝟏 𝐙𝐓ො 𝐲= 𝐙†ො 𝐲 (12) where 𝐙† = 𝐙𝐓𝐙 −𝟏 𝐙𝐓 (13) is the Moore-Penrose pseudo-inverse of 𝐙 Use the pseudo-inverse to find the least-squares solutions to an over-determined system K. Webb ESC 440 70 Coefficient of Determination Goodness of fit characterized by the coefficient of determination: 𝑟2 = 𝑆𝑡−𝑆𝑟 𝑆𝑡 where 𝑆𝑟 is given by (3) 𝑆𝑟= ො 𝐲−𝐙 𝐚𝐓ො 𝐲−𝐙 𝐚 (14) and 𝑆𝑡= ො 𝐲−ത 𝐲𝐓ො 𝐲−ത 𝐲 (15) K. Webb ESC 440 71 General Least-Squares in Python Have 𝑛 measurements ො 𝑦= ො 𝑦0 ො 𝑦1 ⋯ ො 𝑦𝑛−1 𝑇 at 𝑛 known independent variable values 𝑥= 𝑥0 𝑥1 ⋯ 𝑥𝑛−1 𝑇 and a model, defined by 𝑚+ 1 basis functions ො 𝑦= 𝑎0𝑧0 + 𝑎1𝑧1 + ⋯+ 𝑎𝑚𝑧𝑚+ 𝑒 Generate design matrix by evaluating 𝑚+ 1 basis functions at all 𝑛 values of 𝑥 𝐙= 𝑧0 𝑥0 𝑧1 𝑥0 ⋯ 𝑧𝑚𝑥0 𝑧0 𝑥1 𝑧1 𝑥1 ⋯ 𝑧𝑚𝑥1 ⋮ ⋮ ⋱ ⋮ 𝑧0 𝑥𝑛−1 𝑧1 𝑥𝑛−1 ⋯ 𝑧𝑚𝑥𝑛−1 K. Webb ESC 440 72 General Least-Squares in Python Solve for vector of fitting coefficients as the solution to the normal equations 𝐚= 𝐙𝐓𝐙 −𝟏 𝐙𝐓ො 𝐲 Or by using np.linalg.lstsq() a = np.linalg.lstsq(Z, yhat) Result is the same, though the methods are different K. Webb ESC 440 Nonlinear Regression 73 K. Webb ESC 440 74 Nonlinear Regression – minimize() Nonlinear models: Have nonlinear dependence on fitting parameters E.g., 𝑦= 𝛼𝑥𝛽 Two options for fitting nonlinear models to data Linearize the model first, then use linear regression Fit a nonlinear model directly by treating as an optimization problem Want to minimize a cost function Cost function is the sum of the squares of the residuals 𝐽= 𝑆𝑟= ො 𝑦−𝑦2 Find the minimum of 𝐽 – a multi-dimensional optimization Use SciPy’s optimize.minimize() K. Webb ESC 440 75 Nonlinear Regression – minimize() Cost function: 𝐽= ො 𝑦−𝛼𝑒𝛽𝑥2 Find 𝛼 and 𝛽 to minimize 𝐽 Use SciPy’s optimize.minimize() Have noisy data that is believed to be best described by an exponential relationship 𝑦= 𝛼𝑒𝛽𝑥 K. Webb ESC 440 76 Multi-Dimensional Optimization – minimize() Find the minimum of a function of two or more variables opt = minimize(f, x0) f: function to be optimized x0: array of initial values opt: optimizeResult object returned – includes: ◼opt.x: the solution of the optimization (i.e., 𝑥𝑜𝑝𝑡) ◼opt.fun: value of objective function at the optimum (i.e., 𝑓𝑥𝑜𝑝𝑡) ◼opt.nit: number of iterations K. Webb ESC 440 77 Nonlinear Regression – minimize() K. Webb ESC 440 78 Nonlinear Regression – curve_fit() An alternative to minimizing a cost function using scipy.optimize.curve_fit(): popt, pcov = curve_fit(f, x, y, p0=None) f: handle to the fitting function – independent variable must be listed first ◼e.g., f = lamba x, A, B: Aexp(Bx) x: independent variable data y: dependent variable data p0: initial guess for popt - optional popt: best-fit parameters pcov: estimated covariance of popt K. Webb ESC 440 79 Nonlinear Regression – curve_fit() K. Webb ESC 440 Polynomial Interpolation 80 K. Webb ESC 440 81 Polynomial Interpolation Sometimes we know both 𝒙 and 𝒚 values exactly Want a function that describes 𝑦= 𝑓(𝑥) ◼Allows for interpolation between know data points Fit an 𝑛𝑡ℎ-order polynomial to 𝑛+ 1 data points 𝑦= 𝑎0𝑥𝑛+ 𝑎1𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯+ 𝑎𝑛 Polynomial will pass through all points We’ll look at polynomial interpolation using Newton’s polynomial The Lagrange polynomial K. Webb ESC 440 82 Polynomial Interpolation 𝑦= 𝑎0𝑥𝑛+ 𝑎1𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯+ 𝑎𝑛 Can approach similar to linear least-squares regression 𝑦= 𝑎0𝑧0 + 𝑎1𝑧1 + ⋯+ 𝑎𝑛𝑧𝑛 where 𝑧0 = 𝑥𝑛, 𝑧1 = 𝑥𝑛−1, … 𝑧𝑛= 1 For an 𝑛𝑡ℎ-order polynomial, we have 𝑛+ 1 equations with 𝑛+ 1 unknowns In matrix form 𝐲= 𝐙 𝐚 K. Webb ESC 440 83 Polynomial Interpolation Now, unlike for linear regression All 𝑛+ 1 values in 𝐲 are known exactly 𝑛+ 1 equations with 𝑛+ 1 unknown coefficients 𝐙 is square 𝑛+ 1 × 𝑛+ 1 𝑦1 𝑦2 ⋮ 𝑦𝑛+1 = 𝑥1 𝑛 𝑥1 𝑛−1 ⋯ 1 𝑥2 𝑛 𝑥2 𝑛−1 ⋯ 1 ⋮ ⋮ ⋱ ⋮ 𝑥𝑛+1 𝑛 𝑥𝑛+1 𝑛−1 ⋯ 1 𝑎0 𝑎1 ⋮ 𝑎𝑛 Could solve by inverting 𝐙 or by using NumPy’s linalg.solve() a = np.linalg.solve(Z, y) 𝐙 is a Vandermonde matrix Tend to be ill-conditioned The techniques that follow are more numerically robust K. Webb ESC 440 Newton Interpolating Polynomial 84 K. Webb ESC 440 85 Linear Interpolation Fit a line (1st-order polynomial) to two data points using a truncated Taylor series (or simple trigonometry): 𝑓 1 𝑥= 𝑓𝑥1 + 𝑓𝑥2 −𝑓𝑥1 𝑥2 −𝑥1 𝑥−𝑥1 where 𝑓 1(𝑥) is the function for the line fit to the data, and 𝑓𝑥𝑖 are the known data values This is the Newton linear-interpolation formula K. Webb ESC 440 86 Quadratic Interpolation To fit a 2nd-order polynomial to three data points, consider the following form 𝑓 2 𝑥= 𝑏0 + 𝑏1 𝑥−𝑥1 + 𝑏2 𝑥−𝑥1 𝑥−𝑥2 Evaluate at 𝑥= 𝑥1 to find 𝑏0 𝑏0 = 𝑓𝑥1 Back-substitution and evaluation at 𝑥= 𝑥2 and at 𝑥= 𝑥3 will yield the other coefficients 𝑏1 = 𝑓𝑥2 −𝑓𝑥1 𝑥2−𝑥1 and 𝑏2 = 𝑓𝑥3 −𝑓𝑥2 𝑥3−𝑥2 −𝑓𝑥2 −𝑓𝑥1 𝑥2−𝑥1 𝑥3−𝑥1 K. Webb ESC 440 87 Quadratic Interpolation 𝑓 2 𝑥= 𝑏0 + 𝑏1 𝑥−𝑥1 + 𝑏2 𝑥−𝑥1 𝑥−𝑥2 Can still view this as a Taylor series approximation 𝑏0 represents an offset 𝑏1 is slope 𝑏2 is curvature Choice of initial quadratic form (Newton interpolating polynomial) was made to facilitate the development Resulting polynomial would be the same for any initial form of an 𝑛𝑡ℎ-order polynomial Solution is unique K. Webb ESC 440 88 𝑛𝑡ℎ-Order Newton Interpolating Polynomial Extending the quadratic example to 𝑛𝑡ℎ-order 𝑓 𝑛𝑥= 𝑏0 + 𝑏1 𝑥−𝑥1 + ⋯+ 𝑏𝑛𝑥−𝑥1 𝑥−𝑥2 ⋯𝑥−𝑥𝑛 Solve for coefficients as before with back-substitution and evaluation of 𝑓𝑥𝑖 𝑏0 = 𝑓𝑥1 𝑏1 = 𝑓𝑥2, 𝑥1 𝑏2 = 𝑓𝑥3, 𝑥2, 𝑥1 ⋮ 𝑏𝑛= 𝑓𝑥𝑛+1, 𝑥𝑛, … , 𝑥2, 𝑥1 𝑓⋯ denotes a finite divided difference K. Webb ESC 440 89 Finite Divided Differences First finite divided difference 𝑓𝑥𝑖, 𝑥𝑗= 𝑓𝑥𝑖−𝑓𝑥𝑗 𝑥𝑖−𝑥𝑗 Second finite divided difference 𝑓𝑥𝑖, 𝑥𝑗, 𝑥𝑘= 𝑓𝑥𝑖, 𝑥𝑗−𝑓𝑥𝑗, 𝑥𝑘 𝑥𝑖−𝑥𝑘 𝑛𝑡ℎ finite divided difference 𝑓𝑥𝑛+1, 𝑥𝑛, … , 𝑥2, 𝑥1 = 𝑓𝑥𝑛+1, … , 𝑥2 −𝑓𝑥𝑛, … , 𝑥1 𝑥𝑛+1 −𝑥1 Calculate recursively K. Webb ESC 440 90 𝑛𝑡ℎ-Order Newton Interpolating Polynomial 𝑛𝑡ℎ-order Newton interpolating polynomial in terms of divided differences: 𝑓 𝑛𝑥= 𝑓𝑥1 + 𝑓𝑥2, 𝑥1 𝑥−𝑥1 + ⋯ +𝑓𝑥𝑛+1, 𝑥𝑛, … , 𝑥2, 𝑥1 𝑥−𝑥1 𝑥−𝑥2 ⋯𝑥−𝑥𝑛 Divided difference table for calculation of coefficients: Chapra K. Webb ESC 440 91 Newton Interpolating Polynomial – Example K. Webb ESC 440 Lagrange Interpolating Polynomial 92 K. Webb ESC 440 93 Linear Lagrange Interpolation Fit a first-order polynomial (a line) to two known data points: 𝑥1, 𝑓𝑥1 and 𝑥2, 𝑓𝑥2 𝑓 1 𝑥= 𝐿1 𝑥∙𝑓𝑥1 + 𝐿2 𝑥∙𝑓𝑥2 𝐿1 𝑥 and 𝐿2 𝑥 are weighting functions, where 𝐿1 𝑥= ቊ1, 𝑥= 𝑥1 0, 𝑥= 𝑥2 𝐿2 𝑥= ቊ1, 𝑥= 𝑥2 0, 𝑥= 𝑥1 The interpolating polynomial is a weighted sum of the individual data point values K. Webb ESC 440 94 Linear Lagrange Interpolation For linear (1st-order) interpolation, the weighting functions are: 𝐿1 𝑥= 𝑥−𝑥2 𝑥1 −𝑥2 𝐿2 𝑥= 𝑥−𝑥1 𝑥2 −𝑥1 The linear Lagrange interpolating polynomial is: 𝑓 1 𝑥= 𝑥−𝑥2 𝑥1 −𝑥2 𝑓𝑥1 + 𝑥−𝑥1 𝑥2 −𝑥1 𝑓𝑥2 K. Webb ESC 440 95 𝑛𝑡ℎ-Order Lagrange Interpolation Lagrange interpolation technique can be extended to 𝑛𝑡ℎ-order polynomials 𝑓 𝑛𝑥= 𝑖=1 𝑛+1 𝐿𝑖𝑥∙𝑓𝑥𝑖 where 𝐿𝑖𝑥= ෑ 𝑗=1 𝑗≠𝑖 𝑛+1 𝑥−𝑥𝑗 𝑥𝑖−𝑥𝑗 K. Webb ESC 440 96 Lagrange Interpolating Polynomial – Example |
17607 | https://www.sciencedirect.com/science/article/pii/S1058274622008102 | “Nearly off-track lesions” or a short distance from the medial edge of the Hill-Sachs lesion to the medial edge of the glenoid track does not seem to be accurate in predicting recurrence after an arthroscopic Bankart repair in a military population: a case-control study - ScienceDirect
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Journal of Shoulder and Elbow Surgery
Volume 32, Issue 4, April 2023, Pages e145-e152
Online Article
“Nearly off-track lesions” or a short distance from the medial edge of the Hill-Sachs lesion to the medial edge of the glenoid track does not seem to be accurate in predicting recurrence after an arthroscopic Bankart repair in a military population: a case-control study
Author links open overlay panel Lukas P.E.Verweij MD a b c, Theodore P.van Iersel MD c d, Derek F.P.van Deurzen MD, PhD c d, Michel P.J.van den Bekerom MD, PhD c d e, Sebastiaan Floor MD f
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Background
On-track lesions with a short distance from the medial edge of the Hill-Sachs lesion to the medial edge of the glenoid track (nearly off-track) may predispose recurrence after arthroscopic Bankart repair (ABR) in the general population. The aim of this study was to determine if a shorter distance between the medial edge of the Hill-Sachs lesion and the medial edge of the glenoid track could accurately predict recurrence after an ABR in a high-demand military population. It was hypothesized that a shorter distance would not accurately predict recurrence.
Materials and methods
A retrospective monocenter case-control study was performed at the Dutch Central Military Hospital. Patients with an on-track Hill-Sachs lesion who underwent a primary ABR between 2014 and 2019 with a minimal follow-up of 2 years and a preoperative magnetic resonance imaging (MRI) assessment received a questionnaire. The primary outcome was recurrence, defined as a complete dislocation or subluxation. Glenoid bone loss was assessed using a linear-based method on MRI. The distance from the medial edge of the Hill-Sachs lesion to the medial edge of the glenoid track was defined as the distance to dislocation (DTD). A receiver operating characteristic curve was created to determine the predictive value of the DTD for recurrence. Logistic regression was used to determine preoperative risk factors that predispose recurrence. Covariates were selected based on univariable analysis and included gender, body mass index, age at surgery and first dislocation, laterality, smoking habits, overhead shoulder activity during work, preoperative dislocations, sports type and level, bony or labral lesions on MRI, and DTD.
Results
In total, 80 patients with an average follow-up of 4.8±1.9 years completed the questionnaire and were included in the analyses. Seventeen patients (21%) experienced recurrence at the final follow-up. No difference in DTD was observed among patients who experienced recurrence (9 ± 4 mm) compared with patients who did not (9 ± 5 mm; P=.81). The receiver operating characteristic curve demonstrated no predictive power of DTD for recurrence (area under the curve=0.49). Smoking at the time of surgery (odds ratio: 3.9; confidence interval: 1.2-12.7; P=.02) and overhead shoulder movement during work (odds ratio: 9.3; confidence interval: 1.1-78.0; P=.04) were associated with recurrence according to the logistic regression analysis.
Conclusion
A shorter DTD demonstrated no accuracy in predicting recurrence in a military population. Smoking at the time of surgery and overhead shoulder activity during work were associated with recurrence; however, these analyses were underpowered to draw valid conclusions.
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Level of evidence
Level III
Retrospective Cohort Comparison
Prognosis Study
Keywords
Recurrence
arthroscopy
Bankart
shoulder
instability
Literature on shoulder instability is characterized by heterogeneous patient groups and outcomes.39, 40, 41 Nonoperative treatment after a first-time anterior dislocation demonstrates recurrence rates of up to 60% at 25-year follow-up.19Arthroscopic repair of the anterior labrum after a first-time dislocation can reduce the recurrence rate to around 10% in the general and a high-demand military population.5, 6, 7,37 However, even though this type of operative treatment can significantly reduce the recurrence rate, it remains challenging to select patients for optimal treatment.
Several risk factors for recurrence have been identified that may assist clinicians in determining which patients are susceptible for failure after the arthroscopic repair. A meta-analysis of risk factors studies showed that, in the general population, these factors include age, participation in competitive sports, bony lesions, anterior labroligamentous periosteal sleeve avulsion lesions, >1 preoperative dislocations, surgical delay, and an instability severity index score (ISI score)>3.41 However, a high amount of heterogeneity could be found in the meta-analysis, as some risk factor studies found significant differences for specific risk factors and others did not.39,41 This might be due to the fact that most studies are retrospective and susceptible to bias or underpowered to draw valid conclusions, but it may also be explained by the heterogeneous aspect of the studied patient groups. For example, Chan et al10 demonstrated that the ISI score or its individual domains are not associated with recurrence after an arthroscopic Bankart repair (ABR) in a military population, which could be due to high shoulder demand. A risk factor that may predispose recurrence in a military population is the glenoid track concept, as Bottoni et al5 demonstrated that off-track lesions were associated with recurrence in a military population. However, the study was underpowered to draw valid conclusions. The glenoid track concept, which was introduced by Yamamoto et al,42 assesses both humeral and glenoid bone lesions to predict recurrence risk. A lesion is considered off-track when the humeral and glenoid bone lesions engage in external rotation and abduction, increasing the risk of recurrence.13 The risk of an off-track lesion can be estimated with a formula that was designed by Di Giacomo et al, which has been validated in previous studies.13,18,34 Off-track lesions have consistently demonstrated to be a strong predictor for recurrence in multiple studies; however, the strength of the association is unclear.39,41 The meta-analysis evaluating the glenoid track concept in predisposing recurrence by Verweij et al41 demonstrated a high variability in effect (I 2=84%) between studies that evaluated the general population, which may be due to heterogeneous patient groups. Furthermore, the on-track/off-track concept is dichotomous and does not take on-track lesions that are close to being off-track or “nearly off-track” into account. Yamamoto et al43 and Li et al21 further developed the glenoid track concept and proposed to use the glenoid track concept as a continuous variable by calculating a ratio between the Hill-Sachs interval and glenoid track and determining the association with recurrence and patient-reported outcomes. Li et al defined the estimated distance between the medial edge of the Hill-Sachs lesion and the medial edge of the glenoid track as the distance to dislocation (DTD). Yamamoto et al43 demonstrated that a shorter distance is associated with worse Western Ontario Shoulder Index scores, and Li et al21 demonstrated that a shorter distance can effectively predict recurrence. These conflicting results illustrate that the overall effect of a shorter DTD is unclear and might change based on the patient group studied.
To clarify the effect of a shorter distance between the medial edge of the Hill-Sachs lesion and the medial edge of the glenoid track on predicting recurrence, this study aimed to evaluate the effect of a shorter distance in a homogeneous and high-demand patient group. Therefore, the aim of this study was to determine if a shorter distance between the medial edge of the Hill-Sachs lesion and the medial edge of the glenoid track could accurately predict recurrence after an ABR in a high-demand military population. It was hypothesized that a shorter distance would not accurately predict recurrence, as the ISI score was unable to predict recurrence in a military population, and a military population is expected to be less effected by the glenoid track concept due to mid-range instability.
Materials and methods
A retrospective monocenter case-control study was performed at the Dutch Central Military Hospital. This hospital only treats patients who are associated with the Dutch military. Local ethical approval was obtained, and all patients provided informed consent through the questionnaire. This article was written according to the Strengthening the Reporting of Observational Studies in Epidemiology (STROBE) checklist for case-control studies.15
Surgical indication and technique
The surgical indication in our population was a traumatic anterior shoulder dislocation with persistent instability. Surgical treatment was performed arthroscopically under general anesthesia in the lateral decubitus position. A posterior portal in the soft point was used for visualization, and an anterior portal was used for instrumentation. Shoulder stabilization was performed through fixation of the Bankart lesion with between 1 and 3 arthroscopic anchors (GRYPHON Suture Anchors; DePuy Synthes, Raynham, MA, USA). This anchor has the possibility to fixate the labrum with 2 separate sutures, which means a 2-point fixation was possible with a single anchor. Internal rotation immobilization in a shoulder immobilizer was to be followed for 4 weeks, and a standardized rehabilitation protocol was performed. Unlimited range of motion was allowed after 6 weeks. Return to sport was allowed at 6 to 9 months if the patient was painfree, had a full range of motion, and had good strength.
Patient selection
All consecutive patients who underwent a primary ABR between 2014 and 2019 at the Dutch Central Military Hospital for treatment of a traumatic anterior shoulder dislocation were identified through diagnosis and procedure codes. Surgery was performed by 3 orthopedic surgeons who specialize in shoulder pathology for the military. Patients between 18 and 55 years old who had an on-track Hill-Sachs lesion on preoperative magnetic resonance imaging (MRI) scan and a good understanding of the Dutch language were included. Only patients with a minimal follow-up of 2 years who had a preoperative MRI scan available were included in the analyses. Patients receiving surgery other than a primary ABR were excluded. In addition, patients were excluded when the ABR was combined with a tenomyodesis of the infraspinatus muscle (remplissage) or if they did not complete the questionnaire. The “cases” in this case-control study were defined as patients who experienced recurrence at the final follow-up, and the controls were defined as patients who did not experience recurrence at the final follow-up. Recurrence was defined as having experienced a complete dislocation or subluxation at the final follow-up.
Baseline characteristics and imaging
Baseline characteristics included gender, age at surgery, preoperative body mass index (kg/m 2), active service at follow-up, side of the surgery, and duration of follow-up and were extracted from the electronic patient files. The preoperative MRI scan was evaluated to identify bony and labral lesions, such as glenoid fractures, bony Bankart lesions, glenoid bone loss, Hill-Sachs lesions, Bankart lesions, anterior labroligamentous periosteal sleeve avulsion lesions, and Perthes lesions. Lesions were defined according to the definitions by Rutgers et al.32 Glenoid bone loss was assessed using the MRI-based method by Owens et al (Fig.1).3,29 A negative percentage was considered to be 0% bone loss. The Hill-Sachs interval was measured as the largest distance measured from the most medial aspect of the Hill-Sachs lesion to the insertion of the infraspinatus tendon.21 The distance from the medial edge of the Hill-Sachs lesion to the medial edge of the glenoid track was defined as the DTD.21 Glenoid track was calculated as “0.83 × Diameter of the glenoid − Glenoid bone loss” and DTD as “Glenoid track − Hill-Sachs interval.”
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Figure 1. Measurement of glenoid bone loss (A) according to the MRI-based method by Owens et al29 and the Hill-Sachs interval (B) as the largest distance measured from the most medial aspect of the Hill-Sachs lesion to the insertion of the infraspinatus tendon.
Questionnaire and patient-reported outcome measures
An electronic questionnaire was created using Castor EDC (version 2021; Castor, Amsterdam, the Netherlands), which is an online tool to send questionnaires and collect data. Eligible patients received an invitation by email between October 2021 and April 2022. If patients did not respond or complete the questionnaire after confirming participation, they received a maximum of 2 reminders by telephone. The primary outcome was recurrence, which was evaluated through the questionnaire. The questionnaire also included questions regarding age at first dislocation, preoperative dislocations, smoking habits, sports level and type, return to (preinjury) level of sport, overhead shoulder activity during work, patient satisfaction, and a validated Dutch version of the Oxford Shoulder Instability Score (OSIS).23 A minimally important clinical difference for the OSIS was set at 6 points.22 Sports type was categorized according to the shoulder demanding types described by Allain et al,2 comprising (type 1) nonimpact sports, (type 2) high-impact sports, (type 3) overhead sports with hitting movements, and (type 4) sports with overhead movements with sudden stops. If patients participated in multiple sports, the main sport was used to classify the patient.
Statistical analysis
Descriptive statistics were presented as average and standard deviation or median and interquartile range according to the distribution. A receiver operating characteristic (ROC) curve was created to determine the predictive value of the DTD for recurrence through the calculation of the area under the curve (AUC). The accuracy of the DTD according to the AUC values was defined as follows: fail (<0.6), poor (0.6-0.7), acceptable (0.7-0.8), excellent (0.8-0.9), and outstanding (0.9-1.0).24Logistic regression was used to determine preoperative risk factors that predispose recurrence. Covariates were selected based on univariable analyses comparing patients who experienced recurrence with the patients who did not. Independent t-tests were used for normally distributed variables, Mann-Whitney U tests for variables that were not normally distributed, and χ 2 or Fisher’s exact tests—in case of low expected values—were used for categorical variables. According to Austin and Steyerberg,4 a variable had sufficient power to be included in the logistic regression analysis per approximately 10 events. The outcome of the logistic regression was presented as odds ratios (OR) and their respective 95% confidence intervals (CI). A P value of ≤.05 was considered significant in all analyses. If a variable was statistically significant in the univariable analyses, the fragility index was determined. The fragility index is a simplified method to assess the robustness of a significant difference and demonstrates how many events need to be changed to nonevents to lose statistical significance.25 Partial deletion was applied to account for missing data if it was present. The analyses were performed using Statistical Package for Social Sciences (SPSS, version 28.0; IBM Corp., Armonk, NY, USA).
Sample size calculation
The sample size was calculated according to the ROC sample size tool developed by Obuchowski.28 This tool uses the type I error, power, AUC, and allocation ratio to estimate a sample size for case-control studies. The type I error was set at 0.05 and the power at 80%. The AUC was set at 0.73 according to Li et al,21 who evaluated the predictive value of the DTD in the general population. The allocation ratio was set at 4 (1/0.26=3.85) and was based on the recurrence rate in the study by Chan et al,10 who evaluated a military population as well. A minimal sample size of 60 patients was needed to achieve 80% power for the ROC analysis. This included a minimum of 12 cases and 48 controls.
Results
The diagnosis and procedure codes identified 198 patients, of whom 122 were eligible for inclusion. Of the eligible patients, 112 (92%) responded to the email or telephone call and agreed to participate, 7 (6%) were lost to follow-up, and 3 (2%) refused to participate. Patients who refused believed that the study had no additional benefit for them and therefore did not want to participate. After the final reminders, the questionnaire was completed by 80 patients between October 2021 and April 2022, demonstrating a response rate of 66%. These patients all had an on-track lesion on preoperative MRI and could be included in the analyses. The analyzed cohort had an average age at surgery of 26.6±5.9 years and follow-up of 4.8±1.9 years (Table I). The proportion of male patients was 91%.
Table I. Patient characteristics and recurrence
| Empty Cell | All patients (n=80) | No recurrence (n=63) | Recurrence (n=17) | Univariable analysis | Fragility index |
--- ---
| Follow-up (y) | 4.8±1.9 | 4.7±1.8 | 5.4±2.1 | .28 | |
| Active service at follow-up, n (%) | 53 (66) | 42 (67) | 11 (65) | .88 | |
| Male gender, n (%) | 73 (91) | 56 (89) | 17 (100) | .17 | |
| BMI preoperatively | 24.7±2.7 | 24.9±2.6 | 23.9±3.2 | .21 | |
| Age at surgery (y) | 26.6±5.9 | 27.0±5.7 | 25.1±6.5 | .25 | |
| Age at first dislocation (y) | 21.5±7.5 | 21.1±7.8 | 23.1±6.3 | .31 | |
| Surgery right side, n (%) | 41 (51) | 34 (54) | 7 (41) | .35 | |
| Smoking at the time of surgery, n (%) | 27 (34) | 17 (27) | 10 (59) | .01 | 2 |
| Overhead activity during work, n (%) | 56 (70) | 40 (63) | 16 (94) | .01 | 1 |
| Preoperative dislocations, n (%) | | | | .19 | |
| 0-1 | 10 (13) | 10 (16) | 0 (0) | | |
| 2 | 11 (14) | 9 (14) | 2 (12) | | |
| 3-5 | 26 (33) | 22 (35) | 4 (24) | | |
| >5 | 33 (41) | 22 (35) | 11 (65) | | |
| Sports level, n (%) | | | | .51 | |
| Recreational | 41 (51) | 31 (49) | 10 (59) | | |
| Competitive | 35 (44) | 28 (44) | 7 (41) | | |
| Professional | 4 (5) | 4 (6) | 0 (0) | | |
| Sports type according to Allain et al,2 n (%) | | | | .64 | |
| Type 1 | 24 (30) | 18 (29) | 6 (35) | | |
| Type 2 | 32 (40) | 26 (41) | 6 (35) | | |
| Type 3 | 19 (24) | 16 (25) | 3 (18) | | |
| Type 4 | 5 (6) | 3 (5) | 2 (12) | | |
| Bony lesions | | | | | |
| Bony Bankart, n (%) | 31 (39) | 24 (38) | 7 (41) | .82 | |
| Glenoid fracture, n (%) | 1 (1) | 1 (2) | 0 (0) | .79 | |
| Glenoid bone loss (%) | 5±6 | 5±5 | 6±7 | .51 | |
| DTD (mm) | 9±5 | 9±5 | 9±4 | .81 | |
| Hill-Sachs, n (%) | 72 (90) | 56 (89) | 16 (94) | .46 | |
| Hill-Sachs interval (mm) | 12±5 | 12±5 | 12±4 | .64 | |
| Labral lesions, n (%) | | | | | |
| Bankart | 72 (90) | 56 (89) | 16 (94) | .46 | |
| ALPSA | 6 (8) | 5 (8) | 1 (6) | .62 | |
| Perthes | 2 (3) | 2 (3) | 0 (0) | .62 | |
BMI, body mass index; DTD,distance to dislocation; ALPSA,anterior labroligamentous periosteal sleeve avulsion.
Primary outcome: recurrence
The recurrence rate at the final follow-up was 21% (17 recurrences; Table I). There was no recurrence observed in female patients. No difference in DTD was observed among patients who experienced recurrence (9±5 mm) compared with patients who did not (9±4 mm; P=.81). In addition, no difference between the Hill-Sachs interval or glenoid bone loss was observed. The ROC curve demonstrated no predictive power of DTD for recurrence (AUC=0.49; Fig.2). Based on the univariable analyses, smoking at the time of surgery and overhead shoulder activity during work could be included in the logistic regression analysis (Table I). The analysis demonstrated that smoking at the time of surgery (OR: 3.9; CI: 1.2-12.7; P=.02) and overhead shoulder movement during work (OR: 9.3; CI: 1.1-78.0; P=.04) were associated with recurrence. The fragility index was 2 for smoking at the time of surgery and 1 for overhead shoulder activity during work.
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Figure 2. Receiver operating characteristic (ROC) curve of the distance to dislocation as a predictor for recurrence. AUC,area under the curve.
Secondary outcomes
There was no difference observed in return to any and preinjury level of sport when comparing patients who experienced recurrence with the control group (Table II). The OSIS was significantly lower in patients who experienced recurrence. The difference between the medians was 8 points and therefore considered to be clinically relevant.22 Patients who experienced recurrence were less satisfied with the procedure in general and postoperatively compared to controls. No differences in patient satisfaction were observed preoperatively.
Table II. Secondary outcomes
| Empty Cell | All patients (n=80) | No recurrence (n=63) | Recurrence (n=17) | Univariable analysis |
--- --- |
| Return to any level of sport, n (%) | 79 (99) | 62 (98) | 17 (100) | .79 |
| Return to preinjury level of sport, n (%) | 56 (70) | 43 (68) | 13 (76) | .51 |
| OSIS | 44 (38-47.75) | 45 (41-48) | 37 (29-44) | .008 |
| Patient satisfaction (VAS 0-10) | | | | |
| In general | 8 (7-10) | 9 (8-10) | 6 (3-8) | <.001 |
| Preoperatively | 4 (1.25-6) | 4 (2-6) | 4 (1-6.50) | .74 |
| Postoperatively | 8 (7-9.75) | 8 (7-10) | 6 (4-8) | .04 |
OSIS, Oxford Shoulder Instability Score at follow-up; VAS, visual analog scale.
Discussion
The aim of this study was to determine if a shorter DTD could accurately predict recurrence after an ABR in a high-demand military population. The most important finding was that a shorter DTD was not accurate in predicting recurrence in a military population. Smoking at the time of surgery and overhead activity during work were associated with recurrence. However, according to Austin and Steyerberg,4 this analysis was underpowered. In addition, the statistical difference in the univariable analysis demonstrated a low fragility index for both factors, suggesting that the statistical differences are not robust.25
The current study had several limitations. First, the retrospective design made the study susceptible to selection bias and incomplete data collection. For example, some patients were unable to complete the questionnaire, were lost to follow-up, or refused to participate, and tissue quality or the quality of the repair could not be properly extracted from the surgical records. Second, patients might have had difficulty to understand what is considered to be a subluxation, which could influence the amount of recurrences and therefore the primary outcome. Third, the interval to follow-up averaged 5 years, which could cause recall bias. Fourth, according to Austin and Steyerberg,4 there should be 10 events when analyzing a variable, and the outcome of the logistic regression analysis in this study was therefore underpowered. In addition, the fragility index was low, indicating that the statistical differences in the univariable analyses were not robust. In contrast to these limitations, the study also had several important strengths. First, the patient group was relatively homogeneous, which decreased selection bias. Second, the study had sufficient power to create an ROC curve and calculate the AUC. Third, the study had an average long follow-up time, which is desirable for the primary outcome recurrence.31
The current study showed that the DTD is not accurate in predicting recurrence in a military population and smoking at the time of surgery, and overhead activity during work may be associated with recurrence. Trasolini et al39 demonstrated that the glenoid track concept is consistent in predisposing recurrence, and Bottoni et al5 demonstrated that off-track lesions may be predictive for recurrence with 15-year follow-up. The effect of the glenoid track concept or having a shorter DTD seems to be smaller in a military population, which may be due to high-demand or mid-range instability.9 If the effect is considerably smaller compared with other patient groups, a larger sample size is needed to identify the effect. The negative effect of smoking on recurrence was also observed by Novakofski et al27 after nonoperative treatment. Smoking has a negative effect on wound healing and might therefore hamper proper healing of the fixation.35 The relationship with overhead activity during work is unclear, but might be explained by end-range instability.9 Future studies are necessary that distinguish between mid-range and end-range instability.
Glenoid track measurements are primarily influenced by glenoid bone loss and the width of a Hill-Sachs lesion. However, the Hill-Sachs lesion is not a perfect circle, is often oval shaped and is characterized by variations in depth, angles, and position on the posterior side of the humeral head.17,20,43 These variations are probably caused by differences in trauma mechanism and recurrence.12,14,26 Schneider et al33 demonstrated that Hill-Sachs lesion measurements show poor reliability on computed tomography. A poor reliability for Hill-Sachs lesion measurements on CT might explain why they do not correlate with arthroscopic measurements.16 The formula of the glenoid track concept may predispose recurrence, but is probably too simplistic to accurately predict recurrence risk by itself.
In addition to the heterogeneity in patient groups, a high variability in the definition of recurrent instability exists.1 The current study defined recurrence as a complete dislocation or subluxation, but even if a patient did not experience recurrence according to this definition, a patient can still be dissatisfied with treatment due to reasons unrelated to recurrence. Park et al30 demonstrated that predictive factors for objective and subjective failure do not necessarily align. An international consensus on how to define recurrent instability has yet to be reached.
Because of the large heterogeneity in shoulder instability research outcomes, future research should focus on increasing the sample size to produce more robust results in risk factor studies. This may clarify how concepts such as the ISI score and the glenoid track can support decision-making in clinical practice. Furthermore, future studies should try and distinguish between mid-range and end-range instability, as this might be a factor that explains the differences in effect size. van Spanning et al36 published a protocol to increase the sample size by combing existing databases, which will be a more accurate method to estimate the influence of risk factors in the general population. However, considering that shoulder instability research is characterized by heterogeneous outcomes and patient groups, prospective (inter)national cohorts are needed to estimate the effect of risk factors in different patient groups.8,11,38,39,41
Conclusions
A shorter DTD demonstrated no accuracy in predicting recurrence in a military population. Smoking at the time of surgery and overhead shoulder activity during work were associated with recurrence; however, these analyses were underpowered to draw valid conclusions.
Disclaimers
Funding: Lukas P.E. Verweij received a grant from the Amsterdam UMC to complete his PhD. This is a personal grant and is not related to commercial entities (
Conflicts of interest: Derek F.P. van Deurzen is a paid instructor for Wright Medical. The other authors, their immediate families, and any research foundation with which they are affiliated are not receive any further financial payments or other benefits from any commercial entity related to the subject of this article.
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Cited by (0)
The Institutional Review Board (IRB) approved this study (W020.150) and the study received local approval by the Central Military Hospital.
Local ethical approval was obtained and all patients provided informed consent through the questionnaire.
© 2022 Published by Elsevier Inc. on behalf of Journal of Shoulder and Elbow Surgery Board of Trustees.
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Genetic Disorder Reference Sheet: Neurofibromatosis Type 2
Suzanne M. Mahon DNS, RN, AOCN®, AGN-BC, FAAN
September 19, 2023|4 min read
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Genetic Disorder Reference ...
Neurofibromatosis type 2 (NF2) is an autosomal dominant disorder characterized by bilateral vestibular schwannomas that can lead to significant hearing loss or deafness. Symptoms include tinnitus, hearing loss, and balance problems, with an average age of onset of 18–24 years and nearly 100% penetrance.
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Diagnostic Criteria for NF2
Major Criteria
Unilateral vestibular schwannoma
First-degree relative with NF2
Two or more meningiomas
Minor Criteria
Ependymoma
Single meningioma
Ocular changes, including juvenile subcapsular or cortical cataract, retinal hamartoma, epiretinal membrane in those younger than 40 years
Non-vestibular schwannoma
Approximately two-thirds of individuals develop spinal tumors, typically schwannomas. Early manifestations in childhood can include skin tumors and ocular changes (retinal hamartoma, thickened optic nerves, cortical wedge cataracts, third cranial nerve palsy). Other manifestations include schwannomas that develop in other cranial and peripheral nerves, meningiomas, ependymomas, and low-grade astrocytomas. Although many are benign, individuals often have multiple tumors that require frequent monitoring and treatment to prevent debilitating side effects.
NF2 Gene
The NF2 gene provides instructions for making the protein merlin, which is produced in the nervous system in the Schwann cells that surround and insulate neurons. Merlin functions as a tumor suppressor.
As a susceptibility biomarker, NF2 predicts an individual’s risk for various benign and malignant tumors. NF2 is also a prognostic biomarker in that an estimated 50%–75% of patients with NF2 develop meningiomas that are often grade II or III, have a worse prognosis, and have a higher recurrence rate than sporadic meningiomas.
Diagnosis
An NF2 clinical diagnosis is established in a proband with bilateral vestibular schwannomas, a detected germline pathogenic variant, a somatic tumor testing–detected identical NF2 pathogenic variants in two or more anatomically distinct NF2-related tumors, or a combination of the clinical criteria listed in the sidebar (two major criteria or one major and two minor criteria).
Family History
Assessment of family history is important, but approximately 50% of individuals have a de novo pathogenic variant. Of those, 25%–50% have somatic mosaicism, meaning that two or more cell lines with different genetic or chromosomal make-up occur in a single tissue after the zygote has formed. Symptoms of mosaic NF2 are milder and often restricted to a certain area or side of the body.
Offspring of an individual with a germline pathogenic variant have a 50% chance of inheriting the variant; offspring of an individual who has mosaic NF2 may have a lower risk. Once the NF2 pathogenic variant has been identified, cascade testing in other family members is feasible. Testing is usually offered by age 10 so that magnetic resonance imaging (MRI) screening can begin if necessary.
Biomarker Testing
Germline biomarker testing for NF2typically includesnext-generation sequencing (NGS) with multiplex ligation-dependent probe amplification (MLPA) to detect exon to multiexon copy number changes. It may also include high-resolution karyotyping to identify chromosomal rearrangements that NGS/MLPA cannot. The double-pronged approach will detect approximately 96% of germline pathogenic variants and is best coordinated by a genomics professional.
Clinical Management
Early identification of tumors can lead to prompt treatment. Best managed by an interprofessional team with experience in NF2, screening begins at age 10 and includes:
Annual neurologic exam by a provider familiar with NF2,beginning at the time of diagnosis
Annual brain MRI, beginning at age 10 through at least age 40
Annual hearing exam, including brain stem auditory evoked response, beginning at the time of diagnosis
Annual ophthalmology exam, beginning in early childhood
Treatment
Vestibular schwannomas are surgically removed whenever possible, with the procedure selected based on the tumors’ size and precise location. Radiation therapy may be considered for some individuals, especially those who are not surgical candidates, but it should be used with cautionin children because of the substantially higher risk of secondary malignancy. The VEGF inhibitor bevacizumab may also treat rapidly growing schwannomas and has enabled some individuals to regain hearing.
Other NF2 tumors may be benign or malignant. Treatment varies depending on the pathology, size, and location of the tumor. Some tumors may grow slowly and not require active intervention for years.
Retaining a patient’s hearing and hearing rehabilitation are primary management goals. Hearing preservation requires coordination with audiology and speech therapy to assist patients and families in coping with complications, such as by learning to use hearing aids, cochlear implants, and lip reading and sign language.
NF2 has no cure, but promptly identifying and referring families with the disorder to a genetics professional and center with experience managing NF2 can help reduce morbidity and mortality.Families benefit from interprofessional care that includes ongoing psychosocial support and monitoring beginning early in childhood.
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17609 | https://edu.rsc.org/esters/100559.tag | All esters articles
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17610 | https://www.gauthmath.com/solution/1813846025641125/8-Given-sin-A-7-25-A-in-quadrant-II-find-a-sin-2A-b-cos-2A-c-tan-2A-d-quadrant-o | Solved: Given sin A= 7/25 , A in quadrant II find a. sin 2A b. cos 2A c. tan 2A d. quadrant of 2A [Math]
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Given sin A= 7/25 , A in quadrant II find a. sin 2A b. cos 2A c. tan 2A d. quadrant of 2A
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Answer
a. $$-\frac{336}{625}$$−625 336
b. $$\frac{527}{625}$$625 527
c. $$-\frac{336}{527}$$−527 336
d. Quadrant II
Explanation
Find $$\cos A$$cos A
Since A is in quadrant II, $$\cos A$$cos A is negative.
Using the Pythagorean identity, $$\cos^{2} A + \sin^{2} A = 1$$cos 2 A+sin 2 A=1, we get:
$$\cos^{2} A = 1 - \sin^{2} A = 1 - \left(\frac{7}{25}\right)^{2} = \frac{576}{625}$$cos 2 A=1−sin 2 A=1−(25 7)2=625 576
Therefore, $$\cos A = -\sqrt{\frac{576}{625}} = -\frac{24}{25}$$cos A=−625 576=−25 24
Calculate $$\sin 2A$$sin 2 A
Using the double angle formula, $$\sin 2A = 2 \sin A \cos A$$sin 2 A=2 sin A cos A:
$$\sin 2A = 2 \cdot \frac{7}{25} \cdot -\frac{24}{25} = -\frac{336}{625}$$sin 2 A=2⋅25 7⋅−25 24=−625 336
Calculate $$\cos 2A$$cos 2 A
Using the double angle formula, $$\cos 2A = \cos^{2} A - \sin^{2} A$$cos 2 A=cos 2 A−sin 2 A:
$$\cos 2A = \left(-\frac{24}{25}\right)^{2} - \left(\frac{7}{25}\right)^{2} = \frac{527}{625}$$cos 2 A=(−25 24)2−(25 7)2=625 527
Calculate $$\tan 2A$$tan 2 A
Using the identity $$\tan A = \frac{\sin A}{\cos A}$$tan A=cos A sin A, we get:
$$\tan 2A = \frac{\sin 2A}{\cos 2A} = \frac{-\frac{336}{625}}{\frac{527}{625}} = -\frac{336}{527}$$tan 2 A=cos 2 A sin 2 A=625 527−625 336=−527 336
Determine the quadrant of 2A.
Since A is in quadrant II, 2A will be in quadrant II.
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17611 | https://en.wiktionary.org/wiki/egregious | Jump to content
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Contents
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1 English
1.1 Etymology
1.2 Pronunciation
1.3 Adjective
1.3.1 Usage notes
1.3.2 Derived terms
1.3.3 Related terms
1.3.4 Translations
1.4 References
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From Wiktionary, the free dictionary
English
[edit]
WOTD – 28 June 2007
Etymology
[edit]
From Latin ēgregius, from e- (“out of”), + grex (“flock”), + English adjective suffix -ous, from Latin suffix -osus (“full of”); reflecting the positive connotations of "standing out from the flock".
Pronunciation
[edit]
IPA(key): /ɪˈɡɹiː.d͡ʒəs/, /əˈɡɹiː.d͡ʒi.əs/
| | |
--- |
| Audio (US): | (file) |
Rhymes: -iːdʒəs
Adjective
[edit]
egregious (comparative more egregious, superlative most egregious)
Conspicuous, exceptional, outstanding; usually in a negative sense.
: The student has made egregious errors on the examination.
16th century, Christopher Marlowe, Ignoto,
: I cannot cross my arms, or sigh "Ah me," / "Ah me forlorn!" egregious foppery! / I cannot buss thy fill, play with thy hair, / Swearing by Jove, "Thou art most debonnaire!"
c. 1604–1605 (date written), William Shakespeare, “All’s Well, that Ends Well”, in Mr. William Shakespeares Comedies, Histories, & Tragedies […] (First Folio), London: […] Isaac Iaggard, and Ed[ward] Blount, published 1623, →OCLC, [Act II, scene iii]:
: My lord, you give me most egregious indignity.
1834, L[etitia] E[lizabeth] L[andon], chapter XXXI, in Francesca Carrara. […], volume III, London: Richard Bentley, […], (successor to Henry Colburn), →OCLC, page 257:
: Good Heaven! when we observe what egregious nonsense other people talk, what woful follies other people commit, sure we must be tempted to turn upon ourselves and ask—"What do I do that is equally silly?"
22 March 2012, Scott Tobias, AV Club The Hunger Games
: When the goal is simply to be as faithful as possible to the material—as if a movie were a marriage, and a rights contract the vow—the best result is a skillful abridgment, one that hits all the important marks without losing anything egregious.
2014 January 21, Hermione Hoby, “Julia Roberts interview for August: Osage County – 'I might actually go to hell for this ...': Julia Roberts reveals why her violent, Oscar-nominated performance in August: Osage County made her feel 'like a terrible person' [print version: 'I might actually go to hell for this ...' (18 January 2014, p. R4)]”, in The Daily Telegraph (Review):
: She's sitting opposite a window that's gently breezing into her face, wafting her hair into cover-girl perfection ... It's a little moment that seems to encapsulate her appeal: ... her gorgeousness being so egregious that even breezes oblige with their tousle-fanning effects ...
2. Outrageously bad; shocking.
1601, Ben Jonson, Poetaster or The Arraignment: […], London: […] [R. Bradock] for M[atthew] L[ownes] […], published 1602, →OCLC, Act III:
: Tuc[ca]. […] Can thy Author doe it impudently enough? / Hiſt[rio]. O, I warrant you, Captaine: and ſpitefully inough too; he ha's one of the moſt ouerflowing villanous wits, in Rome. He will ſlander any man that breathes; If he diſguſt him. / Tucca. I'le know the poor, egregious, nitty Raſcall; and he haue ſuch commendable Qualities, I'le cheriſh him: […]
Usage notes
[edit]
The negative meaning arose in the late 16th century, probably originating in ironic sarcasm. Before that, it meant outstanding in a good way. Webster also labels “distinguished” as an archaic meaning, and notes that contemporary usage often has an unpleasant connotation (for example, “an egregious error”). It generally precedes such epithets as ass, blunderer, rascal, and rogue. The Italian as well as Spanish cognate egregio has retained a strictly positive sense, as has the Portuguese cognate egrégio.
Derived terms
[edit]
egregiosity
egregiously
egregiousness
unegregious
Related terms
[edit]
egregia cum laude
Translations
[edit]
conspicuous, exceptional, outstanding
| |
| Bulgarian: въпи́ещ (bg) (vǎpíešt), отя́влен (bg) (otjávlen) Catalan: flagrant (ca), conspicu (ca), egregi Czech: náramný (cs), mimořádný (cs), přemrštěný, do očí bijící (cs), do nebe volající (cs), do nebe volající (cs) Dutch: flagrant (nl), ongehoord (nl) Finnish: ennenkuulumaton, räikeä (fi), törkeä (fi) French: flagrant (fr), inouï (fr) German: unerhört (de), ungeheuerlich (de), außergewöhnlich (de) Hungarian: szörnyű (hu) Icelandic: svívirðilegur (is), smánarlegur, erki- Norwegian: flagrant, uhørt Persian: فاحش (fa) (fheš) Polish: rażący (pl) Portuguese: odioso (pt), atroz (pt) Russian: вопию́щий (ru) (vopijúščij), отъя́вленный (ru) (otʺjávlennyj) Spanish: flagrante (es), egregio (es) Swedish: flagrant (sv) |
outrageously bad
| |
| Catalan: dolentíssim Czech: nehorázný (cs), neslýchaný (cs) Finnish: ennenkuulumaton, räikeä (fi), törkeä (fi) German: entsetzlich (de), ungeheuerlich (de), unerhört (de) Italian: oltraggioso m, vergognoso (it) m, disdicevole (it) Portuguese: ultrajante Spanish: craso (es), garrafal (es) |
References
[edit]
“egregious”, in Lexico, Dictionary.com; Oxford University Press, 2019–2022.
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17612 | https://math.stackexchange.com/questions/2355579/pile-splitting-problem-proof-by-induction | Stack Exchange Network
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Pile splitting problem (Proof by induction)
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$\begingroup$
Problem: James has a pile of n stones for some positive integer n ¥ 2. At each step, he chooses one pile of stones and splits it into two smaller piles and writes the product of the new pile sizes on the board. He repeats this process until every pile is exactly one stone.
For example, if James has n = 12 stones, he could split that pile into a pile of size 4 and another pile of size 8. James would then write the number 4 · 8 = 32 on the board. He then decides to split the pile of 4 stones into a pile with 1 stone and a pile with 3 stones and writes 1 · 3 = 3 on the board. He continues this way until he has 12 piles with one stone each.
Prove that no matter how James splits the piles (starting with a single pile of n stones), the sum of the numbers on the blackboard at the end of the procedure is always the same.
Hint: First figure out what the formula for the final sum will be. Then prove it using strong induction.
The formula I came up with is $n(n-1) / $2. It could be totally wrong...
My (Partial) Solution:
1) Base Case: (n=2) James has one pile of 2 stones. Suppose he takes the top-most stone and puts in one pile A, which now has size 1. He takes the remaining stone and places it into another pile B, which now has size 1. Then the product of the sizes is 1. The sum of all the products on his board is 1. Now, if he were to start again, and take the bottom stone first and put it in pile A, he has a pile of size 1, and then takes the top stone and puts it in pile B, he has a pile of size 1, and the product is 1, with sum of all products on the board is 1.
2) Inductive Hypothesis (Strong Induction): Suppose for some $k \geq 2$, $n$ stones can be split in $ 2 \leq n \leq k $ stones and $k-n$ stones.
3) Inductive Step: Consider $n=k+1$. What do I do now?
proof-verification
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induction
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edited Jul 12, 2017 at 1:53
Keith AxelrodKeith Axelrod
asked Jul 12, 2017 at 1:13
Keith AxelrodKeith Axelrod
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$\begingroup$ ...and the title has nothing to do with the process or problem. $\endgroup$
David G. Stork
– David G. Stork
2017-07-12 01:30:49 +00:00
Commented Jul 12, 2017 at 1:30
$\begingroup$ I would love to rephrase this question, but this is exactly as I am looking at it given to me in a class practice problem set. $\endgroup$
Keith Axelrod
– Keith Axelrod
2017-07-12 01:31:39 +00:00
Commented Jul 12, 2017 at 1:31
$\begingroup$ Here's my problem that was answered previously, but there's only two lines of explanation for the proof by induction, and it doesn't show the conventional induction process. math.stackexchange.com/questions/1115105/… $\endgroup$
Keith Axelrod
– Keith Axelrod
2017-07-12 01:33:54 +00:00
Commented Jul 12, 2017 at 1:33
$\begingroup$ When using complete or strong induction, the inductive step is not from k to k+1 .. that's weak induction. What you want to show is that if the inductive hypothesis is true for any $n $\endgroup$
Bram28
– Bram28
2017-07-12 02:44:39 +00:00
Commented Jul 12, 2017 at 2:44
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3 Answers 3
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$\begingroup$
First, a couple of comments on how you think about, and write/present this proof.
When doing induction, it is always a good idea to get very clear on exactly what the claim is that you are trying to prove, and thus what the property is that you want to show all natural numbers have. So in this case, that would be the claim that for any number $n$: if you start with a pile of $n$ stones, then no matter how you keep splitting it, the eventual sum of products will be the same and in fact will be $\frac{n(n-1)}{2}$.
It is also a good idea to try to conceptually understand why you would want to use some specific induction method to prove your claim. In this case, you want to prove your formula by strong induction, since you can split any pile anywhere, and you want to show that no matter what smaller size piles you end up with, the formula always holds.
Now, as a base case you can in fact just use $n=1$, in which case there is noting to do, and so the sum of the products is $0$, which is indeed what you get when plugging in $n=1$ for your formula.
For the inductive 'step', again get clear on what exactly the inductive hypothesis is (indeed, you make it seem like the step is separated, and comes only after the inductive hypothesis, but the inductive hypotheses is part of the step, so I strongly discourage the way you do this in your post). Now, you write: $n$ stones can eb split into $n$ and $k-n$ stones. OK, so first of all, that first $n$ should be $k$, but I understand that was just a typo. But far more importantly, this is not the inductive assumption. In fact, it is not even an interesting claim: of course a pile of $k$ stones can be split in two! The inductive assumption is that the earlier claim is true for all piles of size $n
Step: Let $k$ be some arbitrary number greater than the base case (i.e. $k>1$). The inductive hypothesis is that for any $n
OK, so now we try to show that under that assumption it would follow that the claim is true for a pile of size $k$. Ok, so let's consider a pile of size $k$. Does it matter where we split it? No. Let's split it into piles of size $m$ and $k-m$. By the inductive hypothesis, these two piles will end up with a sum of products of $\frac{m(m-1)}{2}$ and $\frac{(k-m)(k-m-1)}{2}$ respectively, but you also obtained a product of $m(k-m)$ by this very split. Hence, the eventual sum of products will be:
$$\frac{m(m-1)}{2}+\frac{(k-m)(k-m-1)}{2}+m(k-m)=$$
$$\frac{m(m-1)+(k-m)(k-m-1)+2m(k-m)}{2}=$$
$$\frac{m^2-m+k^2-mk-k-mk+m^2+m+2mk-2m^2}{2}=$$
$$\frac{k^2-k}{2}=\frac{k(k-1)}{2}$$
And thus we find the formula is true for any pile of size $k$ as well, no matter where you split it.
Finally, you may want to wrap things up, and say something like:
Since we have proven the base and the step, we have completed the proof by strong induction, and hence the claim is proven.
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edited Jul 12, 2017 at 12:27
answered Jul 12, 2017 at 3:29
Bram28Bram28
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I think the first step would be to try to figure out what the formula for the sum would be. If James has $n$ stones, one way he could work through the process is to remove one stone from the pile each time (putting it into its own 1-stone pile). So at first he would have two piles, one with $n-1$ stones and the other with 1 stone - so a product of $n-1$. Then in the next step he would have three piles, with $n-2$, 1, and 1 stones - a product of $n-2$. And so on, until the sum of his products is
$$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$$
Then, for your induction step, you would assume that for any $2 \leq k \leq n$, the sum of the products at the end of all the pile dividing is $\frac{(k-1)k}{2}$. You would use that to show that if you have $n+1$ stones, the sum of the products would be $\frac{n(n+1)}{2}$. So, you'd have to think about what your piles would look like after the first division, and what conclusions you could draw from that - can you take it from there?
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answered Jul 12, 2017 at 1:57
emmaemma
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$\begingroup$ Whoops, I started working on this before you added in your formula - but obviously I agree with you! $\endgroup$
emma
– emma
2017-07-12 01:58:08 +00:00
Commented Jul 12, 2017 at 1:58
$\begingroup$ Thanks Emma! This makes a lot of sense!! Could we just plug in n+1 into both sides of the equation to prove our inductive step? Or do we need to go about figuring out cases and solving those? $\endgroup$
Keith Axelrod
– Keith Axelrod
2017-07-12 02:18:15 +00:00
Commented Jul 12, 2017 at 2:18
Add a comment |
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The answer is 4 8 15 16 23 42
Joke aside what you actually need is iterated induction (sorta).
Some setup:
The ordering of two piles is irrelevant. A pile of 2 and a pile of 3 is the same as a pile of 3 and a pile of 2. This is important, and we need not replicate these cases when resolving the base case.
Proof:
We will use complete induction. For the base case suppose that $n=2$. It can only be split in one manner and therefore the sum is equal to itself.
For the inductive step suppose that the statement is true for all piles of size $k$, where $1 < k \leq m$. We can therefore define a function on the range of $k$ called $p(k)$, which is defined to return the unique pile sums. We seek to prove the case that $n = m+1$. Let us consider the case where we split it into piles of size $1$ and $m$. The resulting sum for $m$ is $p(m)$ by induction. Therefore the sum for that case is $m + p(m)$.
Now suppose that we split the pile into two piles $s$ and $n-s$, where $1 \leq s < n$. By induction we then have that the pile sum is $ns - s^2 + p(s) + p(n-s)$. However $n = m+1$ and so $ms + s - s^2 + p(s) + p(m+1-s)$. We can identify $p(s) = \sum_{i=2}^{s} i-1$. Therefore, $p(s) + s = s + \sum_{i=2}^{s} i-1 = \sum_{i=2}^{s+1} i-1 = p(s+1)$. This means that the current pile sum is now $ms - s^2 + p(s+1) + p(m+1-s) = m + s(m - s) + p(s) + p(m-s) = m + p(m)$. Therefore, we have proven that all pile sums equal the case of when we split it into a pile of size $1$ and $m$. Therefore, by transitivity all the pile sums for the case $n = m+1$ are equal. This completes the base case.
Therefore, by induction all piles of integer size $n \geq 2$ have a unique pile sum.
Remarks:
You can prove that the pile function is the n-series. I won't do that for you. It is just a matter of pulling out nothing but 1's once you have uniqueness.
Note that this could be written as induction upon the pile splitting and the pile size. I worked around it, but feel free to do that if it is easier for you.
In fact, just answer this how you feel it is best answered by you. This is just an example proof.
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answered Jul 12, 2017 at 2:38
user64742user64742
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17614 | http://103.203.175.90:81/fdScript/RootOfEBooks/E%20Book%20collection%20-%202020/MED/WHITE%20-%20livro.pdf | Fluid Mechanics McGraw-Hill Series in Mechanical Engineering CONSULTING EDITORS Jack P. Holman, Southern Methodist University John Lloyd, Michigan State University Anderson Computational Fluid Dynamics: The Basics with Applications Anderson Modern Compressible Flow: With Historical Perspective Arora Introduction to Optimum Design Borman and Ragland Combustion Engineering Burton Introduction to Dynamic Systems Analysis Culp Principles of Energy Conversion Dieter Engineering Design: A Materials & Processing Approach Doebelin Engineering Experimentation: Planning, Execution, Reporting Driels Linear Control Systems Engineering Edwards and McKee Fundamentals of Mechanical Component Design Gebhart Heat Conduction and Mass Diffusion Gibson Principles of Composite Material Mechanics Hamrock Fundamentals of Fluid Film Lubrication Heywood Internal Combustion Engine Fundamentals Hinze Turbulence Histand and Alciatore Introduction to Mechatronics and Measurement Systems Holman Experimental Methods for Engineers Howell and Buckius Fundamentals of Engineering Thermodynamics Jaluria Design and Optimization of Thermal Systems Juvinall Engineering Considerations of Stress, Strain, and Strength Kays and Crawford Convective Heat and Mass Transfer Kelly Fundamentals of Mechanical Vibrations Kimbrell Kinematics Analysis and Synthesis Kreider and Rabl Heating and Cooling of Buildings Martin Kinematics and Dynamics of Machines Mattingly Elements of Gas Turbine Propulsion Modest Radiative Heat Transfer Norton Design of Machinery Oosthuizen and Carscallen Compressible Fluid Flow Oosthuizen and Naylor Introduction to Convective Heat Transfer Analysis Phelan Fundamentals of Mechanical Design Reddy An Introduction to Finite Element Method Rosenberg and Karnopp Introduction to Physical Systems Dynamics Schlichting Boundary-Layer Theory Shames Mechanics of Fluids Shigley Kinematic Analysis of Mechanisms Shigley and Mischke Mechanical Engineering Design Shigley and Uicker Theory of Machines and Mechanisms Stiffler Design with Microprocessors for Mechanical Engineers Stoecker and Jones Refrigeration and Air Conditioning Turns An Introduction to Combustion: Concepts and Applications Ullman The Mechanical Design Process Wark Advanced Thermodynamics for Engineers Wark and Richards Thermodynamics White Viscous Fluid Flow Zeid CAD/CAM Theory and Practice Fluid Mechanics Fourth Edition Frank M. White University of Rhode Island Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Bogotá Caracas Lisbon London Madrid Mexico City Milan New Delhi Seoul Singapore Sydney Taipei Toronto About the Author Frank M. White is Professor 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 text-books on fluid mechanics and heat transfer.
During 1979–1990 he was editor-in-chief of the ASME Journal of Fluids Engi-neering and then served from 1991 to 1997 as chairman of the ASME Board of Edi-tors 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 General Approach xi Preface The fourth edition of this textbook sees some additions and deletions but no philo-sophical change. The basic outline of eleven chapters and five appendices remains the same. The triad of integral, differential, and experimental approaches is retained and is approached in that order of presentation. The book is intended for an undergraduate course in fluid mechanics, and there is plenty of material for a full year of instruction.
The author covers the first six chapters and part of Chapter 7 in the introductory se-mester. The more specialized and applied topics from Chapters 7 to 11 are then cov-ered at our university in a second semester. The informal, student-oriented style is re-tained and, if it succeeds, has the flavor of an interactive lecture by the author.
Approximately 30 percent of the problem exercises, and some fully worked examples, have been changed or are new. The total number of problem exercises has increased to more than 1500 in this fourth edition. The focus of the new problems is on practi-cal and realistic fluids engineering experiences. Problems are grouped according to topic, and some are labeled either with an asterisk (especially challenging) or a com-puter-disk icon (where computer solution is recommended). A number of new pho-tographs and figures have been added, especially to illustrate new design applications and new instruments.
Professor John Cimbala, of Pennsylvania State University, contributed many of the new problems. He had the great idea of setting comprehensive problems at the end of each chapter, covering a broad range of concepts, often from several different chap-ters. These comprehensive problems grow and recur throughout the book as new con-cepts arise. Six more open-ended design projects have been added, making 15 projects in all. The projects allow the student to set sizes and parameters and achieve good de-sign with more than one approach.
An entirely new addition is a set of 95 multiple-choice problems suitable for prepar-ing for the Fundamentals of Engineering (FE) Examination. These FE problems come at the end of Chapters 1 to 10. Meant as a realistic practice for the actual FE Exam, they are engineering problems with five suggested answers, all of them plausible, but only one of them correct.
Learning Tools Content Changes New to this book, and to any fluid mechanics textbook, is a special appendix, Ap-pendix E, Introduction to the Engineering Equation Solver (EES), which is keyed to many examples and problems throughout the book. The author finds EES to be an ex-tremely attractive tool for applied engineering problems. Not only does it solve arbi-trarily complex systems of equations, written in any order or form, but also it has built-in property evaluations (density, viscosity, enthalpy, entropy, etc.), linear and nonlinear regression, and easily formatted parameter studies and publication-quality plotting. The author is indebted to Professors Sanford Klein and William Beckman, of the Univer-sity of Wisconsin, for invaluable and continuous help in preparing this EES material.
The book is now available with or without an EES problems disk. The EES engine is available to adopters of the text with the problems disk.
Another welcome addition, especially for students, is Answers to Selected Prob-lems. Over 600 answers are provided, or about 43 percent of all the regular problem assignments. Thus a compromise is struck between sometimes having a specific nu-merical goal and sometimes directly applying yourself and hoping for the best result.
There are revisions in every chapter. Chapter 1—which is purely introductory and could be assigned as reading—has been toned down from earlier editions. For ex-ample, the discussion of the fluid acceleration vector has been moved entirely to Chap-ter 4. Four brief new sections have been added: (1) the uncertainty of engineering data, (2) the use of EES, (3) the FE Examination, and (4) recommended problem-solving techniques.
Chapter 2 has an improved discussion of the stability of floating bodies, with a fully derived formula for computing the metacentric height. Coverage is confined to static fluids and rigid-body motions. An improved section on pressure measurement discusses modern microsensors, such as the fused-quartz bourdon tube, micromachined silicon capacitive and piezoelectric sensors, and tiny (2 mm long) silicon resonant-frequency devices.
Chapter 3 tightens up the energy equation discussion and retains the plan that Bernoulli’s equation comes last, after control-volume mass, linear momentum, angu-lar momentum, and energy studies. Although some texts begin with an entire chapter on the Bernoulli equation, this author tries to stress that it is a dangerously restricted relation which is often misused by both students and graduate engineers.
In Chapter 4 a few inviscid and viscous flow examples have been added to the ba-sic partial differential equations of fluid mechanics. More extensive discussion con-tinues in Chapter 8.
Chapter 5 is more successful when one selects scaling variables before using the pi theorem. Nevertheless, students still complain that the problems are too ambiguous and lead to too many different parameter groups. Several problem assignments now con-tain a few hints about selecting the repeating variables to arrive at traditional pi groups.
In Chapter 6, the “alternate forms of the Moody chart” have been resurrected as problem assignments. Meanwhile, the three basic pipe-flow problems—pressure drop, flow rate, and pipe sizing—can easily be handled by the EES software, and examples are given. Some newer flowmeter descriptions have been added for further enrichment.
Chapter 7 has added some new data on drag and resistance of various bodies, notably biological systems which adapt to the flow of wind and water.
xii Preface Supplements EES Software Chapter 8 picks up from the sample plane potential flows of Section 4.10 and plunges right into inviscid-flow analysis, especially aerodynamics. The discussion of numeri-cal methods, or computational fluid dynamics (CFD), both inviscid and viscous, steady and unsteady, has been greatly expanded. Chapter 9, with its myriad complex algebraic equations, illustrates the type of examples and problem assignments which can be solved more easily using EES. A new section has been added about the suborbital X-33 and VentureStar vehicles.
In the discussion of open-channel flow, Chapter 10, we have further attempted to make the material more attractive to civil engineers by adding real-world comprehen-sive problems and design projects from the author’s experience with hydropower proj-ects. More emphasis is placed on the use of friction factors rather than on the Man-ning roughness parameter. Chapter 11, on turbomachinery, has added new material on compressors and the delivery of gases. Some additional fluid properties and formulas have been included in the appendices, which are otherwise much the same.
The all new Instructor’s Resource CD contains a PowerPoint presentation of key text figures as well as additional helpful teaching tools. The list of films and videos, for-merly App. C, is now omitted and relegated to the Instructor’s Resource CD.
The Solutions Manual provides complete and detailed solutions, including prob-lem statements and artwork, to the end-of-chapter problems. It may be photocopied for posting or preparing transparencies for the classroom.
The Engineering Equation Solver (EES) was developed by Sandy Klein and Bill Beck-man, both of the University of Wisconsin—Madison. A combination of equation-solving capability and engineering property data makes EES an extremely powerful tool for your students. EES (pronounced “ease”) enables students to solve problems, especially design problems, and to ask “what if” questions. EES can do optimization, parametric analysis, linear and nonlinear regression, and provide publication-quality plotting capability. Sim-ple to master, this software allows you to enter equations in any form and in any order. It automatically rearranges the equations to solve them in the most efficient manner.
EES is particularly useful for fluid mechanics problems since much of the property data needed for solving problems in these areas are provided in the program. Air ta-bles are built-in, as are psychometric functions and Joint Army Navy Air Force (JANAF) table data for many common gases. Transport properties are also provided for all sub-stances. EES allows the user to enter property data or functional relationships written in Pascal, C, C, or Fortran. The EES engine is available free to qualified adopters via a password-protected website, to those who adopt the text with the problems disk.
The program is updated every semester.
The EES software problems disk provides examples of typical problems in this text.
Problems solved are denoted in the text with a disk symbol. Each fully documented solution is actually an EES program that is run using the EES engine. Each program provides detailed comments and on-line help. These programs illustrate the use of EES and help the student master the important concepts without the calculational burden that has been previously required.
Preface xiii Acknowledgments So many people have helped me, in addition to Professors John Cimbala, Sanford Klein, and William Beckman, that I cannot remember or list them all. I would like to express my appreciation to many reviewers and correspondents who gave detailed suggestions and materials: Osama Ibrahim, University of Rhode Island; Richard Lessmann, Uni-versity of Rhode Island; William Palm, University of Rhode Island; Deborah Pence, University of Rhode Island; Stuart Tison, National Institute of Standards and Technol-ogy; Paul Lupke, Druck Inc.; Ray Worden, Russka, Inc.; Amy Flanagan, Russka, Inc.; Søren Thalund, Greenland Tourism a/s; Eric Bjerregaard, Greenland Tourism a/s; Mar-tin Girard, DH Instruments, Inc.; Michael Norton, Nielsen-Kellerman Co.; Lisa Colomb, Johnson-Yokogawa Corp.; K. Eisele, Sulzer Innotec, Inc.; Z. Zhang, Sultzer Innotec, Inc.; Helen Reed, Arizona State University; F. Abdel Azim El-Sayed, Zagazig University; Georges Aigret, Chimay, Belgium; X. He, Drexel University; Robert Lo-erke, Colorado State University; Tim Wei, Rutgers University; Tom Conlisk, Ohio State University; David Nelson, Michigan Technological University; Robert Granger, U.S.
Naval Academy; Larry Pochop, University of Wyoming; Robert Kirchhoff, University of Massachusetts; Steven Vogel, Duke University; Capt. Jason Durfee, U.S. Military Academy; Capt. Mark Wilson, U.S. Military Academy; Sheldon Green, University of British Columbia; Robert Martinuzzi, University of Western Ontario; Joel Ferziger, Stanford University; Kishan Shah, Stanford University; Jack Hoyt, San Diego State University; Charles Merkle, Pennsylvania State University; Ram Balachandar, Univer-sity of Saskatchewan; Vincent Chu, McGill University; and David Bogard, University of Texas at Austin. The editorial and production staff at WCB McGraw-Hill have been most helpful throughout this project. Special thanks go to Debra Riegert, Holly Stark, Margaret Rathke, Michael Warrell, Heather Burbridge, Sharon Miller, Judy Feldman, and Jen-nifer Frazier. Finally, I continue to enjoy the support of my wife and family in these writing efforts.
xiv Preface 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 14 1.6 Thermodynamic Properties of a Fluid 16 1.7 Viscosity and Other Secondary Properties 22 1.8 Basic Flow-Analysis Techniques 35 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 37 1.10 The Engineering Equation Solver 41 1.11 Uncertainty of Experimental Data 42 1.12 The Fundamentals of Engineering (FE) Examination 43 1.13 Problem-Solving Techniques 44 1.14 History and Scope of Fluid Mechanics 44 Problems 46 Fundamentals of Engineering Exam Problems 53 Comprehensive Problems 54 References 55 Chapter 2 Pressure Distribution in a Fluid 59 2.1 Pressure and Pressure Gradient 59 2.2 Equilibrium of a Fluid Element 61 2.3 Hydrostatic Pressure Distributions 63 2.4 Application to Manometry 70 2.5 Hydrostatic Forces on Plane Surfaces 74 vii Contents 2.6 Hydrostatic Forces on Curved Surfaces 79 2.7 Hydrostatic Forces in Layered Fluids 82 2.8 Buoyancy and Stability 84 2.9 Pressure Distribution in Rigid-Body Motion 89 2.10 Pressure Measurement 97 Summary 100 Problems 102 Word Problems 125 Fundamentals of Engineering Exam Problems 125 Comprehensive Problems 126 Design Projects 127 References 127 Chapter 3 Integral Relations for a Control Volume 129 3.1 Basic Physical Laws of Fluid Mechanics 129 3.2 The Reynolds Transport Theorem 133 3.3 Conservation of Mass 141 3.4 The Linear Momentum Equation 146 3.5 The Angular-Momentum Theorem 158 3.6 The Energy Equation 163 3.7 Frictionless Flow: The Bernoulli Equation 174 Summary 183 Problems 184 Word Problems 210 Fundamentals of Engineering Exam Problems 210 Comprehensive Problems 211 Design Project 212 References 213 Chapter 4 Differential Relations for a Fluid Particle 215 4.1 The Acceleration Field of a Fluid 215 4.2 The Differential Equation of Mass Conservation 217 4.3 The Differential Equation of Linear Momentum 223 4.4 The Differential Equation of Angular Momentum 230 4.5 The Differential Equation of Energy 231 4.6 Boundary Conditions for the Basic Equations 234 4.7 The Stream Function 238 4.8 Vorticity and Irrotationality 245 4.9 Frictionless Irrotational Flows 247 4.10 Some Illustrative Plane Potential Flows 252 4.11 Some Illustrative Incompressible Viscous Flows 258 Summary 263 Problems 264 Word Problems 273 Fundamentals of Engineering Exam Problems 273 Comprehensive Applied Problem 274 References 275 Chapter 5 Dimensional Analysis and Similarity 277 5.1 Introduction 277 5.2 The Principle of Dimensional Homogeneity 280 5.3 The Pi Theorem 286 5.4 Nondimensionalization of the Basic Equations 292 5.5 Modeling and Its Pitfalls 301 Summary 311 Problems 311 Word Problems 318 Fundamentals of Engineering Exam Problems 319 Comprehensive Problems 319 Design Projects 320 References 321 Chapter 6 Viscous Flow in Ducts 325 6.1 Reynolds-Number Regimes 325 6.2 Internal versus External Viscous Flows 330 6.3 Semiempirical Turbulent Shear Correlations 333 6.4 Flow in a Circular Pipe 338 viii Contents 6.5 Three Types of Pipe-Flow Problems 351 6.6 Flow in Noncircular Ducts 357 6.7 Minor Losses in Pipe Systems 367 6.8 Multiple-Pipe Systems 375 6.9 Experimental Duct Flows: Diffuser Performance 381 6.10 Fluid Meters 385 Summary 404 Problems 405 Word Problems 420 Fundamentals of Engineering Exam Problems 420 Comprehensive Problems 421 Design Projects 422 References 423 Chapter 7 Flow Past Immersed Bodies 427 7.1 Reynolds-Number and Geometry Effects 427 7.2 Momentum-Integral Estimates 431 7.3 The Boundary-Layer Equations 434 7.4 The Flat-Plate Boundary Layer 436 7.5 Boundary Layers with Pressure Gradient 445 7.6 Experimental External Flows 451 Summary 476 Problems 476 Word Problems 489 Fundamentals of Engineering Exam Problems 489 Comprehensive Problems 490 Design Project 491 References 491 Chapter 8 Potential Flow and Computational Fluid Dynamics 495 8.1 Introduction and Review 495 8.2 Elementary Plane-Flow Solutions 498 8.3 Superposition of Plane-Flow Solutions 500 8.4 Plane Flow Past Closed-Body Shapes 507 8.5 Other Plane Potential Flows 516 8.6 Images 521 8.7 Airfoil Theory 523 8.8 Axisymmetric Potential Flow 534 8.9 Numerical Analysis 540 Summary 555 Problems 555 Word Problems 566 Comprehensive Problems 566 Design Projects 567 References 567 Chapter 9 Compressible Flow 571 9.1 Introduction 571 9.2 The Speed of Sound 575 9.3 Adiabatic and Isentropic Steady Flow 578 9.4 Isentropic Flow with Area Changes 583 9.5 The Normal-Shock Wave 590 9.6 Operation of Converging and Diverging Nozzles 598 9.7 Compressible Duct Flow with Friction 603 9.8 Frictionless Duct Flow with Heat Transfer 613 9.9 Two-Dimensional Supersonic Flow 618 9.10 Prandtl-Meyer Expansion Waves 628 Summary 640 Problems 641 Word Problems 653 Fundamentals of Engineering Exam Problems 653 Comprehensive Problems 654 Design Projects 654 References 655 Chapter 10 Open-Channel Flow 659 10.1 Introduction 659 10.2 Uniform Flow; the Chézy Formula 664 10.3 Efficient Uniform-Flow Channels 669 10.4 Specific Energy; Critical Depth 671 10.5 The Hydraulic Jump 678 10.6 Gradually Varied Flow 682 10.7 Flow Measurement and Control by Weirs 687 Summary 695 Contents ix Problems 695 Word Problems 706 Fundamentals of Engineering Exam Problems 707 Comprehensive Problems 707 Design Projects 707 References 708 Chapter 11 Turbomachinery 711 11.1 Introduction and Classification 711 11.2 The Centrifugal Pump 714 11.3 Pump Performance Curves and Similarity Rules 720 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 729 11.5 Matching Pumps to System Characteristics 735 11.6 Turbines 742 Summary 755 Problems 755 Word Problems 765 Comprehensive Problems 766 Design Project 767 References 767 Appendix A Physical Properties of Fluids 769 Appendix B Compressible-Flow Tables 774 Appendix C Conversion Factors 791 Appendix D Equations of Motion in Cylindrical Coordinates 793 Appendix E Introduction to EES 795 Answers to Selected Problems 806 Index 813 Hurricane Elena in the Gulf of Mexico. Unlike most small-scale fluids engineering applications, hurricanes are strongly affected by the Coriolis acceleration due to the rotation of the earth, which causes them to swirl counterclockwise in the Northern Hemisphere. The physical properties and boundary conditions which govern such flows are discussed in the present chapter. (Courtesy of NASA/Color-Pic Inc./E.R. Degginger/Color-Pic Inc.) 2 1.1 Preliminary Remarks Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid statics) and the subsequent effects of the fluid upon the boundaries, which may be ei-ther solid surfaces or interfaces with other fluids. Both gases and liquids are classified as fluids, and the number of fluids engineering 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. How-ever, 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 the-ory 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 pos-sible 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, 2, 29].1 This book will present many theoretical re-sults 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 diffi-culty of the basic equations, although the boundary-layer approximation found by Lud-wig 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 ve-locities, to a disorderly, random phenomenon called turbulence. The theory of turbu-lent flow is crude and heavily backed up by experiment (Chap. 6), yet it can be quite serviceable as an engineering estimate. Textbooks now present digital-computer tech-niques for turbulent-flow analysis , but they are based strictly upon empirical as-sumptions regarding the time mean of the turbulent stress field.
Chapter 1 Introduction 3 1Numbered references appear at the end of each chapter.
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 in-formation about specific flows, such as the drag and lift of immersed bodies (Chap. 7).
Fortunately, fluid mechanics is a highly visual subject, with good instrumentation [4, 5, 35], and the use of dimensional analysis and modeling concepts (Chap. 5) is wide-spread. 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 tech-nical 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 deformation; 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 corol-lary, we can say that a fluid at rest must be in a state of zero shear stress, a state of-ten called the hydrostatic stress condition in structural analysis. In this condition, 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 the definition of a fluid above, 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 mol-ecules with strong cohesive forces, tends to retain its volume and will form a free sur-face 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 en-counters confining walls. A gas has no definite volume, and when left to itself with-out confinement, a gas forms an atmosphere which is essentially hydrostatic. The hy-drostatic 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 through A.
Since the block sides are unsupported, element A has zero stress on the left and right sides and compression stress 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 or-der to eliminate shear stress. The walls exert a compression stress of p and reduce Mohr’s circle to a point with zero shear everywhere, i.e., the hydrostatic condition. The liquid retains its volume and forms a free surface in the container. If the walls are re-moved, 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 configura-4 Chapter 1 Introduction tion, pouring out over the lip if necessary. Meanwhile, the gas is unrestrained and ex-pands out of the container, filling all available space. Element A in the gas is also hy-drostatic and exerts a compression stress p on the walls.
In the above discussion, clear decisions could be made about solids, liquids, and gases. Most engineering fluid-mechanics problems deal with these clear cases, i.e., 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 temperature and pres-sure 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 peri-ods. 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 de-voted 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 mix-tures or water with entrapped air bubbles. Specialized textbooks present the analysis 1.2 The Concept of a Fluid 5 Static deflection Free surface Hydrostatic condition Liquid Solid A A A (a) (c) (b) (d) 0 0 A A Gas (1) – p – p p p p = 0 τ θ θ θ 2 1 – = p – = p σ σ 1 τ σ τ σ τ σ 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.
1.3 The Fluid as a Continuum of such two-phase flows . Finally, there are situations where the distinction between a liquid and a gas blurs. This is the case at temperatures and pressures above the so-called critical point of a substance, where only a single phase exists, primarily resem-bling a gas. As pressure increases far above the critical point, the gaslike substance be-comes 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 Tc 647 K and pc 219 atm,2 so that typical problems in-volving 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 Tc 126 K and pc 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 discussed above will be beyond our scope.
We have already used technical terms such as fluid pressure and density without a rig-orous discussion of their definition. As far as we know, fluids are aggregations of mol-ecules, widely spaced for a gas, closely spaced for a liquid. The distance between mol-ecules 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 in-terchange 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 m within a given volume is plotted versus the size of the unit volume. There is a limiting volume below which molecular variations may be important and 6 Chapter 1 Introduction Microscopic uncertainty Macroscopic uncertainty 0 1200 δ δ ≈ 10-9 mm3 Elemental volume Region containing fluid = 1000 kg/m3 = 1100 = 1200 = 1300 (a) (b) ρ ρ ρ ρ ρ δ Fig. 1.2 The limit definition of con-tinuum fluid density: (a) an ele-mental volume in a fluid region of variable continuum density; (b) cal-culated density versus size of the elemental volume.
2One atmosphere equals 2116 lbf/ft2 101,300 Pa.
1.4 Dimensions and Units above which aggregate variations may be important. The density of a fluid is best defined as lim → m (1.1) The limiting volume is about 109 mm3 for all liquids and for gases at atmospheric pressure. For example, 109 mm3 of air at standard conditions contains approximately 3 107 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 proper-ties 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 the 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 path3 are com-parable to, or larger than, the physical size of the system. This requires that the contin-uum 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 pre-clude 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 Chap. 4 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 ex-pressing length. Dimension is a powerful concept about which a splendid tool called dimensional analysis has been developed (Chap. 5), while units are the nitty-gritty, the number which the customer wants as the final answer.
Systems of units have always varied widely from country to country, even after in-ternational agreements have been reached. Engineers need numbers and therefore unit systems, and the numbers must be accurate because the safety of the public is at stake.
You cannot design and build a piping system whose diameter is D and whose length is L. And U.S. engineers have persisted too long in clinging to British systems of units.
There is too much margin for error in most British systems, and many an engineering student has flunked a test because of a missing or improper conversion factor of 12 or 144 or 32.2 or 60 or 1.8. Practicing engineers can make the same errors. The writer is aware from personal experience of a serious preliminary error in the design of an air-craft due to a missing factor of 32.2 to convert pounds of mass to slugs.
In 1872 an international meeting in France proposed a treaty called the Metric Con-vention, 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 foundation of our number system, learned from childhood by all. Problems still remained because 1.4 Dimensions and Units 7 3The mean distance traveled by molecules between collisions.
even the metric countries differed in their use of kiloponds instead of dynes or new-tons, kilograms instead of grams, or calories instead of joules. To standardize the met-ric system, a General Conference of Weights and Measures attended in 1960 by 40 countries proposed the International System of Units (SI). We are now undergoing a painful period of transition to SI, an adjustment which 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 Engi-neers, which prepared a useful booklet 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.4 These dimensions and 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 { }. For example, ac-celeration has the dimensions {LT2}. The most crucial of these secondary dimensions is force, which is directly related to mass, length, and time by Newton’s second law F ma (1.2) From this we see that, dimensionally, {F} {MLT2}. A constant of proportionality is avoided by defining the force unit exactly in terms of the primary units. Thus we define the newton and the pound of force 1 newton of force 1 N 1 kg m/s2 (1.3) 1 pound of force 1 lbf 1 slug ft/s2 4.4482 N In this book the abbreviation lbf is used for pound-force and lb for pound-mass. If in-stead one adopts other force units such as the dyne or the poundal or kilopond or adopts other mass units such as the gram or pound-mass, a constant of proportionality called gc must be included in Eq. (1.2). We shall not use gc in this book since it is not nec-essary in the SI and BG systems.
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.
8 Chapter 1 Introduction 4If electromagnetic effects are important, a fifth primary dimension must be included, electric current {I}, whose SI unit is the ampere (A).
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 Table 1.1 Primary Dimensions in SI and BG Systems Primary Dimensions Part (a) Part (b) Part (c) EXAMPLE 1.1 A body weighs 1000 lbf when exposed to a standard earth gravity g 32.174 ft/s2. (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 stan-dard acceleration gmoon 1.62 m/s2? (c) How fast will the body accelerate if a net force of 400 lbf is applied to it on the moon or on the earth?
Solution Equation (1.2) holds with F weight and a gearth: F W mg 1000 lbf (m slugs)(32.174 ft/s2) or m 3 1 2 0 .1 0 7 0 4 (31.08 slugs)(14.5939 kg/slug) 453.6 kg Ans. (a) The change from 31.08 slugs to 453.6 kg illustrates the proper use of the conversion factor 14.5939 kg/slug.
The mass of the body remains 453.6 kg regardless of its location. Equation (1.2) applies with a new value of a and hence a new force F Wmoon mgmoon (453.6 kg)(1.62 m/s2) 735 N Ans. (b) This problem does not involve weight or gravity or position and is simply a direct application of Newton’s law with an unbalanced force: F 400 lbf ma (31.08 slugs)(a ft/s2) or a 3 4 1 0 .0 0 8 12.43 ft/s2 3.79 m/s2 Ans. (c) This acceleration would be the same on the moon or earth or anywhere.
1.4 Dimensions and Units 9 Secondary dimension SI unit BG unit Conversion factor Area {L2} m2 ft2 1 m2 10.764 ft2 Volume {L3} m3 ft3 1 m3 35.315 ft3 Velocity {LT1} m/s ft/s 1 ft/s 0.3048 m/s Acceleration {LT2} m/s2 ft/s2 1 ft/s2 0.3048 m/s2 Pressure or stress {ML1T2} Pa N/m2 lbf/ft2 1 lbf/ft2 47.88 Pa Angular velocity {T1} s1 s1 1 s1 1 s1 Energy, heat, work {ML2T2} J N m ft lbf 1 ft lbf 1.3558 J Power {ML2T3} W J/s ft lbf/s 1 ft lbf/s 1.3558 W Density {ML3} kg/m3 slugs/ft3 1 slug/ft3 515.4 kg/m3 Viscosity {ML1T1} kg/(m s) slugs/(ft s) 1 slug/(ft s) 47.88 kg/(m s) Specific heat {L2T2 1} m2/(s2 K) ft2/(s2 °R) 1 m2/(s2 K) 5.980 ft2/(s2 °R) Table 1.2 Secondary Dimensions in Fluid Mechanics Part (a) Part (b) 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 con-version factors, as in the following example.
EXAMPLE 1.2 An early viscosity unit in the cgs system is the poise (abbreviated P), or g/(cm s), named after J. L. M. Poiseuille, a French physician who performed pioneering experiments in 1840 on wa-ter flow in pipes. The viscosity of water (fresh or salt) at 293.16 K 20°C is approximately 0.01 P. Express this value in (a) SI and (b) BG units.
Solution [0.01 g/(cm s)] 10 1 0 k 0 g g (100 cm/m) 0.001 kg/(m s) Ans. (a) [0.001 kg/(m s)] 14 1 .5 sl 9 ug kg (0.3048 m/ft) 2.09 105 slug/(ft s) Ans. (b) Note: Result (b) could have been found directly from (a) by dividing (a) by the viscosity con-version factor 47.88 listed in Table 1.2.
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 work5 is the Bernoulli relation, named after Daniel Bernoulli, who published a hy-drodynamics textbook in 1738: p0 p 1 2 V2 gZ (1) where p0 stagnation pressure p pressure in moving fluid V velocity density Z altitude g gravitational acceleration 10 Chapter 1 Introduction 5That’s an awful lot of assumptions, which need further study in Chap. 3.
Part (a) Part (b) Part (c) (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.
Solution We can express Eq. (1) dimensionally, using braces by entering the dimensions of each term from Table 1.2: {ML1T2} {ML1T2} {ML3}{L2T2} {ML3}{LT2}{L} {ML1T2} for all terms Ans. (a) Enter the SI units for each quantity from Table 1.2: {N/m2} {N/m2} {kg/m3}{m2/s2} {kg/m3}{m/s2}{m} {N/m2} {kg/(m s2)} The right-hand side looks bad until we remember from Eq. (1.3) that 1 kg 1 N s2/m.
{kg/(m s2)} {N {m s 2 s / 2 m } } {N/m2} 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 theoretical equa-tions in fluid mechanics.
Introducing BG units for each term, we have {lbf/ft2} {lbf/ft2} {slugs/ft3}{ft2/s2} {slugs/ft3}{ft/s2}{ft} {lbf/ft2} {slugs/(ft s2)} But, from Eq. (1.3), 1 slug 1 lbf s2/ft, so that {slugs/(ft s2)} {l { b f f t s s 2 2 / } ft} {lbf/ft2} 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 example, stan-dard atmospheric pressure is 14.7 lbf/in2 2116 lbf/ft2 101,300 Pa. The pascal is a small unit because the newton is less than 1 4 lbf and a square meter is a very large area.
It is felt nevertheless that the pascal will gradually gain universal acceptance; e.g., re-pair manuals for U.S. automobiles now specify pressure measurements in pascals.
Note that not only must all (fluid) mechanics equations be dimensionally homogeneous, 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 Ex. 1.3, but woe unto 1.4 Dimensions and Units 11 Consistent Units Homogeneous versus Dimensionally Inconsistent Equations 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 1 2 V2 constant where h is the fluid enthalpy and V2/2 is its kinetic energy. Colloquial thermodynamic tables might list h in units of British thermal units per pound (Btu/lb), whereas V is likely used in ft/s. It is completely erroneous to add Btu/lb to ft2/s2. The proper unit for h in this case is ft lbf/slug, which is identical to ft2/s2. The conversion factor is 1 Btu/lb 25,040 ft2/s2 25,040 ft lbf/slug.
All theoretical equations in mechanics (and in other physical sciences) are dimension-ally homogeneous; i.e., each additive term in the equation has the same dimensions.
For example, Bernoulli’s equation (1) in Example 1.3 is dimensionally homogeneous: Each term has the dimensions of pressure or stress of {F/L2}. Another example is the equation from physics for a body falling with negligible air resistance: S S0 V0t 1 2 gt2 where S0 is initial position, V0 is initial velocity, and g is the acceleration of gravity. Each term in this relation has dimensions of length {L}. The factor 1 2 , which arises from inte-gration, is a pure (dimensionless) number, {1}. The exponent 2 is also dimensionless.
However, the reader should be warned that many empirical formulas in the engi-neering literature, arising primarily from correlations of data, are dimensionally in-consistent. Their units cannot be reconciled simply, and some terms may contain hid-den variables. An example is the formula which pipe valve manufacturers cite for liquid volume flow rate Q (m3/s) through a partially open valve: Q CV S G p 1/2 where p 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 CV is the valve flow co-efficient, which manufacturers tabulate in their valve brochures. Since SG is dimen-sionless {1}, we see that this formula is totally inconsistent, with one side being a flow rate {L3/T} and the other being the square root of a pressure drop {M1/2/L1/2T}. It fol-lows that CV must have dimensions, and rather odd ones at that: {L7/2/M1/2}. Nor is the resolution of this discrepancy clear, although one hint is that the values of CV 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 CdAopening p 1/2 where is the liquid density and A the area of the valve opening. The discharge coeffi-cient Cd is dimensionless and changes only slightly with valve size. Please believe—un-til we establish the fact in Chap. 5—that this latter is a much better formulation of the data.
12 Chapter 1 Introduction Convenient Prefixes in Powers of 10 Part (a) Part (b) Meanwhile, we conclude that dimensionally inconsistent equations, though they abound in engineering practice, are misleading and vague and even dangerous, in the sense that they are often misused outside their range of applicability.
Engineering results often are too small or too large for the common units, with too 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 106, we convert this to a concise p 114 MPa (megapascals). Similarly, t 0.000000003 s is a proofreader’s nightmare compared to the equivalent t 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.
EXAMPLE 1.4 In 1890 Robert Manning, an Irish engineer, proposed the following empirical formula for the average velocity V in uniform flow due to gravity down an open channel (BG units): V 1.
n 49 R2/3S1/2 (1) 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 n 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 Introduce dimensions for each term. The slope S, being a tangent or ratio, is dimensionless, de-noted by {unity} or {1}. Equation (1) in dimensional form is T L 1.
n 49 {L2/3}{1} This formula cannot be consistent unless {1.49/n} {L1/3/T}. If n is dimensionless (and it is never listed with units in textbooks), then the numerical value 1.49 must have units. This can be tragic to an engineer working in a different unit system unless the discrepancy is properly doc-umented. In fact, Manning’s formula, though popular, is inconsistent both dimensionally and physically and does not properly account for channel-roughness effects except in a narrow range of parameters, for water only.
From part (a), the number 1.49 must have dimensions {L1/3/T} and thus in BG units equals 1.49 ft1/3/s. By using the SI conversion factor for length we have (1.49 ft1/3/s)(0.3048 m/ft)1/3 1.00 m1/3/s Therefore Manning’s formula in SI becomes V 1 n .0 R2/3S1/2 Ans. (b) (2) 1.4 Dimensions and Units 13 Table 1.3 Convenient Prefixes for Engineering Units Multiplicative factor Prefix Symbol 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 10 deka da 101 deci d 102 centi c 103 milli m 106 micro 109 nano n 1012 pico p 1015 femto f 1018 atto a 1.5 Properties of the Velocity Field Eulerian and Lagrangian Desciptions The Velocity Field with R in m and V in m/s. 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 formu-las are dangerous and should either be reanalyzed or treated as having very limited application.
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.6 This treatment of properties as continuum-field functions distinguishes fluid mechanics from solid mechanics, where we are more likely to be interested in the trajectories of indi-vidual particles or systems.
There are two different points of view in analyzing problems in mechanics. The first view, appropriate to fluid mechanics, is concerned with the field of flow and is called the eulerian method of description. In the eulerian method we compute the pressure field p(x, y, z, t) of the flow pattern, not the pressure changes p(t) which a particle ex-periences as it moves through the field.
The second method, which follows an individual particle moving through the flow, is called the lagrangian description. The lagrangian approach, which is more appro-priate to solid mechanics, will not be treated in this book. However, certain numerical analyses of sharply bounded fluid flows, such as the motion of isolated fluid droplets, are very conveniently computed in lagrangian coordinates .
Fluid-dynamic measurements are also suited to the eulerian system. For example, when a pressure probe is introduced into a laboratory flow, it is fixed at a specific po-sition (x, y, z). Its output thus contributes to the description of the eulerian pressure field p(x, y, z, t). To simulate a lagrangian measurement, the probe would have to move downstream at the fluid particle speeds; this is sometimes done in oceanographic mea-surements, where flowmeters drift along with the prevailing currents.
The two different descriptions can be contrasted in the analysis of traffic flow along a freeway. A certain length of freeway may be selected for study and called the field of flow. Obviously, as time passes, various cars will enter and leave the field, and the identity of the specific cars within the field will constantly be changing. The traffic en-gineer ignores specific cars and concentrates on their average velocity as a function of time and position within the field, plus the flow rate or number of cars per hour pass-ing a given section of the freeway. This engineer is using an eulerian description of the traffic flow. Other investigators, such as the police or social scientists, may be inter-ested in the path or speed or destination of specific cars in the field. By following a specific car as a function of time, they are using a lagrangian description of the flow.
Foremost among the properties of a flow is the velocity field V(x, y, z, t). In fact, de-termining the velocity is often tantamount to solving a flow problem, since other prop-14 Chapter 1 Introduction 6One example where fluid-particle paths are important is in water-quality analysis of the fate of contaminant discharges.
erties follow directly from the velocity field. Chapter 2 is devoted to the calculation of the pressure field once the velocity field is known. Books on heat transfer (for exam-ple, Ref. 10) are essentially devoted to finding the temperature field from known ve-locity fields.
In general, velocity is a vector function of position and time and thus has three com-ponents 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) kw(x, y, z, t) (1.4) The use of u, v, and w instead of the more logical component notation Vx, Vy, and Vz is the result of an almost unbreakable custom in fluid mechanics.
Several other quantities, called kinematic properties, can be derived by mathemati-cally manipulating the velocity field. We list some kinematic properties here and give more details about their use and derivation in later chapters: 1.
Displacement vector: r V dt (Sec. 1.9) 2.
Acceleration: a d d V t (Sec. 4.1) 3.
Volume rate of flow: Q (V n) dA (Sec. 3.2) 4.
Volume expansion rate: 1 d d t V (Sec. 4.2) 5.
Local angular velocity: 1 2 V (Sec. 4.8) We will not illustrate any problems regarding these kinematic properties at present. The point of the list is to illustrate the type of vector operations used in fluid mechanics and to make clear the dominance of the velocity field in determining other flow properties.
Note: The fluid acceleration, item 2 above, is not as simple as it looks and actually in-volves four different terms due to the use of the chain rule in calculus (see Sec. 4.1).
EXAMPLE 1.5 Fluid flows through a contracting section of a duct, as in Fig. E1.5. A velocity probe inserted at section (1) measures a steady value u1 1 m/s, while a similar probe at section (2) records a steady u2 3 m/s. Estimate the fluid acceleration, if any, if x 10 cm.
Solution The flow is steady (not time-varying), but fluid particles clearly increase in velocity as they pass from (1) to (2). This is the concept of convective acceleration (Sec. 4.1). We may estimate the acceleration as a velocity change u divided by a time change t x/uavg: ax ve t l i o m c e ity ch c a h n a g n e ge 40 m/s2 Ans.
A simple estimate thus indicates that this seemingly innocuous flow is accelerating at 4 times (3.0 1.0 m/s)(1.0 3.0 m/s) 2(0.1 m) u2 u1 x/[ 1 2 (u1 u2)] 1.5 Properties of the Velocity Field 15 (1) (2) u1 u2 x E1.5 1.6 Thermodynamic Properties of a Fluid the acceleration of gravity. In the limit as x and t become very small, the above estimate re-duces to a partial-derivative expression for convective x-acceleration: ax,convective lim t→0 u t u u x In three-dimensional flow (Sec. 4.1) there are nine of these convective terms.
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 dis-cussion the three most common such properties 1.
Pressure p 2.
Density 3.
Temperature T These three are constant companions of the velocity vector in flow analyses. Four other thermodynamic properties become important when work, heat, and energy balances are treated (Chaps. 3 and 4): 4.
Internal energy e 5.
Enthalpy h û p/ 6.
Entropy s 7.
Specific heats cp and cv In addition, friction and heat conduction effects are governed by the two so-called trans-port properties: 8.
Coefficient of viscosity 9.
Thermal conductivity k All nine of these quantities are true thermodynamic properties which 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 temper-ature are sufficient to fix the value of all the others: (p, T) h h(p, T) (p, T) (1.5) and so on for every quantity in the list. Note that the specific volume, so important in thermodynamic analyses, is omitted here in favor of its inverse, the density .
Recall that thermodynamic properties describe the state of a system, i.e., a collec-tion of matter of fixed identity which 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: (x, y, z, t), etc.
Recall also that thermodynamics is normally concerned with static systems, whereas fluids are usually in variable motion with constantly changing properties. Do the prop-erties retain their meaning in a fluid flow which 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 m, so that a fluid subjected to sudden changes rapidly ad-16 Chapter 1 Introduction Temperature Specific Weight Density justs itself toward equilibrium. We therefore assume that all the thermodynamic prop-erties listed above exist as point functions in a flowing fluid and follow all the laws and state relations of ordinary equilibrium thermodynamics. There are, of course, im-portant nonequilibrium effects such as chemical and nuclear reactions in flowing flu-ids which are not treated in this text.
Pressure is the (compression) stress at a point in a static fluid (Fig. 1.1). Next to ve-locity, 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 lbf/ft2 101,300 Pa. High-speed (compressible) gas flows (Chap. 9), however, are indeed sensitive to the magnitude of pressure.
Temperature T is a measure of the internal energy level of a fluid. It may vary con-siderably during high-speed flow of a gas (Chap. 9). Although engineers often use Cel-sius or Fahrenheit scales for convenience, many applications in this text require ab-solute (Kelvin or Rankine) temperature scales: °R °F 459.69 K °C 273.16 If temperature differences are strong, heat transfer may be important , but our con-cern here is mainly with dynamic effects. We examine heat-transfer principles briefly in Secs. 4.5 and 9.8.
The density of a fluid, denoted by (lowercase Greek rho), is its mass per unit vol-ume. Density is highly variable in gases and increases nearly proportionally to the pres-sure level. Density in liquids is nearly constant; the density of water (about 1000 kg/m3) increases only 1 percent if the pressure is increased by a factor of 220. Thus most liq-uid flows are treated analytically as nearly “incompressible.” In general, liquids are about three orders of magnitude more dense than gases at at-mospheric pressure. The heaviest common liquid is mercury, and the lightest gas is hy-drogen. Compare their densities at 20°C and 1 atm: Mercury: 13,580 kg/m3 Hydrogen: 0.0838 kg/m3 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 di-mensional analysis (Chap. 5). Other fluid densities are listed in Tables A.3 and A.4 (in App. A).
The specific weight of a fluid, denoted by (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: g (1.6) 1.6 Thermodynamic Properties of a Fluid 17 Pressure Specific Gravity Potential and Kinetic Energies The units of are weight per unit volume, in lbf/ft3 or N/m3. In standard earth grav-ity, g 32.174 ft/s2 9.807 m/s2. Thus, e.g., the specific weights of air and water at 20°C and 1 atm are approximately air (1.205 kg/m3)(9.807 m/s2) 11.8 N/m3 0.0752 lbf/ft3 water (998 kg/m3)(9.807 m/s2) 9790 N/m3 62.4 lbf/ft3 Specific weight is very useful in the hydrostatic-pressure applications of Chap. 2. Spe-cific 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, water (for liquids), and air (for gases): SGgas g a a ir s 1.20 5 g k as g/m3 (1.7) SGliquid l w iq a u te id r 99 8 li k qu g i / d m3 For example, the specific gravity of mercury (Hg) is SGHg 13,580/998 13.6. En-gineers 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 molecu-lar activity and molecular bonding forces. This is commonly denoted as internal en-ergy û. A commonly accepted adjustment to this static situation for fluid flow is to add two more energy terms which arise from newtonian mechanics: the potential energy and kinetic energy.
The potential energy equals the work required to move the system of mass m from the origin to a position vector r ix jy kz against a gravity field g. Its value is mg r, or g r per unit mass. The kinetic energy equals the work required to change the speed of the mass from zero to velocity V. Its value is 1 2 mV2 or 1 2 V2 per unit mass.
Then by common convention the total stored energy e per unit mass in fluid mechan-ics is the sum of three terms: e û 1 2 V2 (g r) (1.8) Also, throughout this book we shall define z as upward, so that g gk and g r gz. Then Eq. (1.8) becomes e û 1 2 V2 gz (1.9) The molecular internal energy û is a function of T and p for the single-phase pure sub-stance, whereas the potential and kinetic energies are kinematic properties.
Thermodynamic properties are found both theoretically and experimentally to be re-lated to each other by state relations which differ for each substance. As mentioned, 18 Chapter 1 Introduction State Relations for Gases we shall confine ourselves here to single-phase pure substances, e.g., water in its liq-uid phase. The second most common fluid, air, is a mixture of gases, but since the mix-ture 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 RT R cp cv gas constant (1.10) Since Eq. (1.10) is dimensionally consistent, R has the same dimensions as specific heat, {L2T2 1}, or velocity squared per temperature unit (kelvin or degree Rank-ine). Each gas has its own constant R, equal to a universal constant divided by the molecular weight Rgas M gas (1.11) where 49,700 ft2/(s2 °R) 8314 m2/(s2 K). Most applications in this book are for air, with M 28.97: Rair 1717 ft2/(s2 °R) 287 m2/(s2 K) (1.12) Standard atmospheric pressure is 2116 lbf/ft2, and standard temperature is 60°F 520°R. Thus standard air density is air (171 2 7 1 ) 1 ( 6 520) 0.00237 slug/ft3 1.22 kg/m3 (1.13) This is a nominal value suitable for problems.
One proves in thermodynamics that Eq. (1.10) requires that the internal molecular energy û of a perfect gas vary only with temperature: û û(T). Therefore the specific heat cv also varies only with temperature: cv T û d d T û cv(T) or dû cv(T) dT (1.14) In like manner h and cp of a perfect gas also vary only with temperature: h û p û RT h(T) cp T h p d d T h cp(T) (1.15) dh cp(T) dT The ratio of specific heats of a perfect gas is an important dimensionless parameter in compressible-flow analysis (Chap. 9) k c c p v k(T) 1 (1.16) 1.6 Thermodynamic Properties of a Fluid 19 Part (a) Part (b) As a first approximation in airflow analysis we commonly take cp, cv, and k to be constant kair 1.4 cv k R 1 4293 ft2/(s2 °R) 718 m2/(s2 K) (1.17) cp k k R 1 6010 ft2/(s2 °R) 1005 m2/(s2 K) Actually, for all gases, cp and cv increase gradually with temperature, and k decreases gradually. Experimental values of the specific-heat ratio for eight common gases are shown in Fig. 1.3.
Many flow problems involve steam. Typical steam operating conditions are rela-tively close to the critical point, so that the perfect-gas approximation is inaccurate.
The properties of steam are therefore available in tabular form , but the error of using the perfect-gas law is sometimes not great, as the following example shows.
EXAMPLE 1.6 Estimate and cp of steam at 100 lbf/in2 and 400°F (a) by a perfect-gas approximation and (b) from the ASME steam tables .
Solution First convert to BG units: p 100 lbf/in2 14,400 lb/ft2, T 400°F 860°R. From Table A.4 the molecular weight of H2O is 2MH MO 2(1.008) 16.0 18.016. Then from Eq. (1.11) the gas constant of steam is approximately R 4 1 9 8 , .
7 0 0 1 0 6 2759 ft2/(s2 °R) whence, from the perfect-gas law, R p T 27 1 5 4 9 ,4 (8 0 6 0 0) 0.00607 slug/ft3 Ans. (a) From Fig. 1.3, k for steam at 860°R is approximately 1.30. Then from Eq. (1.17), cp k k R 1 1 1 .3 .3 0 0 (2 75 1 9) 12,000 ft2/(s2 °R) Ans. (a) From Ref. 13, the specific volume v of steam at 100 lbf/in2 and 400°F is 4.935 ft3/lbm. Then the density is the inverse of this, converted to slugs: 1 v 0.00630 slug/ft3 Ans. (b) This is about 4 percent higher than our ideal-gas estimate in part (a).
Reference 13 lists the value of cp of steam at 100 lbf/in2 and 400°F as 0.535 Btu/(lbm °F).
Convert this to BG units: cp 0.535 Btu/(lbm °R)(32.174 lbm/slug) 13,400 ft lbf/(slug °R) 13,400 ft2/(s2 °R) Ans. (b) 1 (4.935 ft2/lbm)(32.174 lbm/slug) 20 Chapter 1 Introduction Fig. 1.3 Specific-heat ratio of eight common gases as a function of tem-perature. (Data from Ref. 12.) This is about 11 percent higher than our ideal-gas estimate in part (a). The chief reason for the discrepancy is that this temperature and this pressure are quite close to the critical point and sat-uration line of steam. At higher temperatures and lower pressures, say, 800°F and 50 lbf/in2, the perfect-gas law gives and cp of steam within an accuracy of 1 percent.
Note that the use of pound-mass and British thermal units in the traditional steam tables re-quires continual awkward conversions to BG units. Newer tables and disks are in SI units.
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 ide-alized state relation for a liquid is const cp cv const dh cp dT (1.18) Most of the flow problems in this book can be attacked with these simple as-sumptions. Water is normally taken to have a density of 1.94 slugs/ft3 and a spe-cific heat cp 25,200 ft2/(s2 °R). The steam tables may be used if more accuracy is required.
1.6 Thermodynamic Properties of a Fluid 21 Ar Atmospheric pressure H2 CO Air and N2 O2 Steam CO2 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 1000 0 3000 4000 5000 = k cp c 2000 Temperature, °R υ State Relations for Liquids 1.7 Viscosity and Other Secondary Properties Viscosity 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 p p a (B 1) a n B (1.19) where B and n are dimensionless parameters which vary slightly with temperature and pa and a are standard atmospheric values. Water can be fitted approximately to the values B 3000 and n 7.
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 ˆ, 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 ‰. The average density of seawater is 2.00 slugs/ft3. Strictly speaking, seawater has three specific heats, all approximately equal to the value for pure water of 25,200 ft2/(s2 °R) 4210 m2/(s2 K).
EXAMPLE 1.7 The pressure at the deepest part of the ocean is approximately 1100 atm. Estimate the density of seawater at this pressure.
Solution Equation (1.19) holds for either water or seawater. The ratio p/pa is given as 1100: 1100 (3001) a 7 3000 or a 4 3 1 0 0 0 0 1 1/7 1.046 Assuming an average surface seawater density a 2.00 slugs/ft3, we compute 1.046(2.00) 2.09 slugs/ft3 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 sec-tion are primary thermodynamic variables characteristic of any system. There are also certain secondary variables which 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.
When a fluid is sheared, it begins to move at a strain rate inversely proportional to a property called its coefficient of viscosity . Consider a fluid element sheared in one 22 Chapter 1 Introduction Fig. 1.4 Shear stress causes contin-uous shear deformation in a fluid: (a) a fluid element straining at a rate /t; (b) newtonian shear dis-tribution in a shear layer near a wall.
plane by a single shear stress , as in Fig. 1.4a. The shear strain angle will contin-uously grow with time as long as the stress is maintained, the upper surface moving at speed u larger than the lower. Such common fluids as water, oil, and air show a linear relation between applied shear and resulting strain rate t (1.20) From the geometry of Fig. 1.4a we see that tan u y t (1.21) In the limit of infinitesimal changes, this becomes a relation between shear strain rate and velocity gradient d d t d d u y (1.22) From Eq. (1.20), then, the applied shear is also proportional to the velocity gradient for the common linear fluids. The constant of proportionality is the viscosity coeffi-cient d d t d d u y (1.23) Equation (1.23) is dimensionally consistent; therefore has dimensions of stress-time: {FT/L2} or {M/(LT)}. The BG unit is slugs per foot-second, and the SI unit is kilo-grams per meter-second. The linear fluids which follow Eq. (1.23) are called newton-ian fluids, after Sir Isaac Newton, who first postulated this resistance law in 1687.
We do not really care about the strain angle (t) in fluid mechanics, concentrating instead on the velocity distribution u(y), as in Fig. 1.4b. 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, t)—in a viscous fluid. Figure 1.4b illustrates a shear layer, or boundary layer, near a solid wall. The shear stress is proportional to the slope of the 1.7 Viscosity and Other Secondary Properties 23 (a) (b) θ δ δu δt δθ δt τ ∝ δ u = u u = 0 δx τ u(y) y τ = du dy du dy No slip at wall Velocity profile δ y 0 µ θ δ The Reynolds Number 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 tem-perature. 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.
Generally speaking, the viscosity of a fluid increases only weakly with pressure. For example, increasing p from 1 to 50 atm will increase of air only 10 percent. Tem-perature, however, has a strong effect, with increasing with T for gases and decreas-ing for liquids. Figure A.1 (in App. A) shows this temperature variation for various com-mon fluids. It is customary in most engineering work to neglect the pressure variation.
The variation (p, T) for a typical fluid is nicely shown by Fig. 1.5, from Ref. 14, which normalizes the data with the critical-point state ( c, pc, Tc). This behavior, called the principle of corresponding states, is characteristic of all fluids, but the actual nu-merical values are uncertain to 20 percent for any given fluid. For example, values of (T) for air at 1 atm, from Table A.2, fall about 8 percent low compared to the “low-density limit” in Fig. 1.5.
Note in Fig. 1.5 that changes with temperature occur very rapidly near the critical point. In general, critical-point measurements are extremely difficult and uncertain.
As we shall see in Chaps. 5 through 7, the primary parameter correlating the viscous behavior of all newtonian fluids is the dimensionless Reynolds number: Re VL V L (1.24) where V and L are characteristic velocity and length scales of the flow. The second form of Re illustrates that the ratio of to has its own name, the kinematic viscosity: (1.25) It is called kinematic because the mass units cancel, leaving only the dimensions {L2/T}.
24 Chapter 1 Introduction , Ratio , Ratio Fluid kg/(m s)† /(H2) kg/m3 m2/s† /(Hg) Hydrogen 8.8 E–6 00,0001.0 00,000.084 1.05 E–4 00,920 Air 1.8 E–5 0,00002.1 00,001.20 1.51 E–5 00,130 Gasoline 2.9 E–4 00,0033 0,0680 4.22 E–7 00,003.7 Water 1.0 E–3 00,0114 0,0998 1.01 E–6 0000,8.7 Ethyl alcohol 1.2 E–3 0,00135 0,0789 1.52 E–6 000,13 Mercury 1.5 E–3 00,0170 13,580 1.16 E–7 0000,1.0 SAE 30 oil 0.29 033,000 0,0891 3.25 E–4 02,850 Glycerin 1.5 170,000 01,264 1.18 E–3 10,300 †1 kg/(m s) 0.0209 slug/(ft s); 1 m2/s 10.76 ft2/s.
Table 1.4 Viscosity and Kinematic Viscosity of Eight Fluids at 1 atm and 20°C Fig. 1.5 Fluid viscosity nondimen-sionalized by critical-point proper-ties. This generalized chart is char-acteristic of all fluids but is only accurate to 20 percent. (From Ref. 14.) Flow between Plates Generally, the first thing a fluids engineer should do is estimate the Reynolds num-ber 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 de-pend upon flow geometry and will be discussed in Chaps. 5 through 7.
Table 1.4 also lists values of 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 relative to thin liquids such as gasoline, water, and al-cohol. Oil and glycerin still have the highest , but the ratio is smaller. For a given value of V and L in a flow, these fluids exhibit a spread of four orders of magnitude in the Reynolds number.
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. 1.6. The clearance between plates is h, and the fluid is newtonian and does not slip at either plate. If the plates are large, 1.7 Viscosity and Other Secondary Properties 25 0.4 = c r µ Tr Tc = T µ µ 0.8 0.7 0.6 0.5 0.4 0.3 0.2 1 0.9 2 3 4 5 6 7 8 9 10 0.6 0.8 1 2 3 4 5 7 6 8 9 10 Liquid Two-phase region Critical point 0.5 0 pr = 0.2 Low-density limit 1 2 3 5 10 25 Dense gas Fig. 1.6 Viscous flow induced by relative motion between two paral-lel plates.
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 d d u y const which we can integrate to obtain u a by The velocity distribution is linear, as shown in Fig. 1.6, and the constants a and b can be evaluated from the no-slip condition at the upper and lower walls: 0 a b(0) at y 0 V a b(h) at y h Hence a 0 and b V/h. Then the velocity profile between the plates is given by u V h y (1.26) as indicated in Fig. 1.6. 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 following example.
EXAMPLE 1.8 Suppose that the fluid being sheared in Fig. 1.6 is SAE 30 oil at 20°C. Compute the shear stress in the oil if V 3 m/s and h 2 cm.
Solution The shear stress is found from Eq. (1.23) by differentiating Eq. (1.26): d d u y h V (1) 26 Chapter 1 Introduction y h u(y) V Moving plate: u = V Viscous fluid u = V u = 0 Fixed plate u Variation of Viscosity with Temperature Thermal Conductivity From Table 1.4 for SAE 30 oil, 0.29 kg/(m s). Then, for the given values of V and h, Eq. (1) predicts 43 kg/(m s2) 43 N/m2 43 Pa Ans.
Although oil is very viscous, this is a modest shear stress, about 2400 times less than atmos-pheric pressure. Viscous stresses in gases and thin liquids are even smaller.
Temperature has a strong effect and pressure a moderate effect on viscosity. The vis-cosity 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: T T 0 n power law (T/T0) T 3/2 (T S 0 S) Sutherland law where 0 is a known viscosity at a known absolute temperature T0 (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. 3.
Liquid viscosity decreases with temperature and is roughly exponential, aebT; but a better fit is the empirical result that ln is quadratic in 1/T, where T is absolute temperature ln 0 a b T T 0 c T T 0 2 (1.28) For water, with T0 273.16 K, 0 0.001792 kg/(m s), suggested values are a 1.94, b 4.80, and c 6.74, with accuracy about 1 percent. The viscosity of water is tabulated in Table A.1. Curve-fit viscosity formulas for 355 organic liquids are given by Yaws et al. . For further viscosity data, see Refs. 28 and 36.
Just as viscosity relates applied stress to resulting strain rate, there is a property called thermal conductivity k which relates the vector rate of heat flow per unit area q to the vector gradient of temperature T. This proportionality, observed experimentally for fluids and solids, is known as Fourier’s law of heat conduction q kT (1.29a) which can also be written as three scalar equations qx k T x qy k T y qz k T z (1.29b) 0.29 kg/(m s) 0.02 m 1.7 Viscosity and Other Secondary Properties 27 0 (1.27) Nonnewtonian Fluids The minus sign satisfies the convention that heat flux is positive in the direction of de-creasing temperature. Fourier’s law is dimensionally consistent, and k has SI units of joules per second-meter-kelvin. Thermal conductivity k is a thermodynamic property and varies with temperature and pressure in much the same way as viscosity. The ra-tio k/k0 can be correlated with T/T0 in the same manner as Eqs. (1.27) and (1.28) for gases and liquids, respectively.
Further data on viscosity and thermal-conductivity variations can be found in Ref. 11.
Fluids which do not follow the linear law of Eq. (1.23) are called nonnewtonian and are treated in books on rheology . Figure 1.7a compares four examples with a new-tonian fluid. A dilatant, or shear-thickening, fluid increases resistance with increasing applied stress. Alternately, a pseudoplastic, or shear-thinning, fluid decreases resistance with increasing stress. If the thinning effect is very strong, as with the dashed-line curve, the fluid is termed plastic. The limiting case of a plastic substance is one which requires a finite yield stress before it begins to flow. The linear-flow Bingham plastic idealization is shown, but the flow behavior after yield may also be nonlinear. An ex-ample of a yielding fluid is toothpaste, which will not flow out of the tube until a fi-nite stress is applied by squeezing.
A further complication of nonnewtonian behavior is the transient effect shown in Fig. 1.7b. 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 which thins out with time and requires decreasing stress is termed thixotropic. We neglect nonnewton-ian effects in this book; see Ref. 6 for further study.
28 Chapter 1 Introduction Shear stress Shear stress Yield stress Plastic Ideal Bingham plastic Dilatant Newtonian Pseudoplastic Shear strain rate d dt θ 0 0 Time (a) (b) Constant strain rate Rheopectic Common fluids Thixotropic τ τ Fig. 1.7 Rheological behavior of various viscous materials: (a) stress versus strain rate; (b) effect of time on applied stress.
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 dL are exposed normal to the cut and parallel to the surface, where is called the coefficient of surface tension. The dimensions of 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 dA; this requires work to be done of amount dA. Thus the coefficient can also be regarded as the surface energy per unit area of the inter-face, in N m/m2 or ft lbf/ft2.
The two most common interfaces are water-air and mercury-air. For a clean surface at 20°C 68°F, the measured surface tension is 0.0050 lbf/ft 0.073 N/m air-water 0.033 lbf/ft 0.48 N/m air-mercury These are design values and can change considerably if the surface contains contami-nants like detergents or slicks. Generally decreases with liquid temperature and is zero at the critical point. Values of for water are given in Fig. 1.8.
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. 1.9. In Fig. 1.9a, the pressure increase in the interior of a liquid cylinder is balanced by two surface-tension forces 2RL p 2L or p R (1.31) We are not considering the weight of the liquid in this calculation. In Fig. 1.9b, the pres-sure increase in the interior of a spherical droplet balances a ring of surface-tension force R2 p 2R or p 2 R (1.32) We can use this result to predict the pressure increase inside a soap bubble, which has two interfaces with air, an inner and outer surface of nearly the same radius R: pbubble 2 pdroplet 4 R (1.33) Figure 1.9c shows the general case of an arbitrarily curved interface whose principal radii of curvature are R1 and R2. A force balance normal to the surface will show that the pressure increase on the concave side is p (R1 1 R2 1) (1.34) 1.7 Viscosity and Other Secondary Properties 29 Surface Tension (1.30) Fig. 1.8 Surface tension of a clean air-water interface. Data from Table A.5.
Fig. 1.9 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.
Equations (1.31) to (1.33) can all be derived from this general relation; e.g., in Eq. (1.31), R1 R and R2 .
A second important surface effect is the contact angle which appears when a liquid interface intersects with a solid surface, as in Fig. 1.10. The force balance would then involve both and . If the contact angle is less than 90°, the liquid is said to wet the solid; if 90°, the liquid is termed nonwetting. For example, wa-ter wets soap but does not wet wax. Water is extremely wetting to a clean glass sur-face, with 0°. Like , the contact angle is sensitive to the actual physico-chemical conditions of the solid-liquid interface. For a clean mercury-air-glass interface, 130°.
Example 1.9 illustrates how surface tension causes a fluid interface to rise or fall in a capillary tube.
30 Chapter 1 Introduction 0 0.050 0.060 0.070 0.080 10 20 30 40 50 60 70 80 90 100 T, °C , N/m ϒ 2RL ∆p 2R L (a) L L πR2 ∆p 2πR (b) (c) ∆p dA dL2 dL1 dL2 dL1 R2 R1 Fig. 1.10 Contact-angle effects at liquid-gas-solid interface. If 90°, the liquid “wets” the solid; if 90°, the liquid is nonwetting.
EXAMPLE 1.9 Derive an expression for the change in height h in a circular tube of a liquid with surface ten-sion and contact angle , as in Fig. E1.9.
Solution The vertical component of the ring surface-tension force at the interface in the tube must bal-ance the weight of the column of fluid of height h 2R cos R2h Solving for h, we have the desired result h 2 c R os Ans.
Thus the capillary height increases inversely with tube radius R and is positive if 90° (wet-ting liquid) and negative (capillary depression) if 90°.
Suppose that R 1 mm. Then the capillary rise for a water-air-glass interface, 0°, 0.073 N/m, and 1000 kg/m3 is h 0.015 (N s2)/kg 0.015 m 1.5 cm For a mercury-air-glass interface, with 130°, 0.48 N/m, and 13,600 kg/m3, the cap-illary rise is h 0.46 cm When a small-diameter tube is used to make pressure measurements (Chap. 2), these capillary effects must be corrected for.
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/ft2, while that of mercury is only 0.0035 lbf/ft2. If the liquid pressure is greater than the vapor 2(0.48)(cos 130°) 13,600(9.81)(0.001) 2(0.073 N/m)(cos 0°) (1000 kg/m3)(9.81 m/s2)(0.001 m) 1.7 Viscosity and Other Secondary Properties 31 Nonwetting Solid Liquid Gas θ θ Fig. E1.9 θ 2R h Vapor Pressure No-Slip and No-Temperature-Jump Conditions pressure, the only exchange between liquid and vapor is evaporation at the inter-face. 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/ft2, and thus water at normal atmospheric pressure will boil. When the liq-uid pressure is dropped below the vapor pressure due to a flow phenomenon, we call the process cavitation. As we shall see in Chap. 2, if water is accelerated from rest to about 50 ft/s, its pressure drops by about 15 lbf/in2, or 1 atm. This can cause cavitation.
The dimensionless parameter describing flow-induced boiling is the cavitation number Ca (1.35) where pa ambient pressure pv vapor pressure V characteristic flow velocity Depending upon 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. 1.11.
Figure 1.12a 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. 1.12b.
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 surface they contact. All gases are, too, pa pv 1 2 V2 32 Chapter 1 Introduction 100 80 60 40 20 0 pv, kPa 0 20 40 60 80 100 T, °C Fig. 1.11 Vapor pressure of water.
Data from Table A.5.
1.7 Viscosity and Other Secondary Properties 33 Fig. 1.12 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, Pennsylva-nia State University); (b) ugliness: collapsing bubbles erode a pro-peller surface. (Courtesy of Thomas T. Huang, David Taylor Research Center.) Fig. 1.13 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. [From Il-lustrated Experiments in Fluid Me-chanics (The NCFMF Book of Film Notes), National Committee for Fluid Mechanics Films, Education Development Center, Inc., copy-right 1972.] 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 temperature of that surface Vfluid Vwall Tfluid Twall (1.36) 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 (Chap. 6).
Figure 1.13 illustrates the no-slip condition for water flow past the top and bottom sur-faces of a fixed thin plate. The flow past the upper surface is disorderly, or turbulent, while the lower surface flow is smooth, or laminar.7 In both cases there is clearly no slip at the wall, where the water takes on the zero velocity of the fixed plate. The ve-locity profile is made visible by the discharge of a line of hydrogen bubbles from the wire shown stretched across the flow.
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 Vnormal(fluid) Vnormal(solid) (1.37) while the tangential velocity Vt is allowed to be independent of the wall. The analysis is much simpler, but the flow patterns are highly idealized.
34 Chapter 1 Introduction 7Laminar and turbulent flows are studied in Chaps. 6 and 7.
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 be-comes 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 de-fined by a2 p s k p T k c c p v (1.38) 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 aideal gas (kRT)1/2 (1.39) 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 m2/(s2 K)}1/2 343 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 V/a of the flow reaches about 0.3.
The speed of sound of water is tabulated in Table A.5. The speed of sound of air (or any approximately perfect gas) is simply calculated from Eq. (1.39).
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: 1.
Control-volume, or integral analysis (Chap. 3) 2.
Infinitesimal system, or differential analysis (Chap. 4) 3.
Experimental study, or dimensional analysis (Chap. 5) In all cases, the flow must satisfy the three basic laws of mechanics8 plus a thermo-dynamic state relation and associated boundary conditions: 1.
Conservation of mass (continuity) 2.
Linear momentum (Newton’s second law) 3.
First law of thermodynamics (conservation of energy) 4.
A state relation like (p, T) 5.
Appropriate boundary conditions at solid surfaces, interfaces, inlets, and exits In integral and differential analyses, these five relations are modeled mathematically and solved by computational methods. In an experimental study, the fluid itself per-forms this task without the use of any mathematics. In other words, these laws are be-lieved to be fundamental to physics, and no fluid flow is known to violate them.
1.8 Basic Flow-Analysis Techniques 35 1.8 Basic Flow-Analysis Techniques 8In fluids which are variable mixtures of components, such as seawater, a fourth basic law is required, conservation of species. For an example of salt conservation analysis, see Chap. 4, Ref. 16.
A control volume is a finite region, chosen carefully by the analyst, with open bound-aries through which mass, momentum, and energy are allowed to cross. The analyst makes a budget, or balance, between the incoming and outgoing fluid and the resul-tant changes within the control volume. The result is a powerful tool but a crude one.
Details of the flow are normally washed out or ignored in control-volume analyses.
Nevertheless, the control-volume technique of Chap. 3 never fails to yield useful and quantitative information to the engineering analyst.
When the conservation laws are written for an infinitesimal system of fluid in motion, they become the basic differential equations of fluid flow. To apply them to a specific problem, one must integrate these equations mathematically subject to the boundary con-ditions of the particular problem. Exact analytic solutions are often possible only for very simple geometries and boundary conditions (Chap. 4). Otherwise, one attempts numeri-cal integration on a digital computer, i.e., a summing procedure for finite-sized systems which one hopes will approximate the exact integral calculus . Even computer analy-sis often fails to provide an accurate simulation, because of either inadequate storage or inability to model the finely detailed flow structure characteristic of irregular geometries or turbulent-flow patterns. Thus differential analysis sometimes promises more than it delivers, although we can successfully study a number of classic and useful solutions.
A properly planned experiment is very often the best way to study a practical en-gineering flow problem. Guidelines for planning flow experiments are given in Chap.
5. For example, no theory presently available, whether differential or integral, calcu-lus or computer, is able to make an accurate computation of the aerodynamic drag and side force of an automobile moving down a highway with crosswinds. One must solve the problem by experiment. The experiment may be full-scale: One can test a real au-tomobile on a real highway in real crosswinds. For that matter, there are wind tunnels in existence large enough to hold a full-scale car without significant blockage effects.
Normally, however, in the design stage, one tests a small-model automobile in a small wind tunnel. Without proper interpretation, the model results may be poor and mislead the designer (Chap. 5). For example, the model may lack important details such as sur-face finish or underbody protuberances. The “wind” produced by the tunnel propellers may lack the turbulent gustiness of real winds. It is the job of the fluid-flow analyst, using such techniques as dimensional analysis, to plan an experiment which gives an accurate estimate of full-scale or prototype results expected in the final product.
It is possible to classify flows, but there is no general agreement on how to do it.
Most classifications deal with the assumptions made in the proposed flow analysis.
They come in pairs, and we normally assume that a given flow is either Steady or unsteady (1.40a) Inviscid or viscous (1.40b) Incompressible or compressible (1.40c) Gas or liquid (1.40d) As Fig. 1.14 indicates, we choose one assumption from each pair. We may have a steady viscous compressible gas flow or an unsteady inviscid ( 0) incompressible liquid flow. Although there is no such thing as a truly inviscid fluid, the assumption 0 gives adequate results in many analyses (Chap. 8). Often the assumptions overlap: A flow may be viscous in the boundary layer near a solid surface (Fig. 1.13) and effec-36 Chapter 1 Introduction Fig. 1.14 Ready for a flow analy-sis? Then choose one assumption from each box.
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines tively inviscid away from the surface. The viscous part of the flow may be laminar or transitional or turbulent or combine patches of all three types of viscous flow. A flow may involve both a gas and a liquid and the free surface, or interface, between them (Chap. 10). A flow may be compressible in one region and have nearly constant den-sity in another. Nevertheless, Eq. (1.40) and Fig. 1.14 give the basic binary assump-tions of flow analysis, and Chaps. 6 to 10 try to separate them and isolate the basic ef-fect of each assumption.
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 in-stant.
2.
A pathline is the actual path traversed by a given fluid particle.
3.
A streakline is the locus of particles which have earlier passed through a pre-scribed point.
4.
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 eas-ier to generate experimentally. Note that a streamline and a timeline are instantaneous lines, while the pathline and the streakline are generated by the passage of time. The velocity profile shown in Fig. 1.13 is really a timeline generated earlier by a single dis-charge of bubbles from the wire. A pathline can be found by a time exposure of a sin-gle marked particle moving through the flow. Streamlines are difficult to generate ex-perimentally in unsteady flow unless one marks a great many particles and notes their direction of motion during a very short time interval [17, p. 35]. In steady flow the sit-uation 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 1.15a shows a typical set of streamlines, and Fig. 1.15b 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 1.16 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: d u x d v y d w z d V r (1.41) 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 37 Steady Unsteady Inviscid Viscous Incompressible Compressible Gas Liquid If the velocities (u, v, w) are known functions of position and time, Eq. (1.41) can be integrated to find the streamline passing through the initial point (x0, y0, z0, t0). The method is straightforward for steady flows (Example 1.10) but may be laborious for unsteady flow.
The pathline, or displacement of a particle, is defined by integration of the velocity components, as mentioned in Sec. 1.5: Pathline: x u dt y v dt z w dt (1.42) Given (u, v, w) as known functions of position and time, the integration is begun at a specified initial position (x0, y0, z0, t0). Again the integration may be laborious.
Streaklines, easily generated experimentally with smoke, dye, or bubble releases, are very difficult to compute analytically. See Ref. 18 for mathematical details.
38 Chapter 1 Introduction (a) V (b) No flow across streamtube walls Individual streamline Fig. 1.15 The most common method of flow-pattern presenta-tion: (a) Streamlines are every-where tangent to the local velocity vector; (b) a streamtube is formed by a closed collection of stream-lines.
z y dy dx dr x v dz u V V w Fig. 1.16 Geometric relations for defining a streamline.
EXAMPLE 1.10 Given the steady two-dimensional velocity distribution u Kx v Ky w 0 (1) where K is a positive constant, compute and plot the streamlines of the flow, including direc-tions, and give some possible interpretations of the pattern.
Solution Since time does not appear explicitly in Eq. (1), the motion is steady, so that streamlines, path-lines, and streaklines will coincide. Since w 0 everywhere, the motion is two dimensional, in the xy plane. The streamlines can be computed by substituting the expressions for u and v into Eq. (1.41): K dx x K dy y or d x x d y y Integrating, we obtain ln x ln y ln C, or xy C Ans. (2) This is the general expression for the streamlines, which are hyperbolas. The complete pat-tern is plotted in Fig. E1.10 by assigning various values to the constant C. The arrowheads can be determined only by returning to Eqs. (1) to ascertain the velocity component direc-tions, assuming K is positive. For example, in the upper right quadrant (x 0, y 0), u is positive and v is negative; hence the flow moves down and to the right, establishing the ar-rowheads as shown.
1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 0 0 C = –3 –2 –1 C = 0 0 + 1 + 2 C = + 3 – 3 – 2 – 1 x C = 0 +1 +2 +3 y Fig. E1.10 Streamlines for the ve-locity distribution given by Eq. (1), for K 0.
Note that the streamline pattern is entirely independent of constant K. It could represent the impingement of two opposing streams, or the upper half could simulate the flow of a single down-ward stream against a flat wall. Taken in isolation, the upper right quadrant is similar to the flow in a 90° corner. This is definitely a realistic flow pattern and is discussed again in Chap. 8.
Finally note the peculiarity that the two streamlines (C 0) have opposite directions and in-tersect. This is possible only at a point where u v w 0, which occurs at the origin in this case. Such a point of zero velocity is called a stagnation point.
A streakline can be produced experimentally by the continuous release of marked par-ticles (dye, smoke, or bubbles) from a given point. Figure 1.17 shows two examples. The flow in Fig. 1.17b is unsteady and periodic due to the flapping of the plate against the oncoming stream. We see that the dash-dot streakline does not coincide with either the streamline or the pathline passing through the same release point. This is characteristic of unsteady flow, but in Fig. 1.17a the smoke filaments form streaklines which are iden-tical to the streamlines and pathlines. We noted earlier that this coincidence of lines is always true of steady flow: Since the velocity never changes magnitude or direction at any point, every particle which comes along repeats the behavior of its earlier neighbors.
Methods of experimental flow visualization include the following: 1.
Dye, smoke, or bubble discharges 2.
Surface powder or flakes on liquid flows 3.
Floating or neutral-density particles 4.
Optical techniques which detect density changes in gas flows: shadowgraph, schlieren, and interferometer 5.
Tufts of yarn attached to boundary surfaces 6.
Evaporative coatings on boundary surfaces 7.
Luminescent fluids or additives 40 Chapter 1 Introduction Fig. 1.17 Experimental visualization of steady and unsteady flow: (a) steady flow past an airfoil visualized by smoke filaments (C. A. A. SCIENTIFIC—Prime Movers Laboratory Systems); (b) unsteady flow past an oscillating plate with a point bubble release (from an experiment in Ref. 17).
(b) Periodic flapping plate Release point Streamline Pathline Streakline Uniform approach flow (a) The mathematical implications of flow-pattern analysis are discussed in detail in Ref.
18. References 19 and 20 are beautiful albums of photographs. References 21 and 22 are monographs on flow visualization.
Most of the examples and exercises in this text are amenable to direct calculation with-out guessing or iteration or looping. Until recently, only such direct problem assign-ments, whether “plug-and-chug” or more subtle, were appropriate for undergraduate engineering courses. However, the recent introduction of computer software solvers makes almost any set of algebraic relations viable for analysis and solution. The solver recommended here is the Engineering Equation Solver (EES) developed by Klein and Beckman and described in Appendix E.
Any software solver should handle a purely mathematical set of relations, such as the one posed in Ref. 33: X ln (X) Y3, X1/2 1/Y. Submit that pair to any commer-cial solver and you will no doubt receive the answer: X 1.467, Y 0.826. However, for engineers, in the author’s opinion, EES is superior to most solvers because (1) equa-tions can be entered in any order; (2) scores of mathematical formulas are built-in, such as the Bessel functions; and (3) thermophysical properties of many fluids are built-in, such as the steam tables . Both metric and English units are allowed. Equations need not be written in the traditional BASIC or FORTRAN style. For example, X Y 1 0 is perfectly satisfactory; there is no need to retype this as X Y 1.
For example, reconsider Example 1.7 as an EES exercise. One would first enter the reference properties p0 and 0 plus the curve-fit constants B and n: Pz 1.0 Rhoz 2.0 B 3000 n 7 Then specify the given pressure ratio and the curve-fit relation, Eq. (1.19), for the equa-tion of state of water: P 1100Pz P/Pz (B 1) (Rho/Rhoz) ^n B If you request an initial opinion from the CHECK/FORMAT menu, EES states that there are six equations in six unknowns and there are no obvious difficulties. Then request SOLVE from the menu and EES quickly prints out Rho 2.091, the correct answer as seen already in Ex. 1.7. It also prints out values of the other five variables.
Occasionally EES reports “unable to converge” and states what went wrong (division by zero, square root of a negative number, etc.). One needs only to improve the guesses and ranges of the unknowns in Variable Information to assist EES to the solution.
In subsequent chapters we will illustrate some implicit (iterative) examples by us-ing EES and will also assign some advanced problem exercises for which EES is an ideal approach. The use of an engineering solver, notably EES, is recommended to all engineers in this era of the personal computer.
1.10 The Engineering Equation Solver 41 1.10 The Engineering Equation Solver EES Earlier in this chapter we referred to the uncertainty of the principle of corresponding states in discussing Fig. 1.5. Uncertainty is a fact of life in engineering. We rarely know any engineering properties or variables to an extreme degree of accuracy. Therefore, we need to know the uncertainty U of our data, usually defined as the band within which the experimenter is 95 percent confident that the true value lies (Refs. 30, 31).
In Fig. 1.5, we were given that the uncertainty of / c is U 20 percent.
Fluid mechanics is heavily dependent upon experimentation, and the data uncer-tainty is needed before we can use it for prediction or design purposes. Sometimes un-certainty completely changes our viewpoint. As an offbeat example, suppose that as-tronomers reported that the length of the earth year was 365.25 days “give or take a couple of months.” First, that would make the five-figure accuracy ridiculous, and the year would better be stated as Y 365 60 days. Second, we could no longer plan confidently or put together accurate calendars. Scheduling Christmas vacation would be chancy.
Multiple variables make uncertainty estimates cumulative. Suppose a given result P depends upon N variables, P P(x1, x2, x3, . . . , xN), each with its own uncertainty; for example, x1 has uncertainty x1. Then, by common agreement among experimenters, the total uncertainty of P is calculated as a root-mean-square average of all effects: P x P 1 x1 2 x P 2 x2 2 x P N xN 2 1/2 (1.43) This calculation is statistically much more probable than simply adding linearly the various uncertainties xi, thereby making the unlikely assumption that all variables simultaneously attain maximum error. Note that it is the responsibility of the ex-perimenter to establish and report accurate estimates of all the relevant uncertain-ties xi.
If the quantity P is a simple power-law expression of the other variables, for ex-ample, P Const x1 n1x2 n2x3 n3 . . . , then each derivative in Eq. (1.43) is proportional to P and the relevant power-law exponent and is inversely proportional to that variable.
If P Const x1 n1x2 n2x3 n3 . . . , then x P 1 n x 1 1 P , x P 2 n x 2 2 P , x P 3 n x 3 3 P , Thus, from Eq. (1.43), P P n1 x x 1 1 2 n2 x x 2 2 2 n3 x x 3 3 2 1/2 (1.44) Evaluation of P is then a straightforward procedure, as in the following example.
EXAMPLE 1.11 The so-called dimensionless Moody pipe-friction factor f, plotted in Fig. 6.13, is calculated in experiments from the following formula involving pipe diameter D, pressure drop p, density , volume flow rate Q, and pipe length L: f 3 2 2 D Q 5 2L p 42 Chapter 1 Introduction 1.11 Uncertainty of Experimental Data Measurement uncertainties are given for a certain experiment: for D: 0.5 percent, p: 2.0 per-cent, : 1.0 percent, Q: 3.5 percent, and L: 0.4 percent. Estimate the overall uncertainty of the friction factor f.
Solution The coefficient 2/32 is assumed to be a pure theoretical number, with no uncertainty. The other variables may be collected using Eqs. (1.43) and (1.44): U f f 5 D D 2 1 p p 2 1 2 2 Q Q 2 1 L L 2 1/2 [{5(0.5%)}2 (2.0%)2 (1.0%)2 {2(3.5%)}2 (0.4%)2]1/2 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 D been larger than 0.5 percent.
The road toward a professional engineer’s license has a first stop, the Fundamentals of Engineering Examination, known as the FE exam. It was formerly known as the Engineer-in-Training (E-I-T) Examination. This 8-h national test will probably soon be required of all engineering graduates, not just for licensure, but as a student-assessment tool. The 120-problem morning session covers many general studies: Chemistry Computers Dynamics Electric circuits Engineering economics Fluid Mechanics Materials science Mathematics Mechanics of materials Statics Thermodynamics Ethics For the 60-problem afternoon session you may choose chemical, civil, electrical, in-dustrial, or mechanical engineering or take more general-engineering problems for re-maining disciplines. As you can see, fluid mechanics is central to the FE exam. There-fore, 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 or forgot to double or halve something or are missing a factor of , etc. 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, Red for flow in a smooth circular pipe, since (see Chaps. 6 and 7) transition is highly dependent upon 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 1.12 The Fundamentals of Engineering (FE) Examination 43 1.12 The Fundamentals of Engineering (FE) Examination undergraduate-training situation. Every effort has been made to keep the FE exam ques-tions in this text unambiguous.
Fluid flow analysis generates a plethora of problems, 1500 in this text alone! To solve these problems, one must deal with various equations, data, tables, assumptions, unit systems, and numbers. The writer recommends these problem-solving steps: 1.
Gather all the given system parameters and data in one place.
2.
Find, from tables or charts, all needed fluid property data: , , cp, k, , etc.
3.
Use SI units (N, s, kg, m) if possible, and no conversion factors will be neces-sary.
4.
Make sure what is asked. It is all too common for students to answer the wrong question, for example, reporting mass flow instead of volume flow, pressure in-stead of pressure gradient, drag force instead of lift force. Engineers are ex-pected to read carefully.
5.
Make a detailed sketch of the system, with everything clearly labeled.
6.
Think carefully and then list your assumptions. Here knowledge is power; you should not guess the answer. You must be able to decide correctly if the flow can be considered steady or unsteady, compressible or incompressible, one-dimensional, or multidimensional, viscous or inviscid, and whether a control vol-ume or partial differential equations are needed.
7.
Based on steps 1 to 6 above, write out the appropriate equations, data correla-tions, and fluid state relations for your problem. If the algebra is straightforward, solve for what is asked. If the equations are complicated, e.g., nonlinear or too plentiful, use the Engineering Equation Solver (EES).
8.
Report your solution clearly, with proper units listed and to the proper number of significant figures (usually two or three) that the overall uncertainty of the data will allow.
Like most scientific disciplines, fluid mechanics has a history of erratically occurring early achievements, then an intermediate era of steady fundamental discoveries in the eighteenth and nineteenth centuries, leading to the twentieth-century era of “modern practice,” as we self-centeredly term our limited but up-to-date knowledge. Ancient civilizations had enough knowledge to solve certain flow problems. Sailing ships with oars and irrigation systems were both known in prehistoric times. The Greeks produced quantitative information. Archimedes and Hero of Alexandria both postulated the par-allelogram law for addition of vectors in the third century B.C. Archimedes (285–212 B.C.) formulated the laws of buoyancy and applied them to floating and submerged bodies, actually deriving a form of the differential calculus as part of the analysis. The Romans built extensive aqueduct systems in the fourth century B.C. but left no records showing any quantitative knowledge of design principles.
From the birth of Christ to the Renaissance there was a steady improvement in the design of such flow systems as ships and canals and water conduits but no recorded evidence of fundamental improvements in flow analysis. Then Leonardo da Vinci (1452–1519) derived the equation of conservation of mass in one-dimensional steady 44 Chapter 1 Introduction 1.13 Problem-Solving Techniques 1.14 History and Scope of Fluid Mechanics 1.14 History and Scope of Fluid Mechanics 45 flow. Leonardo was an excellent experimentalist, and his notes contain accurate de-scriptions of waves, jets, hydraulic jumps, eddy formation, and both low-drag (stream-lined) and high-drag (parachute) designs. A Frenchman, Edme Mariotte (1620–1684), built the first wind tunnel and tested models in it.
Problems involving the momentum of fluids could finally be analyzed after Isaac New-ton (1642–1727) postulated his laws of motion and the law of viscosity of the linear flu-ids now called newtonian. The theory first yielded to the assumption of a “perfect” or frictionless fluid, and eighteenth-century mathematicians (Daniel Bernoulli, Leonhard Euler, Jean d’Alembert, Joseph-Louis Lagrange, and Pierre-Simon Laplace) produced many beautiful solutions of frictionless-flow problems. Euler developed both the differ-ential equations of motion and their integrated form, now called the Bernoulli equation.
D’Alembert used them to show his famous paradox: that a body immersed in a friction-less fluid has zero drag. These beautiful results amounted to overkill, since perfect-fluid assumptions have very limited application in practice and most engineering flows are dominated by the effects of viscosity. Engineers began to reject what they regarded as a totally unrealistic theory and developed the science of hydraulics, relying almost entirely on experiment. Such experimentalists as Chézy, Pitot, Borda, Weber, Francis, Hagen, Poiseuille, Darcy, Manning, Bazin, and Weisbach produced data on a variety of flows such as open channels, ship resistance, pipe flows, waves, and turbines. All too often the data were used in raw form without regard to the fundamental physics of flow.
At the end of the nineteenth century, unification between experimental hydraulics and theoretical hydrodynamics finally began. William Froude (1810–1879) and his son Robert (1846–1924) developed laws of model testing, Lord Rayleigh (1842–1919) pro-posed the technique of dimensional analysis, and Osborne Reynolds (1842–1912) pub-lished the classic pipe experiment in 1883 which showed the importance of the di-mensionless Reynolds number named after him. Meanwhile, viscous-flow theory was available but unexploited, since Navier (1785–1836) and Stokes (1819–1903) had suc-cessfully added newtonian viscous terms to the equations of motion. The resulting Navier-Stokes equations were too difficult to analyze for arbitrary flows. Then, in 1904, a German engineer, Ludwig Prandtl (1875–1953), published perhaps the most impor-tant paper ever written on fluid mechanics. Prandtl pointed out that fluid flows with small viscosity, e.g., water flows and airflows, can be divided into a thin viscous layer, or boundary layer, near solid surfaces and interfaces, patched onto a nearly inviscid outer layer, where the Euler and Bernoulli equations apply. Boundary-layer theory has proved to be the single most important tool in modern flow analysis. The twentieth-century foundations for the present state of the art in fluid mechanics were laid in a series of broad-based experiments and theories by Prandtl and his two chief friendly competitors, Theodore von Kármán (1881–1963) and Sir Geoffrey I. Taylor (1886– 1975). Many of the results sketched here from a historical point of view will, of course, be discussed in this textbook. More historical details can be found in Refs. 23 to 25.
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 circula-tion. All transportation problems involve fluid motion, with well-developed specialties in aerodynamics of aircraft and rockets and in naval hydrodynamics of ships and sub-marines. Almost all our electric energy is developed either from water flow or from P1.4 A beaker approximates a right circular cone of diameter 7 in and height 9 in. When filled with liquid, it weighs 70 oz.
When empty, it weighs 14 oz. Estimate the density of this liquid in both SI and BG units.
P1.5 The mean free path of a gas, , is defined as the average distance traveled by molecules between collisions. A pro-posed formula for estimating of an ideal gas is 1.26 R T 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?
P1.6 In the {MLT } system, what is the dimensional represen-tation of (a) enthalpy, (b) mass rate of flow, (c) bending mo-ment, (d) angular velocity, (e) modulus of elasticity; (f) Poisson’s ratio?
P1.7 A small village draws 1.5 acre ft/day of water from its reservoir. Convert this average water usage to (a) gallons per minute and (b) liters per second.
P1.8 Suppose we know little about the strength of materials but are told that the bending stress in a beam is pro-portional to the beam half-thickness y and also depends upon the bending moment M and the beam area moment of inertia I. We also learn that, for the particular case M 2900 in lbf, y 1.5 in, and I 0.4 in4, the pre-dicted stress is 75 MPa. Using this information and di-mensional reasoning only, find, to three significant fig-ures, the only possible dimensionally homogeneous formula y f(M, I).
Problems 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 an EES icon (for example, Prob.
2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 1.1 to 1.85 (categorized in the problem list below) are followed by fun-damentals of engineering (FE) exam problems FE3.1 to FE3.10, and comprehensive problems C1.1 to C1.4.
Problem Distribution Section Topic Problems 1.1, 1.2, 1.3 Fluid-continuum concept 1.1–1.3 1.4 Dimensions, units, dynamics 1.4–1.20 1.5 Velocity field 1.21–1.23 1.6 Thermodynamic properties 1.24–1.37 1.7 Viscosity; no-slip condition 1.38–1.61 1.7 Surface tension 1.62–1.71 1.7 Vapor pressure; cavitation 1.72–1.75 1.7 Speed of sound; Mach number 1.76–1.78 1.8,9 Flow patterns, streamlines, pathlines 1.79–1.84 1.10 History of fluid mechanics 1.85 P1.1 A gas at 20°C may be considered rarefied, deviating from the continuum concept, when it contains less than 1012 mol-ecules 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. P1.3, show that a tilted free liquid surface, in contact with an atmosphere at pres-sure pa, 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.
steam flow through turbine generators. All combustion problems involve fluid motion, as do the more classic problems of irrigation, flood control, water supply, sewage dis-posal, 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.
46 Chapter 1 Introduction pa Fluid density P1.3 where Q is the volume rate of flow and p is the pressure rise produced by the pump. Suppose that a certain pump develops a pressure rise of 35 lbf/in2 when its flow rate is 40 L/s. If the input power is 16 hp, what is the efficiency?
P1.14 Figure P1.14 shows the flow of water over a dam. The vol-ume flow Q is known to depend only upon 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?
P1.9 The kinematic viscosity of a fluid is the ratio of viscosity to density, /. What is the only possible dimension-less group combining with velocity V and length L? What is the name of this grouping? (More information on this will be given in Chap. 5.) P1.10 The Stokes-Oseen formula for drag force F on a sphere of diameter D in a fluid stream of low velocity V, density , and viscosity , is F 3 DV 9 1 6 V2D2 Is this formula dimensionally homogeneous?
P1.11 Engineers sometimes use the following formula for the volume rate of flow Q of a liquid flowing through a hole of diameter D in the side of a tank: Q 0.68 D2 g h where g is the acceleration of gravity and h is the height of the liquid surface above the hole. What are the dimen-sions of the constant 0.68?
P1.12 For low-speed (laminar) steady flow through a circular pipe, as shown in Fig. P1.12, the velocity u varies with ra-dius and takes the form u B p (r0 2 r2) where is the fluid viscosity and p is the pressure drop from entrance to exit. What are the dimensions of the con-stant B?
Problems 47 r = 0 r u (r) Pipe wall r = r0 P1.13 The efficiency 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: inpu Q t po p wer P1.12 H B Q Water level Dam P1.15 As a practical application of Fig. P1.14, often termed a sharp-crested weir, civil engineers use the following for-mula for flow rate: Q 3.3BH3/2, with Q in ft3/s and B and H in feet. Is this formula dimensionally homogeneous?
If not, try to explain the difficulty and how it might be con-verted to a more homogeneous form.
P1.16 Algebraic equations such as Bernoulli’s relation, Eq. (1) of Ex. 1.3, are dimensionally consistent, but what about differential equations? Consider, for example, the bound-ary-layer x-momentum equation, first derived by Ludwig Prandtl in 1904: u u x v u y p x gx y where is the boundary-layer shear stress and gx is the component of gravity in the x direction. Is this equation dimensionally consistent? Can you draw a general con-clusion?
P1.17 The Hazen-Williams hydraulics formula for volume rate of flow Q through a pipe of diameter D and length L is given by Q 61.9 D2.63 L p 0.54 P1.14 Q/Across-section and the dimensionless Reynolds number of the flow, Re VavgD/ . Comment on your results.
P1.24 Air at 1 atm and 20°C has an internal energy of approxi-mately 2.1 E5 J/kg. If this air moves at 150 m/s at an al-titude z 8 m, what is its total energy, in J/kg, relative to the datum z 0? Are any energy contributions negligible?
P1.25 A tank contains 0.9 m3 of helium at 200 kPa and 20°C.
Estimate the total mass of this gas, in kg, (a) on earth and (b) on the moon. Also, (c) how much heat transfer, in MJ, is required to expand this gas at constant temperature to a new volume of 1.5 m3?
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 lbf/in2 above the ambient atmosphere. If the tire is at sea level, has a volume of 3.0 ft3, and is at 75°F, estimate the total weight of air, in lbf, inside the tire.
P1.27 For steam at 40 lbf/in2, some values of temperature and specific volume are as follows, from Ref. 13: T, °F 400 500 600 700 800 v, ft3/lbm 12.624 14.165 15.685 17.195 18.699 Is steam, for these conditions, nearly a perfect gas, or is it wildly nonideal? If reasonably perfect, find a least-squares† value for the gas constant R, in m2/(s2 K), estimate the per-cent error in this approximation, and compare with Table A.4.
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 pwater vapor Suppose this wet atmosphere is at 40°C and 1 atm. Cal-culate the density of this 100 percent humid air, and com-pare it with the density of dry air at the same conditions.
P1.29 A compressed-air tank holds 5 ft3 of air at 120 lbf/in2 “gage,” that is, above atmospheric pressure. Estimate the energy, in ft-lbf, required to compress this air from the at-mosphere, assuming an ideal isothermal process.
P1.30 Repeat Prob. 1.29 if the tank is filled with compressed wa-ter instead of air. Why is the result thousands of times less than the result of 215,000 ft lbf in Prob. 1.29?
P1.31 The density of (fresh) water at 1 atm, over the tempera-ture range 0 to 100°C, is given in Table A.1. Fit these val-ues to a least-squares† equation of the form a bT cT2, with T in °C, and estimate its accuracy. Use your for-mula to compute the density of water at 45°C, and com-pare your result with the accepted experimental value of 990.1 kg/m3.
where p is the pressure drop required to drive the flow.
What are the dimensions of the constant 61.9? Can this for-mula be used with confidence for various liquids and gases?
P1.18 For small particles at low velocities, the first term in the Stokes-Oseen drag law, Prob. 1.10, is dominant; hence, F KV, where K is a constant. Suppose a particle of mass m is constrained to move horizontally from the initial po-sition x 0 with initial velocity V0. Show (a) that its ve-locity will decrease exponentially with time and (b) that it will stop after traveling a distance x mV0/K.
P1.19 For larger particles at higher velocities, the quadratic term in the Stokes-Oseen drag law, Prob. 1.10, is dominant; hence, F CV2, where C is a constant. Repeat Prob. 1.18 to show that (a) its velocity will decrease as 1/(1 CV0t/m) and (b) it will never quite stop in a finite time span.
P1.20 A baseball, with m 145 g, is thrown directly upward from the initial position z 0 and V0 45 m/s. The air drag on the ball is CV2, as in Prob. 1.19, where C 0.0013 N s2/m2. Set up a differential equation for the ball motion, and solve for the instantaneous velocity V(t) and position z(t).
Find the maximum height zmax reached by the ball, and compare your results with the classical case of zero air drag.
P1.21 A velocity field is given by V Kxti Kytj 0k, where K is a positive constant. Evaluate (a) and (b) V.
P1.22 According to the theory of Chap. 8, as a uniform stream approaches a cylinder of radius R along the symmetry line AB in Fig. P1.22, the velocity has only one component: u U1 R x2 2 for x R where U is the stream velocity far from the cylinder. Us-ing the concepts from Ex. 1.5, find (a) the maximum flow deceleration along AB and (b) its location.
48 Chapter 1 Introduction EES y x u R B A P1.22 P1.23 Experiment with a faucet (kitchen or otherwise) to determine typical flow rates Q in m3/h, perhaps timing the discharge of a known volume. Try to achieve an exit jet condition which is (a) smooth and round and (b) disorderly and fluctuating.
Measure the supply-pipe diameter (look under the sink). For both cases, calculate the average flow velocity, Vavg † The concept of “least-squares” error is very important and should be learned by everyone.
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 difference between these two val-ues represent (see Chap. 2)?
P1.33 Experimental data for the density of mercury versus pres-sure at 20°C are as follows: p, atm 1 500 1,000 1,500 2,000 , kg/m3 13,545 13,573 13,600 13,625 13,653 Fit this data to the empirical state relation for liquids, Eq. (1.22), to find the best values of B and n for mercury.
Then, assuming the data are nearly isentropic, use these values to estimate the speed of sound of mercury at 1 atm and compare with Table 9.1.
P1.34 If water occupies 1 m3 at 1 atm pressure, estimate the pres-sure required to reduce its volume by 5 percent.
P1.35 In Table A.4, most common gases (air, nitrogen, oxygen, hydrogen) have a specific heat ratio k 1.40. Why do ar-gon 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 The isentropic bulk modulus B of a fluid is defined as the isentropic change in pressure per fractional change in den-sity: B p s What are the dimensions of B? Using theoretical p() re-lations, estimate the bulk modulus of (a) N2O, assumed to be an ideal gas, and (b) water, at 20°C and 1 atm.
P1.37 A near-ideal gas has a molecular weight of 44 and a spe-cific heat cv 610 J/(kg K). What are (a) its specific heat ratio, k, and (b) its speed of sound at 100°C?
P1.38 In Fig. 1.6, if the fluid is glycerin at 20°C and the width be-tween 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 for air at 20°C from Table 1.4, estimate its vis-cosity at 500°C by (a) the power law and (b) the Suther-land law. Also make an estimate from (c) Fig. 1.5. Com-pare with the accepted value of 3.58 E-5 kg/m s.
P1.40 For liquid viscosity as a function of temperature, a sim-plification of the log-quadratic law of Eq. (1.31) is An-drade’s equation , A exp (B/T), where (A, B) are curve-fit constants and T is absolute temperature. Fit this relation to the data for water in Table A.1 and estimate the percent error of the approximation.
P1.41 Some experimental values of the viscosity of argon gas at 1 atm are as follows: T, K 300 400 500 600 700 800 , kg/(m s) 2.27 E-5 2.85 E-5 3.37 E-5 3.83 E-5 4.25 E-5 4.64 E-5 Fit these value to either (a) a power law or (b) the Suther-land law, Eq. (1.30).
P1.42 Experimental values for the viscosity of helium at 1 atm are as follows: T, K 200 400 600 800 1000 1200 , kg/(m s) 1.50 E-5 2.43 E-5 3.20 E-5 3.88 E-5 4.50 E-5 5.08 E-5 Fit these values to either (a) a power law or (b) the Suther-land law, Eq. (1.30).
P1.43 Yaws et al. suggest the following curve-fit formula for viscosity versus temperature of organic liquids: log10 A B T CT DT2 with T in absolute units. (a) Can this formula be criticized on dimensional grounds? (b) Disregarding (a), indicate an-alytically how the curve-fit constants A, B, C, D could be found from N data points ( i, Ti) using the method of least squares. Do not actually carry out a calculation.
P1.44 The values for SAE 30 oil in Table 1.4 are strictly “rep-resentative,” not exact, because lubricating oils vary con-siderably according to the type of crude oil from which they are refined. The Society of Automotive Engineers allows certain kinematic viscosity ranges for all lubricat-ing oils: for SAE 30, 9.3 12.5 mm2/s at 100°C. SAE 30 oil density can also vary 2 percent from the tabulated value of 891 kg/m3. Consider the following data for an ac-ceptable grade of SAE 30 oil: T, °C 0 20 40 60 80 100 , kg/(m s) 2.00 0.40 0.11 0.042 0.017 0.0095 How does this oil compare with the plot in Appendix Fig.
A.1? How well does the data fit Andrade’s equation in Prob. 1.40?
P1.45 A block of weight W 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.
P1.46 Find the terminal velocity of the block in Fig. P1.45 if the block mass is 6 kg, A 35 cm2, 15°, and the film is 1-mm-thick SAE 30 oil at 20°C.
P1.47 A shaft 6.00 cm in diameter is being pushed axially through a bearing sleeve 6.02 cm in diameter and 40 cm long. The clearance, assumed uniform, is filled with oil Problems 49 EES whose properties are 0.003 m2/s and SG 0.88. Es-timate the force required to pull the shaft at a steady ve-locity of 0.4 m/s.
P1.48 A thin plate is separated from two fixed plates by very vis-cous liquids 1 and 2, respectively, as in Fig. P1.48. The plate spacings h1 and h2 are unequal, as shown. The con-tact area is A between the center plate and each fluid.
(a) Assuming a linear velocity distribution in each fluid, derive the force F required to pull the plate at velocity V.
(b) Is there a necessary relation between the two viscosi-ties, 1 and 2?
sured torque is 0.293 N m, what is the fluid viscosity?
Suppose that the uncertainties of the experiment are as fol-lows: L (0.5 mm), M (0.003 N m), (1 percent), and ri or ro (0.02 mm). What is the uncertainty in the measured viscosity?
P1.52 The belt in Fig. P1.52 moves at a steady velocity V and skims the top of a tank of oil of viscosity , as shown. As-suming a linear velocity profile in the oil, develop a sim-ple formula for the required belt-drive power P as a func-tion of (h, L, V, b, ). 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, b 60 cm, and h 3 cm?
50 Chapter 1 Introduction Liquid film of thickness h W V Block contact area A θ P1.45 P1.48 h1 h2 µ1 µ2 F, V P1.49 The shaft in Prob. 1.47 is now fixed axially and rotated in-side the sleeve at 1500 r/min. Estimate (a) the torque (N m) and (b) the power (kW) required to rotate the shaft.
P1.50 An amazing number of commercial and laboratory devices have been developed to measure the viscosity of fluids, as described in Ref. 27. The concentric rotating shaft of Prob.
1.49 is an example of a rotational viscometer. Let the in-ner and outer cylinders have radii ri and ro, respectively, with total sleeve length L. Let the rotational rate be (rad/s) and the applied torque be M. Derive a theoretical relation for the viscosity of the clearance fluid, , in terms of these parameters.
P1.51 Use the theory of Prob. 1.50 (or derive an ad hoc expres-sion if you like) for a shaft 8 cm long, rotating at 1200 r/min, with ri 2.00 cm and ro 2.05 cm. If the mea-L V Oil, depth h Moving belt, width b P1.52 P1.53 A solid cone of angle 2, base r0, and density c is rotat-ing with initial angular velocity 0 inside a conical seat, as shown in Fig. P1.53. The clearance h is filled with oil of viscosity . Neglecting air drag, derive an analytical ex-pression for the cone’s angular velocity (t) if there is no applied torque.
Oil h Base radius r0 ω(t) 2θ P1.53 P1.54 A disk of radius R rotates at an angular velocity inside a disk-shaped container filled with oil of viscosity , as shown in Fig. P1.54. Assuming a linear velocity profile and neglecting shear stress on the outer disk edges, derive a formula for the viscous torque on the disk.
P1.55 The device in Fig. P1.54 is called a rotating disk viscometer . Suppose that R 5 cm and h 1 mm. If the torque required to rotate the disk at 900 r/min is 0.537 N m, what is the viscosity of the fluid? If the uncertainty in each parameter (M, R, h, ) is 1 percent, what is the overall uncertainty in the viscosity?
P1.56 The device in Fig. P1.56 is called a cone-plate viscometer . The angle of the cone is very small, so that sin , and the gap is filled with the test liquid. The torque M to rotate the cone at a rate is measured. Assuming a lin-ear velocity profile in the fluid film, derive an expression for fluid viscosity as a function of (M, R, , ). 8 r L 0 4 Q p Pipe end effects are neglected . Suppose our capillary has r0 2 mm and L 25 cm. The following flow rate and pressure drop data are obtained for a certain fluid: Q, m3/h 0.36 0.72 1.08 1.44 1.80 p, 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 s falls due to gravity inside a tube of diameter D0. The clear-ance, D0 D D, is filled with fluid of density and viscosity . 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, D0 2.04 cm, L 15 cm, with a film of SAE 30 oil at 20°C.
P1.60 For Prob. 1.52 suppose that P 0.1 hp when V 6 ft/s, L 4.5 ft, b 22 in, and h 7/8 in. Estimate the vis-cosity of the oil, in kg/(m s). If the uncertainty in each parameter (P, L, b, h, V) is 1 percent, what is the over-all uncertainty in the viscosity?
P1.61 An air-hockey puck has a mass of 50 g and is 9 cm in di-ameter. 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 lin-ear velocity distribution in the air film, how long will it take the puck to (a) slow down to 1 m/s and (b) stop com-pletely? Also, (c) how far along this extremely long table will the puck have traveled for condition (a)?
P1.62 The hydrogen bubbles which produced the velocity pro-files in Fig. 1.13 are quite small, D 0.01 mm. If the hy-drogen-water interface is comparable to air-water and the water temperature is 30°C estimate the excess pressure within the bubble.
P1.63 Derive Eq. (1.37) by making a force balance on the fluid interface in Fig. 1.9c.
P1.64 At 60°C the surface tension of mercury and water is 0.47 and 0.0662 N/m, respectively. What capillary height changes will occur in these two fluids when they are in contact with air in a clean glass tube of diameter 0.4 mm?
P1.65 The system in Fig. P1.65 is used to calculate the pressure p1 in the tank by measuring the 15-cm height of liquid in the 1-mm-diameter tube. The fluid is at 60°C (see Prob.
1.64). Calculate the true fluid height in the tube and the percent error due to capillarity if the fluid is (a) water and (b) mercury.
Problems 51 R R Clearance h Oil Ω P1.54 Ω Fluid R θ θ P1.57 For the cone-plate viscometer of Fig. P1.56, suppose that R 6 cm and 3°. If the torque required to rotate the cone at 600 r/min is 0.157 N m, what is the viscosity of the fluid? If the uncertainty in each parameter (M, R, , ) is 1 percent, what is the overall uncertainty in the vis-cosity?
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 p is the pressure drop from entrance to exit, the theory of Chap. 6 yields a for-mula for viscosity: P1.56 P1.66 A thin wire ring, 3 cm in diameter, is lifted from a water surface at 20°C. Neglecting the wire weight, what is the force required to lift the ring? Is this a good way to mea-sure surface tension? Should the wire be made of any par-ticular material?
P1.67 Experiment with a capillary tube, perhaps borrowed from the chemistry department, to verify, in clean water, the rise due to surface tension predicted by Example 1.9. Add small amounts of liquid soap to the water, and report to the class whether detergents significantly lower the surface tension.
What practical difficulties do detergents present?
P1.68 Make an analysis of the shape (x) of the water-air inter-face near a plane wall, as in Fig. P1.68, assuming that the slope is small, R1 d2 /dx2. Also assume that the pres-sure difference across the interface is balanced by the spe-cific weight and the interface height, p g . The boundary conditions are a wetting contact angle at x 0 and a horizontal surface 0 as x →. What is the max-imum height h at the wall?
mula for the maximum diameter dmax able to float in the liquid. Calculate dmax 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 between two vertical parallel plates a distance W apart, as in Fig.
P1.70. What will h be for water at 20°C if W 0.5 mm?
52 Chapter 1 Introduction 15 cm p1 P1.65 (x) θ x = 0 y = h y x η P1.68 P1.69 A solid cylindrical needle of diameter d, length L, and den-sity n may float in liquid of surface tension Y. Neglect buoyancy and assume a contact angle of 0°. Derive a for-W h θ P1.70 P1.71 A soap bubble of diameter D1 coalesces with another bub-ble of diameter D2 to form a single bubble D3 with the same amount of air. Assuming an isothermal process, de-rive an expression for finding D3 as a function of D1, D2, patm, and Y.
P1.72 Early mountaineers boiled water to estimate their altitude.
If they reach the top and find that water boils at 84°C, ap-proximately 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 Ca 0.25. At what velocity will cavitation bubbles begin to form on the body? Will the body cavitate if V 30 m/s and the water is cold (5°C)?
P1.74 A propeller is tested in a water tunnel at 20°C as in Fig.
1.12a. The lowest pressure on the blade can be estimated by a form of Bernoulli’s equation (Ex. 1.3): pmin p0 1 2 V2 where p0 1.5 atm and V tunnel velocity. If we run the tunnel at V 18 m/s, can we be sure that there will be no cavitation? If not, can we change the water temperature and avoid cavitation?
P1.75 Oil, with a vapor pressure of 20 kPa, is delivered through a pipeline by equally spaced pumps, each of which in-creases 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?
EES P1.76 An airplane flies at 555 mi/h. At what altitude in the stan-dard atmosphere will the airplane’s Mach number be ex-actly 0.8?
P1.77 The density of 20°C gasoline varies with pressure ap-proximately as follows: p, atm 1 500 1000 1500 , lbm/ft3 42.45 44.85 46.60 47.98 Use these data to estimate (a) the speed of sound (m/s) and (b) the bulk modulus (MPa) of gasoline at 1 atm.
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 Examine the photographs in Figs. 1.12a, 1.13, 5.2a, 7.14a, and 9.10b and classify them according to the boxes in Fig.
1.14.
P1.80 A two-dimensional steady velocity field is given by u x2 y2, v 2xy. Derive the streamline pattern and sketch a few streamlines in the upper half plane. Hint: The differential equation is exact.
P1.81 Repeat Ex. 1.10 by letting the velocity components in-crease linearly with time: V Kxti Kytj 0k Find and sketch, for a few representative times, the in-stantaneous streamlines. How do they differ from the steady flow lines in Ex. 1.10?
P1.82 A velocity field is given by u V cos , v V sin , and w 0, where V and are constants. Derive a formula for the streamlines of this flow.
P1.83 A two-dimensional unsteady velocity field is given by u x(1 2t), v y. Find the equation of the time-varying streamlines which all pass through the point (x0, y0) at some time t. Sketch a few of these.
P1.84 Repeat Prob. 1.83 to find and sketch the equation of the pathline which passes through (x0, y0) at time t 0.
P1.85 Do some reading and report to the class on the life and achievements, especially vis-à-vis fluid mechanics, of (a) Evangelista Torricelli (1608–1647) (b) Henri de Pitot (1695–1771) (c) Antoine Chézy (1718–1798) (d) Gotthilf Heinrich Ludwig Hagen (1797–1884) (e) Julius Weisbach (1806–1871) (f) George Gabriel Stokes (1819–1903) (g) Moritz Weber (1871–1951) (h) Theodor von Kármán (1881–1963) (i) Paul Richard Heinrich Blasius (1883–1970) (j) Ludwig Prandtl (1875–1953) (k) Osborne Reynolds (1842–1912) (l) John William Strutt, Lord Rayleigh (1842–1919) (m) Daniel Bernoulli (1700–1782) (n) Leonhard Euler (1707–1783) Fundamentals of Engineering Exam Problems 53 Fundamentals of Engineering Exam Problems FE1.1 The absolute viscosity of a fluid is primarily a func-tion of (a) Density, (b) Temperature, (c) Pressure, (d) Velocity, (e) Surface tension FE1.2 If a uniform solid body weighs 50 N in air and 30 N in water, its specific gravity is (a) 1.5, (b) 1.67, (c) 2.5, (d) 3.0, (e) 5.0 FE1.3 Helium has a molecular weight of 4.003. What is the weight of 2 m3 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 FE1.4 An oil has a kinematic viscosity of 1.25 E-4 m2/s 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 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) 48 Pa, (c) 96 Pa, (d) 192 Pa, (e) 192 Pa FE1.6 The only possible dimensionless group which combines velocity V, body size L, fluid density , and surface ten-sion coefficient is (a) L/V, (b) VL2/, (c) V2/L, (d) LV2/, (e) LV2/ 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 m2/s.
What is the average shear stress in the oil?
(a) 80 Pa, (b) 100 Pa, (c) 125 Pa, (d) 160 Pa, (e) 200 Pa 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 EES FE1.9 A certain water flow at 20°C has a critical cavitation number, where bubbles form, Ca 0.25, where Ca 2(pa pvap)/V2. If pa 1 atm and 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 FE1.10 A steady incompressible flow, moving through a con-traction section of length L, has a one-dimensional av-erage velocity distribution given by u U0(1 2x/L).
What is its convective acceleration at the end of the con-traction, x L?
(a) U0 2/L, (b) 2U0 2/L, (c) 3U0 2/L, (d) 4U0 2/L, (e) 6U0 2/L 54 Chapter 1 Introduction Comprehensive Problems C1.1 Sometimes equations can be developed and practical prob-lems can be solved by knowing nothing more than the di-mensions of the key parameters in the problem. For ex-ample, consider the heat loss through a window in a building. Window efficiency is rated in terms of “R value” which has units of (ft2 h °F)/Btu. A certain manufac-turer 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 dimen-sions are 3 ft by 5 ft. On a given winter day, the temper-ature 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 t, through a window of area A, with R value R, and temperature difference T. 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 $1.25 per gallon. The propane burner is 80 per-cent efficient. Propane has approximately 90,000 Btu of available energy per gallon. In that same 24-h pe-riod, 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?
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 bot-tom of the blade as a function of skater velocity V, blade length L, water thickness (between the blade and the ice) h, water viscosity , and blade width W.
(b) Suppose an ice skater of total mass m is skating along at a constant speed of V0 when she suddenly stands stiff with her skates pointed directly forward, allowing her-self to coast to a stop. Neglecting friction due to air re-sistance, 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 V0, m, L, h, , and W.
(c) Find x for the case where V0 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 !, are placed in a tank of liquid of known surface tension and contact an-gle , as shown in Fig. C1.3. At the free surface of the liq-uid 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 , find an expression for surface tension in terms of the other variables.
h z g L !
!
C1.3 References 55 C1.4 Oil of viscosity and density drains steadily down the side of a tall, wide vertical plate, as shown in Fig. C1.4.
In the region shown, fully developed conditions exist; that is, the velocity profile shape and the film thickness are independent of distance z along the plate. The vertical ve-locity w becomes a function only of x, and the shear re-sistance from the atmosphere is negligible.
(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 , and the slope of the veloc-ity profile at the wall, (dw/dx)wall, 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 , , (dw/dx)wall, and the gravitational accleration g. Note that, for the coordinate system given, both w and (dw/dx)wall are negative. z x g Oil film Air Plate References 1.
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56 Chapter 1 Introduction Roosevelt Dam in Arizona. Hydrostatic pressure, due to the weight of a standing fluid, can cause enormous forces and moments on large-scale structures such as a dam. Hydrostatic fluid analy-sis is the subject of the present chapter. (Courtesy of Dr. E.R. Degginger/Color-Pic Inc.) 58 2.1 Pressure and Pressure Gradient 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 applica-tions in this chapter are (1) pressure distribution in the atmosphere and the oceans, (2) the design of manometer 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 which 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 easily analyzed in arbitrary (nonrigid-body) motions V(x, y, z, t), 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 cir-cle reduces to a point. In other words, the normal stress on any plane through a fluid element at rest is equal to a unique value 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.
Figure 2.1 shows a small wedge of fluid at rest of size x by z by s and depth b into the paper. There is no shear by definition, but we postulate that the pressures px, pz, and pn may be different on each face. The weight of the element also may be important.
Summation of forces must equal zero (no acceleration) in both the x and z directions.
Fx 0 pxb z pnb s sin (2.1) Fz 0 pzb x pnb s cos 1 2 b x z Chapter 2 Pressure Distribution in a Fluid 59 Fig. 2.1 Equilibrium of a small wedge of fluid at rest.
Pressure Force on a Fluid Element but the geometry of the wedge is such that s sin z s cos x (2.2) Substitution into Eq. (2.1) and rearrangement give px pn pz pn 1 2 z (2.3) These relations illustrate two important principles of the hydrostatic, or shear-free, con-dition: (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,’’ z →0 and Eqs. (2.3) become px pz pn p (2.4) Since is arbitrary, we conclude that the pressure p at a point in a static fluid is inde-pendent of orientation.
What about the pressure at a point in a moving fluid? If there are strain rates in a moving fluid, there will be viscous stresses, both shear and normal in general (Sec.
4.3). In that case (Chap. 4) the pressure is defined as the average of the three normal stresses ii on the element p 1 3 ( xx yy zz) (2.5) The minus sign occurs because a compression stress is considered to be negative whereas p is positive. Equation (2.5) is subtle and rarely needed since the great ma-jority of viscous flows have negligible viscous normal stresses (Chap. 4).
Pressure (or any other stress, for that matter) causes no net force on a fluid element unless it varies spatially.1 To see this, consider the pressure acting on the two x faces in Fig. 2.2. Let the pressure vary arbitrarily p p(x, y, z, t) (2.6) 60 Chapter 2 Pressure Distribution in a Fluid ∆s ∆z O θ ∆x θ px pz x z (up) Element weight: dW = g( b ∆x ∆z) 1 2 Width b into paper pn ρ 1An interesting application for a large element is in Fig. 3.7.
Fig. 2.2 Net x force on an element due to pressure variation.
2.2 Equilibrium of a Fluid Element The net force in the x direction on the element in Fig. 2.2 is given by dFx p dy dz p p x dx dy dz p x dx dy dz (2.7) In like manner the net force dFy involves p/ y, and the net force dFz concerns p/ z. The total net-force vector on the element due to pressure is dFpress i p x j p y k p z dx dy dz (2.8) 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. (2.8) as fpress ∇p (2.9) Thus it is not the pressure but the pressure gradient causing a net force which must be balanced by gravity or acceleration or some other effect in the fluid.
The pressure gradient is a surface force which 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 dFgrav g dx dy dz (2.10) or fgrav g In general, there may also be a surface force due to the gradient, if any, of the vis-cous stresses. For completeness, we write this term here without derivation and con-sider it more thoroughly in Chap. 4. For an incompressible fluid with constant viscos-ity, the net viscous force is fVS 2 x V 2 2 y V 2 2 z V 2 ∇2V (2.11) where VS stands for viscous stresses and is the coefficient of viscosity from Chap.
1. Note that the term g in Eq. (2.10) denotes the acceleration of gravity, a vector act-2.2 Equilibrium of a Fluid Element 61 ( p + dx) dy dz ∂p ∂x x dy dx z p dy dz y dz ing toward the center of the earth. On earth the average magnitude of g is 32.174 ft/s2 9.807 m/s2.
The total vector resultant of these three forces—pressure, gravity, and viscous stress—must either keep the element in equilibrium or cause it to move with acceler-ation a. From Newton’s law, Eq. (1.2), we have a f fpress fgrav fvisc ∇p g ∇2V (2.12) This is one form of the differential momentum equation for a fluid element, and it is studied further in Chap. 4. Vector addition is implied by Eq. (2.12): The acceleration reflects the local balance of forces and is not necessarily parallel to the local-velocity vector, which reflects the direction of motion at that instant.
This chapter is concerned with cases where the velocity and acceleration are known, leaving one to solve for the pressure variation in the fluid. Later chapters will take up the more general problem where pressure, velocity, and acceleration are all unknown.
Rewrite Eq. (2.12) as ∇p (g a) ∇2V B(x, y, z, t) (2.13) where B is a short notation for the vector sum on the right-hand side. If V and a dV/dt are known functions of space and time and the density and viscosity are known, we can solve Eq. (2.13) for p(x, y, z, t) by direct integration. By components, Eq. (2.13) is equivalent to three simultaneous first-order differential equations p x Bx(x, y, z, t) p y By(x, y, z, t) p z Bz(x, y, z, t) (2.14) Since the right-hand sides are known functions, they can be integrated systematically to obtain the distribution p(x, y, z, t) except for an unknown function of time, which remains because we have no relation for p/ t. This extra function is found from a con-dition of known time variation p0(t) at some point (x0, y0, z0). If the flow is steady (in-dependent of time), the unknown function is a constant and is found from knowledge of a single known pressure p0 at a point (x0, y0, z0). If this sounds complicated, it is not; we shall illustrate with many examples. Finding the pressure distribution from a known velocity distribution is one of the easiest problems in fluid mechanics, which is why we put it in Chap. 2.
Examining Eq. (2.13), we can single out at least four special cases: 1.
Flow at rest or at constant velocity: The acceleration and viscous terms vanish identically, and p depends only upon gravity and density. This is the hydrostatic condition. See Sec. 2.3.
2.
Rigid-body translation and rotation: The viscous term vanishes identically, and p depends only upon the term (g a). See Sec. 2.9.
3.
Irrotational motion ( V 0): The viscous term vanishes identically, and an exact integral called Bernoulli’s equation can be found for the pressure distri-bution. See Sec. 4.9.
4.
Arbitrary viscous motion: Nothing helpful happens, no general rules apply, but still the integration is quite straightforward. See Sec. 6.4.
Let us consider cases 1 and 2 here.
62 Chapter 2 Pressure Distribution in a Fluid Fig. 2.3 Illustration of absolute, gage, and vacuum pressure read-ings.
Gage Pressure and Vacuum Pressure: Relative Terms 2.3 Hydrostatic Pressure Distributions 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 am-bient atmosphere. The second case occurs because many pressure instruments are of differential 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: 1.
p pa Gage pressure: p(gage) p pa 2.
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 p1 120,000 Pa absolute. This value may be reported as a gage pressure, p1 120,000 90,000 30,000 Pa gage. (One must also record the atmospheric pressure in the laboratory, since pa changes gradu-ally.) Suppose gage 2 reads p2 50,000 Pa absolute. Locally, this is a vacuum pres-sure and might be reported as p2 90,000 50,000 40,000 Pa vacuum. Occasion-ally, in the Problems section, we will specify gage or vacuum pressure to keep you alert to this common engineering practice.
If the fluid is at rest or at constant velocity, a 0 and ∇2V 0. Equation (2.13) for the pressure distribution reduces to ∇p g (2.15) 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 ∇p expresses the magnitude and direc-tion of the maximum spatial rate of increase of the scalar property p. As a result, ∇p 2.3 Hydrostatic Pressure Distributions 63 Absolute zero reference: p = 0 Pa abs = 90,000 Pa vacuum 120,000 90,000 50,000 0 p (Pascals) High pressure: p = 120,000 Pa abs = 30,000 Pa gage Local atmosphere: p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum Vacuum pressure: p = 50,000 Pa abs = 40,000 Pa vacuum 40,000 30,000 50,000 (Tension) Fig. 2.4 Hydrostatic-pressure distri-bution. 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.
is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.15) 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, i.e., “down.’’ If the fluid is a liquid, its free surface, being at atmospheric pres-sure, will be normal to local gravity, or “horizontal.’’ You probably knew all this be-fore, but Eq. (2.15) is the proof of it.
In our customary coordinate system z is “up.’’Thus the local-gravity vector for small-scale problems is g gk (2.16) where g is the magnitude of local gravity, for example, 9.807 m/s2. For these coordi-nates Eq. (2.15) has the components p x 0 p y 0 p z g (2.17) the first two of which tell us that p is independent of x and y. Hence p/ z can be re-placed by the total derivative dp/dz, and the hydrostatic condition reduces to d d p z or p2 p1 2 1 dz (2.18) Equation (2.18) 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 atmos-pheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a horizon-64 Chapter 2 Pressure Distribution in a Fluid Atmospheric pressure: Depth 1 Depth 2 Water Mercury Free surface a b c d A B C D Effect of Variable Gravity Hydrostatic Pressure in Liquids tal plane and are interconnected by the same fluid, water; therefore all points have the same pressure. The same is true 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 g0 2 (2.19) where r0 is the planet radius and g0 is the surface value of g. For earth, r0 3960 statute mi 6400 km. In typical engineering problems the deviation from r0 extends from the deepest ocean, about 11 km, to the atmospheric height of supersonic transport operation, about 20 km. This gives a maximum variation in g of (6400/6420)2, 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 hy-drostatics. In Example 1.7 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. (2.18) integrates to Liquids: p2 p1 (z2 z1) (2.20) or z1 z2 p 2 p 1 We use the first form in most problems. The quantity 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.
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 pa. When r0 r 2.3 Hydrostatic Pressure Distributions 65 Table 2.1 Specific Weight of Some Common Fluids Specific weight at 68°F 20°C Fluid lbf/ft3 N/m3 Air (at 1 atm) 000.0752 000,011.8 Ethyl alcohol 049.2 007,733 SAE 30 oil 055.5 008,720 Water 062.4 009,790 Seawater 064.0 010,050 Glycerin 078.7 012,360 Carbon tetrachloride 099.1 015,570 Mercury 846 133,100 Fig. 2.5 Hydrostatic-pressure distri-bution in oceans and atmospheres.
The Mercury Barometer we introduce the reference value (p1, z1) (pa, 0), Eq. (2.20) becomes, for p at any (negative) depth z, Lakes and oceans: p pa z (2.21) where is the average specific weight of the lake or ocean. As we shall see, Eq. (2.21) 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 From Table 2.1, take 9790 N/m3. With pa 91 kPa and z 60 m, Eq. (2.21) predicts that the pressure at this depth will be p 91 kN/m2 (9790 N/m3)(60 m) 1 1 00 k 0 N N 91 kPa 587 kN/m2 678 kPa Ans.
By omitting pa we could state the result as p 587 kPa (gage).
The simplest practical application of the hydrostatic formula (2.20) is the barometer (Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and in-verted 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 (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.
66 Chapter 2 Pressure Distribution in a Fluid g 0 +b –h Z p ≈ pa – bγair Free surface: Z = 0, p = pa Air Water p ≈ pa + hγwater Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury column is pro-portional to patm; (b) a modern portable barometer, with digital readout, uses the resonating silicon element of Fig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.) Hydrostatic Pressure in Gases From Fig. 2.6, Eq. (2.20) applies with p1 0 at z1 h and p2 pa at z2 0: pa 0 M(0 h) or h (2.22) At sea-level standard, with pa 101,350 Pa and M 133,100 N/m3 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 wa-ter barometer would be 34 ft high.
Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable in Eq. (2.18) if the integration carries over large pressure changes. It is sufficiently accurate to introduce the perfect-gas law p RT in Eq.
(2.18) g g p RT dp dz pa M 2.3 Hydrostatic Pressure Distributions 67 p1 ≈ 0 (Mercury has a very low vapor pressure.) z1 = h h = pa M z Mercury z2 = 0 p2 ≈ pa (The mercury is in contact with the atmosphere.) pa M γ ρ (a) (b) Separate the variables and integrate between points 1 and 2: 2 1 ln 2 1 (2.23) The integral over z requires an assumption about the temperature variation T(z). One common approximation is the isothermal atmosphere, where T T0: p2 p1 exp (2.24) The quantity in brackets is dimensionless. (Think that over; it must be dimensionless, right?) Equation (2.24) 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 T0 Bz (2.25) Here T0 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 val-ues are assumed to apply from 0 to 36,000 ft: T0 518.69°R 288.16 K 15°C B 0.003566°R/ft 0.00650 K/m (2.26) This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.25) into (2.23) and integrating, we obtain the more accurate relation p pa1 g/(RB) where R g B 5.26 (air) (2.27) 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 m2/(s2 K).
The U.S. standard atmosphere is sketched in Fig. 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, us-ing (a) the exact formula and (b) an isothermal assumption at a standard sea-level temperature of 15°C. Is the isothermal approximation adequate?
Solution Use absolute temperature in the exact formula, Eq. (2.27): p pa1 5.26 (101,350 Pa)(0.8872)5.26 101,350(0.52388) 54,000 Pa Ans. (a) This is the standard-pressure result given at z 5000 m in Table A.6.
(0.00650 K/m)(5000 m) 288.16 K Bz T0 g(z2 z1) RT0 dz T g R p2 p1 dp p 68 Chapter 2 Pressure Distribution in a Fluid Part (a) Is the Linear Formula Adequate for Gases?
If the atmosphere were isothermal at 288.16 K, Eq. (2.24) would apply: p pa exp R g T z (101,350 Pa) exp (101,350 Pa) exp( 0.5929) 60,100 Pa Ans. (b) This is 11 percent higher than the exact result. The isothermal formula is inaccurate in the tro-posphere.
The linear approximation from Eq. (2.20) or (2.21), p z, is satisfactory for liq-uids, which are nearly incompressible. It may be used even over great depths in the ocean. For gases, which are highly compressible, it is valid only over moderate changes in altitude.
The error involved in using the linear approximation (2.21) can be evaluated by ex-panding the exact formula (2.27) into a series 1 n 1 n 2 (2.28) where n g/(RB). Introducing these first three terms of the series into Eq. (2.27) and rearranging, we obtain p pa az1 (2.29) Bz T0 n 1 2 Bz T0 n(n 1) 2!
Bz T0 Bz T0 (9.807 m/s2)(5000 m) 287 m2/(s2 K) 2.3 Hydrostatic Pressure Distributions 69 Fig. 2.7 Temperature and pressure distribution in the U.S. standard at-mosphere. (From Ref. 1.) 60 50 40 30 20 10 0 60 50 40 30 20 10 0 –60 –40 –20 0 +20 20.1 km 11.0 km –56.5°C Troposphere Eq. (2.26) 15°C Temperature, °C Pressure, kPa 40 80 120 Altitude z, km Altitude z, km 1.20 kPa Eq. (2.27) 101.33 kPa Eq. (2.24) Part (b) 2.4 Application to Manometry A Memory Device: Up Versus Down Fig. 2.8 Evaluating pressure changes through a column of multi-ple fluids.
Thus the error in using the linear formula (2.21) is small if the second term in paren-theses in (2.29) is small compared with unity. This is true if z 20,800 m (2.30) We thus expect errors of less than 5 percent if z or z is less than 1000 m.
From the hydrostatic formula (2.20), a change in elevation z2 z1 of a liquid is equiv-alent to a change in pressure (p2 p1)/. Thus a static column of one or more liquids or gases can be used to measure pressure differences between two points. Such a de-vice 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 p5 p1, we add the suc-cessive changes p2 p1, p3 p2, p4 p3, and p5 p4. The intermediate values of p cancel, and we have, for the example of Fig. 2.8, p5 p1 0(z2 z1) w(z3 z2) G(z4 z3) M(z5 z4) (2.31) No additional simplification is possible on the right-hand side because of the dif-ferent 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. (2.20), is mathematically correct but vexing to en-gineers, because it combines two negative signs to have the pressure increase down-ward. When calculating hydrostatic pressure changes, engineers work instinctively by simply having the pressure increase downward and decrease upward. Thus they use the following mnemonic, or memory, device, first suggested to the writer by Professor John 2T0 (n 1)B 70 Chapter 2 Pressure Distribution in a Fluid Known pressure p1 Oil, o ρ Water, w ρ Glycerin, G ρ Mercury, M ρ z = z1 z2 z3 z4 z5 z p2 – p1 = – og(z2 – z1) ρ p3 – p2 = – wg(z3 – z2) ρ p4 – p3 = – Gg(z4 – z3) ρ p5 – p4 = – Mg(z5 – z4) ρ Sum = p5 – p1 Fig. 2.9 Simple open manometer for measuring pA relative to atmos-pheric pressure.
Foss of Michigan State University: pdown pup z (2.32) Thus, without worrying too much about which point is “z1” and which is “z2”, the for-mula simply increases or decreases the pressure according to whether one is moving down or up. For example, Eq. (2.31) could be rewritten in the following “multiple in-crease” mode: p5 p1 0z1 z2 wz2 z3 Gz3 z4 Mz4 z5 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.
Figure 2.9 shows a simple open manometer for measuring pA in a closed chamber relative to atmospheric pressure pa, in other words, measuring the gage pressure. The chamber fluid 1 is combined with a second fluid 2, perhaps for two reasons: (1) to protect the environment from a corrosive chamber fluid or (2) because a heavier fluid 2 will keep z2 small and the open tube can be shorter. One can, of course, apply the basic hydrostatic formula (2.20). Or, more simply, one can begin at A, apply Eq. (2.32) “down” to z1, jump across fluid 2 (see Fig. 2.9) to the same pressure p1, and then use Eq. (2.32) “up” to level z2: pA 1zA z1 2z1 z2 p2 patm (2.33) The physical reason that we can “jump across” at section 1 in that a continuous length of the same fluid connects these two equal elevations. The hydrostatic relation (2.20) 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.
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 typical ap-2.4 Application to Manometry 71 zA, pA A ρ2 ρ1 z1, p1 Jump across Open, pa z2, p2 ≈ pa p = p1 at z = z1 in fluid 2 Fig. 2.10 A complicated multiple-fluid manometer to relate pA to pB.
This system is not especially prac-tical but makes a good homework or examination problem.
plication is to measure pressure change across a flow device, as shown. Derive a formula for the pressure difference pa pb in terms of the system parameters in Fig. E2.3.
Solution Using our “up-down” concept as in Eq. (2.32), start at (a), evaluate pressure changes around the U-tube, and end up at (b): pa 1gL 1gh 2gh 1gL pb or pa pb ( 2 1)gh Ans.
The measurement only includes h, the manometer reading. Terms involving L drop out. Note the appearance of the difference in densities between manometer fluid and working fluid. It is a com-mon student error to fail to subtract out the working fluid density 1—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 Ex. 2.3, because of its popularity in engineering experiments, is some-times considered to be the “manometer formula,” it is best not to memorize it but rather to adapt Eq. (2.20) or (2.32) to each new multiple-fluid hydrostatics problem.
For example, Fig. 2.10 illustrates a multiple-fluid manometer problem for finding the 72 Chapter 2 Pressure Distribution in a Fluid zA, pA z1, p1 z1, p1 Jump across A ρ1 ρ2 z2, p2 z2, p2 ρ3 Jump across Jump across z3, p3 z3, p3 ρ4 B zB, pB Flow device 1 2 (a) L h (b) E2.3 E2.4 difference in pressure between two chambers A and B. We repeatedly apply Eq. (2.20), 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 p1) (p1 p2) (p2 p3) (p3 pB) 1(zA z1) 2(z1 z2) 3(z2 z3) 4(z3 zB) (2.34) The intermediate pressures p1,2,3 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 finally goes up to B.
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.
2.4 Application to Manometry 73 A Water flow 5 cm 4 cm Mercury SAE 30 oil Gage B 6 cm 11 cm Solution First list the specific weights from Table 2.1 or Table A.3: water 9790 N/m3 mercury 133,100 N/m3 oil 8720 N/m3 Now proceed from A to B, calculating the pressure change in each fluid and adding: pA W(z)W M(z)M O(z)O pB or pA (9790 N/m3)( 0.05 m) (133,100 N/m3)(0.07 m) (8720 N/m3)(0.06 m) pA 489.5 Pa 9317 Pa 523.2 Pa pB 87,000 Pa where we replace N/m2 by its short name, Pa. The value zM 0.07 m is the net elevation change in the mercury (11 cm 4 cm). Solving for the pressure at point A, we obtain pA 96,351 Pa 96.4 kPa Ans.
The intermediate six-figure result of 96,351 Pa is utterly fatuous, since the measurements cannot be made that accurately.
Fig. 2.11 Hydrostatic force and center of pressure on an arbitrary plane surface of area A inclined at an angle below the free surface.
In making these manometer calculations we have neglected the capillary-height changes due to surface tension, which were discussed in Example 1.9. These effects cancel if there is a fluid interface, or meniscus, on both sides of the U-tube, as in Fig.
2.9. 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.
A common problem in the design of structures which interact with fluids is the com-putation of the hydrostatic force on a plane surface. If we neglect density changes in the fluid, Eq. (2.20) 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 hy-drostatic 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 liq-uid. The panel plane makes an arbitrary angle 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. (2.20) the pressure there is p pa h.
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 (pa h) dA paA h dA (2.35) The remaining integral is evaluated by noticing from Fig. 2.11 that h sin and, 74 Chapter 2 Pressure Distribution in a Fluid Free surface p = pa θ h(x, y) hCG Resultant force: F = pCG A Side view CP x y CG Plan view of arbitrary plane surface dA = dx dy ξ = h sin θ 2.5 Hydrostatic Forces on Plane Surfaces by definition, the centroidal slant distance from the surface to the plate is CG A 1 dA (2.36) Therefore, since is constant along the plate, Eq. (2.35) becomes F paA sin dA paA sin CGA (2.37) Finally, unravel this by noticing that CG sin hCG, the depth straight down from the surface to the plate centroid. Thus F paA hCGA (pa hCG)A pCGA (2.38) 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 at which it is slanted.
Equation (2.38) can be visualized physically in Fig. 2.12 as the resultant of a lin-ear stress distribution over the plate area. This simulates combined compression and bending 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. (2.38).
However, to balance the bending-moment portion of the stress, the resultant force F does not act 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 yCP, we equate FyCP yp dA y(pa sin ) dA sin y dA (2.39) The term pay dA vanishes by definition of centroidal axes. Introducing CG y, 2.5 Hydrostatic Forces on Plane Surfaces 75 Fig. 2.12 The hydrostatic-pressure force on a plane surface is equal, regardless of its shape, to the resul-tant of the three-dimensional linear pressure distribution on that surface F pCGA.
Pressure distribution p (x, y) Centroid of the plane surface Arbitrary plane surface of area A pav = pCG we obtain FyCP sin CG y dA y2 dA sin Ixx (2.40) where again y dA 0 and Ixx 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 sin pC Ix G x A (2.41) The negative sign in Eq. (2.41) shows that yCP is below the centroid at a deeper level and, unlike F, depends upon angle . If we move the plate deeper, yCP approaches the centroid because every term in Eq. (2.41) remains constant except pCG, which increases.
The determination of xCP is exactly similar: FxCP xp dA x[pa (CG y) sin ] dA sin xy dA sin Ixy (2.42) where Ixy is the product of inertia of the plate, again computed in the plane of the plate. Substituting for F gives xCP sin (2.43) For positive Ixy, xCP is negative because the dominant pressure force acts in the third, or lower left, quadrant of the panel. If Ixy 0, usually implying symmetry, xCP 0 and the center of pressure lies directly below the centroid on the y axis.
Ixy pCGA 76 Chapter 2 Pressure Distribution in a Fluid x y b 2 b 2 L 2 L 2 (a) A = bL Ixx = bL3 12 Ixy = 0 x y R R A = R2 Ixx = R4 4 Ixy = 0 π π (b) x y Ixx = bL3 36 A = 2 Ixy = b(b – 2s)L2 72 b 2 b 2 L 3 2L 3 (c) x y R R A = Ixx = 0.10976R4 Ixy = 0 π 2 4R 3 π (d) s bL R2 Fig. 2.13 Centroidal moments of inertia for various cross sections: (a) rectangle, (b) circle, (c) trian-gle, and (d) semicircle.
Part (a) Part (b) In most cases the ambient pressure pa is neglected because it acts on both sides of the plate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam.
In this case pCG hCG, and the center of pressure becomes independent of specific weight F hCGA yCP Ix h x C s G in A xCP Ix h y C s G in A (2.44) Figure 2.13 gives the area and moments of inertia of several common cross sections for use with these formulas.
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.
2.5 Hydrostatic Forces on Plane Surfaces 77 Gage-Pressure Formulas 15 ft Wall 6 ft 8 ft θ Gate Hinge B A Seawater: 64 lbf/ft3 pa pa Solution By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at eleva-tion 3 ft above point B. The depth hCG is thus 15 3 12 ft. The gate area is 5(10) 50 ft2. Ne-glect pa as acting on both sides of the gate. From Eq. (2.38) the hydrostatic force on the gate is F pCGA hCGA (64 lbf/ft3)(12 ft)(50 ft2) 38,400 lbf 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 Ixy 0 and Ixx b 1 L 2 3 (5 ft) 1 (1 2 0 ft)3 417 ft4 The distance l from the CG to the CP is given by Eq. (2.44) since pa is neglected.
l yCP Ix h x C s G in A 0.417 ft (417 ft4)( 1 6 0 ) (12 ft)(50 ft2) E2.5a The distance from point B to force F is thus 10 l 5 4.583 ft. Summing moments coun-terclockwise about B gives PL sin F(5 l) P(6 ft) (38,400 lbf)(4.583 ft) 0 or P 29,300 lbf Ans. (b) With F and P known, the reactions Bx and Bz are found by summing forces on the gate Fx 0 Bx F sin P Bx 38,400(0.6) 29,300 or Bx 6300 lbf Fz 0 Bz F cos Bz 38,400(0.8) or Bz 30,700 lbf Ans. (c) This example should have reviewed your knowledge of statics.
EXAMPLE 2.6 A tank of oil has a right-triangular panel near the bottom, as in Fig. E2.6. Omitting pa, find the (a) hydrostatic force and (b) CP on the panel.
78 Chapter 2 Pressure Distribution in a Fluid pa 5 m 11 m 30° 6 m 4 m Oil: = 800 kg/m3 ρ pa CG CP 4 m 2 m 8 m 4 m Part (c) E2.6 P A F B Bx Bz θ CP CG l 5 ft L = 10 ft E2.5b 2.6 Hydrostatic Forces on Curved Surfaces Fig. 2.14 Computation of hydro-static force on a curved surface: (a) submerged curved surface; (b) free-body diagram of fluid above the curved surface.
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 corner, as shown. The area is 1 2 (6 m)(12 m) 36 m2 The moments of inertia are Ixx b 3 L 6 3 288 m4 and Ixy b(b 72 2s)L2 72 m4 The depth to the centroid is hCG 5 4 9 m; thus the hydrostatic force from Eq. (2.44) is F ghCGA (800 kg/m3)(9.807 m/s2)(9 m)(36 m2) 2.54 106 (kg m)/s2 2.54 106 N 2.54 MN Ans. (a) The CP position is given by Eqs. (2.44): yCP Ix h x C s G in A 0.444 m xCP Ix h y C s G in A 0.111 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 Fig. E2.6.
The resultant pressure force on a curved surface is most easily computed by separat-ing 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 (72 m4)(sin 30°) (9 m)(36 m2) (288 m4)(sin 30°) (9 m)(36 m2) (6 m)6 m 2(6 m)2 72 (6 m)(12 m)3 36 2.6 Hydrostatic Forces on Curved Surfaces 79 Part (a) Curved surface projection onto vertical plane FV FH FH (a) F1 F1 FH FH FV (b) d c a b e Wair W2 W1 Part (b) 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 2.14b shows a free-body diagram of the column of fluid contained in the ver-tical projection above the curved surface. The desired forces FH and FV are exerted by the surface on the fluid column. Other forces are shown due to fluid weight and hori-zontal 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 F1 exactly balance and are not relevant to the discussion. On the lower, irregular por-tion 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 ver-tical left side of the fluid column. This left-side force can be computed by the plane-surface formula, Eq. (2.38), 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.
Summation of vertical forces on the fluid free body then shows that FV W1 W2 Wair (2.45) 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.
Thus the calculation of FV involves little more than finding centers of mass of a col-umn of fluid—perhaps a little integration if the lower portion abc has a particularly vexing shape.
EXAMPLE 2.7 A dam has a parabolic shape z/z0 (x/x0)2 as shown in Fig. E2.7a, with x0 10 ft and z0 24 ft. The fluid is water, 62.4 lbf/ft3, and atmospheric pressure may be omitted. Compute the 80 Chapter 2 Pressure Distribution in a Fluid z = z0( x x0 ( 2 pa = 0 lbf/ft2 gage FV FH z CP x0 x z0 E2.7a E2.7b forces FH and FV on the dam and the position CP where they act. The width of the dam is 50 ft.
Solution The vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with its centroid halfway down, or hCG 12 ft. The force FH is thus FH hCGAproj (62.4 lbf/ft3)(12 ft)(24 ft)(50 ft) 899,000 lbf 899 103 lbf Ans.
The line of action of FH is below the centroid by an amount yCP h Ix C x G s A in pro j 4 ft Thus FH is 12 4 16 ft, or two-thirds, down from the free surface or 8 ft from the bottom, as might have been evident by inspection of the triangular pressure distribution.
The vertical component FV equals the weight of the parabolic portion of fluid above the curved surface. The geometric properties of a parabola are shown in Fig. E2.7b. The weight of this amount of water is FV ( 2 3 x0z0b) (62.4 lbf/ft3)( 2 3 )(10 ft)(24 ft)(50 ft) 499,000 lbf 499 103 lbf Ans.
1 1 2 (50 ft)(24 ft)3(sin 90°) (12 ft)(24 ft)(50 ft) 2.6 Hydrostatic Forces on Curved Surfaces 81 3z0 5 Area = 2x0z0 3 z0 0 3x0 8 x0 = 10 ft Parabola FV This acts downward on the surface at a distance 3x0/8 3.75 ft over from the origin of coordi-nates. Note that the vertical distance 3z0/5 in Fig. E2.7b is irrelevant.
The total resultant force acting on the dam is F (F2 H F2 V)1/2 [(499)2 (899)2]1/2 1028 103 lbf As seen in Fig. E2.7c, this force acts down and to the right at an angle of 29° tan1 4 8 9 9 9 9 . The force F passes through the point (x, z) (3.75 ft, 8 ft). If we move down along the 29° line un-til we strike the dam, we find an equivalent center of pressure on the dam at xCP 5.43 ft zCP 7.07 ft Ans.
This definition of CP is rather artificial, but this is an unavoidable complication of dealing with a curved surface.
E2.7c 2.7 Hydrostatic Forces in Layered Fluids 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 dis-tribution 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 82 Chapter 2 Pressure Distribution in a Fluid z Resultant = 1028 × 103 1bf acts along z = 10.083 – 0.5555x Parabola z = 0.24x2 CP CG 29° 7.07 ft 5.43 ft 899 3.75 ft x 8 ft 499 0 p2 = p1 – 2g(z2 – z1) z z = 0 z2, p2 p = p1 – 2g(z – z1) p a 1 < 2 Fluid 1 p = p a – 1gz p1 = p a – 1gz1 2 Fluid 2 Plane surface F1= pCG1A1 z1, p1 F2= pCG2A2 ρ ρ ρ ρ ρ ρ ρ Fig. 2.15 Hydrostatic forces on a surface immersed in a layered fluid must be summed in separate pieces.
Part (a) E2.8 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 find the total: F Fi pCGiAi (2.46) Similarly, the centroid of the plate portion in each layer can be used to locate the cen-ter of pressure on that portion yCPi ig p s C in Gi A i i Ixxi xCPi ig p s C in Gi A i i Ixyi (2.47) These formulas locate the center of pressure of that particular Fi 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 Fi can then be found by summing moments about some convenient point such as the surface. The follow-ing example will illustrate.
EXAMPLE 2.8 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 Divide the end panel into three parts as sketched in Fig. E2.8, and find the hydrostatic pressure at the centroid of each part, using the relation (2.38) in steps as in Fig. E2.8: 2.7 Hydrostatic Forces in Layered Fluids 83 pa = 0 z = 0 4 ft 11 ft 16 ft 7 ft 8 ft 6 ft Water (62.4) Oil: 55.0 lbf/ft3 (1) (2) 4 ft Mercury (846) (3) PCG1 (55.0 lbf/ft3)(4 ft) 220 lbf/ft2 pCG2 (55.0)(8) 62.4(3) 627 lbf/ft2 pCG3 (55.0)(8) 62.4(6) 846(2) 2506 lbf/ft2 2.8 Buoyancy and Stability These pressures are then multiplied by the respective panel areas to find the force on each portion: F1 pCG1A1 (220 lbf/ft2)(8 ft)(7 ft) 12,300 lbf F2 pCG2A2 627(6)(7) 26,300 lbf F3 pCG3A3 2506(4)(7) 70,200 lbf F Fi 108,800 lbf Ans. (a) Equations (2.47) can be used to locate the CP of each force Fi, noting that 90° and sin 1 for all parts. The moments of inertia are Ixx1 (7 ft)(8 ft)3/12 298.7 ft4, Ixx2 7(6)3/12 126.0 ft4, and Ixx3 7(4)3/12 37.3 ft4. The centers of pressure are thus at yCP1 1 F gI 1 xx1 1.33 ft yCP2 62 2 .4 6 ( , 1 3 2 0 6 0 .0) 0.30 ft yCP3 0.45 ft This locates zCP1 4 1.33 5.33 ft, zCP2 11 0.30 11.30 ft, and zCP3 16 0.45 16.45 ft. Summing moments about the surface then gives FizCPi FzCP or 12,300(5.33) 26,300(11.30) 70,200(16.45) 108,800zCP or zCP 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 be-low 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.: 1.
A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces.
2.
A floating body displaces its own weight in the fluid in which it floats.
These 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. (2.45) for vertical force, the body experiences a net upward force FB FV(2) FV(1) (fluid weight above 2) (fluid weight above 1) weight of fluid equivalent to body volume (2.48) Alternatively, from Fig. 2.16b, we can sum the vertical forces on elemental vertical slices through the immersed body: FB body (p2 p1) dAH (z2 z1) dAH ()(body volume) (2.49) 1,518,000 108,800 846(37.3) 70,200 (55.0 lbf/ft3)(298.7 ft4) 12,300 lbf 84 Chapter 2 Pressure Distribution in a Fluid Part (b) These are identical results and equivalent to law 1 above.
Equation (2.49) assumes that the fluid has uniform specific weight. The line of ac-tion of the buoyant force passes through the center of volume of the displaced body; i.e., its center of mass is 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 draw-ing. 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.
Equation (2.49) can be generalized to a layered fluid (LF) by summing the weights of each layer of density i displaced by the immersed body: (FB)LF ig(displaced volume)i (2.50) Each displaced layer would have its own center of volume, and one would have to sum moments of the incremental buoyant forces to find the center of buoyancy of the im-mersed 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 lbf/ft3. We may record the weight of a person as 180 lbf and thus estimate the person’s total volume as 3.0 ft3. However, in so doing we are neglecting the buoyant force of the air surrounding the person. At standard con-ditions, the specific weight of air is 0.0763 lbf/ft3; hence the buoyant force is approxi-mately 0.23 lbf. If measured in vacuo, the person would weigh about 0.23 lbf 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, e.g., natural convection of heat and vertical mixing in the ocean, are strongly dependent upon 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 (2.49) is modified to apply to this smaller volume FB ()(displaced volume) floating-body weight (2.51) 2.8 Buoyancy and Stability 85 Surface 1 Surface 2 FV (1) Horizontal elemental area dAH z1 – z2 p1 p2 (a) FV (2) (b) Fig. 2.16 Two different approaches to the buoyant force on an arbitrary immersed body: (a) forces on up-per and lower curved surfaces; (b) summation of elemental vertical-pressure forces.
E2.9 Not only does the buoyant force equal the body weight, but also they are collinear since there can be no net moments for static equilibrium. Equation (2.51) is the math-ematical equivalent of Archimedes’ law 2, previously stated.
EXAMPLE 2.9 A block of concrete weighs 100 lbf in air and “weighs’’ only 60 lbf when immersed in fresh wa-ter (62.4 lbf/ft3). What is the average specific weight of the block?
Solution A free-body diagram of the submerged block (see Fig. E2.9) shows a balance between the ap-parent weight, the buoyant force, and the actual weight Fz 0 60 FB 100 or FB 40 lbf (62.4 lbf/ft3)(block volume, ft3) Solving gives the volume of the block as 40/62.4 0.641 ft3. Therefore the specific weight of the block is block 0 1 .6 0 4 0 1 lb ft f 3 156 lbf/ft3 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 in 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 upon its point. The least disturbance will cause it to seek another equilibrium position which is stable. Engineers must design to avoid floating instabil-86 Chapter 2 Pressure Distribution in a Fluid FB W = 100 lbf 60 lbf Stability Fig. 2.17 Static equilibrium of a floating body.
CG W FB B (Displaced volume) × ( γ of fluid) = body weight Neglect the displaced air up here.
Stability Related to Waterline Area ity. 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 mo-ment which 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 archi-tects , but we can at least outline the basic principle of the static-stability calcula-tion. Figure 2.18 illustrates the computation for the usual case of a symmetric floating body. The steps are as follows: 1.
The basic floating position is calculated from Eq. (2.51). The body’s center of mass G and center of buoyancy B are computed.
2.
The body is tilted a small angle , and a new waterline is established for the body to float at this angle. The new position B of the center of buoyancy is cal-culated. A vertical line drawn upward from B intersects the line of symmetry at a point M, called the metacenter, which is independent of for small angles.
3.
If point M is above G, that is, if the metacentric height M G is positive, a restor-ing moment is present and the original position is stable. If M is below G (nega-tive M G , the body is unstable and will overturn if disturbed. Stability increases with increasing M G .
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 in-volved.
Naval architects have developed the general stability concepts from Fig. 2.18 into a simple computation involving the area moment of inertia of the waterline area about the axis of tilt. The derivation assumes that the body has a smooth shape variation (no discontinuities) near the waterline and is derived from Fig. 2.19.
The y-axis of the body is assumed to be a line of symmetry. Tilting the body a small angle then submerges small wedge Obd and uncovers an equal wedge cOa, as shown.
2.8 Buoyancy and Stability 87 Fig. 2.18 Calculation of the meta-center M of the floating body shown in (a). Tilt the body a small angle . Either (b) B moves far out (point M above G denotes sta-bility); or (c) B moves slightly (point M below G denotes instabil-ity).
Line of symmetry G W FB B M G W B' FB ∆ G W M FB B' Small disturbance angle Small disturbance angle (b) Either Restoring moment or Overturning moment (a) (c) θ ∆θ 88 Chapter 2 Pressure Distribution in a Fluid cOdea Obd cOa Obd cOa Obd cOa waterline x υabOde x dυ x dυ x dυ 0 x (L dA) x (L dA) The new position B of the center of buoyancy is calculated as the centroid of the sub-merged portion aObde of the body: y c a e d x x b M O B G G G dA = x tan dx Tilted floating body Variable-width L(x) into paper Original waterline area B Fig. 2.19 A floating body tilted through a small angle . The move-ment x of the center of buoyancy B is related to the waterline area mo-ment of inertia.
0 x L (x tan dx) xL (x tan dx) tan x2 dAwaterline IO tan where IO is the area moment of inertia of the waterline footprint of the body about its tilt axis O. The first integral vanishes because of the symmetry of the original sub-merged portion cOdea. The remaining two “wedge” integrals combine into IO when we notice that L dx equals an element of waterline area. Thus we determine the de-sired distance from M to B: M B M G G B or M G υ I s O ub G B (2.52) The engineer would determine the distance from G to B from the basic shape and design of the floating body and then make the calculation of IO and the submerged volume υsub. If the metacentric height MG is positive, the body is stable for small disturbances. Note that if G B is negative, that is, B is above G, the body is always stable.
EXAMPLE 2.10 A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as in Fig. E2.10. Determine (a) the metacentric height for a small tilt angle and (b) the range of ratio L/H for which the barge is statically stable if G is exactly at the waterline as shown.
IO υsubmerged x 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, IO b(2L)3/12. Meanwhile, υsub 2LbH. Equation (2.52) predicts M G υ I s O ub G B H 2 3 L H 2 H 2 Ans. (a) The barge can thus be stable only if L2 3H2/2 or 2L 2.45H Ans. (b) The wider the barge relative to its draft, the more stable it is. Lowering G would help also.
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 ex-ample, a ship may float the way we like it, so that we can sit upon the deck, or it may float upside down (capsized). An interesting mathematical approach to floating stabil-ity is given in Ref. 11. The author of this reference points out that even simple shapes, e.g., 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 sym-metry tilted from the vertical.
Floating instability occurs in nature. Living fish generally swim with their plane of symmetry 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 Green-land glacier which protruded into the ocean. The exposed surface is rough, indicating that it has undergone further calving. Icebergs are frozen fresh, bubbly, glacial water of average density 900 kg/m3. Thus, when an iceberg is floating in seawater, whose average density is 1025 kg/m3, approximately 900/1025, or seven-eighths, of its vol-ume 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 8bL3/12 2LbH 2.9 Pressure Distribution in Rigid-Body Motion 89 G B H O L L G E2.10 2.9 Pressure Distribution in Rigid-Body Motion or strain rates, so that the viscous term ∇2V in Eq. (2.13) vanishes, leaving a balance between pressure, gravity, and particle acceleration ∇p (g a) (2.53) The pressure gradient acts in the direction g a, and lines of constant pressure (in-cluding 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. 3.12.
If the center of rotation is at point O and the translational velocity is V0 at this point, the velocity of an arbitrary point P on the body is given by2 V V0 r0 where is the angular-velocity vector and r0 is the position of point P. Differentiat-ing, we obtain the most general form of the acceleration of a rigid body: a ( r0) r0 (2.54) Looking at the right-hand side, we see that the first term is the translational accel-eration; the second term is the centripetal acceleration, whose direction is from point d dt dV0 dt 90 Chapter 2 Pressure Distribution in a Fluid Fig. 2.20 A North Atlantic iceberg formed by calving from a Green-land glacier. These, and their even larger Antarctic sisters, are the largest floating bodies in the world.
Note the evidence of further calv-ing fractures on the front surface.
(Courtesy of S/ oren Thalund, Green-land tourism a/s Iiulissat, Green-land.) 2 For a more detailed derivation of rigid-body motion, see Ref. 4, Sec. 2.7.
P perpendicular toward the axis of rotation; and the third term is the linear accelera-tion due to changes in the angular velocity. It is rare for all three of these terms to ap-ply to any one fluid flow. In fact, 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 which 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 dis-cuss the pressure distribution in a tank of rigidly accelerating water. The following is an example where the water in the tank will reach uniform acceleration rapidly.
EXAMPLE 2.11 A tank of water 1 m deep is in free fall under gravity with negligible drag. Compute the pres-sure at the bottom of the tank if pa 101 kPa.
Solution Being unsupported in this condition, the water particles tend to fall downward as a rigid hunk of fluid. In free fall with no drag, the downward acceleration is a g. Thus Eq. (2.53) for this situation gives ∇p (g g) 0. The pressure in the water is thus constant everywhere and equal to the atmospheric pressure 101 kPa. In other words, the walls are doing no service in sus-taining the pressure distribution which would normally exist.
In this general case of uniform rigid-body acceleration, Eq. (2.53) applies, a having the same magnitude and direction for all particles. With reference to Fig. 2.21, the par-allelogram 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 such that tan1 g ax az (2.55) 2.9 Pressure Distribution in Rigid-Body Motion 91 Uniform Linear Acceleration p = p1 p2 p3 x z ∆ p µg – a ax a az Fluid at rest S –a θ = tan –1 ax g + az g az ax θ Fig. 2.21 Tilting of constant-pressure surfaces in a tank of liquid in rigid-body acceleration.
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 di-rection g a is greater than in ordinary hydrostatics and is given by G where G [a2 x (g az)2]1/2 (2.56) These results are independent of the size or shape of the container as long as the fluid is continuously connected throughout the container.
EXAMPLE 2.12 A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s2. The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigid-body acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate the gage pressure in the corner at point A if the density of coffee is 1010 kg/m3.
Solution The free surface tilts at the angle given by Eq. (2.55) regardless of the shape of the mug. With az 0 and standard gravity, tan1 a g x tan1 9 7 .
.
8 0 1 35.5° If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted sur-face intersects the original rest surface exactly at the centerline, as shown in Fig. E2.12. dp ds 92 Chapter 2 Pressure Distribution in a Fluid Part (a) Part (b) 3 cm ax = 7 m/s2 A ∆z 3 cm 7 cm θ E2.12 Thus the deflection at the left side of the mug is z (3 cm)(tan ) 2.14 cm Ans. (a) This is less than the 3-cm clearance available, so the coffee will not spill unless it was sloshed during the start-up of acceleration.
When at rest, the gage pressure at point A is given by Eq. (2.20): pA g(zsurf zA) (1010 kg/m3)(9.81 m/s2)(0.07 m) 694 N/m2 694 Pa During acceleration, Eq. (2.56) applies, with G [(7.0)2 (9.81)2]1/2 12.05 m/s2. The dis-tance ∆s down the normal from the tilted surface to point A is s (7.0 2.14)(cos ) 7.44 cm Thus the pressure at point A becomes pA G s 1010(12.05)(0.0744) 906 Pa Ans. (b) which is an increase of 31 percent over the pressure when at rest.
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 for the fluid to have attained rigid-body rotation. The fluid acceleration will then be the centripetal term in Eq. (2.54). In the coordinates of Fig.
2.22, the angular-velocity and position vectors are given by k r0 irr (2.57) Then the acceleration is given by ( r0) r2ir (2.58) as marked in the figure, and Eq. (2.53) for the force balance becomes ∇p ir k (g a) (gk r2ir) (2.59) Equating like components, we find the pressure field by solving two first-order partial differential equations r2 (2.60) This is our first specific example of the generalized three-dimensional problem de-scribed by Eqs. (2.14) for more than one independent variable. The right-hand sides of p z p r p z p r 2.9 Pressure Distribution in Rigid-Body Motion 93 z, k r, ir p = pa Ω a = –rΩ2ir –a g g–a Still-water level Axis of rotation p = p1 p2 p3 Fig. 2.22 Development of parabo-loid constant-pressure surfaces in a fluid in rigid-body rotation. The dashed line along the direction of maximum pressure increase is an exponential curve.
Rigid-Body Rotation (2.60) are known functions of r and z. One can proceed as follows: Integrate the first equation “partially,’’ i.e., holding z constant, with respect to r. The result is p 1 2 r22 f(z) (2.61) where the “constant’’ of integration is actually a function f(z).† Now differentiate this with respect to z and compare with the second relation of (2.60): 0 f(z) or f(z) z C (2.62a) where C is a constant. Thus Eq. (2.61) now becomes p const z 1 2 r22 (2.62b) This is the pressure distribution in the fluid. The value of C is found by specifying the pressure at one point. If p p0 at (r, z) (0, 0), then C p0. The final desired dis-tribution is p p0 z 1 2 r22 (2.63) The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressure surface, say, p p1, Eq. (2.63) becomes z a br2 (2.64) Thus the surfaces are paraboloids of revolution, concave upward, with their minimum point 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 cylin-der rotating about its central axis, as in Fig. 2.23. Since the volume of a paraboloid is r22 2g p0 p1 p z 94 Chapter 2 Pressure Distribution in a Fluid Volume = π 2 R2h Still -water level R R Ω h = Ω2R2 2g h 2 h 2 Fig. 2.23 Determining the free-surface position for rotation of a cylinder of fluid about its central axis.
†This is because f(z) vanishes when differentiated with respect to r. If you don’t see this, you should review your calculus.
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 h/2 2R2/(4g), and the edges rise an equal amount.
EXAMPLE 2.13 The coffee cup in Example 2.12 is removed from the drag racer, placed on a turntable, and ro-tated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity which will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition.
Solution The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the distance h/2 in Fig. 2.23. Thus h 2 0.03 m 4 2 g R2 Solving, we obtain 2 1308 or 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. E2.13. The gage pressure here is p0 0, and point A is at (r, z) (3 cm, 4 cm). Equation (2.63) can then be evaluated pA 0 (1010 kg/m3)(9.81 m/s2)(0.04 m) 1 2 (1010 kg/m3)(0.03 m)2(1308 rad2/s2) 396 N/m2 594 N/m2 990 Pa Ans. (b) This is about 43 percent greater than the still-water pressure pA 694 Pa.
Here, as in the linear-acceleration case, it should be emphasized that the paraboloid pressure distribution (2.63) 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 vi-sualize an imaginary free surface extending outside the walls of the container.
EXAMPLE 2.14 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 negligi-ble. Atmospheric pressure is 2116 lbf/ft2. Find the pressure at point A in the rotating condition.
See Fig. E2.14.
2(0.03 m)2 4(9.81 m/s2) 2.9 Pressure Distribution in Rigid-Body Motion 95 Part (a) Part (b) 3 cm A 3 cm 7 cm 3 cm Ω z r 0 E2.13 96 Chapter 2 Pressure Distribution in a Fluid Solution Convert the angular velocity to radians per second: (180 r/min) 18.85 rad/s From Table 2.1 we find for mercury that 846 lbf/ft3 and hence 846/32.2 26.3 slugs/ft3.
At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check this from Eq. (2.64)], 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 at this height, we can calcu-late the constant C in Eq. (2.62b) from the condition pB 2116 lbf/ft2 at (r, z) (10 in, 0): pB 2116 lbf/ft2 C 0 1 2 (26.3 slugs/ft3)( 1 1 0 2 ft)2(18.85 rad/s)2 or C 2116 3245 1129 lbf/ft2 We then obtain pA by evaluating Eq. (2.63) at (r, z) (0, 30 in): pA 1129 (846 lbf/ft3)( 3 1 0 2 ft) 1129 2115 986 lbf/ft2 Ans. This is less than atmospheric pressure, and we can see why if we follow the free-surface pa-raboloid down from point B along the dashed line in the figure. It will cross the horizontal por-tion of the U-tube (where p will be atmospheric) and fall below point A. From Fig. 2.23 the ac-tual drop from point B will be h 2 2 g R2 3.83 ft 46 in Thus pA is about 16 inHg below atmospheric pressure, or about 1 1 6 2 (846) 1128 lbf/ft2 below pa 2116 lbf/ft2, which checks with the answer above. When the tube is at rest, pA 2116 846( 3 1 0 2 ) 4231 lbf/ft2 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 sur-faces, and hence their slope is the negative inverse of the slope computed from Eq.
(2.64): GL where GL stands for gradient line or (2.65) g r2 dz dr 1 r2/g 1 (dz/dr)pconst dz dr (18.85)2( 1 1 0 2 )2 2(32.2) 2 rad/r 60 s/min z Ω r Imaginary free surface 10 in 30 in 0 A B E2.14 2.10 Pressure Measurement 97 Separating the variables and integrating, we find the equation of the pressure-gradient surfaces r C1 exp (2.66) Notice that this result and Eq. (2.64) are independent of the density of the fluid. In the absence of friction and Coriolis effects, Eq. (2.66) defines the lines along which the ap-parent net gravitational field would act on a particle. Depending upon its density, a small particle or bubble would tend to rise or fall in the fluid along these exponential lines, as demonstrated experimentally in Ref. 5. Also, buoyant streamers would align them-selves with these exponential lines, thus avoiding any stress other than pure tension. Fig-ure 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 molecu-lar 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, both for a static fluid and a moving stream. The instrumentation texts in Refs. 7 to 10, 12, and 13 list over 20 designs for pressure measurement instruments.
These instruments may be grouped into four categories: 1.
Gravity-based: barometer, manometer, deadweight piston.
2.
Elastic deformation: bourdon tube (metal and quartz), diaphragm, bellows, strain-gage, optical beam displacement.
3.
Gas behavior: gas compression (McLeod gage), thermal conductance (Pirani gage), molecular impact (Knudsen gage), ionization, thermal conductivity, air piston.
4.
Electric output: resistance (Bridgman wire gage), diffused strain gage, capacita-tive, piezoelectric, magnetic inductance, magnetic reluctance, linear variable dif-ferential transformer (LVDT), resonant frequency.
The gas-behavior gages are mostly special-purpose instruments used for certain scien-tific experiments. The deadweight tester is the instrument used most often for calibra-tions; for example, it is used by the U.S. National Institute for Standards and Tech-nology (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 mea-surements must not disturb the flow. The best way to do this is to take the measure-ment through a static hole in the wall of the flow, as illustrated for the two instruments in Fig. 2.25. 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 mea-suring tube once the pressure has adjusted to a steady value. Thus the flow is almost undisturbed. An oscillating flow pressure, however, can cause a large error due to pos-sible dynamic response of the tubing. Other devices of smaller dimensions are used for dynamic-pressure measurements. Note that the manometers in Fig. 2.25 are arranged to measure the absolute pressures p1 and p2. If the pressure difference p1 p2 is de-2z g 2.10 Pressure Measurement 98 Chapter 2 Pressure Distribution in a Fluid Fig. 2.24 Experimental demonstra-tion with buoyant streamers of the fluid force field in rigid-body rota-tion: (top) fluid at rest (streamers hang vertically upward); (bottom) rigid-body rotation (streamers are aligned with the direction of maxi-mum pressure gradient). (From Ref.
5, courtesy of R. Ian Fletcher.) Fig. 2.25 Two types of accurate manometers for precise measure-ments: (a) tilted tube with eye-piece; (b) micrometer pointer with ammeter detector.
sired, a significant error is incurred by subtracting two independent measurements, and it would be far better to connect both ends of one instrument to the two static holes p1 and p2 so that one manometer reads the difference directly. 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 at-tached to a calibrated dial pointer, as shown. Or the deflection can be used to drive electric-output sensors, such as a variable transformer. Similarly, a membrane or dia-phragm will deflect under pressure and can either be sensed directly or used to drive another sensor.
2.10 Pressure Measurement 99 Flow Flow p2 p1 (a) (b) Fig. 2.26 Schematic of a bourdon-tube device for mechanical mea-surement of high pressures.
Bourdon tube Pointer for dial gage Section AA A A Linkage High pressure Flattened tube deflects outward under pressure Fig. 2.27 The fused-quartz, force-balanced bourdon tube is the most accurate pressure sensor used in commercial applications today.
(Courtesy of Ruska Instrument Corporation, Houston, TX.) 100 Chapter 2 Pressure Distribution in a Fluid An interesting variation of Fig. 2.26 is the fused-quartz, forced-balanced bourdon tube, shown in Fig. 2.27, whose deflection is sensed optically and returned to a zero reference state by a magnetic element whose output is proportional to the fluid pres-sure. The fused-quartz, forced-balanced bourdon tube is reported to be one of the most accurate pressure sensors ever devised, with uncertainty of the order of 0.003 per-cent.
The last category, electric-output sensors, is extremely important in engineering because the data can be stored on computers and freely manipulated, plotted, and an-alyzed. Three examples are shown in Fig. 2.28, the first being the capacitive sensor in Fig. 2.28a. The differential pressure deflects the silicon diaphragm and changes the capacitancce of the liquid in the cavity. Note that the cavity has spherical end caps to prevent overpressure damage. In the second type, Fig. 2.28b, strain gages and other sensors are diffused or etched onto a chip which is stressed by the applied pres-sure. 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 pres-sure. An oscillator excites the element’s resonant frequency and converts it into ap-propriate pressure units. For further information on pressure sensors, see Refs. 7 to 10, 12, and 13.
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 velocity field.
All hydrostatic (Secs. 2.3 to 2.8) and rigid-body (Sec. 2.9) problems are solved in this manner and are classic cases which every student should understand. In arbitrary vis-cous flows, both pressure and velocity are unknowns and are solved together as a sys-tem of equations in the chapters which follow.
Summary Seal diaphragm Filling liquid Sensing diaphragm Cover flange Low-pressure side High-pressure side (a) Fig. 2.28 Pressure sensors with electric output: (a) a silicon dia-phragm whose deflection changes the cavity capacitance (Courtesy of Johnson-Yokogawa Inc.); (b) a sili-con strain gage which is stressed by applied pressure; (c) a microma-chined silicon element which res-onates at a frequency proportional to applied pressure. [(b) and (c) are courtesy of Druck, Inc., Fair-field, CT.] 101 (Druck, Inc., Fairfield, Connecticut) Strain gages Diffused into integrated silicon chip Wire bonding Stitch bonded connections from chip to body plug Etched cavity Micromachined silicon sensor Temperature sensor On-chip diode for optimum temperature performance (c) (b) Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are indicated with an as-terisk, as in Prob. 2.8. Problems labeled with an EES icon (for example, Prob. 2.62), will benefit from the use of the Engi-neering Equation Solver (EES), while problems labeled with a disk icon may require the use of a computer. The standard end-of-chapter problems 2.1 to 2.158 (categorized in the problem list below) are followed by word problems W2.1 to W2.8, fun-damentals of engineering exam problems FE2.1 to FE2.10, com-prehensive problems C2.1 to C2.4, and design projects D2.1 and D2.2.
Problem Distribution Section Topic Problems 2.1, 2.2 Stresses; pressure gradient; gage pressure 2.1–2.6 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 2.48–2.81 2.6 Forces on curved surfaces 2.82–2.100 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 2.152–2.158 2.10 Pressure measurements None P2.1 For the two-dimensional stress field shown in Fig. P2.1 it is found that xx 3000 lbf/ft2 yy 2000 lbf/ft2 xy 500 lbf/ft2 Find the shear and normal stresses (in lbf/ft2) acting on plane AA cutting through the element at a 30° angle as shown.
102 Chapter 2 Pressure Distribution in a Fluid P2.2 For the two-dimensional stress field shown in Fig. P2.1 suppose that xx 2000 lbf/ft2 yy 3000 lbf/ft2 n(AA) 2500 lbf/ft2 Compute (a) the shear stress xy and (b) the shear stress on plane AA.
P2.3 Derive Eq. (2.18) by using the differential element in Fig.
2.2 with z “up,’’no fluid motion, and pressure varying only in the z direction.
P2.4 In a certain two-dimensional fluid flow pattern the lines of constant pressure, or isobars, are defined by the ex-pression P0 Bz Cx2 constant, where B and C are constants and p0 is the (constant) pressure at the origin, (x, z) (0, 0). Find an expression x f(z) for the family of lines which are everywhere parallel to the local pres-sure gradient V p.
P2.5 Atlanta, Georgia, has an average altitude of 1100 ft. On a standard day (Table A.6), pressure gage A in a laboratory experiment reads 93 kPa and gage B reads 105 kPa. Ex-press these readings in gage pressure or vacuum pressure (Pa), whichever is appropriate.
P2.6 Any pressure reading can be expressed as a length or head, h p/ g. What is standard sea-level pressure expressed in (a) ft of ethylene glycol, (b) in Hg, (c) m of water, and (d) mm of methanol? Assume all fluids are at 20°C.
P2.7 The deepest known point in the ocean is 11,034 m in the Mariana Trench in the Pacific. At this depth the specific weight of seawater is approximately 10,520 N/m3. At the surface, 10,050 N/m3. Estimate the absolute pressure at this depth, in atm.
P2.8 Dry adiabatic lapse rate (DALR) is defined as the nega-tive value of atmospheric temperature gradient, dT/dz, when temperature and pressure vary in an isentropic fash-ion. Assuming air is an ideal gas, DALR dT/dz when T T0(p/p0)a, where exponent a (k 1)/k, k cp/cv is the ratio of specific heats, and T0 and p0 are the tempera-ture 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 1)/(kR), where R is the ideal gas constant for air. (b) Calculate the numerical value of DALR for air in units of °C/km.
P2.9 For a liquid, integrate the hydrostatic relation, Eq. (2.18), by assuming that the isentropic bulk modulus, B ( p/ )s, is constant—see Eq. (9.18). Find an expression for p(z) and apply the Mariana Trench data as in Prob. 2.7, using Bseawater from Table A.3.
P2.10 A closed tank contains 1.5 m of SAE 30 oil, 1 m of wa-ter, 20 cm of mercury, and an air space on top, all at 20°C.
The absolute pressure at the bottom of the tank is 60 kPa.
What is the pressure in the air space?
30° A A σxx σxy σyx σyy σxx σyy σyx σxy = = P2.1 P2.16 A closed inverted cone, 100 cm high with diameter 60 cm at the top, is filled with air at 20°C and 1 atm. Water at 20°C is introduced at the bottom (the vertex) to compress the air isothermally until a gage at the top of the cone reads 30 kPa (gage). Estimate (a) the amount of water needed (cm3) and (b) the resulting absolute pressure at the bottom of the cone (kPa).
P2.11 In Fig. P2.11, 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.
Problems 103 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/m3?
1 m 1.5 m 2 m z=0 Gasoline Glycerin A B C Air P2.11 6 cm 12 cm h 8 cm Oil Water P2.12 P2.13 In Fig. P2.13 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?
P2.14 The closed tank in Fig. P2.14 is at 20°C. If the pressure at point A is 95 kPa absolute, what is the absolute pres-sure at point B in kPa? What percent error do you make by neglecting the specific weight of the air?
P2.15 The air-oil-water system in Fig. P2.15 is at 20°C. Know-ing that gage A reads 15 lbf/in2 absolute and gage B reads 1.25 lbf/in2 less than gage C, compute (a) the specific weight of the oil in lbf/ft3 and (b) the actual reading of gage C in lbf/in2 absolute.
1.5 m 1 m h Water Gasoline Liquid, SG = 1.60 P2.13 A Air B Air 4 m 2 m 2 m 4 m Water P2.14 Air Oil Water 1 ft 1 ft 2 ft 2 ft A B C 15 lbf/in2 abs P2.15 P2.19 The U-tube in Fig. P2.19 has a 1-cm ID and contains mer-cury as shown. If 20 cm3 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?
P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56 lbf/ft3. 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 ab-solute? See Fig. P2.21.
P2.18 The system in Fig. P2.18 is at 20°C. If atmospheric pres-sure is 101.33 kPa and the pressure at the bottom of the tank is 242 kPa, what is the specific gravity of fluid X?
P2.17 The system in Fig. P2.17 is at 20°C. If the pressure at point A is 1900 lbf/ft2, determine the pressures at points B, C, and D in lbf/ft2.
104 Chapter 2 Pressure Distribution in a Fluid Air Air Air Water A B C 2 ft D 4 ft 3 ft 2 ft 5 ft P2.17 0.5 m SAE 30 oil Water Fluid X 1 m 2 m 3 m Mercury P2.18 Mercury 10 cm 10 cm 10 cm P2.19 Oil 3-in diameter 1 in 15 in 1-in diameter F 2000 lbf Mercury 80 cm A B h?
Air: 180 kPa abs Water P2.20 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 wa-ter 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 tension ef-fects are negligible, what is the density of the oil, in kg/m3?
P2.24 In Prob. 1.2 we made a crude integration of the density distribution (z) in Table A.6 and estimated the mass of the earth’s atmosphere to be m 6 E18 kg. Can this re-mental observations. (b) Find an expression for the pres-sure at points 1 and 2 in Fig. P2.27b. Note that the glass is now inverted, so the original top rim of the glass is at the bottom of the picture, and the original bottom of the glass is at the top of the picture. The weight of the card can be neglected.
Problems 105 sult 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 atmos-pheric mass?
P2.25 Venus has a mass of 4.90 E24 kg and a radius of 6050 km.
Its atmosphere is 96 percent CO2, but let us assume it to be 100 percent. Its surface temperature averages 730 K, decreasing to 250 K at an altitude of 70 km. The average surface pressure is 9.1 MPa. Estimate the atmospheric pressure of Venus at an altitude of 5 km.
P2.26 Investigate the effect of doubling the lapse rate on atmos-pheric pressure. Compare the standard atmosphere (Table A.6) with a lapse rate twice as high, B2 0.0130 K/m.
Find the altitude at which the pressure deviation is (a) 1 percent and (b) 5 percent. What do you conclude?
P2.27 Conduct an experiment to illustrate atmospheric pressure.
Note: Do this over a sink or you may get wet! Find a drink-ing 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 in-dex 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 experi-Gasoline SG = 0.68 30 cm Water h?
2 cm Air pgage Vent P2.22 8 cm 6 cm Water Oil 10 cm P2.23 Card Top of glass Bottom of glass Card Original top of glass Original bottom of glass 1 G 2 G P2.27a P2.27b (c) Estimate the theoretical maximum glass height such that this experiment could still work, i.e., such that the wa-ter would not fall out of the glass.
P2.28 Earth’s atmospheric conditions vary somewhat. On a cer-tain day the sea-level temperature is 45°F and the sea-level pressure is 28.9 inHg. An airplane overhead registers an air temperature of 23°F and a pressure of 12 lbf/in2. Esti-mate the plane’s altitude, in feet.
P2.29 Under some conditions the atmosphere is adiabatic, p (const)( k), where k is the specific heat ratio. Show that, for an adiabatic atmosphere, the pressure variation is given by p p01 k/(k1) Compare this formula for air at z 5000 m with the stan-dard atmosphere in Table A.6.
P2.30 In Fig. P2.30 fluid 1 is oil (SG 0.87) and fluid 2 is glyc-erin at 20°C. If pa 98 kPa, determine the absolute pres-sure at point A.
(k 1)gz kRT0 P2.34 Sometimes manometer dimensions have a significant ef-fect. In Fig. P2.34 containers (a) and (b) are cylindrical and conditions are such that pa pb. Derive a formula for the pressure difference pa pb when the oil-water interface on the right rises a distance h h, for (a) d D and (b) d 0.15D. What is the percent change in the value of p?
P2.31 In Fig. P2.31 all fluids are at 20°C. Determine the pres-sure difference (Pa) between points A and B.
106 Chapter 2 Pressure Distribution in a Fluid P2.33 In Fig. P2.33 the pressure at point A is 25 lbf/in2. All flu-ids are at 20°C. What is the air pressure in the closed cham-ber B, in Pa?
P2.30 P2.31 P2.32 P2.33 P2.34 ρ1 ρ2 pa A 10 cm 32 cm P2.35 Water flows upward in a pipe slanted at 30°, as in Fig.
P2.35. The mercury manometer reads h 12 cm. Both flu-ids are at 20°C. What is the pressure difference p1 p2 in the pipe?
P2.36 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 of the tube?
P2.37 The inclined manometer in Fig. P2.37 contains Meriam red manometer oil, SG 0.827. Assume that the reservoir 20 cm 40 cm 8 cm 9 cm 14 cm A B Kerosine Air Water Mercury Benzene 18 cm H 35 cm Mercury Water Meriam red oil, SG = 0.827 A B A 4 cm 3 cm 6 cm 8 cm 5 cm 3 cm Air Liquid, SG = 1.45 B Water SAE 30 oil (a) (b) d L h D D Water SAE 30 oil H 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?
with manometer fluid m. One side of the manometer is open to the air, while the other is connected to new tubing which extends to pressure measurement location 1, some height H higher in elevation than the surface of the manometer liquid.
For consistency, let a be the density of the air in the room, t be the density of the gas inside the tube, m be the den-sity of the manometer liquid, and h be the height difference between the two sides of the manometer. See Fig. P2.38.
(a) Find an expression for the gage pressure at the mea-surement point. Note: When calculating gage pressure, use the local atmospheric pressure at the elevation of the mea-surement point. You may assume that h H; i.e., assume the gas in the entire left side of the manometer is of den-sity t. (b) Write an expression for the error caused by as-suming that the gas inside the tubing has the same density as that of the surrounding air. (c) How much error (in Pa) is caused by ignoring this density difference for the fol-lowing conditions: m 860 kg/m3, a 1.20 kg/m3, t 1.50 kg/m3, H 1.32 m, and h 0.58 cm? (d) Can you think of a simple way to avoid this error?
is very large. If the inclined arm is fitted with graduations 1 in apart, what should the angle be if each graduation corresponds to 1 lbf/ft2 gage pressure for pA?
Problems 107 P2.38 An interesting article appeared in the AIAA Journal (vol. 30, no. 1, January 1992, pp. 279–280). The authors explain that the air inside fresh plastic tubing can be up to 25 percent more dense than that of the surroundings, due to outgassing or other contaminants introduced at the time of manufacture.
Most researchers, however, assume that the tubing is filled with room air at standard air density, which can lead to sig-nificant errors when using this kind of tubing to measure pressures. To illustrate this, consider a U-tube manometer h (1) (2) 30 2 m P2.35 P2.39 An 8-cm-diameter piston compresses manometer oil into an inclined 7-mm-diameter tube, as shown in Fig. P2.39.
When a weight W is added to the top of the piston, the oil rises an additional distance of 10 cm up the tube, as shown.
How large is the weight, in N?
P2.40 A pump slowly introduces mercury into the bottom of the closed tank in Fig. P2.40. At the instant shown, the air pressure pB 80 kPa. The pump stops when the air pres-sure rises to 110 kPa. All fluids remain at 20°C. What will be the manometer reading h at that time, in cm, if it is con-nected to standard sea-level ambient air patm?
50 cm 50 cm Oil SG = 0.8 Water SG = 1.0 L P2.36 1 in Reservoir θ D = 5 16 in pA P2.37 h H 1 U-tube manometer m t (tubing gas) a (air) pa at location 1 p1 P2.38 P2.44 Water flows downward in a pipe at 45°, as shown in Fig.
P2.44. The pressure drop p1 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 p1 p2 in lbf/in2? What is the pressure drop due to fric-tion only between 1 and 2 in lbf/in2? Does the manome-ter reading correspond only to friction drop? Why?
108 Chapter 2 Pressure Distribution in a Fluid P2.41 The system in Fig. P2.41 is at 20°C. Compute the pres-sure at point A in lbf/ft2 absolute.
D = 8 cm d = 7 mm Meriam red oil, SG = 0.827 10 cm 15˚ Piston W 8 cm 9 cm Air: pB Water Mercury Pump patm h 2 cm Hg 10 cm P2.39 P2.40 Water Water 5 in 10 in 6 in Mercury A Oil, SG = 0.85 pa = 14.7 lbf/in2 P2.41 h1 pA 1 pB 1 h 2 h1 ρ ρ ρ P2.42 5 ft Flow 1 2 45˚ 6 in Mercury Water P2.44 P2.42 Very small pressure differences pA pB can be measured accurately by the two-fluid differential manometer in Fig.
P2.42. Density 2 is only slightly larger than that of the upper fluid 1. Derive an expression for the proportional-ity between h and pA pB if the reservoirs are very large.
P2.43 A mercury manometer, similar to Fig. P2.35, records h 1.2, 4.9, and 11.0 mm when the water velocities in the pipe are V 1.0, 2.0, and 3.0 m/s, respectively. Determine if these data can be correlated in the form p1 p2 Cf V2, where Cf is dimensionless.
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 cylindrical tank in Fig. P2.47 is being filled with wa-ter at 20°C by a pump developing an exit pressure of 175 kPa. At the instant shown, the air pressure is 110 kPa and H 35 cm. The pump stops when it can no longer raise the water pressure. For isothermal air compression, esti-mate H at that time.
P2.48 Conduct the following experiment to illustrate air pres-sure. Find a thin wooden ruler (approximately 1 ft in EES Problems 109 45 cm 30 cm 15 cm 40 cm patm Air Oil, SG = 0.85 Water Mercury A a karate chop on the portion of the ruler sticking out over the edge of the desk. Record your results. (c) Explain your results.
P2.49 A water tank has a circular panel in its vertical wall. The panel has a radius of 50 cm, and its center is 2 m below the surface. Neglecting atmospheric pressure, determine the water force on the panel and its line of action.
P2.50 A vat filled with oil (SG 0.85) is 7 m long and 3 m deep and has a trapezoidal cross section 2 m wide at the bot-tom and 4 m wide at the top. Compute (a) the weight of oil in the vat, (b) the force on the vat bottom, and (c) the force on the trapezoidal end panel.
P2.51 Gate AB in Fig. P2.51 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 Suppose that the tank in Fig. P2.51 is filled with liquid X, not oil. Gate AB is 0.8 m wide into the paper. Suppose that liquid X causes a force F on gate AB and that the moment of this force about point B is 26,500 N m. What is the specific gravity of liquid X?
P2.45 7 cm 4 cm 6 cm 9 cm 5 cm 12 cm SAE 30 oil Water Fluid X 10 cm 75 cm H 50 cm Air 20˚ C Water Pump Newspaper Ruler Desk P2.46 P2.48 P2.47 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 hang-ing 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. P2.48. (a) Estimate the total force on top of the newspaper due to air pressure in the room. (b) Careful! To avoid potential injury, make sure nobody is standing directly in front of the desk. Perform P2.53 Panel ABC in the slanted side of a water tank is an isosce-les triangle with the vertex at A and the base BC 2 m, as in Fig. P2.53. Find the water force on the panel and its line of action.
110 Chapter 2 Pressure Distribution in a Fluid 8 m 6 m 1.2 m F 40° X 1 m A B 4 m Oil, SG = 0.82 P2.51 P2.54 If, instead of water, the tank in Fig. P2.53 is filled with liq-uid X, the liquid force on panel ABC is found to be 115 kN.
What is the density of liquid X? The line of action is found to be the same as in Prob. 2.53. Why?
P2.55 Gate AB in Fig. 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.
P2.56 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 lbf. For what water depth h is this condition reached?
P2.57 In Fig. P2.55, gate AB is 5 ft wide into the paper. Suppose that the fluid is liquid X, not water. Hinge A breaks when its reaction is 7800 lbf, and the liquid depth is h 13 ft.
What is the specific gravity of liquid X?
P2.58 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.
Water B, C 3 m 4 m A pa Water pa 4 ft B A h P2.53 P2.55 Water 30 cm 3 m m 200 kg h B A P2.58 P2.59 B h Hinge A P L P2.59 Gate AB has length L, 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 . Find an analytic ex-pression for the force P, perpendicular to AB, required to keep the gate in equilibrium in Fig. P2.59.
P2.60 Find the net hydrostatic force per unit width on the rec-tangular gate AB in Fig. P2.60 and its line of action.
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 wa-ter depth h will the force at point B be zero?
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 con-dition, what will be the reading h on the mercury manome-ter on the left side?
P2.62 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.
Problems 111 Water Glycerin A B 1.8 m 1.2 m 2 m 2 m P2.60 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.
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?
P2.66 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 seep-age of water under the dam, could this force tip the dam over? How does your argument change if there is seepage under the dam?
Water B A Glycerin 1 m h 2 m 60° P2.61 Water h B 15 ft 60˚ 10,000 lb Pulley A P2.62 h 50° 2 cm H Water Plug, D = 4 cm Mercury P2.63 EES Water at 20°C A 20 cm B C 1m h P2.64 P2.67 Generalize Prob. 2.66 as follows. Denote length AB as H, length BC as L, and angle ABC as . Let the dam mater-ial have specific gravity SG. The width of the dam is b.
Assume no seepage of water under the dam. Find an an-alytic relation between SG and the critical angle c for which the dam will just tip over to the right. Use your re-lation to compute c for the special case SG 2.4 (con-crete).
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?
P2.69 The water tank in Fig. P2.69 is pressurized, as shown by the mercury-manometer reading. Determine the hydrosta-tic force per unit depth on gate AB.
P2.70 Calculate the force and center of pressure on one side of the vertical triangular panel ABC in Fig. P2.70. Neglect patm.
P2.71 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 112 Chapter 2 Pressure Distribution in a Fluid P A B 5 m Water 3 m Gate: Side view 60 m C A B Water 20˚C 80 m Dam P2.65 P2.66 A P 3 m Gate 50˚ B 1 m Oil, SG = 0.83 2 m P2.68 1 m 5 m 2 m Water, 20˚C Hg, 20˚C A B 80 cm P2.69 C Water 4 ft B A 2 ft 6 ft P2.70 P2.74 In “soft’’ liquids (low bulk modulus ), it may be neces-sary to account for liquid compressibility in hydrostatic calculations. An approximate density relation would be dp d a2 d or p p0 a2( 0) where a is the speed of sound and (p0, 0) are the condi-tions at the liquid surface z 0. Use this approximation to show that the density variation with depth in a soft liq-uid is 0egz/a2 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 incompress-ible result F 0gh2b/2. Would the center of pressure be below the incompressible position z 2h/3?
P2.75 Gate AB in Fig. P2.75 is hinged at A, has width b into the paper, and makes smooth contact at B. The gate has den-sity s and uniform thickness t. For what gate density s, expressed as a function of (h, t, , ), will the gate just be-gin to lift off the bottom? Why is your answer indepen-dent of gate length L and width b?
2.40). What diameter of the sphere is just sufficient to keep the gate closed?
Problems 113 P2.72 The V-shaped container in Fig. P2.72 is hinged at A and held together by cable BC at the top. If cable spacing is 1 m into the paper, what is the cable tension?
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.
A 4 m B Concrete sphere, SG = 2.4 8 m 6 m Water Cable A 3 m 1 m B C 110˚ Water P2.71 P2.72 P2.76 Consider the angled gate ABC in Fig. P2.76, hinged at C and of width b into the paper. Derive an analytic formula for the horizontal force P required at the top for equilib-rium, as a function of the angle .
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 h 8 m. Neglect atmospheric pressure.
P2.78 Repeat Prob. 2.77 to derive an analytic expression for P as a function of h. Is there anything unusual about your solution?
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 A B h Stop 10 ft Tide range Seawater, SG = 1.025 P2.73 A L t h B P2.75 114 Chapter 2 Pressure Distribution in a Fluid gate will open. Neglect atmospheric pressure. Is this result independent of the liquid density?
P2.80 For the closed tank in Fig. P2.80, all fluids are at 20°C, and the airspace is pressurized. It is found that the net outward hydrostatic force on the 30-by 40-cm panel at the bottom of the water layer is 8450 N. Estimate (a) the pressure in the airspace and (b) the reading h on the mercury manometer.
Water A B C 1m h 1m P pa pa P2.77 P2.81 Gate AB in Fig. P2.81 is 7 ft into the paper and weighs 3000 lbf when submerged. It is hinged at B and rests against a smooth wall at A. Determine the water level h at the left which will just cause the gate to open.
Water A B C h 60 cm 40 cm P2.79 P2.82 The dam in Fig. P2.82 is a quarter circle 50 m wide into the paper. Determine the horizontal and vertical compo-nents of the hydrostatic force against the dam and the point CP where the resultant strikes the dam.
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 lbf.
P2.84 Determine (a) the total hydrostatic force on the curved sur-face AB in Fig. P2.84 and (b) its line of action. Neglect at-mospheric pressure, and let the surface have unit width.
Water Water 4 ft 6 ft 8 ft A h B Air SAE 30 oil Water 60 cm 20 cm 80 cm Panel, 30 cm high, 40 cm wide Mercury 2 m 1 atm h P2.80 P2.81 h Specific weight γ θ θ P B A C P2.76 Problems 115 P2.85 Compute the horizontal and vertical components of the hy-drostatic force on the quarter-circle panel at the bottom of the water tank in Fig. P2.85.
Water 20 m 20 m CP pa = 0 P2.82 Water A B F r = 8 ft A B Water at 20° C 1 m x z z = x3 P2.87 The bottle of champagne (SG 0.96) in Fig. P2.87 is un-der pressure, as shown by the mercury-manometer read-ing. Compute the net force on the 2-in-radius hemispher-ical 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?
P2.83 P2.84 Water 6 m 2 m 5 m 2 m P2.85 Water 10 ft 2 ft P2.86 P2.86 Compute the horizontal and vertical components of the hy-drostatic force on the hemispherical bulge at the bottom of the tank in Fig. P2.86.
4 in 2 in 6 in Mercury r = 2 in P2.87 Water B C O R = 6 m 6 m 6 m A P2.88 P2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and is filled 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?
P2.92 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, deter-mine the force induced in each bolt.
P2.93 In Fig. P2.93, a one-quadrant spherical shell of radius R is submerged in liquid of specific gravity and depth h R. Find an analytic expression for the resultant hydro-static force, and its line of action, on the shell surface.
P2.89 The tank in Fig. P2.89 contains benzene and is pressur-ized to 200 kPa (gage) in the air gap. Determine the ver-tical hydrostatic force on circular-arc section AB and its line of action.
116 Chapter 2 Pressure Distribution in a Fluid P2.90 A 1-ft-diameter hole in the bottom of the tank in Fig. P2.90 is closed by a conical 45° plug. Neglecting the weight of the plug, compute the force F required to keep the plug in the hole.
60 cm 60 cm p = 200 kPa Benzene at 20C A B 30 cm P2.89 Air : Water 45˚ cone F 1 ft p = 3 lbf/in2 gage 3 ft 1 ft P2.90 P2.94 The 4-ft-diameter log (SG 0.80) in Fig. P2.94 is 8 ft long into the paper and dams water as shown. Compute the net vertical and horizontal reactions at point C.
Water 4 m 3cm Six bolts 2 m P2.91 2 m 2 m Water Bolt spacing 25 cm P2.92 z , γ R R x z R h ρ P2.93 wall at A. Compute the reaction forces at points A and B.
P2.95 The uniform body A in Fig. P2.95 has width b into the pa-per 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 R?
Problems 117 P2.96 The tank in Fig. P2.96 is 3 m wide into the paper. Ne-glecting atmospheric pressure, compute the hydrostatic (a) horizontal force, (b) vertical force, and (c) resultant force on quarter-circle panel BC.
2ft Log 2ft Water C Water P2.94 P2.98 Gate ABC in Fig. P2.98 is a quarter circle 8 ft wide into the paper. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force.
A R R h O Water P2.95 P2.99 A 2-ft-diameter sphere weighing 400 lbf closes a 1-ft-di-ameter hole in the bottom of the tank in Fig. P2.99. Com-pute the force F required to dislodge the sphere from the hole.
4 m 4 m 6 m Water C B A P2.96 P2.97 Gate AB in Fig. P2.97 is a three-eighths circle, 3 m wide into the paper, hinged at B, and resting against a smooth 4 m Seawater, 10,050 N/m3 2 m 45° A B P2.97 45° 45° r = 4 ft B C Water A P2.98 Water 3 ft 1 ft F 1 ft P2.99 whether his new crown was pure gold (SG 19.3).
Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?
P2.106 It is found that a 10-cm cube of aluminum (SG 2.71) will remain neutral under water (neither rise nor fall) if it is tied by a string to a submerged 18-cm-diameter sphere of buoyant foam. What is the specific weight of the foam, in N/m3?
P2.107 Repeat Prob. 2.62, assuming that the 10,000-lbf weight is aluminum (SG 2.71) and is hanging submerged in the water.
P2.108 A piece of yellow pine wood (SG 0.65) is 5 cm square and 2.2 m long. How many newtons of lead (SG 11.4) should be attached to one end of the wood so that it will float vertically with 30 cm out of the water?
P2.109 A hydrometer floats at a level which is a measure of the specific gravity of the liquid. The stem is of constant di-ameter D, and a weight in the bottom stabilizes the body to float vertically, as shown in Fig. P2.109. If the position h 0 is pure water (SG 1.0), derive a formula for h as a function of total weight W, D, SG, and the specific weight 0 of water.
P2.100 Pressurized water fills the tank in Fig. P2.100. Compute the net hydrostatic force on the conical surface ABC.
118 Chapter 2 Pressure Distribution in a Fluid P2.101 A fuel truck has a tank cross section which is approxi-mately elliptical, with a 3-m horizontal major axis and a 2-m vertical minor axis. The top is vented to the atmos-phere. If the tank is filled half with water and half with gasoline, what is the hydrostatic force on the flat ellipti-cal end panel?
P2.102 In Fig. P2.80 suppose that the manometer reading is h 25 cm. What will be the net hydrostatic force on the com-plete end wall, which is 160 cm high and 2 m wide?
P2.103 The hydrogen bubbles in Fig. 1.13 are very small, less than a millimeter in diameter, and rise slowly. Their drag in still fluid is approximated by the first term of Stokes’ expression in Prob. 1.10: F 3
VD, where V is the rise velocity. Neglecting bubble weight and setting bubble buoyancy equal to drag, (a) derive a formula for the ter-minal (zero acceleration) rise velocity Vterm of the bubble and (b) determine Vterm in m/s for water at 20°C if D 30 m.
P2.104 The can in Fig. P2.104 floats in the position shown. What is its weight in N?
P2.105 It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine 2 m A C B 150 kPa gage 4 m 7 m Water P2.100 Water 3 cm 8 cm D = 9 cm P2.104 P2.110 An average table tennis ball has a diameter of 3.81 cm and a mass of 2.6 g. Estimate the (small) depth at which this ball will float in water at 20°C and sea level standard air if air buoyancy is (a) neglected and (b) included.
P2.111 A hot-air balloon must be designed to support basket, cords, and one person for a total weight of 1300 N. The balloon material has a mass of 60 g/m2. Ambient air is at 25°C and 1 atm. The hot air inside the balloon is at 70°C and 1 atm.
What diameter spherical balloon will just support the total weight? Neglect the size of the hot-air inlet vent.
P2.112 The uniform 5-m-long round wooden rod in Fig. P2.112 is tied to the bottom by a string. Determine (a) the tension h Fluid, SG > 1 W D SG = 1.0 P2.109 P2.116 The homogeneous 12-cm cube in Fig. 2.116 is balanced by a 2-kg mass on the beam scale when the cube is im-mersed in 20°C ethanol. What is the specific gravity of the cube?
in the string and (b) the specific gravity of the wood. Is it possible for the given information to determine the incli-nation angle ? Explain.
Problems 119 Water at 20°C String 1 m 4 m θ D = 8 cm 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 measurements 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 h 18 in?
Wsteel h P2.113 8 m Hinge = 30 2 kg of lead D = 4 cm B P2.114 P2.117 The balloon in Fig. P2.117 is filled with helium and pres-surized to 135 kPa and 20°C. The balloon material has a P2.114 The uniform rod in Fig. P2.114 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 30?
P2.115 The 2-in by 2-in by 12-ft spar buoy from Fig. P2.113 has 5 lbm of steel attached and has gone aground on a rock, as in Fig. P2.115. Compute the angle at which the buoy will lean, assuming that the rock exerts no moments on the spar.
Rock Seawater Wood A B 8 ft θ SG = 0.6 P2.115 12 cm 2 kg P2.116 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 an-gle will the beam float in the 20°C water?
mass of 85 g/m2. Estimate (a) the tension in the mooring line and (b) the height in the standard atmosphere to which the balloon will rise if the mooring line is cut.
120 Chapter 2 Pressure Distribution in a Fluid P2.118 A 14-in-diameter hollow sphere is made of steel (SG 7.85) with 0.16-in wall thickness. How high will this sphere float in 20°C water? How much weight must be added inside to make the sphere neutrally buoyant?
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 with its upper right corner at the surface, as shown.
Determine (a) the angle and (b) the specific gravity of the wood. (Hint: Both the vertical forces and the moments about the beam centroid must be balanced.) Air: 100 kPa at 20°C D = 10 m P2.117 Water 9 ft 5 lbf 4 in × 4 in θ P2.119 P2.121 The uniform beam in Fig. P2.121, of size L by h by b and with specific weight b, floats exactly on its diagonal when a heavy uniform sphere is tied to the left corner, as shown.
Show that this can only happen (a) when b /3 and (b) when the sphere has size D 1/3 Lhb (SG 1) θ Water 1 m A P2.120 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?
L Diameter D γ Width b << L γ b SG > 1 h << L P2.121 Steel block Mercury: SG = 13.56 a b Water P2.122 EES P2.123 In an estuary where fresh water meets and mixes with sea-water, there often occurs a stratified salinity condition with fresh water on top and salt water on the bottom, as in Fig.
P2.123. The interface is called a halocline. An idealization of this would be constant density on each side of the halo-cline as shown. A 35-cm-diameter sphere weighing 50 lbf would “float’’ near such a halocline. Compute the sphere position for the idealization in Fig. P2.123.
P2.124 A balloon weighing 3.5 lbf is 6 ft in diameter. It is filled with hydrogen at 18 lbf/in2 absolute and 60°F and is re-leased. At what altitude in the U.S. standard atmosphere will this balloon be neutrally buoyant?
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 H/L for which it becomes neutrally stable, i.e., 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 2 ft above the waterline. Is it stable?
P2.132 A solid right circular cone has SG 0.99 and floats ver-tically as in Fig. P2.132. Is this a stable position for the cone?
P2.125 Suppose that the balloon in Prob. 2.111 is constructed to have a diameter of 14 m, is filled at sea level with hot air at 70°C and 1 atm, and is released. If the air inside the bal-loon remains constant and the heater maintains it at 70°C, at what altitude in the U.S. standard atmosphere will this balloon be neutrally buoyant?
P2.126 A cylindrical can of weight W, radius R, and height H is open at one end. With its open end down, and while filled with atmospheric air (patm, Tatm), the can is eased down vertically into liquid, of density , which enters and com-presses the air isothermally. Derive a formula for the height h to which the liquid rises when the can is submerged with its top (closed) end a distance d from the surface.
P2.127 Consider the 2-in by 2-in by 10-ft spar buoy of Prob. 2.113.
How many pounds of steel (SG 7.85) should be added at the bottom to ensure vertical floating with a metacen-tric height M G of (a) zero (neutral stability) or (b) 1 ft (reasonably stable)?
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 S 0.88. Deter-mine if this “cubic’’iceberg is stable for the position shown in Fig. P2.128.
Problems 121 Salinity Idealization Halocline SG = 1.025 35°/°° SG = 1.0 0 P2.123 P2.129 The iceberg idealization in Prob. 2.128 may become un-stable if its sides melt and its height exceeds its width. In Water S = 1.0 M?
G B Specific gravity L = S h 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 M G of this cone, in dimensionless form, versus H/R for a range of S 1.
P2.134 When floating in water (SG 1.0), an equilateral trian-gular 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.
P2.128 Water : SG = 1.0 SG = 0.99 P2.132 (a) (b) P2.134 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 [2SG(1 SG)]1/2 R L P2.136 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG 0.5, floating in wa-ter (SG 1). Show that the body will be stable with its axis horizontal if L/R 2.0.
P2.137 A tank of water 4 m deep receives a constant upward ac-celeration az. Determine (a) the gage pressure at the tank bottom if az 5 m2/s and (b) the value of az which 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 be-fore 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 ax in m/s2. (b) Why doesn’t the solution to part (a) depend upon the density of the fluid? (c) Determine the gage pressure at point A if the fluid is glycerin at 20°C.
122 Chapter 2 Pressure Distribution in a Fluid Fig. P2.139 P2.141 P2.140 Suppose that the elliptical-end fuel tank in Prob. 2.101 is 10 m long and filled completely with fuel oil ( 890 kg/m3). Let the tank be pulled along a horizontal road. For rigid-body motion, find the acceleration, and its direction, for which (a) a constant-pressure surface extends from the top of the front end wall to the bottom of the back end and (b) the top of the back end is at a pressure 0.5 atm lower than the top of the front end.
P2.141 The same tank from Prob. 2.139 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.
P2.142 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/s2, compute (a) the water depth on side AB and (b) the water-pressure force on panel AB. Assume no spilling.
P2.143 The tank of water in Fig. P2.143 is full and open to the at-mosphere at point A. For what acceleration ax in ft/s2 will the pressure at point B be (a) atmospheric and (b) zero absolute?
28 cm 100 cm 15 cm V A a?
30° z x A 28 cm 100 cm ax 15 cm P2.144 Consider a hollow cube of side length 22 cm, filled com-pletely with water at 20°C. The top surface of the cube is horizontal. One top corner, point A, is open through a small hole to a pressure of 1 atm. Diagonally opposite to point A is top corner 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 mo-tion.
P2.145 A fish tank 14 in deep by 16 by 27 in is to be carried in a car which may experience accelerations as high as 6 m/s2. What is the maximum water depth which will avoid Water at 20°C 24 cm 9 cm A B P2.142 Water 2ft 2ft 1ft 1ft A B ax pa = 15 lbf/in2 abs P2.143 with the child, which way will the balloon tilt, forward or backward? Explain. (b) The child is now sitting in a car which is stopped at a red light. The helium-filled 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 traffic 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-filled balloon. Conduct a scientific 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 Fig. 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 as-sumed, what is the water surface angle at position A?
spilling in rigid-body motion? What is the proper align-ment 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. In-side the tank, a 10-cm balloon, filled with helium at 130 kPa, is tethered centrally by a string. If the tank acceler-ates to the right at 5 m/s2 in rigid-body motion, at what angle will the balloon lean? Will it lean to the right or to the left?
Problems 123 P2.147 The tank of water in Fig. P2.147 accelerates uniformly by freely rolling down a 30° incline. If the wheels are fric-tionless, what is the angle ? Can you explain this inter-esting result?
Water at 20°C 60 cm 40 cm 20 cm D = 10 cm String 1 atm A He P2.146 θ 30° P2.147 P2.148 A child is holding a string onto which is attached a he-lium-filled balloon. (a) The child is standing still and sud-denly accelerates forward. In a frame of reference moving 10 r/min A θ 1 ft 6 ft P2.149 D ax L 1 2 L L 1 2 h Rest level P2.150 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 be if ax 6 m/s2? Can the scale markings on the tube be linear multiples of ax?
P2.151 The U-tube in Fig. P2.151 is open at A and closed at D.
If accelerated to the right at uniform ax, what acceleration will cause the pressure at point C to be atmospheric? The fluid is water (SG 1.0).
124 Chapter 2 Pressure Distribution in a Fluid 1 ft 1 ft 1 ft D C B A P2.151 A B C 20 cm 10 cm 5 cm 12 cm Ω P2.155 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 lbf/ft2 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?
30 cm 45˚ A C B P2.157 P2.158 It is desired to make a 3-m-diameter parabolic telescope mirror by rotating molten glass in rigid-body motion un-til 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?
P2.152 A 16-cm-diameter open cylinder 27 cm high is full of wa-ter. 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 ex-posed.
P2.153 Suppose the U-tube in Fig. P2.150 is not translated but rather rotated about its right leg at 95 r/min. What will be the level h in the left leg if L 18 cm and D 5 mm?
P2.154 A very deep 18-cm-diameter can contains 12 cm of water overlaid with 10 cm of SAE 30 oil. If the can is rotated in rigid-body motion about its central axis at 150 r/min, what will be the shapes of the air-oil and oil-water interfaces? What will be the maximum fluid pres-sure in the can in Pa (gage)?
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.
EES 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 pres-sure 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 FE2.2 On a sea-level standard day, a pressure gage, moored be-low the surface of the ocean (SG 1.025), reads an ab-solute 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 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 B?
(a) 5.6 kPa, (b) 10.9 kPa, (c) 106.9 kPa, (d) 112.2 kPa, (e) 157.0 kPa 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, al-though the cone contains 67 percent less water. Can you explain the paradox?
W2.2 Can the temperature ever rise with altitude in the real at-mosphere? Wouldn’t this cause the air pressure to increase upward? Explain the physics of this situation.
W2.3 Consider a submerged curved surface which consists of a two-dimensional circular arc of arbitrary angle, arbitrary depth, and arbitrary orientation. Show that the resultant hy-drostatic 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?
Fundamentals of Engineering Exam Problems 125 5 cm 3 cm 4 cm 8 cm A B Oil Mercury SG = 13.56 Water SG = 1 FE2.3 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, 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 at-mosphere, carrying a person/basket of mass M m. Dis-cuss 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?
FE2.4 In Fig. FE2.3, if the oil in region B has SG 0.8 and the absolute pressure at point B is 14 psia, what is the ab-solute pressure at point B?
(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?
(a) 147 kN, (b) 367 kN, (c) 490 kN, (d) 661 kN, (e) 1028 kN FE2.6 In Prob. FE2.5 above, 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 per-cent 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 con-stant of helium is 2077 m2/(s2K) and balloon material weight is neglected, what is the net lifting force of the balloon?
(a) 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 of buoyancy is above its metacenter, (d) metacenter is above its center of buoyancy, (e) metacenter is above its center of gravity 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 126 Chapter 2 Pressure Distribution in a Fluid D h d Zero pressure level To pressure measurement location p1 a (air) pa m C2.1 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 m while the air has den-sity a. Ignore the effects of surface tension. When there is no pressure difference across the manometer, the elevations on both sides are the same, as indicated by the dashed line.
Height 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 p1gage, taking into account the move-ment of the surface of the reservoir. Your equation should give p1gage as a function of h, m, and the physical para-meters in the problem, h, d, D, and gravity constant g. (b) Write an approximate expression for p1gage, neglecting the change in elevation of the surface of the reservoir liq-uid. (c) Suppose h 0.26 m in a certain application. If pa 101,000 Pa and the manometer liquid has a density of 820 kg/m3, estimate the ratio D/d required to keep the error of the approximation of part (b) within 1 percent of the ex-act measurement of part (a). Repeat for an error within 0.1 percent.
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 pa-rameters in the problem. (b) Find the special case of your result in (a) when ptank pa. (c) Suppose H 5.0 cm, pa is 101.2kPa, ptank is 1.82 kPa higher than pa, and SG0 0.85. Calculate h in cm, ignoring surface tension effects and neglecting air density effects.
Oil Water H h Pressurized air tank, with pressure ptank pa C2.2 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 angular velocity and the manometer legs are 7 cm apart. The manometer center is 5.8 m from the axis of rotation. De-termine the height difference h in two ways: (a) approxi-mately, 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 as-suming that the glass is vertical at all times.
C2.2 A prankster has added oil, of specific gravity SG0, to the left leg of the manometer in Fig. C2.2. Nevertheless, the 8.
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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 b h Y L W Circular arc block Fluid: Pivot arm Pivot Counterweight Side view of block face R D2.2 Design Projects D2.1 It is desired to have a bottom-moored, floating system which 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 sys-tem which will provide this force distribution. The system should be practical, i.e., of inexpensive materials and sim-ple construction.
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 com-pare it with the theoretical value for given depth h. The counterweight is arranged so that the pivot arm is hori-zontal 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 formula for W as a function of h in terms of the system parameters. Finally, suggest some appropriate values of Y, L, etc., for a suitable appartus and plot theoretical W ver-sus h for these values.
R 5.80 m (to center of manometer) h Center of rotation 6.00 rpm Water 7.00 cm C2.3 127 128 Table tennis ball suspended by an air jet. The control volume momentum principle, studied in this chapter, requires a force to change the direction of a flow. The jet flow deflects around the ball, and the force is the ball’s weight. (Courtesy of Paul Silverman/Fundamental Photographs) 3.1 Basic Physical Laws of Fluid Mechanics 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 field 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 then ends with 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 my opinion. Statics problems ba-sically 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 me-chanics. This chapter and the next two outline the three basic approaches to the analy-sis of arbitrary flow problems: 1.
Control-volume, or large-scale, analysis (Chap. 3) 2.
Differential, or small-scale, analysis (Chap. 4) 3.
Experimental, or dimensional, analysis (Chap. 5) The three approaches are roughly equal in importance, but control-volume analysis is “more equal,” being the single most valuable tool to the engineer for flow analysis. It gives “engineering” answers, sometimes gross and crude but always useful. In princi-129 Chapter 3 Integral Relations for a Control Volume Systems versus Control Volumes ple, the differential approach of Chap. 4 can be used for any problem, but in practice the lack of mathematical tools and the inability of the digital computer to model small-scale processes make the differential approach rather limited. Similarly, although the dimensional analysis of Chap. 5 can be applied to any problem, the lack of time and money and generality often makes experimentation a limited approach. But a control-volume analysis takes about half an hour and gives useful results. Thus, in a trio of ap-proaches, the control volume is best. Oddly enough, it is the newest of the three. Dif-ferential analysis began with Euler and Lagrange in the eighteenth century, and dimensional analysis was pioneered by Lord Rayleigh in the late nineteenth century, but the control volume, although proposed by Euler, was not developed on a rigorous basis as an analytical tool until the 1940s.
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 be-tween 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.1 This is a law of mechanics and has a very simple mathematical form, called conservation of mass: msyst const or d d m t 0 (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 will begin to accelerate2 F ma m d d V t d d t (mV) (3.2) In Eq. (2.12) we saw this relation applied to a differential element of viscous incom-pressible fluid. In fluid mechanics Newton’s law is called the linear-momentum rela-tion. Note that it is a vector law which implies the three scalar equations Fx max, Fy may, and Fz maz.
Third, if the surroundings exert a net moment M about the center of mass of the system, there will be a rotation effect M d d H t (3.3) where H (r V) m is the angular momentum of the system about its center of 130 Chapter 3 Integral Relations for a Control Volume 1We are neglecting nuclear reactions, where mass can be changed to energy.
2We are neglecting relativistic effects, where Newton’s law must be modified.
mass. Here we call Eq. (3.3) the angular-momentum relation. Note that it is also a vec-tor equation implying three scalar equations such as Mx dHx/dt.
For an arbitrary mass and arbitrary moment, H is quite complicated and contains nine terms (see, e.g., Ref. 1, p. 285). In elementary dynamics we commonly treat only a rigid body rotating about a fixed x axis, for which Eq. (3.3) reduces to Mx Ix d d t (x) (3.4) where x is the angular velocity of the body and Ix is its mass moment of inertia about the x axis. Unfortunately, fluid systems are not rigid and rarely reduce to such a sim-ple relation, as we shall see in Sec. 3.5.
Fourth, if heat dQ 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 ther-modynamics, dQ dW dE or d d Q t d d W t d d E t (3.5) Like mass conservation, Eq. (3.1), this is a scalar relation having only a single com-ponent.
Finally, the second law of thermodynamics relates entropy change dS to heat added dQ and absolute temperature T: dS d T Q (3.6) This is valid for a system and can be written in control-volume form, but there are al-most 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( , T) and e e( , T) for the particular fluid being stud-ied, as in Sec. 1.6.
The purpose of this chapter is to put our four basic laws into the control-volume form suitable for arbitrary regions in a flow: 1.
Conservation of mass (Sec. 3.3) 2.
The linear-momentum relation (Sec. 3.4) 3.
The angular-momentum relation (Sec. 3.5) 4.
The energy equation (Sec. 3.6) 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 3.1 Basic Physical Laws of Fluid Mechanics 131 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 .
fluid forms the environment whose effect on our product we wish to know. For the three examples cited above, 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 prod-uct. 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.
Although thermodynamics is not at all the main topic of this book, it would be a shame if the student did not review at least the first law and the state relations, as dis-cussed, e.g., in Refs. 6 and 7.
In analyzing a control volume, we convert the system laws to apply to a specific re-gion 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 re-gion, and often simple assumptions will be accurate enough (e.g., uniform inlet and/or outlet flows). The flow conditions away from the control volume are then irrelevant.
The technique for making such localized analyses is the subject of this chapter.
All the analyses in this chapter involve evaluation of the volume flow Q or mass flow m ˙ 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 off the normal. Let 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 parallelopiped in Fig. 3.1b: d V dt dA cos (V n) dA dt The integral of d/dt is the total volume rate of flow Q through the surface S Q S (V n) dA S Vn dA (3.7) 132 Chapter 3 Integral Relations for a Control Volume θ S dA 1 V Unit normal n dA θ n V V dt (a) (b) Volume and Mass Rate of Flow 3.2 The Reynolds Transport Theorem We could replace V n by its equivalent, Vn, 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 m ˙ . If density varies over the surface, it must be part of the surface integral m ˙ S (V n) dA S Vn dA If density is constant, it comes out of the integral and a direct proportionality results: Constant density: m ˙ Q To convert a system analysis to a control-volume analysis, we must convert our math-ematics to apply to a specific region rather than to individual masses. This conversion, called the Reynolds transport theorem, can be applied to all the basic laws. Examin-ing 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 control vol-ume is fixed, moving, or deformable. Figure 3.2 illustrates these three cases. The fixed control volume in Fig. 3.2a encloses a stationary region of interest 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, circles around through the surrounding at-mosphere, and slices through the flange bolts and the fluid within the nozzle. This par-ticular control volume exposes the stresses in the flange bolts, which contribute to ap-plied 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.
3.2 The Reynolds Transport Theorem 133 (a) Control surface (c) Control surface V V (b) Control surface V Fig. 3.2 Fixed, moving, and de-formable control volumes: (a) fixed control volume for nozzle-stress analysis; (b) control volume mov-ing at ship speed for drag-force analysis; (c) control volume de-forming within cylinder for tran-sient pressure-variation analysis.
One-Dimensional Fixed Control Volume If V is constant, this relative motion is a steady-flow pattern, which simplifies the analy-sis.3 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 frame of reference.
Figure 3.2c shows a deforming control volume. Varying relative motion at the bound-aries 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 consider the other cases as advanced topics.
As a simple first example, consider a duct or streamtube with a nearly one-dimensional flow V V(x), as shown in Fig. 3.3. The selected control volume is a portion of the duct which happens to be filled exactly by system 2 at a particular instant t. At time t dt, system 2 has begun to move out, and a sliver of system 1 has entered from the left. The shaded areas show an outflow sliver of volume AbVb dt and an inflow volume AaVa dt.
Now let B be any property of the fluid (energy, momentum, etc.), and let dB/dm be the intensive value or the amount of B per unit mass in any small portion of the fluid. The total amount of B in the control volume is thus BCV CV d d d m B (3.8) 134 Chapter 3 Integral Relations for a Control Volume Fig. 3.3 Example of inflow and outflow as three systems pass through a control volume: (a) Sys-tem 2 fills the control volume at time t; (b) at time t dt system 2 begins to leave and system 1 enters.
3A 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.
System 1 System 2 System 3 x, V(x) a b 1 2 3 Control volume fixed in space d in = AaVa dt out = AbVb dt d (a) (b) Section a Section b 1 2 Arbitrary Fixed Control Volume where d is a differential mass of the fluid. We want to relate the rate of change of BCV to the rate of change of the amount of B in system 2 which happens to coincide with the control volume at time t. The time derivative of BCV is defined by the calcu-lus limit d d t (BCV) d 1 t BCV(t dt) d 1 t BCV(t) d 1 t [B2(t dt) ( d)out ( d)in] d 1 t [B2(t)] d 1 t [B2(t dt) B2(t)] ( AV)out ( AV)in The first term on the right is the rate of change of B within system 2 at the instant it occupies the control volume. By rearranging the last line of the above equation, we have the desired conversion formula relating changes in any property B of a local sys-tem to one-dimensional computations concerning a fixed control volume which in-stantaneously encloses the system.
d d t (Bsyst) d d t CV d ( AV)out ( AV)in (3.9) This is the one-dimensional Reynolds transport theorem for a fixed volume. The three terms on the right-hand side are, respectively, 1.
The rate of change of B within the control volume 2.
The flux of B passing out of the control surface 3.
The flux of B passing into the control surface If the flow pattern is steady, the first term vanishes. Equation (3.9) can readily be gen-eralized to an arbitrary flow pattern, as follows.
Figure 3.4 shows a generalized fixed control volume with an arbitrary flow pattern passing through. The only additional complication is that 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 with the local normal to dA. Some elemental areas will have inflow volume (VA cos )in dt, and others will have outflow volume (VA cos )out dt, as seen in Fig. 3.4. Some surfaces might correspond to streamlines ( 90°) or solid walls (V 0) with neither inflow nor outflow. Equation (3.9) generalizes to d d t (Bsyst) d d t CV d CS V cos dAout CS V cos dAin (3.10) 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 control-volume in-tegrals are concerned with the intensive property . Since the control volume is fixed in space, the elemental volumes d do not vary with time, so that the time derivative of the volume integral vanishes unless either or varies with time (unsteady flow).
3.2 The Reynolds Transport Theorem 135 Fig. 3.4 Generalization of Fig. 3.3 to an arbitrary control volume with an arbitrary flow pattern.
Equation (3.10) expresses the basic formula that a system derivative equals the rate of change of B within the control volume plus the flux of B out of the control surface minus the flux of B into the control surface. The quantity B (or ) may be any vector or scalar property of the fluid. Two alternate forms are possible for the flux terms. First we may notice that V cos is the component of V normal to the area element of the control surface. Thus we can write Flux terms CS Vn dAout CS Vn dAin CS dm ˙ out CS dm ˙ in (3.11a) where dm ˙ Vn dA is the differential mass flux through the surface. Form (3.11a) helps visualize what is being calculated.
A second alternate form offers elegance and compactness as advantages. If n is de-fined as the outward normal unit vector everywhere on the control surface, then V n Vn for outflow and V n Vn for inflow. Therefore the flux terms can be rep-resented by a single integral involving V n which accounts for both positive outflow and negative inflow Flux terms CS (V n) dA (3.11b) The compact form of the Reynolds transport theorem is thus d d t (Bsyst) d d t CV d CV (V n) dA (3.12) This is beautiful but only occasionally useful, when the coordinate system is ideally suited to the control volume selected. Otherwise the computations are easier when the flux of B out is added and the flux of B in is subtracted, according to (3.10) or (3.11a).
136 Chapter 3 Integral Relations for a Control Volume System at time t + dt System at time t d A θ n, Unit outward normal to dA Arbitrary fixed control surface CS Vin dA Vout n, Unit outward normal to dA Fixed control volume CV θ d in = Vin dAin cos in dt = –V• n dA dt θ d out = V dAout cos out dt = V • n dA dt out θ Control Volume of Constant Shape but Variable Velocity4 Arbitrarily Moving and Deformable Control Volume5 Control Volume Moving at Constant Velocity The time-derivative term can be written in the equivalent form d d t CV d CV t ( ) d (3.13) for the fixed control volume since the volume elements do not vary.
If the control volume is moving uniformly at velocity Vs, as in Fig. 3.2b, an observer fixed to the control volume will see a relative velocity Vr of fluid crossing the control surface, defined by Vr V Vs (3.14) where V is the fluid velocity relative to the same coordinate system in which the con-trol volume motion Vs is observed. Note that Eq. (3.14) is a vector subtraction. The flux terms will be proportional to Vr, but the volume integral is unchanged because the control volume moves as a fixed shape without deforming. The Reynolds transport the-orem for this case of a uniformly moving control volume is d d t (Bsyst) d d t CV d CS (Vr n) dA (3.15) which reduces to Eq. (3.12) if Vs 0.
If the control volume moves with a velocity Vs(t) which retains its shape, then the vol-ume elements do not change with time but the boundary relative velocity Vr V(r, t) Vs(t) becomes a somewhat more complicated function. Equation (3.15) is un-changed in form, but the area integral may be more laborious to evaluate.
The most general situation is when the control volume is both moving and deforming arbitrarily, as illustrated in Fig. 3.5. The flux of volume across the control surface is again proportional to the relative normal velocity component Vr n, as in Eq. (3.15).
However, since the control surface has a deformation, its velocity Vs Vs(r, t), so that the relative velocity Vr V(r, t) Vs(r, t) is or can be a complicated function, even though the flux 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 de-rivative must be applied after integration. For the deforming control volume, then, the transport theorem takes the form d d t (Bsyst) d d t CV d CS (Vr n) dA (3.16) This is the most general case, which we can compare with the equivalent form for a fixed control volume 3.2 The Reynolds Transport Theorem 137 4This section may be omitted without loss of continuity.
5This section may be omitted without loss of continuity.
Fig. 3.5 Relative-velocity effects between a system and a control volume when both move and de-form. The system boundaries move at velocity V, and the control sur-face moves at velocity Vs.
d d t (Bsyst) CV t ( ) d CS (V n) dA (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 Vr between the fluid system and the control surface. These differences and mathematical subtleties are best shown by examples.
In many applications, the flow crosses the boundaries of the control surface only at cer-tain simplified inlets and exits which are approximately one-dimensional; i.e., the flow properties are nearly uniform over the cross section of the inlet or exit. Then the double-integral flux terms required in Eq. (3.16) reduce to a simple sum of positive (exit) and negative (inlet) product terms involving the flow properties at each cross section CS (Vr n) dA (
i iVriAi)out (
i iVriAi)in (3.18) An example of this situation is shown in Fig. 3.6. There are inlet flows at sections 1 and 4 and outflows at sections 2, 3, and 5. For this particular problem Eq. (3.18) would be CS (Vr n) dA 2 2Vr2A2 3 3Vr3A3 5 5Vr5A5 1 1Vr1A1 4 4Vr4A4 (3.19) 138 Chapter 3 Integral Relations for a Control Volume Vr Vs V n d in = –(Vr • n) dA dt d out = (Vr • n) dA dt n V Vs Vr = V – Vs System at time t + dt CV at time t + dt System and CV at time t One-Dimensional Flux-Term Approximations Fig. 3.6 A control volume with simplified one-dimensional inlets and exits.
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 tabulated be-low. Find the rate of change of energy of the system which occupies the control volume at this instant.
Section Type , kg/m3 V, m/s A, m2 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 The property under study here is energy, and so B E and dE/dm e, the energy per unit mass. Since the control volume is fixed, Eq. (3.17) applies: d d E t syst CV t (e ) d CS e (V n) dA The flow within is steady, so that (e )/t 0 and the volume integral vanishes. The area inte-gral consists of two inlet sections and one outlet section, as given in the table d d E t syst e1 1A1V1 e2 2A2V2 e3 3A3V3 3.2 The Reynolds Transport Theorem 139 CV 1 2 3 4 5 All sections i: Vri approximately normal to area Ai CS Section 2: uniform Vr2, A2, 2, 2, etc.
ρ β E3.1 3 2 1 CV E3.2 Introducing the numerical values from the table, we have d d E t syst (300 J/kg)(800 kg/m3)(2 m2)(5 m/s) 100(800)(3)(8) 150(800)(2)(17) (2,400,000 1,920,000 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.
Note: This problem involves energy, but suppose we check the balance of mass also.
Then B mass m, and B dm/dm unity. Again the volume integral vanishes for steady flow, and Eq. (3.17) reduces to d d m t syst CS (V n) dA 1A1V1 2A2V2 3A3V3 (800 kg/m3)(2 m2)(5 m/s) 800(3)(8) 800(17)(2) (8000 19,200 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).
EXAMPLE 3.2 The balloon in Fig. E3.2 is being filled through section 1, where the area is A1, velocity is V1, and fluid density is 1. The average density within the balloon is b(t). Find an expression for the rate of change of system mass within the balloon at this instant.
Solution It is convenient to define a deformable control surface just outside the balloon, expanding at the same rate R(t). Equation (3.16) applies with Vr 0 on the balloon surface and Vr V1 at the pipe entrance. For mass change, we take B m and dm/dm 1. Equation (3.16) becomes d d m t syst d d t CS d CS (Vr n) dA Mass flux occurs only at the inlet, so that the control-surface integral reduces to the single neg-ative term 1A1V1. The fluid mass within the control volume is approximately the average den-sity times the volume of a sphere. The equation thus becomes d d m t syst d d t b 4 3 R3 1A1V1 Ans.
This is the desired result for the system mass rate of change. Actually, by the conservation law 140 Chapter 3 Integral Relations for a Control Volume 1 R(t) Pipe Average density: b(t) CS expands outward with balloon radius R(t) ρ 3.3 Conservation of Mass (3.1), this change must be zero. Thus the balloon density and radius are related to the inlet mass flux by d d t ( bR3) 4 3 1A1V1 This is a first-order differential equation which could form part of an engineering analysis of balloon inflation. It cannot be solved without further use of mechanics and thermodynamics to relate the four unknowns b, 1, V1, and R. The pressure and temperature and the elastic prop-erties of the balloon would also have to be brought into the analysis.
For advanced study, many more details of the analysis of deformable control vol-umes can be found in Hansen and Potter and Foss .
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, lin-ear momentum, angular momentum, and energy.
For conservation of mass, as discussed in Examples 3.1 and 3.2, B m and dm/dm 1. Equation (3.1) becomes d d m t syst 0 d d t CV d CS (Vr n) dA (3.20) This is the integral mass-conservation law for a deformable control volume. For a fixed control volume, we have CV t d CS (V n) dA 0 (3.21) If the control volume has only a number of one-dimensional inlets and outlets, we can write CV t d i ( iAiVi)out i ( iAiVi)in 0 (3.22) Other special cases occur. Suppose that the flow within the control volume is steady; then /t 0, and Eq. (3.21) reduces to CS (V n) dA 0 (3.23) This states that in steady flow the mass flows entering and leaving the control volume must balance exactly.6 If, further, the inlets and outlets are one-dimensional, we have 3.3 Conservation of Mass 141 6Throughout this section we are neglecting sources or sinks of mass which might be 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.
Incompressible Flow for steady flow i ( iAiVi)in i ( iAiVi)out (3.24) This simple approximation is widely used in engineering analyses. For example, re-ferring to Fig. 3.6, we see that if the flow in that control volume is steady, the three outlet mass fluxes balance the two inlets: Outflow inflow 2A2V2 3A3V3 5A5V5 1A1V1 4A4V4 (3.25) The quantity AV 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 m ˙ 2 m ˙ 3 m ˙ 5 m ˙ 1 m ˙ 4 (3.26) and, in general, the steady-flow–mass-conservation relation (3.23) can be written as i (m ˙ i)out i (m ˙ i)in (3.27) If the inlets and outlets are not one-dimensional, one has to compute m ˙ by integration over the section m ˙ cs cs (V n) dA (3.28) where “cs’’ stands for cross section. An illustration of this is given in Example 3.4.
Still further simplification is possible if the fluid is incompressible, which we may de-fine as having density variations which are negligible in the mass-conservation re-quirement.7As 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. If the fluid is nearly incompressible, /t is negligible and the volume integral in Eq. (3.21) may be neglected, after which the density can be slipped outside the surface integral and divided out since it is nonzero.
The result is a conservation law for incompressible flows, whether steady or unsteady: CS (V n) dA 0 (3.29) If the inlets and outlets are one-dimensional, we have i (ViAi) out i (ViAi)in (3.30) or Qout Qin where Qi ViAi is called the volume flow passing through the given cross section.
142 Chapter 3 Integral Relations for a Control Volume 7Be warned that there is subjectivity in specifying incompressibility. Oceanographers consider a 0.1 percent density variation very significant, while aerodynamicists often neglect density variations in highly compressible, even hypersonic, gas flows. Your task is to justify the incompressible approximation when you make it.
E3.3 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 CS (V n) dA (3.31) Equation (3.31) allows us to define an average velocity Vav which, when multiplied by the section area, gives the correct volume flow Vav Q A A 1 (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: av A 1 dA (3.33) But the mass flow would contain the product of density and velocity, and the average product ( V)av would in general have a different value from the product of the aver-ages ( V)av A 1 (V n) dA avVav (3.34) 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.
EXAMPLE 3.3 Write the conservation-of-mass relation for steady flow through a streamtube (flow everywhere parallel to the walls) with a single one-dimensional exit 1 and inlet 2 (Fig. E3.3).
Solution For steady flow Eq. (3.24) applies with the single inlet and exit m ˙ 1A1V1 2A2V2 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 Q A1V1 A2V2 const or V2 A A 1 2 V1 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.
EXAMPLE 3.4 For steady viscous flow through a circular tube (Fig. E3.4), the axial velocity profile is given approximately by 3.3 Conservation of Mass 143 2 1 Streamtube control volume V • n = 0 V1 V2 E3.4 E3.5 u U01 R r m so that u varies from zero at the wall (r R), or no slip, up to a maximum u U0 at the cen-terline r 0. For highly viscous (laminar) flow m 1 2 , while for less viscous (turbulent) flow m 1 7 . Compute the average velocity if the density is constant.
Solution The average velocity is defined by Eq. (3.32). Here V iu and n i, and thus V n u. Since the flow is symmetric, the differential area can be taken as a circular strip dA 2 r dr. Equa-tion (3.32) becomes Vav A 1 u dA 1 R2 R 0 U01 R r m 2r dr or Vav U0 (1 m) 2 (2 m) Ans.
For the laminar-flow approximation, m 1 2 and Vav 0.53U0. (The exact laminar theory in Chap.
6 gives Vav 0.50U0.) For turbulent flow, m 1 7 and Vav 0.82U0. (There is no exact turbu-lent 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 Consider the constant-density velocity field u V L 0x 0 w V L 0z similar to Example 1.10. Use the triangular control volume in Fig. E3.5, bounded by (0, 0), (L, L), and (0, L), with depth b into the paper. Compute the volume flow through sections 1, 2, and 3, and compare to see whether mass is conserved.
Solution The velocity field everywhere has the form V iu kw. This must be evaluated along each section. We save section 2 until last because it looks tricky. Section 1 is the plane z L with depth b. The unit outward normal is n k, as shown. The differential area is a strip of depth b varying with x: dA b dx. The normal velocity is (V n)1 (iu kw) k w|1 V L 0z zL V0 The volume flow through section 1 is thus, from Eq. (3.31), Q1 0 1 (V n) dA L 0 (V0)b dx V0bL Ans. 1 144 Chapter 3 Integral Relations for a Control Volume n = k (L, L) L z n = –i 0 0 x n = ?
3 1 2 CV Depth b into paper r r = R x u(r) U0 u = 0 (no slip) E3.6 Since this is negative, section 1 is a net inflow. Check the units: V0bL is a velocity times an area; OK.
Section 3 is the plane x 0 with depth b. The unit normal is n i, as shown, and dA b dz. The normal velocity is (V n)3 (iu kw) (i) u|3 V L 0x s0 0 Ans. 3 Thus Vn 0 all along section 3; hence Q3 0.
Finally, section 2 is the plane x z with depth b. The normal direction is to the right i and down k but must have unit value; thus n (1/2 )(i k). The differential area is either dA 2 b dx or dA 2 b dz. The normal velocity is (V n)2 (iu kw) 1 2 (i k) 1 2 (u w)2 1 2 V0 L x V0 L z xz 2 L V0x or 2 L V0z Then the volume flow through section 2 is Q2 0 2 (V n) dA L 0 2 L V0x (2 b dx) V0bL Ans. 2 This answer is positive, indicating an outflow. These are the desired results. We should note that the volume flow is zero through the front and back triangular faces of the prismatic control vol-ume because Vn 0 on those faces.
The sum of the three volume flows is Q1 Q2 Q3 V0bL V0bL 0 0 Mass is conserved in this constant-density flow, and there are no net sources or sinks within the control volume. This is a very realistic flow, as described in Example 1.10 EXAMPLE 3.6 The tank in Fig. E3.6 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 dh/dt. (b) Compute dh/dt if D1 1 in, D2 3 in, V1 3 ft/s, V2 2 ft/s, and At 2 ft2, as-suming water at 20°C.
Solution 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: d d t 0 CV d 1A1V1 2A2V2 0 (1) Now if At is the tank cross-sectional area, the unsteady term can be evaluated as follows: d d t 0 CV d d d t ( wAth) d d t [ aAt(H h)] wAt d d h t (2) 3.3 Conservation of Mass 145 1 2 Tank area At a w H h Fixed CS ρ ρ Part (a) Part (b) 3.4 The Linear Momentum Equation The a 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 d d h t 1A1V1 wA t 2A2V2 Ans. (a) For water, 1 2 w, and this result reduces to d d h t A1V1 At A2V2 Q1 At Q2 (3) The two inlet volume flows are Q1 A1V1 1 4 ( 1 1 2 ft)2(3 ft/s) 0.016 ft3/s Q2 A2V2 1 4 ( 1 3 2 ft)2(2 ft/s) 0.098 ft3/s Then, from Eq. (3), d d h t 0.057 ft/s Ans. (b) Suggestion: Repeat this problem with the top of the tank open.
An illustration of a mass balance with a deforming control volume has already been given in Example 3.2.
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 con-sequence 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; oth-erwise the results will be unrealistic and probably rotten. We shall see in the examples which follow how mass conservation is constantly checked in performing an analysis of other fluid properties.
In Newton’s law, Eq. (3.2), the property being differentiated is the linear momentum mV. Therefore our dummy variable is B mV and dB/dm V, and application of the Reynolds transport theorem gives the linear-momentum relation for a deformable control volume d d t (mV)syst F d d t CV V d CS V (Vr n) dA (3.35) The following points concerning this relation should be strongly emphasized: 1.
The term V is the fluid velocity relative to an inertial (nonaccelerating) coordi-nate system; otherwise Newton’s law must be modified to include noninertial relative-acceleration terms (see the end of this section).
2.
The term F is the vector sum of all forces acting on the control-volume mate-rial considered as a free body; i.e., it includes surface forces on all fluids and (0.016 0.098) ft3/s 2 ft2 146 Chapter 3 Integral Relations for a Control Volume One-Dimensional Momentum Flux Net Pressure Force on a Closed Control Surface solids cut by the control surface plus all body forces (gravity and electromag-netic) acting on the masses within the control volume.
3.
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 Fx d d t CV u d CS u (Vr n) dA (3.36) and similarly, Fy and Fz would involve v and w, respectively. Failure to ac-count 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 Vr V, and F d d t CV V d CS V (V n) dA (3.37) Again we stress that this is a vector relation and that V must be an inertial-frame ve-locity. Most of the momentum analyses in this text are concerned with Eq. (3.37).
By analogy with the term mass flow used in Eq. (3.28), the surface integral in Eq.
(3.37) is called the momentum-flux term. If we denote momentum by M, then M ˙ CS 0 sec V (V n) dA (3.38) Because of the dot product, the result will be negative for inlet momentum flux and positive for outlet flux. If the cross section is one-dimensional, V and are uniform over the area and the integrated result is M ˙ seci Vi( iVniAi) m ˙ iVi (3.39) for outlet flux and m ˙ iVi for inlet flux. Thus if the control volume has only one-dimensional inlets and outlets, Eq. (3.37) reduces to F d d t CV V d (m ˙ iVi)out (m ˙ iVi)in (3.40) This is a commonly used approximation in engineering analyses. It is crucial to real-ize 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 fluxes minus the vector sum of inlet fluxes.
Generally speaking, the surface forces on a control volume are due to (1) forces ex-posed by cutting through solid bodies which protrude through the surface and (2) forces due to pressure and viscous stresses of the surrounding fluid. The computation of pres-sure force is relatively simple, as shown in Fig. 3.7. Recall from Chap. 2 that the ex-ternal 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 Fpress CS p(n) dA (3.41) 3.4 The Linear Momentum Equation 147 Fig. 3.7 Pressure-force computation by subtracting a uniform distribu-tion: (a) uniform pressure, F pa n dA 0; (b) nonuniform pressure, F (p pa)n dA.
Now if the pressure has a uniform value pa all around the surface, as in Fig. 3.7a, the net pressure force is zero FUP pa(n) dA pa n dA 0 (3.42) where the subscript UP stands for uniform pressure. This result is independent of the shape of the surface8 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 sub-tracting any convenient uniform pressure pa and working only with the pieces of gage pressure which remain, as illustrated in Fig. 3.7b. Thus Eq. (3.41) is entirely equiva-lent to Fpress CS (p pa)(n) dA CS pgage(n) dA This trick can mean quite a saving in computation.
EXAMPLE 3.7 A control volume of a nozzle section has surface pressures of 40 lbf/in2absolute at section 1 and atmospheric pressure of 15 lbf/in2absolute at section 2 and on the external rounded part of the nozzle, as in Fig. E3.7a. Compute the net pressure force if D1 3 in and D2 1 in.
Solution We do not have to bother with the outer surface if we subtract 15 lbf/in2 from all surfaces.
This leaves 25 lbf/in2gage at section 1 and 0 lbf/in2 gage everywhere else, as in Fig. E3.7b.
148 Chapter 3 Integral Relations for a Control Volume Closed CS pa n pa pa pa pgage = p – pa Closed CS pgage = 0 pgage pgage pa (a) (b) n pa 8Can you prove this? It is a consequence of Gauss’ theorem from vector analysis.
E3.7 Then the net pressure force is computed from section 1 only F pg1(n)1A1 (25 lbf/in2) 4 (3 in)2i 177i lbf Ans.
Notice that we did not change inches to feet in this case because, with pressure in pounds-force per square inch and area in square inches, the product gives force directly in pounds. More of-ten, though, the change back to standard units is necessary and desirable. Note: This problem computes pressure force only. There are probably other forces involved in Fig. E3.7, e.g., nozzle-wall stresses in the cuts through sections 1 and 2 and the weight of the fluid within the control volume.
Figure E3.7 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. E3.7.
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 3.8 A fixed control volume of a streamtube in steady flow has a uniform inlet flow ( 1, A1, V1) and a uniform exit flow ( 2, A2, V2), as shown in Fig. 3.8. Find an expression for the net force on the control volume.
Solution Equation (3.40) applies with one inlet and exit F m ˙ 2V2 m ˙ 1V1 ( 2A2V2)V2 ( 1A1V1)V1 3.4 The Linear Momentum Equation 149 2 1 2 1 40 lbf/in2 abs 15 lbf/in2 abs Jet exit pressure is atmospheric 25 lbf/in2 gage 15 lbf/in2 abs (a) (b) 15 lbf/in2 abs Flow Flow 0 lbf/in2 gage 0 lbf/in2 gage 0 lbf/in2 gage Pressure Condition at a Jet Exit Fig. 3.8 Net force on a one-dimen-sional streamtube in steady flow: (a) streamtube in steady flow; (b) vector diagram for computing net force.
Fig. 3.9 Net applied force on a fixed jet-turning vane: (a) geometry of the vane turning the water jet; (b) vector diagram for the net force.
The volume-integral term vanishes for steady flow, but from conservation of mass in Example 3.3 we saw that m ˙ 1 m ˙ 2 m ˙ const Therefore a simple form for the desired result is F m ˙ (V2 V1) Ans.
This is a vector relation and is sketched in Fig. 3.8b. The term F represents the net force act-ing on the control volume due to all causes; it is needed to balance the change in momentum of the fluid as it turns and decelerates while passing through the control volume.
EXAMPLE 3.9 As shown in Fig. 3.9a, a fixed vane turns a water jet of area A through an angle without chang-ing its velocity magnitude. The flow is steady, pressure is pa everywhere, and friction on the vane is negligible. (a) Find the components Fx and Fy of the applied vane force. (b) Find ex-pressions for the force magnitude F and the angle between F and the horizontal; plot them versus .
Solution The control volume selected in Fig. 3.9a 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, 150 Chapter 3 Integral Relations for a Control Volume θ Fixed control volume m V2 ΣF = m (V2 – V1) (a) (b) 1 2 V • n = 0 V2 θ V1 .
m = constant m V1 y x V V 1 2 pa θ CV F F θ Fy Fx mV mV φ (a) (b) Part (a) Part (b) 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 m ˙ 2V2 m ˙ 1V1 But the magnitude V1 V2 V as given, and conservation of mass for the streamtube requires m ˙ 1 m ˙ 2 m ˙ AV. The vector diagram for force and momentum change becomes an isosce-les triangle with legs m ˙V and base F, as in Fig. 3.9b. We can readily find the force components from this diagram Fx m ˙V(cos 1) Fy m ˙V sin Ans. (a) where m ˙V AV2 for this case. This is the desired result.
The force magnitude is obtained from part (a): F (Fx 2 Fy 2)1/2 m ˙V[sin2 (cos 1)2]1/2 2m ˙V sin 2 Ans. (b) From the geometry of Fig. 3.9b we obtain 180° tan1 F F y x 90° 2 Ans. (b) 3.4 The Linear Momentum Equation 151 F m V F m V φ φ 0 45˚ 90˚ 135˚ 180˚ 180˚ 1.0 2.0 θ 90˚ E3.9 These can be plotted versus as shown in Fig. E3.9. Two special cases are of interest. First, the maximum force occurs at 180°, that is, when the jet is turned around and thrown back in the opposite direction with its momentum completely reversed. This force is 2m ˙V and acts to the left; that is, 180°. Second, at very small turning angles ( 10°) we obtain approximately F m ˙V 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 di-rection and thereby creates a lift force normal to the basic flow.
EXAMPLE 3.10 A water jet of velocity Vj impinges normal to a flat plate which moves to the right at velocity Vc, as shown in Fig. 3.10a. Find the force required to keep the plate moving at constant veloc-ity if the jet density is 1000 kg/m3, the jet area is 3 cm2, and Vj and Vc are 20 and 15 m/s, re-Fig. 3.10 Force on a plate moving at constant velocity: (a) jet striking a moving plate normally; (b) con-trol volume fixed relative to the plate.
spectively. 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.
Solution The suggested control volume in Fig. 3.10a cuts through the plate support to expose the desired forces Rx and Ry. This control volume moves at speed Vc and thus is fixed relative to the plate, as in Fig. 3.10b. We must satisfy both mass and momentum conservation for the assumed steady-flow pattern in Fig. 3.10b. There are two outlets and one inlet, and Eq. (3.30) applies for mass conservation m ˙ out m ˙ in or 1A1V1 2A2V2 jAj(Vj Vc) (1) We assume that the water is incompressible 1 2 j, and we are given that A1 A2 1 2 Aj.
Therefore Eq. (1) reduces to V1 V2 2(Vj Vc) (2) Strictly speaking, this is all that mass conservation tells us. However, from the symmetry of the jet deflection and the neglect of fluid weight, we conclude that the two velocities V1 and V2 must be equal, and hence (2) becomes V1 V2 Vj Vc (3) For the given numerical values, we have V1 V2 20 15 5 m/s Now we can compute Rx and Ry from the two components of momentum conservation. Equa-tion (3.40) applies with the unsteady term zero Fx Rx m ˙ 1u1 m ˙ 2u2 m ˙ juj (4) where from the mass analysis, m ˙1 m ˙ 2 1 2 m ˙ j 1 2 jAj(Vj Vc). Now check the flow directions at each section: u1 u2 0, and uj Vj Vc 5 m/s. Thus Eq. (4) becomes Rx m ˙ juj jAj(Vj Vc) (5) 152 Chapter 3 Integral Relations for a Control Volume Nozzle Vj p = pa Vc CS Vc Vj – Vc Aj j CS 1 Ry Rx A1 = Aj 1 2 2 A2 = Aj 1 2 (a) (b) Fig. 3.11 Control-volume analysis of drag force on a flat plate due to boundary shear.
For the given numerical values we have Rx (1000 kg/m3)(0.0003 m2)(5 m/s)2 7.5 (kg m)/s2 7.5 N Ans.
This acts to the left; i.e., 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 m ˙ 11 m ˙ 22 m ˙ jj Check directions again: 1 V1, 2 V2, j 0. Thus Ry m ˙ 1(V1) m ˙ 2(V2) 1 2 m ˙ j(V1 V2) (6) But since we found earlier that V1 V2, this means that Ry 0, as we could expect from the symmetry of the jet deflection.9 Two other results are of interest. First, the relative velocity at section 1 was found to be 5 m/s up, from Eq. (3). If we convert this to absolute motion by adding on the control-volume speed Vc 15 m/s to the right, we find that the absolute velocity V1 15i 5j m/s, or 15.8 m/s at an angle of 18.4° upward, as indicated in Fig. 3.10a. Thus the ab-solute jet speed changes after hitting the plate. Second, the computed force Rx 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 flux when Eq. (4) was rewritten for a radial-deflection condition.
EXAMPLE 3.11 The previous example treated a plate at normal incidence to an oncoming flow. In Fig. 3.11 the plate is parallel to the flow. The stream is not a jet but a broad river, or free stream, of uniform velocity V U0i. 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. 3.10, so that 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 . The 3.4 The Linear Momentum Equation 153 9Symmetry can be a powerful tool if used properly. Try to learn more about the uses and misuses of symmetry conditions. Here we doggedly computed the results without invoking symmetry.
y U0 Oncoming stream parallel to plate 0 1 y = h Streamline just outside the shear-layer region 2 Plate of width b 4 Boundary layer where shear stress is significant p = pa y = δ 3 U0 u(y) x L viscous stresses along the wall can sum to a finite drag force on the plate. These effects are il-lustrated in Fig. 3.11. The problem is to make an integral analysis and find the drag force D in terms of the flow properties , U0, and and the plate dimensions L and b.† 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 0 to h to to L and back to the origin 0, as shown in Fig. 3.11. Had we chosen to cut across horizon-tally 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 1.
From (0, 0) to (0, h): a one-dimensional inlet, V n U0 2.
From (0, h) to (L, ): a streamline, no shear, V n 0 3.
From (L, ) to (L, 0): a two-dimensional outlet, V n u(y) 4.
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 in-compressible and steady, Eq. (3.37) applies with no unsteady term and fluxes only across sec-tions 1 and 3: Fx D 0 1 u(V n) dA 0 3 u(V n) dA h 0 U0(U0)b dy 0 u( u)b dy Evaluating the first integral and rearranging give D U0 2bh b 0 u2dy (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 . This is found by applying mass conser-vation, since the control volume forms a streamtube 0 CS (V n) dA 0 h 0 (U0)b dy 0 ub dy or U0h 0 u dy (2) after canceling b and and evaluating the first integral. Introduce this value of h into Eq. (1) for a much cleaner result D b 0 u(U0 u) dyxL Ans. (3) This result was first derived by Theodore von Kármán in 1921.10 It relates the friction drag on 154 Chapter 3 Integral Relations for a Control Volume †The general analysis of such wall-shear problems, called boundary-layer theory, is treated in Sec. 7.3.
10The autobiography of this great twentieth-century engineer and teacher is recommended for its historical and scientific insight.
Momentum-Flux Correction Factor one side of a flat plate to the integral of the momentum defect u(U0 u) across the trailing cross section of the flow past the plate. Since U0 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. To illustrate the magnitude of this drag force, we can use a simple parabolic approximation for the outlet-velocity profile u(y) which simulates low-speed, or laminar, shear flow u U0 2 y y2 2 for 0 y (4) Substituting into Eq. (3) and letting y/ for convenience, we obtain D bU0 2 1 0 (2 2)(1 2 2) d 1 2 5 U0 2b (5) This is within 1 percent of the accepted result from laminar boundary-layer theory (Chap. 7) in spite of the crudeness of the Eq. (4) approximation. This is a happy situation and has led to the wide use of Kármán’s integral theory in the analysis of viscous flows. Note that D increases with the shear-layer thickness , which itself increases with plate length and the viscosity of the fluid (see Sec. 7.4).
For flow in a duct, the axial velocity is usually nonuniform, as in Example 3.4. For this case the simple momentum-flux calculationu (V n) dA m ˙ V AV2 is some-what in error and should be corrected to AV2, where is the dimensionless momentum-flux correction factor, 1.
The factor accounts for the variation of u2across the duct section. That is, we com-pute the exact flux and set it equal to a flux based on average velocity in the duct u2dA m ˙ Vav AVav 2 or A 1 V u av 2 dA (3.43a) Values of can be computed based on typical duct velocity profiles similar to those in Example 3.4. The results are as follows: Laminar flow: u U01 R r2 2 4 3 (3.43b) Turbulent flow: u U01 R r m 1 9 m 1 5 (3.43c) The turbulent correction factors have the following range of values: Turbulent flow: (1 m)2(2 m)2 2(1 2m)(2 2m) 3.4 The Linear Momentum Equation 155 m 1 5 1 6 1 7 1 8 1 9 1.037 1.027 1.020 1.016 1.013 Noninertial Reference Frame11 These are so close to unity that they are normally neglected. The laminar correction may sometimes be important.
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 given as F m ˙ (
2V2 1V1) (3.43d) Note that the basic parameters and vector character of the result are not changed at all by this correction.
All previous derivations and examples in this section have assumed that the coordinate system is inertial, i.e., 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 sys-tem. An example would be coordinates fixed to a rocket during takeoff. A second ex-ample is any flow on the earth’s surface, which is accelerating relative to the fixed stars because of the rotation of the earth. Atmospheric and oceanographic flows experience the so-called Coriolis acceleration, outlined below. It is typically less than 105g, where g is the acceleration of gravity, but its accumulated effect over distances of many kilo-meters 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. 3.12. Then dV/dt will represent a noninertial acceleration which must be added vectorially to a relative acceleration arel to give the absolute ac-celeration ai relative to some inertial coordinate system XYZ, as in Fig. 3.12. Thus ai d d V t arel (3.44) 156 Chapter 3 Integral Relations for a Control Volume 11This section may be omitted without loss of continuity.
Fig. 3.12 Geometry of fixed versus accelerating coordinates.
Particle Vrel = dr dt y x z X Z Y R Noninertial, moving, rotating coordinates Inertial coordinates r Since Newton’s law applies to the absolute acceleration, F mai m d d V t arel or F marel m d d V t (3.45) Thus Newton’s law in noninertial coordinates xyz is equivalent to adding more “force” terms marel to account for noninertial effects. In the most general case, sketched in Fig. 3.12, the term arel contains four parts, three of which account for the angular ve-locity (t) of the inertial coordinates. By inspection of Fig. 3.12, the absolute dis-placement of a particle is Si r R (3.46) Differentiation gives the absolute velocity Vi V d d R t r (3.47) A second differentiation gives the absolute acceleration: ai d d V t d d 2 t R 2 d d t r 2 V ( r) (3.48) By comparison with Eq. (3.44), we see that the last four terms on the right represent the additional relative acceleration: 1.
d2R/dt2is the acceleration of the noninertial origin of coordinates xyz.
2.
(d/dt) r is the angular-acceleration effect.
3.
2 V is the Coriolis acceleration.
4.
( r) is the centripetal acceleration, directed from the particle normal to the axis of rotation with magnitude 2L, where L is the normal distance to the axis.12 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 co-ordinates merely adds inertial terms by integrating the added relative acceleration over each differential mass in the control volume F 0 CV arel dm d d t 0 CV V d 0 CS V (Vr n) dA (3.49) where arel d d 2 t R 2 d d t r 2 V ( r) This is the noninertial equivalent to the inertial form given in Eq. (3.35). To analyze such problems, one must have knowledge of the displacement R and angular velocity of the noninertial coordinates.
If the control volume is nondeformable, Eq. (3.49) reduces to 3.4 The Linear Momentum Equation 157 12A complete discussion of these noninertial coordinate terms is given, e.g., in Ref. 4, pp. 49 – 51.
E3.12 F 0 CVarel dm d d t 0 CV V d CS V (V n) dA (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 M0, and assume a steady exhaust mass flow m ˙ and exhaust ve-locity Ve 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(t) and in-tegrate using the initial condition V 0 at t 0.
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 equation (3.49) becomes Fz arel dm d d t CVw dm ˙ (m ˙ w)e or mg m d d V t 0 m ˙ Ve with m m(t) M0 m ˙ t The term arel dV/dt of the rocket. The control volume integral vanishes because of the steady rocket-flow conditions. Separate the variables and integrate, assuming V 0 at t 0: V 0 dV m ˙ Ve t 0 M0 d t m ˙ t g t 0 dt or V(t) Veln 1 M m ˙ 0 t gt Ans.
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 Ve.
A control-volume analysis can be applied to the angular-momentum relation, Eq. (3.3), 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 ve-locity, the concept of mass moment of inertia is of no help and we have to calculate 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 HO syst (r V) dm (3.51) 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 d d H m O r V 158 Chapter 3 Integral Relations for a Control Volume Accelerating control volume G m Datum V(t) V(t) Ve z g 3.5 The Angular-Momentum Theorem13 13This section may be omitted without loss of continuity.
m ˙ The Reynolds transport theorem (3.16) then tells us that dH dt O syst d d t CV (r V) d CS (r V) (Vr n) dA (3.52) 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 dH dt O MO (r F)O 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 arel in Eq. (3.49) must also be included MO (r F)O CV (r arel) dm (3.53) where the four terms constituting arel 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. (3.52) and (3.53) to obtain (r F)0 CV (r arel) dm d d t CV (r V) d CS (r V) (Vr n) dA (3.54) For a nondeformable inertial control volume, this reduces to M0 t CV (r V) d CS (r V) (V n) dA (3.55) Further, if there are only one-dimensional inlets and exits, the angular-momentum flux terms evaluated on the control surface become CS (r V) (V n) dA (r V)out m ˙ out (r V)in m ˙ in (3.56) Although at this stage the angular-momentum theorem can be considered to be a sup-plementary topic, it has direct application to many important fluid-flow problems in-volving torques or moments. A particularly important case is the analysis of rotating fluid-flow devices, usually called turbomachines (Chap. 11).
EXAMPLE 3.13 As shown in Fig. E3.13a, a pipe bend 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 pa is zero. (a) Find an expression for the torque T which must be resisted by the support at A, in terms of the flow properties at sections 1 and 2 and the distances h1 and h2. (b) Compute this torque if D1 D2 3 in, p1 100 lbf/in2 gage, p2 80 lbf/in2 gage, V1 40 ft/s, h1 2 in, h2 10 in, and 1.94 slugs/ft3.
3.5 The Angular-Momentum Theorem 159 E3.13a Solution The control volume chosen in Fig. E3.13b cuts through sections 1 and 2 and through the sup-port at A, where the torque TA 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 V, r should be taken from point A to sections 1 and 2. Note that the gage pressure forces p1A1 and p2A2 both have moments about A. Equation (3.55) with one-dimen-sional flux terms becomes MA TA r1 (p1A1n1) r2 (p2A2n2) (r2 V2)( m ˙ out) (r1 V1)(m ˙ in) (1) Figure E3.13c shows that all the cross products are associated either with r1 sin 1 h1 or r2 sin 2 h2, the perpendicular distances from point A to the pipe axes at 1 and 2. Remember that m ˙ in m ˙ out from the steady-flow continuity relation. In terms of counterclockwise moments, Eq. (1) then becomes TA p1A1h1 p2A2h2 m ˙ (h2V2 h1V1) (2) Rewriting this, we find the desired torque to be TA h2(p2A2 m ˙ V2) h1(p1A1 m ˙ V1) Ans. (a) (3) 160 Chapter 3 Integral Relations for a Control Volume p1, V1, A1 1 pa = 0 = constant V2, A2, p2 h2 h1 A 2 ρ Part (a) CV V2 TA V1 A r1 r2 p1A1 p2A2 r1 V1 = h1 V1 r2 V2 = h2 V2 θ θ 2 1 r2 r1 V2 V1 h2 = r2 sin h1 = r1 sin θ θ 2 1 E3.13b E3.13c Part (b) counterclockwise. The quantities p1 and p2 are gage pressures. Note that this result is indepen-dent of the shape of the pipe bend and varies only with the properties at sections 1 and 2 and the distances h1 and h2.† The inlet and exit areas are the same: A1 A2 4 (3)2 7.07 in2 0.0491 ft2 Since the density is constant, we conclude from continuity that V2 V1 40 ft/s. The mass flow is m ˙ A1V1 1.94(0.0491)(40) 3.81 slug/s Equation (3) can be evaluated as TA ( 1 1 0 2 ft)[80(7.07) lbf 3.81(40) lbf] ( 1 2 2 ft)[100(7.07) lbf 3.81(40) lbf] 598 143 455 ft lbf counterclockwise Ans. (b) We got a little daring there and multiplied p in lbf/in2 gage times A in in2 to get lbf without changing units to lbf/ft2 and ft2.
EXAMPLE 3.14 Figure 3.13 shows a schematic of a centrifugal pump. The fluid enters axially and passes through the pump blades, which rotate at angular velocity ; the velocity of the fluid is changed from V1 to V2 and its pressure from p1 to p2. (a) Find an expression for the torque TO which must be applied to these blades to maintain this flow. (b) The power supplied to the pump would be P TO. To illustrate numerically, suppose r1 0.2 m, r2 0.5 m, and b 0.15 m. Let the pump rotate at 600 r/min and deliver water at 2.5 m3/s with a density of 1000 kg/m3. Compute the ide-alized torque and power supplied.
Solution The control volume is chosen to be the angular region between sections 1 and 2 where the flow passes through the pump blades (see Fig. 3.13). 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 (3.55) becomes MO TO (r2 V2)m ˙ out (r1 V1)m ˙ in (1) where steady-flow continuity tells us that m ˙ in Vn12r1b m ˙ out Vn2r2b Q The cross product r V is found to be clockwise about O at both sections: r2 V2 r2Vt2 sin 90° k r2Vt2k clockwise r1 V1 r1Vt1k clockwise Equation (1) thus becomes the desired formula for torque TO Q(r2Vt2 r1Vt1)k clockwise Ans. (a) (2a) 3.5 The Angular-Momentum Theorem 161 †Indirectly, the pipe-bend shape probably affects the pressure change from p1 to p2.
Part (a) Fig. 3.13 Schematic of a simplified centrifugal pump.
This relation is called Euler’s turbine formula. In an idealized pump, the inlet and outlet tan-gential velocities would match the blade rotational speeds Vt1 r1 and Vt2 r2. Then the formula for torque supplied becomes TO Q(r2 2 r1 2) clockwise (2b) Convert to 600(2/60) 62.8 rad/s. The normal velocities are not needed here but follow from the flow rate Vn1 2 Q r1b 2(0.
2 2 .5 m m )( 3 0 / .
s 15 m) 13.3 m/s Vn2 2 Q r2b 2(0.
2 5 .
) 5 (0.15) 5.3 m/s For the idealized inlet and outlet, tangential velocity equals tip speed Vt1 r1 (62.8 rad/s)(0.2 m) 12.6 m/s Vt2 r2 62.8(0.5) 31.4 m/s Equation (2a) predicts the required torque to be TO (1000 kg/m3)(2.5 m3/s)[(0.5 m)(31.4 m/s) (0.2 m)(12.6 m/s)] 33,000 (kg m2)/s2 33,000 N m Ans.
The power required is P TO (62.8 rad/s)(33,000 N m) 2,070,000 (N m)/s 2.07 MW (2780 hp) 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.
162 Chapter 3 Integral Relations for a Control Volume 1 2 ω Blade Pump blade shape Width b CV Inflow z,k r1 r2 r O Blade Vn2 Vt2 Vn1 Vt1 Part (b) Fig. 3.14 View from above of a single arm of a rotating lawn sprinkler.
EXAMPLE 3.15 Figure 3.14 shows a lawn-sprinkler arm viewed from above. The arm rotates about O at con-stant angular velocity . The volume flux entering the arm at O is Q, and the fluid is incom-pressible. There is a retarding torque at O, due to bearing friction, of amount TOk. Find an ex-pression for the rotation in terms of the arm and flow properties.
Solution The entering velocity is V0k, where V0 Q/Apipe. Equation (3.55) applies to the control volume sketched in Fig. 3.14 only if V is the absolute velocity relative to an inertial frame. Thus the exit velocity at section 2 is V2 V0i Ri Equation (3.55) then predicts that, for steady flow, MO TOk (r2 V2)m ˙ out (r1 V1)m ˙ in (1) where, from continuity, m ˙ out m ˙ in Q. The cross products with reference to point O are r2 V2 Rj (V0 R)i (R2 RV0)k r1 V1 0j V0k 0 Equation (1) thus becomes TOk Q(R2 RV0)k V R O Q TO R2 Ans.
The result may surprise you: Even if the retarding torque TO is negligible, the arm rotational speed is limited to the value V0/R imposed by the outlet speed and the arm length.
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 dE/dm e. Equation (3.5) can then be written for a fixed control volume as follows:15 d d Q t d d W t d d E t d d t CVe d CS e (V n) dA (3.57) 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: e einternal ekinetic epotential eother 3.6 The Energy Equation 163 y x R ω 2 Absolute outlet velocity V2 = V0i – Rωi CV Retarding torque T0 O V0 = Q Apipe k Inlet velocity 3.6 The Energy Equation14 14This section should be read for information and enrichment even if you lack formal background in thermodynamics.
15The energy equation for a deformable control volume is rather complicated and is not discussed here. See Refs. 4 and 5 for further details.
where eother could encompass chemical reactions, nuclear reactions, and electrostatic or magnetic field effects. We neglect eother here and consider only the first three terms as discussed in Eq. (1.9), with z defined as “up”: e û 1 2 V2 gz (3.58) The heat and work terms could be examined in detail. If this were a heat-transfer book, dQ/dT would be broken down into conduction, convection, and radiation effects and whole chapters written on each (see, e.g., Ref. 3). Here we leave the term un-touched and consider it only occasionally.
Using for convenience the overdot to denote the time derivative, we divide the work term into three parts: W ˙ W ˙ shaft W ˙ press W ˙ viscous stresses W ˙ s W ˙ p W ˙ The work of gravitational forces has already been included as potential energy in Eq.
(3.58). Other types of work, e.g., those due to electromagnetic forces, are excluded here.
The shaft work isolates that portion of the work which is deliberately done by a machine (pump impeller, fan blade, piston, etc.) protruding through the control sur-face into the control volume. No further specification other than W ˙ s is desired at this point, but calculations of the work done by turbomachines will be performed in Chap. 11.
The rate of work W ˙ p done on 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 dW ˙ p (p dA)Vn,in p(V n) dA The total pressure work is the integral over the control surface W ˙ p CS p(V n) dA (3.59) 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 W ˙ s, not to W ˙ p, 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, the in-ternal work terms again being self-canceling, and consists of the product of each vis-cous stress (one normal and two tangential) and the respective velocity component dW ˙ V dA or W ˙ CS V dA (3.60) where 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 which are solid confining walls, V 0 from the viscous no-slip condition; hence W ˙ zero identically.
164 Chapter 3 Integral Relations for a Control Volume One-Dimensional Energy-Flux Terms Surface of a machine. Here the viscous work is contributed by the machine, and so we absorb this work in the term W ˙ s.
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 nnVn dA. Since viscous normal stresses are extremely small in all but rare cases, e.g., 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 the above discussion is that the rate-of-work term in Eq. (3.57) consists essentially of W ˙ W ˙ s CS p(V n) dA CS ( V)SS dA (3.61) where the subscript SS stands for stream surface. When we introduce (3.61) and (3.58) into (3.57), we find that the pressure-work term can be combined with the energy-flux term since both involve surface integrals of V n. The control-volume energy equation thus becomes Q ˙ W ˙ s (W ˙ υ)SS t CV ep d CS (e p ) (V n) dA (3.62) Using e from (3.58), we see that the enthalpy h ˆ û p/ occurs in the control-sur-face integral. The final general form for the energy equation for a fixed control vol-ume becomes Q ˙ W ˙ s W ˙ υ t CV û 1 2 V2 gz d CS h ˆ 1 2 V2 gz (V n) dA (3.63) As mentioned above, the shear-work term W ˙ is rarely important.
If the control volume has a series of one-dimensional inlets and outlets, as in Fig.
3.6, the surface integral in (3.63) reduces to a summation of outlet fluxes minus in-let fluxes CS (h ˆ 1 2 V2 gz) (V n) dA (h ˆ 1 2 V2 gz)outm ˙ out (h ˆ 1 2 V2 gz)inm ˙ in (3.64) where the values of h ˆ, 1 2 V2, and gz are taken to be averages over each cross section.
3.6 The Energy Equation 165 EXAMPLE 3.16 A steady-flow machine (Fig. E3.16) takes in air at section 1 and discharges it at sections 2 and 3. The properties at each section are as follows: Section A, ft2 Q, ft3/s T, °F p, lbf/in2 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 p3 in lbf/in2 absolute and the heat transfer Q ˙ in Btu/s. Assume that air is a perfect gas with R 1715 and cp 6003 ft lbf/(slug °R).
Solution The control volume chosen cuts across the three desired sections and otherwise follows the solid walls of the machine. Therefore the shear-work term W is negligible. We have enough infor-mation to compute Vi Qi/Ai immediately V1 1 0 0 .4 0 250 ft/s V2 1 4 .
0 0 40 ft/s V3 0 5 .2 0 5 200 ft/s and the densities i pi/(RTi) 1 1715 2 ( 0 7 ( 0 14 4) 460) 0.00317 slug/ft3 2 1 3 7 0 1 ( 5 1 ( 4 5 4 6 ) 0) 0.00450 slug/ft3 but 3 is determined from the steady-flow continuity relation: m ˙ 1 m ˙ 2 m ˙ 3 1Q1 2Q2 3Q3 (1) 0.00317(100) 0.00450(40) 3(50) or 50 3 0.317 0.180 0.137 slug/s 3 0.
5 1 0 37 0.00274 slug/ft3 17 1 1 4 5 4 (6 p 6 3 0) p3 21.5 lbf/in2 absolute Ans.
Note that the volume flux Q1 Q2 Q3 because of the density changes.
For steady flow, the volume integral in (3.63) vanishes, and we have agreed to neglect vis-cous work. With one inlet and two outlets, we obtain Q ˙ W ˙ s m ˙ 1(h ˆ1 1 2 V1 2 gz1) m ˙ 2(h ˆ2 1 2 V2 2 gz2) m ˙ 3(h ˆ3 1 2 V3 2 gz3) (2) where W ˙ s is given in hp and can be quickly converted to consistent BG units: W ˙ s 150 hp [550 ft lbf/(s hp)] 82,500 ft lbf/s negative work on system 166 Chapter 3 Integral Relations for a Control Volume (1) (3) (2) Q = ?
150 hp CV E3.16 The Steady-Flow Energy Equation For a perfect gas with constant cp, h ˆ cpT plus an arbitrary constant. It is instructive to sepa-rate the flux terms in Eq. (2) above to examine their magnitudes: Enthalpy flux: cp(m ˙ 1T1 m ˙ 2T2 m ˙ 3T3) [6003 ft lbf/(slug °R)][(0.317 slug/s)(530 °R) 0.180(560) 0.137(660)] 1,009,000 605,000 543,000 139,000 ft lbf/s Kinetic-energy flux: m ˙ 1( 1 2 V1 2) m ˙ 2( 1 2 V2 2) m ˙ 3( 1 2 V3 2) 1 2 [0.317(250)2 0.180(40)2 0.137(200)2] 9900 150 2750 7000 ft lbf/s Potential-energy flux: g(m ˙ 1z1 m ˙ 2z2 m ˙ 3z3) 32.2[0.317(1.0) 0.180(4.0) 0.137(1.5)] 10 23 7 20 ft lbf/s These are typical effects: The potential-energy flux is negligible in gas flows, the kinetic-energy flux is small in low-speed flows, and the enthalpy flux is dominant. It is only when we neglect heat-transfer effects that the kinetic and potential energies become important. Anyway, we can now solve for the heat flux Q ˙ 82,500 139,000 7000 20 49,520 ft lbf/s (3) Converting, we get Q ˙ 778.2 49 ft ,5 2 lb 0 f/Btu 63.6 Btu/s Ans.
For steady flow with one inlet and one outlet, both assumed one-dimensional, Eq. (3.63) reduces to a celebrated relation used in many engineering analyses. Let section 1 be the inlet and section 2 the outlet. Then Q ˙ W ˙s W ˙ m ˙ 1(h ˆ1 1 2 V1 2 gz1) m ˙ 2(h ˆ2 1 2 V2 2 gz2) (3.65) But, from continuity, m ˙ 1 m ˙ 2 m ˙ , and we can rearrange (3.65) as follows: h ˆ1 1 2 V1 2 gz1 (h ˆ2 1 2 V2 2 gz2) q ws wυ (3.66) where q Q ˙ /m ˙ dQ/dm, the heat transferred to the fluid per unit mass. Similarly, ws W ˙ s/m ˙ dWs/dm and wυ W ˙ υ/m ˙ dWυ/dm. Equation (3.66) is a general form of the steady-flow energy equation, which states that the upstream stagnation enthalpy H1 (h ˆ 1 2 V 2 gz)1 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 ws and w are positive if work is done by the fluid on the surroundings.
3.6 The Energy Equation 167 Friction Losses in Low-Speed Flow Each term in Eq. (3.66) 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 engi-neers. The traditional symbol for head is h, which we do not wish to confuse with en-thalpy. Therefore we use internal energy in rewriting the head form of the energy re-lation: p 1 û g 1 V 2g 1 2 z1 p 2 û g 2 V 2g 1 2 z2 hq hs h (3.67) where hq q/g, hs ws/g, and hυ wu/g are the head forms of the heat added, shaft work done, and viscous work done, respectively. The term p/ is called pressure head and the term V2/2g is denoted as velocity head.
A very common application of the steady-flow energy equation is for low-speed flow with no shaft work and negligible viscous work, such as liquid flow in pipes. For this case Eq. (3.67) may be written in the form p 1 V 2g 1 2 z1 p 2 V 2g 2 2 z2 û2 û g 1 q (3.68) The term in parentheses is called the useful head or available head or total head of the flow, denoted as h0. The last term on the right is the difference between the avail-able head upstream and downstream and is normally positive, representing the loss in head due to friction, denoted as hf. Thus, in low-speed (nearly incompressible) flow with one inlet and one exit, we may write p V 2g 2 zin p V 2g 2 zout hfriction hpump hturbine (3.69) Most of our internal-flow problems will be solved with the aid of Eq. (3.69). 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 ht 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 3.17 Gasoline at 20°C is pumped through a smooth 12-cm-diameter pipe 10 km long, at a flow rate of 75 m3/h (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 V2/(2g). (These numbers are quite realistic for liquid flow through long pipelines.) Solution For gasoline at 20°C, from Table A.3, 680 kg/m3, or (680)(9.81) 6670 N/m3. There is no shaft work; hence Eq. (3.69) applies and can be evaluated: 168 Chapter 3 Integral Relations for a Control Volume p in V 2g in 2 zin p out V 2 2 g out zout hf (1) The pipe is of uniform cross section, and thus the average velocity everywhere is Vin Vout Q A ( ( 7 5 / / 4 3 ) 6 ( 0 0 0 .1 ) 2 m m 3/ ) s 2 1.84 m/s Being equal at inlet and exit, this term will cancel out of Eq. (1) above, but we are asked to com-pute the velocity head of the flow for comparison purposes: V 2g 2 2 ( ( 1 9 .8 .8 4 1 m m / / s s ) 2 2 ) 0.173 m Now we are in a position to evaluate all terms in Eq. (1) except the friction head loss: 0.173 m 0 m 10 6 1 6 , 7 3 0 50 N N /m /m 3 2 0.173 m 150 m hf or hf 364.7 15.2 150 199 m Ans.
The friction head is larger than the elevation change z, and the pump must drive the flow against both changes, hence the high inlet pressure. The ratio of friction to velocity head is V2/ h ( f 2g) 0 1 .1 9 7 9 3 m m 1150 Ans.
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 pressure is needed?
Our correlations for hf will lead to the estimate pinlet 24 atm, as stated above.
EXAMPLE 3.18 Air [R 1715, cp 6003 ft lbf/(slug °R)] flows steadily, as shown in Fig. E3.18, through a turbine which produces 700 hp. For the inlet and exit conditions shown, estimate (a) the exit ve-locity V2 and (b) the heat transferred Q ˙ in Btu/h.
(24)(101,350 N/m2) 6670 N/m3 3.6 The Energy Equation 169 1 2 Turbomachine ws = 700 hp ⋅ D1 = 6 in p1 = 150 lb/in2 T1 = 300° F V1 = 100 ft/s D2 = 6 in p2 = 40 lb/in2 T2 = 35° F Q ?
⋅ E3.18 Part (b) Solution The inlet and exit densities can be computed from the perfect-gas law: 1 R p T 1 1 1715 1 ( 5 4 0 6 ( 0 14 4) 300) 0.0166 slug/ft3 2 R p T 2 2 1715 4 ( 0 4 ( 6 1 0 44 ) 35) 0.00679 slug/ft3 The mass flow is determined by the inlet conditions m ˙ 1A1V1 (0.0166) 4 1 6 2 2 (100) 0.325 slug/s Knowing mass flow, we compute the exit velocity m ˙ 0.325 2A2V2 (0.00679) 4 1 6 2 2 V2 or V2 244 ft/s Ans. (a) The steady-flow energy equation (3.65) applies with W ˙ 0, z1 z2, and h ˆ cpT: Q ˙ W ˙s m ˙ (cpT2 1 2 V2 2 cpT1 1 2 V1 2) Convert the turbine work to foot-pounds-force per second with the conversion factor 1 hp 550 ft lbf/s. The turbine work is positive Q ˙ 700(550) 0.325[6003(495) 1 2 (244)2 6003(760) 1 2 (100)2] 510,000 ft lbf/s or Q ˙ 125,000 ft lbf/s Convert this to British thermal units as follows: Q ˙ (125,000 ft lbf/s) 778.
3 2 6 f 0 t 0 s lb /h f/Btu 576,000 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. (3.64) for a given port should be modified by a dimensionless cor-rection factor so that the integral can be proportional to the square of the average velocity through the port port ( 1 2 V2) (V n) dA ( 1 2 Vav 2 )m ˙ where Vav A 1 u dA for incompressible flow 170 Chapter 3 Integral Relations for a Control Volume Part (a) Kinetic-Energy Correction Factor 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, 1 2 u3dA 1 2 Vav 3 A or A 1 V u av 3 dA (3.70) The term 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 com-plete incompressible steady-flow energy equation (3.69), including pumps, turbines, and losses, would generalize to p g 2 g V2 zin p g 2 g V2 zout hturbine hpump hfriction (3.71) where the head terms on the right (ht, hp, hf) are all numerically positive. All additive terms in Eq. (3.71) have dimensions of length {L}. In problems involving turbulent pipe flow, it is common to assume that 1.0. To compute numerical values, we can use these approximations to be discussed in Chap. 6: Laminar flow: u U0 1 R r 2 from which Vav 0.5U0 and 2.0 (3.72) Turbulent flow: u U01 R r m m 1 7 from which, in Example 3.4, Vav (1 m 2 ) U (2 0 m) Substituting into Eq. (3.70) gives (3.73) and numerical values are as follows: Turbulent flow: These values are only slightly different from unity and are often neglected in elemen-tary turbulent-flow analyses. However, should never be neglected in laminar flow.
(1 m)3(2 m)3 4(1 3m)(2 3m) 3.6 The Energy Equation 171 m 1 5 1 6 1 7 1 8 1 9 1.106 1.077 1.058 1.046 1.037 E3.19 EXAMPLE 3.19 A hydroelectric power plant (Fig. E3.19) takes in 30 m3/s of water through its turbine and dis-charges it to the atmosphere at V2 2 m/s. The head loss in the turbine and penstock system is hf 20 m. Assuming turbulent flow, 1.06, estimate the power in MW extracted by the tur-bine.
172 Chapter 3 Integral Relations for a Control Volume Solution We neglect viscous work and heat transfer and take section 1 at the reservoir surface (Fig. E3.19), where V1 0, p1 patm, and z1 100 m. Section 2 is at the turbine outlet. The steady-flow en-ergy equation (3.71) becomes, in head form, p 1 2 1V g 1 2 z1 p 2 2 2V g 2 2 z2 ht hf p a 1 2 .
( 0 9 6 .
( 8 0 1 ) ) 2 100 m p a 1 2 .0 (9 6 .
( 8 2 1 .0 m m /s /s 2 ) ) 2 0 m ht 20 m The pressure terms cancel, and we may solve for the turbine head (which is positive): ht 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 m ˙ ws ( Q)(ght) (998 kg/m3)(30 m3/s)(9.81 m/s2)(79.8 m) 23.4 E6 kg m2/s3 23.4 E6 N m/s 23.4 MW Ans. 7 The turbine drives an electric generator which probably has losses of about 15 percent, so the net power generated by this hydroelectric plant is about 20 MW.
EXAMPLE 3.20 The pump in Fig. E3.20 delivers water (62.4 lbf/ft3) at 3 ft3/s to a machine at section 2, which is 20 ft higher than the reservoir surface. The losses between 1 and 2 are given by hf KV2 2/(2g), Water 30 m3/s z1 = 100 m z2 = 0 m 2 m/s Turbine 1 E3.20 where K 7.5 is a dimensionless loss coefficient (see Sec. 6.7). Take 1.07. Find the horse-power required for the pump if it is 80 percent efficient.
Solution If the reservoir is large, the flow is steady, with V1 0. We can compute V2 from the given flow rate and the pipe diameter: V2 A Q 2 61.1 ft/s The viscous work is zero because of the solid walls and near-one-dimensional inlet and exit. The steady-flow energy equation (3.71) becomes p 1 2 1V g 1 2 z1 p 2 2 2V g 2 2 z2 hs hf Introducing V1 0, z1 0, and hf KV2 2/(2g), we may solve for the pump head: hs p1 p2 z2 (2 K) V 2g 2 2 The pressures should be in lbf/ft2 for consistent units. For the given data, we obtain hs 20 ft (1.07 7.5) 2 ( ( 6 3 1 2 .1 .2 f f t t / / s s ) 2 2 ) 11 20 497 506 ft The pump head is negative, indicating work done on the fluid. As in Example 3.19, the power delivered is computed from P m ˙ ws Qghs (1.94 slug/ft3)(3.0 ft3/s)(32.2 ft/s2)(507 ft) 94,900 ft lbf/s or hp 55 9 0 4, f 9 t 0 0 lb ft f/ (s lb f h /s p) 173 hp (14.7 10.0)(144) lbf/ft2 62.4 lbf/ft3 3 ft3/s (/4)( 1 3 2 ft)2 3.6 The Energy Equation 173 2 1 z1 = 0 Pump hs (negative) Water Machine D2 = 3 in z2 = 20 ft p2 = 10 lbf/in2 p1 = 14.7 lbf/in2 abs 3.7 Frictionless Flow: The Bernoulli Equation We drop the negative sign when merely referring to the “power” required. If the pump is 80 per-cent efficient, the input power required to drive it is Pinput effic P iency 17 0 3 .8 hp 216 hp Ans.
The inclusion of the kinetic-energy correction factor in this case made a difference of about 1 percent in the result.
Closely related to the steady-flow energy equation is a relation between pressure, ve-locity, and elevation 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 der-ivation 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 which are nearly fric-tionless. This section (and, in more detail, Chap. 8) will address the proper use of the Bernoulli relation.
Consider Fig. 3.15, which is an elemental fixed streamtube control volume of vari-able area A(s) and length ds, where s is the streamline direction. The properties ( , V, p) may vary with s and time but are assumed to be uniform over the cross section A.
The streamtube orientation is arbitrary, with an elevation change dz ds sin . Fric-tion on the streamtube walls is shown and then neglected—a very restrictive assump-tion.
Conservation of mass (3.20) for this elemental control volume yields d d t CV d m ˙ out m ˙ in 0 t d dm ˙ where m ˙ AV and d A ds. Then our desired form of mass conservation is dm ˙ d( AV) t A ds (3.74) 174 Chapter 3 Integral Relations for a Control Volume p+ A A + dA τ = 0 , V θ ds CV dz + d V+ dV p + dp dp 0 0 dp dp dFs ≈ 1 2 dp dA dW ≈ g d (a) (b) ρ ρ ρ ρ p S Fig. 3.15 The Bernoulli equation for frictionless flow along a stream-line: (a) forces and fluxes; (b) net pressure force after uniform sub-traction of p.
This relation does not require an assumption of frictionless flow.
Now write the linear-momentum relation (3.37) in the streamwise direction: dFs d d t CV V d (m ˙ V)out (m ˙ V)in t ( V) A ds d(m ˙ V) where Vs V itself because s is the streamline direction. If we neglect the shear force on the walls (frictionless flow), the forces are due to pressure and gravity. The stream-wise gravity force is due to the weight component of the fluid within the control vol-ume: dFs,grav dW sin A ds sin A dz The pressure force is more easily visualized, in Fig. 3.15b, by first subtracting a uni-form value p from all surfaces, remembering from Fig. 3.7 that the net force is not changed. The pressure along the slanted side of the streamtube has a streamwise com-ponent which acts not on A itself but on the outer ring of area increase dA. The net pressure force is thus dFs,press 1 2 dp dA dp(A dA) A dp to first order. Substitute these two force terms into the linear-momentum relation: dFs A dz A dp t ( V) A ds d(m ˙ V) t VA ds V t A ds m ˙ dV V dm ˙ The first and last terms on the right cancel by virtue of the continuity relation (3.74).
Divide what remains by A and rearrange into the final desired relation: V t ds d p V dV g dz 0 (3.75) This is Bernoulli’s equation for unsteady frictionless flow along a streamline. It is in differential form and can be integrated between any two points 1 and 2 on the stream-line: 2 1 V t ds 2 1 d p 1 2 (V2 2 V1 2) g(z2 z1) 0 (3.76) To evaluate the two remaining integrals, one must estimate the unsteady effect V/t and the variation of density with pressure. At this time we consider only steady (V/t 0) incompressible (constant-density) flow, for which Eq. (3.76) becomes p2 p1 1 2 (V2 2 V1 2) g(z2 z1) 0 or p 1 1 2 V1 2 gz1 p 2 1 2 V2 2 gz2 const (3.77) This is the Bernoulli equation for steady frictionless incompressible flow along a streamline.
3.7 Frictionless Flow: The Bernoulli Equation 175 Relation between the Bernoulli and Steady-Flow Energy Equations Equation (3.77) is a widely used form of the Bernoulli equation for incompressible steady frictionless streamline flow. It is clearly related to the steady-flow energy equa-tion for a streamtube (flow with one inlet and one outlet), from Eq. (3.66), which we state as follows: p 1 1 2 V1 2 gz1 p 2 2 2 V2 2 gz2 (û2 û1 q) ws wv (3.78) This relation is much more general than the Bernoulli equation, because it allows for (1) friction, (2) heat transfer, (3) shaft work, and (4) viscous work (another frictional effect).
If we compare the Bernoulli equation (3.77) with the energy equation (3.78), we see that the Bernoulli equation contains even more restrictions than might first be re-alized. The complete list of assumptions for Eq. (3.77) is as follows: 1.
Steady flow—a common assumption applicable to many flows.
2.
Incompressible flow—acceptable if the flow Mach number is less than 0.3.
3.
Frictionless flow—very restrictive, solid walls introduce friction effects.
4.
Flow along a single streamline—different streamlines may have different “Bernoulli constants” w0 p/ V2/2 gz, depending upon flow conditions.
5.
No shaft work between 1 and 2—no pumps or turbines on the streamline.
6.
No heat transfer between 1 and 2—either added or removed.
Thus our warning: Be wary of misuse of the Bernoulli equation. Only a certain lim-ited set of flows satisfies all six assumptions above. The usual momentum or “me-chanical force” derivation of the Bernoulli equation does not even reveal items 5 and 6, which are thermodynamic limitations. The basic reason for restrictions 5 and 6 is that heat transfer and work transfer, in real fluids, are married to frictional effects, which therefore invalidate our assumption of frictionless flow.
Figure 3.16 illustrates some practical limitations on the use of Bernoulli’s equation (3.77). For the wind-tunnel model test of Fig. 3.16a, the Bernoulli equation is valid in the core flow of the tunnel but not in the tunnel-wall boundary layers, the model sur-face boundary layers, or the wake of the model, all of which are regions with high fric-tion.
In the propeller flow of Fig. 3.16b, Bernoulli’s equation is valid both upstream and downstream, but with a different constant w0 p/ V2/2 gz, caused by the work addition of the propeller. The Bernoulli relation (3.77) is not valid near the propeller blades or in the helical vortices (not shown, see Fig. 1.12a) shed down-stream of the blade edges. Also, the Bernoulli constants are higher in the flowing “slipstream” than in the ambient atmosphere because of the slipstream kinetic en-ergy.
For the chimney flow of Fig. 3.16c, Eq. (3.77) is valid before and after the fire, but with a change in Bernoulli constant that is caused by heat addition. The Bernoulli equa-tion is not valid within the fire itself or in the chimney-wall boundary layers.
The moral is to apply Eq. (3.77) only when all six restrictions can be satisfied: steady incompressible flow along a streamline with no friction losses, no heat transfer, and no shaft work between sections 1 and 2.
176 Chapter 3 Integral Relations for a Control Volume 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 h0 z p/ V2/(2g). In frictionless flow with no work or heat transfer, Eq. (3.77), the EGL has constant height. The hydraulic grade line (HGL) shows the height corre-sponding to elevation and pressure head z p/, that is, the EGL minus the velocity head V2/(2g). The HGL is the height to which liquid would rise in a piezometer tube (see Prob. 2.11) attached to the flow. In an open-channel flow the HGL is identical to the free surface of the water.
Figure 3.17 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 p/ and thus out-line the HGL. The pitot stagnation-velocity tubes measure the total head z p/ V2/(2g), 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 extraction (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 in-creases.
3.7 Frictionless Flow: The Bernoulli Equation 177 Invalid Model Valid Valid (a) (b) Valid Valid, new constant Ambient air Invalid Valid, new constant Invalid (c) Valid Fig. 3.16 Illustration of regions of validity and invalidity of the Bernoulli equation: (a) tunnel model, (b) propeller, (c) chimney.
Hydraulic and Energy Grade Lines Fig. 3.17 Hydraulic and energy grade lines for frictionless flow in a duct.
E3.21 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.
EXAMPLE 3.21 Find a relation between nozzle discharge velocity V2and tank free-surface height h as in Fig.
E3.21. Assume steady frictionless flow.
178 Chapter 3 Integral Relations for a Control Volume Flow 1 2 Energy grade line Hydraulic grade line Arbitrary datum (z = 0) z1 2g V22 V12 2g p1 g z2 p2 g Constant Bernoulli head ρ ρ 1 2 V1 EGL HGL V2 Open jet: p2 = pa V12 2g h = z1 – z2 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 V1 and V2.
Mass conservation is usually a vital part of Bernoulli analyses. If A1 is the tank cross section and A2 the nozzle area, this is approximately a one-dimensional flow with constant density, Eq.
(3.30), A1V1 A2V2 (1) Bernoulli’s equation (3.77) gives p 1 1 2 V1 2 gz1 p 2 1 2 V2 2 gz2 But since sections 1 and 2 are both exposed to atmospheric pressure p1 p2 pa, the pressure terms cancel, leaving V2 2 V1 2 2g(z1 z2) 2gh (2) Eliminating V1 between Eqs. (1) and (2), we obtain the desired result: V2 2 1 2g A h 2 2/A1 2 Ans. (3) Generally the nozzle area A2 is very much smaller than the tank area A1, so that the ratio A2 2/A1 2 is doubly negligible, and an accurate approximation for the outlet velocity is V2 (2gh)1/2 Ans. (4) This formula, discovered by Evangelista Torricelli in 1644, states that the discharge velocity equals the speed which 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 characteristic of gravity-driven flows.
Except for the wall boundary layers, the streamlines from 1 to 2 all behave in the same way, and we can assume that the Bernoulli constant h0 is the same for all the core flow. However, the outlet flow is likely to be nonuniform, not one-dimensional, so that the average velocity is only approximately equal to Torricelli’s result. The engineer will then adjust the formula to include a dimensionless discharge coefficient cd (V2)av A Q 2 cd(2gh)1/2 (5) As discussed in Sec. 6.10, the discharge coefficient of a nozzle varies from about 0.6 to 1.0 as a function of (dimensionless) flow conditions and nozzle shape.
Before proceeding with more examples, we should note carefully that a solution by Bernoulli’s equation (3.77) does not require a control-volume analysis, only a selec-tion of two points 1 and 2 along a given streamline. The control volume was used to derive the differential relation (3.75), but the integrated form (3.77) is valid all along 3.7 Frictionless Flow: The Bernoulli Equation 179 the streamline for frictionless flow with no heat transfer or shaft work, and a control volume is not necessary.
EXAMPLE 3.22 Rework Example 3.21 to account, at least approximately, for the unsteady-flow condition caused by the draining of the tank.
Solution Essentially we are asked to include the unsteady integral term involving V/t from Eq. (3.76).
This will result in a new term added to Eq. (2) from Example 3.21: 2 2 1 V t ds V2 2 V 1 2 2gh (1) Since the flow is incompressible, the continuity equation still retains the simple form A1V1 A2V2 from Example 3.21. To integrate the unsteady term, we must estimate the acceleration all along the streamline. Most of the streamline is in the tank region where V/t dV1/dt. The length of the average streamline is slightly longer than the nozzle depth h. A crude estimate for the integral is thus 2 1 V t ds 2 1 d d V t 1 ds d d V t 1 h (2) But since A1 and A2 are constant, dV1/dt (A2/A1)(dV2/dt). Substitution into Eq. (1) gives 2h A A 2 1 d d V t 2 V2 21 A A1 2 2 2 2gh (3) This is a first-order differential equation for V2(t). It is complicated by the fact that the depth h is variable; therefore h h(t), as determined by the variation in V1(t) h(t) h0 t 0 V1 dt (4) Equations (3) and (4) must be solved simultaneously, but the problem is well posed and can be handled analytically or numerically. We can also estimate the size of the first term in Eq. (3) by using the approximation V2 (2gh)1/2 from the previous example. After differentiation, we ob-tain 2h A A 2 1 d d V t 2 A A 2 1 2 V2 2 (5) which is negligible if A2 A1, as originally postulated.
EXAMPLE 3.23 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. E3.23 is called a venturi tube. Find an expression for the mass flux in the tube as a function of the pressure change.
180 Chapter 3 Integral Relations for a Control Volume E3.23 Solution Bernoulli’s equation is assumed to hold along the center streamline p 1 1 2 V1 2 gz1 p 2 1 2 V2 2 gz2 If the tube is horizontal, z1 z2 and we can solve for V2: V2 2 V1 2 2 p p p1 p2 (1) We relate the velocities from the incompressible continuity relation A1V1 A2V2 or V1 2V2 D D 2 1 (2) Combining (1) and (2), we obtain a formula for the velocity in the throat V2 (1 2 p 4) 1/2 (3) The mass flux is given by m ˙ A2V2 A2 1 2 p 4 1/2 (4) This is the ideal frictionless mass flux. In practice, we measure m ˙ actual cd m ˙ ideal and correlate the discharge coefficient cd.
EXAMPLE 3.24 A 10-cm fire hose with a 3-cm nozzle discharges 1.5 m3/min to the atmosphere. Assuming fric-tionless flow, find the force FB exerted by the flange bolts to hold the nozzle on the hose.
Solution We use Bernoulli’s equation and continuity to find the pressure p1 upstream of the nozzle and then we use a control-volume momentum analysis to compute the bolt force, as in Fig. E3.24.
The flow from 1 to 2 is a constriction exactly similar in effect to the venturi in Example 3.23 for which Eq. (1) gave p1 p2 1 2 (V2 2 V1 2) (1) 3.7 Frictionless Flow: The Bernoulli Equation 181 2 1 p1 HGL p2 E3.24 The velocities are found from the known flow rate Q 1.5 m3/min or 0.025 m3/s: V2 A Q 2 ( 0 /4 .0 ) 2 (0 5 .0 m 3 3 m /s )2 35.4 m/s V1 A Q 1 ( 0 / .
4 0 ) 2 ( 5 0.
m 1 3 m /s )2 3.2 m/s We are given p2 pa 0 gage pressure. Then Eq. (1) becomes p1 1 2 (1000 kg/m3)[(35.42 3.22) m2/s2] 620,000 kg/(m s2) 620,000 Pa gage The control-volume force balance is shown in Fig. E3.24b: Fx FB p1A1 and the zero gage pressure on all other surfaces contributes no force. The x-momentum flux is m ˙V2 at the outlet and m ˙V1 at the inlet. The steady-flow momentum relation (3.40) thus gives FB p1A1 m ˙ (V2 V1) or FB p1A1 m ˙ (V2 V1) (2) Substituting the given numerical values, we find m ˙ Q (1000 kg/m3)(0.025 m3/s) 25 kg/s A1 4 D1 2 4 (0.1 m)2 0.00785 m2 FB (620,000 N/m2)(0.00785 m2) (25 kg/s)[(35.4 3.2) m/s] 4872 N 805 (kg m)/s2 4067 N (915 lbf) Ans.
This gives an idea of why it takes more than one firefighter to hold a fire hose at full discharge.
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 182 Chapter 3 Integral Relations for a Control Volume 2 Water: 1000 kg/m3 1 D1 = 10 cm (a) CV pa = 0 (gage) 1 2 FB 1 2 FB p1 0 0 0 x Control volume (b) D2 = 3 cm Summary 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 that 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.
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 equations were attacked “in the large,” i.e., applied to whole regions of a flow. As such, the typ-ical 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 un-known stresses and forces and moments which make the solution for the desired force difficult or impossible. Finally, every attempt should be made to place the control vol-ume 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 which 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 simplify-ing 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: 1.
Is the control volume nondeforming or nonaccelerating?
2.
Is the flow field steady? Can we change to a steady-flow frame?
3.
Can friction be neglected?
4.
Is the fluid incompressible? If not, is the perfect-gas law applicable?
5.
Are gravity or other body forces negligible?
6.
Is there heat transfer, shaft work, or viscous work?
7.
Are the inlet and outlet flows approximately one-dimensional?
8.
Is atmospheric pressure important to the analysis? Is the pressure hydrostatic on any portions of the control surface?
9.
Are there reservoir conditions which 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 those above, one can avoid pulling Bernoulli’s equation off the shelf when it does not apply.
Summary 183 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 an EES icon, for example, Prob. 3.5, will benefit from the use of the Engineering Equation Solver (EES), while fig-ures labeled with a computer disk may require the use of a computer.
The standard end-of-chapter problems 3.1 to 3.182 (categorized in the problem list below) are followed by word problems W3.1 to W3.7, fundamentals of engineering (FE) exam problems FE3.1 to FE3.10, comprehensive problems C3.1 to C3.4, and design project D3.1.
Problem Distribution Section Topic Problems 3.1 Basic physical laws; volume flow 3.1–3.8 3.2 The Reynolds transport theorem 3.9–3.11 3.3 Conservation of mass 3.12–3.38 3.4 The linear momentum equation 3.39–3.109 3.5 The angular momentum theorem 3.110–3.125 3.6 The energy equation 3.126–3.146 3.7 The Bernoulli equation 3.147–3.182 P3.1 Discuss Newton’s second law (the linear-momentum rela-tion) in these three forms: F ma F d d t (mV) F d d t system V d Are they all equally valid? Are they equivalent? Are some forms better for fluid mechanics as opposed to solid me-chanics?
P3.2 Consider the angular-momentum relation in the form MO d d t system (r V) d 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 u C(R2 r2), where R is the tube radius and r R. Integrate u(r) to find the total volume flow Q through the tube.
P3.4 Discuss whether the following flows are steady or un-steady: (a) flow near an automobile moving at 55 mi/h, (b) flow of the wind past a water tower, (c) flow in a pipe as the downstream valve is opened at a uniform rate, (d) river flow over the spillway of a dam, and (e) flow in the ocean beneath a series of uniform propagating surface waves. Elaborate if these questions seem ambiguous.
P3.5 A theory proposed by S. I. Pai in 1953 gives the follow-ing velocity values u(r) for turbulent (high-Reynolds-num-ber) airflow in a 4-cm-diameter tube: r, cm 0 0.25 0.5 0.75 1.0 1.25 1.5 1.75 2.0 u, m/s 6.00 5.97 5.88 5.72 5.51 5.23 4.89 4.43 0.00 Comment on these data vis-à-vis laminar flow, Prob. 3.3.
Estimate, as best you can, the total volume flow Q through the tube, in m3/s.
P3.6 When a gravity-driven liquid jet issues from a slot in a tank, as in Fig. P3.6, an approximation for the exit veloc-ity distribution is u 2 g (h z ) , where h is the depth of the jet centerline. Near the slot, the jet is horizontal, two-dimensional, and of thickness 2L, as shown. Find a general expression for the total volume flow Q issuing from the slot; then take the limit of your result if L h.
184 Chapter 3 Integral Relations for a Control Volume z = +L x z = –L h z Fig. P3.6 Fig. P3.7 P3.7 Consider flow of a uniform stream U toward a circular cylinder of radius R, as in Fig. P3.7. An approximate the-ory for the velocity distribution near the cylinder is devel-oped in Chap. 8, in polar coordinates, for r R: υr U cos 1 R r2 2 υ U sin 1 R r2 2 where the positive directions for radial (υr) and circum-ferential (υ ) velocities are shown in Fig. P3.7. Compute the volume flow Q passing through the (imaginary) sur-face CC in the figure. (Comment: If CC were far upstream of the cylinder, the flow would be Q 2URb.) U R 2R C C R r θ vr vθ Imaginary surface: Width b into paper P3.8 Consider the two-dimensional stagnation flow of Example 1.10, where u Kx and v Ky, with K 0. Evaluate the volume flow Q, per unit depth into the paper, passing through the rectangular surface normal to the paper which stretches from (x, y) (0, 0) to (1, 1).
P3.9 A laboratory test tank contains seawater of salinity S and density . Water enters the tank at conditions (S1, 1, A1, V1) and is assumed to mix immediately in the tank. Tank water leaves through an outlet A2 at velocity V2. 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 Msalt within the tank.
P3.10 Laminar steady flow, through a tube of radius R and length L, is being heated at the wall. The fluid entered the tube at uniform temperature T0 Tw/3. As the fluid exits the tube, its axial velocity and enthalpy profiles are approxi-mated by u U01 R r2 2 h cp 2 Tw 1 R r2 2 cp const (a) Sketch these profiles and comment on their physical realism. (b) Compute the total flux of enthalpy through the exit section.
P3.11 A room contains dust of uniform concentration C dust/ . It is to be cleaned up by introducing fresh air at velocity Vi through a duct of area Ai on one wall and ex-hausting the room air at velocity V0 through a duct A0 on the opposite wall. Find an expression for the instantaneous rate of change of dust mass within the room.
P3.12 Water at 20°C flows steadily through a closed tank, as in Fig. P3.12. At section 1, D1 6 cm and the volume flow is 100 m3/h. At section 2, D2 5 cm and the average ve-locity is 8 m/s. If D3 4 cm, what is (a) Q3 in m3/h and (b) average V3 in m/s?
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 (Q1, Q2, Q3) and tank diameter d. Then, if the water level h is con-stant, determine the exit velocity V2 for the given data V1 3 m/s and Q3 0.01 m3/s.
Problems 185 1 2 3 Water P3.12 P3.13 P3.14 P3.15 P3.13 Water at 20°C flows steadily at 40 kg/s through the noz-zle in Fig. P3.13. If D1 18 cm and D2 5 cm, compute the average velocity, in m/s, at (a) section 1 and (b) sec-tion 2.
2 1 Water D2 = 7 cm Q3 = 0.01 m3/s D1 = 5 cm d 2 3 1 h P3.15 Water, assumed incompressible, flows steadily through the round pipe in Fig. P3.15. The entrance velocity is constant, u U0, and the exit velocity approximates turbulent flow, u umax(1 r/R)1/7. Determine the ratio U0/umax for this flow.
r r = R x = 0 U0 x = L u(r) P3.16 An incompressible fluid flows past an impermeable flat plate, as in Fig. P3.16, with a uniform inlet profile u U0 and a cubic polynomial exit profile u U0 3 2 3 where y Compute the volume flow Q across the top surface of the control volume.
P3.17 Incompressible steady flow in the inlet between parallel plates in Fig. P3.17 is uniform, u U0 8 cm/s, while downstream the flow develops into the parabolic laminar profile u az(z0 z), where a is a constant. If z0 4 cm and the fluid is SAE 30 oil at 20°C, what is the value of umax in cm/s?
P3.19 A partly full water tank admits water at 20°C and 85 N/s weight flow while ejecting water on the other side at 5500 cm3/s. The air pocket in the tank has a vent at the top and is at 20°C and 1 atm. If the fluids are approximately in-compressible, how much air in N/h is passing through the vent? In which direction?
P3.20 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 flux in mL/s and (b) the average outlet velocity in cm/s.
186 Chapter 3 Integral Relations for a Control Volume P3.17 P3.20 P3.18 P3.22 z = z0 umax z = 0 U0 h = 2 mm D = 10 cm 2 1 2 D1 = 3 mm 1 2 Air D1 = 1 cm D2 = 2.5 cm P3.18 An incompressible fluid flows steadily through the rec-tangular duct in Fig. P3.18. The exit velocity profile is given approximately by u umax1 b y2 2 1 h z2 2 (a) Does this profile satisfy the correct boundary condi-tions for viscous fluid flow? (b) Find an analytical expres-sion for the volume flow Q at the exit. (c) If the inlet flow is 300 ft3/min, estimate umax in m/s for b h 10 cm.
P3.21 A dehumidifier brings in saturated wet air (100 percent rel-ative humidity) at 30°C and 1 atm, through an inlet of 8-cm diameter and average velocity 3 m/s. After some of the water vapor condenses and is drained off at the bottom, the somewhat drier air leaves at approximately 30°C, 1 atm, and 50 percent relative humidity. For steady opera-tion, estimate the amount of water drained off in kg/h. (This problem is idealized from a real dehumidifier.) P3.22 The converging-diverging nozzle shown in Fig. P3.22 ex-pands and accelerates dry air to supersonic speeds at the exit, where p2 8 kPa and T2 240 K. At the throat, p1 284 kPa, T1 665 K, and V1 517 m/s. For steady com-pressible flow of an ideal gas, estimate (a) the mass flow in kg/h, (b) the velocity V2, and (c) the Mach number Ma2.
P3.16 y = 0 CV U0 U0 y = δ Q?
Solid plate, width b into paper Cubic Inlet flow L 2h 2b z x, u y 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 cm3/s, 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?
P3.26 A thin layer of liquid, draining from an inclined plane, as in Fig. P3.26, will have a laminar velocity profile u U0(2y/h y2/h2), where U0 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. Esti-mate U0 in ft/s.
Problems 187 y x θ u (y) g h P3.26 P3.27 P3.25 P3.23 P3.24 D1 = 0.75 in D2 = 0.030 in V2 U0 U0 u Width b into paper Dead air (negligible velocity) Exponential curve U + ∆U z L 2 C L P3.24 Water enters the bottom of the cone in Fig. P3.24 at a uni-formly increasing average velocity V Kt. If d is very small, derive an analytic formula for the water surface rise h(t) for the condition h 0 at t 0. Assume incompress-ible flow.
P3.27 The cone frustum in Fig. P3.27 contains incompressible liquid to depth h. A solid piston of diameter d penetrates the surface at velocity V. Derive an analytic expression for the rate of rise dh/dt of the liquid surface.
P3.28 Consider a cylindrical water tank of diameter D and wa-ter depth h. According to elementary theory, the flow rate from a small hole of area A in the bottom of the tank would be Q CA 2 g h , where C 0.61. If the initial water level is h0 and the hole is opened, derive an expression for the time required for the water level to drop to 1 3 h0.
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 ˙ C , where is the air density in the tank and C is a constant. If 0 is the initial density in a tank of volume , derive a formula for the density change (t) after the hole is opened. Apply your formula to the following 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 re-quired for the tank density to drop by 50 percent.
h(t) V = Kt Diameter d Cone Piston V R d h Cone P3.25 As will be discussed in Chaps. 7 and 8, the flow of a stream U0 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 veloc-ity, decaying vertically above the wake according to the variation u U0 U ez/L, where L is the plate height and z 0 is the top of the wake. Find U as a function of stream speed U0.
P3.30 The V-shaped tank in Fig. P3.30 has width b into the pa-per and is filled from the inlet pipe at volume flow Q. De-rive expressions for (a) the rate of change dh/dt and (b) the time required for the surface to rise from h1 to h2.
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 Vs 8 m/s, and the test-section entrance velocity is V1 35 m/s. Assuming incompressible steady flow of air at 20°C, compute (a) V0, (b) V2, and (c) Vf, in m/s.
188 Chapter 3 Integral Relations for a Control Volume P3.33 P3.34 P3.35 Df = 2.2 m D0 = 2.5 m Vf V2 V1 V0 L = 4 m Test section Ds = 0.8 m Uniform suction 1 3 2 Liquid oxygen: 0.5 slug/s Liquid fuel: 0.1 slug/s 1100° F 15 lbf/in2 D2 = 5.5 in 4000° R 400 lbf/in2 Propellant Propellant Combustion: 1500 K, 950 kPa Exit section De = 18 cm pe = 90 kPa V e = 1150 m/s Te = 750 K 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 bel-lows width into the paper, derive an expression for outlet mass flow m ˙ 0 as a function of stroke (t).
P3.32 Water at 20°C flows steadily through the piping junction in Fig. P3.32, entering section 1 at 20 gal/min. The aver-age 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, estimate the exit velocity from the showerhead jets.
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.
P3.35 In contrast to the liquid 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 mo-lecular weight of 28.
P3.30 P3.31 P3.32 20˚ 20˚ Q h h h Stroke L d < < h m0 (t) θ (t) θ (3) (2) (1) d = 4 cm d = 1.5 cm d = 2 cm P3.36 The jet pump in Fig. P3.36 injects water at U1 40 m/s through a 3-in-pipe and entrains a secondary flow of water U2 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 U3 in m/s.
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.
Problems 189 F Plate Dj = 10 cm Vj = 8 m/s F0 0, V0, D0 ρ P3.40 P3.41 P3.36 P3.38 P3.39 Inlet Mixing region Fully mixed U1 D1 = 3 in U2 U3 D2 = 10 in r h(t) V CV CV V(r)?
V0 Fixed circular disk 1 2 30° D1 = 10 cm D2 = 6 cm P3.37 A solid steel cylinder, 4.5 cm in diameter and 12 cm long, with a mass of 1500 g, falls concentrically through a 5-cm-diameter vertical container filled with oil (SG 0.89). Assuming the oil is incompressible, estimate the oil average velocity in the annular clearance between cylin-der and container (a) relative to the container and (b) rel-ative to the cylinder.
P3.38 An incompressible fluid in Fig. P3.38 is being squeezed outward between two large circular disks by the uniform downward motion V0 of the upper disk. Assuming one-dimensional radial outflow, use the control volume shown to derive an expression for V(r).
P3.39 For the elbow duct in Fig. P3.39, SAE 30 oil at 20°C en-ters section 1 at 350 N/s, where the flow is laminar, and exits at section 2, where the flow is turbulent: u1 Vav,11 R r2 1 2 u2 Vav,21 R r 2 1/7 Assuming steady incompressible flow, compute the force, and its direction, of the oil on the elbow due to momen-tum change only (no pressure change or friction effects) for (a) unit momentum-flux correction factors and (b) ac-tual correction factors 1 and 2.
P3.41 In Fig. P3.41 the vane turns the water jet completely around. Find an expression for the maximum jet velocity V0 if the maximum possible support force is F0.
P3.42 A liquid of density flows through the sudden contraction in Fig. P3.42 and exits to the atmosphere. Assume uniform conditions (p1, V1, D1) at section 1 and (p2, V2, D2) at sec-P3.44 When a uniform stream flows past an immersed thick cylinder, a broad low-velocity wake is created downstream, idealized as a V shape in Fig. P3.44. Pressures p1 and p2 are approximately equal. If the flow is two-dimensional and incompressible, with width b into the paper, derive a P3.46 When a jet strikes an inclined fixed plate, as in Fig. P3.46, it breaks into two jets at 2 and 3 of equal velocity V Vjet but unequal fluxes Q at 2 and (1 )Q at section 3, being a fraction. The reason is that for frictionless flow the fluid can exert no tangential force Ft on the plate. The con-dition Ft 0 enables us to solve for . Perform this analy-sis, and find as a function of the plate angle . Why doesn’t the answer depend upon the properties of the jet?
190 Chapter 3 Integral Relations for a Control Volume 1 2 3 θ F t = 0 (1-α)Q, V αQ, V F n , Q, A, V ρ P3.43 Water at 20°C flows through a 5-cm-diameter pipe which has a 180° vertical bend, 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, p1 165 kPa and p2 134 kPa. Ne-glecting pipe weight, determine the total force which the flanges must withstand for this flow.
Water jet D0 = 5 cm W P3.45 P3.46 formula for the drag force F on the cylinder. Rewrite your result in the form of a dimensionless drag coefficient based on body length CD F/( U2bL).
P3.45 In Fig. P3.45 a perfectly balanced weight and platform are supported by a steady water jet. If the total weight sup-ported is 700 N, what is the proper jet velocity?
P3.47 A liquid jet of velocity Vj and diameter Dj strikes a fixed hollow cone, as in Fig. P3.47, and deflects back as a con-ical sheet at the same velocity. Find the cone angle for which the restraining force F 3 2 Aj Vj 2.
P3.48 The small boat in Fig. P3.48 is driven at a steady speed V0 by a jet of compressed air issuing from a 3-cm-diame-ter hole at Ve 343 m/s. Jet exit conditions are pe 1 atm and Te 30°C. Air drag is negligible, and the hull drag is kV0 2, where k 19 N s2/m2. Estimate the boat speed V0, in m/s.
Atmosphere pa p1 2 1 2 1 P3.42 P3.43 P3.44 2 1 U U 2 L L U U L 2 tion 2. Find an expression for the force F exerted by the fluid on the contraction.
EES P3.49 The horizontal nozzle in Fig. P3.49 has D1 12 in and D2 6 in, with inlet pressure p1 38 lbf/in2absolute and V2 56 ft/s. For water at 20°C, compute the horizontal force provided by the flange bolts to hold the nozzle fixed.
tion 2 at atmospheric pressure and higher temperature, where V2 900 m/s and A2 0.4 m2. Compute the hori-zontal test stand reaction Rx needed to hold this engine fixed.
P3.51 A liquid jet of velocity Vj and area Aj strikes a single 180° bucket on a turbine wheel rotating at angular velocity , as in Fig. P3.51. Derive an expression for the power P de-livered to this wheel at this instant as a function of the sys-tem 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?
Problems 191 P3.48 P3.49 P3.50 P3.51 P3.52 Conical sheet θ Jet F Compressed air V0 V e De = 3 cm Hull drag kV0 2 1 2 Open jet Water P a = 15 lbf/in2 abs 1 2 Rx m fuel Combustion chamber Jet Bucket Wheel, radius R Ω θ θ V1 V2 Top view 2w h Side view P3.47 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 A1 0.5 m2 and V1 250 m/s. The fuel-to-air ratio is 1:30. The air leaves sec-P3.52 The vertical gate in a water channel is partially open, as in Fig. P3.52. Assuming no change in water level and a hydrostatic pressure distribution, derive an expression for the streamwise force Fx on one-half of the gate as a func-tion of ( , h, w, , V1). Apply your result to the case of water at 20°C, V1 0.8 m/s, h 2 m, w 1.5 m, and 50°.
P3.53 Consider incompressible flow in the entrance of a circular tube, as in Fig. P3.53. The inlet flow is uniform, u1 U0.
The flow at section 2 is developed pipe flow. Find the wall drag force F as a function of (p1, p2, , U0, R) if the flow at section 2 is (a) Laminar: u2 umax1 R r2 2 (b) Turbulent: u2 umax1 R r 1/7 the power P delivered to the cart. Also find the cart ve-locity for which (c) the force Fx is a maximum and (d) the power P is a maximum.
P3.56 For the flat-plate boundary-layer flow of Fig. 3.11, assume that the exit profile is given by u U0 sin[y/(2)] for water flow at 20°C: U0 3 m/s, 2 mm, and L 45 cm. Estimate the total drag force on the plate, in N, per unit depth into the paper.
P3.57 Laminar-flow theory [Ref. 3 of Chap. 1, p. 260] gives the following expression for the wake behind a flat plate of length L (see Fig. P3.44 for a crude sketch of wake): u U 1 0.6 64 L x 1/2 exp y 4 2 x U where U is the stream velocity, x is distance downstream of the plate, and y 0 is the plane of the plate. Sketch two wake profiles, for umin 0.9U and umin 0.8U. For these two profiles, evaluate the momentum-flux defect, i.e., the difference between the momentum of a uniform stream U and the actual wake profile. Comment on your results.
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 sup-porting cable.
192 Chapter 3 Integral Relations for a Control Volume 1 2 p2 ≈ pa = 101 kPa h Mercury Water Fy Fx Vc = constant θ , Vj, Aj ρ D = 4 m D0 = 4 cm 60˚ 8 m/s Cable p2 , V2, A2 p1, V1, A1 Pressure ≈ p1 Control volume P3.53 P3.54 P3.55 P3.58 P3.59 r = R 1 U0 r x Friction drag on fluid 2 P3.54 For the pipe-flow-reducing section of Fig. P3.54, D1 8 cm, D2 5 cm, and p2 1 atm. All fluids are at 20°C. If V1 5 m/s and the manometer reading is h 58 cm, es-timate the total force resisted by the flange bolts.
P3.55 In Fig. P3.55 the jet strikes a vane which moves to the right at constant velocity Vc on a frictionless cart. Com-pute (a) the force Fx required to restrain the cart and (b) P3.59 When a pipe flow suddenly expands from A1 to A2, as in Fig. P3.59, low-speed, low-friction eddies appear in the corners and the flow gradually expands to A2 downstream.
Using the suggested control volume for incompressible steady flow and assuming that p p1 on the corner annu-lar ring as shown, show that the downstream pressure is given by p2 p1 V1 2 A A 1 2 1 A A 1 2 Neglect wall friction.
P3.60 Water at 20°C flows through the elbow in Fig. P3.60 and exits to the atmosphere. The pipe diameter is D1 10 cm, while D2 3 cm. At a weight flow rate of 150 N/s, the pressure p1 2.3 atm (gage). Neglecting the weight of wa-ter and elbow, estimate the force on the flange bolts at sec-tion 1.
P3.63 The sluice gate in Fig. P3.63 can control and measure flow in open channels. At sections 1 and 2, the flow is uniform and the pressure is hydrostatic. The channel width is b into the paper. Neglecting bottom friction, derive an expression for the force F required to hold the gate. For what condi-tion h2/h1 is the force largest? For very low velocity V 1 2 gh1, for what value of h2/h1 will the force be one-half of the maximum?
Problems 193 F Water Vj = 50 ft/s Dj = 2 in θ 30˚ 30˚ 3 2 1 V1 h1 F A h2 Sluice gate, width b V2 Plate 25 m/s 25 m/s D1 = 6 cm D2 = 4 cm 40° 1 2 P3.60 P3.61 P3.62 P3.63 P3.64 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 30°, evaluate the horizontal force F required to hold the tank stationary.
P3.62 Water at 20°C exits to the standard sea-level atmosphere through the split nozzle in Fig. P3.62. Duct areas are A1 0.02 m2 and A2 A3 0.008 m2. If p1 135 kPa (ab-solute) and the flow rate is Q2 Q3 275 m3/h, compute the force on the flange bolts at section 1.
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, which this water flow causes on the box.
P3.68 The rocket in Fig. P3.68 has a supersonic exhaust, and the exit pressure pe is not necessarily equal to pa. Show that the force F required to hold this rocket on the test stand is F eAeV e 2 Ae(pe pa). Is this force F what we term the thrust of the rocket?
194 Chapter 3 Integral Relations for a Control Volume 0.1 ft3/s 0.2 ft3/s 0.1 ft3/s P3.65 1 2 Water Scale W?
Fuel mf m0 Oxidizer pa ≠ pe pe , Ae ,Ve e F .
.
P3.66 P3.67 P3.68 P3.70 P3.71 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 diame-ters of 6 cm and equal steady volume flows of 300 m3/h.
What should the scale reading W be in N?
P3.67 Gravel is dumped from a hopper, at a rate of 650 N/s, onto a moving belt, as in Fig. P3.67. The gravel then passes off the end of the belt. The drive wheels are 80 cm in diame-ter and rotate clockwise at 150 r/min. Neglecting system friction and air drag, estimate the power required to drive this belt.
P3.69 The solution to Prob. 3.22 is a mass flow of 218 kg/h with V2 1060 m/s and Ma2 3.41. If the conical section 1–2 in Fig. P3.22 is 12 cm long, estimate the force on these conical walls caused by this high-speed gas flow.
P3.70 The dredger in Fig. P3.70 is loading sand (SG 2.6) onto a barge. The sand leaves the dredger pipe at 4 ft/s with a weight flux of 850 lbf/s. Estimate the tension on the moor-ing line caused by this loading process.
P3.72 When immersed in a uniform stream, a thick elliptical cylinder creates a broad downstream wake, as idealized in P3.71 Suppose that a deflector is deployed at the exit of the jet engine of Prob. 3.50, as shown in Fig. P3.71. What will the reaction Rx on the test stand be now? Is this reaction sufficient to serve as a braking force during airplane land-ing?
30˚ 45° 45° Fig. P3.72. The pressure at the upstream and downstream sections are approximately equal, and the fluid is water at 20°C. If U0 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 CD 2F/( U0 2bL).
mentum changes. (b) Show that Fy 0 only if 0.5. (c) Find the values of and for which both Fx and Fy are zero.
P3.76 The rocket engine of Prob. 3.35 has an initial mass of 250 kg and is mounted on the rear of a 1300-kg racing car.
The rocket is fired up, and the car accelerates on level ground.
If the car has an air drag of kV2, where k 0.65 N s2/m2, and rolling resistance cV, where c 16 N s/m, estimate the velocity of the car after it travels 0.25 mi (1320 ft).
P3.77 Water at 20°C flows steadily through a reducing pipe bend, as in Fig. P3.77. Known conditions are p1 350 kPa, D1 25 cm, V1 2.2 m/s, p2 120 kPa, and D2 8 cm.
Neglecting bend and water weight, estimate the total force which must be resisted by the flange bolts.
Problems 195 P3.72 P3.73 P3.74 P3.75 P3.77 L Width b into paper L L U0 U0 U0 2 60° 120° A B F Water 1 2 pa = 100 kPa 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.
Compute 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.
90 R = 15 cm Horizontal plane y x Vertical plane Radial outflow 6 cm z x 1 cm Fx Fy A V, A (1 – )A P3.74 Water at 20°C flows down through a vertical, 6-cm-diam-eter tube at 300 gal/min, as in Fig. P3.74. The flow then turns horizontally and exits through a 90° radial duct seg-ment 1 cm thick, as shown. If the radial outflow is uni-form and steady, estimate the forces (Fx, Fy, Fz) required to support this system against fluid momentum changes.
P3.75 A jet of liquid of density and area A strikes a block and splits into two jets, as in Fig. P3.75. Assume the same ve-locity V for all three jets. The upper jet exits at an angle and area A. The lower jet is turned 90° downward. Ne-glecting fluid weight, (a) derive a formula for the forces (Fx, Fy) required to support the block against fluid mo-P3.78 A fluid jet of diameter D1 enters a cascade of moving blades at absolute velocity V1 and angle 1, and it leaves at absolute velocity V1 and angle 2, 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 parame-ters.
force exerted by the river on the obstacle in terms of V1, h1, h2, b, , and g. Neglect water friction on the river bottom.
P3.81 Torricelli’s idealization of efflux from a hole in the side of a tank is V 2 g h , as shown in Fig. P3.81. The cylin-drical tank weighs 150 N when empty and contains water at 20°C. The tank bottom is on very smooth ice (static fric-tion coefficient 0.01). The hole diameter is 9 cm. For what water depth h will the tank just begin to move to the right?
196 Chapter 3 Integral Relations for a Control Volume P3.78 P3.79 P3.80 P3.81 P3.82 α1 α2 u 1 β V1 D1 Blades V2 2 β Air jet V V V d = 10 cm 85 V2, h2 V1, h1 Width b into paper 30 cm h Water 1 m Static friction V x V Vj P3.79 Air at 20°C and 1 atm enters the bottom of an 85° coni-cal flowmeter duct at a mass flow of 0.3 kg/s, as shown in Fig. P3.79. It is able to support a centered conical body by steady annular flow around the cone, as shown. The air ve-locity at the upper edge of the body equals the entering velocity. Estimate the weight of the body, in newtons.
P3.80 A river of width b and depth h1 passes over a submerged obstacle, or “drowned weir,” in Fig. P3.80, emerging at a new flow condition (V2, h2). Neglect atmospheric pres-sure, and assume that the water pressure is hydrostatic at both sections 1 and 2. Derive an expression for the P3.82 The model car in Fig. P3.82 weighs 17 N and is to be accelerated from rest by a 1-cm-diameter water jet mov-ing at 75 m/s. Neglecting air drag and wheel friction, estimate the velocity of the car after it has moved for-ward 1 m.
P3.83 Gasoline at 20°C is flowing at V1 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 p1 120 kPa, estimate p2 if the wall friction losses are ne-glected.
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.
EES 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, p1 p2 145 kPa. If the wall friction is negligible, estimate the force of the water on the orifice plate.
P3.88 The boat in Fig. P3.88 is jet-propelled by a pump which develops a volume flow rate Q and ejects water out the stern at velocity Vj. If the boat drag force is F kV2, where k is a constant, develop a formula for the steady forward speed V of the boat.
Problems 197 1 2 V1 p1 Vw 5D V2 p2 D Porous section Vw V Vj Pump Q V(t) h Stopper pa P3.84 P3.85 P3.87 P3.88 P3.90 40 cm 90˚ 25 cm 1 cm P3.86 For the water-jet pump of Prob. 3.36, add the following data: p1 p2 25 lbf/in2, and the distance between sec-tions 1 and 3 is 80 in. If the average wall shear stress be-tween sections 1 and 3 is 7 lbf/ft2, estimate the pressure p3. Why is it higher than p1?
P3.87 Figure P3.87 simulates a manifold flow, with fluid removed from a porous wall or perforated section of pipe. Assume incompressible flow with negligible wall friction and small suction Vw V1. If (p1, V1, Vw, , D) are known, derive expressions for (a) V2 and (b) p2.
P3.89 Consider Fig. P3.36 as a general problem for analysis of a mixing ejector pump. If all conditions (p, , V) are known at sections 1 and 2 and if the wall friction is negligible, derive formulas for estimating (a) V3 and (b) p3.
P3.90 As shown in Fig. P3.90, a liquid column of height h is con-fined in a vertical tube of cross-sectional area A by a stop-per. At t 0 the stopper is suddenly removed, exposing 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. 3.90 to include a linear (laminar) average wall shear stress resistance of the form cV, where c is a constant. Find the differential equation for dV/dt and then solve for V(t), assuming for simplicity that the wall area remains constant.
P3.92 A more involved version of Prob. 3.90 is the elbow-shaped tube in Fig. P3.92, with constant cross-sectional area A and diameter D h, L. Assume incompressible flow, neglect P3.92 P3.96 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.
P3.98 As an extension of Example 3.10, let the plate and its cart (see Fig. 3.10a) be unrestrained horizontally, with frictionless wheels. Derive (a) the equation of motion for cart velocity Vc(t) and (b) 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 conditions of Example 3.10 and a cart mass of 2 kg.
P3.99 Suppose that the rocket motor of Prob. 3.34 is attached to a missile which starts from rest at sea level and moves straight up, as in Fig. E3.12. If the system weighs 950 lbf, which includes 300 lbf of fuel and oxidizer, estimate the velocity and height of the missile (a) after 10 s and (b) af-ter 20 s. Neglect air drag.
P3.100 Suppose that the solid-propellant rocket of Prob. 3.35 is built into a missile of diameter 70 cm and length 4 m.
The system weighs 1800 N, which includes 700 N of propellant. Neglect air drag. If the missile is fired verti-cally from rest at sea level, estimate (a) its velocity and height at fuel burnout and (b) the maximum height it will attain.
P3.101 Modify Prob. 3.100 by accounting for air drag on the mis-sile F C D2V2, where C 0.02, is the air density, D is the missile diameter, and V is the missile velocity. Solve numerically for (a) the velocity and altitude at burnout and (b) the maximum altitude attained.
P3.102 As can often be seen in a kitchen sink when the faucet is running, a high-speed channel flow (V1, h1) may “jump” to a low-speed, low-energy condition (V2, h2) as in Fig.
P3.102. The pressure at sections 1 and 2 is approximately hydrostatic, and wall friction is negligible. Use the conti-nuity and momentum relations to find h2 and V2 in terms of (h1, V1).
198 Chapter 3 Integral Relations for a Control Volume V2 V1 pa h L h1 h3 z z Equilibrium position h2 ≈ 0 Liquid–column length L = h1 + h2 + h3 V P3.93 Extend Prob. 3.92 to include a linear (laminar) average wall shear stress resistance of the form cV, where c is a constant. Find the differential equation for dV/dt and then solve for V(t), assuming for simplicity that the wall area remains constant.
P3.94 Attempt a numerical solution of Prob. 3.93 for SAE 30 oil at 20°C. Let h 20 cm, L 15 cm, and D 4 mm. Use the laminar shear approximation from Sec. 6.4: 8 V/D, where is the fluid viscosity. Account for the de-crease in wall area wetted by the fluid. Solve for the time required to empty (a) the vertical leg and (b) the horizon-tal leg.
P3.95 Attempt a numerical solution of Prob. 3.93 for mercury at 20°C. Let h 20 cm, L 15 cm, and D 4 mm. For mercury the flow will be turbulent, with the wall shear stress estimated from Sec. 6.4: 0.005 V2, where is the fluid density. Account for the decrease in wall area wet-ted by the fluid. Solve for the time required to empty (a) the vertical leg and (b) the horizontal leg. Compare with a frictionless flow solution.
P3.96 Extend Prob. 3.90 to the case of the liquid motion in a fric-tionless U-tube whose liquid column is displaced a dis-tance Z upward and then released, as in Fig. P3.96. Ne-glect the short horizontal leg and combine control-volume analyses for the left and right legs to derive a single dif-ferential equation for V(t) of the liquid column.
P3.97 Extend Prob. 3.96 to include a linear (laminar) average wall shear stress resistance of the form 8 V/D, where is the fluid viscosity. Find the differential equation for dV/dt and then solve for V(t), assuming an initial dis-placement z z0, V 0 at t 0. The result should be a damped oscillation tending toward z 0.
P3.103 Suppose that the solid-propellant rocket of Prob. 3.35 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 M0, and its exit properties are m ˙ and Ve relative to the rocket. Set up the differential equation for rocket motion, and solve for the angular velocity (t) of the rod. Neglect gravity, air drag, and the rod mass.
dips h 2.5 cm into a pond. Neglect air drag and wheel friction. Estimate the force required to keep the cart mov-ing.
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 V0, 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.
Problems 199 y x R m, V e, pe = pa ω, ω .
.
P3.104 Water V0 h P3.107 M V 60˚ h Water P3.108 P3.110 P3.102 Hydraulic jump V1 V2 < V1 h2 > h1 h1 P3.105 Extend Prob. 3.104 to the case where the rocket has a lin-ear air drag force F cV, where c is a constant. Assum-ing no burnout, solve for (t) and find the terminal angu-lar velocity, i.e., the final motion when the angular acceleration is zero. Apply to the case M0 6 kg, R 3 m, m 0.05 kg/s, Ve 1100 m/s, and c 0.075 N s/m to find the angular velocity after 12 s of burning.
P3.106 Extend Prob. 3.104 to the case where the rocket has a qua-dratic air drag force F kV2, where k is a constant. As-suming no burnout, solve for (t) and find the terminal angular velocity, i.e., the final motion when the angular acceleration is zero. Apply to the case M0 6 kg, R 3 m, m 0.05 kg/s, Ve 1100 m/s, and k 0.0011 N s2/m2 to find the angular velocity after 12 s of burning.
P3.107 The cart in Fig. P3.107 moves at constant velocity V0 12 m/s and takes on water with a scoop 80 cm wide which P3.109 Apply Prob. 3.108 to the following case: Mtotal 900 kg, b 60 cm, h 2 cm, V0 120 m/s, with the rocket of Prob. 3.35 attached and burning. Estimate V after 3 s.
P3.110 The horizontal lawn sprinkler in Fig. P3.110 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.
R = 6 in d = in 1 – 4 Q θ θ Q Q x T, Ω 2 2 R0 >> Dpipes R0 θ θ θ d = 7 mm R = 15 cm 50˚ 1 2 3 ft B Vrel, 2 θ2 Blade R2 R1 Q T, P,ω b2 Q 2 cm 2 cm 32 cm P3.112 P3.114 P3.115 P3.116 P3.117 P3.111 In Prob. 3.60 find the torque caused around flange 1 if the center point of exit 2 is 1.2 m directly below the flange center.
P3.112 The wye joint in Fig. P3.112 splits the pipe flow into equal amounts Q/2, which exit, as shown, a distance R0 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 .
P3.117 A simple turbomachine is constructed from a disk with two internal ducts which exit tangentially through square holes, as in Fig. P3.117. 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?
200 Chapter 3 Integral Relations for a Control Volume P3.113 Modify Example 3.14 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 I0. Neglecting air drag, find d/dt and integrate to determine the angular velocity (t), as-suming 0 at t 0.
P3.114 The three-arm lawn sprinkler of Fig. P3.114 receives 20°C water through the center at 2.7 m3/h. If collar friction is negligible, what is the steady rotation rate in r/min for (a) 0° and (b) 40°?
P3.115 Water at 20°C flows at 30 gal/min through the 0.75-in-di-ameter double pipe bend of Fig. P3.115. The pressures are p1 30 lbf/in2 and p2 24 lbf/in2. Compute the torque T at point B necessary to keep the pipe from rotating.
P3.116 The centrifugal pump of Fig. P3.116 has a flow rate Q and exits the impeller at an angle 2 relative to the blades, as shown. The fluid enters axially at section 1. Assuming in-compressible flow at shaft angular velocity , derive a for-mula for the power P required to drive the impeller.
P3.118 Reverse the flow in Fig. P3.116, so that the system oper-ates as a radial-inflow turbine. Assuming that the outflow into section 1 has no tangential velocity, derive an ex-pression for the power P extracted by the turbine.
P3.119 Revisit the turbine cascade system of Prob. 3.78, and de-rive a formula for the power P delivered, using the angular-momentum theorem of Eq. (3.55).
P3.120 A centrifugal pump impeller delivers 4000 gal/min of wa-ter at 20°C with a shaft rotation rate of 1750 r/min. Ne-glect losses. If r1 6 in, r2 14 in, b1 b2 1.75 in, Vt1 10 ft/s, and Vt2 110 ft/s, compute the absolute ve-locities (a) V1 and (b) V2 and (c) the horsepower required.
(d) Compare with the ideal horsepower required.
P3.121 The pipe bend of Fig. P3.121 has D1 27 cm and D2 13 cm. When water at 20°C flows through the pipe at 4000 gal/min, p1 194 kPa (gage). Compute the torque re-quired at point B to hold the bend stationary.
P3.124 A rotating dishwasher arm delivers at 60°C to six nozzles, as in Fig. P3.124. The total flow rate is 3.0 gal/min. Each nozzle has a diameter of 1 3 6 in. If the nozzle flows are equal and friction is neglected, estimate the steady rotation rate of the arm, in r/min.
Problems 201 P3.121 P3.122 P3.123 P3.124 P3.125 1 2 V2, p2 = pa B 50 cm 50 cm V1, p1 C , V V h1 h2 V Fn L h3 ρ 75˚ 4 ft Ω 150 ft/s 5 in 40˚ 5 in 6 in Q x y 0 d <R, L Vw Closed valve R L < P3.122 Extend Prob. 3.46 to the problem of computing the center of pressure L of the normal face Fn, as in Fig. P3.122. (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 h1 and the angle between the plate and the oncoming jet 1.
P3.123 The waterwheel in Fig. P3.123 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 r/min will the horsepower developed be a maximum? Assume that there are many buckets on the waterwheel.
P3.125 A liquid of density flows through a 90° bend as shown in Fig. P3.125 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.
P3.126 P3.130 P3.132 P3.127 P3.127 A power plant on a river, as in Fig. P3.127, must elimi-nate 55 MW of waste heat to the river. The river condi-tions upstream are Qi 2.5 m3/s and Ti 18°C. The river is 45 m wide and 2.7 m deep. If heat losses to the atmos-phere and ground are negligible, estimate the downstream river conditions (Q0, T0).
P3.129 Multnomah Falls in the Columbia River Gorge has a sheer drop of 543 ft. Using the steady-flow energy equation, esti-mate the water temperature change in °F caused by this drop.
P3.130 When the pump in Fig. P3.130 draws 220 m3/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.
202 Chapter 3 Integral Relations for a Control Volume Q0, T0 Qi, Ti Power plant Q Q T + ∆T T Water 6 m 2 m Pump Ve De = 5 cm D = 12 cm 1 2 3 Isothermal steady flow P3.126 There is a steady isothermal flow of water at 20°C through the device in Fig. P3.126. Heat-transfer, gravity, and tem-perature effects are negligible. Known data are D1 9 cm, Q1 220 m3/h, p1 150 kPa, D2 7 cm, Q2 100 m3/h, p2 225 kPa, D3 4 cm, and p3 265 kPa. Com-pute the rate of shaft work done for this device and its di-rection.
P3.128 For the conditions of Prob. 3.127, 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 m3/s, through the plant heat exchanger? How will the value of Q affect the downstream conditions (Q0, T0)?
P3.131 When the pump in Fig. P3.130 delivers 25 kW of power to the water, the friction head loss is 4 m. Estimate (a) the exit velocity Ve and (b) the flow rate Q.
P3.132 Consider a turbine extracting energy from a penstock in a dam, as in Fig. P3.132. For turbulent pipe flow (Chap. 6), the friction head loss is approximately hf CQ2, where the constant C depends upon penstock dimensions and the properties of water. Show that, for a given penstock geom-etry and variable river flow Q, the maximum turbine power possible in this case is Pmax 2 gHQ/3 and occurs when the flow rate is Q H /( 3 C ) .
Penstock Q Turbine H P3.133 The long pipe in Fig. P3.133 is filled with water at 20°C.
When valve A is closed, p1 p2 75 kPa. When the valve is open and water flows at 500 m3/h, p1 p2 160 kPa.
What is the friction head loss between 1 and 2, in m, for the flowing condition?
P3.134 A 36-in-diameter pipeline carries oil (SG 0.89) at 1 mil-lion 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. Esti-mate the horsepower which must be delivered to the oil by each pump.
P3.135 The pump-turbine system in Fig. P3.135 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 tur-bine and (b) delivered by the pump.
P3.138Students in the fluid mechanics laboratory at Penn State use a very simple device to measure the viscosity of water as a function of temperature. The viscometer, shown in Fig.
P3.138, consists of a tank, a long vertical capillary tube, a graduated cylinder, a thermometer, and a stopwatch. Because the tube has such a small diameter, the flow remains lami-nar. Because the tube is so long, entrance losses are negligi-ble. It will be shown in Chap. 6 that the laminar head loss through a long pipe is given by hf, laminar (32 LV)/( gd2), where V is the average speed through the pipe. (a) In a given experiment, diameter d, length L, and water level height H are known, and volume flow rate Q is measured with the stopwatch and graduated cylinder. The temperature of the water is also measured. The water density at this tempera-Problems 203 Water level H L Q d P3.138 P3.133 P3.137 P3.135 1 2 Constant-diameter pipe A Water at 20˚C Pump-turbine 1 2 Z1 = 150 ft Z 2 = 25 ft D = 2 in 120 ft/s D = 6 in Pump 6 ft 10 ft P3.136 A pump is to deliver water at 20°C from a pond to an el-evated tank. The pump is 1 m above the pond, and the tank free surface is 20 m above the pump. The head loss in the system is hf cQ2, where c 0.08 h2/m5. If the pump is 72 percent efficient and is driven by a 500-W motor, what flow rate Q m3/h will result?
P3.137 A fireboat draws seawater (SG 1.025) from a submerged pipe and discharges it through a nozzle, as in Fig. P3.137.
The total head loss is 6.5 ft. If the pump efficiency is 75 percent, what horsepower motor is required to drive it?
P3.139 P3.141 P3.142 P3.143 P3.144 ture is obtained by weighing a known volume of water. Write an expression for the viscosity of the water as a function of these variables. (b) Here are some actual data from an ex-periment: T 16.5°C, 998.7 kg/m3, d 0.041 in, Q 0.310 mL/s, L 36.1 in, and H 0.153 m. Calculate the viscosity of the water in kg/(m s) based on these experi-mental data. (c) Compare the experimental result with the published value of at this temperature, and report a per-centage error. (d) Compute the percentage error in the cal-culation of which would occur if a student forgot to in-clude the kinetic energy flux correction factor in part (b) above (compare results with and without inclusion of kinetic energy flux correction factor). Explain the importance (or lack of importance) of kinetic energy flux correction factor in a problem such as this.
P3.139 The horizontal pump in Fig. P3.139 discharges 20°C wa-ter at 57 m3/h. Neglecting losses, what power in kW is de-livered to the water by the pump?
P3.143 The insulated tank in Fig. P3.143 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 initial mass flow rate into the tank is 0.013 kg/s. Assum-ing an ideal gas, estimate the initial rate of temperature rise of the air in the tank.
204 Chapter 3 Integral Relations for a Control Volume D2 = 3 cm D1 = 9 cm 120 kPa 400 kPa Pump z2 = 150 ft z1 = 50 ft D = 6 in Pump Head, ft Flow rate, ft3/s 300 200 100 0 0 1 2 3 4 Pump performance 2 m 25 m Jet Pump 15 m D = 10 cm De = 5 cm P3.140 Steam enters a horizontal turbine at 350 lbf/in2 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 lbm/s, and the heat losses are 7 Btu/lb of steam. If head losses are negli-gible, how much horsepower does the turbine develop?
P3.141 Water at 20°C is pumped at 1500 gal/min from the lower to the upper reservoir, as in Fig. P3.141. Pipe friction losses are approximated by hf 27V2/(2g), where V is the aver-age velocity in the pipe. If the pump is 75 percent effi-cient, what horsepower is needed to drive it?
P3.144 The pump in Fig. P3.144 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 de-livered by the pump?
is 75 percent efficient and is used for the system in Prob.
3.141. Estimate (a) the flow rate, in gal/min, and (b) the horsepower needed to drive the pump.
P3.142 A typical pump has a head which, for a given shaft rota-tion rate, varies with the flow rate, resulting in a pump per-formance curve as in Fig. P3.142. Suppose that this pump Valve Air supply: Tank : = 200 L T1 = 20°C P1 = 1500 kPa P3.145 The large turbine in Fig. P3.145 diverts the river flow un-der a dam as shown. System friction losses are hf 3.5V2/(2g), where V is the average velocity in the supply EES EES pipe. For what river flow rate in m3/s will the power ex-tracted be 25 MW? Which of the two possible solutions has a better “conversion efficiency”?
P3.146 Kerosine at 20°C flows through the pump in Fig. P3.146 at 2.3 ft3/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?
hf 2 V g 2 1 1 A A 1 2 2 See Sec. 6.7 for further details.
P3.151 In Prob. 3.63 the velocity approaching the sluice gate was assumed to be known. If Bernoulli’s equation is valid with no losses, derive an expression for V1 as a function of only h1, h2, and g.
P3.152 A free liquid jet, as in Fig. P3.152, has constant ambient pressure and small losses; hence from Bernoulli’s equa-tion z V2/(2g) is constant along the jet. For the fire noz-zle in the figure, what are (a) the minimum and (b) the maximum values of for which the water jet will clear the corner of the building? For which case will the jet veloc-ity be higher when it strikes the roof of the building?
Problems 205 Turbine D = 4 m z2 = 10 m z3 = 0 m z1 = 50 m Mercury V1 h?
V2 5 ft D2 = 6 in D1 = 3 in Pump Alcohol, SG = 0.79 F V1 –V2 D1 = 5 cm D2 = 2 cm pa = 101 kPa 40 ft V1 = 100 ft/s X 50 ft θ P3.145 P3.153 P3.152 P3.149 P3.146 P3.147 Repeat Prob. 3.49 by assuming that p1 is unknown and us-ing Bernoulli’s equation with no losses. Compute the new bolt force for this assumption. What is the head loss be-tween 1 and 2 for the data of Prob. 3.49?
P3.148 Reanalyze Prob. 3.54 to estimate the manometer reading h if Bernoulli’s equation is valid with zero losses. For the reading h 58 cm in Prob. 3.54, what is the head loss be-tween sections 1 and 2?
P3.149 A jet of alcohol strikes the vertical plate in Fig. P3.149. 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 (b) the absolute pressure at section 1.
P3.150 Verify that Bernoulli’s equation is not valid for the sudden expansion of Prob. 3.59 and that the actual head loss is given by P3.153 For the container of Fig. P3.153 use Bernoulli’s equation to derive a formula for the distance X where the free jet h Free jet H X P3.154 P3.155 P3.157 P3.158 P3.160 leaving horizontally will strike the floor, as a function of h and H. For what ratio h/H will X be maximum? Sketch the three trajectories for h/H 0.4, 0.5, and 0.6.
P3.154 In Fig. P3.154 the exit nozzle is horizontal. If losses are negligible, what should the water level h cm be for the free jet to just clear the wall?
P3.161 A necked-down section in a pipe flow, called a venturi, de-velops a low throat pressure which can aspirate fluid up-ward from a reservoir, as in Fig. P3.161. Using Bernoulli’s 206 Chapter 3 Integral Relations for a Control Volume 40 cm 80 cm h 30 cm Thin wall Jet Jet 1 in 3 in D1 = 10 cm D2 = 6 cm 8 cm h = 3 cm W = 50 kN V D = 6 m 1 P3.155 Bernoulli’s 1738 treatise Hydrodynamica contains many excellent sketches of flow patterns related to his friction-less relation. One, however, redrawn here as Fig. P3.155, seems physically misleading. Can you explain what might be wrong with the figure?
P3.156 A blimp cruises at 75 mi/h through sea-level standard air.
A differential pressure transducer connected between the nose and the side of the blimp registers 950 Pa. Estimate (a) the absolute pressure at the nose and (b) the absolute velocity of the air near the blimp side.
P3.157 The manometer fluid in Fig. P3.157 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.
P3.158 In Fig. P3.158 the flowing fluid is CO2 at 20°C. Neglect losses. If p1 170 kPa and the manometer fluid is Meriam red oil (SG 0.827), estimate (a) p2 and (b) the gas flow rate in m3/h.
P3.159 Our 0.625-in-diameter hose is too short, and it is 125 ft from the 0.375-in-diameter nozzle exit to the garden. If losses are neglected, what is the minimum gage pressure required, inside the hose, to reach the garden?
P3.160 The air-cushion vehicle in Fig. P3.160 brings in sea-level standard air through a fan and discharges it at high veloc-ity through an annular skirt of 3-cm clearance. If the ve-hicle weighs 50 kN, estimate (a) the required airflow rate and (b) the fan power in kW.
equation with no losses, derive an expression for the ve-locity V1 which is just sufficient to bring reservoir fluid into the throat.
P3.162 Suppose you are designing an air hockey table. The table is 3.0 6.0 ft in area, with 1 1 6 -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 which will meet the requirements. Estimate the volumetric flow rate (in ft3/min) and pressure rise (in lb/in2) required of the blower. Hint: Assume that the air is stagnant in the large volume of the manifold under the table surface, and neglect any frictional losses.
P3.163 The liquid in Fig. P3.163 is kerosine at 20°C. Estimate the flow rate from the tank for (a) no losses and (b) pipe losses hf 4.5V2/(2g).
If the pressure at the centerline at section 1 is 110 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.
P3.165 A venturi meter, shown in Fig. P3.165, is a carefully de-signed constriction whose pressure difference is a measure 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 Q where M is the density of the manometer fluid.
2gh( M ) A2 1 ( D 2/ D 1) 4 Problems 207 Water V1 V2, p2 = pa h pa Water D2 D1 P3.161 P3.164 5 ft D = 1 in V pa = 14.7 lbf/in2 abs p = 20 lbf/in2 abs Air: h 1 2 p1 8 cm Open jet 2 5 cm 12 m P3.163 P3.165 P3.167 P3.164 In Fig. P3.164 the open jet of water at 20°C exits a noz-zle into sea-level air and strikes a stagnation tube as shown.
Sea-level air 12 cm (1) Open jet 4 cm H Water P3.166 An open-circuit wind tunnel draws in sea-level standard air and accelerates it through a contraction into a 1-m by 1-m test section. A differential transducer mounted in the test section wall measures a pressure difference of 45 mm of water between the inside and outside. Estimate (a) the test section velocity in mi/h and (b) the absolute pressure on the front nose of a small model mounted in the test sec-tion.
P3.167 In Fig. P3.167 the fluid is gasoline at 20°C at a weight flux of 120 N/s. Assuming no losses, estimate the gage pres-sure at section 1.
P3.168 In Fig. P3.168 both fluids are at 20°C. If V1 1.7 ft/s and losses are neglected, what should the manometer reading h ft be?
P3.168 P3.170 P3.171 P3.172 P3.169 P3.170 If losses are neglected in Fig. P3.170, for what water level h will the flow begin to form vapor cavities at the throat of the nozzle?
P3.171 For the 40°C water flow in Fig. P3.171, estimate the vol-ume flow through the pipe, assuming no losses; then ex-plain what is wrong with this seemingly innocent ques-tion. If the actual flow rate is Q 40 m3/h, compute (a) the head loss in ft and (b) the constriction diameter D which causes cavitation, assuming that the throat divides the head loss equally and that changing the constriction causes no additional losses.
P3.172 The 35°C water flow of Fig. P3.172 discharges to sea-level standard atmosphere. Neglecting losses, for what nozzle diameter D will cavitation begin to occur? To avoid cavi-tation, should you increase or decrease D from this criti-cal value?
208 Chapter 3 Integral Relations for a Control Volume 2 1 D2 = 8 cm D1 = 5 cm pa = 100 kPa Water at 30°C h Open jet 25 m 10 m 5 cm D 3 in 1 1 in 2 10 ft 2 ft Mercury h Water 3 1 2 h L H V2 P3.169 Once it has been started by sufficient suction, the siphon in Fig. P3.169 will run continuously as long as reservoir fluid is available. Using Bernoulli’s equation with no losses, show (a) that the exit velocity V2 depends only upon gravity and the distance H and (b) that the lowest (vac-uum) pressure occurs at point 3 and depends on the dis-tance L H.
P3.173 The horizontal wye fitting in Fig. P3.173 splits the 20°C water flow rate equally. If Q1 5 ft3/s and p1 25 lbf/in2(gage) and losses are neglected, estimate (a) p2, (b) p3, and (c) the vector force required to keep the wye in place.
1 2 3 D 3 in 1 in 6 ft P3.174 In Fig. P3.174 the piston drives water at 20°C. Neglecting losses, estimate the exit velocity V2 ft/s. If D2 is further constricted, what is the maximum possible value of V2?
stream depth h2, and show that two realistic solutions are possible.
P3.178 For the water-channel flow of Fig. P3.178, h1 0.45 ft, H 2.2 ft, and V1 16 ft/s. Neglecting losses and as-suming uniform flow at sections 1 and 2, find the down-stream depth h2; show that two realistic solutions are pos-sible.
Problems 209 P3.173 P3.174 P3.175 P3.176 P3.177 P3.178 1 D1 = 6 in 50° 30° D2 = 3 in D3 = 4 in Water Q1 1 2 Q1 2 3 Water D1 = 8 in D2 = 4 in V2 p a F = 10 lbf p a 10 cm 30 cm 2 m V 1 Water 5 m V 1 V 2 0.7 m h1 h2 V 2 V 1 H h2 h1 V 1 V 2 H P3.175 If the approach velocity is not too high, a hump in the bot-tom of a water channel causes a dip h in the water level, which can serve as a flow measurement. If, as shown in Fig. P3.175, h 10 cm when the bump is 30 cm high, what is the volume flow Q1 per unit width, assuming no losses? In general, is h proportional to Q1?
P3.176 In the spillway flow of Fig. P3.176, the flow is assumed uniform and hydrostatic at sections 1 and 2. If losses are neglected, compute (a) V2 and (b) the force per unit width of the water on the spillway.
P3.177 For the water-channel flow of Fig. P3.177, h1 1.5 m, H 4 m, and V1 3 m/s. Neglecting losses and as-suming uniform flow at sections 1 and 2, find the down-P3.179 A cylindrical tank of diameter D contains liquid to an ini-tial height h0. At time t 0 a 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 expres-sion for the time t0 to drain the entire tank.
P3.180 The large tank of incompressible liquid in Fig. P3.180 is at rest when, at t 0, the valve is opened to the atmos-phere. Assuming h constant (negligible velocities and accelerations in the tank), use the unsteady frictionless Bernoulli equation to derive and solve a differential equa-tion for V(t) in the pipe.
P3.181 Modify Prob. 3.180 as follows. Let the top of the tank be enclosed and under constant gage pressure p0. Repeat the analysis to find V(t) in the pipe.
P3.182 The incompressible-flow form of Bernoulli’s relation, Eq.
(3.77), 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 flu-ids is isentropic flow of an ideal gas, or p C k, where k cp/c. Substitute this relation into Eq. (3.75), integrate, and eliminate the constant C. Compare your compressible result with Eq. (3.77) and comment.
210 Chapter 3 Integral Relations for a Control Volume P3.180 h ≈ constant L Valve V (t) D Word Problems W3.1 Derive a control-volume form of the second law of ther-modynamics. Suggest some practical uses for your rela-tion in analyzing real fluid flows.
W3.2 Suppose that it is desired to estimate volume flow Q 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 z. Why does the incompressible frictionless Bernoulli 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 p0. 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 . Antoine Chézy in 1768 determined that the average velocity of such an open-channel flow should be V CR t an , 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. Ex-plain why.
W3.7 How does a siphon work? Are there any limitations (e.g., how high or how low can you siphon water away from a tank)? Also, how farcould you use a flexible tube to siphon water from a tank to a point 100 ft away?
Fundamentals of Engineering Exam Problems FE3.1 In Fig. FE3.1 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 friction is neglected, what is the gage pressure at sec-tion 1?
(a) 1.4 kPa, (b) 32 kPa, (c) 43 kPa, (d) 29 kPa, (e) 123 kPa FE3.3 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the exit velocity is V2 8 m/s and h (1) (2) Jet 7 cm 4 cm patm = 101 kPa FE3.1 friction is neglected, what is the axial flange force required to keep the nozzle attached to pipe 1?
(a) 11 N, (b) 56 N, (c) 83 N, (d) 123 N, (e) 110 N FE3.4 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 FE3.5 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 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 gal/min, (c) 214 gal/min, (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. Up-stream pressure is 120 kPa. If cavitation occurs in the throat at a flow rate of 155 gal/min, what is the esti-mated fluid vapor pressure, assuming ideal frictionless flow?
(a) 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. Up-stream 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 Comprehensive Problems 211 V 3 cm F = 23 N FE3.5 FE3.6 Comprehensive Problems C3.1 In a certain industrial process, oil of density flows through the inclined pipe in Fig. C3.1. A U-tube manome-ter, with fluid density m, 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 p1 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?
C3.2 A rigid tank of volume 1.0 m3 is initially filled with air at 20°C and p0 100 kPa. At time t 0, a vacuum pump is turned on and evacuates air at a constant volume flow rate Q 80 L/min (regardless of the pressure). As-sume an ideal gas and an isothermal process. (a) Set up a differential equation for this flow. (b) Solve this equation for t as a function of (, Q, p, p0). (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. 3.40 (jet ve-locity 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, Ve Pump 120 cm 70 cm patm d = 4 cm d = 12 cm (1) (2) Water 4 m/s. (Looking from the left, the exit jet is a circular an-nulus 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 phys-ical explanation as to why F has changed.
C3.4 The air flow underneath an air hockey puck is very com-plex, especially since the air jets from the air hockey 212 Chapter 3 Integral Relations for a Control Volume R F h Ve Vj Ve (1) s h (2) m L table impinge on the underside of the puck at various points nonsymmetrically. A reasonable approximation is that at any given time, the gage pressure on the bottom of the puck is halfway between zero (i.e., atmospheric pressure) and the stagnation pressure of the impinging jets. (Stagnation pressure is defined as p0 1 2 Vjet 2 .) (a) 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 of the air. (b) For W 0.05 lbf and d 2.5 in, estimate the required jet veloc-ity in ft/s.
C3.1 C3.3 Design Project D3.1 Let us generalize Probs. 3.141 and 3.142, in which a pump performance curve was used to determine the flow rate be-tween reservoirs. The particular pump in Fig. P3.142 is one of a family of pumps of similar shape, whose dimen-sionless performance is as follows: Head: 6.04 161 n g 2D h p 2 and n Q Dp 3 Efficiency: 70 91,500 3 po p w ow er er to in w p a u t t er 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 va-lidity is 0 0.027. The pump of Fig. P3.142 had Dp 2 ft in diameter and rotated at n 20 r/s (1200 r/min). The solution to Prob. 3.142, namely, Q 2.57 ft3/s and hp 172 ft, corresponds to 3.46, 0.016, 0.75 (or 75 percent), and power to the water gQhp 27,500 ft lbf/s (50 hp). Please check these numerical values before beginning this project.
Now restudy Prob. 3.142 to select a low-cost pump which rotates at a rate no slower than 600 r/min and de-livers no less than 1.0 ft3/s of water. Assume that the cost of the pump is linearly proportional to the power input re-quired. Comment on any limitations to your results.
References 213 References 1. D. T. Greenwood, Principles of Dynamics, Prentice-Hall, En-glewood Cliffs, NJ, 1965.
2. T. von Kármán, The Wind and Beyond, Little, Brown, Boston, 1967.
3. J. P. Holman, Heat Transfer, 7th ed., McGraw-Hill, New York, 1990.
4. A. G. Hansen, Fluid Mechanics, Wiley, New York, 1967.
5. M. C. Potter and J. F. Foss, Fluid Mechanics, Ronald, New York, 1975.
6. G. J. Van Wylen and R. E. Sonntag, Fundamentals of Classical Thermodynamics, 3d ed., Wiley, New York, 1985.
7. W. C. Reynolds and H. C. Perkins, Engineering Thermody-namics, 2d ed., McGraw-Hill, New York, 1977.
Inviscid potential flow past an array of cylinders. The mathematics of potential theory, pre-sented in this chapter, is both beautiful and manageable, but results may be unrealistic when there are solid boundaries. See Figure 8.13b for the real (viscous) flow pattern. (Courtesy of Tecquipment Ltd., Nottingham, England) 214 4.1 The Acceleration Field of a Fluid 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 re-gion 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 viewpoint 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 sys-tem. In either case the results yield the basic differential equations of fluid motion. Ap-propriate 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. However, certain things can be done which have great educational value. First, e.g., as shown in Chap. 5, the equations (even if unsolved) reveal the basic dimensionless parameters which govern fluid motion. Second, as shown in Chap. 6, a great number of useful so-lutions 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 numerical analysis, 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 digital computer. Reference 1 is an example of a textbook devoted entirely to numerical analysis of fluid motion.
In Sec. 1.5 we established the cartesian vector form of a velocity field which varies in space and time: V(r, t) iu(x, y, z, t) j(x, y, z, t) kw(x, y, z, t) (1.4) 215 Chapter 4 Differential Relations for a Fluid Particle This is the most important variable in fluid mechanics: Knowledge of the velocity vec-tor field 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 co-ordinate lines on a glass window in a wind tunnel. This is the eulerian frame of ref-erence, as opposed to the lagrangian frame, which follows the moving position of in-dividual particles.
To write Newton’s second law for an infinitesimal fluid system, we need to calcu-late the acceleration vector field a of the flow. Thus we compute the total time deriv-ative of the velocity vector: a i j k Since each scalar component (u, , w) is a function of the four variables (x, y, z, t), we use the chain rule to obtain each scalar time derivative. For example, But, by definition, dx/dt is the local velocity component u, and dy/dt , and dz/dt w. The total derivative of u may thus be written in the compact form u w (V )u (4.1) Exactly similar expressions, with u replaced by or w, hold for d/dt or dw/dt. Sum-ming these into a vector, we obtain the total acceleration: a u w (V )V (4.2) Local Convective The term V/t is called the local acceleration, which vanishes if the flow is steady, i.e., independent of time. The three terms in parentheses are called the convective ac-celeration, which arises when the particle moves through regions of spatially varying velocity, as in a nozzle or diffuser. Flows which 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 : u w V where i j k The total time derivative—sometimes called the substantial or material derivative— concept may be applied to any variable, e.g., the pressure: u w (V )p (4.3) Wherever convective effects occur in the basic laws involving mass, momentum, or en-ergy, the basic differential equations become nonlinear and are usually more compli-cated than flows which do not involve convective changes.
We emphasize that this total time derivative follows a particle of fixed identity, mak-ing it convenient for expressing laws of particle mechanics in the eulerian fluid-field p t p z p y p x p t dp dt z y x z y x V t V z V y V x V t dV dt u t u z u y u x u t du dt dz dt u z dy dt u y dx dt u x u t du(x, y, z, t) dt dw dt d dt du dt dV dt 216 Chapter 4 Differential Relations for a Fluid Particle 4.2 The Differential Equation of Mass Conservation description. The operator d/dt is sometimes assigned a special symbol such as D/Dt as a further reminder that it contains four terms and follows a fixed particle.
EXAMPLE 4.1 Given the eulerian velocity-vector field V 3ti xzj ty2k find the acceleration of a particle.
Solution First note the specific given components u 3t xz w ty2 Then evaluate the vector derivatives required for Eq. (4.2) i j k 3i y2k zj 2tyk xj This could have been worse: There are only five terms in all, whereas there could have been as many as twelve. Substitute directly into Eq. (4.2): (3i y2k) (3t)(zj) (xz)(2tyk) (ty2)(xj) Collect terms for the final result 3i (3tz txy2)j (2xyzt y2)k Ans.
Assuming that V is valid everywhere as given, this acceleration applies to all positions and times within the flow field.
All the basic differential equations can be derived by considering either an elemental control volume or an elemental system. Here 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 CV d i (iAiVi)out i (iAiVi)in 0 (3.22) The element is so small that the volume integral simply reduces to a differential term CV d dx dy dz t t t dV dt dV dt V z V y V x w t t u t V t 4.2 The Differential Equation of Mass Conservation 217 Fig. 4.1 Elemental cartesian fixed control volume showing the inlet and outlet mass flows on the x faces.
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 consid-ered to be uniformly varying functions of time and position, such as (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 (T/x) dx. For mass conservation, if u is known on the left face, the value of this product on the right face is u (u/x) dx.
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 clut-tering up the drawing. We can list all these six flows as follows: 218 Chapter 4 Differential Relations for a Fluid Particle y z dz x u + ∂ ∂x ( u) dx dy dz Control volume ρ ρ u dy dz ρ dy dx Face Inlet mass flow Outlet mass flow x u dy dz u (u) dx dy dz y dx dz () dy dx dz z w dx dy w (w) dz dx dy z y x Introduce these terms into Eq. (3.22) above and we have dx dy dz (u) dx dy dz () dx dy dz (w) dx dy dz 0 The element volume cancels out of all terms, leaving a partial differential equation in-volving the derivatives of density and velocity (u) () (w) 0 (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 z y x t z y x t Fig. 4.2 Definition sketch for the cylindrical coordinate system.
Cylindrical Polar Coordinates or unsteady, viscous or frictionless, compressible or incompressible.1 However, the equation does not allow for any source or sink singularities within the element.
The vector-gradient operator i j k 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 di-vergence of the vector V (u) () (w) (V) (4.5) so that the compact form of the continuity relation is (V) 0 (4.6) In this vector form the equation is still quite general and can readily be converted to other than cartesian coordinate systems.
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 about the axis. The three independent velocity components are an axial velocity z, a radial velocity r, and a circumferential velocity , which is positive counterclockwise, i.e., in the direction t z y x z y x 4.2 The Differential Equation of Mass Conservation 219 θ r Typical point (r, , z) Base line r z r d d r Typical infinitesimal element Cylindrical axis θ θ z υ υ υ θ dz 1 One case where Eq. (4.4) might need special care is two-phase flow, where the density is discontinu-ous between the phases. For further details on this case, see, e.g., Ref. 2.
Steady Compressible Flow of increasing . In general, all components, as well as pressure and density and other fluid properties, are continuous functions of r, , z, and t.
The divergence of any vector function A(r, , z, t) is found by making the trans-formation of coordinates r (x2 y2)1/2 tan 1 z z (4.7) and the result is given here without proof2 A (rAr) (A ) (Az) (4.8) The general continuity equation (4.6) in cylindrical polar coordinates is thus (rr) ( ) (z) 0 (4.9) There are other orthogonal curvilinear coordinate systems, notably spherical polar co-ordinates, which occasionally merit use in a fluid-mechanics problem. We shall not treat these systems here except in Prob. 4.12.
There are also other ways to derive the basic continuity equation (4.6) which are in-teresting and instructive. Ask your instructor about these alternate approaches.
If the flow is steady, /t 0 and all properties are functions of position only. Equa-tion (4.6) reduces to Cartesian: (u) () (w) 0 Cylindrical: (rr) ( ) (z) 0 (4.10) Since density and velocity are both variables, these are still nonlinear and rather for-midable, but a number of special-case solutions have been found.
A special case which affords great simplification is incompressible flow, where the density changes are negligible. Then /t 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 di-vided out. The result is V 0 (4.11) valid for steady or unsteady incompressible flow. The two coordinate forms are Cartesian: 0 (4.12a) Cylindrical: (rr) ( ) (z) 0 (4.12b) z 1 r r 1 r w z y u x z 1 r r 1 r z y x z 1 r r 1 r t z 1 r r 1 r y x 220 Chapter 4 Differential Relations for a Fluid Particle Incompressible Flow 2 See, e.g., Ref. 3, p. 783.
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 playing a little fast and loose with density approximations. In essence, we wish to slip the density out of the divergence in Eq. (4.6) and approximate a typical term as, e.g., (u) (4.13) This is equivalent to the strong inequality u or (4.14) As we shall see in Chap. 9, the pressure change is approximately proportional to the density change and the square of the speed of sound a of the fluid p a2 (4.15) Meanwhile, if elevation changes are negligible, the pressure is related to the velocity change by Bernoulli’s equation (3.75) p V V (4.16) Combining Eqs. (4.14) to (4.16), we obtain an explicit criterion for incompressible flow: Ma2 1 (4.17) where Ma V/a 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 ve-locity is less than about 100 m/s (330 ft/s). This encompasses a wide variety of air-flows: 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.3 V2 a2 V V u x x u x x 4.2 The Differential Equation of Mass Conservation 221 3An exception occurs in geophysical flows, where a density change is imposed thermally or mechani-cally 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.
Before attempting to analyze the continuity equation, we shall proceed with the der-ivation 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 con-tinuity 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 crash in flames 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 (a1x b1y c1z)i (a2x b2y c2z)j (a3x b3y c3z)k where a1, b1, etc. const, represent an incompressible flow which conserves mass?
Solution Recalling that V ui j wk, we see that u (a1x b1y c1z), etc. Substituting into Eq.
(4.12a) for incompressible continuity, we obtain (a1x b1y c1z) (a2x b2y c2z) (a3x b3y c3z) 0 or a1 b2 c3 0 Ans.
At least two of constants a1, b2, and c3 must have opposite signs. Continuity imposes no re-strictions whatever on constants b1, c1, a2, c2, a3, and b3, which do not contribute to a mass in-crease or decrease of a differential element.
EXAMPLE 4.3 An incompressible velocity field is given by u a(x2 y2) unknown w b where a and b are constants. What must the form of the velocity component be?
Solution Again Eq. (4.12a) applies (ax2 ay2) 0 or 2ax (1) This is easily integrated partially with respect to y (x, y, z, t) 2axy f(x, z, t) Ans.
y b z y x z y x 222 Chapter 4 Differential Relations for a Fluid Particle 4.3 The Differential Equation of Linear Momentum This is the only possible form for which satisfies the incompressible continuity equation. The function of integration f is entirely arbitrary since it vanishes when is differentiated with re-spect to y.† EXAMPLE 4.4 A centrifugal impeller of 40-cm diameter is used to pump hydrogen at 15°C and 1-atm pressure.
What is the maximum allowable impeller rotational speed to avoid compressibility effects at the blade tips?
Solution The speed of sound of hydrogen for these conditions is a 1300 m/s. Assume that the gas ve-locity leaving the impeller is approximately equal to the impeller-tip speed V r 1 2 D Our rule of thumb, Eq. (4.18), neglects compressibility if V 1 2 D 0.3a 390 m/s or 1 2 (0.4 m) 390 m/s 1950 rad/s Thus we estimate the allowable speed to be quite large 310 r/s (18,600 r/min) Ans.
An impeller moving at this speed in air would create shock waves at the tips but not in a light gas like hydrogen.
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 ap-propriate form of the linear-momentum relation is F CV V d (m ˙ iVi)out (m ˙ iVi)in (3.40) Again the element is so small that the volume integral simply reduces to a derivative term CV V d (V) dx dy dz (4.19) 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 which led up to the equation for net mass flux: t t t 4.3 The Differential Equation of Linear Momentum 223 †This is a very realistic flow which simulates the turning of an inviscid fluid through a 60° angle; see Examples 4.7 and 4.9.
Introduce these terms and Eq. (4.19) into Eq. (3.40), and get the intermediate result F dx dy dz (V) (uV) (V) (wV) (4.20) Note that this is a vector relation. A simplification occurs if we split up the term in brackets as follows: (V) (uV) (V) (wV) V (V) u w (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 which instanta-neously occupies the control volume u w (4.2) Thus we have now reduced Eq. (4.20) to F dx dy dz (4.22) 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 vol-ume? Could we have begun the analysis at Eq. (4.22)?
Equation (4.22) points out that the net force on the control volume must be of dif-ferential size and proportional to the element volume. These forces are of two types, body forces and surface forces. Body forces are due to external fields (gravity, mag-netism, electric potential) which act upon the entire mass within the element. The only body force we shall consider in this book is gravity. The gravity force on the differ-ential mass dx dy dz within the control volume is dFgrav g dx dy dz (4.23) where g may in general have an arbitrary orientation with respect to the coordinate system. In many applications, such as Bernoulli’s equation, we take z “up,” and g gk.
dV dt dV dt V z V y V x V t V z V y V x V t t z y x t z y x t 224 Chapter 4 Differential Relations for a Fluid Particle Faces Inlet momentum flux Outlet momentum flux x uV dy dz uV x (uV) dx dy dz y V dx dz V y (V) dy dx dz z wV dx dy wV z (wV) dz dx dy Fig. 4.3 Notation for stresses.
The surface forces are due to the stresses on the sides of the control surface. These stresses, as discussed in Chap. 2, are the sum of hydrostatic pressure plus viscous stresses ij which arise from motion with velocity gradients p xx yx zx ij xy p yy zy (4.24) xz yz p zz The subscript notation for stresses is given in Fig. 4.3.
It is not these stresses but their gradients, or differences, which cause a net force on the differential control surface. This is seen by referring to Fig. 4.4, which shows 4.3 The Differential Equation of Linear Momentum 225 y z x σy y σy z σy x σx y σx z σx x σz x σz z σz y σi j = Stress in j direction on a face normal to i axis y z x σx x dy dz σz x dx dy dy σy x dx dz dx dz (σyx + ∂σyx ∂y dy) dx dz (σxx + ∂σxx ∂x dx) dy dz (σzx + ∂σzx ∂z dz) dx dy Fig. 4.4 Elemental cartesian fixed control volume showing the surface forces in the x direction only.
only the x-directed stresses to avoid cluttering up the drawing. For example, the left-ward force xx dy dz on the left face is balanced by the rightward force xx dy dz on the right face, leaving only the net rightward force (xx/x) dx dy dz on the right face.
The same thing happens on the other four faces, so that the net surface force in the x direction is given by dFx,surf (xx) (yx) (zx) 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 (xx) (yx) (zx) (4.26) In exactly similar manner, we can derive the y and z forces per unit volume on the control surface (xy) (yy) (zy) (xz) (yz) (zz) (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 surf p viscous (4.28) where the viscous force has a total of nine terms: viscous i j k (4.29) Since each term in parentheses in (4.29) represents the divergence of a stress-compo-nent vector acting on the x, y, and z faces, respectively, Eq. (4.29) is sometimes ex-pressed in divergence form d d F viscous ij (4.30) xx yx zx where ij xy yy zy (4.31) xz yz zz zz z yz y xz x zy z yy y xy x zx z yx y xx x dF d dF d dF d z y x p z dFz d z y x p y dFy d z y x p x dFx d z y x 226 Chapter 4 Differential Relations for a Fluid Particle Inviscid Flow: Euler’s Equation 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. Substitut-ing into Eq. (4.22) and utilizing Eq. (4.23), we have the basic differential momentum equation for an infinitesimal element g p ij (4.32) where u w (4.33) 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 acceleration (4.34) Equation (4.32) is so brief and compact that its inherent complexity is almost invisi-ble. 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 mathemati-cal difficulties inherent in the momentum equation: gx u w gy u w (4.35) gz u w 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 com-ponent 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 ve-locity components. The simplest assumption is frictionless flow ij 0, for which Eq.
(4.35) reduces to g 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, (3.75) or (3.77). The complete analysis of inviscid flow fields, using continuity and the Bernoulli relation, is given in Chap. 8.
For a newtonian fluid, as discussed in Sec. 1.7, the viscous stresses are proportional to the element strain rates and the coefficient of viscosity. For incompressible flow, the dV dt w z w y w x w t zz z yz y xz x p z z y x t zy z yy y xy x p y u z u y u x u t zx z yx y xx x p x V z V y V x V t dV dt dV dt 4.3 The Differential Equation of Linear Momentum 227 Newtonian Fluid: Navier-Stokes Equations generalization of Eq. (1.23) to three-dimensional viscous flow is4 xx 2 yy 2 zz 2 xy yx xz zx (4.37) yz zy where is the viscosity coefficient. Substitution into Eq. (4.35) gives the differential momentum equation for a newtonian fluid with constant density and viscosity gx gy d d t (4.38) gz These are the Navier-Stokes equations, named after C. L. M. H. Navier (1785–1836) and Sir George G. Stokes (1819–1903), who are credited with their derivation. They are second-order nonlinear partial differential equations and are quite formidable, but surprisingly many 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.
Equation (4.38) has four unknowns: p, u, , and w. It should be combined with the incompressible continuity relation (4.12) to form four equations in these four unknowns.
We shall discuss this again in Sec. 4.6, which presents the appropriate boundary con-ditions for these equations.
EXAMPLE 4.5 Take the velocity field of Example 4.3, with b 0 for algebraic convenience u a(x2 y2) 2axy w 0 and determine under what conditions it is a solution to the Navier-Stokes momentum equation (4.38). Assuming that these conditions are met, determine the resulting pressure distribution when z is “up” (gx 0, gy 0, gz g).
Solution Make a direct substitution of u, , w into Eq. (4.38): (0) (2a 2a) 2a2(x3 xy2) (1) p x dw dt 2w z2 2w y2 2w x2 p z 2 z2 2 y2 2 x2 p y du dt 2u z2 2u y2 2u x2 p x w y z u z w x x u y w z y u x 228 Chapter 4 Differential Relations for a Fluid Particle 4When compressibility is significant, additional small terms arise containing the element volume ex-pansion rate and a second coefficient of viscosity; see Refs. 4 and 5 for details.
(0) (0) 2a2(x2y y3) (2) ( g) (0) 0 (3) The viscous terms vanish identically (although is not zero). Equation (3) can be integrated partially to obtain p gz f1(x, y) (4) i.e., the pressure is hydrostatic in the z direction, which follows anyway from the fact that the flow is two-dimensional (w 0). Now the question is: Do Eqs. (1) and (2) show that the given velocity field is a solution? One way to find out is to form the mixed derivative 2p/(x y) from (1) and (2) separately and then compare them.
Differentiate Eq. (1) with respect to y 4a2xy (5) Now differentiate Eq. (2) with respect to x [2a2(x2y y3)] 4a2xy (6) Since these are identical, the given velocity field is an exact solution to the Navier-Stokes equation.
Ans.
To find the pressure distribution, substitute Eq. (4) into Eqs. (1) and (2), which will enable us to find f1(x, y) 2a2(x3 xy2) (7) 2a2(x2y y3) (8) Integrate Eq. (7) partially with respect to x f1 1 2 a2(x4 2x2y2) f2(y) (9) Differentiate this with respect to y and compare with Eq. (8) 2a2x2y f 2(y) (10) Comparing (8) and (10), we see they are equivalent if f 2(y) 2a2y3 or f2(y) 1 2 a2y4 C (11) where C is a constant. Combine Eqs. (4), (9), and (11) to give the complete expression for pres-sure distribution p(x, y, z) gz 1 2 a2(x4 y4 2x2y2) C Ans.
(12) This is the desired solution. Do you recognize it? Not unless you go back to the beginning and square the velocity components: f1 y f1 y f1 x x 2p x y 2p x y p z p y 4.3 The Differential Equation of Linear Momentum 229 4.4 The Differential Equation of Angular Momentum u2 2 w2 V2 a2(x4 y4 2x2y2) (13) Comparing with Eq. (12), we can rewrite the pressure distribution as p 1 2 V2 gz C (14) This is Bernoulli’s equation (3.77). That is no accident, because the velocity distribution given in this problem is one of a family of flows which are solutions to the Navier-Stokes equation and which satisfy Bernoulli’s incompressible equation everywhere in the flow field. They are called irrotational flows, for which curl V V 0. This subject is discussed again in Sec. 4.9.
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 vol-ume is MO CV (r V) d CS (r V)(V n) dA (3.55) We shall confine ourselves to an axis O which is parallel to the z axis and passes through the centroid of the elemental control volume. This is shown in Fig. 4.5. Let be the angle of rotation about O of the fluid within the control volume. The only stresses which have moments about O are the shear stresses xy and yx. We can evaluate the moments about O and the angular-momentum terms about O. A lot of algebra is in-volved, and we give here only the result xy yx (xy) dx (yx) dy dx dy dz (dx dy dz)(dx2 dy2) (4.39) Assuming that the angular acceleration d2 /dt2 is not infinite, we can neglect all higher-d2 dt2 1 12 y 1 2 x 1 2 t 230 Chapter 4 Differential Relations for a Fluid Particle Fig. 4.5 Elemental cartesian fixed control volume showing shear stresses which may cause a net an-gular acceleration about axis O.
x y d y d x yx Axis O = Rotation angle yx + ∂ ∂y ( yx) d y x y + ∂ ∂x ( x y) d x θ τ τ τ τ τ τ 4.5 The Differential Equation of Energy6 order differential terms, which leaves a finite and interesting result xy yx (4.40) Had we summed moments about axes parallel to y or x, we would have obtained ex-actly analogous results xz zx yz zy (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 analy-sis, that the shear stresses are symmetric: ij ji. This is the only result of this sec-tion.5 There is no differential equation to remember, which leaves room in your brain for the next topic, the differential energy equation.
We are now so used to this type of derivation that we can race through the energy equa-tion at a bewildering pace. The appropriate integral relation for the fixed control vol-ume of Fig. 4.1 is ˙ Q ˙ Ws ˙ W CV e d CS e (V n) dA (3.63) where ˙ Ws 0 because there can be no infinitesimal shaft protruding into the control volume. By analogy with Eq. (4.20), the right-hand side becomes, for this tiny element, ˙ Q ˙ W (e) (u) () (w) dx dy dz (4.42) where e p/. When we use the continuity equation by analogy with Eq. (4.21), this becomes ˙ Q ˙ W V p dx dy dz (4.43) To evaluate ˙ Q, we neglect radiation and consider only heat conduction through the sides of the element. The heat flow by conduction follows Fourier’s law from Chap. 1 q k T (1.29a) where k is the coefficient of thermal conductivity of the fluid. 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: de dt z y x t p t 4.5 The Differential Equation of Energy 231 5We are neglecting the possibility of a finite couple being applied to the element by some powerful ex-ternal force field. See, e.g., Ref. 6, p. 217.
6This section may be omitted without loss of continuity.
Faces Inlet heat flux Outlet heat flux x qx dy dz qx (qx) dx dy dz y qy dx dz qy (qy) dy dx dz z qz dx dy qz (qz) dz dx dy z y x Fig. 4.6 Elemental cartesian control volume showing heat-flow and viscous-work-rate terms in the x direction.
By adding the inlet terms and subtracting the outlet terms, we obtain the net heat added to the element ˙ Q (qx) (qy) (qz) dx dy dz q dx dy dz (4.44) As expected, the heat flux is proportional to the element volume. Introducing Fourier’s law from Eq. (1.29), we have ˙ Q (k T) dx dy dz (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 ˙ Wυ,LF wx dy dz where wx (uxx υxy wxz) (4.46) (where the subscript LF stands for left face) and a slightly different work on the right face due to the gradient in wx. These work fluxes could be tabulated in exactly the same manner as the heat fluxes in the previous table, with wx replacing qx, etc. After outlet terms are subtracted from inlet terms, the net viscous-work rate becomes ˙ W (uxx xy wxz) (uyx yy wyz) (uzx zy wzz) dx dy dz (V ij) dx dy dz (4.47) We now substitute Eqs. (4.45) and (4.47) into Eq. (4.43) to obtain one form of the dif-ferential energy equation V p (k T) (V ij) where e û 1 2 V2 gz (4.48) A more useful form is obtained if we split up the viscous-work term (V ij) V ( ij) (4.49) de dt z y x z y x 232 Chapter 4 Differential Relations for a Fluid Particle Heat flow per unit area: qx = – k ∂T ∂x wx Viscous work rate per unit area: wx = –(u x x + x y + w x z) dz dy dx qx + ∂ ∂x (qx ) d x wx + ∂ ∂x (wx ) d x τ υτ τ where is short for the viscous-dissipation function.7 For a newtonian incompressible viscous fluid, this function has the form 2 2 2 2 2 2 2 2 2 (4.50) Since all terms are quadratic, viscous dissipation is always positive, so that 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 ij. This will cause the kinetic and potential energies to cancel, leav-ing a more customary form of the general differential energy equation p( V) (k T) (4.51) 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 trans-fer 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 cus-tomary to make the following approximations: dû c dT c, , k, const (4.52) Equation (4.51) then takes the simpler form c k2T (4.53) which involves temperature T as the sole primary variable plus velocity as a secondary variable through the total time-derivative operator u w (4.54) A great many interesting solutions to Eq. (4.53) are known for various flow conditions, 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 c k 2T (4.55) 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. T t T z T y T x T t dT dt dT dt dû dt w x u z z w y u y x w z y u x 4.5 The Differential Equation of Energy 233 7For further details, see, e.g., Ref. 5, p. 72.
4.6 Boundary Conditions for the Basic Equations There are three basic differential equations of fluid motion, just derived. Let us sum-marize them here: Continuity: (V) 0 (4.56) Momentum: g p ij (4.57) Energy: p( V) (k T) (4.58) where is given by Eq. (4.50). In general, the density is variable, so that these three equations contain five unknowns, , V, p, û, 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, T) û û(p, T) (4.59) For example, for a perfect gas with constant specific heats, we complete the system with û c dT c 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? First, if the flow is unsteady, there must be an initial condition or initial spatial distribution known for each variable: At t 0: , V, p, û, T known f(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.
Figure 4.7 illustrates the three most common types of boundaries encountered in fluid-flow analysis: a solid wall, an inlet or outlet, a liquid-gas interface.
First, for a solid, impermeable wall, there is no slip and no temperature jump in a viscous heat-conducting fluid Vfluid Vwall Tfluid Twall solid wall (4.62) The only exception to Eq. (4.62) occurs in an extremely rarefied gas flow, where slip-page can be present .
Second, at any inlet or outlet section of the flow, the complete distribution of ve-locity, 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 im-mersed 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 (x, y, t) (4.64) p RT dû dt dV dt t 234 Chapter 4 Differential Relations for a Fluid Particle Fig. 4.7 Typical boundary condi-tions in a viscous heat-conducting fluid-flow analysis.
Then there must be equality of vertical velocity across the interface, so that no holes appear between liquid and gas: wliq wgas u (4.65) This is called the kinematic boundary condition.
There must be mechanical equilibrium across the interface. The viscous-shear stresses must balance (zy)liq (zy)gas (zx)liq (zx)gas (4.66) Neglecting the viscous normal stresses, the pressures must balance at the interface ex-cept for surface-tension effects pliq pgas (R 1 x R 1 y ) (4.67) which is equivalent to Eq. (1.34). The radii of curvature can be written in terms of the free-surface position Rx 1 Ry 1 x y (4.68) y 1 ( x) 2 ( y) 2 x 1 ( x) 2 ( y) 2 y x t d dt 4.6 Boundary Conditions for the Basic Equations 235 Z Gas Liquid Inlet: known V, p, T Outlet: known V, p, T Solid contact: ( V, T )fluid = ( V, T )wall Solid impermeable wall Liquid-gas interface z = η(x, y, t): pliq = pgas – (R–1 + R–1) x y wliq = wgas = d dt Equality of q and across interface η τ Simplified Free-Surface Conditions Incompressible Flow with Constant Properties 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 (qz)liq (qz)gas (4.69) Neglecting radiation, this is equivalent to k liq k gas (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” which merely exerts pressure upon 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 conditions at the surface pliq pgas wliq liq 0 liq 0 (4.71) In many cases, such as open-channel flow, we can also neglect surface tension, so that pliq patm (4.72) These are the types of approximations which will be used in Chap. 10. The nondi-mensional forms of these conditions will also be useful in Chap. 5.
Flow with constant , , and k is a basic simplification which will be used, e.g., through-out Chap. 6. The basic equations of motion (4.56) to (4.58) reduce to: Continuity: V 0 (4.73) Momentum: g p 2V (4.74) Energy: c k 2T (4.75) Since is constant, there are only three unknowns: p, V, and T. The system is closed.8 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 Vwall (4.76) dT dt dV dt T z V z t 2 y2 2 x2 T z T z 236 Chapter 4 Differential Relations for a Fluid Particle 8For this system, what are the thermodynamic equivalents to Eq. (4.59)?
Inviscid-Flow Approximations Inlet or outlet: Known V, p (4.77) Free surface: p pa w (4.78) Later, entirely at our leisure,9 we can solve for the temperature distribution from Eq.
(4.75), which depends upon velocity V through the dissipation and the total time-derivative operator d/dt.
Chapter 8 assumes inviscid flow throughout, for which the viscosity 0. The mo-mentum equation (4.74) reduces to g p (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 deriva-tive 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: (Vn)fluid (Vn)wall (4.80) In most cases the wall is fixed; therefore the proper inviscid-flow condition is Vn 0 (4.81) There is no condition whatever on the tangential velocity component at the wall in in-viscid flow. The tangential velocity will be part of the solution, and the correct value will appear after the analysis is completed (see Chap. 8).
EXAMPLE 4.6 For steady incompressible laminar flow through a long tube, the velocity distribution is given by z U1 r 0 where U is the maximum, or centerline, velocity and R is the tube radius. If the wall tempera-ture is constant at Tw 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 cr r d d r z 2 (1) dT dr d dr k r dT dr r2 R2 dV dt t 4.6 Boundary Conditions for the Basic Equations 237 9Since temperature is entirely uncoupled by this assumption, we may never get around to solving for it here and may ask you to wait until a course on heat transfer.
4.7 The Stream Function But since r 0 for this flow, the convective term on the left vanishes. Introduce z into Eq. (1) to obtain r 2 d d r z 2 (2) Multiply through by r/k and integrate once: r C1 (3) Divide through by r and integrate once again: T C1 ln r C2 (4) Now we are in position to apply our boundary conditions to evaluate C1 and C2.
First, since the logarithm of zero is , the temperature at r 0 will be infinite unless C1 0 (5) Thus we eliminate the possibility of a logarithmic singularity. The same thing will happen if we apply the symmetry condition dT/dr 0 at r 0 to Eq. (3). The constant C2 is then found by the wall-temperature condition at r R T Tw C2 or C2 Tw (6) The correct solution is thus T(r) Tw 1 Ans.
(7) which is a fourth-order parabolic distribution with a maximum value T0 Tw U2/(4k) at the centerline.
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 simulta-neously for pressure and velocity. The stream function is a clever device which al-lows us to wipe out the continuity equation and solve the momentum equation directly for the single variable .
The stream-function idea works only if the continuity equation (4.56) can be re-duced to two terms. In general, we have four terms: Cartesian: (u) () (w) 0 (4.82a) Cylindrical: (rr) ( ) (z) 0 (4.82b) z 1 r r 1 r t z y x t r4 R4 U2 4k U2 4k U2 4k U2r4 4kR4 U2r4 kR4 dT dr 4U2r2 R4 dT dr d dr k r 238 Chapter 4 Differential Relations for a Fluid Particle 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 com-mon application is incompressible flow in the xy plane 0 (4.83) This equation is satisfied identically if a function (x, y) is defined such that Eq. (4.83) becomes 0 (4.84) Comparison of (4.83) and (4.84) shows that this new function must be defined such that u (4.85) or V i j Is this legitimate? Yes, it is just a mathematical trick of replacing two variables (u and ) by a single higher-order function . The vorticity, or curl V, is an interesting func-tion curl V 2k z k2 where 2 (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 (2 ) (2 ) 2(2 ) (4.87) where / is the kinematic viscosity. This is partly a victory and partly a defeat: Eq. (4.87) is scalar and has only one variable, , but it now contains fourth-order derivatives and probably will require computer analysis. There will be four boundary conditions required on . For example, for the flow of a uniform stream in the x di-rection past a solid body, the four conditions would be At infinity: U 0 At the body: 0 (4.88) Many examples of numerical solution of Eqs. (4.87) and (4.88) are given in Ref. 1.
One important application is inviscid irrotational flow in the xy plane, where z 0.
Equations (4.86) and (4.87) reduce to 2 0 (4.89) This is the second-order Laplace equation (Chap. 8), for which many solutions and an-alytical techniques are known. Also, boundary conditions like Eq. (4.88) reduce to 2 y2 2 x2 x y x y y x x y 2 y2 2 x2 x y x y x y y x y u x 4.7 The Stream Function 239 Geometric Interpretation of At infinity: Uy const (4.90) At the body: 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 by itself to make the stream function im-mortal and always useful to engineers. Even better, though, has a beautiful geomet-ric interpretation: Lines of constant 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 or u dy dx 0 streamline (4.91) Introducing the stream function from Eq. (4.85), we have dx dy 0 d (4.92) Thus the change in is zero along a streamline, or const along a streamline (4.93) Having found a given solution (x, y), we can plot lines of constant to give the streamlines of the flow.
There is also a physical interpretation which relates 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 (V n) dA i j i j ds(1) dx dy d (4.94) y x dx ds dy ds x y y x dy dx u 240 Chapter 4 Differential Relations for a Fluid Particle Fig. 4.8 Geometric interpretation of stream function: volume flow through a differential portion of a control surface.
dQ = ( V • n) dA = d Control surface (unit depth into paper) dy dx V = i u + j v n = dy ds i – dx ds j ds ψ Fig. 4.9 Sign convention for flow in terms of change in stream func-tion: (a) flow to the right if U is greater; (b) flow to the left if L is greater.
Thus the change in across the element is numerically equal to the volume flow through the element. The volume flow between any two points in the flow field is equal to the change in stream function between those points: Q1→2 2 1 (V n) dA 2 1 d 2 1 (4.95) Further, the direction of the flow can be ascertained by noting whether increases or decreases. As sketched in Fig. 4.9, the flow is to the right if U is greater than L, 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 mechan-ics in 1781.
EXAMPLE 4.7 If a stream function exists for the velocity field of Example 4.5 u a(x2 y2) 2axy w 0 find it, plot it, and interpret it.
Solution Since this flow field was shown expressly in Example 4.3 to satisfy the equation of continuity, we are pretty sure that a stream function does exist. We can check again to see if 0 Substitute: 2ax ( 2ax) 0 checks Therefore we are certain that a stream function exists. To find , we simply set u ax2 ay2 (1) 2axy (2) x y y u x 4.7 The Stream Function 241 Flow (a) (b) Flow 2 > 1 ψ ψ 1 ψ 2 < 1 ψ ψ 1 ψ and work from either one toward the other. Integrate (1) partially ax2y f(x) (3) Differentiate (3) with respect to x and compare with (2) 2axy f(x) 2axy (4) Therefore f(x) 0, or f constant. The complete stream function is thus found ax2y C Ans.
(5) To plot this, set C 0 for convenience and plot the function 3x2y y3 (6) for constant values of . The result is shown in Fig. E4.7a to be six 60° wedges of circulating 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. How can the flow be interpreted?
Since there is slip along all streamlines, no streamline can truly represent a solid surface in a viscous flow. However, 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 equation, as we showed in Example 4.5.
3 a y3 3 x ay3 3 242 Chapter 4 Differential Relations for a Fluid Particle = 2a a 0 –2a = 2a a – a –2a x y 60° – a 0 a 2 a The origin is a stagnation point 60° 60° 60° 60° – a ψ = – 2a ψ ψ Flow around a 60° corner Flow around a rounded 60° corner Incoming stream impinging against a 120° corner By allowing the flow to slip as a frictionless approximation, we could let any given stream-line be a body shape. Some examples are shown in Fig. E4.7b.
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.
E4.7a E4.7b 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 (u) () 0 (4.96) 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 u (4.97) Again lines of constant are streamlines of the flow, but the change in is now equal to the mass flow, not the volume flow dm ˙ (V n) dA d or m ˙ 1→2 2 1 (V n) dA 2 1 (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 mo-mentum 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 typ-ical problem (see, e.g., Ref. 5, chap. 7).
Suppose that the important coordinates are r and , with z 0, and that the density is constant. Then Eq. (4.82b) reduces to (rr) ( ) 0 (4.99) After multiplying through by r, we see that this is the same as the analogous form of Eq. (4.84) 0 (4.100) By comparison of (4.99) and (4.100) we deduce the form of the incompressible polar-coordinate stream function r (4.101) Once again lines of constant are streamlines, and the change in is the volume flow Q1→2 2 1. 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).
As a final example, suppose that the flow is three-dimensional (υr, υz) but with no cir-cumferential variations, / 0 (see Fig. 4.2 for definition of coordinates). Such r 1 r r r 1 r r 1 r x y y x 4.7 The Stream Function 243 Incompressible Plane Flow in Polar Coordinates Incompressible Axisymmetric Flow 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. (4.82b) becomes (rr) (z) 0 (4.102) This doesn’t seem to work: Can’t we get rid of the one r outside? But when we real-ize that r and z are independent coordinates, Eq. (4.102) can be rewritten as (rr) (rz) 0 (4.103) By analogy with Eq. (4.84), this has the form 0 (4.104) By comparing (4.103) and (4.104), we deduce the form of an incompressible axisym-metric stream function (r, z) r z (4.105) Here again lines of constant are streamlines, but there is a factor (2 ) in the volume flow: Q1→2 2 ( 2 1). The sign convention on flow is the same as in Fig. 4.9.
EXAMPLE 4.8 Investigate the stream function in polar coordinates U sin 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 , which has units of square meters per second. Note that /(UR) is dimensionless. Rewrite Eq. (1) in dimensionless form sin η 1 η (2) Of particular interest is the special line 0. From Eq. (1) or (2) this occurs when (a) 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.
For any other nonzero value of it is easiest to pick a value of r and solve for : sin (3) In general, there will be two solutions for because of the symmetry about the y-axis. For ex-ample take /(UR) 1.0: /(UR) r/R R/r r R UR R2 r r 1 r z 1 r r z z r z r z r 1 r 244 Chapter 4 Differential Relations for a Fluid Particle E4.8 Guess r/R 3.0 2.5 2.0 1.8 1.7 1.618 Compute 22° 28° 42° 54° 64° 90° 158° 152° 138° 156° 116° This line is plotted in Fig. E4.8 and passes over the circle r R. You have to watch it, though, because there is a second curve for /(UR) 1.0 for small r R below the x-axis: Guess r/R 0.618 0.6 0.5 0.4 0.3 0.2 0.1 Compute 90° 70° 42° 28° 19° 12° 6° 110° 138° 152° 161° 168° 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 equa-tion (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. 8.3). 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 sim-plification. 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 ex-amining Fig. 4.10. Two fluid lines AB and BC, initially perpendicular at time t, move and deform so that at t dt they have slightly different lengths AB and BC and are slightly off the perpendicular by angles d! and d". Such deformation occurs kinemat-ically because A, B, and C have slightly different velocities when the velocity field V 4.8 Vorticity and Irrotationality 245 Streamlines converge, high-velocity region r = R 0 –1 +1 Singularity at origin – 1 2 +1 2 = +1 ψ UR 0 0 0 0 –1 4.8 Vorticity and Irrotationality Fig. 4.10 Angular velocity and strain rate of two fluid lines de-forming in the xy plane.
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 z about the z axis as the average rate of counter-clockwise turning of the two lines ωz d d β t (4.106) But from Fig. 4.10, d! and d" are each directly related to velocity derivatives in the limit of small dt d! lim dt→0 tan 1 dt d" lim dt→0 tan 1 dt (4.107) Combining Eqs. (4.106) and (4.107) gives the desired result: z (4.108) In exactly similar manner we determine the other two rates: x y (4.109) w x u z 1 2 z w y 1 2 u y x 1 2 u y (u/y) dy dt dy (/y) dy dt x (/x) dx dt dx (u/x) dx dt d!
dt 1 2 246 Chapter 4 Differential Relations for a Fluid Particle ∂u ∂y dy dt d A′ Time: t + dt C ′ B′ dα ∂ ∂x dx dt Line 2 Time t V Line 1 A B C dx dy y x 0 ∂ ∂y dy dt dy + ∂u ∂x dx dt dx + υ β υ 4.9 Frictionless Irrotational Flows The vector ix jy kz is thus one-half the curl of the velocity vector i j k (curl V) (4.110) u w Since the factor of 1 2 is annoying, many workers prefer to use a vector twice as large, called the vorticity: 2 curl V (4.111) Many flows have negligible or zero vorticity and are called irrotational curl V 0 (4.112) The next section expands on this idea. Such flows can be incompressible or com-pressible, 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 ˙ #xy (4.113) When multiplied by viscosity , this equals the shear stress xy in a newtonian fluid, as discussed earlier in Eqs. (4.37). Appendix E lists strain-rate and vorticity compo-nents in cylindrical coordinates.
When a flow is both frictionless and irrotational, pleasant things happen. First, the mo-mentum equation (4.38) reduces to Euler’s equation g p (4.114) Second, there is a great simplification in the acceleration term. Recall from Sec. 4.1 that acceleration has two terms (V )V (4.2) A beautiful vector identity exists for the second term : (V )V ( 1 2 V2) V (4.115) where curl V from Eq. (4.111) is the fluid vorticity.
Now combine (4.114) and (4.115), divide by , and rearrange on the left-hand side.
Dot the entire equation into an arbitrary vector displacement dr: V2 V p g dr 0 (4.116) Nothing works right unless we can get rid of the third term. We want ( V) (dr) 0 (4.117) 1 1 2 V t V t dV dt dV dt u y x d" dt d!
dt z y x 1 2 1 2 4.9 Frictionless Irrotational Flows 247 Velocity Potential This will be true under various conditions: 1.
V is zero; trivial, no flow (hydrostatics).
2.
is zero; irrotational flow.
3.
dr is perpendicular to V; this is rather specialized and rare.
4.
dr is parallel to V; we integrate along a streamline (see Sec. 3.7).
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 dr d V2 g dz 0 (4.118) Except for the first term, these are exact differentials. Integrate between any two points 1 and 2 along the streamline: 2 1 ds 2 1 (V 2 2 V 2 1) g(z2 z1) 0 (4.119) 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. (3.76). For incompressible steady flow, it reduces to V2 gz constant along streamline (4.120) The constant may vary from streamline to streamline unless the flow is also irrotational (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.
Irrotationality gives rise to a scalar function $ similar and complementary to the stream function . From a theorem in vector analysis , a vector with zero curl must be the gradient of a scalar function If V 0 then V $ (4.121) where $ $ (x, y, z, t) is called the velocity potential function. Knowledge of $ thus immediately gives the velocity components u w (4.122) Lines of constant $ are called the potential lines of the flow.
Note that $, unlike the stream function, is fully three-dimensional and not limited to two coordinates. It reduces a velocity problem with three unknowns u, , and w to a single unknown potential $; many examples are given in Chap. 8 and Sec. 4.10. The velocity potential also simplifies the unsteady Bernoulli equation (4.118) because if $ exists, we obtain dr () dr d (4.123) $ t t V t $ z $ y $ x 1 2 p 1 2 dp V t dp 1 2 V t 248 Chapter 4 Differential Relations for a Fluid Particle Orthogonality of Streamlines and Potential Lines Generation of Rotationality Equation (4.118) then becomes a relation between $ and p 2 gz const (4.124) This is the unsteady irrotational Bernoulli equation. It is very important in the analy-sis of accelerating flow fields (see, e.g., 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, and $ both ex-ist and the streamlines and potential lines are everywhere mutually perpendicular ex-cept at a stagnation point. For example, for incompressible flow in the xy plane, we would have u (4.125) (4.126) Can you tell by inspection not only that these relations imply orthogonality but also that $ and satisfy Laplace’s equation?10 A line of constant $ would be such that the change in $ is zero d$ dx dy 0 u dx dy (4.127) Solving, we have $const (4.128) Equation (4.128) is the mathematical condition that lines of constant $ and be mu-tually orthogonal. It may not be true at a stagnation point, where both u and are zero, so that their ratio in Eq. (4.128) is indeterminate.
This is the second time we have discussed Bernoulli’s equation under different circum-stances (the first was in Sec. 3.7). 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 ir-rotational, the latter being an easier condition, allowing a universal Bernoulli constant.
The only remaining question is: When is a flow irrotational? In other words, when does a flow have negligible angular velocity? The exact analysis of fluid rotationality under arbitrary conditions is a topic for advanced study, e.g., 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 which 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.
1 (dy/dx) const u dy dx $ y $ x $ y x $ x y 1 2 dp $ t 4.9 Frictionless Irrotational Flows 249 10 Equations (4.125) and (4.126) are called the Cauchy-Riemann equations and are studied in com-plex-variable theory.
Fig. 4.11 Typical flow patterns il-lustrating viscous regions patched onto nearly frictionless regions: (a) low subsonic flow past a body (U a); frictionless, irrotational potential flow outside the boundary layer (Bernoulli and Laplace equa-tions valid); (b) supersonic flow past a body (U % a); frictionless, rotational flow outside the bound-ary layer (Bernoulli equation valid, potential flow invalid).
2. There are entropy gradients caused by curved shock waves (see Fig. 4.11b).
3. There are density gradients caused by stratification (uneven heating) rather than by pressure gradients.
4. 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 negli-gible. 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 condi-tion 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 un-less 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 bound-ary layer. Two examples are shown in Fig. 4.11. Figure 4.11a shows a low-speed 250 Chapter 4 Differential Relations for a Fluid Particle (a) (b) Curved shock wave introduces rotationality Viscous regions where Bernoulli is invalid: Laminar boundary layer Turbulent boundary layer Slight separated flow Wake flow Viscous regions where Bernoulli's equation fails: Laminar boundary layer Turbulent boundary layer Separated flow Wake flow U Uniform approach flow (irrotational) Uniform supersonic approach (irrotational) U subsonic flow past a body. The approach stream is irrotational; i.e., the curl of a con-stant is zero, but viscous stresses create a rotational shear layer beside and downstream of the body. Generally speaking (see Chap. 6), 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 tur-bulent wake extending far downstream. Some sort of laminar or turbulent viscous the-ory 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 () 0 or 2$ 0 (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.
4.11a outside the shear layer.
Figure 4.11b shows a supersonic flow past a 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 the-ory. The shear layers have the same general character as in Fig. 4.11a except that the separation zone is slight or often absent and the wake is usually thinner. Theory of sep-arated flow is presently qualitative, but we can make quantitative estimates of laminar and turbulent boundary layers and wakes.
EXAMPLE 4.9 If a velocity potential exists for the velocity field of Example 4.5 u a(x2 y2) 2axy w 0 find 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: ( V)z 2 z ( 2axy) (ax2 ay2) 2ay 2ay 0 checks Ans.
The flow is indeed irrotational. A potential exists.
To find $(x, y), set u ax2 ay2 (1) 2axy (2) $ y $ x y x u y x 2$ z2 2$ y2 2$ x2 4.9 Frictionless Irrotational Flows 251 E4.9 Integrate (1) $ axy2 f(y) (3) Differentiate (3) and compare with (2) 2axy f(y) 2axy (4) Therefore f 0, or f constant. The velocity potential is $ axy2 C Ans.
Letting C 0, we can plot the $ lines in the same fashion as in Example 4.7. The result is shown in Fig. E4.9 (no arrows on $). For this particular problem, the $ lines form the same pattern as the lines of Example 4.7 (which are shown here as dashed lines) but are displaced 30°. The $ and lines are everywhere perpendicular except at the origin, a stagnation point, where they are 30° apart. We expected trouble at the stagnation point, and there is no general rule for de-termining the behavior of the lines at that point.
Chapter 8 is devoted entirely to a detailed study of inviscid incompressible flows, es-pecially those which possess both a stream function and a velocity potential. As sketched in Fig. 4.11a, inviscid flow is valid away from solid surfaces, and this inviscid pattern is “patched” onto the near-wall viscous layers—an idea developed in Chap. 7. Various body shapes can be simulated by the inviscid-flow pattern. Here we discuss plane flows, three of which are illustrated in Fig. 4.12.
A uniform stream V iU, as in Fig. 4.12a, possesses both a stream function and a ve-locity potential, which may be found as follows: u U 0 x $ y y $ x ax3 3 $ y ax3 3 252 Chapter 4 Differential Relations for a Fluid Particle 2a a 0 –a = –2a y = –2a –a 0 a 2a x a 0 –a –2a = 2a φ φ φ 4.10 Some Illustrative Plane Potential Flows Uniform Stream in the x Direction Fig. 4.12 Three elementary plane potential flows. Solid lines are streamlines; dashed lines are poten-tial lines.
We may integrate each expression and discard the constants of integration, which do not affect the velocities in the flow. The results are Uniform stream iU: Uy $ Ux (4.130) The streamlines are horizontal straight lines (y const), and the potential lines are ver-tical (x const), i.e., orthogonal to the streamlines, as expected.
Suppose that the z-axis were a sort of thin-pipe manifold through which fluid issued at total rate Q uniformly along its length b. Looking at the xy plane, we would see a cylindrical radial outflow or line source, as sketched in Fig. 4.12b. Plane polar coor-dinates are appropriate (see Fig. 4.2), and there is no circumferential velocity. At any radius r, the velocity is r 0 where we have used the polar-coordinate forms of the stream function and the veloc-ity potential. Integrating and again discarding the constants of integration, we obtain the proper functions for this simple radial flow: Line source or sink: m $ m ln r (4.131) where m Q/(2 b) is a constant, positive for a source, negative for a sink. As shown in Fig. 4.12b, the streamlines are radial spokes (constant ), and the potential lines are circles (constant r).
A (two-dimensional) line vortex is a purely circulating steady motion, f(r) only, r 0. This satisfies the continuity equation identically, as may be checked from Eq.
(4.12b). We may also note that a variety of velocity distributions (r) satisfy the -momentum equation of a viscous fluid, Eq. (E.6). We may show, as a problem ex-ercise, that only one function (r) is irrotational, i.e., curl V 0, and that is K/r, where K is a constant. This is sometimes called a free vortex, for which the stream function and velocity may be found: r 0 $ 1 r r K r $ r 1 r $ 1 r r $ r 1 r m r Q 2 rb 4.10 Some Illustrative Plane Potential Flows 253 (a) (b) (c) U m/r K/r Line Source or Sink at the Origin Line Irrotational Vortex Fig. 4.13 Potential flow due to a line source plus an equal line sink, from Eq. (4.133). Solid lines are streamlines; dashed lines are poten-tial lines.
We may again integrate to determine the appropriate functions: K ln r $ K (4.132) where K is a constant called the strength of the vortex. As shown in Fig. 4.12c, the streamlines are circles (constant r), and the potential lines are radial spokes (constant ). Note the similarity between Eqs. (4.131) and (4.132). A free vortex is a sort of re-versed image of a source. The “bathtub vortex,” formed when water drains through a bottom hole in a tank, is a good approximation to the free-vortex pattern.
Each of the three elementary flow patterns in Fig. 4.12 is an incompressible irrotational flow and therefore satisfies both plane “potential flow” equations 2 0 and 2$ 0. Since these are linear partial differential equations, any sum of such basic solutions is also a solution. Some of these composite solutions are quite interesting and useful.
For example, consider a source m at (x, y) ( a, 0), combined with a sink of equal strength m, placed at (a, 0), as in Fig. 4.13. The resulting stream function is simply the sum of the two. In cartesian coordinates, source sink m tan 1 m tan 1 Similarly, the composite velocity potential is $ $source $sink m ln [(x a)2 y2] m ln [(x a)2 y2] 1 2 1 2 y x a y x a 254 Chapter 4 Differential Relations for a Fluid Particle Superposition: Source Plus an Equal Sink Fig. 4.14 Superposition of a sink plus a vortex, Eq. (4.134), simu-lates a tornado.
By using trigonometric and logarithmic identities, these may be simplified to Source plus sink: m tan 1 $ m ln (4.133) These lines are plotted in Fig. 4.13 and are seen to be two families of orthogonal circles, with the streamlines passing through the source and sink and the potential lines encircling them. They are harmonic (laplacian) functions which are exactly analogous in electromagnetic theory to the electric-current and electric-potential pat-terns of a magnet with poles at (a, 0).
An interesting flow pattern, approximated in nature, occurs by superposition of a sink and a vortex, both centered at the origin. The composite stream function and velocity potential are Sink plus vortex: m K ln r $ m ln r K (4.134) When plotted, these form two orthogonal families of logarithmic spirals, as shown in Fig. 4.14. This is a fairly realistic simulation of a tornado (where the sink flow moves up the z-axis into the atmosphere) or a rapidly draining bathtub vortex. At the center of a real (viscous) vortex, where Eq. (4.134) predicts infinite velocity, the actual cir-culating flow is highly rotational and approximates solid-body rotation ≈Cr.
(x a)2 y2 (x a)2 y2 1 2 2ay x2 y2 a2 4.10 Some Illustrative Plane Potential Flows 255 y x Sink Plus a Vortex at the Origin Uniform Stream Plus a Sink at the Origin: The Rankine Half-Body If we superimpose a uniform x-directed stream against an isolated source, a half-body shape appears. If the source is at the origin, the combined stream function is, in polar coordinates, Uniform stream plus source: Ur sin m (4.135) We can set this equal to various constants and plot the streamlines, as shown in Fig.
4.15. A curved, roughly elliptical, half-body shape appears, which separates the source flow from the stream flow. The body shape, which is named after the Scottish engi-neer W. J. M. Rankine (1820–1872), is formed by the particular streamlines m. The half-width of the body far downstream is m/U. The upper surface may be plotted from the relation r (4.136) m( ) U sin 256 Chapter 4 Differential Relations for a Fluid Particle x y U∞ Us(max) = 1.26U∞ = + m = 0 a a π ψ π ψ = – m ψ π It is not a true ellipse. The nose of the body, which is a “stagnation” point where V 0, stands at (x, y) ( a, 0), where a m/U. The streamline 0 also crosses this point—recall that streamlines can cross only at a stagnation point.
The cartesian velocity components are found by differentiation: u U cos sin (4.137) Setting u 0, we find a single stagnation point at 180° and r m/U, or (x, y) ( m/U, 0), as stated. The resultant velocity at any point is V2 u2 2 U21 cos (4.138) where we have substituted m Ua. If we evaluate the velocities along the upper sur-face m, we find a maximum value Us,max 1.26U at 63°. This point is la-beled in Fig. 4.15 and, by Bernoulli’s equation, is the point of minimum pressure on 2a r a2 r2 m r x m r y Fig. 4.15 Superposition of a source plus a uniform stream forms a Rankine half-body.
E4.10 the body surface. After this point, the surface flow decelerates, the pressure rises, and the viscous layer grows thicker and more susceptible to “flow separation,” as we shall see in Chap. 7.
EXAMPLE 4.10 The bottom of a river has a 4-m-high bump which approximates a Rankine half-body, as in Fig. E4.10. The pressure at point B on the bottom is 130 kPa, and the river velocity is 2.5 m/s. Use inviscid theory to estimate the water pressure at point A on the bump, which is 2 m above point B.
4.10 Some Illustrative Plane Potential Flows 257 Solution As in all inviscid theories, we ignore the low-velocity boundary layers which form on solid sur-faces due to the no-slip condition. From Eq. (4.136) and Fig. 4.15, the downstream bump half-height equals a. Therefore, for our case, a (4 m)/ 1.27 m. We have to find the spot where the bump height is half that much, h 2 m a/2. From Eq. (4.136) we may compute r hA a or 90° Thus point A in Fig. E4.10 is directly above the (initially unknown) origin of coordinates (la-beled O in Fig. E4.10) and is 1.27 m to the right of the nose of the bump. With r a/2 and /2 known, we compute the velocity at point A from Eq. (4.138): V2 A U21 cos 1.405U2 or VA 1.185U 1.185(2.5 m/s) 2.96 m/s For water at 20°C, take 998 kg/m2 and & 9790 N/m3. Now, since the velocity and eleva-tion are known at point A, we are in a position to use Bernoulli’s inviscid, incompressible-flow equation (4.120) to estimate pA from the known properties at point B (on the same streamline): zA zB or 2 m 0 Solving, we find pA (13.60 2.45)(9790) 109,200 Pa Ans.
(2.5)2 2(9.81) 130,000 9790 (2.96 m/s)2 2(9.81 m/s2) pA 9790 N/m3 V2 B 2g pB & V2 A 2g pA & 2 2a a/2 a2 ( a/2)2 2 2 a( ) sin Water at 20°C 2.5 m/s B A 0 4 m 2 m 4.11 Some Illustrative Incompressible Viscous Flows If the approach velocity is uniform, this should be a pretty good approximation, since water is relatively inviscid and its boundary layers are thin.
The inviscid flows of Sec. 4.10 do not satisfy the no-slip condition. They “slip” at the wall but do not flow through the wall. To look at fully viscous no-slip conditions, we must attack the complete Navier-Stokes equation (4.74), and the result is usually not at all irrotational, nor does a velocity potential exist. We look here at three cases: (1) flow between parallel plates due to a moving upper wall, (2) flow between parallel plates due to pressure gradient, and (3) flow between concentric cylinders when the in-ner one rotates. Other cases will be given as problem assignments or considered in Chap. 6. Extensive solutions for viscous flows are discussed in Refs. 4 and 5.
Consider two-dimensional incompressible plane (/z 0) viscous flow between par-allel plates a distance 2h apart, as shown in Fig. 4.16. We assume that the plates are very wide and very long, so that the flow is essentially axial, u ' 0 but w 0.
The present case is Fig. 4.16a, where the upper plate moves at velocity V but there is no pressure gradient. Neglect gravity effects. We learn from the continuity equation (4.73) that 0 0 0 or u u(y) only Thus there is a single nonzero axial-velocity component which varies only across the channel. The flow is said to be fully developed (far downstream of the entrance). Sub-stitute u u(y) into the x-component of the Navier-Stokes momentum equation (4.74) for two-dimensional (x, y) flow: u gx or (0 0) 0 0 0 (4.139) d2u dy2 2u y2 2u x2 p x u y u x u x w z y u x 258 Chapter 4 Differential Relations for a Fluid Particle y = +h V y x u( y) y = –h Fixed (a) Fixed (b) u( y) Fixed umax Couette Flow between a Fixed and a Moving Plate Fig. 4.16 Incompressible viscous flow between parallel plates: (a) no pressure gradient, upper plate mov-ing; (b) pressure gradient p/x with both plates fixed.
Flow due to Pressure Gradient between Two Fixed Plates Most of the terms drop out, and the momentum equation simply reduces to 0 or u C1y C2 The two constants are found by applying the no-slip condition at the upper and lower plates: At y h: u V C1h C2 At y h: u 0 C1( h) C2 or C1 and C2 Therefore the solution for this case (a), flow between plates with a moving upper wall, is u y h y h (4.140) This is Couette flow due to a moving wall: a linear velocity profile with no-slip at each wall, as anticipated and sketched in Fig. 4.16a. Note that the origin has been placed in the center of the channel, for convenience in case (b) below.
What we have just presented is a rigorous derivation of the more informally dis-cussed flow of Fig. 1.6 (where y and h were defined differently).
Case (b) is sketched in Fig. 4.16b. Both plates are fixed (V 0), but the pressure varies in the x direction. If w 0, the continuity equation leads to the same conclusion as case (a), namely, that u u(y) only. The x-momentum equation (4.138) changes only because the pressure is variable: (4.141) Also, since w 0 and gravity is neglected, the y- and z-momentum equations lead to 0 and 0 or p p(x) only Thus the pressure gradient in Eq. (4.141) is the total and only gradient: const 0 (4.142) Why did we add the fact that dp/dx is constant? Recall a useful conclusion from the theory of separation of variables: If two quantities are equal and one varies only with y and the other varies only with x, then they must both equal the same constant. Oth-erwise they would not be independent of each other.
Why did we state that the constant is negative? Physically, the pressure must de-crease in the flow direction in order to drive the flow against resisting wall shear stress.
Thus the velocity profile u(y) must have negative curvature everywhere, as anticipated and sketched in Fig. 4.16b.
dp dx d2u dy2 p z p y p x d2u dy2 V 2 V 2h V 2 V 2h d2u dy2 4.11 Some Illustrative Incompressible Viscous Flows 259 The solution to Eq. (4.142) is accomplished by double integration: u C1y C2 The constants are found from the no-slip condition at each wall: At y h: u 0 or C1 0 and C2 Thus the solution to case (b), flow in a channel due to pressure gradient, is u 1 (4.143) The flow forms a Poiseuille parabola of constant negative curvature. The maximum ve-locity occurs at the centerline y 0: umax (4.144) Other (laminar) flow parameters are computed in the following example.
EXAMPLE 4.11 For case (b) above, flow between parallel plates due to the pressure gradient, compute (a) the wall shear stress, (b) the stream function, (c) the vorticity, (d) the velocity potential, and (e) the average velocity.
Solution All parameters can be computed from the basic solution, Eq. (4.143), by mathematical manipulation.
(a) The wall shear follows from the definition of a newtonian fluid, Eq. (4.37): w xy wall yh 2 h 2 1 yh h ( Ans. (a) The wall shear has the same magnitude at each wall, but by our sign convention of Fig. 4.3, the upper wall has negative shear stress.
(b) Since the flow is plane, steady, and incompressible, a stream function exists: u umax1 0 Integrating and setting 0 at the centerline for convenience, we obtain umaxy Ans. (b) At the walls, y h and 2umaxh/3, respectively.
y3 3h2 x y2 h2 y 2umax h dp dx y2 h2 dp dx y x u y h2 2 dp dx y2 h2 h2 2 dp dx h2 2 dp dx y2 2 dp dx 1 260 Chapter 4 Differential Relations for a Fluid Particle Flow between Long Concentric Cylinders (c) In plane flow, there is only a single nonzero vorticity component: z (curl V)z y Ans. (c) The vorticity is highest at the wall and is positive (counterclockwise) in the upper half and negative (clockwise) in the lower half of the fluid. Viscous flows are typically full of vor-ticity and are not at all irrotational.
(d) From part (c), the vorticity is finite. Therefore the flow is not irrotational, and the velocity potential does not exist.
Ans. (d) (e) The average velocity is defined as Vav Q/A, where Q u dA over the cross section. For our particular distribution u(y) from Eq. (4.143), we obtain Vav u dA h h umax1 b dy umax Ans. (e) In plane Poiseuille flow between parallel plates, the average velocity is two-thirds of the maximum (or centerline) value. This result could also have been obtained from the stream function derived in part (b). From Eq. (4.95), Qchannel upper lower umaxh per unit width whence Vav Q/Ab1 (4umaxh/3)/(2h) 2umax/3, the same result.
This example illustrates a statement made earlier: Knowledge of the velocity vector V [as in Eq. (4.143)] is essentially the solution to a fluid-mechanics problem, since all other flow properties can then be calculated.
Consider a fluid of constant (, ) between two concentric cylinders, as in Fig. 4.17.
There is no axial motion or end effect z /z 0. Let the inner cylinder rotate at angular velocity i. Let the outer cylinder be fixed. There is circular symmetry, so the velocity does not vary with and varies only with r.
4 3 2umaxh 3 2umaxh 3 2 3 y2 h2 1 b(2h) 1 A 2umax h2 u y x 4.11 Some Illustrative Incompressible Viscous Flows 261 Fixed Ω i ro vθ r ri Fluid: ρ, µ Fig. 4.17 Coordinate system for incompressible viscous flow be-tween a fixed outer cylinder and a steadily rotating inner cylinder.
Instability of Rotating Inner Cylinder Flow The continuity equation for this problem is Eq. (D.2): (rr) 0 (rr) or rr const Note that does not vary with . Since r 0 at both the inner and outer cylinders, it follows that r 0 everywhere and the motion can only be purely circumferential, (r). The -momentum equation (D.6) becomes (V ) g 2 For the conditions of the present problem, all terms are zero except the last. Therefore the basic differential equation for flow between rotating cylinders is 2 r (4.145) This is a linear second-order ordinary differential equation with the solution C1r The constants are found by the no-slip condition at the inner and outer cylinders: Outer, at r ro: 0 C1ro Inner, at r ri: iri C1ri The final solution for the velocity distribution is Rotating inner cylinder: iri (4.146) The velocity profile closely resembles the sketch in Fig. 4.17. Variations of this case, such as a rotating outer cylinder, are given in the problem assignments.
The classic Couette-flow solution11 of Eq. (4.146) describes a physically satisfying con-cave, two-dimensional, laminar-flow velocity profile as in Fig. 4.17. The solution is mathematically exact for an incompressible fluid. However, it becomes unstable at a relatively low rate of rotation of the inner cylinder, as shown in 1923 in a classic pa-per by G. I. Taylor . At a critical value of what is now called the Taylor number, denoted Ta, Tacrit 1700 (4.147) the plane flow of Fig. 4.17 vanishes and is replaced by a laminar three-dimensional flow pattern consisting of rows of nearly square alternating toroidal vortices. An ex-ri(ro ri)32 i 2 ro/r r/ro ro/ri ri/ro C2 ri C2 ro C2 r r2 d dr d dr 1 r r2 p 1 r r r d dr 1 r 1 r r 1 r 262 Chapter 4 Differential Relations for a Fluid Particle 11Named after M. Couette, whose pioneering paper in 1890 established rotating cylinders as a method, still used today, for measuring the viscosity of fluids.
perimental demonstration of toroidal “Taylor vortices” is shown in Fig. 4.18a, mea-sured at Ta 1.16 Tacrit by Koschmieder . At higher Taylor numbers, the vortices also develop a circumferential periodicity but are still laminar, as illustrated in Fig.
4.18b. At still higher Ta, turbulence ensues. This interesting instability reminds us that the Navier-Stokes equations, being nonlinear, do admit to multiple (nonunique) lami-nar solutions in addition to the usual instabilities associated with turbulence and chaotic dynamic systems.
This chapter complements Chap. 3 by using an infinitesimal control volume to derive the basic partial differential equations of mass, momentum, and energy for a fluid.
These equations, together with thermodynamic state relations for the fluid and appro-4.11 Some Illustrative Incompressible Viscous Flows 263 (a) (b) Fig. 4.18 Experimental verification of the instability of flow between a fixed outer and a rotating inner cylinder. (a) Toroidal Taylor vor-tices exist at 1.16 times the critical speed; (b) at 8.5 times the critical speed, the vortices are doubly peri-odic. (After Koschmieder, Ref. 18.) This instability does not occur if only the outer cylinder rotates.
Summary Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 4.1 to 4.91 (categorized in the problem list below) are followed by word problems W4.1 to W4.10, fun-damentals of engineering exam problems FE4.1 to FE4.3, and com-prehensive problem C4.1.
Problem distribution Section Topic Problems 4.1 The acceleration of a fluid 4.1–4.8 4.2 The continuity equation 4.9–4.25 4.3 Linear momentum: Navier-Stokes 4.26–4.37 4.4 Angular momentum: couple stresses 4.38 4.5 The differential energy equation 4.39–4.42 4.6 Boundary conditions 4.43–4.46 4.7 Stream function 4.47–4.55 4.8 Vorticity, irrotationality 4.56–4.60 4.9 Velocity potential 4.61–4.67 4.10 Plane potential flows 4.68–4.78 4.11 Incompressible viscous flows 4.79–4.91 P4.1 An idealized velocity field is given by the formula V 4txi 2t2yj 4xzk Is this flow field steady or unsteady? Is it two- or three-di-mensional? At the point (x, y, z) ( 1, 1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the acceleration.
P4.2 Flow through the converging nozzle in Fig. P4.2 can be ap-proximated by the one-dimensional velocity distribution u V01 0 w 0 (a) Find a general expression for the fluid acceleration in the nozzle. (b) For the specific case V0 10 ft/s and L 6 in, compute the acceleration, in g’s, at the entrance and at the exit.
2x L 264 Chapter 4 Differential Relations for a Fluid Particle priate boundary conditions, in principle can be solved for the complete flow field in any given fluid-mechanics problem. Except for Chap. 9, in most of the problems to be studied here an incompressible fluid with constant viscosity is assumed.
In addition to deriving the basic equations of mass, momentum, and energy, this chapter introduced some supplementary ideas—the stream function, vorticity, irrota-tionality, and the velocity potential—which will be useful in coming chapters, espe-cially Chap. 8. Temperature and density variations will be neglected except in Chap.
9, where compressibility is studied.
This chapter ended by discussing a few classical solutions for inviscid flows (uni-form stream, source, sink, vortex, half-body) and for viscous flows (Couette flow due to moving walls and Poiseuille flow due to pressure gradient). Whole books [11, 13] are written on the basic equations of fluid mechanics. Whole books [4, 5, 15] are writ-ten on classical solutions to fluid-flow problems. Reference 12 contains 360 solved problems which relate fluid mechanics to the whole of continuum mechanics. This does not mean that all problems can be readily solved mathematically, even with the mod-ern digital-computer codes now available. Often the geometry and boundary conditions are so complex that experimentation (Chap. 5) is a necessity.
V0 u = 3V0 x = L x x = 0 P4.2 P4.3 A two-dimensional velocity field is given by V (x2 y2 x)i (2xy y)j in arbitrary units. At (x, y) (1, 2), compute (a) the accel-erations ax and ay, (b) the velocity component in the direc-tion 40°, (c) the direction of maximum velocity, and (d) the direction of maximum acceleration.
P4.4 Suppose that the temperature field T 4x2 3y3, in arbi-trary units, is associated with the velocity field of Prob. 4.3.
Compute the rate of change dT/dt at (x, y) (2, 1).
P4.5 The velocity field near a stagnation point (see Example 1.10) may be written in the form u U0 and L are constants (a) Show that the acceleration vector is purely radial. (b) For the particular case L 1.5 m, if the acceleration at (x, y) (1 m, 1 m) is 25 m/s2, what is the value of U0?
P4.6 Assume that flow in the converging nozzle of Fig. P4.2 has the form V V0[1 (2x)/L]i. Compute (a) the fluid accel-eration at x L and (b) the time required for a fluid parti-cle to travel from x 0 to x L.
P4.7 Consider a sphere of radius R immersed in a uniform stream U0, as shown in Fig. P4.7. According to the theory of Chap.
8, the fluid velocity along streamline AB is given by V ui U01 i Find (a) the position of maximum fluid acceleration along AB and (b) the time required for a fluid particle to travel from A to B.
R3 x3 U0y L U0x L the time for which the fluid acceleration at x L is zero.
Why does the fluid acceleration become negative after con-dition (b)?
P4.9 A velocity field is given by V (3y2 3x2)i Cxyj 0k.
Determine the value of the constant C if the flow is to be (a) incompressible and (b) irrotational.
P4.10 Write the special cases of the equation of continuity for (a) steady compressible flow in the yz plane, (b) unsteady in-compressible flow in the xz plane, (c) unsteady compress-ible flow in the y direction only, (d) steady compressible flow in plane polar coordinates.
P4.11 Derive Eq. (4.12b) for cylindrical coordinates by consider-ing the flux of an incompressible fluid in and out of the el-emental control volume in Fig. 4.2.
P4.12 Spherical polar coordinates (r, , $) are defined in Fig.
P4.12. The cartesian transformations are x r sin cos $ y r sin sin $ z r cos Problems 265 x = 0 x = L u(x, t) U0 A x = – 4R B y x Sphere R P4.7 P4.8 When a valve is opened, fluid flows in the expansion duct of Fig. 4.8 according to the approximation V iU1 tanh Find (a) the fluid acceleration at (x, t) (L, L/U) and (b) Ut L x 2L P4.8 y x z P θ φ υθ r = constant υφ r r υ P4.12 The cartesian incompressible continuity relation (4.12a) can be transformed to the spherical polar form (r2r) ( sin ) ($) 0 What is the most general form of r when the flow is purely radial, that is, and $ are zero?
P4.13 A two-dimensional velocity field is given by u where K is constant. Does this field satisfy incompressible Kx x2 y2 Ky x2 y2 $ 1 r sin 1 r sin r 1 r2 continuity? Transform these velocities to polar components r and . What might the flow represent?
P4.14 For incompressible polar-coordinate flow, what is the most general form of a purely circulatory motion, (r, , t) and r 0, which satisfies continuity?
P4.15 What is the most general form of a purely radial polar-coordinate incompressible-flow pattern, r r(r, , t) and 0, which satisfies continuity?
P4.16 An incompressible steady-flow pattern is given by u x3 2z2 and w y3 2yz. What is the most general form of the third component, (x, y, z), which satisfies continuity?
P4.17 A reasonable approximation for the two-dimensional in-compressible laminar boundary layer on the flat surface in Fig. P4.17 is u U for y where Cx1/2, C const (a) Assuming a no-slip condition at the wall, find an ex-pression for the velocity component (x, y) for y . (b) Then find the maximum value of at the station x 1 m, for the particular case of airflow, when U 3 m/s and 1.1 cm.
y2 2 2y P4.20 A two-dimensional incompressible velocity field has u K(1 e ay), for x L and 0 y . What is the most general form of (x, y) for which continuity is satisfied and 0 at y 0? What are the proper dimensions for con-stants K and a?
P4.21 Air flows under steady, approximately one-dimensional conditions through the conical nozzle in Fig. P4.21. If the speed of sound is approximately 340 m/s, what is the min-imum nozzle-diameter ratio De/D0 for which we can safely neglect compressibility effects if V0 (a) 10 m/s and (b) 30 m/s?
266 Chapter 4 Differential Relations for a Fluid Particle D0 De V0 Ve P4.21 V = constant x = 0 x = L(t) x (t) u(x, t) ρ P4.18 U y x u(x, y) U Layer thickness (x) 0 U = constant δ u(x, y) P4.17 P4.18 A piston compresses gas in a cylinder by moving at constant speed , as in Fig. P4.18. Let the gas density and length at t 0 be 0 and L0, respectively. Let the gas velocity vary lin-early from u V at the piston face to u 0 at x L. If the gas density varies only with time, find an expression for (t).
P4.19 An incompressible flow field has the cylindrical components Cr, z K(R2 r2), r 0, where C and K are con-stants and r R, z L. Does this flow satisfy continuity?
What might it represent physically?
P4.22 Air at a certain temperature and pressure flows through a contracting nozzle of length L whose area decreases linearly, A A0[1 x/(2L)]. The air average velocity increases nearly linearly from 76 m/s at x 0 to 167 m/s at x L. If the density at x 0 is 2.0 kg/m3, estimate the density at x L.
P4.23 A tank volume contains gas at conditions (0, p0, T0). At time t 0 it is punctured by a small hole of area A. Ac-cording to the theory of Chap. 9, the mass flow out of such a hole is approximately proportional to A and to the tank pressure. If the tank temperature is assumed constant and the gas is ideal, find an expression for the variation of den-sity within the tank.
P4.24 Reconsider Fig. P4.17 in the following general way. It is known that the boundary layer thickness (x) increases mo-notonically and that there is no slip at the wall (y 0). Fur-ther, u(x, y) merges smoothly with the outer stream flow, where u U constant outside the layer. Use these facts to prove that (a) the component (x, y) is positive every-where within the layer, (b) increases parabolically with y very near the wall, and (c) is a maximum at y .
P4.25 An incompressible flow in polar coordinates is given by r K cos 1 K sin 1 Does this field satisfy continuity? For consistency, what b r2 b r2 should the dimensions of constants K and b be? Sketch the surface where r 0 and interpret.
P4.26 Curvilinear, or streamline, coordinates are defined in Fig.
P4.26, where n is normal to the streamline in the plane of the radius of curvature R. Show that Euler’s frictionless mo-mentum equation (4.36) in streamline coordinates becomes V gs (1) V gn (2) Further show that the integral of Eq. (1) with respect to s is none other than our old friend Bernoulli’s equation (3.76).
p n 1 V2 R t p s 1 V s V t Is this also the position of maximum fluid deceleration?
Evaluate the maximum viscous normal stress if the fluid is SAE 30 oil at 20°C, with U 2 m/s and a 6 cm.
P4.32 The answer to Prob. 4.14 is f(r) only. Do not reveal this to your friends if they are still working on Prob. 4.14.
Show that this flow field is an exact solution to the Navier-Stokes equations (4.38) for only two special cases of the function f(r). Neglect gravity. Interpret these two cases phys-ically.
P4.33 From Prob. 4.15 the purely radial polar-coordinate flow which satisfies continuity is r f( )/r, where f is an arbi-trary function. Determine what particular forms of f( ) sat-isfy the full Navier-Stokes equations in polar-coordinate form from Eqs. (D.5) and (D.6).
P4.34 The fully developed laminar-pipe-flow solution of Prob.
3.53, z umax(1 r2/R2), 0, r 0, is an exact so-lution to the cylindrical Navier-Stokes equations (App. D).
Neglecting gravity, compute the pressure distribution in the pipe p(r, z) and the shear-stress distribution (r, z), using R, umax, and as parameters. Why does the maximum shear occur at the wall? Why does the density not appear as a pa-rameter?
P4.35 From the Navier-Stokes equations for incompressible flow in polar coordinates (App. D for cylindrical coordinates), find the most general case of purely circulating motion (r), r z 0, for flow with no slip between two fixed con-centric cylinders, as in Fig. P4.35.
Problems 267 r r = a r = b (r ) No slip θ υ P4.35 Stagnation point (u = 0) 0 a y x P4.31 n s, V θ z x y R Streamline P4.26 P4.27 A frictionless, incompressible steady-flow field is given by V 2xyi y2j in arbitrary units. Let the density be 0 constant and ne-glect gravity. Find an expression for the pressure gradient in the x direction.
P4.28 If z is “up,” what are the conditions on constants a and b for which the velocity field u ay, bx, w 0 is an ex-act solution to the continuity and Navier-Stokes equations for incompressible flow?
P4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid in which the velocity field is known, i.e., u 2xy, y2 x2, w 0. (a) Does this flow satisfy conservation of mass? (b) Find the pressure field, p(x, y) if the pressure at the point (x 0, y 0) is equal to pa.
P4.30 Show that the two-dimensional flow field of Example 1.10 is an exact solution to the incompressible Navier-Stokes equations (4.38). Neglecting gravity, compute the pressure field p(x, y) and relate it to the absolute velocity V2 u2 2. Interpret the result.
P4.31 According to potential theory (Chap. 8) for the flow ap-proaching a rounded two-dimensional body, as in Fig. P4.31, the velocity approaching the stagnation point is given by u U(1 a2/x2), where a is the nose radius and U is the velocity far upstream. Compute the value and position of the maximum viscous normal stress along this streamline.
EES P4.36 A constant-thickness film of viscous liquid flows in lami-nar motion down a plate inclined at angle , as in Fig. P4.36.
The velocity profile is u Cy(2h y) w 0 Find the constant C in terms of the specific weight and vis-cosity and the angle . Find the volume flux Q per unit width in terms of these parameters.
u w 0 If the wall temperature is Tw at both walls, use the incom-pressible-flow energy equation (4.75) to solve for the tem-perature distribution T(y) between the walls for steady flow.
P4.41 The approximate velocity profile in Prob. 3.18 and Fig. P3.18 for steady laminar flow through a duct, was suggested as u umax 1 1 With w 0, it satisfied the no-slip condition and gave a reasonable volume-flow estimate (which was the point of Prob. 3.18). Show, however, that it does not satsify the x-momentum Navier-Stokes equation for duct flow with con-stant pressure gradient p/x 0. For extra credit, explain briefly how the actual exact solution to this problem is ob-tained (see, for example, Ref. 5, pp. 120–121).
P4.42 In duct-flow problems with heat transfer, one often defines an average fluid temperature. Consider the duct flow of Fig.
P4.40 of width b into the paper. Using a control-volume in-tegral analysis with constant density and specific heat, de-rive an expression for the temperature arising if the entire duct flow poured into a bucket and was stirred uniformly.
Assume arbitrary u(y) and T(y). This average is called the cup-mixing temperature of the flow.
P4.43 For the draining liquid film of Fig. P4.36, what are the ap-propriate boundary conditions (a) at the bottom y 0 and (b) at the surface y h?
P4.44 Suppose that we wish to analyze the sudden pipe-expansion flow of Fig. P3.59, using the full continuity and Navier-Stokes equations. What are the proper boundary conditions to handle this problem?
P4.45 Suppose that we wish to analyze the U-tube oscillation flow of Fig. P3.96, using the full continuity and Navier-Stokes equations. What are the proper boundary conditions to han-dle this problem?
P4.46 Fluid from a large reservoir at temperature T0 flows into a circular pipe of radius R. The pipe walls are wound with an electric-resistance coil which delivers heat to the fluid at a rate qw (energy per unit wall area). If we wish to analyze z2 h2 y2 b2 4umaxy(h y) h2 268 Chapter 4 Differential Relations for a Fluid Particle y = h y y = 0 u(y) T(y) x Tw Tw P4.40 h h x z, w P4.37 g y u(y ) x θ h P4.36 P4.37 A viscous liquid of constant and falls due to gravity be-tween two plates a distance 2h apart, as in Fig. P4.37. The flow is fully developed, with a single velocity component w w(x). There are no applied pressure gradients, only gravity. Solve the Navier-Stokes equation for the velocity profile between the plates.
P4.38 Reconsider the angular-momentum balance of Fig. 4.5 by adding a concentrated body couple Cz about the z axis .
Determine a relation between the body couple and shear stress for equilibrium. What are the proper dimensions for Cz? (Body couples are important in continuous media with microstructure, such as granular materials.) P4.39 Problems involving viscous dissipation of energy are depen-dent on viscosity , thermal conductivity k, stream velocity U0, and stream temperature T0. Group these parameters into the di-mensionless Brinkman number, which is proportional to .
P4.40 As mentioned in Sec. 4.11, the velocity profile for laminar flow between two plates, as in Fig. P4.40, is this problem by using the full continuity, Navier-Stokes, and energy equations, what are the proper boundary conditions for the analysis?
P4.47 A two-dimensional incompressible flow is given by the ve-locity field V 3yi 2xj, in arbitrary units. Does this flow satisfy continuity? If so, find the stream function (x, y) and plot a few streamlines, with arrows.
P4.48 Determine the incompressible two-dimensional stream function (x, y) which represents the flow field given in Ex-ample 1.10.
P4.49 Investigate the stream function K(x2 y2), K con-stant. Plot the streamlines in the full xy plane, find any stag-nation points, and interpret what the flow could represent.
P4.50 Investigate the polar-coordinate stream function Kr1/2 sin 1 2 , K constant. Plot the streamlines in the full xy plane, find any stagnation points, and interpret.
P4.51 Investigate the polar-coordinate stream function Kr2/3 sin (2 /3), K constant. Plot the streamlines in all ex-cept the bottom right quadrant, and interpret.
P4.52 A two-dimensional, incompressible, frictionless fluid is guided by wedge-shaped walls into a small slot at the ori-gin, as in Fig. P4.52. The width into the paper is b, and the volume flow rate is Q. At any given distance r from the slot, the flow is radial inward, with constant velocity. Find an ex-pression for the polar-coordinate stream function of this flow.
(2L, 0, 0), (2L, 0, b), (0, L, b), and (0, L, 0). Show the di-rection of Q.
P4.55 In spherical polar coordinates, as in Fig. P4.12, the flow is called axisymmetric if $ 0 and /$ 0, so that r r(r, ) and (r, ). Show that a stream function (r, ) exists for this case and is given by r This is called the Stokes stream function [5, p. 204].
P4.56 Investigate the velocity potential $ Kxy, K constant.
Sketch the potential lines in the full xy plane, find any stag-nation points, and sketch in by eye the orthogonal stream-lines. What could the flow represent?
P4.57 Determine the incompressible two-dimensional velocity po-tential $(x, y) which represents the flow field given in Ex-ample 1.10. Sketch a few potential lines and streamlines.
P4.58 Show that the incompressible velocity potential in plane po-lar coordinates $(r, ) is such that r Further show that the angular velocity about the z-axis in such a flow would be given by 2 z (r ) (r) Finally show that $ as defined above satisfies Laplace’s equation in polar coordinates for incompressible flow.
P4.59 Consider the simple flow defined by V xi yj, in arbi-trary units. At t 0, consider the rectangular fluid element defined by the lines x 2, x 3 and y 2, y 3. Deter-mine, and draw to scale, the location of this fluid element at t 0.5 unit. Relate this new element shape to whether the flow is irrotational or incompressible.
P4.60 Liquid drains from a small hole in a tank, as shown in Fig.
P4.60, such that the velocity field set up is given by r 0, z 0, R2/r, where z H is the depth of the water 1 r r 1 r $ 1 r $ r r 1 r sin 1 r2 sin Problems 269 patm z = 0 z r r = R zC?
z = H P4.60 Slot θ = π /4 vr θ = 0 r P4.52 P4.53 For the fully developed laminar-pipe-flow solution of Prob.
4.34, find the axisymmetric stream function (r, z). Use this result to determine the average velocity V Q/A in the pipe as a ratio of umax.
P4.54 An incompressible stream function is defined by (x, y) (3x2y y3) where U and L are (positive) constants. Where in this chap-ter are the streamlines of this flow plotted? Use this stream function to find the volume flow Q passing through the rec-tangular surface whose corners are defined by (x, y, z) U L2 far from the hole. Is this flow pattern rotational or irrota-tional? Find the depth zC of the water at the radius r R.
P4.61 Investigate the polar-coordinate velocity potential $ Kr1/2 cos 1 2 , K constant. Plot the potential lines in the full xy plane, sketch in by eye the orthogonal streamlines, and interpret.
P4.62 Show that the linear Couette flow between plates in Fig.
1.6 has a stream function but no velocity potential. Why is this so?
P4.63 Find the two-dimensional velocity potential $(r, ) for the polar-coordinate flow pattern r Q/r, K/r, where Q and K are constants.
P4.64 Show that the velocity potential $(r, z) in axisymmetric cylindrical coordinates (see Fig. 4.2) is defined such that r z Further show that for incompressible flow this potential satisfies Laplace’s equation in (r, z) coordinates.
P4.65 A two-dimensional incompressible flow is defined by u where K constant. Is this flow irrotational? If so, find its velocity potential, sketch a few potential lines, and in-terpret the flow pattern.
P4.66 A plane polar-coordinate velocity potential is defined by $ K const Find the stream function for this flow, sketch some stream-lines and potential lines, and interpret the flow pattern.
P4.67 A stream function for a plane, irrotational, polar-coordi-nate flow is C K ln r C and K const Find the velocity potential for this flow. Sketch some stream-lines and potential lines, and interpret the flow pattern.
P4.68 Find the stream function and plot some streamlines for the combination of a line source m at (x, y) (0, a) and an equal line source placed at (0, a).
P4.69 Find the stream function and plot some streamlines for the combination of a counterclockwise line vortex K at (x, y) (a, 0) and an equal line vortex placed at ( a, 0).
P4.70 Superposition of a source of strength m at ( a, 0) and a sink (source of strength m) at (a, 0) was discussed briefly in this chapter, where it was shown that the velocity po-tential function is $ m ln (x a)2 y2 (x a)2 y2 1 2 K cos r Kx x2 y2 Ky x2 y2 $ z $ r A doublet is formed in the limit as a goes to zero (the source and sink come together) while at the same time their strengths m and m go to infinity and minus infinity, re-spectively, with the product a m remaining constant. (a) Find the limiting value of velocity potential for the doublet.
Hint: Expand the natural logarithm as an infinite series of the form ln 2# as # goes to zero. (b) Rewrite your result for $doublet in cylin-drical coordinates.
P4.71 Find the stream function and plot some streamlines for the combination of a counterclockwise line vortex K at (x, y) (a, 0) and an opposite (clockwise) line vortex placed at ( a, 0).
P4.72 A coastal power plant takes in cooling water through a ver-tical perforated manifold, as in Fig. P4.72. The total volume flow intake is 110 m3/s. Currents of 25 cm/s flow past the manifold, as shown. Estimate (a) how far downstream and (b) how far normal to the paper the effects of the intake are felt in the ambient 8-m-deep waters.
3 3 1 # 1 # 270 Chapter 4 Differential Relations for a Fluid Particle 25 cm/s Water Manifold 110 m3/s 8 m P4.72 P4.73 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 120 kPa. What is the nose ra-dius of the half-body? At what tunnel flow velocity will cav-itation bubbles begin to form on the surface of the body?
P4.74 Find the stream function and plot some streamlines for the combination of a uniform stream iU and a clockwise line vortex K at the origin. Are there any stagnation points in the flow field?
P4.75 Find the stream function and plot some streamlines for the combination of a line source 2m at (x, y) (a, 0) and a line source m at ( a, 0). Are there any stagnation points in the flow field?
P4.76 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. 4.76. The jet volume flow is 0.4 m3/s 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?
P4.77 A tornado is simulated by a line sink m 1000 m2/s plus a line vortex K 1600 m2/s. Find the angle between any streamline and a radial line, and show that it is independent of both r and . If this tornado forms in sea-level standard air, at what radius will the local pressure be equivalent to 29 inHg?
P4.78 The solution to Prob. 4.68 (do not reveal!) can represent a line source m at (0, a) near a horizontal wall (y 0). [The other source at (0, a) represents an “image” to create the wall.] Find (a) the magnitude of the maxinun flow velocity along the wall and (b) the point of minimum pressure along the wall. Hint: Use Bernoulli’s equation.
P4.79 Study the combined effect of the two viscous flows in Fig.
4.16. That is, find u(y) when the upper plate moves at speed V and there is also a constant pressure gradient (dp/dx). Is superposition possible? If so, explain why. Plot representa-tive velocity profiles for (a) zero, (b) positive, and (c) neg-ative pressure gradients for the same upper-wall speed V.
P4.80 Oil, of density and viscosity , 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 be-come independent of z and of constant thickness . 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 approximate shape.
(b) Suppose that film thickness and the slope of the ve-locity profile at the wall [w/x]wall are measured with a laser-Doppler anemometer (Chap. 6). Find an expression for oil viscosity as a function of (, , g, [w/x]wall).
P4.81 Modify the analysis of Fig. 4.17 to find the velocity u when the inner cylinder is fixed and the outer cylinder rotates at angular velocity 0. May this solution be added to Eq.
(4.146) 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 ve-locity in a viscous incompressible fluid which is at rest far from the cylinder, as in Fig. P4.82. Make simplifying assumptions and derive the governing differential equation and boundary conditions for the velocity field in the fluid.
Do not solve unless you are obsessed with this problem.
What is the steady-state flow field for this problem?
Problems 271 A S 1.2 m/s 0.4 m3/(s • m) G P4.76 z x g Oil film Air Plate P4.80 r θ r = R Ω υ (r, , t) θ θ P4.82 P4.83 The flow pattern in bearing lubrication can be illustrated by Fig. P4.83, where a viscous oil (, ) is forced into the gap h(x) between a fixed slipper block and a wall moving at ve-locity U. If the gap is thin, h L, it can be shown that the pressure and velocity distributions are of the form p p(x), u u(y), w 0. Neglecting gravity, reduce the Navier-Stokes equations (4.38) to a single differential equation for u(y). What are the proper boundary conditions? Integrate and show that u (y2 yh) U1 where h h(x) may be an arbitrary slowly varying gap width. (For further information on lubrication theory, see Ref. 16.) y h dp dx 1 2 Oil inlet y h0 h(x) u( y) h1 Moving wall U Oil outlet x Fixed slipper block P4.83 P4.84 Consider a viscous film of liquid draining uniformly down the side of a vertical rod of radius a, as in Fig. P4.84. At some distance down the rod the film will approach a ter-minal or fully developed draining flow of constant outer ra-dius b, with z z(r), r 0. Assume that the at-mosphere offers no shear resistance to the film motion.
Derive a differential equation for z, state the proper bound-ary conditions, and solve for the film velocity distribution.
How does the film radius b relate to the total film volume flow rate Q?
P4.87 Suppose in Fig. 4.17 that neither cylinder is rotating. The fluid has constant (, , k, cp). What, then, is the steady-flow solution for (r)? For this condition, suppose that the inner and outer cylinder surface temperatures are Ti and To, respectively. Simplify the differential energy equation ap-propriately for this problem, state the boundary conditions, and find the temperature distribution in the fluid. Neglect gravity.
P4.88 The viscous oil in Fig. P4.88 is set into steady motion by a concentric inner cylinder moving axially at velocity U in-side a fixed outer cylinder. Assuming constant pressure and density and a purely axial fluid motion, solve Eqs. (4.38) for the fluid velocity distribution z(r). What are the proper boundary conditions?
272 Chapter 4 Differential Relations for a Fluid Particle Fully developed region Q Film υ b z r pa a ≈ 0 µ ρ a µ z P4.84 P4.85 A flat plate of essentially infinite width and breadth oscil-lates sinusoidally in its own plane beneath a viscous fluid, as in Fig. P4.85. The fluid is at rest far above the plate. Mak-ing as many simplifying assumptions as you can, set up the governing differential equation and boundary conditions for finding the velocity field u in the fluid. Do not solve (if you can solve it immediately, you might be able to get exempted from the balance of this course with credit).
x y Incompressible viscous fluid u(x, y, z, t)?
Plate velocity: U0 sin ω t P4.85 P4.86 SAE 10 oil at 20°C flows between parallel plates 8 cm apart, as in Fig. P4.86. A mercury manometer, with wall pressure taps 1 m apart, registers a 6-cm height, as shown. Estimate the flow rate of oil for this condition.
SAE 10 oil Q 8 mm 6 cm Mercury 1 m P4.86 Fixed outer cylinder U Oil:ρ, µ vz b vz(r) r a P4.88 P4.89 Modify Prob. 4.88 so that the outer cylinder also moves to the left at constant speed V. Find the velocity distribution z(r). For what ratio V/U will the wall shear stress be the same at both cylinder surfaces?
P4.90 A 5-cm-diameter rod is pulled steadily at 2 m/s through a fixed cylinder whose clearance is filled with SAE 10 oil at 20°C, as in Fig. P4.90. Estimate the (steady) force required to pull the inner rod.
P4.91 Consider two-dimensional, incompressible, steady Couette flow (flow between two infinite parallel plates with the up-per plate moving at constant speed and the lower plate sta-tionary, as in Fig. 4.16a). Let the fluid be nonnewtonian, with its viscous stresses given by EES xx a c yy a c zz a c xy yx 1 2 a c xz zx 1 2 a c yz zy 1 2 a c where a and c are constants of the fluid. Make all the same assumptions as in the derivation of Eq. (4.140). (a) Find the velocity profile u(y). (b) How does the velocity profile for this case compare to that of a newtonian fluid?
w y z w x u z x u y w z y u x Fundamentals of Engineering Exam Problems 273 2 cm 1.5 m 5 cm SAE 10 oil 2 m/s Fixed cylinder Moving rod 2 cm P4.90 Word Problems W4.1 The total acceleration of a fluid particle is given by Eq.
(4.2) in the eulerian system, where V is a known function of space and time. Explain how we might evaluate parti-cle acceleration in the lagrangian frame, where particle po-sition r is a known function of time and initial position, r fcn(r0, t). Can you give an illustrative example?
W4.2 Is it true that the continuity relation, Eq. (4.6), is valid for both viscous and inviscid, newtonian and nonnewtonian, compressible and incompressible flow? If so, are there any limitations on this equation?
W4.3 Consider a CD compact disk rotating at angular velocity . Does it have vorticity in the sense of this chapter? If so, how much vorticity?
W4.4 How much acceleration can fluids endure? Are fluids like astronauts, who feel that 5g is severe? Perhaps use the flow pattern of Example 4.8, at r R, to make some estimates of fluid-acceleration magnitudes.
W4.5 State the conditions (there are more than one) under which the analysis of temperature distribution in a flow field can be completely uncoupled, so that a separate analysis for velocity and pressure is possible. Can we do this for both laminar and turbulent flow?
W4.6 Consider liquid flow over a dam or weir. How might the boundary conditions and the flow pattern change when we compare water flow over a large prototype to SAE 30 oil flow over a tiny scale model?
W4.7 What is the difference between the stream function and our method of finding the streamlines from Sec. 1.9? Or are they essentially the same?
W4.8 Under what conditions do both the stream function and the velocity potential $ exist for a flow field? When does one exist but not the other?
W4.9 How might the remarkable three-dimensional Taylor in-stability of Fig. 4.18 be predicted? Discuss a general pro-cedure for examining the stability of a given flow pattern.
W4.10 Consider an irrotational, incompressible, axisymmetric (/ 0) flow in (r, z) coordinates. Does a stream func-tion exist? If so, does it satisfy Laplace’s equation? Are lines of constant equal to the flow streamlines? Does a velocity potential exist? If so, does it satisfy Laplace’s equation? Are lines of constant $ everywhere perpendicu-lar to the lines?
Fundamentals of Engineering Exam Problems This chapter is not a favorite of the people who prepare the FE Exam. Probably not a single problem from this chapter will appear on the exam, but if some did, they might be like these.
FE4.1 Given the steady, incompressible velocity distribution V 3xi Cyj 0k, where C is a constant, if conservation of mass is satisfied, the value of C should be (a) 3, (b) 3/2, (c) 0, (d) 3/2, (e) 3 FE4.2 Given the steady velocity distribution V 3xi 0j Cyk, where C is a constant, if the flow is irrotational, the value of C should be (a) 3, (b) 3/2, (c) 0, (d) 3/2, (e) 3 FE4.3 Given the steady, incompressible velocity distribution V 3xi Cyj 0k, where C is a constant, the shear stress xy at the point (x, y, z) is given by (a) 3, (b) (3x Cy), (c) 0, (d) C, (e) (3 C) Comprehensive Problem C4.1 In a certain medical application, water at room temperature and pressure flows through a rectangular channel of length L 10 cm, width s 1.0 cm, and gap thickness b 0.30 mm as in Fig. C4.1. The volume flow rate is sinusoidal with amplitude ˆ Q 0.50 mL/s and frequency f 20 Hz, i.e., Q ˆ Q sin (2 ft).
(a) Calculate the maximum Reynolds number (Re Vb/ ) based on maximum average velocity and gap thickness.
Channel flow like this remains laminar for Re less than about 2000. If Re is greater than about 2000, the flow will be tur-bulent. Is this flow laminar or turbulent? (b) In this problem, the frequency is low enough that at any given time, the flow can be solved as if it were steady at the given flow rate. (This is called a quasi-steady assumption.) At any arbitrary instant of time, find an expression for streamwise velocity u as a function of y, , dp/dx, and b, where dp/dx is the pressure gradient required to push the flow through the channel at vol-ume flow rate Q. In addition, estimate the maximum mag-nitude of velocity component u. (c) At any instant of time, find a relationship between volume flow rate Q and pressure gradient dp/dx. Your answer should be given as an expres-sion for Q as a function of dp/dx, s, b, and viscosity . (d) Estimate the wall shear stress, w as a function of ˆ Q, f, , b, s, and time (t). (e) Finally, for the numbers given in the prob-lem statement, estimate the amplitude of the wall shear stress, ˆw, in N/m2.
274 Chapter 4 Differential Relations for a Fluid Particle L s b y x z Q References 1.
D. A. Anderson, J. C. Tannehill, and R. H. Pletcher, Compu-tational Fluid Mechanics and Heat Transfer, 2d ed., McGraw-Hill, New York, 1997.
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G. B. Wallis, One-Dimensional Two-Phase Flow, McGraw-Hill, New York, 1969.
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S. M. Selby, CRC Handbook of Tables for Mathematics, 4th ed., CRC Press Inc., Cleveland, OH, 1976.
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H. Schlichting, Boundary Layer Theory, 7th ed., McGraw-Hill, New York, 1979.
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F. M. White, Viscous Fluid Flow, 2d ed., McGraw-Hill, New York, 1991.
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L. E. Malvern, Introduction to Mechanics of a Continuous Medium, Prentice-Hall, Englewood Cliffs, NJ, 1969.
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J. P. Holman, Heat Transfer, 8th ed., McGraw-Hill, New York, 1997.
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W. M. Kays and M. E. Crawford, Convective Heat and Mass Transfer, 3d ed., McGraw-Hill, New York, 1993.
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G. K. Batchelor, An Introduction to Fluid Dynamics, Cam-bridge University Press, Cambridge, England, 1967.
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L. Prandtl and O. G. Tietjens, Fundamentals of Hydro- and Aeromechanics, Dover, New York, 1957.
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R. Aris, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Prentice-Hall, Englewood Cliffs, NJ, 1962.
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G. E. Mase, Continuum Mechanics, Schaum’s Outline Series, McGraw-Hill, New York, 1970.
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W. F. Hughes and E. W. Gaylord, Basic Equations of Engi-neering Science, Schaum’s Outline Series, McGraw-Hill, New York, 1964.
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G. Astarita and G. Marrucci, Principles of Non-Newtonian Fluid Mechanics, McGraw-Hill, New York, 1974.
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A. Szeri, Tribology: Friction, Lubrication, and Wear, McGraw-Hill, New York, 1980.
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G. I. Taylor, “Stability of a Viscous Liquid Contained be-tween Two Rotating Cylinders,” Philos. Trans. Roy. Soc. Lon-don Ser. A, vol. 223, 1923, pp. 289–343.
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E. L. Koschmieder, “Turbulent Taylor Vortex Flow,” J. Fluid Mech., vol. 93, pt. 3, 1979, pp. 515–527.
C4.1 Wind tunnel test of an F-18 fighter plane model. Testing of models is imperative in the design of complex, expensive fluids-engineering devices. Such tests use the principles of dimensional analysis and modeling from this chapter. (Courtesy of Mark E. Gibson/Visuals Unlimited) 276 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 which might result in tables of output, or even mul-tiple 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.
Chapter 3 presented gross control-volume balances of mass, momentum, and en-ergy which led to estimates of global parameters: mass flow, force, torque, total heat transfer. Chapter 4 presented infinitesimal balances which led to the basic partial dif-ferential equations of fluid flow and some particular solutions. These two chapters cov-ered analytical techniques, which are limited to fairly simple geometries and well-defined boundary conditions. Probably one-third of fluid-flow problems can be attacked in this analytical or theoretical manner.
The other two-thirds of all fluid problems are too complex, both geometrically and physically, to be solved analytically. They must be tested by experiment. Their behav-ior is reported as experimental data. Such data are much more useful if they are ex-pressed in compact, economic form. Graphs are especially useful, since tabulated data cannot be absorbed, nor can the trends and rates of change be observed, by most en-gineering eyes. These are the motivations for dimensional analysis. The technique is traditional in fluid mechanics and is useful in all engineering and physical sciences, with notable uses also seen in the biological and social sciences.
Dimensional analysis can also be useful in theories, as a compact way to present an analytical solution or output from a computer model. Here we concentrate on the pre-sentation of experimental fluid-mechanics data.
Basically, dimensional analysis is a method for reducing the number and complexity of experimental variables which affect a given physical phenomenon, by using a sort of compacting technique. If a phenomenon depends upon n dimensional variables, di-mensional analysis will reduce the problem to only k dimensionless variables, where the reduction n k 1, 2, 3, or 4, depending upon the problem complexity. Gener-ally n k equals the number of different dimensions (sometimes called basic or pri-Chapter 5 Dimensional Analysis and Similarity 277 mary or fundamental dimensions) which govern the problem. In fluid mechanics, the four basic dimensions are usually taken to be mass M, length L, time T, and tempera-ture , or an MLT system for short. Sometimes 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 immersed in a stream of fluid depended only on the body length L, stream velocity V, fluid density , and fluid viscosity , that is, F f(L, V, , ) (5.1) Suppose further that the geometry and flow conditions are so complicated that our in-tegral 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, , ) experimentally.
Generally speaking, it takes about 10 experimental points to define a curve. To find the effect of body length in Eq. (5.1), we have to run the experiment for 10 lengths L.
For each L we need 10 values of V, 10 values of , and 10 values of , making a grand total of 104, or 10,000, experiments. At $50 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 V F 2L2 g VL or CF g(Re) (5.2) i.e., the dimensionless force coefficient F/(V2L2) is a function only of the dimension-less Reynolds number VL/. We shall learn exactly how to make this reduction in Secs. 5.2 and 5.3.
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, , or sep-arately but only the grouping VL/. 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 $500, 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 waste money on computer time to find solutions. It suggests variables which can be discarded; sometimes dimensional analysis will immediately reject 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 which can con-vert data from a cheap, small model to design information for an expensive, large pro-totype. 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 278 Chapter 5 Dimensional Analysis and Similarity the full-scale prototype airplane. There are rules we shall explain for finding scaling laws. When the scaling law is valid, we say that a condition of similarity exists be-tween 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: If Rem Rep then CFm CFp (5.3) where subscripts m and p mean model and prototype, respectively. From the definition of force coefficient, this means that F F m p m p V V m p 2 L L m p 2 (5.4) for data taken where pVpLp/p mVmLm/m. 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 ve-locity 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 analy-sis in the broad view has many subtleties and nuances which only time and 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 crustacean 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 V 30 cm/s. The measured drag on the model is 1.3 N. For sim-ilar conditions, what are the velocity and drag of the actual copepod in water? Assume that Eq. (5.1) applies and the temperature is 20°C.
Solution From Table A.3 the fluid properties are: Water (prototype): p 0.001 kg/(m s) p 998 kg/m3 Glycerin (model): m 1.5 kg/(m s) m 1263 kg/m3 The length scales are Lm 100 mm and Lp 1 mm. We are given enough model data to com-pute the Reynolds number and force coefficient Rem m Vm m Lm 25.3 CFm mV F m m 2Lm 2 1.14 Both these numbers are dimensionless, as you can check. For conditions of similarity, the pro-totype Reynolds number must be the same, and Eq. (5.2) then requires the prototype force co-efficient to be the same 1.3 N (1263 kg/m3)(0.3 m/s)2(0.1 m)2 (1263 kg/m3)(0.3 m/s)(0.1 m) 1.5 kg/(m s) 5.1 Introduction 279 5.2 The Principle of Dimensional Homogeneity Rep Rem 25.3 998V 0.
p 0 (0 0 .
1 001) or Vp 0.0253 m/s 2.53 cm/s Ans.
CFp CFm 1.14 or Fp 7.31 107 N Ans.
It would obviously be difficult to measure such a tiny drag force.
Historically, the first person to write extensively about units and dimensional reasoning in physical relations was Euler in 1765. Euler’s ideas were far ahead of his time, as were those of Joseph Fourier, whose 1822 book Analytical Theory of Heat outlined what is now called the principle of dimensional homogeneity and even developed 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 ex-amples of dimensional analysis. The final breakthrough which established the method as we know it today is generally credited to E. Buckingham in 1914 , whose paper out-lined what is now called the Buckingham pi theorem for describing dimensionless para-meters (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 re-sults equivalent to the pi theorem. Following Buckingham’s paper, P. W. Bridgman pub-lished a classic book in 1922 , outlining the general theory of dimensional analysis. The subject continues to be controversial because there is so much art and subtlety in using di-mensional analysis. Thus, since Bridgman there have been at least 24 books published on the subject [2 to 25]. There will probably be more, but seeing the whole list might make some fledgling authors think twice. Nor is dimensional analysis limited to fluid mechan-ics or even engineering. Specialized books have been written on the application of di-mensional analysis to metrology , astrophysics , economics , building scale models , chemical processing pilot plants , social sciences , biomedical sci-ences , pharmacy , fractal geometry , and even the growth of plants .
In making the remarkable jump from the five-variable Eq. (5.1) to the two-variable Eq. (5.2), we were exploiting a rule which is almost a self-evident axiom in physics.
This rule, the principle of dimensional homogeneity (PDH), can be stated as follows: If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous; i.e., each of its additive terms will have the same dimensions.
All the equations which are derived from the theory of mechanics are of this form. For example, consider the relation which expresses the displacement of a falling body S S0 V0t 1 2 gt2 (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.
Fp 998(0.0253)2(0.001)2 280 Chapter 5 Dimensional Analysis and Similarity Consider Bernoulli’s equation for incompressible flow p 1 2 V2 gz const (5.6) Each term, including the constant, has dimensions of velocity squared, or {L2T 2}.
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. For example, which is it: S 1 2 gt2?
or S 1 2 g2t?
(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 sim-ply exploits it further.
Equations (5.5) and (5.6) also illustrate some other factors that often enter into a di-mensional analysis: Dimensional variables are the quantities which 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 S0, V0, and g, and in Eq. (5.6) they are , 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 1 2 and the exponent 2, both of which came from an integration: t dt 1 2 t2, V dV 1 2 V2. Other common dimensionless constants are and e.
Note that integration and differentiation of an equation may change the dimensions but not the homogeneity of the equation. For example, integrate or differentiate Eq. (5.5): S dt S0t 1 2 V0t2 1 6 gt3 (5.8a) d d S t V0 gt (5.8b) In the integrated form (5.8a) every term has dimensions of {LT}, while in the deriva-tive form (5.8b) every term is a velocity {LT1}.
Finally, there are some physical variables that are naturally dimensionless by virtue of their definition 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). All angles are dimen-sionless (ratio of arc length to radius) and should be taken in radians for this reason.
The motive behind dimensional analysis is that any dimensionally homogeneous equation can be written in an entirely equivalent nondimensional form which is more 5.2 The Principle of Dimensional Homogeneity 281 Ambiguity: The Choice of Variables and Scaling Parameters1 compact. Usually there is more than one method 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 illustrates most of the concepts of dimen-sional analysis. It contains five terms (S, S0, V0, t, g) which we may divide, in our think-ing, into variables and parameters. The variables are the things which 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 upon the variables we wish to know: in this case S0, V0, and g. Almost any engineering study can be subdivided in this manner.
To nondimensionalize our results, we need to know how many dimensions are con-tained among our variables and parameters: in this case, only two, length {L} and time {T}. Check each term to verify this: {S} {S0} {L} {t} {T} {V0} {LT1} {g} {LT2} Among our parameters, we therefore select two to be scaling parameters, used to de-fine dimensionless variables. What remains will be the “basic” parameter(s) whose ef-fect 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, some-thing 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.
For 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.
Option 1: Scaling parameters S0 and V0: the effect of gravity g.
First use the scaling parameters (S0, V0) to define dimensionless () displacement and time. There is only one suitable definition for each:2 S S S 0 t V S 0 0 t (5.9) Substitute these variables into Eq. (5.5) and clean everything up until each term is di-mensionless. The result is our first option: S 1 t 1 2 t2 g V S 0 2 0 (5.10) This result is shown plotted in Fig. 5.1a. There is a single dimensionless parameter , which shows here the effect of gravity. It cannot show the direct effects of S0 and V0, 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 accuracy, would look like Fig. 5.1a.
282 Chapter 5 Dimensional Analysis and Similarity 1 I am indebted to Prof. Jacques Lewalle of Syracuse University for suggesting, outlining, and clarify-ing this entire discussion.
2 Make them proportional to S and t. Do not define dimensionless terms upside down: S0/S or S0/(V0t).
The plots will look funny, users of your data will be confused, and your supervisor will be angry. It is not a good idea.
Fig. 5.1 Three entirely equivalent dimensionless presentations of the falling-body problem, Eq. (5.5): the effect of (a) gravity, (b) initial dis-placement, and (c) initial velocity.
All plots contain the same informa-tion.
Option 2: Scaling parameters V0 and g: the effect of initial displacement S0.
Now use the new scaling parameters (V0, g) to define dimensionless () displace-ment and time. Again there is only one suitable definition: S V S 0 2 g t t V g 0 (5.11) Substitute these variables into Eq. (5.5) and clean everything up again. The result is our second option: S t 1 2 t2 g V S 0 2 0 (5.12) 5.2 The Principle of Dimensional Homogeneity 283 t = V0t S0 S = S S0 10 8 6 4 2 0 0 1 2 3 (c) S = S S0 t = g S0 t √ 5 4 3 2 1 0 1 2 3 (a) t = gt V0 S = gS V02 8 6 4 2 0 0 1 2 3 (b) 1 0.5 0 1 0.5 V0 √gS0 = 2 g S0 V02 = 2 1 0.5 0.2 0 g S0 V02 = 2 0 / This result is plotted in Fig. 5.1b. The same single parameter again appears and here shows the effect of initial displacement, which merely moves the curves upward with-out changing their shape.
Option 3: Scaling parameters S0 and g: the effect of initial speed V0.
Finally use the scaling parameters (S0, g) to define dimensionless () displace-ment and time. Again there is only one suitable definition: S S S 0 t t S g 0 1/2 (5.13) Substitute these variables into Eq. (5.5) and clean everything up as usual. The result is our third and final option: S 1 t 1 2 t2 1 V g 0 S 0 (5.14) This final presentation is shown in Fig. 5.1c. Once again the parameter appears, but we have redefined it upside down, 1/ , so that our display parameter V0 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 and that the increase is proportional to time.
Note that, in all three options, the same parameter appears but has a different meaning: 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(t, ) g V S 0 2 0 (5.15) The reduction 5 3 2 should equal the number of fundamental dimensions involved in the problem {L, T}. This idea led to the pi theorem (Sec. 5.3).
The choice of scaling variables is left to the user, and the resulting dimensionless parameters have differing interpretations. For example, in the dimensionless drag-force formulation, Eq. (5.2), it is now clear that the scaling parameters were , V, and L, since they appear in both the drag coefficient and the Reynolds number. Equation (5.2) can thus be interpreted as the variation of dimensionless force with dimensionless vis-cosity, with the scaling-parameter effects mixed between CF and Re and therefore not immediately evident.
Suppose that we wish to study drag force versus velocity. Then we would not use V as a scaling parameter. We would use (, , L) instead, and the final dimensionless function would become CF F 2 fcn(Re) Re VL (5.16) In plotting these data, we would not be able to discern the effect of or , since they appear in both dimensionless groups. The grouping CF again would mean dimension-284 Chapter 5 Dimensional Analysis and Similarity Some Peculiar Engineering Equations less force, and Re is now interpreted as either dimensionless velocity or size.3 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 vari-ables 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 vari-ables 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 which does not re-solve the process. A typical case is Manning’s open-channel formula, discussed in Ex-ample 1.4: V 1.
n 49 R2/3S1/2 (1) Since V is velocity, R is a radius, and n and S are dimensionless, the formula is not di-mensionally homogeneous. This should be 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. Equation (1) in Example 1.4 (see above) 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 61.9D2.63 d d p x 0.54 (5.17) where D is diameter and dp/dx 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” which cannot be made dimen-sionally homogeneous. The “variables” they relate cannot be analyzed by the dimen-sional-analysis technique. Most of these formulas are raw empiricisms convenient to a small group of specialists. Here are three examples: B 1 2 0 5 0 ,0 00 R (5.18) S 130 1 40 API (5.19) 0.0147DE 3 D .7 E 4 0.26tR 1 t 7 R 2 (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.
5.2 The Principle of Dimensional Homogeneity 285 3 We were lucky to achieve a size effect because in this case L, a scaling parameter, did not appear in the drag coefficient.
5.3 The Pi Theorem Equation (5.20) relates the viscosity of a liquid in DE, or degrees Engler, to its vis-cosity tR in Saybolt seconds. Such formulas have a certain usefulness when commu-nicated between fellow specialists, but we cannot handle them here. Variables like Brinell hardness and Saybolt viscosity are not suited to an MLT dimensional system.
There are several methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. The 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 , meaning a product of variables. The dimensionless groups found from the theorem are power products denoted by 1, 2, 3, etc. The method allows the pis to be found in sequential order without resorting to free expo-nents.
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 only k dimensionless variables or ’s. The reduc-tion j n k equals the maximum number of variables which do not form a pi among themselves and is always less than or equal to the number of dimensions de-scribing the variables.
Take the specific case of force on an immersed body: Eq. (5.1) contains five variables F, L, U, , and described by three dimensions {MLT}. Thus n 5 and j 3. There-fore it is a good guess that we can reduce the problem to k pis, with k n j 5 3 2. And this is exactly what we obtained: two dimensionless variables 1 CF and 2 Re. On rare occasions it may take more pis than this minimum (see Example 5.5).
The second part of the theorem shows how to find the pis one at a time: Find the reduction j, then select j scaling variables which do not form a pi among themselves.4 Each desired pi group will be a power product of these j variables plus one additional variable which is assigned any convenient nonzero exponent. Each pi group thus found is independent.
To be specific, suppose that the process involves five variables 1 f(2, 3, 4, 5) Suppose that there are three dimensions {MLT} and we search around and find that in-deed j 3. Then k 5 3 2 and we expect, from the theorem, two and only two pi groups. Pick out three convenient variables which do not form a pi, and suppose these turn out to be 2, 3, and 4. Then the two pi groups are formed by power prod-ucts of these three plus one additional variable, either 1 or 5: 1 (2)a(3)b(4)c1 M0L0T0 2 (2)a(3)b(4)c5 M0L0T0 Here we have arbitrarily chosen 1 and 5, the added variables, to have unit exponents.
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 1 286 Chapter 5 Dimensional Analysis and Similarity 4 Make a clever choice here because all pis will contain these j variables in various groupings.
contains 1 and only 2 contains 5. 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 vari-ables are missing, dimensional analysis will fail.
2.
List the dimensions of each variable according to {MLT} or {FLT}. A list is given in Table 5.1.
3.
Find j. Initially guess j equal to the number of different dimensions present, and look for j variables which do not form a pi product. If no luck, reduce j by 1 and look again. With practice, you will find j rapidly.
4.
Select j scaling parameters which do not form a pi product. Make sure they please you and have some generality if possible, because they will then appear in every one of your pi groups. Pick density or velocity or length. Do not pick surface tension, e.g., or you will form six different independent Weber-number parameters and thoroughly annoy your colleagues.
5.
Add one additional variable to your j repeating variables, and form a power product. Algebraically find the exponents which make the product dimension-less. Try to arrange for your output or dependent variables (force, pressure drop, torque, power) to appear in the numerator, and your plots will look better. Do 5.3 The Pi Theorem 287 Table 5.1 Dimensions of Fluid-Mechanics Properties Dimensions Quantity Symbol MLT FLT Length L L L Area A L2 L2 Volume L3 L3 Velocity V LT1 LT1 Acceleration dV/dt LT2 LT2 Speed of sound a LT1 LT1 Volume flow Q L3T1 L3T1 Mass flow m ˙ MT1 FTL1 Pressure, stress p, ML1T2 FL2 Strain rate ˙ T1 T1 Angle None None Angular velocity T1 T1 Viscosity ML1T1 FTL2 Kinematic viscosity L2T1 L2T1 Surface tension MT2 FL1 Force F MLT2 F Moment, torque M ML2T2 FL Power P ML2T–3 FLT1 Work, energy W, E ML2T2 FL Density ML–3 FT2L–4 Temperature T Specific heat cp, c L2T21 L2T21 Specific weight ML–2T2 FL3 Thermal conductivity k MLT –31 FT11 Expansion coefficient 1 1 Step 2 Step 1 Step 3 Step 4 Step 5 this sequentially, adding one new variable each time, and you will find all n j k desired pi products.
6.
Write the final dimensionless function, and check your work to make sure all pi groups are dimensionless.
EXAMPLE 5.2 Repeat the development of Eq. (5.2) from Eq. (5.1), using the pi theorem.
Solution Write the function and count variables: F f(L, U, , ) there are five variables (n 5) List dimensions of each variable. From Table 5.1 288 Chapter 5 Dimensional Analysis and Similarity F L U {MLT2} {L} {LT1} {ML3} {ML1T1} Find j. No variable contains the dimension , and so j is less than or equal to 3 (MLT). We in-spect the list and see that L, U, and cannot form a pi group because only contains mass and only U contains time. Therefore j does equal 3, and n j 5 3 2 k. The pi theorem guarantees for this problem that there will be exactly two independent dimensionless groups.
Select repeating j variables. The group L, U, we found in step 3 will do fine.
Combine L, U, with one additional variable, in sequence, to find the two pi products.
First add force to find 1. You may select any exponent on this additional term as you please, to place it in the numerator or denominator to any power. Since F is the output, or dependent, variable, we select it to appear to the first power in the numerator: 1 LaUbcF (L)a(LT1)b(ML3)c( MLT2) M0L0T0 Equate exponents: Length: a b 3c 1 0 Mass: c 1 0 Time: b 2 0 We can solve explicitly for a 2 b 2 c 1 Therefore 1 L2U21F U F 2L2 CF Ans.
This is exactly the right pi group as in Eq. (5.2). By varying the exponent on F, we could have found other equivalent groups such as UL1/2/F1/2.
Finally, add viscosity to L, U, and to find 2. Select any power you like for viscosity. By hindsight and custom, we select the power 1 to place it in the denominator: 2 LaUbc1 La(LT1)b(ML3)c(ML1T1)1 M0L0T0 Equate exponents: Length: a b 3c 1 0 Mass: c 1 0 Time: b 1 0 from which we find a b c 1 Therefore 2 L1U111 U L Re Ans.
We know we are finished; this is the second and last pi group. The theorem guarantees that the functional relationship must be of the equivalent form U F 2L2 g U L Ans.
which is exactly Eq. (5.2).
EXAMPLE 5.3 Reduce the falling-body relationship, Eq. (5.5), to a function of dimensionless variables. Why are there three different formulations?
Solution Write the function and count variables S f(t, S0, V0, g) five variables (n 5) List the dimensions of each variable, from Table 5.1: 5.3 The Pi Theorem 289 S t S0 V0 g {L} {T} {L} LT1} {LT2} There are only two primary dimensions (L, T), so that j 2. By inspection we can easily find two variables which cannot be combined to form a pi, for example, V0 and g. Then j 2, and we expect 5 2 3 pi products. Select j variables among the parameters S0, V0, and g. Avoid S and t since they are the dependent variables, which should not be repeated in pi groups.
There are three different options for repeating variables among the group (S0, V0, g). There-fore we can obtain three different dimensionless formulations, just as we did informally with the falling-body equation in Sec. 5.2. Take each option in turn: 1.
Choose S0 and V0 as repeating variables. Combine them in turn with (S, t, g): 1 S1S0 aV0 b 2 t1S0 cV0 d 3 g1S0 eV0 f Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f). Please allow us to give the results here, and you may check the algebra as an exercise: a 1 b 0 c 1 d 1 e 1 f 2 Ans.
1 S S S 0 2 t V S 0 0 t 3 g V S 0 2 0 Thus, for option 1, we know that S fcn(t, ). We have found, by dimensional analy-sis, the same variables as in Eq. (5.10). But here there is no formula for the functional re-lation — we might have to experiment with falling bodies to establish Fig. 5.1a.
2.
Choose V0 and g as repeating variables. Combine them in turn with (S, t, S0): 1 S1V0 agb 2 t1V0 cgd 3 S0 1V0 egf Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f). Once more allow us to give the results here, and you may check the algebra as an exercise.
a 2 b 1 c 1 d 1 e 1 f 2 Ans.
1 S V Sg 0 2 2 t V tg 0 3 g V S 0 2 0 Thus, for option 2, we now know that S fcn(t, ). We have found, by dimensional analysis, the same groups as in Eq. (5.12). The data would plot as in Fig. 5.1b.
3.
Finally choose S0 and g as repeating variables. Combine them in turn with (S, t, V0): 1 S1S0 agb 2 t1S0 cgd 3 V0 1S0 egf Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f). One more time allow us to give the results here, and you may check the algebra as an exercise: a 1 b 0 c 1 2 d 1 2 e 1 2 f 1 2 Ans.
1 S S S 0 2 t t S g 0 3 V g 0 S 0 Thus, for option 3, we now know that S fcn(t, 1/ ). We have found, by di-mensional analysis, the same groups as in Eq. (5.14). The data would plot as in Fig. 5.1c.
Dimensional analysis here has yielded the same pi groups as the use of scaling parameters with Eq. (5.5). Three different formulations appeared, because we could choose three different pairs of repeating variables to complete the pi theorem.
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 , and the pressure drop per unit tube length dp/dx. Us-ing the pi theorem, find an appropriate dimensionless relationship.
290 Chapter 5 Dimensional Analysis and Similarity Solution Write the given relation and count variables: Q fR, , d d p x four variables (n 4) Make a list of the dimensions of these variables from Table 5.1: 5.3 The Pi Theorem 291 P L I E {L} {MLT2} {L} {L4} {ML1T2} Q R dp/dx {L3T1} {L} {ML1T1} {ML2T2} There are three primary dimensions (M, L, T), hence j 3. By trial and error we determine that R, , and dp/dx cannot be combined into a pi group. Then j 3, and n j 4 3 1. There is only one pi group, which we find by combining Q in a power product with the other three: 1 Rab d d p x c Q1 (L)a(ML1T1)b(ML2T2)c(L3T1) M0L0T0 Equate exponents: Mass: b c 0 Length: a b 2c 3 0 Time: b 2c 1 0 Solving simultaneously, we obtain a 4, b 1, c 1. Then 1 R41 d d p x 1 Q or 1 R4( Q dp /dx) const 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. 6.4 shows that the value of the constant is /8.
EXAMPLE 5.5 Assume that the tip deflection 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, f(P, L, I, 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: 5.4 Nondimensionalization of the Basic Equations There are five variables (n 5) and three primary dimensions (M, L, T), hence j 3. But try as we may, we cannot find any combination of three variables which does not form a pi group.
This is because {M} and {T} occur only in P and E and only in the same form, {MT2}. 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.
With j 2, we select L and E as two variables which cannot form a pi group and then add other variables to form the three desired pis: 1 LaEbI1 (L)a(ML1T2)b(L4) M0L0T0 from which, after equating exponents, we find that a 4, b 0, or 1 I/L4. Then 2 LaEbP1 (L)a(ML1T2)b(MLT2) M0L0T0 from which we find a 2, b 1, or 2 P/(EL2), and 3 LaEb1 (L)a(ML1T2)b(L) M0L0T0 from which a 1, b 0, or 3 /L. The proper dimensionless function is 3 f(2, 1), or L f E P L2 , L I 4 Ans.
(1) This is a complex three-variable function, but dimensional analysis alone can take us no further.
We can “improve” Eq. (1) by taking advantage of some physical reasoning, as Langhaar points out [8, p. 91]. For small elastic deflections, 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 3 must be proportional to 2 and inversely proportional to 1. Thus, for these con-ditions, L (const) E P L2 L I 4 or (const) P E L I 3 (2) This could not be predicted by a pure dimensional analysis. Strength-of-materials theory pre-dicts that the value of the constant is 1 3 .
We could use the pi-theorem method of the previous section to analyze problem after problem after problem, finding the dimensionless parameters which govern in each case. Textbooks on dimensional analysis [for example, 7] do this. An alternate 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 pa-rameters, e.g., Reynolds number, in their proper form and proper position, giving clues to when they are negligible. The boundary conditions must also be nondimensional-ized.
Let us briefly apply this technique to the incompressible-flow continuity and mo-mentum equations with constant viscosity: Continuity: V 0 (5.21a) 292 Chapter 5 Dimensional Analysis and Similarity Momentum: d d V t g p 2V (5.21b) Typical boundary conditions for these two equations are Fixed solid surface: V 0 Inlet or outlet: Known V, p (5.22) Free surface, z : w d d t p pa (Rx 1 Ry 1) We omit the energy equation (4.75) and assign its dimensionless form in the problems (Probs. 5.42 and 5.45).
Equations (5.21) and (5.22) contain the three basic dimensions M, L, and T. All vari-ables p, V, x, y, z, and t can be nondimensionalized by using density and two refer-ence constants which 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.
Now define all relevant dimensionless variables, denoting them by an asterisk: V V U x L x y L y z L z (5.23) t t L U p p U 2 gz All these are fairly obvious except for p, where we have slyly introduced the gravity effect, assuming that z is up. This is a hindsight idea suggested by Bernoulli’s equa-tion (3.77).
Since , U, and L are all constants, the derivatives in Eqs. (5.21) can all be handled in dimensionless form with dimensional coefficients. For example, u x ( ( U Lx u ) ) U L u x 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). The resulting dimensionless equations of motion are: Continuity: V 0 (5.24a) Momentum: d d V t p U L 2(V) (5.24b) The dimensionless boundary conditions are: Fixed solid surface: V 0 Inlet or outlet: Known V, p 5.4 Nondimensionalization of the Basic Equations 293 Dimensionless Parameters Free surface, z : w d d t (5.25) p p U a 2 U gL 2 z U 2L (Rx 1 Ry 1) These equations reveal a total of four dimensionless parameters, one in the momentum equation and three in the free-surface-pressure boundary condition.
In the continuity equation there are no parameters. The momentum equation contains one, generally accepted as the most important parameter in fluid mechanics: Reynolds number Re U L It is named after Osborne Reynolds (1842–1912), a British engineer who first proposed it in 1883 (Ref. 4 of Chap. 6). The Reynolds number is always important, with or with-out a free surface, and can be neglected only in flow regions away from high-velocity gradients, e.g., 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 p U a 2 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 p/(U2). If p involves vapor pressure p, it is called the cavitation number Ca (pa p)/(U2).
The second pressure parameter is much more important: Froude number Fr U gL 2 It is named after William Froude (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 Froude number is the dominant effect in free-surface flows and is totally unimportant if there is no free surface. Chapter 10 investigates Froude number effects in detail.
The final free-surface parameter is Weber number We U 2L 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 Froude. The Weber number is important only if it is of or-der unity or less, which typically occurs when the surface curvature is comparable in size to the liquid depth, e.g., in droplets, capillary flows, ripple waves, and very small hydraulic models. If We is large, its effect may be neglected.
294 Chapter 5 Dimensional Analysis and Similarity Compressibility Parameters Oscillating Flows If there is no free surface, Fr, Eu, and We drop out entirely, except for the possi-bility 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.
In high-speed flow of a gas there are significant changes in pressure, density, and tem-perature which must be related by an equation of state such as the perfect-gas law, Eq. (1.10). These thermodynamic changes introduce two additional dimensionless pa-rameters mentioned briefly in earlier chapters: Mach number Ma U a Specific-heat ratio k c c p 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 compressible-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 bound-ary condition. For example, suppose that the inlet stream is of the form u U cos t Nondimensionalization of this relation results in U u u cos U L t The argument of the cosine contains the new parameter Strouhal number St U L The dimensionless forces and moments, friction, and heat transfer, etc., of such an os-cillating flow would be a function of both Reynolds and Strouhal numbers. This param-eter is named after V. Strouhal, a German physicist who experimented in 1878 with wires singing in the wind.
Some flows which you might guess to be perfectly steady actually have an oscilla-tory pattern which 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. Fig-ure 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 Kármán vortex street, after T. von Kármán, who explained it theoretically in 1912. The shedding occurs in the range 102 Re 107, with an average Strouhal number d/(2 U) 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 sus-pension bridge in 1940, when wind-excited vortex shedding caused resonance with the natural torsional oscillations of the bridge.
5.4 Nondimensionalization of the Basic Equations 295 Fig. 5.2 Vortex shedding from a circular cylinder: (a) vortex street behind a circular cylinder (from Ref. 33, courtesy of U.S. Naval Re-search Laboratory); (b) experimen-tal shedding frequencies (data from Refs. 31 and 32).
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. 5.42. Another important and rather sneaky parameter is the wall-roughness ratio /L (in Table 5.2).5 Slight changes in surface roughness have a strik-296 Chapter 5 Dimensional Analysis and Similarity (a) 0.4 0.3 0.2 0.1 0 10 102 103 104 105 106 107 Re = Ud µ (b) St = d 2 U Data spread ω π ρ 5 Roughness is easy to overlook because it is a slight geometric effect which does not appear in the equations of motion.
Other Dimensionless Parameters ing 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 ef-fects, which dominate most flows. Note that we discovered all these parameters (ex-cept /L) simply by nondimensionalizing the basic equations without actually solving them.
If the reader is not satiated with the 15 parameters given in Table 5.2, Ref. 34 con-tains a list of over 300 dimensionless parameters in use in engineering. See also Ref. 35.
5.4 Nondimensionalization of the Basic Equations 297 Table 5.2 Dimensionless Groups in Fluid Mechanics Qualitative ratio Parameter Definition of effects Importance Reynolds number Re U L V I i n sc e o rt s i i a ty Always Mach number Ma U a S F o l u o n w d s s p p e e e e d d Compressible flow Froude number Fr U gL 2 G In ra e v rt i i t a y Free-surface flow Weber number We U 2L Surfa In ce er t t e ia nsion Free-surface flow Cavitation number Ca p U2 p P I r n e e s r s t u ia re Cavitation (Euler number) Prandtl number Pr k cp C D o is n s d ip u a c t t i i o o n n Heat convection Eckert number Ec c U pT 2 0 Kin E e n ti t c ha e l n p e y rgy Dissipation Specific-heat ratio k c c p Inte E r n n t a h l a e l n p e y rgy Compressible flow Strouhal number St U L M O e s a c n ill s a p ti e o e n d Oscillating flow Roughness ratio L W B a o ll d r y ou le g n h g n t e h ss Turbulent, rough walls Grashof number Gr T g 2 L32 B V u is o c y o a s n i c ty y Natural convection Temperature ratio T T w 0 Heat transfer Pressure coefficient Cp D S yn ta a t m ic ic pr p e r s e s s u s r u e re Aerodynamics, hydrodynamics Lift coefficient CL Dy L n i a f m t f ic or f c o e rce Aerodynamics, hydrodynamics Drag coefficient CD Dy D n r a a m g i f c or f c o e rce Aerodynamics, hydrodynamics D 1 2 U2A L 1 2 U2A p p!
1 2 U2 Wall temperature Stream temperature Fig. 5.3 The proof of practical di-mensional analysis: drag coeffi-cients 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.
Dimensional analysis is fun, but does it work? Yes; if all important variables are in-cluded 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 mea-sured drag on smooth cylinders and spheres. The flow is normal to the axis of the cylin-der, which is extremely long, L/d →!. The data are from many sources, for both liq-uids 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. 5.3a are entirely experimental; the analysis of immersed body drag is one of the weakest areas of modern fluid-mechanics theory. Except for some isolated digital-computer calculations, there is no theory for cylinder and sphere drag except creeping flow, Re 1.
The Reynolds number of both bodies is based upon diameter, hence the notation Red. But the drag coefficients are defined differently: 298 Chapter 5 Dimensional Analysis and Similarity A Successful Application 5 4 3 2 1 0 10 102 103 104 105 106 107 Red = (a) CD Cylinder (two – dimensional) Sphere Cylinder length effect (104 < Re < 105) L/d ∞ 40 20 10 5 3 2 1 1.20 0.98 0.91 0.82 0.74 0.72 0.68 0.64 1.5 104 Red (b) 1.0 0.7 0.5 0.3 105 106 = 0.02 ε − d 0.009 0.007 0.004 0.002 0.0005 Smooth Cylinder: L _ d = ∞ Transition to turbulent boundary layer Ud µ ρ CD CD Part (a) Step 1 CD cylinder (5.26) sphere They both have a factor 1 2 as a traditional tribute to Bernoulli and Euler, and both are based on the projected area, i.e., the area one sees when looking toward the body from upstream. The usual definition of CD is thus CD (5.27) However, one should carefully check the definitions of CD, Re, etc., before using data in the literature. Airfoils, e.g., 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 CD f Red, d , L d (5.28) To describe this function completely would require 1000 or more experiments. Therefore it is customary to explore the length and roughness effects separately 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 deflecting over and under the body.
Figure 5.3b shows the effect of wall roughness for an infinitely long cylinder. The sharp drop in drag occurs at lower Red 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 which is exploited in sports by deliberate dimpling of golf balls to give them less drag at their flight Red 105.
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 5.6 The capillary rise h of a liquid in a tube varies with tube diameter d, gravity g, fluid density , surface tension , and the contact angle . (a) Find a dimensionless statement of this relation.
(b) If h 3 cm in a given experiment, what will h be in a similar case if the diameter and sur-face tension are half as much, the density is twice as much, and the contact angle is the same?
Solution Write down the function and count variables: h f(d, g, , , ) n 6 variables drag 1 2 U2(projected area) drag 1 2 U2 1 4 d2 drag 1 2 U2Ld 5.4 Nondimensionalization of the Basic Equations 299 Part (b) List the dimensions {FLT} from Table 5.2: h d g {L} {L} {LT 2} {FT2L4} {FL1} none Find j. Several groups of three form no pi: , , and g or , g, and d. Therefore j 3, and we expect n j 6 3 3 dimensionless groups. One of these is obviously , which is already dimensionless: 3 Ans. (a) If we had carelessly chosen to search for it by using steps 4 and 5, we would still find 3 .
Select j repeating variables which do not form a pi group: , g, d.
Add one additional variable in sequence to form the pis: Add h: 1 agbdch (FT 2L4)a(LT 2)b(L)c(L) F0L0T 0 Solve for a b 0 c 1 Therefore 1 0g0d1h h d Ans. (a) Finally add , again selecting its exponent to be 1: 2 agbd c (FT 2L4)a(LT 2)b(L)c(FL1) F0L0T0 Solve for a b 1 c 2 Therefore 2 1g1d2 g d2 Ans. (a) The complete dimensionless relation for this problem is thus h d F g d2 , Ans. (a) (1) This is as far as dimensional analysis goes. Theory, however, establishes that h is proportional to . Since occurs only in the second parameter, we can slip it outside h d actual g d2 F1() or h gd F1() Example 1.9 showed theoretically that F1() 4 cos .
We are given h1 for certain conditions d1, 1, 1, and 1. If h1 3 cm, what is h2 for d2 1 2 d1, 2 1 2 1, 2 21, and 2 1? We know the functional relation, Eq. (1), must still hold at condition 2 h d 2 2 F 2 gd 2 2 2, 2 But 2 g 2 d2 2 1 g 1 d1 2 1 2 1 21g( 1 2 d1)2 300 Chapter 5 Dimensional Analysis and Similarity Step 2 Step 3 Step 4 Step 5 Step 6 Geometric Similarity Therefore, functionally, h d 2 2 F 1 g 1 d1 2 , 1 h d 1 1 We are given a condition 2 which is exactly similar to condition 1, and therefore a scaling law holds h2 h1 d d 2 1 (3 cm) 1.5 cm Ans. (b) If the pi groups had not been exactly the same for both conditions, we would have had to know more about the functional relation F to calculate h2.
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 there are certain engineering dif-ficulties which need to be discussed.
First, we have more or less taken for granted that the variables which affect the process can be listed and analyzed. Actually, selection of the important variables re-quires considerable judgment and experience. The engineer must decide, e.g., whether viscosity can be neglected. Are there significant temperature effects? Is surface tension important? What about wall roughness? Each pi group which is retained increases the expense and effort required. Judgment in selecting variables will come through prac-tice and maturity; this book should provide some of the necessary experience.
Once the variables are selected and the dimensional analysis is performed, the ex-perimenter seeks to achieve similarity between the model tested and the prototype to be designed. With sufficient testing, the model data will reveal the desired dimension-less function between variables 1 f(2, 3, . . . k) (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 2m 2p, 3m 3p, etc., Eq. (5.29) guarantees that the desired output 1m will equal 1p. But this is easier said than done, as we now discuss.
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 {L} and must be ensured before any sensible model testing can proceed. A formal definition is as follows: 1 2 d1 d1 5.5 Modeling and Its Pitfalls 301 5.5 Modeling and Its Pitfalls Fig. 5.4 Geometric similarity in model testing: (a) prototype; (b) one-tenth-scale model.
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 pro-totype 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: 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.
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 justified by experimental comparison to show that the prototype behavior was not significantly affected by the discrepancy.
Models which appear similar in shape but which clearly violate geometric similar-ity should not be compared except at your own risk. Figure 5.5 illustrates this point.
302 Chapter 5 Dimensional Analysis and Similarity 10° Vp (a) a Homologous points a Vm 10° 40 m 8 m 0.8 m 4 m (b) 1 m 0.1 m Fig. 5.5 Geometric similarity and dissimilarity of flows: (a) similar; (b) dissimilar.
The spheres in Fig. 5.5a are all geometrically similar and can be tested with a high ex-pectation of success if the Reynolds number or Froude number, etc., 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, etc. 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 equiva-lence 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 independent length and time scales, and no additional parameters are necessary (see Chap. 8 for further details).
Frictionless flows with a free surface, as in Fig. 5.6b, are kinematically similar if their Froude numbers are equal Frm g V L m 2 m g V L p 2 p Frp (5.30) Note that the Froude number contains only length and time dimensions and hence is a purely kinematic parameter which fixes the relation between length and time. From Eq. (5.30), if the length scale is Lm Lp (5.31) 5.5 Modeling and Its Pitfalls 303 Huge sphere (a) (b) Medium sphere Tiny sphere V1 V2 V3 V4 V1 V2 V3 Large 4:1 ellipsoid Medium 3.5:1 ellipsoid Small 3:1 ellipsoid Large sphere Kinematic Similarity Fig. 5.6 Frictionless low-speed flows are kinematically similar: (a) Flows with no free surface are kinematically similar with indepen-dent length- and time-scale ratios; (b) free-surface flows are kinemati-cally similar with length and time scales related by the Froude number.
where is a dimensionless ratio, the velocity scale is V V m p L L m p 1/2 (5.32) and the time scale is T T m p L L m p / / V V m p (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 , then the wave period, propa-gation speed, and particle velocities are related by .
If viscosity, surface tension, or compressibility is important, kinematic similarity is dependent upon 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 sim-304 Chapter 5 Dimensional Analysis and Similarity V∞p Prototype V1p (a) V∞ m = βV∞p V1m = V1p V2 m = βV2 p Model Dm = α Dp Prototype waves: Vp Period Tp Cp Hm = Hp λ m = λ p Cm = C p √ Period Tm = T p √ (b) λp Vm = V p √ Model waves: V2p Dp Hp α α α α α β Dynamic Similarity Fig. 5.7 Dynamic similarity in sluice-gate flow. Model and proto-type yield identical homologous force polygons if the Reynolds and Froude numbers are the same cor-responding values: (a) prototype; (b) model.
ilarity is a first requirement; without it, proceed no further. Then dynamic similarity 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 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.
The perfect dynamic similarity shown in Fig. 5.7 is more of a dream than a reality be-cause true equivalence of Reynolds and Froude numbers can be achieved only by dra-matic 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 re-quires equivalent Froude numbers, Eq. (5.30), and equivalent Reynolds numbers V m m Lm V pL p p (5.34) 5.5 Modeling and Its Pitfalls 305 Ffm Ffp a Fpp Fgp Fip (a) (b) Fpm Fgm Fim a' Discrepancies in Water and Air Testing Fig. 5.8 Reynolds-number extrapo-lation, or scaling, of hydraulic data with equal Froude numbers.
But both velocity and length are constrained by the Froude number, Eqs. (5.31) and (5.32). Therefore, for a given length-scale ratio , Eq. (5.34) is true only if m p L L m p V V m p 3/2 (5.35) For example, for a one-tenth-scale model, 0.1 and 3/2 0.032. Since p is un-doubtedly water, we need a fluid with only 0.032 times the kinematic viscosity of wa-ter to achieve dynamic similarity. Referring to Table 1.4, we see that this is impossi-ble: 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 un-avoidably violated. The Froude number is held constant since it is the dominant parameter 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.
Second, consider aerodynamic model testing in air with no free surface. The im-portant parameters are the Reynolds number and the Mach number. Equation (5.34) should be satisfied, plus the compressibility criterion V am m V ap p (5.36) Elimination of Vm/Vp between (5.34) and (5.36) gives m p L L m p a a m p (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 306 Chapter 5 Dimensional Analysis and Similarity log CD Range of Rem Model data: Range of Rep Power-law extrapolation Uncertainty in prototype data estimate log Re 105 106 107 108 Fig. 5.9 Hydraulic model of a bar-rier-beach inlet at Little River, South Carolina. Such models of ne-cessity violate geometric similarity and do not model the Reynolds number of the prototype inlet.
(Courtesy of U.S. Army Engineer Waterways Experiment Station).
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.
Finally, a serious discrepancy of another type occurs in hydraulic models of natural flow systems such as rivers, harbors, estuaries, and embayments. Such flows have large horizontal dimensions and small relative vertical dimensions. If we were to scale an es-tuary model by a uniform linear length ratio of, say, 11000, the resulting model would be only a few millimeters deep and dominated by entirely spurious surface-tension or Weber-number effects. Therefore such hydraulic models commonly violate geo-metric similarity by “distorting” the vertical scale by a factor of 10 or more. Figure 5.9 shows a hydraulic model of a barrier-beach inlet in South Carolina. The horizontal scale reduction is 1300, but the vertical scale is only 160. Since a deeper channel flows more efficiently, the model channel bottom is deliberately roughened more than the natural chan-nel to correct for the geometric discrepancy. Thus the friction effect of the discrepancy can be corrected, but its effect on, say, dispersion of heat and mass is less well known.
EXAMPLE 5.7 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: p fcn(V, , , L, D). We wish to know how p varies with V. (a) Use the pi theorem to rewrite this function in dimensionless form. (b) Then plot this function, using the following data for three pipes and three fluids: 5.5 Modeling and Its Pitfalls 307 D, cm L, m Q, m3/h p, Pa , kg/m3 , kg/(m s) V, m/s 1.0 5.0 0.3 4,680 680† 2.92 E-4† 1.06 1.0 7.0 0.6 22,300 680† 2.92 E-4† 2.12 1.0 9.0 1.0 70,800 680† 2.92 E-4† 3.54 2.0 4.0 1.0 2,080 998‡ 0.0010‡ 0.88 2.0 6.0 2.0 10,500 998‡ 0.0010‡ 1.77 2.0 8.0 3.1 30,400 998‡ 0.0010‡ 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 V Q/A, A D2/4.
†Gasoline.
‡Water.
§Mercury.
(c) Suppose it is further known that p 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 for-mulation. Plot the dimensionless data in this improved manner and comment upon the results.
Solution There are six variables with three primary dimensions involved {MLT}. Therefore we expect that j 6 3 3 pi groups. We are correct, for we can find three variables which do not form a pi product, for example, (, V, L). Carefully select three (j) repeating variables, but not including p or V, which we plan to plot versus each other. We select (, , D), and the pi theorem guar-antees that three independent power-product groups will occur: 1 abDc p 2 deD fV 3 ghDiL or 1 D 2 2 p 2 V D 3 D L We have omitted the algebra of finding (a, b, c, d, e, f, g, h, i) by setting all exponents to zero M0, L0, T0. Therefore we wish to plot the dimensionless relation D 2 2 p fcn V D , D L Ans. (a) We plot 1 versus 2 with 3 as a parameter. There will be nine data points. For example, the first row in the data above yields D 2 2 p (680 ( ) 2 ( .
0 9 .
2 01 E ) -2 4 (4 )2 680) 3.73 E9 V D (680 2 )( .9 1 2 .06 E ) -( 4 0.01) 24,700 D L 500 The nine data points are plotted as the open circles in Fig. 5.10. The values of L/D are listed for each point, and we see a significant length effect. In fact, if we connect the only two points which have the same L/D ( 200), we could see (and cross-plot to verify) that p increases linearly with L, as stated in the last part of the problem. Since L occurs only in 3 L/D, the function 1 fcn(2, 3) must reduce to 1 (L/D) fcn(2), or simply a function involving only two parameters: 308 Chapter 5 Dimensional Analysis and Similarity Fig. 5.10 Two different correlations of the data in Example 5.7: Open circles when plotting D2 p/2 versus ReD, L/D is a parameter; once it is known that p is propor-tional to L, a replot (solid circles) of D3 p/(L2) versus ReD col-lapses into a single power-law curve.
D L 3 2 p fcn V D flow in a long pipe Ans. (c) We now modify each data point in Fig. 5.10 by dividing it by its L/D value. For example, for the first row of data, D3 p/(L2) (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: D L 3 2 p 0.155 V D 1.75 Ans. (c) All newtonian smooth pipe flows should correlate in this manner. This example is a variation of the first completely successful dimensional analysis, pipe-flow friction, performed by Prandtl’s student Paul Blasius, who published a related plot in 1911. For this range of (turbulent-flow) Reynolds numbers, the pressure drop increases approximately as V1.75.
EXAMPLE 5.8 The smooth-sphere data plotted in Fig. 5.3a represent dimensionless drag versus dimensionless viscosity, since (, 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 To display the effect of velocity, we must not use V as a repeating variable. Instead we choose (, , d) as our j variables to nondimensionalize Eq. (5.1), F fcn(d, V, , ). (See Example 5.2 for an alternate approach to this problem.) The pi groups form as follows: 1 abdcF F 2 2 efdgV Vd Ans. (a) 5.5 Modeling and Its Pitfalls 309 105 ReD 104 Π1 Π3 108 107 106 0.155 ReD1.75 109 1010 1011 Π1 L D = 200 500 100 300 700 400 133 900 200 Fig. 5.11 Cross-plot of sphere-drag data from Fig. 5.3a to isolate diam-eter and velocity.
That is, a 1, b 2, c 0, e 1, f 1, and g 1, by using our power-product techniques of Examples 5.2 to 5.6. Therefore a plot of F/2 versus Re will display the direct effect of ve-locity on sphere drag. This replot is shown as Fig. 5.11. The drag increases rapidly with veloc-ity up to transition, where there is a slight drop, after which it increases more quickly than ever.
If the force is known, we may predict the velocity from the figure.
For water at 20°C, take 998 kg/m3 and 0.001 kg/(m s). For steel, s 7.86water 7840 kg/m3. For terminal velocity, the drag equals the net weight of the sphere in water. Thus F Wnet (s w)g 6 d3 (7840 998)(9.81) 6 (0.01)3 0.0351 N Therefore the ordinate of Fig. 5.11 is known: Falling steel sphere: F 2 3.5 E7 From Fig. 5.11, at F/2 3.5 E7, a magnifying glass reveals that Red 2 E4. Then a crude estimate of the terminal fall velocity is Vd 20,000 or V 2.0 m s Ans. (b) 20,000[0.001 kg/(m s)] (998 kg/m3)(0.01 m) (998 kg/m3)(0.0351 N) [0.001 kg/(m s)]2 310 Chapter 5 Dimensional Analysis and Similarity 1011 1010 109 108 107 106 105 104 103 102 10 1 0.1 1 10 102 103 104 105 106 Re = Vd µ F µ2 = π 8 CD Re2 Transition: ρ ρ Better accuracy could be obtained by expanding the scale of Fig. 5.11 in the region of the given force coefficient. However, there is considerable uncertainty in published drag data for spheres, so the predicted fall velocity is probably uncertain by at least "5 percent.
Note that we found the answer directly from Fig. 5.11. We could use Fig. 5.3a also but would have to iterate between the ordinate and abscissa to obtain the final result, since V is contained in both plotted variables.
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 by the technique of dimensional analysis. Tests and experiments are used both to strengthen existing theories and to provide useful engineering results when the-ory 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 dimen-sional homogeneity. A general technique, the pi theorem, is then presented for sys-tematically finding a set of dimensionless parameters by grouping a list of variables which govern any particular physical process. Alternately, direct application of di-mensional analysis to the basic equations of fluid mechanics yields the fundamental parameters governing flow patterns: Reynolds number, Froude number, Prandtl num-ber, 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 sim-ilarity. 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 311 Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon, for example, Prob. 5.61, will ben-efit from the use of the Engineering Equation Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 5.1 to 5.91 (cat-egorized in the problem list below) are followed by word problems W5.1 to W5.10, fundamentals of engineering exam problems FE5.1 to FE5.10, comprehensive applied problems C5.1 to C5.4, and de-sign projects D5.1 and D5.2.
Problem distribution Section Topic Problems 5.1 Introduction 5.1–5.6 5.2 Choosing proper scaling parameters 5.7–5.9 5.2 The principle of dimensional homogeneity 5.10–5.17 5.3 The pi theorem 5.18–5.41 5.4 Nondimensionalizing the basic equations 5.42–5.47 5.4 Data for spheres and cylinders 5.48–5.57 5.5 Scaling of model data 5.58–5.74 5.5 Froude- and Mach-number scaling 5.75–5.84 5.5 Inventive rescaling of the data 5.85–5.91 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 upon the diameter and average velocity. If d 5 cm and the fluid is kerosine at 20°C, find the volume flow rate in m3/h which causes transition.
P5.2 In flow past a thin flat body such as an airfoil, transition to turbulence occurs at about Re 1 E6, based on the distance x from the leading edge of the wing. If an airplane flies at 450 mi/h at 8-km standard altitude and undergoes transition at the 12 percent chord position, how long is its chord (wing length from leading to trailing edge)?
Summary All quantities have their standard meanings; for example, is density.
P5.11 For a particle moving in a circle, its centripetal acceleration takes the form a fcn(V, R), where V is its velocity and R the radius of its path. By pure dimensional reasoning, rewrite this function in algebraic form.
P5.12 The velocity of sound a of a gas varies with pressure p and density . Show by dimensional reasoning that the proper form must be a (const)(p/)1/2.
P5.13 The speed of propagation C of a capillary wave in deep wa-ter is known to be a function only of density , wavelength #, and surface tension . Find the proper functional rela-tionship, completing it with a dimensionless constant. For a given density and wavelength, how does the propagation speed change if the surface tension is doubled?
P5.14 Consider laminar flow over a flat plate. The boundary layer thickness grows with distance x down the plate and is also a function of free-stream velocity U, fluid viscosity , and fluid density . Find the dimensionless parameters for this problem, being sure to rearrange if neessary to agree with the standard dimensionless groups in fluid mechanics, as given in Table 5.2.
P5.15 It is desired to measure the drag on an airplane whose ve-locity is 300 mi/h. Is it feasible to test a one-twentieth-scale model of the plane in a wind tunnel at the same pressure and temperature to determine the prototype drag coefficient?
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, m2 T temperature difference, K The dimensionless form of h, called the Stanton number, is a combination of h, fluid density , specific heat cp, and flow velocity V. Derive the Stanton number if it is propor-tional to h.
P5.17 In some heat-transfer textbooks, e.g., J. P. Holman, Heat Transfer, 5th ed., McGraw-Hill, 1981, p. 285, simplified for-mulas are given for the heat-transfer coefficient from Prob.
5.16 for buoyant or natural convection over hot surfaces.
An example formula is h 1.42 L T 1/4 where L is the length of the hot surface. Comment on the dimensional homogeneity of this formula. What might be the SI units of constants 1.42 and 1 4 ? What parameters might be missing or hidden?
P5.18 Under laminar conditions, the volume flow Q through a small triangular-section pore of side length b and length L P5.3 An airplane has a chord length L 1.2 m and flies at a Mach number of 0.7 in the standard atmosphere. If its Reynolds number, based on chord length, is 7 E6, how high is it flying?
P5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm-di-ameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather bal-loon moored in sea-level standard air under dynamically similar conditions?
P5.5 An automobile has a characteristic length and area of 8 ft and 60 ft2, respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: V, mi/h 20 40 60 Drag, lbf 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 SAE 10 oil at 20°C flows past an 8-cm-diameter sphere. At flow velocities of 1, 2, and 3 m/s, the measured sphere drag forces are 1.5, 5.3, and 11.2 N, respectively. Estimate the drag force if the same sphere is tested at a velocity of 15 m/s in glycerin at 20°C.
P5.7 A body is dropped on the moon (g 1.62 m/s2) 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 S 0.84. Estimate (a) the initial displacement, (b) the fi-nal displacement, and (c) the time of impact.
P5.8 The Bernoulli equation (5.6) can be written in the form p p0 1 2 V2 gz (1) where p0 is the “stagnation” pressure at zero velocity and elevation. (a) State how many scaling variables are needed to nondimensionalize this equation. (b) Suppose that we wish to nondimensionalize Eq. (1) in order to plot dimen-sionless pressure versus velocity, with elevation as a para-meter. Select the proper scaling variables and carry out and plot the resulting dimensionless relation.
P5.9 Modify Prob. 5.8 as follows. Suppose that we wish to nondi-mensionalize Eq. (1) in order to plot dimensionless pressure versus gravity, with velocity as a parameter. Select the proper scaling variables and carry out and plot the resulting dimensionless relation.
P5.10 Determine the dimension {MLT} of the following quan-tities: (a) u u x (b) 2 1 (p p0) dA (c) cp x 2 T y (d) u t dx dy dz 312 Chapter 5 Dimensional Analysis and Similarity (See Prob. 5.21 for an explanation of this concept.) Rewrite the function in terms of pi groups.
P5.25 When a viscous fluid is confined between two long con-centric cylinders as in Fig. 4.17, the torque per unit length T required to turn the inner cylinder at angular velocity $ is a function of $, cylinder radii a and b, and viscosity .
Find the equivalent dimensionless function. What happens to the torque if both a and b are doubled?
P5.26 A pendulum has an oscillation period T which is assumed to depend upon its length L, bob mass m, angle of swing , 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 pe-riod when it swings at 45°? A similarly constructed pendu-lum, with L 30 cm and m 100 g, is to swing on the moon (g 1.62 m/s2) at 20°. (b) What will be its pe-riod?
P5.27 In studying sand transport by ocean waves, A. Shields in 1936 postulated that the threshold wave-induced bottom shear stress % required to move particles depends upon grav-ity g, particle size d and density p, and water density and viscosity . Find suitable dimensionless groups of this prob-lem, 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 ve-locity V, density , and viscosity . Its center deflection is assumed to be a function of all these variables. (a) Rewrite this proposed function in dimensionless form. (b) Suppose it is known that is independent of , inversely propor-tional to E, and dependent only upon V2, not and V sep-arately. Simplify the dimensionless function accordingly.
Hint: Take L, , and V as repeating variables.
P5.29 When fluid in a pipe is accelerated linearly from rest, it be-gins as laminar flow and then undergoes transition to tur-bulence at a time ttr which depends upon the pipe diameter D, fluid acceleration a, density , and viscosity . Arrange this into a dimensionless relation between ttr and D.
P5.30 In forced convection, the heat-transfer coefficient h, as de-fined in Prob. 5.16, is known to be a function of stream ve-locity U, body size L, and fluid properties , , cp, and k.
Rewrite this function in dimensionless form, and note by name any parameters you recognize. Hint: Take L, , k, and as repeating variables.
P5.31 The heat-transfer rate per unit area q to a body from a fluid in natural or gravitational convection is a function of the temperature difference T, gravity g, body length L, and three fluid properties: kinematic viscosity , conductivity k, and thermal expansion coefficient . Rewrite in dimen-sionless form if it is known that g and appear only as the product g.
is a function of viscosity , pressure drop per unit length p/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 as-sumed to be a function of density , wavelength #, depth h, gravity g, and surface tension . Rewrite this relationship in dimensionless form. What results if is negligible? Hint: Take #, , and g as repeating variables.
P5.20 The power input P to a centrifugal pump is assumed to be a function of the volume flow Q, impeller diameter D, ro-tational rate $, and the density and viscosity of the fluid. Rewrite this as a dimensionless relationship. Hint: Take $, , and D as repeating variables.
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 di-mensions of each variable, some workers list the powers of each primary dimension for each variable in an array: 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 j n k, the desired reduction between original variables and the pi groups. This is a general property of dimensional matri-ces, as noted by Buckingham .
P5.22 When freewheeling, the angular velocity $ of a windmill is found to be a function of the windmill diameter D, the wind velocity V, the air density , the windmill height H as compared to the atmospheric boundary layer height L, and the number of blades N: $ fcnD, V, , H L , N Viscosity effects are negligible. Find appropriate pi groups for this problem and rewrite the function above in dimen-sionless form.
P5.23 The period T of vibration of a beam is a function of its length L, area moment of inertia I, modulus of elasticity E, den-sity , and Poisson’s ratio . Rewrite this relation in di-mensionless form. What further reduction can we make if E and I can occur only in the product form EI? Hint: Take L, , 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 , density , vis-cosity , and speed of sound a of the air. Write out the di-mensional matrix of this function and determine its rank.
Problems 313 F L U M 1 0 0 1 1 L 1 1 1 3 1 T 2 0 1 0 1 Buckingham pi theorem, find an expression for rate of heat loss as a function of the other three parameters in the prob-lem. (b) If the temperature difference T doubles, by what factor does the rate of heat loss increase?
P5.37 The pressure difference p across an explosion or blast wave is a function of the distance r from the blast center, time t, speed of sound a of the medium, and total energy E in the blast. Rewrite this relation in dimensionless form (see Ref.
18, chap. 4, for further details of blast-wave scaling). How does p change if E is doubled?
P5.38 The size d of droplets produced by a liquid spray nozzle is thought to depend upon the nozzle diameter D, jet velocity U, and the properties of the liquid , , and . Rewrite this relation in dimensionless form. Hint: Take D, , and U as repeating variables.
P5.39 In turbulent flow past a flat surface, the velocity u near the wall varies approximately logarithmically with distance y from the wall and also depends upon viscosity , density , and wall shear stress %w. For a certain airflow at 20°C and 1 atm, %w 0.8 Pa and u 15 m/s at y 3.6 mm. Use this information to estimate the velocity u at y 6 mm.
P5.40 Reconsider the slanted-plate surface tension problem (see Fig. C1.1) as an exercise in dimensional analysis. Let the capillary rise h be a function only of fluid properties, grav-ity, bottom width, and the two angles in Fig. C1.1. That is, h fcn(, , g, L, , ). (a) Use the pi theorem to rewrite this function in terms of dimensionless parameters. (b) Ver-ify that the exact solution from Prob. C1.1 is consistent with your result in part (a).
P5.41 A certain axial-flow turbine has an output torque M which is proportional to the volume flow rate Q and also depends upon the density , rotor diameter D, and rotation rate $. How does the torque change due to a doubling of (a) D and (b) $?
P5.42 Nondimensionalize the energy equation (4.75) and its boundary conditions (4.62), (4.63), and (4.70) by defining T T/T0, where T0 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.
P5.43 The differential equation of salt conservation for flowing seawater is S t u S x S y w S z & 2 x S 2 2 y S 2 2 z S 2 where & is a (constant) coefficient of diffusion, with typi-cal units of square meters per second, and S is the salinity in parts per thousand. Nondimensionalize this equation and discuss any parameters which appear.
P5.44 The differential energy equation for incompressible two-di-mensional flow through a “Darcy-type” porous medium is approximately P5.32 A weir is an obstruction in a channel flow which can be cal-ibrated to measure the flow rate, as in Fig. P5.32. The vol-ume flow Q varies with gravity g, weir width b into the pa-per, 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, H).
314 Chapter 5 Dimensional Analysis and Similarity H Q Weir P5.32 P5.33 A spar buoy (see Prob. 2.113) has a period T of vertical (heave) oscillation which depends upon the waterline cross-sectional area A, buoy mass m, and fluid specific weight . 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. 8 of Chap. 1) depends only upon the density , 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 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 Fig. P5.35 depends upon the radius R, rotation rate $, fluid viscosity , and cone angle . Rewrite this relation in di-mensionless form. How does the relation simplify it if it is known that M is proportional to ?
θ Ω θ R Fluid P5.35 P5.36 The rate of heat loss, Q .
loss through a window or wall is a function of the temperature difference between inside and outside T, the window surface area A, and the R value of the window which has units of (ft2 h °F)/Btu. (a) Using EES P5.50 When a micro-organism moves in a viscous fluid, it turns out that fluid density has nearly negligible influence on the drag force felt by the micro-organism. Such flows are called creeping flows. The only important parameters in the prob-lem are the velocity of motion U, the viscosity of the fluid , and the length scale of the body. Here assume the mi-cro-organism’s body diameter d as the appropriate length scale. (a) Using the Buckingham pi theorem, generate an expression for the drag force D as a function of the other parameters in the problem. (b) The drag coefficient dis-cussed in this chapter CD D/( 1 2 U2A) is not appropriate for this kind of flow. Define instead a more appropriate drag coefficient, and call it Cc (for creeping flow). (c) For a spher-ically shaped micro-organism, the drag force can be calcu-lated exactly from the equations of motion for creeping flow.
The result is D 3 Ud. Write expressions for both forms of the drag coefficient, Cc and CD, for a sphere under con-ditions of creeping flow.
P5.51 A ship is towing a sonar array which 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 A 1-in-diameter telephone wire is mounted in air at 20°C and has a natural vibration frequency of 12 Hz. What wind velocity in ft/s will cause the wire to sing? At this con-dition what will the average drag force per unit wire length be?
P5.53 Vortex shedding can be used to design a vortex flowmeter (Fig. 6.32). A blunt rod stretched across the pipe sheds vor-tices whose frequency is read by the sensor downstream.
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 m3/h. How might the meter react to other liquids?
P5.54 A fishnet is made of 1-mm-diameter strings knotted into 2 2 cm squares. Estimate the horsepower required to tow 300 ft2 of this netting at 3 kn in seawater at 20°C. The net plane is normal to the flow direction.
P5.55 The radio antenna on a car begins to vibrate wildly at 500 Hz when the car is driven at 55 mi/h. Estimate the diame-ter of the antenna.
P5.56 A wooden flagpole, of diameter 5 in and height 30 ft, frac-tures at its base in hurricane winds at sea level. If the frac-ture stress is 3500 lbf/in2, estimate the wind velocity in mi/h.
P5.57 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?
cp p x T x cp p y T y k 2 y T 2 0 where is the permeability of the porous medium. All other symbols have their usual meanings. (a) What are the ap-propriate dimensions for ? (b) Nondimensionalize this equation, using (L, U, , T0) as scaling constants, and dis-cuss any dimensionless parameters which arise.
P5.45 In natural-convection problems, the variation of density due to the temperature difference T creates an important buoy-ancy term in the momentum equation (5.30). To first-order accuracy, the density variation would be 0(1 T), where is the thermal-expansion coefficient. The momen-tum equation thus becomes 0 d d V t (p 0gz) 0 T gk 2V where we have assumed that z is up. Nondimensionalize this equation, using Eqs. (5.23), and relate the parameters you find to the list in Table 5.2.
P5.46 The differential equation for compressible inviscid flow of a gas in the xy plane is 2 t ' 2 t (u2 2) (u2 a2) 2 x ' 2 (2 a2) 2 y ' 2 2u x 2' y 0 where ' is the velocity potential and a is the (variable) speed of sound of the gas. Nondimensionalize this relation, using a reference length L and the inlet speed of sound a0 as pa-rameters for defining dimensionless variables.
P5.47 The differential equation for small-amplitude vibrations y(x, t) of a simple beam is given by A 2 t2 y EI 4 x y 4 0 where beam material density A cross-sectional area I area moment of inertia E Young’s modulus Use only the quantities , E, and A to nondimensionalize y, x, and t, and rewrite the differential equation in dimension-less form. Do any parameters remain? Could they be re-moved by further manipulation of the variables?
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. 5.48 is dropped in gasoline at 20°C. Ig-noring its acceleration phase, what will its terminal (con-stant) fall velocity be, from Fig. 5.3a?
Problems 315 tion speed for which the power will not exceed 300 W?
What will the pressure rise be for this condition?
P5.63 The pressure drop per unit length p/L in smooth pipe flow is known to be a function only of the average velocity V, diameter D, and fluid properties and . The following data were obtained for flow of water at 20°C in an 8-cm-diam-eter pipe 50 m long: P5.58 For the steel rod of Prob. 5.57, 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 under “lateral beam vibration.” P5.59 We wish to know the drag of a blimp which will move in 20°C air at 6 m/s. If a one-thirtieth-scale model is tested in water at 20°C, what should the water velocity be? At this velocity, if the measured water drag on the model is 2700 N, what is the drag on the prototype blimp and the power re-quired to propel it?
P5.60 A prototype water pump has an impeller diameter of 2 ft and is designed to pump 12 ft3/s at 750 r/min. A 1-ft-di-ameter model pump is tested in 20°C air at 1800 r/min, and Reynolds-number effects are found to be negligible. For similar conditions, what will the volume flow of the model be in ft3/s? If the model pump requires 0.082 hp to drive it, what horsepower is required for the prototype?
P5.61 If viscosity is neglected, typical pump-flow results from Prob. 5.20 are shown in Fig. P5.61 for a model pump tested in water. The pressure rise decreases and the power required increases with the dimensionless flow coefficient. Curve-fit expressions 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 m3/h. If the pump rotation speed is 30 r/s, find (a) the pressure rise and (b) the power required.
316 Chapter 5 Dimensional Analysis and Similarity Q, m3/s 0.005 0.01 0.015 0.020 p, Pa 5800 20,300 42,100 70,800 U D = 1 cm, L = 60 cm δ = 1 cm?
P5.57 y, 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 Pr es su re ri se Power Pump data (Ω in r/s) = flow coefficient 0 P Ω3D5 ≈ 0.5 + Ω2D2 ≈ 6.0 – 120 ( ( ΩD3 Q ρ ρ 2 3Q ΩD3 ∆p ΩD3 Q P5.61 ω L Stiffness EI M P5.64 P5.65 In turbulent flow near a flat wall, the local velocity u varies only with distance y from the wall, wall shear stress %w, and fluid properties and . The following data were taken in the University of Rhode Island wind tunnel for airflow, 0.0023 slug/ft3, 3.81 E-7 slug/(ft s), and %w 0.029 lbf/ft2: (a) Plot these data in the form of dimensionless u versus di-mensionless y, and suggest a suitable power-law curve fit.
(b) Suppose that the tunnel speed is increased until u 90 ft/s at y 0.11 in. Estimate the new wall shear stress, in lbf/ft2.
EES EES Verify that these data are slightly outside the range of Fig.
5.10. What is a suitable power-law curve fit for the present data? Use these data to estimate the pressure drop for flow of kerosine at 20°C in a smooth pipe of diameter 5 cm and length 200 m if the flow rate is 50 m3/h.
P5.64 The natural frequency of vibration of a mass M attached to a rod, as in Fig. P5.64, depends only upon M and the stiffness EI 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.
P5.62 Modify Prob. 5.61 so that the rotation speed is unknown but D 12 cm and Q 25 m3/h. What is the maximum rota-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. P3.165) varies only with the fluid density, pipe approach velocity, and di-ameter ratio of the meter. A model venturi meter tested in water at 20°C shows a 5-kPa drop when the approach ve-locity is 4 m/s. A geometrically similar prototype meter is used to measure gasoline at 20°C and a flow rate of 9 m3/min. If the prototype pressure gage is most accurate at 15 kPa, what should the upstream pipe diameter be?
P5.72 A one-fifteenth-scale model of a parachute has a drag of 450 lbf when tested at 20 ft/s in a water tunnel. If Reynolds-number effects are negligible, estimate the terminal fall ve-locity at 5000-ft standard altitude of a parachutist using the prototype if chute and chutist together weigh 160 lbf. Ne-glect the drag coefficient of the woman.
P5.73 The yawing moment on a torpedo control surface is tested on a one-eighth-scale model in a water tunnel at 20 m/s, us-ing Reynolds scaling. If the model measured moment is 14 N m, what will the prototype moment be under similar conditions?
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?
P5.75 A one-twelfth-scale model of an airplane is to be tested at 20°C in a pressurized wind tunnel. The prototype is to fly at 240 m/s at 10-km standard altitude. What should the tun-nel pressure be in atm to scale both the Mach number and the Reynolds number accurately?
.
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, lbf 0.016 0.057 0.122 0.208 0.315 0.441 Wave drag, lbf 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 spillway is to be tested by using Froude scaling with a one-thirtieth-scale model. The model flow has an average velocity of 0.6 m/s and a volume flow of 0.05 m3/s. What will the velocity and flow of the prototype be? If the mea-sured force on a certain part of the model is 1.5 N, what will the corresponding force on the prototype be?
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-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 Froude-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?
P5.67 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 char-acteristic dimension dm, such that the ratio dp/dm is some factor f. He then measures the drag on the model at dy-namically similar conditions (also with air at standard at-mospheric 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 Consider flow over a very small object in a viscous fluid.
Analysis of the equations of motion shows that the iner-tial terms are much smaller than the viscous and pressure terms. It turns out, then, that fluid density drops out of the equations of motion. Such flows are called creeping flows. The only important parameters in the problem are the velocity of motion U, the viscosity of the fluid , and the length scale of the body. For three-dimensional bod-ies, like spheres, creeping flow analysis yields very good results. It is uncertain, however, if such analysis can be applied to two-dimensional bodies such as a circular cylinder, since even though the diameter may be very small, the length of the cylinder is infinite for a two-di-mensional flow. Let us see if dimensional analysis can help. (a) Using the Buckingham pi theorem, generate an expression for the two-dimensional drag D2-D as a func-tion of the other parameters in the problem. Use cylinder diameter d as the appropriate length scale. Be careful— the two-dimensional drag has dimensions of force per unit length rather than simply force. (b) Is your result physi-cally plausible? If not, explain why not. (c) It turns out that fluid density cannot be neglected in analysis of creeping flow over two-dimensional bodies. Repeat the dimensional analysis, this time with included as a pa-rameter. Find the nondimensional relationship between the parameters in this problem.
P5.69 A one-sixteenth-scale model of a weir (see Fig. P5.32) has a measured flow rate Q 2.1 ft3/s when the upstream wa-ter height is H 6.3 in. If Q is proportional to weir width b, predict the prototype flow rate when Hproto 3.2 ft.
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: Problems 317 V, ft/s 30 38 48 56 61 F, lbf 1.25 1.95 3.02 4.05 4.81 EES 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?
P5.85 Solve Prob. 5.49, using the modified sphere-drag plot of Fig.
5.11.
P5.86 Solve Prob. 5.49 for glycerin at 20°C, using the modified sphere-drag plot of Fig. 5.11.
P5.87 In Prob. 5.62 it was difficult to solve for $ because it ap-peared in both power and flow coefficients. Rescale the problem, using the data of Fig. P5.61, to make a plot of di-mensionless power versus dimensionless rotation speed. En-ter this plot directly to solve Prob. 5.62 for $.
P5.88 Modify Prob. 5.62 as follows: Let $ 32 r/s and Q 24 m3/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 diameter.
Enter this plot directly to find the desired diameter.
P5.89 Knowing that p is proportional to L, rescale the data of Example 5.7 to plot dimensionless p versus dimensionless diameter. Use this plot to find the diameter required in the first row of data in Example 5.7 if the pressure drop is in-creased to 10 kPa for the same flow rate, length, and fluid.
P5.90 Knowing that p is proportional to L, rescale the data of Example 5.7 to plot dimensionless p versus dimension-less viscosity. Use this plot to find the viscosity required in the first row of data in Example 5.7 if the pressure drop is increased to 10 kPa for the same flow rate, length, and density.
P5.91 Develop a plot of dimensionless p versus dimensionless viscosity, as described in Prob. 5.90. Suppose that L 200 m, Q 60 m3/h, and the fluid is kerosine at 20°C. Use your plot to determine the minimum pipe diameter for which the pressure drop is no more than 220 kPa.
structed by using Froude scaling. What is the minimum scale ratio of the model which will ensure that its minimum We-ber 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 cur-rent speeds are expected?
P5.80 A prototype ship is 35 m long and designed to cruise at 11 m/s (about 21 kn). Its drag is to be simulated by a 1-m-long model pulled in a tow tank. For Froude 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 prototype ship is 400 ft long and has a wetted area of 30,000 ft2. A one-eightieth-scale model is tested in a tow tank according to Froude scaling at speeds of 1.3, 2.0, and 2.7 kn (1 kn 1.689 ft/s). The measured friction drag of the model at these speeds is 0.11, 0.24, and 0.41 lbf, re-spectively. What are the three prototype speeds? What is the estimated prototype friction drag at these speeds if we cor-rect for Reynolds-number discrepancy by extrapolation?
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 lbf/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 318 Chapter 5 Dimensional Analysis and Similarity Word Problems W5.1 In 98 percent of data analysis cases, the “reducing factor” j, which lowers the number n of dimensional variables 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 fol-low for choosing your scaling variables?
W5.4 In an earlier edition, the writer asked the following ques-tion about Fig. 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 Froude and Reynolds numbers together (a ship). Give an example of a flow which 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) which would make the test results dif-ficult 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 (thermodynamics, strength of materials, etc.) which does not satisfy the principle of dimensional homogeneity. Ex-plain what is wrong and whether it can be modified to be homogeneous.
W5.8 Some colleges (e.g., Colorado State University) have en-vironmental wind tunnels which can be used to study, e.g., wind flow over city buildings. What details of scaling might be important in such studies?
W5.9 If the model scale ratio is Lm/Lp, as in Eq. (5.31), and the Weber number is important, how must the model Comprehensive Problems 319 and prototype surface tension be related to for dynamic similarity?
W5.10 For a typical incompressible velocity potential analysis in Chap. 4 we solve 2' 0, subject to known values of '/ n on the boundaries. What dimensionless parameters govern this type of motion?
FE5.6 A football, meant to be thrown at 60 mi/h in sea-level air ( 1.22 kg/m3, 1.78 E-5 N m2), is to be tested us-ing a one-quarter scale model in a water tunnel ( 998 kg/m3, 0.0010 N s/m2). For dynamic similarity, what is the ratio of prototype force to model force?
(a) 3.861, (b) 161, (c) 321, (d) 561, (e) 641 FE5.7 Consider liquid flow of density , viscosity , and veloc-ity U over a very small model spillway of length scale L, such that the liquid surface tension coefficient is im-portant. The quantity U2L/ in this case is important and is called the (a) capillary rise, (b) Froude number, (c) Prandtl num-ber, (d ) Weber number, (e) Bond number FE5.8 If a stream flowing at velocity U past a body of length L causes a force F on the body which depends only upon U, L, and fluid viscosity , then F must be proportional to (a) UL/, (b) U2L2, (c) U/L, (d) UL, (e) UL/ FE5.9 In supersonic wind tunnel testing, if different gases are used, dynamic similarity requires that the model and pro-totype have the same Mach number and the same (a) Euler number, (b) speed of sound, (c) stagnation en-thalpy, (d) Froude number, (e) specific heat ratio FE5.10 The Reynolds number for a 1-ft-diameter sphere moving at 2.3 mi/h through seawater (specific gravity 1.027, vis-cosity 1.07 E-3 N s/m2) is approximately (a) 300, (b) 3000, (c) 30,000, (d) 300,000, (e) 3,000,000 Fundamentals of Engineering Exam Problems FE5.1 Given the parameters (U, L, g, , ) which affect a cer-tain liquid flow problem, the ratio V2/(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 appro-priate tow velocity is (a) 0.19 m/s, (b) 0.35 m/s, (c) 1.31 m/s, (d) 2.55 m/s, (e) 8.35 m/s FE5.3 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, (d) 61,800 N, (e) 275,000 N FE5.4 A tidal estuary is dominated by the semidiurnal lunar tide, with a period of 12.42 h. If a 1500 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 FE5.5 A football, meant to be thrown at 60 mi/h in sea-level air ( 1.22 kg/m3, 1.78 E-5 N s/m2), is to be tested using a one-quarter scale model in a water tunnel ( 998 kg/m3, 0.0010 N s/m2). For dynamic similar-ity, 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 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 %w is a function of density , vis-cosity , average velocity V, pipe diameter d, and wall roughness height . Thus, functionally, we can write %w fcn(, , V, d, ). (a) Using dimensional analysis, rewrite this function in dimensionless form. (b) A certain pipe has d 5 cm and 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 %w, Pa 0.05 0.18 0.37 0.64 0.86 1.25 Plot these data using the dimensionless form obtained in part (a) and suggest a curve-fit formula. Does your plot re-veal the entire functional relation obtained in part (a)?
analysis. Let the vertical velocity be a function only of dis-tance from the plate, fluid properties, gravity, and film thickness. That is, w fcn(x, , , g, ). (a) Use the pi the-orem to rewrite this function in terms of dimensionless pa-rameters. (b) Verify that the exact solution from Prob. 4.80 is consistent with your result in part (a).
C5.4 The Taco Inc. model 4013 centrifugal pump has an impeller of diameter D 12.95 in. When pumping 20°C water at $ 1160 r/min, the measured flow rate Q and pressure rise p are given by the manufacturer as follows: Q, gal/min 200 300 400 500 600 700 p, lb/in2 36 35 34 32 29 23 (a) Assuming that p fcn(, Q, D, $), use the pi theo-rem to rewrite this function in terms of dimensionless pa-rameters 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 p is ex-pected, in lbf/in2?
320 Chapter 5 Dimensional Analysis and Similarity 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 p1 is large and exit diameter d2 is small. In this case, the difference p1 p2 is no longer controlling, and the gas mass flow m ˙ reaches a maximum value which depends upon p1 and d2 and also upon the ab-solute upstream temperature T1 and the gas constant R. Thus, functionally, m ˙ fcn(p1, d2, T1, R). (a) Using dimensional analysis, rewrite this function in dimensionless form. (b) A certain pipe has d2 1 mm. For flow of air, measurements show the following values of mass flow through the nozzle: T1, K 300 300 300 500 800 p1, 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 ob-tained in part (a)?
C5.3 Reconsider the fully developed draining vertical oil-film problem (see Fig. P4.80) as an exercise in dimensional Design Projects D5.1 We are given laboratory data, taken by Prof. Robert Kirch-hoff 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, fac-ing in opposite directions, and glued to thin ( 1 4 -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 $, are as follows: 0.212 0.322 0.458 0.574 U, ft/s $, r/min U, ft/s $, r/min U, ft/s $, r/min U, ft/s $, r/min 18.95 435.00 18.95 225.00 20.10 140.00 23.21 115.00 22.20 545.00 23.19 290.00 26.77 215.00 27.60 145.00 25.90 650.00 29.15 370.00 31.37 260.00 32.07 175.00 29.94 760.00 32.79 425.00 36.05 295.00 36.05 195.00 38.45 970.00 38.45 495.00 39.03 327.00 39.60 215.00 Assume that the angular velocity $ of the device is a func-tion of wind speed U, air density and viscosity , 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 above 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 $ 120 r/min, what should be the proper rod length? What are possible limita-tions of your design? Predict the expected $ (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 ReD 3 E5, which ac-counts for the dimpling of golf balls to increase their dis-tance traveled. Some experimental data for roughened spheres are given in Fig. D5.2. The figure also shows typical golf-ball data. We see that some roughened spheres are better than golf balls in some regions. For the present study, let us neglect the ball’s spin, which causes the very important side-force or Magnus effect (See Fig. 8.11) and assume that the ball is hit without spin and follows the equa-tions of motion for plane motion (x, z): mx ˙˙ F cos mz ˙ ˙ F sin W where F CD 2 4 D2(x ˙ 2 z ˙2) tan1 x z ˙ ˙ The ball has a particular CD(ReD) curve from Fig. D5.2 and is struck with an initial velocity V0 and angle 0. 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 ini-tial 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 golf ball for any conditions? What roughness-effect differ-ences occur between a low-impact duffer and, say, Tiger Woods?
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322 Chapter 5 Dimensional Analysis and Similarity Steam pipe bridge in a geothermal power plant. Pipe flows are everywhere, often occurring in groups or networks. They are designed using the principles outlined in this chapter. (Courtesy of Dr. E. R. Degginger/Color-Pic Inc.) 324 6.1 Reynolds-Number Regimes 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 alternate 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 pip-ing problems.
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 some rather specific digital-computer solutions, and there are a great many experimental data. There is a lot of the-ory 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 agi-tated (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 3 105, where the sharp drop in the drag coefficient appeared. Transition depends upon many effects, e.g., wall roughness (Fig. 5.3b) or fluctuations in the inlet stream, but the primary parame-ter is the Reynolds number. There are a great many data on transition but only a small amount of theory [1 to 3].
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 325 Chapter 6 Viscous Flow in Ducts Fig. 6.1 The three regimes of vis-cous flow: (a) laminar flow at low Re; (b) transition at intermediate Re; (c) turbulent flow at high Re.
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 which damp out quickly (Fig. 6.1a). If transition is occurring, there will be sharp bursts of turbulent fluctuation (Fig. 6.1b) as the increasing Reynolds num-ber causes a breakdown or instability of laminar motion. At sufficiently large Re, the flow will fluctuate continually (Fig. 6.1c) and is termed fully turbulent. The fluctua-tions, 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 frequen-cies. 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 Red,crit 2300. For flow through a 5-cm-diameter pipe, at what velocity will this occur at 20°C for (a) airflow and (b) wa-ter 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 Vd/ 2300. With d known, we introduce the appropriate fluid properties at 20°C from Tables A.3 and A.4: (a) Air: 2300 or V 0.7 (b) Water: 2300 or 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 joined by a liquid sheet. The higher-Reynolds-number turbulent flow (Fig.
6.2b) is unsteady and irregular but, when averaged over time, is steady and predictable.
m s (998 kg/m3)V(0.05 m) 0.001 kg/(m s) Vd m s (1.205 kg/m3)V(0.05 m) 1.80 E-5 kg/(m s) Vd 326 Chapter 6 Viscous Flow in Ducts t u (a) t u (b) t u (c) Small natural disturbances damp quickly Intermittent bursts of turbulence Continuous turbulence Fig. 6.2 Flow issuing at constant speed from a pipe: (a) high-viscosity, low-Reynolds-number, laminar flow; (b) low-viscosity, high-Reynolds-number, turbulent flow. [From Illustrated Experiments in Fluid Mechanics (The NCFMF Book of Film Notes), National Committee for Fluid Mechanics Films, Education Development Center, Inc., copyright 1972.] How did turbulence form inside the pipe? The laminar parabolic flow profile, which is similar to Eq. (4.143), became unstable and, at Red 2300, began to form “slugs” or “puffs” of intense turbulence. A puff has a fast-moving front and a slow-moving rear 6.1 Reynolds-Number Regimes 327 Fig. 6.3 Formation of a turbulent puff in pipe flow: (a) and (b) near the entrance; (c) somewhat down-stream; (d) far downstream. (From Ref. 45, courtesy of P. R. Bandy-opadhyay.) 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 down-stream (Fig. 6.3c) the puff becomes fully turbulent and very active, with helical mo-tions 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 num-ber. If Re UL/, where U is the average stream velocity and L is the “width,” or trans-verse 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 103: laminar, boundary-layer theory useful 103 Re 104: transition to turbulence 104 Re 106: turbulent, moderate Reynolds-number dependence 106 Re : turbulent, slight Reynolds-number dependence These are representative ranges which 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.
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 328 Chapter 6 Viscous Flow in Ducts (a) Flow (b) (c) (d) Historical Outline Fig. 6.4 Experimental evidence of transition for water flow in a 1 4 -in smooth pipe 10 ft long.
mean values of velocity, pressure, force, etc. But turbulence can change the mean val-ues dramatically, e.g., 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 vis-cous flow. He measured water flow in long brass pipes and deduced a pressure-drop law p (const) entrance effect (6.1) This is exactly our laminar-flow scaling law from Example 5.4, but Hagen did not re-alize that the constant was proportional to the fluid viscosity.
The formula broke down as Hagen increased Q beyond a certain limit, i.e., 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 p varies as the sec-ond 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 V 2.2 ft/s the pressure drop is nearly quadratic with V. The actual power p V1.75 seems impossible on dimensional grounds but is easily explained when the dimen-sionless pipe-flow data (Fig. 5.10) are displayed.
In 1883 Osborne Reynolds, a British engineering professor, showed that the change depended upon the parameter Vd/, now named in his honor. By introducing a dye LQ R4 6.1 Reynolds-Number Regimes 329 Turbulent flow ∆p α V 1.75 Laminar flow ∆p α V Transition 0 0.5 1.0 1.5 2.0 2.5 0 20 40 80 100 120 Average velocity V, ft/s Pressure drop ∆p, lbf/ft2 60 6.2 Internal versus External Viscous Flows Fig. 6.5 Reynolds’ sketches of pipe-flow transition: (a) low-speed, laminar flow; (b) high-speed, turbu-lent flow; (c) spark photograph of condition (b). (From Ref. 4.) 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 Red 2100. The flow became fully turbulent, V 2.2 ft/s, at Red 4200. The accepted design value for pipe-flow transition is now taken to be Red,crit 2300 (6.2) 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, Red,crit can be de-layed until much higher values.
Transition also occurs in external flows around bodies such as the sphere and cylin-der 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 which 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 there are no analyses which can simulate the fine-scale random fluctuations of tur-bulent flow.1 Therefore existing turbulent-flow theory is semiempirical, based upon di-mensional analysis and physical reasoning; it is concerned with the mean flow prop-erties only and the mean of the fluctuations, not their rapid variations. The turbulent-flow “theory” presented here in Chaps. 6 and 7 is unbelievably crude yet sur-prisingly effective. We shall attempt a rational approach which places turbulent-flow analysis on a firm physical basis.
Both laminar and turbulent flow may be either internal, i.e., “bounded” by walls, or external and unbounded. This chapter treats internal flows, and Chap. 7 studies exter-nal 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 converges and enters the tube. Viscous boundary layers grow downstream, retarding the axial flow u(r, x) at the wall and thereby accelerating the center-core flow to maintain the in-compressible continuity requirement Q u dA const (6.3) At a finite 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 Le it no longer changes with x and is said to be fully de-veloped, u u(r) only. Downstream of x Le the velocity profile is constant, the wall 330 Chapter 6 Viscous Flow in Ducts Needle Tank Dye filament (a) (b) (c) 1Reference 32 is a computer model of large-scale turbulent fluctuations.
Fig. 6.6 Developing velocity pro-files and pressure changes in the entrance of a duct flow.
shear is constant, and the pressure drops linearly with x, for either laminar or turbu-lent flow. All these details are shown in Fig. 6.6.
Dimensional analysis shows that the Reynolds number is the only parameter af-fecting entrance length. If Le f(d, V, , ) V then g g(Re) (6.4) For laminar flow [2, 3], the accepted correlation is 0.06 Re laminar (6.5) The maximum laminar entrance length, at Red,crit 2300, is Le 138d, which is the longest development length possible.
In turbulent flow the boundary layers grow faster, and Le is relatively shorter, ac-cording to the approximation for smooth walls 4.4 Red 1/6 turbulent (6.6) Some computed turbulent entrance lengths are thus Le d Le d Vd Le d Q A 6.2 Internal versus External Viscous Flows 331 Inviscid core flow 0 Le x Entrance pressure drop Linear pressure drop in fully developed flow region Pressure Entrance length Le (developing profile region) Growing boundary layers Boundary layers merge Developed velocity profile u (r ) u (r, x) x r Fully developed flow region Red 4000 104 105 106 107 108 Le/d 18 20 30 44 65 95 Now 44 diameters may seem “long,” but typical pipe-flow applications involve an L/d value of 1000 or more, in which case the entrance effect may be neglected and a sim-ple analysis made for fully developed flow (Sec. 6.4). This is possible for both lami-nar and turbulent flows, including rough walls and noncircular cross sections.
EXAMPLE 6.2 A 1 2 -in-diameter water pipe is 60 ft long and delivers water at 5 gal/min at 20°C. What fraction of this pipe is taken up by the entrance region?
Solution Convert Q (5 gal/min) 0.0111 ft3/s The average velocity is V Q A 8.17 ft/s From Table 1.4 read for water 1.01 106 m2/s 1.09 105 ft2/s. Then the pipe Reynolds number is Red V d 31,300 This is greater than 4000; hence the flow is fully turbulent, and Eq. (6.6) applies for entrance length L d e 4.4 Re1/6 d (4.4)(31,300)1/6 25 The actual pipe has L/d (60 ft)/[( 1 2 /12) ft] 1440. Hence the entrance region takes up the frac-tion 0.017 1.7% Ans.
This is a very small percentage, so that we can reasonably treat this pipe flow as essentially fully developed.
Shortness can be a virtue in duct flow if one wishes to maintain the inviscid core.
For example, a “long” wind tunnel would be ridiculous, since the viscous core would invalidate the purpose of simulating free-flight conditions. A typical laboratory low-speed wind-tunnel test section is 1 m in diameter and 5 m long, with V 30 m/s. If we take air 1.51 105 m2/s from Table 1.4, then Red 1.99 106 and, from Eq. (6.6), Le/d 49. The test section has L/d 5, which is much shorter than the de-25 1440 Le L (8.17 ft/s)[( 1 2 /12) ft] 1.09 105 ft2/s 0.0111 ft3/s (/4)[( 1 2 /12) ft]2 0.00223 ft3/s 1 gal/min 332 Chapter 6 Viscous Flow in Ducts 6.3 Semiempirical Turbulent Shear Correlations Reynolds’ Time-Averaging Concept velopment length. At the end of the section the wall boundary layers are only 10 cm thick, leaving 80 cm of inviscid core suitable for model testing.
An external flow has no restraining walls and is free to expand no matter how thick the viscous layers on the immersed body may become. Thus, far from the body the flow is nearly inviscid, and our analytical technique, treated in Chap. 7, is to patch an inviscid-flow solution onto a viscous boundary-layer solution computed for the wall region. There is no external equivalent of fully developed internal flow.
Throughout this chapter we assume constant density and viscosity and no thermal in-teraction, so that only the continuity and momentum equations are to be solved for ve-locity and pressure Continuity: 0 Momentum: p g 2V (6.7) subject to no slip at the walls and known inlet and exit conditions. (We shall save our free-surface solutions for Chap. 10.) Both laminar and turbulent flows satisfy Eqs. (6.7). For laminar flow, where there are no random fluctuations, we go right to the attack and solve them for a variety of geometries [2, 3], leaving many more, of course, for the problems.
For turbulent flow, because of the fluctuations, every velocity and pressure term in Eqs.
(6.7) is a rapidly varying random function of time and space. At present our mathe-matics cannot handle such instantaneous fluctuating variables. No single pair of ran-dom functions V(x, y, z, t) and p(x, y, z, t) is known to be a solution to Eqs. (6.7).
Moreover, our attention as engineers is toward the average or mean values of velocity, pressure, shear stress, etc., in a high-Reynolds-number (turbulent) flow. This approach led Osborne Reynolds in 1895 to rewrite Eqs. (6.7) in terms of mean or time-averaged turbulent variables.
The time mean u of a turbulent function u(x, y, z, t) is defined by u T 0 u dt (6.8) where T is an averaging period taken to be longer than any significant period of the fluctuations themselves. The mean values of turbulent velocity and pressure are illus-trated in Fig. 6.7. For turbulent gas and water flows, an averaging period T 5 s is usually quite adequate.
The fluctuation u is defined as the deviation of u from its average value u u u (6.9) also shown in Fig. 6.7. It follows by definition that a fluctuation has zero mean value u T 0 (u u ) dt u u 0 (6.10) 1 T 1 T dV dt w z y u x 6.3 Semiempirical Turbulent Shear Correlations 333 Fig. 6.7 Definition of mean and fluctuating turbulent variables: (a) velocity; (b) pressure.
However, the mean square of a fluctuation is not zero and is a measure of the inten-sity of the turbulence u 2 T 0 u2 dt 0 (6.11) Nor in general are the mean fluctuation products such as u and u p zero in a typi-cal turbulent flow.
Reynolds’ idea was to split each property into mean plus fluctuating variables u u u w w w p p p (6.12) Substitute these into Eqs. (6.7), and take the time mean of each equation. The conti-nuity relation reduces to 0 (6.13) which is no different from a laminar continuity relation.
However, each component of the momentum equation (6.7b), after time averaging, will contain mean values plus three mean products, or correlations, of fluctuating ve-locities. The most important of these is the momentum relation in the mainstream, or x, direction, which takes the form gx u 2 u u w (6.14) The three correlation terms u 2 , u , and u w are called turbulent stresses because they have the same dimensions and occur right alongside the newtonian (lam-inar) stress terms (u /x), etc. Actually, they are convective acceleration terms (which is why the density appears), not stresses, but they have the mathematical effect of stress and are so termed almost universally in the literature.
The turbulent stresses are unknown a priori and must be related by experiment to u z z u y y u x x p x du dt w z y u x 1 T 334 Chapter 6 Viscous Flow in Ducts t u (a) t p (b) u′ u u = u + u′ p′ p = p + p′ p Fig. 6.8 Typical velocity and shear distributions in turbulent flow near a wall: (a) shear; (b) velocity.
geometry and flow conditions, as detailed in Refs. 1 to 3. Fortunately, in duct and boundary-layer flow, the stress u associated with direction y normal to the wall is dominant, and we can approximate with excellent accuracy a simpler streamwise momentum equation gx (6.15) where u lam turb (6.16) Figure 6.8 shows the distribution of lam and turb from typical measurements across a turbulent-shear layer near a wall. Laminar shear is dominant near the wall (the wall layer), and turbulent shear dominates in the outer layer. There is an intermediate re-gion, called the overlap layer, where both laminar and turbulent shear are important.
These three regions are labeled in Fig. 6.8.
In the outer layer turb is two or three orders of magnitude greater than lam, and vice versa in the wall layer. These experimental facts enable us to use a crude but very effective model for the velocity distribution u (y) across a turbulent wall layer.
We have seen in Fig. 6.8 that there are three regions in turbulent flow near a wall: 1. Wall layer: Viscous shear dominates.
2. Outer layer: Turbulent shear dominates.
3. Overlap layer: Both types of shear are important.
From now on let us agree to drop the overbar from velocity u . Let w be the wall shear stress, and let and U represent the thickness and velocity at the edge of the outer layer, y .
For the wall layer, Prandtl deduced in 1930 that u must be independent of the shear-layer thickness u f(, w, , y) (6.17) By dimensional analysis, this is equivalent to u y y p x d u dt 6.3 Semiempirical Turbulent Shear Correlations 335 y y = (x) (x, y) turb lam (a) (b) Viscous wall layer Overlap layer Outer turbulent layer w(x) 0 y U(x) u (x, y) δ τ τ τ τ The Logarithmic-Overlap Law u F u 1/2 (6.18) Equation (6.18) is called the law of the wall, and the quantity u is termed the friction velocity because it has dimensions {LT1}, although it is not actually a flow velocity.
Subsequently, Kármán in 1933 deduced that u in the outer layer is independent of molecular viscosity, but its deviation from the stream velocity U must depend on the layer thickness and the other properties (U u)outer g(, w, , y) (6.19) Again, by dimensional analysis we rewrite this as G (6.20) where u has the same meaning as in Eq. (6.18). Equation (6.20) is called the velocity-defect law for the outer layer.
Both the wall law (6.18) and the defect law (6.20) are found to be accurate for a wide variety of experimental turbulent duct and boundary-layer flows [1 to 3]. They are different in form, yet they must overlap smoothly in the intermediate layer. In 1937 C. B. Millikan showed that this can be true only if the overlap-layer velocity varies logarithmically with y: ln B overlap layer (6.21) Over the full range of turbulent smooth wall flows, the dimensionless constants and B are found to have the approximate values 0.41 and B 5.0. Equation (6.21) is called the logarithmic-overlap layer.
Thus by dimensional reasoning and physical insight we infer that a plot of u versus ln y in a turbulent-shear layer will show a curved wall region, a curved outer region, and a straight-line logarithmic overlap. Figure 6.9 shows that this is exactly the case.
The four outer-law profiles shown all merge smoothly with the logarithmic-overlap law but have different magnitudes because they vary in external pressure gradient. The wall law is unique and follows the linear viscous relation u y (6.22) from the wall to about y 5, thereafter curving over to merge with the logarithmic law at about y 30.
Believe it or not, Fig. 6.9, which is nothing more than a shrewd correlation of ve-locity profiles, is the basis for most existing “theory” of turbulent-shear flows. Notice that we have not solved any equations at all but have merely expressed the streamwise velocity in a neat form.
There is serendipity in Fig. 6.9: The logarithmic law (6.21), instead of just being a short overlapping link, actually approximates nearly the entire velocity profile, except for the outer law when the pressure is increasing strongly downstream (as in a dif-fuser). The inner-wall law typically extends over less than 2 percent of the profile and can be neglected. Thus we can use Eq. (6.21) as an excellent approximation to solve yu u u yu 1 u u y U u u w yu u u 336 Chapter 6 Viscous Flow in Ducts Fig. 6.9 Experimental verification of the inner-, outer-, and overlap-layer laws relating velocity profiles in turbulent wall flow.
nearly every turbulent-flow problem presented in this and the next chapter. Many ad-ditional applications are given in Refs. 2 and 3.
EXAMPLE 6.3 Air at 20°C flows through a 14-cm-diameter tube under fully developed conditions. The cen-terline velocity is u0 5 m/s. Estimate from Fig. 6.9 (a) the friction velocity u, (b) the wall shear stress w, and (c) the average velocity V Q/A.
Solution For pipe flow Fig. 6.9 shows that the logarithmic law, Eq. (6.21), is accurate all the way to the center of the tube. From Fig. E6.3 y R r should go from the wall to the centerline as shown.
At the center u u0, y R, and Eq. (6.21) becomes ln 5.0 (1) Since we know that u0 5 m/s and R 0.07 m, u is the only unknown in Eq. (1). Find the solution by trial and error or by EES u 0.228 m/s 22.8 cm/s Ans. (a) where we have taken 1.51 105 m2/s for air from Table 1.4.
Ru 1 0.41 u0 u 6.3 Semiempirical Turbulent Shear Correlations 337 30 25 20 15 10 5 0 1 y+ = yu 10 102 103 10 4 Linear viscous sublayer, Eq. (6.22) Logarithmic overlap Eq. (6.21) Experimental data u+ = y+ Inn er l ay er Outer law profiles: Strong increasing pressure Flat plate flow Pipe flow Strong decreasing pressure u+ = u u Overlap layer ν E6.3 u ( y) y = R y r r = R = 7 cm u0 = 5 m /s Part (a) Part (b) Assuming a pressure of 1 atm, we have p/(RT) 1.205 kg/m3. Since by definition u (w/)1/2, we compute w u2 (1.205 kg/m3)(0.228 m/s)2 0.062 kg/(m s2) 0.062 Pa Ans. (b) This is a very small shear stress, but it will cause a large pressure drop in a long pipe (170 Pa for every 100 m of pipe).
The average velocity V is found by integrating the logarithmic-law velocity distribution V R 0 u 2r dr (2) Introducing u u[(1/) ln (yu/) B] from Eq. (6.21) and noting that y R r, we can carry out the integration of Eq. (2), which is rather laborious. The final result is V 0.835u0 4.17 m/s Ans. (c) We shall not bother showing the integration here because it is all worked out and a very neat formula is given in Eqs. (6.49) and (6.59).
Notice that we started from almost nothing (the pipe diameter and the centerline velocity) and found the answers without solving the differential equations of continuity and momen-tum. We just used the logarithmic law, Eq. (6.21), which makes the differential equations un-necessary for pipe flow. This is a powerful technique, but you should remember that all we are doing is using an experimental velocity correlation to approximate the actual solution to the problem.
We should check the Reynolds number to ensure turbulent flow Red 38,700 Since this is greater than 4000, the flow is definitely turbulent.
As our first example of a specific viscous-flow analysis, we take the classic problem of flow in a full pipe, driven by pressure or gravity or both. Figure 6.10 shows the geometry of the pipe of radius R. The x-axis is taken in the flow direction and is in-clined to the horizontal at an angle .
Before proceeding with a solution to the equations of motion, we can learn a lot by making a control-volume analysis of the flow between sections 1 and 2 in Fig. 6.10.
The continuity relation, Eq. (3.23), reduces to Q1 Q2 const or V1 V2 (6.23) since the pipe is of constant area. The steady-flow energy equation (3.71) reduces to 1V1 2 gz1 2V2 2 gz2 ghf (6.24) since there are no shaft-work or heat-transfer effects. Now assume that the flow is fully 1 2 p2 1 2 p1 Q2 A2 Q1 A1 (4.17 m/s)(0.14 m) 1.51 105 m2/s Vd 1 R2 Q A 338 Chapter 6 Viscous Flow in Ducts Part (c) 6.4 Flow in a Circular Pipe Fig. 6.10 Control volume of steady, fully developed flow between two sections in an inclined pipe.
developed (Fig. 6.6), and correct later for entrance effects. Then the kinetic-energy cor-rection factor 1 2, and since V1 V2 from (6.23), Eq. (6.24) now reduces to a simple expression for the friction-head loss hf hf z1 z2 z z (6.25) The pipe-head loss equals the change in the sum of pressure and gravity head, i.e., the change in height of the hydraulic grade line (HGL). Since the velocity head is constant through the pipe, hf also equals the height change of the energy grade line (EGL). Re-call that the EGL decreases downstream in a flow with losses unless it passes through an energy source, e.g., as a pump or heat exchanger.
Finally apply the momentum relation (3.40) to the control volume in Fig. 6.10, ac-counting for applied forces due to pressure, gravity, and shear p R2 g(R2) L sin w(2R) L m ˙ (V2 V1) 0 (6.26) This equation relates hf to the wall shear stress z hf (6.27) where we have substituted z L sin from Fig. 6.10.
So far we have not assumed either laminar or turbulent flow. If we can correlate w with flow conditions, we have solved the problem of head loss in pipe flow. Func-tionally, we can assume that w F(, V, , d, ) (6.28) where is the wall-roughness height. Then dimensional analysis tells us that L R 2w g p g p g p g p2 g p1 g 6.4 Flow in a Circular Pipe 339 2 u (r) r r = R 1 x2 – x1 = ∆L p2 x Z2 p1 = p2 + ∆p gx = g sin φ g φ φ Z1 w τ τ (r) Equations of Motion f FRed, (6.29) The dimensionless parameter f is called the Darcy friction factor, after Henry Darcy (1803–1858), a French engineer whose pipe-flow experiments in 1857 first established the effect of roughness on pipe resistance.
Combining Eqs. (6.27) and (6.29), we obtain the desired expression for finding pipe-head loss hf f (6.30) This is the Darcy-Weisbach equation, valid for duct flows of any cross section and for laminar and turbulent flow. It was proposed by Julius Weisbach, a German professor who in 1850 published the first modern textbook on hydrodynamics.
Our only remaining problem is to find the form of the function F in Eq. (6.29) and plot it in the Moody chart of Fig. 6.13.
For either laminar or turbulent flow, the continuity equation in cylindrical coordinates is given by (App. D) 1 r r (rr) 1 r () u x 0 (6.31) We assume that there is no swirl or circumferential variation, / 0, and fully developed flow: u u(r) only. Then Eq. (6.31) reduces to 1 r r (rr) 0 or rr const (6.32) But at the wall, r R, r 0 (no slip); therefore (6.32) implies that υr 0 every-where. Thus in fully developed flow there is only one velocity component, u u(r).
The momentum differential equation in cylindrical coordinates now reduces to u u x d d p x gx 1 r r (r) (6.33) where can represent either laminar or turbulent shear. But the left-hand side vanishes because u u(r) only. Rearrange, noting from Fig. 6.10 that gx g sin : 1 r r (r) d d x (p gx sin ) d d x (p gz) (6.34) Since the left-hand side varies only with r and the right-hand side varies only with x, it follows that both sides must be equal to the same constant.2 Therefore we can inte-grate Eq. (6.34) to find the shear distribution across the pipe, utilizing the fact that 0 at r 0 1 2 r d d x (p gz) (const)(r) (6.35) V2 2g L d d 8w V2 340 Chapter 6 Viscous Flow in Ducts 2Ask your instructor to explain this to you if necessary.
Laminar-Flow Solution Thus the shear varies linearly from the centerline to the wall, for either laminar or tur-bulent flow. This is also shown in Fig. 6.10. At r R, we have the wall shear w 1 2 R p L g z (6.36) which is identical with our momentum relation (6.27). We can now complete our study of pipe flow by applying either laminar or turbulent assumptions to fill out Eq. (6.35).
Note in Eq. (6.35) that the HGL slope d(p gz)/dx is negative because both pres-sure and height drop with x. For laminar flow, du/dr, which we substitute in Eq. (6.35) d d u r 1 2 rK K d d x (p gz) (6.37) Integrate once u 1 4 r2 K C1 (6.38) The constant C1 is evaluated from the no-slip condition at the wall: u 0 at r R 0 1 4 R2 K C1 (6.39) or C1 1 4 R2K/. Introduce into Eq. (6.38) to obtain the exact solution for laminar fully developed pipe flow u 4 1 d d x (p gz)(R2 r2) (6.40) The laminar-flow profile is thus a paraboloid falling to zero at the wall and reaching a maximum at the axis umax 4 R 2 d d x (p gz) (6.41) It resembles the sketch of u(r) given in Fig. 6.10.
The laminar distribution (6.40) is called Hagen-Poiseuille flow to commemorate the experimental work of G. Hagen in 1839 and J. L. Poiseuille in 1940, both of whom established the pressure-drop law, Eq. (6.1). The first theoretical derivation of Eq. (6.40) was given independently by E. Hagenbach and by F. Neumann around 1859.
Other pipe-flow results follow immediately from Eq. (6.40). The volume flow is Q R 0 u dA R 0 umax1 R r2 2 2r dr 1 2 umaxR2 8 R 4 d d x (p gz) (6.42) Thus the average velocity in laminar flow is one-half the maximum velocity 6.4 Flow in a Circular Pipe 341 E6.4 V Q A Q R2 1 2 umax (6.43) For a horizontal tube ( z 0), Eq. (6.42) is of the form predicted by Hagen’s exper-iment, Eq. (6.1): p 8 R L 4 Q (6.44) The wall shear is computed from the wall velocity gradient w d d u r rR 2u R max 1 2 R d d x (p gz) (6.45) This gives an exact theory for laminar Darcy friction factor f 8 V w 2 8(8 V V 2 /d) 6 4 V d or flam R 6 e 4 d (6.46) This is plotted later in the Moody chart (Fig. 6.13). The fact that f drops off with in-creasing Red should not mislead us into thinking that shear decreases with velocity: Eq. (6.45) clearly shows that w is proportional to umax; it is interesting to note that w is independent of density because the fluid acceleration is zero.
The laminar head loss follows from Eq. (6.30) hf,lam 6 4 V d L d 2 V g 2 3 2 g d L 2 V 1 28 gd L 4 Q (6.47) We see that laminar head loss is proportional to V.
EXAMPLE 6.4 An oil with 900 kg/m3 and 0.0002 m2/s flows upward through an inclined pipe as shown in Fig. E6.4. The pressure and elevation are known at sections 1 and 2, 10 m apart. Assuming 342 Chapter 6 Viscous Flow in Ducts 10 m Q,V d = 6 cm p2 = 250,000 Pa p1 = 350,000 Pa, z1 = 0 1 2 40˚ steady laminar flow, (a) verify that the flow is up, (b) compute hf between 1 and 2, and compute (c) Q, (d) V, and (e) Red. Is the flow really laminar?
E6.5 Solution For later use, calculate (900 kg/m3)(0.0002 m2/s) 0.18 kg/(m s) z2 L sin 40° (10 m)(0.643) 6.43 m The flow goes in the direction of falling HGL; therefore compute the hydraulic grade-line height at each section HGL1 z1 p g 1 0 90 3 0 5 ( 0 9 ,0 .8 0 0 0 7) 39.65 m HGL2 z2 p g 2 6.43 90 2 0 5 ( 0 9 ,0 .8 0 0 0 7) 34.75 m The HGL is lower at section 2; hence the flow is from 1 to 2 as assumed.
Ans. (a) The head loss is the change in HGL: hf HGL1 HGL2 39.65 m 34.75 m 4.9 m Ans. (b) Half the length of the pipe is quite a large head loss.
We can compute Q from the various laminar-flow formulas, notably Eq. (6.47) Q 1 2 g 8 d 4 L hf 0.0076 m3/s Ans. (c) Divide Q by the pipe area to get the average velocity V Q R2 0 ( .
0 0 .
0 0 7 3 6 )2 2.7 m/s Ans. (d) With V known, the Reynolds number is Red V d 2 0 .7 .0 (0 0 .
0 0 2 6) 810 Ans. (e) This is well below the transition value Red 2300, and so we are fairly certain the flow is lam-inar.
Notice that by sticking entirely to consistent SI units (meters, seconds, kilograms, newtons) for all variables we avoid the need for any conversion factors in the calculations.
EXAMPLE 6.5 A liquid of specific weight g 58 lb/ft3 flows by gravity through a 1-ft tank and a 1-ft capil-lary tube at a rate of 0.15 ft3/h, as shown in Fig. E6.5. Sections 1 and 2 are at atmospheric pres-sure. Neglecting entrance effects, compute the viscosity of the liquid.
Solution Apply the steady-flow energy equation (6.24), including the correction factor : (900)(9.807)(0.06)4(4.9) 128(0.18)(10) 6.4 Flow in a Circular Pipe 343 1 2 d = 0.004 ft Q = 0.15 ft3/ h 1 ft 1 ft Part (a) Part (b) Part (c) Part (d) Part (e) Turbulent-Flow Solution p g 1 2 1V g 1 2 z1 p g 2 2 2V g 2 2 z2 hf The average exit velocity V2 can be found from the volume flow and the pipe size: V2 A Q 2 Q R2 (0.
1 ( 5 0 /3 .0 6 0 0 2 0) ft f ) t 2 3/s 3.32 ft/s Meanwhile p1 p2 pa, and V1 0 in the large tank. Therefore, approximately, hf z1 z2 2 V 2g 2 2 2.0 ft 2.0 2 ( ( 3 3 .3 2 2 .2 f f t t / / s s ) 2 2 ) 1.66 ft where we have introduced 2 2.0 for laminar pipe flow from Eq. (3.72). Note that hf includes the entire 2-ft drop through the system and not just the 1-ft pipe length.
With the head loss known, the viscosity follows from our laminar-flow formula (6.47): hf 1.66 ft 3 2 g d L 2 V 114,500 or 11 1 4 .6 ,5 6 00 1.45 E-5 slug/(ft s) Ans.
Note that L in this formula is the pipe length of 1 ft. Finally, check the Reynolds number: Red Vd 1650 laminar Since this is less than 2300, we conclude that the flow is indeed laminar. Actually, for this head loss, there is a second (turbulent) solution, as we shall see in Example 6.8.
For turbulent pipe flow we need not solve a differential equation but instead proceed with the logarithmic law, as in Example 6.3. Assume that Eq. (6.21) correlates the lo-cal mean velocity u(r) all the way across the pipe u u ( r) 1 ln (R r)u B (6.48) where we have replaced y by R r. Compute the average velocity from this profile V Q A 1 R2 R 0 u 1 ln (R r)u B2r dr 1 2 u 2 ln R u 2B 3 (6.49) Introducing 0.41 and B 5.0, we obtain, numerically, u V 2.44 ln R u 1.34 (6.50) This looks only marginally interesting until we realize that V/u is directly related to the Darcy friction factor (58/32.2 slug/ft3)(3.32 ft/s)(0.004 ft) 1.45 E-5 slug/(ft s) 32(1.0 ft)(3.32 ft/s) (58 lbf/ft3)(0.004 ft)2 344 Chapter 6 Viscous Flow in Ducts u V V w 2 1/2 8 f 1/2 (6.51) Moreover, the argument of the logarithm in (6.50) is equivalent to R u u V 1 2 Red 8 f 1/2 (6.52) Introducing (6.52) and (6.51) into Eq. (6.50), changing to a base-10 logarithm, and re-arranging, we obtain f 1 1 /2 1.99 log (Red f1/2) 1.02 (6.53) In other words, by simply computing the mean velocity from the logarithmic-law cor-relation, we obtain a relation between the friction factor and Reynolds number for tur-bulent pipe flow. Prandtl derived Eq. (6.53) in 1935 and then adjusted the constants slightly to fit friction data better f 1 1 /2 2.0 log (Red f1/2) 0.8 (6.54) This is the accepted formula for a smooth-walled pipe. Some numerical values may be listed as follows: Red 4000 104 105 106 107 108 f 0.0399 0.0309 0.0180 0.0116 0.0081 0.0059 Thus f drops by only a factor of 5 over a 10,000-fold increase in Reynolds number. Equa-tion (6.54) is cumbersome to solve if Red is known and f is wanted. There are many al-ternate approximations in the literature from which f can be computed explicitly from Red 0.316 Red 1/4 4000 Red 105 H. Blasius (1911) f (6.55) 1.8 log R 6.
e 9 d 2 Ref. 9 Blasius, a student of Prandtl, presented his formula in the first correlation ever made of pipe friction versus Reynolds number. Although his formula has a limited range, it illustrates what was happening to Hagen’s 1839 pressure-drop data. For a horizontal pipe, from Eq. (6.55), hf g p f L d 2 V g 2 0.316 Vd 1/4 L d 2 V g 2 or p 0.158 L3/41/4d5/4V7/4 (6.56) at low turbulent Reynolds numbers. This explains why Hagen’s data for pressure drop begin to increase as the 1.75 power of the velocity, in Fig. 6.4. Note that p varies only slightly with viscosity, which is characteristic of turbulent flow. Introducing Q 1 4 d2V into Eq. (6.56), we obtain the alternate form p 0.241L3/41/4d4.75Q1.75 (6.57) For a given flow rate Q, the turbulent pressure drop decreases with diameter even more sharply than the laminar formula (6.47). Thus the quickest way to reduce required 1 2 Vd 6.4 Flow in a Circular Pipe 345 Effect of Rough Walls pumping pressure is to increase the pipe size, although, of course, the larger pipe is more expensive. Doubling the pipe size decreases p by a factor of about 27 for a given Q. Compare Eq. (6.56) with Example 5.7 and Fig. 5.10.
The maximum velocity in turbulent pipe flow is given by Eq. (6.48), evaluated at r 0 u u m ax 1 ln R u B (6.58) Combining this with Eq. (6.49), we obtain the formula relating mean velocity to max-imum velocity um V ax (1 1.33 f )1 (6.59) Some numerical values are Red 4000 104 105 106 107 108 V/umax 0.790 0.811 0.849 0.875 0.893 0.907 The ratio varies with the Reynolds number and is much larger than the value of 0.5 predicted for all laminar pipe flow in Eq. (6.43). Thus a turbulent velocity profile, as shown in Fig. 6.11, is very flat in the center and drops off sharply to zero at the wall.
It was not known until experiments in 1800 by Coulomb that surface roughness has an effect on friction resistance. It turns out that the effect is negligible for laminar pipe flow, and all the laminar formulas derived in this section are valid for rough walls also.
But turbulent flow is strongly affected by roughness. In Fig. 6.9 the linear viscous sub-layer only extends out to y yu/ 5. Thus, compared with the diameter, the sub-layer thickness ys is only y d s 5 d /u R 1 ed 4.
f 1 1/2 (6.60) 346 Chapter 6 Viscous Flow in Ducts umax V (a) (b) V umax Parabolic curve Fig. 6.11 Comparison of laminar and turbulent pipe-flow velocity profiles for the same volume flow: (a) laminar flow; (b) turbulent flow.
Fig. 6.12 Effect of wall roughness on turbulent pipe flow. (a) The log-arithmic overlap-velocity profile shifts down and to the right; (b) ex-periments with sand-grain rough-ness by Nikuradse show a sys-tematic increase of the turbulent friction factor with the roughness ratio.
For example, at Red 105, f 0.0180, and ys/d 0.001, a wall roughness of about 0.001d will break up the sublayer and profoundly change the wall law in Fig. 6.9.
Measurements of u(y) in turbulent rough-wall flow by Prandtl’s student Nikuradse show, as in Fig. 6.12a, that a roughness height will force the logarithm-law pro-file outward on the abscissa by an amount approximately equal to ln , where u/. The slope of the logarithm law remains the same, 1/, but the shift outward causes the constant B to be less by an amount B (1/) ln .
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. 6.12b. We see that laminar friction is unaffected, but turbulent friction, after an onset point, increases monotonically with the roughness ratio /d. For any given /d, the friction factor becomes constant (fully rough) at high Reynolds numbers. These points of change are certain values of u/: u 5: hydraulically smooth walls, no effect of roughness on friction 5 u 70: transitional roughness, moderate Reynolds-number effect u 70: fully rough flow, sublayer totally broken up and friction independent of Reynolds number For fully rough flow, 70, the log-law downshift B in Fig. 6.12a is B 1 ln 3.5 (6.61) 6.4 Flow in a Circular Pipe 347 0.08 0.06 0.04 0.02 0.01 0.10 103 Red f Eq. (6.55a) Eq. (6.54) = 0.0333 ∋ 0.0163 0.00833 0.00397 0.00198 0.00099 (b) Rough Smooth ≈1n ∋ + ∆B log yu v (a) u u 104 105 106 Red 64 d The Moody Chart and the logarithm law modified for roughness becomes u 1 ln y B B 1 ln y 8.5 (6.62) The viscosity vanishes, and hence fully rough flow is independent of the Reynolds num-ber. If we integrate Eq. (6.62) to obtain the average velocity in the pipe, we obtain u V 2.44 ln d 3.2 or f1 1 /2 2.0 log 3 / .
d 7 fully rough flow (6.63) 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: /d 0.00001 0.0001 0.001 0.01 0.05 f 0.00806 0.0120 0.0196 0.0379 0.0716 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. 6.12b has now been replaced by the Moody chart.
In 1939 to cover the transitionally rough range, Colebrook combined the smooth-wall [Eq. (6.54)] and fully rough [Eq. (6.63)] relations into a clever interpolation for-mula f1 1 /2 2.0 log 3 / .
d 7 R 2 ed .5 f 1 1/2 (6.64) 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 (Fig. 6.13). 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).
Equation (6.64) is cumbersome to evaluate for f if Red is known, although it easily yields to the EES Equation Solver. An alternate explicit formula given by Haaland as f1 1 /2 1.8 log R 6.
e 9 d 3 / .
d 7 1.11 (6.64a) varies less than 2 percent from Eq. (6.64).
The shaded area in the Moody chart indicates the range where transition from lam-inar to turbulent flow occurs. There are no reliable friction factors in this range, 2000 Red 4000. Notice that the roughness curves are nearly horizontal in the fully rough regime to the right of the dashed line.
From tests with commercial pipes, recommended values for average pipe roughness are listed in Table 6.1.
348 Chapter 6 Viscous Flow in Ducts Fig. 6.13 The Moody chart for pipe friction with smooth and rough walls. This chart is identical to Eq.
(6.64) for turbulent flow. (From Ref. 8, by permission of the ASME.) 349 Table 6.1 Recommended Roughness Values for Commercial Ducts 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 Values of (Vd) for water at 60°F (velocity, ft/s × diameter, in) 0.10 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.025 0.02 0.01 0.009 0.008 Friction factor f = h L d V 2 2g 0.015 0.05 0.04 0.03 0.02 0.008 0.006 0.004 0.002 0.001 0.0008 0.0006 0.0004 0.0002 0.0001 0.000,05 0.000,01 Relative roughness ε d 0.015 0.01 103 2(103) 3 4 5 6 8104 8000 2 4 6 8 10 20 40 60 100 200 400 600 800 1000 2000 4000 6000 10,000 20,000 40,000 60,000 100,000 Values of (Vd) for atmospheric air at 60°F 80,000 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 20 40 80 100 200 400 600 800 1000 60 2000 4000 6000 8000 10,000 2(104) 3 4 5 6 8105 2(105) 3 4 5 6 8106 2(106) 3 4 5 6 8 107 2(107) 3 4 5 6 8108 ε d = 0.000,001 ε d = 0.000,005 Reynolds number Re = Vd ν Complete turbulence, rough pipes Transition zone Critical zone Laminar flow f = 64 Re Smooth pipes Recr Laminar flow ( ( EXAMPLE 6.63 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.
Solution One can estimate the Reynolds number of water and air from the Moody chart. Look across the top of the chart to V (ft/s) d (in) 36, and then look directly down to the bottom abscissa to find that Red(water) 2.7 105. The roughness ratio for asphalted cast iron ( 0.0004 ft) is d 0.0008 Find the line on the right side for /d 0.0008, and follow it to the left until it intersects the vertical line for Re 2.7 105. Read, approximately, f 0.02 [or compute f 0.0197 from Eq. (6.64a)]. Then the head loss is hf f L d 2 V g 2 (0.02) 2 0 0 .5 0 2(3 (6 2.
f 2 t/ f s t ) / 2 s2) 4.5 ft Ans.
The pressure drop for a horizontal pipe (z1 z2) is p ghf (62.4 lbf/ft3)(4.5 ft) 280 lbf/ft2 Ans.
Moody points out that this computation, even for clean new pipe, can be considered accurate only to about 10 percent.
EXAMPLE 6.7 Oil, with 900 kg/m3 and 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200-mm-diameter cast-iron pipe. Determine (a) the head loss and (b) the pressure drop if the pipe slopes down at 10° in the flow direction.
Solution First compute the velocity from the known flow rate V Q R2 0 ( .
0 2 .1 m m 3/s )2 6.4 m/s Then the Reynolds number is Red V d (6 0 .4 .0 m 00 / 0 s) 1 (0 m .2 2/ m s ) 128,000 From Table 6.1, 0.26 mm for cast-iron pipe. Then d 0 2 .
0 2 0 6 m m m m 0.0013 0.0004 1 6 2 350 Chapter 6 Viscous Flow in Ducts 3This example was given by Moody in his 1944 paper .
6.5 Three Types of Pipe-Flow Problems Enter the Moody chart on the right at /d 0.0013 (you will have to interpolate), and move to the left to intersect with Re 128,000. Read f 0.0225 [from Eq. (6.64) for these values we could compute f 0.0227]. Then the head loss is hf f L d 2 V g 2 (0.0225) 5 0 0 .2 0 m m 2 ( ( 6 9 .
.
4 81 m m /s / ) s 2 2) 117 m Ans. (a) From Eq. (6.25) for the inclined pipe, hf g p z1 z2 g p L sin 10° or p g[hf (500 m) sin 10°] g(117 m 87 m) (900 kg/m3)(9.81 m/s2)(30 m) 265,000 kg/(m s2) 265,000 Pa Ans. (b) EXAMPLE 6.8 Repeat Example 6.5 to see whether there is any possible turbulent-flow solution for a smooth-walled pipe.
Solution In Example 6.5 we estimated a head loss hf 1.66 ft, assuming laminar exit flow ( 2.0). For this condition the friction factor is f hf L d 2 V g 2 (1.66 ft) 0.0388 For laminar flow, Red 64/f 64/0.0388 1650, as we showed in Example 6.5. However, from the Moody chart (Fig. 6.13), we see that f 0.0388 also corresponds to a turbulent smooth-wall condition, at Red 4500. If the flow actually were turbulent, we should change our kinetic-energy factor to 1.06 [Eq. (3.73)], whence the corrected hf 1.82 ft and f 0.0425. With f known, we can estimate the Reynolds number from our formulas: Red 3250 [Eq. (6.54)] or Red 3400 [Eq. (6.55b)] So the flow might have been turbulent, in which case the viscosity of the fluid would have been R V e d d 1.80(3 3 .3 3 2 0 ) 0 (0.004) 7.2 106 slug/(ft s) Ans.
This is about 55 percent less than our laminar estimate in Example 6.5. The moral is to keep the capillary-flow Reynolds number below about 1000 to avoid such duplicate solutions.
The Moody chart (Fig. 6.13) can be used to solve almost any problem involving fric-tion losses in long pipe flows. However, many such problems involve considerable it-eration 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 (0.004 ft)(2)(32.2 ft/s2) (1.0 ft)(3.32 ft/s)2 6.5 Three Types of Pipe-Flow Problems 351 Type 2 Problem: Find the Flow Rate Red, enter the chart, find f, and hence compute hf. This is one of three fundamental problems which are commonly encountered in pipe-flow calculations: 1. Given d, L, and V or Q, , , and g, compute the head loss hf (head-loss prob-lem).
2. Given d, L, hf, , , and g, compute the velocity V or flow rate Q (flow-rate problem).
3. Given Q, L, hf, , , and g, compute the diameter d of the pipe (sizing problem).
Only problem 1 is 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 3: (a) preparation of a suitable new Moody-type chart (see Prob. 6.62 and 6.73); or (b) the use of solver software, especially the Engineering Equation Solver, known as EES , which gives the answer directly if the proper data are entered. Examples 6.9 and 6.11 include the EES 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 f varies so slowly with Red. Alternately, in the spirit of Example 5.7, we could change the scal-ing variables to (, , d) and thus arrive at dimensionless head loss versus dimension-less velocity. The result is4 fcn(Red) where g L d 3h 2 f f R 2 ed 2 (6.65) Example 5.7 did this and offered the simple correlation 0.155 Red 1.75, which is valid for turbulent flow with smooth walls and Red 1 E5.
A formula valid for all turbulent pipe flows is found by simply rewriting the Cole-brook interpolation, Eq. (6.64), in the form of Eq. (6.65): Red (8 )1/2 log 3 / .
d 7 g L d 3h 2 f (6.66) Given , we compute Red (and hence velocity) directly. Let us illustrate these two ap-proaches with the following example.
EXAMPLE 6.9 Oil, with 950 kg/m3 and 2 E-5 m2/s, flows through a 30-cm-diameter pipe 100 m long with a head loss of 8 m. The roughness ratio is /d 0.0002. Find the average velocity and flow rate.
Direct Solution First calculate the dimensionless head-loss parameter: g L d 3h 2 f 5.30 E7 (9.81 m/s2)(0.3 m)3(8.0 m) (100 m)(2 E-5 m2/s)2 1.775 352 Chapter 6 Viscous Flow in Ducts 4The parameter was suggested by H. Rouse in 1942.
Now enter Eq. (6.66) to find the Reynolds number: Red [8(5.3 E7)]1/2 log 0.0 3 0 .7 02 1 5 .7 .3 75 E 7 72,600 The velocity and flow rate follow from the Reynolds number: V R d ed 4.84 m/s Q V 4 d2 4.84 m s 4 (0.3 m)2 0.342 m3/s Ans.
No iteration is required, but this idea falters if additional losses are present.
Iterative Solution By definition, the friction factor is known except for V: f hf L d 2 V g 2 (8 m) 1 0 0 .3 0 m m 2(9.8 V 1 2 m/s2) or fV2 0.471 (SI units) To get started, we only need to guess f, compute V 0 .4 7 1 /f , then get Red, compute a better f from the Moody chart, and repeat. The process converges fairly rapidly. A good first guess is the “fully rough” value for /d 0.0002, or f 0.014 from Fig. 6.13. The iteration would be as follows: Guess f 0.014, then V 0 .4 7 1 /0 .0 1 4 5.80 m/s and Red Vd/ 87,000. At Red 87,000 and /d 0.0002, compute fnew 0.0195 [Eq. (6.64)].
New f 0.0195, V 0 .4 8 1 /0 .0 1 9 5 4.91 m/s and Red Vd/ 73,700. At Red 73,700 and /d 0.0002, compute fnew 0.0201 [Eq. (6.64)].
Better f 0.0201, V 0 .4 7 1 /0 .0 2 0 1 4.84 m/s and Red 72,600. At Red 72,600 and /d 0.0002, compute fnew 0.0201 [Eq. (6.64)].
We have converged to three significant figures. Thus our iterative solution is V 4.84 m/s Q V 4 d2 (4.84) 4 (0.3)2 0.342 m3/s Ans.
The iterative approach is straightforward and not too onerous, so it is routinely used by engi-neers. Obviously this repetitive procedure is ideal for a personal computer.
Engineering Equation Solver (EES) Solution In EES, one simply enters the data and the appropriate equations, letting the software do the rest. Correct units must of course be used. For the present example, the data could be entered as SI: rho=950 nu=2E-5 d=0.3 L=100 epsod=0.0002 hf=8.0 g=9.81 The appropriate equations are the Moody formula (6.64) plus the definitions of Reynolds num-(2 E-5 m2/s)(72,600) 0.3 m 6.5 Three Types of Pipe-Flow Problems 353 EES ber, volume flow rate as determined from velocity, and the Darcy head-loss formula (6.30): Re Vd/nu Q Vpid^2/4 f (2.0log10(epsod/3.7 2.51/Re/f^0.5))^(2) hf fL/dV^2/2/g EES understands that “pi” represents 3.141593. Then hit “SOLVE” from the menu. If errors have been entered, EES will complain that the system cannot be solved and attempt to explain why.
Otherwise, the software will iterate, and in this case EES prints the correct solution: Q=0.342 V=4.84 f=0.0201 Re=72585 The units are spelled out in a separate list as [m, kg, s, N]. This elegant approach to engi-neering problem-solving has one drawback, namely, that the user fails to check the solution for engineering viability. For example, are the data typed correctly? Is the Reynolds number turbulent?
EXAMPLE 6.10 Work Moody’s problem (Example 6.6) backward, assuming that the head loss of 4.5 ft is known and the velocity (6.0 ft/s) is unknown.
Direct Solution Find the parameter , and compute the Reynolds number from Eq. (6.66): g L d 3h 2 f 7.48 E8 Eq. (6.66): Red [8(7.48 E8)]1/2 log 0.0 3 0 .7 08 1 7 .
.
7 4 7 8 5 E 8 274,800 Then V R d ed (1.1 E-5 0 ) .
( 5 274,800) 6.05 ft/s Ans.
We did not get 6.0 ft/s exactly because the 4.5-ft head loss was rounded off in Example 6.6.
Iterative Solution As in Eq. (6.9) the friction factor is related to velocity: f hf L d 2 V g 2 (4.5 ft) 2 0 0 .5 0 f f t t 2(32.
V 2 2 ft/s2) 0.7 V 2 2 45 or V 0 .7 2 4 5 /f Knowing /d 0.0008, we can guess f and iterate until the velocity converges. Begin with the fully rough estimate f 0.019 from Fig. 6.13. The resulting iterates are f1 0.019: V1 0 .7 2 4 5 /f 1 6.18 ft/s Red1 V d 280,700 (32.2 ft/s2)(0.5 ft)3(4.5 ft) (200 ft)(1.1 E-5 ft2/s)2 354 Chapter 6 Viscous Flow in Ducts Type 3 Problem: Find the Pipe Diameter f2 0.0198: V2 6.05 ft/s Red2 274,900 f3 0.01982: V3 6.046 ft/s Ans.
The calculation converges rather quickly to the same result as that obtained through direct com-putation.
The Moody chart is especially awkward for finding the pipe size, since d occurs in all three parameters f, Red, and /d. Further, it depends upon whether we know the ve-locity or the flow rate. We cannot know both, or else we could immediately compute d 4 Q /( V ) .
Let us assume that we know the flow rate Q. Note that this requires us to redefine the Reynolds number in terms of Q: Red V d 4 d Q (6.67) Then, if we choose (Q, , ) as scaling parameters (to eliminate d), we obtain the func-tional relationship Red 4 d Q fcn L g h 5 f , Q (6.68) and can thus solve d when the right-hand side is known. Unfortunately, the writer knows of no formula for this relation, nor is he able to rearrange Eq. (6.64) into the explicit form of Eq. (6.68). One could recalculate and plot the relation, and indeed an inge-nious “pipe-sizing” plot is given in Ref. 13. Here it seems reasonable to forgo a plot or curve fitted formula and to simply set up the problem as an iteration in terms of the Moody-chart variables. In this case we also have to set up the friction factor in terms of the flow rate: f hf L d V 2g 2 8 2 g L h Q fd 2 5 (6.69) The following two examples illustrate the iteration.
EXAMPLE 6.11 Work Example 6.9 backward, assuming that Q 0.342 m3/s and 0.06 mm are known but that d (30 cm) is unknown. Recall L 100 m, 950 kg/m3, 2 E-5 m2/s, and hf 8 m.
Iterative Solution First write the diameter in terms of the friction factor: f 8 2 8.28d5 or d 0.655f1/5 (1) in SI units. Also write the Reynolds number and roughness ratio in terms of the diameter: (9.81 m/s2)(8 m)d5 (100 m)(0.342 m3/s)2 6.5 Three Types of Pipe-Flow Problems 355 Red 4 ( ( 2 0.
E 34 -5 2 m m 2 3 / / s s ) ) d 21, d 800 (2) d 6 E d -5 m (3) Guess f, compute d from (1), then compute Red from (2) and /d from (3), and compute a bet-ter f from the Moody chart or Eq. (6.64). Repeat until (fairly rapid) convergence. Having no ini-tial estimate for f, the writer guesses f 0.03 (about in the middle of the turbulent portion of the Moody chart). The following calculations result: f 0.03 d 0.655(0.03)1/5 0.325 m Red 2 0 1 .
, 3 8 2 0 5 0 67,000 d 1.85 E-4 Eq. (6.54): fnew 0.0203 then dnew 0.301 m Red,new 72,500 d 2.0 E-4 Eq. (6.54): fbetter 0.0201 and d 0.300 m Ans.
The procedure has converged to the correct diameter of 30 cm given in Example 6.9.
EES Solution For an EES solution, enter the data and the appropriate equations. The diameter is unknown.
Correct units must of course be used. For the present example, the data should use SI units: rho=950 nu=2E-5 L=100 eps=6E-5 hf=8.0 g=9.81 Q=0.342 The appropriate equations are the Moody formula, the definition of Reynolds number, vol-ume flow rate as determined from velocity, the Darcy head-loss formula, and the roughness ratio: Re Vd/nu Q Vpid^2/4 f (2.0log10(epsod/3.7 2.51/Re/f^0.5))^(2) hf fL/dV^2/2/g epsod eps/d Hit Solve from the menu. Unlike Example 6.9, this time EES complains that the system can-not be solved and reports “logarithm of a negative number.” The reason is that we allowed EES to assume that f could be a negative number. Bring down Variable Information from the menu and change the limits of f so that it cannot be negative. EES agrees and iterates to the solution: d 0.300 V 4.84 f 0.0201 Re 72,585 The unit system is spelled out as (m, kg, s, N). As always when using software, the user should check the solution for engineering viability. For example, is the Reynolds number turbulent?
(Yes) 356 Chapter 6 Viscous Flow in Ducts EES 6.6 Flow in Noncircular Ducts5 The Hydraulic Diameter EXAMPLE 6.12 Work Moody’s problem, Example 6.6, backward to find the unknown (6 in) diameter if the flow rate Q 1.18 ft3/s is known. Recall L 200 ft, 0.0004 ft, and 1.1 E-5 ft2/s.
Solution Write f, Red, and /d in terms of the diameter: f 8 2 g L h Q fd 2 5 8 2 0.642d5 or d 1.093f1/5 (1) Red (1 4 .
( 1 1.
E 1 -8 5 f f t3 t2 /s /s ) ) d 136 d ,600 (2) d 0.00 d 04 ft (3) with everything in BG units, of course. Guess f; compute d from (1), Red from (2), and /d from (3); and then compute a better f from the Moody chart. Repeat until convergence. The writer tra-ditionally guesses an initial f 0.03: f 0.03 d 1.093(0.03)1/5 0.542 ft Red 13 0 6 .5 ,6 4 0 2 0 252,000 d 7.38 E-4 fnew 0.0196 dnew 0.498 ft Red 274,000 d 8.03 E-4 fbetter 0.0198 d 0.499 ft Ans.
Convergence is rapid, and the predicted diameter is correct, about 6 in. The slight discrepancy (0.499 rather than 0.500 ft) arises because hf was rounded to 4.5 ft.
In discussing pipe-sizing problems, we should remark that commercial pipes are made only in certain sizes. Table 6.2 lists standard water-pipe sizes in the United States.
If the sizing calculation gives an intermediate diameter, the next largest pipe size should be selected.
If the duct is noncircular, the analysis of fully developed flow follows that of the cir-cular pipe but is more complicated algebraically. For laminar flow, one can solve the exact equations of continuity and momentum. For turbulent flow, the logarithm-law ve-locity profile can be used, or (better and simpler) the hydraulic diameter is an excel-lent approximation.
For a noncircular duct, the control-volume concept of Fig. 6.10 is still valid, but the cross-sectional area A does not equal R2 and the cross-sectional perimeter wetted by the shear stress does not equal 2R. The momentum equation (6.26) thus becomes p A gA L sin w L 0 (32.2 ft/s2)(4.5 ft)d5 (200 ft)(1.18 ft3/s)2 6.6 Flow in Noncircular Ducts 357 Table 6.2 Nominal and Actual Sizes of Schedule 40 Wrought-Steel Pipe Nominal size, in Actual ID, in 1 1 8 0.269 1 1 4 0.364 1 3 8 0.493 1 1 2 0.622 1 3 4 0.824 1 3 4 1.049 1 1 2 1.610 2 3 4 2.067 2 1 2 2.469 3 3 4 3.068 Nominal size within 1 percent for 4 in or larger.
5This section may be omitted without loss of continuity.
or hf g p z g w A / L (6.70) This is identical to Eq. (6.27) except that (1) the shear stress is an average value inte-grated around the perimeter and (2) the length scale A/ takes the place of the pipe radius R. For this reason a noncircular duct is said to have a hydraulic radius Rh, de-fined by Rh A (6.71) This concept receives constant use in open-channel flow (Chap. 10), where the chan-nel cross section is almost never circular. If, by comparison to Eq. (6.29) for pipe flow, we define the friction factor in terms of average shear fNCD 8 V w 2 (6.72) where NCD stands for noncircular duct and V Q/A as usual, Eq. (6.70) becomes hf f 4 L Rh 2 V g 2 (6.73) This is equivalent to Eq. (6.30) for pipe flow except that d is replaced by 4Rh. There-fore we customarily define the hydraulic diameter as Dh 4 A wett 4 ed pe a r r i e m a eter 4Rh (6.74) 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. The fact that Dh equals 4Rh is just one of those things: Chalk it up to an en-gineer’s sense of humor. Note that for the degenerate case of a circular pipe, Dh 4R2/(2R) 2R, as expected.
We would therefore expect by dimensional analysis that this friction factor f, based upon hydraulic diameter as in Eq. (6.72), would correlate with the Reynolds number and roughness ratio based upon the hydraulic diameter f F V Dh , D h (6.75) 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: R 6 e 4 Dh 40% laminar flow f (6.76) fMoodyReDh, D h 15% turbulent flow Now let us look at some particular cases.
cross-sectional area wetted perimeter 358 Chapter 6 Viscous Flow in Ducts Flow between Parallel Plates As shown in Fig. 6.14, flow between parallel plates a distance h apart is the limiting case of flow through a very wide rectangular channel. For fully developed flow, u u(y) only, which satisfies continuity identically. The momentum equation in cartesian coordinates reduces to 0 d d p x gx d d y lam d d u y (6.77) subject to no-slip conditions: u 0 at y h. The laminar-flow solution was given as an example in Eq. (4.143). Here we also allow for the possibility of a sloping chan-nel, with a pressure gradient due to gravity. The solution is u 2 1 d d x ( p gz)(h2 y2) (6.78) If the channel has width b, the volume flow is Q h h u(y)b dy b 3 h 3 d d x ( p gz) or V b Q h 3 h 2 d d x ( p gz) 2 3 umax (6.79) Note the difference between a parabola [Eq. (6.79)] and a paraboloid [Eq. (6.43)]: the average is two-thirds of the maximum velocity in plane flow and one-half in axisym-metric flow.
The wall shear stress in developed channel flow is a constant: w d d u y y h h d d x (p gz) (6.80) This may be nondimensionalized as a friction factor: f 8 V w 2 2 4 V h R 2 e 4 h (6.81) These are exact analytic laminar-flow results, so there is no reason to resort to the hydraulic-diameter concept. However, if we did use Dh, a discrepancy would arise. The hydraulic diameter of a wide channel is 6.6 Flow in Noncircular Ducts 359 u max 2 h Y y y = +h u( y) x y = – h b → ∞ Fig. 6.14 Fully developed flow be-tween parallel plates.
Dh 4 A lim b→ 2 4 b (2 bh 4 ) h 4h (6.82) or twice the distance between the plates. Substituting into Eq. (6.81), we obtain the in-teresting result Parallel plates: flam V 96 (4 h) R 9 e 6 Dh (6.83) Thus, if we could not work out the laminar theory and chose to use the approximation f 64/ReDh, we would be 33 percent low. The hydraulic-diameter approximation is relatively crude in laminar flow, as Eq. (6.76) states.
Just as in circular-pipe flow, the laminar solution above becomes unstable at about ReDh 2000; transition occurs and turbulent flow results.
For turbulent flow between parallel plates, we can again use the logarithm law, Eq. (6.21), as an approximation across the entire channel, using not y but a wall coor-dinate Y, as shown in Fig. 6.14: u u (Y ) 1 ln Y u B 0 Y h (6.84) This distribution looks very much like the flat turbulent profile for pipe flow in Fig.
6.11b, and the mean velocity is V 1 h h 0 u dY u 1 ln hu B 1 (6.85) Recalling that V/u (8/f)1/2, we see that Eq. (6.85) is equivalent to a parallel-plate friction law. Rearranging and cleaning up the constant terms, we obtain f 1 1/2 2.0 log (ReDh f 1/2) 1.19 (6.86) where we have introduced the hydraulic diameter Dh 4h. This is remarkably close to the pipe-friction law, Eq. (6.54). 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 (6.86) can be brought into exact agreement with the pipe law by rewrit-ing it in the form f 1 1/2 2.0 log (0.64 ReDh f 1/2) 0.8 (6.87) Thus the turbulent friction is predicted most accurately when we use an effective di-ameter Deff equal to 0.64 times the hydraulic diameter. The effect on f itself is much less, about 10 percent at most. We can compare with Eq. (6.83) for laminar flow, which predicted Parallel plates: Deff 6 9 4 6 Dh 2 3 Dh (6.88) This close resemblance (0.64Dh versus 0.667Dh) occurs so often in noncircular duct flow that we take it to be a general rule for computing turbulent friction in ducts: Deff Dh 4 A reasonable accuracy 360 Chapter 6 Viscous Flow in Ducts Part (a) Deff(laminar theory) extreme accuracy (6.89) Jones shows that the effective-laminar-diameter idea collapses all data for rectan-gular 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 1.9 slugs/ft3 and (a) 0.00002 ft3/s and (b) 0.002 ft3/s. Assume smooth walls.
Solution The viscosity 3.8 105 slug/(ft s). The spacing is 2h 2.4 in 0.2 ft, and Dh 4h 0.4 ft. The Reynolds number is ReDh V Dh (6 0 .
.
0 00 ft 0 /s 0 ) 2 (0 f .
t 4 2/s ft) 120,000 The flow is therefore turbulent. For reasonable accuracy, simply look on the Moody chart (Fig.
6.13) for smooth walls f 0.0173 hf f D L h V 2g 2 0.0173 1 0 0 .4 0 2 ( ( 6 3 .
2 0 .
) 2 2 ) 2.42 ft Ans. (a) Since there is no change in elevation, p ghf 1.9(32.2)(2.42) 148 lbf/ft2 Ans. (a) This is the head loss and pressure drop per 100 ft of channel. For more accuracy, take Deff 2 3 Dh from laminar theory; then Reeff 2 3 (120,000) 80,000 and from the Moody chart read f 0.0189 for smooth walls. Thus a better estimate is hf 0.0189 1 0 0 .4 0 2 ( ( 6 3 .
2 0 .
) 2 2 ) 2.64 ft and p 1.9(32.2)(2.64) 161 lbf/ft2 Better ans. (a) The more accurate formula predicts friction about 9 percent higher.
Compute 0.0038 slug/(ft s). The Reynolds number is 6.0(0.4)/0.002 1200; there-fore the flow is laminar, since Re is less than 2300.
You could use the laminar-flow friction factor, Eq. (6.83) flam 1 9 2 6 00 0.08 from which hf 0.08 1 0 0 .4 0 2 ( ( 6 3 .
2 0 .
) 2 2 ) 11.2 ft and p 1.9(32.2)(11.2) 684 lbf/ft2 Ans. (b) 96 ReDh 6.6 Flow in Noncircular Ducts 361 Part (b) Flow through a Concentric Annulus Alternately you can finesse the Reynolds number and go directly to the appropriate laminar-flow formula, Eq. (6.79) V 3 h 2 L p or p 684 slugs/(ft s2) 684 lbf/ft2 and hf g p 1.9 6 ( 8 3 4 2.2) 11.2 ft This is one of those—perhaps unexpected—problems where the laminar friction is greater than the turbulent friction.
Consider steady axial laminar flow in the annular space between two concentric cylin-ders, 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. (6.34) d d r r d d u r Kr K d d x (p gz) (6.90) Integrate this twice u 1 4 r2 K C1 ln r C2 The constants are found from the two no-slip conditions u(r a) 0 1 4 a2 K C1 ln a C2 u(r b) 0 1 4 b2 K C1 ln b C2 The final solution for the velocity profile is u 4 1 d d x (p gz)a2 r2 a ln 2 ( b/ b a 2 ) ln a r (6.91) 3(6.0 ft/s)0.0038 slug/(ft s) (0.1 ft)2 362 Chapter 6 Viscous Flow in Ducts Fig. 6.15 Fully developed flow through a concentric annulus.
u(r) u(r) r r = b r = a x The volume flow is given by Q a b u2r dr 8 d d x (p gz)a4 b4 (a ln 2 (a/ b b 2 ) )2 (6.92) The velocity profile u(r) resembles a parabola wrapped around in a circle to form a split doughnut, as in Fig. 6.15. The maximum velocity occurs at the radius r 2 a l 2 n (a b /b 2 ) 1/2 u umax (6.93) This maximum is closer to the inner radius but approaches the midpoint between cylin-ders as the clearance a b becomes small. Some numerical values are as follows: b a 0.01 0.1 0.2 0.5 0.8 0.9 0.99 r a b b 0.323 0.404 0.433 0.471 0.491 0.496 0.499 Also, as the clearance becomes small, the profile approaches a parabolic distribution, as if the flow were between two parallel plates [Eq. (4.143)].
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 f with respect to the head loss, as in Eq. (6.73), f hf D L h 2 V g 2 where V (a2 Q b2) (6.94) The hydraulic diameter for an annulus is Dh 4 2 (a (a 2 b b 2 ) ) 2(a b) (6.95) It is twice the clearance, rather like the parallel-plate result of twice the distance be-tween plates [Eq. (6.82)].
Substituting hf, Dh, and V into Eq. (6.94), we find that the friction factor for lami-nar flow in a concentric annulus is of the form f (6.96) The dimensionless term is a sort of correction factor for the hydraulic diameter. We could rewrite Eq. (6.96) as Concentric annulus: f R 6 e 4 eff Reeff 1 ReDh (6.97) Some numerical values of f ReDh and Deff/Dh 1/ are given in Table 6.3.
For turbulent flow through a concentric annulus, the analysis might proceed by patch-ing 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 di-rectly to the friction factor. According to the general rule proposed in Eq. (6.89), tur-bulent friction is predicted with excellent accuracy by replacing d in the Moody chart (a b)2(a2 b2) a4 b4 (a2 b2)2/ln (a/b) 64 ReDh 6.6 Flow in Noncircular Ducts 363 E6.14 by Deff 2(a b)/ , with values listed in Table 6.3.6 This idea includes roughness also (replace /d in the chart by /Deff). For a quick design number with about 10 percent accuracy, one can simply use the hydraulic diameter Dh 2(a b).
EXAMPLE 6.14 What should the reservoir level h be to maintain a flow of 0.01 m3/s through the commercial steel annulus 30 m long shown in Fig. E6.14? Neglect entrance effects and take 1000 kg/m3 and 1.02 106 m2/s for water.
364 Chapter 6 Viscous Flow in Ducts Table 6.3 Laminar Friction Factors for a Concentric Annulus b/a f ReDh Deff/Dh 1/ 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 1 2 Water L = 30 m Q, V a = 5 cm b = 3 cm h = ?
Solution Compute the average velocity and hydraulic diameter V Q A 1.99 m/s Dh 2(a b) 2(0.05 0.03) m 0.04 m Apply the steady-flow energy equation between sections 1 and 2: p 1 1 2 V1 2 gz1 p 2 1 2 V2 2 gz2 ghf But p1 p2 pa, V1 0, and V2 V in the pipe. Therefore solve for hf f D L h 2 V g 2 z1 z2 2 V g 2 But z1 z2 h, the desired reservoir height. Thus, finally, h 2 V g 2 1 f D L h (1) Since V, L, and Dh are known, our only remaining problem is to compute the annulus friction factor f. For a quick approximation, take Deff Dh 0.04 m. Then ReDh V Dh 1 1 .0 .9 2 9 (0.
1 0 0 4 ) 6 78,000 D h 0.
4 0 0 46 m m m m 0.00115 0.01 m3/s [(0.05 m)2 (0.03 m)2] 6Jones and Leung show that data for annular flow also satisfy the effective-laminar-diameter idea.
Other Noncircular Cross Sections where 0.046 mm has been read from Table 6.1 for commercial steel surfaces. From the Moody chart, read f 0.0232. Then, from Eq. (1) above, h 2 ( ( 1 9 .9 .8 9 1 m m / / s s ) 2 2 ) 1 0.0232 0 3 .0 0 4 m m 3.71 m Crude ans.
For better accuracy, take Deff Dh/ 0.670Dh 2.68 cm, where the correction factor 0.670 has been read from Table 6.3 for b/a 3 5 0.6. Then the corrected Reynolds number and rough-ness ratio are Reeff VD eff 52,300 D eff 0.00172 From the Moody chart, read f 0.0257. Then the improved computation for reservoir height is h 2 ( ( 1 9 .9 .8 9 1 m m / / s s ) 2 2 ) 1 0.0257 0 3 .0 0 4 m m 4.09 m Better ans.
The uncorrected hydraulic-diameter estimate is about 9 percent low. Note that we do not replace Dh by Deff in the ratio L/Dh in Eq. (1) since this is implicit in the definition of friction factor.
In principle, any duct cross section can be solved analytically for the laminar-flow ve-locity distribution, volume flow, and friction factor. This is because any cross section 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.
119–122], Berker , and Olson and Wright [12, pp. 315–317]. Reference 34 is de-voted entirely to laminar duct flow.
In general, however, most unusual duct sections have strictly academic and not com-mercial 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.
For turbulent flow in a duct of unusual cross section, one should replace d by Dh on the Moody chart if no laminar theory is available. If laminar results are known, such as Table 6.4, replace d by Deff [64/(f Re)]Dh for the particular geometry of the duct.
For laminar flow in rectangles and triangles, the wall friction varies greatly, be-ing largest near the midpoints of the sides and zero in the corners. 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 sec-ondary 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 con-tours 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 predic-tion of turbulent secondary flow has yet to be achieved, although numerical models are improving .
6.6 Flow in Noncircular Ducts 365 Table 6.4 Laminar Friction Constants f Re for Rectangular and Triangular Ducts Rectangular Isosceles triangle b/a fReDh , deg fReDh 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 b a 2 Fig. 6.16 Illustration of secondary turbulent flow in noncircular ducts: (a) axial mean-velocity contours; (b) secondary-flow cellular mo-tions. (After J. Nikuradse, disserta-tion, Göttingen, 1926.) EXAMPLE 6.15 Air, with 0.00237 slug/ft3 and 0.000157 ft2/s, is forced through a horizontal square 9-by 9-in duct 100 ft long at 25 ft3/s. Find the pressure drop if 0.0003 ft.
Solution Compute the mean velocity and hydraulic diameter V (0 2 .
5 75 ft3 f / t s )2 44.4 ft/s Dh 4 A 4( 3 8 6 1 i i n n2) 9 in 0.75 ft From Table 6.4, for b/a 1.0, the effective diameter is Deff 56 6 .
4 91 Dh 0.843 ft whence Reeff VD eff 4 0 4 .
.
0 4 0 (0 0 .
1 8 5 4 7 3) 239,000 D eff 0 0 .0 .8 0 4 0 3 3 0.000356 From the Moody chart, read f 0.0177. Then the pressure drop is p ghf g f D L h 2 V g 2 0.00237(32.2)0.0177 0 1 .
0 7 0 5 2 4 (3 4 2 .4 .2 2 ) or p 5.5 lbf/ft2 Ans.
Pressure drop in air ducts is usually small because of the low density.
366 Chapter 6 Viscous Flow in Ducts Midplane (a) (b) 6.7 Minor Losses in Pipe Systems7 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 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; e.g., a partially closed valve can cause a greater pres-sure 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 upon the par-ticular manufacturer’s design, so that the values listed here must be taken as average design estimates [15, 16, 35, 43, 46].
The measured minor loss is usually given as a ratio of the head loss hm p/(g) through the device to the velocity head V2/(2g) of the associated piping system Loss coefficient K V2 h /( m 2g) (6.98) Although K is dimensionless, it unfortunately is not correlated in the literature with the Reynolds 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.
An alternate, and less desirable, procedure is to report the minor loss as if it were an equivalent length Leq of pipe, satisfying the Darcy friction-factor relation hm f L d eq V 2g 2 K V 2g 2 or Leq K f d (6.99) Although the equivalent length should take some of the variability out of the loss data, it is an artificial concept and will not be pursued here.
A single pipe system may have many minor losses. Since all are correlated with V 2/(2g), they can be summed into a single total system loss if the pipe has constant diameter htot hf hm 2 V g 2 f d L K (6.100) Note, however, that we must sum the losses separately if the pipe size changes so that V2 changes. The length L in Eq. (6.100) is the total length of the pipe axis, including any bends.
There are many different valve designs in commercial use. Figure 6.17 shows five 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 globe but with a 90° turn; p 1 2 V 2 6.7 Minor Losses in Pipe Systems 367 7This section may be omitted without loss of continuity.
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.
(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 handbook by Lyons .
Table 6.5 lists loss coefficients K for four types of valve, three angles of elbow fit-368 Chapter 6 Viscous Flow in Ducts (e) h D D D (c) D (d) (e) h D h D (a) (a) D D h (b) Table 6.5 Resistance Coefficients K hm/[V2/(2g)] for Open Valves, Elbows, and Tees Nominal diameter, in Screwed Flanged 1 2 1 2 4 1 2 4 8 20 Valves (fully open): Globe 14 8.2 6.9 5.7 13 8.5 6.0 5.8 5.5 Gate 0.30 0.24 0.16 0.11 0.80 0.35 0.16 0.07 0.03 Swing check 5.1 2.9 2.1 2.0 2.0 2.0 2.0 2.0 2.0 Angle 9.0 4.7 2.0 1.0 4.5 2.4 2.0 2.0 2.0 Elbows: 45° regular 0.39 0.32 0.30 0.29 45° long radius 0.21 0.20 0.19 0.16 0.14 90° regular 2.0 1.5 0.95 0.64 0.50 0.39 0.30 0.26 0.21 90° long radius 1.0 0.72 0.41 0.23 0.40 0.30 0.19 0.15 0.10 180° regular 2.0 1.5 0.95 0.64 0.41 0.35 0.30 0.25 0.20 180° long radius 0.40 0.30 0.21 0.15 0.10 Tees: Line flow 0.90 0.90 0.90 0.90 0.24 0.19 0.14 0.10 0.07 Branch flow 2.4 1.8 1.4 1.1 1.0 0.80 0.64 0.58 0.41 Fig. 6.18a Recent measured loss coefficients for 90° elbows. These values are less than those reported in Table 6.5. [From Ref. 48, cour-tesy of R. D. Coffield.] ting, 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 manufactur-ers, 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 upon fittings manufactured in the 1950s. Modern forged and molded fittings may yield somewhat different loss factors, often less than listed in Table 6.5. An example, shown in Fig. 6.18a, gives very 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 di-ameters 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 upon actual design and manufacturing factors, with Table 6.5 only serving 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 6.18b gives average losses for three valves 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 per-cent. Of all minor losses, valves, because of their complex geometry, are most sensi-tive to manufacturers’ design details. For more accuracy, the particular design and man-ufacturer should be consulted .
6.7 Minor Losses in Pipe Systems 369 0.05 0.1 0.2 0.3 0.5 1.0 2.0 3.0 4.0 Reynolds number (millions) 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 K factor 10% 10% Curve-fit correlation K 1.49 Re0.145 Legend Plastic elbow Metal elbow no. 1 Metal elbow no. 2 Fig. 6.18b Average-loss coefficients for partially open valves (see sketches in Fig. 6.17).
The butterfly valve of Fig. 6.19a is a stem-mounted disk which, when closed, seats against an O-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. How-ever, considerable dynamic torque is needed to close these valves, and losses are high when the valves are nearly closed.
Figure 6.19b shows butterfly-valve loss coefficients as a function of the opening an-gle for turbulent-flow conditions ( 0 is closed). The losses are huge when the 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. 6.19b 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.
370 Chapter 6 Viscous Flow in Ducts 20.00 18.00 16.00 14.00 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.25 0.30 0.40 0.50 0.60 0.70 0.75 0.80 0.90 1.00 Fractional opening h D K Gate Disk Globe (a) Fig. 6.19 Performance of butterfly valves: (a) typical geometry (cour-tesy of Grinnell Corp., Cranston, R.I.); (b) loss coefficients for three different manufacturers.
1000.00 100.00 10.00 1.00 0.10 20 30 40 50 60 70 90 Valve opening angle, degrees (b) K 80 Fig. 6.20 Resistance coefficients for 90° bends.
A bend or curve in a pipe, as in Fig. 6.20, always induces a loss larger than the sim-ple Moody friction loss, due to flow separation at the walls and a swirling secondary flow arising from the centripetal acceleration. The loss coefficients K in Fig. 6.20 are for this additional bend loss. The Moody loss due to the axial length of the bend must be computed separately; i.e., the bend length should be added to the pipe length.
As shown in Fig. 6.21, entrance losses are highly dependent upon entrance geom-etry, 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 downstream reservoir and loses all its velocity head due to viscous dissipation. Therefore K 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) be-tween two sizes of pipe. If the exit is to finite-sized pipe, it is termed a sudden ex-pansion (SE). The losses for both are graphed in Fig. 6.22. For the sudden expansion, the shear stress in the corner 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 KSE 1 D d2 2 2 V2 h /( m 2g) (6.101) Note that K is based on the velocity head in the small pipe. Equation (6.101) is in ex-cellent agreement with experiment.
For the sudden contraction, however, flow separation in the downstream pipe causes the main stream to contract through a minimum diameter dmin, called the vena con-tracta, 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 experimental. It 6.7 Minor Losses in Pipe Systems 371 Secondary flow pattern: d = constant 0.80 0.60 0.30 0.20 K 0.10 0.08 1.00 0 0.0005 0.001 0.002 1 1.5 2 4 3 5 6 7 8 910 (Note: Resistance due to bend length must be added.) R d ε d = 0.01 R 0.40 0.50 0.70 Fig. 6.21 Entrance and exit loss co-efficients: (a) reentrant inlets; (b) rounded and beveled inlets. Exit losses are K 1.0 for all shapes of exit (reentrant, sharp, beveled, or rounded). (From Ref. 37.) (b) 0.6 0.4 0.2 0 0 0.10 0.15 0.20 r d L θ K Sharp-edged V 10° 50° 30° r d L d r d , L d θ = 0 0.1 0.2 0.3 0.4 (a) K 0.02 1.0 0.5 V t l t d = 0 l d 1.0 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1.0 Sudden expansion d V D Eq. (6.102) Eq. (6.101) Sudden contraction: d D D Vena contracta K = hm V2/(2g) d V Fig. 6.22 Sudden expansion and contraction losses. Note that the loss is based on velocity head in the small pipe.
372 Fig. 6.23 Flow losses in a gradual conical expansion region.
fits the empirical formula KSC 0.421 D d2 2 (6.102) up to the value d/D 0.76, above which it merges into the sudden-expansion predic-tion, Eq. (6.101).
If the expansion or contraction is gradual, the losses are quite different. Figure 6.23 shows the loss through a gradual conical expansion, usually called a diffuser . There is a spread in the data, depending upon the boundary-layer conditions in the upstream pipe. A thinner entrance boundary layer, like the entrance profile in Fig. 6.6, gives a smaller loss. Since a diffuser is intendsed to raise the static pressure of the flow, dif-fuser data list the pressure-recovery coefficient of the flow Cp (6.103) The loss coefficient is related to this parameter by K V2 h /( m 2g) 1 Cp (6.104) For a given area ratio, the higher the pressure recovery, the lower the loss; hence large Cp means a successful diffuser. From Fig. 6.23 the minimum loss (maximum recov-ery) occurs for a cone angle 2 equal to about 5°. Angles smaller than this give a large Moody-type loss because of their excessive length. For cone angles greater than 40 to 60°, the loss is so excessive that it would actually be better to use a sudden expansion.
d1 4 d2 4 p2 p1 1 2 V1 2 6.7 Minor Losses in Pipe Systems 373 2θ K V1 V2 Fully developed inlet flow Thin inlet boundary layer 0 20 40 60 80 100 120 140 160 180 0.2 0 0.4 0.6 0.8 1.0 1.2 1.3 Total included diffuser angle 2 , degrees θ K = hm V 2/(2g) 1 = 1 – d 4 1 d 4 2 – Cp E6.16 This unexpected effect is due to gross flow separation in a wide-angle diffuser, as we shall see soon when we study boundary layers. Reference 14 has extensive data on dif-fusers.
For a gradual contraction, the loss is very small, as seen from the following exper-imental values : Contraction cone angle 2, 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, 1.94 slugs/ft3 and 0.000011 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig. E6.16. The rough-ness ratio is /d 0.001. Compute the pump horsepower required.
374 Chapter 6 Viscous Flow in Ducts Sharp entrance Open globe valve Pump 400 ft of pipe, d = ft 12-in bend radius Half-open gate valve Sharp exit Screwed regular 90° elbow z2 = 120 ft z1 = 20 ft 2 1 2 12 Solution Write the steady-flow energy equation between sections 1 and 2, the two reservoir surfaces: p g 1 V 2g 1 2 z1 p g 2 V 2g 2 2 z2 hf hm hp where hp is the head increase across the pump. But since p1 p2 and V1 V2 0, solve for the pump head hp z2 z1 hf hm 120 ft 20 ft 2 V g 2 f d L K (1) Now with the flow rate known, calculate V Q A 9.17 ft/s Now list and sum the minor loss coefficients: 0.2 ft3/s 1 4 ( 1 2 2 ft)2 6.8 Multiple-Pipe Systems8 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) 0.15 Regular 90° elbow (Table 6.5) 0.95 Half-closed gate valve (from Fig. 6.18b) 2.7 Sharp exit (Fig. 6.21) 1.0 K 12.2 Calculate the Reynolds number and pipe-friction factor Red V d 139,000 For /d 0.001, from the Moody chart read f 0.0216. Substitute into Eq. (1) hp 100 ft 2 ( ( 9 3 .1 2 7 .2 f f t t / / s s ) 2 2 ) 12.2 100 ft 84 ft 184 ft pump head The pump must provide a power to the water of P gQhp 1.94(32.2) lbf/ft3(184 ft) 2300 ft lbf/s The conversion factor is 1 hp 550 ft lbf/s. Therefore P 2 5 3 5 0 0 0 4.2 hp Ans.
Allowing for an efficiency of 70 to 80 percent, a pump is needed with an input of about 6 hp.
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 cir-cuits 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 Q1 Q2 Q3 const or V1d 1 2 V2d 2 2 V3d3 2 (6.105) Rule 2 is that the total head loss through the system equals the sum of the head loss in each pipe hA→B h1 h2 h3 (6.106) In terms of the friction and minor losses in each pipe, we could rewrite this as 0.0216(400) 1 2 2 9.17( 1 2 2 ) 0.000011 6.8 Multiple-Pipe Systems 375 8This section may be omitted without loss of continuity.
Fig. 6.24 Examples of multiple-pipe systems: (a) pipes in series; (b) pipes in parallel; (c) the three-reservoir junction problem. hA→B V 2g 1 2 f1 d L 1 1 K1 V 2g 2 2 f2 d L 2 2 K2 V 2g 3 2 f3 d L 3 3 K3 (6.107) and so on for any number of pipes in the series. Since V2 and V3 are proportional to V1 from Eq. (6.105), Eq. (6.107) is of the form hA→B V 2g 1 2 (0 1f1 2 f2 3f3) (6.108) where the i 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 f1, f2, and f3 all depend upon V1 through the Reynolds number. Begin by calculating f1, f2, and f3, assuming fully rough flow, and the solution for V1 will converge with one or two iterations. EES is ideal for this purpose.
EXAMPLE 6.17 Given is a three-pipe series system, as in Fig. 6.24a. The total pressure drop is pA pB 150,000 Pa, and the elevation drop is zA zB 5 m. The pipe data are 376 Chapter 6 Viscous Flow in Ducts 2 1 3 (c) z1 z3 HGL HGL HGL z2 zJ + pJ g ρ 1 3 B (b) 2 A 1 2 3 (a) B A Pipe L, m d, cm , mm /d 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, 1000 kg/m3 and 1.02 106 m2/s. Calculate the flow rate Q in m3/h through the system.
Solution The total head loss across the system is hA→B pA g pB zA zB 10 1 0 5 0 0 ( ,0 9 0 .8 0 1) 5 m 20.3 m From the continuity relation (6.105) the velocities are V2 d d 1 2 2 2 V1 1 9 6 V1 V3 d d 1 3 2 2 V1 4V1 and Re2 V V 2 1 d d 2 1 Re1 4 3 Re1 Re3 2 Re1 Neglecting minor losses and substituting into Eq. (6.107), we obtain hA→B V 2g 1 2 1250f1 2500 1 9 6 2 f2 2000(4)2f3 or 20.3 m V 2g 1 2 (1250f1 7900f2 32,000f3) (1) This is the form which was hinted at in Eq. (6.108). It seems to be dominated by the third pipe loss 32,000f3. Begin by estimating f1, f2, and f3 from the Moody-chart fully rough regime f1 0.0262 f2 0.0234 f3 0.0304 Substitute in Eq. (1) to find V 1 2 2g(20.3)/(33 185 973). The first estimate thus is V1 0.58 m/s, from which Re1 45,400 Re2 60,500 Re3 90,800 Hence, from the Moody chart, f1 0.0288 f2 0.0260 f3 0.0314 Substitution into Eq. (1) gives the better estimate V1 0.565 m/s Q 1 4 d1 2V1 2.84 103 m3/s or Q1 10.2 m3/h Ans.
A second iteration gives Q 10.22 m3/h, a negligible change.
The second multiple-pipe system is the parallel-flow case shown in Fig. 6.24b. Here the loss is the same in each pipe, and the total flow is the sum of the individual flows 6.8 Multiple-Pipe Systems 377 hA→B h1 h2 h3 (6.109a) Q Q1 Q2 Q3 (6.109b) If the total head loss is known, it is straightforward to solve for Qi in each pipe and sum them, as will be seen in Example 6.18. The reverse problem, of determining Qi when hf is known, requires iteration. Each pipe is related to hf by the Moody relation hf f(L/d)(V2/2g) fQ2/C, where C 2gd5/8L. Thus each pipe has nearly quadratic nonlinear parallel resistance, and head loss is related to total flow rate by hf where Ci 2 8 g L d i i 5 (6.109c) Since the fi vary with Reynolds number and roughness ratio, one begins Eq. (6.109c) by guessing values of fi (fully rough values are recommended) and calculating a first estimate of hf. Then each pipe yields a flow-rate estimate Qi (Cihf /fi)1/2 and hence a new Reynolds number and a better estimate of fi. Then repeat Eq. (6.109c) to con-vergence.
It should be noted that both of these parallel-pipe cases—finding either Q or hf — are easily solved by EES if reasonable initial 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. (6.109a) we can solve for each V separately 20.3 m V 2g 1 2 1250f1 V 2g 2 2 2500f2 V 2g 3 2 2000f3 (1) Guess fully rough flow in pipe 1: f1 0.0262, V1 3.49 m/s; hence Re1 V1d1/ 273,000.
From the Moody chart read f1 0.0267; recompute V1 3.46 m/s, Q1 62.5 m3/h. [This prob-lem can also be solved from Eq. (6.66).] Next guess for pipe 2: f2 0.0234, V2 2.61 m/s; then Re2 153,000, and hence f2 0.0246, V2 2.55 m/s, Q2 25.9 m3/h.
Finally guess for pipe 3: f3 0.0304, V3 2.56 m/s; then Re3 100,000, and hence f3 0.0313, V3 2.52 m/s, Q3 11.4 m3/h.
This is satisfactory convergence. The total flow rate is Q Q1 Q2 Q3 62.5 25.9 11.4 99.8 m3/h Ans.
These three pipes carry 10 times more flow in parallel than they do in series.
This example is ideal for EES. One enters the pipe data (Li,di,i); the fluid properties (, ); the definitions Qi (/4)d i 2Vi, Rei Vidi/, and hf fi(Li/di)(V i 2/2g); plus the Colebrook formula (6.74) for each friction factor fi. There is no need to use resistance ideas such as Eq. (6.109c). Specify that fi 0 and Rei 4000. Then, if one enters Q ∑Qi (99.8/3600) m3/s, EES quickly solves for hf 20.3 m. Conversely, if one enters hf 20.3 m, EES solves for Q 99.8 m3/h.
Q2 C i/f i 2 378 Chapter 6 Viscous Flow in Ducts EES 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 Q1 Q2 Q3 0 (6.110) 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 pJ at the junction. In other words, let the HGL at the junction have the elevation hJ zJ p g J where pJ is in gage pressure for simplicity. Then the head loss through each, assum-ing p1 p2 p3 0 (gage) at each reservoir surface, must be such that h1 V 2g 1 2 f1 d L 1 1 z1 hJ h2 V 2g 2 2 f2 d L 2 2 z2 hJ (6.111) h3 V 2g 3 2 f3 d L 3 3 z3 hJ We guess the position hJ and solve Eqs. (6.111) for V1, V2, and V3 and hence Q1, Q2, and Q3, iterating until the flow rates balance at the junction according to Eq. (6.110).
If we guess hJ too high, the sum Q1 Q2 Q3 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 z1 20 m z2 100 m z3 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, z3 hJ 40 m. This saves one calculation (Q3 0) and enables us to get the lay of the land: Reservoir hJ, m zi hJ, m fi Vi, m/s Qi, m3/h Li/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 2000 Q 17.1 Since the sum of the flow rates toward the junction is negative, we guessed hJ too high. Reduce hJ to 30 m and repeat: 6.8 Multiple-Pipe Systems 379 Fig. 6.25 Schematic of a piping network.
Reservoir hJ, m zi hJ, m fi Vi, m/s Qi, m3/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 08.0 Q 12.9 This is positive Q, and so we can linearly interpolate to get an accurate guess: hJ 34.3 m.
Make one final list: Reservoir hJ, m zi hJ, m fi Vi, m/s Qi, m3/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 05.7 0.0321 1.32 06.0 Q 0.7 This is close enough; hence we calculate that the flow rate is 52.4 m3/h toward reservoir 3, bal-anced by 47.1 m3/h away from reservoir 1 and 6.0 m3/h away from reservoir 3.
One further iteration with this problem would give hJ 34.53 m, resulting in Q1 52.8, Q2 47.0, and Q3 5.8 m3/h, so that Q 0 to three-place accuracy. Pedagogically 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 or even a city. This network is quite complex algebraically but follows the same basic rules: 380 Chapter 6 Viscous Flow in Ducts 7 2 1 3 6 4 5 10 12 9 11 8 A B C D E Loop I Loop II Loop III Loop I V F G H I 6.9 Experimental Duct Flows: Diffuser Performance 1. The net flow into any junction must be zero.
2. The net head loss around any closed loop must be zero. In other words, the HGL at each junction must have one and only one elevation.
3. All head losses must satisfy the Moody and minor-loss friction correlations.
By supplying these rules 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 pressure) 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 covered in at least one specialized text . Solution on microcomputers is also a reality. Some ex-plicit numerical algorithms have been developed by Ormsbee and Wood . 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 upon ex-periment to determine the flow properties. As mentioned many times before, experi-ment 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 an estuary (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. 6.7) Chapter 7 will treat a great many more external-flow experiments, especially in Sec.
7.5. Here we shall show data for one type of internal flow, the diffuser.
A diffuser, shown in Fig. 6.26a and b, is an expansion or area increase intended to re-duce velocity in order to recover the pressure head of the flow. Rouse and Ince re-late 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.
Neglecting losses and gravity effects, the incompressible Bernoulli equation pre-dicts that 6.9 Experimental Duct Flows: Diffuser Performance 381 Diffuser Performance Fig. 6.26 Diffuser geometry and typical flow regimes: (a) geometry of a flat-walled diffuser; (b) geom-etry of a conical diffuser; (c) flat-diffuser stability map. (From Ref.
14, by permission of Creare, Inc.) p 1 2 V2 p0 const (6.112) where p0 is the stagnation pressure which 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 Cp p p 0 e t p p t t (6.113) where subscripts 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 sec-tion 2 the exit. Application of Bernoulli’s equation (6.112) to this diffuser predicts that p01 p1 1 2 V1 2 p2 1 2 V2 2 p02 or Cp,frictionless 1 V V 2 1 2 (6.114) Meanwhile, steady one-dimensional continuity would require that Q V1A1 V2A2 (6.115) Combining (6.114) and (6.115), we can write the performance in terms of the area ra-tio AR A2/A1, which is a basic parameter in diffuser design: Cp,frictionless 1 (AR)2 (6.116) A typical design would have AR 51, for which Eq. (6.116) 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.
382 Chapter 6 Viscous Flow in Ducts W1 2θ W2 2 L (a) (b) L D De Throat Exit L W1 100 70 40 20 10 7 4 2 1 1 2 4 7 10 20 40 100 Transitory stall Maximum unsteadiness 2 , degrees Cp max c b Jet flow θ (c) c No stall a a b Bistable steady stall 1 b 2θ Fig. 6.27 Diffuser performance: (a) ideal pattern with good perfor-mance; (b) actual measured pattern with boundary-layer separation and resultant poor performance.
The basic reason for the discrepancy is flow separation, as sketched in Fig. 6.27.
The increasing pressure in the diffuser is an unfavorable gradient (Sec. 7.4), which causes the viscous boundary layers to break away from the walls and greatly reduces the performance. Theories can now predict this behavior (see, e.g., Ref. 20).
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 performance. 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, i.e., highest Cp, 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.
6.9 Experimental Duct Flows: Diffuser Performance 383 Thin boundary layers Low velocity, high pressure (a) Thick boundary layers (b) Backflow High velocity, low pressure “Stalled” flow Separation point 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 de-pend upon the following parameters: 1. Any two of the following geometric parameters: a. Area ratio AR A2/A1 or (De/D)2 b. Divergence angle 2 c. Slenderness L/W1 or L/D 2. Inlet Reynolds number Ret V1W1/ or Ret V1D/ 3. Inlet Mach number Mat V1/a1 4. Inlet boundary-layer blockage factor Bt ABL/A1, where ABL is the wall area blocked, or displaced, by the retarded boundary-layer flow in the inlet (typically Bt 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 b/W1 Even with this formidable list, we have omitted five possible important effects: inlet turbulence, inlet swirl, inlet profile vorticity, superimposed pulsations, and downstream obstruction, all of which occur in practical machinery applications.
The three most important parameters are AR, , 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 diffusers, 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.
The experimental design of a diffuser is an excellent example of a successful at-tempt to minimize the undesirable effects of adverse pressure gradient and flow sepa-ration.
384 Chapter 6 Viscous Flow in Ducts Table 6.6 Maximum Diffuser-Performance Data Flat-walled Conical Inlet blockage —— — — — — — — — — — — — — — — — — — — — Bt Cp,max L/W1 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 Fig. 6.28a Typical performance maps for flat-wall and conical dif-fusers at similar operating condi-tions: (a) flat wall. (From Ref. 14, by permission of Creare, Inc.) Almost all practical fluids engineering problems are associated with the need for an accurate flow measurement. There is a need to measure local properties (velocity, pres-sure, 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 visual-ization techniques were discussed in Sec. 1.7 for low-speed flows, and the special op-tical techniques used in high-speed flows are treated in Ref. 21 of Chap. 1. Flow-mea-surement 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: 6.10 Fluid Meters 385 AS = 1.0 Mat = 0.2 Bt = 0.08 ReDh = 279,000 Flat 5 4.5 4 3.5 3 2 1.75 20° 18° 16° 14° 12° 10° 8° 4 5 6° 6 7 8 9 10 12 14 16 18 20 2θ = 4° L W1 0.60 0.62 0.64 0.66 0.68 0.69 0.70 Cp Transitory stall boundary AR 6.10 Fluid Meters Local-Velocity Measurements Fig. 6.28b Typical performance maps for flat-wall and conical dif-fusers at similar operating condi-tions: (b) conical wall. (From Ref.
14, by permission of Creare, Inc.) 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 films 6. Laser-doppler anemometer (LDA) Some of these meters are sketched in Fig. 6.29.
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 sur-face of a channel flow, small neutrally buoyant spheres mixed with a liquid, or hydro-386 Chapter 6 Viscous Flow in Ducts Mt = 0.2 Bt = 0.09 Red = 120,000 Conical AR 25 16 12 10 8 6 5 4 3 2.5 2 1.75 1.5 2 4 6 8 10 12 16 20 25 2 = 18° θ 16° 14° 12° 10° 8° 6° 5° 4° 0.70 3° 0.68 2° 30 (b) Diffuser length–throat diameter ratio L d 0.66 0.64 0.62 0.60 0.58 0.56 0.54 0.52 0.50 0.48 0.46 0.44 Cp Fig. 6.29 Eight common velocity meters: (a) three-cup anemometer; (b) Savonius rotor; (c) turbine mounted in a duct; (d) free-pro-peller meter; (e) hot-wire anemometer; (f) hot-film anemome-ter; (g) pitot-static tube; (h) laser-doppler anemometer.
gen bubbles. Sometimes gas flows can be estimated from the motion of entrained dust particles. One must establish whether the particle motion truly simulates the fluid mo-tion. Floats are commonly used to track the movement of ocean waters and can be de-signed to move at the surface, along the bottom, or at any given depth . Many of-ficial 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 be used in either gases or liquids, and their rotation rate is approximately proportional to the flow velocity.
The cup anemometer (Fig. 6.29a) 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. 6.29d) meters must 6.10 Fluid Meters 387 Laser (g) (a) (d) (b) (e) (c) (f ) (h) Fine wire: Plated film: θ Focusing optics Flow Receiving optics Photo detector Display Fig. 6.30 Pitot-static tube for com-bined measurement of static and stagnation pressure in a moving stream.
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 by electromagnetic or slip-ring devices for either a continuous or a digital reading of flow velocity. All have the disadvantage of being rel-atively large and thus not representing a “point.” Pitot-static tube.
A slender tube aligned with the flow (Figs. 6.29g and 6.30) can mea-sure local velocity by means of a pressure difference. It has sidewall holes to measure the static pressure ps in the moving stream and a hole in the front to measure the stag-nation pressure p0, where the stream is decelerated to zero velocity. Instead of mea-suring p0 or ps separately, it is customary to measure their difference with, say, a trans-ducer, as in Fig. 6.30.
If ReD 1000, where D is the probe diameter, the flow around the probe is nearly frictionless and Bernoulli’s relation, Eq. (3.77), applies with good accuracy. For in-compressible flow ps 1 2 V2 gzs p0 1 2 (0)2 gz0 Assuming that the elevation pressure difference g(zs z0) is negligible, this reduces to V 2 (p0 ps) 1/2 (6.117) This is the Pitot formula, named after the French engineer 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 substan-tial errors in both the p0 and ps 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-filled tubes leading to the pressure sensors, it is not useful for unsteady-flow mea-surements. It does resemble a point and can be made small enough to measure, e.g., 388 Chapter 6 Viscous Flow in Ducts V θ ps Stagnation pressure 4 to 8 holes Static pressure ≈Free-stream pressure +10% –10% Error 0° 10° 20° Static pressure Stagnation pressure Yaw angle p0 pS Differential pressure transducer 8D θ 0 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 V 1 ft/s in standard air, from Eq. (6.117) we compute p0 p equal to only 0.001 lbf/ft2 (0.048 Pa). This is beyond the resolution of most pressure gages.
Electromagnetic meter.
If a magnetic field 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 re-sistance. 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 amplify-ing 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 fine 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 I2R a b(V)n (6.118) where n 1 3 at very low Reynolds numbers and equals 1 2 at high Reynolds numbers.
The hot wire normally operates in the high-Reynolds-number range but should be cal-ibrated in each situation to find the best-fit a, b, and n. The wire can be operated ei-ther at constant current I, so that resistance R is a measure of V, or at constant resis-tance 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 avail-able, as are do-it-yourself designs . Excellent detailed discussions of the hot wire are given in Refs. 1 and 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. 6.29f). A thin metallic film, usually platinum, is plated onto a relatively thick support which 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, coher-ent monochromatic light which 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 f is proportional to the veloc-ity 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 . Their intersection, 6.10 Fluid Meters 389 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 mea-suring volume scatter the beams; they then pass through receiving optics to a pho-todetector which converts the light to an electric signal. A signal processor then converts electric frequency to a voltage which can be either displayed or stored. If ! is the wave-length of the laser light, the measured velocity is given by V 2 si !
n ( f /2) (6.119) 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 scat-tering 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 capa-ble 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, e.g., the amazingly detailed three-dimensional flow profiles measured by Eckardt in a high-speed centrifugal compressor impeller. Extensive discussions of laser velocimetry are given in Refs. 38 and 39.
EXAMPLE 6.20 The pitot-static tube of Fig. 6.30 uses mercury as a manometer fluid. When it is placed in a wa-ter 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 (2.33), with zA z2, the pressure difference is related to h by p0 ps ("M "w)h Taking the specific weights of mercury and water from Table 2.1, we have p0 ps (846 62.4 lbf/ft3) 8 1 .
2 4 ft 549 lbf/ft2 The density of water is 62.4/32.2 1.94 slugs/ft3. Introducing these values into the pitot-static formula (6.117), we obtain 390 Chapter 6 Viscous Flow in Ducts Volume-Flow Measurements V 1 2 .
( 9 5 4 49 sl l u b g f s /f / t f 2 t3 ) 1/2 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 c. Rotating slotted rings d. Nutating disk 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 mea-sure 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 6.10 Fluid Meters 391 Fig. 6.31 The turbine meter widely used in the oil, gas, and water sup-ply industries: (a) basic design; (b) typical calibration curve for a range of crude oils. (Daniel Indus-tries, Inc., Flow Products Division.) Turbine meter.
The turbine meter, sometimes called a propeller meter, is a freely ro-tating propeller which can be installed in a pipeline. A typical design is shown in Fig.
6.31a. There are flow straighteners upstream of the rotor, and the rotation is 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.
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 51 flow-rate range with 0.25 percent accu-racy. Gas meters need many blades to produce sufficient torque and are accurate to 1 percent.
392 Chapter 6 Viscous Flow in Ducts Rotor supports Turbine rotor Magnetic pulse pickup (a) 0 500 1000 1500 2000 1720 1715 1710 1705 1700 1695 1690 Pulses per cubic meter m3/h (b) 10 – in turbine 0.44 0.38 0.20 v = 0.06 cm2/s Fig. 6.32 A Commercial handheld wind-velocity turbine meter. (Cour-tesy of Nielsen-Kellerman Com-pany.) Since turbine meters are very individualistic, flow calibration is an absolute neces-sity. A typical liquid-meter calibration curve is shown in Fig. 6.31b. Researchers at-tempting to establish universal calibration curves have met with little practical success 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 di-rections. Figure 6.32 illustrates a handheld wind velocity meter which 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 cross-flow sheds alternating vortices at a nearly uniform Strouhal number St fL/U, where U is the approach velocity and L is a characteristic body width. Since L and St are con-stant, this means that the shedding frequency is proportional to velocity f (const)(U) (6.120) 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. 6.33.
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 1001) 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.
6.10 Fluid Meters 393 Fig. 6.34 Ultrasonic flowmeters: (a) pulse type; (b) doppler-shift type (from Ref. 41); (c) a portable noninvasive installation (courtesy of Polysonics Inc., Houston, TX).
394 Chapter 6 Viscous Flow in Ducts (c) Fig. 6.33 A vortex flowmeter. (The Foxboro Company.) A B R T (a) (b) Fig. 6.35 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, Westminster, CA.) Ultrasonic flowmeters.
The sound-wave analog of the laser velocimeter of Fig. 6.29h is the ultrasonic flowmeter. Two examples are shown in Fig. 6.34. The pulse-type flowme-ter is shown in Fig. 6.34a. Upstream piezoelectric transducer A is excited with a short sonic pulse which 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 fA.
The same process is duplicated in the reverse direction from B to A, creating frequency fB. The difference fA fB is proportional to the flow rate. Figure 6.33b 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 which is proportional to the flow rate. Ultrasonic meters are nonintru-sive and can be directly attached to pipe flows in the field (Fig. 6.34c). 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 number. For further details see Ref. 41.
Rotameter.
The variable-area transparent rotameter of Fig. 6.35 has a float which, un-der 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 CdAa Afl 2 o W at ne fl t uid 1/2 (6.121) where Wnet is the float’s net weight in the fluid, Aa Atube Afloat is the annular area between the float and the tube, and Cd is a dimensionless discharge coefficient of or-der 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. 6.35. 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 modern alternative is a mass flowmeter which operates on the principle of the Coriolis accel-eration associated with noninertial coordinates [recall Fig. 3.12 and the Coriolis term 2# V in Eq. (3.48)]. The output of the meter is directly proportional to mass flow.
Figure 6.36 is a schematic of a Coriolis device, to be inserted into a piping system.
The flow enters a double-loop, double-tube arrangement which is electromagnetically vi-brated at a high natural frequency (amplitude 1 mm and frequency 100 Hz). The up flow induces inward loop motion, while the down flow creates outward loop motion, both due to the Coriolis effect. Sensors at both ends register a phase difference which is pro-portional to mass flow. Quoted accuracy is approximately 0.2 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 non-linear. In laminar duct flow, however, Q is linearly proportional to p, as in Eq. (6.44): Q [R4/(8L)] p. Thus a laminar flow sensing element is attractive, since its cal-ibration will be linear. To ensure laminar flow for what otherwise would be a turbu-6.10 Fluid Meters 395 Fig. 6.37 A complete flowmeter system using a laminar-flow ele-ment (in this case a narrow annu-lus). The flow rate is linearly pro-portional to the pressure drop.
(Courtesy of Martin Girard, DH Instruments, Inc.) lent condition, all or part of the fluid is directed into small passages, each of which has a low (laminar) Reynolds number. A honeycomb is a popular design.
Figure 6.37 uses axial flow through a narrow annulus to effect laminar flow. The theory again predicts Q p, as in Eq. (6.92). However, the flow is very sensitive to passage size; for example, halving the annulus clearance increases p more than eight 396 Chapter 6 Viscous Flow in Ducts Electrical connector Microprocessor Self-sealing pressure measurement connection Sintered metallic filter Flange connection O-ring Pressure-equalization chamber O-ring-sealed pressure connection Annular laminar-flow path defined by piston and cylinder Platinum resistance thermometer Piston-centering seat Fig. 6.36 A Coriolis mass flowme-ter. (Courtesy of ABB Instrumenta-tion, Inc.) Fig. 6.38 Velocity and pressure change through a generalized Bernoulli obstruction meter.
times. Careful calibration is thus necessary. In Fig. 6.37 the laminar-flow concept has been synthesized into a complete mass-flow system, with temperature control, differ-ential pressure measurement, and a microprocessor all self-contained. The accuracy of this device is rated at 0.2 percent.
Bernoulli obstruction theory.
Consider the generalized flow obstruction shown in Fig.
6.38. The flow in the basic duct of diameter D is forced through an obstruction of diameter d; the $ ratio of the device is a key parameter $ D d (6.122) After leaving the obstruction, the flow may neck down even more through a vena con-tracta of diameter D2 d, as shown. Apply the Bernoulli and continuity equations for incompressible steady frictionless flow to estimate the pressure change: Continuity: Q 4 D2V1 4 D2 2V2 Bernoulli: p0 p1 1 2 V1 2 p2 1 2 V2 2 6.10 Fluid Meters 397 Horizontal Moody loss Nonrecoverable head loss HGL EGL p1 – p2 D V1 Deadwater region Dividing streamline V2 ≈ V1 D D2 ( ) 2 Vena contracta D2 d = D β Eliminating V1, we solve these for V2 or Q in terms of the pressure change p1 p2: A Q 2 V2 ( 2 1 (p 1 D2 4 p /D 2) 4) 1/2 (6.123) 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 D2/d for use in (6.123). Therefore we assume that D2/D $ and then calibrate the device to fit the relation Q AtVt CdAt 2(p 1 1 $ p2 4 )/ 1/2 (6.124) where subscript t denotes the throat of the obstruction. The dimensionless discharge coefficient Cd accounts for the discrepancies in the approximate analysis. By dimen-sional analysis for a given design we expect Cd f($, ReD) where ReD V 1D (6.125) The geometric factor involving $ in (6.124) is called the velocity-of-approach factor E (1 $4)1/2 (6.126) One can also group Cd and E in Eq. (6.124) to form the dimensionless flow coefficient CdE (1 C $ d 4)1/2 (6.127) Thus Eq. (6.124) can be written in the equivalent form Q At 2(p1 p2) 1/2 (6.128) Obviously the flow coefficient is correlated in the same manner: f($, ReD) (6.129) Occasionally one uses the throat Reynolds number instead of the approach Reynolds number Red V td R $ eD (6.130) Since the design parameters are assumed known, the correlation of from Eq. (6.129) or of Cd from Eq. (6.125) is the desired solution to the fluid-metering problem.
The mass flow is related to Q by m ˙ Q (6.131) and is thus correlated by exactly the same formulas.
Figure 6.39 shows the three basic devices recommended for use by the International Organization for Standardization (ISO) : the orifice, nozzle, and venturi tube.
Thin-plate orifice.
The thin-plate orifice, Fig. 6.39b, can be made with $ in the range of 0.2 to 0.8, except that the hole diameter d should not be less than 12.5 mm. To mea-sure p1 and p2, three types of tappings are commonly used: 398 Chapter 6 Viscous Flow in Ducts Fig. 6.39 International standard shapes for the three primary Bernoulli obstruction-type meters: (a) long radius nozzle; (b) thin-plate orifice; (c) venturi nozzle.
(From Ref. 31 by permission of the International Organization for Standardization.) 1. Corner taps where the plate meets the pipe wall 2. D: 1 2 D taps: pipe-wall taps at D upstream and 1 2 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 6.40 shows the discharge coefficient of an orifice with D: 1 2 D or type 2 taps in the Reynolds-number range ReD 104 to 107 of normal use. Although detailed charts such as Fig. 6.37 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 Cd f($) 91.71$2.5ReD 0.75 1 0.
09$ $ 4 4 F1 0.0337$3F2 (6.132) where f($) 0.5959 0.0312$2.1 0.184$8 The correlation factors F1 and F2 vary with tap position: Corner taps: F1 0 F2 0 (6.133a) 6.10 Fluid Meters 399 3 2 d Flow d 0.6 d d t2 < 13 mm t1 < 0.15 D (a) Flow d D Bevel angle: 45° to 60° Edge thickness: 0.005 D to 0.02 D Plate thickness: up to 0.05 D (b) Ellipse 2 D Flow 0.7d Throat tap ISA 1932 nozzle shape 2 d Conical diffuser: < 15° (c) θ Fig. 6.40 Discharge coefficient for a thin-plate orifice with D: 1 2 D taps, plotted from Eqs. (6.132) and (6.133b).
D: 1 2 D taps: F1 0.4333 F2 0.47 (6.133b) Flange taps: F2 D 1 (in) F1 (6.133c) Note that the flange taps (6.133c), 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. (6.133c) 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.
6.39a and a short-radius type (not shown) called the ISA 1932 nozzle [30, 31]. The flow nozzle, with its smooth rounded entrance convergence, practically eliminates the vena contracta and gives discharge coefficients near unity. The nonrecoverable loss is still large because there is no diffuser provided for gradual expansion.
The ISO recommended correlation for long-radius-nozzle discharge coefficient is Cd 0.9965 0.00653$1/2 R 10 eD 6 1/2 0.9965 0.00653 R 10 e 6 d 1/2 (6.134) The second form is independent of the $ ratio and is plotted in Fig. 6.41. A similar ISO correlation is recommended for the short-radius ISA 1932 flow nozzle D 1 (in) D 2.3 in 0.4333 2.0 D 2.3 in 400 Chapter 6 Viscous Flow in Ducts 0.66 0.65 0.64 0.63 0.62 0.61 0.60 0.59 0.58 104 106 107 ReD 0.7 0.6 0.5 0.4 0.3 0.2 β = 0.8 = d D p1 p2 d Flow 1 2 D Cd 105 D D Fig. 6.41 Discharge coefficient for long-radius nozzle and classical Herschel-type venturi.
Cd 0.9900 0.2262$4.1 (0.000215 0.001125$ 0.00249$4.7) R 10 eD 6 1.15 (6.135) Flow nozzles may have $ values between 0.2 and 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. 6.39c, 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 1.5 105 to 2 106. Its discharge coefficient, shown in Fig. 6.42, is given by the ISO correlation formula Cd 0.9858 0.196$4.5 (6.136) It is independent of ReD within the given range. The Herschel venturi discharge varies with ReD but not with $, as shown in Fig. 6.41. Both have very low net losses.
The choice of meter depends upon the loss and the cost and can be illustrated by the following table: Type of meter Net head loss Cost Orifice Large Small Nozzle Medium Medium Venturi Small Large As so often happens, the product of inefficiency and initial cost is approximately constant.
6.10 Fluid Meters 401 1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 104 105 106 107 108 Cd Red, ReD All values of β Long-radius nozzle (Red) Classical Herschel venturi (ReD) Fig. 6.43 Nonrecoverable head loss in Bernoulli obstruction meters.
(Adapted from Ref. 30.) 402 Chapter 6 Viscous Flow in Ducts 3.0 2.5 2.0 1.5 1.0 0.5 0 0.2 0.3 0.4 0.5 0.6 0.7 0.8 β Thin-plate orifice Flow nozzle Venturi: 15° cone angle 7° cone angle Km = hm Vt2/(2g) 0.3 0.4 0.5 0.6 0.7 0.8 0.92 0.94 0.96 0.98 1.00 β Cd International standards: 0.316 < < 0.775 1.5 × 105 < ReD < 2.0 × 106 β Fig. 6.42 Discharge coefficient for a venturi nozzle.
The average nonrecoverable head losses for the three types of meters, expressed as a fraction of the throat velocity head V t 2/(2g), are shown in Fig. 6.43. The orifice has the greatest loss and the venturi the least, as discussed. The orifice and nozzle simu-late partially closed valves as in Fig. 6.18b, while the venturi is a very minor loss.
When the loss is given as a fraction of the measured pressure drop, the orifice and noz-zle have nearly equal losses, as Example 6.21 will illustrate.
Part (a) Part (b) Part (c) 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.
EXAMPLE 6.21 We want to meter the volume flow of water ( 1000 kg/m3, 1.02 106 m2/s) moving through a 200-mm-diameter pipe at an average velocity of 2.0 m/s. If the differential pressure gage selected reads accurately at p1 p2 50,000 Pa, what size meter should be selected for installing (a) an orifice with D: 1 2 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 $ ratio of the meter. Since the discharge coefficient is a complicated function of $, iteration will be necessary. We are given D 0.2 m and V1 2.0 m/s. The pipe-approach Reynolds number is thus ReD V 1D 1.
( 0 2 2 .0 )(0 1 .
0 2 ) 6 392,000 For all three cases [(a) to (c)] the generalized formula (6.128) holds: Vt V $ 1 2 2(p1 p2) 1/2 (1 C $ d 4)1/2 (1) where the given data are V1 2.0 m/s, 1000 kg/m3, and p 50,000 Pa. Inserting these known values into Eq. (1) gives a relation between $ and : 2 $ .
2 0 2(5 1 0 0 , 0 0 0 00) 1/2 or $2 0 .2 (2) The unknowns are $ (or ) and Cd. Parts (a) to (c) depend upon the particular chart or formula needed for Cd fcn(ReD, $). We can make an initial guess $ 0.5 and iterate to convergence.
For the orifice with D: 1 2 D taps, use Eq. (6.132) or Fig. 6.40. The iterative sequence is $1 0.5, Cd1 0.604, 1 0.624, $2 0.566, Cd2 0.606, 2 0.640, $3 0.559 We have converged to three figures. The proper orifice diameter is d $D 112 mm Ans. (a) For the long-radius flow nozzle, use Eq. (6.134) or Fig. 6.41. The iterative sequence is $1 0.5, Cd1 0.9891, 1 1.022, $2 0.442, Cd2 0.9896, 2 1.009, $3 0.445 We have converged to three figures. The proper nozzle diameter is d $D 89 mm Ans. (b) For the venturi nozzle, use Eq. (6.136) or Fig. 6.42. The iterative sequence is $1 0.5, Cd1 0.977, 1 1.009, $2 0.445, Cd2 0.9807, 2 1.0004, $3 0.447 We have converged to three figures. The proper venturi diameter is d $D 89 mm Ans. (c) 6.10 Fluid Meters 403 Summary These meters are of similar size, but their head losses are not the same. From Fig. 6.43 for the three different shapes we may read the three K factors and compute hm,orifice 3.5 m hm,nozzle 3.6 m hm,venturi 0.8 m The venturi loss is only about 22 percent of the orifice and nozzle losses.
Solution The iteration encountered in this example is ideal for the EES. Input the data in SI units: Rho 1000 Nu 1.02E-6 D 0.2 V 2.0 DeltaP 50000 Then write out the basic formulas for Reynolds number, throat velocity and flow coefficient: Re VD/Nu Vt V/Beta 2 Alpha Cd/(1-Beta 4) 0.5 Vt AlphaSQRT(2DeltaP/Rho) Finally, input the proper formula for the discharge coefficient. For example, for the flow nozzle, Cd 0.9965 0.00653Beta 0.5(1E6/Re) 0.5 When asked to Solve the equation, EES at first complains of dividing by zero. One must then tighten up the Variable Information by not allowing $, , or Cd to be negative and, in particu-lar, by confining $ to its practical range 0.2 $ 0.9. EES then readily announces correct an-swers for the flow nozzle: Alpha 1.0096 Cd 0.9895 Beta 0.4451 This chapter is concerned with internal pipe and duct flows, which are probably the most common problems encountered in engineering fluid mechanics. Such flows are very sensitive to the Reynolds number and change from laminar to transitional to tur-bulent flow as the Reynolds number increases.
The various Reynolds-number regimes are outlined, and a semiempirical approach to turbulent-flow modeling is presented. The chapter then makes 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 are discussed for flow-rate and sizing problems, as well as the application of the Moody chart to noncircular ducts us-ing an equivalent duct “diameter.” The addition of minor losses due to valves, elbows, fittings, and other devices is presented in the form of loss coefficients to be incorpo-rated along with Moody-type friction losses. Multiple-pipe systems are discussed briefly and are 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 is presented as experimental data, since the theory of real diffusers is still not well developed. The chapter ends with a discussion of flowmeters, especially the pitot-static tube and the Bernoulli-obstruction type of meter. Flowmeters also re-quire careful experimental calibration.
404 Chapter 6 Viscous Flow in Ducts EES Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 6.1 to 6.160 (categorized in the problem list below) are followed by word problems W6.1 to W6.5, funda-mentals of engineering exam problems FE6.1 to FE6.15, compre-hensive problems C6.1 to C6.5, and design projects D6.1 and D6.2.
Problem distribution Section Topic Problems 6.1 Reynolds-number regimes 6.1–6.7 6.2 Internal and external flow 6.8–6.10 6.3 Turbulent-shear correlations 6.11–6.18 6.4 Laminar pipe flow 6.19–6.41 6.4 Turbulent pipe flow 6.42–6.77 6.5 Flow-rate and pipe-sizing problems 6.78–6.85 6.6 Noncircular ducts 6.86–6.99 6.7 Minor losses 6.100–6.110 6.8 Series and parallel pipe systems 6.111–6.120 6.8 Three-reservoir and pipe-network systems 6.121–6.130 6.9 Diffuser performance 6.131–6.134 6.10 The pitot-static tube 6.135–6.139 6.10 Flowmeters: The orifice plate 6.140–6.148 6.10 Flowmeters: The flow nozzle 6.149–6.153 6.10 Flowmeters: The venturi meter 6.154–6.160 P6.1 In flow past a sphere, the boundary layer becomes turbu-lent at about ReD 2.5 E5. To what air speed in mi/h does this correspond for a golf ball whose diameter is 1.6 in?
Do the pressure, temperature, and humidity of the air make any difference in your calculation?
P6.2 For a thin wing moving parallel to its chord line, transition to a turbulent boundary layer normally occurs at Rex 2.8 106, where x is the distance from the leading edge [2, 3]. If the wing is moving at 20 m/s, at what point on the wing will transition occur at 20°C for (a) air and (b) water?
P6.3 If the wing of Prob. 6.2 is tested in a wind or water tun-nel, it may undergo transition earlier than Rex 2.8 106 if the test stream itself contains turbulent fluctuations. A semiempirical correlation for this case [3, p. 383] is Rexcrit 1/2 where is the tunnel-turbulence intensity in percent. If V 20 m/s in air at 20°C, use this formula to plot the transi-tion position on the wing versus stream turbulence for between 0 and 2 percent. At what value of is xcrit de-creased 50 percent from its value at 0?
1 (1 13.25 2)1/2 0.00392 2 P6.4 For flow of SAE 10 oil through a 5-cm-diameter pipe, from Fig. A.1, for what flow rate in m3/h would we expect tran-sition to turbulence at (a) 20°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 Ud/ 850, where d is the wire diameter [3, p. 386]. If the sphere diameter is 20 cm and transition is observed at ReD 90,000, what is the diameter of the trip wire in mm?
Problems 405 D Trip wire d U P6.5 P6.6 A fluid at 20°C flows at 850 cm3/s through an 8-cm-di-ameter pipe. Determine whether the flow is laminar or tur-bulent if the fluid is (a) hydrogen, (b) air, (c) gasoline, (d) water, (e) mercury, or (f) glycerin.
P6.7 Cola, approximated as pure water at 20°C, is to fill an 8-oz container (1 U.S. gal 128 fl oz) through a 5-mm-di-ameter tube. Estimate the minimum filling time if the tube flow is to remain laminar. For what cola (water) tempera-ture would this minimum time be 1 min?
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 (p/x) if the pipe is (a) hor-izontal and (b) vertical with the flow up?
P6.9 A light liquid ( 950 kg/m3) flows at an average veloc-ity of 10 m/s through a horizontal smooth tube of diame-ter 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 average wall shear stress, in Pa, and (b) the wall shear stress in the fully developed region of the pipe.
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 pB 260 kPa, VB 3.2 m/s, zB 9.1 m. Which way is the flow go-ing? What is the head loss in meters?
P6.11 Derive the time-averaged x-momentum equation (6.14) by direct substitution of Eqs. (6.12) into the momentum equa-tion (6.7). It is convenient to write the convective acceler-ation as d d u t x (u2) y (u) z (uw) which is valid because of the continuity relation, Eq. (6.7).
P6.12 By analogy with Eq. (6.14) write the turbulent mean-mo-mentum differential equation for (a) the y direction and (b) the z direction. How many turbulent stress terms ap-pear in each equation? How many unique turbulent stresses are there for the total of three directions?
P6.13 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: y, in 0.025 0.035 0.047 0.055 0.065 u, ft/s 51.2 54.2 56.8 57.6 59.1 Estimate (a) the wall shear stress and (b) the velocity u at y 0.22 in.
P6.14 Two infinite plates a distance h apart are parallel to the xz plane with the upper plate moving at speed V, as in Fig.
P6.14. There is a fluid of viscosity 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 ver-sus dimensionless plate velocity. Sketch a typical shape of the profile u(y).
constant [2, 3]. Assuming that turb w near the wall, show that this expression can be integrated to yield the log-arithmic-overlap law, Eq. (6.21).
P6.18 Water at 20°C flows in a 9-cm-diameter pipe under fully de-veloped conditions. The centerline velocity is 10 m/s. Com-pute (a) Q, (b) V, (c) w, and (d) p for a 100-m pipe length.
In Probs. 6.19 to 6.99, neglect minor losses.
P6.19 A 5-mm-diameter capillary tube is used as a viscometer for oils. When the flow rate is 0.071 m3/h, the measured pressure 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?
P6.20 A soda straw is 20 cm long and 2 mm in diameter. It de-livers cold cola, approximated as water at 10°C, at a rate of 3 cm3/s. (a) What is the head loss through the straw?
What is the axial pressure gradient p/x if the flow is (b) vertically up or (c) horizontal? Can the human lung de-liver this much flow?
P6.21 Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in diameter, as in Fig. P6.21. Is there any height H for which the flow might not be laminar? What is the flow rate if H 50 cm? Neglect the tube curvature.
406 Chapter 6 Viscous Flow in Ducts Water at 20˚ C L = 1 m, d = 2 mm H x y V h Fixed u υ P6.14 P6.15 Suppose in Fig. P6.14 that h 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?
P6.16 By analogy with laminar shear, du/dy, T. V. Boussi-nesq in 1877 postulated that turbulent shear could also be related to the mean-velocity gradient turb du/dy, where is called the eddy viscosity and is much larger than . If the logarithmic-overlap law, Eq. (6.21), is valid with w, show that uy.
P6.17 Theodore von Kármán in 1930 theorized that turbulent shear could be represented by turb du/dy where 2y2du/dyis called the mixing-length eddy viscosity and 0.41 is Kármán’s dimensionless mixing-length P6.21 P6.22 SAE 30W oil at 20°C flows through a long, horizontal, 12-cm-diameter tube. At section 1, the fluid pressure is 186 kPa. At section 2, which is 6 m further downstream, the pressure is 171 kPa. If the flow is laminar, estimate (a) the mass flow in kg/s and (b) the Reynolds number.
P6.23 Glycerin at 20°C is to be pumped through a horizontal smooth pipe at 3.1 m3/s. It is desired that (1) the flow be laminar and (2) the pressure drop be no more than 100 Pa/m. What is the minimum pipe diameter allowable?
P6.24 The 6-cm-diameter pipe in Fig. P6.24 contains glycerin at 20°C flowing at a rate of 6 m3/h. Verify that the flow is laminar. For the pressure measurements shown, is the flow up or down? What is the indicated head loss for these pres-sures?
P6.25 To determine the viscosity of a liquid of specific gravity 0.95, you fill, to a depth of 12 cm, a large container which drains through a 30-cm-long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of drain-ing is found to be 1.9 cm3/s. What is your estimate of the fluid viscosity? Is the tube flow laminar?
P6.26 Water at 20°C is flowing through a 20-cm-square smooth duct at a (turbulent) Reynolds number of 100,000. For a “laminar flow element” measurement, it is desired to pack the pipe with a honeycomb array of small square passages (see Fig. P6.36 for an example). What passage width h will ensure that the flow in each tube will be laminar (Reynonds number less than 2000)?
P6.27 An oil (SG 0.9) issues from the pipe in Fig. P6.27 at Q 35 ft3/h. What is the kinematic viscosity of the oil in ft3/s? Is the flow laminar?
P6.32 SAE 30 oil at 20°C flows at 0.001 m3/s through 100 m of 1-cm-diameter pipe and then for another 100 m at an in-creased d 2 cm. The double-pipe system slopes upward at 35° in the flow direction. Estimate (a) the total pressure change and (b) the power required to drive the flow.
P6.33 For the configuration shown in Fig. P6.33, the fluid is ethyl alcohol at 20°C, and the tanks are very wide. Find the flow rate which occurs in m3/h. Is the flow laminar?
Problems 407 2.1 atm A B 12 m 3.7 atm P6.30 P6.24 10 ft L = 6 ft D = 1 2 in Q P6.27 P6.28 In Prob. 6.27 what will the flow rate be, in m3/h, if the fluid is SAE 10 oil at 20°C?
P6.29 Oil, with 890 kg/m3 and 0.06 kg/(m s), is to be pumped through 1 km of straight horizontal pipe with a power input of 1 kW. What is the maximum possible mass flow rate, and corresponding pipe diameter, if laminar flow is to be maintained?
P6.30 A steady push on the piston in Fig. P6.30 causes a flow rate Q 0.15 cm3/s through the needle. The fluid has 900 kg/m3 and 0.002 kg/(m s). What force F is re-quired to maintain the flow?
P6.31 SAE 10 oil at 20°C flows in a vertical pipe of diameter 2.5 cm. It is found that the pressure is constant through-out the fluid. What is the oil flow rate in m3/h? Is the flow up or down?
1.5 cm 3 cm Q F D1 = 0.25 mm D2 = 1 cm P6.33 2 mm 1 m 50 cm 40 cm 80 cm P6.34 For the system in Fig. P6.33, if the fluid has density of 920 kg/m3 and the flow rate is unknown, for what value of vis-cosity will the capillary Reynolds number exactly equal the critical value of 2300?
P6.35 Let us attack Prob. 6.33 in symbolic fashion, using Fig.
P6.35. 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 t0 to drain the upper tank completely. Assume quasi-steady laminar flow.
P6.36 For straightening and smoothing an airflow in a 50-cm-di-ameter duct, the duct is packed with a “honeycomb” of thin straws of length 30 cm and diameter 4 mm, as in Fig.
P6.36. The inlet flow is air at 110 kPa and 20°C, moving at an average velocity of 6 m/s. Estimate the pressure drop across the honeycomb.
P6.37 Oil, with 880 kg/m3 and 0.08 kg/(m s), flows through a pipe with d 2 cm and L 12 m. The wall shear stress is 30 Pa. Estimate (a) the Reynolds number, (b) the total pressure drop, and (c) the power required to drive the fluid.
P6.38 SAE 10 oil at 20°C flows through the 4-cm-diameter ver-tical pipe of Fig. P6.38. For the mercury manometer read-ing h 42 cm shown, (a) calculate the volume flow rate in m3/h and (b) state the direction of flow.
P6.39 Light oil, 880 kg/m3 and 0.015 kg/(m s), flows down a vertical 6-mm-diameter tube due to gravity only.
Estimate the volume flow rate in m3/h if (a) L 1 m and (b) L 2 m. (c) Verify that the flow is laminar.
P6.40 SAE 30 oil at 20°C flows in the 3-cm-diameter pipe in Fig. P6.40, which slopes at 37°. For the pressure mea-surements shown, determine (a) whether the flow is up or down and (b) the flow rate in m3/h.
P6.41 In Prob. 6.40 suppose it is desired to add a pump between A and B to drive the oil upward from A to B at a rate of 3 kg/s. At 100 percent efficiency, what pump power is re-quired?
P6.42 It is clear by comparing Figs. 6.12b and 6.13 that the ef-fects of sand roughness and commercial (manufactured) roughness are not quite the same. Take the special case of commercial roughness ratio /d 0.001 in Fig. 6.13, and replot in the form of the wall-law shift B (Fig. 6.12a) versus the logarithm of u/. Compare your plot with Eq. (6.61).
P6.43 Water at 20°C flows for 1 mi through a 3-in-diameter hor-izontal wrought-iron pipe at 250 gal/min. Estimate the head loss and the pressure drop in this length of pipe.
P6.44 Mercury at 20°C flows through 4 m of 7-mm-diameter glass tubing 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 m2/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.
408 Chapter 6 Viscous Flow in Ducts ρ, µ D Z(t) H d h L SAE 10 oil Mercury D = 4 cm 3 m 42 cm P6.38 pB = 180 kPa 20 m 37° pA = 500 kPa 15 m P6.40 P6.35 6 m/s Thousands of straws 50 cm 30 cm P6.36 EES P6.46 Kerosine at 20°C is pumped at 0.15 m3/s through 20 km of 16-cm-diameter cast-iron horizontal pipe. Compute the input power in kW required if the pumps are 85 percent efficient.
P6.47 Derive Eq. (6.59), showing all steps. The constant 1.33 dates back to Prandtl’s work in 1935 and may change slightly to 1.29 in your analysis.
P6.48 Show that if Eq. (6.49) is accurate, the position in a tur-bulent pipe flow where local velocity u equals average ve-locity V occurs exactly at r 0.777R, independent of the Reynolds number.
P6.49 The tank-pipe system of Fig. P6.49 is to deliver at least 11 m3/h of water at 20°C to the reservoir. What is the maxi-mum roughness height allowable for the pipe?
P6.54 In Fig. P6.52 suppose that the fluid is carbon tetrachloride at 20°C and p1 1100 kPa gage. What pipe diameter, in cm, is required to deliver a flow rate of 25 m3/h?
P6.55 The reservoirs in Fig. P6.55 contain water at 20°C. If the pipe is smooth with L 4500 m and d 4 cm, what will the flow rate in m3/h be for z 100 m?
Problems 409 30 m 60 m 80 m 10 m Smooth pipe: d = 5 cm Q Open jet p1 P6.52 L = 5 m, d = 3 cm 4 m 2 m Water at 20°C P6.49 P6.50 Ethanol at 20°C flows at 125 U.S. gal/min through a hor-izontal cast-iron pipe with L 12 m and d 5 cm. Ne-glecting entrance effects, estimate (a) the pressure gradi-ent dp/dx, (b) the wall shear stress w, and (c) the percent 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 per-cent of the pipe diameter and therefore very difficult to probe with a finite-sized instrument. In an effort to gener-ate a thick sublayer for probing, Pennsylvania State Uni-versity in 1964 built a pipe with a flow of glycerin. As-sume 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 effi-ciency if L 40 ft.
P6.52 The pipe flow in Fig. P6.52 is driven by pressurized air in the tank. What gage pressure p1 is needed to provide a 20°C water flow rate Q 60 m3/h?
P6.53 In Fig. P6.52 suppose P1 700 kPa and the fluid specific gravity is 0.68. If the flow rate is 27 m3/h, estimate the viscosity of the fluid. What fluid in Table A.5 is the likely suspect?
1 2 B L, D, ∆ z ε P6.55 P6.56 Consider a horizontal 4-ft-diameter galvanized-iron pipe simulating the Alaskan pipeline. The oil flow is 70 million U.S. gallons per day, at a density of 910 kg/m3 and vis-cosity of 0.01 kg/(m s) (see Fig. A.1 for SAE 30 oil at 100°C). Each pump along the line raises the oil pressure to 8 MPa, which then drops, due to head loss, to 400-kPa at the entrance to the next pump. Estimate (a) the appro-priate distance between pumping stations and (b) the power required if the pumps are 88 percent efficient.
P6.57 John Laufer (NACA Tech. Rep. 1174, 1954) gave velocity data for 20°C airflow in a smooth 24.7-cm-diameter pipe at Re 5 E5: u/uCL 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 The centerline velocity uCL 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. (6.59).
P6.58 In Fig. P6.55 assume that the pipe is cast iron with L 550 m, d 7 cm, and Z 100 m. If an 80 percent efficient EES pump is placed at point B, what input power is required to deliver 160 m3/h of water upward from reservoir 2 to 1?
P6.59 The following data were obtained for flow of 20°C water at 20 m3/h through a badly corroded 5-cm-diameter pipe which slopes downward at an angle of 8°: p1 420 kPa, z1 12 m, p2 250 kPa, z2 3 m. Estimate (a) the roughness ratio of the pipe and (b) the percent change in head loss if the pipe were smooth and the flow rate the same.
P6.60 J. Nikuradse in 1932 suggested that smooth-wall turbulent pipe flow could be approximated by a power-law profile u u CL R y 1/N where y is distance from the wall and N 6 to 9. Find the best value of N which fits Laufer’s data in Prob. 6.57. Then use your formula to estimate the pipe volume flow, and compare with the measured value of 45 ft3/s.
P6.61 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 m3/h.
What happens if the pipe length is 2 km?
P6.62 The Moody chart, Fig. 6.13, is best for finding head loss (or p) when Q, V, d, and L are known. It is awkward for the second type of problem, finding Q when hf or p is known (see Example 6.9). Prepare a modified Moody chart whose abscissa is independent of Q and V, using /d as a parameter, from which one can immediately read the or-dinate to find (dimensionless) Q or V. Use your chart to solve Example 6.9.
P6.63 A tank contains 1 m3 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 m3/h at this instant.
P6.64 For the system in Fig. P6.63, solve for the flow rate in m3/h if the fluid is SAE 10 oil at 20°C. Is the flow lami-nar or turbulent?
P6.65 In Prob. 6.63 the initial flow is turbulent. As the water 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.
P6.66 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 at r 1 cm.
P6.67 What level h must be maintained in Fig. P6.67 to deliver a flow rate of 0.015 ft3/s through the 1 2 -in commercial-steel pipe?
410 Chapter 6 Viscous Flow in Ducts Q 1 m 1 m3 L = 80 cm D = 4 cm P6.63 Water at 20°C h L = 80 ft D = 1 2 in P6.67 P6.68 Water at 20°C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as shown in Fig.
P6.68. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient, what horsepower pump is needed?
Pump L = 2000 ft 2 1 120 ft P6.68 P6.69 For Prob. 6.68 suppose the only pump available can de-liver 80 hp to the fluid. What is the proper pipe size in inches to maintain the 3 ft3/s flow rate?
P6.70 In Fig. P6.68 suppose the pipe is 6-in-diameter cast iron and the pump delivers 75 hp to the flow. What flow rate Q ft3/s results?
P6.71 It is desired to solve Prob. 6.68 for the most economical pump and cast-iron pipe system. If the pump costs $125 per 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 ft3/s flow rate? Make some simplifying assumptions.
P6.72 The 5-m-long pipe in Fig. P6.72 may be oriented at any angle . If entrance losses are negligible for any (pure Moody friction loss), what is the optimum value of for which the jet height loss h is minimum?
cost of the piping is $75 per centimeter of diameter. The power generated may be sold for $0.08 per kilowatthour.
Find the proper pipe diameter for minimum payback time, i.e., minimum time for which the power sales will equal the initial cost of the system.
P6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m3/h, if the fluid is water at 20°C. Which way is the flow?
Problems 411 10 m Water at 20°C θ D = 3 cm, L = 5 m, cast iron ∆ h 200 kPa gage L = 50 m 15 m P6.78 P6.72 P6.73 The Moody chart, Fig. 6.13, is best for finding head loss (or p) when Q, V, d, and L are known. It is awkward for the third type of problem, finding d when hf (or p) and Q are known (see Example 6.11). Prepare a modified Moody chart whose abscissa is independent of d, using as a parameter nondimensionalized without d, from which one can immediately read the (dimensionless) ordinate to find d. Use your chart to solve Example 6.11.
P6.74 In Fig. P6.67 suppose the fluid is gasoline at 20°C and h 90 ft. What commercial-steel pipe diameter is required for the flow rate to be 0.015 ft3/s?
P6.75 You wish to water your garden with 100 ft of 5 8 -in-diame-ter hose whose roughness is 0.011 in. What will be the de-livery, in ft3/s, if the gage pressure at the faucet is 60 lbf/in2? If there is no nozzle (just an open hose exit), what is the maximum horizontal distance the exit jet will carry?
P6.76 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 m3/h. Sketch the EGL and HGL accurately.
Q Water 20˚ C Turbine 30 m D = 4 cm 10 m D = 6 cm 20 m P6.76 P6.77 Modify Prob. 6.76 into an economic analysis, as follows.
Let the 40 m of wrought-iron pipe have a uniform diam-eter d. Let the steady water flow available be Q 30 m3/h.
The cost of the turbine is $4 per watt developed, and the P6.79 A garden hose is to be used as the return line in a water-fall 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 some-thing supplied by the hose manufacturer. So you devise a simple experiment to measure the roughness. The hose is attached to the drain of an aboveground 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 2.0 104 m3/s for a hose that is 10.0 m long and has an inside diameter of 1.50 cm. Estimate the rough-ness height in mm inside the hose.
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 m3/s?
80 m hp 0 Pump performance Parabola Q 2m3/s P6.80 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 m3/s. (Note that the pump head is now in meters of gasoline.) P6.82 The pump in Fig. P6.80 has its maximum efficiency at a head of 45 m. If it is used to pump ethanol at 20°C through 200 m of commercial-steel pipe, what is the proper pipe diameter for maximum pump efficiency?
P6.83 For the system of Fig. P6.55, let z 80 m and L 185 m of cast-iron pipe. What is the pipe diameter for which the flow rate will be 7 m3/h?
P6.84 It is desired to deliver 60 m3/h of water at 20°C through a horizontal asphalted cast-iron pipe. Estimate the pipe di-ameter which will cause the pressure drop to be exactly 40 kPa per 100 m of pipe length.
P6.85 The pump of Fig. P6.80 is used to deliver 0.7 m3/s of methanol at 20°C through 95 m of cast-iron pipe. What is the proper pipe diameter?
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 1 2 in, connects two reservoirs which differ in surface height by 20 ft. Compute the flow rate in ft3/s through the annulus if the fluid is water at 20°C.
P6.88 Show that for laminar flow through an annulus of very small clearance the flow rate Q is approximately propor-tional to the cube of the clearance a b.
P6.89 An annulus of narrow clearance causes a very large pres-sure drop and is useful as an accurate measurement of vis-cosity. If a smooth annulus 1 m long with a 50 mm and b 49 mm carries an oil flow at 0.001 m3/s, what is the oil viscosity if the pressure drop is 250 kPa?
P6.90 A 90-ft-long sheet-steel duct carries air at approximately 20°C and 1 atm. The duct cross section is an equilateral triangle whose side measures 9 in. If a blower can supply 1 hp to the flow, what flow rate, in ft3/s, will result?
P6.91 Heat exchangers often consist of many triangular pas-sages. Typical is Fig. P6.91, with L 60 cm and an isosceles-triangle cross section of side length a 2 cm and included angle $ 80°. If the average velocity is V 2 m/s and the fluid is SAE 10 oil at 20°C, estimate the pressure drop.
P6.92 A large room uses a fan to draw in atmospheric air at 20°C through a 30-cm by 30-cm commercial-steel duct 12 m long, as in Fig. P6.92. Estimate (a) the air flow rate in m3/h if the room pressure is 10 Pa vacuum and (b) the room pressure if the flow rate is 1200 m3/h. Neglect minor losses.
412 Chapter 6 Viscous Flow in Ducts V β L a P6.91 12 m Room 30 cm by 30 cm patm Fan P6.92 P6.93 Modify Prob. 6.91 so that the angle $ is unknown. For SAE 10 oil at 20°C, if the pressure drop is 120 kPa and the flow rate is 4 m3/h, what is the proper value of the angle $, in degrees?
P6.94 As shown in Fig. P6.94, a multiduct cross section consists of seven 2-cm-diameter smooth thin tubes packed tightly in a hexagonal “bundle” within a single 6-cm-diameter tube. Air, at about 20°C and 1 atm, flows through this sys-tem at 150 m3/h. Estimate the pressure drop per meter.
D = 6 cm d = 2 cm P6.94 P6.95 Modify Prob. 6.94 as follows. Let the seven 2-cm tubes be solid rods, so that the air can only pass through the curved triangular cusped passages. Compute the pressure drop per meter of duct length.
P6.96 Hydrogen, at 20°C and approximately 1 atm, is to be pumped through a smooth rectangular duct 85 m long of aspect ratio 5:1. If the flow rate is 0.7 m3/s and the pres-sure drop is 80 Pa, what should the width and height of the duct cross section be?
P6.97 A wind tunnel has a wooden rectangular section 45 cm by 95 cm by 26 m long. It uses sea-level standard air at 31-m/s average velocity. Estimate the pressure drop and the power required if the fan efficiency is 75 percent.
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?
EES P6.99 Air, approximately at sea-level standard conditions, is to be delivered at 3 m3/s through a horizontal square com-mercial-steel duct. What are the appropriate duct dimen-sions if the pressure drop is not to exceed 90 Pa over a 100-m length?
P6.100 Repeat Prob. 6.92 by including minor losses due to a sharp-edged entrance, the exit into the room, and an open gate valve. If the room pressure is 10 Pa vacuum, by what per-centage is the air flow rate decreased from part (a) of Prob.
6.92?
P6.101 Repeat Prob. 6.67 by including losses due to a sharp en-trance and a fully open screwed swing-check valve. By what percentage is the required tank level h increased?
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-ra-dius elbows, 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 ft3/s.
if the surface of reservoir 1 is 45 ft higher than that of reservoir 2.
P6.104 Reconsider the air hockey table of Prob. 3.162 but with in-clusion of minor losses. The table is 3.0 6.0 ft in area, with 1 1 6 -in-diameter holes spaced every inch in a rectan-gular grid pattern (2592 holes total). The required jet speed from each hole is estimated to be Vjet 50 ft/s. Your job is to select an appropriate blower which will meet the re-quirements. Hint: Assume that the air is stagnant in the large volume of the manifold under the table surface, and assume sharp edge inlets at each hole. (a) Estimate the pressure rise (in lb/in2) required of the blower. (b) Com-pare your answer to the previous calculation in which mi-nor losses were ignored. Are minor losses significant in this application?
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 el-bows, 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 m3/s of water at 20°C into the reservoir?
Problems 413 H W L P6.98 20 ft 6˚ cone Pump P6.102 P6.103 The reservoirs in Fig. P6.103 are connected by cast-iron pipes joined abruptly, with sharp-edged entrance and exit.
Including minor losses, estimate the flow of water at 20°C 1 2 D = 2 in L = 20 ft D = 1 in L = 20 ft 1 in 2 in P6.103 1 Elevation 500 m Sharp exit Open globe 45˚ 45˚ P6.105 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 globe valve is fully open. If the mercury manometer shows a 7-in deflection, what is the flow rate in ft3/s?
P6.107 In Fig. P6.107 the pipe is galvanized iron. Estimate the percentage increase in the flow rate (a) if the pipe entrance is cut off flush with the wall and (b) if the butterfly valve is opened wide.
P6.108 Consider the flow system of Fig. P6.102, including the 6° cone diffuser. Suppose the pump head versus flow rate is approximated by hp 45 125Q2 with hp in ft and Q in ft3/s. Estimate the resulting flow rate, in ft3/s.
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° elbows and an open globe valve, all flanged. If the exit elevation is zero, what horsepower is extracted by the tur-bine when the flow rate is 0.16 ft3/s of water at 20°C?
P6.110 In Fig. P6.110 the pipe entrance is sharp-edged. If the flow rate is 0.004 m3/s, what power, in W, is extracted by the turbine?
414 Chapter 6 Viscous Flow in Ducts 7 in 10 ft Mercury Globe Open jet Water at 20˚C 5 m 6 cm Butterfly valve at 30˚ D = 5 cm, L = 2 m P6.106 P6.107 Open globe Turbine 3 in 6 in 2 in Elevation 100 ft P6.109 Water 40 m Turbine Open globe valve Cast iron: L = 125 m, D = 5 cm P6.110 P6.111 For the parallel-pipe system of Fig. P6.111, each pipe is cast iron, and the pressure drop p1 p2 3 lbf/in2. Com-pute the total flow rate between 1 and 2 if the fluid is SAE 10 oil at 20°C.
L = 200 ft 1 2 D = 3 in D = 2 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 lbf/in2, what will the flow rate be? The fluid is SAE 10 oil at 20°C.
P6.113 The parallel galvanized-iron pipe system of Fig. P6.113 delivers gasoline at 20°C with a total flow rate of 0.036 m3/s. If the pump is wide open and not running, with a loss coefficient K 1.5, determine (a) the flow rate in each pipe and (b) the overall pressure drop.
L1 = 60 m, D1 = 5 cm Q = 0.036 m3/s L2 = 55 m, D2 = 4 cm Pump P6.113 P6.114 Modify Prob. 6.113 as follows. Let the pump be running and delivering 45 kW to the flow in pipe 2. The fluid is gasoline at 20°C. Determine (a) the flow rate in each pipe and (b) the overall pressure drop.
P6.115 In Fig. P6.115 all pipes are 8-cm-diameter cast iron. De-termine the flow rate from reservoir 1 if valve C is (a) closed and (b) open, K 0.5.
EES 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 p1 p2 750 kPa, find the resulting flow rate Q m3/h for water at 20°C. Neglect minor losses.
Pipe Length, m Diameter, cm 1 800 12 2 600 8 3 900 10 The total flow rate is 200 m3/h of water at 20°C. Deter-mine (a) the flow rate in each pipe and (b) the pressure drop across the system.
P6.121 Consider the three-reservoir system of Fig. P6.121 with the following data: L1 95 m L2 125 m L3 160 m z1 25 m z2 115 m z3 85 m All pipes are 28-cm-diameter unfinished concrete ( 1 mm). Compute the steady flow rate in all pipes for water at 20°C.
Problems 415 1 2 Z = 25 m Water at 20°C L = 100 m A C B Valve L = 70 m Z = 0 m 10 m 30 m L = 50 m P6.115 1 2 L = 250 m 150 m 100 m P6.116 P6.117 Modify Prob. 6.116 as follows. Let the flow rate be 45 m3/h of water at 20°C. Determine the overall pressure drop p1 p2 in kPa. Neglect minor losses.
P6.118 For the piping system of Fig. P6.118, all pipes are con-crete with a roughness of 0.04 in. Neglecting minor losses, compute the overall pressure drop p1 p2 in lbf/in2 if Q 20 ft3/s. The fluid is water at 20°C.
D = 8 in L = 1500 ft 2 1 L = 1000 ft D = 12 in D = 15 in L = 1200 ft D = 12 in L = 800 ft P6.118 P6.119 Modify Prob. 6.118 as follows. Let the pressure drop p1 p2 be 98 lbf/in2. Neglecting minor losses, determine the flow rate in m3/h.
P6.120 Three cast-iron pipes are laid in parallel with these di-mensions: Z1 Z2 Z3 L1 L2 L3 P6.121 P6.122 Modify Prob. 6.121 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. 6.121 but are about 5.2 times lower. Can you explain this difference?
P6.123 Modify Prob. 6.121 as follows. All data are the same ex-cept that z3 is unknown. Find the value of z3 for which the flow rate in pipe 3 is 0.2 m3/s toward the junction. (This problem requires iteration and is best suited to a digital computer.) P6.124 The three-reservoir system in Fig. P6.124 delivers water at 20°C. The system data are as follows: D1 8 in D2 6 in D3 9 in L1 1800 ft L2 1200 ft L3 1600 ft All pipes are galvanized iron. Compute the flow rate in all pipes.
P6.125 Modify Prob. 6.124 as follows. Let all data be the same except z3, which is unknown. What value of z3 will cause the flow rate through pipe 3 to be 1.0 ft3/s toward the junc-tion?
EES P6.126 Modify Prob. 6.124 as follows. Let all data be the same except that pipe 1 is fitted with a butterfly valve (Fig.
6.19b). Estimate the proper valve opening angle (in de-grees) for the flow rate through pipe 1 to be reduced to 1.5 ft3/s toward reservoir 1. (This problem requires itera-tion and is best suited to a digital computer.) P6.127 In the five-pipe horizontal network of Fig. P6.127, assume that all pipes have a friction factor f 0.025. For the given inlet and exit flow rate of 2 ft3/s of water at 20°C, deter-mine the flow rate and direction in all pipes. If pA 120 lbf/in2 gage, determine the pressures at points B, C, and D.
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 lbf/in2 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 B, C, and D.
416 Chapter 6 Viscous Flow in Ducts 1 2 3 z1= 20 ft z2= 100 ft z3= 50 ft J P6.124 d = 8 in D A B 4000 ft 3000 ft 2 ft3/s 2 ft3/s 9 in 3 in C 6 in 8 in P6.127 P6.128 Modify Prob. 6.127 as follows. Let the inlet flow rate at A and the exit flow at D be unknown. Let pA pB 100 lbf/in2. 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, deliv-ering water at 20°C. The pressures are known at four points as shown: p1 950 kPa p2 350 kPa p3 675 kPa p4 100 kPa Neglecting minor losses, determine the flow rate in each pipe.
p1 L1 p2 L 2 L 3 a L 4 p 4 p 3 P6.129 C D A B 0.5 ft3/s 2.0 ft3/s 0.5 ft3/s 1.0 ft3/s P6.130 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 per-cent. 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 Prob. 6.131 suppose we are limited by space to a to-tal diffuser length of 10 m. What should the diffuser an-gle, exit diameter, and exit pressure be for maximum re-covery?
P6.133 A wind-tunnel test section is 3 ft square with flow prop-erties V 150 ft/s, p 15 lbf/in2 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 Prob. 6.133 suppose we are limited by space to a to-tal diffuser length of 30 ft. What should the diffuser angle, exit height, and exit pressure be for maximum recovery?
P6.135 A small airplane flying at 5000-m altitude uses a pitot stag-nation probe without static holes. The measured stagna-tion pressure is 56.5 kPa. Estimate the airplane’s speed in mi/h and its uncertainty. 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.
head probe and a static pressure probe, as shown in Fig.
P6.139, a distance h1 apart from each other. Both probes are in the main free stream of the water tunnel, unaffected by 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. (b) Is it critical that dis-tance h1 be measured accurately? (c) How does the ex-pression for velocity V differ from that which would be obtained if a pitot-static probe had been available and used with the same U-tube manometer?
P6.140 Kerosine at 20°C flows at 18 m3/h in a 5-cm-diameter pipe.
If a 2-cm-diameter thin-plate orifice with corner taps is in-stalled, what will the measured pressure drop be, in Pa?
P6.141 Gasoline at 20°C flows at 105 m3/h in a 10-cm-diameter pipe. We wish to meter the flow with a thin-plate orifice and a differential pressure transducer which reads best at about 55 kPa. What is the proper $ 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 up-stream pressure is constant at 400 kPa. What flow rate, in Problems 417 Air 8 cm 20°C 1 atm 40 mm P6.136 P6.137 For the 20°C water flow of Fig. P6.137, use the pitot-sta-tic 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?
2 in Mercury 1 ft P6.137 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 probe, as in Fig. P6.138, thus contami-nating the pitot measurement ridiculously with pipe fric-tion losses. If the pipe flow is air at 20°C and 1 atm and the manometer fluid is Meriam red oil (SG 0.827), es-timate the air centerline velocity for the given manometer reading of 16 cm. Assume a smooth-walled tube.
P6.139 Professor Walter Tunnel needs to measure the flow veloc-ity in a water tunnel. Due to budgetary restrictions, he can-not afford a pitot-static probe, but instead inserts a total D = 6 cm 16 cm 10 m Air P6.138 h1 h2 h3 U-tube manometer ptotal pstatic w m V P6.139 EES gal/min, results without the reducer? What reducer orifice diameter would decrease the flow by 40 percent?
P6.143 A 10-cm-diameter smooth pipe contains an orifice plate with D: 1 2 D taps and $ 0.5. The measured orifice pres-sure drop is 75 kPa for water flow at 20°C. Estimate the flow rate, in m3/h. What is the nonrecoverable head loss?
P6.144 Accurate solution of Prob. 6.143, using Fig. 6.40, requires iteration because both the ordinate and the abscissa of this figure contain the unknown flow rate Q. In the spirit of Example 5.8, rescale the variables and construct a new plot in which Q may be read directly from the ordinate. Solve Prob. 6.143 with your new chart.
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.
proximated by thin-plate orifices. Hint: A momentum con-trol volume may be very useful.
P6.148 A smooth pipe containing ethanol at 20°C flows at 7 m3/h through a Bernoulli obstruction, as in Fig. P6.148. Three piezometer tubes are installed, as shown. If the obstruc-tion is a thin-plate orifice, estimate the piezometer levels (a) h2 and (b) h3.
418 Chapter 6 Viscous Flow in Ducts p = 400 kPa 45 holes, 1.5-mm diameter Flow reducer D = 1.5 cm 1 m h (0) = 2 m Q (t ) h (t ) P6.145 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.
P6.147 Air flows through a 6-cm-diameter smooth pipe which has a 2-m-long perforated section containing 500 holes (di-ameter 1 mm), as in Fig. P6.147. Pressure outside the pipe is sea-level standard air. If p1 105 kPa and Q1 110 m3/h, estimate p2 and Q2, assuming that the holes are ap-20 m 3-cm orifice L = 100 m D = 5 cm D = 6 cm 2 m 500 holes (diameter 1 mm) 1 2 P6.146 P6.147 D = 5 cm 5 m d = 3 cm h3 h2 h1= 1 m P6.148 P6.149 Repeat Prob. 6.148 if the obstruction is a long-radius flow nozzle.
P6.150 Gasoline at 20°C flows at 0.06 m3/s through a 15-cm pipe and is metered by a 9-cm long-radius flow nozzle (Fig.
6.39a). What is the expected pressure drop across the noz-zle?
P6.151 Ethyl alcohol at 20°C flowing in a 6-cm-diameter pipe is metered through a 3-cm long-radius flow nozzle. If the measured pressure drop is 45 kPa, what is the estimated volume flow, in m3/h?
P6.152 Kerosine at 20°C flows at 20 m3/h in an 8-cm-diameter pipe. The flow is to be metered by an ISA 1932 flow noz-zle so that the pressure drop is 7000 Pa. What is the proper nozzle diameter?
P6.142 P6.153 Two water tanks, each with base area of 1 ft2, are con-nected by a 0.5-in-diameter long-radius nozzle as in Fig.
P6.153. If h 1 ft as shown for t 0, estimate the time for h(t) to drop to 0.25 ft.
venturi with a 4-cm throat. The venturi is connected to a mercury manometer whose reading is h 40 cm. Estimate (a) the flow rate, in m3/h, and (b) the total pressure dif-ference between points 50 cm upstream and 50 cm down-stream of the venturi.
P6.159 A modern venturi nozzle is tested in a laboratory flow with water at 20°C. The pipe diameter is 5.5 cm, and the ven-turi 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: Use these data to plot a calibration curve of venturi dis-charge coefficient versus Reynolds number. Compare with the accepted correlation, Eq. (6.134).
P6.160 The butterfly-valve losses in Fig. 6.19b may be viewed as a type of Bernoulli obstruction device, as in Fig. 6.38. The “throat area” At in Eq. (6.125) can be interpreted as the two slivers of opening around the butterfly disk when viewed from upstream. First fit the average loss Kmean ver-sus the opening angle in Fig. 6.19b to an exponential curve.
Then use your curve fit to compute the “discharge coeffi-cient” of a butterfly valve as a function of the opening an-gle. Plot the results and compare them to those for a typ-ical flowmeter.
Problems 419 d = in 1 2 1 ft 2 1 ft 2 h = 1ft 2 ft P6.153 Water d h 5 cm Mercury P6.154 P6.154 Water at 20°C flows through the orifice in Fig. P6.154, which is monitored by a mercury manometer. If d 3 cm, (a) what is h when the flow rate is 20 m3/h and (b) what is Q in m3/h when h 58 cm?
P6.155 It is desired to meter a flow of 20°C gasoline in a 12-cm-diameter pipe, using a modern venturi nozzle. In order for international standards to be valid (Fig. 6.42), what is the permissible range of (a) flow rates, (b) nozzle diameters, and (c) pressure drops? (d) For the highest pressure-drop condition, would compressibility be a problem?
P6.156 Ethanol at 20°C flows down through a modern 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 Modify Prob. 6.156 if the fluid is air at 20°C, entering the venturi at a pressure of 18 lbf/in2. Should a compressibil-ity correction be used?
P6.158 Water at 20°C flows in a long horizontal commercial-steel 6-cm-diameter pipe which contains a classical Herschel d = 3 in 4 in D = 6 in 9 in P6.156 m ˙, kg/s 0.95 1.98 2.99 5.06 8.15 h, mm 3.7 15.9 36.2 102.4 264.4 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 tur-bulent flow near a rough surface which 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 (r) varies linearly from zero at the axis to w at the wall. It is claimed that this is also true, at least in the time mean, for fully developed tur-bulent 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 pres-sure gradient in a porous medium was directly proportional to the volume-averaged velocity V of the fluid: p K V where K is termed the permeability of the medium. This is now called Darcy’s law of porous flow. Can you make a Poiseuille flow model of porous-media flow which verifies Darcy’s law? Meanwhile, as the Reynolds number increases, so that VK1/2/ 1, the pressure drop becomes nonlinear, as was shown experimentally by P. H. Forscheimer as early as 1782. The flow is still decidedly laminar, yet the pres-sure gradient is quadratic: p K V CVV Darcy-Forscheimer law where C is an empirical constant. Can you explain the rea-son for this nonlinear behavior?
W6.5 One flowmeter device, in wide use in the water supply and gasoline distribution industries, is the nutating disk. Look this up in the library, and explain in a brief report how it works and the advantages and disadvantages of typical de-signs.
420 Chapter 6 Viscous Flow in Ducts 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) 2100, (c) 4000, (d) 250,000, (e) 500,000 FE6.2 For flow of water at 20°C through a straight, smooth pipe at 0.06 m3/h, the pipe diameter for which transition to tur-bulence occurs is approximately (a) 1.0 cm, (b) 1.5 cm, (c) 2.0 cm, (d) 2.5 cm, (e) 3.0 cm FE6.3 For flow of oil [ 0.1 kg/(m s), SG 0.9] through a long, straight, smooth 5-cm-diameter pipe at 14 m3/h, the pressure drop per meter is approximately (a) 2200 Pa, (b) 2500 Pa, (c) 10,000 Pa, (d) 160 Pa, (e) 2800 Pa FE6.4 For flow of water at a Reynolds number of 1.03 E6 through a 5-cm-diameter pipe of roughness height 0.5 mm, the ap-proximate Moody friction factor is (a) 0.012, (b) 0.018, (c) 0.038, (d) 0.049, (e) 0.102 FE6.5 Minor losses through valves, fittings, bends, contractions, etc., are commonly modeled as proportional to (a) total head, (b) static head, (c) velocity head, (d) pres-sure drop, (e) velocity FE6.6 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 with its surface elevation at 560 m. If minor losses are neglected, the ex-pected flow rate through the pipe is (a) 0.048 m3/h, (b) 2.87 m3/h, (c) 134 m3/h, (d) 172 m3/h, (e) 385 m3/h FE6.7 If, in Prob. FE6.6 the pipe is rough and the actual flow rate is 90 m3/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 FE6.8 Suppose in Prob. FE6.6 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 (a) 90 m3/h, (b) 579 m3/h, (c) 748 m3/h, (d) 949 m3/h, (e) 1048 m3/h FE6.9 Oil [ 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 (a) 30 kPa, (b) 40 kPa, (c) 50 kPa, (d) 60 kPa, (e) 70 kPa FE6.10 Air at 20°C and approximately 1 atm flows through a smooth 30-cm-square duct at 1500 ft3/min. The expected pressure drop per meter of duct length is (a) 1.0 Pa, (b) 2.0 Pa, (c) 3.0 Pa, (d) 4.0 Pa, (e) 5.0 Pa FE6.11 Water at 20°C flows at 3 m3/h through a sharp-edged 3-cm-diameter orifice in a 6-cm-diameter pipe. Estimate the expected pressure drop across the orifice.
(a) 440 Pa, (b) 680 Pa, (c) 875 Pa, (d) 1750 Pa, (e) 1870 Pa FE6.12 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?
(a) 0.015, (b) 0.017, (c) 0.019, (d) 0.026, (e) 0.032 FE6.13 What is the hydraulic diameter of a rectangular air-venti-lation duct whose cross section is 1 m by 25 cm?
(a) 25 cm, (b) 40 cm, (c) 50 cm, (d) 75 cm, (e) 100 cm FE6.14 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 FE6.15 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?
(a) 0.010, (b) 0.015, (c) 0.020, (d) 0.025, (e) 0.030 Comprehensive Problems 421 C6.2 Comprehensive Problems C6.1 A pitot-static probe will be used to measure the velocity dis-tribution in a water tunnel at 20°C. The two pressure lines from the probe will be connected to a U-tube manometer which uses a liquid of specific 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 which 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 (, ) from a large tank to two other higher-elevation tanks, as shown in Fig.
C6.2. The same pipe of diameter d and roughness is used throughout. All minor losses except through the valve are ne-glected, and the partially closed valve has a loss coefficient Kvalve. Turbulent flow may be assumed with all kinetic en-ergy flux correction coefficients equal to 1.06. The pump net head H is a known function of QA and hence also of VA QA/Apipe; for example, H a bVA 2, where a and b are con-stants. Subscript J refers to the junction point at the tee where branch A splits into B and C. Pipe length LC is much longer than LB. 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, VB, VC, fA, fB, fC, pJ. Write down the eight equations needed to resolve this problem, but do not solve, since an elaborate iteration procedure, or an equation solver such as EES, 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 contin-uous water flow rate Q of 1.39 103 m3/s (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 supplying swimming pool wa-ter for the slide. The pump is 80 percent efficient and will rest fully submerged 1.00 m below the water surface. The rough-ness 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 at-mosphere. The hose outlet is 4.00 m above the water surface.
For fully developed turbulent pipe flow, the kinetic energy flux correction factor is about 1.06. Ignore any minor losses here.
Assume that 998 kg/m3 and 1.00 106 m2/s for this water. Find the brake horsepower (i.e., the actual shaft power in watts) required to drive the pump.
Large tank Large tank Branch A, LA Branch C, LC Branch B, LB Pump VA VB VC Large tank Valve 1 2 3 J C6.4 Suppose you build a house out in the “boonies” where you need to run a pipe to the nearest water supply, which is for-tunately at an elevation of about 1000 m above that of your house. The pipe will be 6.0 km long (the distance to the wa-ter supply), 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, etc., since the length is so long here and major losses dominate. Assume the out-let of the pipe is open to the atmosphere. (b) Which is more important in this problem, the head loss due to elevation dif-ference 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 9.4 104 m3/s, 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 commercial steel. Both ducts are the same length. In the cross sections shown in Fig. C6.5 R 15.0 mm and a 25.0 mm.
(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, as-suming 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 com-pare. (c) If the losses are different for the two cases, explain why. Which duct, if any, is more “efficient”?
422 Chapter 6 Viscous Flow in Ducts Q Pump Sliding board Tube 4.00 m Weee!
Ladder Water 1.00 m C6.3 C6.5 Design Projects D6.1 A hydroponic garden uses the 10-m-long perforated-pipe sys-tem in Fig. D6.1 to deliver water at 20°C. The pipe is 5 cm in diameter and contains a circular hole every 20 cm. A pump delivers water at 75 kPa (gage) at the entrance, while the other end of the pipe is closed. If you attempted, e.g., Prob. 3.125, you know that the pressure near the closed end of a perforated R A b a “manifold’’ is surprisingly high, and there will be too much flow through the holes near that end. One remedy is to vary the hole size along the pipe axis. Make a design analysis, per-haps using a personal computer, to pick the optimum hole-size distribution that will make the discharge flow rate as uniform as possible along the pipe axis. You are constrained to pick hole sizes that correspond only to commercial (numbered) met-ric drill-bit sizes available to the typical machine shop.
D6.2 It is desired to design a pump-piping system to keep a 1-mil-lion-gallon capacity water tank filled. The plan is to use a modified (in size and speed) version of the model 1206 cen-trifugal pump manufactured by Taco Inc., Cranston, Rhode Island. Test data have been provided to us by Taco Inc. for a small model of this pump: D 5.45 in, # 1760 r/min, tested with water at 20°C: Q, gal/min 0 5 10 15 20 25 30 35 40 45 50 55 60 H, ft 28 28 29 29 28 28 27 26 25 23 21 18 15 Efficiency, % 0 13 25 35 44 48 51 53 54 55 53 50 45 The tank is to be filled daily with rather chilly (10°C) ground-water from an aquifer, which is 0.8 mi from the tank and 150 ft lower than the tank. Estimated daily water use is 1.5 million gal/day. Filling time should not exceed 8 h per day.
The piping system should have four “butterfly” valves with variable openings (see Fig. 6.19), 10 elbows of various an-gles, and galvanized-iron pipe of a size to be selected in the design. The design should be economical—both in capital costs and operating expense. Taco Inc. has provided the fol-lowing cost estimates for system components: Pump and motor $3500 plus $1500 per inch of im-peller size Pump speed Between 900 and 1800 r/min Valves $300 $200 per inch of pipe size Elbows $50 plus $50 per inch of pipe size Pipes $1 per inch of diameter per foot of length Electricity cost 10¢ per kilowatthour Your design task is to select an economical pipe size and pump impeller size and speed for this task, using the pump-test data in nondimensional form (see Prob. 5.61) as design data. Write a brief report (5 to 6 pages) showing your cal-culations and graphs.
References 423 10 m 20 cm Pump D6.1 References 1.
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P. W. Runstadler, Jr., et al., “Diffuser Data Book,” Creare Inc.
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R. W. Jepson, Analysis of Flow in Pipe Networks, Ann Ar-bor Pub., Ann Arbor, MI, 1976.
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L. E. Ormsbee and D. J. Wood, “Explicit Pipeline Network Calibration,” J. Water Resources Planning and Management, vol. 112, no. 2, April 1986, pp. 166–182.
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84, 1962, pp. 303–312.
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424 Chapter 6 Viscous Flow in Ducts 426 The turbulent wake behind a bluff body immersed in a stream flow is a subject of the present chapter. This is a digitized video image showing the distribution of tracer-dye concentration in the wake of the body. Compare with Fig. 5.2a of the text, which is a laminar wake. (Courtesy of R. Balachandar, by permission of the American Society of Mechanical Engineers.) 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 lay-ers grow. Although boundary-layer theory (Sec. 7.3) is helpful in understanding exter-nal flows, complex body geometries usually require experimental data on the forces and moments caused by the flow. Such immersed-body flows are commonly encoun-tered in engineering studies: aerodynamics (airplanes, rockets, projectiles), hydrody-namics (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 UL/ is low (Fig. 7.1a), the viscous region is very broad and ex-tends 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 dis-tribution along the plate. Thus, although in principle it should be possible to patch the 427 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.
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 digital com-puter .
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 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 for-mulas for flat-plate flow are x laminar (7.1a) turbulent (7.1b) 0.16 Rex 1/7 5.0 Rex 1/2 428 Chapter 7 Flow Past Immersed Bodies Large viscous displacement effect ReL = 10 U x L Viscous region Inviscid region U U u < U u = 0.99U δ ≈ L x U U u < U Viscous Inviscid region δ L Laminar BL Turbulent BL Small displacement effect ReL = 107 U (a) (b) Fig. 7.2 Illustration of the strong interaction between viscous and in-viscid regions in the rear of blunt-body flow: (a) idealized and defi-nitely false picture of blunt-body flow; (b) actual picture of blunt-body flow.
where Rex Ux/ is called the local Reynolds number of the flow along the plate sur-face. The turbulent-flow formula applies for Rex greater than approximately 106.
Some computed values from Eq. (7.1) are Rex 104 105 106 107 108 (/x)lam 0.050 0.016 0.005 (/x)turb 0.022 0.016 0.011 The blanks indicate that the formula is not applicable. In all cases these boundary lay-ers 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 pressure field 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 dis-crepancy in the viscous-inviscid patching concept. Figure 7.2 shows two sketches of flow past a two- or three-dimensional blunt body. In the idealized sketch (7.2a), 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 7.1 Reynolds-Number and Geometry Effects 429 (a) (b) Thin front boundary layer Beautifully behaved but mythically thin boundary layer and wake Outer stream grossly perturbed by broad flow separation and wake Red = 10 5 Red = 10 5 actual flow (Fig. 7.2b), 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 separates, 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. 7.2b are normally studied experimentally.
Reference 5 is an example of efforts to improve the theory of separated-boundary-layer flows. Reference 6 is a 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 20°C. At what distance x from the leading edge will the boundary-layer thickness be 1 in?
Solution Since we do not know the Reynolds number, we must guess which of Eqs. (7.1) applies. From Table 1.4 for water, 1.09 105 ft2/s; hence U 1.09 20 1 f 0 t /s 5 ft2/s 1.84 106 ft1 With 1 in 1 1 2 ft, try Eq. (7.1a): Laminar flow: x (Ux/ 5 )1/2 or x 2( 5 U 2 /v) 511 ft Now we can test the Reynolds number to see whether the formula applied: Rex U x 1 ( .
2 0 0 9 f t/s 1 ) 0 (5 1 5 1 ft f 2 t) /s 9.4 108 This is impossible since the maximum Rex for laminar flow past a flat plate is 3 106. So we try again with Eq. (7.1b): Turbulent flow: x (U 0 x .
/ 1 6 )1/7 or x 7/6 7/6 (4.09)7/6 5.17 ft Ans.
Test Rex 1 ( .
2 0 0 9 f t/s 1 )( 0 5 .1 5 7 ft f 2 t / ) s 9.5 106 This is a perfectly proper turbulent-flow condition; hence we have found the correct position x on our second try.
( 1 1 2 ft)(1.84 106 ft1)1/7 0.16 (U/v)1/7 0.16 ( 1 1 2 ft)2(1.84 106 ft1) 25 430 Chapter 7 Flow Past Immersed Bodies 7.2 Momentum-Integral Estimates Kármán’s Analysis of the Flat Plate Fig. 7.3 Growth of a boundary layer on a 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 Fig. 7.3.
A shear layer of unknown thickness grows along the sharp flat plate in Fig. 7.3. The no-slip wall condition retards the flow, making it into a rounded profile u(y), which merges into the external velocity U constant at a “thickness” y (x). By utilizing the control volume of Fig. 3.11, we found (without making any assumptions about lam-inar 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 D(x) b (x) 0 u(U u) dy (7.2) where b is the plate width into the paper and the integration is carried out along a ver-tical 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 Kármán , who wrote it in the convenient form of the momentum thickness D(x) bU2 0 U u 1 U u dy (7.3) Momentum thickness is thus a measure of total plate drag. Kármán then noted that the drag also equals the integrated wall shear stress along the plate D(x) b x 0 w(x) dx or d d D x b w (7.4) Meanwhile, the derivative of Eq. (7.3), with U constant, is d d D x bU2 d d x By comparing this with Eq. (7.4) Kármán arrived at what is now called the momentum-integral relation for flat-plate boundary-layer flow 7.2 Momentum-Integral Estimates 431 x y U x = 0 p = pa U w (x) x = L u (x, y) U (x) δ τ w U2 d d x (7.5) It is valid for either laminar or turbulent flat-plate flow.
To get a numerical result for laminar flow, Kármán assumed that the velocity pro-files had an approximately parabolic shape u(x, y) U 2 y y2 2 0 y (x) (7.6) which makes it possible to estimate both momentum thickness and wall shear 0 2 y y2 2 1 2 y y2 2 dy 1 2 5 (7.7) w u y y0 2 U By substituting (7.7) into (7.5) and rearranging we obtain d 15 U dx (7.8) where /. We can integrate from 0 to x, assuming that 0 at x 0, the lead-ing edge 1 2 2 15 U x or x 5.5 U x 1/2 (7.9) This is the desired thickness estimate. It is all approximate, of course, part of Kár-mán’s momentum-integral theory , but it is startlingly accurate, being only 10 per-cent higher than the known exact 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 cf 2 U w 2 1/2 (7.10) Again this estimate, in spite of the crudeness of the profile assumption (7.6) is only 10 percent higher than the known exact laminar-plate-flow solution cf 0.664/Rex 1/2, treated in Sec. 7.4. The dimensionless quantity cf, called the skin-friction coefficient, is analogous to the friction factor f in ducts.
A boundary layer can be judged as “thin” if, say, the ratio /x is less than about 0.1.
This occurs at /x 0.1 5.0/Rex 1/2 or at Rex 2500. For Rex 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 Rex for laminar flow is about 3 106, where measurements on a smooth flat plate show that the flow undergoes transition to a turbulent boundary layer. From 3 106 upward the turbulent Reynolds number may be arbitrarily large, and a practical limit at present is 5 109 for oil supertankers.
0.73 Rex 1/2 1 8 5 Rex 5.5 Rex 1/2 432 Chapter 7 Flow Past Immersed Bodies Displacement Thickness Fig. 7.4 Displacement effect of a boundary layer.
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 (x) to satisfy conservation of mass between the inlet and outlet h 0 Ub dy 0 ub dy h (7.11) The quantity is called the displacement thickness of the boundary layer. To relate it to u(y), cancel and b from Eq. (7.11), evaluate the left integral, and slyly add and subtract U from the right integrand: Uh 0 (U u U) dy U(h ) 0 (u U) dy or 0 1 U u dy (7.12) Thus the ratio of / varies only with the dimensionless velocity-profile shape u/U.
Introducing our profile approximation (7.6) into (7.12), we obtain by integration the approximate result 1 3 x (7.13) These estimates are only 6 percent away from the exact solutions for laminar flat-plate flow given in Sec. 7.4: 0.344 1.721x/Rex 1/2. Since is much smaller than x for large Rex and the outer streamline slope V/U is proportional to , 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 Kár-mán’s momentum-integral theory is effective and useful. Many details of this theory are given in Refs. 1 to 3.
EXAMPLE 7.2 Are low-speed, small-scale air and water boundary layers really thin? Consider flow at U 1 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 20°C.
1.83 Rex 1/2 7.2 Momentum-Integral Estimates 433 y U U Simulated effect 0 y = h x Outer streamline y = h + U u δ h h δ Part (a) Part (b) 7.3 The Boundary-Layer Equations Solution From Table A.3, air 1.61 E-4 ft2/s. The trailing-edge Reynolds number thus is ReL U L 1 ( .
1 61 ft E /s -) 4 (1 ft f 2 t / ) s 6200 Since this is less than 106, 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 0.0634 or, at x 1 ft, 0.0634 ft 0.76 in Ans. (a) From Table A.2 water 1.08 E-5 ft2/s. The trailing-edge Reynolds number is ReL 92,600 This again satisfies the laminar and thinness conditions. The boundary-layer thickness is x 0.0164 or, at x 1 ft, 0.0164 ft 0.20 in Ans. (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 (digital-computer) solutions, (2) experimentation, and (3) boundary-layer theory.
Computational fluid dynamics (CFD) is now well developed and described in ad-vanced texts such as that by Anderson et al. . Thousands of computer solutions and models have been published; execution times, mesh sizes, and graphical pre-sentations 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.
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 which are solved relatively easily and patched onto the outer inviscid-flow field.
5.0 9 2 ,6 0 0 (1 ft/s)(1 ft) 1.08 E-5 ft2/s 5.0 6 2 0 0 x 434 Chapter 7 Flow Past Immersed Bodies Derivation for Two-Dimensional Flow One of the great achievements of boundary-layer theory is its ability to predict the flow separation illustrated in Fig. 7.2b. Before 1904 no one realized that such thin shear layers could cause such a gross effect as flow separation. Unfortunately, even today the-ory cannot accurately predict the behavior of the separated-flow region and its interac-tion with the outer layer. This is the weakness of boundary-layer theory, which we hope will be overcome by intensive research into the dynamics of separated flows .
We consider only steady two-dimensional incompressible viscous flow with the x di-rection along the wall and y normal to the wall, as in Fig. 7.3.1 We neglect gravity, which is important only in boundary layers where fluid buoyancy is dominant [2, sec.
4.13]. From Chap. 4, the complete equations of motion consist of continuity and the x- and y-momentum relations u x y 0 (7.14a) u u x u y p x 2 x u 2 2 y u 2 (7.14b) u x y p y 2 x 2 2 y 2 (7.14c) These should be solved for u, , 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.
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: u (7.15a) Rates of change: u x u y x y (7.15b) Our discussion of displacement thickness in the previous section was intended to jus-tify these assumptions.
Applying these approximations to Eq. (7.14c) results in a powerful simplification p y 0 or 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.
(7.14b) is assumed to be known in advance from Bernoulli’s equation applied to the outer inviscid flow p x d d p x U d d U x (7.17) 7.3 The Boundary-Layer Equations 435 1For 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].
7.4 The Flat-Plate Boundary Layer 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. (7.14b) is negligible due to Eqs. (7.15) 2 x u 2 2 y u 2 (7.18) However, neither term in the continuity relation (7.14a) 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 Continuity: u x y 0 (7.19a) Momentum along wall: u u x u y U d d U x 1 y (7.19b) where u y laminar flow u y u turbulent flow These are to be solved for u(x, y) and (x, y), with U(x) assumed to be a known func-tion from the outer inviscid-flow analysis. There are two boundary conditions on u and one on : At y 0 (wall): u 0 (no slip) (7.20a) At y (x) (outer 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) 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.2 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 con-fine ourselves primarily to flat-plate solutions (Sec. 7.4).
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.
436 Chapter 7 Flow Past Immersed Bodies 2For further mathematical details, see Ref. 2, sec. 2.8.
Laminar Flow For laminar flow past the plate, the boundary-layer equations (7.19) can be solved ex-actly for u and , assuming that the free-stream velocity U is constant (dU/dx 0).
The solution was given by Prandtl’s student Blasius, in his 1908 dissertation from Göt-tingen. With an ingenious coordinate transformation, Blasius showed that the dimen-sionless velocity profile u/U is a function only of the single composite dimensionless variable (y)[U/(x)]1/2: U u f() y v U x 1/2 (7.21) where the prime denotes differentiation with respect to . Substitution of (7.21) into the boundary-layer equations (7.19) reduces the problem, after much algebra, to a sin-gle third-order nonlinear ordinary differential equation for f f 1 2 ff 0 (7.22) The boundary conditions (7.20) become At y 0: f(0) f(0) 0 (7.23a) As y →: f() →1.0 (7.23b) This is the Blasius equation, for which accurate solutions have been obtained only by numerical integration. Some tabulated values of the velocity-profile shape f() u/U are given in Table 7.1.
Since u/U approaches 1.0 only as y →, it is customary to select the boundary-layer thickness as that point where u/U 0.99. From the table, this occurs at 5.0: 99% U x 1/2 5.0 or x Blasius (1908) (7.24) 5.0 Rex 1/2 7.4 The Flat-Plate Boundary Layer 437 y[U/(x)]1/2 u/U y[U/(x)]1/2 u/U 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 1.00000 2.6 0.77246 Table 7.1 The Blasius Velocity Profile [1 to 3] With the profile known, Blasius, of course, could also compute the wall shear and dis-placement thickness cf x (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 w(x) 0.3321 x / 1 2 / 2 1/2U1.5 The wall shear drops off with x1/2 because of boundary-layer growth and varies as ve-locity to the 1.5 power. This is in contrast to laminar pipe flow, where w U and is independent of x.
If w(x) is substituted into Eq. (7.4), we compute the total drag force D(x) b x 0 w(x) dx 0.664b1/2
1/2U1.5x1/2 (7.26) The drag increases only as the square root of the plate length. The nondimensional drag coefficient is defined as CD 2 U D 2 ( b L L ) 2cf (L) (7.27) Thus, for laminar plate flow, CD equals twice the value of the skin-friction coefficient at the trailing edge. This is the drag on one side of the plate.
Kármán pointed out that the drag could also be computed from the momentum re-lation (7.2). In dimensionless form, Eq. (7.2) becomes CD L 2 0 U u 1 U u dy (7.28) This can be rewritten in terms of the momentum thickness at the trailing edge CD 2 L (L) (7.29) Computation of from the profile u/U or from CD gives x laminar flat plate (7.30) Since is so ill defined, the momentum thickness, being definite, is often used to cor-relate data taken for a variety of boundary layers under differing conditions. The ratio of displacement to momentum thickness, called the dimensionless-profile shape fac-tor, is also useful in integral theories. For laminar flat-plate flow H 1 0 .
.
7 6 2 6 1 4 2.59 (7.31) A large shape factor then implies that boundary-layer separation is about to occur.
0.664 Rex 1/2 1.328 ReL 1/2 1.721 Rex 1/2 0.664 Rex 1/2 438 Chapter 7 Flow Past Immersed Bodies Fig. 7.5 Comparison of dimension-less laminar and turbulent flat-plate velocity profiles.
Transition to Turbulence If we plot the Blasius velocity profile from Table 7.1 in the form of u/U versus y/, 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 profiles. Notice how strikingly different in shape they are from the laminar profiles. Instead of decreasing monotonically 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 Rex,tr 3 E6 .
However, for typical commercial surfaces and gusty free streams, a more realistic value is Rex,tr 5 E5.
EXAMPLE 7.3 A sharp flat plate with L 1 m and b 3 m is immersed parallel to a stream of velocity 2 m/s.
Find the drag on one side of the plate, and at the trailing edge find the thicknesses , , and for (a) air, 1.23 kg/m3 and 1.46 105 m2/s, and (b) water, 1000 kg/m3 and 1.02 106 m2/s.
7.4 The Flat-Plate Boundary Layer 439 1.0 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1.0 u U y δ Turbulent 10 5 = Rex 10 6 10 7 Seventh root profile, Eq. (7.39) Exact Blasius profile for all laminar Rex ( Table 7.1) Parabolic approximation, Eq. ( 7.6) Part (b) Part (a) Solution The airflow Reynolds number is V L 1 ( .
2 4 .
6 0 m/ 1 s 0 )( 1 5 .0 m m 2/ ) s 137,000 Since this is less than 3 106, we assume that the boundary layer is laminar. From Eq. (7.27), the drag coefficient is CD (13 1 7 .
, 3 0 2 0 8 0)1/2 0.00359 Thus the drag on one side in the airflow is D CD 1 2 U2bL 0.00359( 1 2 )(1.23)(2.0)2(3.0)(1.0) 0.0265 N Ans. (a) The boundary-layer thickness at the end of the plate is L (137 5 ,0 .0 00)1/2 0.0135 or 0.0135(1.0) 0.0135 m 13.5 mm Ans. (a) We find the other two thicknesses simply by ratios: 1.
5 7 .
2 0 1 4.65 mm 2 .5 9 1.79 mm Ans. (a) Notice that no conversion factors are needed with SI units.
The water Reynolds number is ReL 1.0 2 2 .0 (1.
1 0 0 ) 6 1.96 106 This is rather close to the critical value of 3 106, so that a rough surface or noisy free stream might trigger transition to turbulence; but let us assume that the flow is laminar. The water drag coefficient is CD (1.96 1 .32 1 8 06)1/2 0.000949 and D 0.000949( 1 2 )(1000)(2.0)2(3.0)(1.0) 5.70 N Ans. (b) The drag is 215 times more for water in spite of the higher Reynolds number and lower drag coefficient because water is 57 times more viscous and 813 times denser than air. From Eq.
(7.26), in laminar flow, it should have (57)1/2(813)1/2 7.53(28.5) 215 times more drag.
The boundary-layer thickness is given by L (1.96 5.0 106)1/2 0.00357 or 0.00357(1000 mm) 3.57 mm Ans. (b) By scaling down we have 5.0 ReL 1/2 440 Chapter 7 Flow Past Immersed Bodies Turbulent Flow 1.
5 7 .
2 0 1 1.23 mm 2 .5 9 0.48 mm Ans. (b) The water layer is 3.8 times thinner than the air layer, which reflects the square root of the 14.3 ratio of air to water kinematic viscosity.
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 integal 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: w(x) U2 d d x (7.32) From the definition of cf, Eq. (7.10), this can be rewritten as cf 2 d d x (7.33) Now recall from Fig. 7.5 that the turbulent profiles are nowhere near parabolic. Going back to Fig. 6.9, 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 (6.21) holds all the way across the boundary layer u u 1 ln yu B u w 1/2 (7.34) with, as usual, 0.41 and B 5.0. At the outer edge of the boundary layer, y and u U, and Eq. (7.34) becomes u U 1 ln u B (7.35) But the definition of the skin-friction coefficient, Eq. (7.10), is such that the following identities hold: u U c 2 f 1/2 u Re 1/2 (7.36) Therefore Eq. (7.35) is a skin-friction law for turbulent flat-plate flow c 2 f 1/2 2.44 ln 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: Re 104 105 106 107 cf 0.00493 0.00315 0.00217 0.00158 cf 2 cf 2 7.4 The Flat-Plate Boundary Layer 441 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 cf 0.02 Re 1/6 (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 (x) in terms of (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 U u turb y 1/7 (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: 0 y 1/71 y 1/7 dy 7 7 2 (7.40) We accept this result and substitute Eqs. (7.38) and (7.40) into Kármán’s momentum law (7.33) cf 0.02 Re 1/6 2 d d x 7 7 2 or Re 1/6 9.72 d d x 9.72 d d ( ( R R e e x) ) (7.41) Separate the variables and integrate, assuming 0 at x 0: Re 0.16 Rex 6/7 or x (7.42) Thus the thickness of a turbulent boundary layer increases as x6/7, far more rapidly than the laminar increase x1/2. Equation (7.42) is the solution to the problem, because all other parameters are now available. For example, combining Eqs. (7.42) and (7.38), we obtain the friction variation cf (7.43) Writing this out in dimensional form, we have w,turb (7.44) Turbulent plate friction drops slowly with x, increases nearly as and U2, and is rather insensitive to viscosity.
We can evaluate the drag coefficient from Eq. (7.29) CD 7 6 cf (L) (7.45) 0.031 ReL 1/7 0.0135
1/76/7U13/7 x1/7 0.027 Rex 1/7 0.16 Rex 1/7 442 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.
Then CD is only 16 percent greater than the trailing-edge skin friction [compare with Eq. (7.27) for laminar flow].
The displacement thickness can be estimated from the one-seventh-power law and Eq. (7.12): 0 1 y 1/7 dy 1 8 (7.46) The turbulent flat-plate shape factor is approximately H 1.3 (7.47) These are the basic results of turbulent flat-plate theory.
Figure 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 ef-fect of wall roughness, which is quite strong. The proper roughness parameter here is x/ or L/, by analogy with the pipe parameter /d. In the fully rough regime, CD is in-dependent of the Reynolds number, so that the drag varies exactly as U2 and is inde- 1 8 7 7 2 7.4 The Flat-Plate Boundary Layer 443 0.014 0.012 0.010 0.008 0.006 0.004 0 10 5 10 8 ReL Laminar: Eq. ( 7.27 ) Turbulent smooth Eq. ( 7.45 ) Transition Eq. ( 7.49 ) 200 500 1000 2000 10 4 2 × 10 4 2 × 10 5 0.002 10 6 10 7 10 9 CD 5 × 10 4 L ε = 300 10 6 Fully rough Eq. ( 7.48b) 5000 Part (b) Part (a) pendent of . 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: cf 2.87 1.58 log x 2.5 (7.48a) CD 1.89 1.62 log L 2.5 (7.48b) Equation (7.48b) 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 5 105 ReL 8 107, 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 upon the Reynolds number Retrans where transition begins: CD 1 R 4 e 4 L 0 Retrans 5 105 (7.49a) CD 8 R 7 e 0 L 0 Retrans 3 106 (7.49b) EXAMPLE 7.4 A hydrofoil 1.2 ft long and 6 ft wide is placed in a water flow of 40 ft/s, with 1.99 slugs/ft3 and 0.000011 ft2/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 edge, (c) laminar turbulent flow with Retrans 5 105, and (d) turbulent rough-wall flow with 0.0004 ft.
Solution The Reynolds number is ReL U L ( 0 4 .
0 00 f 0 t/ 0 s) 1 ( 1 1.
f 2 t2 f / t s ) 4.36 106 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), ( L L) (4.36 0 .1 1 6 06)1/7 0.018 or 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.
For fully turbulent smooth-wall flow, the drag coefficient on one side of the plate is, from Eq.
(7.45), CD (4.36 0 .03 1 1 06)1/7 0.00349 0.031 ReL 1/7 0.031 ReL 1/7 444 Chapter 7 Flow Past Immersed Bodies Part (c) Part (d) Then the drag on both sides of the foil is approximately D 2CD( 1 2 U2)bL 2(0.00349)( 1 2 )(1.99)(40)2(6.0)(1.2) 80 lb Ans. (b) With a laminar leading edge and Retrans 5 105, Eq. (7.49a) applies: CD 0.00349 4.3 1 6 4 40 106 0.00316 The drag can be recomputed for this lower drag coefficient: D 2CD( 1 2 U2)bL 72 lbf Ans. (c) Finally, for the rough wall, we calculate L 0.
1 0 .
0 2 04 ft ft 3000 From Fig. 7.6 at ReL 4.36 106, this condition is just inside the fully rough regime. Equa-tion (7.48b) applies: CD (1.89 1.62 log 3000)2.5 0.00644 and the drag estimate is D 2CD( 1 2 U2)bL 148 lbf 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.
The flat-plate analysis of the previous section should give us a good feeling for the be-havior of both laminar and turbulent boundary layers, except for one important effect: flow separation. Prandtl showed that separation like that in Fig. 7.2b is caused by ex-cessive momentum loss near the wall in a boundary layer trying to move downstream against increasing pressure, dp/dx 0, which is called an adverse pressure gradient.
The opposite case of decreasing pressure, dp/dx 0, is called a favorable gradient, where flow separation can never occur. In a typical immersed-body flow, e.g., Fig. 7.2b, 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 (7.19b) at the wall, where u 0, we obtain y wall 2 y u 2 wall U d d U x d d p x or 2 y u 2 wall 1 d d p x (7.50) 7.5 Boundary Layers with Pressure Gradient 445 3This section may be omitted without loss of continuity.
7.5 Boundary Layers with Pressure Gradient3 Fig. 7.7 Effect of pressure gradient on boundary-layer profiles; PI point of inflection.
for either laminar or turbulent flow. Thus in an adverse gradient the second derivative of velocity is positive at the wall; yet it must be negative at the outer layer (y ) 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. 7.7a) the profile 446 Chapter 7 Flow Past Immersed Bodies U u PI (a) Favorable gradient: d U d x > 0 d p d x < 0 No separation, PI inside wall (b) Zero gradient: d U d x = 0 d p d x = 0 No separation, PI at wall PI τ w = 0 (c) Weak adverse gradient: d U d x < 0 d p d x > 0 No separation, PI in the flow (d ) Critical adverse gradient: Zero slope at the wall: Separation (e) Excessive adverse gradient: Backflow at the wall: Separated flow region U u U u U u U u d p d x > 0 PI PI Backflow Fig. 7.8 Boundary-layer growth and separation in a nozzle-diffuser con-figuration.
is very rounded, there is no point of inflection, there can be no separation, and lami-nar profiles of this type are very resistant to a transition to turbulence [1 to 3].
In a zero pressure gradient (Fig. 7.7b), e.g., flat-plate flow, the point of inflection is at the wall itself. There can be no separation, and the flow will undergo transition at Rex no greater than about 3 106, as discussed earlier.
In an adverse gradient (Fig. 7.7c to e), a point of inflection (PI) occurs in the bound-ary layer, its distance from the wall increasing with the strength of the adverse gradi-ent. For a weak gradient (Fig. 7.7c) the flow does not actually separate, but it is vul-nerable to transition to turbulence at Rex as low as 105 [1, 2]. At a moderate gradient, a critical condition (Fig. 7.7d) is reached where the wall shear is exactly zero (u/y 0). This is defined as the separation point ( w 0), because any stronger gradient will actually cause backflow at the wall (Fig. 7.7e): the boundary layer thickens greatly, and the main flow breaks away, or separates, from the wall (Fig. 7.2b).
The flow profiles of Fig. 7.7 usually occur in sequence as the boundary layer pro-gresses along the wall of a body. For example, in Fig. 7.2a, a favorable gradient oc-curs on the front of the body, zero pressure gradient occurs just upstream of the shoul-der, and an adverse gradient occurs successively as we move around the rear of the body.
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 7.5 Boundary Layers with Pressure Gradient 447 Nearly inviscid core flow Boundary layers U(x) Profile point of inflection Separation point w = 0 x Dividing streamline Backflow Separation Nozzle: Decreasing pressure and area Increasing velocity Favorable gradient Throat: Constant pressure and area Velocity constant Zero gradient Diffuser: Increasing pressure and area Decreasing velocity Adverse gradient (boundary layer thickens) U(x) ( x) δ ( x) δ τ Laminar Integral Theory does the throat flow where the pressure gradient is approximately zero. But the ex-panding-area diffuser produces low velocity and increasing pressure, an adverse gra-dient. If the diffuser angle is too large, the adverse gradient is excessive, and the bound-ary layer will separate at one or both walls, with backflow, 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 sep-aration. 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. New techniques are now developed for analyzing the strong inter-action effects caused by separated flows [5, 6].
Both laminar and turbulent theories can be developed from Kármán’s general two-dimensional boundary-layer integral relation , which extends Eq. (7.33) to variable U(x) 1 2 cf d d x (2 H) U d d U x (7.51) where (x) is the momentum thickness and H(x) (x)/ (x) is the shape factor. From Eq. (7.17) negative dU/dx is equivalent to positive dp/dx, that is, an adverse gradient.
We can integrate Eq. (7.51) to determine (x) for a given U(x) if we correlate cf and H with the momentum thickness. This has been done by examining typical velocity profiles of laminar and turbulent boundary-layer flows for various pressure gradients.
Some examples are given in Fig. 7.9, showing that the shape factor H is a good indi-cator of the pressure gradient. The higher the H, the stronger the adverse gradient, and separation occurs approximately at H (7.52) The laminar profiles (Fig. 7.9a) clearly exhibit the S shape and a point of inflection with an adverse gradient. But in the turbulent profiles (Fig. 7.9b) 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 complicated al-gebraically and will be omitted here. The reader is referred to advanced texts [1, 2, 9].
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 , defined as d d U x (7.53) Using a straight-line fit to his correlation, Thwaites was able to integrate Eq. (7.51) in closed form, with the result 2 2 0 U U 0 6 0.
U 45 6 x 0 U5 dx (7.54) 2 v laminar flow turbulent flow 3.5 2.4 w U2 448 Chapter 7 Flow Past Immersed Bodies where 0 is the momentum thickness at x 0 (usually taken to be zero). Separation (cf 0) was found to occur at a particular value of Separation: 0.09 (7.55) Finally, Thwaites correlated values of the dimensionless shear stress S w /(
U) with , and his graphed result can be curve-fitted as follows: S() w U ( 0.09)0.62 (7.56) This parameter is related to the skin friction by the identity S 1 2 cf Re (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 exact digital-com-puter solutions of the laminar-boundary-layer equations (7.19). Complete details of Thwaites’ and other laminar theories are given in Refs. 2 and 3.
As a demonstration of Thwaites’ method, take a flat plate, where U constant, 0, and 0 0. Equation (7.54) integrates to 2 0.4 U 5x 7.5 Boundary Layers with Pressure Gradient 449 1.0 0.8 0.6 0.2 0.2 0.4 0.6 0.8 1.0 Favorable gradients: u U y δ (a) Points of inflection (adverse gradients) 3.5 (Separation) 3.2 2.9 2.7 2.6 (Flat plate) 2.4 2.2 = H = 0.4 θ δ 0 Flat plate Separation 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.2 0.4 0.6 0.8 1.0 u U y δ (b) 0.1 0.3 0.5 0.7 0.9 θ H = = 1.3 δ 2.4 2.3 2.22.1 2.01.9 1.8 1.7 1.6 1.5 1.4 0 Fig. 7.9 Velocity profiles with pressure gradient: (a) laminar flow; (b) turbulent flow with adverse gradients.
Part (a) or x (7.58) This is within 1 percent of Blasius’ exact solution, Eq. (7.30).
With 0, Eq. (7.56) predicts the flat-plate shear to be w U (0.09)0.62 0.225 or cf 2 U w 2 (7.59) This is also within 1 percent of the Blasius result, Eq. (7.25). However, the general ac-curacy 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 use Thwaites’ method to make qualitative assessments of the boundary-layer be-havior.
EXAMPLE 7.5 In 1938 Howarth proposed a linearly decelerating external-velocity distribution U(x) U01 L x (1) as a theoretical model for laminar-boundary-layer study. (a) Use Thwaites’ method to compute the separation point xsep for 0 0, and compare with the exact digital-computer solution xsep/L 0.119863 given by H. Wipperman in 1966. (b) Also compute the value of cf 2 w/(U2) at x/L 0.1.
Solution First note that dU/dx U0/L constant: Velocity decreases, pressure increases, and the pres-sure gradient is adverse throughout. Now integrate Eq. (7.54) 2 x 0 U5 01 L x 5 dx 0.075 U L 0 1 L x 6 1 (2) Then the dimensionless factor is given by d d U x 2U L 0 0.0751 L x 6 1 (3) From Eq. (7.55) we set this equal to 0.09 for separation sep 0.09 0.0751 6 1 or 1 (2.2)1/6 0.123 Ans. (a) xsep L xsep L 2 v 0.45v U6 0(1 x/L)6 0.671 Rex 1/2 0.671 Rex 1/2 450 Chapter 7 Flow Past Immersed Bodies Part (b) 7.6 Experimental External Flows This is less than 3 percent higher than Wipperman’s exact solution, and the computational ef-fort is very modest.
To compute cf at x/L 0.1 (just before separation), we first compute at this point, using Eq. (3) (x 0.1L) 0.075[(1 0.1)6 1] 0.0661 Then from Eq. (7.56) the shear parameter is S(x 0.1L) (0.0661 0.09)0.62 0.099 1 2 cf Re (4) We can compute Re in terms of ReL from Eq. (2) or (3) L 2 2 0 U .0 L 6 / 6 1 0.
R 06 eL 61 or Re 0.257 ReL 1/2 at L x 0.1 Substitute into Eq. (4): 0.099 1 2 cf(0.257 ReL 1/2) or cf ReL U L Ans. (b) We cannot actually compute cf without the value of, say, U0L/.
Boundary-layer theory is very interesting and illuminating and gives us a great quali-tative 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 for the forces on an arbitrary body im-mersed in a stream flowing at an arbitrary Reynolds number. Therefore experimenta-tion 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 chap-ters we shall study data on supersonic airfoils (Chap. 9), open-channel friction (Chap.
10), and turbomachinery performance (Chap. 11).
0.77 ReL 1/2 7.6 Experimental External Flows 451 Fig. 7.10 Definition of forces and moments on a body immersed in a uniform flow.
Drag of Immersed Bodies 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 performs 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 situation is more properly the role of a textbook on aerodynamics [for example, 13]. We shall limit the discussion here to lift and drag.
When the body has symmetry about the lift-drag axis, e.g., airplanes, ships, and cars moving directly into a stream, the side force, yaw, and roll vanish, and the problem re-duces 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 rev-olution 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.4 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 num-ber CD f(Re) (7.60) 452 Chapter 7 Flow Past Immersed Bodies Arbitrary body Lift force Yawing moment Drag force Rolling moment Pitching moment Side force Freestream velocity V 4In 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.
Fig. 7.11 Only the drag force oc-curs if the flow is parallel to both planes of symmetry.
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 Re V L (7.61) 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 de-pending upon the body shape: CD (7.62) The factor 1 2 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, 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 CD,press CD,fric (7.63) drag 1 2 V2A 7.6 Experimental External Flows 453 V Vertical plane of symmetry Horizontal plane of symmetry Principal chord line Doubly symmetric body Drag only if V is parallel to chord line Characteristic Area Friction Drag and Pressure Drag Fig. 7.12 Drag of a streamlined two-dimensional cylinder at Rec 106: (a) effect of thickness ratio on percentage of friction drag; (b) total drag versus thickness when based upon two different areas.
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 cylinder 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, simulating a circular cylin-der, the friction drag is only about 3 percent. Friction and pressure drag are about equal at thickness t/c 0.25. Note that CD in Fig. 7.12b looks quite different when based upon frontal area instead of planform area, planform being the usual choice for this body shape. The two curves in Fig. 7.12b represent exactly the same drag data.
Figure 7.13 illustrates the dramatic effect of separated flow and the subsequent fail-ure of boundary-layer theory. The theoretical inviscid pressure distribution on a circu-lar cylinder (Chap. 8) is shown as the dashed line in Fig. 7.13c: Cp 1 4 sin2 (7.64) where p and V are the pressure and velocity, respectively, in the free stream. The ac-tual laminar and turbulent boundary-layer pressure distributions in Fig. 7.13c are star-tlingly 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 82°, which p p 1 2 V2 454 Chapter 7 Flow Past Immersed Bodies 0.3 0.2 0.1 00 CD 0.2 0.4 0.6 0.8 1.0 Circular cylinder CD based upon frontal area (t b) CD based upon planform area (c b) Width b t V c Thickness ratio t c (b) Flat plate 100 50 0 0 0.2 0.4 0.6 0.8 1.0 (a) 3 Percentage of pressure drag Data scatter Friction drag percent t c Fig. 7.13 Flow past a circular cylinder: (a) laminar separation; (b) turbulent separation; (c) theoretical and actual surface-pressure distri-butions.
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 CD 1.2.
The turbulent boundary layer in Fig. 7.13b is more resistant, and separation is de-layed until 120°, with a resulting smaller wake, higher pressure on the rear, and 75 percent less drag, CD 0.3. This explains the sharp drop in drag at transition in Fig. 5.3.
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°, CD 0.5, while the turbulent flow (Fig. 7.14b) separates at 120°, CD 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 dim-pled 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.
7.6 Experimental External Flows 455 Separation V p∞ Broad wake 82° CD = 1.2 (a) θ Separation V p∞ Narrow wake 120° CD = 0.3 (b) θ 1.0 0.0 – 1.0 – 2.0 – 3.00° 45° 90° 135° 180° (c) θ Turbulent Laminar Inviscid theory θ Cp = 1 – 4 sin2 Cp = p – p∞ ρ V 2/ 2 Fig. 7.14 Strong differences in lam-inar 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, tur-bulent flow induced by patch of nose-sand roughness. (U.S. Navy photograph, Ordnance Test Station, Pasadena Annex.) 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 rec-tangular cylinder (Fig. 7.15a) has rampant separation at all sharp corners and very high drag. Rounding its nose (Fig. 7.15b) reduces drag by about 45 percent, but CD is still high. Streamlining 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 con-trast, the circular cylinder (Fig. 7.15d) has one-eighth the thickness and one-three-hundredth the cross section (c) (Fig. 7.15c), yet it has the same drag. For high-per-formance vehicles and other moving bodies, the name of the game is drag reduction, for which intense research continues for both aerodynamic and hydrodynamic appli-cations [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 CD at very low (creep-ing flow) Re 1.0, while they spread apart at high Reynolds numbers according to 456 Chapter 7 Flow Past Immersed Bodies (a ) (b ) V (a) CD = 1.1 V (c) CD = 2.0 V (b) CD = 0.15 V (d ) 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).
Fig. 7.16 Drag coefficients of smooth bodies at low Mach num-bers: (a) two-dimensional bodies; (b) three-dimensional bodies. Note the Reynolds-number independence of blunt bodies at high Re.
their degree of streamlining. All values of CD 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].
Table 7.2 gives a few data on drag, based on frontal area, of two-dimensional bod-ies of various cross section, at Re 104. 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 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.
7.6 Experimental External Flows 457 10 100 10 1 0.1 0.01 0.001 0.1 1 10 100 10 3 10 4 10 5 10 6 10 7 Re Airfoil Seagull Sailplane CD Stokes' law: 24/Re Disk (a) 100 10 1 0.1 0.01 CD 0.1 1 100 10 3 10 4 10 5 10 6 10 7 Re (b) Smooth flat plate parallel to stream L D = ∞ = 5 Smooth circular cylinder Square cylinder Plate normal to stream Transition Pigeon Vulture 2:1 ellipsoid Airship hull Sphere 458 Chapter 7 Flow Past Immersed Bodies Table 7.2 Drag of Two-Dimensional Bodies at Re 104 Square cylinder: Half tube: Half-cylinder: Equilateral triangle: 2.1 1.6 1.2 2.3 1.2 1.7 1.6 2.0 2.0 1.4 1.0 0.7 Plate: Thin plate normal to a wall: Hexagon: L H Rounded nose section: 0.5 1.16 1.0 0.90 2.0 0.70 4.0 0.68 6.0 0.64 H L Rounded nose section: L/H: 0.1 1.9 0.4 2.3 0.7 2.7 1.2 2.1 2.0 1.8 2.5 1.4 3.0 1.3 6.0 0.9 Elliptical cylinder: Laminar 1.2 0.6 0.35 0.25 Turbulent 0.3 0.2 0.15 0.1 1:1 2:1 4:1 8:1 CD: L/H: CD: CD based CD based CD based on frontal on frontal on frontal Shape area Shape area Shape area Shape CD based on frontal area Flat nose section L/H: CD: L/H: CD: Plate: 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.
7.6 Experimental External Flows 459 L = 20 ft 5 ft/s h = 6 in E7.6 Solution Assume seawater with 1.99 slugs/ft3 and kinematic viscosity 0.000011 ft2/s. With a pil-ing width of 0.5 ft, we have Reh 0 (5 .0 f 0 t 0 /s 0 ) 1 (0 1 .5 ft2 ft / ) s 2.3 105 This is the range where Table 7.2 applies. The worst case occurs when the flow strikes the flat side of the piling, CD 2.1. The frontal area is A Lh (20 ft)(0.5 ft) 10 ft2. The drag is estimated by F CD( 1 2 V2A) 2.1( 1 2 )(1.99 slugs/ft3)(5 ft/s)2(10 ft2) 522 lbf If the flow is uniform, the center of this force should be at approximately middepth. Therefore the bottom bending moment is M0 F 2 L 522(10) 5220 ft lbf Ans.
According to the flexure formula from strength of materials, the bending stress at the bottom would be S M I 0y 251,000 lbf/ft2 1740 lbf/in2 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.
7.16b. Again we can conclude that sharp edges always cause flow separation and high drag which is insensitive to the Reynolds number. Rounded bodies like the ellipsoid have drag which depends upon the point of separation, so that both the Reynolds num-(5220 ft lb)(0.25 ft) 1 1 2 (0.5 ft)4 460 Chapter 7 Flow Past Immersed Bodies Table 7.3 Drag of Three-Dimensional Bodies at Re 104 CD based on Body frontal area Body CD based on frontal area Cube: 1.07 0.81 Cup: 1.4 0.4 1.17 Disk: 1.2 Parachute (Low porosity): Rectangular plate: h b h b/h 1 5 10 20 ∞ 1.18 1.2 1.3 1.5 2.0 L/D: CD: 1 0.64 2 0.68 3 0.72 5 0.74 10 0.82 20 0.91 40 0.98 ∞ 1.20 L D Short cylinder, laminar flow: Porous parabolic dish : Porosity: 0 1.42 0.95 0.1 1.33 0.92 0.2 1.20 0.90 0.3 1.05 0.86 0.4 0.95 0.83 0.5 0.82 0.80 Flat-faced cylinder: Ellipsoid: L /d 0.5 1 2 4 8 1.15 0.90 0.85 0.87 0.99 L /d 0.75 Laminar 0.5 0.47 0.27 0.25 0.2 Turbulent 0.2 0.2 0.13 0.1 0.08 1 2 4 8 d L CD A ≈ 9 ft2 C D A ≈ 1.2 ft2 Average person: U, m/s: CD: 10 1.2 ± 0.2 20 1.0 ± 0.2 30 0.7 ± 0.2 40 0.5 ± 0.2 Pine and spruce trees : L d Cone: 10˚ 0.30 20˚ 0.40 30˚ 0.55 40˚ 0.65 60˚ 0.80 75˚ 1.05 90˚ 1.15 θ θ : CD: CD: CD: CD based on CD based on Body Ratio frontal area Body Ratio frontal area Aerodynamic Forces on Road Vehicles Fig. 7.17 Aerodynamics of automo-biles: (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].
ber and the character of the boundary layer are important. Body length will generally de-crease 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 L/d but friction increases, so that minimum drag occurs at about L/d 2.
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 . Consumer, manufacturer, and government interest has cycled between high speed/high horsepower and lower 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 0.35, based upon the frontal 7.6 Experimental External Flows 461 Fig. 7.18 Drag reduction of a trac-tor-trailer truck: (a) horsepower re-quired to overcome resistance; (b) deflector added to cab reduces air drag by 20 percent. (Uniroyal Inc.) area. Since the frontal area has also decreased sharply, the actual raw drag force on cars has dropped even more than indicated in Fig. 7.17a. The practical minimum, shown tentatively for the year 2000, is CD 0.15 for a tear-shaped vehicle, which can be achieved any time the public is willing to purchase such a shape. Note that basing CD on the frontal area is awkward, since one would need an accurate drawing of the au-tomobile to estimate its frontal area. For this reason, some technical articles simply re-port the raw drag in newtons or pound-force, or the product CDA.
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 vari-ety of flow and geometric effects. Simple changes in part of the shape can have a large influence on aerodynamic forces. Figure 7.17b shows force data by Bearman et al. for an idealized smooth automobile shape with upsweep in the rear of the bottom sec-tion. We see that by simply adding an upsweep angle of 25°, we can quadruple 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, e.g., changes in a windshield shape might interact with downstream 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 which adds separate effects such as front ends, cowls, fenders, windshield, roofs, and rear ends.
Figure 7.18 shows the horsepower required to drive a typical tractor-trailer truck at speeds up to 80 mi/h (117 ft/s or 36 m/s). The rolling resistance increases linearly and the air drag quadratically with speed (CD 1.0). The two are about equally im-portant at 55 mi/h, which is the nominal speed limit in the United States. As shown in Fig. 7.18b, air drag can be reduced by attaching a deflector to the top of the trac-tor. If the angle of the deflector is adjusted to carry the flow smoothly over the top and around the sides of the trailer, the reduction in CD is about 20 percent. Thus, at 55 mi/h the total resistance is reduced 10 percent, with a corresponding reduction in 462 Chapter 7 Flow Past Immersed Bodies Horsepower required 0 20 30 40 50 60 70 80 10 Vehicle speed, mi/h (a) (b) Gross engine horsepower required Air resistance Rolling resistance, hp 550 500 450 400 350 300 250 200 150 100 50 0 E7.7 fuel costs and/or trip time for the trucker. This type of applied fluids engineering can be a large factor in many of the conservation-oriented transportation problems of the future.
EXAMPLE 7.7 A high-speed car with m 2000 kg, CD 0.3, and A 1 m2 deploys a 2-m parachute to slow down from an initial velocity of 100 m/s (Fig. E7.7). Assuming constant CD, brakes free, and no rolling resistance, calculate the distance and velocity of the car after 1, 10, 100, and 1000 s.
For air assume 1.2 kg/m3, and neglect interference between the wake of the car and the para-chute.
7.6 Experimental External Flows 463 d p = 2 m V0 = 100 m/s x Solution Newton’s law applied in the direction of motion gives Fx m d d V t Fc Fp 1 2 V 2(CDcAc CDpAp) where subscript c denotes the car and subscript p the parachute. This is of the form d d V t m K V2 K CDA 2 Separate the variables and integrate V V0 d V V 2 m K t 0 dt or V0 1 V1 m K t Rearrange and solve for the velocity V: V 1 (K V0 /m)V0t K (1) We can integrate this to find the distance traveled: S V 0 ln (1 t) m K V0 (2) Now work out some numbers. From Table 7.3, CDp 1.2; hence CDcAc CDpAp 0.3(1 m2) 1.2 4 (2 m)2 4.07 m2 (CDcAc CDpAp) 2 Other Methods of Drag Reduction Drag of Surface Ships Then m K V0 0.122 s1 Now make a table of the results for V and S from Eqs. (1) and (2): t, s 1 10 100 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. 7.81). Mostly, though, drag is bad and should be reduced. The classical method of drag reduction is stream-lining (Figs. 7.15 and 7.18). For example, nose fairings and body panels have produced motorcycles which can travel over 200 mi/h. More recent research has uncovered other methods which hold great promise, especially for turbulent flows.
1.
Oil pipelines introduce an annular core of water to reduce the pumping power . The low-viscosity water rides the wall and reduces friction up to 60 per-cent.
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 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 air-craft 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.
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 .
Drag reduction is presently an area of intense and fruitful research and applies to many types of airflows and water flows for both vehicles and conduits.
The drag data above, such as Tables 7.2 and 7.3, are for bodies “fully immersed” in a free stream, i.e., with no free surface. If, however, the body moves at or near a free liq-uid surface, wave-making drag becomes important and is dependent upon both the Reynolds number and the Froude number. To move through a water surface, a ship 1 2 (4.07 m2)(1.2 kg/m3)(100 m/s) 2000 kg 464 Chapter 7 Flow Past Immersed Bodies Body Drag at High Mach Numbers 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 Ffric Fwave or CD CD,fric CD,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 sys-tem 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 : high drag if N 1, 3, 5, 7, . . . ; Fr (7.65) low drag if N 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 negligible vari-ation 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. In-troduction of a bulb protrusion on the bow, curve B, greatly reduces the drag. Adding 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 N 4, Fr 0.27, which is a nearly “waveless” con-dition. In this figure CD,wave is defined as 2Fwave/(V2L2) 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 digital 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 above are for nearly incompressible flows, with Mach numbers assumed less than about 0.5. Beyond this value compressibility can be very important, with CD fcn(Re, Ma). As the stream Mach number increases, at some subsonic value Mcrit 1 which depends upon the body’s bluntness and thickness, the local velocity at some point near the body surface will become sonic. If Ma increases beyond Macrit, shock waves form, intensify, and spread, raising surface pressures near the front of the body and therefore increasing the pressure drag. The effect can be dramatic with CD 0.53 N V g L 7.6 Experimental External Flows 465 Fig. 7.19 Wave-making drag on a ship model. (After Inui .) Note: The drag coefficient is defined as CDW 2F/(V2L2).
increasing tenfold, and 70 years ago this sharp increase was called the sonic barrier, implying that it could not be surmounted. Of course, it can be—the rise in CD is fi-nite, 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.5 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 pro-jectiles. Also the Reynolds number (laminar versus turbulent boundary-layer flow) has a large effect below Macrit for spheres and cylinders but becomes unimportant above Ma 1. In contrast, the effect of the Reynolds number is small for airfoils and pro-jectiles and is not shown in Fig. 7.20. A general statement might divide Reynolds- and Mach-number effects as follows: Ma 0.4: Reynolds number important, Mach number unimportant 0.4 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.
9.10b 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 does not eliminate the bow shock. Chapter 9 gives a brief treatment of compressibility.
References 30 and 31 are more advanced textbooks devoted entirely to compressible flow.
466 Chapter 7 Flow Past Immersed Bodies 0.10 0.20 0.30 0.40 0.50 Design speed Potential-flow theory A Main hull (without bulb) B With bow-bulb C With bow and stern-bulbs A B C Fr = V √Lg 0.002 0.001 0 CD, wave 0.60 5There is a slight effect of the specific-heat ratio k which would appear if other gases were tested.
Fig. 7.20 Effect of the Mach num-ber on the drag of various body shapes. (Data from Refs. 23 and 29.) Biological Drag Reduction 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 al-lows it to reconfigure in high winds and thus reduce drag and damage. Tree root sys-tems 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 recon-figure. We saw this in Table 7.3, where tree drag coefficients reduced by 60 per-cent 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. 7.21(a), broad and open in low wind, curls into a conical low-drag shape as wind increases. A compound black walnut leaf group, Fig. 7.21(b), clusters into a low-drag shape at high wind speed. Although drag coefficients were reduced up to 50 per-cent by flexibility, Vogel points out that rigid structures are sometimes just as effec-tive. An interesting recent symposium was devoted entirely to the solid mechanics and fluid mechanics of biological organisms.
Lifting bodies (airfoils, hydrofoils, or vanes) are intended to provide a large force nor-mal to the free stream and as little drag as possible. Conventional design practice has evolved a shape not unlike a bird’s wing, i.e., relatively thin (t/c 0/18) 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 7.6 Experimental External Flows 467 0.0 1.0 2.0 3.0 4.0 Mach number Airfoil Pointed body of revolution Sphere Laminar, Re ≈ 1 E5 Turbulent, Re ≈ 1 E6 CD Laminar, Re ≈ 1 E5 Turbulent, Re ≈ 1 E6 Cylinder in cross flow: 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 Forces on Lifting Bodies Fig. 7.22 Definition sketch for a lifting vane.
Fig. 7.21 Biological adaptation to wind forces: (a) a tulip tree leaf curls into a conical shape at high velocity; (b) black walnut leaves cluster into a low-drag shape as wind increases. (From Vogel, Ref.
33.) 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 . The lift L and the drag D vary with this angle. The dimensionless forces are defined with respect to the planform area Ap bc: Lift coefficient: CL (7.66a) Drag coefficient: CD (7.66b) D 1 2 V2Ap L 1 2 V2Ap 468 Chapter 7 Flow Past Immersed Bodies 5 m/s 10 m/s 20 m/s (a) 5 m/s 10 m/s 20 m/s (b) α Planform area = bc Angle of attack V c = chord b = span t = thickness Lift Drag 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 stream-lines flow smoothly from trailing edge: lift is now 80 percent devel-oped; (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 modern aircraft, Ap c db.
For low-speed flow with a given roughness ratio, CL and CD should vary with and the chord Reynolds number CL f( , Rec) or CD f( , Rec) where Rec Vc/. 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 separation which 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 Fig. 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 stop-ping vortex of opposite (clockwise) sense will form and be shed. During flight, in-creases or decreases in lift will cause incremental starting or stopping vortices, 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. 7.23d, 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-7.6 Experimental External Flows 469 (a) (b) (c) (d ) Fig. 7.24 At high angle of attack, smoke-flow visualization shows stalled flow on the upper surface of a lifting vane. [From Ref. 19, Illus-trated Experiments in Fluid Me-chanics (The NCFMF Book of Film Notes), National Committee for Fluid Mechanics Films, Education Development Center, Inc., copy-right 1972.] ble begins to creep forward on the upper surface.6 At a certain angle 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 hydrody-namicist 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 ex-tended and developed by Prandtl and Kármán and their students, it is now possible to design a low-speed airfoil to satisfy particular surface-pressure distributions and bound-ary-layer characteristics. There are whole families of airfoil designs, notably those de-veloped in the United States under the sponsorship of the NACA (now NASA). Ex-tensive theory and data on these airfoils are contained in Ref. 16. We shall discuss this further in Chap. 8.
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, 470 Chapter 7 Flow Past Immersed Bodies 6For some airfoils the bubble leaps, not creeps, forward, and stall occurs rapidly and dangerously.
Fig. 7.25 Lift and drag of a sym-metric NACA 0009 airfoil of infi-nite span, including effect of a split-flap deflection. Note that roughness can increase CD from 100 to 300 percent.
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 2 sin 2 c h (7.67) where h/c is the maximum camber expressed as a fraction of the chord. The NACA 0009 has zero camber; hence CL 2 sin 0.11 , where 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; e.g., 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 co-efficient. This is a salutary effect, since the prototype will probably be at a higher Reynolds number than the model (107 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) and changes the zero-lift point to 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 gradi-ents over much of the upper surface at low angles. Thus separation is small, and tran-sition to turbulence is delayed; the airfoil retains a good length of laminar flow even 7.6 Experimental External Flows 471 CL α 6 × 10 6 3 × 10 6 –12 – 8 – 4 CD – 8 – 4 0 4 8 12 16 0 4 8 12 16 , deg α , deg α 1.6 1.2 0.8 0.4 With flap at 60° Rec = 6 × 10 6 No flap No flap 0.01 0.02 0.03 0.04 Split flap Rec = 9 × 10 6 With flap at 60° Rec = 3 × 10 6 6 × 10 6 9 × 10 6 Fig. 7.26 Lift-drag polar plot for standard (0009) and a laminar-flow (63-009) NACA airfoil.
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, i.e., 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), AR A b2 p b c (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 ef-fective angle of attack increases, as in Fig. 7.27, by the amount ! C A L R (7.69) When applied to Eq. (7.67), the finite-span lift becomes CL 2 s 1 in ( 2/ AR 2h/c) (7.70) The associated drag increase is !CD CL sin ! CL ! , or CD CD (7.71) CL 2 AR 472 Chapter 7 Flow Past Immersed Bodies 1.2 0.8 0.4 0 0 0.008 0.016 0.024 CL Stall CD Low-drag “bucket” NACA 0009 NACA 63 – 009 0009 with split flap Stall Fig. 7.27 Effect of finite aspect ra-tio on lift and drag of an airfoil: (a) effective angle increase; (b) in-duced drag increase.
where CD is the drag of the infinite-span airfoil, as sketched in Fig. 7.25. These cor-relations 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 L W CL,max( 1 2 Vs 2Ap) or Vs 1/2 (7.72) The stall speed of typical aircraft varies between 60 and 200 ft/s, depending upon the weight and value of CL,max. The pilot must hold the speed greater than about 1.2Vs 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 7.28a shows six such devices whose lift performance is given in 7.28b along with a standard (A) and laminar-flow (B) airfoil. The double-slotted flap achieves CL,max 3.4, and a combination of this plus a leading-edge slat can achieve CL,max 4.0. These are not scientific curiosities; e.g., the Boeing 727 commercial jet aircraft uses a triple-slotted flap plus a leading-edge slat during landing.
Also shown as C in Fig. 7.28b is the Kline-Fogleman airfoil , not yet a reality.
The designers are amateur model-plane enthusiasts who did not know that conventional aerodynamic wisdom forbids a sharp leading edge and a step cutout from the trailing edge. The Kline-Fogleman airfoil has relatively high drag but shows an amazing con-tinual increase in lift out to 45°. In fact, we may fairly say that this airfoil does not stall and provides smooth performance over a tremendous range of flight condi-tions. No explanation for this behavior has yet been published by any aerodynamicist.
This airfoil is under study and may or may not have any commercial value.
Another violation of conventional aerodynamic wisdom is that military aircraft are beginning to fly, briefly, above the stall point. Fighter pilots are learning to make quick 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 67°.
2W CL,maxAp 7.6 Experimental External Flows 473 CL AR = ∞ AR = A p b 2 (a) CD AR = A p b 2 AR = ∞ α ∆ CD ≈ CL π AR 2 (b) CD ∞ β – CL π AR ∆ ≈ α α Fig. 7.29 New aircraft designs do not necessarily look like your typi-cal jetliner. (From Ref. 42.) 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.
The Kline-Fogleman airfoil in Fig. 7.28 is a departure from conventional aerodynamics, but there have been other striking departures, as detailed in a recent article . These new aircraft, conceived presently as small models, have a variety of configurations, as shown in Fig. 7.29: ring-wing, cruciform, flying saucer, and flap-wing. A saucer config-uration (Fig. 7.29c), with a diameter of 40 in, has been successfully flown by radio con-trol, and its inventor, Jack M. Jones, plans for a 20-ft two-passenger version. Another 18-in-span microplane called the Bat (not shown), made by MLB Co., flies for 20 min at 40 mi/h and contains a video camera for surveillance. New engines have been reduced to a 10- by 3-mm size, producing 20 W of power. At the other end of the size spectrum, Boeing and NASA engineers have proposed a jumbo flying-wing jetliner, similar in shape to the stealth bomber, which would carry 800 passengers for a range of 7000 mi.
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.
474 Chapter 7 Flow Past Immersed Bodies Plain flap or aileron Split flap External airfoil flap Slotted flap Double-slotted flap Leading edge slat (a) D E F G H I Optimum but cumbersome combination Kline-Fogleman airfoil α, deg (b) CL 4 3 2 1 0 – 10 10 20 30 40 50° H G F E, D I A B C New Aircraft Designs (a) Ring-wing (b) Cruciform delta (c) Flying saucer (d) Flap-wing dragonfly EXAMPLE 7.8 An aircraft weighs 75,000 lb, has a planform area of 2500 ft2, and can deliver a constant thrust of 12,000 lb. It has an aspect ratio of 7, and CD 0.02. Neglecting rolling resistance, estimate the takeoff distance at sea level if takeoff speed equals 1.2 times stall speed. Take CL,max 2.0.
Solution The stall speed from Eq. (7.72), with sea-level density 0.00237 slug/ft3, is Vs 1/2 1/2 112.5 ft/s Hence takeoff speed V0 1.2Vs 135 ft/s. The drag is estimated from Eq. (7.71) for AR 7 as CD 0.02 0.02 0.0455CL 2 A force balance in the direction of takeoff gives Fs m d d V t thrust drag T kV2 k 1 2 CDAp (1) Since we are looking for distance, not time, we introduce dV/dt V dV/ds into Eq. (1), sepa-rate variables, and integrate S0 0 dS m 2 V0 0 k const or S0 2 m k ln 2 m k ln (2) where D0 kV0 2 is the takeoff drag. Equation (2) is the desired theoretical relation for takeoff distance. For the particular numerical values, take m 75 3 , 2 0 .
0 2 0 2329 slugs CL0 1.39 CD0 0.02 0.0455(CL0)2 0.108 k 1 2 CD0Ap ( 1 2 )(0.108)(0.00237)(2500) 0.319 slug/ft D0 kV0 2 5820 lb Then Eq. (2) predicts that S0 2(0 2 .
3 3 2 1 9 9 s s l l u u g g s /ft) ln 12,0 1 0 2 0 ,0 00 5820 3650 ln 1.94 2420 ft Ans.
A more exact analysis accounting for variable k gives the same result to within 1 percent.
75,000 1 2 (0.00237)(135)2(2500) W 1 2 V0 2 Ap T T D0 T T kV0 2 d(V2) T kV2 CL 2 7 2(75,000) 2.0(0.00237)(2500) 2W CL,maxAp 7.6 Experimental External Flows 475 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 bound-ary layer and wake in the vicinity of the body. Flow outside these “shear layers” is es-sentially inviscid and can be predicted by potential theory and Bernoulli’s equation.
The chapter begins 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 solves the boundary-layer equations to give very accurate formulas for flat-plate flow at high Reynolds numbers. Rough-wall effects are included, and Sec. 7.5 gives a brief introduction to pressure-gradient effects. An ad-verse (decelerating) gradient is seen to cause flow separation, 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. Section 7.6 gives data on drag coefficients of various two- and three-dimensional body shapes. The chapter ends with a brief discussion of lift forces gen-erated by lifting bodies such as airfoils and hydrofoils. Airfoils also suffer flow sepa-ration or stall at high angles of incidence.
476 Chapter 7 Flow Past Immersed Bodies Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 7.1 to 7.124 (categorized in the problem list below) are followed by word problems W7.1 to W7.12, fun-damentals of engineering exam problems FE7.1 to FE7.10, com-prehensive problems C7.1 to C7.4, and design project D7.1.
Problem Distribution Section Topic Problems 7.1 Reynolds-number and geometry 7.1–7.5 7.2 Momentum-integral estimates 7.6–7.12 7.3 The boundary-layer equations 7.13–7.15 7.4 Laminar flat-plate flow 7.16–7.29 7.4 Turbulent flat-plate flow 7.30–7.46 7.5 Boundary layers with pressure gradient 7.47–7.52 7.6 Drag of two-dimensional bodies 7.53–7.63 7.6 Drag of three-dimensional bodies 7.64–7.114 7.6 Lifting bodies—airfoils 7.115–7.124 P7.1 For flow at 20 m/s past a thin flat plate, estimate the dis-tances x from the leading edge at which the boundary-layer thickness will be either 1 mm or 10 cm for (a) air and (b) water at 20°C and 1 atm.
P7.2 Air, equivalent to that at a standard altitude of 4000 m, flows at 450 mi/h past a wing which has a thickness of 18 cm, a chord length of 1.5 m, and a wingspan of 12 m. What is the appropriate value of the Reynolds number for cor-relating the lift and drag of this wing? Explain your se-lection.
P7.3 Equation (7.1b) 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 Rex,crit and turbulent thereafter. Apply this scheme to com-putation 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. (7.1b). Assume Rex,crit 1.2 E6.
P7.4 Air at 20°C and 1 atm flows at 15 m/s past a flat plate with Rex,crit 1 E6. At what point x will the boundary-layer thickness be 8 mm? Why do Eqs. (7.1) seem to fail?
Make a sketch illustrating the discrepancy; then use the ideas in Prob. 7.3 to complete this problem correctly.
P7.5 SAE 30 oil at 20°C flows at 1.8 ft3/s 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.
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” con-cept 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?
Problems 477 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 /x 9.2/Rex 1/2, which is 80 per-cent high. What is the reason for the inaccuracy? [Hint: The answer lies in evaluating the laminar boundary-layer momentum equation (7.19b) at the wall, y 0.] 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) R and (b) R. What are the proper boundary conditions?
2 m /s 3 m 40 × 40 cm square duct Boundary layers Ucore P7.7 P7.8 Air, 1.2 kg/m3 and 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 m2, 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 the parabolic profile, Eq. (7.6), with a more ac-curate sinusoidal profile: U u sin 2 y Compute momentum-integral estimates of cf, /x, /x, and H.
P7.10 Repeat Prob. 7.9, using the polynomial profile suggested by K. Pohlhausen in 1921: U u 2 y 2 y3 3 y4 4 Does this profile satisfy the boundary conditions of lami-nar flat-plate flow?
P7.11 Find the correct form for a cubic velocity-profile polyno-mial u A By Cy2 Dy3 to replace Eq. (7.6) in a flat-plate momentum analysis.
Find the value of / for this profile, but do not pursue the analysis further.
P7.12 The velocity profile shape u/U 1 exp (4.605y/) is a smooth curve with u 0 at y 0 and u 0.99U at y and thus would seem to be a reasonable substitute δ (x) x y r u p ≈ constant R U P7.13 P7.14 Show that the two-dimensional laminar-flow pattern with dp/dx 0 u U0(1 eCy) 0 0 is an exact solution to the boundary-layer equations (7.19).
Find the value of the constant C in terms of the flow pa-rameters. Are the boundary conditions satisfied? What might this flow represent?
P7.15 Discuss whether fully developed laminar incompressible flow between parallel plates, Eq. (4.143) and Fig. 4.16b, represents an exact solution to the boundary-layer equations (7.19) and the boundary conditions (7.20). In what sense, if any, are duct flows also boundary-layer flows?
P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20°C. Compute the total friction drag if the stream is parallel to (a) the long side and (b) the short side.
P7.17 Helium at 20°C and low pressure flows past a thin flat plate 1 m long and 2 m wide. It is desired that the total friction drag of the plate be 0.5 N. What is the appropriate absolute pressure of the helium if U 35 m/s?
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: y, 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 For the Blasius flat-plate problem, Eqs. (7.21) to (7.23), does a two-dimensional stream function " (x, y) exist? If 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 L 50 cm, determine the values of the manometer read-ings h1 and h2, in mm.
P7.18 In Prob. 7.7, when the duct is perfectly square, the core velocity speeds up. Suppose we wish to hold the core ve-locity constant by slanting the upper and lower walls while keeping the front and rear walls parallel. What an-gle of slant will be the best approximation? For this con-dition, what will be the total friction drag on the duct walls?
P7.19 Program a method of numerical solution of the Blasius flat-plate relation, Eq. (7.22), subject to the conditions in (7.23). You will find that you cannot get started with-out knowing the initial second derivative f (0), which lies between 0.2 and 0.5. Devise an iteration scheme which starts at f (0) 0.2 and converges to the correct value. Print out u/U f() and compare with Table 7.1.
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 Meriam red oil, SG 0.827. Use this information to esti-mate the downstream position x of the pitot tube. Assume laminar flow.
so, determine the correct dimensionless form for ", as-suming that " 0 at the wall, y 0.
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) Es-timate the drag on the sheet of plywood if the bound-ary-layer is turbulent (assume the wood is smooth), and compare the result to that of the laminar boundary-layer case.
478 Chapter 7 Flow Past Immersed Bodies h 20 m/s Boundary layer 2 mm x P7.20 P7.23 U Boundary layer h2 2 mm h1 2 mm L L P7.24 P7.25 Modify Prob. 7.24 to the following somewhat more diffi-cult scenario. Let the known data be U 15 m/s and h1 8 mm of water. Use this information to determine (a) L, in cm, (b) h2, in mm.
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 (b)? Ex-plain your results.
P7.27 A thin smooth disk of diameter D is immersed parallel to a uniform stream of velocity U. Assuming laminar flow and using flat-plate theory as a guide, develop an approx-imate formula for the drag of the disk.
P7.28 Flow 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 Fig. P7.28.
The cross section is a by a, and the box length is L. As-suming laminar flat-plate flow and an array of N N boxes, derive a formula for (a) the total drag on the bun-dle of boxes and (b) the effective pressure drop across the bundle.
Problems 479 P7.29 Let the flow straighteners in Fig. P7.28 form an array of 20 20 boxes of size a 4 cm and L 25 cm. If the approach velocity is U0 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. 6.6.
P7.30 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate surface is smooth.
P7.31 Repeat Prob. 7.16 if the fluid is water at 20°C and the plate has an average roughness height of 1.5 mm.
P7.32 A flat plate of length L and height 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 turbu-lent and that the approaching flow is a one-seventh-power law u(y) U0 y 1/7 Using strip theory, derive a formula for the drag coeffi-cient of this plate. Compare this result with the drag of the same plate immersed in a uniform stream U0.
1 3 2 4 1 (a) P7.26 (b) 1 2 3 4 a a L U0 P7.28 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 w 0.0225U2 U 1/4 Show that this formula can be combined with Eqs. (7.33) and (7.40) to derive the following relations for turbulent flat-plate flow: x cf CD These formulas are limited to Rex between 5 105 and 107.
P7.34 A thin equilateral-triangle plate is immersed parallel to a 12 m/s stream of water at 20°C, as in Fig. P7.34. Assum-ing Retr 5 105, estimate the drag of this plate.
P7.35 The solutions to Prob. 7.26 are (a) F 2.83F1-plate and (b) F 2.0F1-plate. Do not reveal these results to your friends.
Repeat Prob. 7.26 assuming the boundary-layer flow is tur-bulent, and comment on the striking increase in numerical values.
0.072 ReL 1/5 0.0577 Rex 1/5 0.37 Rex 1/5 y x U L δ δ y = u(y) P7.32 wall turbulent-flow theory to estimate the position x of the probe, in m.
P7.42 A four-bladed helicopter rotor rotates at n r/min in air with properties (, ). Each blade has chord length C and ex-tends from the center of rotation out to radius R (the hub size is neglected). Assuming turbulent flow from the lead-ing edge, develop an analytical estimate for the power P required to drive this rotor.
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 leading edge, estimate (a) the stream velocity U, (b) the boundary-layer thickness at the element, and (c) the bound-ary-layer velocity u, in m/s, at 5 mm above the element.
P7.36 A ship is 125 m long and has a wetted area of 3500 m2.
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 A wind tunnel has a test section 1 m square and 6 m long with air at 20°C moving at an average velocity of 30 m/s.
It is planned to slant the walls outward slightly to account for the growing boundary-layer displacement thickness on the four walls, thus keeping the test-section velocity con-stant. At what angle should they be slanted to keep V con-stant between x 2 m and x 4 m?
P7.38 Atmospheric boundary layers are very thick but follow for-mulas 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 stand-ing 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 Retr 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 CD 0.01 0.05L/b, where L is the flag length in the flow direction. Test Reynolds numbers ReL were 1 E6 or greater. (a) Explain why, for L/b 1, these drag val-ues are much higher than for a flat plate. Assuming sea-level standard air at 50 mi/h, with area bL 4 m2, find (b) the proper flag dimensions for which the total drag is ap-proximately 400 N.
P7.41 Repeat Prob. 7.20 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-480 Chapter 7 Flow Past Immersed Bodies 2 m 2 m 2 m 12 m/s P7.34 x U Floating element with negligible gap P7.43 P7.44 Extensive measurements of wall shear stress and local ve-locity for turbulent airflow on the flat surface of the Uni-versity of Rhode Island wind tunnel have led to the fol-lowing proposed correlation: y 2 2 w 0.0207 u y 1.77 Thus, if y and u(y) are known at a point in a flat-plate bound-ary layer, the wall shear may be computed directly. If the answer to part (c) of Prob. 7.43 is u 27 m/s, determine whether the correlation is accurate for this case.
P7.45 A thin sheet of fiberboard weighs 90 N and lies on a rooftop, 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 # 0.12, what wind velocity will gen-erate enough fluid friction to dislodge the board?
P7.46 A ship is 150 m long and has a wetted area of 5000 m2.
If it is encrusted with barnacles, the ship requires 7000 hp to overcome friction drag when moving in seawater at 15 kn and 20°C. What is the average roughness of the barna-cles? How fast would the ship move with the same power if the surface were smooth? Neglect wave drag.
EES thin, show that the core velocity U(x) in the diffuser is given approximately by U 1 (2x U t 0 an )/W where W is the inlet height. Use this velocity distribution with Thwaites’ method to compute the wall angle for which laminar separation will occur in the exit plane when diffuser length L 2W. Note that the result is indepen-dent of the Reynolds number.
Problems 481 Fiberboard Roof 2 m 3 m 1 m 1.5 m 2 m U P7.47 As a case similar to Example 7.5, Howarth also proposed the adverse-gradient velocity distribution U U0(1 x2/L2) and computed separation at xsep/L 0.271 by a se-ries-expansion method. Compute separation by Thwaites’ method and compare.
P7.48 In 1957 H. Görtler proposed the adverse-gradient test cases U (1 U x 0 /L)n and computed separation for laminar flow at n 1 to be xsep/L 0.159. Compare with Thwaites’ method, assum-ing 0 0.
P7.49 Based strictly upon your understanding of flat-plate the-ory plus adverse and favorable pressure gradients, explain the direction (left or right) for which airflow past the slen-der airfoil shape in Fig. P7.49 will have lower total (fric-tion pressure) drag.
P7.45 U ?
U ?
P7.49 P7.50 For flow past a cylinder of radius R as in Fig. P7.50, the theoretical inviscid velocity distribution along the surface is U 2U0 sin (x/R), where U0 is the oncoming stream velocity and x is the arc length measured from the nose (Chap. 8). Compute the laminar separation point xsep and sep by Thwaites’ method, and compare with the digital-computer solution xsep/R 1.823 ( sep 104.5°) given by R. M. Terrill in 1960.
P7.51 Consider the flat-walled diffuser in Fig. P7.51, 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 R U0 θ x xsep, θsep P7.50 L U0 x U(x) θ θ Constant width b W P7.51 P7.52 In Fig. P7.52 a slanted upper wall creates a favorable pres-sure gradient on the upper surface of the flat plate. Use Thwaites’ theory to estimate CD F 1 2 U2 0bL P7.52 Flat plate Slanted wall U0 h0 x L, b U(x) 1 2 h0 on the upper plate surface if U0 L/ 105. Compare with Eq. (7.27).
P7.53 From Table 7.2, the drag coefficient of a wide plate nor-mal to a stream is approximately 2.0. Let the stream con-ditions be U and p. 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 A chimney at sea level is 2 m in diameter and 40 m high.
When it is subjected to 50 mi/h storm winds, what is the estimated wind-induced bending moment about the bot-tom of the chimney?
P7.55 A ship tows a submerged cylinder, which is 1.5 m in di-ameter and 22 m long, at 5 m/s in fresh water at 20°C. Es-timate the towing power, in kW, required if the cylinder is (a) parallel and (b) normal to the tow direction.
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 55 mi/h, estimate the force on the sign (a) with no crosswind and (b) with a 10 mi/h crosswind.
how fast can Joe ride into the head wind? (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 1-cm squares. Estimate the drag of 1 m2 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 ft2 of this net?
P7.61 A filter may be idealized as an array of cylindrical fibers normal to the flow, as in Fig. P7.61. 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 !p through a filter of thick-ness L.
482 Chapter 7 Flow Past Immersed Bodies 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 A long cylinder of rectangular cross section, 5 cm high and 30 cm long, is immersed in water at 20°C flowing at 12 m/s parallel to the long side of the rectangle. Es-timate the drag force on the cylinder, per unit length, if the rectangle (a) has a flat face or (b) has a rounded nose.
P7.59 Joe 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, i.e., 0.80 N of force per m/s of speed. The drag area (CDA) of Joe and his bike is 0.422 m2. Joe’s mass is 80 kg and that of the bike is 15 kg. He now encoun-ters a head wind 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, i.e., 60 cm 8 m Phil's Pizza: 555-5748 P7.56 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 The cross section of a cylinder is shown in Fig. P7.63. As-sume that on the front surface the velocity is given by po-tential theory (Sec. 8.4), V 2U sin , from which the surface pressure is computed by Bernoulli’s equation. In R Separated-flow wake θ V = 2 U∞ sin θ p∞ , U∞ P7.63 Filter section U p + ∆ p U p Array of cylinders (fibers) P7.61 the separated flow on the rear, the pressure is assumed equal to its value at 90°. Compute the theoretical drag coefficient and compare with Table 7.2.
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 The drag of a sphere at very low Reynolds numbers ReD 1 was given analytically by G. G. Stokes in 1851: F 3 VD [2, pp. 175–178]. This formula is an exam-ple of Stokes’law of creeping motion, where inertia is neg-ligible. Show that the drag coefficient in this region is CD 24/ReD. A 1-mm-diameter sphere falls in 20°C glycerin at 2.5 mm/s. Is this a creeping motion? Compute (a) the Reynolds number, (b) the drag, and (c) the specific gravity of the sphere.
P7.66 A sphere of density s and diameter D is dropped from rest in a fluid of density and viscosity . Assuming a constant drag coefficient Cd 0, derive a differential equa-tion for the fall velocity V(t) and show that the solution is V 1/2 tanh Ct C 1/2 where S s/ is the specific gravity of the sphere mate-rial.
P7.67 Apply the theory of Prob. 7.66 to a steel sphere of diam-eter 2 cm, dropped from rest in water at 20°C. Estimate the time required for the sphere to reach 99 percent of its terminal (zero-acceleration) velocity.
P7.68 A baseball weighs 145 g and is 7.35 cm in diameter. It is dropped from rest from a 35-m-high tower at approxi-mately sea level. Assuming a laminar-flow drag coeffi-cient, estimate (a) its terminal velocity and (b) whether it will reach 99 percent of its terminal velocity before it hits the ground.
P7.69 Two baseballs from Prob. 7.68 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 A baseball from Prob. 7.68 is batted upward during a game at an angle of 45° and an initial velocity of 98 mi/h. Ne-glect spin and lift. Estimate the horizontal distance trav-eled, (a) neglecting drag and (b) accounting for drag in a numerical (computer) solution with a transition Reynolds number ReD,crit 2.5 E5.
3gCd 0(S 1) 4S2D 4gD(S 1) 3Cd 0 Problems 483 P7.71 A football weighs 0.91 lbf and approximates an ellipsoid 6 in in diameter and 12 in long (Table 7.3). It is thrown upward at a 45° angle with an initial velocity of 80 ft/s.
Neglect spin and lift. Assuming turbulent flow, estimate the horizontal distance traveled, (a) neglecting drag and (b) accounting for drag with a numerical (computer) model.
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 which will ensure that all sediment (SG 2.55) will fall to the bot-tom for particle diameters greater than (a) 1 mm and (b) 100 m.
P7.73 A balloon is 4 m in diameter and contains helium at 125 kPa and 15°C. Balloon material and payload weigh 200 N, not including the helium. Estimate (a) the terminal as-cent velocity in sea-level standard air, (b) the final stan-dard 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.
P7.74 If D 4 m in Prob. 7.73 and the helium remains at 15°C and 125 kPa, estimate the time required for the balloon to rise through the standard atmosphere from sea level to its equilibrium altitude, which is approxi-mately 4000 m.
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, 28 cm Ω Baseball 28 cm Baseball P7.69 D = 50 cm θ U P7.75 EES not including the helium. The helium pressure is 120 kPa.
Estimate the tilt angle if the airstream velocity U is (a) 5 m/s or (b) 20 m/s.
P7.76 Extend Prob. 7.75 to make a smooth plot of tilt angle versus stream velocity U in the range 1 U 12 mi/h.
(A spreadsheet is recommended for this task.) Comment on the effectiveness of this system as an air-velocity in-strument.
P7.77 Modify Prob. 7.75 as follows: Let the fluid be water at 20°C, and let the 50-cm-diameter sphere be solid cork (SG 0.16). Estimate the tilt angle if the water veloc-ity U is 3 m/s.
P7.78 Apply Prob. 7.61 to a filter consisting of 300-
m-diameter fibers packed 250 per square centimeter in the plane of Fig. P7.61. For air at 20°C and 1 atm flowing at 1.5 m/s, estimate the pressure drop if the filter is 5 cm thick.
P7.79 Assume that a radioactive dust particle approximates a sphere of density 2400 kg/m3. How long, in days, will it take such a particle to settle to sea level from an alti-tude of 12 km if the particle diameter is (a) 1 m or (b) 20 m?
P7.80 A heavy sphere attached to a string should hang at an angle when immersed in a stream of velocity U, as in Fig. P7.80. Derive an expression for as a function of the sphere and flow properties. What is if the sphere is steel (SG 7.86) of diameter 3 cm and the flow is sea-level standard air at U 40 m/s? Neglect the string drag.
1.2 ft2 falling feet first [12, p. 313]. What are the mini-mum and maximum terminal speeds that can be achieved by a skydiver at 5000-ft standard altitude?
P7.83 A high-speed car has a drag coefficient of 0.3 and a frontal area of 1 m2. A parachute is to be used to slow this car from 80 to 40 m/s in 8 s. What should the chute diameter be? What distance will be traveled during this decelera-tion? Take m 2000 kg.
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?
484 Chapter 7 Flow Past Immersed Bodies P7.81 A typical U.S. Army parachute has a projected diame-ter of 28 ft. For a payload mass of 80 kg, (a) what ter-minal velocity 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 (c) horizontally? (Neglect the fact that flat shapes are not dynamically stable in free fall.) P7.82 The average skydiver with parachute unopened weighs 175 lbf and has a drag area CDA 9 ft2 spread-eagled and P7.85 An aluminum cylinder (SG 2.7) slides concentrically down a taut 1-mm-diameter wire as shown in Fig. P7.85.
Its length L 8 cm, and its radius R 1 cm. A 2-mm-di-ameter hole down the cylinder center is lubricated by SAE 30 oil at 20°C. Estimate the terminal fall velocity V of the cylinder if ambient air drag is (a) neglected and (b) in-cluded. Assume air at 1 atm and 20°C.
θ D, ρ s U P7.80 P7.84 P7.85 Oil film V R L P7.86 Hoerner [Ref. 12, pp. 3–25] states that the drag coefficient of a flag of 21 aspect ratio is 0.11 based on planform 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 yet avoids breaking the flagpole in hur-ricane (75 mi/h) winds?
P7.87 A tractor-trailer truck has a drag-area CDA 8 m2 bare and 6.7 m2 with an aerodynamic deflector (Fig. 7.18b). Its rolling 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 CD A of 35 ft2. Es-timate the horsepower required to drive the truck at 55 mi/h (a) clean and (b) with the 3- by 6-ft sign in Fig.
P7.88 installed if the rolling resistance is 150 lbf at sea level.
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 $ with wind velocity U in the range 0 U 25 m/s for sea-level stan-dard air.
Problems 485 P7.89 A water tower is approximated by a 15-m-diameter sphere mounted on a 1-m-diameter rod 20 m long. Estimate the bending moment at the root of the rod due to aerodynamic forces during hurricane winds of 40 m/s.
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 lbf?
P7.91 A cup anemometer uses two 5-cm-diameter hollow hemi-spheres connected to 15-cm rods, as in Fig. P7.91. Rod drag is negligible, and the central bearing has a retarding P7.92 A 1500-kg automobile uses its drag area CDA 0.4 m2, plus brakes and a parachute, to slow down from 50 m/s.
Its brakes apply 5000 N of resistance. Assume sea-level standard air. If the automobile must stop in 8 s, what di-ameter parachute is appropriate?
P7.93 A hot-film probe is mounted on a cone-and-rod system in a sea-level airstream of 45 m/s, as in Fig. P7.93. Estimate the maximum cone vertex angle allowable if the flow-in-duced bending moment at the root of the rod is not to ex-ceed 30 N cm.
Eat at Joe's 6 ft 3 ft P7.88 D = 5 cm Ω D = 5 cm U 15 cm 15 cm P7.91 Ho t film 20 cm 3 cm 5-mm dia.
45 m/s P7.93 EES P7.94 A rotary mixer consists of two 1-m-long half-tubes rotat-ing around a central arm, as in Fig. P7.94. Using the drag from Table 7.2, derive an expression for the torque T re-quired to drive the mixer at angular velocity $ in a fluid of density . Suppose that the fluid is water at 20°C and the maximum driving power available is 20 kW. What is the maximum rotation speed $ r/min?
mobile of mass 1500 kg and frontal area 2 m2, the fol-lowing 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 coef-ficient. This problem is well suited for digital-computer analysis but can be done by hand also.
P7.98 A buoyant ball of specific gravity SG 1 dropped into water at inlet velocity V0 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 h as a function of the system properties. How far will a 5-cm-diameter ball with SG 0.5 and CD 0.47 penetrate if it enters at 10 m/s?
486 Chapter 7 Flow Past Immersed Bodies P7.99 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 be-gin to rotate clockwise.
P7.94 D = 5 cm Ω R = 1 m h V0 Diameter D (SG < 1) P7.98 D = 1 cm D = 2 cm U Stop 10 cm 10 cm Hinge 45˚ P7.99 P7.95 An airplane weighing 12 kN, with a drag-area CDA 5 m2, lands at sea level at 55 m/s and deploys a drag para-chute 3 m in diameter. No other brakes are applied. (a) How long will it take the plane to slow down to 30 m/s? (b) How far will it have traveled in that time?
P7.96 A Savonius rotor (Fig. 6.29b) can be approximated by the two open half-tubes in Fig. P7.96 mounted on a cen-tral axis. If the drag of each tube is similar to that in Table 7.2, derive an approximate formula for the rota-tion rate $ as a function of U, D, L, and the fluid prop-erties (, ).
Axis U L L D D Ω P7.96 P7.97 A simple measurement of automobile drag can be found by an unpowered coastdown on a level road with no wind. Assume constant rolling resistance. For an auto-P7.100 In creeping motion or Stokes’ flow at velocity U past a sphere of diameter D, density (fluid inertia) is unim-portant and the drag force is F 3 UD. This formula is valid if ReD 1.0 (see Fig. 7.16b). (a) Verify that Stokes’ formula is equivalent to CD 24/ReD. (b) De-termine the largest diameter raindrop whose terminal fall velocity follows Stokes’ formula in sea-level standard air.
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 upon relative velocities be-tween the iceberg and the fluid, derive an approximate expression for the steady iceberg speed V when driven by wind velocity U.
Problems 487 P7.102 Sand particles (SG 2.7), approximately spherical with diameters from 100 to 250 m, are introduced into an up-ward-flowing stream of water at 20°C. What is the mini-mum water velocity which will carry all the sand particles upward?
P7.103 When immersed in a uniform stream V, a heavy rod hinged at A will hang at Pode’s angle , after an analysis by L. Pode in 1951 (Fig. P7.103). Assume that the cylin-der has normal drag coefficient CDN and tangential coef-ficient CDT which relate the drag forces to VN and VT, re-spectively. Derive an expression for Pode’s angle as a function of the flow and rod parameters. Find for a steel rod, L 40 cm, D 1 cm, hanging in sea-level air at V 35 m/s.
P7.104 Suppose that the body in Fig. P7.103 is a thin plate weighing 0.6 N, of length L 20 cm and width b 4 cm into the paper. For sea-level standard air, plot Pode’s angle versus velocity V in the range 0 V 40 m/s.
Would this device make a good air-velocity meter (anemometer)?
P7.105 A ship 50 m long, with a wetted area of 800 m2, has the hull shape tested in Fig. 7.19. There are no bow or stern bulbs. The total propulsive power available is 1 MW. For seawater at 20°C, plot the ship’s velocity V kn versus power P for 0 P 1 MW. What is the most efficient setting?
P7.106 A smooth steel 1-cm-diameter sphere (W 0.04 N) is fired vertically at sea level at the initial supersonic veloc-ity V0 1000 m/s. Its drag coefficient is given by Fig.
7.20. Assuming that the speed of sound is constant at a 343 m/s, compute the maximum altitude of the projectile (a) by a simple analytical estimate and (b) by a digital-computer program.
P7.107 Repeat Prob. 7.106 if the body is a 9-mm steel bullet (W 0.07 N) which approximates the “pointed body of revolu-tion” in Fig. 7.20.
P7.108 The data in Fig. P7.108 are for the lift and drag of a spin-ning sphere from Ref. 12, pp. 7–20. Suppose that a tennis ball (W 0.56 N, D 6.35 cm) is struck at sea level with initial velocity V0 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 be-fore it strikes the ground.
P7.109 Repeat Prob. 7.108 if the ball is struck with “underspin” (front of the ball rotating upward).
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 lbf 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 deviated 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 veloc-ity of 20 m/s while it is 50 cm above the table, as in Fig. P7.111. 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 deceler-ates during flight.
P7.101 U > > D L L / 8 7L / 8 V Iceberg P7.103 V A θ L, D, ρ s CD N , VN CD T , VT P7.112 Repeat Prob. 7.111 by making a detailed digital-computer solution for the flight path of the ball. Use Fig. P7.108 for lift and drag.
P7.113 An automobile has a mass of 1000 kg and a drag-area CDA 0.7 m2. 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 tan1 0.1 5.71°). How far up the hill will the car come to a stop?
P7.114 Suppose that the car in Prob. 7.113 is placed at the top of the 10 percent grade hill and released from rest to coast down without brakes. What will be its speed, in km/h, af-ter dropping a vertical distance of 20 m?
P7.115 An airplane weighs 180 kN and has a wing area of 160 m2 and a mean chord of 4 m. The airfoil properties are given by Fig. 7.25. If the airplane cruises at 250 mi/h at 3000-m standard altitude, what propulsive power is re-quired to overcome wing drag?
P7.116 The airplane of Prob. 7.115 is designed to land at V0 1.2Vstall, using a split flap set at 60°. What is the proper landing speed in mi/h? What power is required for takeoff at the same speed?
P7.117 Suppose that the airplane of Prob. 7.115 takes off at sea level without benefit of flaps, with CL constant so that the takeoff speed is 100 mi/h. Estimate the takeoff distance if the thrust is 10 kN. How much thrust is needed to make the takeoff distance 1250 m?
P7.118 Suppose that the airplane of Prob. 7.115 is fitted with all the best high-lift devices of Fig. 7.28. What is its mini-mum stall speed in mi/h? Estimate the stopping distance if the plane lands at V0 1.25Vstall with constant CL 3.0 and CD 0.2 and the braking force is 20 percent of the weight on the wheels.
P7.119 An airplane has a mass of 5000 kg, a maximum thrust of 7000 N, and a rectangular wing with aspect ratio 6.0. It takes off at sea level with a 60° split flap whose two-dimensional properties are shown in Fig. 7.25. Assume all lift and all drag are due to the wing. What is the proper wing size if the takeoff distance is to be 1 km?
P7.120 Show that if Eqs. (7.70) and (7.71) are valid, the maxi-mum lift-to-drag ratio occurs when CD 2CD. What are 488 Chapter 7 Flow Past Immersed Bodies P7.111 20 m/s 50 cm 4 m ω?
?
P7.108 EES EES Figure Unavailable (L/D)max and for a symmetric wing when AR 5 and CD 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 d l r i a ft g For a sailplane of mass 200 kg, wing area 12 m2, and as-pect ratio 11, with an NACA 0009 airfoil, estimate (a) the stall speed, (b) the minimum gliding angle, and (c) the maximum 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 CL,max 1.2 and CD 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 attain-able.
P7.123 In prewar days there was a controversy, perhaps apocryphal, about whether the bumblebee has a legitimate aerodynamic right to fly. The average bumblebee (Bombus terrestris) weighs 0.88 g, with a wing span of 1.73 cm and a wing area of 1.26 cm2. 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 The bumblebee can hover at zero speed by flapping its wings. Using the data of Prob. 7.123, 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 sec-ond of such a model wing are needed to support the bee’s weight? (Actual measurements of bees show a flapping rate of 194 Hz.) Fundamentals of Engineering Exam Problems 489 Word Problems W7.1 How do you recognize a boundary layer? Cite some phys-ical properties and some measurements which reveal ap-propriate characteristics.
W7.2 In Chap. 6 the Reynolds number for transition to turbu-lence in pipe flow was about Retr 2300, whereas in flat-plate flow Retr 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 “bound-ary-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 180° to face the stream.
Can you explain this?
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? Ex-plain 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 bi-cycle racing, apply to the material studied in this chapter?
W7.11 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.
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) 7.2 E6, (e) 7.2 E7 FE7.2 If, in Prob. FE7.1, the drag coefficient based upon frontal area is 0.5, what is the drag force on the sphere?
(a) 17 N, (b) 51 N, (c) 102 N, (d) 130 N, (e) 203 N FE7.3 If, in Prob. FE7.1, the drag coefficient based upon frontal area is 0.5, at what terminal velocity will an aluminum sphere (SG 2.7) fall in still water?
(a) 2.3 m/s, (b) 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, (b) 2.6 mm, (c) 5.3 mm, (d) 7.5 mm, (e) 20.2 mm FE7.5 In Prob. FE7.4, for the same flow conditions, what is the wall shear stress at x 60 cm from the leading edge?
(a) 0.053 Pa, (b) 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 upon frontal area, is 1.15. Estimate the wind-induced bending moment at the base of the pole.
(a) 9.7 kN m, (b) 15.2 kN m, (c) 19.4 kN m, (d) 30.5 kN m, (e) 61.0 kN 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 fracture at a base bending moment of 486 kN m, and its drag coef-ficient based upon frontal area is 0.5, what is the approximate maximum allowable wind velocity to avoid fracture?
(a) 50 mi/h, (b) 75 mi/h, (c) 100 mi/h, (d) 125 mi/h, (e) 150 mi/h FE7.8 A dust particle of density 2600 kg/m3, small enough to satisfy Stokes’ drag law, settles at 1.5 mm/s in air at 20°C and 1 atm. What is its approximate diameter?
(a) 1.8 m, (b) 2.9 m, (c) 4.4 m, (d) 16.8 m, (e) 234 m FE7.9 An airplane has a mass of 19,550 kg, a wing span of 20 m, and an average wing chord of 3 m. When flying in air of density 0.5 kg/m3, its engines provide a thrust of 12 kN against an overall drag coefficient of 0.025. What is its approximate velocity?
(a) 250 mi/h, (b) 300 mi/h, (c) 350 mi/h, (d) 400 mi/h, (e) 450 mi/h FE7.10 For the flight conditions of the airplane in Prob. FE7.9 above, what is its approximate lift coefficient?
(a) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, (e) 0.5 490 Chapter 7 Flow Past Immersed Bodies Comprehensive Problems C7.1 Jane wants to estimate the drag coefficient of herself on her bicycle. She measures the projected frontal area to be 0.40 m2 and the rolling resistance to be 0.80 N s/m. The mass of the bike is 15 kg, while the mass of Jane is 80 kg. Jane coasts down a long hill which 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 CD of the rider and bicycle combination.
C7.2 Air at 20°C and 1 atm flows at Vavg 5 m/s between long, smooth parallel heat-exchanger plates 10 cm apart, as in Fig.
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. As-sume all plates are 1 m wide into the paper. Find (a) the pres-sure drop in Pa/m without the small plates present. Then find (b) the number of small plates per meter of channel length which will cause the pressure drop to rise to 10.0 Pa/m.
C7.3 A new pizza store is planning to open. They 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 & V C7.1 high and 5 ft long. The boss (having no feel for fluid me-chanics) 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 Fig. C7.3.) (a) Calculate the drag (in lbf) 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 ft2. For V 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 lbf at 40 mi/h, calcu-late the horsepower 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.
C7.2 Interrupter plates L 1 cm U 5 m/s Design Project D7.1 It is desired to design a cup anemometer for wind speed, similar to Fig. P7.91, with a more sophisticated approach than the “average-torque” method of Prob. 7.91. 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 materi-als—are to be selected through your analysis. Make suit-able assumptions about the instantaneous drag of the cups and rods at any given angle (t) of the system. Compute the instantaneous torque T(t), and find and integrate the in-stantaneous angular acceleration of the device. Develop a complete theory for rotation rate versus wind speed in the range 0 U 50 mi/h. Try to include actual commercial bearing-friction properties.
References 491 C7.3 Air Cup shape L m C7.4 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 the rod L. (a) Set up the differential equation for the os-cillation (t), including different cup drag (air density ) in each direction, and (b) nondimensionalize this equation.
(c) Determine the natural frequency of oscillation for small 1 rad. (d) For the special case L 1 m, D 10 cm, m 50 g, and air at 20°C and 1 atm, with (0) 30°, find (numerically) the time required for the oscillation ampli-tude to drop to 1°.
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492 Chapter 7 Flow Past Immersed Bodies Cylindrical wave pattern produced in a ripple tank. When not modified by the no-slip condi-tion at solid surfaces, waves are nearly inviscid and well represented by the potential theory of this chapter. (Courtesy of Dr. E. R. Degginger/Color-Pic Inc.) 494 8.1 Introduction and Review Motivation.
The basic partial differential equations of mass, momentum, and energy were discussed in Chap. 4. A few solutions were then given for incompressible poten-tial flow in Sec. 4.10 and for incompressible viscous flow in Sec. 4.11. The viscous so-lutions were limited to simple geometries and unidirectional flows, where the difficult nonlinear convective terms were neglected. The potential flows were not limited by such nonlinear terms. Then, in Chap. 7, we found an approximation: patching boundary-layer flows onto an outer inviscid flow pattern. For more complicated viscous flows, we found no theory or solutions, just experimental data.
The purposes of the present chapter are (1) to explore more examples of potential theory and (2) to indicate some flows which can be approximated by computational fluid dynamics (CFD). The combination of these two gives us a good picture of in-compressible-flow theory and its relation to experiment. One of the most important ap-plications of potential-flow theory is to aerodynamics and marine hydrodynamics. First, however, we will review and extend the concepts of Sec. 4.10.
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 (decreas-ing pressure along the surface), and the boundary layers will be attached and thin: In-viscid theory will give excellent results for the outer flow if Re 104. For the inter-nal flow between bodies, the boundary layers will grow and eventually meet, and the inviscid core vanishes. Inviscid theory works well in a “short’’ duct L/D 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.
495 Chapter 8 Potential Flow and Computational Fluid Dynamics 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 (in-creasing 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 analy-sis of separated-flow regions is an active research area at present.
Recall from Sec. 4.9 that if viscous effects are neglected, low-speed flows are irrota-tional, V 0, and the velocity potential exists, such that V or u w (8.1) The continuity equation (4.73), V 0, reduces to Laplace’s equation for : 2 0 (8.2) and the momentum equation (4.74) reduces to Bernoulli’s equation: V2 gz const where V (8.3) Typical boundary conditions are known free-stream conditions Outer boundaries: Known , , (8.4) and no velocity normal to the boundary at the body surface: Solid surfaces: 0 where n is perpendicular to body (8.5) Unlike the no-slip condition in viscous flow, here there is no condition on the tangen-tial surface velocity Vs ∂/∂s, where s is the coordinate along the surface. This ve-locity is determined as part of the solution to the problem.
∂ ∂n ∂ ∂z ∂ ∂y ∂ ∂x 1 2 p ∂ ∂t ∂2 ∂z2 ∂2 ∂y2 ∂2 ∂x2 ∂ ∂z ∂ ∂y ∂ ∂x Fig. 8.1 Patching viscous- and in-viscid-flow regions. Potential the-ory in this chapter does not apply to the boundary-layer regions 496 Chapter 8 Potential Flow and Computational Fluid Dynamics Freestream Inviscid internal core Inviscid external flow Inviscid external flow Boundary layer Boundary layer Boundary layer Boundary layer Fully viscous flow Separation Separation Review of Velocity-Potential Concepts Review of Stream Function Concepts Occasionally the problem involves a free surface, for which the boundary pressure is known and equal to pa, usually a constant. The Bernoulli equation (8.3) then sup-plies a relation at the surface between V and the elevation z of the surface. For steady flow, e.g., Free surface: V2 2 const 2gzsurf (8.6) It should be clear to the reader that this use of Laplace’s equation, with known values of the derivative of 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 the gen-eral theory and its application to fluid mechanics [2 to 4]. There are many analyt-ical techniques, including superposition of elementary functions, conformal mapping, numerical finite differences , numerical finite elements , numerical boundary el-ements , and electric or mechanical analogs now outdated. Having found (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.
Recall from Sec. 4.7 that if a flow is described by only two coordinates, the stream function also exists as an alternate approach. For plane incompressible flow in xy coordinates, the correct form is u ∂ ∂ y ∂ ∂ x (8.7) The condition of irrotationality reduces to Laplace’s equation for also: 2
z 0 ∂ ∂ x ∂ ∂ u y ∂ ∂ x ∂ ∂ x ∂ ∂ y ∂ ∂ y or ∂ ∂ 2 x 2 ∂ ∂ 2 y 2 0 (8.8) The boundary conditions again are known velocity in the stream and no flow through any solid surface: Free stream: Known ∂ ∂ x , ∂ ∂ y (8.9a) Solid surface: body const (8.9b) Equation (8.9b) is particularly interesting because any line of constant in a flow can therefore be interpreted as a body shape and may lead to interesting applica-tions.
For the applications in this chapter, we may compute either or 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 lines, and the other set will be the lines. Both sets of lines are laplacian and could be useful.
8.1 Introduction and Review 497 Fig. 8.2 Streamlines and potential lines are orthogonal and may re-verse roles if results are useful: (a) typical inviscid-flow pattern; (b) same as (a) with roles reversed.
Many solutions in this chapter are conveniently expressed in polar coordinates (r, ).
Both the velocity components and the differential relations for and are then changed, as follows: r (8.10) Laplace’s equation takes the form ∂ ∂ r r ∂ ∂ r ∂ ∂ 2 2 0 (8.11) Exactly the same equation holds for the polar-coordinate form of (r, ).
An intriguing facet of potential flow with no free surface is that the governing equa-tions (8.2) and (8.8) contain no parameters, nor do the boundary conditions. Therefore the solutions are purely geometric, depending only upon the body shape, the free-stream orientation, and—surprisingly—the position of the rear stagnation point.1 There is no Reynolds, Froude, or Mach number to complicate the dynamic similarity. Inviscid flows are kinematically similar without additional parameters—recall Fig. 5.6a.
Recall from Sec. 4.10 that we defined three elementary potential flows which are quite useful: (1) uniform stream in the x direction, (2) line source or sink at the origin, and (3) line vortex at the origin. (Recall Fig. 4.12 for these geometries.) Let us review these special cases here: Uniform stream iU: Uy Ux (8.12a) Line source or sink: m m ln r (8.12b) Line vortex: K ln r K (8.12c) 1 r2 1 r ∂ ∂r ∂ ∂ 1 r ∂ ∂ 1 r ∂ ∂r 498 Chapter 8 Potential Flow and Computational Fluid Dynamics (b) (a) φ1 φ2 φ3 ψ 1 ψ 2 ψ 3 ψ 1 ψ 2 ψ 3 φ1 φ2 φ3 Plane Polar Coordinates 1 The rear stagnation condition establishes the net amount of “circulation’’ about the body, giving rise to a lift force. Otherwise the solution could not be unique. See Sec. 8.4.
8.2 Elementary Plane-Flow Solutions Circulation The source “strength’’m and the vortex “strength’’K have the same dimensions, namely, velocity times length, or {L2/T}.
If the uniform stream is written in plane polar coordinates, it becomes Uniform stream iU: Ur sin Ur cos (8.13) This makes it easier to superimpose, say, a stream and a source or vortex by using the same coordinates. If the uniform stream is moving at angle with respect to the x-axis, i.e., u U cos ∂ ∂ y ∂ ∂ x U sin ∂ ∂ x ∂ ∂ y then by integration we obtain the correct functions for flow at an angle: U(y cos x sin ) U(x cos y sin ) (8.14) These expressions are useful in airfoil angle-of-attack problems (Sec. 8.7).
The line-vortex flow is irrotational everywhere except at the origin, where the vortic-ity V is infinite. This means that a certain line integral called the fluid circulation does not vanish when taken around a vortex center.
With reference to Fig. 8.3, the circulation is defined as the counterclockwise line integral, around a closed curve C, of arc length ds times the velocity component tan-gent to the curve C V cos ds 0 C V ds 0 C (u dx dy w dz) (8.15) From the definition of , V ds ds d for an irrotational flow; hence nor-mally in an irrotational flow would equal the final value of minus the initial value of . Since we start and end at the same point, we compute 0, but not for vortex flow: With K from Eq. (8.12c) there is a change in of amount 2K as we make one complete circle: Path enclosing a vortex: 2K (8.16) 8.2 Elementary Plane-Flow Solutions 499 Closed curve C : Streamline d S α V Γ = οC V cos d s α Fig. 8.3 Definition of the fluid cir-culation .
Fig. 8.4 Intersections of elementary streamlines can be joined to form a combined streamline.
8.3 Superposition of Plane-Flow Solutions Graphical Method of Superposition Alternately the calculation can be made by defining a circular path of radius r around the vortex center, from Eq. (8.15) 0 C ds 2 0 K r r d 2K (8.17) In general, denotes the net algebraic strength of all the vortex filaments contained within the closed curve. In the next section we shall see that a region of finite circu-lation within a flowing stream will be subjected to a lift force proportional to both U and .
It is easy to show, by using Eq. (8.15), that a source or sink creates no circulation.
If there are no vortices present, the circulation will be zero for any path enclosing any number of sources and sinks.
We can now form a variety of interesting potential flows by summing the velocity-potential and stream functions of a uniform stream, source or sink, and vortex. Most of the results are classic, of course, needing only a brief treatment here.
A simple means of accomplishing tot i graphically is to plot the individual stream functions separately and then look at their intersections. The value of tot at each intersection is the sum of the individual values i which cross there. Connecting intersections with the same value of tot creates the desired superimposed flow stream-lines.
A simple example is shown in Fig. 8.4, summing two families of streamlines a and b. The individual components are plotted separately, and four typical intersections are shown. Dashed lines are then drawn through intersections representing the same sum of a b. These dashed lines are the desired solution. Often this graphical method is a quick means of evaluating the proposed superposition before a full-blown numer-ical plot routine is executed.
500 Chapter 8 Potential Flow and Computational Fluid Dynamics Combined streamline Family (a) Family (b) ψ ψ = 1 2 ψ ψ = 1 + ψ ψ ψ = 2 ψ ψ = 1 ψ = 2ψ1 ψ = 2ψ2 ψ ψ = 2 Some Examples from Chap. 4 In Sec. 4.10 we discussed a number of superposition examples.
1.
Source m at ( a, 0) plus an equal sink at (a, 0), Eq. (4.133), and Fig. 4.13: m tan 1 m ln (4.133) The streamlines and potential lines are two families of orthogonal circles as plotted in Fig. 4.13. They resemble a magnet with poles at (x, y) (a, 0).
2.
Sink m plus a vortex K, both at the origin, Eq. (4.134), and Fig. 4.14: m K ln r m ln r K (4.134) The streamlines are logarithmic spirals swirling into the origin, as in Fig. 4.14.
They resemble a tornado or a bathtub vortex.
3.
Uniform stream iU plus a source m at the origin, Eq. (4.135) and Fig. 4.15, the Rankine half-body: Ur sin m Ur cos m ln r (4.135) If the origin contains a source, a plane half-body is formed with its nose to the left, as in Fig. 8.5a. If the origin is a sink, m 0, the half-body nose is to the right, as in Fig. 8.5c. In either case the stagnation point is at a position a m/U away from the origin.
(x a)2 y2 (x a)2 y2 1 2 2ay x2 y2 a2 8.3 Superposition of Plane-Flow Solutions 501 Us (max) = 1.26 U∞ Laminar separation (a) 1.0 0.5 0 (d) s a – 6 Separation = + π m ψ = – π m ψ ψ = 0 U∞ x y y x (c) – 4 – 8 0 – 2 Us U∞ 1.0 0.5 0 (b) s a 2 4 0 8 6 a Us U∞ Fig. 8.5 The Rankine half-body; pattern (c) is not found in a real fluid because of boundary-layer separation: (a) uniform stream plus a source equals a half-body; stagna-tion point at x 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.
Flow Past a Vortex Although the inviscid-flow patterns, Fig. 8.5a and c, are mirror images, their viscous (boundary-layer) behavior is different. The body shape and the velocity along the sur-face are repeated here from Sec. 4.10: V2 U2 1 cos along r (8.18) The computed surface velocities are plotted along the half-body contours in Fig. 8.5b and d as a function of arc length s/a measured from the stagnation point. These plots are also mirror images. However, if the nose is in front, Fig. 8.5b, the pressure gradi-ent there is favorable (decreasing pressure along the surface). In contrast, the pressure gradient is adverse (increasing pressure along the surface) when the nose is in the rear, Fig. 8.5d, and boundary-layer separation may occur.
Application to Fig. 8.5b of Thwaites’ 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. 8.5a is a very realistic picture of streamlines past a half-body nose. In contrast, when applied to the tail, Fig. 8.5c, Thwaites’ method predicts separation at about s/a 2.2, or 110°. Thus, if a half-body is a solid surface, Fig. 8.5c 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 8.1, then Fig. 8.5c is quite realistic and useful. Computations for turbulent boundary-layer the-ory would be similar: separation on the tail, no separation on the nose.
EXAMPLE 8.1 An offshore power plant cooling-water intake sucks in 1500 ft3/s in water 30 ft deep, as in Fig.
E8.1. If the tidal velocity approaching the intake is 0.7 ft/s, (a) how far downstream does the in-take effect extend and (b) how much width L of tidal flow is entrained into the intake?
Solution Recall from Eq. (4.131) that the sink strength m is related to the volume flow Q and the depth b into the paper m 7.96 ft2/s Therefore from Fig. 8.5 the desired lengths a and L are a 11.4 ft Ans. (a) L 2a 2(11.4 ft) 71 ft Ans. (b) Consider a uniform stream U in the x direction flowing past a vortex of strength K with center at the origin. By superposition the combined stream function is stream vortex Ur sin K ln r (8.19) 7.96 ft2/s 0.7 ft/s m U 1500 ft3/s 2(30 ft) Q 2b m( ) U sin 2a r a2 r2 502 Chapter 8 Potential Flow and Computational Fluid Dynamics Boundary-Layer Separation on a Half-Body 0.7 ft /s Top view a ?
Half-body shape Intake 1500 ft3/s L?
E8.1 Fig. 8.6 Flow of a uniform stream past a vortex constructed by the graphical method.
An Infinite Row of Vortices The velocity components are given by r U cos U sin (8.20) The streamlines are plotted in Fig. 8.6 by the graphical method, intersecting the cir-cular streamlines of the vortex with the horizontal lines of the uniform stream.
By setting r 0 from (8.20) we find a stagnation point at 90°, r a K/U, or (x, y) (0, a). This is where the counterclockwise vortex velocity K/r ex-actly cancels the stream velocity U.
Probably 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.
Consider an infinite row of vortices of equal strength K and equal spacing a, as in Fig.
8.7a. This case is included here to illustrate the interesting concept of a vortex sheet.
From Eq. (8.12c), the ith vortex in Fig. 8.7a has a stream function i K ln ri, so that the total infinite row has a combined stream function K i1 ln ri (8.21) It can be shown [2, sec. 4.51] that this infinite sum of logarithms is equivalent to a closed-form function 1 2 K ln cosh cos (8.22) Since the proof uses the complex variable z x iy, i ( 1)1/2, we are not going to show the details here.
2x a 2y a 1 2 K r ∂ ∂r ∂ ∂ 1 r 8.3 Superposition of Plane-Flow Solutions 503 y x The Vortex Sheet Fig. 8.7 Superposition of vortices: (a) an infinite row of equal strength; (b) streamline pattern for part (a); (c) vortex sheet: part (b) viewed from afar.
The streamlines from Eq. (8.22) are plotted in Fig. 8.7b, 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 dif-ferentiating Eq. (8.22) u ya (8.23) where the plus sign applies below the row and the minus sign above the row. This uni-form left and right streaming is sketched in Fig. 8.7c. We stress that this effect is in-duced by the row of vortices: There is no uniform stream approaching the row in this example.
When Fig. 8.7b is viewed from afar, the streaming motion is uniform left above and uniform right below, as in Fig. 8.7c, and the vortices are packed so closely together K a y 504 Chapter 8 Potential Flow and Computational Fluid Dynamics i th vor tex K K K K K K K K x y ( x, y) ri (a) a a a a a a a x y (b) x y (c) u = – π K / a u = + π K / a •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Fig. 8.8 A doublet, or source-sink pair, is the limiting case of Fig.
4.13 viewed from afar. Streamlines are circles tangent to the x-axis at the origin. This figure was prepared using the contour feature of MAT-LAB [34, 35].
The Doublet that they are smudged into a continuous vortex sheet. The strength of the sheet is de-fined as (8.24) and in the general case can vary with x. The circulation about any closed curve which encloses a short length dx of the sheet would be, from Eqs. (8.15) and (8.23), d ul dx uu dx (ul uu) dx dx dx (8.25) where the subscripts l and u stand for lower and upper, respectively. Thus the sheet strength d/dx is the circulation per unit length of the sheet. Thus when a vortex sheet is immersed in a uniform stream, 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, e.g., plate or thin airfoil. This is the ba-sis of the thin-airfoil theory mentioned in Sec. 8.7.
As we move far away from the source-sink pair of Fig. 4.13, the flow pattern begins to resemble a family of circles tangent to the origin, as in Fig. 8.8. This limit of van-ishingly 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 2am re-main constant. Let us call this constant . Then the stream function of a doublet is lim a→0 m tan 1 (8.26) 2am y x2 y2 2amy x2 y2 2ay x2 y2 a2 2K a 2K a 8.3 Superposition of Plane-Flow Solutions 505 We have used the fact that tan 1 as becomes small. The quantity is called the strength of the doublet.
Equation (8.26) can be rearranged to yield x2 y 2 2 so that, as advertised, the streamlines are circles tangent to the origin with centers on the y-axis. This pattern is sketched in Fig. 8.8.
Although the author has in the past laboriously sketched streamlines by hand, this is no longer necessary. Figure 8.8 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. 8.8, 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 from Eq. (8.26), with 1 for conve-nience. The plot would include grid lines, scale markings, and a surrounding box, and 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. 8.8, has no markings but the streamlines themselves. MATLAB is thus a recommended tool and, in addition, has scores of other uses. All this chapter’s problem assignments which call for “sketch the streamlines/potential lines” can be com-pleted 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. (4.133) as a →0 and 2am doublet or x 2 y2 2 (8.27) The potential lines are circles tangent to the origin with centers on the x-axis. Simply turn Fig. 8.8 clockwise 90° to visualize the lines, which are everywhere normal to the streamlines.
The doublet functions can also be written in polar coordinates (8.28) These forms are convenient for the cylinder flows of the next section.
cos r sin r 2 2 x x2 y2 2 2 506 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.9 Flow past a Rankine oval: (a) uniform stream plus a source-sink pair; (b) oval shape and streamlines for m/(Ua) 1.0.
8.4 Plane Flow Past Closed-Body Shapes The Rankine Oval A variety of closed-body external flows can be constructed by superimposing a uni-form stream with sources, sinks, and vortices. The body shape will be closed only if the net source outflow equals the net sink inflow.
A cylindrical shape called a Rankine oval, which is long compared with its height, is formed by a source-sink pair aligned parallel to a uniform stream, as in Fig. 8.9a.
From Eqs. (8.12a) and (4.133) the combined stream function is Uy m tan 1 Ur sin m(1 2) (8.29) When streamlines of constant are plotted from Eq. (8.29), an oval body shape ap-pears, as in Fig. (8.9b). The half-length L and half-height h of the oval depend upon the relative strength of source and stream, i.e., the ratio m/(Ua), which equals 1.0 in Fig. 8.9b. The circulating streamlines inside the oval are uninteresting and not usually shown. The oval is the line 0.
2ay x2 y2 a2 8.4 Plane Flow Past Closed-Body Shapes 507 h +m – m a L (b) (a) U∞ Sou rce θ1 y x (x, y) θ2 Sink +m – m r1 r2 a a Shoulder u max = 1.74 U∞ Flow Past a Circular Cylinder with Circulation 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 pa-rameters are a function of the basic dimensionless parameter m/(Ua), which we can determine from Eq. (8.29): cot 1 1/2 (8.30) 1 As we increase m/(Ua) 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/(Ua) →, L/h →1.0 and umax/U →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 cylin-der, 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 Ur sin K ln r const (8.31) The doublet strength has units of velocity times length squared. For convenience, let Ua2, where a is a length, and let the arbitrary constant in Eq. (8.31) equal K ln a. Then the stream function becomes U sin r K ln (8.32) The streamlines are plotted in Fig. 8.10 for four different values of the dimension-less vortex strength K/(Ua). For all cases the line 0 corresponds to the circle r r a a2 r sin r 2m/(Ua) 1 h2/a2 umax U 2m Ua L a h/a 2m/(Ua) h a 508 Chapter 8 Potential Flow and Computational Fluid Dynamics m/(Ua) h/a L/a L/h umax/U 0.0 0.0 1.0 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 1.000 2.000 Table 8.1 Rankine-Oval Parameters from Eq. (8.30) Fig. 8.10 Flow past a circular cylinder with circulation for values of K/(Ua) of (a) 0, (b) 1.0, (c) 2.0, and (d) 3.0.
a, that is, the shape of the cylindrical body. As circulation 2K increases, the ve-locity becomes faster and faster below the cylinder and slower and slower above it.
The velocity components in the flow are given by r U cos 1 a r2 2 U sin 1 a r2 2 (8.33) The velocity at the cylinder surface r a is purely tangential, as expected r(r a) 0 (r a) 2U sin (8.34) For small K, two stagnation points appear on the surface at angles s where 0, or, from Eq. (8.34), sin s (8.35) Figure 8.10a is for K 0, s 0 and 180°, or doubly symmetric inviscid flow past a cylinder with no circulation. Figure 8.10b is for K/(Ua) 1, s 30 and 150°, and Fig. 8.10c is the limiting case where the two stagnation points meet at the top, K/(Ua) 2, s 90°.
K 2Ua K a K r ∂ ∂r ∂ ∂ 1 r 8.4 Plane Flow Past Closed-Body Shapes 509 K a θ (a) (b) (d) (c) (d) The Kutta-Joukowski Lift Theorem For K 2Ua, Eq. (8.35) is invalid, and the single stagnation point is above the cylinder, as in Fig. 8.10d, at a point y h given by [ (2 4)1/2] 2 (8.36) In Fig. 8.10d, K/(Ua) 3.0, and h/a 2.6.
For the cylinder flows of Fig. 8.10b to d there is a downward force, or negative lift, called the Magnus effect, which is proportional to stream velocity and vortex strength.
We can see from the streamline pattern that the velocity on the top of the cylinder is less and therefore the pressure higher from Bernoulli’s equation; this explains the force.
There is no viscous force, of course, because our theory is inviscid.
The surface velocity is given by Eq. (8.34). From Bernoulli’s equation (8.4), ne-glecting gravity, the surface pressure ps is given by p U2 ps 2U sin 2 or ps p 1 2 U2 (1 4 sin2 4 sin 2) (8.37) where K/(Ua) and p 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 2 0 (ps p) cos ba d where ps p is substituted from Eq. (8.37). But the integral of cos times any power of sin over a full cycle 2 is identically zero. Thus we obtain the (perhaps surpris-ing) result D(cylinder with circulation) 0 (8.38) This is a special case of d’Alembert’s paradox, mentioned in Sec. 1.10: According to inviscid theory, the drag of any body of any shape immersed in a uni-form 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 pro-found effect of the thin viscous boundary layer on the flow pattern in the rear, as in Fig. 7.2b, for example.
The lift force L normal to the stream, taken positive upward, is given by summa-tion of vertical pressure forces L 2 0 (ps p) sin ba d Since the integral over 2 of any odd power of sin is zero, only the third term in the parentheses in Eq. (8.37) contributes to the lift: L U2 ba 2 0 sin2 d U(2K)b 4K aU 1 2 K a 1 2 1 2 K Ua 1 2 h a 510 Chapter 8 Potential Flow and Computational Fluid Dynamics Experimental Lift and Drag of Rotating Cylinders or U (8.39) Notice that the lift is independent of the radius a of the cylinder. Actually, though, as we shall see in Sec. 8.7, the circulation depends upon the body size and orientation through a physical requirement.
Equation (8.39) 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 im-mersed in a uniform stream equals u, where is the total net circulation con-tained within the body shape. The direction of the lift is 90° from the stream di-rection, rotating opposite to the circulation.
The problem in airfoil analysis, Sec. 8.7, is thus to determine the circulation as a function of airfoil shape and orientation.
It is nearly impossible to test Fig. 8.10 by constructing a doublet and vortex with the same center and then letting a stream flow past them. But one physical realization would be a rotating cylinder in a stream. The viscous no-slip condition would cause the fluid in contact with the cylinder to move tangentially at the cylinder peripheral speed a
. A net circulation would be set up by this no-slip mechanism, but it turns out to be less than 50 percent of the value expected from inviscid theory, pri-marily because of flow separation behind the cylinder.
Figure 8.11 shows experimental lift and drag coefficients, based on planform area 2ba, of rotating cylinders. From Eq. (8.38) the theoretical drag is zero, but the actual CD is quite large, more even than the stationary cylinder of Fig. 5.3. The theoretical lift follows from Eq. (8.39) CL (8.40) where s K/a is the peripheral speed of the cylinder.
Figure 8.11 shows that the theoretical lift from Eq. (8.40) is much too high, but the measured lift is quite respectable, much larger in fact than a typical airfoil of the same chord length, e.g., Fig. 7.25. Thus rotating cylinders have practical possibilities. The Flettner rotor ship built in Germany in 1924 employed rotating vertical cylinders which 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 en-ergy costs.
EXAMPLE 8.2 The experimental Flettner rotor sailboat at the University of Rhode Island is shown in Fig. E8.2.
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.
2s U 2 UKb U2 ba L 1 2 U2 (2ba) L b 8.4 Plane Flow Past Closed-Body Shapes 511 Fig. 8.11 Theoretical and experi-mental lift and drag of a rotating cylinder. (From Ref. 22.) 512 Chapter 8 Potential Flow and Computational Fluid Dynamics 10 8 6 4 2 0 8 6 4 2 0 Theory Experiment (Ref. 5) CL , CD CL CD U∞ ω a Theory CD = 0 CL = U∞ ω 2 π a Velocity ratio U∞ aω CL , CD E8.2 (Courtesy of R. C. Lessmann, University of Rhode Island.) The Kelvin Oval Potential-Flow Analogs Solution Convert the rotation rate to 2(220)/60 23.04 rad/s. The wind velocity is 10 kn 16.88 ft/s, so the velocity ratio is 1.71 Entering Fig. 8.11, we read CL 3.3 and CD 1.2. From Table A.6, standard air density is 0.002377 slug/ft3. Then the estimated rotor lift and drag are L CL 1 2 U2 2ba 3.3( 1 2 )(0.002377)(16.88)2(2)(10)(1.25) 27.9 lbf D CD 1 2 U2 2ba L 27.9 10.2 lbf The maximum thrust available is the resultant of these two F [(27.9)2 (10.2)2]1/2 29 lbf Ans. If aligned along the boat’s keel, this thrust will drive the boat at a speed of about 5 kn through the water.
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 U 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.
8.12. The combined stream function is Uy K ln (8.41) The body shape is the line 0, and some of these shapes are shown in Fig. 8.12.
For K/(Ua) 10 the shape is within 1 percent of a Rankine oval (Fig. 8.9) turned 90°, but for small K/(Ua) the waist becomes pinched in, and a figure-eight shape oc-curs at 0.5. For K/(Ua) 0.5 the stream blasts right between the vortices and iso-lates 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. 2 to 4 for further de-tails. A summary of elementary potential flows is given in Table 8.2.
For complicated potential-flow geometries, one can resort to other methods than su-perposition of sources, sinks, and vortices. There are a variety of devices which simu-late solutions to Laplace’s equation.
From 1897 to 1900 Hele-Shaw developed a technique whereby laminar flow be-tween very closely spaced parallel plates simulated potential flow when viewed from above the plates. Obstructions simulate body shapes, and dye streaks represent the streamlines. The Hele-Shaw apparatus makes an excellent laboratory demonstration of potential flow [10, pp. 8–10]. Figure 8.13a illustrates Hele-Shaw (potential) flow x2 (y a)2 x2 (y a)2 1 2 1.2 3.3 CD CL (1.25 ft)(23.04 rad/s) 16.88 ft/s a U 8.4 Plane Flow Past Closed-Body Shapes 513 Fig. 8.12 Kelvin-oval body shapes as a function of the vortex-strength parameter K/(Ua); outer stream-lines not shown.
through an array of cylinders, a flow pattern that would be difficult to analyze just us-ing Laplace’s equation. However beautiful this array pattern may be, it is not a good approximation to real (laminar viscous) array flow. Figure 8.13b shows experimental streakline patterns for a similar staggered-array flow at Re 6400. We see that the in-teracting wakes of the real flow (Fig. 8.13b) cause intensive mixing and transverse mo-tion, not the smooth streaming passage of the potential-flow model (Fig. 8.13a). 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 avail-able, 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.
514 Chapter 8 Potential Flow and Computational Fluid Dynamics U∞ a + K – K 1.0 0.75 0.55 0.5 = 1.5 U∞a K y x Type of flow Potential functions Remarks Stream iU Uy Ux See Fig. 4.12a Line source (m 0) or sink (m 0) m m ln r See Fig. 4.12b Line vortex K ln r K See Fig. 4.12c Half-body Ur sin m Ur cos m ln r See Fig. 8.5 Doublet See Fig. 8.8 Rankine oval Ur sin m(1 2) See Fig. 8.9 Cylinder with circulation U sin r K ln See Fig. 8.10 r a a2 r cos r sin r Table 8.2 Summary of Plane Incompressible Potential Flows Fig. 8.13 Flow past a staggered ar-ray of cylinders: (a) potential-flow model using the Hele-Shaw appara-tus (Tecquipment Ltd., Nottingham, England); (b) experimental streak-lines for actual staggered-array flow at ReD 6400. (From Ref. 36, courtesy of Jack Hoyt, with the per-mission of the American Society of Mechanical Engineers.) Hand-sketching “curvilinear square’’ techniques were also popular. The availability and the simplicity of digital-computer potential-flow methods [5 to 7] have made analog models obsolete.
EXAMPLE 8.3 A Kelvin oval from Fig. 8.12 has K/(Ua) 1.0. Compute the velocity at the top shoulder of the oval in terms of U.
8.4 Plane Flow Past Closed-Body Shapes 515 (a ) (b) 8.5 Other Plane Potential Flows2 Solution We must locate the shoulder y h from Eq. (8.41) for 0 and then compute the velocity by differentiation. At 0 and y h and x 0, Eq. (8.41) becomes ln With K/(Ua) 1.0 and the initial guess h/a 1.5 from Fig. 8.12, we iterate and find the lo-cation h/a 1.5434.
By inspection 0 at the shoulder because the streamline is horizontal. Therefore the shoul-der velocity is, from Eq. (8.41), uyh yh U Introducing K Ua and h 1.5434a, we obtain ushoulder U(1.0 1.84 0.39) 2.45U Ans.
Because they are short-waisted compared with a circular cylinder, all the Kelvin ovals have shoul-der velocity greater than the cylinder result 2.0U from Eq. (8.34).
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)1/2 It turns out that any arbitrary analytic function of this complex variable z has the re-markable property that both its real and its imaginary parts are solutions of Laplace’s equation. If f(z) f(x iy) f1(x, y) i f2(x, y) then ∂ ∂ 2 x f 2 1 ∂ ∂ 2 y f 2 1 0 ∂ ∂ 2 x f 2 2 ∂ ∂ 2 y f 2 2 (8.42) We shall assign the proof of this as a problem. Even more remarkable if you have never seen it before is that lines of constant f1 will be everywhere perpendicular to lines of constant f2: f1C (8.43) We also leave this proof as a problem exercise. This is true for totally arbitrary f(z) as long as this function is analytic; i.e., it must have a unique derivative df/dz at every point in the region.
The net result of Eqs. (8.42) and (8.43) is that the functions f1 and f2 can be inter-preted to be the potential lines and streamlines of an inviscid flow. By long custom we 1 (dy/dx)f2C dy dx K h a K h a ∂ ∂y h/a 1 h/a 1 K Ua h a 516 Chapter 8 Potential Flow and Computational Fluid Dynamics 2 This section may be omitted without loss of continuity.
Uniform Stream at an Angle of Attack Line Source at a Point z0 let the real part of f(z) be the velocity potential and the imaginary part be the stream function f(z) (x, y) i (x, y) (8.44) We try various functions f(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 [4, chap. 5; 10, chap. 5] and a more ad-vanced [2, 3,] level. The method is less important now because of the popularity of digital-computer techniques.
As a simple example, consider the linear function f(z) Uz Ux iUy It follows from Eq. (8.44) that Ux and Uy, which, we recall from Eq.
(8.12a), represents a uniform stream in the x direction. Once you get used to the com-plex variable, the solution practically falls in your lap.
To find the velocities, you may either separate and from f(z) and differentiate or differentiate f directly i i u i (8.45) Thus the real part of df/dz equals u(x, y), and the imaginary part equals (x, y). To get a practical result, the derivative df/dz must exist and be unique, hence the require-ment that f be an analytic function. For Eq. (8.45), df/dz U u, since it is real, and 0, as expected.
Sometimes it is convenient to use the polar-coordinate form of the complex variable z x iy rei r cos ir sin where r (x2 y2)1/2 tan 1 This form is especially convenient when powers of z occur.
All the elementary plane flows of Sec. 8.2 have a complex-variable formulation. The uniform stream U at an angle of attack has the complex potential f(z) Uze i (8.46) Compare this form with Eq. (8.14).
Consider a line source of strength m placed off the origin at a point z0 x0 iy0. Its complex potential is f(z) m ln (z z0) (8.47) This can be compared with Eq. (8.12b), which is valid only for the source at the ori-gin. For a line sink, the strength m is negative.
y x ∂ ∂y ∂ ∂y ∂ ∂x ∂ ∂x df dz 8.5 Other Plane Potential Flows 517 Flow around a Corner of Arbitrary Angle Line Vortex at a Point z0 Fig. 8.14 Streamlines for corner flow, Eq. (8.49) for corner angle of (a) 60°, (b) 90°, (c) 120°, (d) 270°, and (e) 360°.
If a line vortex of strength K is placed at point z0, its complex potential is f(z) iK ln (z z0) (8.48) to be compared with Eq. (8.12c). Also compare to Eq. (8.47) to see that we reverse the meaning of and simply by multiplying the complex potential by i.
Corner flow is an example of a pattern that cannot be conveniently produced by su-perimposing sources, sinks, and vortices. It has a strikingly simple complex represen-tation f(z) Azn Arnein Arn cos n iArn sin n where A and n are constants.
It follows from Eq. (8.44) that for this pattern Arn cos n Arn sin n (8.49) Streamlines from Eq. (8.49) are plotted in Fig. 8.14 for five different values of n.
The flow is seen to represent a stream turning through an angle /n. Patterns in Fig. 8.14d and e are not realistic on the downstream side of the corner, where separa-tion will occur due to the adverse pressure gradient and sudden change of direction. In general, separation always occurs downstream of salient, or protruding corners, except in creeping flows at low Reynolds number Re 1.
Since 360° 2 is the largest possible corner, the patterns for n 1 2 do not repre-sent corner flow. They are peculiar-looking, and we ask you to plot one as a problem.
If we expand the plot of Fig. 8.14a to c to double size, we can represent stagnation flow toward a corner of angle 2 2/n. This is done in Fig. 8.15 for n 3, 2, and 518 Chapter 8 Potential Flow and Computational Fluid Dynamics (a) (a) (b) (c) (d) n = 3 n = 2 n = 3 2 n = 2 3 n = 1 2 (e) Fig. 8.15 Streamlines for stagnation flow from Eq. (8.49) for corner angle 2 of (a) 120°, (b) 180°, and (c) 240°.
1.5. These are very realistic flows; although they slip at the wall, they can be patched to boundary-layer theories very successfully. We took a brief look at corner flows be-fore, in Examples 4.5 and 4.9 and in Probs. 4.49 to 4.51.
We treat this case separately because the Kelvin ovals of Fig. 8.12 failed to degener-ate into a flat plate as K became small. The flat plate normal to a uniform stream is an extreme case worthy of our attention.
Although the result is quite simple, the derivation is very complicated and is given, e.g., in Ref. 2, sec. 9.3. There are three changes of complex variable, or mappings, be-ginning with the basic cylinder-flow solution of Fig. 8.10a. First the uniform stream is rotated to be vertical upward, then the cylinder is squeezed down into a plate shape, and finally the free stream is rotated back to the horizontal. The final result for com-plex potential is f(z) i U(z2 a2)1/2 (8.50) where 2a is the height of the plate. To isolate or , square both sides and separate real and imaginary parts 2 2 U2 (x2 y2 a2) U2 xy 8.5 Other Plane Potential Flows 519 (a) (b) n = 3 n = 2 (c) n = 3 2 Flow Normal to a Flat Plate Fig. 8.16 Streamlines in upper half-plane for flow normal to a flat plate of height 2a: (a) continuous poten-tial-flow theory, Eq. (8.51); (b) ac-tual measured flow pattern; (c) dis-continuous potential theory with k 1.5.
We can solve for to determine the streamlines 4 2U2 (x2 y2 a2) U4 x2y2 (8.51) Equation (8.51) is plotted in Fig. 8.16a, revealing a doubly symmetric pattern of stream-lines which approach very closely to the plate and then deflect up and over, with very high velocities and low pressures near the plate tips.
The velocity s along the plate surface is found by computing df/dz from Eq. (8.50) and isolating the imaginary part plate surface (8.52) Some values of surface velocity can be tabulated as follows: y/a 0.0 0.2 0.4 0.6 0.71 0.8 0.9 1.0 s/U 0.0 0.204 0.436 0.750 1.00 1.33 2.07 y/a (1 y2/a2)1/2 s U 520 Chapter 8 Potential Flow and Computational Fluid Dynamics C L U∞ y a x (a) Broad, low pressure region of separated flow Constant-pressure region Free streamline discontinuity at V = k U∞ C L a x (b) U∞ C L a x (c) U∞ 8.6 Images3 The origin is a stagnation point; then the velocity grows linearly at first and very rapidly near the tip, with both velocity and acceleration being infinite at the tip.
As you might guess, Fig. 8.16a is not realistic. In a real flow the sharp salient edge causes separation, and a broad, low-pressure wake forms in the lee, as in Fig. 8.16b.
Instead of being zero, the drag coefficient is very large, CD 2.0 from Table 7.2.
A discontinuous-potential-flow theory which accounts for flow separation was de-vised by Helmholtz in 1868 and Kirchhoff in 1869. This free-streamline solution is shown in Fig. 8.16c, with the streamline which breaks away from the tip having a con-stant velocity V kU. From Bernoulli’s equation the pressure in the dead-water re-gion behind the plate will equal pr p 1 2 U2 (1 k2) to match the pressure along the free streamline. For k 1.5 this Helmholtz-Kirchoff theory predicts pr p 0.625 U2 and an average pressure on the front pf p 0.375 U2 , giving an over-all drag coefficient of 2.0, in agreement with experiment. However, the coefficient k is a priori unknown and must be tuned to experimental data, so free-streamline theory can be considered only a qualified success. For further details see Ref. 2, sec. 11.2.
The previous solutions have all been for unbounded flows, such as a circular cylinder immersed in a broad expanse of uniformly streaming fluid, Fig. 8.10a. However, many practical problems involve a nearby rigid boundary constraining the flow, e.g., (1) groundwater flow near the bottom of a dam, (2) an airfoil near the ground, simulating landing or takeoff, or (3) a cylinder mounted in a wind tunnel with narrow walls. In such cases the basic unbounded-potential-flow solutions can be modified for wall ef-fects by the method of images.
Consider a line source placed a distance a from a wall, as in Fig. 8.17a. To create the desired wall, an image source of identical strength is placed the same distance be-low the wall. By symmetry the two sources create a plane-surface streamline between them, which is taken to be the wall.
In Fig. 8.17b a vortex near a wall requires an image vortex the same distance be-low but of opposite rotation. We have shaded in the wall, but of course the pattern could also be interpreted as the flow near a vortex pair in an unbounded fluid.
In Fig. 8.17c an airfoil in a uniform stream near the ground is created by an image airfoil below the ground of opposite circulation and lift. This looks easy, but actually it is not because the airfoils are so close together that they interact and distort each other’s shapes. A rule of thumb is that nonnegligible shape distortion occurs if the body shape is within two chord lengths of the wall. To eliminate distortion, a whole series of “corrective’’ images must be added to the flow to recapture the shape of the origi-nal isolated airfoil. Reference 2, sec. 7.75, has a good discussion of this procedure, which usually requires digital-computer summation of the multiple images needed.
Figure 8.17d shows a source constrained between two walls. One wall required only one image in Fig. 8.17a, but two walls require an infinite array of image sources above and below the desired pattern, as shown. Usually computer summation is necessary, but sometimes a closed-form summation can be achieved, as in the infinite vortex row of Eq. (8.22).
8.6 Images 521 3 This section may be omitted without loss of continuity.
Fig. 8.17 Constraining walls can be created by image flows: (a) source near a wall with identical image source; (b) vortex near a wall with image vortex of opposite sense; (c) airfoil in ground effect with image airfoil of opposite circulation; (d) source between two walls requiring an infinite row of images.
EXAMPLE 8.4 For the source near a wall as in Fig. 8.17a, the wall velocity is zero between the sources, rises to a maximum moving out along the wall, and then drops to zero far from the sources. If the source strength is 8 m2/s, how far from the wall should the source be to ensure that the maxi-mum velocity along the wall will be 5 m/s?
Solution At any point x along the wall, as in Fig. E8.4, each source induces a radial outward velocity r m/r, which has a component r cos along the wall. The total wall velocity is thus 522 Chapter 8 Potential Flow and Computational Fluid Dynamics a a (a) (b) (c) (d) E8.4 uwall 2r cos From the geometry of Fig. E8.4, r (x2 a2)1/2 and cos x/r. Then the total wall velocity can be expressed as u This is zero at x 0 and at x →. To find the maximum velocity, differentiate and set equal to zero 0 at x a and umax We have omitted a bit of algebra in giving these results. For the given source strength and max-imum velocity, the proper distance a is a 1.625 m Ans. For x 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. (8.39), the problem in airfoil theory is to determine the net circulation as a function of airfoil shape and free-stream angle of attack .
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 corre-sponding to different values of circulation . Four examples of this nonuniqueness were shown for the cylinder flows in Fig. 8.10. The same is true of the airfoil, and Fig. 8.18 shows three mathematically acceptable “solutions’’ to a given airfoil flow for small (Fig. 8.18a), large (Fig. 8.18b), and medium (Fig. 8.18c) net circulation. You can guess 8 m2/s 5 m/s m umax m a du dx 2mx x2 a2 8.7 Airfoil Theory 523 Wall θ θ a a x r r Source m = 8 m2/s Source m r = m r r 4 This section may be omitted without loss of continuity.
8.7 Airfoil Theory4 The Kutta Condition Fig. 8.18 The Kutta condition prop-erly simulates the flow about an airfoil; (a) too little circulation, stagnation point on rear upper sur-face; (b) too much, stagnation point on rear lower surface; (c) just right, Kutta condition requires smooth flow at trailing edge.
which case best simulates a real airfoil from the earlier discussion of transient-lift de-velopment in Fig. 7.23. It is the case (Fig. 8.18c) 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, approximating most air-foil 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 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 Kutta depends upon flow velocity, angle of attack, and airfoil shape.
The flat plate is the simplest airfoil, having no thickness or “shape,’’ but even its the-ory is not so simple. The problem can be solved by a complex-variable mapping [2, p. 480], but here we shall use a vortex-sheet approach. Figure 8.19a shows a flat plate of length C simulated by a vortex sheet of variable strength (x). The free stream U is at an angle of attack with respect to the plate chord line.
To make the lift “up’’ with flow from left to right as shown, we specify here that the circulation is positive clockwise. Recall from Fig. 8.7c that there is a jump in tan-gential velocity across a sheet equal to the local strength uu ul (x) (8.53) 524 Chapter 8 Potential Flow and Computational Fluid Dynamics (a) (b) (c) Γ < ΓKutta Γ > ΓKutta Γ = ΓKutta Flat-Plate Airfoil Vortex-Sheet Theory Fig. 8.19 Vortex-sheet solution for the flat-plate airfoil; (a) sheet geometry; (b) theoretical pressure coefficient on upper and lower sur-faces; (c) upper-surface velocity with laminar separation points S.
If we omit the free stream, the sheet should cause a rightward flow u 1 2 on the upper surface and an equal and opposite leftward flow on the lower surface, as shown in Fig. 8.19a. The Kutta condition for this sharp trailing edge requires that this veloc-ity difference vanish at the trailing edge to keep the exit flow smooth and parallel (C) 0 (8.54) The proper solution must satisfy this condition, after which the total lift can be com-puted by summing the sheet strength over the whole airfoil. From Eq. (8.39) for a foil of depth b L Ub C 0 (x) dx (8.55) 8.7 Airfoil Theory 525 α U∞ y 0 (a) δ u ≈1 2 γ x = C x δ u ≈1 2 γ CL ≈ area between curves (b) Cpu = – Cpl Cpl = 2 C x – 1) ( 1 2 (c) Uu U∞ 4° 5° 6° x C Separation: S (6°) S (5°) S (4°) S (3°) γ ( x) Cp sin α = 3° α 8 6 4 2 0 – 2 – 4 – 6 – 8 1.2 1.1 1.00 0.2 0.4 0.6 0.8 1.0 An alternate way to compute lift is from the dimensionless pressure coefficient Cp on the upper and lower surfaces Cpu,l 1 (8.56) where the last expression follows from Bernoulli’s equation. The surface velocity squared is given by combining the uniform stream and the vortex-sheet velocity com-ponents from Fig. 8.19a: U2 u,l (U cos u)2 (U sin )2 U2 u,l U2 2U u cos u2 U2 1 (8.57) where we have made the approximations u U and cos in the last expres-sion, assuming a small angle of attack. Equations (8.56) and (8.57) combine to the first-order approximation Cpu,l (8.58) The lift force is the integral of the pressure difference over the length of the airfoil, as-suming depth b L C 0 (pl pu)b dx or CL 1 0 (Cpl Cpu) 2 1 0 d (8.59) Equations (8.55) and (8.59) are entirely equivalent within the small-angle approxima-tions.
The sheet strength (x) is computed from the requirement that the net normal ve-locity (x) be zero at the sheet (y 0), since the sheet represents a solid plate or stream surface. Consider a small piece of sheet dx located at position x0. The ve-locity at point x on the sheet is that of an infinitesimal line vortex of strength d dx d x The total normal velocity induced by the entire sheet at point x is thus sheet C 0 (8.60) Meanwhile, from Fig. 8.19a, the uniform stream induces a constant normal veloc-ity at every point on the sheet given by stream U sin Setting the sum of sheet and stream equal to zero gives the integral equation C 0 2
U sin (8.61) to be solved for (x) subject to the Kutta condition (C) 0 from Eq. (8.54). dx x0 x dx 2
(x0 x) dx 2
(x0 x) d 2
rx0→x x C U dx C L 1 2 U2 bC U 2 u U 2 u U U2 u,l U2 pu,l p 1 2 U2 526 Chapter 8 Potential Flow and Computational Fluid Dynamics Although Eq. (8.61) is quite formidable (and not only for beginners), in fact it was solved long ago by using integral formulas developed by Poisson in the nineteenth cen-tury. The sheet strength which satisfies Eq. (8.61) is (x) 2U sin 1 1/2 (8.62) From Eq. (8.58) the surface-pressure coefficients are thus Cpu,l 2 sin 1 1/2 (8.63) Details of the calculations are given in advanced texts [for example, 11, chap. 5].
The pressure coefficients from Eq. (8.63) are plotted in Fig. 8.19b, showing that the upper surface has pressure continually increasing with x, that is, an adverse gradient.
The upper-surface velocity Uu U u U 1 2 is plotted in Fig. 8.19c for var-ious angles of attack. Above 5° the sheet contribution u is about 20 percent of U so that the small-disturbance assumption is violated. Figure 8.19c also shows sep-aration points computed by Thwaites’ laminar-boundary-layer method, Eqs. (7.54) and (7.55). The prediction is that a flat plate would be extensively stalled on the upper sur-face for 6°, which is approximately correct.
The lift coefficient of the airfoil is proportional to the area between cpl and cpu in Fig. 8.19b, from Eq. (8.59): CL 2 1 0 d 4 sin 1 0 1 1/2 d 2 sin 2 (8.64) This is a classic result which was alluded to earlier in Eq. (7.70) without proof.
Also of interest is the moment coefficient about the leading edge (LE) of the air-foil, taken as positive counterclockwise CMLE 1 0 (Cpl Cpu) d sin CL (8.65) Thus the center of pressure (CP), or position of the resultant lift force, is at the one-quarter-chord point CP (8.66) This theoretical result is independent of the angle of attack.
These results can be compared with experimental results for NACA airfoils in Fig.
8.20). The thinnest NACA airfoil is t/C 0.06, and the thickest is 24 percent, or t/C 0.24. The lift-curve slope dCL/d is within 9 percent of the theoretical value of 2 for all the various airfoil families at all thicknesses. Increasing thickness tends to increase both CL,max and the stall angle. The stall angle at t/C 0.06 is about 8° and would be even less for a flat plate, verifying the boundary-layer separation esti-mates in Fig. 8.19c. Best performance is usually at about the 12 percent thickness point for any airfoil.
1 4 x C 1 4 2 x C x C MLE 1 2 U2 bC2 x C C x x C U C x C x 8.7 Airfoil Theory 527 Fig. 8.20 Lift characteristics of smooth NACA airfoils as a func-tion of thickness ratio, for infinite aspect ratio. (From Ref. 12.) The theory of thick cambered airfoils is covered in advanced texts [for example, 2 to 4]; Ref. 13 has a thorough and comprehensive review of both inviscid and viscous as-pects of airfoil behavior.
Basically the theory uses a complex-variable mapping which transforms the flow about a cylinder with circulation in Fig. 8.10 into flow about a foil shape with circu-lation. 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 528 Chapter 8 Potential Flow and Computational Fluid Dynamics 7 6 2 π 5 2 π t C 65-series 63-, 64-series 4-digit, 5-digit CL max dCL dα stall α 6% 9% 12% 15% 18% 2π (1 + 0.77 t/C) 2.0 1.0 0 t C 6% 9% 12% 15% 18% 20° 10° 0° t C 6% 9% 12% 15% 18% 00 24-63-230-Series: 00 24-63-230-Series: Potential Theory for Thick Cambered Airfoils Fig. 8.21 Characteristics of NACA airfoils: (a) typical thick cambered airfoil; (b) center-of-pressure data; and (c) minimum drag coefficient.
Kutta bCU1 0.77 sin ( ) (8.67) where tan 1 (2h/C) and h is the maximum camber, or maximum deviation of the airfoil midline from its chord line, as in Fig. 8.21a.
The lift coefficient of the infinite-span airfoil is thus CL 21 0.77 sin ( ) (8.68) t C U 1 2 U2 bC t C 8.7 Airfoil Theory 529 Midline Chordline h (a) t C xcp C (b) CD min C 63-X X X 64-X X X 65-X X X 00X X, 14X X 24X X, 44X X 230X X 6% 9% 12% 15% 18% 0.28 0.27 0.26 0.25 0.24 0.23 t C (c) 6% 9% 12% 15% 18% 0.015 0.010 0.005 0 (Smooth) 66-63-64-65 Rough (all) t i g i d -5 , t i g i d -4 t Wings of Finite Span This reduces to Eq. (8.64) when the thickness and camber are zero. Figure 8.20 shows that the thickness effect 1 0.77t/C is not verified by experiment. Some airfoils in-crease lift with thickness, others decrease, and none approach the theory very closely, the primary reason being the boundary-layer growth on the upper surface affecting the airfoil “shape.’’ Thus it is customary to drop the thickness effect from the theory CL 2 sin ( ) (8.69) The theory correctly predicts that a cambered airfoil will have finite lift at zero angle of attack and zero lift (ZL) at an angle ZL tan 1 (8.70) Equation (8.70) overpredicts the measured zero-lift angle by 1° or so, as shown in Table 8.3. The measured values are essentially independent of thickness. The designation XX in the NACA series indicates the thickness in percent, and the other digits refer to cam-ber and other details. For example, the 2415 airfoil has 2 percent maximum camber (the first digit) occurring at 40 percent chord (the second digit) with 15 percent max-imum thickness (the last two digits). The maximum thickness need not occur at the same position as the maximum camber.
Figure 8.21b shows the measured position of the center of pressure of the various NACA airfoils, both symmetric and cambered. In all cases xCP is within 0.02 chord length of the theoretical quarter-chord point predicted by Eq. (8.66). The standard cam-bered airfoils (24, 44, and 230 series) lie slightly forward of x/C 0.25 and the low-drag (60 series) foils slightly aft. The symmetric airfoils are at 0.25.
Figure 8.21c shows the minimum drag coefficient of NACA airfoils as a function of thickness. As mentioned earlier in conjunction with Fig. 7.25, these foils when smooth actually have less drag than turbulent flow parallel to a flat plate, especially the low-drag 60 series. However, for standard surface roughness all foils have about the same minimum drag, roughly 30 percent greater than that for a smooth flat plate.
The results of airfoil theory and experiment in the previous subsection were for two-dimensional, or infinite-span, wings. But all real wings have tips and are therefore of finite span or finite aspect ratio AR, defined by AR (8.71) b C b2 Ap 2h C 530 Chapter 8 Potential Flow and Computational Fluid Dynamics Airfoil series Camber h/C, % Measured ZL, deg Theory , deg 24XX 2.0 2.1 2.3 44XX 4.0 4.0 4.6 230XX 1.8 1.3 2.1 63-2XX 2.2 1.8 2.5 63-4XX 4.4 3.1 5.0 64-1XX 1.1 0.8 1.2 Table 8.3 Zero-Lift Angle of NACA Airfoils Fig. 8.22 Lifting-line theory for a finite wing: (a) actual trailing-vor-tex system behind a wing; (b) sim-ulation by vortex system “bound’’ to the wing; (c) downwash on the wing due to an element of the trail-ing-vortex system.
where b is the span length from tip to tip and Ap is the planform area of the wing as seen from above. The lift and drag coefficients of a finite-aspect-ratio wing depend strongly upon the aspect ratio and slightly upon the planform shape of the wing.
Vortices cannot end in a fluid; they must either extend to the boundary or form a closed loop. Figure 8.22a shows how the vortices which provide the wing circulation bend downstream at finite wing tips and extend far behind the wing to join the start-ing vortex (Fig. 7.23) downstream. The strongest vortices are shed from the tips, but 8.7 Airfoil Theory 531 U• (a) (b) y x Circulation Γ( y) x Wing replaced by “ lifting line” η y, η γ ( ) d = vor tex sheet element η d w = downwash due to γ d η (c) y = 0 y = 1 2 b y = – 1 2 b some are shed from the body of the wing, as sketched schematically in Fig. 8.22b. The effective circulation (y) of these trailing shed vortices is zero at the tips and usually a maximum at the center plane, or root, of the wing. In 1918 Prandtl successfully mod-eled this flow by replacing the wing by a single lifting line and a continuous sheet of semi-infinite trailing vortices of strength (y) d/dy, as in Fig. 8.22c. Each elemen-tal piece of trailing sheet ( ) d induces a downwash, or downward velocity, dw(y), given by dw(y) at position y on the lifting line. Note the denominator term 4 rather than 2 because the trailing vortex extends only from 0 to rather than from to .
The total downwash w(y) induced by the entire trailing vortex system is thus w(y) (1/2)b (1/2)b (8.72) When the downwash is vectorially added to the approaching free stream U, the ef-fective angle of attack at this section of the wing is reduced to eff i i tan 1 (8.73) where we have used a small-amplitude approximation w U.
The final step is to assume that the local circulation (y) is equal to that of a two-dimensional wing of the same shape and same effective angle of attack. From thin-air-foil theory, Eqs. (8.55) and (8.64), we have the estimate CL 2eff or CU eff (8.74) Combining Eqs. (8.72) and (8.74), we obtain Prandtl’s lifting-line theory for a finite-span wing (y) C(y)U (y) (1/2)b (1/2)b (8.75) This is an integrodifferential equation to be solved for (y) subject to the conditions ( 1 2 b) ( 1 2 b) 0. It is similar to the thin-airfoil integral equation (8.61) and even more formidable. Once it is solved, the total wing lift and induced drag are given by L U (1/2)b (1/2)b (y) dy Di U (1/2)b (1/2)b (y)i(y) dy (8.76) Here is a case where the drag is not zero in a frictionless theory because the down-wash causes the lift to slant backward by angle i so that it has a drag component par-allel to the free-stream direction, dDi dL sin i dLi.
The complete solution to Eq. (8.75) for arbitrary wing planform C(y) and arbitrary (d/d ) d y 1 4U Ub 1 2 U2 bC w U w U ( ) d y 1 4 ( ) d 4(y ) 532 Chapter 8 Potential Flow and Computational Fluid Dynamics twist (y) is treated in advanced texts [for example, 11]. It turns out that there is a sim-ple representative solution for an untwisted wing of elliptical planform C(y) C0 1 2 1/2 (8.77) The area and aspect ratio of this wing are Ap (1/2)b (1/2)b C dy bC0 AR (8.78) The solution to Eq. (8.75) for this C(y) is an elliptical circulation distribution of ex-actly similar shape (y) 0 1 2 1/2 (8.79) Substituting into Eq. (8.75) and integrating give a relation between 0 and C0 0 (8.80) where is assumed constant across the untwisted wing.
Substitution into Eq. (8.76) gives the elliptical-wing lift L 1 4 2bC0 U2 /(1 2/AR) or CL (8.81) If we generalize this to a thick cambered finite wing of approximately elliptical plan-form, we obtain CL (8.82) This result was given without proof as Eq. (7.70). From Eq. (8.72) the computed down-wash for the elliptical wing is constant w(y) const (8.83) Finally, the induced drag coefficient from Eq. (8.76) is CDi CL (8.84) This was given without proof as Eq. (7.71).
Figure 8.23 shows the effectiveness of this theory when tested against a nonellipti-cal cambered wing by Prandtl in 1921 . Figure 8.23a and b shows the measured lift curves and drag polars for five different aspect ratios. Note the increase in stall an-gle and drag and the decrease in lift slope as the aspect ratio decreases.
Figure 8.23c shows the lift data replotted against effective angle of attack eff ( )/(1 2/AR), as predicted by Eq. (8.82). These curves should be equivalent to C2 L AR w U 2U 2 AR 2 sin ( ) 1 2/AR 2 1 2/AR C0U 1 2/AR 2y b 4b C0 1 4 2y b 8.7 Airfoil Theory 533 Fig. 8.23 Comparison of theory and experiment for a finite wing: (a) measured lift ; (b) measured drag polar ; (c) lift reduced to infinite aspect ratio; (d) drag polar reduced to infinite aspect ratio.
an infinite-aspect-ratio wing, and they do collapse together except near stall. Their com-mon slope dCL/d is about 10 percent less than the theoretical value 2, but this is consistent with the thickness and shape effects noted in Fig. 8.20.
Figure 8.23d shows the drag data replotted with the theoretical induced drag CDi C2 L /(AR) subtracted out. Again, except near stall, the data collapse onto a single line of nearly constant infinite-aspect-ratio drag CD0 0.01. We conclude that the finite-wing theory is very effective and may be used for design calculations.
The same superposition technique which worked so well for plane flow in Sec. 8.3 is also successful for axisymmetric potential flow. We give some brief examples here.
Most of the basic results carry over from plane to axisymmetric flow with only slight changes owing to the geometric differences. Consider the following related flows: 534 Chapter 8 Potential Flow and Computational Fluid Dynamics α CL AR = 7 5 3 2 1 (a) CD (c) (d ) CD – C 2 L πAR β = – 5° 1 + 2 /AR α + β 1.5 1.0 0.5 0 – 5° 0° 10° 20° CL AR = 7 5 3 2 1 (b) 1.5 1.0 0.5 00 0.2 0.1 CDo ≈ 0.01 CL 1.5 1.0 0.5 0 – 5° 0° 10° 20° CL 1.5 1.0 0.5 00 0.1 0.05 2 π ( + ) α β 1 7 3,2 AR = 5 AR = 2 7 5 3 1 8.8 Axisymmetric Potential Flow5 5 This section may be omitted without loss of continuity.
Fig. 8.24 Spherical polar coordi-nates for axisymmetric flow.
Spherical Polar Coordinates Basic plane flow Counterpart axisymmetric flow Uniform stream Uniform stream Line source or sink Point source or sink Line doublet Point doublet Line vortex No counterpart Rankine half-body cylinder Rankine half-body of revolution Rankine-oval cylinder Rankine oval of revolution Circular cylinder Sphere Symmetric airfoil Tear-shaped body Since there is no such thing as a point vortex, we must forgo the pleasure of studying circulation effects in axisymmetric flow. However, as any smoker knows, there is an axisymmetric ring vortex, and there are also ring sources and ring sinks, which we leave to advanced texts [for example, 3].
Axisymmetric potential flows are conveniently treated in the spherical polar coordi-nates of Fig. 8.24. There are only two coordinates (r, ), and flow properties are con-stant on a circle of radius r sin about the x-axis.
The equation of continuity for incompressible flow in these coordinates is ∂ ∂ r (r2r sin ) ∂ ∂ (r sin ) 0 (8.85) where r and are radial and tangential velocity as shown. Thus a spherical polar stream function6 exists such that r (8.86) In like manner a velocity potential (r, ) exists such that r ∂ ∂ r (8.87) ∂ ∂ 1 r ∂ ∂r 1 r sin ∂ ∂ 1 r2 sin 8.8 Axisymmetric Potential Flow 535 y z r θ νθ νr Properties vary with on a circle about z axis θ Axis of symmetry Properties do not vary on a circle about x axis x 6 It is often called Stokes’ stream function, having been used in a paper Stokes wrote in 1851 on vis-cous sphere flow.
Uniform Stream plus a Point Source Uniform Stream in the x Direction Point Source or Sink Point Doublet These formulas serve to deduce the and functions for various elementary-axisymmetric potential flows.
A stream U in the x direction has components r U cos U sin Substitution into Eqs. (8.86) and (8.87) and integrating give Uniform stream: 1 2 Ur2 sin2 Ur cos (8.88) As usual, arbitrary constants of integration have been neglected.
Consider a volume flux Q issuing from a point source. The flow will spread out radi-ally and at radius r will equal Q divided by the area 4r2 of a sphere. Thus r 0 (8.89) with m Q/(4) for convenience. Integrating (8.86) and (8.87) gives Point source m cos (8.90) For a point sink, change m to m in Eq. (8.90).
Exactly as in Fig. 8.8, place a source at (x, y) ( a, 0) and an equal sink at ( a, 0), taking the limit as a becomes small with the product 2am held constant doublet lim a→0 (m cos source m cos sink) (8.91) We leave the proof of this limit as a problem. The point-doublet velocity potential doublet lim a→0 (8.92) The streamlines and potential lines are shown in Fig. 8.25. Unlike the plane doublet flow of Fig. 8.8, neither set of lines represents perfect circles.
By combining Eqs. (8.88) and (8.90) we obtain the stream function for a uniform stream plus a point source at the origin 1 2 Ur2 sin2 m cos (8.93) From Eq. (8.86) the velocity components are, by differentiation, r U cos U sin (8.94) m r2 cos r2 m rsink m rsource sin2 r m r m r2 Q 4r2 536 Chapter 8 Potential Flow and Computational Fluid Dynamics 2am 2am Fig. 8.25 Streamlines and potential lines due to a point doublet at the origin, from Eqs. (8.91) and (8.92).
Setting these equal to zero reveals a stagnation point at 180° and r a (m/U)1/2, as shown in Fig. 8.26. If we let m Ua2, the stream function can be rewrit-ten as cos 2 sin2 (8.95) The stream surface which passes through the stagnation point (r, ) (a, ) has the value Ua2 and forms a half-body of revolution enclosing the point source, as shown in Fig. 8.26. 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 max-imum velocity and minimum pressure along the half-body surface occur at 70.5°, r a 1 2 Ua2 8.8 Axisymmetric Potential Flow 537 x y Potential lines Stagnation point U∞ 2 a 2 a Source Halfbody Vs max = 1.155 U∞ r a = csc θ 2 θ x a y r Fig. 8.26 Streamlines for a Rankine half-body of revolution.
Fig. 8.27 Streamlines and potential lines for inviscid flow past a sphere.
r a 3 , Vs 1.155U. Downstream of this point there is an adverse gradient as Vs slowly decelerates to U, but boundary-layer theory indicates no flow separation. Thus Eq. (8.95) 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, e.g., similar to Fig. 8.5c, separation is predicted and the inviscid pattern is not realistic.
From Eqs. (8.88) and (8.91), combination of a uniform stream and a point doublet at the origin gives Ur2 sin2 sin2 (8.96) Examination of this relation reveals that the stream surface 0 corresponds to the sphere of radius r a 1/3 (8.97) This is exactly analogous to the cylinder flow of Fig. 8.10a formed by combining a uniform stream and a line doublet.
Letting 1 2 Ua3 for convenience, we rewrite Eq. (8.96) as sin2 (8.98) The streamlines for this sphere flow are plotted in Fig. 8.27. By differentiation from Eq. (8.86) the velocity components are r U cos 1 U sin 2 (8.99) We see that the radial velocity vanishes at the sphere surface r a, as expected. There is a stagnation point at the front (a, ) and the rear (a, 0) of the sphere. The maximum a3 r3 1 2 a3 r3 a r r2 a2 1 2 Ua2 2 U r 1 2 538 Chapter 8 Potential Flow and Computational Fluid Dynamics Uniform Stream plus a Point Doublet U∞ Vmax = 1.5 U∞ Potential lines a S θ Laminar separation at 76° The Concept of Hydrodynamic Mass velocity occurs at the shoulder (a, 1 2 ), where r 0 and 1.5U. The surface-velocity distribution is Vs ra 3 2 U sin (8.100) Note the similarity to the cylinder surface velocity equal to 2U sin from Eq. (8.34) with zero circulation.
Equation (8.100) predicts, as expected, an adverse pressure gradient on the rear ( 90°) of the sphere. If we use this distribution with laminar-boundary-layer theory [for example, 15, p. 298], separation is computed to occur at about 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. (8.100) to be inaccurate even in the front of the sphere. The measured maximum surface velocity is equal only to about 1.3U and oc-curs at about 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 called the added or virtual mass) of the fluid. If the instantaneous body velocity is U(t), the summation of forces must include this effect F (m mh) (8.101) where mh, the hydrodynamic mass, is a function of body shape, the direction of mo-tion, and (to a lesser extent) flow parameters such as the Reynolds number.
According to potential theory [2, sec. 6.4; 3, sec. 9.22], mh depends only on the shape and direction of motion and can be computed by summing the total kinetic en-ergy of the fluid relative to the body and setting this equal to an equivalent body en-ergy KEfluid 1 2 dm V2 rel 1 2 mhU2 (8.102) The integration of fluid kinetic energy can also be accomplished by a body-surface in-tegral involving the velocity potential [16, sec. 11].
Consider the previous example of a sphere immersed in a uniform stream. By sub-tracting out the stream velocity we can replot the flow as in Fig. 8.28, showing the streamlines relative to the moving sphere. Note the similarity to the doublet flow in Fig. 8.25. The relative-velocity components are found by subtracting U from Eqs. (8.99) r The element of fluid mass, in spherical polar coordinates, is dm (2r sin )r dr d When dm and V2 rel 2 r 2 are substituted into Eq. (8.102), the integral can be eval-uated KEfluid 1 3 a3U2 Ua3 sin 2r3 Ua3 cos r3 dU dt 8.8 Axisymmetric Potential Flow 539 Fig. 8.28 Potential-flow streamlines relative to a moving sphere. Com-pare with Figs. 8.25 and 8.27.
8.9 Numerical Analysis or mh(sphere) 2 3 a3 (8.103) 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.
(8.33) after subtracting out the stream velocity. The result is mh(cylinder) a2L (8.104) 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 (FDM) [5, 20] 3.
a. Integral methods with distributed singularities b. The boundary-element method 540 Chapter 8 Potential Flow and Computational Fluid Dynamics U V Fluid particle: d m d ( KE ) = 1 2 d m V 2 The Finite-Element Method The Finite-Difference Method Methods 3a and 3b are closely related, having first been developed on an ad hoc ba-sis by aerodynamicists in the 1960s and then generalized into a multipurpose ap-plied-mechanics technique in the 1970s .
Methods 1 (or FEM) and 2 (or FDM), though strikingly different in concept, are comparable in scope, mesh size, and general accuracy. We concentrate here on the lat-ter method for illustration purposes.
The finite-element method is applicable to all types of linear and nonlinear par-tial differential equations in physics and engineering. The computational domain is di-vided into small regions, usually triangular or quadrilateral. These regions are delin-eated with a finite number of nodes where the field variables—temperature, velocity, pressure, stream function, etc.—are to be calculated. The solution in each region is ap-proximated by an algebraic combination of local nodal values. Then the approximate 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 un-known nodal values. The nodal equations are solved simultaneously, by matrix inver-sion 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 nodel 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 it-eration or matrix inversion. Nodal spacings need not be equal.
Here we illustrate the two-dimensional Laplace equation, choosing for convenience the stream-function form 0 (8.105) subject to known values of along any body surface and known values of ∂ /∂x and ∂ /∂y in the free stream.
Our finite-difference technique divides the flow field into equally spaced nodes, as shown in Fig. 8.29. To economize on the use of parentheses or functional notation, sub-scripts i and j denote the position of an arbitrary, equally spaced node, and i,j denotes the value of the stream function at that node i,j (x0 i x, y0 j y) Thus, i1,j is just to the right of i,j, and i,j1 is just above.
An algebraic approximation for the derivative ∂ /∂x is (x x, y) (x, y) x ∂ ∂x ∂2 ∂y2 ∂2 ∂x2 8.9 Numerical Analysis 541 Fig. 8.29 Definition sketch for a two-dimensional rectangular finite-difference grid.
A similar approximation for the second derivative is The subscript notation makes these expressions more compact ( i1,j i,j) (8.106) ( i1,j 2 i,j i 1,j) These formulas are exact in the calculus limit as x →0, but in numerical analysis we keep x and y finite, hence the term finite differences.
In an exactly similar manner we can derive the equivalent difference expressions for the y direction ( i,j1 i,j) ( i,j1 2 i,j i,j 1) (8.107) The use of subscript notation allows these expressions to be programmed directly into a scientific computer language such as BASIC or FORTRAN.
When (8.106) and (8.107) are substituted into Laplace’s equation (8.105), the result is the algebraic formula 2(1 ) i,j i 1,j i1,j ( i,j 1 i,j1) (8.108) 1 y2 ∂2 ∂y2 1 y ∂ ∂y 1 x2 ∂2 ∂x2 1 x ∂ ∂x (x, y) (x x, y) x (x x, y) (x, y) x 1 x ∂2 ∂x2 542 Chapter 8 Potential Flow and Computational Fluid Dynamics ∆ y ∆ x ∆ x ∆ y ψi, j + 1 ψi, j – 1 ψi –1, j ψi +1, j ψi, j where ( x/ y)2 depends upon the mesh size selected. This finite-difference model of Laplace’s equation states that every nodal stream-function value i,j is a linear com-bination of its four nearest neighbors.
The most commonly programmed case is a square mesh ( 1), for which Eq.
(8.108) reduces to i,j 1 4 ( i,j1 i,j 1 i1,j i 1,j) (8.109) Thus, for a square mesh, each nodal value equals the arithmetic average of the four neighbors shown in Fig. 8.29. The formula is easily remembered and easily pro-grammed. If P(I, J) is a subscripted variable stream function, the BASIC or FORTRAN statement of (8.109) is P(I, J) 0.25 (P(I, J 1) P(I, J 1) P(I 1, J) P(I 1, J)) (8.110) This is applied in iterative fashion sweeping over each of the 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 P(I, J), 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.
Convergence can be speeded up by the successive overrelaxation (SOR) method, discussed by Patankar . The modified SOR form of the iteration is P(I, J) P(I, J) 0.25 A (P(I, J 1) P(I, J 1) P(I 1, J) P(I 1, J) 4 P(I, J)) (8.111) The recommended value of the SOR convergence factor A is about 1.7. Note that the value A 1.0 reduces Eq. (8.111) to (8.110).
Let us illustrate the finite-difference method with an example.
EXAMPLE 8.5 Make a numerical analysis, using x y 0.2 m, of potential flow in the duct expansion shown in Fig. 8.30. The flow enters at a uniform 10 m/s, where the duct width is 1 m, and is assumed to leave at a uniform velocity of 5 m/s, where the duct width is 2 m. There is a straight section 1 m long, a 45° expansion section, and a final straight section 1 m long.
Solution Using the mesh shown in Fig. 8.30 results in 45 boundary nodes and 91 internal nodes, with i varying from 1 to 16 and j varying from 1 to 11. The internal points are modeled by Eq. (8.110).
For convenience, let the stream function be zero along the lower wall. Then since the volume flow is (10 m/s)(1 m) 10 m2/s per unit depth, the stream function must equal 10 m2/s along the upper wall. Over the entrance and exit planes, the stream function must vary linearly to give uniform velocities: Inlet: (1, J) 2 (J 6) for J 7 to 10 Exit: (16, J) J 1 for J 2 to 10 8.9 Numerical Analysis 543 Fig. 8.30 Numerical model of po-tential flow through a two-dimen-sional 45° expansion. The nodal points shown are 20 cm apart.
There are 45 boundary nodes and 91 internal nodes.
All these boundary values must be input to the program and are shown printed in Fig. 8.31.
Initial guesses are stored for the internal points, say, zero or an average value of 5.0 m2/s.
The program then starts at any convenient point, such as the upper left (2, 10), and evaluates Eq.
(8.110) at every internal point, repeating this sweep iteratively until there are no further changes (within some selected maximum change) in the nodal values. The results are the finite-differ-ence simulation of this potential flow for this mesh size; they are shown printed in Fig. 8.31 to three-digit accuracy. The reader should test a few nodes in Fig. 8.31 to verify that Eq. (8.110) is satisfied everywhere. The numerical accuracy of these printed values is difficult to estimate, since there is no known exact solution to this problem. In practice, one would keep decreasing the mesh size to see whether there were any significant changes in nodal values.
This problem is well within the capability of a small personal computer. The values shown in Fig. 8.31 were obtained after 100 iterations, or 6 min of execution time, on a Macintosh SE personal computer, using BASIC.
544 Chapter 8 Potential Flow and Computational Fluid Dynamics (1, 11) y = 2 m (16, 11) (i, j) 10 m /s 5 m /s i j (11, 1) y = 0 m (16, 1) 1 m 1 m 1 m (1, 6) y = 1 m (6, 6) 45° ψ = 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 8.00 8.02 8.04 8.07 8.12 8.20 8.30 8.41 8.52 8.62 8.71 8.79 8.85 8.91 8.95 10.00 9.00 6.00 6.03 6.06 6.12 6.22 6.37 6.58 6.82 7.05 7.26 7.44 7.59 7.71 7.82 7.91 8.00 4.00 4.03 4.07 4.13 4.26 4.48 4.84 5.24 5.61 5.93 6.19 6.41 6.59 6.74 6.88 7.00 2.00 2.02 2.05 2.09 2.20 2.44 3.08 3.69 4.22 4.65 5.00 5.28 5.50 5.69 5.85 6.00 0.00 0.00 0.00 0.00 0.00 1.33 2.22 2.92 3.45 3.87 4.19 4.45 4.66 4.84 5.00 ψ = 0.00 0.00 1.00 1.77 2.37 2.83 3.18 3.45 3.66 3.84 4.00 0.00 0.80 1.42 1.90 2.24 2.50 2.70 2.86 3.00 0.00 0.63 1.09 1.40 1.61 1.77 1.89 2.00 0.00 0.44 0.66 0.79 0.87 0.94 1.00 0.00 0.00 0.00 0.00 0.00 0.00 Fig. 8.31 Stream-function nodal val-ues for the potential flow of Fig.
8.30. Boundary values are known in-puts. Internal nodes are solutions to Eq. (8.110).
Although Fig. 8.31 is the computer solution to the problem, these numbers must be manip-ulated to yield practical engineering results. For example, one can interpolate these numbers to sketch various streamlines of the flow. This is done in Fig. 8.32a. We see that the streamlines are curved both upstream and downstream of the corner regions, especially near the lower wall.
This indicates that the flow is not one-dimensional.
The velocities at any point in the flow can be computed from finite-difference formulas such as Eqs. (8.106) and (8.107). For example, at the point (I, J) (3, 6), from Eq. (8.107), the hor-izontal velocity is approximately u(3, 6) 10.45 m/s and the vertical velocity is zero from Eq. (8.106). Directly above this on the upper wall, we es-timate u(3, 11) 9.65 m/s 10.00 8.07 0.2 (3, 11) (3,10) y 2.09 0.00 0.2 (3, 7) (3, 6) y 8.9 Numerical Analysis 545 p1 V1 (a) ψ = 10 8 6 4 2 0 0.75 One-dimensional approximation Eq. (1) Lower surface Upper surface Cp = p – p1 ρV1 2/ 2 1.0 0.8 0.6 0.4 0.2 0.0 – 0.2 – 0.4 – 0.6 – 0.8 (b) Fig. 8.32 Useful results computed from Fig. 8.31: (a) streamlines of the flow; (b) pressure-coefficient distribution along each wall.
The Boundary-Element Method The flow is not truly one-dimensional in the entrance duct. The lower wall, which contains the diverging section, accelerates the fluid, while the flat upper wall is actually decelerating the fluid.
Another output function, useful in making boundary-layer analyses of the wall regions, is the pressure distribution along the walls. If p1 and V1 are the pressure and velocity at the entrance (I 1), conditions at any other point are computed from Bernoulli’s equation (8.3), neglecting gravity p 1 2 V2 p1 1 2 V2 1 which can be rewritten as a dimensionless pressure coefficient Cp 1 2 This determines p after V is computed from the stream-function differences in Fig. 8.31.
Figure 8.32b shows the computed wall-pressure distributions as compared with the one-dimensional continuity approximation V1A1 V(x)A(x), or Cp(one-dim) 1 2 (1) The one-dimensional approximation, which is rather crude for this large (45°) expansion, lies between the upper and lower wall pressures. One-dimensional theory would be much more ac-curate for a 10° expansion.
Analyzing Fig. 8.32b, we predict that boundary-layer separation will probably occur on the lower wall between the corners, where pressure is strongly rising (highly adverse gradient). There-fore potential theory is probably not too realistic for this flow, where viscous effects are strong.
(Recall Figs. 6.27 and 7.8.) Potential theory is reversible; i.e., when we reverse the flow arrows in Fig. 8.32a, then Fig.
8.32b is still valid and would represent a 45° contraction flow. The pressure would fall on both walls (no separation) from x 3 m to x 1 m. Between x 1 m and x 0, the pressure rises on the lower surface, indicating possible separation, probably just downstream of the corner.
This example should give the reader an idea of the usefulness and generality of numerical analysis of fluid flows.
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, including FORTRAN programs for potential theory and elas-tostatics. There are no interior elements. Rather, all nodes are placed on the boundary of the domain, as in Fig. 8.33. The “element’’ is a small piece of the boundary sur-rounding the node. The “strength’’ of the element can be either constant or variable.
For plane potential flow, the method takes advantage of the particular solution ln (8.112) which satisfies Laplace’s equation, 2 0. Each element i is assumed to have a dif-ferent strength i. Then r represents the distance from that element to any other point in the flow field. Summing all these elemental effects, with proper boundary condi-tions, will give the total solution to the potential-flow problem.
At each element of the boundary, we typically know either the value of or the value of ∂ /∂n, where n is normal to the boundary. (Mixed combinations of and 1 r 1 2 A1 A V V1 p p1 1 2 V2 1 546 Chapter 8 Potential Flow and Computational Fluid Dynamics ∂ /∂n are also possible but are not discussed here.) The correct strengths i 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 (singu-lar) 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 i N j1 j 0 j ds N j1 j 0 j ds i 1 to N (8.113) The integrals, which involve the logarithmic particular solution from Eq. (8.112), are evaluated numerically for each element. Reference 7 recommends and gives a program for gaussian quadrature formulas.
Equations (8.113) contain 2N element values, i and (∂ /∂n)i, of which N are known from the given boundary conditions. The remaining N are solved simultaneously from Eqs. (8.113). Generally this completes the analysis only the boundary solution is computed, and interior points are not studied. In most cases, the boundary velocity and pressure are all that is needed.
We illustrated the method with stream function . Naturally the entire technique also applies to velocity potential , if we are given proper conditions on or ∂/∂n at each boundary element. The method is readily extended to three dimensions .
Reference 7 gives a complete FORTRAN listing for solving Eqs. (8.113) numeri-cally for constant, linear, and quadratic element strength variations. We now use their constant-element-strength program, POCONBE , to take an alternate look at Ex-ample 8.5, which used the finite-difference method.
EXAMPLE 8.6 Solve the duct expansion problem, Example 8.5, using boundary elements. Use the same grid spacing x y 0.2 m for the element sizes.
Solution The boundary nodes are equally spaced, as shown in Fig. 8.34. There are only 45 nodes, whereas there were 91 interior points for the FDM solution of Example 8.5. We expect the same accu-racy for 50 percent fewer nodes. (Had we reduced the grid size to 0.1 m, there would be 90 ∂ ∂n ∂ ∂n 1 2 8.9 Numerical Analysis 547 Element j Node j Domain: 2 ψ = 0 ∆ rj n Element i Node i ds Fig. 8.33 Boundary elements of constant strength in plane potential flow.
One-Dimensional Unsteady Flow Fig. 8.34 Boundary elements corre-sponding to the same grid size as Fig. 8.31. Nodal values of stream function and computed surface ve-locity are shown.
nodes as opposed to 406 interior points a savings of 78 percent.) The program POCONBE asks you to input the location of these 45 nodes. The stream-function values are known all around the boundary: equals 0 on the bottom and 10.0 on the top and is linearly increasing from 0 to 10.0 at entrance and exit. These values of , shown on the outside in Fig. 8.34, are inputted into the program.
Once the input of nodes and element values is complete, the program immediately computes and displays or stores the 45 unknowns, which in this case are the values of ∂ /∂n all around the boundary. These values are shown on the inside of the top and bottom surfaces in Fig. 8.34 and represent the local surface velocity near each element, in m/s. The values of ∂ /∂n at en-trance and exit, which are small fractions representing vertical velocity components, are not shown here.
548 Chapter 8 Potential Flow and Computational Fluid Dynamics 8 6 4 2 9 8 7 6 5 4 3 2 1 9.73 9.68 9.46 9.12 8.68 8.16 7.62 7.10 6.63 6.23 5.88 5.59 5.28 5.54 10.8 10.9 10.1 10.4 10.8 14.1 2.73 11.2 7.14 5.69 4.49 2.23 3.61 4.15 4.48 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Boundary velocities listed inside Stream function values listed outside [In BEM there are no interior nodes.] U = 10.0 m/s U = 5.0 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 5.31 The reader may verify that use of the surface velocities in Fig. 8.34 to compute surface pres-sure coefficients, as in Example 8.5, leads to curves very similar to those shown in Fig. 8.32.
The BEM approach, using the same boundary nodes, has accuracy comparable to that for an FDM computation. For further details see Ref. 7.
Our previous finite-difference model of Laplace’s equation, e.g., Eq. (8.109), 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 different, and they have been met to a large extent, so there are now many textbooks [20, 23 to 27] on (fully viscous) computational fluid dynamics (CFD). This is not a textbook on CFD, but we will address some of the issues in this section.
We begin with a simplified problem, showing that even a single viscous term intro-duces new effects and possible instabilities. Recall (or review) Prob. 4.85, 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 U0(t), as in Fig. 8.35. A uniform vertical grid, of spacing y, has nodes n at which the local velocity un j is to be calculated, where su-Viscous-Flow Computer Models Fig. 8.35 An equally spaced finite-difference mesh for one-dimen-sional viscous flow [Eq. (8.114)].
perscript j denotes the time-step j t. The wall is n 1. If u u(y, t) only and w 0, continuity, V 0, is satisfied and we need only solve the x-momentum Navier-Stokes equation: v (8.114) where !/ . Utilizing the same finite-difference approximations as in Eq. (8.106), we may model Eq. (8.114) algebraically as a forward time difference and a central spa-tial difference: Rearrange and find that we can solve explicitly for un at the next time-step j 1: un j1 (1 2") un j " (un 1 j un1 j ) " (8.115) Thus u at node n at the next time-step j 1 is a weighted average of three previous values, similar to the “four-nearest-neighbors” average in the laplacian model of Eq.
(8.109). Since the new velocity is calculated immediately, Eq. (8.115) is called an ex-plicit model. It differs from the well-behaved laplacian model, however, because it may be unstable. The weighting coefficients in Eq. (8.115) must all be positive to avoid di-vergence. Now " is positive, but (1 2") may not be. Therefore, our explicit viscous flow model has a stability requirement: " # (8.116) Normally one would first set up the mesh size y in Fig. 8.35, after which Eq. (8.116) would limit the time-step t. The solutions for nodal values would then be stable, but not necessarily that accurate. The mesh sizes y and t could be reduced to increase accuracy, similar to the case of the potential-flow laplacian model (8.109).
For example, to solve Prob. 4.85 numerically, one sets up a mesh with plenty of nodes (30 or more y within the expected viscous layer); selects t according to Eq.
1 2 t y2 t y2 u n1 j 2un j u n 1 j y2 un j1 un j t ∂2u ∂y2 ∂u ∂t 8.9 Numerical Analysis 549 y y y y Wall n 1 n 1 n 1 u U0 n An Alternate Implicit Approach (8.116); and sets two boundary conditions for all j: u1 U0 sin t† and uN 0, where N is the outermost node. For initial conditions, perhaps assume the fluid initially at rest: un 1 0 for 2 # n # N 1. Sweeping the nodes 2 # n # N 1 using Eq. (8.115) (an Excel spreadsheet is excellent for this), one generates numerical values of un j 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. 8.115 to demonstrate this. In many finite-difference problems, a stability limitation such as Eq. (8.116) requires an extremely small time-step. To allow larger steps, one can recast the model in an im-plicit fashion by evaluating the second-derivative model in Eq. (8.114) at the next time-step: un j1 t un j This rearrangement is unconditionally stable for any ", but now we have three un-knowns: "un 1 j1 (1 2")un j1 " un1 j1 un j (8.117) This is an implicit model, meaning that one must solve a large system of algebraic equations for the new nodal values at time j 1. Fortunately, the system is narrowly banded, with the unknowns confined to the principal diagonal and its two nearest di-agonals. In other words, the coefficient matrix of Eq. (8.117) is tridiagonal, a happy event. A direct method, called the tridiagonal matrix algorithm (TDMA), is available and explained in most CFD texts [20, 23 to 27]. Appendix A of Ref. 20 includes a complete program for solving the TDMA. If you have not learned the TDMA yet, Eq.
(8.117) converges satisfactorily by rearrangement and iteration: un j1 (8.118) At each time-step j 1, sweep the nodes 2 # n # N 1 over and over, using Eq.
(8.118), until the nodal values have converged. This implicit method is stable for any ", however large. To ensure accuracy, though, one should keep t and y small com-pared to the basic time and length scales of the problem. This author’s habit is to keep t and y small enough that nodal values change no more than 10 percent from one (n, j) to the next.
EXAMPLE 8.7 SAE 30 oil at 20°C is at rest near a wall when the wall suddenly begins moving at a constant 1 m/s. Using the explicit model of Eq. (8.114), estimate the oil velocity at y 3 cm after 1 sec-ond of wall motion.
un j " (un 1 j1 un1 j1) 1 2" un1 j1 2un j1 un 1 j1 y2 550 Chapter 8 Potential Flow and Computational Fluid Dynamics †Finite differences are not analytical; one must set U0 and equal to numerical values.
Steady Two-Dimensional Laminar Flow Solution For SAE 30 oil, from Table A-3, 0.29/891 3.25 E-4 m2/s. For convenience in putting a node exactly at y 3 cm, choose y 0.01 m. The stability limit (8.116) is t/ y2 0.5, or t 0.154 s. Again for convenience, to hit t 1 s on the nose, choose t 0.1 s, or " 0.3255 and (1 2") 0.3491. Then our explicit algebraic model (8.115) for this problem is un j1 0.3491 un j 0.3255(un 1 j un 1 j ) (1) We apply this relation from n 2 out to at least n N 15, to make sure that the desired value of u at n 3 is accurate. The wall no-slip boundary requires u1 j 1.0 m/s constant for all j.
The outer boundary condition is uN 0. The initial conditions are un 1 0 for n $ 2. We then apply Eq. (1) repeatedly for n $ 2 until we reach j 11, which corresponds to t 1 s. This is easily programmed on a spreadsheet such as Excel. Here we print out only j 1, 6, and 11 as follows: j t u1 u2 u3 u4 u5 u6 u7 u8 u9 u10 u11 1 0.000 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 6 0.500 1.000 0.601 0.290 0.107 0.027 0.004 0.000 0.000 0.000 0.000 0.000 11 1.000 1.000 0.704 0.446 0.250 0.123 0.052 0.018 0.005 0.001 0.000 0.000 Note: Units for t and u’s are s and m/s, respectively.
Our numerical estimate is u4 11 u(3 cm, 1 s) 0.250 m/s, which is about 4 percent high— this problem has a known exact solution, u 0.241 m/s . We could improve the accuracy indefinitely by decreasing y and t.
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: Continuity: 0 (8.119a) x momentum: u (8.119b) y momentum u (8.119c) These equations, to be solved for (u, , 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 dom-inant derivatives involve p. This situation has led to several different “pressure-adjust-ment” schemes in the literature [20, 23 to 27], most of which manipulate the continu-ity equation to insert a pressure correction.
A second difficulty in Eq. (8.119b and c) is the presence of nonlinear convective accelerations such as u(∂u/∂x), which create asymmetry in viscous flows. Early at-∂2 ∂y2 ∂2 ∂x2 ∂p ∂y 1 ∂ ∂y ∂ ∂x ∂2u ∂y2 ∂2u ∂x2 ∂p ∂x 1 ∂u ∂y ∂u ∂x ∂ ∂y ∂u ∂x 8.9 Numerical Analysis 551 Commercial CFD Codes tempts, which modeled such terms with a central difference, led to numerical instabil-ity. The remedy is to relate convection finite differences solely to the upwind flow en-tering the cell, ignoring the downwind cell. For example, the derivative ∂u/∂x could be modeled, for a given cell, as (uupwind ucell)/ x. 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. Fig-ure 8.36 illustrates a CFD solution of two-dimensional flow past an NACA 66(MOD) hydrofoil . The gridding in Fig. 8.36a 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 for-mation when surface pressures drop below the local vapor pressure. Figure 8.36b com-pares computed and experimental surface pressure coefficients for an angle of attack of 1°. The dimensionless pressure coefficient is defined as Cp (psurface p)/( V 2/2).
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 ex-tremely effective tool for engineers.
The coming of the third millennium has seen an enormous emphasis on computer ap-plications in nearly every field, fluid mechanics being a prime example. It is now pos-sible, at least for moderately complex geometries and flow patterns, to model on a dig-ital computer, approximately, the equations of motion of fluid flow, with dedicated CFD textbooks available [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 modern 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 are extensive software packages which 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 con-venience is great; the danger is also great. That is, computations are not merely auto-matic, 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 VL/!, 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 conditions and may be inaccurate or unrealistic for others. This is discussed by Freitas , who 552 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.36 CFD results for water flow past an NASA 66(MOD) hy-drofoil [from Ref 28, with permis-sion of the American Society of Mechanical Engineers]: (a) C grid-ding, 262 by 91 nodes; (b) surface pressures at 1°.
compared eight different commercial-code calculations (FLOW-3D, FLOTRAN, STAR-CD, N3S, CFD-ACE, FLUENT, CFDS-FLOW3D, and NISA/3D-FLUID) with exper-imental results for five benchmark flow experiments. Calculations were made by the vendors themselves. Freitas concludes that commercial codes, though promising in gen-eral, can be inaccurate for certain laminar- and turbulent-flow situations. Further re-search is recommended before engineers can truly rely upon such software to give gen-erally accurate fluid-flow predictions.
In spite of the above 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 turbulent flow past a cube mounted on the floor of a channel whose clearance is twice 8.9 Numerical Analysis 553 (a) 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.1 0.2 0.0 0.5 1.0 G Expt.
Comp.
Cp x/C (b) G G G G G G G G G G G G G Fig. 8.37 Flow over a surface-mounted cube creates a complex and perhaps unexpected pattern: (a) experimental oil-streak visualiza-tion of surface flow at Re 40,000 (based on cube height) ( from Ref.
31, courtesy of Robert Martinuzzi, with the permission of the Ameri-can Society of Mechanical Engi-neers); (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 (from Ref.
31, courtesy of Robert Martinuzzi, with the permission of the Ameri-can Society of Mechanical Engi-neers).
554 Chapter 8 Potential Flow and Computational Fluid Dynamics (b) (a) (c) the cube height. Compare Fig. 8.37a, a top view of the experimental surface flow as visualized by oil streaks, with Fig. 8.37b, 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 re-sult when surface shear flows meet an obstacle. We conclude that CFD has a tremen-dous potential for flow prediction.
This chapter has analyzed a highly idealized but very useful type of flow: inviscid, in-compressible, irrotational flow, for which Laplace’s equation holds for the velocity po-tential (8.1) and for the plane stream function (8.7). The mathematics is well devel-oped, 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, (3) use of variable-strength vortex sheets, and (4) numerical analy-sis on a digital computer. Potential theory is especially useful and accurate for thin bodies such as airfoils. The only requirement is that the boundary layer be thin, i.e., that the Reynolds number be large.
For blunt bodies or highly divergent flows, potential theory serves as a first approxi-mation, to be used as input to a boundary-layer analysis. The reader should consult the advanced texts [for example, 2 to 4, 10 to 13] for further applications of potential the-ory. Section 8.9 discusses computational methods for viscous (nonpotential) flows.
Problems 555 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 an EES icon will benefit from the use of the En-gineering Equation Solver (EES), while problems labeled with a computer disk may require the use of a computer. The standard end-of-chapter problems 8.1 to 8.115 (categorized in the problem list below) are followed by word problems W8.1 to W8.7, comprehen-sive problems C8.1 to C8.3, and design projects D8.1 to D8.3.
Problem Distribution Section Topic Problems 8.1 Introduction and review 8.1–8.7 8.2 Elementary plane-flow solutions 8.8–8.17 8.3 Superposition of plane flows 8.18–8.34 8.4 Plane flow past closed-body shapes 8.35–8.59 8.5 The complex potential 8.60–8.71 8.6 Images 8.72–8.79 8.7 Airfoil theory: Two-dimensional 8.80–8.84 8.7 Airfoil theory: Finite-span wings 8.85–8.90 8.8 Axisymmetric potential flow 8.91–8.103 8.8 Hydrodynamic mass 8.104–8.105 8.9 Numerical methods 8.106–8.115 P8.1 Prove that the streamlines (r, ) in polar coordinates from Eqs. (8.10) are orthogonal to the potential lines (r, ).
P8.2 The steady plane flow in Fig. P8.2 has the polar velocity components %r and r 0. Determine the circula-tion around the path shown.
R2 R1 P8.2 P8.3 Using cartesian coordinates, show that each velocity com-ponent (u, , w) of a potential flow satisfies Laplace’s equa-tion 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 (r, )?
P8.5 Consider the two-dimensional velocity distribution u By, Bx, where B is a constant. If this flow possesses a Summary stream function, find its form. If it has a velocity potential, find that also. Compute the local angular velocity of the flow, if any, and describe what the flow might represent.
P8.6 An incompressible flow has the velocity potential 2Bxy, where B is a constant. Find the stream function of this flow, sketch a few streamlines, and interpret the 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?
P8.8 For the velocity distribution of Prob. 8.5, evaluate the cir-culation around the rectangular closed curve defined by (x, y) (1, 1), (3, 1), (3, 2), and (1, 2). Interpret your re-sult, especially vis-à-vis the velocity potential.
P8.9 Consider the two-dimensional flow u Ax, Ay, where A is a constant. Evaluate the circulation around the rectangular closed curve defined by (x, y) (1, 1), (4, 1), (4, 3), and (1, 3). Interpret your result, especially vis-à-vis the velocity potential.
P8.10 A mathematical relation sometimes used in fluid mechan-ics is the theorem of Stokes C V ds A ( & V) n dA where A is any surface and C is the curve enclosing that sur-face. The vector ds is the differential arc length along C, and n is the unit outward normal vector to A. How does this relation simplify for irrotational flow, and how does the re-sulting line integral relate to velocity potential?
P8.11 A power plant discharges cooling water through the man-ifold in Fig. P8.11, which is 55 cm in diameter and 8 m high and is perforated with 25,000 holes 1 cm in diame-ter. Does this manifold simulate a line source? If so, what is the equivalent source strength m?
P8.12 Consider the flow due to a vortex of strength K at the ori-gin. Evaluate the circulation from Eq. (8.15) about the clockwise path from (r, ) (a, 0) to (2a, 0) to (2a, 3/2) to (a, 3/2) and back to (a, 0). Interpret the result.
P8.13 A well-known exact solution to the Navier-Stokes equa-tions (4.38) is the unsteady circulating motion 1 exp r z 0 where K is a constant and is the kinematic viscosity. Does this flow have a polar-coordinate stream function and/or ve-locity potential? Explain. Evaluate the circulation for this motion, plot it versus r for a given finite time, and interpret compared to ordinary line vortex motion.
P8.14 A tornado may be modeled as the circulating flow shown in Fig. P8.14, with r z 0 and (r) such that r r # R v r R Determine whether this flow pattern is irrotational in ei-ther the inner or outer region. Using the r-momentum equation (D.5) of App. D, determine the pressure distri-bution p(r) in the tornado, assuming p p as r →.
Find the location and magnitude of the lowest pressure. R2 r r2 4 t K 2r 556 Chapter 8 Potential Flow and Computational Fluid Dynamics Inlet P8.11 P8.15 Evaluate Prob. 8.14 for the particular case of a small-scale tornado, R 100 m, ,max 65 m/s, with sea-level con-ditions at r . Plot p(r) out to r 400 m.
P8.16 Consider inviscid stagnation flow, Kxy (see Fig.
8.15b) 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)1/2. Give a physical inter-pretation of the flow pattern.
P8.17 Examine the flow of Fig. 8.30 as an analytical (not a nu-merical) problem. Give the appropriate differential equa-tion and the complete boundary conditions for both the stream function and the velocity potential. Is a Fourier-series solution possible?
R θ (r) r P8.14 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 ( a, 0). What is the pattern viewed from afar?
P8.20 Plot the streamlines of the flow due to a line vortex K at (0, a) and a vortex K at (0, a). What is the pattern viewed from afar?
P8.21 Plot the streamlines of the flow due to a line vortex K at (a, 0) and a vortex 2K at ( a, 0). What is the pattern viewed from afar?
P8.22 Plot the streamlines of a uniform stream V iU plus a clockwise line vortex K located at the origin. Are there any stagnation points?
P8.23 Find the resultant velocity vector induced at point A in Fig. P8.23 by the uniform stream, vortex, and line source.
P8.26 Find the resultant velocity vector induced at point A in Fig.
P8.26 by the uniform stream, line source, line sink, and vortex.
Problems 557 P8.24 Line sources of equal strength m Ua, where U is a ref-erence 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 coef-ficient Cp along the wall, where p0 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. (4.134) simulate a tor-nado as in Fig. P8.25. Suppose that the circulation about the tornado is 8500 m2/s and that the pres-sure at r 40 m is 2200 Pa less than the far-field pres-sure. Assuming inviscid flow at sea-level density, es-timate (a) the appropriate sink strength m, (b) the pressure at r 15 m, and (c) the angle at which the streamlines cross the circle at r 40 m (see Fig.
P8.25).
p p0 1 2 U2 U = 8 m /s K = 25 m2 /s m = 15 m2 /s 1.5 m 2 m 1 m A P8.23 40 m β P8.25 P8.27 A counterclockwise line vortex of strength 3K at (x, y) (0, a) is combined with a clockwise vortex K at (0, a). Plot the streamline and potential-line pattern, and find the point of minimum velocity between the two vortices.
P8.28 Sources of equal strength m are placed at the four sym-metric positions (x, y) (a, a), ( a, a), ( a, a), and (a, a). Sketch the streamline and potential-line patterns.
Do any plane “walls’’ appear?
P8.29 A uniform water stream, U 20 m/s and 998 kg/m3, 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 ap-propriate source strength m and (c) the pressure at the nose of the body.
P8.30 Suppose that the total discharge from the manifold in Fig.
P8.11 is 450 m3/s and that there is a uniform ocean cur-rent of 60 cm/s to the right. Sketch the flow pattern from above, showing the dimensions and the region where the cooling-water discharge is confined.
P8.31 A Rankine half-body is formed as shown in Fig. P8.31.
For the stream velocity and body dimension shown, com-pute (a) the source strength m in m2/s, (b) the distance a, (c) the distance h, and (d) the total velocity at point A.
U = 6 m /s m = 12 m2 /s m = – 10 m2 /s K = 9 m2 /s 20° 2 m 1 m A 2 m 1 m P8.26 EES P8.32 Sketch the streamlines, especially the body shape, due to equal line sources m at ( a, 0) and (a, 0) plus a uni-form stream U ma.
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 U ma.
P8.34 Consider three equally spaced sources of strength m placed at (x, y) (0, a), (0, 0), and (0, a). Sketch the result-ing streamlines, noting the position of any stagnation points. What would the pattern look like from afar?
P8.35 Consider three equal sources m in a triangular configura-tion: one at (a/2, 0), one at ( a/2, 0), and one at (0, a).
Plot the streamlines for this flow. Are there any stagnation points? Hint: Try the MATLAB contour command .
P8.36 When a line source-sink pair with m 2 m2/s 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 U 10 m/s, as in Fig. P8.37. Estimate (a) the ve-locity at point A and (b) the location of point B where a particle approaching the stagnation point achieves its max-imum deceleration.
P8.40 Consider a uniform stream U plus line sources m at (x, y) (a, 0) and ( a, 0) and a single line sink 2m at the origin. Does a closed-body shape appear? If so, plot its shape for m/(Ua) equal to (a) 1.0 and (b) 5.0.
P8.41 A Kelvin oval is formed by a line-vortex pair with K 9 m2/s, a 1 m, and U 10 m/s. What are the height, width, and shoulder velocity of this oval?
P8.42 For what value of K/(Ua) does the velocity at the shoul-der of a Kelvin oval equal 4U? What is the height h/a of this oval?
P8.43 Consider water at 20°C flowing at 6 m/s past a 1-m-di-ameter circular cylinder. What doublet strength in m3/s is required to simulate this flow? If the stream pressure is 200 kPa, use inviscid theory to estimate the surface pres-sure at equal to (a) 180°, (b) 135°, and (c) 90°.
P8.44 Suppose that circulation is added to the cylinder flow of Prob. 8.43 sufficient to place the stagnation points at equal to 50° and 130°. What is the required vortex strength K in m2/s? Compute the resulting pressure and surface ve-locity at (a) the stagnation points and (b) the upper and lower shoulders. What will the lift per meter of cylinder width be?
P8.45 What circulation K must be added to the cylinder flow in Prob. 8.43 to place the stagnation point exactly at the up-per shoulder? What will the velocity and pressure at the lower shoulder be then? What value of K causes the lower shoulder pressure to be 10 kPa?
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 in-side 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.
558 Chapter 8 Potential Flow and Computational Fluid Dynamics (0, 3 m) 7 m /s (4m , 0) A x y h + m Source a P8.31 P8.38 A uniform stream U in the x direction combines with a source m at (a, 0) and a sink m at ( a, 0). Plot the re-sulting streamlines and note any stagnation points.
P8.39 Sketch the streamlines of a uniform stream U past a line source-sink pair aligned vertically with the source at a and the sink at a on the y-axis. Does a closed-body shape form?
A B?
2 m 1 m 10 m/s P8.37 P8.47 A circular cylinder is fitted with two surface-mounted pres-sure sensors, to measure pa at 180° and pb at 105°.
The intention is to use the cylinder as a stream velocime-ter. Using inviscid theory, derive a formula for estimating U in terms of pa, pb, , and the cylinder radius a.
P8.48 Wind at U and p flows past a Quonset hut which is a half-cylinder of radius a and length L (Fig. P8.48). The in-ternal pressure is pi. Using inviscid theory, derive an ex-U = 25 m /s D = 2 m p = 50 k Pa (gage) P8.46 EES pression for the upward force on the hut due to the dif-ference between pi and ps.
ft2. 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 out-side the range of Fig. 8.11. The measured lift and drag co-efficients 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 re-quired to drive the rotors.
Problems 559 P8.49 In strong winds the force in Prob. 8.48 can be quite large.
Suppose that a hole is introduced in the hut roof at point A to make pi equal to the surface pressure there. At what angle 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 which 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 U?
U∞ , p∞ A a θ pi ps ( ) θ P8.48 U a/2 Umax?
Bump a U h?
P8.50 P8.55 Assume that the Flettner rotorship 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.
P8.56 The measured drag coefficient of a cylinder in crossflow, based on frontal area DL, is approximately 1.0 for the lam-inar-boundary-layer range (see Fig. 7.16a). Boundary-layer separation occurs near the shoulder (see Fig. 7.13a).
This suggests an analytical model: the standard inviscid-flow solution on the front of the cylinder and constant pres-sure (equal to the shoulder value) on the rear. Use this model to predict the drag coefficient and comment on the results with reference to Fig. 7.13c.
P8.57 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.
P8.58 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.
P8.59 By analogy with Prob. 8.58 plot the streamlines due to counterclockwise line vortices K at (0, 0) and (4a, 0) plus a clockwise vortex K at (a, 0). Again a cylinder ap-pears.
P8.51 Modify Prob. 8.50 as follows. Let the bump be such that Umax 1.5U. Find (a) the upstream elevation h and (b) the height of the bump.
P8.52 The Flettner rotor sailboat in Fig. E8.2 has a water drag coefficient of 0.006 based on a wetted area of 45 ft2. 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 optimum angle between the boat and the wind?
P8.53 Modify Prob. 8.52 as follows. For the same sailboat data, find the wind velocity, in mi/h, which will drive the boat at an optimum speed of 10 kn parallel to its keel.
P8.54 The original Flettner rotor ship was approximately 100 ft long, displaced 800 tons, and had a wetted area of 3500 U∞ ω ω P8.54 P8.60 One of the corner-flow patterns of Fig. 8.15 is given by the cartesian stream function A(3yx2 y3). Which one? Can the correspondence be proved from Eq. (8.49)?
P8.61 Plot the streamlines of Eq. (8.49) in the upper right quad-rant for n 4. How does the velocity increase with x out-ward along the x-axis from the origin? For what corner an-gle and value of n would this increase be linear in x? For what corner angle and n would the increase be as x5?
P8.62 Combine stagnation flow, Fig. 8.14b, with a source at the origin: f(z) Az2 m ln z Plot the streamlines for m AL2, where L is a length scale.
Interpret.
P8.63 The superposition in Prob. 8.62 leads to stagnation flow near a curved bump, in contrast to the flat wall of Fig.
8.14b. Determine the maximum height H of the bump as a function of the constants A and m.
P8.64 Determine qualitatively from boundary-layer theory (Chap.
7) whether any of the three stagnation-flow patterns of Fig.
8.15 can suffer flow separation along the walls.
P8.65 Potential flow past a wedge of half-angle leads to an important application of laminar-boundary-layer theory called the Falkner-Skan flows [15, pp. 242–247]. Let x de-note distance along the wedge wall, as in Fig. P8.65, and let 10°. Use Eq. (8.49) to find the variation of surface velocity U(x) along the wall. Is the pressure gradient ad-verse or favorable?
P8.69. What hyphenated word (originally French) might describe such a flow pattern?
560 Chapter 8 Potential Flow and Computational Fluid Dynamics x θ θ U (x) P8.65 P8.70 Show that the complex potential f U{z 1 4 a coth [(z/a)]} represents flow past an oval shape placed mid-way between two parallel walls y 1 2 a. 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 po-tential 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?
y x y = a ( = 0) ψ Plot the streamlines inside this region P8.69 P8.71 Weir 0 a a y x + m P8.72 P8.66 The inviscid velocity along the wedge in Prob. 8.65 has the analytic form U(x) Cxm, where m n 1 and n is the exponent in Eq. (8.49). Show that, for any C and n, computation of the boundary layer by Thwaites’ method, Eqs. (7.53) and (7.54), leads to a unique value of the Thwaites parameter . Thus wedge flows are called simi-lar [15, p. 244].
P8.67 Investigate the complex potential function f(z) U(z a2/z) and interpret the flow pattern.
P8.68 Investigate the complex potential function f(z) Uz m ln [(z a)/(z a)] and interpret the flow pattern.
P8.69 Investigate the complex potential f(z) A cosh [(z/a)], and plot the streamlines inside the region shown in Fig.
P8.72 Use the method of images to construct 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?
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.
P8.74 A positive line vortex K is trapped in a corner, as in Fig.
P8.74. Compute the total induced velocity vector at point B, (x, y) (2a, a), and compare with the induced veloc-ity when no walls are present.
P8.77 Discuss how the flow pattern of Prob. 8.58 might be in-terpreted 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.
Problems 561 P8.75 The flow past a cylinder very near a wall might be simu-lated by doublet images, as in Fig. P8.75. Explain why the result is not very successful and the cylinder shape be-comes badly distorted.
P8.73 a x y + m 2 a 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 ve-locities at corresponding points A, B and C, D, comparing with a cylinder flow in an infinite expanse of fluid.
B 2a a 0 a 2a x y K V?
P8.74 P8.75 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 at x a. How many images are needed to es-timate this velocity within 1 percent?
2a D U∞ 4a 4a B A C P8.76 2a a y a x U∞ P8.78 + m 2a a x y 0 P8.79 P8.80 The beautiful expression for lift of a two-dimensional air-foil, Eq. (8.69), arose from applying the Joukowski trans-formation, ' z a2/z, where z x iy and ' i.
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 1 unit for convenience, show that (a) a circle with center at the origin and radius 1 will become an ellipse in the ' plane and (b) a circle with center at x ( 1, y 0, and radius (1 () will become an air-foil shape in the ' plane. Hint: The Excel spreadsheet is excellent for solving this problem.
P8.81 A two-dimensional airfoil has 2 percent camber and 10 per-cent thickness. If C 1.75 m, estimate its lift per meter when immersed in 20°C water at 6° and U 18 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 1 4 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 horse-power required to achieve a speed of 15 kn.
P8.83 Two-dimensional lift-drag data for the NACA 2412 airfoil with 2 percent camber (from Ref. 12) may be curve-fitted accurately as follows: CL 0.178 0.109 0.001092 CD 0.0089 1.97 E-4 8.45 E-5 2 1.35 E-5 3 9.92 E-7 4 with in degrees in the range 4° 10°. Com-pare (a) the lift-curve slope and (b) the angle of zero lift with theory, Eq. (8.69). (c) Prepare a polar lift-drag plot and compare with Fig. 7.26.
P8.84 Reference 12 contains inviscid-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: x/c V/U(upper) V/U(lower) 0.0 0.00 0.00 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 Use these data, plus Bernoulli’s equation, to estimate (a) the lift coefficient and (b) the angle of attack if the airfoil is symmetric.
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 lbf and drag of 1.5 lbf. Estimate from wing theory (a) the angle of attack, (b) the minimum drag of the wing and the angle of attack at which it occurs, and (c) the max-imum 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 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 rec-tangular hydrofoil of aspect ratio 8, 2 percent camber, and 12 percent thickness. If the boat travels at 8 m/s and 3.5°, 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 which delivers 50 hp to the water.
P8.88 The Boeing 727 airplane has a gross weight of 125,000 lbf, a wing area of 1200 ft2, and an aspect ratio of 6. It is fitted with two turbofan engines and cruises at 532 mi/h at 30,000-ft standard altitude. Assume for this problem that its airfoil is the NACA 2412 section described in Prob.
8.83. If we neglect all drag except the wing, what thrust is required from each engine for these conditions?
P8.89 The Beechcraft T-34C aircraft has a gross weight of 5500 lbf and a wing area of 60 ft2 and flies at 322 mi/h at 10,000-ft standard altitude. It is driven by a propeller which de-livers 300 hp to the air. Assume for this problem that its airfoil is the NACA 2412 section described in Prob. 8.83, and neglect all drag except the wing. What is the appro-priate aspect ratio for the wing?
P8.90 When moving at 15 m/s in seawater at its maximum lift-to-drag ratio of 18:1, a symmetric hydrofoil, of plan area 3 m2, develops a lift of 120 kN. Estimate from wing the-ory (a) the aspect ratio and (b) the angle of attack in de-grees.
P8.91 If (r, ) in axisymmetric flow is defined by Eq. (8.85) and the coordinates are given in Fig. 8.24, determine what partial differential equation is satisfied by .
P8.92 A point source with volume flow Q 30 m3/s is im-mersed 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, ) on the body surface where the local velocity equals 4.5 m/s.
P8.93 The Rankine half-body of revolution (Fig. 8.26) could simulate the shape of a pitot-static tube (Fig. 6.30). Ac-562 Chapter 8 Potential Flow and Computational Fluid Dynamics EES cording 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 U? Compare your answer with the recommendation x 8D in Fig.
6.30.
P8.94 Determine whether the Stokes streamlines from Eq. (8.86) are everywhere orthogonal to the Stokes potential lines from Eq. (8.87), as is the case for cartesian and plane po-lar 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 m at (a, 0), and a stream U in the x di-rection 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, U, and a.
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 Fig.
P8.98? If so, how would you establish the mathematical 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/(UL2).
Problems 563 y x a a + m – m r θ U∞ P8.95 P8.99 Consider air flowing past a hemisphere resting on a flat surface, as in Fig. P8.99. If the internal pressure is pi, find an expression for the pressure force on the hemisphere. By analogy with Prob. 8.49, at what point A on the hemisphere should a hole be cut so that the pressure force will be zero according to inviscid theory?
U pa = 40 kPa Water at 20˚ C A 80 cm Rankine ovoid P8.97 P8.96 Suppose that a sphere with a single stagnation hole is to be used as a velocimeter. The pressure at this hole is used to compute the stream velocity, but there are errors if the hole is not perfectly aligned with the oncoming stream.
Using inviscid incompressible theory, plot the percent er-ror in stream velocity estimate as a function of misalign-ment angle . At what angle is the error 10 percent? 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 ap-pear at point A. Compute the stream velocity U, ne-glecting surface wave formation, for which cavitation oc-curs.
P8.98 U∞ + m – m y x 0 Point source Line sink of total strength L P8.99 U∞ , p∞ pi 2 a 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 invis-cid 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 con-dition, what will be the pressure at point A on the sphere which is 45° up from the direction of travel?
pa = 101.35 k Pa D = 1 m A V 3 m P8.100 P8.101 Normally by its very nature inviscid theory is incapable of predicting body drag, but by analogy with Fig. 8.16c we can analyze flow approaching a hemisphere, as in Fig.
P8.101. Assume that the flow on the front follows invis-cid sphere theory, Eq. (8.96), and the pressure in the rear equals the shoulder pressure. Compute the drag coefficient and compare with experiment (Table 7.3). What are the defects and limitations of this analysis?
imum thickness. Use these results to derive a formula from the time history U(t) of the cylinder if it is accelerated from rest in a still fluid by the sudden application of a constant force F.
P8.106 Laplace’s equation in plane polar coordinates, Eq. (8.11), is complicated by the variable radius. Consider the finite-difference mesh in Fig. P8.106, with nodes (i, j) equally spaced and r apart. Derive a finite-difference model for Eq. (8.11) similar to the cartesian expression (8.109).
564 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.102 A golf ball weighs 0.102 lbf 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 coeffi-cient 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 (a) without spin and (b) with backspin of 7500 r/min.
P8.103 Modify Prob. 8.102 as follows. Golf balls are dimpled, not smooth, and have higher lift and lower drag (CL 0.2 and CD 0.3 for typical backspin). Using these values, make a computer analysis of the ball trajectory for the initial con-ditions of Prob. 8.102. If time permits, investigate the ef-fect of initial angle for the range 10° 0 50°.
P8.104 Consider a cylinder of radius a moving at speed U 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 rela-tive kinetic energy, and verify the hydrodynamic mass of a cylinder from Eq. (8.104).
P8.105 In Table 7.2 the drag coefficient of a 4:1 elliptical cylin-der in laminar-boundary-layer flow is 0.35. According to Patton , the hydrodynamic mass of this cylinder is hb/4, where b is width into the paper and h is the max-U∞ , p∞ Hemisphere Rear pressure assumed equal to shoulder pressure a P8.101 P8.107 Set up the numerical problem of Fig. 8.30 for an expan-sion of 30°. A new grid system and a nonsquare mesh may be needed. Give the proper nodal equation and boundary conditions. If possible, program this 30° expansion and solve on a digital computer.
P8.108 Consider two-dimensional potential flow into a step con-traction as in Fig. P8.108. The inlet velocity U1 7 m/s, and the outlet velocity U2 is uniform. The nodes (i, j) are labeled in the figure. Set up the complete finite-difference algebraic relations for all nodes. Solve, if possible, on a digital computer and plot the streamlines in the flow.
Still fluid U∞ a P8.104 ∆ r ∆ r ∆θ ∆θ i – 1, j i + 1, j i, j + 1 i, j – 1 rj + 1 rj – 1 νθ i, j νr rj P8.106 P8.108 i = 1 j = 1 2 3 4 5 6 7 8 U1 U2 2 3 4 5 6 7 8 9 10 P8.109 Consider inviscid flow through a two-dimensional 90° bend with a contraction, as in Fig. P8.109. Assume uni-form flow at the entrance and exit. Make a finite-differ-ence computer analysis for small grid size (at least 150 nodes), determine the dimensionless pressure distribution along the walls, and sketch the streamlines. (You may use either square or rectangular grids.) P8.112 In his CFD textbook, Patankar replaces the left-hand sides of Eq. (8.119b and c) with the following two ex-pressions, respectively: ∂ ∂ x (u2) ∂ ∂ y (u) and ∂ ∂ x (u) ∂ ∂ y (2) 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 Repeat Example 8.7 using the implicit method of Eq.
(8.118). Take t 0.2 s and y 0.01 m, which ensures that an explicit model would diverge. Compare your ac-curacy with Example 8.7.
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 1.4845'1/2 0.63' 1.758'2 1.4215'3 0.5075'4 where ' x/C and the maximum thickness tmax occurs at ' 0.3. Use this shape as part of the lower boundary for zero angle of attack. Let the thickness be fairly large, say, tmax 0.12, 0.15, or 0.18. Choose a generous number of nodes ($60), and calculate and plot the velocity distribu-tion V/U along the airfoil surface. Compare with the the-oretical results in Ref. 12 for NACA 0012, 0015, or 0018 airfoils. If time permits, investigate the effect of the bound-ary lengths L1, L2, and L3, which can initially be set equal to the chord length C.
2y tmax Problems 565 P8.110 For fully developed laminar incompressible flow through a straight noncircular duct, as in Sec. 6.6, the Navier-Stokes equations (4.38) reduce to ∂ ∂ 2 y u 2 ∂ ∂ 2 z u 2 const 0 where (y, z) is the plane of the duct cross section and x is along the duct-axis. Gravity is neglected. Using a non-square rectangular grid ( x, y), 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. 8.110 numerically for a rectangular duct of side length b by 2b, using at least 100 nodal points. Eval-uate the volume flow rate and the friction factor, and com-pare with the results in Table 6.4: Q 0.1143 f ReDh 62.19 where Dh 4A/P 4b/3 for this case. Comment on the possible truncation errors of your model.
dp dx b4 !
dp dx 1 !
V1 = 10 m/s V2 5 m 6 m 10 m 10 m 15 m 16 m P8.109 y U∞ L1 x = 0 x = C L2 L3 Airfoil half-contour x P8.114 P8.115 Use the explicit method of Eq. (8.115) to solve Prob. 4.85 numerically for SAE 30 oil at 20°C with U0 1 m/s and M rad/s, where M is the number of letters in your sur-name. (This author will solve the problem for M 5.) When steady oscillation is reached, plot the oil velocity versus time at y 2 cm.
Word Problems W8.1 What simplifications 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 im-portant parameters?
W8.2 In this chapter we superimpose many basic solutions, a con-cept associated with linear equations. Yet Bernoulli’s equa-tion (8.3) is nonlinear, being proportional to the square of the velocity. How, then, do we justify the use of superposi-tion in inviscid-flow analysis?
W8.3 Give a physical explanation of circulation as it relates to the lift force on an immersed body. If the line integral de-fined by Eq. (8.15) is zero, it means that the integrand is a perfect differential—but of what variable?
W8.4 Give a simple proof of Eq. (8.42), namely, that both the real and imaginary parts of a function f(z) are laplacian if z x iy. What is the secret of this remarkable behavior?
W8.5 Figure 8.14 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 corners.
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.
566 Chapter 8 Potential Flow and Computational Fluid Dynamics Comprehensive Problems C8.1 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 irrota-tional 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 in-formation. For full credit, attach a printout of your spread-sheet, with stream function converged and the value of the stream function at each node displayed to four digits of ac-curacy.
cm and (b) the instantaneous boundary-layer thickness (where u 0.99 u). Hint: There is a nonzero pressure gradient in the outer (nearly shear-free) stream, n N, which must be included in Eq. (8.114) and your explicit model.
C8.3 Consider plane inviscid flow through a symmetric diffuser, 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 is 18.5° (L 3h). Set up a nonsquare potential-flow mesh for this problem, and calcu-late and plot (a) the velocity distribution and (b) the pres-sure coefficient along the centerline. Assume uniform inlet and exit flows. Inlet Outlet 5 3.333 1.667 4 3 2 1 0 Wall, 0 Wall, 0 Wall, 5 C8.1 2h 2h L V h C8.3 C8.2 Use an explicit method, similar to but not identical to Eq.
(8.115), 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, u uN at, where a 9 m/s2. At t 1 s, determine (a) the oil velocity at y 1 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 x-y 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 ve-locity 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 varia-tion of dimensionless pressure coefficient, Cp (p p)/( U2/2), 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 coeffi-cient 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 discuss the agreement with prediction. Give reasons for any disagreements.
References 567 a y x C8.4 Design Projects D8.1 In 1927, Theodore von Kármán 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 angle of at-tack. A total of N sources and sinks are distributed along the axis within the body, with strengths mi at positions xi, for i 1 to N. The object is to find the correct distribution of strengths which approximates a given body shape y(x) at a finite num-ber of surface locations and then to compute the approximate surface velocity and pressure. The technique should work for either two-dimensional bodies (distributed line sources) or bod-ies of revolution (distributed point sources).
For our body shape let us select the NACA 0018 airfoil, given by the formula in Prob. 8.114 with tmax 0.18. De-velop the ideas stated above into N simultaneous algebraic equations which can be used to solve for the N unknown line source/sink strengths. Then program your equations for a computer, with N $ 20; solve for mi; compute the surface velocities; and compare with the theoretical velocities for this shape in Ref. 12. Your goal should be to achieve accuracy within 1 percent of the classical 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 distribution which 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 func-tion of body thickness. Especially note the thicknesses where (a) the local pressure is 50 kPa and (b) cavitation first oc-curs on the surface.
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 vec-tor notation; and (b) expanding out the scalar terms and substi-tuting terms from the definition of irrotationality.
References 1.
O. D. Kellogg, Foundations of Potential Theory, Dover, New York, 1969.
2.
J. M. Robertson, Hydrodynamics in Theory and Application, Prentice-Hall, Englewood Cliffs, NJ, 1965.
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L. M. Milne-Thomson, Theoretical Hydrodynamics, 4th ed., Macmillan, New York, 1960.
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I. G. Currie, Fundamental Mechanics of Fluids, 2d ed., McGraw-Hill, New York, 1993.
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S. V. Patankar, Numerical Heat Transfer and Fluid Flow, McGraw-Hill, New York, 1980.
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G. F. Carey and J. T. Oden, Finite Elements: Fluid Mechan-ics, vol. 6, Prentice-Hall, Englewood Cliffs, NJ, 1986.
D8.1 Body shape Typical body point Source Axis x U∞ i = 1 i = N y j m i 7.
C. A. Brebbia and J. Dominquez, Boundary Elements—An In-troductory Course, 2d ed., Computational Mechanics Publi-cations, Southampton, England, and McGraw-Hill, New York, 1991.
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A. D. Moore, “Fields from Fluid Flow Mappers,’’ J. Appl.
Phys., vol. 20, 1949, pp. 790–804.
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H. J. S. Hele-Shaw, “Investigation of the Nature of the Sur-face Resistance of Water and of Streamline Motion under Cer-tain Experimental Conditions,’’ Trans. Inst. Nav. Archit., vol.
40, 1898, p. 25.
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R. H. F. Pao, Fluid Dynamics, Merrill, Columbus, OH, 1967.
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A. M. Kuethe and C.-Y. Chow, Foundations of Aerodynam-ics, 5th ed., Wiley, New York, 1997.
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I. H. Abbott and A. E. von Doenhoff, Theory of Wing Sec-tions, Dover, New York, 1959.
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B. Thwaites (ed.), Incompressible Aerodynamics, Clarendon Press, Oxford, England, 1960.
14.
L. Prandtl, “Applications of Modern Hydrodynamics to Aero-nautics,’’ NACA Rep. 116, 1921.
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F. M. White, Viscous Fluid Flow, 2d ed., McGraw-Hill, New York, 1991.
16.
C. S. Yih, Fluid Mechanics, McGraw-Hill, New York, 1969.
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K. T. Patton, “Tables of Hydrodynamic Mass Factors for Translational Motion,’’ ASME Winter Annual Meeting, Paper 65-WA/UNT-2, 1965.
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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.
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K. H. Huebner, The Finite Element Method for Engineers, 3d ed., Wiley, New York, 1994.
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J. C. Tannehill, D. A. Anderson, and R. H. Pletcher, Compu-tational Fluid Mechanics and Heat Transfer, 2d ed., Taylor and Francis, Bristol, PA, 1997.
21.
J. N. Newman, Marine Hydrodynamics, M.I.T. Press, Cam-bridge, MA, 1977.
568 Chapter 8 Potential Flow and Computational Fluid Dynamics 22.
H. Rouse, Elementary Mechanics of Fluids, Dover, New York, 1978.
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J. H. Ferziger and M. Peric, Computational Methods for Fluid Dynamics, Springer-Verlag, New York, 1996.
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C. Hirsch, Numerical Computation of Internal and External Flows, 2 vols., Wiley, New York, 1990.
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K. A. Hoffmann and S. T. Chiang, CFD for Engineers, 2 vols., Engineering Education System, New York, 1993.
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J. D. Anderson, Computational Fluid Dynamics: The Basics with Applications, McGraw-Hill, New York, 1995.
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C. A. J. Fletcher, Computational Techniques for Fluid Dy-namics, 2 vols., Springer Verlag, New York, 1997.
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M. Deshpande, J. Feng, and C. L. Merkle, “Numerical Mod-eling of the Thermodynamic Effects of Cavitation,” J. Fluids Eng., June 1997, pp. 420–427.
29.
P. A. Libby, An Introduction to Turbulence, Taylor and Fran-cis, Bristol, PA, 1996.
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C. J. Freitas, “Perspective: Selected Benchmarks from Com-mercial CFD Codes,” J. Fluids Eng., vol. 117, June 1995, pp.
208–218.
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480–483.
570 The Concorde 264 supersonic airliner. Flying more than twice as fast as the speed of sound, as discussed in the present chapter, the Concorde is a milestone in commercial aviation. However, this great technical achievement is accompanied by high expense for the traveller. (Courtesy of Don Riepe/Peter Arnold, Inc.) 9.1 Introduction Motivation.
All eight of our previous chapters have been concerned with “low-speed’’ or “incompressible’’ flow, i.e., 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 be-come significant and the flow is termed compressible. Such flows are difficult to obtain in liquids, since high pressures of order 1000 atm are needed to generate sonic veloci-ties. In gases, however, a pressure ratio of only 21 will likely cause sonic flow. Thus compressible gas flow is quite common, and this subject is often called gas dynamics.
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 is made to order for the Engineering Equation Solver (EES) in App. E. Compressible-flow analysis is filled with scores of complicated algebraic equations, most of which are very difficult to manipulate or in-vert. Consequently, for nearly a century, compressible-flow textbooks have relied upon extensive tables of Mach number relations (see App. B) for numerical work. With EES, however, any set of equations in this chapter can be typed out and solved for any vari-able—see part (b) of Example 9.13 for an especially intricate example. With such a tool, App. B serves only as a backup and indeed 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 cri-terion for a nearly incompressible flow was a small Mach number Ma V a 1 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 571 Chapter 9 Compressible Flow 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 mo-mentum 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, , T, V). Thus the general theory of compressible flow is quite com-plicated, and we try here to make further simplifications, especially by assuming a re-versible adiabatic or isentropic flow.
The Mach number is the dominant parameter in compressible-flow analysis, with dif-ferent effects depending upon its magnitude. Aerodynamicists especially make a dis-tinction between the various ranges of Mach number, and the following rough classi-fications are commonly used: Ma 0.3: incompressible flow, where density effects are negligible.
0.3 Ma 0.8: subsonic flow, where density effects are important but no shock waves appear.
0.8 Ma 1.2: transonic flow, where shock waves first appear, dividing sub-sonic 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.
The numerical values listed above 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 super-sonic-flow effects may go against intuition, you should study these differences carefully.
In addition to geometry and Mach number, compressible-flow calculations also depend upon a second dimensionless parameter, the specific-heat ratio of the gas: k c c p (9.1) 572 Chapter 9 Compressible Flow The Mach Number The Specific-Heat Ratio 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 ap-pear in these later chapters of the text.
Recall from Fig. 1.3 that k for the common gases decreases slowly with temperature and lies between 1.0 and 1.7. Variations in k have only a slight effect upon compressible-flow computations, and air, k 1.40, is the dominant fluid of interest. Therefore, although we assign some problems involving, e.g., steam and CO2 and helium, the compressible-flow tables in App. B are based solely upon 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. References 1 to 6, 26, 29, and 33 are intro-ductory, fairly elementary treatments, while Refs. 7 to 14, 27 to 28, 31 to 32, and 35 are advanced. From time to time we shall defer some specialized topic to these texts.
We note in passing that there are at least two flow patterns which depend strongly upon very small density differences, acoustics, and natural convection. Acoustics [9, 14] is the study of sound-wave propagation, which is accompanied by extremely small changes in density, 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.
In principle, compressible-flow calculations can be made for any fluid equation of state, and we shall assign 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 p RT R cp c const k c c p const (9.2) 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 con-stant specific heats.
Recall from Sec. 1.6 that the gas constant is related to a universal constant di-vided by the gas molecular weight Rgas M gas (9.3) where 49,720 ft2/(s2 °R) 8314 m2/(s2 K) For air, M 28.97, and we shall adopt the following property values for air through-out this chapter: R 1717 ft2/(s2 °R) 287 m2/(s2 K) k 1.400 c k R 1 4293 ft2/(s2 °R) 718 m2/(s2 K) (9.4) cp k k R 1 6010 ft2/(s2 °R) 1005 m2/(s2 K) 9.1 Introduction 573 The Perfect Gas Isentropic Process Experimental values of k for eight common gases were shown in Fig. 1.3. From this figure and the molecular weight, the other properties can be computed, as in Eqs. (9.4).
The changes in the internal energy û and enthalpy h of a perfect gas are computed for constant specific heats as û2 û1 c(T2 T1) h2 h1 cp(T2 T1) (9.5) For variable specific heats one must integrate û c dT and h cp dT or use the gas tables . Most modern thermodynamics texts now contain software for evaluat-ing properties of nonideal gases .
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] T ds dh d p (9.6) Introducing dh cp dT for a perfect gas and solving for ds, we substitute T p/R from the perfect-gas law and obtain 2 1 ds 2 1 cp d T T R 2 1 d p p (9.7) If cp is variable, the gas tables will be needed, but for constant cp we obtain the ana-lytic results s2 s1 cp ln T T 2 1 R ln p p 2 1 c ln T T 2 1 R ln 2 1 (9.8) Equations (9.8) are used to compute the entropy change across a shock wave (Sec. 9.5), which is an irreversible process.
For isentropic flow, we set s2 s1 and obtain the interesting power-law relations for an isentropic perfect gas p p 2 1 T T 2 1 k/(k 1) 2 1 k (9.9) These relations are used in Sec. 9.3.
EXAMPLE 9.1 Argon flows through a tube such that its initial condition is p1 1.7 MPa and 1 18 kg/m3 and its final condition is p2 248 kPa and T2 400 K. Estimate (a) the initial temperature, (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 m2/(s2 K) and k 1.67. Therefore estimate its specific heat at constant pressure from Eq. (9.4): 574 Chapter 9 Compressible Flow 9.2 The Speed of Sound cp k k R 1 1 1 .
.
6 6 7 7 (2 08 1 ) 519 m2/(s2 K) The initial temperature and final density are estimated from the ideal gas law, Eq. (9.2): T1 p 1 1 R 454 K Ans. (a) 2 T p 2 2 R 2.98 kg/m3 Ans. (b) From Eq. (9.5) the enthalpy change is h2 h1 cp(T2 T1) 519(400 454) 28,000 J/kg (or m2/s2) Ans. (c) The argon temperature and enthalpy decrease as we move down the tube. Actually, there may not be any external cooling; i.e., the fluid enthalpy may be converted by friction to increased ki-netic energy (Sec. 9.7).
Finally, the entropy change is computed from Eq. (9.8): s2 s1 cp ln T T 2 1 R ln p p 2 1 519 ln 4 4 0 5 0 4 208 ln 0.
1 2 .
4 7 8 E E 6 6 66 400 334 m2/(s2 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 mov-ing subsonically through a tube with large frictional effects (Sec. 9.7).
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 by 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, , T, V 0) at the left, leaving behind at the right a fluid of increased properties (p p, , T T) and a fluid velocity V 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 which would be necessary in Fig. 9.1a, we adopt instead the control volume of Fig. 9.1b, which moves at wave speed C to the left. The wave ap-pears fixed from this viewpoint, and the fluid appears to have velocity C on the left and C V on the right. The thermodynamic properties p, , and T are not affected by this change of viewpoint.
The flow in Fig. 9.1b is steady and one-dimensional across the wave. The continu-ity equation is thus, from Eq. (3.24), AC ( )(A)(C V) or V C (9.10) 248 E3 N/m2 (400 K)[208 m2/(s2 K)] 1.7 E6 N/m2 (18 kg/m3)[208 m2/(s2 K)] 9.2 The Speed of Sound 575 Fig. 9.1 Control-volume analysis of a finite-strength pressure wave: (a) control volume fixed to still fluid at left; (b) control volume moving left at wave speed C.
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, 14] show that the thickness of pressure waves in gases is of order 10 6 ft at atmospheric pressure. Thus we can safely neglect friction and ap-ply the one-dimensional momentum equation (3.40) across the wave Fright m ˙ (Vout Vin) or pA (p p)A (AC)(C V C) (9.11) Again the area cancels, and we can solve for the pressure change p C V (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 C2 p 1 (9.13) The larger the strength / of the wave, the faster the wave speed; i.e., powerful ex-plosion waves move much more quickly than sound waves. In the limit of infinitesi-mal strength →0, we have what is defined to be the speed of sound a of a fluid: a2 p (9.14) 576 Chapter 9 Compressible Flow p ρ T V = 0 C p + ∆ p ρ + ∆ρ T + ∆T ∆V Moving wave of frontal area A (a) p ρ T V = C (b) Fixed wave V = C – ∆V p + ∆ p ρ + ∆ρ T + ∆T Friction and heat transfer effects are confined to wave interior 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 which was equivalent to assuming an isothermal process, the result being 20 percent low for air, for example. He rational-ized the discrepancy as being due to the “crassitude’’ (dust particles, etc.) in the air; the error is certainly understandable when we reflect that it was made 180 years be-fore 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 a p s 1/2 k p T 1/2 (9.15) for any fluid, gas or liquid. Even a solid has a sound speed.
For a perfect gas, From Eq. (9.2) or (9.9), we deduce that the speed of sound is a k p 1/2 (kRT)1/2 (9.16) The speed of sound increases as the square root of the absolute temperature. For air, with k 1.4 and R 1717, an easily memorized dimensional formula is a (ft/s) 49[T (°R)]1/2 a (m/s) 20[T (K)]1/2 (9.17) 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)1/2 968 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 K p s p s (9.18) For example, at standard conditions, the bulk modulus of carbon tetrachloride is 163,000 lbf/in2 absolute, and its density is 3.09 slugs/ft3. Its speed of sound is there-fore [163,000(144)/3.09]1/2 2756 ft/s, or 840 m/s. Steel has a bulk modulus of about 29 106 lbf/in2 absolute and water about 320 103 lbf/in2 absolute, or 90 times less.
For solids, it is sometimes assumed that the bulk modulus is approximately equiv-alent to Young’s modulus of elasticity E, but in fact their ratio depends upon Poisson’s ratio K E 3(1 2) (9.19) The two are equal for 1 3 , which is approximately the case for many common met-als such as steel and aluminum.
9.2 The Speed of Sound 577 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: H2 4,246 1,294 He 3,281 1,000 Air 1,117 340 Ar 1,040 317 CO2 873 266 CH4 607 185 238UF6 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.
9.3 Adiabatic and Isentropic Steady Flow 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 m2/(s2 K), and the given temperature is 300°C 273 573 K. Thus from Eq. (9.16) we estimate aCO (kRT)1/2 [1.40(297)(573)]1/2 488 m/s Ans. As mentioned in Sec. 9.1, the isentropic approximation greatly simplifies a compress-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. (3.66) h1 1 2 V2 1 gz1 h2 1 2 V2 2 gz2 q w (9.20) where point 1 is upstream of point 2. You may wish to review the details of Eq. (3.66) and its development. We saw in Example 3.16 that potential-energy changes of a gas are extremely small compared with kinetic-energy and enthalpy terms. We shall ne-glect the terms gz1 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 h1 1 2 V2 1 h2 1 2 V2 2 const (9.21) The constant in Eq. (9.21) is equal to the maximum enthalpy which the fluid would achieve if brought to rest adiabatically. We call this value h0, the stagnation enthalpy of the flow. Thus we rewrite Eq. (9.21) in the form h 1 2 V2 h0 const (9.22) 578 Chapter 9 Compressible Flow V h0 δ T > δV if Pr < 1 δV Insulated wall Fig. 9.2 Velocity and stagnation-enthalpy distributions near an insu-lated wall in a typical high-speed gas flow.
Mach-Number Relations Isentropic Pressure and Density Relations This should hold for steady adiabatic flow of any compressible fluid outside the bound-ary layer. The wall in Fig. 9.2 could be either the surface of an immersed body or the wall of a duct. We have shown the details of Fig. 9.2; typically the thermal-layer thick-ness T is greater than the velocity-layer thickness V because most gases have a di-mensionless Prandtl number Pr less than unity (see, e.g., 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 1 2 V2 cpT0 (9.23) This establishes the stagnation temperature T0 of an adiabatic perfect-gas flow, i.e., 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 that the velocity achieves a maximum value Vmax (2h0)1/2 (2cpT0)1/2 (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).
The dimensionless form of Eq. (9.23) brings in the Mach number Ma as a parameter, by using Eq. (9.16) for the speed of sound of a perfect gas. Divide through by cpT to obtain 1 2 V cp 2 T T T 0 (9.25) But, from the perfect-gas law, cpT [kR/(k 1)]T a2/(k 1), so that Eq. (9.25) be-comes 1 (k 2a 1 2 )V2 T T 0 or T T 0 1 k 2 1 Ma2 Ma V a (9.26) This relation is plotted in Fig. 9.3 versus the Mach number for k 1.4. At Ma 5 the temperature has dropped to 1 6 T0.
Since a T1/2, the ratio a0/a is the square root of (9.26) a a 0 T T 0 1/2 1 1 2 (k 1)Ma2 1/2 (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 pres-ence of irreversibilities such as friction losses or shock waves.
9.3 Adiabatic and Isentropic Steady Flow 579 Fig. 9.3 Adiabatic (T/T0 and a/a0) and isentropic (p/p0 and /0) prop-erties versus Mach number for k 1.4.
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 p p 0 T T 0 k/(k 1) 1 1 2 (k 1) Ma2 k/(k 1) (9.28a) 0 T T 0 1/(k 1) 1 1 2 (k 1) Ma2 1/(k 1) (9.28b) 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 p0 and 0 are the isentropic stagnation pressure and density, respec-tively, i.e., the pressure and density which the flow would achieve if brought isentrop-ically to rest. In an adiabatic nonisentropic flow p0 and 0 retain their local meaning, but they vary throughout the flow as the entropy changes due to friction or shock waves.
The quantities h0, T0, and a0 are constant in an adiabatic nonisentropic flow (see Sec.
9.7 for further details).
The isentropic assumptions (9.28) are effective, but are they realistic? Yes. To see why, differentiate Eq. (9.22) Adiabatic: dh V dV 0 (9.29) Meanwhile, from Eq. (9.6), if ds 0 (isentropic process), dh d p (9.30) Combining (9.29) and (9.30), we find that an isentropic streamtube flow must be d p V dV 0 (9.31) 580 Chapter 9 Compressible Flow 0.5 0 1 2 3 4 5 Mach number a a0 T T0 ρ p p0 0 ρ 1.0 Relationship to Bernoulli’s Equation Critical Values at the Sonic Point But this is exactly the Bernoulli relation, Eq. (3.75), 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 (a0, T0, p0, 0) 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, , 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 1.4 p p 0 k 2 1 k/(k 1) 0.5283 0 k 2 1 1/(k 1) 0.6339 (9.32) T T 0 k 2 1 0.8333 a a 0 k 2 1 1/2 0.9129 In all isentropic flow, all critical properties are constant; in adiabatic nonisentropic flow, a and T are constant, but p and 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 (kRT)1/2 k 2 k 1 RT0 1/2 (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.
Since the great bulk of our practical calculations are for air, k 1.4, the stagnation-property ratios p/p0, etc., from Eqs. (9.26) to (9.28), are tabulated for this value in Table B.1. The increments in Mach number are rather coarse in this table because the values are only meant as a guide; these equations are now a trivial matter to manipulate on a hand calculator. Thirty years ago every text had extensive compressible-flow tables with Mach-number spacings of about 0.01, so that accurate values could be interpolated.
For k 1.4, the following numerical versions of the isentropic and adiabatic flow formulas are obtained: T T 0 1 0.2 Ma2 0 (1 0.2 Ma2)2.5 p p 0 (1 0.2 Ma2)3.5 (9.34) Or, if we are given the properties, it is equally easy to solve for the Mach number (again with k 1.4) Ma2 5 T T 0 1 5 0 2/5 1 5 p p 0 2/7 1 (9.35) Note that these isentropic-flow formulas serve as the equivalent of the frictionless adi-abatic 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 9.3 Adiabatic and Isentropic Steady Flow 581 Some Useful Numbers for Air 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.
EXAMPLE 9.3 Air flows adiabatically through a duct. At point 1 the velocity is 240 m/s, with T1 320 K and p1 170 kPa. Compute (a) T0, (b) p01, (c) 0, (d) Ma, (e) Vmax, and (f ) V. At point 2 further downstream V2 290 m/s and p2 135 kPa. (g) What is the stagnation pressure p02?
Solution For air take k 1.4, cp 1005 m2/(s2 K), and R 287 m2/(s2 K). With V1 and T1 known, we can compute T01 from Eq. (9.23) without using the Mach number: T01 T1 2 V c 2 1 p 320 320 29 349 K Ans. (a) Then compute Ma1 from the known temperature ratio, using Eq. (9.35): Ma2 1 5 3 3 4 2 9 0 1 0.453 Ma1 0.67 Ans. (d) Alternately compute a1 kR T 1 359 m/s, whence Ma1 V1/a1 240/359 0.67. The stagnation pressure at section 1 follows from Eq. (9.34): p01 p1(1 0.2 Ma2 1)3.5 (170 kPa)[1 0.2(0.67)2]3.5 230 kPa Ans. (b) We need the density from the perfect-gas law before we can compute the stagnation density: 1 R p T 1 1 (2 1 8 7 7 0 ) , ( 0 3 0 2 0 0) 1.85 kg/m3 whence 01 1(1 0.2 Ma2 1)2.5 (1.85)[1 0.2(0.67)2]2.5 2.29 kg/m3 Ans. (c) Alternately, we could have gone directly to 0 p0/(RT0) (230 E3)/[(287)(349)] 2.29 kg/m3.
Meanwhile, the maximum velocity follows from Eq. (9.24): Vmax (2cpT0)1/2 [2(1005)(349)]1/2 838 m/s Ans. (e) and the sonic velocity from Eq. (9.33) is V k 2 k 1 RT0 1/2 1 2 .4 (1 .4) 1 (287)(349) 1/2 342 m/s Ans. (f) At point 2, the temperature is not given, but since we know the flow is adiabatic, the stagnation temperature is constant: T02 T01 349 K. Thus, from Eq. (9.23), T2 T02 2 V c 2 2 p 349 2 ( ( 2 1 9 0 0 0 ) 5 2 ) 307 K Then, although the flow itself is not isentropic, the local stagnation pressure is computed by the local isentropic condition (240 m/s)2 2[1005 m2/(s2 K)] 582 Chapter 9 Compressible Flow 9.4 Isentropic Flow with Area Changes p02 p2 T T 0 2 2 k/(k 1) (135) 3 3 4 0 9 7 3.5 211 kPa Ans. (g) This is 8 percent less than the upstream stagnation pressure p01. Notice that, in this last part, we took advantage of the given information (T02, p2, V2) to obtain p02 in an efficient manner. You may verify by comparison that approaching this part through the (unknown) Mach number Ma2 is more laborious.
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) which 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 d d h x 1 h(x) R(x) (9.36) then the flow is approximately one-dimensional, as in Fig. 9.4b, with V V(x) react-ing to area change A(x). Compressible-flow nozzles and diffusers do not always sat-isfy 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), (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) d d V V d A 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: 9.4 Isentropic Flow with Area Changes 583 y x Wall radius of curvature R(x) (a) (b) V(x) Area A(x) y x h(x) V(x, y) Fig. 9.4 Compressible flow through a duct: (a) real-fluid velocity pro-file; (b) one-dimensional approxi-mation.
Fig. 9.5 Effect of Mach number on property changes with area change in duct flow.
Momentum d p V dV 0 Sound speed: dp a2 d (9.39) Now eliminate dp and d between Eqs. (9.38) and (9.39) to obtain the following rela-tion between velocity change and area change in isentropic duct flow: d V V d A A Ma2 1 1 d V p 2 (9.40) 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 Ma2 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 in-creases, 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 de-crease, which speeds up a subsonic flow and slows down a supersonic flow.
What about the sonic point Ma 1? Since infinite acceleration is physically im-possible, Eq. (9.40) indicates that dV can be finite only when dA 0, that is, a mini-mum 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 ex-panding the gas from a stagnant reservoir. The bulge section fails; the bulge Mach num-ber moves away from a sonic condition rather than toward it.
Although supersonic flow downstream of a nozzle requires a sonic throat, the op-584 Chapter 9 Compressible Flow Duct geometry Subsonic Ma < 1 Supersonic Ma > 1 d A > 0 d V < 0 d p > 0 Subsonic diffuser d V > 0 d p < 0 Supersonic nozzle d A < 0 d V > 0 d p < 0 Subsonic nozzle d V < 0 d p > 0 Supersonic diffuser posite is not true: A compressible gas can pass through a throat section without be-coming sonic.
We can use the perfect-gas and isentropic-flow relations to convert the continuity re-lation (9.37) into an algebraic expression involving only area and Mach number, as fol-lows. Equate the mass flow at any section to the mass flow under sonic conditions (which may not actually occur in the duct) VA VA or A A V V (9.41) Both the terms on the right are functions only of Mach number for isentropic flow.
From Eqs. (9.28) and (9.32) 0 0 k 2 1 1 1 2 (k 1) Ma2 1/(k 1) (9.42) From Eqs. (9.26) and (9.32) we obtain V V (kRT V )1/2 (kR V T)1/2 T T 0 1/2 T T 0 1/2 M 1 a k 2 1 1 1 2 (k 1) Ma2 1/2 (9.43) Combining Eqs. (9.41) to (9.43), we get the desired result A A M 1 a (1/2)(k 1)/(k 1) (9.44) For k 1.4, Eq. (9.44) takes the numerical form A A M 1 a (1 1 0 .
.
7 2 28 Ma2)3 (9.45) 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.
1 1 2 (k 1) Ma2 1 2 (k 1) 9.4 Isentropic Flow with Area Changes 585 A min Subsonic Supersonic (a) A max Ma < 1 Ma > 1 (b) Subsonic: (Supersonic: Subsonic: Supersonic) Ma = 1 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 physi-cal grounds.
Perfect-Gas Relations Fig. 9.7 Area ratio versus Mach number for isentropic flow of a perfect gas with k 1.4.
Figure 9.7 shows that the minimum area which 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.
From Eq. (9.41) the inverse ratio A/A equals V/(V), 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 widened. If the throat is constricted further, the mass flow through the duct must de-crease.
From Eqs. (9.32) and (9.33) the maximum mass flow is m ˙ max AV 0 k 2 1 1/(k 1) A k 2 k 1 RT0 1/2 k1/2 k 2 1 (1/2)(k 1)/(k 1) A0(RT0)1/2 (9.46a) For k 1.4 this reduces to m ˙ max 0.6847A0(RT0)1/2 0.6 (R 8 T 47 0) p 1 0 /2 A (9.46b) 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.
Equation (9.46) gives the maximum mass flow, which occurs at the choking condition (sonic exit). It can be modified to predict the actual (nonmaximum) mass flow at any 586 Chapter 9 Compressible Flow 3.0 2.0 1.0 00 0.5 1.0 1.5 2.0 2.5 Mach number Exact Eq. (9.45) Curve fit Eq. (9.48c) Curve fit Eq. (9.48b) A A Choking The Local Mass-Flow Function Part (a) section where local area A and pressure p are known.1 The algebra is convoluted, so here we give only the final result, expressed in dimensionless form: Mass-flow function A m ˙ p R 0 T 0 k 2 k 1 p p 0 2/ k 1 p p 0 (k 1) /k (9.47) We stress that p and A in this relation are the local values at position x. As p/p0 falls, this function rises rapidly and then levels out at the maximum of Eq. (9.46). A few val-ues may be tabulated here for k 1.4: p/p0 1.0 0.98 0.95 0.9 0.8 0.7 0.6 0.5283 Function 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.
The only cumbersome algebra in these problems is the inversion of Eq. (9.45) to compute the Mach number when A/A is known. If available, EES is ideal for this sit-uation and will yield Ma in a flash. In the absence of EES, the following curve-fitted formulas are suggested; given A/A, they estimate the Mach number within 2 per-cent for k 1.4 if you stay within the ranges listed for each formula: 1 1 0 .7 .2 2 7 8 ( A A / / A A ) 2 1.34 A A (9.48a) 1 0.88ln A A 0.45 1.0 A A 1.34 subsonic flow (9.48b) Ma 1 1.2 A A 1 1/2 1.0 A A 2.9 (9.48c) 216 A A 254 A A 2/3 1/5 2.9 A A supersonic flow (9.48d) Formulas (9.48a) and (9.48d) are asymptotically correct as A/A →, while (9.48b) and (9.48c) are just curve fits. However, formulas (9.48b) and (9.48c) are seen in Fig.
9.7 to be accurate within their recommended ranges.
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, e.g., 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 m2 and V1 180 m/s, p1 500 kPa, and T1 470 K. Compute (a) T0, (b) Ma1, (c) p0, and (d) both A and m ˙. If at section 2 the area is 0.036 m2, compute Ma2 and p2 if the flow is (e) subsonic or (f) supersonic. Assume k 1.4.
Solution A general sketch of the problem is shown in Fig. E9.4. With V1 and T1 known, the energy equa-tion (9.23) gives 9.4 Isentropic Flow with Area Changes 587 1The author is indebted to Georges Aigret, of Chimay, Belgium, for suggesting this useful function.
E9.4 T0 T1 2 V c 2 1 p 470 2 ( ( 1 1 8 0 0 0 ) 5 2 ) 486 K Ans. (a) The local sound speed a1 kR T 1 [(1.4)(287)(470)]1/2 435 m/s. Hence Ma1 V a1 1 1 4 8 3 0 5 0.414 Ans. (b) With Ma1 known, the stagnation pressure follows from Eq. (9.34): p0 p1(1 0.2 Ma2 1)3.5 (500 kPa)[1 0.2(0.414)2]3.5 563 kPa Ans. (c) Similarly, from Eq. (9.45), the critical sonic-throat area is A A 1 (1 1 .7 0 2 .
8 2 M M a a 1 2 1)3 [1 1.7 0 2 .
8 2 ( ( 0 0 .
.
4 4 1 1 4 4 ) )2]3 1.547 or A 1.
A 5 1 47 0 1 .0 .5 5 4 m 7 2 0.0323 m2 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 Eq. (9.46), which remains valid, based on the numerical value of A, whether or not a throat actually exists: m ˙ 0.6847 p0 R A T 0 0.6847 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 p1/p0 500/563 0.889, Eq. (9.47) yields m ˙ 2 ( 0 1 .
.
4 4 ) (0 .8 8 9 )2 /1 .4 [1 ( 0 .8 8 9 )0 .4 /1 .4 ] 0.447 m ˙ 33.4 k s g Ans. (d) Assume subsonic flow corresponds to section 2E in Fig. E9.4. The duct contracts to an area ra-tio A2/A 0.036/0.0323 1.115, which we find on the left side of Fig. 9.7 or the subsonic part of Table B.1. Neither the figure nor the table is that accurate. There are two accurate op-tions. First, Eq. (9.48b) gives the estimate Ma2 1 0.88 ln (1.115)0.45 0.676 (error less than 0.5 percent). Second, EES (App. E) will give an arbitrarily accurate solution with only three statements (in SI units): A2 0.036 Astar 0.0323 A2/Astar (1+0.2Ma2^2)^3/1.2^3/Ma2 2 8 7 (4 8 6 ) 563,000(0.05) (563,000)(0.0323) (2 8 7 )( 4 8 6 ) 588 Chapter 9 Compressible Flow Subsonic Throat Possibly supersonic Assume isentropic flow 2 E A1 = 0.05 m 2 1 2 F A2 = 0.036 m 2 A2 = 0.036 m 2 p1 = 500 k P a T1 = 470 K V1 = 180 m /s Part (b) Part (c) Part (d) Part (e) EES Part (f) Specify that you want a subsonic solution (e.g., limit Ma2 1), and EES reports Ma2 0.6758 Ans. (e) [Ask for a supersonic solution and you receive Ma2 1.4001, which is the answer to part (f).] The pressure is given by the isentropic relation p2 56 1 3 .35 k 8 Pa 415 kPa Ans. (e) Part (e) does not require a throat, sonic or otherwise; the flow could simply be contracting sub-sonically from A1 to A2. This time assume supersonic flow, corresponding to section 2F in Fig. E9.4. Again the area ra-tio is A2/A 0.036/0.0323 1.115, and we look on the right side of Fig. 9.7 or the supersonic part of Table B.1—the latter can be read quite accurately as Ma2 1.40. Again there are two other accurate options. First, Eq. (9.48c) gives the curve-fit estimate Ma2 1 1.2(1.115 1)11/2 1.407, only 0.5 percent high. Second, EES will give a very accurate solution with the same three statements from part (e). Specify that you want a supersonic solution (e.g., limit Ma2 1), and EES reports Ma2 1.4001 Ans. (f) Again the pressure is given by the isentropic relation at the new Mach number: p2 56 3 3 .18 k 3 Pa 177 kPa Ans. (f) 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 p0 200 kPa and T0 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.
Solution The throat area follows from Eq. (9.47), because the throat flow must be sonic to produce a su-personic exit: A 0.00830 m2 D2 or Dthroat 10.3 cm Ans. (a) With the exit Mach number known, the isentropic-flow relations give the pressure and temper-ature: pe 11,700 Pa Ans. (b) Te 222 K Ans. (c) 500 2.25 T0 1 0.2(2.5)2 200,000 17.08 p0 [1 0.2(2.5)2]3.5 1 4 3.0[287(500)]1/2 0.6847(200,000) m ˙ (RT0)1/2 0.6847p0 p0 [1 0.2(1.4001)2]3.5 p0 [1 0.2(0.676)2]3.5 9.4 Isentropic Flow with Area Changes 589 EES 9.5 The Normal-Shock Wave The exit velocity follows from the known Mach number and temperature Ve Mae (kRTe)1/2 2.5[1.4(287)(222)]1/2 2.5(299 m/s) 747 m/s Ans. (d) The exit area follows from the known throat area and exit Mach number and Eq. (9.45): 2.64 or Ae 2.64A 2.64(0.0083 m2) 0.0219 m2 1 4 D2 e or De 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.
A common irreversibility occurring in supersonic internal or external flows is the normal-shock wave sketched in Fig. 9.8. Except at near-vacuum pressures such shock waves are very thin (a few micrometers thick) and approximate a discontinuous change in flow properties. We select a control volume just before and after the wave, as in Fig. 9.8.
The analysis is identical to that of Fig. 9.1; i.e., a shock wave is a fixed strong pres-sure 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: 1V1 2V2 G const (9.49a) p1 p2 2V2 2 1V2 1 (9.49b) Energy: h1 1 2 V2 1 h2 1 2 V2 2 h0 const (9.49c) Perfect gas: (9.49d) Constant cp: h cpT k const (9.49e) Note that we have canceled out the areas A1 A2, which is justified even in a variable 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 (p1, V1, 1, h1, T1) are known, Eqs. (9.49) are five algebraic relations in the five unknowns (p2, V2, 2, h2, T2). 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 s2 s1.
The velocities V1 and V2 can be eliminated from Eqs. (9.49a) to (9.49c) to obtain the Rankine-Hugoniot relation h2 h1 (p2 p1) (9.50) 1 1 1 2 1 2 p2 2T2 p1 1T1 [1 0.2(2.5)2]3 1.728(2.5) Ae A 590 Chapter 9 Compressible Flow Fig. 9.8 Flow through a fixed normal-shock wave.
This contains only thermodynamic properties and is independent of the equation of state. Introducing the perfect-gas law h cpT kp/[(k 1)], we can rewrite this as (9.51) We can compare this with the isentropic-flow relation for a very weak pressure wave in a perfect gas 1/k (9.52) Also, the actual change in entropy across the shock can be computed from the perfect-gas relation ln k (9.53) Assuming a given wave strength p2/p1, we can compute the density ratio and the en-tropy change and list them as follows for k 1.4: 1 2 p2 p1 s2 s1 c p2 p1 2 1 k 1 k 1 1 p2/p1 p2/p1 2 1 9.5 The Normal-Shock Wave 591 Fixed normal shock Isoenergetic T01 = T02 1 2 Isentropic upstream s = s1 Isentropic downstream s = s2 > s1 p02 < p01 Ma 1 > 1 Ma 2 < 1 Thin control volume A1≈ A2 A > A 2 1 2/1 Eq. (9.51) Isentropic 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 s2 s1 c p2 p1 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.2 We see also that weak-shock waves (p2/p1 2.0) are very nearly isentropic.
For a perfect gas all the property ratios across the normal shock are unique functions of k and the upstream Mach number Ma1. For example, if we eliminate 2 and V2 from Eqs. (9.49a) to (9.49c) and introduce h kp/[(k 1)], we obtain (k 1) (9.54) But for a perfect gas 1V2 1/p1 kV2 1/(kRT1) k Ma2 1, so that Eq. (9.54) is equivalent to [2k Ma2 1 (k 1)] (9.55) From this equation we see that, for any k, p2 p1 only if Ma1 1.0. Thus for flow through a normal-shock wave, the upstream Mach number must be supersonic to sat-isfy the second law of thermodynamics.
What about the downstream Mach number? From the perfect-gas identity V2 kp Ma2, we can rewrite Eq. (9.49b) as 1 1 k k M M a a 2 2 1 2 (9.56) which relates the pressure ratio to both Mach numbers. By equating Eqs. (9.55) and (9.56) we can solve for Ma2 2 2 ( k k Ma 1 2 1 ) M ( a k 2 1 1 2 ) (9.57) Since Ma1 must be supersonic, this equation predicts for all k 1 that Ma2 must be subsonic. Thus a normal-shock wave decelerates a flow almost discontinuously from supersonic to subsonic conditions.
Further manipulation of the basic relations (9.49) for a perfect gas gives additional equations relating the change in properties across a normal-shock wave in a perfect gas [2 (k 1) Ma2 1] (9.58) T02 T01 k/(k 1) 1/(k 1) Of additional interest is the fact that the critical, or sonic, throat area A in a duct in-creases across a normal shock k 1 2k Ma2 1 (k 1) (k 1) Ma2 1 2 (k 1) Ma2 1 02 01 p02 p01 2k Ma2 1 (k 1) (k 1)2 Ma2 1 T2 T1 V1 V2 (k 1) Ma2 1 (k 1) Ma2 1 2 2 1 p2 p1 1 k 1 p2 p1 21V2 1 p1 1 k 1 p2 p1 592 Chapter 9 Compressible Flow Mach-Number Relations 2 This is true also for most real gases; see Ref. 14, sec. 7.3.
Fig. 9.9 Change in flow properties across a normal-shock wave for k 1.4.
(1/2)(k 1)/(k 1) (9.59) All these relations are given in Table B.2 and plotted versus upstream Mach number Ma1 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 first 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; i.e., the flow across the shock is adiabatic but non-isentropic. Other basic principles governing the behavior of shock waves can be sum-marized 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, e.g., shock tubes, and in steady flow for certain ranges of the downstream pressure. Figure 9.10a shows a normal shock in a supersonic nozzle. Flow is from left to right. The oblique wave pat-tern to the left is formed by roughness elements on the nozzle walls and indicates that the upstream flow is supersonic. Note the absence of these Mach waves (see Sec. 9.10) in the subsonic flow downstream.
2 (k 1) Ma2 1 2 (k 1) Ma2 2 Ma2 Ma1 A 2 A 1 9.5 The Normal-Shock Wave 593 6 5 4 3 2 1 0 1 1.5 2 2.5 3 3.5 4 Ma1 Ma 2 p2 p1 T2 T1 2 A 1 A V1 V2 = ρ2 ρ1 p02 p01 = ρ02 ρ01 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-corner curva-ture are optical distortions. (Cour-tesy of U.S. Army Ballistic Re-search Laboratory, Aberdeen Proving Ground.) 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 which 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 in-side the shock near the body nose is thus subsonic and at relatively high temperature T2 T1, and convective heat transfer is especially high in this region.
Each nonnormal portion of the bow shock in Fig. 9.10b 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.
594 Chapter 9 Compressible Flow (a ) (b ) Moving Normal Shocks Note the fine-grained turbulent wake structure in the rear of the body in Fig. 9.10b.
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, e.g., Ref. 14, chap. 9, or Ref. 8, chap. 16.
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 sim-ulate a still fluid, we move the shock of Fig. 9.8 to the left at speed V1; 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 V1 V2 following the shock. The ther-modynamic 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) p1, (b) p2, (c) p02, (d) A 2, (e) p03, (f) A 3, (g) p3, (h) T03, and (i) T3.
Solution The reservoir conditions are the stagnation properties, which, for assumed one-dimensional adi-abatic frictionless flow, hold through the throat up to section 1 p01 300 kPa T01 500 K A shock wave cannot exist unless Ma1 is supersonic; therefore the flow must have accelerated through a throat which is sonic At A 1 1 m2 We can now find the Mach number Ma1 from the known isentropic area ratio A A 1 1 2.0 From Eq. (9.48c) Ma1 1 1.2(2.0 1)1/2 2.20 Further iteration with Eq. (9.45) would give Ma1 2.1972, showing that Eq. (9.48c) gives sat-isfactory accuracy. The pressure p1 follows from the isentropic relation (9.28) (or Table B.1) [1 0.2(2.20)2]3.5 10.7 or p1 28.06 kPa Ans. (a) 300 kPa 10.7 p01 p1 2 m2 1 m2 9.5 The Normal-Shock Wave 595 1 2 3 1 m2 2 m2 3 m2 E9.6 The pressure p2 is now obtained from Ma1 and the normal-shock relation (9.55) or Table B.2 [2.8(2.20)2 0.4] 5.48 or p2 5.48(28.06) 154 kPa Ans. (b) In similar manner, for Ma1 2.20, p02/p01 0.628 from Eq. (9.58) and A 2/A 1 1.592 from Eq. (9.59), or we can read Table B.2 for these values. Thus p02 0.628(300 kPa) 188 kPa Ans. (c) A 2 1.592(1 m2) 1.592 m2 Ans. (d) The flow from section 2 to 3 is isentropic (but at higher entropy than the flow upstream of the shock). Thus p03 p02 188 kPa Ans. (e) A 3 A 2 1.592 m2 Ans. (f) Knowing A 3, we can now compute p3 by finding Ma3 and without bothering to find Ma2 (which happens to equal 0.547). The area ratio at section 3 is A A 3 3 1.884 Then, since Ma3 is known to be subsonic because it is downstream of a normal shock, we use Eq. (9.48a) to estimate Ma3 0.330 The pressure p3 then follows from the isentropic relation (9.28) or Table B.1 [1 0.2(0.330)2]3.5 1.078 or p3 174 kPa Ans. (g) Meanwhile, the flow is adiabatic throughout the duct; thus T01 T02 T03 500 K Ans. (h) Therefore, finally, from the adiabatic relation (9.26) 1 0.2(0.330)2 1.022 or T3 489 K Ans. (i) Notice that this type of duct-flow problem, with or without a shock wave, requires straightfor-ward application of algebraic perfect-gas relations coupled with a little thought given to which formula is appropriate for the particular situation.
500 K 1.022 T03 T3 188 kPa 1.078 p03 p3 1 0.27/(1.884)2 1.728(1.884) 3 m2 1.592 m2 1 2.4 p2 p1 596 Chapter 9 Compressible Flow E9.7 Part (a) Part (b) 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 lbf/in2 absolute. Estimate (a) the shock speed C and (b) the air velocity V just inside the shock.
9.5 The Normal-Shock Wave 597 C V POW!
p = 14.7 1bf / in2 abs T = 520° R 200 1bf / in2 abs Solution 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 V1 p1 14.7 lbf/in2 absolute T1 520°R V V1 V2 p2 200 lbf/in2 absolute The speed of sound outside the shock is a1 49T1 1/2 1117 ft/s. We can find Ma1 from the known pressure ratio across the shock 13.61 From Eq. (9.55) or Table B.2 13.61 (2.8 Ma2 1 0.4) or Ma1 3.436 Then, by definition of the Mach number, C V1 Ma1 a1 3.436(1117 ft/s) 3840 ft/s Ans. (a) To find V2, we need the temperature or sound speed inside the shock. Since Ma1 is known, from Eq. (9.58) or Table B.2 for Ma1 3.436 we compute T2/T1 3.228. Then T2 3.228T1 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 V2 2 2cp(T1 T2) V2 1 2(6010)(520 1679) (3840)2 815,000 or V2 903 ft/s Notice that we did this without bothering to compute Ma2, which equals 0.454, or a2 49T2 1/2 2000 ft/s.
1 2.4 200 lbf/in2 absolute 14.7 lbf/in2 absolute p2 p1 9.6 Operation of Converging and Diverging Nozzles Converging Nozzle Finally, the air velocity behind the shock is V V1 V2 3840 903 2940 ft/s Ans. (b) Thus a powerful explosion creates a brief but intense blast wind as it passes.3 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. 9.11a. There is an upstream reser-voir at stagnation pressure p0. The flow is induced by lowering the downstream out-side, or back, pressure pb below p0, resulting in the sequence of states a to e shown in Fig. 9.11b and c.
For a moderate drop in pb to states a and b, the throat pressure is higher than the critical value p which would make the throat sonic. The flow in the nozzle is sub-sonic throughout, and the jet exit pressure pe equals the back pressure pb. The mass flow is predicted by subsonic isentropic theory and is less than the critical value m ˙ max, as shown in Fig. 9.11c.
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, pe pb, and the mass flow equals its maximum value from Eq. (9.46). The flow upstream of the throat is subsonic every-where and predicted by isentropic theory based on the local area ratio A(x)/A and Table B.1.
Finally, if pb is lowered further to conditions d or e below p, the nozzle cannot re-spond further because it is choked at its maximum throat mass flow. The throat re-mains sonic with pe p, and the nozzle-pressure distribution is the same as in state c, as sketched in Fig. 9.11b. The exit jet expands supersonically so that the jet pres-sure can be reduced from p down to pb. The jet structure is complex and multidi-mensional and is not shown here. Being supersonic, the jet cannot send any signal up-stream 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 p0 decreasing and pb 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 p0(t) and pb(t).
EXAMPLE 9.8 A converging nozzle has a throat area of 6 cm2 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.
598 Chapter 9 Compressible Flow 3 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, e.g., Ref. 5, sec. 4.5.
Fig. 9.11 Operation of a converging nozzle: (a) nozzle geometry show-ing characteristic pressures; (b) pressure distribution caused by various back pressures; (c) mass flow versus back pressure.
Solution From Eq. (9.32) for k 1.4 the critical (sonic) throat pressure is 0.5283 or p (0.5283)(120 kPa) 63.4 kPa If the back pressure is less than this amount, the nozzle flow is choked.
For pb 90 kPa p, the flow is subsonic, not choked. The exit pressure is pe pb. The throat Mach number is found from the isentropic relation (9.35) or Table B.1: Ma2 e 5 2/7 1 5 2/7 1 0.4283 Mae 0.654 To find the mass flow, we could proceed with a serial attack on Mae, Te, ae, Ve, and e, hence 120 90 p0 pe p p0 9.6 Operation of Converging and Diverging Nozzles 599 p0 pb ⋅pe Jet boundary (a) 1.0 p0 p p0 p x (b) Sonic point a b Subsonic jet c d e Supersonic jet expansion d e c 1.0 b a p0 p (c) 1.0 p0 pb ⋅ m ⋅ m max 0 0 Part (a) Converging-Diverging Nozzle to compute eAeVe. However, since the local pressure is known, this part is ideally suited for the dimensionless mass-flow function in Eq. (9.47). With pe/p0 90/120 0.75, compute m ˙ Ap R 0 T 0 (0 .7 5 )2 /1 .4 [1 ( 0 .7 5 )0 .4 /1 .4 ] 0.6052 hence m ˙ 0.6052 0.129 kg/s Ans. (a) for pe pb 90 kPa Ans. (a) For pb 45 kPa p, the flow is choked, similar to condition d in Fig. 9.11b. The exit pres-sure is sonic: pe p 63.4 kPa Ans. (b) The (choked) mass flow is a maximum from Eq. (9.46b): m ˙ m ˙ max 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 pres-sure pb is low enough, there will be supersonic flow in the diverging portion and a va-riety of shock-wave conditions may occur, which are sketched in Fig. 9.12b. Let the back pressure be gradually decreased.
For curves A and B in Fig. 9.12b 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, e.g., Table B.1. The exit pressure pe pb, and the jet is subsonic.
For curve C the area ratio Ae/At exactly equals the critical ratio Ae/A for a subsonic Mae in Table B.1. 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 pe pb.
Now jump for a moment to curve H. Here pb is such that pb/p0 exactly corresponds to the critical-area ratio Ae/A for a supersonic Mae in Table B.1. The diverging flow is entirely supersonic, including the jet flow, and pe pb. This is called the design pressure ratio of the nozzle and is the back pressure suitable for operating a supersonic wind tunnel or an efficient rocket exhaust.
Now back up and suppose that pb lies between curves C and H, which is impossi-ble according to purely isentropic-flow calculations. Then back pressures D to F oc-cur 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-diffuser 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 pb.
0.6847(120,000)(0.0006) [287(400)]1/2 0.6847p0Ae (RT0)1/2 (0.0006)(120,000) 2 8 7 (4 0 0 ) 2(1.4) 0.4 600 Chapter 9 Compressible Flow Part (b) Fig. 9.12 Operation of a converg-ing-diverging nozzle: (a) nozzle geometry with possible flow con-figurations; (b) pressure distribution caused by various back pressures; (c) mass flow versus back pressure.
Finally, at back pressure I, pb 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, e.g., Ref. 9, sec. 5.4, for fur-ther details of these off-design jet-flow configurations.
Note that for pb 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 boundary-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, e.g., Ref. 14, pp. 292 and 293, for further details).
9.6 Operation of Converging and Diverging Nozzles 601 p0 pt pb Possible complex jet geometry Throat Possible normal shock Adverse pressure gradient 1.0 p p0 p p0 x Sonic throat Supersonic Shock pe A B D E F G I 1.0 B A Design pressure ratio (a) (b) (c) pb p0 1.0 p p0 m ⋅ m max ⋅ C H 0 0 H F E D C G I Part (a) Part (b) Part (c) EXAMPLE 9.9 A converging-diverging nozzle (Fig. 9.12a) has a throat area of 0.002 m2 and an exit area of 0.008 m2. Air stagnation conditions are p0 1000 kPa and T0 500 K. Compute the exit pressure and mass flow for (a) design condition and the exit pressure and mass flow if (b) pb 300 kPa and (c) pb 900 kPa. Assume k 1.4.
Solution The design condition corresponds to supersonic isentropic flow at the given area ratio Ae/At 0.008/0.002 4.0. We can find the design Mach number either by iteration of the area-ratio for-mula (9.45), using EES, or by the curve fit (9.48d) Mae,design [216(4.0) 254(4.0)2/3]1/5 2.95 (exact 2.9402) The accuracy of the curve fit is seen to be satisfactory. The design pressure ratio follows from Eq. (9.34) [1 0.2(2.95)2]3.5 34.1 or pe,design 100 3 0 4.1 kPa 29.3 kPa Ans. (a) Since the throat is clearly sonic at design conditions, Eq. (9.46b) applies m ˙ design m ˙ max Ans. (a) 3.61 kg/s For pb 300 kPa we are definitely far below the subsonic isentropic condition C in Fig. 9.12b, but we may even be below condition F with a normal shock in the exit, i.e., in condition G, where oblique shocks occur outside the exit plane. If it is condition G, then pe pe,design 29.3 kPa because no shock has yet occurred. To find out, compute condition F by assuming an exit normal shock with Ma1 2.95, that is, the design Mach number just upstream of the shock.
From Eq. (9.55) [2.8(2.95)2 0.4] 9.99 or p2 9.99p1 9.99pe,design 293 kPa Since this is less than the given pb 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 Ans. (b) Also m ˙ m ˙ max 3.61 kg/s Ans. (b) The throat is still sonic and choked at its maximum mass flow.
Finally, for pb 900 kPa, which is up near condition C, we compute Mae and pe for condition C as a comparison. Again Ae/At 4.0 for this condition, with a subsonic Mae estimated from the curve-fitted Eq. (9.48a): Mae(C) 0.147 (exact 0.14655) 1 0.27/(4.0)2 1.728(4.0) 1 2.4 p2 p1 0.6847(106 Pa)(0.002 m2) [287(500)]1/2 0.6847p0At (RT0)1/2 p0 pe 602 Chapter 9 Compressible Flow 9.7 Compressible Duct Flow with Friction4 Then the isentropic exit-pressure ratio for this condition is [1 0.2(0.147)2]3.5 1.0152 or pe 985 kPa The given back pressure of 900 kPa is less than this value, corresponding roughly to condition D in Fig. 9.12b. Thus for this case there is a normal shock just downstream of the throat, and the throat is choked pe pb 900 kPa m ˙ m ˙ max 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.
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, 8, 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 1.
Steady one-dimensional adiabatic flow 2.
Perfect gas with constant specific 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.
Consider the elemental duct control volume of area A and length dx in Fig. 9.13.
The area is constant, but other flow properties (p, , T, h, V) may vary with x. Appli-1000 1.0152 p0 pe 9.7 Compressible Duct Flow with Friction 603 4 This section may be omitted without loss of continuity.
Control volume τ w π D d x V x x + d x V + d V p p + d p T T + d T h h + d h Area A Diameter D ρ + d ρ ρ d x Fig. 9.13 Elemental control volume for flow in a constant-area duct with friction.
Adiabatic Flow cation of the three conservation laws to this control volume gives three differential equations Continuity: V A m ˙ G const or 0 (9.60a) x momentum: pA (p dp)A wD dx m ˙ (V dV V) or dp V dV 0 (9.60b) Energy: h 1 2 V2 h0 cpT0 cpT 1 2 V2 or cp dT V dV 0 (9.60c) Since these three equations have five unknowns—p, , T, V, and w—we need two ad-ditional relations. One is the perfect-gas law p RT or (9.61) To eliminate w as an unknown, it is assumed that wall shear is correlated by a local Darcy friction factor f w 1 8 fV2 1 8 fkp Ma2 (9.62) where the last form follows from the perfect-gas speed-of-sound expression a2 kp/.
In practice, f can be related to the local Reynolds number and wall roughness from, say, the Moody chart, Fig. 6.13.
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 p1, T1, V1, etc., are known, to determine p(x), T(x), etc., 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 fac-tor, by using the definition of Mach number V2 Ma2 kRT or (9.63) By eliminating variables between Eqs. (9.60) to (9.63), we obtain the working relations k Ma2 f (9.64a) f (9.64b) dV V dx D k Ma2 2(1 Ma2) d dx D 1 (k 1) Ma2 2(1 Ma2) dp p dT T 2 d Ma Ma 2 dV V dT T d dp p 4wdx D dV V d 604 Chapter 9 Compressible Flow k Ma2 f (9.64c) f (9.64d) k Ma2 f (9.64e) All these except dp0/p0 have the factor 1 Ma2 in the denominator, so that, like the area-change formulas in Fig. 9.5, subsonic and supersonic flow have opposite effects: dx D 1 1 2 (k 1) Ma2 1 Ma2 d Ma2 Ma2 dx D k(k 1) Ma4 2(1 Ma2) dT T dx D 1 2 d0 0 dp0 p0 9.7 Compressible Duct Flow with Friction 605 Property Subsonic Supersonic p Decreases Increases Decreases Increases V Increases Decreases p0, 0 Decreases Decreases T Decreases Increases Ma Increases Decreases Entropy Increases Increases We have added to the list above 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 above is the Mach number. Whether the inlet flow is subsonic or supersonic, the duct Mach number always tends downstream toward Ma 1 be-cause this is the path along which the entropy increases. If the pressure and density are computed from Eqs. (9.64a) 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. The maximum entropy occurs at Ma 1, so that the second law requires that the duct-flow properties continually ap-proach the sonic point. Since p0 and 0 continually decrease along the duct due to the frictional (nonisentropic) losses, they are not useful as reference properties. Instead, the sonic properties p, , T, p 0, and 0 are the appropriate constant reference quanti-ties in adiabatic duct flow. The theory then computes the ratios p/p, T/T, etc., 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 num-ber to friction. Separate the variables and integrate: L 0 f 1.0 Ma2 d Ma2 (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 ln (9.66) where f is the average friction factor between 0 and L. In practice, an average f is al-ways assumed, and no attempt is made to account for the slight changes in Reynolds (k 1) Ma2 2 (k 1) Ma2 k 1 2k 1 Ma2 k Ma2 f L D 1 Ma2 k Ma4[1 1 2 (k 1) Ma2] dx D Fig. 9.14 Adiabatic frictional flow in a constant-area duct always ap-proaches Ma 1 to satisfy the sec-ond law of thermodynamics. The computed curve is independent of the value of the friction factor.
number along the duct. For noncircular ducts, D is replaced by the hydraulic diameter Dh (4 area)/perimeter as in Eq. (6.74).
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 which never become sonic, for which the so-lution uses the differences in the tabulated “maximum,’’ or sonic, length. For example, the length L required to develop from Ma1 to Ma2 is given by f f D L 1 f D L 2 (9.67) This avoids the need for separate tabulations for short ducts.
It is recommended that the friction factor f be 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. The average friction factor is 0.024.
What length of duct is necessary to accelerate the flow from Ma1 0.1 to Ma2 0.5? What ad-ditional length will accelerate it to Ma3 1.0? Assume k 1.4. L D 606 Chapter 9 Compressible Flow 2.0 0.2 Mach number 0 1.0 3.0 4.0 0.4 0.6 0.8 1.0 1.2 Subsonic duct flow k = 1.4 Maximum entropy at Ma = 1.0 s v c Supersonic duct flow Solution Equation (9.67) applies, with values of f L/D computed from Eq. (9.66) or read from Table B.3: f Ma0.1 Ma0.5 66.9216 1.0691 65.8525 Thus L 55 m Ans. (a) The additional length L to go from Ma 0.5 to Ma 1.0 is taken directly from Table B.2 f Ma0.5 1.0691 or L L Ma0.5 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 be derived from Eqs. (9.64).
Equation (9.64e) can be used to eliminate f dx/D from each of the other relations, giv-ing, for example, dp/p as a function only of Ma and d Ma2/Ma2. For convenience in tabulating the results, each expression is then integrated all the way from (p, Ma) to the sonic point ( p, 1.0). The integrated results are p p M 1 a 2 (k k 1 1 ) Ma2 1/2 (9.68a) V V M 1 a 2 (k k 1 1 ) Ma2 1/2 (9.68b) T T a a 2 2 2 (k k 1 1 ) Ma2 (9.68c) p p 0 0 0 0 M 1 a 2 (k k 1 1 ) Ma2 (1/2)(k 1)/(k 1) (9.68d) All these ratios are also tabulated in Table B.3. For finding changes between points Ma1 and Ma2 which are not sonic, products of these ratios are used. For example, p p 2 1 p p 2 p p 1 (9.69) since p is a constant reference value for the flow.
EXAMPLE 9.11 For the duct flow of Example 9.10 assume that, at Ma1 0.1, we have p1 600 kPa and T1 450 K. At section 2 farther downstream, Ma2 0.5. Compute (a) p2, (b) T2, (c) V2, and (d) p02.
1.0691(0.02 m) 0.024 fL D L D 65.8525(0.02 m) 0.024 f L D f L D 0.024 L 0.02 m L D 9.7 Compressible Duct Flow with Friction 607 Choking due to Friction Solution As preliminary information we can compute V1 and p01 from the given data: V1 Ma1 a1 0.1[(1.4)(287)(450)]1/2 0.1(425 m/s) 42.5 m/s p01 p1(1 0.2 Ma2 1)3.5 (600 kPa)[1 0.2(0.1)2]3.5 604 kPa Now enter Table B.3 or Eqs. (9.68) to find the following property ratios: 608 Chapter 9 Compressible Flow Use these ratios to compute all properties downstream: p2 p1 (600 kPa) 117 kPa Ans. (a) T2 T1 (450 K) 429 K Ans. (b) V2 V1 (42.5 m/s) 208 Ans. (c) p02 p01 (604 kPa) 139 kPa Ans. (d) Note the 77 percent reduction in stagnation pressure due to friction. The formulas are seductive, so check your work by other means. For example, check p02 p2(1 0.2 Ma2 2)3.5.
The theory here predicts that for adiabatic frictional flow in a constant-area duct, no matter what the inlet Mach number Ma1 is, the flow downstream tends toward the sonic point. There is a certain duct length L(Ma1) 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 L?
Then the flow conditions must change, and there are two classifications.
Subsonic inlet.
If L L(Ma1), 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 su-personic duct flow. Even an infinite inlet Mach number will be reduced to sonic con-ditions in only 41 diameters for f 0.02. Some typical numerical values are shown in Fig. 9.15, assuming an inlet Ma 3.0 and f 0.02. For this condition L 26 di-ameters. If L is increased beyond 26D, the flow will not choke but a normal shock will form at just the right place for the subsequent subsonic frictional flow to become sonic exactly at the exit. Figure 9.15 shows two examples, for L/D 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 L/D 63. Further increase in L causes the shock to move upstream of 1.3399 5.8218 p02/p 0 p01/p 0 m s 0.5345 0.1094 V2/V V1/V 1.1429 1.1976 T2/T T1/T 2.1381 10.9435 p2/p p1/p Section Ma p/p T/T V/V p0/p 0 1 0.1 10.9435 1.1976 0.1094 5.8218 2 0.5 2.1381 1.1429 0.5345 1.3399 Fig. 9.15 Behavior of duct flow with a nominal supersonic inlet condition Ma 3.0: (a) L/D 26, flow is supersonic throughout duct; (b) L/D 40 L/D, normal shock at Ma 2.0 with subsonic flow then accelerating to sonic exit point; (c) L/D 53, shock must now occur at Ma 2.5; (d) L/D 63, flow must be entirely subsonic and choked at exit.
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 p0 200 kPa, T0 500 K, and V1 100 m/s. The friction factor is 0.02. Compute (a) the maximum duct length for these conditions, (b) the mass flow if the duct length is 15 m, and (c) the reduced mass flow if L 30 m.
Solution First compute T1 T0 500 500 5 495 K a1 (kRT1)1/2 20(495)1/2 445 m/s Thus Ma1 0.225 For this Ma1, from Eq. (9.66) or interpolation in Table B.3, 11.0 f L D 100 445 V1 a1 1 2 (100 m/s)2 1005 m2/(s2 K) 1 2 V2 1 cp 9.7 Compressible Duct Flow with Friction 609 3.0 2.5 2.0 1.5 1.0 0.5 10 d c a f = 0.020 k = 1.4 Mach number 20 30 40 50 60 D x b 0 Part (a) Part (b) Part (c) The maximum duct length possible for these inlet conditions is L 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 inlet conditions 01 R p T 01 0 1.394 kg/m3 1 1.359 kg/m3 whence m ˙ 1AV1 (1.359 kg/m3) (0.03 m)2 (100 m/s) 0.0961 kg/s Ans. (b) Since L 30 m is greater than L, the duct must choke back until L L, corresponding to a lower inlet Ma1: L L 30 m 20.0 It is difficult to interpolate for fL/D 20 in Table B.3 and impossible to invert Eq. (9.66) for the Mach number without laborious iteration. But it is a breeze for EES to solve Eq. (9.66) for the Mach number, using the following three statements: k 1.4 fLD 20 fLD (1 Ma^2)/k/Ma^2 (k 1)/2/kLN((k 1)Ma^2/(2 (k 1)Ma^2)) Simply specify Ma 1 in the Variable Information menu and EES cheerfully reports Machoked 0.174 (23 percent less) T1,new 497 K a1,new 20(497 K)1/2 446 m/s V1,new Ma1 a1 0.174(446) 77.6 m/s 1,new 1.373 kg/m3 m ˙ new 1AV1 1.373 4 π (0.03)2 (77.6) 0.0753 kg/s (22 percent less) Ans. (c) The adiabatic frictional-flow assumption is appropriate to high-speed flow in short ducts. For flow in long ducts, e.g., natural-gas pipelines, the gas state more closely ap-01 [1 0.2(0.174)2]2.5 T0 1 0.2(0.174)2 0.02(30 m) 0.03 m f L D 4 1.394 1.0255 01 [1 0.2(0.225)2]2.5 200,000 Pa 287(500 K) 11.0(0.03 m) 0.02 (f L/D)D f 610 Chapter 9 Compressible Flow Isothermal Flow with Friction EES Mass Flow for a Given Pressure Drop proximates 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 (9.70) Again it is possible to write all property changes in terms of the Mach number. Inte-gration of the Mach-number–friction relation yields ln (k Ma2) (9.71) which is the isothermal analog of Eq. (9.66) for adiabatic flow.
This friction relation has the interesting result that Lmax becomes zero not at the sonic point but at Macrit 1/k1/2 0.845 if k 1.4. The inlet flow, whether subsonic or supersonic, tends downstream toward this limiting Mach number 1/k1/2. If the tube length L is greater than Lmax from Eq. (9.71), a subsonic flow will choke back to a smaller Ma1 and mass flow and a supersonic flow will experience a normal-shock ad-justment similar to Fig. 9.15.
The exit isothermal choked flow is not sonic, and so the use of the asterisk is in-appropriate. Let 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 prop-erties: p p Ma k1/2 (9.72) The complete analysis and some examples are given in advanced texts [for example, 8, sec. 6.4].
An interesting by-product of the isothermal analysis is an explicit relation between the pressure drop and duct mass flow. This is a common problem which requires numeri-cal iteration for adiabatic flow, as outlined below. In isothermal flow, we may substi-tute dV/V dp/p and V2 G2/[p/(RT)]2 in Eq. (9.63) to obtain f 0 Since G2RT is constant for isothermal flow, this may be integrated in closed form be-tween (x, p) (0, p1) and (L, p2): G2 2 (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 de-rived in several textbooks [5, p. 212; 34, p. 418]: V2 1 (9.74) a2 0[1 (V1/V2)2] kf L/D (k 1) ln (V2/V1) p2 1 p2 2 RT[ f L/D 2 ln (p1/p2)] m ˙ A 2 dp p dx D 2p dp G2RT V V 1 Ma k1/2 1 k Ma2 k Ma2 f Lmax D 9.7 Compressible Duct Flow with Friction 611 Part (a) Part (b) where a0 (kRT0)1/2 is the stagnation speed of sound, constant for adiabatic flow. We assign the proof of this as a problem exercise. This may be combined with continuity for constant duct area V1/V2 2/1, plus the following combination of adiabatic en-ergy and the perfect-gas relation: (9.75) If we are given the end pressures, neither V1 nor V2 will likely be known in advance.
Here, if EES is not available, we suggest only the following simple procedure. Begin with a0 a1 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 V1. Use V1 to improve your estimate of a0, 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/k1/2 for isothermal flow or greater than 1.0 for adiabatic flow. We illustrate both adiabatic and isothermal flow with the following example.
EXAMPLE 9.13 Air enters a pipe of 1-cm diameter and 1.2-m length at p1 220 kPa and T1 300 K. If f 0.025 and the exit pressure is p2 140 kPa, estimate the mass flow for (a) isothermal flow and (b) adiabatic flow.
Solution For isothermal flow Eq. (9.73) applies without iteration: 2 ln 2 ln 3.904 G2 85,700 or G 293 kg/(s m2) Since A (/4)(0.01 m)2 7.85 E-5 m2, 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: 2 1.626 kg/m3 V2 180 m/s or Ma2 0.52 This is well below choking, and the isothermal solution is accurate.
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. A few years ago the author would have done just that, laboriously. However, EES makes handwork and manipulation of equations un-180 347 180 [1.4(287)(300)]1/2 V2 kR T 293 1.626 G 2 140,000 (287)(300) p2 RT (220,000 Pa)2 (140,000 Pa)2 287 m2/(s2 K)(3.904) 220 140 (0.025)(1.2 m) 0.01 m p1 p2 f L D 2a2 0 (k 1)V2 1 2a2 0 (k 1)V2 2 p2 p1 T1 T2 p2 p1 V1 V2 612 Chapter 9 Compressible Flow EES 9.8 Frictionless Duct Flow with Heat Transfer5 necessary, although careful programming and good guesses are required. If we ignore superflu-ous output such as T2 and V2, 13 statements are appropriate. First, spell out the given physical properties (in SI units): k 1.4 P1 220000 P2 140000 T1 300 Next, apply the adiabatic friction relations, Eqs. (9.66) and (9.67), to both points 1 and 2: fLD1(1 Ma1^2)/k/Ma1^2 (k 1)/2/kLN((k 1)Ma1^2/(2 (k-1)Ma1^2)) fLD2(1 Ma2^2)/k/Ma2^2 (k 1)/2/kLN((k 1)Ma2^2/(2 (k-1)Ma2^2)) DeltafLD 0.0251.2/0.01 fLD1 fLD2 DeltafLD Then apply the pressure-ratio formula (9.68a) to both points 1 and 2: P1/Pstar ((k 1)/(2 (k-1)Ma1^2))^0.5/Ma1 P2/Pstar ((k 1)/(2 (k-1)Ma2^2))^0.5/Ma2 These are adiabatic relations, so we need not further spell out quantities such as T0 or a0 unless we want them as additional output.
The above 10 statements are a closed algebraic system, and EES will solve them for Ma1 and Ma2. However, the problem asks for mass flow, so we complete the system: V1 Ma1sqrt(1.4287T1) Rho1 P1/287/T1 Mdot Rho1(pi/40.01^2)V1 If we apply no constraints, EES reports “cannot solve”, because its default allows all variables to lie between and . So we enter Variable Information and constrain Ma1 and Ma2 to lie between 0 and 1 (subsonic flow). EES still complains that it “cannot solve” but hints that “better guesses are needed”. Indeed, the default guesses for EES variables are normally 1.0, too large for the Mach numbers. Guess the Mach numbers equal to 0.8 or even 0.5, and EES still complains, for a subtle reason: Since f
L/D 0.025(1.2/0.01) 3.0, Ma1 can be no larger than 0.36 (see Table B.3). Finally, then, we guess Ma1 and Ma2 0.3 or 0.4, and EES happily re-ports the solution: Ma1 0.3343 Ma2 0.5175 f D L 1 3.935 f D L 2 0.9348 p 67,892 Pa m ˙ 0.0233 kg/s Ans. (b) Though the programming is complicated, the EES approach is superior to hand iteration and, of course, we can save this program for use again with new data.
Heat addition or removal has an interesting effect on a compressible flow. Advanced texts [for example, 8, 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.
9.8 Frictionless Duct Flow with Heat Transfer 613 5 This section may be omitted without loss of continuity.
Fig. 9.16 Elemental control volume for frictionless flow in a constant-area duct with heat transfer. The length of the element is indetermi-nate in this simplified theory.
Consider the elemental duct control volume in Fig. 9.16. Between sections 1 and 2 an amount of heat Q is added (or removed) to each incremental mass m passing through. With no friction or area change, the control-volume conservation relations are quite simple: Continuity: 1V1 2V2 G const (9.76a) x momentum: p1 p2 G(V2 V1) (9.76b) Energy: ˙ Q m ˙ (h2 1 2 V2 2 h1 1 2 V2 1) or q h02 h01 (9.76c) The heat transfer results in a change in stagnation enthalpy of the flow. We shall not spec-ify exactly how the heat is transferred—combustion, nuclear reaction, evaporation, con-densation, 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 the-ory 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 h02 h01 cp(T02 T01) 1/2 (9.77) For a given heat transfer q Q/m or, equivalently, a given change h02 h01, Eqs.
(9.76) and (9.77) can be solved algebraically for the property ratios p2/p1, Ma2/Ma1, etc., 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 T0 and T versus Mach number in the duct. Heating increases T0, and cooling decreases it. The maximum possible T0 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/k1/2 and then decreases. Thus there is a peculiar—or at least unexpected—region where heat-T2 T1 Ma2 Ma1 Ma2 a2 Ma1 a1 V2 V1 p1 1T1 p2 2T2 Q m ˙ Q m ˙ 614 Chapter 9 Compressible Flow Control volume V1, p1, T1, T01 V2, p2, T2, T02 A2 = A1 τ w = 0 q = δQ δm 1 2 Fig. 9.17 Effect of heat transfer on Mach number.
ing (increasing T0) actually decreases the gas temperature, the difference being reflected in a large increase of the gas kinetic energy. For k 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 T0 change on duct-flow properties is as follows: Heating Cooling Subsonic Supersonic Subsonic Supersonic T0 Increases Increases Decreases Decreases Ma Increases Decreases Decreases Increases p Decreases Increases Increases Decreases Decreases Increases Increases Decreases V Increases Decreases Decreases Increases p0 Decreases Decreases Increases Increases s Increases Increases Decreases Decreases T Increases † Decreases Increases up to Ma 1/k1/2 and decreases thereafter.
†Decreases up to Ma 1/k1/2 and increases thereafter.
Probably the most significant item on this list is the stagnation pressure p0, 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 0, T, p, , V, and p 0. The inlet is assumed to be at ar-bitrary Mach number Ma. Equations (9.76) and (9.77) then take the following form: T T 0 0 (9.78a) (k 1) Ma2 [2 (k 1) Ma2] (1 k Ma2)2 9.8 Frictionless Duct Flow with Heat Transfer 615 0 k = 1.4 T0 (max) at Ma = 1.0 T0 T T (max) at Ma = 1 k1/ 2 Cooling Heating 0.5 1 1.5 2 2.5 Mach number Cooling Heating T, T0 Mach-Number Relations Part (a) (9.78b) (9.78c) (9.78d) p p 0 0 k/(k 1) (9.78e) These formulas are all tabulated versus Mach number in Table B.4. The tables are very convenient if inlet properties Ma1, V1, etc., are given but are somewhat cum-bersome if the given information centers on T01 and T02. 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 V1 75 m/s, p1 150 kPa, and T1 300 K. The heat addition by combustion is 900 kJ/kg of mix-ture. Compute (a) the exit properties V2, p2, and T2 and (b) the total heat addition which would have caused a sonic exit flow.
Solution First compute T01 T1 V2 1/(2cp) 300 (75)2/[2(1005)] 303 K. Then compute the change in stagnation temperature of the gas: q cp(T02 T01) or T02 T01 303 K 1199 K We have enough information to compute the initial Mach number: a1 kR T 1 [1.4(287)(300)]1/2 347 m/s Ma1 0.216 For this Mach number, use Eq. (9.78a) or Table B.4 to find the sonic value T0: At Ma1 0.216: 0.1992 or T 0 1521 K Then the stagnation temperature ratio at section 2 is T02/T 0 1199/1521 0.788, which cor-responds in Table B.4 to a Mach number Ma2 0.573.
Now use Table B.4 at Ma1 and Ma2 to tabulate the desired property ratios.
303 K 0.1992 T01 T 0 75 347 V1 a1 900,000 J/kg 1005 J/(kg K) q cp 2 (k 1) Ma2 k 1 k 1 1 k Ma2 (k 1) Ma2 1 k Ma2 V V k 1 1 k Ma2 p p (k 1)2 Ma2 (1 k Ma2)2 T T 616 Chapter 9 Compressible Flow Section Ma V/V p/p T/T 1 0.216 0.1051 2.2528 0.2368 2 0.573 0.5398 1.6442 0.8876 Choking Effects due to Simple Heating The exit properties are computed by using these ratios to find state 2 from state 1: V2 V1 (75 m/s) 385 m/s Ans. (a) p2 p1 (150 kPa) 109 kPa Ans. (a) T2 T1 (300 K) 1124 K Ans. (a) The maximum allowable heat addition would drive the exit Mach number to unity: T02 T 0 1521 K qmax cp(T 0 T01) 1005 J/(kg K) 1.22 E6 J/kg Ans. (b) Equation (9.78a) and Table B.4 indicate that the maximum possible stagnation tem-perature in simple heating corresponds to T 0, 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. For a subsonic inlet there is no theo-retical 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 Ma1 is infinite, there is a finite ratio T01/ T 0 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 su-personic flow only a finite amount of cooling can be allowed before the exit flow ap-proaches infinite Mach number, with T02/T 0 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 de-crease in mass flow?
Solution For q 1400 kJ/kg, the exit will be choked at the stagnation temperature T 0 T01 303 1696 K This is higher than the value T 0 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 T01/ T 0 303/1696 0.1787. From Table B.4 or Eq. (9.78a) for this condition, we read the 1.4 E6 J/kg 1005 J/(kg K) q cp 0.8876 0.2368 T2/T T1/T 1.6442 2.2528 p2/p p1/p 0.5398 0.1051 V2/V V1/V 9.8 Frictionless Duct Flow with Heat Transfer 617 Part (b) Relationship to the Normal-Shock Wave new, lowered entrance Mach number: Ma1,new 0.203. With T01 and p1 known, the other inlet properties follow from this Mach number: T1 301 K a1 kR T 1 [1.4(287)(301)] 1/2 348 m/s V1 Ma1 a1 (0.202)(348 m/s) 70 m/s 1 1.74 kg/m3 Finally, the new lowered mass flow per unit area is 1V1 (1.74 kg/m3)(70 m/s) 122 kg/(s m2) This is 7 percent less than in Example 9.14, due to choking by excess heat addition.
The normal-shock-wave relations of Sec. 9.5 actually lurk within the simple heating rela-tions as a special case. From Table B.4 or Fig. 9.17 we see that for a given stagnation tem-perature less than T 0 there are two flow states which satisfy the simple heating relations, one subsonic and the other supersonic. These two states have (1) the same value of T0, (2) the same mass flow per unit area, and (3) the same value of p V2. 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 Ma1 be supersonic.
To illustrate this point, take Ma1 3.0 and from Table B.4 read T01/T 0 0.6540 and p1/p 0.1765. Now, for the same value T02/T 0 0.6540, use Table B.4 or Eq. (9.78a) to compute Ma2 0.4752 and p2/p 1.8235. The value of Ma2 is exactly what we read in the shock table, Table B.2, as the downstream Mach number when Ma1 3.0.
The pressure ratio for these two states is p2/p1 (p2/p)/(p1/p) 1.8235/0.1765 10.33, which again is just what we read in Table B.2 for Ma1 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 which 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 ap-parent if the flow is supersonic. Figure 9.18 shows a celebrated graphical construction which 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.
m ˙ new A 150,000 (287)(301) p1 RT1 303 1 0.2(0.203)2 T01 1 0.2 Ma2 1 618 Chapter 9 Compressible Flow 9.9 Two-Dimensional Supersonic Flow Mach Waves 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.
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 spher-ical disturbance fronts are shown in Fig. 9.18. The behavior of these fronts is quite dif-ferent according to whether the particle speed is subsonic or supersonic.
In Fig. 9.18a, the particle moves subsonically, U a, Ma U/a 1. The spherical dis-turbances 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 t during the time interval t in which the particle has moved only U t. 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, with-out turning around to look at you, takes to the air and avoids being hit by your car.
At sonic speed, U a, Fig. 9.18b, 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 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 which 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 9.9 Two-Dimensional Suspersonic Flow 619 U > a U δ t (a) Typical pressure disturbance caused by particle passage Limiting Mach wave U = a (c) Zone of silence Zone of action Supersonic Mach wave (b) µ = sin–1 1 Ma U < a a δ t U δ t U δ t a δ t a δ t Fig. 9.19 Supersonic wave pattern emanating from a projectile moving at Ma 2.0. The heavy lines are oblique-shock waves and the light lines Mach waves (Courtesy of U.S.
Army Ballistic Research Labora-tory, Aberdeen Proving Ground.) sin1 U a t t sin1 U a sin1 M 1 a (9.79) 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, 90°; the Mach cone becomes a plane front moving with the particle, in agree-ment with Fig. 9.18b.
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.
The Mach wave need not be a cone: Similar waves are formed by a small distur-bance of any shape moving supersonically with respect to the ambient fluid. For ex-ample, 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 . 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 vis-ible along the body surface downstream of the recompression shock, especially at the rear corner. Their angle is about 30°, indicating a Mach number of about 2.0 along this surface. A more complicated system of Mach waves emanates from the supersonic pro-jectile in Fig. 9.19. The Mach angles change, indicating a variable supersonic Mach 620 Chapter 9 Compressible Flow E9.16 number along the body surface. There are also several stronger oblique-shock waves formed along the surface.
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 an-gle will be somewhat larger than the Mach-wave angle 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 , as seen in Fig. E9.16. Thus 9.9 Two-Dimensional Suspersonic Flow 621 µ BOOM!
Bow wave Ma = ?
5 km 9 km tan 0.5556 or 29.05° Hence, from Eq. (9.79), Ma csc 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 , unlike the normal-shock wave, for which the downstream 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 , and the downstream flow V2 turns at an an-gle which is a function of and state 1 conditions. The upstream flow is always su-personic, but the downstream Mach number Ma2 V2/a2 may be subsonic, sonic, or supersonic, depending upon the conditions.
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 5 km 9 km The Oblique-Shock Wave Fig. 9.20 Geometry of flow through an oblique-shock wave.
just encompassing the wave, we can then derive the following integral relations, can-celing out A1 A2 on each side of the wave: Continuity: 1Vn1 2Vn2 (9.80a) Normal momentum: p1 p2 2V2 n2 1V2 n1 (9.80b) Tangential momentum: 0 1Vn1(Vt2 Vt1) (9.80c) Energy: h1 1 2 V2 n1 1 2 V2 t1 h2 1 2 V2 n2 1 2 V2 t2 h0 (9.80d) We see from Eq. (9.80c) that there is no change in tangential velocity across an oblique shock Vt2 Vt1 Vt const (9.81) Thus tangential velocity has as its only effect the addition of a constant kinetic energy 1 2 V2 t to each side of the energy equation (9.80d). We conclude that Eqs. (9.80) are iden-tical to the normal-shock relations (9.49), with V1 and V2 replaced by the normal com-ponents Vn1 and Vn2. All the various relations from Sec. 9.5 can be used to compute properties of an oblique-shock wave. The trick is to use the “normal’’ Mach numbers in place of Ma1 and Ma2: Man1 Ma1 sin Man2 Ma2 sin ( ) (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 Ma1 replaced by Man1: p p 2 1 k 1 1 [2k Ma2 1 sin2 (k 1)] (9.83a) 2 1 tan ta ( n ) V V n n 1 2 (9.83b) T T 2 1 [2 (k 1) Ma2 1 sin2 ] (9.83c) T02 T01 (9.83d) 2k Ma2 1 sin2 (k 1) (k 1)2 Ma2 1 sin2 (k 1) Ma2 1 sin2 (k 1) Ma2 1 sin2 2 Vn2 a2 Vn1 a1 622 Chapter 9 Compressible Flow β Vn1 > a1 β V1 > a1 Vn2 < a2 θ Deflection angle V2 Vt2 = Vt1 Oblique shock wave Vt1 p p 0 0 2 1 k/(k 1) 1/(k 1) (9.83e) Ma2 n2 (9.83f) All these are tabulated in the normal-shock Table B.2. If you wondered why that table listed the Mach numbers as Man1 and Man2, it should be clear now that the table is also valid for the oblique-shock wave.
Thinking all this over, we realize by 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 con-stant tangential speed Vt. 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 increases with speed Vt up to a maximum and then decreases. From the geom-etry of Fig. 9.20 the deflection angle is given by tan 1 V V n t 2 tan 1 V V n t 1 (9.84) If we differentiate with respect to Vt and set the result equal to zero, we find that the maximum deflection occurs when Vt/Vn1 (Vn2/Vn1)1/2. We can substitute this back into Eq. (9.84) to compute max tan 1 r1/2 tan 1 r 1/2 r (9.85) For example, if Man1 3.0, from Table B.2 we find that Vn1/Vn2 3.8571, the square root of which is 1.9640. Then Eq. (9.85) predicts a maximum deflection of tan 1 1.9640 tan 1 (1/1.9640) 36.03°. The deflection is quite limited even for infinite Man1: From Table B.2 for this case Vn1/Vn2 6.0, and we compute from Eq. (9.85) that max 45.58°.
This limited-deflection idea and other facts become more evident if we plot some of the solutions of Eqs. (9.83). For given values of V1 and a1, 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 Vx and Vy, with x parallel to V1. Such a plot is called a hodograph. The heavy dark line which looks like a fat airfoil is the lo-cus, or shock polar, of all physically possible solutions for the given Ma1. The two dashed-line fishtails are solutions which 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 crosses the polar at two possible solutions: the strong shock, which greatly de-celerates the flow, and the weak shock, which causes a much milder deceleration. 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 finite size, there is a maximum deflection max, shown in Fig. 9.21, which just grazes the upper edge of the polar curve. This verifies the kinematic discussion which led to Eq. (9.85). What happens if a supersonic flow is forced to deflect through an angle greater than max? The answer is illustrated in Fig.
9.22 for flow past a wedge-shaped body.
Vn1 Vn2 (k 1) Ma2 n1 2 2k Ma2 n1 (k 1) k 1 2k Ma2 1 sin2 (k 1) (k 1) Ma2 1 sin2 2 (k 1) Ma2 1 sin2 9.9 Two-Dimensional Suspersonic Flow 623 Fig. 9.21 The oblique-shock polar hodograph, showing double solu-tions (strong and weak) for small deflection angle and no solutions at all for large deflection.
In Fig. 9.22a the wedge half-angle is less than max, and thus an oblique shock forms at the nose of wave angle just sufficient to cause the oncoming supersonic stream to deflect through the wedge angle . Except for the usually small effect of boundary-layer growth (see, e.g., Ref. 19, sec. 7–5.2), the Mach number Ma2 is con-stant along the wedge surface and is given by the solution of Eqs. (9.83). The pres-sure, density, and temperature along the surface are also nearly constant, as predicted 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 Fig. 9.10.
624 Chapter 9 Compressible Flow Vy Weak wave angle β θ Normal shock Strong shock Weak shock θmax V1 Rarefaction shock impossible by second law Vx Mach wave (V2 = V1) Ma1 > 1 Ma 2 Ma 2 (a) θ < max θ Weak shock family above sonic line Sonic line Strong shock family below sonic line Ma < 1 Ma < 1 Ma > 1 (b) θ > max θ Ma1 > 1 Ma > 1 Sonic line Fig. 9.22 Supersonic flow past a wedge: (a) small wedge angle, at-tached oblique shock forms; (b) large wedge angle, attached shock not possible, broad curved detached shock forms.
In Fig. 9.22b the wedge half-angle is greater than max, and an attached oblique shock is impossible. The flow cannot deflect at once through the entire angle max, 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 max.
Fig. 9.23 Oblique-shock deflection versus wave angle for various up-stream Mach numbers, k 1.4: dash-dot curve, locus of max, di-vides strong (right) from weak (left) shocks; dashed curve, locus of sonic points, divides subsonic Ma2 (right) from supersonic Ma2 (left).
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 that partic-ular value of and the given Ma1. 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 analy-sis of detached shock waves is extremely complex, and experimentation is usually needed, e.g., 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 varies with Ma1 and , from Eq. (9.83b).
By using a trigonometric identity for tan ( ) this can be rewritten in the more con-venient form tan (9.86) All possible solutions of Eq. (9.86) for k 1.4 are shown in Fig. 9.23. For deflections max there are two solutions: a weak shock (small ) and a strong shock (large ), as expected. All points along the dash-dot line for max 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) the weak-shock family satisfies the wave-angle relation sin 1 M 1 a1 (9.87) 2 cot (Ma2 1 sin2 1) Ma2 1 (k cos 2) 2 9.9 Two-Dimensional Suspersonic Flow 625 50° 40° 30° 20° 10° 0° 30° 60° 90° k = 1.4 Ma1 = ∞ 10 4 3 2.5 2 1.8 1.6 1.4 1.2 Deflection angle θ Wave angle β 6 Very Weak Shock Waves 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 90°.
Two additional oblique-shock charts are given in App. B, where Fig. B.1 gives the downstream Mach number Ma2 and Fig. B.2 the pressure ratio p2/p1, each plotted as a function of Ma1 and . Additional graphs, tables, and computer programs are given in Refs. 24 and 25.
For any finite the wave angle for a weak shock is greater than the Mach angle .
For small Eq. (9.86) can be expanded in a power series in tan with the following linearized result for the wave angle: sin sin 4 k c os 1 tan (tan2 ) (9.88) For Ma1 between 1.4 and 20.0 and deflections less than 6° this relation predicts 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 p 1 p1 tan (tan2 ) (9.89) k Ma2 1 (Ma2 1 1)1/2 626 Chapter 9 Compressible Flow 2.0 1.0 3.0 p2 – p1 p1 5° 10° k = 1.4 Ma 1 = 10 8 6 4 3 2 Eq. (9.89), Ma 1 = 2 Flow deflection θ 0 0 15° 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.
E9.17 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 computed from Eq. (9.83a). At very small deflections the curves are linear with slopes given by Eq. (9.89).
Finally, 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: s2 cp s1 tan3 (tan4 ) (9.90) The entropy change is cubic in the deflection angle . Thus weak shock waves are very nearly isentropic, a fact which is also used in the next section.
EXAMPLE 9.17 Air at Ma 2.0 and p 10 lbf/in2 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 1.4 compute from exact oblique-shock theory (a) the wave angle , (b) Ma2, and (c) p2. Also use the linearized theory to esti-mate (d) and (e) p2.
Solution With Ma1 2.0 and 10° known, we can estimate 40° 2° from Fig. 9.23. For more (hand calculated) accuracy, we have to solve Eq. (9.86) by iteration. Or we can program Eq.
(9.86) in EES with six statements (in SI units, with angles in degrees): Ma 2.0 k 1.4 Theta 10 Num 2(Ma^2SIN(Beta)^2 1)/TAN(Beta) Denom Ma^2(k COS(2Beta)) 2 Theta ARCTAN(Num/Denom) Specify that Beta 0 and EES promptly reports an accurate result: 39.32° Ans. (a) The normal Mach number upstream is thus Man1 Ma1 sin 2.0 sin 39.32° 1.267 With Man1 we can use the normal-shock relations (Table B.2) or Fig. 9.9 or Eqs. (9.56) to (9.58) to compute Man2 0.8031 1.707 Thus the downstream Mach number and pressure are Ma2 1.64 Ans. (b) p2 (10 lbf/in2 absolute)(1.707) 17.07 lbf/in2 absolute Ans. (c) Notice that the computed pressure ratio agrees with Figs. 9.24 and B.2.
0.8031 sin (39.32° 10°) Man2 sin ( ) p2 p1 (k2 1)Ma6 1 12(Ma2 1 1)3/2 9.9 Two-Dimensional Suspersonic Flow 627 Ma 1 = 2.0 Ma 2 β p1 = 10 lbf / in2 10° EES 9.10 Prandtl-Meyer Expansion Waves For the linearized theory the Mach angle is sin 1 (1/2.0) 30°. Equation (9.88) then estimates that sin sin 30° 0.622 or 38.5° Ans. (d) Equation (9.89) estimates that 1 1.57 or p2 1.57(10 lbf/in2 absolute) 15.7 lbf/in2 absolute 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 which obstructs a supersonic flow and thus decreases its Mach number and velocity. The pre-sent section treats gradual changes in flow angle which are primarily expansive; i.e., 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 that 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 grad-ual changes. The gradual compression of Fig. 9.25a is essentially isentropic, with a 1.4(2)2 tan 10° (22 1)1/2 p2 p1 2.4 tan 10° 4 cos 30° 628 Chapter 9 Compressible Flow Oblique shock Slip line Mach waves Ma decreases Ma > 1 (a) Ma 1 > 1 Ma 2 < Ma 1 (c) Mach waves Ma increases (b) Ma > 1 Ma increases (d) Ma > 1 Oblique shock Mach waves 9.25 Some examples of supersonic expansion and compression: (a) gradual isentropic compression on a concave surface, Mach waves co-alesce farther out to form oblique shock; (b) gradual isentropic ex-pansion on convex surface, Mach waves diverge; (c) sudden compres-sion, nonisentropic shock forms; (d) sudden expansion, centered isentropic fan of Mach waves forms.
smooth increase in pressure along the surface, but the Mach angle decreases 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. 9.25d 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 Fig. 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 sub-sonic, since Mach wave and shock wave patterns cannot exist in subsonic flow.
Consider a small, nearly infinitesimal flow deflection d such as occurs between the first two Mach waves in Fig. 9.25a. From Eqs. (9.88) and (9.89) we have, in the limit, sin 1 M 1 a (9.91a) d p p (Ma k 2 M a 1 2 )1/2 d (9.91b) Since the flow is nearly isentropic, we have the frictionless differential momentum equation for a perfect gas dp V dV kp Ma2 d V V (9.92) Combining Eqs. (9.91a) and (9.92) to eliminate dp, we obtain a relation between turn-ing angle and velocity change d (Ma2 1)1/2 d V V (9.93) This can be integrated into a functional relation for finite turning angles if we can re-late V to Ma. We do this from the definition of Mach number V Ma a or d V V d M M a a d a a (9.94) Finally, we can eliminate da/a because the flow is isentropic and hence a0 is constant for a perfect gas a a0[1 1 2 (k 1) Ma2] 1/2 or d a a (9.95) 1 2 (k 1) Ma d Ma 1 1 2 (k 1) Ma2 9.10 Prandtl-Meyer Expansion Waves 629 The Prandtl-Meyer Perfect-Gas Function Fig,. 9.26 The Prandtl-Meyer su-personic expansion from Eq. (9.99) for k 1.4.
Eliminating dV/V and da/a from Eqs. (9.93) to (9.95), we obtain a relation solely be-tween turning angle and Mach number d d M M a a (9.96) Before integrating this expression, we note that the primary application is to ex-pansions, i.e., increasing Ma and decreasing . Therefore, for convenience, we define the Prandtl-Meyer angle (Ma) which increases when decreases and is zero at the sonic point d d 0 at Ma 1 (9.97) Thus we integrate Eq. (9.96) from the sonic point to any value of Ma 0 d Ma 1 d M M a a (9.98) The integrals are evaluated in closed form, with the result, in radians, (Ma) K1/2 tan 1 Ma2 K 1 1/2 tan 1 (Ma2 1)1/2 (9.99) where K k k 1 1 This is the Prandtl-Meyer supersonic expansion function, which is plotted in Fig. 9.26 (Ma2 1)1/2 1 1 2 (k 1) Ma2 (Ma2 1)1/2 1 1 2 (k 1) Ma2 630 Chapter 9 Compressible Flow 140° 120° 100° 80° 60° 40° 20° 0° 0 1 4 8 12 16 20 Mach number k = 1.4 ω Ma → ∞: = 130.45° ω Part (a) and tabulated in Table B.5 for k 1.4, K 6. The angle changes rapidly at first and then levels off at high Mach number to a limiting value as Ma →: max 2 (K1/2 1) 130.45° if k 1.4 (9.100) Thus a supersonic flow can expand only through a finite turning angle before it reaches infinite Mach number, maximum velocity, and zero temperature.
Gradual expansion or compression between finite Mach numbers Ma1 and Ma2, nei-ther of which is unity, is computed by relating the turning angle to the difference in Prandtl-Meyer angles for the two conditions 1→2 (Ma2) (Ma1) (9.101) The change 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.
EXAMPLE 9.18 Air (k 1.4) flows at Ma1 3.0 and p1 200 kPa. Compute the final downstream Mach num-ber and pressure for (a) an expansion turn of 20° and (b) a gradual compression turn of 20°.
Solution The isentropic stagnation pressure is p0 p1[1 0.2(3.0)2]3.5 7347 kPa and this will be the same at the downstream point. For Ma1 3.0 we find from Table B.5 or Eq. (9.99) that 1 49.757°. The flow expands to a new condition such that 2 1 49.757° 20° 69.757° Linear interpolation in Table B.5 is quite accurate, yielding Ma2 4.32. Inversion of Eq. (9.99), to find Ma when is given, is impossible without iteration. Once again, our friend EES easily handles Eq. (9.99) with four statements (angles specified in degrees): k 1.4 C ((k 1)/(k 1))^0.5 Omega 69.757 Omega CARCTAN((Ma^2–1)^0.5/C) – ARCTAN((Ma^2–1)^0.5) Specify that Ma 1, and EES readily reports an accurate result:6 Ma2 4.32 Ans. (a) The isentropic pressure at this new condition is p2 31.9 kPa Ans. (a) 7347 230.1 p0 [1 0.2(4.32)2]3.5 9.10 Prandtl-Meyer Expansion Waves 631 6The author saves these little programs for further use, giving them names such as Prandtl-Meyer.
EES EES Application to Supersonic Airfoils The flow compresses to a lower Prandtl-Meyer angle 2 49.757° 20° 29.757° Again from Eq. (9.99), Table B.5, or EES we compute that Ma2 2.125 Ans. (b) p2 773 kPa Ans. (b) Similarly, density and temperature changes are computed by noticing that T0 and 0 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 9.27a shows a flat-plate foil at an angle of attack. There is a leading-edge shock on the lower edge with flow deflection , while the upper edge has an ex-pansion fan with increasing Prandtl-Meyer angle . We compute p3 with ex-pansion 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 , and the drag parallel to the stream is D F sin . The dimensionless coefficients CL and CD have the same definitions as in low-speed flow, Eq. (7.66), ex-cept that the perfect-gas-law identity 1 2 V2 1 2 kp Ma2 is very useful here CL CD (9.102) The typical supersonic lift coefficient is much smaller than the subsonic value CL 2 , but the lift can be very large because of the large value of 1 2 V2 at supersonic speeds.
At the trailing edge in Fig. 9.27a, a shock and fan appear in reversed positions and bend the two flows back so that they are parallel in the wake and have the same pres-sure. 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 interesting, but in the theory you ignore the trailing-edge pattern entirely, since it D 1 2 kp Ma2 bC L 1 2 kp Ma2 bC 7347 9.51 p0 [1 0.2(2.125)2]3.5 632 Chapter 9 Compressible Flow Part (b) Fig. 9.27 Supersonic airfoils: (a) flat plate, higher pressure on lower surface, drag due to small downstream component of net pres-sure force; (b) diamond foil, higher pressures on both lower surfaces, additional drag due to body thick-ness.
does not affect the surface pressures: The supersonic surface flow cannot “hear’’ the wake disturbances.
The diamond foil in Fig. 9.27b adds two more wave patterns to the flow. At this particular 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 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 (9.27a) and can be ignored in the calculation. Both lower-surface pressures p2 and p4 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 p5 on the trailing surfaces are lower than their counterparts p2 and p3. The dia-mond drag is greater than the flat-plate drag, but this must be endured in practice to achieve a wing structure strong enough to hold these forces.
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 ReC and moderate supersonic Ma the boundary layers are thin and separation seldom oc-curs, so that the shock expansion theory, although frictionless, is quite successful. Let us look now at an example.
9.10 Prandtl-Meyer Expansion Waves 633 α Ma ∞ p∞ Expansion fan Ma 3 > Ma ∞ p3 < p∞ p03 = p0∞ Oblique shock Ma 2 < Ma ∞ p2 > p∞ p02 < p0∞ Oblique shock Vortex sheet Expansion fan (a) α Ma ∞ p∞ (b) p3 > p∞ p2 > p3 p5 < p3 p4 > p5 p4 < p2 p0∞ E9.19 EXAMPLE 9.19 A flat-plate airfoil with C 2 m is immersed at 8° in a stream with Ma 2.5 and p 100 kPa. Compute (a) CL and (b) CD, 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 expan-sion 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.
634 Chapter 9 Compressible Flow 8° Ma ∞ = 2.5 p∞ = 100 k Pa p0∞ = 1709 k Pa Ma 2 = 2.169 = 1.657 p2 p∞ p 2 = 165.7 k Pa Do not compute ∞ = 39.124° ω = = 8° θ α = 30.01° β p 03 = p0∞ = 1709 k Pa = 30.05 p03 p3 p 3 = 56.85 k Pa Ma 3 = 2.867 ∆ = 8° = ω α 3 = 47.124° ω The important final results are p2 and p3, from which the total force per unit width on the plate is F (p2 p3)bC (165.7 56.85)(kPa)(1 m)(2 m) 218 kN The lift and drag per meter width are thus L F cos 8° 216 kN Ans. (c) D F sin 8° 30 kN Ans. (d) These are very large forces for only 2 m2 of wing area.
From Eq. (9.102) the lift coefficient is CL 0.246 Ans. (a) The comparable low-speed coefficient from Eq. (8.64) is CL 2 sin 8° 0.874, which is 3.5 times larger.
From Eq. (9.102) the drag coefficient is CD 0.035 Ans. (b) From Fig. 7.25 for the NACA 0009 airfoil CD at 8° is about 0.009, or about 4 times smaller.
30 kN 1 2 (1.4)(100 kPa)(2.5)2(2 m2) 216 kN 1 2 (1.4)(100 kPa)(2.5)2(2 m2) Thin-Airfoil Theory 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 para-dox of zero body drag does not occur in supersonic flow.
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 surface de-flection relative to the free stream and condition 1 is the free stream, Ma1 Ma. For the flat-plate airfoil, the total force F is based on p2 p p3 p2 p p p3 p p (Ma k 2 M a2 1)1/2 [ ( )] (9.103) Substitution into Eq. (9.102) gives the linearized lift coefficient for a supersonic flat-plate airfoil CL (Ma2 4 1)1/2 (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 CD CL tan CL (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 yu(x) and the lower profile by yl(x). Then the complete Ackeret drag theory (see, e.g., Ref. 8, 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 y 2 C 1 C 0 d d y x 2 dx (9.106) The final expression for drag [8, p. 442] is CD 2 1 2 (y u 2 y l 2 ) (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 ex-pansion theory. Consider the following example.
4 (Ma2 1)1/2 4 2 (Ma2 1)1/2 (p2 p3)bC 1 2 kp Ma2 bC 9.10 Prandtl-Meyer Expansion Waves 635 Part (a) EXAMPLE 9.20 Repeat parts (a) and (b) of Example 9.19, using the linearized Ackeret theory.
Solution From Eqs. (9.104) and (9.105) we have, for Ma 2.5 and 8° 0.1396 rad, CL 0.244 CD 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: x C CP 0.5 S 2 u C2 Sl (9.108) where Su is the cross-sectional area between the upper surface and the chord and Sl is the area between the chord and the lower surface. For a symmetric airfoil (Sl Su) we obtain xCP at the half-chord point, in contrast with the low-speed airfoil result of Eq.
(8.66), 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 8° and Ma 2.5. Compute CL and CD by (a) shock expansion theory and (b) Ackeret theory. Pinpoint the difference from Example 9.19.
Solution 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 is p2 p3 186.4 65.9 120.5 kPa, and on the rear half it is p4 p5 146.9 48.1 98.1 kPa. The average pressure difference is 1 2 (120.5 98.1) 109.3 kPa, so that the normal force is F (109.3 kPa)(2 m2) 218.6 kN For the chordwise force P the pressure difference on the top half is p3 p5 65.9 48.8 17.1 kPa, and on the bottom half it is p2 p4 186.4 146.9 39.5 kPa. The average dif-ference is 1 2 (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)(1 m) 2.0 kN 4(0.1396)2 (2.52 1)1/2 4(0.1396) (2.52 1)1/2 636 Chapter 9 Compressible Flow E9.21 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: 1 2 kp Ma2 bC 1 2 (1.4)(100 kPa)(2.5)2(2 m2) 875 kN. Thus, finally, shock expansion theory predicts CL 0.247 CD 0.0370 Ans. (a) Meanwhile, by Ackeret theory, CL is the same as in Example 9.20: CL 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) y u 2 y l 2 tan2 2° 0.00122 Then Eq. (9.107) predicts the linearized result CD [(0.1396)2 1 2 (0.00122 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), p3 60 kPa ( 9 per-cent), p4 140 kPa ( 5 percent), and p5 33 kPa ( 6 percent).
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.
4 (2.52 1)1/2 4(0.1396) (2.52 1)1/2 32.4 kN 875 kN 216.2 kN 875 kN 9.10 Prandtl-Meyer Expansion Waves 637 Ma ∞ = 2.5 p∞ = 100 k Pa p0∞ = 1709 k Pa 8° p 3 = 65.9 kPa Ma 3 = 2.770 p 5 = 48.8 kPa Ma 5 = 2.967 p 4 = 146.9 k Pa Ma 4 = 2.238 Chord length = 2 m ω = 4° ∆ 5 = 49.124° ω ω = 4° ∆ 4 = 32.721° ω ∞ = 39.124° ω Ma 2 = 2.086 p 02 = 1668 k Pa p 2 = 186.4 k Pa 2 = 28.721° ω = 10° θ = 31.85° β ω = 6° ∆ 3 = 45.124° ω 0.07 m 4˚ Part (b) 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 num-ber varies from 1.98 just inside the shock to 1.90 at the surface. (Cour-tesy of U.S. Army Ballistic Re-search Center, Aberdeen Proving Ground.) Three-dimensional supersonic flows are highly complex, especially if they con-cern blunt bodies, which therefore contain embedded regions of subsonic and tran-sonic flow, e.g., Fig. 9.10. Some flows, however, yield to accurate theoretical treat-ment 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, 8, chap. 17], and extensive tables of such solutions have been published [22, 23]. There are similar-ities 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 cor-ner. However, 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 k 1.4 and Ma , the maximum cone half-angle for an attached shock is about 57°, com-pared with the maximum wedge angle of 45.6° (see Ref. 23).
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 by the smearing of oil drops placed on the model surface before the test.
638 Chapter 9 Compressible Flow Reusable Hypersonic Launch Vehicles As we shall see in the next chapter, there is an interesting analogy between gas-dynamic shock waves and the surface water waves which form in an open-channel flow. Chapter 11 of Ref. 14 explains how a water channel can be used in an inex-pensive simulation of supersonic-flow experiments.
Having achieved reliable supersonic flight with both military and commercial air-craft, the next step is probably to develop a hypersonic vehicle that can achieve or-bit, yet be retrieved. Presently the United States employs the Space Shuttle, where only the manned vehicle is retrieved, the very expensive giant rocket boosters be-ing lost. In 1996, NASA selected Lockheed-Martin to develop the X-33, the first smaller-scale step toward a retrievable single-stage-to-orbit (SSTO) vehicle, to be called the VentureStar .
The X-33, shown in an artist’s rendering in Fig. 9.30, will be 20 m long, about half the size of the VentureStar, and it will be suborbital. It will take off vertically, rise to a height of 73 km, and coast back to earth at a steep (stressful) angle. Such a flight will test many new plans for the VentureStar , such as metallic tiles, ti-tanium components, graphite composite fuel tanks, high-voltage control actuators, and Rocketdyne’s novel “aerospike” rocket nozzles. If successful, the VentureStar is planned as a standard reusable, low-cost orbital vehicle. VentureStar will be 39 m long and weigh 9.7 MN, of which 88 percent (965 tons!) will be propellant 9.10 Prandtl-Meyer Expansion Waves 639 Fig. 9.29 Wind-tunnel test of the Cobra P-530 supersonic interceptor.
The surface flow patterns are visu-alized by the smearing of oil droplets. (Courtesy of Northrop Corp.) Fig. 9.30 The X-33 is a half-size suborbital test version of the Ven-tureStar, which is planned as an or-bital, low-cost retrievable space ve-hicle. It takes off vertically but then uses its lifting shape to glide back to earth and land horizontally [36, 37]. (Courtesy of Lockheed Martin Corp.) and only 2.7 percent (260 kN) will be payload. The dream is that the X-33 and Ven-tureStar and their progeny will lead to an era of routine, low-cost space travel ap-propriate to the new millennium. This chapter is a brief introduction to a very broad subject, compressible flow, some-times called gas dynamics. The primary parameter is the Mach number Ma V/a, which is large and causes the fluid density to vary significantly. This means that the continuity and momentum equations must be coupled to the energy relation and the equation of state to solve for the four unknowns (p, , T, V).
The chapter reviews the thermodynamic properties of an ideal gas and derives a formula for the speed of sound of a fluid. The analysis is then simplified to one-dimensional steady adiabatic flow without shaft work, for which the stagnation en-thalpy of the gas is constant. A further simplification to isentropic flow enables for-mulas 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. 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, there is a brief study of constant-area duct flow with friction and with heat transfer, both of which lead to choking of the exit flow.
The chapter ends 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 air-foils.
640 Chapter 9 Compressible Flow Summary Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equations Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 9.1 to 9.157 (categorized in the problem list below) are followed by word problems W9.1 to W9.8, funda-mentals of engineering exam problems FE9.1 to FE9.10, compre-hensive problems C9.1 to C9.3, and design projects D9.1 and D9.2.
Problem distribution Section Topic Problems 9.1 Introduction 9.1–9.9 9.2 The speed of sound 9.10–9.18 9.3 Adiabatic and isentropic flow 9.19–9.33 9.4 Isentropic flow with area changes 9.34–9.53 9.5 The normal-shock wave 9.54–9.62 9.6 Converging and diverging nozzles 9.63–9.85 9.7 Duct flow with friction 9.86–9.107 9.8 Frictionless duct flow with heat transfer 9.108–9.115 9.9 Mach waves 9.116–9.121 9.9 The oblique-shock wave 9.122–9.139 9.10 Prandtl-Meyer expansion waves 9.140–9.147 9.10 Supersonic airfoils 9.148–9.157 P9.1 An ideal gas flows adiabatically through a duct. At section 1, p1 140 kPa, T1 260°C, and V1 75 m/s. Farther downstream, p2 30 kPa and T2 207°C. Calculate V2 in m/s and s2 s1 in J/(kg K) if the gas is (a) air, k 1.4, and (b) argon, k 1.67.
P9.2 Solve Prob. 9.1 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 Compressibility effects become important when the Mach number exceeds approximately 0.3. How fast can a two-dimensional cylinder travel in sea-level standard air be-fore compressibility becomes important somewhere in its vicinity?
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 ex-its at saturation conditions. Estimate the exit velocity and temperature.
P9.6 Is it possible for the steam in Prob. 9.5 to continue ac-celerating until it exits with a moisture content of 12 percent? If so, estimate the new exit velocity and tem-perature.
P9.7 Carbon dioxide (k 1.28) enters a constant-area duct at 400°F, 100 lbf/in2 absolute, and 500 ft/s. Farther down-stream the properties are V2 1000 ft/s and T2 900°F.
Compute (a) p2, (b) the heat added between sections, (c) the entropy change between sections, and (d) the mass flow per unit area. Hint: This problem requires the continuity equation.
P9.8 Atmospheric air at 20°C enters and fills an insulated tank which is initially evacuated. Using a control-volume analy-sis from Eq. (3.63), 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 which exhausts at Vexit 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/m3. 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. P9.10 A certain aircraft flies at the same Mach number regard-less of its altitude. Compared to its speed at 12,000-m stan-dard altitude, it flies 127 km/h faster at sea level. Deter-mine its Mach number.
P9.11 At 300°C and 1 atm, estimate the speed of sound of (a) nitrogen, (b) hydrogen, (c) helium, (d) steam, and (e) 238UF6 (k 1.06).
P9.12 Assume that water follows Eq. (1.19) with n 7 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 deep-est 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 From Prob. 1.33, mercury data fit Eq. (1.19) with n 6 and B 41,000. Estimate (a) the bulk modulus and (b) the speed of sound of mercury at 2500 atm and compare with Table 9.1.
P9.14 Assume steady adiabatic flow of a perfect gas. Show that the energy equation (9.21), when plotted as speed of sound versus velocity, forms an ellipse. Sketch this ellipse; label the intercepts and the regions of subsonic, sonic, and su-personic flow; and determine the ratio of the major and minor axes.
P9.15 A weak pressure wave (sound wave) with a pressure change p 40 Pa propagates through air at 20°C and 1 atm. Estimate (a) the density change, (b) the temperature change, and (c) the velocity change across the wave.
P9.16 A weak pressure pulse p propagates through still air. Dis-cuss the type of reflected pulse which occurs and the boundary conditions which must be satisfied when the Problems 641 EES 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. 9.12 as a guide, estimate the distance to the other object.
P9.18 The properties of a dense gas (high pressure and low tem-perature) are often approximated by van der Waals’ equa-tion of state [17, 18]: p a12 where constants a1 and b1 can be found from the critical temperature and pressure a1 9.0 105 lbf ft4/slug2 for air, and b1 0.65 ft3/slug 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 (b) a van der Waals gas. What percent-age higher density does the van der Waals relation predict?
P9.19 The Concorde aircraft flies at Ma 2.3 at 11-km standard altitude. Estimate the temperature in °C at the front stag-nation point. At what Mach number would it have a front stagnation-point temperature of 450°C?
P9.20 A gas flows at V 200 m/s, p 125 kPa, and T 200°C.
For (a) air and (b) helium, compute the maximum pres-sure and the maximum velocity attainable by expansion or compression.
P9.21 Air expands isentropically through a duct from p1 125 kPa and T1 100°C to p2 80 kPa and V2 325 m/s.
Compute (a) T2, (b) Ma2, (c) T0, (d) p0, (e) V1, and ( f) Ma1.
P9.22 Given the pitot stagnation temperature and pressure and the static-pressure measurements in Fig. P9.22, estimate RTc 8pc 27R2T2 c 64pc RT 1 b1 the air velocity V, assuming (a) incompressible flow and (b) compressible flow.
P9.23 A large rocket engine delivers hydrogen at 1500°C and 3 MPa, k 1.41, R 4124 J/(kg K), to a nozzle which ex-its with gas pressure equal to the ambient pressure of 54 kPa. Assuming isentropic flow, if the rocket thrust is 2 MN, what is (a) the exit velocity and (b) the mass flow of hy-drogen? P9.24 For low-speed (nearly incompressible) gas flow, the stag-nation pressure can be computed from Bernoulli’s equa-tion p0 p V2 (a) For higher subsonic speeds, show that the isentropic relation (9.28a) can be expanded in a power series as fol-lows: p0 p V21 Ma2 Ma4 (b) Suppose that a pitot-static tube in air measures the pres-sure difference p0 p and uses the Bernoulli relation, with stagnation density, to estimate the gas velocity. At what Mach number will the error 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 estimate the static pressure in the duct in lbf/in2 ab-solute.
2 k 24 1 4 1 2 1 2 642 Chapter 9 Compressible Flow Air V 100°C 120 k Pa 80 k Pa P9.22 Air at 100°F Mercury 8 in P9.25 P9.26 Show that for isentropic flow of a perfect gas if a pitot-static probe measures p0, p, and T0, the gas velocity can be calculated from V2 2cpT01 (k 1)/k What would be a source of error if a shock wave were formed in front of the probe?
P9.27 In many problems the sonic () properties are more use-ful reference values than the stagnation properties. For isentropic flow of a perfect gas, derive relations for p/p, p p0 T/T, and / as functions of the Mach number. Let us help by giving the density-ratio formula: 1/(k 1) P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea-level standard air through a converging nozzle whose throat diameter is 3 cm. Estimate (a) the mass-flow rate through the nozzle and (b) the Mach number at the throat.
P9.29 Steam from a large tank, where T 400°C and p 1 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 Air flows in a duct of diameter 5 cm. At one section, T0 300°C, p 120 kPa, and m ˙ 0.4 kg/s. Estimate, at this section, (a) V, (b) Ma, and (c) 0.
P9.31 Air flows adiabatically through a duct. At one section V1 400 ft/s, T1 200°F, and p1 35 lbf/in2 absolute, while farther downstream V2 1100 ft/s and p2 18 lbf/in2 ab-solute. Compute (a) Ma2, (b) Umax, and (c) p02/p01.
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 manometer reads h 30 cm. Assuming isentropic flow, compute the pressure (a) in the tank and (b) in the atmos-phere. (c) What is the exit Mach number?
k 1 2 (k 1) Ma.2 P9.36 An air tank of volume 1.5 m3 is initially at 800 kPa and 20°C. At t 0, it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm2.
Estimate (a) the initial mass flow in kg/s, (b) the time re-quired 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 neg-ligible kinetic and potential energy within. Assume criti-cal flow at the exit, and show that both p0 and T0 decrease during blowdown. Set up first-order differential equations for p0(t) and T0(t), and reduce and solve as far as you can.
Problems 643 30°C Air ptank?
pa?
235 m /s Mercury h P9.32 Insulated tank p0 (t) T0 (t) Volume V Ae, Ve, me ⋅ Measurements of tank pressure and temperature P9.37 P9.33 Air flows isentropically from a reservoir, where p 300 kPa and T 500 K, to section 1 in a duct, where A1 0.2 m2 and V1 550 m/s. Compute (a) Ma1, (b) T1, (c) p1, (d) m ˙, and (e) A. Is the flow choked?
P9.34 Steam in a tank at 450°F and 100 lbf/in2 absolute exhausts through a converging nozzle of 0.1-in2 throat area to a 1-atm environment. Compute the initial mass flow (a) for an ideal gas and (b) from the steam tables .
P9.35 Helium, at T0 400 K, enters a nozzle isentropically. At section 1, where A1 0.1 m2, a pitot-static arrangement (see Fig. P9.25) measures stagnation pressure of 150 kPa and static pressure of 123 kPa. Estimate (a) Ma1, (b) mass flow m ˙, (c) T1, and (d) A.
P9.38 Prob. 9.37 makes an ideal senior project or combined lab-oratory and computer problem, as described in Ref. 30, sec. 8.6. In Bober and Kenyon’s lab experiment, the tank had a volume of 0.0352 ft3 and was initially filled with air at 50 lb/in2 gage and 72°F. Atmospheric pressure was 14.5 lb/in2 absolute, and the nozzle exit diameter was 0.05 in.
After 2 s of blowdown, the measured tank pressure was 20 lb/in2 gage and the tank temperature was 5°F. Compare these values with the theoretical analysis of Prob. 9.37.
P9.39 Consider isentropic flow in a channel of varying area, from section 1 to section 2. We know that Ma1 2.0 and desire that the velocity ratio V2/V1 be 1.2. Estimate (a) Ma2 and (b) A2/A1. (c) Sketch what this channel looks like. For ex-ample, does it converge or diverge? Is there a throat? P9.40 Air, with stagnation conditions of 800 kPa and 100°C, ex-pands isentropically to a section of a duct where A1 20 cm2 and p1 47 kPa. Compute (a) Ma1, (b) the throat area, and (c) m ˙. At section 2 between the throat and section 1, the area is 9 cm2. (d) Estimate the Mach number at sec-tion 2.
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 10x2 with A in cm2 and x in m. It is desired to plot the com-plete family of isentropic pressures p(x) in this nozzle, for the range of inlet pressures 1 p(0) 100 kPa. Indicate those inlet pressures which 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.
P9.42 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 (pa 100 kPa absolute and Ta 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, i.e., one-dimensional isentropic 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 duct with T0 300°C.
At two sections with identical areas of 25 cm2, the pres-sures are p1 120 kPa and p2 60 kPa. Determine (a) the mass flow, (b) the throat area, and (c) Ma2.
P9.44 In Prob. 3.34 we knew nothing about compressible flow at the time, so we merely assumed exit conditions p2 and T2 and computed V2 as an application of the continuity equation. 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, compute the actual exit velocity, pressure, and temperature according to one-dimensional theory. If pa 14.7 lbf/in2 absolute, compute the thrust from the analysis of Prob.
3.68. This thrust is entirely independent of the stagnation temperature (check this by changing T0 to 2000°R if you like). Why?
P9.45 At a point upstream of the throat of a converging-diverg-ing nozzle the properties are V1 200 m/s, T1 300 K, and p1 125 kPa. If the exit flow is supersonic, compute, from isentropic theory, (a) m ˙ and (b) A1. The throat area is 35 cm2.
P9.46 If the author did not falter, the results of Prob. 9.43 are (a) 0.671 kg/s, (b) 23.3 cm2, and (c) 1.32. Do not tell your friends who are still working on Prob. 9.43. Consider a control volume which encloses the nozzle between these two 25-cm2 sections. If the pressure outside the duct is 1 atm, determine the total force acting on this section of nozzle.
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 m2, 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 m2 (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) Ve, (b) Vp, and (c) m ˙ e.
644 Chapter 9 Compressible Flow A(x) p (x)?
p0 p 0 0 1 m 2 m x P9.41 Insulated F Vp Air at 20°C Dp = 12 cm De = 3 mm pa = 1 atm Ve, me ⋅ P9.48 P9.49 Air expands through a nozzle and exits supersonically. The throat area is 10 cm2, and the throat pressure is 100 kPa.
Find the pressure on either side of the throat where the duct area is 24 cm2.
P9.50 Argon expands isentropically in a converging nozzle whose entrance conditions are D1 10 cm, p1 150 kPa, T1 100°C, and m ˙ 1 kg/s. The flow discharges smoothly to an ambient pressure of 101 kPa. (a) What is the exit di-ameter of the nozzle? (b) How much further can the am-bient pressure be reduced before it affects the inlet mass flow?
P9.51 Air, at stagnation conditions of 500 K and 200 kPa, flows through a nozzle. At section 1, where the area is 12 cm2, EES the density is 0.32 kg/m3. Assuming isentropic flow, (a) find the mass flow. (b) Is the flow choked? If so, estimate A. Also estimate (c) p1 and (d) Ma1.
P9.52 A converging-diverging nozzle exits smoothly to sea-level standard atmosphere. It is supplied by a 40-m3 tank ini-tially 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 cm2. P9.53 Air flows steadily from a reservoir at 20°C through a noz-zle of exit area 20 cm2 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) Ve, (b) Mae, and (c) p0 if pa 101 kPa.
P9.58 Argon (Table A.4) approaches a normal shock with V1 700 m/s, p1 125 kPa, and T1 350 K. Estimate (a) V2 and (b) p2. (c) What pressure p2 would result if the same velocity change V1 to V2 were accomplished isentropi-cally?
P9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows through a nozzle. At section 1, where the area is 15 cm2, 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 su-personic flow, a normal shock will stand in front of the probe. Suppose the probe reads p0 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 Repeat Prob. 9.56 except this time let the odd coincidence be that the static pressure downstream of the shock exactly equals the throat pressure. What is the area where the shock stands?
P9.62 An atomic explosion propagates into still air at 14.7 lbf/in2 absolute and 520°R. The pressure just inside the shock is 5000 lbf/in2 absolute. Assuming k 1.4, what are the speed C of the shock and the velocity V just inside the shock?
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 cm2, as shown. Estimate (a) the pres-sure in the tank and (b) the mass flow.
Problems 645 Vacuum tank 1 cm2 2 cm2 3 cm2 Sea-level air P9.63 10 cm2 14 cm2 Sea-level air Air at 100°C Shock P9.57 Air 20°C 135 N Plate A = 20 cm e 2 P9.53 P9.54 For flow of air through a normal shock the upstream con-ditions are V1 600 m/s, T01 500 K, and p01 700 kPa. Compute the downstream conditions Ma2, V2, T2, p2, and p02.
P9.55 Air, supplied by a reservoir at 450 kPa, flows through a con-verging-diverging nozzle whose throat area is 12 cm2. A nor-mal shock stands where A1 20 cm2. (a) Compute the pres-sure just downstream of this shock. Still farther downstream, at A3 30 cm2, estimate (b) p3, (c) A 3, and (d) Ma3.
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 cm2. 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.
P9.64 Air in a large tank at 100°C and 150 kPa exhausts to the atmosphere through a converging nozzle with a 5-cm2 throat area. Compute the exit mass flow if the atmospheric pressure is (a) 100 kPa, (b) 60 kPa, and (c) 30 kPa.
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?
P9.66 In Prob. 9.65 what would be the mercury-manometer read-ing h if the nozzle were operating exactly at supersonic de-sign conditions?
P9.67 In Prob. 9.65 estimate the complete range of manometer readings h for which the flow through the nozzle is en-tirely isentropic, except possibly in the exit plane.
P9.68 Air in a tank at 120 kPa and 300 K exhausts to the at-mosphere through a 5-cm2-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?
P9.69 With reference to Prob. 3.68, show that the thrust of a rocket engine exhausting into a vacuum is given by F where Ae exit area Mae exit Mach number p0 stagnation pressure in combustion chamber Note that stagnation temperature does not enter into the thrust.
P9.70 Air, at stagnation temperature 100°C, expands isentropi-cally through a nozzle of 6-cm2 throat area and 18-cm2 exit area. The mass flow is at its maximum value of 0.5 kg/s. Estimate the exit pressure for (a) subsonic and (b) supersonic exit flow.
P9.71 For the nozzle of Prob. 9.70, allowing for nonisentropic flow, what is the range of exit tank pressures pb for which (a) the diverging nozzle flow is fully supersonic, (b) the exit flow is subsonic, (c) the mass flow is independent of pb, (d) the exit plane pressure pe is independent of pb, and (e) pe pb?
P9.72 Suppose the nozzle flow of Prob. 9.70 is not isentropic but instead has a normal shock at the position where area is 15 cm2. Compute the resulting mass flow, exit pressure, and exit Mach number.
P9.73 Air flows isentropically in a converging-diverging nozzle with a throat area of 3 cm2. At section 1, the pressure is 101 kPa, the temperature is 300 K, and the velocity is 868 m/s. (a) Is the nozzle choked? Determine (b) A1 and (c) the mass flow. Suppose, without changing stagnation con-ditions or A1, the (flexible) throat is reduced to 2 cm2. As-p0Ae(1 k Ma2 e) 1 k 2 1 Ma2 e k/(k 1) suming shock-free flow, will there be any change in the gas properties at section 1? If so, compute new p1, V1, and T1 and explain.
P9.74 The perfect-gas assumption leads smoothly to Mach-num-ber relations which are very convenient (and tabulated).
This is not so for a real gas such as steam. To illustrate, let steam at T0 500°C and p0 2 MPa expand isen-tropically through a converging nozzle whose exit area is 10 cm2. Using the steam tables, find (a) the exit pressure and (b) the mass flow when the flow is sonic, or choked.
What complicates the analysis?
P9.75 A double-tank system in Fig. P9.75 has two identical con-verging nozzles of 1-in2 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. 14, pp. 288–290, has a good discussion.) 646 Chapter 9 Compressible Flow Air 100 lbf / in2 abs 520° R 10 lbf / in2 abs 1 2 3 P9.75 100°C 300 k Pa A t = 10 cm2 h Mercury Ae =30 cm2 P9.65 0.2 cm2 0.3 cm2 p1 = 150 k Pa, T1 = 20°C p2 = 100 k Pa A B mB?
⋅ mA?
⋅ P9.78 P9.76 A large reservoir at 20°C and 800 kPa is used to fill a small insulated tank through a converging-diverging nozzle with 1-cm2 throat area and 1.66-cm2 exit area. The small tank has a volume of 1 m3 and is initially at 20°C and 100 kPa.
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 su-personic nozzle with an exit area 5 times its throat area.
The exit Mach number is 3.8. What is the specific-heat ra-tio of the gas? What might this gas be? If p0 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 re-versed. For the given air properties on either side, compute the mass flow through each hole and explain why they are different.
EES P9.79 Air with p0 300 kPa and T0 500 K flows through a converging-diverging nozzle with throat area of 1 cm2 and exit area of 3 cm2 into a receiver tank. The mass flow is 195.2 kg/h. For what range of receiver pressure is this mass flow possible?
P9.80 A sea-level automobile tire is initially at 32 lbf/in2 gage pressure and 75°F. When it is punctured with a hole which resembles a converging nozzle, its pressure drops to 15 lbf/in2 gage in 12 min. Estimate the size of the hole, in thousandths of an inch. The tire volume is 2.5 ft2.
P9.81 Helium, in a large tank at 100°C and 400 kPa, discharges to a receiver through a converging-diverging nozzle de-signed to exit at Ma 2.5 with exit area 1.2 cm2. Com-pute (a) the receiver pressure and (b) the mass flow at de-sign conditions. (c) Also estimate the range of receiver pressures for which mass flow will be a maximum.
P9.82 Air at 500 K flows through a converging-diverging nozzle with throat area of 1 cm2 and exit area of 2.7 cm2. When the mass flow is 182.2 kg/h, a pitot-static probe placed in the exit plane reads p0 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 down-stream 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 lbf. The chamber pressure and temperature are 600 lbf/in2 absolute and 4000°R, respectively. The exhaust gases ap-proximate k 1.38 with a molecular weight of 26. Esti-mate (a) the exit Mach number and (b) the throat diame-ter.
P9.84 Air flows through a duct as in Fig. P9.84, where A1 24 cm2, A2 18 cm2, and A3 32 cm2. A normal shock stands at section 2. Compute (a) the mass flow, (b) the Mach number, and (c) the stagnation pressure at section 3.
(a) the throat pressure and (b) the stagnation pressure in the upstream supply tank.
P9.86 Air enters a 3-cm-diameter pipe 15 m long at V1 73 m/s, p1 550 kPa, and T1 60°C. The friction factor is 0.018.
Compute V2, p2, T2, and p02 at the end of the pipe. How much additional pipe length would cause the exit flow to be sonic?
P9.87 Air enters a duct of L/D 40 at V1 170 m/s and T1 300 K. The flow at the exit is choked. What is the aver-age friction factor in the duct for adiabatic flow?
P9.88 Air enters a 5- by 5-cm square duct at V1 900 m/s and T1 300 K. The friction factor is 0.02. For what length duct will the flow exactly decelerate to Ma 1.0? If the duct length is 2 m, will there be a normal shock in the duct? If so, at what Mach number will it occur?
P9.89 Air flows adiabatically in a 5-cm-diameter tube with f 0.025. At section 1, V1 75 m/s, T1 350 K, and p1 300 kPa. How much further down the tube will (a) the pressure be 156 kPa, (b) the temperature be 343 K, and (c) the flow reach the choking point?
P9.90 Air, supplied at p0 700 kPa and T0 330 K, flows through a converging nozzle into a pipe of 2.5-cm diame-ter which exits to a near vacuum. If f 0.022, what will be the mass flow through the pipe if its length is (a) 0 m, (b) 1 m, and (c) 10 m?
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.
Problems 647 1 2 3 Air Normal shock Ma1 = 2.5 p1 = 40 k Pa T1 = 30°C P9.84 L = 9 m, D = 6 cm f = 0.025 De = 5 cm Nozzle Pa = 100 k Pa 1 2 Air at 100°C P9.91 P9.85 A large tank delivers air through a nozzle of 1-cm2 throat area and 2.7-cm2 exit area. When the receiver pressure is 125 kPa, a normal shock stands in the exit plane. Estimate P9.92 Modify Prob. 9.91 as follows. Let the pressure in the tank be 700 kPa, and let the nozzle be choked. Determine (a) Ma2 and (b) the mass flow.
P9.93 Air flows adiabatically in a 3-cm-diameter duct. The av-erage friction factor is 0.015. If, at the entrance, V 950 m/s and T 250 K, how far down the tube will (a) the Mach number be 1.8 or (b) the flow be choked?
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 1.4 and the data of Prob. 9.86, draw a Fanno 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.
P9.95 Helium (Table A.4) enters a 5-cm-diameter pipe at p1 550 kPa, V1 312 m/s, and T1 40°C. The friction fac-tor is 0.025. If the flow is choked, determine (a) the length of the duct and (b) the exit pressure.
P9.96 Derive and verify the adiabatic-pipe-flow velocity relation of Eq. (9.74), which is usually written in the form ln P9.97 By making a few algebraic substitutions, show that Eq.
(9.74), or the relation in Prob. 9.96, may be written in the density form 2 1 2 2 2 2 ln 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, f 64/Re, may occur in cap-illary tubes. Consider air, at stagnation conditions of 100°C and 200 kPa, entering a tube 3 cm long and 0.1 mm in di-ameter. 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 f , compute the resulting mass flow.
1 2 f L D 2k k 1 1 V2 2 1 V2 1 a2 0 k V2 V1 k 1 k f L D P9.101 How do the compressible-pipe-flow formulas behave for small pressure drops? Let air at 20°C enter a tube of di-ameter 1 cm and length 3 m. If f 0.028 with p1 102 kPa and p2 100 kPa, estimate the mass flow in kg/h for (a) isothermal flow, (b) adiabatic flow, and (c) incom-pressible 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 f , estimate the mass flow through this system. Is the flow choked?
648 Chapter 9 Compressible Flow Pe = 30 kPa L = 1 m D = 5 cm 550 kPa 100°C L = 1.2 m D = 3 cm Converging nozzle P9.102 D = 4 cm L = 20 m 250 kPa ∆ p Parabola Te = 200°C m ⋅ m ⋅ 0.4 kg / s Pe = 101 k Pa P9.99 P9.100 Modify Prob. 9.99 as follows. Find the length of 4-cm-di-ameter pipe for which the pump pressure rise will be ex-actly 200 kPa.
P9.103 Natural gas, with k 1.3 and a molecular weight of 16, is to be 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 as-tronaut through an umbilical tube 12 m long and 2 cm in diameter. The exit pressure in the tube is 40 kPa. If the de-sired mass flow is 90 kg/h and f 0.025, what should be the pressure in the tank?
P9.105 Air enters a 5-cm-diameter pipe at p1 200 kPa and T1 350 K. The downstream receiver pressure is 74 kPa. The friction factor is 0.02. If the exit is choked, what is (a) the length of the pipe and (b) the mass flow? (c) If p1, T1, and preceiver stay the same, what pipe length will cause the mass flow to increase by 50 percent over (b)? Hint: In part (c) the exit pressure does not equal the receiver pressure.
P9.106 Air at 300 K flows through a duct 50 m long with f 0.019. What is the minimum duct diameter which can carry the flow without choking if the entrance velocity is (a) 50 m/s, (b) 150 m/s, and (c) 420 m/s?
P9.107 A fuel-air mixture, assumed equivalent to air, enters a duct combustion chamber at V1 104 m/s and T1 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 is added during combustion?
EES P9.108 What happens to the inlet flow of Prob. 9.107 if the com-bustion yields 1500 kJ/kg heat addition and p01 and T01 remain the same? How much is the mass flow reduced?
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 m2, 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 noz-zle exit velocity, and (c) the thrust produced by the engine.
P9.110 Compressible pipe flow with heat addition, Sec. 9.8, as-sumes constant momentum (p V2) and constant mass flow but variable stagnation enthalpy. Such a flow is of-ten called Rayleigh flow, and a line representing all pos-sible property changes on a temperature-entropy chart is called a Rayleigh line. Assuming air passing through the flow state p1 548 kPa, T1 588 K, V1 266 m/s, and A 1 m2, 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. 9.110 a Fanno line (see Prob. 9.94) for stagnation enthalpy equal to the value as-sociated with state 1 in Prob. 9.110. 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-à-vis Table B.2.
P9.112 Air enters a duct subsonically at section 1 at 1.2 kg/s. When 650 kW of heat is added, the flow chokes at the exit at p2 95 kPa and T2 700 K. Assuming frictionless heat addition, estimate (a) the velocity and (b) the stagnation pressure at section 1.
P9.113 Air enters a constant-area duct at p1 90 kPa, V1 520 m/s, and T1 558°C. It is then cooled with negligible fric-tion until it exits at p2 160 kPa. Estimate (a) V2, (b) T2, and (c) the total amount of cooling in kJ/kg.
P9.114 We have simplified things here by separating friction (Sec.
9.7) from heat addition (Sec. 9.8). Actually, they often oc-cur together, and their effects must be evaluated simulta-neously. Show that, for flow with friction and heat trans-fer in a constant-diameter pipe, the continuity, momentum, and energy equations may be combined into the following differential equation for Mach-number changes: where dQ is the heat added. A complete derivation, in-cluding many additional combined effects such as area change and mass addition, is given in chap. 8 of Ref. 8.
f dx D k Ma2 [2 (k 1) Ma2] 2(1 Ma2) dQ cpT 1 k Ma2 1 Ma2 d Ma2 Ma2 P9.115 Air flows subsonically in a duct with negligible friction.
When heat is added in the amount of 948 kJ/kg, the pres-sure drops from p1 200 to p2 106 kPa. Estimate (a) Ma1, (b) T1, and (c) V1, assuming T01 305 K.
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 An observer at sea level does not hear an aircraft flying at 6000-m standard altitude until 15 s after it has passed over-head. Estimate the aircraft speed in m/s.
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.
Problems 649 V Particle 3 m 8 m P9.118 V Particle 3 m 8 m 8 m P9.119 3 m 6 m P9.120 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 standard air. From the two disturbance spheres shown, estimate (a) the position of the particle at this instant and (b) the tem-perature in °C at the front stagnation point of the particle.
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 disturbance cone of half-angle 17° is created. Estimate (a) the Mach number, (b) the velocity, and (c) the stagnation temperature of the stream.
P9.122 Supersonic air takes a 5° compression turn, as in Fig.
P9.122. Compute the downstream pressure and Mach num-ber and the wave angle, and compare with small-distur-bance theory.
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 and (b) the wave angle caused by a scratch at b?
650 Chapter 9 Compressible Flow Ma 1 = 3 p1 = 100 k Pa 5° Ma2, p2 P9.122 a b Ma > 1 30° 50° θ φ P9.129 Ma = 3 p = 10 lbf / in2 abs A B 16° 16° P9.132 Ma = 3.0 p = 100 k Pa δ 12 cm P9.128 P9.123 Modify Prob. 9.122 as follows. Let the 5° total turn be in the form of five separate compression turns of 1° each.
Compute the final Mach number and pressure, and com-pare the pressure with an isentropic expansion to the same final Mach number.
P9.124 Determine the validity of the following alternate relation for the pressure ratio across an oblique shock wave: If necessary, your proof (or disproof) may be somewhat tentative and heuristic.
P9.125 Show that, as the upstream Mach number approaches in-finity, the Mach number downstream of an attached oblique-shock wave will have the value Ma2 P9.126 Consider airflow at Ma1 2.2. Calculate, to two decimal places, (a) the deflection angle for which the downstream flow is sonic and (b) the maximum deflection angle.
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 an-gle lies halfway between 1 and 2 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 for which the horizontal component of the total pressure force on the nose is 35 kN/m of depth into the paper.
k 1 2k sin2 ( ) cot sin 2 cos 2 k cot sin 2 cos 2 k p2 p1 P9.130 Modify Prob. 9.129 as follows. If the wave angle is 42°, determine (a) the shock-wave angle and (b) the deflection angle.
P9.131 In Fig. P9.128, assume that the approach stream tempera-ture is 20°C. For what wedge half-angle will the stream temperature along the wedge surface be 200°C?
P9.132 Air flows at Ma 3 and p 10 lbf/in2 absolute toward a wedge of 16° angle at zero incidence in Fig. 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 B?
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 Ma1 2.5 and p1 100 kPa, compute Ma3, p3, and the angle .
P9.137 Generalize Prob. 9.136 into a computer study as follows.
Assuming weak shocks, find and plot all combinations of and h in Fig. P9.136 for which the canceled or “swal-lowed’’ reflected shock is possible.
P9.138 The supersonic nozzle of Fig. P9.138 is overexpanded (case G of Fig. 9.12) with Ae/At 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 pr in kPa?
Problems 651 Shocks (1 m, 1 m) (0, 0) Ma P9.133 Ma 1 = 2.2 2 3 12° max?
θ P9.139 Ma 1 = 2.5 Ma 2 Ma 3 40° φ P9.134 Ma 1 = 3.0 2 3 10 Air: p1 = 100 k Pa ° P9.135 1 m Ma = 3.5 Shock Shock φ h P9.136 Air 4° Jet edge pr?
P9.138 P9.135 A bend in the bottom of a supersonic duct flow induces a shock wave which reflects from the upper wall, as in Fig. P9.135. Compute the Mach number and pressure in region 3.
P9.136 Figure P9.136 is a special application of Prob. 9.135. 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 number in a channel (a supersonic diffuser). If the bend angle is 10°, find (a) the downstream width h and (b) the downstream Mach number. Assume a weak shock wave.
P9.139 Airflow at Ma 2.2 takes a compression turn of 12° and then another turn of angle in Fig. P9.139. What is the maximum value of for the second shock to be attached?
Will the two shocks intersect for any less than max?
P9.140 The solution to Prob. 9.122 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 pres-sure, and compare with small-disturbance theory.
EES P9.142 A supersonic airflow at Ma1 3.2 and p1 50 kPa un-dergoes a compression shock followed by an isentropic ex-pansion 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 Ma1 3.2 passes through a 25° oblique-shock deflection. What isentropic expansion turn is required to bring the flow back to (a) Ma1 and (b) p1?
P9.144 Consider a smooth isentropic compression turn of 20°, as shown in Fig. P9.144. The Mach waves thus generated will form a converging fan. Sketch this fan as accurately as pos-sible, using at least five equally spaced waves, and demon-strate how the fan indicates the probable formation of an oblique-shock wave.
8°? Why does the drag coefficient not have the simple parabolic form CD K 2 in this range?
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 an-gles are 4°, as in Fig. P9.151. Compute the lift and drag coefficient at equal to (a) 0° and (b) 6°.
652 Chapter 9 Compressible Flow Ma ∞ = 2.5 4° 4° P9.151 Ma1 = 3 p1 = 100 k Pa 5° Ma 2, p2 P9.141 Ma = 3.0 Start Mach waves Circular-arc turn Finish 20° P9.144 pa = 10 kPa φ φ Jet edge Jet edge Ma 2 Ma 2 P9.147 P9.145 Air at Ma1 2.0 and p1 100 kPa undergoes an isen-tropic expansion to a downstream pressure of 50 kPa. What is the desired turn angle in degrees?
P9.146 Helium, at 20°C and V1 2010 m/s, undergoes a Prandtl-Meyer expansion until the temperature is 50°C. Estimate the turn angle in degrees.
P9.147 A converging-diverging nozzle with a 4:1 exit-area ratio and p0 500 kPa operates in an underexpanded condition (case I of Fig. 9.12b) as in Fig. P9.147. The receiver pres-sure is pa 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 of the edge of the jet be? Assume k 1.4 as usual.
P9.148 Repeat Example 9.19 for an angle of attack of 6°. Is the lift coefficient linear with angle in this range of 0° 8°? Is the drag coefficient parabolic with in this range?
P9.149 Repeat Example 9.21 for an angle of attack of 2°. Is the lift coefficient linear with angle in this range of 0° P9.152 A supersonic airfoil has a parabolic symmetric shape for upper and lower surfaces yu,l 2t such that the maximum thickness is t at x 1 2 C. Compute the drag coefficient at zero incidence by Ackeret theory, and compare with a symmetric double wedge of the same thickness.
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 m2 and (b) the thrust required in N.
P9.154 A symmetric supersonic airfoil has its upper and lower sur-faces defined by a sine-wave shape: y sin where t is the maximum thickness, which occurs at x C/2. Use Ackeret theory to derive an expression for the drag coefficient at zero angle of attack. Compare your re-sult with Ackeret theory for a symmetric double-wedge air-foil of the same thickness.
P9.155 For the sine-wave airfoil shape of Prob. 9.154, with Ma 2.5, k 1.4, t/C 0.1, and 0°, plot (without com-x C t 2 x2 C2 x C EES puting the overall forces) the pressure distribution p(x)/p along the upper surface for (a) Ackeret theory and (b) an oblique shock plus a continuous Prandtl-Meyer expansion.
P9.156 A thin circular-arc airfoil is shown in Fig. P9.156. The leading edge is parallel to the free stream. Using linearized (small-turning-angle) supersonic-flow theory, derive a for-mula for the lift and drag coefficient for this orientation, and compare with Ackeret-theory results for an angle of attack tan 1 (h/L).
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 the two 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 wa-ter 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, p1 to p2. Describe its behavior on an appropriately labeled Mollier chart for (a) frictionless flow Fundamentals of Engineering Exam Problems 653 Circular-arc foil TE L h LE Ma > 1 P9.156 P9.157 Prove from Ackeret theory that for a given supersonic air-foil shape with sharp leading and trailing edges and a given thickness, the minimum-thickness drag occurs for a sym-metric double-wedge shape.
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 dur-ing 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 ac-tually 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?
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 For 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 For 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 FE9.3 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 FE9.4 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 FE9.5 In Prob. FE9.3, if the area at section 1 is 0.15 m2, 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 FE9.6 For steady isentropic flow, what is the maximum possible mass flow through the duct in Fig. FE9.6?
(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 FE9.6 Throat area 0.05 m2 Exit Tank: 400 K, 300 kPa FE9.7 If the exit Mach number in Fig. FE9.6 is 2.2, what is the exit area? (a) 0.10 m2, (b) 0.12 m2, (c) 0.15 m2 , (d) 0.18 m2, (e) 0.22 m2 FE9.8 If there are no shock waves and the pressure at one duct section in Fig. FE9.6 is 55.5 kPa, what is the velocity at that section? (a) 166 m/s, (b) 232 m/s, (c) 554 m/s, (d) 706 m/s, (e) 774 m/s FE9.9 If, in Fig. FE9.6, there is a normal shock wave at a sec-tion where the area is 0.07 m2, what is the air density just upstream of that shock? (a) 0.48 kg/m3, (b) 0.78 kg/m3, (c) 1.35 kg/m3, (d) 1.61 kg/m3, (e) 2.61 kg/m3 FE9.10 In Prob. FE9.9, what is the Mach number just downstream of the shock wave? (a) 0.42, (b) 0.55, (c) 0.63, (d) 1.00, (e) 1.76 654 Chapter 9 Compressible Flow 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 travels from tank a to tank b, where tank a is much larger than tank b. (a) Find the area at the exit Ae and the back pressure pb which will allow the system to operate at de-sign 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), i.e., 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 cm2, the air pressure is 16 kPa and the temperature is 250 K. Just 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. Throat area 0.07 m2 Tank a Tank b T 500 K p 1.00 MPa Air (k 1.4) Volume huge Volume 100,000 L T 20.0 C Ae, Ve, Mae Shock waves 430 K 50 C9.1 C9.3 Design Projects D9.1 It is desired to select a rectangular wing for a fighter air-craft. The plane must be able (a) to take off and land on a 4500-ft-long sea-level runway and (b) to cruise supersoni-cally at Ma 2.3 at 28,000-ft altitude. For simplicity, as-sume a wing with zero sweepback. Let the aircraft maxi-mum weight equal (30 n)(1000) lbf, 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. Mak-ing suitable assumptions about the effect of finite aspect ra-tio on wing lift and drag for both subsonic and supersonic flight, select a wing of minimum 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, gradually 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 , 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 will the supersonic wave-drag coefficient CD, based on frontal area, be exactly 0.5? (b) Suppose that 20°. Is there a free-stream Mach number for which the wave-drag coefficient CD, based on frontal area, will be exactly 0.5? (c) Investigate the percent increase in CD from (a) and (b) due to including boundary-layer fric-tion drag in the calculation.
References 655 D9.2 h p , Ma References 1.
A. Y. Pope, Aerodynamics of Supersonic Flight, 2d ed., Pit-man, New York, 1958.
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A. B. Cambel and B. H. Jennings, Gas Dynamics, McGraw-Hill, New York, 1958.
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Mech., vol. 13, no. 2, 1946, p. A-91.
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656 Chapter 9 Compressible Flow 658 The Lehigh River, White Haven, Pennsylvania.
Open channel flows are everywhere, often rough and turbulent, as in this photo. They are analyzed by the methods of the present chapter. (Courtesy of Dr. E. R. Degginger/Color-Pic Inc.) 10.1 Introduction Motivation.
The duct flows of Chap. 6 were driven by a pressure difference between the ends of the duct. Such channels are closed and full of fluid, either gas or liquid.
By contrast, an open-channel flow is liquid only and is not full; i.e., there is always a free surface exposed to ambient pressure. The basic balance of forces is between grav-ity (fluid weight) and friction.
Practical open-channel problems almost always concern water as the relevant fluid.
The flow is generally turbulent, due to its large scale and small kinematic viscosity, and is three-dimensional, sometimes unsteady, and often surprisingly complex due to geometric effects. This chapter presents some simple engineering theories and corre-lations for steady flow in straight channels of simple geometry. Many of the concepts from steady duct flow—hydraulic diameter, friction factor, head losses—apply also to open channels.
Simply stated, open-channel flow is the flow of a liquid in a conduit with a free sur-face. 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 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.1 But the free surface complicates the analysis because its shape is a priori unknown: The depth profile changes with conditions and must be computed as part of the problem, espe-cially in unsteady problems involving wave motion.
Before proceeding, we remark, as usual, that whole books have been written on open-channel hydraulics [1 to 4]. There are also specialized texts devoted to wave mo-tion [5 to 7] and to engineering aspects of coastal free-surface flows [8, 9]. This chap-ter is only an introduction to broader and more detailed treatments.
Chapter 10 Open-Channel Flow 1Surface tension is rarely important because open channels are normally quite large and have a very large Weber number.
659 The One-Dimensional Approximation 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 dis-tribution. 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 max-imum velocity is near the surface, and the velocity profile is nearly logarithmic from the bottom to the free surface, as in Eq. (6.84). 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 ve-locity occurring near the outer radius of the bend. Curved natural channels are subject to strong bottom erosion and deposition effects.
660 Chapter 10 Open-Channel Flow 2.0 1.5 1.0 0.5 Triangular channel 2.0 1.5 1.0 0.5 Trapezoidal channel 1.5 1.0 0.5 2.0 2.5 Pipe 2.0 1.5 1.0 0.5 Shallow ditch 2.0 1.5 1.0 0.5 2.5 Natural irregular channel 1.5 1.0 0.5 2.0 2.5 Narrow rectangular section Fig. 10.1 Measured isovelocity contours in typical straight open-channel flows. (From Ref. 3.) Fig. 10.2 Geometry and notation for open-channel flow: (a) side view; (b) cross section. All these parameters are constant in uniform flow.
With the advent of the supercomputer, it is possible to make numerical simulations of complex flow patterns such as in Fig. 10.1 . 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, p1 p2 pa, and we have, for steady flow, V 2g 2 1 z1 V 2g 2 2 z2 hf (10.2) where z denotes the total elevation of the free surface, which includes the water depth y (see Fig. 10.2a) plus the height of the (sloping) bottom. The friction head loss hf is analogous to head loss in duct flow from Eq. (6.30): hf f x2 D h x1 V 2 2 a g v Dh hydraulic diameter 4 P A (10.3) where f is the average friction factor (Fig. 6.13) between sections 1 and 2. Since chan-nels are irregular in shape, their “size” is taken to be the hydraulic diameter, with P the wetted perimeter—see Fig. 10.2b. Actually, open-channel formulas typically use the hydraulic radius Rh 1 4 Dh A P (10.4) The local Reynolds number of the channel would be Re VDh/, which is usually highly turbulent (1 E5). The only commonly occurring laminar channel flows are the thin sheets which form as rainwater drains from crowned streets and airport runways.
The wetted perimeter P (see Fig. 10.2b) 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 (10.5) not 2b 2h.
10.1 Introduction 661 Horizontal θ V x (a) A (b) Rh = A P S = tan θ y y b0 P Flow Classification by Depth Variation 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 alternate 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 (hence the velocity in steady flow) remains 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 yn, 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 ex-amples of this method of classification are shown in Fig. 10.3. The classes can be sum-marized as follows: 1.
Uniform flow (constant depth and slope) 2.
Varied flow a.
Gradually varied (one-dimensional) b.
Rapidly varied (multidimensional) 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 potential theory.
A second and very interesting classification is by dimensionless Froude number, which for a rectangular or very wide channel takes the form Fr V/(gy)1/2, where y is the 662 Chapter 10 Open-Channel Flow Flow Classification by Froude Number: Surface Wave Speed GVF RVF GVF Uniform flow GVF RVF GVF Aerated region Fig. 10.3 Open-channel flow classi-fied by regions of rapidly varying flow (RVF), gradually varying flow (GVF), and uniform-flow depth profiles.
Fig. 10.4 Analysis of a small sur-face wave propagating into still shallow water; (a) moving wave, nonsteady frame; (b) fixed wave, inertial frame of reference.
water depth. The three flow regimes are 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 pursue the analogy in Sec. 10.4.
The Froude-number denominator (gy)1/2 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 y propagating at speed c into still liquid. To achieve a steady-flow in-ertial frame of reference, we fix the coordinates on the wave as in Fig. 10.4b, 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.
For the control volume of Fig. 10.4b, the one-dimensional continuity relation is, for channel width b, cyb (c V)(y y)b or V c y y y (10.7) This is analogous to Eq. (9.10); the velocity change V induced by a surface wave is small if the wave is “weak,” y y. If we neglect bottom friction in the short distance across the wave in Fig. 10.4b, the momentum relation is a balance between the net hy-drostatic pressure force and momentum 1 2 gb[(y y)2 y2] cby(c V c) or g1 y c V (10.8) 1 2 y y 10.1 Introduction 663 Still water c y (a) δ y δV (b) c δy pa = 0 Fixed wave c – δV Control volume τw ≈ 0 g y ρ g( y + δy) ρ Fig. 10.5 Flow under a sluice gate accelerates from subcritical to criti-cal to supercritical flow and then jumps back to subcritical flow.
10.2 Uniform Flow; the Chézy Formula This is analogous to Eq. (9.12). By eliminating V between Eqs. (10.7) and (10.8) we obtain the desired expression for wave propagation speed c2 gy1 y y 1 (10.9) The “stronger” the wave height y, the faster the wave speed c, by analogy with Eq.
(9.13). In the limit of an infinitesimal wave height y →0, the speed becomes c2 0 gy (10.10) This is the surface-wave equivalent of fluid sound speed a, and thus the Froude num-ber in channel flow Fr V/c0 is the analog of the Mach number.
As in gas dynamics, a channel flow can accelerate from subcritical to critical to su-percritical 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 flow “shocks” back to subcritical flow because the downstream “receiver” height is too high to main-tain supercritical flow. Note the similarity with the nozzle gas flows of Fig. 9.12.
The critical depth yc [Q2/(b2g)]1/3 is sketched as a dashed line in Fig. 10.5 for ref-erence. Like the normal depth yn, yc 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. 10.
Uniform flow can occur in long straight runs of constant slope and constant channel cross section. The water depth is constant at y yn, and the velocity is constant at V V0. Let the slope be S0 tan , where is the angle the bottom makes with the horizontal, con-sidered positive for downhill flow. Then Eq. (10.2), with V1 V2 V0, becomes hf z1 z2 S0L (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 that the Darcy-Weisbach relation, Eq. (6.30), holds hf f D L h V 2g 2 0 Dh 4Rh (10.12) 1 2 y y 664 Chapter 10 Open-Channel Flow Subcritical Sluice gate Subcritical Hydraulic jump Supercritical 1/ 3 yc = Q2 b2g The Manning Roughness Correlation with Dh 4A/P used to accommodate noncircular channels. The geometry and nota-tion for open-channel flow analysis are shown in Fig. 10.2.
By combining Eqs. (10.11) and (10.12) we obtain an expression for flow velocity in uniform channel flow V0 8 f g 1/2 Rh 1/2S0 1/2 (10.13) For a given channel shape and bottom roughness, the quantity (8g/f)1/2 is constant and can be denoted by C. Equation (10.13) becomes V0 C(RhS0)1/2 Q CA(RhS0)1/2 (10.14) These are called the Chézy formulas, first developed by the French engineer Antoine Chézy in conjunction with his experiments on the Seine River and the Courpalet Canal in 1769. The quantity C, called the Chézy coefficient, varies from about 60 ft1/2/s for small rough channels to 160 ft1/2/s for large smooth channels (30 to 90 m1/2/s in SI units).
Over the past century a great deal of hydraulics research has been devoted to the correlation of the Chézy 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. 3, chap. 5. Here we confine our treatment to Manning’s corre-lation, the most popular.
The most fundamentally sound approach to the Chézy formula is to use Eq. (10.13) with f estimated from the Moody friction-factor chart, Fig. 6.13. Indeed, the open chan-nel 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. (6.64): f 2.0 log 14. 8Rh 2 (10.15a) A special case, for rocky channel beds, is recommended in Ref. 2: f 1.2 2.03 log d R 84 h % 2 (10.15b) where d84% is the size for which 84 percent of the rocks are smaller (the largest rocks dominate the friction in the channel). Note that d84% and are not equal, being an overall average size. 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 Chézy coefficient C increased approximately as the sixth root of the channel size. He proposed the simple formula C 8 f g 1/2 R n h 1/6 (10.16) where n is a roughness parameter. Since the formula is clearly not dimensionally con-sistent, it requires a conversion factor which changes with the system of units used: 10.2 Uniform Flow; the Chézy Formula 665 Part (a) 1.0 SI units 1.486 BG units (10.17) Recall that we warned about this awkwardness in Example 1.4. You may verify that is the cube root of the conversion factor between the meter and your chosen length scale: In BG units, (3.2808 ft/m)1/3 1.486. The Manning formula for uniform-flow velocity is thus V0 (m/s) 1 n .0 [Rh (m)]2/3S0 1/2 (10.18) V0 (ft/s) 1.4 n 86 [Rh (ft)]2/3S0 1/2 The channel slope S0 is dimensionless, and n is taken to be the same in both systems.
The volume flow rate simply multiplies this result by the area: Uniform flow: Q V0A n ARh 2/3S0 1/2 (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.
Since Manning’s sixth-root size variation is not exact, real channels can have a vari-able n depending upon the water depth. The Mississippi River near Memphis, Ten-nessee, 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.
EXAMPLE 10.1 A finished-concrete 8-ft-wide rectangular channel has a bed slope of 0.5° and a water depth of 4 ft. Predict the uniform flow rate in ft3/s.
Solution From Table 10.1, for finished concrete, n 0.012. The slope S0 tan 0.5° 0.00873. For depth y 4 ft and width b 8 ft, the geometric properties are A by (8 ft)(4 ft) 32 ft2 P b 2y 8 2(4) 16 ft Rh A P 3 1 2 6 f f t t 2 2.0 ft Dh 4Rh 8.0 ft From Manning’s formula (10.19) in BG units, the estimated flow rate is Q 1.4 n 86 ARh 2/3S0 1/2 1 0 .
.
4 0 8 1 6 2 (32 ft2)(2.0 ft)2/3(0.00873)1/2 590 ft3/s Ans.
666 Chapter 10 Open-Channel Flow An interesting discussion of the history and “dimensionality” of Manning’s formula is given in Ref. 3, pp. 98–99.
Normal-Depth Estimates Considering the uncertainty in n ( 17 percent), it would be more realistic to report this esti-mate as Q 600 100 ft3/s. An alternate estimate, using the Moody formula (10.15) with 0.0032 ft from Table 10.1, would give Q 540 ft3/s.
With water depth y known, the computation of Q in Example 10.1 was quite straight-forward. However, if Q is given, the computation of the normal depth yn may require iteration or trial and error. Since the normal depth is a characteristic flow parameter, this is an important type of problem.
EXAMPLE 10.2 The asphalt-lined trapezoidal channel in Fig. E10.2 carries 300 ft3/s of water under uniform-flow conditions when S 0.0015. What is the normal depth yn?
10.2 Uniform Flow; the Chézy Formula 667 Table 10.1 Experimental Values of Manning’s n Factor Average roughness height 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 Cement, 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 ?
?
A more complete list is given in Ref. 3, pp. 110–113.
Uniform Flow in a Partly Full Circular Pipe Solution From Table 10.1, for asphalt, n 0.016. The area and hydraulic radius are functions of yn, which is unknown b0 6 ft 2yn cot 50° A 1 2 (6 b0)yn 6yn y2 n cot 50° P 6 2W 6 2yn csc 50° From Manning’s formula (10.19) with a known Q 300 ft3/s, we have 300 0 1 .
.
0 4 1 9 6 (6yn y2 n cot 50°) 2/3 (0.0015)1/2 or (6yn y2 n cot 50°)5/3 83.2(6 2yn csc 50°)2/3 One can iterate this formula laboriously and eventually find yn 4.6 ft. However, it is a perfect candidate for EES. Instead of manipulating and programming the final formula, one might sim-ply evaluate each separate part of the Chézy equation (in English units, with angles in degrees): P 6 2yn/sin(50) A 6yn yn^2/tan(50) Rh A/P 300 1.49/0.016ARh^(2/3)0.0015^0.5 Hit Solve from the menu bar and EES complains of “negative numbers to a power”. Go back to Variable Information on the menu bar and make sure that yn is positive. EES then immediately solves for P 17.95 A 45.04 Rh 2.509 yn 4.577 ft Ans.
Generally, EES is ideal for open-channel-flow 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 ra-dius R and the angle up to the free surface, the geometric properties are A R2 sin 2 2 P 2R Rh R 2 1 sin 2 2 6yn y2 n cot 50° 6 2yn csc 50° 668 Chapter 10 Open-Channel Flow b0 50° W 6 ft yn E10.2 Fig. 10.6 Uniform flow in a partly full circular channel: (a) geometry; (b) velocity and flow rate versus depth.
10.3 Efficient Uniform-Flow Channels The Manning formulas (10.19) predict a uniform flow as follows: V0 n R 2 1 sin 2 2 2/3 S0 1/2 Q V0R2 sin 2 2 (10.20) For a given n and slope S0, we may plot these two relations versus in Fig. 10.6b.
There are two different maxima, as follows: Vmax 0.718 n R2/3S0 1/2 at 128.73° and y 0.813D (10.21) Qmax 2.129 n R8/3S0 1/2 at 151.21° and y 0.938D As shown in Fig. 10.6b, 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 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 Rh for a given flow area and discharge. Since Rh A/P, maximizing Rh for given A is the same as minimizing the wetted perimeter 10.3 Efficient Uniform-Flow Channels 669 R (a) 0.8 0.6 0.4 0.2 1.0 0.0 0.0 0.2 0.4 0.6 0.8 1.0 y D R R θ y (b) V Vmax Q Qmax Fig. 10.7 Geometry of a trapezoidal channel section.
Best Trapezoid Angle P. There is no general solution for arbitrary cross sections, but an analysis of the trape-zoid section will show the basic results.
Consider the generalized trapezoid of angle in Fig. 10.7. For a given side angle , the flow area is A by y2 cot (10.22) The wetted perimeter is P b 2W b 2y(1 2)1/2 (10.23) Eliminating b between (10.22) and (10.23) gives P A y y 2y(1 2)1/2 (10.24) To minimize P, evaluate dP/dy for constant A and and set equal to zero. The result is A y2[2(1 2)1/2 ] P 4y(1 2)1/2 2y Rh 1 2 y (10.25) The last result is very interesting: For any angle , the most efficient cross section for uniform flow occurs when the hydraulic radius is half the depth.
Since a rectangle is a trapezoid with 0, the most efficient rectangular section is such that A 2y2 P 4y Rh 1 2 y b 2y (10.26) To find the correct depth y, these relations must be solved in conjunction with Man-ning’s flow-rate formula (10.19) for the given discharge Q.
Equations (10.25) are valid for any value of . What is the best value of for a given depth and area? To answer this question, evaluate dP/d from Eq. (10.24) with A and y held constant. The result is 2 (1 2)1/2 cot 3 1 1/2 or 60° (10.27) Thus the very best trapezoid section is half a hexagon.
Similar calculations with a circular channel section running partially full show best efficiency for a semicircle, y 1 2 D. In fact, the semicircle is the best of all possible 670 Chapter 10 Open-Channel Flow W y b θ y α θ = cot α 10.4 Specific Energy; Critical Depth channel sections (minimum wetted perimeter for a given flow area). The percentage improvement over, say, half a hexagon is very slight, however.
EXAMPLE 10.3 What are the best dimensions for a rectangular brick channel designed to carry 5 m3/s of water in uniform flow with S0 0.001?
Solution From Eq. (10.26), A 2y2 and Rh 1 2 y. Manning’s formula (10.19) in SI units gives, with n 0.015 from Table 10.1, Q 1 n .0 ARh 2/3S0 1/2 or 5 m3/s 0 1 .0 .0 15 (2y2) 1 2 y 2/3 (0.001)1/2 which can be solved for y8/3 1.882 m8/3 y 1.27 m Ans.
The proper area and width are A 2y2 3.21 m2 b A y 2.53 m Ans.
It is constructive to see what flow rate a half-hexagon and semicircle would carry for the same area of 3.214 m2.
For the half-hexagon (HH), with 1/31/2 0.577, Eq. (10.25) predicts A y2 HH[2(1 0.5772)1/2 0.577] 1.732y2 HH 3.214 or yHH 1.362 m, whence Rh 1 2 y 0.681 m. The half-hexagon flow rate is thus Q 0 1 .0 .0 15 (3.214)(0.681)2/3(0.001)1/2 5.25 m3/s or about 5 percent more than that for the rectangle.
For a semicircle, A 3.214 m2 D2/8, or D 2.861 m, whence P 1 2 D 4.494 m and Rh A/P 3.214/4.484 0.715 m. The semicircle flow rate will thus be Q 0 1 .0 .0 15 (3.214)(0.715)2/3(0.001)1/2 5.42 m3/s or about 8 percent more than that of the rectangle and 3 percent more than that of the half-hexagon.
As suggested by Bakhmeteff in 1911, the specific energy E is a useful parameter in channel flow E y 2 V g 2 (10.28) 10.4 Specific Energy; Critical Depth 671 Fig. 10.8 Specific-energy consider-ations: (a) illustration sketch; (b) depth versus E from Eq.
(10.29), showing minimum specific energy occurring at critical depth.
where y is the water depth. It is seen from Fig. 10.8a 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 for the same specific energy.
Consider the possible states at a given location. Let q Q/b Vy be the discharge per unit width of a rectangular channel. Then, with q constant, Eq. (10.28) becomes E y 2 q g 2 y2 q Q b (10.29) Figure 10.8b 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 dE/dy 0 at constant q, we find that Emin occurs at y yc q g 2 1/3 b Q 2 2 g 1/3 (10.30) The associated minimum energy is Emin E(yc) 3 2 yc (10.31) The depth yc corresponds to channel velocity equal to the shallow-water wave prop-agation speed C0 from Eq. (10.10). To see this, rewrite Eq. (10.30) as q2 gyc 3 (gyc)yc 2 Vc 2yc 2 (10.32) By comparison it follows that the critical channel velocity is Vc (gyc)1/2 C0 Fr 1 (10.33) For E Emin no solution exists in Fig. 10.8b, and thus such a flow is impossible phys-ically. For E Emin two solutions are possible: (1) large depth with V Vc, called sub-critical, and (2) small depth with V Vc, called supercritical. In subcritical flow, dis-turbances can propagate upstream because wave speed C0 V. In supercritical flow, waves are swept downstream: Upstream is a zone of silence, and a small obstruction in the flow will create a wedge-shaped wave exactly analogous to the Mach waves in 672 Chapter 10 Open-Channel Flow Horizontal EGL h f E V 2 2 g y y yc E min Supercritical Subcritical Critical Constant q y = E (b) E 0 (a) Rectangular Channels Part (a) Part (b) Nonrectangular Channels Fig. 9.18c.2 The angle of these waves must be sin1 c V 0 sin1 (gy V )1/2 (10.34) The wave angle and the depth can thus be used as a simple measurement of supercrit-ical-flow velocity.
Note from Fig. 10.8b that small changes in E near Emin 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.
EXAMPLE 10.4 A wide rectangular clean-earth channel has a flow rate q 50 ft3/(s ft). (a) What is the criti-cal depth? (b) What type of flow exists if y 3 ft?
Solution From Table 10.1, n 0.022 and 0.12 ft. The critical depth follows from Eq. (10.30): yc q g 2 1/3 3 5 2 0 .
2 2 1/3 4.27 ft Ans. (a) If the actual depth is 3 ft, which is less than yc, the flow must be supercritical.
Ans. (b) If the channel width varies with y, the specific energy must be written in the form E y 2 Q gA 2 2 (10.35) The critical point of minimum energy occurs where dE/dy 0 at constant Q. Since A A(y), Eq. (10.35) yields, for E Emin, d d A y g Q A 2 3 (10.36) But dA b0 dy, where b0 is the channel width at the free surface. Therefore Eq. (10.36) is equivalent to Ac b0 g Q2 1/3 (10.37a) Vc A Q c g b A 0 c 1/2 (10.37b) For a given channel shape A(y) and b0(y) and a given Q, Eq. (10.37) has to be solved by trial and error or by EES to find the critical area Ac, from which Vc can be computed.
10.4 Specific Energy; Critical Depth 673 2This is the basis of the water-channel analogy for supersonic gas-dynamics experimentation [14, chap. 11].
Critical Uniform Flow: The Critical Slope Part (a) Part (b) By comparing the actual depth and velocity with the critical values, we can deter-mine the local flow condition y yc, V Vc: subcritical flow (Fr 1) y yc, V Vc: supercritical flow (Fr 1) If a critical channel flow is also moving uniformly (at constant depth), it must corre-spond to a critical slope Sc, with yn yc. This condition is analyzed by equating Eq.
(10.37a) to the Chézy (or Manning) formula: Q2 g b A 0 c 3 C2Ac 2RhSc n2 2 Ac 2Rh 4/3Sc or Sc n 2b 2g 0R A h c c 4/3 b P 0 8 f b P 0 (10.38) where 2 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, b0 yc, the formula reduces to Wide rectangular channel: Sc n 2y 2g c 1/3 8 f This is a special case, a reference point. In most channel flows yn yc. For fully rough turbulent flow, the critical slope varies between 0.002 and 0.008.
EXAMPLE 10.5 The 50° triangular channel in Fig. E10.5 has a flow rate Q 16 m3/s. Compute (a) yc, (b) Vc, and (c) Sc if n 0.018.
Solution This is an easy cross section because all geometric quantities can be written directly in terms of depth y P 2y csc 50° A y2 cot 50° (1) Rh 1 2 y cos 50° b0 2y cot 50° The critical-flow condition satisfies Eq. (10.37a) gAc 3 b0Q2 or g(yc 2 cot 50°)3 (2yc cot 50°)Q2 yc g co 2 t Q 2 2 50° 1/5 9.8 2 1 ( ( 1 0 6 .8 ) 3 2 9)2 1/5 2.37 m Ans. (a) With yc known, from Eqs. (1) we compute Pc 6.18 m, Rhc 0.760 m, Ac 4.70 m2, and b0c 3.97 m. The critical velocity from Eq. (10.37b) is Vc Q A 4 1 .
6 70 m m 3/s 2 3.41 m/s Ans. (b) n2g 2Rhc 1/3 674 Chapter 10 Open-Channel Flow y y cot 50° y csc 50° 50° E10.5 Frictionless Flow over a Bump With n 0.018, we compute from Eq. (10.38) a critical slope Sc 2 g R n h 1 2 / P 3b0 1 9 .
.
0 8 ( 1 0 ( .
0 7 .
6 0 0 1 ) 8 1 ) / 2 3 ( ( 6 3 .
.
1 9 8 7 ) ) 0.00542 Ans. (c) A rough analogy to compressible gas flow in a nozzle (Fig. 9.12) is open-channel flow over a bump, as in Fig. 10.9a. The behavior of the free surface is sharply different ac-cording 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. 10.9a are related by continuity and momentum: V1y1 V2y2 V 2g 2 1 y1 V 2g 2 2 y2 h Eliminating V2 between these two gives a cubic polynomial equation for the water depth y2 over the bump: y3 2 E2y2 2 V 2 2 1 g y2 1 0 where E2 V 2g 2 1 y1 h (10.39) This equation has one negative and two positive solutions if h is not too large. Its be-havior is illustrated in Fig. 10.9b and depends upon whether condition 1 is on the up-per or lower leg of the energy curve. The specific energy E2 is exactly h less than the approach energy E1, and point 2 will lie on the same leg of the curve as E1. A sub-10.4 Specific Energy; Critical Depth 675 Part (c) y1 V1 y2 V2 Bump Subcritical approach Supercritical approach flow ∆ h (a) Water depth y1 y2 yc (b) Ec E2 E1 Specific energy 2.0 ∆ h max ∆ h Supercritical bump Subcritical bump 2 1 Fig. 10.9 Frictionless two-dimen-sional flow over a bump: (a) defini-tion sketch showing Froude-number dependence; (b) specific-energy plot showing bump size and water depths.
critical approach, Fr1 1, will cause the water level to decrease at the bump. Super-critical approach flow, Fr1 1, causes a water-level increase over the bump.
If the bump height reaches hmax E1 Ec, as illustrated in Fig. 10.9b, the flow at the crest will be exactly critical (Fr 1). If h hmax, 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 (h 0): Sub-critical approach flow will cause a water-level rise and supercritical flow a fall in depth.
Point 2 will be hto the right of point 1, and critical flow cannot occur.
EXAMPLE 10.6 Water flow in a wide channel approaches a 10-cm-high bump at 1.5 m/s and a depth of 1 m. Es-timate (a) the water depth y2 over the bump and (b) the bump height which will cause the crest flow to be critical.
Solution First check the approach Froude number, assuming C0 g y : Fr1 0.479 (subcritical) For subcritical approach flow, if h 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 h 0.1 m, the spe-cific-energy levels must be E1 V 2g 2 1 y1 2 ( ( 1 9 .
.
5 8 ) 1 2 ) 1.0 1.115 E2 E1 h 1.015 m This physical situation is shown on a specific-energy plot in Fig. E10.6. With y1 in meters, Eq.
(10.39) takes on the numerical values y3 2 1.015y2 2 0.115 0 There are three real roots: y2 0.859 m, 0.451 m, and 0.296 m. The third (negative) so-lution is physically impossible. The second (smaller) solution is the supercritical condition for 1.5 m/s (9 .8 1 m /s 2) (1 .0 m ) V1 g y1 676 Chapter 10 Open-Channel Flow Part (a) 0.90 E Ec = 0.918 m E2 = 1.015 m E1 = 1.115 m Supercritical 0.451 m 0.612 m y2 = 0.859 m Subcritical bump y1 = 1.0 m 1.2 1 2 y ∆ h = 0.1 m 1.0 0.8 0.6 0.4 1.00 1.10 1.20 E10.6 Flow under a Sluice Gate E2 and is not possible for this subcritical bump. The first solution is correct: y2(subcritical) 0.859 m Ans. (a) The surface level has dropped by y1 y2 h 1.0 0.859 0.1 0.041 m. The crest ve-locity is V2 V1y1/y2 1.745 m/s. The Froude number at the crest is Fr2 0.601. Flow down-stream of the bump is subcritical. These flow conditions are shown in Fig. E10.6.
For critical flow in a wide channel, with q Vy 1.5 m2/s, from Eq. (10.31), E2,min Ec 3 2 yc 3 2 q g 2 1/3 3 2 ( 9 1 .
.
8 5 1 m m 2/ / s s ) 2 2 1/3 0.918 m Therefore the maximum height for frictionless flow over this particular bump is hmax E1 E2,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 y1 y2 h 0.191 m.
A sluice gate is a bottom opening in a wall, as sketched in Fig. 10.10a, 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 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. 9.12b).
For free discharge, friction may be neglected, and since there is no bump (h 0), Eq. (10.39) applies with E1 E2: y3 2 V 2g 2 1 y1 y2 2 V 2 2 1 g y2 1 0 (10.40) Given subcritical upstream flow (V1, y1), this cubic equation has only one positive real solution: supercritical flow at the same specific energy, as in Fig. 10.10b. The flow rate varies with the ratio y2/y1; we ask, as a problem exercise, to show that the flow rate is a maximum when y2/y1 2 3 .
The free discharge, Fig. 10.10a, contracts to a depth y2 about 40 percent less than the gate’s gap height, as shown. This is similar to a free orifice discharge, as in Fig.
6.38. 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 CdHb 2 g y1 where Cd (10.41) in the range H/y1 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 sit-uation, and experimental discharge correlations are necessary [2, 15]. See Prob. 10.77.
0.61 1 0 .6 1 H /y 1 10.4 Specific Energy; Critical Depth 677 Part (b) Fig. 10.10 Flow under a sluice gate passes through critical flow: (a) free discharge with vena contracta; (b) spe-cific energy for free discharge; (c) dissipative flow under a drowned gate.
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 en-ergy dissipator and is a feature of stilling-basin and spillway applications . Figure 10.11 shows the jump formed at the bottom of a dam spillway in a model test. It is very important 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 ef-fectively and have application to sewage and water treatment designs.
678 Chapter 10 Open-Channel Flow V1, y1 Gate Vena contracta V2, y2 (a) (b) E1= E2 E Supercritical 1 2 Subcritical y1 y V1, y1 (c) Dissipation High tailwater V2, y2 Gate 10.5 The Hydraulic Jump Fig. 10.11 Hydraulic jump formed on a spillway model for the Karna-fuli Dam in East Pakistan. (Cour-tesy of the St. Anthony Falls Hy-draulic Laboratory, University of Minnesota.) Classification The principal parameter affecting hydraulic-jump performance is the upstream Froude number Fr1 V1/(gy1)1/2. The Reynolds number and channel geometry have only a secondary effect. As detailed in Ref. 16, the following ranges of operation can be out-lined, as illustrated in Fig. 10.12: Fr1 1.0: Jump impossible, violates second law of thermodynamics.
Fr1 1.0 to 1.7: Standing-wave, or undular, jump about 4y2 long; low dissipa-tion, less than 5 percent.
Fr1 1.7 to 2.5: Smooth surface rise with small rollers, known as a weak jump; dissipation 5 to 15 percent.
10.5 The Hydraulic Jump 679 (a) 2 1 (b) V1 y1 V2 (c) (d ) (e) y2 Fig. 10.12 Classification of hy-draulic jumps: (a) Fr 1.0 to 1.7: undular jumps; (b) Fr 1.7 to 2.5: weak jump; (c) Fr 2.5 to 4.5: os-cillating jump; (d) Fr 4.5 to 9.0: steady jump; (e) Fr 9.0: strong jump. (Adapted from Ref. 16.) Fr1 2.5 to 4.5: Unstable, oscillating jump; each irregular pulsation creates a large wave which can travel downstream for miles, damaging earth banks and other structures. Not recommended for design conditions. Dissipation 15 to 45 percent.
Fr1 4.5 to 9.0: Stable, well-balanced, steady jump; best performance and ac-tion, insensitive to downstream conditions. Best design range.
Dissipation 45 to 70 percent.
Fr1 9.0: Rough, somewhat intermittent strong jump, but good perfor-mance. Dissipation 70 to 85 percent.
Further details can be found in Ref. 16 and Ref. 3, chap. 15.
A jump which 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 that the clas-sic 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. 10.4b, where the change in depth y is not neglected. If V1 and y1 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 in-terpret C and y in Fig. 10.4b as upstream conditions V1 and y1, with C V and y y being the downstream conditions V2 and y2, as in Fig. 10.12b. Equation (10.9) be-comes V2 1 1 2 gy1( 1) (10.42) where y2/y1. Introducing the Froude number Fr1 V1/(gy1)1/2 and solving this qua-dratic equation for , we obtain 2 y y 1 2 1 (1 8 Fr2 1)1/2 (10.43) With y2 thus known, V2 follows from the wide-channel continuity relation V2 V y 1 2 y1 (10.44) Finally, we can evaluate the dissipation head loss across the jump from the steady-flow energy equation hf E1 E2 y1 V 2g 2 1 y2 V 2g 2 2 Introducing y2 and V2 from Eqs. (10.43) and (10.44), we find after considerable alge-braic manipulation that hf (y2 4 y1y y 2 1)3 (10.45) Equation (10.45) shows that the dissipation loss is positive only if y2 y1, which is a requirement of the second law of thermodynamics. Equation (10.43) then requires that 680 Chapter 10 Open-Channel Flow Theory for a Horizontal Jump Part (a) Part (b) Part (c) Part (e) Part (d) Fr1 1.0; that is, the upstream flow must be supercritical. Finally, Eq. (10.44) shows that V2 V1 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 horizontal channels. For the theory of prismatic or sloping channels see advanced texts [for example, 3, chaps.
15 and 16].
EXAMPLE 10.7 Water flows in a wide channel at q 10 m3/(s m) and y1 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 V1 y q 1 10 1 m .2 3/ 5 (s m m) 8.0 m/s The upstream Froude number is therefore Fr1 (gy V 1 1 )1/2 [9.81( 8 1 .
.
0 25)]1/2 2.285 From Fig. 10.12 this is a weak jump. The depth y2 is obtained from Eq. (10.43): 2 y y 1 2 1 [1 8(2.285)2]1/2 5.54 or y2 1 2 y1(5.54) 1 2 (1.25)(5.54) 3.46 m Ans. (a) From Eq. (10.44) the downstream velocity is V2 V y 1 2 y1 8.0 3 ( .
1 4 .
6 25) 2.89 m/s Ans. (b) The downstream Froude number is Fr2 (gy V 2 2 )1/2 [9.81 2 (3 .8 .4 9 6)]1/2 0.496 Ans. (c) As expected, Fr2 is subcritical. From Eq. (10.45) the dissipation loss is hf ( 4 3 ( .4 3 6 .4 6)( 1 1 .
.
2 2 5 5 ) ) 3 0.625 m Ans. (d) The percentage dissipation relates hf to upstream energy E1 y1 V 2g 2 1 1.25 2 ( ( 8 9 .
.
0 8 ) 1 2 ) 4.51 m Hence Percentage loss (100) E h 1 f 100 4 (0 .5 .6 1 25) 14 percent Ans. (e) 10.5 The Hydraulic Jump 681 Part (f) Part (g) 10.6 Gradually Varied Flow3 Basic Differential Equation The power dissipated per unit width is Power gqhf (9800 N/m3)10 m3/(s m) 61.3 kW/m Ans. (f) Finally the mass flow rate is m ˙ q (1000 kg/m3)[10 m3/(s m)] 10,000 kg/(s m), and the temperature rise from the steady-flow energy equation is Power dissipated m ˙cp T or 61,300 W/m [10,000 kg/(s m)][4200 J/(kg K)] T from which T 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 po-sition, as in Fig. 10.3. An approximate analysis is possible if the flow is gradually var-ied, e.g., 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 bot-tom friction losses included. The two unknowns for steady flow are velocity V(x) 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 which enter the steady-flow energy equation are shown, and the balance between x and x dx is 2 V g 2 y S0 dx S dx 2 V g 2 d 2 V g 2 y dy or d d y x d d x 2 V g 2 S0 S (10.46) where S0 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).
To eliminate the velocity derivative, differentiate the continuity relation d d Q x 0 A d d V x V d d A x (10.47) 682 Chapter 10 Open-Channel Flow 3This section may be omitted without loss of continuity.
Fig. 10.13 Energy balance between two sections in a gradually varied open-channel flow.
Classification of Solutions But dA b0 dy, where b0 is the channel width at the surface. Eliminating dV/dx be-tween Eqs. (10.46) and (10.47), we obtain d d y x 1 V g 2 A b0 S0 S (10.48) Finally, recall from Eq. (10.37) that V2b0/(gA) is the square of the Froude number of the local channel flow. The final desired form of the gradually varied flow equation is d d y x 1 S0 Fr S 2 (10.49) This equation changes sign according as the Froude number is subcritical or super-critical and is analogous to the one-dimensional gas-dynamic area-change formula (9.40).
The numerator of Eq. (10.49) changes sign according as S0 is greater or less than S, which is the slope equivalent to uniform flow at the same discharge Q S S0n D f h 2 V g 2 R V hC 2 2 n 2 2 R V h 4 2 /3 (10.50) where C is the Chézy coefficient. The behavior of Eq. (10.49) thus depends upon the relative magnitude of the local bottom slope S0(x), compared with (1) uniform flow, y yn, and (2) critical flow, y yc. As in Eq. (10.38), the dimensional parameter 2 equals 1.0 for SI units and 2.208 for BG units.
It is customary to compare the actual channel slope S0 with the critical slope Sc for the same Q from Eq. (10.38). There are five classes for S0, giving rise to twelve distinct types of solution curves, all of which are illustrated in Fig. 10.14: 10.6 Gradually Varied Flow 683 Horizontal EGL S d x V 2 2 g V 2 2 g V 2 2 g + d HGL y V S0 d x x d x Bottom slope S0 V + d V y + d y τ w x + d x slope S Fig. 10.14 Gradually varied flow for five classes of channel slope, showing the twelve basic solution curves.
Slope class Slope notation Depth class Solution curves S0 Sc Steep yc yn S-1, S-2, S-3 S0 Sc Critical yc yn C-1, C-3 S0 Sc Mild yc yn M-1, M-2, M-3 S0 0 Horizontal yn H-2, H-3 S0 0 Adverse yn imaginary A-2, A-3 684 Chapter 10 Open-Channel Flow (a) Steep S0 > Sc yc yn Fr < 1 Fr > 1 Fr > 1 y (x) y (0) S – 1 S – 2 S – 3 (b) Critical S0 = Sc yn = yc Fr < 1 Fr > 1 C – 1 C – 3 (c) Mild S0 < Sc yn yc Fr < 1 Fr< 1 Fr > 1 M – 1 M – 2 M – 3 (d ) Horizontal S0 = 0 yn = ∞ yc Fr < 1 Fr > 1 H – 2 H – 3 (e) Adverse S0 < 0 yn = imaginary yc Fr < 1 Fr > 1 A – 2 A – 3 The solution letters S, C, M, H, and A obviously denote the names of the five types of slope. The numbers 1, 2, 3 relate to the position of the initial point on the solution curve with respect to the normal depth yn and the critical depth yc. In type 1 solutions, the ini-tial point is above both yn and yc, and in all cases the water-depth solution y(x) becomes even deeper and farther away from yn and yc. In type 2 solutions, the initial point lies between yn and yc, and if there is no change in S0 or roughness, the solution tends as-ymptotically toward the lower of yn or yc. In type 3 cases, the initial point lies below both yn and yc, 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, S0 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. 3, chap. 9; see also Ref. 18, sec. 12.7.
The basic relation for gradually varied flow, Eq. (10.49), is a first-order ordinary dif-ferential equation which can be easily solved numerically. For a given constant-volume flow rate Q, it may be written in the form d d y x (10.51) subject to an initial condition y y0 at x x0. It is assumed that the bottom slope S0(x) and the cross-sectional shape parameters (b0, P, A) are known everywhere along the channel.
Then one may solve Eq. (10.51) for local water depth y(x) by any standard numerical method.
The author uses an Excel spreadsheet for a personal computer. Step sizes x may be se-lected so that each change y is limited to no greater than, say, 1 percent. The solution curves are generally well behaved unless there are discontinuous changes in channel para-meters. Note that if one approaches the critical depth yc, the denominator of Eq. (10.51) ap-proaches zero, so small step sizes are required. It helps physically to know what type solu-tion curve (M-1, S-2, etc.) you are proceeding along, but this is not mathematically necessary.
EXAMPLE 10.8 Let us extend the data of Example 10.4 to compute a portion of the profile shape. Given is a wide channel with n 0.022, S0 0.0048, and q 50 ft3/(s ft). If y0 3 ft at x 0, how far along the channel x L does it take the depth to rise to yL 4 ft? Is the 4-ft depth position up-stream or downstream in Fig. E10.8a?
Solution In Example 10.4 we computed yc 4.27 ft. Since our initial depth y 3 ft is less than yc, we know the flow is supercritical. Let us also compute the normal depth for the given slope S0 by setting q 50 ft3/(s ft) in the Chézy formula (10.19) with Rh yn: q n ARh 2/3S0 1/2 1 0 .
.
4 0 8 2 6 2 [yn(1 ft)]yn 2/3(0.0048)1/2 50 ft3/(s ft) Solve for: yn 4.14 ft S0 n2Q2/(2A2Rh 4/3) 1 Q2b0/(gA3) 10.6 Gradually Varied Flow 685 Numerical Solution Thus both y(0) 3 ft and y(L) 4 ft are less than yn, which is less than yc, so we must be on an S-3 curve, as in Fig. 10.14a. For a wide channel, Eq. (10.51) reduces to d d y x S0 1 n2q q 2 2 / / ( ( gy 2y 3) 10/3) with y(0) 3 ft The initial slope is y(0) 0.00494, and a step size x 5 ft would cause a change y (0.00494)(5 ft) 0.025 ft, less than 1 percent. We therefore integrate numerically with x 5 ft to determine when the depth y 4 ft is achieved. Tabulate some values: x, ft 0 50 100 150 200 230 y, ft 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 so-lution curve y(x) is shown as the bold line in Fig. E10.8b.
0.0048 (0.022)2(50)2/(2.208y10/3) 1 (50)2/(32.2y3) 686 Chapter 10 Open-Channel Flow yc = 4.27 ft yn = 4.14 ft y0 = 3 ft y L = 4 ft S0 = 0.0048 x = 0 x = L L = ?
5 4 3 2 1 0 0 50 100 150 200 250 230 ft x y yc yn Present example Other S – 3 solution curves Present example solution curve E10.8a E10.8b 10.7 Flow Measurement and Control by Weirs 687 For little extra effort we can investigate the entire family of S-3 solution curves for this prob-lem. Figure E10.8b 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 yn 4.14 ft.
The solution curves in Fig. 10.14 are somewhat simplistic, since they postulate constant bottom slopes. In practice, channel slopes can vary greatly, S0 S0(x), and the solution curves can cross between two regimes. Other parameter changes, such as A(x), b0(x), and n(x), can cause interesting composite-flow profiles. Some examples are shown in Fig. 10.15.
Figure 10.15a shows a change 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 [3, 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 which would move away from uniform steep flow.
Figure 10.15b shows a mild slope which becomes even milder. The water depth moves smoothly along an M-1 curve to the new (higher) uniform flow. There is no sin-gular point, so gradually varied theory is adequate.
Figure 10.15c illustrates a steep slope which becomes less steep. The depth will change smoothly along an S-3 curve to approach the new (higher) uniform flow. There is no singular point. Compare with Fig. 10.15b.
Figure 10.15d shows a steep slope which changes to mild. It is generally impossi-ble to revert back smoothly from supercritical to subcritical flow without dissipation.
A hydraulic jump will form whose position depends upon downstream conditions. Two cases are illustrated: (1) a jump to an S-1 curve to reach a high downstream normal depth and (2) a change to an M-3 curve and then a jump to a lower downstream depth.
Figure 10.15e illustrates a free overfall with a mild slope. This acts as a control sec-tion to the upstream flow, which then forms an M-2 curve and accelerates to critical 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 computation of y(x).
This is the type of flow which 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 Refs. 1, 3, and 18.
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-chan-nel flowmeter. We used a weir as an example of dimensional analysis in Prob. 5.32.
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 Some Illustrative Composite-Flow Profiles 10.7 Flow Measurement and Control by Weirs discharge q per unit width is proportional to g1/2H3/2 but with somewhat different co-efficients. The short-crested (or thin-plate) weir nappe should be ventilated to the at-mosphere; i.e., it should spring clear of the weir crest. Unventilated or drowned nappes are more difficult to correlate and depend upon tailwater conditions. (The spillway of Fig. 10.11 is a sort of unventilated weir.) 688 Chapter 10 Open-Channel Flow Fig. 10.15 Some examples of com-posite-flow profiles.
(a) Mild M – 2 yn1 Critical flow S – 2 yc yn2 Steep Mild Steep Steep M – 1 yn1 Milder S – 3 yn2 yc yc yn2 Mild S – 1 yn, high yc yn, low (b) (c) (d) (e) yc yn M – 2 Critical flow Free overfall M – 3 yn1 Less steep yn1 Jump Fig. 10.16 Flow over wide, well-ventilated weirs: (a) sharp-crested; (b) broad-crested.
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. 10.122.
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; i.e., Bernoulli’s equation is used with no losses: V 2g 2 2 H h V 2g 2 1 H or V2(h) 2 g h V 2 1 where h is the vertical distance down to point 2, as shown in Fig. 10.16a. If we accept for the moment, without proof, that the flow over the crest draws down to hmin H/3, the volume flow q Q/b over the crest is approximately q crest V2 dh H H/3 (2gh V2 1)1/2 dh 2 3 2 g H V 2g 2 1 3/2 H 3 V 2g 2 1 3/2 (10.52) 10.7 Flow Measurement and Control by Weirs 689 Analysis of Sharp-Crested Weirs V1 1 H Y h 2 Weir Nappe Ventilation ≈ H 1 3 (b) H Weir yc Y V1 L (a) Analysis of Broad-Crested Weirs Experimental Weir Discharge Coefficients Normally the upstream velocity head V2 1/(2g) is neglected, so this expression reduces to Sharp-crested theory: q 0.81( 2 3 )(2g)1/2H3/2 (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. 10.16b can be analyzed more accurately because it cre-ates a short run of nearly one-dimensional critical flow, as shown. Bernoulli’s equation from upstream to the weir crest yields V 2g 2 1 Y H V 2g c 2 Y yc If the crest is very wide into the paper, Vc 2 gyc from Eq. (10.33). Thus we can solve for yc 2 3 H V 3g 2 1 2 3 H 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 yc 3 2 3 2 g H V 2g 2 1 3/2 (10.54) Again we may usually neglect the upstream velocity head V2 1/(2g). The coefficient 1/3 0.577 is about right, but experimental data are preferred.
Theoretical weir-flow formulas may be modified experimentally as follows. Eliminate the numerical coefficients 2 3 and 2 , for which there is much sentimental attachment in the literature, and reduce the formula to Qweir Cdbg H V 2g 2 1 3/2 Cdbg H3/2 (10.55) where b is the crest width and Cd 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. 19.
An accurate ( 2 percent) composite correlation for wide ventilated sharp crests is recommended as follows : Wide sharp-crested weir: Cd 0.564 0.0846 H Y for H Y 2 (10.56) The Reynolds numbers V1H/ for these data varied 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. 10.16b is considerably more sensitive to geometric parameters, including the surface roughness of the crest. If the leading-edge nose is rounded, R/L 0.05, available data [19, chap. 7] may be correlated as follows: Round-nosed broad-crested weir: Cd 0.5441 H / / L L 3/2 (10.57) 1 3 690 Chapter 10 Open-Channel Flow Part (a) where L 0.001 0.2
/ L The chief effect is due to turbulent boundary-layer displacement-thickness growth on the crest as compared to upstream head H. The formula is limited to H/L 0.7, /L 0.002, and V1H/ 3 E5. If the nose is round, there is no significant effect of weir height Y, at least if H/Y 2.4.
If the broad-crested weir has a sharp leading edge, thus commonly called a rectan-gular weir, the discharge may depend upon the weir height Y. However, in a certain range of weir height and length, Cd is nearly constant: Sharp-nosed Cd 0.462 for 0.08 H L 0.33 broad-crested weir: (10.58) 0.22 H Y 0.56 Surface roughness is not a significant factor here. For H/L 0.08 there is large scatter (10 percent) in the data. For H/L 0.33 and H/Y 0.56, Cd increases up to 10 percent due to each parameter, and complex charts are needed for the discharge coefficient [19, chap. 6].
EXAMPLE 10.9 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 cement broad crest 1.2 m long. Neglect V2 1/(2g).
Solution We are given Y 1 m and H Y 1.6 m, hence H 0.6 m. Since H b, we assume that the weir is “wide.” For a sharp crest, Eq. (10.56) applies: Cd 0.564 0.0846 0 1 .6 m m 0.615 Then the discharge is given by the basic correlation, Eq. (10.55): Q Cdbg H3/2 (0.615)(4 m)(9 .8 1 m /s 2) (0.6 m)3/2 3.58 m3/s Ans. (a) We check that H/Y 0.6 2.0 for Eq. (10.56) to be valid. From continuity, V1 Q/(by1) 3.58/[(4.0)(1.6)] 0.56 m/s, giving a Reynolds number V1H/ 3.4 E5.
For a round-nosed broad-crested weir, Eq. (10.57) applies. For an unfinished cement surface, read 2.4 mm from Table 10.1. Then the displacement thickness is L 0.001 0.2
/ L 0.001 0.2 0.0 1 0 .2 24 m m 1/2 0.00994 Then Eq. (10.57) predicts the discharge coefficient: Cd 0.5441 0.6 0.
m 00 / 9 1 9 .2 4 m 3/2 0.528 10.7 Flow Measurement and Control by Weirs 691 Part (b) Other Thin-Plate Weir Designs The estimated flow rate is thus Q Cdbg H3/2 0.528(4 m)(9 .8 1 m 2/ s) (0.6 m)3/2 3.07 m3/s Ans. (b) Check that H/L 0.5 0.7 as required. The approach Reynolds number is V1H/ 2.9 E5, just barely below the recommended limit in Eq. (10.57).
Since V1 0.5 m/s, V2 1/(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.
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 10.2a 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 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.
692 Chapter 10 Open-Channel Flow Thin-plate weir Flow-rate correlation Q 0.564 0.0846 H Y bg1/2H3/2 (a) Full-width rectangle.
Q 0.581(b 0.1H)g1/2H3/2 H 0.5Y (b) Rectangle with side contractions.
Q 0.44 tan 2 g1/2H5/2 20° 100° (c) V notch.
Table 10.2 Thin-Plate Weirs for Flow Measurement L > 2b Plate H Y b Plate b H Y θ Plate H Y Backwater Curves Table 10.2b 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 [19, 20] is to reduce the effective weir width by 0.1H, 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 (H 75 mm) and small slot widths (b 30 cm) are not recommended. See Refs. 19 and 20 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 proportional to H5/2, rather than a power of 3 2 . Application of Bernoulli’s equation to the triangular open-ing, in the spirit of Eq. (10.52), leads to the following ideal flow rate for a V notch: V notch: Qideal 8 15 2 tan 2 g1/2H5/2 (10.59) where 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 QV notch Cd tan 2 g1/2H5/2 Cd 0.44 for 20° 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 (Re 0 W .9 e)1/6 (10.61) where Re g1/2H3/2/ and We gH2/, with being the coefficient of surface ten-sion. Liquids other than water may be used with this formula, as long as Re 300/tan ( /2)3/4 and We 300.
A number of other thin-plate weir designs—trapezoidal, parabolic, circular arc, and U-shaped—are discussed in Ref. 21, which also contains considerable data on broad-crested weirs.
A weir is a flow barrier which not only alters the local flow over the weir but also mod-ifies 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 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 as studied, e.g., by the U.S. Army Corps of Engineers .
EXAMPLE 10.10 A rectangular channel 8 m wide, with a flow rate of 30 m3/s, encounters a 4-m-high sharp-edged dam, as shown in Fig. E10.10a. Determine the water depth 2 km upstream if the channel slope is S0 0.0004 and n 0.025.
10.7 Flow Measurement and Control by Weirs 693 E10.10a Solution First determine the head H produced by the dam, using sharp-crested full-width weir theory, Eq.
(10.56): Q 30 m3/s Cdbg1/2H3/2 0.564 0.0846 4 H m (8 m)(9.81 m/s2)1/2H3/2 Since the term 0.0846H/4 in parentheses is small, we may proceed by iteration or EES to the solution H 1.59 m. Then our initial condition at x 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): yc b Q 2 2 g 1/3 (8 m (3 )2 0 (9 m .8 3 1 /s) m 2 /s2) 1/3 1.13 m Since y(0) is greater than yc, the flow upstream is subcritical. Finally, for reference purposes, es-timate the normal depth from the Chézy equation (10.19): Q 30 m3/s n byRh 2/3S0 1/2 0 1 .0 .0 25 (8 m)yn 8 8y 2 n yn 2/3 (0.0004)1/2 By trial and error or EES solve for yn 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 val-ues y(0), yc, and yn are shown in Fig. E10.10b.
Since y(0) yn yc, 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: d d y x 1.0 A 8y n 0.025 Rh 8 8y 2y b0 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 x 100 m. The complete solution curve is shown in Fig. E10.10b. The desired solution value is At x 2000 m: y 5.00 m Ans.
S0 n2Q2/(2A2Rh 4/3) 1 Q2b0/(gA3) 694 Chapter 10 Open-Channel Flow Backwater curve (M – 1) H (From weir theory) Q X x = 0 x = – 2000 m S0 = 0.0004, b = 8 m Manning's n = 0.025 y ?
Y = 4 m Q = 30 m 3/s yn = 3.20 m yc = 1.13 m Dam Summary Thus, even 2 km upstream, the dam has produced a “backwater” which is 1.8 m above the nor-mal depth which would occur without a dam. For this example, a near-normal depth of, say, 10 cm greater than yn, 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.
This chapter is an introduction to 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 Chézy equation (10.19). Straight prismatic channels can be optimized to find the cross sec-tion which 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 which 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.
The analysis of gradually varied flow leads to a differential equation (10.51) which 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 695 E10.10b 6 5 4 3 2 1 0 – 2000 – 1500 – 1000 – 500 0 x, m y, m M – 1 Solution curve y ≈ 5.00 m at x = –2000 m yn = 3.20 m yc = 1.13 m Weir 5.59 4.0 Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equations Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 10.1 to 10.128 (categorized in the prob-lem list below) are followed by word problems W10.1 to W10.13, fundamentals of engineering exam problems FE10.1 to FE10.7, comprehensive problems C10.1 to C10.3, and design projects D10.1 and D10.2.
Problem distribution Section Topic Problems 10.1 Introduction: Froude number, wave speed 10.1–10.10 10.2 Uniform flow: The Chézy formula 10.11–10.36 10.3 Efficient uniform-flow channels 10.37–10.46 10.4 Specific energy: Critical depth 10.47–10.58 10.4 Flow over a bump 10.59–10.68 10.4 Sluice-gate flow 10.69–10.78 10.5 The hydraulic jump 10.79–10.96 10.6 Gradually varied flow 10.97–10.112 10.7 Weirs and flumes 10.113–10.123 10.7 Backwater curves 10.124–10.128 P10.1 The formula for shallow-water wave propagation speed, Eq. (10.9) or (10.10), is independent of the physical properties of the liquid, i.e., density, viscosity, or sur-face tension. Does this mean that waves propagate at the same speed in water, mercury, gasoline, and glyc-erin? Explain.
P10.2 A shallow-water wave 1 cm high propagates into still wa-ter of depth 1.1 m. Compute (a) the wave speed c and (b) the induced velocity V.
P10.3 Narragansett Bay is approximately 21 (statute) mi long and has an average depth of 42 ft. Tidal charts for the area indicate a time delay of 30 min between high tide at the mouth of the bay (Newport, Rhode Island) and its head (Providence, Rhode Island). Is this delay correlated with the propagation of a shallow-water tidal-crest wave through the bay? Explain.
P10.4 The water-channel 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?
696 Chapter 10 Open-Channel Flow P10.4 P10.5 Water flows rapidly in a channel 15 cm deep. Piercing the surface with a pencil point creates a wedgelike down-stream wave of included angle 35°. Estimate the veloc-ity V of the water.
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. P10.6. From this information estimate (a) the Froude number and (b) the stream velocity.
P10.7 Pebbles dropped successively at the same point, into a water channel flow of depth 65 cm, create two circular V 4 m 9 m 6 m P10.6 V 9 m 4 m 3 m P10.7 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) which propagates south across the Pacific Ocean. If the average ocean depth is 4 km and seawater density is 1025 kg/m3, estimate the time of arrival of this tsunami in Hilo, Hawaii.
P10.9 Equation (10.10) is for a single disturbance wave. For pe-riodic small-amplitude surface waves of wavelength and period T, inviscid theory [5 to 9] predicts a wave propagation speed c2 0 2 g tanh 2 y where y is the water depth and surface tension is ne-glected. (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 . (d) Also for what ratio y/ is the wave speed within 1 percent of limit (c)?
P10.10 If surface tension is included in the analysis of Prob.
10.9, the resulting wave speed is [5 to 9] c2 0 2 g 2 tanh 2 y (a) Determine if this expression is affected by the Reynolds number, Froude number, or Weber number. De-rive the limiting values of this expression for (b) y and (c) y . (d) Finally determine the wavelength crit for a minimum value of c0, assuming that y .
P10.11 A rectangular channel is 2 m wide and contains water 3 m deep. If the slope is 0.85° and the lining is corru-gated metal, estimate the discharge for uniform flow.
P10.20 A circular corrugated-metal storm drain is flowing half full over a slope 4 ft/mi. Estimate the normal discharge if the drain diameter is 8 ft.
P10.21 A sewer pipe has the shape of Fig. P10.21 and is con-structed of unfinished concrete. The slope is 3 ft/mi. Plot the normal flow rate as a function of depth over the full range 0 y 7 ft, and determine the maximum dis-charge and maximum velocity.
P10.12 (a) For laminar draining of a wide thin sheet of water on pavement sloped at angle , as in Fig. P4.36, show that the flow rate is given by Q gbh 3 3 sin 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 f 24/Re, where Re Vavh/.
P10.13 The laminar-draining flow from Prob. 10.12 may undergo transition to turbulence if Re 500. If the pavement slope is 0.0045, what is the maximum sheet thickness, in mm, for which laminar flow is ensured?
P10.14 The Chézy formula (10.18) is independent of fluid den-sity 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 finished-concrete channel of Fig. P10.15 is designed for a flow rate of 6 m3/s at a normal depth of 1 m. De-termine (a) the design slope of the channel and (b) the percentage of reduction in flow if the surface is asphalt.
Problems 697 EES EES 1 m 3 m Proposed barrier P10.16 In Prob. 10.15, for finished concrete, determine the per-centage of reduction in flow if the channel is divided in the center by the proposed barrier in Fig. P10.15. How does your estimate change if all surfaces are clay tile?
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.
30° 2 m 30° P10.18 Modify Prob. 10.17 as follows. Determine the normal depth for which the flow rate will be 8 m3/s.
P10.19 Modify Prob. 10.17 as follows. Let the surface be clean earth, which erodes if V exceeds 1.5 m/s. What is the maximum depth to avoid erosion?
P10.15 P10.17 y 2 ft 3 ft 2 ft P10.21 P10.22 A trapezoidal aqueduct (Fig. 10.7) has b 5 m and 40° and carries a normal flow of 60 m3/s of water when y 3.2 m. For clay tile surfaces, estimate the required elevation drop in m/km.
P10.23 For the aqueduct of Prob. 10.22, if the slope is 0.0004 and the discharge is 40 m3/s, use the Moody-chart for-mulation (10.15a) to estimate the normal depth.
P10.24 A riveted-steel channel slopes at 1:500 and has a V shape with an included angle of 80°. Find the normal depth if the flow rate is 900 m3/h.
P10.25 The equilateral-triangle channel in Fig. P10.25 has con-stant slope S0 and constant Manning factor n. Find Qmax and Vmax. Then, by analogy with Fig. 10.6b, plot the ra-tios Q/Qmax and V/Vmax as a function of y/a for the com-plete range 0 y/a 0.866.
a a y a P10.25 P10.26 Water flows in a 6-m-wide rectangular channel lined with rocks whose dominant size is 20 cm. The channel slope is 0.003. Assuming uniform flow and the rock-friction correlation of Eq. (10.15b), estimate (a) the flow rate if the depth is 2 m and (b) the depth if the flow rate is P10.33 Five of the sewer pipes from Prob. 10.32 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 S0 0.002 is designed to carry 230 ft3/s of water in uniform flow. There is an argument over whether the channel width should be 4 or 8 ft. Which design needs fewer bricks? By what per-centage?
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 y1 20 ft, y2 5 ft, b1 40 ft, b2 100 ft, n1 0.020, n2 0.040, with a slope of 0.0002. Estimate the discharge in ft3/s.
15 m3/s. (c) In part (a) what is the equivalent value of Manning’s roughness factor n for the same flow rate?
Does your result agree with Table 10.1?
P10.27 A circular unfinished-cement water channel has a slope of 1:600 and a diameter of 5 ft. Estimate the normal dis-charge in gal/min for which the average wall shear stress is 0.15 lbf/ft2, and compare your result to the maximum possible discharge for this channel.
P10.28 Show that, for any straight prismatic channel in uniform flow, the average wall shear stress is given by av gRhS0 If you happen to spot this result early, you can use it in solving Prob. 10.27.
P10.29 Suppose that the trapezoidal channel of Fig. P10.17 con-tains sand and silt which we wish not to erode. Accord-ing to an empirical correlation by A. Shields in 1936, the average wall shear stress crit required to erode sand par-ticles of diameter dp is approximated by 0.5 where s 2400 kg/m3 is the density of sand. If the slope of the channel in Fig. P10.17 is 1 900 and n 0.014, de-termine 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 A storm drain has the cross section shown in Fig. P10.31 and is laid on a slope of 1.5 m/km. If it is constructed of brickwork, find the normal discharge when it is exactly half full of water. crit ( s )g dp 698 Chapter 10 Open-Channel Flow 90˚ 45˚ R=1m P10.31 P10.32 A 2-m-diameter clay tile sewer pipe runs half full on a slope of 0.25°. Compute the normal flow rate in gal/min.
y2 n 2 b 2 y1 n 1 b 1 y1 b 2 y2 n 2 P10.35 P10.36 The Blackstone River in northern Rhode Island normally flows at about 25 m3/s and resembles Fig. P10.35 with a clean-earth center channel, b1 20 m and y1 3 m. The bed slope is about 2 ft/mi. The sides are heavy brush with b2 150 m. During hurricane Carol in 1955, a record flow rate of 1000 m3/s was estimated. Use this information to estimate the maximum flood depth y2 during this event.
P10.37 The answer to Prob. 10.34 is that the 8-ft width is 29 per-cent more efficient than the 4-ft width and is almost op-timum. Verify this result, using the “best-efficiency” con-cepts of Sec. 10.3.
P10.38 A rectangular channel has b 3 m and y 1 m. If n and S0 are the same, what is the diameter of a semicircular channel that will have the same discharge? Compare the two wetted perimeters.
P10.39 A trapezoidal channel has n 0.022 and S0 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 ft3/s of water? What is the discharge of a semicircular channel of the same cross-sectional area and the same S0 and n?
P10.40 Using the geometry of Fig. 10.6a, prove that the most ef-ficient circular open channel (maximum hydraulic radius for a given flow area) is a semicircle.
P10.41 Determine the most efficient value of for the V-shaped channel of Fig. P10.41.
EES Fig. P10.50. If the channel surface is painted steel and the depth is 35 cm, determine (a) the Froude number, (b) the critical depth, and (c) the critical slope for uniform flow.
P10.42 Suppose that the side angles of the trapezoidal channel in Prob. 10.39 are reduced to 15° to avoid earth slides. If the bottom flat width is 8 ft, (a) determine the normal depth and (b) compare the resulting wetted perimeter with the solution P 24.1 ft from Prob. 10.39. (Do not reveal this answer to friends still struggling with Prob. 10.39.) P10.43 What are the most efficient dimensions for a riveted-steel rectangular channel to carry 4.8 m3/s at a slope of 1 900?
P10.44 What are the most efficient dimensions for a half-hexa-gon cast-iron channel to carry 15,000 gal/min on a slope of 0.16°?
P10.45 What is the most efficient depth for an asphalt trapezoidal channel, with sides sloping at 45°, to carry 3 m3/s on a slope of 0.0008?
P10.46 It is suggested that a channel which affords minimum erosion has a half-sine-wave shape, as in Fig. P10.46. The local depth takes the form h(z) h0 sin (z/b). For uni-form-flow conditions, determine the most efficient ratio h0/b for this channel shape.
Problems 699 θ θ y P10.41 z Half sine wave z = b b 2 b 2 h0 h (z) P10.46 P10.47 Replot Fig. 10.8b in the form of q versus y for constant E. Does the maximum q occur at the critical depth?
P10.48 A wide, clean-earth river has a flow rate q 150 ft3/(s ft).
What is the critical depth? If the actual depth is 12 ft, what is the Froude number of the river? Compute the critical slope by (a) Manning’s formula and (b) the Moody chart.
P10.49 Find the critical depth of the brick channel in Prob. 10.34 for both the 4- and 8-ft widths. Are the normal flows sub-or supercritical?
P10.50 A pencil point piercing the surface of a rectangular chan-nel flow creates a wedgelike 25° half-angle wave, as in 25° P10.50 P10.51 Modify Prob. 10.50 as follows. Let the water be flowing in an unfinished-cement half-full circular channel of di-ameter 60 cm. Determine (a) the Froude number, (b) the flow velocity, and (c) the critical slope.
P10.52 Water flows full in an asphalt half-hexagon channel of bottom width W. The flow rate is 12 m3/s. Estimate W if the Froude number is exactly 0.60.
P10.53 For the river flow of Prob. 10.48, find the depth y2 which has the same specific energy as the given depth y1 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 m3/s. Compute (a) the critical depth, (b) the critical velocity, and (c) the critical slope for uni-form flow.
P10.55 A trapezoidal channel resembles Fig. 10.7 with b 1 m and 50°. The water depth is 2 m, and the flow rate is 32 m3/s. If you stick your fingernail in the surface, as in Fig. P10.50, what half-angle wave might appear?
P10.56 A riveted-steel triangular duct flows partly full as in Fig.
P10.56. If the critical depth is 50 cm, compute (a) the critical flow rate and (b) the critical slope.
1 m 1 m 1 m P10.56 P10.57 For the triangular duct of Prob. 10.56, if the critical flow rate is 1.0 m3/s, compute (a) the critical depth and (b) the critical slope.
P10.58 A circular corrugated-metal channel is half full and in uniform flow at a slope S0 0.0037. Estimate the Froude number of the flow.
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 de-EES P10.65 Program and solve the differential equation of “friction-less flow over a bump,” from Prob. 10.62, for entrance conditions V0 1 m/s and y0 1 m. Let the bump have the convenient shape h 0.5hmax[1 cos (2x/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 hmax 15 cm (Prob. 10.63), for which the maximum Froude number is 0.425. If more time is available, it is instructive to ex-amine a complete family of surface profiles for hmax 1 cm up to 35 cm (which is the solution of Prob. 10.64).
P10.66 In Fig. P10.62 let V0 6 m/s and y0 1 m. If the max-imum bump height is 35 cm, estimate (a) the Froude num-ber over the top of the bump and (b) the maximum in-crease in the water-surface level.
P10.67 In Fig. P10.62 let V0 6 m/s and y0 1 m. If the flow over the top of the bump is exactly critical (Fr 1.0), determine the bump height hmax.
P10.68 Modify Prob. 10.65 to have a supercritical approach con-dition V0 6 m/s and y0 1 m. If you have time for only one case, use hmax 35 cm (Prob. 10.66), for which the maximum Froude number is 1.47. If more time is available, it is instructive to examine a complete family of surface profiles for 1 cm hmax 52 cm (which is the solution to Prob. 10.67).
P10.69 Given is the flow of a channel of large width b under a sluice gate, as in Fig. P10.69. Assuming frictionless steady flow with negligible upstream kinetic energy, de-rive a formula for the dimensionless flow ratio Q2/(y3 1b2g) as a function of the ratio y2/y1. Show by dif-ferentiation that the maximum flow rate occurs at y2 2y1/3.
pression in the water surface results. If the minimum wa-ter depth over the bump is 50 cm, compute (a) the ve-locity over the bump and (b) the flow rate per meter of width.
700 Chapter 10 Open-Channel Flow V1 y2 = 50 cm 10 cm bump 0.02° P10.59 P10.60 Modify Prob. 10.59 as follows. Again assuming uniform subcritical approach flow (V1, y1), find (a) the flow rate and (b) y2 for which the Froude number Fr2 at the crest of the bump is exactly 0.8.
P10.61 Modify Prob. 10.59 as follows. Again assuming uniform subcritical approach flow (V1, y1), 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 d d y x 1 dh V / 2 d / x (gy) Is the drawdown of the water surface realistic in Fig.
P10.62? Explain under what conditions the surface might rise above its upstream position y0.
V0 y0 V ( x) y ( x) h ( x) Bump P10.62 P10.63 In Fig. P10.62 let V0 1 m/s and y0 1 m. If the max-imum bump height is 15 cm, estimate (a) the Froude num-ber over the top of the bump and (b) the maximum de-pression in the water surface.
P10.64 In Fig. P10.62 let V0 1 m/s and y0 1 m. If the flow over the top of the bump is exactly critical (Fr 1.0), determine the bump height hmax.
V2 y2 Gate y1 V1 P10.69 P10.70 In Fig. P10.69 let y1 90 cm and V1 65 cm/s. Esti-mate (a) y2, (b) Fr2, and (c) the flow rate per unit width.
P10.71 In Fig. P10.69 let y1 95 cm and y2 50 cm. Estimate the flow rate per unit width if the upstream kinetic en-ergy is (a) neglected and (b) included.
P10.72 Water approaches the wide sluice gate of Fig. P10.72 at V1 0.2 m/s and y1 1 m. Accounting for upstream ki-Problems 701 Figure P10.77 shows data from Ref. 3 on drowned ver-tical sluice gates. Use this chart to repeat Prob. 10.73, and plot the estimated flow rate versus y2 in the range 0 y2 110 cm.
(1) (2) (3) 5 P10.72 netic energy, estimate at the outlet, section 2, the (a) depth, (b) velocity, and (c) Froude number.
P10.73 In Fig. P10.69 suppose that y1 1.2 m and the gate is raised so that its gap is 15 cm. Estimate the resulting flow rate per unit width.
P10.74 With respect to Fig. P10.69, show that, for frictionless flow, the upstream velocity may be related to the water levels by V1 where K y1/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 t 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.
2g(y1 y2) K2 1 Gate closed Gate raised to a gap of 10 cm 1 m 3 m P10.75 P10.76 In Prob. 10.75 estimate what gap height would cause the tank level to drop from 1 m to 50 cm in exactly 1 min.
Assume free outflow.
P10.77 Equation (10.41) for the discharge coefficient is for free (nearly frictionless) outflow. If the outlet is drowned, as in Fig. 10.10c, there is dissipation and Cd drops sharply.
0.6 0.5 0.4 0.3 0.2 0.1 Cd y1 H y2 H= 0 2 4 6 8 10 12 14 16 Drowned tailwater Fig. 10.10c Free outflow 2 3 4 5 6 7 8 P10.77 (From Ref. 3, p. 509.) P10.78 Repeat Prob. 10.75 if the gate is drowned at y2 40 cm.
P10.79 Show that the Froude number downstream of a hydraulic jump will be given by Fr2 81/2 Fr1/[(1 8 Fr2 1)1/2 1]3/2 Does the formula remain correct if we reverse subscripts 1 and 2? Why?
P10.80 Water, flowing horizontally in a wide channel of depth 30 cm, undergoes a hydraulic jump whose energy dissi-pation is 71 percent. Estimate (a) the downstream depth and (b) the volume flow rate per meter of width.
P10.81 Water flows in a wide channel at q 25 ft3/(s ft), y1 1 ft, and then undergoes a hydraulic jump. Compute y2, V2, Fr2, hf, the percentage dissipation, and the horsepower dissipated per unit width. What is the critical depth?
P10.82 The flow downstream of a wide hydraulic jump is 7 m deep and has a velocity of 2.2 m/s. Estimate the upstream (a) depth and (b) velocity and (c) the critical depth of the flow.
P10.83 A wide-channel flow undergoes a hydraulic jump from 40 to 140 cm. Estimate (a) V1, (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 y1 10 ft and all losses are neglected except the dissipation in the jump, calculate y2 and y3 and the per-centage of dissipation, and sketch the flow to scale with the EGL included. The channel is horizontal and wide.
P10.85 In Prob. 10.72 the exit velocity from the sluice gate is 4.33 m/s. If there is a hydraulic jump just downstream of section 2, determine the downstream (a) velocity, (b) EES trol-volume sketch of a sloping jump to show why this is so. The sloped-jump chart given in Chow’s figure 15-20 may be approximated by the following curve fit: 2 y y 1 2 [(1 8 Fr2 1)1/2 1]e3.5S0 where 0 S0 0.3 are the channel slopes for which data are available. Use this correlation to modify your solu-tion to Prob. 10.89. If time permits, make a graph of y2/y1 ( 20) versus Fr1 ( 15) for various S0 ( 0.3).
P10.92 At the bottom of an 80-ft-wide spillway is a horizontal hydraulic jump with water depths 1 ft upstream and 10 ft downstream. Estimate (a) the flow rate and (b) the horsepower dissipated.
P10.93 Water in a horizontal channel accelerates smoothly over a bump and then undergoes a hydraulic jump, as in Fig.
P10.93. If y1 1 m and y3 40 cm, estimate (a) V1, (b) V3, (c) y4, and (d) the bump height h.
depth, (c) Froude number, and (d) percent dissipation.
Neglect the effect of the nonhorizontal bottom (see Prob.
10.91).
P10.86 A bore is a hydraulic jump which propagates upstream into a still or slower-moving fluid, as in Fig. 10.4a. Sup-pose that the still water is 2 m deep and the water behind the bore is 3 m deep. Estimate (a) the propagation speed of the bore and (b) the induced water velocity.
P10.87 A tidal bore may occur when the ocean tide enters an es-tuary against an oncoming river discharge, such as on the Severn River in England. Suppose that the tidal bore is 10 ft deep and propagates at 13 mi/h upstream into a river which is 7 ft deep. Estimate the river current in kn.
P10.88 At one point in a rectangular channel 7 ft wide, the depth is 2 ft and the flow rate is 200 ft3/s. If a hydraulic jump occurs nearby, determine (a) whether it is upstream or downstream of this point and (b) the percentage of dis-sipation in the jump.
P10.89 Water 30 cm deep is in uniform flow down a 1° unfin-ished-concrete slope when a hydraulic jump occurs, as in Fig. P10.89. If the channel is very wide, estimate the wa-ter depth y2 downstream of the jump.
702 Chapter 10 Open-Channel Flow V1 = 2 ft / s y2 Jump y1 y3 P10.84 y1 = 30 cm Jump y2?
Unfinished concrete, 1° slope P10.89 P10.90 Modify Prob. 10.89 as follows. Suppose that y2 1 m and y1 30 cm but the channel slope is not equal to 1°.
Determine the proper slope for this condition.
P10.91 No doubt you used the horizontal-jump formula (10.43) to solve Probs. 10.89 and 10.90, which is reasonable since the slope is so small. However, Chow [3, p. 425] points out that hydraulic jumps are higher on sloped channels, due to “the weight of the fluid in the jump.” Make a con-Jump 1 2 3 4 h P10.93 P10.94 Modify Prob. 10.93 as follows. Let the bump height be 20 cm and the subcritical approach velocity be V1 1.5 m/s. Determine (a) y2, (b) the supercritical flow V3, and (c) y4.
P10.95 A 10-cm-high bump in a wide horizontal water channel creates a hydraulic jump just upstream and the flow pat-tern in Fig. P10.95. Neglecting losses except in the jump, for the case y3 30 cm, estimate (a) V4, (b) y4, (c) V1, and (d) y1.
Jump 1 2 3 4 Bump: h = 10 cm P10.95 P10.96 Show that the Froude numbers on either side of a wide hydraulic jump are related by the simple relation Fr2 Fr1(y1/y2)3/2.
no hydraulic jump, compute from gradually varied the-ory the downstream distance where y 2.0 ft.
P10.106 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 b 4 m, L 250 m, the initial depth is y(0) 1.05 m, and the flow rate is 7 m3/s. (b) Then, if your computer program is running well, determine the initial depth y(0) for which the exit flow will be exactly critical.
P10.97 A brickwork rectangular channel 4 m wide is flowing at 8.0 m3/s 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 m3/s per meter of width on a slope of 0.75°. Is this a mild, criti-cal, or steep slope? What type of gradually varied solu-tion curve are we on if the local water depth is (a) 1 m, (b) 2 m, and (c) 3 m?
P10.99 A clay tile V-shaped channel of included angle 60° is flowing at 1.98 m3/s on a slope of 0.33°. Is this a mild, critical, or steep slope? What type of gradually varied so-lution curve are we on if the local water depth is (a) 1 m, (b) 2 m, and (c) 3 m?
P10.100 If bottom friction is included in the sluice-gate flow of Prob. 10.84, the depths (y1, y2, y3) 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 var-ied flow.
P10.101 Consider the gradual change from the profile beginning at point a in Fig. P10.101 on a mild slope S01 to a mild but steeper slope S02 downstream. Sketch and label the curve y(x) expected.
Problems 703 a ?
yn2 yc Mild Mild but steeper yn1 yc P10.101 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 which are expected for gradually varied flow.
P10.103 A circular painted-steel channel is running half full at 1 m3/s and is laid out on a slope of 5 m/km. Is this a mild or steep slope, of type 1, 2, or 3? Take R 50 cm.
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 which are expected for gradually varied flow.
P10.105 In Prob. 10.84 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 a b Steep Steeper yc yn1 yn2 P10.102 a b Steep 50% increase in channel width yc1 yn1 yc2 yn2 P10.104 b x x = 0 x = L 2 b P10.106 P10.113 Figure P10.113 shows a channel contraction section of-ten called a venturi flume [19, p. 167], because mea-surements of y1 and y2 can be used to meter the flow rate.
Show that if losses are neglected and the flow is one-di-mensional and subcritical, the flow rate is given by Q 1/2 Apply this to the special case b1 3 m, b2 2 m, and y1 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.
2g(y1 y2) 1/(b2 2y2 2) 1/(b2 1y2 1) P10.107 A clean-earth wide-channel flow is flowing up an adverse slope with S0 0.002. If the flow rate is q 4.5 m3/(s m), use gradually varied theory to compute the distance for the depth to drop from 3.0 to 2.0 m.
P10.108 Illustrate Prob. 10.104 with a numerical example. Let the channel be rectangular with a width b1 10 m for 0 x 100 m, expanding to b2 15 m for 100 x 250 m. The flow rate is 27 m3/s, and n 0.012. Com-pute the water depth at x 250 m for initial depth y(0) equal to (a) 75 cm and (b) 5 cm. Compare your results with the discussion in Prob. 10.104. Let S0 0.005.
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 yc and the edge the flow is rapidly varied and does not sat-isfy gradually varied theory. Suppose that the flow rate is q 1.3 m3/(s m) and the surface is unfinished ce-ment. Use Eq. (10.51) to estimate the water depth 300 m upstream as shown.
704 Chapter 10 Open-Channel Flow 300 m S0 = 0.06° y?
yc P10.109 P10.110 We assumed frictionless flow in solving the bump case, Prob. 10.65, for which V2 1.21 m/s and y2 0.826 m over the crest when hmax 15 cm, V1 1 m/s, and y1 1 m. However, if the bump is long and rough, friction may be important. Repeat Prob. 10.65 for the same bump shape, h 0.5hmax[1 cos (2x/L)], to compute conditions (a) at the crest and (b) at the end of the bump, x L. Let hmax 15 cm and L 100 m, and assume a clean-earth surface.
P10.111 Modify Prob. 10.110 as follows. Keep all other data the same, and find the bump length L for which the flow first becomes critical somewhere along the bump surface.
P10.112 The clean-earth channel in Fig. P10.112 is 6 m wide and slopes at 0.3°. Water flows at 30 m3/s 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?
Reservoir 30 m3/s 2 m 3 m L P10.112 b1 b2 y1 y2 Top view Side view P10.113 P10.114 Investigate the possibility of choking in the venturi flume of Fig. P10.113. Let b1 4 ft, b2 3 ft, and y1 2 ft.
Compute the values of y2 and V1 for a flow rate of (a) 30 ft3/s and (b) 35 ft3/s. Explain your vexation.
P10.115 Gradually varied theory, Eq. (10.49), neglects the effect of width changes, db/dx, 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 rec-tangular section with b b(x), Eq. (10.49) should be modified as follows: d d y x S0 S V2/(gb) 1 Fr2 P10.122 In 1952 E. S Crump developed the triangular weir shape shown in Fig. P10.122 [19, 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 de-sign is that it has a unique discharge correlation up to near-drowning conditions, H2/H1 0.75: Q Cdbg1/2H1 V 2g 2 1 kh 3/2 where Cd 0.63 and kh 0.3 mm The term kh is a low-head loss factor. Suppose that the weir is 3 m wide and has a crest height Y 50 cm. If the water depth upstream is 65 cm, estimate the flow rate in gal/min.
Investigate a criterion for reducing this relation to Eq.
(10.49).
P10.116 Investigate the possibility of frictional effects in the venturi flume of Prob. 10.113, part (a), for which the frictionless solution is Q 9.88 m3/s. Let the contraction be 3 m long and the measurements of y1 and y2 be at positions 3 m up-stream and 3 m downstream of the contraction, respectively.
Use the modified gradually varied theory of Prob. 10.115, with n 0.018 to estimate the flow rate.
P10.117 A full-width weir in a horizontal channel is 5 m wide and 80 cm high. The upstream depth is 1.5 m. Estimate the flow rate for (a) a sharp-crested weir and (b) a round-nosed broad-crested weir.
P10.118 Using a Bernoulli-type analysis similar to Fig. 10.16a, show that the theoretical discharge of the V-shaped weir in Fig. P10.118 is given by Q 0.7542g1/2 tan H5/2 Problems 705 α H α P10.118 P10.119 Data by A. T. Lenz for water at 20°C (reported in Ref.
19) show a significant increase of discharge coefficient of V-notch weirs (Fig. P10.118) at low heads. For 20°, some measured values are as follows: H, ft 0.2 0.4 0.6 0.8 1.0 Cd 0.499 0.470 0.461 0.456 0.452 Determine if these data can be correlated with the Reynolds and Weber numbers vis-à-vis Eq. (10.61). If not, suggest another correlation.
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 m3/s 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 ?
P10.121 Water flow in a rectangular channel is to be metered by a thin-plate weir with side contractions, as in Table 10.1b, 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 ap-propriate length for the weir width b?
Flow 2 m Y P10.120 Flow 1:2 slope Hydraulic jump 1:5 slope H2 Y H1 P10.122 The Crump weir [19, chap. 4] P10.123 The Crump weir calibration in Prob. 10.122 is for mod-ular flow, i.e., when the flow rate is independent of down-stream tailwater. When the weir becomes drowned, the flow rate decreases by the following factor: Q Qmod f where f 1.0350.817 H H 1 2 4 0.0647 for 0.70 H 2/H 1 0.93, where H denotes H1 V2 1/(2g) kh for brevity. The weir is then double-gaged to measure both H1 and H2. Suppose that the weir crest is 1 m high and 2 m wide. If the measured upstream and downstream water depths are 2.0 and 1.9 m, respectively, estimate the flow rate in gal/min. Comment on the pos-sible uncertainty of your estimate.
slope of 0.15°. If the V-notch weir has 30° and Y 50 cm, estimate, from gradually varied theory, the water depth 100 m upstream.
P10.127 A horizontal gravelly earth channel 2 m wide contains a full-width Crump weir (Fig. P10.122) 1 m high. If the weir is not drowned, estimate, from gradually varied the-ory, the flow rate for which the water depth 100 m up-stream will be 2 m.
P10.128 A rectangular channel 4 m wide is blocked by a broad-crested weir 2 m high, as in Fig. P10.128. The channel is horizontal for 200 m upstream and then slopes at 0.7° as shown. The flow rate is 12 m3/s, and n 0.03. Com-pute the water depth y at 300 m upstream from gradually varied theory.
P10.124 Water flows at 600 ft3/s 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?
706 Chapter 10 Open-Channel Flow Backwater curve 10 ft 15 ft L = ?
P10.124 P10.125 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 rectangular channel 100 ft wide with a flow rate of 800 ft3/s. Estimate the water-depth 2 mi upstream of the dam if S0 0.001.
P10.126 Suppose that the rectangular channel of Fig. P10.120 is made of riveted steel and carries a flow of 8 m3/s on a y?
y ( x) Slope 0.7° 12 m 3/s 200 m 100 m P10.128 Word Problems W10.1 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, ei-ther 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 Chézy equation (10.13)?
W10.4 A shallow-water wave propagates at the speed c0 (gy)1/2. What makes it propagate? That is, what is the bal-ance 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.
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.
W10.10 Cite some similarities, and perhaps some dissimilarities, between a weir and a Bernoulli obstruction flowmeter from Sec. 6.7.
W10.11 Is a bump, Fig. 10.9a, 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 de-sign 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. 10.113?
EES FE10.4 For the channel of Prob. FE10.1, if the water depth is 2 m and the uniform-flow rate is 24 m3/s, what is the ap-proximate 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. FE10.1, if Manning’s roughness factor n 0.020 and Q 29 m3/s, what is the normal depth yn?
(a) 1 m, (b) 1.5 m, (c) 2 m, (d) 2.5 m, (e) 3 m FE10.6 For the channel of Prob. FE10.1, if Q 24 m3/s, what is the critical depth yc?
(a) 1.0 m, (b) 1.26 m, (c) 1.5 m, (d) 1.87 m, (e) 2.0 m FE10.7 For the channel of Prob. FE10.1, if Q 24 m3/s 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 Fundamentals of Engineering Exam Problems The FE Exam is fairly light on open-channel problems in the gen-eral (morning) session, but it plays a big part in the specialized civil engineering (afternoon) exam.
FE10.1 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 For the channel of Prob. FE10.1, the most efficient wa-ter 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. FE10.1 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 m3/s, (b) 18 m3/s, (c) 36 m3/s, (d) 40 m3/s, (e) 53 m3/s Design Projects 707 Comprehensive Problems C10.1 February 1998 saw the failure of the earthen dam im-pounding California Jim’s Pond in southern Rhode Island.
The resulting flood raised temporary havoc in the nearby village 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 re-quired for the pond to drain to a depth of 2 ft.
C10.2 A circular, unfinished concrete drainpipe is laid on a slope of 0.0025 and is planned to carry from 50 to 300 ft3/s of runoff water. Design constraints are that (1) the water depth should be no more than three-fourths of the diameter and (2) the flow should always be subcritical. What is the ap-propriate pipe diameter to satisfy these requirements? If no commercial pipe is exactly this calculated size, should you buy the next smallest or the next largest pipe? C10.3 Extend Prob. 10.72, whose solution was V2 4.33 m/s. Use gradually varied theory to estimate the water depth 10 m downstream at section (3) for (a) the 5° unfinished-concrete slope shown in Fig. P10.72. (b) Repeat your calculation for an upward (adverse) slope of 5°. (c) When you find that part (b) is impossible with gradually varied theory, explain why and repeat for an adverse slope of 1°.
Design Projects D10.1 A straight weedy-earth channel has the trapezoidal shape of Fig. 10.7, with b 4 m and 35°. The channel has a con-stant bottom slope of 0.001. The flow rate varies seasonally from 5 up to 10 m3/s. It is desired to place a sharp-edged weir across the channel so that the water depth 1 km up-stream remains at 2.0 m 10 percent throughout the year.
Investigate 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 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 ft3/s, with a 30-year predicted flood rate of 1200 ft3/s. 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 conditions. You, as project engineer, have to predict how high the dam crest can be raised and still meet this re-quirement.
13.
B. A. Bakhmeteff, Hydraulics of Open Channels, McGraw-Hill, New York, 1932.
14.
P. A. Thompson, Compressible-Fluid Dynamics, McGraw-Hill, New York, 1972.
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E. F. Brater, Handbook of Hydraulics, 6th ed., McGraw-Hill, New York, 1976.
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U.S. Bureau of Reclamation, “Research Studies on Stilling Basins, Energy Dissipators, and Associated Appurtenances,” Hydraulic Lab. Rep. Hyd-399, June 1, 1955.
17.
“Friction Factors in Open Channels, Report of the Commit-tee on Hydromechanics,” ASCE J. Hydraul. Div., March 1963, pp. 97–143.
18.
R. M. Olson and S. J. Wright, Essentials of Engineering Fluid Mechanics, 5th ed., Harper & Row, New York, 1990.
19.
P. Ackers et al., Weirs and Flumes for Flow Measurement, Wiley, New York, 1978.
20.
M. G. Bos, J. A. Replogle, and A. J. Clemmens, Flow Mea-suring Flumes for Open Channel Systems, Wiley, New York, 1984.
21.
M. G. Bos, Long-Throated Flumes and Broad-Crested Weirs, Martinus Nijhoff (Kluwer), Dordrecht, The Netherlands, 1985.
22.
U.S. Army Corps of Engineers, HEC-2 Water Surface Pro-files, Hydrologic Engineering Center, 609 Second Street, Davis, CA.
23.
W. Rodi, Turbulence Models and Their Application in Hy-draulics, Brookfield Publishing, Brookfield, VT, 1984.
References 1.
R. H. French, Open-Channel Hydraulics, McGraw-Hill, New York, 1985.
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M. H. Chaudhry, Open Channel Flow, Prentice-Hall, Upper Saddle River, NJ, 1993.
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Ven Te Chow, Open Channel Hydraulics, McGraw-Hill, New York, 1959.
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F. M. Henderson, Open Channel Flow, Macmillan, New York, 1966.
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B. Kinsman, Wind Waves: Their Generation and Propagation on the Ocean Surface, Prentice-Hall, Englewood Cliffs, NJ, 1964; Dover, New York, 1984.
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M. J. Lighthill, Waves in Fluids, Cambridge University Press, London, 1978.
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C. C. Mei, The Applied Dynamics of Ocean Surface Waves, Wiley, New York, 1983.
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A. T. Ippen, Estuary and Coastline Hydrodynamics, Mc-Graw-Hill, New York, 1966.
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R. M. Sorenson, Basic Coastal Engineering, Wiley, New York, 1978.
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J. M. Robertson and H. Rouse, “The Four Regimes of Open Channel Flow,” Civ. Eng., vol. 11, no. 3, March 1941, pp.
169–171.
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R. W. Powell, “Resistance to Flow in Rough Channels,” Trans. Am. Geophys. Union, vol. 31, no. 4, August 1950, pp.
575–582.
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R. Manning, “On the Flow of Water in Open Channels and Pipes,” Trans. I.C.E. Ireland, vol. 20, 1891, pp. 161–207.
708 Chapter 10 Open-Channel Flow 710 Wind-energy generator farm, Altamont Pass, Cali-fornia. Windmills have been used for power for more than two thousand years. These multi-bladed horizontal-axis wind turbines (HAWT’s) are among the most efficient of windpower designs, as discussed in this chapter. (Courtesy of Kevin Schafer/Peter Arnold, Inc.) 11.1 Introduction and Classification Motivation.
The most common practical engineering application for fluid mechanics is the design of fluid machinery. The most numerous types are machines which add energy to the fluid (the pump family), but also important are those which extract en-ergy (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 per-formance of fluid machines. The emphasis will be upon nearly incompressible flow, i.e., liquids or low-velocity gases. Basic flow principles are discussed, but not the de-tailed construction of the machine.
Turbomachines divide naturally into those which add energy (pumps) and those which 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 Ro-mans in 70 B.C., and Babylonian windmills date back to 700 B.C. .
Machines which deliver liquids are simply called pumps, but if gases are involved, three different terms are in use, depending upon 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 (PDPs) 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 many mechanical designs are in wide use. The text by Warring gives an excellent sum-mary of PDPs. A brief classification of PDP designs is as follows: 711 Chapter 11 Turbomachinery Classification of Pumps 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 PDPs 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 PDPs.
It is rare for such devices to be run backward, so to speak, as turbines or energy ex-tractors, the steam engine (reciprocating piston) being a classic exception.
Since PDPs 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 rea-son. Sturdy construction is required, and complete shutoff would cause damage if pres-sure-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 mo-mentum 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, 11, 14, 31].
Dynamic pumps generally provide a higher flow rate than PDPs and a much stead-ier discharge but are ineffective in handling high-viscosity liquids. Dynamic pumps also generally need priming; i.e., if they are filled with gas, they cannot suck up a liq-uid from below into their inlet. The PDP, on the other hand, is self-priming for most 712 Chapter 11 Turbomachinery Fig. 11.1 Schematic design of posi-tive-displacement pumps: (a) recip-rocating piston or plunger, (b) ex-ternal gear pump, (c) double-screw pump, (d) sliding vane, (e) three-lobe pump, ( f ) double circumfer-ential piston, (g) flexible-tube squeegee.
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, a PDP can operate up to very high pressures (300 atm) but typically produces low flow rates (100 gal/min).
The relative performance (p 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 con-11.1 Introduction and Classification 713 Plunger Suction pipe Packing Discharge pipe Discharge check valve Liquid cylinder Suction check valve Motion Suction Discharge (a) (b) (c) (d) (e) ( f) (g) Fig. 11.2 Comparison of perfor-mance curves of typical dynamic and positive-displacement pumps at constant speed.
Fig. 11.3 Cutaway schematic of a typical centrifugal pump.
stant 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 displacement or the speed.
The reliable constant-speed discharge from PDPs has led to their wide use in meter-ing flows .
The dynamic pump, by contrast in Fig. 11.2, provides a continuous constant-speed variation of performance, from near-maximum p at zero flow (shutoff conditions) to zero p at maximum flow rate. High-viscosity fluids sharply degrade the performance 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], fans [17 to 20], compressors [21 to 23], turbines [24 to 28], and PDPs [35 to 38]. There are several useful handbooks [29 to 32], and at least two elementary textbooks [33, 34] have a comprehensive dis-cussion 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 714 Chapter 11 Turbomachinery 0 Pressure rise or head increase Dynamic pump Positive displacement pump Discharge µ Low µ High µ Low µ High 1 2 Casing Impeller Expanding area scroll 11.2 The Centrifugal Pump Basic Output Parameters 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 dough-nut-shaped diffuser, or scroll, section of the casing decelerates the flow and further in-creases 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.
The blades may be open, i.e., separated from the front casing only by a narrow clear-ance, or closed, i.e., 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.
Assuming steady flow, the pump basically increases the Bernoulli head of the flow be-tween point 1, the eye, and point 2, the exit. From Eq. (3.67), neglecting viscous work and heat transfer, this change is denoted by H: H z2 z1 hs hf (11.1) where hs is the pump head supplied and hf the losses. The net head H is a primary out-put 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 V1 are about the same, z2 z1 is no more than a meter or so, and the net pump head is essentially equal to the change in pressure head H (11.2) The power delivered to the fluid simply equals the specific weight times the discharge times the net head change Pw gQH (11.3) This is traditionally called the water horsepower. The power required to drive the pump is the brake horsepower1 bhp T (11.4) where is the shaft angular velocity and T the shaft torque. If there were no losses, Pw and brake horsepower would be equal, but of course Pw is actually less, and the ef-ficiency of the pump is defined as (11.5) The chief aim of the pump designer is to make as high as possible over as broad a range of discharge Q as possible.
gQH T Pw bhp p g p2 p1 g V2 2g p g V2 2g p g 11.2 The Centrifugal Pump 715 1 Conversion factors may be needed: 1 hp 550 ft lbf/s 746 W.
Elementary Pump Theory The efficiency is basically composed of three parts: volumetric, hydraulic, and me-chanical. The volumetric efficiency is (11.6) where QL is the loss of fluid due to leakage in the impeller-casing clearances. The hy-draulic efficiency is h 1 (11.7) where hf has three parts: (1) shock loss at the eye due to imperfect match between in-let flow and the blade entrances, (2) friction losses in the blade passages, and (3) cir-culation loss due to imperfect match at the exit side of the blades.
Finally, the mechanical efficiency is m 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 hm (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 predic-tive 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) com-plex digital-computer models which 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 com-putational 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-dimen-sional flow and combine idealized fluid-velocity vectors through the impeller with the angular-momentum theorem for a control volume, Eq. (3.55).
The idealized velocity diagrams are shown in Fig. 11.4. The fluid is assumed to en-ter the impeller at r r1 with velocity component w1 tangent to the blade angle 1 plus circumferential speed u1 r1 matching the tip speed of the impeller. Its absolute entrance velocity is thus the vector sum of w1 and u1, shown as V1. Similarly, the flow exits at r r2 with component w2 parallel to the blade angle 2 plus tip speed u2 r2, with resultant velocity V2.
We applied the angular-momentum theorem to a turbomachine in Example 3.14 (Fig. 3.13) and arrived at a result for the applied torque T T Q(r2Vt2 r1Vt1) (11.10) Pf bhp hf hs Q Q QL 716 Chapter 11 Turbomachinery Fig. 11.4 Inlet and exit velocity diagrams for an idealized pump impeller.
where Vt1 and Vt2 are the absolute circumferential velocity components of the flow.
The power delivered to the fluid is thus Pw T Q(u2Vt2 u1Vt1) (11.11) or H (u2Vt2 u1Vt1) These are the Euler turbomachine equations, showing that the torque, power, and ideal head are functions only of the rotor-tip velocities u1,2 and the absolute fluid tangential velocities Vt1,2, 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 V2 u2 w2 2uw cos w cos u Vt or uVt 1 2 (V2 u2 w2) (11.12) Substituting this into Eq. (11.11) gives H [(V2 2 V2 1) (u2 2 u2 1) (w2 2 w2 1)] (11.13) 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 z const (11.14) This is the Bernoulli equation in rotating coordinates and applies to either two- or three-dimensional ideal incompressible flow.
r22 2g w2 2g p g 1 2g 1 g Pw gQ 11.2 The Centrifugal Pump 717 V2 Vt 2 w2 Vn 2 Impeller w1 V1 Vt 1 Vn1 r1 r2 Blade ω 2 β 2 α u 1 = r1 ω 1 β 1 α u 2 = r2 ω Part (a) For a centrifugal pump, the power can be related to the radial velocity Vn Vt tan and the continuity relation Pw Q(u2Vn2 cot 2 u1Vn1 cot 1) (11.15) where Vn2 and Vn1 and where b1 and b2 are the blade widths at inlet and exit. With the pump parameters r1, r2, 1, 2, and known, Eqs. (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 1 90° Vn1 V1 (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: r1 4 in, r2 7 in, 1 30°, 2 20°, speed 1440 r/min. Estimate (a) the design-point discharge, (b) the water horsepower, and (c) the head if b1 b2 1.75 in.
Solution The angular velocity is 2 r/s 2(1440/60) 150.8 rad/s. Thus the tip speeds are u1 r1 150.8(4/12) 50.3 ft/s and u2 r2 150.8(7/12) 88.0 ft/s. From the inlet-velocity diagram, Fig. E11.1a, with 1 90° for design point, we compute Vn1 u1 tan 30° 29.0 ft/s whence the discharge is Q 2r1b1Vn1 (2) (29.0) (8.87 ft3/s)(60 s/min) gal/ft3 3980 gal/min Ans. (a) (The actual pump produces about 3500 gal/min.) The outlet radial velocity follows from Q Vn2 16.6 ft/s This enables us to construct the outlet-velocity diagram as in Fig. E11.1b, given 2 20°. The tangential component is Vt2 u2 Vn2 cot 2 88.0 16.6 cot 20° 42.4 ft/s 2 tan1 21.4° 16.6 42.4 8.87 2( 1 7 2 )(1.75/12) Q 2r2b2 1728 231 1.75 12 4 12 Q 2r1b1 Q 2r2b2 718 Chapter 11 Turbomachinery Part (b) E11.1b E11.1a V2 20° 16.6 ft /s 2 α 88.0 ft /s V1 90° 30° u1 = 50.3 ft/s Effect of Blade Angle on Pump Head Fig. 11.5 Theoretical effect of blade exit angle on pump head versus discharge.
The power is then computed from Eq. (11.11) with Vt1 0 at the design point Pw Qu2Vt2 (1.94 slugs/ft3)(8.87 ft3/s)(88.0 ft/s)(42.4 ft/s) 117 hp Ans. (b) (The actual pump delivers about 125 water horsepower, requiring 147 bhp at 85 percent effi-ciency.) Finally, the head is estimated from Eq. (11.11) H 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, 6, 8, and 31].
The simple theory above can be used to predict an important blade-angle effect. If we neglect inlet angular momentum, the theoretical water horsepower is Pw Qu2Vt2 (11.17) where Vt2 u2 Vn2 cot 2 Vn2 Then the theoretical head from Eq. (11.11) becomes H Q (11.18) The head varies linearly with discharge Q, having a shutoff value u2 2/g, where u2 is the exit blade-tip speed. The slope is negative if 2 90° (backward-curved blades) and positive for 2 90° (forward-curved blades). This effect is shown in Fig. 11.5 and is accurate only at low flow rates.
u2 cot 2 2r2b2g u2 2 g Q 2r2b2 64,100 ft lbf/s (62.4 lbf/ft3)(8.87 ft3/s) Pw gQ 64,100 ft lbf/s 550 11.2 The Centrifugal Pump 719 Head H Unstable: Can cause pump surge Discharge Q 2 > 90° (Forward-curved) β 2 = 90° (Radial blades) β 2 < 90° (Backward-curved) β Part (c) 11.3 Pump Performance Curves and Similarity Rules The measured shutoff head of centrifugal pumps is only about 60 percent of the the-oretical value H0 2r2 2 /g. With the advent of the laser-doppler anemometer, re-searchers can now make detailed three-dimensional flow measurements inside pumps and can even animate the data into a movie .
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 problem in gas-compressor operation. For this reason a backward-curved or radial blade design is gen-erally 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 ft3/min for gases). The dependent variables, or “out-put,’’ are taken to be head H (pressure rise p for gases), brake horsepower (bhp), and efficiency .
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 Qmax. The pos-itive-slope part of the head is shown dashed; as mentioned earlier, this region can be unstable and can cause hunting for the operating point.
720 Chapter 11 Turbomachinery Positive slope may be unstable for certain system loss curves Best efficiency point (BEP) or design point Head Horsepower Efficiency Effect of cavitation or gas entrainment on liquid heads 0 Flow rate Q Q Qmax 0 Fig. 11.6 Typical centrifugal pump performance curves at constant im-peller-rotation speed. The units are arbitrary.
Measured Performance Curves Net Positive-Suction Head The efficiency is always zero at no flow and at Qmax, and it reaches a maximum, perhaps 80 to 90 percent, at about 0.6Qmax. This is the design flow rate Q or best ef-ficiency point (BEP), max. 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 max, so that a wide range of efficient operation is achieved. However, some designs simply do not achieve flat efficiency curves. Note that is not independent of H and P but rather is calculated from the relation in Eq. (11.5), gQH/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 BEP, especially for radial-tipped and forward-curved blades. This is considered unde-sirable because a much larger motor is then needed to provide high flow rates. Back-ward-curved blades typically have their horsepower level off above BEP (“nonover-loading’’ type of curve).
Figure 11.7 shows actual performance data for a commercial centrifugal pump. Figure 11.7a 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 oper-ating range near the BEP. Everything is plotted raw, of course [feet, horsepower, gal-lons per minute (1 U.S. gal 231 in3)] since it is to be used directly by designers. Fig-ure 11.7b 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. 11.7 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 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 ) due to the large change in viscosity (Reynolds num-ber). Again this should become clear with the similarity rules.
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 first occur. The NPSH is defined as NPSH (11.19) where pi and Vi are the pressure and velocity at the pump inlet and p is the vapor pres-sure 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.
p g V i 2 2g pi g 11.3 Pump Performance Curves and Similarity Rules 721 Fig. 11.7 Measured-performance curves for two models of a cen-trifugal water pump: (a) basic cas-ing with three impeller sizes; (b) 20 percent larger casing with three larger impellers at slower speed.
(Courtesy of Ingersoll-Rand Corpo-ration, Cameron Pump Division.) If the pump inlet is placed at a height Zi above a reservoir whose free surface is at pressure pa, we can use Bernoulli’s equation to rewrite NPSH as NPSH Zi hfi (11.20) where hfi is the friction-head loss between the reservoir and the pump inlet. Knowing pa and hfi, we can set the pump at a height Zi which will keep the right-hand side greater than the “required’’ NPSH plotted in Fig. 11.7.
pv g pa g 722 Chapter 11 Turbomachinery 50 Total head, ft 700 600 500 400 300 200 U.S. gallons per minute × 1000 (a) 40 20 NPSH, ft Total head, ft 400 300 250 200 150 0 4 8 12 16 20 24 28 U.S. gallons per minute × 1000 (b) NPSH 350 100 0 4 8 12 16 20 24 28 30 25 20 10 NPSH, ft 15 800 n = 1170 r/min 364 3 -in dia.
32-in dia.
65% 72% 78% 82% 85% 88% 87% 3500 bhp 3000 bhp 87% 2500 bhp 2000 bhp 1500 bhp n = 710 r/min 2 411-in dia.
38-in dia.
28-in dia.
NPSH 60% 72% 80% 84% 86% 88% 35-in dia.
89% 86% 84% 88% 1500 bhp 1250 bhp 1000 bhp Deviations from Ideal Pump Theory 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 deteriora-tion 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, e.g., the 36.75-in-diameter pump at 1170 r/min in Fig. 11.7a. The the-oretical shutoff head is H0(ideal) 1093 ft 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. This is a sharp dropoff and is indicative of nonrecov-erable 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 direc-tion, especially significant at high flow rates These are complicated three-dimensional-flow effects and hence are difficult to pre-dict. Although, as mentioned, numerical (CFD) techniques are becoming more impor-tant , modern performance prediction is still a blend of experience, empirical cor-relations, 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 reser-voir whose surface is at 14.7 lbf/in2 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, p 0.26 lbf/in2 absolute, SG 1.0 and (b) 200°F, p 11.52 lbf/in2 absolute, SG 0.9635?
Solution For either case read from Fig. 11.7a at 24,000 gal/min that the required NPSH is 40 ft. For this case g 62.4 lbf/ft3. From Eq. (11.20) it is necessary that NPSH pa g p Zi hfi or 40 ft Zi 6.0 or Zi 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 g 62.4(0.9635) 60.1 lbf/ft3. Equation (11.20) applies again with the higher p 40 ft Zi 6.0 (14.7 11.52)(144) 60.1 (14.7 0.26)(144) 62.4 [1170(2/60) rad/s]2[(36.75/2)/(12) ft]2 32.2 ft/s2 2r2 2 g 11.3 Pump Performance Curves and Similarity Rules 723 Part (a) Part (b) Dimensionless Pump Performance or Zi 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 de-pendent upon discharge Q, impeller diameter D, and shaft speed n, at least. Other pos-sible parameters are the fluid density , viscosity , and surface roughness . Thus the performance curves in Fig. 11.7 are equivalent to the following assumed functional re-lations:2 gH f1(Q, D, n, , , ) bhp f2(Q, D, n, , , ) (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 (Prob. 5.20). For each function in Eq.
(11.21) there are seven variables and three primary dimensions (M, L, and T); hence we expect 7 3 4 dimensionless pis, and that is what we get. You can verify as an exercise that appropriate dimensionless forms for Eqs. (11.21) are g1 , , (11.22) g2 , , The quantities nD2/ and /D are recognized as the Reynolds number and roughness ratio, respectively. Three new pump parameters have arisen: Capacity coefficient CQ Head coefficient CH (11.23) Power coefficient CP Note that only the power coefficient contains fluid density, the parameters CQ and CH being kinematic types.
Figure 11.7 gives no warning of viscous or roughness effects. The Reynolds num-bers are from 0.8 to 1.5 107, 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 rough-ness ratio have a constant effect, so that Eqs. (11.23) reduce to, approximately, CH CH(CQ) CP CP(CQ) (11.24) bhp n3D5 gH n2D2 Q nD3 D nD2 Q nD3 bhp n3D5 D nD2 Q nD3 gH n2D2 724 Chapter 11 Turbomachinery 2 We adopt gH as a variable instead of H for dimensional reasons.
Fig. 11.8 Nondimensional plot of the pump performance data from Fig. 11.7. These numbers are not representative of other pump de-signs.
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 im-pellers 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; e.g., a factor-of-3 or more viscosity increase causes a clearly visible effect on CH and CP.
The efficiency is already dimensionless and is uniquely related to the other three.
It varies with CQ also (CQ) (11.25) We can test Eqs. (11.24) and (11.25) from the data of Fig. 11.7. The impeller diame-ters 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, CH, and CP are computed with n in r/s, Q in ft3/s (gal/min 2.23 103), H and D in ft, g 32.2 ft/s2, 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 CHS(CQ) (11.26) g(NPSH) n2D2 CHCQ CP 11.3 Pump Performance Curves and Similarity Rules 725 7 6 5 4 3 2 1 CH CQ 0.9 0.8 0.7 0.6 0.7 0.6 0.5 0.4 0.3 CP CHS η D = 38 in D = 32 in CH CHS CP η 0 0 0.05 0.1 0.15 0.2 0.25 0.8 1.0 Part (a) Part (b) The coefficients CP and CHS are seen to correlate almost perfectly into a single func-tion of CQ, while and CH 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 de-velops slightly more head and is more efficient. The overall effect is a resounding vic-tory for dimensional analysis.
The best-efficiency point in Fig. 11.8 is approximately CQ 0.115 CP 0.65 (11.27) max 0.88: 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 CH(0) 6.0, and by extrapolation the shutoff power is CP(0) 0.25 and the maximum discharge is CQ,max 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 n2D2, nD3, and n3D5 for two pumps in Fig. 11.7 we can see readily why the large pump had the same discharge but less power and head: Discharge Head Power D, ft n, r/s nD3, ft3/s n2D2/g, ft n3D5/550, hp Fig. 11.7a 32/12 1170/60 370 0.84 3527 Fig. 11.7b 38/12 710/60 376 0.44 1861 Ratio — — 1.02 0.52 0.53 Discharge goes as nD3, which is about the same for both pumps. Head goes as n2D2 and power as n3D5 for the same (water), and these are about half as much for the larger pump. The NPSH goes as n2D2 and is also half as much for the 38-in pump.
EXAMPLE 11.3 A pump from the family of Fig. 11.8 has D 21 in and n 1500 r/min. Estimate (a) discharge, (b) head, (c) pressure rise, and (d) brake horsepower of this pump for water at 60°F and best ef-ficiency.
Solution In BG units take D 21/12 1.75 ft and n 1500/60 25 r/s. At 60°F, of water is 1.94 slugs/ft3. The BEP parameters are known from Fig. 11.8 or Eqs. (11.27). The BEP discharge is thus Q CQnD3 0.115(25)(1.75)3 (15.4 ft3/s)(448.8) 6900 gal/min Ans. (a) Similarly, the BEP head is H 300-ft water Ans. (b) 5.0(25)2(1.75)2 32.2 CHn2D2 g 726 Chapter 11 Turbomachinery Part (c) Part (d) Part (a) Part (c) Part (d) Part (b) Similarity Rules Since we are not given elevation or velocity-head changes across the pump, we neglect them and estimate p gH 1.94(32.2)(300) 18,600 lbf/ft2 129 lbf/in2 Ans. (c) Finally, the BEP power is P CPn3D5 0.65(1.94)(25)3(1.75)5 590 hp Ans. (d) EXAMPLE 11.4 We want to build a pump from the family of Fig. 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.
Solution 3000 gal/min 6.68 ft3/s and 1200 r/min 20 r/s. At BEP we have Q CQnD3 6.68 ft3/s (0.115)(20)D3 or D 1/3 1.43 ft 17.1 in Ans. (a) The maximum Q is related to Q by a ratio of capacity coefficients Qmax 6000 gal/min Ans. (b) From Fig. 11.8 we estimated the shutoff head coefficient to be 6.0. Thus H(0) 152 ft Ans. (c) Finally, from Eq. (11.27), the NPSH at BEP is approximately NPSH 9.4 ft Ans. (d) Since this a small pump, it will be less efficient than the pumps in Fig. 11.8, 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: 3 2 2 D2 D1 n2 n1 H2 H1 D2 D1 n2 n1 Q2 Q1 0.37(20)2(1.43)2 32.2 CHSn2D2 g 6.0(20)2(1.43)2 32.2 CH(0)n2D2 g 3000(0.23) 0.115 QCQ,max CQ 6.68 0.115(20) 323,000 ft lbf/s 550 11.3 Pump Performance Curves and Similarity Rules 727 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.
3 5 (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 geo-metrically similar family. A graphic display of these rules is given in Fig. 11.9, show-ing the effect of speed and diameter changes on pump performance. In Fig. 11.9a the size is held constant and the speed is varied 20 percent, while Fig. 11.9b 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. 11.9b) is even more dramatic, especially for power, which varies as D5. 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 1 2, 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 max-imum 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 : 1/4 (11.29a) Flow-rate changes : 0.32 (11.29b) Anderson’s formula (11.29b) 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 Q1 Q2 0.94 2 0.94 1 D1 D2 1 2 1 1 D2 D1 n2 n1 2 1 P2 P1 728 Chapter 11 Turbomachinery Q Q H, bhp D = 10 = constant H n = 10 bhp n = 12 n = 8 0 (a) bhp H D = 12 D = 10 D = 8 (b) n = 10 = constant H, bhp 0 Effect of Viscosity 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed Part (a) for turbines if the constant 0.94 is replaced by 0.95. The formulas in Eq. (11.29) as-sume the same value of surface roughness for both machines—one could micropolish 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: /water 1.0 10.0 100 1000 max, % 85 76 52 11 Beyond about 300water 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 formi-dable device, able to deliver very high heads and reasonable flow rates with excellent efficiency. It can match many system requirements. But basically the centrifugal pump is a high-head, low-flow machine, whereas there are many applications requiring 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. (11.27): H 25 ft CH g n2D2 5.
3 0 2 n .
2 2 D2 Q 100,000 gal/min 222.8 ft3/s CQnD3 0.115nD3 The two unknowns are n and D. Solve simultaneously for D 12.4 ft n 1.03 r/s 62 r/min Ans. (a) If you wish to avoid algebraic manipulation, simply program the above two simultaneous equa-tions in EES, using English units: 25 5.0n^2D^2/32.2 222.8 0.115nD^3 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 729 EES Part (b) Specify in Variable Information that n and D are positive, and EES promptly returns the correct solution: D 12.36 ft and n 1.027 r/s.
The most efficient horsepower is then, from Eq. (11.27), bhp CPn3D5 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.
There are other dynamic-pump designs which do provide low head and high discharge.
For example, there is a type of 38-in, 710 r/min pump, e.g., with the same input parame-ters as Fig. 11.7b, 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 veloc-ity 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 Ns or N s.
0.65(1.94)(1.03)3(12.4)5 550 730 Chapter 11 Turbomachinery 1.0 0.9 0.8 0.7 0.6 500 1000 2000 4000 5000 10,000 15,000 Axial flow Ns r/min (gal/min)1/2/(H, ft)3/4 (a) Low Specific speed High Centrifugal Mixed-flow propeller 1000 2000 4000 5000 10,000–15,000 (b) η max 500 Centrifugal pump Mixed flow 0 H, bhp H Q bhp 10 4 10 3 100 1.0 = 10.0 µ µ water Fig. 11.10 Effect of viscosity on centrifugal pump performance.
Fig. 11.11 (a) Optimum efficiency and (b) vane design of dynamic-pump families as a function of specific speed.
The 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 CH, 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: N s (11.30a) Lazy but common: Ns (11.30b) In other words, practicing engineers do not bother to change n to revolutions per sec-ond 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 Ns 17,182N s Note that Ns is applied only to BEP; thus a single number characterizes an entire fam-ily of pumps. For example, the family of Fig. 11.8 has N s (0.115)1/2/(5.0)3/4 0.1014, Ns 1740, regardless of size or speed.
It turns out that the specific speed is directly related to the most efficient pump de-sign, as shown in Fig. 11.11. Low Ns means low Q and high H, hence a centrifugal pump, and large Ns implies an axial pump. The centrifugal pump is best for Ns between 500 and 4000, the mixed-flow pump for Ns between 4000 and 10,000, and the axial-flow pump for Ns above 10,000. Note the changes in impeller shape as Ns increases.
If we use NPSH rather than H in Eq. (11.30), the result is called suction specific speed Rigorous: N ss (11.31a) Lazy: Nss (11.31b) 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 N ss 0.47 Nss 8100 In the absence of test data, this relation can be used, given n and Q, to estimate the minimum required NPSH.
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.
(r/min)(gal/min)1/2 [NPSH (ft)]3/4 nQ1/2 (g NPSH)3/4 (r/min)(gal/min)1/2 [H (ft)]3/4 n(Q)1/2 (gH)3/4 C1/2 Q C3/4 H 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 731 Suction Specific Speed Axial-Flow Pump Theory Fig. 11.12 Analysis of an axial-flow pump: (a) basic geometry; (b) stator blades and exit-velocity diagram; (c) rotor blades and exit-velocity diagram.
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. Fig-ure 11.12b shows the stator blades and their exit-velocity diagram. Since the stator is fixed, ideally the absolute velocity V1 is parallel to the trailing edge of the blade. After vectorially subtracting the rotor tangential velocity u from V1, we obtain the velocity w1 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 rel-ative velocity w2 is parallel to the blade trailing edge, while the absolute velocity V2 should be designed to enter smoothly the next row of stator blades.
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, u1 u2, and one-dimensional continuity requires that the axial-velocity component remain constant Vn1 Vn2 Vn const Q A 732 Chapter 11 Turbomachinery Rotor w2 V2 Vn2 Vt2 (c) 2 β 2 β 2 α ω u = r u Stator w1 V1 Vn 1 Vt 1 (b) u 1 α 1 α 1 β Stator Flow Rotor (a) , n ω r Performance of an Axial-Flow Pump From the geometry of the velocity diagrams, the normal velocity (or volume flow) can be directly related to the blade rotational speed u: u rav Vn1(tan 1 tan 1) Vn2(tan 2 tan 2) (11.32) Thus the flow rate can be predicted from the rotational speed and the blade angles.
Meanwhile, since Vt1 Vn1 cot 1 and Vt2 u Vn2 cot 2, Euler’s relation (11.11) for the pump head becomes gH uVn(cot 2 cot 1) u2 uVn(cot 1 cot 2) (11.33) the preferred form because it relates to the blade angles 1 and 2. The shutoff or no-flow head is seen to be H0 u2/g, just as in Eq. (11.18) for a centrifugal pump. The blade-angle parameter cot 1 cot 2 can be designed to be negative, zero, or posi-tive, 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 com-plexity 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 pro-peller pump is shown in Fig. 11.13. Note, as expected, the higher CQ and lower CH compared with Fig. 11.8. The head curve drops sharply with discharge, so that 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 sud-11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 733 4 3 2 1 0 0 0.2 0.4 0.6 0.8 CQ CP CH η CH, CP 1.0 0.8 0.6 0.4 0.2 η 0 Fig. 11.13 Dimensionless perfor-mance curves for a typical axial-flow pump, Ns 12,000. Con-structed from data given by Stepanoff for a 14-in pump at 690 r/min.
Fig. 11.14 Optimum efficiency of pumps versus capacity and specific speed. (Adapted from Refs. 4 and 31.) denly 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, CH 1.07, CP 0.70, and max 0.84.
From this we compute N s (0.55)1/2/(1.07)3/4 0.705, Ns 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 D 46 in and n 430 r/min, with bhp 750; this is a much more reasonable design solution, with im-provements still possible at larger-Ns conditions.
Specific speed is such an effective parameter that it is used as an indicator of both per-formance and efficiency. Figure 11.14 shows a correlation of the optimum efficiency of a pump as a function of the specific speed and capacity. Because the dimensional parameter Q is a rough measure of both size and Reynolds number, increases with Q. When this type of correlation was first published by Wislicenus in 1947, it be-came 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.
Figure 11.15 shows the effect of specific speed on the shape of the pump perfor-mance curves, normalized with respect to the BEP point. The numerical values shown are representative but somewhat qualitative. The high-specific-speed pumps (Ns 10,000) have head and power curves which drop sharply with discharge, implying over-load or start-up problems at low flow. Their efficiency curve is very narrow.
734 Chapter 11 Turbomachinery 1.0 0.8 0.6 0 100 300 ∞ η max 100 300 1000 3000 10,000 30,000 Ns 0.2 0.4 Q = 5 gal/min 10 30 1000 10,000 Pump Performance versus Specific Speed Fig. 11.15 Effect of specific speed on pump performance curves.
A low-specific-speed pump (Ns 600) has a broad efficiency curve, a rising power curve, and a head curve which “droops’’ at shutoff, implying possible surge or hunt-ing problems.
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 firm 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 exper-imental and computational results for a centrifugal pump diffuser. A photograph of the device is shown in Fig. 11.16a. It is made of clear Perspex, so that laser measurements 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. 11.16b. The computations used a turbulence formulation called the k- 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.
The ultimate test of a pump is its match with the operating-system characteristics. Phys-ically, the system head must match the head produced by the pump, and this intersec-tion should occur in the region of best efficiency.
The system head will probably contain a static-elevation change z2 z1 plus fric-tion losses in pipes and fittings Hsys (z2 z1) K (11.34) fL D V2 2g 11.5 Matching Pumps to System Characteristics 735 3 2 1 600 H H 0 1 2 Q Q 0 1 2 Q Q 3 2 1 0 1 2 Q Q η 1.0 0.2 0.6 600 4000 Ns = 10,000 bhp bhp Ns = 10,000 4000 Ns = 10,000 4000 600 Computational Fluid Dynamics 11.5 Matching Pumps to System Characteristics Fig. 11.16 Turbomachinery design now involves both experimentation and computational fluid dynamics (CFD): (a) a centrifugal impeller and diffuser (Courtesy of K. Eisele and Z. Zhang, Sulzer Innotec Ltd.); (b) a three-dimensional CFD model grid for this system. (From Ref. 56 by permission of the American So-ciety of Mechanical Engineers.) where K denotes minor losses and V is the flow velocity in the principal pipe. Since V is proportional to the pump discharge Q, Eq. (11.34) represents a system-head curve Hs(Q). Three examples are shown in Fig. 11.17: a static head Hs a, static head plus laminar friction Hs a bQ, and static head plus turbulent friction Hs a cQ2.
The intersection of the system curve with the pump performance curve H(Q) defines 736 Chapter 11 Turbomachinery Impeller Diffuser (b) (a) Fig. 11.17 Illustration of pump op-erating points for three types of system-head curves.
the operating point. In Fig. 11.17 the laminar-friction operating point is at maximum efficiency while the turbulent and static curves are off design. This may be unavoid-able 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 pos-sible because commercial pumps have only certain discrete sizes and speeds. Let us il-lustrate 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-ID pipe with friction factor f 0.030.
(a) What will the operating point and efficiency be? (b) To what speed should the pump be changed to operate at the BEP?
Solution For reservoirs the initial and final velocities are zero; thus the system head is Hs z2 z1 120 ft 0.030(1500 ft) 1 1 6 2 ft V2 2g fL D V2 2g 11.5 Matching Pumps to System Characteristics 737 Operating points Q1 Q2 Q3 Q Pump curves 1 2 3 Turbulent friction Laminar friction Static head Pump H(Q) System curves H(Q) Pump (Q) η Η,η Part (a) From continuity in the pipe, V Q/A Q/[ 1 4 ( 1 1 6 2 ft)2], and so we substitute for V above to get Hs 120 0.269Q2 Q in ft3/s (1) Since Fig. 11.7a uses thousands of gallons per minute for the abscissa, we convert Q in Eq. (1) to this unit: Hs 120 1.335Q2 Q in 103 gal/min (2) We can plot Eq. (2) on Fig. 11.7a and see where it intersects the 32-in pump-head curve, as in Fig. E11.6. A graphical solution gives approximately H 430 ft Q 15,000 gal/min Pumps Combined in Parallel 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 Hpump 490 0.26Q2 Q in 103 gal/min (3) Equations (2) and (3) must match at the operating point: 490 0.26Q2 120 1.335Q2 or Q2 232 Q 15.2 103 gal/min 15,200 gal/min Ans. (a) H 490 0.26(15.2)2 430 ft Ans. (a) To move the operating point to BEP, we change n, which changes both Q n and H n2.
From Fig. 11.7a, at BEP, H 386 ft; thus for any n, H 386(n/1170)2. Also read Q 20 103 gal/min; thus for any n, Q 20(n/1170). Match H to the system characteristics, Eq. (2), H 386 2 120 1.33520 2 Ans. (b) which gives n2 0. Thus it is impossible to operate at maximum efficiency with this particular system and pump.
If a pump provides the right head but too little discharge, a possible remedy is to com-bine two similar pumps in parallel, i.e., 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. 11.18. 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 QAB will be less than the separate operating discharges QA QB but certainly greater than either one.
n 1170 n 1170 490 120 0.26 1.335 738 Chapter 11 Turbomachinery Hpump Operating point 15,000 gal/min 490 ft H Q 430 ft 120 ft Hs E11.6 Part (b) Pumps Combined in Series 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 g(QAB)(HAB)/(550 bhpAB).
If pumps A and B are not identical, as in Fig. 11.18, 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. 11.19, 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 11.5 Matching Pumps to System Characteristics 739 H Operating points B A A + B Pump A Pump B Combined in parallel System curve QA QB 0 Q H A A + B 0 Q HB HA System curve Combined in series Pump A Pump B Operating points B Fig. 11.18 Performance and operat-ing points of two pumps operating singly and combined in parallel.
Fig. 11.19 Performance of two pumps combined in series.
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, i.e., re-quires 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 com-bined power is the sum of brake horsepower for A and B at the operating point flow rate. The combined efficiency is 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 lbf/in2 absolute. Figure 11.20 shows a section of a seven-stage centrifugal propane compressor which develops 300 lbf/in2 rise at 40,000 ft3/min and 35,000 bhp.
Most of the discussion in this chapter concerns incompressible flow, that is, negligible change in fluid density. Even the pump of Fig. 11.7, which can produce 600 ft of head at 1170 r/min, will only increase standard air pressure by 46 lbf/ft2, about a 2 percent change in density. The picture changes at higher speeds, p n2, and multiple stages, where very large changes in pressure and density are achieved. Such devices are called compressors, as in Fig. 11.20. The concept of static head, H p/g, becomes inap-propriate, since varies. Compressor performance is measured by (1) the pressure ra-tio across the stage p2/p1 and (2) the change in stagnation enthalpy (h02 h01), where h0 h 1 2 V2 (see Sec. 9.3). Combining m stages in series results in pfinal/pinitial (p2/p1)m. 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].
Compressor efficiency, from inlet condition 1 to final outlet f, is defined by the change in gas enthalpy, assuming an adiabatic process: comp 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 ˙(RT0)1/2/(D2p0), which will reach a maximum when choking occurs. For further de-tails, see Refs. 21 to 23. Tf T01 T0f T01 hf h01 h0f h01 g(QAB)(HAB) 550 bhpAB 740 Chapter 11 Turbomachinery Multistage Pumps Compressors 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 1 2 Q at the same 1170 r/min speed. The system curve is the same, and the balance-of-head relation becomes H 490 0.26( 1 2 Q)2 120 1.335Q2 or Q2 Q 16,300 gal/min Ans.
This is only 7 percent more than a single pump. Each pump delivers 1 2 Q 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.
490 120 1.335 0.065 11.5 Matching Pumps to System Characteristics 741 Fig. 11.20 Cross section of a seven-stage centrifugal propane compressor which delivers 40,000 ft3/min at 35,000 bhp and a pres-sure rise of 300 lbf/in2. Note the second inlet at stage 5 and the varying impeller designs. (Courtesy of DeLaval-Stork V.O.F., Centrifugal Compressor Division.) 11.6 Turbines Reaction Turbines EXAMPLE 11.8 Suppose the elevation change in Example 11.6 is raised from 120 to 500 ft, greater than a sin-gle 32-in pump can supply. Investigate using 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 sys-tem-head curve is replaced by 500. The balance of heads becomes H 2(490 0.26Q2) 500 1.335Q2 or Q2 1 9 .3 8 3 0 5 5 0 0 .
0 52 Q 16.1 103 gal/min Ans.
The operating head is 500 1.335(16.1)2 845 ft, or 97 percent more than that for a single pump in Example 11.5. Each pump is operating at 16.1 103 gal/min, which from Fig. 11.7a is 83 percent efficient, a pretty good match to the system. To pump at this operating point re-quires 4100 bhp, or about 2050 bhp for each pump.
A turbine extracts energy from a fluid which possesses high head, but it is fatuous to say a turbine is a pump run backward. Basically there are two types, reaction and im-pulse, 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 im-peller 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 more efficient at low-power settings. Figure 11.21 shows sketches of runner designs for Fran-cis 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 11.22 shows a radial tur-bine runner. Again assume one-dimensional frictionless flow through the blades. Ad-justable inlet guide vanes are absolutely necessary for good efficiency. They bring the inlet flow to the blades at angle 2 and absolute velocity V2 for minimum “shock’’ or 742 Chapter 11 Turbomachinery Idealized Radial Turbine Theory Fig. 11.21 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, Nsp 29.
Fig. 11.22 Inlet- and outlet-velocity diagrams for an idealized radial-flow reaction turbine runner.
directional-mismatch loss. After vectorially adding in the runner tip speed u2 r2, the outer blade angle should be set at angle 2 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. (3.55), to Fig.
11.22 (see Example 3.14 for a similar case) yields an idealized formula for the power P extracted by the runner: P T Q(r2Vt2 r1Vt1) Q(u2V2 cos 2 u1V1 cos 1) (11.35) 11.6 Turbines 743 (a) (b) (c) Nsp = 20 Nsp = 60 Nsp = 140 0.4 0.3 0.2 0.1 0 2 3 CQ CP (d) 10.0 9.0 1.0 0.8 0.6 0.4 0.2 0.0 CQ CH CH η η 1 Adjustable guide vane Blade Runner Vt 2 Vn 2 w2 V2 r2 r1 u1 w1 V1 ω β2 2 α ω u 2 = r 2 β1 1 α Power Specific Speed where Vt2 and Vt1 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 Vn2 V2 sin 2 is proportional to the flow rate Q. If the flow rate changes and the runner speed u2 is constant, the vanes must be ad-justed to a new angle 2 so that w2 still follows the blade surface. Thus adjustable in-let vanes are very important to avoid shock loss.
Turbine parameters are similar to those of a pump, but the dependent variable is the out-put brake horsepower, which depends upon the inlet flow rate Q, available head H, im-peller speed n, and diameter D. The efficiency is the output brake horsepower divided by the available water horsepower gQH. The dimensionless forms are CQ, CH, 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 independent variable: CH CH(CP) CQ CQ(CP) (CP) (11.36) Figure 11.21d shows typical performance curves for a small Francis radial turbine. The maximum efficiency point is called the normal power, and the values for this particu-lar turbine are max 0.89 CP 2.70 CQ 0.34 CH 9.03 A parameter which compares the output power with the available head, independent of size, is found by eliminating the diameter between CH and CP. It is called the power specific speed: Rigorous form: N sp (11.37a) Lazy but common: Nsp (11.37b) For water, 1.94 slugs/ft3 and Nsp 273.3N sp. The various turbine designs divide up nicely according to the range of power specific speed, as follows: Turbine type Nsp 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 Nsp, like Ns for pumps, is defined only with respect to the BEP and has a single value for a given turbine family. In Fig. 11.21d, Nsp 273.3(2.70)1/2/(9.03)5/4 29, regardless of size.
(r/min)(bhp)1/2 [H (ft)]5/4 n(bhp)1/2 1/2(gH)5/4 C P 1/2 C H 5/4 bhp gQH 744 Chapter 11 Turbomachinery Impulse Turbines 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 en-gineering 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. 11.23. The reversible pump-and-turbine design of Fig. 11.23d 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 hu-man scale, as shown in Fig. 11.24. The economic advantages of small-scale model test-ing are evident from this photograph of the Francis turbine units at Grand Coulee Dam.
For high head and relatively low power, i.e., low Nsp, not only would a reaction tur-bine require too high a speed but also the high pressure in the runner would require a massive casing thickness. The impulse turbine of Fig. 11.25 is ideal for this situation.
Since Nsp 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 Vj. The jet strikes the buckets and imparts a momentum change similar to that in our control-volume analysis for a moving vane in Example 3.10 or Prob. 3.51. The buckets have an elliptical split-cup shape, as in Fig. 11.25b. They are named Pelton wheels, after Lester A. Pelton (1829–1908), who produced the first efficient design.
From Example 3.10 the force and power delivered to a Pelton wheel are theoretically F Q(Vj u)(1 cos ) (11.38) P Fu Qu(Vj u)(1 cos ) where u 2nr is the bucket linear velocity and r is the pitch radius, or distance to the jet centerline. A bucket angle 180° gives maximum power but is physically impractical. In practice, 165°, or 1 cos 1.966 or only 2 percent less than maximum power.
From Eq. (11.38) the theoretical power of an impulse turbine is parabolic in bucket speed u and is maximum when dP/du 0, or u 2nr 1 2 Vj (11.39) For a perfect nozzle, the entire available head would be converted to jet velocity Vj (2gH)1/2. Actually, since there are 2 to 8 percent nozzle losses, a velocity coefficient C is used Vj C (2gH)1/2 0.92 C 0.98 (11.40) By combining Eqs. (11.36) and (11.40), the theoretical impulse turbine efficiency becomes 2(1 cos )(C ) (11.41) where peripheral-velocity factor Maximum efficiency occurs at 1 2 C 0.47.
u (2gH)1/2 11.6 Turbines 745 Fig. 11.23 Large-scale turbine designs depend upon 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 Allis-Chalmers Fluid Products Company.) 746 Chapter 11 Turbomachinery (a ) (b ) (c ) (d ) Fig. 11.24 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 of Allis-Chalmers Fluid Products Company.) Fig. 11.25 Impulse turbine: (a) side view of wheel and jet; (b) top view of bucket; (c) typical velocity diagram.
Figure 11.26 shows Eq. (11.41) plotted for an ideal turbine ( 180°, Cv 1.0) and for typical working conditions ( 160°, Cv 0.94). The latter case predicts max 85 percent at 0.47, but the actual data for a 24-in Pelton wheel test are somewhat less efficient due to windage, mechanical friction, backsplashing, and nonuni-11.6 Turbines 747 r ω n, π u = 2 nr Vj β ≈ 165˚ Vj α u w β Split bucket Needle valve (a) (c) (b) Fig. 11.26 Efficiency of an impulse turbine calculated from Eq. (11.41): solid curve ideal, 180°, Cv 1.0; dashed curve actual, 160°, Cv 0.94; open circles data, Pelton wheel, diameter 2 ft.
form bucket flow. For this test max 80 percent, and, generally speaking, an impulse turbine is not quite as efficient as the Francis or propeller turbines at their BEPs.
Figure 11.27 shows the optimum efficiency of the three turbine types, and the im-portance of the power specific speed Nsp as a selection tool for the designer. These ef-ficiencies are optimum and are obtained in careful design of large machines.
The water power available to a turbine may vary due to either net-head or flow-rate changes, both of which are common in field installations such as hydroelectric plants.
The demand for turbine power also varies from light to heavy, and the operating re-sponse is a change in the flow rate by adjustment of a gate valve or needle valve (Fig.
11.25a). As shown in Fig. 11.28, all three turbine types achieve fairly uniform effi-ciency as a function of the level of power being extracted. Especially effective is the adjustable-blade (Kaplan-type) propeller turbine, while the poorest is a fixed-blade pro-peller. The term rated power in Fig. 11.28 is the largest power delivery guaranteed by the manufacturer, as opposed to normal power, which is delivered at maximum effi-ciency.
For further details of design and operation of turbomachinery, the readable and in-teresting treatment in Ref. 33 is especially recommended. The feasibility of micro-hydropower is discussed in . See also Refs. 27 and 28.
748 Chapter 11 Turbomachinery 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 η u (2gH)1/2 = φ 0 1.0 0.9 η Nsp 10 100 1000 0.8 1 Impulse Francis Propeller Fig. 11.27 Optimum efficiency of turbine designs.
Fig. 11.28 Efficiency versus power level for various turbine designs at constant speed and head.
EXAMPLE 11.9 Investigate the possibility of using (a) a Pelton wheel similar to Fig. 11.26 or (b) the Francis turbine family of Fig. 11.21d to deliver 30,000 bhp from a net head of 1200 ft.
Solution From Fig. 11.27, the most efficient Pelton wheel occurs at about Nsp 4.5 or n 183 r/min 3.06 r/s From Fig. 11.26 the best operating point is 0.47 or D 13.6 ft Ans. (a) This Pelton wheel is perhaps a little slow and a trifle large. You could reduce D and increase n by increasing Nsp to, say, 6 or 7 and accepting the slight reduction in efficiency. Or you could use a double-hung, two-wheel configuration, each delivering 15,000 bhp, which changes D and n by the factor 21/2: Double wheel: n (183)21/2 260 r/min D 9.6 ft Ans. (a) 13.6 21/2 D(3.06 r/s) [2(32.2)(1200)]1/2 (r/min)(30,000 bhp)1/2 (1200 ft)1.25 11.6 Turbines 749 1.0 0.9 0.8 0.7 0.6 20 40 60 80 100 Rated power, percent Kaplan (adjustable blade) Impulse Francis Fixed-blade propeller 10° 20° η 0.5 0 Part (a) Part (b) The Francis wheel of Fig. 11.21d must have Nsp 29 or n 1183 r/min 19.7 r/s Then the optimum power coefficient is CP 2.70 or D5 412 D 3.33 ft 40 in Ans. (b) This is a faster speed than normal practice, and the casing would have to withstand 1200 ft of water or about 520 lbf/in2 internal pressure, but the 40-in size is extremely attractive. Francis turbines are now being operated at heads up to 1500 ft.
Wind energy has long been used as a source of mechanical power. The familiar four-bladed windmills of Holland, England, and the Greek islands have been used for cen-turies to pump water, grind grain, and saw wood. Modern research concentrates on the ability of wind turbines to generate electric power. Koeppl stresses the potential for propeller-type machines. Spera gives a detailed discussion of the technical and economic feasibility of large-scale electric power generation by wind. See also Refs.
46, 48, and 50 to 52.
Some examples of wind turbine designs are shown in Fig. 11.29. The familiar Amer-ican multiblade farm windmill (Fig. 11.29a) is of low efficiency, but thousands are in use as a rugged, reliable, and inexpensive way to pump water. A more efficient design is the propeller mill in Fig. 11.29b, similar to the pioneering Smith-Putnam 1250-kW two-bladed system which operated on Grampa’s Knob, 12 mi west of Rutland, Ver-mont, from 1941 to 1945. The Smith-Putnam design broke because of inadequate blade strength, but it withstood winds up to 115 mi/h and its efficiency was amply demon-strated .
The Dutch, American multiblade, and propeller mills are examples of horizontal-axis wind turbines (HAWTs), which are efficient but somewhat awkward in that they require extensive bracing and gear systems when combined with an electric generator.
Thus a competing family of vertical-axis wind turbines (VAWTs) has been proposed which simplifies gearing and strength requirements. Figure 11.29c shows the “egg-beater’’ VAWT invented by G. J. M. Darrieus in 1925, now used in several govern-ment-sponsored demonstration systems. To minimize centrifugal stresses, the twisted blades of the Darrieus turbine follow a troposkien curve formed by a chain anchored at two points on a spinning vertical rod.
An alternative VAWT, simpler to construct than the troposkien, is the straight-bladed Darrieus-type turbine in Fig. 11.29d. This design, proposed by Reading University in England, has blades which pivot due to centrifugal action as wind speeds increase, thus limiting bending stresses.
30,000(550) (1.94)(19.7)3D5 P n3D5 (r/min)(30,000 bhp)1/2 (1200 ft)1.25 750 Chapter 11 Turbomachinery Wind Turbines Fig. 11.29 Wind turbine designs: (a) the American multiblade farm HAWT; (b) propeller HAWT (Courtesy of Grumman Aerospace Corp.); (c) the Darrieus VAWT (Courtesy of National Research Council, Canada); (d) modified straight-blade Darrieus VAWT (Courtesy of Reading University— Nat’l Wind Power Ltd.).
11.6 Turbines 751 (a ) (b ) (d ) (c ) Idealized Wind Turbine Theory Fig. 11.30 Idealized actuator-disk and streamtube analysis of flow through a windmill.
The ideal, frictionless efficiency of a propeller windmill was predicted by A. Betz in 1920, using the simulation shown in Fig. 11.30. The propeller is represented by an ac-tuator disk which creates across the propeller plane a pressure discontinuity of area A and local velocity V. The wind is represented by a streamtube of approach velocity V1 and a slower downstream wake velocity V2. The pressure rises to pb just before the disk and drops to pa just after, returning to free-stream pressure in the far wake. To hold the propeller rigid when it is extracting energy from the wind, there must be a leftward force F on its support, as shown.
A control-volume–horizontal-momentum relation applied between sections 1 and 2 gives Fx F m ˙ (V2 V1) A similar relation for a control volume just before and after the disk gives Fx F (pb pa)A m ˙ (Va Vb) 0 Equating these two yields the propeller force F (pb pa)A m ˙ (V1 V2) (11.42) Assuming ideal flow, the pressures can be found by applying the incompressible Bernoulli relation up to the disk From 1 to b: p 1 2 V2 1 pb 1 2 V2 From a to 2: pa 1 2 V2 p 1 2 V 2 2 Subtracting these and noting that m ˙ AV through the propeller, we can substitute for pb pa in Eq. (11.42) to obtain pb pa 1 2 (V2 1 V 2 2) V(V1 V2) or V 1 2 (V1 V2) (11.43) 752 Chapter 11 Turbomachinery Streamtube passing through propeller Wind V1, p∞ Swept area A pb pa V Wake V2, p∞ p∞ p∞ F pb pa p Fig. 11.31 Estimated performance of various wind turbine designs as a function of blade-tip speed ratio.
(From Ref. 53.) Continuity and momentum thus require that the velocity V through the disk equal the average of the wind and far-wake speeds.
Finally, the power extracted by the disk can be written in terms of V1 and V2 by combining Eqs. (11.42) and (11.43) P FV AV2(V1 V2) 1 4 A(V 2 1 V2 2)(V1 V2) (11.44) For a given wind speed V1, we can find the maximum possible power by differentiat-ing P with respect to V2 and setting equal to zero. The result is P Pmax 2 8 7 AV 3 1 at V2 1 3 V1 (11.45) which corresponds to V 2V1/3 through the disk.
The maximum available power to the propeller is the mass flow through the pro-peller times the total kinetic energy of the wind Pavail 1 2 m ˙ V2 1 1 2 AV3 1 Thus the maximum possible efficiency of an ideal frictionless wind turbine is usually stated in terms of the power coefficient CP (11.46) Equation (11.45) states that the total power coefficient is Cp,max 1 2 6 7 0.593 (11.47) This is called the Betz number and serves as an ideal with which to compare the ac-tual performance of real windmills.
Figure 11.31 shows the measured power coefficients of various wind turbine de-signs. The independent variable is not V2/V1 (which is artificial and convenient only in P 1 2 AV3 1 11.6 Turbines 753 0.6 0.5 0.4 0.3 0.2 0.1 0 1 2 3 4 5 6 7 8 American multiblade Savonius rotor Ideal, propeller type Grumman windstream (Fig. 11.29 b) Dutch, four-arm Darrieus VAWT High-speed HAWT Ideal Betz number ω Speed ratio r/V1 Cp Fig. 11.32 World availability of land-based wind energy: estimated annual electric output in kWh/kW of a wind turbine rated at 11.2 m/s (25 mi/h). (From Ref. 54.) the ideal theory) but the ratio of blade-tip speed r to wind speed. Note that the tip can move much faster than the wind, a fact disturbing to the laity but familiar to en-gineers in the behavior of iceboats and sailing vessels. The Darrieus has the many ad-vantages of a vertical axis but has little torque at low speeds and also rotates more slowly at maximum power than a propeller, thus requiring a higher gear ratio for the generator. The Savonius rotor (Fig. 6.29b) has been suggested as a VAWT design be-cause it produces power at very low wind speeds, but it is inefficient and susceptible to storm damage because it cannot be feathered in high winds.
As shown in Fig. 11.32, there are many areas of the world where wind energy is an attractive alternative. Greenland, Newfoundland, Argentina, Chile, New Zealand, Ice-land, Ireland, and the United Kingdom have the highest prevailing winds, but Australia, e.g., with only moderate winds, has the potential to generate half its electricity with wind turbines . In addition, since the ocean is generally even windier than the land, there are many island areas of high potential for wind power. There have also been proposals for ocean-based floating windmill farms. The inexhaustible availability of the winds, coupled with improved low-cost turbine designs, indicates a bright fu-ture for this alternative.
754 Chapter 11 Turbomachinery Under 750 750 – 2250 2250 – 3750 3750 – 5000 Over 5000 P11.3 A PDP can deliver almost any fluid, but there is always a limiting very high viscosity for which performance will deteriorate. Can you explain the probable reason?
P11.4 Figure 11.2c shows a double-screw pump. Sketch a sin-gle-screw pump and explain its operation. How did Archimedes’ screw pump operate?
P11.5 What type of pump is shown in Fig. P11.5? How does it operate?
Problems Most of the problems herein are fairly straightforward. More dif-ficult or open-ended assignments are labeled with an asterisk. Prob-lems labeled with an EES icon will benefit from the use of the En-gineering Equations Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 11.1 to 11.103 (categorized in the prob-lem list below) are followed by word problems W11.1 to W11.10, comprehensive problems C11.1 to C11.3, and design project D11.1.
Problem distribution Section Topic Problems 11.1 Introduction and classification 11.1–11.14 11.2 Centrifugal pump theory 11.15–11.21 11.3 Pump performance and similarity rules 11.22–11.41 11.3 Net positive-suction head 11.42–11.44 11.4 Specific speed: Mixed- and axial-flow pumps 11.45–11.62 11.5 Matching pumps to system characteristics 11.63–11.73 11.5 Pumps in parallel or series 11.74–11.81 11.5 Pump instability 11.82–11.83 11.6 Reaction and impulse turbines 11.84–11.99 11.6 Wind turbines 11.100–11.103 P11.1 Describe the geometry and operation of a peristaltic pos-itive-displacement pump which occurs in nature.
P11.2 What would be the technical classification of the fol-lowing turbomachines: (a) a household fan, (b) a wind-mill, (c) an aircraft propeller, (d) a fuel pump in a car, (e) an eductor, ( f ) a fluid-coupling transmission, and (g) a power plant steam turbine?
Problems 755 Turbomachinery design is perhaps the most practical and most active application of the principles of fluid mechanics. There are billions of pumps and turbines in use in the world, and thousands of companies are seeking improvements. This chapter discusses both positive-displacement devices and, more extensively, rotodynamic machines. With the centrifugal pump as an example, the basic concepts of torque, power, head, flow rate, and efficiency are developed for a turbomachine. Nondimensionalization leads to the pump similarity rules and some typical dimensionless performance curves for ax-ial and centrifugal machines. The single most useful pump parameter is found to be the specific speed, which delineates the type of design needed. An interesting design application is the theory of pumps combined in series and in parallel.
Turbines extract energy from flowing fluids and are of two types: impulse turbines, which convert the momentum of a high-speed stream, and reaction turbines, where the pressure drop occurs within the blade passages in an internal flow. By analogy with pumps, the power specific speed is important for turbines and is used to classify them into impulse, Francis, and propeller-type designs. A special case of reaction turbine with unconfined flow is the wind turbine. Several types of windmills are discussed and their relative performances compared.
Summary P11.5 P11.6 Figure P11.6 shows two points a half-period apart in the operation of a pump. What type of pump is this? How does it work? Sketch your best guess of flow rate versus time for a few cycles.
P11.7 A piston PDP has a 5-in diameter and a 2-in stroke and operates at 750 r/min with 92 percent volumetric effi-ciency. (a) What is its delivery, in gal/min? (b) If the pump delivers SAE 10W oil at 20°C against a head of 50 ft, what horsepower is required when the overall efficiency is 84 percent?
P11.8 A centrifugal pump delivers 550 gal/min of water at 20°C when the brake horsepower is 22 and the efficiency is 71 percent. (a) Estimate the head rise in ft and the pressure rise in lbf/in2. (b) Also estimate the head rise and horse-power if instead the delivery is 550 gal/min of gasoline at 20°C.
P11.9 Figure P11.9 shows the measured performance of the Vickers model PVQ40 piston pump when delivering SAE 10W oil at 180°F ( 910 kg/m3). Make some general observations about these data vis-à-vis Fig. 11.2 and your intuition about the behavior of piston pumps.
P11.10 Suppose that the piston pump of Fig. P11.9 is used to de-liver 15 gal/min of water at 20°C using 20 brake horse-power. Neglecting Reynolds-number effects, use the fig-ure to estimate (a) the speed in r/min and (b) the pressure rise in lbf/in2.
P11.11 A pump delivers 1500 L/min of water at 20°C against a pressure rise of 270 kPa. Kinetic- and potential-energy changes are negligible. If the driving motor supplies 9 kW, what is the overall efficiency?
P11.12 In a test of the centrifugal pump shown in Fig. P11.12, the following data are taken: p1 100 mmHg (vacuum) and p2 500 mmHg (gage). The pipe diameters are D1 12 cm and D2 5 cm. The flow rate is 180 gal/min of light oil (SG 0.91). Estimate (a) the head developed, in meters; and (b) the input power required at 75 percent efficiency.
P11.13 A 20-hp pump delivers 400 gal/min of gasoline at 20°C with 75 percent efficiency. What head and pressure rise result across the pump?
P11.14 A pump delivers gasoline at 20°C and 12 m3/h. At the inlet p1 100 kPa, z1 1 m, and V1 2 m/s. At the exit p2 500 kPa, z2 4 m, and V2 3 m/s. How much power is required if the motor efficiency is 75 percent?
P11.15 A lawn sprinkler can be used as a simple turbine. As shown in Fig. P11.15, flow enters normal to the paper in the center and splits evenly into Q/2 and Vrel leaving each nozzle. The arms rotate at angular velocity and do work on a shaft. Draw the velocity diagram for this turbine.
Neglecting friction, find an expression for the power de-livered to the shaft. Find the rotation rate for which the power is a maximum.
756 Chapter 11 Turbomachinery Flow out Flow in Flow out Flow in Check valve A B A B P11.6 (2) 65 cm (1) R Q R ω Q 2 , Vrel Q 2 , Vrel θ Vc ρ , V, A P11.12 P11.15 P11.16 For the “sprinkler turbine’’ of Fig. P11.15, let R 18 cm, with total flow rate of 14 m3/h of water at 20°C. If the noz-zle exit diameter is 8 mm, estimate (a) the maximum power delivered in W and (b) the appropriate rotation rate in r/min.
P11.17 A centrifugal pump has d1 7 in, d2 13 in, b1 4 in, b2 3 in, 1 25°, and 2 40° and rotates at 1160 r/min. If the fluid is gasoline at 20°C and the flow enters the blades radially, estimate the theoretical (a) flow rate in gal/min, (b) horsepower, and (c) head in ft.
P11.18 A jet of velocity V strikes a vane which moves to the right at speed Vc, as in Fig. P11.18. The vane has a turning angle .
P11.18 EES Derive an expression for the power delivered to the vane by the jet. For what vane speed is the power maximum?
P11.19 A centrifugal pump has r2 9 in, b2 2 in, and 2 35° and rotates at 1060 r/min. If it generates a head of 180 ft, determine the theoretical (a) flow rate in gal/min and (b) horsepower. Assume near-radial entry flow.
P11.20 Suppose that Prob. 11.19 is reversed into a statement of the theoretical power Pw 153 hp. Can you then com-pute the theoretical (a) flow rate and (b) head? Explain and resolve the difficulty which arises.
P11.21 The centrifugal pump of Fig. P11.21 develops a flow rate of 4200 gal/min of gasoline at 20°C with near-radial absolute Problems 757 15 30 45 60 0 20 40 60 80 100 0 19 38 57 76 95 500 1000 1500 2000 0 100 80 60 40 20 0 Volumetric efficiency, percent Speed, r/min Delivery, L/min Overall efficiency, percent Input power, kW Pump displacement: 41 cm3/r 210 bar 140 bar 70 bar 35 bar 35 bar 70 bar 140 bar 210 bar 210 bar 140 bar 70 bar 35 bar 210 bar - 3000 lb/in2 140 bar - 2000 lb/in2 70 bar - 1000 lb/in2 35 bar - 500 lb/in2 Fig. P11.9 Performance of the model PVQ40 piston pump deliv-ering SAE 10W oil at 180°F. (Courtesy of Vickers Inc., PDN/PACE Division.) 30° 1750 r / min 2 in 4 in 3 in P11.21 inflow. Estimate the theoretical (a) horsepower, (b) head rise, and (c) appropriate blade angle at the inner radius.
P11.22 A 37-cm-diameter centrifugal pump, running at 2140 r/min with water at 20°C, produces the following per-formance data: Q, m3/s 0.0 0.05 0.10 0.15 0.20 0.25 0.30 H, m 105 104 102 100 95 85 67 P, kW 100 115 135 171 202 228 249 (a) Determine the best efficiency point. (b) Plot CH ver-sus CQ. (c) If we desire to use this same pump family to deliver 7000 gal/min of kerosine at 20°C at an input power of 400 kW, what pump speed (in r/min) and impeller size (in cm) are needed? What head will be developed?
P11.23 If the 38-in-diameter pump of Fig. 11.7b is used to de-liver 20°C kerosine at 850 r/min and 22,000 gal/min, what (a) head and (b) brake horsepower will result?
P11.24 Figure P11.24 shows performance data for the Taco, Inc., model 4013 pump. Compute the ratios of measured shut-off head to the ideal value U2/g for all seven impeller sizes. Determine the average and standard deviation of this ratio and compare it to the average for the six im-pellers in Fig. 11.7.
P11.25 At what speed in r/min should the 35-in-diameter pump of Fig. 11.7b be run to produce a head of 400 ft at a dis-charge of 20,000 gal/min? What brake horsepower will be required? Hint: Fit H(Q) to a formula.
P11.26 Determine if the seven Taco, Inc., pump sizes in Fig.
P11.24 can be collapsed into a single dimensionless chart 758 Chapter 11 Turbomachinery 0 100 200 300 400 500 600 700 800 Flow, gal / min Curves based on clear water with specific gravity of 1.0 100 80 60 40 20 0 Head, ft 12.95 in 12.50 in 12.00 in 11.50 in 11.00 in 10.50 in 10.00 in 10 15 20 25 30 35 40 45 L/s 50 30 25 20 15 10 5 0 3 4 5 6 NPSH, ft 50% 60% 65% 70% 74% 76% 78%79% 80% 79% 78% 74% 70% 65% 60% 50% 10 bhp 7.5 bhp 5 bhp 76% Model 4013 FM Series 1160 RPM Curve No. 806 Min. Imp. Dia. 10.0 Sizes 5 x 4 x13 aco 5 Fig. P11.24 Performance data for a centrifugal pump. (Courtesy of Taco, Inc., Cranston, Rhode Island.) of CH, CP, and versus CQ, as in Fig. 11.8. Comment on the results.
P11.27 The 12-in pump of Fig. P11.24 is to be scaled up in size to provide a head of 90 ft and a flow rate of 1000 gal/min at BEP. Determine the correct (a) impeller diameter, (b) speed in r/min, and (c) horsepower required.
P11.28 Tests by the Byron Jackson Co. of a 14.62-in-diameter centrifugal water pump at 2134 r/min yield the follow-ing data: Q, ft3/s 0 2 4 6 8 10 H, ft 340 340 340 330 300 220 bhp 135 160 205 255 330 330 What is the BEP? What is the specific speed? Estimate the maximum discharge possible.
P11.29 If the scaling laws are applied to the pump of Prob.
11.28 for the same impeller diameter, determine (a) the speed for which the shutoff head will be 280 ft, (b) the speed for which the BEP flow rate will be 8.0 ft3/s, and (c) the speed for which the BEP conditions will require 80 hp.
P11.30 A pump from the same family as Prob. 11.28 is built with D 18 in and a BEP power of 250 hp for gasoline (not water). Using the scaling laws, estimate the resulting (a) speed in r/min, (b) flow rate at BEP, and (c) shutoff head.
P11.31 Figure P11.31 shows performance data for the Taco, Inc., model 4010 pump. Compute the ratios of measured shut-off head to the ideal value U2/g for all seven impeller sizes. Determine the average and standard deviation of this ratio, and compare it to the average of 0.58 0.02 for the seven impellers in Fig. P11.24. Comment on your results.
Problems 759 Curves based on clear water with specific gravity of 1.0 Head, ft 10.40 in 10.00 in 9.50 in 9.00 in 8.50 in 8.00 in 7.70 in 10 L/s 10 12 14 16 NPSH, ft 50% 60%65% 70% 74% 78% 80% 82% 83% 82% 76% 74% 65% 60% 50% 10 bhp 15 bhp 30 bhp 140 120 100 80 60 40 20 18 20 22 80% 78% 70% 20 bhp 25 bhp Head, m 40 35 30 25 20 15 10 0 Flow, gal / min 125 250 375 500 625 750 875 1000 1125 1250 20 30 40 50 60 70 Model 4010 CM & FM Series 1760 RPM Curve No.756 Min. Imp. Dia. 7.70 Sizes 5 x 4 x10 aco Fig. P11.31 Performance data for a family of centrifugal pump impellers. (Courtesy of Taco, Inc., Cranston, Rhode Island.) EES P11.32 Determine if the seven Taco, Inc., impeller sizes in Fig.
P11.31 can be collapsed into a single dimensionless chart of CH, CP, and versus CQ, as in Fig. 11.8. Comment on the results.
P11.33 Clearly the maximum efficiencies of the pumps in Figs.
P11.24 and P11.31 decrease with impeller size. Compare max for these two pump families with both the Moody and the Anderson correlations, Eqs. (11.29). Use the cen-tral impeller size as a comparison point.
P11.34 You are asked to consider a pump geometrically similar to the 9-in-diameter pump of Fig. P11.31 to deliver 1200 gal/min at 1500 r/min. Determine the appropriate (a) im-peller diameter, (b) BEP horsepower, (c) shutoff head, and (d) maximum efficiency. The fluid is kerosine, not water.
P11.35 An 18-in-diameter centrifugal pump, running at 880 r/min with water at 20°C, generates the following per-formance data: Q, gal/min 0.0 2000 4000 6000 8000 10,000 H, ft 92 89 84 78 68 50 P, hp 100 112 130 143 156 163 Determine (a) the BEP, (b) the maximum efficiency, and (c) the specific speed. (d) Plot the required input power versus the flow rate. P11.36 Plot the dimensionless performance curves for the pump of Prob. 11.35 and compare with Fig. 11.8. Find the ap-propriate diameter in inches and speed in r/min for a geo-metrically similar pump to deliver 400 gal/min against a head of 200 ft. What brake horsepower would be re-quired?
P11.37 The efficiency of a centrifugal pump can be approximated by the curve fit aQ bQ3, where a and b are con-stants. For this approximation, (a) what is the ratio of Q at BEP to Qmax? If the maximum efficiency is 88 per-cent, what is the efficiency at (b) 1 3 Qmax and (c) 4 3 Q?
P11.38 A 6.85-in pump, running at 3500 r/min, has the follow-ing measured performance for water at 20°C: Q, gal/min 50 100 150 200 250 300 350 400 450 H, ft 201 200 198 194 189 181 169 156 139 , % 29 50 64 72 77 80 81 79 74 (a) Estimate the horsepower at BEP. If this pump is rescaled in water to provide 20 bhp at 3000 r/min, de-termine the appropriate (b) impeller diameter, (c) flow rate, and (d) efficiency for this new condition.
P11.39 The Allis-Chalmers D30LR centrifugal compressor de-livers 33,000 ft3/min of SO2 with a pressure change from 14.0 to 18.0 lbf/in2 absolute using an 800-hp motor at 3550 r/min. What is the overall efficiency? What will the flow rate and p be at 3000 r/min? Estimate the diame-ter of the impeller.
P11.40 The specific speed Ns, as defined by Eqs. (11.30), does not contain the impeller diameter. How then should we size the pump for a given Ns? Logan suggests a pa-rameter called the specific diameter Ds, which is a di-mensionless combination of Q, gH, and D. (a) If Ds is proportional to D, determine its form. (b) What is the re-lationship, if any, of Ds to CQ, CH, and CP? (c) Esti-mate Ds for the two pumps of Figs. 11.8 and 11.13.
P11.41 It is desired to build a centrifugal pump geometrically similar to that of Prob. 11.28 to deliver 6500 gal/min of gasoline at 20°C at 1060 r/min. Estimate the resulting (a) impeller diameter, (b) head, (c) brake horsepower, and (d) maximum efficiency.
P11.42 An 8-in model pump delivering 180°F water at 800 gal/min and 2400 r/min begins to cavitate when the inlet pressure and velocity are 12 lbf/in2 absolute and 20 ft/s, respectively. Find the required NPSH of a prototype which is 4 times larger and runs at 1000 r/min.
P11.43 The 28-in-diameter pump in Fig. 11.7a at 1170 r/min is used to pump water at 20°C through a piping system at 14,000 gal/min. (a) Determine the required brake horsepower. The average friction factor is 0.018. (b) If there is 65 ft of 12-in-diameter pipe upstream of the pump, how far below the sur-face should the pump inlet be placed to avoid cavitation?
P11.44 The pump of Prob. 11.28 is scaled up to an 18-in diam-eter, operating in water at best efficiency at 1760 r/min.
The measured NPSH is 16 ft, and the friction loss be-tween the inlet and the pump is 22 ft. Will it be sufficient to avoid cavitation if the pump inlet is placed 9 ft below the surface of a sea-level reservoir?
P11.45 Determine the specific speeds of the seven Taco, Inc., pump impellers in Fig. P11.24. Are they appropriate for centrifugal designs? Are they approximately equal within experimental uncertainty? If not, why not?
P11.46 The answer to Prob. 11.40 is that the dimensionless “spe-cific diameter” takes the form Ds D(gH)1/4/Q1/2, evaluated at the BEP. Data collected by the author for 30 different pumps indicate, in Fig. P11.46, that Ds corre-lates well with specific speed Ns. Use this figure to esti-mate the appropriate impeller diameter for a pump which delivers 20,000 gal/min of water and a head of 400 ft when running at 1200 r/min. Suggest a curve-fit formula to the data. Hint: Use a hyperbolic formula.
P11.47 A typical household basement sump pump provides a discharge of 5 gal/min against a head of 15 ft. Estimate (a) the maximum efficiency and (b) the minimum horse-power required to drive such a pump at 1750 r/min.
760 Chapter 11 Turbomachinery EES P11.48 Compute the specific speeds for the pumps in Probs.
11.28, 11.35, and 11.38 plus the median sizes in Figs.
P11.24 and P11.31. Then determine if their maximum ef-ficiencies match the values predicted in Fig. 11.14.
P11.49 Data collected by the author for flow coefficient at BEP for 30 different pumps are plotted versus specific speed in Fig. P11.49. Determine if the values of C Q for the five pumps in Prob. 11.48 also fit on this correlation. If so, suggest a curve-fitted formula for the data.
P11.52 An axial-flow fan operates in sea-level air at 1200 r/min and has a blade-tip diameter of 1 m and a root diameter of 80 cm. The inlet angles are 1 55° and 1 30°, while at the outlet 2 60°. Estimate the theoretical val-ues of the (a) flow rate, (b) horsepower, and (c) outlet angle 2.
P11.53 If the axial-flow pump of Fig. 11.13 is used to deliver 70,000 gal/min of 20°C water at 1170 r/min, estimate (a) the proper impeller diameter, (b) the shutoff head, (c) the shutoff horsepower, and (d) p at best efficiency.
P11.54 The Colorado River Aqueduct uses Worthington Corp.
pumps which deliver 200 ft3/s at 450 r/min against a head of 440 ft. What types of pump are these? Estimate the impeller diameter.
P11.55 We want to pump 70°C water at 20,000 gal/min and 1800 r/min. Estimate the type of pump, the horsepower re-quired, and the impeller diameter if the required pressure rise for one stage is (a) 170 kPa and (b) 1350 kPa.
P11.56 A pump is needed to deliver 40,000 gal/min of gasoline at 20°C against a head of 90 ft. Find the impeller size, speed, and brake horsepower needed to use the pump families of (a) Fig. 11.8 and (b) Fig. 11.13. Which is the better design?
P11.57 Performance data for a 21-in-diameter air blower running at 3550 r/min are as follows: p, inH2O 29 30 28 21 10 Q, ft3/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 the performance compare with Fig. 11.8? What are C Q, C H, and C P?
Problems 761 500 0 1000 1500 2000 2500 3000 3500 0 2 4 6 8 10 12 14 16 18 20 Ns Ds Data from 30 different pump designs Fig. P11.46 Specific diameter at BEP for 30 commer-cial pumps.
0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 0 500 1000 1500 2000 2500 3000 3500 NS C Q Data from 30 different pump designs 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 500 1000 1500 2000 2500 3000 3500 Ns C P Data from 30 different pump designs Fig. P11.49 Flow coefficient at BEP for 30 commercial pumps.
P11.50 Data collected by the author for power coefficient at BEP for 30 different pumps are plotted versus specific speed in Fig. P11.50. Determine if the values of C P for the five pumps in Prob. 11.48 also fit on this correlation. If so, suggest a curve-fitted formula for the data.
P11.51 An axial-flow pump delivers 40 ft3/s of air which enters at 20°C and 1 atm. The flow passage has a 10-in outer radius and an 8-in inner radius. Blade angles are 1 60° and 2 70°, and the rotor runs at 1800 r/min. For the first stage compute (a) the head rise and (b) the power required.
Fig. P11.50 Power coefficient at BEP for 30 commer-cial pumps.
P11.58 The Worthington Corp. model A-12251 water pump, op-erating at maximum efficiency, produces 53 ft of head at 3500 r/min, 1.1 bhp at 3200 r/min, and 60 gal/min at 2940 r/min. What type of pump is this? What is its efficiency, and how does this compare with Fig. 11.14? Estimate the impeller diameter.
P11.59 Suppose it is desired to deliver 700 ft3/min of propane gas (molecular weight 44.06) at 1 atm and 20°C with a single-stage pressure rise of 8.0 inH2O. Determine the appropriate size and speed for using the pump families of (a) Prob. 11.57 and (b) Fig. 11.13. Which is the bet-ter design?
P11.60 A 45-hp pump is desired to generate a head of 200 ft when running at BEP with 20°C gasoline at 1200 r/min.
Using the correlations in Figs. P11.49 and P11.50, de-termine the appropriate (a) specific speed, (b) flow rate, and (c) impeller diameter.
P11.61 A mine ventilation fan, running at 295 r/min, delivers 500 m3/s of sea-level air with a pressure rise of 1100 Pa. Is this fan axial, centrifugal, or mixed? Estimate its diame-ter 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. 11.61 had a diameter of 20 ft [20, p. 339]. What would be the proper diameter for the pump family of Fig. 11.14 to pro-vide 500 m3/s at 295 r/min and BEP? What would be the resulting pressure rise in Pa?
P11.63 The 36.75-in pump in Fig. 11.7a at 1170 r/min is used to pump water at 60°F from a reservoir through 1000 ft of 12-in-ID galvanized-iron pipe to a point 200 ft above the reservoir surface. What flow rate and brake horse-power will result? If there is 40 ft of pipe upstream of the pump, how far below the surface should the pump in-let be placed to avoid cavitation?
P11.64 In Prob. 11.63 the operating point is off design at an ef-ficiency of only 77 percent. Is it possible, with the sim-ilarity rules, to change the pump rotation speed to deliver the water near BEP? Explain your results.
P11.65 The 38-in pump of Fig. 11.7a is used in series to lift 20°C water 3000 ft through 4000 ft of 18-in-ID cast-iron pipe.
For most efficient operation, how many pumps in series are needed if the rotation speed is (a) 710 r/min and (b) 1200 r/min?
P11.66 It is proposed to run the pump of Prob. 11.35 at 880 r/min to pump water at 20°C through the system in Fig. P11.66.
The pipe is 20-cm-diameter commercial steel. What flow rate in ft3/min will result? Is this an efficient application?
P11.67 The pump of Prob. 11.35, running at 880 r/min, is to pump water at 20°C through 75 m of horizontal galva-nized-iron pipe. All other system losses are neglected.
Determine the flow rate and input power for (a) pipe di-ameter 20 cm and (b) the pipe diameter found to yield maximum pump efficiency.
P11.68 A 24-in pump is homologous to the 32-in pump in Fig.
11.7a. At 1400 r/min this pump delivers 12,000 gal/min of water from one reservoir through a long pipe to an-other 50 ft higher. What will the flow rate be if the pump speed is increased to 1750 r/min? Assume no change in pipe friction factor or efficiency.
P11.69 The pump of Prob. 11.38, 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 which results if the pipe diameter is 3 in.
P11.70 The pump of Prob. 11.28, operating at 2134 r/min, is used with 20°C water in the system of Fig. P11.70. (a) If it is operating at BEP, what is the proper elevation z2? (b) If z2 225 ft, what is the flow rate if d 8 in.?
762 Chapter 11 Turbomachinery 4 m Pump 8 m 3 m 20 m 12 m P11.66 z1 = 100 ft Pump z 2 1500 ft of cast-iron pipe P11.70 P11.71 The pump of Prob. 11.38, running at 3500 r/min, deliv-ers water at 20°C through 7200 ft of horizontal 5-in-di-ameter 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 which causes the pump to op-erate at BEP. (c) If the latter condition holds continuously for 1 year, estimate the energy cost at 10 ¢/kWh.
P11.72 Performance data for a small commercial pump are as follows: Q, gal/min 0 10 20 30 40 50 60 70 H, ft 75 75 74 72 68 62 47 24 EES This pump supplies 20°C water to a horizontal 5 8 -in-diameter garden hose ( 0.01 in) which is 50 ft long.
Estimate (a) the flow rate and (b) the hose diameter which would cause the pump to operate at BEP.
P11.73 The piston pump of Fig. P11.9 is run at 1500 r/min to deliver SAE 10W oil through 100 m of vertical 2-cm-di-ameter wrought-iron pipe. If other system losses are ne-glected, estimate (a) the flow rate, (b) the pressure rise, and (c) the power required.
P11.74 The 32-in pump in Fig. 11.7a is used at 1170 r/min in a system whose head curve is Hs (ft) 100 1.5Q2, with Q in thousands of gallons of water per minute. Find the discharge 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. 11.7b are installed in paral-lel for the system of Fig. P11.75. Neglect minor losses.
For water at 20°C, estimate the flow rate and power re-quired if (a) both pumps are running and (b) one pump is shut off and isolated.
P11.80 It is proposed to use one 32- and one 28-in pump from Fig.
11.7a in parallel to deliver water at 60°F. The system-head curve is Hs 50 0.3Q2, with Q in thousands of gallons per minute. What will the head and delivery be if both pumps run at 1170 r/min? If the 28-in pump is reduced be-low 1170 r/min, at what speed will it cease to deliver?
P11.81 Reconsider the system of Fig. P6.68. Use the Byron Jack-son pump of Prob. 11.28 running at 2134 r/min, no scal-ing, to drive the flow. Determine the resulting flow rate between the reservoirs. What is the pump efficiency?
P11.82 The S-shaped head-versus-flow curve in Fig. P11.82 oc-curs in some axial-flow pumps. Explain how a fairly flat system-loss curve might cause instabilities in the opera-tion of the pump. How might we avoid instability?
Problems 763 z 1 = 200 ft 1 statute mile of cast-iron pipe, 24-in diameter Two pumps z2 = 300 ft P11.75 P11.76 Two 32-in pumps from Fig. 11.7a are combined in par-allel to deliver water at 60°F through 1500 ft of hori-zontal pipe. If f 0.025, what pipe diameter will ensure a flow rate of 35,000 gal/min for n 1170 r/min?
P11.77 Two pumps of the type tested in Prob. 11.22 are to be used at 2140 r/min to pump water at 20°C vertically up-ward 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?
P11.78 Suppose that the two pumps in Fig. P11.75 are modified to be in series, still at 710 r/min. What pipe diameter is required for BEP operation?
P11.79 Two 32-in pumps from Fig. 11.7a are to be used in se-ries at 1170 r/min to lift water through 500 ft of vertical cast-iron pipe. What should the pipe diameter be for most efficient operation? Neglect minor losses.
H Q 0 H Q 0 P11.82 P11.83 P11.83 The low-shutoff head-versus-flow curve in Fig. P11.83 occurs in some centrifugal pumps. Explain how a fairly flat system-loss curve might cause instabilities in the op-eration of the pump. What additional vexation occurs when two of these pumps are in parallel? How might we avoid instability?
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, num-EES ber, and size of turbine which 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?
P11.85 Turbines at the Conowingo Plant on the Susquehanna River each develop 54,000 bhp at 82 r/min under a head of 89 ft. What type of turbines are these? Estimate the flow rate and impeller diameter.
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 ft3/s for a head of 30 ft.
What type of turbines are these? How does their perfor-mance compare with Fig. 11.21?
P11.87 An idealized radial turbine is shown in Fig. P11.87. The absolute flow enters at 30° and leaves radially inward.
The flow rate is 3.5 m3/s of water at 20°C. The blade thickness is constant at 10 cm. Compute the theoretical power developed at 100 percent efficiency.
P11.92 At a proposed turbine installation the available head is 800 ft, and the water flow rate is 40,000 gal/min. Dis-cuss the size, speed, and number of turbines which might be suitable for this purpose while using (a) a Pelton wheel and (b) a Francis wheel.
P11.93 Figure P11.93 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, 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, includ-ing idealized velocity vector diagrams.
764 Chapter 11 Turbomachinery V2 b = 10 cm 30° V1 40 cm 70 cm 135 r / min P11.87 P11.88 A certain turbine in Switzerland delivers 25,000 bhp at 500 r/min under a net head of 5330 ft. What type of tur-bine is this? Estimate the approximate discharge and size.
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 diam-eter is 4 in.
P11.90 An idealized radial turbine is shown in Fig. P11.90. The absolute flow enters at 25° with the blade angles as shown. The flow rate is 8 m3/s of water at 20°C. The blade thickness is constant at 20 cm. Compute the theo-retical power developed at 100 percent efficiency.
P11.91 The flow through an axial-flow turbine can be idealized by modifying the stator-rotor diagrams of Fig. 11.12 for energy absorption. Sketch a suitable blade and flow arrangement and the associated velocity vector diagrams.
For further details see chap. 8 of Ref. 25.
b = 20 cm 1.2 m 35° 25° V2 Vr 2 80 r / min Vr 1 30° 0.8 m P11.90 Flow Flow P11.93 P11.94 A simple cross-flow turbine, Fig. P11.93, 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 rev/min.
P11.95 One can make a theoretical estimate of the proper diam-eter for a penstock in an impulse turbine installation, as in Fig. P11.95. Let L and H be known, and let the tur-bine 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 H/3, (b) the optimum jet veloc-ity is (4gH/3)1/2, and (c) the best nozzle diameter is Dj [D5/(2 fL)]1/4, where f is the pipe-friction factor.
(11.20), the empirical criterion given by Wislicenus for cavitation is Nss 11,000 Use this criterion to compute how high z1 z2, the im-peller eye in Fig. P11.98, can be placed for a Francis tur-bine with a head of 300 ft, Nsp 40, and pa 14 lbf/in2 absolute before cavitation occurs in 60°F water.
P11.100 One of the largest wind generators in operation today is the ERDA/NASA two-blade propeller HAWT in San-dusky, Ohio. The blades are 125 ft in diameter and reach maximum power in 19 mi/h winds. For this condition es-timate (a) the power generated in kW, (b) the rotor speed in r/min, and (c) the velocity V2 behind the rotor.
P11.101 A Darrieus VAWT in operation in Lumsden, Saskatchewan, that is 32 ft high and 20 ft in diameter sweeps out an area of 432 ft2. Estimate (a) the maximum power and (b) 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 A very large Darrieus VAWT was constructed by the U.S.
Department of Energy near Sandia, New Mexico. It is 60 ft high and 30 ft in diameter, with a swept area of 1200 ft2. If the turbine is constrained to rotate at 90 r/min, use Fig. 11.31 to plot the predicted power output in kW versus wind speed in the range V 5 to 40 mi/h.
(r/min)(gal/min)1/2 [NPSH (ft)]3/4 Word Problems 765 Impulse wheel Dj Vj Reservoir Penstock: L, D H P11.95 P11.96 Apply the results of Prob. 11.95 to determining the opti-mum (a) penstock diameter and (b) nozzle diameter for the data of Prob. 11.92 with a commercial-steel penstock of length 1500 ft.
P11.97 Consider the following nonoptimum version of Prob.
11.95: H 450 m, L 5 km, D 1.2 m, Dj 20 cm.
The penstock is concrete, 1 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. P11.98. Explain at least two advantages in us-ing a draft tube.
P11.99 Turbines can also cavitate when the pressure at point 1 in Fig. P11.98 drops too low. With NPSH defined by Eq.
1 2 P11.98 Word Problems W11.1 We know that an enclosed rotating bladed impeller will impart energy to a fluid, usually in the form of a pres-sure rise, but how does it actually happen? Discuss, with sketches, the physical mechanisms through which an im-peller actually transfers energy to a fluid.
W11.2 Dynamic pumps (as opposed to PDPs) have difficulty moving highly viscous fluids. Lobanoff and Ross suggest the following rule of thumb: D (in) 0.015/water, where D is the diameter of the discharge pipe. For example, SAE 30W oil ( 300water) should EES require at least a 4.5-in outlet. Can you explain some rea-sons 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 liq-uid temperature?
W11.4 For nondimensional fan performance, Wallis sug-gests that the head coefficient should be replaced by FTP/(n2D2), 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 sub-ject, and report to the class about its effect on pump per-formance.
W11.8 Explain why some pump performance curves may lead to unstable operating conditions.
W11.9 Why are Francis and Kaplan turbines generally consid-ered unsuitable for hydropower sites where the available head exceeds 1000 ft?
W11.10 Do some reading on the performance of the free propeller that is used on small, low-speed aircraft. What dimen-sionless parameters are typically reported for the data?
How do the performance and efficiency compare with those for the axial-flow pump?
766 Chapter 11 Turbomachinery C11.1 Comprehensive Problems C11.1 The net head of a little aquarium pump is given by the man-ufacturer as a function of volume flow rate as listed below: Q, m3/s H, mH2O 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 pump water from the lower reservoir to the upper reservoir as shown in Fig. C11.1? 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.
C11.2 Reconsider Prob. 6.68 as an exercise in pump selection. Se-lect an impeller size and rotational speed from the Byron Jackson pump family of Prob. 11.28 which will deliver a flow rate of 3 ft3/s to the system of Fig. P6.68 at minimum input power. Calculate the horsepower required.
C11.3 Reconsider Prob. 6.77 as an exercise in turbine selection.
Select an impeller size and rotational speed from the Fran-cis turbine family of Fig. 11.21d which will deliver maxi-mum power generated by the turbine. Calculate the turbine power output and remark on the practicality of your design.
C11.4 A pump provides a net head H which is dependent on the volume flow rate Q, as follows: H a bQ2, where a 80 m and b 20 s2/m5. The pump delivers water at 20°C through a horizontal 30-cm-diameter cast-iron pipe which is 120 m long. The pressures at the inlet and exit of the sys-tem are the same. Neglecting minor losses, calculate the ex-pected volume flow rate in gal/min.
C11.5 In Prob. 11.23, estimate the efficiency of the pump in two ways: (a) Read it directly from Fig. 11.7b (for the dynam-ically similar water pump); and (b) Calculate it from Eq.
(11.5) for the actual kerosene flow. Compare your results and discuss any discrepancies.
Pump Q Q 0.80 m References 767 Design Project D11.1 To save on electricity costs, a town water supply system uses gravity-driven flow from five large storage tanks during the day and then refills these tanks from 10 p.m. to 6 a.m. at a cheaper night rate of 7 ¢/kWh. The total resupply 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 which can be controlled at any desirable angle. Select a suitable pump family from one of the six data sets in this chapter: Figs. 11.8, P11.24, and P11.31 plus Probs. 11.28, 11.35, and 11.38. Assume ideal similarity (no Reynolds-num-ber or pump roughness effects). The goal is to determine pump and pipeline sizes which achieve minimum total cost over a 5-year period. Some suggested cost data are 1. Pump and motor: $2500 plus $1500 per inch of pipe size 2. Valves: $100 plus $100 per inch of pipe size 3. Pipelines: 50¢ 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.
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K. Eisele et al., “Flow Analysis in a Pump Diffuser: Part 1, Measurements; Part 2, CFD,” J. Fluids Eng,, vol. 119, De-cember 1997, pp. 968–984.
768 Chapter 11 Turbomachinery Fig. A.1 Absolute viscosity of com-mon fluids at 1 atm.
Appendix A Physical Properties of Fluids Absolute viscosity µ, N ⋅ s / m2 0.5 0.4 0.3 0.2 0.1 0.06 0.04 0.03 0.02 0.01 6 4 3 2 1 × 10 – 3 1 × 10 – 4 1 × 10 – 5 5 – 20 0 20 40 60 80 100 120 Temperature, °C Aniline Mercury Kerosine Air 6 4 3 2 6 4 3 2 Hydrogen Carbon dioxide Helium Benzene Gasoline (SG 0.68) Water Ethyl alcohol Carbon tetrachloride Crude oil (SG 0.86) Castor oil SAE 30 oil Glycerin SAE 10 oil 769 Fig. A.2 Kinematic viscosity of common fluids at 1 atm.
770 Appendix A Kinematic viscosity ν, m2 / s 1 × 10 – 3 8 6 4 3 2 1 × 10 – 4 1 × 10 – 5 1 × 10 – 6 1 × 10 – 7 –20 0 20 40 60 80 100 120 Temperature, °C 8 6 4 3 2 8 6 4 3 2 8 6 4 3 2 Mercury Gasoline (SG 0.68) Carbon tetrachloride Water Ethyl alcohol Kerosine Benzene Crude oil (SG 0.86) Carbon dioxide SAE 30 oil Air and oxygen Hydrogen SAE 10 oil Glycerin Helium Table A.1 Viscosity and Density of Water at 1 atm Suggested curve fits for water in the range 0 T 100°C: (kg/m3) 1000 0.0178T°C 4°C1.7 0.2% ln 0 1.704 5.306z 7.003z2 z 2 T 73 K K 0 1.788 E-3 kg/(m s) Physical Properties of Fluids 771 T, °C , kg/m3 , N s/m2 , m2/s T, °F , slug/ft3 , lb s/ft2 , ft2/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 Table A.2 Viscosity and Density of Air at 1 atm T, °C , kg/m3 , N s/m2 , m2/s T, °F , slug/ft3 , lb s/ft2 , ft2/s 40 1.520 1.51 E-5 0.99 E-5 40 2.94 E-3 3.16 E-7 1.07 E-4 0 1.290 1.71 E-5 1.33 E-5 32 2.51 E-3 3.58 E-7 1.43 E-4 20 1.200 1.80 E-5 1.50 E-5 68 2.34 E-3 3.76 E-7 1.61 E-4 50 1.090 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: R p T Rair 287 J/(kg K) Power law: 0 T T 0 0.7 Sutherland law: 0 T T 0 3/2 T T 0 S S Sair 110.4 K with T0 273 K, 0 1.71 E-5 kg/(m s), and T in kelvins.
Table A.3 Properties of Common Liquids at 1 atm and 20°C (68°F) Molecular Specific-heat Power-law Gas weight R, m2/(s2 K) g, N/m3 , N s/m2 ratio exponent n† H2 2.016 4124 00.822 9.05 E-6 1.41 0.68 He 4.003 2077 01.630 1.97 E-5 1.66 0.67 H2O 18.020 0461 07.350 1.02 E-5 1.33 1.15 Ar 39.944 0208 16.300 2.24 E-5 1.67 0.72 Dry air 28.960 0287 11.800 1.80 E-5 1.40 0.67 CO2 44.010 0189 17.900 1.48 E-5 1.30 0.79 CO 28.010 0297 11.400 1.82 E-5 1.40 0.71 N2 28.020 0297 11.400 1.76 E-5 1.40 0.67 O2 32.000 0260 13.100 2.00 E-5 1.40 0.69 NO 30.010 0277 12.100 1.90 E-5 1.40 0.78 N2O 44.020 0189 17.900 1.45 E-5 1.31 0.89 Cl2 70.910 0117 28.900 1.03 E-5 1.34 1.00 CH4 16.040 0518 06.540 1.34 E-5 1.32 0.87 †The power-law curve fit, Eq. (1.27), /293K (T/293)n, fits these gases to within 4 percent in the range 250 T 1000 K. The temperature must be in kelvins.
772 Appendix A Bulk modulus, Viscosity Liquid , kg/m3 , kg/(m s) , N/m p, N/m2 N/m2 parameter C† Ammonia 13,608 2.20 E-4 2.13 E-2 9.10 E5 — 1.05 Benzene 13,881 6.51 E-4 2.88 E-2 1.01 E4 1.43 E9 4.34 Carbon tetrachloride 31,590 9.67 E-4 2.70 E-2 1.20 E4 9.65 E8 4.45 Ethanol 13,789 1.20 E-3 2.28 E-2 5.73 E3 9.03 E8 5.72 Ethylene glycol 31,117 2.14 E-2 4.84 E-2 1.23 E1 — 11.7 Freon 12 31,327 2.62 E-4 — — — 1.76 Gasoline 13,680 2.92 E-4 2.16 E-2 5.51 E4 9.58 E80 3.68 Glycerin 31,260 1.49 6.33 E-2 1.43 E-2 4.34 E90 28.0 Kerosine 13,804 1.92 E-3 2.83 E-2 3.11 E3 1.63 E90 5.56 Mercury 13,550 1.56 E-3 4.84 E-1 1.13 E-3 2.55 E10 1.07 Methanol 13,791 5.98 E-4 2.25 E-2 1.34 E4 8.33 E80 4.63 SAE 10W oil 13,870 1.04 E-1‡ 3.63 E-2 — 1.31 E90 15.7 SAE 10W30 oil 13,876 1.7 E-1‡ — — — 14.0 SAE 30W oil 13,891 2.9 E-1‡ 3.53 E-2 — 1.38 E90 18.3 SAE 50W oil 13,902 8.6 E-1‡ — — — 20.2 Water 13,998 1.00 E-3 7.28 E-2 2.34 E3 2.19 E90 Table A.1 Seawater (30%) 31,025 1.07 E-3 7.28 E-2 2.34 E3 2.33 E90 7.28 In contact with air.
†The viscosity-temperature variation of these liquids may be fitted to the empirical expression 2 0°C exp C 2 T 93 K K 1 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.
Table A.4 Properties of Common Gases at 1 atm and 20°C (68°F) Table A.5 Surface Tension, Vapor Pressure, and Sound Speed of Water z, m T, K p, Pa , kg/m3 a, m/s 0500 291.41 107,508 1.2854 342.2 0,0000 288.16 101,350 1.2255 340.3 0,0500 284.91 095,480 1.1677 338.4 01,000 281.66 089,889 1.1120 336.5 01,500 278.41 084,565 1.0583 334.5 02,000 275.16 079,500 1.0067 332.6 02,500 271.91 074,684 0.9570 330.6 03,000 268.66 070,107 0.9092 328.6 03,500 265.41 065,759 0.8633 326.6 04,000 262.16 061,633 0.8191 324.6 04,500 258.91 057,718 0.7768 322.6 05,000 255.66 054,008 0.7361 320.6 05,500 252.41 050,493 0.6970 318.5 06,000 249.16 047,166 0.6596 316.5 06,500 245.91 044,018 0.6237 314.4 07,000 242.66 041,043 0.5893 312.3 07,500 239.41 038,233 0.5564 310.2 08,000 236.16 035,581 0.5250 308.1 08,500 232.91 033,080 0.4949 306.0 09,000 229.66 030,723 0.4661 303.8 09,500 226.41 028,504 0.4387 301.7 10,000 223.16 026,416 0.4125 299.5 10,500 219.91 024,455 0.3875 297.3 11,000 216.66 022,612 0.3637 295.1 11,500 216.66 020,897 0.3361 295.1 12,000 216.66 019,312 0.3106 295.1 12,500 216.66 017,847 0.2870 295.1 13,000 216.66 016,494 0.2652 295.1 13,500 216.66 015,243 0.2451 295.1 14,000 216.66 014,087 0.2265 295.1 14,500 216.66 013,018 0.2094 295.1 15,000 216.66 012,031 0.1935 295.1 15,500 216.66 011,118 0.1788 295.1 16,000 216.66 010,275 0.1652 295.1 16,500 216.66 009,496 0.1527 295.1 17,000 216.66 008,775 0.1411 295.1 17,500 216.66 008,110 0.1304 295.1 18,000 216.66 007,495 0.1205 295.1 18,500 216.66 006,926 0.1114 295.1 19,000 216.66 006,401 0.1029 295.1 19,500 216.66 005,915 0.0951 295.1 20,000 216.66 005,467 0.0879 295.1 22,000 218.60 004,048 0.0645 296.4 24,000 220.60 002,972 0.0469 297.8 26,000 222.50 002,189 0.0343 299.1 28,000 224.50 001,616 0.0251 300.4 30,000 226.50 001,197 0.0184 301.7 40,000 250.40 000,287 0.0040 317.2 50,000 270.70 000,080 0.0010 329.9 60,000 255.70 000,022 0.0003 320.6 70,000 219.70 000,006 0.0001 297.2 Physical Properties of Fluids 773 T, °C , N/m p, kPa a, m/s 0 0.0756 0.611 1402 10 0.0742 1.227 1447 20 0.0728 2.337 1482 30 0.0712 4.242 1509 40 0.0696 7.375 1529 50 0.0679 12.34 1542 60 0.0662 19.92 1551 70 0.0644 31.16 1553 80 0.0626 47.35 1554 90 0.0608 70.11 1550 100 0.0589 101.3 1543 120 0.0550 198.5 1518 140 0.0509 361.3 1483 160 0.0466 617.8 1440 180 0.0422 1,002 1389 200 0.0377 1,554 1334 220 0.0331 2,318 1268 240 0.0284 3,344 1192 260 0.0237 4,688 1110 280 0.0190 6,412 1022 300 0.0144 8,581 920 320 0.0099 11,274 800 340 0.0056 14,586 630 360 0.0019 18,651 370 374 0.019 22,090 0 Critical point.
Table A.6 Prop-erties of the Standard Atmosphere Table B.1 Isentropic Flow of a Perfect Gas, k 1.4 Appendix B Compressible-Flow Tables Ma p/p0 /0 T/T0 A/A 0.0 1.0 1.0 1.0 0.02 0.9997 0.9998 0.9999 28.9421 0.04 0.9989 0.9992 0.9997 14.4815 0.06 0.9975 0.9982 0.9993 9.6659 0.08 0.9955 0.9968 0.9987 7.2616 0.1 0.9930 0.9950 0.9980 5.8218 0.12 0.9900 0.9928 0.9971 4.8643 0.14 0.9864 0.9903 0.9961 4.1824 0.16 0.9823 0.9873 0.9949 3.6727 0.18 0.9776 0.9840 0.9936 3.2779 0.2 0.9725 0.9803 0.9921 2.9635 0.22 0.9668 0.9762 0.9904 2.7076 0.24 0.9607 0.9718 0.9886 2.4956 0.26 0.9541 0.9670 0.9867 2.3173 0.28 0.9470 0.9619 0.9846 2.1656 0.3 0.9395 0.9564 0.9823 2.0351 0.32 0.9315 0.9506 0.9799 1.9219 0.34 0.9231 0.9445 0.9774 1.8229 0.36 0.9143 0.9380 0.9747 1.7358 0.38 0.9052 0.9313 0.9719 1.6587 0.4 0.8956 0.9243 0.9690 1.5901 0.42 0.8857 0.9170 0.9659 1.5289 0.44 0.8755 0.9094 0.9627 1.4740 0.46 0.8650 0.9016 0.9594 1.4246 0.48 0.8541 0.8935 0.9559 1.3801 0.5 0.8430 0.8852 0.9524 1.3398 0.52 0.8317 0.8766 0.9487 1.3034 0.54 0.8201 0.8679 0.9449 1.2703 0.56 0.8082 0.8589 0.9410 1.2403 0.58 0.7962 0.8498 0.9370 1.2130 0.6 0.7840 0.8405 0.9328 1.1882 0.62 0.7716 0.8310 0.9286 1.1656 0.64 0.7591 0.8213 0.9243 1.1451 0.66 0.7465 0.8115 0.9199 1.1265 0.68 0.7338 0.8016 0.9153 1.1097 0.7 0.7209 0.7916 0.9107 1.0944 0.72 0.7080 0.7814 0.9061 1.0806 Ma p/p0 /0 T/T0 A/A 0.74 0.6951 0.7712 0.9013 1.0681 0.76 0.6821 0.7609 0.8964 1.0570 0.78 0.6690 0.7505 0.8915 1.0471 0.8 0.6560 0.7400 0.8865 1.0382 0.82 0.6430 0.7295 0.8815 1.0305 0.84 0.6300 0.7189 0.8763 1.0237 0.86 0.6170 0.7083 0.8711 1.0179 0.88 0.6041 0.6977 0.8659 1.0129 0.9 0.5913 0.6870 0.8606 1.0089 0.92 0.5785 0.6764 0.8552 1.0056 0.94 0.5658 0.6658 0.8498 1.0031 0.96 0.5532 0.6551 0.8444 1.0014 0.98 0.5407 0.6445 0.8389 1.0003 1.0 0.5283 0.6339 0.8333 1.0000 1.02 0.5160 0.6234 0.8278 1.0003 1.04 0.5039 0.6129 0.8222 1.0013 1.06 0.4919 0.6024 0.8165 1.0029 1.08 0.4800 0.5920 0.8108 1.0051 1.1 0.4684 0.5817 0.8052 1.0079 1.12 0.4568 0.5714 0.7994 1.0113 1.14 0.4455 0.5612 0.7937 1.0153 1.16 0.4343 0.5511 0.7879 1.0198 1.18 0.4232 0.5411 0.7822 1.0248 1.2 0.4124 0.5311 0.7764 1.0304 1.22 0.4017 0.5213 0.7706 1.0366 1.24 0.3912 0.5115 0.7648 1.0432 1.26 0.3809 0.5019 0.7590 1.0504 1.28 0.3708 0.4923 0.7532 1.0581 1.3 0.3609 0.4829 0.7474 1.0663 1.32 0.3512 0.4736 0.7416 1.0750 1.34 0.3417 0.4644 0.7358 1.0842 1.36 0.3323 0.4553 0.7300 1.0940 1.38 0.3232 0.4463 0.7242 1.1042 1.4 0.3142 0.4374 0.7184 1.1149 1.42 0.3055 0.4287 0.7126 1.1262 1.44 0.2969 0.4201 0.7069 1.1379 1.46 0.2886 0.4116 0.7011 1.1501 774 Ma p/p0 /0 T/T0 A/A 2.56 0.0533 0.1232 0.4328 2.7891 2.58 0.0517 0.1205 0.4289 2.8420 2.6 0.0501 0.1179 0.4252 2.8960 2.62 0.0486 0.1153 0.4214 2.9511 2.64 0.0471 0.1128 0.4177 3.0073 2.66 0.0457 0.1103 0.4141 3.0647 2.68 0.0443 0.1079 0.4104 3.1233 2.7 0.0430 0.1056 0.4068 3.1830 2.72 0.0417 0.1033 0.4033 3.2440 2.74 0.0404 0.1010 0.3998 3.3061 2.76 0.0392 0.0989 0.3963 3.3695 2.78 0.0380 0.0967 0.3928 3.4342 2.8 0.0368 0.0946 0.3894 3.5001 2.82 0.0357 0.0926 0.3860 3.5674 2.84 0.0347 0.0906 0.3827 3.6359 2.86 0.0336 0.0886 0.3794 3.7058 2.88 0.0326 0.0867 0.3761 3.7771 2.9 0.0317 0.0849 0.3729 3.8498 2.92 0.0307 0.0831 0.3696 3.9238 2.94 0.0298 0.0813 0.3665 3.9993 2.96 0.0289 0.0796 0.3633 4.0763 2.98 0.0281 0.0779 0.3602 4.1547 3.0 0.0272 0.0762 0.3571 4.2346 3.02 0.0264 0.0746 0.3541 4.3160 3.04 0.0256 0.0730 0.3511 4.3990 3.06 0.0249 0.0715 0.3481 4.4835 3.08 0.0242 0.0700 0.3452 4.5696 3.1 0.0234 0.0685 0.3422 4.6573 3.12 0.0228 0.0671 0.3393 4.7467 3.14 0.0221 0.0657 0.3365 4.8377 3.16 0.0215 0.0643 0.3337 4.9304 3.18 0.0208 0.0630 0.3309 5.0248 3.2 0.0202 0.0617 0.3281 5.1210 3.22 0.0196 0.0604 0.3253 5.2189 3.24 0.0191 0.0591 0.3226 5.3186 3.26 0.0185 0.0579 0.3199 5.4201 3.28 0.0180 0.0567 0.3173 5.5234 3.3 0.0175 0.0555 0.3147 5.6286 3.32 0.0170 0.0544 0.3121 5.7358 3.34 0.0165 0.0533 0.3095 5.8448 3.36 0.0160 0.0522 0.3069 5.9558 3.38 0.0156 0.0511 0.3044 6.0687 3.4 0.0151 0.0501 0.3019 6.1837 3.42 0.0147 0.0491 0.2995 6.3007 3.44 0.0143 0.0481 0.2970 6.4198 3.46 0.0139 0.0471 0.2946 6.5409 3.48 0.0135 0.0462 0.2922 6.6642 3.5 0.0131 0.0452 0.2899 6.7896 3.52 0.0127 0.0443 0.2875 6.9172 3.54 0.0124 0.0434 0.2852 7.0471 3.56 0.0120 0.0426 0.2829 7.1791 3.58 0.0117 0.0417 0.2806 7.3135 3.6 0.0114 0.0409 0.2784 7.4501 3.62 0.0111 0.0401 0.2762 7.5891 Table B.1 (Cont.) Isentropic Flow of a Perfect Gas, k 1.4 Ma p/p0 /0 T/T0 A/A 1.48 0.2804 0.4032 0.6954 1.1629 1.5 0.2724 0.3950 0.6897 1.1762 1.52 0.2646 0.3869 0.6840 1.1899 1.54 0.2570 0.3789 0.6783 1.2042 1.56 0.2496 0.3710 0.6726 1.2190 1.58 0.2423 0.3633 0.6670 1.2344 1.6 0.2353 0.3557 0.6614 1.2502 1.62 0.2284 0.3483 0.6558 1.2666 1.64 0.2217 0.3409 0.6502 1.2836 1.66 0.2151 0.3337 0.6447 1.3010 1.68 0.2088 0.3266 0.6392 1.3190 1.7 0.2026 0.3197 0.6337 1.3376 1.72 0.1966 0.3129 0.6283 1.3567 1.74 0.1907 0.3062 0.6229 1.3764 1.76 0.1850 0.2996 0.6175 1.3967 1.78 0.1794 0.2931 0.6121 1.4175 1.8 0.1740 0.2868 0.6068 1.4390 1.82 0.1688 0.2806 0.6015 1.4610 1.84 0.1637 0.2745 0.5963 1.4836 1.86 0.1587 0.2686 0.5910 1.5069 1.88 0.1539 0.2627 0.5859 1.5308 1.9 0.1492 0.2570 0.5807 1.5553 1.92 0.1447 0.2514 0.5756 1.5804 1.94 0.1403 0.2459 0.5705 1.6062 1.96 0.1360 0.2405 0.5655 1.6326 1.98 0.1318 0.2352 0.5605 1.6597 2.0 0.1278 0.2300 0.5556 1.6875 2.02 0.1239 0.2250 0.5506 1.7160 2.04 0.1201 0.2200 0.5458 1.7451 2.06 0.1164 0.2152 0.5409 1.7750 2.08 0.1128 0.2104 0.5361 1.8056 2.1 0.1094 0.2058 0.5313 1.8369 2.12 0.1060 0.2013 0.5266 1.8690 2.14 0.1027 0.1968 0.5219 1.9018 2.16 0.0996 0.1925 0.5173 1.9354 2.18 0.0965 0.1882 0.5127 1.9698 2.2 0.0935 0.1841 0.5081 2.0050 2.22 0.0906 0.1800 0.5036 2.0409 2.24 0.0878 0.1760 0.4991 2.0777 2.26 0.0851 0.1721 0.4947 2.1153 2.28 0.0825 0.1683 0.4903 2.1538 2.3 0.0800 0.1646 0.4859 2.1931 2.32 0.0775 0.1609 0.4816 2.2333 2.34 0.0751 0.1574 0.4773 2.2744 2.36 0.0728 0.1539 0.4731 2.3164 2.38 0.0706 0.1505 0.4688 2.3593 2.4 0.0684 0.1472 0.4647 2.4031 2.42 0.0663 0.1439 0.4606 2.4479 2.44 0.0643 0.1408 0.4565 2.4936 2.46 0.0623 0.1377 0.4524 2.5403 2.48 0.0604 0.1346 0.4484 2.5880 2.5 0.0585 0.1317 0.4444 2.6367 2.52 0.0567 0.1288 0.4405 2.6865 2.54 0.0550 0.1260 0.4366 2.7372 Compressible-Flow Tables 775 776 Appendix B Table B.1 (Cont.) Isentropic Flow of a Perfect Gas, k 1.4 Ma p/p0 /0 T/T0 A/A 3.64 0.0108 0.0393 0.2740 7.7305 3.66 0.0105 0.0385 0.2718 7.8742 3.68 0.0102 0.0378 0.2697 8.0204 3.7 0.0099 0.0370 0.2675 8.1691 3.72 0.0096 0.0363 0.2654 8.3202 3.74 0.0094 0.0356 0.2633 8.4739 3.76 0.0091 0.0349 0.2613 8.6302 3.78 0.0089 0.0342 0.2592 8.7891 3.8 0.0086 0.0335 0.2572 8.9506 3.82 0.0084 0.0329 0.2552 9.1148 3.84 0.0082 0.0323 0.2532 9.2817 3.86 0.0080 0.0316 0.2513 9.4513 3.88 0.0077 0.0310 0.2493 9.6237 3.9 0.0075 0.0304 0.2474 9.7990 3.92 0.0073 0.0299 0.2455 9.9771 3.94 0.0071 0.0293 0.2436 10.1581 3.96 0.0069 0.0287 0.2418 10.3420 3.98 0.0068 0.0282 0.2399 10.5289 4.0 0.0066 0.0277 0.2381 10.7188 4.02 0.0064 0.0271 0.2363 10.9117 4.04 0.0062 0.0266 0.2345 11.1077 4.06 0.0061 0.0261 0.2327 11.3068 4.08 0.0059 0.0256 0.2310 11.5091 4.1 0.0058 0.0252 0.2293 11.7147 4.12 0.0056 0.0247 0.2275 11.9234 4.14 0.0055 0.0242 0.2258 12.1354 4.16 0.0053 0.0238 0.2242 12.3508 4.18 0.0052 0.0234 0.2225 12.5695 4.2 0.0051 0.0229 0.2208 12.7916 4.22 0.0049 0.0225 0.2192 13.0172 4.24 0.0048 0.0221 0.2176 13.2463 4.26 0.0047 0.0217 0.2160 13.4789 4.28 0.0046 0.0213 0.2144 13.7151 4.3 0.0044 0.0209 0.2129 13.9549 4.32 0.0043 0.0205 0.2113 14.1984 Ma p/p0 /0 T/T0 A/A 4.34 0.0042 0.0202 0.2098 14.4456 4.36 0.0041 0.0198 0.2083 14.6965 4.38 0.0040 0.0194 0.2067 14.9513 4.4 0.0039 0.0191 0.2053 15.2099 4.42 0.0038 0.0187 0.2038 15.4724 4.44 0.0037 0.0184 0.2023 15.7388 4.46 0.0036 0.0181 0.2009 16.0092 4.48 0.0035 0.0178 0.1994 16.2837 4.5 0.0035 0.0174 0.1980 16.5622 4.52 0.0034 0.0171 0.1966 16.8449 4.54 0.0033 0.0168 0.1952 17.1317 4.56 0.0032 0.0165 0.1938 17.4228 4.58 0.0031 0.0163 0.1925 17.7181 4.6 0.0031 0.0160 0.1911 18.0178 4.62 0.0030 0.0157 0.1898 18.3218 4.64 0.0029 0.0154 0.1885 18.6303 4.66 0.0028 0.0152 0.1872 18.9433 4.68 0.0028 0.0149 0.1859 19.2608 4.7 0.0027 0.0146 0.1846 19.5828 4.72 0.0026 0.0144 0.1833 19.9095 4.74 0.0026 0.0141 0.1820 20.2409 4.76 0.0025 0.0139 0.1808 20.5770 4.78 0.0025 0.0137 0.1795 20.9179 4.8 0.0024 0.0134 0.1783 21.2637 4.82 0.0023 0.0132 0.1771 21.6144 4.84 0.0023 0.0130 0.1759 21.9700 4.86 0.0022 0.0128 0.1747 22.3306 4.88 0.0022 0.0125 0.1735 22.6963 4.9 0.0021 0.0123 0.1724 23.0671 4.92 0.0021 0.0121 0.1712 23.4431 4.94 0.0020 0.0119 0.1700 23.8243 4.96 0.0020 0.0117 0.1689 24.2109 4.98 0.0019 0.0115 0.1678 24.6027 5.0 0.0019 0.0113 0.1667 25.0000 Man1 Man2 p2/p1 V1/V2 2/1 T2/T1 p02/p01 A 2/A 1 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.02 0.9805 1.0471 1.0334 1.0132 1.0000 1.0000 1.04 0.9620 1.0952 1.0671 1.0263 0.9999 1.0001 1.06 0.9444 1.1442 1.1009 1.0393 0.9998 1.0002 1.08 0.9277 1.1941 1.1349 1.0522 0.9994 1.0006 1.1 0.9118 1.2450 1.1691 1.0649 0.9989 1.0011 1.12 0.8966 1.2968 1.2034 1.0776 0.9982 1.0018 1.14 0.8820 1.3495 1.2378 1.0903 0.9973 1.0027 1.16 0.8682 1.4032 1.2723 1.1029 0.9961 1.0040 1.18 0.8549 1.4578 1.3069 1.1154 0.9946 1.0055 1.2 0.8422 1.5133 1.3416 1.1280 0.9928 1.0073 1.22 0.8300 1.5698 1.3764 1.1405 0.9907 1.0094 1.24 0.8183 1.6272 1.4112 1.1531 0.9884 1.0118 Table B.2 Normal-Shock Relations for a Perfect Gas, k 1.4 Compressible-Flow Tables 777 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4 Man1 Man2 p2/p1 V1/V2 2/1 T2/T1 p02/p01 A 2/A 1 1.26 0.8071 1.6855 1.4460 1.1657 0.9857 1.0145 1.28 0.7963 1.7448 1.4808 1.1783 0.9827 1.0176 1.3 0.7860 1.8050 1.5157 1.1909 0.9794 1.0211 1.32 0.7760 1.8661 1.5505 1.2035 0.9758 1.0249 1.34 0.7664 1.9282 1.5854 1.2162 0.9718 1.0290 1.36 0.7572 1.9912 1.6202 1.2290 0.9676 1.0335 1.38 0.7483 2.0551 1.6549 1.2418 0.9630 1.0384 1.4 0.7397 2.1200 1.6897 1.2547 0.9582 1.0436 1.42 0.7314 2.1858 1.7243 1.2676 0.9531 1.0492 1.44 0.7235 2.2525 1.7589 1.2807 0.9476 1.0552 1.46 0.7157 2.3202 1.7934 1.2938 0.9420 1.0616 1.48 0.7083 2.3888 1.8278 1.3069 0.9360 1.0684 1.5 0.7011 2.4583 1.8621 1.3202 0.9298 1.0755 1.52 0.6941 2.5288 1.8963 1.3336 0.9233 1.0830 1.54 0.6874 2.6002 1.9303 1.3470 0.9166 1.0910 1.56 0.6809 2.6725 1.9643 1.3606 0.9097 1.0993 1.58 0.6746 2.7458 1.9981 1.3742 0.9026 1.1080 1.6 0.6684 2.8200 2.0317 1.3880 0.8952 1.1171 1.62 0.6625 2.8951 2.0653 1.4018 0.8877 1.1266 1.64 0.6568 2.9712 2.0986 1.4158 0.8799 1.1365 1.66 0.6512 3.0482 2.1318 1.4299 0.8720 1.1468 1.68 0.6458 3.1261 2.1649 1.4440 0.8639 1.1575 1.7 0.6405 3.2050 2.1977 1.4583 0.8557 1.1686 1.72 0.6355 3.2848 2.2304 1.4727 0.8474 1.1801 1.74 0.6305 3.3655 2.2629 1.4873 0.8389 1.1921 1.76 0.6257 3.4472 2.2952 1.5019 0.8302 1.2045 1.78 0.6210 3.5298 2.3273 1.5167 0.8215 1.2173 1.8 0.6165 3.6133 2.3592 1.5316 0.8127 1.2305 1.82 0.6121 3.6978 2.3909 1.5466 0.8038 1.2441 1.84 0.6078 3.7832 2.4224 1.5617 0.7948 1.2582 1.86 0.6036 3.8695 2.4537 1.5770 0.7857 1.2728 1.88 0.5996 3.9568 2.4848 1.5924 0.7765 1.2877 1.9 0.5956 4.0450 2.5157 1.6079 0.7674 1.3032 1.92 0.5918 4.1341 2.5463 1.6236 0.7581 1.3191 1.94 0.5880 4.2242 2.5767 1.6394 0.7488 1.3354 1.96 0.5844 4.3152 2.6069 1.6553 0.7395 1.3522 1.98 0.5808 4.4071 2.6369 1.6713 0.7302 1.3695 2.0 0.5774 4.5000 2.6667 1.6875 0.7209 1.3872 2.02 0.5740 4.5938 2.6962 1.7038 0.7115 1.4054 2.04 0.5707 4.6885 2.7255 1.7203 0.7022 1.4241 2.06 0.5675 4.7842 2.7545 1.7369 0.6928 1.4433 2.08 0.5643 4.8808 2.7833 1.7536 0.6835 1.4630 2.1 0.5613 4.9783 2.8119 1.7705 0.6742 1.4832 2.12 0.5583 5.0768 2.8402 1.7875 0.6649 1.5039 2.14 0.5554 5.1762 2.8683 1.8046 0.6557 1.5252 2.16 0.5525 5.2765 2.8962 1.8219 0.6464 1.5469 2.18 0.5498 5.3778 2.9238 1.8393 0.6373 1.5692 2.2 0.5471 5.4800 2.9512 1.8569 0.6281 1.5920 2.22 0.5444 5.5831 2.9784 1.8746 0.6191 1.6154 2.24 0.5418 5.6872 3.0053 1.8924 0.6100 1.6393 2.26 0.5393 5.7922 3.0319 1.9104 0.6011 1.6638 2.28 0.5368 5.8981 3.0584 1.9285 0.5921 1.6888 2.3 0.5344 6.0050 3.0845 1.9468 0.5833 1.7144 2.32 0.5321 6.1128 3.1105 1.9652 0.5745 1.7406 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4 Man1 Man2 p2/p1 V1/V2 2/1 T2/T1 p02/p01 A 2/A 1 2.34 0.5297 6.2215 3.1362 1.9838 0.5658 1.7674 2.36 0.5275 6.3312 3.1617 2.0025 0.5572 1.7948 2.38 0.5253 6.4418 3.1869 2.0213 0.5486 1.8228 2.4 0.5231 6.5533 3.2119 2.0403 0.5401 1.8514 2.42 0.5210 6.6658 3.2367 2.0595 0.5317 1.8806 2.44 0.5189 6.7792 3.2612 2.0788 0.5234 1.9105 2.46 0.5169 6.8935 3.2855 2.0982 0.5152 1.9410 2.48 0.5149 7.0088 3.3095 2.1178 0.5071 1.9721 2.5 0.5130 7.1250 3.3333 2.1375 0.4990 2.0039 2.52 0.5111 7.2421 3.3569 2.1574 0.4911 2.0364 2.54 0.5092 7.3602 3.3803 2.1774 0.4832 2.0696 2.56 0.5074 7.4792 3.4034 2.1976 0.4754 2.1035 2.58 0.5056 7.5991 3.4263 2.2179 0.4677 2.1381 2.6 0.5039 7.7200 3.4490 2.2383 0.4601 2.1733 2.62 0.5022 7.8418 3.4714 2.2590 0.4526 2.2093 2.64 0.5005 7.9645 3.4937 2.2797 0.4452 2.2461 2.66 0.4988 8.0882 3.5157 2.3006 0.4379 2.2835 2.68 0.4972 8.2128 3.5374 2.3217 0.4307 2.3218 2.7 0.4956 8.3383 3.5590 2.3429 0.4236 2.3608 2.72 0.4941 8.4648 3.5803 2.3642 0.4166 2.4005 2.74 0.4926 8.5922 3.6015 2.3858 0.4097 2.4411 2.76 0.4911 8.7205 3.6224 2.4074 0.4028 2.4825 2.78 0.4896 8.8498 3.6431 2.4292 0.3961 2.5246 2.8 0.4882 8.9800 2.6636 2.4512 0.3895 2.5676 2.82 0.4868 9.1111 3.6838 2.4733 0.3829 2.6115 2.84 0.4854 9.2432 3.7039 2.4955 0.3765 2.6561 2.86 0.4840 9.3762 3.7238 2.5179 0.3701 2.7017 2.88 0.4827 9.5101 3.7434 2.5405 0.3639 2.7481 2.9 0.4814 9.6450 3.7629 2.5632 0.3577 2.7954 2.92 0.4801 9.7808 3.7821 2.5861 0.3517 2.8436 2.94 0.4788 9.9175 3.8012 2.6091 0.3457 2.8927 2.96 0.4776 10.0552 3.8200 2.6322 0.3398 2.9427 2.98 0.4764 10.1938 3.8387 2.6555 0.3340 2.9937 3.0 0.4752 10.3333 3.8571 2.6790 0.3283 3.0456 3.02 0.4740 10.4738 3.8754 2.7026 0.3227 3.0985 3.04 0.4729 10.6152 3.8935 2.7264 0.3172 3.1523 3.06 0.4717 10.7575 3.9114 2.7503 0.3118 3.2072 3.08 0.4706 10.9008 3.9291 2.7744 0.3065 3.2630 3.1 0.4695 11.0450 3.9466 2.7986 0.3012 3.3199 3.12 0.4685 11.1901 3.9639 2.8230 0.2960 3.3778 3.14 0.4674 11.3362 3.9811 2.8475 0.2910 3.4368 3.16 0.4664 11.4832 3.9981 2.8722 0.2860 3.4969 3.18 0.4654 11.6311 4.0149 2.8970 0.2811 3.5580 3.2 0.4643 11.7800 4.0315 2.9220 0.2762 3.6202 3.22 0.4634 11.9298 4.0479 2.9471 0.2715 3.6835 3.24 0.4624 12.0805 4.0642 2.9724 0.2668 3.7480 3.26 0.4614 12.2322 4.0803 2.9979 0.2622 3.8136 3.28 0.4605 12.3848 4.0963 3.0234 0.2577 3.8803 3.3 0.4596 12.5383 4.1120 3.0492 0.2533 3.9483 3.32 0.4587 12.6928 4.1276 3.0751 0.2489 4.0174 3.34 0.4578 12.8482 4.1431 3.1011 0.2446 4.0877 3.36 0.4569 13.0045 4.1583 3.1273 0.2404 4.1593 3.38 0.4560 13.1618 4.1734 3.1537 0.2363 4.2321 3.4 0.4552 13.3200 4.1884 3.1802 0.2322 4.3062 778 Appendix B Compressible-Flow Tables 779 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4 Man1 Man2 p2/p1 V1/V2 2/1 T2/T1 p02/p01 A 2/A 1 3.42 0.4544 13.4791 4.2032 3.2069 0.2282 4.3815 3.44 0.4535 13.6392 4.2178 3.2337 0.2243 4.4581 3.46 0.4527 13.8002 4.2323 3.2607 0.2205 4.5361 3.48 0.4519 13.9621 4.2467 3.2878 0.2167 4.6154 3.5 0.4512 14.1250 4.2609 3.3151 0.2129 4.6960 3.52 0.4504 14.2888 4.2749 3.3425 0.2093 4.7780 3.54 0.4496 14.4535 4.2888 3.3701 0.2057 4.8614 3.56 0.4489 14.6192 4.3026 3.3978 0.2022 4.9461 3.58 0.4481 14.7858 4.3162 3.4257 0.1987 5.0324 3.6 0.4474 14.9533 4.3296 3.4537 0.1953 5.1200 3.62 0.4467 15.1218 4.3429 3.4819 0.1920 5.2091 3.64 0.4460 15.2912 4.3561 3.5103 0.1887 5.2997 3.66 0.4453 15.4615 4.3692 3.5388 0.1855 5.3918 3.68 0.4446 15.6328 4.3821 3.5674 0.1823 5.4854 3.7 0.4439 15.8050 4.3949 3.5962 0.1792 5.5806 3.72 0.4433 15.9781 4.4075 3.6252 0.1761 5.6773 3.74 0.4426 16.1522 4.4200 3.6543 0.1731 5.7756 3.76 0.4420 16.3272 4.4324 3.6836 0.1702 5.8755 3.78 0.4414 16.5031 4.4447 3.7130 0.1673 5.9770 3.8 0.4407 16.6800 4.4568 3.7426 0.1645 6.0801 3.82 0.4401 16.8578 4.4688 3.7723 0.1617 6.1849 3.84 0.4395 17.0365 4.4807 3.8022 0.1589 6.2915 3.86 0.4389 17.2162 4.4924 3.8323 0.1563 6.3997 3.88 0.4383 17.3968 4.5041 3.8625 0.1536 6.5096 3.9 0.4377 17.5783 4.4156 3.8928 0.1510 6.6213 3.92 0.4372 17.7608 4.5270 3.9233 0.1485 6.7348 3.94 0.4366 17.9442 4.5383 3.9540 0.1460 6.8501 3.96 0.4360 18.1285 4.5494 3.9848 0.1435 6.9672 3.98 0.4355 18.3138 4.5605 4.0158 0.1411 7.0861 4.0 0.4350 18.5000 4.5714 4.0469 0.1388 7.2069 4.02 0.4344 18.6871 4.5823 4.0781 0.1364 7.3296 4.04 0.4339 18.8752 4.5930 4.1096 0.1342 7.4542 4.06 0.4334 19.0642 4.6036 4.1412 0.1319 7.5807 4.08 0.4329 19.2541 4.6141 4.1729 0.1297 7.7092 4.1 0.4324 19.4450 4.6245 4.2048 0.1276 7.8397 4.12 0.4319 19.6368 4.6348 4.2368 0.1254 7.9722 4.14 0.4314 19.8295 4.6450 4.2690 0.1234 8.1067 4.16 0.4309 20.0232 4.6550 4.3014 0.1213 8.2433 4.18 0.4304 20.2178 4.6650 4.3339 0.1193 8.3819 4.2 0.4299 20.4133 4.6749 4.3666 0.1173 8.5227 4.22 0.4295 20.6098 4.6847 4.3994 0.1154 8.6656 4.24 0.4290 20.8072 4.6944 4.4324 0.1135 8.8107 4.26 0.4286 21.0055 4.7040 4.4655 0.1116 8.9579 4.28 0.4281 21.2048 4.7135 4.4988 0.1098 9.1074 4.3 0.4277 21.4050 4.7229 4.5322 0.1080 9.2591 4.32 0.4272 21.6061 4.7322 4.5658 0.1062 9.4131 4.34 0.4268 21.8082 4.7414 4.5995 0.1045 9.5694 4.36 0.4264 22.0112 4.7505 4.6334 0.1028 9.7280 4.38 0.4260 22.2151 4.7595 4.6675 0.1011 9.8889 4.4 0.4255 22.4200 4.7685 4.7017 0.0995 10.0522 4.42 0.4251 22.6258 4.7773 4.7361 0.0979 10.2179 4.44 0.4247 22.8325 4.7861 4.7706 0.0963 10.3861 4.46 0.4243 23.0402 4.7948 4.8053 0.0947 10.5567 4.48 0.4239 23.2488 4.8034 4.8401 0.0932 10.7298 Table B.2 (Cont.) Normal-Shock Relations for a Perfect Gas, k 1.4 780 Appendix B Man1 Man2 p2/p1 V1/V2 2/1 T2/T1 p02/p01 A 2/A 1 4.5 0.4236 23.4583 4.8119 4.8751 0.0917 10.9054 4.52 0.4232 23.6688 4.8203 4.9102 0.0902 11.0835 4.54 0.4228 23.8802 4.8287 4.9455 0.0888 11.2643 4.56 0.4224 24.0925 4.8369 4.9810 0.0874 11.4476 4.58 0.4220 24.3058 4.8451 5.0166 0.0860 11.6336 4.6 0.4217 24.5200 4.8532 5.0523 0.0846 11.8222 4.62 0.4213 24.7351 4.8612 5.0882 0.0832 12.0136 4.64 0.4210 24.9512 4.8692 5.1243 0.0819 12.2076 4.66 0.4206 25.1682 4.8771 5.1605 0.0806 12.4044 4.68 0.4203 25.3861 4.8849 5.1969 0.0793 12.6040 4.7 0.4199 25.6050 4.8926 5.2334 0.0781 12.8065 4.72 0.4196 25.8248 4.9002 5.2701 0.0769 13.0117 4.74 0.4192 26.0455 4.9078 5.3070 0.0756 13.2199 4.76 0.4189 26.2672 4.9153 5.3440 0.0745 13.4310 4.78 0.4186 26.4898 4.9227 5.3811 0.0733 13.6450 4.8 0.4183 26.7133 4.9301 5.4184 0.0721 13.8620 4.82 0.4179 26.9378 4.9374 5.4559 0.0710 14.0820 4.84 0.4176 27.1632 4.9446 5.4935 0.0699 14.3050 4.86 0.4173 27.3895 4.9518 5.5313 0.0688 14.5312 4.88 0.4170 27.6168 4.9589 5.5692 0.0677 14.7604 4.9 0.4167 27.8450 4.9659 5.6073 0.0667 14.9928 4.92 0.4164 28.0741 4.9728 5.6455 0.0657 15.2284 4.94 0.4161 28.3042 4.9797 5.6839 0.0647 15.4672 4.96 0.4158 28.5352 4.9865 5.7224 0.0637 15.7902 4.98 0.4155 28.7671 4.9933 5.7611 0.0627 15.9545 5.0 0.4152 29.0000 5.0000 5.8000 0.0617 16.2032 Table B.3 Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4 Ma f L/D p/p T/T / V/V p0/p 0 0.0 1.2000 0.0 0.02 1778.4500 54.7701 1.1999 0.0219 28.9421 0.04 440.3520 27.3817 1.1996 0.0438 14.4815 0.06 193.0310 18.2508 1.1991 0.0657 9.6659 0.08 106.7180 13.6843 1.1985 0.0876 7.2616 0.1 66.9216 10.9435 1.1976 0.1094 5.8218 0.12 45.4080 9.1156 1.1966 0.1313 4.8643 0.14 32.5113 7.8093 1.1953 0.1531 4.1824 0.16 24.1978 6.8291 1.1939 0.1748 3.6727 0.18 18.5427 6.0662 1.1923 0.1965 3.2779 0.2 14.5333 5.4554 1.1905 0.2182 2.9635 0.22 11.5961 4.9554 1.1885 0.2398 2.7076 0.24 9.3865 4.5383 1.1863 0.2614 2.4956 0.26 7.6876 4.1851 1.1840 0.2829 2.3173 0.28 6.3572 3.8820 1.1815 0.3043 2.1656 0.3 5.2993 3.6191 1.1788 0.3257 2.0351 0.32 4.4467 3.3887 1.1759 0.3470 1.9219 0.34 3.7520 3.1853 1.1729 0.3682 1.8229 0.36 3.1801 3.0042 1.1697 0.3893 1.7358 0.38 2.7054 2.8420 1.1663 0.4104 1.6587 0.4 2.3085 2.6958 1.1628 0.4313 1.5901 0.42 1.9744 2.5634 1.1591 0.4522 1.5289 Compressible-Flow Tables 781 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4 Ma f L/D p/p T/T / V/V p0/p 0 0.44 1.6915 2.4428 1.1553 0.4729 1.4740 0.46 1.4509 2.3326 1.1513 0.4936 1.4246 0.48 1.2453 2.2313 1.1471 0.5141 1.3801 0.5 1.0691 2.1381 1.1429 0.5345 1.3398 0.52 0.9174 2.0519 1.1384 0.5548 1.3034 0.54 0.7866 1.9719 1.1339 0.5750 1.2703 0.56 0.6736 1.8975 1.1292 0.5951 1.2403 0.58 0.5757 1.8282 1.1244 0.6150 1.2130 0.6 0.4908 1.7634 1.1194 0.6348 1.1882 0.62 0.4172 1.7026 1.1143 0.6545 1.1656 0.64 0.3533 1.6456 1.1091 0.6740 1.1451 0.66 0.2979 1.5919 1.1038 0.6934 1.1265 0.68 0.2498 1.5413 1.0984 0.7127 1.1097 0.7 0.2081 1.4935 1.0929 0.7318 1.0944 0.72 0.1721 1.4482 1.0873 0.7508 1.0806 0.74 0.1411 1.4054 1.0815 0.7696 1.0681 0.76 0.1145 1.3647 1.0757 0.7883 1.0570 0.78 0.0917 1.3261 1.0698 0.8068 1.0471 0.8 0.0723 1.2893 1.0638 0.8251 1.0382 0.82 0.0559 1.2542 1.0578 0.8433 1.0305 0.84 0.0423 1.2208 1.0516 0.8614 1.0237 0.86 0.0310 1.1889 1.0454 0.8793 1.0179 0.88 0.0218 1.1583 1.0391 0.8970 1.0129 0.9 0.0145 1.1291 1.0327 0.9146 1.0089 0.92 0.0089 1.1011 1.0263 0.9320 1.0056 0.94 0.0048 1.0743 1.0198 0.9493 1.0031 0.96 0.0021 1.0485 1.0132 0.9663 1.0014 0.98 0.0005 1.0238 1.0066 0.9833 1.0003 1.0 0.0000 1.0000 1.0000 1.0000 1.0000 1.02 0.0005 0.9771 0.9933 1.0166 1.0003 1.04 0.0018 0.9551 0.9866 1.0330 1.0013 1.06 0.0038 0.9338 0.9798 1.0492 1.0029 1.08 0.0066 0.9133 0.9730 1.0653 1.0051 1.1 0.0099 0.8936 0.9662 1.0812 1.0079 1.12 0.0138 0.8745 0.9593 1.0970 1.0113 1.14 0.0182 0.8561 0.9524 1.1126 1.0153 1.16 0.0230 0.8383 0.9455 1.1280 1.0198 1.18 0.0281 0.8210 0.9386 1.1432 1.0248 1.2 0.0336 0.8044 0.9317 1.1583 1.0304 1.22 0.0394 0.7882 0.9247 1.1732 1.0366 1.24 0.0455 0.7726 0.9178 1.1879 1.0432 1.26 0.0517 0.7574 0.9108 1.2025 1.0504 1.28 0.0582 0.7427 0.9038 1.2169 1.0581 1.3 0.0648 0.7285 0.8969 1.2311 1.0663 1.32 0.0716 0.7147 0.8899 1.2452 1.0750 1.34 0.0785 0.7012 0.8829 1.2591 1.0842 1.36 0.0855 0.6882 0.8760 1.2729 1.0940 1.38 0.0926 0.6755 0.8690 1.2864 1.1042 1.4 0.0997 0.6632 0.8621 1.2999 1.1149 1.42 0.1069 0.6512 0.8551 1.3131 1.1262 1.44 0.1142 0.6396 0.8482 1.3262 1.1379 1.46 0.1215 0.6282 0.8413 1.3392 1.1501 1.48 0.1288 0.6172 0.8344 1.3520 1.1629 1.5 0.1361 0.6065 0.8276 1.3646 1.1762 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4 782 Appendix B Ma f L/D p/p T/T / V/V p0/p 0 1.52 0.1433 0.5960 0.8207 1.3770 1.1899 1.54 0.1506 0.5858 0.8139 1.3894 1.2042 1.56 0.1579 0.5759 0.8071 1.4015 1.2190 1.58 0.1651 0.5662 0.8004 1.4135 1.2344 1.6 0.1724 0.5568 0.7937 1.4254 1.2502 1.62 0.1795 0.5476 0.7869 1.4371 1.2666 1.64 0.1867 0.5386 0.7803 1.4487 1.2836 1.66 0.1938 0.5299 0.7736 1.4601 1.3010 1.68 0.2008 0.5213 0.7670 1.4713 1.3190 1.7 0.2078 0.5130 0.7605 1.4825 1.3376 1.72 0.2147 0.5048 0.7539 1.4935 1.3567 1.74 0.2216 0.4969 0.7474 1.5043 1.3764 1.76 0.2284 0.4891 0.7410 1.5150 1.3967 1.78 0.2352 0.4815 0.7345 1.5256 1.4175 1.8 0.2419 0.4741 0.7282 1.5360 1.4390 1.82 0.2485 0.4668 0.7218 1.5463 1.4610 1.84 0.2551 0.4597 0.7155 1.5564 1.4836 1.86 0.2616 0.4528 0.7093 1.5664 1.5069 1.88 0.2680 0.4460 0.7030 1.5763 1.5308 1.9 0.2743 0.4394 0.6969 1.5861 1.5553 1.92 0.2806 0.4329 0.6907 1.5957 1.5804 1.94 0.2868 0.4265 0.6847 1.6052 1.6062 1.96 0.2929 0.4203 0.6786 1.6146 1.6326 1.98 0.2990 0.4142 0.6726 1.6239 1.6597 2.0 0.3050 0.4082 0.6667 1.6330 1.6875 2.02 0.3109 0.4024 0.6608 1.6420 1.7160 2.04 0.3168 0.3967 0.6549 1.6509 1.7451 2.06 0.3225 0.3911 0.6491 1.6597 1.7750 2.08 0.3282 0.3856 0.6433 1.6683 1.8056 2.1 0.3339 0.3802 0.6376 1.6769 1.8369 2.12 0.3394 0.3750 0.6320 1.6853 1.8690 2.14 0.3449 0.3698 0.6263 1.6936 1.9018 2.16 0.3503 0.3648 0.6208 1.7018 1.9354 2.18 0.3556 0.3598 0.6152 1.7099 1.9698 2.2 0.3609 0.3549 0.6098 1.7179 2.0050 2.22 0.3661 0.3502 0.6043 1.7258 2.0409 2.24 0.3712 0.3455 0.5989 1.7336 2.0777 2.26 0.3763 0.3409 0.5936 1.7412 2.1153 2.28 0.3813 0.3364 0.5883 1.7488 2.1538 2.3 0.3862 0.3320 0.5831 1.7563 2.1931 2.32 0.3911 0.3277 0.5779 1.7637 2.2333 2.34 0.3959 0.3234 0.5728 1.7709 2.2744 2.36 0.4006 0.3193 0.5677 1.7781 2.3164 2.38 0.4053 0.3152 0.5626 1.7852 2.3593 2.4 0.4099 0.3111 0.5576 1.7922 2.4031 2.42 0.4144 0.3072 0.5527 1.7991 2.4479 2.44 0.4189 0.3033 0.5478 1.8059 2.4936 2.46 0.4233 0.2995 0.5429 1.8126 2.5403 2.48 0.4277 0.2958 0.5381 1.8192 2.5880 2.5 0.4320 0.2921 0.5333 1.8257 2.6367 2.52 0.4362 0.2885 0.5286 1.8322 2.6865 2.54 0.4404 0.2850 0.5239 1.8386 2.7372 2.56 0.4445 0.2815 0.5193 1.8448 2.7891 2.58 0.4486 0.2781 0.5147 1.8510 2.8420 Compressible-Flow Tables 783 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4 Ma f L/D p/p T/T / V/V p0/p 0 2.6 0.4526 0.2747 0.5102 1.8571 2.8960 2.62 0.4565 0.2714 0.5057 1.8632 2.9511 2.64 0.4604 0.2682 0.5013 1.8691 3.0073 2.66 0.4643 0.2650 0.4969 1.8750 3.0647 2.68 0.4681 0.2619 0.4925 1.8808 3.1233 2.7 0.4718 0.2588 0.4882 1.8865 3.1830 2.72 0.4755 0.2558 0.4839 1.8922 3.2440 2.74 0.4791 0.2528 0.4797 1.8978 3.3061 2.76 0.4827 0.2498 0.4755 1.9033 3.3695 2.78 0.4863 0.2470 0.4714 1.9087 3.4342 2.8 0.4898 0.2441 0.4673 1.9140 3.5001 2.82 0.4932 0.2414 0.4632 1.9193 3.5674 2.84 0.4966 0.2386 0.4592 1.9246 3.6359 2.86 0.5000 0.2359 0.4552 1.9297 3.7058 2.88 0.5033 0.2333 0.4513 1.9348 3.7771 2.9 0.5065 0.2307 0.4474 1.9398 3.8498 2.92 0.5097 0.2281 0.4436 1.9448 3.9238 2.94 0.5129 0.2256 0.4398 1.9497 3.9993 2.96 0.5160 0.2231 0.4360 1.9545 4.0763 2.98 0.5191 0.2206 0.4323 1.9593 4.1547 3.0 0.5222 0.2182 0.4286 1.9640 4.2346 3.02 0.5252 0.2158 0.4249 1.9686 4.3160 3.04 0.5281 0.2135 0.4213 1.9732 4.3989 3.06 0.5310 0.2112 0.4177 1.9777 4.4835 3.08 0.5339 0.2090 0.4142 1.9822 4.5696 3.1 0.5368 0.2067 0.4107 1.9866 4.6573 3.12 0.5396 0.2045 0.4072 1.9910 4.7467 3.14 0.5424 0.2024 0.4038 1.9953 4.8377 3.16 0.5451 0.2002 0.4004 1.9995 4.9304 3.18 0.5478 0.1981 0.3970 2.0037 5.0248 3.2 0.5504 0.1961 0.3937 2.0079 5.1210 3.22 0.5531 0.1940 0.3904 2.0120 5.2189 3.24 0.5557 0.1920 0.3872 2.0160 5.3186 3.26 0.5582 0.1901 0.3839 2.0200 5.4201 3.28 0.5607 0.1881 0.3807 2.0239 5.5234 3.3 0.5632 0.1862 0.3776 2.0278 5.6286 3.32 0.5657 0.1843 0.3745 2.0317 5.7358 3.34 0.5681 0.1825 0.3714 2.0355 5.8448 3.36 0.5705 0.1806 0.3683 2.0392 5.9558 3.38 0.5729 0.1788 0.3653 2.0429 6.0687 3.4 0.5752 0.1770 0.3623 2.0466 6.1837 3.42 0.5775 0.1753 0.3594 2.0502 6.3007 3.44 0.5798 0.1736 0.3564 2.0537 6.4198 3.46 0.5820 0.1718 0.3535 2.0573 6.5409 3.48 0.5842 0.1702 0.3507 2.0607 6.6642 3.5 0.5864 0.1685 0.3478 2.0642 6.7896 3.52 0.5886 0.1669 0.3450 2.0676 6.9172 3.54 0.5907 0.1653 0.3422 2.0709 7.0471 3.56 0.5928 0.1637 0.3395 2.0743 7.1791 3.58 0.5949 0.1621 0.3368 2.0775 7.3135 3.6 0.5970 0.1616 0.3341 2.0808 7.4501 3.62 0.5990 0.1590 0.3314 2.0840 7.5891 3.64 0.6010 0.1575 0.3288 2.0871 7.7305 3.66 0.6030 0.1560 0.3262 2.0903 7.8742 Table B.3 (Cont.) Adiabatic Frictional Flow in a Constant-Area Duct for k 1.4 784 Appendix B Ma f L/D p/p T/T / V/V p0/p 0 3.68 0.6049 0.1546 0.3236 2.0933 8.0204 3.7 0.6068 0.1531 0.3210 2.0964 8.1691 3.72 0.6087 0.1517 0.3185 2.0994 8.3202 3.74 0.6106 0.1503 0.3160 2.1024 8.4739 3.76 0.6125 0.1489 0.3135 2.1053 8.6302 3.78 0.6143 0.1475 0.3111 2.1082 8.7891 3.8 0.6161 0.1462 0.3086 2.1111 8.9506 3.82 0.6179 0.1449 0.3062 2.1140 9.1148 3.84 0.6197 0.1436 0.3039 2.1168 9.2817 3.86 0.6214 0.1423 0.3015 2.1195 9.4513 3.88 0.6231 0.1410 0.2992 2.1223 9.6237 3.9 0.6248 0.1397 0.2969 2.1250 9.7990 3.92 0.6265 0.1385 0.2946 2.1277 9.9771 3.94 0.6282 0.1372 0.2923 2.1303 10.1581 3.96 0.6298 0.1360 0.2901 2.1329 10.3420 3.98 0.6315 0.1348 0.2879 2.1355 10.5289 4.0 0.6331 0.1336 0.2857 2.1381 10.7188 Table B.4 Frictionless Duct Flow with Heat Transfer for k 1.4 Ma T0/T 0 p/p T/T / V/V p0/p 0 0.0 0.0 2.4000 0.0 0.0 1.2679 0.02 0.0019 2.3987 0.0023 0.0010 1.2675 0.04 0.0076 2.3946 0.0092 0.0038 1.2665 0.06 0.0171 2.3800 0.0205 0.0086 1.2647 0.08 0.0302 2.3787 0.0362 0.0152 1.2623 0.1 0.0468 2.3669 0.0560 0.0237 1.2591 0.12 0.0666 2.3526 0.0797 0.0339 1.2554 0.14 0.0895 2.3359 0.1069 0.0458 1.2510 0.16 0.1151 2.3170 0.1374 0.0593 1.2461 0.18 0.1432 2.2959 0.1708 0.0744 1.2406 0.2 0.1736 2.2727 0.2066 0.0909 1.2346 0.22 0.2057 2.2477 0.2445 0.1088 1.2281 0.24 0.2395 2.2209 0.2841 0.1279 1.2213 0.26 0.2745 2.1925 0.3250 0.1482 1.2140 0.28 0.3104 2.1626 0.3667 0.1696 1.2064 0.3 0.3469 2.1314 0.4089 0.1918 1.1985 0.32 0.3837 2.0991 0.4512 0.2149 1.1904 0.34 0.4206 2.0657 0.4933 0.2388 1.1822 0.36 0.4572 2.0314 0.5348 0.2633 1.1737 0.38 0.4935 1.9964 0.5755 0.2883 1.1652 0.4 0.5290 1.9608 0.6151 0.3137 1.1566 0.42 0.5638 1.9247 0.6535 0.3395 1.1480 0.44 0.5975 1.8882 0.6903 0.3656 1.1394 0.46 0.6301 1.8515 0.7254 0.3918 1.1308 0.48 0.6614 1.8147 0.7587 0.4181 1.1224 0.5 0.6914 1.7778 0.7901 0.4444 1.1141 0.52 0.7199 1.7409 0.8196 0.4708 1.1059 0.54 0.7470 1.7043 0.8469 0.4970 1.0979 0.56 0.7725 1.6678 0.8723 0.5230 1.0901 0.58 0.7965 1.6316 0.8955 0.5489 1.0826 0.6 0.8189 1.5957 0.9167 0.5745 1.0753 Compressible-Flow Tables 785 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4 Ma T0/T 0 p/p T/T / V/V p0/p 0 0.62 0.8398 1.5603 0.9358 0.5998 1.0682 0.64 0.8592 1.5253 0.9530 0.6248 1.0615 0.66 0.8771 1.4908 0.9682 0.6494 1.0550 0.68 0.8935 1.4569 0.9814 0.6737 1.0489 0.7 0.9085 1.4235 0.9929 0.6975 1.0431 0.72 0.9221 1.3907 1.0026 0.7209 1.0376 0.74 0.9344 1.3585 1.0106 0.7439 1.0325 0.76 0.9455 1.3270 1.0171 0.7665 1.0278 0.78 0.9553 1.2961 1.0220 0.7885 1.0234 0.8 0.9639 1.2658 1.0255 0.8101 1.0193 0.82 0.9715 1.2362 1.0276 0.8313 1.0157 0.84 0.9781 1.2073 1.0285 0.8519 1.0124 0.86 0.9836 1.1791 1.0283 0.8721 1.0095 0.88 0.9883 1.1515 1.0269 0.8918 1.0070 0.9 0.9921 1.1246 1.0245 0.9110 1.0049 0.92 0.9951 1.0984 1.0212 0.9297 1.0031 0.94 0.9973 1.0728 1.0170 0.9480 1.0017 0.96 0.9988 1.0479 1.0121 0.9658 1.0008 0.98 0.9997 1.0236 1.0064 0.9831 1.0002 1.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.02 0.9997 0.9770 0.9930 1.0164 1.0002 1.04 0.9989 0.9546 0.9855 1.0325 1.0008 1.06 0.9977 0.9327 0.9776 1.0480 1.0017 1.08 0.9960 0.9115 0.9691 1.0632 1.0031 1.1 0.9939 0.8909 0.9603 1.0780 1.0049 1.12 0.9915 0.8708 0.9512 1.0923 1.0070 1.14 0.9887 0.8512 0.9417 1.1063 1.0095 1.16 0.9856 0.8322 0.9320 1.1198 1.0124 1.18 0.9823 0.8137 0.9220 1.1330 1.0157 1.2 0.9787 0.7958 0.9118 1.1459 1.0194 1.22 0.9749 0.7783 0.9015 1.1584 1.0235 1.24 0.9709 0.7613 0.8911 1.1705 1.0279 1.26 0.9668 0.7447 0.8805 1.1823 1.0328 1.28 0.9624 0.7287 0.8699 1.1938 1.0380 1.3 0.9580 0.7130 0.8592 1.2050 1.0437 1.32 0.9534 0.6978 0.8484 1.2159 1.0497 1.34 0.9487 0.6830 0.8377 1.2264 1.0561 1.36 0.9440 0.6686 0.8269 1.2367 1.0629 1.38 0.9391 0.6546 0.8161 1.2467 1.0701 1.4 0.9343 0.6410 0.8054 1.2564 1.0777 1.42 0.9293 0.6278 0.7947 1.2659 1.0856 1.44 0.9243 0.6149 0.7840 1.2751 1.0940 1.46 0.9193 0.6024 0.7735 1.2840 1.1028 1.48 0.9143 0.5902 0.7629 1.2927 1.1120 1.5 0.9093 0.5783 0.7525 1.3012 1.1215 1.52 0.9042 0.5668 0.7422 1.3095 1.1315 1.54 0.8992 0.5555 0.7319 1.3175 1.1419 1.56 0.8942 0.5446 0.7217 1.3253 1.1527 1.58 0.8892 0.5339 0.7117 1.3329 1.1640 1.6 0.8842 0.5236 0.7017 1.3403 1.1756 1.62 0.8792 0.5135 0.6919 1.3475 1.1877 1.64 0.8743 0.5036 0.6822 1.3546 1.2002 1.66 0.8694 0.4940 0.6726 1.3614 1.2131 1.68 0.8645 0.4847 0.6631 1.3681 1.2264 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4 786 Appendix B Ma T0/T 0 p/p T/T / V/V p0/p 0 1.7 0.8597 0.4756 0.6538 1.3746 1.2402 1.72 0.8549 0.4668 0.6445 1.3809 1.2545 1.74 0.8502 0.4581 0.6355 1.3870 1.2692 1.76 0.8455 0.4497 0.6265 1.3931 1.2843 1.78 0.8409 0.4415 0.6176 1.3989 1.2999 1.8 0.8363 0.4335 0.6089 1.4046 1.3159 1.82 0.8317 0.4257 0.6004 1.4102 1.3324 1.84 0.8273 0.4181 0.5919 1.4156 1.3494 1.86 0.8228 0.4107 0.5836 1.4209 1.3669 1.88 0.8185 0.4035 0.5754 1.4261 1.3849 1.9 0.8141 0.3964 0.5673 1.4311 1.4033 1.92 0.8099 0.3895 0.5594 1.4360 1.4222 1.94 0.8057 0.3828 0.5516 1.4408 1.4417 1.96 0.8015 0.3763 0.5439 1.4455 1.4616 1.98 0.7974 0.3699 0.5364 1.4501 1.4821 2.0 0.7934 0.3636 0.5289 1.4545 1.5031 2.02 0.7894 0.3575 0.5216 1.4589 1.5246 2.04 0.7855 0.3516 0.5144 1.4632 1.5467 2.06 0.7816 0.3458 0.5074 1.4673 1.5693 2.08 0.7778 0.3401 0.5004 1.4714 1.5924 2.1 0.7741 0.3345 0.4936 1.4753 1.6162 2.12 0.7704 0.3291 0.4868 1.4792 1.6404 2.14 0.7667 0.3238 0.4802 1.4830 1.6653 2.16 0.7631 0.3186 0.4737 1.4867 1.6908 2.18 0.7596 0.3136 0.4673 1.4903 1.7168 2.2 0.7561 0.3086 0.4611 1.4938 1.7434 2.22 0.7527 0.3038 0.4549 1.4973 1.7707 2.24 0.7493 0.2991 0.4488 1.5007 1.7986 2.26 0.7460 0.2945 0.4428 1.5040 1.8271 2.28 0.7428 0.2899 0.4370 1.5072 1.8562 2.3 0.7395 0.2855 0.4312 1.5104 1.8860 2.32 0.7364 0.2812 0.4256 1.5134 1.9165 2.34 0.7333 0.2769 0.4200 1.5165 1.9476 2.36 0.7302 0.2728 0.4145 1.5194 1.9794 2.38 0.7272 0.2688 0.4091 1.5223 2.0119 2.4 0.7242 0.2648 0.4038 1.5252 2.0451 2.42 0.7213 0.2609 0.3986 1.5279 2.0789 2.44 0.7184 0.2571 0.3935 1.5306 2.1136 2.46 0.7156 0.2534 0.3885 1.5333 2.1489 2.48 0.7128 0.2497 0.3836 1.5359 2.1850 2.5 0.7101 0.2462 0.3787 1.5385 2.2218 2.52 0.7074 0.2427 0.3739 1.5410 2.2594 2.54 0.7047 0.2392 0.3692 1.5434 2.2978 2.56 0.7021 0.2359 0.3646 1.5458 2.3370 2.58 0.6995 0.2326 0.3601 1.5482 2.3770 2.6 0.6970 0.2294 0.3556 1.5505 2.4177 2.62 0.6945 0.2262 0.3512 1.5527 2.4593 2.64 0.6921 0.2231 0.3469 1.5549 2.5018 2.66 0.6896 0.2201 0.3427 1.5571 2.5451 2.68 0.6873 0.2171 0.3385 1.5592 2.5892 2.7 0.6849 0.2142 0.3344 1.5613 2.6343 2.72 0.6826 0.2113 0.3304 1.5634 2.6802 2.74 0.6804 0.2085 0.3264 1.5654 2.7270 2.76 0.6781 0.2058 0.3225 1.5673 2.7748 Compressible-Flow Tables 787 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4 Ma T0/T 0 p/p T/T / V/V p0/p 0 2.78 0.6761 0.2030 0.3186 1.5693 2.8235 2.8 0.6738 0.2004 0.3149 1.5711 2.8731 2.82 0.6717 0.1978 0.3111 1.5730 2.9237 2.84 0.6696 0.1953 0.3075 1.5748 2.9752 2.86 0.6675 0.1927 0.3039 1.5766 3.0278 2.88 0.6655 0.1903 0.3004 1.5784 3.0813 2.9 0.6635 0.1879 0.2969 1.5801 3.1359 2.92 0.6615 0.1855 0.2934 1.5818 3.1914 2.94 0.6596 0.1832 0.2901 1.5834 3.2481 2.96 0.6577 0.1809 0.2868 1.5851 3.3058 2.98 0.6558 0.1787 0.2835 1.5867 3.3646 3.0 0.6540 0.1765 0.2803 1.5882 3.4245 3.02 0.6522 0.1743 0.2771 1.5898 3.4854 3.04 0.6504 0.1722 0.2740 1.5913 3.5476 3.06 0.6486 0.1701 0.2709 1.5928 3.6108 3.08 0.6469 0.1681 0.2679 1.5942 3.6752 3.1 0.6452 0.1660 0.2650 1.5957 3.7408 3.12 0.6435 0.1641 0.2620 1.5971 3.8076 3.14 0.6418 0.1621 0.2592 1.5985 3.8756 3.16 0.6402 0.1602 0.2563 1.5998 3.9449 3.18 0.6386 0.1583 0.2535 1.6012 4.0154 3.2 0.6370 0.1565 0.2508 1.6025 4.0871 3.22 0.6354 0.1547 0.2481 1.6038 4.1602 3.24 0.6339 0.1529 0.2454 1.6051 4.2345 3.26 0.6324 0.1511 0.2428 1.6063 4.3101 3.28 0.6309 0.1494 0.2402 1.6076 4.3871 3.3 0.6294 0.1477 0.2377 1.6088 4.4655 3.32 0.6280 0.1461 0.2352 1.6100 4.5452 3.34 0.6265 0.1444 0.2327 1.6111 4.6263 3.36 0.6251 0.1428 0.2303 1.6123 4.7089 3.38 0.6237 0.1412 0.2279 1.6134 4.7929 3.4 0.6224 0.1397 0.2255 1.6145 4.8783 3.42 0.6210 0.1381 0.2232 1.6156 4.9652 3.44 0.6197 0.1366 0.2209 1.6167 5.0536 3.46 0.6184 0.1351 0.2186 1.6178 5.1435 3.48 0.6171 0.1337 0.2164 1.6188 5.2350 3.5 0.6158 0.1322 0.2142 1.6198 5.3280 3.52 0.6145 0.1308 0.2120 1.6208 5.4226 3.54 0.6133 0.1294 0.2099 1.6218 5.5188 3.56 0.6121 0.1280 0.2078 1.6228 5.6167 3.58 0.6109 0.1267 0.2057 1.6238 5.7162 3.6 0.6097 0.1254 0.2037 1.6247 5.8173 3.62 0.6085 0.1241 0.2017 1.6257 5.9201 3.64 0.6074 0.1228 0.1997 1.6266 6.0247 3.66 0.6062 0.1215 0.1977 1.6275 6.1310 3.68 0.6051 0.1202 0.1958 1.6284 6.2390 3.7 0.6040 0.1190 0.1939 1.6293 6.3488 3.72 0.6029 0.1178 0.1920 1.6301 6.4605 3.74 0.6018 0.1166 0.1902 1.6310 6.5739 3.76 0.6008 0.1154 0.1884 1.6318 6.6893 3.78 0.5997 0.1143 0.1866 1.6327 6.8065 3.8 0.5987 0.1131 0.1848 1.6335 6.9256 3.82 0.5977 0.1120 0.1830 1.6343 7.0466 3.84 0.5967 0.1109 0.1813 1.6351 7.1696 Table B.4 (Cont.) Frictionless Duct Flow with Heat Transfer for k 1.4 788 Appendix B Ma T0/T 0 p/p T/T / V/V p0/p 0 3.86 0.5957 0.1098 0.1796 1.6359 7.2945 3.88 0.5947 0.1087 0.1779 1.6366 7.4215 3.9 0.5937 0.1077 0.1763 1.6374 7.5505 3.92 0.5928 0.1066 0.1746 1.6381 7.6816 3.94 0.5918 0.1056 0.1730 1.6389 7.8147 3.96 0.5909 0.1046 0.1714 1.6396 7.9499 3.98 0.5900 0.1036 0.1699 1.6403 8.0873 4.0 0.5891 0.1026 0.1683 1.6410 8.2269 Table B.5 Prandtl-Meyer Supersonic Expansion Function for k 1.4 Ma , deg Ma , deg Ma , deg Ma , deg 1.00 0.0 1.05 0.49 3.05 50.71 5.05 77.38 7.05 91.23 1.10 1.34 3.10 51.65 5.10 77.84 7.10 91.49 1.15 2.38 3.15 52.57 5.15 78.29 7.15 91.75 1.20 3.56 3.20 53.47 5.20 78.73 7.20 92.00 1.25 4.83 3.25 54.35 5.25 79.17 7.25 92.24 1.30 6.17 3.30 55.22 5.30 79.60 7.30 92.49 1.35 7.56 3.35 56.07 5.35 80.02 7.35 92.73 1.40 8.99 3.40 56.91 5.40 80.43 7.40 92.97 1.45 10.44 3.45 57.73 5.45 80.84 7.45 93.21 1.50 11.91 3.50 58.53 5.50 81.24 7.50 93.44 1.55 13.38 3.55 59.32 5.55 81.64 7.55 93.67 1.60 14.86 3.60 60.09 5.60 82.03 7.60 93.90 1.65 16.34 3.65 60.85 5.65 82.42 7.65 94.12 1.70 17.81 3.70 61.60 5.70 82.80 7.70 94.34 1.75 19.27 3.75 62.33 5.75 83.17 7.75 94.56 1.80 20.73 3.80 63.04 5.80 83.54 7.80 94.78 1.85 22.16 3.85 63.75 5.85 83.90 7.85 95.00 1.90 23.59 3.90 64.44 5.90 84.26 7.90 95.21 1.95 24.99 3.95 65.12 5.95 84.61 7.95 95.42 2.00 26.38 4.00 65.78 6.00 84.96 8.00 95.62 2.05 27.75 4.05 66.44 6.05 85.30 8.05 95.83 2.10 29.10 4.10 67.08 6.10 85.63 8.10 96.03 2.15 30.43 4.15 67.71 6.15 85.97 8.15 96.23 2.20 31.73 4.20 68.33 6.20 86.29 8.20 96.43 2.25 33.02 4.25 68.94 6.25 86.62 8.25 96.63 2.30 34.28 4.30 69.54 6.30 86.94 8.30 96.82 2.35 35.53 4.35 70.13 6.35 87.25 8.35 97.01 2.40 36.75 4.40 70.71 6.40 87.56 8.40 97.20 2.45 37.95 4.45 71.27 6.45 87.87 8.45 97.39 2.50 39.12 4.50 71.83 6.50 88.17 8.50 97.57 2.55 40.28 4.55 72.38 6.55 88.47 8.55 97.76 2.60 41.41 4.60 72.92 6.60 88.76 8.60 97.94 2.65 42.53 4.65 73.45 6.65 89.05 8.65 98.12 2.70 43.62 4.70 73.97 6.70 89.33 8.70 98.29 2.75 44.69 4.75 74.48 6.75 89.62 8.75 98.47 2.80 45.75 4.80 74.99 6.80 89.90 8.80 98.64 2.85 46.78 4.85 75.48 6.85 90.17 8.85 98.81 2.90 47.79 4.90 75.97 6.90 90.44 8.90 98.98 2.95 48.78 4.95 76.45 6.95 90.71 8.95 99.15 3.00 49.76 5.00 76.92 7.00 90.97 9.00 99.32 Compressible-Flow Tables 789 Fig. B.1 Mach number downstream of an oblique shock for k 1.4.
4.0 3.0 2.0 1.0 0 2.0 3.0 4.0 1.0 Ma 2 Ma1 Mach line = 0° Weak shock Strong shock = β 20° 25 30 35 40 45 55 50 60 65 70 75 85 90 80 Normal shock 5 10 15 20 25 30 35 θ β Ma1 Ma 2 θ θ 790 Appendix B Fig. B.2 Pressure ratio downstream of an oblique shock for k 1.4.
θ Weak shock Strong shock β 80° Normal shock 5 10 15 20 25 θ 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Ma1, p1 p2 70 90 75 65 60 55 50 45 40 35 30 25 20 β 30° θ = = p2 p1 1.0 1.0 Ma1 2.0 3.0 4.0 791 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 ft3 0.028317 m3 1 mi 5280 ft 1609.344 m 1 U.S. gal 231 in3 0.0037854 m3 1 nautical mile (nmi) 6076 ft 1852 m 1 L 0.001 m3 0.035315 ft3 1 yd 3 ft 0.9144 m 1 U.S. fluid ounce 2.9574 E-5 m3 1 angstrom (Å) 1.0 E-10 m 1 U.S. quart (qt) 9.4635 E-4 m3 Mass Area 1 slug 32.174 lbm 14.594 kg 1 ft2 0.092903 m2 1 lbm 0.4536 kg 1 mi2 2.78784 E7 ft2 2.59 E6 m2 1 short ton 2000 lbm 907.185 kg 1 acre 43,560 ft2 4046.9 m2 1 tonne 1000 kg 1 hectare (ha) 10,000 m2 Velocity Acceleration 1 ft/s 0.3048 m/s 1 ft/s2 0.3048 m/s2 1 mi/h 1.466666 ft/s 0.44704 m/s 1 kn 1 nmi/h 1.6878 ft/s 0.5144 m/s Mass flow Volume flow 1 slug/s 14.594 kg/s 1 gal/min 0.002228 ft3/s 0.06309 L/s 1 lbm/s 0.4536 kg/s 1 106 gal/day 1.5472 ft3/s 0.04381 m3/s Pressure Force 1 lbf/ft2 47.88 Pa 1 lbf 4.448222 N 16 oz 1 lbf/in2 144 lbf/ft2 6895 Pa 1 kgf 2.2046 lbf 9.80665 N 1 atm 2116.2 lbf/ft2 14.696 lbf/in2 1 U.S. (short) ton 2000 lbf 101,325 Pa 1 dyne 1.0 E-5 N 1 inHg (at 20°C) 3375 Pa 1 ounce (avoirdupois) (oz) 0.27801 N 1 bar 1.0 E5 Pa Appendix C Conversion Factors Energy Power 1 ft lbf 1.35582 J 1 hp 550 ft lbf/s 745.7 W 1 Btu 252 cal 1055.056 J 778.17 ft lbf 1 ft lbf/s 1.3558 W 1 kilowatt hour (kWh) 3.6 E6 J Specific weight Density 1 lbf/ft3 157.09 N/m3 1 slug/ft3 515.38 kg/m3 1 lbm/ft3 16.0185 kg/m3 1 g/cm3 1000 kg/m3 Viscosity Kinematic viscosity 1 slug/(ft s) 47.88 kg/(m s) 1 ft2/h 0.000025806 m2/s 1 poise (P) 1 g/(cm s) 0.1 kg/(m s) 1 stokes (St) 1 cm2/s 0.0001 m2/s Temperature scale readings TF 9 5 TC 32 TC 5 9 (TF 32) TR TF 459.69 TK TC 273.16 where subscripts F, C, R, and K 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/(h ft °R) 1.7307 W/(m K) 1 Btu/(lb °R) 4186.8 J/(kg 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.
792 Appendix C The equations of motion of an incompressible newtonian fluid with constant , k, and cp are given here in cylindrical coordinates (r, , z), which are related to cartesian co-ordinates (x, y, z) as in Fig. 4.2: x r cos y r sin z z (D.1) The velocity components are r,
, and z. The equations are: Continuity: 1 r r (rr) 1 r (
) z (z) 0 (D.2) Convective time derivative: V r r 1 r z z (D.3) Laplacian operator: 2 1 r r r r r 1 2 2 2 z 2 2 (D.4) The r-momentum equation: t r (V )r 1 r 2 1 p r gr 2r r 2 r r 2 2 (D.5) The -momentum equation: t (V ) 1 r r 1 r p g 2 r 2 r 2 2 r (D.6) The z-momentum equation: t z (V )z 1 p z gz 2z (D.7) Appendix D Equations of Motion in Cylindrical Coordinates 793 The energy equation: cp T t (V )T k2T [2(2 rr 2 2 zz) 2 z 2 rz 2 r
] (D.8) where rr r r 1 r r zz z z
z 1 r z z (D.9) rz z r r z r 1 r r
r Viscous stress components: rr 2rr 2 zz 2zz (D.10) r r
z
z rz rz Angular-velocity components: r 1 r z z z r r z (D.11) z 1 r r (r
) 1 r r 794 Appendix D Overview Background Information EES (pronounced “ease”) is an acronym for Engineering Equation Solver. The basic function provided by EES is the numerical solution of nonlinear algebraic and differ-ential equations. In addition, EES provides built-in thermodynamic and transport prop-erty functions for many fluids, including water, dry and moist air, refrigerants, and combustion gases. Additional property data can be added by the user. The combina-tion of equation solving capability and engineering property data makes EES a very powerful tool.
A license for EES is provided to departments of educational institutions which adopt this text by WCB/McGraw-Hill. If you need more information, contact your local WCB/McGraw-Hill representative, call 1-800-338-3987, or visit our website at www.mhhe.com. A commercial version of EES can be obtained from: F-Chart Software 4406 Fox Bluff Rd Middleton, WI 53562 Phone: (608)836-8531 Fax: (608)836-8536 The EES program is probably installed on your departmental computer. In addition, the license agreement for EES allows students and faculty in a participating educa-tional department to copy the program for educational use on their personal computer systems. Ask your instructor for details.
To start EES from the Windows File Manager or Explorer, double-click on the EES program icon or on any file created by EES. You can also start EES from the Windows Run command in the Start menu. EES begins by displaying a dialog window which shows registration information, the version number, and other infor-mation. Click the OK button to dismiss the dialog window.
Detailed help is available at any point in EES. Pressing the F1 key will bring up a Help window relating to the foremost window. (See Fig. E.1.) Clicking the Contents Appendix E Introduction to EES 795 Fig. E.1 EES Help index.
button will present the Help index shown below. Clicking on an underlined word (shown in green on color monitors) will provide help relating to that subject.
EES commands are distributed among nine pull-down menus as shown below. Many of the commands are accessible with the speed button palette that appears below the menu bar. A brief summary of their functions follows. (A tenth pull-down menu, which is made visible with the Load Textbook command described below, provides ac-cess to problems from this text.) 796 Appendix E The System menu appears above the File menu. The System menu is not part of EES but rather is a feature of the Windows operating system. It holds commands which allow window moving, resizing, and switching to other ap-plications.
The File menu provides commands for loading, merging, and saving work files; libraries; and printing. The Load Textbook command in this menu reads the problem disk developed for this text and creates a new menu to the right of the Help menu for easy access to EES problems accompanying this text.
The Edit menu provides the editing commands to cut, copy, and paste informa-tion.
The Search menu provides Find and Replace commands for use in the Equations window.
The Options menu provides commands for setting the guess values and bounds of variables, the unit system, default information, and program preferences. A command is also provided for displaying information on built-in and user-supplied functions.
The Calculate menu contains the commands to check, format, and solve the equation set.
The Tables menu contains commands to set up and alter the contents of the parametric and lookup tables and to do linear regression on the data in these tables. The parametric table, which is similar to a spreadsheet, allows the equation set to be solved repeatedly while varying the values of one or more variables. The lookup table holds user-supplied data which can be interpolated and used in the solution of the equation set.
The Plot menu provides commands to modify an existing plot or prepare a new plot of data in the parametric, lookup, or array tables. Curve-fitting capability is also provided.
The Windows menu provides a convenient method of bringing any of the EES windows to the front or to organize the windows.
The Help menu provides commands for accessing the on-line help documentation.
A basic capability provided by EES is the solution of a set of nonlinear algebraic equations. To demonstrate this capability, start EES and enter this simple example prob-lem in the Equations window.
Introduction to EES 797 Text is entered in the same manner as for any word processor. Formatting rules are as follows: 1.
Uppercase and lowercase letters are not distinguished. EES will (optionally) change the case of all variables to match the manner in which they first appear.
2.
Blank lines and spaces may be entered as desired since they are ignored.
3.
Comments must be enclosed within braces { } or within quotation marks “ ”.
Comments may span as many lines as needed. Comments within braces may be nested, in which case only the outermost set of braces is recognized. Comments within quotes will also be displayed in the Formatted Equations window.
4.
Variable names must start with a letter and consist of any keyboard characters except ( ) ‘/ ^ { } : " or ;. Array variables are identified with square braces around the array index or indices, e.g., X[5,3]. The maximum variable length is 30 characters.
5.
Multiple equations may be entered on one line if they are separated by a semi-colon (;). The maximum line length is 255 characters.
6.
The caret symbol (^) or is used to indicate raising to a power.
7.
The order in which the equations are entered does not matter.
8.
The position of knowns and unknowns in the equation does not matter.
If you wish, you may view the equations in mathematical notation by selecting the Formatted Equations command from the Windows menu.
798 Appendix E Select the Solve command from the Calculate menu. A Dialog window will ap-pear indicating the progress of the solution. When the calculations are completed, the button will change from Abort to Continue.
Click the Continue button. The solution to this equation set will then be displayed.
Let us now solve Prob. 6.55 from the text, for a cast-iron pipe, to illustrate the capa-bilities of the EES program. This problem, without EES, would require iteration for Reynolds number, velocity, and friction factor, a daunting task. State the problem: 6.55 Reservoirs 1 and 2 contain water at 20°C. The pipe is cast iron, with L 4500 m and D 4 cm. What will be the flow rate in m3/h if z 100 m?
This is a representative problem in pipe flow (see Fig. E.2), and, being water in a rea-sonably large (noncapillary) pipe, it will probably be turbulent (Re 4000). The steady-flow energy equation (3.71) may be written between the surfaces of reservoirs 1 and 2: p g 1 V 2g 1 2 z1 p g 2 V 2g 2 2 z2 hf where hf f D L V 2 p g ipe 2 Since p1 p2 patm and V1 V2 0, this relation simplifies to z f D L 2 V g 2 (1) where V Q/A is the velocity in the pipe. The friction factor f is a function of Reynolds number and pipe roughness ratio, if the flow is turbulent, from Eq. (6.64): f 1 1 /2 2.0 log10 3 / .
D 7 R 2 e .5 f 1 1/2 if Re 4000 (2) Finally, we need the definitions of Reynolds number and volume flow rate: Re V D (3) and Q V 4 D2 (4) where and are the fluid density and viscosity, respectively.
Introduction to EES 799 A Pipe Friction Example Problem 1 2 z L, D, Fig. E.2 Sketch of the flow system.
Fig. E.3 Unit Selection dialog window.
Fig. E.4 Equations window.
There are a total of 11 variables involved in this problem: (L, D, z, , g, , , V, Re, f, Q). Of these, seven can be specified at the start (L, D, z, , g, , ), while four (V, Re, f, Q) must be calculated from relations (1) to (4) above. These four equations in four unknowns are well posed and solvable but only by laborious iteration, exactly what EES is designed to do.
Start EES or select the New command from the File menu if you have already been using the program. A blank Equations window will appear. Our recommenda-tion is to always set the unit system immediately: Select Unit System from the Options menu (Fig. E.3). We select SI and Mass units and trig Degrees, although we do not actually have trigonometric functions this time. We select kPa for pres-sure and Celsius for temperature, which will be handy for using the EES built-in physical properties of water.
800 Appendix E Now, onto the blank screen, enter the equations for this problem (Fig. E.4), of which five are known input values, two are property evaluations, and four are the relations (1) to (4) from above.
There are several things to notice in Fig. E.4. First, quantities in quotes, such as “m,” are for the user’s benefit and ignored by EES. Second, we changed Eps and D to me-ters right away, to keep the SI units consistent. Third, we called on EES to input the viscosity and density of water at 20°C and 1 atm, a procedure well explained in the Help menu. For example, viscosity(water,T 20,P 101) meets the EES requirement that temperature (T) and pressure (P) should be input in °C and kPa—EES will then evaluate in kg/(m s). Finally, note that EES recognizes pi to be 3.141593.
In Fig. E.4 we used only one built-in function, log10. There are many such func-tions, found by scrolling down the Function Information command in the Options menu.
Having entered the equations, check the syntax by using the Check/Format com-mand in the Calculate menu. If you did well, EES will report that the 11 equations in 11 unknowns look OK. If not, EES will guess at what might be wrong. If OK, why not go for it? Hit the Solve command in the Options menu. EES reports “logarithm of a neg-ative number—try setting limits on the variables”. We might have known. Go to the Vari-able Information command in the Options menu. Abox, listing the 11 variables, will appear (Fig. E.5). All default EES “guesses” are unity; all default limits are to , which is too broad a range. Enter (as already shown in Fig. E.5) guesses for f 0.02 and Re 10,000, while V 1 and Q 1 seem adequate, and other variables are fixed. Make sure that f, Re, V, and Q cannot be negative. The “display” columns normally say “A”, au-tomatic, satisfactory for most variables. We have changed “A” to “F” (fixed decimal) for Q and V to make sure they are displayed to four decimal places. The “units” column is nor-mally blank—type in the correct units and they will be displayed in the solution.
Our guesses and limits are excellent, and the Solve command now iterates and reports success: “max residual 2E-10”, a negligible error. (The default runs for 100 Introduction to EES 801 Fig. E.5 Variable Information window with units and guess values entered.
Parametric Studies with Tabular Input iterations, which can be modified by the Stop Criteria command in the Op-tions menu.) Hit Continue, and the complete solution is displayed for all vari-ables (Fig. E.6).
802 Appendix E This is the correct solution to Prob. 6.55: this cast-iron pipe, when subjected to a 100-m elevation difference, will deliver Q 3.17 m3/h of water. EES did all the iteration.
One of the most useful features of EES is its ability to provide parametric studies. For example, suppose we wished to know how varying z changed the flow rate Q. First comment out the equation that reads DELTAZ 100 by enclosing it within braces. (If you select the equation and press the right mouse button, EES will automatically en-ter the braces.) Select the New Parametric Table command in the Options menu. A dialog will be displayed (Fig. E.7) listing all the variables in the problem.
Highlight what you wish to vary: z. Also highlight variables to be calculated and tab-ulated: V, Q, Re, and f.
Fig. E.6 The Solutions window for Prob. 6.55.
Fig. E.7 New Parametric Table window showing selected variables (V is not shown).
Fig. E.8 Parametric Table window.
Fig. E.9 Solve Table Dialog.
Introduction to EES 803 Clearly the parametric table operates much like a spreadsheet. Select Solve Table from the Calculate menu, and the Solve Table dialog window will appear (Fig. E.9).
These are satisfactory default values; the author has changed nothing. Hit the OK button and the calculations will be made and the entire parametric table filled out, as in Fig. E.10.
The flow rates can be seen in Fig. E.10, but as always, in the author’s experience, a plot is more illuminating. Select New Plot window from the Plot menu. The New Plot window dialog (Fig. E.11) will appear. Choose z as the x-axis and Q as the y-axis.
We added grid lines. Click the OK button, and the desired plot will appear in the Plot window (Fig. E.12). We see a nonlinear relationship, roughly a square-root type, and learn that flow rate Q is not linearly proportional to head difference z.
The plot appearance in Fig. E.12 can be modified in several ways. Double-click the mouse in the plot rectangle to see some of these options.
Click the OK button and the new table will be displays (Fig. E.8). Enter 10 values of z that cover the range of interest—we have selected the linear range 10 z 500 m.
Fig. E.10 Parametric Table window after calculations are completed.
804 Appendix E Fig. E.11 New Plot Setup dialog window.
Introduction to EES 805 Fig. E.12 Plot window for flow rate versus elevation difference.
A problems disk developed for EES has been included with this textbook. Place the disk in the disk drive, and then select the Load Textbook command in the File menu. Use the Windows Open File command to open the textbook problem in-dex file which, for this book, is named WHITE.TXB. A new menu called Fluid Mechanics will appear to the right of the Help menu. This menu will provide ac-cess to all the EES problem solutions developed for this book organized by chapter.
As an example, select Chap. 6 from the Fluid Mechanics menu. A dialog win-dow will appear listing the problems in Chap. 6. Select Problem 6.55—Flow Between Reservoirs. This problem is a modification (smooth instead of cast-iron pipe) of the problem you just entered. It provides a diagram window in which you can enter z.
Enter other values, and then select the Solve command in the Calculate menu to see their effect on the flow rate. At this point, you should explore. Try whatever you wish. You can’t hurt anything.
The on-line help (invoked by pressing F1) will provide details for the EES commands.
EES is a powerful tool that you will find very useful in your studies.
Loading a Textbook File 1.68 h (/g)1/2 cot 1.70 h 2 cos /(gW) 1.72 z 4800 m 1.74 Cavitation occurs for both (a) and (b) 1.76 z 7500 m 1.78 (a) 25°C; (b) 4°C 1.80 x2y y3/3 constant 1.82 y x tan constant 1.84 x x0{ln (y/y0) ln2 (y/y0)} Chapter 2 2.2 xy 289 lb/ft2, AA 577 lb/ft2 2.4 x Const e2Cz/B 2.6 (a) 30.3 ft; (b) 30.0 in; (c) 10.35 m; (d) 13,100 mm 2.8 DALR 9.77°C/km 2.10 10,500 Pa 2.12 8.0 cm 2.14 74,450 Pa with air; 75,420 Pa without air 2.16 (a) 21,526 cm3; (b) 137 kPa 2.18 1.56 2.20 14 lbf 2.22 0.94 cm 2.24 psealevel 117 kPa, mexact 5.3 E18 kg 2.26 (a) 2580 m; (b) 5410 m 2.28 4400 400 ft 2.30 101,100 Pa 2.32 22.6 cm 2.34 p h[ water(1 d2/D2) oil(1 d2/D2)] 2.36 25° 2.38 (a) p1,gage (m a)gh (t a)gH 2.40 21.3 cm 2.42 pA pB (2 1)gh 2.44 (a) 171 lb/ft2; (b) 392 lb/ft2; manometer reads friction loss Chapter 1 1.2 1.3 E44 molecules 1.4 1.63 slug/ft3, 839 kg/m3 1.6 (a) {L2/T2}; (b) {M/T} 1.8 1.00 My/I 1.10 Yes, all terms are {ML/T2} 1.12 {B} {L1} 1.14 Q Const B g1/2H3/2 1.16 All terms are {ML2T2} 1.18 V V0emt/K 1.20 zmax 64.2 m at t 3.36 s 1.22 (a) 0.372U 2/R; (b) x 1.291 R 1.24 e 221,000 J/kg 1.26 Wair 0.71 lbf 1.28 wet 1.10 kg/m3, dry 1.13 kg/m3 1.30 W1-2 21 ft lbf 1.32 (a) 76 kN; (b) 501 kN 1.34 1300 atm 1.36 (a) BN2O 1.33 E5 Pa; (b) Bwater 2.13 E9 Pa 1.38 1380 Pa, ReL 28 1.40 A 0.0016 kg/(m s), B 1903 K 1.42 /200K (T K/200 K)0.68 1.44 Data 50 percent higher; Andrade fit varies 50 percent 1.46 V 15 m/s 1.48 F (1/h1 2/h2)AV 1.50 M(ro ri)/(2ri 3L) 1.52 P 73 W 1.54 M R4/h 1.56 3M sin /(2R3) 1.58 0.040 kg/(m s), last 2 points are turbulent flow 1.60 0.88 0.023 kg/(m s) 1.62 28,500 Pa 1.64 (a) 0.023 m; (b) 0.069 m 1.66 F 0.014 N 806 Answers to Selected Problems 2.46 1.45 2.48 F 39700 N 2.50 (a) 524 kN; (b) 350 kN; (c) 100 kN 2.52 0.96 2.54 879 kg/m3 2.56 16.08 ft 2.58 0.40 m 2.60 Fnet 23,940 N at 1.07 m above B 2.62 10.6 ft 2.64 1.35 m 2.66 F 1.18 E9 N, MC 3.13 E9 N m counterclockwise, no tipping 2.68 18,040 N 2.70 4490 lbf at 1.44 ft to right and 1.67 ft up from point B 2.72 33,500 N 2.74 F ba20{(a2/g)[exp(gh/a2) 1] h} 2.76 P ( /24)h2b(3 csc2 ) 2.78 P R3/4 2.80 (a) 58,800 Pa; (b) 0.44 m 2.82 FH 97.9 MN, FV 153.8 MN 2.84 FH 4895 N, FV 7343 N 2.86 FH 0, FV 6800 lbf 2.88 FH 176 kN, FV 31.9 kN, yes 2.90 467 lbf 2.92 Fone bolt 11,300 N 2.94 Cx 2996 lb, Cz 313 lbf 2.96 (a) 940 kN; (b) 1074 kN; (c) 1427 kN 2.98 FH 7987 lbf, FV 2280 lbf 2.100 FH 0, FV 297 kN 2.102 124 kN 2.104 5.0 N 2.106 4310 N/m3 2.108 12.6 N 2.110 h (a) 7.05 mm; (b) 7.00 mm 2.112 (a) 39 N; (b) 0.64 2.114 0.636 2.116 19100 N/m3 2.118 (a) draft 7.24 in; (b) 25 lbf 2.120 34.3° 2.122 a/b 0.834 2.124 6850 m 2.126 h/H Z (Z2 1 )1/2, d/H, Z (2 )/2, pa/(gH) 2.128 Yes, stable if S 0.789 2.130 Slightly unstable, MG 0.007 m 2.132 Stable if R/h 3.31 2.134 (a) unstable; (b) stable 2.136 MG L2/(3R) 4R/(3) 0 if L 2R 2.138 2.77 in deep; volume 10.8 fluid ounces 2.140 ax (a) 1.96 m/s2 (deceleration); (b) 5.69 m/s2 (deceleration) 2.142 (a) 16.3 cm; (b) 15.7 N 2.144 (a) ax 319 m/s2; (b) no effect, pA pB 2.146 Leans to the right at 27° 2.148 Leans to the left at 27° 2.150 5.5 cm; linear scale OK 2.152 (a) 224 r/min; (b) 275 r/min 2.154 (a) both are paraboloids; (b) pB 2550 Pa (gage) 2.157 77 r/min, minimum pressure halfway between B and C 2.158 10.57 r/min Chapter 3 3.2 r position vector from point O 3.6 Q (2b/3)(2g)1/2[(h L)3/2 (h L)3/2] 3.8 Q K per unit depth 3.10 (/3)R2U0cpTw 3.12 (a) 44 m3/h; (b) 9.6 m/s 3.14 dh/dt (Q1 Q2 Q3)/(d2/4) 3.16 Qtop 3U0b/8 3.18 (b) Q 16bhumax/9 3.20 (a) 7.97 mL/s; (b) 1.27 cm/s 3.22 (a) 0.06 kg/s; (b) 1060 m/s; (c) 3.4 3.24 h [3Kt2d2/(8 tan2 )]1/3 3.26 Q 2U0bh/3 3.28 t (1 1/3 )h0 1/2/(2CA2 g ) 3.30 (a) dh/dt Q/(2hb cot 20°) 3.32 Vhole 6.1 m/s 3.34 V2 4660 ft/s 3.36 U3 6.33 m/s 3.38 V V0r/(2h) 3.40 500 N to the left 3.42 F (p1 pa)A1 1A1V1 2[(D1/D2)2 1] 3.44 F U2Lb/3 3.46 (1 cos )/2 3.48 V0 2.27 m/s 3.50 102 kN 3.52 F WhV1 2[1/(1 sin ) 1] to the left 3.54 163 N 3.56 2.45 N/m 3.58 40 N 3.60 2100 N 3.62 3100 N 3.64 980 N 3.66 8800 N 3.70 91 lbf 3.72 Drag 4260 N 3.74 Fx 0, Fy 17 N, Fz 126 N 3.76 77 m/s 3.80 F (/2)gb(h1 2 h2 2) h1bV1 2(h1/h2 1) 3.82 25 m/s 3.84 23 N 3.86 274 kPa Answers to Selected Problems 807 3.88 V [2 2Vj]1/2, Q/2k 3.90 dV/dt g 3.92 dV/dt gh/(L h) 3.94 h 0 at t 70 s 3.96 d2Z/dt2 2gZ/L 0 3.100 (a) 507 m/s and 1393 m; (b) 14.5 km 3.102 h2/h1 1 2 1 2 [1 8V1 2/(gh1)]1/2 3.104 (Ve/R) ln (1 m ˙ t/M0) 3.106 final 75 rad/s 3.108 (a) V V0/(1 CV0t/M), C bh(1 cos ) 3.110 (a) 0.113 ft lbf; (b) 250 r/min 3.112 T m ˙ R0 2 3.114 (a) 414 r/min; (b) 317 r/min 3.116 P Qr2[r2 Q cot 2/(2r2b2)] 3.118 P QuVn(cot 1 cot 2) 3.120 (a) 22 ft/s; (b) 110 ft/s; (c) 710 hp 3.122 L h1 (cot )/2 3.124 41 r/min 3.126 15.5 kW (work done on the fluid) 3.128 1.07 m3/s 3.130 34 kW 3.134 4500 hp 3.136 5.6 m3/h 3.138 gd4(H L )/(128L Q) 2Q/(16L ) 3.140 1640 hp 3.142 (a) 1150 gal/min; (b) 67 hp 3.144 26 kW 3.146 h 3.6 ft 3.148 hf 0.21 m 3.152 (a) 85.9°; (b) 55.4° 3.154 h 0.133 m 3.156 (a) 102 kPa; (b) 88 mi/h 3.158 (a) 169.4 kPa; (b) 209 m3/h 3.160 (a) 31 m3/s; (b) 54 kW 3.162 Q 166 ft3/min, p 0.0204 lbf/in2 3.164 (a) 5.25 kg/s; (b) 2.9 cm 3.166 (a) 60 mi/h; (b) 1 atm 3.168 h 1.08 ft 3.170 h 1.76 m 3.172 D 0.132 ft 3.174 (a) 5.61 ft/s; (b) further constriction reduces V2 3.176 (a) 9.3 m/s; (b) 68 kN/m 3.178 h2 2.03 ft (subcritical) or 0.74 ft (supercritical) 3.180 V Vf tanh (Vft/2L), Vf (2gh)1/2 3.182 kp/[(k 1)] V2/2 gz constant Chapter 4 4.2 (a) du/dt (2V0 2/L)(1 2x/L) 4.4 At (2, 1), dT/dt 125 units 4.6 (a) 6V0 2/L; (b) L ln 3/(2V0) 4.8 (a) 0.0196 V2/L; (b) at t 1.05 L/U 4.12 If 0, r r2 fcn (, ) 4.14 fcn(r) only 4.16 y2 3x2y fcn(x, z) 4.18 0L0/(L0 Vt) 4.20 0 const, {K} {L/T}, {a} {L1} 4.22 xL 1.82 kg/m3 4.28 Exact solution for any a or b 4.30 p const (K2/2)(x2 y2) 4.32 f1 C1r; f2 C2/r 4.34 p p(0) 4umaxz/R2 4.36 C g sin /(2) 4.38 Cz yx xy 4.42 Tmean (∫uT dy)/(∫u dy) 4.48 Kxy const 4.50 Inviscid flow around a 180° turn 4.52 4Q/(b) 4.54 Q ULb 4.60 Irrotational, z0 H 2R2/(2g) 4.62 Vy2/(2h) const 4.66 K sin /r 4.68 m tan1[2xy/(x2 y2 a2)] 4.70 cos /r2, 2am 4.72 (a) 8.8m; (b) 55 m 4.74 Uy K ln r 4.76 (a) 0.106 m from A; (b) 0.333 m above the wall 4.78 (a) Vwall,max m/L ; (b) pmin at x L 4.80 (a) w (g/2)(2x x2) 4.82 Obsessive result: R2/r 4.84 z (gb2/2) ln (r/a) (g/4)(r2 a2) 4.86 Q 0.0031 m3/(s m) 4.88 z U ln (r/b)/[ln (a/b)] 4.90 F 3.34 N Chapter 5 5.2 1.21 m 5.4 V 1.55 m/s, F 1.3 N 5.6 F 450 N 5.10 (a) {ML2T2}; (b) {MLT2} 5.14 /x fcn (Ux/) 5.16 Stanton number h/(Vcp) 5.18 Q/[( p/L)b4] const 5.20 P/(3D5) fcn[Q/(D3), D2/] 5.22 D/V fcn(N, H/L) 5.24 F/(V2L2) fcn(, VL/, L/D, V/a) 5.26 (a) indeterminate; (b) T 2.75 s 5.28 /L fcn[L/D, VD/, E/(V2)] 5.30 hL/k fcn(UL/, cp/k) 5.32 Q/(bg1/2H3/2) const 5.34 khydrogen 0.182 W/(m K) 808 Answers to Selected Problems 5.36 (a) QlossR/(A ) constant 5.38 d/D fcn(UD/, U2D/Y) 5.40 h/L fcn(gL2/Y, , ) 5.44 (a) {} {L2} 5.48 F 0.17 N; (doubling U quadruples F) 5.50 (a) F/(UL ) constant 5.52 U 5 ft/s, F 0.003 lbf/ft 5.54 Power 7 hp 5.56 V 128 ft/s 87 mi/h 5.58 V 2.8 m/s 5.60 Prototype power 157 hp 5.62 max 26.5 r/s; p 22,300 Pa 5.64 aluminum 0.77 Hz 5.66 (a) V 27 m/s; (b) z 27 m 5.68 (a) F/(U) constant; (b) No, not plausible 5.70 F 87 lbf 5.72 V 25 ft/s 5.74 Prototype moment 88 kN m 5.76 Drag 107,000 lbf 5.78 Weber no. 100 if Lm/Lp 0.0090 5.80 (a) 1.86 m/s; (b) 42,900; (c) 254,000 5.82 Speeds: 19.6, 30.2, and 40.8 ft/s; Drags: 14,600; 31,800; and 54,600 lbf 5.84 Vm 39 cm/s; Tm 3.1 s; Hm 0.20 m 5.88 At 340 W, D 0.109 m 5.90 pD/(V2L) 0.155(VD/)1/4 Chapter 6 6.2 (a) x 2.1 m; (b) x 0.14 m 6.4 (a) 39 m3/h; (b) 1.3 m3/h 6.6 (a) laminar; (b) laminar 6.8 (a) 3600 Pa/m; (b) 13,400 Pa/m 6.10 (a) from A to B; (b) hf 7.8 m 6.18 (a) 0.054 m3/s; (b) 8.5 m/s; (c) 122 Pa; (d) 542 kPa 6.20 (a) 0.204 m; (b) 19,800 Pa/m; (c) 9980 Pa/m 6.22 (a) 39 kg/s; (b) 1430 6.24 Head loss 25 m 6.26 4 mm 6.28 Q 0.31 m3/h 6.30 F 4 N 6.32 (a) 127 MPa; (b) 127 kW 6.34 0.000823 kg/(m s) 6.36 p 65 Pa 6.38 (a) 19.3 m3/h; (b) flow is up 6.40 (a) flow is up; (b) 1.86 m3/h 6.44 hf 10.5 m, p 1.4 MPa 6.46 Input power 11.2 MW 6.48 r/R 1 e3/2 6.50 (a) 4000 Pa/m; (b) 50 Pa; (c) 46 percent 6.52 p1 2.38 MPa 6.54 D 0.118 m 6.56 (a) 188 km; (b) 27 MW 6.58 Power 870 kW 6.64 Q 19.6 m3/h (laminar, Re 1450) 6.66 (a) 56 kPa; (b) 85 m3/h; (c) u 3.3 m/s at r 1 cm 6.68 Power 204 hp 6.70 Q 2.21 ft3/s 6.72 Optimum 90° (0.7 m rise) 6.74 D 0.52 in 6.76 Q 15 m3/h 6.78 Q 25 m3/h (to the left) 6.80 Q 0.905 m3/s 6.82 D 0.394 m 6.84 D 0.104 m 6.86 (a) 3.0 m/s; (b) 0.325 m/m; (c) 2770 Pa/m 6.90 Q 19.6 ft3/s 6.92 (a) 1530 m3/h; (b) 6.5 Pa (vacuum) 6.94 260 Pa/m 6.96 Cross section 0.106 m by 0.531 m 6.98 Approximately 128 squares 6.102 (a) 5.55 hp; (b) 5.31 hp with 6° cone 6.104 p 0.0305 lbf/in2 6.106 Q 0.0296 ft3/s 6.108 Q 0.22 ft3/s 6.110 840 W 6.112 Q 0.0151 ft3/s 6.114 (a) Q1 0.0167 m3/s, Q2 0.0193 m3/s, p 774 kPa 6.116 Q 0.027 m3/s 6.118 p 131 lbf/in2 6.120 Q1 0.0109 m3/s, Q2 0.0264 m3/s, Q3 0.0183 m3/s 6.122 Increased /d and L/d are the causes 6.124 Q1 2.09 ft3/s, Q2 1.61 ft3/s, Q3 0.49 ft3/s 6.126 opening 35° 6.128 QAB 3.47, QBC 2.90, QBD 0.58, QCD 5.28, QAC 2.38 ft3/s (all) 6.130 QAB 0.95, QBC 0.24, QBD 0.19, QCD 0.31, QAC 1.05 ft3/s (all) 6.132 2 6°, De 2.0 m, pe 224 kPa 6.134 2 10°, We 8.4 ft, pe 2180 lbf/ft2 6.136 (a) 25.5 m/s, (b) 0.109 m3/s, (c) 1.23 Pa 6.138 46.7 m/s 6.140 p 273 kPa 6.142 Q 18.6 gal/min, dreducer 0.84 cm 6.144 Q 54 m3/h 6.146 (a) 0.00653 m3/s; (b) 100 kPa 6.148 (a) 1.58 m; (b) 1.7 m 6.150 p 27 kPa 6.152 D 4.12 cm 6.154 h 59 cm Answers to Selected Problems 809 6.156 Q 0.924 ft3/s 6.158 (a) 49 m3/h; (b) 6200 Pa Chapter 7 7.2 Rec 1.5 E7 7.4 d 8 mm lies in the transition region 7.6 H 2.5 (versus 2.59 for Blasius) 7.8 Approximately 0.08 N 7.12 Does not satisfy ∂2u/∂y2 0 at y 0 7.14 C 0/ const 0 (wall suction) 7.16 (a) F 181 N; (b) 256 N 7.18 0.16°; Fdrag 0.024 N 7.20 x 0.91 m 7.22 ( xU)1/2 f(!) 7.24 h1 9.2 mm; h2 5.5 mm 7.26 Fa 2.83 F1, Fb 2.0 F1 7.28 (a) Fdrag 2.66 N2(L)1/2U3/2a 7.30 (a) F 72 N; (b) 79 N 7.32 F 0.0245 1/7 L6/7 U0 13/7 7.34 F 725 N 7.36 7.2 m/s 14 kn 7.38 (a) 7.6 m/s; (b) 6.2 m/s 7.40 L 3.53 m, b 1.13 m 7.42 P4 blades 0.0321/7(C)6/7 20/7 R27/7 7.44 Accurate to about 6 percent 7.46 9 mm, U 11.2 m/s 22 kn 7.48 Separation at x/L 0.158 (1 percent error) 7.50 Separation at x/R 1.80 rad 103.1° 7.52 CD (ReL )1/2 2.67 (by numerical integration) 7.54 Moment 200,000 N m 7.56 (a) 10 N; (b) 80 N 7.58 (a) 3200 N/m; (b) 2300 N/m 7.60 Tow power 140 hp 7.62 Square side length 0.83 m 7.64 t1000–2000m 202 s 7.68 (a) 34 m/s; (b) no, only 67 percent of terminal velocity at impact 7.70 (a) 642 ft; (b) 425 ft 7.72 (a) L 6.3 m; (b) 120 m 7.78 p 100 Pa 7.80 72° 7.82 Vmin 138 ft/s; (b) Vmax 377 ft/s 7.84 V 9 m/s 7.86 Approximately 3.05 m by 6.1 m 7.88 (a) 62 hp; (b) 86 hp 7.90 Voverturn 145 ft/s 99 mi/h 7.94 Torque (CD/4)2DR4, max 85 r/min 7.96 avg 0.21 U/D 7.98 (b) h 0.18 m 7.100 (b) Dmax 78 m 7.106 (a) 300 m; (b) 380 m 7.108 xball 13 m 7.110 y 1.9 ft 7.114 Vfinal 18.3 m/s 66 km/h 7.116 (a) 87 mi/h; (b) 680 hp 7.118 (a) 21 m/s; (b) 360 m 7.120 (L/D)max 21; 4.8° 7.122 (a) 6.7 m/s; (b) 13.5 m/s 26 kn 7.124 crude theory 340 r/s Chapter 8 8.2 " (R2 2 R1 2) 8.4 No, 1/r is not a proper two-dimensional potential 8.6 B(y2 x2) 8.8 " 4B 8.12 " 0 8.14 Irrotational outer, rotational inner; minimum p p 2R2 at r 0 8.18 From afar: a single source 4m 8.20 Vortex near a wall (see Fig. 8.17b) 8.22 Same as Fig. 8.6 except upside down 8.24 Cp {2(x/a)/[1 (x/a)2]}2, Cp,min 1.0 at x a 8.26 Vresultant 9.4 m/s at 47° 8.28 Creates a source in a square corner 8.34 Two stagnation points, at x a/3 8.36 U 12.9 m/s, 2L 53 cm, Vmax 22.5 m/s 8.42 K/(U
a) 0.396, h/a 1.124 8.44 K 4.6 m2/s; (a) 218 kPa; (b) 214 kPa at upper shoulder, 6 kPa at lower shoulder (cavitation) 8.46 F1-bolt 5000 N 8.50 h 3a/2, Umax 5U/4 8.52 Vboat 10.2 ft/s with wind at 44° 8.54 Fparallel 6700 lbf, Fnormal 2700 lbf, power 560 hp (very approximate) 8.56 CD 2.67 (too high, incorrect prear) 8.60 This is Fig. 8.15a, flow in a 60° corner 8.62 Stagnation flow near a “bump” 8.64 All favorable gradients: no separation 8.66 0.45m/(5m 1) if U Cxm 8.68 Flow past a Rankine oval 8.70 Applied to wind-tunnel “blockage” 8.72 Adverse gradient for x a 8.74 VB,total (8Ki 4Kj)/(15a) 8.78 Need an infinite array of images 8.82 (a) 4.5 m/s; (b) 1.13; (c) 1.26 hp 8.84 (a) 0.21; (b) 1.9° 8.86 (a) 26 m; (b) 8.7; (c) 1600 N 8.88 Thrust1-engine 2900 lbf 8.90 (a) 4.0; (b) 4.8° 8.92 (a) 0.77 m; (b) V 4.5 m/s at (r, ) (1.81, 51°) and (1.11, 88°) 8.94 Yes, they are orthogonal 810 Answers to Selected Problems 8.98 Yes, a closed teardrop shape appears 8.100 V 14.1 m/s, pA 115 kPa 8.102 (a) 1250 ft; (b) 1570 ft (crudely) Chapter 9 9.2 (a) V2 450 m/s, s 515 J/(kg K); (b) V2 453 m/s, s 512 J/(kg K) 9.4 About 50 m/s 9.6 Exit at about T2 54°C and V2 1445 m/s 9.8 410 K 9.10 Ma 0.78 9.12 (a) 2.13 E9 Pa and 1460 m/s; (b) 2.91 E9 Pa and 1670 m/s; (c) 2645 m/s 9.18 (a) 930 ft/s; (b) 878 ft/s 9.20 (a) air: 144 kPa and 995 m/s; (b) helium: 128 kPa and 2230 m/s 9.22 (a) 267 m/s; (b) 286 m/s 9.24 (b) at Ma 0.576 9.28 (a) 0.17 kg/s; (b) 0.90 9.30 (a) 262 m/s; (b) 0.563; (c) 0.905 kg/m3 9.32 (a) 141 kPa; (b) 101 kPa; (c) 0.706 9.34 (a) 0.00424 slug/s; (b) 0.00427 slug/s 9.40 (a) 2.50; (b) 7.6 cm2; (c) 1.27 kg/s; (d) Ma2 1.50 9.42 (a) Ma 0.90, T 260 K, V 291 m/s 9.44 Ve 5680 ft/s, pe 15.7 psia, Te 1587°R, thrust 4000 lbf 9.46 Rx 8 N (to the left) 9.48 (a) 313 m/s; (b) 0.124 m/s; (c) 0.00331 kg/s 9.50 (a) Dexit 5.8 cm 9.52 (a) 5.9 cm2; (b) 773 kPa 9.54 Ma2 0.648, V2 279 m/s, T2 461°K, p2 458 kPa, p02 607 kPa 9.56 At about A1 24.7 cm2 9.58 (a) 306 m/s; (b) 599 kPa; (c) 498 kPa 9.60 Upstream: Ma 1.92, V 585 m/s 9.62 C 19,100 ft/s, Vinside 15,900 ft/s 9.64 (a) 0.150 kg/s; (b, c) 0.157 kg/s 9.66 h 1.09 m 9.68 patm 92.6 kPa; max flow 0.140 kg/s 9.70 (a) 388 kPa; (b) 19 kPa 9.72 Mass flow 0.5 kg/s, pe 185 kPa, Mae 0.407 9.74 (a) 1.096 MPa; (b) 2.24 kg/s 9.76 tshocks 23 s; tchoking-stops 39 s 9.78 Case A: 0.0071 kg/s; B: 0.0068 kg/s 9.80 A 2.4 E-6 ft2 or Dhole 0.021 in 9.82 Ve 110 m/s, Mae 0.67 (yes) 9.84 (a) 0.96 kg/s; (b) 0.27; (c) 435 kPa 9.86 V2 107 m/s, p2 371 kPa, T2 330 K, p02 394 kPa 9.88 L 2 m, yes, a shock at Ma2 2.14 9.90 (a) 0.764 kg/s; (b) 0.590 kg/s; (c) 0.314 kg/s 9.92 (a) 0.45; (b) 2.04 kg/s 9.98 (a) 430; (b) 0.12; (c) 0.00243 kg/h 9.100 Lpipe 69 m 9.102 Flow is choked at 0.69 kg/s 9.104 ptank 99 kPa 9.106 (a) 0.031 m; (b) 0.53 m; (c) 26 m 9.108 Mass flow drops by about 32 percent 9.112 (a) 105 m/s; (b) 215 kPa 9.116 Vplane 2640 ft/s 9.118 V 204 m/s, Ma 0.6 9.120 P is 3 m ahead of the small circle, Ma 2.0, Tstag 518 K 9.122 23.13°, Ma2 2.75, p2 145 kPa 9.126 (a) 25.9°; (b) 26.1° 9.128 wedge 15.5° 9.130 (a) 57.87°; (b) 21.82° 9.132 (a) pA 18.0 psia; (b) pB 121 psia 9.134 Ma3 1.02, p3 727 kPa, 42.8° 9.136 (a) h 0.40 m; (b) Ma3 2.43 9.138 pr 21.7 kPa 9.140 Ma2 2.75, p2 145 kPa 9.142 (a) Ma2 2.641, p2 60.3 kPa; (b) Ma2 2.299, p2 24.1 kPa 9.146 9.47° (helium) 9.148 CL 0.184 (approximately linear), CD 0.0193 (approximately parabolic) 9.150 (a) 4.10°; (b) drag 2150 N/m 9.152 Parabolic shape has 33 percent more drag Chapter 10 10.2 (a) C 3.31 m/s; (b) V 0.030 m/s 10.4 These are piezometer tubes (no flow) 10.6 (a) Fr 3.8; (b) Vcurrent 7.7 m/s 10.8 ttravel 6.3 h 10.10 crit 2(/g)1/2 10.14 Flow must be fully rough turbulent (high Re) for Chézy to be valid 10.16 20 percent less flow, independent of n 10.18 yn 0.993 m 10.20 Q 74 ft3/s 10.22 S0 0.00038 (or 0.38 m/km) 10.24 yn 0.56 m 10.26 (a) 17.8 m3/s; (b) 1.79 m 10.30 t 32 min 10.32 74,000 gal/min 10.34 If b 4 ft, y 9.31 ft, P 22.62 ft; if b 8 ft, y 4.07 ft, P 16.14 ft 10.36 y2 3.6 m 10.38 Dsemicircle 2.67 m (16 percent less perimeter) 10.42 P 41.3 ft (71 percent more than Prob. 10.39) 10.44 Hexagon side length b 2.12 ft 10.46 Best h0/b 0.53 0.03 Answers to Selected Problems 811 10.48 (a) 0.00634; (b) 0.00637 10.50 (a) 2.37; (b) 0.62 m; (c) 0.0026 10.52 W 2.06 m 10.54 (a) 1.98 m; (b) 3.11 m/s; (c) 0.00405 10.56 (a) 1.02 m3/s; (b) 0.0205 10.58 Fr 0.628R1/6, R in meters 10.60 (a) 0.052 m3/(m s); (b) 0.0765 m 10.64 hmax 0.35 m 10.66 (a) 1.47; (b) y2 1.19 m 10.70 (a) 0.15 m; (b) 3.2; (c) 0.59 m3/(s m) 10.72 (a) 0.046 m; (b) 4.33 m/s; (c) 6.43 10.76 H 0.011 m 10.78 t 8.6 s (crude analysis) 10.80 (a) 3.83 m; (b) 4.83 m3/(s m) 10.82 (a) 0.88 m; (b) 17.6 m/s; (c) 2.89 m 10.84 y2 0.82 ft; y3 5.11 ft; 47 percent 10.86 (a) 6.07 m/s; (b) V 2.03 m/s 10.88 (a) downstream; (b) 5.7 percent 10.90 0.0207 (or 1.19°) 10.92 (a) 3370 ft3/s; (b) 7000 hp 10.94 (a) 0.61 m; (b) 3.74 m/s; (c) 0.89 m 10.98 (a) steep S-3; (b) S-2; (c) S-1 10.106 No entry depth leads to critical flow 10.108 (a, b) Both curves reach y yn 0.5 m at x 250 m 10.110 (a) ycrest 0.782 m; (b) y(L) 0.909 m 10.112 M-1 curve, with y 2 m at L 214 m 10.114 Vexing! Flow chokes at Q 17 m3/s 10.116 Q 9.51 m3/s 10.120 Y 0.64 m, 34° 10.122 5500 gal/min 10.124 M-1 curve, y 10 ft at x 3040 ft 10.126 At x 100 m, y 2.81 m 10.128 At 300 m upstream, y 2.37 m Chapter 11 11.6 This is a diaphragm pump 11.8 (a) H 112 ft and p 49 lb/in2; (b) H 112 ft (of gasoline); P 15 hp 11.10 (a) 1300 r/min; (b) 2080 lbf/in2 11.12 (a) 11.3 m; (b) 1520 W 11.14 1870 W 11.16 (a) 1450 W; (b) 1030 r/min 11.18 Vvane (1/3)Vjet for max power 11.20 (a) 2 roots: Q 7.5 and 38.3 ft3/s; (b) 2 roots; H 180 ft and 35 ft 11.22 (a) BEP 92 percent at Q 0.22 m3/s 11.26 Correlation is “fair,” not geometrically similar 11.28 BEP at about 6 ft3/s; Ns 1430, Qmax 12 ft3/s 11.30 (a) 1700 r/min; (b) 8.9 ft3/s; (c) 330 ft 11.32 Correlation “fair,” not geometrically similar 11.34 (a) 11.5 in; (b) 28 hp; (c) 100 ft; (d) 78 percent 11.36 D 9.8 in, n 2100 r/min 11.38 (a) 18.5 hp; (b) 7.64 in; (c) 415 gal/min; (d) 81 percent 11.40 (a) Ds D(gH)1/4/Q1/2 11.42 NPSHproto 23 ft 11.44 No cavitation, required depth is only 5 ft 11.46 Ds C/Ns, C 7800 7 percent 11.52 (a) 7.97 m3/s; (b) 14.6 kW; (c) 28.3° 11.54 Centrifugal pumps, D 7.2 ft 11.56 (a) D 5.67 ft, n 255 r/min, P 700 hp; (b) D 1.76 ft, n 1770 r/min, P 740 hp 11.58 Centrifugal pump, ! 67 percent, D 0.32 ft 11.60 (a) 623; (b) 762 gal/min; (c) 1.77 ft 11.62 D 18.7 ft, p 1160 Pa 11.64 No speed is able to get to BEP 11.66 Q 1240 ft3/min 11.68 Qnew 15,300 gal/min 11.70 (a) 212 ft; (b) 5.8 ft3/s 11.72 (a) 10 gal/min; (b) 1.3 in 11.74 (a) 14.9; (b) 15.9; (c) 20.7 kgal/min 11.76 Dpipe 1.70 ft 11.78 Dpipe 1.67 ft, P 2000 hp 11.80 Q32 22,900 gal/min; Q28 8400 gal/min, H 343 ft for both 11.84 Two turbines: (a) D 9.6 ft; (b) D 3.3 ft 11.86 Nsp 70, hence Francis turbines 11.88 Q 52 ft3/s, D 10.5 ft 11.90 P 800 kW 11.92 Pelton and Francis wheels both OK 11.94 (a) 71 percent; (b) Nsp 19 11.96 (a) 1.68 ft; (b) 0.78 ft 11.100 (a) 190 kW; (b) 24 r/min; (c) 9.3 ft/s 11.102 Q 29 gal/min 812 Answers to Selected Problems Index A Acceleration of a particle, 15, 90, 216 centripetal, 90, 157 convective, 216 Coriolis, 157 local, 216 Ackeret airfoil theory, 635–636 Acoustics, 573 Actuator disk theory, 752–753 Added mass, 539–540 Adiabatic flow, 578–579 atmospheric lapse rate, 102, 105 with friction, 604–606, 776–780 Adverse pressure gradient, 430, 445–448, 501 Aerodynamic forces and moments, 452–453 NACA designs, 471, 528–530, 565 Air-cushion vehicle, 206 Airfoil description, 468, 529 Airfoil theory, 523–534 finite-span, 530–534 supersonic flow, 632–637 thick-cambered, 528–530 thin-plate, 524–528 Andrade’s equation, 49 Anemometer cup, 387, 485 hot-wire and film, 387, 389 Angle of attack definition, 468 in inviscid flow, 499, 517 Angular momentum theorem, 130, 158–159, 230 Angular velocity of a fluid, 246–247 Annulus flow in, 362–364 laminar friction factors, 364 Answers to selected problems, 806–812 Archimedes’ laws of buoyancy, 44, 84 Area, body reference, 453 Area change in a duct, 583–587 Aspect ratio of a diffuser, 384 of a wing, 472, 473, 530 Atmosphere isothermal, 68 U. S. standard, 69, 773 Automobile drag forces, 461–463 Average velocity, 143, 26 in pipe flow, 144, 341, 344, 346 Avogadro’s number, 46 Axial-flow pumps, 730–734 Axisymmetric potential flow, 534–540 B Backwater curve, 693–695 Barometer, 66–67 Basic equations (see Differential equations of flow) Basic laws of fluid motion, 35, 129–131 Bend losses, 371 813 Bump, channel flow over, 675–676 Butterfly valve, 370 Buoyancy, 84–86 Buoyant force, 85 C Cambered airfoil, 468, 471, 528–530 Capillary effect, 31, 299–300 Cauchy-Riemann equations, 249 Cavitation, 32–33, 552 of a pump, 720, 721 of a turbine, 765 Cavitation erosion, 33 Cavitation number, 32, 294, 297 Center of buoyancy, 85, 88 Center of mass, 80 Center of pressure, 75–76 of an airfoil, 527, 636 Centrifugal pump, 161–162, 200, 714–718 dimensionless coefficients, 724–726 performance curves, 720–723, 758, 759 similarity rules, 727–729 Centripetal acceleration, 90, 157 Centroid, 75 of various cross-sections, 76 Channel flow (see Open channel flow) Chézy coefficient, 665 Chézy formulas, 664–666 Choked flow, 586, 598, 740 due to friction, 608–609 due to heat transfer, 617 in an open channel, 676 Chord line, 452, 468 Circular section open channel, 668–669 pipe flow, 338–357 Circulation, 499 at airfoil trailing edge, 524 on a cylinder, 509, 511 Classification of flow, 36–37 Colebrook pipe-friction formula, 348 Complex-variable potential theory, 516–521 Bernoulli, Daniel, 10, 174 Bernoulli constant, 176, 248 Bernoulli obstruction meters, 397–404 Bernoulli’s equation, 10, 174–177, 230, 248–249 outside a boundary layer, 435 compared to the energy equation, 176, 580 for irrotational flow, 230, 249, 496 for isentropic flow, 580–581 limitations and assumptions, 176, 177 in rotating coordinates, 717 for unsteady flow, 175, 248, 249, 496 Betz number, 753 BG units, 8 Bingham-plastic fluid, 28 Blasius flat-plate solution, 437–439, 478 Blasius pipe friction formula, 345 Blowdown analysis, 598, 643 Blower, 711 Blunt-body flows, 429 Body forces, 61, 224 Bore in a channel, 702 Boundary conditions, 34, 234–236 for a boundary layer, 436 free surface or interface, 234–236, 497 at an inlet or outlet, 234, 496 for inviscid flow, 496, 497 kinematic, 235 no-slip, 24, 34, 234 Boundary-element method, 546–548 Boundary layer, 23, 45, 250, 431, 496, 578 displacement thickness, 433, 438, 443 equations of, 434–436 on a flat plate, 153–155, 266 momentum thickness, 431, 438 with pressure gradient, 445–450 with rough walls, 443–444 separation, 435, 447, 447–449 shape factor, 438, 443, 448 thickness, 428, 442 transition, 298, 432, 439 Bourdon tube gage, 99–100 Brinkman number, 268 Broad-crested weir, 689, 690 Buckingham pi theorem, 280, 286–288 Bulk modulus, 49, 102, 577 of various liquids, 772 814 Index Index 815 Composite channel flows, 687–688 Compressibility criterion, 35, 221, 571 Compressible flow, 220, 315, 571 with area change, 583–587, 774–776 with friction, 603–613, 780–784 with heat transfer, 613–618, 784–788 tables, 774–788 Compressor, 711, 740–741 Computational fluid dynamics, 3, 434, 465, 540–555, 735–736 commercial codes, 552–553 Concentric cylinder flows, 261–263 Cone flow, supersonic, 638 Conformal mapping, 516, 562 Conical diffuser, 373, 386 Conjugate depths, 699 Conservation laws, 35, 130–131 for angular momentum, 130, 158–159 for energy, 131, 163–165, 231 for linear momentum, 130, 146 for mass, 130, 141–146 for salt or species, 35, 315 Consistent units, 11–13 Contact angle, 30–31 Continuity, equation of, 218 cylindrical polar form, 219–220, 793 incompressible flow, 220 spherical polar form, 265 turbulent flow, 334 Continuum, 6–7 Contraction losses, 372, 374 Control section of a channel, 687, 693 Control surface, 136 Control volume, 36, 133 arbitrary but fixed, 135–136 deformable, 137–138, 140 differential-sized, 218 guidelines for selection, 183 moving, 133, 137, 152 one-dimensional, 134–135 Convective acceleration, 15–16 Converging-diverging nozzle, 600–603 Converging nozzle, 598–600 Conversion factors, 8–9, 791–792 Coriolis acceleration, 156, 157, 250 Corner flow, inviscid, 242, 518–519 Correlations, turbulent, 334 Corresponding states, law of, 24 Couette flow between cylinders, 261–263 instability of, 262–263 between plates, 25–26, 258–259 nonnewtonian, 272 Couple, concentrated, 268 Creeping motion, 25, 298, 315, 317, 328, 483 past a sphere, 47, 457, 483 Critical channel flow, 671–674 Critical depth, 664, 672 Critical Reynolds number, 329–330 Critical slope of a channel, 674 Critical sonic-point properties, 581, 605 Critical state, 6, 24 Crossflow turbine, 764 Crump weir, 705 Cup anemometer, 387, 485 Cup-mixing temperature, 268 Curl of a vector, 247 Current meter, 387, 389 Curved surface, force on, 79–82 Curvilinear coordinate system, 267 Cylinder array of, 515 in inviscid flow, 245, 508–510 rotating, 512 in viscous flow, 261–263, 295–296, 298, 455 Cylindrical coordinates, 219 equations of motion, 793–794 D da Vinci, Leonardo, 44, 143 d’Alembert paradox, 45, 510 Darcy friction factor, 340, 342, 344, 604 Darcy’s law of porous flow, 420 Darcy-Weisbach equation, 340, 664 Decimal prefixes, 13 Deformable control volume, 133, 137 Deformation of a fluid element, 245–247 Del operator, 216, 219 816 Index Density definition of, 6, 17 of various fluids, 771–772 Detached shock wave, 624 Diaphragm transducer, 101 Differential equations of flow, 36, 215, 682, 793 angular momentum, 230 continuity or mass, 217–221 cylindrical coordinates, 793–794 energy, 231–233 incompressible, 220, 228, 236–237 linear momentum, 62, 223–227 Diffuser flows, 313, 381–385, 447 head loss, 373 performance maps, 385, 386 separation and stall, 383, 447–448 stability map, 382 subsonic versus supersonic, 584 Digital computer (see Computational fluid dynamics) Dilatant fluid, 28 Dimensional analysis, 7, 12, 277 of the basic equations, 292–294 of the boundary conditions, 293–294 of pipe flow, 307–309 pitfalls of, 305–307 of turbomachines, 724–726, 744 Dimensional homogeneity, 11–12, 280 nonhomogeneity, 285–286 Dimensional matrix, 313 Dimensionless groups, list of, 297 Dimensions, 7–13, 278 list of, 8–9, 287 Discharge coefficient, 12, 179, 181, 398 flow nozzle, 401 orifice plate, 399–400 sluice gate, 677 venturi, 401, 402 weir, 690–692 Displacement thickness, 433 for a flat plate, 433, 438, 443 Dissipation function, 233 of a hydraulic jump, 680 Divergence of a vector, 220, 226 Dot product, 133 Doublet line, 270, 505–506 point, 536 Downwash on a wing, 531, 532 Draft tube, 765 Drag, 452–467 biological adaptation, 467–468 induced, 472, 533 Drag coefficient, 195, 297, 453, 468, 632 of airfoils, 457, 468, 471–473 of a cylinder, 298, 455, 457 rotating, 511–512 on a flat plate, 438, 442–443 at high Mach numbers, 465–466 of road vehicles, 461–463 of a sphere, 298, 456, 457 spinning, 487, 488 of surface ships, 464–465 of three-dimensional bodies, 457, 460 of two-dimensional bodies, 457–458 Drag reduction, 456, 462, 464 Drowned channel flow, 678, 701 Duct flow, 325, 603 compressible, with friction, 603–613, 780 with heat transfer, 613–618, 784 Dynamic similarity, 304–305 E Eckert number, 297 Eddy viscosity, 406 Effective duct diameter, 360 Efficiency, 47, 715 of an open channel, 669–671 of a turbomachine, 715, 734 volumetric, 716, 757 of wind turbines, 753 Elbows, losses in, 368, 369 Elliptical wing, 533 Energy, 18, 131, 163 Energy equation, 163–165, 231–233 steady flow, 167–168 Energy flux, 165, 231 Energy grade line, 177–178, 339 672 Index 817 Engineering Equation Solver (EES), 41 Enthalpy, 19, 165, 574 Entrance length, 331 Entrance losses, 331, 371, 372 Entrance region, 330–331 Entropy, of an ideal gas, 574 Entropy change, 574 across a normal shock, 591 across a weak oblique shock, 627 Equations of motion (see Differential equa-tions of flow) Equilibrium of forces, 61–62 Equivalent length, minor losses, 367 Erosion of particles, 313, 698 Euler, Leonhard, 45, 174, 294 Euler number, 294, 297 Euler turbine equations, 161–162, 717 Eulerian description, 14, 216 Euler’s equation, 227, 237, 247 Exit pipe loss, 371, 372 Expansion losses, 372–373 Explicit numerical model, 549 External flow, 333, 427, 451–467 F Falling-body problem, 12 dimensional analysis of, 282–285, 289 Fanno line, 648 Favorable pressure gradient, 430, 445–447, 495, 502 Film of fluid draining down an inclined plane, 268 down a vertical cylinder, 272 down a vertical plate, 271 Finite-difference method, 541–543 Finite-element method, 541 Finite-span wings, 472–473, 530–534 First law of thermodynamics, 131, 163 Fittings, losses in, 368 Flap, airfoil, 471, 472, 474 Flat-plate flow,153–155, 428, 431–433 Blasius solution, 437–439, 478 integral theory, laminar, 432 turbulent, 441–444 normal to the stream, 457–459, 519–521 with rough walls, 443–444 Flettner rotorship, 511–512, 559 Floating element shear measurement, 480 Flooding of channels, 698 Flow between plates, 25–27, 258–260, 359– 360 Flow coefficient of a meter, 398 Flow meters (see Fluid meters) Flow net, 497 Flow nozzle, 399, 400–401 Flow straighteners, 479 Flow visualization, 40, 426, 470, 515, 554 Fluctuation, turbulent, 326, 333–334 Fluid, definition of, 4 Fluid meters, 385–404 Coriolis type, 395, 396 electromagnetic, 389 flow nozzle, 399, 400–401 head losses, 402 hot-wire, hot-film, 387, 389 laminar-flow element, 396–397 laser-doppler, 387, 389–390 obstruction type, 397–404 orifice plate, 398–400 pitot-static tube, 387, 388–389 rotameter, 395 Savonius rotor, 387, 486 turbine type, 392–393 ultrasonic, 394–395 venturi meter, 399, 401–402 volume flow type, 391 vortex type, 393–394 Fluid properties, 769–773 Force coefficient, 278, 310 Forces hydrostatic, 74–84 on a turning vane, 150–151 Fourier’s law of conduction, 27, 231 Francis turbine, 742 Free-body concept, 77, 133 Free overfall, 687, 688 Free-streamline theory, 520–521 Free-surface flows, 4–5, 236, 326, 497, 659 818 Index Free vortex, 253 Friction drag, 154, 453 Friction losses, 168 Friction factor, 42, 340, 342, 604 compressible flow, 606 Friction factor—Cont.
laminar pipe-flow, 342 noncircular ducts, 364, 365 rocky channels, 665 turbulent pipe-flow, 345, 348 Friction velocity, 336 Frictionless flow, 227, 495, 613 Frontal area, 453, 462 Froude, William, 45, 294 Froude number, 294, 297, 465–466, 662, 679, 683 Froude scaling laws, 303–304 Fully developed flow, 258, 330–331 Fully rough flow in channels, 665 on a flat plate, 443–444 in pipes, 347, 348 Fundamentals of Engineering (FE) Exam, 43 G Gage pressure, 63, 77, 148 Gages, pressure, 97–101 Gas constant, 19, 573 of various gases, 772 Gas dynamics (see Compressible flow) Geometric similarity, 301–303 violation of, 302, 303, 307 Glide angle, 489 Gradient operator, 61, 216, 219 Gradual contraction loss, 374 Gradual expansion loss, 373 Gradually varied flow, 662, 682–687 classification, 683–685 effect of width changes, 704 Grashof number, 297 Gravity acceleration of, 9, 64 variation with radius, 65 Gravity force on an element, 61, 224 Grid, numerical, 542, 549, 553, 564 H Hagen, G. L. H., 329 Hagen-Poiseuille flow, 341 Half-body, plane, 256–258, 501–502 axisymmetric, 537–538 Halocline, 120 Hazen-Williams formula, 47, 285 Head loss, 168, 340, 661 of a hydraulic jump, 680 minor, 367–375 in pipe flow, 340 Heat addition, flow with, 613–618, 784–788 Heat conduction equation, 233 Heat flux through an element, 232 Heat transfer coefficient, 312 Hele-Shaw flow, 513–514 Herschel-type venturi, 401 High-lift devices, 474 History of fluid mechanics, 44–46, 280 Hodograph for an oblique shock, 623, 624 Homologous points, 302, 727 Honeycomb flow straightener, 408 Horseshoe vortex, 555 Hot-wire or hot-film anemometer, 387, 389 Hydraulic diameter, 358, 363, 661 Hydraulic efficiency, 716 Hydraulic grade line, 177–178, 339, 659 Hydraulic jump, 198–199, 664, 678–681, 702 classification, 679 sloping, 702 Hydraulic model, 307 Hydraulic radius, 13, 358, 661 Hydraulically smooth wall, 347 Hydrodynamic mass, 539–540 Hydrogen bubble technique, 34 Hydrometer, 118 Hydrostatic condition, 4, 59, 62, 63 in gases, 67–69 in liquids, 65–66 Index 819 Hydrostatic forces on curved surfaces, 79–83 in layered fluids, 82–84 laboratory apparatus, 127 on plane surfaces, 74–79 Hydrostatic pressure distribution, 63–65 Hypersonic flow, 572, 639 I Icebergs, 89–90 Ideal gas (see Perfect-gas law) Images, 521–522 Implicit numerical model, 550 Impulse turbines, 745–749 Incompressible flow, 17, 142–143, 220, 236, 572 Induced drag, 533 Inertial coordinate system, 156 Initial conditions, 234 Integral equations (see Control volume) Intensity of turbulence, 334, 405 Interface, 29 Internal energy, 18, 574 Internal flow, 330 Inviscid flow, 36, 496 Irrotational flow, 230, 247–249, 496 frictionless, 247, 579–582 Isentropic flow, 579–582, 529 with area change, 583–587, 774 compared to Bernoulli’s equation, 580–581 tables, 774–776 Isentropic process, 574 Isothermal duct flow, 610–611 Isovelocity contours, 660 J Jet exit pressure condition, 149 Jet flow, laminar and turbulent, 326–327 Jet pump, 189, 712 Jet-turning vane, 150–151 Joukowski transformation, 562 K Kaplan turbine, 742, 746, 749 Kármán momentum-integral relation, 154, 431 Kármán vortex street, 295–296 Kelvin oval, 513–514 Kinematic properties, 15 Kinematic similarity, 303–304, 498 Kinematic viscosity, 24 of various fluids, 24, 770–772 Kinetic energy, 18, 164 correction factor, 170–171 Kline-Fogleman airfoil, 473, 474 Kutta condition, 523–524 Kutta-Joukowski lift theorem, 510–511 L Lagrangian description, 14 Laminar flow, 34, 326, 327, 551 in a concentric annulus, 346–364 between parallel plates, 25–27, 258–260, 359–360 in a pipe, 341–344 between rotating cylinders, 261–263 Laplace’s equation, 239, 251, 496, 497, 516, 793 numerical simulation, 541–543 in polar coordinates, 498, 564, 793 Lapse rate, 68 Large-eddy simulation, 554 Laser-Doppler anemometer, 387, 389 Law-of-the-wall, 336 Lawn sprinkler analysis, 163 Layered fluids, 70, 82–84 Lift definition of, 452, 468 in flow past a cylinder, 510–511 820 Index Lift coefficient of airfoils, 470–474, 527, 529–530, 533 supersonic, 632–637 definition, 297, 468, 632 maximum, 473, 528 of a rotating cylinder, 511–512 of a rotating sphere, 488 Lift-drag polar plot, 472 Lifting line theory, 532 Lifting vane, 150–151, 469 Linear momentum, 130, 146–158, 223–227 Liquids versus gases, 4–6 Local acceleration, 216 Local mass-flow function, 586–587 Logarithmic velocity profile, 336, 344 Loss minor, 367–375 in pumps, 723 Lubricating oil properties, 769, 770, 772 Lubrication theory, 271 M Mach angle, 619 Mach cone, 619 Mach number, 35, 221, 295, 297, 306, 572, 579, 592 effect on body drag, 467 Mach waves, 594, 618–621, 628 analogy to water waves, 663, 673, 681 Magnus effect, 510 Manifold flow, 422–423 Manning, Robert, 13, 665 Manning roughness factor, 13, 285, 665 for various channels, 667 Manometer, 70–73, 97, 99 two-fluid differential, 72, 108 Mass, units of, 8 Mass flow, 133, 142, 586, 599, 611 in choked flow, 586 MATLAB contouring, 506 Mean free path of a gas, 7, 46 Meniscus, 74 Metacenter, 87 Metacentric height, 87, 88 Meter (see Fluid meters) Minor losses in pipe flow, 367–375 Mixing-length theory, 406 Model-testing principles, 278, 301–307 pitfalls and discrepancies, 296–297 Mohr’s circle, 4–5, 59 Molecular weight, 19, 573 of various gases, 772 Moment of inertia, 76, 88, 131 for various areas, 76 Momentum angular, 130, 158, 230 linear, 130, 146–148, 223–227 Momentum flux, 147, 224 correction factor, 155–156 Momentum integral theory, 155, 431–433, 448 Momentum thickness, 431 for a flat plate, 438, 442 Thwaites’ parameter, 448 Moody chart, 349, 443, 606, 665 Moody pump-size formula, 728 Moving shock wave, 595–596 Multiple-pipe systems, 375–381 N NACA airfoils, 470–471, 528, 567 Nappe, 687, 689 Natural convection, 312, 315, 573 Navier-Stokes equations, 45, 228, 789 nonuniqueness of, 263 Net positive suction head, 721–722 Network, piping, 380–381 Neutral buoyancy, 86, 386 Newton, Sir Isaac, 23, 45, 577 Newtonian fluid, 23, 227–228 Newton’s second law, 8, 130, 146 for a fluid element, 62, 216, 224 in noninertial coordinates, 156–158 No-slip condition, 24, 34, 234, 259, 262, 340 No-temperature-jump condition, 34, 234 Noncircular duct flow, 357–366 Index 821 Nondimensionalization (see Dimensional analysis) Noninertial coordinate system, 156–158 Nonnewtonian fluids, 28, 272 Nonwetting liquid, 30, 31 Normal channel depth, 662, 667 Normal shock wave, 590–595, 618 tables, 776–780 Normal stresses, 225 Nozzle flow, 598–601 analogy with a sluice gate, 664, 677 choked, 586, 600 converging-diverging, 600–603 design conditions, 600–601 Nozzle flow—Cont.
subsonic versus supersonic, 584 Numerical analysis, 3, 434, 540–555 instability, 549 of inviscid flow, 540–548 of open-channel flow, 685–687 of pumps, 735–736 of viscous flow, 548–555 O Oblique shock wave, 620, 621–628, 789–790 reflection of, 651 One-dimensional approximation, 134–135, 138, 147, 165, 583, 660–661 One-seventh power-law, 439, 442, 479 Open channel flow, 236, 659 analogy with gas dynamics, 663, 664, 673, 677, 681 classification of, 662–664 critical flow, 671–674 gradually varied flow, 682–687 most efficient section, 669–671 over weirs, 687–693 Orifice plate, 398–400 Orthogonality conditions, 249, 497–498, 516 Outer layer, turbulent, 335–336 Overlap layer, 335–336, 441 Overrelaxation, 543 P Parallel plates, 26, 258–259, 359–360 Pascal unit, 9, 11 Pascal’s law, 71 Pathline, 37–38 Pelton wheel turbine, 745, 747 Perfect-gas law, 19, 35, 67, 573, 585 Permeability of porous media, 315, 420 Physical properties of fluids, 769–773 Pi theorem, 286–288 Piezometer, 103 Pipe flow, 328, 338–357 bend loss, 371 compressible, 604–613, 780–788 flow rate determination, 352–355 head loss or pressure drop, 307, 339, 351 laminar, 341–344 minor losses, 367–375 in a network, 380–381 noncircular, 357–366 in parallel, 376–378 with rough walls, 346–349 in series, 375–376 sizing problem, 355–357 turbulent, 344–348 Pipe standard sizes, 357 Pipelines, 168, 610 Pitching moment, 452, 527 Pitot-static tube, 387, 388, 642 Planform area, 299, 453, 468 Pode’s angle, 487 Poiseuille, J. L. M., 10, 341 Poiseuille flow, 260, 341–342 Poisson’s ratio, 577 Polar coordinates, 220, 243, 498, 499 Polar drag plot, 472 Positive-displacement pump, 711–714, 757 Potential energy, 18, 164 Potential flow, 252–257, 497 analog methods, 513–515 axisymmetric, 534–540 complex variable, 516–521 numerical analysis, 540–548 Potential lines, 248, 498 Potential vortex, 253, 498 822 Index Power coefficient, 724, 753, 761 Power-law correlation, for pipe flow, 309, 345 for velocity profile, 155, 439, 442, 479 for viscosity, 27, 772 Power product method, 286 Power specific speed, 744 Prandtl, Ludwig, 2, 45, 434, 629 flat-plate formulas, 479 lifting line theory, 532 Prandtl-Meyer angle, 630, 788 Prandtl-Meyer expansion waves, 628–631, 788 Prandtl number, 297, 579 Prefixes for units, 13 Pressure, 17, 59–61 absolute versus gage, 63 at a point, 60 stagnation, 312, 580 vacuum, 63 vapor, 31–32, 772, 773 Pressure coefficient, 297, 454, 526, 546 Pressure condition at a jet exit, 149 Pressure distribution, 62, 89 hydrostatic, 63–65 in irrotational flow, 249 in a nozzle, 599, 601 in rigid-body translation, 91–93 in rotating rigid-body motion, 93–97 Pressure drop in pipes, 339, 345 Pressure drag, 453 Pressure force on a control volume, 147–148 on a curved surface, 79–82 on an element, 60–61 on a plane surface, 74–79 Pressure gradient, 61, 96, 259, 341 adverse and favorable, 430, 445–448, 501 Pressure head, 65, 102, 168 Pressure measurement, 97–101 Pressure recovery of a diffuser, 373, 382, 385, 386 Pressure transducers, 99–101 Primary dimensions, 8, 278 Principle of corresponding sates, 24–25 Principle of dimensional homogeneity, 10–11, 280 Problem-solving techniques, 44 Product of inertia, 76 Propeller turbine, 742, 746, 749 Properties of fluids, 769–773 Propulsion, rocket, 158 Prototype, 36, 278 Pseudoplastic fluid, 28 Pump-system matching, 735–740 Pump-turbine system, 203, 746 Pumps, 47, 711 axial-flow, 729–733 centrifugal, 161–162, 714–718 dimensionless, 724 effect of blade angle, 719 effect of viscosity, 729, 730 multistage, 740 net positive-suction head, 721–722 in parallel, 738–739 performance curves, 204, 212, 714, 720– 722, 733–735, 758, 759 positive-displacement, 711–714 in series, 739–740 similarity rules, 727–728 size effects, 728 R Radius of curvature, 29, 235 Rankine half-body, plane, 256–258, 501–502 axisymmetric, 537–538 Rankine oval, plane, 507–508 axisymmetric, 563 Rankine-Hugoniot relations, 590 Rapidly varied channel flow, 662, 687 Rarefaction shock, 593, 624 Rayleigh line, 649 Reaction turbines, 742 Rectangular duct flow, 365, 366 Relative roughness, 349 Relative velocity, 137–138, 152 Reversible adiabatic flow (see Isentropic flow) Reynolds, Osborne, 45, 294, 330 Reynolds number, 24, 278, 294, 297, 325, 427 for an airfoil, 469 local, 429 Reynolds pipe-flow experiment, 330 Reynolds time-averaging concept, 333–334 Index 823 Reynolds transport theorem, 133–141 Rheology, 5, 28 Rheopectic fluid, 28 Rigid-body fluid motion, plane, 89–97 Rocket motion, 158 Rolling moment, 452 Rotameter, 395 Rotating cylinder, 512 sphere, 488 Rotationality, generation of, 249–250 Rough-wall effects on channels, 665–667 on cylinder drag, 298 on a flat plate, 443–444 on pipe flow, 346–348 on pumps, 726 sand-grain tests, 347 on sphere drag, 321, 456 Roughness of commercial pipes, 349 of open channels, 667 S Salinity, 22 Sandgrain roughness, 347 Savonius rotor, 387, 486 Saybolt viscosity, 286 Scaling laws, 278–279, 304, 306 Scaling parameters, 282 Schedule-40 pipe sizes, 357 Seawater properties, 22, 772 Second law of thermodynamics, 131, 233, 606, 624, 679, 680 Secondary dimensions, 8–9 Secondary flow, 365–366 Separated flow, 429, 455, 456, 502 Separation bubble, 469 Separation point on an airfoil, 525 on a cylinder, 455 definition of, 447 in a diffuser, 447 in a laminar boundary layer, 449, 501–502 on a sphere, 456 Shaft work, 164 Shape factor, 438, 443, 448 Sharp-crested weir, 689–690 Shear stresses, 4–5, 23, 225 turbulent, 334 Shear work, 164 Shock-expansion theory, 632–637 Shock polar, 623, 624 Shock-tube wind tunnel, 598 Shock wave, 250, 591, 521 detached, 624 linearized, 626 moving, 595–596 normal, 590–595, 776–780 oblique, 621–628, 789–790 rarefaction, 593, 624 strong versus weak, 624–626 Shut-off head of a pump. 719–720 SI units, 7–8 Silicon resonance transducer, 67, 101 Similarity, 279, 301 dynamic, 304–305 geometric, 301–303 violations, 302, 303, 307 kinematic, 303–304, 498 for pumps, 727–728 Sink line, 253, 498, 563 point, 536 Siphon, 208, 406 Skin friction coefficient, 432, 438, 441, 442, 449 Slip conditions in inviscid flow, 237 Sluice gate, 664, 677–678 drowned, 678, 701 Smoke-flow visualization, 40, 470 Soap bubble, 29 Sonic boom, 620, 621 Sonic point, 581, 605 Source line, 253, 498, 517 point, 536 Spar buoy, 119 Specific diameter, 760–761 Specific energy, 671–672 Specific gravity, 12, 18 Specific heat, 19–20, 573 824 Index Specific-heat ratio, 19, 295, 297, 572 of common gases, 21, 772 Specific speed, 730–731, 734, 735 Specific weight, 17 of common fluids, 65, 772 Speed of sound, 35, 221, 575–577 in the atmosphere, 773 of a perfect gas, 35, 577 of various materials, 577 of water, 577, 769 Sphere inviscid flow, 265, 538–539 viscous flow, 298, 321, 456, 457, 483, 488 Spherical droplet, 29 Spherical polar coordinates, 265, 535–536 Stability of floating bodies, 86–89 Stability of a pump, 720, 763 Stability map of a diffuser, 382 Stagnation density, 580 Stagnation enthalpy, 167, 578, 614 Stagnation point, 38, 249, 252, 256, 519 plane flow near, 39–40, 265 Stagnation pressure, 312, 580 Stagnation properties, 578–580 Stagnation speed of sound, 579 Stagnation temperature, 579 Stall angle of attack, 528 Stall speed, 473 Stalled airfoil, 470 Standard atmosphere, 69, 773 Stanton number, 312 Starting vortex, 469 State, equation of, 16, 18, 67, 131, 234, 573 for gases, 18–20, 573–574 for liquids, 21–22 van der Waals, 642 Static-pressure measurement, 97 Steady-flow energy equation, 167–168, 578, 661 Stokes flow past a sphere, 47, 457, 483, 487 Stokes’ stream function, 269, 535 Stopping vortex, 469 Strain rate, 23, 247 Stratified flow, 221, 250 Streakline, 37–38 Stream function, 238–244, 497 axisymmetric flow, 243–244, 535 compressible flow, 243 geometric interpretation, 240–241 irrotational flow, 239, 497 polar coordinates, 243 of Stokes, 269, 535 Streamline, 37–40, 240, 498 Streamline coordinates, 267 Streamlining of bodies, 456 Streamtube, 38, 143, 174 Stress gradients, 225 Stress tensor, 225, 227, 794 symmetry condition, 231 Strouhal number, 295, 296, 297 Subcritical channel flow, 663, 672 Subsonic flow, 572 Substantial derivative, 216 Suction specific speed, 731 Sudden expansion or contraction, 192–193, 371–372 Supercritical channel flow, 663, 672–673 Superposition of potential flows, 254–257, 500–501 Supersonic airfoil theory, 632–637 Supersonic flow, 149, 572, 618 Surface forces, 61, 225 Surface tension, 29–31, 235, 693 of air-water, 30, 773 of various interfaces, 29, 772 Sutherland-law viscosity formula, 27, 771 Swallow float, 86 System, 16, 130 System-matching of pumps, 735–740 Systems of units, 7 T Tainter gate, 115 Takeoff analysis for aircraft, 475 Taylor number, 262 Taylor vortices, 263 Tee-junction losses, 368 Temperature definition, 17 rise due to dissipation, 238 Terminal velocity, 309, 483 Index 825 Thermal conductivity, 27, 231 Thermodynamic properties, 16, 131, 771–772 Thickness drag, supersonic, 635 Thin-airfoil theory, 524–527, 635 Thixotropic fluid, 28 Three-dimensional flow, 535–540, 637 compressible, 637–640 Three-reservoir pipe junction, 376, 379 Throat in a duct, 585 Thwaites’ integral method, 448–450 Time-averaging of turbulence, 333–334 Timeline, 37 Tornado flow model, 255, 501, 556 Torricelli’s formula, 179 Total head, 168, 177 Trailing vortex, 469, 531–532 Transition to turbulence, 326–330 in a boundary layer, 299, 432, 439 on a flat plate, 405, 439 in a jet exit stream, 327 in pipe flow, 326, 328 in sphere flow, 310, 405 Transitional roughness, 347 Transonic flow, 572 Transport properties, 16 Trapezoidal channel, 668, 670–671 Triangular duct flow, 365, 366 Tri-diagonal matrix, 550 Trip wire, 405 Troposphere, 68 Tube bundle, 412 Turbines, 742–749 efficiency, 744, 748–749 impulse, 745–749 reaction, 742 windmills, 750–754 Turbomachine classification, 711–714 Turbulent flow, 3, 34, 376 on a flat plate, 441–444 fluctuations 326, 333–334 historical details, 328–330 intensity, 405 intermittency, 326 numerical models, 552–553 in a pipe, 328, 344–348 Turbulent puff, 328 Turbulent shear flow, 333–337 logarithmic overlap layer, 335–337 wall and outer layers, 335–336 Turbulent stresses, 334 Two-phase flow, 6, 219 U Ultrasonic flowmeter, 394 Uncertainty of data, 42–43 Uncoupling of velocity and temperature, 236–237 Uniform channel flow, 662, 664–667 Uniform stream, 252–253, 498, 517, 536 Unit normal vector, 132, 136 U. S. Standard Atmosphere, 69, 773 Units, 7–10 Universal gas constant, 19, 573 Unsteady Bernoulli equation, 175, 248, 249, 496 Unsteady flow, 36, 40, 549 Upwind differencing, 552 V V-notch weir, 692–693, 705 Vacuum pressure, 63 Valve flows, 12, 367–370 Van der Waals’ equation, 642 Vane flow, 150–151 Vapor pressure, 31–32 of various fluids, 772 of water, 32, 773 Varied flow, 662, 682–687 Vector differentiation, 215–217 Velocity-defect law, 336 Velocity diagrams, 717, 732, 743 Velocity field, 14–15, 215 Velocity gradient, 23, 246 Velocity head, 168, 367 Velocity measurement, 385–390 Velocity of approach factor, 398 Velocity potential, 248, 496, 535 826 Index Velocity profile, 23, 34, 439, 449, 535 for the Blasius solution, 437, 439 Vena contracta, 391, 392, 397, 398, 678 VentureStar spacecraft, 639–640 Venturi flume, 704 Venturi meter, 180–181, 207, 399, 401–402 Virtual mass, 539–540 Viscometer, 50, 51, 203 Viscosity, 22–24 formula for liquids, 27 generalized chart, 25 Sutherland and power-law, 27, 772 of various fluids, 24, 769, 771–772 Viscous dissipation, 233 Viscous flow analysis, 258–263 Viscous force on an element, 226 Viscous stresses, 23, 225, 228 symmetry condition, 231 Viscous sublayer, 337–347 Viscous work, 164 Visualization of flow, 40 Volume expansion rate, 15 Volume flow, 132, 142 measurement of, 391–404 Von Kármán, Theodore, 45, 154, 406, 431, 567 Vortex, line, 253–254, 498, 503, 518 infinite row, 503–504 potential, 253, 498 starting and stopping, 469 trailing, 531–532 Vortex flowmeter, 393–394 Vortex shedding, 295–296 Vortex sheet, 504–505 for a thin airfoil, 524–527 Vorticity, 247 W Wake flow, 190, 195, 250, 426, 429, 455 Wall roughness, 339–340 Wall shear stress, 341, 342, 438, 442 Waterline area, 88 Wave drag of ships, 464–466 of supersonic bodies, 635 Wave motion, 494, 576, 663–664, 672 periodic, 696 Weber number, 294, 297, 693 Weirs, 47, 687–693 broad-crested, 689, 690 Crump type, 705 drowned, 196, 705 inviscid flow model, 560 sharp-crested, 689–690 Wetted area, 453, 465 Wetted perimeter, 358, 661 Wind turbines, 750–754 performance of, 753 typical designs, 751 world energy distribution, 754 Wing theory, two-dimensional, 523–534 finite span, 472, 530–534 Work, 164 due to viscous stresses, 232 X X-33 spacecraft, 639–640 Y Yawing moment, 452 Young’s modulus, 577 Z Zero-lift airfoil angle, 530 Zones of action and silence, 620, 672 StudyGuide for Fluid Mechanics Preface The following materials are provided as a study guide for the text Fluid Mechanics by Frank White. A brief summary of the key concepts and theory is presented for each chapter along with the final form of basic equations (without detailed derivations) used in the various analyses being presented. In most cases, a detailed explanation for the physical significance of each term in a fundamental governing equation is given (e.g., linear momentum, pg. III-8) to assist the student in identifying when a given term should be included in the analysis. Example problems are provided for major sections. In each case, the starting general equation used in the solution is given followed by any necessary simplifications and the resulting complete solution. Where appropriate, the control volume and coordinate system used in the analysis are shown with the problem schematic. In many cases, an explanation is given with the final numerical answer to help the student understand the engineering significance of the answer (e.g., forces on curved surfaces, pg. II–18). In selected cases, computer based solutions to example problems are provided as an example to the student in the use of computer based problem solving techniques (e.g., parallel pipe sections, pg. VI-23). For problems areas involving multiple steps in the solution, a summary of the steps used in a typical problem solution sequence is provided and enclosed in a boxed border (e.g., rigid body motion, pg. II-22). Areas where the author’s experience has shown that mistakes in the analysis can easily occur are noted as Key Points (e.g., laminar flat plate boundary layer, pg. VII-5) throughout the material. Finally, the author of this study guide appreciates the opportunity to contribute to the instructional materials provided with one of the leading texts in the area of fluid mechanics and to collaborate with an educator with whom he has long has the highest respect and had the privilege to further his education in fluid mechanics while a student at Georgia Tech. Jerry R. Dunn, P.E. Associate Professor, Department of Mechanical Engineering Texas Tech University I-1 I. FLUID MECHANICS I.1 Basic Concepts & Definitions: Fluid Mechanics - Study of fluids at rest, in motion, and the effects of fluids on boundaries. Note: This definition outlines the key topics in the study of fluids: (1) fluid statics (fluids at rest), (2) momentum and energy analyses (fluids in motion), and (3) viscous effects and all sections considering pressure forces (effects of fluids on boundaries). Fluid - A substance which moves and deforms continuously as a result of an applied shear stress. The definition also clearly shows that viscous effects are not considered in the study of fluid statics. Two important properties in the study of fluid mechanics are: Pressure and Velocity These are defined as follows: Pressure - The normal stress on any plane through a fluid element at rest. Key Point: The direction of pressure forces will always be perpendicular to the surface of interest. Velocity - The rate of change of position at a point in a flow field. It is used not only to specify flow field characteristics but also to specify flow rate, momentum, and viscous effects for a fluid in motion. I-2 I.4 Dimensions and Units This text will use both the International System of Units (S.I.) and British Gravitational System (B.G.). A key feature of both is that neither system uses gc. Rather, in both systems the combination of units for mass acceleration yields the unit of force, i.e. Newton’s second law yields S.I. 1 Newton (N) = 1 kg m/s2 B.G. 1 lbf = 1 slug ft/s2 This will be particularly useful in the following: Concept Expression Units momentum mV & kg/s m/s = kg m/s2 =N slug/s ft/s = slug ft/s2 = lbf manometry ρ g h kg/m3m/s2m = (kg m/s2)/ m2 =N/m2 slug/ft3ft/s2ft = (slug ft/s2)/ft2 = lbf/ft2 dynamic viscosity µ N s /m2 = (kg m/s2) s /m2 = kg/m s lbf s /ft2 = (slug ft/s2) s /ft2 = slug/ft s Key Point: In the B.G. system of units, the mass unit is the slug and not the lbm. and 1 slug = 32.174 lbm. Therefore, be careful not to use conventional values for fluid density in English units without appropriate conversions, e.g., ρw = 62.4 lb/ft3 For this case the manometer equation would be written as ∆P = ρ g gc h I-3 Example: Given: Pump power requirements are given by p W & = fluid densityvolume flow rategpump head = ρ Q g hp For ρ = 1.928 slug/ft3, Q = 500 gal/min, and hp = 70 ft, Determine: The power required in kW. p W & = 1.928 slug/ft3 500 gal/min1 ft3/s /448.8 gpm32.2 ft/s2 70 ft p W & = 4841 ft–lbf/s 1.355810-3 kW/ft–lbf/s = 6.564 kW Note: We used the following: 1 lbf = 1 slug ft/s2 to obtain the desired units Recommendation: In working with problems with complex or mixed system units, at the start of the problem convert all parameters with units to the base units being used in the problem, e.g. for S.I. problems, convert all parameters to kg, m, & s; for BG problems, convert all parameters to slug, ft, & s. Then convert the final answer to the desired final units. 1.5 Properties of the velocity Field Two important properties in the study of fluid mechanics are Pressure and Velocity The basic definition for velocity has been given previously, however, one of its most important uses in fluid mechanics is to specify both the volume and mass flow rate of a fluid. I-4 Volume flow rate: = V n dA V dA cs Q n cs ⋅ = ⋅ = ⋅ = ⋅ = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & where Vn is the normal component of velocity at a point on the area across which fluid flows. Key Point: Note that only the normal component of velocity contributes to flow rate across a boundary. Mass flow rate: m = n cs cs V n d A V d A ρ ρ ρ ρ ρ ρ ρ ρ ⋅ = ⋅ = ⋅ = ⋅ = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & NOTE: While not obvious in the basic equation, Vn must also be measured relative to any flow area boundary motion, i.e., if the flow boundary is moving, Vn is measured relative to the moving boundary. This will be particularly important for problems involving moving control volumes in Ch. III. I-5 1.6 Thermodynamic Properties All of the usual thermodynamic properties are important in fluid mechanics P - Pressure (kPa, psi) T- Temperature (oC, oF) ρ ñ Density (kg/m3, slug/ft3) Alternatives for density γ - specific weight = weight per unit volume (N/m3, lbf/ft3) γ = ρ g H2O: γ = 9790 N/m3 = 62.4 lbf/ft3 Air: γ = 11.8 N/m3 = 0.0752 lbf/ft3 S.G. - specific gravity = ρ / ρ (ref) where: ρ (ref) = ρ (water at 1 atm, 20˚C) for liquids = 998 kg/m3 = ρ (air at 1 atm, 20˚C) for gases = 1.205 kg/m3 Example: Determine the static pressure difference indicated by an 18 cm column of fluid (liquid) with a specific gravity of 0.85. ∆P = ρ g h = S.G. γ h = 0.85 9790 N/m3 0.18 m = 1498 N/m2 = 1.5 kPa I.7 Transport Properties Certain transport properties are important as they relate to the diffusion of momentum due to shear stresses. Specifically: µ ≡ coefficient of viscosity (dynamic viscosity) {M / L t } ν ≡ kinematic viscosity ( µ / ρ ) { L 2 / t } I-6 This gives rise to the definition of a Newtonian fluid. Newtonian fluid: A fluid which has a linear relationship between shear stress and velocity gradient. τ = µ dU dy The linearity coefficient in the equation is the coefficient of viscosity µ . Flows constrained by solid surfaces can typically be divided into two regimes: a. Flow near a bounding surface with 1. significant velocity gradients 2. significant shear stresses This flow region is referred to as a "boundary layer." b. Flows far from bounding surface with 1. negligible velocity gradients 2. negligible shear stresses 3. significant inertia effects This flow region is referred to as "free stream" or "inviscid flow region." An important parameter in identifying the characteristics of these flows is the Reynolds number = Re = ρV L µ This physically represents the ratio of inertia forces in the flow to viscous forces. For most flows of engineering significance, both the characteristics of the flow and the important effects due to the flow, e.g., drag, pressure drop, aerodynamic loads, etc., are dependent on this parameter. II-1 II. Fluid Statics From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant along a horizontal plane, 2. Pressure at a point is independent of orientation, 3. Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth. These concepts are key to the solution of problems in fluid statics, e.g. 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no bearing on the pressure at a point in a static fluid. 3. Vertical depth is a key dimension in determining pressure change in a static fluid. If we were to conduct a more general force analysis on a fluid in motion, we would then obtain the following: ∇P = ρ g −a { } + µ ∇ 2V Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch. II inertial effects (ρ a), Ch. III viscous effects ( µ∇ 2V ) Chs VI & VII Note: For problems involving the effects of both (1) fluid statics and (2) inertial effects, it is the net v g −v a acceleration vector that controls both the magnitude and direction of the pressure gradient. II-2 This equation can be simplified for a fluid at rest (ie., no inertial or viscous effects) to yield ∇ p = ρ g ∂p ∂x = 0 ; ∂p ∂y = 0 ; ∂p ∂z = dp dz = −ρ g P2 −P 1 = −ρ gd Z 1 2 ∫ For liquids and incompressible fluids, this integrates to P1 – P2 = -ρg (Z2 – Z1) Note: Z2 – Z1 is positive for Z2 above Z1. but P2 – P1 is negative for Z2 above Z1. 2 P 1 P2 x y z Free surface Pressure = Pa h = Z - Z 2 1 Z Z 1 We can now define a new fluid parameter useful in static fluid analysis: γ = ρg ≡ specific weight of the fluid With this, the previous equation becomes (for an incompressible, static fluid) P2 – P1 = - γ (Z2 – Z1) The most common application of this result is that of manometry. II-3 Consider the U-tube, multi- fluid manometer shown on the right. If we first label all intermediate points between A & a, we can write for the overall pressure change PA - Pa = (PA- P1) + (P1 - P2) + (P2 - Pa ) This equation was obtained by adding and subtracting each intermediate pressure. The total pressure difference now is expressed in terms of a series of intermediate pressure differences. Substituting the previous result for static pressure difference, we obtain PA - PB = - ρ g(ZA- Z1) – ρ g (Z1 – Z2) – ρ g (Z2 - ZB ) Again note: Z positive up and ZA > Z1 , Z1 < Z2 , Z2 < Za . In general, follow the following steps when analyzing manometry problems: 1. On manometer schematic, label points on each end of manometer and each intermediate point where there is a fluid-fluid interface: e.g., A – 1 – 2 - B 2. Express overall manometer pressure difference in terms of appropriate intermediate pressure differences. PA - PB = (PA- P1) + (P1 – P2) + (P2 - PB ) 3. Express each intermediate pressure difference in terms of appropriate product of specific weight elevation change (watch signs) PA - PB = - ρ g(zA- z1) – ρ g (z1 – z2) – ρ g (z2 - zB ) 4. Substitute for known values and solve for remaining unknowns. II-4 When developing a solution for manometer problems, take care to: 1. Include all pressure changes 2. Use correct ∆Z and γ with each fluid 3. Use correct signs with ∆ Z. If pressure difference is expressed as PA – P1, the elevation change should be written as ZA – Z1 4. Watch units. Manometer Example: Given the indicated manometer, determine the gage pressure at A. Pa = 101.3 kPa. The fluid at A is Meriam red oil no. 3. ρgw = 9790 N/m3 ρg A = S.G.ρgw = 0.839790 N/m3 ρg A = 8126 N/m3 ρgair = 11.8 N/m3 1 2 a A 1 10 cm 18 cm S.G. = .83 H 0 2 With the indicated points labeled on the manometer, we can write PA - Pa = (PA- P1) + (P1 – P2) + (P2 - Pa ) Substituting the manometer expression for a static fluid, we obtain PA - Pa = - ρgA(zA- z1) – ρgw(z1 – z2) – ρga(z2 - za ) Neglect the contribution due to the air column. Substituting values, we obtain PA - Pa = - 8126 N/m3 0.10 m – 9790 N/m3 -0.18 = 949.6 N/m2 Note why: (zA- z1) = 0.10 m and (z1 – z2) = -0.18 m, & did not use Pa Review the text examples for manometry. II-5 Hydrostatic Forces on Plane Surfaces Consider a plane surface of arbitrary shape and orientation, submerged in a static fluid as shown: If P represents the local pressure at any point on the surface and h the depth of fluid above any point on the surface, from basic physics we can easily show that the net hydrostatic force on a plane surface is given by (see text for development): F = PdA A ∫ = Pcg A The basic physics says that the hydrostatic force is a distributed load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on one side of a plane surface submerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure acts normal to a surface, the direction of the resultant force will always be normal to the surface. Note: In most cases since it is the net hydrostatic force that is desired and the contribution of atmospheric pressure Pa will act on both sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is obtained by F = ρ gh cgA F = PcgA II-6 Pcg is now the gage pressure at the centroid of the area in contact with the fluid. Therefore, to obtain the net hydrostatic force F on a plane surface: 1. Determine depth of centroid hcg for the area in contact with the fluid 2. Determine the (gage) pressure at the centroid Pcg 3. Calculate F = PcgA. The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must be taken when dealing with layered fluids. The required procedure is essentially that the force on the plane area in each layer of fluid must be determined individually for each layer using the steps listed above. We must now determine the effective point of application of F. This is commonly called the “center of pressure - cp” of the hydrostatic force. Define an x – y coordinate system whose origin is at the centroid, c.g, of the area. The location of the resultant force is determined by integrating the moment of the distributed fluid load on the surface about each axis and equating this to the moment of the resultant force. Therefore, for the moment about the x axis: F y cp = y P d A ∫ A Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), for Ixx defined as the ≡ moment of inertia, or 2nd moment of area we obtain Ycp = −ρgsinθ I xx PcgA ≤ 0 Therefore, the resultant force will always act at a distance ycp below the centroid of the surface ( except for the special case of sin θ = 0 ). II-7 a PROPERTIES OF PLANE SECTIONS Geometry Centroid Moment of Inertia I x x Product of Inertia I x y Area b 1 h b b L y x L 2R b L b y x Fluid Specific Weight Seawater Glycerin Mercury Carbon .0752 57.3 62.4 49.2 11.8 8,996 9,790 7,733 64.0 78.7 846.
99.1 10,050 12,360 133,100 15,570 0 0 0 h 0 y x y x x y R y x y x s R Air Oil Water Ethyl 3 1bf /ft N /m3 bL3 12 b L ⋅ b + b 1 ( ) h 2 a = 4 R 3π π 16 − 4 9π R 4 1 8 − 4 9π R 4 π R 2 4 a = L 3 bL 3 36 b b −2s ( )L 2 72 1 2 b ⋅L b 3 , L 3 bL 3 36 − b 2L 2 72 b ⋅L 2 π R 2 2 R 4 π 8 − 8 9π 0 ,a = 4R 3π π R 4 4 0, 0 π R 2 b 2 , L 2 a ) = h b + 2b 1 ( 3 b + b1 ( ) ) 3 b 2 + 4bb 1 + b 1 2 ( 36 b + b 1 ( ) N /m3 3 1bf/ft II-8 Proceeding in a similar manner for the x location, and defining Ixy = product of inertia, we obtain X cp = − ρgsinθ I xy PcgA where Xcp can be either positive or negative since Ixy can be either positive or negative. Note: For areas with a vertical plane of symmetry (e.g., squares, circles, isosceles triangles, etc.) through the centroid, i.e. the ( y - axis), the center of pressure is located directly below the centroid along the plane of symmetry, i.e., Xcp = 0. Key Points: The values Xcp and Ycp are both measured with respect to the centroid of the area in contact with the fluid. Xcp and Ycp are both measured in the plane of the area; i.e., Ycp is not necessarily a vertical dimension, unless θ = 90o. Special Case: For most problems where (1) we have a single, homogeneous fluid ( i.e., not applicable to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific weight γ cancels in the equation for Ycp and Xcp and we have the following simplified expressions: F = ρ g h cgA Ycp = −I xx sinθ h cgA Xcp = − Ixy sinθ h cgA However, for problems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > Patm , these simplifications do not occur and the original, basic expressions for F , Ycp , and Xcp must be used; i.e., take care to use the approximate expressions only for cases where they apply. The basic equations always work. II-9 Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid 2. The centroid is used to determine the magnitude of the force. This is not the location of the resultant force 3. The location of the resultant force will be at the center of pressure which will be at a location Ycp below the centroid and Xcp as specified previously 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate b. Horizontal force at wall - A c. Hinge reactions - B 8’ θ Seawater • c .g.
hc.g.
A B 64 lbf/ft3 15’ 6’ 9’ a. By geometry: θ = tan-1 (6/8) = 36.87o Neglect Patm Since plate is rectangular, hcg = 9 ft + 3ft = 12 ft A = 10 x 5 = 50 ft2 Pcg = γ hcg = 64 lbf/ft3 12 ft = 768 lbf/ft2 ∴ Fp = Pcg A = 768 lbf/ft2 50 ft2 = 38,400 lbf II-10 b. Horizontal Reaction at A Must first find the location, c.p., for Fp ycp = −ρ gsinθ Ixx P cg A = −Ixx sinθ hcg A For a rectangular wall: Ixx = bh3/12 Ixx = 5 103/12 = 417 ft4 Note: The relevant area is a rectangle, not a triangle. θ • c.g.
• c.p.
Bx Bz P Fw 8 ft 6 ft y c.p.
θ Note: Do not overlook the hinged reactions at B. 4 2 417 0.6 0.417 12 50 cp ft y ft ft ft = − = − = − = − = − = − = − = − 4 2 417 0.6 0.417 12 50 cp ft y ft ft ft = − = − = − = − = − = − = − = − below c.g. xcp = 0 due to symmetry 0 B M = = = = ∑ ∑ ∑ ∑ (5 0.417) 38,400 6 0 P − ⋅ − = − ⋅ − = − ⋅ − = − ⋅ − = P = 29,330 lbf ← ← ← ← θ • c.g.
• c.p.
Bx Bz P Fw 8 ft 6 ft y c.p.
θ II-11 c. F x ∑ =0, B x + Fsinθ −P= 0 Bx + 38,4000.6 - 29,330 = 0 Bx = 6290 lbf → → → → F z ∑=0, B z −Fcosθ =0 Bz = 38,400 0.8 = 30,720 lbf ↑ ↑ ↑ ↑ Note: Show the direction of all forces in final answers. Summary: To find net hydrostatic force on a plane surface: 1. Find area in contact with fluid. 2. Locate centroid of that area. 3. Find hydrostatic pressure Pcg at centroid, typically = γ γ γ γ hcg ( generally neglect Patm ). 4. Find force F = Pcg A. 5. Location will not be at c.g., but at a distance ycp below centroid. ycp is in the plane of the area. Review all text examples for forces on plane surfaces. II-12 Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate separate x and y components of net hydrostatic force. Consider curved surface, a-b. Force balances in x & y directions yields Fh = FH Fv = Wair + W1 + W2 From this force balance, the basic rules for determining the horizontal and vertical component of forces on a curved surface in a static fluid can be summarized as follows: Horizontal Component, Fh 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. The horizontal force will act through the c.p. (not the centroid) of the projected area. b a cp hcg Fh ycp a’ b’ Projected vertical plane Curved surface II-13 Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid: 1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid - c.g. of the vertical plane. 4. Determine the depth of the centroid - hcg of the vertical plane. 5. Determine the pressure - Pcg = g hcg at the centroid of the vertical plane. 6. Calculate Fh = Pcg A, where A is the area of the projection of the curved surface into the vertical plane, ie., the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane , not the centroid. Get the picture? All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane. Vertical Component - Fv The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid is important in that there may not always actually be fluid directly above the surface. ( See graphic that follows.) This effective column of fluid is specified by identifying the column of fluid that would be required to cause the pressure at each location on the surface. II-14 Thus to identify the effective volume - Veff: 1. Identify the curved surface in contact with the fluid. 2. Identify the pressure at each point on the curved surface. 3. Identify the height of fluid required to develop the pressure. 4. These collective heights combine to form Veff. b a V eff P P P Fluid above the surface a b V eff P P P fluid No fluid actually above surface These two examples show two typical cases where this concept is used to determine Veff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The vertical direction will be the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1. Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column of fluid - Fv = ρgVeff. 4. The location of Fv is through the centroid of Veff. II-15 Finding the Location of the Centroid A second problem associated with the topic of curved surfaces is that of finding the location of the centroid of Veff. Recall: Centroid = the location where the first moment of a point area, volume, or mass equals the first moment of the distributed area, volume, or mass, e.g. xcgV 1 = xdV V 1 ∫ This principle can also be used to determine the location of the centroid of complex geometries. For example: If Veff = V1 + V2 then xcgVeff = x1V1 + x2V2 or VT = V1 + Veff xTVT = x1V1 + xcgVeff b a 2 V V 1 a b V1 V eff fluid Note: In the figures shown above, each of the x values would be specified relative to a vertical axis through b since the cg of the quarter circle is most easily specified relative to this axis. II-16 Example: Gate AB holds back 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment about B. • • 15 ft A B Water H F F V Width - W a. Horizontal component γ = ρg = 62.4 lbf/ft3 Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = Pcg Ap Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis. • • A B a b h cg Pcg The curved surface projects onto plane a - b and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: hcg = 7.5, Pcg = γhcg = 62.4 lbf/ft3 7.5 ft = 468 lbf/ft2 Fh = Pcg A = 468 lbf/ft2 15 ftW= 7020 W lbf Location: Ixx = bh3/12 = W 153 /12 = 281.25 W ft4 ycp = −Ixxsinθ hcg A = −281.25W ft4sin90o 7.5 ft15W ft2 = −2.5 ft The location is 2.5 ft below the c.g. or 10 ft below the surface, 5 ft above the bottom. II-17 b. Vertical force: Rule: Fv equals the weight of the effective column of fluid above the curved surface. A • c.g.
C B • b F v • Q: What is the effective volume of fluid above the surface? What volume of fluid would result in the actual pressure distribution on the curved surface? Vol = A - B - C Vrec = Vqc + VABC, VABC = Vrec - Vqc VABC = Veff = 152 W - π 152/4W = 48.29 W ft3 Fv = ρg Veff = 62.4 lbf/ft3 48.29 ft3 = 3013 lbf Note: Fv is directed upward even though the effective volume is above the surface. c. What is the location? Rule: Fv will act through the centroid of the “effective volume causing the force. A • c.g.
C B • b F v • We need the centroid of volume A-B-C. How do we obtain this centroid? Use the concept which is the basis of the centroid, the “first moment of an area.” II-18 Since: Arec = Aqc + AABC Mrec = Mqc + MABC MABC = Mrec - Mqc Note: We are taking moments about the left side of the figure, ie., point b. WHY? (The c.g. of the quarter circle is known to be 4 R/ 3 π w.r.t. b.) xcg A = xrec Arec - xqc Aqc xcg {152 - π152/4} = 7.5152 - {415/3/π} π152/4 xcg = 11.65 ft { distance to rt. of b to centroid } Q: Do we need a y location? Why? d. Calculate the moment about B needed for equilibrium. 0 B M = = = = ∑ ∑ ∑ ∑ clockwise positive. MB +5F h + 15−xv ( )F v = 0 ( ( ( ( ) ) ) ) 5 7020 15 11.65 3013 0 B M W W + × + − = + × + − = + × + − = + × + − = ( ( ( ( ) ) ) ) 5 7020 15 11.65 3013 0 B M W W + × + − = + × + − = + × + − = + × + − = a P g y G g ρ ρ ρ ρ ≠ ≠ ≠ ≠ ≠ ≠ ≠ ≠ MB +35,100W +10,093.6W = 0 MB = −45,194W ft −lbf Why negative? The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilibrium. Always review your answer (all aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand about the problem. Begin to think like an engineer. II-19 Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of buoyancy. The basic principle was discovered by Archimedes. It can be easily shown that (see text for detailed development) the buoyant force Fb is given by: Fb = ρ g Vb where Vb is the volume of the fluid displaced by the submerged body and ρ g is the specific weight of the fluid displaced. V b Patm Fb Thus, the buoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the buoyant force is: Through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on buoyancy. II-20 Pressure distribution in rigid body motion All of the problems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid body motion, where the entire fluid mass moves and accelerates uniformly (as a rigid body). The container of fluid shown below is accelerated uniformly up and to the right as shown. From a previous analysis, the general equation governing fluid motion is ∇ P = ρ( g −a ) + µ ∇ 2 V For rigid body motion, there is no velocity gradient in the fluid, therefore µ∇ 2V = 0 The simplified equation can now be written as ∇ P = ρ( g −a ) = ρG where G = g −a ≡ the net acceleration vector acting on the fluid. II-21 This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid body motion: ∇ P = f {fluid density & the net acceleration vector- G = g −a } ∇ P acts in the vector direction of G = g −a Lines of constant pressure are perpendicular to G . The new orientation of the free surface will also be perpendicular to G . The equations governing the analysis for this class of problems are most easily developed from an acceleration diagram. Acceleration diagram: For the indicated geometry: 1 tan x z a g a θ θ θ θ − − − − = = = = + + + + 1 tan x z a g a θ θ θ θ − − − − = = = = + + + + dP ds = ρG where G = a x 2 + (g + a z ) 2 { } 1 2 and P2 −P 1 = ρG(s 2 −s1 ) Note: P2 −P 1 ≠ρ g z2 −z1 ( ) and s2 – s1 is not a vertical dimension a -a g G θ θ Free surface P 2 P 1 s ax az Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G . II-22 In analyzing typical problems with rigid body motion: 1. Draw the acceleration diagram taking care to correctly indicate –a, g, and θ, the inclination angle of the free surface. 2. Using the previously developed equations, solve for G and θ. 3. If required, use geometry to determine s2 – s1 (the perpendicular distance from the free surface to a given point) and then the pressure at that point relative to the surface using P2 – P1 = ρ G (s2 – s1) . Key Point: Do not use ρg to calculate P2 – P1, use ρ G. Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. Mug is accelerated to the right with ax = 7 m/s2 . Assuming rigid body motion. ρc = 1010 kg/m3, Determine: a. Will the coffee spill? b. Pg at “a & b”. c. Fnet on left wall. a. First draw schematic showing original orientation and final orientation of the free surface. a b 7 cm 10 cm ∆ z ax θ 6 cm ρc = 1010 kg/m3 ax = 7m/s2 az = 0 g = 9.8907 m/s2 Have a new free surface angle θ where 1 tan x z a g a θ θ θ θ − − − − = = = = + + + + θ = tan−1 7 9.807 =35.5° ∆z = 3 tan 35.5 = 2.14 cm a -a g G θ II-23 hmax = 7 + 2.14 = 9.14 cm < 10 cm ∴ Will not spill. b. Pressure at “ a & b.” Pa = ρ G ∆ sa G = {a2x + g2}.5 = { 72 + 9.8072}.5 G = 12.05 m/s2 ∆ sa = {7 + z} cos θ ∆ sa = 9.14 cm cos 35.5 = 7.44 cm Pa = 1010 kg/m312.05m/s20.0744 m Pa = 906 (kg m/s2)/m2 = 906 Pa Note: a P g y G g ρ ρ ρ ρ ≠ ≠ ≠ ≠ ≠ ≠ ≠ ≠ a b 7 cm 10 cm ∆ z ax θ 6 cm θ ∆ sa Q: How would you find the pressure at b, Pb? c. What is the force on the left wall? We have a plane surface, what is the rule? Find cg, Pcg, F = Pcg. A Vertical depth to cg is: zcg = 9.14/2 = 4.57 cm ∆scg = 4.57 cos 35.5 = 3.72 cm Pcg = ρ G ∆scg Pcg = 1010 kg/m312.05 m/s2 0.0372 m Pcg = 452.7 N/m2 F = Pcg A = 452.7 N/m20.09140.06m2 F = 2.48 N ← a b 7 cm 10 cm ∆ z ax θ 6 cm θ • cg ∆ scg II-24 What is the direction? Horizontal, perpendicular to the wall; i.e., Pressure always acts normal to a surface. Q: How would you find the force on the right wall? III-1 III. Control Volume Relations for Fluid Analysis From consideration of hydrostatics, we now move to problems involving fluid flow with the addition of effects due to fluid motion, e.g. inertia and convective mass, momentum, and energy terms. We will present the analysis based on a control volume (not differential element) formulation, e.g. similar to that used in thermodynamics for the first law. Basic Conservation Laws: Each of the following basic conservation laws is presented in its most fundamental, fixed mass form. We will subsequently develop an equivalent expression for each law that includes the effects of the flow of mass, momentum, and energy (as appropriate) across a control volume boundary. These transformed equations will be the basis for the control volume analyses developed in this chapter. Conservation of Mass: Defining m as the mass of a fixed mass system, the mass for a control volume V is given by m sys = ρdV sys ∫ The basic equation for conservation of mass is then expressed as dm dt sys = 0 The time rate of change of mass for the control volume is zero since at this point we are still working with a fixed mass system. Linear Momentum: Defining P sys as the linear momentum of a fixed mass, the linear momentum of a fixed mass control volume is given by: III-2 P sys = mV = V ρdV sys ∫ where V is the local fluid velocity and dV is a differential volume element in the control volume. The basic linear momentum equation is then written as F = ∑ dP dt sys = d mV ( ) dt sys Moment of Momentum: Defining H as the moment of momentum for a fixed mass, the moment of momentum for a fixed mass control volume is given by H sys = r × V ρdV sys ∫ where r is the moment arm from an inertial coordinate system to the differential control volume of interest. The basic equation is then written as M sys = r × ∑ F = ∑ dH dt sys Energy: Defining E sys as the total energy of an element of fixed mass, the energy of a fixed mass control volume is given by E sys = eρ dV sys ∫ III-3 where e is the total energy per unit mass ( includes kinetic, potential, and internal energy ) of the differential control volume element of interest. The basic equation is then written as sys d E d t Q W = = = = − − − − & & (Note: written on a rate basis) It is again noted that each of the conservation relations as previously written applies only to fixed, constant mass systems. However, since most fluid problems of importance are for open systems, we must transform each of these relations to an equivalent expression for a control volume which includes the effect of mass entering and/or leaving the system. This is accomplished with the Reynolds transport theorem. Reynolds Transport Theorem We define a general, extensive property ( an extensive property depends on the size or extent of the system) Bsys where Bsys = β ρd V sys ∫ Bsys could be total mass, total energy, total momentum, etc., of a system. and Bsys per unit mass is defined as β or β = dB dm Thus, β is the intensive equivalent of Bsys . Applying a general control volume formulation to the time rate of change of Bsys , we obtain the following (see text for detailed development): III-4 e i e e e e i i i i sys cv A A d B dV V d A V d A d t t β ρ β ρ β ρ β ρ β ρ β ρ β ρ β ρ β ρ β ρ β ρ β ρ ∂ ∂ ∂ ∂ = + − = + − = + − = + − ∂ ∂ ∂ ∂ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ↓ ↓ ↓ ↓ System rate Rate of Rate of B Rate of B of change change of leaving c.v. entering c.v. of B B in c.v. ↓ ↓ transient term convective terms where B is any conserved quantity, e.g. mass, linear momentum, moment of momentum, or energy. We will now apply this theorem to each of the basic conservation equations to develop their equivalent open system, control volume forms. Conservation of mass For conservation of mass, we have that B = m and β = 1 From the previous statement of conservation of mass and these definitions, Reynolds transport theorem becomes 0 e i e e e i i i cv A A dV V d A V d A t ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ∂ ∂ ∂ ∂ + − = + − = + − = + − = ∂ ∂ ∂ ∂∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ or 0 e i e e e i i i cv A A dV V d A V d A t ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ∂ ∂ ∂ ∂ + − = + − = + − = + − = ∂ ∂ ∂ ∂∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ↓ ↓ ↓ Rate of change Rate of mass Rate of mass of mass in c.v., leaving c.v., entering c.v., ↓ ↓ ↓ = 0 for steady-state e m & i m & III-5 This can be simplified to 0 e i cv d m m m d t + − = + − = + − = + − = ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ & & Note that the exit and inlet velocities Ve and Vi are the local components of fluid velocities at the exit and inlet boundaries relative to an observer standing on the boundary. Therefore, if the boundary is moving, the velocity is measured relative to the boundary motion. The location and orientation of a coordinate system for the problem are not considered in determining these velocities. Also, the result of V e ⋅dA e and V i ⋅dA i is the product of the normal velocity component times the flow area at the exit or inlet, e.g. Ve,n dAe and Vi,n dAi Special Case: For incompressible flow with a uniform velocity over the flow area, the previous integral expressions simplify to: cs m V d A AV ρ ρ ρ ρ ρ ρ ρ ρ = = = = = = = = ∫ ∫ ∫ ∫ & Conservation of Mass Example Water at a velocity of 7 m/s exits a stationary nozzle with D = 4 cm and is directed toward a turning vane with θ = 40o, Assume steady-state. Determine: a. Velocity and flow rate entering the c.v. b. Velocity and flow rate leaving the c.v. III-6 a. Find V1 and 1 m & Recall that the mass flow velocity is the normal component of velocity measured relative to the inlet or exit area. Thus, relative to the nozzle, V(nozzle) = 7 m/s and since there is no relative motion of point 1 relative to the nozzle, we also have V1 = 7 m/s ans. From the previous equation: cs m V d A AV ρ ρ ρ ρ ρ ρ ρ ρ = = = = = = = = ∫ ∫ ∫ ∫ & = 998 kg/m37 m/sπ0.042/4 1 m & = 8.78 kg/s ans. b. Find V2 and 2 m & Determine the flow rate first. Since the flow is steady state and no mass accumulates on the vane: 1 m & = 2 m & , 2 m & = 8.78 kg/s ans. Now: 2 m & = 8.78 kg/s = ρ A V)2 Since ρ and A are constant, V2 = 7 m/s ans. Key Point: For steady flow of a constant area, incompressible stream, the flow velocity and total mass flow are the same at the inlet and exit, even though the direction changes. or alternatively: Rubber Hose Concept: For steady flow of an incompressible fluid, the flow stream can be considered as a rubber hose and if it enters a c.v. at a velocity of V, it exits at a velocity V, even if it is redirected. III-7 Problem Extension: Let the turning vane (and c.v.) now move to the right at a steady velocity of 2 m/s (other values remain the same); perform the same calculations. Therefore: Given: Uc = 2 m/s VJ = 7 m/s For an observer standing at the c.v. inlet (point 1) V1 = VJ – Uc = 7 – 2 = 5 m/s 1 m & = ρ1 V1 A1 = 998 kg/m35 m/sπ0.042/4 = 6.271 kg/s Note: The inlet velocity used to specify the mass flow rate is again measured relative to the inlet boundary but now is given by VJ – Uc . Exit: 1 m & = 2 m & = 6.271 kg/s Again, since ρ and A are constant, V2 = 5 m/s. Again, the exit flow is most easily specified by conservation of mass concepts. Note: The coordinate system could either have been placed on the moving cart or have been left off the cart with no change in the results. Key Point: The location of the coordinate system does not affect the calculation of mass flow rate which is calculated relative to the flow boundary. It could have been placed at Georgia Tech with no change in the results. Review material and work examples in the text on conservation of mass. III-8 Linear Momentum For linear momentum, we have that B = P = mV and β = V From the previous statement of linear momentum and these definitions, Reynolds transport theorem becomes ( ( ( ( ) ) ) ) e i e e e i i i cv A A sys d mV F V dV V V d A V V d A d t t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ∂ ∂ ∂ ∂ = = + ⋅ − ⋅ = = + ⋅ − ⋅ = = + ⋅ − ⋅ = = + ⋅ − ⋅ ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ or e i e i cv A A F V dV V d m V d m t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ∂ ∂ ∂ ∂ = + − = + − = + − = + − ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & ↓ ↓ ↓ ↓ = the ∑ ∑ ∑ ∑of the = the rate of = the rate of = the rate of external forces change of momentum momentum acting on the c.v. momentum leaving the entering the c.v. in the c.v. c.v. = body + point + = 0 for distributed, e.g. steady-state (pressure) forces and where V is the vector momentum velocity relative to an inertial reference frame. Key Point: Thus, the momentum velocity has magnitude and direction and is measured relative to the reference frame (coordinate system) being used for the problem. The velocities in the mass flow terms i m & and e m & are scalars, as noted previously, and are measured relative to the inlet or exit boundary. Always clearly define a coordinate system and use it to specify the value of all inlet and exit momentum velocities when working linear momentum problems. III-9 For the 'x' direction, the previous equation becomes , , e i x x x e e x i i cv A A F V dV V d m V d m t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ∂ ∂ ∂ ∂ = + − = + − = + − = + − ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & Note that the above equation is also valid for control volumes moving at constant velocity with the coordinate system placed on the moving control volume. This is because an inertial coordinate system is a nonaccelerating coordinate system which is still valid for a c.s. moving at constant velocity. Example: A water jet 4 cm in diameter with a velocity of 7 m/s is directed to a stationary turning vane with θ = 40o. Determine the force F necessary to hold the vane stationary. Governing equation: , , e i x x x e e x i i cv A A F V dV V d m V d m t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ∂ ∂ ∂ ∂ = + − = + − = + − = + − ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & Since the flow is steady and the c.v. is stationary, the time rate of change of momentum within the c.v. is zero. Also with uniform velocity at each inlet and exit and a constant flow rate, the momentum equation becomes b e e i i F m V m V − = − − = − − = − − = − & & Note that the braking force, Fb, is written as negative since it is assumed to be in the negative x direction relative to positive x for the coordinate system. III-10 From the previous example for conservation of mass, we can again write cs m V d A AV ρ ρ ρ ρ ρ ρ ρ ρ = = = = = = = = ∫ ∫ ∫ ∫ & = 998 kg/m37 m/sπ0.042/4 1 m & = 8.78 kg/s and V1 = 7 m/s and for the exit: 2 m & = 8.78 kg/s and V2 = 7 m/s inclined 40û above the horizontal. Substituting in the momentum equation, we obtain -Fb = 8.78 kg/s 7 m/s cos 40o - 8.78 kg/s 7 m/s and -Fb = - 14.4 kg m/s2 or Fb = 14.4 N ← ← ← ← ans. Note: Since our final answer is positive, our original assumption of the applied force being to the left was correct. Had we assumed that the applied force was to the right, our answer would be negative, meaning that the direction of the applied force is opposite to what was assumed. Modify Problem: Now consider the same problem but with the cart moving to the right with a velocity Uc = 2 m/s. Again solve for the value of braking force Fb necessary to maintain a constant cart velocity of 2 m/s. Note: The coordinate system for the problem has now been placed on the moving cart. III-11 The transient term in the momentum equation is still zero. With the coordinate system on the cart, the momentum of the cart relative to the coordinate system is still zero. The fluid stream is still moving relative to the coordinate system, however, the flow is steady with constant velocity and the time rate of change of momentum of the fluid stream is therefore also zero. Thus The momentum equation has the same form as for the previous problem (However the value of individual terms will be different.) b e e i i F m V m V − = − − = − − = − − = − & & 1 m & = ρ1 V1 A1 = 998 kg/m35 m/sπ0.042/4 = 6.271 kg/s = 2 m & Now we must determine the momentum velocity at the inlet and exit. With the coordinate system on the moving control volume, the values of momentum velocity are V1 = VJ – Uc = 7 – 2 = 5 m/s and V2 = 5 m/s inclined 40o The momentum equation ( x - direction ) now becomes -Fb = 6.271 kg/s 5 m/s cos 40o - 6.271 kg/s 5 m/s and -Fb = - 7.34 kg m/s2 or Fb = 7.34 N ← ← ← ← ans. Question: What would happen to the braking force Fb if the turning angle had been > 90o, e.g., 130 o? Can you explain based on your understanding of change in momentum for the fluid stream? Review and work examples for linear momentum with fixed and non-accelerating (moving at constant velocity) control volumes. Accelerating Control Volume The previous formulation applies only to an inertial coordinate system, i.e., fixed or moving at constant velocity (non-accelerating). III-12 We will now consider problems with accelerating control volumes. For these problems we will again place the coordinate system on the accelerating control volume, thus making it a non-inertial coordinate system. For coordinate systems placed on an accelerating control volume, we must account for the acceleration of the c.s. by correcting the momentum equation for this acceleration. This is accomplished by including the term as shown below: e i cv cv e i cv cv A A F a d m V dV V d m V d m t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ∂ ∂ ∂ ∂ − = + − − = + − − = + − − = + − ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & ↓ integral sum of the local c.v. (c.s.) acceleration the c.v. mass The added term accounts for the acceleration of the control volume and allows the problem to be worked with the coordinate system placed on the accelerating c.v. Note: Thus, all vector (momentum) velocities are then measured relative to an observer (coordinate system) on the accelerating control volume. For example, the velocity of a rocket as seen by an observer (c.s.) standing on the rocket is zero and the time rate of change of momentum is zero in this reference frame even if the rocket is accelerating. Accelerating Control Volume Example A turning vane with θ = 60 o accelerates from rest due to a jet of water (VJ = 35 m/s, AJ = 0.003 m2 ). Assuming the mass of the cart mc, is 75 kg and neglecting drag and friction effects, find: a. Cart acceleration at t = 0. b. Uc as a f(t) III-13 Starting with the general equation shown above, we can make the following assumptions: 1. ∑ ∑ ∑ ∑ Fx = 0, no friction or body forces. 2. The jet has uniform velocity and constant properties. 3. The entire cart accelerates uniformly over the entire control volume. 4. Neglect the relative momentum change of the jet stream that is within the control volume. With these assumptions, the governing equation simplifies to , , c c e x e i x i a m m V m V − = − − = − − = − − = − & & We thus have terms that account for the acceleration of the control volume, for the exit momentum, and for the inlet momentum (both of which change with time.) Mass flow: As with the previous example for a moving control volume, the mass flow terms are given by: i e m m m = = = = = = = = = = = = & & & ρ AJ (VJ – Uc) Note that since the cart accelerates, Uc is not a constant but rather changes with time. Momentum velocities: Ux,i= VJ - Uc Ux,e = (VJ - Uc ) cos θ Substituting, we now obtain - ac mc = ρ AJ (VJ – Uc)2 cos θ - ρ AJ (VJ – Uc) Solving for the cart acceleration, we obtain ac = ρ AJ 1 −cosθ ( ) VJ −Uc ( ) 2 mc III-14 Substituting for the given values at t = 0, i.e., Uc = 0, we obtain ac (t = 0) = 24.45 m/s2 = 2.49 g’s Note: The acceleration at any other time can be obtained once the cart velocity Uc at that time is known. To determine the equation for cart velocity as a function of time, the equation for the acceleration must be written in terms of Uc (t) and integrated. dUc dt = ρ AJ 1−cosθ ( )VJ −Uc ( ) 2 mc Separating variables, we obtain dUc VJ −Uc ( ) 2 0 Uc(t) ∫ = ρ AJ 1−cosθ ( ) mc 0 t ∫ dt Completing the integration and rearranging the terms, we obtain a final expression of the form Uc VJ = VJ bt 1+ VJ bt where b = ρ AJ 1−cosθ ( ) mc Substituting for known values, we obtain VJ b = 0.699 s-1 Thus the final equation for Uc is give by Uc VJ = 0.699t 1+0.699t III-15 The final results are now given as shown below: t Uc/VJ Uc ac (s) (m/s) (m/s2) 0 0.0 0.0 24.45 2 0.583 20.0 4.49 5 0.757 27.2 1.22 10 0.875 30.6 0.39 15 0.912 31.9 0.192 ∞ 1.0 35 0.0 Uc vs t 0 5 10 15 20 25 30 35 0 5 10 15 20 t(s) Note that the limiting case occurs when the cart velocity reaches the jet velocity. At this point, the jet can impart no more momentum to the cart, the acceleration is now zero, and the terminal velocity has been reached. Review the text example on accelerating control volumes. Moment of Momentum (angular momentum) For moment of momentum we have that B = H = r × m V ( ) and β = r ×V From the previous equation for moment of momentum and these definitions, Reynolds transport theorem becomes III-16 e i e i cv A A M r V dV r V d m r V d m t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ∂ ∂ ∂ ∂ = × + × − × = × + × − × = × + × − × = × + × − × ∑ ∑ ∑ ∑ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & ↓ ↓ ↓ ↓ = the ∑ ∑ ∑ ∑ of all = the rate of = the rate of = the rate of external change of mom- moment of moment of moments ent of momentum momentum momentum acting in the c.v. = 0 leaving entering on the c.v. for steady state the c.v. the c.v. For the special case of steady-state, steady-flow and uniform properties at any exit or inlet, the equation becomes e e i i M m r V m r V = × − × = × − × = × − × = × − × ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ & & For moment of momentum problems, we must be careful to correctly evaluate the moment of all applied forces and all inlet and exit momentum flows, with particular attention to the signs. Moment of Momentum Example: A small lawn sprinkler operates as indicated. The inlet flow rate is 9.98 kg/min with an inlet pressure of 30 kPa. The two exit jets direct flow at an angle of 40o above the horizontal. For these conditions, determine the following: a. Jet velocity relative to the nozzle. b. Torque required to hold the arm stationary. c. Friction torque if the arm is rotating at 35 rpm. d. Maximum rotational speed if we neglect friction. 160 mm D = 5 mm J III-17 a. R = 160 mm, DJ = 5 mm, Therefore, for each of the two jets: QJ = 0.5 9.98 kg/min/998 kg/m3 = 0.005 m3/min AJ = π π π π π π π π 0.00252 = 1.96310-5 m2 VJ = 0.005 m3/min / 1.96310-5 m2 /60 s/min VJ = 4.24 m/s relative to the nozzle exit ans. b. Torque required to hold the arm stationary. First develop the governing equations and analysis for the general case of the arm rotating. With the coordinate system at the center of rotation of the arm, a general velocity diagram for the case when the arm is rotating is shown in the adjacent schematic. R V cos θ J o ω rω + Taking the moment about the center of rotation, the moment of the inlet flow is zero since the moment arm is zero for the inlet flow. The basic equation then becomes ( ( ( ( ) ) ) ) 0 2 cos e J T m R V R α ω α ω α ω α ω = − = − = − = − & Note that the net momentum velocity is the difference between the tangential component of the jet exit velocity and the rotational speed of the arm. Also note that the direction of positive moments was taken as the same as for VJ and opposite to the direction of rotation. For a stationary arm R ω = 0. We thus obtain for the stationary torque III-18 To = 2 ρ QJ R VJ cos α 3 3 1min 2 998 .005 4.24 cos4.160 min 60 o o kg m m T m s m = = = = To = 0.0864 N m clockwise. ans. A resisting torque of 0.0864 N m must be applied in the clockwise direction to keep the arm from rotating in the counterclockwise direction. c. At ω = 30 rpm, calculate the friction torque Tf 1min 30 2 min 60 s rev rad rad rev ω π π ω π π ω π π ω π π = = = = = = = = 3 3 1min 2 998 0.005 0.16 4.24 cos40 .16 min 60 s o o kg m m rad T m m s m π π π π = − = − = − = − ans. Note; The resisting torque decreases as the speed increases. d. Find the maximum rotational speed. The maximum rotational speed occurs when the opposing torque is zero and all the moment of momentum goes to the angular rotation. For this case, VJ cos θ – Rω = 0 4.2 193.8 4 / cos40 cos s 20.3 193.8 0.16 J rad rpm m s V rad rpm R m s θ θ θ θ ω ω ω ω • • • • = = = = = = = = = = = = = = = = = = = = ans. Review material and examples on moment of momentum. III-19 Energy Equation (Extended Bernoulli Equation) For energy, we have that B = E = e ρd V cv ∫ and β = e = u + 1 2 V 2 + g z From the previous statement of conservation of energy and these definitions, Reynolds transport theorem becomes: e i e e e e i i i i cv A A sys d E Q W e dV e V d A e V d A d t t ∂ ∂ ∂ ∂ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ∂ ∂ ∂ ∂ − = = + ⋅ − ⋅ − = = + ⋅ − ⋅ − = = + ⋅ − ⋅ − = = + ⋅ − ⋅ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & & After extensive algebra and simplification (see text for detailed development), we obtain: P1 −P2 ρ g = V2 2 −V1 2 2 g + Z2 −Z1 + hf, 1 −2 − h p ↓ ↓ ↓ ↓ ↓ Pressure drop Pressure Pressure Pressure Pressure from 1 – 2, drop due to drop due to drop due to drop due to in the flow acceleration elevation frictional mechanical direction of the fluid change head loss work on fluid Note: this formulation must be written in the flow direction from 1 - 2 to be consistent with the sign of the mechanical work term and so that hf,1-2 is always a positive term. Also note the following: ❑ The points 1 and 2 must be specific points along the flow path ❑ Each term has units of linear dimension, e.g., ft or meters, and z2 – z1 is positive for z2 above z1 ❑ The term hf,1-2 is always positive when written in the flow direction and for internal, pipe flow includes pipe or duct friction losses and fitting or piping component (valves, elbows, etc.) losses, III-20 ❑ The term hp is positive for pumps and fans ( i.e., pumps increase the pressure in the flow direction) and negative for turbines (turbines decrease the pressure in the flow direction) ❑ For pumps: hp = ws g where ws = the useful work per unit mass to the fluid Therefore: p s w g h = = = = and f s p W mw Qg h ρ ρ ρ ρ = = = = = = = = & & where f W & = the useful power delivered to the fluid and f p p W W η η η η = = = = & & where ηp is the pump efficiency Example Water flows at 30 ft/s through a 1000 ft length of 2 in diameter pipe. The inlet pressure is 250 psig and the exit is 100 ft higher than the inlet. Assuming that the frictional loss is given by 18 V2/2g, Determine the exit pressure. 1 2 100 ft 250 psig 2 Given: V1 = V2 = 30 ft/s, L = 1000 ft, Z2 – Z1 = 100 ft, P1 = 250 psig Also, since there is no mechanical work in the process, the energy equation simplifies to III-21 P 1 −P 2 ρg = Z2 −Z1 + hf P 1 −P 2 ρg = 100 ft + 18302 ft2 / s 2 64.4 ft / s2 = 351.8 ft P1 – P2 = 62.4 lbf/ft3 351.8 ft = 21,949 psf = 152.4 psi P2 = 250 – 152.4 = 97.6 psig ans. Problem Extension A pump driven by an electric motor is now added to the system. The motor delivers 10.5 hp. The flow rate and inlet pressure remain constant and the pump efficiency is 71.4 %, determine the new exit pressure. Q = AV = π π π π π π π π (1/12)2 ft2 30 ft/s = 0.6545 ft3/s Wf = ηp Wp= ρ Q g hp 3 3 0.714 10.5 550 / / 101 62.4 / 0.6545 / p hp ft lbf s hp h ft lbm ft ft s − − − − = = = = = = = = The pump adds a head increase equal to 101 ft to the system and the exit pressure should increase. Substituting in the energy equation, we obtain P 1 −P 2 ρg = 100 ft + 18302 ft2 / s 2 64.4 ft / s2 −101 ft = 250.8 ft P1 – P2 = 62.4 lbf/ft3 250.8 ft = 15,650 psf = 108.7 psi P2 = 250 – 108.7 = 141.3 psig ans. Review examples for the use of the energy equation IV - 1 Ch. IV Differential Relations for a Fluid Particle This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common boundary conditions are presented that allow exact solutions to be obtained. Two of the most common simplifications are 1). steady flow and 2). incompressible flow. The Acceleration Field of a Fluid A general expression of the flow field velocity vector is given by: V (r ,t) = ˆ i u x, y,z,t ( ) + ˆ j v x, y,z,t ( ) + ˆ k w x, y, z,t ( ) One of two reference frames can be used to specify the flow field characteristics: eulerian – the coordinates are fixed and we observe the flow field characteristics as it passes by the fixed coordinates. lagrangian - the coordinates move through the flow field following individual particles in the flow. Since the primary equation used in specifying the flow field velocity is based on Newton’s second law, the acceleration vector is an important solution parameter. In cartesian coordinates, this is expressed as a = d V d t = ∂V ∂t + u∂V ∂x + v∂V ∂y + w∂V ∂z = ∂V ∂t + V ⋅∇ ( )V total local convective The acceleration vector is expressed in terms of three types of derivatives: Total acceleration = total derivative of velocity vector = local derivative + convective derivative of velocity vector IV - 2 Likewise, the total derivative (also referred to as the substantial derivative ) of other variables can be expressed in a similar form, e.g., d P d t = ∂P ∂t + u ∂P ∂x + v∂P ∂y + w ∂P ∂z = ∂P ∂t + V ⋅∇ ( )P Example 4.1 Given the eulerian velocity-vector field V = 3t ˆ i + x z ˆ j + t y 2 ˆ k find the acceleration of the particle. For the given velocity vector, the individual components are u = 3 t v = x z w = t y2 Evaluating the individual components, we obtain ∂V ∂t = 3 i + y2 k ∂V ∂x = z j ∂V ∂y = 2 t y k ∂V ∂z = x j Substituting, we obtain d V d t = ( 3 i + y2 k) + (3 t) (z j) + (x z) (2 t y k) + (t y2) (x j) After collecting terms, we have d V d t = 3 i + (3 t z + t x y2) j + ( 2 x y z t + y2) k ans. IV - 3 The Differential Equation of Conservation of Mass If we apply the basic concepts of conservation of mass to a differential control volume, we obtain a differential form for the continuity equation in cartesian coordinates ∂ρ ∂t + ∂ ∂x ρu ( )+ ∂ ∂y ρ v ( )+ ∂ ∂z ρ w ( )= 0 and in cylindrical coordinates ∂ρ ∂t + 1 r ∂ ∂r r ρvr ( )+ 1 r ∂ ∂θ ρ vθ ( )+ ∂ ∂z ρvz ( )= 0 Steady Compressible Flow For steady flow, the term ∂ ∂t = 0 and all properties are function of position only. The previous equations simplify to Cartesian: ∂ ∂x ρu ( )+ ∂ ∂y ρv ( )+ ∂ ∂z ρw ( )= 0 Cylindrical: 1 r ∂ ∂r r ρ vr ( )+ 1 r ∂ ∂θ ρvθ ( )+ ∂ ∂z ρ vz ( )= 0 Incompressible Flow For incompressible flow, density changes are negligible, ρ = const., and ∂ρ ∂t = 0 In the two coordinate systems, we have Cartesian: ∂u ∂x + ∂v ∂y + ∂w ∂z = 0 IV - 4 Cylindrical: 1 r ∂ ∂r r vr ( )+ 1 r ∂ ∂θ vθ ( )+ ∂ ∂z vz ( )= 0 Key Point: It is noted that the assumption of incompressible flow is not restricted to fluids which cannot be compressed, e.g. liquids. Incompressible flow is valid for (1) when the fluid is essentially incompressible (liquids) and (2) for compressible fluids for which compressibility effects are not significant for the problem being considered. The second case is assumed to be met when the Mach number is less than 0.3: Ma = V/c < 0.3 Gas flows can be considered incompressible The Differential Equation of Linear Momentum If we apply Newton’s Second Law of Motion to a differential control volume we obtain the three components of the differential equation of linear momentum. In cartesian coordinates, the equations are expressed in the form: Inviscid Flow: Euler’s Equation If we assume the flow is frictionless, all of the shear stress terms drop out. The resulting equation is known as Euler’s equation and in vector form is given by: ρg - ∇P = ρ dV d t IV - 5 where d V dt is the total or substantial derivative of the velocity discussed previously and ∇P is the usual vector gradient of pressure. This form of Euler’s equation can be integrated along a streamline to obtain the frictionless Bernoulli’s equation ( Sec. 4.9). The Differential Equation of Energy The differential equation of energy is obtained by applying the first law of thermodynamics to a differential control volume. The most complex element of the development is the differential form of the control volume work due to both normal and tangential viscous forces. When this is done, the resulting equation has the form ρ d u dt + P ∇⋅V ( ) = ∇⋅k∇T ( ) + Φ where Φ is the viscous dissipation function. The term for the total derivative of internal energy includes both the transient and convective terms seen previously. Two common assumptions used to simplify the general equation are: 1. du ≈ Cv dT and 2. Cv, µ, k, ρ ≈ constants With these assumptions, the energy equation reduces to ρCv d T d t = k ∇ 2 T + Φ It is noted that the flow-work term was eliminated as a result of the assumption of constant density, ρ, for which the continuity equation becomes ∇⋅V = 0 ,thus eliminating the term P ∇⋅V ( ) . We now have the three basic differential equations necessary to obtain complete flow field solutions of fluid flow problems. IV - 6 Boundary Conditions for the Basic Equations In vector form, the three basic governing equations are written as Continuity: ∂ρ ∂t + ∇⋅ρV ( )= 0 Momentum: ρ d V dt = ρ g −∇P + ∇⋅τi j Energy: ρ d u dt + P ∇⋅V ( ) = ∇⋅k∇T ( ) + Φ We have three equations and five unknowns: ρ, V, P, u, and T ; and thus need two additional equations. These would be the equations of state describing the variation of density and internal energy as functions of P and T, i.e., ρ = ρ (P,T) and u = u (P,T) Two common assumptions providing this information are either: 1. Ideal gas: ρ = P/RT and du = Cv dT 2. Incompressible fluid: ρ = constant and du = C dT Time and Spatial Boundary Conditions Time Boundary Conditions: If the flow is unsteady, the variation of each of the variables (ρ, V, P, u, and T ) must be specified initially, t = 0, as functions of spatial coordinates e.g. x,y,z. Spatial Boundary Conditions: The most common spatial boundary conditions are those specified at a fluid – surface boundary. This typically takes the form of assuming equilibrium (e.g., no slip condition – no property jump) between the fluid and the surface at the boundary. IV - 7 This takes the form: Vfluid = Vwall Tfluid = Twall Note that for porous surfaces with mass injection, the wall velocity will be equal to the injection velocity at the surface. A second common spatial boundary condition is to specify the values of V, P, and T at any flow inlet or exit. Example 4.6 For steady incompressible laminar flow through a long tube, the velocity distribution is given by vz =U 1 −r 2 R 2 vr = 0 vθ = 0 where U is the maximum or centerline velocity and R is the tube radius. If the wall temperature is constant at Tw and the temperature T = T(r) only, find T(r) for this flow. For the given conditions, the energy equation reduces to ρCv vr d T dr = k r d d r r d T d r + µ d vz d r 2 Substituting for vz and realizing the vr = 0, we obtain k r d d r r d T d r = −µ d vz d r 2 = −4U 2 µ r 2 R 4 Multiply by r/k and integrate to obtain IV - 8 d T d r = −µU 2r 3 k R 4 + C1 Integrate a second time to obtain T = −µU 2r 4 4 k R 4 + C1 ln r + C2 Since the term, ln r, approaches infinity as r approaches 0, C1 = 0. Applying the wall boundary condition, T = Tw at r = R, we obtain for C2 C2 = Tw + µ U 2 4 k The final solution then becomes T r ( )= Tw + µ U 2 4k 1 −r 4 R 4 The Stream Function The necessity to obtain solutions for multiple variables in multiple governing equations presents an obvious mathematical challenge. However, the stream function, Ψ , allows the continuity equation to be eliminated and the momentum equation solved directly for the single variable, Ψ . The use of the stream function works for cases when the continuity equation can be reduced to only two terms. For example, for 2-D, incompressible flow, continuity becomes ∂u ∂x + ∂v ∂y = 0 IV - 9 Defining the velocity components to be u = ∂Ψ ∂y and v = −∂Ψ ∂x which when substituted into the continuity equation yields ∂ ∂x ∂Ψ ∂y + ∂ ∂y −∂Ψ ∂x = 0 and continuity is automatically satisfied. Geometric interpretation of Ψ Ψ Ψ Ψ It is easily shown that lines of constant Ψ Ψ Ψ Ψ are flow streamlines. Since flow does not cross a streamline, for any two points in the flow we can write Q 1→2 = V ⋅n ( ) 1 2 ∫ d A = d Ψ 1 2 ∫ = Ψ2 −Ψ 1 Thus the volume flow rate between two points in the flow is equal to the difference in the stream function between the two points. Steady Plane Compressible Flow In like manner, for steady, 2-D, compressible flow, the continuity equation is ∂ ∂x ρ u ( )+ ∂ ∂y ρv ( )= 0 For this problem, the stream function can be defined such that ρu = ∂Ψ ∂y and ρ v = −∂Ψ ∂x IV - 10 As before, lines of constant stream function are streamlines for the flow, but the change in stream function is now related to the local mass flow rate by ( ( ( ( ) ) ) ) 2 2 1 2 2 1 1 1 m V n d A d ρ ρ ρ ρ − − − − = ⋅ = Ψ = Ψ −Ψ = ⋅ = Ψ = Ψ −Ψ = ⋅ = Ψ = Ψ −Ψ = ⋅ = Ψ = Ψ −Ψ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ & Vorticity and Irrotationality The concept of vorticity and irrotationality are very useful in analyzing many fluid problems. The analysis starts with the concept of angular velocity in a flow field. Consider three points, A, B, & C, initially perpendicular at time t, that then move and deform to have the position and orientation at t + dt. The lines AB and BC have both changed length and incurred angular rotation dα and dβ relative to their initial positions. Fig. 4.10 Angular velocity and strain rate of two fluid lines deforming in the x-y plane We define the angular velocity ωz about the z axis as the average rate of counter-clockwise turning of the two lines expressed as ω z = 1 2 dα dt −d β d t IV - 11 Applying the geometric properties of the deformation shown in Fig. 4.10 and taking the limit as ∆t → 0, we obtain ω z = 1 2 d v d x −d u d y In like manner, the angular velocities about the remaining two axes are ω x = 1 2 d w d y −d v d z ω y = 1 2 d u d z −d w d x From vector calculus, the angular velocity can be expressed as a vector with the form ω ω ω ω = i ω x + j ω y + k ωz = 1/2 the curl of the velocity vector, e.g. ω = 1 2 curl V ( )= 1 2 i j k ∂ ∂x ∂ ∂y ∂ ∂z u v w The factor of 2 is eliminated by defining the vorticity, ξ , as follows: ξ ξ ξ ξ = 2 ω ω ω ω = curl V Frictionless Irrotational Flows When a flow is both frictionless and irrotational, the momentum equation reduces to Euler’s equation given previously by ρg - ∇P = ρ dV d t IV - 12 As shown in the text, this can be integrated along the path, ds, of a streamline through the flow to obtain ∂V ∂t 1 2 ∫ ds + d P ρ 1 2 ∫ + 1 2 V2 2 −V 1 2 ( )+ g z2 −z1 ( )= 0 For steady, incompressible flow this reduces to P ρ + 1 2 V 2 + gz = constant along a streamline V-1 V. Modeling, Similarity, and Dimensional Analysis To this point, we have concentrated on analytical methods of solution for fluids problems. However, analytical methods are not always satisfactory due to: (1) limitations due to simplifications required in the analysis, (2) complexity and/or expense of a detailed analysis. The most common alternative is to: Use experimental test & verification procedures. However, without planning and organization, experimental procedures can : (a) be time consuming, (b) lack direction, (c) be expensive. This is particularly true when the test program necessitates testing at one set of conditions, geometry, and fluid with the objective to represent a different but similar set of conditions, geometry, and fluid. Dimensional analysis provides a procedure that will typically reduce both the time and expense of experimental work necessary to experimentally represent a desired set of conditions and geometry. It also provides a means of "normalizing" the final results for a range of test conditions. A normalized (non-dimensional) set of results for one test condition can be used to predict the performance at different but fluid dynamically similar conditions ( including even a different fluid). The basic procedure for dimensional analysis can be summarized as follows: V-2 1. Compile a list of relevant variables (dependent & independent) for the problem being considered, 2. Use an appropriate procedure to identify both the number and form of the resulting non-dimensional parameters. This procedure is outlined as follows for the Buckingham Pi Theorem Definitions: n = the number of independent variables relevant to the problem j’ = the number of independent dimensions found in the n variables j = the reduction possible in the number of variables necessary to be considered simultaneously k = the number of independent pi terms that can be identified to describe the problem, k = n - j Summary of Steps: 1. List and count the n variables involved in the problem. 2. List the dimensions of each variable using {MLTΘ} or {FLTΘ}. Count the number of basic dimensions ( j’) for the list of variables being considered. 3. Find j by initially assuming j = j’ and look for j repeating variables which do not form a pi product. If not successful, reduce j by 1 and repeat the process. 4. Select j scaling, repeating variables which do not form a pi product. 5. Form a pi term by adding one additional variable and form a power product. Algebraically find the values of the exponents which make the product dimensionless. Repeat the process with each of the remaining variables. 6. Write the combination of dimensionless pi terms in functional form: Πk = f( Π1, Π2, …Πi) Consider the following example for viscous pipe flow. The relevant variables for this problem are summarized as follows: ∆P = pressure drop ρ = density V = velocity D = diameter µ = viscosity ε = roughness L = length Seven pipe flow variables: {∆P ρ, V, D, µ, ε, L } dependent independent V-3 Use of the Buckingham Pi Theorem proceeds as follows: 1. Number of independent variables: n = 7 2. List the dimensions of each variable ( use m L t Θ ): variables ∆P ρ V D µ ε L dimensions mL-1t-2 mL-3 Lt-1 L mL-1t-1 L L The number of basic dimensions is j’ = 3. 3. Choose j = 3 with the repeating variables being ρ, V, and D. They do not form a dimensionless pi term. No combination of the 3 variables will eliminate the mass dimension in density or the time dimension in velocity. 4. This step described in the above step. The repeating variables again are ρ, V, and D and j = 3. Therefore, k = n – j = 7 – 3 = 4 independent Π terms. 5. Form the Π terms: Π1 = ρa Vb Dc µ−1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-1 )−1 In order for the Π term to have no net dimensions, the sum of the exponents for each dimension must be zero. Therefore, we have: mass: a - 1 = 0 , a = 1 time: - b + 1 = 0, b = 1 length: -3a + b + c + 1 = 0, c = 3 – 1 – 1 = 1 We therefore have Π1 = ρ V D /µ = Re = Reynolds number Repeating the process by adding the roughness ε V-4 Π2 = ρa Vb Dc ε1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: a = 0 , a = 0 time: - b = 0, b = 0 Length: -3a + b + c + 1 = 0, c = – 1 Π2 = ε / D Roughness ratio Repeat the process by adding the length L. Π3 = ρa Vb Dc L1 = (mL-3)a ( Lt-1)b Lc ( L )1 Solving: mass: a = 0 , a = 0 time: - b = 0, b = 0 length: -3a + b + c + 1 = 0, c = – 1 Π3 = L / D length-to-diameter ratio These three are the independent Π terms. Now obtain the dependent Π term by adding ∆P Π4 = ρa Vb Dc ∆P1 = (mL-3)a ( Lt-1)b Lc ( mL-1t-2 )1 Solving: mass: a + 1 = 0 , a = -1 time: - b - 2 = 0, b = -2 length: -3a + b + c - 1 = 0, c = 0 Π4 = ∆P / ρ V2 Pressure coefficient V-5 Application of the Buckingham Pi Theorem to the previous list of variables yields the following non-dimensional combinations: ∆P ρV2 = f ρVD µ , L D, ε D or { { { { } } } } p C f Re,L,ε ε ε ε = = = = Thus, a non-dimensional pressure loss coefficient for viscous pipe flow would be expected to be a function of (1) the Reynolds number, (2) a non-dimensional pipe length, and (3) a non-dimensional pipe roughness. This will be shown to be exactly the case in Ch. VI, Viscous Internal Flow. A list of typical dimensionless groups important in fluid mechanics is given in the accompanying table. From these results, we would now use a planned experiment with data analysis techniques to get the exact form of the relationship among these non - dimensional parameters. The next major step is concerned with the design and organization of the experimental test program Two key elements in the test program are: design of the model specification of the test conditions, particularly when the test must be performed at conditions similar, but not the same as the conditions of interest. Similarity and non-dimensional scaling The basic requirement is in this process to achieve 'similarity' between the 'experimental model and its test conditions' and the 'prototype and its test conditions' in the experiment. V-6 Table 5.2 Dimensional Analysis and Similarity Parameter Definition Qualitative ratio of effects Importance Reynolds number RE = ρUL µ Inertia Viscosity Always Mach number MA = U A Flow speed Sound speed Compressible flow Froude number Fr = U2 gL Inertia Gravity Free-surface flow Weber number We = ρ U2L γ Inertia Surfacetension Free-surface flow Cavitation number (Euler number) Ca = p - p v ρU 2 Pressure Inertia Cavitation Prandtl number Pr = Cpµ k Dissipation Conduction Heat convection Eckert number Ec = U 2 cpT o Kinetic energy Enthalpy Dissipation Specific-heat ratio γ = cp cv Enthalpy Internal energy Compressible flow Strouhal number St = ω L U Oscillation Mean speed Oscillating flow Roughness ratio ε L Wall roughness Body length Turbulent,rough walls Grashof number Gr = β ∆TgL3ρ2 µ 2 Buoyancy Viscosity Natural convection Temperature ratio Tw To Walltemperature Stream temperature Heat transfer Pressure coefficient Cp = p −p∞ 1/2ρU 2 Staticpressure Dynamic pressure Aerodynamics, hydrodynamics Lift coefficient CL = L 1/2ρU 2A Lift force Dynamic force Aerodynamics hydrodynamics Drag coefficient CD = D 1/ 2ρ U 2A Lift force Dynamic force Aerodynamics, hydrodynamics V-7 In this context, “similarity” is defined as Similarity: All relevant dimensionless parameters have the same values for the model & the prototype. Similarity generally includes three basic classifications in fluid mechanics: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity Geometric similarity In fluid mechanics, geometric similarity is defined as follows: Geometric Similarity All linear dimensions of the model are related to the corresponding dimensions of the prototype by a constant scale factor SFG Consider the following airfoil section (Fig. 5.4): Fig. 5.4 Geometric Similarity in Model Testing For this case, geometric similarity requires the following: V-8 SFG = rm rp = Lm Lp = Wm Wp = ⋅⋅⋅ In addition, in geometric similarity, All angles are preserved. All flow directions are preserved. Orientation with respect to the surroundings must be same for the model and the prototype, ie., Angle of attack )m = angle of attack )p Kinematic Similarity In fluid mechanics, kinematic similarity is defined as follows: Kinematic Similarity The velocities at 'corresponding' points on the model & prototype are in the same direction and differ by a constant scale factor SFk. Therefore, the flows must have similar streamline pattterns Flow regimes must be the same. These conditions are demonstrated for two flow conditions, as shown in the following kinematically similar flows (Fig. 5.6). Fig. 5.6a Kinematically Similar Low Speed Flows V-9 Fig. 5.6b Kinematically Similar Free Surface Flows The conditions of kinematic similarity are generally met automatically when geometric and dynamic similarity conditions are satisfied. Dynamic Similarity In fluid mechanics, dynamic similarity is typically defined as follows: Dynamic Similarity This is basically met if model and prototype forces differ by a constant scale factor at similar points. This is illustrated in the following figure for flow through a sluice gate (Fig. 5.7). Fig. 5.7 Dynamic Similarity for Flow through a Sluice Gate V-10 This is generally met for the following conditions: 1. Compressible flows: model & prototype Re, Ma, are equal Rem = Rep, Mam= Map , γm = γp 2. Incompressible flows a. No free surface Rem = Rep b. Flow with a free surface Rem = Rep , Frm = Frp Note: The parameters being considered, e.g., velocity, density, viscosity, diameter, length, etc., generally relate to the flow, geometry, and fluid characteristics of the problem and are considered to be independent variables for the subject problem. The result of achieving similarity by the above means is that relevant non - dimensional dependent variables, e.g., CD, Cp, Cf, or Nu, etc., are then equal for both the model and prototype. This result would then indicate how the relevant dependent results, e.g. drag force, pressure forces, viscous forces, are to be scaled for the model to the prototype. Equality of the relevant non-dimensional independent variables Re, Ma, x/L, etc., indicates how the various independent variables of importance should be scaled. V-11 An example of this scaling is shown as follows: Example: The drag on a sonar transducer prototype is to be predicted based on the following wind tunnel model data and prototype data requirements. Determine the model test velocity Vm necessary to achieve similarity and the expected prototype force Fp based on the model wind tunnel test results. Given: Prototype Model sphere sphere D 1 ft 6 in V 5 knots unknown? F ? 5.58 lbf ρ 1.98 slugs ft 3 0.00238 slugs ft 3 ν 1.4 10-5 ft 2 s 1.56 10-4 ft 2 s From dimensional analysis: CD = f R e { } F D2 ρ V2 = f VD υ For the prototype, the actual operating velocity and Reynolds number are: Prototype: 1 5 6080 8.44 3600 p na mi ft hr ft V hr na mi s s ⋅ ⋅ ⋅ ⋅ = = = = = = = = ⋅ ⋅ ⋅ ⋅ 1 5 6080 8.44 3600 p na mi ft hr ft V hr na mi s s ⋅ ⋅ ⋅ ⋅ = = = = = = = = ⋅ ⋅ ⋅ ⋅ V-12 R e p = VD υ p = 8.44ft s 1ft 1.4 10 −5 ft 2 s = 6.0310 5 Equality of Reynolds number then yields the required model test velocity of R e m = R e p = VD υ m = > Vm = 188ft s Based on actual test results for the model, i.e. measured Fm, equality of model and prototype drag coefficients yields ∴CD p = C Dm = > Fp = Fm ρ p ρm Vp 2 Vm 2 D p 2 D m 2 = 37.4 lbf Note: All fluid dynamic flows and resulting flow characteristics are not Re dependent. Example: The drag coefficient for bluff bodies with a fixed point of separation; e.g., radar antennae, generally have a constant, fixed number for CD which is not a function of Re. CD = const . ≠f R e ( ) VI-1 VI. VISCOUS INTERNAL FLOW To date, we have considered only problems where the viscous effects were either: a. known: i.e. - known FD or hf b. negligible: i.e. - inviscid flow This chapter presents methodologies for predicting viscous effects and viscous flow losses for internal flows in pipes, ducts, and conduits. Typically, the first step in determining viscous effects is to determine the flow regime at the specified condition. The two possibilities are: a. Laminar flow b. Turbulent flow The student should read Section 6.1 in the text, which presents an excellent discussion of the characteristics of laminar and turbulent flow regions. For steady flow at a known flow rate, these regions exhibit the following: Laminar flow: A local velocity constant with time, but which varies spatially due to viscous shear and geometry. Turbulent flow: A local velocity which has a constant mean value but also has a statistically random fluctuating component due to turbulence in the flow. Typical plots of velocity time histories for laminar flow, turbulent flow, and the region of transition between the two are shown below. Fig. 6.1 (a) Laminar, (b)transition, and (c) turbulent flow velocity time histories. VI-2 Principal parameter used to specify the type of flow regime is the Reynolds number - Re = ρV D µ = V D ν V - characteristic flow velocity D - characteristic flow dimension µ - dynamic viscosity υ - kinematic viscosity = µ ρ We can now define the Recr ≡ critical or transition Reynolds number Recr ≡ Reynolds number below which the flow is laminar, above which the flow is turbulent While transition can occur over a range of Re, we will use the following for internal pipe or duct flow: Recr ≅2300 =ρVD µ cr = VD υ cr Internal Viscous Flow A second classification concerns whether the flow has significant entrance region effects or is fully developed. The following figure indicates the characteristics of the entrance region for internal flows. Note that the slope of the streamwise pressure distribution is greater in the entrance region than in the fully developed region. Typical criteria for the length of the entrance region are given as follows: Laminar: Le D ≅0.06 Re VI-3 Turbulent L e D ≅4.4Re1/ 6 where: Le = length of the entrance region Note: Take care in neglecting entrance region effects. In the entrance region, frictional pressure drop/length > the pressure drop/length for the fully developed region. Therefore, if the effects of the entrance region are neglected, the overall predicted pressure drop will be low. This can be significant in a system with short tube lengths, e.g., some heat exchangers. Fully Developed Pipe Flow The analysis for steady, incompressible, fully developed, laminar flow in a circular horizontal pipe yields the following equations: U r ( ) = −R2 4µ dP dx 1 −r2 R2 VI-4 U Umax = 1 −r2 R2 , Umax = 2V avg and Q = A Vavg = π π π π R2 Vavg Key Points: Thus for laminar, fully developed pipe flow (not turbulent): a. The velocity profile is parabolic. b. The maximum local velocity is at the centerline (r = 0). c. The average velocity is one-half the centerline velocity. d. The local velocity at any radius varies only with radius, not on the streamwise (x) location ( due to the flow being fully developed). Note: All subsequent equations will use the symbol V (no subscript) to represent the average flow velocity in the flow cross section. Darcy Friction Factor: We can now define the Darcy friction factor f as: f ≡ D L ∆P f ρ V 2 2 where ∆Pf = the pressure drop due to friction only. The general energy equation must still be used to determine total pressure drop. Therefore, we obtain ∆P f = ρ ghf = f L D ρ V 2 2 and the friction head loss hf is given as h f = f L D V 2g 2 Note: The definitions for f and hf are valid for either laminar or turbulent flow. However, you must evaluate f for the correct flow regime, laminar or turbulent. VI-5 Key Point: It is common in industry to define and use a “fanning” friction factor ff . The fanning friction factor differs from the Darcy friction factor by a factor of 4. Thus, care should be taken when using unfamiliar equations or data since use of ff in equations developed for the Darcy friction factor will result in significant errors (a factor of 4). Your employer will not be happy if you order a 10 hp motor for a 2.5 hp application. The equation suitable for use with ff I s h f = 4 ff L D V 2g 2 Laminar flow: Application of the results for the laminar flow velocity profile to the definition of the Darcy friction factor yields the following expression: f = 64 Re laminar flow only (Re < 2300) Thus with the value of the Reynolds number, the friction factor for laminar flow is easily determined. Turbulent flow: A similar analysis is not readily available for turbulent flow. However, the Colebrook equation, shown below, provides an excellent representation for the variation of the Darcy friction factor in the turbulent flow regime. Note that the equation depends on both the pipe Reynolds number and the roughness ratio, is transcendental, and cannot be expressed explicitly for f . f = −2log 2.51 Re f 1/ 2 + ε / D 3.7 turbulent flow only (Re > 2300) VI-6 where ε = nominal roughness of pipe or duct being used. (Table 6.1, text) (Note: Take care with units for ε; ε /D must be non-dimensional). A good approximate equation for the turbulent region of the Moody chart is given by Haaland’s equation: f = −1.8log 6.9 Re + ε / D 3.7 1.11 −2 Note again the roughness ratio ε/D must be non-dimensional in both equations. Graphically, the results for both laminar and turbulent flow pipe friction are represented by the Moody chart as shown below. Typical roughness values are shown in the following table: VI-7 Table 6.1 Average roughness values of commercial pipe Haaland’s equation is valid for turbulent flow (Re > 2300) and is easily set up on a computer, spreadsheet, etc. Key fluid system design considerations for laminar and turbulent flow a. Most internal flow problems of engineering significance are turbulent, not laminar. Typically, a very low flow rate is required for internal pipe flow to be laminar. If you open your kitchen faucet and the outlet flow stream is larger than a kitchen match, the flow is probably turbulent. Thus, check your work carefully if your analysis indicates laminar flow. b. The following can be easily shown: Laminar flow: ∆Pf ~ µ,L,Q,D−4 { } { { { { } } } } 2 4 ~ , , , f W L Q D µ µ µ µ − − − − & Turbulent flow: ∆Pf ~ ρ 3/ 4, µ 1/ 4, L, Q 1.75,D −4.75 { } { { { { } } } } 3/ 4 1/ 4 2.75 4.75 ~ , , , , f W L Q D ρ µ ρ µ ρ µ ρ µ − − − − & Thus both pressure drop and pump power are very dependent on flow rate and pipe/conduit diameter. Small changes in diameter and/or flow rate can significantly change circuit pressure drop and power requirements. VI-8 Example ( Laminar flow): Water, 20oC flows through a 0.6 cm tube, 30 m long, at a flow rate of 0.34 liters/min. If the pipe discharges to the atmosphere, determine the supply pressure if the tube is inclined 10o above the horizontal in the flow direction. 10o 1 2 L = 30 m D = 0.6 cm L Water Properties: Energy Equation ρ = 998 kg/m3 ρg = 9790 N/m3 ν = 1.005 E-6 m2/s P 1 −P 2 ρg = V2 2 −V 1 2 2 g +Z2 −Z1 + hf −hp which for steady-flow in a constant diameter pipe with P2 = 0 gage becomes, P 1 ρ g = Z2 −Z1 + hf = L sin 10o + hf V = Q A = 0.34E −3 m 3 / min1min/ 60s π 0.3/100 ( ) 2 m 2 = 0.2 m / s Re = V D υ = 0.20.006 1.005E −6 =1197 → laminar flow f = 64 Re = 64 1197 = 0.0535 hf = f L D V 2 2g = 0.0535 30m 0.006m 0.2 2 29.807m / s2 = 0.545m P 1 ρg = 30sin10Þ+ 0.545 = 5.21 + 0.545 = 5.75m gravity friction total head head head loss P1 = 9790 N/m35.75 m = 56.34 kN/m3 (kPa) ~ 8.2 psig ans. VI-9 Example: (turbulent flow) Oil, ρ = 900 kg/m3, ν = 1 E-5 m2/s, flows at 0.2 m3/s through a 500 m length of 200 mm diameter, cast iron pipe. If the pipe slopes downward 10o in the flow direction, compute hf, total head loss, pressure drop, and power required to overcome these losses. L 10 o L = 500 m D = 200 mm 1 2 The energy equation can be written as follows where ht = total head loss. P 1 −P 2 ρg = ht = V2 2 −V 1 2 2g +Z2 −Z1 + hf −hp which reduces to ht = Z2 −Z1 + hf V = Q A = 0.2m 3 / s π .1 ( ) 2 m 2 = 6.4 m / s Table 6.1, cast iron, ε = 0.26 mm Re = V D υ = 6.4.2 1E −5 = 128,000 →turbulent flow, ε D = 0.26 200 = 0.0013 Since flow is turbulent, use Haaland’s equation to determine friction factor (check your work using the Moody chart). f = −1.8log 6.9 Re + ε / D 3.7 1.11 −2 , f = −1.8log 6.9 128,000 + 0.0013 3.7 1.11 −2 f = 0.02257 hf = f L D V 2 2g = 0.02257 500m 0.2m 6.4 2 2 9.807m / s 2 =116.6m ans. ht = Z2 – Z1 + hf = - 500 sin 0 + 116.6 = - 86.8 + 116.6 = 29.8 m ans. VI-10 Note that for this problem, there is a negative gravity head loss (i.e. a head increase) and a positive frictional head loss resulting in the net head loss of 29.8 m. ∆P = ρ ght = 900kg / m 3 9.807m/ s 2 29.8m = 263kPa ans. 3 2 0.2 / 273,600 / 54.7 t W Qg h Q P m s N m kw ρ ρ ρ ρ = = ∆ = = = = ∆ = = = = ∆ = = = = ∆ = = & ans. Note that this is not necessarily the power required to drive a pump, as the pump efficiency will typically be less than 100%. These problems are easily set up for solution in a spreadsheet as shown below. Make sure that the calculation for friction factor includes a test for laminar or turbulent flow with the result proceeding to the correct equation. Always verify any computer solution with problems having a known solution. FRICTIONAL HEAD LOSS CALCULATION All Data are entered in S.I. Units e.g. (m, sec., kg), except as noted, ε Ex.
6.7 Input Data Calculated Results L = 500 (m ) V = 6.37 (m/sec) D = 0.2 (m) Re = 127324 ε = 0.26 (mm) e/D = 0.0013 ρ = 900 (kg/m^3) f= 0.02258 ν = 1.00E-05 (m^2/sec) hf = 116.62 Q = 0.2 (m^3/sec) sum Ki = 0.00 D1 = 0.08 (m) hm = 0.00 (m) D2 = 0.08 (m) d KE = 0.00 (m) ht = 29.62 (m) d Z = -87 (m) (m) P1-P2 = 261.43 (kPa) VI-11 Solution Summary: To solve basic pipe flow frictional head loss problem, use the following procedure: 1. Use known flow rate to determine Reynolds number. 2. Identify whether flow is laminar or turbulent. 3. Use appropriate expression to determine friction factor (w ε/D if necessary). 4. Use definition of hf to determine friction head loss. 5. Use general energy equation to determine total pressure drop. Unknown Flow Rate and Diameter Problems Problems involving unknown flow rate and diameter in general require iterative/ trial & error solutions due to the complex dependence of Re, friction factor, and head loss on velocity and pipe size. Unknown Flow Rate: For the special case of known friction loss hf a closed form solution can be obtained for the problem of unknown Q. The solution proceeds as follows: Given: Known values for D, L, hf, ρ, and µ calculate V or Q. Define solution parameter: ς = 1 2 f ReD 2 = gD3 hf Lυ 2 Note that this solution does not contain velocity and the parameter ζ can be calculated from known values for D, L, hf, ρ, and µ. The Reynolds number and subsequently the velocity can be determined from ζ and the following equations: Turbulent: ReD = −8ς ( ) 1/ 2 log ε / D 3.7 + 1.775 ζ VI-12 Laminar: ReD = ς 32 and laminar to turbulent transition can be assumed to occur approximately at ζ = 73,600 (check Re at end of calculation to confirm). Note that this procedure is not valid (except perhaps for initial estimates) for problems involving significant minor losses where the head loss due only to pipe friction is not known. For these problems a trial and error solution using a computer is best. Example 6.9 Oil, with ρ = 950 kg/m3 and ν = 2 E-5 m2/s, flows through 100 m of a 30 cm diameter pipe. The pipe is known to have a head loss of 8 m and a roughness ratio ε/D = 0.0002. Determine the flow rate and oil velocity possible for these conditions. Without any information to the contrary, we will neglect minor losses and KE head changes. With these assumptions, we can write: ς = g D 3 hf Lυ 2 = 9.807m / s2 0.303 m3 8.0 m 100m 2 E−5m2 / s ( ) 2 = 5.3E7> 73,600; turbulent ReD = −85.3E7 ( ) 1/ 2 log 0.0002 3.7 + 1.775 5.3E7 = 72,600 checks, turbulent ReD = V D υ , V = 72,6002 E −5m2 / s 0.3m = 4.84 m s ans. Q = AV = π .152 m24.84 m/s = 0.342m3/s ans. This is the maximum flow rate and oil velocity that could be obtained through the given pipe and given conditions (hf = 8 m). VI-13 Note that this problem could have also been solved using a computer based trial and error procedure in which a value is assumed for the fluid flow rate until a flow rate is found which results in the specified head loss. Note also that with this procedure, the problem being solved can include the effects of minor losses, KE, and PE changes with no additional difficulty. Unknown Pipe Diameter: A similar difficulty arises for problems involving unknown pipe difficulty, except a closed form, analytical solution is not available. Again, a trial and error solution is appropriate for use to obtain the solution and the problem can again include losses due to KE, PE, and piping components with no additional difficulty. Non-Circular Ducts: For flow in non-circular ducts or ducts for which the flow does not fill the entire cross-section, we can define the hydraulic diameter Dh as Dh = 4 A P where A = cross-sectional area of actual flow, P = wetted perimeter, i.e. the perimeter on which viscous shear acts A P Cross sectional area -Perimeter - With this definition, all previous equations for the Reynolds number Re friction factor f and head loss hf are valid as previously defined and can be used on both circular and non-circular flow cross sections. Minor Losses In addition to frictional losses for a length L of pipe, we must also consider losses due to various fittings (valves, unions, elbows, tees, etc.). These losses are expressed as hm = Ki V 2 2g VI-14 where hm = the equivalent head loss across the fitting or flow component. V = average flow velocity for the pipe size of the fitting Ki = the minor loss coefficient for given flow component; valve, union, etc. See Sec. 6.7, Table 6.5, 6.6, Fig. 6.19, 6.20, 6.21, 6.22, etc. Table 6.5 shows minor loss K values for several common types of valves, fully open, and for elbows and tees. Table 6.5 Minor loss coefficient for common valves and piping components Figure 6.18 shows minor loss K values for several types of common valves. Note that the K valves shown here are for the indicated fractional opening. Also, fully open values may not be consistent with values indicated in Table 6.5 for fully open valves or for the valve of a particular manufacturer. In general, use specific manufacturer’s data when available. VI-15 Fig. 6.18 Average loss coefficients for partially open valves Note that exit losses are K ≅ 1 for all submerged exits, e.g., fluid discharged into a tank at a level below the fluid surface. Also, for an open pipe discharge to the atmosphere, there is no loss coefficient when the energy equation is written only to the end of the pipe. In general, do not take point 1 for an analysis to be in the plane of an inlet having an inlet loss. You do not know what fraction of the inlet loss to consider. Fig. 6.21 Entrance and exit loss coefficients (a) reentrant inlets; (b) rounded and beveled inlets VI-16 Fig. 6.22 Sudden contraction and expansion losses. Note that the losses shown in Fig. 6.22 do not represent losses associated with pipe unions or reducers. These must be found in other sources in the literature. Also note that the loss coefficient is always based on the velocity in the smaller diameter (d) of the pipe, irrespective of the direction of flow. Assume that this is also true for reducers and similar area change fittings. These and other sources of data now provide the ability to determine frictional losses for both the pipe and other piping/duct flow components. The total frictional loss now becomes hf = f L D V 2g 2 + ∑Ki V 2 2g or hf = f L D + ∑Ki V 2g 2 These equations would be appropriate for a single pipe size (with average velocity V). For multiple pipe/duct sizes, this term must be repeated for each pipe size. Key Point: The energy equation must still be used to determine the total head loss and pressure drop from all possible contributions. VI-17 Example 6.16 Water, ρ = 1.94 slugs/ft3 and ν = 1.1 E-5 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2–in diameter pipe with ε/D = 0.001 having the indicated minor losses. Compute the pump horsepower (motor size) required. Writing the energy equation between points 1 and 2 (the free surfaces of the two reservoirs), we obtain P 1 −P 2 ρ g = V 2 2 −V 1 2 2g + Z2 −Z1 + hf −hp For this problem, the pressure (P1 = P2) and velocity (V1 = V2 = 0) head terms are zero and the equation reduces to hp = Z2 −Z1 + hf = Z2 −Z1 + f L D + ∑Ki V 2g 2 For a flow rate Q = 0.2 ft3/s we obtain VI-18 V = Q A = 0.2 ft3 / s π 1/12 ( )2 ft2 = 9.17 ft / s With ε/D = 0.001 and Re = V D ν = 9.17 ft / s 2/12 ( ) ft 1.1E−5 ft2 / s =139,000 the flow is turbulent and Haaland’s equation can be used to determine the friction factor: f = −1.8log 6.9 139,000 + .001 3.7 1.11 −2 = 0.0214 the minor losses for the problem are summarized in the following table: Note: The loss for a pipe bend is not the same as for an elbow fitting. If there were no tank at the pipe discharge and point 2 were at the pipe exit, there would be no exit loss coefficient. However, there would be an exit K.E. term. Loss element Ki Sharp entrance (Fig. 6.21) 0.5 Open globe valve (Table 6.5) 6.9 12 " bend, R/D = 12/6 = 2 (Fig. 6.19) 0.15 Threaded, 90Þ, reg. elbow, (Table 6.5) 0.95 Gate valve, 1/2 closed (Fig. 6.18) 2.7 Submerged exit (Fig. 6.20) 1 Ki = 12.2 Substituting in the energy equation we obtain h p = 120 −20 ( ) + 0.0214 400 2/12 9.17 2 64.4 + 12.2 9.17 2 64.4 h p =100 + 67.1+ 15.9 = 183ft VI-19 Note the distribution of the total loss between static, pipe friction, and minor losses. The power required to be delivered to the fluid is give by P f = ρ Qghp =1.94 slug ft 3 32.2 ft s 2 0.2 ft 3 s 183 ft = 2286 ft lbf P f = 2286ftlbf 550ftlbf/s/hp=4.2hp If the pump has an efficiency of 70 %, the power requirements would be specified by 4.2 6 0.70 p hp hp w = = = = = = = = & Solution Summary: To solve basic pipe flow pressure drop problem, use the following procedure: 1. Use known flow rate to determine Reynolds number. 2. Identify whether flow is laminar or turbulent. 3. Use appropriate expression to find friction factor (with ε/D if necessary). 4. Use definition of hf to determine friction head loss. 5. Tabulate and sum minor loss coefficients for piping components. 6a. Use general energy equation to determine total pressure drop, or 6b. Determine pump head requirements as appropriate. 7. Determine pump power and motor size if required. VI-20 Multiple-Pipe Systems Basic concepts of pipe system analysis apply also to multiple pipe systems. However, the solution procedure is more involved and can be iterative. Consider the following: a. Multiple pipes in series b. Multiple pipes in parallel Series Pipe System: The indicated pipe system has a steady flow rate Q through three pipes with diameters D1, D2, & D3. Two important rules apply to this problem. a b 1 2 3 1. The flow rate is the same through each pipe section. For incompressible flow, this is expressed as Q1 = Q2 = Q3 = Q or D1 2V1 = D2 2V2 = D3 2V3 2. The total frictional head loss is the sum of the head losses through the various sections. hf ,a−b = hf ,1 + hf ,2 + hf ,3 hf ,a−b = f L D + Ki ∑ D1 V 1 2 2g + f L D + Ki ∑ D2 V 2 2 2g + f L D + Ki ∑ D3 V 3 2 2g Note: Be careful how you evaluate the transitions from one section to the next. In general, loss coefficients for transition sections are based on the velocity of the smaller section. VI-21 Example: Given a pipe system as shown in the previous figure. The total pressure drop is Pa – Pb = 150 kPa and the elevation change is Za – Zb = – 5 m. Given the following data, determine the flow rate of water through the section. Pipe L (m) D (cm) e (mm) e/D 1 100 8 0.24 0.003 2 150 6 0.12 0.02 3 80 4 0.2 0.005 The energy equation is written as where hf is given by the sum of the total frictional losses for three pipe sections. With no pump; hp is 0, Zb - Za = - 5 m and ht = 15.3 m for ∆P = 150 kPa P a −P b ρ g = V b 2 −Va 2 2 g + Zb −Za + h f −hp ht = P a −P b ρ g = 150,000 N / m2 9790 N / m2 = 15.3m Since the flow rate Q and thus velocity is the only remaining variable, the solution is easily obtained from a spreadsheet by assuming Q until ∆P = 150 kPa. Fluid 1 2 3 ρ(kg/m^3) = 1000 L(m) 100 150 80 ν(m^2/s) = 1.02E-06 D(m) = 0.08 0.06 0.04 ε(mm) = 0.24 0.12 0.20 inlet & exit ε/D= 0.003 0.002 0.005 dZ (m) = -5 Da(m) = 0.08 V(m/s)= 0.56 1.00 2.25 Db(m) = 0.04 Re = 44082.8 58777.1 88165.6 Assume f= 0.02872 0.02591 0.03139 Q (m^3/s)= 0.00283 hf = 0.58 3.30 16.18 Va(m/s)= 0.56 Ki 0 0 0 Vb(m/s)= 2.25 hm= 0 0 0 dKE(m) = 0.24 hf (net) = 20.08 hf(calc) = 0.58 3.30 16.18 Actual Calculated Pa - Pb (kPa) 150 150.00 Q(m^3/hr) = 10.17 VI-22 Thus it is seen that a flow rate of 10.17 m3/hr produces the indicated head loss through each section and a net total ∆P = 150 kPa. A solution can also be obtained by writing all terms explicitly in terms of a single velocity, however, the algebra is quite complex (unless the flow is laminar), and an iterative solution is still required. All equations used to obtained the solution are the same as those presented in previous sections. Parallel Pipe Systems A flow rate QT enters the indicated parallel pipe system. The total flow splits and flows through 3 pipe sections, each with different diameters and lengths. Two basic rules apply to parallel pipe systems; a b 1 2 3 QT QT 1. The total flow entering the parallel section is equal to the sum of the flow rates through the individual sections, 2. The total pressure drop across the parallel section is equal to the pressure drop across each individual parallel segment. Note that if a common junction is used for the start and end of the parallel section, the velocity and elevation change is also the same for each section. Thus, the flow rate through each section must be such that the frictional loss is the same for each and the sum of the flow rates equals the total flow. For the special case of no kinetic or potential energy change across the sections, we obtain: ht = ( hf + hm)1 = ( hf + hm)2 =( hf + hm)3 and QT = Q1 + Q2 + Q3 VI-23 Again, the equation used for both the pipe friction and minor losses is the same as previously presented. The flow and pipe dimensions used for the previous example are now applied to the parallel circuit shown above. Example: A parallel pipe section consists of three parallel pipe segments with the lengths and diameters shown below. The total pressure drop is 150 kPa and the parallel section has an elevation drop of 5 m. Neglecting minor losses and kinetic energy changes, determine the flow rate of water through each pipe section. The solution is iterative and is again presented in a spreadsheet. The net friction head loss of 20.3 m now occurs across each of the three parallel sections. Fluid 1 2 3 ρ(kg/m^3) = 1000 L(m) 100 150 80 ν(m^2/s) = 1.02E-06 D(m) = 0.08 0.06 0.04 ε(mm) = 0.24 0.12 0.20 inlet & exit ε/D= 0.003 0.002 0.005 dZ (m) = -5 Q(m^3/hr) = 62.54 25.95 11.41 Da(m) = 0.08 V(m/s)= 3.46 2.55 2.52 Db(m) = 0.04 Re = 271083.5 149977.8 98919.7 f= 0.02666 0.02450 0.03129 Q (m^3/hr)= 99.91 hf = 20.30 20.30 20.30 Assume Ki = 0 0 0 Q1 (m^3/s)= 62.54 hm= 0.00 0.00 0.00 Q2(m^3/s)= 25.95 hf,net(m) = 20.30 20.30 20.30 hf + ∆z= 15.30 15.30 15.30 Pa - Pb (kPa) 150.13 ht(m) = 15.31 Q(m^3/hr) = 62.54 25.95 11.41 Total Flow, Qt(m^3/hr) = 99.91 The strong effect of diameter can be seen with the smallest diameter having the lowest flow rate, even though it also has the shortest length of pipe. VII-1 VII. Boundary Layer Flows The previous chapter considered only viscous internal flows. Viscous internal flows have the following major boundary layer characteristics: An entrance region where the boundary layer grows and dP/dx ≠ constant, A fully developed region where: • The boundary layer fills the entire flow area. • The velocity profiles, pressure gradient, and τw are constant; i.e., they are not equal to f(x), • The flow is either laminar or turbulent over the entire length of the flow , i.e., transition from laminar to turbulent is not considered. However, viscous flow boundary layer characteristics for external flows are significantly different as shown below for flow over a flat plate: y x δ(x) turbulent laminar xcr laminar to turbulent transition edge of boundary layer free stream U∞ Fig. 7.1 Schematic of boundary layer flow over a flat plate For these conditions, we note the following characteristics: • The boundary layer thickness δ grows continuously from the start of the fluid-surface contact, e.g., the leading edge. It is a function of x, not a constant. • Velocity profiles and shear stress τ are f(x,y). • The flow will generally be laminar starting from x = 0. • The flow will undergo laminar-to-turbulent transition if the streamwise dimension is greater than a distance xcr corresponding to the location of the transition Reynolds number Recr. • Outside of the boundary layer region, free stream conditions exist where velocity gradients and therefore viscous effects are typically negligible. VII-2 As it was for internal flows, the most important fluid flow parameter is the local Reynolds number defined as Rex = ρU∞x µ = U∞x υ where ρ = fluid density µ = fluid dynamic viscosity ν = fluid kinematic viscosity U∞ = characteristic flow velocity x = characteristic flow dimension It should be noted at this point that all external flow applications will not use a distance from the leading edge x and the characteristic flow dimension. For example, for flow over a cylinder, the diameter will be used as the characteristic dimension for the Reynolds number. Transition from laminar to turbulent flow typically occurs at the local transition Reynolds number which for flat plate flows can be in the range of 500,000 ≤Recr ≤3,000,00 With xcr = the value of x where transition from laminar to turbulent flow occurs, the typical value used for steady, incompressible flow over a flat plate is Recr = ρU∞xcr µ = 500,000 Thus for flat plate flows for which: x < xcr the flow is laminar x ≥xcr the flow is turbulent The solution to boundary layer flows is obtained from the reduced “Navier – Stokes” equations, i.e., Navier-Stokes equations for which boundary layer assumptions and approximations have been applied. VII-3 Flat Plate Boundary Layer Theory Laminar Flow Analysis For steady, incompressible flow over a flat plate, the laminar boundary layer equations are: Conservation of mass: ∂u ∂x + ∂v ∂y = 0 'X' momentum: u ∂u ∂x + v ∂u ∂y = −1 ρ d p d x + 1 ρ ∂ ∂y µ ∂u ∂y 'Y' momentum: −∂p ∂y = 0 The solution to these equations was obtained in 1908 by Blasius, a student of Prandtl's. He showed that the solution to the velocity profile, shown in the table below, could be obtained as a function of a single, non-dimensional variable η defined as η = y U ∞ υ x 1/2 with the resulting ordinary differential equation: ′ ′ ′ f + 1 2 f ′ ′ f = 0 and ′ f η ( )= u U ∞ Boundary conditions for the differential equation are expressed as follows: at y = 0, v = 0 → f (0) = 0 ; y component of velocity is zero at y = 0 at y = 0 , u = 0 →′ f 0 ( )= 0; x component of velocity is zero at y = 0 VII-4 The key result of this solution is written as follows: ∂ 2 f ∂η 2 y=0 = 0.332 = τ w µU∞U∞/ υ x With this result and the definition of the boundary layer thickness, the following key results are obtained for the laminar flat plate boundary layer: Local boundary layer thickness δ x ( )= 5x Re x Local skin friction coefficient: (defined below) Cfx = 0.664 Re x Total drag coefficient for length L ( integration of τw dA over the length of the plate, per unit area, divided by 0.5 ρ U∞ 2 ) CD = 1.328 Rex where by definition Cfx = τw x ( ) 1 2 ρU∞ 2 and CD = F D / A 1 2 ρ U∞ 2 With these results, we can determine local boundary layer thickness, local wall shear stress, and total drag force for laminar flow over a flat plate. Example: Air flows over a sharp edged flat plate with L = 1 m, a width of 3 m and U∞ = 2 m/s . For one side of the plate, find: δ(L), Cf (L), τw(L), CD, and FD. Air: ρ = 1.23 kg/m3 ν = 1.46 E-5 m2/s First check Re: Re L = U∞L υ = 2m / s 2.15m 1.46E −5m2 / s = 294,520 < 500,000 Key Point: Therefore, the flow is laminar over the entire length of the plate and calculations made for any x position from 0 - 1 m must be made using laminar flow equations. VII-5 Boundary layer thickness at x = L: δ L ( ) = 5 L ReL = 52.15m 294,520 = 0.0198m =1.98cm Local skin friction coefficient at x = L: Cf L ( ) = 0.664 ReL = 0.664 294,520 = 0.00122 Surface shear stress at x = L: τ w =1/ 2 ρU∞ 2 Cf = 0.51.23kg / m 3 2 2 m 2 / s 2 0.00122 τ w = 0.0030 N / m 2 Pa ( ) Drag coefficient over total plate, 0 – L: CD L ( )= 1.328 ReL = 1.328 294,520 = 0.00245 Drag force over plate, 0 – L: F D =1/ 2ρU∞ 2 CD A = 0.51.23kg / m 3 2 2 m 2 / s 2 0.0024522.15m 2 F D = 0.0259 N Two key points regarding this analysis: 1. Each of these calculations can be made for any other location on the plate by simply using the appropriate x location for anyx L ≤ ≤ ≤ ≤ . 2. Be careful not to confuse the calculation for Cf and CD. Cf is a local calculation at a particular x location (including x = L) and can only be used to calculate local shear stress, not drag force. CD is an integrated average over a specified length (including any x ≤L) and can only be used to calculate the average shear stress and the integrated force over the length. VII-6 Turbulent Flow Equations While the previous analysis provides an excellent representation of laminar, flat plate boundary layer flow, a similar analytical solution is not available for turbulent flow due to the complex nature of the turbulent flow structure. However, experimental results are available to provide equations for key flow field parameters. A summary of the results for boundary layer thickness and local and average skin friction coefficient for a laminar flat plate and a comparison with experimental results for a smooth, turbulent flat plate are shown below. Laminar Turbulent δ x ( )= 5x Re x δ x ( ) = 0.37x Rex .2 Cfx = 0.664 Rex Cfx = 0.0592 Rex .2 CD = 1.328 Re L CD = 0.074 ReL .2 for turbulent flow over entire plate, 0 – L, i.e. assumes turbulent flow in the laminar region. where Cfx = τw 1 2 ρU∞ 2 local drag coefficient based on local wall shear stress (laminar or turbulent flow region). and CD = total drag coefficient based on the integrated force over the length 0 to L CD = F / A 1 2 ρU∞ 2 = 1 2 ρU∞ 2 A ( ) −1 τ w 0 L ∫ x ( )w dx A careful study of these results will show that in general, boundary layer thickness grows faster for turbulent flow and wall shear and total friction drag are greater for turbulent flow than for laminar flow given the same Reynolds number. VII-7 It is noted that the expressions for turbulent flow are valid only for a flat plate with a smooth surface. Expressions including the effects of surface roughness are available in the text. Combined Laminar and Turbulent Flow y x U δ(x) turbulent laminar xcr laminar to turbulent transition edge of boundary layer free stream Flat plate with both laminar and turbulent flow sections For conditions (as shown above) where the length of the plate is sufficiently long that we have both laminar and turbulent sections: Local values for boundary layer thickness and wall shear stress for either the laminar or turbulent sections are obtained from the expressions for δ(x) and Cfx for laminar or turbulent flow as appropriate for the given region. The result for average drag coefficient CD and thus total frictional force over the laminar and turbulent portions of the plate is given by (assuming a transition Re of 500,000) CD = 0.074 ReL .2 −1742 ReL Calculations assuming only turbulent flow can be made typically for two cases 1. when some physical situation (a trip wire) has caused the flow to be leading from the leading edge or 2. if the total length L of the plate is much greater than the length xcr of the laminar section such that the total flow can be considered turbulent from x = 0 to L. Note that this will overpredict the friction drag force since turbulent drag is greater than laminar. With these results, a detailed analysis can be obtained for laminar and/or turbulent flow over flat plates and surfaces that can be approximated as a flat plate. VII-8 Example: Water flows over a sharp flat plate 2.55 m long, 1 m wide, with U∞ = 2 m/s. Estimate the error in FD if it is assumed that the entire plate is turbulent. Water: ρ = 1000 kg/m3 ν = 1.02 E- m2/s Reynolds number: ReL = U∞L υ = 2m / s2.55m 1.02E −6m 2 / s = 5E6 > 500,000 with Recr = 500,000 ⇒xcr = 0.255m ( or 10% laminar) a. Assume that the entire plate is turbulent CD = 0.074 ReL .2 = 0.074 5E6 ( ) .2 = 0.00338 F D = 0.5ρU∞ 2 CD A = 0.51000 kg m 3 2 2 m 2 s 2 0.003382.55m 2 F D = 17.26 N This should be high since we have assumed that the entire plate is turbulent and the first 10% is actually laminar. b. Consider the actual combined laminar and turbulent flow: CD = 0.074 ReL .2 −1742 ReL = 0.00338 −1742 5 E6 = 0.00303 Note that the CD has decreased when both the laminar and turbulent sections are considered. F D = 0.5ρU∞ 2 CD A = 0.51000 kg m 3 2 2 m 2 s 2 0.003032.55m 2 VII-9 F D = 15.46 N {Lower than the fully turbulent value} Error = 17.26 −15.46 15.46 100 =11.6% high Question: Since xcr = 0.255 m, what would your answers represent if you had calculated the Re, CD, and FD using x = xcr = 0.255 m? Answer: You would have the value of the transition Reynolds number and the drag coefficient and drag force over the laminar portion of the plate (assuming you used laminar equations). If you used turbulent equations, you would have red marks on your paper. VII-10 Von Karman Integral Momentum Analysis While the previous results provide an excellent basis for the analysis of flat plate flows, complex geometries and boundary conditions make analytical solutions to most problems difficult. An alternative procedure provides the basis for an approximate solution which in many cases can provide excellent results. The key to practical results is to use a reasonable approximation to the boundary layer profile, u(x,y). This is used to obtain the following: a. Boundary layer mass flow: 0 m ubdy δ δ δ δ ρ ρ ρ ρ = = = = ∫ ∫ ∫ ∫ & where b is the width of the area for which the flow rate is being obtained. b. Wall shear stress: τ w = µ d u d y y=0 You will also need the streamwise pressure gradient d P d x for many problems. The Von Karman integral momentum theory provides the basis for such an approximate analysis. The following summarizes this theory. Displacement thickness: Consider the problem indicated in the adjacent figure: A uniform flow field with velocity U∞ approaches a solid surface. As a result of viscous shear, a boundary layer velocity profile develops. Streamline δ y=h +δ U y x 0 h u Simulated effect U h U ∞ ∞ ∞ VII-11 A viscous boundary layer is created when the flow comes in contact with the solid surface. Key point: Compared to the uniform velocity profile approaching the solid surface, the effect of the viscous boundary layer is to displace streamlines of the flow outside the boundary layer away from the wall. With this concept, we define δ = displacement thickness δ = distance the solid surface would have to be displaced to maintain the same mass flow rate as for non-viscous flow. From the development in the text, we obtain δ = 1 −u U∞ 0 δ ∫ dy Therefore, with an expression for the local velocity profile we can obtain δ = f(δ) Example: Given: u U∞ = 2 y δ − y δ 2 determine an expression for δ = f(δ) Note that for this assumed form for the velocity profile: 1. At y = 0, u = 0 correct for no slip condition 2. At y = δ, u = U∞ correct for edge of boundary layer 3. The form is quadratic To simplify the mathematics, let η = y/δ, at y = 0, η = 0 ; at y = δ , η = 1; dy = δ dη Therefore: u U∞ = 2η −η2 VII-12 Substituting: δ = 1 −2η + η2 ( ) 0 1 ∫ δ dη = δ η −2η2 2 + η3 3 0 1 which yields δ = 1 3δ Therefore, for flows for which the assumed quadratic equation approximates the velocity profile, streamlines outside of the boundary layer are displaced approximately according to the equation δ = 1 3δ This closely approximates flow for a flat plate. Key Point: When assuming a form for a velocity profile to use in the Von Karman analysis, make sure that the resulting equation satisfies both surface and free stream boundary conditions as well as has a form that approximates u(y). Momentum Thickness: The second concept used in the Von Karman momentum analysis is that of momentum thickness - θ The concept is similar to that of displacement thickness in that θ is related to the loss of momentum due to viscous effects in the boundary layer. Consider the viscous flow regions shown in the adjacent figure. Define a control volume as shown and integrate around the control volume to obtain the net change in momentum for the control volume. VII-13 If D = drag force on the plate due to viscous flow, we can write - D = ∑ ( momentum leaving c.v. ) - ∑ ( momentum entering c.v. ) Completing an analysis shown in the text, we obtain D = ρU∞ 2 θ θ = u U∞ 0 δ ∫ 1 −u U∞ d y Using a drag coefficient defined as CD = D/A 1 2 ρ U∞ 2 We can also show that CD = 2θ L ( ) L where: θ(L) is the momentum thickness evaluated over the length L. Thus, knowledge of the boundary layer velocity distribution u = f(y) allows the drag coefficient to be determined. Momentum integral: The final step in the Von Karman theory applies the previous control volume analysis to a differential length of surface. Performing an analysis similar to the previous analysis for drag D we obtain τ w ρ = δ U∞ dU∞ d x + d d x U∞ 2 θ ( ) This is the momentum integral for 2-D, incompressible flow and is valid for laminar or turbulent flow. VII-14 where δ U∞ dU∞ d x = −δ ρ d P d x Therefore, this analysis also accounts for the effect of freestream pressure gradient. For a flat plate with non-accelerating flow, we can show that P = const., U∞= const., dU∞ d x = 0 Therefore, for a flat plate, non-accelerating flow, the Von Karman momentum integral becomes τ w ρ = d d x U∞ 2 θ ( )= U∞ 2 dθ d x From the previous analysis and the assumed velocity distribution of u U∞ = 2 y δ − y δ 2 = 2η −η2 The wall shear stress can be expressed as τ w = µ d u d y w = 2U∞ 2 δ −2 y δ 2 y=0 = 2 µU∞ δ (A) Also, with the assumed velocity profile, the momentum thickness θ can be evaluated as θ = u U∞ 0 δ ∫ 1 −u U∞ d y or VII-15 θ = 2η −η2 ( ) 0 δ ∫ 1 −2η + η2 ( ) δ dη = 2δ 15 We can now write from the previous equation for τw τ w = ρ U∞ 2 dθ d x = 2 15 ρ U∞ 2 dδ d x Equating this result to Eqn. A we obtain τ w = 2 15 ρ U∞ 2 dδ d x = 2 µU∞ δ or δ dδ = 15µ ρU∞ d x which after integration yields δ = 30 µ x ρU∞ 1/2 or δ = 5.48 Rex Note that the this result is within 10% of the exact result from Blasius flat plate theory. Since for a flat plate, we only need to consider friction drag (not pressure drag), we can write Cfx = τw x ( ) 1 2 ρU∞ 2 = 2 µU∞ δ 1 1 2 ρ U∞ 2 Substitute for δ to obtain Cfx = 2µ U∞ 5.48 Re 1 2 ρU∞ 2 = 0.73 Rex VII-16 Exact theory has a numerical constant of 0.664 compared with 0.73 for the previous result. It is seen that the von Karman integral theory provides the means to determine approximate expressions for δ, τw, and Cf using only an assumed velocity profile. Solution summary: 1. Assume an analytical expression for the velocity profile for the problem. 2. Use the assumed velocity profile to determine the solution for the displacement thickness for the problem. 3. Use the assumed velocity profile to determine the solution for the momentum thickness for the problem. 4. Use the previous results and the von Karman integral momentum equation to determine the solution for the drag/wall shear for the problem. VII-17 Bluff Body, Viscous Flow Characteristics ( Immersed Bodies) In general, a body immersed in a flow will experience both externally applied forces and moments as a result of the flow about its external surfaces. The typical terminology and designation of these forces and moments are given in the diagram shown below. The orientation of the axis for the drag force is typically along the principal body axis, although in certain applications, this axis is aligned with the principal axis of the free stream, approach velocity U. Since in many cases the drag force is aligned with the principal axis of the body shape and not necessarily aligned with the approaching wind vector. Review all data carefully to determine which coordinate system is being used: body axis coordinate system or a wind axis coordinate system. These externally applied forces and moments are generally a function of a. Body geometry b. Body orientation c. Flow conditions VII-18 These forces and moments are also generally expressed in the form of a non-dimensional force/moment coefficient, e.g. the drag coefficient: CD = FD/A 1 2 ρ U∞ 2 It is noted that it is common to see one of three reference areas used depending on the application: 1. Frontal (projected) area: Used for thick, stubby, non-aerodynamic shapes, e.g., buildings, cars, etc. 2. Planform (top view, projected) area: Used for flat, thin shapes, e.g., wings, hydrofoils, etc. 3. Wetted area: The total area in contact with the fluid. Used for surface ships, barges, etc. The previous, flat plate boundary layer results considered only the contribution of viscous surface friction to drag forces on a body. However, a second major (and usually dominant) factor is pressure or form drag. Pressure drag is drag due to the integrated surface pressure distribution over the body. Therefore, in general , the total drag coefficient of a body can be expressed as C D = CD,press + CD,friction or CD = FD,total/A 1 2 ρ U∞ 2 = FD,press./A 1 2 ρU ∞ 2 + FD,friction/A 1 2 ρ U∞ 2 Which factor, pressure or friction drag, dominates depends largely on the aerodynamics (streamlining) of the shape and to a lesser extent on the flow conditions. VII-19 Typically the most important factor in the magnitude and significance of pressure or form drag is the boundary layer separation and resulting low pressure wake region associated with flow around non - aerodynamic shapes. Consider the two shapes shown below: High Pressure Low Pressure Wake boundary layer separation large pressure drag Low pressure drag no separated flow region The flow around the streamlined airfoil remains attached, producing no boundary layer separation and comparatively small pressure drag. However, the flow around the less aerodynamic circular cylinder separates, resulting in an area of high surface pressure on the front side and low surface pressure on the back side and thus significant pressure drag. This effect is shown very graphically in the following figures from the text. VII-20 Fig. 7.12 Drag of a 2-D, streamlined cylinder The previous figure shows the effect of streamlining and aerodynamics on the relative importance of friction and pressure drag. While for a thin flat plate (t/c = 0), all the drag is due to friction with no pressure drag, for a circular cylinder (t/c = 1), only 3% of the drag is due to friction with 97% due to pressure. Likewise for most bluff, non-aerodynamic bodies, pressure (also referred to as form drag) is the dominant contributor to the total drag. However, the magnitude of the pressure (and therefore the total) drag can also be changed by reducing the size of the low pressure wake region. One way to do this is to change the flow conditions from laminar to turbulent. This is illustrated in the following figures from the text for a circular cylinder. Fig. 7.13 Circular cylinder with (a) laminar separation and (b) turbulent separation Note that for the cylinder on the left, the flow is laminar, boundary layer separation occurs at 82o and the CD is 1.2, whereas for the cylinder on the right, the flow is turbulent and separation is delayed and occurs at 120o. The drag coefficient CD is 0.3, a factor of 4 reduction due to a smaller wake region and reduced pressure drag. VII-21 It should also be pointed out that the friction drag for the cylinder on the right is probably greater (turbulent flow conditions) than for the cylinder on the left (laminar flow conditions). However, since pressure drag dominates, the net result is a significant reduction in the total drag. The pressure distribution for laminar and turbulent flow over a cylinder is shown in Fig. 7.13c to the right. The front-to-rear pressure difference is greater for laminar flow, thus greater drag. Finally, the effect of streamlining on total drag is shown very graphically with the sequence of modifications in Fig. 7.15. Two observations can be made: (1) As body shape changes from a bluff body with fixed points of separation to a more aerodynamic shape, the effect of pressure drag and the drag coefficient will decrease. Fig. 7.15 The effect of streamlining on total drag (2) The addition of surface area from (a) to (b) and (b) to (c) increases the friction drag, however, since pressure drag dominates, the net result is a reduction in the drag force and the CD . VII-22 The final two figures show results for the drag coefficient for two and three dimensional shapes with various geometries. Table 7.2 CD for Two-Dimensional Bodies at Re ≥ 104 First note that all values in Table 7.2 are for 2-D geometries, that is, the bodies are very long (compared to the cross-section dimensions) in the dimension perpen-dicular to the page. Key Point: Non – aerodynamic shapes with fixed points of separations (sharp corners) have a single value of CD, irrespective of the value of the Reynolds number, e.g. square cylinder, half-tube, etc. Aerodynamic shapes generally have a reduction in CD for a change from laminar to turbulent flow as a result of the shift in the point of boundary layer separation, e.g. elliptical cylinder. VII-23 Table 7.3 Drag of three-dimensional bodies at Re ≥ 104 The geometries in Table 7.3 are all 3-D and thus are finite perpendicular to the page. Similar to the results from the previous table, bluff body geometries with fixed points of separation have a single CD, whereas aerodynamic shapes such as slender bodies of revolution have individual values of CD for laminar and turbulent flow. VII-24 In summary, one must remember that broad generalizations such as saying that turbulent flow always increases drag, drag coefficients always depend on Reynolds number, or increasing surface area increases drag are not always valid. One must consider carefully all effects (viscous and pressure drag) due to changing flow conditions and geometry. Example: A square 6-in piling is acted on by a water flow of 5 ft/s that is 20 ft deep. Estimate the maximum bending stress exerted by the flow on the bottom of the piling. Water: ρ = 1.99 slugs/ft3 ν = 1.1 E – 5 ft2/s Assume that the piling can be treated as 2-D and thus end effects are negligible. Thus for a width of 0.5 ft, we obtain: Re = 5 ft / s.5 ft 1.1E −5 ft2 / s = 2.3E 5 In this range, Table 7.2 applies for 2-D bodies and we read CD = 2 .1. The frontal area is A = 200.5 = 10 ft2 F D = 0.5ρU∞ 2 CD A = 0.51.99 slug ft 3 5 2 ft 2 s 2 2.110 ft 2 =522lbf For uniform flow, the drag should be uniformly distributed over the total length with the net drag located at the mid-point of the piling. Thus, relative to the bottom of the piling, the bending moment is given by Mo = F 0.5 L = 522 lbf 10 ft = 5220 ft-lbf VII-25 From strength of materials, we can write σ = Mo c I = 5220 ft −lbf 0.25 ft 1 12 0.5 ft 0.5 3 ft 3 = 251,000 psf = 1740 psi where c = distance to the neutral axis, I = moment of inertia = b h3/12 Question: Since pressure acts on the piling and increases with increasing depth, why wasn’t a pressure load considered? Answer: Static pressure does act on the piling, but it acts uniformly around the piling at every depth and thus cancels. Dynamic pressure is considered in the drag coefficients of Tables 7.2 and 7.3 and does not have to be accounted for separately. VIII - 1 Ch. VIII Potential Flow and Computational Fluid Dynamics Review of Velocity-Potential Concepts This chapter presents examples of problems and their solution for which the assumption of potential flow is appropriate. For low speed flows where viscous effects are neglected, the flow is irrotational and ∇× V = 0 V = ∇φ u = ∂φ ∂x v = ∂φ ∂y w = ∂φ ∂z The continuity equation , ∇⋅V = 0 , now reduces to ∇ 2 V = ∂ 2 φ ∂x 2 + ∂ 2 φ ∂y 2 + ∂ 2 φ ∂z 2 = 0 The momentum equation reduces to Bernoulli’s equation: ∂φ ∂t + P ρ + 1 2 V 2 + gz = const Review of Stream Function Concepts For plane incompressible flow in x-y coordinates a stream function exists such that u = ∂Ψ ∂y and v = −∂Ψ ∂x The condition of irrotationality reduces to Lapace’s equation for Ψ and ∂ 2 Ψ ∂x 2 + ∂ 2 Ψ ∂y 2 = 0 VIII - 2 Elementary Plane-Flow Solutions Three useful plane-flow solutions that are very useful in developing more complex solutions are: Uniform stream, iU, in the x direction: Ψ = U y φ = U x Line source or sink: Ψ = mθ φ = mln r Line vortex: Ψ = −K ln r φ = Kθ In these expressions, the source strength, ‘m’ and vortex strength, ‘ K ‘, have the dimensions of velocity times length, or [L2/t]. If the uniform stream is written in plane polar coordinates, we have Uniform stream, iU: Ψ = U r sinθ φ = U rcosθ For a uniform stream moving at an angle, a , relative to the x-axis, we can write u = Ucosα = ∂Ψ ∂y = ∂φ ∂x v = Usinα = −∂Ψ ∂x = ∂φ ∂y After integration, we obtain the following expressions for the stream function and velocity potential: Ψ = U y cosα −xsinα ( ) φ =U x cosα + ysinα ( ) Circulation The concept of fluid circulation is very useful in the analysis of certain potential flows, in particular those useful in aerodynamics analyses. Consider Figure 8.3 shown below: VIII - 3 We define the circulation, Γ , as the counterclockwise line integral of the arc length, ds times the velocity component tangent to the closed curve, C, e.g. Γ = V cosα d s c ∫ = V ⋅ds c ∫ Γ = u dx + vdy + wdz ( ) c ∫ For most flows, this line integral around a closed path, starting and stoping at the same point, yields Γ = 0. However, for a vortex flow for which φ = K θ the integral yields Γ = 2 π K An equivalent calculation can by made by defining a circular path of radius r around the vortex center to yield Γ = vθ c ∫ d s = K r 0 2π ∫ r dφ = 2π K Superposition of Potential Flows Due to the mathematical character of the equations governing potential flows, the principle of superposition can be used to determine the solution of the flow which results from combining two individual potential flow solutions. Several classic examples of this are presented as follows: VIII - 4 1. Source m at ( -a,0) added to an equal sink at (+a, 0). ψ = −mtan −1 2a y x 2 + y 2 −a 2 φ = 1 2 mln x + a ( ) 2 + y2 x −a ( ) 2 + y2 The streamlines and potential lines are two families of orthogonal circles (Fig. 4.13). 2. Sink m plus a vortex K, both at the origin. ψ = mθ −K ln r φ = mln r + Kθ The streamlines are logarithmic spirals swirling into the origin (Fig. 4.14). They resemble a tornado or a bathtub vortex. 3. Uniform steam i U∞ plus a source m at the origin (Fig. 4.15), the Ranking half body. If the origin contains a source, a plane half-body is formed with its nose to the left as shown below. If the origin contains a sink, m < 0, the half-body nose is to the right.. For both cases, the stagnation point is at a position a = m / U∞ away from the origin. VIII - 5 Example 8.1 An offshore power plant cooling water intake has a flow rate of 1500 ft3/s in water 30 ft deep as in Fig. E8.1. 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 of tidal flow in entrained into the intake? The sink strength is related to the volume flow, Q and water depth by m = Q 2π b = 1500 ft 3 / s 2π 30 ft =7.96 ft 2 / s The lengths a and L are given by a = m U∞ = 7.96 ft 2 / s 0.7 ft / s = 11.4 ft L = 2π a = 2π 11.4 ft = 71 ft Flow Past a Vortex Consider a uniform stream, U∞ flowing in the x direction past a vortex of strength K with the center at the origin. By superposition the combined stream function is ψ = ψ stream + ψ vortex = U∞rsinθ −K ln r The velocity components of this flow are given by vr = 1 r ∂ψ ∂θ = U∞cosθ vθ = −∂ψ ∂r = −U∞sinθ + K r Setting vr and vθ = 0, we find the stagnation point at θ = 90û, r = a = K/ U∞ or (x,y) = (0,a). VIII - 6 An Infinite Row of Vortices Consider an infinite row of vortices of equal strength K and equal spacing a. A single vortex, i , has a stream function given by Fig. 8.7 Superposition of vortices ψ i = −K ln r i i=1 ∞ ∑ This infinite sum can also be expressed as ψ =−1 2 K ln 1 2 cosh 2π y a −cosh 2π x a The resulting left and right flow above and below the row of vortices is given by u = ∂ψ ∂y y >a = ± π K a VIII - 7 Plane flow past Closed-Body Shapes Various types of external flows over a closed-body can be constructed by superimposing a uniform stream with sources, sinks, and vortices. Key Point: The body shape will be closed only if the net source of the outflow equals the net sink inflow. Two examples of this are presented below. The Rankine Oval A Rankine Oval is a cylindrical shape which is long compared to its height. It is formed by a source-sink pair aligned parallel to a uniform stream. The individual flows used to produce the final result and the combined flow field are shown in Fig. 8.9. The combined stream function is given by ψ = U∞y −m tan −1 2a y x 2 + y 2 −a 2 or ψ = U∞rsinθ + m θ1 −θ2 ( ) Fig. 8.9 The Rankine Oval The oval shaped closed body is the streamline, ψ = 0. Stagnation points occur at the front and rear of the oval, x = ± L, y = 0. Points of maximum velocity and minimum pressure occur at the shoulders, x = 0, y = ± h. Key geometric and flow parameters of the Rankine Oval can be expressed as follows: VIII - 8 h a = cot h / a 2m / U∞a ( ) L a = 1+ 2m U∞a 1/ 2 umax U∞ =1 + 2m / U∞a ( ) 1 + h 2 / a 2 As the value of the parameter m / U∞a ( ) is increased from zero, the oval shape increases in size and transforms from a flat plate to a circular cylinder at the limiting case of m / U∞a ( )= ∞. Specific values of these parameters are presented in Table 8.1 for four different values of the dimensionless vortex strength, K / U∞a ( ). Table 8.1 Rankine-Oval Parameters m / U∞a ( ) h / a L / a L / h umax /U∞ 0.0 0.0 1.0 ∞ 1.0 0.01 0.31 1.10 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 4.458 1.033 1.968 10.0 14.130 14.177 1.003 1.997 ∞ ∞ ∞ 1.000 2.000 Flow Past a Circular Cylinder with Circulation It is seen from Table 8.1 that as source strength m becomes large, the Rankine Oval becomes a large circle, much greater in diameter than the source-sink spacing 2a. Viewed, from the scale of the cylinder, this is equivalent to a uniform stream plus a doublet. To add circulation, without changing the shape of the cylinder, we place a vortex at the doublet center. For these conditions the stream function is given by VIII - 9 ψ = U∞sinθ r −a 2 r −K ln r a Typical resulting flows are shown in Fig. 8.10 for increasing values of non-dimensional vortex strength K / U∞a. Fig. 8.10 Flow past a cylinder with circulation for values of K / U∞a of (a) 0, (b) 1.0, (c) 2.0, and (d) 3.0 Again the streamline ψ = 0 is corresponds to the circle r = a. As the counter-clockwise circulation Γ = 2π K increases, velocities below the cylinder increase and velocities above the cylinder decrease. In polar coordinates, the velocity components are given by vr = 1 r ∂ψ ∂θ = U∞cosθ 1 −a 2 r 2 vθ = −∂ψ ∂r = −U∞sinθ 1 + a 2 r 2 + K r For small K, two stagnation points appear on the surface at angles θs or for which VIII - 10 sinθs = K 2U∞a Thus for K = 0, θs = 0 and 180o. For K / U∞a = 1, θs = 30 and 150o . Figure 8.10c is the limiting case for which with K / U∞a = 2, θs = 90o and the two stagnation points meet at the top of the cylinder. The Kutta-Joukowski Lift Theorem The development in the text shows that from inviscid flow theory, The lift per unit depth of any cylinder of any shape immersed in a uniform stream equals to ρU∞Γ where Γ is the total net circulation contained within the body shape. The direction of the lift is 90o from the stream direction, rotating opposite to the circulation. This is the well known Kutta-Joukowski lift theorem. IX-1 IX. Compressible Flow Compressible flow is the study of fluids flowing at speeds comparable to the local speed of sound. This occurs when fluid speeds are about 30% or more of the local acoustic velocity. Then, the fluid density no longer remains constant throughout the flow field. This typically does not occur with fluids but can easily occur in flowing gases.
Two important and distinctive effects that occur in compressible flows are (1) choking where the flow is limited by the sonic condition that occurs when the flow velocity becomes equal to the local acoustic velocity and (2) shock waves that introduce discontinuities in the fluid properties and are highly irreversible.
Since the density of the fluid is no longer constant in compressible flows, there are now four dependent variables to be determined throughout the flow field. These are pressure, temperature, density, and flow velocity. Two new variables, temperature and density, have been introduced and two additional equations are required for a complete solution. These are the energy equation and the fluid equation of state. These must be solved simultaneously with the continuity and momentum equations to determine all the flow field variables.
Equations of State and Ideal Gas Properties: Two equations of state are used to analyze compressible flows: the ideal gas equation of state and the isentropic flow equation of state. The first of these describe gases at low pressure (relative to the gas critical pressure) and high temperature (relative to the gas critical temperature). The second applies to ideal gases experiencing isentropic (adiabatic and frictionless) flow. The ideal gas equation of state is = P R T In this equation, R is the gas constant, and P and T are the absolute pressure and absolute temperature respectively. Air is the most commonly incurred compressible flow gas and its gas constant is Rair = 1717 ft2/(s2-oR) = 287 m2/(s2-K).
IX-2 Two additional useful ideal gas properties are the constant volume and constant pressure specific heats defined as Cv = d u d T and Cp = d h d T where u is the specific internal energy and h is the specific enthalpy. These two properties are treated as constants when analyzing elemental compressible flows.
Commonly used values of the specific heats of air are: cv = 4293 ft2/(s2-oR) = 718 m2/(s2-K) and cp = 6010 ft2/(s2-oR) = 1005 m2/(s2-K). Additional specific heat relationships are R = Cp −Cv and k = C p Cv The specific heat ratio k for air is 1.4.
When undergoing an isentropic process (constant entropy process), ideal gases obey the isentropic process equation of state: P k = constant Combining this equation of state with the ideal gas equation of state and applying the result to two different locations in a compressible flow field yields P2 P 1 = T 2 T 1 k/ k−1 ( ) = 2 1 k Note: The above equations may be applied to any ideal gas as it undergoes an isentropic process.
Acoustic Velocity and Mach Number The acoustic velocity (speed of sound) is the speed at which an infinitesimally small pressure wave (sound wave) propagates through a fluid. In general, the acoustic velocity is given by IX-3 a 2 = P The process experienced by the fluid as a sound wave passes through it is an isentropic process. The speed of sound in an ideal gas is then given by a = k RT The Mach number is the ratio of the fluid velocity and speed of sound, Ma = V a This number is the single most important parameter in understanding and analyzing compressible flows.
Mach Number Example: An aircraft flies at a speed of 400 m/s. What is this aircraft’s Mach number when flying at standard sea-level conditions (T = 289 K) and at standard 15,200 m (T = 217 K) atmosphere conditions?
At standard sea-level conditions, a = k RT = 1.4 ( ) 287 ( ) 289 ( ) = 341m /s and at 15,200 m, a = 1.4 ( ) 287 ( ) 217 ( ) = 295m /s. The aircraft Mach numbers are then sea −level : Ma = V a = 400 341 =1.17 15,200 m : Ma = V a = 400 295 =1.36 Note: Although the aircraft speed did not change, the Mach number did change because of the change in the local speed of sound.
IX-4 Ideal Gas Steady Isentropic Flow When the flow of an ideal gas is such that there is no heat transfer (i.e., adiabatic) or irreversible effects (e.g., friction, etc.), the flow is isentropic. The steady-flow energy equation applied between two points in the flow field becomes h1 + V 1 2 2 = h2 + V 2 2 2 = ho = constant where h0 is called the stagnation enthalpy that remains constant throughout the flow field. Observe that the stagnation enthalpy is the enthalpy at any point in an isentropic flow field where the fluid velocity is zero or very nearly so.
The enthalpy of an ideal gas is given by h = Cp T over reasonable ranges of temperature. When this is substituted into the adiabatic, steady-flow energy equation, we see that ho = Cp T o = constant and To T =1 + k −1 2 Ma 2 Thus, the stagnation temperature To remains constant throughout an isentropic or adiabatic flow field and the relationship of the local temperature to the field stagnation temperature only depends upon the local Mach number.
Incorporation of the acoustic velocity equation and the ideal gas equations of state into the energy equation yields the following useful results for steady isentropic flow of ideal gases.
To T =1 + k −1 2 Ma 2 ao a = T o T 1 / 2 = 1 + k−1 2 Ma2 1 / 2 P o P = To T k/ k−1 ( ) = 1 + k −1 2 Ma 2 k / k−1 ( ) o = To T 1/ k−1 ( ) = 1+ k −1 2 Ma2 1/ k−1 ( ) IX-5 The values of the ideal gas properties when the Mach number is 1 (i.e., sonic flow) are known as the critical or sonic properties and are given by To T = 1+ k −1 2 ao a = T o T 1 / 2 = 1 + k −1 2 1 / 2 P o P = To T k / k−1 ( ) = 1+ k −1 2 k / k−1 ( ) o = To T 1/ k−1 ( ) = 1 + k −1 2 1/ k−1 ( ) given by Isentropic Flow Example: Air flowing through an adiabatic, frictionless duct is supplied from a large supply tank in which P = 500 kPa and T = 400 K. What are the Mach number Ma. the temperature T, density _, and fluid V at a location in this duct where the pressure is 430 kPa?
The pressure and temperature in the supply tank are the stagnation pressure and temperature since the velocity in this tank is practically zero. Then, the Mach number at this location is Ma = 2 k −1 P o P k−1 ( )/ k −1 Ma = 2 0.4 500 430 0.4/1.4 −1 Ma = 0.469 and the temperature is given by IX-6 T = To 1 + k −1 2 Ma 2 T = 400 1 + 0.2 0.469 ( ) 2 T = 383K The ideal gas equation of state is used to determine the density, = P R T = 430,000 287 ( ) 383 ( ) = 3.91kg / m 3 Using the definition of the Mach number and the acoustic velocity, V = Ma k RT = 0.469 1.4 ( ) 287 ( ) 383 ( ) =184 m / s Solving Compressible Flow Problems Compressible flow problems come in a variety of forms, but the majority of them can be solved by 1. Use the appropriate equations and reference states (i.e., stagnation and sonic states) to determine the Mach number at all the flow field locations involved in the problem.
2. Determine which conditions are the same throughout the flow field (e.g., the stagnation properties are the same throughout an isentropic flow field).
3. Apply the appropriate equations and constant conditions to determine the necessary remaining properties in the flow field.
4. Apply additional relations (i.e., equation of state, acoustic velocity, etc.) to complete the solution of the problem.
Most compressible flow equations are expressed in terms of the Mach number.
You can solve these equations explicitly by rearranging the equation, by using tables, or by programming them with spreadsheet or EES software.
IX-7 Isentropic Flow with Area Changes All flows must satisfy the continuity and momentum relations as well as the energy and state equations. Application of the continuity and momentum equations to a differential flow (see textbook for derivation) yields: d V V = 1 Ma 2 −1 d A A This result reveals that when Ma < 1 (subsonic flow) velocity changes are the opposite of area changes. That is, increases in the fluid velocity require that the area decrease in the direction of the flow. For supersonic flow (Ma > 1), the area must increase in the direction of the flow to cause an increase in the velocity.
Changes in the fluid velocity dV can only be finite in sonic flows (Ma = 1) when dA = 0. The effect of the geometry upon velocity, Mach number, and pressure is illustrated in Figure 1 below.
Figure 1 Combining the mass flow rate equation ˙ m = A V = constant with the preceding isentropic flow equations yields = 2 k +1 1+ k −1 2 Ma 2 1 / k−1 ( ) IX-8 V V = 1 Ma 2 k +1 1 + k −1 2 Ma 2 1 / 2 A A = 1 Ma 1 + 0.5 k −1 ( )Ma 2 0.5 k +1 ( ) k+1 ( )/ 2 k−1 ( ) [ ] where the sonic state (denoted with ) may or may not occur in the duct. If the sonic condition does occur in the duct, it will occur at the duct minimum or maximum area. If the sonic condition occurs, the flow is said to be choked since the mass flow rate ˙ m = A V = A V is the maximum mass flow rate the duct can accommodate without a modification of the duct geometry.
Review Example 9.4 of the textbook.
Normal Shock Waves Under the appropriate conditions, very thin, highly irreversible discontinuities can occur in otherwise isentropic compressible flows. These discontinuities are known as shock waves which when they are perpendicular to the flow velocity vector are called normal shock waves. A normal shock wave in a one-dimensional flow channel is illustrated in Figure 2.
Figure 2 IX-9 Application of the second law of thermodynamics to the thin, adiabatic normal shock wave reveals that normal shock waves can only cause a sharp rise in the gas pressure and must be supersonic upstream and subsonic downstream of the normal shock. Rarefaction waves that result in a decrease in pressure and increase in Mach number are impossible according to the second law.
Application of the conservation of mass, momentum, and energy equations along with the ideal gas equation of state to a thin, adiabatic control volume surrounding a normal shock wave yields the following results.
Ma2 2 = k −1 ( )Ma1 2 + 2 2k Ma1 2 −k −1 ( ) , Ma1 >1 P 2 P 1 = 1 + k Ma1 2 1 + k Ma2 2 2 1 = V 1 V2 = k +1 ( ) Ma 1 2 k −1 ( ) Ma 1 2 + 2 To1 = T o 2 T2 T 1 = 2 + k −1 ( )Ma1 2 [ ] 2k Ma 1 2 − k −1 ( ) k +1 ( ) 2Ma1 2 P o2 P o1 = o2 o1 = k +1 ( ) Ma1 2 2 + k−1 ( )Ma1 2 k / k−1 ( ) k + 1 2k Ma 1 2 −k −1 ( ) 1/ k−1 ( ) A2 A 1 = Ma2 Ma1 2 + k −1 ( )Ma1 2 2 + k −1 ( )Ma2 2 k+1 ( )/ 2 k−1 ( ) [ ] When using these equations to relate conditions upstream and downstream of a normal shock wave, keep the following points in mind: IX-10 1. Upstream Mach numbers are always supersonic while downstream Mach numbers are subsonic.
2. Stagnation pressures and densities decrease as one moves downstream across a normal shock wave while the stagnation temperature remains constant.
3. Pressures increase greatly while temperature and density increase moderately across a shock wave in the downstream direction.
4. The effective throat area increases across a normal shock wave in the downstream direction.
5. Shock waves are very irreversible causing the specific entropy downstream of the shock wave to be greater than the specific entropy upstream of the shock wave.
Moving normal shock waves such as those caused by explosions, spacecraft reentering the atmosphere, and others can be analyzed as stationary normal shock waves by using a frame of reference that moves at the speed of the shock wave in the direction of the shock wave.
Converging-Diverging Nozzle Example: Also see Example 9.6 of textbook Air is supplied to the converging-diverging nozzle shown here from a large tank where P = 2 Mpa and T = 400 K. A normal shock wave in the diverging section of this nozzle forms at a point P o1 = P o2 = 2 MPa where the upstream Mach number is 1.4. The ratio of the nozzle exit area to the throat area is 1.6. Determine (a) the Mach number downstream of the shock wave, (b) the Mach number at the nozzle exit, (c) the pressure at the nozzle exit, and (d) the temperature at the nozzle exit.
This flow is isentropic from the supply tank (1) to just upstream of the normal shock (2) and also from just downstream of the shock (3) to the exit (4). Stagnation temperatures do not change in isentropic flows or across shock waves, To1 = T o 2 = To3 = To4 = 400 K . Stagnation pressures do not change in isentropic flows, P o1 = P o2 = 2 MPa and P o3 = P o4 , but stagnation pressures change across shocks, P o2 > P o3 .
IX-11 Based upon the Mach number at 2 and the isentropic relations, A2 At = A3 A t = A 2 At = 1 Ma2 1+ 0.2 Ma2 2 ( ) 3 1.728 =1.115 The normal shock relations can be used to work across the shock itself. The answer to (a) is then: Ma3 = k −1 ( )Ma2 2 + 2 2k Ma2 2 −k −1 ( ) 1 / 2 = 0.4 ( ) 1.4 ( ) 2 + 2 2 1.4 ( ) 1.4 ( ) 2 −0.4 1 / 2 = 0.740 Continuing to work across the shock, P o4 = P o3 = P o2 k + 1 ( )Ma2 2 2 + k −1 ( ) Ma2 2 k/ k−1 ( ) k + 1 2 k Ma2 2 −k −1 ( ) 1/ k−1 ( ) P o4 = P o3 = 2 2.4 ( ) 0.74 ( ) 2 2 + 0.4 ( ) 0.74 ( ) 2 3.5 2.4 2 1.4 ( ) 0.74 ( ) 2 −0.4 2.5 = 1.92 MPa A3 A2 = Ma3 Ma2 2 + k −1 ( )Ma2 2 2 + k −1 ( )Ma3 2 k+1 ( )/ 2 k−1 ( ) [ ] =1.044 Now, we know A4/At, and the flow is again isentropic between states 3 and 4.
Writing an expression for the area ratio between the exit and the throat, we have A4 At = 1.6 = A4 A4 A4 A3 A3 A2 A2 A t = A4 A4 1 ( ) 1.044 ( ) 1.115 ( ) Solving for A4 A4 we obtain A4 A4 =1.374 Using a previously developed equation for choked, isentropic flow, we can write IX-12 A4 A4 =1.374 = 1 Ma 1 + 0.5 k −1 ( )Ma 2 0.5 k +1 ( ) k+1 ( )/ 2 k−1 ( ) [ ] or 1.374 = 1 Ma4 1+ 0.2Ma4 2 ( ) 3 1.728 The solution of this equation gives answer (b) Ma4 = 0.483.
Now that the Mach number at 4 is known, we can proceed to apply the isentropic relations to obtain answers (c) and (d).
P 4 = P o4 1 + 0.5 k −1 ( )Ma4 2 [ ] k/ k−1 ( ) = 1.92 MPa 1 + 0.2 0.483 ( ) 2 [ ] 3.5 =1.637MPa T4 = T o4 1 + 0.5 k −1 ( )Ma4 2 = 400 K 1 + 0.2 0.483 ( ) 2 = 382 K Note: Observe how the sonic area downstream from the shock is not the same as upstream of the shock. Also, observe the use of the area ratios to determine the Mach number at the nozzle exit.
The following steps can be used to solve most one-dimensional compressible flow problems.
1. Clearly identify the flow conditions:e.g., isentropic flow, constant stagnation temperature, constant stagnation pressure, etc.
2. Use the flow condition relationships, tables, or software to determine the Mach number at major locations in the flow field.
3. Once the Mach number is known at the principal flow locations, one can proceed to use the flow relations, tables, or software to determine other flow properties such as fluid velocity, pressure, and temperature. This may require the reduction of property ratios to the product of several ratios, as was done with the area ratio in the above example to obtain the answer.
IX-13 Operation of Converging-Diverging Nozzles A converging-diverging nozzle like that shown in Figure 3 can operate in several different modes depending upon the ratio of the discharge and supply pressure Pd/Ps. These modes of operation are illustrated on the pressure ratio – axial position diagram of Figure 3.
Figure 3 Mode (a) The flow is subsonic throughout the nozzle, supply, and discharge chambers. Without friction, this flow is also isentropic and the isentropic flow equations may be used throughout the nozzle.
Mode (b) The flow is still subsonic and isentropic throughout the nozzle and chambers. The Mach number at the nozzle throat is now unity.
At the throat, the flow is sonic, the throat is choked, and the mass flow rate through the nozzle has reached its upper limit. Further reductions in the discharge tank pressure will not increase the mass flow rate any further.
Mode (c) A shock wave has now formed in the diverging section of the nozzle. The flow is subsonic before the throat, same as mode (b), the throat is choked, same as mode (b), and the flow is supersonic IX-14 and accelerating between the throat and just upstream of the shock.
The flow is isentropic between the supply tank and just upstream of the shock. The flow downstream of the shock is subsonic and decelerating. The flow is also isentropic downstream of the shock to the discharge tank. The flow is not isentropic across the shock.
Isentropic flow methods can be applied upstream and downstream of the shock while normal shock methods are used to relate conditions upstream to those downstream of the shock.
Mode (d) The normal shock is now located at the nozzle exit. Isentropic flow now exists throughout the nozzle. The flow at the nozzle exit is subsonic and adjusts to flow conditions in the discharge tank, not the nozzle. Isentropic flow methods can be applied throughout the nozzle.
Mode (e) A series of two-dimensional shocks are established in the discharge tank downstream of the nozzle. These shocks serve to decelerate the flow. The flow is isentropic throughout the nozzle, same as mode (d).
Mode (f) The pressure in the discharge tank equals the pressure predicted by the supersonic solution of the nozzle isentropic flow equations. The pressure ratio is known as the supersonic design pressure ratio.
Flow is isentropic everywhere in the nozzle, same as mode (d) and (e), and in the discharge tank.
Mode (g) A series of two-dimensional shocks are established in the discharge tank downstream of the nozzle. These shocks serve to decelerate the flow. The flow is isentropic throughout the nozzle, same as modes (d), (e), and (f).
Review Example 9.9 of the textbook.
IX-15 Adiabatic, Constant Duct Area Compressible Flow with Friction When compressible fluids flow through insulated, constant-area ducts, they are subject to Moody-like pipe-friction which can be described by an average Darcy-Weisbach friction factor f . Application of the conservation of mass, momentum, and energy principles as well as the ideal gas equation of state yields the following set of working equations.
f L D = 1−Ma2 k Ma 2 + k +1 2k ln k+ 1 ( )Ma 2 2 + k −1 ( ) Ma 2 P P = 1 Ma k +1 ( ) 2 + k −1 ( ) Ma2 1 / 2 = V V = 1 Ma 2 + k −1 ( )Ma 2 k +1 1 / 2 T T = a a2 = k +1 ( ) 2 + k −1 ( )Ma2 P P o = o o = 1 Ma 2 + k −1 ( ) Ma2 k +1 k+1 ( )/ 2 k−1 ( ) [ ] where the asterisk state is the sonic state at which the flow Mach number is one.
This state is the same throughout the duct and may be used to relate conditions at one location in the duct to those at another location. The length of the duct enters these calculations by f ∆L D = f L D 1 − f L D 2 Thus, given the length ∆L of the duct and the Mach number at the duct entrance or exit, the Mach number at the other end (or location) of the duct can be determined.
IX-16 Compressible Flow with Friction Example: Air enters a 0.01-m-diameter duct ( f = 0.05) with Ma = 0.05. The pressure and temperature at the duct inlet are 1.5 MPa and 400 K. What are the (a) Mach number, (b) pressure, and (c) temperature in the duct 50 m from the entrance?
At the duct entrance, with f = 0.05, D = 0.01 m, and Ma = 0.05, we obtain f L D 1 = 1−Ma 2 k Ma 2 + k +1 2k ln k +1 ( ) Ma 2 2 + k −1 ( ) Ma 2 1 f L D 1 = 1−0.05 2 1.4 0.05 ( ) 2 + 2.4 2.8 ln 2.4 ( )0.05 2 2 + 0.4 ( )0.05 2 1 = 280 Then, at the duct exit we obtain f L D 2 = f L D 1 −f ∆L D = 280 −0.05 ( )50 0.01 = 30 We can not write for the duct exit that f L D 2 = 30 = 1−Ma 2 k Ma 2 + k +1 2k ln k+ 1 ( ) Ma 2 2 + k −1 ( ) Ma 2 2 or 30 = 1−Ma2 2 1.4 Ma2 2 + 2.4 2.8 ln 2.4 Ma2 2 2 + 0.4 Ma2 2 The solution of the second of these equations gives answer (a) Ma2 = 0.145.
Writing the following expression for pressure ratios yields for (b), P 2 = P 1 P 2 P 2 P 2 P 1 P 1 P 1 IX-17 P 2 = 1.5 ( ) 1 Ma2 k +1 ( ) 2 + k −1 ( ) Ma2 2 1 / 2 1 ( )Ma1 1 2 + k −1 ( ) Ma 1 2 k +1 1 / 2 P 2 = 1.5 ( ) 1 0.145 2.4 2+ 0.4 ( )0.145 2 1 / 2 1 ( ) 0.05 1 2 + 0.4 ( )0.05 2 2.4 1 / 2 = 0.516 Application of the temperature ratios yields answer (c), T 2 = T 1 T1 T 1 T 2 T 1 T 2 T2 = 400 2 + k −1 ( )Ma1 2 2 + k −1 ( )Ma2 2 = 400 2 + 0.4 ( )0.05 2 2 + 0.4 ( )0.145 2 = 399 This example demonstrates what happens when the flow at the inlet to the duct is subsonic, the Mach number increases as the duct gets longer. When the inlet flow is supersonic, the Mach number decreases as the duct gets longer. A plot of the specific entropy of the fluid as a function of the duct Mach number (length) is presented in Figure 4 for both subsonic and supersonic flow.
Figure 4 These results clearly illustrate that the Mach number in the duct approaches unity as the length of the duct is increased. Once the sonic condition exists at the duct exit, the flow becomes choked. This figure also demonstrates that the flow can never proceed from subsonic to supersonic (or supersonic to subsonic) flow, as this would result in a violation of the second law of thermodynamics.
IX-18 Other compressible flows in constant area ducts such as isothermal flow with friction and frictionless flow with heat addition may be analyzed in a similar manner using the equations appropriate to each flow. Many of these flows also demonstrate choking behavior.
Oblique Shock Waves Bodies moving through a compressible fluid at speeds exceeding the speed of sound create a shock system shaped like a cone. The half-angle of this shock cone is given by = sin −1 1 Ma This angle is known as the Mach angle. The interior of the shock cone is called the zone of action. Inside the zone of action, it is possible to hear any sounds produced by the moving body. Outside the Mach cone, in what is known as the zone of silence, sounds produced by the moving body cannot be heard.
An oblique shock wave at angle with respect to the approaching compressible fluid whose Mach number is supersonic is shown in Figure 5. Observe that the streamlines (parallel to the velocity vector) have been turned by the deflection angle by passing through the oblique shock wave.
Figure 5 IX-19 This flow is readily analyzed by considering the normal velocity components V n1 = V 1sin and Vn2 = V 2 sin − ( ) and the tangential components V t1 and Vt2. Application of the momentum principle in the tangential direction (along which there are no pressure changes) verifies that 12 = tt VV V t1 = Vt2 By defining the normal Mach numbers as Man1 = Vn1 a1 = Ma1sin and Man2 = V n 2 a2 = Ma2 sin − ( ) The simultaneous solution of the conservation of mass, momentum, and energy equations in the normal direction along with the ideal gas equation of state are the same as those of the normal shock wave with Ma1 replaced with Man1 and Ma2 replaced with Man2. In this way, all the results developed in the normal shock wave section can be applied to two-dimensional oblique shock waves.
Oblique Shock Example: A two-dimensional shock wave is created at the leading edge of an aircraft flying at Ma = 1.6 through air at 70 kPa, 300 K. If this oblique shock forms a 55o angle with respect to the approaching air, what is (a) the Mach number of the flow after the oblique shock (this is not the normal Mach number) and (b) the streamline deflection angle ?
The velocity of the fluid upstream of the oblique shock wave is V 1 = Ma1 a1 = Ma1 k R T = 1.6 1.4 ( ) 287 ( ) 300 ( ) = 556m / s whose components are V n1 = V1sin = 556sin55 = 455m / s V t1 = Vt2 = V 1 cos = 556cos55 = 319m / s IX-20 The upstream normal Mach number is then Man1 = Ma1sin =1.6sin55 = 1.311 and the downstream normal Mach number is Man2 = k −1 ( )Man1 2 + 2 2k Man1 2 −k −1 ( ) 1 / 2 = 0.4 ( ) 1.311 ( ) 2 + 2 2 1.4 ( ) 1.311 ( ) 2 −0.4 1 / 2 = 0.780 and the downstream temperature is T 2 = T 1 k −1 ( )Man1 2 + 2 [ ]2 k Man1 2 −k −1 ( ) k +1 ( ) 2 Man1 2 T 2 = 300 0.4 ( )1.311 2 + 2 [ ]2 1.4 ( )1.311 2 −0.4 2.4 ( ) 21.311 2 = 359 K Now, the downstream normal velocity is V n2 = Man2 a2 = Man2 k R T 2 = 0.780 1.4 ( ) 287 ( ) 359 ( ) = 296m /s and the downstream fluid velocity is V 2 = V n2 2 + V t 2 2 = 296 2 + 319 2 = 435m / s and the downstream Mach number is Ma2 = V2 a2 = 435 1.4 ( ) 287 ( ) 359 ( ) =1.15 According to the geometry of Figure 5, IX-21 = −tan −1 Vn2 V t2 = 55 −tan −1 296 319 =12.1 Other downstream properties can be calculated in the same way as the downstream temperature by using the normal Mach numbers in the normal shock relations.
Prandl-Meyer Expansion Waves The preceding section demonstrated that when the streamlines of a supersonic flow are turned into the direction of the flow an oblique compression shock wave is formed. Similarly, when the streamlines of a supersonic flow are turned away from the direction of flow as illustrated in Figure 6, an expansion wave system is established. Unlike shock waves (either normal or oblique) which form a strong discontinuity to change the flow conditions, expansion waves are a system of infinitesimally weak waves distributed in such a manner as required to make the required changes in the flow conditions.
Figure 6 The Mach waves that accomplish the turning of supersonic flows form an angle with respect to the local flow velocity equal to the Mach angle = sin −1 1 /Ma ( ) and are isentropic. Application of the governing conservation equations and equation of state to an infinitesimal turning of the supersonic flow yields IX-22 − Ma ( ) = Ma ( ) = k +1 k −1 1 / 2 tan −1 k −1 ( ) Ma 2 −1 ( ) k +1 1 / 2 −tan −1 Ma 2 −1 ( ) 1 / 2 where Ma ( ) is the Prandl-Meyer expansion function. The overall change in the flow angle as a supersonic flow undergoes a Prandl-Meyer expansion is then ∆ = Ma1 ( ) − Ma2 ( ) where 1 refers to the upstream condition and 2 refers to the downstream condition.
The flow through a Prandl-Meyer expansion fan is isentropic flow. The isentropic flow equations can then be used to relate the fluid properties upstream and downstream of the expansion fan.
Example: Air at 80 kPa, 300 K with a Mach number of 1.5 turns the sharp corner of an airfoil as shown here. Determine the angles of the initial and final Mach waves, and the downstream pressure and temperature of this flow.
The initial angle between the flow velocity vector and the Prandtl-Meyer fan is the Mach angle.
1 = sin −1 1 Ma 1 = sin −1 1 1.5 = 41.8 0 IX-23 The upstream Prandtl-Meyer function is Ma1 ( ) = k +1 k −1 1 / 2 tan −1 k −1 ( ) Ma1 2 −1 ( ) k +1 1 / 2 −tan −1 Ma1 2 −1 ( ) 1 / 2 Ma1 ( ) = 2.4 0.4 1 / 2 tan −1 0.4 ( ) 1.5 2 −1 ( ) 2.4 1 / 2 −tan −1 1.5 2 −1 ( ) 1 / 2 Ma1 ( ) =11.90 0 The downstream Prandtl-Meyer function is then Ma2 ( ) = Ma 1 ( ) −∆= 11.9 0 −10 0 =1.90 0 Solving the Prandtl-Meyer function gives the downstream Mach number Ma2 =1.13. The downstream Mach angle is then 2 = 62.2 0 . According to the geometry of the above figure, 2 = 2 −∆ = 62.2 0 −10 0 = 52.2 0 Since T0 and P0 remain constant, the isentropic flow relations yield T 2 = T 1 T01 T 1 T2 T02 = T 1 1+ k −1 2 Ma1 2 1+ k −1 2 Ma2 2 = 300 1+ 0.2 1.5 ( ) 2 1+ 0.2 1.13 ( ) = 346K P 2 = P 1 P 01 P 1 P 2 P 02 = P 1 1+ k −1 2 Ma1 2 1+ k −1 2 Ma2 2 k/ k−1 ( ) = 80 1+ 0.2 1.5 ( ) 2 1+ 0.2 1.13 ( ) 3.5 =132 MPa Students are encouraged to examine the flow visualization photographs in Ch 9. X-1 Ch. 10 Open-Channel Flow Previous internal flow analyses have considered only closed conduits where the fluid typically fills the entire conduit and may be either a liquid or a gas. This chapter considers only partially filled channels of liquid flow referred to as open-channel flow. Open-Channel Flow: Flow of a liquid in a conduit with a free surface. Open-channel flow analysis basically results in the balance of gravity and friction forces. One Dimensional Approximation While open-channel flow can, in general, be very complex ( three dimensional and transient), one common approximation in basic analyses is the One-D Approximation: The flow at any local cross section can be treated as uniform and at most varies only in the principal flow direction. This results in the following equations. Conservation of Mass (for ρ = constant) Q = V(x) A(x) = constant Energy Equation V 1 2 2 g + Z1 = V 2 2 2g + Z2 + hf X-2 The equation in this form is written between two points ( 1 – 2 ) on the free surface of the flow. Note that along the free surface, the pressure is a constant, is equal to local atmospheric pressure, and does not contribute to the analysis with the energy equation. The friction head loss hf is analogous to the head loss term in duct flow, Ch. VI, and can be represented by h f = f x2 −x1 D h Vavg 2 2g where P = wetted perimeter Dh = hydraulic diameter = 4A P Note: One of the most commonly used formulas uses the hydraulic radius: Rh = 1 4Dh = A P Flow Classification by Depth Variation The most common classification method is by rate of change of free-surface depth. The classes are summarized as 1. Uniform flow (constant depth and slope) 2. Varied flow a. Gradually varied (one-dimensional) b. Rapidly varied (multidimensional) Flow Classification by Froude Number: Surface Wave Speed A second classification method is by the dimensionless Froude number, which is a dimensionless surface wave speed. For a rectangular or very wide channel we have Fr = V co = V gy ( ) 1/2 where y is the water depth and co = the speed of a surface wave as the wave height approaches zero. There are three flow regimes of incompressible flow. These have analogous flow regimes in compressible flow as shown below: X-3 Incompressible Flow Compressible Flow Fr < 1 subcritical flow Fr = 1 critical flow Fr > 1 supercritical flow Ma < 1 subsonic flow Ma = 1 sonic flow Ma > 1 supersonic flow Hydraulic Jump Analogous to a normal shock in compressible flow, a hydraulic jump provides a mechanism by which an incompressible flow, once having accelerated to the supercritical regime, can return to subcritical flow. This is illustrated by the following figure. Fig. 10.5 Flow under a sluice gate accelerates from subcritical to critical to supercritical and then jumps back to subcritical flow. The critical depth yc = Q b 2g 1/3 is an important parameter in open-channel flow and is used to determine the local flow regime (Sec. 10.4). Uniform Flow; the Chezy Formula Uniform flow 1. Occurs in long straight runs of constant slope 2. The velocity is constant with V = Vo 3. Slope is constant with So = tan θ X-4 From the energy equation with V1 = V2 = Vo, we have h f = Z1 −Z2 = SoL Since the flow is fully developed, we can write from Ch. VI h f = f L Dh V o 2 2g and Vo = 8g f 1/2 Rh 1/2So 1/2 For fully developed, uniform flow, the quantity 8g f 1/2 is a constant and can be denoted by C. The equations for velocity and flow rate thus become Vo = C Rh 1/2So 1/2 and Q = CA Rh 1/2 So 1/2 The quantity C is called the Chezy coefficient, and varies from 60 ft1/2/s for small rough channels to 160 ft1/2/s for large rough channels (30 to 90 m1/2/s in SI). The Manning Roughness Correlation The friction factor f in the Chezy equations can be obtained from the Moody chart of Ch. VI. However, since most flows can be considered fully rough, it is appropriate to use Eqn 6.64: fully rough flow: f ≈2.0log 3.7Dh ε −2 However, most engineers use a simple correlation by Robert Manning: S.I. Units Vo m/s ( )≈α n Rh m ( ) [ ] 2/3 So 1/2 B.G. Units Vo ft/s ( )≈α n Rh ft ( ) [ ] 2/3 So 1/2 X-5 where n is a roughness parameter given in Table 10.1 and is the same in both systems of units and α is a dimensional constant equal to 1.0 in S.I. units and 1.486 in B.G. units. The volume flow rate is then given by Uniform flow Q = Vo A ≈α n A Rh 2/3 So 1/2 Table 10.1 Experimental Values for Manning’s n Factor X-6 Example 10.1 Given: Rectangular channel, finished concrete, slope = 0.5˚ water depth: y = 4 ft, width: b = 8 ft Find: Volume flow rate (ft3/s) 4 ft 8 ft cross-section θ For the given conditions: n = 0.012 So = tan 0.5˚ = 0.0873 A = b y = (8 ft) (4 ft) = 32 ft2 P = b + 2 y = 8 + 2 (4) = 16 ft Rh = A P = 32 ft 2 16ft = 2 ft D h = 4Rh = 8ft Using Manning’s formula in BG units, we obtain for the flow rate Q ≈1.486 0.012 32ft 2 ( ) 2ft ( ) 2/3 0.00873 ( ) 1.2 ≈590 ft 3/s ans. Alternative Problem The previous uniform problem can also be formulated where the volume flow rate Q is given and the fluid depth is unknown. For these conditions, the same basic equations are used and the area A and hydraulic radius Rh are expressed in terms of the unknown water depth yn. The solution is then obtained using iterative or systematic trial and error techniques that are available in several math analysis/ math solver packages such as EES (provided with the text) or Mathcad ®. X-7 Uniform Flow in a Partly Full, Circular Pipe Fig. 10.6 shows a partly full, circular pipe with uniform flow. Since frictional resistance increases with wetted perimeter, but volume flow rate increases with cross sectional flow area, the maximum velocity and flow rate occur before the pipe is completely full. For this condition, the geometric properties of the flow are given by the equations below. Fig. 10.6 Uniform Flow in a Partly Full, Circular Channel A = R 2 θ −sin2θ 2 P = 2Rθ Rh = R 2 1−sin2θ 2θ The previous Manning formulas are used to predict Vo and Q for uniform flow when the above expressions are substituted for A, P, and Rh. Vo ≈α n R 2 1−sin2θ 2θ 2 /3 So 1/2 Q = Vo R 2 θ −sin2θ 2 These equations have respective maxima for Vo and Q given by X-8 Vmax = 0.718α n R 2/3 So 1/2 at θ =128.73Þ and y = 0.813D Q max = 2.129α n R 8/3 So 1/2 at θ = 151.21Þ and y = 0.938D Efficient Uniform Flow Channels A common problem in channel flow is that of finding the most efficient low-resistance sections for given conditions. This is typically obtained by maximizing Rh for a given area and flow rate. This is the same as minimizing the wetted perimeter. Note: Minimizing the wetted perimeter for a given flow should minimize the frictional pressure drop per unit length for a given flow. It is shown in the text that for constant value of area A and α = cot θ, the minimum value of wetted perimeter is obtained for A = y 2 2 1+α 2 ( ) 1/ 2 −α [ ] P = 4 y 1+α 2 ( ) 1/ 2 −2α y Rh = 1 2 y Note: For any trapezoid angle, the most efficient cross section occurs when the hydraulic radius is one-half the depth. For the special case of a rectangle (a = 0, q = 90˚), the most efficient cross section occurs with A = 2y 2 P = 4 y Rh = 1 2 y b = 2y X-9 Best Trapezoid Angle The general equations listed previously are valid for any value of α. For a given, fixed value of area A and depth y the best trapezoid angle is given by 1/2 1 = cot = 3 or α θ θ α θ θ α θ θ α θ θ = = = = 60o Example 10.3 What are the best dimensions for a rectangular brick channel designed to carry 5 m3/s of water in uniform flow with So = 0.001? Taking n = 0.015 from Table 10.1, A = 2 y2 , and Rh = 1/2 y ; Manning’s formula is written as Q ≈1.0 n ARh 2/3 So 1/2 or 5m 3 / s = 1.0 0.015 2 y 2 ( ) 1 2 y 2 / 3 0.001 ( ) 1/ 2 This can be solved to obtain y 8/ 3 =1.882 m 8 / 3 or y = 1.27m The corresponding area and width are A = 2y 2 = 3.21m 2 and b = A y = 2.53m Note: The text compares these results with those for two other geometries having the same area. X-10 Specific Energy: Critical Depth One useful parameter in channel flow is the specific energy E, where y is the local water depth. Defining a flow per unit channel width as q = Q/b we write E = y + V 2 2 g E = y + q 2 2g y 2 Fig. 10.8b is a plot of the water depth y vs. the specific energy E. The water depth for which E is a minimum is referred to as the critical depth yc. Fig. 10.8 Specific Energy Illustration Emin occurs at y = yc = q 2 g 1/3 = Q 2 b 2 g 1/3 The value of Emin is given by Emin = 3 2 yc At this value of minimum energy and minimum depth we can write Vc = gyc ( ) 1/2 = Co and Fr =1 X-11 Depending on the value of Emin and V, one of several flow conditions can exist. For a given flow, if E < Emin E = Emin E > Emin , V < Vc E > Emin , V > Vc No solution is possible Flow is critical, y = yc, V = Vc Flow is subcritical, y > yc ,disturbances can propagate upstream as well as downstream Flow is supercritical, y < yc , disturbances can only propagate downstream within a wave angle given by µ = sin -1 Co V = sin -1 gy ( ) 1/ 2 V Nonrectangular Channels For flows where the local channel width varies with depth y, critical values can be expressed as A c = bo Q 2 g 1/3 and Vc = Q A c = g Ac bo 1/2 where bo = channel width at the free surface. These equations must be solved iteratively to determine the critical area Ac and critical velocity Vc. For critical channel flow that is also moving with constant depth (yc), the slope corresponds to a critical slope Sc given by X-12 Sc = n 2 gA c α 2 bo Rh,c and α = 1. for S I units and 2.208 for B. G. units Example 10.5 Given: a 50˚, triangular channel has a flow rate of Q = 16 m3/s. Compute: (a) yc, (b) Vc, (c) Sc for n = 0.018 a. For the given geometry, we have P = 2 ( y csc 50˚) A = 2[ y (1/2 y cot 50 ˚)] Rh = A/P = y/2 cos 50˚ bo = 2 ( y cot 50˚) For critical flow, we can write g A c 3 = bo Q 2 or g yc 2 cot50Þ ( )= 2 yc cot 50Þ ( )Q 2 yc = 2.37 m ans. b. With yc, we compute Pc = 6.18 m Ac = 4.70 m2 bo,c = 3.97 m The critical velocity is now V c = Q A = 16m 3 / s 4.70m = 3.41 m / s ans. c. With n = 0.018, we compute the critical slope as Sc = g n 2 P α 2 bo Rh 1/3 = 9.81 0.018 ( ) 2 6.18 ( ) 1.0 2 3.97 ( ) 0.76 ( ) 1 /3 = 0.0542 X-13 Frictionless Flow over a Bump Frictionless flow over a bump provides a second interesting analogy, that of compressible gas flow in a nozzle. The flow can either increase or decrease in depth depending on whether the initial flow is subcritical or supercritical. The height of the bump can also change the results of the downstream flow. Fig. 10.9 Frictionless, 2-D flow over a bump Writing the continuity and energy equations for two dimensional, frictionless flow between sections 1 and 2 in Fig. 10.10, we have V 1 y1 = V 2 y2 and V 1 2 2g + y1 = V2 2 2g + y2 + ∆h Eliminating V2, we obtain y2 3 −E2 y2 2 + V 1 2 y1 2 2 g = 0 where E2 = V 1 2 2g + y1 −∆h The problem has the following solutions depending on the initial flow condition and the height of the jump: X-14 Key Points: 1. The specific energy E2 is exactly ∆h less than the approach energy E1. 2. Point 2 will lie on the same leg of the curve as point 1. 3. For Fr < 1, subcritical approach 4. For Fr > 1, supercritical approach 5. For bump height equal to ∆hmax = E1 - Ec 6. For ∆h > ∆hmax The water level will decrease at the bump. Flow at point 2 will be subcritical. The water level will increase at the bump. Flow at point 2 will be supercritical. Flow at the crest will be exactly critical (Fr =1). No physically correct, frictionless solutions are possible. Instead, the channel will choke and typically result in a hydraulic jump. Flow under a Sluice Gate A sluice gate is a bottom opening in a wall as shown below in Fig. 10.10a. For free discharge through the gap, the flow smoothly accelerates to critical flow near the gap and the supercritical flow downstream. Fig. 10.10 Flow under a sluice gate This is analogous to the compressible flow through a converging-diverging nozzle. For a free discharge, we can neglect friction. Since this flow has no bump (∆h = 0) and E1 = E2, we can write X-15 y2 3 − V 1 2 2g + y1 y2 2 + V 1 2 y1 2 2g = 0 This equation has the following possible solutions. Subcritical upstream flow and low to moderate tailwater (downstream water level) Subcritical upstream flow and high tailwater One positive, real solution. Supercritical flow at y2 with the same specific energy E2 = E1. Flow rate varies as y2/y1. Maximum flow is obtained for y2/y1 = 2/3. The sluice gate is drowned or partially drowned (analogous to a choked condition in compressible flow). Energy dissipation will occur downstream in the form of a hydraulic jump and the flow downstream will be subcritical. The Hydraulic Jump The hydraulic jump is an irreversible, frictional dissipation of energy which provides a mechanism for super-critical flow to transition (jump) to subcritical flow analogous to a normal shock in compressible flow. The development of the theory is equivalent to that for a strong fixed wave (Sec 10.1) and is summarized for a hydraulic jump in the following section. X-16 Theory for a Hydraulic Jump If we apply the continuity and momentum equations between points 1 and 2 across a hydraulic jump, we obtain 2 y2 y1 = −1 + 1+ 8 Fr 1 2 ( ) 1 /2 which can be solved for y2. We obtain V2 from continuity: V 2 = V 1 y1 y2 The dissipation head loss is obtained from the energy equation as hf = E1 −E2 = V 1 2 2 g + y1 − V 2 2 2g + y2 or hf = y2 −y1 ( ) 3 4 y1 y2 Key points: 1. Since the dissipation loss must be positive, y2 must be > y1. 2. The initial Froude number Fr1 must be > 1 (supercritical flow). 3. The downstream flow must be subcritical and V2 < V1. Example 10.7 Water flows in a wide channel at q = 10 m3/(s m) and y1 = 1.25 m. If the flow undergoes a hydraulic jump, compute: (a) y2, (b) V2, (c) Fr2, (d) hf, (e) the percentage dissipation, (f) power dissipated/unit width, and (g) temperature rise. X-17 a. The upstream velocity is V 1 = q y1 = 10 m3 / s⋅m ( ) 1.25m = 8.0 m / s The upstream Froude number is Fr 1 = V 1 g y1 ( ) 1/ 2 = 8.0 9.81 1.25 ( ) [ ] 1/ 2 = 2.285 This is a weak jump and y2 is given by 2 y2 y1 = −1 + 1+ 8 2.285 ( ) 2 ( ) 1/ 2 = 5.54 and y2 =1/ 2 y1 5.54 ( ) = 3.46 m b. The downstream velocity is V 2 = V 1 y1 y2 = 8.0 1.25 ( ) 3.46 = 2.89m / s c. The downstream Froude number is Fr 2 = V2 g y2 ( ) 1/ 2 = 2.89 9.81 3.46 ( ) [ ] 1/ 2 = 0.496 and Fr2 is subcritical as expected. d. The dissipation loss is given by hf = y2 −y1 ( ) 3 4 y1 y2 = 3.46 −1.25 ( ) 3 4 3.46 ( ) 1.25 ( ) = 0.625m e. The percentage dissipation is the ratio of hf/E1. E1 = V 1 2 2g + y1 =1.25 + 8.0 2 2 9.81 ( ) = 4.51m X-18 The percentage loss is thus given by % Loss = hf E1 100 = 0.625 4.51 100 = 14% f. The power dissipated per unit width is Power = ρ Q g hf = 9800 M/m310 m3/(s m) 0.625 m = 61.3 kw/m g. Using Cp = 4200 J/kg K, the temperature rise is given by p Power dissipated mC T = ∆ = ∆ = ∆ = ∆ & or 61,300 W/m = 10,000 kg/s m 4200 J/kg K ∆T ∆T = 0.0015˚K negligible temperature rise XI - 1 XI. Turbomachinery This chapter considers the theory and performance characteristics of the mechanical devices associated with the fluid circulation. General Classification: Turbomachine - A device which adds or extracts energy from a fluid. Adds energy: Pump Extracts energy: Turbine In this context, a pump is a generic classification that includes any device that adds energy to a fluid, e.g. fans, blowers, compressors. We can classify pumps by operating concept: 1. Positive displacement 2. Dynamic (momentum change) General Performance Characteristics Positive Displacement Pumps 1. Delivers pulsating or periodic flow (cavity opens, fluid enters, cavity closes, decreasing volume forces fluid out exit opening. 2. Not sensitive to wide viscosity changes. 3. Delivers a moderate flow rate. 4. Produces a high pressure rise. 5. Small range of flow rate operation (fixed pump speed). Dynamic Pumps 1. Typically higher flow rates than PD’s. 2. Comparatively steady discharge. 3. Moderate to low pressure rise. 4. Large range of flow rate operation. 5. Very sensitive to fluid viscosity. XI - 2 Typical Performance Curves (at fixed impeller speed) Fig. 11.2 Performance curves for dynamic and positive displacement pumps Centrifugal Pumps Most common turbomachine used in industry. Includes the general categories of (a) liquid pumps, (b) fans, (c) blowers, etc. They are momentum change devices and thus fall within the dynamic classification. Typical schematic shown as Fig. 11.3 Cutaway schematic of a typical centrifugal pump XI - 3 Writing the energy equation across the device and solving for hp – hf ,we have H = hp −hf = P2 −P 1 ρ g + V2 2 −V 1 2 2g + Z2 −Z1 where H is the net useful head delivered to the fluid, the head that results in pressure, velocity, and static elevation change. Since for most pumps (not all), V1 = V2 and ∆Z is small, we can write H ≅∆P ρg Since friction losses have already been subtracted, this is the ideal head delivered to the fluid. Note that velocity head has been neglected and can be significant at large flow rates where pressure head is small. The ideal power to the fluid is given by Pw = ρ Q g H The pump efficiency is given by η = Pw BHP = ρQgH BHP = ρQ gH ω T where BHP = shaft power necessary to drive the pump ω = angular speed of shaft T = torque delivered to pump shaft Note that from the efficiency equation, pump efficiency is zero at zero flow rate Q and at zero pump head,H. XI - 4 Basic Pump Theory Development of basic pump theory begins with application of the integral conservation equation for moment-of-momentum previously presented in Ch. III. Applying this equation to a centrifugal pump with one inlet, one exit, and uniform properties at each inlet and exit, we obtain e i T r x V r x V e i m m = − = − = − = − & & where T is the shaft torque needed to drive the pump V i , V e are the absolute velocities at the inlet and exit of the pump This is used to determine the change of angular momentum across the device. Fig. 11.4 Inlet and exit velocity diagrams for an idealized impeller Since the velocity diagram is key to the analysis of the device, we will discuss the elements in detail. XI - 5 1. At the inner radius r1 have two velocity components: a. the circumferential velocity due to the impeller rotation u1 = r1ω blade tip speed at inner radius b. relative flow velocity tangent to the blade w1 tangent to the blade angle β1 These combine to yield the absolute inlet velocity V1 at angle α1 u1 w1 V 1 V n1 V t1 α1 β1 The absolute velocity can be resolved into two absolute velocity components: 1. Normal ( radial ) component: V n1 = V 1 sinα1 = w1 sinβ1 Note that for ideal pump design, V n1 = V 1 and α1 = 90 o 2. Absolute tangential velocity: V t1 = V 1 cosα1 = u1 - w1 cos β1 again, ideally V t1 = 0 It is also important to note that V n1 is use to determine the inlet flow rate, i.e., Q = A1V n1 = 2π r1 b1 Vn1 where b1 is the inlet blade width XI - 6 Likewise for the outer radius r2 we have the following: a. the circumferential velocity due to the impeller rotation u2 = r2 ω blade tip speed at outer radius b. relative flow velocity tangent to the blade w 2 tangent to the blade angle β2 These again combine to yield the absolute outlet velocity V2 at angle α2 u2 w2 V 2 V t2 α2 β2 V n2 The exit absolute velocity can also be resolved into two absolute velocity components: 1. Normal ( radial ) component: V n2 = V 2 sinα2 = w2 sinβ2 = Q 2π r2 b2 Note that Q is the same as for the inlet flow rate 2. Absolute tangential velocity: V t2 = V 2 cosα 2 = u2 - w2 cos β2 V t2 = u2 -Vn2 tan β2 = u2 -Q 2π r2 b2 tan β2 where Q = A1V n1 = 2π r1 b1 Vn1 = A2Vn2 = 2π r2 b2 V n2 Again, each of the above expressions follows easily from the velocity diagram, and the student should draw and use the diagram with each pump theory problem. XI - 7 We can now apply moment - of – momentum equation. T = ρQ r2 Vt2 −r1Vt1 { } (again Vt1 is zero for the ideal design) For a sign convention, we have assumed that Vt1 and Vt2 are positive in the direction of impeller rotation. The “ ideal” power supplied to the fluid is given by Pw = ω T = ρQ ω r2 Vt2 −ω r1 Vt1 { } or Pw = ω T = ρQ u2 Vt2 −u1 Vt1 { }= ρ Qg H Since these are ideal values, the shaft power required to drive a non-ideal pump is given by BHP = Pw ηp The head delivered to the fluid is H = ρQ u2 Vt2 −u1 V t1 { } ρQ g = u2 V t2 −u1 Vt1 { } g For the special case of purely radial inlet flow H = u2 V t2 g XI - 8 From the exit velocity diagram, substituting for Vt2 we can show that H = u2 2 g − ω Q 2π b2 gtan β2 has the form C1 - C2 Q where: C1 = u2 2 g shutoff head, the head produced at zero flow, Q = 0 Example: A centrifugal water pump operates at the following conditions: speed = 1440 rpm, r1 = 4 in, r2 = 7 in, β1 = 30o, β2 = 20o, b1 = b2 = 1.75 in Assuming the inlet flow enters normal to the impeller (zero absolute tangential velocity): find: (a) Q, (b) T, (c) Wp, (d) hp, (e) ∆P ω = 1440 rev min 2π 60 = 150.8 rad s Calculate blade tip velocities: u1 = r1ω = 4 12 ft150.8 rad s = 50.3 ft s u2 = r2 ω = 7 12 ft150.8 rad s = 88 ft s Since design is ideal, at inlet α1 = 90o, Vt1 = 0 Vn1 = U1 tan 300 = 50.3 tan 30o = 29.04 ft/s Q = 2π r 1 b1 V n1 30Þ 30Þ 90Þ V = V 1 1 n w1 u1 r 1 • XI - 9 Q = 2π 4 12 ft1.75ft29.04 ft s = 8.87 ft 3 s Q = 8.87 ft 3 s 60 s min 7.48 gal ft3 = 3981 gal min Repeat for the outlet: Vn2 = Q 2π r2 b2 = 8.87 ft3 s 2π 7 12 ft1.75 12 ft Vn2 =16.6 ft s w 2 = Vn2 sin20 o = 16.6ft/s sin20 o = 48.54 ft s 20Þ 20Þ u2 r 2 • w2 V 2 α2 V t 2 = u2 - w 2 cos β2 = 88 −48.54cos20 o = 42.4 ft s We are now able to determine the pump performance parameters. Since for the centrifugal pump, the moment arm r1 at the inlet is zero, the momentum equation becomes Ideal moment of momentum delivered to the fluid: T = ρQ r2 Vt2 { }= 1.938slug ft3 8.87 ft 3 s 7 12 ft 42.4 ft s = 425.1ft −lbf Ideal power delivered to the fluid: P = ω T =150.8 rad s 425.1ft −lbf = 64,103ft −lbf s =116.5hp XI - 10 Head produced by the pump (ideal): H = P ρ gQ = 64,103ft −lbf/s 62.4 lbf ft3 8.87 ft3 s = 115.9 ft Pressure increase produced by the pump: ∆P = ρgH = 62.4 ft 3 s 115.9ft = 7226psf = 50.2psi Pump Performance Curves and Similarity Laws Pump performance results are typically obtained from an experimental test of the given pump and are presented graphically for each performance parameter. • Basic independent variable - Q {usually gpm or cfm } • Dependent variables typically H – head pressure rise, in some cases ∆P BHP – input power requirements (motor size) η – pump efficiency • These typically presented at fixed pump speed and impeller diameter Typical performance curves appear as XI - 11 Fig. 11.6 Typical Centrifugal Pump Performance Curves at Fixed Pump Speed and diameter These curves are observed to have the following characteristics: 1. hp is approximately constant at low flow rate. 2. hp = 0 at Qmax. 3. BHP is not equal to 0 at Q = 0. 4. BHP increases monotonically with the increase in Q. 5. ηp = 0 at Q = 0 and at Qmax. 6. Maximum pump efficiency occurs at approximately Q = 0.6 Qmax . This is the best efficiency point BEP. At any other operating point, efficiency is less, pump head can be higher or lower, and BHP can be higher or lower. 7. At the BEP, Q = Q, hp = hp, BHP = BHP. Measured Performance Data Actual pump performance data will typically be presented graphically as shown in Fig. 11.7. Each graph will usually have curves representing the pump head vs flow rate for two or more impeller diameters for a given class/model of pumps having a similar design. The graphs will also show curves of constant efficiency and constant pump power (BHP) for the impeller diameters shown. All curves will be for a fixed pump impeller speed. XI - 12 Fig. 11.7 Measured performance curves for two models of a centrifugal water pump XI - 13 How to Read Pump Performance Curves Care must be taken to correctly read the performance data from pump curves. This should be done as follows: (1) For a given flow rate Q (2) Read vertically to a point on the pump head curve h for the impeller diameter D of interest. (3) All remaining parameters ( efficiency & BHP) are read at this point; i.e., graphically interpolate between adjacent curves for BHP to obtain the pump power at this point. Note that the resulting values are valid only for the conditions of these curves: (1) pump model and design, (2) pump speed – N, (3) impeller size – D, (4) fluid (typically water) Thus for the pump shown in Fig. 11.7a with an impeller diameter D = 32 in, we obtain the following performance at Q = 20,000 gpm: Q = 20,000 gpm, D = 32 in, N = 1170 rpm H ≅ 385 ft, BHP ≅ 2300 bhp, ηp ≅ 86.3 % Note that points that are not on an h vs. Q curve are not valid operating points. Thus for Fig. 11.7b, the conditions Q = 22,000 gpm, BHP = 1500 bhp, hp = 250 ft do not correspond to a valid operating point because they do not fall on one of the given impeller diameter curves. However, for the same figure, the point Q = 20,000 gpm, BHP = 1250 bhp is a valid point because it coincidentally also falls on the D = 38 in impeller curve at hp = 227 ft. XI - 14 Net Positive Suction Head - NPH One additional parameter is typically shown on pump performance curves: NPSH = head required at the pump inlet to keep the fluid from cavitating. NPSH is defined as follows: NPSH = Pi ρg + V i 2 2g −Pv ρg where Pi = pump inlet pressure Pv = vapor pressure of fluid Considering the adjacent figure, write the energy equation between the fluid surface and the pump inlet to obtain the following: z=0 zi Pi Pa Pump inlet NPSH = Pi ρg + V i 2 2g −Pv ρg = Pa ρg −Zi −hf,a−i −Pv ρg For a pump installation with this configuration to operate as intended, the right-hand-side of the above equation must be > the NPSH value for the operating flow rate for the pump. Example: A water supply tank and pump are connected as shown. Pa = 13.6 psia and the water is at 20 o C with Pv = 0.34 psia. The system has a friction loss of 4.34 ft. Will the NPSH of the pump of Fig. 11.7a at 20,000 gpm work? 10 ft i a XI - 15 Applying the previous equation we obtain NPSH = Pa ρg −Zi −hf,a−i −Pv ρg NPSH = 13.6−0.34 ( )lbf/in2 144in2/ft2 62.4lbf/ft3 −−10ft ( ) −4.34ft NPSH = 36.26ft The pump will work because the system NPSH as shown in Fig. 11.7a is 30 ft which provides a 6.3 ft safety margin. Conversely, the pump could be located as close as 3.7 ft below the water surface and meet NPSH requirements. Pump Similarity Laws Application of the dimensional analysis procedures of Ch. V will yield the following three dimensionless performance parameters: Dimensionless flow coefficient: CQ = Q ω D3 Dimensionless head coefficient: CH = gH ω 2 D2 Dimensionless power coefficient: CP = BHP ρω 3 D5 where ω is the pump speed in radians/time and other symbols are standard design and operating parameters with units that make the coefficients dimensionless. How are these used? These terms can be used to estimate design and performance changes between two pumps of similar design. XI - 16 Stated in another way: If pumps 1 and 2 are from the same geometric design family and are operating at similar operating conditions, the flow rates, pump head, and pump power for the two pumps will be related according to the following expressions: Q2 Q1 = N2 N1 D2 D1 3 H2 H1 = N2 N1 2 D2 D1 2 BHP2 BHP 1 = ρ2 ρ1 N 2 N1 3 D2 D1 5 Use to predict the new flow rate for a design change in pump speed N and impeller diameter D. Used to predict the new pump head H for a design change in pump speed, N and impeller diameter D. Used to predict the new pump power BHP for a design change in fluid, ρ, pump speed N and impeller diameter D. Example It is desired to modify the operating conditions for the 38 in diameter impeller pump of Fig. 11.7b to a new pump speed of 900 rpm and a larger impeller diameter of 40 in. Determine the new pump head and power for the new pump speed at the BEP. Q(gpm) H(ft) BEP 2 BEP1 • • XI - 17 For the D = 38 in impeller of Fig. 11.7b operating at 710 rpm, we read the best efficiency point (BEP) values as Q = 20,000 gpm, H = 225 ft, BHP = 1250 hp Applying the similarity laws for N2 = 900 rpm and D2 = D1 = 38 in, we obtain Q2 Q1 = N2 N1 D2 D1 3 = 900 710 40 38 3 = 1.478 Q2 = 20,0001.478 = 29,570 gpm ans. H2 H1 = N2 N1 2 D2 D1 2 = 900 710 2 40 38 2 =1.78 H2 = 2251.78 = 400.5 ft ans. BHP2 BHP 1 = ρ2 ρ1 N2 N1 3 D2 D1 5 = 1 ( ) 900 710 3 40 38 5 = 2.632 BHP2 = 3290 hp ans. Thus, even small changes in the speed and size of a pump can result in significant changes in flow rate, head, and power. It is noted that every point on the original 38 in diameter performance curve exhibits a similar translation to a new operating condition. The similarity laws are obviously useful to predict changes in the performance characteristics of an existing pump or to estimate the performance of a modified pump design prior to the construction of a prototype. XI - 18 Matching a Pump to System Characteristics The typical design/sizing requirement for a pump is to select a pump which has a pump head which matches the required system head at the design/operating flow rate for the piping system. Key Point hp = hsys at Qdes. It is noted that pump selection should occur such that the operating point of the selected pump should occur on the pump curve near or at the BEP. From the energy equation in Ch. VI, the system head is typically expressed as hsys = P2 −P 1 ρg + V2 2 −V1 2 2g + Z2 −Z1 + f L D + ∑Ki V 2g 2 Thus the selection of a pump for a piping system design should result in a pump for which the pump head hp at the design flow rate Qdes is equal ( or very close) to the head requirements hsys of the piping system at the same flow rate, and this should occur at or near the point of maximum efficiency for the chosen pump. Q(gpm) • Qdes Hdes hp hsys p η Other operating and performance requirements (such as NPSH) are obviously also a part of the selection criteria for a pump. XI - 19 Pumping Systems: Parallel and Series Configurations For some piping system designs, it may be desirable to consider a multiple pump system to meet the design requirements. Two typical options include parallel and series configurations of pumps. Specific performance criteria must be met when considering these options. Given a piping system which has a known design flow rate and head requirements, Qdes, hdes. The following pump selection criteria apply. Pumps in Parallel: Assuming that the pumps are identical, each pump must provide the following: Q(pump) = 0.5 Qdes h(pump) = hdes Pumps in Series: Assuming that the pumps are identical, each pump must provide the following: Q (pump) = Qdes h(pump) = 0.5 hdes For example, if the design point for a given piping system were Qdes = 600 gpm, and hsys = 270 ft, the following pump selection criteria would apply: 1. Single pump system Q(pump) = 600 gpm, hp = 270 ft 2. Parallel pump system Q(pump) = 300 gpm, hp = 270 ft for each of the two pumps 3. Series pump system Q(pump) = 600 gpm, hp = 135 ft for each of the two pumps |
17615 | https://brainly.com/question/14126664 | [FREE] Lead thiocyanate, Pb(SCN)₂, has a Ksp value of 2.00 \times 10^{-5}. a) Calculate the molar solubility of - brainly.com
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Lead thiocyanate, Pb(SCN)₂, has a Ksp value of 2.00×1 0−5.
a) Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold.
b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.
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the molar solubility of lead thiocyanate (Pb(SCN)₂) in pure water is approximately: 1.26 × 10⁻² M
, we can qualitatively state that the molar solubility of Pb(SCN)₂ in 0.500 M KSCN will be lower than 1.26 × 10⁻² M obtained in pure water (part a).
Solubility Equilibrium:
Pb(SCN)₂ (s) ⇌ Pb²⁺ (aq) + 2 SCN⁻ (aq)
a) Molar Solubility in Pure Water:
Write the Solubility Product Expression (Ksp):
Ksp = [Pb²⁺][SCN⁻]²
Let x be the molar solubility of Pb(SCN)₂. This means the concentration of Pb²⁺ will also be x, and the concentration of SCN⁻ will be 2x (since two SCN⁻ ions dissociate from each Pb(SCN)₂ molecule).
Substitute x into the Ksp expression:
Ksp = (x)(2x)² = 4x³
Plug in the given Ksp value:
2.00 × 10⁻⁵ = 4x³
Solve for x (consider using logarithms or calculators):
x ≈ 1.26 × 10⁻² M
Therefore, the molar solubility of lead thiocyanate in pure water is approximately 1.26 × 10⁻² M.
b) Molar Solubility in 0.500 M KSCN:
Common Ion Effect: The presence of KSCN (which dissociates into K⁺ and SCN⁻) creates a common ion (SCN⁻) with Pb(SCN)₂. This suppresses the dissociation of Pb(SCN)₂ and reduces its solubility compared to pure water.
Modified Equilibrium Expression: We can still use the Ksp expression, but we need to consider the initial concentration of SCN⁻ from KSCN:
Ksp = [Pb²⁺][SCN⁻]² = x( [SCN⁻] from Pb(SCN)₂ + [SCN⁻] from KSCN )²
[SCN⁻] from KSCN is already 0.500 M.
Substitute x and 0.500 M into the expression and solve for x. The calculation will involve solving a cubic equation, which may require numerical methods or software.
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The molar solubility of lead thiocyanate, Pb(SCN)₂, in pure water is approximately 1.71 × 10⁻² M. In the presence of 0.500 M KSCN, the molar solubility decreases to about 8.00 × 10⁻⁵ M due to the common ion effect. This illustrates how the solubility of a salt can be influenced by the concentration of one of its ions in solution.
Explanation
To address the question about the molar solubility of lead thiocyanate, Pb(SCN)₂, we will approach it in two parts: a) calculating the molar solubility in pure water and b) calculating the molar solubility in a solution of KSCN.
a) Molar Solubility in Pure Water:
Write the Solubility Reaction: Lead thiocyanate dissociates in water according to the following equilibrium:
P b(SCN)2(s)⇌P b(a q)2++2 SC N(a q)−
Set Up the Ksp Expression: The solubility product constant (Ksp) can be expressed as:
Ks p=[P b 2+][SC N−]2
Let x be the molar solubility of Pb(SCN)₂. In this case, the concentration of P b 2+ will be x, and the concentration of SC N− will be 2 x due to the dissociation of one formula unit of Pb(SCN)₂ producing two SCN⁻ ions.
Substituting into the Ksp Expression: Substituting these into the Ksp expression gives:
Ks p=(x)(2 x)2=4 x 3
Plug in the Given Ksp Value: We know that Ksp = 2.00×1 0−5. So:
2.00×1 0−5=4 x 3
Solving for x:
x 3=4 2.00×1 0−5=5.00×1 0−6
x=3 5.00×1 0−6≈1.71×1 0−2 M
Therefore, the molar solubility of lead thiocyanate in pure water is approximately 1.71×1 0−2 M.
b) Molar Solubility in 0.500 M KSCN:
Common Ion Effect: In this scenario, adding KSCN introduces SCN⁻ ions into the solution, which suppresses the solubility of Pb(SCN)₂ due to the common ion effect.
Modified Ksp Expression: We still use the Ksp expression, but now we need to account for the initial concentration of SCN⁻ provided by KSCN:
Ks p=[P b 2+][SC N−]2=x(0.500+2 x)2
Assuming x is Small: For diluted solutions, we can assume that x is small compared to 0.500 M, thus:
Ks p≈x(0.500)2
2.00×1 0−5=x(0.500)2
2.00×1 0−5=x(0.25)
Solving for x:
x=0.25 2.00×1 0−5=8.00×1 0−5 M
Thus, the molar solubility of Pb(SCN)₂ in a 0.500 M KSCN solution is approximately 8.00×1 0−5 M.
Examples & Evidence
For example, the addition of potassium nitrate to a saturated solution of lead(II) nitrate would similarly decrease the solubility of lead(II) iodide due to the presence of common ions. This principle is widely observed in salt solubility studies.
This response is based on established principles of chemical equilibrium and solubility product calculations commonly found in advanced chemistry courses.
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(a) The molar solubility of lead thiocyanate in pure water is 1.71 × 10⁻² M. (b) In a 0.500 M KSCN solution, the molar solubility decreases to 8.00 × 10⁻⁵ M.
(a) Let's start by calculating the molar solubility of lead thiocyanate, Pb(SCN)₂, in pure water:
Write the dissociation equation:
Pb(SCN)₂(s) ⇌ Pb²⁺(aq) + 2 SCN⁻(aq)
Set up the expression for the solubility product constant (Ksp):
K s p=[Pb 2+][SCN−]2
Given Ksp = 2.00 × 10⁻⁵
Let the molar solubility be s, where s is the concentration of Pb²⁺ and 2s is the concentration of SCN−:
K s p=s(2 s)2=4 s 3
2.00×1 0−5=4 s 3
s 3=4 2.00×1 0−5=5.00×1 0−6
s=(5.00×1 0−6)1/3≈1.71×1 0−2 M
Therefore, the molar solubility of lead thiocyanate in pure water is 1.71 × 10⁻² M.
(b) Now, calculate the molar solubility of lead thiocyanate in 0.500 M KSCN:
In this case, the initial [SCN⁻] is 0.500 M due to KSCN.
Using the Ksp expression:
K s p=[Pb 2+][SCN−]2
2.00×1 0−5=(s)(0.500+2 s)2
Assuming s is much smaller than 0.500, we simplify to:
2.00×1 0−5=(s)(0.500+2 s)2
s≈0.25 2.00×1 0−5=8.00×1 0−5 M
Therefore, the molar solubility of lead thiocyanate in 0.500 M KSCN is 8.00 × 10⁻⁵ M.
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### Free Chemistry solutions and answers
Community Answer Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold. Lead thiocyanate, Pb(SCN)2, has a Ksp value of .
Community Answer 1 Lead thiocyanate, Pb(SCN)₂, has a [tex]K_{sp}[/tex] value of 2.00×10⁻⁵. 1. Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold. Express your answer with the appropriate units. 1.71×10⁻² M Common-Ion Effect Consider the dissolution of AB(s) : AB(s)⇌A (aq) + B⁻(aq) Le Châtelier's principle tells us that an increase in either [A ] or [B⁻] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A or B⁻ ions. 2. Calculate the molar solubility of lead thiocyanate in 0.800 M KSCN.
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Community Answer Calculate the molar solubility of lead thiocyanate in 1.00 M KSCN.
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17616 | https://www.youtube.com/watch?v=HH8yUIeQtDA | Minimising Without Calculus
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We minimise 5x^2 + y^2 + 2z^2 - 2xy + 4xz + 6z without using calculus. Our approach relies on repeatedly completing the square.
00:00 Intro
00:50 Completing the square
02:43 Terms with x
04:11 Terms with z
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Transcript:
Intro okay so we're going to solve this problem where we need to find the minimum possible value of this expression with our three variables x y and Zed which will have to be real numbers and the idea here is instead of using calculus we're going to use a completing the square style argument to get this into a slightly different form so at the moment we've got 5 x^2 y^2 and 2 z^ 2 we know that all of these are positive but this doesn't necessarily help us with finding the minimum of the function CU we could choose x y and Zed all zero but that doesn't necessarily mean that we've minimized the remaining terms there so the idea is to use a completing the square kind of argument to somehow factorize these remaining terms into some other squared terms so that we're left with a sum of some squared terms then perhaps just a constant at the end which can tell us what the minimum of our function is going to be so if we look at each of the Completing the square different variables which one are we going to start with well X appears in three different terms Z appears in three different terms but Y is only actually in these two terms the y^2 and the - 2x y so if we just focus on these two terms the only ones containing y at the moment we've got y^2 - 2x Y and if we wanted to complete the square for this expression treating it as a polinomial in y we could just take half of this coefficient - 2x so we'd have y - x all s but then when we expand this we get y^2 - 2x y but we also have a plus x^2 so we just need to take away X2 so that we get this y^2 - 2x y so we can say then that our y^ 2 - 2x y term can be written as y - x all 2 - x^2 then this is quite helpful now because when we if we treat this as just f of x y and Zed we'll call it so our original expression there we can take our y terms we've combined them into a single y term and then we've gained an extra X term so f of x y and Zed we have our y - x all s and we've got this - x^ 2 but remember we've also got a plus 5 x^2 so we just take away 1 x^2 and we're left with + 4 x^2 then we've also got still + 2 z^ 2 we've got rid of our - 2x y that's been absorbed into our y - x all 2 and we've still got + 4 x z and plus 6 Z so you can see now once we've dealt with this first term there's only actually the 4 x^2 and the 4X Z left which contain X so I think that would be a sensible one to look at next then after that we can deal with our remaining Zed terms CU Zed appears in actually three of our remaining terms there so just focusing on our X terms Terms with x we've got 4x^2 + 4 x z and if we want to treat this like a polinomial in X and complete the square we'd need to first of all to get the 4x^2 part we'd need 2X or squ then we need to think what do we multiply this by twice to get back to 4X Z so we would actually just need to multiply that by Zed twice so 2 x Z gives us 2 x z and we get that again we get back to 4 x z so when we expand the brackets here we get 4x^2 + 4x Z then we also have a plus z^ s term so just like before we can take away z s so we can see that our original expression these two terms involve X is equivalent to 2x + Z 2 - z^ 2 so this is really useful now because we can substitute this into our original Expressions if we keep calling this f of x y and Zed we've got y - x all 2 then taking our two x terms we can write these as + 2x + Z all squar then we've got minus z^ 2 we've also got + 2 z^ 2 so we've just got + one lot of z^ s now and the only remaining piece is this + 6 Z so you can see now we're really close to expressing this as a sum of things entirely squared so if we Terms with z complete the square for our z^ s + 6 Z term we first of all just keep the first two terms y - x^2 + 2x + Z all 2 but then completing the square here we just do half of the six so Z + 3 all s but then we need to take away 3 squar we just take away 9 you can see now we've expressed our entire function as just the sum of three different things squared - 9 so then the only possibility to make this as small as possible is to make each of these three squares equal to zero it turns out this is possible because we could take in order for Z + 3 to be equal to zero we need Zed is minus 3 then looking at 2x+ Zed to be zero if Zed is minus 3 we need 2x to be equal to three so X would have to be equal to 3 / 2 and finally for y - x to be Z you need y equal to X so y would also have to be equal to 3 over2 so this gives us a solution then the minimum of our function is obtained for these different values of x y and Zed and the minimum value then of our function is - 9 |
17617 | https://www.quora.com/How-many-ways-are-there-to-write-2-n-as-the-sum-of-powers-of-2 | How many ways are there to write [math]2^n[/math] as the sum of powers of [math]2[/math]? - Quora
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How many ways are there to write 2 n 2 n as the sum of powers of 2 2?
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Pablo Emanuel
MSc in Math and IMO medalist · Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 844 answers and 5.4M answer views
·5y
Based on the question comments, we should ignore the order of the factors. So, each way of writing 2 n 2 n as the sum of powers of 2 2 is uniquely determined by the number of factors equal to 2 i 2 i, for all i between 0 0 and n n.
In other words, we can represent each solution by a (n+1)(n+1)-tuple, starting with the number of factors 2 n 2 n and ending with the number of factors 2 0 2 0. For instance, the solutions for 2 2 2 2 are
(1,0,0)(1,0,0)
(0,2,0)(0,2,0)
(0,1,2)(0,1,2)
(0,0,4)(0,0,4)
It’s pretty straightforward to see that we can always get to all solutions starting with (1,0,0,⋯,0)(1,0,0,⋯,0) and building a graph where, on each edge, we replace one unit
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Based on the question comments, we should ignore the order of the factors. So, each way of writing 2 n 2 n as the sum of powers of 2 2 is uniquely determined by the number of factors equal to 2 i 2 i, for all i between 0 0 and n n.
In other words, we can represent each solution by a (n+1)(n+1)-tuple, starting with the number of factors 2 n 2 n and ending with the number of factors 2 0 2 0. For instance, the solutions for 2 2 2 2 are
(1,0,0)(1,0,0)
(0,2,0)(0,2,0)
(0,1,2)(0,1,2)
(0,0,4)(0,0,4)
It’s pretty straightforward to see that we can always get to all solutions starting with (1,0,0,⋯,0)(1,0,0,⋯,0) and building a graph where, on each edge, we replace one unit in a position i i with 2 2 units at the position i+1 i+1. For instance, from (0,1,2,0)(0,1,2,0) we’d be able to go to (0,0,4,0)(0,0,4,0) or (0,1,1,2)(0,1,1,2).
Now, assume that we found all the f(n)f(n) solutions for 2 n 2 n. The first observation is that, if we just append a 0 0 at the end of each tuple, automatically they become solutions for 2 n+1 2 n+1 (we’re just multiplying all the factors by 2).
The other observation is that the only new nodes in our graph for 2 n+1 2 n+1 will come from replacing one unit at position n n with 2 2 units at position n+1 n+1.
The third observation is that, no two of this new edges will generate the same new node.
Putting all of this together, we just need to look at the solutions that don’t have 0 0 in the last position, and apply this recursive algorithm to generate all solutions.
Let’s do it for the first few n n’s (I’m highlighting in bold the items generated at that level, which are the ones interesting for the next level)
n=0 n=0
(1)
n=1 n=1
(1,0)
(0,2)
n=2 n=2
(1,0,0)
(0,2,0)
(0,1,2)
(0,0,4)
n=3 n=3
(1,0,0,0)
(0,2,0,0)
(0,1,2,0)
(0,1,1,2)
(0,1,0,4)
(0,0,4,0)
(0,0,3,2)
(0,0,2,4)
(0,0,1,6)
(0,0,0,8)
Looking just at the last position, what we see is
(1)→(2)→(2,4)→(2,4,2,4,6,8)→⋯(1)→(2)→(2,4)→(2,4,2,4,6,8)→⋯
What we’re doing is replacing each element k k of our sequence with all 2∗i 2∗i where i i goes from 1 1 to k k.
Also, if we call s(n)s(n) the number of elements of that sequence at step n n, we have
f(n)=s(n)+f(n−1)(1)(1)f(n)=s(n)+f(n−1)
The next observation is that we don’t care about the order of the elements in each of these sequences. We only need to know how many 2s, how many 4s, etc, we have in the sequence. So, in summary, the sequence we care about can be written (starting at n = 1) as
(1)(1) [i.e. one 2] →(1,1)→(1,1) [one 2, one 4] →(2,2,1,1)→(2,2,1,1) [two 2, two 4, one 6, one 8]
And s(n)s(n) is simply the sum of the numbers of the n th n th tuple.
The algorithm to generate the (n+1)(n+1)-th tuple from the n n-th tuple is pretty simple
The new number of 2’s and 4’s is simply the sum of all the numbers in the previous tuple. The number of 6’s and 8’s is the sum of all the numbers of the previous tuple except the first. The number of 10’s and 12’s is the sum except the first two items, etc.
i.e., our sequence would continue as
(2,2,1,1)→(6,6,4,4,2,2,1,1)→(26,26,20,20,14,14,10,10,6,6,4,4,2,2,1,1)→⋯(2,2,1,1)→(6,6,4,4,2,2,1,1)→(26,26,20,20,14,14,10,10,6,6,4,4,2,2,1,1)→⋯
Putting all of this together, we come up with this short python script to calculate the first 25 values of f f.
import numpy as np
s = np.zeros(27)
s = 1
s = 1
s = 2
f = np.zeros(27)
f = 1
arr = [1,1]
for i in range(3,26):
arr2 = []
S = s[i-1]
arr2.append(S)
arr2.append(S)
for k in arr:
S = S-k
if(S == 0):
break
arr2.append(S)
arr2.append(S)
s[i] = sum(arr2)
arr = arr2
for i in range(1,26):
f[i] = f[i-1] + s[i]
ratio = f[i] / f[i-1]
if(i>1):
ratio2 = (f[i] / f[i-1])/(f[i-1]/f[i-2])
else:
ratio2 = 1
print ("s[" + str(i) + "] = " + str(s[i]) + ", f[" + str(i) + "] = " + str(f[i]) + ", ratio = " + str(ratio) + ", ratio2 = " + str(ratio2))
Which yields the following results
s = 1.0, f = 2.0, ratio = 2.0, ratio2 = 1
s = 2.0, f = 4.0, ratio = 2.0, ratio2 = 1.0
s = 6.0, f = 10.0, ratio = 2.5, ratio2 = 1.25
s = 26.0, f = 36.0, ratio = 3.6, ratio2 = 1.44
s = 166.0, f = 202.0, ratio = 5.611111111111111, ratio2 = 1.5586419753086418
s = 1626.0, f = 1828.0, ratio = 9.049504950495049, ratio2 = 1.6127830604842661
s = 25510.0, f = 27338.0, ratio = 14.955142231947484, ratio2 = 1.652592303530302
s = 664666.0, f = 692004.0, ratio = 25.312897797936937, ratio2 = 1.6925882352267438
s = 29559718.0, f = 30251722.0, ratio = 43.71610857740707, ratio2 = 1.7270290002502218
s = 2290267226.0, f = 2320518948.0, ratio = 76.70700358809327, ratio2 = 1.7546622076910166
s = 314039061414.0, f = 316359580362.0, ratio = 136.3313928699711, ratio2 = 1.777300461403148
s = 77160820913242.0, f = 77477180493604.0, ratio = 244.9022735614625, ratio2 = 1.796374763038203
s = 3.431739276248976e+16, f = 3.4394869942983364e+16, ratio = 443.9354881508985, ratio2 = 1.8127046421211976
s = 2.7859502236825854e+19, f = 2.789389710676884e+19, ratio = 810.9900445330586, ratio2 = 1.8268195856813192
s = 4.157581110633654e+22, f = 4.160370500344332e+22, ratio = 1491.4984752470318, ratio2 = 1.8391082422051033
s = 1.1474658165418717e+26, f = 1.1478818535919061e+26, ratio = 2759.0856475328387, ratio2 = 1.8498749367315717
s = 5.88765612737581e+29, f = 5.888804009229402e+29, ratio = 5130.148186246164, ratio2 = 1.8593653266376555
s = 5.642056933022321e+33, f = 5.642645813423244e+33, ratio = 9581.989491549797, ratio2 = 1.8677802557904546
s = 1.0138681878302956e+38, f = 1.01392461428843e+38, ratio = 17968.957255413992, ratio2 = 1.87528459212573
s = 3.428771169943382e+42, f = 3.428872562404811e+42, ratio = 33817.82544860287, ratio2 = 1.8820137956760767
s = 2.189326046230645e+47, f = 2.1893603349562692e+47, ratio = 63850.73504804681, ratio2 = 1.888079265921124
s = 2.6470464793327483e+52, f = 2.6470683729360976e+52, ratio = 120906.01673337481, ratio2 = 1.8935728248452375
s = 6.0762816989879806e+57, f = 6.07630816967171e+57, ratio = 229548.5916342969, ratio2 = 1.898570458577786
s = 2.6545023583174967e+63, f = 2.6545084346256665e+63, ratio = 436862.0485502933, ratio2 = 1.9031353903764126
s = 2.211829250988816e+69, f = 2.2118319054972507e+69, ratio = 833235.9681536136, ratio2 = 1.907320562449105
I conjecture the second order ratio f(n)/f(n−1)f(n−1)/f(n−2)f(n)/f(n−1)f(n−1)/f(n−2) converges, and I’d be very happy if it converged to exactly 2 2, but, at this point I can’t see any obvious way to prove it. If anyone can see why it should or shouldn’t be true, I’d love to know.
EDIT: I actually see why the second-order ratio would converge to 2 2. Our sequence at step n+1 n+1 is a set of 2 n 2 n numbers, the first (and greatest) of which is s(n)s(n). Assume that, for big n n, the average of those numbers became a fixed fraction of s(n)s(n) i.e. α s(n)α s(n). That would mean that s(n+1)=2 n α s(n)s(n+1)=2 n α s(n), i.e.
s(n+1)s(n)=α 2 n s(n+1)s(n)=α 2 n
or
s(n+1)/s(n)s(n)/s(n−1)=α 2 n α 2 n−1=2 s(n+1)/s(n)s(n)/s(n−1)=α 2 n α 2 n−1=2
Actually, what seems to be the case is that the α n α n actually converges to 0 0, but the α n α n−1→1 α n α n−1→1, which is what we need.
alpha(2) = 0.75 , ratio = 0.75
alpha(3) = 0.5416666666666666 , ratio = 0.7222222222222222
alpha(4) = 0.39903846153846156 , ratio = 0.7366863905325444
alpha(5) = 0.3060993975903614 , ratio = 0.7670924662505444
alpha(6) = 0.24513760762607625 , ratio = 0.8008431560330364
alpha(7) = 0.20355559094472755 , ratio = 0.8303727564120787
alpha(8) = 0.17372281482353544 , ratio = 0.853441627504632
alpha(9) = 0.15132682171667708 , ratio = 0.8710820272535429
alpha(10) = 0.1339052349134517 , ratio = 0.8848744286994735
alpha(11) = 0.11997291648020549 , ratio = 0.8959538927492177
alpha(12) = 0.10858191525495921 , ratio = 0.9050535607582257
alpha(13) = 0.0990989661621674 , ratio = 0.9126654832848999
alpha(14) = 0.09108512081531195 , ratio = 0.9191329066567512
alpha(15) = 0.08422656532790153 , ratio = 0.9247016919336681
alpha(16) = 0.078292966661883 , ratio = 0.9295519336100375
alpha(17) = 0.07311140015404986 , ratio = 0.9338182377197392
alpha(18) = 0.06854947140259744 , ratio = 0.9376030449172061
alpha(19) = 0.06450406553551997 , ratio = 0.9409856008470375
alpha(20) = 0.06089365306241435 , ratio = 0.9440281408135817
alpha(21) = 0.05765290752420168 , ratio = 0.9467802410393906
alpha(22) = 0.05472886328987777 , ratio = 0.9492819293962506
alpha(23) = 0.05207812210197664 , ratio = 0.9515659374494739
alpha(24) = 0.04966478797651845 , ratio = 0.9536593481475285
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Gil Gautard
Studied Electronics and Communication Engineering (Graduated 1969) · Author has 309 answers and 136.7K answer views
·5y
2^n = 2^(n-1) + 2^(n-1) = 2^(n-2) + 2^(n-2) + 2^(n-2) + 2^(n-2) =
2^n = 8 times 2^(n-3) = … = 2^n times 1.
Thus we have n+1 ways of writing 2^n
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Ragu Rajagopalan
Passionate Maths solver ;Reviving knowledge after 3 decades · Author has 10K answers and 7.5M answer views
·May 10
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How do I find a type of sum ∞∑n=1 2 n n 2 n!∑n=1∞2 n n 2 n!?
S=∞∑n=1 2 n⋅n 2 n!=x=2∞∑n=1 x n⋅n(n−1)!=m=(n−1)∞∑m=0 x m+1⋅(m+1)m!S=∑n=1∞2 n⋅n 2 n!=⏟x=2∑n=1∞x n⋅n(n−1)!=⏟m=(n−1)∑m=0∞x m+1⋅(m+1)m!
Note :e x=∞∑m=0 x m m!⟹x⋅e x=∞∑m=0 x m+1 m!Note :e x=∑m=0∞x m m!⟹x⋅e x=∑m=0∞x m+1 m!
Differentiating both sides with respet to 'x' :Differentiating both sides with respet to 'x' :
(x+1)⋅e x=∞∑m=0(m+1)⋅x m m!(x+1)⋅e x=∑m=0∞(m+1)⋅x m m!
\text{Multiplying throughout wit\text{Multiplying throughout wit
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S=∞∑n=1 2 n⋅n 2 n!=x=2∞∑n=1 x n⋅n(n−1)!=m=(n−1)∞∑m=0 x m+1⋅(m+1)m!S=∑n=1∞2 n⋅n 2 n!=⏟x=2∑n=1∞x n⋅n(n−1)!=⏟m=(n−1)∑m=0∞x m+1⋅(m+1)m!
Note :e x=∞∑m=0 x m m!⟹x⋅e x=∞∑m=0 x m+1 m!Note :e x=∑m=0∞x m m!⟹x⋅e x=∑m=0∞x m+1 m!
Differentiating both sides with respet to 'x' :Differentiating both sides with respet to 'x' :
(x+1)⋅e x=∞∑m=0(m+1)⋅x m m!(x+1)⋅e x=∑m=0∞(m+1)⋅x m m!
Multiplying throughout with 'x' :Multiplying throughout with 'x' :
x⋅(x+1)⋅e x=∞∑m=0(m+1)⋅x m+1 m!=S x⋅(x+1)⋅e x=∑m=0∞(m+1)⋅x m+1 m!⏟=S
⟹S x=2=2⋅(2+1)⋅e 2=6 e 2⟹S x=2=2⋅(2+1)⋅e 2=6 e 2
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How many ways are there to express 2 n−1 2 n−1 as the sum of powers of 2,2, where order doesn't matter?
What are the number of ways of choosing k k numbers from 1,2,3,...,n 1,2,3,...,n such that their sum is n(n+1)4 n(n+1)4?
Let f(n)f(n) be the number of ways to write n n as a sum of powers of 2,2, where we keep track of the order of the summation. For example, f(4)=6 f(4)=6 because 4 4 can be written as 4,2+2,2+1+1,1+2+1,1+1+2,4,2+2,2+1+1,1+2+1,1+1+2, and 1+1+1+1.1+1+1+1. How do I find the smallest n>2013 n>2013 for which f(n)f(n) is odd?
Let f(n)f(n) be the number of ways to write n n as a sum of powers of 2,2, where we keep track of the order of the summation. For example, f(4)=6 f(4)=6 because 4 4 can be written as 4,2+2,2+1+1,1+2+1,1+1+2,4,2+2,2+1+1,1+2+1,1+1+2, and 1+1+1+1.1+1+1+1. How do I find the smallest n n greater than 2013 2013 for which f(n)f(n) is odd?
How many numbers in the set {1,2,3,...,2 n−1}{1,2,3,...,2 n−1} can be expressed as a sum of k k or less powers of 2, with k≤n k≤n?
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·
10mo
To find the number of ways to write 2 n 2 n as the sum of powers of 2, we can express 2 n 2 n in the form:
2 n=2 k 1+2 k 2+⋯+2 k m 2 n=2 k 1+2 k 2+⋯+2 k m
where k 1,k 2,…,k m k 1,k 2,…,k m are non-negative integers subject to the condition that k 1>k 2>⋯>k m k 1>k 2>⋯>k m. This ensures that each power of 2 is used at most once in the sum.
Key Insight
The key insight here is that each way of writing 2 n 2 n corresponds to a unique binary representation of numbers less than 2 n 2 n. Specifically, to form 2 n 2 n as a sum of distinct powers of 2, we can think of this in terms of binary digits:
Every distinct sum of powers of 2 corresp
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To find the number of ways to write 2 n 2 n as the sum of powers of 2, we can express 2 n 2 n in the form:
2 n=2 k 1+2 k 2+⋯+2 k m 2 n=2 k 1+2 k 2+⋯+2 k m
where k 1,k 2,…,k m k 1,k 2,…,k m are non-negative integers subject to the condition that k 1>k 2>⋯>k m k 1>k 2>⋯>k m. This ensures that each power of 2 is used at most once in the sum.
Key Insight
The key insight here is that each way of writing 2 n 2 n corresponds to a unique binary representation of numbers less than 2 n 2 n. Specifically, to form 2 n 2 n as a sum of distinct powers of 2, we can think of this in terms of binary digits:
Every distinct sum of powers of 2 corresponds to a selection of bits in the binary representation of 2 n 2 n.
Each bit in the binary representation can either be included or not included, leading to a combinatorial explosion of combinations.
Combinatorial Argument
For 2 n 2 n, we are essentially looking at all the combinations of the powers of 2 from 2 0 2 0 to 2 n−1 2 n−1 that can sum up to 2 n 2 n.
The number 2 n 2 n has a binary representation that consists of a 1 followed by n n zeros, which means it can be formed by summing any combination of powers of 2 from 2 0 2 0 to 2 n−1 2 n−1.
Counting the Combinations
The problem can be reformulated as counting the number of partitions of n n using distinct parts. Each valid partition corresponds to a unique way of choosing powers of 2 that add up to 2 n 2 n.
The number of ways to partition n n into distinct parts can be calculated using the Fibonacci numbers, where the number of partitions of n n into distinct parts is given by F n+1 F n+1, where F k F k is the k k-th Fibonacci number.
Thus, the number of ways to express 2 n 2 n as a sum of distinct powers of 2 is given by:
F n+1 F n+1
where F 0=0 F 0=0, F 1=1 F 1=1, F 2=1 F 2=1, F 3=2 F 3=2, F 4=3 F 4=3, F 5=5 F 5=5, and so on, following the Fibonacci sequence.
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Apoorv Khandelwal
Math enthusiast. · Author has 179 answers and 685.5K answer views
·8y
Related
What are the number of ways of choosing k k numbers from 1,2,3,...,n 1,2,3,...,n such that their sum is n(n+1)4 n(n+1)4?
Note that it is only possible for a subset of {1,…,n}{1,…,n} to sum to n(n+1)4 n(n+1)4 if n≡3 mod 4 n≡3 mod 4 or n≡0 mod 4 n≡0 mod 4 , since n(n+1)4 n(n+1)4 is an integer only for these conditions. The number of ways to choose the relevant subsets is given in A063865 - OEIS. No one appears to have found a closed form for this, but there is an asymptotic approximation: √6 π n−3 2 2 n 6 π n−3 2 2 n. I computed the following comparison for the first few n n with the below Python script.
!/usr/bin/python
import itertools
import math
def main():
maxN = 24
print "n:\tf(n):\tapprox f(n):\t% error:
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Note that it is only possible for a subset of {1,…,n}{1,…,n} to sum to n(n+1)4 n(n+1)4 if n≡3 mod 4 n≡3 mod 4 or n≡0 mod 4 n≡0 mod 4 , since n(n+1)4 n(n+1)4 is an integer only for these conditions. The number of ways to choose the relevant subsets is given in A063865 - OEIS. No one appears to have found a closed form for this, but there is an asymptotic approximation: √6 π n−3 2 2 n 6 π n−3 2 2 n. I computed the following comparison for the first few n n with the below Python script.
!/usr/bin/python
import itertools
import math
def main():
maxN = 24
print "n:\tf(n):\tapprox f(n):\t% error:"
for n in range(1, maxN + 1):
mod4 = n % 4
if mod4 == 3 or mod4 == 0:
realVal = f(n)
approximateVal = approximateF(n)
print str(n) + "\t" + str(realVal) + "\t" + str(approximateVal) + "\t" + str(percentError(realVal, approximateVal))
def percentError(expected, actual):
return (actual - expected) 100 / expected
def f(n):
doubleN = n + 0.0
target = doubleN (doubleN + 1) / 4
targetMap = dict()
targetMap = 1
inputList = range(1, n + 1)
toReturn = 0
for k in inputList:
for comb in itertools.combinations(inputList, k):
thisSum = getSum(comb)
if thisSum == target:
toReturn += 1
if thisSum not in targetMap:
targetMap[thisSum] = 0
targetMap[thisSum] += 1
for val in range(doubleN (doubleN + 1) / 2 + 1):
print str(val) + ": " + str(targetMap[val])
return toReturn
def approximateF(n):
return math.sqrt(6.0 / math.pi) (n (-1.5)) (2 n)
def getSum(thisComb):
toReturn = 0
for num in thisComb:
toReturn += num
return toReturn
if name == 'main':
main()
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Is there a way to check if someone has a dating profile?
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Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. This tool checks a bunch of dating apps and websites to see if that person has a profile—either now or in the past.
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Abdelhadi Nakhal
Hydrogeologist engineer · Author has 1.3K answers and 428.7K answer views
·Feb 21
Related
How many integers from 1 1 to 1000 1000 can be written as the sum of three nonnegative integer powers of two?
N.B: I work with the hypotheisis of differents powers.
1000=2 9+2 8+2 7+2 2=(1110000100)2 1000=2 9+2 8+2 7+2 2=(1110000100)2
So the greatest integer is 2 9+2 8+2 7 2 9+2 8+2 7
The others are: 2 a+2 b+2 c 2 a+2 b+2 c with: 0≤c<b<a≤9 0≤c<b<a≤9
Their number is computed like this:
for c=0 c=0
cases for:b number of choices for:a 1 8 2 7⋯⋯8 1 cases for:b number of choices for:a 1 8 2 7⋯⋯8 1
n(c=0)=t 8 n(c=0)=t 8
For c=1 c=1
cases for:b number of choices for:a 2 7 3 6⋯⋯8 1 cases for:b number of choices for:a 2 7 3 6⋯⋯8 1
n(c=1)=t 7 n(c=1)=t 7
We have now a good pattern:
n=7∑c=0 n(c)=t 8+t 7+⋯+t 1 n=∑c=0 7 n(c)=t 8+t 7+⋯+t 1
We compute by hand t k t k (you use summation technics)
\boxed{n=122\boxed{n=122
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N.B: I work with the hypotheisis of differents powers.
1000=2 9+2 8+2 7+2 2=(1110000100)2 1000=2 9+2 8+2 7+2 2=(1110000100)2
So the greatest integer is 2 9+2 8+2 7 2 9+2 8+2 7
The others are: 2 a+2 b+2 c 2 a+2 b+2 c with: 0≤c<b<a≤9 0≤c<b<a≤9
Their number is computed like this:
for c=0 c=0
cases for:b number of choices for:a 1 8 2 7⋯⋯8 1 cases for:b number of choices for:a 1 8 2 7⋯⋯8 1
n(c=0)=t 8 n(c=0)=t 8
For c=1 c=1
cases for:b number of choices for:a 2 7 3 6⋯⋯8 1 cases for:b number of choices for:a 2 7 3 6⋯⋯8 1
n(c=1)=t 7 n(c=1)=t 7
We have now a good pattern:
n=7∑c=0 n(c)=t 8+t 7+⋯+t 1 n=∑c=0 7 n(c)=t 8+t 7+⋯+t 1
We compute by hand t k t k (you use summation technics)
n=122 n=122
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Let A=333 333 333.A=333 333 333.What's the sum of all digits of the number C C?
How many numbers in the set {1,2,3,...,2 n−1}{1,2,3,...,2 n−1} can be expressed as a sum of k k or less powers of 2, with k≤n k≤n?
How do you find the sum n∑k=0 k 2(n k)∑k=0 n k 2(n k) ?
What are all the possible 4-number combinations of the numbers 1, 4, 2, and 5 if repetition is not allowed?
What are the combinations of 5 numbers among 1 to 35?
Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Yair Livne
, Master's Mathematics, Hebrew University of Jerusalem (2007) and
Michael Lamar
, PhD in Applied Mathematics · Author has 8.7K answers and 171.8M answer views
·7y
Related
In how many ways can 2 n 2 n be written as the sum of the square of 4 4 integers?
First of all notice that 0 0 is an integer, so 0 2 0 2 is the square of an integer; therefore, whenever we manage to write anything as a sum of one, two or three squares, we can make it a sum of four squares by adding zeros.
Next, whenever we use some square like 4 2 4 2, we can clearly also use (−4)2(−4)2 instead. The question is about sums of squares of integers, and integers may be negative. But should we regard 4 2+4 2 4 2+4 2 and 4 2+(−4)2 4 2+(−4)2 as different representations? This is unclear. So what I’ll do is count representations by sums of squares of nonnegative integers, and we’ll do the needed counting to accom
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First of all notice that 0 0 is an integer, so 0 2 0 2 is the square of an integer; therefore, whenever we manage to write anything as a sum of one, two or three squares, we can make it a sum of four squares by adding zeros.
Next, whenever we use some square like 4 2 4 2, we can clearly also use (−4)2(−4)2 instead. The question is about sums of squares of integers, and integers may be negative. But should we regard 4 2+4 2 4 2+4 2 and 4 2+(−4)2 4 2+(−4)2 as different representations? This is unclear. So what I’ll do is count representations by sums of squares of nonnegative integers, and we’ll do the needed counting to accommodate signs in the end.
So let’s see:
1=1 2 1=1 2. One way.
2=1 2+1 2 2=1 2+1 2. One way.
4=2 2=1 2+1 2+1 2+1 2 4=2 2=1 2+1 2+1 2+1 2. Two ways.
8=2 2+2 2 8=2 2+2 2. One way.
16=4 2=2 2+2 2+2 2+2 2 16=4 2=2 2+2 2+2 2+2 2. Two ways.
32=4 2+4 2 32=4 2+4 2. One way.
64=8 2=4 2+4 2+4 2+4 2 64=8 2=4 2+4 2+4 2+4 2. Two ways.
At this point we should feel convinced that we get the pattern. The case of 2 0=1 2 0=1 is an outlier. For other powers 2 n 2 n there is one way when n n is odd and two ways when n n is even. From now on, we set aside the case n=0 n=0 since it is trivial, and focus on n>0 n>0.
Looking at our little table, we observe that the ways we can write a power of 2 2 as a sum of four squares are essentially the same in all cases:
When n n is odd, we write 2 n 2 n as the sum of two of its halves, as in 8=4+4 8=4+4 or 32=16+16 32=16+16. Each of those halves is an even power of 2 2, so it’s a square.
When n n is even, 2 n 2 n is itself a square, which gives us one representation, and it’s also the sum of four of its quarters, which is another. For example 64=64=16+16+16+16 64=64=16+16+16+16.
It is clear that these representations are always available to us. But why are they the only ones? How do we know that there isn’t some other way of representing, say, 2 23 2 23 as a sum of four squares in some completely different way?
The key is this: if N N is an even number, the number of representations of N N and of 4 N 4 N is the same.
Given a representation of N N as a sum of four squares, say
N=a 2+b 2+c 2+d 2 N=a 2+b 2+c 2+d 2,
we can produce a representation of 4 N 4 N simply like so:
4 N=(2 a)2+(2 b)2+(2 c)2+(2 d)2 4 N=(2 a)2+(2 b)2+(2 c)2+(2 d)2.
That’s kind of obvious. The interesting part is the opposite: any representation of 4 N 4 N must involve only even squares, and therefore can be halved to produce a representation of N N.
Why can’t there be any odd squares involved? Ah, simple: squares of odd numbers leave a remainder of 1 1 upon division by 8 8, and squares of even numbers leave a remainder of 0 0 or 4 4. When you add four squares, if any odd ones are involved, the sum total cannot be 0 0 mod 8 8, but it needs to be: 4 N 4 N is divisible by 8 8 when N N is even.
Therefore, representations of 4 N 4 N as sums of four squares must involve only even squares, and so each one is simply a doubling of a representation of N N. In other words, N N and 4 N 4 N have the same number of representations, which explains why our pattern has a period of 2 2. Once we’ve determined the number of representations of 2 2 and 4 4, the rest just follows.
Now, what about those signs? Well, for odd powers of 2 2 we see that we actually have 3 3 representations, or 4 4 if we care about the order, and for even ones we have 6 6 representations, or 17 17 if we care about order.
Even better, suppose we bring in the 0 0’s as well, and we take into account the order of all the summands. So now, the representations 4 2+4 2+0 2+0 2 4 2+4 2+0 2+0 2 is counted differently from 0 2+4 2+0 2+4 2 0 2+4 2+0 2+4 2.
This may seem strange, but it yields a very nice and clean result: all powers of 2 2 have exactly 24 24 representations. You can check that this is so for both the odd-power and even-power cases we analyzed.
In fact, a famous and wonderful theorem of Jacobi tells you exactly how many representations exist for any natural number as a sum of four squares, when counted in this specific way. The number is this:
r 4(N)=8∑m∣N,4∤m m r 4(N)=8∑m∣N,4∤m m
In words, the number of ways to represent N N as a sum of four squares is eight times the sum of those divisors of N N that aren’t divisible by 4 4. When N=2 n N=2 n, there are just two such divisors, 1 1 and 2 2, and their sum is 3 3, and multiplying by eight gives 24 24, just like we’ve found.
Math works.
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Richard P
MMath in Mathematics, Churchill College, Cambridge (Graduated 2011) · Author has 774 answers and 301K answer views
·Feb 21
Related
How many integers from 1 1 to 1000 1000 can be written as the sum of three nonnegative integer powers of two?
Let the integer be n (so 1 ≤ n ≤ 1000), and write it in binary notation. The integer n can be written as the sum of three distinct nonnegative integer powers of two precisely when the binary notation of n contains exactly three “1” digits.
Since n ≤ 1000 < 1024 = 2^10, we have at most 10 binary digits. Include leading zeros where necessary so that all representations have exactly 10 digits.
Now note that if n ≥ 15 64 = 960, its binary representation starts with at least 4 “1” digits. Indeed, if n > 7 128 = 896, its binary representation starts with at least 3 “1” digits, with at least one
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Let the integer be n (so 1 ≤ n ≤ 1000), and write it in binary notation. The integer n can be written as the sum of three distinct nonnegative integer powers of two precisely when the binary notation of n contains exactly three “1” digits.
Since n ≤ 1000 < 1024 = 2^10, we have at most 10 binary digits. Include leading zeros where necessary so that all representations have exactly 10 digits.
Now note that if n ≥ 15 64 = 960, its binary representation starts with at least 4 “1” digits. Indeed, if n > 7 128 = 896, its binary representation starts with at least 3 “1” digits, with at least one further “1” digit in the rest of the binary representation. So no integer greater than 896 can be written in the required form.
So we can choose three of the 10 binary digits, and set them to “1” and the rest of the binary digits to zero. This can be done in 10!/(3! 7!) = 120 ways, giving 120 integers which can be represented as the sum of three distinct nonnegative integer powers of two. The lowest such integer is 7 = 1 + 2 + 4 and the highest is 896 = 512 + 256 + 128.
If we do not require the three powers of two to be distinct, then we also allow the value of n to be expressible as the sum of two distinct nonnegative integer powers of two (simply divide the larger power into two equal halves, e.g. 10 = 4+4+2, or 24 = 8+8+8), or as a single power of two (e.g. 8 = 4+2+2). However the two cases of n=1 and n=2 cannot be so expressed. So we find that there are 10!/(2! 8!) = 45 integers with two binary “1” digits, and 8 single powers of 2 which can be expressed in the right form.
So in all there are 120 + 45 + 8 = 173 integers n, with 1 ≤ n ≤ 1000 which can be represented as the sum of three nonnegative integer powers of two. Of these, 120 can be represented as the sum of three distinct nonnegative integer powers of two.
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Enrico Gregorio
Associate professor in Algebra · Upvoted by
Alon Amit
, Lover of math. Also, Ph.D. and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 17.7K answers and 15.4M answer views
·Updated May 11
Related
How do I find a type of sum ∞∑n=1 2 n n 2 n!∑n=1∞2 n n 2 n!?
Start by rewriting your series as
2∑n≥1 2 n−1 n(n−1)!=2∑n≥0 2 n(n+1)n!2∑n≥1 2 n−1 n(n−1)!=2∑n≥0 2 n(n+1)n!
which is a special case of
f(x)=2∑n≥0(n+1)x n n!=2∑n≥1 x n(n−1)!+2∑n≥0 x n n!f(x)=2∑n≥0(n+1)x n n!=2∑n≥1 x n(n−1)!+2∑n≥0 x n n!
Thus we can rewrite
f(x)=2 x∑n≥0 x n n!+2∑n≥0 x n n!=2 x e x+2 e x=(2 x+2)e x f(x)=2 x∑n≥0 x n n!+2∑n≥0 x n n!=2 x e x+2 e x=(2 x+2)e x
For x=2 x=2 we have
f(2)=6 e 2 f(2)=6 e 2
If you just want to show that the series is convergent, you can apply the ratio test:
2 n+1(n+1)2/(n+1)!2 n n 2/n!=2 n+1(n+1)2 n 2 2 n+1(n+1)2/(n+1)!2 n n 2/n!=2 n+1(n+1)2 n 2
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Mark Bradley
Economist
·9mo
What are the dumbest financial mistakes most Americans make?
Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one
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Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up.
Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one of the biggest mistakes and easiest ones to fix.
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Ellis Cave
40+ years as an Electrical Engineer · Author has 7.7K answers and 4.3M answer views
·Feb 21
Related
How many integers from 1 1 to 1000 1000 can be written as the sum of three nonnegative integer powers of two?
Brute force solution, using the J programming language:
+/m=.1e3>s=.+/"1 c3=.(3 comb 10){p2=.2^1 to 10
84
The answer is that there are 84 integers from 1 to 1000 that can be written as the sum of three unique powers of two.
List each of the 84 sums, along with its three powers of 2:
(m#s),.m#c3
14 2 4 8
22 2 4 16
38 2 4 32
70 2 4 64
134 2 4 128
262 2 4 256
518 2 4 512
26 2 8 16
42 2 8 32
74 2 8 64
138 2 8 128
266 2 8 256
522 2 8 512
50 2 16 32
82 2 16 64
146 2 16 128
274 2 16 256
530 2 16 512
98 2 32 64
162 2 32 128
290 2 32 256
546 2 32 512
194 2 64 128
322 2 64 256
578 2 64 512
386 2 128 256
642 2 128 512
770 2 256 512
28 4
Continue Reading
Brute force solution, using the J programming language:
+/m=.1e3>s=.+/"1 c3=.(3 comb 10){p2=.2^1 to 10
84
The answer is that there are 84 integers from 1 to 1000 that can be written as the sum of three unique powers of two.
List each of the 84 sums, along with its three powers of 2:
(m#s),.m#c3
14 2 4 8
22 2 4 16
38 2 4 32
70 2 4 64
134 2 4 128
262 2 4 256
518 2 4 512
26 2 8 16
42 2 8 32
74 2 8 64
138 2 8 128
266 2 8 256
522 2 8 512
50 2 16 32
82 2 16 64
146 2 16 128
274 2 16 256
530 2 16 512
98 2 32 64
162 2 32 128
290 2 32 256
546 2 32 512
194 2 64 128
322 2 64 256
578 2 64 512
386 2 128 256
642 2 128 512
770 2 256 512
28 4 8 16
44 4 8 32
76 4 8 64
140 4 8 128
268 4 8 256
524 4 8 512
52 4 16 32
84 4 16 64
148 4 16 128
276 4 16 256
532 4 16 512
100 4 32 64
164 4 32 128
292 4 32 256
548 4 32 512
196 4 64 128
324 4 64 256
580 4 64 512
388 4 128 256
644 4 128 512
772 4 256 512
56 8 16 32
88 8 16 64
152 8 16 128
280 8 16 256
536 8 16 512
104 8 32 64
168 8 32 128
296 8 32 256
552 8 32 512
200 8 64 128
328 8 64 256
584 8 64 512
392 8 128 256
648 8 128 512
776 8 256 512
112 16 32 64
176 16 32 128
304 16 32 256
560 16 32 512
208 16 64 128
336 16 64 256
592 16 64 512
400 16 128 256
656 16 128 512
784 16 256 512
224 32 64 128
352 32 64 256
608 32 64 512
416 32 128 256
672 32 128 512
800 32 256 512
448 64 128 256
704 64 128 512
832 64 256 512
896 128 256 512
Upvote ·
Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
James McElhatton Ph.D. (Glasgow, 1976)
, B.Sc. Mathematics & Chemistry, University of Malta (1967) · Author has 8.7K answers and 171.8M answer views
·2y
Related
How do you find all n∈N n∈N which can be written as a sum of 2 2 odd composite numbers?
Observe the numbers 9 9, 25 25 and 35 35. They are all odd and composite, but they leave different remainders upon division by 3 3: the remainders are 0 0, 1 1 and 2 2 respectively. This means that you can subtract one of these numbers from any n n to obtain a multiple of 3 3.
With that, given any number n n:
If n n is divisible by 3 3, write n=9+(n−9)n=9+(n−9)
If n n is 1 1 mod 3 3 write n=25+(n−25)n=25+(n−25)
If n n is 2 2 mod 3 3 write n=35+(n−35)n=35+(n−35).
The n−X n−X part is always divisible by 3 3, so it’s composite as long as it is greater than 3 3. And if n n is even, both summands are odd.
Therefore, the numbers representable as a sum of two odd composite numbers ar
Continue Reading
Observe the numbers 9 9, 25 25 and 35 35. They are all odd and composite, but they leave different remainders upon division by 3 3: the remainders are 0 0, 1 1 and 2 2 respectively. This means that you can subtract one of these numbers from any n n to obtain a multiple of 3 3.
With that, given any number n n:
If n n is divisible by 3 3, write n=9+(n−9)n=9+(n−9)
If n n is 1 1 mod 3 3 write n=25+(n−25)n=25+(n−25)
If n n is 2 2 mod 3 3 write n=35+(n−35)n=35+(n−35).
The n−X n−X part is always divisible by 3 3, so it’s composite as long as it is greater than 3 3. And if n n is even, both summands are odd.
Therefore, the numbers representable as a sum of two odd composite numbers are precisely all even numbers which are at least 40 40, as well as 18,24,30,34 18,24,30,34 and 36 36. You can directly check the finite list of small even numbers to see that they are not representable in this way.
And of course, odd numbers aren’t the sum of two odd numbers, so the answer is complete.
Upvote ·
99 97
9 4
9 1
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) and
Siddhant Grover
, MSc in Statistics & BSc Mathematics, Hindu College, University of Delhi (2022) · Author has 8.3K answers and 20.4M answer views
·Updated 4y
Related
How calculate n∑k=0(3 n 3 k)∑k=0 n(3 n 3 k) solution is 2(−1)n+8 n 3.2(−1)n+8 n 3. ?
Remark: Jan van Delden has already given a great and well-motivated solution to this post (if you have not read his first, please do so now). Please treat this post as an extension of his solution.
The series in question (as well as those that Jan van Delden give at the end of his answer) can be addressed are special cases of the following theorem:
Multisection Formula: Let ω=e 2 π i/n ω=e 2 π i/n and fix l∈{0,1,…,n−1}l∈{0,1,…,n−1}.
If f(x)=∑∞k=0 a k x k f(x)=∑k=0∞a k x k, then
∑k≡l mod n a k x k=1 n n−1∑s=0 ω−l s f(ω s x),∑k≡l mod n a k x k=1 n∑s=0 n−1 ω−l s f(ω s x),
or equivalen
Continue Reading
Remark: Jan van Delden has already given a great and well-motivated solution to this post (if you have not read his first, please do so now). Please treat this post as an extension of his solution.
The series in question (as well as those that Jan van Delden give at the end of his answer) can be addressed are special cases of the following theorem:
Multisection Formula: Let ω=e 2 π i/n ω=e 2 π i/n and fix l∈{0,1,…,n−1}l∈{0,1,…,n−1}.
If f(x)=∑∞k=0 a k x k f(x)=∑k=0∞a k x k, then
∑k≡l mod n a k x k=1 n n−1∑s=0 ω−l s f(ω s x),∑k≡l mod n a k x k=1 n∑s=0 n−1 ω−l s f(ω s x),
or equivalently,
∞∑j=0 a n j+l x n j+l=1 n n−1∑s=0 ω−l s f(ω s x).∑j=0∞a n j+l x n j+l=1 n∑s=0 n−1 ω−l s f(ω s x).
Proof: To show this, note that
n−1∑s=0 ω−l s f(ω s x)=n−1∑s=0 ω−l s∞∑k=0 a k(ω s x)k=∞∑k=0 a k x k n−1∑s=0 ω(k−l)s.∑s=0 n−1 ω−l s f(ω s x)=∑s=0 n−1 ω−l s∑k=0∞a k(ω s x)k=∑k=0∞a k x k∑s=0 n−1 ω(k−l)s.
However, since ω,ω 2,…ω n−1 ω,ω 2,…ω n−1 are all n n-th roots of unity and satisfy
x n−1 x−1=x n−1+x n−2+…+1=0,x n−1 x−1=x n−1+x n−2+…+1=0,
it follows that
1+ω m+ω 2 m+…+ω(n−1)m={0 if n∤m,n if n∣m.1+ω m+ω 2 m+…+ω(n−1)m={0 if n∤m,n if n∣m.
Then, ∑n−1 s=0 ω(k−l)s∑s=0 n−1 ω(k−l)s is nonzero (and equals n n) precisely when k≡l mod n k≡l mod n. Therefore, we obtain
n−1∑s=0 ω−l s f(ω s x)=∑k≡l mod n n⋅a k x k,∑s=0 n−1 ω−l s f(ω s x)=∑k≡l mod n n⋅a k x k,
which is equivalent to the result that we wanted to prove. ■◼
Back to the original question. Given all of the 3s that occur in the series, we suspect that this comes from a trisection of the Binomial Theorem (the fact that series is finite only simplifies matters; we need not worry about convergence).
Since we know that
(1+x)3 n=3 n∑k=0(3 n k)x k,(1+x)3 n=∑k=0 3 n(3 n k)x k,
applying the Multisection Formula (with n=3 n=3 in its statement) yields
∑k≡0 mod 3(3 n k)x k=1 3[(1+x)3 n+(1+ω x)3 n+(1+ω 2 x)3 n],∑k≡0 mod 3(3 n k)x k=1 3[(1+x)3 n+(1+ω x)3 n+(1+ω 2 x)3 n],
where ω=e 2 π i/3=−1+i√3 2 ω=e 2 π i/3=−1+i 3 2.
Then, letting x=1 x=1 yields
∑k≡0 mod 3(3 n k)=1 3(2 3 n+(1+ω)3 n+(1+ω 2)3 n).∑k≡0 mod 3(3 n k)=1 3(2 3 n+(1+ω)3 n+(1+ω 2)3 n).
Since ω 2+ω+1=0 ω 2+ω+1=0, this readily yields (after re-indexing the left side)
n∑k=0(3 n 3 k)=2⋅(−1)n+8 n 3,∑k=0 n(3 n 3 k)=2⋅(−1)n+8 n 3,
as required.
Final remarks: The Multisection Formula takes care of the other two ‘trisections’ with no real extra effort.
For instance,
∑k≡1 mod 3(3 n k)x k=1 3[(1+x)3 n+ω−1(1+ω x)3 n+ω−2(1+ω 2 x)3 n].∑k≡1 mod 3(3 n k)x k=1 3[(1+x)3 n+ω−1(1+ω x)3 n+ω−2(1+ω 2 x)3 n].
Letting x=1 x=1 and simplifying yields
n−1∑k=0(3 n 3 k+1)=8 n−(−1)n 3.∑k=0 n−1(3 n 3 k+1)=8 n−(−1)n 3.
Analogously,
∑k≡2 mod 3(3 n k)x k=1 3[(1+x)3 n+ω−2(1+ω x)3 n+ω−4(1+ω 2 x)3 n].∑k≡2 mod 3(3 n k)x k=1 3[(1+x)3 n+ω−2(1+ω x)3 n+ω−4(1+ω 2 x)3 n].
and letting x=1 x=1 yields
n−1∑k=0(3 n 3 k+2)=8 n−(−1)n 3.∑k=0 n−1(3 n 3 k+2)=8 n−(−1)n 3.
Upvote ·
99 27
9 1
9 3
Sergei Michailov
Retired System Analyst (2009–present) · Author has 5.5K answers and 3.5M answer views
·7y
Related
How many 2-digit numbers can be written as the sum of exactly six different powers of 2, including 2^0?
The answer is 2. The numbers are:
32+16+8+4+3+1 = 63 and
64+16+8+4+2+1 = 95
The eligible powers of 2 are :
64, 32, 16, 8, 4, 2, 1 (128 is too much because it has three digits).
Case 1. We do not include 64 in the sum, then we have only one sum which is
32+16+8+4+3+1 = 63. This case is good.
Case 2. We do include 64.
Case 2.1. Then we can not include 32 because 64+32 = 96. 16, 8, and 4 are out of picture because they produce the sum of 100 or more before we choose the remaining three numbers. What is left is 2 and 1. But then we have 4 powers of 2, not 6.
Case 2.2. We have 64+16. With the remaining four
Continue Reading
The answer is 2. The numbers are:
32+16+8+4+3+1 = 63 and
64+16+8+4+2+1 = 95
The eligible powers of 2 are :
64, 32, 16, 8, 4, 2, 1 (128 is too much because it has three digits).
Case 1. We do not include 64 in the sum, then we have only one sum which is
32+16+8+4+3+1 = 63. This case is good.
Case 2. We do include 64.
Case 2.1. Then we can not include 32 because 64+32 = 96. 16, 8, and 4 are out of picture because they produce the sum of 100 or more before we choose the remaining three numbers. What is left is 2 and 1. But then we have 4 powers of 2, not 6.
Case 2.2. We have 64+16. With the remaining four digits we have 64+16+8+4+2+1 = 95. This is the second good case.
All other cases have less than six powers of 2.
Upvote ·
9 1
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Contents
(Top)
1 Uses
2 History
3 Properties
3.1 Propagation modes and cutoff frequencies
3.2 Impedance matching
4 Electromagnetic waveguides
4.1 Radio-frequency waveguides
4.2 Optical waveguides
5 Acoustic waveguides
6 Infinite tubes
7 Sound synthesis
8 See also
9 Notes
10 References
11 External links
Waveguide
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From Wikipedia, the free encyclopedia
Structure that guides waves efficiently
A waveguide is a structure that guides waves by restricting the transmission of energy to one direction. Common types of waveguides include acoustic waveguides which direct sound, optical waveguides which direct light, and radio-frequency waveguides which direct electromagnetic waves other than light like radio waves.
Without the physical constraint of a waveguide, waves would expand into three-dimensional space and their intensities would decrease according to the inverse square law.
There are different types of waveguides for different types of waves. The original and most common meaning is a hollow conductive metal pipe used to carry high frequency radio waves, particularly microwaves. Dielectric waveguides are used at higher radio frequencies, and transparent dielectric waveguides and optical fibers serve as waveguides for light. In acoustics, air ducts and horns are used as waveguides for sound in musical instruments and loudspeakers, and specially-shaped metal rods conduct ultrasonic waves in ultrasonic machining.
The geometry of a waveguide reflects its function; in addition to more common types that channel the wave in one dimension, there are two-dimensional slab waveguides which confine waves to two dimensions. The frequency of the transmitted wave also dictates the size of a waveguide: each waveguide has a cutoff wavelength determined by its size and will not conduct waves of greater wavelength; an optical fiber that guides light will not transmit microwaves which have a much larger wavelength. Some naturally occurring structures can also act as waveguides. The SOFAR channel layer in the ocean can guide the sound of whale song across enormous distances. Any shape of waveguide can support EM waves, however irregular shapes are difficult to analyse. Commonly used waveguides are rectangular or circular in cross-section.
Uses
[edit]
The uses of waveguides for transmitting signals were known even before the term was coined. The phenomenon of sound waves guided through a taut wire have been known for a long time, as well as sound through a hollow pipe such as a cave or medical stethoscope. Other uses of waveguides are in transmitting power between the components of a system such as radio, radar or optical devices. Waveguides are the fundamental principle of guided wave testing (GWT), one of the many methods of non-destructive evaluation.
Specific examples:
Optical fibers transmit light and signals for long distances with low attenuation and a wide usable range of wavelengths.
In a microwave oven a waveguide transfers power from the magnetron, where waves are formed, to the cooking chamber.
In a radar, a waveguide transfers radio frequency energy to and from the antenna, where the impedance needs to be matched for efficient power transmission (see below).
Rectangular and circular waveguides are commonly used to connect feeds of parabolic dishes to their electronics, either low-noise receivers or power amplifier/transmitters.
Waveguides are used in scientific instruments to measure optical, acoustic and elastic properties of materials and objects. The waveguide can be put in contact with the specimen (as in a medical ultrasonography), in which case the waveguide ensures that the power of the testing wave is conserved, or the specimen may be put inside the waveguide (as in a dielectric constant measurement, so that smaller objects can be tested and the accuracy is better.
A transmission line is a commonly used specific type of waveguide.
History
[edit]
| |
| This section duplicates the scope of other articles, specifically Waveguide (electromagnetism)#History. Please discuss this issue and help introduce a summary style to the section by replacing the section with a link and a summary or by splitting the content into a new article. (November 2020) |
The first structure for guiding waves was proposed by J. J. Thomson in 1893, and was first experimentally tested by Oliver Lodge in 1894. The first mathematical analysis of electromagnetic waves in a metal cylinder was performed by Lord Rayleigh in 1897.: 8 For sound waves, Lord Rayleigh published a full mathematical analysis of propagation modes in his seminal work, "The Theory of Sound". Jagadish Chandra Bose researched millimeter wavelengths using waveguides, and in 1897 described to the Royal Institution in London his research carried out in Kolkata.
The study of dielectric waveguides (such as optical fibers, see below) began as early as the 1920s, by several people, most famous of which are Rayleigh, Sommerfeld and Debye. Optical fiber began to receive special attention in the 1960s due to its importance to the communications industry.
The development of radio communication initially occurred at the lower frequencies because these could be more easily propagated over large distances. The long wavelengths made these frequencies unsuitable for use in hollow metal waveguides because of the impractically large diameter tubes required. Consequently, research into hollow metal waveguides stalled and the work of Lord Rayleigh was forgotten for a time and had to be rediscovered by others. Practical investigations resumed in the 1930s by George C. Southworth at Bell Labs and Wilmer L. Barrow at MIT. Southworth at first took the theory from papers on waves in dielectric rods because the work of Lord Rayleigh was unknown to him. This misled him somewhat; some of his experiments failed because he was not aware of the phenomenon of waveguide cutoff frequency already found in Lord Rayleigh's work. Serious theoretical work was taken up by John R. Carson and Sallie P. Mead. This work led to the discovery that for the TE01 mode in circular waveguide losses go down with frequency and at one time this was a serious contender for the format for long-distance telecommunications.: 544–548
The importance of radar in World War II gave a great impetus to waveguide research, at least on the Allied side. The magnetron, developed in 1940 by John Randall and Harry Boot at the University of Birmingham in the United Kingdom, provided a good power source and made microwave radar feasible. The most important centre of US research was at the Radiation Laboratory (Rad Lab) at MIT but many others took part in the US, and in the UK such as the Telecommunications Research Establishment. The head of the Fundamental Development Group at Rad Lab was Edward Mills Purcell. His researchers included Julian Schwinger, Nathan Marcuvitz, Carol Gray Montgomery, and Robert H. Dicke. Much of the Rad Lab work concentrated on finding lumped element models of waveguide structures so that components in waveguide could be analysed with standard circuit theory. Hans Bethe was also briefly at Rad Lab, but while there he produced his small aperture theory which proved important for waveguide cavity filters, first developed at Rad Lab. The German side, on the other hand, largely ignored the potential of waveguides in radar until very late in the war. So much so that when radar parts from a downed British plane were sent to Siemens & Halske for analysis, even though they were recognised as microwave components, their purpose could not be identified.
At that time, microwave techniques were badly neglected in Germany. It was generally believed that it was of no use for electronic warfare, and those who wanted to do research work in this field were not allowed to do so.
— H. Mayer, wartime vice-president of Siemens & Halske
German academics were even allowed to continue publicly publishing their research in this field because it was not felt to be important.: 548–554: 1055, 1057
Immediately after World War II waveguide was the technology of choice in the microwave field. However, it has some problems; it is bulky, expensive to produce, and the cutoff frequency effect makes it difficult to produce wideband devices. Ridged waveguide can increase bandwidth beyond an octave, but a better solution is to use a technology working in TEM mode (that is, non-waveguide) such as coaxial conductors since TEM does not have a cutoff frequency. A shielded rectangular conductor can also be used and this has certain manufacturing advantages over coax and can be seen as the forerunner of the planar technologies (stripline and microstrip). However, planar technologies really started to take off when printed circuits were introduced. These methods are significantly cheaper than waveguide and have largely taken its place in most bands. However, waveguide is still favoured in the higher microwave bands from around Ku band upwards.: 556–557: 21–27, 21–50
Properties
[edit]
Propagation modes and cutoff frequencies
[edit]
A propagation mode in a waveguide is one solution of the wave equations, or, in other words, the form of the wave. Due to the constraints of the boundary conditions, there are only limited frequencies and forms for the wave function which can propagate in the waveguide. The lowest frequency in which a certain mode can propagate is the cutoff frequency of that mode. The mode with the lowest cutoff frequency is the fundamental mode of the waveguide, and its cutoff frequency is the waveguide cutoff frequency.: 38
Propagation modes are computed by solving the Helmholtz equation alongside a set of boundary conditions depending on the geometrical shape and materials bounding the region. The usual assumption for infinitely long uniform waveguides allows us to assume a propagating form for the wave, i.e. stating that every field component has a known dependency on the propagation direction (i.e. ). More specifically, the common approach is to first replace all unknown time-varying fields (assuming for simplicity to describe the fields in cartesian components) with their complex phasors representation , sufficient to fully describe any infinitely long single-tone signal at frequency , (angular frequency ), and rewrite the Helmholtz equation and boundary conditions accordingly. Then, every unknown field is forced to have a form like , where the term represents the propagation constant (still unknown) along the direction along which the waveguide extends to infinity. The Helmholtz equation can be rewritten to accommodate such form and the resulting equality needs to be solved for and , yielding in the end an eigenvalue equation for and a corresponding eigenfunction for each solution of the former.
The propagation constant of the guided wave is complex, in general. For a lossless case, the propagation constant might be found to take on either real or imaginary values, depending on the chosen solution of the eigenvalue equation and on the angular frequency . When is purely real, the mode is said to be "below cutoff", since the amplitude of the field phasors tends to exponentially decrease with propagation; an imaginary , instead, represents modes said to be "in propagation" or "above cutoff", as the complex amplitude of the phasors does not change with .
Impedance matching
[edit]
In circuit theory, the impedance is a generalization of electrical resistance in the case of alternating current, and is measured in ohms (). A waveguide in circuit theory is described by a transmission line having a length and characteristic impedance.: 2–3, 6–12: 14 In other words, the impedance indicates the ratio of voltage to current of the circuit component (in this case a waveguide) during propagation of the wave. This description of the waveguide was originally intended for alternating current, but is also suitable for electromagnetic and sound waves, once the wave and material properties (such as pressure, density, dielectric constant) are properly converted into electrical terms (current and impedance for example).: 14
Impedance matching is important when components of an electric circuit are connected (waveguide to antenna for example): The impedance ratio determines how much of the wave is transmitted forward and how much is reflected. In connecting a waveguide to an antenna a complete transmission is usually required, so an effort is made to match their impedances.
The reflection coefficient can be calculated using: , where (Gamma) is the reflection coefficient (0 denotes full transmission, 1 full reflection, and 0.5 is a reflection of half the incoming voltage), and are the impedance of the first component (from which the wave enters) and the second component, respectively.
An impedance mismatch creates a reflected wave, which added to the incoming waves creates a standing wave. An impedance mismatch can be also quantified with the standing wave ratio (SWR or VSWR for voltage), which is connected to the impedance ratio and reflection coefficient by: , where are the minimum and maximum values of the voltage absolute value, and the VSWR is the voltage standing wave ratio, which value of 1 denotes full transmission, without reflection and thus no standing wave, while very large values mean high reflection and standing wave pattern.
Electromagnetic waveguides
[edit]
Radio-frequency waveguides
[edit]
Main articles: Waveguide (radio frequency) and Transmission line
Waveguides can be constructed to carry waves over a wide portion of the electromagnetic spectrum, but are especially useful in the microwave and optical frequency ranges. Depending on the frequency, they can be constructed from either conductive or dielectric materials. Waveguides are used for transferring both power and communication signals.: 1–3: xiii–xiv
Optical waveguides
[edit]
Main article: Waveguide (optics)
Waveguides used at optical frequencies are typically dielectric waveguides, structures in which a dielectric material with high permittivity, and thus high index of refraction, is surrounded by a material with lower permittivity. The structure guides optical waves by total internal reflection. An example of an optical waveguide is optical fiber.
Other types of optical waveguide are also used, including photonic-crystal fiber, which guides waves by any of several distinct mechanisms. Guides in the form of a hollow tube with a highly reflective inner surface have also been used as light pipes for illumination applications. The inner surfaces may be polished metal, or may be covered with a multilayer film that guides light by Bragg reflection (this is a special case of a photonic-crystal fiber). One can also use small prisms around the pipe which reflect light via total internal reflection —such confinement is necessarily imperfect, however, since total internal reflection can never truly guide light within a lower-index core (in the prism case, some light leaks out at the prism corners).
Acoustic waveguides
[edit]
Main article: Waveguide (acoustics)
An acoustic waveguide is a physical structure for guiding sound waves. Sound in an acoustic waveguide behaves like electromagnetic waves on a transmission line. Waves on a string, like the ones in a tin can telephone, are a simple example of an acoustic waveguide. Another example are pressure waves in the pipes of an organ. The term acoustic waveguide is also used to describe elastic waves guided in micro-scale devices, like those employed in piezoelectric delay lines and in stimulated Brillouin scattering.
Infinite tubes
[edit]
Main article: Wave equation
A waveguide (or tube) impose a boundary condition on the wave equation such that the wave function must be equal to zero on the boundary and that the allowed region is finite in all but one dimension. An infinitely long cylinder is an example. Mathematically, any tube with a bulge, where the width of the tube increases, admits at least one non-propagating bound state. Using the variational principles, Jeffrey Goldstone and Robert Jaffe have shown a tube of constant width with a twist admits a bound state.
Sound synthesis
[edit]
Main article: Digital waveguide synthesis
Sound synthesis uses digital delay lines as computational elements to simulate wave propagation in tubes of wind instruments and the vibrating strings of string instruments.
See also
[edit]
Circular polarization
Earth–ionosphere waveguide
Linear polarization
Orthomode transducer
Polarization
Flap attenuator
Notes
[edit]
^ Institute of Electrical and Electronics et al. 1997.
^ Payne & Webb 1971.
^ Olisa, Khan & Starr 2021.
^ Baker-Jarvis 1990.
^ EETech Media.
^ McLachlan 1964.
^ Rayleigh 1894.
^ Emerson 1997a.
^ Emerson 1997b, Reprint.
^ a b c Balanis 1989.
^ Oliner 2006, Reprint.
^ a b Oliner 2006.
^ Levy & Cohn 1984.
^ Han & Hwang 2012.
^ a b Cronin 1995.
^ Pozar 2012.
^ Ramo, Whinnery & Van Duzer 1994.
^ Marcuvitz 1951.
^ Beranek & Mellow 2012, Characteristic Impedance.
^ a b c Khare & Nema 2012.
^ Beranek & Mellow 2012, Pressure and density effects.
^ Zhang, Krooswyk & Ou 2015, Reflection coefficient.
^ Okamoto 2010.
^ Herres.
^ "Light Pipe Technologies". Archived from the original on March 14, 2007.
^ Saxe 1989.
^ Goldstone & Jaffe 1992.
^ Smith 1996.
References
[edit]
Baker-Jarvis, James (1990). Transmission/Reflection and short-Circuit Line Permittivity Measurements (PDF). Boulder, Colorado: National Institute of Standards and Technology.
Balanis, Constantine A. (1989). Engineering Electromagnetics. Wiley. ISBN 978-0-471-62194-2. Archived from the original on May 14, 2009.
Beranek, Leo Leroy; Mellow, Tim (2012). Acoustics: Sound Fields and Transducers. Academic Press. ISBN 978-0-12-391421-7.
Cronin, N. J. (1995). Microwave and Optical Waveguides. CRC Press. p. 38. ISBN 978-0-7503-0216-6.
EETech Media, LLC. "Waveguides". All About Circuits. Retrieved December 31, 2023.
Emerson, D.T. (1997a). "The work of Jagadis Chandra Bose: 100 years of mm-wave research". 1997 IEEE MTT-S International Microwave Symposium Digest. Vol. 2. pp. 553–556. doi:10.1109/MWSYM.1997.602853. ISBN 0-7803-3814-6. S2CID 9039614.
Emerson, D.T. (1997b). "The work of Jagadis Chandra Bose: 100 years of millimeter-wave research". IEEE Transactions on Microwave Theory and Techniques. 45 (12): 2267–2273. Bibcode:1997ITMTT..45.2267E. doi:10.1109/22.643830. ISBN 9780986488511.
Goldstone, J.; Jaffe, R. L. (1992). "Bound states in twisting tubes". Physical Review B. 45 (24): 14100–14107. Bibcode:1992PhRvB..4514100G. doi:10.1103/PhysRevB.45.14100. PMID 10001530.
Han, C.C.; Hwang, Y. (2012). "Satellite attenas". In Lo, Y. T.; Lee, S. W. (eds.). Antenna Handbook: Volume III Applications. Springer Science & Business Media. ISBN 978-1-4615-2638-4.
Herres, David. "Optical Fiber: A waveguide for light and internal reflection". Test & Measurement Tips. Retrieved January 1, 2024.
Institute of Electrical and Electronics, Engineers; Radatz, Jane; Standards Coordinating Committee, Terms and Definitions; IEEE Computer Society, Standards Coordinating Committee (1997). The IEEE Standard Dictionary of Electrical and Electronics Terms (6th ed.). New York, New York: Institute of Electrical and Electronics Engineers. ISBN 978-1-55937-833-8.
Khare, Rashmi; Nema, Rajesh (2012). "Review of Impedance Matching Networks for Bandwidth Enhancement". International Journal of Emerging Technology and Advanced Engineering. 2 (1): 92–96.
Levy, R.; Cohn, S.B. (1984). "A History of Microwave Filter Research, Design, and Development". IEEE Transactions on Microwave Theory and Techniques. 32 (9): 1055–1067. Bibcode:1984ITMTT..32.1055L. doi:10.1109/TMTT.1984.1132817.
Lo, Y. T.; Lee, S. W. (December 6, 2012). Antenna Handbook: Volume III Applications. Springer Science & Business Media. ISBN 978-1-4615-2638-4.
McLachlan, Norman William (1964). Theory and Application of Mathieu Functions, By N.W. Mclachlan. New York, New York: Dover.
Marcuvitz, Nathan (1951). Waveguide Handbook. IET. ISBN 978-0-86341-058-1. {{cite book}}: ISBN / Date incompatibility (help)
Okamoto, Katsunari (2010). Fundamentals of Optical Waveguides. Elsevier. ISBN 978-0-08-045506-8.
Oliner, Arthur A. (January 30, 2006). "The evolution of electromagnetic waveguides: from hollow metallic guides to microwave integrated circuits". In Sarkar, T. K.; Mailloux, Robert; Oliner, Arthur A.; Salazar-Palma, Magdalena; Sengupta, Dipak L. (eds.). History of Wireless. John Wiley & Sons. pp. 543–566. ISBN 978-0-471-78301-5.
Olisa, Samuel Chukwuemeka; Khan, Muhammad A.; Starr, Andrew (2021). "Review of Current Guided Wave Ultrasonic Testing (GWUT) Limitations and Future Directions". Sensors. 21 (3): 811. Bibcode:2021Senso..21..811O. doi:10.3390/s21030811. PMC 7865912. PMID 33530407.
Payne, R.; Webb, D. (1971). "Orientation by means of long range acoustic signaling in baleen whales". Annals of the New York Academy of Sciences. 188 (1): 110–141. Bibcode:1971NYASA.188..110P. doi:10.1111/j.1749-6632.1971.tb13093.x. ISSN 0077-8923. PMID 5288850. S2CID 42324742.
Pozar, David M. (2012). Microwave Engineering. John Wiley & Sons. ISBN 978-0-470-63155-3.
Ramo, Simon; Whinnery, John R.; Van Duzer, Theodore (1994). Fields and Waves in Communication Electronics. New York: John Wiley and Sons. pp. 321–324. ISBN 978-0-471-58551-0.
Rayleigh, John William Strutt Baron (1894). The Theory of Sound. Macmillan.
Saxe, Steven G. (1989). "Prismatic film light guides: Performance and recent developments". Solar Energy Materials. 19 (1): 95–109. doi:10.1016/0165-1633(89)90026-9. ISSN 0165-1633.
Smith, Julius O. (1996). "Physical Modeling Synthesis Update". Computer Music Journal. 20 (2): 44–56. doi:10.2307/3681331. ISSN 0148-9267. JSTOR 3681331.
Zhang, Hanqiao; Krooswyk, Steven; Ou, Jeff (2015). "Chapter 1 - Transmission line fundamentals". High Speed Digital Design. Morgan Kaufmann. pp. 1–26. ISBN 978-0-12-418663-7.
External links
[edit]
Look up waveguide in Wiktionary, the free dictionary.
Wikimedia Commons has media related to Waveguides.
Electromagnetic Waves and Antennas: Waveguides Archived March 29, 2024, at the Wayback Machine Sophocles J. Orfanidis, Department of Electrical and Computer Engineering, Rutgers University
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Volume 22, 2020, e00945
Case report Multiple cranial nerve palsies in malignant external otitis: A rare presentation of a rare condition
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Abstract
Malignant external otitis (MEO) is a rare inflammatory and infectious condition, typically caused by Pseudomonas aeruginosa, that mainly affects diabetic or immunocompromised elderly patients and is associated with severe morbidity and mortality. It begins in the external auditory canal and rapidly progresses through the skull base, leading to osteomyelitis and may result in cranial neuropathy, especially of the facial nerve. Here we describe a rare neurological presentation of MEO in a 65-year old diabetic man, who presented with an 8-month progressing left otitis externa and evolved with ipsilateral proptosis, ophthalmoplegia, blindness, facial palsy, hearing loss and contralateral evolvement of the temporal bone with hearing impairment. He was initially treated with oral ciprofloxacin and after one week was transferred to our tertiary hospital, where antibiotic therapy was switched to meropenem and vancomycin due to the severity of the case and to the hospitals microbiological profile. The patient underwent left canal wall-up mastoidectomy with insertion of ear ventilation tube bilaterally, with good recovery of right ear hearing capacity, but with no improvements of neurological deficits nor left hearing function. All microbiological tests performed were negative, and this was interpreted as a possible consequence of the early use of antibiotics. Unfortunately, the patient was infected by Sars-CoV-2 during hospitalization and passed away after ten days of COVID-19 intensive care unit internment.
Keywords
Malignant external otitis
Multiple cranial neuropathy
Ophthalmoplegia
Osteomyelitis
Diabetes mellitus
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Vieta's Formulas
In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments.
Contents
1 Statement
2 Proof
3 Problems
3.1 Introductory
3.2 Intermediate
4 Advanced
5 See also
Statement
Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots.
Vieta’s formulas then state that This can be compactly summarized as for some such that .
Proof
Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant .
Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof.
Problems
Here are some problems with solutions that utilize Vieta's quadratic formulas:
Introductory
2005 AMC 12B Problem 12
2007 AMC 12A Problem 21
2010 AMC 10A Problem 21
2003 AMC 10A Problem 18
2021 AMC 12A Problem 12
Intermediate
2017 AMC 12A Problem 23
2003 AIME II Problem 9
2008 AIME II Problem 7
2021 Fall AMC 12A Problem 23
2019 AIME I Problem 10
Advanced
2020 AIME I Problem 14
See also
Polynomial
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17621 | https://math.stackexchange.com/questions/380582/finding-poles-indicating-their-order-and-then-computing-their-residues | Skip to main content
Finding poles, indicating their order and then computing their residues
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I really don't understand the concept behind finding poles in Complex Analysis and I can't find anything on the internet or in books that helps me grasp the concept...
The following are past exam questions that I'm looking at but don't know where to go with with them in order to find their poles, indicate their order to then compute their residues.. Any help would be greatly appreciated...
(i)f(z)=sinz(z−1)sinhz
(ii)g(z)=sinzz(z2+4)
complex-analysis
singularity
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The general, intuitive idea of poles is that they are points where evaluating your function would entail dividing by zero. The order of the pole is the exponent in the factor that is going to zero in the denominator. It's best to start with some simple examples, such as rational functions:
f(z)=(z+1)(z−2)(z+1)(z−1)(z−3)2
Notice that the denominator goes to zero at z=−1,1,3. For z=−1, however, there's also a copy of (z+1) on the top, so this is a pole of order zero, or a removable singularity, so it normally doesn't count. At z=1, we have one copy of (z−1) in the denominator, so it's a pole of order 1. For z=3, we have (z−3) with multiplicity 2, so it's a pole of order 2.
Now let's look at a slightly more interesting example:
f(z)=zsinz
As we all know, sin(z)=0 when z=nπ, where n is an integer, and furthermore, all these zeros of sin(z) are single roots, so naturally we might think that f(z) has poles of order 1 at z=nπ for every n. However, the z in the numerator cancels out the zero at z=0 in the denominator, so in fact f(z) in this case has a pole of order 1 for z=nπ for every nonzero n, while for n=0, there won't be a pole (there is instead a removable singularity).
So the general strategy can be described as this:
Identify all the zeros in the denominator, along with their multiplicities.
Identify any zeros in the numerator that are also zeros in the denominator, along with their multiplicities.
Each zero in the denominator is a pole whose order is given by taking the multiplicity in the denominator and subtracting away the multiplicity in the numerator. If the order isn't positive, then it actually isn't a pole.
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answered May 3, 2013 at 20:12
Christopher A. WongChristopher A. Wong
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Thank you so much, your explanation is so much clearer than anything else i've found!!! I fully understand the simpler example but im still a little confused on the harder example, like the examples I have given.. For (ii) would there be a pole of z=2i with order 1, z=-2i with order 1, then another at z=0 with order 1.. But im not really sure if that is all. Then the zero for the numerator isn't also a zero in the denominator, so that isnt a pole?
Mathsstudent147
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Mathsstudent147
05/03/2013 20:30:02
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May 3, 2013 at 20:30
Correct. There are simple poles at z=±2i, and since sin0=0 z=0 will be a removable singularity.
mscook
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mscook
05/03/2013 20:34:35
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May 3, 2013 at 20:34
I got my z=0 with order 1 from the z part of the denomiator, as z=0 would make the denomiator zero.. is this right?
Mathsstudent147
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Mathsstudent147
05/03/2013 20:37:48
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May 3, 2013 at 20:37
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Yes, but at z=0 the numerator is 0 also (sin0=0), and they both have the same order. So the zero in the numerator "cancels" with the zero in the denominator, and it will be a removable singularity and not a pole.
mscook
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mscook
05/03/2013 20:57:50
Commented
May 3, 2013 at 20:57
Clear, concise explanation, thanks!
user29553
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user29553
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(i) The poles of f(z)=sinz(z−1)sinhz will be the poles of sinz along with the zeroes of (z−1)sinhz). Since sin(z) is analytic it will have no poles.
Thus the poles will by z=1 and the zeros of sinhz, which are z=nπi for z∈Z. and they will all be simple.
At z=1, we can use the formula Res(f,c)=limz→c(z−c)f(z). Thus Res(f,1)=limz→1sinzsinhz=sin1sinh1
For the z=nπi poles, we want to use the formula Res(gh,c)=g(c)h′(c), so Res(f,nπi)=sin(nπi)(nπi−1)cosh(nπi)=sinh(nπ)nπi−1
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why does a different formula for the residue have to be used to z=nπi?
Mathsstudent147
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05/03/2013 20:39:06
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May 3, 2013 at 20:39
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Either formula would work for any of the poles (they hold for all poles of order 1), however I think using them this way is far easier. For z=1 the (z−1) cancels out with the first method, so all you have to do is plug in 1. For the nπi poles, the limit method would be pretty untractable, but the derivative is easy.
mscook
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mscook
05/03/2013 20:49:52
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May 3, 2013 at 20:49
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Really they are actually just 2 different ways of approaching the same exact method, use the first approach if it's easier to cancel the (z−c) term and use the second if it's easier to take the derivative of the denominator.
mscook
–
mscook
05/03/2013 20:55:12
Commented
May 3, 2013 at 20:55
If it is written that find poles of function f (z)= 1/(z-1) for |z|>6 then what should we consider z=1 is pole or not?
Gilll
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Gilll
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Combinational Logic
Combinational circuits are built of five basic logic gates:
AND gate - output is 1 if BOTH inputs are 1
OR gate - output is 1 if AT LEAST one input is 1
XOR gate - output is 1 if ONLY one input is 1
NAND gate - output is 1 if AT LEAST one input is 0
NOR gate - output is 1 if BOTH inputs are 0
There is a sixth element in digital logic, the inverter (sometimes called a NOT gate). Inverters aren't truly gates, as they do not make any decisions. The output of an inverter is a 1 if the input is a 0, and vise versa.
A few things of note about the above image:
Usually, the name of the gate is not printed; the symbol is assumed to be sufficient for identification.
The A-B-Q type terminal notation is standard, although logic diagrams will usually omit them for signals which are not inputs or outputs to the system as a whole.
Two input devices are standard, but you will occasionally see devices with more than two inputs. They will, however, only have one output.
Digital logic circuits are usually represented using these six symbols; inputs are on the left and outputs are to the right. While inputs can be connected together, outputs should never be connected to one another, only to other inputs. One output may be connected to multiple inputs, however.
Truth Tables
The descriptions above are adequate to describe the functionality of single blocks, but there is a more useful tool available: the truth table. Truth tables are simple plots which explain the output of a circuit in terms of the possible inputs to that circuit. Here are truth tables describing the six main elements:
Truth tables can be expanded out to an arbitrary scale, with as many inputs and outputs as you can handle before your brain melts. Here's what a four-input circuit and truth table look like:
Written Boolean Logic
It is, of course, useful to be able to write in a simple mathematical format an equation representing a logical operation. To that end, there are mathematical symbols for the unique operations: AND, OR, XOR, and NOT.
A AND B should be written as AB (or sometimes A • B)
A OR B should be written as A + B
A XOR B should be written as A ⊕ B
NOT A should be written as A' or A
You'll note that there are two missing elements on that list: NAND and NOR. Typically, those are simply represented by complementing the appropriate representation:
A NAND B is written as (AB)' , (A • B)' , or (AB)
A NOR B is written as (A + B)' or (A + B)
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Mathematics Word Problems
Sphere 3D
Distance Formula
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Analytic Geometry
5
What is the minimal distance between any point on the sphere (x-2) ^2 +(y-1) ^2 +(z-3) ^2=1 and any point on the sphere (x+2) ^2 +(y-2) ^2 +(z-4) ^2 =4?
Dave Benson
trying to make maths easy.
·
Author has 6.1K answers and 2.1M answer views
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1y
Distance between centres is:√{(x₂-x₁)²+(y₂-y₁)²+(z₂-z₁)²} & centres: (2,1,3) & (-2,2,4)
Distance between centres is √{(2+2)²+(1–2)²+(3–4)²} = √18 & then minus radii 2 & 1
minimal distance between any point on the sphere = √18 -3 Answer
Related questions
What is the distance of point P (2,3,4) from the line (x+3) /2= (y-2) /3= z/2 "measured perpendicular" to Z axis?
What is the equation of the sphere which passes through the points (1,-3,4); (1,-5,2); (1, -3,0) and whose centre lies on the plane x+y+z=0?
What is the equation of the circle on the sphere x^2+y^2 + z^2 = 49 whose centre is at the point (2, -1, 3)?
What is the equation of the sphere which has its centre at the origin and which touches the line 2(x+1) =2-y=z+3?
How should we solve it? P (2, 2, 3) point (x -2) ^ 2 + (y -1) ^ 2 + (z -1) ^ 2 = 1/5 is a point outside the sphere. This point is closest to which point on the sphere?
Ravi Sharma
Former Group A Officer From Indian Railways (1973–2009)
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Author has 15.1K answers and 3.8M answer views
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1y
COORDINATES OF CENTER
SPHERE 1 (2, 1, 3)
SPHERE 2 (—2, 2, 4)
DISTANCE BETWEEN CENTERS
✓{(2+2)²+(1–2)²+(3–4)²}=✓18= 3✓2
RADIUS OF SPHERES
SPHERE 1 = 1
SPHERE 2= 2
MINIMUM DISTANCE BETWEEN POINTS ON SPHERES= 3✓2—(1+2)= 3✓2–3= 3(✓2–1)= 1.242
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987)
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Author has 6.2K answers and 1.7M answer views
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1y
1st sphere: center at (2, 1, 3), radius 1.
2nd sphere: center at (-2, 2, 4), radius 2
The distance between centers
sqrt(4^2 + 1^2 + 1^2) = 3sqrt(2) ≈ 4.243
The minimum distance is 4.243 - 1 - 2 ≈ 1.243
Simon Bridge
Scientist
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Author has 86.2K answers and 38.8M answer views
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1y
Unless the sphere’s overlap, you are looking for the distance between their centers less the sum of their radii.
So what is the distance between points (2,1,3) and (-2, 2, 4) minus 3?
Related questions
How do you show that whether the plane x+2y-z=6 is a tangent plane of the sphere x^2+y^2+z^2=6 and the line x-1=(y-2) /2= -(z+1) is a normal line of the sphere x^2+y^2 +z^2=6?
What is the distance from the point (4, 3, 1) to the line x=0, y=3+2t, z=1+3t?
A sphere has centre (1,-1,0) and radius = 1. What is the point on the sphere closest to the plane -2x+3y+z=6? What is the distance between this point and the plane?
What is the distance between 2 points (2,-1, 3) and (-2, 1, 3)?
How do you find the points on the line x+2 /3 = y+1 /2 = z-3 /2 which are at a distance 3√2 from the point (1,2,3)?
Matthew John Stevens
Bachelors in Electrical and Electronics Engineering & Mathematics, University of Pretoria (Graduated 2021)
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Updated 5y
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How should we solve it? P (2, 2, 3) point (x -2) ^ 2 + (y -1) ^ 2 + (z -1) ^ 2 = 1/5 is a point outside the sphere. This point is closest to which point on the sphere?
Ok, the previous answer of Mr. Sánchez is correct. Here is my solotion to find the closest point on the sphere to P. As he has stated, the closest point to P would be the point on the sphere that lies on the line through the center of the sphere and point P. Let us call this point point B. The vector that points to point P will be called [math]\boldsymbol{P}[/math]. I will call the vector that points to the origin of the sphere, the vector [math]\boldsymbol{O}[/math]. The point at the center of the sphere will be called O. The radius of the sphere will be called the distance [math]r[/math].
NB: I think it is best you draw pictur
Ok, the previous answer of Mr. Sánchez is correct. Here is my solotion to find the closest point on the sphere to P. As he has stated, the closest point to P would be the point on the sphere that lies on the line through the center of the sphere and point P. Let us call this point point B. The vector that points to point P will be called [math]\boldsymbol{P}[/math]. I will call the vector that points to the origin of the sphere, the vector [math]\boldsymbol{O}[/math]. The point at the center of the sphere will be called O. The radius of the sphere will be called the distance [math]r[/math].
NB: I think it is best you draw pictures of the vectors as you follow the solution.
So from the equation of the sphere, one can see that [math]r^2 = \frac{1}{5}[/math], which leads to the fact that [math]r = \frac{1}{\sqrt{5}}[/math]
We can also see from the equation of the sphere, that [math]\boldsymbol{O} = \langle 2, 1, 1 \rangle[/math]. And from the problem statement we see that [math]\boldsymbol{P} = \langle 2, 2, 3 \rangle[/math]
Now, if we want to find the point on the line that goes from the center of the sphere to the point P, we will have to find a unit vector that points to P from point O, a vector that points from point O to point P is can be calculated as [math]\boldsymbol{P} -\boldsymbol{O}[/math], which I will call [math]\vec{OP}[/math].
Thus:
[math]\vec{OP} = \langle 2, 2, 3 \rangle - \langle 2, 1, 1 \rangle[/math].
= [math]\langle 0, 1, 2 \rangle[/math].
[math]\vec{OP}[/math] is not a unit vector, but it points from point O to point P. To now find the unit vector that points from point O to point P, which I will call [math]\boldsymbol{u}_{\vec{OP}}, [/math]one must divide [math]\vec{OP}[/math] by its magnitude, i.e.
[math]\boldsymbol{u}_{\vec{OP}} = \frac{\vec{OP}}{\| \vec{OP} \|}[/math].
So:
[math]\boldsymbol{u}_{\vec{OP}} = \frac{\langle 0, 1, 2 \rangle}{\sqrt{0^2 + 1^2 + 2^2}}[/math]
Thus:
[math]\boldsymbol{u}_{\vec{OP}} = \langle 0, \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle[/math].
The goal here is to find a vector that points from point O to the point that is closest on the sphere to point P. This means that vector that we want to find is a vector that has length equal to the radius of the sphere, and points from O to P, the tip of this vector will be on the point on the sphere that is closest to point P if the tail of this vector is placed at point O, since it points in the same direction as [math]\vec{OP}[/math].
We have found the unit vector that points from point O to point P, [math]\boldsymbol{u}_{\vec{OP}}[/math], and since it is a unit vector, its length is equal to 1. So if we want to find a vector that has length [math]r[/math], and points in the same direction as [math]\vec{OP}[/math], we simply have to multiply [math]\boldsymbol{u}_{\vec{OP}} [/math]with [math]r[/math]. So this vector, which we will call [math]\boldsymbol{A}[/math], is calculated as:
[math]\boldsymbol{A} = r\boldsymbol{u}_{\vec{OP}}[/math]
So
[math]\boldsymbol{A} = \frac{1}{\sqrt{5}} \langle 0, \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle[/math].
So
[math]\boldsymbol{A} = \langle 0, \frac{1}{5}, \frac{2}{5} \rangle[/math].
Now, the resultant vector of the sum of [math]\boldsymbol{O} [/math] and [math] \boldsymbol{A}[/math] will point to the point on the sphere which is closest to point P. Let us call this vector [math]\boldsymbol{B}[/math]
So:
[math]\boldsymbol{B} = \boldsymbol{O} + \boldsymbol{A}[/math]
Thus [math]\boldsymbol{B} = \langle 2, 1, 1 \rangle + \langle 0, \frac{1}{5}, \frac{2}{5} \rangle.[/math]
[math]= \langle 2, \frac{6}{5}, \frac{7}{5} \rangle[/math]
As stated, this vector [math]\boldsymbol{B} [/math] points to the point on the sphere that is closest to point P, in other words, the tip of its head is on the point on the sphere that is closest to point P. This means that point B, the point on the sphere that is closest to point P, is equal to math.[/math]
So anwer: B = math[/math].
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm
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Author has 9.3K answers and 6.2M answer views
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1y
Related
What is the shortest distance between the point (-1,5) and y^2=x?
Given a parabola having equation [math]y^2=x[/math]
Given a point [math]A(-1,5)[/math] on parabola
Let [math]B(k^2,k)[/math] be a point on parabola such that distance [math]AB[/math] is the shortest
Let [math]AB=p[/math]
[math]p^2= (-1-k^2)^2+(5-k)^2[/math]
[math]p^2= k^4+3k^2-10k+26[/math]
Differentiating w.r.t. [math]k[/math],
[math]2p \cdot \dfrac{dp}{dk}= 4k^3+6k-10[/math]
For the distance to be maximum or minimum, the condition is that
[math]\dfrac{dp}{dk}= 0[/math]
[math]2p \cdot 0 = 4k^3+6k-10[/math]
[math]4k^3+6k-10=0[/math]
There are [math]3[/math] roots. One is integer and two are complex
[math]k=1[/math]
Now, [math]k=1 \implies k^2=1[/math]
math[/math] is the closest point
[math]p=\sqrt{ (-1-1^2)^2+(5-1)^2 } [/math]
[math]p = 2\sqrt{5}[/math]
Note: We must ensure that second derivative of [math]p[/math] is positive in order to conclude
Given a parabola having equation [math]y^2=x[/math]
Given a point [math]A(-1,5)[/math] on parabola
Let [math]B(k^2,k)[/math] be a point on parabola such that distance [math]AB[/math] is the shortest
Let [math]AB=p[/math]
[math]p^2= (-1-k^2)^2+(5-k)^2[/math]
[math]p^2= k^4+3k^2-10k+26[/math]
Differentiating w.r.t. [math]k[/math],
[math]2p \cdot \dfrac{dp}{dk}= 4k^3+6k-10[/math]
For the distance to be maximum or minimum, the condition is that
[math]\dfrac{dp}{dk}= 0[/math]
[math]2p \cdot 0 = 4k^3+6k-10[/math]
[math]4k^3+6k-10=0[/math]
There are [math]3[/math] roots. One is integer and two are complex
[math]k=1[/math]
Now, [math]k=1 \implies k^2=1[/math]
math[/math] is the closest point
[math]p=\sqrt{ (-1-1^2)^2+(5-1)^2 } [/math]
[math]p = 2\sqrt{5}[/math]
Note: We must ensure that second derivative of [math]p[/math] is positive in order to conclude that [math]k[/math] so obtained gives minimum value
Dwaraganathan Rengasamy
Studied IIT-JEE 2019 Aspirant & Mathematics
·
Author has 94 answers and 435.4K answer views
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6y
Related
What are the maximum and minimum values of a point (3,4,12) from the sphere x^2+y^2+z^2=1 using the Lagrange method?
I think you are asking the maximum and minimum distances of the point math[/math] from the sphere [math]x^2 + y^2 + z^2 = 1[/math] .
The whole thing Lagrange multiplier is about is…
…this!
The inverted triangle represents partial derivative of the function.
And lamda is some constant.
As we are supposed to find minimum and maximum values of [math]\sqrt{(x-3)^2 + (y-4)^2 + (z-12)^2}[/math]
Let [math]f(x,y,z) = (x-3)^2 + (y-4)^2 + (z-12)^2[/math]
[math]g(x) = x^2 + y^2 + z^2 - 1[/math] is the given constraint.
Using the Lagrange multiplier now,
[math]\nabla _{x,y,z} f = \lambda \nabla _{x,y,z} g[/math]
[math]\implies \frac{\partial f(x,y,z)}{\partial x} = \lambda \frac{\partial[/math]
I think you are asking the maximum and minimum distances of the point math[/math] from the sphere [math]x^2 + y^2 + z^2 = 1[/math] .
The whole thing Lagrange multiplier is about is…
…this!
The inverted triangle represents partial derivative of the function.
And lamda is some constant.
As we are supposed to find minimum and maximum values of [math]\sqrt{(x-3)^2 + (y-4)^2 + (z-12)^2}[/math]
Let [math]f(x,y,z) = (x-3)^2 + (y-4)^2 + (z-12)^2[/math]
[math]g(x) = x^2 + y^2 + z^2 - 1[/math] is the given constraint.
Using the Lagrange multiplier now,
[math]\nabla _{x,y,z} f = \lambda \nabla _{x,y,z} g[/math]
[math]\implies \frac{\partial f(x,y,z)}{\partial x} = \lambda \frac{\partial g(x,y,z)}{\partial x}[/math]
[math]\frac{\partial ((x-3)^2 + (y-4)^2 + (z-12)^2)}{\partial x} = \lambda \frac{\partial (x^2 + y^2 + z^2 - 1)}{\partial x}[/math]
[math]2(x-3) = \lambda (2x)[/math]
[math]x(\lambda - 1) = -3[/math]
[math]x = \frac{3}{1 - \lambda}[/math]
[math]\frac{x}{3} = \frac{1}{1 - \lambda}[/math]
Similarly,
[math]\frac{y}{4} = \frac{1}{1 - \lambda}[/math]
[math]\frac{z}{12} = \frac{1}{1 - \lambda}[/math]
Replacing [math]\lambda[/math] with a new constant,
[math]\frac{x}{3} = \frac{y}{4} = \frac{z}{12} = k[/math]
[math]x = 3k[/math]
[math]y = 4k[/math]
[math]z = 12k[/math]
Substitute the values in the constrained equation to get the value of the constant.
math^2 + (4k)^2 + (12k)^2 = 1[/math]
Solving this,
[math]k = ± \frac{1}{13}[/math]
[math]\implies x = ± \frac{3}{13}[/math]
[math]y = ± \frac{4}{13}[/math]
[math]z = ± \frac{12}{13}[/math]
[math]f_{min} = (\frac{3}{13}-3)^2 + (\frac{4}{13}-4)^2 + (\frac{12}{13}-12)^2[/math]
[math]= \frac{156^2}{13^2} = 12^2[/math]
[math]f_{max} = (- \frac{3}{13}-3)^2 + (- \frac{4}{13}-4)^2 + (- \frac{12}{13}-12)^2[/math]
[math]= \frac{182^2}{13^2} = 14^2[/math]
Minimum distance [math]= \sqrt{f_{min}} = 12[/math]
Maximum distance [math]= \sqrt{f_{max}} = 14[/math]
If you are just about to find the answer to problem by any method,
You may notice that nearest and farthest points on a sphere from any point lie on the line joining that point and the centre of the sphere.
In this case, the line joining centre of sphere math[/math] and given point math[/math]. The line is nothing but [math]\frac{x}{3} = \frac{y}{4} = \frac{z}{12}[/math].
We directly arrive at the required relation to solve without any differentiation.
Lagrange methods are universal tools to solve various kind of math problems. But you don’t actually need a hammer to open boxes,… While you have small pretty keys.
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Gustavo Sánchez
M.S. in Physics, University of Buenos Aires (Graduated 1986)
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Author has 3K answers and 2.2M answer views
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5y
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How should we solve it? P (2, 2, 3) point (x -2) ^ 2 + (y -1) ^ 2 + (z -1) ^ 2 = 1/5 is a point outside the sphere. This point is closest to which point on the sphere?
Poits for which
math^2+(y-1)^2+(z-1)^2 < 1/5[/math]
are inside the sphere, if
math^2+(y-1)^2+(z-1)^2 = 1/5[/math]
are on the sphere, and if
math^2+(y-1)^2+(z-1)^2 > 1/5[/math]
are outside the sphere.
In particular P(2,2,3) gives
math^2+(2-1)^2+(3-1)^2 = 0+1+4 = 5 > 1/5[/math],
hence P is outside.
The closest point on the sphere is located on the line passing through the center and P. Using
[math]x’ = x-2[/math]
[math]y’ = y-1[/math]
[math]z’ = z-1[/math]
the center has coordinates (0,0,0), and P has coordinates (0,1,2). The line passing by the origin and P is
math = \lambda (0,1,2)[/math]
and putting
[math]x’^2+y’^2+z’^2 = 1/5[/math]
we have
[math]\lambda^2+4\lambda^2 = 1/5[/math]
and therefore
[math]\lamb[/math]
Poits for which
math^2+(y-1)^2+(z-1)^2 < 1/5[/math]
are inside the sphere, if
math^2+(y-1)^2+(z-1)^2 = 1/5[/math]
are on the sphere, and if
math^2+(y-1)^2+(z-1)^2 > 1/5[/math]
are outside the sphere.
In particular P(2,2,3) gives
math^2+(2-1)^2+(3-1)^2 = 0+1+4 = 5 > 1/5[/math],
hence P is outside.
The closest point on the sphere is located on the line passing through the center and P. Using
[math]x’ = x-2[/math]
[math]y’ = y-1[/math]
[math]z’ = z-1[/math]
the center has coordinates (0,0,0), and P has coordinates (0,1,2). The line passing by the origin and P is
math = \lambda (0,1,2)[/math]
and putting
[math]x’^2+y’^2+z’^2 = 1/5[/math]
we have
[math]\lambda^2+4\lambda^2 = 1/5[/math]
and therefore
[math]\lambda = \sqrt{1/25} = 1/5.[/math]
Hence the coordinates of the closest point, Q, are
[math]x’ = 0[/math]
[math]y’ = 1/5[/math]
[math]z’ = 2/5[/math]
or
[math]x = 2[/math]
[math]y = 6/5[/math]
[math]z = 7/5,[/math]
i.e. Q(2,6/5,7/5).
Pranesh Muddebihal
Studied at Indian Institute of Science, Bangalore (IISc)
·
Author has 199 answers and 498.3K answer views
·
6y
Related
What is maximum and minimum distance of the point (3,4,12) of the sphere x^2+y^2+z^2=4?
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present)
·
Author has 6.2K answers and 4.3M answer views
·
1y
Related
What is the shortest distance between the point (-1,5) and y^2=x?
Since only the upper half of the parabola is relevant to this exercise, we can let the point on the parabola be
(x, √x)
, which point establishes a
normal
to the curve of the parabola, thus the shortest distance from
(-1, 5)
to
(x, √x)
. By differentiating
y^2 = x
implicitly
with respect to
x
, we get the slope of any tangent to the parabola. The
negative reciprocal
of this slope is the slope of the normal that we seek, and we will establish this slope, thus the point where the normal lies, by tying together the points
(x, √x)
and
(-1, 5)
. Study carefully the math below, and note that the graph b
Since only the upper half of the parabola is relevant to this exercise, we can let the point on the parabola be (x, √x), which point establishes a normal to the curve of the parabola, thus the shortest distance from (-1, 5) to (x, √x). By differentiating y^2 = x implicitly with respect to x, we get the slope of any tangent to the parabola. The negative reciprocal of this slope is the slope of the normal that we seek, and we will establish this slope, thus the point where the normal lies, by tying together the points (x, √x) and (-1, 5). Study carefully the math below, and note that the graph below confirms our conclusion nicely.
Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm
·
Author has 9.3K answers and 6.2M answer views
·
Jul 25
Related
What is the shortest distance between a point in the circle x^2+y^2=1 and the line y=x+2?
Given equation of circle
[math]x^2+y^2=1[/math]
This circle has center [math]O(0,0)[/math] and radius [math]1[/math]
Given equation of line,
[math]y=x+2[/math]
Let [math]A[/math] and [math]B[/math] respectively be the points on the line and circle such that [math]AB[/math] is the smallest distance.
[math]OA= \dfrac{1(0)-1(0)+2}{\sqrt{1^2+(-1)^2}}=\dfrac{2}{\sqrt{2}}= \sqrt{2}[/math]
[math]AB= OA-OB = \sqrt{2}-1[/math]
[math]AB= \sqrt{2}-1[/math]
Let a circle be defined having center at origin such that line is tangential to this circle.
We write,
Abscissa of [math]A[/math] is the double root . Let it be [math]\alpha[/math]. By Vieta’s formula,
[math]\alpha+\alpha= -2 \implies \alpha=-1[/math]
[math]\alpha^2 [/math]
Given equation of circle
[math]x^2+y^2=1[/math]
This circle has center [math]O(0,0)[/math] and radius [math]1[/math]
Given equation of line,
[math]y=x+2[/math]
Rewriting,
[math]x-y+2=0[/math]
Let [math]A[/math] and [math]B[/math] respectively be the points on the line and circle such that [math]AB[/math] is the smallest distance.
[math]OB=1[/math]
[math]OA= \dfrac{1(0)-1(0)+2}{\sqrt{1^2+(-1)^2}}=\dfrac{2}{\sqrt{2}}= \sqrt{2}[/math]
[math]AB= OA-OB = \sqrt{2}-1[/math]
[math]AB= \sqrt{2}-1[/math]
Aliter
Let a circle be defined having center at origin such that line is tangential to this circle.
[math]x^2+y^2=k[/math]
We write,
[math]x^2+(x+2)^2=k[/math]
[math]x^2+2x+2-\frac{k}{2}=0[/math]
Abscissa of [math]A[/math] is the double root . Let it be [math]\alpha[/math]. By Vieta’s formula,
[math]\alpha+\alpha= -2 \implies \alpha=-1[/math]
[math]\alpha^2 = 2-\frac{k}{2}[/math]
math^2= 2-\frac{k}{2} \implies k=2[/math]
[math]OA^2=2 \implies OA=\sqrt{2}[/math]
[math]AB= OA-OB= \sqrt{2}-1[/math]
Viktor T. Toth
IT pro, part-time physicist -- patreon.com/vttoth
·
Upvoted by
Tom Heine
, BS Aerospace and Aeronautical Engineering & Mathematics, Western Michigan University (2002) and
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 10.1K answers and 172.7M answer views
·
10y
Related
Why doesn't
math ^2 [/math]
equal
[math] x^2+y^2 [/math]
?
Others provided you with the algebra and explained that exponentiation does not distribute over addition, but here is a way to think about this in terms of geometry:
This is a square with sides that have lengths of
[math]x+y[/math]
. Clearly, the area of the square is
math^2[/math]
. But it is also equal to the sum of the areas of the shapes inside: A square with area
[math]x^2[/math]
, another square with area
[math]y^2[/math]
, and two rectangles, with area
[math]xy[/math]
. The total comes to
[math]x^2 + y^2 + 2xy[/math]
.
Others provided you with the algebra and explained that exponentiation does not distribute over addition, but here is a way to think about this in terms of geometry:
This is a square with sides that have lengths of
[math]x+y[/math]
. Clearly, the area of the square is
math^2[/math]
. But it is also equal to the sum of the areas of the shapes inside: A square with area
[math]x^2[/math]
, another square with area
[math]y^2[/math]
, and two rectangles, with area
[math]xy[/math]
. The total comes to
[math]x^2 + y^2 + 2xy[/math]
.
Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
·
Author has 6.2K answers and 7M answer views
·
5y
Related
What is the minimum distance between xy=4 and x^2+y^2=4?
X×Y = 4 is a diagonal hyperbola, whose transverse axis is line Y = X . Vertices of this hyperbola are intersection points with transverse axis.
Obviously, transverse axis intersects hyperbola at (2,2) and (-2,-2)
X^2 + Y^2 = 4 is a circle with centre at origin. It's intersection with Y = X is,
2X^2 = 4 => X = +/- √2 Intersection coordinates are (√2,√2) and (-√2,-√2)
Minimum distance = √[(2-√2)^2 + (2-√2)^2] =>
√[2(2-√2)^2)] => √2(2-√2) => 2(√2 - 1).
X×Y = 4 is a diagonal hyperbola, whose transverse axis is line Y = X . Vertices of this hyperbola are intersection points with transverse axis.
Obviously, transverse axis intersects hyperbola at (2,2) and (-2,-2)
X^2 + Y^2 = 4 is a circle with centre at origin. It's intersection with Y = X is,
2X^2 = 4 => X = +/- √2 Intersection coordinates are (√2,√2) and (-√2,-√2)
Minimum distance = √[(2-√2)^2 + (2-√2)^2] =>
√[2(2-√2)^2)] => √2(2-√2) => 2(√2 - 1).
Related questions
What is the distance of point P (2,3,4) from the line (x+3) /2= (y-2) /3= z/2 "measured perpendicular" to Z axis?
What is the equation of the sphere which passes through the points (1,-3,4); (1,-5,2); (1, -3,0) and whose centre lies on the plane x+y+z=0?
What is the equation of the circle on the sphere x^2+y^2 + z^2 = 49 whose centre is at the point (2, -1, 3)?
What is the equation of the sphere which has its centre at the origin and which touches the line 2(x+1) =2-y=z+3?
How should we solve it? P (2, 2, 3) point (x -2) ^ 2 + (y -1) ^ 2 + (z -1) ^ 2 = 1/5 is a point outside the sphere. This point is closest to which point on the sphere?
How do you show that whether the plane x+2y-z=6 is a tangent plane of the sphere x^2+y^2+z^2=6 and the line x-1=(y-2) /2= -(z+1) is a normal line of the sphere x^2+y^2 +z^2=6?
What is the distance from the point (4, 3, 1) to the line x=0, y=3+2t, z=1+3t?
A sphere has centre (1,-1,0) and radius = 1. What is the point on the sphere closest to the plane -2x+3y+z=6? What is the distance between this point and the plane?
What is the distance between 2 points (2,-1, 3) and (-2, 1, 3)?
How do you find the points on the line x+2 /3 = y+1 /2 = z-3 /2 which are at a distance 3√2 from the point (1,2,3)?
What is the equation of the sphere passing through (1,-3,4), (1-5,2), and (1,-3,0) if these points have a center on the plane x+y+z?
What is the distance between point (1, 5, 3) from the plane 4x+y+8z+33=0?
Is it true that 4 is the maximum number of points on a sphere where (1) the points are all equidistant from each other, and (2) the sum of the distances between different pairs of points is the most possible? Or is it 5?
What is the shortest and longest distance from the point (1,2,-1) to the sphere is x2+y2+Z2=24?
How do you show that the plane 2x-2y+z +12=0 touches the sphere x^2+y^2-2x-4y+2z=3, and how do you find the point of contact?
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17624 | https://pubmed.ncbi.nlm.nih.gov/19557682/ | The evolution and maintenance of Hox gene clusters in vertebrates and the teleost-specific genome duplication - PubMed
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. 2009;53(5-6):765-73.
doi: 10.1387/ijdb.072533km.
The evolution and maintenance of Hox gene clusters in vertebrates and the teleost-specific genome duplication
Shigehiro Kuraku1,Axel Meyer
Affiliations Expand
Affiliation
1 Lehrstuhl fur Zoologie und Evolutionsbiologie, Department of Biology, University of Konstanz, Germany.
PMID: 19557682
DOI: 10.1387/ijdb.072533km
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The evolution and maintenance of Hox gene clusters in vertebrates and the teleost-specific genome duplication
Shigehiro Kuraku et al. Int J Dev Biol.2009.
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. 2009;53(5-6):765-73.
doi: 10.1387/ijdb.072533km.
Authors
Shigehiro Kuraku1,Axel Meyer
Affiliation
1 Lehrstuhl fur Zoologie und Evolutionsbiologie, Department of Biology, University of Konstanz, Germany.
PMID: 19557682
DOI: 10.1387/ijdb.072533km
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Abstract
Hox genes are known to specify spatial identities along the anterior-posterior axis during embryogenesis. In vertebrates and most other deuterostomes, they are arranged in sets of uninterrupted clusters on chromosomes, and are in most cases expressed in a "colinear" fashion, in which genes closer to the 3-end of the Hox clusters are expressed earlier and more anteriorly and genes close to the 5-end of the clusters later and more posteriorly. In this review, we summarize the current understanding of how Hox gene clusters have been modified from basal lineages of deuterostomes to diverse taxa of vertebrates. Our parsimony reconstruction of Hox cluster architecture at various stages of vertebrate evolution highlights that the variation in Hox cluster structures among jawed vertebrates is mostly due to secondary lineage-specific gene losses and an additional genome duplication that occurred in the actinopterygian stem lineage, the teleost-specific genome duplication (TSGD).
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Pattern formation today.Chuong CM, Richardson MK.Chuong CM, et al.Int J Dev Biol. 2009;53(5-6):653-8. doi: 10.1387/ijdb.082594cc.Int J Dev Biol. 2009.PMID: 19557673 Free PMC article.Review.
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Hox clusters as models for vertebrate genome evolution.Hoegg S, Meyer A.Hoegg S, et al.Trends Genet. 2005 Aug;21(8):421-4. doi: 10.1016/j.tig.2005.06.004.Trends Genet. 2005.PMID: 15967537 Review.
The "fish-specific" Hox cluster duplication is coincident with the origin of teleosts.Crow KD, Stadler PF, Lynch VJ, Amemiya C, Wagner GP.Crow KD, et al.Mol Biol Evol. 2006 Jan;23(1):121-36. doi: 10.1093/molbev/msj020. Epub 2005 Sep 14.Mol Biol Evol. 2006.PMID: 16162861
Estimation of Hox gene cluster number in lampreys.Sharman AC, Holland PW.Sharman AC, et al.Int J Dev Biol. 1998 May;42(4):617-20.Int J Dev Biol. 1998.PMID: 9694633
The bithorax complex of Drosophila an exceptional Hox cluster.Maeda RK, Karch F.Maeda RK, et al.Curr Top Dev Biol. 2009;88:1-33. doi: 10.1016/S0070-2153(09)88001-0.Curr Top Dev Biol. 2009.PMID: 19651300 Review.
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17625 | https://practicalneurology.com/diseases-diagnoses/headache-pain/dizziness-and-vertigo/30119/ | Media formats available:
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A stepwise and careful history helps identify the many causes of vertigo and dizziness. In the next issue, Part 2 covers the oculomotor and vestibular examinations.
First, Characterize Symptoms
Definitions
In 2009, the Barany Society published the first consensus classification of vestibular symptoms. Internal vertigo is a false or distorted sensation of self-motion including spinning, swaying, bobbing, tilting, bouncing, and sliding. External vertigo is a false or distorted sensation of the surroundings, excluding bidirectional motion, which is known as oscillopsia. The feeling of being unstable without a particular direction preference while sitting, standing, or walking is unsteadiness.Dizziness is a nonmotion sensation of disrupted spatial orientation. Purposefully, the definitions do not suggest a particular disease pathophysiology.1 Orthostatic hypotension and benign paroxysmal positional vertigo (BPPV), for instance, can both induce vertigo or dizziness,2,3 although the term vertigo will be used throughout this article to describe either symptom. Patients may have more than a single symptom at a time. For example, a common combination of symptoms in vestibular neuritis includes vertigo (due to semicircular canal imbalance) and oscillopsia (due to horizontal jerk nystagmus). Patients may also have symptoms that transition from one to another over time; for example, acute vertigo to chronic unsteadiness.
Direction
In some vestibular disorders (eg, vestibular paroxysmia), patients have directionally specific spinning that may be better recognized in vertigo than in external vertigo.4 Spinning vertigo that changes direction during a single event, is unique to Ménière’s disease and related to the phases of the attack—excitatory, inhibitory, or recovery.5 Understanding the direction of vertigo can occasionally help lateralize the disorder or better understand the pathophysiology.
Duration
Vertigo spells are brief, usually lasting seconds in patients with BPPV, vestibular paroxysmia, and cardiac arrhythmias. Patients with Ménière’s disease, vestibular migraine (VM), or transient ischemic attacks (TIAs) often present with vertigo spells lasting minutes to hours. In patients with with vestibular neuritis or central vestibular lesions from stroke or demyelination, vertigo lasts days to weeks. Patients with bilateral vestibular loss (BVL), uncompensated unilateral vestibular loss (UVL), chronic intoxication, or persistent postural perceptual dizziness (PPPD) often have months to years of symptoms.
In episodic conditions, asking the patient how long the specific vestibular symptom lasted (ie, dizziness or vertigo), rather than how long an attack lasted, can give a better estimate of duration. For example, a patient with BPPV may overestimate attack duration (eg, 5 minutes) because of persisting vegetative symptoms (eg, nausea, vomiting, and sweating) even when the spinning associated with BPPV lasted less than 1 minute.6
Second, Categorize Symptoms
Vestibular disorders can be grouped by presentation into acute, episodic, and chronic vestibular syndromes (AVS, EVS, and CVS, respectively). Patients with AVS present with more than 24 hours of continuous vertigo (lasting days to weeks and monophasic) with nausea/vomiting, imbalance, head motion intolerance, spontaneous nystagmus (eg, stroke or vestibular neuritis). Patients with EVS have similar symptoms and signs as AVS, lasting seconds to hours (eg, Ménière’s disease, VM). Patients with CVS have constant vestibular symptoms for weeks to years (eg, bilateral vestibular loss).
A convenient strategy is to employ a 2-layer approach to acute and episodic clinical syndromes (Figures 1-4).1
First Layer: Are symptoms provoked or unprovoked?
Second Layer: Do vestibular symptoms occur in isolation or are there additional neurologic or audiologic symptoms?
Core clinical syndromes commonly overlap and evolve as do symptoms (Figure 4, Table).7
Third, Identify Symptom Triggers
Head Motion, Position, Orthostatic Change, or Exertion
Head Motion. In a patient with a vestibular disorder (peripheral or central), a normal head movement can lead to a faulty estimation of movement. Symptoms are experienced during or time-locked with the head movement.
Position. Attacks of vertigo and dizziness in patients with BPPV or central positional nystagmus occur in response to changes in the gravitational vector during movements such as looking down-then-up, bending over, or rolling over in bed. Symptoms are triggered by the head movement.
Orthostatic Change. Triggered by changing from a seated to standing position or from lying to seated, orthostatic vertigo and may be related to a neurologic condition (eg, multiple system atrophy), medications (eg, antihypertensives), hypovolemia, or presyncope. If symptoms are triggered by standing up from sitting when there is no change in spatial orientation of the head with respect to gravity (eg, standing from a chair without moving the head), orthostatic hypotension is favored. Symptoms triggered by going from sitting to lying or rolling over in bed (both cause a change in gravitational vector), in contrast, favor BPPV.
Exertion. Exertion-related vertigo may be caused simply by activities leading to head movements that trigger positional or head motion-induced symptoms. If vertigo occurs with exertion when the head is stationary, however, cardiopulmonary disorders should be considered. Additionally, hyperventilation and changes in cerebrospinal fluid (CSF) pH that occur with strenuous activity may also lead to 8th cranial nerve hyperexcitability in vestibular paroxysmia or with acoustic neuroma.
Head Turning and Eye Position
Head Turning. Some patients with vestibular paroxysmia may experience short spells of vertigo with head turn (to the right or left) in the upright position, because they may have more neurovascular contact with certain head positions. Rotational vertebral artery occlusion syndrome should be considered when protracted head turn induces stereotypical vertigo attacks.4
Eye Position. A particular eye position may bring on symptoms. For example, monocular oscillopsia can be induced by down and medial gaze in superior oblique myokymia.
Sound, Valsalva, and External Ear Pressure Changes
Patients with a third-window syndrome (superior canal dehiscence syndrome [SCDS], perilymph fistula, or enlarged vestibular aqueduct syndrome) have inner ear bony structure deficits. Changes in intracranial or middle ear pressure or loud sound (Tulio phenomenon) often lead to inappropriate excitation or inhibition of a semicircular canal, causing vertigo and nystagmus. In SCDS, excitatory stimuli for the anterior (also known as superior) canal include Valsalva against pinched nostrils (eg, blowing the nose), positive pressure in the external auditory canal (EAC) (eg, inserting a wet finger into the EAC) or a loud sound. The resultant nystagmus will be downbeat-torsional (top poles beating toward the affected ear). Inhibitory stimuli include Valsalva against a closed glottis (eg, heavy objects lifting, coughing, straining, and laughing) or negative pressure in the EAC (eg, pulling out a wet finger from the EAC). The resultant nystagmus will be upbeat-torsional (top poles beating toward the unaffected ear).9 A history of head trauma including barotrauma and blast injury is a known risk factor.10 In comparison to patients with third-window syndromes, vertigo spells in Chiari malformation or situational and vasovagal syncope may be driven by closed-glottis Valsalva maneuvers but are not triggered by loud sound or middle ear pressure changes.
Complex Visual Environments, Passive Self-Motion, or Illusions of Passive Self-Motion
Loss or distortion of vestibular input may cause increased reliance on visual information for balance, which is thought to be the genesis of most visual vertigo. Complex visual environments include patterned wallpaper or carpets or a busy grocery store. In patients who overrely on visual input, there is commonly impaired compensation for moving scenes, (eg, passive self-motion on a car, bus, train, or plane) or illusions of passive self-motion (eg, video games on large screens, 3D movies, virtual reality headsets or looking at traffic) that can cause spatial disorientation. Abnormal visual dependency can cause anxiety specific to open spaces or shopping centers and ultimately agoraphobia. Visual tasks that require fixation on small target (eg, mobile devices or reading a book) may be a trigger in patients with PPPD.11
Walking on Uneven Surfaces or in the Dark
Understanding that normal balance relies upon visual, proprioceptive, and vestibular inputs can help the clinician establish which system(s) are impaired depending on the specific condition(s) that worsen balance. Asking about situations with down-regulation of visual cues (walking in the dark) and disrupted somatosensory cues (walking on uneven surfaces) can have localizing value.
A history of oscillopsia while walking (ie, head movement dependent) will favor BVL,12 whereas an abnormal general neurologic exam in a patient with imbalance may suggest a nonvestibular etiology (eg, diminished vibration/proprioception sensation and hyporeflexia in polyneuropathy). In elderly patients, imbalance is often multifactorial, and related to a combination of orthopedic, neuropathic, visual, proprioceptive, and vestibular impairments. Recovery from vestibular neuritis may be suboptimal because of combined abnormal visual dependency and anxiety, dispite objective vestibular function tests showing recovery of function.13
Fourth, Assess Associated Symptoms
Vegetative Symptoms: Nausea and Vomiting
When it’s not clear whether a patient with acute prolonged or episodic symptoms has a vestibular or nonvestibular etiology, the presence of vegetative symptoms suggest a vestibular disorder.
Auditory Symptoms: Deafness, Tinnitus and Aural Fullness
Ischemic auditory symptoms. New unilateral hearing loss in a patient with AVS is concerning for anterior inferior cerebellar artery (AICA) ischemia. The inner ear is particularly susceptible to ischemic injury because it is supplied by an end artery, the internal auditory artery (IAA). The most common mechanism of IAA territory infarction is thrombotic stenosis of the parent vessel, usually the AICA, or the origin of the AICA in the basilar artery. Spells of vertigo associated with auditory symptoms (eg, tinnitus or unilateral hearing loss) that last for minutes can represent AICA TIAs that precede stroke. Patients with transient vestibular or auditory symptoms and vascular risk factors should have head and neck vascular imaging and brain MRI.14
Nonischemic auditory symptoms. Clinically, labyrinthitis resembles vestibular neuritis but can be differentiated by the presence of acute unilateral hearing loss. In patients with acute or chronic middle ear infection or meningitis, abrupt audiovestibular symptoms are concerning for bacterial labyrinthitis. Abrupt-onset audiovestibular symptoms with ipsilesional peripheral facial paresis and vesicular rash (may involve the auricle, EAC, and tympanic membrane) is concerning for herpes zoster (Ramsay-Hunt syndrome).
Head injuries, especially those including temporal bone trauma, can cause hearing loss with or without vestibular symptoms. In patients with a recent temporal bone fracture, barotrauma, or stapes surgery, co-occurrence of episodic vertigo and unilateral hearing loss may suggest a perilymphatic fistula.10 Supranormal bone thresholds and a low-frequency conductive hearing loss in the presence of normal tympanometry are associated audiologic findings of SCDS.9
Progressive peripheral facial nerve weakness with auditory symptoms (eg, tinnitus and hearing loss) suggests a neoplastic process in the middle ear (eg, glomus body tumor), internal auditory canal, or cerebellopontine angle (eg, vestibular and facial schwannoma or metastasis).15,16
Fluctuating sensorineural hearing loss in the low-to-medium frequency (> 30 dB, < 2000 Hz) range with tinnitus and aural fullness occur in association with spontaneous episodic vertigo in patients with Ménière’s disease. With recurrent attacks, hearing loss is often progressive.5
Cogan’s syndrome has similar presentation to Ménière’s disease (episodic vertigo with fluctuating hearing loss) with a more fulminant course and interstitial keratitis.17 Other inflammatory disorders that may involve the inner ear include polyarteritis nodosa, systemic lupus, Sjögren syndrome, Vogt-Koyanagi-Harada syndrome, granulomatosis with polyangiitis, and relapsing polychondritis.18 Patients with presyncope may experience muffled sounds. Patients with VM may have a mild feeling of aural fullness and tinnitus.19
Migraine Headache
Episodes of VM typically last 5 minutes to 72 hours and are accompanied by a variety of sensations including dizziness, vertigo, and/or unsteadiness that are often aggravated by head movements and visual stimulation. Associated migraine features may include headache (unilateral onset, throbbing or pounding in quality, moderate to severe in intensity and worsening by routine physical activity), phonophobia (bilateral sound discomfort), photophobia and visual aura, and nausea.19 Patients with migraine may experience associated migraine features before, during, or after vestibular symptoms. Commonly, patients are overwhelmed by vestibular symptoms and devalue the associated headache or nonheadache migraine features during the acute phase.19
Neurologic Symptoms
Neurologic symptoms of diplopia, dysarthria, numbness, weakness, hiccups, vision loss, or clouded consciousness should be assessed. Although stroke is the most common cause of central AVS, multiple sclerosis and Wernicke’s encephalopathy also should be considered.
In addition to nonischemic causes of a lower motor neuron facial palsy discussed, the presence of vertigo and a 7th nerve palsy should also raise suspicion for an AICA territory infarction involving the root entry zone and 7th nerve fascicle.14
Inflammatory disorders (eg, granulomatosis with polyangiitis, lupus erythematosus, or sarcoidosis) and infectious diseases (eg, tuberculosis, fungal infection, or syphilis) are often subacute and may present as focal (inflammatory infiltrate or abscess) or diffuse (multiple cranial neuropathy) neurologic deficits. Autoimmune encephalitis (paraneoplastic or non-paraneoplastic) also has a subacute temporal profile.
Autophony
A patient with SCDS may report hearing an echo of their own voice, footsteps, or movements of the jaws and eyes while chewing or looking around, respectively. Autophony is also a core clinical symptom in patients with a patulous eustachian tube; however, these patients will note an echo of their own breath sounds that patients with SCDS will not.9
Anxiety
Anxiety is a common comorbidity of all vestibular disorders and may adversely influence recovery. Of note, spells of dizziness or vertigo may result from a panic attack, agoraphobia, traumatic stress disorders, or generalized anxiety. Patients with anxiety-related dizziness are less likely to report external vertigo or significant vegetative symptoms.
Last,in Cases of Isolated Vertigo
The most common diagnoses of AVS are vestibular neuritis and stroke (about 5%-10% of cases), which can be indistinguishable with history and general neurologic examination. Even MR with diffusion-weighted imaging (DWI) in the first 24 to 48 hours of an attack may be falsely negative in 6% to 21% of strokes.20 Approximately 29% of patients with vertigo caused by a posterior circulation stroke reported a preceding isolated episode of vertigo that lasted minutes.21 Transient isolated vertigo attacks are potentially a warning sign for strokes, and like any other transient neurologic attack, vascular imaging (CT angiography or MRA) should be considered.22 Because there is potential for acute MRI results to be false negative, CT-perfusion may be helpful in some cases if posterior fossa ischemia is suspected.20
Conclusion
By approaching the vestibular history methodically, an accurate diagnosis can usually be made, or the differential diagnosis at least narrowed substantially. Using symptom type alone to diagnose dizzy or vertiginous patients is an ineffective approach and often leads to the wrong diagnosis and inappropriate testing. Once timing and triggers are understood, categorization into acute, episodic, or chronic vestibular syndromes can be done and a targeted exam chosen. That examination will then confirm the diagnosis and/or inform further audiovestibular testing or neuroimaging. The approach to the vestibular and ocular motor examination will be discussed in Part 2 in the next issue of Practical Neurology.
Bisdorff A, Von Brevern M, Lempert T, Newman-Toker DE. Classification of vestibular symptoms: towards an international classification of vestibular disorders. J Vestib Res. 2009;19(1-2):1-13.
Choi JH, Seo JD, Kim MJ, et al. Vertigo and nystagmus in orthostatic hypotension. Eur J Neurol. 2015;22(4):648-655.
Von Brevern M, Bertholon P, Brandt T, et al. Benign paroxysmal positional vertigo: diagnostic criteria. J Vestib Res. 2015;25(3-4):105-117.
Strupp M, Lopez-Escamez JA, Kim JS, et al. Vestibular paroxysmia: diagnostic criteria. J Vestib Res. 2016;26(5-6):409-415.
Lopez-Escamez JA, Carey J, Chung WH, et al. Diagnostic criteria for Menière’s disease. J Vestib Res. 2015;25(1):1-7.
Bronstein AM, Lempert T. Symptoms and syndromes in the patient with dizziness or unsteadiness. In Bronstein AM (ed): Oxford Textbook of Vertigo and Imbalance. Oxford, United Kingdom: Oxford University Press; 2013: 354.
Bisdorff A. Vestibular symptoms and history taking. Handb Clin Neurol. 2016;137:83-90.
Edlow JA, Newman-Toker DE, Savitz SI. Diagnosis and initial management of cerebellar infarction. Lancet Neurol. 2008;7(10):951-964.
Ward BK, Carey JP, Minor LB. Superior canal dehiscence syndrome: lessons from the first 20 years. Front Neurol. 2017;8:177.
Minor LB. Labyrinthine fistulae: pathobiology and management. Curr Opin Otolaryngol Head Neck Surg. 2003;11(5):340-346.
Staab JP, Eckhardt-Henn A, Horii A, et al. Diagnostic criteria for persistent postural-perceptual dizziness (PPPD): consensus document of the Committee for the Classification of Vestibular Disorders of the Bárány Society. J Vestib Res. 2017;27(4):191-208.
Strupp M, Kim JS, Murofushi T, et al. Bilateral vestibulopathy: diagnostic criteria consensus document of the Classification Committee of the Bárány Society. J Vestib Res. 2017;27(4):177-189.
Cousins S, Kaski D, Cutfield N, et al. Predictors of clinical recovery from vestibular neuritis: a prospective study. Ann Clin Transl Neurol. 2017;4(5):340-346.
Choi KD, Lee H, Kim JS. Ischemic syndromes causing dizziness and vertigo. Handb Clin Neurol. 2016;137:317-40.
Lees KA, Tombers NM, Link MJ, et al. Natural history of sporadic vestibular schwannoma: a volumetric study of tumor growth. Otolaryngol Head Neck Surg. 2018;159(3):535-542.
West N, Sass H, Møller MN, Cayé-Thomasen P. Facial nerve schwannomas presenting with vestibular dysfunction: a case series. Acta Neurochir (Wien). 2018;160(12):2315-2319.
Gluth MB, Baratz KH, Matteson EL, Driscoll CL. Cogan syndrome: a retrospective review of 60 patients throughout a half century. Mayo Clin Proc. 2006;81(4):483-488.
Girasoli L, Cazzador D, Padoan R, et al. Update on vertigo in autoimmune disorders, from diagnosis to treatment. J Immunol Res. 2018;2018:5072582.
Lempert T, Olesen J, Furman J, et al. Vestibular migraine: diagnostic criteria. J Vestib Res. 2012;22(4):167-172.
Choi JH, Oh EH, Park MG, et al. Early MRI-negative posterior circulation stroke presenting as acute dizziness. J Neurol. 2018;265(12):2993-3000.
Grad A, Baloh RW. Vertigo of vascular origin. Clinical and electronystagmographic features in 84 cases. Arch Neurol. 1989;46(3):281-284.
Kattah JC. Use of HINTS in the acute vestibular syndrome. an overview. Stroke Vasc Neurol. 2018;3(4):190-196.
Disclosure
The authors have no financial or other relationships relevant to this content to disclose.
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Authors (Current Affiliations)
Daniel R. Gold, DO
Departments of Neurology, Ophthalmology, Otolaryngology—Head and Neck Surgery, Neurosurgery, and Emergency Medicine
The Johns Hopkins University School of Medicine Baltimore, MD
Ari A. Shemesh, MD
Department of Neurology
The Johns Hopkins University School of Medicine Baltimore, MD
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17626 | https://math.stackexchange.com/questions/886832/how-to-go-from-this-equation-to-the-equation-of-an-hyperbola | graphing functions - How to go from this equation to the equation of an hyperbola - Mathematics Stack Exchange
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How to go from this equation to the equation of an hyperbola
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I've seen that x∗y=1 x∗y=1 graphs an hyperbola, but I am struggling to get that equation to the form x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1. How can I do this? Ultimately, what I want is to be able to graph it by hand by looking at the equation.
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asked Aug 4, 2014 at 4:02
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The formula x 2/a 2−y 2/b 2=1 x 2/a 2−y 2/b 2=1 is for hyperbolas whose axes are aligned with the coordinate axes. x y=1 x y=1 does not have this feature, so it doesn't fit that equation form.
See this page for more details.
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answered Aug 4, 2014 at 4:03
BatmanBatman
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I think we need a suitable transformation here.Mikasa –Mikasa 2014-08-04 04:12:10 +00:00 Commented Aug 4, 2014 at 4:12
Well, rotating the coordinate system 45 degrees would make the hyperbola fit this form, but that doesn't really answer the question (as it doesn't actually fit the form in the original coordinate system).Batman –Batman 2014-08-04 04:20:02 +00:00 Commented Aug 4, 2014 at 4:20
Right. I really meant what @lab did.Mikasa –Mikasa 2014-08-04 05:29:04 +00:00 Commented Aug 4, 2014 at 5:29
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Consider the following fact: we say that an algebraic curve f(x,y)=0 f(x,y)=0 is a blow-up of the algebraic curve g(x,y)=0 g(x,y)=0 if g(0,0)=0 g(0,0)=0 and f(x,y)−g(x,y)f(x,y)−g(x,y) is constant. Hence the hyperbola whose equation is x y=1 x y=1 is a blow-up of the double line x y=0 x y=0 (union of the line x=0 x=0 and the line y=0 y=0) while the hyperbola having equation x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1 is a blow-up of the double line y=±b a y=±b a. The double lines we got are just the asymptotes of our hyperbolas: in the first case, they are parallel to the coordinate axis, in the second one, they are not.
Anyway, by having the asymptotes and a point on we can graph any hyperbola.
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answered Aug 4, 2014 at 4:19
Jack D'AurizioJack D'Aurizio
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As the nature of a Curve is invariant under Rotation of axes
Using Rotation of axes, x=h cos θ−k sin θ,y=h sin θ+k cos θ x=h cosθ−k sinθ,y=h sinθ+k cosθ on the Rectangular Hyperbola
x 2−y 2=a 2,x 2−y 2=a 2,
(h 2−k 2)(cos 2 θ)−h k sin 2 θ=a 2(h 2−k 2)(cos2 θ)−h k sin2 θ=a 2
Set sin 2 θ=±1 sin2 θ=±1
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answered Aug 4, 2014 at 4:27
lab bhattacharjeelab bhattacharjee
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17627 | https://math.fandom.com/wiki/Geometric_sequence | Skip to content
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A geometric sequence is a sequence in the form
Where r is a common ratio. If , the sequence is convergent, and has a defined limit. If , the sequence is divergent, and will have an undefined limit. An example of a convergent sequence is
which will converge to 0.
A geometric series is the sum of a geometric sequence. The sum of the series is equal to
If a geometric series is infinite and convergent, the formula will still apply (again, if , the series is convergent), but since
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Cubes as the sum of odd integers
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$\begingroup$
It is well known that$1^3=1$ $2^3=3+5$ $3^3=7+9+11$ $4^3=13+15+17+19$ $5^3=21+23+25+27+29$ and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array \begin{align} \begin{matrix} 1\ 3 & 5\ 7 & 9 & 11\ 13& 15 & 17 & 19\ &&&&\ddots \end{matrix} \end{align} and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$. \begin{align} \begin{matrix} 1\ 3 & 5\ 7 & 9 & 11\ 13& 15 & 17 & 19\ \vdots \ A_{(i-1)1}&...&A_{(i-1)(i-1)}\ A_{i1}&...&A_{ij} &...&A_{ii} \end{matrix} \end{align} It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration. \begin{align} A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\ &=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\ &=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\ &...\ &=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\ &=(i-1)i-1. \end{align} Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!
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asked Jul 17, 2020 at 16:01
JMMJMM
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$\begingroup$ If your goal was to elude induction, your iteration approach on $A_{(i-1)(i-1)}$ is formally done using induction. However your proof seems good to me $\endgroup$
Gabrielek
– Gabrielek
2020-07-17 16:58:13 +00:00
Commented Jul 17, 2020 at 16:58
$\begingroup$ Do you know of any methods which successfully elude induction? $\endgroup$
JMM
– JMM
2020-07-17 17:49:51 +00:00
Commented Jul 17, 2020 at 17:49
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See my answer to this question.
It is a little appreciated fact that every power $k\ge 2$ of any positive integer $n$ can be expressed as the sum of exactly $n$ consecutive odd numbers, viz: $$n^k=\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$ So $n$ consecutive odd numbers can be found that sum to $n^3$ for any $n$. $$n^3=\sum_{i=\frac{n^{2}-n}{2}+1}^{\frac{n^{2}+n}{2}}(2i-1)$$ If you compute the starting and ending values for the summation for any particular $n$, you get exactly the numbers in your exposition.
The general formula specifies $n$ consecutive odd numbers with an average value of $n^{k-1}$ summing to $n\cdot n^{k-1}=n^k$, and is not dependent on induction in any particular case.
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answered Jul 18, 2020 at 1:04
Keith BackmanKeith Backman
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$\begingroup$ Awesome! This is exactly the kind of answer I'm looking for! +1 my friend :) $\endgroup$
JMM
– JMM
2020-07-18 03:50:14 +00:00
Commented Jul 18, 2020 at 3:50
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There is a simple way to find the odd numbers that sum up to a cube. A cube can be written as $n^3=n.n^2$. So a cube $n^3$ is a sum of $n$ squares. There are two cases to consider: odd and even $n$. I will simply provide two examples to illustrate the method.
$5^3= 5^2 + 5^2 + 5^2 + 5^2 + 5^2$ $5^3 =(5^2-4) + (5^2-2) + 5^2 + (5^2+2) + (5^2+4)$ $5^3= 21 + 23 + 25 + 27 + 29$
In other words, we subtract or add $2,4,...$ from $n^2$ to get the corresponding odd number while keeping the square in the middle.
For even numbers, we use the same principle which says that a cube $n^3$ is a sum of $n$ squares.
$4^3= 4^2 + 4^2 + 4^2 + 4^2$ $4^3= (4^2-3) + (4^2-1) + (4^2 +1) + (4^2+3)$ $4^3= 13 + 15 + 17 + 19$
The only difference between odd and even case is the fact that for odd numbers we keep the middle square and we subtract or add $2,4,6,...$ from the other squares to get an odd number. For even number, we do not keep a square but we subtract or add $1,3,5,...$ to get an odd number.
Edit #1 Dec 29 2022
Here's a method that doesn't rely on induction. It is based on the properties of the square of a triangular number $T_{n}$ and few other properties that will be pointed out when they are used. We know that $T_{n}^2 = 1^3 + 2^3 + 3^3 + ... + n^3$. I think it's easier to take an example and show how to get the odd numbers that sum up to the given cube. Example: $4^3= 13 + 15 +17 +19$ At this point we introduce the triangular number $T_{4}=4(4+1)/2=10$. We consider the square $T_{4}^2=10^2=100$. The sum of the (equal) factors of $T_{4}^2$ is $s=10+10=20$. If we consider $4^3=4\cdot16=64$ we see that the sum of the factors of $4^3$ is $t=4+16=20$, it is the same as that of $T_{4}^2$. We know that there are $9$ integers that have the same sum of factors as $T_{4}^2$ and they are: $19,36,51,64,75,84,91,96,99$. Note that $4^3=64$ is one of them. These factors are $1+19=2+18=3+17=...=9+11$. It turned out that $19$ is the forth and last odd number that add up to $4^3=64$. At this point we can just go backward and find the other $3$ odd numbers by subtracting $2$ from $19$ then $2$ from $17$ then $2$ from $15$ and then $2$ from $15$ to get $4^3= 13 + 15 + 17 + 19= 64$. We can do better. We know that the ending number of $(n-1)^3$ and the starting number of $n^3$ differ by $2$ as shown by the data provided by the OP. So it's enough to consider $T_{3}=T_{4}-4=6$ and square it and consider the sum of factors equal to $s=6+6=1+11$. So we know that the the last odd number of the three odd numbers that add up to $3^3=27$ is $11$ since $3^3= 7 + 9 + 11$. Therefore we know that the first odd number of the four that add up to $4^3$ is $11+2=13$. So now we have the both the starting odd number and the ending one and we can write: $4^3 = 13 + 15 + 17 + 19$ In fact the starting and ending odd numbers are given by $2T_{n-1}+1$ and $2T_{n}-1$. In our case $2T_{3} + 1 =2\cdot6 + 1=13$ and $2T_{4}-1=2\cdot10 - 1=19$.
In fact the starting and ending odd number sandwich the square $n^2$ in the multiplication table and are listed in the diagonal above and below the main diagonal of the squares. In our case we have $12$, $4^2=16$, $20$. It's enough to remember to add $1$ to the smaller number and subtract $1$ from the larger number. The starting odd number has the same difference of factors as $n\cdot(n^2)$ and the ending odd number of the sum has the same sum of factors as $n\cdot(n^2)$. In our case $16-4=12= 13-1$ and $16+4=20= 1 + 19$.
Finally, another way to get $4^3$ is to add up $4+8+12+16+12 +8+4=64$, the inverted L in the multiplication table starting with $4$ and ending with $4$. This is valid for any integer $n^3$.
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edited Dec 29, 2022 at 20:33
answered Jul 21, 2020 at 13:42
user25406user25406
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$\begingroup$ Hey this is a really cool way of looking at it! Thanks a lot for the insight! :) $\endgroup$
JMM
– JMM
2020-07-21 13:45:48 +00:00
Commented Jul 21, 2020 at 13:45
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The OP's claim can be summarized by saying that $n^3$ can be expressed as the sum of $n$ consecutive odd numbers starting with $n(n+1) - (2n - 1)$ and ending with $n(n+1) - 1$.
This is easy to show by a direct proof, using a result for $n^2$ that is well-known and easy to prove by induction.
Take the series of odd numbers in reverse order, from largest to smallest. Then we are saying that: \begin{eqnarray} n^3 & = & \big[(n^2 + n - 1) + (n^2 + n - 3) + \ldots + (n^2 + n - (2n - 1))\big] \ & = & n (n^2 + n) - [1 + 3 + \ldots + 2n - 1] \end{eqnarray} This in turn gives:
\begin{equation} n^2 = 1 + 3 + \ldots + 2n - 1 \end{equation}
which is an exercise in elementary proof by induction.
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answered Dec 22, 2020 at 14:27
Shrisha RaoShrisha Rao
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$\begingroup$ Terrific insight. Best answer yet! $\endgroup$
JMM
– JMM
2020-12-22 14:30:55 +00:00
Commented Dec 22, 2020 at 14:30
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The proposed identity says that $n^3 =\sum_{k=0}^{n-1}(n(n-1)+1+2k) $.
(Figuring out how to write this is the hard part.)
The right side is $\sum_{k=0}^{n-1}(n(n-1)+1+2k) =n(n(n-1)+1)+2\sum_{k=0}^{n-1}k =n^3-n^2+n+n(n-1) =n^3 $.
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answered Jul 18, 2020 at 3:56
marty cohenmarty cohen
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There are identitity's given below:
For even cube:
$p^3=(p^2-p+1)+\cdots+(p^2-5)+(p^2-3)+(p^2-1)+(p^2+1)+(p^2+3)+(p^2+5)+\cdots+(p^2+p-1)$
For odd cube:
$q^3=(q^2-q+1)+\cdots+(q^2-6)+(q^2-4)+(q^2-2)+(q)^2+(q^2+2)+(q^2+4)+(q^2+6)+\cdots+(q^2+p-1)$
For, $p=8$ we get:
$8^3=(57+59+61+63+65+67+69+71)$
For, $q=9$ we get:
$9^3=(73+75+77+79+81+83+85+87+89)$
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edited Jul 18, 2020 at 14:02
Integrand
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answered Jul 17, 2020 at 20:18
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Benchmark Fractions – Definition with Examples
Home » Math Vocabulary » Benchmark Fractions – Definition with Examples
What Are Benchmark Fractions?
Benchmark Fractions Definition
How to Compare Fractions Using Benchmark Fractions
Solved Examples
Practice Problems
Frequently Asked Questions
What Are Benchmark Fractions?
Benchmark fractions are fractions that are easy to identify and visualize. Let’s understand this using a real-life example. What do you do when you are trying to find a new address or when you explore a new street? You try to identify a benchmark near it or a nearby location you are familiar with and then you find the new address comparing it against the benchmark point.
Similarly, when comparing or ordering fractions becomes difficult, we can take help from benchmark fractions. Let’s brush up some basic concepts first!
A fraction can be defined as a part of a whole. The denominator represents the total number of equal parts of a whole and is written at the bottom. The numerator represents the number of parts taken from the whole and is written at the top.
$\frac{3}{5}$ represents 3 parts out of 5 equal parts. 3 is the numerator, and 5 is the denominator.
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Benchmark Fractions Definition
In math, benchmark fractions can be defined as fractions that we can use when measuring, comparing, or ordering other fractions. They are used as a “benchmark” for other common factors.
Common benchmark fractions examples: $0, 1,\frac{1}{4} , \frac{1}{2}$ , etc.
These fractions are most helpful when the fractions are to be compared or to be placed on a number line. Here’s a benchmark fractions chart placed on a number line, which can help you compare fractions:
Let’s understand how the fraction strip is helpful. Let’s compare the unlike fractions $\frac{3}{5}$ and $\frac{7}{8}$ easily using this chart.
Observe the one-fifth strip and the one-eighth strip. The 3 parts of the one-fifth strip represent 35and the 7 parts of the one-eighth strip $\frac{7}{8}$. Comparing the region occupied by the shaded parts, we get $\frac{3}{5} \lt \frac{7}{8}$.
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How to Compare Fractions Using Benchmark Fractions
The benchmark chart can be used to compare two or more fractions, considering the length of the corresponding fractions. The process is comparable to using fraction strips for approximating measurements.
The most common benchmark fraction is $\frac{1}{2}$. It is right in the middle of zero and one. It can also be written in the equivalent fraction form as $\frac{2}{4}, \frac{3}{6}, \frac{4}{8}$, and so on.
If the numerator is half of the denominator, then the fraction is equivalent to $\frac{1}{2}$.
If a fraction is equivalent to $\frac{1}{2}$, its position on the number line will be the same as $\frac{1}{2}$.
Example 1: Let’s find out if the fraction $\frac{5}{12}$ is less than $\frac{1}{2}$ or greater than $\frac{1}{2}$ or equals $\frac{1}{2}$.
What’s the equivalent fraction of $\frac{1}{2}$ closer to $\frac{5}{12}$? It’s $\frac{6}{12}$.
So, $\frac{6}{12} = \frac{1}{2}$
It is clear that $\frac{5}{12} \lt \frac{6}{12}$ since the denominator is the same and $5 \lt 6$.
So, what do we know? $\frac{5}{12}$ is less than $\frac{6}{12}$ and $\frac{6}{12}$ equals $\frac{1}{2}$.
We can easily conclude that $\frac{5}{12}$ is less than $\frac{1}{2}$.
We know that $\frac{4}{16} = \frac{1}{4}$.
Also, $\frac{5}{16}$ is greater than $\frac{4}{16}$.
So, we can say that $\frac{5}{16}$ is greater than $\frac{1}{4}$.
Using Benchmark Fractions for Rounding Fractions
If the numerator is much smaller than the denominator, we round the given fraction to 0
Example: $\frac{1}{9}, \frac{1}{5}$ can be rounded to 0 since they are much closer to 0 on a number line.
If the numerator is almost half of the denominator, then we round the given fraction to $\frac{1}{2}$.
Example: $\frac{5}{11}, \frac{2}{6}, \frac{5}{8}$ can be rounded to $\frac{1}{2}$.
If the numerator and denominator are almost equal, then we round the given fraction to 1.
Example: $\frac{10}{11}, \frac{5}{6}$ can be rounded to 1.
Let’s Sing!
When comparing or ordering two different fractions,
Let benchmark fractions be your guiding action!
Just begin by putting the fractions on a number line.
Then, choose the benchmark to compare—it’ll be fine!
Solved Examples
1. Compare $\frac{5}{8}$ and $\frac{1}{2}$.
Solution: We already know that $\frac{4}{8}$ is an equivalent fraction of $\frac{1}{2}$.
Since 5 is greater than 4, we can say that $\frac{5}{8}$ is greater than $\frac{4}{8}$.
So, $\frac{5}{8}$ is greater than $\frac{1}{2}$.
2. Compare whether $\frac{2}{6}$ is less or more than $\frac{2}{3}$.
Solution: $\frac{2}{6}$ is an equivalent fraction of $\frac{1}{3}$.
So, $\frac{2}{6} = \frac{1}{3}$.
Since $\frac{1}{3}$ is less than $\frac{2}{3}, \frac{2}{6}$ is less than $\frac{2}{3}$.
3. Which is greater: $\frac{4}{8}$ or $\frac{6}{12}$?
Solution: $\frac{4}{8}$ can also be written as $\frac{1}{2}$.
$\frac{6}{12}$ can also be written as $\frac{1}{2}$.
So, this means both fractions are equal.
Practice Problems
Benchmark Fractions - Definition with Examples
Attend this quiz & Test your knowledge.
1
Which fraction is closer to $0: \frac{1}{2} ,\frac{1}{4}$ or $\frac{3}{4}$ ?
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{3}{4}$
All of the above
CorrectIncorrect
Correct answer is: $\frac{1}{4}$$\frac{1}{2}$ is the midpoint of 0 and 1. $\frac{1}{4}$ is exactly in the middle of 0 and $\frac{1}{2}$. $\frac{3}{4}$ is exactly in the middle of $\frac{1}{2}$ and 1. So, $\frac{1}{4}$ is closer to 0.
2
Compare the fractions $\frac{1}{4}$ and $\frac{3}{12}$.
$\frac{1}{4} \gt \frac{3}{12}$
$\frac{1}{4} \lt \frac{3}{12}$
Both are equal
Neither
CorrectIncorrect
Correct answer is: Both are equalOn simplifying $\frac{3}{12}$, we get $\frac{1}{4}$. Therefore,$\frac{1}{4} = \frac{3}{12}$
3
Which is greater among $\frac{12}{18}, \frac{14}{21}$, and $\frac{18}{27}$?
$\frac{12}{18}$
$\frac{14}{21}$
$\frac{18}{27}$
All are equal
CorrectIncorrect
Correct answer is: All are equal$\frac{12}{18}$ can be simplified as $\frac{2}{3}$. $\frac{14}{21}$ can be simplified as $\frac{2}{3}$. $\frac{18}{27}$ can be simplified as $\frac{2}{3}$. This means all three are equal.
4
Compare $0, \frac{2}{6}$ and $\frac{6}{9}$.
$\frac{2}{6} \gt \frac{6}{9} \gt 0$
$\frac{6}{9} \gt \frac{2}{6} \gt 0$
$\frac{2}{6} \gt \frac{6}{9}$
$0 \gt \frac{6}{9} \gt \frac{2}{6}$
CorrectIncorrect
Correct answer is: $\frac{6}{9} \gt \frac{2}{6} \gt 0$Both the fractions are greater than 0 since they are positive. $\frac{2}{6}$ can be simplified as $\frac{1}{3}$. $\frac{6}{9}$ can also be written as $\frac{2}{3}$. Since $\frac{2}{3} \gt \frac{1}{3}$, we can say that $\frac{6}{9} \gt \frac{2}{6}$.
Frequently Asked Questions
What is the purpose of using fraction strips?
Fraction strips are a practical way for students to grasp the fundamentals of fractions. Fraction strips simplify studying the division of fractions into halves, thirds, quarters, etc.
For example, these strips can be cut into various sizes, so it’s possible to work on multiple problems. They are especially useful for understanding addition, subtraction, and multiplication.
What is the easiest way of comparing fractions?
Converting fractions to decimal numbers makes it much easier to determine which fraction is larger. The fraction with the larger decimal value is the greater the fraction.
What does comparing fractions mean?
You compare the two fractions to determine if one is less than, greater than, or equal to the other.
What are benchmark numbers?
Benchmark numbers are numbers that are used to compare other numbers or quantities. They also help in estimation problems. Common example of benchmark numbers is multiples of 10 or 100.
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17631 | https://www.whitman.edu/mathematics/calculus_online/section09.01.html | We have seen how integration can be used to find an area between a
curve and the x-axis. With very little change we can find some areas
between curves; indeed, the area between a curve and the x-axis may
be interpreted as the area between the curve and a second "curve''
with equation y=0. In the simplest of cases, the idea is quite easy
to understand.
Example 9.1.1 Find the area below f(x)=−x2+4x+3 and above
g(x)=−x3+7x2−10x+5 over the interval 1≤x≤2. In
figure 9.1.1 we show the two curves together, with
the desired area shaded, then f alone with the area under f
shaded, and then g alone with the area under g shaded.
| | | |
---
| 0,0 1 2 3 10 | 0,0 1 2 3 10 | 0,0 1 2 3 10 |
Figure 9.1.1. Area between curves as a difference of areas.
It is clear
from the figure that the area we want is the area under f minus the
area under g, which is to say
∫21f(x)dx−∫21g(x)dx=∫21f(x)−g(x)dx.
It doesn't matter whether we compute the two integrals on the left and
then subtract or compute the single integral on the right. In this
case, the latter is perhaps a bit easier:
∫21f(x)−g(x)dx=∫21−x2+4x+3−(−x3+7x2−10x+5)dx=∫21x3−8x2+14x−2dx=x44−8x33+7x2−2x∣∣∣21=164−643+28−4−(14−83+7−2)=23−563−14=4912.
□
It is worth examining this problem a bit more. We have seen one way to
look at it, by viewing the desired area as a big area minus a small
area, which leads naturally to the difference between two
integrals. But it is instructive to consider how we might find the
desired area directly. We can approximate the area by dividing the
area into thin sections and approximating the area of each section by
a rectangle, as indicated in
figure 9.1.2.
The area of a typical rectangle is
Δx(f(xi)−g(xi)), so the total area is approximately
∑i=0n−1(f(xi)−g(xi))Δx.
This is exactly the sort of sum that turns into an integral in the
limit, namely the integral
∫21f(x)−g(x)dx.
Of course, this is the integral we actually computed above, but we
have now arrived at it directly rather than as a modification of the
difference between two other integrals. In that example it really
doesn't matter which approach we take, but in some cases this second
approach is better.
0,0
1
2
3
10
Figure 9.1.2. Approximating area between curves with rectangles.
Example 9.1.2 Find the area below f(x)=−x2+4x+1 and above
g(x)=−x3+7x2−10x+3 over the interval 1≤x≤2; these are the
same curves as before but lowered by 2. In
figure 9.1.3 we show the two curves
together. Note that the lower curve now dips below the x-axis. This
makes it somewhat tricky to view the desired area as a big area minus
a smaller area, but it is just as easy as before to think of
approximating the area by rectangles. The height of a typical
rectangle will still be f(xi)−g(xi), even if g(xi) is
negative. Thus the area is
∫21−x2+4x+1−(−x3+7x2−10x+3)dx=∫21x3−8x2+14x−2dx.
This is of course the same integral as before, because the region
between the curves is identical to the former region—it has just
been moved down by 2.
□
0,0
1
2
3
10
Figure 9.1.3. Area between curves.
Example 9.1.3 Find the area between f(x)=−x2+4x and
g(x)=x2−6x+5 over the interval 0≤x≤1; the
curves are shown in figure 9.1.4. Generally we
should interpret "area'' in the usual sense, as a necessarily
positive quantity. Since the two curves cross, we need to compute two
areas and add them. First we find the intersection point of the
curves:
−x2+4x0x=x2−6x+5=2x2−10x+5=10±100−40−−−−−−−√4=5±15−−√2.
The intersection point we want is x=a=(5−15−−√)/2. Then
the total area is
∫a0x2−6x+5−(−x2+4x)dx+∫1a−x2+4x−(x2−6x+5)dx=∫a02x2−10x+5dx+∫1a−2x2+10x−5dx=2x33−5x2+5x∣∣∣a0+−2x33+5x2−5x∣∣∣1a=−523+515−−√,
after a bit of simplification.
□
0,0
1
5
Figure 9.1.4. Area between curves that cross.
Example 9.1.4 Find the area between f(x)=−x2+4x and
g(x)=x2−6x+5; the
curves are shown in figure 9.1.5. Here we
are not given a specific interval, so it must be the case that there
is a "natural'' region involved. Since the curves are both parabolas,
the only reasonable interpretation is the region between the two
intersection points, which we found in the previous example:
5±15−−√2.
If we let a=(5−15−−√)/2 and b=(5+15−−√)/2,
the total area is
∫ba−x2+4x−(x2−6x+5)dx=∫ba−2x2+10x−5dx=−2x33+5x2−5x∣∣∣ba=515−−√.
after a bit of simplification.
□
0,0
1
2
3
4
5
5
−5
Figure 9.1.5. Area bounded by two curves.
Exercises 9.1
Find the area bounded by the curves.
You can use
Sage
to help check your work.
Ex 9.1.1
y=x4−x2 and y=x2 (the part to the right of the y-axis)
(answer)
Ex 9.1.2
x=y3 and x=y2
(answer)
Ex 9.1.3
x=1−y2 and y=−x−1
(answer)
Ex 9.1.4
x=3y−y2 and x+y=3
(answer)
Ex 9.1.5
y=cos(πx/2) and y=1−x2 (in the first quadrant)
(answer)
Ex 9.1.6
y=sin(πx/3) and y=x (in the first quadrant)
(answer)
Ex 9.1.7
y=x−−√ and y=x2
(answer)
Ex 9.1.8
y=x−−√ and y=x+1−−−−√, 0≤x≤4
(answer)
Ex 9.1.9
x=0 and x=25−y2
(answer)
Ex 9.1.10
y=sinxcosx and y=sinx, 0≤x≤π
(answer)
Ex 9.1.11
y=x3/2 and y=x2/3
(answer)
Ex 9.1.12
y=x2−2x and y=x−2
(answer)
The following three exercises expand on the geometric interpretation
of the hyperbolic functions. Refer to section 4.11
and particularly to figure 4.11.2 and
exercise 6
in section 4.11.
Ex 9.1.13
Compute ∫x2−1−−−−−√dx using the substitution
u=arccoshx, or x=coshu;
use exercise 6
in section 4.11.
Ex 9.1.14
Fix t>0.
Sketch the region R in the right half plane bounded by the curves
y=xtanht, y=−xtanht, and x2−y2=1. Note well: t is fixed,
the plane is the x-y plane.
Ex 9.1.15
Prove that the area of R is t. |
17632 | https://www.cuemath.com/questions/how-do-you-know-if-an-absolute-value-equation-has-no-solution/ | How do you know if an absolute value equation has no solution?
Absolute value functions are always greater than or equal to zero.
Answer: There are 0 solutions to the equation |-5 - 4x| + 9 = -40
There is one major red flag of an equation with an absolute value that has no solution.
Explanation:
An absolute value expression can never be less than zero. That is a fully reduced absolute value expression must be greater than or equal to zero.
If on simplifying an absolute value expression, the value on the other side of the equation has a negative sign i.e. it is a negative number then it has no solutions.
Lets verify this with an example,
⇒ |-5 - 4x| + 9 = -40
First, we have to isolate the absolute value expression, so we'll subtract 9 from both sides.
⇒ |-5 - 4x| = -49
We observe that LHS is positive while RHS is negative, which contradicts.
Since the number on the right-hand side comes out to be negative, thus the absolute function has 0 solutions possible. |
17633 | https://www.teachoo.com/subjects/cbse-maths/class-9th/ch14-9th-statistics/ | Statistics Class 9 - NCERT Solutions, MCQ, Practice Ques [2026 Exams]
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Class 9
Chapter 12 Class 9 Statistics
Chapter 12 Class 9 Statistics
Master Chapter 12 Class 9 Statistics with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Updated for 2026 Exams NCERT Book.
Get solutions of all NCERT Questions of Chapter 12 Class 9 Statistics. All exercise questions and examples are solved with detailed explanations of each and every question.
In this chapter, we will learn
The difference between primary and secondary data
Representing raw data as ungrouped and grouped frequency distribution table
Representing data as a bar graph
Drawing a histogram - with uneven class interval
Drawing frequency polygon - with and without a histogram
Finding mean, median and mode of raw data
Click on a NCERT Exercise below, or start the chapter from the concepts given below.
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Representing raw data as ungrouped and grouped frequency distribution table
Representing data as a bar graph
Drawing a histogram - with uneven class interval
Drawing frequency polygon - with and without a histogram
Finding mean, median and mode of raw data
Click on a NCERT Exercise below, or start the chapter from the concepts given below.
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17634 | https://achievethecore.org/coherence-map/6/28/288/288 | Measurement And Data
Solve Problems Involving Measurement And Conversion Of Measurements From A Larger Unit To A Smaller Unit.
Supporting Cluster
4.MD.A.1
Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two- column table. For example, know that 1 ft is 12 times as long as 1 in. Express the length of a 4 ft snake as 48 in. Generate a conversion table for feet and inches listing the number pairs (1, 12), (2, 24), (3, 36), ...
Operations And Algebraic Thinking
Use The Four Operations With Whole Numbers To Solve Problems.
Major Cluster
4.OA.A.2
Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.
See Glossary, Table 2
Number And Operations-Fractions
Apply And Extend Previous Understandings Of Multiplication And Division To Multiply And Divide Fractions.
Major Cluster
5.NF.B.3
Interpret a fraction) as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers) leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models) or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
Number And Operations-Fractions
Apply And Extend Previous Understandings Of Multiplication And Division To Multiply And Divide Fractions.
Major Cluster
5.NF.B.5
Interpret multiplication as scaling (resizing), by:
Number And Operations-Fractions
Apply And Extend Previous Understandings Of Multiplication And Division To Multiply And Divide Fractions.
Major Cluster
5.NF.B.6
Solve real world problems involving multiplication of fractions) and mixed numbers, e.g., by using visual fraction models) or equations to represent the problem.
Operations And Algebraic Thinking
Write And Interpret Numerical Expressions.
Additional Cluster
5.OA.A.2
Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them. For example, express the calculation “add 8 and 7, then multiply by 2” as 2 × (8 + 7). Recognize that 3 × (18932 + 921) is three times as large as 18932 + 921, without having to calculate the indicated sum or product.
Operations And Algebraic Thinking
Analyze Patterns And Relationships.
Additional Cluster
5.OA.B.3
Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule “Add 3” and the starting number 0, and given the rule “Add 6” and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so.
Ratios And Proportional Relationships
Understand Ratio Concepts And Use Ratio Reasoning To Solve Problems.
Major Cluster
6.RP.A.1
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
Example Task
6.RP The Escalator, Assessment Variation
Provided by Illustrative Mathematics
Task
Ty took the escalator to the second floor. The escalator is 12 meters long, and he rode the escalator for 30 seconds. Which statements are true? Select all that apply.
He traveled 2 meters every 5 seconds.
Every 10 seconds he traveled 4 meters.
He traveled 2.5 meters per second.
He traveled 0.4 meters per second.
Every 25 seconds, he traveled 7 meters.
Solution
This is a one-point item.
(a), (b) and (d) are all correct.
Download Example Task
Progressions
It is important for students to focus on the meaning of the terms “for every,” “for each,” “for each 1,” and “per” because these equivalent ways of stating ratios and rates are at the heart of understanding the structure in these tables, providing a foundation for learning about proportional relationships in Grade 7.
Please reference page 5 in the Progression document
Download Progressions PDF
Tasks
Bag of Marbles Games at Recess
Assessments
Ratios and Rates Mini-Assessment
Prior Grade-Level Tasks
The tasks below align to standards that are critical building blocks to this standard. Students do not necessarily need to be successful with these tasks to be able to engage with this standard, but these tasks can help teachers ground themselves in the skills and knowledge that students are bringing to the current unit of study and gather valuable information to use when teaching this standard. These tasks can be used as homework, warm-ups, in one-on-one work with students, or in PLC discussions. Because of their close ties to the work of this standard, the standards and the tasks below are a good place to start to:
review just before the current unit of study,
surface content that deserves additional support while addressing this standard,
focus the time on readiness for the current unit of study, and/or
provide visibility into a student’s thinking about a related task and determine how best to bring the student into the current unit of study.
6.EE.C.9: Families of Triangles 5.OA.3 Sidewalk Patterns 4.MD.A.1 Who is the Tallest? 4.OA.A.5 Inside Mathematics Bikes & Trikes 3.OA.D.9 Patterns in the Multiplication Table
NWEA Assessment Item Illustrating 6.RP.A.1 Smarter Balanced Assessment Item Illustrating 6.RP.A.1
Focus in Grade 6
Ratios And Proportional Relationships
Understand Ratio Concepts And Use Ratio Reasoning To Solve Problems.
Major Cluster
6.RP.A.2
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger."
Expectations for unit rates in this grade are limited to non-complex fractions.
Ratios And Proportional Relationships
Understand Ratio Concepts And Use Ratio Reasoning To Solve Problems.
Major Cluster
6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams), double number line diagrams), or equations.
Send Feedback |
17635 | https://www.teacherspayteachers.com/browse/independent-work/worksheets/free?search=solving%20unit%20rate%20problems | Solving Unit Rate Problems
Use unit rates with fractions to solve multi-step problems
Unit Rate and Proportion Word Problems
Get more with resources under $5
Find the Unit Rate from Word Problems | Solve & Color Math Activity or Practice
Ratios Lesson - Solve Unit Rate Problems (includes unit price & constant speed)
Solve Unit Rate Word Problems Worksheet L16 6th Grade iReady Math Exit Tickets
Practice Solving "Finding" Unit Rates word Problems, worksheets with Answers
Unit Rates Word Problems Worksheet - Includes answer key
Multistep Problems Using Unit Rate
Comparing Cost Using Unit Rate
Ontario Grade 8 Math Unit - Integers, Square Roots, and Order of Operations
Guided Unit Rates with Fractions Practice: Perfect Purple Paint
Unit Rates with Proportions
Customary Units of Measurement
ratios and unit rates
Proportions Worksheet
FREE Unit Rates: Which is the better value?
Thanksgiving Rates Freebie | Unit Rates | Fall | Math 6 | Math 7
Solving Story Problems Stations - Better Buy or More for Your Money Using Rates
Integer Operations - Lesson 1-1 - Pre-Algebra - (7Honors/ 8th)
Grade 7 Common Core Math: Ratios & Proportional Relationships 7.RP.A.2 #3
7th Grade Common Core Aligned Proportion Equation Worksheet
Metacognition Unit Self Reflection
(FREE) Volume of CUBE & CUBOID Printable Worksheets
Product Cost for store items
New Year: Math Review + Goal Setting
Fractions: Computing the Total Purchase Cost (Interactive Digital Worksheet)
Error Analysis: Reading Graphs in the Coordinate Plane
Decimals |
17636 | https://math.stackexchange.com/questions/3753876/transforming-constraints-into-linear-inequality | optimization - Transforming constraints into linear inequality - Mathematics Stack Exchange
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Transforming constraints into linear inequality
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I want to model the following two constraints in terms of LP, but after trying various ways without success, I wonder if it is possible at all?
Given x x and y i j y i j are binary variables. We need the following two constraints:
If x=1 x=1, then ∑n i=1 y i j≤1∑i=1 n y i j≤1 for any j=1,2,…,n j=1,2,…,n
If x=0 x=0, then ∑n i=1 y i j=2∑i=1 n y i j=2 for any j=1,2,…,n j=1,2,…,n.
Can someone please help me write these two constraints in terms of linear inequality? It seems so simple yet surprisingly difficult to me. Any help would really be appreciated.
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asked Jul 11, 2020 at 23:38
ghjkghjk
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Answer:
⎧⎩⎨⎪⎪⎪⎪⎪⎪∑i=1 n y i j≤2−x∑i=1 n y i j≥2−2 x for every j=1,2,…,n for every j=1,2,…,n{∑i=1 n y i j≤2−x for every j=1,2,…,n∑i=1 n y i j≥2−2 x for every j=1,2,…,n
Motivation:
What you want can be rewritten as:
If x=1 x=1, then 0≤∑i=1 n y i j≤1 0≤∑i=1 n y i j≤1 for every j=1,2,…,n j=1,2,…,n;
If x=0 x=0, then 2≤∑i=1 n y i j≤2 2≤∑i=1 n y i j≤2 for every j=1,2,…,n j=1,2,…,n.
Therefore, we just need to find two linear functions f,g f,g such that f(1)=0 f(1)=0, f(0)=2 f(0)=2, g(1)=1 g(1)=1, g(0)=2 g(0)=2. This is easy to do, using slope-intercept or whichever is your favourite way to compute equations for straight lines.
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answered Jul 11, 2020 at 23:55
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Thank you so much for your help! Brilliant.ghjk –ghjk 2020-07-12 06:02:35 +00:00 Commented Jul 12, 2020 at 6:02
can we rewrite it the other way around? I meant, if now we have the constraint is the summation, and what we want is x=0 x=0 if ∑n i=1 y i j≤1∑i=1 n y i j≤1 and x=1 x=1 if ∑n i=1 y i j=2∑i=1 n y i j=2, how can we do this? x x and y i j y i j are binary. Is the answer simply as: ∑n i=1 y i j−1≥x∑i=1 n y i j−1≥x?_ ghjk –ghjk 2020-07-13 07:11:34 +00:00 Commented Jul 13, 2020 at 7:11
Do you allow that ∑i=1 n y i j>2∑i=1 n y i j>2, or must it be either =2=2 or ≤1≤1?J. C. –J. C. 2020-07-13 14:40:29 +00:00 Commented Jul 13, 2020 at 14:40
Good point. I prefer it to be either 2 2 or ≤1≤1, but if ∑n i=1 y i j>2∑i=1 n y i j>2, then x=1 x=1 as well.ghjk –ghjk 2020-07-13 15:31:32 +00:00 Commented Jul 13, 2020 at 15:31
This is probably the answer: ∑n i=1 y i j=2⌊x⌋∑i=1 n y i j=2⌊x⌋ghjk –ghjk 2020-07-13 15:50:26 +00:00 Commented Jul 13, 2020 at 15:50
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17637 | https://math.stackexchange.com/questions/1968175/if-z-and-w-are-two-complex-numbers-prove-that-zw-zw | If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|. - Mathematics Stack Exchange
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If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
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If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
My Attempt;
If z is given as x + iy, and w = u + iv, then |z| = √(x^2 + y^2), = r, say; and |w| = √(u^2 + v^2), = p, say Then we can also view z as r cis θ, and w as p cis φ (i.e., p (cos φ + i sin φ) in polar form; so that θ and φ can be determined from x, y, u, and v -- to within 2π.) Now if z + w = q, say, consider the triangle formed by z, w, and q. The angle opposite q is π - |φ - θ| . So we have |q|^2 = |z|^2 + |w|^2 + 2 |
Is this correct. Please somebody check it. And if there is any other method... then plaese do it.
z| |w| cos (θ + φ), which is < |z + w|^2, since cos x
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asked Oct 14, 2016 at 11:30
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Perhaps writing the expression explicitly it can be clearer:
⎧⎩⎨z=a+b i w=x+i y⟹|z+w|=(a+x)2+(b+y)2−−−−−−−−−−−−−−−√≤?a 2+b 2−−−−−−√+x 2+y 2−−−−−−√{z=a+b i w=x+i y⟹|z+w|=(a+x)2+(b+y)2≤?a 2+b 2+x 2+y 2
Well, now square both sides:
a 2+2 a x+x 2+b 2+2 b y+y 2≤?a 2+b 2+x 2+y 2+2(a 2+b 2)(x 2+y 2)−−−−−−−−−−−−−−−√⟺a 2+2 a x+x 2+b 2+2 b y+y 2≤?a 2+b 2+x 2+y 2+2(a 2+b 2)(x 2+y 2)⟺
⟺2(a x+b y)≤?2(a 2+b 2)(x 2+y 2)−−−−−−−−−−−−−−−√⟺⟺2(a x+b y)≤?2(a 2+b 2)(x 2+y 2)⟺
⟺a 2 x 2+2 a b x y+b 2 y 2≤a 2 x 2+a 2 y 2+b 2 x 2+b 2 y 2⟺⟺a 2 x 2+2 a b x y+b 2 y 2≤a 2 x 2+a 2 y 2+b 2 x 2+b 2 y 2⟺
2 a b x y≤a 2 y 2+b 2 x 2⟺a 2 y 2+b 2 x 2−2 a b x y≥0⟺2 a b x y≤a 2 y 2+b 2 x 2⟺a 2 y 2+b 2 x 2−2 a b x y≥0⟺
⟺(a y−b x)2≥0✓⟺(a y−b x)2≥0✓
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answered Oct 14, 2016 at 11:38
DonAntonioDonAntonio
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@DonAntonii, How did you say that it has been proved? Where is your conclusion?Ger Wyn –Ger Wyn 2016-10-14 17:27:33 +00:00 Commented Oct 14, 2016 at 17:27
@GerWyn Observe the double implications and reach the wanted inequality. I thought it was pretty clear... You understand now?DonAntonio –DonAntonio 2016-10-14 17:29:57 +00:00 Commented Oct 14, 2016 at 17:29
But if you state the conclusion, it would be easier for the OP to know it easily.Ger Wyn –Ger Wyn 2016-10-14 17:34:36 +00:00 Commented Oct 14, 2016 at 17:34
@GerWyn I think anyone doing complex analysis can very easily follow the above: it's like A⟺B⟺C⟺D A⟺B⟺C⟺D , so if D D is clear, and right, then also A A is .DonAntonio –DonAntonio 2016-10-14 17:36:48 +00:00 Commented Oct 14, 2016 at 17:36
I was just asking if you could show the conclusions, too. As you wish.Ger Wyn –Ger Wyn 2016-10-14 17:48:36 +00:00 Commented Oct 14, 2016 at 17:48
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0Spivak, Ch. 25, "Complex Numbers": Prove |z+w|≤|z|+|w||z+w|≤|z|+|w|, z z and w w complex numbers when z=λ w z=λ w. Must we consider separate cases?
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How to solve a system of linear inequalities which satisfies most of the given constraints?
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I need to find a solution to a system of linear inequalities of the form
```
a11x1 + a12x2 +... a1ixi >= b1
a21x1 + a22x2 +... a2ixi >= b2
```
and so on and in case no solution exists, then return a solution x which satisfies most of the constraints. If multiple values satisfy give any.
Any ideas on how this can be done?
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edited Jun 30, 2015 at 8:26
mattmilten
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asked Jun 29, 2015 at 11:24
marvel308marvel308
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Formulating this as an optimization problem is probably not the big issue here. But I guess it's not possible to formulate this as a problem with smooth constraints. So the real problem is probably that problems of this kind are very hard to optimize.
– cel
Commented
Jun 29, 2015 at 11:43
There exists an exponential solution , but that is inefficient . Can it be done in polynomial time ?
– marvel308
Commented
Jun 29, 2015 at 12:02
If I had to guess I would say that this problem is equal to optimizing a L0 norm, and this is proven to be np-hard. But that's just my intuition, I cannot say for sure.
– cel
Commented
Jun 29, 2015 at 12:11
1
You could add a non-negative slack variable to every inequality and then minimize the sum of all slack variables. If your optimal objective function value then is 0 you have found a feasible solution for your x variables. If not, then at least some of your slacks may be 0, though this won't yield a solution that satisfies most constraints, as you have required.
– mattmilten
Commented
Jun 29, 2015 at 21:39
What size of problem are we talking about? how many x's and how many constraints? There are several options you might want to try.
– Ioannis
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Jul 19, 2015 at 1:38
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You need to have an objective function also, you can solve them in Excel Solver. But that wont be suitable for big equations. For bigger equations, you can use a Java Library (LpSolve). And as mentioned above, you can add slack variables in every equation with high penalty cost in the objective function.
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answered Jul 18, 2015 at 17:52
Amit MadanAmit Madan
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17639 | https://www.mometrix.com/academy/piecewise-functions/ | How to find the Piecewise Functions (Video & Practice Questions)
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Piecewise Functions
How to find the Piecewise Functions (Video & Practice Questions)
On this page
Function – Definition
Absolute Value Functions
Other Piecewise Functions
Naming Piecewise Functions
Piecewise Function Practice Questions
[x] Transcript - [x] Practice
Hi, and welcome to this video about piecewise functions! In this video, we will explore:
What piecewise functions are
How piecewise functions are defined
And how piecewise functions can be used
Function – Definition
First, let’s look at the definition of a function. A function is a relationship where a single output is assigned to each input.
Many functions belong to function families because their equations and graphs all have similar characteristics.
For example, functions in the linear family have equations that resemble f(x)=m x+b f(x)=m x+b and their graphs are straight lines. Functions in the quadratic family have equations that look like f(x)=a x 2 f(x)=a x 2, and their graphs are parabolas.
Piecewise functions are not considered a function family on their own. As the name suggests, they are functions comprised of pieces of other functions.
Absolute Value Functions
The first piece we’re going to look at is the absolute value function.
Functions in the absolute value family have equations that resemble f(x)=|x|f(x)=|x| and their graphs all have a characteristic “V” shape. This is the graph and a section of the table of values of f(x)=|x|f(x)=|x|.
Instead of a single V, f(x)f(x) can also be visualized as pieces of two linear functions. On the left, f(x)=−x f(x)=−x, and on the right, f(x)=x f(x)=x.
As we “read” the graph from left to right, we are on the function f(x)=−x f(x)=−x until x=0 x=0. At that point, the function definition changes to f(x)=x f(x)=x.
The domain of the absolute value function is all real numbers. Normally, both f(x)=−x f(x)=−x and f(x)=x f(x)=x also have domains of all real numbers, but if we were to graph them together, the graph would look like this and we would no longer have a function.
So each piece needs to be defined on a section of its domain in order to define a piecewise function. If the left function is only defined for negative x x-values and the right is only defined for positive x x-values (and we put the 0 into one of them—more on that in a minute), we can define this as a single piecewise function.
One way to visualize this is to graph both linear functions and erase the sections that are not part of the absolute value function.
Other Piecewise Functions
Let’s take a quick break and practice describing a couple of piecewise functions in terms of what their pieces look like and where those pieces are defined.
Integer Function
First up is the greatest integer function, f(x)=⌊x⌋f(x)=⌊x⌋. This is also called the floor function or stair step function.
This function is made up of pieces of constant functions that are 1 unit wide.
Sawtooth Function
Next, we have the sawtooth function, f(x)=x–⌊x⌋f(x)=x–⌊x⌋. This function is also called the castle rim function.
This function is made up of pieces of parallel linear functions that are one unit long
Naming Piecewise Functions
There are two criteria for naming piecewise functions:
Reading the pieces from top-to-bottom in the list corresponds to reading the graph from left-to-right. So the first piece in the list is the left piece on the graph.
The domains of the pieces must “add up to” the domain of the entire function.
Let’s start by rewriting our absolute value function in this form.
Here’s the list of expressions that define each piece from left to right:
f(x)={−x x f(x)={−x x
Now let’s take a minute to consider the domain of each piece.
f(x)=|x|f(x)=|x| has a domain of all real numbers. We can input any number we want and get a single output, but we have to be careful with our piecewise definition when we consider x=0 x=0. Why? Because 0 is a defined point on both pieces, but we only need to include it once.
If we write the function like this:
f(x)={−x,x,if x<0 if x>0 f(x)={−x,if x<0 x,if x>0
Neither piece includes 0 and the domain is incomplete.
If we write it like this:
f(x)={−x,x,if x≤0 if x≥0 f(x)={−x,if x≤0 x,if x≥0
We are not defining a function, because we’re saying that f(x)f(x) equals both −x−x and x x at x=0 x=0 (even though the output would technically be the same in this case).
Since 0 is in the domain of both pieces, we simply choose which piece to put it in. Both of these equations correctly define the function.
All of our examples so far have depicted piecewise functions with domains of all real numbers. The pieces do not need to connect and they do not need to extend to plus or minus infinity.
Take a look at this function and try to define it with an equation (and yes that single point is a part of it!):
f(x)=⎧⎩⎨⎪⎪⎪⎪x,1,x−−√,x,−7<x≤−3 x=−1 0≤x<4 6<x≤7 f(x)={x,−7<x≤−3 1,x=−1 x,0≤x<4 x,6<x≤7
Like other functions, piecewise functions can be used to tell stories. This is the story of a car journey Bob took last week:
d(t)=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪50 t,150,25 t+62.5,250,−62.5 t,0≤t≤3 3<t<3.5 3.5≤t<7.5 7.5≤t<8 8≤t≤12 d(t)={50 t,0≤t≤3 150,3<t<3.5 25 t+62.5,3.5≤t<7.5 250,7.5≤t<8−62.5 t,8≤t≤12
Let’s see how many of these review questions we can answer about Bob’s journey before we go:
How long did the journey last? 12 hrs
What’s the farthest distance Bob drove from his home? 250 mi
How long did it take him to get there? 7.5 hrs
What was Bob doing from 3-3.5 hrs? not moving
When did Bob turn toward home? at the 8 hr mark
When was Bob driving the fastest? from hr 8 to hr 12 he traveled 62.5 mph
What was Bob’s average speed from 0 to 8 hrs? 31.25 mph
Thanks for watching, and happy studying!
Piecewise Function Practice Questions
Question #1:
The following piecewise function shows the speed of a car as a function of time. Daniel says that the flat top portion of the graph shows that the car came to a stop. Is Daniel correct?
Daniel is correct.
Daniel is incorrect.
[x] Show Answer
Answer:
Each “piece” of the graph shows the car traveling at a different speed. The first piece of the graph shows the car gradually increasing speed as time elapses. The second piece of the graph shows the car maintaining a steady speed as time elapses. The third piece of the graph shows the car steadily reducing speed.
Even though the flat piece of the graph seems to show the car stopping, the speed is maintained as time is passing.
[x] Hide Answer
Question #2:
What kind of piecewise function is graphed below?
Linear
Quadratic
Step
Absolute Value
[x] Show Answer
Answer:
An absolute value function will form a V shape. This type of function is essentially two “pieces” of linear functions.
[x] Hide Answer
Question #3:
Define the following piecewise function.
h(x)=2 if x≤3 h(x)=2 if x≤3 x if x>−1 x if x>−1
h(x)=2 if x≤1 h(x)=2 if x≤1 x if x>1 x if x>1
h(x)=2 if x≤9 h(x)=2 if x≤9 x if x>3 x if x>3
h(x)=1 if x≤1 h(x)=1 if x≤1 x if x>−3 x if x>−3
[x] Show Answer
Answer:
The graph shows that when x x is less than or equal to 1 1, then h(x)h(x) is equal to 2 2. For example, when x x is equal to 0 0, h(x)h(x) equals 2 2. This is defined as h(x)=2 if x>1 h(x)=2 if x>1.
When x is greater than 1 1, then h(x)h(x) is equal to x x. For example, when x x is equal to 4 4, h(x)h(x) is also equal to 4 4. This is defined as h(x)=x if x>1 h(x)=x if x>1.
These two pieces define the piecewise function for the graph:
h(x)=2 if x≤1 h(x)=2 if x≤1
x if x>1 x if x>1
[x] Hide Answer
Question #4:
The graph below shows the relationship between time in minutes (x x-axis) and water in gallons (y y-axis). The graph shows the amount of water that was used (in gallons) when George gave his new puppy a bath.
When did George turn the water faucet off? How much water was in the bath when he bathed the puppy?
George turned the water off at 30 minutes. There were 24 gallons of water in the tub when he bathed the puppy.
George turned the water off at 12 minutes. There were 18 gallons of water in the tub when he bathed the puppy.
George turned the water off at 6 minutes. There were 24 gallons of water in the tub when he bathed the puppy.
George turned the water off at 24 minutes. There were 6 gallons of water in the tub when he bathed the puppy.
[x] Show Answer
Answer:
The amount of water in gallons stops increasing when George turns the water faucet off. The x x-axis shows us that when 6 6 minutes have passed, the amount of water stops increasing. This means that George turns off the water after 6 6 minutes, and begins the puppy bath. The y y-axis indicates that at this time the water is currently at 24 24 gallons.
[x] Hide Answer
Question #5:
Define the following piecewise function.
f(x)=x+2 5≤x≤−1 f(x)=x+2 5≤x≤−1−x 1<x≤3−x 1<x≤3
f(x)=x+5 7≤x≤1 f(x)=x+5 7≤x≤1−x 1<x≤3−x 1<x≤3
f(x)=x+1−4≤x≤1 f(x)=x+1−4≤x≤1−x 1<x≤3−x 1<x≤3
f(x)=x+3−5≤x≤1 f(x)=x+3−5≤x≤1−x−4<x≤2−x−4<x≤2
[x] Show Answer
Answer:
The graphed line on the left side represents x x-values that are greater than or equal to −4−4, and less than or equal to 1 1. This region can be described as −4≤x≤1−4≤x≤1. In this region, the f(x)f(x)-value, or y y-value, is always one more than the x x-value. For example, when x x is 1 1, y y is 2 2. When x x is −3−3, y y is −2−2. This means that in the region of −4≤x≤1−4≤x≤1, y y is always one more than x x. This is defined as f(x)=x+1−4≤x≤1 f(x)=x+1−4≤x≤1.
The graphed line on the right side represents x x-values that are greater than 1 1, and less than or equal to 3 3. This region can be described as 1<x≤3 1<x≤3. In this region, the f(x)f(x)-value, or the y y-value, is always the x x-value negated. For example, when x x is 2 2, y y is −2−2. When x x is 3 3, y y is −3−3. This means that in the region of 1<x≤3 1<x≤3, y y is always the negative value of x x. This is defined as f(x)=−x 1<x≤3 f(x)=−x 1<x≤3.
When both “pieces” are put together, they define the piecewise function as
f(x)=x+1−4≤x≤1 f(x)=x+1−4≤x≤1
−x 1<x≤3−x 1<x≤3
[x] Hide Answer
Return to Algebra II Videos
707921
by Mometrix Test Preparation | Last Updated: August 12, 2025
On this page
Function – Definition
Absolute Value Functions
Other Piecewise Functions
Naming Piecewise Functions
Piecewise Function Practice Questions
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Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE’s) deal with functions of one variable, which can often be thought …Show more
Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE’s) deal with functions of one variable, which can often be thought of as time.Show less
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17641 | https://testbook.com/question-answer/the-sum-of-3-digit-numbers-abc-bca-and-cab-is-alw--624719dab225957c3e1efbf2 | [Solved] The sum of 3-digit numbers abc, bca and cab is always divisi
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The sum of 3-digit numbers abc, bca and cab is always divisible by:
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Option 3 : 37
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Given:
The 3-digit numbers abc, bca and cab
Calculation:
Generic forms are:
abc = 100a + 10b + c
bca = a + 100b + 10c
cab = 10a + b + 100c
So, (100a + 10b + c) + (a + 100b + 10c) + (10a + b + 100c)
= 111a + 111b + 111c
= 111 (a + b + c)
= 37× 3(a + b + c)
∴The sum of 3-digit numbers abc, bca and cab is always divisible by 37
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More Divisibility and Remainder Questions
Q1.What is the number that when divided by 7, 9 and 12 leaves the same remainder 1 in each case ?
Q2.When a number divided by 114 it leaves a remainder 14. What will be the remainder, when the same number divided by 19.
Q3.If the 4-digit number 48ab is divisible by 2, 5 and 7, then what is the value of (10a - b) ?
Q4.Which of the following numbers is NOT divisible by 4?
Q5.Which of the following number is divisible by 6?
Q6.The greatest number of four digits which when divided by 12, 16 and 24 leave remainders 2, 6 and 14 respectively is ?
Q7.What should be added to 1459 so that it is exactly divisible by 12?
Q8.A Question is given below followed by two Statements I and II. Question: If n is an integer greater than 1, is 3n - 2n divisible by 35? Statement I: n is divisible by 15. Statement II: n is divisible by 18. Which one of the following is correct in respect of the above Question and the Statements?
Q9.A Question is given below followed by two Statements I and II. Question: What is the remainder of [(x + 1)N /x], wherein x and N are natural numbers? Statement-l: x = 5, and N is divisible by 7. Statement-II: x = 17, and N is divisible by 20. Which of the following is correct in respect of the above Question and the Statements?
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Q1.If a shopkeeper gives a candy for Rs. 5 and also a candy for 2 wrappers of the same candy, then for Rs. 40 how many candies can a child eat?
Q2.In a series 1, 2, 3, 4, 5......, multiples of 2, 3, 4 and 5 are deleted from the series. In the remaining series, what will be the first number greater than 50?
Q3.The Lowest Common Multiple (LCM) of two numbers is 48. The numbers are in the ratio of 3 : 2. What is the sum of the numbers?
Q4.What is the number that when divided by 7, 9 and 12 leaves the same remainder 1 in each case ?
Q5.A set of natural numbers has to be formed that consists of six numbers, and the last number is 30. The First number of the set is the highest root of the given equation, and the second number of the set is (n + 4) - 2n, where 'n' is the difference between the roots of the given equation. The third number of the set is (n + 1)2- (n + 1). The fourth number of the set is (n + 2)2- (n + 2). The fifth number of the set is (n + 1)3– (n + 5). Note: x2- 14x + 48 = 0 Which of the statement/s is/are correct. I. The product of second and fifth number is perfect square. II. The average of last four number is 17. III. If the seventh number of the set is sixth number + n3, then the resultant number is a factor of 114.
Q6.When a number divided by 114 it leaves a remainder 14. What will be the remainder, when the same number divided by 19.
Q7.LCM and GCD of two numbers are 63 and 21 respectively. If one of them is 21, then find the second number?
Q8.The sum of two numbers is 384 HCF of the numbers is 48 the difference of the number is
Q9.If the 4-digit number 48ab is divisible by 2, 5 and 7, then what is the value of (10a - b) ?
Q10.The LCM of two numbers is 48. The numbers are in the ratio of 2∶ 3. Find the sum of the number.
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17642 | https://www.statpearls.com/point-of-care/31223 | Vitamin D | Treatment & Management | Point of Care
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Vitamin D
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Author: Krati ChauhanAuthor: Mahsa ShahrokhiEditor: Martin R. HueckerUpdated: 4/9/2023 9:13:09 AM
Indications
Vitamin D is labeled as the "sunshine vitamin," as it is produced in the skin on sun exposure. Vitamin D is required to maintain the serum calcium concentration within the normal physiologic range for musculoskeletal health.
The Endocrine Society, the National and International Osteoporosis Foundation, and the American Geriatric Society define vitamin D deficiency as the level of 25-hydroxyvitamin (25 OH D) of less than 30 ng/mL. The Endocrine Society recommends a preferred range of 40 to 60 ng/mL. In contrast, the National Institute of Health defines vitamin D deficiency as less than 20 ng/ml. Some authorities define insufficiency as 12 to 19 ng/mL and deficiency as less than 12 ng/mL. To maintain this level, the Endocrine Society recommends an intake of 400 to 1000 International Units (IU) daily for infants less than one year, 600 to 1000 IU for children and adolescents from 1 to 18 years, and 1500 to 2000 IU for all adults.
Vitamin D deficiency in children causes rickets and prevents children from reaching their peak bone mass and genetically determined height. In adults, vitamin D deficiency results in abnormal mineralization of the collagen matrix in bone, referred to as osteomalacia. This collagen matrix is weak, does not provide adequate structural support, and increases the risk of fracture. This abnormally mineralized matrix pushes the periosteum, a highly innervated structure, outward and results in aching bones, a common complaint in vitamin D deficient individuals. Vitamin D deficiency also results in muscle weakness and muscle pain. Patients complain of generalized bone and muscle pain. Around 40% to 60% of patients with generalized myalgias and bone pain have vitamin D deficiency.
Vitamin D deficiency (level < 30 ng/mL) and insufficiency (level between 20 to 30 ng/mL) are a problem across the globe. Pregnant women, African Americans, Hispanics, obese adults, and children are at high risk for vitamin D deficiency. In the United States, 50% of children ages 1 to 5 and 70% of children ages 6 to 11 have vitamin D deficiency. It is attributed to an increase in the incidence of obesity, a decrease in milk consumption, and the use of sun protection.
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Mechanism of Action
Vitamin D is a hormone obtained through dietary consumption and skin production. Ultraviolet B (UVB) radiation, wavelength (290 to 315 nm), converts 7-dehydrocholesterol in the skin to previtamin D. This previtamin D undergoes heat isomerization and is converted to vitamin D. Vitamin D from the skin and diet is metabolized in the liver to 25-hydroxyvitamin D (25 OH D), and 25-hydroxyvitamin D is useful in assessing vitamin D status. In the kidneys, 25 hydroxyvitamin D converts to the biologically active form: 1,25-dihydroxy vitamin D (1,25 (OH)) by the enzyme 25-hydroxyvitamin D-1 alpha-hydroxylase (CYP27B1). Renal production of 1,25-dihydroxy vitamin is under the regulation of parathyroid, calcium, and phosphorus levels.
1,25-dihydroxy vitamin D binds to the vitamin D receptor, a hormone receptor present at the nucleus inside the cell. Gene transcription is modified through the binding of vitamin D to its receptor, resulting in the activation of certain genes and suppression of others. It stimulates intestinal calcium and phosphorus absorption. In the absence of vitamin D, approximately 10 to 15% of dietary calcium and 60% of phosphorus are absorbed. In the presence of vitamin D, this percentage of absorption is increased to 30% to 40% for calcium and 80% for phosphorus. In the kidneys, 1,25-dihydroxy vitamin promotes calcium reabsorption.
Vitamin D has a physiologic function outside calcium metabolism. Vitamin D receptor is present in the small intestine, colon, T and B lymphocytes, mononuclear cells, brain, and skin. It stimulates insulin production, modulates the function of activated T and B lymphocytes, prevents inflammatory bowel diseases, and affects myocardial contractility.
Topical 1,25-dihydroxy vitamin D has utility in the treatment of psoriasis. It reduces scaling and erythema in psoriasis. Keratinocytes in the skin, which function abnormally in psoriasis, have vitamin D receptors, and vitamin D inhibits their proliferation and induces differentiation.
Administration
Vitamin D administration can be oral, or the skin can make it via UV exposure. A serum level of 25-hydroxyvitamin D (25 OH D) of at least 30 ng/ml (78 nmoL/L) is required to maintain the physiologic function of vitamin D. Recommendations are to use 25-hydroxyvitamin D (25 OH D) as a measure of vitamin D status as it has a half-life of 2 weeks; whereas, 1,25-dihydroxy vitamin D (1,25 (OH)), the biologically active form, has a serum half-life of < 4 hours and should not be used to measure vitamin D status.
Factors that alter the amount of UVB radiation reaching the skin change the cutaneous production of vitamin D. Melanin in the skin absorbs UVB radiation and prevents the conversion of 7-dehydrocholesterol to vitamin D. Hence, individuals with increased skin pigmentation have decreased cutaneous production of vitamin D and require a longer duration of exposure to UVB radiation to produce vitamin D. Sunscreen, which also absorbs UVB radiation, decreases cutaneous production of vitamin D. A sunscreen with a sun protection factor (SPF) of 8 reduces cutaneous production of UVB by > 95% and a sunscreen with an SPF of 15 will reduce this to > 98%.
During winter, sun rays enter at a more oblique angle, and the ozone layer absorbs a higher amount of UVB radiation. Hence, less UVB radiation reaches the skin. For this reason, during the winter months, there is a decrease in the production of vitamin D.Similarly, at latitudes greater than 37 degrees, there is a decrease in the UVB radiation reaching the skin, which reduces vitamin D production. In the early morning and the evening, the sun's rays enter at an oblique angle, and the skin produces very little UVB.
Vitamin D is fat-soluble and stored in body fat. In obese individuals, a greater amount of vitamin D is stored in fat, and less is available for biological functions. Hence obese people require larger units of vitamin D supplementation to maintain an adequate serum level of vitamin D.
Very few foods are a natural source of vitamin D. These include oily fish such as salmon, mackerel, and sardines. Foods fortified with vitamin D are milk and orange juice (100 units per 8 ounce serving) and some bread and cereals. An important source of oral vitamin D is vitamin D supplements, which are available both over the counter and through prescription. These are available in strengths of 1000 IU, 2000 IU, 5000 IU, and 50,000 IU, which are available only through prescription.
Monitoring
The recommendation is to check the level of the circulating form of vitamin D(25-hydroxyvitamin D) at least twice a year. Once in spring, which will reflect low levels after the winter, and once in fall, which will reflect higher levels after the summer, and the dose should be adjusted accordingly.
Toxicity
Vitamin D intoxication is extremely rare. Vitamin D intoxication from sun exposure does not occur as the skin destroys excess vitamin D. The only way a person may get vitamin D toxicity is by ingestion of extremely high doses of vitamin D for a prolonged period. Concentrations over 150 ng/mL (325 nmoL/L) may result in vitamin D intoxication and are associated with hypercalcemia. Some symptoms associated with vitamin D toxicity and hypercalcemia include constipation, polydipsia, polyuria, and confusion.
Enhancing Healthcare Team Outcomes
All interprofessional healthcare team members, including clinicians, mid-level practitioners, nurses, pharmacists, and dieticians, need to be aware of vitamin D deficiency (level < 30 ng/mL) and insufficiency (level between 20 to 30 ng/mL) are a problem across the globe. Pregnant women, African Americans, Hispanics, obese adults, and children are at high risk for vitamin D deficiency. In the United States, 50% of children ages 1 to 5 and 70% of children ages 6 to 11 have vitamin D deficiency. Experts attribute this fact to an increase in the incidence of obesity, a decrease in milk consumption, and the use of sun protection. The interprofessional healthcare team needs to examine all these factors when assessing the patient.
Very few foods are a natural source of vitamin D. These include oily fish such as salmon, mackerel, and sardines. Foods fortified with vitamin D are milk and orange juice, and some bread and cereals. An important source of oral vitamin D is vitamin D supplements, which are available both over the counter and through prescription. These are available in strengths of 1000 IU, 2000 IU, 5000 IU, and 50,000 IU, which are available only through prescription.By engaging in interprofessional collaboration, the healthcare team can ensure patients are adequately supplied with this vital nutrient and drive their patients to better health. [Level 5]
References
Stamm E, Acchini A, Da Costa A, Besse S, Christou F, Launay C, Balmer P, Humbert M, Nguyen S, Major K, Bosshard W, Büla C. [Year in review : geriatrics]. Revue medicale suisse. 2019 Jan 9:15(N° 632-633):50-52 [PubMed PMID: 30629369]
. . :(): [PubMed PMID: 30628983]
Level 1 (high-level) evidence
Hernigou P, Auregan JC, Dubory A. Vitamin D: part II; cod liver oil, ultraviolet radiation, and eradication of rickets. International orthopaedics. 2019 Mar:43(3):735-749. doi: 10.1007/s00264-019-04288-z. Epub 2019 Jan 9 [PubMed PMID: 30627846]
Tang H,Li D,Li Y,Zhang X,Song Y,Li X, Effects of Vitamin D Supplementation on Glucose and Insulin Homeostasis and Incident Diabetes among Nondiabetic Adults: A Meta-Analysis of Randomized Controlled Trials. International journal of endocrinology. 2018; [PubMed PMID: 30627160]
Level 1 (high-level) evidence
Fink C, Peters RL, Koplin JJ, Brown J, Allen KJ. Factors Affecting Vitamin D Status in Infants. Children (Basel, Switzerland). 2019 Jan 8:6(1):. doi: 10.3390/children6010007. Epub 2019 Jan 8 [PubMed PMID: 30626163]
Häusler D, Weber MS. Vitamin D Supplementation in Central Nervous System Demyelinating Disease-Enough Is Enough. International journal of molecular sciences. 2019 Jan 8:20(1):. doi: 10.3390/ijms20010218. Epub 2019 Jan 8 [PubMed PMID: 30626090]
Nair R, Maseeh A. Vitamin D: The "sunshine" vitamin. Journal of pharmacology & pharmacotherapeutics. 2012 Apr:3(2):118-26. doi: 10.4103/0976-500X.95506. Epub [PubMed PMID: 22629085]
Maurya VK, Aggarwal M. Factors influencing the absorption of vitamin D in GIT: an overview. Journal of food science and technology. 2017 Nov:54(12):3753-3765. doi: 10.1007/s13197-017-2840-0. Epub 2017 Sep 23 [PubMed PMID: 29085118]
Level 3 (low-level) evidence
Wacker M, Holick MF. Sunlight and Vitamin D: A global perspective for health. Dermato-endocrinology. 2013 Jan 1:5(1):51-108. doi: 10.4161/derm.24494. Epub [PubMed PMID: 24494042]
Level 3 (low-level) evidence
Awadh AA, Hilleman DE, Knezevich E, Malesker MA, Gallagher JC. Vitamin D supplements: The pharmacists' perspective. Journal of the American Pharmacists Association : JAPhA. 2021 Jul-Aug:61(4):e191-e201. doi: 10.1016/j.japh.2021.02.002. Epub 2021 Feb 10 [PubMed PMID: 33674204]
Level 3 (low-level) evidence
Marcinowska-Suchowierska E, Kupisz-Urbańska M, Łukaszkiewicz J, Płudowski P, Jones G. Vitamin D Toxicity-A Clinical Perspective. Frontiers in endocrinology. 2018:9():550. doi: 10.3389/fendo.2018.00550. Epub 2018 Sep 20 [PubMed PMID: 30294301]
Level 3 (low-level) evidence
Bassatne A, Chakhtoura M, Saad R, Fuleihan GE. Vitamin D supplementation in obesity and during weight loss: A review of randomized controlled trials. Metabolism: clinical and experimental. 2019 Mar:92():193-205. doi: 10.1016/j.metabol.2018.12.010. Epub 2019 Jan 4 [PubMed PMID: 30615949]
Level 1 (high-level) evidence
Teymoori-Rad M, Shokri F, Salimi V, Marashi SM. The interplay between vitamin D and viral infections. Reviews in medical virology. 2019 Mar:29(2):e2032. doi: 10.1002/rmv.2032. Epub 2019 Jan 6 [PubMed PMID: 30614127]
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17644 | https://www.vedantu.com/chemistry/henderson-hasselbalch-equation | Chemistry
Henderson-Hasselbalch Equation Explained
Henderson-Hasselbalch Equation Explained
Reviewed by:
Ritika Singla
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How to Calculate Buffer pH Using the Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation is essential in chemistry and helps students understand buffer solutions, acid-base equilibrium, and the pH control of many real-life solutions.
This equation is a cornerstone for both practical labs and theoretical concepts, making it vital for a strong foundation in physical chemistry.
What is Henderson-Hasselbalch Equation in Chemistry?
A Henderson-Hasselbalch equation refers to a mathematical relationship that calculates the pH of a buffer solution using the concentrations of a weak acid (or base) and its conjugate partner.
This concept appears in buffer solutions, acid-base equilibrium, and chemical equilibrium, making it a foundational part of your chemistry syllabus.
Molecular Formula and Composition
The Henderson-Hasselbalch equation can be written as:
pH = pKa + log([A-]/[HA])
Here, [A-] is the concentration of the conjugate base, [HA] is the concentration of the weak acid, and pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka). It is used to estimate the pH of mixtures, especially in buffer solutions.
Preparation and Synthesis Methods
To make buffer solutions for applying the Henderson-Hasselbalch equation, you can mix a weak acid (like acetic acid) with its salt (like sodium acetate), or a weak base (like ammonia) with its salt (like ammonium chloride). Ensure the concentrations of acid and salt are similar for best buffer action.
Physical Properties of Henderson-Hasselbalch Buffers
Buffer solutions prepared for pH control are usually clear, colorless or weakly colored liquids. Their key feature is stable pH value. The solution’s pH depends on the chosen acid/base pair and their concentration ratio, not on appearance, odor or boiling point.
Chemical Properties and Reactions
Buffers made using the Henderson-Hasselbalch equation resist pH change when small amounts of acid or base are added.
If an acid is added, the base component neutralizes it; if a base is added, the acid component neutralizes it.
This stability is crucial for chemical and biological systems.
Frequent Related Errors
Using the Henderson-Hasselbalch equation for strong acids or bases instead of weak ones.
Getting the logarithm term upside down: log([base]/[acid]) – make sure the conjugate base is on top.
Applying the equation when buffer concentrations are too low or too diluted.
Neglecting significant temperature changes, which can affect pKa and final pH.
Uses of Henderson-Hasselbalch Equation in Real Life
The Henderson-Hasselbalch equation is widely used in laboratories to prepare standardized buffer solutions, in acid-base analysis, and for medical purposes such as maintaining blood pH. It is also essential in pharmaceutical formulations and food processing.
Relation with Other Chemistry Concepts
The Henderson-Hasselbalch equation ties closely with concepts like pH and pOH pKa and pKb, and equilibrium, reinforcing the importance of acid-base balance in chemical reactions, buffer action, and titration curves.
Step-by-Step Reaction Example
Write the dissociation equation for acetic acid in water:
CH3COOH ⇌ H+ + CH3COO-
Express the acid dissociation constant (Ka):
Ka = [H+][CH3COO-] / [CH3COOH]
Take negative log on both sides to convert Ka to pKa:
-log Ka = -log([H+][A-]/[HA])
Rearranged, this gives:
pH = pKa + log([A-]/[HA])
Example Calculation:
If [CH3COO-] = 0.5 M, [CH3COOH] = 0.2 M, pKa = 4.7
pH = 4.7 + log(0.5/0.2) = 4.7 + log(2.5) = 4.7 + 0.40 = 5.10 (rounded).
Lab or Experimental Tips
Always use freshly prepared solutions and measure concentrations accurately. The closer your acid and conjugate base concentrations, the more effective your buffer will be. Vedantu educators recommend using a pH meter for precise measurements in experiments involving the Henderson-Hasselbalch equation.
Try This Yourself
Write the Henderson-Hasselbalch equation for a basic buffer (use pOH and pKb).
Calculate the pH of a buffer solution with 0.4 M weak acid and 0.4 M salt, given pKa = 5.0.
Find a real-life example where the Henderson-Hasselbalch equation is critical (for example: blood pH regulation).
Final Wrap-Up
We explored the Henderson-Hasselbalch equation—its formula, derivation, real-life uses, and practical examples. Mastering this equation helps in understanding buffers, acid-base chemistry, and biological systems. For guided learning and more chemistry topics, join live sessions and download resources from Vedantu.
FAQs on Henderson-Hasselbalch Equation Explained
What is the Henderson-Hasselbalch equation?
The Henderson-Hasselbalch equation is a chemical formula used to estimate the pH of a buffer solution consisting of a weak acid and its conjugate base. The equation is:
pH = pKa + log([A-] / [HA])
Where:
• pH: measure of solution acidity
• pKa: acid dissociation constant (logarithmic scale)
• [A-]: concentration of conjugate base
• [HA]: concentration of weak acid
What does the Henderson-Hasselbalch equation explain?
The equation explains the relationship between the pH of a buffer solution and the ratio of the concentrations of the weak acid and its conjugate base.
Main uses:
• Quick calculation of buffer pH
• Predicting how buffer composition affects pH
• Explaining the pH stability of biological and chemical systems
How do you use the Henderson-Hasselbalch equation to calculate pH?
To calculate pH with the Henderson-Hasselbalch equation:
Identify the pKa of the weak acid from data or tables.
Measure or determine the concentrations of the conjugate base [A-] and weak acid [HA].
Substitute values into:
pH = pKa + log([A-] / [HA])
Calculate the log and sum to find the pH.
When can you not use the Henderson-Hasselbalch equation?
The Henderson-Hasselbalch equation should not be used:
• When using strong acids or strong bases
• If buffer component concentrations are too low (<0.001 M)
• When the ratio [A-]/[HA] is extremely high or low
• During titrations past the buffer region
• For solutions without a true buffer system
What are the applications of the Henderson-Hasselbalch equation?
Key applications include:
• Calculating blood pH in medical studies
• Designing laboratory and industrial buffers
• Analyzing pH in biochemical reactions
• Supporting acid-base equilibrium calculations in exams and practicals
How is the Henderson-Hasselbalch equation derived?
The equation is derived from the acid dissociation constant (Ka) of a weak acid:
Start with Ka = [H+][A-]/[HA]
Rearrange to solve for [H+]
Take the negative logarithm to introduce pH and pKa
Result: pH = pKa + log([A-] / [HA])
How does the Henderson-Hasselbalch equation apply to basic buffers?
For basic buffers, use a similar equation with base components:
• pOH = pKb + log([B+] / [BOH])
• Calculate pOH, then find pH using pH = 14 - pOH
• Applicable for weak base and its salt with a strong acid
Why might your calculated pH differ from experimental results using the Henderson-Hasselbalch equation?
Calculated pH may differ due to:
• Ionic strength changes
• Temperature variations
• Non-ideal behavior of solutions
• Impurities or measurement errors
• Dilution effects at low concentrations
What is the significance of pKa in the Henderson-Hasselbalch equation?
pKa indicates the strength of a weak acid.
In the equation:
• Lower pKa = stronger acid
• pKa helps determine the ideal buffer region (greatest capacity when pH ≈ pKa)
• Used to compare different acids for buffer selection
Give an example of solving a buffer pH using the Henderson-Hasselbalch equation.
Example: Calculate pH for 0.2 M acetic acid (pKa = 4.76) and 0.1 M sodium acetate.
Solution:
pH = 4.76 + log(0.1/0.2)
= 4.76 + log(0.5)
= 4.76 + (-0.301)
= 4.46
How does dilution affect buffer pH using this equation?
Dilution generally does not change the pH significantly, as the ratio [A-]/[HA] remains constant. However:
• Buffer capacity decreases
• Extreme dilution may invalidate buffer action and the equation's accuracy
What is the buffer region in context of the Henderson-Hasselbalch equation?
The buffer region is the pH range where a buffer can effectively resist changes in acidity or basicity.
• Typically, within ±1 pH unit of the pKa value
• Buffer works best when [A-] ≈ [HA]
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17645 | https://www.reddit.com/r/learnmath/comments/6dv16u/gr_11_find_the_exact_value_of_cos120_without_a/ | [GR 11] Find the exact value of cos120 without a calculator : r/learnmath
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badboyzpwns
[GR 11] Find the exact value of cos120 without a calculator
cos120 = cos(180 - 120)
cos120 = cos60 or -(1/2)
I don't get the first step at all, could someone clarify it's significance? I understand the second step though.
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17646 | http://www.mathisfun.com/equal-less-greater.html | Equal, Less and Greater Than Symbols
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Equal, Greater or Less Than
As well as the familiar equals sign (=) it is also very useful to show if something is not equal to (≠) greater than (>) or less than (<)
)
These are the important signs to know:
=When two values are equal
we use the "equals" sign example: 2+2=4
≠When two values are definitely not equal
we use the "not equal to" sign example: 2+2≠9
<When one value is smaller than another
we use a "less than" sign example: 3<5
When one value is bigger than another
we use a "greater than" sign example: 9>6
Less Than and Greater Than
The "less than" sign and the "greater than" sign look like a "V" on its side, don't they?
To remember which way around the "<" and ">" signs go, remember that the wide open side faces the larger number:
BIG > small
small < BIG
The "small" end always points to the smaller number, like this:
Greater Than Symbol: BIG > small
Example:
10 > 5
"10 is greater than 5"
Or the other way around:
5 < 10
"5 is less than 10"
Do you see how the symbol "points at" the smaller value?
... Or Equal To ...
Sometimes we know a value is smaller, but may also be equal to!
Example, a jug can hold up to 4 cups of water.
So how much water is in it?
It could be 4 cups or it could be less than 4 cups: So until we measure it, all we can say is "less than or equal to" 4 cups.
To show this, we add an extra line at the bottom of the "less than" or "greater than" symbol like this:
The "less than or equal to" sign:≤
The "greater than or equal to" sign:≥
All The Symbols
Below is a summary of all the symbols:
| Symbol | Words | Example Use |
---
| = | equals | 1 + 1 = 2 |
| ≠ | not equal to | 1 + 1 ≠ 3 |
| | | |
| > | greater than | 5 > 2 |
| < | less than | 7 < 9 |
| | | |
| ≥ | greater than or equal to | marbles ≥ 1 |
| ≤ | less than or equal to | dogs ≤ 3 |
Why Use Them?
Because there are things we do not know exactly ...
... but can still say something about.
So we have ways of saying what we do know (which may be useful!)
Example: John had 10 marbles, but lost some. How many has he now?
Answer: He must have less than 10:
Marbles < 10
If John still has some marbles we can also say he has greater than zero marbles:
Marbles > 0
But if we thought John could have lost all his marbles we would say
Marbles ≥ 0
In other words, the number of marbles is greater than or equal to zero.
Combining
We can sometimes say two (or more) things on the one line:
Example: Becky starts with $10, buys something and says "I got change, too". How much did she spend?
Answer: Something greater than $0 and less than $10 (but NOT $0 or $10):
"What Becky Spends" > $0
"What Becky Spends" < $10
This can be written down in just one line:
$0 < "What Becky Spends" < $10
That says that $0 is less than "What Becky Spends" (in other words "What Becky Spends" is greater than $0) and what Becky Spends is also less than $10.
Notice that ">" was flipped over to "<" when we put it before what Becky spends. Always make sure the small end points to the small value.
Changing Sides
We saw in that previous example that when we change sides we flipped the symbol as well.
This:Becky Spends > $0(Becky spends greater than $0)
is the same as this:$0 < Becky Spends($0 is less than what Becky spends)
Just make sure the small end points to the small value!
Here is another example using "≥" and "≤":
Example: Becky has $10 and she is going shopping. How much will she spend (without using credit)?
Answer: Something greater than, or possibly equal to, $0 and less than, or possibly equal to, $10:
Becky Spends ≥ $0
Becky Spends ≤ $10
This can be written down in just one line:
$0 ≤ Becky Spends ≤ $10
A Long Example: Cutting Rope
Here is an interesting example I thought of:
Example: Sam cuts a 10m rope into two. How long is the longer piece? How long is the shorter piece?
Answer: Let us call the longer length of rope "L", and the shorter length "S"
L must be greater than 0m (otherwise it isn't a piece of rope), and also less than 10m:
L > 0
L < 10
So:
0 < L < 10
That says that L (the Longer length of rope) is between 0 and 10 (but not 0 or 10)
The same thing can be said about the shorter length "S":
0 < S < 10
But I did say there was a "shorter" and "longer" length, so we also know:
S < L
(Do you see how neat mathematics is? Instead of saying "the shorter length is less than the longer length", we can just write "S < L")
We can combine all of that like this:
0 < S < L < 10
That says a lot:
0 is less that the short length, the short length is less than the long length, the long length is less than 10.
Reading "backwards" we can also see:
10 is greater than the long length, the long length is greater than the short length, the short length is greater than 0.
It also lets us see that "S" is less than 10 (by "jumping over" the "L"), and even that 0<10 (which we know anyway), all in one statement.
NOW, there is one more thing!
If Sam tried really hard he might be able to cut the rope EXACTLY in half, so each half is 5m, but we know he didn't because we said there was a "shorter" and "longer" length, so we also know:
S < 5
and
L > 5
We can put that into our very neat statement here:
0 < S < 5 < L < 10
And IF we thought the two lengths MIGHT be exactly 5 we could change that to
0 < S ≤ 5 ≤ L < 10
An Example Using Algebra
OK, this example may be complicated if you don't know Algebra, but I thought you might like to see it anyway:
Example: What is x+3, when we know that x is greater than 11?
If x > 11 , then x+3 > 14
(Imagine that "x" is the number of people at your party. If there are more than 11 people at your party, and 3 more arrive, then there must be more than 14 people at your party now.)
Mathopolis:Q1)Q2)Q3)Q4)Q5)Q6)Q7)Q8)Q9)Q10)
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17647 | https://dwest.web.illinois.edu/regs/combnull.html | Complexity of the Combinatorial Nullstellensatz (2002)
Originator: Noga Alon
(presented by Dan Schreiber - REGS 2009)
Background:
Let F be a field, and let ƒ be a polynomial in variables
x1,...,xn over F whose degree is
∑i=1n ti.
The Combinatorial Nullstellensatz [Al] states that if the coefficient of
∏i=1n xiti
in ƒ is nonzero, and S1,...,Sn are subsets of
F with |Si| > ti for all i, then
there exists s∈∏Si such that ƒ(s)≠0.
The proof of the Combinatorial Nullstellensatz is non-constructive. The
computational problem NULLSTELLENSATZ is the following: given a polynomial
ƒ and sets S1,...,Sn satisfying the
conditions of the theorem, find a vector s∈∏Si
such that ƒ(s)≠0.
Definitions:
In order to study the complexity of NULLSTELLENSATZ, we provide informal
definitions of various complexity classes of problems.
P is the class of decision problems that can be solved by an
algorithm whose running time is bounded by a polynomial function of the size of
the input. A decision problem can be viewed as a 0,1-function ("Boolean"
function) on binary strings.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
Background:
Let F be a field, and let ƒ be a polynomial in variables
x1,...,xn over F whose degree is
∑i=1n ti.
The Combinatorial Nullstellensatz [Al] states that if the coefficient of
∏i=1n xiti
in ƒ is nonzero, and S1,...,Sn are subsets of
F with |Si| > ti for all i, then
there exists s∈∏Si such that ƒ(s)≠0.
The proof of the Combinatorial Nullstellensatz is non-constructive. The
computational problem NULLSTELLENSATZ is the following: given a polynomial
ƒ and sets S1,...,Sn satisfying the
conditions of the theorem, find a vector s∈∏Si
such that ƒ(s)≠0.
Definitions:
In order to study the complexity of NULLSTELLENSATZ, we provide informal
definitions of various complexity classes of problems.
P is the class of decision problems that can be solved by an
algorithm whose running time is bounded by a polynomial function of the size of
the input. A decision problem can be viewed as a 0,1-function ("Boolean"
function) on binary strings.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
The proof of the Combinatorial Nullstellensatz is non-constructive. The
computational problem NULLSTELLENSATZ is the following: given a polynomial
ƒ and sets S1,...,Sn satisfying the
conditions of the theorem, find a vector s∈∏Si
such that ƒ(s)≠0.
Definitions:
In order to study the complexity of NULLSTELLENSATZ, we provide informal
definitions of various complexity classes of problems.
P is the class of decision problems that can be solved by an
algorithm whose running time is bounded by a polynomial function of the size of
the input. A decision problem can be viewed as a 0,1-function ("Boolean"
function) on binary strings.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
Definitions:
In order to study the complexity of NULLSTELLENSATZ, we provide informal
definitions of various complexity classes of problems.
P is the class of decision problems that can be solved by an
algorithm whose running time is bounded by a polynomial function of the size of
the input. A decision problem can be viewed as a 0,1-function ("Boolean"
function) on binary strings.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
P is the class of decision problems that can be solved by an
algorithm whose running time is bounded by a polynomial function of the size of
the input. A decision problem can be viewed as a 0,1-function ("Boolean"
function) on binary strings.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
NP is the class of decision problems for which "yes" answers can
be confirmed in time that is polynomial in the size of the input. In terms of
the corresponding Boolean function ƒ, a "verifier" function
V in P and a polynomial g are required such that whenever
x is a binary string, ƒ(x)=1 if and only if there exists
y∈{0,1}g(|x|) such that V(x,y)=1.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
FP is the class of binary relations R such that there exists a
polynomial time algorithm that, given x, outputs y for which
R(x,y) holds. In modeling decision problems, x and y can
be viewed as the input and the "certificate", respectively, such that
R(x,y)=1 if and only if y is a string that the verifier can use
to show that the answer to the decision problem is "yes".
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
FNP is the class of binary relations R such that there exists
a polynomial time algorithm that, given both x and y, determines
whether R(x,y) holds. When restricted to certificates of decision
problems, FNP is roughly the same as NP.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
TFNP is the class of binary relations R such that (1) there
exists a polynomial time algorithm that, given both x and y,
determines whether R(x,y) holds, and (2)
for every x, there exists y such that R(x,y) holds.
Here "T" stands for "Total" and for decision problems indicates that the
answer is always "yes" and is verified by an appropriate string.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
NULLSTELLENSATZ belongs to TFNP, because the theorem guarantees a "yes"
answer (that is, existence of s such that ƒ(s)≠0),
and the correctness of s can be verified in polynomial time.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
Question 1:[Al] Is there a polynomial-time algorithm for solving
NULLSTELLENSATZ?
Comments:
Alon says 'it seems likely that such algorithms do exist.'
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
In studying a class of problems, complexity theorists often seek problems in
the class that, if solvable in polynomial time, yield polynomial-time
algorithms for all problems in the class. Such problems are called
complete for the class. There are reasons to believe that no problems
are TFNP-complete.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
Papadimitriou [P] developed a framework by which we might analyze the
complexity of subclasses of problems of TFNP based on the type of argument that
is used to prove totality. One such lemma is the parity argument, which
states that every graph has an even number of odd degree nodes. We denote by
PPA the subclass of TFNP whose totality is proved via the parity
argument on undirected graphs.
A special case of Chévalley's Theorem considers polynomials
p1,...,pn in n variables over the finite
field F2. It states that if
n>∑i=1m deg(pi) and
(0,...,0) is a common zero of p1,...,pn,
then there is another common zero (a1,...,an).
Given polynomials p1,...,pn satisfying the
conditions, let CHEVALLEY-MOD-2 be the computational problem of finding a
nonzero solution.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532.
CHEVALLEY-MOD-2 is in PPA, as we can prove totality by analyzing the
degrees of an appropriate bipartite graph and invoking the parity argument. In
general, to show membership in PPA one must prove totality either by
directly using the parity argument or by using another theorem whose
computational version has previously been shown to be in PPA.
Question 2: Is NULLSTELLENSATZ in PPA?
Comment: One can use the Combinatorial Nullstellensatz to prove
Chévalley's Theorem, but this does not show that NULLSTELLENSATZ is in
PPA. One would have to work in the other direction and prove the
Nullstellensatz from Chévalley's Theorem!
References:
[Al] Noga Alon, Combinatorial Nullstellensatz. Combinatorics, Probability and Computing 8 (1999), 7-29.
[P] Christos Papadimitriou, On the complexity of the parity argument and other inefficient proofs of existence. Journal of Computer and System Sciences 3 (1994), 498-532. |
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Searching a sorted array to find the $k$ closest values to a target value $T$
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Let $A$ be a sorted array of $N$ values. I am interested in finding the index $j$ such that the elements $A_j, A_{j + 1}, ..., A_{j + k - 1}$ have the $k$ closest values to the given target value $T$. Assume that $k$ is comparable, in magnitude, to $N$; i.e. we don't have $k \ll N$.
One can naively do this by first (binary) searching for the closest element, and then successively probing the left/right neighbors of that element to arrive at the result. This method has a time complexity of $O(\log N + k)$.
I suspect one can modify the original binary search algorithm to arrive at the result in $O(\log N + \log k)$ time. I tried searching online for such a variant of the binary search algorithm, without success. I'd appreciate any/all pointers to sources I might have missed, and also original attempts at formulating such a variant.
Note: It turns out there is an elegant binary search cousin that computes the result in $O(\log(N - k))$ time. See Ivan Smirnov's answer for a discussion.
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edited Jul 1, 2017 at 20:48
iheapiheap
asked Jun 29, 2017 at 9:16
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$\begingroup$ After finding the value closest to $T$, try performing another binary search (of a somewhat different kind) to find the $k$ closest values. The binary search will take an additional $O(\log k)$ essentially since you are searching among $2k$ values. What you are searching for is an interval of size $k$ around the value you found in the first binary search, and you are looking for the interval whose endpoints are closest to $T$. $\endgroup$
Yuval Filmus
– Yuval Filmus
2017-06-29 10:50:12 +00:00
Commented Jun 29, 2017 at 10:50
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You can do it in two binary searches. For simplicity I assume that all numbers are distinct and, even more, absolute differences between $T$ and all other elements are distinct too. The solution is easily adapted if it is not the case.
What means that a selected subrange is optimal? That means that we can't move it neither right ($|A_{j+k} - T| > |A_{j} - T|$) nor left ($|A_{j+k-1} - T| < |A_{j-1}|$). That looks like a binary search as well!
Look at the function $h(j) = \mathbb{1}{|A_{j+k} - T| > |A_{j} - T|}$ (that is, $h(j)$ is 1 iff the answer does not become better if we move the subrange to the right). Now you have to find minimum $j$ s.t. $h(j) = 1$. As the function is monotonous, it can be done by a binary search by $j$ in $O(\log n)$ time. (Actually, it is $O(\log (n-k))$ because $j < n - k$).
If there are identical values in the array, we need some case handling. Whenever $|A_{j+k}-T| = |A_j-T|$, there are three cases possible: Either $A_{j} = A_{j+k}$ and the segment is to the left of $T$; in this case we should move right. Or, $A_{j} = A_{j+k}$ and the segment is to the right of $T$, and we should move left. Or the segment contains $T$, in this case we've found the optimal answer.
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edited Jul 2, 2017 at 0:26
iheap
17555 bronze badges
answered Jun 29, 2017 at 10:54
Ivan SmirnovIvan Smirnov
96466 silver badges1313 bronze badges
$\endgroup$
10
$\begingroup$ Thanks for the answer! This is in line with what I was thinking. I've actually come up with another algorithm that ~~is asymptotically slightly better; i.e. it~~ runs in $O(\log(N - k))$ time. The difference only matters in cases where $k$ is close to $N$. I will add another answer that explains that approach. $\endgroup$
iheap
– iheap
2017-06-30 18:56:08 +00:00
Commented Jun 30, 2017 at 18:56
$\begingroup$ @iheap You are welcome! By the way, my algorithm in fact also works in $O(\log (N-k))$ time because $j $\endgroup$
Ivan Smirnov
– Ivan Smirnov
2017-06-30 18:57:35 +00:00
Commented Jun 30, 2017 at 18:57
$\begingroup$ Great. In my approach, I was searching for the optimal $j$ that minimizes $\max(|A_{j + k} - T|, |A_{j} - T|)$. This can be done in logarithmic time if all elements are distinct. If not, the algorithm might spend linear time, though. Does your method always take logarithmic time without the all-distinct assumption? $\endgroup$
iheap
– iheap
2017-06-30 19:05:50 +00:00
Commented Jun 30, 2017 at 19:05
$\begingroup$ @iheap It should, if you replace $<$ with $\leq$ in one inequality and carefully think about invariants. To make your life easier try adding $-\infty$ to the beginning of the array and $+\infty$ to the end so you don't think about corner cases. $\endgroup$
Ivan Smirnov
– Ivan Smirnov
2017-06-30 19:09:14 +00:00
Commented Jun 30, 2017 at 19:09
1
$\begingroup$ @iheap Let's try the following fix. We know that $A_j \leq T \leq A_{j+k}$. So whenever $|A_{j+k}-T| = |A_j-T|$, there are three cases: either $A_j = A_{j+k}$ and the segment is to the left of $T$; in this case we should move right. Or $A_j = A_{j+k}$ and the segment is to the right of $T$, and we should move left. Or the segment contains $T$, in this case we're done. I think that it fixes the binary search. $\endgroup$
Ivan Smirnov
– Ivan Smirnov
2017-07-01 10:52:50 +00:00
Commented Jul 1, 2017 at 10:52
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A Venn diagram is a graphical representation that shows similarities and differences between items, concepts or data. They utilize shapes, often circles, to illustrate relationships and compare data.
Venn diagrams are certainly one of the most well-known types of diagrams. This design, featuring a series of simple circles or ellipses or rectangles that overlap is ubiquitous. The overlapping region of a Venn diagram indicates similarities while others showcase differences. Although the Venn diagram was created to describe mathematically logical relationships, it’s now used to show all types of relationships.
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17650 | https://emcrit.org/ibcc/svcs/ | Internet Book of Critical Care (IBCC)
Online Medical Education on Emergency Department (ED) Critical Care, Trauma, and Resuscitation
You are here: Home / IBCC / Superior vena cava syndrome (SVCS)
Superior vena cava syndrome (SVCS)
by Josh Farkas
CONTENTS
Causes
Signs & symptoms
Diagnosis: CT scan findings
Grading system to define severity
Management:
Range of treatment options
Management of critical SVC syndrome
Pathophysiology & anatomy
Questions & discussion
abbreviations used in the pulmonary section: 5
ABPA: Allergic bronchopulmonary aspergillosis 📖
AE-ILD: Acute exacerbation of ILD 📖
AEP: Acute eosinophilic pneumonia 📖
AFB: Acid Fast Bacilli
AIP: Acute interstitial pneumonia (Hamman-Rich syndrome) 📖
ANA: Antinuclear antibody 📖
ANCA: Antineutrophil cytoplasmic antibodies 📖
ARDS: Acute respiratory distress syndrome 📖
ASS: Antisynthetase Syndrome 📖
BAL: Bronchoalveolar lavage 📖
BiPAP: Bilevel positive airway pressure 📖
CEP: Chronic eosinophilic pneumonia 📖
CF: Cystic fibrosis 📖
COP: Cryptogenic organizing pneumonia 📖
CPAP: Continuous positive airway pressure 📖
CPFE: Combined pulmonary fibrosis and emphysema 📖
CTD-ILD: Connective tissue disease associated interstitial lung disease 📖
CTEPH: Chronic thromboembolic pulmonary hypertension 📖
DAD: Diffuse alveolar damage 📖
DAH: Diffuse alveolar hemorrhage 📖
DIP: Desquamative interstitial pneumonia 📖
DLCO: Diffusing capacity for carbon monoxide 📖
DRESS: Drug reaction with eosinophilia and systemic symptoms 📖
EGPA: Eosinophilic granulomatosis with polyangiitis 📖
FEV1: Forced expiratory volume in 1 second 📖
FVC: Forced vital capacity 📖
GGO: Ground glass opacity 📖
GLILD: Granulomatous and lymphocytic interstitial lung disease 📖
HFNC: High flow nasal cannula 📖
HP: Hypersensitivity pneumonitis 📖
IPAF: Interstitial pneumonia with autoimmune features 📖
IPF: Idiopathic pulmonary fibrosis 📖
IVIG: Intravenous immunoglobulin 📖
LAM: Lymphangioleiomyomatosis 📖
LIP: Lymphocytic interstitial pneumonia 📖
MAC: Mycobacterium avium complex 📖
MCTD: Mixed connective tissue disease 📖
NIV: Noninvasive ventilation (including CPAP or BiPAP) 📖
NSIP: Nonspecific interstitial pneumonia 📖
NTM: Non-tuberculous mycobacteria 📖
OHS: Obesity hypoventilation syndrome 📖
OP: Organizing pneumonia 📖
OSA: Obstructive sleep apnea 📖
PAP: Pulmonary alveolar proteinosis 📖
PE: Pulmonary embolism 📖
PFT: Pulmonary function test 📖
PLCH: Pulmonary Langerhans cell histiocytosis 📖
PPFE: Pleuroparenchymal fibroelastosis 📖
PPF: Progressive pulmonary fibrosis 📖
PVOD/PCH Pulmonary veno-occlusive disease/pulmonary capillary hemangiomatosis 📖
RB-ILD: Respiratory bronchiolitis-associated interstitial lung disease 📖
RP-ILD: Rapidly progressive interstitial lung disease 📖
TNF: tumor necrosis factor
UIP: Usual Interstitial Pneumonia 📖
--- causes of SVC syndrome (back to contents) ---
malignancy (~70%)
Non-small cell lung cancer (~50% of malignant etiologies).
Small cell lung cancer (~25% of malignant etiologies).
Lymphoma (~10% of malignant etiologies).
Others:
Thymoma or thyroid malignancy.
Primary germ cell neoplasms.
Mesothelioma.
Esophageal cancer.
Solid tumors with mediastinal lymph node metastasis (e.g., breast cancer).
device-related thrombosis (~25%)
Pacemaker/defibrillator.
Hemodialysis catheter.
Central venous catheter.
Chronic central venous port.
lymphadenopathy
Infection (e.g., histoplasmosis, tuberculosis).
Sarcoidosis.
mediastinal fibrosis
Radiation fibrosis.
Idiopathic fibrosis.
other
Retrosternal thyroid.
Aortic aneurysm.
Mediastinal hematoma.
Spontaneous thrombosis due to thrombophilia or Behcet disease. (33103626)
--- signs & symptoms (back to contents) ---
Severity relates to how rapidly occlusion develops, because a gradually developing occlusion may lead to collateral venous drainage.
edema & plethora (redness)
Facial and neck edema (including periorbital/conjunctival edema).
Arm swelling.
Plethora (redness) of the face and neck.
venous distension
Distended neck veins.
Distended chest veins (these develop gradually, so prominently dilated subcutaneous veins on the chest may imply a more subacute disease course).
pulmonary involvement (due largely to airway edema)
Stridor, hoarseness.
Dyspnea, sometimes with orthopnea.
Tongue swelling.
Cough.
cerebral perfusion affected (including cerebral edema)
Headache (may be worse lying flat or bending over). (30037444)
Blurry vision.
Dizziness, especially when leaning forward. (33357528)
Altered mental status (including confusion, obtundation).
impaired cardiac output
This is uncommon, but may be more likely in some situations:
Acute SVC thrombosis.
Pre-existing heart failure.
Coexisting pulmonary emboli due to thrombus formation in the SVC.
Mass compresses both the SVC and also the heart and/or pulmonary arteries. (33103626)
--- CT scan findings in SVC syndrome (back to contents) ---
If SVC syndrome is suspected, this should be communicated with the radiology department, since multiphase imaging could improve diagnostic yield. (33103626)
direct evidence of obstruction
Extrinsic compression vs. thrombosis.
indirect evidence
Collateral vessel dilation implies a more clinically relevant occlusion.
Collateral venous drainage into the portal venous system may rarely cause contrast enhancement to occur near the falciform ligament (“CT quadrate lobe hot spot sign”). (36062219)
--- Yu grading system for the severity of SVC syndrome (back to contents) ---
Grade 0 (asymptomatic, ~10%)
Radiographic superior vena cava obstruction in the absence of symptoms.
Grade 1 (mild, ~25%): symptoms may include
Edema in head or neck (vascular distention).
Cyanosis and/or plethora.
Grade 2 (moderate, ~50%)
Edema in head or neck with functional impairment, such as:
Mild dysphagia.
Cough.
Mild-moderate impairment of the head, jaw, or eyelid movements.
Visual disturbances caused by ocular edema.
Grade 3 (severe, ~10%): one or more of the following
Mild/moderate cerebral edema (e.g., headache, dizziness).
Mild/moderate laryngeal edema.
Diminished cardiac reserve (e.g., syncope after bending or coughing).
Grade 4 (life-threatening, ~5%): one or more of the following
Significant cerebral edema (e.g., confusion, obtundation).
Significant laryngeal edema (e.g., stridor).
Significant hemodynamic compromise:
Syncope without precipitating factors.
Hypotension.
Renal insufficiency. (18670297; 📄)
--- management: range of different treatment options for SVC syndrome (back to contents) ---
interventional radiology 🏆
stenting
Basics:
Time to improvement: ~0-72 hours.
Stenting is increasingly considered the standard of care for both benign and malignant etiologies. (33357528)
Stenting is used as a front-line therapy for patients with severe SVC syndrome (e.g., cerebral or laryngeal edema, or postural syncope). (33357528)
Advantages:
High success rate.
Fastest resolution of symptoms.
Can be combined with other modalities (e.g., radiation, chemotherapy).
Disadvantages:
Major complications are rare, but can occur: (33103626)
Stent migration, re-occlusion, or malposition.
Pericardial tamponade if SVC perforation occurs below the level of the azygos vein).
Hemothorax.
Pulmonary edema (due to increased venous return).
Lower long-term durability compared to surgery, so may be less ideal for patients with long life expectancy.
May not be an option in patients with pacemakers/defibrillators (stenting can entrap the pacer lead).
thrombectomy
Thrombus may be removed using aspiration thrombectomy and/or catheter-directed thrombolysis.
angioplasty
Angioplasty without stent insertion may be utilized for patients with pacemaker/defibrillator-related SVC stenosis.
tissue biopsy
Tissue biopsy may be performed to determine the diagnosis in patients with lymphadenopathy or mass compressing the superior vena cava.
surgery
Basics:
Time to improvement: ~0-72 hours.
Often involves the use of a saphenous vein graft to bypass the SVC. (33357528)
Advantages:
Could be considered in SVC syndrome from benign etiology (e.g., extensive venous thrombosis).
Could be considered in SVC syndrome that isn't amenable to stenting.
Might be more durable in some situations. However, saphenous vein bypass grafting remains vulnerable to re-occlusion.
Disadvantages:
Invasive surgery.
May involve prolonged intubation.
Higher rate of complications as compared to stenting (e.g., pulmonary embolism, deep venous thrombosis, mediastinal hematoma).
chemotherapy
Chemotherapy is an option for chemosensitive tumors (e.g., small cell lung carcinoma, non-Hodgkin lymphoma, germ cell tumor).
Time to improvement: ~1-2 weeks (depending on malignancy).
Advantages:
Provides treatment of systemic malignancy.
Disadvantages:
Delayed relief of symptoms.
radiation
Radiation is an option for radiosensitive tumors (including non-small-cell lung carcinoma and low grade lymphomas that may not be highly chemosensitive).
Time to improvement: ~3-30 days.
Advantages:
In some cases, may be used to provide definitive treatment of the malignancy (e.g., definitive chemoradiation for non-small-cell lung carcinoma).
Disadvantages:
Delayed relief of symptoms.
Ineffective in up to 20% of patients. (33357528)
Radiotherapy may exacerbate SVC obstruction by causing edema. (36062219)
Complications may include SVC perforation, inhibition of collateral venous development.
Radiation prior to tissue diagnosis of malignancy may subsequently obscure the diagnosis. (33357528)
--- management of critical SVC syndrome (back to contents) ---
general supportive care
Elevation of the head of the bed.
Intubation may be needed for patients with severe airway obstruction and/or obtundation.
diuresis
Gentle diuresis may be considered to reduce venous pressure, but this isn't supported by high-quality data.
anticoagulation
If there is a thrombus within the SVC, systemic anticoagulation with a heparin infusion is generally utilized.
Factors that may suggest the presence of thrombosis: (33103626)
SVC syndrome associated with indwelling devices (e.g., central lines, dialysis catheters, cardiac devices).
More acute onset of symptoms.
Radiological evidence of thrombosis (e.g., pulmonary embolism).
steroid
Steroid is a niche therapy that may be considered for carefully selected patients.
Potential indications for steroid:
(1) Steroid may shrink any reactive lymphadenopathy or lymphoma (with the drawback that this may compromise subsequent tissue diagnosis of lymphoma).
(2) Prophylaxis against radiation-induced edema.
definitive therapy
Interventional radiology is generally the preferred therapy for critical SVC syndrome, since it may allow for immediate resolution (unlike chemotherapy or radiotherapy, which can take weeks to have an effect).
--- pathophysiology & anatomy (back to contents) ---
mechanisms of SVC syndrome may include:
One or more of the following may be involved:
External compression or direct invasion of the SVC.
Intrinsic stenosis of the SVC (e.g., following indwelling catheter or pacemaker/defibrillator placement).
Thrombosis of SVC (acute thrombosis may occur superimposed on compression or stenosis, leading to acute deterioration).
basic anatomy of the SVC
The SVC has three major tributaries: the right and left brachiocephalic veins and the azygos vein.
anatomic types of SVC obstruction
Brachiocephalic vein obstruction (pink panel above):
Right brachiocephalic vein may empty via the azygos network.
Upper SVC, above the azygos vein (yellow panel above):
Azygos system functions as a collateral drainage network, draining blood into the (remaining) SVC.
At the level of the azygos vein (blue panel above):
This may be the most problematic location for an occlusion.
Both the SVC and the Azygos system are blocked.
Lower SVC, below the azygos vein (purple panel above):
Blood flows backwards through the Azygos vein to drain into the IVC.
Less severe symptoms occur, because blood can be rerouted around the obstruction via a relatively robust venous network.
--- questions & discussion (back to contents) ---
To keep this page small and fast, questions & discussion about this post can be found on another page here.
Guide to emoji hyperlinks
= Link to online calculator.
= Link to Medscape monograph about a drug.
= Link to IBCC section about a drug.
= Link to IBCC section covering that topic.
= Link to FOAMed site with related information.
📄 = Link to open-access journal article.
= Link to supplemental media.
References
25229918 Kumar B, Hosn NA. Images in clinical medicine. Superior vena cava syndrome. N Engl J Med. 2014 Sep 18;371(12):1142. doi: 10.1056/NEJMicm1311911 [PubMed]
30037444 Zimmerman S, Davis M. Rapid Fire: Superior Vena Cava Syndrome. Emerg Med Clin North Am. 2018 Aug;36(3):577-584. doi: 10.1016/j.emc.2018.04.011 [PubMed]
33103626 Klein-Weigel PF, Elitok S, Ruttloff A, Reinhold S, Nielitz J, Steindl J, Hillner B, Rehmenklau-Bremer L, Wrase C, Fuchs H, Herold T, Beyer L. Superior vena cava syndrome. Vasa. 2020 Oct;49(6):437-448. doi: 10.1024/0301-1526/a000908 [PubMed]
33357528 Azizi AH, Shafi I, Shah N, Rosenfield K, Schainfeld R, Sista A, Bashir R. Superior Vena Cava Syndrome. JACC Cardiovasc Interv. 2020 Dec 28;13(24):2896-2910. doi: 10.1016/j.jcin.2020.08.038 [PubMed]
36062219 Quencer KB. Superior Vena Cava Syndrome: Etiologies, Manifestations, and Treatments. Semin Intervent Radiol. 2022 Aug 31;39(3):292-303. doi: 10.1055/s-0042-1753480 [PubMed]
Books:
Shah, P. L., Herth, F. J., Lee, G., & Criner, G. J. (2018). Essentials of Clinical pulmonology. In CRC Press eBooks.
Shepard, JO. (2019). Thoracic Imaging The Requisites (Requisites in Radiology) (3rd ed.). Elsevier.
Walker C & Chung JH (2019). Muller’s Imaging of the Chest: Expert Radiology Series. Elsevier.
Palange, P., & Rohde, G. (2019). ERS Handbook of Respiratory Medicine. European Respiratory Society.
Rosado-De-Christenson, M. L., Facr, M. L. R. M., & Martínez-Jiménez, S. (2021). Diagnostic imaging: chest. Elsevier.
Murray & Nadel: Broaddus, V. C., Ernst, J. D., MD, King, T. E., Jr, Lazarus, S. C., Sarmiento, K. F., Schnapp, L. M., Stapleton, R. D., & Gotway, M. B. (2021). Murray & Nadel’s Textbook of Respiratory Medicine, 2-Volume set. Elsevier.
Fishman's: Grippi, M., Antin-Ozerkis, D. E., Cruz, C. D. S., Kotloff, R., Kotton, C. N., & Pack, A. (2023). Fishman’s Pulmonary Diseases and Disorders, Sixth Edition (6th ed.). McGraw Hill / Medical.
The Internet Book of Critical Care is an online textbook written by Josh Farkas (@PulmCrit), an associate professor of Pulmonary and Critical Care Medicine at the University of Vermont.
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Magnesium - 12 MgYour user agent does not support the HTML5 Audio element. 🔊
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🇬🇧Magnesium
🇺🇦Магній
🇨🇳鎂
🇳🇱Magnesium
🇫🇷Magnésium
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Magnesium - 12 Mg: the essentials
▸▸ Mg Essentials
Physical properties
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Name: magnesium
Symbol: Mg
Atomic number: 12
Relative atomic mass (A r): 24.305 ±0.002. Range: [24.304, 24.307]
Standard state: solid at 298 K
Appearance: silvery white
Classification: Metallic
Group in periodic table: 2
Group name: Alkaline earth metal
Period in periodic table: 3
Block in periodic table: s
Shell structure: 2.8.2
CAS Registry: 7439-95-4
Magnesium atoms have 12 electrons and the shell structure is 2.8.2. The ground state electronic configuration of neutral magnesium is[Ne].3s2 and the term symbol of magnesium is1S0.
Magnesium: description Your user agent does not support the HTML5 Audio element. 🔊
Magnesium is a grayish-white, fairly tough metal. Magnesium is the eighth most abundant element in the earth's crust although not found in it's elemental form. It is a Group 2 element (Group IIA in older labelling schemes). Group 2 elements are called alkaline earth metals. Magnesium metal burns with a very bright light.
Magnesium is an important element for plant and animal life. Chlorophylls are porphyrins based upon magnesium. The adult human daily requirement of magnesium is about 0.3 g day-1.
Magnesium tarnishes slightly in air, and finely divided magnesium readily ignites upon heating in air and burns with a dazzling white flame. Normally magnesium is coated with a layer of oxide, MgO, that protects magnesium from air and water.
Magnesium: physical properties
Density of solid: 1738 kg m-3
Molar volume: 14.00 cm 3
Thermal conductivity: 160 W m‑1 K‑1
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Magnesium: heat properties
Melting point: 923 [650°C (1202°F)]K
Boiling point: 1363 [1090°C (1994°F)]K
Enthalpy of fusion: 20.5 kJ mol-1
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Magnesium: atom sizes
Atomic radius (empirical): 150 pm
Molecular single bond covalent radius: 139 (coordination number 2)ppm
van der Waals radius: 251 ppm
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Magnesium: electronegativities
Pauling electronegativity: 1.31 (Pauling units)
Allred Rochow electronegativity: 1.23 (Pauling units)
Mulliken-Jaffe electronegativity: 1.37 (sp orbital)
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Magnesium: orbital properties
First ionisation energy: 737.75 kJ mol‑1
Second ionisation energy: 1450.68 kJ mol‑1
Third ionisation energy: 7732.68 kJ mol‑1
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Magnesium: abundances
Universe: 600000 ppb by weight
Crustal rocks: 29000000 ppb by weight
Human: 270000 ppb by weight
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Magnesium: crystal structure
The solid state structure of magnesium is: hcp (hexagonal close-packed).
More crystallographic data...
Magnesium: biological data
Human abundance by weight: 270000 ppb by weight
Magnesium is an important element for plants and animals. Chlorophylls (responsible for the green colour of plants) are compounds knonw as porphyrins and are based upon magnesium. Magnesium is required for the proper working of some enzymes. The adult daily requirement of magnesium is about 0.3 g day-1.
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Magnesium: uses
Uses...
Magnesium: reactions
Reactions of magnesium as the element with air, water, halogens, acids, and bases where known.
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Magnesium: binary compounds
Binary compounds with halogens (known as halides), oxygen (known as oxides), hydrogen (known as hydrides), and other compounds of magnesium where known.
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Magnesium: compound properties
Bond strengths; lattice energies of magnesium halides, hydrides, oxides (where known); and reduction potentials where known.
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Magnesium: history
Magnesium was discovered by Sir Humphrey Davy in 1755 at England. Origin of name: from the Greek word "Magnesia", a district of Thessaly.
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Magnesium: isotopes
Isotope abundances of magnesium with the most intense signal set to 100%.
The Magnesium isotopes Mg-25 and Mg-26 are used to study the absorption and metabolism of Mg in the human body and they are also used for heart disease studies. Mg-25 is also used for the production of the radioisotope Na-22.
More isotope and NMR data...
Magnesium: isolation
Isolation: magnesium can be made commercially by several processes and would not normally be made in the laboratory because of its ready availability. There are massive amounts of magnesium in seawater. This can be recovered as magnesium chloride, MgCl 2 through reaction with calcium oxide, CaO.
CaO + H 2 O → Ca 2+ + 2OH-
Mg 2+ + 2OH- → Mg(OH)2
Mg(OH)2 + 2HCl → MgCl 2 + 2H 2 O
Electrolysis of hot molten MgCl 2 affords magnesium as a liquid whih is poured off and chlorine gas.
cathode: Mg 2+(l) + 2e- → Mg
anode: Cl-(l) → 1/2 Cl 2 (g) + e-
The other methos used to produce magnesium is non electrolytic and involves dolomite, [MgCa(CO 3)2], an important magnesium mineral. This is "calcined" by heating to form calcined dolomite, MgO.CaO, and this reacted with ferrosilicon alloy.
2[MgO.CaO] + FeSi → 2Mg + Ca 2 SiO 4 + Fe
The magnesium may be distilled out from this mixture of products.
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17652 | https://www.sciencedirect.com/science/article/pii/S0166218X03002737 | Sets in ℤn with distinct sums of pairs - ScienceDirect
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1. Introduction
2. Theoretical results
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Volume 138, Issues 1–2, 29 March 2004, Pages 99-106
Sets in ℤ n with distinct sums of pairs☆
Author links open overlay panel Harri Haanpää a, Antti Huima a, Patric Östergård b
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Abstract
A subset S={s 1,…,s k} of an abelian group G is called an S t-set of size k if all sums of t different elements in S are distinct. A function with applications in coding theory, v γ(k) denotes the order of the smallest cyclic group in which an S 2-set of size k exists. A lower bound for v γ(k) is given in this study, and exact values of v γ(k) are obtained for k⩽15. For the related problem in which all sums of any two, not necessarily distinct, elements in S are required to be different, values of the corresponding function v δ(k) for each k⩽14 are given.
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Keywords
Additive base
Backtrack search
Difference set
Packing
1. Introduction
This work considers packing problems in cyclic groups. A subset S of an abelian group where |S|=k is an S t-set of size k if all sums of t different elements in S are distinct in the group. For the group operations, we use additive notation throughout the paper. See , for open problems in additive number theory related to S t-sets and similar configurations.
Two central functions in the study of S t-sets are v(k) and v γ(k), which give the order of the smallest abelian group and cyclic group, respectively, in which an S 2-set of size k exists. Since cyclic groups are abelian, clearly v(k)⩽v γ(k).
One motivation for studying v(k), v γ(k), and S t-sets is that they have applications in coding theory , , . A constant weight error-correcting code is a set of binary vectors of length k and weight w such that the Hamming distance between any two vectors is at least d. Given k, w, and d, the maximum size of such a code is denoted by A(k,d,w). In [5, Theorem 16] it is shown that A(k,6,w)≥(k w)/v(k).
In this paper we consider v γ(k), which is called v ℤ(k) in . In , values of v γ(k) and corresponding S 2-sets are presented for k⩽10. The (computer-aided) methods for obtaining these results are, however, not discussed in . We present an exhaustive search method for determining v γ(k) and finding corresponding S 2-sets. We determine v γ(k) for k⩽15.
With a slight modification, our method allows us to also consider v δ(k), the order of the smallest cyclic group in which there exists a k-subset such that the sums of any two elements of the subset, not necessarily different, are distinct. The values of v δ(k) and corresponding sets are given for k⩽10 and k=12 in . In , constructions are developed to determine lower bounds for the maximum cardinality of such a subset in a given cyclic group ℤ n; also, the maximum cardinality is determined by a computer search for n⩽134, whereby v δ(k) is determined for k⩽12. Using our method we determine v δ(k) and the corresponding subsets for k⩽14.
Some theoretical results on v γ(k) and v δ(k) are presented in Section 2, and a computational method for finding their values is discussed in Section 3. The search results are documented in Section 4. We list the values of v γ(k) for k⩽15 and v δ(k) for k⩽14 along with the corresponding k-element sets.
2. Theoretical results
In this section, we review some lower bounds for v δ(k) and present a lower bound for v γ(k). These bounds can be used to limit the scope of the computer search.
An (n,k,λ)-difference set is a k-subset of a group of order n such that the list of non-zero differences contains each non-zero group element exactly λ times. When the group is cyclic, we have a cyclic difference set . The problems we are considering are closely connected to the theory of difference sets with λ=1, since x 1+x 2=x 3+x 4⇔x 1−x 4=x 3−x 2. When, in calculating v δ(k), we allow taking sums of equal elements, the relation is straightforward and the theory of cyclic difference sets is directly applicable.
By a simple volume argument, we know that v δ(k)⩾k(k−1)+1: from a k-subset, k(k−1) differences can be formed, none of which is zero and no two of which may be equal. When k−1 is a prime power, a cyclic difference set construction by Singer shows that the bound is sharp . The series of values of k that are not settled by this result is thus 7,11,13,15,16,…. Also, when k is a prime power, another cyclic difference set construction by Bose shows that v δ(k)⩽k 2−1. In it is shown that v δ(7)=48, and our calculations show that the bound by Bose is sharp also for k=11 and 13.
We next present a volume argument for v γ(k). In that case some values of differences may occur more than once, since if x 3−x 2=x 2−x 1, then the right-hand side of x 1+x 3=x 2+x 2 is not the sum of two distinct elements, and thus does not contradict the assumption of distinct sums of pairs. Our idea for adapting the volume argument for v γ(k) involves finding upper bounds for the number of difference values that may occur more than once.
Let S={s 1,…,s k} be a set with distinct sums of pairs in a cyclic group ℤ n. For each d≠0 in ℤ n, we construct the undirected graph G d by taking as the edge set E(G d) the set of those pairs of elements in S whose difference equals ±d, and as the vertex set V(G d) the endpoints of the edges.
Following the usual convention, let P i denote a path of length i, and C i a cycle of length i.
Lemma 1
Every non-empty graph G d is isomorphic to P 1, P 2, or C 3.
Proof
By definition, each vertex of G d has degree at least one. Each vertex of G d has degree at most two: no vertex s can be adjacent to a vertex other than s+d or s−d. Every pair of edges {s i 1,s i 2}, {s i 3,s i 4} in G d, ordered so that s i 1−s i 2=s i 3−s i 4, must have a common endpoint, or s i 1+s i 4=s i 2+s i 3 will contradict the assumption of distinct sums of pairs. Since every vertex has degree at least one, and every pair of edges must have an endpoint in common, the graph must be connected. A connected graph with highest degree at most two can only be a path or a cycle. Paths of more than two edges and cycles with more than three edges contain a pair of edges without a common endpoint. Hence, if G d is non-empty, it must be isomorphic to P 1, P 2, or C 3.□
Let p be the number of graphs G d isomorphic to P 2, and c the number of graphs G d isomorphic to C 3.
Lemma 2
p+3 c⩽2 k.
Proof
The vertices that correspond to a given element of S can have degree two in at most two graphs G d. Otherwise, among the (at least three) graphs there would have to be two graphs, say G d and G d′, where d≠−d′. Then {s−d′,s−d,s+d,s+d′} would be a set of four distinct elements in S, and the assumption of distinct sums of pairs would be contradicted by (s−d′)+(s+d′)=(s−d)+(s+d). From this, and noting that the graphs G d isomorphic to P 2 have one vertex of degree two whereas those isomorphic to C 3 have three, the lemma follows.□
Theorem 3
v γ(k)⩾k(k−3).
Proof
Let us calculate an upper bound for the total number of times the different values may occur as the difference of two elements of S. The difference value d occurs once if d≠−d and G d is isomorphic to P 1; twice if either G d is isomorphic to P 1 and d=−d, which is possible for at most one value in ℤ n, or if G d is isomorphic to P 2; and three times if G d is isomorphic to C 3. The number of difference values that occur at least once is then at most n−1, the number of difference values occurring at least twice is at most 1+p+c, and the number of difference values occurring three times is c. The total, then, is at most n+p+2 c. From a k-element subset, k(k−1) differences can be formed; thus k(k−1)⩽n+p+2 c=n−c+p+3 c⩽n+2 k (by Lemma 2), and the theorem follows.□
This bound has the same asymptotic behavior as a similar bound in [6, Lemma 6], but it is slightly stronger.
3. Computer search
3.1. A backtrack algorithm
The natural way to approach this problem is to consider backtrack algorithms. In a backtrack search, there are two main ways of pruning the search tree. First, a search branch may be pruned as soon as it is clear that a configuration cannot lead to a desired solution, and, second, of search branches leading to equivalent configurations one may prune all but one.
Before the search, we fix n, the order of the cyclic group. For equivalence of sets, we use the following definition, which is used in for difference sets. Two sets S 1,S 2⊆ℤ n are equivalent if S 2=S 1 a+b:={s a+b|s∈S 1}, where b is any group element and a is co-prime with n. For the purposes of canonicity testing, ℤ n is to be interpreted as a ring, i.e., addition and multiplication are carried out modulo n. It is not difficult to see that if all sums of pairs in S 1 are different, then this also holds for all sums of pairs in S 2. Hence, only one set in each equivalence class needs to be considered in the search. Since a given set may have as many as nφ(n) different equivalent sets, where φ is the Euler totient function, a considerable speedup may be achieved if equivalence testing can be carried out fast.
All one-element sets are clearly equivalent, so we may start from {0}. Throughout the search, we only need examine the search branches that correspond to the set S′ if no set equivalent to S′ has been considered earlier in the search. Since the algorithm searches all possible sets in lexicographical order, this is the case precisely when S′ is the lexicographically first member of its equivalence class. Such a set is called the canonical representative of its equivalence class. Pruning the search when sums of pairs occur more than once or when S′ is not in canonical form we get the following algorithm. At line 2, if a bound such as Theorem 3 shows that no S 2-set with |S|+1 elements can exist, one may stop the search.
functionrecsearch(S,p)
if|S|>m a x kthen
larger distinct-sum set found; store; m a x k←|S|
end if
fori=pton-1do
S′←S∪{i}
if sums of pairs occur at most once then
ifS′ is a canonical representative then
recsearch(S′,i+1)
end if
end if
end for
return
function search
m a x k←1
recsearch({0},1)
return
Our backtrack search method with isomorph pruning is an orderly algorithm. Any set S with at least two elements that is in canonical form may be reached in the search from another set in canonical form. Thus, the algorithm searches all sets with distinct sums of pairs that are in canonical form.
The implementation of the two tests in lines 7 and 8 of our algorithm is crucial for its performance. These will be discussed in the following subsections.
3.2. Testing occurrences of sums
In order to check whether an element c can be added to S without violating the distinct sums condition, we need to calculate the new sums that are possible to obtain after adding the element c to the set S and verify that none of them are obtainable from the elements of S prior to adding c to the set. When calculating v γ(k), we restrict ourselves to sums of two distinct elements; in that case the new sums are obtained by adding c to each element of S. In calculating v δ(k) we also have sums of two identical elements, and we need to take also the additional new sum c+c into account.
As for testing occurrences of sums, we obtained the best performance with bitmask representations. These were also used recently in a search for Golomb rulers , another similar problem in additive number theory.
In the bitmask representation, the search set S is represented as a bit sequence that contains n bits. The i th bit is 1 if and only if i∈S (the 0th bit is leftmost). Similarly, the set of sums that can be composed from the elements in S is represented as a v-bit sequence with 1s in the corresponding positions.
The crucial point of this representation is that if S is represented as a bit sequence s, the set S+c is represented as s⋙c, where ⋙ denotes right rotation by c bits, allowing us to calculate very quickly the new sums introduced by adding the element c to the set S.
Assume that all sums of pairs in S are different on line 6 of our algorithm. Let s be the bit sequence representing S and u the sequence representing the sums that can be composed from the elements of S. To perform the test in line 7, it is enough to calculate (1)(s⋙i)AND u where the AND operation is carried out bitwise. This can be done very fast. If the value of (1) is 0, then the test is passed. In calculating v δ(k), the new sums are most conveniently calculated by substituting the bitmap s′, corresponding to the set S′, for s in (1).
Another benefit of the bitmask representation is that it can be used to compare the lexicographical ordering of two sets very efficiently, also when several machine words are used for their representation in the actual implementation.
3.3. Testing canonicity
The efficiency of our algorithm very much depends on efficient testing of canonicity of S′ in line 8. The most straightforward, but also the most time-consuming way to do this is to compute all nφ(n) equivalent sets corresponding to the allowable φ(n) different values of a and n different values of b, find the lexicographically first one of them and compare it to S′. To speedup this direct approach, we may use some additional information that we have about the canonical representative.
We know that 0 is always in the canonical representative of any non-empty set. For any set with at least two elements, the second smallest element in the canonical representative is c=min x 1,x 2∈S′gcd(x 2-x 1,n). It is therefore sufficient to only check such choices of a and b that f(x)=(a x+b)mod n maps a pair (x 1,x 2) in S′ to (0,c). The canonicity test is then performed by finding all such choices of a and b, applying the corresponding mapping to S′, and checking whether the result lexicographically precedes S′. In this manner it is only necessary to compare O(k 2) sets. Of course, if one of the mappings gives a set that lexicographically precedes S′, we may stop testing right away—we already know that S′ is not in canonical form.
4. Computational results
We implemented the algorithm described in Section 3 in C. For best possible performance, canonicity testing was not performed on all levels in the search tree. For example, with k=12, the test should be done with at most about 7 (depending somewhat on the value of n) elements in S′.
Generally, the program finds the maximum subset with distinct sums fairly quickly. Verifying that no larger subset with distinct sums exists takes significantly longer. To determine the value of v γ(k), and analogously v δ(k), we have to calculate a lower bound for v γ(k) (e.g., from Theorem 3), and run our algorithm for all consecutive values of n starting from the lower bound until we find a packing that gives the value of v γ(k). This holds only for the modular versions of packing and covering problems, whereas, for example, the Golomb ruler problem can be solved in a more direct way (see [6, Lemma 4] and the comments thereafter).
We used the program to compute v γ(k) for k⩽15. Our new results on v γ(k) for 11⩽k⩽15 improve on the bounds given for v(k) in [3, Table 5]. Our results on v γ(k) for 11⩽k⩽15 are thus also the best-known upper bounds on v(k). The results are presented in Table 1. We similarly calculated the values of v δ(k) and the corresponding lexicographically first maximum sets for k⩽14. Our results are summarized in Table 2. The value of v δ(13) is new. The k-element subsets given in Table 1, Table 2 are unique up to equivalence, with the exceptions of v γ(7) and v γ(11), for which the lexicographically first k-subsets of each equivalence class is given.
Table 1. Values of v γ(k) and the corresponding sets for k⩽15
| k | v γ(k) | The corresponding sets |
---
| 1 | 1 | {0} |
| 2 | 2 | {0,1} |
| 3 | 3 | {0,1,2} |
| 4 | 6 | {0,1,2,4} |
| 5 | 11 | {0,1,2,4,7} |
| 6 | 19 | {0,1,2,4,7,12} |
| 7 | 28 | {0,1,2,4,8,15,20}, |
| | | {0,1,2,5,9,17,23} |
| 8 | 40 | {0,1,5,7,9,20,23,35} |
| 9 | 56 | {0,1,2,4,7,13,24,32,42} |
| 10 | 72 | {0,1,2,4,7,13,23,31,39,59} |
| 11 | 96 | {0,1,2,4,10,16,30,37,50,55,74}, |
| | | {0,1,2,4,11,21,40,52,70,75,83}, |
| | | {0,1,2,4,13,26,34,40,50,55,78}, |
| | | {0,1,2,4,16,22,27,35,52,59,69} |
| 12 | 114 | {0,1,4,14,22,34,39,66,68,77,92,108} |
| 13 | 147 | {0,1,2,4,7,29,40,54,75,88,107,131,139} |
| 14 | 178 | {0,1,2,4,16,51,80,98,105,111,137,142,159,170} |
| 15 | 183 | {0,1,2,14,18,21,27,52,81,86,91,128,139,161,169} |
Table 2. Values of v δ(k) and corresponding sets for k⩽15
| k | v δ(k) | Lexicographically first set |
---
| 1 | 1 | {0} |
| 2 | 3 | {0,1} |
| 3 | 7 | {0,1,3} |
| 4 | 13 | {0,1,3,9} |
| 5 | 21 | {0,1,4,14,16} |
| 6 | 31 | {0,1,3,8,12,18} |
| 7 | 48 | {0,1,3,15,20,38,42} |
| 8 | 57 | {0,1,3,13,32,36,43,52} |
| 9 | 73 | {0,1,3,7,15,31,36,54,63} |
| 10 | 91 | {0,1,3,9,27,49,56,61,77,81} |
| 11 | 120 | {0,1,3,20,31,35,45,53,58,74,114} |
| 12 | 133 | {0,1,3,12,20,34,38,81,88,94,104,109} |
| 13 | 168 | {0,1,3,11,30,34,46,83,103,108,121,147,162} |
| 14 | 183 | {0,1,3,16,23,28,42,76,82,86,119,137,154,175} |
The maximum subsets of cyclic groups of small order may be obtained electronically from the WWW page 〈
The computations were run in a heterogeneous, non-dedicated network of PCs, using the autoson distributed batch system . For determining v γ(15), Theorem 3 gives us the lower bound v γ(15)⩾180. For each 180⩽n⩽182 an exhaustive search shows that v γ(15)≠n, as ℤ n contains no 15-element subset. Each of these searches takes an estimated two to three days on a 1.4 GHz AMD Athlon PC. It takes 15 min to find the 15-element subset of ℤ 183 listed in Table 1, which shows that v γ(15)⩽183, and by combining these results we get v γ(15)=183.
Verifying that the values given in Table 1, Table 2 are upper bounds of v γ(k) and v δ(k) is straightforward. Verifying that they are minimal is much harder. However, for v γ(k), two independent implementations gave the same results for k⩽13. Additionally, our computations confirm the computational results in and the computational results on v γ(k) and v δ(k) in .
While for most values of k the best-known upper bound for v(k) is derived from a subset in a cyclic group, for k∈{6,7,9} the bounds in show that for those values v(k)<v γ(k). Exhaustively searching abelian groups of small order for maximum subsets with distinct sums of pairs would be a natural way to continue this research.
Acknowledgements
The authors wish to thank professor Alexander Pott for his most helpful comments.
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2009, Discrete Mathematics Show abstract We consider the following generalization of the concept of harmonious graphs. Given a graph G=(V,E) and a positive integer t≥|E|, a function h̃:V(G)→Z t is called a t-harmonious labeling of G if h̃ is injective for t≥|V| or surjective for t<|V|, and h̃(v)+h̃(w)≠h̃(x)+h̃(y) for all distinct edges v w,x y∈E(G). Then the smallest possible t such that G has a t-harmonious labeling is named the harmonious order of G. We determine the harmonious order of some non-harmonious graphs, such as complete graphs K n (n≥5), complete bipartite graphs K m,n (m,n>1), even cycles C n, some powers of paths P n k, disjoint unions of triangles n K 3 (n even). We also present some general results concerning harmonious order of the Cartesian product of two given graphs or harmonious order of the disjoint union of copies of a given graph. Furthermore, we establish an upper bound for harmonious order of trees.
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17653 | https://www.quora.com/What-is-the-reason-for-arranging-elements-in-the-periodic-table-according-to-atomic-number-instead-of-weight-or-density | What is the reason for arranging elements in the periodic table according to atomic number instead of weight or density? - Quora
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What is the reason for arranging elements in the periodic table according to atomic number instead of weight or density?
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Clive Holloway
Former Full Professor at York University (Canada) (1968–2003) · Author has 2.9K answers and 2.2M answer views
·1y
Because the weight depends on both the number of protons and neutrons in the atomic nucleus, whereas the chemistry depends on the number of electrons (= number of protons). Since H is the only atom that has no neutrons, it would be the only atom that you could guarantee was in the right place. And in this special case, the H isotopes with one (D)and two (T) neutrons, have measurably different properties because they are twice and three times heavier the H. Even so, most of the elements follow their atomic weight also, except a few important exceptions.
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Why are elements arranged in periodic table based on their atomic number and not atomic mass?
How can you find an atomic number without the periodic table?
Did Mendeleev arrange elements in order of increasing atomic weight instead of increasing atomic number?
As you go down the periodic table, why is the atomic weight no longer exactly twice the atomic number for heavier elements?
How do I learn atomic numbers in periodic table 😢?
Steve Geo
Ph.D. in Chemistry&Organic Chemistry, University of Iowa (Graduated 1964) · Author has 2.8K answers and 582.8K answer views
·1y
Dmitri Mendeleev originally ordered the elements in order of atomic weight. This resulted in an almost perfect match, but there were some anomalies that stuck out. Later investigators who knew about atomic numbers found that ordering the elements by atomic number gave a perfect fit.
Upvote ·
Shashank Tripathy
Student ( currently in standard 12). (2002–present) · Author has 146 answers and 348.7K answer views
·7y
Originally Answered: Why are elements arranged in periodic table based on their atomic number and not atomic mass? ·
It can be easily understood by studying mendelev’s periodic table’s demerits.
Since the atomic mass of the elemnts does not increases in a regular manner , so it is not possible to predict that how many elements can be later discovered between two succesive elements. Further the isotopes of elements were also placed in seperate groups although they had similiar properties. Lastly, it could not be explained why some elemnts having less mass came after the elemnts that had greater mass like Co and Ni.( Co has a mass of 58.9 while Ni has mass of 58.7)
However , the mordern periodic table cleared al
Continue Reading
It can be easily understood by studying mendelev’s periodic table’s demerits.
Since the atomic mass of the elemnts does not increases in a regular manner , so it is not possible to predict that how many elements can be later discovered between two succesive elements. Further the isotopes of elements were also placed in seperate groups although they had similiar properties. Lastly, it could not be explained why some elemnts having less mass came after the elemnts that had greater mass like Co and Ni.( Co has a mass of 58.9 while Ni has mass of 58.7)
However , the mordern periodic table cleared all the anomilities . In the mordern periodic table elemmets are arranged according to their atomic mass. Thus the position of isotopes was not needed to be speciefied( as they had the same atomic number so they all had the same place). Further atomic number increases by a fixed number that is 1 , so no two elemts can be discovered between any two elements( although their isotopes can be discovered but they too would have the same place as the main element). Lastly position of Co and Ni was also specified as Co has atomic number-27 while Ni has atomic number-28 , hence Co comes before Ni although Co has a greter maas than Ni.
That is why atmoic number was considered a mkre fundamental charecteristic than atomic mass, and hence the mordern periodic table has been made according to atomic number of elemnts.
Upvote ·
9 9
Bernd Leps
Former scientific official; retired · Author has 5.6K answers and 1.2M answer views
·2y
Originally Answered: Why do we use the periodic table instead of just listing all the elements by their atomic numbers? ·
The trick is, that the periodic table is the list of elements by their atomic numbers.
The long list but is broken into lines (called “periods”), so that you get some vertical columns.
In each column, the elements share some structural properties and therefore are chemically related.
If we would use a single line list, we would not obtain this additional information.
Upvote ·
9 1
Related questions
More answers below
Why do we use the periodic table instead of just listing all the elements by their atomic numbers?
What is an atomic number on the periodic table?
Which element in the periodic table hasn't been arranged according to its increasing atomic number?
Why is the atomic number important in the placement of particular elements in the periodic table?
Why are there only a certain number of elements in the periodic table?
Adi
Ph.D in Rare Earth&Organometallic Chemistry, Imperial College London (Graduated 2014) · Author has 2.2K answers and 4.7M answer views
·7y
Originally Answered: Why is the periodic table arranged according to the atomic numbers and not the atomic mass? ·
All elements have isotopes. These isotopes have different numbers of neutrons and therefore have different masses. These neutrons have a negligible impact on chemical reactivity in most cases.
You will notice that even though we give protons and neutrons a mass of 1, all masses are non-integer numbers. The classic example is chlorine. Chlorine exists on Earth as a 75:25 mix of 35Cl and 37Cl. This averages out at an atomic mass of 35.5, which is listed.
Roughly speaking the elements are mostly in mass order anyway.
Upvote ·
9 1
John Darling
BS in Chemistry&Mathematics, Bachelor of Science Degrees · Author has 105 answers and 209.2K answer views
·6y
Originally Answered: Why is the periodic table based on atomic numbers? ·
The periodic table is just a tool humans use to classify the elements we have found in nature. The first versions were based on atomic mass, but atomic number was eventually determined to provide a more useful arrangement.
A table based on atomic number best represents and categorizes the elements based on their recurring chemical and physical properties.
Upvote ·
9 1
Nawaal Saleem
Knows Spanish · Upvoted by
Malcolm Sargeant
, Degree level applied chemistry + 20yr experience in corrosion prevention and water treatment
·4y
Originally Answered: Why are elements arranged in periodic table based on their atomic number and not atomic mass? ·
It is also that if you look at the periodic table, arranging it based on the elements’ atomic masses would mean that potassium would come at argon’s place in the noble gas group. Potassium’s atomic mass is 39 while argon’s is 40. This is a serious error as potassium is highly reactive with its own distinctive properties and is a completely different element from the other noble gases. So arranging the periodic table according to the atomic numbers was deemed suitable. Other reasons may also count.
Upvote ·
9 7
Alvaro Wang
Studied Environmental Chemistry at University of São Paulo (USP) · Author has 107 answers and 378.8K answer views
·7y
Originally Answered: Why are elements arranged in periodic table based on their atomic number and not atomic mass? ·
There might be some reasons why we do not arrange elements by mass, the main reason is the existence of isotopes.
There are many isotopes from different elements which are isobars, rendering such arrangement useless. I am not sure and I am quite lazy to check, but I wouldn´t be surprised if some isotopes from different elements would actually end up switching places, messing with the awesome periodic properties.
Even if you arrange them by the average mass, you would have overlap, for instance Lawrencium and Dubnium (262) (source: Chemical elements of the periodic table sorted by Atomic Mass).
On
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There might be some reasons why we do not arrange elements by mass, the main reason is the existence of isotopes.
There are many isotopes from different elements which are isobars, rendering such arrangement useless. I am not sure and I am quite lazy to check, but I wouldn´t be surprised if some isotopes from different elements would actually end up switching places, messing with the awesome periodic properties.
Even if you arrange them by the average mass, you would have overlap, for instance Lawrencium and Dubnium (262) (source: Chemical elements of the periodic table sorted by Atomic Mass).
On the other hand, arranging them by atomic number gives a neat integral number, keeps the periodic properties intact, and chemists happy.
Upvote ·
Tomas De Paulis
Studied Chemistry at Stockholm University (Graduated 1978) · Author has 478 answers and 287.3K answer views
·6y
Related
Why is the periodic table organized by atomic number rather than alphabetically?
It is called the periodic table for a reason. The atomic number denotes the number of protons in the atomic nucleus. The chemical properties of each element is defined by its electrons. Because the number of electrons in a neutral element must be matched by the same number of protons, the atomic number will explain the chemistry of the element. The column position of an element in the periodic table tells how many electrons have to be added or taken away to have a complete 8–electron outer shell. This, in turn tells what kinds of salt or combinations are possible. Once each shell is filled the
Continue Reading
It is called the periodic table for a reason. The atomic number denotes the number of protons in the atomic nucleus. The chemical properties of each element is defined by its electrons. Because the number of electrons in a neutral element must be matched by the same number of protons, the atomic number will explain the chemistry of the element. The column position of an element in the periodic table tells how many electrons have to be added or taken away to have a complete 8–electron outer shell. This, in turn tells what kinds of salt or combinations are possible. Once each shell is filled the next shell is available, resulting in similar chemistry as those of the previous shell. An alphabetical table would make no sense, what so ever.
Upvote ·
9 6
Torbjörn Alm
Studied Chemistry&Data at Stockholm University (Graduated 1966) · Author has 163 answers and 17.5K answer views
·2y
Originally Answered: What is the reason elements are placed according to their atomic number and not according to their mass number or chemical properties? ·
The atomic number defines the number of electrons. It is the number of electrons in the outmost shell that defines the chemical properties. There are especially heavier elements, that have isotopes mith mass either lower or higher.
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Tom Kent
Scientific software designer, botany, photography · Upvoted by
Roger Mercer
, PhD in Chemistry from the University of Toronto. Director of Florida State University College of Medicine Tra… and
Malcolm Sargeant
, Degree level applied chemistry + 20yr experience in corrosion prevention and water treatment · Author has 608 answers and 1.4M answer views
·12y
Originally Answered: Why is the Periodic Table arranged by atomic number and not atomic mass? ·
Here is bit more detail. Atomic number is the number of protons, while atomic mass is the number of protons and neutrons (and presumably the nearly weightless electrons). The chemical behavior of the element is determined by the number of protons, not so much because of the protons per se, but because in a neutral atom, there are a matching number of electrons. The electrons are arranged in orbitals or shells, and those in the outer shell are responsible for chemical reactivity at earthly temperatures.
The number of neutrons in the atom has little effect on chemical behavior, because they do no
Continue Reading
Here is bit more detail. Atomic number is the number of protons, while atomic mass is the number of protons and neutrons (and presumably the nearly weightless electrons). The chemical behavior of the element is determined by the number of protons, not so much because of the protons per se, but because in a neutral atom, there are a matching number of electrons. The electrons are arranged in orbitals or shells, and those in the outer shell are responsible for chemical reactivity at earthly temperatures.
The number of neutrons in the atom has little effect on chemical behavior, because they do not influence the number of outer shell electrons. Many atoms have multiple isotopes -- for example, hydrogen has no neutrons, while its isotopes deuterium and tritium have one or two, respectively. Each of these isotopes have a different atomic mass. So when we say "the atomic mass of hydrogen is x," we need to take into account the relative natural occurrence of each isotope to produce an average value. (Or we need to talk about the atomic mass of each isotope separately, but that would be a table of the isotopes.)
So atomic mass is not a natural way to order a periodic table, since periodic tables were (and are) principally aimed at capturing the chemical relationships between elements.
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99 28
9 1
Daniel Iyamuremye
Former Senior Lecturer (Retired) (2000–2018) · Author has 11.3K answers and 1.9M answer views
·10mo
Originally Answered: Why are elements arranged in periodic table based on their atomic number and not atomic mass? ·
The first version of Mendeleev's Periofic Table was arranged based on their atomic mass. This led to the situation where a few number of elements were placed in group of elements with very different properties.
Latter on, after the discovery of the atomic structure, it was observed that atomic number was a better organizing criteria.
Upvote ·
John Finagin
Retired science teacher · Upvoted by
Malcolm Sargeant
, Degree level applied chemistry + 20yr experience in corrosion prevention and water treatment · Author has 220 answers and 642.3K answer views
·9y
Originally Answered: Why is the Periodic Table arranged by atomic number and not atomic mass? ·
Atomic number is the number of protons in the nucleus of each element's atoms. That number is unique to each element. Atomic mass is determined by the number of protons and neutrons combined. An electron's mass is 1/1840th of a proton's, so to all intents and purposes it has a minimal effect on atomic mass.
The number of neutrons can vary, creating two or more isotopes of the same element. Carbon, for example has an atomic number of 6. That means it has 6 protons. But it has two isotopes, Carbon-12 with 6 protons and 6 neutrons, and Carbon-14 with 6 protons and 8 neutrons.
Dmitri Mendeleev, wh
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Atomic number is the number of protons in the nucleus of each element's atoms. That number is unique to each element. Atomic mass is determined by the number of protons and neutrons combined. An electron's mass is 1/1840th of a proton's, so to all intents and purposes it has a minimal effect on atomic mass.
The number of neutrons can vary, creating two or more isotopes of the same element. Carbon, for example has an atomic number of 6. That means it has 6 protons. But it has two isotopes, Carbon-12 with 6 protons and 6 neutrons, and Carbon-14 with 6 protons and 8 neutrons.
Dmitri Mendeleev, who first came up with the idea of a periodic table in 1871 to group the behaviour of elements, was a keen player of patience, the solitaire playing card game. Think of him laying out the elements like playing cards and looking for patterns, so predictions about newly discovered elements could be made.
He actually started out with atomic weights and the table worked, but there were some anomalies. It was the English physicist Henry Moseley in 1913 who proposed that using atomic number was a much more reliable way of arranging the table. To be fair to Mendeleev our understanding of the basic structure of the atom wasn't understood until Ernest Rutherford came up with the model in 1911.
Poor Henry Moseley was killed at Gallipoli in 1915. Just think what other insights he might have had should he have lived.
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Related questions
Why are elements arranged in periodic table based on their atomic number and not atomic mass?
How can you find an atomic number without the periodic table?
Did Mendeleev arrange elements in order of increasing atomic weight instead of increasing atomic number?
As you go down the periodic table, why is the atomic weight no longer exactly twice the atomic number for heavier elements?
How do I learn atomic numbers in periodic table 😢?
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Why are elements arranged in periodic table based on their atomic number and not atomic mass?
How can you find an atomic number without the periodic table?
Did Mendeleev arrange elements in order of increasing atomic weight instead of increasing atomic number?
As you go down the periodic table, why is the atomic weight no longer exactly twice the atomic number for heavier elements?
How do I learn atomic numbers in periodic table 😢?
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Which element in the periodic table hasn't been arranged according to its increasing atomic number?
Why is the atomic number important in the placement of particular elements in the periodic table?
Why are there only a certain number of elements in the periodic table?
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17654 | https://physics.stackexchange.com/questions/552965/torque-and-axis-of-rotation-for-couple | newtonian mechanics - Torque and Axis of Rotation for couple - Physics Stack Exchange
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Torque and Axis of Rotation for couple
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I was recently instructed by my instructor that the axis of rotation in the case of a couple always passes through the center of mass and is parallel to torque,Why is this true?
newtonian-mechanics
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edited May 19, 2020 at 4:21
Qmechanic♦
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asked May 18, 2020 at 14:14
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In mechanics, a couple refers to two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action.
A better term is force couple or pure moment. Its effect is to create rotation without translation or, more generally, without any acceleration of the centre of mass.
Your one statement is correct that the axis of rotation is parallel to the torque but it passes through center of mass is not always correct.
In the diagram above, a couple is applied to a disk of diameter D D. That is, a force F F is applied to opposite sides of the disk. The torque do to the couple is:
τ=F×d 1+F×d 2 τ=F×d 1+F×d 2
where d 1 d 1 and d 2 d 2 are the distance to some (arbitrary) point O O.
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answered May 18, 2020 at 17:02
SarGeSarGe
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Could you provide an example for the same? good read tho user245838 –user245838 2020-05-19 09:05:04 +00:00 Commented May 19, 2020 at 9:05
Example for what??SarGe –SarGe 2020-05-19 09:06:55 +00:00 Commented May 19, 2020 at 9:06
that axis may not pass through com user245838 –user245838 2020-05-19 09:08:05 +00:00 Commented May 19, 2020 at 9:08
The axis passes through the point where the object is hinged. If it is not hinged, the axis passes through center of mass.SarGe –SarGe 2020-05-19 09:15:24 +00:00 Commented May 19, 2020 at 9:15
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For any couple, the axis of rotation is always through the COM and is truly parallel to the torque. Torque is a vector whose direction is determined by the right hand rule. Depending on the orientation, the torque can be pointing either up/down or into/out of the reference plane. And this direction will always be parallel to the rotation axis. To understand why an object including a couple would rotate about this point, you must remember that rotation also has it's own inertia. Every point/axis is not created equal when it comes to rotation. This implies that any axis at all has it's own inertia (how easily you can rotate the object about the axis). It turns out objects have least inertia for rotation about their COM . For this reason, an object would naturally prefer to rotate about this point than any other point since it is easier for them. Except rotation is forced about any other axis, every object would tend to swing about this unique point. You may as well consider the parallel axis theorem I p=I c m+M d 2 I p=I c m+M d 2 and the implication for this explanation. Good luck!
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edited May 19, 2020 at 18:30
answered May 18, 2020 at 20:57
EMMANUEL CHIDERAEMMANUEL CHIDERA
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The direction of torque is given by the right hand rule and the statement that the rotation axis is parallel to torque is correct.ModCon –ModCon 2020-05-19 05:59:24 +00:00 Commented May 19, 2020 at 5:59
@ModCon I admit. I just thought about it.EMMANUEL CHIDERA –EMMANUEL CHIDERA 2020-05-19 06:39:00 +00:00 Commented May 19, 2020 at 6:39
Could it be said by the right hand thumb rule that the direction of movement of palm w.r.t the thumb is actually w.r.t . the axis( which passes through the thumb)?user245838 –user245838 2020-05-19 09:07:05 +00:00 Commented May 19, 2020 at 9:07
1 @JoeSantino Yes! It is correct to imagine the direction the thumb points as the rotation axis. Remember the thumb points in the direction of the torque which we have said is also parallel (same direction) as the axis of rotation. So yes!EMMANUEL CHIDERA –EMMANUEL CHIDERA 2020-05-19 09:31:12 +00:00 Commented May 19, 2020 at 9:31
1 But @Joe Santino's question says that it always passes through COM, which is wrong.SarGe –SarGe 2020-05-21 11:17:26 +00:00 Commented May 21, 2020 at 11:17
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17655 | https://math.stackexchange.com/questions/4079732/when-does-the-cauchy-schwarz-inequality-become-an-equality | vector spaces - When does the Cauchy-Schwarz inequality become an equality? - Mathematics Stack Exchange
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When does the Cauchy-Schwarz inequality become an equality?
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For two vectors |f⟩|f⟩ and |g⟩|g⟩, the Cauchy-Schwarz inequality becomes an equality when
|h⟩=|f⟩−⟨f|g⟩⟨g|g⟩|g⟩|h⟩=|f⟩−⟨f|g⟩⟨g|g⟩|g⟩
is a null vector. This happens when |f⟩|f⟩ is linearly dependent on |g⟩|g⟩ i.e. |f⟩=α|g⟩|f⟩=α|g⟩ where α α is a complex number.
But if we substitute this into the expression of |h⟩|h⟩ above we get,
|h⟩=α|g⟩−α∗⟨g|g⟩⟨g|g⟩|g⟩=(α−α∗)|g⟩|h⟩=α|g⟩−α∗⟨g|g⟩⟨g|g⟩|g⟩=(α−α∗)|g⟩
which is nonzero unless α α is real.
What is wrong with my argument?
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edited Mar 27, 2021 at 20:41
RobPratt
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SolidificationSolidification
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You probably want ⟨g|f⟩⟨g|f⟩ in the numerator of the coefficient.Andrew D. Hwang –Andrew D. Hwang 2021-03-27 21:18:38 +00:00 Commented Mar 27, 2021 at 21:18
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Your vector |h⟩|h⟩ is meant to be the coprojection of f f onto the span of g g. For it to be that, when you take the inner product of f f with g g, f f must appear in the linear position, not the conjugate-linear position. Which position that actually is varies somewhat among authors, but generally in bra-ket notation the inner product is linear in the kets, i.e. in the second position, which seems to be how you're doing it.
In other words you should have had |h⟩=|f⟩−⟨g|f⟩⟨g|g⟩|g⟩|h⟩=|f⟩−⟨g|f⟩⟨g|g⟩|g⟩, and then the problem disappears.
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answered Mar 27, 2021 at 23:32
IanIan
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Please see Proof 3 here @Ian en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Proofs z, u, and v are respectively h, f, and g in my notation.Solidification –Solidification 2021-03-28 02:10:49 +00:00 Commented Mar 28, 2021 at 2:10
@mithusengupta123 Yes, as they say there, they are using the convention that ⟨⋅,⋅⟩⟨⋅,⋅⟩ is linear in the first argument instead of the second. Physicists usually use the opposite convention.Ian –Ian 2021-03-28 02:40:33 +00:00 Commented Mar 28, 2021 at 2:40
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The Cauchy-Schwarz inequality does become an equality when your condition is satisfied (provided the vectors are of the proposed form |f⟩=α|g⟩|f⟩=α|g⟩), but not only when it is. The equality is trivial if |f⟩|f⟩ or |g⟩|g⟩ are null vectors, so let us assume they aren't. The condition we need is
⟨f|f⟩=⟨f|g⟩⟨g|f⟩⟨g|g⟩⟨f|f⟩=⟨f|g⟩⟨g|f⟩⟨g|g⟩
which only requires |f⟩=α|g⟩|f⟩=α|g⟩ just as you stated (to prove this condition from the above equation you may write |f⟩−α|g⟩=|u⟩|f⟩−α|g⟩=|u⟩ and then prove that |u⟩|u⟩ is zero). In fact, if you let ⟨g|⟨g| act on the vectors from your first equation, you will find that ⟨g|h⟩=⟨g|f⟩−⟨f|g⟩⟨g|h⟩=⟨g|f⟩−⟨f|g⟩, so |h⟩|h⟩ is only a null vector if ⟨g|f⟩⟨g|f⟩ is a real number, which is why your assumption leads to real α α.
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edited Mar 27, 2021 at 23:49
answered Mar 27, 2021 at 23:23
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No, Cauchy-Schwarz is indeed an equality precisely when f f and g g are parallel, in which case the coprojection of one onto the other will be zero (if it is defined). The OP just made an error in their definition of the coprojection.Ian –Ian 2021-03-27 23:32:47 +00:00 Commented Mar 27, 2021 at 23:32
@Ian yes. But because of that error the condition he wrote will only lead to the correct result if α α is real. It should be ⟨g|f⟩⟨g|f⟩ rather than ⟨f|g⟩⟨f|g⟩. They will be the same thing when that inner product is real, which only happens when α α is.Othin –Othin 2021-03-27 23:38:59 +00:00 Commented Mar 27, 2021 at 23:38
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A specific (but not exclusive) case for when Cauchy-Schwarz has an equality is when f=g f=g. By Cauchy-Schwarz,
|⟨f,f⟩|≤∥f∥⋅∥f∥=∥f∥2|⟨f,f⟩|≤‖f‖⋅‖f‖=‖f‖2
But recall that for any inner product ⟨⋅,⋅⟩⟨⋅,⋅⟩ in a vector space, the naturally accompanying norm ∥⋅∥‖⋅‖ is defined via ∥f∥2=⟨f,f⟩‖f‖2=⟨f,f⟩; hence,
|⟨f,f⟩|≤∥f∥⋅∥f∥=∥f∥2=⟨f,f⟩|⟨f,f⟩|≤‖f‖⋅‖f‖=‖f‖2=⟨f,f⟩
where we can drop the |⋅||⋅| on the LHS since f f points in the same direction as itself, resulting in one particular case of equality.
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answered May 21, 2024 at 4:02
Aram NazaryanAram Nazaryan
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Species Habitat
Silver hake are widely distributed and are found throughout the Northwest Atlantic Ocean from Newfoundland to South Carolina. Two stocks have been identified in the United States – a northern stock inhabiting the waters of the Gulf of Maine and Northern Georges Bank and a southern stock inhabiting the waters of Southern Georges Bank and the Mid-Atlantic Bight. Silver hake are nocturnal predators and spend their days resting on sandy, muddy, and pebbly ocean floors. From dusk to midnight, they will move up in the water column to feed.
Silver hake have been observed at temperatures ranging from 36-63° F (2-17° C) and are commonly found between 45-50° F (7-10° C). They are found at depths ranging from 36-1,640 feet (11-500 meters) with older, larger whiting preferring deeper waters. The fish will migrate in response to seasonal changes in water temperatures, moving into shallow, warmer waters in the spring where they will spawn in the late spring and early summer months. During the summer, portions of both the northern and southern stocks can be found on Georges Bank. As the water temperature cools in the autumn months, whiting will move offshore into deeper waters. During winter, fish in the northern stock will move to deep basins in the Gulf of Maine while the southern stock will move to the outer continental shelf and slope waters.
The spatial distribution of silver hake is highly correlated with the position of the Gulf Stream. Changes in distribution are in response to changes in temperature on the continental shelf, which responds to circulation of the North/South paths of the Gulf Stream. Silver hake shift geographically to remain within their preferred temperature range, and the alteration in global climate is showing a population shift northwards
Science & Management:
Wild Science: Because silver hake shift geographically, the alteration in global climate is showing a population shift northwards within their preferred temperature range. The change in temperature on the continental shelf also changes ocean circulation and overall temperatures, and ultimately the path of the Gulf Stream, which may further affect populations and migrations. Spawning is also affected by temperature, with fluctuations influencing spawning peaks and juvenile growth. Research is needed to further understand the long-term impacts of global climate change on silver hake populations.
More research is also needed in relation to silver hake migrations, biology, and population dynamics. Habitat studies typically include location, depth, temperature, and sometimes salinity levels. Future habitat studies should also include bottom type as a habitat consideration. Food habits analysis is needed to determine predator-prey relationships, distribution, and abundance.
Management:
NOAA Fisheries and the New England Fishery Management Council manage the silver hake fishery in the US under the Northeast Multispecies Fishery Management Plan (FMP) for Small Mesh Multispecies. As vessels participating in this fishery use small mesh to target silver hake and other species, the fishery is managed as a series of exemptions under Northeast Multispecies FMP. The FMP establishes:
Permitting requirements
Possession limits
A cap on groundfish bycatch that vessels can take
Seasonal and spatial closure throughout the Gulf of Maine and Georges Bank for fishers using small mesh
Silver hake are managed as two stocks in the Northwest Atlantic – a northern stock that ranges from the Gulf of Maine and Northern Georges Bank and a southern stock that ranges from Southern Georges Bank to Cape Hatteras. According to a 2014 stock assessment neither the northern nor southern stock are overfished nor subject to overfishing.
Fisheries and Oceans Canada (DFO) manage the silver hake fishery in Canada. DFO conducts research surveys and based on the results of those surveys, sets annual total allowable catch (TAC) limits.
Wild
Impact on Stock
Silver hake are found in the Atlantic Ocean from South Carolina in the United States to Newfoundland in Canada. They are primarily targeted in the U.S. Atlantic. There are two silver hake stocks: a northern one in the Gulf of Maine and northern Georges Bank, and one in southern Georges Bank and the Mid-Atlantic Bight coastal region. A Seafood Watch report from 2016 noted that, according to the most recent assessment, both stocks are not overfished and overfishing is not occurring.
Habitat Impacts
Silver hake are migratory and also move up through the water column to feed. They are mainly fished using small-mesh multispecies trawlers. In some areas, fishermen use modified gear to catch silver hake, which reduces contact with the seafloor. The fish prefer resting on sandy and muddy ocean bottoms during the day, a habitat that isn’t heavily impacted by bottom trawls, according to Seafood Watch.
Bycatch
The mid-water trawls that target silver hake result in little bycatch, FishWatch reported in 2015. Silver hake are part of the small-mesh multispecies fishery in the United States, which can unintentionally catch squid, Atlantic butterfish, and Atlantic mackerel. Overall, Seafood Watch called bycatch a moderate conservation concern in the fishery.
Management Effectiveness
In the United States, silver hake are managed by NOAA Fisheries and the New England Fishery Management Council under the Northeast Multispecies Fishery Management Plan for small-mesh species. Measures include permitting requirements, a cap on groundfish bycatch as well as seasonal and spatial limitations. Seafood Watch called the management effective.
Farmed
| Origin | Harvest Method | Sustainability Ratings |
---
| Canada | Wild-caught | |
| Unassessed Origin | Unassessed Fishing Methods | |
| USA - Gulf of Maine | Bottom Trawls | |
| USA - Mid-Atlantic | Bottom Trawls | |
| USA - North Georges Bank | Bottom Trawls | |
| USA - South Georges Bank | Bottom Trawls | |
| Name | Country | State/Province |
---
| Catanese Classic Seafood | United States | Ohio |
| Channel Fish Processing Company, Inc. | United States | Massachusetts |
| John Nagle Co. | United States | Massachusetts |
| Maine Shellfish Company | United States | Maine |
| Monterey Fish Market | United States | California |
| Profish Ltd. | United States | District of Columbia |
| Raw Seafoods | United States | Massachusetts |
| Samuels & Son Seafood Company, Inc. | United States | Pennsylvania |
| Sea Star Seafoods | Canada | Nova Scotia |
| Sea to Table, USA | United States | New York |
| True Fin | United States | Maine |
| Walden Local, Inc. | United States | Massachusetts |
Acknowledgements
Environmental Defense Fund
NOAA Fisheries
Seafood Watch Program
SeafoodSource
Last Updated: 8/7/2020
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17659 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6404699/ | Approach to the Diagnosis and Management of Subarachnoid Hemorrhage - PMC
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West J Emerg Med
. 2019 Feb 28;20(2):203–211. doi: 10.5811/westjem.2019.1.37352
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Approach to the Diagnosis and Management of Subarachnoid Hemorrhage
Evie Marcolini
Evie Marcolini, MD
University of Vermont Medical Center, Department of Emergency Medicine, Burlington, Vermont
Find articles by Evie Marcolini
,✉, Jason Hine
Jason Hine, MD
†Tufts University School of Medicine, Department of Emergency Medicine, Boston, Massachusetts
Find articles by Jason Hine
†
Author information
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University of Vermont Medical Center, Department of Emergency Medicine, Burlington, Vermont
†Tufts University School of Medicine, Department of Emergency Medicine, Boston, Massachusetts
✉
Address for Correspondence: Evie Marcolini, MD, University of Vermont College of Medicine, Department of Surgery, Division of Emergency Medicine, 111 Colchester Avenue, Burlington, VT 05401. Email: evie.marcolini@yale.edu.
✉
Corresponding author.
Received 2018 Dec 19; Revised 2019 Jan 9; Accepted 2019 Jan 16; Issue date 2019 Mar.
Copyright: © 2019 Marcolini et al.
This is an open access article distributed in accordance with the terms of the Creative Commons Attribution (CC BY 4.0) License. See:
PMC Copyright notice
PMCID: PMC6404699 PMID: 30881537
Abstract
Headache is one of the most common reasons for presentation to the emergency department (ED), seen in up to 2% of patients.1 Most are benign, but it is imperative to understand and discern the life-threatening causes of headache when they present. Headache caused by a subarachnoid hemorrhage (SAH) from a ruptured aneurysm is one of the most deadly, with a median case-fatality of 27–44%.2 Fortunately, it is also rare, comprising only 1% of all headaches presenting to the ED.3 On initial presentation, the one-year mortality of untreated SAH is up to 65%.4 With appropriate diagnosis and treatment, mortality can be reduced to 18%.5 The implications are profound: Our careful assessment leading to the detection of a SAH as the cause of headache can significantly decrease our patients’ mortality. If this were an easy task, the 12% reported rate of missed diagnosis would not exist.6 We have multiple tools and strategies to evaluate the patient with severe headache and must understand the strengths and limitations of each tool. Herein we will describe the available strategies, as well as the ED management of the patient with SAH.
INTRODUCTION
A 50-year-old female was preparing her children for school when she experienced a headache severe enough to make her lie down on the sofa. She managed to get the children off to school, but the headache did not abate. She was used to headaches, as she had migraines periodically that were controlled with over-the-counter medications, but this one was different and much more intense. She took a couple of acetaminophen, and when the pain was not relieved she brought herself to the emergency department (ED).
Headache is one of the most common reasons for presentation to the ED, seen in up to 2% of patients.1 Most are benign, but it is imperative to understand and discern the life-threatening causes of headache when they present. Headache caused by a subarachnoid hematoma (SAH) from a ruptured aneurysm is one of the most deadly, with a median case-fatality of 27–44%.2 Fortunately, it is also rare, comprising only 1% of all headaches presenting to the ED.3 On initial presentation, the one-year mortality of untreated SAH is up to 65%.4 With appropriate diagnosis and treatment, mortality can be reduced to 18%.5
The implications are profound: Our careful assessment leading to the detection of a SAH as the cause of headache can significantly decrease our patients’ mortality. If this were an easy task, the 12% misdiagnosis rate would not exist.6 We have multiple tools and strategies to evaluate the patient with severe headache, and must understand the strengths and limitations of each tool.
Pathophysiology
Eighty-five percent of cases of atraumatic SAH result from a ruptured aneurysm.7 Alternate etiologies include perimesencephalic hemorrhage, which has a benign course, as well as arteriovenous malformations, dural arteriovenous fistula, arterial dissection, mycotic aneurysm, and cocaine abuse. The prevalence of aneurysms in the general population is roughly 2–5%,8 greater in those with family history of aneurysms, and/or personal history of Ehlers-Danlos or polycystic kidney disease. Not all aneurysms are dangerous. Factors associated with the risk of rupture include hypertension, tobacco use, excessive alcohol use, sympathomimetic drugs, Black race, Hispanic ethnicity, and aneurysmal size > 10 millimeters (mm).9 Aneurysmal SAH is more common in women and in patients 40–60 years old.
Aneurysms typically present at cerebral artery bifurcation points in both anterior or posterior regions. Aneurysmal pathophysiology has been theorized to involve congenital weakness in the vessel wall, or degenerative changes resulting in destruction of elasticity of the vessel wall at points of high turbulence such as bifurcations.10
Classification
There are several systems of classification for SAH. The Hunt and Hess score and World Federation of Neurological Surgeons grading system are both used to predict patient outcome, and the Fisher grade helps to predict vasospasm. Given the retrospective derivation of these scales and little if any assessment of intra- and interobserver variability, no single scale can be recommended over others.11 In terms of patient-centered outcomes and prognosis, specific scores were not seen to perform any better than the Glasgow Coma Scale (GCS).12
The classification systems do, however, help highlight an important concept of spectrum bias. As we delve into the diagnosis of SAH, it is important to note that some patients with SAH, for example Hunt and Hess grades I and II patients, are more commonly missed because symptoms are milder, and they may have smaller aneurysms with less subarachnoid blood. These patients do not necessarily do better or have less morbidity with rupture or re-rupture.
Diagnosis
The diagnosis of SAH should be considered in any patient with a severe and sudden onset or rapidly escalating headache. With such a large number of patients presenting to the ED with a chief complaint of headache, differentiating those with a benign cause from those with an emergent etiology such as SAH can be difficult. Deciding which patients require a workup for SAH is often the most challenging part of the emergency physician’s care, in part due to the low frequency and high acuity of the illness.
Classic teaching characterizes the headache of SAH as a “thunderclap headache,” which is defined as a sudden, severe headache often described as the worst of the patient’s life.14 The headache is typically a sudden onset, which is commonly characterized as occurring within a few minutes, although research parameters include headache that reaches maximum intensity within one hour. Symptoms that increase the likelihood of a subarachnoid bleed as the cause of headache include exertional onset, syncope, vomiting, neck pain, and seizures.15 Focal neurologic deficits, meningismus, and/or retinal hemorrhage may be present, but up to 50% of SAH patients have a normal neurologic exam.16 Recent research has attempted to shed light on which elements of the history and physical exam are correlated with and discriminating for the diagnosis of SAH.
Perry et. al published the Ottawa SAH Rule in 2013 after prospectively assessing 2131 adult patients with a non-traumatic headache that reached maximum intensity within one hour (Figure 1).17 Subjects were excluded if they had a pattern of similar headaches, had papilledema, or focal neurologic deficits on exam, or had a prior history of aneurysm, SAH, neoplasm, or hydrocephalus. Of the 2,131 patients investigated, 132 were ultimately diagnosed with SAH, giving a prevalence of 6.2%. The authors describe a decision rule with 100% sensitivity, although the specificity is at best 15% (Table). By this rule, any one criterion suggests that the patient should get a full workup. The low specificity, however, can have the deleterious effect of increasing the number of patients who undergo full workups, and are subsequently exposed to unnecessary radiation, procedures, and perhaps invasive procedures. While the merits of the Ottawa decision rule can be argued, it has helped delineate which historical elements, signs, and symptoms are statistically correlated with a confirmed diagnosis of SAH. Given that one of the most difficult elements of a SAH diagnosis is determining in whom a workup is needed, these data can inform the clinician’s process of determining pretest probability, even if the rule is not used in its entirety.
Figure 1.
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Ottawa subarachnoid hemorrhage decision rule.
Table.
Hunt and Hess grading for subarachnoid hemorrhage.13
| Grade | Criteria | Survival |
:---:
| I | Asymptomatic or mild headache with slight nuchal rigidity | 70% |
| II | Moderate to severe headache, nuchal rigidity, no neurological deficit other than cranial nerve palsy | 60% |
| III | Drowsiness, confusion, or mild focal deficit | 50% |
| IV | Stupor, moderate to severe hemiparesis, possibly early decerebrate rigidity or vegetative disturbance | 20% |
| V | Deep coma, decerebrate rigidity, moribund appearance | 10% |
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Diagnostic Tools
Computed Topography
When a clinical suspicion for SAH exists based on history and physical exam, non-contrast computed tomography (CT) is the first diagnostic tool. It is also valuable in excluding other pathologies such as intracranial hemorrhage, malignancy, or abscess.
Timing of Computed Tomography
At the onset of the bleed, subarachnoid blood is the most readily visible on CT, but it becomes more difficult to appreciate as red blood cell (RBC) degradation progresses. Advances in neuroimaging have increased the sensitivity of non-contrast CT, raising questions regarding the need for lumbar puncture (LP) in the face of a negative CT.
A meta-analysis published in 2016 attempted to answer the question of CT sensitivity with relation to time from symptom onset.18 The analysis, which included five studies, assessed patients with a thunderclap headache and normal neurologic exam. While the results carry many of the limitations of a meta-analysis, a conservative statistical analysis showed that a non-contrast CT completed within six hours of headache onset had a sensitivity of 98.7% with confidence intervals 97.1%–99.4%. The authors took into consideration the following criteria: patient must have a hematocrit > 30% and an isolated thunderclap headache without seizure, syncope, or neck pain; and the CT image must be third generation or newer, of high quality, read by an attending-level radiologist, and evaluated with the indication for imaging being thunderclap headache or concern for SAH. If these criteria are met, many consider a negative head CT within six hours to be a “rule-out” study given the sensitivity and confidence intervals.
Lumbar Puncture
If non-contrast head CT is not definitive (time to study, patient elements [i.e., severe anemia], interpretation limitations [i.e., trainee radiologist, motion artifact], etc) the next recommended diagnostic tool is the LP. In these instances the LP is looking for two elements that raise the concern for SAH: 1) RBCs; and 2) xanthochromia (bilirubin in cerebrospinal fluid [CSF]).
Given the sensitivity of the CT discussed above, shared decision-making should be conducted with regard to LP. In particular, with sensitivity of near 99% for an adequate study if completed within six hours, and meeting the criteria outlined above (Dubosh), patients should be made aware of the low diagnostic utility of LP if completed after a CT.19 In this setting, risks (adverse events and false positives) generally outweigh benefits and LP is advised against. There are rare instances in which the clinical scenario so strongly suggests SAH that even an adequate negative CT completed within six hours is unable to rule out SAH and should be followed by LP. If the imaging is completed after the six-hour timeframe, the sensitivity of CT drops to 85.7%. In these cases, the diagnostic utility of LP increases as the probability of SAH after negative CT also increases. In such cases, LP is indicated. It should also be noted that, in keeping with the low prevalence of this disease, one recent study showed a roughly 0.4% of LPs revealed aneurysms.20 Shared decision-making is still recommended, as with any invasive procedure.
Red Blood Cells
Intact RBCs will be seen early in the course of the disease and decrease as the cells break down and are resorbed. Fitting the pathophysiology, the presence of RBCs in the fourth tube of CSF is thought to represent SAH. Unfortunately, a LP is often a technically difficult procedure and rates of “traumatic tap,” or introduction of erythrocytes by local trauma and needle manipulation can approach 30%.21 This complicates the diagnosis of SAH by RBC results. Because differentiating between a true SAH and a traumatic tap is of the utmost clinical importance, authors have researched criteria to help differentiate the two.
Perry et al. published data comparing LP results in patients with SAH (by research gold-standard confirmation) to those without that final diagnosis, most notably patients with a traumatic tap without concurrent SAH.21 In this analysis, the researchers found that setting a cutoff of 2,000 × 10 6 RBCs per liter (L) in the final CSF tube combined with no xanthochromia, irrespective of RBCs in the first tube, captured all patients with a final diagnosis of SAH while excluding most patients with a traumatic tap. Patients were considered to have a SAH if they had any of the following: CT head positive for blood in the subarachnoid space; xanthochromia on LP; or RBCs of 2,000 × 10 6 in the final tube of CSF with an aneurysm on CT angiography (CTA) that required neuro-intervention or resulted in death. To our knowledge, this fourth-tube cutoff for diagnosis of SAH has not yet been incorporated into professional society guidelines. Generally, it is believed that a traumatic tap produces a lower RBC count and possibly a more rapidly diminishing count from tube one to four.22 Multiple authors have shown that the approach of comparing the first and fourth tubes is unreliable, in light of the fact that traumatic tap and SAH are independent entities.21,23
Xanthochromia
True xanthochromia is pathognomonic for SAH. This is valuable when there is high clinical suspicion and RBC count is not sufficiently elevated to differentiate from a traumatic tap diagnostic. Xanthochromia is detected either by visual inspection of the CSF tube vs a tube of water, or by spectrophotometry. RBCs that have shed into CSF from SAH will ultimately break down and release oxyhemoglobin, which then converts to bilirubin in vivo, interpreted as xanthochromia, or literally “yellow color.” It should be noted that blood from a traumatic tap can produce oxyhemoglobin when exposed to natural light, which can produce a yellow color, but since it is outside the body it will not produce bilirubin.24 Protecting the specimen from light will minimize the conversion of RBCs to oxyhemoglobin. Alternatively, spectrophotometry can differentiate the oxyhemoglobin of traumatic tap from the bilirubin of SAH. Visual inspection, however, is still used in most institutions.
Timing
As with CT, controversy and practice variations exist with respect to timing of the LP. However, given the timing of RBC breakdown, the presence of any xanthochromia is delayed and most conservative estimates state an “up to 12 hours” timeframe.25–27 In pursuit of xanthochromia, some have historically advocated a delayed LP approach, typically 12 hours from ictus, but the literature supporting this approach used spectrophotometry, which is not available in most labs in the United States.28 No literature supports waiting for 12 hours to perform LP.29 Given the flow of ED care and desire to expeditiously diagnose SAH, it is reasonable to obtain CT and then immediate LP (if needed), with attention to xanthochromia supplemented by the RBC cutoff criteria as needed. If the LP shows > 2000 × 10 6 RBCs in tube four, standard practice is to follow this with a CTA to assess for aneurysm. Conversely, if the cell count is < 2000 × 10 6/L and no xanthochromia is seen, then SAH is ruled out.
In the Perry study four out of five sites used visual inspection for xanthochromia, and 39% of all LPs were done within 12 hours of headache onset. Considering this, and the fact that results were confirmed with blood in subarachnoid space on CT, xanthochromia or RBCs in the final tube, and an aneurysm by cerebral angiography requiring neurovascular intervention or resulting in death, we believe that visual inspection is not only the most-often used modality to determine xanthochromia, but is reasonable for this purpose. Often, there is lack of clarity on exact time of ictus, and more importantly we have no pathophysiologic data showing a standard timeframe for the processes of RBC degradation into xanthochromia. Given the pathognomonic characteristics of xanthochromia, these authors (JH + EM) recommend that CSF samples be analyzed for both RBC count and xanthochromia regardless of timing of LP.
Computed Tomography Angiography
Over the last decade, CTA of the brain has become part of the discussion in ruling out SAH. As a non-invasive means of highlighting vascular anatomy and detecting aneurysms, CTA has many advantages. Much like non-contrast head CT, advances in neuroimaging have shown CTA to have a sensitivity of up to 98% and a specificity of 100% for aneurysms in patients with known SAH. These statistics are derived from a small data set (n= 65) where CTA results were compared to gold standards of digital subtraction angiography or surgical findings.30,31
Some propose CTA as an alternative to LP after a negative non-contrast CT.19,31 With the prevalence of aneurysms estimated to be ~2–5% in the general population,8 there is a concern for incidental findings and false positives. An aneurysm found on CTA may be incidental and unrelated to the cause of headache. For example, a patient with a moderate pretest probability of SAH on presentation at 12 hours of symptoms is generally not thought able to be ruled out by non-contrast head CT given its sensitivity of ~85%. Some advocate that if the patient has a CTA that is negative for aneurysm after a negative non-contrast head CT, this is accepted as being conclusively negative for SAH.19 In addition, if it is not possible to perform LP for any reason, such as coagulopathy, the CTA could be used, with acknowledgment and consideration for its limitations.
Based on best available literature, a CTA without findings of aneurysm when coupled with a negative non-contrast head CT has a post-test probability of disease of < 1%.31 This percentage is important because it falls below most clinicians’ test threshold, which is the probability of disease below which no further investigation is required. However there is one confounding factor in this suggested algorithm (Figure 2). The sensitivity of CTA is 92.3% for aneurysms < 4mm,32 and in contrast to pathologies where the size of the lesion correlates with the severity of disease (i.e., pulmonary embolus), a small, ruptured cerebral aneurysm can still lead to significant morbidity and mortality.
Figure 2.
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Algorithmic assessment for SAH in patient with sudden onset severe headache.
HA, headache; SAH, subarachnoid hemorrhage; NCHCT, non-contrast head computed tomagraphy; LP, lumbar puncture; CTA, computed tomography angiography; MRI, magnetic resonance imaging.
1 With criteria met for Perry study [Perry et al. BMJ 2011]47
2 Patient factors include anticoagulation status, patient willingness to undergo LP, history of lumbar spinal fusion or other surgery, and time from ictus (with longer time favoring MRI)
3 Caveat for this strategy includes the potential to miss aneurysms < 4 millimeters
4 MRI is an acceptable diagnostic at > 24 hours from ictus, prior to this sensitivity is lacking.
This is the recommended strategy by AHA/ASA, ACEP, and these authors
Recommended to decrease the false positive rate of CTA.
Collectively, for patients in whom a CT is completed at > 6 hours, a CT-CTA approach is pursued by some, but has limitations, most notably, the finding of incidental aneurysms and inability to detect small culprit aneurysms. If the CTA is positive for aneurysm, completing an LP at that time to determine incidental vs symptomatic could be considered. Limitations to this approach include radiation dose to patient, contrast dye exposure, and detriments to department flow of such an algorithm. As noted above, this approach of CT-CTA carries a low sensitivity for small but symptomatic aneurysms.19 If this approach is used, the limitations and risk of false positive results should be discussed with the patient in a shared decision-making process.
Magnetic Resonance Imaging
Magnetic resonance imaging (MRI) can be used to assess for SAH, with certain limitations. The challenge with using MRI for SAH is that the blood is combined with CSF that has a high oxygen concentration, thus delaying the transition of blood products to a deoxyhemoglobin state that is better imaged with MRI.33 Since there are no data showing a discrete timeframe for the use of MRI, the decision to use MRI to assess for blood should be used in consultation with radiology and neurology or neurosurgery. The combination of fluid-attenuated inversion recovery and susceptibility-weighted imaging has been shown to be 100% sensitive for SAH, although most cases were imaged greater than 24 hours after the ictus of headache.34 If MRI is negative for SAH, LP is still recommended.1,35,36 Magnetic resonance angiogram is 95% sensitive for aneurysms > 3 mm.37 With all of these limitations, MR imaging is not recommended as a primary imaging modality, but may be useful in certain atypical cases, in particular in patients with a long delay from ictus to presentation.
Summary of Available Diagnostic Tools
Many tools are available to assess for SAH including non-contrast CT, LP, CTA, and MRI. Understanding the potentially high mortality in the case of a missed SAH should mandate a diagnostic strategy with the highest sensitivity possible, which is currently accepted to be non-contrast CT followed, if negative, by LP.1,31,34 This is the algorithm supported by both the American Heart Association (AHA) and American College of Emergency Physicians (ACEP). This strategy, of course, should take into account the previously described limitations of the LP. While CT/LP remains the most accepted rule-out method, other approaches do exist. Many practitioners have accepted the recent literature showing non-contrast CT to be an acceptable stand-alone study if completed within six hours.18 If using any of the other tools described above, we must appreciate and work within the known limitations of each method.
ED Management
Once the diagnosis of SAH is established, the most important time-sensitive goals include confirmation of airway security and stabilization of hemodynamics. Intubation should be undertaken in the setting of low Glasgow Coma Scale Score or inability to protect the airway, but care should be taken to mitigate increases in mean arterial pressure (MAP) during the intubation process. This can be accomplished through careful choice of sedation agents for rapid sequence intubation and push-dose vasoactive agents if blood pressure does become elevated. Cardiac monitoring is important, as patients with devastating brain injury are at risk for neurocardiogenic stunning.38
The next priorities are to reduce systolic blood pressure (BP) and reverse anticoagulation to mitigate the risk of aneurysm re-rupture. Specific BP goals are unclear and need to be weighed against the risk of ischemia or infarction with hypotension. Guidelines recommend targeting BP < 160 systolic,35 although many consider lower targets of 140–150. Nicardipine (5 milligrams per hour (mg/h) intravenous (IV), may increase by 2.5 mg/h q5–15 minutes (min); Max: 15 mg/h), labetalol (40–80 mg IV q10 min, start 20 mg IV × 1; Max 300 mg/total dose; Alt: 2 mg/min IV), and clevidipine (4–6 mg/h IV, start 1–2 mg/h IV, double rate q 90 seconds until near BP goal, then increase. By smaller increments q5–10 min; max:32 mg/h) are effective agents, often used in infusion form to avoid hypotension. In the setting of bradycardia, hydralazine may also be used. Nitroprusside and nitroglycerin should be avoided due to their significant vasodilatory effect and the risk of increasing intracranial pressure (ICP).
Reversing anticoagulation should be accomplished as soon as possible. Vitamin K antagonists can be reversed with phytonadione (vitamin K) and 4-factor prothrombin complex concentrate (PCC) or fresh frozen plasma. PCC is preferable as it has a more rapid onset, does not need to be thawed or blood-type matched, and can be infused rapidly with less volume and risk of fluid overload.39 Antiplatelet agents should be reversed with platelet infusion, and desmopressin should be considered.40 The utility of platelet administration has been questioned recently after a recent trial showed increased mortality with platelet infusion for patients taking antiplatelet therapy.41 This trial, however, studied patients with spontaneous intracerebral hemorrhage, a different pathophysiology than SAH, and generalization of the results is not directly applicable.
Direct thrombin inhibitors such as dabigatran can be reversed with idarucizimab, which is United States Food and Drug Administration (FDA) approved and widely available. Andexanet alpha, an antidote for Factor Xa inhibitors (apixaban, edoxaban, rivaroxaban) is FDA approved for reversal of major bleeding with apixaban and rivaroxaban and available on a limited basis (Young).39 If the patient with SAH is taking any Factor Xa inhibitor, including unfractionated heparin or fondaparinux, PCC is recommended as a first-line agent for reversal, unless Andexanet alpha is indicated and available.
Regardless of anticoagulation mechanism, a pre-approved institutional protocol should be in place for rapid utilization, with input from hematology, blood bank, emergency medicine, and neurosurgery in order to most efficiently reverse anticoagulation. Other strategies to reduce risk of aneurysmal re-rupture are targeted toward controlling pain, nausea, and valsalva effect by treatment with analgesics, antiemetics, and stool softeners as needed. Fentanyl is a very effective and easily titratable analgesic, and is quickly titrated off to facilitate neurologic exams. Nimodipine, a calcium channel blocker used to improve outcome in SAH patients can be started in the ED, with caution given to the patient’s ability to swallow and the potential to inappropriately reduce BP.35 Other best practices include arterial-line BP monitoring, crystalloid to target euvolemia, and head of bed at 30° to protect against aspiration and to allow jugular venous outflow for ICP protection.
Many patients with SAH will require ventriculostomy drainage, either for hydrocephalus or periprocedurally to help with ICP complications. Antiepileptic medications may be recommended if the neurologic exam is poor, or the amount of blood is significant, portending risk of clinical or subclinical seizure. There has not been a definitive study to recommend any specific antiepileptic agent, as each has therapeutic benefits and risks and is ideally tailored by the patient’s profile. The most common agents are phenytoin (load 10–20mg/kilogram [kg] IV max: 50mg/min), fosphenytoin (10–20 phenytoin sodium equivalent (PE)/kilogram (kg) IV; infuse slowly over 30 min; max: 150mg PE/min) and levetiracetam (15–20mg/kg over 30 min).
These therapeutic modalities should be discussed with the admitting neurointensivist or neurosurgery team. Continuous electroencephalogram (EEG) monitoring may be started in the intensive care unit (ICU).
The ultimate therapeutic goal, once a bleeding aneurysm is identified, is to secure it surgically by coiling or clipping. While coiling is the preferred method,42 since it is less invasive than open surgical clipping, data are inconclusive as to whether long-term outcomes are better with either procedure, but guidelines suggest that coiling should be performed if both are possible.43,44 In some cases, tortuous vascular anatomy or other contraindications to coiling make open surgery necessary. Earlier treatment and securing the aneurysm is associated with lower risk of rebleeding.44 In the event that surgical treatment is delayed, antifibrinolytics such as aminocaproic acid may be used for a short period of time to mitigate the risk of re-rupture. Tranexamic acid and prothrombin complex concentrates have not been studied in this setting. This treatment modality is not backed by evidence, invokes risk of thrombosis, and is best discussed with the neurosurgery team.44,45
Once the aneurysm is secured, the greatest risk to patient outcome is that of vasospasm and delayed cerebral infarction (DCI). Many strategies are employed to assess for vasospasm, including hourly neurologic exam, strict euvolemia, continuous EEG, transcranial Doppler, permissive or induced hypertension, electrolyte monitoring and CT or direct angiography. If vasospasm is detected, timely treatment is paramount to decrease the risk of associated DCI. Treatment can be catheter-directed calcium channel blocker administration, such as nicardipine or verapamil, or vessel angioplasty.46
Our patient had a non-contrast CT 10 hours after onset of headache, which was negative for blood but positive for mild hydrocephalus. Hydrocephalus presents in 20–30% of SAH patients, and is generally thought to be a result of fibrotic changes associated with inflammatory reaction to blood at the arachnoid granulation. This can be suggestive of a pathologic process but is not diagnostic of SAH. The patient then consented for a LP, which showed RBCs in tubes 1 (14,000×10^6) through tube 4 (13,500×10^6). She was started on a nicardipine infusion for BP management and was given fentanyl for pain control. Neurosurgery was consulted, and she was admitted to the neuro ICU for hourly neurologic examinations and preparation for coiling the next day.
Many centers have access to neurosurgical coiling capability, as part of a comprehensive stroke center designation, but in some areas surgical clipping may be the only available procedure. Coiling is typically preferred, having shown better outcomes in the long run, but in some cases patient anatomy (tortuous vessels or plaque in carotid arteries) may preclude this procedure and the patient may need to be transferred to another center.42 Ventriculostomy placement prior to transfer will depend on the presence and severity of hydrocephalus, and should be discussed with the neurosurgery team.
The assessment and treatment of SAH is a dynamic and changing field, with the advent of advanced imaging, better understanding of pathophysiology and improved surgical techniques. SAH is rare but can be a devastating occurrence. Understanding the pathophysiology, demographics and risk factors helps to accurately evaluate the patient who presents to the ED with sudden-onset severe headache. The diagnostic strategy is key to decide which patients warrant full workup.
CT followed by LP is the standard diagnostic strategy, as per guidelines from ACEP, AHA and ASA, but many other advanced imaging options have come to the fore, making it important to understand the benefits and limitations of each diagnostic tool. Diagnostic sensitivity is critical, as a missed diagnosis of SAH can lead to increased mortality if the aneurysm re-bleeds. Shared decision-making can ensure that each patient understands risks and benefits. Disease recognition and prompt diagnosis is the primary responsibility of the emergency physician, while patient-specific management decisions are best made in a multi-disciplinary fashion.
CONCLUSION
Despite advances in the diagnosis and treatment of aneurysmal subarachnoid hemorrhage, mortality remains high. We are indebted to scholars who have contributed to the growing body of knowledge around aneurysmal SAH and appreciate that there is much more work to be done for this devastating disease.
Footnotes
Section Editor: Edward P. Sloan, MD, MPH
Full text available through open access at
Conflicts of Interest: By the West JEM article submission agreement, all authors are required to disclose all affiliations, funding sources and financial or management relationships that could be perceived as potential sources of bias. No author has professional or financial relationships with any companies that are relevant to this study. There are no conflicts of interest or sources of funding to declare.
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Primary School
Primary 1
P1.1 Numbers to 100
P1.1.1 Counting to 10
P1.1.2 Add & Subtract 1 to 10
P1.1.3 Digit
P1.1.4 Add & Subtract Numbers to 100
Addition
Addition & Subtraction Checkpoint
Addition of Length
P1.2 Ordinal Numbers
Primary 2
P2.1 Numbers to 1000
P2.1.1 Counting to 1000
P2.1.2 Spelling Numbers to 1000
P2.1.3 Ordering numbers to 1000
P2.2 Addition & Subtraction to 1000
P.2.2.3 Addition and Subtraction to 1000
P2.3 Multiplication and Division (2,3,4,5, and 10)
P2.4 Lengths & Mass
P2.4.1 Lengths
P2.6 Fractions Part 1
P2.6.1 What is a Fraction?
P2.6.3 Ordering Fractions
P2.7 Time
Clock Activity 2
P2.8 2D & 3D Shapes
P2.8.1 2D Shapes (2)
P2.10 Pictograms
Primary 3
P3.1 Numbers to 10 000
P3.3 Multiplication
P3.4 Division
P3.6 Time
P3.7 Fractions (Part 2)
P3.7.1 Equivalent Fractions
P3.9 Angles
P3.11 Perimeter & Area
P3.12 Bar Graphs
Workbook Tasks
P3. Revision
Mental Sums Quiz
Arithmetic Game
Mathjong Game
Primary 4
P4.1 Numbers to 100 000
P4.2 Rounding & Estimation
P4.3 Factors & Multiples
P4.5 Fractions
4.7 Decimals
P4.7.1 Decimals
4.7.3 Convert Fractions to Decimals
P4.9 Angles
P4.9.4 8 Point Compass
Cardinal Quiz
P4.5.2 Improper Fractions
Primary 5
P5.2 Order of Operations
P5.6 Rate
P5.7 Percentage
P5.8 Triangles
P5.8.1 Area of Triangle
P5.9 Volume
P5.9.1 Volume of Cube / Cuboid
P5.10 Angles
P5.10.3 Vertically Opposite Angles
P5.11 Quadrilaterals
Primary 6
6.4 Percentage
6.6 Angles
6.7 Circles
6.7.1 Estimating Pi
6.7.2 Circumference of Circle
6.7.3 Area of Circle Proofs
6.8 Volume of a Cuboid
Sec 1
1.1 Factors & Multiples
1.1.1 Natural Numbers
1.1.2 Factors
Amicable & Perfect Numbers
1.1.3 Prime Numbers
1.1.4 Prime Factorisation
1.1.5 Highest Common Factor
Euclidean Algorithm
1.1.6 Lowest Common Multiple
1.1.7 Squares and Cubes
(Olympiad) Trailing Zeroes
(Olympiad) Divisibility Tests
1.2 Real Numbers
1.2.1 Negative Numbers
Poem: A Positive Reminder
1.2.2 Integers
Palindromic Numbers
1.2.3 Rational Numbers
1.2.4 Irrational Numbers
1.2.5 BODMAS on Real Numbers
1.3 Approximation & Estimation
1.3.1 Rounding
1.3.2 Decimal Places
1.3.3 Significant Figures
1.4 Introduction to Algebra
1.4.0 History of Algebra
1.4.1 Algebraic Expressions
1.4.2 Like & Unlike Terms
1.4.3 Algebraic Expressions
1.4.4 Expand Algebraic Expressions
1.4.5 Factorise Algebraic Expressions
1.4.6 Linear Equations
1.4.7 Linear Inequalities
1.5 Introduction to Geometry
1.5.1 Definitions of Point & Lines
1.5.2 Angle Properties
1.5.3 Parallel Lines & Transversals
1.5.4 Line of Symmetry
1.5.5 Rotational Symmetry
1.5.6 Loci & Construction
1.5.6.1 More Loci
1.5.6.2. Voronoi Diagrams
NUSH Loci Problems
Challenging Questions (Geometry I)
(Enrichment) Centres of a Triangle
1.6 Triangles, Quadrilaterals & Polygons
1.6.1 Special Triangles
1.6.2 Triangle Properties
1.6.3 Types of Quadrilaterals
1.6.4 Types of Polygons
1.6.5 Polygon Properties
Checkpoint (Geometry)
Challenging Question (Polygon)
(Olympiad) Useful Property (2 Squares)
(Olympiad) Area of Dodecagon
(Olympiad) Viviani's Theorem
1.7 Ratio, Rate & Speed
1.7.1 Ratio
1.7.3 Speed
1.8 Percentages
1.8.1 Calculating Percentages
1.9 Number Patterns
1.10 Coordinate Geometry
1.10.0 Basic Definitions
1.10.1 Coordinates System
1.10.2 Gradient of a Straight Line
Optical Illusion
1.10.3 Equation of a Straight Line
1.10.3.1 Intercept & General Form
1.10.4 Drawing a Straight Line Graph
1.10.5 Real World Context
1.11 Perimeters and Areas of Plane Figures
1.11.1 Squares & Rectangles
1.11.2 Parallelograms
1.11.3 Rhombus
1.11.4 Trapezium
1.12 Volumes & Areas of Solids
1.12.1 Prism
1.12.2 Cylinders
(BL) Prisms and Cylinders
Nets of Cubes (all 11)
Cube Rotation & Shadows
(Enrichment) Platonic Solids
Truncated Hexahedron
Nets of Polyhedrons
(Enrichment) Net of a 4D Hypercube
(Enrichment) Olympiad Questions
Sec 2
2.1 Proportions
2.1.1 Direct Proportions
2.1.2 Inverse Proportions
2.1.3 Map Scales
2.2 Algebraic Expansion & Factorisation
2.2.1 Quadratic Identities
2.2.2 Quadratic Expansion
2.2.3 Quadratic Factorisation
2.2.4 Group Factorisation
(Olympiad) Mental Square Roots
2.3 Simple Algebraic Fractions
2.3.1 Simplify Algebraic Fraction
2.3.2 Product of Algebraic Fractions
2.3.3 Sum of Algebraic Fractions
2.3.4 Change Subject
2.4 Quadratic Equations
2.4.1 Solving Quadratic Equations (Factorisation)
2.5 Simultaneous Equations
2.5.2 Elimination Method
2.5.3 Graphical Solution
Cow & Chicken
2.6 Congruency & Similarity
2.7 Pythagoras' Theorem
2.7.1 Proof of Pythagoras' Theorem
2.7.2 Application of Pythagoras' Theorem
2.7.3 Converse of Pythagoras' Theorem
2.7.4 Beauty of Pythagoras' Theorem
(Olympiad) Corollaries & Challenging Problems
2.8 Trigonometric Ratios
2.8.1 Trigonometric Ratios
2.8.2 Inverse Trigonometric Functions
Challenging Questions
2.9 Mensuration of Pyramids, Cones & Spheres
2.9.1 Surface Area of Pyramids
2.9.2 Volume of Pyramids
2.9.3 Surface Area of a Cone
2.9.5 Surface Area of Sphere
2.9.6 Volume of Sphere
NUSH Y1 Mensuration Problems
2.10 Statistics Part 2
2.10.2 Stem & Leaf Diagrams
2.10.3 Histogram (Ungrouped)
Histogram vs Bar Chart
2.11 Probability
2.11.1 Sample Space
Sec 3
3.1 Indices
3.1.1 Laws of Indices
3.1.2 Index Law of Addition & Subtraction
3.1.3 Index Power Law
3.1.4 Index Distribution Law
3.1.5 Zero Index Law
3.1.6 Negative Index Law
3.1.7 Fractional Index Law
3.1.8 Solve Index Equations
3.1.9 Standard Form
Prefixes
3.1.10 Compound Interest
CPF
3.1.11 Exponential Graph
TDD Challenge
Password Math
3.2 Quadratics Equations
3.2.1 Completing the Square
Completing the Square Method
Graph of Complete-the-Square Form
3.2.2 Quadratic Formula
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17662 | https://www.quora.com/What-is-the-simplest-method-for-finding-a-remainder-in-modular-arithmetic | Something went wrong. Wait a moment and try again.
Modular Arithmetic
Operations (math)
Arithmetic Function
Remainder Problems (math)
Number Theory
Discrete Arithmetic
5
What is the simplest method for finding a remainder in modular arithmetic?
MUHAMM SHAHZAIB
The simplest method for finding a remainder in modular arithmetic is to use the modulus operator. In essence, if you want to find the remainder of ( a ) divided by ( n ), you would calculate ( a \mod n ). Here’s the basic formula:
This means you subtract from ( a ) the largest multiple of ( n ) that is less than or equal to ( a ). The result is the remainder when ( a ) is divided by ( n ).
For example, to find the remainder of ( 17 ) divided by ( 5 ), you calculate ( 17 \mod 5 ) which gives ( 2 ), because ( 17 - (3 \times 5) = 2 ).
For larger numbers or more complex calculat
The simplest method for finding a remainder in modular arithmetic is to use the modulus operator. In essence, if you want to find the remainder of ( a ) divided by ( n ), you would calculate ( a \mod n ). Here’s the basic formula:
remainder=a−([a/n]×n)
This means you subtract from ( a ) the largest multiple of ( n ) that is less than or equal to ( a ). The result is the remainder when ( a ) is divided by ( n ).
For example, to find the remainder of ( 17 ) divided by ( 5 ), you calculate ( 17 \mod 5 ) which gives ( 2 ), because ( 17 - (3 \times 5) = 2 ).
For larger numbers or more complex calculations, you might use additional properties of congruences, such as the Chinese Remainder Theorem or Fermat’s Little Theorem, to simplify the process before applying the modulus operator.
Related questions
How do you use modular arithmetic to find the remainder of $12^ {157}} $ when divided by $10 (modular arithmetic, math)?
What is the method for finding the smallest value of x in modular arithmetic?
What is the history of modular arithmetic?
What is the process for finding the order of a number in modular arithmetic?
What is the method for finding the remainder of a number when divided by another number using modular arithmetic?
Goh Kim Tee
Former Tutor at Private (non-agency) (2008–2017)
·
Author has 4.2K answers and 2.4M answer views
·
7y
How do I use modular arithmetic to find the remainder of a number^X power?
TQVM A2A ,the honour is mine.
e.g. 31^32/33
① No-theorem method ...advantage:no traps,nothing to memorise
= (33–2)^32
= (-2)^32 M32
≡ 2^32 M3
= 2²×(2^5)^6
=4×(33–1)^6
≡ 4×(-1)^6
≡ 4×1
≡ 4 M33
★★method
(ii)
(a) 31 ≡ 33–2 ≡ [-2] M33
(b) 31^5
= [-2]^5M33
≡ -32M33
≡ -(31)–1
≡ -[-2]-1
≡ 1 M33☜
(c)
31^30
=(31^5)^6
≡1 M33
(d) 31^32
≡ 31²×31^30
≡ [-2]²× M33
≡ (4)M33
② Fermat ‘s LITTLE Theorem FLT :
a^(n-1)≡ 1 (mod n) 【memorise 3,4,5→3^4≡ 1mod5】
Note the DIVISOR "n" MUST be PRIME ! , before anything else.
JUST as important, don’t be CONFUSED :
BASE "a" NEED NOT BE PRIME ! ,but MUST BE CO-PRIME with the must-be prime
TQVM A2A ,the honour is mine.
e.g. 31^32/33
① No-theorem method ...advantage:no traps,nothing to memorise
★method(i)
31^32
= (33–2)^32
= (-2)^32 M32
≡ 2^32 M3
= 2²×(2^5)^6
=4×(33–1)^6
≡ 4×(-1)^6
≡ 4×1
≡ 4 M33
★★method
(ii)
(a) 31 ≡ 33–2 ≡ [-2] M33
(b) 31^5
= [-2]^5M33
≡ -32M33
≡ -(31)–1
≡ -[-2]-1
≡ 1 M33☜
(c)
31^30
=(31^5)^6
≡1 M33
(d) 31^32
≡ 31²×31^30
≡ [-2]²× M33
≡ (4)M33
② Fermat ‘s LITTLE Theorem FLT :
a^(n-1)≡ 1 (mod n) 【memorise 3,4,5→3^4≡ 1mod5】
Note the DIVISOR "n" MUST be PRIME ! , before anything else.
JUST as important, don’t be CONFUSED :
BASE "a" NEED NOT BE PRIME ! ,but MUST BE CO-PRIME with the must-be prime divisor "n" 【 no common factor except 1 】
As it is FLT not applicable for 31^32/33 【33 is not prime】
③ ONE way to overcome "non-prime divisor" is via EULER’S TOTIENT FUNCTION φ(n) THEOREM【ETFT】 .
THEN a^{φ(n)} ≡ 1 mod n
BUT THEN a & n must be co-prime, another stumblimg block :
φ(n)=φ(33)=φ(prime p×prime q)
=n×{(p-1)/p}{(q-1)/q}
=(p-1)/(q-1)
=φ(3×11)
2×10=20
∴ 31^20≡1 mod33
31^32=31^12×31^20≡31^12
As you can see ETFT only reduce the power ...
31^12
= (33-2)^12
≡ 2²×2^10
= 4×(2^5)²
= 4×32²
= 4×(33–1)²
≡ 4 m33
④ ANOTHER way to bypass the non-prime/co-prime problem is the" factor out" rule :【not applicable for 31^32/33,as 31 is prime】
e.g. 44^32M33
={(4×11)×44^31(MOD3×11)}
≡ 11{4×44^31(M3)}
【FLT/ETFT→44^2 ≡1M3←44^30=(44²)^15 ≡1M3】
≡ 11{4×44×44^30(mod3)}
=11{4×44(M3)}
≡ 11{(3+1)(42+2)(M3)}
≡ 11{1×2M(3)}
≡ {2×11M(3×11)}
≡ 22 M33 ...beautiful answer
ADDED :
⑤ Pattern cycles method.
31^1≡(33–2)≡-2 M33
31²≡(-2)²≡4M33
31³ ≡(-2)³≡ -8 M33≡
31^4≡(-2^)4≡16
31^5=(-2)^5=-32=-32+33≡1M33☜【end of first cycle of order 5】
HENCE ,the pattern is :
31^15≡31^10≡31^5≡1M33≡31^30
31^11≡31^1≡31^6≡(-2) M33≡31^31
31^12≡31²≡31^7≡4M33≡31^32☜♦
31^13≡31³≡31^8≡-8M33
31^14≡31^4≡31^9≡16M33
WRITE OUT THE PATTERN:
{31^1 , 31² , 31³ , 31^4 , 31^5 }
{31^6 , 31^7 , 31^8 , 31^9 , 31^10}
{...............................,31^14 , 31^15}
{..................................31^29 , 31^30}
{31^21 , 31^32 , 31^33 , 31^34 , 31^35 }
{ -2 , 4 , (-8) , 16 , 1}
{ -2 , 4 , (-8) , 16 , 1}.........
∴31^(5n) ≡ 1 (mod33),n∈N
∴ 31^30 ≡ 1 M33
31^32 ≡ 4 M33
PS :FOR EXAM ,it’s difficult to present this type of answer...highly NOT recommended .
⑥ Continuos squaring method (used in ①)
e.g. 7^(1,234) mod100≡last two digits of 7^8
7^4≡(7²)²=49²=(50–1)²=2,500–100+1≡1 (mod100)
∴7^(4n)≡1 M100 , n∈ N7
7^(1,232)=(7^4)^(308)≡ 1 M100
7^(1,234)≈7²×7^(1,232)≡ 49M100
Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994)
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 10.6K answers and 23.6M answer views
·
10mo
Related
Why division is not allowed as a modular arithmetic operation?
Division appears in modular arithmetic as multiplication by the multiplicative inverse.
That’s what division usually is, e.g. 2/7=2×17.
That of course can be viewed as an algebra multiplication problem.
x=2/7 is the solution to 7x=2 in the rational numbers.
Similarly, we might like to solve the congruence
7x≡2mod11
That means we’re looking for the integers x that when multiplied by 7 give a remainder of 2 when divided by 11.
We want to divide both sides by 7, meaning we want to multiply by the multiplicative inverse, aka the reciprocal, of 7(mod11). There’s only 11 numb
Division appears in modular arithmetic as multiplication by the multiplicative inverse.
That’s what division usually is, e.g. 2/7=2×17.
That of course can be viewed as an algebra multiplication problem.
x=2/7 is the solution to 7x=2 in the rational numbers.
Similarly, we might like to solve the congruence
7x≡2mod11
That means we’re looking for the integers x that when multiplied by 7 give a remainder of 2 when divided by 11.
We want to divide both sides by 7, meaning we want to multiply by the multiplicative inverse, aka the reciprocal, of 7(mod11). There’s only 11 numbers to try, 0 through 10. For small moduli, the most expedient way to find multiplicative inverses is to guess them. See if you can figure out the inverse of 7 before continuing.
The theorem is which is mod arithmetic on the integers from to is a field for prime which in particular means every non-zero element has a multiplicative inverse. For composite finite rings , we get a multiplication group, so multiplicative inverses, for the elements relatively prime to . The elements with common factors with won’t have reciprocals.
We seek the inverse of 7 mod 11. That’s the solution to the congruence
which is the equation which is called a Linear Diophantine Equation as we require and to be integers.
There’s a long established theory of these going back to Euclid; the workhorse of the field is Euclid’s GCD algorithm, which gives us the simple continued fraction expansion of 7/11; we find:
The numerators and the fraction bars are boilerplate, so we abbreviate that:
We can twiddle the last partial quotient to get an even number of them,
The convergents are what we get stopping early,
The magic of convergents is the cross product which basically means is the best possible approximation to for the denominator size.
We only needed the last one, the penultimate convergent to give us the reciprocal of 7 mod 11, namely 8, because it says:
That lets us solve our congruence,
Answer:
That’s 5, 16, 27, 38, etc.
Check:
In the case where we can just divide without remainders, feel free:
we can immediately write
just like if we were clever enough we could write
Carlos Eŭ Th
Triple IMO bronze medalist
·
Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 5.9K answers and 4.4M answer views
·
7y
Related
What is the simplest way to compute the remainder
in
when
is large and
is small using a computer?
If you have a language such as Python, which can handle large integers, limited only to its resources, then you can write:
r = l % s
A similarly capable language that used Pascal-like syntax would be
r := l MOD s
Now, if
is very large, you don’t have a language that operates with large integers, but have
written as a text string (in base 10), and
is small enough.
If
is coprime with 10, then find
for Euler’s totien function. Now, divide in groups of
digits, right to left. Add those chunks. Repeat until you have
digits or less. Then operate with the result.
2. If
s
is not coprime
If you have a language such as Python, which can handle large integers, limited only to its resources, then you can write:
r = l % s
A similarly capable language that used Pascal-like syntax would be
r := l MOD s
Now, if is very large, you don’t have a language that operates with large integers, but have written as a text string (in base 10), and is small enough.
If is coprime with 10, then find for Euler’s totien function. Now, divide in groups of digits, right to left. Add those chunks. Repeat until you have digits or less. Then operate with the result.
If s is not coprime with 10, it means that for coprime with 10. Have and do as above, and have the reminder modulo . ().
If Then take the last digits and find the reminder modulo . ().
If Then take the last digits and find the reminder modulo . ().
If , then the reminder modulo are the last digits. ().
If . Then the last digits are the reminder modulo . (). Take the next digits and find the reminder modulo . (). With and find . Analogously if .
With this reminders you can calculate the reminder.
Note that this concept is similar if is represented as a string in any other base, just that you have to check if is coprime or not with such base. If has a binary representation, you can use base 16 (nibbles), base 256 (bytes) or even base 2 (bits).
This is if you want to program that. There are already open libraries that deal with large integers for C, C++ and other languages. Find them and check their syntax.
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Related questions
What is the method for finding the value of
math! [/math]
in modular arithmetic?
Why is modular arithmetic useful?
What is the method for finding the period of modular arithmetic?
Where is modular arithmetic applied?
How is modular arithmetic used in cryptography?
Gary Russell
Former Professor at University of Iowa (1996–2025)
·
Author has 6K answers and 3M answer views
·
Updated Mar 10
Related
How do you use modular arithmetic to find the remainder of $12^ {157}} $ when divided by $10 (modular arithmetic, math)?
Note that
12^157 (mod 10) = 2^157 (mod 10)
Also, in mod 10,
x^5 = x (mod 10) for all x
Now, note that
157 = (31)(5) + 2
So,
2^157 = (2^31)(2^2) (mod 10) = 2^33 (mod 10) = (2^6)(2^3) (mod 10) = 2^9 (mod 10) = 2^5 (mod 10) = 2 (mod 10)
So,
12^157 = 2 (mod 10)
The remainder is 2.
UPDATE : A METHOD USING MODULAR ARITHMETIC THEOREMS
Note that
12^157 (mod 10) = 2^157 (mod 10)
Also, note that 10 = (2)(5) and both are coprime. We compute 2^157 mod 2 and 2^157 mod 5. Then, we combine the answers.
By inspection,
2^157 mod 2 = 0 mod 2
Now, 5 is coprime with 2 and 5 is prime. By Fermat’s Little Theorem,
2^4k = 1 mod 5
Also,
Note that
12^157 (mod 10) = 2^157 (mod 10)
Also, in mod 10,
x^5 = x (mod 10) for all x
Now, note that
157 = (31)(5) + 2
So,
2^157 = (2^31)(2^2) (mod 10) = 2^33 (mod 10) = (2^6)(2^3) (mod 10) = 2^9 (mod 10) = 2^5 (mod 10) = 2 (mod 10)
So,
12^157 = 2 (mod 10)
The remainder is 2.
UPDATE : A METHOD USING MODULAR ARITHMETIC THEOREMS
Note that
12^157 (mod 10) = 2^157 (mod 10)
Also, note that 10 = (2)(5) and both are coprime. We compute 2^157 mod 2 and 2^157 mod 5. Then, we combine the answers.
By inspection,
2^157 mod 2 = 0 mod 2
Now, 5 is coprime with 2 and 5 is prime. By Fermat’s Little Theorem,
2^4k = 1 mod 5
Also,
157 = 1 mod 4 = 4k + 1
And so,
2^157 mod 5 = 2^(4k+1) mod 5 = 2 mod 5
Because 2 and 5 are coprime, we can combine the boldface equations using the Chinese Remainder Theorem. We write
2^157 = r mod 10
r = 2p = 5q + 2
5q = -2 mod 2 = 0 mod 2
q = 0, r = 2 and
2^157 mod 10 = 2 mod10
12^157 mod 10 = 2^157 mod 10 = 2 mod 10
as before.
Harald Overbeek
loves mental calculation
·
Author has 185 answers and 626.4K answer views
·
9y
Related
What are some well-known uses of modular arithmetic?
Checking answers of calculations.
Nine proof or casting out nines:
One of the most useful ways of using modular arithmetic is by checking the answers of long additions, subtractions, multiplications, divisions, etc., etc.
The check is done, not by repeating the calculation, but by doing a reduced version of the same calculation. it is sometimes called the nine proof. An eleven proof also exists.
An example will make this clear.
Let's do a multiplication:
17 X 31 = 527
Now we check it by reducing the numbers to a single digit:
17 = 1+7 = 8
31 = 3 + 1 = 4
527 = 5 + 2 + 7 = 14. 14 = 1+4 = 5
Now do the m
Checking answers of calculations.
Nine proof or casting out nines:
One of the most useful ways of using modular arithmetic is by checking the answers of long additions, subtractions, multiplications, divisions, etc., etc.
The check is done, not by repeating the calculation, but by doing a reduced version of the same calculation. it is sometimes called the nine proof. An eleven proof also exists.
An example will make this clear.
Let's do a multiplication:
17 X 31 = 527
Now we check it by reducing the numbers to a single digit:
17 = 1+7 = 8
31 = 3 + 1 = 4
527 = 5 + 2 + 7 = 14. 14 = 1+4 = 5
Now do the multiplication again, using the single digits:
8 X 4 = 32. 3 + 2 = 5
The answer to the original multiplication, 527, reduced is 5. The reduced calculation also leads to 5, so the answer is correct.
This is an example of modulo 9 arithmetic. 17 modulo 9 = 8, because 17/9 = 1 rest 8.
31 modulo 9 = 4, because we can subtract 27, a multiple of nine and that leave a rest of 4.
The modulo nine of a number can be calculated by adding the digits of a number together. If the result is higher than 9 as was the case in 527, repeat the process until we have a number between zero and nine.
Eleven proof:
The same calculation can be done using modulo 11 arithmetic:
17 mod 11 = 6 (subtract 11 from 17)
31 mod 11 = 9 (subtract 22 from 31)
6 X 9 = 54 and 54 mod 11 = 10 (subtract 44 from 54)
527 mod 11 = 10 (527 - 440 = 87 and 87 - 77 = 10)
17 reduced is 6 and 31 reduced is 9. 6 X 9 = 54 and 54 reduces is 10.
Since 99 is a factor of 11, we can calculate the modulo of 527 as 27+5 = 32 and 32 - 22 = 10.
This is not as easy as the modulo 9 calculation. However; with some practice the calculations go quick.
Clock calculations:
If the time is 11 o' clock and you need to be somewhere in 4 hours, you know that that meeting is at 3 o' clock.
You have added 4 to 11 to get 15 and then did a modulo 12 to arrive at 3.
Date calculations:
If the date is 29th of January and you have an appointment in 4 days, you know it is on the 2nd of February.
You have added 4 to 29 to arrive at 33 and then did a modulo 31, for Januari has 31 days.
Other regular uses:
Checksums in bank account and IBAN numbers. Finds typo's in typed in numbers.
Checksums in files. Finds out if a file is transferred well form one place to another.
etc., etc.
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Grant Molnar
PhD in Mathematics & Number Theory, Dartmouth College (Graduated 2023)
·
7y
Related
Where is modular arithmetic applied?
What time is it?
7:00 eh? And there are 24 hours in a day? So yesterday it was -17:00?
No? Why not?
To avoid uselessly large numbers that tell us nothing about what time of day it is, our calendars system uses modular arithmetic on the hours of the day. You add 24 hours, and you get the same time. (You add 12 hours, and you switch from AM to PM or back, if you use that sort of time) So you are using modular arithmetic every time you look at a clock.
“Maybe the clock makers used modular arithmetic, but I just check what time it is.” I could debate that, but alright. Let’s do something else then.
Wha
What time is it?
7:00 eh? And there are 24 hours in a day? So yesterday it was -17:00?
No? Why not?
To avoid uselessly large numbers that tell us nothing about what time of day it is, our calendars system uses modular arithmetic on the hours of the day. You add 24 hours, and you get the same time. (You add 12 hours, and you switch from AM to PM or back, if you use that sort of time) So you are using modular arithmetic every time you look at a clock.
“Maybe the clock makers used modular arithmetic, but I just check what time it is.” I could debate that, but alright. Let’s do something else then.
What is 4+7?
11, right! If you didn’t have that memorized, you compute it effectively by doing modular arithmetic. 4+7 is 1 mod 10, so you should have a 1 in the 1’s place, but it is also between 10 and 20, so the answer is 11. Doing basic arithmetic quickly goes quicker with modular arithmetic as a trick up your sleeve.
But modular arithmetic is used in more exotic ways too. Suppose you are a computer chip designer, and you have built a chip that has three interchanable processing points and a lot of requests to use each of them. You want to divide those requests up evenly, but be able to recall where any given request went without wasting a lot of memory storing that information.
What do you do?
You consider those requests’ labels mod 3, and then send each one to the appropriate processing center. If you need to figure out where you sent a request, the info is sitting there where in the label, no extra storage required. Modular arithmetic is fast to compute, which is a major boon to computer scientists.
Or how about this! You want to withdraw some money from the bank online. The bank doesn’t mind you having the money, but it doesn’t want someone else stealing your money. But your communication goes over public channels. What do you do? The classic solution is to use the RSA algorithm, which allows you to publicly send information to the bank that no one outside the bank can decode, using a public key. Explaining how the algorithm works would take us too far afield, but what your computer does is take the information you want to send, raise it to a power the bank gave you but reduce mod N for an N the bank gave you as well, and then pass the resulting garbled nonsense over the wire (the inverse log problem is hard, and so is factoring N, and that’s where the security of this algorithm comes from). Then the bank raises your answer to another power and reduces mod N, and your request materializes clearly for them. Modular arithmetic lies at the heart of security for national and international finance.
None of that is why I care about it though. As a number theorist, modular arithmetic gives me a way to verify a given polynomial equation has no solutions. It lets me easily get p-adic information about all manner of power series. In my most recent paper, it told me what prime levels I could prove to have Zagier duality using a particular method I uncovered. It also lets me construct a variety of fundamental examples of groups and rings. Modular arithmetic is one of the most basic and important tools a pure mathematician has in their toolbox.
Jonas
Upvoted by
Justin Rising
, PhD in statistics · Author has 2.4K answers and 6.5M answer views
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6y
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Why do computers find modular arithmetic easy?
Here is how computers store numbers:
This can store numbers from 0 to 9,999. What happens if you add 1? You should get 10,000, right?
But of course what you get is 0,000. Which happens to be congruent to 10,000 modulo 10,000. The first digit gets just lost.
This counter doesn’t find modulo arithmetic easy. It can only do modulo arithmetic, and only modulo 10,000. It can’t do anything else.
Computers work roughly the same way, but of course with binary digits instead of decimal digits.
Here is how computers store numbers:
This can store numbers from 0 to 9,999. What happens if you add 1? You should get 10,000, right?
But of course what you get is 0,000. Which happens to be congruent to 10,000 modulo 10,000. The first digit gets just lost.
This counter doesn’t find modulo arithmetic easy. It can only do modulo arithmetic, and only modulo 10,000. It can’t do anything else.
Computers work roughly the same way, but of course with binary digits instead of decimal digits.
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Aarav Agrawal
Studied at Atomic Energy Central School, No. 4, Mumbai (Graduated 2023)
·
Author has 92 answers and 45.5K answer views
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4y
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What is the remainder when 5^18 is divided by 19 using modular arithmetic?
Sorry for bad handwriting, but your answer should be 1 according to this
Sorry for bad handwriting, but your answer should be 1 according to this
Aaron Welles
BS in Mathematics
·
Author has 54 answers and 113.4K answer views
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9y
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Why division is not allowed as a modular arithmetic operation?
Think of 'division' as 'inverse multiplication'. To divide by a number by [math]a[/math] we multiply the number by the multiplicative inverse of [math]a[/math].
So what's a multiplicative inverse? For any number [math]a[/math], its multiplicative inverse, [math]a'[/math], is the number for which [math]a \cdot a' = 1[/math]. In other words, multiplying a number by its inverse results in [math]1.[/math]
Example: Consider the real numbers with normal addition, multiplication, etc. Let's try dividing a number [math]r[/math] by [math]5[/math]. Instead of carrying out division we can multiply [math]r[/math] by the multiplicative inverse of [math]5[/math], which is [math]1/5[/math], because [math] 5 \cdot \frac{1}{5} = 1[/math]. This gives us this obvious
Think of 'division' as 'inverse multiplication'. To divide by a number by [math]a[/math] we multiply the number by the multiplicative inverse of [math]a[/math].
So what's a multiplicative inverse? For any number [math]a[/math], its multiplicative inverse, [math]a'[/math], is the number for which [math]a \cdot a' = 1[/math]. In other words, multiplying a number by its inverse results in [math]1.[/math]
Example: Consider the real numbers with normal addition, multiplication, etc. Let's try dividing a number [math]r[/math] by [math]5[/math]. Instead of carrying out division we can multiply [math]r[/math] by the multiplicative inverse of [math]5[/math], which is [math]1/5[/math], because [math] 5 \cdot \frac{1}{5} = 1[/math]. This gives us this obvious identity:
[math]r\div 5 = r \cdot \frac{1}{5}[/math].
Unfortunately, finding the multiplicative inverse in modular arithmetic is not so obvious, and, in fact, some numbers may not have inverses, while the others do. And numbers without inverses might divide some numbers, but not others. This is why you can't, in general, divide in modular arithmetic.
Example 1: Though 3 has no inverse (mod 12), 3 divides 9.
[math]\quad 9\div 3=3[/math] (mod 12).
Example 2: [math]7\div 3[/math] mod 12 cannot be computed. There is no number [math]x[/math] such that [math]3\cdot x = 7[/math] mod 12.
If an inverse exists for a number [math]a[/math], however, then we can divide any number by [math]a[/math].
Condition: The multiplicative inverse of a number [math]a[/math] exists (modulo [math]m[/math]), if and only if [math]a[/math] and [math]m[/math] are coprime (i.e., if gcd([math]a[/math], [math]m[/math]) = [math]1[/math]).
Where that condition comes from is not obvious, but consider the special case when [math]m[/math] is prime. Then gcd([math]a[/math],[math]m[/math])=[math]1[/math] for all [math]a[/math]. That is, if your modding by a prime number then every (non-zero) element will have a multiplicative inverse. And so you can divide at will. (Except by zero! [math]0[/math] has no inverse!).
Example: Let's mod by 7 and compute the inverses:
[math]\begin{align} \text{inverse of }0&= \text{NONE},\ \text{inverse of }1&= 1, \quad\text{(1 is its own inverse)}\ \text{inverse of }2&= 4, \quad\text{(since } 2\cdot 4=8=1 \text{ mod 7)}\ \text{inverse of }3&= 5, \quad\text{(since } 3\cdot 5=15=1 \text{ mod 7)}\ \text{inverse of }4&= 2,\ \text{inverse of }5&= 3,\ \text{inverse of }6&= 6, \quad\text{(since } 6\cdot 6=36=1 \text{ mod 7)}\ \end{align}[/math]
Which happens to give a neat property that multiplying a number by [math]6[/math] and dividing it by [math]6[/math] give the same result mod 7. And, since every non-zero element has an inverse, this is a Division ring, in which you can divide any number by any other non-zero number.
Links:
Modular multiplicative inverse
Division ring
Henry Zhu
B.S. in Computer Science, University of California, Davis (Graduated 2022)
·
Upvoted by
Gilbert Doan
, Master (Unattempted) Mathematics, San Jose State University (2018) · Author has 290 answers and 778.9K answer views
·
Updated 5y
Related
What is modular arithmetic?
Modular arithmetic concentrates on using arithmetic with numbers that involve modulus operators.
The foundation of modular arithmetic is congruence. If [math]a, b \in \mathbb{Z}[/math] (a and b are integers) and [math]m \in \mathbb{N}[/math] (m is a natural number), we can state:
[math]a \equiv b (mod\enspace m)[/math] if [math]m|(a-b)[/math] ([math]m[/math] is a divisor of math[/math])
In other words, [math]a[/math] is congruent to [math]b (mod\enspace n)[/math].
For example:
[math]20 \equiv 4 (mod\enspace 4)[/math], because [math]4|(20–4)[/math]
There are three properties in modular arithmetic that depend on congruence, given the following constraints: [math]a, b, c, d \in \mathbb{Z}[/math] ([math]a[/math], [math]b[/math], [math]c[/math], and [math]d[/math] are integers) and [math]m \i[/math]
Modular arithmetic concentrates on using arithmetic with numbers that involve modulus operators.
The foundation of modular arithmetic is congruence. If [math]a, b \in \mathbb{Z}[/math] (a and b are integers) and [math]m \in \mathbb{N}[/math] (m is a natural number), we can state:
[math]a \equiv b (mod\enspace m)[/math] if [math]m|(a-b)[/math] ([math]m[/math] is a divisor of math[/math])
In other words, [math]a[/math] is congruent to [math]b (mod\enspace n)[/math].
For example:
[math]20 \equiv 4 (mod\enspace 4)[/math], because [math]4|(20–4)[/math]
There are three properties in modular arithmetic that depend on congruence, given the following constraints: [math]a, b, c, d \in \mathbb{Z}[/math] ([math]a[/math], [math]b[/math], [math]c[/math], and [math]d[/math] are integers) and [math]m \in \mathbb{N}[/math] (m is a natural number).
If [math]a \equiv b (mod\enspace m)[/math] and [math]c \equiv d (mod\enspace m)[/math]
[math]a + c \equiv b + d (mod\enspace m)[/math]
[math]a c \equiv b d (mod\enspace m)[/math]
[math]a k \equiv b k (mod\enspace m)[/math]
These basic properties allow us to solve all sorts of different modular arithmetic problems—problems that seem daunting at first.
For example: Find the remainder when [math]3^{254}[/math] is divided by [math]7[/math]
In this problem, we start out by finding the first non-zero power congruent to [math]1 (mod\enspace 7)[/math]. In this case, it is [math]3^6 = 729 \equiv 1 (mod\enspace 7)[/math]. We can multiply this expression by itself through the second property to get:
[math]3^6 3^6 \equiv 1 1 (mod\enspace 7)[/math]
which simplifies to:
[math]3^{12} \equiv 1 (mod\enspace 7)[/math]
We can continuously multiply this expression by [math]3^6 \equiv 1 (mod\enspace 7)[/math] to get the following expression:
[math]3^{252} \equiv 1 (mod\enspace 7)[/math]
The following congruence will help tie the knot for us to solve this problem:
[math]3^2 \equiv 2 (mod\enspace 7)[/math]
Multiplying [math]3^{252} \equiv 1 (mod\enspace 7)[/math] and [math]3^2 \equiv 2 (mod\enspace 7)[/math] through the second property gives us
[math]3^{252} 3^2 \equiv 1 2 (mod\enspace 7) [/math]
[math]3^{254} \equiv 2 (mod\enspace 7)[/math]
Thus, the remainder when [math]3^{254}[/math] is divided by [math]7[/math] is 2.
Bernard Montaron
PhD in Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980)
·
Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 3.2K answers and 2.1M answer views
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2y
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What is the definition of modular arithmetic? What is an example of a problem that can be solved using modular arithmetic?
Modular arithmetic is the arithmetic of integers modulo [math]m[/math], some integer generally positive and greater than 1, where integers [math]n[/math] are grouped in [math]m[/math] classes represented by the remainder [math]r[/math], called the residue, in the Euclidean division of [math]n[/math] by [math]m[/math], i.e. [math]n=r+qm[/math] with [math]0\leq r\leq m-1[/math].
All integers having the same residue [math]r[/math] modulo [math]m[/math] are considered equivalent, and belong to the equivalence class represented by [math]r[/math]. An addition and multiplication table can be defined modulo [math]m[/math] , for example math+1=0[/math].
There is a huge number of applications of modular arithmetic. I’ll give two examples.
First, how do you prove th
Modular arithmetic is the arithmetic of integers modulo [math]m[/math], some integer generally positive and greater than 1, where integers [math]n[/math] are grouped in [math]m[/math] classes represented by the remainder [math]r[/math], called the residue, in the Euclidean division of [math]n[/math] by [math]m[/math], i.e. [math]n=r+qm[/math] with [math]0\leq r\leq m-1[/math].
All integers having the same residue [math]r[/math] modulo [math]m[/math] are considered equivalent, and belong to the equivalence class represented by [math]r[/math]. An addition and multiplication table can be defined modulo [math]m[/math] , for example math+1=0[/math].
There is a huge number of applications of modular arithmetic. I’ll give two examples.
First, how do you prove that the equations [math]x^3+y^3+z^3=4+9k[/math] or [math]x^3+y^3+z^3=5+9k[/math] have no solutions in integers math[/math] for any integer [math]k[/math] ?
This is done by calculating the residues of a cube modulo 9, it’s easy to see that whatever the integer [math]x[/math] is, the only possible residues of [math]x^3[/math] modulo 9 are -1 (i.e. 8), 0, 1. Hence, the only possible residues of [math]x^3+y^3+z^3[/math] modulo 9 are [math]0,\pm 1, \pm 2, \pm 3[/math] i.e. [math]0, 1, 2, 3, 6, 7, 8[/math]. Since neither 4 or 5 belong to this set, this proves the result.
My second example, is how do you prove that there is an infinity of integers that cannot be expressed as a sum of less than 15 powers of 4 of integers?
This is proved by considering the residues of [math]x^4[/math] modulo 16. If [math]x[/math] is even, then [math]x^4[/math] is a multiple of 16, and the residue is 0 modulo 16. If [math]x[/math] is odd, i.e. [math]x=2k+1[/math] then [math]x^2=4k(k+1)+1[/math] and
and since [math]k(k+1)[/math] is always even, we obtain the residue 1 modulo 16.
This implies directly that an integer of the form [math]16k+15[/math] cannot be expressed as a sum of less than 15 powers of 4, which completes the proof.
Madeline Griswold
Mathematician
·
Upvoted by
Gilbert Doan
, Master (Unattempted) Mathematics, San Jose State University (2018)
·
7y
Related
What is modular arithmetic?
The study of math that has to do with cycles of numbers or remainders.
The most commonly used example of modular arithmetic is the clock (a mod 12 system).
Using modular arithmetic, you can question what time will it be after 67 hours. This is written as 12 mod 67. In order to calculate this you divide 67 by 12 and the remainder is the answer. 12 mod 67 is 7
Modular arithmetic is applicable to many different areas. Our current counting system is a base 10 system, but using another system (such as base 12) involves modular arithmetic. Additionally, much of cryptography (code breaking) involves mod
The study of math that has to do with cycles of numbers or remainders.
The most commonly used example of modular arithmetic is the clock (a mod 12 system).
Using modular arithmetic, you can question what time will it be after 67 hours. This is written as 12 mod 67. In order to calculate this you divide 67 by 12 and the remainder is the answer. 12 mod 67 is 7
Modular arithmetic is applicable to many different areas. Our current counting system is a base 10 system, but using another system (such as base 12) involves modular arithmetic. Additionally, much of cryptography (code breaking) involves modular arithmetic.
And my favorite part of it is that while it sounds super smart and professional, the basic concept is merely finding the remainder.
Related questions
How do you use modular arithmetic to find the remainder of $12^ {157}} $ when divided by $10 (modular arithmetic, math)?
What is the method for finding the smallest value of x in modular arithmetic?
What is the history of modular arithmetic?
What is the process for finding the order of a number in modular arithmetic?
What is the method for finding the remainder of a number when divided by another number using modular arithmetic?
What is the method for finding the value of
in modular arithmetic?
Why is modular arithmetic useful?
What is the method for finding the period of modular arithmetic?
Where is modular arithmetic applied?
How is modular arithmetic used in cryptography?
What is the explanation for the sum of all remainders being zero after division by four in modular arithmetic?
How do you determine the remainder when (modular arithmetic, math)?
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Matchstick Puzzles
Subject: Mathematics
Age range: 7-11
Resource type: Worksheet/Activity
Mammathematicians
4.56 30 reviews
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These are really fun starter activities for students, especially those who find maths difficult and struggle to enjoy it. This activity is perfect for the run up to the holidays and can be used with any class, primary or secondary. Several of the puzzles also have multiple solutions, so this isn’t just a “who can finish first activity”.
This presentation contains over 3 0 puzzles of varying difficulty, each one with its own animated solution.
If you do download this presentation, please leave feedback on how I can improve them. Or if you would like me to increase the number of puzzles, feel free to post a link to the puzzle you’d like me to put in.
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Such a cool resource. Nice one! Definitely going to use these with my students. They'll love it! Thanks for posting it.
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aimee_lisa
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Wow. that animation at the beginning of the PowerPoint! That must have taken ages. Well done. The puzzles are good and there are a lot of different ones, all with solutions (thank goodness). My only criticism is that it runs a bit slow, not sure why that is - but it isn't just my computer.
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pjackson2009
7 years ago
Yes, the animation did take a long time. But it was fun to make. It i my view that activities should come with solutions where possible. It saves everyone so much time! Sorry that it is running a bit slow. I'm not sure why that is. Perhaps saving the PowerPoint as two separate documents will help (cut and paste half the slides to another one).
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17664 | https://www.youtube.com/watch?v=c6bx_fn6W2Q | How To Find, Calculate The Original List Price Of A Product From A Series Discount Explained
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In this video we discuss how to find the original list price of a product from a series discount in business. We go through an example and cover the formula used to do the calculation
Transcript/notes
In business, a trade discount is the amount a manufacturer reduces the list price of a product, when it sells to a reseller.
Series discounts, also called chain discounts are basically multiple trade discounts. They are often written as a number, a forward slash then another number, such as 20/10. This example means there is a 20% trade discount applied to the list price, then a separate 10% discount is applied to the 20% discounted price.
In a series discount, the net cost equivalent is the complements of each of the discounts in the series times each other. As an example, in our 20/10 series discount, (show 20% and 10%), the complement of 20%, the first discount is 1 minus .2, which equals .8. And the complement of 10%, the second discount is 1 minus .1, which equals .9. Now we multiply these complements, which equals .72, and that is the net cost equivalent.
If we are given the net cost of a product, which is the cost of the product after all the series discounts are applied, and we know the series discount, such as 20/10, we can find the original list price of the product.
The formula for this is, the list price equals, the net cost divided by the net cost equivalent.
As an example, lets say that a given product has a net cost of $243.20 after a series discount of 20/20. What is the original list price of the product?
To answer this, we first need to calculate the net cost equivalent. The net cost equivalent equals, the complement of 20, which is 1 minus .2, so, .8, times the complement of the second discount, also 20, so, again, 1 minus .2, or .8.
Now we have .8 times .8, which equals .64 as the net cost equivalent.
Now we can use the formula. The list price equals, $243.20, the net cost, divided by .64, the net cost equivalent, and this calculates to $380 as the original list price of the product.
Timestamps
0:00 What are trade and series discounts
0:27 Net cost equivalent explained
0:55 Finding the original list price
1:13 Example of finding the list price
1 comments
Transcript:
What are trade and series discounts In business, a trade discount is the amount a manufacturer reduces the list price of a product, when it sells to a reseller. Series discounts, also called chain discounts are basically multiple trade discounts. They are often written as a number, a forward slash then another number, such as 20/10. This example means there is a 20% trade discount applied to the list price, then a separate 10% discount is applied to the 20% discounted price. Net cost equivalent explained In a series discount, the net cost equivalent is the complements of each of the discounts in the series times each other. As an example, in our 20/10 series discount, the complement of 20%, the first discount is 1 minus .2, which equals .8. And the complement of 10%, the second discount is 1 minus .1, which equals .9. Now we multiply these complements, which equals .72, and that is the net cost equivalent. Finding the original list price If we are given the net cost of a product, which is the cost of the product after all the series discounts are applied, and we know the series discount, such as 20/10, we can find the original list price of the product. The formula for this is, the list price equals, the net cost divided by the net cost equivalent. Example of finding the list price As an example, lets say that a given product has a net cost of $243.20 after a series discount of 20/20. What is the original list price of the product? To answer this, we first need to calculate the net cost equivalent. The net cost equivalent equals, the complement of 20, which is 1 minus .2, so, .8, times the complement of the second discount, also 20, so, again, 1 minus .2, or .8. Now we have .8 times .8, which equals .64 as the net cost equivalent. And now we can use the formula. The list price equals, $243.20, the net cost, divided by .64, the net cost equivalent, and this calculates to $380 as the original list price of the product. Alright my friends, hopefully you got something out of this video, I do have more videos right there for you, till next time, I am outta here. |
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If a,b are relatively prime proof and ab is a perfect square, so are a, b
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Prove that if a and b are relatively prime integers and ab is a perfect square so are a and b. Show by counterexample that the relatively prime condition is necessary.
I dont know how to start this proof. Also the second "counterexample" part is messing me up. Thanks for any help!
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edited Dec 28, 2019 at 15:00
Martin Sleziak
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asked Jun 7, 2014 at 15:46
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Is 2×18 a perfect square?
– egreg
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Jun 7, 2014 at 15:48
Counterexample: Let's pick a=b=2. Then ab=4 which is a perfect square, and hence so are a and b. Hm... Something seems off there
– Sten
Commented
Jun 7, 2014 at 15:48
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@egreg and Sten, (2 and 18) and (2 and 2) are not relatively prime.
– Rocket Man
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Jun 7, 2014 at 15:48
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@AJStas - That's the counterexample part
– Sten
Commented
Jun 7, 2014 at 15:51
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Minor comment: The theorem as stated is false, for (−4)(−9) is a perfect square, but neither −4 nor −9 is a perfect square. Easy fix: specify that a and b are positive integers.
– André Nicolas
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Jun 7, 2014 at 16:22
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Counterexample (hint): take a=b (hint: not any a will do)
Proof (hint): write ab=c2 and consider the decomposition of c into prime factors; each of these occurs either in a or in b but not in both.
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edited Jun 7, 2014 at 15:59
answered Jun 7, 2014 at 15:49
VladimirVladimir
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– Vladimir
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Jun 7, 2014 at 15:51
I admitted fault above. OOPS! :)
– Rocket Man
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Jun 7, 2014 at 15:53
@Vladimir this doesn't work for just any a,b as a counterexample, for instance a=b=4.
– DanZimm
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Jun 7, 2014 at 15:54
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Sure this doesn't; I should have written "hint" there, too
– Vladimir
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Jun 7, 2014 at 15:57
@Vladimir just bustin chops, realize now my comment sounds jerkish, sorry!
– DanZimm
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First, you must restrict to a,b∈N else it is false, e.g. (−1)(−4)=22 but −4 is not a square.
Theorem a,b coprime, ab=n2⇒a,b are squares, if a,b,n∈N.
Proof By induction on n. Clear if n=1. Else n>1, so some prime p∣n so p2∣n2=ab, thus p2∣a or p2∣b, by a,b coprime. Wlog p2∣b, thus ab=n2⇒ a(b/p2)=(n/p)2 by canceling p2. Since n/p<n, by induction there are c,d∈N such that a=c2, b/p2=d2, so b=(pd)2. QED
Remark The key property used is that p2∣ab⇒p2∣a or p2∣b, if a,b are coprime. This is an immediate consequence of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations), or equivalent well-known properties, e.g. one can iterate Euclid's Lemma p∣ab⇒p∣a or p∣b. The same holds true for k'th powers of primes, so the proof generalizes from squares to k'th powers and, further, from Z to any UFD (e.g. F[x], a polynomial ring over a field). In the general case one must allow for unit factors, i.e. the result is that a=uc2, b=u−1d2 for some c,d and some unit (invertible) u. We eliminated u=±1∈Z by requiring a,b>0.
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answered Jun 7, 2014 at 18:16
Bill DubuqueBill Dubuque
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See this answer for a proof using gcd laws.
– Bill Dubuque
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Here is some help on the proof, I see there is already help for the counterexample.
If a and b are relatively prime then consider the prime factorization of each
a=pa11⋯pakk
b=qb11⋯qbmm
since they are relatively prime they can share no prime factors. That is pi≠qj for any i and j. Now
ab=pa11⋯pakkqb11⋯qbmm
is a perfect square. What does this say about the exponents? (Hint: Think about parity)
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answered Jun 7, 2014 at 15:55
John MachacekJohn Machacek
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Does that mean exponents cannot be equal?
– user02398409
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Jun 7, 2014 at 15:56
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The square root is the same as 12 power. That is taking the square root cuts your exponents in half. When I cut my exponents in half I still need them to be integers. Half of an integer only still an integer if it is what?
– John Machacek
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Jun 7, 2014 at 16:00
So only the exponents can be evens
– user02398409
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Correct, very good!
– John Machacek
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Jun 7, 2014 at 16:02
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A perfect square has only prime factors with an even exponent, like 86436=22⋅32⋅74.
If a and b have no common prime factor, then all their own exponents must be even. Like 9604=22⋅74 and 9=32.
But when a and b have common prime factors with an odd exponent, the product will have an even exponent as well. Like 1372=22⋅73 and 63=32⋅7.
Beware that you can also have non-relatively-primes a and b giving a perfect square and being perfect squares themselves, like 196=22⋅72 and 441=32⋅72.
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Consider a=pα11pα22⋯ pαrr and b=qβ11qβ22⋯ qβss. Since gcd(a,b)=1 we have {p1,p2,⋯,pr}∩{q1,q2,⋯,qs}=ϕ. Now ab=pα11pα22⋯ pαrr×qβ11qβ22⋯ qβss and ab is a perfect square hence all exponents of p′is and q′js are even for 1≤i≤r, 1≤j≤s. Hence a and b are also square since exponent are even.
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answered Jun 8, 2014 at 11:31
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2017 AMC 12A Problems/Problem 23
Contents
[hide]
1 Problem
2 Solution
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6 (Fast)
8 Solution 7
9 Solution 8
10 Solution 9 (Risky)
11 Solution 10
12 Solution 11
13 Note
14 Video Solution 1
15 Video Solution 2
16 Video Solution 3 by Punxsutawney Phil
17 See Also
Problem
For certain real numbers , , and , the polynomial has three distinct roots, and each root of is also a root of the polynomial What is ?
Solution
Let and be the roots of . Let be the additional root of . Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:
Thus .
Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:
Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:
It follows that . But so
Now we can factor in terms of as
Then and
Hence .
Solution 2
Since all of the roots of are distinct and are roots of , and the degree of is one more than the degree of , we have that
for some number . By comparing coefficients, we see that . Thus,
Expanding and equating coefficients we get that
The third equation yields , and the first equation yields . So we have that
Solution 3
Let the roots of be and the roots of be . Then by Vietas, so . Again by Vietas, . Finally, .
Solution 4
must have four roots, three of which are roots of . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of and are the same, we know that
where is the fourth root of . Substituting and expanding, we find that
Comparing coefficients with , we see that
(Solution 1.1 picks up here.)
Let's solve for and . Since , , so . Since , , and . Thus, we know that
Taking , we find that
Solution 5
A faster ending to Solution 1 is as follows. We shall solve for only and . Since , , and since , . Then,
Solution 6 (Fast)
Let the term be the linear term that we are solving for in the equation . Now, we know that must have , because only . In addition, we know that, by distributing, . Therefore, , and all the other variables are quickly solved for.
Solution 7
We notice that the constant term of and the constant term in . Because can be factored as (where is the unshared root of , we see that using the constant term, and therefore . Now we once again write out in factored form:
.
We can expand the expression on the right-hand side to get:
Now we have .
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:
and finally,
.
We know that is the sum of its coefficients, hence . We substitute the values we obtained for and into this expression to get .
Solution 8
Let and be the roots of . Let be the additional root of . Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:
Thus .
Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:
Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:
It follows that . But so
Now we can factor in terms of as
Then and
Hence .
Solution 9 (Risky)
Let the roots of be , , and . Let the roots of be , , , and . From Vieta's, we have: The fourth root is . Since , , and are common roots, we have: Let : Note that This gives us a pretty good guess of .
Solution 10
First off, let's get rid of the term by finding . This polynomial consists of the difference of two polynomials with common factors, so it must also have these factors. The polynomial is , and must be equal to . Equating the coefficients, we get equations. We will tackle the situation one equation at a time, starting the terms. Looking at the coefficients, we get . The solution to the previous is obviously . We can now find and . , and . Finally , Solving the original problem, .
Solution 11
Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of eventually brings us the final step minus after we multiply by . Now we equate coefficients of same-degree terms. This gives us . We are interested in finding , which equals . ~skyscraper
Note
Note that for any polynomial is simply the sum of the coefficients of the polynomial.
Video Solution 1
Video Solution 2
~ pi_is_3.14
Video Solution 3 by Punxsutawney Phil
See Also
2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23Followed by
Problem 25
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All AMC 10 Problems and Solutions
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22Followed by
Problem 24
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All AMC 12 Problems and Solutions
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17667 | https://alexanderrem.weebly.com/uploads/7/2/5/6/72566533/polynomials.pdf | Winter Camp 2011 Polynomials Alexander Remorov Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic poly-nomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
(University of Toronto Math Competition 2010) Solution: The standard approach would be to write f(x) = ax2 + bx + c and play around with the coefficients of f(x)f(x + 1). It is doable, but quite messy. Let us look at the roots. Let f(x) = a(x −r)(x −s), then: f(x)f(x + 1) = a2 · (x −r)(x −s + 1) · (x −s)(x −r + 1) = = a2([x2 −(r + s −1)x + rs] −r)([x2 −(r + s −1)x + rs] −s) and we are done by setting g(x) = a2(x −r)(x −s), h(x) = x2 −(r + s −1)x + rs.
Warm-up Problem 2: The polynomial f(x) = xn + a1xn−1 + a2xn−2 + · · · + an−1x + an = 0 with integer non-zero coefficients has n distinct integer roots. Prove that if the roots are pairwise coprime, then an−1 and an are coprime.
(Russian Math Olympiad 2004) Solution: Assume gcd(an−1, an) ̸= 1, then both an−1 and an are divisible by some prime p. Let the roots of the polynomial be r1, r2, · · · , rn. Then r1r2 · · · rn = (−1)nan. This is divisible p, so at least one of the roots, wolog r1, is divisible by p. We also have: r1r2 · · · rn−1 + r1r3r4 · · · rn−1 + · · · + r2r3 · · · rn = (−1)n−1an−1 ≡0 mod p All terms containing r1 are divisible by p, hence r2r3 · · · rn is divisible by p. Hence gcd(r1, r2r3 · · · rn) is divisible by p contradicting the fact that the roots are pairwise coprime. The result follows.
1 Algebra Fundamental Theorem of Algebra: A polynomial P(x) of degree n with complex coefficients has n complex roots. It can be uniquely factored as: P(x) = a(x −r1)(x −r2) · · · (x −rn) Vieta’s Formulas: Let P(x) = anxn + an−1xn−1 + · · · + a1x + a0 with complex coefficients have roots r1, r2, · · · , rn. Then: n X i=1 ri = (−1)1 an−1 an ; X i |a1 + a2 + · · · + an|. Prove that P(x) is irreducible (that is, cannot be factored into two polynomials with integer coefficients of degree at least 1).
2 Winter Camp 2011 Polynomials Alexander Remorov 4. (Russia 2003) The side lengths of a triangle are the roots of a cubic equation with rational coefficients. Prove that the altitudes are the roots of a degree six equation with rational coefficients.
5. (Russia 1997) Does there exist a set S of non-zero real numbers such that for any positive integer n there exists a polynomial P(x) with degree at least n, all the roots and all the coefficients of which are from S?
6. (Putnam 2010) Find all polynomials P(x), Q(x) with real coefficients such that P(x)Q(x + 1) −P(x + 1)Q(x) = 1.
7. (IMO SL 2005) Let a, b, c, d, e, f be positive integers. Suppose that S = a + b + c + d + e + f divides both abc + def and ab + bc + ca −de −ef −fd. Prove that S is composite.
8. (USAMO 2002) Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients can be written as the average of two monic polynomials of degree n with n real roots.
9. (Iran TST 2010) Find all two-variable polynomials P(x, y) such that for any real numbers a, b, c: P(ab, c2 + 1) + P(bc, a2 + 1) + P(ca, b2 + 1) = 0 10. (China TST 2007) Prove that for any positive integer n, there exists exactly one polynomial P(x) of degree n with real coefficients, such that P(0) = 1 and (x + 1)(P(x))2 −1 is an odd function. (A function f(x) is odd if f(x) = −f(−x) for all x).
2 Number Theory By Z[x] we denote all the polynomials of one variable with integer coefficients. Arguably the most useful property when it comes to polynomials and integers is: If P(x) ∈Z[x], and a, b are integers, then (a −b)|(P(a) −P(b)) Recall that polynomial in Z[x] is irreducible over the integers if it cannot be factored into two polynomials with integer coefficients.
Eisenstein’s Criterion: Let P(x) = anxn + an−1xn−1 + ... + a1x + a0 ∈Z[x] be a polynomial and p be a prime dividing a0, a1, ..., an−1, such that p ∤an and p2 ∤a0. Then P(x) is irreducible.
Proof: Assume P(x) = Q(x)R(x), where Q(x) = bkxk + bk−1xk−1 + ... + b1x + b0, R(x) = clxl + cl−1xl−1 + ... + c1x + c0. Then b0c0 is divisible by p but not p2. Wolog p|b0, p ∤c0. Since p|a1 = b1c0 + b0c1 it follows that p|b1. Since p|a2 = b2c0 + b1c1 + b0c2 it follows that p|b2. By induction it follows that p|bk which implies that p|an, a contradiction.
Lemma [Schur] Let P(x) ∈Z[x] be a non-constant polynomial. Then there are infinitely many primes dividing at least one of the non-zero terms in the sequence P(1), P(2), P(3), ....
Proof: Assume first that P(0) = 1. There exists an integer M such that P(n) ̸= 1 for all n > M (or else P(x)−1 has infinitely many roots and therefore is constant). We also have P(n!) ≡1( mod n!), and by taking arbitrarily large integers n we can generate arbitrarily large primes dividing P(n!).
3 Winter Camp 2011 Polynomials Alexander Remorov If P(0) = 0, the result is obvious. Otherwise consider Q(x) = P(xP(0)) P(0) and apply the same line of reasoning to Q(x); the result follows.
For polynomials in Z[x] it is often useful to work modulo a positive integer k. If P(x) = Pn i=0 aixi ∈ Z[x] and k is a positive integer we call P(x) = Pn i=0 aixi the reduction of P(x) (mod k), where ai = ai (mod k). Some useful facts about reduced polynomials: 1. Let P(x), Q(x), R(x), S(x) ∈Z[x], such that P(x) = (Q(x) + R(x))S(x). Then P(x) = (Q(x) + R(x))(S(x)).
2. Let p be a prime and P(x) ∈Z[x]. Then the factorization of P(x) is unique modulo p (more formally, in Fp[x] up to permutation.) Note that this result does not hold when p is not a prime. For example, x = (2x + 3)(3x + 2) mod 6 if x, 2x + 3, 3x + 2 are prime.
Also remember that all roots of P(x) modulo p are in the set {0, 1, . . . , p −1}.
2.1 Warm-Up 1.
(a) Let p be a prime number. Prove that P(x) = xp−1 + xp−2 + ... + x + 1 is irreducible.
(b) Prove Eisenstein’s Criterion by considering a reduction modulo p.
2. (Iran 2007) Does there exist a sequence of integers a0, a1, a2, ... such that gcd(ai, aj) = 1 for i ̸= j, and for every positive integer n, the polynomial n X i=0 aixi is irreducible?
3.
(a) (Bezout) Let P(x), Q(x) be polynomials with integer coefficients such that P(x), Q(x) do not have any roots in common. Prove that there exist polynomials A(x), B(x) and an integer N such that A(x)P(x) + B(x)Q(x) = N.
(b) Let P(x), Q(x) be monic non-constant irreducible polynomials with integer coefficients.
For all sufficiently large n, P(n) and Q(n) have the same prime divisors. Prove that P(x) ≡Q(x).
2.2 Problems 1. (a) (USAMO 1974)Let a, b, c be three distinct integers. Prove that there does not exist a polynomial P(x) with integer coefficients such that P(a) = b, P(b) = c, P(c) = a.
(b) (IMO 2006) Let P(x) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Let Q(x) = P(P(. . . P(P(x)) . . .)), where the polynomial P is composed k times. Prove that there are at most n integers t such that Q(t) = t.
2. (Romania TST 2007) Let P(x) = xn + an−1xn−1 + · · · + a1x + a0 be a polynomial of degree n ≥3 with integer coefficients such that P(m) is even for all even integers m. Furthermore, a0 is even, and ak + an−k is even for k = 1, 2, ..., n −1. Suppose P(x) = Q(x)R(x) where Q(x), R(x) are polynomials with integer coefficients, deg Q ≤deg R, and all coefficients of R(x) are odd. Prove that P(x) has an integer root.
3. (USA TST 2010) Let P(x) be a polynomial with integer coefficients such that P(0) = 0 and gcd(P(0), P(1), P(2), . . .) = 1. Prove that there are infinitely many positive integers n such that gcd(P(n) −P(0), P(n + 1) −P(1), P(n + 2) −P(2), . . .) = n.
4 Winter Camp 2011 Polynomials Alexander Remorov 4. (Iran TST 2004) Let P(x) be a polynomial with integer coefficients such that P(n) > n for every positive integer n. Define the sequence xk by x1 = 1, xi+1 = P(xi) for i ≥1. For every positive integer m, there exists a term in this sequence divisible by m. Prove that P(x) = x + 1.
5. (China TST 2006) Prove that for any n ≥2, there exists a polynomial P(x) = xn+an−1xn−1+ ... + a1x + a0 such that: (a) a0, a1, ..., an−1 all are non-zero.
(b) P(x) is irreducible.
(c) For any integer x, |P(x)| is not prime.
6. (Russia 2006) A polynomial (x+1)n −1 is divisible by a polynomial P(x) = xk + ak−1xk−1 + · · · + a1x + a0 of even degree k, such that all of its coefficients are odd integers. Prove that n is divisible by k + 1.
7. (USAMO 2006) For an integer m, let p(m) be the greatest prime divisor of m. By convention, we set p(±1) = 1 and p(0) = ∞. Find all polynomials f with integer coefficients such that the sequence {p f n2 −2n}n≥0 is bounded above. (In particular, f n2 ̸= 0 for n ≥0.) 8. Find all non-constant polynomials P(x) with integer coefficients, such that for any relatively prime integers a, b, the sequence {f(an+b)}n≥1 contains an infinite number of terms and any two of which are relatively prime.
9. (IMO SL 2009) Let P(x) be a non-constant polynomial with integer coefficients. Prove that there is no function T from the set of integers into the set of integers such that the number of integers x with T n(x) = x is equal to P(n) for every positive integer n, where T n denotes the n-fold application of T.
10. (USA TST 2008) Let n be a positive integer. Given polynomial P(x) with integer coefficients, define its signature modulo n to be the (ordered) sequence P(1), . . . , P(n) modulo n. Of the nn such n-term sequences of integers modulo n, how many are the signature of some polynomial P(x) if: (a) n is a positive integer not divisible by the square of a prime.
(b) n is a positive integer not divisible by the cube of a prime.
5 Winter Camp 2011 Polynomials Alexander Remorov 3 Hints to Selected Problems 3.1 Algebra 2. Difference of squares.
3. Prove that for any complex root r of P(x), we have |r| > 1.
4. Use Heron’s formula to prove the square of the area is a rational number.
5. Look at the smallest and the largest numbers in S by absolute value. Use Vieta’s thoerem.
7. The solution involves polynomials.
8. A polynomial has n real roots iffit changes sign n + 1 times. Define one of the polynomials as kQ(x) where Q(x) has n roots and k is a constant.
9. Prove that P(x, y) is divisible by x2(y −1).
10. Let P(x) = Q(x) + R(x) where Q is an even function and R is an odd function.
3.2 Number Theory 2. Reduce modulo 2. Prove that deg R(x) = 1.
3. Let P(x) = xkQ(x) with Q(x) ̸= 0. Consider prime n = pk where p is prime.
4. Prove that xk+1 −xk|xk+2 −xk+1.
5. Reduce modulo 2.
6. Use Eisenstein’s Criterion.
7. Look at the irreducibe factors of f. Prove they are of form 4x −k2.
8. What can you say about gcd(n, f(n))?
9. For k ∈N, look at ak, the number of integers x, such that k is the smallest integer for which T k(x) = x.
10. First solve the problem if n is a prime.
6 |
17668 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_(Morris)/04%3A_Design_Theory/19%3A_Designs_and_Codes/19.04%3A_Using_the_Parity-Check_Matrix_For_Decoding | 19.4: Using the Parity-Check Matrix For Decoding - Mathematics LibreTexts
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19: Designs and Codes
4: Design Theory
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19.4: Using the Parity-Check Matrix For Decoding
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19.4: Using the Parity-Check Matrix For Decoding
Last updated Jul 13, 2021
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19.3: Using the Generator Matrix For Encoding
19.5: Codes From Designs
Page ID 60176
Joy Morris
University of Lethbridge
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. General Method.
Notation
A binary linear code is oftype (n,k)(or we say C is an (n,k) code) if its generator matrix G=[I k A] is an n×k matrix. In other words, G encodes messages of length k as codewords of length n, which means that the number of check bits is n−k. We usually use r to denote the number of check bits, so r=n−k. Then A is an r×k matrix.
Exercise 19.4.1
How many codewords are there in a binary linear code of type (n,k)?
Definition: Parity-Check Matrix
If G=[I k A] is the generator matrix of a binary linear code C, and A is an r×k matrix (so C is of type (k+r,k)), then the parity-check matrix of C is
(19.4.1)P=[A I r].
Example 19.4.1
1) For the code C of Example 19.3.2, the matrix A is 2×3, so r=2. Therefore, the parity-check matrix of C is
P=[A I r]=[A I 2]=[1 0 1 1 0 0 1 1 0 1].
2) For a single parity check-bit, as in Example 19.3.1, we have A=[1 1 1]. This is a 1×3 matrix, so r=1. Therefore, the parity-check matrix of the code is
P=[A I r]=[A I 1]=[1 1 1 1]
(since I 1=).
Exercise 19.4.2
Suppose the generator matrix of the binary linear code C is
[1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 0 1 0 1 1 1]
What is the parity-check matrix of this code?
The parity-check matrix can be used to check whether a message we received is a valid codeword:
Proposition 19.4.1
A column vector x is a codeword if and only if Px=0.
Proof
(⇒) Since x is a codeword, we have x=Gm for some (k-bit) message m. This means that
(19.4.2)x=Gm=[I k A]m=[m Am]
Then
(19.4.3)Px=[A I r][m Am]=[Am+Am]=[2Am]≡0(mod 2).
(⇐) Suppose Px=0. Write x=[m y], where
m is the first k rows of x, and
y is the remaining r=n−k rows of x.
Then
(19.4.4)0=Px=[A I r][m y]=[Am+y].
This means y=−Am=Am(mod 2), so
(19.4.5)x=[m y]=[m Am]=[I−k A]m=Gm
so x∈C.
Example 19.4.3
Here is a simple illustration of Proposition 19.4.1. For the code in which every codeword is required to have an even number of 1 s, Example 19.4.1(2) tells us that the parity-check matrix is P=[1 1 1 1]. Hence, for any 4-bit string x 1x 2x 3x 4, we have
Px=[1 1 1 1][x 1 x 2 x 3 x 4]=[x 1+x 2+x 3+x 4].
This is 0(mod 2) if and only if there are an even number of 1 s in x, which is what it means to say that x is a codeword.
Example 19.4.4
Use the parity-check matrix to determine whether each of these words is in the code C of Example 19.3.2:
11111,10101,00000,11010.
Solution
From Example 19.4.1(1), we know that the parity-check matrix of this code is
P=[1 0 1 1 0 0 1 1 0 1].
We have:
P=[1 1 1 1 1]=[1·1+0·1+1·1+1·1+0·1 0·1+1·1+1·1+0·1+1·1]=[1 1]≠[0 0], so 11111 is not a codeword.
P=[1 0 1 0 1]=[1·1+0·0+1·1+1·0+0·1 0·1+1·0+1·1+0·0+1·1]=[0 0], so 10101 is a codeword.
P=[0 0 0 0 0]=[1·0+0·0+1·0+1·0+0·0 0·0+1·0+1·0+0·0+1·0]=[0 0], so 00000 is a codeword.
P=[1 1 0 1 0]=[1·1+0·1+1·0+1·1+0·0 0·1+1·1+1·0+0·1+1·0]=[0 1]≠[0 0], so 11010 is not a codeword.
(These answers can be verified by looking at the list the elements of \mathcal{C} in the solution of Example 19.3.2.)
It is evident from the parity-check matrix whether a code corrects every single-bit error:
Theorem 19.4.1
A binary linear code C can correct every single-bit error if and only if the columns of its parity-check matrix are all distinct and nonzero.
Proof
Suppose a codeword x is transmitted, but the ith bit gets changed, so a different string y is received. Let ei be the string that is all 0 s, except that the i th bit is 1, so y=x+e i. Then
(19.4.6)P y=P(x+e i)=P x+Pe i=0+Pe i=P ei
is the i th column of P.
Therefore, if all the columns of P are nonzero, then P y is nonzero, so the receiver can detect that there was an error. If, in addition, all of the columns of P are distinct, then P y is equal to the i th column of P, and not equal to any other column, so the receiver can conclude that the error is in the ith bit. Changing this bit corrects the error.
Conversely, if either the i th column of P is zero, or the i th column is equal to the j th column, then either Pe i=0 or Pe i=Pe j. Therefore, when the codeword 00...0 is sent, and an error changes the i th bit, resulting in the message e i being received, either Pe i=0, so the receiver does not detect the error (and erroneously concludes that the message e i is what was sent), or cannot tell whether the error is in the i th bit (and message 0 was sent) or the error is in the j th bit (and message e i+e j was sent). In either case, this is a single-bit error that cannot be corrected.
Exercise 19.4.3
The parity-check matrix of the binary linear code C is
P=[0 1 1 1 0 1 0 1 0 1].
Can C correct all single-bit errors?
The proof of Theorem 19.4.1 shows how to correct any single-bit error (when it is possible):
General Method.
Assume the word y has been received. Calculate P y.
If P y=0, then y is a codeword. Assume there were no errors, so y is the codeword that was sent.
Now suppose P y≠0.
If P y is equal to the i th column of P, then let x=y+e i. (In other words, create x by changing the i th bit of y from 0 to 1 or vice-versa.) Then x is a codeword. Assume it is the codeword that was sent.
If P y is not equal to any of the columns of P, then at least two of the bits of y are wrong. Do not try to correct the error.
Example 19.4.5
Suppose the parity-check matrix of a binary linear code is
P=[1 1 1 1 0 0 1 0 1 0 1 0 1 1 0 0 0 1].
Decode each of the following received words:
111000,101001,001101.
Solution
Let P be the given parity-check matrix. Then:
P=[1 1 1 0 0 0]=[1 0 0]. This is the 4 th column of P, so changing the 4 th bit corrects the error. This means that the received word 111000 decodes as 1111―00.
P=[1 0 1 0 0 1]=[0 0 0]. This is 0, so there is no error. This means that the received word 101001 decodes as 101001.
P=[0 0 1 1 0 1]=[0 1 1]. This is not any of the columns of P, so there are at least two errors. Therefore, we cannot decode the received word 001101.
Exercise 19.4.4
1) The parity-check matrix of a certain binary linear code is
P=[1 0 1 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1].
(a) Decode each of the following received words: 10001111, 11110000, 01111101.
(b) Find the generator matrix of the code.
2) The parity check matrix of a certain binary linear code is
P=[1 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1].
(a) Can the code correct all single-bit errors?
(b) Decode each of the following received words: 001001, 110011, 000110
Example 19.4.6
Let
P=[1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 0 1 0 1 0 1 1 0 1 1].
This is a 4×11 matrix whose columns list all of the binary vectors of length 4 that have at least two 1 s. The corresponding 4×15 parity-check matrix P=[A I 4] lists all 2 4−1=15 nonzero binary vectors of length 4 (without repetition), so the resulting binary linear code can correct all single-bit errors.
The corresponding generator matrix G=[I 11 A]. is a 15×11 matrix, so it takes an 11-bit message, and adds only 15−11=4 check bits. This is much more efficient than the triplerepetition code of Example 19.1.3, which would have to add 22 check bits to detect every single-bit error in an 11-bit message.
Note
Generalizing Example 19.4.6, a binary linear code is called a Hamming code if the columns of its parity-check matrix P=[A I r] are a list of all the 2 r−1 nonzero binary vectors of length r (in some order, and without repetition). Every Hamming code can correct all single-bit errors. Because of their high efficiency, Hamming codes are often used in real-world applications. But they only correct single-bit errors, so other binary linear codes (which we will not discuss) need to be used in situations where it is likely that more than one bit is wrong.
Exercise 19.4.5
1) Explain how to make a binary linear code of type (29,24) that corrects all single-bit errors.
2) Explain why it is impossible to find a binary linear code of type (29,25) that corrects all single-bit errors.
3) For each k≤20, find the smallest possible number r of check bits in a binary linear code that will let you send k-bit messages and correct all single-bit errors. (That is, for each k, we want a code of type (n,k) that corrects all single-bit errors, and we want r=n−k to be as small as possible.)
4) What is the smallest possible number r of check bits in a binary linear code that will let you send 100-bit messages and correct all single-bit errors?
This page titled 19.4: Using the Parity-Check Matrix For Decoding is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Joy Morris.
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17670 | https://magma.maths.usyd.edu.au/magma/handbook/text/195 | Documentation
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Residue Class Rings
The ring Z/mZ consists of representatives for the residue classes
of integers modulo m > 1.
This online help node and the nodes below it describe
the operations in Magma for such rings and their elements.
At any stage during a session, Magma will have at most one copy of
Z/mZ present, for any m>1. In other words, different names for the
same residue class ring will in fact be different references to the
same structure. This saves memory and avoids confusion about different
but isomorphic structures.
If m is a prime number, the ring Z/mZ forms a field; however,
Magma has special functions for dealing with finite fields.
The operations described here should not be used for
finite field calculations: the implementation of finite field
arithmetic in Magma
takes full advantage of the special structure of finite fields
and leads to superior performance.
Contents
Creation
Coercion
Elementary Invariants
Structure Operations
Ring Predicates and Booleans
Homomorphisms
Creation
In addition to the general quotient constructor, a number
of abbreviations are provided for computing residue class rings.
quo : RngInt, RngInt -> RngIntRes
Given the ring of integers Z, and an ideal I, create the
residue class ring modulo the ideal.
Note that Z/I does not give this residue ring, but rather,
in compatibility with Z as a number field order, returns
the ideal quotient, namely Z itself
(see Section Z as a Number Field Order above,
and the example in Sections Ideals of Z).
quo : RngInt, RngIntElt -> RngIntRes
Given the ring of integers Z, and an integer m ≠0, create
the residue class ring Z/mZ.
ResidueClassRing(m) : RngIntElt -> RngIntRes, Map
IntegerRing(m) : RngIntElt -> RngIntRes
Integers(m) : RngIntElt -> RngIntRes
RingOfIntegers(m) : RngIntElt -> RngIntRes
Given an integer greater than zero, create the residue class
ring Z/mZ and also returns the map from Z into Z/mZ.
ResidueClassField(p) : RngIntElt -> FldFin, Map
Given a prime integer p construct the residue class field Fp and the map from
Z into Fp.
ResidueClassRing(Q) : RngIntEltFact -> RngIntRes
IntegerRing(Q) : RngIntEltFact -> RngIntRes
Integers(Q) : RngIntEltFact -> RngIntRes
Create the residue class ring Z/mZ, where m is the integer
corresponding to the factorization sequence Q. This is more
efficient than creating the ring by m alone, since the
factorization Q will be stored so it can be reused later.
Example RngIntRes_residue-ring (H20E2)
We construct a residue ring having modulus the largest prime not
exceeding 216.
```
p := PreviousPrime(2^16);
p;
65521
R := ResidueClassRing(p);
Residue class ring of integers modulo 65521
```
Now we try to find an element x in R such that x3 = 23.
```
exists(t){x : x in R | x^3 eq 23};
true
t;
12697
```
Coercion
Automatic
coercion takes place between Z/mZ and Z so that a binary operation
like + applied to an element of Z/mZ and an integer will result in a
residue class from Z/mZ.
Using !, elements from a prime field GF(p) can be coerced
into Z/pZ, and elements from Z/pZ can be coerced into GF(pr).
Also, transitions between Z/mZ and Z/nZ can be made using !
provided that m divides n or n divides m. In cases where there
is a choice -- such as when an element r from Z/mZ is coerced into
Z/nZ with m dividing n -- the result will be the residue class
containing the representative for r.
Example RngIntRes_Coercion (H20E3)
```
r := ResidueClassRing(3) ! 5;
r;
2
ResidueClassRing(6) ! r;
2
```
So the representative 2 of 5 mod 3
is mapped to the residue class 2 mod 6,
and not to 5 mod 6.
Elementary Invariants
Characteristic(R) : RngIntRes -> RngIntResElt
# R : RngIntRes -> RngIntResElt
Modulus(R) : RngIntRes -> RngInt
Given a residue class ring R=Z/mZ, this function returns the common
modulus m for the elements of R.
FactoredModulus(R) : RngIntRes -> RngIntEltFact
Given a residue class ring R=Z/mZ, this function returns the
factorization of the common modulus m for the elements of R.
Structure Operations
Category(R) : RngIntRes -> Cat
Parent(R) : RngIntRes -> PowerStructure
PrimeRing(R) : RngIntRes -> RngIntRes
Center(R) : RngIntRes -> RngIntRes
AdditiveGroup(R) : RngIntRes -> GrpAb, Map
Given R=Z/mZ, create the abelian group of integers modulo m under addition.
This returns the finite additive abelian group A (of order m) together
with a map from A to the ring Z/mZ, sending A.1 to 1.
MultiplicativeGroup(R) : RngIntRes -> GrpAb, Map
UnitGroup(R) : RngIntRes -> GrpAb, Map
Given R=Z/mZ, create the multiplicative group of R as an abelian group.
This returns an (additive) abelian group A of order φ(m), together
with a map from A to R.
sub< R | n > : RngIntRes, RngIntResElt -> RngIntRes
Given R, the ring of integers modulo m or an ideal of it, and an element
n of R create the ideal of R generated by n.
Set(R) : RngIntRes -> SetEnum
Create the enumerated set consisting of the elements of the residue class
ring R.
Ring Predicates and Booleans
IsCommutative(R) : RngIntRes -> BoolElt
IsUnitary(R) : RngIntRes -> BoolElt
IsFinite(R) : RngIntRes -> BoolElt
IsOrdered(R) : RngIntRes -> BoolElt
IsField(R) : RngIntRes -> BoolElt
IsEuclideanDomain(R) : RngIntRes -> BoolElt
IsPID(R) : RngIntRes -> BoolElt
IsUFD(R) : RngIntRes -> BoolElt
IsDivisionRing(R) : RngIntRes -> BoolElt
IsEuclideanRing(R) : RngIntRes -> BoolElt
IsPrincipalIdealRing(R) : RngIntRes -> BoolElt
IsDomain(R) : RngIntRes -> BoolElt
R eq R : RngIntRes, Rng -> BoolElt
R ne R : RngIntRes, Rng -> BoolElt
Homomorphisms
Ring homomorphisms with domain Z/mZ are completely determined
by the image of 1. As usual,
we require our homomorphisms to map 1 to 1. Therefore, the general
homomorphism constructor with domain Z/mZ needs no arguments.
hom< R -> S | > : RngIntRes, Rng -> Map
Given a residue class ring R, and a ring S,
create a homomorphism from R to S, determined by f(1R) = 1S.
Note that it is the responsibility of the user that the map defines
a homomorphism!
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V2.28, 28 February 2025
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How to check if expression contains a Complex expression?
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Asked 14 years, 3 months ago
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Is there a way to check if an expression contains complex expressions / imaginary numbers?
The documentation says that you can't check if an expression contains I because of how it is interpreted. I have also tried ImaginaryQ[expr_] := expr != Conjugate[expr] and Simplify[expr] =!= Simplify[Conjugate[expr]], but it does not yield accurate results. I have also tried to use MemberQ[expr, Complex], but that does not seem to work either.
I posted some examples into a notebook:
math
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edited Jun 29, 2011 at 16:38
Charles
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asked Jun 29, 2011 at 16:34
eacousineaueacousineau
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1
Do you need to check for expressions like (-1)^(1/3) or just explicit Complex[a, b] == a + I b objects which Yoda's answer addresses?Simon –Simon 2011-06-29 22:16:55 +00:00 Commented Jun 29, 2011 at 22:16
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How about
javascript
ImaginaryQ[expr_] := ! FreeQ[expr, _Complex]
Using it on two of your examples:
```javascript
imExpr = a Sin[a + 2 I];
ImaginaryQ@imExpr
( True )
reExpr = a Sin[a^2 + a];
ImaginaryQ@reExpr
( False )
```
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answered Jun 29, 2011 at 17:02
abcdabcd
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eacousineau
eacousineauOver a year ago
That works perfectly! But why is the form argument for FreeQ a pattern, _Complex, instead of just a symbol, Complex?
2011-07-04T03:58:37.793Z+00:00
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abcdOver a year ago
Complex is the head for all complex numbers & expressions. So we use a pattern _Complex to capture anything with that Head. If you used the symbol Complex instead, this example: FreeQ[{a,b,Complex},Complex] will return False, eventhough Complex is not a complex number... FreeQ[{a,b,Complex},_Complex] returns True as it should.
2011-07-04T04:06:03.167Z+00:00
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To be clear as to why MemberQ[expr,Complex] will not necessarily return True for reals (and may or may not return True for complex expressions). MemberQ is not asking if something is a member of the set of reals or anything like that.
MemberQ[expr,form] returns True if one of the elements of level 1 of expr matches form. Level 1 is what you get second from top if you do TreeForm. Also, by default, MemberQ does not look at heads. Thus:
javascript
l = List[1 + I];
MemberQ[l, Complex, Heads -> True]
MemberQ[List@l, Complex, Heads -> True]
(
->
True
False
)
(the Heads->True part is to make MemberQ also look at heads of expressions). To understand why, look at TreeForm@l and Treeform[List@l]:
Thus, there is a Complex at the first level in the first case, and no Complex at level 1 in the second. This is why we get True and False above. One can use
javascript
MemberQ[List@l, Complex, -1, Heads -> True]
(
-> True
)
to match on all levels.
Finally, to see that MemberQ really is a structural question, try MemberQ[1 + Exp[3I], Complex, Heads -> True] which gives False even though the first argument is obviously complex.
So to sum up, MemberQ has little to do with mathematics; it's a construct to test patterns in lists (or any expression, the head does not matter).
In any case, if one is going to use structural tests, FreeQ is the easiest way, while Elementis the way to do this with mathematical tests.
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edited May 23, 2017 at 12:30
CommunityBot
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answered Jun 29, 2011 at 18:53
aclacl
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acl
aclOver a year ago
@yoda sure, I wouldn't use Element either as I think of it as an assertion rather than a test. I was commenting on the answer stating that MemberQ[expr, Complex] should not be used because that should give True for real numbers, which I felt reflected a lack of distinction between structural and mathematical operations (ie, MemberQ is not about being or not being a member of a mathematical set)
2011-06-29T19:03:11.347Z+00:00
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CharlesOver a year ago
@yoda: If Mathematica were smart enough, Not[Element[a + b I, Reals]] would return something like b = 0. I would think Element[a + I b, Complexes] would return True. (All this is assuming that we tell Math'ca that a and b are Reals.)
2011-06-29T19:04:57.447Z+00:00
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abcdOver a year ago
Yes of course. I meant to post that to the other answer. I've done that and removed this :)
2011-06-29T19:05:24.197Z+00:00
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aclOver a year ago
@Charles Thanks! By the way, what you want is more what Reduce does than Elements; thus Reduce[a + bI \[Element] Reals, {a, b}] or Reduce[a + bI \[Element] Reals && a \[Element] Reals && b \[Element] Reals, {a, b}]
2011-06-29T19:12:58.763Z+00:00
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eacousineauOver a year ago
@acl I agree with Charles - great explanation!
2011-07-04T04:04:59.967Z+00:00
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I would not use MemberQ[expr, Complex] because that should give True for real numbers. Or rather, Element[expr, Complexes] would—I'm not sure what, if anything, your version would do. What about
javascript
Not[Element[expr, Reals]]
or
javascript
Im[expr] != 0
?
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edited Jun 29, 2011 at 18:49
answered Jun 29, 2011 at 16:37
CharlesCharles
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acl
aclOver a year ago
MemberQ[expr, Complex] does not return True for reals because it is a structural question, not a mathematical one
2011-06-29T17:50:14.06Z+00:00
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CharlesOver a year ago
@yoda: Thanks, corrected. Mathematica is something like my 10th language and I get confused every so often.
2011-06-29T18:50:16.263Z+00:00
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abcdOver a year ago
@Charles: Yes, that's a mistake I make often too :) For me it's more common with Real because, Real is a valid head in Mathematica and the syntax coloring won't highlight it, whereas I catch Imag instantly because it's undefined and shows up as blue.
2011-06-29T18:54:41.213Z+00:00
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aclOver a year ago
@yoda, @Charles and how about Trace vs Tr, especially with something along the lines of f[i_] := i^2; Trace[Table[f[i], {i, 1, 10}, {j, 1, 10}]] (page of stuff) vs f[i_] := i^2; Tr[Table[f[i], {i, 1, 10}, {j, 1, 10}]] (385)...
2011-06-29T18:58:16.29Z+00:00
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abcdOver a year ago
@Charles: Btw, the problem with Element is for undefined symbols, it's more of an assertion than a test. For e.g. Element[a + I b, Complexes]. Mathematically, yes it shouldn't return either True or False, because we haven't defined a and b. However, I assume OP wanted to only check for the presence of a I in the expression. Similarly, Im[expr] also doesn't always return true/false (e.g., check with OP's first example imExpr)
2011-06-29T19:04:19.327Z+00:00
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17672 | https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:sqrt-eq/v/extraneous-solutions-of-radical-equations | Square-root equations intro (video) | Khan Academy
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Algebra 2
Course: Algebra 2>Unit 10
Lesson 2: Square-root equations
Intro to square-root equations & extraneous solutions
Square-root equations intro
Intro to solving square-root equations
Square-root equations intro
Solving square-root equations
Solving square-root equations: one solution
Solving square-root equations: two solutions
Solving square-root equations: no solution
Square-root equations
Math>
Algebra 2>
Equations>
Square-root equations
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Square-root equations intro
FL.BEST.Math: MA.912.AR.7.1
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About About this video Transcript
Sal gives an example of how an extraneous solution arises when solving 2x-1=√(8-x).
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caldwelljt 2 years ago Posted 2 years ago. Direct link to caldwelljt's post “I'm curious, why isn't it...” more I'm curious, why isn't it a rule that when squaring we default to comparing only to an absolute value? Stepping back for a minute, I'm wondering if doing so requires some sort of larger math principle or this elusive "proof" thing I've not really grasped (i.e. I know what a proof is, but is that's what's necessary and/or does it hurt us once we go further into math).... I can sort of see how that whatever we do here has to hold up even if we layer tons of other math (and other types of math) on top of it, so that, say, this is just some sub-component of a much larger string of equations or just a large equation, that doing so doesn't prevent us from suddenly being unable to solve something that SHOULD be solvable?
Ok, so I hope that makes sense to someone. Whenever I ask questions like this I feel like my math "footing" isn't secure and I'm not truly grasping some fundamental "why/how" and/or that someone/somewhere/somehow is going to connect some of these larger principle dots for me later and that I just need to go through the routine of learning the how first and the why will come later... other times I feel like I MISSED something...
help?
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Marvyn Greco 10 months ago Posted 10 months ago. Direct link to Marvyn Greco's post “how did he do that? turni...” more how did he do that? turning 4x^2 -3x-7=0 into x=sqrt9-447 all over 8
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kubleeka 10 months ago Posted 10 months ago. Direct link to kubleeka's post “Sal applied the quadratic...” more Sal applied the quadratic formula. If you have an equation that looks like
ax²+bx+c=0
then the solutions are
x=(-b±√(b²-4ac))/2a
This is covered in Unit 9 of Khan Academy's Algebra section.
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siena a year ago Posted a year ago. Direct link to siena's post “how is 14/8 a solution if...” more how is 14/8 a solution if it doesnt fit back into the original equation
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “Why do you think it won't...” more Why do you think it won't work? Sal didn't try it.
I'm going to reduce the fraction to start:
x = 14/8 = 7/4
Substitute into original equation: 2(7/4)-1=√(8-7/4)
Simplify both sides
-- Left side: 2(7/4)-1 = 7/2-1 = 7/2-2/2 = 5/2
-- Right side: √(8-7/4) = √(32/4-7/4) = √(25/4) = 5/2
x=7/4 is a valid solution. It makes the 2 sides of the equation equal.
Hope this helps.
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allen.m.muresan a year ago Posted a year ago. Direct link to allen.m.muresan's post “Is the extraneous solutio...” more Is the extraneous solution like the "dark matter" of algebra?
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Double_Mass 3 months ago Posted 3 months ago. Direct link to Double_Mass's post “No, because we know what ...” more No, because we know what they are and where they come from; they are the answers to problems that have the same answer after being squared as the problem that the solution is extraneous to, but not before.
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beauty840725 a year ago Posted a year ago. Direct link to beauty840725's post “Hi, In the quiz, in the p...” more Hi, In the quiz, in the process of the solving the problem, it says
8y^2+2y-3=0,
=(4y+3)(2y-1)=0
I thought (y+3)(y-1)=0. why 4y and 2y is there?
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “This is the same situatio...” more This is the same situation as in your other question. You need to use factoring by grouping. Your factorings need to create 8y^2 rather than just y^2 as the first term. If you multiply your 2 factors: (y+3)(y-1), they create y^2+2y-3 which is not the original equation. Use the link I gave you in your other question to learn about factoring by grouping.
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Lola.Oretade 7 months ago Posted 7 months ago. Direct link to Lola.Oretade's post “Can both answers be extra...” more Can both answers be extraneous solutions?
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Kim Seidel 7 months ago Posted 7 months ago. Direct link to Kim Seidel's post “It could be possible, but...” more It could be possible, but it would also be an indication that the 2 sides of the original equation were not equal in the first place. So, it wasn't a true equation.
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G-First 7 months ago Posted 7 months ago. Direct link to G-First's post “If anyone asks: why is th...” more If anyone asks: why is the equation not equivalent when it is squared?
To answer this question I would make an easy equation
2+2=1+3
(simply)
4=4
imagine if the answer didn't look clear so we squared it by 2
4^2=4^2
(simply)
16=16
16=16 would look like the right answer but it is actually 4=4 and 16=16 is not the same as 4=4 also the original equation (2+2=1+3) does not equal to 16=16 but we got 16=16 from the original equation.
The reason why I wrote this example is because in algebra when you square a equation you alter the equation a little bit this is one of the reasons why not all
of the solutions are correct.
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FractalShield 5 years ago Posted 5 years ago. Direct link to FractalShield's post “Is there any way to just ...” more Is there any way to just eyeball an extraneous solution? Like maybe a pattern? or do we always have to test it out?
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David Severin 5 years ago Posted 5 years ago. Direct link to David Severin's post “I do not know any shortcu...” more I do not know any shortcut, but it should not take long to determine if what is under the square root is positive or negative.
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james.walkerusa 2 years ago Posted 2 years ago. Direct link to james.walkerusa's post “Why can't -3 be a solutio...” more Why can't -3 be a solution to square-root of 9? (-3)squared is 9.
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FreeRadical 2 years ago Posted 2 years ago. Direct link to FreeRadical's post “-3 is a solution to the s...” more -3 is a solution to the sqrt(9).
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Isabella Park 7 years ago Posted 7 years ago. Direct link to Isabella Park's post “What is the exact definit...” more What is the exact definition of an extraneous solution?
Is substitution(just plugging it in) the simplest way of checking for extraneous solutions?
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Video transcript
[Voiceover] So let's say we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side of the equation. So, we might say, "Well, let's just get rid of the radical. "Let's square both sides of this equation." So we might say that this is the same thing as two-x minus one squared is equal to the square root of eight minus x, eight minus x squared, and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel, "Okay we did legitimate operations. "We did the same thing to both sides. "That these are equivalent equations." But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it, you're losing information. So, for example, this would be true even if the original equation were two-x... Let me, this in a different color. Even if the original equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to this right-hand side, and not the yellow one, then we would call that an extraneous solution. So, let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, we are going to get four-x squared minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the quadratic formula, here. So the solutions are going to be... X is going to be equal to negative b, so three, plus or minus the square root of b-squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three plus or minus the square root of... Let's see, four times four is 16 times seven. 16 times seven is gonna be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven. The two, four. So, it's 112 plus nine. So, 121, that worked out nicely. So, plus or minus the square root of 121, all of that over eight. Well, that is equal to three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14-eights. Or, if we subtract 11, three minus 11 is negative eight. Negative eight divided by eight is negative one. So we have to think about... You might say, "Okay, I found two solutions "to the radical equation." But remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure that they're legitimate, or maybe one of these is an extraneous solution. In fact, one is very likely a solution to this radical equation which wasn't our original goal. So, let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one. So that would be negative two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine. This is a positive square root. This is not true. So this, right over here, that is an extraneous solution. Extraneous. Extraneous solution. It is a solution to this one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three. So, it checks out for this one. So, this one right over here is the extraneous solution. This one right over here is gonna be the actual solution for our original equation, and you can test it out on your own. In fact, I encourage you to do so.
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17673 | https://codental.uobaghdad.edu.iq/wp-content/uploads/sites/14/2021/06/Development-of-muscular-system-converted.pdf | Embryology Lect16 Development of muscular system Dr.Enas Fadhil The formation of the muscular system begins about 4 th week of embryonic development. The beginning cells are called Myoblasts. Most of our muscles develops from the mesodermal germ layer. Except some smooth muscle tissues (pupil, sweat glands and mammary gland differentiate from ectoderm) The muscular system consist of; 1. Skeletal musculature 2. Cardiac musculature 3. Smooth musculature Skeletal muscles are derived from paraxial mesoderm This forms; Somites from the occipital to the sacral regions Somitomeres in the head Smooth muscles differentiate from splanchnic mesoderm surrounding the gut and its derivatives. Cardiac muscles are derived from splanchnic mesoderm surrounding the heart tube Striated skeletal musculature Musculature of the head, axial skeleton and body wall are formed by Somites and somitomeres From the occipital region caudally, somites form and differentiate into; ►Sclerotome ►Dermatome ►Two muscle-forming regions One in the dorsolateral region of the somite provides progenitor cells for limb and body wall musculature (hypomeric) The other in the dorsalmedial region forms the myotome (epimeric musculature). • Precursor cells, the myoblasts, fuse and form long, multinucleated muscle fibers • Myofibrils soon appear in the cytoplasm, and by the end of the third month, cross- striations appear in skeletal muscle • A similar process occurs in the seven somitomeres in the head region rostral to the occipital somites. Patterning of muscle Patterns of muscle formation are controlled by connective tissue into which myoblasts migrate In the head region these connective tissues are derived from neural crest cells; in cervical and occipital regions they differentiate from somatic mesoderm; and In the body wall and limbs they originate from somatic mesoderm . Derivatives of precursors muscle cells By the end of the 5th week prospective muscle cells are collected into two parts: Epimere (small dorsal portion) – innervated by the dorsal primary ramus Hypomere (larger ventral part) – innervated by the ventral primary ramus Myoblasts of the epimeres form the extensor muscles of the vertebral column, and those of the hypomeres give rise to muscles of the limbs and body wall. Transverse section through the thoracic region of a 5-week embryo Myoblasts from cervical hypomeres form the geniohyoid, and prevertebral muscles. Those from thoracic segments split into three layers, which in the thorax are represented by; External Intercostal Internal Intercostal Innermost Intercostal In the abdominal wall these three muscle layers consist of the external oblique, the internal oblique, and the transversus abdominis muscles. • Myoblasts from the hypoblast of lumbar segments form the quadrates lumborum muscle • Those from sacral and coccygeal regions form the pelvic diaphragm and striated muscles of the anus. A ventral longitudinal column arises at the ventral tip of the hypomeres. This column is represented by the rectus abdominis muscle and the infrahyoid musculature. Head musculature All voluntary muscles of the head region are derived from paraxial mesoderm (somitomeres and somites); Including muscle of the tongue, eye (except that of the iris, which is derived from optic cup ectoderm), and that associated with the pharyngeal (visceral) arches. Patterns of muscle formation in the head are directed by connective tissue elements (Neural crest cells) . Limb musculature Connective tissue dictates the pattern of muscle formation in the limb Derived from the somatic mesoderm The mesenchyme is derived from dorsolateral cells of the somites that migrate into the limb bud to form the muscles. With elongation of the limb buds, the muscle tissue splits into flexor and extensor components . The upper limb buds lie opposite the lower five cervical and upper two thoracic segments, and the lower limb buds lie opposite the lower four lumbar and upper two sacral segments. |
17674 | https://www.bccdc.ca/health-info/diseases-conditions/staphylocococcus-aureus | Staphylocococcus aureus (food poisoning)
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Staphylocococcus aureus (food poisoning)
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Staphylococcal food poisoning is an acute intoxication that occurs when food contaminated with enterotoxin produced by this bacterium is consumed. Although precise data regarding the exact number of cases is lacking, staphylococcal food poisoning is considered to be among the most common causes of gastroenteritis worldwide. The presence of staphylococcal enterotoxin in food is usually due to cross contamination of ready to eat food with either raw food or, most likely, contamination from a food handler that is carrying Staphylococcus aureus. To prevent food becoming contaminated, good food safety practices should be followed, such as practicing frequent hand washing, wearing gloves when handling foods, and ensuring separation between the areas where raw and cooked products are handled.
Definition
Illness caused by Staphylococcus aureus is an acute intoxication that develops after the ingestion of food contaminated with the enterotoxin produced by this bacterium. S. aureus is also associated to other health problems ranging from skin infections to severe invasive infections of the lungs or the heart. Drug resistant strains are common, especially in hospital settings. This summary refers solely to staphylococcal food poisoning.
Symptoms
Symptoms include:
vomiting
nausea
diarrhea, usually watery but sometimes with blood
cramps
other symptoms may include mild fever, weakness, dizziness and chills.
Symptoms usually start 1 to 10 hours after exposure and go away in 1 to 2 days. In some cases, the illness may be more severe. If you have serious symptoms, you should see your doctor.
Causes
For staphylococcal food poisoning to occur following the ingestion of a given food, two conditions are necessary. First, S. aureus has to be present in the food; second, foods stored at incorrect temperatures and time allow growth of this pathogen and the production of enterotoxin.
Although S. aureus can be found in food-producing animals and raw foods, humans are considered the main reservoir for this pathogen. S. aureus can be present in healthy individuals, usually on the skin and mucous membranes, for example in the nasal cavity. Food can become contaminated during preparation if the food handler is a carrier of S.aureus and this is transferred to the food through direct contact with contaminated skin or by coughing and sneezing.
The growth of S. aureus in food to a sufficient level as to allow enterotoxin production is possible only under certain conditions. For example, it needs temperatures of between 7°C and 48°C to be able to grow, with an optimum temperature of 37°C. Enterotoxin production will only take place once the levels of this microorganism are large (greater than 10,000 S. aureus per gram of food).
The foods that have been most frequently implicated in cases of staphylococcal food poisoning are poultry and cooked meat products such as ham or corned beef. Other foods implicated were milk and milk products, canned food and bakery products.
Complications
Staphylococcal food poisoning is usually self-limiting, resolving within one or two days. In a small percentage of cases it can be more severe, especially in infants, the elderly and immunocompromised patients.
Tests and diagnosis
Staphylococcal food poisoning can be confirmed if the enterotoxin or large numbers of S. aureus are found in the food. S. aureus can also be detected in stool samples from patients.
Treatment and drugs
Most people recover without treatment. The administration of fluids is recommended if the diarrhoea or vomiting is severe. Antibiotics are not indicated as the symptoms are caused by the enterotoxin and not the bacteria.
Prevention
Control measures should be applied first to avoid contaminating the food with S. aureus and also to prevent growth and the formation of enterotoxin in the food.
To avoid contaminating the food with S. aureus, handle and prepare food safely:
Ensure raw foods of animal origin are obtained following good hygienic practices, to reduce the possibility of S. aureus contamination.
Food handlers should use appropriate protective clothing (e.g. gloves) and thoroughly wash hands.
Food handlers with skin lesions should have them properly covered prior to handling food. If this is not possible, they should not work while handling food until the lesions have healed.
Avoid cross-contamination by keeping work surfaces clean and ensuring separation between areas where raw and cooked foods are manipulated.
To prevent growth of S. aureus and the formation of enterotoxin:
Ensure food is maintained either at a temperature above 60°C or refrigerated below 4°C.
Cool cooked foods that will not be immediately consumed to below 4°C within 6 hours.
When reheating food, ensure that the temperature reaches at least 74°C.
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17675 | https://jimmymaths.com/number-pattern-formulas/ | How to Derive a Number Pattern Formula
In this revision note, you will learn how to derive a number pattern formula so that you will be proficient in solving Number Pattern O-Level Math questions.
Number patterns are hidden all around us, and learning to identify and utilise them can unlock numerous benefits such as the prediction of outcomes and the forming of relationships between various objects and variables.
Building upon our earlier understanding of number sequences (learnt in primary school), we will learn in this chapter new types of sequences, tools to find their general formulas as well as their practical applications in everyday life.
Before you read on, you might want to download this entire revision notes in PDF format to print it out, or to read it later. This will be delivered to your email inbox.
Click here to download
Number Pattern Definitions
Consider the following whole numbers:
The above set of numbers 2, 7, 12, 17, 22, … forms a number sequence.
The numbers in the number sequence are called terms of the sequence. The 1st, 2nd, 3rd, 4th and 5th terms are 2, 7, 12, 17, and 22 respectively. We use T1 to represent the 1st term, T2 to represent the 2nd term, T3 to represent the 3rd term and so on.
The numbers in the number sequence are guided by a specific rule: begin with 2, then add 5 to the current term to obtain the next term.
The general term of the above number sequence, Tn , is represented by the algebraic expression 5n −3, where n is the ordinal numerical value of the term. For instance, to obtain the 100th term in the number sequence above, we let n = 100, then T100 = 5(100) − 3 = 497.
Hence the general term, Tn = 5 − 3, is a number pattern formula to enable us to accurately predict any term in the sequence without having to list out all the previous terms (1st to 99th).
Obtaining the Number Pattern Formula
To obtain the number pattern formula of a number sequence, you need to apply some key skills below:
Identifying patterns: This is the most crucial skill. You need to observe and analyze the sequence to identify the underlying pattern that governs the change from one term to the next. This can involve looking for constant differences, ratios, or other recurring features.
Understanding different types of sequences: Familiarity with various types of sequences, such as arithmetic, geometric, quadratic, and Fibonacci, is essential. Each type has its specific pattern and formula for generating the terms.
Applying mathematical concepts: You need to apply mathematical concepts like arithmetic operations, powers of numbers, and even algebra to express the pattern in a mathematical form.
Inductive reasoning: You need to generalise the observed patterns to form a rule that applies to all terms in the sequence, including those not explicitly given.
Translating patterns into formulas: Once you have identified the pattern and rule, you need to translate it into a concise and accurate formula using mathematical notation.
Checking and verifying: It’s crucial to verify your formula by testing it on several terms in the sequence and ensuring it accurately predicts the values.
The above steps can be demonstrated in the given number sequence:
2, 7, 12, 17, 22, …
Steps | Actions To Take
1. Identifying patterns | Observe that the numbers are increasing.The numbers are increasing by a constant amount of 5.
2.Understanding different types of sequences | Based on the above initial observations, this is an arithmetic sequence.[We will learn more about this type of sequence and other sequences in later part of the notes.]
3.Applying mathematical concepts | Increasing by 5 is the same as +5.
4.Inductive reasoning | We can represent each term of the number sequence as follows:T1= 2T2= 2 + 5T3= 2 + 5 + 5T4= 2 + 5 + 5 + 5T5= 2 + 5 + 5 + 5 + 5
5.Translating patterns into formulas | Observe that the pattern always begin with 2, and we progressively add 5 to the current term to obtain the next one.The number of 5s in the current term is always one less than the ordinal numerical value. For instance, we add one 5 to the 2ndterm, we add two 5s to the 3rdterm, we add three 5s to the 4thterm and so on.Therefore, to obtain thenthterm, we begin with 2 and then we add(n– 1) number of 5s. This translates to:Tn= 2 + (n – 1)5= 5n – 3
6.Checking and verifying | Once we have the general term Tn= 5n – 3, we mustcheckthat the formula can indeed give usallthe correct terms.We progressively substitutenwith 1, 2 and 3, into the formula and cross check with the given terms in the number sequence.T1= 5(1) – 3 = 2[✔]T2= 5(2) – 3 = 7[✔]T3= 5(3) – 3 = 12[✔]Check for at least the first 3 terms to be sure that we have the correct formula.
Steps
Actions To Take
Identifying patterns
Observe that the numbers are increasing.
The numbers are increasing by a constant amount of 5.
Based on the above initial observations, this is an arithmetic sequence.
[We will learn more about this type of sequence and other sequences in later part of the notes.]
We can represent each term of the number sequence as follows:
T1 = 2
T2 = 2 + 5
T3 = 2 + 5 + 5
T4 = 2 + 5 + 5 + 5
T5 = 2 + 5 + 5 + 5 + 5
Observe that the pattern always begin with 2, and we progressively add 5 to the current term to obtain the next one.
The number of 5s in the current term is always one less than the ordinal numerical value. For instance, we add one 5 to the 2nd term, we add two 5s to the 3rd term, we add three 5s to the 4th term and so on.
Therefore, to obtain the nth term, we begin with 2 and then we add (n – 1) number of 5s. This translates to:
Tn = 2 + (n – 1)5
= 5n – 3
Once we have the general term Tn = 5n – 3, we must check that the formula can indeed give us all the correct terms.
We progressively substitute n with 1, 2 and 3, into the formula and cross check with the given terms in the number sequence.
T1 = 5(1) – 3 = 2 [✔]
T2 = 5(2) – 3 = 7 [✔]
T3 = 5(3) – 3 = 12 [✔]
Check for at least the first 3 terms to be sure that we have the correct formula.
Number Pattern Formula of Different Sequences
Arithmetic Sequences
An arithmetic sequence (also known as a linear sequence) is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference, d, is called the common difference.
Number Pattern Formula for Arithmetic Sequences: Tn = a + (n – 1)d
where n is the ordinal numerical value of the term, a is the first term and d is the common difference between any two consecutive terms.
Examples:
Applications to real life:
Arithmetic Sequences Example
A certain number sequence is given as:
3, 7, 11, 15, …
(a) Write down the next two terms of the sequence.
(b) Write down an expression ,in terms of n, for the nth term of the sequence.
(c) Hence
(i) find the 45th term of the number sequence.
(ii) determine if 400 is in the number sequence.
Solution:
(a) 19, 23
(b) Tn = 3 + (n − 1)(4) = 4n − 1
(c)(i) T45 = 4(45) − 1 = 179
(ii) Assume 400 is in the number sequence. Then Tn = 400 for a positive integer n.
4n − 1 = 400
4n = 399
n = 99.75 (not a positive integer)
Since n is not a positive integer for Tn = 400, 400 is not in the number sequence.
Geometric Sequences
A geometric sequence is a sequence of numbers where each term is obtained by multiplying the previous term by a constant value. This constant value is called the common ratio, r.
Number Pattern Formula for Geometric Sequence: Tn = arn–1,
where n is the ordinal numerical value of the term, a is the first term and r is the common ratio between any two consecutive terms.
Examples:
Applications to real life:
Geometric Sequences Example
A certain number sequence is given as:
4, 20, 100, 500, …
(a) Write down the next two terms of the sequence.
(b) Write down an expression ,in terms of n, for the nth term of the sequence.
(c) Hence
(i) find the 20th term of the number sequence, giving your answer in standard form.
(ii) explain why the difference between any two consecutive terms is always exactly divisible by 2.
Solution:
(a) 2 500, 12 500
(b) Tn = 4(5)n – 1
(c)(i) T20 = 4(5)20−1 = 7.63 × 1013
(ii) Tn+1 − Tn = 4(5)n+1 − 4(5)n
= 4(5 n+1 − 5n)
= (2)(2)(5n+1 − 5n)
Since the difference between any two consecutive terms is a multiple of 2, the difference is always exactly divisibly by 2.
Harmonic Sequences
A harmonic sequence is a sequence of numbers where each term is the reciprocal of the corresponding term in an arithmetic sequence. In simple terms, it’s a sequence of reciprocals of an arithmetic sequence.
Number Pattern Formula for Harmonic Sequences: Tn =1/ (a+(n−1)d),
where n is the ordinal numerical value of the term, a is the denominator of the first term, and d is the common difference between the denominators of any two consecutive terms.
Examples:
Harmonic sequence: 1, 1/4, 1/7, 1/10, 1/13, 1/16, … (a = 1, d = 3)
Harmonic sequence: 1, 2/3, 1/2, 2/5, 1/3, 2/7, 1/4, … (a = 1, d =1/2)
Applications to real life:
Square Numbers
A square number is a positive integer obtained by squaring an integer. In simpler terms, it’s the product of a number multiplied by itself.
For example, the sequence 1, 4, 9, 16, 25,… is a sequence of square numbers.
Number Pattern Formula for Square Numbers: Tn = n2,
where n is the ordinal numerical value of the term.
As a result, we can also rewrite each square number as a sum of the number of objects in each row:
T1 =1 T2= 1+ 3 = 4 T3 = 1+ 3+ 5 = 9 T4 = 1 +3 + 5 + 7 = 16
Applications to real life:
Sequence of Triangular Numbers
A triangular number is a positive integer that represents the number of objects arranged in an equilateral or right-angled isosceles triangle.
Each additional row adds one more object than the previous row, creating a triangular shape, like the examples shown below.
Hence, the sequence 1, 3, 6, 10, 15, 21,… is a sequence of triangular numbers. We can also rewrite each triangular number as a sum of the number of objects that form the triangle:
T1 = 1 T2 = 1 + 2 = 3 T3 = 1 + 2 + 3 = 6 T4 = 1 + 2 + 3 + 4 = 10 T5 = 1 + 2 + 3 + 4 + 5 = 15 T6 =1 + 2 + 3 + 4 + 5 + 6 = 21
Number Pattern Formula for Triangular Numbers: Tn =n(n + 1)/2,
where n is the ordinal numerical value of the term.
Applications to real life:
Geometry: calculating areas and volumes of certain shapes.
Fibonacci Sequence
The Fibonacci sequence is a sequence of numbers where each number is the sum of the two preceding ones. Starting with 0 and 1, the sequence continues:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
Number Pattern Formula for Fibonacci Sequence: Tn = Tn –1 + Tn – 2,
where n is the ordinal numerical value of the term.
Unique Properties:
Applications to real life:
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17676 | https://www.mathway.com/popular-problems/Trigonometry/338997 | Expand the Trigonometric Expression cos(45+x) | Mathway
Enter a problem...
[x]
Trigonometry Examples
Popular Problems
Trigonometry
Expand the Trigonometric Expression cos(45+x)
cos(45+x)cos(45+x)
Step 1
Rearrange terms.
cos(x+45)cos(x+45)
Step 2
Double negation
cos(x−(−45))cos(x-(-45))
Step 3
Apply the difference of angles identity cos(x−y)=cos(x)cos(y)+sin(x)sin(y)cos(x-y)=cos(x)cos(y)+sin(x)sin(y).
cos(x)cos(−45)+sin(x)sin(−45)cos(x)cos(-45)+sin(x)sin(-45)
Step 4
Simplify each term.
Tap for more steps...
Step 4.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
cos(x)cos(45)+sin(x)sin(−45)cos(x)cos(45)+sin(x)sin(-45)
Step 4.2
The exact value of cos(45)cos(45) is √2 2 2 2.
cos(x)√2 2+sin(x)sin(−45)cos(x)2 2+sin(x)sin(-45)
Step 4.3
Combine cos(x)cos(x) and √2 2 2 2.
cos(x)√2 2+sin(x)sin(−45)cos(x)2 2+sin(x)sin(-45)
Step 4.4
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.
cos(x)√2 2+sin(x)(−sin(45))cos(x)2 2+sin(x)(-sin(45))
Step 4.5
The exact value of sin(45)sin(45) is √2 2 2 2.
cos(x)√2 2+sin(x)(−√2 2)cos(x)2 2+sin(x)(-2 2)
Step 4.6
Combine sin(x)sin(x) and √2 2 2 2.
cos(x)√2 2−sin(x)√2 2 cos(x)2 2-sin(x)2 2
cos(x)√2 2−sin(x)√2 2 cos(x)2 2-sin(x)2 2
c o s(4 5+x)c o s(4 5+x)
cos(45+x)x cos(45+x)x cos(45+x)x 2 cos(45+x)x 2 cos(45+x)x 3 cos(45+x)x 3
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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17677 | https://convertlive.com/es/u/convertir/kil%C3%B3metros/a/metros | Kilómetros a Metros
Kilómetros = Metros
Convertir de Kilómetros a Metros. Escriba la cantidad que desea convertir y presione el botón convertir (↻).
Pertenece en la categoría
Longitud
| | | |
---
| 1 Kilómetros = 1000 Metros | 10 Kilómetros = 10000 Metros | 2500 Kilómetros = 2500000 Metros |
| 2 Kilómetros = 2000 Metros | 20 Kilómetros = 20000 Metros | 5000 Kilómetros = 5000000 Metros |
| 3 Kilómetros = 3000 Metros | 30 Kilómetros = 30000 Metros | 10000 Kilómetros = 10000000 Metros |
| 4 Kilómetros = 4000 Metros | 40 Kilómetros = 40000 Metros | 25000 Kilómetros = 25000000 Metros |
| 5 Kilómetros = 5000 Metros | 50 Kilómetros = 50000 Metros | 50000 Kilómetros = 50000000 Metros |
| 6 Kilómetros = 6000 Metros | 100 Kilómetros = 100000 Metros | 100000 Kilómetros = 100000000 Metros |
| 7 Kilómetros = 7000 Metros | 250 Kilómetros = 250000 Metros | 250000 Kilómetros = 250000000 Metros |
| 8 Kilómetros = 8000 Metros | 500 Kilómetros = 500000 Metros | 500000 Kilómetros = 500000000 Metros |
| 9 Kilómetros = 9000 Metros | 1000 Kilómetros = 1000000 Metros | 1000000 Kilómetros = 1000000000 Metros |
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17678 | https://www.1066.co.nz/Mosaic%20DVD/alphametics/example1.pdf | How to Solve Each Alphametic puzzle is constructed from up to 10 letters. Each letter represents a number between 0 and 9. Two letters can not represent the same letter. The objective is to determine the number used to exchange for each letter. The only obvious clue is that a word can not start with a zero. The following is an example to illustrate how to solve a simple Alphametic using logic. 1 2 3 4 A N + C A N N E E D (Column) A good first step is to look for the number 0. Since the conventions outline that no word may start with 0, we know that neither A, C or N are 0. Next we attempt to establish whether any columns add up to more than 10 (e.g. 4 + 7 = 11), which indicates this column carries over to the next column. From the example, column 1 contains only the letter N. For this to happen, the addition from the preceding column must have generated a carry, therefore N = 1. From column 4, if N + N = D, this means that D = 2. From column 2, C = 9 as any number less than 9 would not have produced a carry to column 1. By the same reasoning, column 3 must also produce a carry. Using this we can determine that E = 0 from column 2. Finally, in column 3, A + A = 0. This means that A = 5 as we have already determined that column 4 had no carry. We have now replaced all the letters with numbers, providing the solution: A=5, C=9, D=2, E=0, N=1. Tips and Tricks The following is a selection of tips and tricks to help in solving Alphametics using logic. There are many more tips and tricks that you will discover as you progress through this book. A + B A If this occurs in the right end column, then B is clearly the number 0 as any number plus 0 leaves that number unchanged. A X + B Y A Z If the previous situation occurs but not in the right end column, B must be either 0 or 9 depending on whether the preceding column has a carry. A + A B If this occurs in the right end column, then B is an even number and A is not 0. A X + Y B Z This arrangement tells you that B is 1 more than A. A + B C If it can be shown that this column has no carry, the smallest A or B can be is 1 and the smallest C can be is 3. In addition neither A or B can be 9. Using Solving Aids Each puzzle in this book has two pages; the first page contains the puzzle and the second contains several aids to help in solving the puzzle. These solving aids can reduce the complexity of the problem by working through the logic of the problem visually, rather than in your head. When you start out it is a good idea to use these as it will help you to understand the process for solving the puzzles. You use the first solving aid to help record what each letter could be and what it is not. As you solve the puzzle you will discover the certain letters can not logically be certain numbers, cross these out on the table (shown here in grey). Eventually there will be only one number associated with each letter. You will have solved the puzzle. The second solving aid is also used to eliminate numbers, by associating letters in a sum. It helps in applying your logic while solving a puzzle. When you discover a range of values for a letter, use the table to see the effect on other letters. For example, the third tip (A + A = B) is shown in the first table. If later we determine that A is less than 5, the second table shows you that B must be 2, 4, 6, or 8 and has no carry second table shows you that B must be 2, 4, 6, or 8 and has no carry. A D E G M N P R S T 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 ( 0 , 1 ) c ar ry A 0 1 2 3 4 5 6 7 8 9 A 0 1 2 3 4 5 6 7 8 9 B 0 1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 1 6 17 18 1 9 ( 0 , 1 ) ca rr y A 0 1 2 3 4 5 6 7 8 9 A 0 1 2 3 4 5 6 7 8 9 B 0 1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 1 6 17 18 1 9 |
17679 | https://content.next.westlaw.com/Glossary/PracticalLaw/I8d74c59cef2a11e28578f7ccc38dcbee | Glossary | Practical Law - Legal Resources & Know-How for Professionals
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17680 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11633866/ | Elevated Middle Cerebral Artery Peak Systolic Velocity in Non-Anemic Fetuses: Providing a Better Understanding of Enigmatic Middle Cerebral Artery Peak Systolic Velocity - PMC
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Fetal Diagn Ther
. 2024 Jul 18;51(6):550–558. doi: 10.1159/000540342
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Elevated Middle Cerebral Artery Peak Systolic Velocity in Non-Anemic Fetuses: Providing a Better Understanding of Enigmatic Middle Cerebral Artery Peak Systolic Velocity
Saja Anabusi
Saja Anabusi
a Fetal Medicine Unit, Department of Obstetrics and Gynaecology, Mount Sinai Hospital and University of Toronto, Toronto, ON, Canada
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a, Tim Van Mieghem
Tim Van Mieghem
a Fetal Medicine Unit, Department of Obstetrics and Gynaecology, Mount Sinai Hospital and University of Toronto, Toronto, ON, Canada
b Ontario Fetal Centre, Toronto, ON, Canada
Find articles by Tim Van Mieghem
a,b, Greg Ryan
Greg Ryan
a Fetal Medicine Unit, Department of Obstetrics and Gynaecology, Mount Sinai Hospital and University of Toronto, Toronto, ON, Canada
b Ontario Fetal Centre, Toronto, ON, Canada
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a,b, Shiri Shinar
Shiri Shinar
a Fetal Medicine Unit, Department of Obstetrics and Gynaecology, Mount Sinai Hospital and University of Toronto, Toronto, ON, Canada
b Ontario Fetal Centre, Toronto, ON, Canada
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a,b,✉
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a Fetal Medicine Unit, Department of Obstetrics and Gynaecology, Mount Sinai Hospital and University of Toronto, Toronto, ON, Canada
b Ontario Fetal Centre, Toronto, ON, Canada
✉
Correspondence to: Shiri Shinar, shiri.shinar@sinaihealth.ca
✉
Corresponding author.
Received 2024 Jan 18; Accepted 2024 Jun 29; Issue date 2024 Dec.
© 2024 The Author(s). Published by S. Karger AG, Basel
This article is licensed under the Creative Commons Attribution 4.0 International License (CC BY) ( Usage, derivative works and distribution are permitted provided that proper credit is given to the author and the original publisher.
PMC Copyright notice
PMCID: PMC11633866 PMID: 39025054
Abstract
Introduction
Our aim was to investigate the incidence, comorbidities, and outcomes of fetuses with an elevated middle cerebral artery peak systolic velocity (MCA-PSV) >1.5 multiples of median (MoM), despite normal hemoglobin (Hgb) levels on fetal blood sampling (FBS).
Methods
A single-center observational retrospective cohort study of all fetuses undergoing FBS and MCA-PSV >1.5 MoM. Only those fetuses with no or mild anemia were included. Indications for Doppler assessment, associated anomalies, and neonatal outcomes were collected.
Results
Overall, 383 fetuses had an MCA-PSV >1.5 MoM and underwent FBS. Twenty-three (6%) fetuses met our inclusion criteria and had no or only mild anemia. Associations with elevated MCA-PSV were elucidated in 12 of the 23 cases (52.2%) and included mild anemia (n = 2), intracranial hemorrhage (n = 3), genetic disease (n = 1), idiopathic nonimmune hydrops (NIH, n = 1), hypoxic-ischemic encephalopathy (n = 1), maternal and or fetal acidosis (n = 3), and fetal growth restriction (n = 1). Favorable perinatal outcomes were observed in truly unexplained 11 cases with no additional anomalies (47.8%).
Conclusion
Elevated MCA-PSV >1.5 MoM with normal Hgb levels is seen in 6% of pregnancies undergoing FBS and is often associated with other significant maternal or fetal problems. Those with unexplained and isolated MCA-PSV elevation have normal outcomes.
Keywords: Middle cerebral artery peak systolic velocity, Fetal anemia, Fetal Doppler
Introduction
An elevated peak systolic velocity (PSV) in the middle cerebral artery (MCA) is a well-documented indicator of fetal anemia in cases of fetal red cell alloimmunization, with the MCA-PSV increasing as anemia severity progresses . In the absence of previous intrauterine transfusion (IUT), a value of MCA-PSV greater than 1.5 multiples of median (MoM) has a sensitivity of 100% (95% CI: 86–100%) and a false-positive rate of 12% for the diagnosis of moderate to severe fetal anemia [2–6]. Its high reliability allows to avoid unnecessary fetal blood sampling (FBS) in up to 70% of cases or rhesus alloimmunization .
Aside from fetal anemia, several other causes of elevated MCA-PSV have been documented. High values may also be caused by fetal activity or fetal growth restriction (FGR) with cerebral vasodilatation and redistribution of cardiac output . In addition, a cross-sectional study in patients with gestational diabetes demonstrated a significantly higher MCA-PSV when compared with nondiabetic controls . Finally, the reliability of MCA-PSV measurements following multiple transfusions has been questioned. In a recent meta-analysis , the accuracy of MCA-PSV >1.5 MoM for the prediction of moderate to severe anemia, defined as a hemoglobin (Hgb) deficit of more than 70 g/L, decreased significantly even after a single IUT and continued to decline with increasing numbers of IUTs.
Despite these known causes for “false-positive” MCA-PSV elevation and its recognized limitations in diagnosing anemia, some fetuses with repeatedly high MCA-PSV measurements without a clear explanation will still undergo diagnostic FBS while investigating other potential causes. The aim of our study was to describe such a cohort and document the phenotypes and outcomes of fetuses presenting with an elevated MCA-PSV >1.5 MoM, who had normal or near normal Hgb on FBS.
Methods
We retrospectively reviewed all fetuses who underwent FBS due to elevated MCA-PSV >1.5 MoM at the fetal medicine unit of Mount Sinai Hospital, Toronto, Canada, between 2007 and 2021. Only fetuses with no or mild anemia, defined as a Hgb deviation of equal to or less than 20 g/L from a gestational age (GA)-adjusted mean on FBS, were included in this study. This definition was chosen as it stems from a relatively large cohort and is GA-adjusted .
In our unit, MCA Doppler velocimetry is routinely measured in high-risk pregnancies with alloimmunization, FGR, suspected antepartum hemorrhage, and monochorionic twins and when clinical or ultrasound findings could be associated with fetal anemia (such as fetal hydrops, decreased fetal movements, and polyhydramnios). MCA-PSV measurements were obtained using a transabdominal approach using either Philips iU-22 (Philips Healthcare, PA, USA) or Voluson E10 (GE Healthcare, Zipf, Austria) ultrasound machines as previously described . An axial section of the fetal head was obtained, and color flow mapping was used to identify the circle of Willis and the MCA. The pulsed-wave Doppler gate was placed on the proximal one-third of the MCA, and the angle of insonation was less than 20°. Attention was taken to avoid any unnecessary pressure on the fetal head and the mechanical and thermal indices were kept below 1. At least three consecutive waveforms, in the absence of fetal body or breathing movements, were recorded, and the highest peak of the Doppler waveform was considered as the PSV (cm/s) . Measurements >1.5 MoM were considered elevated . MCA pulsatility index (MCA-PI) was also obtained. When the cause of the MCA-PSV elevation was unclear, a further workup was initiated including maternal blood for Kleihauer Betke test, Hgb electrophoresis, indirect Coombs, and TORCH screen including parvovirus serologies and measurements were typically repeated at least 24 h later to confirm the finding and reduce the risk of false-positive results. Patients with persistently elevated and often progressively increasing MCA-PSV were offered diagnostic FBS.
All fetal blood samples were obtained within 24 h of an elevated MCA-PSV measurement. FBS was done as previously described using an aseptic technique under local anesthesia with maternal sedation. The puncture site was either at the placental cord insertion or in the intrahepatic portion of the umbilical vein, based on accessibility.
In our cohort, to investigate alternative causes of elevated MCA-PSV, we examined maternal demographics and pregnancy characteristics, ultrasound findings, genetic testing results when available, FBS results (Hgb, hematocrit, platelet count, and pH when available), and perinatal outcomes. For the latter, we collected 5-min Apgar scores, neonatal Hgb levels, and any neonatal adverse events prior to discharge from hospital. Cases that despite this evaluation for potential cause, remained with no identifiable reason for elevated MCA-PSV Doppler velocimetry, were defined as unexplained. The study was approved by our Ethics Review Board (REB #21-0276-C).
Statistical Analysis
In this descriptive study, continuous variables are presented as mean ± SD or median (interquartile range), as appropriate. Categorical variables are expressed as n (%).
Results
During the 14-year study period, 383 fetuses had an MCA-PSV >1.5 MoM and underwent FBS. Only 23 fetuses (6%) met the study’s inclusion criteria, with no or only mild anemia on initial sampling. Maternal and pregnancy characteristics for this cohort are presented in Table 1.
Table 1.
Maternal and pregnancy characteristics in cases of unexplained elevated MCA-PSV
| | n = 23 |
--- |
| Maternal age, years | 31.9±5.5 |
| Prepregnancy BMI, kg/m 2 | 28.6±7.0 |
| Nulliparity | 8 (34.7) |
| Known maternal chronic disease† | 7 (30.4) |
| Maternal ethnicity |
| Caucasian | 14 (60.8) |
| Black | 1 (4.3) |
| Asian | 6 (26) |
| Missing | 2 (8.6) |
| ART, n (%) | 0 |
| GA at FBS, weeks | 32.5±2.2 |
| Fetal Hgb, g/L | 132.8±15.5 |
| Fetal Hct, % | 40.5±5.8 |
| Fetal PLT, ×10 9/L | 177.3±87.2 |
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Data are reported as mean ± SD or n (%) as appropriate.
BMI, body mass index; ART, assisted reproductive technology; MCA, middle cerebral artery; PSV, peak systolic velocity; Hgb, hemoglobin; FBS, fetal blood sampling; Hct, hematocrit; PLT, platelets.
†Diabetes mellitus, hypertension, and hypothyroidism.
Indications for MCA-PSV Measurement
The most common indication for MCA-PSV measurement in our study cohort was abnormal fetal surveillance (atypical/abnormal fetal heart rate monitoring, abnormal biophysical profile, or fetal/placental abnormalities; n = 11, 47.8%). The fetal/placental abnormalities group included cases of nonimmune hydrops of unknown etiology (n = 1), trisomy 21 (n = 1), Noonan syndrome (n = 1), FGR (n = 1), suspected twin anemia polycythemia sequence (n = 2) in monochorionic diamniotic twins, maternal cytomegalovirus seroconversion in pregnancy with unknown fetal status (n = 1), and abnormal biophysical profile score in a case followed for placenta accreta spectrum (n = 1). Three pregnancies presented with non-reassuring fetal status on cardiotocography. Of these, two were associated with maternal metabolic acidosis at time of the MCA-PSV sampling (diabetic ketoacidosis in one and pancreatitis in another).
The second most common indication of MCA-PSV sampling was red cell alloimmunization (n = 5, 21.7%). Despite an elevated MCA-PSV, the fetal Hgb was either normal (n = 3) or consistent with mild anemia only (n = 2). These cases accounted for 2.8% of the 176 cases of alloimmunization in the 383 fetuses that underwent FBS. None of these five cases had an IUT prior to FBS. Four fetuses (17.4%) underwent MCA-PSV measurements because of maternal antepartum hemorrhage, and in three cases (13%), the reason for initial MCA-PSV measurement was unclear.
Final Diagnosis and Perinatal Outcome
Indications of Doppler assessment and perinatal findings are presented in Table 2, and neonatal outcomes are presented in Table 3 and Figure 1. An underlying condition associated with the finding of elevated MCA-PSV >1.5 MoM was eventually elucidated in 12 of the 23 cases, defined as explained elevated MCA-PSV (52.2%) and included mild anemia (n = 2), intracranial hemorrhage (n = 3), postnatal diagnosis of a rare genetic condition (n = 1) (Table 2, case 2), idiopathic nonimmune hydrops (NIH, n = 1), metabolic fetal acidosis (maternal and fetal n = 3), fetal hypoxic-ischemic encephalopathy (HIE, n = 1) (Table 2, case 11), and FGR (n = 1). In the cases where an underlying explanation other than mild anemia was identified, outcomes varied. Poorest outcomes (neonatal death [n = 5] or termination of pregnancy due to expected poor prognosis [n = 1]) were observed in cases with intracranial hemorrhage (n = 3), HIE (n = 1), NIH (n = 1), or a genetic abnormality (n = 1). A favorable outcome was observed in the 11 neonates (47.8%) with no identifiable cause for elevated MCA-PSV (Table 3). This included one fetus with trisomy 21, but no hematologic abnormality to explain the elevated MCA-PSV. Of note, the MCA-PI did not differ between the cases where the cause of elevated MCA-PSV was elucidated and those where the cause remained unknown, making MCA redistribution an unlikely explanation for elevated MCA-PSV Doppler velocimetry.
Table 2.
Antenatal and postnatal findings in fetuses with elevated MCA-PSV
| | Reason for referral | Specific anomaly/risk factor for fetal anemia | GA at FBS | MCA-PSV, cm/s | MCA-PSV MoM | Fetal Hgb, g/L | Approximate mean Hgb for GA, g/L | Delta | Hct, % | PLT | Final diagnosis | Outcome |
--- --- --- --- --- ---
| 1 | Abnormal fetal surveillance | Echogenic liver, splenomegaly, and CMV seroconversion in pregnancy | 32.7 | 80 | 1.75 | 142 | 142 | 0 | 0.41 | 261 | Unexplained | Live birth without major sequelae |
| 2 | Abnormal fetal surveillance | Splenomegaly, hepatomegaly | 33.7 | 90 | 1.9 | 176 | 144 | Above mean | 0.52 | 38 | Noonan-like syndrome | NND |
| JMML associated with a de novo mutation in the CBL gene |
| 3 | Abnormal fetal surveillance | AVSD and pericardial effusion | 35.1 | 95.3 | 1.85 | 125 | 144 | 19 | 0.36 | 144 | Unexplained | Live birth |
| Trisomy 21 |
| 4 | Abnormal fetal surveillance | Maternal pancreatitis | 33.3 | 112 | 2.37 | 157 | 142 | Above mean | 0.49 | 212 | Maternal and fetal acidosis | Live birth without major sequelae |
| Abnormal fetal tracing |
| 5 | Abnormal fetal surveillance | Maternal DKA | 32.1 | 97 | 2.17 | 135 | 140 | 5 | 0.4 | 227 | Maternal DKA and fetal acidosis | Live birth without major sequelae |
| Suspected fetal hydrops |
| 6 | Abnormal fetal surveillance | Polyhydramnios | 32.7 | 93 | 2 | 129 | 142 | 13 | 0.42 | 77 | Fetal acidosis of unknown etiology | Live birth without major sequelae |
| 7 | Abnormal fetal surveillance | FGR | 29.6 | 99 | 2.49 | 125 | 136 | 11 | 0.38 | 264 | FGR | Live birth without major sequelae |
| 8 | Abnormal fetal surveillance | Hydrops | 30 | 60 | 1.5 | 116 | 135 | 19 | 0.34 | 94 | NIH of unknown etiology | NND |
| 9 | Abnormal fetal surveillance | Suspected TAPS | 26.6 | 72 | 2.11 | 146 | 130 | Above mean | 0.46 | N/A | Periventricular hemorrhagic infarct diagnosed postnatally | NND |
| 10 | Abnormal fetal surveillance | TAPS and mild VM with suspected IVH | 26.3 | 67 | 1.97 | 152 | 128 | Above mean | 0.58 | 125 | Mid VM and IVH | NND |
| 11 | Abnormal fetal surveillance | BPP 2/8 | 30.1 | 107 | 2.62 | 116 | 136 | 20 | 0.34 | 95 | HIE | NND |
| 12 | Alloimmunization | Anti-D | 34 | 89 | 1.85 | 127 | 144 | 13 | 0.38 | 204 | Mild anemia | Live birth without major sequelae |
| 13 | Alloimmunization | Anti-D | 30.5 | 70 | 1.67 | 137 | 137 | 0 | 0.36 | 114 | Unexplained | Live birth without major sequelae |
| 14 | Alloimmunization | Anti-D | 32.7 | 75 | 1.63 | 136 | 142 | 6 | 0.39 | 180 | Unexplained | Live birth without major sequelae |
| 15 | Alloimmunization | Anti-C and anti-E | 34.1 | 75 | 1.53 | 125 | 144 | 19 | 0.42 | N/A | Mild anemia | Live birth without major sequelae |
| 16 | Alloimmunization | Anti-Fya | 29 | 80 | 2.1 | 134 | 134 | 0 | 0.42 | 341 | Unexplained | Live birth without major sequelae |
| 17 | Suspected APH | Positive BK post-MVA | 31.7 | 66 | 1.51 | 135 | 140 | 5 | 0.39 | 298 | Unexplained | Live birth without major sequelae |
| 18 | Suspected APH | APH | 35.4 | 100 | 1.93 | 125 | 136 | 11 | 0.35 | 166 | Unexplained | Live birth without major sequelae |
| 19 | Suspected APH | Post-MVA | 32.7 | 71 | 1.72 | 128 | 142 | 14 | 0.4 | 168 | Periventricular hemorrhagic infarct | TOP |
| 20 | Suspected APH | APH | 32.8 | 83 | 1.8 | 152 | 141 | | 0.46 | 164 | Unexplained | Live birth without major sequelae |
| 21 | No indication for follow-up | Previous stillbirth | 34.1 | 91 | 1.85 | 128 | 144 | 16 | 0.38 | 142 | Unexplained | Live birth without major sequelae |
| 22 | No indication for follow-up | Maternal diabetes routine follow-up | 32.1 | 98 | 2.19 | 150 | 140 | | 0.47 | 225 | Unexplained | Live birth without major sequelae |
| 23 | No indication for follow-up | | 32.5 | 80 | 1.75 | 128 | 141 | 13 | 0.37 | 212 | Unexplained | Live birth without major sequelae |
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FBS, fetal blood sample; MCA, middle cerebral artery; PSV, peak systolic velocity; Hgb, hemoglobin; Hct, hematocrit; PLT, platelets; JMML, juvenile myelomonocytic leukemia; CBL, Casitas B-lineage lymphoma; NND, neonatal death; AVSD, atrioventricular septal defect; DKA, diabetic ketoacidosis; FGR, fetal growth restriction; NIH, nonimmune hydrops; TAPS, twin anemia polycythemia sequence; IVH, intraventricular hemorrhage; VM, ventriculomegaly; BPP, biophysical profile; HIE, hypoxic-ischemic encephalopathy; BK, Kleihauer Betke; MVA, motor vehicle accident; APH, antepartum hemorrhage; TOP, termination of pregnancy; CMV, cytomegalovirus.
Explained elevated MCA-PSV.
Table 3.
Perinatal outcomes of fetuses with elevated MCA-PSV despite normal hemoglobin
| | n = 23 |
--- |
| GA at delivery, weeks | 34.1±3.4 |
| Cesarean delivery | 15 (65.2) |
| Birthweight, g | 2,682±1,138 |
| Female sex | 11 (47.8) |
| 5′ Apgar score (IQR) | 9 (8.9) (n = 19) |
| Termination of pregnancy | 1 (4.4) |
| Early neonatal death | 5 (22) |
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Data are reported as mean ± SD, n (%) or median and IQR as appropriate.
IQR, interquartile range.
Fig. 1.
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Outcomes stratified by indication for FBS.
Discussion
We here present a population of fetuses with high MCA-PSV Doppler despite a normal or near normal Hgb level. Overall, this population comprised 6% of all pregnancies presenting with an elevated MCA-PSV >1.5 MoM that underwent FBS. In our cohort, the leading causes of abnormally high MCA-PSV in fetuses without moderate to severe anemia included intracranial hemorrhage, fetal acidosis, and mild anemia. When no associated findings were seen, neonatal outcomes were favorable. However, when an association with an underlying condition other than mild anemia or transient acidosis was detected, prognosis was generally poor with high rates of neonatal death.
The reported performance rate of MCA-PSV varies in the literature and depends on multiple factors, including fetal activity, diabetes, FGR, and previous blood transfusions [7–10]. Some have also associated it with ethnicity, creating reference ranges that are population-specific [18–20]. Mari et al. , in their seminal paper, reported a sensitivity of 100% with a false-positive rate of 12%. In comparison, a recent meta-analysis reported an 86% pooled sensitivity and 72% pooled specificity . We, on the other hand, found a lower false-positive rate of 6%. This lower rate may be explained by performance of the MCA assessment almost exclusively only when indicated and repeating the assessment multiple times. Moreover, it is not common practice in our unit to perform FBS due to an elevated MCA-PSV when the index of suspicion is very low. In these cases, we generally follow closely and perform a FBS only in the absence of an alternative diagnosis and when the measurements are persistently elevated on subsequent follow-up visits.
A rather frequent explanation for elevated MCA-PSV with normal Hgb in our cohort was the presence of an intracranial hemorrhage. Our group has previously demonstrated that in cases of acute cerebral injury, as occurs with single demise in monochorionic twins, MCA-PSV of the survivor co-twin may be elevated . It is possible that in the three hemorrhagic cases in our cohort, transient fetal hypovolemia ensued, which resulted in cerebral HIE and altered cerebral blood flow . This would also explain why the MCA-PSV was elevated in the case of postnatal detection of HIE (Table 2, case 11). Another possible explanation for the elevated MCA-PSV in these cases may be a pressure effect from a focal bleed. This may cause disruption of the cerebral tissue and edema as well as affect cerebral blood flow regulation, altering flow velocity in that region. Such changes have been documented postnatally . More research is required to establish a direct causal relationship between cerebral tissue disruption and elevated MCA-PSV Doppler. Given these findings, a detailed neurosonogram may be worthwhile in these cases of unexplained elevated MCA-PSV.
Maternal acidosis can result in fetal acidosis and acidemia, triggering hemodynamic modifications aimed at protecting areas essential to fetal survival, such as the brain. This selective vasodilatation, known as fetal brain sparing or centralization [24, 25], results in altered Doppler flow to the brain, mainly affecting the pulsatility index. Our finding that maternal metabolic acidosis resulted in elevated MCA-PSV is contradictory to the results by Liao et al. , which did not find a correlation between fetal acidemia at birth MCA-PSV Doppler. We hypothesize that the maternal acidosis lowers the maternal blood oxygen affinity, promoting higher availability of placental oxygen transport to the fetal system (the Bohr effect). In addition, fetal acidemia also decreases the oxygen affinity, thereby promoting oxygen availability to tissues . In order to maintain oxygen delivery to the brain, the fetal circulation may undergo adaptive changes affecting the MCA-PSV .
Not surprisingly, despite no significant anemia in fetuses in this cohort, those with high MCA-PSV and associated anomalies (NIH, three intracranial hemorrhages, and genetic syndrome [case 2]) had poor outcomes and a high risk of demise. This finding is in line with findings from a previous study that demonstrated a valuable role for MCA-PSV in prediction of impending fetal death in growth-restricted fetuses. Thus, in our practice, growth-restricted fetuses who show persistent and progressive elevation in the MCA-PSV measurement would undergo further assessment through FBS to determine their Hgb status. On the other hand, in cases where no anomaly was detected after thorough prenatal and postnatal investigations, outcomes were favorable with healthy neonates. This finding aligns with Rosello et al. , who showed no correlation between MCA-PSV MoM in the absence of anemia and neonatal Hgb and pH at time of delivery.
It should be noted that in three of our fetuses with unexplained elevated MCA-PSV, no clear indication for this assessment was found. Despite lack of clinical suspicion, a fetal blood sample was carried out with a normal Hgb as a result. This emphasizes that when no indication for MCA-PSV sampling exists, it should be discouraged, as it may lead to unnecessary fetal interventions that carry a risk to the pregnancy.
One of the strengths of this study is the large number of FBS performed in a single tertiary center and analyzed by the same laboratory. In each case included in the study, several FBS were meticulously collected in each procedure to ensure diagnostic certainty regarding fetal anemia. Fetal Hgb levels were adjusted for GA to ensure accurate diagnosis of fetal anemia and thus MCA-PSV test accuracy. Additionally, the assessment for fetal anemia was done within a similar timeframe as was the MCA-PSV reading, allowing for a reliable correlation between the two and thereby comprising the study population in which we were interested. Finally, the comprehensive nature of data collection incorporating neonatal outcomes and postnatal investigations from a single tertiary care center further enhances the overall robustness of the study.
This study has some notable limitations, owing mostly to its retrospective design and different physician practices affected by ease and individual comfort level with FBS. The latter may have resulted in some physicians performing a fetal blood sample despite a low index of suspicion of fetal anemia and values just above 1.5 MoM, while others may have acted only on higher values and a higher index of suspicion. This may have affected our reported false-positive rate of MCA-PSV for anemia detection. An additional limitation is that this study spanned over a 14-year time period. During this time, genetic investigations changed significantly from routine karyotyping to chromosomal microarray and whole exome sequencing. Thus, is it possible that older cases with an unknown cause for elevated MCA-PSV may have been explained though next-generation sequencing that was not available at that time?
In conclusion, in experienced fetal medicine centers, the false-positive rate of MCA-PSV for detection of fetal anemia may be lower than previously reported. Although approximately half of the cases with an elevated MCA-PSV without significant fetal anemia resulted in normal outcome, the other 50% were associated with an increased risk of perinatal morbidities largely related to and driven by findings detectable by prenatal ultrasonography, antenatal fetal surveillance, and/or genetic testing. As such, in non-anemic fetuses, a thorough workup and continued or increased close fetal surveillance would seem indicated. Further research in a larger fetal population is recommended to confirm these findings and establish appropriate clinical investigative strategies.
Statement of Ethics
The study was approved by Mount Sinai Health System Ethics Review Board (REB #21-0276-C). This study was granted an exemption from requiring written informed consent by the Mount Sinai Health System Ethics Review Board, as it involved retrospective analysis of anonymized data and did not pose any risk to participants’ confidentiality or privacy.
Conflict of Interest Statement
The authors have no conflicts of interest to declare.
Funding Sources
The authors received no financial support for the research, authorship, and/or publication of this article.
Author Contributions
Saja Anabusi and Shiri Shinar contributed to the study design, data collection, data analysis, and writing of the manuscript and its revisions. Tim Van Mieghem contributed to the study design and writing of the manuscript. Greg Ryan contributed to the data collection and writing of the manuscript.
Funding Statement
The authors received no financial support for the research, authorship, and/or publication of this article.
Data Availability Statement
The data that support the findings of this study are not publicly available due to their containing information that could compromise the privacy of research participants and are integral to future research plans. However, they are available from the corresponding author, SS, upon reasonable request. Requests for data access can be directed to Shiri Shinar at shiri.shinar@sinaihealth.ca.
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The data that support the findings of this study are not publicly available due to their containing information that could compromise the privacy of research participants and are integral to future research plans. However, they are available from the corresponding author, SS, upon reasonable request. Requests for data access can be directed to Shiri Shinar at shiri.shinar@sinaihealth.ca.
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Who Is at Fault in a Motorcycle and Pedestrian Accident?
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Motorcycle Accident
By Allen Clardy
Motorcycle and pedestrian accidents present complex legal challenges, often leaving those involved with significant injuries and unanswered questions about liability. Determining who is at fault in such incidents requires a thorough understanding of traffic laws, the circumstances of the accident, and the behaviors of both parties involved.
As a personal injury lawyer, it is crucial to analyze each case meticulously to establish fault and seek justice for the injured party.
Understanding Liability in Motorcycle and Pedestrian Accidents
The Basics of Fault Determination
Fault in motorcycle and pedestrian accidents is determined based on the principles of negligence. Negligence occurs when an individual fails to exercise reasonable care, resulting in harm to another person. To establish negligence, four elements must be proven:
Duty of care:The defendant owed a duty of care to the plaintiff.
Breach of duty: The defendant breached that duty of care.
Causation: The breach caused the plaintiff’s injuries.
Damages: The plaintiff suffered actual damages (physical, emotional, financial) as a result.
Common Causes of Motorcycle and Pedestrian Accidents
Accidents involving motorcycles and pedestrians can happen due to various reasons, including:
Distracted Driving: Motorcyclists not paying attention to the road, often due to mobile phone use, can fail to see pedestrians.
Speeding: High speeds reduce reaction times and increase the severity of collisions.
Impaired Driving: Alcohol or drug impairment can lead to reckless behavior and accidents.
Pedestrian error: Pedestrians crossing roads improperly or unexpectedly entering traffic can cause accidents.
Poor visibility: Weather conditions, low light, or blind spots can obscure the view of both motorcyclists and pedestrians.
Legal Obligations of Motorcyclists and Pedestrians in South Carolina
Both motorcyclists and pedestrians have legal responsibilities to ensure their own and others’ safety. Understanding these obligations is critical in determining fault.
Motorcyclists’ Responsibilities
Obey traffic laws:Motorcyclists must adhere to all traffic signals, speed limits, and road signs.
Yield to pedestrians: When crossing at marked crosswalks or intersections, motorcyclists must yield to pedestrians.
Drive sober: Operating a motorcycle under the influence of alcohol or drugs is illegal and highly dangerous.
Avoid distractions: Staying focused on the road and avoiding distractions like mobile phones is essential.
Wear a helmet: Riders under the age of 21 are required to wear helmets.Helmets must be approved by the Department of Public Safety and meet specific safety standards. Passengers under the age of 21 must also wear helmets.
Remain in a single lane: Lane splitting (riding between lanes of traffic) is illegal in South Carolina.
Have working lights: Motorcycles must have a working headlight, tail light, and brake light. Headlights must be used at all times, day and night.
Hold a valid motorcycle license: Motorcyclists must have a valid motorcycle endorsement on their driver’s license.
Additional laws can be read in the SCDPS’ South Carolina Motorcycle Laws article 29.
Pedestrians’ Responsibilities
Use crosswalks: Pedestrians should cross streets at designated crosswalks and follow pedestrian signals.
Yield to vehicles:Pedestrians must yield to vehicles when crossing streets at points without marked crosswalks or signals.
Stay alert: Avoiding distractions, such as texting while walking, ensures pedestrians are aware of their surroundings.
Obey traffic signals:Pedestrians must adhere to “walk” and “don’t walk” signals at intersections.
Avoid impairment: Just as with drivers, pedestrians impaired by alcohol or drugs may not act safely.
Walk on the left:When sidewalks are not available, pedestrians should walk on the left side of the roadway facing traffic.
Determining Fault in Specific Scenarios
The phrase “pedestrians have the right of way” isn’t always accurate. This doesn’t give pedestrians the right to be ignorant. They share equal responsibility for helping to ensure that roads are safe. In the scenario where a pedestrian is crossing traffic against safety signals or is walking in restricted areas, they can be deemed negligent. As a result of this, they can also be held responsible for an accident.
Scenario 1: Motorcyclist Failing to Yield
If a motorcyclist fails to yield to a pedestrian at a marked crosswalk and an accident occurs, the motorcyclist is likely to be found at fault. Traffic laws clearly state that vehicles must yield the right-of-way to pedestrians at crosswalks. In this scenario, the motorcyclist breached their duty of care by not yielding, leading to the pedestrian’s injuries.
Scenario 2: Pedestrian Jaywalking
Conversely, if a pedestrian jaywalks (crosses the street outside of a designated crosswalk) and is struck by a motorcyclist, the pedestrian may be found at fault. Jaywalking is illegal and puts pedestrians at risk, as drivers, including motorcyclists, are not expecting someone to cross at that point. The pedestrian’s action of crossing the road unsafely constitutes a breach of their duty of care.
Scenario 3: Shared Fault
In some cases, both the motorcyclist and the pedestrian may share fault. For instance, if a pedestrian crosses at a crosswalk but does not look for oncoming traffic, and a motorcyclist is speeding, both parties may be considered negligent. The motorcyclist for speeding and the pedestrian for not ensuring the road was clear before crossing. In such cases, comparative negligence rules may apply, where the fault is apportioned between both parties.
Scenario 4: Poor Visibility Conditions
Accidents occurring under poor visibility conditions, such as fog or heavy rain, add another layer of complexity. If a motorcyclist hits a pedestrian in these conditions, the determination of fault will consider whether the motorcyclist was driving appropriately for the weather conditions. Similarly, if the pedestrian did not take precautions, such as wearing visible clothing or using a flashlight, their negligence may be considered.
Investigating the Accident
Gathering Evidence
A thorough investigation is critical in determining fault in motorcycle and pedestrian accidents. Key steps include:
Collecting witness statements: Eyewitnesses can provide valuable insights into how the accident occurred.
Reviewing surveillance footage:Nearby cameras may have captured the incident, offering visual evidence.
Examining police reports: Law enforcement documentation can provide an unbiased account of the accident and initial determinations of fault.
Accident reconstruction: Experts can recreate the accident scenario to better understand the dynamics involved.
Analyzing Evidence
Evidence must be meticulously analyzed to build a strong case. This involves:
Assessing damage:Examining the extent and nature of damage to both the motorcycle and the pedestrian.
Evaluating medical reports: Understanding the injuries sustained and how they correlate with the accident details.
Comparing statements:Cross-referencing witness statements with physical evidence to identify inconsistencies or corroborations.
Legal Considerations and Comparative Negligence
Pure Comparative Negligence
In states that follow pure comparative negligence laws, each party can recover damages proportionate to their degree of fault. For example, if a pedestrian is found to be 30% at fault and the motorcyclist 70%, the pedestrian can still recover 70% of the total damages awarded.
Modified Comparative Negligence
Some states follow modified comparative negligence rules, which bar recovery if a party is found to be more than 50% at fault. In these states, if a pedestrian is 51% responsible for the accident, they would not be able to recover any damages.
Contributory Negligence
A few states adhere to contributory negligence laws, where any fault on the part of the plaintiff completely bars recovery. Thus, if a pedestrian is found to be even 1% at fault, they would be unable to recover damages from the motorcyclist.
The Role of Insurance Companies
Filing Claims
Both motorcyclists and pedestrians may need to file claims with insurance companies following an accident. Motorcyclists typically have liability insurance that may cover injuries to pedestrians. Pedestrians may also have personal injury protection (PIP) coverage under their health insurance policies.
Insurance Adjusters
Insurance adjusters will investigate the claim to determine the extent of liability and damages. They will evaluate medical records, witness statements, and other evidence. It is crucial to be cautious when dealing with insurance adjusters, as their goal is often to minimize the payout.
Settlements
Insurance companies may offer settlements to resolve claims quickly. It is important to consult with a motorcycle injury lawyerbefore accepting any settlement offers to ensure that the compensation is fair and covers all damages, including medical expenses, lost wages, and pain and suffering.
Seeking Legal Assistance
Why Hire a Personal Injury Lawyer?
Navigating the complexities of fault determination and insurance claims can be challenging. A motorcycle injury lawyer can:
Provide legal knowledge: Understand the nuances of traffic laws and negligence principles.
Conduct thorough investigations: Gather and analyze evidence to build a strong case.
Negotiate with insurance companies: Ensure fair compensation for the injured party.
Represent in court: Advocate for the injured party’s rights if a fair settlement cannot be reached.
Steps to Take After an Accident
Seek medical attention: Prioritize health and get immediate medical care for any injuries.
Report the accident: File a police report to document the incident.
Gather information: Collect contact information from witnesses and take photographs of the accident scene.
Consult a lawyer: Seek legal advice from a motorcycle injury lawyer to understand your rights and options.
Determining who is at fault in a motorcycle and pedestrian accident requires a careful analysis of the actions of both parties involved. Understanding traffic laws, gathering and analyzing evidence, and navigating insurance claims are crucial steps in seeking justice and compensation. By adhering to their respective legal responsibilities, both motorcyclists and pedestrians can contribute to safer roadways and reduce the risk of such accidents.
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Allen Clardy has been an attorney proudly serving South Carolinians since 1995. A lifelong Upstate resident, Allen is a trusted legal advocate known for his client-first approach and decades of success representing injured individuals. Allen started The Clardy Law Firm in 2010, and it has grown into a respected statewide practice with a reputation for personalized, results-driven representation. Allen has been recognized as a South Carolina Super Lawyer, Best Lawyer by Best Lawyers in America®, The Greenville Business magazine's Legal Elite, and Greenville TALK® magazine's Top Lawyer. Allen is also honored to have won a Greer Citizen award on two occasions for his work with the Toys for Tots campaign by the Greer Police Department. He is also an active supporter of the Greenville Humane Society, The PENFED foundation that supports our veterans, and other causes that give back to the community he’s proud to call home.
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17682 | https://www.scielo.cl/pdf/proy/v36n1/art10.pdf | Jensen’s and the quadratic functional equations with an endomorphism KH. Sabour IBN Tofail University, Morocco and S. Kabbaj IBN Tofail University, Morocco Received : October 2016. Accepted : October 2016 Proyecciones Journal of Mathematics Vol. 36, No 1, pp. 187-194, March 2017.
Universidad Cat´ olica del Norte Antofagasta - Chile Abstract We determine the solutions f : S →H of the generalized Jensen’s functional equation f(x + y) + f(x + ϕ(y)) = 2f(x), x, y ∈S, and the solutions f : S →H of the generalized quadratic functional equation f(x + y) + f(x + ϕ(y)) = 2f(x) + 2f(y), x, y ∈S, where S is a commutative semigroup, H is an abelian group (2-torsion free in the first equation and uniquely 2-divisible in the second) and ϕ is an endomorphism of S.
Subjclass : Primary 39B52.
Keywords : Functional equation, Jensen, quadratic, additive func-tion, semigroup.
188 KH. Sabour and S. Kabbaj 1. Set up, notation and terminology Throughout the paper we work in the following framework and with the following notation and terminology. We use it without explicit mentioning.
S is a commutative semigroup [a set equipped with an associative com-position rule (x, y) 7→x + y], ϕ : S →S is an endomorphism and (H, +) denotes an abelian group with neutral element 0. We say that H is 2-torsion free if [h∈H and 2h = 0] ⇒h = 0. H is said to be uniquely 2-divisible if for any h ∈H the equation 2x = h has exactly one solution x ∈H.
A function A : S →H is said to be additive if A(x + y) = A(x) + A(y) for all x, y ∈S.
We recall that the Cauchy difference Cf of a function f : S →H is defined by Cf(x, y) := f(x + y) −f(x) −f(y), x, y ∈S.
2. Introduction Let σ ∈Hom(S, S) satisfy σ2 = id. In , Sinopoulos determined the general solution f : S →H, where H is 2-torsion free, of Jensen’s functional equation f(x + y) + f(x + σ(y)) = 2f(x), x, y ∈S, (2.1) and the general solution f : S →H, where H is uniquely 2-divisible, of the quadratic functional equation f(x + y) + f(x + σ(y)) = 2f(x) + 2f(y), x, y ∈S.
(2.2) The purpose of the present paper is to solve the functional equations (2.1) and (2.2) on semigroups without using the condition σ2 = id. More precisely, we solve the functional equations f(x + y) + f(x + ϕ(y)) = 2f(x), x, y ∈S, (2.3) f(x + y) + f(x + ϕ(y)) = 2f(x) + 2f(y), x, y ∈S, (2.4) where ϕ is an endomorphism of S. By elementary methods we show that our formulas for the solutions of (2.3) (resp.
(2.4)) on semigroups are the same as those of (2.1) (resp. (2.2)), so that our results constitute a natural extension of earlier results of, e.g., , from Jensen’s (resp. the quadratic) functional equation with an involutive automorphism to that Jensen’s and the quadratic functional equations with an ...
189 with an endomorphism. Note that the equation (2.3) has been solved in under the assumption that ϕ is surjective. Here we remove this restriction.
A similar functional equation that has been studied is f(x + y) + f(x + ϕ(y)) = 2f(x)f(y), x, y ∈S, (2.5) where f : S →C is the function to determine. Eq. (2.5) was solved in a more general framework (see ).
3. The generalized Jensen’s functional equation In this section, we solve the functional equation (2.3) by expressing its solutions in terms of additive functions.
Lemma 3.1. Let f : S →H be a solution of the functional equation (2.3).
Then f(x + ϕ2(y)) = f(x + y) for all x, y ∈S.
(3.1) Proof.
Replacing y by ϕ(y) in (2.3), we get that f(x + ϕ(y)) + f(x + ϕ2(y)) = 2f(x).
Using this equation and (2.3), we obtain (3.1). 2 Lemma 3.2. Let S be a semigroup (not necessarily abelian), φ : S →S an endomorphism. If f : S →H and Φ : S × S →H satisfy f(xy) + f(φ(y)x) = Φ(x, y) for all x, y ∈S, (3.2) then 2f(xyz) = Φ(x, yz) −Φ(φ(z)x, y) + Φ(xy, z) for all x, y, z ∈S.
(3.3) Proof.
Making the substitutions (x, yz), (φ(z)x, y), and (xy, z) in (3.2), we get respectively f(xyz) + f(φ(yz)x) = Φ(x, yz), f(φ(z)xy) + f(φ(yz)x) = Φ(φ(z)x, y) f(xyz) + f(φ(z)xy)) = Φ(xy, z).
Subtracting the middle identity from the sum of the other two we get (3.3).
2 190 KH. Sabour and S. Kabbaj Theorem 3.3. Suppose that H is 2-torsion free.
The general solution f : S →H of the functional equation (2.3) is f = A + c, where A : S →H is an additive map such that A ◦ϕ = −A, and where c ∈H is a constant.
Proof.
The method used here is closely related to and inspired by the one in [11, Proof of Theorem 2]. Assume that f : S →H is a solution of (2.3). Replacing y by y + ϕ(y) in (2.3) and using Lemma 3.1, we get f(x + y + ϕ(y)) = f(x).
(3.4) Using lemma 3.2 with Φ(x, y) := 2f(x), we find after division by 2 that f(x + y + z) = f(x) −f(ϕ(z) + x) + f(x + y) = f(x) −[f(x + z) + f(ϕ(z) + x)] + f(x + z) + f(x + y) = f(x + y) + f(x + z) −f(x).
Setting here z = ϕ(x) and using (3.4), we get f(y) + f(x) = f(x + ϕ(x)) + f(x + y).
(3.5) Interchanging x and y in (3.5), we get that f(x + ϕ(x)) = f(y + ϕ(y)) for all x, y ∈S. So f(x + ϕ(x)) is a constant, say c. By using (3.5), we infer that the function A(x) := f(x) −c is additive. Substituting f into (2.3) we see that A ◦ϕ = −A.
The other direction of the proof is trivial to verify. 2 As an immediate consequence of Theorem 3.3, we have the following result.
Corollary 3.4 (2, Theorem 3.2). Suppose that H is 2-torsion free and let σ, τ ∈Hom(S, S) such that σ2 = τ2 = id. The general solution f : S → H of the functional equation f(x+σ(y)) + f(x + τ(y)) = 2f(x), x, y ∈S, is f = A+c, where A : S →H is an additive map such that A◦τ = −A◦σ, and where c ∈H is a constant.
Proof.
The proof follows on putting ϕ = τ ◦σ in Theorem 3.3. 2 4. The generalized quadratic functional equation In this section, we generalize Sinopoulos’s result [11, Theorem 3] on semi-groups by solving the functional equation (2.4).
Jensen’s and the quadratic functional equations with an ...
191 The following lemma lists pertinent basic properties of any solution f : S →H of (2.4).
Lemma 4.1. Suppose that H is 2-torsion free and let f : S →H be a solution of the functional equation (2.4).
(a) f ◦ϕ = f.
(b) For all x, y, z ∈S, we have f(x + y + z) = f(x + y) + f(x + z) +f(y + z) −f(x) −f(y) −f(z).
(4.1) (c) Cf : S × S →H is a symmetric, bi-additive map satisfying Cf(x, ϕ(y)) = −Cf(x, y) for all x, y ∈S.
(d) Let A : S →H be A(x) := f(x + ϕ(x)), x ∈S. Then A ◦ϕ = A and A is additive.
(e) 2f(x) = Cf(x, x) + A(x) for all x ∈S.
Proof.
(a) Let us first observe that f ◦ϕ is a solution of (2.4). We next replace x by ϕ(x) in (2.4) we find that f(ϕ(x) + y) + f(ϕ(x) + ϕ(y)) = 2f(ϕ(x)) + 2f(y).
Adding this equation to (2.4), we get [f(x + y) + f(ϕ(x) + y)] + [f(x + ϕ(y)) + f(ϕ(x) + ϕ(y))] = 2f(x) + 2f(ϕ(x)) + 4f(y).
Using (2.4) and the fact that H is 2-torsion free, we obtain [f(x) + f(y)] + [f(x) + f(ϕ(y))] = f(x) + f(ϕ(x)) + 2f(y), i.e.
f(x) −f ◦ϕ(x) = f(y) −f ◦ϕ(y) for all x, y ∈S.
From this last equation we infer that f −f ◦ϕ is a constant in H, say c.
Using the fact that f −f ◦ϕ is a solution of (2.4) and that H is 2-torsion free, we see that c = 0.
192 KH. Sabour and S. Kabbaj (b) Putting Φ(x, y) := 2f(x) + 2f(y) in lemma 3.2, we get after division by 2 that f(x + y + z) = f(x) + f(y + z) −f(ϕ(z) + x) −f(y) + f(x + y) + f(z) = f(x) + f(y + z) −[f(x + z) + f(ϕ(z) + x)] + f(x + z) −f(y) + f(x + y) + f(z) = f(x + y) + f(x + z) + f(y + z) −f(x) −f(y) −f(z), we get (4.1).
(c) That Cf is symmetric and bi-additive follows immediately from the very definition of Cf and (4.1). Let x, y ∈S be arbitrary. By help of (2.4) and (a), we get that Cf(x, ϕ(y)) = f(x + ϕ(y)) −f(x) −f ◦ϕ(y) = 2f(x) + 2f(y) −f(x + y) −f(x) −f ◦ϕ(y) = f(x) + f(y) −f(x + y) = −Cf(x, y).
(d) A is ϕ-even, because A(ϕ(x)) = f(ϕ(x) + ϕ2(x)) = 2f ◦ϕ(x + ϕ(x)) = f(x + ϕ(x)) = A(x) for all x ∈S.
Next, let x, y ∈S be arbitrary. By help of (4.1) and (a), we find A(x + y) = f((x + ϕ(x)) + y + ϕ(y)) = [f(x + ϕ(x) + y) + f(x + ϕ(x) + ϕ(y))] + f(y + ϕ(y)) −f(x + ϕ(x)) −f(y) −f ◦ϕ(y) = 2f(x + ϕ(x)) + 2f(y) + f(y + ϕ(y)) −f(x + ϕ(x)) −f(y) −f ◦ϕ(y) = f(x + ϕ(x)) + f(y + ϕ(y)) = A(x) + A(y).
(e) Using (2.4), we obtain Cf(x, x) + A(x) = f(x + x) + f(x + ϕ(x)) −2f(x) = 4f(x) −2f(x) = 2f(x) for all x ∈S.
The second main theorem of the present paper reads as follows.
Jensen’s and the quadratic functional equations with an ...
193 Theorem 4.2. Suppose that H is uniquely 2-divisible. The general solu-tion f : S →H of the functional equation (2.4) is f(x) = Q(x, x) + A(x), x ∈S, where Q : S × S →H is an arbitrary symmetric, bi-additive map such that Q(x, ϕ(y)) = −Q(x, y) for all x, y ∈S, and where A : S →H is an arbitrary additive function such that A ◦ϕ = A.
Proof.
That all solutions of (2.4) have this form is a consequence of Lemma 4.1 and the fact that H is uniquely 2-divisible. Conversely, simple computations based on the properties of Q and A, show that the indicated functions are solutions. 2 Acknowledgement. We wish to express our thanks to the referees for useful comments.
References B. Fadli, D. Zeglami and S. Kabbaj, On a Gajda’s type quadratic equation on a locally compact abelian group, Indagationes Math. 26, pp. 660-668, (2015).
B. Fadli, A. Chahbi, Iz. EL-Fassi and S. Kabbaj, On Jensen’s and the quadratic functional equations with involutions, Proyecciones (Antofa-gasta) 35 (2), pp. 213-223, (2016).
B. Fadli, S. Kabbaj, Kh. Sabour and D. Zeglami, Functional equations on semigroups with an endomorphism, Acta Math. Hungar. (2016), DOI 10.1007/s10474-016-0635-9.
B. Fadli, D. Zeglami and S. Kabbaj, A variant of Jensen’s functional equation on semigroups, Demonstratio Math., to appear.
P. de Place Friis and H. Stetkær, On the quadratic functional equation on groups, Publ. Math. Debrecen 69, pp. 65-93, (2006).
P. Kannappan, Functional equations and inequalities with applica-tions, Springer, New York, (2009).
194 KH. Sabour and S. Kabbaj C.T. Ng, Jensen’s functional equation on groups, Aequationes Math.
39, pp. 85-99, (1990).
C.T. Ng, Jensen’s functional equation on groups, II, Aequationes Math. 58, pp. 311-320, (1999).
C.T. Ng, Jensen’s functional equation on groups, III, Aequationes Math. 62, pp. 143-159, (2001).
Th. M. Rassias, Inner Product Spaces and Applications, Pitman Re-search Notes in Mathematics Series, Addison Wesley Longman Ltd, 376, (1997).
P. Sinopoulos, Functional equations on semigroups, Aequationes Math.
59, pp. 255-261, (2000).
H. Stetkær, Functional equations on abelian groups with involution, Aequationes Math. 54, pp. 144-172, (1997).
H. Stetkær, On Jensen’s functional equation on groups, Aequationes Math. 66, pp. 100-118, (2003).
H. Stetkær, Functional Equations on Groups, World Scientific Pub-lishing Co, Singapore, (2013).
KH. Sabour Department of Mathematics, Faculty of Sciences, IBN Tofail University, BP: 14000. Kenitra, Morocco e-mail : khsabour2016@gmail.com and S. Kabbaj Department of Mathematics, Faculty of Sciences, IBN Tofail University, BP: 14000. Kenitra, Morocco e-mail : samkabbaj@yahoo.fr |
17683 | https://missmastalio.wordpress.com/2020/02/25/classroom-mashup-03-quick-desmos-activities/ | Classroom Mashup 03: Quick Desmos Activities – Mastalio. Math. Mavericks.
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Classroom Mashup 03: Quick Desmos Activities
First, I’m renaming my “Friday Mashup” series, because who wants to be limited to posting that only on Fridays? It’s a quick rundown of things I want to share that aren’t quite big enough for their own blog post. Here are previous ones:
01 | 02
Today’s is about two quick Desmos Activity Builders that I created, one for my Algebra 1 class and one for Algebra 2.
Polygraph: Polynomial Functions
If you haven’t used a Desmos Polygraph in your classroom yet, you’re missing out! It’s the most surefire way to force students to reckon with their vocabulary understanding. There’s no way for them to pretend here. If they don’t understand it, they have to ask for help or they literally cannot play the game. There exists a Polygraph for pretty much every type of function, but most of them are the graphs of those functions.
My Algebra 1 students are learning terminology for degrees of polynomials and the number of terms that it has, and in past years I have realized that this vocabulary does not stick at all. We finish a whole unit on solving quadratics and they don’t know what the word quadratic actually means! So I made a polygraph, but of polynomial EXPRESSIONS instead of graphs. In Desmos, you can insert a picture into their graphs, so I typed out the equations, screengrabbed them, and inserted them into graphs, turning the axes and gridlines off.
Here you can see an example game from two of my students.
(Sidenote: You can sign up for a free teacher account with Equatio and easily type mathematical notation or create graphs without having to open a Google Doc or Word Doc, for purposes of screengrabbing. I use it constantly to insert equations on things that don’t have a built in equation editor, or for things like this)
Matching Factored Form to Standard Form Polynomials
My Algebra 2 students are just starting to factor polynomial expressions, in preparation for solving polynomial equations. I have a traditional worksheet assignment for them on this topic, but I wanted a bit of practice first that was more guided. I thought initially of doing a question stack, but I didn’t want to do 10 questions and didn’t want to spend the time creating a new template for fewer questions. My next thought was matching, but I realized I didn’t really want to print that out on cards, and then my brain remembered that card sorts are an option on Desmos Activity Builder!
Each polynomial expression has a matching factored form card, which students can drag on top of each other to match together. The answer key is even set so that you can put the teacher screen on the board for students to check their work if you want! I then simply added a graph screen onto the activity, since my students are factoring using x-intercepts on a graph to get them started, so that they could do their graphing without leaving the activity or needing multiple tabs. If you don’t want this screen, you could make a copy of the activity and delete that screen!
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I haven’t used this one with students yet, but I think it will be a good quick check of their understanding and a confidence builder before we do more open ended factoring work.
Let me know if you use either of these Desmos Activities, or if you have ideas to improve them (or if you improve them yourself!)
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17684 | https://www.askiitians.com/forums/11-grade-physics-others/what-is-the-relation-between-compressibility-and-b_470582.htm | Flag 11 grade physics others> What is the relation between compressibil...
What is the relation between compressibility and bulk modulus?
What is the relation between compressibility and bulk modulus?
Aniket Singh , 6 Months ago
Askiitians Tutor Team
Last Activity: 6 Months ago
The compressibility of a substance refers to its ability to undergo compression or decrease in volume when subjected to an external force. It is a measure of how much a substance can be compressed under a given pressure. On the other hand, the bulk modulus is a material property that describes the resistance of a substance to uniform compression or volume change.
The relation between compressibility (β) and bulk modulus (K) can be defined mathematically as:
β = 1 / K
Here, β represents the compressibility and K represents the bulk modulus. This equation implies an inverse relationship between compressibility and bulk modulus. In other words, as the bulk modulus increases, the compressibility decreases, and vice versa.
This relationship can be understood intuitively: substances with higher bulk modulus are less compressible because they exhibit greater resistance to changes in volume when subjected to external forces. Conversely, substances with lower bulk modulus are more compressible as they allow for easier volume changes under external pressures.
It's important to note that the bulk modulus and compressibility are related to each other, but they are not the same thing. The bulk modulus describes the material's response to compression, while compressibility measures the actual degree of compression a substance undergoes.
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17685 | https://journal.chestnet.org/article/S0012-3692(16)58836-9/pdf | Nitroprusside Therapyin Post-Open Heart Hypertensives A Ritual Tryst with Cyanide Death?
To the Editor:
Patel et al (Chest 1986; 89:663-67) confirmed our initial fears about therapy with intravenous nitroprusside (NP) in open heart patients.' Extrapolating from their experience, about 3,000 patients may have cyanide toxicity and 1,000 cyanide deaths will occur from NP therapy in the postoperative open heart period in the critical care setting (even after appropriate recognition and use of antidote) in the U.S. alone this year, assuming 180,000 coronary bypass operations are done and NP is used to treat hypertension. Not stated in their report is the total number of patients who received NP, of which seven patients developed cyanide toxicity with close to 50 percent mor-tality. One of the two patients we initially reported' died a month later of congestive heart and respiratory failure. Of note, the patient was seen and assessed by astute clinical staff, Fellows, and critical care nurses in the acute care setting only a few minutes before she was found by chance to be gasping for breath but arousable. Clinical assessment and laboratory values of NP on suspicion for blood gas, electrolytes, lactic acid and mixed venous Po. were within normal limits, except for elevated cyanide levels were assessed just prior to cessation. With immediate cessation of NP infusion and start of intravenous hydralazine, the patient made remarkable recovery with
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peripheral and pulmonary vascular resistances, which were elevated during NP infusion. No sodium thiosulphate was given. Apart from hypertension, these patients in the immediate postop erative period have stable vital systems functions. With the use of intravenous NP a small but unpredictable percentage may be exposed to cyanide toxicity within six hours or days, even when half the recommended doses are used.' Such patients have close to 50 percent mortality as reported by Patel et aI, even when detected and treated. Nitroprusside as a factor in the causation of death and multisystem failure in these patients is at present empirical. Our own clinical experience, however, parallels that of Patel et al. Intravenous nitroglycerine (NTG) therapy, on the other hand, has beneficial physiologic effects enhancing collateral circulation and reducing infarct size. 3,4 Tachyphalaxis to NTG can be reversed with Nvactycysteine." Methemoglobinemia is corrected with intravenous methylene blue. Alcoholic intoxication can be avoided by using alcohol-free NTG. Prophylactic NTG infusion may even reduce the incidence of postoperative hypertension" and, if given with in-travenous magnesium, may have added benefit of reducing ar-rhythmtas.?" No other toxic or adverse effects ofNTG are reported. Other vasodilators may be used in conjunction with NTG. Cyanid e in expe rime ntal syste ms gen er ates free radic al injury to endothelial cells which may be operative in multisystem failure." Sodium nitroprusside does not blunt catecholamine response in postoperative cardiac surgery patients." Catecholamines may also be involved in oxidant stress. Cyanide has also been implicated with cytotoxic inhibition, perhaps a factor in sepsis. Converting enzyme inhibitors on the other hand blunt sympa-thetic response in patients with congestive failure." If this response holds true in postoperative cardiac surgery patients, then a combina-tion of nonsulphated converting enzyme Inhibitors" and nitro-glycerine offers a physiologic approach to the treatment of postoperative hypertension without the attendant and unpredicta-ble risks of cyanide toxicity when nitroprusside is used. With our initial clinical suspicions and the further information now provided by Patel et aI, use of NP therapy in postoperative open heart patients for treatment of hypertension needs urgent reevalua-tion.
796
Sarvotham S. S., M.D., F.R.C.S.(C), Chagrin Valley Medical Center Chagrin Falls, Ohio
REFERENCES 1 Sarvotham SS, Patel C, Venus B, Mathu M. IV nitroglycerin vs nitroprusside treatment of myocardial ischemia. Crit Care Med 1985; 13:212-13 2 Pasch T, Schulx V, Hoppelshauser G. Nitroprusside-induced formation of cyanide and detoxication with thiosulfate during deliberate hypotension. J Cardiovasc Pharmacol1983; 5:77-84 3 Vatner SF, Pagani M, Manders WT, Pasipoularides AD. Alpha adrenergic vasoconstriction and nitroglycerine vasodilation of large coronary arteries in the conscious dog. J Clin Invest 1980; 65:5-14 4 Mudge GH Jr, Grossman W, Mills RM Jr, Lesch M, Braunwald E. Reflex increase in coronary vascular resistance in patients with ischemic heart disease. N Engl J Med 1976; 295:1333-37 5 Packer M, Lcc WH, Kessler P, Medina N, Yushak M. Induction of nitrate tolerance in human heart failure by continuous in-travenous infusion of nitroglycerin and reversal of tolerance by N-
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6 Gallagher JD, Moore RA, Jose AB, Botros SB, Clark DL. Prophylactic nitroglycerin infusions during coronary artery by-pass surgery. Anesthesiology 1986; 64:785-89 7 Sarvotham SS. Letter to the editor. Resident Staff Phys 1985; 31:14-15 8 Rasmussen HS, Norregard P, Lindeneg 0, McNair P, Backer V,
Balslev S. Intravenous magnesium in acute myocardial infarc-tion. Lancet 1986; 1:234-35 9 DiCarlo LA, Morady F, deBuitleir BM, Krol RB, Schurig L, Annesley TM. Effects of magnesium sulfate on cardiac conduc-tion and re iractonnesss in humans. JA (; (; l \Jisb; t : l;1bb-b~
10 Ward WG, ReifRL. High dose intravenous nitroglycerin during cardiopulmonary resuscitation for refractory cardiac arrest. Am J Cardiol1984; 53:1725 11 Ma yer DB, Feld L Miletich DL Albrecht RF. The efficacy of magnesium sulfate in the treatment of halothan epinephrine induced cardiac arrhythmias. Anesthesiology 1985; 63:3A 12 Weiss SJ, Young J, LoBuglio AF, Slivka A, Nimeh NF. Role of hydrogen peroxide in neutrophil-mediated destruction of cultured endothelial cells. J Clin Invest 68:714-721, 1981. 13 Corr lA, Grounds RM, Brown MJ, Whitwam JG. Plasma Cate-cholamine changes during cardiopulmonary bypass: a ran-domised double blind comparison of trimetaphan camsylate and sodium nitroprusside. BR Heart J 56:89-93, 1986. 14 Daly P, Rouleau JL, Cousineau D, Burgess JH, Chatterjee K. Effects of captopril and a combination of hydralazine and isosorbide dinitrate on myocardial sympathetic tone in patients with severe congestive heart failure. BR Heart J 56:152-7, 1986. 15 Arrick BA, Griffo W, Cohn Z, Nathan C. Hydrogen peroxide from cellular metabolism of cystine. J Clin Invest 76:567-574, 1985.
CPAP with Minimal Work of Breathing
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We refer to the response ofMathru et al (Chest, 1986;90:151)to our recent letter to Chest: concerning the use of weighted reservoir bag circuits to provide CPAP with minimal work of breathing. While not wishing to prolong the debate unduly, their response demands com-ment. They question our recent study which demonstrated the advantages of a weighted reservoir bag system and low gas flow rates in providing CPAP with minimal work of breathing. Unfortunately, they have misread the paper and have misrepresented our work. They cite some of our data which compares a CPAP circuit with a weighted and an unweighted reservoir bag. They state that it is "dif-ficult to relate to the clinical situation" because we used "a 75 Urn in continuous gas flow" and "inspired with a giant syringe at 1.25 Usec
and 1.4 Usec from weighted and unweighted bags respectively." This is not so. A "giant syringe" was not used to characterize the systems. Each circuit was examined by measuring airway pressure, flow and |
17686 | https://www.chimica-online.it/download/unita-di-misura-della-lunghezza-d-onda.htm | Unità di misura della lunghezza d'onda
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Unità di misura della lunghezza d'onda
Qual è l'unità di misura della lunghezza d'onda?
Un'onda elettromagnetica è una forma di energia che si propaga sia nel vuoto che in qualsiasi mezzo materiale trasparente (acqua, aria, vetro, ecc.); è generata da una carica elettrica che subisce una accelerazione ed è dovuta alla contemporanea propagazione di un campo elettrico (E) e di un campo magnetico (B) oscillanti nel tempo.
Le onde elettromagnetiche sono quindi radiazioni che si propagano nello spazio a causa delle perturbazioni di un campo elettromagnetico. Esse sono trasversali rispetto alla direzione di propagazione; il vettore campo elettrico E, il vettore campo magnetico H e la direzione di avanzamento dell'onda costituiscono tre assi mutuamente ortogonali.
Nella propagazione dell'onda elettromagnetica il campo elettrico e il campo magnetico sono entrambi sinusoidali e tra loro perpendicolari.
Siccome l'interazione con la materia riguarda unicamente la componente del campo elettrico E, un'onda elettromagnetica viene descritta solitamente come l'oscillazione sinusoidale del campo elettrico.
In questa lezione non ci soffermeremo sulle proprietà delle onde elettromagnetiche ma focalizzeremo la nostra attenzione unicamente sull'unità di misura della lunghezza d'onda, una delle proprietà caratteristiche delle onde elettromagnetiche.
Che cos'è la lunghezza d'onda?
La lunghezza d'onda di un'onda elettromagnetica, solitamente indicata con la lettera λ (si legga "lambda"), rappresenta la distanza fra due punti massimi (creste) o due minimi (avvallamenti) dell'onda elettromagnetica, o più in generale la distanza tra due punti successivi in fase fra loro nel profilo spaziale (il profilo spaziale di un'onda elettromagnetica è il profilo dell'onda che si ottiene fissando il tempo e studiando come varia l'onda nello spazio).
Lunghezza d'onda.
SI tratta dunque di una lunghezza e come tale viene solitamente espressa in metri o in uno dei suoi multipli o sottomultipli: chilometri (1 km = 1000 m), centimetri (1 cm = 0,1 m), micrometri (1 µm = 10-6 m), nanometri (1 nm = 10-9 m) o picometri (1 pm = 10-12 m).
A seconda della lunghezza d'onda dell'onda elettromagnetica è preferita una unità di misura piuttosto che un'altra: nel caso dei raggi gamma, aventi valori di lunghezza d'onda molto piccoli, si preferisce l'utilizzo dei picometri mentre per quanto riguarda le microonde si preferisce l'utilizzo dei centimetri o dei metri; viceversa, per onde elettromagnetiche che ricadono nel campo del visibile la lunghezza d'onda viene solitamente espressa in nanometri (1 nm = 10-9 m).
Talvolta è possibile che le lunghezza d'onda di onde elettromagnetiche ad elevata frequenza vengano espresse in un'unità di misura non facente parte del Sistema Internazionale, l'angstrom: un angstrom corrisponde a 10-10 metri (1 Å = 10-10 m).
Esercizi sulle onde elettromagnetiche
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17687 | https://library.fiveable.me/key-terms/physical-chemistry-i/w-=-nrtlnv2v1 | W = nrtln(v2/v1) - (Physical Chemistry I) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Physical Chemistry I
W = nrtln(v2/v1)
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key term - W = nrtln(v2/v1)
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Definition
The equation $$w = nRT \ln(\frac{v_2}{v_1})$$ represents the work done by an ideal gas during an isothermal expansion or compression process. In this expression, 'w' is the work, 'n' is the number of moles of gas, 'R' is the ideal gas constant, and 'T' is the absolute temperature. The ratio $$\frac{v_2}{v_1}$$ indicates the change in volume, showcasing how the work is related to both the amount of gas and the volume it occupies during the thermodynamic process.
5 Must Know Facts For Your Next Test
This equation specifically applies to isothermal processes where temperature remains constant throughout the expansion or compression of the gas.
In this equation, if the volume increases (i.e., $$v_2 > v_1$$), the work done is positive, indicating that the system has done work on its surroundings.
Conversely, if the volume decreases (i.e., $$v_2 < v_1$$), the work done is negative, indicating that work is done on the system by its surroundings.
The natural logarithm function $$\ln$$ highlights how the ratio of volumes affects the calculation of work, emphasizing non-linear relationships in thermodynamic changes.
Understanding this equation helps bridge concepts of mechanical work and thermal energy changes in gases under ideal conditions.
Review Questions
How does the equation w = nRTln(v2/v1) relate to the concept of isothermal processes?
The equation $$w = nRT \ln(\frac{v_2}{v_1})$$ specifically describes work done during an isothermal process where temperature remains constant. This means that as a gas expands or compresses while keeping its temperature stable, the work done can be calculated using this equation. The presence of absolute temperature (T) ensures that this relationship holds true across different volumes, maintaining consistency in thermodynamic principles.
Discuss how changes in volume impact the sign and magnitude of work done as represented by w = nRTln(v2/v1).
In the equation $$w = nRT \ln(\frac{v_2}{v_1})$$, changes in volume directly affect both the sign and magnitude of work done. If volume increases ($$v_2 > v_1$$), the natural logarithm yields a positive value, indicating that work is done by the gas on its surroundings. Conversely, if volume decreases ($$v_2 < v_1$$), it results in a negative value for work, which signifies that work is done on the gas by its surroundings. Thus, understanding how these relationships impact thermodynamic processes is crucial.
Evaluate how this equation bridges mechanical work and thermal energy within the context of ideal gas behavior.
The equation $$w = nRT \ln(\frac{v_2}{v_1})$$ effectively illustrates how mechanical work done by or on a gas relates to thermal energy changes in an ideal gas scenario. As a gas undergoes isothermal expansion or compression, energy transfer occurs through work while maintaining a constant temperature. This highlights a key principle in thermodynamics: that energy conservation can manifest in multiple forms—mechanical through work and thermal through temperature stability—demonstrating interconnectivity between these fundamental concepts.
Related terms
Isothermal Process:
A thermodynamic process in which the temperature remains constant as the system expands or compresses.
Ideal Gas Law:An equation of state for an ideal gas, given by $$PV = nRT$$, relating pressure (P), volume (V), temperature (T), and number of moles (n).
Thermodynamics:
The branch of physics that deals with heat, work, and the transformation of energy in physical and chemical processes.
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17688 | https://faculty.cst.temple.edu/~dfuter/research/determinants.pdf | Journal of Knot Theory and Its Ramifications Vol. 19, No. 6 (2010) 765–782 c ⃝World Scientific Publishing Company DOI: 10.1142/S021821651000811X ALTERNATING SUM FORMULAE FOR THE DETERMINANT AND OTHER LINK INVARIANTS OLIVER T. DASBACH Department of Mathematics Louisiana State University Baton Rouge, LA 70803, USA kasten@math.lsu.edu DAVID FUTER Department of Mathematics Temple University Philadelphia, PA 19122, USA dfuter@temple.edu EFSTRATIA KALFAGIANNI Department of Mathematics Michigan State University East Lansing, MI 48824, USA kalfagia@math.msu.edu XIAO-SONG LIN Department of Mathematics University of California, Riverside Riverside, CA 92521, USA NEAL W. STOLTZFUS Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA stoltz@math.lsu.edu Accepted 31 July 2008 ABSTRACT A classical result states that the determinant of an alternating link is equal to the number of spanning trees in a checkerboard graph of an alternating connected projection of the link. We generalize this result to show that the determinant is the alternating sum of the number of quasi-trees of genus j of the dessin of a non-alternating link.
∗We regretfully inform you that Xiao-Song Lin passed away on the 14th of January, 2007.
765 766 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus Furthermore, we obtain formulas for coefficients of the Jones polynomial by counting quantities on dessins. In particular, we will show that the jth coefficient of the Jones polynomial is given by sub-dessins of genus less or equal to j.
Keywords: Knots; knot determinant; Jones polynomial; Turaev genus; Bollob´ as– Riordan–Tutte polynomial; dessin d’enfant.
Mathematics Subject Classification 2010: 57M25, 57M27 1. Introduction A classical result in knot theory states that the determinant of an alternating link is given by the number of spanning trees in a checkerboard graph of an alter-nating, connected link projection (see e.g. ). For non-alternating links, one has to assign signs to the trees and count the trees with signs, where the geometric meaning of the signs is not apparent. Ultimately, these theorems are reflected in Kauffman’s spanning tree expansion for the Alexander polynomial (see [14,18]) as well as Thistlethwaite’s spanning tree expansion for the Jones polynomial ; the determinant is the absolute value of the Alexander polynomial as well as of the Jones polynomial at −1.
The first purpose of this paper is to show that the determinant theorem for alternating links has a very natural, topological/geometrical generalization to non-alternating links, using the framework that we developed in : Every link diagram induces an embedding of the link into the neighborhood of an orientable surface, its Turaev surface, such that the projection is alternating on that surface. Now the two checkerboard graphs are graphs embedded on surfaces, i.e. dessins d’enfant (aka. combinatorial maps or ribbon graphs), and these two graphs are dual to each other. The minimal genus of all surfaces coming from that construction is the Turaev genus of the link. However, as in one does not need the reference to the surface to construct the dessin directly from the diagram and to compute its genus. The Jones polynomial can then be considered as an evaluation of the Bollob´ as–Riordan–Tutte polynomial of the dessin . Alternating non-split links are precisely the links of Turaev genus zero. Our determinant formula recovers the classical determinant formula in that case.
For a connected link projection of higher Turaev genus, we will show that the determinant is given as the alternating sum of the number of spanning quasi-trees of genus j, as defined below, in the dessin of the link projection. Thus, the sign has a topological/geometrical interpretation in terms of the genus of sub-dessins.
In particular, we will show that for Turaev genus 1 projections the determinant is the difference between the number of spanning trees in the dessin and the number of spanning trees in the dual of the dessin. The class of Turaev genus one knots and links includes for example all non-alternating pretzel knots.
Every link can be represented as a dessin with one vertex, and we will show that with this representation the numbers of j-quasi-trees arise as coefficients of the characteristic polynomial of a certain matrix assigned to the dessin. In particular, we will obtain a new determinant formula for the determinant of a link which comes Alternating Sum Formulae for the Determinant and Other Link Invariants 767 solely from the Jones polynomial. Recall that the Alexander polynomial — and thus every evaluation of it — can be expressed as a determinant in various ways. The Jones polynomial, however, is not defined as a determinant.
The second purpose of the paper is to develop dessin formulas for coefficients of the Jones polynomial. We will show that the jth coefficient is completely deter-mined by sub-dessins of genus less or equal to j and we will give formulas for the coefficients. Again, we will discuss the simplifications in the formulas if the dessin has one vertex. Starting with the work of the first and fourth author , the coef-ficients of the Jones polynomial have recently gained a new significance because of their relationship to the hyperbolic volume of the link complement. Under certain conditions, the coefficients near the head and the tail of the polynomial give linear upper and lower bounds for the volume. In [9,10] this was done for alternating links and in [11,12] it was generalized to a larger class of links.
The paper is organized as follows: Section 2 recalls the pertinent results of .
In Sec. 3, we develop the alternating sum formula for the determinant of the link.
Section 4 shows a duality result for quasi-trees and its application to knots of Turaev genus one. In Sec. 5, we look at the situation when the dessin has one vertex. Section 6 shows results on the coefficients of the Jones polynomial within the framework of dessins.
2. The Dessin D’enfant Coming from a Link Diagram We recall the basic definitions of : A dessin d’enfant (combinatorial map, oriented ribbon graph) can be viewed as a multi-graph (i.e. loops and multiple edges are allowed) equipped with a cyclic order on the edges at every vertex. Isomorphisms between dessins are graph isomorphisms that preserve the given cyclic order of the edges.
Equivalently, dessins correspond to graphs embedded on an orientable surface such that every region in the complement of the graph is a disk. We call the regions the faces of the dessins. Thus, the genus g(D) of a dessin D with v(D) vertices, e(D) edges, f(D) faces and k components is determined by its Euler characteristic: χ(D) = v(D) −e(D) + f(D) = 2k −2g(D).
For each Kauffman state of a (connected) link diagram, a dessin is constructed as follows: Given a link diagram P(K) of a link K we have, as in Fig. 1, an A-splicing and a B-splicing at every crossing. For any state assignment of an A or B at each crossing we obtain a collection of non-intersecting circles in the plane, together with embedded arcs that record the crossing splice. Again, Fig. 1 shows this situation locally. In particular, we will consider the state where all splicings are A-splicings.
The collection of circles will be the set of vertices of the dessin.
To define the desired dessin associated to a link diagram, we need to define an orientation on each of the circles resulting from the A or B splicings, according to a given state assignment. We orient the set of circles in the plane by orienting each 768 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus Fig. 1.
Splicings of a crossing, A-graph and B-graph.
component clockwise or anti-clockwise according to whether the circle is inside an odd or even number of circles, respectively. Given a state assignment s : E →{A, B} on the crossings (the eventual edge set E(D) of the dessin), the associated dessin is constructed by first resolving all the crossings according to the assigned states and then orienting the resulting circles according to a given orientation of the plane.
The vertices of the dessin correspond to the collection of circles and the edges of the dessin correspond to the crossings. The orientation of the circles defines the orientation of the edges around the vertices. We will denote the dessin associated to state s by D(s). Of particular interest for us will be the dessins D(A) and D(B) coming from the states with all-A splicings and all-B splicings. For alternating projections of alternating links D(A) and D(B) are the two checkerboard graphs of the link projection. In general, we showed in that D(A) and D(B) are dual to each other.
We will need several different combinatorial measurements of the dessin: Definition 2.1. Denote by v(D), e(D) and f(D) the number of vertices, edges and faces of a dessin D. Furthermore, we define the following quantities: k(D) = the number of connected components of D, g(D) = 2k(D) −v(D) + e(D) −f(D) 2 , the genus of D, n(D) = e(D) −v(D) + k(D), the nullity of D.
The following spanning sub-dessin expansion was obtained in by using results of . A spanning sub-dessin is obtained from the dessin by deleting edges. Thus, it has the same vertex set as the dessin.
Theorem 2.2. Let ⟨P⟩∈Z[A, A−1] be the Kauffman bracket of a connected link projection diagram P and D := D(A) be the dessin of P associated to the all-A-splicing. The Kauffman bracket can be computed by the following spanning sub-dessin H expansion: A−e(D)⟨P⟩= A2−2v(D)(X −1)−k(D) !
H⊂D (X −1)k(H)Y n(H)Zg(H) Alternating Sum Formulae for the Determinant and Other Link Invariants 769 under the following specialization: {X →−A4, Y →A−2δ, Z →δ−2} where δ := (−A2 −A−2).
Theorem 2.2, after substitution, yields the following sub-dessin expansion for the Kauffman bracket of P.
Corollary 2.3.
⟨P⟩= !
H⊂D Ae(D)−2e(H) " −A2 −A−2#f(H)−1 .
3. Dessins Determine the Determinant The determinant of a link is ubiquitous in knot theory. It is the absolute value of the Alexander polynomial at −1 as well as the Jones polynomial at −1. Furthermore, it is the order of the first homology group of the double branched cover of the link complement. For other interpretations, see e.g. .
We find the following definition helpful: Definition 3.1. Let D be a connected dessin that embeds into a surface S. A spanning quasi-tree of genus j or spanning j-quasi-tree in D is a sub-dessin H of D with v(H) vertices and e(H) edges such that H is connected and spanning and (1) H is of genus j.
(2) S −H has one component, i.e. f(H) = 1.
(3) H has e(H) = v(H) −1 + 2j edges.
In particular, the spanning 0-quasi-trees are the regular spanning trees of the graph. Note that by Definition 2.1 either two of the three conditions in Definition 3.1 imply the third one.
Theorem 2.2 now leads to the following formula for the determinant det(K) of a link K.
Theorem 3.2. Let P be a connected projection of the link K and D := D(A) be the dessin of P associated to the all-A splicing. Suppose D is of genus g(D).
Furthermore, let s(j, D) be the number of spanning j-quasi-trees of D.
Then det(K) = $ $ $ $ $ $ g(D) !
j=0 (−1)j s(j, D) $ $ $ $ $ $ .
Proof. Recall that the Jones polynomial JK(t) can be obtained from the Kauffman bracket, up to a sign and a power of t, by the substitution t := A−4.
770 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus By Theorem 2.2, we have for some power u = u(D): ± JK(A−4) = Au !
H⊂D (X −1)k(H)−1Y n(H)Zg(H) = Au !
H⊂D A−2−2e(H)+2v(H)δf(H)−1.
(3.1) We are interested in the absolute value of JK(−1). Thus, δ = 0 and, since k(H) ≤f(H): |JK(−1)| = $ $ $ $ $ $ !
H⊂D,f(H)=1 A−2−2e(H)+2v(H) $ $ $ $ $ $ (3.2) = $ $ $ $ $ $ !
H⊂D,f(H)=1 A−4g(H) $ $ $ $ $ $ .
(3.3) Collecting the terms of the same genus and setting A−4 := −1 proves the claim.
Remark 3.3. For genus j = 0 we have s(0, D) is the number of spanning trees in the dessin D. Recall that a link has Turaev genus zero if and only if it is alternating.
Thus, in particular, we recover the well-known theorem that for alternating links the determinant of a link is the number of spanning trees in a checkerboard graph of an alternating connected projection.
Theorem 3.2 is a natural generalization of this theorem for non-alternating link projections.
Example 3.4. Figure 2 shows the non-alternating 8-crossing knot 821, as drawn by Knotscape ( and Fig. 3 the all-A associated dessin.
The dessin in Fig. 3 is of genus 1. Thus, the only quasi-trees are of genus 0 and of genus 1. The quasi-trees of genus 0 are the spanning trees of the dessin. The dessin in the example contains 9 spanning trees, i.e. s(0, D) = 9. A quasi-tree of genus 1 must have 4 edges. Furthermore, it must contain either of the two loops, otherwise it would not be of genus 1. Two of the remaining three edges must form a cycle which interlinks with that loop. A simple count yields 24 of these and thus the determinant of the knot is 24 −9 = 15.
4. Duality The following theorem is a generalization of the result that for planar graphs the spanning trees are in one-one correspondence to the spanning trees of the dual graphs: Theorem 4.1. Let D = D(A) be the dessin of all-A splicings of a connected link projection of a link L. Suppose D is of genus g(D) and D∗is the dual of D.
Alternating Sum Formulae for the Determinant and Other Link Invariants 771 Fig. 2.
The eight-crossing knot 821 with its all-A splicing projection diagram.
Fig. 3.
All-A splicing dessin for 821.
772 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus We have: The j-quasi-trees of D are in one-one correspondence to the (g(D)−j)-quasi-trees of D∗. Thus s(j, D) = s(g(D) −j, D∗).
Proof. Let H be a spanning j-quasi-tree in D. Denote by D −H the sub-dessin of D obtained by removing the edges of H from D, and by (D −H)∗the sub-dessin of the dual D∗obtained by removing the edges dual to the edges in H. From f(H) = 1, it follows that (D −H)∗is connected and spanning. Furthermore, f((D −H)∗) = 1.
We have: v(H) −e(H) + f(H) = v(D) −e(H) + 1 = 2 −2j v(D) −e(D) + f(D) = 2 −2g(D).
Thus, v((D −H)∗) −e((D −H)∗) + f((D −H)∗) = f(D) −(e(D) −e(H)) + 1 = 2 −2g(D) −v(D) + e(H) + 1 = 2 −2(g(D) −j).
Hence, (D −H)∗is a (g(D) −j)-quasi-tree in D∗.
Recall that the Turaev genus zero links are precisely the alternating links. The following corollary generalizes to the class of Turaev genus one links the aforemen-tioned, classical interpretation of the determinant of connected alternating links as the number of spanning trees in its checkerboard graph.
Corollary 4.2. Let D = D(A) be the all-A dessin of a connected link projection of a link L and D∗its dual. Suppose D is of Turaev genus one. Then det(L) = |#{spanning trees in D} −#{spanning trees in D∗}|.
We apply Corollary 4.2 to compute the determinants of non-alternating pretzel links. The Alexander polynomial as well as the Jones polynomial, and consequently the determinant is invariant under mutations (see, e.g. ). Hence, it is sufficient to consider the case of K(p1, . . . , pn, −q1, . . . , −qm) pretzel links, as depicted in Fig. 4.
We assume that the links are non-alternating, i.e. n ≥1 and m ≥1.
Example 4.3. Consider the pretzel link K(p1, . . . , pn, −q1, . . . , −qm), where n ≥1, m ≥1 and pi, qi > 0 for all i. The determinant of K(p1, . . . , pn, −q1, . . . , −qm), is det(K(p1, . . . , pn, −q1, . . . , −qm)) = $ $ $ $ $ $ n % i=1 pi m % j=1 qj n !
i=1 1 pi − m !
j=1 1 qj $ $ $ $ $ $ .
Alternating Sum Formulae for the Determinant and Other Link Invariants 773 Fig. 4.
The K(p1, . . . , pn, −q1, . . . , −qm) pretzel link.
Fig. 5.
The all-A splicings of the K(p1, . . . , pn, −q1, . . . , −qm) pretzel link.
Proof. Figure 5 shows the all-A splicing diagram of these links. The all-A dessin D = D(A) has v(D) = n + m !
j=1 (qj −1) = n −m + m !
j=1 qj vertices and e(D) = n i=1 pi + m j=1 qj edges. For the numbers of faces, we have to count the vertices in the all-B dessin. We compute: f(D) = m −n + n !
i=1 pi.
Now, we get for the Euler characteristic: χ(D) = v(D) −e(D) + f(D) = 0 and thus the Turaev genus is one.
774 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus It remains to compute the difference between the number of spanning trees in the dessin and the number of spanning trees in its dual. This is a simple counting argument.
Remark 4.4. The class of Turaev genus one knots and links is quite rich. For example, it contains all non-alternating Montesinos links. It also contains all semi-alternating links (whose diagrams are constructed by joining together two alternat-ing tangles, and thus have exactly two over-over crossing arcs and two under-under arcs).
5. Dessins with One Vertex 5.1. Link projection modifications Here we show that every knot/link admits a projection with respect to which the all-A dessin has one vertex. Such dessins are useful for computations.
Lemma 5.1. Let ˜ P be a projection of a link L with corresponding all-A dessin ˜ D.
Then ˜ P can be modified by Reidemeister moves to new a projection P such that the corresponding dessin D = D(A) has one vertex. Furthermore, we have: (1) e(˜ D) + 2v( ˜ D) −2 = e(D), (2) g(˜ D) + v( ˜ D) −1 = g(D).
Proof. For a connected projection of the link L, consider the collection of circles that we obtain by an all-A splicing of the crossings. If there is only one circle we are done. Otherwise, one can perform a Reidemeister move II near a crossing on two arcs that lie on two neighbor circles as in Fig. 6.
The new projection will have one circle less in its all-A splicing diagram. Also two crossings were added and a new face was created. If the link projection is non-connected, one can transform it by Reidemeister II moves into a connected link projection. It is easy to check that the genus behaves as predicted. The claim follows.
Remark 5.2. Dessins with one vertex are equivalent to Manturov’s “d-diagrams” . Note that the procedure of using just Reidemeister moves of type II is similar in spirit to Vogel’s proof of the Alexander theorem [21, 3].
Fig. 6.
Reduction of the number of vertices by a Reidemeister II move.
Alternating Sum Formulae for the Determinant and Other Link Invariants 775 5.2. The determinant of dessins with one vertex Dessins with one vertex can also described as chord diagrams. The circle of the chord diagram corresponds to the vertex and the chords correspond to the edges.
In our construction, the circle of the chord diagram is the unique circle of the state resolution, and the chords correspond to the crossings. The cyclic orientation at the vertex induces the order of the chords around the circle. For each chord diagram D, one can assign an intersection matrix [2,7] as follows: Fix a base point on the circle, disjoint from the chords and number the chords consecutively.
The intersection matrix is given by: IM(D)ij = + sign(i −j), if the ith chord and the jth chord intersect, 0, else.
Recall that the number of spanning j-quasi trees in D was denoted by s(j, D).
Now Theorem 5.3. For a dessin D with one vertex, the characteristic polynomial of IM(D) satisfies : det(IM(D) −xI) = (−1)m ⌊m 2 ⌋ !
j=0 s(j, D)xm−2j, where m = e(D) is the number of chords, i.e. the number of edges in the dessin.
In particular det(D) = | det(IM(D) − √ −1I)|.
Proof. The result follows from combining Theorem 3.2 and a result of Bar-Natan and Garoufalidis . Bar-Natan and Garoufalidis use chord diagrams to study weight systems coming from Vassiliev invariant theory, thus in a differ-ent setting than we do. However, by for a chord diagram D the determinant det(IM(D)) is either 0 or 1 and, translated in our language, it is 1 precisely if f(D) = 1.
Furthermore, since f(D) −1 and the number of edges have the same parity, we know that det(IM(D)) = 0 for an odd number of edges.
The matrix IM(D) has zeroes on the diagonal. Thus, the coefficient of xm−j in det(IM(D) −xI) is (−1)m−j times the sum over the determinants of all j × j submatrices that are obtained by deleting m −j rows and the m −j corresponding columns in the matrix IM(D). Those submatrices are precisely IM(H) for H a sub-dessin of D with j edges. In particular, the determinant of IM(H) is zero for j odd.
For j even we know that det(IM(H)) = 1 if f(H) = 1 and 0 otherwise. Since for 1-vertex dessins D the genus 2g(D) = e(D) −f(D) + 1 those H with f(H) = 1 are precisely the j/2-quasi-trees. This, together with Theorem 3.2 implies the claim.
776 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus Fig. 7.
The (p, q)-twist knot and its all-A splicing dessin in chord diagram form.
Example 5.4. The (p, q)-twist knots as in Fig. 7 have an all-A-dessin with one vertex.
The figure-8 knot is given as the (2, 3)-twist knot. Its intersection matrix is IM(D) = 0 0 −1 −1 −1 0 0 −1 −1 −1 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 .
The characteristic polynomial of IM(D) is −6x3−x5. In particular, the determinant of the figure-8 knot is 6 −1 = 5.
5.3. The Jones polynomial at t = −2 By work of Jaeger, Vertigan and Welsh , evaluating the Jones polynomial is #P-hard at all points, except at eight points: All fourth and sixth roots of unity.
In particular, the determinant arises as one of these exceptional points. However, letting computational complexity aside, Theorem 2.2 gives an interesting formula in terms of the genus for yet another point: t = A−4 = −2.
Lemma 5.5. Let P be the projection of a link K with dessin D = D(A) such that D has one vertex. Then, the Kauffman bracket at t = A−4 := −2 evaluates to ⟨P⟩= Ae(D) !
H⊂D (A−4)g(H).
Proof. By Corollary 2.3, we have the following sub-dessin expansion for the Kauff-man bracket of P: ⟨P⟩= !
H⊂D Ae(D)−2e(H) " −A2 −A−2#f(H)−1 .
The term A−2(−A2 −A−2) = (−1 −A−4) Alternating Sum Formulae for the Determinant and Other Link Invariants 777 is 1 at t = A−4 = −2 and, with v(D) = v(H) for all spanning sub-dessin H of D, the claim follows.
6. Dessins and the Coefficients of the Jones Polynomial Let P be a connected projection of a link L, with corresponding all-A dessin D := D(A) and let ⟨P⟩= !
H⊂D Ae(D)−2e(H)(−A2 −A−2)f(H)−1 (6.1) denote the spanning sub-dessin expansion of the Kauffman bracket of P as obtained earlier. Let H0 ⊂D denote the spanning sub-dessin that contains no edges (so v(H0) = v(D) and e(H0) = 0) and let M := M(P) and m := m(P) denote the maximum and minimum powers of A that occur in the terms that lead to ⟨P⟩. We have M(P) ≤e(D) + 2v(D) −2, and the exponent e(D) + 2v(D) −2 is realized by H0; see [8, Lemma 7.1]. Let aM denote the coefficient of the extreme term Ae(D)+2v(D)−2 of ⟨P⟩. Below we will give formulae for aM; similar formulae can be obtained for the lowest coefficient, say am, if one replaces the the all-A dessin with the all-B dessin in the statements below.
We should note that aM is not, in general, the first non-vanishing coefficient of the Jones polynomial of L. Indeed, the exponent e(D) + 2v(D) −2 as well as the expression for aM we obtain below, depends on the projection P and it is not, in general, an invariant of L. In particular, aM might be zero and, for example, we will show that this is the case in Example 6.2.
The following theorem extends and recovers results of Bae and Morton, and Manch´ on [1,16] within the dessin framework.
Theorem 6.1. We have (1) For l ≥0, let aM−l denote the coefficient of Ae(D)+2v(D)−2−4l in the Kauffman bracket ⟨P⟩. Then, the term aM−l only depends on spanning sub-dessins H ⊂D of genus g(H) ≤l.
(2) The highest term is given by aM = !
H⊂D, g(H)=0=k(H)−v(D) (−1)v(D)+e(H)−1.
(6.2) In particular, if D does not contain any loops then aM = (−1)v(D)−1 and the only contribution comes from H0.
Proof. The contribution of a spanning H ⊂D to ⟨P⟩is given by XH := Ae(D)−2e(H)(−A2 −A−2)f(H)−1.
(6.3) 778 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus A typical monomial of XH is of the form Ae(D)−2e(H)+2f(H)−2−4s, for 0 ≤s ≤f(H) −1.
For a monomial to contribute to aM−l, we must have e(D) −2e(H) + 2f(H) −2 −4s = e(D) + 2v(D) −2 −4l, (6.4) or f(H) = v(D) + e(H) + 2s −2l.
(6.5) Now we have 2g(H) = 2k(H) −v(D) + e(H) −f(H) = 2k(H) −2v(D) + 2l −2s, or g(H) = k(H) −v(D) + l −s. But since v(D) ≥k(H) (every component must have a vertex) and s ≥0, we conclude that l = g(H) + v(D) −k(H) + s ≥g(H), as desired. Now to get the claims for aM: Note that for a monomial of XH to contribute to aM, we must have g(H) = k(H) −v(D) −s (6.6) which implies that s = g(H) = 0 and v(D) = k(H). It follows that H contributes to aM if and only if all of the following conditions are satisfied: (1) f(H) = v(D) + e(H).
(2) k(H) = v(D). Thus, H consists of k := k(H) components each of which has exactly one vertex and either H has no edges or every edge is a loop.
(3) g(H) = 0.
(4) The contribution of H to aM is (−1)f(H)−1.
This finishes the proof of the theorem.
Example 6.2. The all-A dessin of Fig. 3 contains one sub-dessin with no edges, two sub-dessins with exactly one loop and one sub-dessin of genus zero with two loops. Thus aM = 0.
A connected link projection is called A-adequate if and only if the all-A dessin D(A) contains no loops; alternating links admit such projections. We consider two edges as equivalent if they connect the same two vertices. Let e′ = e′(D(A)) denote the number of equivalence classes of edges.
The following is an extension in to the class of adequate links of a result in for alternating links. We will give the dessin proof for completeness, since it shows a subtlety when dealing with dessins in our context: Not all dessins can occur as a dessin of a link diagram.
Corollary 6.3. For A-adequate diagrams aM−1 = (−1)v(e′ −v + 1).
Alternating Sum Formulae for the Determinant and Other Link Invariants 779 Proof. With the notation and setting of the proof of Theorem 6.1, we are looking to calculate the coefficient of the power Ae(D)+2v(D)−6. The analysis in the proof of Theorem 6.1 implies that a spanning sub-dessin H ⊂D contributes to aM−1 if it satisfies one of the following: (1) v(H) = k(H) and g(H) = 1.
(2) v(H) = k(H) and g(H) = 0.
(3) v(H) = k(H) + 1 and g(H) = 0.
Since the link is adequate D(A) contains no loops and we cannot have any H as in (1). Furthermore, the only H with the properties of (2) is the sub-dessin H0 that contains no edges. Finally, the only case that occurs in (3) consists of those sub-dessins H1 that are obtained from H0 by adding edges between a pair of vertices.
The dessin is special since it comes from a link diagram. Each vertex in the dessin represents a circle in the all-A splicing diagram of the link and each edge represents an edge there. Because these edges do not intersect, H1 must have genus 0.
Note that any sub-dessin H′ ⊂H1 is either H0 or is of the sort described in (3).
We will call H1 maximal if it is not properly contained in one of the same type with more edges. Thus, there are e′ maximal H1 for D(A). The contribution of H1 to aM−1 is (−1)v(D)−3+e(H1). Thus, the contribution of all H′ ⊂H1 that are not H0 is e(H1) !
j=1 .e(H1) j / (−1)v(D)−3+j = (−1)v(D).
Thus, the total contribution in aM−1 of all such terms is (−1)ve′.
To finish the proof, observe that the contribution of H0 comes from the second term of the binomial expansion XH0 := Ae(D)(−A2 −A−2)f(H0)−1.
(6.7) Since f(H0) = v this later contribution is (−1)v−1(v −1).
The expression in Theorem 6.1 becomes simpler, and the lower order terms easier to express, if the dessin D has only one vertex. By Lemma 5.1 the projection P can always be chosen so that this is the case.
Corollary 6.4. Suppose P is a connected link projection such that D = D(A) has one vertex. Then, aM−l = g(H)=l !
H⊂D, g(H)=0 (−1)e(H) .e(H) −2g(H) l −g(H) / .
(6.8) In particular, aM = !
H⊂D, g(H)=0 (−1)e(H) (6.9) 780 O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin & N. W. Stoltzfus and aM−1 = !
H⊂D, g(H)=1 (−1)e(H) + !
H⊂D, g(H)=0 (−1)e(H)e(H).
(6.10) Proof. For a 1-vertex dessin D, we have k(D) = v(D) = 1 and, thus 2g(D) = e(D) −f(D) + 1.
Now, Eq. (6.1) simplifies to ⟨P⟩= !
H⊂D Ae(D)−2e(H)+2f(H)−2 " −1 −A−4#f(H)−1 = !
H⊂D Ae(D)−4g(H) " −1 −A−4#e(H)−2g(H) .
The claim follows from collecting the terms.
Parallel edges, i.e. neighboring edges that are parallel in the chord diagram, in a dessin are special since they correspond to twists in the diagram. It is useful to introduce weighted dessins: Collect all edges, say µ(c) −1 edges parallel to a given edge c and replace this set by c weighted with weight µ(c). Note, that ˜ D has the same genus as D.
Corollary 6.5. For a knot projection with a 1-vertex dessin D and weighted dessin ˜ D, we have : ⟨P⟩= !
˜ H⊂˜ D Ae(D)−4g(˜ H)(−1 −A−4)−2g(˜ H) % c∈˜ H (−1 −A−4µ(c)).
Proof. For a given edge c collect in ⟨P⟩= !
H⊂D Ae(D)−4g(H) " −1 −A−4#e(H)−2g(H) all terms where H contains an edge parallel to c. This sub-sum is !
H⊂D, H contains edge parallel to c Ae(D)−4g(H) " −1 −A−4#e(H)−2g(H) = !
H⊂D, ˜ H=H−{edges parallel to c}∪c µ(c) !
j=1 .µ(c) j / Ae(D)−4g(H) " −1 −A−4#e(˜ H)−1+j−2g(H) = !
H⊂D, ˜ H=H−{edges parallel to c}∪c Ae(D)−4g(H)(−1 −A−4)e(˜ H)−1−2g(H)(−1 −A−4µ(c)) The claim follows by repeating this procedure for each edge c.
Alternating Sum Formulae for the Determinant and Other Link Invariants 781 Example 6.6. The (p, q)-twist knot is represented by the weighted, 1-vertex dessin with two intersecting edges, one with weight p and one with weight q. By Corollary 6.5, its Kauffman bracket is: A−p−q⟨P⟩= 1 + (−1 −A−4p) + (−1 −A−4q) + A−4(−1 −A−4)−2(−1 −A−4p)(−1 −A−4q).
Remark 6.7. Corollary 6.4 implies the following for the first coefficient aM. Sup-pose D is a 1-vertex, genus 0 dessin with at least one edge. Then, every sub-dessin also has genus 0.
Thus, !
H⊂D, g(H)=0 (−1)e(H) = e(D) !
j=0 .e(D) j / (−1)j = 0.
For an arbitrary dessin, let H1, . . . , Hn be the maximal genus 0 sub-dessins of D. Define a function φ on dessins which is 1 if the dessin contains no edges and 0 otherwise.
Then aM = !
i φ(Hi) − !
i,j, i<j φ(Hi ∩Hj) + !
i,j,k, i<j<k φ(Hi ∩Hj ∩Hk) −· · · Acknowledgments We thank James Oxley and the referee for helpful comments. The first author was supported in part by NSF grants DMS-0806539 and DMS-0456275 (FRG), the second author by NSF grant DMS-0353717 (RTG), the third author by NSF grants DMS-0805942 and DMS-0456155 (FRG) and the fifth author by NSF grant DMS-0456275 (FRG).
References Y. Bae and H. R. Morton, The spread and extreme terms of Jones polynomials, J.
Knot Theory Ramifications 12(3) (2003) 359–373.
D. Bar-Natan and S. Garoufalidis, On the Melvin-Morton-Rozansky conjecture, Invent. Math. 125(1) (1996) 103–133.
J. S. Birman and T. E. Brendle, Braids: A survey, in Handbook of Knot Theory, (Elsevier B. V., Amsterdam, 2005), pp. 19–103.
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B. Bollob´ as and O. Riordan, A polynomial of graphs on surfaces, Math. Ann. 323(1) (2002) 81–96.
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O. T. Dasbach, D. Futer, E. Kalfagianni, X.-S. Lin and N. W. Stoltzfus, The Jones polynomial and graphs on surfaces, J. Combin. Theory Ser. B 98(2) (2008) 384–399.
O. T. Dasbach and X.-S. Lin, On the head and the tail of the colored Jones polyno-mial, Compos. Math. 142(5) (2006) 1332–1342.
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D. Futer, E. Kalfagianni and J. S. Purcell, Symmetric links and Conway sums: Volume and Jones polynomial, Math. Res. Lett. 16(2) (2009) 233–253.
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17689 | https://www.omnicalculator.com/physics/ideal-gas-law | Ideal Gas Law Calculator
This ideal gas law calculator will help you establish the properties of an ideal gas subject to pressure, temperature, or volume changes. Read on to learn about the characteristics of an ideal gas, how to use the ideal gas law equation, and the definition of the ideal gas constant.
We also recommend checking out our combined gas law calculator for further understanding of the basic thermodynamic processes of ideal gases.
What is an ideal gas?
An ideal gas is a special case of any gas that fulfills the following conditions:
The gas consists of a large number of molecules that move around randomly.
All molecules are point particles (they don't take up any space).
The molecules don't interact except for colliding.
All collisions between the particles of the gas are perfectly elastic (visit our conservation of momentum calculator to learn more).
The particles obey Newton's laws of motion.
Ideal gas law equation
The properties of an ideal gas are all summarized in one formula of the form:
p⋅V=n⋅R⋅T
where:
p – Pressure of the gas, measured in Pa;
V – Volume of the gas, measured in m³;
n – Amount of substance, measured in moles;
R – Ideal gas constant; and
T – Temperature of the gas, measured in kelvins.
To find any of these values, simply enter the other ones into the ideal gas law calculator.
For example, if you want to calculate the volume of 40 moles of a gas under a pressure of 1013 hPa and at a temperature of 250 K, the result will be equal to:
V = nRT/p = 40 × 8.31446261815324 × 250 / 101300 = 0.82 m³.
Ideal gas constant
The gas constant (symbol R) is also called the molar or universal constant. It is used in many fundamental equations, such as the ideal gas law.
The value of this constant is 8.31446261815324 J/(mol·K).
The gas constant is often defined as the product of Boltzmann's constant k (which relates the kinetic energy and temperature of a gas) and Avogadro's number (the number of atoms in a mole of substance):
R=NAk=(6.02214076×1023/mol)⋅(1.38064852×10−23 J/K)=8.3144626 J/(mol⋅K)
You might find this air pressure at altitude calculator useful, too.
FAQs
When can I use the ideal gas law?
You can apply the ideal gas law for every gas at a density low enough to prevent the emergence of strong intermolecular forces. In these conditions, every gas is more or less correctly modeled by the simple equation PV = nRT, which relates pressure, temperature, and volume.
What is the formula of the ideal gas law?
The formula of the ideal gas law is:
PV = nRT
where:
P — Pressure, in pascal;
V — Volume in cubic meters;
n — Number of moles;
T — Temperature in kelvin; and
R — Ideal gas constant.
Remember to use consistent units! The value commonly used for R, 8.314... J/mol·K refers to the pressure measured exclusively in pascals.
What is the pressure of 0.1 moles of a gas at 50 °C in a cubic meter?
268.7 Pa, or 0.00265 atm. To find this result:
Convert the temperature into kelvin:
T [K] = 273.15 + 50 = 323.15 K.
2. Compute the product of temperature, the number of moles, and the gas constant: nRT = 0.1 mol × 323.15 K × 8.3145 J/mol·K = 268.7 J (that is, energy).
3. Divide by the volume. In this case, the volume is 1, hence:
P = 268.7 Pa.
What are the three thermodynamics laws that can be identified in the ideal gas law?
The ideal gas law has four parameters. One of them is the number of moles which is a bit outside the scope of thermodynamics. The other three are pressure, temperature, and volume. We can identify three laws by fixing, in turn, each one of the three:
Fixing the temperature, we find the isothermal transformation (or Boyle's law): PV = k.
Fixing the volume, we find the isochoric transformation (Charles's law): P/T = k.
Fixing the pressure, we have the isobaric transformation (Gay-Lussac's law): V/T = k.
How do I calculate the temperature of a gas given moles, volume and pressure?
To calculate the temperature of a gas given the pressure and the volume, follow these simple steps:
Calculate the product of pressure and volumes. Be sure you're using consistent units: a good choice is pascals and cubic meters.
Calculate the product of the number of moles and the gas constant. If you used pascals and cubic meters, the constant is R = 8.3145 J/mol·K.
Divide the result of step 1 by the result of step 2: the result is the temperature (in kelvin): T = PV/nR
Additional parameters
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17690 | https://esther.rice.edu/selfserve/!bwzkpsyl.v_viewDoc?term=202210&crn=12265&type=SYLLABUS | MECH 590 Aerospace Propulsion
Fall 2021
Instructor : Professor P. Rodi, ( Patrick.Rodi@Rice.edu , GRB W200F)
Tech. TA : N/A
Class Hours : Monday/Wednesday/Friday, 11:00am – 11:50am, using Zoom via Canvas
Office Hours : Monday/Wednesday/Friday, 10:00am - 10:45am or by appointment (using Zoom)
Course URL:
Course Outcomes Expected :Students completing the course should be able to: 1. Derive and apply the basic supersonic one- and two-dimensional shock and expansion-wave relations. 2. Derive and apply the governing thermodynamic equations of turbine powered air-breathing propulsion. 3. Perform basic engine cycle analysis and quantify expected engine performance for both ideal and real turbine powered jet engines. 4. Derive and apply the basic governing equations for chemical fueled rocket engines. 5. Perform basic engine cycle analysis and quantify expected engine performance for chemical fueled rocket engines. 6. Derive and Apply The Rocket Equation and perform rocket performance analysis for single-stage and multi-stage rockets.
Textbooks :1. Mechanics and Thermodynamics of Propulsion by Hill and Peterson (required text, either 1 st or 2 nd
edition), AIAA. ( 2. Supplementary material from Modern Compressible Flow by Anderson, McGraw-Hill. 3. Auxiliary publication Aerothermodynamics of Gas Turbine and Rocket Propulsion , by G. C. Oates, AIAA. 4. Auxiliary publication Elements of Gas Turbine Propulsion , by J.D. Mattingly, McGraw-Hill, Inc.. 5. Auxiliary publication Rocket Propulsion Elements , by G. P. Sutton & O. Biblarz, Wiley-Interscience.
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Distribution : Engineering science, 3 credits. The course emphasizes theoretical concepts and analysis, but includes some design and synthesis techniques for propulsion systems that can be useful for larger design projects.
ADA Notice : Any student with a disability requiring accommodations in this course is encouraged to contact the instructor after class or during office hours. Syllabus
Syllabus & Introduction Conservation equations, review of thermodynamics, compressibility. 2. One-Dimensional Supersonic Flow One-dimensional flow equations, normal shock relations, Hugoniot equation, heat addition, friction. 3. Two-Dimensional Supersonic Flow Oblique shocks, detached shocks, Prandtl-Meyer expansion waves, shock-expansion theory, quasi one-dimensional flow, conical flow. 4. Boundary Layer Mechanics and Heat Transfer Viscous boundary layer, laminar boundary layer, turbulent boundary layer, boundary layer heat transfer. 5. Thermodynamics of Aircraft Jet Engines Basic equations, thrust and efficiency, the ideal ramjet, the ideal turbojet, typical engine performance. 6. Aerothermodynamics of Inlets, Combustors, and Nozzles Subsonic inlets and diffusers, supersonic inlets, combustion time and distances, exhaust nozzle flowfields and performance. 7. Jet Engine Turbomachines: Axial Compressors Angular momentum, single-stage axial compressor, multi-stage axial compressor, instabilities and unsteady flow, starting. 8. Jet Engine Turbomachines: Centrifugal Compressors and Axial Turbines The centrifugal compressor, the axial turbine, turbine and compressor matching, stresses in compressors and turbines 9. Scramjets, and Detonation Engines Scramjet operation, deflagration vs. detonation, Pulse Detonation Engines, Rotary Detonation Engines. 10. Performance of Rocket Vehicles Basic equations, for thrust and specific impulse, single-stage rockets, multi-stage rockets, the rocket equation, high thrust space missions, low thrust space missions. 11. Chemical Rockets: An Introduction Basic equations for chemical rockets, analysis of an ideal rocket. 12. Chemical Rockets: Propellants and Combustion Liquid, solid, and hybrid propellants, combustion chambers, combustion instabilities. 13. Chemical Rockets: Expansion in Nozzles Basic equations of rocket exhaust nozzles, nozzle shape, effects of friction and heat transfer, the effect of back pressure on nozzle flow. 14. Chemical Rockets: Thrust Chambers Introduction, rocket heat transfer, rocket construction, liquid-propellant pressurization, selection of combustion pressure, ignition, thrust vectoring |
17691 | https://www.youtube.com/watch?v=-deXwDCodws | Solve Ambiguous Case Triangles FAST with Law of Sines!
Angie Teaches Math
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Posted: 7 May 2024
How to identify and solve for 2 possible triangles when using the law of sines--this is the LOS ambiguous case! Learn about the angle side side case and how to find both possible triangles when solving non-right triangles using the law of sines.
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0:00 Are there 2 triangles? Ambiguous Case
1:54 Triangle 1 with LOS
8:59 Triangle 2 with LOS
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Are there 2 triangles? Ambiguous Case how do I know that this triangle with two known sides and one angle is our ambiguous case well I'm going to start by thinking of the angle I'm going to put that 34.5 de down this way next I'm going to put a = 10 it could have been on either side of angle B but I'm going to make it easier by putting it up there on the top that leaves me with side B now side B has to be across from angle B so let's go ahead and draw it in here now you'll know notice that the given information is in order an angle a side and then a side so this is my angle side side case which results in two possible triangles here's the easy way to remember the initials of angle side and side happen to spell s so this is the S case two triangles you're welcome let's go ahead and finish working on this one because there's a second triangle we need to identify we don't know the length of our bottom side so I could actually have taken side B and swung it to the left and drawn it this way this gives me another triangle now let's go ahead and just start by looking at the larger of the two triangles we're going to call that triangle one in this triangle we need to find angles A and C and side C notice that angle a here is an acute angle that's going to become important in our second triangle the smaller of the triangles we still need to find angles A and C and side C but in this triangle they're going to be very different values so let's go ahead and give them subscripts of two and I want you to take a look at that angle a it is larger than 90° so it's an obtuse angle let's go back to that first triangle and Triangle 1 with LOS the law of signs as I'm applying the law of signs I want to pick up a side and an angle across from each other so I want to pick up angle B and side B and the only other piece of information that I know is side A so I need to pick up side a and angle a let's go ahead and put this together in our law of signs so I've got the S of a over side a which is 10 is equal to the sign of our angle B which is 34.5 5 over side B which happens to be six now I'm solving for angle a so I want to start by isolating sin a let's multiply that 10 on the other side so I get sin a by itself so sin a is equal to 10 the S of 34.5 all divided by six now in order to get a all by itself I really need to apply a sin inverse here so the sin inverse is going to be applied to both sides the left and the right hand side so s inverse of as I do this the S inverse and the sign cross each other out and I do end up with one version anyway of angle a and I can get that by putting this into my calculator now there's one little problem here and that's the range or the possible outputs that sin inverse will give you and it will only give you a value between 90° or negative pi Hales if you're working in radians up to 90° so it's only going to give you acute angles and that's exactly what we're going to get out of our calculator the very first thing I want you to do is make sure you're in degrees so I'm going to click on mode and my radian and degree line is the fourth line down I am in degrees and I know because it's highlighted if you're not Arrow over to it and then hit enter so it's highlighted let's quit here by hitting second followed by mode and I'm ready to type this in so I want to grab that s inverse it's the second of s I'm using my TI calculator here and then I'm going to type it in exactly like I see it so 10 the sign of 34.5 parentheses and then divided by six close our parentheses and we end up with just like we expected an angle measure between 90 and positive 90° so we get that a is approximately I'm going to round that to one decimal place and call that 707 de and 70.7 would live right about here but I know that there's a second value that has this same s value which means it would have the same y value so it would live over here I'm doing my best to draw everything nice and straight okay so it would live right there so to find that second value I would take this would be for a sub 2 this would be for my second triangle so we're going to hang on to this for a minute it would be 180 minus that 70. 7° and if I go ahead and do 180 minus 70.7 I end up with what's going to be that obtuse angle for the second triangle of 109 109.3 now I'm going to hang on to that 109.3 de for triangle number two I really want to finish up here triangle number one with that 70.7 de so this angle right here is 70 .7 de back up in our triangle we still need to find angle C and side C let's do angle C first I know that the sum of all of my angles A + B + C is equal to 180° so to get angle C I'm just going to take 180 and subtract those other two angle measures so 180 minus my 34.5 minus my 70.7 is going to give me angle C and that leaves me with 74.8 so that is angle C so angle C is 74.8 De all we've got left now is side C we want to use the law of signs one more time as I'm looking at that law of signs I've got of course still angle B and side B and going to use that as one of my fractions and then for the other fraction I need to use angle C and side C so I can solve for that c so we've got the sign of let's do our B's first 34.5 divided by side B which is 6 is equal to the sign of angle C and that is 74.8 divided by side C which is what we are looking for now I need to do some cross multiplying cross multiplying cross multiplying and I end up with C the sign of 34.5 is equal to 6 the other sign so 6 times the S of 74.8 I need to divide off that s of 34.5 that's going to leave me with C so s of 3 4.5 / by the S of 34.5 okay so if I cross these signs out C is going to equal or be approximate because I've been rounding here that value let's get this into our calculator 6 the S of 74.8 parentheses divided by the S of 30 4.5 and I get that side C which is going to be if I'm rounding it's going to be 10.2 so C is approximately 10.2 let me go back up to that triangle and I can erase a bunch of this now because I've got everything and I can fill in that last value of 10.2 now we're ready for for triangle Triangle 2 with LOS number two let's go back up and grab that second angle a so a sub 2 was 180 minus 70.7 which gives us that obtuse angle greater than 90° of 109.3 I'm going to add that angle measure to our triangle here 109.3 and we're going to find the other angles in exact ly the same order that we finished off triangle number one so that means that I want to start with finding angle measure C to find angle measure number c and this is C sub 2 right my other angle C that's going to be 180 minus my other two angle measures so I've got a 34.5 and I've got a 109.3 and we get angle C's measure of 36.2 so C sub it's about cuz I'm rounding but I'm just going to put equals um 36.2 de let's go ahead and add that to our triangle so 36.2 de which just leaves side C and just like we did before we've already laid the ground workor just like we did before we're going to use the law of signs here and to use the law of signs I'm going to pick up that given information angle B and side B is one of my frac FRS and I'm looking for side C so I need to pick up angle C and side C for my other fraction okay let's put that together I'm going to start with my B's so the sign of 34.5 divided by side B which is six is equal to the sign of angle C we just found that that was 36.2 all divided by side C we'll call that c sub 2 let's go ahead and cross multiply remember your goal here is to solve for C so I end up with C sub 2 times that s of [Music] 34.5 is equal to 6 the other sign so the sign of 36.2 all I need to do is to divide off the sign of 34.5 34.5 do the same same thing on the other side again it feels super familiar same steps different values so this gives me c sub 2 so my second side C is equal to what I've got here let's get this into our calculator so 6 the S of 36.2 close the parentheses divided by the sign of 34.5 34.5 enter you don't need those parentheses and I get 6.3 if I'm rounding so about 6.3 and that's that last value that we needed for triangle number two you are doing so good law of cosiness is next [Applause] |
17692 | https://www.pbslearningmedia.org/resource/our20-math-7512/interpreting-negative-numbers/ | Interpreting Negative Numbers | PBS LearningMedia
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In this video lesson, students investigate different ways to represent addition of signed numbers on a number line. They review what they know about negative numbers including placing them on the number line, comparing and ordering them, and interpreting them in the contexts of temperature and elevation.
Using the context of temperature helps students make sense of the addition equations. Students see that an increase in temperature can be represented as adding a positive value and a decrease in temperature can be represented as adding a negative value. When students use quantitative contexts like temperature to aid in abstract reasoning about numeric expressions with signed numbers, they engage in MP2.
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Unit 5, Lessons 1 & 2: Interpreting Negative NumbersSpanish Student Activity: Unit 5, Lessons 1 & 2: Interpreting Negative Numbers
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17693 | https://www.merriam-webster.com/dictionary/isosceles%20triangle | Est. 1828
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isosceles triangle
noun
mathematics
: a triangle in which two sides have the same length
Examples of isosceles triangle in a Sentence
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The triangles vertices are formed by the stars Deneb, Vega, and Altair, forming a celestial isosceles triangle thatd make Euclid proud, that spans the constellations Cygnus, Lyra, and Aquila. —Tom Hawking, Popular Science, 31 July 2025 The trio will appear to form a broad isosceles triangle with Venus marking the vertex angle, while Aldebaran and the moon form the base of the triangle. —Joe Rao, Space.com, 18 July 2025 The only obstacle is that this hunk is her best friend Lady Danburys (Adjoa Andoh) estranged brother, forming an odd isosceles triangle of Regency relations. —Devon Ivie, Vulture, 14 June 2024
Rhymes for isosceles triangle
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“Isosceles triangle.” Merriam-Webster.com Dictionary, Merriam-Webster, Accessed 28 Sep. 2025.
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17694 | https://math.stackexchange.com/questions/2240535/extend-a-fourier-function-to-an-odd-function | Extend a fourier function to an odd function - Mathematics Stack Exchange
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Extend a fourier function to an odd function
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Given is a function f(x)= 1-x with 0 < x < 1 that should be extended on the interval [-1, 1] to get an odd function. Outside this interval it should be continued with a period of 2.
If it would be an even function, I would extend it to f(x) = 1-|x| as even functions are symmetric around the y-axis. Would this be correct?
Now I am not sure, how to extend it to an odd function. I remember that odd functions should be point symmetric around the origin, and I think that in this case it will stay the same: f(x)= 1-x. Can this be correct?
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asked Apr 18, 2017 at 17:41
mrs fouriermrs fourier
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You need 1−x 1−x on (0,1)(0,1) so on (−1,0)(−1,0) the result is −1−x−1−x. There is a discontinuity at 0 0, where any odd function must be 0 0. By periodicity, the function should vanish at all even integers. You can work out the function elsewhere; note that, except at discontinuities, the gradient is always −1−1.
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answered Apr 18, 2017 at 17:47
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Thanks a lot. This really makes sense. :)mrs fourier –mrs fourier 2017-04-22 16:28:19 +00:00 Commented Apr 22, 2017 at 16:28
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17695 | https://nebusresearch.wordpress.com/2012/04/06/the-difference-of-two-triangles/ | Skip to content
Joseph Nebus's work in progress.
The Difference Of Two Triangles
[ Trapezoid Week continues! ]
Yesterday I set out a diagram, showing off one example of a trapezoid, with which I mean to show one way to get the formula for a trapezoid’s area. The approach being used here is to find two triangles so that the difference in area between the two is the area of the trapezoid. This can often be a convenient way of finding the area of something: find simple shapes to work with so that the area we want is the sum or the difference of these easy areas. Later on I mean to do this area as the sum of simple shapes.
For now, though, I have the trapezoid set up so its area will be the difference of two triangle areas. The area of a triangle is a simple enough formula: it’s one-half the length of the base times the height. We’ll see much of that formula.
In the diagram I put the longer base, connecting points A and B, of length b1, at the bottom, and the shorter base, connecting points D and E, of length b2, on top. And I extended the two non-parallel legs so that they eventually come together at a point, labelled C. The trapezoid is the figure bounded by the points A, B, E, and D, or to put it succinctly, ABED. The distance between the line AB and the line DE is the altitude a. And the perpendicular distance between line AB and the point C is the first height, h1. The perpendicular height between the line DE and the point C is the second height, h2. That’s all background.
There are two triangles in the figure. The bigger one is triangle ΔABC, and its area is (1/2) b1 h1, half its base times its height. The smaller triangle is triangle ΔDEC, and its area is (1/2) b2 h2, half its base times its height. The trapezoid ABED is the part of triangle ΔABC not already in the triangle ΔDEC; its area has to be the difference between them. That is, the trapezoid has area (1/2) b1 h1 – (1/2) b2 h2. This is perfectly true, although it requires we figure out whta h1 and h2 are, which might be work. What we’ll do is try to get rid of h1 and h2.
Here’s one piece that lets us get rid of h2 at least. (It’ll return, but, we’ll get rid of it for good shortly after). The height h1 is the distance between the parallel bases, which we were calling a, plus the distance between the shorter base DE and the point C, h2. So we can either replace h1 with a + h2, or we can replace h2 with h1 – a. Either way will work, but, I want to use the second expression. By replacing h2 where it appears in the area for the trapezoid, we get the equation:
ABED = (1/2) b1 h1 – (1/2) b2 (h1 – a)
That second part of the above line we can expand, according to the distributive property. According to the distributive property, for any three numbers, s (x + y) is the same number as s x + s y. So we can change the way we write the equation — though not its value, or its truth — into this:
ABED = (1/2) b1 h1 – (1/2) b2 h1 + (1/2) b2 a
Now we have the area as the sum of three terms, the middle one negative. This is about where I got stuck in class, by the way, as I couldn’t think of how to get rid of the h1 there. At best I could shuffle between h1 and h2, without eliminating both together.
The insight that I needed, and that came to me after class, was to realize that the triangles ΔABC and ΔDEC are similar. That has a precise mathematical definition: it means the triangles have the same interior angles. If they differ at all, it’s only in their size. If I were to show a picture of ΔABC by itself, and to show a zoomed-in picture of ΔDEC by itself, there’d be no way of telling them apart, other than the labels.
Why this is important is because of a neat property regarding similar triangles. If you have two similar triangles, then, the ratios of corresponding pieces of the two triangles will be equal. That is, for example, the ratio of the length of AB to the length of AC will be the same as the length of DE to the length of DC. More immediately usefully, the ratio of the length of AB, b1, to the height of ΔABC, h1, is equal to the ratio of the length of DE, b2, to the height of ΔDEC, h2.
Put in a more familiar equation form, h1 / b1 = h2 / b2. Or we can rewrite that in a couple of ways; the one I am going to use is: b2 = b1 h2 / h1.
Now I’ll put this expression for b2 in to the equation just where I got stuck, and before long, both the h2 and the h1 terms will melt away. Starting with the expression where I got stuck:
ABED = (1/2) b1 h1 – (1/2) b2 h1 + (1/2) b2 a
Replacing b2 as I intended gives me this:
ABED = (1/2) b1< h1 – (1/2) (b1 h2 / h1 ) h1 + (1/2) b2 a
In the middle term, we have a quantity which is divided by h1 only to be multiplied by h1 again. As long as h1 isn’t equal to zero, this division and multiplication come out to the same thing as multiplying the rest of the quantity by 1, which is just leaving the quantity alone. And we can be fairly sure that whatever h1 is, it isn’t equal to zero. So:
ABED = (1/2) b1 h1 – (1/2) b1 h2 + (1/2) b2 a
Now I’m going to call on the distributive property again, combining the first and second terms again:
ABED = (1/2) b1 ( h1 – h2 ) + (1/2) b2 a
That’s a lot of work not to have got rid of h1 or h2 … except, that we do know something about h1 minus h2. That difference is the distance between the two parallel bases, that is, a. So the area of the trapezoid is:
ABED = (1/2) b1 a + (1/2) b2 a
And — you may have seen this coming — with yet another round of the distributive property we get:
ABED = (1/2) ( b1 + b2 ) a
So we’ve gotten exactly the formula we should have gotten.
This is one way of showing the formula’s true. It’s not the best. I’m not sure this is even the best way of doing it by the difference-between-triangles, particularly as it seems to need to remember this bit about similar triangles. It’s not an obscure property of similar triangles, but it isn’t as obvious as the area for the triangles. The proof also does make one hard-to-remove assumption: we’re supposing that the legs AE and BE, if we keep drawing those lines out, eventually come together in a point. That is, the proof is not going to work on parallelograms or on rectangles, which at least last time around I had wanted to include as trapezoids.
So, how could this be done better?
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Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him. View all posts by Joseph Nebus
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17696 | https://www.quora.com/What-is-modular-arithmetic-and-are-there-any-real-world-uses | What is modular arithmetic, and are there any real world uses? - Quora
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What is modular arithmetic, and are there any real world uses?
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Keith Ramsay
Ph.D. in mathematics · Author has 5.4K answers and 8.2M answer views
·4y
Modular arithmetic does have a number of practical applications, but for a mathematician, giving examples is a little like giving practical applications of some basic tool like a lathe. To us, modular arithmetic is quite a basic tool, especially in number theory. I’ll try to give some good examples at the end.
Modular arithmetic is an example of defining a new concept by abstraction from an old one, namely integer arithmetic. We define what is known as an “equivalence relation” on the integers, and define arithmetic on the “equivalence classes”.
Two integers a a and b b are said to be congruent modu
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Modular arithmetic does have a number of practical applications, but for a mathematician, giving examples is a little like giving practical applications of some basic tool like a lathe. To us, modular arithmetic is quite a basic tool, especially in number theory. I’ll try to give some good examples at the end.
Modular arithmetic is an example of defining a new concept by abstraction from an old one, namely integer arithmetic. We define what is known as an “equivalence relation” on the integers, and define arithmetic on the “equivalence classes”.
Two integers a a and b b are said to be congruent modulo an integer n n if a−b a−b is divisible by n n. The integer n n is known as the “modulus”. We write this as a≡b(mod n)a≡b(mod n). This counts as an equivalence relation because it satisfies three properties. First, for any integer a a, we have that a a is congruent to itself mod n n. Second, if a a is congruent to b b modulo n n, then b b is congruent to a a mod n n. Third, if a a is congruent to b b and b b is congruent to c c modulo n n, then a a is congruent to c c modulo n n. They follow from the facts that a−a=0 a−a=0 is divisible by any integer, if a−b a−b is divisible by n n then so is b−a=−(a−b)b−a=−(a−b), and if a−b a−b and b−c b−c are divisible by n n, then so is (a−b)+(b−c)=a−c(a−b)+(b−c)=a−c.
Once we know that a relationship is an equivalence relation, it follows that the domain (the integers) is partitioned into equivalence classes, consisting in all of the integers congruent to some one given integer. So for example, modulo 2 2 all the even integers are congruent to each other, and all the odd integers are congruent to each other, but none of either one is congruent to any of the other.
The way “sum” (and product and other operations) is defined for congruence classes is by saying that the sum of two congruence classes A A and B B is the congruence class which the sums a+b a+b where a a is in A A and b b is in B B fall. The only way that this makes sense is if the congruence classes are not empty, and if there is only one congruence class that the sums a+b a+b belong to. Congruence classes (like any equivalence class) always have an element, because a congruence class is the set of all integers congruent to some given integer, which is always a member. The other thing we have to show is that if a a is congruent to a′a′ and b b is congruent to b′b′ modulo n n, then a+b a+b is congruent to a′+b′a′+b′. For addition, that’s pretty simple since (a+b)−(a′+b′)=(a−a′)+(b−b′)(a+b)−(a′+b′)=(a−a′)+(b−b′) which is divisible by n n if a−a′a−a′ and b−b′b−b′ are. To show that the product of two congruence classes is well-defined is a little bit less simple, but follows by a b−a′b′=(a b−a b′)+(a b′−a′b′)=a(b−b′)+(a−a′)b′a b−a′b′=(a b−a b′)+(a b′−a′b′)=a(b−b′)+(a−a′)b′ being divisible by n n if a−a′a−a′ and b−b′b−b′ are.
One practical application of modular arithmetic is the RSA (Rivest-Shamir-Adelman) cryptosystem, where you use a modulus n n which can be written as n=p q n=p q where p p and q q are two primes each requiring many digits to write out in decimal (and as I recall, they prefer that (p−1)/2(p−1)/2 and (q−1)/2(q−1)/2 also be prime to make the system harder to crack). The modulus n n is published, as well as an integer a a which is used an exponent. Anybody can do arithmetic modulo n n, so they can take a piece of a message encoded as an integer m m, 0<m<n 0<m<n, and raise it to the a power modulo n n. What they can’t currently do is to reverse that encryption process. If you know what p p and q q are, you can find a second exponent b b with the property that m a b m a b is congruent to m m modulo n n, which lets you decrypt the message. Quantum computing seems likely to undermine the security of RSA once quantum computers having enough qubits to store several integers as big as n n are available, because there is in theory an algorithm for quantum computers that would let one factor n n into p q p q in a reasonable amount of time.
Some popular books about number theory like to give clock time as an example of modular arithmetic. The hours on a 12 12 or 24 24 hour clock are essentially the integers modulo 12 12 or 24 24, so if you know something will take 16 hours and figure out what time of day it will be when it is done, you are basically doing modular arithmetic.
Computers typically are digital and use bits for storage and processing. Arithmetic modulo 2 2 is often referred to in discussions of computers under a different name. One way to describe it is as taking the “exclusive or” or XOR of two bits. One thinks of 0 0 as representing “false” and 1 1 as representing “true”. The sum a+b a+b of two bits modulo 2 2 is under that representation the same as “a a or b b, but not both” (a a exclusive-or b b).
Modular arithmetic is used in a lot of check-sums. For example, the ISBN (international standard book number) of a book satisfies a congruence relation on the digits in it. It’s harder then accidentally to make a typo in the book number without it causing the congruence to be false (letting one know that a mistake has been made).
I’m not terribly familiar with all the practical applications of finite fields. Mathematicians in algebra (number theorists especially) also regard finite fields as a basic tool although we don’t tend to focus on direct practical applications. The integers modulo p p where p p is a prime is one simple example of a finite field, and other finite fields can be thought of as extensions of such a “prime” subfield. I know they are also used in cryptography. I once saw a job advertisement from the NSA (National Security Agency of the U.S.) looking for mathematicians with good familiarity with Galois theory and finite fields. Naturally they’re not going to tell everybody exactly what they are doing with them.
Some computations are more easily done by first doing them modulo some large number n n and then working backward to what the actual answer is.
It was only several years after I had heard of modular arithmetic that I read that it could be extended to rational numbers whose denominator has no common factor greater than 1 1 with the modulus n n.
The integers modulo n are often denoted Z/(n)Z/(n). Sometimes one sees it written just as Z n Z n but that conflicts with the notation used in number theory where that is used to denote the n n-adic numbers (where n n usually is a prime).
A lot of people like to think of the congruence classes in terms of a set of representatives, one taken from each congruence class, typically {0,1,2,…,n−1}{0,1,2,…,n−1}. Pure mathematicians tend to shy away from committing to any one such set of representatives as if it were “canonical”, although it can be convenient in some cases to make such a selection. Sometimes though it’s more useful to use a different range, like {−n/2+1,…,n/2}{−n/2+1,…,n/2} when n n is even or {−(n−1)/2,…,(n−1)/2}{−(n−1)/2,…,(n−1)/2} when n is odd. Most of the time, though, picking a representative out of the congruence class is just irrelevant, as all members are equally suited (for many purposes).
Modular arithmetic is a special case of reducing a ring by an ideal (which is short for the older term, “ideal divisor”). If we back up to what we did when we defined arithmetic modulo n n in the integers, we made equivalence classes on which the arithmetic operations are defined. To generalize this, we consider a more general algebraic structure on which operations like addition and multiplication are defined, known as a “ring”. An example of a ring would be the complex numbers of the form a+b√5 i a+b 5 i where a a and b b are integers (denoted Z[√5 i]Z[5 i]).
If we analyze what we did by reducing modulo n n, we can discover that it hinges on what elements we decide to make equivalent to 0 0. In order to make addition well-defined on the equivalence classes, a a and b b have to be equivalent when a−b a−b is equivalent to 0 0, and vice-versa. Call the set of elements which are considered equivalent to 0 0 the set I I (for “ideal”). To make addition well-defined, again we need for the sum of any two elements in I I to be an element of I I. To make multiplication well-defined, we need for the product of an element in I I with any other element still to be in I I.
The only ideals in the integers are the sets I I of the form {k n:k∈Z}{k n:k∈Z}., in other words the integers divisible by some given integer n n. The congruence classes are then just the integers modulo n n. (In the special case of n=0 n=0, modern authors tend to say that 0 0 is the one integer divisible by 0 0, although some authors, especially of older texts, have defined 0 0 not to be divisible by 0 0.) This is partly why modular arithmetic is important for the study of integers. The set of multiples of a given element is always an ideal. An ideal of that form is called a principal idea. A ring like the integers where every ideal is principal is called a principal ideal domain (PID).
Some rings however have other ideals. The ring Z[√5 i]Z[5 i] defined above, for example, has some additional ones. It fails unique factorization because (2)(3)=(1−√5 i)(1+√5 i)(2)(3)=(1−5 i)(1+5 i). There is an ideal consisting of all the complex numbers of the form a+b√5 i a+b 5 i where a a is congruent to b b modulo 2 2. This idea is non-principal, and it acts like the “missing” common divisor between 2 2 and 1+√5 i 1+5 i., which don’t have common divisors in this ring besides 1 1 and −1−1.
Algebraic number theory is pervaded with talk of ideals in rings, as in the example just above.
Algebraic geometry is pervaded with talk of ideals in rings. The set of polynomials which are 0 0 everywhere on some curve or surface is an ideal in the polynomials. For example, the polynomials in x x and y y which are 0 0 everywhere on the unit circle x 2+y 2=1 x 2+y 2=1 constitute an ideal, which happens to be the multiples of x 2+y 2–1 x 2+y 2–1.
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What are some practical applications of modular arithmetic?
What are some well-known uses of modular arithmetic?
Why is modular arithmetic useful?
Other than telling time, what are some applications of modular arithmetic?
What is modular arithmetic?
Henry Zhu
B.S. in Computer Science, University of California, Davis (Graduated 2022) · Upvoted by
Gilbert Doan
, Master (Unattempted) Mathematics, San Jose State University (2018) · Author has 290 answers and 779.4K answer views
·Updated 5y
Originally Answered: What is modular arithmetic? ·
Modular arithmetic concentrates on using arithmetic with numbers that involve modulus operators.
The foundation of modular arithmetic is congruence. If a,b∈Z a,b∈Z (a and b are integers) and m∈N m∈N (m is a natural number), we can state:
a≡b(m o d m)a≡b(m o d m) if m|(a−b)m|(a−b) (m m is a divisor of (a−b)(a−b))
In other words, a a is congruent to b(m o d n)b(m o d n).
For example:
20≡4(m o d 4)20≡4(m o d 4), because 4|(20–4)4|(20–4)
There are three properties in modular arithmetic that depend on congruence, given the following constraints: a,b,c,d∈Z a,b,c,d∈Z (a a, b b, c c, and d d are integers) and m\i m\i
Continue Reading
Modular arithmetic concentrates on using arithmetic with numbers that involve modulus operators.
The foundation of modular arithmetic is congruence. If a,b∈Z a,b∈Z (a and b are integers) and m∈N m∈N (m is a natural number), we can state:
a≡b(m o d m)a≡b(m o d m) if m|(a−b)m|(a−b) (m m is a divisor of (a−b)(a−b))
In other words, a a is congruent to b(m o d n)b(m o d n).
For example:
20≡4(m o d 4)20≡4(m o d 4), because 4|(20–4)4|(20–4)
There are three properties in modular arithmetic that depend on congruence, given the following constraints: a,b,c,d∈Z a,b,c,d∈Z (a a, b b, c c, and d d are integers) and m∈N m∈N (m is a natural number).
If a≡b(m o d m)a≡b(m o d m) and c≡d(m o d m)c≡d(m o d m)
a+c≡b+d(m o d m)a+c≡b+d(m o d m)
a∗c≡b∗d(m o d m)a∗c≡b∗d(m o d m)
a∗k≡b∗k(m o d m)a∗k≡b∗k(m o d m)
These basic properties allow us to solve all sorts of different modular arithmetic problems—problems that seem daunting at first.
For example: Find the remainder when 3 254 3 254 is divided by 7 7
In this problem, we start out by finding the first non-zero power congruent to 1(m o d 7)1(m o d 7). In this case, it is 3 6=729≡1(m o d 7)3 6=729≡1(m o d 7). We can multiply this expression by itself through the second property to get:
3 6∗3 6≡1∗1(m o d 7)3 6∗3 6≡1∗1(m o d 7)
which simplifies to:
3 12≡1(m o d 7)3 12≡1(m o d 7)
We can continuously multiply this expression by 3 6≡1(m o d 7)3 6≡1(m o d 7) to get the following expression:
3 252≡1(m o d 7)3 252≡1(m o d 7)
The following congruence will help tie the knot for us to solve this problem:
3 2≡2(m o d 7)3 2≡2(m o d 7)
Multiplying 3 252≡1(m o d 7)3 252≡1(m o d 7) and 3 2≡2(m o d 7)3 2≡2(m o d 7) through the second property gives us
3 252∗3 2≡1∗2(m o d 7)3 252∗3 2≡1∗2(m o d 7)
3 254≡2(m o d 7)3 254≡2(m o d 7)
Thus, the remainder when 3 254 3 254 is divided by 7 7 is 2.
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Madeline Griswold
Mathematician · Upvoted by
Gilbert Doan
, Master (Unattempted) Mathematics, San Jose State University (2018)
·7y
Originally Answered: What is modular arithmetic? ·
The study of math that has to do with cycles of numbers or remainders.
The most commonly used example of modular arithmetic is the clock (a mod 12 system).
Using modular arithmetic, you can question what time will it be after 67 hours. This is written as 12 mod 67. In order to calculate this you divide 67 by 12 and the remainder is the answer. 12 mod 67 is 7
Modular arithmetic is applicable to many different areas. Our current counting system is a base 10 system, but using another system (such as base 12) involves modular arithmetic. Additionally, much of cryptography (code breaking) involves mod
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The study of math that has to do with cycles of numbers or remainders.
The most commonly used example of modular arithmetic is the clock (a mod 12 system).
Using modular arithmetic, you can question what time will it be after 67 hours. This is written as 12 mod 67. In order to calculate this you divide 67 by 12 and the remainder is the answer. 12 mod 67 is 7
Modular arithmetic is applicable to many different areas. Our current counting system is a base 10 system, but using another system (such as base 12) involves modular arithmetic. Additionally, much of cryptography (code breaking) involves modular arithmetic.
And my favorite part of it is that while it sounds super smart and professional, the basic concept is merely finding the remainder.
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Lukas Schmidinger
I have graduate CS and my studies included math courses. · Author has 27.7K answers and 14.9M answer views
·4y
What is modular arithmetic, and are there any real world uses?
Basically it looks at the other result of whole number division: the reminder (well at least with non-negative integers).
Or more mathematically a mod b≡c a mod b≡c
means that there is an integer d d such that b d+c=a b d+c=a
normally d=⌊a b⌋b d=⌊a b⌋b
where ⌊a b⌋⌊a b⌋ is the largest integer less or equal than a b a b thus d d is the largest integer multiple of d d less or equal than a a and c c is the gap between that.
On a number bases this is great because of the following rules:
(a±b)mod(a±b)mod
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What is modular arithmetic, and are there any real world uses?
Basically it looks at the other result of whole number division: the reminder (well at least with non-negative integers).
Or more mathematically a mod b≡c a mod b≡c
means that there is an integer d d such that b d+c=a b d+c=a
normally d=⌊a b⌋b d=⌊a b⌋b
where ⌊a b⌋⌊a b⌋ is the largest integer less or equal than a b a b thus d d is the largest integer multiple of d d less or equal than a a and c c is the gap between that.
On a number bases this is great because of the following rules:
(a±b)mod c≡((a mod c)±(b mod c))mod c(a±b)mod c≡((a mod c)±(b mod c))mod c
(a×b)mod c≡((a mod c)×(b mod c))mod c(a×b)mod c≡((a mod c)×(b mod c))mod c
(a b)mod c≡((a mod c)b)mod c(a b)mod c≡((a mod c)b)mod c
It allows to answer questions to answer questions like this: Lukas Schmidinger's answer to How would I prove that (1990^1990) - 1 can be divided by 1991 (feel free to use modular arithmetic)?
But to more practical uses, well I think I will quote the Wikipedia article on the topic fully:
In theoretical mathematics, modular arithmetic is one of the foundations of number theory, touching on almost every aspect of its study, and it is also used extensively in group theory, ring theory, knot theory, and abstract algebra. In applied mathematics, it is used in computer algebra, cryptography, computer science, chemistry and the visual and musical arts.
A very practical application is to calculate checksums within serial number identifiers. For example, International Standard Book Number (ISBN) uses modulo 11 (for 10 digit ISBN) or modulo 10 (for 13 digit ISBN) arithmetic for error detection. Likewise, International Bank Account Numbers (IBANs), for example, make use of modulo 97 arithmetic to spot user input errors in bank account numbers. In chemistry, the last digit of the CAS registry number (a unique identifying number for each chemical compound) is a check digit, which is calculated by taking the last digit of the first two parts of the CAS registry number times 1, the previous digit times 2, the previous digit times 3 etc., adding all these up and computing the sum modulo 10.
In cryptography, modular arithmetic directly underpins public key systems such as RSA and Diffie–Hellman, and provides finite fields which underlie elliptic curves, and is used in a variety of symmetric key algorithms including Advanced Encryption Standard (AES), International Data Encryption Algorithm (IDEA), and RC4. RSA and Diffie–Hellman use modular exponentiation.
In computer algebra, modular arithmetic is commonly used to limit the size of integer coefficients in intermediate calculations and data. It is used in polynomial factorization, a problem for which all known efficient algorithms use modular arithmetic. It is used by the most efficient implementations of polynomial greatest common divisor, exact linear algebra and Gröbner basis algorithms over the integers and the rational numbers. As posted on Fidonet in the 1980's and archived at Rosetta Code, modular arithmetic was used to disprove Euler's sum of powers conjecture on a Sinclair QLmicrocomputer using just one-fourth of the integer precision used by a CDC 6600supercomputer to disprove it two decades earlier via a brute force search.
In computer science, modular arithmetic is often applied in bitwise operations and other operations involving fixed-width, cyclic data structures. The modulo operation, as implemented in many programming languages and calculators, is an application of modular arithmetic that is often used in this context. The logical operator XOR sums 2 bits, modulo 2.
In music, arithmetic modulo 12 is used in the consideration of the system of twelve-tone equal temperament, where octave and enharmonic equivalency occurs (that is, pitches in a 1∶2 or 2∶1 ratio are equivalent, and C-sharp is considered the same as D-flat).
The method of casting out nines offers a quick check of decimal arithmetic computations performed by hand. It is based on modular arithmetic modulo 9, and specifically on the crucial property that 10 ≡ 1 (mod 9).
Arithmetic modulo 7 is used in algorithms that determine the day of the week for a given date. In particular, Zeller's congruence and the Doomsday algorithm make heavy use of modulo-7 arithmetic.
More generally, modular arithmetic also has application in disciplines such as law (e.g., apportionment), economics (e.g., game theory) and other areas of the social sciences, where proportional division and allocation of resources plays a central part of the analysis.
Modular arithmetic - Wikipedia
In short it is a great and easy check system.
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Where is modular arithmetic applied?
How do you teach modular arithmetic?
What is a simple way of explaining modular arithmetic?
How is modular arithmetic used in cryptography?
What are some examples of real world applications of modular arithmetic?
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Modular arithmetic is a system of arithmetic for integers, where numbers wrap around upon reaching a certain value known as the modulus. The notation a mod n a mod n represents the remainder when a a is divided by n n. For example, 7 mod 5=2 7 mod 5=2 because when you divide 7 by 5, the remainder is 2.
Key Concepts
Congruence: In modular arithmetic, we say that two numbers a a and b b are congruent modulo n n if they give the same remainder when divided by n n. This is written as:
a≡b mod n a≡b mod n
For example, 10≡2 mod 8 10≡2 mod 8 because both 10 and 2 leave a remainder of 2 when divided by 8.
2. Operations: You can per
Continue Reading
Modular arithmetic is a system of arithmetic for integers, where numbers wrap around upon reaching a certain value known as the modulus. The notation a mod n a mod n represents the remainder when a a is divided by n n. For example, 7 mod 5=2 7 mod 5=2 because when you divide 7 by 5, the remainder is 2.
Key Concepts
Congruence: In modular arithmetic, we say that two numbers a a and b b are congruent modulo n n if they give the same remainder when divided by n n. This is written as:
a≡b mod n a≡b mod n
For example, 10≡2 mod 8 10≡2 mod 8 because both 10 and 2 leave a remainder of 2 when divided by 8.
2. Operations: You can perform addition, subtraction, and multiplication in modular arithmetic, and the results are also taken modulo n n. For instance:
(3+5)mod 7=1(3+5)mod 7=1
(6−4)mod 5=2(6−4)mod 5=2
(3×4)mod 5=2(3×4)mod 5=2
Real-World Uses
Cryptography: Modular arithmetic is fundamental in various encryption algorithms, such as RSA, which relies on the properties of large prime numbers and modular exponentiation.
Computer Science: It is widely used in hash functions, random number generation, and algorithms that require wrapping around values, such as circular buffers.
Clock Arithmetic: A common everyday example is how we tell time. For instance, if it is 10 o'clock and you add 5 hours, you arrive at 3 o'clock, not 15. This is effectively (10+5)mod 12=3(10+5)mod 12=3.
Error Detection: Checksums and hash codes often use modular arithmetic to verify the integrity of data. For example, the checksum might be the sum of data bytes modulo 256.
Game Development: In games, modular arithmetic can be used to cycle through levels or manage inventory systems where items wrap around after reaching a certain count.
Conclusion
Modular arithmetic is a versatile mathematical concept with applications across various fields, making it an essential tool in both theoretical and practical contexts. Whether in computer science, cryptography, or everyday life, its principles help manage cyclical and repetitive scenarios efficiently.
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Robert Langner
Author has 3.8K answers and 1.7M answer views
·4y
Telling time is one real world use. If it’s 11 now, what time will it be in 2 hours? 1. You calculate time mod 12.
Another use is in cryptocurrencies, or in general all Internet security. It’s based on number theory and modular arithmetic. Without modular arithmetic there’s no Internet commerce or bitcoin.
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Trym Bruset
Data scientist and topologer living in Norway · Author has 326 answers and 738.8K answer views
·11y
Originally Answered: Number Theory: What is modular arithmetic? ·
Intuitively: Clock face arithmetic.
Given a module (say 12), we can preform arithmetic on a group whose elements are equivalence classes.
The equivalence classes in modular arithmetic is a set of integers whose remainder are equal when divided by the module. So 7 and 19 are in the same equivalence class modulo 12, since both leave remainder 7. (07 and 19 are on the same place on a clock face).
The problem with composite modules (like 12=223) is that they have zero divisors, that is, nonzero elements a and b, such that ab=0.
On a clock a and b can be 4 and 3, so if you count 4, 3 times, you'
Continue Reading
Intuitively: Clock face arithmetic.
Given a module (say 12), we can preform arithmetic on a group whose elements are equivalence classes.
The equivalence classes in modular arithmetic is a set of integers whose remainder are equal when divided by the module. So 7 and 19 are in the same equivalence class modulo 12, since both leave remainder 7. (07 and 19 are on the same place on a clock face).
The problem with composite modules (like 12=223) is that they have zero divisors, that is, nonzero elements a and b, such that ab=0.
On a clock a and b can be 4 and 3, so if you count 4, 3 times, you're back at 0 without passing through all elements.
Now having zero divisors causes some problems when you're doing arithmetic, perhaps most notably:
The cancellation laws fails, so given ab=ac we can no longer conclude b=c.
Multiplicative inverses fail, so for an element a, there is no element a' so that aa'=1.
So if you want all the good stuff associated with a finite field, make sure your module is prime.
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Bernard Montaron
PhD in Mathematics&Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980) · Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 3.2K answers and 2.1M answer views
·2y
Originally Answered: What is the definition of modular arithmetic? What is an example of a problem that can be solved using modular arithmetic? ·
Modular arithmetic is the arithmetic of integers modulo m m, some integer generally positive and greater than 1, where integers n n are grouped in m m classes represented by the remainder r r, called the residue, in the Euclidean division of n n by m m, i.e. n=r+q m n=r+q m with 0≤r≤m−1 0≤r≤m−1.
All integers having the same residue r r modulo m m are considered equivalent, and belong to the equivalence class represented by r r. An addition and multiplication table can be defined modulo m m , for example (m−1)+1=0(m−1)+1=0.
There is a huge number of applications of modular arithmetic. I’ll give two examples.
First, how do you prove th
Continue Reading
Modular arithmetic is the arithmetic of integers modulo m m, some integer generally positive and greater than 1, where integers n n are grouped in m m classes represented by the remainder r r, called the residue, in the Euclidean division of n n by m m, i.e. n=r+q m n=r+q m with 0≤r≤m−1 0≤r≤m−1.
All integers having the same residue r r modulo m m are considered equivalent, and belong to the equivalence class represented by r r. An addition and multiplication table can be defined modulo m m , for example (m−1)+1=0(m−1)+1=0.
There is a huge number of applications of modular arithmetic. I’ll give two examples.
First, how do you prove that the equations x 3+y 3+z 3=4+9 k x 3+y 3+z 3=4+9 k or x 3+y 3+z 3=5+9 k x 3+y 3+z 3=5+9 k have no solutions in integers (x,y,z)(x,y,z) for any integer k k ?
This is done by calculating the residues of a cube modulo 9, it’s easy to see that whatever the integer x x is, the only possible residues of x 3 x 3 modulo 9 are -1 (i.e. 8), 0, 1. Hence, the only possible residues of x 3+y 3+z 3 x 3+y 3+z 3 modulo 9 are 0,±1,±2,±3 0,±1,±2,±3 i.e. 0,1,2,3,6,7,8 0,1,2,3,6,7,8. Since neither 4 or 5 belong to this set, this proves the result.
My second example, is how do you prove that there is an infinity of integers that cannot be expressed as a sum of less than 15 powers of 4 of integers?
This is proved by considering the residues of x 4 x 4 modulo 16. If x x is even, then x 4 x 4 is a multiple of 16, and the residue is 0 modulo 16. If x x is odd, i.e. x=2 k+1 x=2 k+1 then x 2=4 k(k+1)+1 x 2=4 k(k+1)+1 and
x 4=(4 k(k+1)+1)2=16 k 2(k+1)2+8 k(k+1)+1 x 4=(4 k(k+1)+1)2=16 k 2(k+1)2+8 k(k+1)+1
and since k(k+1)k(k+1) is always even, we obtain the residue 1 modulo 16.
This implies directly that an integer of the form 16 k+15 16 k+15 cannot be expressed as a sum of less than 15 powers of 4, which completes the proof.
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Kevin Smyth
Bsc. (hons) in Mathematics, Chelsea College of Science and Technology, University of London (Graduated 1969) · Author has 4K answers and 1.4M answer views
·2y
Originally Answered: What is the definition of modular arithmetic? What is an example of a problem that can be solved using modular arithmetic? ·
Put simply, modular arithmetic creates a finite arithmetic field of integers with a maximum number M, such that M + 1 = 0. So, the numbers “wrap around”.
One everyday fairly recent use of modular arithmetic is an electronic computer. Your desktop computer, handy and iPad, etc. all have finite space to store data and, naturally, a finite data bus width. In the 1990s, a typical computer data bus was 16 bits. Some counts might be able to reach that and then wrap around to zero. It was thought there would be a problem with some old programs when 1999 changed to 2,000. It turned out that the mainly
Continue Reading
Put simply, modular arithmetic creates a finite arithmetic field of integers with a maximum number M, such that M + 1 = 0. So, the numbers “wrap around”.
One everyday fairly recent use of modular arithmetic is an electronic computer. Your desktop computer, handy and iPad, etc. all have finite space to store data and, naturally, a finite data bus width. In the 1990s, a typical computer data bus was 16 bits. Some counts might be able to reach that and then wrap around to zero. It was thought there would be a problem with some old programs when 1999 changed to 2,000. It turned out that the mainly first generation of computer programmers was more conscientious and cleverer than given credit for. The millennium problem proved to be vastly overblown.
One unavoidable demonstration of modular arithmetic is the clock on the wall. In fact, when children are first introduced to modular arithmetic, it is often called “clock” arithmetic. A clock has two modular fields. The hour hand is modular 12 (there are also special cases of modular 24). The minute and second hands are both modular 60. The reason for 12 and 60 seems that our ancestors first performed arithmetic with fractions, and 12 and 60 have both a reasonable number of factors.
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Harald Overbeek
loves mental calculation · Author has 185 answers and 626.6K answer views
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What are some well-known uses of modular arithmetic?
Checking answers of calculations.
Nine proof or casting out nines:
One of the most useful ways of using modular arithmetic is by checking the answers of long additions, subtractions, multiplications, divisions, etc., etc.
The check is done, not by repeating the calculation, but by doing a reduced version of the same calculation. it is sometimes called the nine proof. An eleven proof also exists.
An example will make this clear.
Let's do a multiplication:
17 X 31 = 527
Now we check it by reducing the numbers to a single digit:
17 = 1+7 = 8
31 = 3 + 1 = 4
527 = 5 + 2 + 7 = 14. 14 = 1+4 = 5
Now do the m
Continue Reading
Checking answers of calculations.
Nine proof or casting out nines:
One of the most useful ways of using modular arithmetic is by checking the answers of long additions, subtractions, multiplications, divisions, etc., etc.
The check is done, not by repeating the calculation, but by doing a reduced version of the same calculation. it is sometimes called the nine proof. An eleven proof also exists.
An example will make this clear.
Let's do a multiplication:
17 X 31 = 527
Now we check it by reducing the numbers to a single digit:
17 = 1+7 = 8
31 = 3 + 1 = 4
527 = 5 + 2 + 7 = 14. 14 = 1+4 = 5
Now do the multiplication again, using the single digits:
8 X 4 = 32. 3 + 2 = 5
The answer to the original multiplication, 527, reduced is 5. The reduced calculation also leads to 5, so the answer is correct.
This is an example of modulo 9 arithmetic. 17 modulo 9 = 8, because 17/9 = 1 rest 8.
31 modulo 9 = 4, because we can subtract 27, a multiple of nine and that leave a rest of 4.
The modulo nine of a number can be calculated by adding the digits of a number together. If the result is higher than 9 as was the case in 527, repeat the process until we have a number between zero and nine.
Eleven proof:
The same calculation can be done using modulo 11 arithmetic:
17 mod 11 = 6 (subtract 11 from 17)
31 mod 11 = 9 (subtract 22 from 31)
6 X 9 = 54 and 54 mod 11 = 10 (subtract 44 from 54)
527 mod 11 = 10 (527 - 440 = 87 and 87 - 77 = 10)
17 reduced is 6 and 31 reduced is 9. 6 X 9 = 54 and 54 reduces is 10.
Since 99 is a factor of 11, we can calculate the modulo of 527 as 27+5 = 32 and 32 - 22 = 10.
This is not as easy as the modulo 9 calculation. However; with some practice the calculations go quick.
Clock calculations:
If the time is 11 o' clock and you need to be somewhere in 4 hours, you know that that meeting is at 3 o' clock.
You have added 4 to 11 to get 15 and then did a modulo 12 to arrive at 3.
Date calculations:
If the date is 29th of January and you have an appointment in 4 days, you know it is on the 2nd of February.
You have added 4 to 29 to arrive at 33 and then did a modulo 31, for Januari has 31 days.
Other regular uses:
Checksums in bank account and IBAN numbers. Finds typo's in typed in numbers.
Checksums in files. Finds out if a file is transferred well form one place to another.
etc., etc.
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David Dirkse
Former Customer Engineer (1968–1993)
·2y
Originally Answered: What is the definition of modular arithmetic? What is an example of a problem that can be solved using modular arithmetic? ·
modular arithmetic counts in remainders
in case of:
modulus 10: we take the lowest digit of the number
modulus 2: we only count even or odd
modulus 5: 1 mod 5 = 6 mod 5 = 11 mod 5 = 16 mod 5 = 1
Example: we have a rectangle with 5 columns and 10 rows. making 50 fields counted n=0..49
So for field n we calculate the column = n mod 5.
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Quoss Wimblik
Hobbyist · Author has 378 answers and 90.3K answer views
·4y
Originally Answered: What is modular arithmetic? ·
When you divide two numbers you can get a Quotient and a Remainder. In the Mod function with Modular Arithmetic they deal just with the remainder.
Modular arithmetic helps in many fields and is a well studied aspect of math’s and computing.
Mod functions can be represented as working like clocks as well but it’s the same thing as working with the remainder which works like clock.
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Tom Musgrove
Android developer, machine learning, and researcher · Author has 4.2K answers and 10.1M answer views
·11y
Originally Answered: What is the significance of modular arithmetic? ·
One obvious usage is 'clock arithmetic' ie what time will it be a certain number of hours later. It also has numerous uses in a number of fields of mathematics, particularly number theory. It also has practical applications for cryptography.
Modular arithmetic
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Finding the phase shift of a cosine function, given the graph
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Here is what I have found so far:
Vertical Displacement = 2 units down (-2)
Amplitude = 4
Period = 2pi/3
I am now trying to find the phase shift. I moved the 'working x-axis' 2 units down, in accordance with the vertical displacement of -2. Since the cosine function starts at 1 on the y-axis, how do I find the phase shift. If the x-axis is at -2, then the point where y=1 is a tiny fraction of a unit away from the y-axis. So how do I find the phase shift?
algebra-precalculus
trigonometry
graphing-functions
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asked Dec 1, 2014 at 22:52
McBMcB
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1 Note that the simple cosine graph intersects the y-axis at (0,1).Pakquebchsoflwty –Pakquebchsoflwty 2014-12-01 22:55:03 +00:00 Commented Dec 1, 2014 at 22:55
Noted. @Pakquebchsoflwty When my x-axis is at -2, y=1 (relative to the working axis) is very close to (0,1). How do I determine the phase shift from there?McB –McB 2014-12-01 22:57:06 +00:00 Commented Dec 1, 2014 at 22:57
2 You are not looking for where y=1. You are looking for where y=1×4−2 y=1×4−2 You need to take into account the amplitude and vertical shift of the function. So, where does y=2 that will give you the least possible shift?turkeyhundt –turkeyhundt 2014-12-01 22:57:44 +00:00 Commented Dec 1, 2014 at 22:57
1 Actually, -1. The phase shift is the opposite of what many people intuitively think. I always think of it like this. Cosine is a maximum at cos(0)cos(0) So in this case, we want a maximum at y=1, so cos(x−1)cos(x−1) fits the bill.turkeyhundt –turkeyhundt 2014-12-01 23:03:29 +00:00 Commented Dec 1, 2014 at 23:03
1 well yeah 1 unit. but what are the units here guys? pi/6. so it's pi/6 shift. not 1!wild_child –wild_child 2016-09-22 07:17:07 +00:00 Commented Sep 22, 2016 at 7:17
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The simple function 4 cos(x)4 cos(x) will have a relative maximum at (0,4)(0,4). However, your function seems to be 4 cos(x−1)−2 4 cos(x−1)−2. The phase shift I believe is ω ϕ ω ϕ. In this case, ω=1 ω=1 and ϕ=1 ϕ=1. So the phase shift is 1 1=1 1 1=1.
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answered Dec 2, 2014 at 4:05
SalmonKillerSalmonKiller
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0How to find the phase shift of this cosine graph?
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FAQ
Thematic maps are a powerful way to visualize specific types of data across geographic areas — such as population density, climate zones, traffic intensity, or access to public services. They help turn raw numbers into spatial insights, making it easier to spot patterns and support data-driven decisions.
For example, urban planners use thematic maps to optimize transport routes, public health agencies track disease outbreaks, and journalists illustrate voting trends by region. These maps often combine thematic data with a geographic base layer to provide clear context and orientation.
In this guide, you'll learn what thematic maps are, explore their main types, and discover real-world examples and use cases across different industries.
What Are Thematic Maps?
A simple thematic maps definition is: maps that focus on visualizing a specific theme, topic, or dataset across a geographic area. Unlike general-purpose maps that show physical or political features, thematic maps highlight patterns or variations in data — such as income levels, air pollution, or school performance.
The goal of a thematic map is to communicate a clear message about the distribution of a particular variable. These maps are widely used in fields like urban planning, environmental science, public health, and business analysis.
Practical examples include:
Population density maps that help city planners identify areas for housing or infrastructure development (see World Population History).
Air quality maps showing pollution levels across different neighborhoods (e.g., EPA AirNow).
Choropleth election maps used by media to display voting outcomes by region (see U.S. Census Bureau Median Household Income Map).
Crime rate maps helping law enforcement allocate resources more effectively.
Retail analysis maps visualizing store performance or customer concentration (often used in GIS-driven market research).
By combining data with geography, thematic maps turn complex statistics into intuitive visuals — making trends and relationships easier to understand and act upon.
Types of Thematic Maps + Examples and Use Cases
There are many types of thematic maps, each suited to different kinds of data and storytelling goals. Below are the most common types, along with practical examples and real-world use cases:
Choropleth Maps
Choropleth maps use color shading to represent data values across geographic regions — for example, darker colors may indicate higher population density or income. These maps are commonly used when data is tied to administrative boundaries such as countries, states, or districts.
Results of the 2025 German federal election. Main map shows single-member constituencies (Direktmandate); crosshatching indicates seats forfeited under the new electoral law. Bottom left shows proportional list seats by state. Image by Erinthecute, licensed under CC BY-SA 4.0.
Examples
A map showing COVID-19 infection rates by district, with darker shades indicating higher case counts.
A national unemployment map highlighting joblessness levels by state or region.
A median household income map using color gradients to show economic disparities between counties.
A voter turnout map where intensity of shading represents the percentage of registered voters who participated in an election.
You can see a live example built with Leaflet using Geoapify services. To get accurate geographic boundaries for your own data, check out the Boundaries API. If you’re looking for a full explanation of how choropleth maps work and when to use them, this guide covers the details.
Pros
Simple to read and understand.
Great for comparing regions at a glance.
Widely supported across mapping tools.
Cons
Can be misleading when area sizes differ greatly.
Needs careful color classification to avoid misinterpretation.
Not suitable for exact counts or small areas with large values.
For large regions or detailed maps, boundary data can be heavy — you may need to simplify polygons to avoid performance issues when loading on the web.
Bivariate Maps
Bivariate maps are thematic maps that display two variables simultaneously on the same map. This is typically done by combining two color scales — such as hue and saturation — so that each area reflects a combination of values for both variables.
This type of map is useful when you want to understand the relationship or interaction between two datasets across geographic space — for example, comparing income and education, or population density and access to healthcare.
Bivariate vs. Choropleth maps
A choropleth map shows a single variable (e.g., unemployment rate) using a single color scale.
A bivariate map, in contrast, combines two variables in a grid of color combinations, allowing both to be visualized at once — but with added complexity in interpretation.
Bivariate choropleth map showing average wait times and number of yellow taxi requests in NYC by zone (June 2023). Author: Lucoooo, created with QGIS. Licensed under CC BY-SA 4.0.
Examples
A map showing median income and high school graduation rate in each county, using blended color schemes.
A housing map combining rental cost and commute time to identify areas with poor affordability.
A map comparing internet access and poverty levels to highlight digital divides.
Pros
Enables multivariate spatial analysis in a single visual.
Helps reveal correlations or contrasts between two related datasets.
Saves space and avoids splitting attention across two maps.
Cons
Harder to read and interpret, especially for general audiences.
Requires a clear and intuitive legend to be useful.
Color combinations may be challenging for accessibility (e.g., color blindness).
Isoline Maps
Isoline maps use continuous lines to connect points with equal values of a specific variable. These lines — known as isolines — help visualize how that variable changes across space. Common types include isotherms (equal temperature), isobars (equal atmospheric pressure), and contour lines (equal elevation).
When isolines are drawn from precise point measurements (like weather stations), it's called an isoline map. If the values represent averaged data across regions (like population density), it's referred to as an isopleth map — a closely related but distinct type.
Water depth around the island of Rhodes visualized with contour-like isolines. Map data © OpenSeaMap, under the Open Database License (ODbL).
Examples
A topographic map showing terrain elevation with closely spaced contour lines in mountainous areas.
A weather map using isobars to indicate high and low-pressure zones.
A wind speed map where isotachs connect regions with equal wind velocity.
A noise map illustrating equal decibel levels around an airport or highway.
If you're interested in travel-based isolines, Geoapify Isochrone Maps offer an intuitive way to show how far someone can travel within a set time or distance from a point. You can also use the Isoline API to generate isochrones (travel time areas) and isodistances (travel distance areas) using various transport modes and constraints.
Pros
Ideal for displaying gradual, continuous changes in spatial data.
Great for highlighting patterns, boundaries, and transitional zones.
Widely used in scientific, environmental, and engineering applications.
Cons
Requires dense and accurate point data for reliable results.
Can be challenging to interpret for non-experts.
May become visually cluttered in areas with rapid or extreme data variation.
Results can be too rough or generalized if data is sparse or unevenly distributed.
Isopleth Maps
Isopleth maps are used to represent continuous phenomena that vary across space — such as population density, rainfall, or air pollution. These maps use shaded regions or filled contours to illustrate areas with similar values, helping to visualize gradients and transitions in the data.
Isopleths are often derived from interpolated data and show values that are distributed throughout an area, not just at specific points. They're commonly used when data is aggregated (e.g., per square kilometer) or smoothed over a surface.
Isopleth vs. Isoline
Isolines are lines that connect points of equal value — such as elevation or temperature. These maps typically show only the lines.
Isopleths use those isolines to create filled regions between the lines, often with a color gradient to represent value ranges.
In short: Isolines show the lines, while isopleths show the regions between them.
Estimated tsunami travel times from the 1755 Lisbon earthquake (red: 1–4h, yellow: 5–6h, green: 7–14h, blue: 15–21h); source: NOAA NGDC, public domain.
Examples
A population density map highlighting urban-to-rural transitions using smoothed color gradients.
A median household income map showing gradual economic variation across neighborhoods.
A crop productivity map identifying high- and low-yield agricultural zones.
A climate index map visualizing comfort zones based on temperature and humidity.
A disease vulnerability map estimating relative risk across a country based on healthcare data and demographics.
Pros
Ideal for visualizing smooth spatial patterns in continuous or averaged data.
Makes it easier to identify regions with similar characteristics.
Works well with statistical, environmental, and demographic data.
Cons
May hide localized extremes due to averaging or smoothing.
Needs careful interpolation to avoid distortions.
Often confused with isoline maps, especially when visual styles are similar.
Less precise for point-specific data, as it generalizes over areas.
Heat Maps
Heat maps are thematic maps that use color intensity or gradients to represent the density or intensity of a phenomenon across a surface. They are typically raster-based and do not rely on administrative boundaries — making them ideal for visualizing data that occurs irregularly or continuously in space.
Unlike choropleth or dot density maps, heat maps interpolate values based on proximity and concentration, resulting in smoothed visual gradients that highlight hotspots and cold zones.
Mean wind speed in China from the Global Wind Atlas. Source: Global Wind Atlas, licensed under CC BY 4.0.
Examples
A crime heat map of a city, with red areas indicating high incident concentration.
A retail footfall map showing customer density around shopping centers.
An air quality heat map based on sensor networks across a region.
A bike-sharing activity map displaying usage intensity across urban stations.
Pros
Very effective at showing concentration or intensity without requiring boundaries.
Ideal for point-based or location event data (e.g., GPS, sensors, logs).
Visually intuitive and engaging for general audiences.
Cons
Can distort actual values — intensity may be influenced by clustering or scaling method.
Not suitable for exact or comparative analysis (values are not precise).
May over-smooth or exaggerate hotspots depending on the chosen parameters.
Dot-Density Maps
Dot-density maps use dots to represent a quantity of a variable within a defined area. Each dot corresponds to a specific number of occurrences (e.g., one dot = 1,000 people), and the distribution of dots visually conveys how the data is spread across space.
These maps are especially useful for showing absolute counts and distributions rather than ratios or densities. They don’t show exact locations of individual events or people, but rather illustrate spatial patterns based on data aggregated by regions.
Dot distribution map showing places of worship in Kerala, India — one of the most religiously diverse states in the country. Source: Survey of India Open Data. Author: Planemad, licensed under CC BY-SA 4.0.
Examples
A U.S. population map, where each dot represents 1,000 people, dispersed to reflect urban and rural densities.
A crop distribution map, with dots showing wheat farms across agricultural zones.
A disease outbreak map, using dots to represent reported cases by region.
A refugee movement map, displaying displaced individuals by host regions.
Pros
Great for visualizing absolute numbers and spatial distribution.
Easy to understand and interpret visually.
Does not require complex classification or interpolation.
Cons
Can be misleading if dots overlap heavily or are randomly placed within large areas.
Not ideal for small value differences — hard to detect subtle variation.
Dot placement is usually randomized within a polygon, not precise.
Size Theme / Graduated Symbol Maps
Graduated symbol maps (also known as size theme maps) use symbols of varying size — typically circles — to represent quantitative values at specific geographic locations. The size of each symbol is scaled proportionally (or semi-proportionally) to the data value, making it easy to compare magnitudes across locations.
These maps are especially effective for point-based data, such as city populations, event counts, or production volumes. They’re ideal when you want to highlight absolute values rather than ratios or densities.
Terminology note:
“Graduated symbol map” is the standard term in cartography and GIS tools, while “size theme map” is a more informal name that refers to the same concept.
Graduated symbol map showing antimicrobial use in 2020 (red circles) and projected increase by 2030 (dark red rings). Source: PLOS Global Public Health, licensed under CC BY 4.0.
Examples
A city population map, where larger circles indicate more residents.
An earthquake magnitude map with symbols sized according to Richter scale values.
An airport map showing passenger volumes with graduated circles.
A refugee map where symbol size represents the number of displaced people hosted in each region.
Pros
Excellent for comparing absolute values between locations.
Easy to understand and visually engaging.
Works well on both small and large geographic scales.
Cons
Can be difficult to interpret exact values without a clear legend.
Overlapping symbols may obscure data in densely populated areas.
Not suitable for normalized data (e.g., rates or percentages).
Cartograms
Cartograms are thematic maps in which geographic regions are resized or distorted based on a specific variable, rather than their actual land area. The shape and relative position of regions may be preserved (in some types) or altered to prioritize data representation over geographic accuracy.
Cartograms are powerful for showing comparative quantities in a visually impactful way — for example, emphasizing countries with larger populations, economies, or carbon emissions regardless of their physical size.
Cartogram showing market capitalisation of publicly listed firms in Switzerland, aggregated by canton of registered office. Author: Sdnegel, licensed under CC BY-SA 4.0
Types of Cartograms
Contiguous cartograms: Maintain shared borders between regions but distort shape (e.g., squashed or stretched countries).
Non-contiguous cartograms: Maintain original shapes but allow regions to shrink or expand independently, creating space between them.
Dorling cartograms: Replace regions with scaled circles, often positioned to resemble the original geography.
Examples
A world population cartogram where densely populated countries (like India or China) appear much larger.
A US election results cartogram scaled by number of voters or electoral votes instead of state size.
A carbon footprint map where countries are resized according to their CO₂ emissions.
A global vaccination coverage map showing doses administered, not landmass.
Pros
Highly effective at drawing attention to disparities in data.
Excellent for showing comparative values (e.g., who contributes most or least).
Memorable and eye-catching for communication and storytelling.
Cons
Distorts geographic accuracy, which can confuse some viewers.
Harder to interpret spatial relationships like distance or adjacency.
May be less familiar to general audiences than standard map types.
Tools & Platforms to Create Thematic Maps
From beginner-friendly editors to advanced GIS platforms.
You can build thematic maps with a wide range of tools — whether you're a developer, analyst, educator, or journalist. Below are both no-code and low-code platforms suited for different needs and experience levels.
No-Code Mapping Tools (No Programming Required)
These tools let you upload your data and visually build thematic maps — ideal for users who don’t want to write code.
Recommended Platforms:
Mapifator
A no-code thematic map builder by Geoapify. Create interactive maps with custom icons, filters, and storytelling features.
Datawrapper
Great for journalists and educators. Supports choropleths, symbol maps, and more with just a CSV file and a few clicks.
Flourish
Powerful visualization tool with templates for choropleths, tile maps, symbol maps, and animated storytelling — all no-code.
ArcGIS Online
ESRI’s browser-based platform for creating interactive maps using templates and data layers — no code required for most use cases.
CARTO Builder
A cloud GIS tool offering an intuitive drag-and-drop interface for building choropleth and bivariate maps. Advanced users can add SQL queries or scripts, but basics require no code.
Best for:
Educators, journalists, NGOs, marketers, and analysts.
Low-Code / Advanced Tools (For Developers and Analysts)
These tools provide more power and flexibility but assume some level of coding or GIS expertise.
Recommended Platforms:
QGIS
Open-source GIS software for professional users. Supports all thematic map types but has a learning curve and scripting options (Python, SQL).
Mapbox Studio
Vector styling tool for creating custom basemaps and map layers. Styling requires understanding of Mapbox expressions.
Kepler.gl
High-performance web-based map builder by Uber. Excellent for large datasets and time-based visualizations. Some technical knowledge recommended.
Tableau
Visual analytics platform with strong mapping support — great for symbol and filled maps. Advanced use cases involve calculated fields or GIS integration.
Power BI
Business analytics tool with mapping capabilities. Supports shape maps, ArcGIS integration, and custom visuals — works best with structured spatial data.
Leaflet and MapLibre
JavaScript libraries for building fully custom interactive maps. Requires coding, but offer maximum control and integration with Geoapify services.
Best for:
Developers, GIS analysts, data visualization experts, and projects requiring full customization or interactivity.
Power Your Maps with Geoapify APIs
Combine your preferred tool with Geoapify services for better data and map functionality:
Map Tiles – Add clean, customizable base layers.
Boundaries API – Retrieve administrative and regional polygons for choropleths and region-based maps.
Isoline API – Generate travel-time and distance zones for isochrones, accessibility, and logistics maps.
Place Details API – Get detailed POI metadata and original boundary geometry to enrich your thematic maps.
Conclusion
Thematic maps are more than just visuals — they are powerful tools for understanding data in a spatial context. Whether you're analyzing population trends, exploring environmental impacts, or building interactive dashboards, choosing the right map type helps turn raw data into clear insights.
With a growing ecosystem of no-code and developer-friendly tools, it's easier than ever to create professional, meaningful maps for any audience or use case.
Ready to build your own thematic map?
Start by exploring our Map Tiles as a base layer, and enhance your maps with APIs like the Boundaries API and Isoline API. Whether you're coding or not, Geoapify provides everything you need to bring your data to life.
→ Explore Geoapify Tools and start mapping today!
FAQ
Which type of thematic map is best for visualizing population density?
Population density is typically visualized using choropleth maps for regional comparisons or dot-density maps when you want to show detailed spatial distribution.
How can I improve map performance when loading large datasets?
To improve performance, use simplified geometry (e.g., reduced polygon detail) when rendering large boundary datasets. You can simplify features in a GIS tool or request simplified results from APIs such as the Geoapify Boundaries API.
Do I need to provide attribution when using Geoapify APIs or map tiles?
Yes, attribution is required, especially on the Free plan. Include a visible mention like: Powered by Geoapify. Attribution to OpenStreetMap is also required where applicable.
Can I visualize my own data on Geoapify map tiles?
Yes, you can overlay your own or third-party data on top of Geoapify map tiles using libraries like Leaflet, MapLibre, or QGIS. This allows you to build fully customized thematic maps with your data.
How can I create a thematic map without programming?
You can use no-code tools like Mapifator, QGIS, or Datawrapper. These platforms let you upload your data and build interactive maps using visual editors, without writing a single line of code.
Can I use Leaflet or MapLibre to build custom thematic maps?
Yes, both Leaflet and MapLibre are excellent open-source JavaScript libraries for building custom, interactive thematic maps. You can combine them with Geoapify APIs and Map Tiles for full flexibility.
Where can I get country, state, or city boundaries for choropleth maps?
Use the Geoapify Boundaries API to get administrative and political boundary polygons for countries, regions, cities, and more. It’s perfect for generating choropleth and region-based maps.
How can I generate travel-time zones for accessibility or service area maps?
Geoapify offers an Isoline API that allows you to generate isochrones (travel-time areas) and isodistances based on driving, walking, biking, and public transit.
How can I enrich my thematic map with details about specific places?
Use the Place Details API to retrieve additional information about places, such as POI metadata, categories, and original boundary geometry. This is helpful for adding context or interactivity to your map.
Can I combine multiple Geoapify APIs in one map?
Yes, and it's encouraged! For example, you can combine Map Tiles for your base layer, Boundaries API for regions, Isoline API for accessibility zones, and Place Details API for interactive popups — all within the same map interface.
Geo dataData visualization |
17699 | https://www.doubtnut.com/qna/185017902 | If k(x)=5x+2, what is the vlaue of k(4)−k(1) ?
15
17
19
21
The correct Answer is:A
To solve the question, we need to find the value of k(4)−k(1) given that k(x)=5x+2. 1. Calculate k(4): k(4)=5⋅4+2 =20+2 =22 2. Calculate k(1): k(1)=5⋅1+2 =5+2 =7 3. Subtract k(1) from k(4): k(4)−k(1)=22−7 =15 Final Answer: The value of k(4)−k(1) is 15. ---
To solve the question, we need to find the value of k(4)−k(1) given that k(x)=5x+2.
Calculate k(4):
k(4)=5⋅4+2
=20+2
=22
Calculate k(1):
k(1)=5⋅1+2
=5+2
=7
Subtract k(1) from k(4):
k(4)−k(1)=22−7
=15
Final Answer:
The value of k(4)−k(1) is 15.
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