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6100	https://open.oregonstate.education/anatomy2e/chapter/bone-classification/	6.2 Bone Classification – Anatomy & Physiology 2e Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Book Contents Navigation Contents An Introduction to the Human Body 1.0 Introduction: Human Body 1.1 How Structure Determines Function 1.2 Structural Organization of the Human Body 1.3 Homeostasis 1.4 Anatomical Terminology 1.5 Medical Imaging The Chemical Level of Organization 2.0 Introduction: Chemical Level of Organization 2.1 Elements and Atoms: The Building Blocks of Matter 2.2 Chemical Bonds 2.3 Chemical Reactions 2.4 Inorganic Compounds Essential to Human Functioning 2.5 Organic Compounds Essential to Human Functioning The Cellular Level of Organization 3.0 Introduction: Cellular Level of Organization 3.1 The Cell Membrane 3.2 The Cytoplasm and Cellular Organelles 3.3 The Nucleus and DNA Replication 3.4 Protein Synthesis 3.5 Cell Growth and Division 3.6 Cellular Differentiation The Tissue Level of Organization 4.0 Introduction: Tissue Level of Organization 4.1 Types of Tissues 4.2 Epithelial Tissue 4.3 Connective Tissue 4.4 Muscle Tissue 4.5 Nervous Tissue 4.6 Tissue Injury and Aging The Integumentary System 5.0 Introduction: Integumentary System 5.1 Layers of the Skin 5.2 Accessory Structures of the Skin 5.3 Functions of the Integumentary System 5.4 Diseases, Disorders, and Injuries of the Integumentary System Bone Tissue and the Skeletal System 6.0 Introduction: Bone Tissue and Skeletal System 6.1 The Functions of the Skeletal System 6.2 Bone Classification 6.3 Bone Structure 6.4 Bone Formation and Development 6.5 Fractures: Bone Repair 6.6 Exercise, Nutrition, Hormones, and Bone Tissue 6.7 Calcium Homeostasis: Interactions of the Skeletal System and Other Organ Systems Axial Skeleton 7.0 Introduction: Axial Skeleton 7.1 Divisions of the Skeletal System 7.2 The Skull 7.3 The Vertebral Column 7.4 The Thoracic Cage 7.5 Embryonic Development of the Axial Skeleton The Appendicular Skeleton 8.0 Introduction: Appendicular Skeleton 8.1 The Pectoral Girdle 8.2 Bones of the Upper Limb 8.3 The Pelvic Girdle and Pelvis 8.4 Bones of the Lower Limb 8.5 Development of the Appendicular Skeleton Joints 9.0 Introduction: Joints 9.1 Classification of Joints 9.2 Fibrous Joints 9.3 Cartilaginous Joints 9.4 Synovial Joints 9.5 Types of Body Movements 9.6 Anatomy of Selected Synovial Joints 9.7 Development of Joints Muscle Tissue 10.0 Introduction: Muscle Tissue 10.1 Overview of Muscle Tissues 10.2 Skeletal Muscle 10.3 Muscle Fiber Excitation, Contraction, and Relaxation 10.4 Nervous System Control of Muscle Tension 10.5 Types of Muscle Fibers 10.6 Exercise and Muscle Performance 10.7 Smooth Muscle Tissue 10.8 Development and Regeneration of Muscle Tissue The Muscular System 11.0 Introduction: Muscular System 11.1 The Roles of Agonists, Antagonists, and Synergists 11.2 The Organization of Muscle Fascicles and Their Role in Generating Force 11.3 The Criteria Used to Name Skeletal Muscles 11.4 Axial Muscles of the Head, Neck, and Back 11.5 Axial Muscles of the Abdominal Wall and Thorax 11.6 Muscles of the Pectoral Girdle and Upper Limbs 11.7 Appendicular Muscles of the Pelvic Girdle and Lower Limbs The Nervous System and Nervous Tissue 12.0 Introduction: Nervous System and Nervous Tissue 12.1 Structure and Function of the Nervous System 12.2 Nervous Tissue 12.3 The Function of Nervous Tissue 12.4 Communication Between Neurons 12.5 The Action Potential The Peripheral Nervous System 13.0 Introduction: Peripheral Nervous System 13.1 Sensory Receptors 13.2 Ganglia and Nerves 13.3 Spinal and Cranial Nerves 13.4 Ventral Horn Output and Reflexes The Central Nervous System 14.0 Introduction: Central Nervous System 14.1 Embryonic Development 14.2 Blood Flow, the Meninges, and Cerebrospinal Fluid Production and Circulation 14.3 The Brain 14.4 The Spinal Cord 14.5 Sensory and Motor Pathways 14.6 Testing the Spinal Nerves (Sensory and Motor Exams) 14.7 The Cranial Nerve Exam The Special Senses 15.0 Introduction: Special Senses 15.1 Taste 15.2 Smell 15.3 Hearing 15.4 Equilibrium 15.5 Vision The Autonomic Nervous System 16.0 Introduction: Autonomic Nervous System 16.1 Divisions of the Autonomic Nervous System 16.2 Autonomic Reflexes and Homeostasis 16.3 Central Control 16.4 Drugs That Affect the Autonomic System The Endocrine System 17.0 Introduction: Endocrine System 17.1 An Overview of the Endocrine System 17.2 Hormones 17.3 The Pituitary Gland and Hypothalamus 17.4 The Thyroid Gland 17.5 The Parathyroid Glands 17.6 The Adrenal Glands 17.7 The Pineal Gland 17.8 Gonadal and Placental Hormones 17.9 The Pancreas 17.10 Organs with Secondary Endocrine Functions 17.11 Development and Aging of the Endocrine System The Cardiovascular System: Blood 18.0 Introduction: Cardiovascular System: Blood 18.1 Functions of Blood 18.2 Production of the Formed Elements 18.3 Erythrocytes 18.4 Leukocytes and Platelets 18.5 Hemostasis 18.6 Blood Typing The Cardiovascular System: The Heart 19.0 Introduction: Cardiovascular System: The Heart 19.1 Heart Anatomy 19.2 Cardiac Muscle and Electrical Activity 19.3 Cardiac Cycle 19.4 Cardiac Physiology 19.5 Development of the Heart The Cardiovascular System: Blood Vessels and Circulation 20.0 Introduction: Cardiovascular System: Blood Vessels and Circulation 20.1 Structure and Function of Blood Vessels 20.2 Blood Flow, Blood Pressure, and Resistance 20.3 Capillary Exchange 20.4 Homeostatic Regulation of the Vascular System 20.5 Circulatory Pathways 20.6 Development of Blood Vessels and Fetal Circulation The Lymphatic and Immune System 21.0 Introduction: Lymphatic and Immune System 21.1 Anatomy of the Lymphatic and Immune Systems 21.2 Barrier Defenses and the Innate Immune Response 21.3 The Adaptive Immune Response: T Lymphocytes and Their Functional Types 21.4 The Adaptive Immune Response: B Lymphocytes and Antibodies 21.5 The Immune Response against Pathogens 21.6 Diseases Associated with Depressed or Overactive Immune Responses 21.7 Transplantation and Cancer Immunology The Respiratory System 22.0 Introduction: Respiratory System 22.1 Organs and Structures of the Respiratory System 22.2 The Lungs 22.3 The Process of Breathing 22.4 Gas Exchange 22.5 Transport of Gases 22.6 Modifications in Respiratory Functions 22.7 Embryonic Development of the Respiratory System The Digestive System 23.0 Introduction: Digestive System 23.1 Overview of the Digestive System 23.2 Digestive System Processes and Regulation 23.3 The Mouth, Pharynx, and Esophagus 23.4 The Stomach 23.5 Accessory Organs in Digestion: The Liver, Pancreas, and Gallbladder 23.6 The Small and Large Intestines 23.7 Chemical Digestion and Absorption: A Closer Look Metabolism and Nutrition 24.0 Introduction: Metabolism and Nutrition 24.1 Overview of Metabolic Reactions 24.2 Carbohydrate Metabolism 24.3 Lipid Metabolism 24.4 Protein Metabolism 24.5 Metabolic States of the Body 24.6 Energy and Heat Balance 24.7 Nutrition and Diet The Urinary System 25.0 Introduction: Urinary System 25.1 Internal and External Anatomy of the Kidney 25.2 Microscopic Anatomy of the Kidney: Anatomy of the Nephron 25.3 Physiology of Urine Formation: Overview 25.4 Physiology of Urine Formation: Glomerular Filtration 25.5 Physiology of Urine Formation: Tubular Reabsorption and Secretion 25.6 Physiology of Urine Formation: Medullary Concentration Gradient 25.7 Physiology of Urine Formation: Regulation of Fluid Volume and Composition 25.8 Urine Transport and Elimination 25.9 The Urinary System and Homeostasis Fluid, Electrolyte, and Acid-Base Balance 26.0 Introduction: Fluid, Electrolyte, and Acid-Base Balance 26.1 Body Fluids and Fluid Compartments 26.2 Water Balance 26.3 Electrolyte Balance 26.4 Acid-Base Balance 26.5 Disorders of Acid-Base Balance The Sexual Systems 27.0 Introduction: Sexual Systems 27.1 Anatomy of Sexual Systems 27.2 Development of Sexual Anatomy 27.3 Physiology of the Female Sexual System 27.4 Physiology of the Male Sexual System 27.5 Physiology of Arousal and Orgasm Development and Inheritance 28.0 Introduction: Development and Inheritance 28.1 Fertilization 28.2 Embryonic Development 28.3 Fetal Development 28.4 Maternal Changes During Pregnancy, Labor, and Birth 28.5 Adjustments of the Infant at Birth and Postnatal Stages 28.6 Lactation 28.7 Patterns of Inheritance Appendix A. Anatomical Terminology Appendix B. Chemical Abbreviations and Equations Appendix C. Tables by Chapter Appendix D. Activity Overview Appendix E. Content Mapping Glossary Accessibility and Versioning Anatomy & Physiology 2e 6.2 Bone Classification Learning Objectives By the end of this section, you will be able to: Classify bones according to their shapes Describe the function of each category of bones The 206 bones that compose the adult skeleton are divided into five categories based on their shapes (Figure 6.2.1). Like other structure/function relationships in the body, their shapes and their functions are related such that each categorical shape of bone has a distinct function. Figure 6.2.1 – Classifications of Bones: Bones are classified according to their shape. Long Bones A long bone is one that is cylindrical in shape, being longer than it is wide. Keep in mind, however, that the term describes the shape of a bone, not its size. Long bones are found in the upper limbs (humerus, ulna, radius) and lower limbs (femur, tibia, fibula), as well as in the hands (metacarpals, phalanges) and feet (metatarsals, phalanges). Long bones function as rigid bars that move when muscles contract. Short Bones A short bone is one that is cube-like in shape, being approximately equal in length, width, and thickness. The only short bones in the human skeleton are in the carpals of the wrists and the tarsals of the ankles. Short bones provide stability and support as well as some limited motion. Flat Bones The term flat boneis somewhat of a misnomer because, although a flat bone is typically thin, it is also often curved. Examples include the cranial (skull) bones, the scapulae (shoulder blades), the sternum (breastbone), and the ribs. Flat bones serve as points of attachment for muscles and often protect internal organs. Irregular Bones An irregular bone is one that does not have any easily characterized shape and therefore does not fit any other classification. These bones tend to have more complex shapes, like the vertebrae that support the spinal cord and protect it from compressive forces. Many bones of the face, particularly the jaw bones that contain teeth, are classified as irregular bones. Sesamoid Bones A sesamoid bone is a small, round bone that forms in tendons (sesamo- = “sesame” and -oid = “resembling”). Tendons are a dense connective tissue that connect bones to muscles and sesamoid bones form where a great deal of pressure is generated in a joint. The sesamoid bones protect tendons by helping them overcome excessive forces but also allow tendons and their attached muscles to be more effective. Sesamoid bones vary in number and placement from person to person but are typically found in tendons associated with the feet, hands, and knees. The patellae (singular = patella) are the only sesamoid bones found in common with every person. Table 6.1 reviews bone classifications with their associated features, functions, and examples. Table 6.1: Bone Classification\| Bone classification \| Features \| Function(s) \| Examples \| --- --- \| \| Long \| Cylinder-like shape, longer than it is wide \| Movement, support \| Femur, tibia, fibula, metatarsals, humerus, ulna, radius, metacarpals, phalanges \| \| Short \| Cube-like shape, approximately equal in length, width, and thickness \| Provide stability, support, while allowing for some motion \| Carpals, tarsals \| \| Flat \| Thin and curved \| Points of attachment for muscles; protectors of internal organs \| Sternum, ribs, scapulae, cranial bones \| \| Irregular \| Complex shape \| Protect internal organs, movement, support \| Vertebrae, facial bones \| \| Sesamoid \| Small and round; embedded in tendons \| Protect tendons from excessive forces, allow effective muscle action \| Patellae \| Chapter Review Bones can be classified according to their shapes. Long bones, such as the femur, are longer than they are wide. Short bones, such as the carpals, are approximately equal in length, width, and thickness. Flat bones are thin, but are often curved, such as the ribs. Irregular bones such as those of the face have no characteristic shape. Sesamoid bones, such as the patellae, are small and round, and are located in tendons. Review Questions Critical Thinking Questions What are the structural and functional differences between a tarsal and a metatarsal? RevealStructurally, a tarsal is a short bone, meaning its length, width, and thickness are about equal, while a metatarsal is a long bone whose length is greater than its width. Functionally, the tarsal provides limited motion, while the metatarsal acts as a rigid bar against which muscle can act. What are the structural and functional differences between the femur and the patella? RevealStructurally, the femur is a long bone, meaning its length is greater than its width, while the patella, a sesamoid bone, is small and round. Functionally, the femur acts as a rigid bar for movement, while the patella protects the patellar tendon from excessive forces. Glossary flat bone thin and curved bone; serves as a point of attachment for muscles and protects internal organs irregular bone bone of complex shape; protects internal organs from compressive forces long bone cylinder-shaped bone that is longer than it is wide; functions as a lever sesamoid bone small, round bone embedded in a tendon; protects the tendon from compressive forces short bone cube-shaped bone that is approximately equal in length, width, and thickness; provides limited motion tendon a dense connective tissue that connect bones to muscles definition cylinder-shaped bone that is longer than it is wide; functions as a lever ×Close definition cube-shaped bone that is approximately equal in length, width, and thickness; provides limited motion ×Close definition thin and curved bone; serves as a point of attachment for muscles and protects internal organs ×Close definition bone of complex shape; protects internal organs from compressive forces ×Close definition small, round bone embedded in a tendon; protects the tendon from compressive forces ×Close definition a dense connective tissue that connect bones to muscles ×Close definition Previous/next navigation Previous: Functions of the Skeletal System Next: Bone Structure Back to top License Anatomy & Physiology 2e Copyright © 2025 by Lindsay M. Biga, Staci Bronson, Sierra Dawson, Amy Harwell, Robin Hopkins, Joel Kaufmann, Mike LeMaster, Philip Matern, Katie Morrison-Graham, Kristen Oja, Devon Quick, Jon Runyeon, and OpenStax is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted. Share This Book Pressbooks Powered by Pressbooks Pressbooks User Guide \|Pressbooks Directory \|Contact Pressbooks on YouTubePressbooks on LinkedIn
6101	https://www.tiktok.com/@freegcsemathsteacher/video/7146091508939197701?lang=en	Sign up \| TikTok TikTok Log in TikTok Search For You Explore Following Upload LIVE Profile More Log in Company Program Terms & Policies © 2025 TikTok 38.4K217 7796 1357 00:08 / 00:50 freegcsemathsteacher FreeGCSEMathsTeacher · 2022-9-22 Follow more #maths#mathematics#algebra#algebra1#gcsemaths#gcsemathsrevision Oh No (Instrumental) - Kreepa 217 comments Log in to comment fransarylibros Why expanding? Don't they teach that (a+b)²=a²+2ab+b² ? 2022-9-23 Reply 242 View 11 replies ShahnawazAnsari yeah (a+b)2 formula goes well with it. 2022-9-23 Reply 2 View 1 reply Hex u don't know ANYTHING about this 2023-1-14 Reply 0 Sep Tembre is this FOIL? I have a vague memory of it 2023-7-15 Reply 2 View 1 reply L.e.y.a00 This also works with the box method 2022-9-25 Reply 2 View 1 reply jcp0027 Perfect 🥰 2022-9-22 Reply 3 View 1 reply Maria Bertoldi to complicated 2022-9-24 Reply 0 View 1 reply Inday 🇵🇭 why you write ur X like that. 😅😅😅😅😅😅 2022-9-23 Reply 0 gumballdad Question like terms doesn’t suppose to be cancel out just a question no hate 2022-9-22 Reply 0 hnasif.mv 😅😅😅 RIGHT 2022-9-22 Reply 2 View 1 reply Em i love this type of math its so easy 2022-10-15 Reply 0 Daily Reminder for expansion, we can use pascals triangle 2022-9-23 Reply 1 View 1 reply C.Ianis (a+b)la puterea 2 = a la 2 + 2ab +b la 2. 2022-9-22 Reply 33 Gaelle_RG Thank u 2022-9-23 Reply 2 View 1 reply Carmela FOIL method 2022-9-23 Reply 1 View 1 reply kebab thanks 2022-9-24 Reply 0 𝒸𝒽𝓁𝑜𝑒💐 how do people physically and mentally understand algebra 2023-5-19 Reply 0 ᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠ you can also use smile method HAHAHA 2022-9-23 Reply 0 View 1 reply itsdeep213 Don’t the like terms get cancelled out though 2022-9-24 Reply 1 View 3 replies You may like Log in Log in Introducing keyboard shortcuts! Go to previous video Go to next video Like video Mute / unmute video Play / Pause Skip forward Skip backward Full screen Choose your interests Animals Comedy Travel Food Sports Beauty & Style Art Gaming Science & Education Dance Skip Done By continuing, you agree to TikTok’s Terms of Service and confirm that you have read TikTok’s Privacy Policy. Already have an account? Log in Unmute
6102	https://www.youtube.com/watch?v=PhQzxjEH2dQ	Sign analysis of first derivative to determine where f(x) is increasing or decreasing (Part 1) Math surgery 1530 subscribers Description 1382 views Posted: 26 Oct 2017 Transcript: the first derivative can be used to determine whether a function is increasing or decreasing and you're going to learn in this video how to find the intervals where the function is either increasing or decreasing so start with example one find the interval on which f of X is increasing and the intervals on which it's decreasing so the first step is determine the derivative or just the first derivative which is f dash X and we have ready work this out in a previous video but this is just straightforward so 6x squared minus 6x minus 12 it's a polynomial it's easy to depreciate and then step two we're going to find the critical points okay now um step two we need to find the critical points and these are basically the points where the first derivative is equal to zero so find the critical points and these are the points where F dash X 0 okay so we can find this points by L just write this properly equal to 0 ok so we can find this point by equating this function to 0 and then that's solving for x we already done this in the previous video and the values of X were X is equal to negative 1 and X is equal to 2 were calculated in the previous video okay step three evaluate the derivative between or on either side of the critical points so we have our critical points but we need to investigate what's happening between them or on either side of these guys so step three a one-way the derivative between or on either side of the critical points so we want to find what's the derivative between and on either side of these guys here which means we need to draw something like this so basically these are values of x by the way and we have the critical points values of X we have X is equal to negative 1 1 and X is equal to 2 ok but what I'm interested in is the science of the first derivative okay I'm interested in the science of the first derivative okay because I know that at x equals negative 1 the first derivative is zero at x equals 2 to the first derivative is zero but what happens between them and on either side so what I'm going to do is to choose what we call test points so I'll just choose any point on the left ok so any point on the left any point here in between and any point and these points are called test points on the Left I can adjust negative two and I may want to find out if I play black negative two into the first derivative okay here what do I get in fact if you plug negative two there you will end up with 24 so this will be equal to 24 how about if I plug in x equals to 0 because 0 is between negative 1 & 2 um if I plug in F - 0 and that would just be if you plug in 0 you'll end up with negative 12 okay how about if I plug in and number to the right of 2 for example 3 and if you plug in 3 what do you end up with if you plug in 3 you'll end up with 24 and you can try that in a calculator to check that ok 24 now just remember that this points here ok these points here they have a special name okay they're just called test points okay because we're using them to test the science of the sides of the first derivative and here we can see that the first derivative is positive here it's negative and here it's positive that's all we need to know okay and for example if this puzzle is positive we know that the function is increasing if it's negative before then we know that the function is decreasing so that leads us to UM step 4 so step four will be find the intervals where f of X is freezing is increasing or decreasing so find the intervals where f of X is increasing or decreasing okay we can go ahead and say all right so we're I will this function be increasing okay where is the first derivative greater than zero and we know that the function will be increasing sorry the function be treating when the first derivative is greater than 0 so and we just need to inspect on the graph where is it greater than zero and that will be when X is less than negative 1 and when X is greater than 2 okay and then when is the function decreasing and we know that a function is decreasing when the first derivative is less than 0 and we just need to investigate okay here when is the first derivative less than 0 when um X between negative 1 & 2 so between and that's how you determine the intervals where the function is either increasing or decreasing alright let's look at the example to find the intervals on which this function is increasing and the intervals on which it's decreasing so step one just like what you did previously is to determine the first derivative determine the derivative now we have already done this in the previous video and our first derivative was 6 X of x squared minus 1 squared so this was completed the previous video ok step 2 find the critical points so step 2 and find the critical points and these are the points where the first derivative is equal to 0 or the first derivative does not exist which means the vertical asymptotes so these are the points where the first derivative is equal to 0 or the first derivative okay we have already done this in the previous video and in fact we had our critical points as X is equal to 1 x equals negative 1 and X goes to 0 so calculated in the previous video okay so we have our critical points what's the next the next is to evaluate between or on either side of the critical points so step three would be evaluate the derivative value of the first derivative between on either side of the critical points okay so which means we need to draw a sign diagram so something that looks like that what we did previously and this is for X values and this X values are basically the key we are going to mark on this x one is the critical points so we have zero we have negative one and we have sorry I get c1 should be on the left we have negative one and we have one so by the way just remember that these guys here they are simple okay and this is also an asymptote which means the derivative does not exist at this points and maybe I may just want to fish to the left so X now what I'm interested in is the sign of the first derivative and to find the signs of the first derivative I'm going to use test points so I'll pick a test point here this one here this one here and the test point there okay so to the left of negative one I can pick negative two so okay I need to find the value of that so negative two if you plug negative two into the first derivative what do you end up with you'll end up with negative 4 over 3 so that would be negative 4 over 3 and you can check this on your calculator in between here plug in what's between there negative 1/2 if you plug in negative 1/2 in the first derivative what do you end up with okay that you are plugging in you end up with negative 16 over 3 and you can continue the same process here plug in 1/2 that would be positive 16 over 3 okay and then plug in here 2 which will be 4 over 3 now let's now mark in the signs so this is negative that's negative that's positive and that's positive ok now we need to determine where this function is increasing or decreasing so that leads us to step 4 so step 4 find the intervals where f of X is increasing or decreasing by my intervals where f of X is increasing or decreasing and we know that f of X is increasing when when f dash X is greater than zero so you know that this is zero and just by inspecting on the graph greater than zero that will be between 0 & 1 when X is between 0 & 1 you cannot say when X is greater than 0 because we have an asymptote which means the function does not exist there so between 0 & 1 and when X is greater than 1 okay decreasing when f dash X is less than zero when is it less than zero to the left of negative 1 and between negative 1 & 0 so that will be between negative 1 & 0 and when X is less than negative 1
6103	http://pcwww.liv.ac.uk/~karpenk/JournalUDT/vol11/no2/06BaGr.pdf	uniform distribution theory DOI: 10.1515/udt-2016–0017 Unif. Distrib. Theory 11 (2016), no.2, 151–161 SPATIAL EQUIDISTRIBUTION OF BINOMIAL COEFFICIENTS MODULO PRIME POWERS Guy Barat—Peter J. Grabner † Dedicated to the memory of Pierre Liardet ABSTRACT. The spatial distribution of binomial coeﬃcients in residue classes modulo prime powers is studied. It is proved inter alia that empirical distribution of the points (k, m)p−m with 0 ≤k ≤n < pm and n k ≡a (mod p)s (for (a, p) = 1) for m →∞tends to the Hausdorﬀmeasure on the “p-adic Sierpi´ nski gasket”, a fractals studied earlier by von Haeseler, Peitgen, and Skordev. Communicated by Jean Louis Verger-Gaugry 1. Introduction and Results Binomial coeﬃcients and their number theoretic properties are the subject of a vast number of investigations. For instance, D. Singmaster have studied divisibility properties and proved that for any integer m almost all binomial coeﬃcients are divisible by m in the following sense lim N→∞ 2 N(N + 1)# k, n ; 0 ≤k ≤n < N ∧m \| n k = 1. After this it is natural to ask what happens for the remaining set of density 0, or how the binomial coeﬃcients behave after dividing out the highest possible power of m. For prime p, the ﬁrst question has been answered independently 2010 M athem ati cs Subj ect C l as s i f i cati on: Primary 11B65; Secondary 11A63. K eyw or ds: Binomial coeﬃcients, equidistribution. † This author is supported by the Austrian Science Fund FWF projects F5503 (part of the Special Research Program (SFB) “Quasi-Monte Carlo Methods: Theory and Applications”) and W1230 (Doctoral Program “Discrete Mathematics” ). 151 G. BARAT—P. J. GRABNER in and , namely the binomial coeﬃcients not divisible by p are evenly distributed in the prime residue classes modulo p lim N→∞ # k, n ; 0 ≤k ≤n < N ∧ n k ≡a (mod p) # k, n ; 0 ≤k ≤n < N ∧ n k ̸≡0 (mod p) = 1 p −1 for (a, p) = 1. The methods used in these two papers are rather diﬀerent: in multiplicative characters are used, whereas in polynomial congruences over ﬁnite ﬁelds are applied. Both methods are based on ´ E. Lucas’ congruence n k ≡ n0 k0 n1 k1 · · · nL kL (mod p), where n = PL ℓ=0 nℓpℓand k = PL ℓ=0 kℓpℓare the respective p-adic digital expan-sions of n and k (with possible leading zeroes in the expansion of k). The result was extended to prime powers in using a generalisation of Lucas’ congru-ence due to A. Granville , see Theorem 3. The second question was addressed in the same paper, where it was shown that the p-free parts of the binomial co-eﬃcients are uniformly distributed in Z∗ p (p-adic integers). Notice that the p-free part of an integer n is given by n(p) = np−vp(n), where vp denotes the p-adic valuation. The result reads as lim N→∞ 2 N(N + 1)# ( k, n ; 0 ≤k ≤n < N ∧ n k (p) ≡a (mod ps) ) = 1 φ(ps) = 1 ps−1(p −1) (1) for all s ∈N and all a not divisible by p; φ denotes Euler’s totient as usual. Furthermore, the number of binomial coeﬃcients up to row N which have p-adic valuation j, has been studied by L. Carlitz . This is based on E. E. Kum-mer’s result , which states that the p-valuation of n k equals the number of carries in the subtraction n −k performed in base p. In Carlitz’ result could be reﬁned to a precise asymptotic formula; recently L. Spiegelhofer and M. Wallner found expressions for # ( k, n ; 0 ≤k ≤n ∧vp n k ! = j ) in terms of the numbers of certain blocks occurring in the digital expansion of n. Also recently, F. Greinecker could prove that binomial coeﬃcients, Stirling numbers, and more generally number schemes satisfying generalisations of Lucas’ congruence are spatially uniformly distributed modulo p in a sense that we will make precise below. 152 SPATIAL EQUIDISTRIBUTION OF BINOMIAL COEFFICIENTS MODULO PRIME POWERS The present paper exhibits detailed properties of the distribution of binomial coeﬃcients in residue classes modulo prime powers in the respects introduced above. The ﬁrst theorem states that p-free parts of n k are uniformly distributed in residue classes modulo ps and simultaneously spatially with respect to two--dimensional Lebesgue measure λ2 (restricted to the triangle {(x, y) ∈R2; 0 ≤x ≤y ≤1} and normalised). Theorem 1 . Let p be a prime, s ≥1, and (a, p) = 1. Then for any λ2-continuity set A we have lim m→∞ 2 pm(pm + 1)# ( k, n ; 0≤k≤n<pm∧ n k (p) ≡a (mod ps)∧(k, n)p−m ∈A ) = λ2(A) φ(ps) . (2) Let µ be the log p(p+1) 2 log p -dimensional Hausdorﬀmeasure restricted to the “p-adic Sierpi´ nski gasket”—the attractor of an iterated function system given in and described below. The second theorem states that for given j the binomial coeﬃcients with prescribed p-valuation equal to j exhibit a similar behaviour as well; their p-free parts are uniformly distributed modulo ps, whereas they are spatially uniformly distributed with respect to µ. Theorem 2 . Let p be a prime, s ≥1, j ≥0, and (a, p) = 1. Then for any µ-continuity set A we have lim m→∞ # k, n ; 0≤k≤n 0, some non-negative matrix B, and the matrix norm ∥· ∥ induced by the maximum norm on the vector space. Assume that B has 1 as a simple dominating eigenvalue. Then the sequence of matrices PK(t) = B p−Kt B p−(K−1)t · · · B p−1t converges to a limit P(t) for all t; P(t) is continuous at t = 0. In the sequel ∥·∥will always denote the matrix norm induced by the maximum norm. Remark 2 . It is not stated in [6, Lemma 5] but immediately follows from the proof that if B(t) depends continuously on t, then the convergence to P(t) as stated in the Lemma is uniform in t on compact subsets of Rd and P(t) is continuous on Rd. Furthermore, the relation P(t) = P(0)P(t) holds. Since the matrix P(0) has rank 1 by the Perron-Frobenius theorem, this implies that the matrix P(t) has rank at most 1. 154 SPATIAL EQUIDISTRIBUTION OF BINOMIAL COEFFICIENTS MODULO PRIME POWERS The second lemma is a generalisation of Lemma 1. Lemma 2 . Let A(t) and B(t) be matrix functions mapping vectors t ∈Rd continuously to square matrices satisfying ∥B(t) −B(0)∥≤C∥t∥ for ∥t∥≤T, (7) ∥A(t) −A(0)∥≤C∥t∥ for ∥t∥≤T, (8) \|[B(t)]i,j\| ≤[B(0)]i,j for i, j and all t, (9) \|[A(t)]i,j\| ≤[A(0)]i,j for i, j and all t (10) for some C, T > 0. Furthermore, assume that B = B(0) has 1 as simple domi-nating eigenvalue and set A = A(0). Deﬁne P (j) K inductively by setting P (0) K (t) = B p−Kt B p−(K−1)t · · · B p−1t , P (0) 0 (t) = I and P (j) K (t) = K X m=1 P (0) K−m(p−mt)A(p−mt)P (j−1) m−1 (t). (11) Then P (j)(t) = lim K→∞ P (j) K (t) K j = P(0)A jP(t), (12) where P(t) is the limit given in Lemma 1. The convergence is uniform on com-pact subsets of Rd. Remark 3 . Notice that P (j) K (t) is the sum of all products of matrices A(·) and B(·) containing exactly j matrices A(·). In our application, this will reﬂect the combinatorial structure of j carries. P r o o f. We proceed by induction on j to prove lim K→∞ P (j) K (t) K j = P(0)A P (j−1)(t). From Lemma 1 and Remark 2 we have that PK(t) K converges uniformly to P(t) on compact subsets of Rd, which is the case j = 0. 155 G. BARAT—P. J. GRABNER For the step j −1 →j we use the recursion formula (11). We split the range of summation into m < √ K, √ K ≤m < K − √ K and m ≥K − √ K and estimate the ﬁrst and the last sum X m< √ K P (0) K−m(p−mt)A(p−mt)P (j−1) m−1 (t) =O  X m< √ K mj−1  =O(Kj/2) X K− √ K≤m≤K P (0) K−m(p−mt)A(p−mt)P (j−1) m−1 (t) = O(Kj−1√ K) = O(Kj−1/2). For the middle sum, we write PK−m(p−mt) = P(0) + o(1), A(p−mt) = A + o(1), and P (j−1) m−1 (t) = m −1 j −1 P (j−1)(t) + o(mj−1) (which hold uniformly on compact subsets of Rd) and insert these to obtain X √ K≤m<K− √ K P (0) K−m(p−mt)A(p−mt)P (j−1) m−1 (t) = X √ K≤m ν mod p, (14) where κ mod p denotes the non-negative remainder in the Euclidean division κ/p. The states (·, ·, 0) represent the situation that no carry occurred in the subtrac-tion of the least signiﬁcant digits, whereas (·, ·, 1) encode the situation that a carry occurred. A similar automaton was used in [13, 14] in the study of the p-adic Sierpi´ nski gasket. For a given pair of integers (n, k) we start at the state n mod ps−1, k mod ps−1 ; technically, we would have to add extra states, which emulate reading the ﬁrst s −1 digits. For t = (t1, t2) ∈R2 we deﬁne the marked transition matrix Mχ(t) of the automaton deﬁned above by Mχ(t),(ν2,κ2,η2)=                                                χ (±1)η1 (n!)p (k!)p((n−k−η1)!)p e(εt1+δt2) if n−k−η1 ≥0 and ν2 = ⌊ν1 p ⌋+ εps−2 and κ2 = ⌊κ1 p ⌋+ δps−2 and [ [(κ1 mod p) + η1 > (ν1 mod p)] ] = η2, χ (±1)η1 (n!)p (k!)p((ps−1+n−k−η1)!)p e(εt1+δt2) if n−k−η1 <0 and ν2 = ⌊ν1 p ⌋+ εps−2 and κ2 = ⌊κ1 p ⌋+ δps−2 and [ [(κ1 mod p) + η1 > (ν1 mod p)] ] = η2, 0, otherwise, where, for short, we denote n = ν1 + εps−1 and k = κ1 + δps−1. The value (±1) is chosen according to Theorem 3. Here and later on we use Iverson’s notation [ [A] ], which is 1, if A is true, and 0 otherwise. Then we have S(1) χ (m, t) = v(p−mt)T Mχ p−(m−s)t Mχ p−(m−s−1)t · · · Mχ(p−1t)w, (15) where w denotes the column vector with all entries (1 −η). 157 G. BARAT—P. J. GRABNER The vector v(t) is given by v(t) = e (νt1 + κt2) . We write 1 for the vector with all entries 1 and observe that v(0) = 1. We notice that the automaton deﬁned by the transition function (14) has exactly p2 transitions emanating from each state. The marked transition matrix Mχ0(0) (χ0 being the principal character) marks each of these transitions by 1; thus this matrix has exactly p2 entries 1 per line and is a Perron-Frobenius ma-trix with dominating eigenvalue p2. From (13) the sum S(1) χ (m, t) has pm(pm+1) 2 summands. Thus we divide (15) by pm(pm+1) 2 and let m tend to inﬁnity. For the principal character this results in ˆ λ(t) = lim m→∞ pm(pm + 1) 2 −1 Sχ0(m, t) = 1 T P(t)w, (16) where P(t) = lim m→∞ pm(pm + 1) 2 −1 Mχ0 p−(m−s)t Mχ0 p−(m−s−1)t · · · Mχ0(p−1t) is a convergent inﬁnite matrix product applying Lemma 1 with B(t) = 1 p2 Mχ0(t). The limit ˆ λ(t) is the Fourier transform of the two-dimensional Lebesgue mea-sure restricted to the triangle {(x, y) ∈R2 ; 0 ≤x ≤y ≤1}, normalised to total measure 1. For non-principal characters χ, at least one non-zero entry of Mχ(0) diﬀers from 1, because n 1 (p) =n(p) implies that the character is evaluated at all prime residue classes (mod ps); thus we have ∥Mχ(0)∥< p2. From this we conclude, using the continuity of Mχ(t) at t = 0, lim m→∞ pm(pm + 1) 2 −1 Sχ(m, t) = 1 T lim m→∞ pm(pm + 1) 2 −1 Mχ p−(m−s)t Mχ p−(m−s−1)t · · · Mχ(p−1t)w = 0. Summing up, we have lim m→∞ pm(pm + 1) 2 −1 X 0≤k≤n<pm n k (p) ≡a (mod ps) e (nt1 + kt2)p−m = 1 φ(ps) X χ χ(a) lim m→∞ pm(pm + 1) 2 −1 S(1) χ (m, t) = 1 φ(ps) ˆ λ(t), which ﬁnishes the proof of Theorem 1 by Levy’s continuity theorem. □ 158 SPATIAL EQUIDISTRIBUTION OF BINOMIAL COEFFICIENTS MODULO PRIME POWERS P r o o f o f T h e o r e m 2. In order to obtain the desired result, we study the exponential sum S(2) χ (m, j, t1, t2) = X 0≤k≤n<pm pj∥ n k χ p−j n k ! e (kt1 + nt2)p−m . (17) The underlying automaton is the same as in the proof of Theorem 1, but we have to take into account the number of carries in the subtraction. This is done by a new marking of the transitions using the following two matrices B(t),(ν2,κ2,η2) = ( 2 p(p+1)Mχ0(t),(ν2,κ2,η2) if η1 = 0, 0 if η1 = 1, A(t),(ν2,κ2,η2) = ( Mχ0(t),(ν2,κ2,η2) if η1 = 1, 0 if η1 = 0. The matrix B(t) encodes all transitions without carry, whereas A(t) encodes a carry in the subtraction of the least signiﬁcant digit. Notice that we normalise B(t) to satisfy the assumption of Lemma 2. We obtain Sχ0(m, j, t) = p(p + 1) 2 m min(j,s−1) X ℓ=0 vℓ(p−mt)T P (j−ℓ) m−s (t)w, (18) where the vectors vℓencode starting blocks of s −1 digits containing ℓcarries in the subtraction: vℓ(t)(ν,κ,η) =      e(νt1 + κt2) if η = [ [ν < κ] ] and there are exactly ℓcarries in the subtraction ηps + ν −κ, 0, otherwise. Applying Lemma 2 yields lim m→∞ Sχ0(m, j, t) m j p(p+1) 2 m = v0(0)T P(0)A jP(t)w. (19) We now observe that v0(0) is a vector with non-negative entries and (P(0)A)j is a matrix with non-negative entries. Thus v0(0)T P(0)A j is a vector with non-negative entries. Using the relation P(t) = P(0)P(t), we can rewrite the limit (19) as v0(0)T P(0)A jP(0)P(t)w; the vector v0(0)TP(0)A jP(0) is now proportional to the left Perron-Frobenius eigenvalue of B, because P(0) is a matrix of rank 1 with this eigenvector. Thus the limit (19) is proportional to vTP(t)w for any non-negative vector v. 159 G. BARAT—P. J. GRABNER We set ˆ µ(t) = lim m→∞ Sχ0(m, 0, t) p(p+1) 2 m . Then by the above argument, we have ˆ µ(t) = lim m→∞ Sχ0(m, j, t) Sχ0(m, j, 0). From we infer that the sets Qm = ( (n, k)p−m ; 0 ≤k ≤n < pm and p ̸ \| n k ) tend to the attractor Q of the iterated function system deﬁned by Fa,b(x, y) = x + a p , y + b p , where 0 ≤b ≤a < p (the “p-adic Sierpi´ nski triangle”). Furthermore, the measures µm = 1 p(p+1) 2 m X x∈Qm δx tend to the Hausdorﬀmeasure of dimension s = log p(p+1) 2 log p restricted to Q and normalised to total mass 1. This implies that ˆ µ(t) is the Fourier transform of this measure. In order to sieve out the binomial coeﬃcients with p−jn k ≡a (mod ps) we consider the sum 1 φ(ps) X χ χ(a)Sχ(m, j, t) and observe, using the same arguments as in the proof of Theorem 1, that Sχ(m, j, t) (for χ ̸= χ0) is of smaller order of magnitude than Sχ0(m, j, t). This ﬁnishes the proof of Theorem 2. □ REFERENCES BARAT, G.—GRABNER, P. J.: Digital functions and distribution of binomial coeﬃ-cients, J. London Math. Soc. 64 (2001), 523–547. BARBOLOSI, D.—GRABNER, P. J.: Distribution des coeﬃcients multinomiaux et q-binomiaux modulo p, Indag. Math. 7 (1996), 129–135. CARLITZ, L.: The number of binomial coeﬃcients divisible by a ﬁxed power of a prime, Rend. Circ. Matem. Palermo 16 (1967), 299–320. 160 SPATIAL EQUIDISTRIBUTION OF BINOMIAL COEFFICIENTS MODULO PRIME POWERS DAVIS, K. S.—WEBB, W.: Lucas congruence for prime powers, European J. Combin. 11 (1990), 229–233. GARFIELD, R.—WILF, H. S.: The distribution of the binomial coeﬃcients modulo p, J. Number Theory 41 (1992), 1–5. GRABNER, P. J.—HEUBERGER, C.—PRODINGER, H.: Counting optimal joint digit expansions, Integers 5 (2005), no. 3, A09, 19 pages (electronic). GRANVILLE, A.: Arithmetic properties of binomial coeﬃcients. I. Binomial coeﬃcients modulo prime powers, Organic mathematics (Burnaby, BC, 1995), Amer. Math. Soc., Providence, RI, 1997, pp. 253–276. GREINECKER, F.: Spatial equidistribution of combinatorial number schemes, J. Fractal Geom. (2016) (to appear). KUMMER, E. E.: ¨ Uber die Erg¨ anzungss¨ atze zu den allgemeinen Reciprocit¨ atsgesetzen, J. reine angew. Math. 44 (1852), 93–146. LUCAS, ´ E: Sur les congruences des nombres eul´ eriens et des coeﬃcients diﬀ´ erentiels des fonctions trigonom´ etriques, suivant un module premier, Bull. Soc. Math. France 6 (1878), 49–54. SINGMASTER, D.: Notes on binomial coeﬃcients, I—A generalization of Lucas’ con-gruence, II—The least n such that pe divides an r-nomial coeﬃcient of rank n, III—Any integer divides almost all binomial coeﬃcients, J. London Math. Soc. 8 (1974), 545–548, 549–554, 555–560. SPIEGELHOFER, L.—WALLNER, M.: Divisibility of binomial coeﬃcients by powers of primes, arXiv:1604.07089, 2016. VON HAESELER, F.—PEITGEN, H.-O.—SKORDEV, G.: Pascal’s triangle, dynamical systems and attractors, Ergodic Theory Dynam. Systems 12 (1992), no. 3, 479–486. , Cellular automata, matrix substitutions and fractals, Ann. Math. Artiﬁcial Intelligence 8 (1993), 345–362. (Theorem proving and logic programming (1992).) Received May 10, 2016 Accepted June 14, 2016 Guy Barat Peter J. Grabner Institut f¨ ur Analysis und Zahlentheorie Technische Universit¨ at Graz Kopernikusgasse 24 8010 Graz AUSTRIA E-mail: guy.barat@tugraz.at peter.grabner@tugraz.at 161
6104	https://www.media4math.com/library/definition-polynomial-concepts-difference-squares	Definition--Polynomial Concepts--Difference of Squares \| Media4Math Skip to main content Enable accessibility for low vision Open the accessibility menu Skip to content Home Search Subscribe ContentNewest ResourcesLesson Plan LibraryContent ShowcaseAnimated Math Clip ArtSAT Test PrepMath Fluency CentersMath Visual GlossaryStudent Tutorials Standards AlignmentsStandards Alignments (K-12)Standards Reports Getting StartedEducatorsTutorsParents Partners Shop@Media4Math Persistent Menu Library Classroom Log in Register Register to SaveSubscribe to DownloadPreviewAdd to Slideshow (Subscribers Only) If you are a subscriber, please log in. Algebra>>Polynomials>>Factoring Polynomials Resource Related Resources Display Title Definition--Polynomial Concepts--Difference of Squares Difference of Squares Topic Polynomials Definition The difference of squares is a polynomial expression of the form a² - b², which can be factored as (a + b)(a - b). Description The difference of squares is a fundamental concept in algebra, particularly in the study of polynomials. This special factoring technique is widely used to simplify expressions and solve equations. Understanding the difference of squares pattern allows students to quickly recognize and factor certain quadratic expressions, making complex algebraic manipulations more manageable. The importance of the difference of squares extends beyond basic algebra. It is a crucial tool in advanced mathematics, including calculus and number theory. In practical applications, this concept is used in various fields such as physics, engineering, and computer science, where it helps in simplifying calculations and modeling real-world phenomena. Mastering the difference of squares enhances problem-solving skills and provides a foundation for understanding more complex polynomial factorizations. For a complete collection of terms related to polynomials click on this link: Polynomials Collection Loading… An error occurred while loading related resources. No related resources found. \| Common Core Standards \| CCSS.MATH.CONTENT.HSA.APR.A.1, CCSS.MATH.CONTENT.HSA.APR.B.2, CCSS.MATH.CONTENT.HSA.APR.C.5, CCSS.MATH.CONTENT.HSA.APR.C.4, CCSS.MATH.CONTENT.HSA.APR.B.3, CCSS.MATH.CONTENT.HSF.IF.C.7.C \| \| Grade Range \| 8 - 12 \| \| Curriculum Nodes \| Algebra •Polynomials •Factoring Polynomials \| \| Copyright Year \| 2021 \| \| Keywords \| polynomials, monomials, definitions, glossary term, difference of squares \| The Media4Math Definitions Library A Visual Glossary for Your Students Vocabulary is an important part of the math curriculum. In fact, many students struggle with math concepts because they lack the mastery of key vocabulary. Textbook instruction or examples often rely on these key terms and without a proper grounding in the relevant vocabulary, students will continue to struggle. With that in mind, Media4Math has developed an extensive glossary of key math terms. Each definition is a downloadable image that can easily be incorporated into a lesson plan. Furthermore, each definition includes a clear explanation and a contextual example of the term. To see the complete collection of these terms, click on this link. Math vocabulary doesn't consist of isolated terms. In fact, for any given concept there are clusters of vocabulary terms that students need to learn in order to better understand the concept. The Media4Math glossary consists of clusters of such terms. This is a summary of these clusters. Click on each link to see that collection of terms and definitions. 3D Geometry Coordinate Systems Fraction Concepts Linear Systems Polygon Concepts Rationals and Radicals Triangle Concepts Angle Concepts Early Elementary Definitions Functions Concepts Linear Functions Concepts Polynomial Concepts Ratios, Proportions, and Percents Trigonometry Charts and Graphs Equation Concepts Geometry Math Properties Primes and Composites Sequences and Series Variables and Unknowns Circles Exponential Concepts Geometric Theorems Measures of Central Tendency Quadratics Concepts Slope Closure Property Factors and Multiples Inequalities Place Value Concepts Quadrilateral Concepts Statistics and Probability Instructional Suggestions Here are some idea for how to use this library of vocabulary terms: Creating Connections As you introduce a new topic, for example Slope, go to the corresponding collection of definitions by linking on one of the collections above. Each definition includes an example of the term. Have students research one or more of these terms. Have a group of students research these terms and begin making connections. The idea is to encourage students to start using these terms as they begin discussing the main concept. To continue this example, let's look at the collection of terms under slope. Clicking on the link reveals that there are 17 terms under the category of slope. As you can see, this is more than just a simple definition of a single term. As students analyze these definitions, they begin to see common terms: ratio, rise over run, change in coordinates, and so forth. Working in teams, students can begin to build connections among these terms. Encourage them make connections among these related terms, creating a graphic similar to this: With an activity like this, students begin to use math vocabulary but, more important, tie it to math concepts. Creating Word Games Once students are familiar with the collection of terms have them create word search or crossword puzzles using these terms. There are many free online tools for creating such puzzles. In addition, Media4Math has a collection of word games and vocabulary puzzles to allow students to further practice their vocabulary skills. To see the current collection of puzzles see the links below: Word Search Puzzles Crossword Puzzles The Illustrated Math Dictionary Subscribers to Media4Math also get access to The Illustrated Math Dictionary. This ebook brings together math definitions and related multimedia resources. This ebook brings together the glossary terms for concepts like Linear Functions, Quadratic Functions, and Polynomial functions. Each term has an audio component, along with related resources.The Illustrated Math Dictionary is more than just a vocabulary tool. Use it for instruction or review. Additional Resources The What Works Clearinghouse has a number of Practice Guides that focus on evidence-based practices that will help struggling students. In particular, the Practice Guide entitled, Assisting Students Struggling with Mathematics: Intervention in the Elementary Grades, emphasizes the importance of math vocabulary. In particular, make a not of the following point: Mathematical language is academic language that conveys mathematical ideas.This includes vocabulary, terminology, and language structures used when thinking about, talking about, and writing about mathematics. Mathematical language conveys a more precise understanding of mathematics than the conversational or informal language used every day to communicate with others. Footer menu About Privacy Policy Blog Contact Us Accessibility Advertise with Us! Follow © Media4Math. All rights reserved
6105	https://www.studocu.com/en-us/messages/question/3076761/squeeze-theorem-to-find-the-limit-of-the-sequence-cos1n-11n	[Solved] Squeeze Theorem to find the limit of the sequence cos1n 11n - Single-Variable Calculus I (MATH 1231) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent Squeeze Theorem to find the limit of the sequencecos(1/n) -1/(1/n) Single-Variable Calculus I (MATH 1231) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist The George Washington University Single-Variable Calculus I Question Squeeze Theorem to find the limit of the sequence cos1n 11n The George Washington University Single-Variable Calculus I Question ### Josh 2 years ago Squeeze Theorem to find the limit of the sequence cos(1/n) -1/(1/n) Like 0 Related documents Msth Class Notes: Exploring Transcendental Functions and Inverses Introduction to Philosophy Lecture notes None Unit 1 notes 1 - CALC 1 Single-Variable Calculus I Lecture notes None Calculus Cheat Sheet: Limits and Integrals Overview Single-Variable Calculus Ii Lecture notes 67%(6) Single-Variable Calculus Ii Lecture slides, lecture 11-1 Single-Variable Calculus Ii Lecture slides None Answer Created with AI 2 years ago Answer The Squeeze Theorem, also known as the Sandwich Theorem, is a useful tool in calculus for finding the limit of a function or sequence. It states that if a function or sequenc... 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Single-Variable Calculus I (MATH 1231) 1 year ago Two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results. a=16, b=25, A=60° Single-Variable Calculus I (MATH 1231) 1 year ago A ball is thrown upward and outward from a height of 5 feet. The height of the ball, f(x), in feet, can be modeled by f(x)=-0.1x^2+0.8x+5 where x is the ball's horizontal distance, in feet, from where it was thrown. a. What is the maximum height of the ball and how far from where it was thrown does this occur? b. How far does the ball travel horizontally before hitting the ground? Single-Variable Calculus I (MATH 1231) 2 years ago Use the comparison test to determine whether the following series converge.∑∞. (𝑛+1√−𝑛)/𝑛𝑛=1 Single-Variable Calculus I (MATH 1231) 2 years ago For each of the following series, if the divergence test applies, either state that lim𝑛→∞𝑎𝑛 does not exist or find lim𝑛→∞𝑎𝑛. If the divergence test does not apply, state why.𝑎𝑛=𝑛/√(3n^2+2n+1)𝑎𝑛=2^𝑛+3^𝑛/10^(𝑛/2) Single-Variable Calculus I (MATH 1231) 2 years ago Using sigma notation, write the following expressions as infinite series.1-1/2+1/3−1/4+...In the following exercises, compute the general term 𝑎𝑛 of the series with the given partial sum 𝑆𝑛. 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6106	https://www.sciencedirect.com/science/article/abs/pii/S0363018813000716	Skip to article My account Sign in Access through your organization Purchase PDF Patient Access Article preview Abstract Introduction Section snippets References (35) Cited by (68) Current Problems in Diagnostic Radiology Volume 43, Issue 2, MarchApril 2014, Pages 55-67 Congenital Cystic Neck Masses: Embryology and Imaging Appearances, With Clinicopathological Correlation Author links open overlay panel, , , , , , rights and content Congenital cystic masses of the neck are uncommon and can present in any age group. Diagnosis of these lesions can be sometimes challenging. Many of these have characteristic locations and imaging findings. The most common of all congenital cystic neck masses is the thyroglossal duct cyst. The other congenital cystic neck masses are branchial cleft cyst, cystic hygroma (lymphangioma), cervical thymic and bronchogenic cysts, and the floor of the mouth lesions including dermoid and epidermoid cysts. In this review, we illustrate the common congenital cystic neck masses including embryology, clinical findings, imaging features, and histopathological findings. Introduction Congenital cystic masses of the neck are uncommon but important pediatric neck lesions. Many of these have characteristic locations and imaging findings. The most common congenital cystic neck mass is thyroglossal duct cyst (TGD).1 The other common lesions are second branchial cleft cyst and cystic hygroma (lymphangioma). The uncommon cystic lesions include first and third branchial cleft cysts, fourth branchial apparatus anomaly, cervical thymic and bronchogenic cysts, and the floor of the mouth lesions including dermoid and epidermoid cysts. Anomalies of the branchial apparatus could also present as sinuses and fistulae in addition to cysts. In this article, we discuss the common congenital cystic neck masses in an orderly fashion, including embryology and pathogenesis, clinical findings, imaging features, and histopathological findings. Section snippets Embryology and Pathogenesis The thyroid gland develops in the region of foramen cecum of the tongue during the third gestational week. The thyroid anlage descends down to reach the thyroid bed anterior to laryngeal cartilages through the TGD by the seventh week of gestation, and TGD begins to involute by 8-10 weeks of gestation.2 If any segment of the TGD fails to involute, then the persistent secretory activity from the epithelial lining owing to repeated infection or inflammation would give rise to TGD cyst.1 Clinical Findings Most TGD Embryology and Pathogenesis Branchial apparatus include 6 arches (mesoderm) interfaced by 4 clefts (ectoderm) and pouches (endoderm) on each side at the end of fourth week of embryonic life (Fig 4).2 The fifth arch is rudimentary and does not contribute to formation of any of the adult structure. Incomplete obliteration of I, II, III, or IV branchial apparatus can lead to respective branchial remnants like isolated cyst (most common), fistula, or sinus (rare). Second branchial cleft cyst accounts for>90% of all branchial Embryology and Pathogenesis There are 4 types of lymphangiomas in the neck, capillary, cavernous, cystic, and venolymphatic forms. Of all these forms, cystic hygroma (cystic lymphangioma) is the most common form of lymphangioma and constitutes approximately 5% of all benign tumors of infancy and childhood.13 Cystic hygroma is thought to arise from an early sequestration of the embryonic lymphatic channels.15, 16 This sequestration occurs more commonly in the developing jugular lymph sac pair than in the other 4 embryonic Embryology and Pathogenesis The dermoid cysts are more common than epidermoid cysts in the head and neck region. The dermoid cysts typically occur along embryonic lines of fusion suggesting that they exist because of entrapment of epithelial elements during development. Lateral eyebrow is the most common location for head and neck dermoid, and the floor of the mouth is the second most common location.20 Clinical Findings Dermoid cysts usually manifest during the second and third decades of life; however, the epidermoid cysts appear much Embryology and Pathogenesis The third branchial pouch forms the thymus and inferior parathyroid glands during fetal development. The thymus descends down to the mediastinum through thymopharyngeal duct, traveling lateral to the thyroid gland. The cervical thymic cyst arises along this tract (Fig 16). The exact pathogenesis of cervical thymic cysts remains controversial even today. Two mechanisms (congenital and acquired) are proposed. Most authorities favor the congenital persistence of the thymopharyngeal duct remnants Embryology and Pathogenesis Cervical bronchogenic cysts are extremely rare. Anomalous foregut development is suspected to play a role in the development of bronchogenic cyst, but the reason why these cysts reach an aberrant position in the neck remains unclear.28 They have been reported in infants as well as in adults and occur in males approximately 3 times as often as in females.14, 29 Clinical Findings Extrathoracic bronchogenic cyst usually is located in the suprasternal notch, presternum, shoulder, neck, base of the tongue, Conclusion Clinical evaluation including both history and physical examination is important in the evaluation of a suspected congenital neck mass. Most common presentation includes fluctuant slow-growing neck masses, and sometimes presents with repeated infections, especially the branchial cleft cysts. US and CT are the most common imaging modalities used with MR evaluation for more complex lesions (Table 2). A systematic approach based on the knowledge of embryology and anatomy of the cervical region References (35) D.S. Foley et al. ### Thyroglossal duct and other congenital midline cervical anomalies ### Semin Pediatr Surg (2006) S.A. Deane et al. ### Surgery for thyroglossal duct and branchial cleft anomalies ### Am J Surg (1978) W.P. Work ### Cysts and congenital lesions of the parotid gland ### Otolaryngol Clin North Am (1977) R.L. Telander et al. ### Thyroglossal and branchial cleft cysts and sinuses ### Surg Clin North Am (1977) R.C. King et al. ### Dermoid cyst in the floor of the mouth. Review of the literature and case reports ### Oral Surg Oral Med Oral Pathol (1994) C.J. Howell ### The sublingual dermoid cyst. Report of five cases and review of the literature ### Oral Surg Oral Med Oral Pathol (1985) F. Tovi et al. ### The aberrant cervical thymus. Embryology, pathology, and clinical implications ### Am J Surg (1978) C.W. Wagner et al. ### Respiratory complications in cervical thymic cysts ### J Pediatr Surg (1988) M. Ibrahim et al. ### Congenital cystic lesions of the head and neck ### Neuroimaging Clin N Am (2011) A.E. Obiechina et al. ### Coexisting congenital sublingual dermoid and bronchogenic cyst ### Br J Oral Maxillofac Surg (1999) R.H. Allard ### The thyroglossal cyst ### Head Neck Surg (1982) - P.S. Mohan et al. ### Thyroglossal duct cysts: A consideration in adults ### Am Surg (2005) - D.B. Hawkins et al. ### Cysts of the thyroglossal duct ### Laryngoscope (1982) - D.T. Wadsworth et al. ### Thyroglossal duct cysts: Variability of sonographic findings ### Am J Roentgenol (1994) - T.J. Vogl et al. ### Cystic masses in the floor of the mouth: Value of MR imaging in planning surgery ### Am J Roentgenol (1993) - W.P. Work ### Newer concepts of first branchial cleft defects ### Laryngoscope (1972) - H. Bailey ### The diagnosis of branchial cyst; with a note upon its removal ### Br Med J (1928) Cited by (68) Congenital Neck Masses 2018, Clinics in Perinatology ### Confirmation of etiology in fetal hydrops by sonographic evaluation of fluid allocation patterns 2015, European Journal of Obstetrics and Gynecology and Reproductive Biology Citation Excerpt : The majority of fetal hydrops due to aneuploidy (44/85 (51.8%)) could be secured at this early time of pregnancy. Aneuploid fetuses with trisomy 13, 18 and 21 as well as fetuses with Turner's syndrome may present frequently (>60%) with cystic hygroma (and hydrops fetalis) during fetal ultrasound [13,14] (Fig. 1). Fetal ascites may not be a general issue in aneuploidies, but may be sometimes present in cases with (transient) myeloproliferative disorders. To evaluate patterns of fluid allocations in different etiologies of hydrops fetalis. This report is a retrospective cohort study on 20,395 fetal sonographic evaluations in a single tertiary center from 2000 to 2014. Special emphasis was placed on the exact description of the distinct fluid allocation sites in each fetus. Postmortem/postnatal records were evaluated additionally. Mean follow up of the surviving neonates was 34 days (1060 days). There seem to be distinctive patterns of fluid allocation in some etiologies leading to fetal hydrops including aneuploidies and Parvovirus B19 related infections. Due to the allocation patterns of fluid filled sites in fetuses with hydrops fetalis the spectrum of possible etiologies may be narrowed already during initial ultrasound scan. It can contribute substantially to diagnostic accuracy as well as to parental counseling. This knowledge may also help to omit delay in diagnostic routines. ### Cervical lymphatic malformations: Prenatal characteristics and ex utero intrapartum treatment 2019, Prenatal Diagnosis ### Imaging of the sublingual and submandibular spaces 2018, Insights into Imaging ### Thyroglossal Duct Cyst Carcinomas: A Clinicopathologic Series of 22 Cases with Staging Recommendations 2017, Head and Neck Pathology ### Update from the 4th Edition of the World Health Organization Classification of Head and Neck Tumours: What Is New in the 2017 WHO Blue Book for Tumors and Tumor-Like Lesions of the Neck and Lymph Nodes 2017, Head and Neck Pathology View all citing articles on Scopus View full text Copyright © 2014 Elsevier, Inc. Published by Mosby, Inc. All rights reserved.
6107	https://medlineplus.gov/druginfo/meds/a604006.html	Memantine: MedlinePlus Drug Information Skip navigation An official website of the United States government Here’s how you know Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. National Library of Medicine Menu Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia About MedlinePlus Show Search Search MedlinePlus GO About MedlinePlus What's New Site Map Customer Support Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia Español You Are Here: Home → Drugs, Herbs and Supplements → Memantine URL of this page: Memantine pronounced as (mem' an teen) Why is this medication prescribed? How should this medicine be used? Other uses for this medicine What special precautions should I follow? What special dietary instructions should I follow? What should I do if I forget a dose? What side effects can this medication cause? What should I know about storage and disposal of this medication? In case of emergency/overdose What other information should I know? Brand names Why is this medication prescribed? Expand Section Memantine is used to treat the symptoms of Alzheimer's disease (AD; a brain disease that slowly destroys the memory and the ability to think, learn, communicate and handle daily activities). Memantine is in a class of medications called NMDA receptor antagonists. It works by decreasing abnormal activity in the brain. Memantine may improve the ability to think and remember or may slow the loss of these abilities in people who have AD. However, memantine will not cure AD or prevent the loss of these abilities at some time in the future. How should this medicine be used? Expand Section Memantine comes as a tablet, a solution (liquid), and an extended-release (long-acting) capsule to take by mouth. The solution and tablet are usually taken once or twice a day with or without food. The capsule is taken once a day with or without food. Follow the directions on your prescription label carefully, and ask your doctor or pharmacist to explain any part you do not understand. To help you remember to take memantine, take it at around the same time(s) every day. Take memantine exactly as directed. Do not take more or less of it or take it more often than prescribed by your doctor. Swallow the extended-release capsules whole; do not chew, divide, or crush them. If you are unable to swallow the extended-release capsules, you can carefully open a capsule and sprinkle the contents on a spoonful of applesauce. Swallow this mixture immediately without chewing it. Do not save this mixture to use at a later time. If you are taking the oral solution, follow the manufacturer's directions to measure your dose using the oral syringe that is supplied with the medication. Slowly squirt the medication from the syringe into a corner of your mouth and swallow it. Do not mix the medication with any other liquid. After you take your medication, follow the manufacturer's directions to re-seal the bottle and clean the oral syringe. Ask your pharmacist or doctor if you have any questions about how to use this medication. Your doctor will probably start you on a low dose of memantine and gradually increase your dose, not more than once every week. Memantine helps to control the symptoms of Alzheimer's disease but does not cure it. Continue to take memantine even if you feel well. Do not stop taking memantine without talking to your doctor. Ask your doctor or pharmacist for a copy of the manufacturer's information for the patient. Other uses for this medicine Expand Section This medication may be prescribed for other uses; ask your doctor or pharmacist for more information. What special precautions should I follow? Expand Section Before taking memantine, tell your doctor and pharmacist if you are allergic to memantine, any other medications, or any of the ingredients in memantine tablets, capsules, and oral solution. Ask your pharmacist or check the manufacturer's patient information for a list of the ingredients. tell your doctor and pharmacist what prescription and nonprescription medications, vitamins, nutritional supplements, and herbal products you are taking while taking memantine. Your doctor may need to change the doses of your medications or monitor you carefully for side effects. the following nonprescription products may interact with memantine: dextromethorphan (Robitussin); sodium bicarbonate (Soda Mint, baking soda); cimetidine (Tagamet); randitidine; nicotine. Be sure to let your doctor and pharmacist know that you are taking these medications before you start taking memantine. Do not start any of these medications while taking memantine without discussing with your healthcare provider. tell your doctor if you have or have a urinary tract infection now or if you develop one during your treatment with memantine and if you have or have ever had seizures, difficulty urinating, or kidney or liver disease. tell your doctor if you are pregnant, plan to become pregnant, or are breast-feeding. If you become pregnant while taking memantine, call your doctor. if you are having surgery, including dental surgery, tell the doctor or dentist that you are taking memantine. What special dietary instructions should I follow? Expand Section Unless your doctor tells you otherwise, continue your normal diet. What should I do if I forget a dose? Expand Section Take the missed dose as soon as you remember it. However, if it is almost time for the next dose, skip the missed dose and continue your regular dosing schedule. Do not take a double dose to make up for a missed one. If you forget to take memantine for several days, call your doctor before you start to take the medication again. What side effects can this medication cause? Expand Section Memantine may cause side effects. Tell your doctor if any of these symptoms are severe or do not go away: dizziness confusion aggression depression headache sleepiness diarrhea constipation nausea vomiting weight gain pain anywhere in your body, especially your back cough Some side effects can be serious. If you experience either of the following symptoms, call your doctor immediately or get emergency medical help: shortness of breath hallucination (seeing things or hearing voices that do not exist) Memantine may cause other side effects. Call your doctor if you have any unusual problems while taking this medication. If you experience a serious side effect, you or your doctor may send a report to the Food and Drug Administration's (FDA) MedWatch Adverse Event Reporting program online ( or by phone (1-800-332-1088). What should I know about storage and disposal of this medication? Expand Section Keep this medication in the container it came in, tightly closed, and out of reach of children. Store it at room temperature and away from excess heat and moisture (not in the bathroom). Dispose of unneeded medications in a way so that pets, children, and other people cannot take them. Do not flush this medication down the toilet. Use a medicine take-back program. Talk to your pharmacist about take-back programs in your community. Visit the FDA's Safe Disposal of Medicines website for more information. Keep all medication out of sight and reach of children as many containers are not child-resistant. Always lock safety caps. Place the medication in a safe location – one that is up and away and out of their sight and reach. In case of emergency/overdose Expand Section In case of overdose, call the poison control helpline at 1-800-222-1222. Information is also available online at If the victim has collapsed, had a seizure, has trouble breathing, or can't be awakened, immediately call emergency services at 911. Symptoms of overdose may include: restlessness slowed movements agitation weakness slowed heartbeat confusion dizziness unsteadiness double vision hallucination (seeing things or hearing voices that do not exist) sleepiness loss of consciousness vomiting lack of energy sense that you or your surroundings are spinning What other information should I know? Expand Section Keep all appointments with your doctor. Do not let anyone else take your medication. Ask your pharmacist any questions you have about refilling your prescription. Keep a written list of all of the prescription and nonprescription (over-the-counter) medicines, vitamins, minerals, and dietary supplements you are taking. Bring this list with you each time you visit a doctor or if you are admitted to the hospital. You should carry the list with you in case of emergencies. Brand names Expand Section Namenda® Namenda® Titration Pak¶ Namenda XR® Namzaric®(as a combination product containing Donepezil, Memantine) ¶ This branded product is no longer on the market. Generic alternatives may be available. Last Revised - 04/15/2016 Browse Drugs and Medicines Learn how to cite this page Was this page helpful? Yes No Thank you for your feedback! American Society of Health-System Pharmacists, Inc. Disclaimer AHFS® Patient Medication Information™. © Copyright, 2025. The American Society of Health-System Pharmacists®, 4500 East-West Highway, Suite 900, Bethesda, Maryland. All Rights Reserved. Duplication for commercial use must be authorized by ASHP. 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6108	https://news.ycombinator.com/item?id=15946418	\| \| \| \| --- \| \| \| \| --- \| \| Hacker Newsnew \| past \| comments \| \| \| \| \| login \| \| \| \| \| \| pdkl95 on Dec 17, 2017 \| parent \| context \| favorite \| on: Matrices from a geometric perspective 3Blue1Brown has an outstanding series of short videos, "Essence of linear algebra", which covers vectors, matrix transformations, and the related math. What I particularly like about these videos is that the concepts are introduced first without the numbers and calculations (that stuff is covered later). The concepts are first introduced as abstract animations to emphasize the "shape" of what the vector, matrix, etc really represent. \| \| \| \| --- \| \| \| \| \| edanm on Dec 17, 2017 \| next [] Amazing series! Open question I still have - what is the "geometric interpretation" of the transpose operation? A^T? Considering the fact that the transpose shows up all the time, I'm very surprised that I've never seen a good explanation of how I should be visualizing it. \| \| \| \| \| \| macawfish on Dec 18, 2017 \| parent \| next [] Graph theory offers one nice illustration of the transpose (for square matrices). Here's a directed graph: ¬---------¬ n n n It can be represented by this adjacency matrix: X = [ 0 1 0 ] [ 0 0 1 ] [ 1 0 0 ] See how there is a 1 for each row i and column j where there is an edge from node i to node j? Okay, now look at the transpose, Xáµ: Xáµ = [ 0 0 1 ] [ 1 0 0 ] [ 0 1 0 ] If we interpret this matrix as an adjacency matrix, its graph looks like this: ¬---------¬ n n n Taking the transpose of a directed graph's adjacency matrix just reverses the direction of the edges! P.S: some more cool things about adjacency matrices: Imagine that a vector represents a node or set of nodes on a graph. Like this: n = [ 1 ], n = [ 0 ], n = [ 0 ], [ 0 ] [ 1 ] [ 0 ] [ 0 ] [ 0 ] [ 1 ] Watch what happens when we multiply X with n: Xn = [ 0 0 1 ] [ 1 ] [ 0 ] [ 1 0 0 ] [ 0 ] = [ 1 ] = n [ 0 1 0 ] [ 0 ] [ 0 ] So the adjacency matrix is not only an index of edges, it's also a little machine that can push nodes around on a graph. You can even do it with two at a time: X(n + n) = [ 0 0 1 ] [ 1 ] [ 0 ] [ 1 0 0 ] [ 1 ] = [ 1 ] = n + n [ 0 1 0 ] [ 0 ] [ 1 ] If you're interested in diving further into algebraic representations of graphs, check out the Leavitt Path algebra: \| \| \| \| \| \| thethirdone on Dec 18, 2017 \| root \| parent \| next [] Notably adjacency matrices are orthogonal so A^T = A^-1. This extends to probabilistic transitions as well. \| \| \| \| \| \| macawfish on Dec 18, 2017 \| root \| parent \| next [] Well, the adjacency matrix I used as an example is orthogonal, but they don't have to be. Any matrix with 1's and 0's can be interpreted as an unweighted adjacency matrix for an undirected graph (if the matrix is symmetric) or directed graph (if it's not symmetric). For example, here's an adjacency matrix that's not an orthogonal matrix: [ 1 0 ] [ 1 1 ] \| \| \| \| \| \| obastani on Dec 18, 2017 \| parent \| prev \| next [] It's a bit difficult to explain the visualization without pictures, but I'll give it a shot. The transpose is really about converting the matrix to operate on a different vector space, namely, the dual space. In particular, the dual space of a vector space V is the vector space of "linear functionals", which are linear functions \phi: V -> R A linear functional on R^2 looks like a gradient (the "gradient fill" gradient, not a calculus gradient). These gradients are in one-to-one correspondence to vectors in R^2. In particular, given a vector w \in R^2, the direction of the gradient is along the direction of v, and the speed with which the gradient is changing corresponds to the magnitude of w. The precise mathematical correspondence is that (i) given a vector w \in R^2, the function f_w(v) = is a linear function (here, <,> is the inner/dot product), and (ii) every linear function has this form. Now, note that f_w is exactly multiplication by the transpose w^T of w! In particular, f_w(v) = w^T v More generally, for any linear map A : V -> W, the adjoint A of A is defined to be the linear map from the dual space of W to the dual space of V that satisfies = The transpose A^T is the adjoint of A when V is a finite-dimensional real vector space: = (A^T w)^T v = w^T A v = In summary, you can try to visualize A^T as a linear map "acting" on dual vectors. For example, let v \in R^2 and let w be a dual vector (i.e., a gradient), and suppose that A rotates v clockwise by 90 degrees. To preserve the inner product , A^T rotates w counter-clockwise by 90 degrees. \| \| \| \| \| \| pandaman on Dec 18, 2017 \| root \| parent \| next [] And a practical example of this is checking transformed points against a view frustum. Instead of transforming points into the view space a transposed matrix allows you to transform the frustum into the object space and check untransformed points against it. This works only on non-perspective transforms, of course, but the view transform should not be perspective anyways. To visualize this, take a simplest case of 2D space and non-homogenous coordinates. A simple frustum would be an angle made by two rays from the origin. You can see that rotating this space is as same as rotating the frustum in the opposite direction (though in this case transposed matrix is the same as inverse) but stretching the space opens/closes the frustum depending on in which direction it pulls its normals. \| \| \| \| \| \| thethirdone on Dec 18, 2017 \| parent \| prev \| next [] > Open question I still have - what is the "geometric interpretation" of the transpose operation? A^T? AFAIK, there is not a solid geometric interpretation. Part of the trouble is that the transpose can change the shape of the matrix. For example, for a vector the transpose turns it into a matrix outputting a single number. These two objects seem to be rather incomparable geometrically. The best intuition I have for transposition is that it represents a time reversal (but not necessarily an inverse). In the case of a vector, you have to think of it as a linear transformation that maps 1 to that vector. The transpose instead maps that vector to its length squared. I have more vague intuition for the why its length squared and how it relates to projections, but its hard to put into words. With rotation matrices, this time reversal results in the inverse. So essentially rotation/skewing is reversed, but scaling is not. \| \| \| \| \| \| akalin on Dec 18, 2017 \| parent \| prev \| next [] I don't have a complete answer, but a first step would be to think of the coordinate-free concept behind transposes, which is the adjoint: \| \| \| \| \| \| jacobolus on Dec 18, 2017 \| root \| parent \| next [] You want to start with: \| \| \| \| \| \| Crye on Dec 17, 2017 \| prev \| next [] I was hoping this was the top comment. Linear algebra would have been much more interesting if I had watch those videos first. He 3B1B such an amazing teacher! \| \| \| \| \| \| jjcc on Dec 18, 2017 \| prev \| next [] It remind me 3B1B either when I read this much earlier blog. 3B1B is really really wonderful tutor. Liner algebra is only one series among others. \| \| \| \| \| \| lexpar on Dec 17, 2017 \| prev [] Can't overstate how great his stuff is. \| \| \| \| Guidelines \| FAQ \| Lists \| API \| Security \| Legal \| Apply to YC \| Contact \|
6109	https://literacylearn.com/context-clue-types-examples/	Skip to content All Resources \| Grade 2 \| Grade 3 \| Reading Resources \| Vocabulary 6 Context Clue Types with Examples & FREE Mini-Book This post may contain affiliate links. As an Amazon affiliate, we earn from qualifying purchases. Email Facebook X Learn how to teach the 6 types of context clues with examples of each one. This is an essential strategy for understanding and learning new vocabulary! Get a fun and free mini-book that clearly shows the different types of context clues, signal words, and examples for your students to use! Get a new freebie every week! Skip Ahead Toggle What Are Context Clues? Context cluesare words around the unknown word which give hints as to the word’s meaning. In order to understand what you’re reading, you must know word meanings. Occasionally, you’ll find a word you can decode (or read accurately) but are unsure of its meaning. Strong readers will use context clues to determine the meanings of difficult words. 🤔 Think about this word: petulantly. Have you heard this word before? Do you know what it means? If not, try reading the sentence below and try to infer what the word means using context: The child crossed his arms, frowned, and pouted petulantly when his ice cream cone fell on the ground. Notice how the other words within the sentence can help you figure out the meaning of petulantly? If you look closely, you’ll notice two kinds of context clues: Example context clue: “The child crossed his arms.” Synonym context clues: “frowned” and “pouted.” The various types of context clues help you to infer the meaning of the word: “In an angry or sulky way.” You can almost visualize the child’s face; If it were an emoji, it might be this one: 😖 By explicitly teaching the 6 types of context clues, students can learn to use context to help determine the meaning of a tricky or unknown word. This improves reading comprehension and will be an incredibly powerful tool in a child’s word-learning toolbox! 6 Types of Context Clues According to the Vocabulary Handbook (2021), there are 6 types of context clues: Definition: The meaning of the word is stated in the sentence. Synonym: Another word with a similar meaning is stated in the sentence. Antonym: A word with an opposite meaning is stated in the sentence. Example: An example of the definition is included within the sentence. Appositive: The word’s meaning is included and set apart by commas. General: Clues are given over the course of multiple sentences. 🚦 Each type has signal words or punctuation that help the context clue types stand out. 🔍 Context Clue Types Examples It’s super helpful to teach kids the types of context clues using examples. Below are the types and a few examples of each, with signal words or punctuation underlined. Definition Type The garden had flowers growing in abundance, meaning the flowers were growing everywhere. Meander means to follow a winding or complicated course. A mesa is a flat-topped hill with steep sides. Appositive Type The dog was really enormous, or gigantic, and took up the whole couch. Sophia had a mischievous, or playful, smile on her face when she hid behind the door and scared her brother. The teacher was elated, or extremely happy, when all the students aced their spelling test. Synonym Type The jubilant crowd went crazy when the team won the championship game as they joyfully cheered. In the heat of the summer, the kids were thankful for the enormous tree. It was so big that it shaded the entire playground. Sarah’s impeccable manners impressed everyone at the dinner party. Her sister, Clara, had flawless manners, too. Antonym Type The weather was scorching hot, but the pool water was refreshingly cool. Jeremy’s room was a chaotic mess, unlike his sister’s room which was neat and organized. The movie’s beginning was very dull; however, as it went on, it became exciting and action-packed. Example Type The farmer planted different crops, such as corn, beans, and potatoes. The zoo had an exhibit with large canines, including wolves, wild dogs, and coyotes. My sister has a variety of pets. For example, a dog, a cat, two rats, a bunny, a horse, a snake, and a goldfish. General Type The children galivanted around in the first snowfall of the season. They spent hours building snowmen, having snowball fights, making snow angels, and enjoying the winter weather. My dad’s business was going through a merger. Dad said he was joining forces with another business and they would work together from now on. He was excited about the new team he would be working with. The sweet aroma of freshly baked bread, cookies, and cakes filled the air at the bakery. Mom and I couldn’t help but buy a box full of bakery treats after taking a sniff. 👨🏽‍🏫 How to Teach Word-Learning Strategies It’s going to be very important to teach kids explicitly about context clues as outlined above. But it is also imperative to teach them to use morphemic (word part) analysis. This is looking for word part clues within the unfamiliar words. Kids should ask themselves: Does it have a prefix? Does it have a suffix? What is the meaning of the base word? Does it have a Greek or Latin root? You also want to make sure your students know how to use a dictionary. More importantly, that they know how to use what they find in a dictionary in a meaningful way! Research from the National Reading Panel has shown that these three instructional strategies, when used together, are most effective to help kids discover word meaning. As children get older and read more complex texts, they will continue to see unknown words and will need to use context clues, morphemic analysis, and dictionaries to learn new vocabulary. Using other vocabulary-building strategies, like semantic mapping, is important too! ✨ So, the best way to teach these word-learning strategies is to explicitly teach them! That’s why we created our context clue resources, to help make your teaching easy and effective. When teaching context clues, teach clues one at a time, and give kids opportunities to practice. 👍🏼 If you don’t know much about morphemic analysis, I’d highly recommend IMSE’s Morphology course. I took it and learned a ton! You can read more about this in this post on the best OG trainings. The Mini-Book We’re providing you with a FREE context clue mini-book that you can print and use with your students. Download the color OR black and white version below! 📖 This mini-book is PERFECT for kids to use when reading! 🔖 They can keep it close and use it as a bookmark when reading on their own. 💛 Students love putting the book together (you may see them start to make other books following this method too). You can show our video for step-by-step instructions on how to make the mini-book. It also pairs perfectly with our Context Clue Anchor Chart. You can print the anchor chart, hang it up, and now your students have multiple resources to help them learn the meanings of new words. Related Posts Science of Reading Morphology Poster 3 Sounds of Suffix -ED All About Orthographic Mapping Download & Print We’d love to hear about your experience using these resources! Please leave a comment below or tag us on Instagram @literacylearn! Context Clues Mini Book (B&W) 1 file(s) 0.00 KB Download Context Clues Mini Book (Color) 1 file(s) 0.00 KB Download DOWNLOAD TERMS: All of our resources and printables are designed for personal use only in homes and classrooms. Each teacher must download his or her own copy. You may not:Save our files to a shared drive, reproduce our resources on the web, or make photocopies for anyone besides your own students. To share with others, please use the social share links provided or distribute thelink to the blog post so others can download their own copies. Your support in this allows us to keep making free resources for everyone! Please see our Creative Credits page for information about the licensed clipart we use. If you have any questions or concerns regarding our terms, please email us. Thank you! Get a new freebie every week! Related 2 Comments This is an EXCELLENT source! Thank you so much for providing it for FREE! I appreciate what you do for us so much. 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6110	https://math.answers.com/math-and-arithmetic/How_do_you_convert_143_base_5_to_decimal_form	0 Subjects>Math>Math & Arithmetic How do you convert 143 base 5 to decimal form? Anonymous ∙ 13y ago Updated: 11/29/2024 Well, honey, to convert 143 base 5 to decimal form, you simply multiply each digit by 5 raised to the power of its position from right to left (starting at 0). So, it's 1 x 5^2 + 4 x 5^1 + 3 x 5^0, which equals 25 + 20 + 3, giving you a decimal form of 48. Hope that clears things up for ya! BettyBot ∙ 10mo ago What else can I help you with? Trending Questions How do you get hoopa in x and y? How do you write 36.9 in standard form? Different ways to say or write 1 quarter? How many sq ft on area of 18 ft by 18 ft? How many times does 7 go into 365? What is the awnser to 12x4? How many perpendicular faces does a hexagonal have? If y varies directly with x and y is 42 when x is 4 what is the constant of variation expressed in decimal form? You cosigned for your father on his house and now he asked you to sign a quitclaim deeding the property to his name only As a cosigner are you entitled to any compensation even though you paid zero? What is the numeral for twenty one thousandths? What is next number 1806? How many syllables are in the word now? What are synonym for retort? What Is Point Of Inflection? What does a sheet of 3 8 x 4 x 8 plywood weight? What is 475 times 2? How do you write 7650 as a product of its prime factor? How much is 8.99? What is Pi to the second decimal? Four less than a number equals ten? Resources Leaderboard All Tags Unanswered Top Categories Algebra Chemistry Biology World History English Language Arts Psychology Computer Science Economics Product Community Guidelines Honor Code Flashcard Maker Study Guides Math Solver FAQ Company About Us Contact Us Terms of Service Privacy Policy Disclaimer Cookie Policy IP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
6111	https://eprint.iacr.org/2017/754.pdf	Long-Term Secure Time-Stamping using Preimage-Aware Hash Functions⋆ (Full Version) Ahto Buldas1,2, Matthias Geihs3, and Johannes Buchmann3 1 Tallinn University of Technology, Tallinn, Estonia. 2 Cybernetica AS, Tallinn, Estonia. 3 Darmstadt University of Technology, Darmstadt, Germany. Keywords: Long-Term Security, Timestamps, Preimage Aware Hash Functions Abstract. Commonly used digital signature schemes have a limited life-time because their security is based on computational assumptions that will potentially break in the future when more powerful computers are available. In 1993, Bayer et al. proposed a method for prolonging the life-time of a digital signature by time-stamping the signature together with the signed document. Based on their idea long-term timestamp schemes have been developed that generate renewable timestamps. To minimize the risk of a design failure that aﬀects the security of these schemes, it is important to formally analyze their security. However, many of the proposed schemes have not been subject to a formal security analysis yet. In this paper, we address this issue by formally analyzing the secu-rity of a hash-based long-term timestamp scheme that is based on the ideas of Bayer et al. Our analysis shows that the security level of this scheme degrades cubic over time, a security loss that needs to be taken into account when the scheme is used in practice. 1 Introduction 1.1 Motivation More and more information is generated and stored in digital form. In many cases it is important to ensure the integrity of this information. For example, in the case of electronic health records, it is indispensable that unintentional changes to the health records can be detected. Most commonly, integrity of such sensitive information is protected using digital signature schemes. However, digital signature schemes used today can provide security only for a limited time period. Their security is based on computational assumptions, which means that they provide security only as long as the computational resources of an attacker ⋆This work has been co-funded by the DFG as part of project S6 within the CRC 1119 CROSSING. A short version of this paper is published in the proceedings of ProvSec 2017 . are insuﬃcient to solve a given computational problem. The widely used RSA digital signature scheme , for example, is broken if an attacker can ﬁnd the prime factors of large integers. Using current computers, this seems to be infeasible. However, large integers can eﬃciently be factored using a quantum computer. To mitigate the security risk, in 1993 Bayer et al. proposed a method for prolonging the security period of a digital signature beyond the validity of the corresponding digital signature scheme. Their idea is to timestamp the signature together with the signed document in order to prove that a signature for that document was known when the corresponding signature scheme was still considered secure. Timestamps can be viewed as a digital signature themselves and have a limited security period as well. In order to protect the validity of the initial signature over a long time period, timestamps must therefore also be renewed periodically. To understand the security of long-term integrity protection using timestamp renewal, a security model must be provided for such a scheme. Unfortunately, Bayer et al. did not provide a security model for their scheme. In fact, the security of long-term timestamp schemes has not been formally analyzed until recently. In , Geihs et al. analyze the security of long-term timestamp schemes that use digital signature schemes and rely on trusted timestamp services. Their analysis is done in the random oracle model and shows that the security level of such long-term timestamp schemes degrades over time. An alternative method for time-stamping uses hash functions and relies on the availability of a trusted public repository. The security of long-term timestamp schemes based on this time-stamping method has not been studied yet. 1.2 Contribution In this work we analyze the security of long-term timestamp schemes that use hash-based timestamps. We present a security model that is based on the ideal primitive model proposed by Dodis et al. . The ideal primitive model is a more reﬁned model compared to the random oracle model used by Geihs et al. . Based on the ideal primitive model, we deﬁne the notion of extractable time-stamping, which engages important aspects of existing security notions for timestamp schemes and at the same time appears very suitable for analyzing the security of long-term timestamp schemes. By our security analysis we establish a bound on the security level of hash-based long-term timestamp schemes that degrades cubic over time. Furthermore, we provide a framework for obtaining numeric estimations of the security loss in a practical scenario. Such a frame-work is valuable for engineers who design systems that rely on long-term digital evidence. 1.3 Organization In Section 2, we brieﬂy discuss fundamentals on cryptographic security proofs, recall the deﬁnition of preimage-aware hash functions, and ﬁnally describe time-2 stamp schemes and long-term timestamp schemes in more detail. In Section 3, we deﬁne what it means for a hash-based timestamp scheme to be extractable. We show how to construct an extractable timestamp scheme using a preimage aware hash function and provide a bound on the security level of this construction. Then, in Section 4, we deﬁne what it means for a long-term timestamp scheme to be extractable and construct a hash-based long-term timestamp scheme that uses hash-based timestamp schemes for protection renewal. We prove a bound on the security level of this construction in terms of the security level of the used timestamp schemes. In Section 5, we use the results of our security analysis to evaluate the security level of hash-based long-term time-stamping in a practical scenario. In Section 6, we conclude our work and propose research directions for future work. 2 Preliminaries 2.1 Computational Security Most commonly used cryptographic schemes provide computational security. A cryptographic primitive is computationally secure if for a probabilistic adversary that is given a certain amount of computational resources the probability to break the security of the scheme is negligible. Such an adversary can be thought of as a program that runs on a computing machine, formally modeled, for exam-ple, as a Turing Machine or a Quantum Turing Machine . The resources of such an adversary are measured in terms of the number of operation steps performed by the machine. Let p be an integer. By the class of p-step adversaries we mean all programs that halt after at most p steps. Deﬁnition 1. For ϵ : N →[0, 1], we say a cryptographic scheme is ϵ(p)-secure, if any p-step adversary breaks the scheme with probability at most ϵ(p). We remark that in the traditional view of cryptography the step count of an adversary is sometimes also referred to as the computation time of the ad-versary. In particular, no distinction between real time and computation steps is usually made. Furthermore, the class of computing machines considered is usually assumed to be the class of Turing Machines. We will see in Section 4 that for analyzing long-term security of cryptographic schemes it is useful to distinguish between real time and computational steps and to consider that the computational technology used by an adversary may change over time. 2.2 Hash Functions A hash function H : {0, 1}∗→{0, 1}n is a function that converts bitstrings of arbitrary length to n-bit digests. By a hash chain c, we mean a sequence c, c, . . . , c[m] of 2n-bit strings and an m-bit string ι that is called the shape of c. The bits of ι are denoted by ι, . . . , ι[m]. Every c[k] consists of two n-bit halfs denoted by c[k]0 and c[k]1. By x c ⇝r we mean that: 3 – H(c) = r – H(c[k + 1]) = c[k]ι[k], for every k ∈{1, . . . , m −1} – H(x) = c[m]ι[m] For example, a hash chain can be seen as a path through a hash tree from a leaf to the root (Figure 1). r c0 c0 . . . . . . c1 c0 c1 c1 . . . . . . Fig. 1. A hash tree that contains hash chain c = [c0∥c1, c0∥c1, c0∥c1] of shape ι = [0, 1, 0], where for any x with H(x) = c0, x c ⇝r. Preimage Awareness. Informally, a hash function H is called preimage aware (PrA) if whenever somebody ﬁrst outputs as hash value y and later comes up with a preimage x, H(x) = y, then it must have known x when outputting y. The notion of preimage aware hash functions was formalized by Dodis et al. for hash functions HP that use an ideal primitive P. The primitive is ideal in the sense that it can only be called via an oracle P that records all calls made to P in an advice string adv. More formally, for HP to be preimage aware, there must exist an eﬃcient algorithm E (the so-called extractor) which when given y and the list adv of calls to the ideal primitive P, outputs x such that HP (x) = y, or ⊥ if the extraction failed. The adversary tries to ﬁnd x and y so that E(adv, y) ̸= x and y = HP (x). Deﬁnition 2 (Preimage Aware). Let ϵ : N3 →[0, 1]. A function HP is ϵ-secure preimage aware (PrA) if for every pE, pA, and q, there is a pE-step extractor E, such that for every pA-step adversary A that makes at most q calls to Ex, AdvPrA P,H(A, E) = Pr h ExpPrA P,H(A, E) = 1 i ≤ϵ(pE, pA, q) . 4 Algorithm 1: The Preimage Awareness (PrA) experiment ExpPrA P,H(A, E). x ←AP,Ex; y ←HP (x); if y ∈Q and V[y] ̸= x then return 1; else return 0; oracle P(m): z ←P(m); adv ←adv\|\|(m, z); return z; oracle Ex(y): x ←E(y, adv); Q ←Q ∪{y}; V[y] ←x; return x; 2.3 Time-Stamping Digital time-stamping was ﬁrst proposed by Haber and Stornetta . It is used to prove that a given data object existed at a certain point in time. In the following we describe a hash-based timestamp scheme based on their ideas [11, 1] and survey various security models that have been proposed for such a scheme. Scheme Description. The following hash-based timestamp scheme is associ-ated with a hash function H and a set of allowed hash chain shapes S. It uses a trusted repository Rep that accepts hash value queries. If Rep receives a hash value query r, it publishes r so that everybody can verify that r existed at this point in time t. The time-stamping procedure is divided into rounds. During each round, a timestamp server receives a set of bitstrings {x1, . . . , xn} from clients. At the end of each round it runs algorithm Stamp to generate timestamps for these bitstrings and returns the timestamps to the clients. Algorithm Verify is used to verify timestamps. Stamp: On input of bitstrings x1, . . . , xn (n ≤\|S\|), a hash tree is computed from leaves x1, . . . , xn. Let r be the root of that hash tree and ci be the hash chain corresponding to the path from leaf xi to the root r (cf. Figure 1). The timestamp server publishes the root hash r at the repository Rep and for i ∈{1, . . . , n}, sends ci as the response to request xi. Hash chain ci is also called a timestamp for bitstring xi. Verify: On input x, hash chain c, and a hash value r published at the repository, it is checked that c has allowed shape, shape(c) ∈S, and c is a hash chain from x to r, x c ⇝r. The algorithm outputs 1 if these conditions hold, otherwise the algorithm outputs 0. Security Model. Intuitively, security of a timestamp scheme means that an adversary cannot back-date any x, i.e., generate an x and a hash chain c such that Verify(x, c, r) = 1 for an r published at Rep before the generation of x. Such a condition is formalized, for example, in [7, 5], where a two-stage adversary A = (A1, A2) is considered. At the ﬁrst phase of the attack, A1 stores hashes into the repository Rep in an arbitrary way. At the second stage, A2 presents 5 an unpredictable x, a hash chain c, and selects an r published at Rep so that Verify(x, c, r) = 1. The unpredictability of x is essential as otherwise x could have been pre-computed by A1 before r is published and hence x could in fact be older than r. The security deﬁnitions of [7, 5] model the future as a computationally eﬃ-cient stochastic process which may not be the case in the real world. There are no arguments against the future documents having arbitrary distributions. Addi-tionally, the success of A is deﬁned as the average over such a distribution and it might still be easy to backdate ﬁxed documents. Having such arguments in mind, extraction-based security deﬁnitions for time-stamping have been explored in . Intuitively, such conditions say that whenever A1 publishes a hash r to Rep and later A2 outputs a document x and a hash chain c with Verify(x, c, r) = 1, then x must have been “known” by A1 when r was stored in Rep. Formally, this is expressed by assuming the existence of an extraction algorithm E that depends on A1 and outputs a set of bitstrings X such that if A2 outputs (x, c) with Verify(x, c, r) = 1 for some r ∈Rep, then x ∈X with overwhelming probability. In Sections 3 and 4 we propose extraction-based security deﬁnitions for time-stamp schemes and long-term timestamp schemes in the ideal primitive model. We then analyze the security of the hash-based long-term timestamp scheme described in Section 2.4 based on these deﬁnitions. 2.4 Long-Term Time-Stamping We describe a long-term timestamp scheme that is based on the idea of Bayer et al. to extend the lifetime of a digital signature by using timestamps . Here we assume that timestamp schemes TS = {TSi}i are available for usage over time. Each timestamp scheme TSi ∈TS is associated with a start time ts i and a breakage time tb i. The start time deﬁnes when the scheme becomes available and the breakage time deﬁnes after which time the timestamps created using this scheme are not considered valid anymore. Additionally, we assume the ex-istence of a repository Rep that is used for publishing root hash values. The long-term timestamp scheme is deﬁned by algorithm Stamp for creating an ini-tial timestamp, algorithm Renew for renewing a timestamp, and algorithm Verify for verifying a timestamp. Stamp: This algorithm gets as input a timestamp scheme identiﬁer i and a se-quence of bitstrings x1, . . . , xn. It creates timestamps for the bitstrings using scheme TSi by computing (r, c1, . . . , cn) ←TSi.Stamp(x1, . . . , xn). Then, the root hash r is published together with identiﬁer i at the repository Rep. Let t be the time when r was published. For j ∈{1, . . . , n}, the algorithm responds to request xj with timestamp Tj = [(i, cj, r, t)]. Renew: This algorithm gets as input a timestamp scheme identiﬁer i′ and a sequence of bitstrings and timestamps (x1, T1), . . . , (xn, Tn). The algorithm renews the timestamps using scheme TSi′ as follows. First, it computes new timestamps (r, c1, . . . , cn) ←TSi′.Stamp(x1∥T1, . . . , xn∥Tn). Then, it pub-lishes the root hash r together with the timestamp scheme identiﬁer i′ at 6 the repository Rep. Let t be the time when r is published. For j ∈{1, . . . , n}, the algorithm sends (i′, cj, r, t) as the response to request (xj, Tj). The client receiving (i′, cj, r, t) updates its timestamp by appending (i′, cj, r, t) to Tj. Verify: This algorithm takes as input a bitstring x, a long-term timestamp T = (C1∥. . . ∥Cn), where Cj = (ij, cj, rj, tj), a time t, a reference R to the trusted repository Rep, and a set of admissible timestamp schemes TS = {TSi}i. For j ∈{1, . . . , n}, it is veriﬁed that TSij.Verify((x∥C1∥. . . ∥Cj−1), cj, rj) = 1, (ij, rj) ∈R[tj], and tb ij > tj+1. The algorithm outputs 1 if these conditions hold, otherwise it outputs 0. 3 Extractable Time-Stamping In the following we deﬁne extractable time-stamping for timestamp schemes as described in Section 2.3. Informally, extractability of a timestamp scheme TS means that if a root hash r is published at the repository at time t and later some-one comes up with a bitstring x and a hash chain c such that TS.Verify(x, c, r) = 1, then x must have been known at time t. Our notion of extractable time-stamping is reminiscent of PrA hash functions and knowledge-binding com-mitments . After the deﬁnition of extractable timestamp schemes, we analyze the security of the timestamp scheme from Section 2.3 instantiated using a PrA hash function. 3.1 Deﬁnition More formally, extractability of a timestamp scheme TSP with an ideal primitive P is deﬁned using an experiment ExpExTs (Algorithm 2). In this experiment an adversary A publishes root hashes at the repository. The adversary also uses the ideal primitive P and queries to P are recorded in an advice string adv. The deﬁnition of extractable time-stamping requires the existence of an extractor E with the following properties. Whenever A publishes a root hash r, the extractor E extracts from r and the advice adv, a set of supposedly timestamped bitstrings X. At the end, A outputs a bitstring x, a timestamp c, and a root hash r. It wins if c is valid for x and r, r was published, and x was not extracted. Deﬁnition 3 (Extractable Time-Stamping). Let ϵ : N3 →[0, 1]. A time-stamp scheme TSP using ideal primitive P is ϵ-secure extractable (ExTs) if for all integers pE, pA, and qE, there is a pE-step extractor E, such that for every pA-step adversary A that makes at most q calls to Rep, AdvExTs P,TS (A, E) = Pr h ExpExTs P,TS (A, E) = 1 i ≤ϵ(pE, pA, q) . 3.2 Security Analysis We analyze the security of the hash-based timestamp scheme from Section 2.3. We ﬁrst recall various properties of hash chain shapes which are useful for analyzing the security of hash-based timestamp schemes. 7 Algorithm 2: The extractable time-stamping experiment ExpExTs P,TS (A, E). (x, c, r) ←AP,Rep; if TS.Verify(x, c, r) = 1, r ∈R, and x ̸∈L[r] then return 1; else return 0; oracle P(m): z ←P(m); adv ←adv\|\|(m, z); return z; oracle Rep(r): X ←E(adv, r); R ←R ∪{r}; L[r] ←X; return X; Deﬁnition 4. We say that a timestamp scheme associated with allowed shapes S is N-bounded if \|S\| ≤N. Deﬁnition 5. An N-bounded timestamp scheme is said to be shape-compact, if the length of allowed hash chains does not exceed 2 log2 N. Next we proof a bound on the security of the hash-based timestamp scheme from Section 2.3 if instantiated as N-bounded and shape compact. Theorem 1. The timestamp scheme from Section 2.3 instantiated as N-bounded and shape compact and with an ϵ-secure PrA hash function HP is ϵ′-secure ex-tractable with ϵ′(pE, pA, q) = ϵ α · pE 2N log2 N , β · (pA + 2qN log2 N), 2qN log2 N , for some small constants α and β. A detailed proof of Theorem 1 can be found in Appendix A. The log2 N term can be eliminated if a more eﬃcient tree-extractor is used. Theorem 2. The timestamp scheme from Section 2.3 instantiated as N-bounded and shape compact and with an ϵ-secure PrA hash function HP is ϵ′-secure ex-tractable with ϵ′(pE, pA, q) = ϵ α · pE 2N , β · (pA + 2Nq), 2Nq , for some small constants α and β. Proof (Sketch). The list extractor used in the proof of Theorem 1, extracts a separate hash chain for each leaf of the hash tree. This means that each of the inner nodes of the tree are extracted many times. The eﬃciency of the extraction can be improved by avoiding redundant extraction of hash chains that partially overlap. This is what the tree extractor does and it leads to the improved security bound. 8 4 Extractable Long-Term Time-Stamping In this section we propose a security model for the hash-based long-term time-stamp scheme described in Section 2.4. First, we deﬁne a model of time and an adversary model suitable for our security analysis. Then, we deﬁne what it means for a long-term timestamp scheme to be extractable. Finally, we prove a security bound for the long-term timestamp from Section 2.4 based on the security level of the timestamp schemes used for timestamp renewal. 4.1 Model of Real Time and Computation To model the breakage of a hash function or a hash-based timestamp scheme, we cannot use the traditional (timeless) security because it does not make sense to say that before tb i the hash function is secure but after tb i it is not secure anymore. However, in reality hash functions that have been considered secure in the past may appear insecure at a later point because the computational power of real world adversaries typically increases over time as well as new attacking algorithms, which were not known before tb i, may be discovered. In order to model adversaries that increase their abilities over time, we assume that the class Mt of computing machines available at a time t widens when t increases, i.e., Mt ⊂Mt′ for t < t′. Using our approach we can model, for example, that at some point quantum computers become available.4 Model of Real Time. To realize our adversary model we must be able to set in relation computation time with other events (e.g., the arise of a new computational technology) happening at given points in time. Therefore, we require a model of real time in contrast to conventional security models, where time is commonly considered the same as the number of steps performed by the adversary. In recent literature, various methods for modeling real time have been pro-posed. Schwenk and Geihs et al. , for example, model real time by deﬁn-ing a global clock that advances whenever the adversary performs work. Another model of real time has been proposed by Canetti et al. . They work in a com-putational framework that supports concurrency and deﬁne a global clock that runs concurrently to all other processes and ticks at a deﬁned rate. For our paper we follow the time formalism used by Schwenk and Geihs et al. . That is, we use a global clock Clock that holds state time, initialized to 0. Here, we give the adversary the control over advancing time. It may do so by calling Clock(t) as an oracle and if t > time, the clock is advanced to time = t (Algorithm 3). We remark that by advancing time, the adversary also burns computation power and triggers events in the security experiment. 4 We remark that while our model considers quantum computing, quantum commu-nication is not considered. 9 Model of Computation. As described in the previous section, we consider adversaries AClock that are associated with a global clock Clock. We bound the computational power of adversary A with respect to the time deﬁned by Clock. For ρ : N →N, we say A is ρ-step-bounded if at any time t, it performed less than ρ(t) computation steps. We say A is ρ-call-bounded if at any time t, it performed less than ρ(t) oracle calls. To allow for modeling that computational technology progress over time, we deﬁne an adversary AClock as a sequence (AClock 0 , AClock 1 , AClock 2 , . . .) of machines such that At ∈Mt, where Mt is the class of computing machines available at time t. Executing AClock at time = t means executing the component AClock t . The adversary AClock t then runs until it calls Clock(t′) after which the control is given to AClock t′ . Here, AClock t′ gets access to the internal state of AClock t . An extractors E is also deﬁned as a sequence (E0, E1, E2, . . .) such that Et ∈Mt, but the components do not have access to the clock-oracle. Calling an extractor E at time = t means calling Et. 4.2 Security Deﬁnition We deﬁne extractable long-term time-stamping using an experiment ExpExLTs (Algorithm 3). Similar to the deﬁnitions of PrA hash functions and extractable time-stamping without renewal, our deﬁnition of extractable long-term time-stamping uses an ideal primitive P which can only be called via an oracle P that records all calls to P in an advice string adv. The experiment also involves a global clock Clock as described in Section 4.1. The experiment involves an adversary A and an extractor E. The adversary A may publish root hash values r at the repository Rep at any time by calling Rep(r). When Rep is called with root hash r at time t, it records r associated with t in a global table R (i.e., R[t] ←R[t] ∪{r}, where initially R[t] = {}). Additionally, the extractor E on input adv and r extracts a set of bitstrings X that is stored associated with time t in a table L (i.e., L[t] ←L[t] ∪X, where initially L[t] = {}). The goal of the adversary A is to produce (x, T, t) such that T is a valid long-term timestamp for bitstring x and time t, and x was not extracted at time t (i.e., Verify(x, c, t, R) = 1 and x ̸∈L[t]). Deﬁnition 6 (Extractable Long-Term Time-Stamping). Let M describe the available machines classes and TS describe the available timestamp schemes. Let ϵ : N4 →[0, 1]. A long-term timestamp scheme LTSP , which uses an ideal primitive P, is ϵ-secure extractable (for M and TS) if for all bounds ρE, ρA, and q, there is a ρE-bounded extractor E ∈M, such that for every ρA-step-bounded and q-call-bounded adversary A ∈M, and for every time t: AdvExLTs P,LTS,TS(A, E, t) = Pr h ExpExLTs P,LTS,TS(A, E, t) = 1 i ≤ϵ(ρE, ρA, q, t) . 10 Algorithm 3: The extractable long-term time-stamping experiment ExpExLTs P,LTS,TS(A, E, t∗). (x, T, t) ←AClock,P,Rep; if LTS.Verify(x, T, t, R, TS) = 1, x ̸∈L[t], time ≤t∗then return 1; else return 0; oracle Clock(t): if t > time then time ←t; oracle P(m): z ←P(m); adv ←adv\|\|(m, z); return z; oracle Rep(r): X ←E(adv, r); t ←time; R[t] ←R[t]∥r; L[t] ←L[t]∥X; return X; 4.3 Security Analysis Before we analyze the security of the long-term timestamp scheme described in Section 2.4, we adapt the notion of extractable time-stamping from Section 3 to the long-term setting where the class of computing machines available changes over time. Deﬁnition 7 (Extractable Time-Stamping (Reﬁned)). Let ME and MA be classes of machines and ϵ : N3 →[0, 1]. We say a non-renewable timestamp scheme TS is ϵ-secure extractable for adversaries of MA and extractors of ME if for all integers pE, pA, and qE, there exists a pE-step extractor E ∈ME, such that for every pA-step adversary A ∈MA that makes at most q calls to Rep: AdvExTs P,TS (A, E) ≤ϵ(pE, pA, q) . We now prove a bound on the security level of the long-term timestamp scheme described in Section 2.4 in terms of the security level of the available timestamp schemes. Theorem 3. Let M describe the available computing machine classes and TS = {TSP i }i describe the available timestamp schemes, which use an ideal primitive P. If for every i, TSP i is ϵi-secure extractable for adversaries of Mtb i and extrac-tors of Mts i , then the long-term timestamp scheme described in Section 2.4 is ϵ-secure extractable with ϵ(ρE, ρA, q, t) = X i∈{i:tb i≤t} ϵi α · ρE(tb i), β · ρA(tb i) + q(tb i)ρE(tb i) , q(tb i) , for some small constants α and β. A detailed proof of Theorem 3 can be found in Appendix B. 11 5 Evaluation We evaluate which protection level the long-term timestamp scheme described in Section 2.4 provides in a practical scenario. For our evaluation we consider a scenario where data is protected over a time period of Y years. The security level of the long-term timestamp scheme is evaluated in terms of the security level of the hash functions that are used to instantiate the available timestamp schemes. Here, we assume that all used hash functions have the same security level during their validity period, where by security level we mean a bound on the success probability of an adversary. 5.1 Scenario We assume that a set TS = {TSP i }i of available hash-based timestamp schemes, where for each i, HP i is the hash function used by TSP i . We assume that the PrA-security of a hash function derives from the ratio of the adversary power pA and the extractor power pE, and is inﬂuenced by the number of repository calls q and a base security level δ. Concretely, we assume that each hash function Hi is ϵ-secure PrA until its breakage time tb i with ϵ(pE, pA, q) = pA pE qδ. Furthermore, we assume that each timestamp scheme TSi is N-bounded and shape compact, which means that each timestamp round up to N timestamps are generated. For our practical security analysis we neglect the constants α and β derived in Theorem 2 and Theorem 3 as they are in most cases close to 1. By Theorem 2 we obtain that each timestamp scheme TSi is ϵ′-secure ex-tractable until time tb i with ϵ′(pE, pA, q) ≤ϵ pE 2N , pA + 2Nq, 2Nq = pA + 2Nq pE (2N)2qδ . Furthermore, by Theorem 3 we obtain that the long-term timestamp scheme is ϵ′′-secure long-term extractable with ϵ′′(ρE, ρA, q, t) ≤ X i∈It ϵ′ ρE(tb i), ρA(tb i) + q(tb i)ρE(tb i), q(tb i) ≤ X i∈It ρA(tb i) + q(tb i)ρE(tb i) + 2Nq(tb i) ρE(tb i) (2N)2q(tb i)δ ≤ X i∈It ρA(tb i) ρE(tb i) + 2N ρE(tb i) + 1 q(tb i) (2N)2q(tb i)δ . We assume that the adversary and the extractor have the same computation power, i.e., ρA(t) ρE(t) = 1, and we observe that any reasonable extractor E extracts at least 2N bitstrings before the breakage time of a scheme, i.e., ρE(tb i) ≥2N. Let time be denoted in years and assume that each year at most L new time-stamp schemes become available, and at most R root hashes are published at the repository, i.e., \|It\| = \|{i : tb i ≤t}\| ≤tL and q(t) ≤tR. We obtain the following bound on the security level of the long-term timestamp scheme: ϵ′′(ρE, ρB, q, t) ≤12t3(NR)2Lδ . 12 5.2 Results In Figure 2 we show the security level of the long-term timestamp scheme for diﬀerent time spans Y and parameters N, L, R, and δ. The default parameters are Y = 100, L = 10, N = 232, R = 365, and δ = 2−192. By the upper left graph of Figure 2, we observe a cubic security loss over time for the case that the security level of the used hash function remains constant. Using a base security level of δ = 2−192 for the hash function, the security level of the long-term timestamp scheme after 1 year is 2−104, after 10 years it drops to 2−94, and after 100 years it drops to 2−84. There is a linear loss in security for increasing the number L of short-term timestamp schemes used per year, as depicted in the upper right graph. Allowing a larger number R of hash values to be published at the repository results in a quadratic security loss, as can be seen in the middle left graph. The number N of timestamps that can be issued per root hash is an important factor because in practice it may be large. The security level decreases quadratically when N is increased, as can be seen in the middle right graph. Using N = 216 results in security level 2−116 after 100 years, while using N = 248 results in security level 2−52. Finally, the bottom graph shows how the security of the long-term timestamp scheme depends linear on the base security level δ of the used hash functions. 6 Conclusions and Future Work We have formally analyzed the security of hash-based long-term time-stamping as proposed by Bayer et al. . We prove that the security level of the discussed scheme degrades cubic over time if the security level of the used cryptographic primitives remains constant. This shows that long-lived systems need to be de-signed using a certain security margin. Our analysis provides a framework for analyzing the security level of similar schemes. For future work it would be interesting to see whether the security bound proved by us can be improved and the security loss over time can be reduced. It would also be interesting to establish a security model for long-term time-stamping that does not rely on idealized assumptions such as a random oracle or an ideal primitive. References 1. Bayer, D., Haber, S., Stornetta, W.S.: Improving the eﬃciency and reliability of digital time-stamping. In: Capocelli, R., De Santis, A., Vaccaro, U. (eds.) Sequences II: Methods in Communication, Security, and Computer Science. pp. 329–334. Springer New York, New York, NY (1993) 2. van Breugel, F., Chechik, M. (eds.): CONCUR 2008 - Concurrency Theory, 19th International Conference, CONCUR 2008, Toronto, Canada, August 19-22, 2008. Proceedings, Lecture Notes in Computer Science, vol. 5201. Springer (2008) 13 50 100 85 90 95 Y −log2(ϵ′′) 100 101 102 103 80 85 L −log2(ϵ′′) 102 103 104 105 106 70 80 90 R −log2(ϵ′′) 20 30 40 50 60 80 100 120 log2(N) −log2(ϵ′′) 200 300 400 0 100 200 300 −log2(δ) −log2(ϵ′′) Fig. 2. Evaluation of the security level ϵ′′ of long-term time-stamping when run for Y years, and with parameters L, R, N, and δ. Here, L is the number of new short-term timestamp schemes per year, R is the number of published root hashes per short-term scheme, N is the number of documents covered by one root hash, and δ is the base security level of the hash functions. 14 3. Buldas, A., Geihs, M., Buchmann, J.A.: Long-term secure time-stamping using preimage-aware hash functions (short version). In: Provable Security: 11th Interna-tional Conference, ProvSec 2017, Xi’an, China, October 23-25, 2017, Proceedings. Springer International Publishing (2017) 4. Buldas, A., Laanoja, R.: Security proofs for hash tree time-stamping using hash functions with small output size. In: Boyd, C., Simpson, L. (eds.) Information Security and Privacy - 18th Australasian Conference, ACISP 2013, Brisbane, Aus-tralia, July 1-3, 2013. Proceedings. Lecture Notes in Computer Science, vol. 7959, pp. 235–250. Springer (2013) 5. Buldas, A., Laur, S.: Do broken hash functions aﬀect the security of time-stamping schemes? In: Zhou, J., Yung, M., Bao, F. (eds.) Applied Cryptography and Network Security, 4th International Conference, ACNS 2006, Singapore, June 6-9, 2006, Proceedings. Lecture Notes in Computer Science, vol. 3989, pp. 50–65 (2006) 6. Buldas, A., Laur, S.: Knowledge-binding commitments with applications in time-stamping. In: Okamoto, T., Wang, X. (eds.) Public Key Cryptography - PKC 2007, 10th International Conference on Practice and Theory in Public-Key Cryptogra-phy, Beijing, China, April 16-20, 2007, Proceedings. Lecture Notes in Computer Science, vol. 4450, pp. 150–165. Springer (2007) 7. Buldas, A., Saarepera, M.: On provably secure time-stamping schemes. In: Lee, P.J. (ed.) Advances in Cryptology - ASIACRYPT 2004, 10th International Conference on the Theory and Application of Cryptology and Information Security, Jeju Island, Korea, December 5-9, 2004, Proceedings. Lecture Notes in Computer Science, vol. 3329, pp. 500–514. Springer (2004) 8. Deutsch, D.: Quantum theory, the church-turing principle and the universal quan-tum computer. Proceedings of the Royal Society of London A: Mathematical, Phys-ical and Engineering Sciences 400(1818), 97–117 (1985) 9. Dodis, Y., Ristenpart, T., Shrimpton, T.: Salvaging merkle-damg˚ ard for practical applications. In: Joux, A. (ed.) Advances in Cryptology - EUROCRYPT 2009, 28th Annual International Conference on the Theory and Applications of Cryptographic Techniques, Cologne, Germany, April 26-30, 2009. Proceedings. Lecture Notes in Computer Science, vol. 5479, pp. 371–388. Springer (2009) 10. Geihs, M., Demirel, D., Buchmann, J.A.: A security analysis of techniques for long-term integrity protection. In: 14th Annual Conference on Privacy, Security and Trust, PST 2016, Auckland, New Zealand, December 12-14, 2016. pp. 449– 456. IEEE (2016) 11. Haber, S., Stornetta, W.S.: How to time-stamp a digital document. In: Menezes, A., Vanstone, S.A. (eds.) Advances in Cryptology - CRYPTO ’90, 10th Annual International Cryptology Conference, Santa Barbara, California, USA, August 11-15, 1990, Proceedings. Lecture Notes in Computer Science, vol. 537, pp. 437–455. Springer (1990) 12. Merkle, R.C.: A certiﬁed digital signature. In: Brassard, G. (ed.) Advances in Cryp-tology - CRYPTO ’89, 9th Annual International Cryptology Conference, Santa Barbara, California, USA, August 20-24, 1989, Proceedings. Lecture Notes in Com-puter Science, vol. 435, pp. 218–238. Springer (1989) 13. Rivest, R.L., Shamir, A., Adleman, L.M.: A method for obtaining digital signatures and public-key cryptosystems. Commun. ACM 21(2), 120–126 (1978) 14. Schwenk, J.: Modelling time for authenticated key exchange protocols. In: Kuty-lowski, M., Vaidya, J. (eds.) Computer Security - ESORICS 2014 - 19th European Symposium on Research in Computer Security, Wroclaw, Poland, September 7-11, 2014. Proceedings, Part II. Lecture Notes in Computer Science, vol. 8713, pp. 277–294. Springer (2014) 15 15. Turing, A.M.: On computable numbers, with an application to the entschei-dungsproblem. Proceedings of the London mathematical society 2(1), 230–265 (1937) A Proof of Theorem 1 An N-bounded shape compact timestamp scheme using an ϵ-secure PrA hash function HP is ϵ′-secure extractable with ϵ′(pE, pA, q) = ϵ α · pE 2N log2 N , β · (pA + 2qN log2 N), 2qN log2 N , for some small constants α and β. Proof. Let S be the set of allowed shapes associated with an N-bounded shape compact timestamp scheme TS and let E be an extractor for HP that extracts a preimage given a hash value and the ideal primitive calls. Such an extractor exists if HP is PrA-secure. Using E as a black box, we construct a list extractor L (Algorithm 4), which extracts timestamped bitstrings from published root hash values. Having as input a P-query string adv and a bitstring r (a hash value), the list extractor L extracts for every allowed hash-chain shape ι ∈S, the corresponding hash chain in a top-down way, starting from r. Due to the shape-compactness, the step count of L is bounded by α·pE·N·maxι∈S length(ι) ≤ 2 · α · pE · N log2(N), for some small constant α and if pE is a bound on the step count of E. Algorithm 4: List extractor L(adv, r). for ι ∈S do x ←E(adv, r), c ←[], k ←0; while x ̸= ⊥and k < length(ι) do k ←k + 1; x ←E(adv, xι[k]) (where x = x0∥x1); L[r] ←L[r] ∪{x}; return L; Let AP,Rep be an adversary that participates in the ExTs-experiment denoted by ExpExTs P,TS (A, L). We construct an adversary BP,Ex (Algorithm 5) that partici-pates in the PrA-experiment ExpPrA P,H(B, E) as follows. The adversary BP,Ex runs algorithm AP,Rep, where calls to the repository Rep are simulated using algorithm REx (Algorithm 6). 16 Algorithm 5: PrA adversary BP,Ex. (x, c, r) ←AP,REx; ι ←shape(c); if ∃k : R[c[k]ι[k]] = 1, V[c[k]ι[k]] ̸= c[k + 1] then return c[k + 1]; else return x; Algorithm 6: Rep simulator REx(r). for ι ∈S do x ←Ex(r), k ←0; while x ̸= ⊥and k < length(ι) do k ←k + 1; x ←Ex(xι[k]); L[r] ←L[r] ∪{x}; return L[r]; Note that whenever A succeeds in the ExTs-experiment, it ﬁnds (x, c, r) such that x c ⇝r, x ∈R, and x ̸∈L[r]. This means there must be k such that HP (c[k+ 1]) = c[k]ι[k] (where ι is the shape of c), but the extractor E failed to extract c[k + 1] from c[k]ι[k]. Therefore, after simulating A, the adversary B obtains the hash chain c and ﬁnds the smallest k, such that R[c[k]ι[k]] = 1 and V[c[k]ι[k]] ̸= c[k + 1], and outputs c[k + 1], or, if no such k exists, outputs x. We observe that B succeeds in the PrA experiment whenever A succeeds in the ExTs experiment, AdvExTs P,TS (A, L) ≤AdvPrA P,H(B, E) . The step count of BP,Ex is O(pA + 2qRN log2 N), where pA is the step count of A and qR is the number of calls to Rep. It follows that AdvExTs P,TS (A, L) ≤AdvPrA P,H(B, E) ≤ϵ(pE, α · (pA + 2qRN log2 N), 2qRN log2 N) , for some small constant α and where pE is a bound on the step count of E. The step count of L is pL = O(2pEN log2 N). We obtain that the timestamp scheme constructed from HP is ϵ′-secure extractable with ϵ′(pL, pA, qR) = ϵ β · pL 2N log2 N , α · (pA + 2qRN log2 N), 2qRN log2 N , for some small constant β. ⊓ ⊔ B Proof of Theorem 3 Let M describe the available computing machine classes and TS = {TSP i }i de-scribe the available timestamp schemes, which use ideal primitive P. For every i, assume TSi is ϵi-secure extractable for adversaries of Mtb i and extractors of Mts i . Then the long-term timestamp scheme described in Section 2.4 is ϵ-secure extractable with ϵ(ρE, ρA, q, t) = X i∈{i:tb i≤t} ϵi α · ρE(tb i), β · (ρA(tb i) + q(tb i)ρE(tb i)), q(tb i) , for some small constants α and β. 17 Proof. Let Li ∈Mts i be the list extractor for TSi that exists because TSi is ϵi-secure extractable for adversaries of Mtb i and extractors of Mts i . Using the Li’s as black boxes, we construct a long-term list extractor L ∈M in the following way (Algorithm 7). Having as input an ideal primitive advice string adv and a bitstring r, the list extractor L decomposes r into a timestamp scheme identiﬁer i and a hash value r′, checks if scheme i can be used at the current time, and if so, extracts a set of bitstrings X from r′ using list extractor Li. The step count of L in this run is at most the step count of Li (plus a small simulation overhead). Algorithm 7: Long-term extractor L(adv, r) r = (i, r′); if ts i ≤time < tb i then return Li(adv, r′); else return ⊥; Let A be any adversary that participates in the ExLTs experiment. For every i, we construct an adversary B(i) (Algorithm 8) that participates in the ExTs experiment as follows.5 The adversary BP,Rep (i) simluates a run of AClock,P,R where the clock Clock is simulated using Algorithm 10 and the repository R is simu-lated using Algorithm 9. The simulation is performed only while time does not exceed the breakage time tb i. The adversary B(i) tries to ﬁnd a moment t′′ < tb i when A submits (j, r′′) to the repository to renew a tuple (x′, i, c′, r′, t′) valid with scheme TSi and A submitted (i, r′) to the repository, but x′ has not been extracted. This violates the ExLTs-condition for TSi. 5 We use the notation B(i) to distinguish between B(i) and the time-components B = (B0, B1, . . .) of an adversary. 18 Algorithm 8: ExLTs adversary BP,Rep (i) . (x, c, t) ←AClock,P,R while time < tb i until ∃j, x′, c′, r′, t′, r′′, t′′ such that – (j, r′′) ∈R[t′′] and ts j < t′′ < tb i, tb j, – (x′, i, c′, r′, t′) ∈L[t′′], – Si.Verify(x′, c′, r′) = 1, – (i, r′) ∈R[t′] and ts i < t′ < tb i, – x′ ̸∈L[t′]; if ∃such (j, x′, c′, r′, t′, r′′, t′′) then return x′; else return ⊥; Algorithm 9: Repository simulator R(r). r = (j, r′); t ←time; if ts j ≤t < tb j then if j=i then X ←Rep(r′); else X ←Lj(adv, r′); else X ←⊥; R[t] ←R[t]∥r; L[t] ←L[t]∥X; return X; Algorithm 10: Clock simulator Clock(t). if t > time then time ←t; Deﬁne It = {i : tb i ≤t}. We observe that whenever A is successful until time t, one of {Bi : tb i ≤t} is successful, AdvExLTs P,LTS (A, L, t) ≤ X i∈It AdvExTs P,TSi(B(i), Li) . Assume A is ρA-step-bounded and q-call-bounded, and assume L is ρL-bounded. Then, for each i, the step count of B(i) is O(ρB(tb i)+q(tb i)ρL(tb i)). We obtain that for every t, AdvExLTs P,LTS (A, L, t) ≤ X i∈It AdvExTs P,TSi(B(i), Li) ≤ X i∈It ϵi α · ρL(tb i), β · (ρB(tb i) + q(tb i)ρL(tb i)), q(tb i) , for some small constants α and β. 19
6112	https://webbook.nist.gov/cgi/cbook.cgi?ID=C7664417	Ammonia Jump to content National Institute of Standards and Technology NIST Chemistry WebBook, SRD 69 Home Search Name Formula IUPAC identifier CAS number More options NIST Data SRD Program Science Data Portal Office of Data and Informatics About FAQ Credits More documentation Ammonia Formula: H 3 N Molecular weight: 17.0305 IUPAC Standard InChI:InChI=1S/H3N/h1H3 Copy IUPAC Standard InChIKey:QGZKDVFQNNGYKY-UHFFFAOYSA-N Copy CAS Registry Number: 7664-41-7 Chemical structure: This structure is also available as a 2d Mol file or as a computed3d SD file View 3d structure (requires JavaScript / HTML 5) Isotopologues: Ammonia-d3 Other names: Ammonia gas; Nitro-Sil; Spirit of Hartshorn; NH3; Ammonia, anhydrous; Anhydrous ammonia; Aromatic Ammonia, Vaporole Permanent link for this species. Use this link for bookmarking this species for future reference. Information on this page: Notes Other data available: Gas phase thermochemistry data Phase change data Reaction thermochemistry data: reactions 1 to 50, reactions 51 to 100, reactions 101 to 146 Henry's Law data Gas phase ion energetics data Ion clustering data IR Spectrum Mass spectrum (electron ionization) Vibrational and/or electronic energy levels Gas Chromatography Fluid Properties Data at other public NIST sites: Electron-Impact Ionization Cross Sections (on physics web site) Gas Phase Kinetics Database X-ray Photoelectron Spectroscopy Database, version 5.0 Options: Switch to calorie-based units Data at NIST subscription sites: NIST / TRC Web Thermo Tables, "lite" edition (thermophysical and thermochemical data) NIST / TRC Web Thermo Tables, professional edition (thermophysical and thermochemical data) NIST subscription sites provide data under the NIST Standard Reference Data Program, but require an annual fee to access. The purpose of the fee is to recover costs associated with the development of data collections included in such sites. Your institution may already be a subscriber. Follow the links above to find out more about the data in these sites and their terms of usage. Notes Go To:Top Data from NIST Standard Reference Database 69: NIST Chemistry WebBook The National Institute of Standards and Technology (NIST) uses its best efforts to deliver a high quality copy of the Database and to verify that the data contained therein have been selected on the basis of sound scientific judgment. However, NIST makes no warranties to that effect, and NIST shall not be liable for any damage that may result from errors or omissions in the Database. Customer support for NIST Standard Reference Data products. © 2025 by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. Copyright for NIST Standard Reference Data is governed by the Standard Reference Data Act. Privacy Statement Privacy Policy Security Notice Disclaimer (Note: This site is covered by copyright.) Accessibility Statement FOIA Contact Us
6113	https://www.regiscollege.edu/blog/public-health/what-is-epidemiology-in-public-health	Skip to main content What Is Epidemiology in Public Health? Blog All Posts Careers for Veterans College Advice Completing Your Degree Dental Hygiene Education Health Sciences Marketing and Communications Medical Imaging Nursing Occupational Therapy Online Learning Public Health Regis College’s MPH Program: A Commitment to Real-World Impact Earning a Master’s in Public Health Online Interdisciplinary Public Health Career Paths What is Health Equity and Why Does it Matter? What Is a Master’s in Public Health? Public Health Skills Professionals Need to Make an Impact Real-World Impact: Regis College’s Public Health Practicums Understanding Public Health Policy and How You Can Influence It Public Health Degree Courses: What’s the Curriculum? BA in Public Health vs BS: What’s the Difference? BA in Public Health Salary: What Is Your Earning Potential? What Is Epidemiology in Public Health? Current Public Health Challenges in the Greater Boston Area How to Become a Public Health Research Assistant How to Become a Public Health Program Coordinator How to Become a Community Health Worker How to Become a Health Educator Public Health Career Outlook: Professions and Salaries Top Public Health Companies to Hiring in Massachusetts What Can You Do With a Bachelor's in Public Health? Is a Public Health Degree Worth It? The Future of Public Health: Nine Trends You Should Know About Speech-Language Pathology One of the great things about public health as a field of study is that earning a public health degree will prepare you for a wide variety of careers. Some people study public health with the intention of becoming a community health worker and directly interacting with the communities they serve. Others dream of becoming public health program coordinators who design public health initiatives. And others want to become public health educators or researchers, who also play crucial roles in improving the health of their communities. Want to learn more about Public Health Careers? Download Our Free Guide! One public health career that has been in the limelight in recent years, largely due to the coronavirus pandemic, is epidemiology. Are you interested in potentially becoming an epidemiologist? Below, we take a closer look at what epidemiology is, what an epidemiologist does, and explore the relationship between epidemiology and public health. What is Epidemiology? Epidemiology is, at its heart, the science and study of disease. It is a branch of public health that studies disease and how it spreads amongst human and non-human populations, with the goal of using that understanding to help control and limit the spread of pathogens and other negative health problems. The Centers for Disease Control and Prevention (CDC) define epidemiology as “the method used to find the causes of health outcomes and diseases in populations. By definition, epidemiology is the study (scientific, systematic, and data-driven) of the distribution (frequency, pattern) and determinants (causes, risk factors) of health-related states and events (not just diseases) in specified populations (neighborhood, school, city, state, country, global).” As mentioned, though epidemiology is often focused on disease, it also includes other health concerns such as: Environmental exposures (e.g., lead, heavy metals, air pollutants) Foodborne illnesses (e.g., salmonella, E. coli) Injuries (e.g., homicide, suicide, or domestic violence) Non-infectious diseases (e.g., cancer, birth defects) Natural disasters (e.g., hurricanes, earthquakes) And more What Does an Epidemiologist Do? An epidemiologist is a public health worker who is responsible for investigating patterns in illness and injury. How they do this will depend on the populations they serve and the health challenges they are studying. That being said, epidemiologists will often: Conduct research Collect data Analyze that data Identify patterns and their underlying causes Communicate with other public health workers and policy makers Inform public health programs and initiatives The end goal of the epidemiologist is to reduce negative health outcomes while increasing positive health outcomes in the communities that they serve. How Much Does an Epidemiologist Make? Exactly how much an epidemiologist makes will depend on a number of factors, including the geographic location in which they work, the specific type of organization acting as their employer, and their level of experience. That being said, the average yearly salary of an epidemiologist in the United States is $107,110 according to Salary.com. Most epidemiologists will earn between $91,853 and $128,308 per year, though higher salaries are possible with increased experience. How to Become and Epidemiologist While there are many entry-level epidemiology positions available to those pursuing a degree in public health (i.e., contact tracing, census work, and data entry level positions), you will need to earn a master’s degree if you hope to become an epidemiologist who is in charge of an effort. Most typically, this will be a master’s degree in public health (MPH). During the course of earning your degree, you will gain a thorough understanding of the best research practices and protocols, informing you on everything that you do as an epidemiologist. Knowing that a master’s degree is required, you will of course need to first earn a bachelor’s degree. Earning a bachelor’s degree in public health can be an excellent choice, as it will directly prepare you for a career in epidemiology and empower you to work entry-level public health positions. This is not the only option though. Other potential undergraduate degrees might include biology or statistics. Epidemiology In Today’s Healthcare Epidemiology plays a vital role in today’s healthcare system by providing valuable information for healthcare managers and policymakers. It helps professionals predict health needs, understand health conditions, and identify relationships between the demand and need for healthcare services. This perspective is crucial for: Monitoring health conditions Assessing the impacts of programs Forming the scientific basis for policy making It’s also essential for controlling infectious diseases and integrating epidemiology into technology assessment and epidemiological research. For example, AI-powered machine learning is used for pandemic preparedness and response through more accurate projections in public health. It also identifies emerging areas beyond infectious disease management, such as the impacts of pandemics on mental health or chronic conditions. In addition, another technology regarded as GeoAI, combines spatial science, machine learning, and high-performance computing to extract knowledge from spatial big data. It is used in environmental epidemiology to model exposure and offers numerous advantages, like incorporating large amounts of data and scalability across different geographic areas. The Science of Public Health Epidemiologists play a crucial role in the field of public health, collecting and analyzing data about injury, disease, and other negative health outcomes and using that data to identify trends. Ultimately, the work of epidemiologists helps to control and reduce negative health outcomes and improve the health of their populations. As such, epidemiology is a crucial part of the field of public health, and a worthwhile career to consider. Related Blogs Earning a Master’s in Public Health Online Discover the benefits of pursuing an online Master's in Public Health (MPH) at Regis College. Regis College’s MPH Program: A Commitment to Real-World Impact Explore how Regis’s curriculum extends beyond the classroom to empower students with real-world impact through practical applications. Interdisciplinary Public Health Career Paths Discover unique public health career paths that extend beyond traditional clinical roles and make a meaningful impact on global health. Published Date
6114	https://www.youtube.com/watch?v=VVz0iFYhl88	Transformations of the Logarithmic Function Aidan Potts 210 subscribers 11 likes Description 591 views Posted: 4 Feb 2021 In this video I outline how to apply transformations to the logarithmic function using the mapping rule. I review the typical transformations for a function which are vertical stretch/compression, horizontal stretch/compression, vertical translation, and horizontal translation. I go over what form the function should be in to ensure accurate transformations of various points. Using the mapping rule, we take points on the parent function graph and transform those points to get the transformed graph. We go over two example problems that involve transforming the logarithmic graph. This series is tailored to cover all things related to logarithms. This involves, properties of logarithms, logarithm rules and laws, transformations of log functions, solving logarithmic equations and more! This content is a part of the Grade 12 Advanced Functions Curriculum (MH4FU). Please SHARE this video with a friend that would benefit and LIKE this video and SUBSCRIBE to this channel to see more videos just like this! Let me know in the comments below what type of video you want me to do next! Timestamps: 0:00 – Intro 0:43 – Review of Transformations 2:02 – Review of Mapping Rule 4:35 – Mapping Rule applied to Logarithm Function 7:06 – Example Problem #1 11:39 – Example Problem #2 17:07 – Conclusion 3 comments Transcript: Intro hey everyone welcome back to the channel so this is gonna be part two on all things related to logarithms and today we're going to be going over transformations of the logarithmic function by the end of this video you'll understand what each transformation does to the log function we'll also review the mapping rule and we'll learn how to graph these log functions we'll also go over two example problems where we practice what we've learned so far if you guys like this content i'd really appreciate you liking this video and subscribing to the channel to see more videos just like this alright let's get started [Applause] [Music] all right so at this point we've Review of Transformations transformed lots of different graphs we've done reciprocal and rational function transformations you could do transformations on sine graphs and trigonometric graphs and here this is actually no different so we're always going to see this form here this form of y is equal to a times the function k x minus d plus c so what that is is the proper form that allows us to understand what these transformations do and note carefully that this k is separate from this x here so we want we don't want them to switch like smushed together we're going to go over that in one of the example problems but everything stays the same so the absolute value of a so this value right here that represents the vertical stretch or compression the k value here that represents in the absolute value of it represents the horizontal stretch and compression you'll note this reflection here so if there's a negative in sign in front of the a or the k that's going to reflect the graph so if we have a negative sign in front of the a that's going to be a reflection over the x-axis if it's or if it's in front of the k variable then that's actually a reflection over the y-axis lastly we have the transformations involving translation so that's d so that's horizontal translation so shifting the graph left and right and we also have c which is vertical translation or shifting the graph up and down now i want to review the mapping rule so Review of Mapping Rule the mapping rule is saying that x comma y is going to be transformed by like the coordinates x comma y so a point of x comma y can be transformed to be x over k plus d that's the x coordinate now comma a times y plus c so this is basically just breaking down this transformed function and again we want the form of y is equal to a f k x minus d plus c and this f here can just be replaced with whatever function we're dealing with it could be replaced with a sine function uh you know a quadratic function a trigonometric function and now we're going to be applying it to the log function before we do that i want to just kind of review the mapping rule and how it's being used to find out the graphs of transform transformed functions so here if we look at this parent function of x squared right this is going to say that negative 2 negative 2 squared that's going to be 4. negative 1 negative 1 squared that's going to be 1. then 0 0 squared is 0 and so on and so forth now we're going to apply a transformation and you can note that this really represents the a the k the d value and the c value so it's of the form like we talked about before this a f k x minus d plus c and in this case we're just applying the quadratic function to it so what we're going to do for this mapping rule is going to we're going to look at the points on the parent function first and then we're going to apply these transformations both to the x coordinate and the y coordinate to get our transform graph so here we're going to have the first point of the transformed point is going to be well we'll have negative 2 negative 2 over k so in this case it's it's 2 as well and then it's going to be plus 4 because that's the d value that's a horizontal translation and it makes sense because the x coordinate represents horizontal on the graph right the horizontal kind of situation the x coordinate and the k value that represents horizontal compression and the d value represents horizontal translation so it makes sense that when we're applying these transformations the horizontal parts go to the x values and the vertical parts go to the y values so now for the y point what we're going to do is going to be 4 which is our y value multiplied by our a value which is 3 and then plus our c value which is three as well so here if we solve for this point it's gonna be negative two over two that's gonna be negative one plus three that's gonna be three next we're gonna have four times three is twelve plus three well that's gonna be fifteen and you can see that i did this previously but that is the point we get right there and that's kind of how you solve for these transformed graphs and find out what this graph is going to look like so Mapping Rule applied to Logarithm Function now we're going to apply it to the logarithmic function so now it's important to note all of our transformed variables or all of our kind of expressions here so our a value is 2 we're going to have a k value of 2 a d value of 4 and a c value of 5. once we have all that and we kind of check to make sure this is of the form that we're used to seeing so that y is equal to a f k x minus d plus c form and it is so now we can use the mapping rule to find out what these transformed points are going to have and once we have the points we can obviously just connect the points to get the graph so what we're going to do is first we're going to do 0.01 divided by because that's our x value that's our x value of our parent function divided by k so that's 2 and that's gonna be plus four so this comma and then we can do our y value so it's going to be negative two multiplied by a which is two and then we're going to plus the c value which is five so here we'll get the point of zero point so point zero one divided by two that's point zero zero five plus four this is going to be four point zero zero five comma negative four plus one that's gonna be one so we get the value of 4.005 comma 1. we can now apply the same strategy to the next point on the parent function so again we're going to look at 0.1 so we're going to be 0.1 divided by 2 plus 4. that's going to give us 4.05 next we're going to do negative 1 times 2 and then plus our c value which is 5 that's going to give us the value 3. so here we get 4.05 comma 3. again that's another point on our transform graph we can keep going and say 1 divided by 2 plus 4 well that's going to give us 4.5 here and then now it's going to be 2 times 0 because that's our parent point plus 5. so this will be 4.5 comma 5. next we can do 10 divided by 2 plus 4 this is 9 and then we're going to do 1 times 2 plus 5 this is 9 comma 7 that's another point there finally we can do 100 divided by 2 and plus 4 so we get 54 and then it's going to be 2 times 2 plus 5. so that's going to be 54 comma 9. and now what we've done is we've taken our transformed graph and we've just related it to our parent function and applied all transformation to each of the points so once we have all the points we can just connect them and that will be the graph of our transformed function Example Problem #1 okay so moving on to our example problem here we're asked to sketch the graph and the graph is log base 10 of negative 2x minus 4. and again i mentioned in the earlier part of this video that we want these x's and y or the k and x to be separate right we don't want them to be kind of smooshed together we have to have it always of this form so that's the first thing you check for when you look for a transformed graph so this is actually going to turn into and i'm just going to separate the k and the x variable now so this is log base 10 and this is going to be i'm going to factor out the negative 2 so then we have x plus 2 and that's going to be our new graph and this actually meets the form that we're looking for so once we have that form then we can just pull out the parameters we need so now i'm just going to write down all of the variables we have so we're going to have a is equal to well a is just equal to one our k value is equal to negative two and that negative represents a reflection in the y axis then we're going to have d well that actually equals negative two because of this positive usually a positive d is of the form negative so minus like two but in this case it's plus two so we know that that's actually a horizontal shift to the left and finally we have c and that is equal to zero because there's no vertical translation so now to apply the mapping rule what we're going to need to do is look at the parent points of our function so in this example the parent function is still y is equal to log x base 10. so we can actually just apply the same func or the same points we had in our previous example or kind of our our first step to understand what the mapping rule does to the logarithmic function and use that those points to now find the transformed function so here i'm going to take our a values k values d values and c values and apply the mapping rule so again i'm going to have 0.01 divided by well now k is negative 2 so we have to account for that negative and then it's going to be plus well actually it's going to be minus d because of this negative 2 here comma and then a y so negative two times one plus c and c is zero so that is going to be our point here this is gonna be point zero point zero one divided by negative two minus two so we get negative two point zero zero five comma negative two so that would be our first point here so now looking at our next point we're gonna have point one divided by two or actually negative two so it's gonna be negative point zero five and then it'll be minus two so here we'll get negative two point zero five and then if we look at our y variable our y point it's going to be negative one times one because a value is one and then plus zero so it's gonna be negative one like there now if we continue this function this is going to be one divided by two um so that's times negative one minus two so we get negative two point five then we're gonna have this is gonna be point zero because two times or 1 times 0 plus 0. lastly we'll have 10 divided by negative 2 minus 2 so this is going to be negative 7 and we're going to have 1 times 1 so this will be a value of 1 here our last point is going to be negative 50 minus 2 so it'll be negative 52 and then it'll be 2 times so this will be point number 2. now we have all of our points that we need to graph this function so now we're just going to draw a graph using all these points so now we can draw negative two point zero zero five negative two so we're gonna get a value that's probably right around there and then the next value is gonna be negative two point zero four negative one so it's going to be like right there it's gonna be very close to the same point next we have negative two point five comma zero so that will be right there um we have negative seven or sorry negative seven seven comma one so that's right here so this function is going to look something where we go through all four of these points and we could graph the other one it's just uh very far in the distance so be not very to scale here but it's gonna look something like that and our typical log function is of this form right here so we usually have a vertical asymptote like this right at zero we shifted this vertical asymptote two units right so it's gonna get very close to this value of negative two but never touch it and we also reflected it over the y-axis so i think you could see from these transformations that this blue graph you know it kind of makes sense it passes like the gut test all right so now we have an example Example Problem #2 problem where it's a word problem and it's talking about very similar principle that we've been doing but it's a little bit hidden in kind of the paragraph here so it says function f of x equals log of 10 or log of x base 10 has a point 10 comma 1 on its graph if f x is stretched vertically by a factor of 3 reflecting the x-axis horizon and then horizontally stretched by a factor of 2 horizontally translated 5 units to the right and vertically translated 2 units up determine the equation of the transform function the coordinates of image point transform from 10 comma 1 and lastly the domain and range of the function so basically throughout this paragraph you want to sift through the kind of important information so we're looking at this parent function here this is what we've been using before log 10 or log of x base 10 and then here we're looking for so it's vertically stretched by a factor of 3. so a is equal to 3 because a is responsible for vertical stretching and compression then is reflected in the x-axis so reflected in the x-axis that seems like a negative value that's going to be put on the negative a so it's now negative three so we also see that it's reflected in the x-axis so that's going to be a negative a value so now a is negative three it's then horizontally stretched by a factor of two and remember that k corresponds to horizontal stretching compression in this case we have k is equal to one half because it is stretched usually k is compressed so when it's stretched we know that it's the reciprocal of this factor it's then horizontally translated five units to the right so that would be a d value equal to 5 and then vertically translated 2 units up so that would be c is equal to 2. so now determine the equation of the transformed function well once we have all these variables this is actually pretty easy to do because we're just going to plug in these variables wherever we you know see this function or see the variable so we have y is equal to well our parent function i'll write that down first is log 10 that's the base that's the parent function right there now our a value is negative 3 right so that's where we put that then we have here our k value our k value is 1 over 2 then we have x minus d so x minus 5 in this case and then plus c which is plus 2 in this case so that would be our answer to a because we've just looked through and sifted through all of the important information to show this transformed function as an equation the next thing they ask is the coordinates of image point the coordinates of the image point transformed from 10 comma 1. so what they're saying is take 10 comma 1 of the parent function right the point on the parent function 10 comma 1 and then apply these transformations to see what where that point would be in this transform function so here this is just the mapping rule again so we're going to take 10 divided by k which in this case we had k to be one half so it's divided by one half or 0.5 and then it is plus 5 that is d comma then we have negative three that's our a value times the y value which is one plus two so we're gonna get ten divided by point five plus five so we're gonna get 25 comma negative 3 plus 2 that's going to be negative 1. that would be our answer to part b this is the transformed image now they talk about the domain and range of a function so we can say the domain typically for a logarithmic function there's a vertical asymptote right so the vertical asymptote says that x will never be that value and also there's no negative values on the log function right if we think about doing log base 10 of i don't know negative 3. again we have to talk about what that means we're saying to what power must i raise 10 to get negative 3 and you could put that power to anything you want it could be a negative number a very high number low number whatever you want you'll never get negative three so the domain is already restricted on the parent function for the logarithmic graph so here we just need to see that okay well if we've translated it five units to the right then we know that the domain is going to be x or it's going to be x is greater than or not equal to just greater than 5 and then we can do x represents or belongs to all real numbers so this would be the domain here because again we're looking at this function and we know the domain function is already restricted for its parent function now we just need to apply the transformations to that function so that is the domain right there the range is actually pretty easy because even though these graphs you know taper off pretty quickly if you look at this graph here this blue graph it's getting like it seems like there might be a horizontal asymptote just because of how much it continually like it basically stagnates becomes flat but it never actually becomes flat it'll always keep increasing in the y direction just like the exponential function will always keep increasing in the x direction but it's just going to keep going up at a very small rate each time in the x direction so we actually can say that the range of a logarithmic function is just y belongs to all real numbers and that would be our answer to that one Conclusion so i hope you guys got some value out of this video if you did i'd really appreciate you liking this video and subscribing to the channel to see more videos just like this do you guys have any ideas of what content you want me to cover next let me know in the comments down below thanks again and i'll see you guys in the next video
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6116	https://math.stackexchange.com/questions/2278991/there-is-an-even-prime-greater-than-2	proof verification - There is an even prime greater than $2$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more There is an even prime greater than 2 2 [closed] Ask Question Asked 8 years, 4 months ago Modified8 years, 4 months ago Viewed 2k times This question shows research effort; it is useful and clear -4 Save this question. Show activity on this post. Closed. This question needs details or clarity. It is not currently accepting answers. Want to improve this question? As written, this question is lacking some of the information it needs to be answered. If the author adds details in comments, consider editing them into the question. Once there's sufficient detail to answer, vote to reopen the question. Closed 8 years ago. Improve this question (NOTE: I am new to proof construction. Don't panic if your heart beat increase.) Proof 1: suppose x x is even and prime, then there is k k in N N such that x=2 k x=2 k But x is only divisible by itself and not 2. x 2!=k x 2!=k x x can not be even. Proof 2: x x is prime x=p q x=p q then either p=1 p=1 or q=1 q=1 x x is even x=2 k x=2 k p q=2 k p q=2 k (2)k p q=1(2)k p q=1 which is false proof-verification proof-writing prime-numbers Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited May 15, 2017 at 14:43 DivineCoderDivineCoder asked May 13, 2017 at 8:56 DivineCoderDivineCoder 105 5 5 bronze badges 5 The contradiction tells you exactly that it cannot exist. I'm sorry but I didn't get your point. Could you explain better what are you asking for?Ender Wiggins –Ender Wiggins 2017-05-13 09:05:26 +00:00 Commented May 13, 2017 at 9:05 It isn't clear (at least for me) what you're even trying to prove. Are you trying to prove the claim is false or else to prove it is true? This is the first thing that must be crystal clear before even reading your "proof".DonAntonio –DonAntonio 2017-05-13 09:06:12 +00:00 Commented May 13, 2017 at 9:06 I have a hard time following this. It sort of looks like you are arguing that if x>2 x>2 is both even and prime, then you get a contradiction.Mankind –Mankind 2017-05-13 09:07:03 +00:00 Commented May 13, 2017 at 9:07 @Mankind You are right, this is exactly what I am trying to do DivineCoder –DivineCoder 2017-05-13 10:06:42 +00:00 Commented May 13, 2017 at 10:06 Your mistake (after the recent edit) is in the line x=2 p q x=2 p q. All you know is that x=2 k x=2 k for some number k k. There aren't two numbers p p and q q to deal with. Please read the two day old correct answer below.Ethan Bolker –Ethan Bolker 2017-05-15 13:52:03 +00:00 Commented May 15, 2017 at 13:52 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Here's a proof that 2 2 is the only even prime. Let a∈N a∈N be prime and even. (We know this is okay to do, as there is at least one even prime) Assume a>2 a>2 Since a a is even, it can be written as a=2 k a=2 k for some k∈N k∈N. By assumption, k>1 k>1. a 2=k a 2=k, which implies that a a is divisible by 2 2. Since a a is prime, its only divisors are 1 1 and a a. However this is contradicted by the fact that 2\|a 2\|a and a>2 a>2. Thus the assumption that there is an even prime larger than 2 2 is false. NOTE: This proof was much more verbose than necessary, but it seemed you were in need a fully stated proof by contradiction argument. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 13, 2017 at 9:34 answered May 13, 2017 at 9:25 infinitylordinfinitylord 5,156 2 2 gold badges 25 25 silver badges 47 47 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. What? Whatever you're trying to do, it gets completely lost because you're not explaining what you're trying to accomplish, but just stating some things. And for the mathematical content, I gave up after: If x>n x>n then x n x n does not belong to N N. That's simply not true: 4>2 4>2 and 4 2=2∈N 4 2=2∈N. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 13, 2017 at 9:07 Arthur 205k 14 14 gold badges 186 186 silver badges 332 332 bronze badges answered May 13, 2017 at 9:07 Henrik supports the communityHenrik supports the community 7,077 13 13 gold badges 28 28 silver badges 34 34 bronze badges 2 I am to prove that the claim is false.DivineCoder –DivineCoder 2017-05-13 09:34:09 +00:00 Commented May 13, 2017 at 9:34 @user2240414 Make that perfectly clear in the body of your question before it gets closed...DonAntonio –DonAntonio 2017-05-13 09:51:09 +00:00 Commented May 13, 2017 at 9:51 Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions proof-verification proof-writing prime-numbers See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Prove that the order of U(n)U(n) is even when n>2 n>2. 3The Prime Numbers Set is infinite. Is this proof correct? 2Proof that odd + odd = even 2Determine if n is a prime. 2Relevance of prime being divisble by 4 k+1 4 k+1 in proof that 'There are infinitely many primes of the shape 4 k+3 4 k+3' 4Possible proof of Fermat's Last Theorem for prime exponents greater than 2 0Prove that every integer n > 1 is either prime or can be expressed as a product of primes. (Is proof valid?) 2There does not exist an even prime greater than two: is there an intuitionistic proof? Hot Network Questions How do you create a no-attack area? Why multiply energies when calculating the formation energy of butadiene's π-electron system? 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6117	https://kolibri.teacherinabox.org.au/modules/en-wikipedia_for_schools-static/wp/t/Tetrahedron.htm	Tetrahedron Related subjects: Mathematics Did you know... SOS Children, an education charity, organised this selection. Sponsoring children helps children in the developing world to learn too. \| Regular Tetrahedron \| \| --- \| \| (Click here for rotating model) \| \| \| Type \| Platonic solid \| \| Elements \| F = 4, E = 6 V = 4 (χ = 2) \| \| Faces by sides \| 4{3} \| \| Schläfli symbol \| {3,3} and s{2,2} \| \| Wythoff symbol \| 3 \| 2 3 \| 2 2 2 \| \| Coxeter–Dynkin \| \| \| Symmetry \| Td, A3, [3,3], (332) \| \| Rotation group \| T, [3,3]+, (332) \| \| References \| U01, C15, W1 \| \| Properties \| Regular convex deltahedron \| \| Dihedral angle \| 70.528779° = arccos(1/3) \| \| 3.3.3 ( Vertex figure) \| Self-dual ( dual polyhedron) \| \| Net \| \| A tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex. A regular tetrahedron is one in which the four triangles are regular, or "equilateral," and is one of the Platonic solids. The tetrahedron is one kind of pyramid, the second most common type; a pyramid has a flat base, and triangular faces above it, but the base can be of any polygonal shape, not just square or triangular. Like all convex polyhedra, a tetrahedron can be folded from a single sheet of paper. Formulas for regular tetrahedron For a regular tetrahedron of edge length : \| \| \| --- \| \| Surface area \| \| \| Volume \| \| \| Height \| \| \| Angle between an edge and a face \| (approx. 55°) \| \| Angle between two faces \| (approx. 71°) \| \| Angle between the segments joining the centre and the vertices \| (approx. 109.471°) \| \| Solid angle at a vertex subtended by a face \| (approx. 0.55129 steradians) \| \| Radius of circumsphere \| \| \| Radius of insphere that is tangent to faces \| \| \| Radius of midsphere that is tangent to edges \| \| \| Radius of exspheres \| \| \| Distance to exsphere centre from a vertex \| \| Note that with respect to the base plane the slope of a face () is twice that of an edge (), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof). Volume of any tetrahedron The volume of any tetrahedron is given by the pyramid volume formula: where A is the area of the base and h the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apexes to the opposite faces are inversely proportional to the areas of these faces. For a tetrahedron with vertices a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is (1/6)·\|det(a−b, b−c, c−d)\|, or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so where a, b, and c represent three edges that meet at one vertex, and is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 1/6 of the volume of any parallelepiped which shares with it three converging edges. It should be noted that the triple scalar can be represented by the following determinants: : or where is expressed as a row or column vector etc. : Hence : where etc. : which gives If we are given only the distances between the vertices of any tetrahedron, then we can compute its volume using the formula: If the determinant's value is negative this means we can not construct a tetrahedron with the given distances between the vertices. Distance between the edges Any two opposite edges of a tetrahedron lie on two skew lines. If the closest pair of points between these two lines are points in the edges, they define the distance between the edges; otherwise, the distance between the edges equals that between one of the endpoints and the opposite edge. Three dimensional properties of a generalized tetrahedron As with triangle geometry, there is a similar set of three dimensional geometric properties for a tetrahedron. A generalised tetrahedron has an insphere, circumsphere, medial tetrahedron and exspheres. It has respective centers such as incenter, circumcenter, excenters, Spieker centre and points such as a centroid. However there is, generally, no orthocenter in the sense of intersecting altitudes. There is an equivalent sphere to the triangular nine point circle that is the circumsphere of the medial tetrahedron. However its circumsphere does not, generally, pass through the base points of the altitudes of the reference tetrahedron. To resolve these inconsistencies, a substitute centre known as the Monge point that always exists for a generalized tetrahedron is introduced. This point was first identified by Gaspard Monge. For tetrahedra where the altitudes do intersect, the Monge point and the orthocenter coincide. The Monge point is define as the point where the six midplanes of a tetrahedron intersect. A midplane is defined as a plane that is orthogonal to an edge joining any two vertices that also contains the centroid of an opposite edge formed by joining the other two vertices. An orthogonal line dropped from the Monge point to any face is coplanar with two other orthogonal lines to the same face. The first is an altitude dropped from a corresponding vertex to the chosen face. The second is an orthogonal line to the chosen face that passes through the orthocenter of that face. This orthogonal line through the Monge point lies mid way between the altitude and the orthocentric orthogonal line. The Monge point, centroid and circumcenter of a tetrahedron are colinear and form the Euler line of the tetrahedron. However, unlike the triangle, the centroid of a tetrahedron lies at the midpoint of its Monge point and circumcenter. There is an equivalent sphere to the triangular nine point circle for the generalized tetrahedron. It is the circumsphere of its medial tetrahedron. It is a twelve point sphere centered at the circumcenter of the medial tetrahedron. By definition it passes through the centroids of the four faces of the reference tetrahedron. It passes through four substitute Euler points that are located at a distance of 1/3 of the way from M, the Monge point, toward each of the four vertices. Finally it passes through the four base points of orthogonal lines dropped from each Euler point to the face not containing the vertex that generated the Euler point. If T represents this twelve point center then it also lies on the Euler line, unlike its triangular counterpart, the center lies 1/3 of the way from M, the Monge point towards the circumcenter. Also an orthogonal line through T to a chosen face is coplanar with two other orthogonal lines to the same face. The first is an orthogonal line passing through the corresponding Euler point to the chosen face. The second is an orthogonal line passing through the centroid of the chosen face. This orthogonal line through the twelve point center lies mid way between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelve point centre lies at the mid point of the corresponding Euler point and the orthocenter for that face. The radius of the twelve point sphere is 1/3 of the circumradius of the reference tetrahedron. If OABC forms a generalized tetrahedron with a vertex O as the origin and vectors and represent the positions of the vertices A, B and C with respect to O, then the radius of the insphere is given by: and the radius of the circumsphere is given by: which gives the radius of the twelve point sphere: where: The vector position of various centers are given as follows: The centroid The circumcenter The Monge point The Euler line relationships are: It should also be noted that: and: Geometric relations A tetrahedron is a 3- simplex. Unlike in the case of other Platonic solids, all vertices of a regular tetrahedron are equidistant from each other (they are in the only possible arrangement of four equidistant points). A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual. A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are : (+1, +1, +1); : (−1, −1, +1); : (−1, +1, −1); : (+1, −1, −1). For the other tetrahedron (which is dual to the first), reverse all the signs. The volume of this tetrahedron is 1/3 the volume of the cube. Combining both tetrahedra gives a regular polyhedral compound called the stella octangula, whose interior is an octahedron. Correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e., rectifying the tetrahedron). The above embedding divides the cube into five tetrahedra, one of which is regular. In fact, 5 is the minimum number of tetrahedra required to compose a cube. Inscribing tetrahedra inside the regular compound of five cubes gives two more regular compounds, containing five and ten tetrahedra. Regular tetrahedra cannot tessellate space by themselves, although it seems likely enough that Aristotle reported it was possible. However, two regular tetrahedra can be combined with an octahedron, giving a rhombohedron which can tile space. However, there is at least one irregular tetrahedron of which copies can tile space. If one relaxes the requirement that the tetrahedra be all the same shape, one can tile space using only tetrahedra in various ways. For example, one can divide an octahedron into four identical tetrahedra and combine them again with two regular ones. (As a side-note: these two kinds of tetrahedron have the same volume.) The tetrahedron is unique among the uniform polyhedra in possessing no parallel faces. Related polyhedra Truncated tetrahedron Two tetrahedra in a cube Compound of five tetrahedra Compound of ten tetrahedra Intersecting tetrahedra An interesting polyhedron can be constructed from five intersecting tetrahedra. This compound of five tetrahedra has been known for hundreds of years. It comes up regularly in the world of origami. Joining the twenty vertices would form a regular dodecahedron. There are both left-handed and right-handed forms which are mirror images of each other. The isometries of the regular tetrahedron The proper rotations and reflections in the symmetry group of the regular tetrahedron The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron (see above, and also animation, showing one of the two tetrahedra in the cube). The symmetries of a regular tetrahedron correspond to half of those of a cube: those which map the tetrahedrons to themselves, and not to each other. The tetrahedron is the only Platonic solid that is not mapped to itself by point inversion. The regular tetrahedron has 24 isometries, forming the symmetry group Td, isomorphic to S4. They can be categorized as follows: T, isomorphic to alternating group A4 (the identity and 11 proper rotations) with the following conjugacy classes (in parentheses are given the permutations of the vertices, or correspondingly, the faces, and the unit quaternion representation): identity (identity; 1) rotation about an axis through a vertex, perpendicular to the opposite plane, by an angle of ±120°: 4 axes, 2 per axis, together 8 ((1 2 3), etc.; (1±i±j±k)/2) rotation by an angle of 180° such that an edge maps to the opposite edge: 3 ((1 2)(3 4), etc.; i,j,k) reflections in a plane perpendicular to an edge: 6 reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to −x): the rotations correspond to those of the cube about face-to-face axes The isometries of irregular tetrahedra The isometries of an irregular tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3-dimensional point group is formed. An equilateral triangle base and isosceles (and non-equilateral) triangle sides gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C3v, isomorphic to S3. Four congruent isosceles (non-equilateral) triangles gives 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D2d. Four congruent scalene triangles gives 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-group V4 ≅ Z22, present as the point group D2. Two pairs of isomorphic isosceles (non-equilateral) triangles. This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C2v, isomorphic to V4. Two pairs of isomorphic scalene triangles. This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C2 isomorphic to Z2. Two unequal isosceles triangles with a common base. This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group Cs isomorphic to Z2. No edges equal, so that the only isometry is the identity, and the symmetry group is the trivial group. A law of sines for tetrahedra and the space of all shapes of tetrahedra A corollary of the usual law of sines is that in a tetrahedron with vertices O, A, B, C, we have One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface. Putting any of the four vertices in the role of O yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity. One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is a half-circle. What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be a half-circle. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by this sine law further reduce the number of degrees of freedom, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional. Computational uses Complex shapes are often broken down into a mesh of irregular tetrahedra in preparation for finite element analysis and computational fluid dynamics studies. Applications and uses The ammonium ion is tetrahedral Several of the Tetrahedron scoring pieces and goals from the 2005 FIRST Robotics Competition game Packaging The company Tetrapak's Tetra Classic is in the shape of a tetrahedral. Chemistry The tetrahedron shape is seen in nature in covalent bonds of molecules. For instance in a methane molecule (CH4) the four hydrogen atoms lie in each corner of a tetrahedron with the carbon atom in the centre. For this reason, one of the leading journals in organic chemistry is called Tetrahedron. Ammonium is another example. Angle from the centre to any two vertices is , or approximately 109.47°. Electronics If each edge of a tetrahedron were to be replaced by a one ohm resistor, the resistance between any two vertices would be 1/2 ohm. Symbolism The tetrahedron represents the classical element fire. Games Especially in roleplaying, this solid is known as a d4, one of the more common polyhedral dice. Tetrahedrons constructed of 1 1/4" PVC pipe, which were known as 'tetras', were used as the main scoring object on the 2005 FIRST Robotics Competition game Triple Play. The object of the game was to stack these 'tetras' onto larger tetrahedron goals which here placed in a 3×3 matrix. Some Rubik's Cube-like puzzles are tetrahedral, such as the Pyraminx and Pyramorphix. Pyramid head from the Silent Hill games has a tetrahedral on top of his head. Fiction In the Xeelee Sequence of science fiction books by author Stephen Baxter, a blue-green tetrahedron is the symbol of free humanity. The Triforces of the Legend of Zelda cartoon series are green and red tetrahedrons. The triforce's representation in the actual game series (starting in A Link to the Past) is that of an unfolded tetrahedron. The story arc of the fan series "Star Trek: Hidden Frontier" evolves around gigantic ancient artifacts, which later become a centre part of the series. The artifacts are referred to as Tetrahedrons and have the shape of such a geometrical body. In the series, the Tetrahedrons possess the ability to produce a great deal of energy and are of unknown material and origin, however, they appear to be several million years old and it is outlined that these devices have been built by an ancient civilization. Retrieved from " Subjects Art Business Studies Citizenship Countries Design and Technology Everyday life Information Technology Language and literature Title Word Index
6118	https://puzzling.stackexchange.com/questions/99355/two-truths-and-a-lie-a-logic-puzzle	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Two Truths and a Lie: a logic puzzle Ask Question Asked Modified 1 year, 8 months ago Viewed 9k times 5 $\begingroup$ I'm crafting an answer to the game Two Truths and a Lie: each person playing the game says three statements (usually about themselves, though I'm ignoring that requirement for this), two of which are true, and one of which is false. I think I've got three statements for which there is exactly one consistent solution to which statement is false. Here they are: Either this statement is false and #2 is true, or this statement and #3 are both true. If this statement is true, then #1 is false and #3 is true. It is false that the preceding two statements are both true. I believe there is only one correct solution, (i.e. exactly one of the statements must be false and the others true to maintain consistency) but I'd like to make sure. :) logical-deduction liars Share Improve this question edited Jun 25, 2020 at 6:59 melfnt 5,17022 gold badges1414 silver badges6060 bronze badges asked Jun 25, 2020 at 3:03 Ashton WiersdorfAshton Wiersdorf 16111 silver badge44 bronze badges $\endgroup$ Add a comment \| 3 Answers 3 Reset to default 5 $\begingroup$ I don't believe there's a correct solution. First, regarding statement 2: Without even looking at the other statements, statement 2 can't be false because of Curry's paradox. Since every statement in your problem must be true or false, it must be true. Next, regarding statement 1: Since we already established statement 2 to be true, statement 1 must be false. But this means the clause "this statement is false and #2 is true" is true, so statement 1 must be true. This is a contradiction and the puzzle is thus unsolvable. Finally, regarding statement 3: Doesn't matter. There wouldn't be a solution no matter what statement 3 said. Share Improve this answer answered Jun 25, 2020 at 5:07 Joseph Sible-Reinstate MonicaJoseph Sible-Reinstate Monica 1,9441212 silver badges2525 bronze badges $\endgroup$ 6 $\begingroup$ I don't think this is a case of Curry's Paradox, (thanks for the link!) but let me check my discrete math textbook just in case. rot13(lbh nffhzr ahzore gjb gb or snyfr, ubj vf gung n pbagenqvpgvba?) $\endgroup$ Ashton Wiersdorf – Ashton Wiersdorf 2020-06-25 06:04:21 +00:00 Commented Jun 25, 2020 at 6:04 $\begingroup$ @AshtonWiersdorf rot13(Jurer qb V nffhzr ahzore gjb gb or snyfr?) $\endgroup$ Joseph Sible-Reinstate Monica – Joseph Sible-Reinstate Monica 2020-06-25 06:09:20 +00:00 Commented Jun 25, 2020 at 6:09 1 $\begingroup$ You are correctthis does have a problem. I forgot that you can rewrite P Q as ¬ P ¨ Q, so it is indeed always true. Thank you so much for catching that! $\endgroup$ Ashton Wiersdorf – Ashton Wiersdorf 2020-06-25 06:21:10 +00:00 Commented Jun 25, 2020 at 6:21 1 $\begingroup$ Wow. Yup. Curry's Paradox exactly. You are correct on all accounts. ð Any suggestions on how to fix it so it's still tricky? $\endgroup$ Ashton Wiersdorf – Ashton Wiersdorf 2020-06-25 06:31:33 +00:00 Commented Jun 25, 2020 at 6:31 1 $\begingroup$ @AshtonWiersdorf: It isn't tricky in the first place. #1 gives you all the information you need to solve the puzzle without even reading the other 2 statements. But perhaps statement 2 can just say: "#1 is false and #3 is true" $\endgroup$ musefan – musefan 2020-06-25 12:06:07 +00:00 Commented Jun 25, 2020 at 12:06 \| Show 1 more comment 1 $\begingroup$ Answer: This puzzle does work, the solution is that Statement 2 is the lie. For my explanation, I will write like one would write boolean expresions in java. If I wrote "1" that means "1 is true". Same idea with "2" or "3". "!" means not, so "!1" means "1 is not true". "&&" means "and", "\|\|" means "or". Parenthesis work like in math. "if(x){y}" means "if x is true, y must be true" "==" means "is equal to" (I hope that was obvious...), this is not java but "--->" means "simplifies to". Also "und" means "undefined" Translating the problem to java gives the following:Statement 1: ((!1 && 2) \|\| (1 && 3)) == 1Statement 2: if(2){!1 && 3} == 2 Statement 3: !(1 && 2) == 3 Game rules: if(1 && 2){!3}, if(1 && 3){!2}, if(2 && 3){!1}, if(!3){1 && 2}, if(!2){1 && 3}, if(!1){2 && 3} The best strategy is to assume each statement is false and use proof by contradiction to see if it works out or not. Step 1: Assume Statement 1 is false: Because of game rules, 2 && 3Lets falsify statment 1: (!1 && 2) \|\| (1 && 3) == 1(!1 && 2) \|\| (1 && 3) == false---> (true && true)\|\|(false && true) == false---> true \|\| false == false---> true == false This is clearly impossible so !1 == und. 1 must be true. Step 2: Assume Statement 3 is false (I'm skipping 2 for a reason): Because of game rules, 1 && 2Lets falsify statment 3: !(1 && 2) == 3 !(1 && 2) == false---> 1 \|\| 2 == true; This is because of some law that I forgot the name of, but if you work it out its true. I'll give the first comment with the name of the law a shoutout, thanks in advance.true \|\| true = true; substituted 1 and 2 because of game rulesThis obviously checks out, but before saying 3 can be false, we must check if 1 and 2 can be true. I have already proven that 1 must be true. Can two be true?if(2){!1 && 3} == trueund && 3 == true1 cannot be true, so 2 can't be true either. Therefore, 3 must be true. Step 3: Assume Statement 2 is false: Because of game rules, 1 && 3Lets falsify statment 2: if(2){!1 && 3} == 2if(2){!1 && 3 == true} == false ---> if(2){und && 3 == true} == false ---> if(2){und == true} == false This clearly works as undefined can never equal true. 2 must be false. All this work finally shows that: The only possible solution is that Statement 2 is the lie, and Statements 1 & 3 are both true. This was a great puzzle. I actually switched my answer 7 times in the more fun puzzle solve "Is it possible to use these three statments in a two truths and a lie?" when compared to "Which statment is a lie?" Thanks for the great puzzle, I had a lot of fun solving it! Share Improve this answer edited Jun 25, 2020 at 4:40 answered Jun 25, 2020 at 3:19 AnkitAnkit 1,30577 silver badges2222 bronze badges $\endgroup$ 7 $\begingroup$ accidentally hit post... give me a sec to write my explanation. $\endgroup$ Ankit – Ankit 2020-06-25 03:20:19 +00:00 Commented Jun 25, 2020 at 3:20 $\begingroup$ I don't think this is right, because rot13(fgngrzrag gjb orvat snyfr yrnqf gb pbagenqvpgvba). $\endgroup$ Joseph Sible-Reinstate Monica – Joseph Sible-Reinstate Monica 2020-06-25 05:08:18 +00:00 Commented Jun 25, 2020 at 5:08 3 $\begingroup$ This is the answer I was going for, but on further reflection, @JosephSible-ReinstateMonica's answer is the correct one, I believe. I'll rewrite this. Bit of a meta question: what's the best way to submit edits? Should I close this question and open another? Or just add UPDATE at the bottom and let you guys update your answers? (I'm new to this bit of SE) $\endgroup$ Ashton Wiersdorf – Ashton Wiersdorf 2020-06-25 06:22:46 +00:00 Commented Jun 25, 2020 at 6:22 1 $\begingroup$ @AshtonWiersdorf I think it is better if you create a new question. $\endgroup$ justhalf – justhalf 2020-06-25 08:58:00 +00:00 Commented Jun 25, 2020 at 8:58 1 $\begingroup$ @justhalf Rather than "undefined", perhaps it should say "There is as yet insufficient data for a meaningful answer" $\endgroup$ Chronocidal – Chronocidal 2020-06-25 12:16:42 +00:00 Commented Jun 25, 2020 at 12:16 \| Show 2 more comments 0 $\begingroup$ 1 is false Because statement 1 uses either Implying there are only two possibilities: 1-F 2=T implying (3=T) or 1=T 3=T implying (2=F). Another option could be theyre all false..but it leaves no room for that therefore falsifying itself in its rigidity. Either this statement is false and #2 is true, or this statement and #3 are both true. If this statement is true, then #1 is false and #3 is true. Assuming there is indeed one false answer this validates my prior reasoning thus being true It is false that the preceding two statements are both true. Assuming there is indeed one false answer this validates my prior reasoning thus also being true. Share Improve this answer answered Oct 18, 2023 at 10:00 CorinaCorina 1 $\endgroup$ 2 1 $\begingroup$ I don't understand your reasoning for 1 being false. What's the contradiction in there being two possibilities? $\endgroup$ Rand al'Thor – Rand al'Thor ♦ 2023-10-18 10:48:17 +00:00 Commented Oct 18, 2023 at 10:48 $\begingroup$ but then #1 is false and #2 is true, with means (by the first part of the OR condition) that #1 is true $\endgroup$ Alois Christen – Alois Christen 2023-10-18 11:24:18 +00:00 Commented Oct 18, 2023 at 11:24 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions logical-deduction liars See similar questions with these tags. 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6119	https://manual.cfturbo.com/en/md_dimensions_francisturbine.html	Impeller > Main dimensions > Francis Turbine > Dimensions Dimensions <<Click to Display Table of Contents>> Navigation:Impeller>Main dimensions>Francis Turbine> Dimensions In the panel Shaft, the required shaft diameter is computed. ¢Shaft/ Hub The hub diameter is determined by the user and depends on the shaft-hub connection. The main dimensions can be seen on Main dimensions panel. They can be recomputed by pressing the Calculate-button. The computation is based on "Euler's Equation of Turbomachinery", on the continuity equation and the relations for the velocity triangles as well as on the parameters given in the tab sheets Setup and Parameters. Individual main dimensions can be calculated separately using the button inside the value field. You may accept the proposed values or you can modify them slightly, e.g. to meet a certain normalized diameter. In case the checkbox Automatic is activated a new calculation will accomplished after any change of parameter. Then the manual alteration of the main dimensions is not possible. Neighboring components In specific cases the dimensions of the neighboring components at inlet and/ or outlet can be used to get exactly matching geometry. This feature is available only for explicitly uncoupled components or side-by-side impellers. Information In the right panel of any tab sheet an information panel is situated, which holds the computed variables in accordance to the actual state of design, the resulting Meridional section as well as the Cordier-Diagramm with the location of the best point. These three sections can be chosen by the appropriate soft buttons in the heading. In the Value section the following variables are displayed for information which result from calculated or determined main dimensions: Flow properties at: impeller inlet impeller outlet static pressure p total pressure p t volume flow Q mass flow ṁ swirl s Velocity triangles at: impeller inlet impeller outlet velocities u, c m, c m, c u, c, w u, w flow angles α, β Work coefficient Flow coefficient Meridional flow coefficient Specific diameter Inlet velocity Inlet velocity (net) Outlet velocity Outlet velocity (net) Outlet width ratio Meridional velocity ratio Reynolds number with d 1 = highest diameter at inlet with b 1 = width at inlet with d 2 = highest diameter at outlet with b 2 = width at outlet The Meridional preview is until now based on the main dimensions only. The Cordier diagram can be used for checking the impeller diameter d 2. See Cordier. The Velocity triangles are the result of a mid-span calculation and are based on the design point and the main dimensions. Possible warnings \| Problem \| Possible solutions \| --- \| \| Inlet/ Outlet: Pressure and temperature not within the liquid phase of the fluid. Check inlet condition and geometry. \| \| If the fluid model = CoolProp, fluid properties will be calculated as function of pressure and temperature. In case the local pressure and temperature are not above the vapor pressure curve, the fluid state is not valid. \| Increase main dimensions (width, diameter etc.) or change Global setup (e.g., decrease mass flow, increase pressure, decrease temperature). \| \| In/outlet cross section ratio is unusual. Meridional contour should be defined with an appropriate in/outlet width. \| \| Inlet and outlet cross section area are not in the same order of magnitude. The ratio A 2/A 1 is outside the range of 0.1 to 10. This can result in high level of meridional acceleration or deceleration. \| You may change the in/ outlet geometry or ignore the warning because the real cross sections will be estimated based on the meridional contour in subsequent design steps. \| URL of this page:
6120	https://arxiv.org/pdf/math-ph/0011021	%PDF-1.4 %Çì¢ %%Invocation: path/gs -P- -dSAFER -dCompatibilityLevel=1.4 -q -P- -dNOPAUSE -dBATCH -sDEVICE=pdfwrite -sstdout=? -sOutputFile=? -P- -dSAFER -dCompatibilityLevel=1.4 ? 5 0 obj <> stream xÕZK³µ&ÝÉ:ûYÎI8²Ôz'6P87qØ×p\_6¾o¿¯õIçÌÅN!Ë>cÔjuýuKgjü§ü^n¤°ÓÍ³JíSù9½îmî}&ÓÑã ^L;£i"¦ËÍ\|üüïO¿ýÝåñóÝÍù=)¤¾¥i"3=:~¶G\_mÞ;Ú\| wO'å)kx¦¬±9éu<ßgó;ÛE'Lðóx6DkæãíN ££¡ùI~ ^Í·[Riïç/ù)hrós~]º/Ñà½I 2j~w»³Bv,\Õ¯1yp¤©'i]Q]OgÃ zç0ôc²æ¿Ä ¼¶ØRp2«¬TýÑ) LoX+É¬,<êÈiF¸W"m:W[ WyP¢ ¤)ÅN©²³5à}¸ù¯á´2t¦ÊOs-Nh&ÎSÅ»Æ<]l7.J}ëøH!&l:ÍbÌ¦ø &"ï´ÎÞHëÚ<üíVÝ:{0,ûÙYQÀn oï0Ã\_É»Ñ½É´vJjÚÆ+Eø#eG&?¤Åöf'/¢?) 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6121	https://study.com/academy/lesson/order-of-operations-with-fractions.html	Order of Operations with Fractions \| PEMDAS, Examples & Practice - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses Math for Kids Order of Operations with Fractions \| PEMDAS, Examples & Practice Contributors: Anita Dunn, Mark Boster Author Author: Anita Dunn Show more Instructor Instructor: Mark Boster Learn about the order of operations with fractions. Discover how to use PEMDAS with fractions, improper fractions, and mixed numbers, and examine a few examples. Updated: 11/21/2023 Table of Contents What is the Order of Operations with Fractions? Understanding PEMDAS with Fractions Order of Operations Fractions Examples Lesson Summary Show FAQ Do you add or subtract fractions first? That depends on the problem. In the order of operations, addition and subtraction are both on the same level of importance. Addition and subtraction should be carried out as you move from left to right. Do you multiply or divide fractions first? That depends on the problem. When using the order of operations, multiplication and division are on the same level of importance. Therefore, multiplication and division should be done as they occur when moving from left to right. Do you use PEMDAS with fractions? Yes. PEMDAS is the acronym often used for the order of operations. The order of operations should be used for fractions, the same as they are used for whole numbers. Create an account LessonTranscript VideoQuizCourse Click for sound 3:39 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. 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Your next lesson will play in 10 seconds 0:04 Why Order Matters 1:31 Some Practice 2:57 Lesson Summary QuizCourseView Video Only Save Timeline 51K views Recommended lessons and courses for you Related LessonsRelated Courses ##### Brackets in Math \| Definition, Types & Examples 4:35 ##### How to Simplify Expressions with Integers 5:12 ##### Order of Operations With Signed Numbers 6:34 ##### Solving Problems With More Than One Operation: Lesson for Kids 4:06 ##### PEMDAS \| Meaning, Rule & Examples 9:07 ##### Order of Operations in Math \| Steps & Examples 5:50 ##### Number Model \| Definition, Types & Examples 4:48 ##### Showing Students How to Correct Basic Calculation Errors 5:54 ##### Mental Math: Multiplication and Division 4:49 ##### Mental Math: Addition and Subtraction 4:34 ##### Multiplication & Division Fact Families \| Overview & Uses 3:01 ##### Fact Families: Addition & Subtraction \| Definition & Examples 3:32 ##### Mental Math \| Definition, Process & Examples 5:42 ##### Integer Operations \| Rules & Examples 6:53 ##### Multiplying & Dividing Negative Numbers 4:50 ##### The Relationship Between Multiplication & Division 3:50 ##### Multiplying & Dividing Rational Numbers \| Process & Examples 6:19 ##### Missing Quantities in Multiplication & Division Problems: Lesson for Kids 3:31 ##### Arithmetic \| Operators, Properties & Sample Problems 5:02 ##### Arithmetic with Whole Numbers 9:43 ##### MEGA Elementary Education Study Guide and Test Prep ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - Number & Quantity: High School Standards ##### Common Core Math - Algebra: High School Standards ##### Common Core Math - Statistics & Probability: High School Standards ##### Common Core Math - Geometry: High School Standards ##### Common Core Math - Functions: High School Standards ##### Common Core Math Grade 8 - Expressions & Equations: Standards ##### Common Core Math Grade 8 - Functions: Standards ##### MEGA Elementary Education Study Guide and Test Prep ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - Number & Quantity: High School Standards ##### Common Core Math - Algebra: High School Standards ##### Common Core Math - Statistics & Probability: High School Standards ##### Common Core Math - Geometry: High School Standards ##### Common Core Math - Functions: High School Standards ##### Common Core Math Grade 8 - Expressions & Equations: Standards ##### Common Core Math Grade 8 - Functions: Standards What is the Order of Operations with Fractions? ----------------------------------------------- Imagine a pizza. If that pizza is divided into equal pieces, slices of pizza have been created. Similarly, if a whole unit is broken into individual parts, then fractions of that whole unit have been created. A fraction is written in the form a b where a and b are whole numbers, and b is not equal to zero. The top number, the number represented by a, is the numerator, while the bottom number, represented by b, is the denominator. Figure 1: Circle broken into fractional parts. Recall that a mathematical operation includes things like addition, subtraction, multiplication, and division. A mathematical expression will sometimes contain various different operations. The order in which the operations are carried out will often change the solution to the expression. For example, consider the following expression. 2+3∗5. If one first carries out the addition and then the multiplication this expression simplifies as follows. 2+3∗5=5∗5=25. Meanwhile, if instead multiplication is done first followed by addition the expression simplifies to a different number. 2+3∗5=2+15=17. Clearly, this is not ideal. Each expression should be able to be simplified to one solution, and one solution only. This is why the order of operations is so important. The order of operations is like a recipe for a cake. The order in which ingredients are added to a cake batter will change the outcome, and putting the ingredients in the batter in the wrong order will sometimes make the cake not turn out correctly. In the same way, the order of operations needs to be followed so that the expression is solved correctly. The order of operations will be discussed in more detail in the next section, but looking at the expression in the example above: multiplication should be done before addition, so 17 is the correct answer. Understanding PEMDAS with Fractions ----------------------------------- The order of operations is often remembered by either the acronym PEMDAS or the statement "Please excuse my dear aunt Sally." Order of operations: Perform any operations within parenthesis or other grouping symbols such as square brackets. Simplify fractions with exponents or square roots Multiply or divide, moving left to right Add or subtract, moving left to right. Note that the acronym stands for P - parenthesis, E - exponents, MD - multiply or divide, AS - add or subtract. Note that even though M comes before D in the acronym, this does not necessarily mean that multiplication always comes before division. As the order of operations steps notes, multiplication and division should be carried out from left to right. For example: 4÷2×3=2×3=6 Note that the order of operations stays the same, whether the numbers in question are whole numbers, fractions, mixed numbers, or even decimals. No matter what, the same steps need to be carried out in the order given. Mistakes When Using the Order of Operations to Simplify Fractions A few types of mistakes are common when simplifying fractions. It is important to perform each step carefully without skipping ahead to the next step before it should be done. Example: Simplify (1 2+1 3)×6 7. Using the order of operations, first, complete the operation that is located within the parenthesis. Note that this involves adding two fractions with unlike denominators. Therefore, a common denominator needs to be found. The least common denominator of 2 and 3 is 6. (1 2+1 3)×6 7=(1×3 2×3+1×2 3×2)×6 7=(3 6+2 6)×6 7=5 6×6 7 Now, the multiplication can be completed. Note that the denominator of the first fraction, 5 6 and the numerator of the second fraction, 6 7 are the same number, namely 6. Therefore, one can cancel the numerator and the denominator and get the following. 5 6×6 7=5 7 So what could go wrong with this example? Say that instead of completing the operation within the parenthesis first, the person completing the problem instead looks at the multiplication part of the problem and notes that some canceling could be done. Note that this is the incorrect method to solve this problem. (1 2+1 3)×6 7=(1 2+1⧸3 1)×⧸6 2 7 Now say that the person realizes that the addition needs to be completed, except now the numbers being added are 1 2+1 1, which, after finding the common denominator 2, becomes: (1 2+1 1)×2 7=3 2×6 7=18 14=9 7 Clearly, this is not the same as 5 7, which is the correct answer. So what happened? By canceling the numerator and the denominator before the addition was carried out, the order of operations was not followed. Instead, part of the multiplication was handled, and then the parenthesis, and then the rest of the multiplication was completed. This is why the incorrect answer was found. Order of Operations Fractions Examples -------------------------------------- Let's go over some order of operations practice problems. Example 1 Simplify: 1 2+5÷2 3 Note that there are no parenthesis or exponents in this problem. Therefore, the next step would be to complete any multiplication and division. Recall that when dividing fractions, first flip the second fraction and change the division to multiplication. 1 2+5÷2 3=1 2+5×3 2. To more easily complete this multiplication, change the 5 into a fraction by putting a 1 in the denominator. 1 2+5 1×3 2=1 2+15 2 Now any addition or subtraction can be completed. Note that the fractions already have a common denominator. 1 2+15 2=16 2=8 Example 2 Simplify: (1 5×10 7)2÷3 5 First, anything found within the parenthesis needs to be completed. (1 5×10 7)2÷3 5=(1⧸5 1×⧸10 2 7)2÷3 5=(2 7)2÷3 5. Next comes the exponent. (2 7)2÷3 5=4 49÷3 5. Now, comes any multiplication or division. In this case, division is the only operation left. Again, to divide a fraction, flip the second fraction and change the operation to multiplication. 4 49÷3 5=4 49×5 3=20 147 Example 3 Simplify: 3 5÷1 4×1 2 Note that there are no parenthesis or exponents in this problem. It may be tempting to start with the multiplication, because in PEMDAS the M comes before the D. But in actuality, in order of operations multiplication and division are on the same level of importance. Therefore, multiplication and division should be completed as they occur moving left to right. Thus, in this example division should be done before multiplication. 3 5÷1 4×1 2=3 5×4 1×1 2=12 5×1 2=12 10=6 5 Lesson Summary -------------- Fractions show portions of a whole and are any number that can be represented by a b where a and b are whole numbers and b is not zero. The order of operations refers to the order that operations should be done in order to solve the problem. Order of operations should be used when solving expressions involving fractions and mixed numbers, just as they are used for expressions that include only whole numbers. The order of operations is often remembered with the acronym PEMDAS, which stands for parenthesis, exponents, multiplication and division (which should be completed left to right), and addition and subtraction (which should also be completed left to right). Video Transcript Why Order Matters Matter, really, does, what in order go you it? Wait, none of that made any sense! Let's put that sentence in order! 'Does it really matter what order you go in?' Clearly, of course it does! Try to eat a banana before you peel it or a watermelon before you cut it. Well, order really does matter when adding, subtracting, multiplying, or dividing fractions. Just like we put our socks on before our shoes, when solving equations, even ones that contain fractions, order matters. Look at Equation 1 below. You can see it does matter in what order you solve the equation. Example 1 When it was solved the first time, the multiplication was done before the addition. The second time it was solved, addition was done before multiplication! So what order should you go in? Just like when you are solving equations with whole numbers, solving equations with fractions has the same order of operations. The order of operations is the order in which you solve the problem. If numbers are in parentheses, you do them first. Second, you solve for any numbers with exponents. Then, you go from left to right and do the multiplication and division. Finally, you go back and do the addition and subtraction - again, from left to right. Sounds like a lot, right? Well, there is a very popular mnemonic device to help you remember the order of operations: PEMDAS (Please Excuse My Dear Aunt Sally) helps us remember: P - Parentheses E - Exponents M & D - Multiplication and Division (left to right) A & S - Addition and Subtraction (left to right) Some Practice How about trying There is both addition and multiplication in this equation, so what's the first step? Remember your PEMDAS. There are no parentheses or exponents, so multiplication comes first. As we can see: So now you have the first part done. So now you can go ahead and continue with the addition part of the equation. As we can see: And that's our answer! Let's try another one with: A little different this time. What do you do first? It's always whatever's inside the parentheses. So, therefore: Now, outside the parentheses: Not too difficult. Sometimes the equations look tricky, but after you know the order of operations, they aren't that bad at all. Check out the Order of Operations with Fractions diagram: It's important that you remember this order. Maybe looking at the bottom where is tells you in symbols what to do will help you remember it. If all else fails, you always have PEMDAS and your 'Dear Aunt Sally!' Lesson Summary Let's take a couple of moments to review what we've learned about the order of operations we use when working with fractions. The order of operations for mathematical equations, including fractions, is the set of rules which determine the order in which you solve the problem. PEMDAS (Please Excuse My Dear Aunt Sally) is a popular mnemonic device used to help you remember the correct order. It stands for: P = Parentheses E = Exponents M & D =Multiplication and Division (left to right) A & S = Addition and Subtraction (left to right) Using PEMDAS, you should always remember how to solve a fraction question when facing one on a test. Register to view this lesson Are you a student or a teacher? 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Try it now Math for Kids 23 chapters 325 lessons Chapter 1 Numbers for Elementary School Ordinal Numbers: Lesson for Kids 3:30 min Binary Numbers: Lesson for Kids Triangular Numbers: Lesson for Kids Prime Numbers: Lesson for Kids Whole Numbers: Lesson for Kids 2:57 min Pi Lesson for Kids Facts & History Composite Number Lesson for Kids Fibonacci Numbers in Nature: Lesson for Kids Fibonacci Sequence Lesson for Kids: Numbers & Facts Number Line: Lesson for Kids Place Value Games for Kids Place Value: Lesson for Kids 2:48 min Roman Numerals: Lesson for Kids 5:59 min Rounding Numbers: Lesson for Kids Digit in Math \| Meaning & Examples 3:30 min 9/4 as a Mixed Number: How-to & Steps 2:49 min Place Value and Patterns with Rounding Number Line for Addition & Subtraction \| Overview & Examples 2:57 min How to Use a Number Line Counting on the Number Line Numbers 1-50: Lesson for Kids 3:17 min Numbers 1-20: Lesson for Kids Numbers 1-100: Lesson for Kids 2:53 min Numbers 1-30: Lesson for Kids How to Count: Lesson for Kids Chapter 2 Geometry for Elementary School Surface Area Lesson for Kids 3:58 min Types of Angles: Lesson for Kids 3:17 min Types of Triangles: Lesson for Kids Geometric Shapes: Lesson for Kids Isosceles Triangle Lesson for Kids How to Use a Protractor: Lesson for Kids 3-D Shapes: Lesson for Kids 2:59 min Pentagon Shapes: Lesson for Kids 2:16 min Polygon Shapes: Lesson for Kids Diamond Shapes: Lesson for Kids 2:42 min Measuring Angles: Lesson for Kids Circle Definition: Lesson for Kids 3:10 min Circumference Lesson for Kids: Definition & Formula Cubes: Lesson for Kids Properties of a Circle: Lesson for Kids 2:21 min Rectangle Definition: Lesson for Kids Triangle Lesson for Kids: Definition & Facts 3:06 min 2-D Shapes: Lesson for Kids How to Find the Area of a Triangle: Lesson for Kids 3:01 min How to Find the Volume of a Cylinder: Lesson for Kids Pentagon Shape Facts: Lesson for Kids 2:39 min Vertices: Lesson for Kids 3:22 min What is a Regular Polygon? - Lesson for Kids 3:01 min Pythagorean Theorem Proof: Lesson for Kids Pythagorean Theorem Lesson for Kids: Definition & Examples 4:06 min Perimeter of a Pentagon \| Overview, Formula & Examples 2:43 min Pentagon Polygon \| Definition, Types & Examples 3:31 min Geometry Definition: Lesson for Kids Names of Polygons 3:18 min What is a Venn Diagram? - Lesson for Kids 3:32 min Triangle Sum Theorem \| Definition, Proof & Diagrams Facts About Right Angles 3:09 min Octagon Lesson for Kids: Definition & Facts 2:38 min Hexagon Lesson for Kids: Definition & Facts 2:29 min Perimeter Lesson for Kids: Definition & Examples 3:42 min 4-Sided Polygons \| Overview & Shapes 3:55 min Dodecahedron \| Definition, Faces & Examples 2:46 min 3-4-5 Triangle \| Definition, Rules & Examples 4:11 min Triangle Sum Theorem \| Definition & Examples Dodecahedron \| Definition, Properties & Examples How to Calculate the Area of a Hexagon 3:38 min How to Divide an Angle into Two Equal Angles What is a Congruent Angle? - Lesson for Kids Chapter 3 Fractions for Elementary School Comparing Fractions: Lesson for Kids Equivalent Fractions: Lesson for Kids 3:21 min How to Add Fractions: Lesson for Kids Fractions Games for Kids Fractions to Decimals: Lesson for Kids 4:13 min Fractions: Lesson for Kids 2:39 min Numerator & Denominator Lesson for Kids 3:16 min Decimals: Lesson for Kids 3:42 min Multiplying Fractions by Whole Numbers: Lesson for Kids 3:04 min How to Simplify Fractions: Lesson for Kids Improper Fractions: Lesson for Kids Benchmark Fractions \| Definition, Uses & Examples 3:48 min Adding Mixed Numbers \| Steps & Examples Multiplying Compound Fractions Simplifying Compound Fractions Dividing Improper Fractions \| Keep-Change-Flip Method & Examples 4:21 min Adding Compound Fractions Subtracting Compound Fractions How to Subtract Fractions with Variables \| Examples 5:55 min Multiplying Fractions with Common Denominators \| Steps & Examples 5:05 min Subtracting Fractions with Like Denominators Reducing Fractions: Rules & Practice Definition of Simplest Form: Lesson for Kids Adding Fractions with Unlike Denominators \| Overview & Examples 4:51 min Adding & Subtracting Fractions with Unlike Denominators 3:50 min Subtracting Fractions \| Rules & Examples Rules for Multiplying Fractions \| Steps & Examples 4:03 min Rules for Dividing Fractions Plotting Fractions on a Number Line \| Methods & Examples Locating Fractions on a Number Line How to Add & Subtract Two Fractions with Like Denominators How to Find Equivalent Fractions on a Number Line 5:22 min How to Subtract Mixed Fractions with Unlike Denominators Chapter 4 Math Basics for Elementary School Subtraction: Lesson for Kids Expanded Form: Lesson for Kids 3:26 min Logic Problems: Lesson for Kids How to Do Long Multiplication: Lesson for Kids What is Multiplication? - Lesson for Kids Addition: Lesson for Kids 3:30 min Multiplication Properties Lesson for Kids Math Card Games for Kids Telling Time: Activities & Games for Kids Telling Time: Lesson for Kids 3:32 min Counting Money: Lesson for Kids Telling Time to the Minute: Lesson for Kids Divisibility Rules for 10: Lesson for Kids Divisibility Rules for 2: Lesson for Kids 2:51 min Divisibility Rules for 5: Lesson for Kids Division Lesson for Kids: Definition & Method 3:04 min Long Division Steps: Lesson for Kids 3:44 min Estimation: Lesson for Kids 3:44 min Lattice Method of Multiplication \| Definition, Steps & Examples 3:50 min Multiplication Methods & Types 3:52 min Ways to Divide & Types of Division \| Steps & Examples 3:47 min Theorem \| Meaning, Types & Examples 3:11 min Subtraction with Regrouping \| Overview & Examples 3:01 min Dividing a Whole Number by a Decimal \| Steps & Examples 4:46 min Regrouping in Addition \| Definition & Examples 3:09 min Finding the Missing Addend 3:06 min Metric vs. Imperial System \| Units & Measurement 3:46 min History of the Metric System: Lesson for Kids How Money is Made: Lesson for Kids 3:54 min Chinese Multiplication Method \| Overview & Examples 3:46 min Column Addition Method \| Concept & Examples 4 Digit Subtraction with Regrouping 4:19 min How to Multiply Using Expanded Form 3:52 min Writing Numbers in Expanded Form \| Overview & Examples 3:58 min Famous Mathematicians: Lesson for Kids Writing Decimals in Expanded Form \| Steps & Examples 3:30 min Length Lesson for Kids: Definition & Measurement 3:12 min 4 Digit Addition with Regrouping Expanded Notation Method for Division 4:37 min Expanded Notation Method for Multiplication 3:23 min Multiplication & Division Fact Families \| Overview & Uses 3:01 min Fact Families: Addition & Subtraction \| Definition & Examples 3:32 min 4 Digit by 1 Digit Multiplication Metric Conversions: Lesson for Kids 3:22 min How to Divide by Double Digit Numbers 5:58 min Two-Step Math Word Problems 4:03 min How to Borrow in Math Math Word Problems: Lesson for Kids How to Learn Times Tables Box Method Multiplication \| Definition, Steps & Examples 2:29 min 4 Digit by 2 Digit Multiplication 4:06 min Equivalent in Math \| Definition, Numbers & Fractions 4:13 min 1st Grade Math Vocabulary: Lesson for Kids Chapter 5 Statistics for Elementary School Algorithm Lesson for Kids Arrays: Lesson for Kids Probability Lesson for Kids: Examples & Definition 3:52 min What Is Range in Math? - Lesson for Kids 3:11 min Line Graphs: Lesson for Kids 3:39 min Different Types of Graphs: Lesson for Kids 3:37 min Statistics: Lesson for Kids Five Number Summary \| Definition, Conditions & Calculation Chapter 6 Number Properties for Elementary School Associative Property Lesson for Kids 3:55 min Commutative Property: Lesson for Kids 3:45 min What is a Factor? - Lesson for Kids 3:38 min What is a Prime Factor? - Lesson for Kids 3:07 min What are Factors & Multiples? 3:29 min Chapter 7 Algebra for Elementary School History of Algebra: Lesson for Kids Math Formulas: Lesson for Kids Distributive Property: Lesson for Kids Algebra Lesson for Kids: Facts & Rules 4:18 min Foil Method in Math \| Definition & Examples 3:26 min Chapter 8 Math Patterns for Elementary School Fractals Lesson for Kids Fractal Art \| Meaning, Types & Uses Semi-Regular Tessellation \| Definition, Types & Examples Tessellation Shapes, Patterns & Examples 3:39 min Patterns in Math \| Overview, Rule & Types 3:36 min Pascal's Triangle Lesson for Kids: Definition & History Chapter 9 History of Math for Elementary School Al Khwarizmi Lesson for Kids: Biography & Facts Abacus Lesson for Kids: History & Uses Pythagoras Lesson for Kids: Biography & Facts Chapter 10 Math Terms for Elementary School Definition of Like Denominators Attribute in Math \| Definition, Shape & Examples What Is Breadth in Math? 3:01 min Chord in Math \| Definition, Theorems & Calculation Cluster in Math \| Overview & Examples 2:36 min Difference in Math \| Overview, Subtraction & Examples 2:17 min What is Dimension in Math? \| Concept and Examples 2:09 min Edges in Math \| Definition, Identification & Examples 2:20 min Elements of a Set \| Definition & Examples 3:06 min Equation Lesson for Kids: Definition & Examples 3:49 min Exponent Definition: Lesson for Kids Face in Math \| Overview & Shapes Function in Math \| Definition, Types & Examples 2:15 min Infinity in Math \| Definition, Symbol & Signs 2:52 min Integers Lesson for Kids Interval in Math \| Definition & Examples 3:40 min Median Definition: Lesson for Kids What Does Mode Mean in Math? - Lesson for Kids Net in Math \| Definition & Examples Outcome in Math \| Definition & Events Point in Math \| Definition, Uses & Examples 2:14 min Proper Factor in Mathematics \| Overview, Facts & Examples Proportion Lesson for Kids: Definition & Examples Quantity in Math \| Definition, Uses & Examples 2:50 min Quotient Lesson for Kids: Definition & Rule Revolution in Math \| Definition, Types & Angles 5th Grade Math Vocabulary: Lesson for Kids 4th Grade Math Vocabulary: Lesson for Kids 3rd Grade Math Vocabulary: Lesson for Kids 2nd Grade Math Vocabulary: Lesson for Kids Chapter 11 Working with Numbers for Elementary School Standard Algorithm for Addition 3:32 min Standard Algorithm in Math \| Meaning & Examples 2:53 min Standard Algorithm for Division 3:25 min Absolute Value: Lesson for Kids Ascending Order Definition & Examples 3:07 min Decomposing in Math \| Definition, Process & Examples 3:24 min Descending Order \| Definition, Numbers & Examples 3:35 min The Fundamental Theorem of Arithmetic 4:03 min Golden Mean Definition, Uses & Examples Power of a Number \| Overview & Examples 3:38 min Logical Thinking & Reasoning Questions: Lesson for Kids 2:51 min Vertical Addition & Subtraction 3:39 min Scale Factor Lesson for Kids: Definition & Examples Square Roots: Lesson for Kids Chapter 12 Types of Numbers for Elementary School What is an Abundant Number? What is a Base Number? 2:32 min Cardinal Numbers \| Overview, Definition & Examples 3:06 min Consecutive Numbers \| Definition, Example & Formulas 3:22 min Finding the Sum of Consecutive Numbers 4:10 min What is a Cubed Number? Deficient Numbers \| Definition, Properties & Examples What Are Figurate Numbers? - Definition & Examples Opposite Numbers \| Definition & Facts 2:53 min Rectangular Numbers \| Definition, Properties & Types 3:05 min Like & Unlike Terms Is Zero an Integer? 2:17 min What is 0? \| Definition & Types 2:37 min Sieve of Eratosthenes: Lesson for Kids What Are Twin Prime Numbers? 3:22 min Chapter 13 Measurements for Elementary School Acre \| Definition, Area & Measurement 3:10 min Capacity Lesson for Kids: Definition & Facts 2:25 min Centigrade Definition, Conversion & Facts Centimeter Definition, Symbol & Conversion 2:06 min Cubic Centimeter \| Definition & Conversion 3:20 min Cubic Meter \| Definition, Formula & Conversion 4:53 min Hectare Definition, History & Conversion 4:56 min Hectometer \| Overview, Conversion & Examples 3:03 min Grams & Kilograms: Lesson for Kids 2:43 min Kilometer \| Definition, Measurement & Examples 3:23 min Meter in Math \| Definition, Conversion & Examples 3:43 min Milliliter Definition, Abbreviation & Conversion 2:58 min Millimeter \| Meaning, Conversion & Measurement 3:12 min Ounce \| oz Meaning & Examples 2:43 min Pint Definition, Measurement & Conversion 2:13 min Pound \| Definition & Measurement 3:21 min Quart \| Definition, Measurement & Conversion 2:43 min Volume & Capacity: Lesson for Kids 4:19 min International Date Line Lesson for Kids: Definition & Facts Chapter 14 Working with Data for Elementary School Axis Definition: Lesson for Kids Box & Whisker Plot: Lesson for Kids What is a Carroll Diagram? - Definition & Examples What is a Class Interval? - Definition & Example 3:45 min How to Find a Class Interval 3:00 min What is a Column Graph? - Definition & Example 2:42 min Frequency Definition: Lesson for Kids 2:27 min Frequency Distribution in Statistics \| Definition & Examples 3:57 min Frequency Distribution in Statistics \| Table & Examples 3:05 min Frequency Histogram \| Parts & Calculation 3:12 min Frequency Histogram \| Definition, Purpose & Examples 3:57 min Histogram Lesson for Kids Ogive \| Definition, Graph & Examples 3:33 min Ratios Lesson for Kids: Definition & Examples Sets in Math \| Symbols, Definition & Examples 3:02 min Chapter 15 Representing Numbers for Elementary School Bar Notation Overview & Examples \| What Does a Line Over a Number Mean? 2:54 min Base Ten Blocks \| Definition, Names & Examples 3:57 min Base Ten System, Chart & Examples 3:21 min Fractional Notation \| Conversion & Examples 3:57 min Hexadecimal \| Definition, System & Examples Creating Graphs: Lesson for Kids What is a Sector Graph? Stem-and-Leaf Plots with Decimals \| Definition, Steps & Examples 4:00 min Stem and Leaf Plot \| Definition, Steps & Examples 3-Digit Stem-and-Leaf Plots 3:40 min Stem-and-Leaf Display \| Plot, Graph & Diagram 3:38 min Chapter 16 Types of Data for Elementary School Categorical Data Lesson for Kids: Definition & Examples Census Lesson for Kids Discrete Data in Math \| Definition, Representation & Examples 3:27 min Population Lesson for Kids: Definition & Facts Qualitative Data \| Definition & Examples 3:22 min Triple Venn Diagram Overview, Uses & Examples 2:56 min Chapter 17 Math Strategies for Elementary School Jump Strategy Definition, Uses & Steps Solving a 3x3 Magic Square \| Overview, Formula & Examples 3:29 min Magic Squares in Math \| History & Examples 3:05 min Trigonometry: Lesson for Kids Chapter 18 Fraction Operations for Elementary School Multiply Fractions with Different Denominators \| Steps & Examples 3:38 min Dividing Fractions with Unlike Denominators \| Overview & Examples 4:56 min Comparing Fractions with Unlike Denominators 3:35 min How to Add Fractions & Whole Numbers 3:33 min Viewing now Order of Operations with Fractions \| PEMDAS, Examples & Practice 3:39 min Up next Solving One-Step Equations with Fractions \| Methods & Examples Read next lesson Chapter 19 Shapes for Elementary School Vertex Angle of an Isosceles Triangle \| Definition & Examples 3:50 min Vertex Angles of a Kite: Lesson for Kids Geometry in Nature \| Shapes, Types & Examples 3:13 min Perimeter & Area of a Square \| Definition, Formula & Relationship 3:10 min Geometric Shapes in The Real World \| Types & Examples 2:36 min Octagon Shape \| Area & Angles 3:29 min Right Angle Shapes \| Overview, Types & Examples 2:53 min Surface Area of Composite Figures 4:27 min Rotational & Radial Symmetry: Lesson for Kids Cross Section Overview & Examples 2:07 min Chapter 20 Negative Numbers for Elementary School Zero Pair in Math \| Definition & Examples 2:46 min History of Negative Numbers \| Invention & Uses 3:07 min Subtraction of Positive & Negative Numbers 4:34 min Chapter 21 Decimals for Elementary School Long Division with Decimals \| Steps & Examples 3:43 min Dividing Decimals by Whole Numbers Comparing & Ordering Fractions, Decimals & Percents 4:16 min Dividing Decimals by Decimals Chapter 22 Lines & Angles for Elementary School Line Segment: Lesson for Kids 2:54 min Angles Formed by Intersecting Lines \| Overview, Types & Examples 3:42 min Coincident Lines Definition, Conditions & Examples 2:46 min What is an Oblique Line? 2:35 min Chapter 23 Multiplication for Elementary School Three Digit Multiplication Three Digit Multiplication Word Problems 5:19 min Multiplication with Lines \| Steps & Examples 5:51 min Related Study Materials Order of Operations with Fractions \| PEMDAS, Examples & Practice LessonsCoursesTopics ##### Brackets in Math \| Definition, Types & Examples 4:35 ##### How to Simplify Expressions with Integers 5:12 ##### Order of Operations With Signed Numbers 6:34 ##### Solving Problems With More Than One Operation: Lesson for Kids 4:06 ##### PEMDAS \| Meaning, Rule & Examples 9:07 ##### Order of Operations in Math \| Steps & Examples 5:50 ##### Number Model \| Definition, Types & Examples 4:48 ##### Showing Students How to Correct Basic Calculation Errors 5:54 ##### Mental Math: Multiplication and Division 4:49 ##### Mental Math: Addition and Subtraction 4:34 ##### Multiplication & Division Fact Families \| Overview & Uses 3:01 ##### Fact Families: Addition & Subtraction \| Definition & Examples 3:32 ##### Mental Math \| Definition, Process & Examples 5:42 ##### Integer Operations \| Rules & Examples 6:53 ##### Multiplying & Dividing Negative Numbers 4:50 ##### The Relationship Between Multiplication & Division 3:50 ##### Multiplying & Dividing Rational Numbers \| Process & Examples 6:19 ##### Missing Quantities in Multiplication & Division Problems: Lesson for Kids 3:31 ##### Arithmetic \| Operators, Properties & Sample Problems 5:02 ##### Arithmetic with Whole Numbers 9:43 ##### MEGA Elementary Education Study Guide and Test Prep ##### Math Review for Teachers: Study Guide & Help ##### Common Core Math - 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6122	https://www.chem.fsu.edu/chemlab/chm1045/empirical.html	Empirical and Molecular Formulas The content that follows is the substance of lecture 9. In this lecture we cover the relationship between Empirical and Molecular formulas and the calculations used to determine one from the other. The Empirical Formula An Empirical formula is the chemical formula of a compound that gives the proportions (ratios) of the elements present in the compound but not the actual numbers or arrangement of atoms. This would be the lowest whole number ratio of the elements in the compound. For Example: In order to determine the Empirical formula for a compound or molecule, we need to know the mass percentages of the the elements in the compound. Once we have this information we can convert it to moles to determine the ratios between the elements. A simple rhyme can be used to remember the process: Percent to Mass Mass to Mole Divide by Small Multiply 'til Whole For Example: NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet and find the molecular formula. (The molar mass of NutraSweet is 294.30 g/mol) Start with the number of grams of each element, given in the problem. \| \| \| --- \| \| \| If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given. \| Convert the mass of each element to moles using the molar mass from the periodic table. Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. \| \| \| --- \| \| \| If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple. \| Now, we can find the molecular formula by finding the mass of the empirical formula and setting up a ratio: Some more detailed instructions: Here are some practice problems with answers:
6123	https://www.ebsco.com/research-starters/geography-and-cartography/maps-and-mathematical-cartography	Skip to main navigation Research Starters Home Maps and mathematical cartography Maps are visual representations of the Earth's features, encompassing both natural and artificial elements, typically displayed on a plane with specific scales and projections. The field of mathematical cartography focuses on the systematic representation of the Earth's surface through various map projections, which aim to balance the inherent distortions that arise when translating three-dimensional reality into two dimensions. Maps serve to convey information about spatial relationships, often using symbols to categorize and enhance legibility, while their scale can vary depending on the area represented, leading to approximations in distance measurements. Historically, maps have evolved from ancient clay representations to sophisticated tools that support navigation and scientific analysis. The classification of maps into general reference, thematic, and charts reflects their diverse purposes, from providing general geographic information to depicting specific themes. The development of mathematical cartography has been significantly influenced by advancements in calculus and the contributions of key figures such as Lambert and Gauss. Today, technology, including geographic information systems and computer software, democratizes map-making, enabling a wider audience to create customized maps and model complex systems, reinforcing the integral role of mathematics in both geography and cartography. Published in: 2024 By: Gaspar, Joaquim Alves Go to EBSCOhost and sign in to access more content about this topic. On this Page History Mathematical Cartography Bibliography Subject Terms Maps History of cartography Cartography Maps and mathematical cartography SUMMARY: Scales and projections are used to display geographic features on maps. The word “map” is the name given to any representation of the Earth’s features—natural and artificial—usually on a plane using a given scale and map projection. In scientific and mathematics applications, the term “map” is more broadly interpreted. The purpose of a map is to register and transmit information about those features and the spatial relations between them. A common characteristic of all maps is that they are reduced and conventional representations of reality, which makes them significantly different from an aerial photograph. While an aerial photograph depicts all the physical objects that a sensor could detect and register, a map is a selection of natural and artificial objectsvisible and invisiblechosen to fit the cartographer’s purpose and the limits imposed by the available space. These objects are represented on maps in a conventional way by means of symbols; this is not the case with photographs, in which they are depicted by the visual image they present when viewed from above by the sensor. The symbols on a map are designed to categorize features by type and to optimize the document’s legibility. Very often, their size is not proportional to the size of the objects they represent. For example, roads are symbolized by lines of variable thickness and pattern, often much larger than the corresponding width of the actual roads, since representing them to exact scale would often make them too thin, even invisible. In other cases, such as with cities, features are symbolized by punctual symbols whose color and shape depend on the classification scheme chosensuch as administrative status or population. Maps are usually classified in three main categoriesgeneral reference maps, thematic maps, and charts. A general reference map depicts generic geographic information of various types considered useful to a large spectrum of users. This information may include topography, political and administrative borders, and land cover. The best example of a general reference map is the topographic map. A thematic map, on the other hand, represents the geographic distribution of a specific theme or group of themes such as geological features, population, or air temperature. A chart is a special type of map designed to support navigation, either maritime with nautical charts or aerial with aerial charts. History Maps were first made by the ancient civilizations of Europe and the Middle East several centuries before the Common Era. One of the oldest known is a Babylonian clay map of the world c. 600 B.C.E., now kept in the British Museum. Though it is documented in the testimony of Ptolemy of Alexandria (c. 90–169 C.E.) and others that maps were drawn in Greece as early as the seventh century B.C.E., none are known to have survived. However, several medieval manuscript maps have survived that represent the ecumene (the known inhabited part of the world around the Mediterranean basin). Few had any practical purpose, and most were symbolic representations inspired by religion and myth rather than by reality. In his Geography, published for the first time in the second century C.E., Ptolemy describes three map projections in detail and presents a list of more than 8000 places in the ecumene, defined by their latitudes and longitudes. This list permitted others to redraw the maps that may have accompanied the original text once the work was translated into Latin and disseminated throughout Europe during the fifteenth century. The publication of several editions of Geography did much to bring about the rebirth of scientific cartography. By this time, nautical charts had already been used to navigate in the Mediterranean for at least two centuries. And while terrestrial cartography quickly adopted the geographic coordinates and map projections proposed by Ptolemy, nautical charts remained based on the magnetic directions and estimated distances observed by pilots at sea. Still, these representations were of astonishing accuracy and detail compared with the traditional maps of the time. It is now known that the first nautical charts, commonly known as “portolan charts,” were constructed in the first half of the thirteenth century, probably in Genoa, after the introduction of the magnetic compass and the adoption of the decimal system in Europe. This basic model continued to be used in nautical cartography for a long time, though much improved by the introduction of astronomical navigation during the fifteenth century. The resulting modality, based on observed latitudes and magnetic directions, became known as the “latitude chart”or “plane chart”and played a fundamental role in the discoveries and maritime expansion periods. In 1569, an important world map specifically conceived for supporting maritime navigation was constructed by the Flemish cartographer Gerard Kremer (1512–1594), better known by the Latinized name of “Gerardus Mercator.” Contrary to traditional portolan charts, this map was based on the latitudes and longitudes of places and represented all rhumb lines (lines of constant course) as straight segments making true angles with the meridians. Though Mercator did not explain how the planisphere was made, a geometric method was most likely used. The mathematics of the projection is not trivial and its formalization had to wait until after calculus was developed, more than one century later. As for its full adoption as a navigational tool, that did not occur until the middle of the eighteenth century, when the marine chronometer was invented and longitudes could finally be determined at sea. Mathematical Cartography Maps may depict only a small part of the whole surface of the Earth. The word “scale” means the quotient between a length measured on a map and the corresponding distance measured on the Earth’s surface. Because it is not possible to represent the spherical surface of the Earth in a plane without distorting the relative position of the places (and thus, the shape of all objects), the scale of a map is not constant, always varying from place to place and, in the generality of cases, also with the direction. In large-scale maps, like the plant of a city or the topographic map of a small region, these distortions can be ignored and the scale considered constant for most practical purposes. That is not the case when a large area of the Earth’s surface is represented, like in a planisphere or a map of a whole continent. Here shapes may be strongly deformed and the scale varies significantly from place to place. Measurements made on those maps with the purpose of evaluating distances between places, using their graphical or numerical scale, are only approximations, as the scale strictly applies only to certain parts of the maps— like the central meridian or parallel—and their use in the other regions may lead to very large errors. “Map projection” refers to any systematic way of representing the surface of the Earth on a plane. The process consists of two independent steps. First, one has to replace the irregular topographic surface, with all its mountains and valleys, with a simpler geometrical model, usually a sphere or an ellipsoid where a system of geographic coordinates (latitude and longitude) is established. Second, one has to project that model onto a plane surface. This step may be accomplished by some geometric construction or by a mathematical function that transforms each pair of geographic coordinates latitude (j) and longitude (l) into a pair of Cartesian coordinates x and y, defined on the plane. Depending on the purpose of the map, there are many different map projections to choose from. Knowing that none of them conserves the relative position of all places on the surface of the Earth, the choice is usually driven by the type of geometric property one wants to preserve. For example, equivalent or equal-area projections conserve the relative areas of all objects and are typically used in political maps. Conformal projections conserve the angles around any point on the map (the scale does not vary with direction), as well as the shape of small objects, and are utilized in nautical charts and topographic maps. Equidistant projections conserve the scale of certain lines and are used whenever one wants to preserve distances measured along those lines. This is the case of the azimuthal equidistant projection, where distances measured from the center of the projection along all great circles are conserved. This property is useful, for example, for quickly determining the distance of any place in the world measured from a chosen location. However, it is not possible for a map projection to have all these properties at the same time, and the conservation of some properties is usually accompanied by significant distortions of the others. A significant example is the Mercator projection (which is conformal), where all rhumb lines are represented by straight segments making true angles with the meridians. However, the scale increases with latitude in this projection, strongly affecting the proportion of the areas. The branch of cartography dealing with map projections is known as “mathematical cartography.” Though some map projections have been well known since remote antiquity, when they were often used for representing the sky, a more formal approach became possible only after the development of calculus. The most important contributions in the formalization of mathematical cartography were those of Johann Heinrich Lambert (1728–1777), Joseph-Louis Lagrange (1736–1813), Carl Friedrich Gauss (1777–1855) and Nicolas Auguste Tissot (1824–1897). Computers and geographic information systems have made it possible for previously unforeseen numbers of users to produce good-quality maps tailored to their specific needs and at a reasonable cost. They also allow scientists and mathematicians to map increasingly complex systems and concepts, such as the universe and the World Wide Web. They can also often render in three dimensions and beyond. In mathematics, maps can be used to alternatively express functions or connect mathematical objects. In conceiving those systems, as well as in acquiring the geographic data necessary to construct the representations within, mathematics continues to play a fundamental role. Bibliography Brown, Lloyd A. The Story of Maps. Dover Publications, 1977. Bugayevskiy, Lev, and John Snyder. Map Projections. A Reference Manual. Taylor & Francis, 1995. Ehrenberg, Ralph E. Mapping the World: An Illustrated History of Cartography. National Geographic, 2005. Snyder, John. Flattening the Earth. University of Chicago Press, 1993. Zuravicky, Orli. Map Math: Learning about Latitude and Longitude Using Coordinate Systems. Rosen Publishing, 2005.
6124	https://www.chegg.com/homework-help/questions-and-answers/social-surplus-maximized-competitive-equilibrium-consumer-surplus-maximized-ballocative-pr-q33135400	Solved When social surplus is maximized in competitive \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Business Economics Economics questions and answers When social surplus is maximized in competitive equilibrium, A.consumer surplus is maximized. B.allocative and productive efficiency are achieved. C. marginal social benefit is greater than marginal social cost. D. All of these options are correct. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: When social surplus is maximized in competitive equilibrium, A.consumer surplus is maximized. B.allocative and productive efficiency are achieved. C. marginal social benefit is greater than marginal social cost. D. All of these options are correct. When social surplus is maximized in competitive equilibrium, A.consumer surplus is maximized. B.allocative and productive efficiency are achieved. C. marginal social benefit is greater than marginal social cost. D. All of these options are correct. There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! To start, focus on understanding the relationship between total surplus and competitive equilibrium, where total surplus is the sum of consumer surplus and producer surplus. Step 1 The social surplus (or total surplus) is the sum of ... View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. Cookie NoticeYour Privacy ChoicesDo Not Sell My InfoGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
6125	https://www.studocu.com/row/messages/question/8537779/give-equation-3-find-mpl-mpk-and-mrts	[Solved] give equation 3 find MPL MPK and MRTS - economics (14444) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent give equation 3 find MPL MPK and MRTS economics (14444) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Aksum University economics Question give equation 3 find MPL MPK and MRTS Aksum University economics Question ### tariku 1 year ago give equation 3 find MPL MPK and MRTS Like 0 Answer Created with AI 1 year ago To find the marginal product of labor (MPL), marginal product of capital (MPK), and the marginal rate of technical substitution (MRTS) for a production function, we can start with the general production function: js Q = f(K, L) Where: Q = quantity of output K = quantity of capital L = quantity of labor The marginal product of labor (MPL) is the derivative of the production function with respect to labor: js MPL = ∂Q/∂L The marginal product of capital (MPK) is the derivative of the production function with respect to capital: js MPK = ∂Q/∂K The marginal rate of technical substitution (MRTS) is the ratio of the marginal products of labor and capital: js MRTS = MPL/MPK By finding the partial derivatives of the production function with respect to labor and capital, you can calculate MPL, MPK, and MRTS. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: economics 14444 Aksum University 30 Documents Go to course 57 Assessment of Micro and Small Enterprises in Community Development - MSc Thesis economics 100%(6) 42 Challenges and Prospects of Small Business Enterprises in AHSEA TOWN: A Research Study economics 100%(5) 35 Calc I - Intro to Calculus for Economics Students economics 100%(2) 1 Aksum University Shire Campus economics 100%(1) Discover more from: economics 14444Aksum University30 Documents Go to course 57 Assessment of Micro and Small Enterprises in Community Development - MSc Thesis economics 100%(6) 42 Challenges and Prospects of Small Business Enterprises in AHSEA TOWN: A Research Study economics 100%(5) 35 Calc I - Intro to Calculus for Economics Students economics 100%(2) 1 Aksum University Shire Campus economics 100%(1) 1 Title not provided. Please upload a document with a title for me to evaluate and generate a new one.economics 100%(1) 28 MICROECONOMICS II: Oligopoly and Game Theory Concepts Explained economics None Related Answered Questions 5 months ago According to the Keynesian school of thought the 1930s great depression of US and western Europe was because of: Insufficient production levels especially agricultural products Failure of the financial sector Aggregate demand deficiency Bad weather conditions that strongly affect agricultural productions All of the above 2.Which of the following is/are true about the new Keynesian macroeconomists? A.New Keynesians argued existence of market imperfections B.They assumed complete and continuous wage and price flexibility C.New Keynesians have strongly believed that markets clear via market economics (14444) 1 year ago Interprate the results Y(L,K)=L+2k economics (14444) 1 year ago Give equation 3 find MPL MPK MRTS economics (14444) 2 years ago Question 2 Suppose you receive the following data about the South African labour market situation in 2022: Broad unemployment rate 40% Number of people younger than 15 years or older than 64 years 20 000 000 Number of discouraged work-seekers 1 000 000 Number of people of the working-age that were neither willing, nor able to work 4 000 000 Total number of people in the population 50 000 000 Calculate the following. Round your answers to the nearest integer. a) The number of narrowly unemployed people. b) The narrow participation rate. c) The narrow unemployment rate. d) The number of employed people. economics (14444) 2 years ago Calculate the values of the two interest rates on which Nomandia's feasible sets are based. Round your answers to once decimal. economics (14444) Ask AI Home My Library Discovery Discovery Universities High Schools Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English Rest of the World Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English Rest of the World Studocu is not affiliated to or endorsed by any school, college or university. 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6126	https://german.stackexchange.com/questions/27398/what-is-the-reason-why-the-translation-of-speed-is-geschwindigkeit	word usage - What is the reason why the translation of “Speed” is “Geschwindigkeit”? - German Language Stack Exchange Join German Language By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community German Language helpchat German Language Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What is the reason why the translation of “Speed” is “Geschwindigkeit”? Ask Question Asked 9 years, 9 months ago Modified9 years, 9 months ago Viewed 3k times This question shows research effort; it is useful and clear 13 Save this question. Show activity on this post. I am not a native speaker of German and I’ve been learning the language for some months. Recently, I read some texts about cars and I noticed the word Geschwindigkeit, which means speed in English. I noticed also that this word is used in the speedometer of the cars. I wonder why the translation for speed is Geschwindigkeit (from the adjective geschwind + ig + keit) and not another German word that is closer or more related like Schnelligkeit (from the adjective schnell + ig + keit). Is there a specific reason why Geschwindigkeit is used instead of Schnelligkeit to indicate the speed of a vehicle? Is Schnelligkeit uncommon or wrong to say speed? word-usage etymology Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Dec 30, 2015 at 18:22 Wrzlprmft♦ 22k 8 8 gold badges 74 74 silver badges 139 139 bronze badges asked Dec 30, 2015 at 18:14 Alberto SolanoAlberto Solano 297 3 3 silver badges 13 13 bronze badges Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 16 Save this answer. Show activity on this post. From their etymology both adjectives geschwind, and schnell used to have a common meaning of being strong, and agile in Old High German, presumably coming from usage for describing a man fighting. Over time both words are still used for speed, and we do have both derived nouns Geschwindigkeit, and Schnelligkeit, where the agility aspect is a bit better preserved in Schnelligkeit. For speed as a physical unit and for all the derivates from this meaning we only use Geschwindigkeit in modern German. The English speed is much closer to physics because it shares its etymology with Old High German spuot (speed, progress). This meaning only has a remnant in the modern German verb sich sputen. It may come from a different cultural approach where speed in the English language was preserved for something moving fast where in German speaking regions an additional connotation was also put on strength. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 31, 2015 at 7:35 answered Dec 30, 2015 at 21:39 TakkatTakkat 70.8k 30 30 gold badges 172 172 silver badges 424 424 bronze badges 1 Thank you very much for your explanation. That's what I wanted to understand.Alberto Solano –Alberto Solano 2016-01-03 10:57:59 +00:00 Commented Jan 3, 2016 at 10:57 Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. I am not a native speaker, but studied in Germany and still fairly fluent. It is like in English, with slight differences in word meanings and what is the "correct" term. Schnelligkeit means more speediness, where Geschwindigkeit refers to velocity, sort of a quality versus a quantity thing. We would never talk about rapidness of a car but rather its acceleration. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Dec 31, 2015 at 2:39 user unknown 23.6k 4 4 gold badges 50 50 silver badges 97 97 bronze badges answered Dec 31, 2015 at 2:23 BillBill 51 1 1 bronze badge 1 Actually, the acceleration (0 to 60 in X seconds) is different from its (current) speed, but I will upvote you nonetheless for a good answer :-) Welcome aboard Mawg –Mawg 2020-02-19 05:28:39 +00:00 Commented Feb 19, 2020 at 5:28 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. To answer your question: The German translation of the english word »speed« is »Geschwindigkeit«, because »Geschwindigkeit« is the German word for what is called »speed« in English. But I guess you wanted to know why this word exists in German. Try another point of view: Why does there exist such a word like »speed« in English? I'm a German native speaker, and for me it's a miracle why there is such a word like »speed« in English for something that has such a natural name like »Geschwindigkeit« in German. The more natural english word would be »fastness« because it derives the same way from the adjective »fast« as the German »Geschwindigkeit« derives from the adjective »geschwind«. Adding the suffix -keit (as well as -heit and some other suffixes) to an adjective is a very common way to create a noun from an adjective, comparable to the english suffix -ness: heiter -> Heiterkeit / happy - happiness einsam -> Einsamkeit / lonely -> loneliness höflich -> Höflichkeit / polite -> politeness - geschwind -> Geschwindigkeit / fast ->speed ??? The adjective »geschwind« is a synonym for »schnell«. For the etymology of both words and the etymology of »speed« look at Takkats answer. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:42 CommunityBot 1 answered Dec 31, 2015 at 6:10 Hubert SchölnastHubert Schölnast 130k 20 20 gold badges 221 221 silver badges 430 430 bronze badges 1 But of course there are many other cases where the quality is not named directly after the measure: height not tallness, weight not heaviness, distance not remoteness, wealth not richness, temperature not hotness. In all those cases both words exist, but their connotations are subtly different.Michael Kay –Michael Kay 2017-12-12 23:04:12 +00:00 Commented Dec 12, 2017 at 23:04 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I guess that there is no particular reason. Theoretically it would make sense to say "Schnelligkeit" instead of "Geschwindigkeit" to describe the speed of a vehicle. But it just is not used like that. From my understanding, "Geschwind" is just the "older" less "modern" adjective so I think the word "Geschwindigkeit" has naturalized and it stayed that way - it's just a convention. Using "Schnelligkeit" would not be wrong, but it would sound weird. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Dec 30, 2015 at 19:32 geruetzelgeruetzel 196 2 2 silver badges 5 5 bronze badges 1 9 Well, Schnelligkeit would particularly sound like you wanted to express fastness, as a contrast to Langsamkeit, i.e. slowness. Geschwindigkeit is rather perceived as a "neutral" way to express any speed, fast or slow (even though it cannot be explained ethymologically, given that geschwind on its own means rather fast or quick again).O. R. Mapper –O. R. Mapper 2015-12-30 20:22:10 +00:00 Commented Dec 30, 2015 at 20:22 Add a comment\| Your Answer Thanks for contributing an answer to German Language Stack Exchange! 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6127	https://en.wikipedia.org/wiki/Oxonium_ion	Jump to content Search Contents (Top) 1 Alkyloxonium 2 Oxocarbenium ions 3 Gold-stabilized species 4 Relevance to natural product chemistry 5 See also 6 References Oxonium ion العربية Català Čeština Deutsch Español Euskara فارسی Français Nederlands 日本語 Nordfriisk Norsk bokmål Polski Português Simple English Slovenčina Српски / srpski Suomi தமிழ் Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Cation containing an oxygen atom with 3 bonds and 1+ formal charge In chemistry, an oxonium ion is any cation containing an oxygen atom that has three bonds and 1+ formal charge. The simplest oxonium ion is the hydronium ion (H3O+). Alkyloxonium [edit] Hydronium is one of a series of oxonium ions with the formula RnH3−nO+. Oxygen is usually pyramidal with an sp3 hybridization. Those with n = 1 are called primary oxonium ions, an example being protonated alcohol (e.g. methanol). In acidic media, the oxonium functional group produced by protonating an alcohol can be a leaving group in the E2 elimination reaction. The product is an alkene. Extreme acidity, heat, and dehydrating conditions are usually required. Other hydrocarbon oxonium ions are formed by protonation or alkylation of alcohols or ethers (R−C−+O−R1R2). Secondary oxonium ions have the formula R2OH+, an example being protonated ethers. Tertiary oxonium ions have the formula R3O+, an example being trimethyloxonium. Tertiary alkyloxonium salts are useful alkylating agents. For example, triethyloxonium tetrafluoroborate (Et3O+)(BF−4), a white crystalline solid, can be used, for example, to produce ethyl esters when the conditions of traditional Fischer esterification are unsuitable. It is also used for preparation of enol ethers and related functional groups. \| \| \| \| \| general pyramidaloxonium ion \| skeletal formula of thetrimethyloxonium cation \| ball-and-stick modelof trimethyloxonium \| space-filling modelof trimethyloxonium \| Oxatriquinane and oxatriquinacene are unusually stable oxonium ions, first described in 2008. Oxatriquinane does not react with boiling water or with alcohols, thiols, halide ions, or amines, although it does react with stronger nucleophiles such as hydroxide, cyanide, and azide. Oxocarbenium ions [edit] Another class of oxonium ions encountered in organic chemistry is the oxocarbenium ions, obtained by protonation or alkylation of a carbonyl group e.g. R−C=+O−R′ which forms a resonance structure with the fully-fledged carbocation R−+C−O−R′ and is therefore especially stable: Gold-stabilized species [edit] An unusually stable oxonium species is the gold complex tris[triphenylphosphinegold(I)]oxonium tetrafluoroborate, [(Ph3PAu)3O][BF4], where the intramolecular aurophilic interactions between the gold atoms are believed responsible for the stabilisation of the cation. This complex is prepared by treatment of Ph3PAuCl with Ag2O in the presence of NaBF4: : 3 Ph3PAuCl + Ag2O + NaBF4 → [(Ph3PAu)3O]+[BF4]− + 2 AgCl + NaCl It has been used as a catalyst for the propargyl Claisen rearrangement. Relevance to natural product chemistry [edit] Complex bicyclic and tricyclic oxonium ions have been proposed as key intermediates in the biosynthesis of a series of natural products by the red algae of the genus Laurencia. Several members of these elusive species have been prepared explicitly by total synthesis, demonstrating the possibility of their existence. The key to their successful generation was the use of a weakly coordinating anion (Krossing's anion, [Al(pftb)4]−, pftb = perfluoro-tert-butoxy) as the counteranion. As shown in the example below, this was executed by a transannular halide abstraction strategy through the reaction of the oxonium ion precursor (an organic halide) with the silver salt of the Krossing's anion Ag[Al(pftb)4]•CH2Cl2, generating the desired oxonium ion with simultaneous precipitation of inorganic silver halides. The resulting oxonium ions were characterized comprehensively by nuclear magnetic resonance spectroscopy at low temperature (−78 °C) with support from density functional theory computation. These oxonium ions were also demonstrated to directly give rise to multiple related natural products by reacting with various nucleophiles, such as water, bromide, chloride, and acetate. See also [edit] Acylium ion, a type of oxonium ion with the structure R–C≡O+ Onium ion, a +1 cation derived by protonation of a hydride (includes oxonium ions) Pyrylium, a subtype of oxonium ion Sulfonium, a sulfur analog that can be chiral References [edit] ^ March, Jerry (2007), Advanced Organic Chemistry: Reactions, Mechanisms, and Structure (4th ed.), New York: Wiley, p. 497 ^ Olah, George A. (1998). Onium Ions. John Wiley & Sons. p. 509. ISBN 9780471148777. ^ Olah, George A. (1993). "Superelectrophiles". Angew. Chem. Int. Ed. Engl. 32 (6): 767–788. doi:10.1002/anie.199307673. ^ Raber, Douglas J.; Gariano Jr, Patrick; Brod, Albert O.; Gariano, Anne L.; Guida, Wayne C. (1977). "Esterification Of Carboxylic Acids With Trialkyloxonium Salts: Ethyl And Methyl 4-acetoxybenzoates". Org. Synth. 56: 59. doi:10.15227/orgsyn.056.0059. ^ Struble, Justin R.; Bode, Jeffrey W. (2010). "Synthesis Of A N-mesityl Substituted Aminoindanol-derived Triazolium Salt". Org. Synth. 87: 362. doi:10.15227/orgsyn.087.0362. ^ Hegedus, Lous S.; Mcguire, Michael A.; Schultze, Lisa M. (1987). "1,3-Dimethyl-3-methoxy-4-phenylazetidinone". Org. Synth. 65: 140. doi:10.15227/orgsyn.065.0140. ^ Schmidbaur, Hubert (2000). "The Aurophilicity Phenomenon: A Decade of Experimental Findings, Theoretical Concepts and Emerging Application". Gold Bulletin. 33 (1): 3–10. doi:10.1007/BF03215477. ^ Schmidbaur, Hubert (1995). "Ludwig Mond Lecture: High-Carat Gold Compounds". Chem. Soc. Rev. 24 (6): 391–400. doi:10.1039/CS9952400391. ^ Bruce, M. I.; Nicholson, B. K.; Bin Shawkataly, O.; Shapley, J. R.; Henly, T. (1989). "Synthesis of Gold-Containing Mixed-Metal Cluster Complexes". In Kaesz, Herbert D. (ed.). Inorganic Syntheses. Vol. 26. John Wiley & Sons, Inc. pp. 324–328. doi:10.1002/9780470132579.ch59. ISBN 9780470132579. ^ Sherry, Benjamin D.; Toste, F. Dean (2004). "Gold(I)-Catalyzed Propargyl Claisen Rearrangement" (PDF). Journal of the American Chemical Society. 126 (49): 15978–15979. CiteSeerX 10.1.1.604.7272. doi:10.1021/ja044602k. ISSN 0002-7863. PMID 15584728. ^ a b Sam Chan, Hau Sun; Nguyen, Q. Nhu N.; Paton, Robert S.; Burton, Jonathan W. (2019-10-09). "Synthesis, Characterization, and Reactivity of Complex Tricyclic Oxonium Ions, Proposed Intermediates in Natural Product Biosynthesis". Journal of the American Chemical Society. 141 (40). A full list of references encompassing the contributions from Braddock, Snyder, Murai, Suzuki, Fukuzawa, Burton, Kim, and Fox are available inside.: 15951–15962. doi:10.1021/jacs.9b07438. ISSN 0002-7863. PMID 31560524. S2CID 203580092. ^ Krossing, Ingo (2001). "The Facile Preparation of Weakly Coordinating Anions: Structure and Characterisation of Silverpolyfluoroalkoxyaluminates AgAl(ORF)4, Calculation of the Alkoxide Ion Affinity". Chemistry – A European Journal. 7 (2): 490–502. doi:10.1002/1521-3765(20010119)7:2<490::aid-chem490>3.0.co;2-i. ISSN 1521-3765. PMID 11271536. ^ Wang, Bin-Gui; Gloer, James B.; Ji, Nai-Yun; Zhao, Jian-Chun (March 2013). "Halogenated Organic Molecules of Rhodomelaceae Origin: Chemistry and Biology". Chemical Reviews. 113 (5): 3632–3685. doi:10.1021/cr9002215. ISSN 0009-2665. PMID 23448097. ^ Zhou, Zhen-Fang; Menna, Marialuisa; Cai, You-Sheng; Guo, Yue-Wei (2015-02-11). "Polyacetylenes of Marine Origin: Chemistry and Bioactivity". Chemical Reviews. 115 (3): 1543–1596. doi:10.1021/cr4006507. ISSN 0009-2665. PMID 25525670. ^ Wanke, Tauana; Philippus, Ana Cláudia; Zatelli, Gabriele Andressa; Vieira, Lucas Felipe Oliveira; Lhullier, Cintia; Falkenberg, Miriam (2015-11-01). "C15 acetogenins from the Laurencia complex: 50 years of research – an overview". Revista Brasileira de Farmacognosia. 25 (6): 569–587. doi:10.1016/j.bjp.2015.07.027. ISSN 0102-695X. Retrieved from " Categories: Oxonium compounds Oxycations Hidden categories: Articles with short description Short description is different from Wikidata Oxonium ion Add topic
6128	https://openstax.org/books/college-algebra-2e/pages/1-5-factoring-polynomials	Skip to ContentGo to accessibility pageKeyboard shortcuts menu College Algebra 2e 1.5 Factoring Polynomials College Algebra 2e1.5 Factoring Polynomials Search for key terms or text. Learning Objectives In this section, you will: Factor the greatest common factor of a polynomial. Factor a trinomial. Factor by grouping. Factor a perfect square trinomial. Factor a difference of squares. Factor the sum and difference of cubes. Factor expressions using fractional or negative exponents. Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1. Figure 1 The area of the entire region can be found using the formula for the area of a rectangle. The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of units2. The other rectangular region has one side of length and one side of length giving an area of units2. So the region that must be subtracted has an area of units2. The area of the region that requires grass seed is found by subtracting units2. This area can also be expressed in factored form as units2. We can confirm that this is an equivalent expression by multiplying. Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions. Factoring the Greatest Common Factor of a Polynomial When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, is the GCF of and because it is the largest number that divides evenly into both and The GCF of polynomials works the same way: is the GCF of and because it is the largest polynomial that divides evenly into both and When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables. Greatest Common Factor The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials. How To Given a polynomial expression, factor out the greatest common factor. Identify the GCF of the coefficients. Identify the GCF of the variables. Combine to find the GCF of the expression. Determine what the GCF needs to be multiplied by to obtain each term in the expression. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by. Example 1 Factoring the Greatest Common Factor Factor Solution First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of , and is . (Note that the GCF of a set of expressions in the form will always be the exponent of lowest degree.) And the GCF of , and is . Combine these to find the GCF of the polynomial, . Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that , and Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by. Analysis After factoring, we can check our work by multiplying. Use the distributive property to confirm that Try It #1 Factor by pulling out the GCF. Factoring a Trinomial with Leading Coefficient 1 Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial has a GCF of 1, but it can be written as the product of the factors and Trinomials of the form can be factored by finding two numbers with a product of and a sum of The trinomial for example, can be factored using the numbers and because the product of those numbers is and their sum is The trinomial can be rewritten as the product of and Factoring a Trinomial with Leading Coefficient 1 A trinomial of the form can be written in factored form as where and Q&A Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. How To Given a trinomial in the form factor it. List factors of Find and a pair of factors of with a sum of Write the factored expression Example 2 Factoring a Trinomial with Leading Coefficient 1 Factor Solution We have a trinomial with leading coefficient and We need to find two numbers with a product of and a sum of In the table below, we list factors until we find a pair with the desired sum. \| Factors of \| Sum of Factors \| --- \| \| \| \| \| \| 14 \| \| \| \| \| \| 2 \| Now that we have identified and as and write the factored form as Analysis We can check our work by multiplying. Use FOIL to confirm that Q&A Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter. Try It #2 Factor Factoring by Grouping Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial can be rewritten as using this process. We begin by rewriting the original expression as and then factor each portion of the expression to obtain We then pull out the GCF of to find the factored expression. Factor by Grouping To factor a trinomial in the form by grouping, we find two numbers with a product of and a sum of We use these numbers to divide the term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression. How To Given a trinomial in the form factor by grouping. List factors of Find and a pair of factors of with a sum of Rewrite the original expression as Pull out the GCF of Pull out the GCF of Factor out the GCF of the expression. Example 3 Factoring a Trinomial by Grouping Factor by grouping. Solution We have a trinomial with and First, determine We need to find two numbers with a product of and a sum of In the table below, we list factors until we find a pair with the desired sum. \| Factors of \| Sum of Factors \| --- \| \| \| \| \| \| 29 \| \| \| \| \| \| 13 \| \| \| \| \| \| 7 \| So and Analysis We can check our work by multiplying. Use FOIL to confirm that Try It #3 Factor ⓐ ⓑ Factoring a Perfect Square Trinomial A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term. We can use this equation to factor any perfect square trinomial. Perfect Square Trinomials A perfect square trinomial can be written as the square of a binomial: How To Given a perfect square trinomial, factor it into the square of a binomial. Confirm that the first and last term are perfect squares. Confirm that the middle term is twice the product of Write the factored form as Example 4 Factoring a Perfect Square Trinomial Factor Solution Notice that and are perfect squares because and Then check to see if the middle term is twice the product of and The middle term is, indeed, twice the product: Therefore, the trinomial is a perfect square trinomial and can be written as Try It #4 Factor Factoring a Difference of Squares A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied. We can use this equation to factor any differences of squares. Differences of Squares A difference of squares can be rewritten as two factors containing the same terms but opposite signs. How To Given a difference of squares, factor it into binomials. Confirm that the first and last term are perfect squares. Write the factored form as Example 5 Factoring a Difference of Squares Factor Solution Notice that and are perfect squares because and The polynomial represents a difference of squares and can be rewritten as Try It #5 Factor Q&A Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored. Factoring the Sum and Difference of Cubes Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial. Similarly, the difference of cubes can be factored into a binomial and a trinomial, but with different signs. We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example. The sign of the first 2 is the same as the sign between The sign of the term is opposite the sign between And the sign of the last term, 4, is always positive. Sum and Difference of Cubes We can factor the sum of two cubes as We can factor the difference of two cubes as How To Given a sum of cubes or difference of cubes, factor it. Confirm that the first and last term are cubes, or For a sum of cubes, write the factored form as For a difference of cubes, write the factored form as Example 6 Factoring a Sum of Cubes Factor Solution Notice that and are cubes because Rewrite the sum of cubes as Analysis After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check. Try It #6 Factor the sum of cubes: Example 7 Factoring a Difference of Cubes Factor Solution Notice that and are cubes because and Write the difference of cubes as Analysis Just as with the sum of cubes, we will not be able to further factor the trinomial portion. Try It #7 Factor the difference of cubes: Factoring Expressions with Fractional or Negative Exponents Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, can be factored by pulling out and being rewritten as Example 8 Factoring an Expression with Fractional or Negative Exponents Factor Solution Factor out the term with the lowest value of the exponent. In this case, that would be Try It #8 Factor Media Access these online resources for additional instruction and practice with factoring polynomials. Identify GCF Factor Trinomials when a Equals 1 Factor Trinomials when a is not equal to 1 Factor Sum or Difference of Cubes 1.5 Section Exercises Verbal 1. If the terms of a polynomial do not have a GCF, does that mean it is not factorable? Explain. A polynomial is factorable, but it is not a perfect square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF? 3. How do you factor by grouping? Algebraic For the following exercises, find the greatest common factor. 4. 5. 6. 7. 8. 9. For the following exercises, factor by grouping. 10. 11. 12. 13. 14. 15. For the following exercises, factor the polynomial. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. For the following exercises, factor the polynomials. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. Real-World Applications For the following exercises, consider this scenario: Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of m2, as shown in the figure below. The length and width of the park are perfect factors of the area. 51. Factor by grouping to find the length and width of the park. A statue is to be placed in the center of the park. The area of the base of the statue is Factor the area to find the lengths of the sides of the statue. 53. At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is Factor the area to find the lengths of the sides of the fountain. For the following exercise, consider the following scenario: A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd. as shown in the figure below. The flagpole will take up a square plot with area yd2. Find the length of the base of the flagpole by factoring. Extensions For the following exercises, factor the polynomials completely. 55. 56. 57. 58. 59. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Jay Abramson Publisher/website: OpenStax Book title: College Algebra 2e Publication date: Dec 21, 2021 Location: Houston, Texas Book URL: Section URL: © Jun 16, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
6129	https://www.sciencedirect.com/science/article/pii/S2950198924000278	Skip to article My account Sign in View PDF JAAD Reviews Volume 2, December 2024, Pages 1-12 Clinical review Lichen planus or not? A review of conditions with lichenoid nail changes Author links open overlay panel, , rights and content Under a Creative Commons license Open access Nail lichen planus is associated with numerous characteristic nail matrix and nail bed findings, which can collectively be referred to as lichenoid nail changes. Individual lichenoid changes like dorsal pterygium and trachyonychia may present in many dermatologic, genetic, and autoimmune diseases. A few conditions, including graft versus host disease, systemic amyloidosis, and dyskeratosis congenita, may rarely manifest with multiple lichenoid nail changes together, often making them difficult to discern from true nail lichen planus. The accurate clinical differentiation and diagnosis of an underlying condition are key to the successful management of lichenoid nail changes. This review uses evidence-based data and clinical expertise to summarize specific nail findings and diagnostic/differentiating features that can help to guide the clinical assessment, diagnosis, and management of lichenoid nail changes associated with a variety of diseases. Key words alopecia areata amyloidosis atopic dermatitis autoimmune hemolytic anemia cicatricial pemphigoid Darier disease dyskeratosis congenita epidermolysis bullosa graft versus host disease hereditary punctuate palmoplantar keratoderma Huriez syndrome ichthyosis vulgaris idiopathic atrophy of the nails IgA deficiency immune dysregulation polyendocrinopathy enteropathy X-linked syndrome (IPEX) incontientia pigmenti Kindler syndrome lichen nitidus lichen planus lichen striatus lupus erythematosus Marfan syndrome nail nail lichen planus nail matrix nail plate nail trauma onycholysis onychomatricoma onychorrhexis onychotillomania paraneoplastic pemphigus pemphigus foliaceus pemphigus vulgaris porokeratosis primary biliary cirrhosis psoriasis pterygium reflex sympathetic dystrophy sarcoidosis SÃ©zary syndrome Steven-Johnson syndrome subungual hyperkeratosis systemic amyloidosis toxic epidermal necrolysis trachyonychia vitiligo Abbreviations used DC dyskeratosis congenita GVHD graft versus host disease LN lichen nitidus LP lichen planus LS lichen striatus Cited by (0) : Funding sources: None. : IRB approval status: Not applicable. © 2024 The Author(s). Published by Elsevier Inc. on behalf of the American Academy of Dermatology.
6130	https://www.vedantu.com/neet/full-form-of-iupac	Full Form Science Medicine Full Form OF IUPAC Full Form of IUPAC: International Union of Pure and Applied Chemistry Download PDF Courses Class 11 NEET Course (2023-25) Class 12 NEET Course (2023-24) NEET Repeater Course (2023-24) Class 8 NEET Foundation Course Class 9 NEET Foundation Course Class 10 NEET Foundation Course Exam Info Important Dates (Exam Dates) Eligibility Criteria Application Form Form Correction Centres (Exam Centres) Admit Card Reservation Criteria Slot Booking Neet Syllabus Neet Physics Syllabus Neet Chemistry Syllabus Neet Biology Syllabus Weightage Pattern Study Material Important Questions Revision Notes Formulas Difference between Chapter specific previous year question and answer Question Papers Previous Year Paper Solutions Question Paper Analysis Mock Test Sample paper Answer Key Cut off Neet Colleges Government College Private College Results Percentage Vs Marks Rank List Counselling Marks and Rank Neet Question Answers Neet Physics Question Answers Neet Chemistry Question Answers Neet Biology Question Answers News Videos FAQs What is the Full Form of IUPAC? The full form of IUPAC is the International Union of Pure and Applied Chemistry . It is a globally recognized organization responsible for standardizing chemical nomenclature, terminology, and measurement units. Established in 1919 , IUPAC plays a crucial role in developing systematic naming conventions for chemical compounds, ensuring consistency in scientific communication worldwide. It also contributes to advancing research, setting industry standards, and promoting education in chemistry. History of IUPAC: The International Union of Pure and Applied Chemistry (IUPAC) was established in 1919 to create a standardized system for chemical nomenclature, terminology, and measurement units. It was formed by chemists from different countries to overcome inconsistencies in naming chemical compounds and ensure uniform scientific communication worldwide. Over the years, IUPAC has played a crucial role in updating the periodic table, defining atomic weights, and setting global chemistry standards. It has also contributed to advancing research, industry regulations, and education, making it a key organization in the development of chemical sciences. Importance of IUPAC Standardizes chemical nomenclature to ensure global consistency. Provides a systematic naming system for chemical compounds. Prevents confusion in scientific communication and publications. Contributes to research, industry regulations, and education. Establishes international measurement units and symbols. Plays a key role in classifying new elements and updating the periodic table. Helps in setting safety guidelines and industry standards. Supports advancements in chemical sciences and innovations. Role of IUPAC Establishes standardized chemical nomenclature for global consistency. Develops and updates systematic naming rules for chemical compounds. Regulates symbols, units, and terminology in chemistry. Classifies new elements and maintains the periodic table. Supports scientific research, education, and industry applications. Ensures safety guidelines and best practices in chemical sciences. Promotes international collaboration and knowledge exchange. Advances innovations in pure and applied chemistry. Interesting Facts about IUPAC Founded in 1919, IUPAC was created to bring uniformity to chemical terminology worldwide. It is responsible for naming all newly discovered chemical elements in the periodic table. IUPAC introduced the official systematic naming of organic and inorganic compounds. The organization plays a key role in defining atomic weights and isotopic compositions. IUPAC revised the Periodic Table layout, ensuring accuracy in element classification. It organizes international conferences and educational programs to promote chemistry. Conclusion: The International Union of Pure and Applied Chemistry (IUPAC) plays a vital role in standardizing chemical nomenclature, terminology, and scientific measurements worldwide. Its contributions to research, education, and industry regulations ensure consistency and accuracy in chemical sciences. As a globally recognized authority, IUPAC continues to advance innovations, update the periodic table, and promote scientific collaboration, making it indispensable in the field of chemistry. If you are preparing for MBBS, here are a few NEET study materials that will help you in preparation for your Exam. Essential Study Materials for NEET UG Success \| Study Materials for NEET 2025 \| \| \| S. No \| Study Materials Links \| \| 1 \| NEET Revision Notes \| \| 2 \| NEET Important Questions \| \| 3 \| NEET Question Papers \| \| 4 \| NEET Sample Papers \| \| 5 \| NEET Syllabus \| \| 6 \| NEET Mock Test Series \| \| 7 \| NEET Formulas \| \| 8 \| NEET Difference Between \| \| 9 \| NEET Most Important Chapters \| \| 10 \| NEET Question and Answers \| \| 11 \| NEET UG Preparation Books \| \| 12 \| NEET Chapter-Wise Weightage \| FAQs on Full Form of IUPAC: International Union of Pure and Applied Chemistry What is IUPAC naming in chemistry? IUPAC naming is a standardized system for naming organic and inorganic compounds, ensuring global consistency in chemical nomenclature. What is the IUPAC Formula? The IUPAC formula represents a chemical compound using a systematic approach, following specific naming rules for elements and functional groups. What is the scientific name of IUPAC? The scientific name of IUPAC is the International Union of Pure and Applied Chemistry. What is 5 in IUPAC? In IUPAC nomenclature, "pent-" is the prefix used for compounds with five carbon atoms in their structure. What is the IUPAC name of alcohol? The IUPAC name for alcohols follows the "-ol" suffix, such as ethanol (C₂H₅OH) for ethyl alcohol. Who found IUPAC? IUPAC was established in 1919 by chemists from various countries to standardize chemical nomenclature and measurements. Where is IUPAC located? IUPAC’s headquarters is in Zürich, Switzerland, while its secretariat is based in Research Triangle Park, North Carolina, USA. Why is IUPAC important in chemistry? IUPAC ensures global consistency in chemical nomenclature, terminology, and measurement units, helping scientists communicate effectively across different regions. How does IUPAC name organic compounds? IUPAC follows a systematic naming approach based on the number of carbon atoms, functional groups, and molecular structure, ensuring clarity and uniformity. What are the major contributions of IUPAC? IUPAC has contributed to naming chemical elements, defining atomic weights, updating the periodic table, and setting international standards for chemistry research and education. 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6131	https://www.nhs.uk/mental-health/conditions/agoraphobia/overview/	Skip to main content Overview - Agoraphobia Agoraphobia is a fear of being in situations where escape might be difficult or that help wouldn't be available if things go wrong. Many people assume agoraphobia is simply a fear of open spaces, but it's actually a more complex condition. Someone with agoraphobia may be scared of: travelling on public transport visiting a shopping centre leaving home If someone with agoraphobia finds themselves in a stressful situation, they'll usually experience the symptoms of a panic attack, such as: rapid heartbeat rapid breathing (hyperventilating) feeling hot and sweaty feeling sick They'll avoid situations that cause anxiety and may only leave the house with a friend or partner. They'll order groceries online rather than going to the supermarket. This change in behaviour is known as avoidance. Read more about the symptoms of agoraphobia. What causes agoraphobia? Agoraphobia can develop as a complication of panic disorder, an anxiety disorder involving panic attacks and moments of intense fear. It can arise by associating panic attacks with the places or situations where they occurred and then avoiding them. Not all people with agoraphobia have a history of panic attacks. In these cases, their fear may be related to issues like a fear of crime, terrorism, illness or being in an accident. Read more about the possible causes of agoraphobia. Diagnosing agoraphobia Speak to your GP if you think you may be affected by agoraphobia. It should be possible to arrange a telephone consultation if you don't feel ready to visit your GP in person. Your GP will ask you to describe your symptoms, how often they occur, and in what situations. It's very important you tell them how you've been feeling and how your symptoms are affecting you. Your GP may ask you the following questions: Do you find leaving the house stressful? Are there certain places or situations you have to avoid? Do you have any avoidance strategies to help you cope with your symptoms, such as relying on others to shop for you? It can sometimes be difficult to talk about your feelings, emotions, and personal life, but try not to feel anxious or embarrassed. Your GP needs to know as much as possible about your symptoms to make the correct diagnosis and recommend the most appropriate treatment. Read more about diagnosing agoraphobia. Treating agoraphobia Lifestyle changes may help, including taking regular exercise, eating more healthily, and avoiding alcohol, drugs and drinks that contain caffeine, such as tea, coffee and cola. Self-help techniques that can help during a panic attack include staying where you are, focusing on something that's non-threatening and visible, and slow, deep breathing. If your agoraphobia fails to respond to these treatment methods, see your GP. You can also refer yourself directly for talking therapies, including cognitive behavioural therapy (CBT), without seeing your GP. Read more about talking therapies on the NHS If you're under 18, or want to get help for someone under 18, find out how to get mental health support for children and young people. Medication may be recommended if self-help techniques and lifestyle changes aren't effective in controlling your symptoms. You'll usually be prescribed a course of selective serotonin reuptake inhibitors (SSRIs), which are also used to treat anxiety and depression. In severe cases of agoraphobia, medication can be used in combination with other types of treatment, such as CBT and relaxation therapy. Read more about treating agoraphobia. Page last reviewed: 31 October 2022 Next review due: 31 October 2025
6132	https://ntrs.nasa.gov/api/citations/19780004170/downloads/19780004170.pdf	19780004170 NASA-RP-1009 78N12113 An introduction to orbit dynamics and its application to satellite-based earth monitoring systems NASA Reference Publication 1009 An Introduction to Orbit Dynamics and Its Application to Satellite-Based Earth Monitoring Missions David R. Brooks Langley Research Center Hampton, Virginia National Aeronautics and Space Administration Scientific and Technical Informalion Office 1977 PREFACE This report provides, by analysis and example, an appreciation of the long-term behavior of orbiting satellites at a level of complexity suitable for the initial phases of planning Earth monitoring missions. The basic orbit dynamics of satellite motion are covered in detail. Of particular interest are orbit plane precession, Sun-synchronous orbits, and establishment of conditions for repetitive surface coverage. Orbit plane precession relative to the Sun is shown to be the driving factor in observed patterns of surface illumination, on which are superimposed effects of the seasonal motion of the Sun relative to the equator. Considerable attention is given to the special geometry of Sun-synchronous orbits, orbits whose precession rate matches the average apparent precession rate of the Sun. The solar and orbital plane motions take place within an inertial framework, and these motions are related to the longitude-latitude coordinates of the satellite ground track through appropriate spatial and temporal coordinate systems. It is shown how orbit parameters can be chosen to give repetitive coverage of the same set of longitude-latitude values on a daily or longer basis. Several potential Earth monitoring missions which illus-trate representative applications of the orbit dynamics are described. The interactions between orbital properties and the resulting coverage and illumina-tion patterns over specific sites on the Earth's surface are shown to be at the heart of satellite mission planning. A requirement for global surface coverage is seen to be a potentially severe constraint on satellite systems, depending on the mission objectives. In some contexts, this requirement strongly sug-gests multiple-satellite systems, whereas in less restrictive situations, "global coverage" can be redefined such that a single satellite can provide adequate data. The approach taken in the examples stresses the need for consid-ering mission output as a whole, treating the entire data base as an entity rather than a series of separate measurements. This is consistent with use of this type of data as input to large-scale, long-term activities which can best justify the use of satellite systems. iii CONTENTS PREFACE .................................. iii I. INTRODUCTION ............................ I II. DYNAMICS OFEARTH-ORBITING SATELLITES ................ 2 Coordinate Systems in Space .................... 2 Coordinate Systems for Time and the Relationship Between Longitude and Right Ascension .................. 5 Equations for the Propagation of Satellite Orbits ......... 9 Sun-SynchronousOrbits ...................... 15 III. ESTABLISHING PATTERNS OFSURFACE COVERAGE FROM A SATELLITE ..... 20 Ground Tracks for Circular Orbits ................. 20 Designing Orbits for Repetitive Longitude-Latitude Coverage .... 22 Solar Illumination Considerations in Satellite Orbit Analysis . 25 Pointing Angle Geometries for Locating the Sun .......... 28 The Geometry of Sunrise and Sunset as Viewed From a Satellite . 30 IV. SOME EXAMPLES OFGEOGRAPHICAL COVERAGE ANDSOLAR ILLUMINATION PATTERNS FORLONG-DURATION EARTH MONITORING MISSIONS ....... 33 Establishing a Needfor Repetitive Coverage Patterns ....... 33 Yearly Illumination Patterns Over a Single Site .......... 34 Yearly Variations in Illumination Angle as a Function of Latitude ............................ 37 Solar Illumination Effects on Global Surface Coverage ....... 38 Tangent-Point Distributions For Solar Occultation Experiments 41 V. SYMBOLS ............................... 47 APPENDIX A - "BASIC" LANGUAGE ALGORITHMS FORCALCULATING ORBIT PARAMETERS ............................... APPENDIX B - ORBITELEMENTS FORTHEEARTH'S MOTION AROUND THESUN..... APPENDIX C - DETERMINING THECONDITIONS UNDER WHICH THESUNIS OCCULTED BY THEEARTH RELATIVE TOA SATELLITE HAVING FIXEDORBIT ELEMENTS ................................ REFERENCES ................................ TABLES .................................. 51 55 56 59 60 v I. INTRODUCTION Since the beginning of the space age, satellites have been employed to examine our own planet. There is currently an abundanceof missions - specula-tive, planned, and operational - to exploit the capabilities of a variety of space-based sensing systems in order to understand, monitor, control, and uti-lize the Earth and its atmosphere. Oneof the initial steps in designing such missions is, invariably, to gain an understanding of howa satellite orbit can best be matched to the requirements for meeting one or more objectives. This matching involves not just selecting the best initial or instantaneous orbit parameters but also attempting to maximize the useful return of data over the desired range of conditions for the entire life of the mission. The purpose of this report is to provide, by analysis and example, an appreciation of the long-term behavior of orbiting satellites at a level of complexity suitable for the initial phases of planning Earth monitoring missions. Section II of the report, "Dynamics of Earth-Orbiting Satellites," deals with the basic orbit dynamics required to depict the motion of satellites in their orbits. The time scale of interest for Earth observation ranges typi-cally from a few months to a few years. This meansthat the major effects of the Earth's nonsymmetrical gravitational field must be accounted for but some less significant orbital perturbations can be ignored at this stage of analy-sis. With these limits in mind, it is possible to achieve a sufficiently rep-resentative and self-consistent picture of satellite motion which allows an analyst to study and comparevarious orbit options while avoiding the complexi-ties and loose ends of higher order orbit dynamics. A subsection of Section II deals with Sun-synchronous orbits, which have important implications becauseof their special geometrical relationship to the Sun's apparent motion about the Earth. Section III, "Establishing Patterns of Surface Coverage From a Satellite," deals with viewing the Earth's surface and the ways in which orbits can be modi-fied to produce particular patterns of coverage. The problems of keeping track of the Sun's apparent motion with respect to a satellite are examined in detail. This involves such requirements as looking at (or not looking at) the Sunfor measurementsor calibration as well as accounting for variations in lighting con-ditions beneath the path of a satellite during the entire course of a mission. Section IV, "SomeExamplesof Geographical Coverage and Solar Illumination Patterns for Long-Duration Earth Monitoring Missions," presents sometypical applications of the orbit analyses of the previous sections. Several differ-ent types of missions are considered, each with its own sampling and coverage requirements. The limitations of a single satellite in satisfying someglobal coverage requirements are illustrated, and the alternative of multiple-satellite systems is briefly discussed. The general aim of the analytic sections of this report is to present a tutorial approach to orbit dynamics around which a mission analyst, starting with no previous knowledge of orbital mechanics, can construct a set of self-contained and self-consistent computing tools. The requirements for the exact form of such tools will vary with circumstance, so that no attempt has been made here to formulate specific computer programs, although some BASIC language algo-rithms have been given in appendix A for a few standard calculations. In the applications section (Section IV) the aim has been not to exhaust the possibil-ities of analysis for each topic of interest, but to provide, through typical situations, a feeling for the potential and limitations of space-based Earth monitoring missions. In particular, Section IV is intended to provide some insight into several types of coverage patterns and how they develop over seasonal and yearly time periods. II. DYNAMICS OF EARTH-ORBITING SATELLITES In this section, the equations required to propagate a satellite along its orbit are considered. First, the necessary coordinate systems are derived for locating a satellite in time and space relative to the Earth and Sun. Then, first-order perturbations are introduced to take into account the well-known oblateness of the Earth, the result of which is that orbital paths are not fixed in inertial space but precess about the Earth. Coordinate Systems in Space As outlined in the Introduction, analysis for Earth observations involves locations on the Earth's surface (the satellite ground track, for example) and also the Sun's position relative to a satellite in space or its viewing point on the Earth's surface or elsewhere. Thus, it is natural to assume that two basic coordinate systems for locating points will be required. One is a system of locating points on the Earth's surface -longitudes and latitudes -and the other describes the apparent motion of the Sun through space relative to the Earth. The longitude-latitude L-k system is described in figure I. The equator forms the x-y plane, and the x-axis is arbitrarily defined by a vector in the Z f/ i \ Greenwich / meridian/ / / i Figure I.-Definition of a longitude-latitude coordinate system. equatorial plane which passes through the samemeridian that passes through Greenwich, England. This meridian has a longitude of 0° by definition. The latitude of a point P is defined as the angle between the equator and P, mea-sured along the meridian passing through P. Longitude is defined as the angle between the Greenwich meridian and the meridian passing through P. Latitude is usually defined as having values between 90° (North Pole) and -90° (South Pole). Longitude usually takes positive values between 0° and 180 ° for points east of Greenwich and between 0o and -180° for points west of Greenwich, although it is often convenient in computer applications to define longitude between 0° and 360 ° measuredeast from Greenwich. The Earth-Sun geometry is illustrated in figure 2. The ecliptic plane is defined by the Earth's motion around the Sun. The x-axis of an Earth-centered / Ecliptic S-'--/ / n°rth p°le // -''' A / o . ox / Equatorial plane Figure 2.-Earth's motion around the Sun. Cartesian coordinate system is defined by the vector pointing from the Earth to the Sun at the instant of the vernal equinox, that is, the instant at which the subsolar point crosses the equator moving from south to north. This Earth-centered geometry is shown in figure 3. The celestial sphere is a fictitious globe of apparently infinite radius which contains the "fixed "I stars against which the Sun, the planets, and the satellite appear to move as viewed from the Earth. The projection of the Earth's equator onto the celestial sphere is called the equinoctial and the ecliptic is defined by the Sun's trace as it IAs used herein, "fixed "means that the motion of stars against the celes-tial sphere is too small to be noticed on the time scales of importance in this analysis. North Pole Figure 3.-Celestial sphere and right-ascensionmdeclination coordinate system. appears to move among the stars. The vernal equinox represents a particular location on the celestial sphere, determined by the intersection of the equinoc-tial and the ecliptic, as the Sun moves from south to north. From the signs of the zodiac, this intersection is called the first point of Aries and is repre-sented by the symbol for the ram's head T. This definite point in the heavens is now located about 15 ° west of the constellation Pleiades. (The directions east and west on the celestial sphere are defined in the same way as an observer would identify these directions on a globe model of the Earth.) The reference planes and directions thus established depend on the motion of the Earth about the Sun and about its own axis. The Earth can be thought of as an orbiting gyroscope which maintains the reference systems and their inter-relationships. For many purposes, this is a sufficiently accurate mental pic-ture, but in fact the gyroscope analogy suffers from some irregularities. Chiefly because of the small unsymmetrical forces of the Sun and the Moon on the Earth's equatorial bulge and secondarily because of the perturbing effects of the other planets, the Earth's axis precesses in space at the rate of about 50.25 seconds of arc in a year, advancing the time of the vernal equinox by about 20 minutes per year. This results in a slow but steady westward motion in the location of the first point of Aries, with a period of about 25 725 years. An additional secondary effect is the variation of the obliquity (incli-nation) of the Earth's equator to the ecliptic plane. The obliquity is cur-rently about 23°.44 and is decreasing at a rate of about 47 seconds of arc per century. Detailed discussions of these effects are given in reference I. 4 It is easy to see from figure 3 that the position of a point P on the celestial sphere can be defined by two angles relative to the vernal equinox vector and the equinoctial. The right ascension _ is measured along the equi-noctial positive to the east from T, and the declination _ is measured north from the equinoctial. The dizzying change in perspective from figure 2 to fig-ure 3 is actually simply accomplished by vector transformations. Starting with an Earth ephemeris (a table of or an analytic expression for Earth positions relative to the Sun as a function of time), the negative of the Earth's position vector in heliocentric space is the Sun's position in a geocentric ecliptic sys-tem. Since the ecliptic and equatorial systems share the same x-axis (that is, T), the transformation into the equatorial (equinoctial) system involves only a vector rotation about T. Note now that the L-k and -6 coordinate systems both have the equatorial plane as their basis, with only the difference in x-axis to distinguish them. Note too, that values of right ascension and declination, although defined against the celestial sphere, are as suitable for defining points on the Earth's surface as are longitudes and latitudes. For this analy-sis, declination and latitude are, in fact, equivalent and can be used inter-changeably. Finally, note that the right ascension-declination system is an inertial coordinate system in space as perceived by an observer having the van-tage point implied by figure 2. That is, although the origin of the coordinate system rotates around the Sun with the Earth, the directions of the three axes remain fixed in space. From the vantage point of an observer on Earth, an iner-tial system is one whose axes are pointed at fixed positions on the celestial sphere. The discussion of the previous paragraph shows that this inertial frame-work is only an approximation, as could eventually be determined (and has been) by patient observation of the Earth-Sun system as shown in figure 2. However, for present purposes, the inertial framework is a useful and very good approxi-mation for analysis and will henceforth be accepted as valid. The relationship between the inertial (right-ascension--declination) and rotating (longitude-latitude) systems depends on the location of the Greenwich meridian relative to T, the specification of which requires definition of a coordinate system for time. This problem is dealt with in the next section, which digresses long enough to present the necessary background information and ends with the equa-tions which allow transformation back and forth between longitude and right ascensi on. Coordinate Systems for Time and the Relationship Between Longitude and Right Ascension The definition of a time-measuring system requires an agreed-upon starting point and a set of periodic events which can be used to measure intervals of time. Fundamental to an understanding of the physical sciences is an acceptance of the existence of a uniform time which corresponds to the time variable in dynamical equations. However, in practice, it is necessary to admit that mea-sures of time exist only as matters of definition, that these definitions are imperfect ones applied to particular situations, and that discrepancies will always exist at some level of application. If the Earth-Sun-celestial sphere system is used as the means of defining a time system, there are two conventional ways to proceed -sidereal time and 5 solar time. Sidereal time is governed by the rotation of the Earth relative to the stars. If the instant of passage of a particular star over a meridian is noted, the star will cross the samemeridian again in exactly one sidereal day, by definition. This definition is complicated by the motion of the vernal equi-nox, but for most purposes the problems are not worth considering. Solar time is governed by the rotation of the Earth relative to the Sun. If the instant of passage of the Sun over a meridian is noted, the Sunwill cross the samemeridian again in exactly one solar day, by definition. This observable solar day is the basis of what is called apparent solar time. It is not hard to see the difficulties in applying this definition to practical measurements. It is complicated by the slightly elliptical motion of the Earth around the Sun and by the fact that the Sun does not movein the equatorial plane, with the result that the observed length of solar days is only approxi-mately constant. To circumvent this problem, "meansolar time" is defined rela-tive to a fictitious "meansun" which movesalong the equinoctial (the projec-tion of the equator on the celestial sphere) at a constant rate equal to the average apparent rate of motion of the real Sun. Although the meansolar time system is the basis of all civil time measurements- the time by which people set their watches - it is still full of inconsistencies because of the variable rotation rate of the Earth, the fact that the fictitious meansun cannot be observed, and the fact that the accepted description of the fictitious meansun is based on another time system (ephemeris time), which dependson the internal consistency of the gravitational theories describing the entire solar system. However, at somepoint it is necessary to cease worrying about discrepancies and accept meansolar time as a precise and uniform method of relating nonsimul-taneous events. Hereinafter, "time" meansmeansolar time unless stated other-wise. A "day" is 86 400 meansolar seconds, by definition, and a "second" is always a meansolar second. The meaning of a "year" is complicated by the variety of choices available. Thus, a sidereal year is based on the motion of the Earth relative to the stars -the cycle time for astronomical observation. A year defined relative to the Sun is based on the passage of the Earth from one vernal (or autumnal) equinox to the next. This interval of time is called a tropical year - the cycle time for Earth seasons. It is nearly constant, but not quite, because of the interaction of the planets in the solar system. It is slowly decreasing linearly with time at a rate of 0.53 sec/century (ref. I); a current approximate length, good for the pertinent time frame, is 365.24220 days. This discussion makesclear that calendar systems are bound to be complicated, as they seek to maintain seasonal relationships by using units (days) which are not exact subdivisions of the tropical (seasonal) year. Within the framework of meansolar days, there are two time systems of interest in the bookkeeping sense. The first is universal time (UT), a unit of which is, for all practical purposes, equal to the corresponding unit of mean solar time. Universal time is the practical basis of civil time. It is expressed in hours, minutes, and seconds or in fractions of a meansolar day. The zero reference for a universal time day is Greenwich midnight, at the start of a calendar day. Hence, Jan. 1.0 is the start of calendar day Jan. I, at Greenwichmidnight. The other system of interest, and an important one for analysis, is the Julian date (J.D.) system, which is a method of counting meansolar days consec-utively. Its origin was set long enough ago that all recorded astronomical events can, in principle, be assigned an unambiguousJulian date. Since the system was devised by astronomers for their own purposes, the Julian day starts at Greenwich noon so that all astronomical observations during a night of view-ing activity have the sameJulian date. Onestandard reference epoch for Julian date tables is Jan. 0.5, 1900 (UT). This is Greenwich noon on calendar day Dec. 31, 1899, or Julian date 241 5020.0. Julian dates are used extensively for keeping track of days in preference to the more cumbersome calendar system. A Julian year is exactly 365.25 days, the average length of the calendar year according to the Julian calendar, and, advantageously, is an exact decimal frac-tion. A Julian century is 36 525 days. Table I, from reference I, lists Julian dates for the first day of each month in the last half of the 20th century. Note that the calendar day starts at a Julian date of XXXXXXX.5, so that Jan. 1.0, 1950, is J.D. 243 3282.5. Returning now to the relationship between right ascension and longitude, it can be expressed in terms of a constant related to a particular epoch, the Earth's rotational rate, and somemeasureof time. Reference I gives the fol-lowing equation for the right ascension of the Greenwichmeridian at 12 midnight (0 hours UT): _g,o = 99°'6909833 + 36000°'7689T + 0°'00038708T2 (I) where T is the time in Julian centuries from Jan. 0.5, 1900: T = J.D. -241 5020.0 36 525 (2) The appearance of a second-order term in T is due to the accumulated effects of small secular and periodic variations in the Earth's rotational rate and cannot be accounted for in terms of the previous discussion; this part of equa-tion (I) must just be accepted as fact. Note that although equation (I) appears to give ag,o as a continuous function of T, it does not actually do so. The value of T must correspond to a time of O h UT, that is, to Julian dates of the form XXX XXXX.5. A detailed discussion of equation (I) and its relationship to the mean solar time system may be found in reference I. At this point, it is useful to reconsider the concept of sidereal time, that is, time relative to a "fixed" x-axis -the vernal equinox. In this sense, the angular position of any meridian is equivalent to some fraction of a side-real day, with 360 ° = 24 sidereal hours. Thus, sidereal time and right ascen-sion are equivalent measurements of angular position. To determine the sidereal time (right ascension) of any meridian at any time, it is only necessary to know the rotational rate of the Earth @. Then a = eg,o + (t - to)e + Lm-ag + Lm (3) where t -to is the universal time of interest and Lm is the longitude of the meridian. The Sun's apparent sidereal rate is 360 ° in 365.2422 mean Solar days, so that the Earth must rotate 360 + 360/365.2422 = 360.9856473 degrees 7 per mean solar day or 0.25068447 degree per mean solar minute. An example (from ref. 2) will make clear the necessary computations: What is the local sidereal time !right ascension ) of a point at 298°.2213 east longitude, at 10h15m30 s UT on Oct. 12, 1962? (The superscripts show the conventional notations for hours, minutes, and seconds.) Convert to Julian days: 243 7938.0 +0.5 +11.0 243 7949.5 J.D., Oct. 0.5, 1962 (from table I) Add 0.5 day to obtain J.D., Oct. 1.0, 1962 Add 11 days to obtain J.D., Oct. 12.0, 1962 J.D., O h UT, Oct. 12, 1962 From equations (I) and (2), T = 243 7949.5 -241 5020.0 = 0.62777549 Julian century 36 525 g,o : 99°'69098 + 280°'40033 + 0°'0001525 = 20°'0915 10h15m30 s : 615.5 min And from equation (3), eg = 20o.0915 + (615.5 -0)(0°.25068447) = 174°.3878 Thus, = 174o.3878 + 298°.2213 = 112o.6091 Note the convention of expressing sidereal "time" in terms of an angle. It could be expressed as 112o.6091 = (24)(112.6091/360) hours = 7h30m26 s, but the time units are not the same as mean solar time unitsl Therefore, to avoid con-fusion, the angular notation is used. A final topic of practical interest in a discussion of time systems is the matter of converting back and forth between the Julian date system and a calendar-clock system. This is a somewhat tedious procedure for hand calcula-tions but is easily taken care of in computerized orbit analysis with the aid of transformation subroutines. It is necessary only to keep in mind that clock times derived from Julian dates are Greenwich times, so that the local clock hour, if that is desired, depends on the time zone of the point of interest. There are 24 such zones, 15 ° wide, one of which is centered about the Greenwich meridian, ±7.5 ° • For the purposes of analysis, all time zones are considered to follow meridians, so that local geographical and "daylight savings" fluctua-tions, such as are common over populated areas of the Earth, are ignored. For example, eastern standard time is 5 hours behind Greenwich mean time, being in the time zone centered at -75 °. Therefore, local noon EST occurs at J.D. XXX XXXX.O + 5/24 = XXX XXXX.208333. Equations for the Propagation of Satellite Orbits In an unperturbed spherically symmetric gravitational field, orbiting objects follow the classical laws of Keplerian dynamics. The position of a satellite In a Cartesian coordinate system can be specified completely either by a set of six orbit elements or by six vector components -a position vector plus a velocity vector. (For a basic review of elliptic orbital motion, see ref. 3.) Figure 4 shows the orientation of an orbit in the right ascension-z ra ( rp r Equator Y r + r a a= P 2 rp = a(1-e) r a = a(1 + e) p = a(1 -e 2) x T Figure 4.-Defining the position and velocity of a satellite in orbit around the Earth. declination system previously discussed in detail. Two of the orbit elements, right ascension of the ascending node _ and inclination i, position the orbi-tal plane in space. The remaining angular elements, argument of perigee m and true anomaly f, specify the angular position of a satellite at point P along its orbital path relative to the x-y (equatorial) plane. The size and shape of the orbital path are determined by the semimaJor axis a and eccentricity e. (Many orbits of interest for Earth monitoring have semimajor axes between about 6600 and 8000 km and are circular, or nearly so, with eccentricities in the range 0.0 to 0.1.) The semimajor axis a is the average of the perigee and apogee rp and ra (minimum and maximum radii relative to the source of the gravitational field), as shown in figure 4. The eccentricity e can be thought of as a dimensionless measure of the departure from a circular orbit of radius a, re]ating a to rp and ra as shown. The semilatus rectum p is a derived quantity which will be used in the following analysis; it is measured through the gravitational center along a line perpendicular to a line from peri-gee to apogee and is equal to the semimaJor axis for a circular orbit. Knowl-edge of all six Keplerian elements -a, e, i, , , and f -is equivalent to knowing the satellite's state vector (position and velocity), which may be obtained in terms of orbit elements as follows (ref. 3): rx = [cos f(cos _ cos _ -cos i sin _ sin ) + sin f(-sin _ cos -cos i sin _ cos _ P I + e cos f (4a) ry = [cos f(cos _ sin _ + cos i cos _ sin ) + sin f(-sin _ sin + cos i cos _ cos _ P I + e cos f (4b) P = Lrc°s f sin i sin _ + sin f sin i cos _71-r z f + e COS (4c) _ [(cos + rx : f e)(-sin _ cos _ cos i sin _ cos ) -sin f(cos _ cos Vr -cos i sin _ sin )] (4d) _y : _cos f + e)(-sin _ cos _ + cos i cos _ cos ) -sin f (cos m sin + cos i cos _ sin )] (4e) rz : _p___cos f + e)(sin i cos _ -sin f sin i sin _ (4f) The gravitational constant _ is equal to 398 601.2 km3/sec 2 for the Earth; it is this quantity which takes into account the mass associated with a gravita-tional source and, hence, determines the magnitude of the velocities associated with orbits around a particular body. The inverse of the transformation in equa-tions (4), from state vector to orbit elements, and many other relationships referring to Keplerian orbits and their elements are summarized in reference 3. Because of the Earth's oblateness, not all the Keplerian elements are con-stant for Earth satellites. In particular, the orientation of the orbital plane and the argument (location) of the perigee m precess as a result of the effects of the bulge around the Earth's equator. These precessions, which can be understood in terms of gyroscopic analogies, were thoroughly discussed in the late 1950's. Reference 3 contains an extensive bibliography. The perturbations in _ and _ are significant, amounting to several degrees per day, and there-fore must be accounted for. The usual treatment of gravitational potential involves expansion of the potential in a series. The following discussion utilizes the first-order theory of Kozai (ref. 4), involving the so-called J2 term in the gravitational repre-sentation. In this formulation I0 = 3 J2 r@-2 M(2 - 2.5 sin 2 i) 2 p2 : _ 3 J2 r_2 _ cos i 2 p2 where J2 = 1.6238235 × 10-3 (a dimensionless constant) 2 (5a) (5b) r@ Earth's mean equatorial radius, 6378.145 km perturbed mean motion of the satellite in its orbit, rad/sec The perturbed mean motion M (in radians per second) is based on a modification of the unperturbed (Keplerian) value, which is Mo : 2__ (6) T o where To : 2_aV (7) Then, according to Kozai, M = MO_ + _ J2 r_2 _ -e2_ - _ sin2 ]--(8) The mean anomaly at any time is, by definition, just the constant value of multiplied by the time interval from perigee. It is equal to the true anomaly f only for circular orbits, since according to Kepler's laws of orbital motion, the time derivative of true anomaly f is a function of position in an orbit when e # 0, being larger (faster) at perigee than at apogee. (If this were not so, orbiting objects in elliptical orbits could not sweep out equal areas in equal time, as required by Kepler's laws.) By definition, the time required for the mean anomaly to proceed from 0 to 2w tad (0 ° to 360o), that is, from perigee to perigee, is called the anomalistic period TA: A -Z. (9) There is another equally valid definition of period brought about by the motion of the perigee -the nodal period T N. Figure 5 illustrates the geometry. According to this figure, 2 (10) TN - _ + 11 Initial __ condition: o)o , Anomalistic per iod Perigee Node to perigee to node Nodal period Figure 5.-Anomalistic and nodal orbit periods. For perigee precession in the direction shown, the anomalistic period is a lit-tle longer than the nodal period. Note that the nodal period is the one by which equator crossings of an object in a circular orbit would be counted. That is, if a satellite started at its ascending node on the equator, it would cross again after one nodal period. Note, too, that TN as defined only describes the observable behavior of objects in circular orbits, or the mean motion for elliptical orbits. That is, the nodal crossing times of an object in an elliptical orbit - what might be called the observed nodal periods - are not constant but vary with the location of the perigee. Such an object does not cross a given latitude after a length of time TN" Therefore, a distinc-tion must in general be made between the actual position (true anomaly) andthe "average" position described by the preceding equations for the mean anomaly. The relation between the two involves the solution of a transcendental equation, Kepler's equation, a discussion of which can be found in any textbook on orbital mechanics. Briefly, M": E - e sin E (11) where M and E must be expressed in radians and E is called the eccen-tric anomaly; E can be directly related to true anomaly in a number of ways (ref. 3). For example, f : cos-l{ cos E-e I - e cos E/ (12) There are many ways to solve equation (11) iteratively for E. For example, an easily programed "graphical" solution is given in reference 5. In many cases, a series expanslon to the required accuracy can be done without iteration to get directly from M to f. (See ref. 3.) Equations (5) can be used to propagate Earth orbit elements in the absence of additional perturbations. Of course, a consideration of atmospheric drag would introduce such perturbations -specifically, a secular (nonperiodic) change in semimaJor axis. There are.other sources which could be considered too, in addition to those due to the neglected higher order gravitational terms. 12 One example would be the effects of solar radiation, which are known to cause substantial orbit perturbations on hollow balloon satellites. These effects will be ignored for analysis of Earth observation mission, as they will add little to understanding patterns of surface coverage and other pertinent relationships. At any time t after some initial orbit configuration, (13a) (13b) A useful formulation gives the changes A_ and A_ anomaly step AM: A_ = (1.6238235 x I0-3)r_ 2 (2 -2.5 sin 2 i) AM p2 corresponding to a mean-(14a) (-1.6238235 x 10-3)rm2.cos i p2 (14b) Expressing AM in degrees gives , and A_ units of degrees. The time required for a step AM is t : TA AM 360 (15) Note that for AM : 360 ° , the time required for a step will be one anomalis-tic period. Often it is desirable to use nodal periods to generate conditions at successive crossings of a particular latitude. Then AM should be modified: AM N : AM T-- N (16) TA Thus, in a computer program for generating orbits, AM = 360 ° could he input to produce one orbit revolution, and if desired, it could be modified in the pro-gram by an input flag to generate nodal instead of anomalistic revolutions. Of course, any fraction or multiple of either anomalistic or nodal revolutions can be generated with equations (14) to (16) by the appropriate choice of AM. As a final comment on the generation and propagation of orbits with the first-order equations, note that the concept of a perigee must be retained even for circular orbits. It must be assigned an initial value and its motion must be accounted for with equation (14a). This follows from figure 5, which shows that from the point of view of an observer on the Earth, the precession of the perigee in a circular orbit is part of, and cannot be distinguished from, the motion of the satellite around its orbit. Thus, if an observer timed successive equator crossings of a satellite in a circular orbit, he would measure the nodal period, which can only be related to the orbit dynamics through both the mean motion and the perigee precession. The distinction would become clear to the observer only if the satellite were in an eccentric orbit, in which case he 13 would note that the altitude at each crossing and the time between crossings undergo a periodic cycle with a period equal to the time for the perigee to precess 360 ° . In order to illustrate the magnitude of first-order effects on typical Earth orbits, table 2 lists Keplerian (unperturbed) and perturbed periods for a range of satellite altitudes h and inclinations i, and table 3 lists pre-cession rates _ and _ as a function of inclination over the same range of altitudes. For the usually small values of e associated with Earth observa-tion studies, the restriction of tables 2 and 3 to circular orbits does not pre-vent the values from being representative, as can be seen by examining the con-tribution of e (through p) in equations (5). The information in tables 2 and 3 is also plotted in figures 6 and 7, respectively. 7000 6800 /" 6600 ./ 6400 / 6200 / °' / sec 6000 / 5800 / / 5600 / 5400 / 52OO 2O0 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 Alti_de, km (a) Keplerian period. Figure 6.-Keplerian period and deviations from the Keplerian period TO as a function of orbit altitude and inclination. 14 10 -2 h = 1500 k I TN -TO -6:2:::::: ::__or -10 / TA -To, h = 200 km sec TN - To, h = 200 km TN - ¢o, h = 1500 km TA - TO, h = 200 km 'A - To' h = 1500 krn 14 h-I o0 7 -18 rh_ 200 km -22 --/ -26 0 10 20 30 40 50 60 70 80 90 i, deg (b) Deviations from a Keplerian period. Figure 6.-Concluded. Sun-Synchronous Orbits For many types of Earth observation it would be desirable to view the same surface regions repeatedly under constant lighting conditions. If it were possi-ble to establish an orbit whose nodal precession matched the solar precession, then the Sun-orbit plane geometry would be fixed, with the lighting conditions at a particular latitude dependent only on the north-south movement of the Sun with the seasons. The best available way to accomplish this is to set the orbi-tal precession rate equal tothe average solar precession rate. Recall that the Sun precesses 360 ° in I tropical year of 365.2422 days. Therefore, from equa-tion (5b), the orbital precession rate, in degrees per day, is 360 : 0.9856473 : _ : (-1.6238235 x 10-3)r@2_ cos i 365.2422 p2 Substituting for M from equation (8) and converting the units of M from radians per second to degrees per day yields: 0.9856473 : -2.0645874 x 1014 cos iF1 + 6.605833 x 104 ) a3.5(i _ e2)2 [ ( _ e2_3/2 (I -1.5 sin 2 i.j (17) 15 Precession rate, deg/day 18 16 14 12 10 6 4 . h, km 00 \ \ \ o\ \ 0 \ \ \ \ 15oo /// / -6 _ // 200 / -8 f I -I0 0 10 20 30 40 50 60 70 80 90 100 i , deg Figure 7.-Precession rates for perigee and right ascension of the ascending node for perturbed circular orbits, as a function of inclination. 16 Note that because of the minus sign appearing in the expression for , the Sun-synchronous inclination will always be greater than 90 °. For many purposes it may be sufficient to ignore the bracketed term in equation (17). As an example, for a = 7000 km and e = O, the Sun-synchronous inclination i s is found to be 97°.874474 with this term ignored. Iterating once by substituting this value for i in the bracketed term and recomputing yield i s = 97°.879528. The sin-gle iteration procedure gives a Sun-synchronous orbit precession which matches the Sun's position after I year to better than 10 -5 deg. The variation of Sun-synchronous orbit inclination for circular orbits up to an altitude of 1600 km is given in figure 8. i s , deg 103 102 i01 100 99 98 97j 96 200 400 600 800 I000 1200 1400 1600 h, km Figure 8.-Variation of Sun-synchronous orbit inclination for circular orbits as a function of altitude h. It is interesting to consider how well a Sun-synchronous orbit actually maintains a constant relationship to the Sun throughout the year. To do this, first assume a fictitious sun 2 that travels at a constant rate along the ecliptic projection on the celestial sphere (fig. 3). This would mean that the ficti-tious sun travels around the celestial sphere at a constant rate of 0.9856473 deg/day. Figure 9 shows how to compare the right ascension of the ascending node of a satellite orbit with the right ascension of the fictitious subsolar 2This "fictitious sun" is an ad hoc mathematical entity for the purposes of this discussion, and is not the same as the fictitious mean sun which forms the basis for the mean solar time system. (See p. 6 and ref. I.) 17 x Celestial / X Fictiti ous subsolar point Eclip ht ascension of satellite [_-" I-.LL:-----'..-.. ascending node after time t Equinoc tial_..------" _c T c = 0.9856473t tan _ = cos 23°.44 tan c c a c = right ascension of fictitious subsolar point after time t Figure 9.-Relationship between a Sun-synchronous orbit and the subsolar point for a fictitious sun which moves at a constant rate along the ecliptic. point. It assumes that a Sun-synchronous orbit is started at the vernal equi-nox, so that the subsolar point is initially in the orbital plane. For constant motion of this fictitious sun, the right ascensions must be equal at each equi-nox (where the ecliptic intersects the equinoctial) and at each solstice (where the ecliptic reaches its highest or lowest point relative to the equinoctial). For the fictitious sun in its circular orbit, the vernal equinox, summer sol-stice, autumnal equinox, and winter solstice occur at exactly I/4-year intervals. In figure 10 the dashed line gives the resulting difference between the right ascension of the ascending node of a Sun-synchronous orbit and that of the fictitious subsolar point for one-fourth of the yearly cycle. The solid line in figure 10 shows the difference between the same satellite quantity and the right ascension of the real subsolar point throughout I year. Also shown in the figure is the actual subsolar declination 6® as the real Sun passes through the summer solstice, autumnal equinox, and winter solstice. The points at which _ = 0 define the equinoxes, and the maximum and minimum values of 6 define The summer and winter solstices, respectively. The three separate o_dinates for Q in figure 10 show when these events occur during the year relative to I/4, I/2, and 3/4 of a year, as denoted by the three tick marks on the time axis. The greater than expected lagging of the real subsolar point is due to the eccentricity of the Earth's orbit around the Sun; the tick marks show 18 o I +o +o o oo 0 .H 0 \ \ o,I . o o I J _d 0 0 C_J 0 P4 I I I c o • _ 0 co L. 0 _ 0 . r-4 co 0 o •-- o m o ! c_ • H o o o c - I o . O 0 0 eJ 0 ...I _ -0 _ 0 goo ,el 19 that the autumnal equinox occurs more than 6 months after the vernal equinox, implying that the Earth moves through its aphelion (farthest point from the Sun) during this period. During the time between autumn and the next spring, the Sun ,catches up" as the Earth moves through its perihelion (closest point to the Sun). Thus the satellite leads the Sun for most of the year, by as much as about 6° . Although these data are typical for the time periods of interest, they are, strictly speaking, a function of time, since the perihelion of the Earth is precessing around the Sun at a rate of approximately 1.7 degrees per century. Hence the year for which the calculations have been made (1981) is specified. III. ESTABLISHING PATTERNS OF SURFACE COVERAGE FROM A SATELLITE The underlying assumption for this section is that any Earth observation mission requires some type of repetitive surface coverage pattern to achieve its goals. To establish these patterns, it is necessary to couple the motion of a satellite in inertial space (that is, with respect to the inertial right-ascension--declination coordinate system) with its motion over the surface of the rotating Earth. For a randomly chosen orbit, there is no reason to expect the satellite ground track (that is, the longitude-latitude history) of the sub-satellite point to be repetitive. This is because, in general, the satellite period bears no useful relationship to the Earth's rotational period unless it is specifically selected for a particular purpose. Ground Tracks for Circular Orbits Several typical ground tracks are shown in figure 11 for circular orbits with a semimajor axis of 7000 km and varying inclinations. One-fourth of the I 8o I I_. ..-..---85 ......... .----,.80 70 J' _ -60 ,_r 50 50 = . ,,... ..--71 ... 2o 0 10 20 30 40 50 60 70 80 90 Longitude, deg Figure 11 .- Ground tracks for circular orbits, through one-fourth of a nodal period. 2O track is shown- the track over one-fourth of a nodal period. The satellite does not cover a longitude of 90° during this time becauseof the rotation of the Earth and, secondarily, the orbital precession terms. The longitude-latitude history relative to an initial point on the equator can be calculated as a function of mean-anomalystep AM (in degrees) according to the following equations: sin-IFsin i sin (AM+ OAM)7 (18) 1 t J AL = tan-1[cos i tan(AM + A_AM _ + A_AM -AM _A(360.9856473) (360)(86 400) (19) Note that these equations work only for circular orbits, or the mean motion for elliptical orbits, for the reasons discussed earlier in reference to the defini-tion of nodal period (eq. (10)). Since orbits for coverage of the Earth's sur-face are commonly circular anyhow, for a variety of reasons, this is not a par-ticularly restrictive limitation. The longitudes and latitudes are referenced to a fictitious spherical earth, even though the dynamical equations of the orbit reflect the fact that the Earth is oblate. Transformation of longitude-latitude data from a sphere to points on an oblate spheroidal surface is easily achieved (ref. 6) but is not considered necessary for the present objective of understanding and modifying patterns of surface coverage. The cumulative effect of the apparent westward drift of the ground track is further illustrated in figure 12, which shows the longitude-latitude trace of 9O 7O 5O 3O 10 -10 _a -30 -5o -7o -9o TN la ='7000 km ' i 33 ° 29 28 27 26 25 24 ¥_3 22 21 I 20 19 18 / 0 60 120 180 240 300 Longi_de, deg Figure 12.-Ground track for I revolution of a circular orbit and starting longitudes for subsequent orbits. 36O 21 a _ircular orbit with i = 63 ° and a = 7000 km. (The reason for the choice of these particular orbit elements will be made clear in a subsequent section. It is immediately clear, however, that the 63 ° inclination covers, in latitude at least, most of the inhabited areas of the Earth.) The longitudes at which suc-cessive orbits cross the equator are shown in figure 12 by the numbered marks along the axis. The westward movement of each successive orbital track is, for any orbit, an obvious consequence of the Earth's rotation. The gaps in longi-tude between successive orbits constitute one of the basic restrictions in obtain-ing surface coverage. There is not much that can be done to alter the magnitude of the gaps, as can be seen from table 4. This shows, as a function of altitude, the change in longitude AL between successive orbital passes over the same lat-itude. To simplify construction of the table, the orbital precession effects, which are considerably less than I degree per orbit, have been neglected. Hence the quantity AL is the rotation of the Earth under a hypothetical unperturbed (Keplerian) circular orbit during one Keplerian period. More exact calculations would require the use of nodal periods and inclusion of the precession effects. Table 4 also illustrates the range of orbital periods available for Earth obser-vation. In principle, any orbital period between about 16 and I or less orbits per day can be obtained; however, for reasons of sensor resolution and launch system capability, most Earth observation orbits will remain at below about 1500 km, or a period of less than about 7000 see. The lower limit on period is set by the lowest altitude which is practical in terms of coping with atmo-spheric drag -about 200 km. The last nine columns of table 4 show the longitu-dinal displacement in kilometers at nine different latitudes for a spherical earth whose radius is equal to 6378.145 km (the mean equatorial radius of the Earth). Designing Orbits for Repetitive Longitude-Latitude Coverage The satellite ground track and successive equator crossings presented in figure 12 at least suggest that the longitude-latitude history will not repeat itself in any useful way; that is, there is no discernible repetitive coverage pattern for this particular orbit. According to table 4, the ground track can-not possibly repeat itself any more often than once a day, as it takes about I day for the Earth's Surface to rotate once under the orbital trace. Except for that restriction, however, it is easy to redesign the orbit to pass over the same longitude-latitude points every day, every 2 days, and so forth. Again, because of the general differences between mean and true anomaly, this discus-sion is valid only for circular orbits, or the "mean" location of satellites in elliptical orbits. Consider the geometry shown in figure 13. At some given time a satellite passes over longitude L, which is shown on the equator but could be at any lati-tude. For repetitive coverage of L, the point L must undergo n revolutions under the orbit trace in the time required for the satellite to undergo m nodal revolutions (where n and m are integers!. The point L moves with respect to the orbit plane at an angular rate e - , and the satellite's orbital rate is M + . (Recall that e is the Earth's rotational rate.) Therefore, (m)(360) = (n)(360) (20) 22 _ North AS Orbit trace I 1VI + ¢b "\ \ _ I_x' Figure 13.-Earth-orbit geometry for defining the repetition factor Q. The repetition factor Q is defined as the number of satellite revolutions divided by the number of revolutions of the point L: Q_m:+ n e- (21) Note that except for the _ and _ terms, Q would be exactly equal to the number of orbits per day; however, because of the need to maintain repeated cov-erage over a long period of time, it is necessary to retain these small terms when designing for a particular repetition factor. An integer value of Q, for example, Q = 14/I, means that the longitude and latitude points covered by a particular ground track will be exactly duplicated 14 orbits later, the next day. What is lost to gain this advantage is that other longitudes, those not covered during the first day, will never be covered If Q is given the value 29 = 14_, the initial longitude coverage will be duplicated on every other 2 2 day after launch, with longitude points halfway in between being covered on the alternate days. Sun-synchronous orbits present a special case in defining Q. Since the Earth's rotation rate e is 360 ° plus the average.solar precession in one day, by definition, and the orbit is designed so that _ is the average solar pre-cession in one day, the synchronous repetition factor is Qs - + _ (22) 360 where the units of M and _ are degrees per day. That is, for Sun-synchronous orbits, Qs is numerically identical with the number of nodal revolutions per day. 23 The range of circular orbit altitudes required to produce specified values of Q (convertible to semimajor axis, in the average sense discussed previ-ously) is shown in figure 14, which also illustrates the small inclination 1700 h, km 1600 _ 1500 _ 1400 __= 90° 13oo \ 1200 0£ II00 1000 9O0 800 700 6OO 500 400 3OO 15 15½ 16 2OO 12 12½ 13 13 14 14_ Q Figure 14.-Circular orbit altitudes h to achieve specified repetition factors. 24 dependence (through 6). An extensive listing of circular orbit parameters for particular values of Q is given in table 5. Repetition rates of I to 10 days and 2 or 3 weeks are considered. Table 6 gives the same data for Sun-synchronous orbits, in which case both semimajor axis and inclination must have a particular value to satisfy two independent constraints: the correct precession rate, dependent on the satellite-Sun geometry, and repetitive ground coverage, depen-dent on the satellite-Earth geometry. Appendix A presents several computing algorithms written in BASIC language for the Wang 2200 computer system. These can be used to calculate many useful orbit parameters based on the first-order perturbations discussed in the preced-ing sections. In particular, orbit parameters which produce orbits having desired repetition factors can be generated; the necessary computations are very tedious to do by hand to the accuracy required for internally self-consistent, long-term orbit propagation. Solar Illumination Considerations in Satellite Orbit Analysis Previously it has been shown that the apparent motion of the Sun about the Earth is not coupled in any discernible way with the Earth's rotation. As a result, the specification of orbit parameters for a particular type of repeti-tive ground coverage must be made independently of what the solar illumination conditions will be during the specified repeat cycle. Even for Sun'synchronous orbits, which maintain a "fixed" geometry with respect to the Sun, figure 4 serves as a reminder that the resulting illumination geometry is not really fixed at all and actually varies in a more complex way than might be expected. As a basis for consideration of solar illumination patterns, table 7 pre-sents a summary of solar position information, starting at the beginning of a calendar year and continuing through the vernal equinox of the following year. The calculations are based on equations for the Earth's motion around the Sun, presented in appendix B. Note that figure 10 can be obtained from these data. It should be emphasized that the large number of significant figures shown in the position data is merely for the sake of internal consistency when working with the data in other contexts (as in computer programs) and need not necessar-ily represent actual solar positions to the indicated accuracy. As an illustration of the relationship between the Sun's apparent motion in its nearly inertial framework and the Earth's rotation about its own axis, con-sider the following problem: It is well known that local clock noon does not in general correspond to high noon -the instant at which the subsolar point crosses an observer's meridian. For an arbitrary date, say May I, 1981, find the actual local clock time at which high noon occurs for an observer at a longitude L m of -70 ° . The first step in the solution is to locate both the Sun and the observer in a common coordinate system. The latitude (declination) does not matter, as the location of high noon only involves lining up the observer's meridian with the subsolar point. The right ascension of the observer is cal-culated from equations (I) to (3). The Julian date for O h UT, May I, 19BI, is 244 4725.5. In the following calculations, note the practice of removing inte-ger multiples of 360 ° from the angular values, and of converting negative angles to their positive equivalents between 0 ° and 360o: 25 T = 244 4725.5 -241 5020.0 = 0.813292 Julian century 36 525 ag,o : 990.6909833 + 36000°.7689T + 0o.00038708T 2 : 218o.838139 mobs : g,o + Lm : 148°'838139 The right ascension of the Sun at O h UT is given in table 7: eQ : 38°.127349 The observer's right ascension changes at a rate equal to the sidereal rotation of the Earth. The corresponding motion of the subsolar point is not constant but changes with the seasons and the position of the Earth in its heliocentric orbit. The rate can be determined to sufficient accuracy by examining the change in solar right ascension as given in table 7. The average of the 5-day changes prior to and following May I, 1981, is Finally, the time at which the Sun crosses the observer's meridian is given by 148°.838139 + 360.9856473t = 38o.127349 + 0.9545191t 360.0311282t = -110°.71079 = 2490.28921 t = 0.69241016 day Therefore, high noon occurs at J.D. 244 4726.19241016 = May I, 1981, 16h37m04 s UT, or 11h37m04 s EST (local clock time). Also of interest in table 7 is the variation of solar declination 6 as a function of time. Clearly, this seasonal variation has a dominant effect on solar illumination during any Earth observing mission. Figure 15 illustrates the range of Sun elevation angles available throughout the year as a function of latitude. (A geometrical description of the elevation angle is given in the following section.) Variation of Sun elevation angle _ with time is shown for summer, winter, and the equinoxes in table 8. The time system used in this table is an hour-angle (B) system; the elevation angle _ is given at any lati-tude X by : 90 ° -cos-1(cos 6® cos X cos B ÷ sin 6® sin X) (23) where B is the equatorial angle from the subsatellite meridian to the subsolar meridian. The reference (0 hours) is local high noon (B = 0°). For example, at a relative time of +2 hours or -2 hours (10 a.m. or 2 p.m., respectively, in a "clock" terminology), the Sun is on a meridian +30 ° or -30 ° away from the local meridian. The use of this relative time system serves as a warning to exercise care when dealing with orbit timing. A "9:00 a.m." orbit could mean that a 26 90 7O 5O 30 Sun elevation 10 angle, deg -10 -30 -5O -70 -90 -90 -70 -50 -30 -10 10 30 50 70 90 Subsatellite latitude, deg Figure 15.-Range of Sun elevation angles as a function of subsatellite latitude, with subsolar declination (time of year) as a parameter. satellite crosses the equator at 9:00 a.m. local time, but such a description might just as well be loosely used to mean that at the equator crossing, the subsolar meridian is 45 ° to the east of the subsatellite meridian. These condi-tions are not, in general, exactly equivalentl The reason for belaboring these distinctions will be apparent in a later example, where care is taken to sepa-rate actual Sun lighting conditions at the Earth's surface from the intuitive feelings about sunlight that an observer from the northern temperate latitudes tends to associate with a given local clock time. As an example of the use of figure 15 or equation (23) to provide limiting values for Sun lighting conditions, consider an observer at 50 ° latitude. The range of available elevation angles is bounded by the intersection of the edges of the rectangle appropriate to the subsolar declination at the date of observa-tion with a vertical line at 50 ° . The subsolar declination is directly related to the date of observation. Using table 7, for example, the observer could relate a particular date to the corresponding subsolar declination for use with 27 figure 15, interpolating as required in the table and figure. Suppose the date is around June 20, near the summer solstice. Then the subsolar declination is at its maximum of 23o.44; there is an appropriate boundary indicated in fig-ure 15 for this value which may be used without interpolation. The maximum and minimum values of _ occur for 8 = 0 ° and 8 = 180 ° , respectively (0 and 12 hours). Thus, either approximately from figure 15 or more accurately from equation (23), the maximum and minimum values of _ at the time of the summer solstice are found to be 63o.44 and -16°.56, respectively, for an observer at 50 ° latitude. Before examining solar illumination patterns during long-duration missions, it is first necessary to establish the reference systems for specifying solar position from the ground and from a satellite. Following that, some special attention will be given to the problem of solar occultation, since the observa-tion of sunset and sunrise relative to a satellite -at the beginning and end of periods of darkness -has particular significance for many types of atmo-spheric profile measurements. Pointing Angle Geometries for Locating the Sun To account for the Sun's location relative to a satellite or its subpoint on the Earth's surface, it is necessary to have three pieces of information: (I) a reference system, (2) the solar elevation or zenith angle, and (3) the direction from which the Sun is shining. The geometry selected for locating the Sun from the Earth's surface is shown in figure 16. The point P is the Z y x T Figure 16.-Pointing angles for the Sun relative to the Earth's surface. 28 subsatellite point, _ is the unit vector to the satellite, and __ is a unit vector in the direction of the Sun. The zenith angle _ is measured downward from the local vertical , and the elevation angle _ is measured upward from the local horizon. The angle n, measured counterclockwise from the parallel of latitude through P (that is, from a unit vector pointing east from P), defines th direction from which the Sun is shining. Note that P is a point of right ascension and declination, so that a sub-satellite point given in terms of longitude and latitude must be transformed to the proper coordinate system. Note, too, that since the Sun appears to be infi-nitely far away, _ has essentially constant components regardless of the location of P. Th_s, P could Just as well be at the satellite, in which case n and _ or _ could serve as pointing angles to the Sun from the satellite. However, for this purpose other reference systems, such as those described in the following paragraph, are more convenient. Figure 17 illustrates a coordinate system for defining pointing angles relative to a satellite. The unit vector _ is normal to the orbit, in the z x F T yy I_ = (sin 9sin i, -cos sin i, cos i) Figure 17.-Definition of a Cartesian coordinate system for pointing relative to a satellite. direction of the angular momentum vector. For circular orbits the heading vec-tor H is in the same direction as the velocity vector. These two vectors, with , form a Cartesian coordinate system. 29 A ^ One way to find the Sun relative to H, N, and _ is to rotate first about _ and then about N. Figure 18(a) shows the pointing angles A and B which result. Note that A is identical with the sun elevation angle . Another useful way to find the Sun is to rotate first about 9, with results as shown in figure 18(b); I is rotation in the plane of satellite motion and 0 is rotation out of plane. i i i _0 / f / / _0 .// r <:> I= tan-l( A = 90 ° -cos -I(@ • ,). O : 90 ° -cos-I(N • ®). Figure 18.-Pointing angles for locating the Sun relative to a satellite. For noncircular orbits it may be useful to consider a coordinate system which retains N, replaces H with the unit velocity vector @, and replaces with a new unit vector @ x N to give a Cartesian system. The Geometry of Sunrise and Sunset as Viewed From a Satellite As previously mentioned, the times during which the Sun rises and sets on the Earth's horizon as viewed from a satellite have a special significance for many types of atmospheric observations. The usual astronomical definitions of sunrise or sunset, based on the appearance of the solar image to an observer on the ground, are not the same as employed here. It is perfectly satisfactory for the present to define sunrise or sunset as the instant at which the center of the solar image (as located by a unit vector _ in the right-ascension--declination system) is coincident with the horiz'Von of a spherical earth of radius 6378.145 km. During sunrise or sunset it is, in principle, possible to obtain vertical profiles of many different atmospheric constituents which atten-uate the Sun's radiation. Of interest during these measurements is the tangent point on the Earth (or on a concentric spherical shell) at which the measurement is made. The geometry is shown in sketch (a). A vector rT from the Earth's 3O V /0 -S Satellite Sketch (a) center is constructed perpendicular to the line from the satellite to the Sun (along ). The path from the satellite to the point of intersection of T along ( is given by the vector 3. The tangent latitude and longitude are the coordinates of the point at which _T pierces the Earth's surface. Depend-ing on the orbit parameters (and hence the apparent sunrise or sunset rate) and the time required for measurements, the tangent point may or may not change sig-nificantly during the course of a series of measurements made as the Sun's image passes through the atmosphere. Thus, although the problem of interpreting solar occultation measurements is, per se, beyond the scope of this paper, the orbital information necessary for taking these effects into account can easily be supplied. Since _ and __ are known, the triangular geometry for finding _ and _T is straightforward: (24) _T = g + _ (25) The apparent sunrise or sunset rate along _T is given by R : _ • eT (26) which can conveniently be expressed as the rate, in kilometers per second, at which the solar image appears to move vertically with respect to the .horizon at the tangent point, or as an angular rate by using the magnitude of . To provide a feeling for the magnitude of the sunrise and sunset rates, assume that the Sun is just rising directly in front of and below a satellite in a circular orbit with a radius of 7000 km. In terms of the angles defined in 31 figure 18, B and 0 are zero and A and I are negative. The geometry is given in sketch (b). v F (7000 k v F T (6378.145 km) Sketch (b) For this simple case: S = \/70002 -6378.1452 = 2884 km W and I = A = -[90 ° -cos-I(s/7000)] = -24o.3 The Keplerian period of this orbit T o is 5828.5 sec, so that (to a good approx-imation even for the perturbed case): v : 2_(7000/5828.5) : 7.546 km/sec The projection of this velocity onto T is v • rT, given by R : v sin A : 3.11 km/sec : 0.062 deg/sec It is easy to see from this case that the Sun's image appears to pass through 50 km or so of atmosphere in less than 20 sec. At the other extreme, however, when the satellite is travelling along the terminator, as when the orbit is entering a period of no occultation for several orbits, the sunrise or sunset rate approaches zero. Then vertical profiles of the atmosphere become more difficult to obtain because the tangent point as previously defined undergoes large excursions across the Earth's surface. The equations for determining the time of sunrise and sunset relative to perigee in an inertially fixed orbit are worked out in appendix C. Strictly speaking, this procedure must be an iterative one, since the orbit actually moves in inertial space between the instant of perigee passage and the calcu-lated time of entry into and exit from a period of darkness. As a practical matter, the iteration is usually not worth the effort. With appendix C, it is 32 possible to propagate an orbit in steps of one anomalistic period, from perigee to perigee, computing the sunrises and sunsets at each step and determining those periods of continuous solar exposure. This information has obvious impor-tance for the scientific objectives of any mission as well as for the engineer-ing design of the satellite and its systems. IV. SOME EXAMPLES OF GEOGRAPHICAL COVERAGE AND SOLAR ILLUMINATION PATTERNS FOR LONG-DURATION EARTH MONITORING MISSIONS It should be clear from the preceding sections that the discussions of orbit analysis were intended to encompass mission situations extending for a year or so -long enough for orbits to precess completely through several cycles, for the Sun to go through an annual illumination cycle, and for repeti-tive coverage of locations on the Earth's surface. The dynamic relationships between a satellite and the targets of its sensors are often decisive factors in assessing the worth of a mission. It is not so much a question of establishing a particular set of viewing conditions at a particular time as it is a question of assessing and optimizing the total output from all measurement opportunities over the nominal course of the mission. This section presents some representa-tive patterns of coverage based on actual orbit analysis experience with several proposed Earth observation missions. The examples are intended to illustrate some important types of coverage and their associated problems. They include (I) yearly illumination patterns over a single site, (2) yearly variations in illumination angle as a function of latitude, (3) solar illumination effects on global surface coverage, and (4) tangent-point distributions for solar occultation experiments. Establishing a Need for Repetitive Coverage Patterns Each of the following examples requires, in some form, the establishment of repeatable patterns; therefore, it is appropriate at this point to consider the general necessity for repetition. The quantities or features of interest for measuring or viewing from a spacecraft have one property in common: they all vary in at least three dimensions -longitude, latitude, and time. If they did not, there would be little justification for viewing them from spaceborne platforms. In order to separate the contributions of each dimension, it is clearly necessary to have repetition - repetition of two quantities to isolate the third. Repeated measurements over the same location (longitude and lati-tude) to isolate a dependence on time is one important example. Note that other desirable repetitive pairings -longitude or latitude with time to isolate the dependence on a space dimension - are not possible with a single satellite, which can repeat itself in space but not in time. Thus, it must be sufficient to match repetition in space coordinates with later but presumably identical conditions in the time coordinate. If the time dependence is obvious or domi-nating, as _n a diurnal cycle, identical conditions may be easy to establish. Otherwise, interpretation of a set of sequential measurements is one of the major problems in utilizing spacecraft data. 33 A less obvious reason for repetitive measurements is the need to determine error associated with measurements. Assuming that the errors resulting from making and interpreting measurements are randomly distributed, multiple measure-ments under the same conditions are required to isolate error from the other factors influencing the value of the measured quantity. The extent to which this can be done depends on the feasibility of substituting equivalent values in the time dimension for identical ones. Spatially distributed quantities are not exempt from similar considerations, as it is often necessary to group nearly identical spatial coordinates with the assumption that variations of interest exist only over spatial scales that are large compared with the differences among such a grouping of available data points. Yearly Illumination Patterns Over a Single Site A straightforward application of the previous analyses is to establish a satellite orbit that gives repetitive coverage over a selected site or group of sites, determine one or more parameters of interest, and study their behavior over the course of a year. This type of parametric study has practical applica-tions in Earth resource management; pollution monitoring; land-use planning; and many other research, survey, or inventory tasks. The example in this section is based on a 1973 study of orbit requirements for observing the eastern coast of the continental United States (ref. 7). In this study, an orbit was selected that gives good coverage of the entire east coast and its performance was assessed by reference to a single site at Norfolk, Virginia, which is roughly midway along the Atlantic Coast. The orbit inclination of 63 ° was selected by minimizing the off-nadir point-ing required to cover 25 east coast sites of interest. This orbit has been noted previously, in the section "Establishing Patterns of Surface Coverage From a Satellite," as covering in latitude most of the inhabited areas of the Earth. A ground track is plotted in figure 12; however, it is a "random" orbit in the sense that its other elements were not selected to give a particular repetition factor for regular coverage of a selected site. For a circular orbit, this means just changing the semimajor axis to the appropriate value. This is easily done with the algorithms described in appendix A. Table 9 gives a list of orbit parameters for a 63 ° circular orbit which gives daily coverage over a particular site (Q = 15). The list includes the starting conditions which will cause the satellite to pass over Norfolk at the desired time on its initial orbit (local clock noon). Note that the anomalistic and nodal periods are nearly identical, as the inclination of 63 ° is very close to the value at which perigee precession is zero. (See fig. 7.) The ground track of this orbit is plotted for the first 15 revolutions in figure 19, as seen from a vantage point in space directly over Norfolk. The parameter of interest in reference 7, and of general interest for any similar mission, is the variation of Sun elevation angle during the repeated passages over the site. Clearly, some of the passes will take place in darkness because of the nodal precession of the orbit. The precession rate relative to the Sun is about 4! deg/day; therefore, it would be reasonable to expect a com-2 plete illumination cycle to take about 80 days. "On top" of this cycle should 34 la" = 28_7.371 kra Q= 15 L o =-760.289 k o = 36°.853 ." / .........i..... Figure 19.-Ground track history of a 63° orbit started at Norfolk, Virginia, with a l-day repeat cycle. Q = 15. be an elevation-angle envelope resulting from the seasonal variation of the Sun's declination. The actual variation in Sun elevation angle over Norfolk at the time of satellite passage is illustrated in figure 20. The points show the elevation angle for an orbit initially passing over Norfolk on Jan. I, 1981, at 17h UT (12:00 noon EST). Bear in mind that the curve is really not continuous but is a set of 365 discrete points, one for each pass over Norfolk during the year. The solid lines show the maximum and minimum available Sun elevation 35 8O 6O Maximum available elevation an Minimum available elevation angle a = 6887.371 km i = 63 ° Q= 15 to = 12:00 noon EST, Jan. 1, 1981 Norfolk, Virginia -8o I l I 0 30 60 90 120 150 210 240 270 300 330 360 Time, days from t o Figure 20.-Sun elevation angle over Norfolk, Virginia, for a satellite in a 63 ° circular orbit. Lo = -760.289 ;to 360.853 angles at this site as calculated from equation (23). It is interesting to consider how little control can be exercised over the resulting illumination pattern. By starting the orbit at a different time of day, the patterns can be shifted within the envelope of available elevation angles; except for this, there are no available parameters for altering the pattern in any significant way. A different inclination will speed up or slow down the illumination cycle without altering its basic behavior. A different value of Q will alter the coverage of other places, all of which will have their own latitude-dependent illumination patterns similar to those in figure 20. The price for repetitive coverage over more sites is, as always, less frequent coverage of each site. The resulting trade-off is a standard study for any Earth-oriented mission with multiple objectives, each of which may have to be compromised for the total optimization of the mission. There is also the possibility of using a Sun-synchronous orbit. Then, most of the rest of the east coast coverage will be lost because orbits at synchro-nous inclination no longer follow the coastline. In this case, the initial tim-ing of the orbit is important, as it can be used to regulate the Sun elevation pattern for the rest of the year, within the allowable envelope. If it is desired to view the nadir at the maximum Sun elevation angle (not necessarily a desirable situation), this condition can be approximately met by starting the orbit at local clock noon. Indeed, the elevation angle for such an orbit started at Norfolk would follow the top solid line in figure 20 and would be indistin-guishable from it on the scale of this figure. Curves for other local times would have the same shape as the maximum-minimum envelopes in figure 20 but 36 would be displaced within these envelopes by an amount proportional to the local time, between noon and midnight. Yearly Variations in Illumination Angle as a Function of Latitude The specific example of a 63° orbit discussed in the previous section can easily be generalized to predict the illumination conditions encountered by any orbit. The envelope of available Sun elevation angles apparent in figure 20 is but one of a family of such curves which are a function of latitude and time. Figure 21 gives 10 such yearly curves for latitudes from 0o to 90° in 10° incre--"---C""-'---I I I I I I i i I I f i 1 I " "'= 120 150 180 210 240 270 300 330 360 Time, days Figure 21 .-Available Sun elevation angle as a function of time at various latitudes, ments. The data may be calculated with the aid of equation (23); they are an extension of the data listed in table 8. It is within these envelopes, and of course their corresponding southern hemisphere equivalents, that the Sun eleva-tion angles vary. As usual, Sun-synchronous orbits present a special case, as their orbital planes do not precess on the average relative to the Sun. Assume that a satel-lite in a Sun-synchronous orbit has been started on the equator at the point and time of the vernal equinox. The Sun elevation angles observed as the satel-lite passes several different latitudes are recorded in figure 22. These curves, for a "high noon" configuration, closely resemble the elevation-angle envelopes of figure 21, but small differences are detectable, especially at lower lati-tudes. These differences arise from the peculiarities of the Sun's apparent motion around the Earth, as previously discussed in conjunction with figure 10. It is often of interest to adjust the initial geometry of Sun-synchronous orbits away from high noon to accommodate sensor_ - photographic, for example - for which longer shadows might be desired to provide contrast in the observed sur-face features. 37 Nadir latitude, deg :-.. 80 • J---4- f \ \ \ //i \ / / // v / 60 50 ' 40 , 30. 20 / / f 0 40 80 120 160 200 240 280 320 360 (T) Time from launch, days Figure 22.-History of observed Sun elevation angle for a Sun-synchronous orbit started at the location and time of the vernal equinox. Solar Illumination Effects on Global Surface Coverage One possible line of background justification for the examples in this sec-tion is the desire to monitor, on a global scale, the radiation balance of the Earth. It is a problem which has widespread appeal because of the importance of the radiation balance in predicting long-term trends in the Earth's weather and climate. The present state of knowledge of the Earth's radiation processes is such that significant advances require accurate measurements on a global scale of the emitted and reflected radiation, in particular. At least one of these, the reflected radiation, presents an exceedingly complex problem in data inter-pretation and requires not just global coverage on a longitude-latitude grid of 5° or so but repetitive coverage of the same locations over a wide range of Sun geometries. This particular problem has been explored in an analysis of orbit requirements for Earth radiation monitoring (ref. 8). It can be viewed more generally as a problem that imposes the most severe possible requirements on the ability to extract adequate data from Earth-orbiting satellites. For the moment, consider the performance of a single nadir-viewing satel-lite in covering longitude, latitude, and time (or equivalently, illumination angle). At the outset it can be stated that coverage of the longitude-latitude grid, within the latitude limits of the orbit, is easy to obtain. In fact, by appropriately specifying the repetition factor Q, it is possible to get exten-sive coverage in these two dimensions over as fine a grid as required. The problem lies in the time coverage, for reasons which are made qualitatively clear by considering the relative slowness with which a satellite orbit moves with respect to the Sun or, put another way, the slowness with which the satel-38 lite orbit plane changes its position relative to the plane of the solar termina-tor. It has already been demonstrated in previous sections that it is possible to cover a particular site at most on a once per day basis, thereby giving up a fine longitude-latitude grid in favor of a coarse one (with longitude points roughly 25 ° apart according to table 4). The illumination on subsequent passes will change daily (recall the coverage of Norfolk, Virginia, detailed in fig. 20), but the coverage does not even approach the requirements for estab-lishing diurnal changes in radiation patterns. The best that can be hoped for is that over a sufficiently long period of time, say a month, adequate coverage will be achieved to allow extrapolation in both space and time to the desired quanti-ties. This process requires extensive modeling of (for example) the radiation environment and an appreciation for the orbit constraints; it poses a problem sufficiently complex to have defied satisfactory solution up to the present time. Since it is really the Sun elevation angle which is of interest for the radiation problem, rather than clock time per se, the single satellite coverage is first expressed in terms of an hour-angle system relative to the satellite, as a function of latitude. (It is the same system previously defined for use in eq. (23), where 8 is the hour angle.) For each part of table 10, the globe has been divided into thirty-six 5 ° latitude strips and the horizontal scale is divided into twenty-four "l-hour" segments. Thus, each step on the horizontal scale represents 15 ° of separation between the subsatellite and subsolar meri-dians. The contents of a latitude-time "box" are increased by one each time a measurement at the satellite nadir point is assumed to have been made within the space and time boundaries of that box. It is assumed that measurements are made from orbit at a constant rate, which amounts to one measurement every 5° of mean anomaly. The results of 30 000 such measurements (about 30 days) are given in tables 10(a), (b), and (c), for 50 ° , 80 °, and Sun-synchronous orbits, respec-tively, at an altitude of 800 km. To make the orbit geometry as easy as possible to visualize, the initial conditions are very special ones: in each case the satellite starts on the equa-tor at the location and time of a vernal equinox so that the first measurement records an hour angle of zero. The boxing algorithm truncates hour-angle values rather than rounding them off, so that the hour-angle equivalent time of "zero" might go into either the first or last box, depending on its precise value. It can be seen that with the given initial condition, the hour angle will then be 6 (a zenith angle of 90 °) at a mean anomaly of 90o; 12, at 180°; 18, at 270o; and 0(24) again, at 360 ° . The imprecision of ignoring the distinction between anomalistic and nodal revolutions is of no noticeable consequence for making this point. The change in the coverage pattern over 30 days is due to orbital plane precession relative to the Sun and, to a smaller extent over this short period of time, the seasonal motion of the Sun into the northern hemisphere. For the 50 ° orbit this relative precession rate is about -5.2 deg/day; for the 80 ° orbit it is about -2.1 deg/day; and for the Sun-synchronous orbit it is, by design, zero. Eventually, the nonsynchronous satellites will make measurements at all hour angles for each available latitude. The question is whether or not the necessary extrapolations can be made to determine diurnal variations and aver-age out or otherwise account for their effects. The difficulty in doing this over the required time scale (say, within 30 days) provides the argument in 39 favor of multiple satellites, which will cause the latitude-time grid to be filled in much more quickly than will a single satellite. The trade-off between the cost of additional satellites and increased mission capability is at the heart of the justification process for this type of spaceborne monitoring pro-gram. One attempt at presenting the options for multiple satellite missions may be found in reference 8, where the ability of various satellite systems to fill in the latitude-time grid in 30 days is examined. A two-satellite config-uration of one 50 ° and one 80 ° orbit was recommended in this study for "zonal" coverage over latitude bands, but such a system still may not be able to pro-vide total latitude-time coverage as quickly as desired for accurate interpre-tation of radiation data. Another solution to this problem is to use multiple Sun-synchronous satellites to give continuous coverage of a particular local illumination condition. Four such satellites might be sufficient to allow con-fident modeling of the diurnal radiation cycle. However, the expense of such an extensive system has so far been an insurmountable obstacle to its implementation. There is another way of looking at the latitude-illumination grid. It can be argued that not only is clock time an insufficient measure of illumination coverage, but relative solar time (hour angle) by itself is not much better, as it still does not unambiguously specify the illumination conditions at the nadir point. 3 Thus, an additional parameter of real interest is the solar zenith angle. These data are shown in table 11 for the same three orbits previously used - 50 °, 80 °, and Sun-synchronous - and may be compared with the corresponding parts of table 10. The pattern of zenith-angle coverage can best be visualized by first considering the Sun-synchronous orbit from which is viewed a dawn or dusk condi-tion (zenith angle of 90 ° ) near the South or North Pole, respectively. The other orbits initially have the same type of pattern, but they precess away from this geometry. In 30 days the 50 ° orbit has precessed about -156 ° relative to the Sun, and thus has viewed dawn and dusk conditions over every available latitude. The 80 ° orbit will fill in its "hole in the middle" in a similar fashion if given enough time -it has precessed only about -63 ° in 30 days. The foregoing discussion of the problem of radiation-balance coverage may appear quite discouraging from the point of view of the mission analyst, who apparently cannot exercise sufficient control over orbital properties to achieve the desired results with reasonable satellite systems. However, the definition of global coverage is to some extent a flexible one which can be expected to relax considerably under different circumstances. (Such a situation will be dealt with in the following section.) In the present case, it is clear that sensors with sufficiently wide fields of view will allow global coverage in some sense. Then, the problem becomes not one of obtaining sufficient data but one of interpreting the available information in an unambiguous way. The implied trade-off between flight hardware for measurements and computer soft-ware for analysis is an important consideration in mission planning but lies outside the scope of this analysis. 3Interpretation of the reflections of solar energy by the Earth's surface requires such an unambiguous two-parameter specification of solar position. 4O Tangent-Point Distributions For Solar Occultation Experiments The three previous sections have dealt with someof the coverage patterns resulting from a nadir-looking satellite system. Another basic type of sensing involves looking not at the Earth but at the Sun under specialized conditions. It has long been recognized that in studies of atmospheric constituents, it is profitable to use the Sun as a radiation source. It is easy to visualize a process whereby solar radiation received by a satellite when the Sun is well above the horizon is comparedwith the radiation attenuated by the atmosphere as the Sun rises or sets through the Earth's atmosphere. In keeping with previ-ous general discussions, it is important to examine not just one or a few of these measurementsin isolation, but to view a set of such measurementsin the context of a total mission. The data to be presented here are applicable to any solar occultation experiment; they were generated originally in support of a proposed mission for measuring halogens in the stratosphere (ref. 9). The question in this case is a typical one: Is the coverage resulting from a flight project (assumedto last I year) sufficient in space and time to permit useful interpretation of the underlying distributions of the variable of interest? The particular example of stratospheric halogens allows an immediate but qualified "yes" to this question, as do manyother proposed experiments, since information of any kind about the distribution of manystratospheric constituents is virtually nonexistent. How-ever, the problem of justifying the return of the mission relative to its cost is another obstacle which must be overcome. Becauseof the paucity of worldwide stratospheric data, the definition of global coverage can indeed be relaxed, as suggested in the previous section. It may at first be sufficient merely to establish average levels over various parts of the globe, taking note of only the coarsest features of the underlying fac-tors which drive the distributions. But it is at this point that the qualifica-tion of the "yes" becomesapparent: What is "coarse"? Howwell can time aver-ages be made? Can spatial and temporal distributions be separated, and to what extent? To help answer these questions, several pertinent sets of solar occulta-tion data are presented in figure 23. In figure 23(a) the distribution of tan-gent latitudes during I year is presented for a 70° orbit, chosen because it is approximately the lowest orbit inclination at this moderate altitude (a = 6978 km) which will allow somepolar coverage. The curves appear to be continuous, but in reality they are madeup of the discrete contributions of the sunrises and sunsets viewed from the satellite throughout the year: 9600 separate occurrences at the rate of about 14.9 per day. The orbit is deliber-ately started during a period of total solar exposure, and its subsequent pre-cession through cycles of occultation and exposure is dependent in a complex way on the time of year of the launch and the precession rate of the orbit rela-tive to the Sun. The small dots represent sunrises; the crosses, sunsets, a convention which also applies to the rest of figure 23. Note that in contrast to nadir-looking coverage, the latitude limits of the tangent point are not restricted to ±70 ° , but extend northward and southward depending on the solar position. The length of the cycles can be altered, similarly to surface illu-mination cycles previously discussed, by choosing orbits with faster or slower precession rates. However, since the precession rate is mostly a function of inclination, this choice cannot be madeindependently of the coverage in 41 • Sunrise + Sunset 9O 6O il I} -30 0 30 60 90 120 150 180 210 240 270 300 330 360 Time from launch, days (a) Tangent latitude on the Earth's surface at sunrise or sunset. nset 120 _ 0 -120 0 30 60 90 120 150 180 210 240 Time from launch, days ? 270 300 330 360 (b) Pointing (yaw) angle to the Sun relative to the plane of satellite motion at sunrise or sunset. Figure 23.-Solar occultation data during a l-year flight for a 70° orbit at h = 600 km. 42 3.5 2.5 1.5 o .5 -.5 oF-4 -1.5 > -2.5, Sunrise 4. --1 Sunset k i { 30 60 90 120 150 180 210 240 270 300 Time from launch, days (c) Apparent vertical rate of the solar image through the atmosphere at sunrise and sunset. 330 360 Figure 23.-Concluded. latitude. Also, the exact shape of the cycles and the location of the periods of total sunlight are determined by the timing and initial geometry of the launch. Therefore, figure 23 must be considered to be only representative in details, while retaining generality in overall cyclic patterns. Figure 23(b) illustrates another important facet of solar occultation exper-iments which must be considered in mission design: tracking the Sun. The quan-tity plotted is a "yaw" angle, previously defined as B, which locates the solar image on the Earth's horizon relative to the direction of satellite motion. The constantly changing satellite-Sun geometry causes an expected variation of B over a range of ±180 °, a characteristic feature of Sun-tracking measurements. A question clearly framed by the figure is: What level of sophistication is warranted in designing a Sun tracker? The problem of finding the Sun when it is "up" and following it as it sets is easier than predicting where it will be just as it rises -a differentiation which carries with it the prospect of los-ing half of all the occultation opportunities. The sunrise-sunset symmetry evident in the pointing requirements relative to 900 and -90 o is encouraging, however, with the possibility of utilizing a sunset to predict the value of B for the following sunrise. A third cyclic pattern of interest during the mission is the variation in the vertical rate R at which the Sun rises or sets. A useful approximation is to consider the value of R at the horizon as characteristic of a particu-43 lar measurement opportunity, even though the rate is not actually constant dur-ing the entire residence time of the solar image within the vertical limits of the atmosphere. This quantity is plotted in figure 23(c). It is clear from previous discussions that the maximum values of IRI occur when the Sun is ahead of or behind the satellite (B = 0 ° or ±180 ° ) and that the minimum values occur at B = 90 ° or -90 ° . Whereas the significance of this number depends on the nature of the experiment being performed, it is easy to see that vertical profiles of the atmosphere over a well-defined spot on the Earth can best be achieved when IRI is as large as possible. At the other extreme the longitude-latitude coordinates can vary considerably during the passage of the solar image through the stratosphere. Without going into details of data reduction, it is reasonable to conclude that the distribution of R over the life of a mission will affect the overall accuracy of the resulting global distributions. The distribution of measurements of IRI in 0.5 km/sec intervals is given in fig-ure 24 for the 9600 measurements of this l-year mission. 3000 Total: 9600 measurements 2500 2OOO Number of measurements 1500 1000 500 0 .5 1.0 1.5 2.0 2.5 3.0 3.5 [R[, km/sec Figure 24.-Distribution of apparent rate of the solar image vertically through the atmosphere as viewed from a satellite during sunrise or sunset. Up to now, no mention has been made of longitude coverage; in the previous section it has been assumed to be less critical than latitude coverage. Tangent latitude is plotted against tangent longitude in figure 25, where it is clear that in some sense the coverage in longitude is uniform at least over a time span of I year; that is, there are no obvious gaps in this plot. It is useful 44 90 Sunset 60 30 - 0 _ -30 -60 0 30 60 90 120 150 180 210 240 270 300 330 3G0 Tangent longitude, deg Sunrise 90 -. ,....-, ... ¢ ,.,.. ... , ....,., .,. -...• , ",,,,., ",. :, ..::, ,.,.9., ,:"t:...,:,, .r. ,;,...-., ,-.. • •.' •.....-" -..-." -.': : " "'-.' "./ .-'.•. ." ... .• m;. .. •,..:....'-.-..-':..... ..;-• -•%".". '.. .'. I..';.. _ .',,; • .,,,. , .-,..,, ,--...'., ..... -,, ".,..-'.-.....'C ...... -.... .-. .... ,.... ".,.. ".-.... -.-,,. "" " • i • "" "' ° • • L. -t... • t. %% 'klk.. .P %. ".-. . ,. . {:--" .... ...... . . ., ..... . .. • ., . .. ... •. . -. • ... -.. ,. . . • , • -• -. . ..o... ..'... •.'•.. .°;. • . .t.. .,,. -.. , -',, • -. ".-. ",-• , "., . • •,'. • "-;. • .< :... • :., . : ".': ,.:,:':,. ,.',:., :.:, ..;,..'..,...,.:.., ,',.,,: :.,.,-,¢ ,,,,'--.,.'-,--..:.,,,-., ,..,,-., 60 .:,.;.;..;..,•.,...;.,.; .... '',..'.;,-...•'%.,, "&." "4.-'..'T':4:."..%". N.-. _2.-..."'i.""" ._m." .. ._r. % .-w.. "....-..... .. "'iLn . %. ;" .• •" ._a "1 .JA I' o.% .m.[l" ;.e b .. -... • ;'-".'' "''' ", • ".. • .". ",';'" " :L"" • •o'." ".'" "," ".". ._4 ,,.. • _ 30-,-". -,". ." • . " . • . ." • • . • . " .,. .,. ., :. ' 0 .,. • . ." . : • •. • ., .• • ,• , ,. ,• , . . :" ,• ;" . ,., , ... • .. .4-) __ •. .. .. . .. _ • • ; ... .. . -• • • . • . • . . . • .•. . 0 ". " .... • ........ ,'" ., •" .." "" .... -30 c_ -60 -9O ...... • ;. -;.. ..... . , •.. ,. :,. %. ,.. -. :,,-".' :. • :. • ".,. " ..,,. -,,. "". "k .. . ,,." . .,... .. ,,... ." ,,.. ." -.' • .." ''.. " ",,,. . . , .. .. . .. _ .. .. .o .. :. :. • .. _ , • ,.•. . ..., ,... ... ...... : .,: • . . • . • • -. % ° • ; • . • .. . ...... . I.'. ." . ." ' _,,," . ." . .,." . ;" , .. ., ,-. .; ,,. ,",., ,., • , •. , ,,,, , .-,, -: • • .'.m --'"" t.''"' ." ;" m '• .e _ .: . ..-"-%. ..:',I, . .'.....,'...,, . ....%. °......... ... ....'. .;,.... . ,.,. . ... .... ,a. • .:,,..-.'¢:..: ....,. .:..',,: "._, . ," :a,r • , • ",. , "-'.¢.. ;,;'. • :.,<. . ,,,,. , ..,, ..., .:. .: .r.: _ ., .. ..,, ..,. .. • , "e,, "-,¢, " ,¢. " "_ ., • • • _ , , " • -" -" -" ., ,, , lr""l" .."7'". ".'7": .." ",. ..... -". "" -• .'' ,,'--'" -. • "" .." "" ." "" -'." ..-.'" . .,.o'" • .',, • ..:..,,-•:, :.;,".../;.../.o..s. .,.. .:-.., ,.:. 4,:.:.:,. _ ":,. :.. ",,.:-- .-':5 " ",::,-'"," . [,;'"I'" ""'I'":" ": I'" "I I r 'I I ,,1 I I 0 30 60 90 120 150 180 210 240 270 300 330 360 Tangent longitude, deg Figure 25.-Distribution of tangent longitude and latitude on the Earth's surface at sunrise or sunset after a -year, solar occultation mission for a 70 °, 600-km orbit• 45 to look at smaller time intervals, however, to see how the longitude-latitude coverage is built up. In figure 26, about 400 orbits from the beginning of the A-Sunset O Sunrise 90 I 3 , 4-" 4-4"4-4. 44. -4-..,. 4-4. 4.4. 4.4. .4.4. ..i. 4. F-. 4..4." @, .4-4-.4--4-..4-4-4-4--4-4-.4-4-.4-4-.4-30- +15 "14 +13 "12 11 10 9 " 8 7 6 5 +4 • 3 2 01 q)l __ 07 06 05 04 03 02 O9 O8 1 l 1 I l 1 l l I 0 30 60 90 120 150 180 210 240 270 300 330 360 Tangent longitude, deg Figure 26.-Distribution of tangent longitude and latitude at sunrise and sunset for about 400 orbits during the first period of occultation. first period of occultation are considered. The first 15 occulted orbits are numbered, starting from the point where the Sun just reaches the horizon so that sunrise and sunset occur at essentially the same time and place. It can be seen that the longitude points are fairly regularly spaced throughout a day - a time interval during which the latitude changes relatively little. This is encouraging from the point of view of taking averages over bands of latitude, a procedure which is commonly used for modeling the stratosphere. The ability of th_s mission to achieve yearly averages of a single strato-spheric quantity as a function of latitude is shown in figure 27. The solid line gives the actual yearly averages of a dimensionless parameter F, produced by a hypothetical stratospheric model which varies in longitude, latitude, and time and has random components to account for measurement and analysis errors and natural random fluctuations. (It can be thought of as representing a total vertical burden of some stratospheric constituent.) A l-year mission which extracts 9600 measurements from this model is flown, and the resulting 36 lati-tude band averages are calculated with standard statistical procedures for grouping and weighting the available data. Although a detailed statistical analysis of the measurements is beyond the scope of this report, the lack of agreement in figure 27 between "theory" (perfect sampling) and "experiment" (available sampling) near both poles can readily be attributed to an insuffi-cient number of temporally spaced measurements near the poles, as is evident in figure 23(a). In any event, the solar occultation technique has obvious potential for space-based stratospheric measurements and may allow global model-ing in certain restricted circumstances. 46 4O F 30 --O A: atUsa_r (ePe_fs_:u lat_Sampling) .O 20 I ] ] I I I I I I I ] ] ] I ] ] ] -90 -60 -30 0 30 60 90 Latitude, deg Figure 27.-Actual and measured (simulated) yearly averages of a hypo-thetical stratospheric constituent as a function of latitude. a A,B C h ^ H V. SYMBOLS semimajor axis, km pointing angles to Sun relative to a satellite, deg (see fig. 18(a)) location of a fictitious sun along the celestial ecliptic, deg (see fig. 9) Julian days since Jan. 1.0, 1900 = d/10000 eccentricity eccentric anomaly, rad true anomaly, deg or rad dimensionless parameter used for evaluating ability of a solar occul-tation mission to extract information on global distribution of stratospheric constituents altitude above surface of a fictitious spherical earth of radius r_, km unit heading vector (see fig. 17) 47 i I,O J2 J.D. L m,n M P P Q r,r re R -9 S t T .9 V,V x,y,z 8 inclination of an orbit, deg pointing angles to Sun relative to a satellite, deg (see fig. 18(b)) term in expansion of Earth's gravitational potential (see eqs. (5)) Julian date or day longitude, deg integers (see eq. (20)) mean anomaly, deg or tad unit vector normal to an orbit (see fig. 17) semilatus rectum, km point on Earth's surface, along a satellite orbit, or on the celestial sphere orbit repetition factor satellite distance from center of Earth or a position vector, km Earth's mean equatorial radius, 6378.145 km rate at which solar image appears to rise or set on the horizon as viewed from a satellite; km/sec or deg/sec vector in direction of Sun as viewed from a satellite, with length as defined in sketch (a), km time; days, min, or sec, as defined in context Julian centuries since Jan. 1.0, 1900 speed, km/sec, or a velocity vector unit vector pointing toward Sun, defined in right-ascension--declination coordinate system Cartesian coordinates right ascension, deg hour angle of subsolar meridian relative to a satellite meridian, deg declination, deg obliquity of Earth's equator to the ecliptic, deg 48 n 6 T T solar zenith angle, deg (see fig. 16) direction from which Sun is shining relative to Earth's surface, deg (see fig. 16) rotation rate of Earth, 0.25068447 deg/min or 360.9856473 deg/day latitude, deg Earth's gravitational constant, 398601.2 kmS/sec 2 Sun elevation angle, deg (see fig. 16) period, sec or orbits/day referring to vernal equinox (as when indicating direction of the first point of Aries) argument of perihelion, deg right ascension of ascending node, deg Subscripts: A anomalistic periods g Greenwich meridian m meridian AM evaluated for a mean-anomaly step size of AM N nodal periods o initial or unperturbed (Keplerian) quantities s Sun-synchronous quantities Earth ® Sun obs observer x,y,z components of a position vector Superscripts: h,m,s hours, minutes, seconds d/dt degrees 49 A dimensioned vector unit vector 5O APPENDIX A "BASIC" LANGUAGE ALGORITHMS FOR CALCULATING ORBIT PARAMETERS Many of the equations of previous sections on orbital dynamics may easily be incorporated into small computer programs for calculating relevant orbit properties. Table AI gives a BASIC language algorithm written for the Wang 2200 computer system. With an input of semimaJor axis, eccentricity, and inclination, plus the desired mean-anomaly step size, the precession rates and other data are calculated. The Sun-synchronous inclination is also calculated; all the orbit data can be obtained for synchronous orbits by setting I = S and rerunning the program. The last two lines of printout will show that the inclination for syn-chronous orbits as calculated from just one iteration does, in fact, give very nearly the desired result. An example of the output is given in the table. Table A2 lists another useful BASIC algorithm. With inputs of Q (in terms of an integer plus a fraction), eccentricity, inclination, and a guess at the semimaJor axis, a simple iterative technique causes the semimajor axis to con-verge to the correct value for the desired Q. For Sun-synchronous orbits the inclination is calculated internally and adjusted to maintain the synchronous precession rate as semimajor axis is varied. This algorithm will generate data such as those presented in tables 5 and 6. It can be modified to vary eccentri-city instead of semimajor axis, but simultaneous freedom of these two variables would lead to an infinite number of solutions for each value of Q. 51 APPENDIX A TABLE AI.-"BASIC" ALGORITHM FORGENERATING ORBITPARAMETERS ONTHEWANG 2200 SYSTEM 10SELECT ]) 20XI=66058. 33128 30X2=6. 283185308 40](3=398601.2 50X4=57.29577951 60X5=.9856473 701NPUT "lEAN ANOIALY 80 INPUT "UNPERTURBED 90INPUT "ECCENTRICITY", E 1001NPUT "INCLINATION" , I IIOPRINT "D ",D 120PRINT "A0" ,A0 130PRINT "E ",E 140PRINT "I ", I 150PO--(A0(I-EE)) !2 160T0=X2AOSQR(AO/X3) 170110=X 2/T0 180'II=M0" (I+XI/P 190DI=XI/POMI ( 200D2=-XI/P0MI 210TI=X2/MIIS0/ 220T2=X2/(MI+OI) 230S0=-A0 !i. 5P0 240S=S0/ (I+XI/PO 250SO--ARCCOS (SO) 260S=ARCCOS (S) 270D3=XI/ 28004=-XI 290D5=II0 300D6=-XI 340PRINT 35OPRINT 36OPR INT 370PRINT 330PRINT 390PRINT 400PRINT 410PRINT 420PRINT 430PRINT 440PRINT 450PRINT STEP",D SEHI:IAJOR AXIS", A0 OSQR(I-EE)(I-I.5SIN(1)!2))X486400 2-2.5"SIN(I)!2) cos(z) #PI.86400 .180/#PI.36400 X5/_1/SQR(X3)/X4/86400 SQR(I-EE)(I-I.5SIN(ARCCOS(S0))!2)) P0"(2-2.5SIN (I) !2)D /POCOS (I)D I I+XI/P0SQR(I-EE)(I-I.5SIN(S) !2)) P0D5COS (S) X4"36400 310DT=D6365. 2422 320Q:(:II+DI)/(.7292115061E-4ISO/#PI86400-D2) "UNPERTURBED PERIOD,SEC ",T0 "UNPERTURBED MEAN :IOTION,RAD/SEC ",MO ":lEAN IOTION,DEG/DAY ",MI "PERIGEE RATE,DEG/DAY ",DI "NODE RATE,DEG/DAY ",D2 "PERIGEE STEP,DEG/MEAN ANOMALY STEP ",D3 "NODE STEP,DEG/>IEAN ANO:'IALY STEP ",D4 "ANOMALISTIC PERIOD ",TI "NODAL PERIOD ",T2 "REPETITION FACTOR Q ",Q "UNPERTURBED SYNCHRONOUS INCLINATION "_SO "SYNCHRONOUS INCLINATION ",S 455MI=MO(I+XI/P0SQR(I-EE)(I-I.5SIN(S) !2))X4"86400 460DI=XI/POMI(2-2.5SIN(S) !2) 470QI= (MI+DI)/360 480PRINT "SYNCIIRONOUS REPETITION FACTOR ",QI 490PRINT "DALLY SYNCHRONOUS ORBIT PRECESSION ",D6 500PRINT "YEARLY SYNCHRONOUS ORBIT PRECESSION ",D7 52 APPENDIX A TABLE AI.-Concluded D 360 AO 7000 E 0 I 60 UNPERTURBED PERIOD,SEC 5828.5110951 UNPERTURBED MEAN !OTION,RAD/SEC 1.07800863E-03 MEAN_IOTION,DEG/DAY 5335.626538 PERIGEE RATE,DEG/DAY .8991392485551 NODERATE,I)EG/DAY -3.596556994221 PERIGEE STEP,DEG/MEANANOI_ALYSTEP 6.06658144E-02 NODE STEP,DEG/MEANANOIIALY STEP -.2426632577633 ANO:_ALIST IC PERIOD 5829.493459208 NODALPERIOD 5828.511260928 REPETITION FACTORQ 14.63737447681 UNPERTURBED SYNCHRONOUS INCLINATION 97.87448384351 SYNCIIRONOU S INCLINATION 97.87952788402 SYNCHRONOUS REPETITION FACTOR 14.80520669859 DALLY SYNCHRONOUS ORBIT PRECESSION .9856473476633 YEARLY SYNCHRONOUS ORBIT PRECESSION 360.0000056847 (Set I=S and rerun -RUNIO -to get synchronous orbit data) D 36O A0 7000 E 0 I 97.87952788402 UNPERTURBED PERIOD,SEC 5828.5110951 UNPERTURBED IEAN :[OTION,RAD/SEC I®07800863E-03 MEAN_OTION,DEG/DAY 5333.131479134 PERIGEE RATE,DEG/DAY -3.25706764226 NODERATE,DEG/DAY .9856473476636 PERIGEE STEP,DEG/MEANANO_ALY STEP -.2198603870542 NODESTEP,DEG/_IEANANOMALYSTEP 6.65337141E-02 ANO_IALISTIC PERIOD 5832.22073668 NODALPERIOD 5835.78478641 REPETITION FACTORQ 14.80520831492 UNPERTURBED SYNCIIRONOUS INCLINATION 97.87448384351 SYNCHRONOUS INCLINATION 97.87952788402 SYNCHRONOUS REPETITION FACTOR 14.80520669859 DALLY SYNCHRONOUS ORBIT PRECESSION .9856473476633 YEARLY SYNCIIRONOUS ORBIT PRECESSION 360.0000056847 53 APPENDIX A TABLE A2.-"BASIC" ALGORITHM FOR GENERATING ORBITS WITH SPECIFIED REPETITION FACTORS ON THE WANG 2200 SYSTEM i0 PRIHT 20 SELECT D 30 XI=66058.33128:X2=6.283185308:X3=398601.2:X4=57.29577951 40 INPUT "REPETITION FACTOR: INTEGER,NUM,DENO_I",Q,N,U 50 FO--Q+N/U 60 II_PUT "GUESS AT SE\|IMAJOR AXIS",AI 70 A2=AI+IO 80 INPUT "ECCENTRICITY", E 90 INPUT "INCLINATION-USE 999 FOR SUN SYNC",I i00 IF I=999THEN 130 Ii0 11=1:12=1:13=1 120 PRINT 130 PRII{TUSING 390,F0,N,U; 140 PI=(AI(I-EE)) !2:P2=(A2(I-EE)) !2 150 TI=X2AISQR(AI/X3) :T2=X2A2SQR(A2/X3) 160 MI=X2/TI:M2=X2/T2:IF I[]999THEN 190 170 GOSUB 'I(AI,PI):II=S 180 GOSUB 'I(A2,P2):I2=S 190 SI=_II(I+XI/PISQR(I-EE)(I-I.5SI_4(II) E2))X486400 200 S2--M2(I+XI/P2SQR(I-EE)(I-I.5SIN(12) !2))X4"86400 210 DI=XI/PISI(2-2.5SIN(II) !2) :D2=XI/P2S2(2-2.5SIN(12) !2) 220 CI---XI/PISICOS(II) :C2=-XI/P2SICOS(12) 230 F1 =(SI+DI)/(.729211506E-4IS0/#PI86400-Ci)-FO 240 F2=(S2+D2)/(.72921iSO6E-4I80/#PI86400-C2)-FO 250 IF ABS(FI)[IE-6 TIIEN 360:IF ABS(F2)[IE-6 THEN 370 260 A3=(A2-AI)F2/ (FI-F2)+A2 270 P3 =(A3(I-EE)) !2:T3=X2A3SQR(A3/X3):A3=X2/T3:IF I[]999THEN 280 GOSUB 'I(A3,P3):I3=S 290 S3=II3(I+XI/P3SQR(I-EE)(I-I 5"SIl{(13)!2))X4"84600 300 D3=XI/P3S3(2-2.5SIN(13) !2) :C3=-XI/P3S3COS(13) 310 F3= ($3+D3) / (. 7 292115061E-4IS0/#PI86400-C3)-F0 320 IF ABS(F3)[IE-6 TIIEN 340 330 AI=A2 :A2=A3 :GOTO 140 340 A=A3:F=F3+F0 350 GOTO 380 360 A=AI:F=FI+F0:GOTO 380 370 A=A2 :F=F2+F0 380 I=I3:PRI_[TUSI:G 400,A+.0000005;I+.0000005 390,'## ##/## 400Z ####.###### ###. #####' 410 GOTO 40 420 END 430DEFFN'I(A,P) 440 SO=-A!I.5Pe.9856473/XI/SQR(X3)/X4/86400 450 S=S0/(I+XI/PSQR(I-EE)(I-I.5SIN(ARCCOS(S0))!2)) 460 S=ARCCOS(S) 470 RETU Rq 290 (Sample case for a = 7000, e = O, i -- 60, Q = 13 1/2) 13 II 2 7396.373144 60.000000 (i = 999) 13 i/ 2 7445.166714 q9.793197 54 APPENDIX B ORBIT ELEMENTS FOR THE EARTH'S MOTION AROUND THE SUN The equations for the Earth's orbit elements can be obtained by inverting the equations for the apparent motion of the Sun obtained from astronomical observation. The time-varying orbit elements of the Sun can be found in refer-ence I, and inverted orbit elements for the Earth, or more specifically, mean elements for the gravitational center of the Earth-Moon system (barycenter) are given explicitly in reference 6: d = J.D. -241 5020.0 D = d/10000 T = d/36 525 a = 1.00000023 AU e@ = 0.01675104 - 0.0000418T -0.000000126T 2 = 101o.220833 + 0o.0000470684d + 0o.0000339D 2 + 0o.00000007D3 M_ = 358o.475845 + 0°.985600267d - 0°.0000112D 2 - 0°.O0000007D 3 e_ = 23°.452294 -0°.0035626D -0°.000000123D 2 + 0°.0000000103D 3 Reference 6 gives the value of I AU (astronomical unit) as 149 597 893 ± 5 km. The true anomaly of the Earth can be obtained from implementation of any of the many available iterative solutions to Kepler's equation, or by expansion (ref. 3, eq. (6-15)): f : M + 2e sin M + 5e-- 2 sin 2M + e3(13 sin 3M -3 sin M) + e4(I03 sin 4M 4 12 96 -44 sin 2M) + e5 (I097 sin 5M -645 sin 3M + 50 sin M) + e6 (1223 sin 6M 960 960 -902 sin 4M + 85 sin 2M) + . In this expression, both f and M must be in radians. 55 APPENDIX C DETERMINING THE CONDITIONS UNDER WHICH THE SUN IS OCCULTED BY THE EARTH RELATIVE TO A SATELLITE HAVING FIXED ORBIT ELEMENTS Solution of the occultation problem involves determining the points on its orbit at which a satellite enters or leaves the Earth's shadow. The geometry is shown in sketch (CI). The symbols used for this appendix form a self-Sketch (CI) contained set, and they are not all defined in or consistent with the symbol list in the body of the report. All the vector quantities are expressed in a system whose x-y plane lies in the plane of the satellite orbit, with x and y oriented as indicated in sketch (CI). Such a coordinate system is called the PQW system. Vectors can be rotated into the PQW system by standard procedures. (See, for example, ref. 2.) A0 to A4 constants as defined below vector perpendicular to the terminator, extending from the terminator to the satellite, km e eccentricity 56 f P F ^ % % APPENDIX C true anomaly, deg semilatus rectum, km radius vector to satellite, km unit vector to the Sun Earth radius vector to the terminator, km unit vector in direction of the perigee of the satellite orbit unit vector at right angles to _ along the semilatus rectum ^ angle between r® and 3, deg The geometric constraint is the requirement that at a point where the satellite enters or leaves the Earth's shadow, r® d = -d = -(r 2 r@2) I/2 • -: r cos That is, __ and d are antiparallel. This restriction can be written in another way ^ _ _ ^ rQ • d : ® ( -F_) : • _ : rr® ( cos f + 9 sin f) : r cos ¢ The x-and y-components of the unit vector to the Sun are r(Dx = ® " _ r(Dy : r(D" 9 Then, rGx cos f + r(Dy sin f : -I/2 Since r : p/(1 + e cos f), P ( cos f + F p2 (I + e cos f) rGx roy sin f) + k(1 + e cos f)2 This can be rewritten in terms of cos f: A0 cos 4 f + AI cos3 f + A2 cos 2 f + A 3 cos f + A 4 = 0 The constants are (---) 4e4 (--)2(r_y + (r_x+ r_y) 2 A0 = -2 -r_x)e2 =0 57 A, r2 APPENDIX C A2 = 6 - 2 --r_y) A3 = 4(_)4e- 4(_)2(I- ry)e 4 2 The fourth-order equation may be solved by standard procedures. The two spuri-ous roots may be rejected by noting that Ir@ × I = r sin @ = I@I and A r@ • /r < 0 for occultation to occur. 58 REFERENCES I. Explanatory Supplement to the Astronomical Ephemeris and the American Ephemeris and Nautical Almanac. H.M. Naut. Alm. Off., 1961. 2. Escobal, Pedro Ramon: Methods of Orbit Determination. John Wiley & Sons, Inc., c.1965. 3. Orbital Flight Handbook. Part I -Basic Techniques and Data. NASA SP-33, Pt. I, 1963. 4. Kozai, Yoshihide: The Motion of a Close Earth Satellite. Astron. J., vol. 64, no. 9, Nov. 1959, pp. 367-377. 5. McCuskey, S. W.: Introduction to Celestial Mechanics. Addison-Wesley Pub. Co., Inc., c.1963. 6. Melbourne, William G.; Mulholland, J. Derral; Sjogren, William L.; and Sturms, Francis M., Jr.: Constants and Related Information for Astrody-namic Calculations, 1968. Tech. Rep. No. 32-1306 (Contract NAS 7-100), Jet Propulsion Lab., California Inst. Technol., July 15, 1968. 7. Harrison, Edwin F.; and Green, Richard N.: Orbit Analysis for Coastal Zone Oceanography Observations. AIAA Paper No. 73-207, Jan. 1973. 8. Harrison, Edwin F.; Brooks, David R.; and Gibson, Gary G.: Mission Analysis To Define Satellite Orbits for Earth Radiation Budget Measurements. AIAA Paper 76-811, Aug. 1976. 9. Russell, James M., III; Park, Jae H.; and Drayson, S. Roland: Global Monitor-ing of Stratospheric Halogen Compounds From a Satellite Using Gas Filter Spectroscopy in the Solar Occultation Mode. Appl. Opt., vol. 16, no. 3, Mar. 1977, pp. 607-612. 59 TABLE I.-JULIAN DAY NUMBER, 1950-1999, OF DAY COF_4ENCING AT GREENWICH NOON ON: Year Jan. 0.5 a Feb, 0,5 Mar. 0.5 Apr. 0,5 May0.5 June0.5 JulyO.5 Aug. 0.5 Sept. 0.50ct, O.5 Nov, 0,5 Dec. 0. 1950 243 3282 3313 3341 3372 3402 3433 3463 3494 3525 3555 3586 3616 1951 3647 3678 3706 3737 3767 3798 3828 3859 3890 3920 3951 3961 1952 4012 4043 4072 4103 4133 4164 4194 4225 4256 4286 4317 4347 1953 4378 4409 4437 4468 4498 4529 4559 4590 4621 4651 4682 4712 1954 4743 4774 4802 4833 4863 4894 4924 4955 4986 5016 5047 5077 1955 243 5108 5139 5167 5198 5228 5259 5289 5320 5351 5381 5412 5442 1956 5473 5504 5533 5564 5594 5625 5655 5686 5717 5747 5778 5808 1957 5839 5870 5898 5929 5959 5990 6020 6051 6082 6112 6143 6173 1958 6204 6235 6263 6294 6324 6355 6385 6416 6447 6477 6508 6538 1959 6569 6600 6628 6659 6689 6720 6750 6781 6812 6842 6873 6903 1960 243 6934 6965 6994 7025 7055 7086 7116 7147 7178 7208 7239 7269 1961 7300 7331 7359 7390 7420 7451 7481 7512 7543 7573 7604 7634 1962 7665 7696 7724 7750 7785 7816 7846 7877 7908 7938 7969 7999 1963 8030 8061 8089 8120 8150 8181 8211 8242 8273 8303 8334 8364 1964 8395 8426 8455 8486 8516 8547 8577 8608 8639 8669 8700 8730 1965 243 8761 8792 8820 8851 8881 8912 8942 8973 9004 9034 9065 9095 1966 9126 9157 9185 9216 9246 9277 9307 9338 9369 9399 9430 9460 1967 9491 9522 9550 9581 9611 9642 9672 9703 9734 9764 9795 9825 1968 9856 9887 9916 9947 9977 0008 _0038 _0069 50100 _0130 _0161 0191 1969 244 0222 0253 0281 0312 0342 0373 0403 0434 0465 0495 0526 0556 1970 244 0587 0618 0646 0677 0707 0738 0768 0799 0830 0860 0891 0921 1971 0952 0983 IOll i042 1072 1103 i133 I164 I195 1225 1256 1286 1972 1317 1348 1377 1408 1438 1469 1499 1530 1561 1591 1622 1652 1973 1683 1714 1742 1773 1803 1834 1864 1895 1926 1956 1987 2017 1974 2048 2079 2107 2138 2168 2199 2229 2260 2291 2321 2352 2382 1975 244 2413 2444 2472 2503 2533 2564 2594 2625 2656 2686 2717 2747 1976 2778 2809 2838 2869 2899 2930 2960 2991 3022 3052 3083 3113 1977 3144 3175 3203 3234 3264 3295 3325 3356 3387 34_7 3448 3478 1978 3509 3540 3568 3599 3629 3660 3690 3721 3752 3782 3813 3843 1979 3874 3905 3933 3964 3994 4025 4055 4086 4117 4147 4178 4208 1980 244 4239 4270 4299 4330 4360 4391 4421 4452 4483 4513 4544 4574 1981 4605 4636 4664 4695 4725 4756 4786 4817 4848 4878 4909 4939 1982 4970 5001 5029 5060 5090 5121 5151 5182 5213 5243 5274 5304 1983 5335 5366 5394 5425 5455 5486 5516 5547 5578 5608 5639 5669 1984 5700 5731 5760 5791 5821 5852 5882 5913 5944 5974 6005 6035 1985 244 6066 6097 6125 6156 6186 6217 6247 6278 6309 6339 6370 6400 1986 6431 6462 6490 6521 6551 6582 6612 6643 6674 6704 6735 6765 1987 6796 6827 6855 6886 6916 6947 6977 7008 7039 7069 7100 7130 1988 7161 7192 7221 7252 7282 7313 7343 7374 7405 7435 7466 7496 1989 7527 7558 7586 7617 7647 7678 7708 7739 7770 7800 7831 7861 1990 244 7892 7923 7951 7982 8012 8043 8073 8104 8135 8165 8196 8226 1991 8257 8288 8316 8347 8377 8408 8438 8469 8500 8530 8561 8591 1992 8622 8653 8682 8713 8743 8774 8804 8835 8866 8896 8927 8957 1993 8988 9019 9047 9078 9108 9139 9169 9200 9231 9261 9292 9322 1994 9353 9384 9412 9443 9473 9504 9534 9565 9596 9626 9657 9687 1995 244 9718 9749 9777 9808 9838 9869 9899 9930 9961 9991 0022 _0052 1996 245 0083 0114 0143 0174 0204 0235 0265 0296 0327 0357 0388 0418 1997 0449 0480 0508 0539 0569 0600 0630 0661 0692 0722 0753 0783 1998 0814 0845 0873 0904 0934 0965 0995 I026 i057 1087 Ill8 i148 1999 245 I179 1210 1238 1269 1299 1330 1360 1391 1422 1452 1483 1513 2000 245 1544 1575 1604 1635 1665 1696 1726 1757 1788 1818 1849 1879 ajan. 0.5 = Greenwioh noon (12 h) UT, Dec. 31. 60 TABLE 2.-KEPLERIAN AND PERTURBED ORBITAL PERIODS AS A FUNCTION OF ALTITUDE AND INCLINATION h, To, TA' TN' TA " TO, and TN - To,a see, fop i, deE, of -km sec 0 I0 20 30 40 50 60 70 80 90 200 5309.6 5301.6 5301.9 5303.0 5304.6 5306.6 5308.7 5310.7 5312.3 5313.3 5313.7 -3.I -7.7 -6.7 -5.1 -3.1 -i.0 1.0 2.6 3.7 4.1 5285.4 5286.4 5289.2 5293.5 5290.7 5304.4 5309.6 5314.0 5316.8 5317.8 -24.2 -23.3 -20.5 -16.2 -10.9 -5.3 0.O 4.3 7.1 3.1 300 5431.2 5423.1 5423.5 5424.6 5426.2 5428.1 5430.2 5432.2 5433.8 5434.8 5435.2 -8.0 -7.7 -6.6 -5.0 -3.1 -I.0 1.0 2.6 3.7 4.0 5407.1 5408.1 5410.9 5415.1 5420.4 5425.9 5431.2 5435.5 5438.3 5439.2 -24.1 -23.1 -20.3 -16 .i -i0.8 -5.2 0.0 4.3 7.1 8.1 400 5553.6 5545.7 5546.0 5547.1 5548.6 5550.6 5552.7 5554.6 5556.2 5557.3 5557.6 -8.0 -7.6 -6.6 -5.0 -3.0 -1.0 1.0 2.6 3.6 4.0 5529.8 5530.7 5533.5 5537.7 5542.9 5548.4 5553.6 5557.9 5560.7 5561.6 -23.9 -22.9 -20.2 -15.9 -10.7 -5.2 0.0 4.3 7.0 8.0 500 5677.0 5669.1 5669.4 5670.5 5672.0 5674.0 5676.0 5678.0 5679.6 5680.6 5680.9 -7.9 -7.6 -6.5 -4.9 -3.0 -0.9 1.0 2.6 3.6 4.0 5653.3 5654.2 5657.0 5661.2 5666.3 5671.8 5677.0 5681.2 5684.0 5684.9 -23.7 -22.8 -20.0 -15.8 -10.7 -5.2 0.0 4.2 7.0 7.9 600 5801.2 5793.4 5793.7 5794.8 5796.3 5798.2 5800.3 5302.2 5803.8 5304.8 5805.2 -7.9 -7.5 -6.5 -4.9 -3.0 -0.9 1.0 2.6 3.6 3.9 5777.7 5778.6 5781.4 5785.5 5790.6 5796.1 5801,2 5305.4 5808.2 5809.1 -23.5 -22.6 -19.9 -15.7 -I0.6 -5.1 0.0 4.2 6.9 7.9 700 5926.4 5918.6 5918.9 5919.9 5921.5 5923.4 5925.4 5927.4 5928.9 5929.9 5930.3 -7.8 -7.5 -6.4 -4.9 -3.0 -0.9 1.0 2.5 3.6 3.9 5903.0 5904.0 5906.6 5910.3 5915.9 5921.3 5926.4 5930.5 5933.3 5934.2 -23.4 -22.4 -19.7 -15.6 -10.5 -5.1 0.0 4.2 6.9 7.3 800 6052.4 6044.7 6045.0 6046.0 6047.6 6049.5 6051.5 6053.4 6054.9 6055.9 6056.3 -7.7 -7.4 -6.4 -4.8 -2.9 -0.9 1.0 2.5 3.5 3.9 6029.2 6030.1 6032.8 6036.9 6:)42.0 6047.4 6052.4 6056.5 6059.2 6060.2 -23.2 -22.3 -19.6 -15.5 -I0.4 -5.1 0.0 4.1 6.8 7.3 9006179.3 6171.6 6172.0 6173.0 6174.5 6"176.4 6173.4 6180.3 6181.8 6132.8 6183.2 -7.7 -7.3 -6.3 -4.8 -2.9 -0.9 1.0 2.5 3.5 3.9 6156.3 6157.2 6159.9 6164.0 6169.0 6174.3 6179.3 6183.4 6186.1 6187.0 -23.1 -22.1 -19.5 -15.4 -10.4 -5.0 0.0 4.1 6.8 7.7 1000!6307.1 6299.5 6299.8 6300.8 6302.3 6304.2 6306.2 6308.1 6309.6 6310.6 6311.0 -7.6 -7.3 -6.3 -4.8 -2.9 -0.9 1.0 2.5 3.5 3.8 6284.2 6285.1 6287.8 6291.8 6296.8 6302.1 6307.1 6311.2 6313.9 6314.8 -22.9 -22.0 -19.3 -15.3 -10.3 -5.0 0.0 4.1 6.7 7.7 1100!6435.3 6428.2 6428.5 6429.5 6431.0 6432.9 6434.9 6436.7 6438.3 6439.2 6439.6 -7.6 -7.3 -6.3 -4.7 -2.9 -0.9 1.0 2.5 3.5 3.8 6413.0 6414.0 6416.6 6420.6 6425.6 6430.8 6435.8 6439.8 6442.5 6443.4 -22.7 -21.8 -19.2 -15.2 -10.2 -5.0 0.0 4.0 6.7 7.6 1200 6565.3 6557.8 6558.1 6559.1 6560.6 6562.4 6564.4 6566.2 6567.8 6568.7 6569.1 -7.5 -7.2 -6.2 -4.7 -2.9 -0.9 0.9 2.5 3.4 3.8 6542.7 6543.6 6546.2 6550.2 6555.1 6560.4 6565.3 6569.3 6572.0 6572.9 -22.6 -21.7 -19.1 -15.1 -i0.2 -4.9 0.0 4.0 6.6 7.6 1300 6695.7 6688.2 6688.5 6639.5 6691.0 6692.8 6694.8 6696.6 6698.1 6699.1 6699.4 -7.5 -7.2 -6.2 -4.7 -2.9 -0.9 0.9 2.4 3.4 3.8 6673.2 6674.1 6676.7 6630.7 6685.6 6690.3 6695.7 6699.7 6702.3 6703.2 -22.4 -21.5 -19.0 -15.0 -10.1 -4.9 0.0 4.0 6.6 7.5 1400 6826.9 6819.5 6819.8 6820.8 6822.3 6824.1 6826.0 6827.8 6329.3 6330.3 6830.6 -7.4 -7.1 -6.1 -4.7 -2,8 -0.9 0.9 2.4 3.4 3.7 6804.6 6805.5 6808.1 6812.0 6816.9 6822.1 6826.9 6830.9 6833.5 6834.4 -22.3 -21.4 -18.8 -14.9 -i0.0 -4.9 0.0 4.0 6.6 7.5 1500 6959.0 6951.6 6951.9 6952.9 6954.4 6956.2 6953.1 6959.9 6961.4 6962.4 6962.7 -7.4 -7.1 -6.1 -4.6 -2.8 -0.9 0.9 2.4 3.4 3.7 6936.8 6937.7 6940.3 6944.2 6949.0 6954.2 6959.0 6962.9 6965.5 6966.4 -22.2 -21.3 -18.7 -14.8 -10.0 -4.8 0.0 3.9 6.5 7.4 akt each value of altitude h, the first llne of data is TA; the second line, _A - To; the third line, TN; and the fourth line, z N -TO, all in seconds. 61 Z O h.-I E-q Z H ,J O Z H r t.'l :3 E- I--I E- ,-1 B_ 0 Z 0 H Z r.. H m: 0 .] o r_ Cz: E_ gg 0 r_ z Q i-4 e_ ! r. ,-1 O _ _ _ _ _ _O _ _ O O O el ee em el em ee ee ee ee el me O I I I I I I I II II I I II • • • • el ee le • ee ee ee I I I I I I I II II I I i. i. • .i • el le • _ I _ I I I I I I I I I I I I o _ _ _ _ _ _ _ _ _ o _ o • e e el ee oe i_ el e_ ee el ee O_ O_ _ h_ _ _ O_ NN O NO NO II _1 _1 I I I I I II II I I I "3 _O NO 0 NI NI I I I I I II II I I Z O ee el ie el et el o le el el el _ e • e e el el ee el _l eo el NI NI HI I I I I II II I I . _ .e • • • !! i _i el .e el el NI NI NI NI I I I II II I I o _ , o _ _ e. • e • ie e • e NI NI HI HI I I I II II I I O • el ee _e ee e el el ee el • 1 NI NI NI I I I II II I I o o (N .-I u'l ¢'] _0PI O u"l IrlU .-0 --..'r .--I u',Q O u"1 ',DO 0_, u'l I ,,-I I ,-I I vii I I I I I I I -,-I "El 62 H Ix:; 0 H a: ,-I r.r.1 r..) H rJ r_ O Z O I.-i [-4 O r_ 0 m ,-.1 ),-t H C3 Z O ,-1 I ,-3 E-4 J ., . 0 0 ,,,T ',O O0 C _ , _ O0 CD 0,1 -o" . CO C O0 CO 0 0", 0' 0' 0' O_ 0 C) C C C 63 TABLE 5.-CIRCULAR ORBIT PARAMETERS FOR SPECIFIED REPETITION FACTORS Orbit altitude, km, for inclination, i, deE, of -Q 0 10 20 30 40 50 60 70" 80 90 12 1616.2 1616.5 1617.6 1619.6 1622.7 1627.1 1633.0 1640.6 1649.9 1660.7 12 1/ 2 1395.9 1396.3 1397.5 1399.8 1403.3 1408.2 1414.7 1423.O 1433.O 1444.6 12 I/ 3 1467.7 1468.1 1469.3 1471.5 1474.8 1479.5 1485.9 1493.9 1503.7 1515.0 12 2/ 3 1325.6 1326.0 1327.3 1329.7 1333.3 1338.3 1345.1 1353.6 1363.8 1375.7 12 i/ 4 1504.2i 1504.6 1505.8 1507.9 1511.2 1515.8 1522.0 1530.O 1539.6 1550.8 12 3/ 4'1291.01291.4 1292.8 1295.2 1298.8 1304.O 1310.8 1319.4 1329.8 1341.8 12 i/ 5 1526.3 1526.7 1527.8 1529.9 1533.2 1537.8 1543.9 1551.8 1561.4 1572.5 12 4/ 5 1270.4 1270.8 1272.2 1274.6 1278.3 1283.6 1290.4 1299.1 1309.6 1321.7 12 I/ 6 1541.1 1541.5 1542.6 1544.7 1547.9 1552.5 155_.6 1566.4 1575.9 1587.0 12 5/ 6 1256.7 1257.2 1258.6 1261.0 1264.8 1270.0 1276.9 1285.7 1296.2 1308.3 12 i/ 7 1551.7 1552.1 1553.2 1555.3 1558.5 1563.0 1569.1 1576.9 1586.4 1597.4 12 6/ 7 1247.0 1247.5 1248.9 1251.3 1255.1 1260.4 1267.3 1276.111286.6 1298.8 12 i/ 3 1559.7 1560.1 1561.2 1563.3 1566.5 1571.O 1577.1 1584.8 1594.3 1605.3 12 7/ 3 1239.8 1240.2 1241.6 1244.1 1247.9 1253.1 1260.1 1268.9:1279.5 1291.7 12 I/ 9 1565.9 1566.3 156;.4 1569.5 1572.7 1577.2 1583.2 1591.O 1600.4 1611.4 12 8/ 9 1234.1 1234.6 1236.0 1233.5 1242.2 1247.5 1254.5 1263.3i1274.O 1286.2 12 I/I0 1570.9 1571.3 1572.4 1574.5!1577.6 1532.1 15SJ.2 1595.911605.3 161b.3 12 O/lO 1229.6 1230.1 1231.5 1234.O 1237.8 1243.1 1250.1 1258.911269.5 1281.8 12 i/II 1575.0 1575.4 1576.5 157_.5 1531.7 1536.2 1592.2 1599.9 1609. 1620.3 12 I0/ii 1225.9 1226.4 1227.8 1230.3 1234.1 1239.411246.4 1255.3 1265. 127S.2 12 1/12 1578.4 1578.8 1579.9 1581.9 1585.1 1589.611595.6 1603.3 1612.7 1623.6 12 11/1211222.9 1223.3 1224.7 1227.2 1231.O 1236.4 1243.4 1252.2 1262.9 1275.2 12 1/13 1581.3 1581.7 1582.8 1584.8 1588.O 1592.5 1598.5 1606.2 1615.6 1626.5 12 12/13 1220.3 1220.7 1222.1 1224.7 1228.5 1233.8 1240.811249.7 1260.4 1272.6 12 1/14 1583.8 1584.1 1585.3 1587.3 1590.4 1594.9 1600.9 1608.6 1618.O 1628.9 12 13/14 1218.1 1218.5 1219.9 1222.4 1226.3 12_1.6 1238.6 1247.5 1258.2 1270.5 12 1/15 1535.9 1586.3 1587.4 1589.4 1592.6 1597.1 1603.1 1610.7 1620.1 1631.0 12 14/15 1216.1 1216.61218.O 1220.5 1224.3 1229.7 1236.7 1245.6 1256.3 1268.6 12 1/16 1587.8 1588.211589.3 1591.3 1594.5 1598.9 1604.9 1612.6!1622.0 1632.9 12 15/16 1214.4 1214.9 1216.3 1218.8 1222.7 1228.O 1235.1 1243.9 1254.6 1267.0 12 1/17 1589.5 15G9.8 1591.0 1593.0 1596.1 1600.6 1606.6 1614.2 1623.6 1634.5 12 16/17 1213.0 1213.4 1214.8 1217.4 1221.2 1226.5 1233.6 1242.5 1253.2 1265.5 12 I/I_ 1591.O 1591.3 1592.4 1594.4 1597.6 1602.0 1608.0 1615.7 1625.1 1636.0 12 17/18 1211.6 1212.1 1213.5 L216.1 1219.9 1225.2!1232.3 1241.2 1251.9 1264.2 12 1/19 1592.3 1592.6 1593.7 1595.8 1598.9 1603.411609.4 1617.0 1626.4'1637.2 12 18/19,1210.5 1210.9 1212.4.1214.9 1218.7 1224.1 1231.1 1240.0 1250.7 1263.1 12 1/20 1593.5 1593.S 1594.9 1597.O 16OO.1 1604.5 1610.5 1618.2 1627.5 163S.4 12 19/20 1209.4 1209.9 1211.3 1213.8 1217.7 1223.0 1230.1 1239.0 1249.7 1262.0 12 1/21 1594.5 1594.9 1596.0 1598.0 1601.2 1605.6 1611.6 1619.3 1628.6 1639.5 12 20/21 1208.5 1208.9 1210.3 1212.9 1216.7 1222.1 1229.1 1238.0 1248.7 1261.1 64 TABLE 5.-Continued Orbit altitude, , for inclination, ir de_1 of -Q 0 10 20 30 40 50 60 70 80 90 13 1189.3 1189.8 1191.2 1193.8 1197.7 1203.1 1210.2 1219.2 1230.0 1242.4 13 I/ 2 995.2 995.7 997.4 1000.2 1004.5 1010.5 1018.2 1027.9 1039.5 1052.8 13 i/ 3 1058.6 1059.1 1060.7 1063.4 1067.6 1073.4 1080.9 1090.4 1101.7 1114.7 13 2/ 3 933.0 933.6 935.3 938.2 942.7 948.8 956.8 966.7 978.6 992.1 13 i/ 4 1090.8 1091.3 1092.8 1095.5 1099.6 1105.3 1112.7 1122.1 1133.2 1146.1 13 3/ 4 902.4 903.0 904.7 907.7 912.2 918.4 926.5 936.6 945.6 962.3 13 I/ 5 1110.3 1110.8 1112.3 1115.0 ii19.0 1124.6 1132.0 i141.3ii152.3 1165.1 13 4/ 5 884.2 884.7 386.5 889.5 894.1 900.4 908.5 918.7 930.7 944.5 13 I/ 6 1123.3 1123.8 1125.3 1128.0 1132.0 1137.6 1144.9 1154.1 1165.1 1177.8 13 5/ 6 872.1 872.7 874.4 877.5 882.1 888.4 896.6 906.8 918.9 932.7 13 I/ 7 i132.7 1133.1 1134.6 1137.3 1141.3 1146.8 1154.1 1163.3 1174.3 1187.0 13 6/ 7 863.5 864.0 865.8 868.9 873.5 879.8 888.1 898.3 910.4 924.3 13 i/ 8 1139.7 1140.2 1141.7 1144.3 1148.3 I153.8 1161.1 1170.2 1181.2 1193.9 13 7/ 3 857.0 857.6 859.4 862.5 867.1 873.41 881.7 891.9 904.1 918.0 13 i/ 9 1145.2 1145.7 1147.1 1149.8 1153.7 1159.2 1166.5 1175.6 1186.6 1199.2 13 8/ 9 852.0 352.6 854.4 857.5 862.1 868.5 876.8 857.0 899.2 913.2 13 I/i0 1149.6 1150.0 1151.5 1154.1 1158.1 1163.6 1170.9 i180.0 1190.9 1203.5 13 9/10 848.0 848.6 850.4 853.5 858.1 864.5 872.8 883.1 895.3 909.3 13 I/II 1153.2 1153.6 1155.1 1157.7 1161.7 1167.2 1174.4 I183.51194.4 1207.0 13 I0/ii 844.8 845.3 347.1 850.2 854.9 861.3 869.6 879.9 892.1 906.1 13 1/12 1156.2 1156.6 1158.1 1160.7 1164.7 1170.2 1177.4 1186.5 1197.4 1209.9 13 11/12 842.0 _42.6 844.4 847.5 852.2 858.6 866.9 877.2 889.5 903.4 13 1/13 1158.7 1159.2 1160.6 1163.2 1167.2 1172.7 1179.9 1189.0 1199.9 1212.4 13 12/13 _39.7 G40.3 842.1 845.2 849.9 856.3 864.6 874.9 887.2 901.2 13 1/14 1160.9 1161.3 1162.8 1165.4 1169.3 I174. 1182.1 1191.1 1202.O 1214.6 13 13/14 837.8 838.4 840.Iu 843.3 847.9 854.3 862.7 873.0 885.3 899.3 13 1/15 1162.8 1163.2 1164.7 1167.3 1171.211176.7 1183.9 1193.0 1203.9 1216.4 13 14/15 836.1 836.6 838.4 841.6 846.2 852.7 861.0 871.3 883.6 897.6! 13 1/16 1164.4 1164.9 1166.4 1169.0 1172.9 1178.4 1185.6 1194.6 1205.5 1218.0 13 15/16 834.6 835.2 836.9 840.1 844.8 851.2 859.5 369.9 882.2i 896.2 13 1/17 1165.9 1166.4 1167.8 I170.4 1174.3 1179.8 I187.0 i196.1 1206.911219.5 13 16/17 333.3 333.8 835.6 838.8 843.4 849.9i 358.2 863.6 880.9 894_9 13 1/18 1167.2 I167.7 I169.1 1171.7 1175.6 1181.1 1188.3 1197.3 1208.2 1220.7 13 17/13 832.1 832.7 834.5 837.6 842.3 848.7 857.1 867.4 879.7 893.8 13 1/19 1168.3 1168.8 1170.3 1172.9 I176.8 1182.2 i189.4 1198.5 1209.4!1221.9 13 18/19 831.0 831.6 833.4 836.6 841.2 847.7 856.0 866.4 878.7 892.7 13 1/20 L169.4 1169.9 1171.3 I173.9 1177.8 I183.3 I_90.5 1199.5 1210,4 1222.9 13 19/20 830.1 830.7 832.5 835.6 840.3 046.7 855.1 865.5 877.8 891.8 13 1/21 1170.3!1170.8 1172.3 1174.9 1178.8 1184.2 1191.4 1200.5 1211.311223.8 13 20/21 829.3 829.8 831.6 834.8 839.5 845.9 354.3 864.6 877.0 891.0 65 TABLE 5.-Continued Orbit altitude, km, for inclination, i, deg, of -Q 0 10 20 30 40 50 60 70 80 90 14 812.3 812.9 814.7 817.9 322.6 829.1 837.5 348.0 860.3 74.5 14 I/ 2 639.5 640.1 642.2 645.7 650.9 657.9 667.0 678.2 691.5 706.5 14 1/ 3 696.0 696.7 698.6 702.0 707.0 713.9 722.3 733.7 746.7 761.4 14 2/ 3 534.0 584.7 586.7 590.4 595.7 603.0 612.3 623.8 637.3 652.6 14 1/ 4 724.7 725.3 727.2 730.6 735.5 742.3 751.1 761.9 774.7 789.3 14 3/ 4 556.6 557.3 559.4 563.1 568.5 575.9 535.3 596.9 610.6 626.1 14 1/ 5 742.0 742.6 744.5 747.8 752.7 759.4 763.1 778.9 791.6 806.1 14 4/ 5 540.3 541.0 543.1 546.8 552.3 559.7 569.3 580.9 594.7 610.3 14 1/ 6 753.6 754.2 756.1 759.4 764.3 770.9 779.6 790.3 803.0 817.4 14 5/ 6 529.5 330.2 532.3 536.1 541.6 549.0 553.6 570.3 584.1 599.3 14 1/ 7 761.9 762.5 764.4 767.7 772.5 779.2 7°7.8 798.5 311.1 325.5 14 6/ 7 521.8 522.5 524.6 528.4 533.9 541.4 351.0 562.8 576.6 592.3 14 1/ 8 768.2 763.8 770.6 773.9 773.7 735.4 794.0 804.6 817.2 831.5 14 7/ 3 516.0 516.7 518.9 522.6 528.2 535.7 545.3 557.1 571.0 586.7 14 1/ 9 773.0 773.6 775.5 778.8 733.6 790.2 793.3 309.4 022.0 836.3 14 8/ 9 511.5 512.2 514.4 518.2 523.7 531.3 540.9 552.7 566.6 532.4 14 1/10 776.9 777.5 779.4 732.6 787.5 794.1 302.6 313.2 825.8 340.1 14 9/I0 507.9 508.6 510.8 514.6 520.2 527.7 537.4 549.2 563.2 573.9 14 1/11 780.1 780.7 782.6 785.3 790.6 797.2 805.8 316.4 828.9 843.2 14 I0/11 505.0 505.7 507.9 511.7 517.3 524.8 534.5 546.4 560.3 576.1 14 1/12 782.8 783.4 785.3 788.5 793.3 799.9 803.4 819.0 831.5 845.3 14 11/12 502.6 503.3 505.5 509.3 514.8 522.4 532.1 544.0 557.9 573.7 14 1/13 785.1 785.7 787.5 790.7 795.5 802.1 810.6 321.2 833.7 848.0 14 12/13 500.5 501.2 503.4 507.2 512.8 520.4 530.1 542.0 555.9 571.7 14 1/14 737.0 787.6 789.4 792.7 797.5 804.0 812.6 323.1 835.6 349.9 14 13/14 498.8 499.5 501.7 505.5 511.0 518.6 528.3 540.2 554.2 570.0 14 1/15 788.7 789.3 791.1 794.3 799.1 805.7 814.2 824.8 837.3 851.5 14 14/15 497.2 497.9 500.1 503.9 509.5 517.1 526.3 538.7 552.7 568.5 14 1/16 790.1 790.7 792.6 795.8 800.6 307 .2 315.7 826.2 838.7 352.9 14 15/16 495.9 496.6 498.8 502.6 508.2 515. 525.5 537.4 551.4 567.2 14 1/17 791.4 792.0 793.9 797.1 801.9 808.4 816.9 827.5 840.0 354.2 i4 16/17 494.7 495.4 497.6 501.4 507.0 514.6 524.4 536.3 550.3 566.1 14 1/18 792.6 793.2 795.0 798.2 803_0 809.6 818.1 828.6 841.1 355.3 14 17/18 493.7 494.4 496.6 500.4 506.0 513.6 523.3 535.2 549.3 565.1 14 1/19 793.6 794.2 796.1 799.3 304.1 810.6 819.1 829.6 042.1 856.3 14 18/19 492.7 493.4 495.6 499.5 505.1 512.7 522.4 534.3 548.3 564.2 14 1/20 794.6 795.2 797.0 800.2 805.0 811.5 820.0 830.5 843.0 357.2 14 19/20 491.9 492.6 494.8 498.6 504.2 511.3 521.6 533.5 547.5 563.3 14 1/21 795.4 796.0 797.8 801.0 805.8 812.4 820.9 831.4 343.8 358.0 14 20/21 491.1 491.8 494.0 497.9 503.5 511.1 520.3 532.8 546.8 562.6 66 TABLE 5.-Concluded Orbit altitude, km, for inclination, i, deg, of -Q 0 10 20 30 40 50 60 70 80 90 13 475.9 476.0 47S.9 492.7 4,;g.4 496.0 505.9 517.9 532.0 547.9 15 i/ 2 320.g 321.6 324.0 328.2 334.3 342.6 353.2 366.0 381.0 397.S 15 I/ 3 371.6 372.4 374.7 378.8 3G4.g 392.9 403.2 415.7 430.4 447.0 15 2/ 3 270. 271.6 274.1 27g.4 234.8 293.2 304.0 317.2 332.4 349.6 15 I/ 4 397.3 39S.i 400.4 404.4 410.4 418.3 428.5 440.9 455.5 471.8 15 3/ 4 246.1 246.9 249.5 253.9 260.3 268.9 279.3 293.0 303.5 325.8 15 I/ 5 412.9 413.6 415.9 419.9 425.8 433.7 443.8 456.1 470.6 486.9 15 4/ 5 231.4 232.2 234.8 239.2 245.7 254.3 265.3 278.7 294.2 311.6 15 I/ 6 423.3 424.1 426.3 430.3 436.1 444.0 454.1 466.3 480.7 497.0 15 5/ 6 221.6 222.5 225.1 229.5 236.0 244.7 255.7 269.1 284.7 302.2 15 I/ 7 430.G 431.5 433.8 437.7 443.5 451.4 461.4 473.6 438.0 504.2 15 6/ 7 214.7 215.5 218.1 222.6 229.1 237.8 248.9 262.3 27.0 295.5 15 i/ 8 436.4 437.1 439.4 443.3 449.1 456.9 466.9 479.1 493.4 509.6 15 7/ 3 209.5 210.3 212.9 217.4 223.9 232.7 243.8 257.3 272.9 290.5 15 i/ 9 440.7 441.5 443.7 447.7 453.4 461.2 471.2 483.4 497.7 513.8 15 3/ 9 205.4 206.3 203.9 213.4 219.9 228.7 239.8 253.3 269.0 286.6 15 1/10 444.2 445.0 447.2 451.2 456.9 464.7 474.7 486.8 501.1 517.2 15 9/10 202.2 203.0 205.7 210.2 216.7 225.5 236.7 250.2 265.9 283.5 15 1/11 447.1 447.8 450.i 454.0 459.8 467.5 477.5 489.6 503.9 520.0 15 10/II 199.6 200.4 203.0 207.5 214.1 222.9 234.1 247.6 263.3 281.0 15 1/12 449.5 450.2 452.5 456.4 462.1 469.9 479.8 492.0 506.2 522.3 15 11/12 197.4 19G.2 200.8 205.3 211.9 220.7 231.9 245.4 261.2 278.8 15 1/13 451.5 452.3 454.5 458.4 464.1 471.9 481.8 494.0 508.2 524.3 15 12/13 195.5 196.4 199.0 203.5 210.1 218.9 230.1 243.6 259.4 277.0 15 1/14 453.3 454.0 456.2 460.i 465.9 473.6 483.5 495.7 509.9 525.9 15 13/14 193.9 194.8 197.4 201.9 208.5 217.3 228.5 242.1 257.8 275.5 15 1/15 454.8 455.5 457.7 461.6 467.4 475.1 435.0 497.1 511.4 527.4 15 14/15 192.5 193.4 196.0 200.5 207.1 216.0 227.2 240.7 256.5 274.2 15 1/16 456.1 456.8 459.1 463.0 468.7 476.4 486.3 498.4 512.6 523.7 15 15/16 191.3 192.2 194.8 199.3 205.9 214.8 226.0 239.6 255.3 273.0 15 1/17 '457.2 458.0 460.2 464.1 469.8 477.6 487.5 499.6 513.3 529.8 15 16/17 190.3 191.1 193.8 198..3 204.9 213.7 224.9 233.5 254.3 272.0 15 1/18 458.3 459.0 461.3 465.1 470.9 478.6 488.5 500.6 514.8 530.3 15 17/18 189.3 190.2 192.8 197.3 203.9 212.8 224.0 237.6 253.4 271.1 15 1/19 459.2 459.9 462.2 466.1 471.8 479.5 489.4 501.5 515.7 531.7 15 18/19 188.5 189.3 192.0 196.5 203.1 212.0 223.2 236.8 252.6 270.3 15 1/20 460.0 460.8 463.0 466.9 472.6 480.3 490.2 502.3 516.5 532.5 15 19/20 187.7 188.6 191.2 195.7 202.4 211.2 222.4 236.0 251.8 269.5 15 1/21 460.8 461.5 463.8 467.6 473.4 431.1 491.0 503.0 517.2 533.2 15 20/21 187.0 187.9 190.5 195.1 201.7 210.5 221.8 235.3 251.2 268.9 67 TABLE 6.- CIRCULAR SUN-SYNCHRONOUS ORBIT PARAMETERS FOR SPECIFIED REPETITION FACTORS Q 12 12 1/ 2 12 1/ 3 12 2/ 3 12 II 4 12 3/ 4 12 I/ 5 12 4/ 5 12 i/ 6 12 5/ 6 12 i/ 7 12 61 7 12 11 8 12 7/ 8 12 11 9 12 3/ 9 12 1/lO 12 9/10 12 1/14 12 13/14 12 1121 12 20/21 13 13 11 2 13 11 3 13 21 3 13 ii 4 13 31 4 13 i/ 5 13 4/ 5 13 11 6 13 5/ 6 13 1/ 7 13 61 7 13 1/ 8 13 7/ 8 13 1/ 9 13 81 9 13 1110 13 9/10 13 1114 13 13/14 13 1121 13 20121 as, kln is, deg 8054.630 7837.948 7908.546 7768.903 7944.445 1734.9A6 7966.181 7714.749 7980.754 7701. 357 7991.204 7691.827 7999.064 7684.698 8005. 190 7679.165 8010.100 7674.746 8022.761 7663.412 8033. 349 7653.999 7635. 259 7445.167 7507.209 7384. 389 7538.717 7354.462 7557.781 7336.650 7570.558 7324.835 7579.717 102.944 I01. 748 102.128 i01. 385 102.325 101.210 102.445 101.106 102.526 101 .O38 102.585 100.990 102.629 100.954 102.663 i00.926 102.691 lOO .904 102. 762 i00.84 7 102.822 100.800 100. 706 99.793 100.085 99.514 100.235 99,378 100. 327 99,298 100. 389 99.246 IO0.433 7316 7586 7310 7591 7305 7596 7301 7607 7291 7616.632 7283,023 .425 99.208 .604 100.467 .133 99.180 .972 100.493 .248 99.159 .273 100.514 .347 99.141 .361 I00.569 .338 99.097 100.614 99.061 68 TABLE 6.-Concluded 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 16 Q as, km i s , deg i/ 2 1/ 3 2/ 3 1/ 4 3/ 4 1/ 5 4/ 5 1/ 6 5/ 6 1/ 7 6/ 7 1/ 8 71 8 I/ 9 8/ 9 1/10 9/10 1/14 13114 i121 20/21 1/ 2 1/ 3 7266.465 7098.096 7153.130 7044.107 7181.049 7017.493 7197.932 7001.646 7209.242 6991.130 7217.348 6983.643 7223.442 6978.0t_0 7228.191 6973.691 7231.996 6970.216 7241.802 6961.301 7250 .O00 6953.893 6939.136 6788.770 6837.983 98.988 98.275 98.504 98.056 98.621 97.949 98.693 97.886 98.741 97.844 98.776 91.815 98.802 97.793 98.822 97.776 98.839 97.762 98.881 97.727 98.917 97.698 97.641 97.074 97.256 21 3 6740 11 4 6862 31 4 6716 1/ 5 6878 41 5 6702 1/ 6 6888 5/ 6!6692 1/ 76895 6/ 7 6686 1/ 8 6900 7/ 8 6681 1/ 9 6905 8/ 9 6677 i/i0 6908 9/10 6674 1/14 6917 13/14 6666 1/21 6924 20/21 6659 6646. .430 96.899 .926 97.350 .58O 96.813 .001 97.407 .371 96.763 .097 97.445 .940 96.729 .331 97.473 .224 96.706 .769 97.493 .197 96.688 .005 97.5i0 .294 96.674 .400 97.523 .176 96.663 .147 97.556 .175 96.635 ,457 97.584 .525 96.612 276 96.566 69 TABLE 7.-SOLAR POSITION DATA AS A FUNCTION OF TIME J.D. Calendar Days from Days from x-, y-, and z-components of ;a _44 xxxx.5 day 1981T Jan. 1.0, 1981 right ascension and declination 4605.5 4610.5 4615.5 4620.5 4625.5 4630.5 4635.5 4640.5 4645.5 4650.5 4655.5 4660.5 4665.5 4670.5 4675.5 4680.5 4685.5 4690.5 4695.5 4700.5 4705.5 4710.5 4715.5 4720.5 4725.5 4730.5 4735.5 4740.5 4745.5 4750.5 81/ 1/ 1/ 0 81/ 11 6/ 0 gl/ l/ill 0 _1/ 1/16/ 0 31/ 11211 0 _11 11261 n _11 11311 o _11 21 51 o '_11 21101 o 811 21151 0 _1/ 2/20/ 0 81I 21251 0 _11 31 21 o 8II 3I 71 o 81/ 3/12/ 0 811 31171 0 811 31221 0 811 11271 0 all 41 II 0 811 41 61 0 _II 41111 0 _ll 41161 o 811 41211 0 81/ 4/26/ 0 81/ 51 1/ n _1/ 51 / o 11 51111 o _11_51161 0 811 51211 0 811 5/261 0 -78.6992631 -73.6992631 -68.6992631 -67.6902631 -58.6Q92631 -Sq.6gg2631 -4R.6992631 -43.6gg2631 -3R.69g2631 -33.6992631 -2n.6992631 -23.6992631 -I.6902631 -13.6992631 -8.6992631 -3.6992631 1.3007369 6.3007369 11.3007369 16.3007369 21.3007369 26.3007369 31.30_7368 36.300736g 41.3007369 46.3007369 51.3007369 56.3007369 61.3007369 66.3007369 0 5 i0 15 20 25 30 35 40 45 50 55 60 65 70 75 80 35 gO 95 I00 105 II0 115 120 125 130 135 140 145 0.1821703 -0.9021130 -0.3911599 -78.583370 -23.026691 0.2687963 -0.8836995 -0.3831757 -73.081722 -22.530532 0.3532844 -0.8583031 -0.3721637 -67.627542 -21.849120 0.4349470 -0.8261368 -0.3582162 -62.233954 -2o.g90690 0.5131237 -0.7874736 -0.3414516 -56.911432 -19.965343 0.5871874 -0.7426430 -0.3220129 -51.667547 -18.7R4701 0.6565507 -0.6920272 -0.3000657 -46.506911 -17.461549 0_72_6715 -0.6360570 -0.2757967 -41.43127" -16.009503 0.7780581 -0.5757062 -0.2494116 -36.439765 -14.442697 0.8_12727 -0.5099864 -0.2211320 -31.579148 -12.77553_ 0.376Q352 -0.44OQ415 -0.1911938 -26.694212 -11.022466 0.9157256 -0.36R6411 -0.15g8441 -21.928099 -9.197850 0.9473857 -0.2936752 -0.1273386 -17.222644 -7.315830 0.9717197 -0.2166476 -0.0939392 -12.568691 -5.390267 0.9885947 -0.1381703 -0.0599111 -7.956371 -3.434713 0.9979401 -0.0588577 -0.0255209 -3.37534g -1.462399 0.9997459 0.0206790 0.0089665 1.184954 0.513750 0.9940616 0.0998369 0.0432896 5.735175 2.481089 0.980q934 0.1780259 0.0771926 10.285819 4.427217 0.9607014 0.2546732 0.1104271 14.847094 6.339941 0.9333973 0.3292274 0.1427541 19.428741 8.207247 0.8993399 0.4011617 0.1739450 24.039855 10.017274 0.85R8_27 0.4699771 0.2037836 2R.6R8705 11.75_304 n.812219_ 0.5352049 0.2320666 33.382532 13.418772 0.75q8_13 0.5g64090 0.2586048 38.127349 14.987297 0.7022320 0.6531877 0.2832242 42.927723 16.452734 0.6397148 0.7051751 0.3057661 47.786575 17.804262 0.5727992 0.7520423 0.3260878 52.704987 19.031497 0.5019768 0.7934983 0.3440633 57.682053 20.124629 0.4277578 0.8292908 0.3595830 62.7147R8 21.074589 aRespeetively (top line). bin degrees (bottom line). 7O TABLE 7.-Continued J.D. 244 xxxx .5 4755.5 4760.5 4765.5 4770.5 4775.5 4780.5 4785.5 4790.5 4795.5 4800.5 4805.5 4810.5 4815.5 482O .5 4825.5 4_30.5 4a35.5 4840.5 4845.5 4850.5 4855.5 4860.5 4865,5 4870.5 4875.5 4880.5 4885.5 4890.5 4895.5 4900.5 Calendar day 811 51311 0 811 61 51 o 811 61101 0 311 61151 0 811 6120/ 0 81/ 6125/ 0 81/ 613o/ o 811 7/ 5/ o 81/ 71101 0 811 7/15/ 0 811 7/20/ o 81/ 7/25/ 0 81/ 7/30/ 0 811 81 41 0 81/ 3/ 0/ 0 I/ 8/14/ 0 all 8/19/ 0 :81/ 81241 o 811 81201 0 81/ 91 31 o 81/ 9/ 8/ o Sll 91131 0 811 91181 0 811 91231 o 81/ 9/28/ 0 81/lO/ 3/ o 81/10/ 8/ 0 811101131 0 811101181 0 81110123/ 0 Days from 1981T 71.3007360 76.3007369 81.3007369 86.3007369 91.3007369 96.3007369 101.3007369 I06.3007360 111.3007369 116.3007369 121.3007369 126.3007369 131.3007369 136.3007369 141.3007360 146.1007369 151.3007360 156.3007369 161.3007360 166.3007360 171,3007369 176.3007369 181.3007369 186.3007369 191.3007369 196.3007369 201.3007369 20.3007369 211.3007369 216.3007369 Days from Jan. 1.0, 1981 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 2"0 295 and z-components of a X- y-, right ascension and declinatio_ u 0.3506685 0.8592n50 0.3725542 67.7_8107 71.373232 0.2712473 0.8830692 0,3829014 72.924913 22.513521 0.1900419 0.9007454 0.3905658 7_.086287 22.080713 0.1076068 0.91213_I .3955057 81.271794 21.207518 0.0244908 0.9171809 0.3976061 88.469887 23,414235 -0.05_7201 0.9158823 0.39712QI 93.66_3q3 23.398829 -0.1414958 0.9082146 0.3938130 98.855048 23.1_1968 -0.2232742 0,8943046 0._877729 104.018036 22.815Q95 -0.3035090 0.8741872 0.3790499 109.146491 22.274847 -0.3816632 0.8480146 0.3677014 114.230930 21.573926 -0.4572113 0.8159553 0.3538003 119.263594 20.719940 -0.5296418 0.7782138 0.3374355 12.238676 19.720709 -0.5984601 0.7350298 0.3187108 129.152429 18.584977 -0.6631906 0.6866775 0.2977450 134.003190 17.322218 -0.7233793 0.6334647 0.2746718 139.791302 15.942462 -0.7785067 0.5757321 0.2&963_8 143.518081 14.456144 -0.828_401 0.5138519 0.2228074 148.190143 12.87_081 -0.8725366 0.4482265 0.1943521 152.810199 II.20687_ -0.9105459 0.3792875 0.1644599 157.385851 9.465863 -0.9421629 0.3074934 0.1333298 161.924887 7.662054 -0.9671206 0.2033281 0.1011716 166.435989 5.806641 -0.9851929 0.15729_7 0.0682051 170.928554 3.910901 -0.9961975 0.0799329 0.0346590 175.412528 1.986217 -0.9999981 0.0017759 0.0007700 179.898247 0.044119 -0.9965074 -0.0766122 -0.0332192 -175.603707 -1.903671 -0.9856891 -0.1546602 -0.0670610 -171.082670 -3.845198 -0.9675600 -0.2317897 -0.1005044 -166.528047 -5.768222 -0.9421917 -0.3074189 -0.1332974 -161.929496 -7.660183 -0.9097123 -0.3809676 -0.1651883 -157.277130 -9.508174 -0.8703065 -0.4518614 -0.1959279 -152.561749 -11.298939 aRespectively (top line). bin degrees (bottom line). 71 TABLE 7.- Concluded J.D. Calendar Days from Daysfrom 244xxxx.5 day 19_IT Jan. 1.0, 1981 4905.5 4910.5 4915.5 4920.5 4925.5 4930.5 4935.5 4940.5 4945.5 495O.5 4055.5 4960.5 4965.5 4970.5 4975.5 4980.5 4985.5 4990.5 4995.5 5000.5 5005.5 5010.5 5015.5 5020.5 5O25.5 5030.5 5035.5 5040.5 5045.5 5050.5 81/10/28/ 0 81/ii/ 2/ 0 81/iii 7/ 0 81/11/12/ 0 81111/17/ 0 81/Ii/22/ 0 81/11/27/ 0 81/12/ 2/ o 811121 7/ o 81/12/12/ n 81112117/ 0 81/12/22/ 0 81112127/ 0 221.3007369 226.3007369 231. 3007369 236.3007369 241. 3007369 246.3007369 251.3O07369 256.3007369 261.3q07369 266.3007369 271.3007369 276.3007_69 281.3007369 82/ 1/ 1/ 0 82/ 11 61 0 821 1/11/ 0 821 11161 0 821 1/21/ 0 82/ 1/26/ 0 821 11311 o !82/ 2/ 5/ o s2/ 2/10/ 0 82/ 21151 0 82/ 2/20/ O 82/ 2/25/ O s_l 31 21 0 82/ 31 7/ o 821 31121 o 821 3/17/ q 82/ 31221 0 286.3007369 291.30O7369 296.3007160 301. 3007369 306.3007369 311. 3007369 316.3007369 321. 3007369 326. 3007369 331.3O07369 336. 3007369 341. 3007369 346. 3007369 351.3o07369 356.3007360 361. 3007369 366. 3007369 30O 305 310 315 320 325 330 335 340 345 350 355 360 365 370 375 38O 385 390 395 400 405 410 415 420 425 430 435 440 445 aRespectively (top line). bin degrees (bottom line). ^ a x-, y-, and z-components of xc_ right ascension and declinatiofi u -0.8242168 -0.5195372 -0.2252723 -147.775111 -13.018892 -0.7717430 -0.5834481 -0.2529841 -142.910226 -14.654172 -0.7132410 -0.6430690 -0.2788359 -137.961694 -16.190740 -0.6491219 -0.6979018 -0.3026114 -132.926046 -17.614522 -0.5798494 -0.7474810 -0.3241090 -127.802092 -18.911607 -0.5050370 -0.7913788 -0.3431431 -122.591224 -20.068490 -0.4279443 -0.8292102 -0.35q5469 -117.297654 -21.072375 -0.3464723 -0.8606381 -0.3731740 -111.92_537 -21.911505 -0.2621588 -0.8853772 -0.3839000 ;106.493941 -22.575527 -0.1756723 -0.90319R0 -0.301_2_0 -101.006626 -23.055840 -0.007705 _ -0.q139303 -0.3962815 -95.4P1630 -23.345q26 O.OOl0301 -0.0174653 -0.397_143 -89.935665 -23.441613 o.0808142 -0.9137579 -0.3962067 -84.306367 -23.341258 0.1770221 -0.902fl273 -0.3914671 -78.851454 -23.0_5823 0.2646340 -0.8847572 -0.3R36319 -73.347fl86 -22.558835 0.3492_19 -0.8596054 -0.3727650 -67.891082 -21.886248 0.4310573 -0.8278510 -0.3599576 -62.494286 -21.036198 0.5094182 -0.7894969 -0.3423268 -57.168107 -20.018702 0.5836956 -0.7449571 -0.3230143 -51.920262 -18.845316 0.6533003 -0.6946124 -0.3011847 -46.755505 -17.528777 0.7176877 -0.6388912 -0.2770239 -41.67572_ -16.082665 0.7763635 -0.5782654 -0.2507365 -36.680157 -14.521100 0.828_872 -0.5132449 -0.2225435 -31.765689 -12.858472 0.8748758 -0.4443720 -0.1926802 -26.927193 -11.109239 0.9140064 -0.3722156 -0.1613930 -22.157883 -9.207763 0.9_60176 -0.2973645 -0.1289375 -17.449650 -7.408200 0.9707107 -0.2204221 -0.0055752 -12.793379 -5.484429 0.9879498 -0.1420002 -0.0615714 -8.179231 -3.530015 0.9976610 -0.0627131 -0.0271Q24 -3.596808 -1.558205 0.99O8317 0.0168276 0.0072964 0.964222 0.418060 72 TABLE 8.-AVAILABLE SUNELEVATION ANGLE AS A FUNCTION OF LATITUDE FOR THE WINTER AND SUMMER SOLSTICES AND THE EQUINOXES Sea son deg Winter solstice 0 i0 20 30 4O 50 60 7O 8O 90 Vernal or autumnal equinox 0 i0 20 30 4O 50 6O 7O 8O 90 Summer solstice o i0 20 30 40 50 60 70 80 oO Hour angle, B, deg ..... Sun elevation angle, _, deg, at local time relative to high noon, sidereal hours, of -66.6 56.6 46.6 36.6 26.6 16.6 6.6 -3.5 -13.5 -23.4 52.6 45.5 37.6 29.3 20.7 11.9 3.t7 -5.9 -14.7 -23.4 27 22 17 11.4 5.5 -0.6 -6.6 -12.5 -18.2 -23.4 6 .3 0.0 .5 -4.0 .2 -7.8 -11.5 -14.8 -17.7 -20.2 -22.0 -23.1 -23.4 -27.3 -31.4 -3t,.6 -36.6 -37.4 -36.8 -35.0 -32.1 -2_q.1 -23.4 I0 -52.6 -58.4 -62.0 -62.5 -59.8 -54.6 -47.9 -40.2 -32.0 -23.4 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 i0.0 0.0 60.0 58.5 54.5 48.6 41.6 33.8 25.7 17.2 8.6 0.0 30.0 29.5 28.0 25.7 22.5 18.7 14.5 9.8 5.0 0.0 66.6 76,6 86.6 83.5 73.5 63.5 53.5 43.4 33.5 23.4 52.6 58.4 62.0 62.5 59.8 54.6 47.9 40.2 32.0 23.4 30 27 31 34 36 37 36 35 32 28 23 6O .3 .4 .6 .6 .4 .8 .0 .I .I .4 0.0 4.0 7.8 ii .5 14.8 17.7 20.2 22.0 23.1 23.4 9O 0.0 -30.0 q. 0 -2q. 5 0.0 -28.0 0.0 -25.7 0.0 -22.5 0.0 -18.7 0.0 -14.5 0.0 -9.8i 0.0 -5.0 0.0 0.0 -27.3 -22.5 -17.2 -11.4 -5.5 0.6 6.6 12.5 18.2 23.4 120 -60.0 -5S.5 -54.5 -43.6 -41.6 -33.8 -25.7 -17.2 -8.6 0.0 -52.6 -45.5 -37.6 -20.3 -20.7 -11.9 -3.0 5.9 14.7 23.4 150 12 -66.6 -76.6 -86.6 -83.5 -73.5 -63.5 -53.5 -43.4 -33.5 -23.4 -90.0 -80.0 -70.0 -60.0 -50.0 -40.0 -30.0 -20.0 -10.0 0.0 -66.6 -56.6 -46.6 -36.6 -26.6 -16.6 -6.6 3.5 13.5 23.4 180 73 TABLE 9.-ORBIT PARAMETERS FOR A CIRCULAR ORBIT WHICH PASSES OVER THE SAME SITES ON THE EAST COAST ONCE PER DAY, WITH NORFOLK, VIRGINIA, AS INITIAL CONDITION Inclination, deg ............................. 63.0 Eccentricity ............................... 0.0 Semimajor axis, km .......................... 6887.371 Anomalistic period, sec ....................... 5689.92 Nodal period, see .......................... 5689.80 Mean motion: 5446.5085 deg/day ............................. Anomalistic revolutions per day ................. 15..184746 Perigee precession rate: deg/orbit ............................ deg/day ............................. 0.OO7654 0.116232 Nodal precession rate: deg/orbit ............................ deg/day ............................. -0.227599 -3.456029 Repetition factor, Q .......................... 15.0 Initial conditions to pass over Norfolk, Virginia (L = -76.289 ° , at 12:00 noon EST on Jan I, 1981: Longitude of ascending node: -98°.741 J.D. = 244 4605.2083333 = 42o.309 k = 36.853 °) 74 v 04 o 0"3 o o o v ! o oJ o 0J I o 0o o It ,--I o ,..-i o ¢, o o z _ @ -,-t , oooooooooo__oooooooo _ ooooooooo_~oooooooo oooooooo$__oooooooo® oooooooo_o_ $___oooooooo N 00000000_ ____ooooooooo 00000000 OOOOOOOOOOOO ooooooooo_ooooooooooooo_ _ 00000000__0_ 0000000000000000 oooooooo__OOOOOOOOOOOOOoooooo_ OOOOOOOO_O_O_O_OOOOOOOOO0oOOOOO oooooooo_ .....__NNOOOOOOOOOOOOO_N_ OOOOOOOO_OOOOOOOOOOO O0000000m__O00000000 N OOOOOOO_OOOOOOOOO OOOOOOOON__O_OOOOOOOO OOOOOOOO_N Nm_ __o_oooooooo o _O0_O0000_OONO0000000 N OOOOOOOOOOO_O_O_NOOOOOOOO OOOOOOOOOOOOO_O__N_OOOOOOOO OOOOOOOOOOOOOOOOm__N_OOOOOOOO OOO_OOOO_OOOOOOOOOO_NO_OOOOOOOO _O_OmO_O_O_O_O_O_O_O_O_O_O_0_O_O_O_0 IIIIIIIIIIIIIIIII Illlllllllllllll 75 I ?0 J o 0 oo cO 0 o co v o C cO =T o4 2-0 cO r_ QJ = ; m _ i_ [ 00000_000 N_OO_N __ __N ooooo_o__ .... o__o_oo OO_m_O000oo0ooooo_ooooooooo00, 00_0_000000000000000000000000_000000 O0_mO000000000000_O00000000000000000 00_000000000000000_00000_0000_0_0000 00_000000000_0_00_3000000000000000' 00_0000000000000000000000000000000 0o__0o0o0o0oo0o00_00o0o0_00o0_ N .... oo_ _ oooo ,o oo_oo ooo_ 0 00 _ O000_00 .00300 O00300, _ 'n _ _ _0 O0 ooOoOooO,_OOOO0OOO0OOO0OOOOOOOOO_o°m IIIlilllllllllll ?6 o I. o .-3 CO I 0 L 0 g - o In o o o m .,-i co .o o L ,o z c_ oJ 04 0 OJ aO o4 0 cO u_ . _P ,J O00000000_N_O__O0000000 O000000(0 _0 O00 _00000000000 _D 0_0 _ 0000 Oo 000o000000 _0000 :)000000000000 0_000: 0 0o0000o _00o000000000000000000 )0_0 _ O0 O000000 D, 000000000000 ,_00000000_0 00300000000_0000000000000000000000_0 00_00_00_30_0_000_00000_00_00030_0_0 000000000000_000000000000000000000_0 0 00_000000000000000_00000000000000_0 .... ooo ......... oo°+oo°o .......... +oo 90)_000000000000900000000000000_000 Oo O0 o- 0 _ _ _0 _ 00000 (_00000000 O0 _00000 0 o -+r ' oo 00000o000000000 O0000000000000 " c0 _0 o(.)_0 _0_0 .. _ 000000000 , o 00 O000000 .+) o 0 o 00 0 _ooo00,)0000_0000000000000000000000o 0_03000000000000000-_0000000000000000 O0OOOOOOOOOOO030OO0O0OOOOOOOOOOOO0OO 0_000000_000_000000_0000000000000000 0 0"o. -- -_00000 O0 O00 ') _ 0000 O0000 O000 000000 --4+--I 00_0000000000_000000000000000000000 0_000000000)000000000000000000000 00000__0 _ .... 000_000000000000000 ill\|\|iililiilill ?? z == X 0_ Z 0 o° o o ,L o a0 o v o o oo _ o ° o I 7-N II1 I-4 o r b,-u b-0 t_ oJ L_ r_ I r--0 -0 OJ . o e-g o Z _ m 0 ('4 m O00000000000000000_O000000QO0000000 oooooooooooo _N__'o_gZgO0000000 oooooooo_o_e__o_oooooooo O0OOOOO0_m_NmN__4_o_ooooooooo 0o00o09o_o_0_o___ooooooo0o oo00ooooo_oomJ_OO_o__ooooooooOo I ooooooOo_o_oo.%o_oooooooooooo ooooooooooo_m__ooooooo_oooo_o O000000000000N%O00._o00000000 0000000000000000_0000000000000000_0 I ...... _ 0 _0_0 _ 0 0 _ 0 _ 0_0 '_0 llillllll l llll IIIII_\|III\|IIIII! ?8 .,4 o I. N °o l_i cO s i:i °o cO L ¢ o 100000ooo000 DOOOOO--4_q,O0000000000000000 ...4,"t ua I ,o I L_ 0 U_ u_ u_ t'--o', g u_ t_ ,,1-.. -- .-. .- r- :o 4" 0 ' _ ", _ _ 4 _ 79 o L_ LG Lm r_ Lm I b-rM 0 --" o -Lr_ , Oa hi) • u3 •,-4 _ _ s m L t_ 0 m _ .o m e--; 0 • , 0 _ t---i _ o m -=t ! o : m i . c_ ,..1 - G _J < - {] o o L) < 0 O : <J _b 0 0 ) 0 ) 0 0 () ] • 0 0 0 :] ) 0 0 m '] _ 0 rJ_ D_,D :. )0 ):D:),D :,1{:' ) ' J ) )O.D.D .J:D JO-- 0. " _t r---,a -n o) o 9o 0 u,.] 0oo ,D,D 0 - P- qr L) 0 00_m L) 0 O0 0 <m 0 0 <J 0 0 I I I I I I I I ! I ! I ! I I ! o_o_o_m_o_o_o_o_o_o_o_o_o_o_o_o ............... °...... ,, ,, ,,TT ,, , Y_Y_,,, 8O I, Relx_t No. 2. Government Accession No. NASA RP-IO09 41 Title and Subtitle AN INTRODUCTION TO ORBIT DYNAMICS AND ITS APPLICATION TO SATELLITE-BASED EARTH MONITORING MISSIONS 7. Author(s} David R. Brooks 9. Performing Organization Name and Addre_ NASA Langley Research Center Hampton, VA 23665 12. S_nsoring Agency Name and Addr_s _ationaIAeronautics and Space Administration Washington, DC 20546 3. Recipient's Catalog No. 5. Report Date November 1977 6. Performing Organization Code 8. Performing Orgamzation Report No. L-11710 10. Work Unit NO. 683-75-33-02 '11. Contract or Grant No. 13. Type of Report and Period Covered Reference Publication 14. Sponsoring Agency Code 15. Supplementary Notes 16. Abstract A self-contained tutorial treatment is given to the long-term behavior of satel-lites at a level of complexity suitable for the initial planning phases of Earth monitoring missions. First-order perturbation theory is used to describe in detail the basic orbit dynamics of satellite motion around the Earth and relative to the Sun. Surface coverage capabilities of satellite orbits are examined. Several examples of simulated observation and monitoring missions are given to illustrate representative applications of the theory. The examples stress the need for devis-ing ways of maximizing total mission output in order to make the best possible use of the resultant data base as input to those large-scale, long-term Earth monitor-ing activities which can best justify the use of satellite systems. 17. Key Words (Suggested by Authoris)) Orbit dynamics Remote sensing Satellite monitoring 18. Dislribution Statement Unclassified -Unlimited Subject Category 13 19. Security Cla_if. (of this report] 20. Security Classif. (of this page} 21. No, of Pages 22. Price" Unclassified Unclassified 84 $5. O0 For sale by the National Technical Information Serwce, Springfield, Virgima 22161 NASA-Langley, 1977
6133	https://lizaab.com/2022/06/22/mathematics-of-meetings/	Mathematics of Meetings Written by Liz Aab My profession is to go to meetings and write documents, I often tell my 6 year old. A lot of work happens at meetings. So if one wants to accelerate progress at work, getting a new product launched, ramping up sales, finishing a report, whatever, there are two ways to go faster: make each meeting more effective, and/or have the next meeting soon. (I got this idea while swimming, where there are only two things that affect your lap time: stroke length (how far you propel yourself with each stroke) and stroke rate (time between strokes).) Many articles and techniques have been written about the subtle interpersonal techniques around how to have an effective meeting (stroke length). Not so much has been written about meeting frequency (stroke rate). Fortunately, meeting frequency is simple math: reduce the wait time to next meeting, and your initiative or organization moves faster. The target metric (or Key Performance Indicator, KPI): Average Time to Next Meeting. So one Friday night, I pulled out my trusty loved-hated Excel and tried to model out what are the key variables that affect this time-to-next-meeting. I was surprised to find a few very important variables that massively slow an organization. Here they are: 1. The number of people needed at the meeting. We all know that more people means slower progress, but I was shocked to see in numbers how much a few extra people delay the next available slot — like Covid, it’s exponential. Say I have hour-long meetings randomly throughout an 8 hour work day, and usually spend half my day (4 of the 8 slots) in meetings. I want to find one slot where both of us are free. Maybe at 9am I am free, but not my colleague; 10am my colleague is free but not me; 11am we are both in meetings; and noon we are both free. In the model, it takes us an average of nearly 4 hours to meet up — so still same day, no biggie. Add a third person, and on average it will take us 10 slots to meet up, so by tomorrow morning. Add a fourth, and now it’s 20 slots (within 2-3 days); to have 9 people meet, it’ll take 534 slots, or nearly 10 weeks. Top tip to speed up work: have max 3 people in a team that need to meet. 2. What proportion of each day a person spends in meetings I assumed here all of us had already booked 50% their day with meetings (4 of 8 slots). Some people rarely have meetings, and can be scheduled in any day. But if you are trying to schedule in someone whose main job is to have meetings, for instance a manager or senior executive, they may spend 80-90 or even 100% of their time in meetings. You don’t need an Excel model to figure out that scheduling a meeting with someone whose calendar is already 100% full for weeks means you won’t be able to schedule a meeting at all. Top tip: cut down the number of meetings you commit to, and avoid including people who are always in meetings (such as senior managers). 2b. …especially when >85% of your day is in meetings You will know either from trying to fly during the (pre-pandemic) Christmas holiday, that when people and planes are operating near 100% of capacity (or really anywhere above 85%), your waiting time explodes from minutes to hours. When we want to go faster in business we tend to load people up to 100% (or 150%) of the time they have; loading people’s calendars more than 85% with meetings hits a real tipping point (Here’s a picture from queuing theory) — you end up needing to wait a LOT longer to have that next meeting. The fix: resist adding more work until you and the many stakeholders involved have finished the first. Celebrate those who operate at less than 85% of their capacity — who have some down time. Culturally, it can be hard to tell a boss (or boss’s boss) you’re too busy for them; but the fastest teams are the best at pushing back on taking on more work. 3. How many slots there are per day. There are 2 ways of increasing the number of slots per day : longer working hours, and/or have shorter slots. If you are looking for a slot, it’s a lot easier to half the meeting length (e.g. from 1 hour to 30 minutes) than to double the number of hours worked (from 8 hours to 16 hours).Tech teams famously use the 10 minute standing meeting to address this problem. In general, I’ve often wondered Outlook/Google calendar had 20, 40, 60 minute default meetings rather than 30 and 60 minute ones, if we’d all meet up faster. Working remotely has helped put more slots into a day — no more travel time to and from meetings. But days of back-to-back virtual meetings mean we, as humans, need longer slots: if you have too short, task-focused meetings only, you destroy human connection; and often don’t let the important stuff that really matters come to light. 4. How correlated their schedules are. In my model, I’ve randomly assigned a calendar slot to empty or full — there’s no correlation between my calendar and my colleague’s. But that’s not usually the case. One way teams solve the Time to Next Meeting problem is by coordinating schedules well in advance and on a recurring basis — the standing Tuesday 9am team meeting, the monthly senior management meeting, the annual offsite conference. Of course, if your stakeholders’ calendars are now full of recurring standing meetings, go back to top tip 2 above! I’ve also assumed that calendars can’t change. But often they do: you can propose a time and the other person can move something around to make it work. How helpful would Outlook, Google Calendars, Calendly and other calendar programs be if they knew where there was some flexibility in the slot, and could adjust others’ calendars for you! One common ‘calendar correlation problem’ large corporates face is using specialists : legal, financial, technical, central services, etc. Often their calendars are truly uncorrelated to the teams they support. The main fixes I have found are switching to one-to-ones with the experts, giving up on meetings entirely and switching to email, using specialists in a very short timeframe (e.g. all within the same week when they might have a lull from other projects) or abandoning specialist support entirely. If you have other solutions, let me know. 5. How much the team’s working hours overlap. Having worked in global jobs spanning multiple time zones my entire career, I know working across time zones kills any hope of that next meeting happening quickly. I chose to live in London in part because it straddles the Americas-Europe-Asia time span. But if you need a team with people from London, the US (hopefully not California), Bangalore and Singapore, you likely have 1 (awkward) hour a day to meet, at best — 7 am California (preferably not a Monday) / 3pm London / 10 pm Singapore (preferably not a Friday). It’s not just time zones — it’s holidays too. My first summer working in Europe, I was advised to ‘make sure you get your deliverables done by June, before the July and August holiday period.’ Having moved to London from China, where all the nations holidays were synchronized around (then) 3 major holiday weeks, and in America, where people didn’t take more than a few days holiday at a time, it was a learning curve: some schools (and their parents) break for a few weeks holiday in July, others in August, Germany has a week long holiday sometime in October, others take a few weeks off for Eid, in addition to the Christmas, New Year and Chinese New Year holidays I already had learned to work around. In my model, I didn’t assume any holidays — once you add those in, the time-to-meeting explodes. The solutions I have seen are to synchronize shut downs (e.g. everyone takes off the week after Christmas), reduce participants from teams that have different holiday schedules, pre-plan projects with the holiday calendar in mind and ask people to work during their holidays (ouch). So that’s it: it’s just math. It’s quite amazing that there’s a single metric that you can measure to figure out if you work in a fast organization/team/project, or a slow one: ask your colleagues how long it takes on average to book the next meeting. Is it same day, next day, 2-3 days, >1 week, or >2 weeks? Not everyone wants a fast organization, and a lot of mistakes are made by going too fast. There are good reasons for having a slower organization, where the tradeoffs are worthwhile: maybe you make huge, multi-year investments, with safety critical operations, and massive risks. In that case, it’s worth it. But if you are in a business or organization which doesn’t have these constraints, especially where the market is changing quickly or you have aggressive growth targets to hit, then consider a few ‘quick’ fixes: Delegate power and authority to small (3 person) teams Keep everyone working in the same time zone, on the same correlated schedules; avoid jobs or reporting lines that span time zones Keep everyone at least 15% below what they can do (think: a day free a week) Shorten the default meeting time to 20, 40 or 60 minutes Use specialists with uncorrelated schedules sparingly, in one-to-ones, in bursts, or by email Don’t require all decisions or input to be made, live, by people whose job it is to primarily attend meetings Does this resonate with you? Any other variables you noticed? Any other fixes you’ve found that work? And if you’ve made it this far, let me know — it means I found someone else who finds it interesting to think about the mathematics of meetings! We should be friends. Yours, Liz If you enjoyed my writing, you may enjoy my recently published novel Turning Forward, available in eBook and paperback on Amazon Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading… More posts The 6pm moon The Bravery of Trees Mathematics of Meetings Food goes in, what goes out? Reblog Subscribe Subscribed Aabservations Already have a WordPress.com account? Log in now. Aabservations Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar
6134	https://michaelminn.net/tutorials/data/2024-ei-statistical-review.html	Home / Tutorials Energy Institute Statistical Review of World Energy This is a polygon dataset representing 2023 energy data from the 2024 Energy Institute Statistical Review of World Energy. This is a continuation of an annual review that had been published since 1952 by the energy company bp. All values are expressed in exajoules (EJ) for comparability. Reserves are converted to EJ using conversion factors of 0.043 EJ per MM tonnes oil equivalent (MMTOE) and 1.019 EJ per trillion cubic feet (TCF) natural gas. Although the heat content of coal varies widely by quality, the US average 0.0223 EJ per MM tonnes (19.180 million Btu per short ton) is used for simplicity. Data is for 2023 except where noted. Reserves numbers are from 2020 because the methodology and timing for reserves calculation was under review at the time the report was published. There are four types of numbers depending on the resource: Reserves for fossil fuels represent the estimated heat value for quantities of a fuel resource in the ground within a country's borders "...which geological and engineering data demonstrate with reasonable certainty to be recoverable in future years...under current economic and operating conditions" (bp 2019). Production for fossil fuels represents the estimated heat value of the fuel produced each year. Generation for non-fossil fuel generated electricity represents the heat equivalent in fossil fuel that would be needed to generate that same amount of electricity each year. Generation and consumption heat values for non-fossil fuels do not align because the conversion factor for electricity to heat value (calorific) equivalents adjusts for fossil-fuel generating plant losses (around 59.6%) so the numbers are comparable to fossil-fuels. Consumption is the estimated heat value of a resource consumed within the borders of the country each year. Primary consumption is total energy consumed within a country in a year measured from primary fossil fuel or renewable energy sources. Country boundaries are from the Natural Earth large scale (1:10m Cultural Vectors) polygons for world countries (Admin 0 - Countries). French Guiana and Antarctica have been removed to facilitate cleaner mapping in Mercator or equirectantular projections. 2023 population data is from the World Bank ( and is provided to facilitate per capita normalization. Fields Name: Short country name from the Energy Institute data. ISO_A3: Three-digit ISO alpha 3 code for the country. Total_Population Primary_Consumption_EJ Oil_Consumption_EJ Gas_Consumption_EJ Coal_Consumption_EJ Nuclear_Consumption_EJ Hydro_Consumption_EJ Solar_Consumption_EJ Wind_Consumption_EJ Oil_Reserves_EJ (2020) Oil_Production_EJ Gas_Reserves_EJ (2020) Gas_Production_EJ Coal_Reserves_EJ (2020) Coal_Production_EJ CO2_MM_Tonnes
6135	https://rechneronline.de/pi/reuleaux-triangle.php	Reuleaux Triangle - Geometry Calculator Geometry \| Forms \| Contact & PrivacyGeometricCalculatorsGerman: Geometrierechner, Formen 1DLine, Circular Arc, Parabola, Helix, Koch Curve2DRegular Polygons: Equilateral Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, Nonagon, Decagon, Hendecagon, Dodecagon, Hexadecagon, N-gon, Polygon Ring Other Polygons: Triangle, Right Triangle, Isosceles Triangle, IR Triangle, 1/2 EL Triangle, Golden Triangle, Quadrilateral, Rectangle, Golden Rectangle, Rhombus, Equidiagonal Rhombus, Parallelogram, Kite, 60-90-120 Kite, Half Square Kite, Right Kite, Trapezoid, Right Trapezoid, Isosceles Trapezoid, Tri-equilateral Trapezoid, Obtuse Trapezoid, Cyclic Quadrilateral, Tangential Quadrilateral, Arrowhead, Concave Quadrilateral, Crossed Rectangle, Antiparallelogram, House-Shape, Symmetric Pentagon, Diagonally Bisected Octagon, Cut Rectangle, Concave Pentagon, Concave Regular Pentagon, Stretched Pentagon, Straight Bisected Octagon, Stretched Hexagon, Symmetric Hexagon, Semi-regular Hexagon, Parallelogon, Concave Hexagon, Arrow-Hexagon, Rectangular Hexagon, L-Shape, Sharp Kink, T-Shape, Square Heptagon, Truncated Square, Stretched Octagon, Frame, Open Frame, Grid, Cross, X-Shape, H-Shape, Threestar, Fourstar, Pentagram, Hexagram, Unicursal Hexagram, Oktagram, Star of Lakshmi, Double Star Polygon, Polygram, The Hat, Polygon Round Forms: Circle, Semicircle, Circular Sector, Circular Segment, Circular Layer, Circular Central Segment, Round Corner, Circular Corner, Circle Tangent Arrow, Drop Shape, Crescent, Pointed Oval, Two Circles, Lancet Arch, Knoll, Elongated Semicircle, Annulus, Semi-Annulus, Annulus Sector, Annulus Segment, Annulus stripe, Curved Rectangle, Cash, Rounded Polygon, Rounded Rectangle, Ellipse, Semi-Ellipse, Elliptical Segment, Elliptical Sector, Elliptical Ring, Stadium, Half Stadium, Stadium Segment, Spiral, Log. Spiral, Reuleaux Triangle, Cycloid, Double Cycloid, Astroid, Hypocycloid, Cardioid, Epicycloid, Parabolic Segment, Heart, Tricorn, Pointed Semicircle, Interarc Triangle, Circular Arc Triangle, Interarc Quadrangle, Intercircle Quadrangle, Circular Arc Quadrangle, Circular Arc Polygon, Claw, Half Yin-Yang, Arbelos, Salinon, Bulge, Lune, Three Circles, Polycircle, Round-Edged Polygon, Rose, Gear, Oval, Egg-Profile, Lemniscate, Squircle, Circular Square, Digon, Spherical Triangle 3DPlatonic Solids: Tetrahedron, Cube, Octahedron, Dodecahedron, Icosahedron Archimedean Solids: Truncated Tetrahedron, Cuboctahedron, Truncated Cube, Truncated Octahedron, Rhombicuboctahedron, Truncated Cuboctahedron, Icosidodecahedron, Truncated Dodecahedron, Truncated Icosahedron, Snub Cube, Rhombicosidodecahedron, Truncated Icosidodecahedron, Snub Dodecahedron Catalan Solids: Triakis Tetrahedron, Rhombic Dodecahedron, Triakis Octahedron, Tetrakis Hexahedron, Deltoidal Icositetrahedron, Hexakis Octahedron, Rhombic Triacontahedron, Triakis Icosahedron, Pentakis Dodecahedron, Pentagonal Icositetrahedron, Deltoidal Hexecontahedron, Hexakis Icosahedron, Pentagonal Hexecontahedron Johnson Solids: Pyramids, Cupolae, Rotunda, Elongated Pyramids, Gyroelongated Pyramids, Bipyramids, Elongated Bipyramids, Gyroelongated Square Dipyramid, Gyrobifastigium, Disheptahedron, Snub Disphenoid, Sphenocorona, Disphenocingulum Other Polyhedrons: Cuboid, Square Pillar, Triangular Pyramid, Square Pyramid, Regular Pyramid, Pyramid, Square Frustum, Regular Frustum, Frustum, Bent Pyramid, Regular Bipyramid, Bipyramid, Bifrustum, Frustum-Pyramid, Ramp, Right Wedge, Wedge, Half Tetrahedron, Rhombohedron, Parallelepiped, Regular Prism, Prism, Oblique Prism, Anticube, Antiprism, Prismatoid, Trapezohedron, Disphenoid, Corner, General Tetrahedron, Wedge-Cuboid, Half Cuboid, Skewed Cuboid, Ingot, Skewed Three-Edged Prism, Cut Cuboid, Truncated Cuboid, Obtuse Edged Cuboid, Elongated Dodecahedron, Truncated Rhombohedron, Obelisk, Bent Cuboid, Hollow Cuboid, Hollow Pyramid, Hollow Frustum, Star Pyramid, Stellated Octahedron, Small Stellated Dodecahedron, Great Stellated Dodecahedron, Great Dodecahedron, Great Icosahedron Round Forms: Sphere, Hemisphere, Quarter Sphere, Spherical Corner, Cylinder, Cut Cylinder, Oblique Cylinder, Bent Cylinder, Elliptic Cylinder, Generalized Cylinder, Cone, Truncated Cone, Oblique Circular Cone, Elliptic Cone, Truncated Elliptic Cone, General Cone, General Truncated Cone, Bicone, Truncated Bicone, Pointed Pillar, Rounded Cone, Elongated Hemisphere, Drop, Spheroid, Ellipsoid, Semi-Ellipsoid, Spherical Sector, Spherical Cap, Spherical Segment, Spherical Central Segment, Double Calotte, Rounded Disc, Double Sphere, Spherical Wedge, Half Cylinder, Diagonally Halved Cylinder, Cylindrical Wedge, Cylindrical Sector, Cylindrical Segment, Flat End Cylinder, Half Cone, Conical Sector, Conical Wedge, Spherical Shell, Half Spherical Shell, Spherical Shell Cap, Cylindrical Shell, Cut Cylindrical Shell, Oblique Cylindrical Shell, Hollow Cone, Truncated Hollow Cone, Spherical Ring, Torus, Spindle Torus, Toroid, Torus Sector, Toroid Sector, Arch, Reuleaux-Tetrahedron, Capsule, Half Capsule, Capsule Segment, Double Point, Anticone, Truncated Anticone, Sphere-Cylinder, Lens, Concave Lens, Barrel, Egg Shape, Paraboloid, Hyperboloid, Oloid, Steinmetz Solids, Solid of Revolution 4DTesseract, HypersphereDiscover more Calculators Calculator Haushaltsrechner Zahl in Wort Gesundheitsrechner-Apps Ingenieur-Tools Zahlen umwandeln mathematischen Mathematik-Nachhilfe Wirtschaftskurse Anzeige Reuleaux Triangle Calculator ================================================================================== Calculations at a Reuleaux triangle. This is an equilateral triangle with the side length a, where around each vertex a circle with the radius a is drawn. The three resulting circular segments are added to the equilateral triangle. The Reuleaux triangle is also the intersection of those three circles. It is the simplest form of a curve of constant width, this width is also a. Its two-dimensional equivalent is the Reuleaux tetrahedron. Enter one value and choose the number of decimal places. Then click Calculate. Radius, side length (a): Arc length (l): Perimeter (p): Area (A): Round to decimal places. Formulas: l = a π / 3 p = a π A = ( π - √3 ) a² / 2 pi: π = 3.141592653589793... Radius, length and perimeter have the same unit (e.g. meter), the area has this unit squared (e.g. square meter). A shape that has the same thickness or width everywhere is called a curve of constant width. If you put this shape between two parallel lines so that they touch the curve of constant width, you can rotate it within these lines so that the contact is maintained. There is never a gap between the shape and the lines or the shape extends beyond these lines. The center of the curve of constant width does not have to be exactly in the middle of the lines and can change its position by rotating it. The only curve of constant width with a fixed center is the circle, which is the trivial case of a curve of constant width and the one with the largest area per width. The one with the smallest area per width is the Reuleaux triangle. In between are the regular Reuleaux polygons, which get closer and closer to a circle as the number of corners increases. A Reuleaux polygon is only for an odd number of corners a curve of constant width. The Reuleaux triangle was already known before Franz Reuleaux, and Leonardo da Vinci used it to map the earth. Franz Reuleaux researched in the field of gear theory and used this form for mechanical applications, therefore this form was named after him. © Jumk.de Webprojects \| Online Calculators Discover more Calculator Calculators Graphen zeichnen Wirtschaft Rechner Technik-Gadgets Mathematisch Geographie-Rechner Geometrie-Formelsammlungen Online-Rechner Physik-Rechner Anzeige Discover more Calculators Calculator Technik-Gadgets Taschenrechner Statistik-Kurse Zeitmanagement-Apps Geometrie Rechner Physik-Lehrbücher Physik-Rechner Geographie-Rechner Anzeige ↑ up
6136	https://math.stackexchange.com/questions/13094/significance-of-sqrtnan	algebra precalculus - Significance of $\sqrt[n]{a^n} $?! - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Significance of a n−−√n a n n?! Ask Question Asked 14 years, 10 months ago Modified1 year, 11 months ago Viewed 6k times This question shows research effort; it is useful and clear 35 Save this question. Show activity on this post. There is a formula given in my module: a n−−√n={a\|a\|if n is odd if n is even a n n={a if n is odd\|a\|if n is even I don't really understand the differences between them, kindly explain with an example. algebra-precalculus radicals absolute-value faq Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 17, 2018 at 19:59 F.A. 1,009 2 2 gold badges 12 12 silver badges 21 21 bronze badges asked Dec 5, 2010 at 9:58 QuixoticQuixotic 22.9k 34 34 gold badges 134 134 silver badges 225 225 bronze badges 0 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 50 Save this answer. Show activity on this post. (−1)2−−−−−√=1–√=1=\|−1\|(−1)2=1=1=\|−1\| because in order to make the square root an unambiguous operation, we agree that the square root of a nonnegative number x x is always the (unique) nonnegative number r r such that r 2=x r 2=x. But with cubic roots there is no problem: −1−−−√3=−1−1 3=−1, because every real number has a unique cubic root. The same is true with 4th, 6th, 8th, 10th, etc. powers, since a n=(−a)n a n=(−a)n, and the 4th, 6th, 8th, 10th, etc. roots are defined to be the unique nonnegative real number that "works", so that they are unambiguous. That is, there are two numbers which when squared will give you the value 2 2 2 2: both 2 2 and −2−2. There are two numbers that when taken to the fourth power will give you (−6)4(−6)4: both −6−6 and 6 6. And so on. Generally, both a a and −a−a will, when raised to an even n n th power, give the same answer: a n=(−a)n a n=(−a)n. And we agree that a square root (fourth root, sixth root, etc.) will always be the nonnegative answer, so the n n th root of a n a n will be \|a\|\|a\| when n n is even. (Don't let the big −− in "−a−a" fool you; that does not mean that −a−a is negative, it just means the additive inverse of whatever a a is; if a a is positive, then −a−a is negative, but if a a is negative, say a=−3 a=−3, then −a−a is positive, −a=−(−3)=3−a=−(−3)=3. Repeat after me: the proper way to pronounce "-a" is not "negative a", the proper pronunciation is "minus a"). But if n n is odd, then every number has a unique n n th root. In particular, the only number that when cubed gives 2 3 2 3 is 2 2; the only number which, when raised to the fifth power, gives (−6)5(−6)5, is −6−6. There is no longer the problem that both 6 6 and −6−6 are possible answers, so we can simply say that the cubic root of (−2)3(−2)3 is −2−2, the fifth root of 7 5 7 5 is 7 7, etc. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 26, 2023 at 1:28 J. W. Tanner 64.1k 4 4 gold badges 44 44 silver badges 89 89 bronze badges answered Dec 5, 2010 at 10:01 Arturo MagidinArturo Magidin 419k 60 60 gold badges 864 864 silver badges 1.2k 1.2k bronze badges 8 5 I presume that's a typo in your first expression... :)J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-05 10:02:58 +00:00 Commented Dec 5, 2010 at 10:02 4 @Tretwick Marian: i i is the number one imagines at 4am in the morning... the time here. I should really go to bed now.Arturo Magidin –Arturo Magidin 2010-12-05 10:07:36 +00:00 Commented Dec 5, 2010 at 10:07 5 I remain envious of you, Arturo; I'm not that eloquent at 4 A.M. ;P J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-05 10:09:10 +00:00 Commented Dec 5, 2010 at 10:09 2 ...and if you catch yourself saying "negative a" again, the penance is to stand in the corner and keep repeating "minus a" until you are parched. :P J. M. ain't a mathematician –J. M. ain't a mathematician 2010-12-05 10:18:12 +00:00 Commented Dec 5, 2010 at 10:18 5 @Arturo: Of course, by "square roots are always nonnegative" you mean something more like "x−−√x is defined to be the nonnegative square root of a nonnegative number x x."Jonas Meyer –Jonas Meyer 2010-12-05 13:16:08 +00:00 Commented Dec 5, 2010 at 13:16 \|Show 3 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus radicals absolute-value faq See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 13Square root of a number squared is equal to the absolute value of that number 2Question regarding the square root of a squared number. 0What is x 2−−√x 2? 1Evaluating (cos θ)2−−−−−−√(cos⁡θ)2 at θ=π θ=π. 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6137	https://critcareedu.com.au/acute-pericarditis/	Critcareedu Critcareedu Remifentanyl Vs Fentanyl Dr Swapnil Pawar Analgesics Dr Swapnil Pawar Effects of the sudden and sustained increase in LV Afterload Dr Swapnil Pawar Remifentanyl Vs Fentanyl Dr Swapnil Pawar Empiric Antimicrobials used in Hospital Dr Swapnil Pawar Analgesics Dr Swapnil Pawar Navigate by tag Most loved podcasts Remifentanyl Vs Fentanyl September 14, 2024 Empiric Antimicrobials used in Hospital September 13, 2024 Analgesics August 1, 2024 ICU Fellowship Snippet – Type 1 Respiratory failure in immunocompramised patient July 5, 2024 SWITCH Trial June 13, 2024 Oxygen Delivery Devices June 13, 2024 Venous Admixture and effect of Supplemental oxygen May 16, 2024 ICU Fellowship Snippet – Asthma Management May 8, 2024 medical students Acute Pericarditis Dr Swapnil Pawar July 19, 2021 1 Pericarditis Blog written by – Dr Andrew Lam play_arrow Acute Pericarditis Dr Swapnil Pawar Acute Pericarditis Dr Swapnil Pawar High-Risk Features that May Require Admission: Fever > 38°C The sub-Acute course of symptoms, progressing over several days to weeks Evidence of cardiac tamponade ECG Findings in Acute Pericarditis Acute Pericarditis – Widespread concave ST elevation (V2-6) and (I, II, aVL, aVF) Aetiology of Pericarditis Indeterminate: Assumed to be of viral or immune origin Infectious: Viral – Coxsackie, echovirus, adenovirus, EBV, CMV, influenza, HIV Bacterial – M. Tuberculosis, Staphylococcus, Streptococcus, Haemophilus, Salmonella, Chlamydia Fungal – Histoplasmosis, Aspergillus Parasitic – Echinococcus, Toxoplasma, Amoebic Malignant: Metastatic breast or lung cancer, melanoma, leukemia, lymphoma. Cardiac: Post-infarction, Dressler’s syndrome Metabolic: Uraemia Radiotherapy Drug-Induced (rare): Isoniazid, hydralazine, anticoagulants, thrombolytics, phenytoin, penicillins References: Burns, E., Buttner, R. (2021). Pericarditis. Life in the Fastlane. Retrieved 5th July 2021 from Hoit, B.D. (2021) Etiology of pericardial disease. UpToDate. Retrieved 5th July 2021 from Imazio, M. (2021) Acute pericarditis: Clinical presentation, diagnostic evaluation and diagnosis. UpToDate. Retrieved 5th July 2021 from Author Dr Swapnil Pawar list Archive Previous episode Gastrointestinal Acute Mesenteric Ischemia Dr Swapnil Pawar July 14, 2021 play_arrow Acute Mesenteric Ischemia Dr Swapnil Pawar Mesenteric Ischemia Blog written by – Dr Hyerim Suh Types of mesenteric ischaemic Primary Investigations to order Bloods (Including FBC, EUC, coagulation panel, […] Read more trending_flat Similar episodes Featured Empiric Antimicrobials used in Hospital Dr Swapnil Pawar September 13, 2024 Opioids Analgesics Dr Swapnil Pawar August 1, 2024 Critcareedu ©Allrights reserved.
6138	https://community.the-hospitalist.org/content/dermatomyositis-cancer-screening-guidelines-get-real-world-validation	Dermatomyositis Cancer Screening Guidelines Get Real-World Validation \| MDedge Skip to main content Dermatomyositis Cancer Screening Guidelines Get Real-World Validation Breadcrumb Home User login Username Password Reset your password Article Type News Changed Thu, 10/03/2024 - 13:06 Author(s) Tara Haelle Newly issued guidelines for cancer screening in patients with dermatomyositis had 100% sensitivity in a single institution’s cohort, though most of the cancers found would have been detected with standard cancer screenings recommended for the general population, according to aresearch letterpublished in JAMA Dermatology. “These early results emphasize the continued need to refine risk assessment and cancer screening for patients with dermatomyositis while balancing resource use and outcomes,” concluded Caroline J. Stone and her colleagues at the Department of Dermatology, Perelman School of Medicine, University of Pennsylvania, Philadelphia. dermatology.cdlib.org/CC BY-SA 3.0/wikimedia Patients with dermatomyositis have approximately a 4.7 times greater risk for cancer than those without it, according to a2016 meta-analysis. Despite the well-established link between cancer and dermatomyositis, cancer in people with idiopathic inflammatory myopathies is commonly diagnosed at a later stage and is the leading cause of death in people with these conditions. Guidelines First Presented in 2022 and Published in 2023 A wide variability in screening practices eventually led the International Myositis Assessment & Clinical Studies Group (IMACS) to present the first evidence-based and consensus-based guidelines for cancer screening of patients with idiopathic inflammatory myopathies, including those with dermatomyositis, at the 2022 annual meeting of theAmerican College of Rheumatologyand publish them in 2023 in Nature Reviews Rheumatology. The guidelines advise low-risk patients to undergo basic cancer screening with routine blood and urine studies, liver function tests, plain chest radiography, and age- and sex-appropriate cancer screening. Intermediate- and high-risk patients are recommended to undergo enhanced screening that can include mammography, Pap tests, endoscopy/colonoscopy, pelvic and transvaginal ultrasonography, prostate-specific antigen or cancer antigen 125 blood tests, fecal occult blood tests, and CT of the neck, thorax, abdomen, and pelvis. But because the guidelines are new, little evidence exists regarding their validation in real-world cohorts. Researchers, therefore, assessed the IMACS guidelines in 370 patients, aged 18-80 years, who visited the University of Pennsylvania rheumatology-dermatology specialty clinic between July 2008 and January 2024. All participants had dermatomyositis and at least 3 years of follow-up and were an average 48 years old. The vast majority were women (87%) and White participants (89%). Most (68.6%) had myositis-specific autoantibody test results, one of the factors included in the guidelines for determining whether the patient should be classified as low, intermediate, or high risk. Other factors for risk stratification included myositis subtype, age at disease onset, and clinical features. About half (49.2%) had classic dermatomyositis, 42.4% had amyopathic dermatomyositis, 3.8% had juvenile dermatomyositis, 3.2% had hypomyopathic dermatomyositis, 0.8% had antisynthetase syndrome, and 0.5% had immune-mediated necrotizing myopathy. Just over half the patients (54%) were classified as high risk, while 37.3% were classified as intermediate risk and 8.9% as low risk using the guidelines. Among the 18 patients (4.9%) with paraneoplastic dermatomyositis, 15 were classified as high risk and 3 as intermediate risk. Of the patients diagnosed with cancer, 55% of cases were diagnosed about a year before their dermatomyositis diagnosis. In three patients, symptoms “suggestive of cancer at the time of dermatomyositis diagnosis, including lymphadenopathy and unexplained weight loss,” led to diagnostic testing that found an underlying cancer. In the eight patients diagnosed with cancer after their dermatomyositis diagnosis, 75% of the cancers were identified during the first year of follow-up and 25% in the second year. Five were identified based on basic cancer screening and three on enhanced screening. A total of 11 patients (3%) developed intravenous contrast allergies, and no other adverse events were reported to be associated with cancer screening, but the study was not designed to capture other types of adverse screening effects, such as cost, quality of life, or risk from radiation exposure. The most common neoplasm identified was breast cancer, found in nine (50%) of the patients using mammography. Two patients had lung cancer identified with chest radiography and two had ovarian cancer identified with abdominal radiography and CT. The remaining five patients included one each with bladder cancer, papillary thyroid cancer, renal cell carcinoma, non-Hodgkin lymphoma, and adenocarcinoma with unknown primary. The sensitivity of the guidelines in detecting cancer related to dermatomyositis was 100%, though the authors noted that the “IMACS risk-stratification scheme may overestimate cancer risk and encourage enhanced screening protocols of unclear benefit.” Most of the cancers found after dermatomyositis diagnosis were detected with routine age- and sex-related screening that already falls under basic cancer screening recommendations for the general population. Nonetheless, 90% of the participants fell into the intermediate- and high-risk groups, warranting a more comprehensive and costly enhanced screening protocol. Will the Guidelines Lead to Overscreening? The 4.9% cancer prevalence is considerably lower than the typical 15%-25% prevalence among patients with dermatomyositis, but the findings, regardless, suggest the guidelines will lead to overscreening, wrote Andrea D. Maderal, MD, University of Miami Miller School of Medicine in Florida, and Alisa Femia, MD, New York University Grossman School of Medicine, New York City, in anaccompanying editorial. Given that the median age in patients with cancer in the study was 58 years — 18 years older than the age cutoff for high-risk criteria — one way to refine the guidelines may be to increase the age for the high-risk category, they suggested. “While these guidelines led to many ultimately unnecessary screening tests based on currently recommended designations of intermediate-risk and high-risk patients, these guidelines reflect a more conservative approach to screening than was previously performed,” Dr. Maderal and Dr. Femia wrote. Jeff Gehlhausen, MD, PhD, an assistant professor of dermatology at Yale School of Medicine, New Haven, Connecticut, said he is not concerned about overscreening in patients, however, and is “very enthusiastic” about the findings. “Patients are very anxious for good reason,” given the typical cancer prevalence of 25% in this population, he said in an interview. “I think therein lies the challenge — with that risk, what is ‘enough’ screening?” Yet this “incredibly impressive” study “provides real insights into the applicability of the IMACS screenings to our dermatomyositis management,” including relevance to his own patients. “Their findings are instructive for how to better evaluate these patients in a more mindful fashion,” he said, and they are particularly welcome, given how widely variable practice has historically been before the guidelines were issued. “This question has been an outstanding one for decades, and nearly every doctor has a different answer,” Dr. Gehlhausen said. “The introduction of the guidelines alone are now much more actionable with this study, and that’s why it’s such an important one for our community.” Benedict Wu, DO, PhD, director of Inpatient Dermatology and an assistant professor at Montefiore Einstein and a member of the Montefiore Einstein Comprehensive Cancer Center in New York City, similarly regarded the findings as reassuring, though he was surprised at the low prevalence of cancer in the patients. “The most reassuring finding was that the detection of most malignancies was possible by using routine age- and sex-related screening combined with basic cancer screening,” Wu said in an interview. “Basic cancer screening can reduce costs while keeping patients safe.” He also found it reassuring that all the paraneoplastic dermatomyositis was in intermediate- or high-risk patients, and while he does not see the IMACS guidelines as overestimating cancer risk, he does think “the risk stratification and recommended screening tests could be revised to be less ‘aggressive.’ ” The overall low rate of cancer in the group “calls into question the need for stringent and annual cancer screening,” he said. “In this large cohort of patients, the fact that malignancy was detected within 2 years of dermatomyositis diagnosis will help guide us with long-term screening recommendations.” Despite the study’s small size and single-center design, the demographics of the patients nearly represents exactly what is found in the United States more broadly, Wu noted. He also drew attention to how many patients lacked the myositis antibody profile performed, and he agreed with the authors that more extensive and prospective studies need to be conducted. He also emphasized the need to keep in mind that “the primary goal of dermatomyositis management should focus on controlling/reducing the disease burden.” The research was funded by the National Institutes of Health and the US Department of Veterans Affairs. The authors had no disclosures. Dr. Maderal reported personal fees from argenx. No disclosures were noted for Dr. Gehlhausen and Dr. Wu. A version of this article appeared on Medscape.com. Publications MDedge Rheumatology Rheumatology News Dermatology News Oncology Practice MDedge Dermatology MDedge Hematology and Oncology Topics Lupus & Connective Tissue Diseases Autoimmune Diseases Rare Diseases Preventive Care Sections From the Journals Latest News Author(s) Tara Haelle Author(s) Tara Haelle Newly issued guidelines for cancer screening in patients with dermatomyositis had 100% sensitivity in a single institution’s cohort, though most of the cancers found would have been detected with standard cancer screenings recommended for the general population, according to aresearch letterpublished in JAMA Dermatology. “These early results emphasize the continued need to refine risk assessment and cancer screening for patients with dermatomyositis while balancing resource use and outcomes,” concluded Caroline J. Stone and her colleagues at the Department of Dermatology, Perelman School of Medicine, University of Pennsylvania, Philadelphia. dermatology.cdlib.org/CC BY-SA 3.0/wikimedia Patients with dermatomyositis have approximately a 4.7 times greater risk for cancer than those without it, according to a2016 meta-analysis. Despite the well-established link between cancer and dermatomyositis, cancer in people with idiopathic inflammatory myopathies is commonly diagnosed at a later stage and is the leading cause of death in people with these conditions. Guidelines First Presented in 2022 and Published in 2023 A wide variability in screening practices eventually led the International Myositis Assessment & Clinical Studies Group (IMACS) to present the first evidence-based and consensus-based guidelines for cancer screening of patients with idiopathic inflammatory myopathies, including those with dermatomyositis, at the 2022 annual meeting of theAmerican College of Rheumatologyand publish them in 2023 in Nature Reviews Rheumatology. The guidelines advise low-risk patients to undergo basic cancer screening with routine blood and urine studies, liver function tests, plain chest radiography, and age- and sex-appropriate cancer screening. Intermediate- and high-risk patients are recommended to undergo enhanced screening that can include mammography, Pap tests, endoscopy/colonoscopy, pelvic and transvaginal ultrasonography, prostate-specific antigen or cancer antigen 125 blood tests, fecal occult blood tests, and CT of the neck, thorax, abdomen, and pelvis. But because the guidelines are new, little evidence exists regarding their validation in real-world cohorts. Researchers, therefore, assessed the IMACS guidelines in 370 patients, aged 18-80 years, who visited the University of Pennsylvania rheumatology-dermatology specialty clinic between July 2008 and January 2024. All participants had dermatomyositis and at least 3 years of follow-up and were an average 48 years old. The vast majority were women (87%) and White participants (89%). Most (68.6%) had myositis-specific autoantibody test results, one of the factors included in the guidelines for determining whether the patient should be classified as low, intermediate, or high risk. Other factors for risk stratification included myositis subtype, age at disease onset, and clinical features. About half (49.2%) had classic dermatomyositis, 42.4% had amyopathic dermatomyositis, 3.8% had juvenile dermatomyositis, 3.2% had hypomyopathic dermatomyositis, 0.8% had antisynthetase syndrome, and 0.5% had immune-mediated necrotizing myopathy. Just over half the patients (54%) were classified as high risk, while 37.3% were classified as intermediate risk and 8.9% as low risk using the guidelines. Among the 18 patients (4.9%) with paraneoplastic dermatomyositis, 15 were classified as high risk and 3 as intermediate risk. Of the patients diagnosed with cancer, 55% of cases were diagnosed about a year before their dermatomyositis diagnosis. In three patients, symptoms “suggestive of cancer at the time of dermatomyositis diagnosis, including lymphadenopathy and unexplained weight loss,” led to diagnostic testing that found an underlying cancer. In the eight patients diagnosed with cancer after their dermatomyositis diagnosis, 75% of the cancers were identified during the first year of follow-up and 25% in the second year. Five were identified based on basic cancer screening and three on enhanced screening. A total of 11 patients (3%) developed intravenous contrast allergies, and no other adverse events were reported to be associated with cancer screening, but the study was not designed to capture other types of adverse screening effects, such as cost, quality of life, or risk from radiation exposure. The most common neoplasm identified was breast cancer, found in nine (50%) of the patients using mammography. Two patients had lung cancer identified with chest radiography and two had ovarian cancer identified with abdominal radiography and CT. The remaining five patients included one each with bladder cancer, papillary thyroid cancer, renal cell carcinoma, non-Hodgkin lymphoma, and adenocarcinoma with unknown primary. The sensitivity of the guidelines in detecting cancer related to dermatomyositis was 100%, though the authors noted that the “IMACS risk-stratification scheme may overestimate cancer risk and encourage enhanced screening protocols of unclear benefit.” Most of the cancers found after dermatomyositis diagnosis were detected with routine age- and sex-related screening that already falls under basic cancer screening recommendations for the general population. Nonetheless, 90% of the participants fell into the intermediate- and high-risk groups, warranting a more comprehensive and costly enhanced screening protocol. Will the Guidelines Lead to Overscreening? The 4.9% cancer prevalence is considerably lower than the typical 15%-25% prevalence among patients with dermatomyositis, but the findings, regardless, suggest the guidelines will lead to overscreening, wrote Andrea D. Maderal, MD, University of Miami Miller School of Medicine in Florida, and Alisa Femia, MD, New York University Grossman School of Medicine, New York City, in anaccompanying editorial. Given that the median age in patients with cancer in the study was 58 years — 18 years older than the age cutoff for high-risk criteria — one way to refine the guidelines may be to increase the age for the high-risk category, they suggested. “While these guidelines led to many ultimately unnecessary screening tests based on currently recommended designations of intermediate-risk and high-risk patients, these guidelines reflect a more conservative approach to screening than was previously performed,” Dr. Maderal and Dr. Femia wrote. Jeff Gehlhausen, MD, PhD, an assistant professor of dermatology at Yale School of Medicine, New Haven, Connecticut, said he is not concerned about overscreening in patients, however, and is “very enthusiastic” about the findings. “Patients are very anxious for good reason,” given the typical cancer prevalence of 25% in this population, he said in an interview. “I think therein lies the challenge — with that risk, what is ‘enough’ screening?” Yet this “incredibly impressive” study “provides real insights into the applicability of the IMACS screenings to our dermatomyositis management,” including relevance to his own patients. “Their findings are instructive for how to better evaluate these patients in a more mindful fashion,” he said, and they are particularly welcome, given how widely variable practice has historically been before the guidelines were issued. “This question has been an outstanding one for decades, and nearly every doctor has a different answer,” Dr. Gehlhausen said. “The introduction of the guidelines alone are now much more actionable with this study, and that’s why it’s such an important one for our community.” Benedict Wu, DO, PhD, director of Inpatient Dermatology and an assistant professor at Montefiore Einstein and a member of the Montefiore Einstein Comprehensive Cancer Center in New York City, similarly regarded the findings as reassuring, though he was surprised at the low prevalence of cancer in the patients. “The most reassuring finding was that the detection of most malignancies was possible by using routine age- and sex-related screening combined with basic cancer screening,” Wu said in an interview. “Basic cancer screening can reduce costs while keeping patients safe.” He also found it reassuring that all the paraneoplastic dermatomyositis was in intermediate- or high-risk patients, and while he does not see the IMACS guidelines as overestimating cancer risk, he does think “the risk stratification and recommended screening tests could be revised to be less ‘aggressive.’ ” The overall low rate of cancer in the group “calls into question the need for stringent and annual cancer screening,” he said. “In this large cohort of patients, the fact that malignancy was detected within 2 years of dermatomyositis diagnosis will help guide us with long-term screening recommendations.” Despite the study’s small size and single-center design, the demographics of the patients nearly represents exactly what is found in the United States more broadly, Wu noted. He also drew attention to how many patients lacked the myositis antibody profile performed, and he agreed with the authors that more extensive and prospective studies need to be conducted. He also emphasized the need to keep in mind that “the primary goal of dermatomyositis management should focus on controlling/reducing the disease burden.” The research was funded by the National Institutes of Health and the US Department of Veterans Affairs. The authors had no disclosures. Dr. Maderal reported personal fees from argenx. No disclosures were noted for Dr. Gehlhausen and Dr. Wu. A version of this article appeared on Medscape.com. Newly issued guidelines for cancer screening in patients with dermatomyositis had 100% sensitivity in a single institution’s cohort, though most of the cancers found would have been detected with standard cancer screenings recommended for the general population, according to aresearch letterpublished in JAMA Dermatology. “These early results emphasize the continued need to refine risk assessment and cancer screening for patients with dermatomyositis while balancing resource use and outcomes,” concluded Caroline J. Stone and her colleagues at the Department of Dermatology, Perelman School of Medicine, University of Pennsylvania, Philadelphia. dermatology.cdlib.org/CC BY-SA 3.0/wikimedia Patients with dermatomyositis have approximately a 4.7 times greater risk for cancer than those without it, according to a2016 meta-analysis. Despite the well-established link between cancer and dermatomyositis, cancer in people with idiopathic inflammatory myopathies is commonly diagnosed at a later stage and is the leading cause of death in people with these conditions. Guidelines First Presented in 2022 and Published in 2023 A wide variability in screening practices eventually led the International Myositis Assessment & Clinical Studies Group (IMACS) to present the first evidence-based and consensus-based guidelines for cancer screening of patients with idiopathic inflammatory myopathies, including those with dermatomyositis, at the 2022 annual meeting of theAmerican College of Rheumatologyand publish them in 2023 in Nature Reviews Rheumatology. The guidelines advise low-risk patients to undergo basic cancer screening with routine blood and urine studies, liver function tests, plain chest radiography, and age- and sex-appropriate cancer screening. Intermediate- and high-risk patients are recommended to undergo enhanced screening that can include mammography, Pap tests, endoscopy/colonoscopy, pelvic and transvaginal ultrasonography, prostate-specific antigen or cancer antigen 125 blood tests, fecal occult blood tests, and CT of the neck, thorax, abdomen, and pelvis. But because the guidelines are new, little evidence exists regarding their validation in real-world cohorts. Researchers, therefore, assessed the IMACS guidelines in 370 patients, aged 18-80 years, who visited the University of Pennsylvania rheumatology-dermatology specialty clinic between July 2008 and January 2024. All participants had dermatomyositis and at least 3 years of follow-up and were an average 48 years old. The vast majority were women (87%) and White participants (89%). Most (68.6%) had myositis-specific autoantibody test results, one of the factors included in the guidelines for determining whether the patient should be classified as low, intermediate, or high risk. Other factors for risk stratification included myositis subtype, age at disease onset, and clinical features. About half (49.2%) had classic dermatomyositis, 42.4% had amyopathic dermatomyositis, 3.8% had juvenile dermatomyositis, 3.2% had hypomyopathic dermatomyositis, 0.8% had antisynthetase syndrome, and 0.5% had immune-mediated necrotizing myopathy. Just over half the patients (54%) were classified as high risk, while 37.3% were classified as intermediate risk and 8.9% as low risk using the guidelines. Among the 18 patients (4.9%) with paraneoplastic dermatomyositis, 15 were classified as high risk and 3 as intermediate risk. Of the patients diagnosed with cancer, 55% of cases were diagnosed about a year before their dermatomyositis diagnosis. In three patients, symptoms “suggestive of cancer at the time of dermatomyositis diagnosis, including lymphadenopathy and unexplained weight loss,” led to diagnostic testing that found an underlying cancer. In the eight patients diagnosed with cancer after their dermatomyositis diagnosis, 75% of the cancers were identified during the first year of follow-up and 25% in the second year. Five were identified based on basic cancer screening and three on enhanced screening. A total of 11 patients (3%) developed intravenous contrast allergies, and no other adverse events were reported to be associated with cancer screening, but the study was not designed to capture other types of adverse screening effects, such as cost, quality of life, or risk from radiation exposure. The most common neoplasm identified was breast cancer, found in nine (50%) of the patients using mammography. Two patients had lung cancer identified with chest radiography and two had ovarian cancer identified with abdominal radiography and CT. The remaining five patients included one each with bladder cancer, papillary thyroid cancer, renal cell carcinoma, non-Hodgkin lymphoma, and adenocarcinoma with unknown primary. The sensitivity of the guidelines in detecting cancer related to dermatomyositis was 100%, though the authors noted that the “IMACS risk-stratification scheme may overestimate cancer risk and encourage enhanced screening protocols of unclear benefit.” Most of the cancers found after dermatomyositis diagnosis were detected with routine age- and sex-related screening that already falls under basic cancer screening recommendations for the general population. Nonetheless, 90% of the participants fell into the intermediate- and high-risk groups, warranting a more comprehensive and costly enhanced screening protocol. Will the Guidelines Lead to Overscreening? The 4.9% cancer prevalence is considerably lower than the typical 15%-25% prevalence among patients with dermatomyositis, but the findings, regardless, suggest the guidelines will lead to overscreening, wrote Andrea D. Maderal, MD, University of Miami Miller School of Medicine in Florida, and Alisa Femia, MD, New York University Grossman School of Medicine, New York City, in anaccompanying editorial. Given that the median age in patients with cancer in the study was 58 years — 18 years older than the age cutoff for high-risk criteria — one way to refine the guidelines may be to increase the age for the high-risk category, they suggested. “While these guidelines led to many ultimately unnecessary screening tests based on currently recommended designations of intermediate-risk and high-risk patients, these guidelines reflect a more conservative approach to screening than was previously performed,” Dr. Maderal and Dr. Femia wrote. Jeff Gehlhausen, MD, PhD, an assistant professor of dermatology at Yale School of Medicine, New Haven, Connecticut, said he is not concerned about overscreening in patients, however, and is “very enthusiastic” about the findings. “Patients are very anxious for good reason,” given the typical cancer prevalence of 25% in this population, he said in an interview. “I think therein lies the challenge — with that risk, what is ‘enough’ screening?” Yet this “incredibly impressive” study “provides real insights into the applicability of the IMACS screenings to our dermatomyositis management,” including relevance to his own patients. “Their findings are instructive for how to better evaluate these patients in a more mindful fashion,” he said, and they are particularly welcome, given how widely variable practice has historically been before the guidelines were issued. “This question has been an outstanding one for decades, and nearly every doctor has a different answer,” Dr. Gehlhausen said. “The introduction of the guidelines alone are now much more actionable with this study, and that’s why it’s such an important one for our community.” Benedict Wu, DO, PhD, director of Inpatient Dermatology and an assistant professor at Montefiore Einstein and a member of the Montefiore Einstein Comprehensive Cancer Center in New York City, similarly regarded the findings as reassuring, though he was surprised at the low prevalence of cancer in the patients. “The most reassuring finding was that the detection of most malignancies was possible by using routine age- and sex-related screening combined with basic cancer screening,” Wu said in an interview. “Basic cancer screening can reduce costs while keeping patients safe.” He also found it reassuring that all the paraneoplastic dermatomyositis was in intermediate- or high-risk patients, and while he does not see the IMACS guidelines as overestimating cancer risk, he does think “the risk stratification and recommended screening tests could be revised to be less ‘aggressive.’ ” The overall low rate of cancer in the group “calls into question the need for stringent and annual cancer screening,” he said. “In this large cohort of patients, the fact that malignancy was detected within 2 years of dermatomyositis diagnosis will help guide us with long-term screening recommendations.” Despite the study’s small size and single-center design, the demographics of the patients nearly represents exactly what is found in the United States more broadly, Wu noted. He also drew attention to how many patients lacked the myositis antibody profile performed, and he agreed with the authors that more extensive and prospective studies need to be conducted. He also emphasized the need to keep in mind that “the primary goal of dermatomyositis management should focus on controlling/reducing the disease burden.” The research was funded by the National Institutes of Health and the US Department of Veterans Affairs. The authors had no disclosures. Dr. Maderal reported personal fees from argenx. No disclosures were noted for Dr. Gehlhausen and Dr. Wu. A version of this article appeared on Medscape.com. 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6139	https://www.geogebra.org/m/VHQzeMX9	Google Classroom GeoGebra Classroom Home Resources Profile Classroom App Downloads Entering Vectors Author:GeoGebra Institute of MEI Topic:Vectors Vectors can be created using the Vector or Vector from Point tools in the third menu. Vector joins two points A and B with the vector AB. Vector from Point will translate an existing vector so that it starts from a point. Alternatively enter a set of coordinates with a lower-case name will create the object as a position vector, e.g. v=(2,3,-1) Line[A,v] will create a a line through point A in the direction v. Planes can be entered directly in Cartesian form: e.g. 2x+3y-z=2. New Resources רישום חופשי The Red Chair Untitled 拼砌四邊形 - 工作紙 Ringed Polyhedra Discover Resources Area and Perimeter of Right Triangle Step response of a spring-mass-dashpot system Olimpiada Tulcea. สมการเชิงเส้น perpendicular lines to an oval and an ellipse Discover Topics Cylinder Quadratic Functions Algebra Linear Functions General Quadrilateral
6140	https://www.tuhh.de/ti3/batra/publications/NPE8n04.pdf	NEW POINT ESTIMATES FOR NEWTON’S METHOD PRASHANT BATRA ∗ Abstract. The Newton iteration is customarily used for (sequential) approximation of zeros of diﬀerentiable functions. Beside the classical Kantorovich theory there exist convergence criteria which only involve data at one point, i.e. point estimates. The suﬃcient conditions ensure immediate quadratic convergence to a single zero and have been frequently used by diﬀerent authors to design robust, fast and eﬃcient root-ﬁnding methods for polynomials. In this paper a suﬃcient condition for the simultaneous convergence of the one-dimensional Newton iteration for polynomials will be given. The new condition involves only n point evaluations of the Newton correction and the minimum mutual distance of approximations to ensure ’simultaneous’ quadratic convergence to the pairwise distinct n roots. To establish the new convergence condition a new error estimate for Newton’s method will be proven. This estimate is applicable if n (rough) approximations of the pairwise distinct roots are well separated. The resulting error estimate is independent of n. Mathematics Subject Classiﬁcations(1991): 65H05, ? Keywords: Polynomial roots, simultaneous methods, Newton iteration, convergence theorems, practical conditions for convergence, point estimates. 1. Introduction. The well known work of Ostrowski and Kantorovich yields convergence criteria for a twice diﬀerentiable function f, whenever Df(z0)−1 · f(z0) and D2f(z) · Df(z0)−1 can be suitably estimated on a circle around z0 . Starting afresh, Myong-Hi Kim and Steve Smale independently derived suﬃcient convergence conditions for the Newton method from data at one point, see or . Before we formulate these conditions for holomorphic functions, two deﬁnitions (of Smale’s) are appropriate. Definition 1.1. Given a holomorphic function f : C →C. If for all Newton iterates zk of z0 ∈C : \|zk+1 −zk\| ≤( 1 2)2k−1\|z1 −z0\|, k = 0, 1, · · ·, then the element z0 ∈C is called an approximate zero (of f). The convergence of Newton’s method from an approximate zero is obviously quadratic. An existence criterion for approximate zeros can be established via a simultaneous estimation of the Newton corrrection and certain coeﬃcients of a Taylor expansion. For this purpose deﬁne for a holomorphic function f the following. Definition 1.2. α(z, f) := \|f(z)/f ′(z)\| supk>1 \| f (k)(z) k!f ′(z)\|1/(k−1). If α(z0, f) is suitably small, the starting point z0 is an approximate zero, as shown by Kim and Smale. The condition involves only evaluation of functions at a single point. Using the majorant sequence technique, Xing-Hua Wang and Dan-Fu Han established a sharp result which yields the following theorem. Theorem 1.3. Given a holomorphic function f and z0 ∈C, suppose α(z0, f) < 3 −2 √ 2. (1.1) Then z0 is an approximate zero. For a polynomial, this suﬃcient condition involves only computable data. Based on such results, new algorithms for polynomial root approximation have been devised, see for example , . These methods apply the Newton iteration sequentially for each root. On the other hand, there exist several methods for the simultaneous ∗Inst. f. Informatik III, Technical University Hamburg-Harburg, Eißendorfer Str. 38, 21071 Ham-burg, Germany (batra@tu-harburg.de). 1 approximation of polynomial zeros which are based on variants of Newton’s method in Cn. In this paper, a practical condition will be established which guarantees the convergence of Newton’s method from n pairwise diﬀerent approximations to n pairwise diﬀerent zeros. Before presenting the details we sketch the way to derive such conditions. Given a polynomial P(z) = an ·Q(z −ζi) of degree n, consider the formal identity (derived in Section 3 as (3.2) ) α(z, P) = \| P(z) P ′(z)\| · max k>1 \| n X i1<i2<···<ik i1,i2,···,ik=1 1 (z −ζi1) · · · (z −ζik) P(z) P ′(z)\|1/(k−1). (1.2) If z = z0 is close to ζ1, and well separated from the other zeros ζ2, · · · , ζn, condition (1.1) holds true, i.e. the Newton iteration converges starting with z = z0. How to guarantee the separation? Given z1 with P(z1)·P ′(z1) ̸= 0, there is the well-known estimate for the distance to the closest root, ζ1 say: P ′(z1) P(z1) = n X i=1 1 z1 −ζi ≤ n \|z1 −ζ1\| or \|z1 −ζ1\| ≤n · P(z1) P ′(z1) . (1.3) Combining Theorem 1.3, (1.2) and (1.3) it is possible to prove a suﬃcient convergence condition like max i P(zi) P ′(zi) < mini̸=j \|zi −zj\| 5 · n2 (1.4) for the convergence from the n points zi to the n pairwise distinct zeros of P(z) . A second line of investigation starts from good practical error estimates for simul-taneous methods. For example, it has been shown that the Durand-Kerner iteration zi+1 := zi − P (zi) an·Q j̸=i(zi−zj) is convergent if max i P(zi) an · Q j̸=i(zi −zj) < mini̸=j \|zi −zj\| 2 · n (1.5) holds true, see . As the Durand-Kerner iteration is Newton’s iteration for Vi` ete’s system of equations , it seems natural to look for weaker conditions than (1.4) in the one-dimensional case as well. The suﬃciency of condition (1.5) for the Durand-Kerner method has been established using a quite sharp error estimate from . A new error estimate will be used here to establish a new suﬃcient convergence for Newton’s method. This condition improves on (1.4) by O(n), from 5n2 to 8n in the denominator. For the general case, the estimation (1.3) is sharp. Generally it cannot be used to determine the number of zeros in the disc Ii := {z ∈C : \|z −zi\| ≤n · P (zi) P ′(zi) }. But if the discs Ii are mutually disjoint, each disc contains exactly one zero, and one may ask whether there exist better estimates in this case. The calculation necessary to determine whether two discs Ii, Ij are disjoint can trivially be modiﬁed to yield a relative measure of separation. The measure of separation can be used to improve the estimation (1.3). This leads to new inclusion discs ˆ Ii ⊂Ii for the roots. Recursive application of this idea will lead to the following result in Section 2. 2 Theorem 1.4. Given a polynomial P(z) of degree n ≥3. Given n values zi with P ′(zi) ̸= 0. If maxi P (zi) P ′(zi) < mini̸=j \|zi−zj\| C·n , C > 2, then each circle \|z −zi\| < C C−1 · P (zi) P ′(zi) contains a zero. This estimate will be derived by completely elementary means. The new error estimate is used in Section 3 to prove the following new convergence condition which merely involves data from Newton’s method. Theorem 1.5. Given a polynomial P(z) of degree n ≥3. Given n values zi with P ′(zi) ̸= 0. Assume that max i P(zi) P ′(zi) < mini̸=j \|zi −zj\| 8 · n . (1.6) Then the following holds true. i) The Newton iteration converges for each zi, and convergence is quadratic. ii) There is exactly one zero in Ki := {z ∈C : \|z −(zi −P (zi) P ′(zi))\| ≤\| P (zi) P ′(zi)\|}. Remark. Given n approximations zi, to check the general convergence criterion (1.1) the n2 values P (k)(zi) k!P ′(zi) for k = 0, 2, · · · , n; i = 1, · · · , n have to be calculated. The new criterion only involves the evaluation of n Newton corrections P (zi) P ′(zi) and improves on (1.4) by O(n). A theorem similar to Theorem 1.5 could be derived by assuming that a certain isolation of a single zero is already guaranteed. Estimations of that kind have, for example, been applied in the analysis of Pan’s algorithm for the improvement of Weyl’s quadtree construction as stated in . 2. A special error estimate. We restate the very well known estimation (1.3) from the introduction: Lemma 2.1. Given a polynomial P(z) of degree n. Given z0 ∈C with P ′(z0) ̸= 0. The circle \|z −z0\| ≤n · \| P (z0) P ′(z0)\| contains at least one root of P(z). Without further assumption the inclusion radius n · \| P (z0) P ′(z0)\| of Lemma 2.1 is op-timal, but not in the situation of Theorem 1.5. If the set of roots can be separated into at least two clusters and if the relative size of separation is known, this informa-tion can be used directly in the proof of Lemma 2.1. This observation leads to the following theorem. Theorem 2.2. Given P (z) = an · Qn i=1(z −ζi) ∈C[z], an ̸= 0, and z1 with P ′(z1) ̸= 0. Assume that the roots ζ1, · · · , ζk (k ≤n) lie in D1 := {z ∈C : \|z −z1\| ≤ n · \| P (z1) P ′(z1)\|}, and all other roots in the exterior of D1, i.e., 1.) ζi ∈D1 for 1 ≤i ≤k, 2.) ∃S > 1 : \|z1 −ζi\| ≥S · n · \| P (z1) P ′(z1)\| for k + 1 ≤i ≤n. Then the circle \|z −z1\| ≤ k n −(n −k)/S · n · \| P(z1) P ′(z1)\| contains a root. Proof. The cases n = 1 or P(z1) = 0 are trivial. Assume n > 1 and P(z1) ̸= 0. The roots are assumed to be numbered such that ζ1 is the root with minimal distance to z1, i.e. \|z1 −ζ1\| = min i \|z1 −ζi\|. Deﬁne δm := Snk+(n−k)δm−1 Sn = k + n−k Sn δm−1 and 3 δ0 := n. First, it will be shown by induction that the circle \|z −z1\| ≤δm · \| P(z1) P ′(z1)\| (2.1) contains a root for every m. The claimed inclusion (2.1) is guaranteed for m = 0 by Lemma 2.1. Assume that (2.1) holds true for m, i.e. \|z1 −ζ1\| ≤δm · \| P (z1) P ′(z1)\|. Remember that the roots are numbered such, that for all i with 2 ≤i ≤k it holds \|z1 −ζi\| ≥\|z1 −ζ1\|. By assumption 2.) it is \|z1 −ζi\| ≥Sn · \| P (z1) P ′(z1)\| for k + 1 ≤i ≤n, and therefore \|z1 −ζi\| ≥Sn δm · \|z1 −ζ1\|. These estimates together show \|P ′(z1) P(z1) \| = \| n X i=1 1 z1 −ζi \| ≤ k X i=1 \| Sn Sn(z1 −ζ1)\| + n X i=k+1 \| δm Sn(z1 −ζ1)\| = Snk + (n −k)δm Sn 1 \|z1 −ζ1\|. Therefore, \|z1 −ζ1\| ≤Snk+(n−k)δm Sn · \| P (z1) P ′(z1)\| = δm+1 · \| P (z1) P ′(z1)\|. The inclusion (2.1) is established by induction. As n−k Sn < 1, the sequence {δm} converges to the ﬁxpoint ˆ δ := Snk Sn−(n−k) > 0. To obtain estimates suitable for the named applications, Theorem 2.2 will be ap-plied for all zeros simultaneously. The improved inclusion implicitly causes a stronger separation, and this leads to recursive improvement. Proposition 2.3. Given P(z) = an · Qn i=1(z −ζi) ∈C[z] of degree n ≥3, and n values zi with P ′(zi) ̸= 0. Assume that for some C > 2, max i \| P(zi) P ′(zi)\| < min i̸=j \|zi −zj\| C · n . Let E0 := 1 and ρ0 := C −1, and deﬁne two sequences by Eν+1 := 1 n −n−1 ρνEν , ρν+1 := ρ0 + 1 Eν+1 −1. (2.2) With suitable enumeration of the roots, the following holds true for all ν: i) ζi ∈{z ∈C : \|z −zi\| ≤Eν · n · \| P (zi) P ′(zi)\|} for all i. ii) ζj ∈{z ∈C : \|z −zi\| ≥ρν · Eν · n · \| P (zi) P ′(zi)\|} for all j ̸= i. The proof by induction is based on the following result. Lemma 2.4. Given P(z) = an · n Q i=1 (z −ζi) ∈C[z] of degree n ≥3. Given n values zi with P ′(zi) ̸= 0. Assume there exist E > 0, ρ ≥1 with ρ · E > 1, such that 1.) ζi ∈{z ∈C : \|z −zi\| ≤E · n · \| P (zi) P ′(zi)\|} for each i, 2.) ζj ∈{z ∈C : \|z −zi\| ≥ρ · E · n · \| P (zi) P ′(zi)\|} for each j ̸= i. Deﬁne ˆ δ := ρ·E·n ρ·E·n−(n−1). Then the following inclusion holds true: ζi ∈{z ∈C : \|z −zi\| ≤ˆ δ · \| P (zi) P ′(zi)\|} for all i. 4 Proof. For any index i, the assumptions of Theorem 2.2 are satisﬁed for k = 1: 1.) by Lemma 2.1, 2.) with S := ρE. An induction argument yields a sequence of radii with ﬁxpoint ˆ δ as in the proof of Theorem 2.2. Proof. [of Proposition 2.3] By assumption, n ≥3, ρ > 1 and ρ0 · E0 > 1. Expanding the deﬁning formulas, it is obvious that the sequence {Eν} is monotonically decreasing and the sequence {ρν} is monotonically increasing. Moreover, ρν · Eν > 1. By Lemma 2.1 at least one root is contained in each circle \|z −zi\| ≤n·\| P (zi) P ′(zi)\|. These circles are mutually disjoint, and each contains exactly one zero, which is obvious from min u̸=v \|zu −zv\| > (C −1) · n · \| P(zi) P ′(zi)\| + n · \| P(zj) P ′(zj)\| > n · \| P(zi) P ′(zi)\| + n · \| P(zj) P ′(zj)\| , (2.3) for arbitrary i, j. Hence, for each approximation zi the assumptions of Lemma 2.4 are satisﬁed with E := E0, ρ := ρ0, therefore i), ii) hold true for ν = 0. Assume, that the claimed estimates hold for ν = ν0. Then, with a ﬁxed enumeration, the assumptions of Lemma 2.4 are satisﬁed with E := Eν, ρ := ρν > 1 because of ρν · Eν > 1. This leads to the inclusion estimate ζi ∈{z ∈C : \|z −zi\| ≤Eν+1 · n · \| P(zi) P ′(zi)\|}. (2.4) The exclusion region in ii) is a trivial consequence. The limits of the sequences {Eν}, {ρν} in Proposition 2.3 yield new error estimates as well. Theorem 1.4 will be a simple consequence of the following result. Theorem 2.5. Given P(z) = an · Qn i=1(z −ζi) ∈C[z] of degree n ≥3, and n values zi with P ′(zi) ̸= 0. Assume that for some C > 2, max i \| P(zi) P ′(zi)\| < min i̸=j \|zi −zj\| C · n . With ρ0 := C −1 and ˆ E = 1 n + n−1 n · 2 ρ0·n+2·√ ( ρ0·n 2 )2−(n−1) the following holds true (for a suitable enumeration of the zi): i) The root ζi lies in the circle \|z −zi\| ≤ˆ E · n · \| P(zi) P ′(zi)\|, ii) and for j ̸= i, the root ζj lies in \|z −zi\| ≥ h C −ˆ E i · n · \| P(zi) P ′(zi)\|, where iii) ˆ E < C C−1. Proof. Deﬁne the two sequences {ρν}, {Eν} as in (2.2). We ﬁnd, that ρν · Eν > 1, and that both sequences are monotonic. By Proposition 2.3 we have the following properties of the sequences. With suitable ﬁxed enumeration it holds for each index i = 1, · · · , n and every iteration index ν ≥0: 5 1.) Each root ζi is isolated in the circle \|z −zi\| ≤Eν · n · \| P (zi) P ′(zi)\|. 2.) The roots ζj, j ̸= i, lie in the region \|z −zi\| ≥ρν · Eν · n · \| P (zi) P ′(zi)\|. If P(zi) = 0 for some i, then i), ii) hold true for that index, according to the assump-tion and Lemma 2.1. Otherwise, we obtain a strictly monotonic sequence of inclusion radii ˆ E·n·\| P (zi) P ′(zi)\| with a non-zero limit. Therefore, the sequences Eν, ρν are bounded. Passing to the limits ˆ E and ˆ ρ retains the inclusion-exclusion properties from 1.), 2.). It remains to calculate the limits. The recursion formula gives ˆ E = ˆ ρ · ˆ E ˆ ρ · ˆ E · n −(n −1) , (2.5) ˆ ρ = ρ0 + (1 −ˆ E) ˆ E = C −ˆ E ˆ E . (2.6) Rearrangement of (2.5) shows ˆ E = ˆ ρ + (n −1) ˆ ρ · n . (2.7) Substitution in (2.6) shows ˆ ρ = ρ0 + (1 −ˆ ρ+(n−1) ˆ ρ·n ) ˆ ρ+(n−1) ˆ ρ·n = ˆ ρ · ρ0 · n + ˆ ρ · n −ˆ ρ −(n −1) ˆ ρ + (n −1) . This gives the quadratic equation ˆ ρ 2 −ˆ ρ · ρ0 · n = −(n −1). Therefore, ˆ ρ is either of ρ0·n 2 ± q ( ρ0·n 2 )2 −(n −1). Moreover, ρ0 · n 2 − r (ρ0 · n 2 )2 −(n −1) < ρ0. As the sequence {ρν} is monotonically increasing, its limit ˆ ρ is given by ˆ ρ = ρ0 · n 2 + r (ρ0 · n 2 )2 −(n −1) . Substituting this in (2.7) shows ˆ E = 1 n + n −1 n · 1 ˆ ρ = 1 n + n −1 n · 2 ρ0 · n + 2 · q ( ρ0·n 2 )2 −(n −1) . (2.8) Passing in 1.), 2.) from Eν, ρν to ˆ E, ˆ ρ respectively yields i), ii). The inequality ( 1 n + n−1 n · 2 ρ0·n+2·√ ( ρ0·n 2 )2−(n−1)) · n < C C−1 is easily veriﬁed for ρ0 = C −1, which yields iii). 6 3. Simultaneous convergence of Newton’s method. Given a polynomial P of degree n, the function α(z, f)\|f=P reads α(z, P) = \| P(z) P ′(z)\| max k=2,···,n \|P (k)(z) k!P ′(z)\|1/(k−1). Recall that by Theorem 1.3 the Newton iteration converges quadratically starting from z0, if α(z0, P) < 3 −2 √ 2. The suﬃcient convergence condition α(z0, P) < 3−2 √ 2 involves for a polynomial of degree n only n point evaluations of rational functions, i.e., the condition only relies on ’attainable data’. But the essential values P (k)(z) k!P ′(z) which have to be calculated, are not necessary in the actual computation. A new suﬃcient convergence condition for Newton’s method will be established which only involves information ’naturally’ available from the Newton iteration. As before, denote the zeros of P(z) by ζ1, · · · , ζn. To employ the new estimate from Theorem 1.4, a diﬀerent expression for α(z, P) is useful. Suppose P(z) ̸= 0. Then \|P (k)(z) k!P ′(z)\| = \| 1 k!P (k)(z) · P(z)−1 · P(z) P ′(z)\| = \| 1 k! n X i1=1 n X i2=1 i2̸=i1 · · · n X ik=1 ∀m<k:ik̸=im n Y j=1 j̸=i1,i2,···,ik (z −ζj) n Y j=1 (z −ζj) −1 P(z) P ′(z)\| = \| 1 k! · k! n X i1,i2,···,ik=1 i1<i2<···<ik n Y j=1 j̸=i1,i2,···,ik (z −ζj) n Y j=1 (z −ζj)−1 P(z) P ′(z)\| = \| n X i1,i2,···,ik=1 i1<i2<···1 \|P (k)(z) k!P ′(z)\|1/(k−1) = \| P(z) P ′(z)\| · max k>1 \| n X i1<i2<···<ik i1,i2,···,ik=1 1 (z −ζi1) · · · (z −ζik) P(z) P ′(z)\|1/(k−1). (3.2) If approximations zi of the roots ζi are given, lower estimates of \|zi −ζj\|, j = 1, · · · , n facilitate the estimation of α(zi, P) from (3.2). Consider a ﬁxed index i. Choosing a suitable numeration of the approximations zj, assume \|zi−ζi\| = minj \|zj −ζi\|. Then, a lower estimate of \|zi−ζi\| in the non-trivial case P ′(zi) · P(zi) ̸= 0 can be obtained from lower estimates of \|zi −ζj\|, j ̸= i and an upper estimate of \|zi −ζi\| as follows. Using the relation (zi −ζi) · P ′(zi) P(zi) = n X j=1 zi −ζi zi −ζj or (zi −ζi) = 1 + n X j=1 j̸=i zi −ζi zi −ζj · P(zi) P ′(zi) 7 gives \|zi −ζi\| ≥ 1 −(n −1)\|zi −ζi\| minj̸=i \|zi −ζj\| · \| P(zi) P ′(zi)\|, (3.3) which can be utilized for a lower estimate. The well-known Lemma 2.1 shows \|zi −ζi\| ≤n · \| P (zi) P ′(zi)\|. So it is quite obvious, that conditions like (1.4) can be used to estimate α(z, P) by estimating suitably the distances \|zi −ζj\| in (3.2). The use of Theorem 1.4 improves the estimates involved, so we are already prepared to prove Theorem 1.5. Proof. [of Theorem 1.5] Assume P(zi) ̸= 0. Let N := maxj P (zj) P ′(zj) . By Theo-rem 1.4 with n ≥3 (and C = 8), a ﬁxed enumeration can be chosen such that the following estimates hold true. \|zi −ζi\| < 8 7 · \| P(zi) P ′(zi)\| ≤8 7 · N for each i, (3.4) \|zi −ζj\| ≥\|zi −zj\| −\|zj −ζj\| > 8(n −1 7)N for each j ̸= i. (3.5) These estimates together with (3.3) yield \|zi −ζi\| > 1 −(n −1) 8 7 · N 8(n −1 7) · N · \| P(zi) P ′(zi)\| > 6 7 · \| P(zi) P ′(zi)\|. (3.6) To prove convergence, α(zi, P) will be estimated for each i ∈{1, · · · , n}. It is useful to split the sum in (3.2) dependent on i: n X i1<i2<···<ik i1,i2,···,ik=1 1 (z −ζi1)(z −ζi2) · · · (z −ζik) = n X i1<i2<···<ik i1,i2,···,ik=1 i∈{i1,···,ik} 1 (z −ζi1) · · · (z −ζik) + n X i1<i2<···1 ( ′ X + ′′ X ) P(z) P ′(z) 1/(k−1) . (3.8) For ﬁxed i and k ≥2, the sum P′ (i ∈{i1, · · · , ik}) comprises exactly n−1 k−1 ≤ (n −1)k−1 terms, the complementary sum P′′ comprises n−1 k < (n −1)k terms accordingly. The convergence of Newton’s method can be deduced after appropiate estimation of α(zi, P). Let µ := minj̸=i \|zi −ζj\|. From (3.7) with (3.4), (3.5) and (3.6) it follows ′ X + ′′ X < (n −1)k−1 · 1 \|zi −ζi\|( 1 µ)k−1 + (n −1)k · ( 1 µ)k 8 < \| P ′(zi) 6/7 · P(zi)\|( (n −1) 8(n −1 7)N )k−1 + ( n −1 8(n −1 7)N )k < \| P ′(zi) 6/7 · P(zi)\|( 1 8 · N )k−1 + ( 1 8 · N )k , ′ X + ′′ X! P(zi) P ′(zi) < 7 6( 1 8 · N )k−1 + ( 1 8 · N )k · \| P(zi) P ′(zi)\| ≤7 6( 1 8 · N )k−1 + 1 8 · ( 1 8 · N )k−1 = 7 6 + 1 8 ( 1 8 · N )k−1. (3.9) Hence, α(zi, P) may be estimated with (3.8), (3.9) as α(zi, P) < P(zi) P ′(zi) · max k>1 1 8 · N k−17 6 + 1 8 1/(k−1) ≤1 8 · 7 6 + 1 8 . (3.10) This yields α(zi, P) < 3−2 √ 2, and Theorem 1.3 implies that zi is an approximate zero. By Deﬁnition 1.1, the quadratic convergence is guaranteed. As P∞ k=0(1/2)2k−1 = 5/3, all Newton iterates of zi lie in {z ∈C : \|z −zi\| ≤5/3 · \| P (zi) P ′(zi)\|}. By (3.4) there is a zero in {z ∈C : \|z −zi\| ≤8/7 · \| P (zi) P ′(zi)\|}. The inclusion region with center zi −P (zi) P ′(zi) stated in Theorem 1.5, ii) is a simple consequence. These circles are pairwise disjoint. Remark. A diﬀerent proof shows that the new criterion (1.6) can be relaxed to the domain of linear convergence. 4. Conclusion. A suﬃcient criterion for quadratic convergence of Newton’s method has been established using a new error estimate. This criterion involves only n Newton corrections, and it improves by O(n) on suﬃcient conditions directly computed from the point estimate theory of Kim and Smale. REFERENCES Aberth, O. Iteration methods for ﬁnding all zeros of a polynomial simultaneously. Math. of Computation, 27:339–344, 1973. Batra, P. Absch¨ atzungen und Iterationsverfahren f¨ ur Polynom-Nullstellen. PhD thesis, Tech-nical University of Hamburg, 1998. Batra, P. Improvement of a convergence condition for the Durand-Kerner iteration. Journal of Computational and Applied Mathematics, 96:117–125, 1998. Carstensen, C. Anwendungen von Begleitmatrizen. ZAMM, 71(6):T 809 – T 812, 1991. Kerner, I. O. Ein Gesamtschrittverfahren zur Berechnung der Nullstellen von Polynomen. Numerische Mathematik, 8:290–294, 1966. Kim, M.-H. Computational Complexity of the Euler-Type Algorithms for the Roots of Complex Polynomials. PhD thesis, City University of New York, 1985. Kim, M.-H. On approximate zeros and rootﬁnding algorithms for a complex polynomial. Math-ematics of Computation, 51:707–719, 1988. Kim, M.-H. ; Sutherland, S. Polynomial root-ﬁnding algorithms and branched covers. SIAM J. Comput., 23(2):415–436, 1994. Ortega, J. M. ; Rheinboldt, W.C. Iterative solution of nonlinear equations in several variables. Academic Press, 1970. Ostrowski, A. Solution of Equations in Euclidean and Banach Spaces. Academic Press, 1973. 9 Pan, V. Y. On approximating complex polynomial zeros: Modiﬁed quadtree (Weyl’s) construc-tion and improved Newton’s iteration. Technical Report 2894, INRIA, Sophia-Antipolis, 1996. Smale, S. Algorithms for solving equations. In Proceedings of the International Congress of Mathematicians,Berkley, Ca.,USA,1986, pages 172–195. AMS, Providence,R.I. Smale, S. Newton’s method estimates from data at one point. In The merging of disciplines, pages 185–196. Springer-Verlag, New York, 1986. Wang, X.-H. ; Han, D.-F. On dominating sequence method in the point estimate and Smale theorem. Science in China, 33(2):135–144, 1990. 10
6141	https://www.cedars-sinai.org/health-library/diseases-and-conditions/p/post-polio-syndrome.html	For Patients For Providers Health Sciences University 1-800-CEDARS-1 English عربى 简体中文繁體中文 فارسي עִברִית 日本語 한국어 Русский Español Tagalog For Patients Home Get Care 24/7 Virtual Care Urgent Care Primary Care Pediatric Primary Care Routine Gynecology Virtual Second Opinion Browse All Find a Doctor Locations See All Locations Hospitals Primary Care Urgent Care Facilities Emergency Rooms Surgery Centers Medical Offices Imaging Facilities Labs Specialties Cancer Digestive & Liver Diseases Ear, Nose & Throat Heart Lung Neurology & Neurosurgery Obstetrics & Gynecology Orthopaedics Pediatrics at Guerin Children's Spine Surgery Transplant Urology Women's Health Browse All Records & Billing Medical Records Request Insurance & Billing Pay Your Bill Advanced Healthcare Directive Initiate a Request Help Paying Your Bill Sign In For Patients For Providers Health Sciences University 1-800-CEDARS-1 Post-Polio Syndrome Not what you're looking for? Start New Search ABOUT CAUSES DIAGNOSIS TREATMENT NEXT STEPS What is post-polio syndrome? Post-polio syndrome (PPS) is a disorder of the nerves and muscles. It happens in some people many years after they have had polio. PPS may cause new muscle weakness that gets worse over time, pain in the muscles and joints, and tiredness. People with PPS often feel exhausted. Polio is a contagious disease caused by the polio virus. It can spread through body fluids. It most often strikes young children. Severe polio may lead to paralysis and breathing problems. Polio vaccines have eliminated wild poliovirus in the United States. But there have been recent outbreaks in Afghanistan, Pakistan, Liberia, and Madagascar. The CDC states that the best way to keep people safe from the polio virus is by maintaining a high level of community immunity (protection) through vaccines. PPS happens only in those who had polio. It starts an average of 35 years after the polio infection. It causes the muscles to weaken and may eventually make it hard to breathe due to the damage done to the muscles involved in breathing. What causes post-polio syndrome? The cause of PPS is still not clear. Some experts think it may have to do with the way that the nerves have to regrow branches. This may overwork the nerve cells after a bout of polio. Another idea under investigation is that the virus “sleeps” in the nervous system. It reactivates later, causing PPS. It is also possible that the immune system somehow gets mixed up and attacks the body’s own nerves. Researchers are still trying to learn more about the possible causes of PPS. What are the symptoms of post-polio syndrome? People who survive polio are later at risk for PPS. PPS affects your nerves and muscles. Symptoms usually start between 20 and 40 years after the original polio illness. But they may appear anywhere from 10 to 70 years after. Muscle weakness may be the main symptom. This weakness may affect one side of your body more than the other. In general, symptoms of PPS may include: Progressive weakness (common). Tiredness (fatigue) (common). Pain in the muscles and joints (common). Reduced muscle mass (atrophy). Trouble swallowing. Breathing problems. Sleep disorders. Sensitivity to cold temperatures. You may find that your symptoms get more noticeable as the nerves and muscles continue to decline over time. Specially designed exercise programs and physical therapy may help improve some of the muscle weakness. How is post-polio syndrome diagnosed? Your health care provider will ask about your health history, your recent symptoms, and other health conditions. The provider typically does a physical exam and tests your muscle strength. You may need testing that includes: Blood tests to rule out other causes for your muscle weakness. Electromyography (EMG) to measure the electrical activity of the muscles. Muscle biopsy to look for signs of damage in the muscle cells. MRI or CT scans. Your health care provider may make a diagnosis of PPS if you had polio in the past, have new muscle weakness and other symptoms that last for a year, and have no other cause for your symptoms. You may first see your primary health care provider and then be referred to a specialist, such as a neurologist who specializes in neuromuscular diseases. How is post-polio syndrome treated? There is no cure for PPS, but there are things you can do to help you manage the condition. The goal of treatment for PPS is to reduce its impact on your daily life. You may use a cane or walker to save your energy and muscle strength. Make sure you get plenty of rest. People with PPS who have new muscle weakness and tiredness (fatigue) may improve muscle strength with specially designed, low-intensity, muscle-strengthening exercise programs known as non-fatiguing exercises. You may do these exercises in brief cycles, in which short repeats of exercise alternate with periods of rest in between. It's very important for people with PPS to not exert too much effort. Exercising in warm temperatures and in water may improve your well-being. Along with the exercise program, other supportive therapies that you may use include: Assistive devices, such as lightweight braces, canes, walkers, scooters, and wheelchairs. Medicines to ease pain. Medicines to ease fatigue. Physical therapy to keep as much mobility as possible. Occupational therapy to help with ways to adapt. Speech-language therapy if needed for trouble swallowing. Assisted breathing with a positive pressure breathing machine, if needed, especially at night. Emotional and mental health support. You may get care from a variety of providers in addition to your neurologist. These might include physical therapists, occupational therapists, speech-language therapists, and respiratory specialists. You can work with your medical team to design the best plan of care for your situation. Key points about post-polio syndrome PPS is a disorder that happens in as many as half of people many years after they had polio. The symptoms include progressive muscle weakness, pain in the muscles and joints, and tiredness. Some people with PPS may eventually need a machine to help with breathing. Prescribed special exercise programs, assistive devices, physical therapy, occupational therapy, and medicines may all help you manage the symptoms of PPS. PPS is rarely life-threatening, but the symptoms can significantly interfere with a person's ability to function independently. Next steps Here are some tips to help you get the most from a visit to your health care provider: Know the reason for your visit and what you want to happen. Before your visit, write down questions you want answered. Bring someone with you to help you ask questions and remember what your provider tells you. At the visit, write down the name of a new diagnosis and any new medicines, treatments, or tests. Also, write down any new instructions your provider gives you. Know why a new medicine or treatment is prescribed and how it will help you. Also, know what the side effects are and when they should be reported. Ask if your condition can be treated in other ways. Know why a test or procedure is recommended and what the results could mean. Know what to expect if you do not take the medicine or have the test or procedure. If you have a follow-up appointment, write down the date, time, and purpose for that visit. Know how you can contact your provider if you have questions. Medical Reviewer: Shaziya Allarakha MD Medical Reviewer: Esther Adler Medical Reviewer: Marianne Fraser MSN RN © 2000-2025 The StayWell Company, LLC. All rights reserved. This information is not intended as a substitute for professional medical care. Always follow your healthcare professional's instructions. Not what you're looking for? Start New Search Looking for a Physician? Choose a doctor and schedule an appointment. Find a Doctor Looking for Virtual Care? Connect with providers for illnesses like flus, UTIs and rashes (ages 3+) or chronic conditions like asthma, weight management and diabetes (ages 18+). Get Care Now Want More Information? Cedars-Sinai has a range of comprehensive treatment options. See Our Programs Need Help? 1-800-CEDARS-1 (1-800-233-2771) Monday‑Friday, 8 am‑5 pm PT Schedule a Callback Expert Care for Life™ Starts Here Adult Primary Care Pediatric Primary Care Urgent Care Need Help? 1-800-CEDARS-1 (1-800-233-2771) Schedule a Callback Looking for a Physician Choose a doctor and schedule an appointment. Find a Doctor
6142	https://www.geeksforgeeks.org/dsa/maximum-points-of-intersection-n-lines/	Maximum points of intersection n lines You are given n straight lines. You have to find a maximum number of points of intersection with these n lines. Examples: Input : n = 4 Output : 6 Input : n = 2 Output :1 Approach : As we have n number of line, and we have to find the maximum point of intersection using this n line. So this can be done using the combination. This problem can be thought of as a number of ways to select any two lines among n line. As every line intersects with others that are selected. So, the total number of points = nC2 Below is the implementation of the above approach: ```` // CPP program to find maximum intersecting // points include using namespace std; define ll long int // nC2 = (n)(n-1)/2; ll countMaxIntersect(ll n) { return (n) (n - 1) / 2; } // Driver code int main() { // n is number of line ll n = 8; cout << countMaxIntersect(n) << endl; return 0; } // Java program to find maximum intersecting // points public class GFG { // nC2 = (n)(n-1)/2; static long countMaxIntersect(long n) { return (n) (n - 1) / 2; } // Driver code public static void main(String args[]) { // n is number of line long n = 8; System.out.println(countMaxIntersect(n)); } // This code is contributed by ANKITRAI1 } Python3 program to find maximum intersecting points nC2 = (n)(n-1)/2 def countMaxIntersect(n): return int(n(n - 1)/2) Driver code if name=='main': n is number of line n = 8 print(countMaxIntersect(n)) this code is contributed by Shashank_Sharma // C# program to find maximum intersecting // points using System; class GFG { // nC2 = (n)(n-1)/2; public static long countMaxIntersect(long n) { return (n) (n - 1) / 2; } // Driver code public static void Main() { // n is number of line long n = 8; Console.WriteLine(countMaxIntersect(n)); } } // This code is contributed by Soumik PHP // PHP program to find maximum intersecting // points // nC2 = (n)(n-1)/2; function countMaxIntersect($n) { return ($n) ($n - 1) / 2; } // Driver code // n is number of line $n = 8; echo countMaxIntersect($n) . "\n"; // This code is contributed by ChitraNayal ? // Javascript program to find maximum intersecting // points // nC2 = (n)(n-1)/2; function countMaxIntersect(n) { return (n) (n - 1) / 2; } // Driver code // n is number of line var n = 8; document.write( countMaxIntersect(n) ); ```` Time Complexity: O(1) Auxiliary Space: O(1) S Explore DSA Fundamentals Logic Building Problems Analysis of Algorithms Data Structures Array Data Structure String in Data Structure Hashing in Data Structure Linked List Data Structure Stack Data Structure Queue Data Structure Tree Data Structure Graph Data Structure Trie Data Structure Algorithms Searching Algorithms Sorting Algorithms Introduction to Recursion Greedy Algorithms Graph Algorithms Dynamic Programming or DP Bitwise Algorithms Advanced Segment Tree Binary Indexed Tree or Fenwick Tree Square Root (Sqrt) Decomposition Algorithm Binary Lifting Geometry Interview Preparation Interview Corner GfG160 Practice Problem GeeksforGeeks Practice - Leading Online Coding Platform Problem of The Day - Develop the Habit of Coding Thank You! What kind of Experience do you want to share?
6143	https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:torque-and-rotational-dynamics/xf557a762645cccc5:newton-s-second-law-in-rotational-form/v/more-on-moment-of-inertia	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
6144	https://www.chegg.com/homework-help/questions-and-answers/7-express-coefficients-f-left-y-x-1-x-2-x-3-x-4-right-left-y-x-1-x-3-x-2-x-4-right-left-y--q112633364	Solved 7. Express the coefficients of \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Advanced Math Advanced Math questions and answers 7. Express the coefficients of f=(y−x1x2−x3x4)(y−x1x3−x2x4)(y−x1x4−x2x3)∈Q[x1,x2,x3,x4][y] in terms of the elementary symmetric polynomials. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 7. Express the coefficients of f=(y−x1x2−x3x4)(y−x1x3−x2x4)(y−x1x4−x2x3)∈Q[x1,x2,x3,x4][y] in terms of the elementary symmetric polynomials. Show transcribed image text There are 4 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 f=(y−x 1 x 2−x 3 x 4)(y−x 1 x 3−x 2 x 4)(y−x 1 x 4−x 2 x 3)∈Q[x 1,x 2,x 3,x 4][y] observe that, the roots of f are α 1:=(x 1 x 2+x 3 x 4)α 2:=(x 1 x 3+x 2 x 4)α 3:=(x 1 x 4+x 2 x 3) View the full answer Step 2 UnlockStep 3 UnlockStep 4 UnlockAnswer Unlock Previous questionNext question Transcribed image text: Express the coefficients of f=(y−x 1x 2−x 3x 4)(y−x 1x 3−x 2x 4)(y−x 1x 4−x 2x 3)∈Q[x 1,x 2,x 3,x 4][y] in terms of the elementary symmetric polynomials. Not the question you’re looking for? Post any question and get expert help quickly. 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6145	https://worksheetgenius.com/calc/fraction-of-number/what-is-225-490-of-499/	What is 225/490 of 499? Calculating the fraction of a whole number is a very useful skill to learn that helps students to understand the nature of numbers and their interactions. In this article, we'll explain how to calculate 225/490 of 499 with step-by-step-examples. Looking for fractions of whole number worksheets? Click here to see all of our free fraction of a number worksheets. Calculating the Answer as a Number The simple rule to remember whenever you want to find a fraction of an amount is to divide the amount by the fraction's denominator, and then multiply that answer by the fraction. Using this rule, you'll be able to work out the fractional amount of the original number. Let's work it out together. First we take the amount, 499, and divide it by the denominator, 490: Next, we take the answer, 1.0183673469388, and multiply it by the numerator, 225: As you can see, the answer to the question "what is 225/490 of 499?" as a number is 229.13265306122. Increase or Decrease an Amount by a Fraction What if you wanted to increase or decrease 499 by 225/490? Once you have calculated the answer above, 229.13265306122, you deduct that amount from the whole number to decrease it by 225/490 and you add it to increase: Increase = 499 + 229.13265306122 = 728.13265306122 Decrease = 499 - 229.13265306122 = 269.86734693878 Calculating the Answer as a Fraction Sometimes, you might want to show your answer as another fraction. In that case, we can do the following. First, we take the whole number and turn it into a fraction by using 1 as the fraction denominator: Now that we have two fractions, we can multiply the numerators and denominators together to get our answer in fraction form: We can simplify this new fraction down to lower terms. To do that we need to know something in math which is called the GCD, or greatest common divisor. I won't go through the steps of finding the GCD here as we'll cover it in a future article, but for now all you need to know is that the greatest common divisor of 112275 and 490 is 5. Using the GCD, we can divide the new numerator (112275) and the denominator (490) by 5 to simplify the fraction: 112275 ÷ 5 = 22455 490 ÷ 5 = 98 Finally, we can put the fraction answer together: Now you might have noticed that the fraction we have has a numerator that is larger than the denominator. This is called a mixed or improper fraction and means that there is a whole number involved. We can simplify this down to a mixed number. We'll put together a blog post on converting improper fractions to a mixed number to explain those steps in more detail, but for the purposes of this article, we'll go ahead and just give you the mixed number answer: 229 13/98 Hopefully this article helps you to understand how you can work with fractions of whole numbers and work this out quickly for yourself whenever you need it. Practice Fraction of Number Worksheets Like most math problems, finding the fraction of a number is something that will get much easier for you the more you practice the problems and the more you practice, the more you understand. Whether you are a student, a parent, or a teacher, you can create your own fractions of a whole number worksheets using our fraction of a number worksheet generator. This completely free tool will let you create completely randomized, differentiated, fraction of a number problems to help you with your learning and understanding of fractions. Practice Fractions of a Number Using Examples If you want to continue learning about how to calculate the fraction of a whole number, take a look at the quick calculations and random calculations in the sidebar to the right of this blog post. We have listed some of the most common fractions in the quick calculation section, and a selection of completely random fractions as well, to help you work through a number of problems. Each article will show you, step-by-step, how to work out the fraction of any whole number and will help students to really learn and understand this process. Calculate Another Fraction of a Number Enter your fraction in the A and B boxes, and your whole number in the C box below and click "Calculate" to calculate the fraction of the number. Link to Us / Reference this Page Please use the tool below to link back to this page or cite/reference us in anything you use the information for. Your support helps us to continue providing content! What is 225/490 of 499? "What is 225/490 of 499?". WorksheetGenius.com. Accessed on September 29, 2025. "What is 225/490 of 499?". WorksheetGenius.com, Accessed 29 September, 2025 What is 225/490 of 499?. WorksheetGenius.com. Retrieved from Quick Fraction of a Whole Number Calculations Random Fraction of a Whole Number Calculations © 2007-2025 Worksheet Genius 🧠. All rights reserved.
6146	https://www.scribd.com/document/155521650/Lesson-Plan-Inverse-Proportion	Lesson Plan Inverse Proportion \| PDF \| Formula \| Lesson Plan Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(5)100% found this document useful (5 votes) 5K views 1 page Lesson Plan Inverse Proportion This lesson plan aims to teach students about inverse proportion. The learning objectives are for students to be able to derive a formula to model inverse proportion and use it to calculate … Full description Uploaded by Jonathan Robinson AI-enhanced description Go to previous items Go to next items Download Save Save Lesson Plan Inverse Proportion For Later Share 100%100% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Lesson Plan Inverse Proportion For Later You are on page 1/ 1 Search Fullscreen mr-mathematics.com Figure 1 Lesson Planning Sheet Title: Inverse Proportion Learning Objectives: By the end of the lesson:  All students should be able to derive a formula in the form 1/x to model inverse proportion.  Most students should be able to derive a formula to model inverse proportion and use it to calculate unknown values.  Some students should be able to derive and use a formula to model when one unit is inversely proportion to the square or cube of another. Keywords : Model, Inverse proportion, Varies indirectly, Constant of Proportionality, Derive, Formula Learning Activities Starter/Introduction Students are challenged to use solve a simple inverse proportion problem using their powers of intuition. It is important to have the class describe their method in order to develop their mathematical reaso ning. Have students work in pairs to challenge each other and promote discussion. A pair of students could demo nstrate the solution to their peers at the front of the class.Development To model an inverse proportion it is important to ensure students have a secure understanding of reciprocal functions. Discuss that two measurements are considered to be inverse proportion, or vary indirectly when, as one unit increases the other decreases.The notation for modelling inverse proportion is shown to the right in figure 1.Use the x and y values in the table to highlight that the y value decreases as the x increases.When the problem in given in words rather than as a table it is important for the students to identify key vocabulary in order to derive the equation. Underline key words and phrases such as ‘square of’, ‘inversely proportional’ or ‘varies indirectly’. The interactive Excel file can be u sed to provide additional ex amples. Use mini-whiteboards to assess and feedback. When ready, the students should be able to work through the prob lems on the third slide inde pendently. Feedback solutions throughout to maintain pace.Plenary The plenary is used to consolidate student’s understanding between direct and inverse proportion. Ideally, the class would recognise that   cannot be the relationship since one increases when the o ther decreases. Understanding that k is constant between the different x and y values indicates that the correct relationship is modelled by   √  Have solutions presented for assessment and feedback. Resources: Mini-whiteboards Interactive Excel File Differentiation More able:  Students could solve problems in a real life context rather than using data presented in a table.Less Able  Students may benefit from modelling inverse proportions in the form    rather than     . adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Grade Level 11 - Direct & Inverse Proportions Lesson Plan No ratings yet Grade Level 11 - Direct & Inverse Proportions Lesson Plan 4 pages Statistical Inspector No ratings yet Statistical Inspector 22 pages Matrix Algebra: Operations & Applications No ratings yet Matrix Algebra: Operations & Applications 89 pages Matteo Petrera - Mathematical Physics I - Dynamical Systems and Classical Mechanics - Lecture Notes-Logos Verlag Berlin (2013) No ratings yet Matteo Petrera - Mathematical Physics I - Dynamical Systems and Classical Mechanics - Lecture Notes-Logos Verlag Berlin (2013) 268 pages Properties of Real Numbers 100% (1) Properties of Real Numbers 54 pages Full Many-Sorted Algebras For Deep Learning and Quantum Technology 1st Edition Giardina Ph.D. PDF All Chapters 100% (3) Full Many-Sorted Algebras For Deep Learning and Quantum Technology 1st Edition Giardina Ph.D. PDF All Chapters 62 pages 798 Books Doubtnut Question Bank No ratings yet 798 Books Doubtnut Question Bank 27 pages Resp 7 No ratings yet Resp 7 44 pages Chapter 1 Linear-Programming No ratings yet Chapter 1 Linear-Programming 52 pages Ferguson Part2 No ratings yet Ferguson Part2 96 pages Mathematical Induction No ratings yet Mathematical Induction 10 pages Volume Calculation for Solid Figures No ratings yet Volume Calculation for Solid Figures 54 pages 1 s2.0 S0895717710005613 Main No ratings yet 1 s2.0 S0895717710005613 Main 10 pages Grade 8 Module No ratings yet Grade 8 Module 9 pages Introduction To MATLAB For Engineers, Third Edition: Numerical Methods For Calculus and Differential Equations No ratings yet Introduction To MATLAB For Engineers, Third Edition: Numerical Methods For Calculus and Differential Equations 47 pages Basic Mathematics Diploma Course No ratings yet Basic Mathematics Diploma Course 7 pages XI Real Numbers-1 No ratings yet XI Real Numbers-1 9 pages Duality No ratings yet Duality 28 pages Lesson Plan Trig Intro 0% (1) Lesson Plan Trig Intro 1 page Conservative Property of Potential Field No ratings yet Conservative Property of Potential Field 12 pages PEUN Assignment 2 No ratings yet PEUN Assignment 2 7 pages 13 - Exponential Function No ratings yet 13 - Exponential Function 14 pages Monday 24 May 2021: Further Pure Mathematics No ratings yet Monday 24 May 2021: Further Pure Mathematics 6 pages Difference Equations No ratings yet Difference Equations 3 pages Revision Test MEP Year 9 100% (1) Revision Test MEP Year 9 6 pages Chapter-11 Direct and Inverse Proportion 100% (1) Chapter-11 Direct and Inverse Proportion 2 pages Efficient Power Flow via Newton's Method No ratings yet Efficient Power Flow via Newton's Method 12 pages Antenna Arrays: Advanced Problems No ratings yet Antenna Arrays: Advanced Problems 14 pages Mathdlp No ratings yet Mathdlp 10 pages A Grasping Force Optimization Algorithm For Multiarm Robots With Multifingered Hands No ratings yet A Grasping Force Optimization Algorithm For Multiarm Robots With Multifingered Hands 13 pages Lesson Plan Pie Charts 90% (10) Lesson Plan Pie Charts 2 pages Solving Word Problem Involving Direct Proportion in Different Context Such As Distance, Rate and Time Using Appropriate Strategies and Tools Grade 6 Lesson Plan No ratings yet Solving Word Problem Involving Direct Proportion in Different Context Such As Distance, Rate and Time Using Appropriate Strategies and Tools Grade 6 Lesson Plan 5 pages Introduction To The Practice of Statistics: Section 4.3 Homework Answers No ratings yet Introduction To The Practice of Statistics: Section 4.3 Homework Answers 7 pages Advanced Math Olympiad Problems No ratings yet Advanced Math Olympiad Problems 4 pages Algorithm Complexity Basics No ratings yet Algorithm Complexity Basics 9 pages Off To A Good Start! 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6147	http://mathgardenblog.blogspot.com/2013/07/radical-axis.html	Math Garden: Radical axis and radical center Typesetting math: 100% Math for the curious Pages Home About Radical axis and radical center I n previous post, we have learned about the concept of power of a point with respect to a circle. The power of a point P with respect to a circle centered at O and of radius r is defined by the following formula P(P,(O))=P U→×P V→=P O 2−r 2=(P x−O x)2+(P y−O y)2−r 2, here, U and V are two intersection points of the circle (O) with an arbitrary line passing though P. Power of a point: P(P,(O))=P U→×P V→=P O 2−r 2=(P x−O x)2+(P y−O y)2−r 2. The value of the power of a point gives us information about relative position of the point with respect to the circle. If the power of the point P is a positive number then P is outside the circle, if it is a negative number then P is inside the circle, and if it is equal to zero then P is on the circle. Today, we will look at application of the power of a point concept. We will use two main tools: radical axis and radical center. Radical axis is often used to prove that a certain number of points lie on the same straight line, and radical center is used to prove a certain number of lines meet at a common point. S uppose we have two circles (O 1,r 1) and (O 2,r 2) intersecting at two points I and J as in the following figure. If P is a point on the line I J then clearly P has equal power with respect to these two circles P(P,(O 1))=P I→×P J→=P(P,(O 2)). Upon more careful observation though, we see that if the point P does not lie on the line I J then the powers of the point P with respect to these two circles will not be the same. For instance, in the following figure, we can see that P(P,(O 1))=P I→×P A→≠P I→×P B→=P(P,(O 2)). So we can conclude that P has equal power to the two circles if and only if the point P lies on the line I J. N ow let us look at a different scenario. Suppose we have two circles (O 1,r 1) and (O 2,r 2) which do not have intersections. Is it possible to construct a point P that has equal power to these two circles? Yes, there is a simple method. Please have a ruler and a compass ready to draw with me. Let us draw a circle (O) so that it cuts the circle (O 1) at two points A, B, and cuts the circle (O 2) at two points C, D.Let us take the intersection point P of the two lines A B and C D. As we have previously pointed out, since P lies on A B, P has equal power to (O) and (O 1). Since P lies on C D, P has equal power to (O) and (O 2). Therefore, P has equal power to the two circles (O 1) and (O 2)! Let us now draw a lot of circles, with different positions of (O), can you observe any pattern in the intersection points P? Can you see that P is moving around on a line? We have reason now to suspect that the locus of all the points that have equal power to two circles is a straight line. This line is called the radical axis of the two circles. Radical axis Let us now state the theorem on radical axis. There are many ways to prove this theorem. We will show two proofs. One proof is using vector and the other makes use of coordinate system. Theorem on radical axis.Given two circles(O 1,r 1) and (O 2,r 2) with two distinct centers(O 1≠O 2). Then the locus of all the point P such that P has equal power to the two circles is a straight line. This line is called the radical axis of the two circles (O 1) and (O 2). The radical axis is perpendicular to the line O 1 O 2 connecting the two centers. Using the power of a point formula P(P,(O 1))=P O 2 1−r 2 1, P(P,(O 2))=P O 2 2−r 2 2, we can see that P(P,(O 1))=P(P,(O 2)) is equivalent to P O 2 2−P O 2 1=r 2 2−r 2 1=c. A proof using vector. c=P O 2 2−P O 2 1=(P O 2→+P O 1→)(P O 2→−P O 1→)=2 P M→O 1 O 2→, here, M is the midpoint of O 1 O 2. Draw the perpendicular line P H to O 1 O 2, we have c=P O 2 2−P O 2 1=2 P M→O 1 O 2→=2(P H→+H M→)O 1 O 2→=2 H M→O 1 O 2→. This implies that the point H on O 1 O 2 is determined uniquely by the formula H M¯¯¯¯¯¯¯¯¯¯=c 2 O 1 O 2¯¯¯¯¯¯¯¯¯¯¯¯. Therefore, the locus of the point P is the line passing through H and perpendicular to O 1 O 2, and the theorem is proved. A proof using coordinate system. c=P O 2 2−P O 2 1=[(P x−O 2 x)2+(P y−O 2 y)2]−[(P x−O 1 x)2+(P y−O 1 y)2] =2(O 1 x−O 2 x)P x+2(O 1 y−O 2 y)P y+(O 2 2 x+O 2 2 y−O 2 1 x−O 2 1 y). This means that the locus of the point P is the straight line determined by the following equation (O 1 x−O 2 x)x+(O 1 y−O 2 y)y+1 2(O 2 2 x+O 2 2 y−O 2 1 x−O 2 1 y−c)=0. The gradient of this line shows that it is perpendicular to O 1 O 2. Thus, the theorem is proved. W e have shown two proofs of the radical axis theorem. So now we know that the locus of all points that have equal power to two circles is a straight line and this line is called the radical axis of the two circles. A radical axis is perpendicular to the line connecting the centers of the two circles. In a special case, if the two circles intersect at two points then the radical axis is the line connecting these two intersection points. Using radical axis we can prove that certain points lie on the same straight line. The strategy is to show that each of these points has equal power to two certain circles and then conclude that they all lie on the radical axis of the two circles. Radical center L et us now move on to three circles. Suppose that we have three circles(O 1), (O 2) and (O 3), and we want to find a point P such that it has equal power to the three circles P(P,(O 1))=P(P,(O 2))=P(P,(O 3)). Now, if we want P to have equal power to (O 1) and (O 2) then P must be on the radical axis of the two circles (O 1) and (O 2). Similarly, we can see that P must be on the radical axis of the two circles (O 2) and (O 3), and also on the radical axis of the two circles (O 3) and (O 1). Thus, the three radical axes concurrent at the point P. This point is called the radical center of the three circles (O 1), (O 2) and (O 3). From the radical axis theorem, we know that the radical axis is perpendicular to the line connecting the two centers of the circles. So if the centers O 1, O 2, O 3 are collinear then all three radical axes are perpendicular to the line O 1 O 2 O 3. In this case, the three radical axes are parallel. Let us state the theorem on radical center. Theorem on radical center.Given three circles (O 1), (O 2) and (O 3) whose centers are three distinct points. If the three centers O 1, O 2, O 3 lie on a straight line then the radical axes of the three pairs of circles {(O 1),(O 2)}, {(O 2),(O 3)} and {(O 3),(O 1)} are parallel to each other; If the three points O 1, O 2, O 3 are not collinear then the radical axes of the three pairs of circles {(O 1),(O 2)}, {(O 2),(O 3)} and {(O 3),(O 1)} meet at a single point and this point is called the radical center. The radical center has equal power to the three circles. Application R adical axis is a useful tool to prove the collinearity of points. To prove that a certain number of points lie on a straight line, we may try to show that each of these points has equal power to two circles and then draw the conclusion that they all lie on the radical axis of the circles. Similarly, we can use radical center to prove the concurrency of lines. The strategy is to show that each of these lines is the radical axis of a pair of circles and then conclude that they meet at the radical center. Let us look at some exercises. Problem 1.Given a triangle A B C. Construct outside of the triangle A B C three isosceles triangles B A′C, C B′A and A C′B whose bases are B C, C A and A B, respectively. Prove that the lines passing through A, B, C, respectively, and perpendicular to B′C′, C′A′, A′B′ are concurrent. Solution. Using A′ as the center, draw a circle passing the two points B and C. Using B′ as the center, draw a circle passing the two points C and A. And using C′ as the center, draw a circle passing the two points A and B. The line passing through A perpendicular to B′C′ clearly is the radical axis of the two circles (B′) and (C′). Similarly, the line passing through B perpendicular to C′A′ is the radical axis of the two circles (C′) and (A′), and the line passing through C perpendicular to A′B′ is the radical axis of the two circles (A′) and (B′). Therefore, the three lines must concurrent at the radical center of the three circles (A′), (B′), (C′), and the problem is proved. Problem 2(IMO 1995). Let A, B, C, D be four distinct points on a line, in that order. The circles with diameters A C and B D intersect at X and Y. The line X Y meets B C at Z. Let P be a point on the line X Y other than Z. The line C P intersects the circle with diameter A C at C and M, and the line B P intersects the circle with diameter B D at B and N. Prove that the lines A M, D N, X Y are concurrent. Solution. To prove that A M, D N, X Y are concurrent, we take T as the intersection point of A M and X Y and then show that the point T is on the line D N. Since T is on X Y, T has equal power to the two circles with diameters B D and A C. Since T is on A M, T has equal power to the two circles with diameters A C and A P. Since T is on P Z, T has equal power to the two circles with diameters A P and D P. It follows that T has equal power to the two circles with diameters B D and D P. The line D N is the radical axis of these two circles, therefore, T is on D N. Thus, the problem is proved. T oday we have learned about the power of a point with respect to a circle and the concept of radical axis and radical center of circles. Let us stop here for now. In the next post we will use this tool to prove Pascal's theorem for the circle case. Hope to see you there. Homework. Use Pythagorean Theorem to find the locus of all the points P such that P O 2 2−P O 2 1=c. (Hint: Draw the line P H perpendicular to O 1 O 2 and then use Pythagorean Theorem.) Suppose that A 1 A 2, B 1 B 2, C 1 C 2, D 1 D 2 are tangent to the two circles (O 1) and (O 2) as in the following figure. Prove that the midpoints A, B, C, D of A 1 A 2, B 1 B 2, C 1 C 2, D 1 D 2 lie on a straight line. Given a square A B C D, using its four vertices as center, draw four circles of equal radius. Take a point P lying outside of these circles. Through P draw two lines such that the first line intersects with the circle (A) at two points A 1, A 2, and intersects with the circle (C) at two points C 1, C 2; the second line intersects with the circle (B) at two points B 1, B 2, and intersects with the circle (D) at two points D 1, D 2. Prove that P A 1×P A 2+P C 1×P C 2=P B 1×P B 2+P D 1×P D 2. (IMO 2000). Suppose A B is tangent to the circles C A M N and N M B D, M lies between C and D on the line C D, and C D is parallel to A B. The chords N A and C M meet at P; the chords N B and M D meet at Q. The rays C A and D B meet at E. Prove that P E=Q E. Hint: Prove that N M intersects A B at the midpoint of A B Prove that M is the midpoint of P Q Prove that A B is a midline of the triangle E C D Prove that E M is perpendicular to C D Prove that E P Q is an isosceles triangle Labels: Cartesian coordinate, circle, collinear, concurrent, geometry, plane geometry, power of a point, radical axis, radical center Newer PostOlder PostHome Support Math Garden on facebook Blog Archive ►2016(7) ►December(1) ►October(1) ►May(1) ►April(1) ►March(2) ►February(1) ►2015(12) ►December(1) ►November(1) ►October(1) ►July(1) ►May(2) ►April(4) ►March(1) ►January(1) ►2014(12) ►December(1) ►November(3) ►August(1) ►July(1) ►June(1) ►April(1) ►March(1) ►February(2) ►January(1) ▼2013(26) ►October(3) ►September(2) ►August(2) ▼July(2) Radical axis and radical center Power of a point to a circle ►June(3) ►May(3) ►April(3) ►March(3) ►February(3) ►January(2) ►2012(36) ►December(1) ►November(7) ►October(3) ►September(6) ►August(5) ►July(4) ►June(6) ►May(4) Vietnamese Version A facebook friendship problem Caveman's multiplication When I look in your eyes Hexagrammum Mysticum Theorem Pythagorean Theorem 1 = 2012 = 2013 Fibonacci numbers and a tile matching puzzle James' geometry art James' question Vianney's awesome square! 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6148	https://www.fq.math.ca/Papers/57-1/wadsanthat12092018.pdf	DISTRIBUTION OF CYCLE LENGTHS OF A QUADRATIC MAP OVER FINITE FIELDS OF CHARACTERISTIC 2 ATSANON WADSANTHAT AND CHATCHAWAN PANRAKSA Abstract. The map x 7→x2+x deﬁned on a ﬁxed ﬁnite ﬁeld of characteristic 2 is investigated as a dynamical system. The map is known to be a linear map. Its nilpotent points form a subﬁeld, and periodic cycles are somewhat uniform. A general upper bound for the cycle lengths is given in terms of the Carmichael function of the ﬁeld degree. 1. Introduction This work stems from our earlier study of maps of the form x 7→x2 + bx deﬁned over ﬁnite ﬁelds of characteristic 2 for general b in the ﬁeld . It was found that the case b = 1 is an exception compared with others. For instance, because the map looks simpler than others, an explicit formula for its iterations can always be given. The iteration formula provides a convenient ground for the description of orbits of points, and hence the behavior of the map. It turns out that binary expansion and primality play a large role in determining the orbits. The map is an example of a ﬁnite linear dynamical system, which has been studied before. The set of periodic points of such a dynamical system is fully described in . The crucial property used is that the given system preserves an algebraic operation, which is the addition of vectors in a vector space in that case. Other operation-preserving maps were studied, such as the squaring map x 7→x2 over ﬁnite rings in , , and , and the more general x 7→xn in . Because basic algebraic structures can be realized as modules, and that the given map is linear, the previous studies used Fitting’s lemma and our discussion will do the same. Fitting’s lemma states that, for a given module endomorphism, a simultaneously Artinian and Noetherian module can be written as a direct sum of two certain submodules. These two submodules, within respective contexts, are the sets of nilpotent and periodic points. This decomposition allows for describing the maps with ease; the sets can be investigated separately. Within our context, however, more can be said on the sets. The set of nilpotent points can be identiﬁed quickly. Moreover, the set of periodic points is found to be largely uniform, especially if the degree of the ﬁeld is prime or a power of 2. Before proceeding, let us establish some notations and recall a few deﬁnitions. Unless stated otherwise, N will denote a positive integer, possibly with some conditions, and q = 2N. The symbol Fq represents the ﬁnite ﬁeld of q elements, so it will be of characteristic 2 and degree N. The function f : Fq →Fq is given by f (x) = x2 + x. For any positive integer n and any x ∈Fq, fn (x) is the nth iteration of x. A given point x ∈Fq is termed nilpotent (respectively, periodic) if there exists an n such that fn (x) = 0 (respectively, fn (x) = x). We make a few observations as well. The function f is a linear one because for all x, y ∈Fq, f (x + y) = (x + y)2 + (x + y) = f (x) + f (y) . (1.1) Thus, by Fitting’s lemma , there exists a positive integer n such that Fq = Ker fn ⊕Im fn, (1.2) FEBRUARY 2019 35 THE FIBONACCI QUARTERLY where the ﬁrst and second summands can be proved to be the sets of nilpotent and periodic points, respectively. Hence, nilpotent points form an additive subgroup of Fq, and periodic points form another one. The two main results of this paper are to characterize the set of nilpotent points of f, and primes in terms of its cycle lengths. First, if N = 2kM is the degree of the ﬁeld, and M is odd, then the nilpotent points form a subﬁeld of degree 2k. Second, if N is odd, then N is prime if and only if the nonzero cycles of x 7→x2 + x are equal in length. These statements are provided here, and proved later near the end of Sections 3 and 4, respectively. Theorem 1.1. Let N = 2kM, where k is nonnegative and M is odd. Then, the set of nilpotent points of f is a subﬁeld of Fq of degree 2k. Theorem 1.2. Suppose that N is an odd number. The nonzero periodic cycles of x 7→x2 + x on Fq are of equal length if and only if N is prime. 2. General Iteration Identities In this section, a closed form for fn (x) in terms of powers of x is given. It is shown that the formula involves, in large part, the binary expansion of the number n of iterations. One of these formulas can be used in obtaining an upper bound for periodic cycle lengths. Lemma 2.1. Let n = 2k be a positive integer. Then for all x ∈Fq, fn (x) = x2n + x. (2.1) Proof. The idea is to use induction on the exponent k. If n = 20 = 1, then for all x ∈Fq, f1 (x) = x220 + x = x2 + x, which holds by deﬁnition of f. Suppose that the formula holds for a ﬁxed k and all x ∈Fq. Then for n = 2k+1 and all x ∈Fq, f2k+1 (x) = f2k f2k (x) = f2k x22k + x = x22k + x 22k + x22k + x = x22k+1 + x. By induction, it is concluded that (2.1) holds for all powers of 2 and all x ∈Fq. □ Proposition 2.2. Let n = 2r + 2s for some diﬀerent r, s ∈N. Then for all x ∈Fq, fn (x) = x22r+2s + x22r + x22s + x. (2.2) 36 VOLUME 57, NUMBER 1 DISTRIBUTION OF CYCLE LENGTHS Proof. Let r, s ∈N be given diﬀerent integers, and let x ∈Fq be arbitrary. Consider f2r+2s (x) = f2r f2s (x) = f2r x22s + x = x22s + x 22r + x22s + x = x22r+2s + x22r + x22s + x. Since r, s, and x are arbitrary, the proof is complete. □ Proposition 2.3. Let n ∈N have the binary expansion n = m X k=1 2rk (2.3) where rk ∈N for k = 1, 2, . . . , m. Then for all x ∈Fq, fn (x) = X x2c, (2.4) where c ranges over {0, 1}-linear combinations from {2r1, 2r2, . . . , 2rm}. Proof. Suppose that (2.4) holds for every x ∈Fq and for positive integers with at most m terms in the right side of (2.3). Let x ∈Fq be arbitrary and consider when n has m + 1 terms in its binary expansion, say n = m+1 X k=1 2rk = 2rm+1 + m X k=1 2rk. (2.5) Let n′ = n −2rm+1. Then n′ has m terms in its binary expansion, and f2m+1 fn′ (x) = f2m+1 X x2c = X x2c+2m+1 + X x2c, where c runs through all {0, 1}-linear combinations from {2r1, 2r2, . . . , 2rm}. Any {0, 1}-linear combination d from {2r1, 2r2, . . . , 2rm, 2rm+1} either contains the term 2rm+1 or not. If it does, then x2d is found in the ﬁrst sum. Otherwise, it is in the second sum. Hence, the sums can be written as one sum X x2d where d runs through all {0, 1}-linear combinations from {2r1, 2r2, . . . , 2rm+1}. The proof is complete. □ 3. Nilpotent Points Recall that a point x ∈Fq is nilpotent if there exists an n ∈N such that fn (x) = 0. The set of nilpotent points is known to be a subspace of the vector space Fq, which means that the sum of two nilpotent points is again nilpotent. For the map f, however, nilpotent points are more limited in choice, as we shall show that they form a subﬁeld of Fq. If N = 2kM, where M is odd and k is nonnegative, the subﬁeld of nilpotent points has degree 2k. Lemma 3.1. Let x, y ∈Fq be nilpotent points of f. Then, xy is a nilpotent point of f. FEBRUARY 2019 37 THE FIBONACCI QUARTERLY Proof. Let x, y ∈Fq be nilpotent points of f. By deﬁnition, there exists an n ∈N such that fn (x) = fn (y) = 0. Let m be such that 2m ≥n. Taking iterations of f yields f2m (x) = f2m (y) = 0. By Lemma 2.1, x22m = x, (3.1) y22m = y. (3.2) Multiplying both equations yields (xy)22m = xy f2m (xy) = 0. By deﬁnition, xy is nilpotent. □ Proof of Theorem 1.1. Let F be the subﬁeld of Fq of degree 2k. Then for every x ∈F, x22k = x. (3.3) This implies that f2k (x) = 0, so x is a nilpotent point of f. Conversely, let x be a nilpotent point of f. By deﬁnition, there exists an n ∈N such that fn (x) = 0. Let m ∈N be such that 2m ≥n. On the one hand, if m ≤k, then f2k (x) = 0, and immediately x ∈F. On the other hand, if m > k, then x satisﬁes x2N = x and x22m = x. By the division algorithm, there exist unique d1 and r1 such that 2m = Nd1 + r1 and 0 ≤r1 < N. It must be the case that x22m = x x2Nd1+r1 = x x2r1 = x. By the Euclidean algorithm, the process can be applied until an equation of the form x2r = x, (3.4) where r = gcd (2m, N) = 2k is derived. Therefore, x22k = x, and x ∈F. □ As a corollary, we identify N for which f is a ﬁxed point system. A mapping is termed as such when all its periodic points are ﬁxed ones. Corollary 3.2. The map f is a ﬁxed point system if and only if N = 2k for some nonnegative integer k. Proof. Suppose that N = 2k for some nonnegative integer k. Theorem 1.1 dictates that the subﬁeld of nilpotent points of f must be of degree 2k, implying that every point x ∈Fq is nilpotent. Hence, every point is eventually mapped to 0, which is the sole ﬁxed point of f, and f is a ﬁxed point system. Conversely, suppose that f is a ﬁxed point system. Since the ﬁeld Fq being ﬁnite, there exist an n such that fn (x) = 0 for all x ∈Fq. There exists an m ∈N such that 2m ≥n. This implies f2m (x) = 0 for all x ∈Fq, so x22m = x. Choosing k to be the least possible such m yields the assertion. □ Corollary 3.3. If N is odd, then f has only two nilpotent points, namely 0 and 1. Theorem 1.1 and its corollaries intuitively suggest that if N = 2kM is the degree of the given ﬁeld, then the orbits are largely unaﬀected by its subﬁeld of degree 2k. In this sense, N can be assumed to be odd, which corresponds to k = 0. 38 VOLUME 57, NUMBER 1 DISTRIBUTION OF CYCLE LENGTHS 4. Periodic Cycles In this section, we investigate the cycle structure of f by giving a crude upper bound for the maximal cycle length of f, and show how the primality of N plays a part in determining cycles. To be speciﬁc, assuming that N is odd, all periodic cycles of f except one are of the same length if and only if N is a prime. A few observations are in order. The map f has exactly one ﬁxed point, namely 0, which is obtained from solving f (x) = x. Thus, no other points are ﬁxed points of f, and we call a periodic cycle nonzero if it is not that ﬁxed point. We also recall that the Carmichael function λ (N) is the least positive integer such that for all a relatively prime to N, aλ(N) −1 is divisible by N. For odd prime power P = pr, λ (pr) = pr −pr−1. Lemma 4.1. Let N = 2kM, where k is nonnegative and M is odd. For all x in Fq, f2λ(M)+k (x) = f2k (x) , (4.1) where λ (M) is the Carmichael function of M. Proof. Let x ∈Fq be arbitrary. By the deﬁnition of λ, 2λ(M) = Mc + 1 for some positive integer c. This implies 2λ(M)+k = 2kMc + 2k = Nc + 2k. (4.2) Consider f2λ(M)+k (x) = x22λ(M)+k + x = x2Nc+2k + x = x22k + x = f2k (x) . The proof is complete. □ Thus, if N = 2kM and M is odd, Lemma 4.1 shows that the trees attached to each periodic point are of height 2k, and the cycles are of length 2λ(N)+k −2k. Since these facts are crucial to the overall description of f as a dynamical system, they are recorded as a corollary. Corollary 4.2. Let N = 2kM, where k is a nonnegative integer and M is an odd number. Then, the following conditions hold. (1) For every x ∈Fq, there exists a nonnegative integer m such that m ≤2k and fm (x) is a periodic point. (2) Every periodic point x ∈Fq of f has period 2λ(N)+k −2k. The bound in Corollary 4.2, however, might be ineﬀective in some cases, one of which is highlighted below. It concerns diﬀerent powers of 2, and Mersenne numbers in particular. Proposition 4.3. Suppose that the binary expansion of N is a geometric series, namely N = 2a + 2a+r + 2a+2r + . . . + 2a+(k−1)r. (4.3) Then for all x ∈Fq, f2a+kr (x) = f2a (x) . (4.4) FEBRUARY 2019 39 THE FIBONACCI QUARTERLY Proof. Let x ∈Fq be arbitrary. Note that (2r −1) N = 2a+kr −2a. Consider f2a+kr (x) = x22a+kr + x = x2(2r−1)N+2a + x = x22a + x = f2a (x) . Noticing that x is arbitrary completes the proof. □ In particular, Proposition 4.3 applies to numbers N of the form 2m + 1 and 2m −1, which include Mersenne and Fermat primes. Noticing that their binary expansions are geometric series proves the following corollaries. Corollary 4.4. Let N = 2r −2s for some nonnegative integers r, s with r > s + 1. Then for all x ∈Fq, f2r (x) = f2s (x) . (4.5) Corollary 4.5. Let N = 2r + 2s for some nonnegative integers r, s with r > s + 1. Then for all x ∈Fq, f22r (x) = f22s (x) . (4.6) Lemma 4.1, Proposition 4.3, and its two corollaries, give an upper bound for cycle lengths. In general, the map f on a ﬁnite ﬁeld of degree N = 2kM, where k is nonnegative and M is odd, has cycle lengths not longer than 2λ(M)+k −2k. If N has a binary expansion which is a geometric series, the cycle length is a multiple of N, likely between N and N2 −2N. In case of N = 2r −2s = 2s (2r−s −1) and r > s + 1, it is inferred from Corollary 4.4 that the lengths of periodic cycles cannot be greater than 2r −2s = N. Theorem 1.1 indicates that the subspace of nilpotent points is of dimension 2s. Consequently, the subspace of periodic points must be of dimension k = N −2s. Periodic cycles of maximal length must be at least k iterations long by virtue of linear independence. Thus, the length of such cycles must be N. It is stressed that, for N = 2r −2s with r > s + 1, the bound from Corollary 4.4 is exact for maximal cycles. It remains to prove Theorem 1.2, which demonstrates how primality of an odd N aﬀects the cycle lengths. Speciﬁcally, if the degree N of the ﬁeld is an odd prime, then all periodic cycles, except the sole ﬁxed point, have the same length. The converse also holds. Proof of Theorem 1.2. Assume that N is odd. Let {C1, C2, . . . , Ck} be the enumeration of non-ﬁxed periodic cycles of the map, with lengths c1, c2, . . ., ck, respectively. The sum of these lengths is equal to the number of nonzero periodic points, which is 2N−1 −1. By Lemma 4.1, for each i = 1, 2, . . . , k, there exists an integer di such that cidi = 2λ(N) −1. Counting the number of periodic points and considering the deﬁnition of di yield k X j=1 1 dj = 2N−1 −1 2λ(N) −1. (4.7) Dividing N −1 by λ (N) gives positive integers s and t such that N −1 = sλ (N) + t and 0 ≤λ (N). If N is prime, then λ (N) = N −1 and the right side of (4.7) is 1. If N is composite, then λ (N) ≤N −3 and the right side of (4.7) is at least 4. Note that λ (N) = N −2 is impossible because N −2 is odd but λ (N) is always even. 40 VOLUME 57, NUMBER 1 DISTRIBUTION OF CYCLE LENGTHS Deﬁne µ = k X j=1 cj dj = 1 2λ(N) −1 k X j=1 c2 j. (4.8) The Cauchy-Schwarz inequality implies µ2 ≤   k X j=1 1 dj     k X j=1 c2 j dj  . (4.9) Consider the expression s = k X j=1 (cj −µ)2 dj = k X j=1 c2 j dj −  2 − k X j=1 1 dj  µ2. (4.10) Preparations are done at this point. On the one hand, suppose that cj = µ for all j. Then s = 0, and equality holds in (4.9). An appropriate substitution yields µ2 =   k X j=1 1 dj    2 − k X j=1 1 dj  µ2. This implies Pk j=1 1 dj = 1, so λ (N) = N −1. Therefore, N is prime. On the other hand, suppose that N is prime. Then λ (N) = N −1, so from 2N−1 −1 = c1d1 = c1 + c2 + . . . + ck, d1 = 1 + c2 c1 + c3 c1 + . . . + ck c1 . To interpret this, if the nonzero periodic points are grouped into cycles of length c1, then all those points are exhausted. In this way, there are c2 c1 cycles of length c2, c3 c1 cycles of length c3, and so on. Thus, c1 divides each of c2, c3, . . ., ck. More generally, for any i, j ∈{1, 2, 3, . . . , k}, ci divides cj. It follows that ci = cj for any pair i, j ∈{1, 2, 3, . . . , k}. As an implication, di = k for all i. Hence, for each i, ci = 2N−1 −1 k = k X j=1 cj k = µ. □ The cycle lengths of f for N = 2, 3, 4, . . . , 15 are enumerated in Table 1 below. It shows how the primality and binary expansion of N regulate the existence and uniformity of cycles. In particular, it veriﬁes our theorems in Sections 3 and 4, and illustrates how they apply for each N. For N = 3, 5, 7, 11, 13, there is only one cycle length. In contrast, for N = 9, 15, there are at least two diﬀerent lengths. FEBRUARY 2019 41 THE FIBONACCI QUARTERLY N Upper bound Obtained from corollary Nonzero cycles 2 1 3.2 None 3 3 4.4 1 cycle of length 3 4 1 3.2 None 5 15 4.5 1 cycle of length 15 6 6 4.4 1 cycle of length 3 2 cycles of length 6 7 7 4.4 9 cycles of length 7 8 1 3.2 None 9 63 4.5 1 cycle of length 3 4 cycles of length 63 10 30 4.2 1 cycle of length 15 8 cycles of length 30 11 1023 4.2 3 cycles of length 341 12 12 4.4 1 cycle of length 3 2 cycles of length 6 20 cycles of length 12 13 4095 4.2 5 cycles of length 819 14 14 4.4 9 cycles of length 7 288 cycles of length 14 15 15 4.4 1 cycle of length 3 3 cycles of length 5 1091 cycles of length 15 TABLE 1. Cycle lengths of f on Fq as N ranges from 2 to 15. All examples in Table 1 are calculated using Sage computer algebra software, version 6.4.1 . The code for each N, ran case-by-case, is given in Figure 1. sage : #Set degree N and order q , vary N on each run sage : N=6; q=2ˆN sage : #Build f i n i t e f i e l d sage : a=var ( ‘ a ’ ) ; F= GF(q , ‘ a ’ ) sage : #Create digraph of f sage : C =F. l i s t () sage : G =DiGraph ( [C, lambda i , j : F( i ˆ2+ i)==F( j ) ] ) sage : #List c y c l e s and count lengths sage : l1= G. a l l s i m p l e c y c l e s () sage : l2 =[ len ( l )−1 f o r l in l1 ] sage : #Count c y c l e s sage : from i t e r t o o l s import groupby sage : [ ( len ( l i s t (m) ) , l ) f o r l ,m in groupby ( l2 ) ] FIGURE 1. Code for ﬁnding cycle lengths of f for each N. 42 VOLUME 57, NUMBER 1 DISTRIBUTION OF CYCLE LENGTHS FIGURE 2. Direct graph of f on F64 ∼ = F2[a]/ a6 + a4 + a3 + a + 1 . 5. Conclusion and Discussion For the speciﬁc map x 7→x2 +x deﬁned on ﬁnite ﬁelds of characteristic 2, the nilpotent and cycle structure are obtained in terms of degree of the ﬁeld. By Theorem 1.1, the set of nilpotent points is the maximal subﬁeld of degree 2k. The degree N characterizes the uniformity of cycle lengths. By Theorem 1.2, except the ﬁxed point, periodic cycles are of the same length if and only if N is prime. It is likely that the bound from Corollary 4.2 can be made sharper by considering the order of 2 in the multiplicative group (Z/nZ)× or considering a geometric sum. It might be possible to use Corollary 4.4 and Theorem 1.2 in conjunction to derive a test for Mersenne primes, or Lemma 4.1 to estimate λ (N), using only bitwise operations. With a so-called normal basis for Fq, the map x 7→x2 + x is a cyclic permutation of bits, followed by an exclusive or. References E. L. Blanton Jr., S. P. Hurd, and J. S. McCranie, On a digraph deﬁned by squaring modulo n, The Fibonacci Quarterly, 30.4 (1992), 322–334. R. Burton, Elementary Number Theory, 7th ed., McGraw-Hill, New York, 2011. H. Griﬃn, Elementary Theory of Numbers, McGraw-Hill, New York, 1954. FEBRUARY 2019 43 THE FIBONACCI QUARTERLY R. A. Hern´ andez Toledo, Linear ﬁnite dynamical systems, Comm. Algebra, 33.9 (2005), 2977–2989. S. Lang, Algebra, 3rd ed., Addison-Wesley Publishing Company, Massachusetts, 1993. R. Lidl and H. Niederreiter, Finite Fields, 2nd ed., Cambridge University Press, Cambridge, England, UK, 1997. C. Lucheta, E. Miller, and C. Reiter, Digraph from powers modulo p, The Fibonacci Quarterly, 34.3 (1996), 226–239. T. D. Rogers, The graph of the square mapping on the prime ﬁelds, Discrete Math., 148 (1996), 317–324. SageMath, the Sage Mathematics Software System (Version 6.4.1), The Sage Developers, 2014, http: //www.sagemath.org. T. Vasiga and J. Shallit, On the iteration of certain quadratic maps over GF(p), Discrete Math., 277 (2004), 219–240. W. R. Wade, An Introduction to Analysis, 4th ed., Pearson, New Jersey, 2010. A. Wadsanthat, C. Panraksa, and W. Kositwattanarerk, Linear maps given by quadratic polynomials, East-West J. Math., preprint. MSC2010: 11B39, 33C05 Department of Mathematics, Faculty of Science, Mahidol University, Ratchathewi, Bangkok, 10240 Thailand E-mail address: atsanon.wad@student.mahidol.ac.th Mahidol University International College, Mahidol University, Salaya, Nakhon Pathom, 73170 Thailand E-mail address: chatchawan.pan@mahidol.edu 44 VOLUME 57, NUMBER 1
6149	https://www.mathpages.com/home/kmath689/kmath689.htm	\| \| \| Intersecting Circles \| \| \| \| The bonds that unite another person to ourself exist only in our mind. \| \| Marcel Proust \| \| \| \| A “circle” is ordinarily defined as the locus of points (on a plane) equi-distant from a given point called the “center”. Algebraically, in terms of a system of Cartesian coordinates “x” and “y”, we typically express this by defining a circle as the locus of points with coordinates (x,y) such that \| \| \| \| \| \| \| \| where (x0,y0) are the real coordinates of the center, and R is the real radius of the circle, i.e., the metrical distance from the center to each point of the circle. Usually we tacitly restrict our attention to points with purely real coordinates, but without this restriction the algebraic definition actually implies a more elaborate structure, consisting of all the “points” with complex coordinates that satisfy the above equation. \| \| \| \| As an example, suppose the center of the circle is at the origin of the coordinates, so x0 = y0 = 0, and consider the set of complex numbers x = xr + ixi and y = yr + iyi that satisfy the equation \| \| \| \| \| \| \| \| where R, xr, xi, yr, yi are all purely real-valued numbers. This implies \| \| \| \| \| \| \| \| Obviously if the imaginary components of x and y are both zero, the left hand condition is automatically satisfied and the right hand condition reduces to the usual equation for a real circle centered on the origin. However, if we allow xi and yi to have non-zero values, we have solutions for all sets of values of xr,yr outside the original circle, i.e., such that the sum of the squares of xr and yr exceeds R2. The locus of these solutions could be regarded as the algebraically complete “circle”. \| \| \| \| To visualize a complete circle, we can solve the left hand condition for yi, insert this into the right hand condition, and solve for xr to give \| \| \| \| \| \| \| \| If xi = 0 this equation simply gives x as a function of y for an ordinary real circle. For any other given value of xi this equation gives xr as a function of yr for lines of constant xi. Since xr must be real-valued, we know the values of yr must lie within the range \| \| \| \| \| \| \| \| A plot of the lines of constant xi for a circle of radius R = 1 is shown below. Note that since xi appears only squared in the equation for these loci, they correspond to both positive and negative values of xi. \| \| \| \| \| \| \| \| As an aside, these contours are remarkably similar to the stream lines in potential fluid flow around a sphere or cylinder. A contour plot of the upper surface (i.e., the positive values of xi) viewed from the side is shown below. \| \| \| \| \| \| \| \| In the same way that we found the loci of constant xi as a function of xr and yr, we can also solve for the loci of constant values of yi, and by symmetry it is the same as those shown above, except rotated 90 degrees. Thus a complete depiction of an algebraically complete circle is as shown below. \| \| \| \| \| \| \| \| In the three-dimensional space representing the coordinate components xr, yr, xi, we see that an algebraically complete circle can be described as a double-covering surface that extends from the real circle out to infinity. Likewise in the space of xr, yr, yi the circle is a double-covering surface. However, these two surfaces are not independent, because the relation xrxi + yryi = 0 implies that the signs of xi and yi are linked. At any given point on the xr,yr plane, there are two possible values of xi (of equal magnitude and opposite signs), and two possible values of yi, but this doesn’t represent four possible combinations, because we have yi = -(xr/yr)xi. For example, if xr/yr is positive, then yi and xi must have opposite signs. Hence there are only two complex “points on the circle” for any given values of xr and yr. (Of course, these two “points” merge into one on the real circle, where the two parts of each surface meet.) \| \| \| \| Incidentally, the flat “basin” inside the real circle (as depicted in the countour plot above) is the region where the computed value of xi (and also yi) would be imaginary, which means the value of x = xr + ixi and y = yr + iyi would be purely real. This implies that the resulting values of x and y are on the real circle, so the values of xi and yi are just such as to map each point (xr,yr) inside the circle to a point on the circle. Specifically, it maps the point (xr,yr) to the point with the coordinates \| \| \| \| \| \| \| \| where r2 = xr2 + yr2. It’s easy to verify that these points do indeed lie on the circle of radius R, and the second terms are perpendicular to the first terms, so it follows that each point inside the circle is mapped to the circle along the line perpendicular to the line from the center of the circle, as shown below. \| \| \| \| \| \| \| \| Now we consider the intersections between two circles, given the (purely real) radii and the coordinates of the centers. Let r1 and r2 denote the radii of the two circles, and let (x1,y1) and (x2,y2) denote the coordinates of their centers (in terms of a Cartesian coordinate system). For convenience we also let L denote the distance between the centers. It is straightforward to show that the intersection of these two circles consists of two points, with the coordinates \| \| \| \| \| \| \| \| where A is the area of the triangle with edge lengths r1, r2, L, as given by Heron’s formula \| \| \| \| \| \| \| \| As a simple example, consider two circles of radii 5 and 3 units respectively, with centers located a distance of 4 units apart, as depicted for a typical configuration in the figure below. \| \| \| \| \| \| \| \| Since in this example we have r12 – r22 = L2, the first two terms of the equations for the x and y coordinates of the intersection points are simply x2 and y2 respectively, so the equations reduce to \| \| \| \| \| \| \| \| where we have put Δx = x2 – x1 and Δy = y2 – y1. Also, the area of the triangle with edge lengths L, r1, r2 in this case is simply Lr2/2 (which can be confirmed by Heron’s formula), so we have \| \| \| \| \| \| \| \| By similar triangles we have r2/L = dx/Δy = dy/Δx, so we make these substitutions to arrive at the expected result \| \| \| \| \| \| \| \| This merely confirms that equations (1) give the real intersection points for real circles that overlap in real space. \| \| \| \| However, those equations also apply to any two circles, even if they don’t overlap in real space. For example, consider two identical circles, each of radius r1 = r2 = 2 units, with centers separated by a distance of L = 5 units. \| \| \| \| \| \| \| \| These circles clearly have no real points of contact, but the algebraically complete circles do have two points of intersection in the full complex space. The area of the triangle with edge lengths 5,2,2 is imaginary, given by Heron’s formula as (15/4)i. Since r1 = r2, the second terms in equations (1) are zero, so the real parts of the coordinates of the intersection points are given by the first terms, which represents simply the midpoint between the centers of the circles. The imaginary parts of the intersection coordinates are proportional to the real components of the vector between the centers. For this example we have \| \| \| \| \| \| \| \| Thus if we rotate the configuration, keeping the radii and distance between centers the same, the vector between the centers rotates and the vector consisting of xi and yi rotates, but they are 90 degrees out of phase, since xi is proportional to the y component of the vector between the centers, and yi is proportional to the x component. The magnitude of the imaginary vector is a constant (2A/L)i. For a specific numerical example, suppose the centers of the two circles are located at (12,7) and (16,10). Inserting these values into the above equations, we get the coordinates of the two points of intersection \| \| \| \| \| \| \| \| It can easily be confirmed that each of these points is a distance from 2 units from each of the centers. \| \| \| \| If the radii of the circles are unequal, the real parts of x∩ and y∩ are no longer at the mid-point between the centers, but it still falls on the line between the centers, shifted toward the smaller circle. For example, if the radius of the second circle in the preceding figure was reduced from 2 units to just 1, the real parts of the intersection coordinates would shift from the midpoint (14, 17/2) to (14 + 12/50, 17/2 + 9/50) as shown below. \| \| \| \| \| \| \| \| It’s interesting to consider the case when the radii of the circles shrink toward zero, in which case the area approaches A = (L2/4)i and the coordinates of the points of intersection approach \| \| \| \| \| \| \| \| In this limiting case the squared distances from the origin of the coordinate system to these two intersection points are \| \| \| \| \| \| \| \| Hence if we let d1 and d2 denote the distances from the origin to the centers of the (vanishingly small) circles, and if we let δ1 and δ2 denote the distances from the origin to the points of intersection of those circles, we have δ1δ2 = d1d2. \| \| \| \| Return to MathPages Main Menu \|
6150	https://dokumen.pub/vertebrate-life-9thnbsped-9780321773364.html	Vertebrate Life [9th ed.] 9780321773364 - DOKUMEN.PUB Anmelden Registrierung Deutsch English Español Português Français Dom Najlepsze kategorie CAREER & MONEY PERSONAL GROWTH POLITICS & CURRENT AFFAIRS SCIENCE & TECH HEALTH & FITNESS LIFESTYLE ENTERTAINMENT BIOGRAPHIES & HISTORY FICTION Najlepsze historie Najlepsze historie Dodaj historię Moje historie Home Vertebrate Life [9th ed.] 9780321773364 Vertebrate Life [9th ed.] 9780321773364 3,599 379 38MB English Pages 729 Year 2013 Report DMCA / Copyright DOWNLOAD FILE Polecaj historie ###### Vertebrate Motoneurons 107 85 47MB Read more ###### Evolution of Vertebrate Design 9780226220635 The Evolution of Vertebrate Design is a solid introduction to vertebrate evolution, paleontology, vertebrate biology, an 436 69 10MB Read more ###### Great Transformations in Vertebrate Evolution 9780226268392 How did flying birds evolve from running dinosaurs, terrestrial trotting tetrapods evolve from swimming fish, and whales 849 139 28MB Read more ###### Vertebrate Faunal Remains from Grasshopper Pueblo, Arizona 9781949098891, 9780915703210 Grasshopper Pueblo is a large fourteenth-century community in the forested Mogollon highlands of central Arizona. 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Heiser Categories Biology Zoology _Table of contents : Cover......Page 1 Title Page......Page 2 Copyright Page......Page 3 About the Authors......Page 4 Contents......Page 6 Preface......Page 14 PART I: Vertebrate Diversity, Function, and Evolution......Page 16 1.1 The Vertebrate Story......Page 17 1.3 Phylogenetic Systematics......Page 23 1.4 The Problem with Fossils: Crown and Stem Groups......Page 26 1.5 Evolutionary Hypotheses......Page 27 1.6 Earth History and Vertebrate Evolution......Page 30 Discussion Questions......Page 31 Additional Information......Page 32 2.1 Vertebrates in Relation to Other Animals......Page 34 2.2 Definition of a Vertebrate......Page 39 2.3 Basic Vertebrate Structure......Page 40 Discussion Questions......Page 60 Additional Information......Page 61 3.1 Reconstructing the Biology of the Earliest Vertebrates......Page 62 3.2 Extant Jawless Fishes......Page 65 3.3 The Importance of Extant Jawless Vertebrates in Understanding Ancient Vertebrates......Page 71 3.5 The Basic Gnathostome Body Plan......Page 72 3.6 The Transition from Jawless to Jawed Vertebrates......Page 77 3.7 Extinct Paleozoic Jawed Fishes......Page 81 Summary......Page 83 Additional Information......Page 84 PART II: Non-Amniotic Vertebrates: Fishes and Amphibians......Page 86 4.1 The Aquatic Environment......Page 87 4.2 Water and the Sensory World of Fishes......Page 92 4.4 Exchange of Water and Ions......Page 99 4.5 Responses to Temperature......Page 105 4.6 Body Size and Surface/Volume Ratio......Page 109 Summary......Page 111 Additional Information......Page 112 5.1 Chondrichthyes—The Cartilaginous Fishes......Page 114 5.3 The Paleozoic Chondrichthyan Radiation......Page 117 5.4 The Early Mesozoic Elasmobranch Radiation......Page 120 5.5 Extant Lineages of Elasmobranchs......Page 121 5.6 Batoidea: Skates and Rays......Page 131 5.7 Holocephali—The Little Known Chondrichthyans......Page 133 Discussion Questions......Page 134 Additional Information......Page 136 6.1 The Origin of Bony Fishes......Page 137 6.2 Evolution of the Actinopterygii......Page 141 6.3 Extant Actinopterygii—Ray-Finned Fishes......Page 146 6.4 Locomotion in Water......Page 152 6.5 Actinopterygian Reproduction......Page 156 6.6 The Adaptable Fishes—Teleosts in Contrasting Environments......Page 162 6.7 Conservation of Fishes......Page 165 6.8 Sarcopterygii—The Lobe-Finned Fishes......Page 168 Discussion Questions......Page 171 Additional Information......Page 172 7.1 Earth History, Changing Environments, and Vertebrate Evolution......Page 173 7.2 Continental Geography of the Paleozoic......Page 174 7.3 Paleozoic Climates......Page 177 7.4 Paleozoic Terrestrial Ecosystems......Page 178 7.5 Paleozoic Extinctions......Page 180 Additional Information......Page 181 8.1 Support and Locomotion on Land......Page 182 8.2 Eating on Land......Page 190 8.5 Pumping Blood Uphill......Page 193 8.6 Sensory Systems in Air......Page 197 8.8 Controlling Body Temperature in a Changing Environment......Page 199 Summary......Page 202 Additional Information......Page 203 9.1 Tetrapod Origins......Page 204 9.2 Radiation and Diversity of Non-Amniotic Paleozoic Tetrapods......Page 212 9.3 Amniotes......Page 216 Summary......Page 223 Additional Information......Page 224 10.1 Amphibians......Page 226 10.2 Diversity of Life Histories of Amphibians......Page 237 10.3 Amphibian Metamorphosis......Page 252 10.4 Exchange of Water and Gases......Page 253 10.5 Poison Glands and Other Defense Mechanisms......Page 258 10.7 Why Are Amphibians Vanishing?......Page 261 Discussion Questions......Page 265 Additional Information......Page 266 PART III: Sauropsida: Turtles, Lepidosaurs, and Archosaurs......Page 268 11 Synapsids and Sauropsids: Two Approaches to Terrestrial Life......Page 269 11.1 Taking Advantage of the Opportunity for Sustained Locomotion......Page 270 11.2 Increasing Gas Exchange: The Trachea and Lungs......Page 274 11.3 Transporting Oxygen to the Muscles: Structure of the Heart......Page 280 11.4 Taking Advantage of Wasted Energy: Endothermy......Page 281 11.5 Getting Rid of Wastes: The Kidneys and Bladder......Page 286 11.6 Sensing and Making Sense of the World: Eyes, Ears, Tongues, Noses, and Brains......Page 296 Discussion Questions......Page 299 Additional Information......Page 300 12.1 Everyone Recognizes a Turtle......Page 302 12.2 But What Is a Turtle? Phylogenetic Relationships of Turtles......Page 305 12.3 Turtle Structure and Function......Page 306 12.4 Ecology and Behavior of Turtles......Page 313 12.5 Reproductive Biology of Turtles......Page 315 12.6 Hatching and the Behavior of Baby Turtles......Page 317 12.7 Conservation of Turtles......Page 321 Summary......Page 322 Additional Information......Page 323 13.1 The Lepidosaurs......Page 325 13.2 Radiation of Sphenodontids and the Biology of Tuatara......Page 326 13.3 Radiation of Squamates......Page 327 13.4 Ecology and Behavior of Squamates......Page 336 13.5 Behavioral Control of Body Temperatures by Ectotherms......Page 353 13.6 Temperature and Ecology of Squamates......Page 357 Summary......Page 361 Additional Information......Page 362 14.1 Vertebrates and Their Environments......Page 364 14.2 Dealing with Dryness—Ectotherms in Deserts......Page 365 14.3 Coping with Cold—Ectotherms in Subzero Conditions......Page 371 14.4 The Role of Ectothermic Tetrapods in Terrestrial Ecosystems......Page 374 Discussion Questions......Page 377 Additional Information......Page 378 15 Geography and Ecology of the Mesozoic Era......Page 379 15.2 Mesozoic Terrestrial Ecosystems......Page 380 15.3 Mesozoic Climates......Page 383 15.4 Mesozoic Extinctions......Page 384 Additional Information......Page 385 16.1 The Mesozoic Fauna......Page 386 16.3 Marine Lineages......Page 387 16.4 Semiaquatic and Terrestrial Diapsids: Crocodylomorpha......Page 394 16.5 The First Evolution of Flight: Pterosauria......Page 397 16.6 Dinosaurs......Page 399 16.7 Terrestrial Herbivores: Ornithischian and Sauropod Saurischian Dinosaurs......Page 403 16.8 Terrestrial Carnivores: Theropod Dinosaurs......Page 412 16.9 The Second Evolution of Flight: Birds......Page 416 Discussion Questions......Page 419 Additional Information......Page 420 17.1 Early Birds and Extant Birds......Page 422 17.2 The Structure of Birds......Page 425 17.3 Wings and Flight......Page 430 17.4 The Hindlimbs......Page 437 17.5 Feeding and Digestion......Page 440 17.6 Sensory Systems......Page 444 17.7 Social Behavior......Page 447 17.8 Mating Systems......Page 449 17.9 Oviparity......Page 450 17.10 Nests......Page 452 17.11 Orientation and Navigation......Page 455 17.12 Migration......Page 456 Summary......Page 459 Additional Information......Page 460 PART IV: Synapsida: The Mammals......Page 462 18.1 The Origin of Synapsids......Page 463 18.2 Diversity of Nonmammalian Synapsids......Page 465 18.3 Evolutionary Trends in Synapsids......Page 470 18.4 The First Mammals......Page 477 18.5 The Radiation of Mesozoic Mammals......Page 481 Summary......Page 483 Additional Information......Page 484 19 Geography and Ecology of the Cenozoic Era......Page 486 19.1 Cenozoic Continental Geography......Page 487 19.2 Cenozoic Terrestrial Ecosystems......Page 488 19.3 Cenozoic Climates......Page 490 19.4 Cenozoic Extinctions......Page 492 Additional Information......Page 493 20.1 Major Lineages of Mammals......Page 495 20.2 Mammalian Ordinal Diversity......Page 496 20.3 Features Shared by All Mammals......Page 503 20.4 Features That Differ Among Mammalian Groups......Page 515 20.5 Cenozoic Mammal Evolution......Page 518 Additional Information......Page 526 21.1 Mammalian Reproduction......Page 528 21.2 Some Extreme Placental Mammal Reproductive Specializations......Page 533 21.3 Are Placental Mammals Reproductively Superior to Marsupials?......Page 534 21.4 Specializations for Feeding......Page 535 21.5 Locomotor Specializations......Page 542 21.6 Evolution of Aquatic Mammals......Page 545 Discussion Questions......Page 550 Additional Information......Page 551 22.1 Endothermic Thermoregulation......Page 552 22.2 Endotherms in the Cold......Page 555 22.3 Avoiding Cold and Sharing Heat......Page 556 22.4 Facultative Heterothermy......Page 558 22.5 Migration to Avoid Cold......Page 562 22.6 Endotherms in the Heat......Page 564 Discussion Questions......Page 572 Additional Information......Page 573 23.1 Social Behavior......Page 574 23.2 Population Structure and the Distribution of Resources......Page 575 23.3 Advantages of Sociality......Page 579 23.4 Body Size, Diet, and the Structure of Social Systems......Page 580 23.5 Horns and Antlers......Page 585 23.6 Primate Societies......Page 588 Additional Information......Page 594 24.1 Primate Origins and Diversification......Page 596 24.2 Origin and Evolution of the Hominoidea......Page 605 24.3 Origin and Evolution of Humans......Page 609 24.4 Derived Hominins (the Genus Homo)......Page 615 24.5 Evolution of Human Characteristics......Page 620 24.6 How Many Species of Humans Were Contemporaneous?......Page 625 Discussion Questions......Page 626 Additional Information......Page 627 25 The Impact of Humans on Other Species of Vertebrates......Page 629 25.1 Humans and the Pleistocene Extinctions......Page 630 25.2 Humans and Recent Extinctions......Page 632 25.3 Global Climate Change and Vertebrates......Page 635 25.4 Organismal Biology and Conservation......Page 639 25.5 Captive Breeding......Page 643 25.6 The Paradoxes of Conservation......Page 646 Summary......Page 647 Additional Information......Page 648 Appendix......Page 650 A......Page 672 B......Page 673 C......Page 674 D......Page 675 E......Page 676 H......Page 677 I......Page 678 M......Page 679 P......Page 680 S......Page 682 V......Page 684 Z......Page 685 Credits......Page 686 A......Page 694 B......Page 696 C......Page 697 D......Page 700 E......Page 701 F......Page 703 H......Page 704 K......Page 706 L......Page 707 M......Page 708 O......Page 710 P......Page 711 R......Page 713 S......Page 714 T......Page 717 V......Page 719 Z......Page 720_ Citation preview F. HARVEY POUGH Rochester Institute of Technology CHRISTINE M. JANIS Brown University JOHN B. HEISER Cornell University VERTEBRATE LIFE NINTH EDITION Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Editor-in-Chief: Beth Wilbur Executive Director of Development: Deborah Gale Development and Project Editor: Crystal Clifton, Progressive Publishing Alternatives Managing Editor: Mike Early Production Project Manager: Camille Herrera Production Management: Rose Kernan, Cenveo Publisher Services Compositor: Cenveo Publisher Services Interior and Cover Designer: Emily Friel, Integra Software Services, Inc. Illustrators: Dartmouth Publishing, and Precision Graphics Art Development Editor: Jennifer Kane Associate Director of Image Management: Travis Amos Supervisor Publishing Production, Photos: Donna Kalal Photo Researcher: Maureen Spuhler Photo Clearance: Q2A/Bill Smith Manager of Permissions, Text: Beth Wollar Permissions Assistant, Text: Tamara Efsen Manufacturing Buyer: Michael Penne Marketing Manager: Lauren Harp Cover Photo Credit: Rod Williams / Nature Picture Library Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on the appropriate page within the text and on p. C-1. Copyright ©2013, 2009, 2005 Pearson Education, Inc. All rights reserved. Manufactured in the United States of America. 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[Library of Congress Cataloging-in-Publication Data on File] 1 2 3 4 5 6 7 8 9 10—EB—16 15 14 13 12 www.pearsonhighered.com ISBN 10: 0-321-77336-5 ISBN 13: 978-0-321-77336-4 (Student edition) About the Authors F. Harvey Pough is a Professor of Biology at the Rochester Institute of Technology. He began his biological career at the age of fourteen when he and his sister studied the growth and movements of a population of eastern painted turtles in Rhode Island. His research now focuses on organismal biology, blending physiology, morphology, behavior, and ecology in an evolutionary perspective. Undergraduate students regularly participate in his research and are coauthors of many of his publications. He especially enjoys teaching undergraduates and has taught introductory biology and courses in vertebrate zoology, functional ecology, herpetology, environmental physiology, and animal behavior. He has published more than a hundred papers reporting the results of field and laboratory studies of turtles, snakes, lizards, frogs, and tuatara that have taken him to Australia, New Zealand, Fiji, Mexico, Costa Rica, Panama, and the Caribbean, as well as most parts of the United States. He is a Fellow of the American Association for the Advancement of Science and a past President of the American Society of Ichthyologists and Herpetologists. Christine M. Janis is a Professor of Biology at Brown University, where she teaches comparative anatomy and vertebrate evolution and is a recipient of the Elizabeth Leduc Prize for Distinguished Teaching in the Life Sciences. A British citizen, she obtained her bachelor's degree at Cambridge University and then crossed the pond to get her Ph.D. at Harvard University. She is a vertebrate paleontologist with a particular interest in mammalian evolution and faunal responses to climatic change. She first became interested in vertebrate evolution after seeing the movie Fantasia at the impressionable age of seven. That critical year was also the year that she began riding lessons, and she has owned at least one horse since the age of twelve. Many years later, she is now an expert on ungulate (hoofed mammal) evolution and has recently expanded her interests to the evolution of the Australian mammal fauna, especially the kangaroos. She is a Fellow of the Paleontological Society and is currently President of the Society for the Study of Mammalian Evolution. She attributes her life history to the fact that she has failed to outgrow either the dinosaur phase or the horse phase. John B. Heiser was born and raised in Indiana and completed his undergraduate degree in biology at Purdue University. He earned his Ph.D. in ichthyology from Cornell University for studies of the behavior, evolution, and ecology of coral reef fishes, research that he continues today with colleagues specializing in molecular biology. For fifteen years, he was Director of the Shoals Marine Laboratory operated by Cornell University and the University of New Hampshire on the Isles of Shoals in the Gulf of Maine. While at the Isles of Shoals, his research interests focused on opposite ends of the vertebrate spectrum—hagfish and baleen whales. He enjoys teaching vertebrate morphology, evolution, and ecology, both in the campus classroom and in the field, and is a recipient of the Clark Distinguished Teaching Award from Cornell University. His hobbies are natural history, travel and nature photography, and videography, especially underwater using scuba. He has pursued his natural history interests on every continent and all the world's major ocean regions. Because of his experience, he is a popular ecotourism leader, having led Cornell Adult University groups to the Caribbean, Sea of Cortez, French Polynesia, Central America, the Amazon, Borneo, Antarctica, and Spitsbergen in the High Arctic. iii Brief Contents PART I Vertebrate Diversity, Function, and Evolution 1 1 The Diversity, Classification, and Evolution of Vertebrates 2 2 Vertebrate Relationships and Basic Structure 19 3 Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates PART II Non-Amniotic Vertebrates: Fishes and Amphibians 4 5 6 7 8 9 10 Living in Water 72 Radiation of the Chondrichthyes 99 Dominating Life in Water: The Major Radiation of Fishes Geography and Ecology of the Paleozoic Era 158 Living on Land 167 Origin and Radiation of Tetrapods 189 Salamanders, Anurans, and Caecilians 211 122 PART III Sauropsida: Turtles, Lepidosaurs, and Archosaurs 11 12 13 14 15 16 17 Synapsids and Sauropsids: Two Approaches to Terrestrial Life Turtles 287 The Lepidosaurs: Tuatara, Lizards, and Snakes 310 Ectothermy: A Low-Cost Approach to Life 349 Geography and Ecology of the Mesozoic Era 364 Mesozoic Diapsids: Dinosaurs, Crocodilians, Birds, and Others Avian Specializations 407 PART IV Synapsida: The Mammals 18 19 20 21 22 23 24 25 Glossary Credits Index iv 447 The Synapsida and the Evolution of Mammals 448 Geography and Ecology of the Cenozoic Era 471 Mammalian Diversity and Characteristics 480 Mammalian Specializations 513 Endothermy: A High-Energy Approach to Life 537 Body Size, Ecology, and Sociality of Mammals 559 Primate Evolution and the Emergence of Humans 581 The Impact of Humans on Other Species of Vertebrates 614 Appendix A-1 G-1 C-1 I-1 71 253 254 371 47 Contents Preface xv PART I Vertebrate Diversity, Function, and Evolution 1 The Diversity, Classification, and Evolution of Vertebrates 1.1 The Vertebrate Story 2 1.2 Classification of Vertebrates 8 1.3 Phylogenetic Systematics 8 1.4 The Problem with Fossils: Crown and Stem Groups 1.5 Evolutionary Hypotheses 12 1.6 Earth History and Vertebrate Evolution 15 Summary 16 Discussion Questions 16 Additional Information 17 2 Vertebrate Relationships and Basic Structure 2.1 Vertebrates in Relation to Other Animals 2.2 Definition of a Vertebrate 24 2.3 Basic Vertebrate Structure 25 Summary 45 Discussion Questions 45 Additional Information 46 3 1 2 11 19 19 Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates 47 3.1 3.2 3.3 Reconstructing the Biology of the Earliest Vertebrates 47 Extant Jawless Fishes 50 The Importance of Extant Jawless Vertebrates in Understanding Ancient Vertebrates 56 3.4 The Radiation of Paleozoic Jawless Vertebrates—“Ostracoderms” 3.5 The Basic Gnathostome Body Plan 57 3.6 The Transition from Jawless to Jawed Vertebrates 62 3.7 Extinct Paleozoic Jawed Fishes 66 Summary 68 Discussion Questions 69 Additional Information 69 57 v PART II Non-Amniotic Vertebrates: Fishes and Amphibians 4 Living in Water 71 72 4.1 The Aquatic Environment 72 4.2 Water and the Sensory World of Fishes 77 4.3 The Internal Environment of Vertebrates 84 4.4 Exchange of Water and Ions 84 4.5 Responses to Temperature 90 4.6 Body Size and Surface/Volume Ratio 94 Summary 96 Discussion Questions 97 Additional Information 97 5 Radiation of the Chondrichthyes 99 5.1 Chondrichthyes—The Cartilaginous Fishes 99 5.2 Evolutionary Diversification of Chondrichthyes 102 5.3 The Paleozoic Chondrichthyan Radiation 102 5.4 The Early Mesozoic Elasmobranch Radiation 105 5.5 Extant Lineages of Elasmobranchs 106 5.6 Batoidea: Skates and Rays 116 5.7 Holocephali—The Little Known Chondrichthyans 118 Summary 119 Discussion Questions 119 Additional Information 121 6 Dominating Life in Water: The Major Radiation of Fishes 122 6.1 The Origin of Bony Fishes 122 6.2 Evolution of the Actinopterygii 126 6.3 Extant Actinopterygii—Ray-Finned Fishes 131 6.4 Locomotion in Water 137 6.5 Actinopterygian Reproduction 141 6.6 The Adaptable Fishes—Teleosts in Contrasting Environments 6.7 Conservation of Fishes 150 6.8 Sarcopterygii—The Lobe-Finned Fishes 153 Summary 156 Discussion Questions 156 Additional Information 157 7 Geography and Ecology of the Paleozoic Era 7.1 7.2 7.3 7.4 vi Contents 147 158 Earth History, Changing Environments, and Vertebrate Evolution Continental Geography of the Paleozoic 159 Paleozoic Climates 162 Paleozoic Terrestrial Ecosystems 163 158 7.5 Paleozoic Extinctions 165 Additional Information 166 8 9 10 Living on Land 167 8.1 Support and Locomotion on Land 167 8.2 Eating on Land 175 8.3 Reproduction on Land 178 8.4 Breathing Air 178 8.5 Pumping Blood Uphill 178 8.6 Sensory Systems in Air 182 8.7 Conserving Water in a Dry Environment 184 8.8 Controlling Body Temperature in a Changing Environment Summary 187 Discussion Questions 188 Additional Information 188 Origin and Radiation of Tetrapods 184 189 9.1 Tetrapod Origins 189 9.2 Radiation and Diversity of Non-Amniotic Paleozoic Tetrapods 9.3 Amniotes 201 Summary 208 Discussion Questions 209 Additional Information 209 Salamanders, Anurans, and Caecilians 197 211 10.1 Amphibians 211 10.2 Diversity of Life Histories of Amphibians 222 10.3 Amphibian Metamorphosis 237 10.4 Exchange of Water and Gases 238 10.5 Poison Glands and Other Defense Mechanisms 243 10.6 Mimicry 246 10.7 Why Are Amphibians Vanishing? 246 Summary 250 Discussion Questions 250 Additional Information 251 PART III Sauropsida: Turtles, Lepidosaurs, and Archosaurs 11 253 Synapsids and Sauropsids: Two Approaches to Terrestrial Life 254 11.1 Taking Advantage of the Opportunity for Sustained Locomotion 255 11.2 Increasing Gas Exchange: The Trachea and Lungs 259 Contents vii 12 13 14 15 11.3 Transporting Oxygen to the Muscles: Structure of the Heart 265 11.4 Taking Advantage of Wasted Energy: Endothermy 266 11.5 Getting Rid of Wastes: The Kidneys and Bladder 271 11.6 Sensing and Making Sense of the World: Eyes, Ears, Tongues, Noses, and Brains Summary 284 Discussion Questions 284 Additional Information 285 Turtles 287 12.1 Everyone Recognizes a Turtle 287 12.2 But What Is a Turtle? Phylogenetic Relationships of Turtles 12.3 Turtle Structure and Function 291 12.4 Ecology and Behavior of Turtles 298 12.5 Reproductive Biology of Turtles 300 12.6 Hatching and the Behavior of Baby Turtles 302 12.7 Conservation of Turtles 306 Summary 307 Discussion Questions 308 Additional Information 308 The Lepidosaurs: Tuatara, Lizards, and Snakes Contents 310 13.1 The Lepidosaurs 310 13.2 Radiation of Sphenodontids and the Biology of Tuatara 311 13.3 Radiation of Squamates 312 13.4 Ecology and Behavior of Squamates 321 13.5 Behavioral Control of Body Temperatures by Ectotherms 338 13.6 Temperature and Ecology of Squamates 342 Summary 346 Discussion Questions 347 Additional Information 347 Ectothermy: A Low-Cost Approach to Life 349 14.1 Vertebrates and Their Environments 349 14.2 Dealing with Dryness—Ectotherms in Deserts 350 14.3 Coping with Cold—Ectotherms in Subzero Conditions 356 14.4 The Role of Ectothermic Tetrapods in Terrestrial Ecosystems 359 Summary 362 Discussion Questions 362 Additional Information 363 Geography and Ecology of the Mesozoic Era 15.1 Mesozoic Continental Geography 365 15.2 Mesozoic Terrestrial Ecosystems 366 viii 290 364 281 15.3 Mesozoic Climates 368 15.4 Mesozoic Extinctions 369 Additional Information 370 16 Mesozoic Diapsids: Dinosaurs, Crocodilians, Birds, and Others 371 16.1 The Mesozoic Fauna 371 16.2 Characteristics of Diapsids 372 16.3 Marine Lineages 372 16.4 Semiaquatic and Terrestrial Diapsids: Crocodylomorpha 379 16.5 The First Evolution of Flight: Pterosauria 382 16.6 Dinosaurs 384 16.7 Terrestrial Herbivores: Ornithischian and Sauropod Saurischian Dinosaurs 16.8 Terrestrial Carnivores: Theropod Dinosaurs 397 16.9 The Second Evolution of Flight: Birds 401 Summary 404 Discussion Questions 404 Additional Information 405 17 Avian Specializations 388 407 17.1 Early Birds and Extant Birds 407 17.2 The Structure of Birds 410 17.3 Wings and Flight 415 17.4 The Hindlimbs 422 17.5 Feeding and Digestion 425 17.6 Sensory Systems 429 17.7 Social Behavior 432 17.8 Mating Systems 434 17.9 Oviparity 435 17.10 Nests 437 17.11 Orientation and Navigation 440 17.12 Migration 441 Summary 444 Discussion Questions 445 Additional Information 445 PART IV Synapsida: The Mammals 18 447 The Synapsida and the Evolution of Mammals 448 18.1 The Origin of Synapsids 448 18.2 Diversity of Nonmammalian Synapsids 450 18.3 Evolutionary Trends in Synapsids 455 Contents ix 18.4 The First Mammals 462 18.5 The Radiation of Mesozoic Mammals Summary 468 Discussion Questions 469 Additional Information 469 19 466 Geography and Ecology of the Cenozoic Era 471 19.1 Cenozoic Continental Geography 472 19.2 Cenozoic Terrestrial Ecosystems 473 19.3 Cenozoic Climates 475 19.4 Cenozoic Extinctions 477 Additional Information 479 20 Mammalian Diversity and Characteristics 480 20.1 Major Lineages of Mammals 480 20.2 Mammalian Ordinal Diversity 481 20.3 Features Shared by All Mammals 488 20.4 Features That Differ Among Mammalian Groups 20.5 Cenozoic Mammal Evolution 503 Summary 511 Discussion Questions 511 Additional Information 511 21 Mammalian Specializations 500 513 21.1 Mammalian Reproduction 513 21.2 Some Extreme Placental Mammal Reproductive Specializations 21.3 Are Placental Mammals Reproductively Superior to Marsupials? 21.4 Specializations for Feeding 520 21.5 Locomotor Specializations 527 21.6 Evolution of Aquatic Mammals 530 Summary 535 Discussion Questions 535 Additional Information 536 22 Endothermy: A High-Energy Approach to Life 22.1 Endothermic Thermoregulation 537 22.2 Endotherms in the Cold 540 22.3 Avoiding Cold and Sharing Heat 541 22.4 Facultative Heterothermy 543 22.5 Migration to Avoid Cold 547 22.6 Endotherms in the Heat 549 Summary 557 x Contents 537 518 519 Discussion Questions 557 Additional Information 558 23 Body Size, Ecology, and Sociality of Mammals 559 23.1 Social Behavior 559 23.2 Population Structure and the Distribution of Resources 560 23.3 Advantages of Sociality 564 23.4 Body Size, Diet, and the Structure of Social Systems 565 23.5 Horns and Antlers 570 23.6 Primate Societies 573 Summary 579 Discussion Questions 579 Additional Information 579 24 Primate Evolution and the Emergence of Humans 581 24.1 Primate Origins and Diversification 581 24.2 Origin and Evolution of the Hominoidea 590 24.3 Origin and Evolution of Humans 594 24.4 Derived Hominins (the Genus Homo) 600 24.5 Evolution of Human Characteristics 605 24.6 How Many Species of Humans Were Contemporaneous? Summary 611 Discussion Questions 611 Additional Information 612 25 The Impact of Humans on Other Species of Vertebrates 610 614 25.1 Humans and the Pleistocene Extinctions 615 25.2 Humans and Recent Extinctions 617 25.3 Global Climate Change and Vertebrates 620 25.4 Organismal Biology and Conservation 624 25.5 Captive Breeding 628 25.6 The Paradoxes of Conservation 631 Summary 632 Discussion Questions 633 Additional Information 633 Appendix Glossary Credits Index A-1 G-1 C-1 I-1 Contents xi This page intentionally left blank Preface T he theme of Vertebrate Life is organismal biology—that is, how the anatomy, physiology, ecology, and behavior of animals interact to produce organisms that function effectively in their environments and how lineages of organisms change through evolutionary time. The ninth edition emphasizes advances in our understanding of vertebrates since the eighth edition was published. Several topics have been greatly expanded. New to This Editions • Molecular biology. Molecular studies have produced • • • • new information about phylogenetic relationships that illuminates events as distant as the origin of jawless vertebrate lineages and as recent as the separation of the clouded leopards on Borneo, Sumatra, and the Asian mainland. Fossil evidence. Newly described fossils have expanded our understanding of the evolutionary diversity of vertebrate lineages, particularly of dinosaurs (including birds) and our own human lineage. Climate change. The ever-increasing evidence of global climate change has important implications for the biology and conservation of vertebrates, and new information about atmospheric conditions during the Paleozoic and Mesozoic eras sheds light on vertebrate diversification. Conservation. As the pace of extinction quickens, specific situations raise acute concerns: the global decline in amphibian populations, part of which can be traced to the worldwide spread of a fungal infection; the threats posed to fisheries by fish farming and transgenic fishes; and the difficulty of preserving large animals that require huge home ranges, and especially the problems associated with large predators, such as tigers, that sometimes eat people. Access to information. The expansion of electronic databases and the accessibility of online resources give students increased access to the primary literature and authoritative secondary sources, and we have added more citations of printed and online journals and of websites to encourage students to explore these sources. • Discussion questions. We have found that openended discussion questions are an effective way to increase active learning in class meetings, and we have added discussion questions in this edition. (And, because we sometimes find ourselves wondering what response the author expected when we use questions from textbooks, we have provided answers to our questions in the Instructor Resource Center at www.pearsonhighered.com.) • Lists of derived characters accompanying the cladograms. The extensive legends accompanying the cladograms provide important information, but they break the flow of text in the chapters. In this edition we have moved the legends to an appendix. • Images from the text. The figures from the text can be downloaded in jpg and PowerPoint formats from the Instructor Resource Center at www.pearsonhighered.com. Amid all of these changes, the element of Vertebrate Life that has always been most important to the authors has remained constant: We are biologists because we care enormously about what we do and the animals we work with. We are deeply committed to passing on the fascination and sheer joy that we have experienced to new generations of biologists and to providing information and perspectives that will help them with the increasingly difficult task of ensuring that the enormous vigor and diversity of vertebrate life do not vanish. Acknowledgments We are very grateful to the excellent production team assembled by Pearson for this edition: Editor in Chief, Beth Wilbur; our outstanding Project Editor, Crystal Clifton; Photo Researcher, Maureen Spuhler; and Nesbitt Graphics. Their mastery of every step on the complex path from a manuscript to a bound copy of a book has been enormously comforting to the authors. We are especially pleased by the return of Jennifer Kane as the artist for this edition. Jennifer first met Vertebrate Life when she was a student, and she brings that perspective to her work. Jennifer combines the ability to render anatomical information accurately xiii with an empathy for vertebrates that allows her to produce drawings so lifelike that they appear ready to walk off the page. Writing a book with a scope as broad as this one requires the assistance of many people. We list below the colleagues who generously provided comments, suggestions, and photographs and who responded to our requests for information. Mike Barton, Centre College Robin Beck, University of New South Wales Kayla Bell, Rochester Institute of Technology Michael Bell, Stony Brook University Daniel Blackburn, Trinity College Elizabeth Brainerd, Brown University Larry Buckley, Rochester Institute of Technology Gabbie Burns, Rochester Institute of Technology Jennifer Burr, Nazareth College Nathan Cawley, Rochester Institute of Technology Jenny Clack, University of Cambridge Andy Derocher, University of Alberta Phillip Donoghue, University of Bristol Robert E. Espinoza, California State University, Northridge Colleen Farmer, University of Utah David Fastovsky, University of Rhode Island Patricia Freeman, University of Nebraska Gene Helfman, University of Georgia Chris Hess, Butler University Morna Hilderbrand, Rochester Institute of Technology Jill Holiday, University of Florida Anne Houtman, Rochester Institute of Technology Rebecca Hull, Rochester Institute of Technology Christopher Irish, Rochester Institute of Technology Tom Kemp, Oxford University Haeja Kessler, Rochester Institute of Technology xiv Preface Jamie Kraus, Rochester Institute of Technology Kyle Mara, Temple University Kaitlyn Moranz, Rochester Institute of Technology Phillip Motta, University of South Florida Henry Mushinksy, University of South Florida Warren Porter, University of Wisconsin Marilyn Renfree, University of Melbourne Tony Russell, University of Calgary Roger Seymour, University of Adelaide Joan Sharp, Simon Fraser University Adrienne Sherman, Rochester Institute of Technology Rick Shine, University of Sydney Susan Smith, Rochester Institute of Technology Harshita Sood, Rochester Institute of Technology Ben Speers-Roesch, University of British Columbia Karen Steudal, University of Wisconsin C. Richard Tracy, University of Nevada, Reno Jason Treberg, University of Manitoba Hugh Tyndale-Biscoe, Canberra, Australia David Wake, University of California, Berkeley John Waud, Rochester Institute of Technology Lisa Whitenack, Allegheny College Amy Yuhas, Rochester Institute of Technology We especially appreciate the work of reviewers of this text: Christine Dahlin, University of Pittsburgh at Johnstown; Margaret B. Fusari, University of California, Santa Cruz; Marion Preest, Keck Science Department of The Claremont Colleges. We are deeply indebted to Adwoa Boateng and Morna Hilderbrand at the Wallace Center, Rochester Institute of Technology. Their skills and assistance were essential to our extensive review of the enormous literature of vertebrate biology. PART Vertebrate Diversity, Function, and Evolution T I he more than 63,000 living species of vertebrates inhabit nearly every part of Earth, and other kinds of vertebrates that are now extinct lived in habitats that no longer exist. Increasing knowledge of the diversity of vertebrates was a product of the European exploration and expansion that began in the ﬁfteenth and sixteenth centuries. In the middle of the eighteenth century, Swedish naturalist Carolus Linnaeus developed a binominal classiﬁcation to catalog the varieties of animals and plants. Despite some problems in reﬂecting evolutionary relationships, the Linnaean system remains the basis for naming organisms today. A century later, Charles Darwin and Alfred Russel Wallace explained the diversity of plants and animals as the product of natural selection and evolution. In the early twentieth century, their work was coupled with the burgeoning information about mechanisms of genetic inheritance. This combination of genetics and evolutionary biology, known as the New Synthesis, or neo-Darwinism, continues to be the basis for understanding the mechanics of evolution. Methods of classifying animals also changed during the twentieth century; and classiﬁcation, which began as a way of trying to organize the diversity of organisms, has become a powerful tool for generating testable hypotheses about evolution. Vertebrate biology and the fossil record of vertebrates have been at the center of these changes in our view of life. Comparative studies of the anatomy, embryology, behavior, and physiology of living vertebrates have often supplemented the fossil record. These studies reveal that evolution acts by changing existing traits. All vertebrates share basic characteristics that are the products of their common ancestry, and the process of evolution can be analyzed by tracing the modiﬁcations of these characters. Thus, an understanding of vertebrate form and function is basic to understanding the evolution of vertebrates and the ecology and behavior of living species. 1 1 CHAPTER The Diversity, Classification, and Evolution of Vertebrates E volution is central to vertebrate biology because it provides a principle that organizes the diversity we see among living vertebrates and helps to ﬁt extinct forms into the context of living species. Classiﬁcation, initially a process of attaching names to organisms, has become a method of understanding evolution and planning strategies for conservation. Current views of evolution stress natural selection operating at the level of individuals as the predominant mechanism that produces changes in a population over time. The processes and events of evolution are intimately linked to the changes that have occurred on Earth during the history of vertebrates. These changes have resulted from the movements of continents and the eﬀects of those movements on climates and geography. In this chapter, we present an overview of the scene, the participants, and the events that have shaped the biology of vertebrates. 1.1 The Vertebrate Story Mention “animal” and most people will think of a vertebrate. Vertebrates are abundant and conspicuous parts of people’s experience of the 2 natural world. Vertebrates are also very diverse: The more than 63,000 extant (currently living) species of vertebrates range in size from ﬁshes weighing as little as 0.1 gram when fully mature to whales weighing over 100,000 kilograms. Vertebrates live in virtually all the habitats on Earth. Bizarre ﬁshes, some with mouths so large they can swallow prey larger than their own bodies, cruise through the depths of the sea, sometimes luring prey to them with glowing lights. Fifteen kilometers above the ﬁshes, migrating birds ﬂy over the crest of the Himalayas, the highest mountains on Earth. The behaviors of vertebrates are as diverse and complex as their body forms and habitats. Vertebrate life is energetically expensive, and vertebrates get the energy they need from food they eat. Carnivores eat the ﬂesh of other animals and show a wide range of methods of capturing prey. Some predators search the environment to ﬁnd prey, whereas others wait in one place for prey to come to them. Some carnivores pursue their prey at high speeds, and others pull prey into their mouths by suction. M a n y vertebrates swallow their prey intact, sometimes while it is alive and struggling, but other vertebrates have very speciﬁc methods of dispatching prey. Venomous snakes inject complex mixtures of toxins, and cats (of all sizes, from house cats to tigers) kill their prey with a distinctive bite on the neck. Herbivores eat plants. Plants cannot run away when an animal approaches, so they are easy to catch, but they are hard to chew and digest and frequently contain toxic compounds. Herbivorous vertebrates show an array of specializations to deal with the diﬃculties of eating plants. Elaborately sculptured teeth tear apart tough leaves and expose the surfaces of cells, but the cell walls of plants contain cellulose, which no vertebrate can digest. Herbivorous vertebrates rely on microorganisms living in their digestive tracts to digest cellulose. In addition, these endosymbionts (organisms that live inside another organism) detoxify the chemical substances that plants use to protect themselves. Reproduction is a critical factor in the evolutionary success of an organism, and vertebrates show an astonishing range of behaviors associated with mating and reproduction. In general, males court females and females care for the young, but these roles are reversed in many species of vertebrates. At the time of birth or hatching, some vertebrates are entirely self-suﬃcient and never see their parents, whereas other vertebrates (including humans) have extended periods of obligatory parental care. Extensive parental care is found in seemingly unlikely groups of vertebrates—ﬁshes that incubate eggs in their mouths, frogs that carry their tadpoles to water and then return to feed them, and birds that feed their nestlings a ﬂuid called crop milk that is very similar in composition to mammalian milk. The diversity of living vertebrates is enormous, but the species now living are only a small proportion of the species of vertebrates that have existed. For each living species, there may be more than a hundred extinct species, and some of these have no counterparts among living forms. For example, the dinosaurs that dominated Earth for 180 million years are so entirely diﬀerent from living animals that it is hard to reconstruct the lives they led. Even mammals were once more diverse than they are now. The Pleistocene epoch saw giants of many kinds—ground sloths as big as modern rhinoceroses and raccoons as large as bears. The number of species of terrestrial vertebrates probably reached its maximum in the middle Miocene epoch, 12 to 14 million years ago, and has been declining since then. The story of vertebrates is fascinating. Where they originated, how they evolved, what they do, and how they work provide endless intriguing details. In preparing to tell this story, we must introduce some basic information, including what the diﬀerent kinds of ver- tebrates are called, how they are classiﬁed, and what the world was like as the story of vertebrates unfolded. Major Extant Groups of Vertebrates Two major groups of vertebrates are distinguished on the basis of an innovation in embryonic development: the appearance of three membranes formed by tissues that come from the embryo itself. One of these membranes, the amnion, surrounds the embryo, and animals with this structure are called amniotes. The division between nonamniotes and amniotes corresponds roughly to aquatic and terrestrial vertebrates, although many amphibians and a few ﬁshes lay non-amniotic eggs in nests on land. Among the amniotes, we can distinguish two major evolutionary lineages—the sauropsids (reptiles, including birds) and the synapsids (mammals). These lineages separated from each other in the Late Devonian period, before vertebrates had developed many of the characters we see in extant species. As a result, synapsids and sauropsids represent parallel but independent origins of basic characters such as lung ventilation, kidney function, insulation, and temperature regulation. Figure 1–1 shows the major kinds of vertebrates and the approximate numbers of living species. In the following sections, we brieﬂy describe the diﬀerent kinds of vertebrates. Non-Amniotes The embryos of non-amniotes are enclosed and protected by membranes that are produced by the reproductive tract of the female. This is the condition seen among the invertebrate relatives of vertebrates, and it is retained in the non-amniotes: the ﬁshes and amphibians. Hagfishes and Lampreys—Myxiniformes and Petromyzontiformes Lampreys and hagﬁshes are elongate, limb- less, scaleless, and slimy and have no internal bony tissues. They are scavengers and parasites and are specialized for those roles. Hagﬁshes are marine, living on the seabed at depths of 100 meters or more. In contrast, many species of lampreys are migratory, living in oceans and spawning in rivers. Hagﬁshes and lampreys are unique among living vertebrates because they lack jaws; this feature makes them important in the study of vertebrate evolution. They have traditionally been grouped as agnathans (Greek a = without and gnath = jaw) or cyclostomes (Greek cyclo = round and stoma = mouth), but they are probably not closely related to each other and instead represent two independent (i.e., separate) evolutionary lineages. Sharks, Rays, and Ratfishes—Chondrichthyes The name Chondrichthyes (Greek chondro = cartilage and ichthyes = The Vertebrate Story 3 Myxiniformes and Petromyzontiformes 119 Chondrichthyes 1,201 Mammalia 5,488 m Mam als Aves ~10,000 Bir Actinopterygii 30,554 ds Fishes ibians es Crocodilia 23 Lepidosauria Amph til p Re Snakes 3,315 Anura 5,996 Lizards 5,461 Testudinia 313 Gymnophiona 186 Actinistia + Dipnoi 8 Caudata 619 Figure 1–1 Diversity of vertebrates. Areas in the diagram correspond to approximate numbers of living species in each group. (These are estimates, and the numbers change frequently as new species are described.) Common names are in the center circle, and formal names for the groups are on the outer circle. The two major lineages of extant vertebrates are the Actinopterygii (ray-ﬁnned ﬁshes) and the Sarcopterygii (lobe-ﬁnned ﬁshes), each of which includes more than 30,000 extant species. (The Sarcopterygii includes the lineages Actinistia, Dipnoi, Caudata, Anura, Gymnophiona, Testudinia, Lepidosauria, Crocodilia, Aves, and Mammalia.) ﬁsh) refers to the cartilaginous skeletons of these ﬁshes. Extant sharks and rays form a group called the Neoselachii (Greek neo = new and selach = shark), but the two kinds of ﬁshes diﬀer in body form and habits. Sharks have a reputation for ferocity that most species would have difﬁculty living up to. Some sharks are small (15 centimeters or less); and the largest species, the whale shark (which grows to 10 meters), is a ﬁlter feeder that subsists on plankton it strains from the water. Rays are mostly bottom feeders; they are dorsoventrally ﬂattened and swim with undulations of their extremely broad pectoral ﬁns. 4 CHAPTER 1 The second group of chondrichthyans, the ratﬁshes or chimaerans, gets its name, Holocephalii (Greek holo = whole and cephal = the head), from the single gill cover that extends over all four gill openings. These are bizarre marine animals with long, slender tails and bucktoothed faces that look rather like rabbits. They live on the seaﬂoor and feed on hard-shelled prey, such as crustaceans and mollusks. Bony Fishes—Osteichthyes Bony ﬁshes, the Osteich- thyes (Greek osteo = bone and ichthyes = ﬁsh), are so The Diversity, Classification, and Evolution of Vertebrates diverse that any attempt to characterize them brieﬂy is doomed to failure. Two broad categories can be recognized: the ray-ﬁnned ﬁshes (actinopterygians; Greek actino = ray and ptero = wing or ﬁn) and the lobe-ﬁnned or ﬂeshy-ﬁnned ﬁshes (sarcopterygians; Greek sarco = ﬂesh). The ray-ﬁnned ﬁshes have radiated extensively in fresh and salt water. More than 30,500 species of ray-ﬁnned ﬁshes have been named, and several thousand additional species may await discovery. A single project, the Census of Marine Life, is describing 150 to 200 previously unknown species of ray-ﬁnned ﬁshes annually. Two major groups can be distinguished among actinopterygians. The Chondrostei (bichirs, sturgeons, and paddleﬁshes) are survivors of an early radiation of bony ﬁshes. Bichirs are swamp- and river-dwellers from Africa; they are known as African reed ﬁsh in the aquarium trade. Sturgeons are large ﬁshes with protrusible, toothless mouths that are used to suck food items from the bottom. Sturgeons are the source of caviar—eggs are taken from the female before they are laid. Of course, this kills the female sturgeon, and many species have been driven close to extinction by overﬁshing. Paddleﬁshes (two species, one in the Mississippi drainage of North America and another nearly extinct species in the Yangtze River of China) have a paddlelike snout with organs that locate prey by sensing electrical ﬁelds. The Neopterygii, the modern radiation of ray-ﬁnned ﬁshes, can be divided into three lineages. Two of these— the gars and the bowﬁns—are relicts of earlier radiations. These ﬁshes have cylindrical bodies, thick scales, and jaws armed with sharp teeth. They seize prey in their mouths with a sudden rush or gulp, and they lack the specializations of the jaw apparatus that allow later bony ﬁshes to use more complex feeding modes. The third lineage of neopterygians, the Teleostei, includes almost 30,500 species of ﬁshes covering every imaginable combination of body size, habitat, and habits. Most of the ﬁshes that people are familiar with are teleosts—the trout, bass, and panﬁsh that anglers seek; the sole (a kind of ﬂounder) and swordﬁsh featured by seafood restaurants; and the salmon and tuna whose by-products ﬁnd their way into canned catfood. Modiﬁcations of the body form and jaw apparatus have allowed many teleosts to be highly specialized in their swimming and feeding habits. In one sense, only eight species of lobe-ﬁnned ﬁshes survive: the six species of lungﬁshes (Dipnoi) found in South America, Africa, and Australia and the two species of coelacanths (Actinistia), one from deep waters oﬀ the east coast of Africa and a second species recently discovered near Indonesia. These are the living ﬁshes most closely related to terrestrial vertebrates, and a more accurate view of sarcopterygian diversity includes their terrestrial descendants—amphibians, mammals, turtles, lepidosaurs (the tuatara, lizards, and snakes), crocodilians, and birds. From this perspective, bony ﬁshes include two major evolutionary radiations—one in the water and the other on land, with each containing more than 30,000 living species. Salamanders, Frogs, and Caecilians—Urodela, Anura, and Gymnophiona These three groups of vertebrates are popularly known as amphibians (Greek amphi = double and bios = life) in recognition of their complex life histories, which often include an aquatic larval form (the larva of a salamander or caecilian and the tadpole of a frog) and a terrestrial adult. All amphibians have bare skins (i.e., lacking scales, hair, or feathers) that are important in the exchange of water, ions, and gases with their environment. Salamanders are elongate animals, mostly terrestrial, and usually with four legs; anurans (frogs, toads, and tree frogs) are short-bodied, with large heads and large hind legs used for walking, jumping, and climbing; and gymnophians (caecilians) are legless aquatic or burrowing animals. Amniotes An additional set of membranes associated with the embryo appeared during the evolution of vertebrates. They are called fetal membranes because they are derived from the embryo itself rather than from the reproductive tract of the mother. The amnion is one of these membranes, and vertebrates with an amnion are called amniotes. In general, amniotes are more terrestrial than non-amniotes; but there are also secondarily aquatic species of amniotes (such as sea turtles and whales) as well as many species of salamanders and frogs that spend their entire lives on land despite being non-amniotes. However, many features distinguish non-amniotes (ﬁshes and amphibians) from amniotes (mammals and reptiles, including birds), and we will use the terms to identify which of the two groups is being discussed. By the Permian, amniotes were well established on land. They ranged in size from lizardlike animals a few centimeters long through cat- and dog-size species to the cow-size parieasaurs. Some were herbivores; others were carnivores. In terms of their physiology, however, we can infer that they retained ancestral characters. They had scale-covered skins without an insulating layer of hair or feathers, a simple kidney that could not produce highly concentrated urine, simple lungs, and a heart in which the ventricle was not divided by a septum. Early in their evolutionary history, terrestrial vertebrates split into two lineages—the synapsids (now represented by mammals) and the sauropsids (the The Vertebrate Story 5 reptiles, including birds). Terrestrial life requires lungs to extract oxygen from air, hearts that can separate oxygen-rich arterial blood from oxygen-poor venous blood, kidneys that can eliminate waste products while retaining water, and insulation and behaviors to keep body temperature stable as the external temperature changes. These features evolved in both lineages; but, because synapsids and sauropsids evolved terrestrial specializations independently, the lungs, hearts, kidneys, and body coverings of synapsids and sauropsids are diﬀerent. Sauropsid Amniotes Extant sauropsids are the animals we call reptiles: turtles, the scaly reptiles (tuatara, lizards, and snakes), crocodilians, and birds. Extinct sauropsids include the forms that dominated the world during the Mesozoic era—dinosaurs and pterosaurs (flying reptiles) on land and a variety of marine forms, including ichthyosaurs and plesiosaurs, in the oceans. Turtles—Testudinia Turtles (Latin testudo = a turtle) are probably the most immediately recognizable of all vertebrates. The shell that encloses a turtle has no exact duplicate among other vertebrates, and the morphological modiﬁcations associated with the shell make turtles extremely peculiar animals. They are, for example, the only vertebrates with the shoulders (pectoral girdle) and hips (pelvic girdle) inside the ribs. Tuatara, Lizards, and Snakes—Lepidosauria These three kinds of vertebrates can be recognized by their scalecovered skin (Greek lepido = scale and saur = lizard) as well as by characteristics of the skull. The tuatara, a stocky-bodied animal found only on some islands near New Zealand, is the sole living remnant of an evolutionary lineage of animals called Sphenodontida, which was more diverse in the Mesozoic. In contrast, lizards and snakes (which are highly specialized lizards) are now at the peak of their diversity. Alligators and Crocodiles—Crocodilia These impressive animals, which draw their name from crocodilus, the Greek word for crocodile, are in the same evolutionary lineage (the Archosauria) as dinosaurs and birds. The extant crocodilians are semiaquatic predators, with long snouts armed with numerous teeth. They range in size from the saltwater crocodile, which has the potential to grow to a length of 7 meters, to dwarf crocodiles and caimans that are less than a meter long. Their skin contains many bones (osteoderms; Greek osteo = bone and derm = skin) that lie beneath the scales and provide a kind of armor plating. Crocodilians are noted for the parental care they provide for their eggs and young. 6 CHAPTER 1 Birds—Aves The birds (Latin avis = a bird) are a lineage of dinosaurs that evolved ﬂight in the Mesozoic. Feathers are characteristic of extant birds, and feathered wings are the structures that power a bird’s ﬂight. Recent discoveries of dinosaur fossils with traces of feathers show that feathers evolved before ﬂight. This oﬀset between the times that feathers and ﬂight appeared illustrates an important principle: the function of a trait in an extant species is not necessarily the same as its function when it ﬁrst appeared. In other words, current utility is not the same as evolutionary origin. The original feathers were almost certainly structures that were used in courtship displays, and their modiﬁcation as airfoils, for streamlining, and as insulation in birds is a secondary event. Synapsid Amniotes The synapsid lineage contains the three kinds of extant mammals: the monotremes (prototheria; the platypus and echidna), marsupials (metatherians), and placentals (eutherians). Extinct synapsids include forms that diversiﬁed in the Paleozoic era—pelycosaurs and therapsids—and the rodentlike multituberculates of the late Mesozoic. Mammals—Mammalia The living mammals (Latin mamma = a teat) can be traced to an origin in the late Paleozoic, from some of the earliest fully terrestrial vertebrates. Extant mammals include about 5500 species, most of which are placental mammals. Both placentals and marsupials possess a placenta, a structure that transfers nutrients from the mother to the embryo and removes the waste products of the embryo’s metabolism. Placentals have an extensive system of placentation and a long gestation period, whereas marsupials have a short gestation period and give birth to very immature young that continue their development attached to a nipple, often in a pouch on the mother’s abdomen. Marsupials dominate the mammalian fauna only in Australia. Kangaroos, koalas, and wombats are familiar Australian marsupials. The strange monotremes, the platypuses and the echidnas, are mammals whose young are hatched from eggs. All mammals, including monotremes, feed their young with milk. New Species New species of vertebrates are described weekly—this is why we use the words “approximately” and “about” when we cite the numbers of species. More than 300 new species of ﬁshes are described annually. In 2002, a molecular analysis of the relationships of rhacophorid frogs on Madagascar increased the number of species from 18 to more than 100. The Diversity, Classification, and Evolution of Vertebrates Among terrestrial mammals rodents led the way with 174 new species described between 1993 and 2008, and bats were second with 94 species. Most rodents are small and bats are nocturnal, so it is not surprising to ﬁnd new species in those groups. It is, however, something of a surprise to ﬁnd that primates, with 55 new species, followed bats in the list because primates are mostly diurnal and large enough to be conspicuous elements of a fauna; other newly identiﬁed species are also large. • In 2005, a new species of mangabey monkey as big as (a) a medium-size dog was discovered in Tanzania. Subsequent study showed that it was so diﬀerent from related species that a new genus, Rungwecebus, was created for it. In addition to being large, this monkey has a loud vocalization and occurs in forest adjacent to cultivated areas. In fact, the type specimen was captured in a trap set by a farmer in a ﬁeld of maize. • Three new species of whales have been described since 2000—two rorquals (Balaenoptera omurai from the Indo-Paciﬁc region and B. edeni, which has a worldwide distribution) and a new right whale (Eubalaena japonica) from the North Paciﬁc. These animals are from 11 to 17 meters long. • In 2001 the largest extant land mammal, the African elephant, was shown to consist of two distinct species: the African savannah elephant (Loxodonta africana) and the African forest elephant (Loxodonta cyclotis). (b) Many of these new species occur in isolated populations and some are on the brink of extinction. For example, 221 of the 408 new species of terrestrial mammals that were described between 1993 and 2008 occur in isolated populations, and 34 of those new species are considered to be at risk of extinction. This situation places a biologist who discovers a new species in a paradoxical situation—on one hand, zoological convention calls for killing an individual of the new species and depositing it in a museum to serve as the holotype— that is, the single specimen on which the species is based in the original publication. On the other hand, even one individual may be a signiﬁcant loss from a small population. The descriptions of two new species of primates and a lizard have been based on photographs, measurements, and tissue samples from living individuals that were released instead of being killed (Figure 1–2). (The validity of designating a living holotype rests on interpretation of article 73.1.4 of the 1999 edition of the International Code of Zoological Nomenclature, and this practice has generated a lively controversy.) (c) Figure 1–2 Three new species of vertebrates that were described on the basis of living holotypes. (a) The highland mangabey, Rungwecebus kipunji. The species was initially described on the basis of this photograph. Later, another individual that had been killed by a farmer became available for study. (b) The blonde capuchin monkey, Cebus queirozi. These are two of the images from the series of photographs of an anesthetized monkey in the description of the species. (The monkey was released after it recovered consciousness.) (c) The pink Galápagos land iguana, Conolophus marthae. An electronic identiﬁcation tag has been implanted in this individual. When it dies a natural death, its remains will be deposited in a museum. The Vertebrate Story 7 1.2 Classification of Vertebrates The diversity of vertebrates (more than 63,000 living species and perhaps a hundred times that number of species now extinct) makes the classiﬁcation of vertebrates an extraordinarily diﬃcult task. Yet classiﬁcation has long been at the heart of evolutionary biology. Initially, classiﬁcation of species was seen as a way of managing the diversity of organisms, much as an ofﬁce ﬁling system manages the paperwork of the oﬃce. Each species could be placed in a pigeonhole marked with its name; when all species were in their pigeonholes, the diversity of vertebrates would have been encompassed. This approach to classiﬁcation was satisfactory as long as species were regarded as static and immutable: once a species was placed in the ﬁling system, it was there to stay. Acceptance of the fact that species evolve has made that kind of classiﬁcation inadequate. Now biologists must express evolutionary relationships among species by incorporating evolutionary information in the system of classiﬁcation. Ideally, a classiﬁcation system should not only attach a label to each species but also encode the evolutionary relationships between that species and other species. Modern techniques of systematics (the evolutionary classiﬁcation of organisms) have become methods for generating testable hypotheses about evolution. Classification and Names Our system of naming species is pre-Darwinian. It traces back to methods established by the naturalists of the seventeenth and eighteenth centuries, especially those of Carl von Linné, a Swedish naturalist, better known by his Latin pen name, Carolus Linnaeus. The Linnaean system employs binominal nomenclature to designate species and arranges species into hierarchical categories (taxa, singular taxon) for classiﬁcation. Binominal Nomenclature The scientiﬁc naming of species became standardized when Linnaeus’s monumental work, Systema Naturae (The System of Nature), was published in sections between 1735 and 1758. Linnaeus attempted to give an identifying name to every known species of plant and animal. His method assigns a binominal (two-word) name to each species. Familiar examples include Homo sapiens for human beings (Latin hom = human and sapien = wise), Passer domesticus for the house sparrow (Latin passer = sparrow and domesticus = belonging to 8 CHAPTER 1 the house), and Canis familiaris for the domestic dog (Latin canis = dog and familiaris = of the family). Why use Latin words? Latin was the early universal language of European scholars and scientists. It has provided a uniform usage that scientists, regardless of their native language, continue to recognize worldwide. The same species may have diﬀerent colloquial names, even in the same language. For example, Felis concolor (Latin for “the uniformly colored cat”) is known in various parts of North America as cougar, puma, mountain lion, American panther, painter, and catamount. In Central and South America it is called león colorado, onça-vermelha, poema, guasura, or yaguá-pitá. But biologists of all nationalities recognize the name Felis concolor as referring to a speciﬁc kind of cat. Hierarchical Groups Linnaeus and other naturalists of his time developed what they called a natural system of classiﬁcation. The species is the basic level of biological classiﬁcation, but the deﬁnition of a species has been contentious, partly because criteria that have been used to identify extant species (e.g., reproductive isolation from other species) don’t work for fossil species and don’t always correspond to genetic diﬀerences. Similar species are grouped together in a genus (plural genera), based on characters that deﬁne the genus. The most commonly used characters were anatomical because they can be most easily preserved in museum specimens. Thus all doglike species—various wolves, coyotes, and jackals— were grouped together in the genus Canis because they all share certain anatomical features, such as an erectile mane on the neck and a skull with a long, prominent sagittal crest on the top from which massive temporal (jaw-closing) muscles originate. Linnaeus’s method of grouping species was functional because it was based on anatomical (and to some extent on physiological and behavioral) similarities and diﬀerences. Linnaeus lived before there was any knowledge of genetics and the mechanisms of inheritance, but he used characters that we understand today are genetically determined biological traits that generally express the degree of genetic similarity or diﬀerence among groups of organisms. Genera are placed in families, families in orders, orders in classes, and animal classes in phyla (singular phylum). 1.3 Phylogenetic Systematics All methods of classifying organisms, even preLinnaean systems, are based on similarities among the included species, but some similarities are more The Diversity, Classification, and Evolution of Vertebrates signiﬁcant than others. For example, nearly all vertebrates have paired limbs, but only a few kinds of vertebrates have mammary glands. Consequently, knowing that the species in question have mammary glands tells you more about the closeness of their relationship than knowing that they have paired limbs. You would thus give more weight to the presence of mammary glands than to paired limbs. A way to assess the relative importance of diﬀerent characteristics was developed in the mid-twentieth century by Willi Hennig, who introduced a method of determining evolutionary relationships called phylogenetic systematics (Greek phyla = tribe and genesis = origin). An evolutionary lineage is a clade (from cladus, the Greek word for a branch), and phylogenetic systematics is also called cladistics. Cladistics recognizes only groups of organisms that are related by common descent. The application of cladistic methods has made the study of evolution rigorous. The groups of organisms recognized by cladistics are called natural groups, and they are linked in a nested series of ancestor-descendant relationships that trace the evolutionary history of the group. Hennig’s contribution was to insist that these groups can be identiﬁed only on the basis of derived characters. “Derived” means “diﬀerent from the ancestral condition.” A derived character is called an apomorphy (Greek apo = away from [i.e., derived from] and morph = form, which is interpreted as “away from the ancestral condition”). For example, the feet of terrestrial vertebrates have distinctive bones—the carpals, tarsals, and digits. This arrangement of foot bones is diﬀerent from the ancestral pattern seen in lobe-ﬁnned ﬁshes, and all lineages of terrestrial vertebrates had that derived pattern of foot bones at some stage in their evolution. (Many groups of terrestrial vertebrates—horses, for example—have subsequently modiﬁed the foot bones, and some, such as snakes, have lost the limbs entirely. The signiﬁcant point is that those evolutionary lineages include species that had the derived terrestrial pattern.) Thus, the terrestrial pattern of foot bones is a shared derived character of terrestrial vertebrates. In cladistic terminology, shared derived characters are called synapomorphies (Greek syn = together, so synapomorphy can be interpreted as “together away from the ancestral condition”). Of course, organisms also share ancestral characters— that is, characters that they have inherited unchanged from their ancestors. These are called plesiomorphies (Greek plesi = near in the sense of “similar to the ancestor”). Terrestrial vertebrates have a vertebral column, for example, that was inherited from lobe-ﬁnned ﬁshes. Hennig called shared ancestral characters symplesiomorphies (sym, like syn, is a Greek root that means “together”). Symplesiomorphies tell us nothing about degrees of relatedness. The principle that only shared derived characters can be used to determine evolutionary relationships is the core of cladistics. The conceptual basis of cladistics is straightforward, although applying cladistic criteria to real organisms can become very complicated. To illustrate cladistic classiﬁcation, consider the examples presented in Figure 1–3. Each of the three cladograms (diagrams showing hypothetical sequences of branching during evolution) illustrates a possible evolutionary relationship for the three taxa (plural of taxon, which means a species or group of species), identiﬁed as 1, 2, and 3. To make the example a bit more concrete, we can consider three characters: the number of toes on the front foot, the skin covering, and the tail. For this example, let’s say that in the ancestral character state there are ﬁve toes on the front foot, and in the derived state there are four toes. We’ll say that the ancestral state is a scaly skin, and the derived state is a lack of scales. As for the tail, it is present in the ancestral state and absent in the derived state. Figure 1–3 shows the distribution of those three character states in the three taxa. The animals in taxon 1 have ﬁve toes on the front feet, lack scales, and have a tail. Animals in taxon 2 have ﬁve toes, scaly skins, and no tails. Animals in taxon 3 have four toes, scaly skins, and no tails. How can we use this information to decipher the evolutionary relationships of the three groups of animals? Notice that the derived number of toes occurs only in taxon 3, and the derived tail condition (absent) is found in taxa 2 and 3. The most parsimonious phylogeny (i.e., the evolutionary relationship requiring the fewest number of changes) is represented by Figure 1–3(a). Only three changes are needed to produce the derived character states: 1. In the evolution of taxon 1, scales are lost. 2. In the evolution of the lineage including taxon 2 + taxon 3, the tail is lost. 3. In the evolution of taxon 3, a toe is lost from the front foot. The other two phylogenies shown in Figure 1–3 are possible, but they would require tail loss to occur independently in taxon 2 and in taxon 3. Any change in a structure is an unlikely event, so the most plausible phylogeny is the one requiring the fewest changes. The second and third phylogenies each require four evolutionary changes, so they are less parsimonious than the ﬁrst phylogeny we considered. Phylogenetic Systematics 9 (a) 1 2 3 Scales→ no scales 5 toes→ 4 toes Tail→ no tail (b) 2 Tail→ no tail 1 Scales→ no scales 3 Tail→ no tail 5 toes→ 4 toes (c) 3 1 Tail→ no tail Scales→ no scales 2 Tail→ no tail 5 toes→ 4 toes Figure 1–3 Three cladograms showing the possible evolutionary relationships of three taxa. Bars connect derived characters (apomorphies). The black bar shows a shared derived character (a synapomorphy) of the lineage that includes taxa 2 and 3. Colored bars represent two independent origins of the same derived character state that must be assumed to have occurred if there was no apomorphy in the most recent common ancestor of taxa 2 and 3. The labels identify changes from the ancestral character state to the derived condition. Cladogram (a) requires a total of three changes from the ancestral condition to explain the distribution of characters in the extant taxa, whereas cladograms (b) and (c) require four changes. Because cladogram (a) is more parsimonious (i.e., requires the smallest number of changes), it is considered to be the most likely sequence of changes. A phylogeny is a hypothesis about the evolutionary relationships of the groups included. Like any scientiﬁc hypothesis, it can be tested when new data become available. If it fails that test, it is falsiﬁed; that is, it is rejected, and a diﬀerent hypothesis (a diﬀerent cladogram) takes its place. The process of testing hypotheses and replacing those that are falsiﬁed is a continuous one, and changes in the cladograms in suc10 CHAPTER 1 cessive editions of this book show where new information has generated new hypotheses. The most important contribution of phylogenetic systematics is that it enables us to frame testable hypotheses about the sequence of events during evolution. So far we have avoided a central issue of phylogenetic systematics: How do scientists know which character state is ancestral (plesiomorphic) and which is The Diversity, Classification, and Evolution of Vertebrates Batrachia 7. Lissamphibia Archosauria 11. Diapsida 10. Sauropsida Eutheria (placentals) Metatheria (marsupials) Monotremata (platypus, spiny anteaters) Aves (birds) Crocodilia (alligators and crocodiles) Anura (frogs) Caudata (salamanders) Gymnophiona (caecilians) Dipnoi (lungfishes) Actinistia (coelacanths) Actinopterygii (ray-finned fishes) Chondrichthyes (sharks, skates, rays, ratfishes) Petromyzontiformes (lampreys) Myxiniformes (hagfishes) Outgroups (tunicates and cephalochordates) Evolutionary lineages must have a single evolutionary origin; that is, they must be monophyletic (Greek mono = one, single) and include all the descendants of that Lepidosauria (tuatara, lizards, snakes) 1.4 The Problem with Fossils: Crown and Stem Groups Testudinia (turtles) ancestor. The cladogram depicted in Figure 1–4 is a hypothesis of the evolutionary relationships of the major living groups of vertebrates. A series of dichotomous branches extends from the origin of vertebrates to the groups of extant vertebrates. Cladistic terminology assigns names to the lineages originating at each branch point. This process produces a nested series of groups, starting with the most inclusive. For example, the Gnathostomata includes all vertebrate animals that have jaws; that is, every taxon above number 2 in Figure 1–4 is included in the Gnathostomata; every taxon above number 3 is included in the Osteichthyes (bony ﬁshes); and so on. Because the lineages are nested, it is correct to say that humans are both gnathostomes and osteichthyans. After number 6, the cladogram divides into Lissamphibia and Amniota, and humans are in the Amniota lineage. The cladogram divides again above number 9 into two lineages, the Sauropsida and Synapsida lineages. Humans are in the Eutheria, which is in the synapsid lineage. derived (apomorphic)? That is, how can we determine the direction (polarity) of evolutionary transformation of the characters? For that, we need additional information. Increasing the number of characters we are considering can help, but comparing the characters we are using with an outgroup that consists of the closest relatives of the ingroup (i.e., the organisms we are studying) is the preferred method. A well-chosen outgroup will possess ancestral character states compared to the ingroup. For example, lobe-ﬁnned ﬁshes are an appropriate outgroup for terrestrial vertebrates. Theria Synapsida (including Mammalia) 9. Amniota Tetrapoda 5. Rhipidistia 4. Sarcopterygii ? Osteichthyes 2. Gnathostomata 1. Vertebrata Figure 1–4 Phylogenetic relationships of extant vertebrates. This diagram depicts the probable relationships among the major groups of extant vertebrates. Note that the cladistic groupings are nested progressively; that is, all placental mammals are therians, all therians are synapsids, all synapsids are amniotes, all amniotes are tetrapods, and so on. In a phylogenetic classiﬁcation lineages can be named at each branching point, although it is not necessary to do so. The dashed line and question mark indicate uncertainty about the branching sequence for hagﬁshes. The Problem with Fossils: Crown and Stem Groups 11 This method of tracing ancestor-descendant relationships allows us to decipher evolutionary pathways that extend from fossils to living groups, but a diﬃculty arises when we try to ﬁnd names for groups that include fossils. The derived characters that deﬁne the extant groups of vertebrates did not necessarily appear all at the same time. On the contrary, evolution usually acts by gradual and random processes, and derived characters appear in a stepwise fashion. The extant members of a group have all of the derived characters of that group because that is how we deﬁne the group today; but, as you move backward through time to fossils that are ancestral to the extant species, you encounter forms that have a mosaic of ancestral and derived characters. The further back in time you go, the fewer derived characters the fossils have. What can we call the parts of lineages that contain these fossils? They are not included in the extant groups because they lack some of the derived characters of those groups, but the fossils in the lineage are more closely related to the extant group than they are to animals in other lineages. The solution to this problem lies in naming two types of groups: crown groups and stem groups. The crown groups are deﬁned by the extant species, the ones that have all the derived characters. The stem groups are the extinct forms that preceded the point at which the ﬁrst member of the crown group branched oﬀ. Basically, stem groups contain fossils with some derived characters, and crown groups contain extant species plus those fossils that have all of the derived characters of the extant species. Stem groups are paraphyletic (Greek para = beside, beyond); that is, they do not contain all of the descendants of the ancestor of the stem group plus the crown group because the crown group is excluded by deﬁnition. 1.5 Evolutionary Hypotheses Phylogenetic systematics is based on the assumption that organisms in a lineage share a common heritage, which accounts for their similarities. Because of that common heritage, we can use cladograms to ask questions about evolution. By examining the origin and signiﬁcance of characters of living animals, we can make inferences about the biology of extinct species. For example, the phylogenetic relationship of crocodilians, dinosaurs, and birds is shown in Figure 1–5. We know that both crocodilians and birds display extensive parental care of their eggs and young. Some fossilized dinosaur nests contain remains of baby dinosaurs, suggesting that at least some dinosaurs may also have cared for their young. Is that a plausible inference? Obviously there is no direct way to determine what sort of parental care dinosaurs had. The intermediate lineages in the cladogram (pterosaurs and dinosaurs) are extinct, so we cannot observe their reproductive behavior. However, the phylogenetic diagram in Figure 1–5 provides an indirect way to approach the question by examining the lineage that includes the closest living relatives of dinosaurs, crocodilians and Archosauria Dinosauria Saurischia Aves Lepidosauria Phytosaurs Crocodilians Pterosaurs Ornithischian dinosaurs Saurischian dinosaurs Most recent common ancestor of crocodilians and birds Figure 1–5 Using a cladogram to make inferences about behavior. The cladogram shows the relationships of the Archosauria, the evolutionary lineage that includes living crocodilians and birds. (Phytosaurs were crocodile-like animals that disappeared at the end of the Triassic, and pterosaurs were the ﬂying reptiles of the Jurassic and Cretaceous.) Both extant groups— crocodilians and birds—display extensive parental care of eggs and young. The most parsimonious explanation of this situation assumes that parental care is an ancestral character of the archosaur lineage. 12 CHAPTER 1 The Diversity, Classification, and Evolution of Vertebrates Birds birds. Crocodilians are more basal than pterosaurs and dinosaurs and birds are more derived; together crocodilians and birds form what is called an extant phylogenetic bracket. Both crocodilians and birds, the closest living relatives of the dinosaurs, do have parental care. Looking at living representatives of more distantly related lineages (outgroups), we see that parental care is not universal among ﬁshes, amphibians, or sauropsids other than crocodilians. The most parsimonious explanation of the occurrence of parental care in both crocodilians and birds is that it had evolved in that lineage before the crocodilians separated from dinosaurs + birds. (We cannot prove that parental care did not evolve separately in crocodilians and in birds, but one change to parental care is more likely than two changes.) Thus, the most parsimonious hypothesis is that parental care is a derived character of the evolutionary lineage containing crocodilians + dinosaurs + birds (the Archosauria). That means we are probably correct when we interpret the fossil evidence as showing that dinosaurs did have parental care. Figure 1–5 also shows how cladistics has made talking about restricted groups of animals more complicated than it used to be. Suppose you wanted to refer to just the two lineages of animals that are popularly known as dinosaurs—ornithischians and saurischians. What could you call them? Well, if you call them dinosaurs, you’re not being phylogenetically correct, because the Dinosauria lineage includes birds. So if you say dinosaurs, you are including ornithischians + saurischians + birds, even though any seven-year-old would understand that you are trying to restrict the conversation to extinct Mesozoic animals. In fact, there is no correct name in cladistic terminology for just the animals popularly known as dinosaurs. That’s because cladistics recognizes only monophyletic lineages, and a monophyletic lineage includes an ancestral form and all its descendants. The most recent common ancestor of ornithischians, saurischians, and birds in Figure 1–5 lies at the intersection of the lineage of ornithischians with saurischians + birds, so Dinosauria is a monophyletic lineage. If birds are omitted, however, all the descendants of the common ancestor are no longer included; and ornithischians + saurischians minus birds does not ﬁt the deﬁnition of a monophyletic lineage. It would be called a paraphyletic group. The stem groups discussed in the previous section are paraphyletic because they do not include all of the descendants of the fossil forms. Biologists who are interested in how organisms live often want to talk about paraphyletic groups. After all, the dinosaurs (in the popular sense of the word) differed from birds in many ways. The only correct way of referring to the animals popularly known as dinosaurs is to call them nonavian dinosaurs, and you will ﬁnd that and other examples of paraphyletic groups later in the book. Sometimes even this construction does not work because there is no appropriate name for the part of the lineage you want to distinguish. In this situation we will use quotation marks (e.g., “ostracoderms”) to indicate that the group is paraphyletic. Another important term is sister group. The sister group is the monophyletic lineage most closely related to the monophyletic lineage being discussed. In Figure 1–5, for example, the lineage that includes crocodilians + phytosaurs is the sister group of the lineage that includes pterosaurs + ornithischians + saurischians + birds. Similarly, pterosaurs are the sister group of ornithischians + saurischians + birds, ornithischians are the sister group of saurischians + birds, and saurischians are the sister group of birds. Determining Phylogenetic Relationships We’ve established that the derived characters systematists use to group species into higher taxa must be inherited through common ancestry. That is, they are homologous (Greek homo = same) similarities. In principle, that notion is straightforward; but in practice, the determination of common ancestry can be complex. For example, birds and bats have wings that are modiﬁed forelimbs, but the wings were not inherited from a common ancestor with wings. The evolutionary lineages of birds (Sauropsida) and bats (Synapsida) diverged long ago, and wings evolved independently in the two groups. This process is called convergent evolution. Parallel evolution describes the situation in which species that have diverged relatively recently develop similar specializations. The long hind legs that allow the North American kangaroo rats and the African jerboa to jump are an example of parallel evolution in these two lineages of rodents. A third mechanism, reversal, can produce similar structures in distantly related organisms. Sharks and cetaceans (porpoises and whales) have very similar body forms, but they arrived at that similarity from diﬀerent directions. Sharks retained an ancestral aquatic body form, whereas cetaceans arose from a lineage of terrestrial mammals with well-developed limbs that returned to an aquatic environment and reverted to the aquatic body form. Convergence, parallelism, and reversal are forms of homoplasy (Greek homo = same and plas = form, shape). Homoplastic similarities do not indicate common ancestry. Indeed, they complicate the process of deciphering evolutionary relationships. Convergence Evolutionary Hypotheses 13 Bornean Clouded Leopard Sumatran Clouded Leopard Mainland Clouded Leopard Mainland Clouded Leopard Leopard Snow Leopard Lion Jaguar Sumatran Clouded Leopard Tiger Bornean Clouded Leopard Domestic cat (a) (b) (c) (d) Figure 1–6 An example of the value of phylogenetic analyses in conservation. (a) A cladogram of clouded leopards and (b) a map showing the geographic locations of surviving populations. The clouded leopard found on the Asian mainland, Neofelis nebulosa (c), is as distant genetically from the species of clouded leopard found on Sumatra and Borneo, Neofelis diardi (d), as lions are from tigers. The island forms of Neofelis diardi are more closely related to each other than either is to the mainland form. Nonetheless, the genetic diﬀerences that distinguish the leopards on Sumatra from those on Borneo are large enough to be separated in captive breeding programs for the two forms. and parallelism give an appearance of similarity (as in the wings of birds and bats) that is not the result of common evolutionary origin. Reversal, in contrast, conceals similarity (e.g., between cetaceans and their four-legged terrestrial ancestors) that is the result of common evolutionary origin. Phylogeny and Conservation Combining genetic analysis with cladistic analyses can provide an important tool for biologists concerned with conservation (Figure 1–6). For example, some of the new species of mammals described in section 1.1 14 CHAPTER 1 were identiﬁed by comparing their DNA with the DNA of related species. When a genetic diﬀerence is large, it means that the two forms have been reproductively isolated from each other and have followed diﬀerent evolutionary pathways. From a conservationist’s perspective, lineages that have evolved substantial genetic diﬀerences are Evolutionarily Signiﬁcant Units (ESUs), and management plans should protect the genetic diversity of ESUs. For example, a genetic study published in 2007 revealed that the clouded leopards on the islands of Borneo and Sumatra (Neofelis diardi) and those on the The Diversity, Classification, and Evolution of Vertebrates Asian mainland (Neofelis nebulosa) separated between 1.4 and 2.8 million years ago, and the three forms have been following independent evolutionary pathways since then. The genetic mainland form and the island forms are genetically diﬀerent. Furthermore, the island populations are reproductively isolated from each other, and the clouded leopards on Borneo and Sumatra are genetically distinct. Thus, the three forms represent three ESUs, and conservation plans should treat the mainland species and the two island species separately. Before this study the three forms were grouped together, and a portion of the genetic diversity of clouded leopards was lost through interbreeding in captivity. 1.6 Earth History and Vertebrate Evolution Since their origin in the early Paleozoic, vertebrates have been evolving in a world that has changed enormously and repeatedly. These changes have aﬀected vertebrate evolution both directly and indirectly. Understanding the sequence of changes in the positions of continents, and the signiﬁcance of those positions regarding climates and interchange of faunas, is central to understanding the vertebrate story. These events are summarized inside the front cover of the book, and Chapters 7, 15, and 19 give details. The history of Earth has occupied three geological eons: the Archean, Proterozoic, and Phanerozoic. Only the Phanerozoic, which began about 542 million years ago, contains vertebrate life, and it is divided into three geological eras: the Paleozoic (Greek paleo = ancient and zoo = animal), Mesozoic (Greek meso = middle), and Cenozoic (Greek cen = recent). These eras are divided into periods, which can be further subdivided in a variety of ways, such as the subdivisions called epochs within the Cenozoic era from the Paleocene to the Recent. Movement of landmasses, called continental drift, has been a feature of Earth’s history at least since the Proterozoic, and the course of vertebrate evolution has been shaped extensively by continental movements. By the early Paleozoic, roughly 540 million years ago, a recognizable scene had appeared. Seas covered most of Earth as they do today, large continents ﬂoated on Earth’s mantle, life had become complex, and an atmosphere of oxygen had formed, signifying that the photosynthetic production of food resources had become a central phenomenon of life. The continents still drift today—North America is moving westward and Australia northward at the speed of approximately 4 centimeters per year (about the rate at which ﬁngernails grow). Because the movements are so complex, their sequence, their varied directions, and the precise timing of the changes are diﬃcult to summarize. When the movements are viewed broadly, however, a simple pattern unfolds during vertebrate history: fragmentation, coalescence, fragmentation. Continents existed as separate entities over 2 billion years ago. Some 300 million years ago, all of these separate continents combined to form a single landmass known as Pangaea, which was the birthplace of terrestrial vertebrates. Persisting and drifting northward as an entity, this huge continent began to break apart about 150 million years ago. Its separation occurred in two stages: ﬁrst into Laurasia in the north and Gondwana in the south, and then into a series of units that have drifted and become the continents we know today. The complex movements of the continents through time have had major eﬀects on the evolution of vertebrates. Most obvious is the relationship between the location of landmasses and their climates. At the end of the Paleozoic, much of Pangaea was located on the equator, and this situation persisted through the middle of the Mesozoic. Solar radiation is most intense at the equator, and climates at the equator are correspondingly warm. During the late Paleozoic and much of the Mesozoic, large areas of land enjoyed tropical conditions. Terrestrial vertebrates evolved and spread in these tropical regions. By the end of the Mesozoic, much of Earth’s landmass had moved out of equatorial regions; and, by the mid-Cenozoic, most terrestrial climates in the higher latitudes of the Northern and Southern Hemispheres were temperate instead of tropical. A less obvious eﬀect of the position of continents on terrestrial climates comes from changes in patterns of oceanic circulation. For example, the Arctic Ocean is now largely isolated from the other oceans, and it does not receive warm water via currents ﬂowing from more equatorial regions. High latitudes are cold because they receive less solar radiation than do areas closer to the equator, and the Arctic Basin does not receive enough warm water to oﬀset the lack of solar radiation. As a result, the Arctic Ocean is permanently frozen, and cold climates extend well southward across the continents. The cooling of climates in the Northern Hemisphere at the end of the Eocene epoch, around 34 million years ago, may have been a factor leading to the extinction of archaic mammals, and it is partly the result of changes in oceanic circulation at that time. Another factor that inﬂuences climates is the relative levels of the continents and the seas. At some Earth History and Vertebrate Evolution 15 periods in Earth’s history, most recently in the late Mesozoic and again in the ﬁrst part of the Cenozoic, shallow seas ﬂooded large parts of the continents. These epicontinental seas extended across the middle of North America and the middle of Eurasia during the Cretaceous period and early Cenozoic. Water absorbs heat as air temperature rises, and then releases that heat as air temperature falls. Thus, areas of land near large bodies of water have maritime climates—they do not get very hot in summer or very cold in winter, and they are usually moist because water that evaporates from the sea falls as rain on the land. Continental climates, which characterize areas far from the sea, are usually dry with cold winters and hot summers. The draining of the epicontinental seas at the end of the Cretaceous probably contributed to the demise of the dinosaurs by making climates in the Northern Hemisphere more continental. In addition to changing climates, continental drift has formed and broken land connections between the continents. Isolation of diﬀerent lineages of vertebrates on diﬀerent landmasses has produced dramatic examples of the independent evolution of similar types of organisms, such as the diversiﬁcation of mammals in the midCenozoic, a time when Earth’s continents reached their greatest separation during the history of vertebrates. Much of evolutionary history appears to depend on whether a particular lineage of animals was in the right place at the right time. This random element of evolution is assuming increasing prominence as more detailed information about the times of extinction of old groups and radiation of new groups suggests that competitive replacement of one group by another is not the usual mechanism of large-scale evolutionary change. The movements of continents and their eﬀects on climates and the isolation or dispersal of animals are taking an increasingly central role in our understanding of vertebrate evolution. On a continental scale, the advance and retreat of glaciers in the Pleistocene caused homogeneous habitats to split and merge repeatedly, isolating populations of widespread species and leading to the evolution of new species. Summary The more than 63,000 species of living vertebrates span a size range from less than a gram to more than 100,000 kilograms. They live in habitats extending from the bottom of the sea to the tops of mountains. This extraordinary diversity is the product of more than 500 million years of evolution, and the vast majority of species fall into one of the two major divisions of bony fishes (Osteichthyes)—the aquatic ray-finned fishes (Actinopterygii) and the primarily terrestrial lobe-finned fishes and tetrapods (Sarcopterygii), each of which contains more than 25,000 extant species. Phylogenetic systematics, usually called cladistics, classifies animals on the basis of shared derived character states. Natural evolutionary groups can be defined only by these derived characters; retention of ancestral characters does not provide information about evolutionary lineages. Application of this principle produces groupings of animals that reflect evolutionary history as accurately as we can discern it and forms a basis for making hypotheses about evolution and for designing management plans that conserve the genetic diversity of evolutionary lineages. Earth has changed dramatically during the half-billion years of vertebrate history. Continents were fragmented when vertebrates first appeared; coalesced into one enormous continent, Pangaea, about 300 million years ago; and began to fragment again about 150 million years ago. This pattern of fragmentation, coalescence, and fragmentation has resulted in isolation and renewed contact of major groups of vertebrates on a worldwide scale. Discussion Questions 1. Why don’t phylogenetic (cladistic) classiﬁcations have a ﬁxed number of hierarchical categories like those in a Linnaean classiﬁcation? 16 CHAPTER 1 What aspect of evolution does a phylogenetic classiﬁcation represent more clearly than a Linnaean classiﬁcation does? The Diversity, Classification, and Evolution of Vertebrates What is the meaning of an Evolutionarily Signiﬁcant Unit (ESU) in conservation biology? 4. Tetrapoda (node 6 in Figure 1–4) is a crown group, whereas Tetrapodomorpha is the corresponding stem group. What organisms are included in Tetrapoda? In Tetrapodomorpha? What is the diﬀerence between parallel and convergent evolution? 6. What is the signiﬁcance of an extant phylogenetic bracket? Additional Information Anderson, R. P., and C. O. Handley, Jr. 2001. A new species of three-toed sloth (Mammalia: Xenarthra) from Panamá, with a review of the genus Bradypus. Proceedings of the Biological Society of Washington 114:1–33. Budd, G. 2001. Climbing life’s tree. 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World-wide genetic diﬀerentiation of Eubalaena: Questioning the number of right whale species. Molecular Ecology 9:1793–1802. Rovero, F., et al. 2008. A new species of giant sengi or elephant-shrew (genus Rhynchocyon) highlights the exceptional biodiversity of the Udzungwa Mountains of Tanzania. Journal of Zoology 274:126–133. Sangster, G. 2009. Increasing numbers of bird species result from taxonomic progress, not taxonomic inﬂation. Proceedings of the Royal Society B 276:3185–3191. Singleton, M. 2009. The phenetic aﬃnities of Rungwecebus kipunji. Journal of Human Evolution 56:25–42. Swiderski, D. L. (ed.). 2001. Beyond reconstruction: Using phylogenies to test hypotheses about vertebrate evolution. American Zoologist 41:485–607. Templeton, A. R. 2010. The diverse applications of cladistic analysis of molecular evolution, with special reference to nested clade analysis. International Journal of Molecular Sciences 11:124–139. Timm, R. M., et al. 2005. What constitutes a proper description? Science 309:2163–2164. Wado, S., et al. 2003. A newly discovered species of living baleen whale. Nature 426:278–281. Additional Information 17 Wilting, A., et al. 2007. Clouded leopard phylogeny revisited: Support for species recognition and population division between Borneo and Sumatra. Frontiers in Zoology 4:15. content/4/1/15 Zinner, D., et al. 2009. Is the new primate genus Rungwecebus a baboon? PLoS ONE 4(3): e4859. doi:10.1371/ journal.pone.0004859. Websites Census of Marine Life The Census of Marine Life is a growing global network of researchers in more than 80 nations engaged in a 10-year initiative to assess and explain the diversity, distribution, and abundance of marine life in the oceans—past, present, and future. www.coml.org Discover Life Species identiﬁcation and images. 18 CHAPTER 1 International Code of Zoological Nomenclature The International Commission on Zoological Nomenclature sets the procedures for naming species of animals; its goal is to promote “standards, sense, and stability for animal names in science”. hosted-sites/iczn/code/ Journey into Phylogenetic Systematics Explanation of the methods, uses, and implications of cladistic classiﬁcation. clad/clad1.html Tree of Life web project When complete, the Tree of Life web project will have a page for every species of organism, both living and extinct; web pages follow the evolutionary branching patterns of the lineages. phylogeny.html The Diversity, Classification, and Evolution of Vertebrates 2 I n this chapter, we explain the structures that are characteristic of vertebrates, discuss the relationship of vertebrates to other members of the animal kingdom, and describe the systems that make vertebrates functional animals. We need an understanding of the fundamentals of vertebrate anatomy, physiology, and development to appreciate the changes that have occurred during their evolution and to trace homologies between basal vertebrates and more derived ones. 2.1 Vertebrates in Relation to Other Animals Vertebrates are a diverse and fascinating group of animals. Because we are vertebrates ourselves, that statement may seem chauvinistic, but vertebrates are remarkable in comparison with most other animal groups. Vertebrates are the subphylum Vertebrata of the phylum Chordata. At least 30 other animal phyla have been named, but only the phylum Arthropoda (insects, crustaceans, spiders, etc.) rivals the vertebrates in diversity of forms and habitat. And it is only in the phylum Mollusca (snails, clams, and squid) that we ﬁnd animals (such as octopus and squid) that approach the very large size of some vertebrates and also have a capacity for complex learning. The tunicates (subphylum Urochordata) and cephalochordates (subphylum Cephalochordata) are placed with vertebrates in the phylum Chordata. Within the chordates, CHAPTER Vertebrate Relationships and Basic Structure cephalochordates and vertebrates resemble each other anatomically, but molecular characteristics show that tunicates are more closely related to vertebrates than are cephalochordates. Characteristics of Chordates Chordates are united by several shared derived features, which are seen in all members of the phylum at some point in their lives: • A notochord (a dorsal stiﬀening rod that gives the phylum Chordata its name) • A dorsal hollow nerve cord • A segmented, muscular postanal tail (i.e., extending beyond the gut region) • An endostyle (a ciliated, glandular groove on the floor of the pharynx that secretes mucus for trapping food particles during filter feeding; generally homologous with the thyroid gland of vertebrates, an endocrine gland involved with regulating metabolism) Chordates are also characterized by a pharynx (throat region) containing gill slits. Nonvertebrate 19 chordates use the gill slits for ﬁlter feeding, and aquatic vertebrates (ﬁshes) use them for respiration. Some other deuterostomes (the larger grouping to which chordates belong) also have gills, however, and gills may be a primitive feature for chordates as a group. Chordates also share some nervous system features: they all have structures in the brain corresponding to a pineal eye and hormone-regulating pituitary. Chordate Relationships Although chordates are all basically bilaterally symmetrical animals (i.e., one side is the mirror image of the other), they have a left-to-right asymmetry within the body—for example, the position of the heart on the left side and most of the liver on the right side. (Rare human individuals have a condition termed “situs inversus,” in which the positions of the major body organs are reversed.) The relationship of chordates to other kinds of animals is revealed by anatomical, biochemical, and Characteristics of Deuterostomes The chordates, echinoderms, and two other phyla (hemichordates and xenoturbellids) are linked as deuterostomes (Greek deutero = second and stoma = mouth) by several unique embryonic features, such as the way in which their eggs cleave into daughter cells after fertilization, their larval form, and some other features discussed later. Hemichordates are a small, Xenoturbellida Chordata Ambulacraria Olfactores Vertebrata embryonic characters as well as by the fossil record. Figure 2–1 shows the relationships of animal phyla. Vertebrates superﬁcially resemble other active animals, such as insects, in having a distinct head end, jointed legs, and bilateral symmetry. However, and perhaps surprisingly, developmental and molecular data show that the phylum Chordata is closely related to the phylum Echinodermata (starﬁshes, sea urchins, and the like), which are marine forms without distinct heads and with pentaradial (ﬁvefold and circular) symmetry as adults. Hemichordata (acorn worms, pterobranchs) Echinodermata (starfish, sea urchins, etc.) Arthropoda Annelida Cephalochordata Urochordata (amphioxus) (tunicates) Plus many other more obscure phyla Deuterostomata Protostomata Flatwormlike animals Cnidarians (anemones, corals, etc.) Sponges Coelomata Coelom (body cavity within mesoderm) Bilataria Three cell layers organs bilateral symmetry movement as adult Cell layers and tissues nervous system with neurons Tissue layers: Ectoderm Mesoderm Endoderm Arrows indicate direction of food passage/water flow. Metazoa Collagen, heterotrophy, early embryo forms hollow ball of cells (blastula), sex cells formed in special organs—sperm have whiplike tails Figure 2–1 A simplified phylogeny of the animal kingdom (metazoans). There are at least 30 phyla today (Chordata, Echinodermata, Annelida, etc. are phyla). Possibly as many as 15 additional phyla are known from the early Paleozoic era; they became extinct at the end of the Cambrian period. 20 CHAPTER 2 Mollusca Vertebrate Relationships and Basic Structure poorly known phylum of marine animals containing the earthwormlike acorn worms and the fernlike pterobranchs. Xenoturbellids are small marine wormlike forms (only two species are known), which have recently been identiﬁed as deuterostomes by molecular analysis. Hemichordates were long considered the sister group of chordates because both groups have pharyngeal slits, and hemichordates also have features of the pharynx that can be interpreted as the precursor to an endostyle. However, we now consider these to be primitive features of the larger group (deuterostomes) in which hemichordates and chordates are placed. Although modern echinoderms lack pharyngeal slits, some extinct echinoderms appear to have had them. (The diversity of extinct echinoderms is much greater than that of the living forms.) Furthermore, early echinoderms were bilaterally symmetrical, meaning that the ﬁvefold symmetry of modern echinoderms is probably a derived character of that lineage. Molecular characteristics currently unite echinoderms and hemichordates as the Ambulacraria, and xenoturbellids are more closely related to these phyla than to the chordates. Relationships of Deuterostomes To consider how deuterostomes are related to other animals, we will start at the bottom of the tree and work upward (see Figure 2–1). All animals (metazoans) are multicellular and share some features of reproduction and embryonic development: they have motile sperm with whiplike tails, the embryo initially forms as a hollow ball of cells (the blastula), and sex cells form in special organs. All metazoans also have the structural protein collagen, which in humans forms the matrix of many of our tissues and organs, including the nose and ears. Animals more derived than sponges have a nervous system, and their bodies are made of distinct layers of cells, or germ layers, that are laid down early in development at a stage called gastrulation. Gastrulation occurs when the hollow ball of cells forming the blastula folds in on itself, producing two distinct layers of cells and a gut with an opening to the outside at one end. The outer layer of cells is the ectoderm (Greek ecto = outside and derm = skin), and the inner layer forms the endoderm (Greek endo = within). Jellyﬁshes and related animals have only these two layers of body tissue, making them diploblastic (Greek diplo = two and blast = a bud or sprout). Animals more derived than jellyﬁshes and their kin add an additional, middle cell layer of mesoderm (Greek mesos = middle), making them triploblastic (Greek triplo = three). Triploblasts have a gut that opens at both ends (i.e., with a mouth and an anus) and are bilaterally symmetrical with a distinct anterior (head) end at some point in their life. The mesoderm forms the body’s muscles, and only animals with a mesoderm are able to be motile as adults. (Larval forms do not need muscles because they are small enough to be powered by hairlike cilia on the body surface.) The coelom, an inner body cavity that forms as a split within the mesoderm, is another derived character of most, but not all, triploblastic animals. Coelomate animals (i.e., animals with a coelom) are split into two groups on the basis of how the mouth and anus form. When the blastula folds in on itself to form a gastrula, it leaves an opening to the outside called the blastopore (Latin porus = a small opening). During the embryonic development of coelomates, a second opening develops. In the lineage called protostomes (Greek proto = ﬁrst and stome = mouth), the blastopore (which was the ﬁrst opening in the embryo) becomes the mouth, whereas in deuterostomes the second opening becomes the mouth and the blastopore becomes the anus. Chordates, hemichordates, and echinodems are deuterostomes, whereas mollusks (snails, clams, and squid), arthropods (insects, crustaceans, and spiders), annelids (earthworms), and many other phyla are protostomes (see Figure 2–1). Extant Nonvertebrate Chordates The two groups of extant nonvertebrate chordates are small marine animals. More types of nonvertebrate chordates may have existed in the past, but such softbodied animals are rarely preserved as fossils. Some possible Early Cambrian chordates are described at the end of this section. Urochordates Present-day tunicates (subphylum Uro- chordata) are marine animals that ﬁlter particles of food from the water with a basketlike perforated pharynx. There are about 3000 living species, and all but 100 or so are sedentary as adults, attaching themselves to the substrate either singly or in colonies. Adult tunicates (also known as sea squirts, or ascideans) bear little apparent similarity to cephalochordates and vertebrates. Most tunicate species have a brief freeswimming larval period lasting a few minutes to a few days, after which the larvae metamorphose into sedentary adults attached to the substrate, although some species remain motile as adults. Tunicate larvae have a notochord, a dorsal hollow nerve cord, and a muscular postanal tail that moves in a ﬁshlike swimming pattern (Figure 2–2). It was long believed that the earliest chordates would have been sessile as adults (like most Vertebrates in Relation to Other Animals 21 (a) Free-swimming larval tunicate (Urochordata) Inlet Nervous system Outlet Notochord Tail Adhesive surfaces Pharynx (b) Sessile adult tunicate (Urochordata) Inlet Outlet from well-developed adult atrium Pharynx Tail remnant Notochord remnant Anus Reproductive cells Reduced nervous system Stomach Adhesive surface (c) The lancelet, amphioxus (Cephalochordata) (posterior myomeres removed) Wheel organ Pharynx Velum Nerve cord Myomeres Notochord Cecum Midgut Fin-ray boxes Intestine Anus Buccal cirri Anterior end of notochord Atriopore 10 mm Endostyle Atrial wall Gonads Fin-ray boxes Figure 2–2 Nonvertebrate chordates. Tunicates have a free-swimming larva (a) that metamorphoses into a sessile adult (b), whereas amphioxus (c) is free-swimming throughout its life. 22 CHAPTER 2 Vertebrate Relationships and Basic Structure other deuterostomes) and that cephalochordates and vertebrates evolved from an ancestor that resembled a tunicate larva. However, it now seems more likely that a sessile adult stage (Figure 2–2b) is a derived character for tunicates and that the living species that remain free-swimming as adults most resemble the ancestral chordate. The ancestral chordate (and indeed, the ancestral deuterostome) was probably a free-swimming wormlike creature that used gill slits for ﬁlter feeding. Cephalochordates The subphylum Cephalochordata contains about 27 species, all of which are small, superﬁcially ﬁshlike marine animals usually less than 5 centimeters long. The best-known cephalochordate is the lancelet (Branchiostoma lanceolatum), more commonly known as amphioxus (Greek amphi = both and oxy = sharp). Amphioxus means “sharp at both ends,” an appropriate term for an animal in which the front and rear ends are nearly the same shape because it lacks a distinct head. Lancelets are widely distributed in marine waters of the continental shelves and are usually burrowing, sedentary animals as adults, although the adults of a few species retain an active, free-swimming behavior. A notable characteristic of amphioxus is its ﬁshlike locomotion produced by myomeres—blocks of striated muscle ﬁbers arranged along both sides of the body and separated by sheets of connective tissue. Sequential contraction of the myomeres bends the body from side to side, resulting in forward or backward propulsion. The notochord acts as an incompressible elastic rod, extending the full length of the body and preventing the body from shortening when the myomeres contract. While the notochord of vertebrates ends midway through the head region, the notochord of amphioxus extends from the tip of the snout to the end of the tail, projecting well beyond the region of the myomeres at both ends. This anterior elongation of the notochord apparently is a specialization that aids in burrowing. Figure 2–2c shows some details of the internal structure of amphioxus. Amphioxus and vertebrates diﬀer in the use of the pharyngeal slits. Amphioxus has no gill tissue associated with these slits; its body is small enough for oxygen uptake and carbon dioxide loss to occur by diﬀusion over the body surface, and the gill slits are used for ﬁlter feeding. Water is moved over the gill slits by cilia on the gill bars between the slits, aided by the features of the buccal (mouth region, Latin bucc = cheek) cirri and the wheel organ, while the velum is a ﬂap helping to control the one-way ﬂow of water. In addition to the internal body cavity, or coelom, amphioxus has an external body cavity called the atrium, which is also seen in tunicates and hemichordates—and thus is probably an ancestral deuterostome feature— but is absent from vertebrates. (This atrium is not the same as the atrium of the vertebrate heart; the word atrium [plural atria] comes from the Latin term for an open space.) The atrium of amphioxus is formed by outgrowths of the body wall (metapleural folds), which enclose the body ventrally. (Imagine yourself wearing a cape and then extending your arms until there is a space between the cape and your body— that space would represent the position of the atrium in amphioxus.) The atrium opens to the outside world via the atriopore, an opening in front of the anus. The atrium appears to work in combination with the beating of the cilia on the gill bars and the wheel organ in the head to control passage of substances through the pharynx and is probably functionally associated with the ancestral chordate character of using the gill slits for ﬁlter feeding. Cephalochordates have several anatomical features that are shared with vertebrates but absent from tunicates. In addition to the myomeres, amphioxus has a vertebrate-like tail ﬁn, a circulatory system similar to that of vertebrates, with a dorsal aorta and a ventral heartlike structure that forces blood through the gills, and specialized excretory cells called podocytes. Olfactores—Tunicates Plus Vertebrates Cephalochordates were long considered to be the sister group of vertebrates on the basis of the shared anatomical characters described above. However, some (but not all) molecular analyses have placed tunicates as the sister group of vertebrates. (The group that includes vertebrates plus tunicates is the Olfactores.) If tunicates are the sister group of vertebrates, the sessile nature of most adult tunicates must be a derived character of the lineage, and both ancestral tunicates and ancestral chordates likely were mobile. A sister-group relationship of vertebrates and tunicates is now generally accepted, as we show in Figure 2–1. Morphology still strongly supports a vertebrate/cephalochordate association, however, and the features that cephalochordates share with vertebrates may simply be ancestral chordate features. Cambrian Chordates The best-known early chordatelike animal is Pikaia, which looks a little like an amphioxus. About 100 specimens of this animal are known from the Middle Cambrian Burgess Shale in British Columbia. Pikaia is probably not a true cephalochordate: although it has an obvious notochord, the myomeres are straight rather than V-shaped and there is no evidence of gills. Recently spectacular fossils of soft-bodied animals have been found in the Early Cambrian Chengjiang Vertebrates in Relation to Other Animals 23 Dorsal fin Notochord Brain Spinal cord Myomere Pharynx "Tail" Tail fin Eye Endostyle Anus Visceral arch skeleton Ventral aorta Atriopore Gonads Intestine Ventral fin Heart 1 mm Figure 2–3 The yunnanozoan Haikouella, from the Early Cambrian Chenjiang Fauna of southern China, an early chordate or chordate relative. formation in southern China. This 522-million-year-old deposit is at least 17 million years older than the Burgess Shale. The Chengjiang deposit includes the earliestknown true vertebrates (described in Chapter 3) and some intriguing fossils that may be early chordates. (The ﬂattened “road kill-like” nature of the Chengjiang specimens makes it diﬃcult to interpret their structure.) The most vertebrate-like member of the Chengjiang Fauna is Haikouella, which is known from more than 300 individuals (Figure 2–3). Haikouella has the chordate features of myomeres, a notochord, and a pharynx apparently enclosed in an atrium, as well as derived features that suggest that it is the sister group to vertebrates. These features include a large brain, clearly deﬁned eyes, thickened branchial bars (that appear to be made of a type of cartilage similar to that of lamprey larvae), and an upper lip like that of larval lampreys. The endostyle and tentacles surrounding the mouth suggest that this animal was a suspension feeder, like amphioxus. Haikouella appears to have developed a vertebratelike muscular pharynx. The thickened branchial bars appear stout enough to support both gill tissue and pharyngeal muscles, suggesting that the gills were used for respiration as well as for feeding. 2.2 Definition of a Vertebrate The term vertebrate is derived from the vertebrae that are serially arranged to make up the spinal column, or backbone, of vertebrate animals. In ourselves, as in other land vertebrates, the vertebrae form around the notochord during development and also encircle the nerve cord. The bony vertebral column replaces the original notochord after the embryonic period. In many 24 CHAPTER 2 fishes the vertebrae are made of cartilage rather than bone. All vertebrates have the uniquely derived feature of a cranium, or skull, which is a bony, cartilaginous, or ﬁbrous structure surrounding the brain. Vertebrates also have a prominent head containing complex sense organs. Although many of the genes that determine head development in vertebrates are present in amphioxus, the anterior portion of the vertebrate head does seem to be a new feature of vertebrates. Hagﬁsh have been considered to lack vertebrae, but new developmental evidence suggests homologs to the ventral portion of the vertebrae in the tail of jawed ﬁshes. The structures corresponding to vertebrae in lampreys are segmental cartilaginous rudiments (arcualia) ﬂanking the nerve cord. Fully formed vertebrae, with a centrum (plural centra) surrounding the notochord, are found only in gnathostomes (jawed vertebrates— see Chapter 3), and many jawed ﬁshes retain a functional notochord as adults. Because of the apparent lack of vertebrae in hagﬁshes, some people have preferred the term Craniata to Vertebrata for the subphylum. However, we continue to use the familiar term vertebrate in this book, and the term Craniata is redundant if hagﬁshes are indeed the sister group of lampreys (see Chapter 3). Unique Embryonic Features of Vertebrates Several embryonic features may account for many of the diﬀerences between vertebrates and other chordates. The two primary ones are duplication of the Hox gene complex (homeobox genes) and the appearance of a new embryonic tissue, the neural crest. Vertebrate Relationships and Basic Structure Hox Genes Hox genes regulate the expression of a hierarchical network of other genes that control the process of development from front to back along the body. Vertebrates have more Hox genes than other groups of animals: Jellyﬁshes (and possibly also sponges) have one or two Hox genes, the common ancestor of protostomes and deuterostomes probably had seven, and more derived metazoans have up to thirteen. However, vertebrates are unique in having undergone duplications of the entire Hox complex. The ﬁrst duplication event seems to have occurred at the start of vertebrate evolution because amphioxus and tunicates have a single Hox cluster, whereas the living jawless vertebrates have two. A second duplication event had taken place by the evolution of gnathostomes, because all jawed vertebrates have at least four clusters. Finally, an additional duplication event occurred in both teleost ﬁshes (derived bony ﬁshes) and frogs. Interactions among genes modify the eﬀects of those genes, and more genes allow more interactions that probably produce more complex structures. The doubling and redoubling of the Hox gene sequence during vertebrate evolution is believed to have made the structural complexity of vertebrates possible. Neural Crest Neural crest is a new tissue in embryological development that forms many novel structures in vertebrates, especially in the head region (a more detailed description is given in section 2.3). The evolution of the neural crest is the most important innovation in the origin of the vertebrate body plan. Neural-crest tissue is a fourth germ layer that is unique to vertebrates and is on a par with ectoderm, endoderm, and mesoderm. Neural-crest cells originate at the lateral boundary of the neural plate, the embryonic structure that makes the nerve cord, and migrate throughout the body to form a variety of structures, including pigment cells. A similar population of cells, with a similar genetic expression, can be found in amphioxus, but here the cells do not migrate and do not change into diﬀerent cell types. Recently cells resembling migratory neuralcrest cells have been identiﬁed in the larval stage of one tunicate species, where they differentiate into pigment cells. These cells in tunicates may represent a precursor to the vertebrate neural crest. (This appears to be one morphological feature that tunicates share with vertebrates, but it is not seen in cephalochordates.) If the Cambrian chordate Haikouella has been correctly interpreted as having eyes and a muscular pharynx, these features would imply the presence of neural crest in this animal. Placodes Another new type of embryonic tissue in vertebrates, which is similar to neural crest but probably has a diﬀerent origin, forms the epidermal thickenings (placodes) that give rise to the complex sensory organs of vertebrates, including the nose, eyes, and inner ear. Some placode cells migrate caudally to contribute, along with the neural-crest cells, to the lateral line system and to the cranial nerves that innervate it. MicroRNAs The appearance of many microRNAs is a genetic innovation in vertebrates that may contribute to their anatomical complexity. MicroRNAs are noncoding RNA sequences 22 bases long that have been added to the genomes of metazoans throughout their evolutionary history. MicroRNAs regulate the synthesis of proteins by binding to complementary base sequences of messenger RNAs. The phylum Chordata is characterized by the addition of two new microRNAs, another three are shared by vertebrates and tunicates, and all vertebrates possess an additional 41 unique microRNAs. (All vertebrate lineages have independently acquired yet more microRNAs of their own, and mammals in particular have a great number of novel microRNAs.) MicroRNAs are involved in regulating the development of some derived vertebrate structures, including the liver and the kidney. Brains The brains of vertebrates are larger than the brains of primitive chordates, and they have three parts—the forebrain, midbrain, and hindbrain. The brain of amphioxus appears simple, but genetic studies show that amphioxus has all the genes that code for the vertebrate brain with the exception of those directing formation of the front part of the vertebrate forebrain, the telencephalon (the portion of the brain that contains the cerebral cortex, the area of higher processing in vertebrates). The presence of the genes in amphioxus combined with the absence of a complex brain reinforces the growing belief that the diﬀerences in how genes are expressed in different animals are as important as diﬀerences in what genes are present. Other unique vertebrate features include a multilayered epidermis and blood vessels that are lined by endothelium. 2.3 Basic Vertebrate Structure This section serves as an introduction to vertebrate anatomical structure and function. The heart of this section is in Table 2–1 and Figure 2–4, which contrast the basic vertebrate condition with that of a nonvertebrate chordate such as amphioxus. These same systems will Basic Vertebrate Structure 25 Amphioxus-like nonvertebrate chordate Cerebral vesicle V-shaped segmental myomeres Notochord Nerve cord Pigment spot Simple caudal fin Cloaca Gut (unmuscularized) Endostyle Atrium Gill slit Cecum Pharynx Single-chambered (unmuscularized) "heart" (a) Atriopore Hypothetical primitive vertebrate Tripartite brain Lateral line Notochord Nerve cord Cranium Figure 2–4 Generalized chordate structure. (a) A generalized amphioxus nonvertebrate chordate; (b) a hypothetical ancestral vertebrate. (Note that the myomeres actually extend all the way down the body; see Figure 2–10.) 26 CHAPTER 2 Liver Cranial sense Thyroid organs: nose, eye, Gill slit Gill arch inner ear with gill (cartilage) tissue Pharynx (b) (muscularized) be further discussed for more derived vertebrates in later chapters; our aim here is to provide a general introduction to the basics of vertebrate anatomy. More detail can be found in books listed at the end of the chapter. At the whole-animal level, an increase in body size and increased activity distinguish vertebrates from nonvertebrate chordates. Early vertebrates generally had body lengths of 10 centimeters or more, which is about an order of magnitude larger than the bodies of nonvertebrate chordates. Because of their relatively large size, vertebrates need specialized systems to carry out processes that are accomplished by diﬀusion or ciliary action in smaller animals. Vertebrates are also more active animals than other chordates, so they need organ systems that can carry out physiological processes at a greater rate. The transition from nonvertebrate chordate to vertebrate was probably related to the adoption of a more actively predaceous mode of life, as evidenced by the features of the vertebrate head W-shaped segmental myomeres Caudal fin with dermal fin rays Kidney Archinephric duct Pancreatic tissue Three-chambered heart Gut (muscularized) (largely derived from neural-crest tissue) that would enable suction feeding with a muscular pharynx, and a bigger brain and more complex sensory organs for perceiving the environment. Vertebrates are characterized by mobility, and the ability to move requires muscles and a skeleton. Mobility brings vertebrates into contact with a wide range of environments and objects in those environments, and a vertebrate’s external protective covering must be tough but ﬂexible. Bone and other mineralized tissues that we consider characteristic of vertebrates had their origins in this protective integument. Embryonic Development Studying embryos can show how systems develop and how the form of the adult is related to functional and historical constraints during development. Scientists no longer adhere to the biogenetic law that “ontogeny recapitulates phylogeny” (i.e., the idea that the embryo Vertebrate Relationships and Basic Structure Table 2–1 Comparison of features in nonvertebrate chordates and ancestral vertebrates Generalized Nonvertebrate Chordate (based on features of the living cephalochordate amphioxus) Ancestral Vertebrate (based on features of the living jawless vertebrates—hagfishes and lampreys) Brain and Head End Notochord extends to tip of head (may be derived condition). Head extends beyond tip of notochord. No cranium (skull). Cranium—skeletal supports around brain, consisting of capsules surrounding the main parts of the brain and their sensory components plus underlying supports. Simple brain (= cerebral vesicle), no specialized sense organs (except photoreceptive frontal organ, probably homologous with the vertebrate eye). Tripartite brain and multicellular sense organs (eye, nose, inner ear). Poor distance sensation (although the skin is sensitive). Improved distance sensation: in addition to the eyes and nose, also have a lateral line system along the head and body that can detect water movements (poorly developed lateral line system on the head is found only in hagfishes). No electroreception. Electroreception may be an ancestral vertebrate feature (but absent in hagfishes, possibly lost). Pharynx and Respiration Gill arches used for filter feeding (respiration is by diffusion over the body surface). Gill arches (= pharyngeal arches) support gills that are used primarily for respiration. Numerous gill slits (up to 100 on each side). Fewer gill slits (6 to 10 on each side), individual gills with highly complex internal structure (gill filaments). Pharynx not muscularized (except in wall of atrium, or external body cavity). Pharynx with specialized (branchiomeric) musculature. Water moved through pharynx and over gills by ciliary action. Water moved through pharynx and over gills by active muscular pumping. Gill arches made of collagen-like material Gill arches made of cartilage (allows for elastic recoil—aids in pumping). Feeding and Digestion Gut not muscularized: food passage by means of ciliary action. Gut muscularized: food passage by means of muscular peristalsis. Digestion of food is intracellular: individual food particles taken into cells lining gut. Digestion of food is extracellular: enzymes poured onto food in gut lumen, then breakdown products absorbed by cells lining gut. No discrete liver and pancreas: structure called the midgut cecum or diverticulum is probably homologous to both. Discrete liver and pancreatic tissue. Heart and Circulation Ventral pumping structure (no true heart, just contracting regions of vessels; = sinus venosus of vertebrates). Also accessory pumping regions elsewhere in the system. Ventral pumping heart only (but accessory pumping regions retained in hagfishes). Three-chambered heart (listed in order of blood flow): sinus venosus, atrium, and ventricle. No neural control of the heart to regulate pumping. Neural control of the heart (except in hagfishes). Circulatory system open: large blood sinuses; capillary system not extensive. Circulatory system closed: without blood sinuses (some remain in hagfishes and lampreys); extensive capillary system. Blood not specifically involved in the transport of respiratory gases (O2 and CO2 mainly transported via diffusion). No red blood cells or respiratory pigment. Blood specifically involved in the transport of respiratory gases. Red blood cells containing the respiratory pigment hemoglobin (binds with O2 and CO2 and aids in their transport). (continued) Basic Vertebrate Structure 27 Table 2–1 Comparison of features in nonvertebrate chordates and ancestral vertebrates (Continued) Generalized Nonvertebrate Chordate Excretion and Osmoregulation No specialized kidney. Coelom filtered by solenocytes (flame cells) that work by creating negative pressure within cell. Cells empty into the atrium (false body cavity) and then to the outside via the atriopore. Body fluids same concentration and ionic composition as seawater. No need for volume control or ionic regulation. Ancestral Vertebrate Specialized glomerular kidneys: segmental structures along dorsal body wall; works by ultrafiltration of blood. Empty to the outside via the archinephric ducts leading to the cloaca. Body fluids more dilute than seawater (except for hagfishes). Kidney important in volume regulation, especially in freshwater environment. Monovalent ions regulated by the gills (also the site of nitrogen excretion), divalent ions regulated by the kidney. Support and Locomotion Notochord provides main support for body muscles. Notochord provides main support for body muscles, vertebral elements around nerve cord at least in all vertebrates except hagfishes. Myomeres with simple V shape. Myomeres with more complex W shape. No lateral fins; no median fins besides tail fin. Initially no lateral fins. Caudal (tail) fin has dermal fin rays. Dorsal fins present in all except hagfishes. faithfully passes through its ancestral evolutionary stages in the course of its development) proposed by the nineteenth-century embryologist Haeckel. Nevertheless, embryology can provide clues about the ancestral condition and about homologies between structures in diﬀerent animals. The development of vertebrates from a single fertilized cell (the zygote) to the adult condition will be summarized only brieﬂy. This is important background information for many studies, but a detailed treatment is beyond the scope of this book. Note, however, that there is an important distinction in development between vertebrates and invertebrates: invertebrates develop from cell lineages whose fate is predetermined, but vertebrates are much more ﬂexible in their development and use inductive interactions between developing structures to determine the formation of diﬀerent cell types and tissues. We saw earlier that all animals with the exception of sponges are formed of distinct tissue layers, or germ layers. The fates of germ layers have been very conservative throughout vertebrate evolution. The outermost germ layer, the ectoderm, forms the adult superﬁcial layers of skin (the epidermis); the linings of the most anterior and most posterior parts of the digestive tract; and the nervous system, including most of the sense organs (such as the eye and the ear). The innermost layer, the endoderm, forms the rest of the digestive tract’s lining as well as the lining of glands associated with the gut—including the liver 28 CHAPTER 2 and the pancreas—and most respiratory surfaces of vertebrate gills and lungs. Endoderm also forms the taste buds and the thyroid, parathyroid, and thymus glands. The middle layer, the mesoderm, is the last of the three layers to appear in development, perhaps reﬂecting the fact that it is the last layer to appear in animal evolution. It forms everything else: muscles, skeleton (including the notochord), connective tissues, and circulatory and urogenital systems. A little later in development, there is a split within the originally solid mesoderm layer, forming a coelom or body cavity. The coelom is the cavity containing the internal organs. In mammals it is divided into the pleural cavity (around the lungs), the peritoneal cavity (around the viscera), and the pericardial cavity (around the heart). In other animals, which either lack lungs entirely or lack a diaphragm separating the pleural cavity from the peritoneal cavity, these two cavities are united into the pleuroperitoneal cavity. These coelomic cavities are lined by thin sheets of mesoderm—the peritoneum (around the pleural or peritoneal cavity) and the pericardium (around the heart). The gut is suspended in the peritoneal cavity by sheets of peritoneum called mesenteries. Neural crest forms many of the structures in the anterior head region, including some bones and muscles that were previously thought to be formed by mesoderm. Neural crest also forms almost all of the peripheral nervous system (i.e., that part of the nervous Vertebrate Relationships and Basic Structure Ectoderm Hindbrain Auditory placode Somites Nephrotome Nerve cord Midbrain Notochord Dorsal mesentery Outer layer of lateral plate (forms outer peritoneum and appendicular skeleton) Gut Pharynx Gut endoderm Forebrain Coelom Body wall Inner layer of lateral plate (forms gut muscles, heart muscles, blood and blood vessels, connective tissue, and inner peritoneum) Optic cup Stomadeum Pharyngeal Heart clefts Nasal placode Unsegmented lateral plate mesoderm Ectoderm Ventral mesentery Figure 2–5 Three-dimensional view of a portion of a generalized vertebrate embryo at the developmental stage (called the pharyngula) when the developing gill pouches appear. The ectoderm is stripped oﬀ the left side, showing segmentation of the mesoderm in the trunk region and pharyngeal development. The stomadeum is the developing mouth. system outside of the brain and the spinal cord) and contributes to portions of the brain. Some structures in the body that are new features of vertebrates are also formed from neural crest. These include the adrenal glands, pigment cells in the skin, secretory cells of the gut, and smooth muscle tissue lining the aorta. Figure 2–5 shows a stage in early embryonic development in which the ancestral chordate feature of pharyngeal pouches in the head region makes at least a ﬂeeting appearance in all vertebrate embryos. In ﬁshes the grooves between the pouches (the pharyngeal clefts) perforate to become the gill slits, whereas in land vertebrates these clefts disappear in later development. The linings of the pharyngeal pouches give rise to half a dozen or more glandular structures often associated with the lymphatic system, including the thymus gland, parathyroid glands, carotid bodies, and tonsils. The dorsal hollow nerve cord typical of vertebrates and other chordates is formed by the infolding and subsequent pinching oﬀ and isolation of a long ridge of ectoderm running dorsal to the developing notochord. The notochord itself appears to contain the developmental instructions for this critical embryonic event, which is probably why the notochord is retained in the embryos of vertebrates (such as us) that no longer have the complete structure in the adult. The cells that will form the neural crest arise next to the developing nerve cord (the neural tube) at this stage. Slightly later in development, these neural-crest cells disperse laterally and ventrally, ultimately settling and diﬀerentiating throughout the embryo. Embryonic mesoderm becomes divided into three distinct portions, as shown in Figure 2–5, with the result that adult vertebrates are a strange mixture of segmented and unsegmented components. The dorsal (upper) part of the mesoderm, lying above the gut and next to the nerve cord, forms an epimere, a series of thick-walled segmental buds (somites), which extends from the head end to the tail end. The ventral (lower) part of the mesoderm, surrounding the gut and containing the coelom, is thin-walled and unsegmented and is called the lateral plate mesoderm (or hypomere). Small segmental buds linking the somites and the lateral plate are called nephrotomes (the mesomere or the intermediate mesoderm). The nervous system also follows this segmented versus unsegmented pattern, as will be discussed later. The segmental somites will eventually form the dermis of the skin, the striated muscles of the body that are used in locomotion, and portions of the skeleton (the vertebral column, ribs, and portions of the back of the skull). Some of these segmental muscles later Basic Vertebrate Structure 29 migrate ventrally from their originally dorsal (epaxial) position to form the layer of striated muscles on the underside of the body (the hypaxial muscles), and from there they form the muscles of the limbs in tetrapods (four-footed land vertebrates). The lateral plate forms all the internal, nonsegmented portions of the body, such as the connective tissue, the blood vascular system, the mesenteries, the peritoneal and pericardial linings of the coelomic cavities, and the reproductive system. It also forms the smooth muscle of the gut and the cardiac (heart) muscle. The nephrotomes form the kidneys (which are elongated segmental structures in the ancestral vertebrate condition), the kidney drainage ducts (the archinephric ducts), and the gonads. Some exceptions exist to this segmented versus nonsegmented division of the vertebrate body. The locomotory muscles, both axial (within the trunk region) and appendicular (within the limbs), and the axial skeleton are derived from the somites. Curiously, however, the limb bones are mostly derived from the lateral plate, as are the tendons and ligaments of the appendicular muscles, even though they essentially form part of the segmented portion of the animal. The explanation for this apparent anomaly may lie in the fact that limbs are add-ons to the basic limbless vertebrate body plan, as seen in the living jawless vertebrates. The boundary of the complex interaction between the somite-derived muscles and structures derived from the outer layer of the lateral plate in the embryo is known as the lateral somitic frontier. This area is involved in the switching on and oﬀ of regulatory genes and is thus of prime importance in evolutionary change. Other peculiarities are found in the expanded front end of the head of vertebrates, which has a complex pattern of development and does not follow the simple segmentation of the body. The head mesoderm contains only somites (no lateral plate), which give rise to the striated eye muscles and branchiomeric muscles powering the pharyngeal arches (gills and jaws). Within the brain, the anteriormost part of the forebrain (the front of the telencephalon) and the midbrain are not segmented, but the hindbrain shows segmental divisions during development (rhombomeres). Adult Tissue Types There are several kinds of tissue in vertebrates: epithelial, connective, vascular (i.e., blood), muscular, and nervous. These tissues are combined to form larger units called organs, which often contain most or all of the ﬁve basic tissue types. Connective Tissue A fundamental component of most animal tissues is the ﬁbrous protein collagen. Collagen 30 CHAPTER 2 is primarily a mesodermal tissue: in addition to the softer tissues of organs, it forms the organic matrix of bone and the tough tissue of tendons and ligaments. Vertebrates have a unique type of ﬁbrillar collagen that may be responsible for their ability to form an internal skeleton. Collagen is stiﬀ and does not stretch easily. In some tissues, collagen is combined with the protein elastin, which can stretch and recoil. Another important ﬁbrous protein, seen only in vertebrates, is keratin. While collagen forms structures within the mesoderm, keratin is primarily an ectodermal tissue. Keratin is mainly found in the epidermis (outer skin) of tetrapods, making structures such as hair, scales, feathers, claws, horns, and beaks; it also forms the horny toothlike structures of the living jawless vertebrates. The Integument The external covering of vertebrates, the integument, is a single organ, making up 15 to 20 percent of the body weight of many vertebrates and much more in armored forms. It includes the skin and its derivatives, such as glands, scales, dermal armor, and hair. The skin protects the body and receives information from the outside world. The major divisions of the vertebrate skin are the epidermis (the superﬁcial cell layer derived from embryonic ectoderm) and the unique vertebrate dermis (the deeper cell layer of mesodermal and neural-crest origin). The dermis extends deeper into a subcutaneous tissue (hypodermis) that is derived from mesoderm and overlies the muscles and bones. The epidermis forms the boundary between a vertebrate and its environment and is of paramount importance in protection, exchange, and sensation. It often contains secretory glands and may play a signiﬁcant role in osmotic and volume regulation. The dermis, the main structural layer of the skin, includes many collagen ﬁbers that help to maintain its strength and shape. The dermis contains blood vessels, and blood ﬂow within these vessels is under neural and hormonal control (e.g., as in human blushing, when the vessels are dilated and blood rushes to the skin). The dermis also houses melanocytes (melanincontaining pigment cells that are derived from the neural crest) and smooth muscle ﬁbers, such as the ones in mammals that produce skin wrinkling around the nipples. In tetrapods, the dermis houses most of the sensory structures and nerves associated with sensations of temperature, pressure, and pain. The hypodermis, or subcutaneous tissue layer, lies between the dermis and the fascia overlying the muscles. This region contains collagenous and elastic ﬁbers and is the area in which subcutaneous fat is stored by Vertebrate Relationships and Basic Structure birds and mammals. The subcutaneous striated muscles of mammals, such as those that enable them to ﬂick the skin to get rid of a ﬂy, are found in this area. Mineralized Tissues Vertebrates have a unique type of mineral called hydroxyapatite, a complex compound of calcium and phosphorus. Hydroxyapatite is more resistant to acid than is calcite (calcium carbonate), which forms the shells of mollusks. The evolution of this unique calcium compound in vertebrates may be related to the fact that vertebrates rely on anaerobic metabolism during activity, producing lactic acid that lowers blood pH. A skeleton made of hydroxyapatite may be more resistant to acidiﬁcation of the blood during anaerobic metabolism than is the calcite that forms the shells of mollusks. Vertebrate mineralized tissues are composed of a complex matrix of collagenous ﬁbers, cells that secrete a proteinaceous tissue matrix, and crystals of hydroxyapatite. The hydroxyapatite crystals are aligned on the matrix of collagenous ﬁbers in layers with alternating directions, much like the structure of plywood. This combination of cells, ﬁbers, and minerals gives bone its complex latticework appearance that combines strength with relative lightness and helps to prevent cracks from spreading. Six types of tissues can become mineralized in vertebrates, and each is formed from a diﬀerent cell lineage in development. • Mineralized cartilage. Cartilage is an important structural tissue in vertebrates and many invertebrates but is not usually mineralized. Mineralized cartilage occurs naturally only in jawed vertebrates, where it forms the main mineralized internal skeletal tissue of sharks. (Sharks and other cartilaginous ﬁshes appear to have secondarily lost true bone.) Some fossil jawless vertebrates also had internal calciﬁed cartilage, probably evolved independently from the condition in sharks. • Bone. The internal skeleton of bony fishes and tetrapods is formed by bone. Bone may replace cartilage in development, as it does in our own skeletons, but bone is not simply cartilage to which minerals have been added. Rather, it is composed of diﬀerent types of cells—osteocytes (Greek osteo = bone and cyte = cell), which are called osteoblasts (Greek blasto = a bud) while they are actually making the bone; in contrast, chondrocytes form cartilage. The cells that form bone and cartilage are derived from the mesoderm, except in the region in the front of the head, where they are derived from neural-crest tissue. Bone and mineralized cartilage are both about 70 percent mineralized. • Enamel and dentine. The other types of mineralized tissues are found in the teeth and in the mineralized exoskeleton of ancestral vertebrates. The enamel and dentine that form our teeth are the most mineralized of the tissues—enamel is about 96 percent mineralized, and dentine is about 90 percent mineralized. This high degree of mineralization explains why teeth are more likely to be found as fossils than are bones. The cells that form dentine (odontoblasts) are derived from neural-crest tissue, and those that form enamel (amyloblasts) are derived from the ectoderm. • Enameloid. Enameloid resembles enamel in its degree of hardness and its position on the outer layer of teeth or dermal scales, but it is produced by mesodermal cells. Enameloid is the enamel-like tissue that was present in ancestral vertebrates and is found today in cartilaginous ﬁshes. Both enamel and enameloid may have evolved independently on a number of occasions. • Cementum. Cementum is a bonelike substance that fastens the teeth in their sockets in some vertebrates, including mammals, and may grow to become part of the tooth structure itself. Bone Bone remains highly vascularized even when it is mineralized (ossiﬁed). This vascularization allows bone to remodel itself. Old bone is eaten away by specialized blood cells (osteoclasts, from the Greek clast = broken), which are derived from the same cell lines as the macrophage white blood cells that engulf foreign bacteria in the body. Osteoblasts enter behind the osteoclasts and deposit new bone. In this way, a broken bone can mend itself and bones can change their shape to suit the mechanical stresses imposed on an animal. This is why exercise builds up bone and why astronauts lose bone in the zero gravity of space. Mineralized cartilage is unable to remodel itself because it does not contain blood vessels. There are two main types of bone in vertebrates: dermal bone, which, as its name suggests, is formed in the skin without a cartilaginous precursor; and endochondral bone, which is formed in cartilage. Dermal bone (Figure 2–6) is the earliest type of vertebrate bone ﬁrst seen in the fossil jawless vertebrates called ostracoderms, which are described in Chapter 3. Only in the bony ﬁshes and tetrapods is the endoskeleton composed primarily of bone. In these vertebrates, the endoskeleton is initially laid down in cartilage and is replaced by bone later in development. Dermal bone originally was formed around the outside of the body, like a suit of armor (ostracoderm Basic Vertebrate Structure 31 Dentine tubercles Spongy acellular bone (a) Basal acellular bone Enamel The Skeletomuscular System Dentine Pulp cavity Cementum Periodontal ligament Enamel organ Dermal papilla (b) Replacement tooth © 1998 The McGraw-Hill Companies, Inc. Figure 2–6 Organization of vertebrate mineralized tissues. (a) Three-dimensional block diagram of dermal bone from an extinct jawless vertebrate (heterostracan ostracoderm). (b) Section through a developing tooth (shark scales are similar). means “shell-skinned”), forming a type of exoskeleton. We think of vertebrates as possessing only an endoskeleton, but most of our skull bones are dermal bones, and they form a shell around our brains. The endoskeletal structure of vertebrates initially consisted of only the braincase and was originally formed from cartilage. Thus, the condition in many early vertebrates was a bony exoskeleton and a cartilaginous endoskeleton (Figure 2–7). Teeth Teeth form from a type of structure called a dermal papilla, so they form only in the skin, usually over dermal bones. When the tooth is fully formed, it erupts through the gum line. Replacement teeth may start to develop to one side of the main tooth even before its eruption. The basic structure of the teeth of jawed vertebrates is like the structure of odontodes, which were the original toothlike components of the original 32 CHAPTER 2 vertebrate dermal armor, and odontodes are homologous with teeth and the dermal denticles of cartilaginous ﬁshes. Teeth are composed of an inner layer of dentine and an outer layer of enamel or enameloid around a central pulp cavity (Figure 2–6). Shark scales (dermal denticles) have a similar structure. There has been considerable controversy about whether dental tissues are always derived from the ectoderm, or whether they can form from the endoderm, as seen in pharyngeal teeth in some ﬁshes. Recent experimental studies have shown that the critical issue in tooth development is the neural-crest precursor that forms the dentine, and that either ectoderm or endoderm may be co-opted to form the outer layers. The basic endoskeletal structural features of chordates are the notochord, acting as a dorsal stiﬀening rod running along the length of the body, and some sort of gill skeleton that keeps the gill slits open. The cranium surrounding the brain was the ﬁrst part of the vertebrate skeleton to evolve. Next the dermal skeleton of external plates and the axial skeleton (vertebrae, ribs, and median ﬁn supports) were added, and still later the appendicular skeleton (bones of the limb skeleton and limb girdles) evolved. The Cranial Skeleton The skull, or cranium, is formed by three basic components: the chondrocranium (Greek chondr = cartilage [literally “gristle”] and cran = skull) surrounding the brain; the splanchnocranium (Greek splanchn = viscera) forming the gill supports; and the dermatocranium (Greek derm = skin) forming in the skin as an outer cover that was not present in the earliest vertebrates. The splanchnocranial components of the vertebrate skeleton are known by a confusing variety of names. In general they can be called gill arches because they support the gill tissue and muscles. The anterior elements of the splanchnocranium are specialized into nongill-bearing structures in all extant vertebrates, such as the jaws of gnathostomes. Other names for these structures are pharyngeal arches (because they form in the pharynx region) and branchial arches (which is just a fancy way of saying gill arches because the Greek word branchi means gill). Yet another name for these structures is visceral arches, because the splanchnocranium is also known as the visceral skeleton. (Still another name associated with the pharyngeal region, aortic arches, refers not to the gill skeleton but to the segmental arteries that supply the gill arches.) Vertebrate Relationships and Basic Structure The dermal skeleton or exoskeleton (a) The endodermal skeleton or endoskeleton Axial Cranial Figure 2–7 Vertebrate skeletons. (b) Appendicular We will call these structures pharyngeal arches when we are discussing the embryonic elements of their development, and gill arches in adults, especially for those arches that actually do bear gill tissue (i.e., arches 3–7). The vertebrate chondrocranium and splanchnocranium are formed primarily from neural-crest tissue, although a splanchnocranium-equivalent formed by endodermal tissue is present in cephalochordates and hemichordates. Thus, a structure with the same function as the vertebrate splanchnocranium preceded the origin of vertebrates and of neural-crest tissue, although only vertebrates have a true splanchnocranium (i.e., one that is derived from neural-crest tissue). The chondrocranium and splanchnocranium are formed from cartilage in the ancestral vertebrate condition, but they are made of endochondral bone in the adults of some bony ﬁshes and most tetrapods. The dermatocranium is made from dermal bone, which is formed in a membrane rather than in a cartilaginous precursor. (Because it forms in a membrane it is sometimes called membrane bone.) The dermatocranium is cartilaginous only as a secondary condition in some ﬁshes, such as sturgeons, where ossiﬁcation of the dermatocranium has been lost. Figure 2–8 shows a diagrammatic representation (a) the originally dermal bone exoskeleton and (b) the originally cartilaginous bone endoskeleton. (The animal depicted is an extinct bony ﬁsh.) of the structure and early evolution of the vertebrate cranium, and Figure 2–9 (see page 36) illustrates three vertebrate crania in more detail. The Cranial Muscles There are two main types of stri- ated muscles in the head of vertebrates: the extrinsic eye muscles and the branchiomeric muscles. Six muscles in each eye rotate the eyeball in all vertebrates except hagﬁshes, in which their absence may represent secondary loss. Like the striated muscles of the body, these muscles are innervated by somatic motor nerves. The branchiomeric muscles are associated with the splanchnocranium and are used to suck water into the mouth during feeding and respiration. Branchiomeric muscles are innervated by cranial nerves that exit from the dorsal part of the spinal cord (unlike striated muscles, which are innervated by motor nerves that exit from the ventral part of the spinal cord). The reason for this diﬀerence is not clear, but it emphasizes the extent to which the vertebrate head diﬀers in its structure and development from the rest of the body. The Axial Skeleton and Musculature The notochord is the original “backbone” of all chordates, although it is never actually made of bone. The notochord has a core Basic Vertebrate Structure 33 Lateral Views (e) Osteichthyan Cross-sectional Views (at level of dotted line in lateral views) Dermal operculum covering gills (connects to shoulder girdle) Jaws encased in dermal bone Chondrocranium Original upper jaw (d) Chondrichthyan Chondrocranium Gill slit that was here was squeezed out and now forms the spiracle Gill arch 1 (= mandibular arch) forms jaws and holds teeth Gill arch 2 (= hyoid arch) forms jaw and tongue support Dermal bone upper jaw Dermal bone lower jaw Dermal bone also forms palate Hyoid elements Gular bones Chondrocranium Chondrocranium and splanchnocranium may now be made of bone Upper jaw Gill arches are now more complexly hinged Teeth Lower jaw Hyoid elements Dermatocranium lost (c) Ostracoderm Original lower jaw Chondrocranium (cartilaginous) Dermal shield surrounds rest of cranium Dermal head shield (bony) Evolution of jaws Splanchnocranial elements may form oral cartilages in this region Splanchnocranium (cartilaginous) Chondrocranium distorted up and back (b) Lamprey Otic capsule Evolution of dermal bone Anterior gill arches (1 and 2) form supports for oral hood and tongue Parachordal Posterior gill arches (3-7 plus 3 additional) form complex branchial basket Lingual cartilage Notochord Branchial basket (a) Basic vertebrate cranium Optic capsule (surrounds eyes and midbrain) Otic capsule (surrounds inner ear and hindbrain) Otic capsule Nasal capsule (surrounds nose and forebrain) Parachordals (posterior underlying support) Parachordal Notochord (extends to front of otic capsule) Gill arch 2 Anterior trabecula 1 2 3 4 5 6 7 (anterior underlying support) Gill openings in here Zigzag shape of gill arches allows muscles to change the volume of the pharynx Chondrocranium Ventral view of basic primitive vertebrate Optic capsule Otic capsule Notochord Splanchnocranium Dermatocranium 34 CHAPTER 2 Vertebrate Relationships and Basic Structure Nasal capsule Parachordal Anterior trabecula Figure 2–8 Diagrammatic view of the form and early evolution of the cranium of vertebrates. The ancestral condition (a) was to have a chondrocranium formed from the paired sensory capsules, one pair for each part of the tripartite brain, with the underlying support provided by paired anterior trabeculae (at least in jawed vertebrates) and parachordals ﬂanking the notochord posteriorly. The splanchnocranium was probably ancestrally made up of seven pairs of pharyngeal arches supporting six gill openings, without any anterior specializations. In the lamprey (b), the mandibular (second segment arch, but termed arch 1 because there is no arch in the ﬁrst segment) pharyngeal arch becomes the velum and other supporting structures in the head, and the remainder of the splanchnocranium forms a complex branchial basket on the outside of the gills (possibly in association with the unique mode of tidal gill ventilation). Above the level of the lamprey, the chondrocranium and splanchnocranium are surrounded with a dermatocranium of dermal bone, as ﬁrst seen in ostracoderms (c). In gnathostomes (d,e) the pharyngeal arches of the mandibular (second) and hyoid (third) head segments become modiﬁed to form the jaws and jaw supports. The dermatocranium is lost in chondrichthyans (d). In osteichthyans (e) the dermatocranium forms in a characteristic pattern, including a bony operculum covering the gills and aiding in ventilation in bony ﬁshes. The basics of this pattern are still seen in us. of large, closely spaced cells packed with incompressible ﬂuid-ﬁlled vacuoles wrapped in a complex ﬁbrous sheath that is the site of attachment for segmental muscles and connective tissues. The notochord ends anteriorly just posterior to the pituitary gland and continues posteriorly to the tip of the ﬂeshy portion of the tail. The original form of the notochord is lost in adult tetrapods, but portions remain as components of the intervertebral discs between the vertebrae. The axial muscles are composed of myomeres that are complexly folded in three dimensions so that each one extends anteriorly and posteriorly over several body segments (Figure 2–10 on page 37). Sequential muscle blocks overlap and produce undulation of the body when they contract. In amphioxus, myomeres have a simple V shape, whereas in vertebrates they have a W shape. The myomeres of jawed vertebrates are divided into epaxial (dorsal) and hypaxial (ventral) portions by a sheet of ﬁbrous tissue called the horizontal septum. The segmental pattern of the axial muscles is clearly visible in ﬁshes. It is easily seen in a piece of raw or cooked ﬁsh where the ﬂesh ﬂakes apart in zigzag blocks, each block representing a myomere. (This pattern is similar to the fabric pattern of interlocking V shapes known as herringbone, although “herring muscle” would be a more accurate description.) In tetrapods, the pattern is less obvious, but the segmental pattern can be observed on the six-pack stomach of body builders, where each ridge represents a segment of the rectus abdominis muscle (a hypaxial muscle of tetrapods). Locomotion Many small aquatic animals, especially lar- vae, move by using cilia to beat against the water. However, ciliary propulsion works only at very small body sizes. Adult chordates use the serial contraction of segmental muscle bands in the trunk and tail for locomo- tion, a feature that possibly ﬁrst appeared as a startle response in larvae. The notochord stiﬀens the body so it bends from side to side as the muscles contract. (Without the notochord, contraction of these muscles would merely compress the body like an accordion.) Most ﬁshes still use this basic type of locomotion. The paired ﬁns of jawed ﬁshes are generally used for steering, braking, and providing lift–not for propulsion except in some specialized ﬁshes such as skates and rays that have winglike pectoral ﬁns and in some derived bony ﬁshes (teleosts) such as seahorses and coral reef ﬁshes. Energy Acquisition and Support of Metabolism Food must be processed by the digestive system into molecules small enough to pass through the walls of the intestine, then transported by the circulatory system to the body tissues. Oxygen is required for this process; the respiratory and the circulatory systems are closely intertwined with those of the digestive system. Feeding and Digestion Feeding includes getting food into the mouth, mechanical processing (“chewing” in the broad sense—although today only mammals truly chew their food), and swallowing. Digestion includes the breakdown of complex compounds into small molecules that are absorbed across the wall of the gut and transported to the tissues. Vertebrate ancestors probably ﬁltered small particles of food from the water, as amphioxus and larval lampreys still do. Most vertebrates are particulate feeders; that is, they take in their food as bite-sized pieces rather than as tiny particles. Vertebrates move the food through the gut by rhythmical muscular contractions (peristalsis), and digest it by secreting digestive enzymes produced by the liver and the pancreas Basic Vertebrate Structure 35 (a) Chondrocranium and splanchnocranium of a lamprey Nasal capsule Position of first gill pouch Arcualia (vertebral rudiment) Gill slit Optic capsule Otic capsule Cartilages around buccal funnel Notochord Lingual cartilage Pericardial cartilage Branchial arch Hypobranchial rod Horizontal bar (b) Chondrocranium and splanchnocranium of a shark Position of spiracle Otic capsule Hyomandibular Position of first gill pouch Pharyngobranchial Optic capsule Rostrum Nasal capsule Epibranchial Position of fifth gill pouch Ceratobranchial Palatoquadrate Mandibular arch (upper and lower jaw) Hyoid arch Mandibular cartilage Hypobranchial Basibranchial (c) Dermatocranium of a primitive generalized (basal) bony fish Dermal roof bones Orbit Dermal bones of the supracleithral and opercular series Naris Palatoquadrate cartilage Mandibular cartilage (articular) Figure 2–9 The crania of three vertebrates. (a) the chondrocranium and splanchnocranium of a lamprey compared to (b) the chondrocranium and splanchnocranium of a living cartilaginous vertebrate (a shark), and (c) the dermatocranium of a generalized bony ﬁsh. 36 CHAPTER 2 Dermal lower jaw bones Dermal gular bones Splanchnocranium Chondrocranium Dermatocranium Vertebrate Relationships and Basic Structure © 1994 Cengage Learning, Inc. Myomeres V-shaped (a) Amphioxus Myomeres W-shaped Lamprey (b) Horizontal septum Shark (dogfish) Myomeres more complexly folded (c) Epaxial muscle Red muscle Hypaxial muscle (d) Bony fish (perch) Figure 2–10 Chordate body muscles (myomeres). (a) amphioxus (nonvertebrate chordate), (b) lamprey (jawless vertebrate), and jawed vertebrates, (c) shark, and (d) bony ﬁsh. into the gut. The pancreas also secretes the hormones insulin and glucagon, which are involved in the regulation of glucose metabolism and blood-sugar levels. In the primitive vertebrate condition, there is no stomach, no division of the intestine into small and large portions, and no distinct rectum. The intestine empties to the cloaca, which is the shared exit for the urinary, reproductive, and digestive systems in all vertebrates except therian mammals. Respiration and Ventilation Ancestral chordates probably relied on oxygen absorption and carbon dioxide loss by diﬀusion across a thin skin (cutaneous respiration). This is the mode of respiration of amphioxus, which is small and sluggish. Cutaneous respiration is important for many vertebrates (especially modern amphibians), but the combination of large body size and high levels of activity make specialized gas-exchange structures essential for most vertebrates. Gills are eﬀective in water, whereas lungs work better in air. Both gills and lungs have large surface areas that allow oxygen to diﬀuse from the surrounding medium (water or air) into the blood. Cardiovascular System Blood carries oxygen and nutrients through the vessels to the cells of the body, removes carbon dioxide and other metabolic waste products, and stabilizes the internal environment. Blood also carries hormones from their sites of release to their target tissues. Blood is a ﬂuid tissue composed of liquid plasma, red blood cells (erythrocytes) that contain the ironrich protein hemoglobin, and several diﬀerent types of white blood cells (leukocytes) that are part of the Basic Vertebrate Structure 37 Carotid arteries Mesenteric arteries Dorsal aorta Efferent gill arteries Heart Ventricle Sinus venosus HEAD Segmental arteries GUT TRUNK GILLS Ventral aorta LIVER Atrium Hepatic vein Common cardinal vein Anterior cardinal veins KIDNEY Hepatic portal vein Renal portal vein Posterior cardinal veins Oxygenated blood Deoxygenated blood Figure 2–11 Diagrammatic plan of vertebrate cardiovascular circuit. All vessels are paired on the left and right sides of the body except for the midline ventral aorta and dorsal aorta. Note that the cardinal veins actually run dorsally in the real animal, ﬂanking the carotid arteries (anterior cardinals) or the dorsal aorta (posterior cardinals). immune system. Cells specialized to promote clotting of blood (called platelets or thrombocytes) are present in all vertebrates except mammals, in which they are replaced by noncellular platelets. Vertebrates have closed circulatory systems; that is, the arteries and veins are connected by capillaries. Arteries carry blood away from the heart, and veins return blood to the heart (Figure 2–11). Blood pressure is higher in the arterial system than in the venous system, and the walls of arteries have a layer of smooth muscle that is absent from veins. The following features are typical of vertebrate circulatory systems: • Capillary beds. Interposed between the smallest arteries (arterioles) and the smallest veins (venules) are the capillaries, which are the sites of exchange between blood and tissues. Their walls are only one cell layer thick; so diﬀusion is rapid, and capillaries pass close to every cell. Collectively the capillaries provide an enormous surface area for the exchange of gases, nutrients, and waste products. Arteriovenous anastomoses connect some arterioles directly to venules, allowing blood to bypass a capillary bed, and normally only a fraction of the capillaries in a tissue have blood ﬂowing through them. • Portal vessels. Blood vessels that lie between two capillary beds are called portal vessels. The hepatic 38 CHAPTER 2 portal vein, seen in all vertebrates, lies between the capillary beds of the gut and the liver (see Figure 2–11). Substances absorbed from the gut are transported directly to the liver, where toxins are rendered harmless and some nutrients are processed or removed for storage. Most vertebrates also have a renal portal vein between the veins returning from the tail and posterior trunk and the kidneys (see Figure 2–11). The renal portal system is not well developed in jawless vertebrates and has been lost in mammals. • The heart. The vertebrate heart is a muscular tube folded on itself and is constricted into three sequential chambers: the sinus venosus, the atrium, and the ventricle. Our so-called four-chambered heart lacks a distinct sinus venosus, and the original atrium and ventricle have been divided into left and right chambers. The sinus venosus is a thin-walled sac with few cardiac muscle ﬁbers. Suction produced by muscular contraction draws blood anteriorly into the atrium, which has valves at each end that prevent backﬂow. The ventricle is thick-walled, and the muscular walls have an intrinsic pulsatile rhythm, which can be speeded up or slowed down by the nervous system. Contraction of the ventricle forces the blood into the ventral aorta. Mammals no longer have a distinct structure identiﬁable as the sinus venosus; Vertebrate Relationships and Basic Structure rather, it is incorporated into the wall of the right atrium as the sinoatrial node, which controls the basic pulse of the heartbeat. • The aorta. The basic vertebrate circulatory plan consists of a heart that pumps blood into the single midline ventral aorta. Paired sets of aortic arches (originally six pairs) branch from the ventral aorta (Figure 2–12). One member of each pair supplies the left side and the other the right side. In the original vertebrate circulatory pattern, which is retained in ﬁshes, the aortic arches lead to the gills, where the blood is oxygenated and returns to the dorsal aorta. The dorsal aorta is paired above the gills, and the vessels from the most anterior arch run forward to the head as the carotid arteries. Behind the gill region, the two vessels unite into a single dorsal aorta that carries blood posteriorly. is known as an opisthonephric kidney. The compact bean-shaped kidney seen in adult amniotes (the metanephric kidney) includes only the metanephros, drained by a new tube, the ureter, derived from the basal portion of the archinephric duct. The basic units of the kidney are microscopic structures called nephrons. Vertebrate kidneys work by ultraﬁltration: high blood pressure forces water, ions, and small molecules through tiny gaps in the capillary walls. Nonvertebrate chordates lack true kidneys. Amphioxus has excretory cells called solenocytes associated with the pharyngeal blood vessels that empty individually into the false body cavity (the atrium). The eﬄuent is discharged to the outside via the atriopore. The solenocytes of amphioxus are thought to be homologous with the podocytes of the vertebrate nephron, which are the cells that form the wall of the renal capsule. The dorsal aorta is ﬂanked by paired cardinal veins that return blood to the heart (see Figure 2–11). Anterior cardinal veins (the jugular veins) draining the head and posterior cardinal veins draining the body unite on each side in a common cardinal vein that enters the atrium of the heart. In lungﬁshes and tetrapods, the posterior cardinal veins are essentially replaced by a single midline vessel, the posterior vena cava. Blood is also returned separately to the heart from the gut and liver via the hepatic portal system. The Gonads—Ovaries and Testes Although the gonads are derived from the mesoderm, the gametes (eggs and sperm) are formed in the endoderm and then migrate up through the dorsal mesentery (see Figure 2–5) to enter the gonads. The archinephric duct drains urine from the kidney to the cloaca and from there to the outside world. In jawed vertebrates, this duct is also used for the release of sperm by the testes. Reproduction is the means by which gametes are produced, released, and combined with gametes from a member of the opposite sex to produce a fertilized zygote. Vertebrates usually have two sexes, and sexual reproduction is the norm—although unisexual species occur among ﬁshes, amphibians, and lizards. The gonads are paired in jawed vertebrates but are single in the jawless ones: it is not clear which represents the ancestral vertebrate condition. The gonads usually lie on the posterior body wall behind the peritoneum (the lining of the body cavity); it is only among mammals that the testes are found outside the body in a scrotum. The gonads (ovaries in females, testes in males) also produce hormones, such as estrogen and testosterone. In living jawless vertebrates, which probably represent the ancestral vertebrate condition, there is no special tube or duct for the passage of the gametes. Rather, the sperm or eggs erupt from the gonad and move through the coelom to pores that open to the base of the archinephric ducts. In jawed vertebrates, however, the gametes are always transported to the cloaca via specialized paired ducts (one for each gonad). In males, sperm are released directly into the archinephric ducts that drain the kidneys in non-amniotes and embryonic amniotes. In females, the egg is still released into the coelom but is then transported via a new structure, the oviduct. The oviducts produce Excretory and Reproductive Systems Although the func- tions of the excretory and reproductive systems are entirely diﬀerent, both systems are formed from the nephrotome or intermediate mesoderm, which forms the embryonic nephric ridge (Figure 2–13 on page 41). The kidneys are segmental, whereas the gonads (ovaries in females and testes in males) are unsegmented. The Kidneys The kidneys dispose of waste products, pri- marily nitrogenous waste from protein metabolism, and regulate the body’s water and minerals—especially sodium, chloride, calcium, magnesium, potassium, bicarbonate, and phosphate. In tetrapods the kidneys are responsible for almost all these functions, but in ﬁshes and amphibians the gills and skin also play important roles (see Chapter 4). The kidney of ﬁshes is a long, segmental structure extending the entire length of the dorsal body wall. In all vertebrate embryos, the kidney is composed of three portions: pronephros, mesonephros, and metanephros (see Figure 2–13). The pronephros is functional only in the embryos of living vertebrates and possibly in adult hagﬁshes. The kidney of adult ﬁshes and amphibians includes the mesonephric and metanephric portions and Basic Vertebrate Structure 39 (d) Generalized gnathostome Arch 1 lost L R Gill slit 1 turned into spiracle 1 Gill slit 3 2 Ventral aorta Subclavian artery (to forelimb) 4 Aortic arch (c) Lamprey R Arch 1 lost L Carotid artery 5 6 Conus arteriosus (fourth heart chamber) 1 2 Gill slit 1 lost Ventricle 3 Carotid artery Atrium 4 Ventral aorta 5 Gill slit Sinus venosus 6 Aortic arch 7 3 gill slits and aortic arches added posteriorly 8 9 Dorsal aorta Iliac artery (to hindlimb) End of body in all gnathostomes Caudal artery (to tail) Ventricle Atrium Sinus venosus Dorsal aorta (b) Hypothetical first vertebrate R L Carotid artery 1 Gill slit 2 3 Aortic arch 4 Ventral aorta 5 6 (a) Protovertebrate R 1 Gill slit 2 L Carotid artery (takes blood to head) 3 Aortic arch Atrium Ventricle Sinus venosus Dorsal aorta 4 5 6 Ventral aorta (under gills, blood runs forward) Single heart chamber (= sinus venosus) 40 CHAPTER 2 Paired dorsal aortae (above gills, blood runs back toward body) Cardinal veins returning blood from body and head Dorsal aorta (to body) Vertebrate Relationships and Basic Structure Oxygenated blood Deoxygenated blood Figure 2–12 Diagrammatic view of the form and early evolution of the heart and aortic arches of vertebrates. The view is from the ventral side of the animal. In the protovertebrate (a) and the earliest vertebrate condition (b), there were probably six pairs of aortic arches, just as are seen in the embryos of all living vertebrates, although arch 1 is never seen in the adults. Protovertebrates used these arches for feeding, not for respiration, so the blood is not shown as picking up oxygen in this illustration. In lampreys (c), additional aortic arches are added posteriorly to accommodate more gill openings. In gnathostomes (d), the subclavian and iliac arteries are added to the main circulatory system, supplying the forelimbs and hindlimbs, respectively. A fourth chamber is also added to the heart, the conus arteriosus, which damps out the pulsatile component of the blood ﬂow. In jawed vertebrates, elongated portions of the neurons, the axons, are encased in a fatty insulating coat, the myelin sheath, that increases the conduction velocity of the nerve impulse. The axons are generally collected like wires in a cable, forming a nerve. Information enters the neuron via short processes called dendrites. The brain and spinal cord are known as the central nervous system (CNS), and the nerves running between the CNS and the body are known as the peripheral nervous system (PNS). substances associated with the egg, such as the yolk or the shell. The oviducts can become enlarged and fused in various ways to form a single uterus or paired uteri in which eggs are stored and young develop. Vertebrates may deposit eggs that develop outside the body or retain the eggs within the mother’s body until embryonic development is complete. Shelled eggs must be fertilized in the oviduct before the shell and albumen are deposited. Many viviparous vertebrates and vertebrates that lay shelled eggs have some sort of intromittent organ—such as the pelvic claspers of sharks and the penis of amniotes—by which sperm are inserted into the female’s reproductive tract. The Spinal Cord The nerves of the PNS are segmentally arranged, exiting from either side of the spinal cord between the vertebrae. The spinal cord receives sensory inputs, integrates them with other portions of the CNS, and sends impulses that cause muscles to contract and glands to alter their secretion. The spinal cord has considerable autonomy in many vertebrates. Even complex movements such as swimming are controlled by the spinal cord rather than the brain, and ﬁshes continue coordinated swimming movements when the brain is severed from the spinal cord. Our familiar knee-jerk Coordination and Integration The nervous and endocrine systems respond to conditions inside and outside an animal, and together they control the actions of organs and muscles, coordinating them so they work in concert. General Features of the Nervous System Individual cells called neurons are the basic units of the nervous system. Nephric ridge Archinephric duct extending by proliferation Archinephric duct derived from tubules Developing lateral plate Pronephric tubules Nephrotomes Developing somites M o Pr ph ne s ro o es s ro ph e n Op a- s et ro M ph ne s ro ph e n ho ist Figure 2–13 Kidney development in a generalized vertebrate embryo, showing the nephrotome regions. Basic Vertebrate Structure 41 reaction is produced by the spinal cord as a reﬂex arc. Development of more complex connections within the spinal cord and between the spinal cord and the brain has been a trend in vertebrate evolution. The Somatic and Visceral Nervous Systems Vertebrates are unique in having a dual type of nervous system: the somatic nervous system (known as the voluntary nervous system) and the visceral nervous system (called the involuntary nervous system). This dual nervous system mimics the dual pattern of development of the mesoderm. The somatic nervous system innervates the structures derived from the segmented portion (the somites), including the striated muscles that we can move consciously (e.g., the limb muscles), and relays information from sensation that we are usually aware of (e.g., from temperature and pain receptors in the skin). The visceral nervous system innervates the smooth and cardiac muscles that we usually cannot move consciously (e.g., the gut and heart muscles) and relays information from sensations that we are not usually aware of, such as the receptors monitoring the levels of carbon dioxide in the blood. Each spinal nerve complex is made up of four types of ﬁbers: somatic sensory ﬁbers coming from the body wall, somatic motor ﬁbers running to the body, visceral sensory ﬁbers coming from the gut wall and blood vessels, and visceral motor ﬁbers running to the muscles and glands of the gut and to the blood vessels of both the gut and peripheral structures like the skin. The motor portion of the visceral nervous system is known as the autonomic nervous system. In more derived vertebrates, such as mammals, this system becomes divided into two portions: the sympathetic nervous system (usually acting to speed things up) and the parasympathetic nervous system (usually acting to slow things down). Cranial Nerves Vertebrates also have nerves that emerge directly from the brain; these cranial nerves (10 pairs in the ancestral vertebrate condition, 12 in amniotes) are identiﬁed by Roman numerals. Some of these nerves, such as the ones supplying the nose (the olfactory nerve, I) or the eyes (the optic nerve, II), are not really nerves at all, but tracts—that is, parts of the CNS. Somatic motor ﬁbers in cranial nerves innervate the muscles that move the eyeballs and the branchiomeric muscles that power the jaws and gill arches, and sensory nerves convey information from the head, including the sense of taste. The special sensory nerves that supply the lateral line in ﬁshes are also derived from the cranial nerves. The vagus nerve (cranial nerve X) ramiﬁes through all but the most posterior part of the trunk, carrying 42 CHAPTER 2 the visceral motor nerve supply to various organs. People who break their necks may be paralyzed from the neck down (i.e., lose the function of their skeletal muscles) but still may retain their visceral functions (workings of the gut, heart, etc.) because the vagus nerve is independent of the spinal cord and exits above the break. Brain Anatomy and Evolution All chordates have some form of a brain, as a thickening of the front end of the notochord. The brain of all vertebrates is a tripartite (three-part) structure (Figure 2–14), and the telencephalon (front part of the forebrain) and the olfactory receptors are probably true new features in vertebrates. In the ancestral condition, the forebrain is associated with the sense of smell, the midbrain with vision, and the hindbrain with balance and detection of vibrations (hearing, in the broad sense). These portions of the brain are associated with the nasal, optic, and otic capsules of the chondrocranium, respectively (see Figure 2–8). The vertebrate brain has three parts: the forebrain, midbrain, and hindbrain. The Forebrain The forebrain has two parts: • The anterior region of the adult forebrain, the telencephalon, develops in association with the olfactory capsules and coordinates inputs from other sensory modalities. The telencephalon becomes enlarged in diﬀerent vertebrate groups and is known as the cerebrum or cerebral hemispheres. Tetrapods developed an area in the cerebrum called the neocortex or neopallium, which is the primary seat of sensory integration and nervous control. Bony ﬁshes also evolved a larger, more complex telencephalon, but by a completely diﬀerent mechanism. Sharks and, perhaps surprisingly, hagﬁshes independently evolved relatively large forebrains, although a large cerebrum is primarily a feature of birds and mammals. • The posterior region of the forebrain is the diencephalon, which contains structures that act as a major relay station between sensory areas and the higher brain centers. The pituitary gland, an important endocrine organ, is a ventral outgrowth of the diencephalon. The ﬂoor of the diencephalon (the hypothalamus) and the pituitary gland form the primary center for neural-hormonal coordination and integration. Another endocrine gland, the pineal organ, is a median dorsal outgrowth of the diencephalon that is a photoreceptor. Many early tetrapods had a hole in the skull over the pineal gland to admit light, and this condition persists in some reptiles (e.g., the tuatara and many lizards). Vertebrate Relationships and Basic Structure MYELENCEPHALON Forebrain (prosencephalon) Midbrain (mesencephalon) Hindbrain (rhombencephalon) METENCEPHALON DIENCEPHALON TELENCEPHALON Lateral view of the exterior of the brain Cerebellum Lateral view of a section cut through the midline of the brain Tectum Pineal gland Cortex Ventricle IV Medulla oblongata Figure 2–14 The generalized vertebrate brain. The Midbrain The midbrain develops in conjunction with the eyes and receives input from the optic nerve, although in mammals the forebrain has taken over much of the task of vision. The Hindbrain The hindbrain has two portions: • The posterior portion, the myelencephalon, or medulla oblongata, controls functions such as respiration and acts as a relay station for receptor cells from the inner ear. • The anterior portion of the hindbrain, the metencephalon, develops an important dorsal outgrowth, the cerebellum—present as a distinct structure only in jawed vertebrates among living forms. The cerebellum coordinates and regulates motor activities whether they are reflexive (such as maintenance of posture) or directed (such as escape movements). The Sense Organs We think of vertebrates as having ﬁve senses—taste, touch, sight, smell, and hearing— but this list does not reﬂect the ancestral condition, nor does it include all the senses of living vertebrates. Complex, multicellular sense organs that are formed from epidermal placodes and tuned to the sensory worlds of the species that possess them are derived features of many vertebrate lineages. Pituitary gland Hypothalamus Olfactory bulb © 1998 The McGraw-Hill Companies, Inc. Chemosensation: Smell and Taste The senses of smell and taste both involve the detection of dissolved molecules by specialized receptors. We think of these two senses as being closely interlinked; for example, our sense of taste is poorer if our sense of smell is blocked because we have a cold. However, the two senses are actually very diﬀerent in their innervation. Smell is a somatic sensory system—sensing items at a distance, with the sensations being received in the forebrain. Taste is a visceral sensory system—sensing items on direct contact, with the sensations being received initially in the hindbrain. Vision The receptor ﬁeld of the vertebrate eye is arrayed in a hemispherical sheet, the retina, which originates as an outgrowth of the brain. The retina contains two types of light-sensitive cells, cones and rods, which are distinguished from each other by morphology, photochemistry, and neural connections. Electroreception The capacity to perceive electrical im- pulses generated by the muscles of other organisms is also a form of distance reception, but one that works only in water. Electroreception was probably an important feature of early vertebrates and is seen today primarily in ﬁshes and monotreme mammals. Many extant ﬁshes produce electrical discharge for communication with other individuals or for protection from predators. Basic Vertebrate Structure 43 Balance and Orientation Originally the structures in the inner ear (the vestibular apparatus) detected an animal’s position in space, and these structures retain that function today in both aquatic and terrestrial vertebrates. The basic sensory cell in the inner ear is the hair cell, which detects the movement of ﬂuid resulting from a change of position or the impact of sound waves. The vestibular apparatus (one on either side of the animal) is enclosed within the otic capsule of the skull and consists of a series of sacs and tubules containing a ﬂuid called endolymph. The lower parts of the vestibular apparatus, the sacculus and utriculus, house sensory organs called maculae, which contain tiny crystals of calcium carbonate resting on hair cells. Sensations from the maculae tell the animal which way is up and detect linear acceleration. The upper part of the vestibular apparatus contains the semicircular canals. Sensory areas located in swellings at the end of each canal (ampullae) detect angular acceleration through cristae, hair cells embedded in a jellylike substance, by monitoring the displacement of endolymph during motion. Jawed vertebrates have three semicircular canals on each side of the head, hagﬁshes have one, and lampreys and fossil jawless vertebrates have two (Figure 2–15). We often fail to realize the importance of our own vestibular senses because we usually depend on vision to Lamprey (2 semicircular canals) Posterior SSC Posterior ampulla Endolymphatic duct Anterior SSC Anterior ampulla Lagena (a) Sacculus Utriculus Generalized gnathostome (shark) (3 semicircular canals) Posterior SSC Horizontal SSC Endolymphatic duct Anterior SSC Utriculus © 1998 The McGraw-Hill Companies, Inc. Figure 2–15 Anatomy of the vestibular apparatus in fishes. The lamprey (a) has two semicircular canals (SSC), whereas gnathostomes (represented by a shark, b) have three semicircular canals. 44 Detection of Water Vibration Fishes and aquatic amphib- ians have a structure running along the body on either side, called the lateral line. Within this system, hair cells are aggregated into neuromast organs, which detect the movement of water around the body, and information is then fed back to the vestibular apparatus for integration (see Chapter 4). Hearing The inner ear is also used for hearing (recep- tion of sound waves) by tetrapods and by a few derived ﬁshes. In tetrapods only, the inner ear contains the cochlea (organ of hearing, also known as the lagena in nonmammalian tetrapods). The cochlea and vestibular apparatus together are known as the membranous labyrinth. Sound waves are transmitted to the cochlea, where they create waves of compression that pass through the endolymph. These waves stimulate the auditory sensory cells, which are variants of the basic hair cell. The Endocrine System The endocrine system transfers information from one part of the body to another via the release of a chemical messenger (hormone) that produces a response in the target cells. The time required for an endocrine response ranges from seconds to hours. Hormones are produced in discrete endocrine glands, whose primary function is hormone production and excretion (e.g., the pituitary, thyroid, thymus, and adrenals), and by organs with other major bodily functions—such as the gonads, kidneys, and gastrointestinal tract. Endocrine secretions are predominantly involved in controlling and regulating energy use, storage, and release, as well as in allocating energy to special functions at critical times. The trend in the evolution of vertebrate endocrine glands has been consolidation from scattered clusters of cells or small organs in ﬁshes to larger, better-deﬁned organs in amniotes. The Immune System Sacculus (b) determine our position. We can sometimes be fooled, however, as when sitting in a stationary train or car and thinking that we are moving, only to realize from the lack of input from our vestibular system that it is the vehicle next to us that is moving. CHAPTER 2 Vertebrates have adaptive immunity, a type of immune system diﬀerent from that of invertebrates. While all animals have innate, speciﬁcally genetically encoded responses to pathogens that are ﬁxed and unchanging, vertebrates additionally have evolved lymphocytes (a type of white blood cell), which provide a system of adjustable antigen recognition. The adaptive immune systems of jawless and jawed vertebrates are somewhat Vertebrate Relationships and Basic Structure different. While gnathostomes generate lymphocyte receptors via immunoglobulin gene segments, lampreys and hagﬁshes employ leucine-rich repeat molecules. In addition, lampreys and hagﬁshes lack a thymus gland, which produces lymphocytes in gnathostomes. The agnathan condition is probably the ancestral condition for vertebrates, as it more closely resembles the mode of antipathogen responses in invertebrates. Summary Vertebrates are large, active members of the phylum Chordata, a group of animals whose other members, tunicates and cephalochordates (amphioxus), are small and sluggish or entirely sessile as adults. Chordates share with many other derived animal phyla the features of being bilaterally symmetrical, with a distinct head and tail end. Both embryological and molecular evidence show that chordates are related to other sessile marine animals, such as echinoderms. Chordates are distinguished from other animals by the presence of a notochord, a dorsal hollow nerve chord, a muscular postanal tail, and an endostyle (which is homologous to the vertebrate thyroid gland). Vertebrates appear to be most closely related to tunicates among nonvertebrate chordates, and most of the differences in structure and physiology between vertebrates and other chordates reflect an evolutionary change to larger body sizes, greater levels of activity, and predation rather than filter feeding. Vertebrates have the unique features of an expanded head with multicellular sense organs and a cranium housing an enlarged, tripartite brain. The features that distinguish vertebrates from other chordates appear to be related to two critical embryonic innovations: a doubling of the Hox gene complex and the development of neural-crest tissue. The diverse activities of vertebrates are supported by a complex morphology. Study of embryology can throw light on the genetic basis and developmental pathways underlying these structures. In particular, neural-crest cells, which are unique to vertebrates, are responsible for many of their derived characters, especially those of the new anterior portion of the head. An adult vertebrate can be viewed as a group of interacting anatomical and physiological systems involved in protection, support and movement, acquisition of energy, excretion, reproduction, coordination, and integration. These systems underwent profound changes in function and structure at several key points in vertebrate evolution. The most important transition was from the prevertebrate condition—as represented today by the cephalochordate amphioxus—to the vertebrate condition, shown by hagfishes and lampreys. Other important transitions, to be considered in later chapters, include the shift from jawless to jawed vertebrates and from fish to tetrapod. Discussion Questions 1. Suppose new molecular data showed that tunicates and vertebrates are sister taxa. What difference would this make to our assumptions about the form of the original chordate animal? What additional features might this animal have possessed? 2. We noted that many features typical of vertebrates, such as a distinct head and limbs of some sort for locomotion, are seen in invertebrates such as insects and crustaceans, and even in cephalopods (squid and octopuses). What characteristic do these animals share with vertebrates that might have led to the independent evolution of such structures? 3. Why would evidence of sense organs in the head of a fossil animal (such as Haikouella) suggest a close relationship with vertebrates—that is, which critical vertebrate feature would have to be present? Vertebrates have been thought of as “dual animals,” consisting of both segmented and unsegmented portions. How is this duality reflected in their embryonic development and the structure of the nervous system? 5. Among the extant vertebrates, only the bony fishes (Osteichthyes) possess bone. (a) Why, then, do we make the assumption that the ancestors of cartilaginous fishes must have had bone that was subsequently lost? (b) Why don’t we think this is true for the lampreys and hagfishes? 6. Amphioxus obtains its oxygen by diffusion over the body surface. Why aren’t vertebrates generally able to do this—that is, why do most aquatic vertebrates rely on gills for gas exchange? Discussion Questions 45 Additional Information Donoghue, P. C. J., and M. A. Purnell. 2009. The evolutionary emergence of vertebrates from among their spineless relatives. Evolution Education Outreach 2:204–212. Fraser, G. J., et al. 2010. The odontode explosion: The origin of tooth-like structures in vertebrates. Bioessays 32:808–817. Gai, Z. et al. 2011. Fossil jawless fish from China foreshadows early jawed vertebrate anatomy. Nature 476:324–327. Garcia-Fernàndez, J., and E. Benito-Gutiérrez. 2009. It’s a long way from amphioxus: Descendants of the earliest chordate. BioEssays 31:665–675. Guo, P., et al. 2009. Dual nature of the adaptive immune system in lampreys. Nature 459:796–801. Huysseune, A., et al. 2009. Evolutionary and developmental origins of the vertebrate dentition. Journal of Anatomy 214:465–476. Janvier, J. 2011. Comparative anatomy: All vertebrates do have vertebrae. Current Biology 21:R661–R663. Kuraku, S. 2011. Hox Gene clusters of early vertebrates: Do they serve as reliable markers for genome evolution? Genomics, Proteomics & Bioinformatics 9:97–103. Lowe, C. B. et al. 2011. Three periods of regulatory innovation during vertebrate evolution. Science 333:1019–1024. Manning, G., and E. Scheeﬀ. 2010. How the vertebrates were made: Selective pruning of a double-duplicated genome. BMC Biology 8:144. Ouilon, S. et al. 2011. Evolution of repeated structures along the body axis of jawed vertebrates, insights from the Scyliorhinus canicula Hox code. Evolution & Development 13:247–259. 46 CHAPTER 2 Sansom R. S., et al. 2011. Decay of vertebrate characters in hagﬁsh and lamprey (Cyclostomata) and the implications for the vertebrate fossil record. Proceedings of the Royal Society B 278:1150–1157. Shubin, N. 2009. Your Inner Fish: A Journey into the 3.5-Billion-Year History of the Human Body. New York: Vintage Books. Sire, H.-Y., et al. 2009. Origin and evolution of the integumentary skeleton in non-tetrapod vertebrates. Journal of Anatomy 214:409–440. Websites Metazoan diversity: The Cambrian explosion .html Introduction to the Urochordata Introduction to the Cephalochordata The amphioxus song (annotated) Haikouella and other Chengjiang animals Xidazoon-Haikouella/XidazoonHaikouella.htm Hox genes and metazoan evolution .php Evolution of the circulatory system .html Vertebrate Relationships and Basic Structure 3 T he earliest vertebrates represented an important advance over the nonvertebrate chordate filter feeders. The most conspicuous new feature of these early vertebrates was a distinct head end, containing a tripartite brain enclosed by a cartilaginous cranium (skull) and complex sense organs. They used the newly acquired pharyngeal musculature that powered the gill skeleton to draw water into the mouth and over the gills, which were now used for respiration rather than for filter feeding. Early vertebrates were active predators rather than sessile filter feeders. Many of them also had external armor made of bone and other mineralized tissues. We know a remarkable amount about the anatomy of some of these early vertebrates because the internal structure of their bony armor reveals much of their soft anatomy. Gnathostomes represent an advance in the vertebrate body plan for high levels of activity and predation. Jaws themselves are homologous with the structures that form the gill arches and probably first evolved as devices to improve the strength and effectiveness of gill ventilation. Later the jaws were modified for seizing and holding prey. In this chapter we trace the earliest steps in the radiation of vertebrates, beginning more than 500 million years ago, and discuss the biology of both the Paleozoic agnathans (the ostracoderms) and the extant forms (hagfishes and lampreys). We also consider the transition from the jawless condition to the jawed one and the biology of some of the early types of jawed fishes that did not survive the Paleozoic era (placoderms and acanthodians). CHAPTER Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates 3.1 Reconstructing the Biology of the Earliest Vertebrates The Earliest Evidence of Vertebrates Until very recently our oldest evidence of vertebrates consisted of fragments of the dermal armor of the jawless vertebrates colloquially known as ostracoderms. These animals were very different from any vertebrate alive today. They were essentially fishes encased in bony armor (Greek ostrac = shell and derm = skin), quite unlike the living jawless vertebrates that lack bone completely. Bone fragments that can be assigned definitively to vertebrates are known from the Ordovician period, some 480 million years ago, although there are some pieces of mineralized material that are tentatively believed to represent vertebrates from the Late Cambrian period, around 500 million years ago. This was about 80 million years before whole-body vertebrate fossils became abundant, in the Late Silurian period. 47 Early Cambrian Vertebrates Recent finds of early soft-bodied vertebrates from the Early Cambrian of China extend the fossil record back by another 40 million years or so, to around 520 million years ago. Two Early Cambrian vertebrates, Myllokunmingia and Haikouichthys (although they may be the same species), are found in the Chengjiang Fauna, the fossil deposit that also yielded the possible vertebrate relative Haikouella (see Chapter 2). These specimens are small, fish-shaped, and about 3 centimeters long (Figure 3–1a), and Haikouichthys is now known from hundreds of well-preserved individuals. A third type of early vertebrate, Zhongjianichthys, from the same deposit has a more eel-like body form. Evidence of a notochord and myomeres marks these animals as chordates, and the presence of a cranium (nasal capsules and possibly otic capsules) and paired sensory structures (probably representing eyes) at the head end marks them as vertebrates because these structures are formed from neural-crest tissue (see Chapter 2). These animals also have additional vertebrate features: a dorsal fin and a ribbonlike ventral fin (but without the fin rays seen in other vertebrates); six to seven gill pouches with evidence of filamentous gills and a branchial skeleton; myomeres that are probably W-shaped rather than V-shaped; and segmental structures flanking the notochord that may represent lampreylike arcualia. However, unlike ostracoderms, the Early Cambrian vertebrates lack any evidence of bone or mineralized scales. Additionally, evidence for serial segmental gonads, as in amphioxus, emphasizes their ancestral condition—in derived vertebrates the gonads are nonsegmental. Ordovician Vertebrates The next good evidence of early vertebrates is from the Early Ordovician. Several sites have yielded bone fragments, suggesting that by this time vertebrates Dorsal fin Myomeres Mouth Gill pouches Ventral fin 10 mm (a) Head shield Bony trunk scales Mouth Eye (b) 10 mm Gill openings Figure 3–1 Some of the earliest vertebrates. (a) The Early Cambrian Myllokunmingia from China. (b) The Ordovician pteraspid ostracoderm Astraspis from North America. 48 CHAPTER 3 Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates had diversified and had a worldwide distribution. The earliest complete fossils of vertebrates are from the Late Ordovician of Bolivia, Australia, North America, and Arabia. These were torpedo-shaped jawless fishes, ranging from 12 to 35 centimeters in length, with a bony external armor. The head and gill region were encased by many small, close-fitting, polygonal bony plates, and the body was covered by overlapping bony scales. The bony head shield shows the presence of sensory canals, special protection around the eye, and—in the reconstruction of the North American Astraspis (Figure 3–1b)—as many as eight gill openings on each side of the head. The Ordovician was also a time for great radiation and diversification among marine invertebrates, following the extinctions at the end of the Cambrian. The early radiation of vertebrates involved both jawed and jawless groups, although the first evidence of jawed fishes is in the Middle Ordovician, a little later than the first ostracoderms. By the Late Silurian (about 400 million years ago) both complete fossils of ostracoderms and fragmentary fossils of gnathostomes are known from diverse fossil assemblages worldwide. The Origin of Bone and Other Mineralized Tissues Mineralized tissues composed of hydroxyapatite are a major new feature of vertebrates. Enamel (or enameloid) and dentine, which occur primarily in the teeth among living vertebrates, are at least as old as bone and were originally found in intimate association with bone in the dermal armor of ostracoderms. However, mineralized tissues did not appear at the start of vertebrate evolution and are lacking in the extant jawless vertebrates. The Earliest Mineralized Tissues The origin of vertebrate mineralized tissues remains a puzzle: the earliestknown tissues were no less complex in structure than the mineralized tissues of living vertebrates. The basic units of mineralized tissue appear to be odontodes, little toothlike elements formed in the skin. They consist of projections of dentine, covered in some cases with an outer layer of enameloid, with a base of bone. Our own teeth are very similar to these structures (see Figure 2–6 in Chapter 2). Odontode-like structures are seen today as the tiny sharp denticles in the skin of sharks—and the larger scales, plates, and shields on the heads of many ostracoderms and early bony fishes are interpreted as aggregations of odontodes. Note that these bony elements would not have been external to the skin like a snail’s shell. Rather, they were formed within the dermis of the skin and overlain by a layer of epidermis, as with our own skull bones. The original condition for vertebrate bone is to lack cells in the adult form; this type of acellular bone is also known as aspidin. Cellular bone is found only in gnathostomes and in some derived ostracoderms (osteostracans). What could have been the original selective advantage of mineralized tissues in vertebrates? The detailed structure of the tissues forming the bony armor points to a function more complex than mere protection. Suggestions about the function of mineralized tissues include storage and/or regulation of minerals such as calcium and phosphorus, and insulation around electroreceptive organs (like those of living sharks) that enhanced detection of prey. A rigid bony head shield might also have served a hydrodynamic function in the earlier, finless vertebrates, providing stability while swimming in the open water. Although we think of a skeleton as primarily supportive or protective, our own bone also serves as a store of calcium and phosphorus. These hypotheses about the initial advantage of bone and other mineralized tissues—protection, electroreception, and mineral storage—are not mutually exclusive. All of them may have been involved in the evolutionary origin of these complex tissues. The Problem of Conodonts Curious microfossils known as conodont elements are widespread and abundant in marine deposits from the Late Cambrian to the Late Triassic periods. Conodont elements are small (generally less than 1 millimeter long) spinelike or comblike structures composed of apatite (the particular mineralized calcium compound that is characteristic of vertebrate hard tissues). They originally were considered to be the partial remains of marine invertebrates, but recent studies of conodont mineralized tissues have shown that they are similar in their microstructure to dentine, which is a uniquely vertebrate tissue. Thus, conodont elements are now considered to be the toothlike elements of true vertebrates. This interpretation has been confirmed by the discovery of impressions of complete conodont animals with vertebrate features (such as a notochord, a cranium, myomeres, and large eyes) and with conodont elements arranged within the pharynx in a complex apparatus (Figure 3–2). The paleontological community is still debating how conodont teeth fit into the evolutionary picture, although the general consensus is that they represent true vertebrates. However, some workers argue that various features of conodonts are not vertebrate-like; Reconstructing the Biology of the Earliest Vertebrates 49 10 mm (a) Ant erio SS S S S r S S S S environments. Under what conditions did the first vertebrates evolve? The vertebrate kidney is very good at excreting excess water while retaining biologically important molecules and ions. That is what a freshwater fish must do, because its body fluids are continuously being diluted by the osmotic inflow of water and it must excrete that water to regulate its internal concentration (see Chapter 4). Thus, the properties of the vertebrate kidney suggest that vertebrates evolved in freshwater. Despite the logic of that inference, however, a marine origin of vertebrates is now widely accepted. Osmoregulation is complex and fishes use cells in the gills as well as the kidney to control their internal fluid concentration. Probably the structure of the kidney is merely fortuitously suited to freshwater. Two lines of evidence support the hypothesis of a marine origin of vertebrates: • The earliest vertebrate fossils are found in marine sediments. Pb • All nonvertebrate chordates and deuterostome Pa (b) Figure 3–2 Conodonts. (a) Clydagnathus. (b) Close-up of the feeding apparatus (conodont elements) inside the head of Idiognathodus. The anterior S elements appear to be modified for grasping, and the posterior P elements for crushing. for example, conodont myomeres have the V shape of nonvertebrate chordates. The ability to form mineralized tissues in the skin is a feature of all vertebrates from ostracoderms onward, but the production of teeth, or toothlike structures, as seen in conodonts, may have evolved independently several different times among vertebrates, as discussed later in the chapter. Accepting conodonts as vertebrates has changed our ideas about early vertebrate interrelationships, particularly the importance—in the vertebrate phylogeny— of having mineralized tissues. The toothlike structures of conodonts and the bony dermal skeletons of ostracoderms are now thought to make these animals more derived vertebrates than the soft-bodied, jawless fishes living today (Figure 3–3). The Environment of Early Vertebrate Evolution By the Late Silurian, ostracoderms and early jawed fishes were abundant in both freshwater and marine 50 CHAPTER 3 invertebrate phyla are exclusively marine forms, and they have body fluids with approximately the same osmolal concentration as their surroundings. Hagfishes also have concentrated body fluids, and these high body-fluid concentrations probably represent the original vertebrate condition. 3.2 Extant Jawless Fishes The extant jawless vertebrates—hagfishes and lampreys—once were placed with the ostracoderms in the class “Agnatha” because they lack the gnathostome features of jaws and two sets of paired fins. They also have other ancestral features, such as lack of specialized reproductive ducts, and neither has mineralized tissues. However, it is now clear that the “Agnatha” is a paraphyletic assemblage, and that ostracoderms are actually more closely related to gnathostomes than are living jawless vertebrates. Hagfishes and lampreys have often been linked as cyclostomes because they have round, jawless mouths, but this grouping may also be paraphyletic, since lampreys and gnathostomes share derived features that hagfishes lack (see Figure 3–3 and Figure 3–4 on page 52). Because both hagfishes and lampreys appear to be more distantly related to gnathostomes than were the armored ostracoderms of the Paleozoic, we will look at them before considering the extinct agnathans. The fossil record of the extant jawless vertebrates is sparse. Lampreys are known from the Late Devonian period— Priscomyzon, a short-bodied form, from South Africa; Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates Recent Cambrian 488 SARCOPTERYGII ACANTHODII PLACODERMI OSTEOSTRACI PITURIASPIDA GALEASPIDA THELODONTI ANASPIDA HETEROSTRACI ASTRASPIDA 444 ARANDASPIDA 416 STEM CHONDRICHTHYANS “Cephalaspida” ACTINOPTERYGII HOLOCEPHALI “Ostracoderms” 359 Myllokunmingia/Haikouichthys PALEOZOIC Devonian Carboniferous 299 Ordovician Silurian PETROMYZONTIFORMES MYXINIFORMES Permian 251 ELASMOBRANCHII 201.5 CONODONTA Triassic 0.01 15. 12. Osteichthyes Chondrichthyes ? 13. Eugnathostomata 11. Gnathostomata ? ? Myopterygii Pteraspida 4. 542 5. Millions of years Figure 3–3 Phylogenetic relationships of early vertebrates. This diagram depicts the probable relationships among the major groups of “fishes,” including living and extinct jawless vertebrates and the earliest jawed vertebrates. The black lines show relationships only; they do not indicate times of divergence or the unrecorded presence of taxa in the fossil record. Lightly shaded bars indicate ranges of time when the taxon is believed to be present, because it is known from earlier and later times, but is not recorded in the fossil record during this interval. The hatched bar shows probable occurrence based on limited evidence. Only the best-corroborated relationships are shown and question marks indicate uncertainty about relationships. The numbers at the branch points indicate derived characters that distinguish the lineages—see the Appendix for a list of these characters. Extant Jawless Fishes 51 GNATHOSTOMES (jawed vertebrates) Pla co rm s”† de co “O str a Ea ve rly C rte am bra b tes rian Ha † gfi sh es La mp rey s Co no do nt s † CYCLOSTOMES OSTEICHTHYANS (bony vertebrates) de rm s† Ho l o (ch ce rat ime pha fis ras lan he or s Ela s) s (sh mo ark bra s a nc nd hs Ac ray an s) tho dia ns Ac † (ra tinop y-f inn teryg ed ian fis s he s) CHONDRICHTHYANS (cartilaginous fishes) ? Sa (lo rcop an be-fi tery d t nn gia etr ed ns ap fis o d he s) s “AGNATHANS” (jawless vertebrates) Endochondral bone Teeth on jaws Jaws, pelvic fins Pectoral fins ? Dermal bone Mineralized tissues W-shaped myomeres Distinct head with cranium Figure 3–4 Simplified cladogram of vertebrates. Only living taxa and major extinct groups are shown. Quotation marks indicate paraphyletic groups. An asterisk indicates possibly paraphyletic groups. A dagger indicates extinct groups. the Late Carboniferous period—Hardistiella from Montana and Mayomyzon from Illinois; and the Early Cretaceous period—Mesomyzon from southern China, the first known freshwater form. All these fossil lampreys appear to have been specialized parasites similar to the living forms. Myxinikela (an undisputed hagfish) and a second possible hagfish relative, Gilpichthys, have been found in the same Carboniferous deposits as Mayomyzon. Hagfishes—Myxiniformes There are about 75 species of hagfishes in two major genera (Eptatretus and Myxine). Adult hagfishes (Figure 3–5) are elongated, scaleless, pinkish to purple in color, and about half a meter in length. Hagfishes are entirely marine, with a nearly worldwide distribution except for the polar regions. They are primarily deepsea, cold-water inhabitants. They are the major scavengers of the deep-sea floor, drawn in large numbers by their sense of smell to carcasses. Structural Characteristics Large mucous glands that open through the body wall to the outside are a unique feature of hagfishes. These so-called slime glands secrete enormous quantities of mucus and tightly 52 CHAPTER 3 coiled proteinaceous threads. The threads straighten on contact with seawater to entrap the slimy mucus close to the hagfish’s body. An adult hagfish can produce enough slime within a few minutes to turn a bucket of water into a gelatinous mess. This obnoxious behavior is apparently a deterrent to predators. When danger has passed, the hagfish makes a knot in its body and scrapes off the mass of mucus, then sneezes sharply to blow its nasal passage clear. Hagfishes lack any trace of vertebrae, and their internal anatomy shows many additional unspecialized features. For example, the kidneys are simple, and there is only one semicircular canal on each side of the head. Hagfishes have a single terminal nasal opening that connects with the pharynx via a broad tube, and the number of gill openings on each side (1 to 15) varies with the species. The eyes are degenerate or rudimentary and covered with a thick skin. The mouth is surrounded by six tentacles that can be spread and swept to and fro by movements of the head when the hagfish is searching for food. Two horny plates in the mouth bear sharp toothlike structures made of keratin rather than mineralized tissue. These tooth plates lie to each side of a protrusible tongue and spread apart when the tongue is protruded. When the Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates (a) Lateral view External gill opening 10 mm (b) Sagittal section of head region Nostril Olfactory sac Brain Nasopharyngeal duct Velum Notochord Spinal cord Excurrent branchial duct Barbel Mouth Branchial pouch Teeth on tongue Tongue muscle Pharynx Internal openings to branchial pouches Figure 3–5 Hagfishes. tongue is retracted, the plates fold together and the teeth interdigitate in a pincerlike action. Hagfishes have large blood sinuses and very low blood pressure. In contrast to all other vertebrates, hagfishes have accessory hearts in the liver and tail regions in addition to the true heart near the gills. These hearts are aneural, meaning that their pumping rhythm is intrinsic to the hearts themselves rather than coordinated via the central nervous system. In all these features, hagfishes resemble the condition seen in amphioxus—although, like other vertebrates, their blood does have red blood cells containing hemoglobin, and the true heart has three chambers. Feeding Hagfishes attack dead or dying vertebrate prey. Once attached to the flesh, they can tie a knot in their tail and pass it forward along their body until they are braced against their prey and can tear off the flesh in their pinching grasp. They often begin by eating only enough outer flesh to enter the prey’s coelomic cavity, where they dine on soft parts. Some recent, but controversial, research suggests that hagfishes can actually absorb dissolved organic nutrients through their skin and gill tissues. Reproduction In most species, female hagfishes out- number males by a hundred to one; the reason for this strange sex ratio is unknown. Examination of the gonads suggests that at least some species are hermaphroditic, but nothing is known of mating. The yolky eggs, which are oval and more than a centimeter long, are encased in a tough, clear covering that is secured to the sea bottom by hooks. The eggs are believed to hatch into small, completely formed hagfishes, bypassing a larval stage. Unfortunately, almost nothing is known of the embryology or early life history of any hagfish because few eggs have been available for study. However, some hagfish embryos have recently been examined, and it has been determined that hagfishes possess neural crests like other vertebrates. Hagfishes and Humans We still know very little about hagfish ecology: we do not know how long hagfishes live; how old they are when they first begin to reproduce; exactly how, when, or where they breed; where the youngest juveniles live; what the diets and energy requirements of free-living hagfishes are; or virtually any of the other information needed for good management of commercially exploited populations of hagfishes. And, strangely enough, hagfishes do have an economic Extant Jawless Fishes 53 importance for humans. Almost all so-called eel-skin leather products are made from hagfish skin. Worldwide demand for this leather has eradicated economically harvestable hagfish populations in Asian waters and in some sites along the West Coast of North America. Lampreys—Petromyzontiformes There are around 40 species of lampreys in two major genera (Petromyzon and Lampetra); the adults of different species range in size from around 10 centimeters up to 1 meter. Structural Characteristics Although lampreys are similar to hagfishes in size and shape (Figure 3–6), they have many features that are lacking in hagfishes but shared with gnathostomes. Traditionally it has been assumed that only lampreys had structures homologous with the vertebrae of jawed vertebrates: minute cartilaginous elements called arcualia, homologous with the neural arches of vertebrae. However, recent work has shown that hagfishes also have vertebral rudiments in the ventral portion of their tails (homologs of the hemal arches). Lampreys are unique among living vertebrates in having a single nasal opening situated on the top of the head, combined with a duct leading to the hypophysis (pituitary) and known as a nasohypophysial opening. Development of this structure involves distortion of the front of the head, and its function is not known. Several groups of ostracoderms had an apparently similar structure, which evidently evolved convergently in those groups. The eyes of lampreys are large and well developed, as is the pineal body, which lies under a pale spot just posterior to the nasal opening. In contrast to hagfishes, lampreys have two semicircular canals on each side of the head—a condition shared with the extinct ostracoderms as well as in gnathostomes. In addition, the heart is innervated by the parasympathetic nervous system (the vagus nerve, X), as in gnathostomes, but not in hagfishes. Chloridetransporting cells in the gills and well-developed kidneys regulate ions, water, and nitrogenous wastes, as well as overall concentration of body fluids, allowing lampreys to exist in a variety of salinities. Lampreys have seven pairs of gill pouches that open to the outside just behind the head. In most other fishes and in larval lampreys, water is drawn into the mouth and then pumped out over the gills in continuous or flow-through ventilation. Adult lampreys spend much of their time with their suckerlike mouths affixed to the bodies of other fishes, and during this time they cannot ventilate the gills in a flow-through fashion. Instead, 54 CHAPTER 3 they use a form of tidal ventilation by which water is both drawn in and expelled through the gill slits. A flap called the velum prevents water from flowing out of the respiratory tube into the mouth. The lampreys’ mode of ventilation is not very efficient at oxygen extraction, but it is a necessary compromise given their specialized mode of feeding. Feeding Most adult lampreys are parasitic on other fishes, although some small, freshwater species have nonfeeding adults. The parasitic species attach to the body of another vertebrate (usually a bony fish that is larger than the lamprey) by suction and rasp a shallow, seeping wound through the integument of the host. The round mouth is located at the bottom of a large fleshy funnel (the oral hood), the inner surface of which is studded with keratinized conical spines. The oral hood, which appears to be a hypertrophied upper lip, is a unique derived structure in lampreys. The protrusible tonguelike structure is also covered with spines, and together these structures allow tight attachment and rapid abrasion of the host’s integument. This tongue is not homologous with the tongue of gnathostomes because the tongue muscle is innervated by a different cranial nerve (the trigeminal nerve, V, rather than the hypoglossal nerve, XII). An oral gland secretes an anticoagulant that prevents the victim’s blood from clotting. Feeding is probably continuous when a lamprey is attached to its host. The bulk of an adult lamprey’s diet consists of body fluids of fishes. The digestive tract is straight and simple, as one would expect for an animal with a diet as rich and easily digested as blood and tissue fluids. Lampreys generally do not kill their hosts, but they do leave a weakened animal with an open wound. At sea, lampreys feed on several species of whales and porpoises in addition to fishes. Swimmers in the Great Lakes, after having been in the water long enough for their skin temperature to drop, have reported initial attempts by lampreys to attach to their bodies. Reproduction Lampreys are primarily found in northern latitude temperate regions, although a few species are known from southern temperate latitudes. Nearly all lampreys are anadromous; that is, they live as adults in oceans or big lakes and ascend rivers and streams to breed. Some of the most specialized species live only in freshwater and do all of their feeding as larvae, with the adults acting solely as a reproductive stage in the life history of the species. Little is known of the habits of adult lampreys because they are generally observed only during reproductive activities or when captured with their host. Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates (a) Lateral view of an adult Dorsal fin Nasohypophyseal opening Eye Oral hood 10 mm Gill openings (b) Sagittal section of head region Nasohypophyseal Olfactory Nasal duct sac opening Mouth Pineal eye Hypophyseal pouch Brain Spinal cord Notochord Dorsal aorta Esophagus Ventral aorta Oral hood Tongue muscle Velum Respiratory tube Internal openings to branchial ducts (c) Larval lamprey (ammocoete) Region of mandibular arch Otic capsule Brain Gill pouch Eye spot Spinal cord Notochord Nostril Upper lip Liver Heart Velum Gill opening Endostyle Figure 3–6 Lampreys. Extant Jawless Fishes 55 Female lampreys produce hundreds to thousands of eggs, about a millimeter in diameter and devoid of any specialized covering such as that found in hagfishes. Male and female lampreys construct a nest by attaching themselves by their mouths to large rocks and thrashing about violently. Smaller rocks are dislodged and carried away by the current. The nest is complete when a pit is rimmed upstream by large stones, downstream by a mound of smaller stones and sand that produces eddies. Water in the nest is oxygenated by this turbulence but does not flow strongly in a single direction. The female attaches to one of the upstream rocks, laying eggs, and the male wraps around her, fertilizing the eggs as they are extruded, a process that may take two days. Adult lampreys die after breeding once. The larvae hatch in about two weeks. The larvae are radically different from their parents and were originally described as a distinct genus, Ammocoetes (see Figure 3–6c). This name has been retained as a vernacular name for the larval form. A week to ten days after hatching, the tiny 6- to 10-millimeter-long ammocoetes leave the nest. They are wormlike organisms with a large, fleshy oral hood and nonfunctional eyes hidden deep beneath the skin. Currents carry the ammocoetes downstream to backwaters and quiet banks, where they burrow into the soft mud or sand and spend three to seven years as sedentary filter feeders. The protruding oral hood funnels water through the muscular pharynx, where food particles are trapped in mucus and carried to the esophagus. An ammocoete may spend its entire larval life in the same bed of sediment, with no major morphological or behavioral change until it is 10 centimeters or longer and several years old. Adult life is usually no more than two years, and many lampreys return to spawn after one year. Lampreys and Humans During the past hundred years, humans and lampreys have increasingly been at odds. Although the sea lamprey, Petromyzon marinus, seems to have been indigenous to Lake Ontario, it was unknown from the other Great Lakes of North America before 1921. From the 1920s to the 1950s, lampreys expanded rapidly across the entire Great Lakes basin, and by 1946 they inhabited all the Great Lakes. Lampreys were able to expand unchecked until sporting and commercial interests became alarmed at the reduction of economically important fish species, such as lake trout, burbot, and lake whitefish. Chemical lampricides as well as electrical barriers and mechanical weirs at the mouths of spawning streams 56 CHAPTER 3 have been employed to bring the Great Lakes lamprey populations down to their present level. 3.3 The Importance of Extant Jawless Vertebrates in Understanding Ancient Vertebrates The fossil record of the first vertebrates reveals little about their pre-Silurian evolution, and it yields no undisputed clues about the evolution of vertebrate structure from the condition in nonvertebrate chordates. Hagfishes and lampreys may provide examples of the early agnathous radiation, but do hagfishes really represent a less derived type of vertebrate than the lamprey? This question is important for understanding the biology of the first vertebrates: Were the first vertebrates as lacking in derived characters as living hagfishes appear to be, or were they somewhat more complex animals? Anatomical Evidence On balance, anatomical features of hagfishes appear to represent a more ancestral condition than those in lampreys, but the interpretation of many of these features is controversial. • Some aspects of the anatomy of hagfishes, notably aspects of the brain and neuroanatomy, do appear to be truly ancestral. • Other apparently basal features, such as the rudimentary eyes, may represent a secondary loss of more derived characters. • Some characters, such as the very simple kidney, the lack of innervation of the heart (and the presence of amphioxus-like accessory hearts), and the body fluids that are the same concentration as seawater, are simply hard to interpret. Molecular Evidence The majority of molecular studies link hagfishes and lampreys as sister taxa, but we cannot get molecules from early nonvertebrate chordates (such as the Cambrian Haikouella) or from early vertebrates (such as the Cambrian vertebrate Haikouichthys and the huge diversity of ostracoderms) for comparison. Missing data of this sort can bias the computer programs that create phylogenies by introducing a statistical artifact known as “long branch attraction” that does not represent the true phylogenetic relationship. We do not Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates claim here that the molecular findings in the case of lampreys and hagfishes are artifacts, but the difference in the results obtained from morphological data versus molecular data is troubling. Recent molecular studies have focused on small portions of the genome called microRNAs, which are particularly prominent in vertebrates, and which appear to have been accumulating in vertebrate genomes over time (see Chapter 2). Studies of hagfishes and lampreys show that these two types of cyclostomes have four unique families of microRNAs, suggesting that they are in fact closely related. However, microRNAs apparently may be lost over time, so the debate is likely to continue. 3.4 The Radiation of Paleozoic Jawless Vertebrates— “Ostracoderms” “Ostracoderms” is a paraphyletic assemblage because some more derived types of ostracoderms are clearly more closely related to the gnathostomes (jawed vertebrates) than others (see Figure 3–3). Our interpretation of exactly how different lineages of ostracoderms are related to one another and to living vertebrates has changed considerably since the latter part of the twentieth century. Ostracoderms are clearly more derived than extant agnathans: Ostracoderms had dermal bone, and impressions on the underside of the dorsal head shield suggest that they had derived (i.e., gnathostome-like) features—a cerebellum in the hindbrain and an olfactory tract connecting the olfactory bulb with the forebrain. (Living agnathans lack a distinct cerebellum, and their olfactory bulbs are incorporated within the rest of the forebrain rather than placed more anteriorly and linked to the head via the olfactory tract [cranial nerve I].) Characters of Ostracoderms Most ostracoderms are characterized by the presence of a covering of dermal bone, usually in the form of an extensive armored shell, or carapace, but sometimes in the form of smaller plates or scales (e.g., anaspids), and some are relatively naked (e.g., thelodonts). Ostracoderms ranged in length from about 10 centimeters to more than 50 centimeters. Although they lacked jaws, some apparently had various types of movable mouth plates that have no analogues in any living vertebrates. These plates were arranged around a small, circular mouth that appears to have been located farther forward in the head than the larger, more gaping mouth of jawed vertebrates. Most species of ostracoderms probably ate small, soft-bodied prey. Most ostracoderms had some sort of midline dorsal fin, and although many heterostracans and anaspids had some sort of anterior, paired, finlike projections, only the more derived osteostracans had true pectoral fins, with an accompanying pectoral girdle and endoskeletal fin supports. As in living jawless vertebrates, the notochord must have been the main axial support throughout adult life. Figure 3–7 depicts some typical ostracoderms. Agnathans and Gnathostomes During the Late Silurian and the Devonian, most major known groups of extinct agnathans coexisted with early gnathostomes, and it is highly unlikely that ostracoderms were pushed into extinction by the radiation of gnathostomes after 50 million years of coexistence. Jawless and jawed vertebrates appear to represent two different basic types of animals that probably exploited different types of resources. The initial reduction of ostracoderm diversity at the end of the Early Devonian may be related to a lowering of global sea levels, with the resulting loss of coastal marine habitats. The extinction of the ostracoderms in the Late Devonian occurred at the same time as mass extinctions among many marine invertebrates, which suggests that their demise was not due to gnathostome competition. Gnathostomes also suffered in the Late Devonian mass extinctions, and the placoderm lineage that dominated the Devonian period became extinct at its end. 3.5 The Basic Gnathostome Body Plan Gnathostomes are considerably more derived than agnathans, not only in their possession of jaws but also in many other ways. Jaws allow a variety of new feeding behaviors, including the ability to grasp objects firmly, and along with teeth enable the animal to cut food to pieces small enough to swallow or to grind hard foods. New food resources became available when vertebrates evolved jaws: Herbivory was now possible, as was taking bite-sized pieces from large prey items, and many gnathostomes became larger than contemporary jawless vertebrates. A grasping, movable jaw also permits manipulation of objects: The Basic Gnathostome Body Plan 57 The heterostracan Pteraspis (Pteraspida) (Late Silurian) Mouth The psammosteid heterostracan (Pteraspida) Drepanaspis (dorsal view) (Early Devonian) 10 mm Gill opening Hypocercal tail (a) Gill opening Mouth (b) The anaspid Pharyngolepis (Cephalaspida) (Late Silurian) 10 mm Eye Orbital plate Hypocercal tail Gill openings (c) Lateral spine 10 mm Anal spine Late Silurian thelodonts Phlebolepis (d) 10 mm The osteostracan Hemicyclaspis (Cephalaspida) (Late Devonian) Loganellia (a fork-tail) Dorsal scales = anterior dorsal fin Posterior dorsal fin 10 mm Heterocercal tail Galeaspid (Cephalaspida) from the Early Devonian Hypochordal lobe ? sensory field area Ventrolateral ridge (e) Pectoral fin 10 mm Pituriaspid (Cephalaspida) from the Devonian (f) (g) 58 CHAPTER 3 10 mm Figure 3–7 Ostracoderm diversity: Pteraspida and common Cephalaspida. Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates jaws are used to dig holes, to carry pebbles and vegetation to build nests, and to grasp mates during courtship and juveniles during parental care. However, it seems that the likely origin of jaws was for more efficient gill ventilation rather than predation, as will be discussed later. Gnathostome Biology Gnathostomes are first known with certainty from the Early Silurian, but isolated sharklike scales suggest that they date back to the Middle Ordovician. The difference between gnathostomes and agnathans is traditionally described as the presence of jaws that bear teeth and two sets of paired fins or limbs (pectoral and pelvic). However, the gnathostome body plan (Figure 3–8) reveals that they are characterized by many other features, implying that gnathostomes represent a basic step up in level of activity and complexity from the jawless vertebrates. Interestingly, the form of the gnathostome lower jaw seems to be most variable in 9 Nerve cord 5 6 Dorsal fin the Silurian, when jawed fishes were not very common. By the Devonian, when the major radiation of jawed vertebrates commenced, the variety of jaw forms had stabilized. These derived features include improvements in locomotor and predatory abilities and in the sensory and circulatory systems. Just as jawless vertebrates show a duplication of the Hox gene complex in comparison to nonvertebrate chordates (see Chapter 2), living jawed vertebrates show evidence of a second Hox duplication event. Gene duplication would have resulted in a greater amount of genetic information, perhaps necessary for building a more complex type of animal. However, as can be seen in Figure 3–3, a number of extinct ostracoderm taxa lie between living agnathans and gnathostomes, and many features seen today only in gnathostomes might have been acquired in a steplike fashion throughout ostracoderm evolution. Paired pectoral fins, for example, are usually considered to be a gnathostome character, but some osteostracans had evolved paired fins. Segmental axial muscles Neural arch Notochord 4 Hemal arch 1 2 3 8 7 Caudal fin Anal fin Pelvic fins Pectoral fins Figure 3–8 Generalized jawed vertebrate (gnathostome) showing derived features compared to the jawless vertebrate (agnathan) condition. Legend: 1. Jaws (containing teeth) formed from the mandibular gill arch. 2. Gill skeleton consists of jointed branchial arches and contains internal gill rakers that stop particulate food from entering the gills. Gill musculature is also more robust. 3. Hypobranchial musculature allows strong suction in inhalation and suction feeding. 4. Two distinct olfactory tracts, leading to two distinct nostrils. 5. Original first gill slit squeezed to form the spiracle, situated between the mandibular and hyoid arches. 6. Three semicircular canals in the inner ear (addition of horizontal canal). 7. Addition of a conus arteriosus to the heart, between the ventricle and the ventral aorta. (Note that the position of the heart is actually more anterior than shown here, right behind the most posterior gill arch.) 8. Horizontal septum divides the trunk muscle into epaxial (dorsal) and hypaxial (ventral) portions. It also marks the position of the lateral line canal, containing the neuromast sensory organs. 9. Vertebrae now have centra (elements surrounding the notochord) and ribs, but note that the earliest gnathostomes have only neural and hemal arches, as shown in the posterior trunk. The Basic Gnathostome Body Plan 59 Jaws and Teeth Extant gnathostomes have teeth on their jaws, but teeth must have evolved after the jaws were in place because early members of the most basal group of jawed fishes—the placoderms—appear to have lacked true teeth (see section 3.6). Another reason to dissociate the evolution of jaws from the evolution of teeth is the presence of pharyngeal toothlike structures in various jawless vertebrates. These are most notable in the conodonts previously discussed (see Figure 3–2) but also have been reported in thelodonts, which are poorly known ostracoderms that lack a well-mineralized skeleton. Developmental studies support the notion that the first teeth were not oral structures but were located in the pharynx. Bony fishes and tetrapods have teeth embedded in the jawbones (Figure 3–9). However, because teeth form from a dermal papilla, they can be embedded only in dermal bones, and cartilaginous fishes such as sharks and rays lack dermal bone. The teeth of cartilaginous fishes form within the skin, resulting in a tooth whorl that rests on the jawbone but is not actually embedded in it. This condition is probably the ancestral one for all gnathostomes more derived than placoderms because it is also seen in the extinct acanthodians (see section 3.6). Adding jaws and hypobranchial muscles (which are innervated by spinal nerves) to the existing branchiomeric muscles (innervated by cranial nerves) allowed vertebrates to add powerful suction to their feeding mechanisms (Figure 3–10a). The cranium of gnathostomes has also been elongated both anteriorly and posteriorly compared with the agnathan condition. segmental muscles, providing increased anchorage for axial muscles. There is now a clear distinction between the epaxial and hypaxial blocks of the axial muscles, which are divided by a horizontal septum made of thin fibrous tissue that runs the length of the animal. The lateral line canal—containing the organs that sense vibrations in the surrounding water—lies in the plane of this septum, perhaps Replacement teeth Skin Mandible (a) Tooth whorl Replacement tooth (b) Pleurodont Replacement tooth (c) Acrodont Vertebrate and Ribs Progressively more complex ver- tebrae are another gnathostome feature. Gnathostome vertebrae initially consisted of arches flanking the nerve cord dorsally (the neural arches, which are homologous with the arcualia of lampreys) with matching arches below the notochord (the hemal arches, which may be present in the tail only—see the posterior portion of the trunk in Figure 3–8). More derived gnathostomes had a vertebral centrum or central elements with attached ribs (Figure 3–10b). Still more complete vertebrae support the notochord and eventually replace it as a supporting rod for the axial muscles used for locomotion (mainly in tetrapods). Well-developed centra were not a feature of the earliest jawed fishes and are unknown in the two extinct groups of fishes—placoderms and acanthodians. Ribs are another new feature in gnathostomes. They lie in the connective tissue between successive 60 CHAPTER 3 Replacement tooth (d) Thecodont © 1998 The McGraw-Hill Companies, Inc. Figure 3–9 Gnathostome teeth. (a) Tooth whorl of a chondrichthyan. (b–d) Teeth embedded in the dermal bones of the jaw as seen in osteichthyan fishes and tetrapods. (b) Pleurodont, the basal condition: teeth set in a shelf on the inner side of the jawbone, seen in some bony fishes and in modern amphibians and some lizards. (c) Acrodont condition: teeth fused to the jawbone, seen in most bony fishes and in some reptiles (derived independently). (d) Thecodont condition: teeth set in sockets and held in place by peridontal ligaments, seen in archosaurian reptiles and mammals (derived independently). Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates Adductor mandibulae Neural arch Hypobranchials Spinal cord Cucullaris Notochord Pectoral-fin elevator Centrum elements Rib (b) Ventral constrictors Branchiomeric muscle (cranially innervated) Pectoral fin depressor Spinally innervated muscles (a) Figure 3–10 Some gnathostome specializations. (a) Ventral view of a dogfish, showing both branchiomeric muscles and the new, spinally innervated, hypobranchial muscles. (b) View of generalized gnathostome vertebral form, showing central elements and ribs. reflecting improved integration between locomotion and sensory feedback. In the inner ear there is a third (horizontal) semicircular canal, which may reflect an improved ability to navigate and orient in three dimensions. What About Soft Tissues? Features such as jaws and ribs can be observed in fossils, so we know that they are unique to gnathostomes. However, we cannot know for sure whether other new features within the soft anatomy characterize gnathostomes alone or whether they were adopted somewhere within the ostracoderm lineage. • We can surmise that some features of the nervous system that are seen only in gnathostomes among living vertebrates were acquired by the earliest ostracoderms. For example, impressions on the inner surface of the dermal head shield reveal the presence of a cerebellum in the brain and olfactory tracts. • We can see that no ostracoderm possessed a third semicircular canal. • A new feature of the nervous system in gnathostomes is the insulating sheaths of myelin on the nerve fibers, which increase the speed of nerve impulses. • The heart of gnathostomes has an additional small chamber in front of the pumping ventricle, the conus arteriosus, which acts as an elastic reservoir that smoothes out the pulsatile nature of the flow of blood produced by contractions of a more powerful heart. (Some workers consider that lampreys, but not hagfishes, have a small conus arteriosus despite their weak hearts and low blood pressures.) • Living agnathans lack a stomach, but some derived ostracoderms may have had a stomach. Gnathostomes also have a type of cartilage different from the cartilage seen in the hagfishes and lampreys. (See Chapter 2 and the legend to Figure 3–3 in the Appendix for additional gnathostome characters.) • One particularly important gnathostome feature is in the reproductive system, where the gonads now have their own specialized ducts leading to the cloaca (see Chapter 2). The Origin of Fins Guidance of a body in three-dimensional space is complicated. Fins act as hydrofoils, applying pressure to the surrounding water. Because water is practically incompressible, force applied by a fin in one direction against the water produces a thrust in the opposite direction. A tail fin increases the area of the tail, giving more thrust during propulsion, and allows the fin to exert the force needed for rapid acceleration. Rapid adjustments of the body position in the water may be The Basic Gnathostome Body Plan 61 Pitch Pitch Yaw Roll Yaw Figure 3–11 An early jawed fish (the acanthodian, Climatius). Views from the side and front illustrate pitch, yaw, and roll and the fins that counteract these movements. especially important for active, predatory fishes such as the early gnathostomes, and the unpaired fins in the midline of the body (the dorsal and anal fins) control the tendency of a fish to roll (rotate around the body axis) or yaw (swing to the right or left) (Figure 3–11). The paired fins (pectoral and pelvic fins) can control the pitch (tilt the fish up or down) and act as brakes, and they are occasionally specialized to provide thrust during swimming, like the enlarged pectoral fins of skates and rays. The basic form of the pectoral and pelvic fins, as seen today in chondrichthyans and basal bony fishes, is for a tribasal condition. This means that three main elements within the fin articulate with the limb girdle within the body, and molecular studies back up the anatomical ones in showing that this type of fin represents the ancestral gnathostome condition. The paired fins of gnathostomes also share patterns of development, involving particular Hox genes, with the median fins of lampreys. Fins have non-locomotor functions as well. Spiny fins are used in defense, and they may become systems to inject poison when combined with glandular secretions. Colorful fins are used to send visual signals to potential mates, rivals, and predators. Even before the gnathostomes appeared, fishes had structures that served the same purpose as fins. Many ostracoderms had spines or enlarged scales derived from dermal armor that acted like immobile fins. Some anaspids had long finlike sheets of tissue running along the flanks, and osteostracans had pectoral fins. 62 CHAPTER 3 3.6 The Transition from Jawless to Jawed Vertebrates In Chapter 2 we saw that the branchial arches were a fundamental feature of the vertebrate cranium, providing support for the gills. It has long been known that vertebrate jaws are made of the same material as the skeletal elements that support the gills (cartilage derived from the neural crest), and they clearly develop from the first (mandibular) arch of this series in vertebrates. (Note that while the word “mandible” commonly refers to the lower jaw only, the term “mandibular arch” includes both upper and lower jaws.) It is helpful at this stage to envisage the vertebrate head as a segmented structure, with each branchial arch corresponding to a segment (Figure 3–12). The mandibular arch that forms the gnathostome jaw is formed within the second head segment; jaw supports are formed from the hyoid arch of the third head segment; and the more posterior branchial arches that form the gill supports (arches 3 through 7) are formed in head segments 4 through 8. The branchial arch numbering does not match the numbering of the head segments because no evidence exists for a branchial arch functioning as a gill support structure in the first (premandibular) segment at any point in vertebrate history. No living vertebrate has a pair of gill-supporting branchial arches in such an anterior position as the Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates Ancestral agnathous condition 1st gill slit Braincase 1st gill arch (a) 2nd gill arch Gnathostome condition Lateral view Palatoquadrate (upper jaw = 1st gill arch) Spiracle (= 1st gill slit) Anterior view (form of gill arches 3-7) Pharyngeobranchial Notochord Pharyngeobranchial Epibranchial Ceratobranchial Epibranchial Hypobranchial Ceratobranchial Mandible (lower jaw = 1st gill arch) (b) Hyoid arch (= 2nd gill arch) Hypobranchial Basibranchial Basibranchial Figure 3–12 Evolution of the vertebrate jaw from anterior branchial arches. Colored shading indicates splanchnocranium elements (branchial arches and their derivatives). jaws as shown in Figure 3-12(a). However, all vertebrates have some structure in this position (i.e., in the second head segment) that appears to represent the modification of an anterior pair of pharyngeal arches; these are the jaws in gnathostomes and velar cartilages (i.e., the structures that support the velum) in lampreys and hagfishes. Thus it has been proposed that the common ancestor of living vertebrates had an unmodified pair of branchial arches in this position, with a fully functional gill slit lying between the first and second arches, and that living jawless and jawed vertebrates are both divergently specialized from this condition. There is no trace of a gill slit between arches 1 and 2 in living jawless vertebrates; but, in many living cartilaginous fishes, and in a few bony fishes, there is a small hole called the spiracle in this position, which is now used for water intake. Figure 3–12 summarizes the differences in the gill arches between jawed vertebrates and their presumed jawless ancestor and illustrates the major components of the hinged gnathostome gill arches (as seen in arches 3 through 7). Evolution of Gills of Early Vertebrates Because ostracoderms are now viewed as “stem gnathostomes,” any understanding of the origin of gnathostomes must encompass the view that at some point a jawless vertebrate was transformed into a jawed one. However, for much of the last century, researchers considered jawless and jawed vertebrates as two separate evolutionary radiations. This was because of the apparent nonhomology of their branchial arches: at least in living jawless vertebrates (the situation is less clear for the fossil ones), the gill arches lie lateral to (that is, external to) the gill structures, whereas in jawed vertebrates they lie medially (internal to the gills). Additionally, extant jawless vertebrates have pouched gills with small, circular openings that are different from the flatter, more lens-shaped openings between the gills of gnathostomes. However, in the early twenty-first century a veritable explosion of studies on vertebrate head development (especially studies of lampreys) showed that this difference in gill arch position may be produced simply by a switch in developmental timing. The Transition from Jawless to Jawed Vertebrates 63 In addition, hagfishes and lampreys both have some evidence of internal branchial elements in the form of velar cartilages, and some living sharks have evidence of small external branchial cartilages. That observation raises the possibility that the earliest vertebrates may have had both internal and external gill arches, and that is a condition from which both the cyclostome and gnathostome conditions could be derived. The external branchial basket of lampreys, which squeezes the pharyngeal area from the outside, may be essential for the specialized tidal ventilation of adult lampreys. In contrast, the strengthened (and jointed) internal branchial arches in gnathostomes may be related to the more powerful mode of gill ventilation used by these vertebrates. In any event, the difference in gill skeletons that was once thought to be an insurmountable problem for evolving jawed vertebrates from jawless ones is now inconsequential. However, there is one aspect of the agnathan anatomy that did need to change before jaws could develop: the nasohypophyseal duct (see Figure 3–6b) that connects to the pituitary in the brain. In lampreys this duct obstructs the lateral growth of the neuralcrest tissue that forms the gnathostome jaw, so this connection had to be broken before jaws could evolve. In gnathostomes the paired olfactory sacs are widely separated and no longer connect to the hypophyseal pouch. This separation has now been identified in a derived ostracoderm, one of the galeaspids. Thus, the rearrangement of cranial anatomy necessary for jaw evolution was acquired in jawless fishes closely related to gnathostomes. Stages in the Origin of Jaws In recent years, molecular developmental biology has provided fresh insights into the issue of the origin of jaws. It is worth noting that some of the controversies described above relate to a couple of misconceptions about the probable processes of evolution: • First, we should consider that evolutionary transformation does not involve changing one adult structure into another adult structure; instead, morphology is altered via modifications of gene expression and shifts in developmental timing. • Second, the fact that the lamprey lies below the position of gnathostomes on the cladogram does not mean that it represents a basal vertebrate condition. On the contrary, lampreys, with their bizarrely hypertrophied upper lip, are highly derived in their own right. The same genes are expressed in the mandibular segments of lampreys and gnathostomes, indicating that 64 CHAPTER 3 the structures in this position are likely to be homologous. The lamprey velum and velar cartilages, composed of mandibular segment tissue, are highly specialized structures, and the lamprey upper lip is a strange mixture of material from the mandibular (second) segment and the premandibular (first) segment. Thus, lampreys clearly do not represent a generalized ancestral condition from which gnathostomes might have been derived. Why Evolve Jaws? Although the complex evolutionary hypotheses about the roles of internal and external branchial arches described above may be valid, a relatively simple change in the direction of the streams of neuralcrest tissue that form the arches could also account for the difference in position of jaws in the adult. What is perhaps of more interest to evolutionary morphologists than how such a shift in branchial arch position could have occurred is the reason why jaws evolved at all. The notion that derived predatory vertebrates should convert gill arches into toothed jaws has been more or less unquestioned for decades. It is a common assumption that jaws are superior devices for feeding, and thus more derived vertebrates were somehow bound to obtain them. However, this simplistic approach does not address the issue of how the evolutionary event might actually have taken place: What use would a protojaw be prior to its full transformation? And even if early vertebrates needed some sort of superior mouth anatomy, why modify a branchial gill-supporting arch, which initially was located some distance behind the mouth opening? Why not just modify the existing cartilages and plates surrounding the mouth? There is nothing about those structures that prevents them from being modified. Quite the contrary, in fact—the living agnathans have specialized oral cartilages, and various ostracoderms apparently had oral plates. Additionally, as we noted earlier, gnathostome teeth must have evolved after jaws evolved, so the first jaws were toothless; and of what use could a toothless jaw be? (A movie entitled Gums wouldn’t sell out a theater.) Jon Mallatt has proposed a novel explanation of the origin of jaws based on the hypothesis that jaws were initially important for gill ventilation rather than predation. Living gnathostomes are more active than jawless vertebrates and have greater metabolic demands, and features of the earliest known ones suggest that this was the condition from the start of their evolutionary history. One derived gnathostome feature associated with such high activity is the powerful mechanism for pumping water over the gills. Gnathostomes have a characteristic series Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates (a) Upper-lip skeleton Upper lip Mouth opening Lower lip Gill opening Hyoid arch Mandibular arch Other branchial arches (b) (c) Otic capsule Upper lip First gill opening Hyoid arch Hyoid arch Enlarged mandibular arch Mandibular arch (d) (e) Spiracle Upper lip First gill opening Hyoid arch Jaw = Mandibular arch Hyoid arch Jaw = Mandibular arch Figure 3–13 Proposed evolution of vertebrate jaws. (a) Jawless gnathostome ancestor. The ‘upper lip skeleton’ is derived from the mandibular arch and is considered to be the precursor of the upper jaw. (b) Early pre-gnathostome with jointed branchial arches. (c) Late pre-gnathostome with enlarged mandibular arch now employed to firmly close mouth during ventilation. (d) Early gnathostome with mandibular arch now used as a feeding jaw. (e) More derived gnathostome (e.g., a shark) with teeth added to jaw. of internal branchial muscles (with cranial nerve innovation) as well as the new, external hypobranchials (with spinal nerve innervation). These muscles not only push water through the pharynx in exhalation but also suck water into the pharynx during inhalation. Gnathostome fishes can generate much stronger suction than agnathans, and powerful suction is also a way to draw food into the mouth. Living agnathans derive a certain amount of suction from their pumping velum, but this pump mostly pushes (rather than sucks) water, and its action is weak. Mallatt proposed that the mandibular branchial arch enlarged into protojaws because it played an essential role in forceful ventilation—rapidly closing and opening the entrance to the mouth (Figure 3–13). During strong exhalation, as the pharynx squeezed water back across the gills, water was kept from exiting via the mouth by bending the mandibular arch sharply shut. Next, during forceful inspiration, the mandibular arch was rapidly straightened to reopen the mouth and allow water to enter. To accommodate the forces of the powerful muscles that bent The Transition from Jawless to Jawed Vertebrates 65 the arch (the adductor mandibulae, modified branchial muscles) and straightened it (the hypobranchial muscles), the mandibular arch enlarged and became more robust. The advantage of using the mandibular arch would be that the muscles controlling it are of the same functional series as the other ventilatory muscles, and their common origin would ensure that all of the muscles were controlled by the same nerve circuits. In contrast, the muscles of the more anterior oral cartilages would not have been coupled with the musculature of the pharyngeal arches. Probably that would not have mattered if feeding had been the original function of jaws, whereas it would matter if the jaws had to coordinate their movements with the pharyngeal arches that were responsible for ventilation. This line of reasoning may explain why the mandibular arch, rather than the more anterior oral cartilages, became the jaws of gnathostomes. 3.7 Extinct Paleozoic Jawed Fishes With jaws that can grasp prey, muscles that produce powerful suction, and other features indicative of higher levels of activity, gnathostomes were able to enter ecological niches unavailable to agnathan vertebrates. We have numerous fossils of the entire bodies of gnathostomes (rather than fragments such as teeth and scales) from Devonian sediments. At this point, gnathostomes can be divided into four distinctive clades: two extinct groups—placoderms and acanthodians, and two groups that survive today—chondrichthyans (cartilaginous fishes) and osteichthyans (bony vertebrates): • Placoderms were highly specialized, armored fishes that appear to be basal to other gnathostomes. • The acanthodians, or “spiny sharks,” were small, more generalized fishes. Although acanthodians have been traditionally grouped with the bony fishes, new studies suggest that the different species occupied a diversity of positions in the vertebrate phylogeny. • The cartilaginous fishes, which include sharks, rays, and ratfishes, evolved distinctive specializations of small dermal scales, internal calcification, jaw and fin mobility, and reproduction. • Bony fishes evolved endochondral bone in their internal skeleton, a distinctive dermal head skeleton that included an operculum covering the gills, and an internal air sac forming a lung or a swim bladder. The bony fishes include the ray-finned fishes (actinopterygians), which comprise the majority of living fishes, and the lobe-finned fishes (sarcopterygians). 66 CHAPTER 3 Only a few lobe-finned fishes survive today (lungfishes and coelacanths), but they were more diverse in the Paleozoic. (Furthermore, sarcopterygian fishes are the group that gave rise to the tetrapods, and from this perspective there are as many extant species of sarcopterygians as of actinopterygians.) If we take another step back, Osteichthyes includes tetrapods, and we are ourselves highly modified fishes. Bony fishes by themselves constitute a paraphyletic group because their common ancestor is also the ancestor of tetrapods, and the same is true of the lobe-finned fishes. Before studying the extant groups of jawed fishes, we turn to the placoderms and acanthodians to examine the variety of early gnathostomes. Figures 3–3 and 3–4 show the interrelationships of gnathostome fishes. Living and extinct groups of chondrichthyans are discussed in Chapter 5, and osteichthyans are discussed in Chapter 6. Placoderms—The Armored Fishes As the name placoderm (Greek placo = plate and derm = skin) implies, placoderms were covered with a thick bony shield over the anterior one-third to one-half of their bodies. Unlike the ostracoderms, the bony shield of placoderms was divided into separate head and trunk portions, linked by a mobile joint that allowed the head to be lifted up during feeding (Figure 3–14a). The endoskeleton was mineralized by perichondral bone ossification around the rim of the bone (perichondral bone). (Ossification throughout the entire bone [endochondral bone] appears to be limited to osteichthyans.) Many (though not all) researchers consider that placoderms are not a distinctive clade, but rather represent a paraphyletic assemblage at the stem of gnathostome evolution. This would mean that some placoderm species were more closely related to the living groups of gnathostomes than were others. History of Placoderms Placoderms are known from the Early Silurian to the end of the Devonian and were the most diverse vertebrates of their time in terms of number of species, types of morphological specializations, and body sizes, with the largest reaching a length of 8 meters, the size of a great white shark. Like the ostracoderms, they suffered massive losses in the Late Devonian extinctions; but, unlike any ostracoderm group, a few placoderm lineages continued for another 5 million years, until the very end of the period. Ancestral placoderms were primarily marine, but a great many lineages became adapted to freshwater and estuarine habitats. Biology of Placoderms Placoderms have no modern analogues, and their massive external armor makes Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates (a) The arthrodire placoderm Coccosteus Head/trunk joint Lateral view Dorsal view 50 mm Frontal view 50 mm Tooth plates (b) The antiarch placoderm Bothriolepis Pectoral fins encased in armor (c) The giant Late Devonian arthrodire placoderm Dunkleosteus 20 mm (d) The Early Devonian Climatius, a generalized acanthodian Pectoral spine Branchiostegal rays 200 mm 20 mm Pelvic spine Anal spine Intermediate spines Figure 3–14 Extinct Paleozoic fishes: placoderms and acanthodians. interpreting their lifestyle difficult, although they were apparently primarily inhabitants of deep water. Most placoderms lacked true teeth; their toothlike structures (tooth plates, see Figure 3–14a) were actually projections of the dermal jawbones that were subject to wear and breakage without replacement. However, some later placoderms did possess true teeth, possibly evolved convergently with teeth of other gnathostomes. It is impossible to know whether placoderms had many of the soft-anatomy characters of living gnathostomes, but it has been determined that, like their extant relatives, they had myelinated nerve sheaths. Structures preserved in one placoderm were initially interpreted as being lungs, but this notion has now been disproved: these supposed “lungs” appear to be portions of the digestive tract that were filled with sediment after death. More than half of the known placoderms, nearly 200 genera, belonged to the predatory arthrodires (Greek arthros = a joint and dira = the neck). As their Extinct Paleozoic Jawed Fishes 67 name suggests, they had specializations of the joint between the head shield and the trunk shield, allowing an enormous head-up gape, probably increasing both predatory and respiratory efficiency. There was a large diversity of placoderms, many of them flattened, bottom-dwelling forms, some even resembling modernday skates and rays. The antiarchs, such as Bothriolepis, looked rather like armored catfishes (Figure 3–14b). Their pectoral fins were also encased in the bony shield, so that their front fins looked more like those of a crab. Perhaps the best-known placoderm is Dunkleosteus, a voracious, 8-meter-long predatory arthrodire (Figure 3–14c); biomechanical studies have shown that this fish had an extraordinarily powerful bite and rapid gape expansion for suction of prey items. A recent study of placoderms from the Late Devonian of Australia that preserve some soft-tissue structure has provided much more information about these fishes. For example, one way in which they are less derived than other gnathostomes is that their segmental muscles (myomeres) resemble those of lampreys in being only weakly W-shaped and not distinctly separated into epaxial and hypaxial portions. Perhaps the most interesting finding was evidence in one form, appropriately named Materpiscis (Latin mater = mother and piscis = fish), that had embryos and a structure interpreted as an umbilical cord preserved within the body cavity. This evidence of viviparity in at least some placoderms matches the observation that some placoderms also have claspers on their pelvic fins, resembling the pelvic claspers in male modern cartilaginous fishes that are used for internal fertilization (see Chapter 5). We can infer from this that the placoderms, like living chondrichthyans, had internal fertilization and probably complex courtship behaviors. Acanthodians Acanthodians are so named because of the stout spines (Greek acantha = spine) anterior to their well-developed dorsal, anal, and paired fins. Most researchers now consider these fishes to be an array of early forms, some of which were more closely related to the Osteichthyes, others to the Chondrichthyes, and still others perhaps less derived than any extant gnathostome. A recent analysis of Acanthodes indicates that the form of the chondrichthyan braincase is basal for gnathostomes. History of Acanthodians Acanthodians lived from the Late Ordovician through the Early Permian periods, with their major diversity in the Early Devonian. The earliest forms were marine, but by the Devonian they were predominantly a freshwater group. Acanthodians had a basic fusiform fish shape with a heterocercal tail fin (i.e., with the upper lobe larger than the lower lobe; Figure 3–14d). This tail shape in modern fishes is associated with living in the water column, rather than being bottom-dwelling like the placoderms. Biology of Acanthodians Acanthodians were usually not more than 20 centimeters long, although some species reached lengths of 2 meters. They had paired fins, lots of them—some species had as many as six pairs of ventrolateral fins in addition to the pectoral and pelvic fins that gnathostomes have (see Figure 3–14d). Most acanthodians had large heads with wide-gaping mouths, and the teeth (in the species that had teeth—some species were toothless) formed a sharklike tooth whorl. The acanthodids, the only group to survive into the Permian, were elongate, toothless, and with long gill rakers. They were probably plankton-eating filter feeders. Summary Fossil evidence indicates that vertebrates evolved in a marine environment. Jawless vertebrates are first known from the Early Cambrian, and there is evidence that the first jawed vertebrates (gnathostomes) evolved as long ago as the Middle Ordovician. The first vertebrates would have been more active than their ancestors, with a switch from filter feeding to more active predation and with a muscular pharyngeal pump for gill ventilation. The first mineralized tissues were seen in the teeth of conodonts, enigmatic animals that have only recently been considered true vertebrates. Bone is a feature of many early vertebrates, although it was absent from the Early Cambrian forms and is also absent in the living jawless vertebrates—the hagfishes 68 CHAPTER 3 and lampreys. Bone is first found with accompanying external layers of dentine and enamel-like tissue in the dermal armor of early jawless fishes called the ostracoderms. Current explanations for the original evolutionary use of bone include protection, a store for calcium and phosphorus, and housing for electroreceptive sense organs. Among the living jawless fishes, hagfishes are apparently less derived in their anatomy than the lampreys and all other known vertebrates, but molecular data now suggest that they form a clade with lampreys, which would mean that the apparently ancestral anatomical features represent secondary loss of derived characters. Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates Lampreys, in turn, are less derived than the extinct armored ostracoderms. Ostracoderms are widely known from the Silurian and Early Devonian, and none survived past the end of the Devonian. Ostracoderms were not a unified evolutionary group: some forms (Cephalaspida) were more closely related to gnathostomes than were other forms, sharing with gnathostomes the feature of a pectoral fin. Ostracoderms and gnathostomes flourished together for 50 million years; thus there is little evidence to support the idea that jawed vertebrates outcompeted and replaced jawless ones. Just as the evolution of vertebrates from nonvertebrate chordates represented an advance in basic anatomy and physiology, so did the evolution of jawed vertebrates from jawless ones. Jaws may have evolved initially to improve gill ventilation rather than to bite prey. In addition to jaws, gnathostomes have a number of derived anatomical features (such as true vertebrae, ribs, and a complete lateral line sensory system), suggesting a sophisticated and powerful mode of locomotion and sensory feedback. The early radiation of jawed fishes, first known in detail from the fossil record of the Late Silurian, included four major groups. Two groups, the chondrichthyans (cartilaginous fishes) and osteichthyans (bony fishes), survive today. Osteichthyans were the forms that gave rise to tetrapods in the Late Devonian. The other two groups, placoderms and acanthodians, are now extinct. Placoderms did not survive past the Devonian, but acanthodians survived almost until the end of the Paleozoic. Placoderms were armored fishes, superficially like the ostracoderms in their appearance. They were the most diverse fishes of the Devonian and included large predatory forms. Placoderms are considered to be basal to all other gnathostomes, perhaps forming a paraphyletic stem group. Acanthodians are more generalized fishes than the placoderms were. Acanthodians were originally considered to be the sister group to the bony fishes, but are now considered to be a generalized basal gnathostome assemblage. Discussion Questions 1. Did jawed vertebrates outcompete the jawless ones? Does the fossil record provide any evidence? 2. How did the realization that conodont animals were vertebrates change our ideas about the pattern of vertebrate evolution? 3. How is the body-fluid composition of hagfishes different from that of other vertebrates? How does this difference figure in the debate about whether vertebrates arose in fresh or salt water? 4. Ventilating the gills by taking water in through the gill openings, as well as using the gill openings for water ejection, is not a very efficient way of doing things. Why, then, do adult lampreys ventilate their gills this way? 5. We usually think of jaws as structures that evolved for biting. What might have been a different original use, and what is the evidence for this? 6. Recent fossil evidence shows a placoderm with a developing embryo inside her body. How might we have been able to speculate that placoderms were viviparous in the absence of such evidence? Additional Information Ahlberg, P. E. 2009. Birth of the jawed vertebrates. Nature 457:1094–1095. Ahlberg, P., et al. 2009. Pelvic claspers confirm chondrichthyan-like internal fertilization in arthrodires. Nature 460:888–889. Anderson, P. S. L., and M. W. Westneat. 2009. A biomechanical model of feeding kinematics for Dunkleosteus terrelli (Arthrodira, Placodermi). Paleobiology 35: 251–269. Anderson, P. S. L., et al. 2011. Initial radiation of jaws demonstrated stability despite faunal and environmental changes. Nature 476:206–209. Brazeau, M. D. 2009. The braincase and jaws of a Devonian ‘acanthodian’ and modern gnathostome origins. Nature 457:305–308. Cerny, R., et al. 2010. Evidence for the prepattern/ cooption model of vertebrate jaw evolution. Proceedings of the National Academy of Sciences 107:17262–17267. Coates, M. I. 2009. Beyond the age of fishes. Nature 458:413–414. Davis, S. P. et al. 2012. Acanthodes and shark-like conditions in the last common ancestor of modern gnathostomes. Nature 486:247-251. Donoghue, P. C. J., and M. A. Purnell. 2009. The evolutionary emergence of vertebrates from among their spineless relatives. Evolution Education Outreach 2:204–212. Friedman, M., and M. D. Brazeau. 2010. A reappraisal of the origin and basal radiation of the Osteichthyes. Journal of Vertebrate Paleontology 30:36–56. Gai, Z., et al. 2011. Fossil jawless fish from China foreshadows early jawed vertebrate anatomy. Nature 476:324–327. Additional Information 69 Gillis, J. A., and N. H. Shubin. 2009. The evolution of gnathostome development: Insights from chondrichthyan embryology. Genesis 47:825–841. Glover, C. N., et al. 2011. Adaptations to in situ feeding: Novel nutrient acquisition pathways in an ancient vertebrate. Proceedings of the Royal Society of London B 278:3096–3101. Goujet, D. 2011. “Lungs” in Placoderms, a persistent palaeobiological myth related to environmental preconceived interpretations. Comptes Rendus Palevol 10: 323–329. Janvier, J. 2011. Comparative anatomy: All vertebrates do have vertebrea. Current Biology 21:R661-R663. Janvier, P. 2008. Early jawless vertebrates and cyclostome origins. Zoological Science 25:1045–1056. Long, J. A. 2011. The Rise of Fishes, 2nd ed. Baltimore, MD: Johns Hopkins University Press. Long, J. A., et al. 2008. Live birth in the Devonian Period. Nature 453:650–653. Mallatt, J. 2008. Origin of the vertebrate jaw: Neoclassical ideas versus newer, development-based ideas. Zoological Science 25:990-998. Near, T. J. 2009. Conflict and resolution between phylogenies inferred from molecular and phenotypic data sets for hagfish, lampreys, and gnathostomes. Journal of Experimental Zoology (Mol Dev Evol) 312B:749–761. Richardson, M. K., et al. 2010. Developmental anatomy of lampreys. Biological Reviews 85:1–33. 70 CHAPTER 3 Shubin, N. 2009. Your Inner Fish: A Journey into the 3.5-Billion-Year History of the Human Body. New York: Random House. Turner, S., et al. 2010. False teeth: Conodont-vertebrate phylogenetic relationships revisited. Geodiversitas 32:545–594. Young, G. V. 2010. Placoderms (armored fish): Dominant vertebrates of the Devonian Period. Annual Review of Earth and Planetary Sciences 38:523–550. Zhu, M., et al. 2009. The oldest articulated osteichthyan reveals mosaic gnathostome characters. Nature 458:469– 474. Websites Information about living agnathans .html Information about basal gnathostomes Information about placoderms fifthb.html placodermi.html Information about acanthodians .html Early Vertebrates: Jawless Vertebrates and the Origin of Jawed Vertebrates PART II Non-Amniotic Vertebrates: Fishes and Amphibians V ertebrates originated in the sea, and more than half of the species of living vertebrates are the products of evolutionary lineages that have never left an aquatic environment. Water now covers 73 percent of Earth’s surface (the percentage has been both higher and lower in the past) and provides habitats extending from deep oceans, lakes, and mighty rivers to fast-ﬂowing streams and tiny pools in deserts. Fishes have adapted to all these habitats, and there are more than 37,000 species of extant cartilaginous and bony ﬁshes. Life in water poses challenges for vertebrates but oﬀers many opportunities. Aquatic habitats are some of the most ecologically productive on Earth, and energy is plentifully available in many of them. Some aquatic habitats (coral reefs are an example) have enormous structural complexity, whereas others (like the open ocean) have virtually none. The diversity of ﬁshes reﬂects specializations for this variety of habitats. The diversity of ﬁshes and the habitats in which they live has oﬀered unparalleled scope for variations in life history. Some ﬁshes produce millions of eggs that are released into the water to drift and develop on their own, other species of ﬁshes produce a few eggs and guard both the eggs and the young, and numerous ﬁshes give birth to young that require no parental care. Males of some species of ﬁshes are larger than females; in others the reverse is true. Some species have no males at all, and a few species of ﬁshes change sex partway through life. Feeding mechanisms have been a central element in the evolution of ﬁshes, and the specializations of modern ﬁshes range from species that swallow prey longer than their own bodies to species that rapidly extend their jaws like a tube to suck up minute invertebrates from tiny crevices. In the Devonian period, vertebrates entered a new world as ﬁshlike forms emerged onto the land and occupied terrestrial environments. In this part of the book, we consider the evolution of this extraordinary array of vertebrates. 71 4 CHAPTER Living in Water A lthough life evolved in water and the earliest vertebrates were aquatic, the physical properties of water create some diﬃculties for aquatic animals. To live successfully in open water, a vertebrate must adjust its buoyancy to remain at a selected depth and force its way through a dense medium to pursue prey or to escape its own predators. Heat ﬂows rapidly between an animal and the water around it, and it is diﬃcult for an aquatic vertebrate to maintain a body temperature that is diﬀerent from the water’s temperature. (That phenomenon was dramatically illustrated when the Titanic sank— in the cold water of the North Atlantic, most of the victims died from hypothermia rather than by drowning.) Ions and water molecules move readily between the external environment and an animal’s internal body ﬂuids, so maintaining a stable internal environment can be diﬃcult. On the plus side, ammonia is extremely soluble in water so disposal of nitrogenous waste products is easier in aquatic environments than on land. The concentration of oxygen in water is lower than it is in air, however, and the density of water imposes limits on the kinds of gas-exchange structures that can be eﬀective. Despite these challenges, many vertebrates are entirely aquatic. Fishes, and especially the bony ﬁshes, have diversiﬁed into an enormous array of sizes and ways of life. In this chapter we will examine some of the challenges of living in water and the ways aquatic vertebrates have responded to them. 4.1 The Aquatic Environment Seventy-three percent of the surface of Earth is covered by freshwater or salt water. Most 72 of this water is held in the ocean basins, which are populated everywhere by vertebrates. Freshwater lakes and rivers hold a negligible amount of the water on Earth—about 0.01 percent. This is much less than the amount of water tied up in the atmosphere, ice, and groundwater, but freshwater habitats are exceedingly rich biologically, and nearly 40 percent of all bony fishes live in freshwater. Water and air are both ﬂuids at biologically relevant temperatures and pressures, but they have diﬀerent physical properties that make them drastically diﬀerent environments for vertebrates to live in (Table 4–1). In air, for example, gravity is an important force acting on an animal, but ﬂuid resistance to movement (air resistance) is trivial for all but the fastest birds. In water, the opposite relationship holds—gravity is negligible, but ﬂuid resistance to movement is a major factor with which aquatic vertebrates must contend and most ﬁshes are streamlined. Although each major clade of aquatic vertebrates solved environmental challenges in somewhat diﬀerent ways, the basic specializations needed by all aquatic vertebrates are the same. Table 4–1 The physical properties of freshwater and air at 20°C. Most of these properties change with temperature and atmospheric pressure, and some are aﬀected by the presence of solutes as well. Property Freshwater Air Comparison Density 1 kg . liter-1 0.0012 kg . liter-1 Water is about 800 times as dense as air. Dynamic viscosity 1 mPa . s 0.018 mPa . s Water is about 55 times as viscous as air. Oxygen content 6.8 ml . liter 209 ml . liter The oxygen content of freshwater decreases from about 15 ml . liter-1 at 0°C to about 5 ml . liter-1 at 40°C. Seawater contains less oxygen than freshwater—5.2 ml . liter-1 at 20°C. Heat capacity 4.18 kJ . kg-1 . °K-1 0.0012 kJ . kg-1 . °K-1 The heat capacity of water is about 3500 times that of air. Heat conductivity 0.58 W . m-1 . °K-1 0.024 W . m-1 . °K-1 Water conducts heat about 24 times as fast as air. -1 -1 Obtaining Oxygen in Water—Gills Most aquatic vertebrates have gills, which are specialized structures where oxygen and carbon dioxide are exchanged. Teleosts are derived ray-ﬁnned ﬁshes, and this group includes the majority of species of extant freshwater and marine ﬁshes. The gills of teleosts are enclosed in pharyngeal pockets called the opercular cavities (Figure 4–1). The ﬂow of water is usually unidirectional—in through the mouth and out through the gills. Flaps just inside the mouth and ﬂaps at the margins of the gill covers (opercula, singular operculum) of bony ﬁshes act as valves to prevent backﬂow. The respiratory surfaces of the gills are delicate projections from the lateral side of each gill arch. Two columns of gill ﬁlaments extend from each gill arch. The tips of the ﬁlaments from adjacent arches meet when the ﬁlaments are extended. As water leaves the buccal cavity, it passes over the ﬁlaments. Gas exchange takes place at the numerous microscopic projections from the ﬁlaments called secondary lamellae. The pumping action of the mouth and opercular cavities (buccal pumping) creates a positive pressure across the gills so that the respiratory current is only slightly interrupted during each pumping cycle. Some ﬁlter-feeding ﬁshes and many pelagic ﬁshes—such as mackerel, certain sharks, tunas, and swordﬁshes— have reduced or even lost the ability to pump water across the gills. These ﬁshes create a respiratory current by swimming with their mouths open, a method known as ram ventilation, and they must swim continuously. Many other ﬁshes rely on buccal pumping when they are at rest and switch to ram ventilation when they are swimming. The arrangement of blood vessels in the gills maximizes oxygen exchange. Each gill ﬁlament has two arteries, an aﬀerent vessel running from the gill arch to the ﬁlament tip and an eﬀerent vessel returning blood to the arch. Each secondary lamella is a blood space connecting the afferent and efferent vessels (Figure 4–2 on page 75). The direction of blood flow through the lamellae is opposite to the direction of water ﬂow across the gill. This arrangement, known as countercurrent exchange, assures that as much oxygen as possible diﬀuses into the blood. Pelagic ﬁshes such as tunas, which sustain high levels of activity for long periods, have skeletal tissue reinforcing the gill ﬁlaments, large gill exchange areas, and a high oxygen-carrying capacity per milliliter of blood compared with sluggish bottom-dwelling fishes, such as toadfishes and flat fishes (Table 4–2 on page 76). Obtaining Oxygen from Air—Lungs and Other Respiratory Structures Although the vast majority of ﬁshes depend on gills to extract dissolved oxygen from water, ﬁshes that live in water with low oxygen levels cannot obtain enough oxygen via gills alone. These ﬁshes supplement the oxygen they get from their gills with additional oxygen obtained from the air via lungs or accessory air respiratory structures. The accessory surfaces used to take up oxygen from air include enlarged lips that are extended just above the water surface and a variety of internal structures into which air is gulped. The anabantid ﬁshes of tropical Asia (including the bettas and gouramies seen in pet stores) have vascularized chambers in the rear of the head, called labyrinths. Air is sucked into the mouth and transferred to the labyrinth, where gas exchange takes place. Many of these ﬁshes are facultative air breathers; that is, they switch oxygen uptake from their gills to accessory respiratory structures when the The Aquatic Environment 73 LATERAL VIEW OF HEAD Gill filaments Water flow Gill arch Efferent artery (to dorsal aorta) Mouth Secondary lamellae Buccal cavity Gill arch Gill filaments (primary lamellae) Gill skeleton Gill slit Gill filament ORAL CAVITY Operculum Water flow Opercular cavity DIAGRAMMATIC HORIZONTAL SECTION THROUGH HEAD (a) Gill arch Opercular cavity Afferent artery (from ventral aorta) DETAIL OF GILL FILAMENTS (b) Figure 4–1 Anatomy of bony fish gills. (a) Position of gills in head and general ﬂow of water; (b) countercurrent ﬂow of water (colored arrows) and blood (black arrows) through the gills. level of oxygen in the water becomes low. Others, like the electric eel and some of the snakeheads, are obligatory air breathers. The gills alone cannot meet the respiratory needs of these ﬁshes, even if the water is saturated with oxygen, and they drown if they cannot reach the surface to breathe air. We think of lungs as being the respiratory structures used by terrestrial vertebrates, as indeed they are, but lungs ﬁrst appeared in ﬁshes and preceded the evolution of tetrapods by millions of years. Lungs develop embryonically as outpocketings (evaginations) of the pharyngeal region of the digestive tract, originating from its ventral or dorsal surface. The lungs of bichirs (a group of air-breathing ﬁshes from Africa), lungﬁshes, and tetrapods originate from the ventral surface of the gut, whereas the lungs of gars (a group of primitive bony ﬁshes) and the lungs of the derived bony ﬁshes known as teleosts originate embryonically from its dorsal surface. 74 CHAPTER 4 Living in Water Lungs used for gas exchange need a large surface area, which is provided by ridges or pockets in the wall. This structure is known as an alveolar lung, and it is found in gars, lungﬁshes, and tetrapods. Increasing the volume of the lung by adding a second lobe is another way to increase the surface area, and the lungs of lungﬁshes and tetrapods consist of two symmetrical lobes. (Bichirs have non-alveolar lungs with two lobes, but one lobe is much smaller than the other; gars have single-lobed alveolar lungs.) Adjusting Buoyancy Holding a bubble of air inside the body changes the buoyancy of an aquatic vertebrate, and bichirs and teleost ﬁshes use the lungs and swim bladders to regulate their position in the water. Air-breathing aquatic vertebrates (whales, dolphins, seals, and penguins, for example) can adjust their buoyancy by altering the volume of air in their lungs when they dive. Water entering gill (low oxygen) Water leaving gill (low oxygen) Water current Blood leaving gill (high oxygen) Secondary lamellae Cross section of gill filament Blood entering gill (low oxygen) (a) Blood flow through secondary lamellae Water current 15 40 70 100 Percent saturation Diffusion gradient 5 (b) 30 60 90 Percent saturation Blood flow Water current 50 50 65 100 Percent saturation Diffusion gradient 50 (c) 50 35 Percent saturation 5 Blood flow Figure 4–2 Countercurrent exchange in the gills of bony fishes. (a) The direction of water ﬂow across the gill opposes the ﬂow of blood through the secondary lamellae. Blood cells are separated from oxygen-rich water only by the thin epithelial cells of the capillary wall, as shown in the cross section of a secondary lamella. (b) Countercurrent ﬂow maintains a diﬀerence in oxygen concentration (a diﬀusion gradient) between blood and water for the full length of the lamella and results in a high oxygen concentration in the blood leaving the gills. (c) If water and blood ﬂowed in the same direction, the diﬀerence in oxygen concentration and the diﬀusion gradient would be high initially, but would drop to zero as the concentration of oxygen equalized. No further exchange of oxygen would occur, and the blood leaving the gills would have a low oxygen concentration. Bony Fishes Many bony ﬁshes are neutrally buoyant (i.e., have the same density as water). These ﬁshes do not have to swim to maintain their vertical position in the water column. The only movement they make when at rest is backpedaling of the pectoral ﬁns to counteract the forward thrust produced by water as it is ejected from the gills and a gentle undulation of the tail ﬁn to keep them level in the water. Fishes capable of hovering in the water like this usually have welldeveloped swim bladders. The swim bladder is located between the peritoneal cavity and the vertebral column (Figure 4–3). The bladder occupies about 5 percent of the body volume of marine teleosts and 7 percent of the volume of freshwater teleosts. The diﬀerence in volume corresponds to the diﬀerence in density of salt water and freshwater—salt water is denser, so a smaller swim bladder is suﬃcient. The swim bladder wall, which has smooth walls composed of interwoven collagen ﬁbers without blood vessels, is virtually impermeable to gas. Neutral buoyancy produced by a swim bladder works as long as a ﬁsh remains at one depth, but if a ﬁsh swims vertically up or down, the hydrostatic pressure that the surrounding water exerts on the bladder changes, which in turn changes the volume of the bladder. For example, when a ﬁsh swims deeper, the additional pressure of the water column above it compresses the gas in its swim bladder, making the bladder smaller and reducing the buoyancy of the ﬁsh. When the ﬁsh swims toward the surface, water pressure decreases, the swim bladder expands, and the ﬁsh becomes more buoyant. To maintain neutral buoyancy, a ﬁsh must adjust the volume of gas in its swim bladder as it changes depth. A bony ﬁsh regulates the volume of its swim bladder by secreting gas into the bladder to counteract the increased external water pressure when it swims down and removing gas when it swims up. Primitive teleosts— such as bony tongues, eels, herrings, anchovies, salmon, and minnows—retain a connection, the pneumatic duct, between the gut and swim bladder (see Figure 4–3a). These ﬁshes are called physostomous (Greek phys = bladder and stom = mouth), and goldﬁsh are a familiar example of this group. Because they have a connection between the gut and the swim bladder, they can gulp air at the surface to ﬁll the bladder and can burp gas out to reduce its volume. The pneumatic duct is absent in adult teleosts from more derived clades, a condition termed physoclistous (Greek clist = closed). Physoclists regulate the volume of the swim bladder by secreting gas from the blood into the bladder. Both physostomes and physoclists have a gas gland, which is located in the anterior ventral ﬂoor of the swim bladder (see Figure 4–3b). Underlying the gas gland is an area with many capillaries arranged to give countercurrent ﬂow of blood entering The Aquatic Environment 75 Table 4–2 Anatomical and physiological characteristics of three types of fishes Oxygen Consumption (ml O2 . g–1 . h–1) Gill Area (mm2 . g body mass–1) Oxygen Capacity (ml O2 . 100 ml blood–1) High, swims continuously 0.73 1160 14.8 Porgy (Stenotomus) Intermediate 0.17 506 7.3 Toadfish (Opsanus) Sluggish, bottom dweller 0.11 197 6.2 Species of Fishes Activity Mackerel (Scomber) and leaving the area. This structure, which is known as a rete mirabile (“wonderful net,” plural retia mirabilia), moves gas (especially oxygen) from the blood to the gas bladder. It is remarkably eﬀective at extracting oxygen from the blood and releasing it into the swim bladder, even when the pressure of oxygen in the bladder is many times higher than its pressure in blood. Gas secretion occurs in many deep-sea ﬁshes despite the hundreds of atmospheres of gas pressure within the bladder. The gas gland secretes oxygen by releasing lactic acid and carbon dioxide, which acidify the blood in the rete mirabile. Acidiﬁcation causes hemoglobin to release oxygen into solution (the Bohr and Root eﬀects). Because of the anatomical relations of the rete mirabile, which folds back upon itself in a countercurrent multiplier arrangement, oxygen released from the hemoglobin accumulates and is retained within the rete until its pressure exceeds the oxygen pressure in Swim bladder Hemoglobin saturation Normal Stomach Pneumatic duct Intestines Dorsal aorta To heart (c) Liv e r Constrictor muscles Gas gland (b) Oxygen pressure Ovale Rete mirabile with parallel vessels dissected apart Hemoglobin saturation (a) Bohr effect Normal (d) Figure 4–3 Swim bladder of bony fishes. (a) The swim bladder is in the coelomic cavity just beneath the vertebral column. This is a physostomous ﬁsh, in which the swim bladder retains its ancestral connection to the gut via the pneumatic duct. (b) The vascular connections of a physoclistous swim bladder, which has lost its connection to the gut. (c) The Bohr eﬀect is a reduction in the aﬃnity of hemoglobin for oxygen in the presence of acid. By creating a Bohr eﬀect, the gas gland causes hemoglobin to release oxygen (i.e., to bind less oxygen). (d) The Root eﬀect is a reduction in the maximum amount of oxygen that hemoglobin can bind. By creating a Root eﬀect, the gas gland prevents oxygen in the gland from binding to hemoglobin in the blood. As a result, the oxygen pressure in the gas gland rises, and oxygen is released into the swim bladder. 76 CHAPTER 4 Living in Water Root effect Oxygen pressure the swim bladder. At this point oxygen diﬀuses into the bladder, increasing its volume. The maximum multiplication of gas pressure that can be achieved is proportional to the length of the capillaries of the rete mirabile, and deep-sea ﬁshes have very long retia. A large Root eﬀect is characteristic only of the blood of ray-ﬁnned ﬁshes, and it is essential for the function of the gas gland. Physoclistous ﬁshes have no connection between the swim bladder and the gut, so they cannot burp to release excess gas from the bladder. Instead, physoclists open a muscular valve, called the ovale, located in the posterior dorsal region of the bladder adjacent to a capillary bed. The high internal pressure of oxygen in the bladder causes it to diffuse into the blood of this capillary bed when the ovale sphincter is opened. Cartilaginous Fishes Sharks, rays, and ratﬁshes do not have swim bladders. Instead, these ﬁshes use the liver to create neutral buoyancy. The average tissue densities of sharks with their livers removed are heavier than water—1.06 to 1.09 grams per milliliter compared to about 1.025 grams per milliliter for seawater. The liver of a shark, however, is well known for its high oil content (shark-liver oil). Shark-liver tissue has a density of only 0.95 gram per milliliter, which is lighter than water, and the liver may contribute as much as 25 percent of the body mass. A 4-meter tiger shark (Galeocerdo cuvieri) weighing 460 kilograms on land may weigh as little as 3.5 kilograms in the sea. Not surprisingly, bottom-dwelling sharks, such as nurse sharks, have livers with fewer and smaller oil vacuoles in their cells, and these sharks are negatively buoyant. Nitrogen-containing compounds in the blood of cartilaginous ﬁshes also contribute to their buoyancy. Urea and trimethylamine oxide in the blood and muscle tissue provide positive buoyancy because they are less dense than an equal volume of water. Chloride ions, too, are lighter than water and provide positive buoyancy, whereas sodium ions and protein molecules are denser than water and are negatively buoyant. Overall these solutes provide positive buoyancy. Deep-Sea Fishes Many deep-sea ﬁshes have deposits of light oil or fat in the gas bladder, and others have reduced or lost the gas bladder entirely and have lipids distributed throughout the body. These lipids provide static lift, just like the oil in shark livers. Because a smaller volume of the bladder contains gas, the amount of secretion required for a given vertical descent is less. Nevertheless, a long rete mirabile is needed to secrete oxygen at high pressures, and the gas gland in deepsea ﬁshes is very large. Fishes that migrate over large vertical distances depend more on lipids such as wax esters than on gas for buoyancy, whereas their close relatives that do not undertake such extensive vertical movements depend more on gas for buoyancy. Air-Breathing Divers Air in the lungs of air-breathing aquatic vertebrates reduces their density. Unlike most ﬁshes, air-breathing vertebrates must return to the surface at intervals, so they do not hover at one depth in the water column. Deep-diving animals, such as elephant seals and some whales and porpoises, face a different problem, however. These animals dive to depths of 1000 meters or more and are subjected to pressures more than 100 times higher than at the surface. Under those conditions, nitrogen would be forced from the air in the lungs into solution in the blood and carried to the tissues at high pressure. When the animal rose toward the surface, the nitrogen would be released from solution. If the animal moved upward too fast, the nitrogen would form bubbles in the tissues—this is what happens when human deep-sea divers get “the bends” (decompression sickness). Specialized diving mammals avoid the problem by allowing the thoracic cavity to collapse as external pressure rises. Air is forced out of the lungs as they collapse, reducing the amount of nitrogen that diﬀuses into the blood. Even these specialized divers would have problems if they made repeated deep dives, however; a deep dive is normally followed by a period during which the animal remains near the surface and makes only shallow dives until the nitrogen level in its blood has equilibrated with the atmosphere. Despite these specializations, the bones of sperm whales, which dive to depths of 2000 m, contain areas of dead tissue caused by the bends. 4.2 Water and the Sensory World of Fishes Water has properties that inﬂuence the behaviors of ﬁshes and other aquatic vertebrates. Light is absorbed by water molecules and scattered by suspended particles. Objects become invisible at a distance of a few hundred meters even in the very clearest water, whereas distance vision is virtually unlimited in clear air. Fishes supplement vision with other senses, some of which can operate only in water. The most important of these aquatic senses is mechanical and consists of detecting water movement via the lateral line system. Small currents of water can stimulate the sensory organs of the lateral line because water is dense and Water and the Sensory World of Fishes 77 viscous. Electrical sensitivity is another sensory mode that depends on the properties of water and does not operate in air. In this case it is the electrical conductivity of water that is the key. Even vision is diﬀerent in water and air because of the diﬀerent refractive properties of the two media. Vision Vertebrates generally have well-developed eyes, but the way an image is focused on the retina is diﬀerent in terrestrial and aquatic animals. Air has an index of refraction of 1.00, and light rays bend as they pass through a boundary between air and a medium with a diﬀerent refractive index. The amount of bending is proportional to the diﬀerence in indices of refraction. Water has a refractive index of 1.33, and the bending of light as it passes between air and water causes underwater objects to appear closer to an observer in air than they really are. The corneas of the eyes of terrestrial and aquatic vertebrates have an index of refraction of about 1.37, so light is bent as it passes through the air-cornea interface. As a result, the cornea of a terrestrial vertebrate plays a substantial role in focusing an image on the retina. This relationship does not hold in water, however, because the refractive index of the cornea is too close to that of water for the cornea to have much eﬀect in bending light. The lens plays the major role in focusing light on the retina of an aquatic vertebrate, and ﬁshes have spherical lenses with high refractive indices. The entire lens is moved toward or away from the retina to focus images of objects at diﬀerent distances from the ﬁsh. Terrestrial vertebrates have ﬂatter lenses, and muscles in the eye change the shape of the lens to focus images. Aquatic mammals such as whales and porpoises have spherical lenses like those of ﬁshes. Chemosensation: Taste and Odor Fishes have taste-bud organs in the mouth and around the head and anterior ﬁns. In addition, olfactory organs on the snout detect soluble substances. Sharks and salmon can detect odors at concentrations of less than 1 part per billion. Sharks, and perhaps bony ﬁshes, compare the time of arrival of an odor stimulus on the left and right sides of the head to locate the source of the odor. Homeward-migrating salmon are directed to their stream of origin from astonishing distances by a chemical signature from the home stream that was permanently imprinted when they were juveniles. Plugging the nasal olfactory organs of salmon destroys their ability to home. 78 CHAPTER 4 Living in Water Touch Mechanical receptors detect touch, sound, pressure, and motion. Like all vertebrates, ﬁshes have an internal ear (the labyrinth organ, not to be confused with the organ of the same name that assists in respiration in anabantid ﬁshes) that detects changes in speed and direction of motion. Fishes also have gravity detectors at the base of the semicircular canals that allow them to distinguish up from down. Most terrestrial vertebrates also have an auditory region of the inner ear that is sensitive to sound-pressure waves. These diverse functions of the labyrinth depend on basically similar types of sense cells, the hair cells (Figure 4–4). In ﬁshes and aquatic amphibians, clusters of hair cells and associated support cells form neuromast organs that are dispersed over the surface of the head and body. In jawed ﬁshes, neuromast organs are often located in a series of canals on the head, and one or more canals pass along the sides of the body onto the tail. This surface receptor system of ﬁshes and aquatic amphibians is referred to as the lateral line system. Lateral line systems are found only in aquatic vertebrates because air is not dense enough to stimulate the neuromast organs. Amphibian larvae have lateral line systems, and permanently aquatic species of amphibians, such as African clawed frogs and mudpuppies, retain lateral lines throughout their lives. Terrestrial species of amphibians lose their lateral lines when they metamorphose into adults, however, and terrestrial vertebrates that have secondarily returned to the water, such as whales and porpoises, do not have lateral line systems. Detecting Water Displacement Neuromasts of the lateral line system are distributed in two conﬁgurations—within tubular canals or exposed in epidermal depressions. Many kinds of ﬁshes have both arrangements. Hair cells have a kinocilium placed asymmetrically in a cluster of kinocilia. Hair cells are arranged in pairs with the kinocilia positioned on opposite sides of adjacent cells. A neuromast contains many such hair-cell pairs. Each neuromast has two aﬀerent nerves: one transmits impulses from hair cells with kinocilia in one orientation, and the other carries impulses from cells with kinocilia positions reversed by 180 degrees. This arrangement allows a ﬁsh to determine the direction of displacement of the kinocilia. All kinocilia and microvilli are embedded in a gelatinous secretion, the cupula (Latin cupula = a small tub). Displacement of the cupula causes the kinocilia to bend. The resultant deformation either excites or Pore in skin surface Canal Lateral line Neuromast organ Pores in skin surface Canal Neuromast organ Nerve (a) Cupula Hair-cell receptor potential Hyperpolarization Depolarization Kinocilia Excitation Recording sites Static discharge Inhibition Nerve impulse discharge (b) Figure 4–4 Lateral line systems. (a) Semidiagrammatic representations of the two conﬁgurations of lateral line organs in ﬁshes. (b) Hair-cell deformations and their eﬀect on hair-cell transmembrane potential (receptor potential) and aﬀerent nerve-cell discharge rates. Deﬂection of the kinocilium (dark line) in one direction (the right in this diagram) depolarizes the cell and increases the discharge rate (excitation). Deﬂection of the kinocilium in the opposite direction (to the left in the diagram) hyperpolarizes the cell and reduces the discharge rate (inhibition). inhibits the neuromast’s nerve discharge. Each haircell pair, therefore, signals the direction of cupula displacement. The excitatory output of each pair has a maximum sensitivity to displacement along the line joining the kinocilia, and falling oﬀ in other directions. The net eﬀect of cupula displacement is to increase the ﬁring rate in one aﬀerent nerve and to decrease it in the other nerve. These changes in lateral line nerve ﬁring rates thus inform a ﬁsh of the direction of water currents on diﬀerent surfaces of its body. Several surface-feeding ﬁshes and African clawed frogs provide vivid examples of how the lateral line Water and the Sensory World of Fishes 79 organs act under natural conditions. These animals ﬁnd insects on the water surface by detecting surface waves created by their prey’s movements. Each neuromast group on the head of the killiﬁsh, Aplocheilus lineatus, provides information about surface waves coming from a diﬀerent direction (Figure 4–5). The groups of neuromasts have overlapping stimulus ﬁelds, allowing the ﬁsh to determine the precise (a) Freestanding organs Nasal organs Supraorbital canal Postorbital canal (b) Left nasal field Left orbital field Right supraorbital field Figure 4–5 Distribution of the lateral line canal organs. (a) The dorsal surface of the head of the killiﬁsh Fundulus notatus. (b) The sensory ﬁelds of the head canal organs in a diﬀerent species of killiﬁsh, Aplocheilus lineatus. The wedgeshaped areas indicate the ﬁelds of view for each group of canal organs. Note that ﬁelds overlap on opposite sides as well as on the same side of the body, allowing the lateral line system to localize the source of a water movement. 80 CHAPTER 4 Living in Water location of the insect. Removing a neuromast group from one side of the head disturbs the directional response to stimuli, showing that a ﬁsh combines information from groups on both sides of the head to interpret water movements. The large numbers of neuromasts on the heads of some ﬁshes might be important for sensing vortex trails in the wakes of adjacent ﬁshes in a school. Many of the ﬁshes that form extremely dense schools (herrings, atherinids, mullets) lack lateral line organs along the ﬂanks and retain canal organs only on the head. These well-developed cephalic canal organs concentrate sensitivity to water motion in the head region, where it is needed to sense the turbulence into which the ﬁsh is swimming, and the reduction of ﬂank lateral line elements would reduce noise from turbulence beside the ﬁsh. Electrical Discharge Unlike air, water conducts electricity, and the torpedo ray of the Mediterranean, the electric catﬁsh from the Nile River, and the electric eel of South America can discharge enough electricity to stun prey animals and deter predators. The weakly electric knifeﬁshes (Gymnotidae) of South America and the elephant ﬁshes (Mormyridae) of Africa use electrical signals for courtship and territorial defense. All of these electric ﬁshes use modiﬁed muscle tissue to produce the electrical discharge. The cells of such modiﬁed muscles, called electrocytes, are muscle cells that have lost the capacity to contract and are specialized for generating an ion current ﬂow (Figure 4–6). When at rest, the membranes of muscle cells and nerve cells are electrically charged, with the intracellular ﬂuids about 84 millivolts more negative than the extracellular ﬂuids. The imbalance is primarily due to sodium ion exclusion. When the cell is stimulated, sodium ions ﬂow rapidly across the smooth surface, sending its potential to a positive 67 millivolts. Only the smooth surface depolarizes; the rough surface remains at –84 millivolts, so the potential diﬀerence across the cell is 151 millivolts (from –84 to +67 millivolts). Because electrocytes are arranged in stacks like the batteries in a ﬂashlight, the potentials of many layers of cells combine to produce high voltages. The South American electric eel has up to 10,000 layers of cells and can generate potentials in excess of 600 volts. Most electric ﬁshes are found in tropical freshwaters of Africa and South America. Few marine forms can generate specialized electrical discharges—among marine cartilaginous ﬁshes, only the torpedo ray (Torpedo), the ray genus Narcine, and some skates are electric; and +− −+ +− −+ + +K − −+ − 84 mV +− − + − 84 mV Na+ high +− −+ outside cell Resting (a) electrocyte −+ Na+ K+ high inside cell −+ −+ −+ − 84 mV −+ −+ −+ K+ −+ + 67 mV −+ Na+ − + Active electrocyte (b) (c) At the peak of nerve stimulation At rest Tail +− +− +− +− −+ + −+ −+ −+ +− +− +− +− −+ −+ −+ −+ −+ +− +− +− +− −+ −+ −+ −+ −+ Space outside cells Electrocyte Resting electrocytes (d) −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ − − − − − + Active electrocytes −+ −+ −+ −+ Spinal motor neurons + − Head −+ −+ −+ −+ Positive current flow during discharge © 1994 American Institute of Physics Figure 4–6 Weakly electric fishes. Some ﬁshes use transmembrane potentials of modiﬁed muscle cells to produce a discharge. In this diagram the smooth surface is on the left and the rough surface on the right. Only the smooth surface is innervated. (a) At rest, K+ (potassium ion) is maintained at a high internal concentration and Na+ (sodium ion) at a low internal concentration by the action of a Na+/K+ cell-membrane pump. Permeability of the membrane to K+ exceeds the permeability to Na+. As a result, K+ diﬀuses outward faster than Na+ diﬀuses inward (arrow) and sets up the –84-mV resting potential. (b) When the smooth surface of the cell is stimulated by the discharge of the nerve, Na+ diﬀuses into the cell and K+ diﬀuses out of the smooth surface, changing the net potential to +67 mV. The rough surface does not depolarize and retains a –84-mV potential, creating a potential diﬀerence of 151 mV across the cell. (c) A weakly electric South American gymnotid, showing the location of electrocytes along the sides of the body. (d) By arranging electrocytes in series so that the potentials of individual cells are summed, some electric ﬁshes can generate very high voltages. Electric eels, for example, have 10,000 electrocytes in series and produce potentials in excess of 600 volts. among marine teleosts, only the stargazers (family Uranoscopidae) produce specialized discharges. Electroreception by Sharks and Rays The high conductivity of seawater makes it possible for sharks to detect the electrical activity that accompanies muscle contractions of their prey. Sharks have structures known as the ampullae of Lorenzini on their heads, and rays have them on the pectoral ﬁns as well. The ampullae are sensitive electroreceptors (Figure 4–7). The canal connecting the receptor to the surface pore is ﬁlled with an electrically conductive gel, and the wall of the canal is nonconductive. Because the canal runs for some distance beneath the epidermis, the sensory cell can detect a diﬀerence in electrical potential between the tissue in which it lies (which reﬂects the adjacent epidermis and environ- ment) and the distant pore opening. Thus, it can detect electric ﬁelds, which are changes in electrical potential in space. Electroreceptors of sharks respond to minute changes in the electric field surrounding an animal. They act like voltmeters, measuring differences in electrical potentials across the body surface. Ampullary organs are remarkably sensitive, with thresholds lower than 0.01 microvolt per centimeter, a level of detection achieved by only the best voltmeters. Sharks use their electrical sensitivity to detect prey. All muscle activity generates electrical potential: motor nerve cells produce extremely brief changes in electrical potential, and muscular contraction generates changes of longer duration. In addition, a steady potential issues from an aquatic organism as a result of Water and the Sensory World of Fishes 81 livolt per centimeter—well above the level that can be detected by ampullary organs. In addition, ocean currents generate electrical gradients as large as 0.5 millivolt per centimeter as they carry ions through Earth’s magnetic ﬁeld. Nostril Gelfilled canal Nerves (a) (b) Sensory cell Epidermis Surface pore Figure 4–7 Ampullae of Lorenzini. (a) Distribution of the ampullae on the head of a spiny dogﬁsh, Squalus acanthias. Open circles represent the surface pores; the black dots are positions of the sensory cells. (b) A single ampullary organ consists of a sensory cell connected to the surface by a pore ﬁlled with a substance that conducts electricity. Electrolocation by Teleosts Unusual arrangements of electrocytes are present in several species of freshwater ﬁshes that do not produce electric shocks. In these ﬁshes—which include the knifeﬁshes (Gymnotidae) of South America and the elephant ﬁshes (Mormyridae) of Africa—the discharge voltages are too small to be of direct defensive or oﬀensive value. These weakly electric ﬁshes are mostly nocturnal and usually live in turbid waters where vision is limited to short distances even in daylight; they use their discharges for electrolocation and social communications. When a ﬁsh discharges its electric organ, it creates an electric ﬁeld in its immediate (a) the chemical imbalance between the organism and its surroundings. A shark can locate and attack a hidden ﬁsh by relying only on this electrical activity (Figure 4–8). Sharks may use electroreception for navigation as well as for locating prey. The electromagnetic ﬁeld at Earth’s surface produces tiny voltage gradients, and a swimming shark could encounter gradients as large as 0.4 mil- Figure 4–8 Electrolocation capacity of sharks. (a) A shark can locate a live ﬁsh concealed from sight beneath the sand. (b) The shark can still detect the ﬁsh when it is covered by an agar shield that blocks olfactory cues but allows the electrical signal to pass. (c) The shark follows the olfactory cues (displaced by the agar shield) when the live ﬁsh is replaced by chopped bait that produces no electrical signal. (d) The shark is unable to detect a live ﬁsh when it is covered by a shield that blocks both olfactory cues and the electrical signal. (e) The shark attacks electrodes that give oﬀ an electrical signal duplicating a live ﬁsh without producing olfactory cues. These experiments indicate that when the shark was able to detect an electrical signal, it used that to locate the ﬁsh—and it was also capable of homing in on a chemical signal when no electrical signal was present. This dual system allows sharks to ﬁnd both living and dead food items. 82 CHAPTER 4 Living in Water Point of attack Live flounder (b) Agar shield No vertical olfactory cue (c) Chopped bait No electrical cues Displaced olfactory cue (d) Electrical insulation No attack (e) Live electrodes Each accurately attacked vicinity (Figure 4–9). Because of the high energy costs of maintaining a continuous discharge, electric ﬁshes produce a pulsating discharge. Most weakly electric teleost ﬁshes pulse at rates between 50 and 300 cycles per second, but the knifeﬁshes of South America reach 1700 cycles per second, which is the most rapid continuous ﬁring rate known for any vertebrate muscle or nerve. The electric ﬁeld from even weak discharges may extend outward for a considerable distance in freshwater because electrical conductivity is relatively low. The electric ﬁeld the ﬁsh creates will be distorted by the presence of electrically conductive and resistant objects. Rocks are highly resistive, whereas other ﬁshes, (a) (b) © 2006 The Company of Biologists Ltd. invertebrates, and plants are conductive. Distortions of the ﬁeld cause a change in the distribution of electrical potential across the ﬁsh’s body surface. An electric ﬁsh detects the presence, position, and movement of objects by sensing where on its body maximum distortion of its electric ﬁeld occurs. The skin of weakly electric teleosts contains special sensory receptors: ampullary organs and tuberous organs. These organs detect tonic (steady) and phasic (rapidly changing) discharges, respectively. Electroreceptors of teleosts are modiﬁed lateral line neuromast receptors. Like lateral line receptors, they have double innervation—an aﬀerent channel that sends impulses to the brain and an eﬀerent channel that causes inhibition of the receptors. During each electric organ discharge, an inhibitory command is sent to the electroreceptors, and the ﬁsh is rendered insensitive to its own discharge. Between pulses, electroreceptors report distortion in the electric ﬁeld or the presence of a foreign electric ﬁeld to the brain. Electric organ discharges vary with habits and habitat. Species that form groups or live in shallow, narrow streams generally have discharges with high frequency and short duration. These characteristics reduce the chances of interference from the discharges of neighbors. Territorial species, in contrast, have long electric organ discharges. Electric organ discharges vary from species to species. In fact, some species of electric ﬁshes were ﬁrst identiﬁed by their electric organ discharges, which were recorded by placing electrodes in water that was too murky for any ﬁshes to be visible. During the breeding season, electric organ discharges distinguish immature individuals, females with eggs, and sexually active males. Other Electroreceptive Vertebrates Electrogenesis and (c) Figure 4–9 Weakly electric fishes. An electric ﬁeld surrounds a weakly electric ﬁsh. Electroreceptors in the skin allow a ﬁsh to detect the presence of nearby objects by sensing distortion of the lines of electrical force. (a) Nonconductive objects, such as rocks, spread the ﬁeld and diﬀuse potential diﬀerences along the body surface. (b) Conductive objects, such as another ﬁsh, concentrate the ﬁeld on the skin of the ﬁsh. (c) When two electric ﬁsh swim close enough to each other to create interference between their electric ﬁelds, they change the frequencies of their discharges. electroreception are not restricted to a single group of aquatic vertebrates, and monotremes (the platypus and the echidna, early oﬀshoots oﬀ the main mammalian lineages that still lay eggs) use electroreception to detect prey. Electrosensitivity was probably an early feature of vertebrate evolution. The brain of the lamprey responds to electric ﬁelds, and it seems likely that the earliest vertebrates had electroreceptive capacity. All ﬁshlike vertebrates of lineages that evolved before the neopterygians have electroreceptor cells. These cells, which have a prominent kinocilium, ﬁre when the environment around the kinocilium is negative relative to the cell. Their impulses pass to the midline region of the posterior third of the brain. Electrosensitivity was apparently lost in neopterygians, and teleosts have at least two separate new evolutions of electroreceptors. Electrosensitivity of teleosts is Water and the Sensory World of Fishes 83 distinct from that of other vertebrates: teleost electroreceptors lack a kinocilium and ﬁre when the environment is positive relative to the cell, and nerve impulses are sent to the lateral portions of the brain rather than to the midline. 4.3 The Internal Environment of Vertebrates Seventy to eighty percent of the body mass of most vertebrates is water, and the chemical reactions that release energy or synthesize new chemical compounds take place in an aqueous environment. The body ﬂuids of vertebrates contain a complex mixture of ions and other solutes. Some ions are cofactors that control the rates of metabolic processes; others are involved in the regulation of pH, the stability of cell membranes, or the electrical activity of nerves. Metabolic substrates and products must diﬀuse from sites of synthesis to the sites of utilization. Almost everything that happens in the body tissues of vertebrates involves water, and maintaining the concentrations of water and solutes within narrow limits is a vital activity. Water sounds like an ideal place to live for an animal that itself consists mostly of water, but in some ways an aquatic environment can be too much of a good thing. Freshwater vertebrates—especially ﬁshes and amphibians— face the threat of being ﬂooded with water that ﬂows into them from their environment, and saltwater vertebrates must prevent the water in their bodies from being sucked out into the sea. Temperature, too, is a critical factor for living organisms because chemical reactions are temperature sensitive. In general, the rates of chemical reactions increase as temperature increases, but not all reactions have the same sensitivity to temperature. A metabolic pathway is a series of chemical reactions in which the product of one reaction is the substrate for the next, yet each of these reactions may have a diﬀerent sensitivity to temperature, so a change in temperature can mean that too much or too little substrate is produced to sustain the next reaction in the series. To complicate the process of regulation of substrates and products even more, the chemical reactions take place in a cellular milieu that itself is changed by temperature because the viscosity of plasma membranes is also temperature sensitive. Clearly, the smooth functioning of metabolic pathways is greatly simpliﬁed if an organism can limit the range of temperatures its tissues experience. Water temperature is more stable than air temperature because water has a much higher heat capacity 84 CHAPTER 4 Living in Water than air. The stability of water temperature simpliﬁes the task of maintaining a constant body temperature, as long as the body temperature the animal needs to maintain is the same as the temperature of the water around it. An aquatic animal has a hard time maintaining a body temperature diﬀerent from water temperature, however, because water conducts heat so well. Heat ﬂows out of the body if an animal is warmer than the surrounding water and into the body if the animal is cooler than the water. In the following sections, we discuss in more detail how and why vertebrates regulate their internal environments and the special problems faced by aquatic animals. 4.4 Exchange of Water and Ions An organism can be described as a leaky bag of dirty water. That is not an elegant description, but it accurately identiﬁes the two important characteristics of a living animal—it contains organic and inorganic substances dissolved in water, and this ﬂuid is enclosed by a permeable body surface. Exchange of matter and energy with the environment is essential to the survival of the organism, and much of that exchange is regulated by the body surface. Water molecules and ions pass through the skin quite freely, whereas larger molecules move less readily. The signiﬁcance of this diﬀerential permeability is particularly conspicuous in the case of aquatic vertebrates, but it applies to terrestrial vertebrates as well. Vertebrates use both active and passive exchange to regulate their internal concentrations in the face of varying external conditions. The Vertebrate Kidney An organism can tolerate only a narrow range of concentrations of the body ﬂuids and must eliminate waste products before they reach harmful levels. The molecules of ammonia that result from the breakdown of protein are especially important because they are toxic. Vertebrates have evolved superb capacities for controlling water balance and excreting wastes, and the kidney plays a crucial role in these processes. The adult vertebrate kidney consists of hundreds to millions of tubular nephrons, each of which produces urine. The primary function of a nephron is removing excess water, salts, waste metabolites, and foreign substances from the blood. In this process, the blood is ﬁrst ﬁltered through the glomerulus, a structure unique to Lumen of the glomerular capsule Capillary bed of glomerulus Afferent arteriole Proximal convoluted tubule Smooth muscle Efferent arteriole Ultrafiltrate Endothelial cell (capillary cell) Figure 4–10 Detail of a typical mammalian glomerulus. Blood pressure forces an ultraﬁltrate of the blood through the walls of the capillary into the lumen of the glomerular capsule. The blood ﬂow to each glomerulus is regulated by smooth muscles that can close oﬀ the aﬀerent and eﬀerent arterioles to adjust the glomerular ﬁltration rate of the kidney as a whole. The ultraﬁltrate, which consists of water, ions, and small molecules, passes from the glomerular capsule into the proximal convoluted tubule, where the process of adding and removing speciﬁc substances begins. vertebrates (Figure 4–10). Each glomerulus is composed of a leaky arterial capillary tuft encapsulated within a sievelike ﬁlter. Arterial blood pressure forces ﬂuid into the nephron to form an ultrafiltrate, composed of blood minus blood cells and larger molecules. The ultraﬁltrate is then processed to return water and essential metabolites (glucose, amino acids, and so on) to the general circulation. The ﬂuid that remains after this processing is urine. Regulation of Ions and Body Fluids The salt concentrations in the body ﬂuids of many marine invertebrates are similar to those in seawater, as are those of hagﬁshes (Table 4–3). It is likely that the ﬁrst vertebrates also had ion levels similar to those in seawater. In contrast, salt levels are greatly reduced in the blood of all other vertebrates. In the context of body ﬂuids, a solute is a small molecule that is dissolved in water or blood plasma. Salt ions, urea, and some small carbohydrate molecules are the solutes primarily involved in the regulation of body ﬂuid concentrations. The presence of solutes lowers the potential activity of water. Water moves from areas of high potential to areas of lower potential; therefore, water ﬂows from a dilute solution (one with a high water potential) to a more concentrated solution (with a lower water potential). This process is called osmosis. Seawater has a solute concentration of approximately 1000 millimoles per kilogram of water (mmol . kg–1). Most marine invertebrates and hagﬁshes have body ﬂuids that are in osmotic equilibrium with seawater; that is, they are isosmolal to seawater. Body ﬂuid concentrations in marine teleosts and lampreys are between 300 and 350 mmol . kg–1. Therefore, water ﬂows outward from their blood to the sea (i.e., from a region of high water potential to a region of lower water potential). Cartilaginous ﬁshes retain urea and other nitrogen-containing compounds, raising the osmolality of their blood slightly above that of seawater so water ﬂows from the sea into their bodies. These osmolal Exchange of Water and Ions 85 Table 4–3 Sodium, chloride and osmolality of the blood of vertebrates and marine invertebrates. Concentrations are expressed in millimoles per kilogram of water; all values are rounded to the nearest 5 units. Osmolality (mmol . kg–1) Na+ Cl- ~1000 475 550 0.3 >0.4 >0.5 >0.6 >0.7 >0.8 >0.9 U © AAAS pper panel: The geographic distribution of amphibian species and the threats they face. (a) Species richness of frogs: Frogs are the most diverse group of amphibians, and in tropical regions of South America, Africa, and Southeast Asia more than 100 species of frogs can be found in the same habitats. (b) Global climate change: In most parts of the world climate change is predicted to make the environment unsuitable for at least 50 percent of the species of frogs in the region, and in some areas 100 percent of the species will be affected. (c) Chytridiomycosis: From 50 to 100 percent of frog species are expected to be threatened by chytridiomycosis, with frog communities in South America, western Europe, and southern Australia feeling the greatest impact. (d) Habitat loss: The greatest impact of habitat loss is predicted to occur in South America, Africa, and Asia where logging and agriculture are destroying native forests. ower panel: Extinctions of Mexican Sceloporus lizards as a result of climate change. (e) In 2009 a survey of 200 sites that had been occupied by species of Sceloporus in 1975 found that 12 percent of the populations were extinct. Extinctions were grouped in the areas most affected by rising temperatures in northern and central Mexico, and populations at high altitudes showed the greatest levels of extinction. Viviparous species had higher rates of extinction than did oviparous species. (f) By 2080 the temperature increase predicted by global climate models will lead to widespread extinctions of populations of Sceloporus in Mexico, and again viviparous species would feel the greatest impact. L Geological Time Scale Eon Era Period Position of continents on the globe Epoch Holocene (Recent) Pleistocene CENOZOIC QUATERNARY Neogene TERTIARY MESOZOIC PALEOZOIC PHANEROZOIC PRE CAMBRIAN JURASSIC 2.6 5.3 33.9 CRETACEOUS 0.01 23 Paleogene Pliocene, Miocene Approximate time since the beginning of each interval in millions of years before the present Oligocene, Eocene, Paleocene Subdivisions exist of all periods, at least into Early and Late portions, and sometimes with a Middle portion. Within these divisions there are further subdivisions, similar to the epochs of the Tertiary and Quaternary Periods. This level of detail is not shown here. 55.8 65.5 145.5 201.6 TRIASSIC 251 PERMIAN 299 CARBONIFEROUS 359 DEVONIAN 416 SILURIAN 444 ORDOVICIAN 488 CAMBRIAN 542 PROTEROZOIC 2,500 ARCHEAN (INCLUDING HADEAN) 4,527 Major Geological, Climatic, and Biological Events nt There were extensive and repeated periods of glaciation in the Northern Hemisphere, and the Neogene midlatitude savanna faunas became extinct. The genus Homo appeared and hominins expanded throughout the Old World. Near the end of the interval, hominins reached the New World. There was an extinction of many large mammals, especially in the New World and in Australia. The remaining carnivorous flightless birds also became extinct. Cooler and more arid climates persisted, resulting from mountain uplift and the formation of the Isthmus of Panama near the end of the interval. The Arctic ice cap had formed by the end of the interval. The first grasslands spread in the middle latitudes. Modern families of mammals and birds radiated, and marine mammals and birds diversified in the oceans. The first hominins were seen near the end of the interval. Global climate was warm in the early part of the interval, with forests above the Arctic Circle, but later temperatures fell in the higher latitudes, with the formation of the Antarctic ice cap. Mammals diversified into larger body sizes and a greater variety of adaptive types, including predators and herbivores. Mammal radiations included archaic forms, now extinct, and the earliest members of living orders. Giant carnivorous flightless birds were common as predators. Further separation of the continents occurred, including the breakup of the southern continent, Gondwana. Teleost fishes radiated, and marine reptiles flourished. Angiosperms first appeared and rapidly diversified to become the dominant land plants by the end of the period. Dinosaurs remained the dominant tetrapods, but small mammals, including the first therians, diversified. Birds and pterosaurs coexisted, and the first snakes appeared. A major mass extinction at the end of the period, defining the end of the Mesozoic, claimed dinosaurs, pterosaurs, and marine reptiles, as well as many marine invertebrates. The world continent began to break up, with the formation of the Atlantic Ocean. Marine invertebrates began to take on a modern aspect with the diversification of predators, modern sharks and rays appeared, and marine reptiles diversified. Conifers and other gymnosperms were the dominant terrestrial vegetation, and insects diversified. Dinosaurs diversified while mammals remained small and relatively inconspicuous. The first birds, lizards, and salamanders were seen at the end of the period. The world continent was relatively high, with few shallow seas. No evidence of glaciation existed, and the interior of the continent was arid. Conifers replaced seed fern terrestrial vegetation in the later part of the period. Non-mammalian synapsids declined, while archosaurian reptiles (including dinosaur ancestors) diversified. Remaining large nonamniote tetrapods were all specialized aquatic forms. First appearances by the end of the period included true mammals, dinosaurs, pterosaurs, marine reptiles, crocodiles, lepidosaurs, froglike amphibians, and teleost fishes. A single world continent, Pangaea, was formed at the end of the period. Glaciation ceased early in the period. The large terrestrial nonamniote tetrapods declined and the amniotes radiated. Amniote diversification included the ancestors of modern reptiles and the ancestors of mammals, the non-mammalian synapsids being the dominant large terrestrial tetrapods. The first herbivorous tetrapods evolved. The largest known mass extinction event occurred on both land and sea at the end of the period. It coincided with low levels of atmospheric O2 and marked the end of the Paleozoic. There was a major glaciation in the second half of the period, with low atmospheric levels of CO2 and high levels of O2. Coal swamps were prevalent in the then-tropical areas of North America and Europe. Major radiation of insects, including flying forms. Diversification of jawed fishes, including sharklike forms and primitive bony fishes, and first appearance of modern types of jawless fishes. Extensive radiation of non-amniote tetrapods, with the appearance of the first amniotes (including the earliest mammal-like reptiles) by the late part of the period. There was major mountain building in North America and Europe. Major freshwater basins, containing the first tetrapods, formed in equatorial regions at the end of the period. About the same time there were the first forests with tall trees on land, and terrestrial arthropods diversified. Both jawed and jawless fishes diversified, but both experienced major extinctions toward the end of the period, with the disappearance of the ostracoderms, the armored jawless fishes. The extensive shallow seas continued, but on dry land there was the first evidence of vascular plants and arthropods. Jawless fishes radiated, and jawed fishes (sharklike forms) were now definitely known. There were widespread shallow seas over the continents, and the global climate was equable until a sharp glaciation at the end of the period. First evidence of complex plants on land. Major radiation of marine animals, including the first well-known jawless fishes and fragmentary evidence of jawed fishes. Continental masses of the late Proterozoic now broken up into smaller blocks, covered by shallow seas. Explosive radiation of animals at the beginning of the period, with first appearance of forms with shells or other hard coverings. First appearance of chordates and great diversification of arthropods, including trilobites. First vertebrates appeared early in the period. Formation of large continental masses. Oxygen first appears in the atmosphere. First eukaryotic organisms appeared around 2 billion years ago. Major diversification of life at 1 billion years ago, with multicellular organisms, including algae. First animals appeared around 600 million years ago, just after a major glaciation. Formation of the Earth. Major bombardment of the Earth by extraterrestrial bodies, precluding formation of life until 4 billion years ago (first fossils known at 3.8 billion years ago). Small continents. Hydrosphere definite at 3.8 billion years, atmosphere without free oxygen. Latin and Greek Lexicon Many biological names and terms are derived from Latin (L) and Greek (G). Learning even a few dozen of these roots is a great aid to a biologist. The following terms are often encountered in a vertebrate biology. The words are presented in the spelling and form in which they are most often encountered; this is not necessarily the original form of the word in its etymologically pure state. An example of how a root is used in vertebrate biology can often be found by referring to the subject index. Remember, however, that some of these roots may be used as suffixes or otherwise embedded in technical words and will require further searching to discover an example. Additional information can be found in a reference such as the Dictionary of Word Roots and Combining Forms, by Donald J. Borror (Palo Alto, Calif.: Mayfield Publishing Co.) a, ab (L) away from a, an (G) not, without acanth (G) thorn actin (G) a ray ad (L) toward, at, near aeros (G) the air aga (G) very much, too much aistos (G) unseen al, alula (L) a wing allant (G) a sausage alveol (L) a pit ambl (G) blunt ammos (G) sand amnion (G) a fetal membrane amphi, ampho (G) both, double amplexus (L) an embracing ampulla (L) a jug or flask ana (G) up, upon, through anat (L) a duck angio (G) a reservoir, vessel ankylos (G) crooked, bent anomos (G) lawless ant, anti (G) against ante (L) before anthrac (G) coal apat (G), illusion, error aphanes (G) invisible, unknown apo, ap (G) away from, separate apsid (G) an arch, loop aqu (L) water arachne (G) a spider arch (G) beginning, first in time argenteus (L) silvery arthr (G) a joint ascidion (G) a little bag or bladder aspid (G) a shield asteros (G) a star atri, atrium (L) an entrance-room audi (L) to hear austri, australis (L) southern avis (L) a bird baen (G) to walk or step bas (G) base, bottom batrachos (G) a frog benthos (G) the sea depths bi, bio (G) life bi, bis (L) two blast (G) bud, sprout brachi (G) arm brachy (G) short branchi (G) a gill or fin buce (L) the check cal (G) beautiful calie (L) a cup capit (L) head carn (L) flesh caud (L) tail cene, ceno (G) new, recent cephal (G) head cer, cerae (G) a horn cerc (G) tail chir, cheir (G) hand choan (G) funnel, tube chondr (G) grit, gristle chord (G) guts, a string chorio (G) skin, membrane chrom (G) color clist (G) closed cloac (L) a sewer coel (G) hollow cornu (L) a horn cortic, cortex (L) bark, rind costa (L) a rib cran (G) the skull creta (L) chalk cretio (L) separate crini (L) the hair cten (G) a comb cut, cutis (L) the skin cyn (G) a dog cytos (G) a cell dactyl (G) a finger de (L) down, away from dectes (G) a biter dendro (G) a tree dent, dont (L) a tooth derm (G) skin desmos (G) a chain, tie, or band deuteros (G) secondary di, dia (G) through, across di, diplo (G) two, double din, dein (G) terrible, powerful dir (G) the neck disc (G) a disk dory (G) a spear draco (L) a dragon drepan (G) a sickle dromo (G) running duc (L) to lead dur (L) hard e, ex (L) out of, from, without echinos (G) a prickly being eco, oikos (G) a house ect (G) outside edaphos (G) the soil or bottom eid (G) form, appearance elasma (G) a thin plate eleuthero (G) free, not bound elopos (G) a kind of sea fish embolo (G) like a peg or stopper embryon (G) a fetus emys (G) a freshwater turtle end (G) within enter (G) bowel, intestine eos (G) the dawn or beginning ep (G) on, upon equi (L) a horse ery (G) to drag or draw erythr (G) red eu, ev (G) good, true eury (G) broad extra (L) beyond, outside falc (L) a sickle, scythe fenestra (L) a window fer (L) a carrier of fil (L) a thread fossa (L) a ditch fundus (L) bottom, foundation galeos (G) a shark gallus (L) poultry gaster (G) the belly genos (G) birth genus (L) a race or stock geo (G) the Earth gephyr (G) a bridge gerrhron (G) made of wicker-work glob (L) a ball glom, glomer (L) a ball of yarn gloss (G) the tongue glyco (G) sweet gnath (G) the jaw gony (G) the knee gorgo (L) a female monster of terrible aspect gracil (L) slender grapho (G) intricate or puzzling grapto (G) inscribed or painted gul (L) the throat gymn (G) naked gyr (G) round, a circle haem (G) blood hal (G) the sea hemi (G) half hepat (G) the liver herp (G) to creep hetero (G) other, different hipp (G) a horse hist (G) web, tissue hol (G) whole, entire homo (G) alike hyp, hypo (G) under, beneath hyper (G) above, beyond hyps (G) high, height ichthy (G) fish in (L) in, into, not, without infra (L) below inter (L) between, among intr (L) inside is, iso (G) similar, equal kin (G) movement labyrinthos (G) a tortuous passageway lecith (G) yolk lepis (G) a scale leptos (G) thin, small, weak lingu (L) tongue liss (G) smooth loph (G) crest, ridge lumin, lumen (L) a light lut (L) yellow lychnos (G) a lamp macr (G) long, large magn (L) great, large mamm (L) a breast mast (G) a breast meatus (L) a passage medull (L) marrow, pith mela (G) black mer (G) a part mes (G) middle meta, met (G) next to micr (G) small mon (G) single morph (G) shape mys (G) muscle, mouse nect (G) swimming neo (G) new, recent nephro (G) kidney nomen (L) name noto (G) the back odont, don (G) tooth oligo (G) few, small, little omni (L) all opisth (G) behind ops (G) appearance orinth (G) a bird oro (L) the mouth orth (G) straight osmos (G) pushing or thrusting oste (G) bone oxys, oxus (G) sharp, pointed palae, paleo (G) ancient par, para (G) beside parous (L) to beget percul (L) a cover, lid peri (G) around, near phag (G) to eat phil (G) loving, friend phor (G) to bear phot (G) light phyl (G) a tribe, race phyll (G) leaf physo (G) an air sac platy (G) broad pleur (G) a rib, the side ppneuma, uma (G) breathing pod (G) a foot poikilos (G) variable poly (G) many por (L) a passage post (L) after, behind prim (L) first pro (G) before, in front of prot (G) first, primary pseud (G) false pter, pteron (G) wing pyg (G) the rump pyxi (G) a box ram (L) a branch rept (L) to crawl retor (L) backward rhaci (G) backbone rhin (G) a nose rhynch (G) a beak, snout sarc (G) flesh saur (G) a lizard scler (G) hard scopos (G) a watchman som (G) the body speci (L) kind, sort sperma (G) seed, sperm sphen (G) a wedge splanchn (G) viscera spondyl (G) vertebra squam (L) a scale steg (G) a roof sten (G) narrow, straight stom (G) mouth styl (G) a pillar sub (L) under, below suchus (G) crocodile super (L) above, over syn, sym (G) together tele (G) perfect, entire tethy (G) a sea goddess tetr (G) four thec (G) a case theri (G) a wild animal therm (G) heat tom (G) a cut, slice top (G) a place trem (G) a hole tri (G) three trich (G) a hair trop (G) a turn, change troph (G) food, feeding tryma (G) a hole ulna (L) the elbow ultim (L) the farthest, last unguis, ungula (L) claw, hoof, nail ura, oura (G) tail ureo (G) urine urs (L) a bear vas (L) a vessel velum (L) a veil or covering ventr (L) the belly vertebra (L) a joint vest (L) a coat vibrissa (L) a nostril hair viscera (L) the internal organs vivus, vivi (L) alive vor (L) to devour xanthos (G) yellow colored xen (G) a stranger or foreigner zoo (G) an animal zyg (G) a coupling or linkage Report "Vertebrate Life [9th ed.] 9780321773364" × Close Submit Contact information Michael Browner info@dokumen.pub Address: 1918 St.Regis, Dorval, Quebec, H9P 1H6, Canada. 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6151	https://www.doubtnut.com/pcmb-questions/122398	The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OIf the relative molecular mass of carbon dioxide is 44 what is the mass of carbon dioxide produced by burning two moles of octane ? More from this Exercise From equation we know that 2moles of octane produces 16 moles of CO2 . Mass of CO2 produced = 16 x 44 = 704 Related Solutions The equation for the burning of octane is 2 C 8 H 18 + 25 O 2 → 16 C O 2 + 18 H 2 O If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ? The equation for the burning of octane is 2C_(8)H_(18) + 25O_(2) to 16CO_(2) + 18H_(2)O How many moles of carbon dioxide are produced when one mole of octane burns ? कार्बन डाइऑक्साइड का सापेक्ष आणविक द्रव्यमान ____________ है। The equation for the burning of octane is 2 C 8 H 18 + 25 O 2 → 16 C O 2 + 18 H 2 O What volume at S.T .P. is occupied by the number of moles of carbon dioxide produced when one mole of octane burns ? Oxygen: Burn :: Carbon-dioxide : ? గాలిలో ఒక మోల్ కార్బన్ ను మండించినప్పుడు వెలువడే కార్బన్ డై ఆక్సైడ్ భారాన్ని లెక్కకట్టండి. Determine the amount of carbon dioxide produced in this case 1mol Carbon is burned in the air. Molecular mass of carbon dioxide is Molecular mass of carbon dioxide is………… . कार्बन डाइऑक्साइड वस्तुओं को जलाने में सहायक है। Carbon dioxide added by biomass burning is In one mole of ethanol (C_2H_5OH) completely burns to carbon dioxide and water the weight of carbon dioxide formed is about : Translate the following statements into chemical equation and balance the equations : carbon disulphide burns in air to give carbon dioxide and sulphur dioxide. ઉત્પન્ન થયેલ કાર્બન ડાયોક્સાઇડનું પ્રમાણ ગણો . જ્યારે, 2 મોલ કાર્બનને 16 g ડાયઑક્સિજનમાં બાળવામાં આવે છે. Multiply the amount of carbon dioxide produced. Whereas, 1 mole of carbon is burned in the air. Multiply the amount of carbon dioxide produced. Whereas, 1 mole of carbon is burned in 16 g of dioxin. এক কেজি মোল কার্বন ডাইঅক্সাইডের ভর কত ? निम्नलिखित के लिए संतुलित रासायनिक समीकरण लिखें- कार्बन डाइसल्फाइड वायु में जलकर कार्बन डाइऑक्साइड तथा सल्फर डाइऑक्साइड बनाती है। कार्बन डाइऑक्साइड में कार्बन का द्रव्यमान क्या है? 3.0g of carbon burns to 8.00g of oxygen to produce 11.00g of carbon dioxide. How many grams of carbon dioxide will be formed when 3.00g of carbon is burned in 50.00g of oxygen? Recommended Content What energy conversions take place in the following when they are work... The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18... What do you understand by the term colour? Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
6152	https://pmc.ncbi.nlm.nih.gov/articles/PMC11648870/	Permafrost instability negates the positive impact of warming temperatures on boreal radial growth - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Proc Natl Acad Sci U S A . 2024 Dec 2;121(50):e2411721121. doi: 10.1073/pnas.2411721121 Search in PMC Search in PubMed View in NLM Catalog Add to search Permafrost instability negates the positive impact of warming temperatures on boreal radial growth Raquel Alfaro-Sánchez Raquel Alfaro-Sánchez a Northern Forestry Centre, Canadian Forest Service, Natural Resources Canada, Edmonton, AB T6H 3S5, Canada b Department of Biology, Wilfrid Laurier University, Waterloo, ON N2L 3C5, Canada Find articles by Raquel Alfaro-Sánchez a,b,1, Andrew D Richardson Andrew D Richardson c Center for Ecosystem Science and Society, Northern Arizona University, Flagstaff, AZ 86011 d School of Informatics, Computing, and Cyber Systems, Northern Arizona University, Flagstaff, AZ 86011 Find articles by Andrew D Richardson c,d, Sharon L Smith Sharon L Smith e Geological Survey of Canada, Natural Resources Canada, Ottawa, ON K1A 0G1, Canada Find articles by Sharon L Smith e, Jill F Johnstone Jill F Johnstone f YukonU Research Centre, Yukon University, Whitehorse, YT Y1A 5K4, Canada g Institute of Arctic Biology, University of Alaska Fairbanks, Fairbanks, AK 99775 Find articles by Jill F Johnstone f,g, Merritt R Turetsky Merritt R Turetsky h Renewable and Sustainable Energy Institute, Department of Ecology and Evolutionary Biology, University of Colorado Boulder, Boulder, CO 80309-0027 Find articles by Merritt R Turetsky h, Steven G Cumming Steven G Cumming i Department of Wood and Forest Sciences, Laval University, Quebec City, QC G1V 0A6, Canada Find articles by Steven G Cumming i, James M Le Moine James M Le Moine j LI-COR Biosciences, Lincoln, NE 68504 Find articles by James M Le Moine j, Jennifer L Baltzer Jennifer L Baltzer b Department of Biology, Wilfrid Laurier University, Waterloo, ON N2L 3C5, Canada Find articles by Jennifer L Baltzer b Author information Article notes Copyright and License information a Northern Forestry Centre, Canadian Forest Service, Natural Resources Canada, Edmonton, AB T6H 3S5, Canada b Department of Biology, Wilfrid Laurier University, Waterloo, ON N2L 3C5, Canada c Center for Ecosystem Science and Society, Northern Arizona University, Flagstaff, AZ 86011 d School of Informatics, Computing, and Cyber Systems, Northern Arizona University, Flagstaff, AZ 86011 e Geological Survey of Canada, Natural Resources Canada, Ottawa, ON K1A 0G1, Canada f YukonU Research Centre, Yukon University, Whitehorse, YT Y1A 5K4, Canada g Institute of Arctic Biology, University of Alaska Fairbanks, Fairbanks, AK 99775 h Renewable and Sustainable Energy Institute, Department of Ecology and Evolutionary Biology, University of Colorado Boulder, Boulder, CO 80309-0027 i Department of Wood and Forest Sciences, Laval University, Quebec City, QC G1V 0A6, Canada j LI-COR Biosciences, Lincoln, NE 68504 1 To whom correspondence may be addressed. Email: raquel.alfarosanchez@nrcan-rncan.gc.ca. Edited by Eric Rignot, University of California Irvine, Irvine, CA; received June 12, 2024; accepted October 1, 2024 Received 2024 Jun 12; Accepted 2024 Oct 1; Issue date 2024 Dec 10. Copyright © 2024 the Author(s). Published by PNAS. This article is distributed under Creative Commons Attribution-NonCommercial-NoDerivatives License 4.0 (CC BY-NC-ND). PMC Copyright notice PMCID: PMC11648870 PMID: 39621910 Significance In boreal forests, short growing seasons, cold temperatures, and the presence of frozen ground (permafrost) limit tree growth. Climate warming can potentially enhance growth by increasing air and soil temperatures. Here, we found reduced tree growth since the 1980 s at southern latitudes and increased growth at higher, colder latitudes, at least until the 2000 s. Our results showed that recent permafrost warming at the highest latitudes, where permafrost is more prevalent in the landscape, has caused significant stress in tree growth due to ground destabilization. Trees growing in unstable permafrost used their nutrient uptake to remain upright instead of increasing their growth. These findings indicate that boreal forests will not become more productive with climate warming and the resulting permafrost thaw. Keywords: climate warming, tree leaning, reaction wood, stemwood non-structural carbohydrates, permafrost Abstract Climate warming can alleviate temperature and nutrient constraints on tree growth in boreal regions, potentially enhancing boreal productivity. However, in permafrost environments, warming also disrupts the physical foundation on which trees grow, leading to leaning trees or “drunken” forests. Tree leaning might reduce radial growth, undermining potential benefits of warming. Here, we found widespread radial growth reductions in southern latitude boreal forests since the 1980s. At mid latitudes, radial growth increased from ~1980 to ~2000 but showed recent signs of decline afterward. Increased growth was evident since the 1980 s at higher latitudes, where radial growth appears to be temperature limited. However, recent changes in permafrost stability, and the associated increased frequency of tree leaning events, emerged as a significant stressor, leading to reduced radial growth in boreal trees at the highest latitudes, where permafrost is extensive. We showed that trees growing in unstable permafrost sites allocated more nonstructural carbohydrate reserves to offset leaning which compromised radial growth and potential carbon uptake benefits of warming. This higher allocation of resources in drunken trees is needed to build the high-density reaction wood, rich in lignin, that is required to maintain a vertical position. With continued climate warming, we anticipate widespread reductions in radial growth in boreal forests, leading to lower carbon sequestration. These findings enhance our understanding of how climate warming and indirect effects, such as ground instability caused by warming permafrost, will affect boreal forest productivity in the future. The boreal biome houses 30 to 40% of terrestrial carbon stocks (1) and provides critical ecosystem services, such as wood production and carbon storage, that have impacts at local, regional, and global scales (2). About 80% of the boreal biome lies within the permafrost region (3), with the majority in the zone of discontinuous permafrost, where permafrost is patchy in the landscape. Boreal ecosystems are warming at three to four times the global mean rate due to Arctic amplification (4, 5). Permafrost temperatures are also increasing (6), particularly in colder permafrost at higher latitudes (7), and it is virtually certain that permafrost thaw will increase and permafrost extent will decrease under continued global warming (8). However, it is unknown how these combined effects of climate change will impact growth trajectories in boreal trees and whether these changes in growth might offset or compound predicted carbon losses from warming soils. In the context of warming temperatures, there is a fundamental debate around the degree to which carbon assimilation via photosynthesis (i.e., carbon source limitation) versus direct environmental limitations to cambial cell development (i.e., growth or carbon sink limitation) controls radial growth (9). In cold-adapted plants, the capacity for wood tissue formation becomes marginal at temperatures below 5 °C (10–12), pointing toward the carbon sink limitation hypothesis as the main driver of growth, i.e., that low temperatures constrain tree growth despite sufficient photosynthesis. Additionally, widespread permafrost thaw may benefit the functioning of overlying forests, mainly due to warmer soils and deeper permafrost tables (13, 14) that support greater soil resource availability and rooting volume (15–17). Moreover, the release of previously inaccessible resources from thawing permafrost may enhance tree productivity (16, 18), at least in the short term (19)—although more research is needed as previous studies have focused mostly on experimentally induced permafrost thaw rather than direct field observations, and not on mature trees (20, 21). Consequently, trees at higher latitudes are expected to increase their radial growth primarily as a response to higher air and soil temperatures, extended growing season lengths, and greater thaw depths. Analysis of satellite imagery reveals greening primarily in the coldest margins of the boreal biome (22). Ultimately, an overall greening may enhance carbon uptake, relieving or even offsetting the loss of carbon that will occur as permafrost thaws, as seen in the Tibetan Plateau (23). However, warmer margins of the boreal biome are exhibiting reductions in carbon sequestration and plant growth and increases in mortality, known as “boreal browning” (22, 24–26). Browning could result from an increase in evapotranspirative demand or reduced soil moisture, leading to drought stress in trees. Projected increases in water loss via evapotranspiration could force the trees to reduce their photosynthetic carbon uptake by partial stomatal closure. Decreased photosynthesis leads to carbon depletion in sink tissue and a reduction in tree growth as a mechanism to reduce the risk of carbon starvation and hydraulic failure (27). Browning could also result from drought responses owing to changes in rooting-zone soil moisture conditions following thaw (15, 28–30). Another extensive, yet often overlooked, disturbance affecting tree growth that is common in permafrost environments is ground instability caused by seasonal and long-term changes in ice-rich permafrost, which causes trees to lean off-vertical (31–34). In permafrost environments of boreal North America, the widespread conifer black spruce [Picea mariana (Mill.) B.S.P.] dominates forested peatlands (35), where it occupies extensive regions of leaning trees known as “drunken” forests. Radial growth in leaning trees is altered due to the formation of reaction wood, which counteracts the tree tilting associated with the physical destabilization of permafrost. Therefore, periods of increased frequency of tilting in boreal conifers growing in areas with frost-susceptible sediments can often be attributed to changes in active layer (seasonally thawed) thickness (ALT) (32, 33). A better understanding of how the complex abiotic interactions that northern boreal trees experience may alter radial growth and carbon reserves requires combining physiological approaches with tree ring and environmental data (36–39). Nonstructural carbohydrates (NSCs) provide insights into the conditions under which growth is constrained by the environment versus actively reduced by the plant and how the mobilization of carbon stores affects productivity of the trees (40). Thus, the study of NSC (mainly sugars and starch) distribution across stemwood tree rings can help elucidate whether reserves deep in the stem are metabolically active and available to support functional processes and how stemwood NSC distribution varies with abiotic factors, such as increased temperatures, drought stress, or the physical destabilization observed on permafrost environments. The cold conditions of near-surface permafrost sites are a well-documented example of stress in trees that lead to elevated concentrations of NSCs (41) at the expense of radial growth, but this conservative approach may be susceptible to change with warming. Impacts of other stressful environmental conditions such as the increased destabilization of permafrost tables on tree productivity and resource allocation remain unexplored. Here, we evaluated the effects of permafrost and climate variability and the occurrence of tree leaning events on black spruce radial growth (tree-ring widths, TRW) across a 650 km climatic and permafrost gradient in the Northwest Territories, Canada. We used permafrost monitoring sites that provided historic records of seasonal thaw to investigate the link between the occurrence of tree leaning events and permafrost table stability (ground stability). To gain insight into the availability of carbon reserves in black spruce trees, we examined radial patterns of stemwood NSCs, and their relationship with TRW, in trees growing over contrasting permafrost table depth and stability conditions. Results Effects of Near-Surface Permafrost Presence and Climate Variability on Radial Growth Trajectories. To determine how the presence or absence of near-surface permafrost and climate variability affected TRW, we analyzed black spruce radial growth patterns in 109 sites from 1940 to 2020. Study sites spanned the low subarctic Taiga Plains in the North (N), the mid and high boreal Taiga Plains in the South (S), and the Taiga Shield (hereafter N Plains, S Plains, and Shield; Fig. 1 A and SI Appendix, Fig. S1 andTable S1A). Over the period from 1940–2020, no significant differences in TRW were observed in the Plains when comparing sites with near-surface permafrost (permafrost table shallower than 2 m, i.e., within the rooting zone—and potentially impacting trees growing over them) and sites with no permafrost or deep permafrost (permafrost table deeper than 2 m, i.e., below the rooting zone). On the Shield, TRW was significantly higher in sites with no or deep permafrost than in sites with near-surface permafrost since ~1980 s (Fig. 1 B and SI Appendix, Tables S4 and S5). Fig. 1. Open in a new tab Location map, radial growth trajectories, and climate-growth sensitivity. (A) Map of sampling sites within three ecoregions, the low subarctic Northern Taiga Plains ecoregion (N Plains), the high-mid boreal Southern Taiga Plains ecoregion (S Plains), and the Taiga Shield ecoregion (Shield). Different symbols indicate whether sites had near-surface permafrost (permafrost table shallower than 2 m, i.e., within the rooting zone), as well as annual records of ALT and NSC data. Points have been jittered for better visualization. Permafrost extent information obtained from ref. 42. Mapping was created using ArcGIS Pro 3.2 (B) Predicted TRW and 95% CI as a function of calendar year for the two-level permafrost factor, i.e., sites with near-surface permafrost or with no or permafrost deeper than two meters for each ecoregion (SI Appendix, Table S4). Other terms in the model are held constant at their medians. (C) Comparison of combined AIC values per ecoregion obtained in linear models (LMs) between individual TRW residuals with seasonal (sp, spring; sm, summer; at, autumn; and wt, winter) temperature (Tave, average air temperature; Tmin, minimum air temperature, and Tmax, maximum air temperature) and moisture (PPT, precipitation; CMD, climate moisture deficit) variables. For each seasonal variable, we used current year and previous year values (indicated with a p after the season abbreviation) from spring of the previous year to fall of the current year, and all seasonal combinations within that time frame (SI Appendix, Table S2). The lowest AIC values correspond to the most influential variable and seasonal window for TRW for each ecoregion (Methods). TRW exhibited a pattern of widespread reductions at southern latitudes and increased growth at northern latitudes. Specifically, in the southernmost ecoregion, the S Plains, TRW decreased from ~1950 to 2020. At mid latitudes, in the Shield ecoregion, TRW increased since ~1980 but showed signs of decline in the last two decades. In the northernmost ecoregion, the N Plains, TRW increased from ~1980 to 2020 (Methods; Fig. 1 B and SI Appendix, Tables S4). TRW was most strongly associated with summer minimum air temperature (Tmin) at higher latitudes (i.e., N Plains) and with summer drought, as measured by CMD, in mid and southern latitudes (i.e., Shield and S Plains ecoregions) (Methods; Fig. 1 C). Higher temperatures and lower moisture deficit enhanced TRW, particularly in sites with no or deep permafrost in the Shield. We observed a nonlinear response of TRW to Tmin in the N Plains and a nonlinear response of TRW to CMD in all three ecoregions, suggesting a deceleration in growth with continued increase in temperature or moisture deficit, respectively (Methods; SI Appendix, Fig. S2 andTable S4). Effect of Permafrost Table Stability on Radial Growth Trajectories. We investigated how permafrost table stability affected radial growth in 15 long-term permafrost monitoring sites in the N Plains. These sites provided annual records of ALT (Methods) from 2008 to 2018–2019. The permafrost monitoring sites showed consistent increases in ALT (i.e., permafrost thaw) over the period studied (median thaw rate in the near-surface permafrost sites 5.3 cm y−1), although some plots also showed permafrost aggradation (SI Appendix, Fig. S3). We converted the site-level ALT rates into absolute rates of permafrost change to obtain a measurement of permafrost table stability, where higher absolute rates indicated greater ground instability, owing either permafrost thaw or aggradation processes (Methods). We found significant reductions in TRW with greater absolute rates of permafrost change over the period of 2008–2020 (Fig. 2 A and SI Appendix, Table S6 A). Fig. 2. Open in a new tab Permafrost changes effects on radial growth and frequency of leaning events. Predicted TRW and 95% CI as a function of (A) absolute rate of permafrost change (SI Appendix, Table S6 A) and as a function of (B) calendar year for the four levels of permafrost factor, i.e., sites with deep stable, near-surface stable, deep unstable, or near-surface unstable permafrost (SI Appendix, Table S6 B). Predicted number of leaning events and 95% CI as a function of (C) absolute rate of permafrost change (SI Appendix, Table S8 B) and as a function of (D) the four levels of permafrost factor (SI Appendix, Table S8 C). Predicted TRW and 95% CI as function of (E) number of leaning events (SI Appendix, Table S9). For all predicted variables, other terms in the models are held constant at their medians. Black inset ticks in the x-axis on panels A, C, and E indicate the frequency of data. Vertical dashed lines in panels A and C indicate the threshold between stable and unstable permafrost tables (Methods). Lowercase letters in panel D indicate significant differences obtained in post hoc analyses. Considering that higher absolute thaw or aggradation rates may have greater impact on trees when occurring within the rooting zone, we classified the permafrost monitoring sites into four permafrost categories, based on their permafrost table depth and stability (Methods): i) deep stable, ii) near-surface stable, iii) deep unstable, or iv) near-surface unstable permafrost sites. Over the period of 2008–2020, greater TRW was observed in the permafrost monitoring sites with deep stable or near-surface stable permafrost (Fig. 2 B and SI Appendix, Table S6 B). In addition, higher air temperatures were associated with greater TRW only in the permafrost monitoring sites with deep stable permafrost, suggesting a deceleration in growth in sites with unstable or near-surface permafrost (SI Appendix, Fig. S4 andTable S6B). No significant differences in TRW were found with respect to permafrost depth conditions (i.e., near-surface stable or unstable vs deep stable or unstable permafrost) in the permafrost monitoring sites (SI Appendix, Tables S6 B and S7). This result is consistent with our findings for the N Plains over the period from 1940 to 2020, where no significant differences were observed based on permafrost conditions at the time of sampling (SI Appendix, Table S4 A). Frequency of Leaning Events as a Surrogate Variable for Historical Changes in Permafrost. We investigated whether the number of leaning events per tree could serve as a proxy for historical changes in permafrost in sites lacking instrumental data on seasonal thaw rates over time. The total number of leaning events per tree was estimated based on their annual lean intensity for all trees across the latitudinal gradient with basal disk samples (Methods; SI Appendix, Table S1 A). We found that older trees experienced a higher number of leaning events (Spearman correlations r s = 0.62, t = 19.1, df = 568, P< 0.001). On average, 81% of the trees exhibited at least one leaning event, and the presence or absence of leaning events in the trees was not driven by permafrost depth conditions (SI Appendix, Table S8 A). The total number of leaning events per tree in the Shield was significantly greater in sites with near-surface permafrost conditions, but no significant differences were observed in the Plains (SI Appendix, Fig. S5 andTable S8A). However, we found that the number of leaning events in the N Plains permafrost monitoring sites significantly increased with rates of permafrost change, i.e., in sites with unstable permafrost conditions (Fig. 2 C and D and SI Appendix, Table S8 B and C), indicating the potential of this variable as a proxy for determining historical site permafrost table stability. We subsequently examined the impact of leaning events on TRW across the entire latitudinal gradient, where 94 out of 109 sites had no historical information about changes in permafrost. We found that TRW was significantly reduced with increased frequency of leaning events (Methods; Fig. 2 E and SI Appendix, Tables S9). Sensitivity analyses examining the interaction between climate stress and the occurrence of leaning events on TRW revealed a cumulative negative effect on growth with increased tree leaning and climate stress in the Plains—warming in the N Plains or moisture deficit in the S Plains. However, no significant interaction effect was found in the Shield ecoregion (SI Appendix, Fig. S6and Table S10). NSCs to Unravel the Mechanism Underlying Radial Growth Response to Changes in Permafrost. To gain insight into the availability of carbon reserves in black spruce trees growing in sites experiencing changes in permafrost conditions and their relationship with TRW, we examined radial NSC (sugars) concentrations in the outermost 13 individual rings (from 2008 to 2020) in a subset of trees from the 15 long-term permafrost monitoring sites (Methods). Sugar content was highest in the outermost ring, i.e., the 1-y-old ring (1.87 ± 0.36%), and declined radially toward the inner rings (0.3 ± 0.16% in the 13-y-old ring). The reduction in sugar concentration in the 13-y-old ring (the oldest ring analyzed) relative to the 1-y-old ring content was on average 83 ± 9% (SI Appendix, Fig. S7 A). We ran Pearson correlations between 1-y-old ring TRW (the outermost, last full tree ring measured) and radial content of sugars (1 to 13 y old) to investigate whether reserves deep in the stem are metabolically active and available to support functional processes such as radial growth. We found significant negative correlations between 1-y-old ring TRW and 3- to 11-y-old sugar content after Bonferroni corrections (Methods; Fig. 3 and SI Appendix, Table S11). Maximum negative correlations were found between 1-y-old ring TRW and 5- to 8-y-old sugar content (r ~ −0.81), pointing toward a higher demand for sugars to build growth from those rings. This suggests that deeper, and presumably older, sugar reserves were used to support recent growth. Fig. 3. Open in a new tab Relationships between radial growth and stemwood sugar concentrations. Relationship between 1-y-old ring TRW [log-transformed] and 1 to 13-y-old ring sugar concentrations. Dashed lines indicate nonsignificant Pearson correlations for a Bonferroni critical value of P< 0.00385 (Methods; SI Appendix, Table S11). Points are raw tree sugar concentrations color-coded to indicate the four levels of permafrost factor. Sugar content in the 1-y-old ring showed a significant positive relationship with absolute site-level rates of permafrost change (SI Appendix, Table S12), in contrast to the negative relationship between 1-y-old ring TRW and absolute rates of permafrost change (Fig. 2 A and SI Appendix, Table S6A). This result suggests that in unstable permafrost sites (sites with higher absolute rate of permafrost change; Fig. 2 C), more new sugars were stored and therefore not available for 1-y-old ring TRW, while in stable permafrost sites, new sugars were more available to trees for growth. We then calculated the percentage of sugar depletion in stemwood rings from 2- to 13-y-old relative to sugar content in the 1-y-old ring and compared the percentages of sugar depletion per tree across the four levels of permafrost conditions (Methods). We found a significant steeper depletion in sites with deep stable permafrost, than in near-surface stable or unstable permafrost sites (Fig. 4 and SI Appendix, Table S13). Specifically, sugar concentrations decreased from the 1-y-old to the 2-y-old ring by 43.7 ± 7% in sites with deep stable permafrost, compared to a decrease of 19.5 ± 11.9% in deep unstable sites, 17.3 ± 11.6% in near-surface stable sites or 12.4 ± 10.2% in near-surface unstable sites. Differences in the percentage of sugar depletion among the four permafrost condition levels were more notable between the 2nd- and the 5th-y-old rings (Fig. 4 and SI Appendix, Table S13). Fig. 4. Open in a new tab Stemwood sugar depletion for contrasting permafrost conditions. Boxplots for 2 to 13-y-old stemwood sugar depletion relative to 1-y-old ring (%) for four levels of permafrost factor, i.e., sites with deep stable, near-surface stable, deep unstable, or near-surface unstable permafrost. Lowercase letters indicate significant differences among the four-level permafrost factor for each stemwood ring obtained in post hoc analyses. The colored areas indicate the stemwood rings with significant differences in the post hoc test (no significant differences were found among the four-level permafrost factor for stemwood rings 6 to 13). Circles are raw percentage of stemwood ring sugar depletion relative to 1-y-old ring (%) color-coded to indicate the four levels of permafrost factor. Discussion While rising levels of atmospheric CO 2 make more carbon available for photosynthesis, additional factors or conditions are required to allow trees to take advantage of this extra carbon through enhanced tree growth (43). In cold-adapted conifers, the allocation of assimilated carbon to structural investment, i.e., tissue growth, is hindered when air temperatures fall below 5 °C (12). Since the 1980 s, warming has relaxed temperature constraints on radial growth at high latitudes (39, 44). This response is evident in our sites within the N Plains and the Shield (until the last two decades), two ecoregions with very low mean annual temperatures but no significant reductions in PPT observed since 1940 (SI Appendix, Figs. S8–S9 These conditions likely sustain adequate soil moisture, meaning an increase in air temperature may enhance tree growth rather than leading to drought stress (45). In contrast, we attribute the observed decrease in TRW at our most southerly sites in the S Plains to the significant increase in summer air temperatures, creating the potential for drought stress (46). From 1940 to 1980 s, black spruce radial growth showed less constraints for growth in the S Plains than at higher latitudes. However, since 1980 s, the species appears to have struggled to acclimate adequately to the excessive warmth recorded at these southern latitudes (Fig. 1 B). Overall, we found a lack of variation in TRW and the number of leaning events linked to the presence or absence of near-surface permafrost at the time of sampling in the Plains sites. This is probably attributable to changes in site permafrost status over the lifespan of the trees. Trends observed in northern environments over recent decades point to shorter-term periods of cooling and permafrost aggradation, superimposed on longer-term trends of warming and permafrost thaw/degradation (7, 8, 13). Indeed, changes in permafrost in our long-term permafrost monitoring sites since 2008 were predominantly unidirectional, characterized by a general increase in ALT (i.e., permafrost thaw) that aligns with these trends observed across northern environments in recent decades (19, 47, 48). Therefore, it is more likely that sites with permafrost within the rooting zone at the time of sampling had experienced these shallower permafrost conditions over the lifespan of the trees, and sites with no permafrost at the time of sampling, may have had shallower permafrost in the recent past that had degraded, but it will depend on when stands were established. Significant differences in TRW and number of leaning events with the presence or absence of near-surface permafrost at the time of sampling in the Shield point to more stable permafrost conditions in that ecoregion, at least since 1980 s. This is because the Shield generally has coarser sediments, resulting in less ground ice compared to the fine-grained frost-susceptible material found overlying the sedimentary basin in the Plains (49). Lower tree growth in Shield near-surface permafrost sites was most likely linked to the challenges posed by the continued presence of permafrost over time, as cold and poorly drained soils limit tree growth (45), and the growth stress associated to seasonal thaw-driven tree leaning. The higher number of leaning events detected in Shield near-surface permafrost sites when compared with Shield sites with deep or no permafrost suggests that black spruce trees experienced higher number of leaning events in those stable permafrost sites. Seasonal and long-term changes in ALT affect the foundation upon which trees grow. Continuous heaving and settlement cause the trees to frequently lean requiring corrective growth responses to maintain a vertical position. Tree response to leaning leaves a distinct signature in the rings of tree stems, characterized by the formation of reaction wood. Reaction wood (compression wood in black spruce trees) is characterized by higher lignin content that is triggered by the gravitational imbalance in leaning trees (34, 50). The production of such dense lignified wood requires greater allocation of carbon, i.e., higher sugar concentrations (51), which can adversely affect tree growth rates (52). By examining the radial patterns in NSC concentrations and their relationship with wood formation (TRW), we were able to assess the physiological mechanisms underlying the response of radial growth across different conditions of permafrost depths and stability. During the peak growing season, sugar concentrations in the stemwood were highest in the outermost rings and declined radially toward the inner rings for all trees (53–55), and the opposite pattern was found for starch concentrations (SI Appendix, Fig. S7). In boreal and temperate biomes, a strong depletion of starch is common during the growing season due to the high demand for carbohydrates to support growth and respiration (56–58). In contrast, soluble sugars serve important immediate physiological functions (e.g., osmotic regulation), with concentrations maintained above a critical threshold (59–61). This requirement to maintain relatively high sugar concentrations in forming rings appears to be particularly crucial for woody species in cold climates (57). Every year, trees mobilize toward the inner rings the sugar reserves not used for growth or functional services. The significant correlations found between 1-y-old ring TRW and deeper sugar reserves suggest that trees can access reserves that are metabolically available on interannual timescales (62) (Fig. 3). Higher sugar reserves (lower rate of sugar depletion) were found in trees growing on near-surface or unstable permafrost sites (Fig. 4). This pattern is consistent with a conservative resource use strategy characterized by increased investment in sugar storage for future years and reduced new carbon sink. Lower radial growth and higher accumulation of carbon stored as NSCs in trees growing on near-surface permafrost may result from low rooting zone temperature impacting root function or nutrition during the growing season (10). However, we found that ground instability within the rooting zone in actively thawing permafrost environments appears to be even a more significant stressor affecting sugar reserves and consequently radial growth (Figs. 2 C–E and 4). These results highlight the importance of permafrost stability in modulating radial growth patterns, particularly in environments with widespread thaw processes. In contrast, the sharp depletion in sugars near the cambium in trees growing under less stressful environmental conditions (i.e., sites with deeper and stable permafrost tables) suggest a less conservative strategy, as trees are allocating proportionally more of their recently fixed carbon to growth in the outermost rings rather than to storage (63). Therefore, trees growing in sites with deep stable permafrost showed large, high-turnover carbohydrate pools, most likely fueling growth from a combination of assimilates from current and previous years (54, 64). Future climate scenarios for the boreal region predict warmer and drier conditions (65), which may exacerbate the risk of carbon starvation and hydraulic failure (27), particularly at southern latitudes. A potential saturation in radial growth with continued warming may cause a reduction in photosynthetic C uptake (decreased photosynthesis), driving long-term growth declines, and enhancing mortality risk (45, 46, 66). This negative feedback would intensify in permafrost environments due to climate-driven changes in permafrost. Permafrost does not need to completely disappear for drought stress to occur; rather, the active layer only needs to become thick and well-drained enough to dry out, altering the rooting-zone soil conditions. However, this will depend on factors such as topography, material type, and drainage characteristics. We suggest that any potential positive effect on tree growth from increased soil nutrients induced by warming-induced permafrost thaw seemed to be overshadowed by the stress in growth caused by ground instability in the plant rooting zone. Permafrost table instability drives trees to lean off-vertical repeatedly, significantly reducing radial growth in favor of more NSC reserves. However, longer monitoring is necessary to determine whether trees can recover from the physical disruption—or if they will succumb to it—once the ground is stabilized again, either because all the permafrost thaws, progresses below the rooting zone, or transitions into stable seasonal thaw. Methods Study Area and Field Sampling. Wood samples and permafrost data were collected in 109 sites dominated by black spruce [P. mariana (Mill.) B.S.P], in low subarctic and high-mid boreal forests in the Northwest Territories, Canada. Most of the sites were situated in the extensive discontinuous permafrost zone, while a few extend into the continuous and the sporadic permafrost zones (42) (Fig. 1). Sites were stratified by level III ecoregions: the Northern and Southern Taiga Plains, henceforth N Plains and S Plains, and the Taiga Shield, henceforth Shield (67, 68). The Plains ecoregions are characterized by undulating glacial tills derived from Cretaceous sedimentary deposits, peatlands in wetter areas often underlain by permafrost, or glaciolacustrine sediments, particularly in the Mackenzie Valley (69). The Shield features hilly pre-Cambrian bedrock sparsely covered with thin layers of till, or lacustrine or aeolian deposits of clay, sand, or gravel (67, 68). Ninety-four out of the 109 sites were established during summer fieldwork campaigns from 2016 to 2022 across the three ecoregions, in areas with no record of stand-replacing fire in recent history [>19 y postfire (70, 71)]. Fifteen out of 109 sites were established in the N Plains ecoregion in summer 2021, near to boreholes instrumented with thermistor strings, which allowed the study of historical changes in permafrost conditions, hereafter permafrost monitoring sites. The permafrost monitoring sites provided information on the texture of surficial materials (SI Appendix, Table S1 B and see ref. 72 for more details). Most of the permafrost monitoring sites with shallow permafrost conditions were likely frost-susceptible, as they were underlain by peat or finer-grained sediments like silt or clay, which have high ice/moisture content and are prone to frost heave. In contrast, sites with deep permafrost were generally not frost-susceptible, as they were typically underlain by coarser sediments such as sand and gravel (even when there was a thin layer of peat at the surface). The permafrost monitoring sites also showed no record of recent stand-replacing fires. Stand-replacing crown fires are the most common fire regime in North America’s boreal forests. However, a significant proportion of low-severity fires also occur within fire perimeters (73). We found evidence of fire-scarred trees at ten out of fifteen permafrost monitoring sites across various permafrost conditions (SI Appendix, Table S1 B and Ref. 74; see section 3 for permafrost conditions), indicating the widespread occurrence of low-severity fires since the last stand-replacing fire. In all sampled sites, we established one to three monitoring plots in representative landscape areas to collect wood samples and permafrost, soils, vegetation and forests stand attributes as part of a broader ecological research project in the NWT (70, 71). Each plot covered an area of 60 m 2 and consisted of two parallel 30-m transects running from south to north, positioned 2 m apart. Details about the sampling design can be found in refs. 70, 71. In the permafrost monitoring sites, the monitoring plots were established 30 to 50 m away from the boreholes to prevent disturbance in their measurements. Wood Samples. In total, we collected wood samples from 961 live black spruce trees across the 109 sites, sampling 387 cores using increment borers, and 574 disks cutting the tree stems at the base of the tree (9 ± 4 trees sampled per site). Tree sampling focused on selecting black spruce trees that were most representative of the 60 m 2 monitoring plots at each site. However, wood samples were collected outside of these monitoring plots to avoid disturbance, as they were established as long-term permanent plots. As a result, our sampling design included a range of trees beyond just the dominant ones, which enhances the robustness of our inferences about the tree population. Wood samples were used to determine tree age and to analyze climate-growth relationships, patterns in tree growth, reaction-wood dynamics (tree leaning events), and, in a subset of trees, stemwood NSC determination. Therefore, cores and disks were collected perpendicular to the stem axis, close to the base, where age estimation is most accurate and reaction wood reaches its maximum (34). The frequency of leaning events in trees growing in permafrost regions can inform about historical or current permafrost thaw dynamics. However, tree leaning events can only be assessed from disks, which represents the 57% of the tree-ring samples collected across the climatic gradient. All cores and disks were processed following standard dendrochronological methods (75). Samples were scanned and annual Tree Ring Widths (TRW) were measured using CooRecorder (76). Cross-dating of individual series was checked using CooRecorder and COFECHA programs (77). In tree cores, usually two perpendicular radii of the trees were measured (1.8 ± 0.6 radius measured per tree). In the tree disks, at least two directions of the wood disk were measured, including the maximum and minimum wood radius, and often one or two more measurements in perpendicular directions to the major and minor axis (2.8 ± 0.6 radius measured per tree). Radii measurements were averaged to obtain the TRW for a given year for each tree and for studying tree leaning events in tree disks. Asymmetry in the width of concentric rings on a tree stem is an indicator of tree growth correcting for departures from a vertical orientation (lean). The annual intensity of tree leaning was defined as the ratio of maximum annual tree ring width to average annual tree ring width (modified from ref. 34), and was calculated by dividing the absolute maximum annual tree ring width from all the measured radius, by the average annual ring width of the rest of measurements. Tree leaning events were identified when the annual intensity of tree leaning was > 2. Then, the number of leaning events was counted per tree. Stemwood NSC analyses were performed on an extra disk taken from each black spruce tree sampled at the 15 permafrost monitoring sites. The extra wood disks cut per tree were immediately frozen in the field on dry ice and then transported frozen to cold storage. Due to the time-consuming and costly nature of determining NSC concentrations, we analyzed a subset of disks, ensuring the selected trees covered a similar age range to minimize potential confounding effects and across contrasting permafrost conditions (see next section). The subset of frozen disks was first freeze-dried, and then annual rings from the period 2008–2020 were carefully separated by razor blade under a stereomicroscope. The samples were transferred to the University of Northern Arizona at Flagstaff, AZ, USA where concentrations of starch and sugars were determined, as percentages of dry weight. We followed protocols of ref. 78 for sugars and of ref. 79 for starch, as described in ref. 54. Sugar concentration was determined for 230 annual rings in 18 trees. Starch concentration was determined only for 36 annual rings in 15 trees instead that in the 230 annual rings, as our preliminary data showed very low concentrations of starch (0.038 ± 0.037 %; mean ± SD), and therefore there were not used for data analyses. Permafrost Data. Presence of near-surface permafrost and permafrost table depth. Information about the presence of near-surface permafrost and permafrost table depth was collected for 94 out of the 109 sites at the time of sampling using a 2 m length graduated steel rod with a handle, also known as frost probe. To do so, we pushed the steel rod into the soil down to the point of firm resistance, against the frost table or rocky ground. Probes typically cannot penetrate rocky ground and make a “ping” or scraping sound when encountering rock, rather than the “thump” when hitting permafrost. Sites that encountered frozen ground during the summer field campaigns were revisited at the end of the summer or early fall (late September and early October) to verify the presence of permafrost and to obtain the maximum thaw depth (80). We probed three to five times along the 30-meter east transect of each plot established per site to obtain a more accurate estimate of near-surface permafrost presence and an average measurement of thaw depth. Sites were classified as near-surface permafrost sites if frozen ground was detected at the end of summer or early fall within the rooting zone (considered up to a depth of 2 m). The remaining sites were classified as sites with no permafrost in the upper two meters. We used the 2-meter permafrost table depth threshold for near-surface permafrost classification because it is unlikely that changes in permafrost table conditions below the plant rooting zone significantly impact black spruce forests, given the typical shallow depth of the rooting zone in this species (19). Presence of near-surface permafrost, permafrost table depth, and permafrost table stability in permafrost monitoring sites. For the 15 permafrost monitoring sites, information about permafrost table depth was collected from the nearby boreholes instrumented with thermistor strings (Data available in ref. 81). The spacing between sensors is 0.5 m in the upper 2 m, increasing to usually 1 to 2 m at greater depths. ALT was estimated as the depth at which the maximum annual ground temperature was 0 °C (see refs. 19, 72, 82 for specific details about the data recording methods at the permafrost monitoring sites). The instrumentation provided continuous records of annual variation in (seasonally thawed) ALT from 2008 to 2018–2019 [the end of the time series varied per site; (81)]. Extreme increases or decreases in ALT within the time series at each site are due to small temperature fluctuations around 0 °C across the temperature profile, which are within the measurement error of the thermistors, and sometimes to larger spacing between thermistors (> 2 m), which affects the accuracy of the interpolation used to determine the thaw depth. We classified permafrost monitoring sites as sites with near-surface permafrost, i.e., within the rooting zone, when maximum ALT values were shallower than 2 m depth for at least one year of the studied period. Permafrost monitoring sites with maximum active layers thicker than 2 m were classified as sites with deep permafrost. To determine whether permafrost monitoring sites had stable or unstable permafrost tables, we used the annual ALT values in LMs as a function of year to test for trends over the studied period at all 15 sites. Only one site showed significant evidence of permafrost aggradation, or decrease of the ALT (slope coefficient −57.4 cm y−1). In seven sites, the trend was positive (slope coefficients ranged from 7.40 to 56.6 cm y−1), consistent with increase permafrost thaw or increasing of the ALT. Seven other sites showed no significant trends in ALT (SI Appendix, Fig. S3). Sites with significant absolute trends in ALT from 2008 to 2018–2019 were classified as sites with unstable permafrost and the remaining sites as sites with stable permafrost. We then defined a four-level permafrost factor to classify the depth and stability of the permafrost table in the 15 permafrost monitoring sites: sites with deep and stable permafrost table (n sites = 2), sites with near-surface and stable permafrost table (n sites = 5), sites with deep and unstable permafrost table (n sites = 3), and sites with near-surface and unstable permafrost table (n sites = 5). Climate Data. We obtained historical time series of five climate variables (SI Appendix, Table S2) for the 109 sampling sites, using the ClimateNA v7.10 software (83). The historical data covered the period 1940–2020, the longest period for which consistent regional data were available ( Seasonal means were calculated for the historical, normal, and future climatic variables following previous northern dendrochronology studies, e.g., ref. 36 where seasons were defined as winter (November–March), spring (April–May), Summer (June–August), and Autumn (September-October). We also obtained seasonal values of the same five variables for the 30-y normal period 1971–2000, and future values for 2011–2040, downscaled from the Coupled Model Intercomparison Project Phase 5 [CMIP5; (84)] under an intermediate emissions scenario, RCP4.5. Temporal trends of the most influential seasonal climate variables for tree growth, obtained in climate-growth sensitivity analyses (see next section), were examined across regions by fitting linear and nonlinear regression models for each site between climate variables and year (Main results section). The temporal trend analyses covered the period from 1940 to 2020. Climate Variable Selection for Tree Growth Analyses. For climate-growth analyses, we selected 798/961 trees from 104/109 sites having at least 31 continuous years of growth rings with paired annual climate and individual TRW data for the period 1940–2020. Climate variables included in these analyses were seasonal values from spring to winter at a lag of 1 y, and spring to autumn at lag 0, for the temperature variables average temperature, minimum temperature, maximum temperature, and moisture variables PPT and CMD (SI Appendix, Table S2). We used generalized additive models (GAM) using functions gam and s of mgcv (85) R package to detrend the individual TRW series and all the site-level time-series seasonal climate variables. Detrending the tree growth series removes low-frequency variation related to nonclimatic drivers, such as growth, tree aging, and competition effects (86). Detrending the climate variables enhances the sensitivity of tree growth to interannual climate variations, a crucial consideration given observed trends in mean temperature over recent decades (87). We used LM to test for associations between detrended, matched series of TRWs and climate variables. Separate models were fit using seasonal values from spring to winter from the current and previous year for a total of 44 seasonal covariates per tree (SI Appendix, Table S2). For each tree and covariate, we compared models with linear and quadratic adjustments selecting the best fit by Akaike information criterion (AIC), and then we selected the most sensitive temperature and PPT model by tree, i.e., the model with the lowest AIC. We then ranked the two groups of variables by selection frequency within ecoregions. Statistical Analyses. In total, we built 11 statistical models including generalized additive mixed models (GAMM), Linear Mixed Effects Models (LMMs), general linear models (GLM), and LMs to analyze black spruce tree growth, leaning events, and NSCs (See next sections). GAMMs were fit using function gamm and s of mgcv (85) R package; LMMs, GLMs, and LMs were fit using function glmmTMB of glmmTMB R package (88). Specifications of the initial model formulas can be found below and summarized in SI Appendix, Table S3. Tree growth analyses. We fit GAMM 1 to analyze the interannual variation in individual black spruce TRW as a response to climate, site permafrost status, while controlling for age and size effects. Models were fit separately for each ecoregion to account for systematic differences, such as surficial geology or parent material. We used natural logarithm of annual TRW as the response variable because of the skewed distributions of their values. Time series replication covered trees younger than 200 y old, from 1940 to 2020 for the Plains and from 1940 to 2017 for the Shield. This time series provided sufficient sample replication over time and within age classes, along with more reliable climate data. GAMM1 included as fixed effects the two-level permafrost factor (i.e., sites with near-surface permafrost vs no/deep permafrost), the primary climatic effects on tree growth obtained through climate-growth sensitivity analyses (See previous section, i.e., Summer Tmin and Summer CMD), annual cumulative basal diameter (reconstructed diameter based on tree rings) and cambial age (ring number from the pith) and calendar year to account for any temporal trends not explained by the climate variables (39, 86, 89). Primary climate drivers were included with quadratic terms to account for potential nonlinear responses of black spruce to climate (39, 66, 90). We introduced second interactions between first-order linear terms for climate drivers and permafrost factor, to explore whether tree growth patterns exhibited any associations with specific permafrost status of the sites at the time of sampling. Smoothing terms “s” were used for the cumulative diameter, cambial age, and calendar year variables. Smoothing terms represent cubic regression splines with a determined degree of smoothness (85). Random effects included tree identity and site to account for repeated measures and spatial variability. To account for temporal autocorrelation within trees, akin to traditional detrending methods, we used the continuous autocorrelation function corCAR1 (86) of mgcv (85) R package. Post hoc comparison of pairwise intercept differences between permafrost conditions within ecoregions was assessed using Wald tests, with the wald_gam function from the itsadug package (91). We specified GAMM 2 to investigate the effects of absolute rates of permafrost change from 2008 to 2020 on TRW in the N Plains permafrost monitoring sites. GAMM 2 included as fixed effects the absolute rate of permafrost change, the annual cumulative diameter, and cambial age with smooth terms. Climatic variables and calendar year were not included in GAMM 2 to avoid collinearity effects between the rate of permafrost change with climatic variables and year. GAMM 3 was specified to test for the effects of the four levels of permafrost factor from 2008 to 2020 on TRW in the N Plains permafrost monitoring sites. GAMM 3 included the same fixed and random effects than GAMM1 but included the four-level permafrost factor instead of the two-level factor. We specified GAMM 8 to investigate the effects of the frequency of leaning events on TRW. Other fixed effects, such as climatic variables, were not included in GAMM 8 because our goal was to explore the generalized effect of the frequency of leaning events on tree growth using the longest available full time series, from 1723 to 2020. GAMM 8 included as fixed effects the number of leaning events per tree, the annual cumulative diameter, and cambial age with smooth terms. GAMM 9 included the same fixed effects as GAMM 8, along with the number of leaning events per tree, annual cumulative diameter, cambial age with smooth terms, and climatic variables. As a result, the time series replication for this model spans from 1940 to 2020, aligning with the time series of climatic variables. GAMMs 2, 3 and 8 and 9 used natural logarithm of annual TRW as the response variable and included the same random and autocorrelation effects as GAMM 1. Lean events analyses. To analyze the relationship between the number of leaning events per tree and site rates of permafrost change, we run Spearman correlations between the total number of leaning events per tree and the absolute site-level rates of permafrost change from 2008 to 2018–2019 obtained in the 15 N Plains monitoring sites. Our sample design, which included all trees within the monitoring plots—regardless of height or diameter, and beyond just the dominant trees—resulted in a very broad tree age distribution (SI Appendix, Fig. S1 B). To determine whether the frequency of leaning varied with tree age, we run Spearman correlations between the total number of leaning events per tree and tree age, across the full climatic gradient. Results indicated a significant correlation (see the Results section). Consequently, we included age as a fixed predictor in the subsequent lean event analyses. We built GLM 4 to model the probability of the presence/absence of leaning events per tree using a binomial response (presence or absence of leaning events per tree) and GLM 5 to model the number of leaning events per tree using a negative binomial response as a function of tree age and the two-level permafrost factor. Models were fit separately for each ecoregion to account for systematic differences, such as surficial geology or parent material. We specified GLM 6 to investigate the effects of absolute rates of permafrost change from 2008 to 2020 on the number of leaning events in the N Plains permafrost monitoring sites. GLM 6 included as fixed effects tree age and the absolute rate of permafrost change. GLM 7 was specified to test for the effects of the four levels of permafrost factor from 2008 to 2020 on the number of leaning events in the N Plains permafrost monitoring sites. GLM 7 included as fixed effects tree age and the four-level permafrost factor. Significant differences (P< 0.05) among levels of the two or four-level permafrost factor were examined using Tukey–Kramer Tukey post hoc tests with the lsmeans R package (92). NSC analyses. NSCs were measured in the stemwood of black spruce trees growing over the four levels of permafrost factor in the N Plains permafrost monitoring sites. NSC concentrations were analyzed in stemwood samples divided into the first 13 individual rings, i.e., from 2008 to 2020 during the middle of the growing season (samples collected at the end of July 2021 when NSCs were still mobilized for growth). As all samples were collected during the same phenological stage for all trees, and no significant latitudinal or elevational gradients were considered for these analyses (Fig. 1), our data can be used to compare stemwood NSC concentrations among trees exposed by contrasting environmental conditions (93), i.e., the four levels of permafrost factor. We analyzed the relationship between sugars and absolute ALT trends in LM 10, while controlling for age effects. The response variable for LM 10 was 1-y-old sugars. We analyzed 1-y-old sugars, instead of the mean sugar values from 2008 to 2018–2019 as we found a significant decrease in sugar concentrations (and a significant increase in starch) toward the pith (Results; SI Appendix, Fig. S7). Next, we used Pearson correlations to explore the relationship between 1-y-old TRW and stemwood sugar content for each ring analyzed, i.e., 1 to 13 y old for sugars. We applied a Bonferroni correction, dividing our significant level α = 0.05 by the number of tests (13 for sugars) to get the Bonferroni critical value of P< 0.00385. Significant correlations were then considered for these Bonferroni critical value. We avoid running separated Pearson correlation sets for the four levels of permafrost factor due to the limited number of NSC samples analyzed per year. LMM 11 was specified to test for differences in the percentage of sugar depletion in stemwood rings from 2- to 13-y-old relative to new sugar content (1-y-old ring) among the four levels of permafrost factor. Time series replication covered the 2008–2020 period. We calculated the percentage of sugar depletion relative to new sugar content following this equation: Sugars relative to 1-y-old ring (%) = where x represents the stemwood rings from 2 to 13 y old. LMM 11 included as fixed predictors the interactions between tree age and the four-level permafrost factor, and between the stemwood ring—as a factor—and the four levels of permafrost factor. Tree identity was included as random effect. Significant differences (P< 0.05) among levels of the four-level permafrost factor for each stemwood ring were examined using Tukey–Kramer Tukey post hoc tests with the lsmeans R package (92). Model selection and model validation. Collinearity among fixed effects was explored in all the statistical models, using function vifstep of usdm (94) R package. We compared Akaike’s information criterion (AIC) of candidate models using the logarithmic transformed and untransformed size and age variables against using smooth terms, after rerunning them using the maximum likelihood (ML) method. We selected the candidate models with lower AIC score. The final models were fitted using the restricted maximum likelihood method (95). Significance of the quadratic and interaction terms in the models was assessed comparing the AIC between models containing and excluding each quadratic or interaction term using the ML method. Nonsignificant quadratic and interaction terms were manually removed from the models. Model validation for GAMMs was visually assessed by plotting the residuals against the fitted values. The model validation for LMMs, GLM, and LMs was conducted using the simulate Residuals function available in the DHARMa package (96). All statistics were carried out in R version 4.1.2 (97). All graphs were created using ggplot2 (98) or base R (97). Supplementary Material Appendix 01 (PDF) pnas.2411721121.sapp.pdf (1.7MB, pdf) Acknowledgments We thank numerous students and technicians for field and lab assistance. We thank the GNWT Aurora Research Institute (Research License 15879). The Wilfrid Laurier University—GNWT Partnership Agreement was instrumental in providing logistical support and laboratory space. The permafrost monitoring in the Mackenzie River Valley is supported by Natural Resources Canada. RAS was supported by María Zambrano MZ2021 postdoctoral research program and the British Ecological Society (Small Research 2021—SR21/1291). Field sampling was supported by the Government of the Northwest Territories Environmental Studies Research Fund and Cumulative Impacts Monitoring Program (Project 170), Natural Science and Engineering Research Council Northern Supplement Funding, Polar Continental Shelf Program, and Northern Student Training Program. NSERC Discovery Grant support (J.F.J., J.L.B., and S.C.), Canada Research Chairs funding (J.L.B.), and in-kind support to J.F.J. from the Bonanza Creek LTER [funded by the US NSF (DEB-2224776) and the USDA Forest Service, Pacific Northwest Research Station] supported this work. The NSC analyses in this paper were conducted while R.A.-S. was a visiting scientist in the Center for Ecosystem Science and Society at Northern Arizona. We thank Emma Sherwood for her support with ArcMap and H. Brendan O’Neill from the Geological Survey of Canada, Natural Resources Canada, for providing comments on this manuscript. Author contributions R.A.-S., S.L.S., J.F.J., M.R.T., S.G.C., and J.L.B. designed research; R.A.-S., A.D.R., and J.M.L.M. performed research; R.A.-S., S.L.S., S.G.C., and J.L.B. analyzed data; and R.A.-S., S.L.S., J.F.J., S.G.C., and J.L.B. wrote the paper. Competing interests The authors declare no competing interest. Footnotes This article is a PNAS Direct Submission. 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6153	https://sites.math.washington.edu/~novik/publications/cs-survey.pdf	A tale of centrally symmetric polytopes and spheres Isabella Novik∗ Department of Mathematics University of Washington Seattle, WA 98195-4350, USA novik@math.washington.edu May 3, 2018 Abstract This paper is a survey of recent advances as well as open problems in the study of face num-bers of centrally symmetric simplicial polytopes and spheres. The topics discussed range from neighborliness of centrally symmetric polytopes and the upper bound theorem for centrally symmetric simplicial spheres to the generalized lower bound theorem for centrally symmetric simplicial polytopes and the lower bound conjecture for centrally symmetric simplicial spheres and manifolds. 1 Introduction The goal of this paper is to survey recent results related to face numbers of centrally symmetric simplicial polytopes and spheres. To put things into perspective, we start by discussing simplicial polytopes and spheres without a symmetry assumption. The classical theorems of Barnette and McMullen , known as the Lower Bound Theorem (LBT, for short) and the Upper Bound Theorem (or UBT), assert that among all d-dimensional simplicial polytopes with n vertices a stacked polytope simultaneously minimizes all the face numbers while the cyclic polytope simultaneously maximizes all the face numbers. Furthermore, the same results continue to hold in the generality of (d −1)-dimensional simplicial spheres (see and ). It is also worth mentioning that in the class of simplicial spheres of dimension d −1 ≥3 with n vertices, the (boundary complexes of the) stacked polytopes are the only minimizers . On the other hand, the maximizers are precisely the ⌊d/2⌋-neighborly spheres — a set that, in addition to the (boundary complex of the) cyclic polytope, includes many other simplicial spheres, see, for instance, a recent paper . In fact, much more is known: the celebrated g-theorem of Billera–Lee and Stanley provides a complete characterization of all possible f-vectors of simplicial polytopes. The equally celebrated g-conjecture posits that the same characterization is valid for the f-vectors of simplicial spheres. (The f-vector encodes the number of faces of each dimension of a simplicial complex.) In ∗Research is partially supported by NSF grants DMS-1361423 and DMS-1664865, and by Robert R. & Elaine F. Phelps Professorship in Mathematics. This material is based on work supported by the National Science Foun-dation under Grant No. DMS-1440140 while the author was in residence at the Mathematical Sciences Research Institute in Berkeley CA, during the Fall 2017 semester. 1 other words, at least conjecturally, the f-vector cannot diﬀerentiate between simplicial polytopes and simplicial spheres. Let us now restrict our world and consider all centrally symmetric (cs, for short) simplicial polytopes and all centrally symmetric simplicial spheres. What is known about the f-vectors of these objects? Are there cs analogs of the LBT and UBT, and ultimately of the g-theorem and g-conjecture, respectively? Surprisingly very little is known, and not for the lack of eﬀort. There is an analog of the LBT for cs simplicial polytopes established by Stanley as well as a characterization of minimizers , and while it is plausible that the same results continue to hold for all cs simplicial spheres, the proofs remain elusive. Still, the most mysterious and fascinating side of the story comes from trying to understand the upper bounds: in contrast with the situation for simplicial polytopes and spheres without a symmetry assumption, the existing upper-bound type results and conjectures indicate striking diﬀerences between f-vectors of cs polytopes and those of cs spheres. In other words, cs spheres and polytopes do not look alike f-wise, even though their non-cs counterparts do! For instance, the (appropriately deﬁned) neighborliness of cs polytopes is quite restrictive , and a cs d-dimensional polytope with more than 2d vertices cannot be even 2-neighborly; yet, by a result of Jockusch , there exist cs 2-neighborly simplicial spheres of dimension 3 with any even number n ≥8 of vertices. Thus, for a suﬃciently large n, the maximum number of edges that a cs 4-dimensional polytope with n vertices can have diﬀers from the maximum number of edges that a cs simplicial sphere of dimension 3 with n vertices can have; moreover, the former quantity (or even its asymptotic behavior) is unknown at present. This indicates how very far we are from even posing a plausible (sharp) Upper Bound Conjecture for cs d-dimensional polytopes with d ≥4. At the same time, Adin and Stanley (unpublished) provided upper bounds on face numbers of cs simplicial spheres of dimension d −1 with n vertices; these bounds are attained by a cs ⌊d/2⌋-neighborly sphere of dimension d −1 with n vertices assuming such a sphere exist. Aside from being of intrinsic interest, additional motivation to better understand the f-vectors of cs polytopes arises from the recently discovered tantalizing connections (initiated by Donoho and his collaborators, see, for instance, [22, 23]) between cs polytopes with many faces and seem-ingly unrelated areas of error-correcting codes and sparse signal reconstruction. Furthermore, any cs convex body in Rd is the unit ball of a certain norm on Rd. As a result, methods used in the study of face numbers of cs complexes, at present, involve not only commutative algebra (via the study of associated Stanley–Reisner rings) but also a wealth of techniques from geometric analysis. Yet, many problems remain notoriously diﬃcult. The goal of this paper is thus to survey the few known results on f-vectors of cs simplicial polytopes and spheres, showcase several existing techniques, and, most importantly, present many open problems. It is our hope that collecting such problems in one place will catalyze progress in this fascinating ﬁeld. For a quick preview of what is known and what is not, see the following table. non-cs cs simplicial polytopes simplicial spheres same? simplicial polytope simplicial spheres same? UBT √ √ yes no plausible conjecture known bounds conjecturally sharp no LBT √ √ yes √ conjecture conjecturally yes GLBT √ conjecture conjecturally yes √ conjecture conjecturally yes The rest of the paper is structured as follows. In Section 2, we set up notation and recall basic 2 deﬁnitions pertaining to simplicial polytopes and spheres. Section 3 is devoted to neighborliness of cs polytopes. This leads to discussion of upper-bound type results and problems on face numbers of cs polytopes, see Section 4. Section 5 deals with neighborliness and upper-bound type results for cs simplicial spheres. Section 6 takes us into the algebraic side of the story: there we present a quick review of the Stanley–Reisner ring — the major algebraic tool in the study of face numbers, sketch the proof of the classical UBT along with Adin–Stanley’s variant of this result for cs spheres, and prepare the ground for the following sections. Sections 7 and 8 are concerned with the lower-bound type results and conjectures. Speciﬁcally, in Section 7 we discuss a cs analog of the Generalized LBT for cs simplicial polytopes — a part of the story that is most well understood, while in Section 8 we consider a natural conjectural cs analog of the LBT for spheres, manifolds, and pseudomanifolds. We close with a few concluding remarks in Section 9. 2 Preliminaries We start with outlining basic deﬁnitions and notation we will use throughout the paper. A polytope is the convex hull of a set of ﬁnitely many points in Rd. One example is the (geometric) d-simplex deﬁned as the convex hull of an arbitrary set of d+1 aﬃnely independent points in Rd. A (proper) face of any convex body K (e.g., a polytope) is the intersection of K with a supporting aﬃne hyperplane; see, for example, Chapter II of . A polytope P is called simplicial if all of its (proper) faces are simplices. The dimension of a polytope P is the dimension of the aﬃne hull of P. We refer to d-dimensional polytopes as d-polytopes, and to i-dimensional faces as i-faces. A polytope P ⊂Rd is centrally symmetric (cs, for short) if P = −P; that is, x ∈P if and only if −x ∈P. An important example of a cs polytope is the d-dimensional cross-polytope C∗ d = conv(±p1, ±p2, . . . , ±pd), where p1, p2, . . . , pd are points in Rd whose position vectors form a basis for Rd. If these position vectors form the standard basis of Rd, denoted {e1, . . . , ed}, then the resulting polytope is the unit ball of the ℓ1-norm; we refer to this particular instance of the cross-polytope as the regular cross-polytope. A simplicial complex ∆on a (ﬁnite) vertex set V = V (∆) is a collection of subsets of V that is closed under inclusion; an example is the (abstract) simplex on V , V := {F : F ⊆V }. The elements of ∆are called faces and the maximal under inclusion faces are called facets. The dimension of a face F ∈∆is dim F = \|F\| −1, and the dimension of ∆is dim ∆:= max{dim F : F ∈∆}. The k-skeleton of ∆, Skelk(∆), is a subcomplex of ∆consisting of all faces of dimension ≤k. The f-vector of a simplicial complex ∆is f(∆) := (f−1(∆), f0(∆), . . . , fdim ∆(∆)), where fi = fi(∆) denotes the number of i-faces of ∆; the numbers fi are called the f-numbers of ∆. Each simplicial complex ∆admits a geometric realization ∥∆∥that contains a geometric i-simplex for each i-face of ∆. A simplicial complex ∆is a simplicial sphere (simplicial ball, or simplicial manifold, respectively) if ∥∆∥is homeomorphic to a sphere (ball, or closed manifold, respectively). If P is a simplicial d-polytope, then the empty set along with the collection of the vertex sets of all the (proper) faces of P is a simplicial sphere of dimension d −1 called the boundary complex of P; it is denoted by ∂P. While it follows from Steinitz’ theorem that every simplicial 2-sphere can be realized as the boundary complex of a simplicial 3-polytope, for d ≥4, there are many more combinatorial types of simplicial (d −1)-spheres than those of boundary complexes of simplicial d-polytopes, see [38, 57, 65]. A simplicial complex ∆is centrally symmetric (or cs) if it is equipped with a simplicial involution φ : ∆→∆such that for every non-empty face F ∈∆, φ(F) ̸= F. We refer to F and 3 φ(F) as antipodal faces. For instance, the boundary complex of any cs simplicial polytope P is a cs simplicial sphere with the involution φ induced by the map φ(v) = −v on the vertices of P. A simplicial complex ∆is k-neighborly if every set of k of its vertices forms a face of ∆. Equivalently, a simplicial complex ∆with n vertices is k-neighborly if its (k−1)-skeleton coincides with the (k −1)-skeleton of the (n −1)-simplex. Since in a cs complex, a vertex and its antipode can never form an edge, this deﬁnition of neighborliness requires a suitable adjustment for cs complexes. We say that a cs simplicial complex ∆is k-neighborly if every set of k of its vertices, no two of which are antipodes, forms a face of ∆. Equivalently, a cs simplicial complex ∆on 2m vertices is k-neighborly if its (k −1)-skeleton coincides with the (k −1)-skeleton of ∂C∗ m. In particular, if ∆is a k-neighborly simplicial complex, then fj−1(∆) = f0(∆) j for all j ≤k, while if ∆is a cs k-neighborly simplicial complex, then fj−1(∆) = 2jf0(∆)/2 j for all j ≤k. It is worth mentioning that similar deﬁnitions apply to general (i.e., not necessarily simplicial) polytopes. Speciﬁcally, a polytope P is k-neighborly if every set of k of its vertices forms the vertex set of a face of P; a cs polytope P is k-neighborly if every set of k of its vertices, no two of which are antipodes, forms the vertex set of a face of P. In the next two sections we work with general cs polytopes. 3 How neighborly can a cs polytope be? Our story begins with the cyclic polytope, Cd(n), which is deﬁned as the convex hull of n ≥d + 1 distinct points on the moment curve {(t, t2, . . . , td) ∈Rd : t ∈R} or on the trigonometric moment curve {(cos t, sin t, cos 2t, sin 2t, . . . , cos kt, sin kt) ∈R2k : t ∈R}, assuming d = 2k. Both types of cyclic polytopes were investigated by Carath´ eodory and later by Gale , who, in particular, showed that the two types are combinatorially equivalent (assuming d is even) and independent of the choice of points. In fact, the two types are projectively equivalent, see [84, Exercise 2.21]. These polytopes were also rediscovered by Motzkin [53, 32] and many others; see books [11, 84] for more information on these amazing polytopes. The properties of the cyclic polytope Cd(n) that will be important for us here are: it is a (non cs) simplicial d-polytope with n vertices; furthermore, it is ⌊d/2⌋-neighborly for all n ≥d + 1. The existence of cyclic polytopes motivated several questions, among them: do there exist cs d-polytopes (apart from the cross-polytope) that are ⌊d/2⌋-neighborly or at least “highly” neighborly? This section discusses the state-of-aﬀairs triggered by this question. It became apparent from the very beginning that the answer is likely to be both interesting and complicated: several works from the sixties indicated that in contrast to the general case, the neighborliness of cs polytopes might be rather restricted. Speciﬁcally, Gr¨ unbaum showed in 1967 [31, p. 116] that while there exists a cs 2-neighborly 4-polytope with 10 vertices, no cs 4-polytope with 12 or more vertices can be 2-neighborly. This observation was extended by McMullen and Shephard who proved that while there exists a cs ⌊d/2⌋-neighborly d-polytope with 2(d + 1) vertices, a cs d-polytope with at least 2(d + 2) vertices cannot be more than ⌊(d + 1)/3⌋-neighborly. A cs ⌊d/2⌋-neighborly d-polytope with 2(d + 1) vertices is easy to construct: for instance, conv ± e1, ±e2, . . . , ±ed, ± Pd i=1 ei does the job. On the other hand, to show that a cs d-polytope with 2(d + 2) vertices can only be ⌊(d + 1)/3⌋-neighborly and to construct cs polytopes achieving this bound, McMullen and Shephard introduced and studied cs transformations of cs polytopes — a cs analog of celebrated Gale transforms. (See [31, Section 4 5.4] and [84, Chapter 6] for a detailed description of Gale transforms and their applications.) Let k(d, n) denote the largest integer k such that there exists a cs d-polytope with 2(n + d) vertices that is k-neighborly. Inﬂuenced by their result, McMullen and Shephard conjectured that, in fact, k(d, n) ≤⌊(d + n −1)/(n + 1)⌋for all n ≥3. Their conjecture was subsequently refuted by Halsey and then by Schneider , but only for d ≫n. Namely, Schneider showed that lim inf d→∞ k(d, 3) d + 3 ≥1 −2−1/2 and lim inf d→∞ k(d, n) d + n ≥0.2390 for all n. On the other hand, Burton proved that a cs d-polytope with a suﬃciently large number of vertices (≈(d/2)d/2) indeed cannot be even 2-neighborly. Burton’s proof is surprisingly short and simple: it relies on John’s ellipsoid theorem, see for instance [8, Chapter 3], along with a quantitative version of the observation that any suﬃciently large ﬁnite subset of the Euclidean unit sphere contains two points that are very close to each other. To the best of the author’s knowledge, McMullen–Shephard’s values k(d, 1) = ⌊d 2⌋and k(d, 2) = ⌊d+1 3 ⌋(see ) remain the only known exact values of k(d, n) for all d. In partic-ular, k(4, 1) = 2 and k(4, 2) = 1, while k(5, 1) = k(5, 2) = 2. Furthermore, the main result of a very recent preprint implies that k(5, 3) = k(5, 4) = 2, but it appears unknown whether k(5, 5) equals 2 or 1. On the other hand, the asymptotics of k(d, n) is now well understood: Theorem 3.1. There exist absolute constants C1, C2 > 0 independent of d and n such that C1d 1 + log n+d d ≤k(d, n) ≤1 + C2d 1 + log n+d d . Theorem 3.1 is due to Linial and Novik . A dual version of the lower bound part of this theorem was also independently established by Rudelson and Vershynin . Two extreme, and hence particularly interesting, cases of Theorem 3.1 deserve a special mention: the case of k(d, n) proportional to d and the case of k(d, n) = 1. Donoho proved that there exists ρ > 0 such that for large d, the orthogonal projection of the 2d-dimensional regular cross-polytope onto a d-dimensional subspace of R2d, chosen uniformly at random, is at least ⌊ρd⌋-neighborly with high probability and provided numerical evidence that ρ ≥0.089. In other words, for large d, k(d, d) ≥ρd. (The estimates from guarantee that ρ ≥1/400.) As for the other extreme, Theorem 3.1 shows that the largest number of vertices in a cs 2-neighborly d-polytope is eΘ(d). In fact, the following more precise result is known, see [46, Theorem 1.2] and [12, Theorem 3.2]. Theorem 3.2. A cs 2-neighborly d-polytope has at most 2d vertices, i.e., k(d, 2d−1 + 1 −d) = 1. On the other hand, for any even d ≥6, there exists a cs 2-neighborly d-polytope with 2( √ 3 d/3−1) vertices. While this paper awaited its publication, the author proved in that for every d ≥2, there exists a cs 2-neighborly d-polytope with 2d−1 + 2 vertices. Our discussion of k(d, n) is summarized in a table below. 5 n k(d, n) 1 d 2 2 d+1 3 d proportional to d ≤2d−2 + 1 −d ≥2 ≥2d−1 + 1 −d 1 We devote the rest of this section to pointing out some of the main ideas used in the proofs. In particular, the proof of the fact that a cs 2-neighborly d-polytope has at most 2d vertices is so short that we cannot avoid the temptation to provide it here. For a polytope P ⊂Rd and a vector a ∈Rd, deﬁne Pa := P + a to be the translation of P by a, where “+” denotes the Minkowski addition. The ﬁrst step in the proof is the following simple observation from : Lemma 3.3. Let P ⊂Rd be a cs d-polytope, and let u and v be vertices of P, so that −v is also a vertex of P. If the polytopes Pu and Pv have intersecting interiors then the vertices u and −v are not connected by an edge. Consequently, if P is a cs 2-neighborly polytope with vertex set V , then the polytopes {Pv : v ∈V } have pairwise disjoint interiors. Proof: The assumption that int (Pu) ∩int (Pv) ̸= ∅implies that there exist x, y ∈int (P) such that x + u = y + v, or equivalently, that (y −x)/2 = (u −v)/2. Since P is centrally symmetric, and x, y ∈int (P), the point q := (y −x)/2 is an interior point of P. As q is also the barycenter of the line segment connecting u and −v, this line segment is not an edge of P. □ The rest of the proof that a cs 2-neighborly d-polytope has at most 2d vertices utilizes a volume trick that goes back to Danzer and Gr¨ unbaum . If P ⊂Rd is a cs 2-neighborly d-polytope with vertex set V , then by Lemma 3.3, the polytopes {Pv : v ∈V } have pairwise disjoint interiors. Therefore, Vol [ v∈V Pv = X v∈V Vol (Pv) = \|V \| · Vol (P). On the other hand, since for v ∈V , Pv = P + v ⊂2P, it follows that S v∈V Pv ⊆2P, and so Vol [ v∈V Pv ≤Vol (2P) = 2d · Vol (P). Comparing these two equations yields \|V \| ≤2d, as desired. The proof of the upper bound part of Theorem 3.1 relies on a more intricate application of the same volume trick. Let P be a cs k-neighborly d-polytope, where k = 2s for some integer s. We say that a family F of (s −1)-faces of P is good if every two elements F ̸= G of F satisfy the following conditions: F and G share at most s/2 vertices, while F and −G have no common vertices. To obtain an upper bound on the size of the vertex set V of P in terms of d and k, one ﬁrst observes that if F is a good family, F ̸= G are in F, and bF and bG are the barycenters of F and G, then the polytopes P + 2bF and P + 2bG have disjoint interiors (cf. Lemma 3.3). One then uses a simple counting argument to show that there is a relatively large (in terms of d, s, and \|V \|) good family F. Since the translates {P + 2bF : F ∈F} of P have pairwise disjoint interiors and are all contained in 3P, the volume trick yields a desired upper bound on \|F\|, and hence also on \|V \|; see the proof of Theorem 1.1 in for more details. 6 The proof of the lower bound in Theorem 3.1 is based on studying the cs transforms of cs polytopes introduced in and on “high-dimensional paradoxes” such as Kaˇ sin’s theorem and its generalization due to Garnaev and Gluskin . Speciﬁcally, Kaˇ sin’s theorem asserts that there is an absolute constant C (for instance, 32 does the job, see [8, Lecture 4]) such that for every d, R2d has a d-dimensional subspace, Ld, with the following property: the ratio of the ℓ2-norm to the ℓ1-norm of any nonzero vector x ∈Ld is in the interval [ 1 √ 2d, C √ 2d]; we refer to such a subspace as a Kaˇ sin subspace. Via cs transforms, d-dimensional subspaces of R2d correspond to cs d-polytopes with 4d vertices; furthermore, a careful analysis of cs transforms shows that the polytopes corresponding to Kaˇ sin subspaces are k-neighborly with k proportional to d. More generally, a theorem due to Garnaev and Gluskin quantiﬁes the extent to which an n-dimensional subspace of Rn+d can be “almost Euclidean” (meaning that the ratio of the ℓ2-norm to the ℓ1-norm of non-zero vectors remains within certain bounds, more precisely, it is ≤˜ C q 1+log((n+d)/d) d for some absolute constant ˜ C). Via cs transforms, n-dimensional subspaces of Rn+d correspond to cs d-polytopes with 2(n+d) vertices, and the “almost Euclidean” subspaces correspond to cs polytopes with neighborliness given by the lower bound in Theorem 3.1, see for technical details. The proof of the Garnaev–Gluskin result and hence also of the lower bound part of Theorem 3.1 is probabilistic in nature: it does not give an explicit construction of neighborly cs polytopes, but rather shows that they form a set of positive probability in a certain probability space. Indeed it is an interesting open question to ﬁnd an explicit construction for highly neighborly cs polytopes that meet the lower bound in Theorem 3.1. We discuss some known explicit constructions (for instance that of a cs 2-neighborly d polytope with ≈ √ 3 d vertices) in the next section. It would also be extremely interesting to shed some light on the exact values of k(d, n): Problem 3.4. Determine the exact values of k(d, n), or at least ﬁnd the value of n0(d) := min{n : k(d, n) = 1}, that is, ﬁnd the number n starting from which a cs d-polytope with 2(d+n) vertices cannot be even 2-neighborly. 4 Towards an upper bound theorem for cs polytopes The fame of the cyclic polytope comes from the Upper Bound Theorem (UBT, for short) conjec-tured by Motzkin and proved by McMullen . It asserts that among all d-polytopes with n vertices, the cyclic polytope Cd(n) simultaneously maximizes all the face numbers. The goal of this section is to summarize several upper-bound type results and problems for cs polytopes motivated by the UBT. What is the maximum number of k-faces that a centrally symmetric d-polytope with N vertices can have? While our discussion in the previous section suggests that at present we are very far from even posing a plausible conjecture, certain asymptotic results on the maximum possible number of edges are known. Speciﬁcally, the following generalization of Theorem 3.2 holds. This result is in sharp contrast with the fact that f1(Cd(n)) = n 2 as long as d ≥4. Theorem 4.1. Let d ≥4. If P ⊂Rd is a cs d-polytope on N vertices, then f1(P) ≤(1 −2−d)N2 2 . 7 On the other hand, there exist cs d-polytopes with N vertices (for an arbitrarily large N) and at least 1 −3−⌊d/2−1⌋ N 2 ≈(1 − √ 3 −d) · N2 2 edges. The ﬁrst part of Theorem 4.1 was established by Barvinok and Novik in [14, Proposition 2.1]; its proof relies on an extension of the argument discussed in the previous section to obtain an upper bound on the number of vertices that a cs 2-neighborly d-polytope can have, and more speciﬁcally on Lemma 3.3. The idea, very roughly, is as follows: each of the N vertex-translates of P, Pu for u ∈V , has the same volume as P, and all of them are contained in the polytope 2P, whose volume is 2dVol (P). Hence, on average, an interior point of 2P belongs to N/2d (out of N) sets int (Pu). Therefore, on average, int (Pu) intersects with the interiors of at least N/2d −1 other vertex-translates of P. Consequently, by Lemma 3.3, the average degree of a vertex of P in the graph of P is ≤N(1 −2−d). This yields the desired upper bound on the number of edges of P. The second part of Theorem 4.1 is due to Barvinok, Lee, and Novik . The proof is based on an explicit construction whose origins can be traced to work of Smilansky . To discuss this part, we start by recalling that the cyclic polytope Cd(n) is deﬁned as the convex hull of n points on the moment curve or, if d = 2k is even, on the trigonometric moment curve Tk : R →R2k, where Tk(t) = (cos t, sin t, cos 2t, sin 2t, . . . , cos kt, sin kt). In an attempt to come up with a cs analog of the cyclic polytope, Smilansky (in the case of k = 2), and Barvinok and Novik (in the case of arbitrary k) considered the symmetric moment curve, Uk : R →R2k, deﬁned by Uk(t) = (cos t, sin t, cos 3t, sin 3t, . . . , cos(2k −1)t, sin(2k −1)t) . Since Uk(t) = Uk(t + 2π) for all t ∈R, from this point on, we think of Uk as deﬁned on the unit circle S = R/2πZ. The name symmetric moment curve is explained by an observation that for all t ∈S, t and t + π form a pair of opposite points and Uk(t + π) = −Uk(t). A bicyclic polytope is then deﬁned as the convex hull of {Uk(t) : t ∈X}, where X ⊂S is a ﬁnite subset of S; we will also assume that X is a cs subset of S. The papers (in the case of k = 2), and along with (in the case of arbitrary k) study the edges of bicyclic polytopes. In particular, when k = 2, the following result established in (see also ) holds. Recall that a face of a convex body K is the intersection of K with a supporting hyperplane. Theorem 4.2. Let Γ ⊂S be an open arc of length 2π/3 and let t1, t2 ∈Γ. Then the line segment conv(U2(t1), U2(t2)) is an edge of the 4-dimensional convex body B2 := conv(U2(t) : t ∈S). One immediate consequence is Corollary 4.3. Let X = {0, π/2, π, 3π/2} ⊂S, let s ≥2 be an integer, and let Xs be a cs subset of S obtained from X by replacing each τ ∈X with a cluster of s points all of which lie on a small arc containing τ. Then the polytope Qs := conv(U2(t) : t ∈Xs) is a cs 4-polytope that has N := 4s vertices and at least 1 2 · N( 3 4N −1) ≈3 4 N 2 edges. Indeed, it follows from Theorem 4.2 that each vertex of Qs is connected by an edge to all other vertices of Qs except possibly those coming from the opposite cluster, yielding the result. 8 Denote by fmax(d, N; k−1) the maximum possible number of (k−1)-faces that a cs d-polytope with N vertices can have. From the above discussion we infer that 3 4 · N2 2 −O(N) ≤fmax(4, N; 1) ≤15 16 · N2 2 . These are currently the best known bounds on the maximum possible number of edges that a cs 4-polytope with N vertices can have. Perhaps somewhat surprisingly, for k > 2, bicyclic polytopes do not have a record number of edges. However, the symmetric moment curve is used in the following construction that produces cs polytopes with the largest known to-date number of edges. Let m ≥1 be an integer. Deﬁne Φm(t) : S →R2m+2 by Φm(t) := (U1(t), U1(3t), U1(32t), . . . , U1(3m(t)) = (cos t, sin t, cos 3t, sin 3t, cos 9t, sin 9t, . . . , cos(3mt), sin(3mt)). Parts 1 and 2 of the following result complete the proofs of Theorem 3.2 and Theorem 4.1, respectively. Theorem 4.4. Fix integers m ≥2 and s ≥2. Let Am be a set of 2(3m −1) equally spaced points on S, and let Am,s be a cs subset of S obtained from Am by replacing each τ ∈Am with a cluster of s points, all of which lie on a very small arc containing τ. Then 1. the polytope Pm := conv(Φm(Am)) is a cs 2-neighborly polytope of dimension 2(m + 1) that has 2(3m −1) vertices, 2. the polytope Pm,s := conv(Φm(Am,s)) is a cs polytope of dimension 2(m + 1) that has N := 2s(3m −1) vertices and more than (1 −3−m) N 2 edges. The assumption m ≥2 is only needed to guarantee that the dimension of Pm is exactly 2(m + 1) rather than ≤2(m + 1). To prove that Pm is cs 2-neighborly and Pm,s has many edges for all m ≥1, it is enough to show that each vertex of Pm,s is connected by an edge to all other vertices of Pm,s except possibly those coming from the opposite cluster. This can be done by induction on m. Since Φ1 = U2, the case of m = 1 is simply Corollary 4.3. For the inductive step, one relies on some standard facts about faces of polytopes along with an observation that the composition of Φm : R →R2(m+1) with the projection of R2(m+1) onto R2m that forgets the ﬁrst two coordinates is the curve t →Φm−1(3t), while the composition of Φm with the projection of R2(m+1) onto R4 that forgets all but the ﬁrst four coordinates is the curve Φ1; see [12, Section 3] for more details. Finally, to extend the construction of Theorem 4.4 to odd dimensions, consider the bipyramid over polytopes Pm and Pm,s. To summarize, 1 −3−⌊d/2−1⌋ N 2 ≤fmax(d, N; 1) ≤ 1 −2−d · N2 2 . (4.1) It is, however, worth pointing out that in view of the main result of , fmax(d, N; 1) might be closer to the right-hand side of eq. (4.1) than to the left one. To extend the above discussion to higher-dimensional faces, we need to look more closely at the curve Uk and its convex hull. 9 One crucial feature of the convex hull Mk = conv(Tk(t) : t ∈S) ⊂R2k of the trigonometric moment curve is that it is k-neighborly, that is, for any p ≤k distinct points t1, . . . , tp ∈S, the convex hull conv(Tk(t1), . . . , Tk(tn)) is a face of Mk; see, for example, Chapter II of . While the convex hull of Uk is not k-neighborly, the following theorem, which is the main result of , shows that it is locally k-neighborly (cf. Theorem 4.2). Theorem 4.5. For every positive integer k there exists a number π 2 < αk < π such that for an arbitrary open arc Γ ⊂S of length αk and arbitrary distinct p ≤k points t1, . . . , tp ∈Γ, the set conv(Uk(t1), . . . , Uk(tp)) is a face of Bk := conv(Uk(t) : t ∈S). The gap between the currently known upper and lower bounds on fmax(d, N; k −1) for k > 2 is much worse than the gap for k = 2 illustrated by eq. (4.1). Indeed, we have: Theorem 4.6. Let 3 ≤k ≤d/2. Then 1 −k2 23/20k22k−d N k ≤fmax(d, N; k −1) ≤ 1 −2−d N N −1 N k . The proof of the upper bound part of this theorem follows easily from the ﬁrst part of Theorem 4.1 together with (1) the well-known perturbation trick that reduces the situation to cs simplicial polytopes, and (2) the standard double-counting argument that relates the number of edges to the number of (k −1)-faces in any simplicial complex with N vertices, see [14, Proposition 2.2] for more details. The proof of the lower bound part can be found in [12, Section 4]. It utilizes a construction that is somewhat along the lines of the construction of Theorem 4.4, but quite a bit more involved: the desired polytope is obtained as the convex hull of a carefully chosen set of points on the curve Ψk,m : S →R2k(m+1) deﬁned by Ψk,m(t) := (Uk(t), Uk(3t), Uk(32t), . . . , Uk(3mt)). (Note that Φm is essentially Ψ2,m: in Ψ2,m(t) every coordinate except for the ﬁrst two and the last two shows up twice; to obtain Φm from Ψ2,m simply leave one copy of each repeated coordinate.) To choose an appropriate set of points one uses the notion of a k-independent family of subsets of {1, 2, . . . , m} and a deterministic construction from of a large k-independent family. Finally, to show that the resulting polytope has many (k −1)-faces, one relies heavily on the local neighborliness of the convex hull of Uk discussed in Theorem 4.5 along with some standard results on faces of polytopes. As the discussion above indicates, at present our understanding of the maximum possible number of faces of a given dimension that a cs d-polytope with a given number of vertices can have is rather limited even for d = 4. In particular, the following questions are wide open: Problem 4.7. Does the limit limN→∞ fmax(d,N;1) (N 2) exist and, if so, what is its value? Or better yet, what is the actual value of fmax(d, N; 1)? Similarly, what is limN→∞ fmax(d,N;k−1) (N k) for 2 < k ≤ d/2? Can we, at least, establish better lower and upper bounds than those given in Theorem 4.6? In fact, it should be stressed that for d ≥6 and N ≥2(d + 2), we do not even know if in the class of cs d-polytopes with N vertices there is a polytope that simultaneously maximizes all the face numbers. (For d = 4, 5, a cs simplicial polytope that maximizes the number of edges automatically maximizes the rest of face numbers.) 10 5 Towards an upper bound theorem for cs simplicial spheres McMullen’s Upper Bound Theorem was extended by Stanley to the class of all simplicial spheres. More precisely, Stanley proved that among all simplicial (d−1)-spheres with n vertices, the boundary complex of the cyclic polytope Cd(n) simultaneously maximizes all the face numbers. (This result was extended even further to some classes of simplicial manifolds and even certain pseudomanifolds with isolated singularities, see [58, 34, 62]. In fact, already in 1964, Victor Klee proved that the UBT holds for all (d −1)-dimensional Eulerian complexes that have at least O(d2) vertices. Whether the UBT holds for all Eulerian complexes remains an open question.) The situation with the face numbers of cs spheres versus face numbers of cs polytopes appears to be drastically diﬀerent. On one hand, as we saw in Sections 3 and 4, a cs 4-polytope P with N ≥12 vertices cannot be 2-neighborly; furthermore, such a P has at most 15 16 · N2 2 edges. On the other hand, it is a result of Jockusch that Theorem 5.1. For every even number N ≥8, there exists a cs 2-neighborly simplicial sphere JN of dimension 3 with N vertices. In particular, JN has N 2 −N 2 = N2 2 −N edges. Jockusch’s proof, see , is by inductive construction. The initial sphere, J8, is the boundary complex of the cross-polytope C∗ 4 with the involution φ induced by the map φ(v) = −v on the vertex set of C∗ 4. The cs sphere (JN+2, φ) is obtained from the cs sphere (JN, φ) by the following procedure: 1. Find a subcomplex BN in JN such that (i) BN is a 3-ball, (ii) BN and φ(BN) share no common facets, (iii) every vertex of JN is a vertex of BN, and (iv) every edge of BN lies on the boundary of BN, i.e., BN has no interior vertices and no interior edges. 2. Let vN+1 and vN+2 be two new vertices. Cut out the interior of BN from JN, and cone the boundary of the resulting hole with vN+1. Similarly, cut out the interior of φ(BN) from JN, and cone the boundary of the resulting hole with vN+2. Extending φ to JN+2 by letting φ(vN+1) = vN+2 and φ(vN+2) = vN+1 provides us with a free involution on JN+2. Furthermore, choosing BN in a way that is speciﬁed in Part (1) of the construction guarantees that the cs sphere JN+2 is 2-neighborly (provided JN was 2-neighborly). To allow for this inductive construction, an essential part of Jockusch’s proof is devoted to deﬁning BN in a way that ensures that the resulting simplicial sphere JN+2 has a subcomplex BN+2 with the same properties. The suspension of a simplicial complex ∆, Σ(∆), is the join of ∆with the 0-dimensional sphere, that is, Σ(∆) = {F, F ∪{u0}, F ∪{w0} : F ∈∆}, where u0 and w0 are two new vertices. Observe that if ∆is cs, then Σ(∆) is cs (with u0 and w0 being antipodes), and if ∆is a (d −1)-sphere, then Σ(∆) is a d-sphere; furthermore, if ∆is cs k-neighborly, then so is Σ(∆). In particular, for every even N ≥8, Σ(JN) is a cs 2-neighborly simplicial 4-sphere with N + 2 vertices. Jockusch’s results lead to the following problem on higher-dimensional cs spheres. Problem 5.2. Let d > 5 and let N ≥2d be an even number. Is there a cs ⌊d/2⌋-neighborly simplicial (d −1)-sphere with N vertices? The boundary complex of C∗ d is the unique cs d-neighborly simplicial (d −1)-sphere with 2d vertices. The boundary complex of conv(±e1, . . . , ±ed, ± Pd i=1 ei) is a cs ⌊d/2⌋-neighborly 11 simplicial (d−1)-sphere with 2(d+1) vertices ; furthermore, in the case of an odd d, [16, Section 6.2] provides a construction of many cs ⌊d/2⌋-neighborly simplicial (d −1)-spheres with 2(d + 1) vertices. Lutz [47, Chapter 4] found (by a computer search) several cs 3-neighborly simplicial 5-spheres with 16 = 2(6 + 2) vertices; his examples have vertex-transitive cyclic symmetry. Suspensions of Lutz’s examples are cs 3-neighborly simplicial 6-spheres with 18 vertices. For all other values of d and N, Problem 5.2 remains wide open. It is also worth pointing out that Pfeiﬂe [64, Chapter 10] investigated possible neigborliness of cs star-shaped spheres — a class of objects that contains all boundary complexes of cs simplicial polytopes and is contained in the class of all cs simplicial spheres. In analogy with McMullen–Shephard’s result about cs d-polytopes with 2(d + 2) vertices, he proved that for all even d ≥4 and for all odd d ≥11, there are no cs ⌊d/2⌋-neighborly star-shaped spheres of dimension d −1 with 2(d + 2) vertices. One of the main reasons for trying to resolve Problem 5.2 comes from the following cs analog of the Upper Bound Theorem; this result is due to Adin and Stanley (unpublished). Theorem 5.3. Among all cs simplicial (d −1)-spheres with N vertices, a cs ⌊d/2⌋-neighborly (d −1)-sphere with N vertices simultaneously maximizes all the face numbers, assuming such a sphere exists. We sketch the proof of Theorem 5.3 in the next section; as in the classic non-cs case, it relies on the theory of Stanley–Reisner rings and on the Dehn–Sommerville relations. To close this section, we posit a weaker, and hence potentially more approachable version, of Problem 5.2. Let ∆be a (d −1)-dimensional simplicial complex and F a face of ∆. The link of F in ∆is lk∆(F) := {G ∈∆ : F ∪G ∈∆, F ∩G = ∅}. (Thus, the link of F describes the local structure of ∆around F.) We say that ∆is pure if all facets of ∆have dimension d −1; equivalently, for any face F ∈∆, the link of F is (d −\|F\| −1)-dimensional. Furthermore, we say that ∆is Eulerian if it is pure and the link of any face F ∈∆(including the empty face) has the same Euler characteristic as Sd−\|F\|−1 — the (d −\|F\| −1)-dimensional sphere. In particular, the class of Eulerian complexes includes all simplicial spheres, all odd-dimensional simplicial manifolds as well as all even-dimensional manifolds whose Euler characteristic is two. Eulerian complexes were introduced by Victor Klee in . Problem 5.4. Let d > 5 and let N ≥2d be an even number. Is there a cs ⌊d/2⌋-neighborly (d −1)-dimensional Eulerian complex with N vertices? 6 The algebraic side of the story: Stanley–Reisner rings We now switch to the algebraic side of the story and present a quick review of the major algebraic tool in the study of face numbers of simplicial complexes — the Stanley–Reisner ring. Along the way, we outline the proof of Theorem 5.3 as well as prepare the ground for our discussion of the lower bound theorems in the next section. The main reference to this material is Stanley’s book . Let ∆be a simplicial complex with vertex set V and let K be an inﬁnite ﬁeld of an arbitrary characteristic. Consider the polynomial ring S := K[xv : v ∈V ] with one variable for each vertex in ∆. The Stanley–Reisner ideal of ∆is the following squarefree monomial ideal I∆:= (xv1xv2 · · · xvk : {v1, v2, . . . , vk} / ∈∆) . 12 The Stanley–Reisner ring (or face ring) of ∆is the quotient K[∆] := S/I∆. Since I∆is a monomial ideal, the quotient ring K[∆] is graded by degree. The deﬁnition of I∆ensures that, as a K-vector space, each graded piece of K[∆], denoted K[∆]i, has a natural basis of monomials whose supports correspond to faces of ∆. Stanley’s and Hochster’s insight (independently from each other) in deﬁning this ring [73, 35] was that algebraic properties of K[∆] reﬂect many combinatorial and topological properties of ∆. For instance, if ∆is (d −1)-dimensional then the Krull dimension of K[∆] is d; in fact, the Hilbert series of K[∆], i.e., Hilb(K[∆], t) := P∞ i=0 dimK K[∆]i · ti, is given by Hilb(K[∆], t) = d X i=0 fi−1(∆)ti (1 −t)i = Pd i=0 fi−1(∆)ti(1 −t)d−i (1 −t)d . The last equation leads to the following deﬁnition: if ∆is a (d −1)-dimensional simplicial complex, then the h-vector of ∆, h(∆) = (h0(∆), h1(∆), . . . , hd(∆)), is the vector whose entries satisfy Pd i=0 hi(∆)ti = Pd i=0 fi−1(∆)ti(1 −t)d−i; equivalently, d X i=0 hi(∆)td−i = d X i=0 fi−1(∆)(t −1)d−i. (6.1) In particular, h0(∆) = 1, h1(∆) = f0(∆) −d, and h2(∆) = f1(∆) −(d −1)f0(∆) + d 2 . The following immediate consequences of eq. (6.1) are worth mentioning: knowing the f-numbers of ∆is equivalent to knowing its h-numbers. Moreover, since the f-numbers are non-negative integer combinations of the h-numbers, any upper/lower bounds on the h-numbers of ∆automatically imply upper/lower bounds on the f-numbers of ∆. Let ∆be a (d−1)-dimensional simplicial complex. A sequence of linear forms, θ1, θ2, . . . , θd ∈ S is called a linear system of parameters (or l.s.o.p.) for K[∆] if the ring K[∆]/(Θ) is a ﬁnite-dimensional K-vector space; here (Θ) := (θ1, . . . , θd). It is well-known that if K is an inﬁnite ﬁeld, then K[∆] admits an l.s.o.p. An l.s.o.p. for K[∆] is a regular sequence if θi+1 is a non-zero-divisor on K[∆]/(θ1, . . . , θi) for all 0 ≤i < d. We say that ∆is Cohen–Macaulay (over K), or K-CM for short, if every l.s.o.p. for K[∆] is a regular sequence. Assume now that ∆is K-CM, and θ1, . . . , θd is an l.s.o.p. for K[∆]. Then θ1 is a non-zero-divisor on K[∆], and so the following sequence of K-vector spaces 0 →K[∆]i−1 ·θ1 − →K[∆]i − → K[∆]/(θ1) i →0 (6.2) is exact (for all i ≥0). Thus, dimK(K[∆]/(θ1))i = dimK K[∆]i −dimK K[∆]i−1 for all i. Mul-tiplying by ti and summing up the resulting equations, we obtain that Hilb(K[∆]/(θ1), t) = (1 −t) Hilb(K[∆], t). Iterating this argument for θ2, . . . , θd leads to the following result due to Stanley [73, Section 4]. Theorem 6.1. Let ∆be a (d−1)-dimensional K-CM complex and let θ1, . . . , θd be an l.s.o.p. for K[∆]. Then Hilb(K[∆]/(Θ), t) = (1 −t)d Hilb(K[∆], t) = Pd i=0 hi(∆)ti. (In particular, the h-numbers of CM complexes are non-negative.) If ∆⊆Γ are simplicial complexes (say, on the same vertex set V ), then I∆⊇IΓ, and so there is a natural surjection K[Γ] →K[∆]. Furthermore, if dim ∆= dim Γ, then any l.s.o.p. θ1, . . . , θd 13 for K[Γ] is also an l.s.o.p. for K[∆], and the induced graded homomorphism K[Γ]/(Θ) →K[∆]/(Θ) is surjective. This observation together with Theorem 6.1 yields the following special case of [76, Theorem 2.1]: Corollary 6.2. Let ∆⊆Γ be simplicial complexes. Assume that both ∆and Γ are K-CM and have the same dimension. Then hi(∆) ≤hi(Γ) for all i. The reason CM complexes are relevant to our discussion is that, by a result of Reisner , ∆is a K-CM complex if and only if ∆is pure and for every face F of ∆(including the empty face), all but the top homology group (computed over K) of the link of F vanish. In particular, all simplicial spheres and balls are CM over any ﬁeld; furthermore, the j-skeleton of a K-CM complex is K-CM for all j. Another very important result about simplicial spheres is Dehn–Sommerville relations estab-lished by Klee : if ∆is a simplicial (d −1)-sphere, then hi(∆) = hd−i(∆) for all 0 ≤i ≤d. (In fact, Klee showed that these relations hold for all Eulerian complexes.) We are now ready to close this section with a sketch of the proof of Stanley’s UBT and of its cs analog – Theorem 5.3. Following the custom, if P is a simplicial polytope, we denote by h(P) the h-vector of the boundary complex of P. Let ∆be a simplicial (d −1)-sphere with vertex set V , \|V \| = n. Let V be the simplex on V , let Γ = Skeld−1(V ), and let Cd(n) be the cyclic polytope. Then ∆⊆Γ are both CM complexes of the same dimension. Hence, by Corollary 6.2, hi(∆) ≤hi(Γ) for all 0 ≤i ≤d. Furthermore, since Cd(n) is ⌊d/2⌋-neighborly, ∂Cd(n) and Γ have the same (⌊d/2⌋−1)-skeleton, and so hi(Γ) = hi(Cd(n)) for all 0 ≤i ≤d/2. (Indeed, hi is determined by f0, f1, . . . , fi−1.) We conclude that hi(∆) ≤hi(Cd(n)) for all 0 ≤i ≤d/2. Dehn–Sommerville relations, applied to ∆and ∂Cd(n), then yield that hi(∆) ≤hi(Cd(n)) also holds for all d/2 ≤i ≤d. Thus, hi(∆) ≤hi(Cd(n)) for all 0 ≤i ≤d, and the Upper Bound Theorem, asserting that fj(∆) ≤fj(Cd(n)) for all 1 ≤j ≤d−1, follows. Similarly, if ∆is a cs simplicial (d −1)-sphere with N = 2m vertices, then ∆is a full-dimensional subcomplex of Γ := Skeld−1(∂C∗ m) (under any bijection from the vertex set of ∆ to that of ∂C∗ m that takes antipodal vertices to antipodal vertices). Further, if S is a cs ⌊d/2⌋-neighborly simplicial (d −1)-sphere with N vertices, then Skel⌊d/2⌋−1(Γ) = Skel⌊d/2⌋−1(S). Ap-plying the same argument as above to ∆, Γ, and S, we conclude that hi(∆) ≤hi(S) for all 0 ≤i ≤d, and hence that fj(∆) ≤fj(S) for all 1 ≤j ≤d −1. This completes the proof of Theorem 5.3. Note that whether a cs ⌊d/2⌋-neighborly simplicial (d−1)-sphere with N = 2m vertices exists or not, the h-numbers (and hence also f-numbers) such a sphere would have are well-deﬁned: hi(S) = hd−i(S) = hi(Skeld−1(C∗ m)) for all i ≤d/2. Thus, independently of the existence of such a sphere, Theorem 5.3 provides upper bounds on face numbers of cs simplicial (d −1)-spheres with N vertices. However, if a cs ⌊d/2⌋-neighborly simplicial (d −1)-sphere with N vertices exists, then these bounds are tight. Note also that the proof of Theorem 5.3 almost does not use the central symmetry assumption: instead, it relies on a much weaker condition, namely, on the existence of a free involution φ : V (∆) →V (∆) on the vertex set of ∆such that {v, φ(v)} is not an edge of ∆for all v ∈V (∆). We refer our readers to for more details on the h-vectors of simplicial spheres and manifolds possessing this weaker property and to for a complete characterization of h-vectors of CM complexes with this property. 14 7 The generalized lower bound theorem for cs polytopes Our ultimate dream is to ﬁnd a cs analog of the g-theorem for cs simplicial polytopes. While at the moment it is completely out of reach (we do not even have a plausible upper bound conjecture, let alone a complete characterization!), establishing the lower-bound type results is a necessary part of the program. To this end, in this section we discuss a cs analog of the Generalized Lower Bound Theorem for cs simplicial polytopes. We start by reviewing the classical Lower Bound Theorem (LBT, for short) and the Generalized Lower Bound Theorem (GLBT, for short). To state these results, we need a few deﬁnitions. A triangulation of a simplicial d-polytope P is a simplicial d-ball B whose boundary, ∂B, coincides with ∂P. A simplicial d-polytope P is called (r−1)-stacked (for some 1 ≤r ≤d) if there exists a triangulation B of P such that Skeld−r(B) = Skeld−r(∂P), i.e., all “new” faces of this triangulation have dimension > d −r. Note that the simplices are the only 0-stacked polytopes, and that 1-stacked polytopes — usually referred to as stacked polytopes — are polytopes that can be obtained from the d-simplex by repeatedly attaching (shallow) d-simplices along facets. In contrast with the cyclic polytopes, two stacked d-polytopes with n vertices may not have the same combinatorial type. However, they do have the same face numbers. Theorem 7.1. (LBT) Let P be a simplicial d-polytope with d ≥4. Then h1(P) ≤h2(P), with equality if and only if P is stacked. Theorem 7.2. (GLBT) Let P be a simplicial d-polytope. Then 1 = h0(P) ≤h1(P) ≤h2(P) ≤· · · ≤h⌊d/2⌋(P). Furthermore, hr(P) = hr−1(P) for some r ≤d/2 if and only if P is (r −1)-stacked. Since fi−1 is a non-negative linear combination of h0, h1, . . . , hi, one immediate corollary of Theorem 7.1 is that among all simplicial d-polytopes with n vertices, a stacked polytope has the smallest number of edges. In fact, Theorem 7.1 together with a well-known reduction due to McMullen, Perles, and Walkup implies that among all simplicial d-polytopes with n vertices, stacked polytopes simultaneously minimize all the face numbers, and that for d ≥4, stacked polytopes are the only minimizers, see [37, Section 5]. The diﬀerences between consecutive h-numbers are known as the g-numbers: g0(P) := 1 and gr(P) := hr(P) −hr−1(P) for 1 ≤r ≤d/2. Non-negativity of g2 for simplicial polytopes of dimension 4 and higher was established by Barnette , while Billera and Lee proved that the equality g2(P) = 0 (for d ≥4) holds if and only if P is stacked. The ﬁrst part of the GLBT was proved by Stanley (a more elementary proof of this part is due to McMullen [49, 50]). The second part was conjectured by McMullen and Walkup, , and proved only recently by Murai and Nevo . Since the d-dimensional cross-polytope has the minimal number of faces among all cs simplicial d-polytopes and since hr(C∗ d) = d r > d r−1 = hr−1(C∗ d) for all 1 ≤r ≤d/2, one expects that for this class of polytopes, the inequalities gr(P) ≥0 of the GLBT can be considerably strengthened. Indeed, the following result of Stanley provides a cs version of the inequality part of the GLBT. This result settled an unpublished conjecture by Bj¨ orner. Theorem 7.3. Let P be a cs simplicial d-polytope. Then gr(P) ≥ d r − d r −1 = gr(C∗ d) for all r ≤d/2. 15 Stanley’s proof of the ﬁrst part of Theorem 7.2 is based on the theory of toric varieties associated with polytopes. Here is a very rough sketch of the argument. If P ⊂Rd is a simplicial d-polytope, then a slight perturbation of the vertices of P does not change its combinatorial type, and so we may assume without loss of generality that P has rational vertices; further, by translating P, we may also assume that the origin is in the interior of P. Let V be the vertex set of P. For each v ∈V , let (v1, v2, . . . , vd) denote the coordinates of v in Rd, and deﬁne θj := P v∈V vjxv ∈R[xv : v ∈V ] for j = 1, 2, . . . , d. Consider R[∂P] — the Stanley–Reisner ring of the boundary complex of P over R. Then θ1, . . . , θd is an l.s.o.p. of R[∂P] (this follows, for instance, from [77, Theorem III.2.4]), and so by Theorem 6.1, dimR (R[∂P]/(Θ))i = hi(P). On the other hand, R[∂P]/(Θ) is isomorphic to the singular cohomology ring of the toric variety XP corresponding to P (see [20, Theorem 10.8] and [26, Section 5.2] for more details). Let ω = P v∈V xv. The toric variety XP is known to satisfy the hard Lefschetz theorem, which implies that for all i ≤d/2, the multiplication map ·ω : (R[∂P]/(Θ))i−1 →(R[∂P]/(Θ))i is injective, and hence that hi−1(P) ≤hi(P) for all i ≤d/2, as desired. Extending Theorem 7.2 to all simplicial spheres is a major open problem in the ﬁeld: it is a part of the celebrated g-conjecture. The main obstacle is that all known proofs of the g-theorem for simplicial polytopes, see and [49, 50], rely heavily on such tools as toric varieties and the hard Lefschetz theorem or polytopal algebras and the Hodge–Riemann–Minkowski quadratic inequalities between mixed volumes, that is, tools that are (at least at present) available only for convex polytopes. To prove Theorem 7.3 a few additional ideas are needed. Let P be a cs simplicial d-polytope or, more generally, let (∆, φ) be a cs CM complex over R with vertex set V . Then the involution φ induces the map σ : xv →xφ(v) on the set of variables of S = R[xv : v ∈V ], which, in turn, extends to a unique automorphism σ on S. Furthermore, since σ(I∆) = I∆, the map σ gives rise to an automorphism on R[∆], which we also denote by σ. The main insight of is that this allows us to equip R[∆] with a ﬁner grading by N × Z/2Z: indeed, deﬁne R∆ := {f ∈R[∆]i : σ(f) = f} and R∆ := {f ∈R[∆]i : σ(f) = −f}. Then R[∆]i = R∆⊕R∆ as vector spaces while R∆·R∆ ⊆R∆ for all i, j ∈N and ϵ1, ϵ2 ∈{±1}. It is also not hard to see that dimR R∆ = dimR R∆ = 1 2 dimR R[∆]i for all i ≥1, and that R∆ = R[∆]0 = R. One can use the Kind–Kleinschmidt criterion [77, Theorem III.2.4] to show that R[∆] has an l.s.o.p. θ1, . . . , θd such that θj ∈R∆ for all j. (For instance, for a cs simplicial polytope P and ∆= ∂P, the l.s.o.p. described in the proof of Theorem 7.2 does the job.) Analyzing the exact sequences as in the proof of Theorem 6.1, see eq. (6.2), but using our ﬁner grading and, in particular, that ·θj maps the degree (i−1, ±1) pieces of R[∆]/(θ1, . . . , θj−1) to the degree (i, ∓1) pieces, then yields that dimR(R[∆]/(Θ))(i,−1) = 1 2 hi(∆) − d i for all 0 ≤i ≤d; (7.1) see for more details. This establishes the following result of Stanley : Theorem 7.4. Let ∆be a cs R-CM complex of dimension d −1. Then hi(∆) ≥ d i for all i. We now consider the case of ∆= ∂P, where P is a cs simplicial d-polytope (with rational vertices), and θ1, . . . , θd are deﬁned using the coordinates of vertices of P as in the sketch of the 16 proof of Theorem 7.2; in particular, θj ∈R∆. Observe that ω = P v∈V xv ∈R∆ and hence that ·ω maps (R[∆]/(Θ))(i−1,−1) to (R[∆]/(Θ))(i,−1). As the map ·ω : (R[∆]/(Θ))i−1 → (R[∆]/(Θ))i is injective for all i ≤d/2, its restriction to (R[∆]/(Θ))(i−1,−1) is also injective. This, along with eq. (7.1), completes the proof of Theorem 7.3. It is worth pointing out that Theorem 7.3 was extended by Adin [3, 4] to the classes of all rational simplicial polytopes with a ﬁxed-point-free linear symmetry of prime and prime-power orders. Furthermore, A’Campo-Neuen extended Theorem 7.3 to toric g-numbers of all cs polytopes (including non-rational non-simplicial ones). The Murai–Nevo proof of the equality case of Theorem 7.2 involves a beautiful blend of tools such as Alexander duality, Stanley–Reisner rings, the Cohen–Macaulay property, as well as generic initial ideals, and, in particular, Green’s crystalization principle [30, Proposition 2.28]. A diﬀerent proof of both parts of Theorem 7.2, including a sharper version of the inequality part, was very recently obtained by Adiprasito [5, Cor. 6.5 and §7]. His proof relies on Lee’s generalized stress spaces . When does equality hold in Theorem 7.3? That is, for a ﬁxed r ≤d/2, is there a characteri-zation of cs simplicial d-polytopes with gr = d r − d r−1 ? In a recent preprint , Klee, Nevo, Novik, and Zheng provide such a characterization in the r = 2 case and a conjectural character-ization in the r > 2 case. Both statements strongly parallel the equality cases of Theorems 7.1 and 7.2. If P is a cs simplicial d-polytope with d ≥4, then we can apply to P the symmetric stacking operation: this operation repeatedly attaches (shallow) simplices along antipodal pairs of facets. Note that symmetric stacking preserves both central symmetry and g2. Theorem 7.5. Let P be a cs simplicial d-polytope with d ≥4. Then g2(P) = d 2 −d if and only if P is obtained from C∗ d by symmetric stacking. Conjecture 7.6. Let P be a cs simplicial d-polytope, and assume that gr(P) = d r − d r−1 for some 3 ≤r ≤⌊d/2⌋. Then there exists a unique polytopal complex C in Rd with the following properties: (i) one of the faces of C is the cross-polytope C∗ d, all other faces of C are simplices that come in antipodal pairs, (ii) C is a “cellulation” of P, that is, S C∈C C = P, and (iii) each element C ∈C of dimension ≤d −r is a face of P. We mention that [41, Conjecture 8.6] provides a more detailed version of the above conjecture and we refer our readers to Ziegler’s book [84, Section 8.1] for the deﬁnition of a polytopal complex. Conjecture 7.6, if true, would imply the following weaker statement that is also wide open at present. (The r = 2 case does hold; this is immediate from Theorem 7.5.) Conjecture 7.7. Let P be a cs simplicial d-polytope. If gr(P) = gr(C∗ d) for some 3 ≤r ≤ ⌊d/2⌋−1, then gr+1(P) = gr+1(C∗ d). One consequence of Theorems 7.3 and 7.5, along with the fact that the f-numbers of spheres are nonnegative linear combinations of the g-numbers, is that among all cs simplicial d-polytopes with n vertices, a polytope obtained from C∗ d by symmetric stacking simultaneously minimizes all the face numbers; furthermore, if d ≥4, then such polytopes are the only minimizers. While we will not discuss here the proof of Theorem 7.5, it is worth pointing out that it relies on the rigidity theory of frameworks (some ingredients of this theory are brieﬂy outlined in the next section), and, in particular, on (1) the theorem of Whiteley asserting that the graph of a simplicial d-polytope (d ≥3) with its natural embedding in Rd is inﬁnitesimally rigid, and 17 on (2) the fact that if P is a cs simplicial d-polytope with g2(P) = d 2 −d, then all stresses on P must be symmetric (i.e., assign the same weight to each edge and its antipode.) This latter observation follows from work of Stanley and Lee , and also from more recent work of Sanyal, Werner, and Ziegler [68, Theorem 2.1]. To summarize: in contrast with the upper-bound type results, an analog of the GLBT for cs simplicial polytopes (at least its inequality part developed in Theorem 7.3) is a very well-understood part of the story. Furthermore, in the case of the LBT we even have a characterization of the minimizers (see Theorem 7.5). The part that is still missing is a characterization of cs simplicial d-polytopes with gr = d r − d r−1 for 3 ≤r ≤⌊d/2⌋. Conjecture 7.6 proposes such a characterization. 8 The lower bound conjecture for cs spheres and manifolds The ﬁnal part of our story concerns a conjectural analog of the LBT for cs spheres (manifolds and even normal pseudomanifolds) — a necessary step in our quest for a cs analog of the g-conjecture. To this end, it is worth recalling that in the world of simplicial complexes without a symmetry assumption, where at present the GLBT (Theorem 7.2) is only known to hold for the class of simplicial polytopes, works of Walkup , Barnette , Kalai , Fogelsanger , and Tay show that the LBT (Theorem 7.1) holds much more generally: Theorem 8.1. Let ∆be a simplicial complex of dimension d −1 ≥3. Assume further that ∆ is a connected simplicial manifold (or even a normal pseudomanifold). Then g2(∆) ≥0, with equality if and only if ∆is the boundary complex of a stacked polytope. In view of this result, we strongly suspect that Stanley’s inequality on the g2-number of cs polytopes (see Theorem 7.3) as well as the characterization of the minimizers given in Theo-rem 7.5 continue to hold in the generality of cs simplicial spheres or perhaps even cs normal pseudomanifolds. This conjecture is however wide open at present. Conjecture 8.2. Let ∆be a cs simplicial complex of dimension d −1 ≥3. Assume further that ∆is a simplicial sphere (or a connected simplicial manifold or even a normal pseudomanifold). Then g2(∆) ≥ d 2 −d. Furthermore, equality holds if and only if ∆is the boundary complex of a cs d-polytope obtained from the cross-polytope C∗ d by symmetric stacking. In the rest of this short section we discuss one potential approach to attacking this conjecture: via the rigidity theory of frameworks. The (now wide) use of this theory in the study of face numbers of simplicial complexes was pioneered by Kalai in his celebrated proof of Theorem 8.1 for simplicial manifolds . Below we review several results and deﬁnitions pertaining to this fascinating theory. We refer our readers to Asimow and Roth , for a friendly introduction to this subject. Let G = (V, E) be a ﬁnite graph. A map ρ : V →Rd is called a d-embedding of G if aﬀ{ρ(v) : v ∈V } = Rd. The graph G, together with a d-embedding ρ, is called a d-framework. An inﬁnitesimal motion of a d-framework (G, ρ) is a map µ : V (G) →Rd such that for any edge {u, v} in G, d dt t=0 (ρ(u) + tµ(u)) −(ρ(v) + tµ(v)) 2 = 0. 18 An inﬁnitesimal motion µ of (G, ρ) is trivial if d dt t=0 (ρ(u) + tµ(u)) −(ρ(v) + tµ(v)) 2 = 0 holds for every two vertices u, v of G. (Trivial inﬁnitesimal motions correspond to a start of an isometric motion of Rd.) We say that a d-framework (G, ρ) is inﬁnitesimally rigid if every inﬁnitesimal motion µ of (G, ρ) is trivial. A stress on a d-framework (G, ρ) is an assignment of weights ω = (ωe : e ∈E(G)) to the edges of G such that for each vertex v equilibrium holds: X u : {u,v}∈E(G) ω{u,v}(ρ(v) −ρ(u)) = 0. We denote the space of all stresses on (G, ρ) by S(G, ρ). The stresses on (G, ρ) correspond to the elements in the kernel of the transpose of a certain f1(G) × d f0(G) matrix Rig(G, ρ) known as the rigidity matrix of (G, ρ). The relevance of rigidity theory to the Lower Bound Theorem is explained by the following fundamental fact that is an easy consequence of the Implicit Function Theorem (see and ). Theorem 8.3. Let (G, ρ) be a d-framework. Then the following statements are equivalent: • (G, ρ) is inﬁnitesimally rigid; • rank Rig(G, ρ) = d f0(G) − d+1 2 ; • dimR S(G, ρ) = f1(G) −d f0(G) + d+1 2 . A graph G is called generically d-rigid if there exists a d-embedding ρ of G such that (G, ρ) is inﬁnitesimally rigid; in this case, the set of inﬁnitesimally rigid d-embeddings of G is an open dense subset of the set of all d-embeddings. Recall that if ∆is a (d −1)-dimensional simplicial complex, then g2(∆) = h2(∆) −h1(∆) = f1(∆) −d f0(∆) + d+1 2 . The last condition of Theorem 8.3 then implies that if the graph (i.e., 1-skeleton) of ∆is generically d-rigid, then g2(∆) ≥0. Two basic but very useful results in rigidity theory are the gluing lemma ([7, Theorem 2] and [83, Lemma 11.1.9]) and the cone lemma . The gluing lemma asserts that if two graphs G1 and G2 are generically d-rigid and share at least d vertices, then their union G1 ∪G2 is also generically d-rigid, while the cone lemma posits that G is generically d-rigid if and only if the graph of the cone over G is generically (d + 1)-rigid. Additionally, by Gluck’s result , the graph of any simplicial 2-sphere is generically 3-rigid. Now, if ∆is a simplicial (d −1)-manifold (with d ≥4) and F is a (d −4)-face of ∆, then the link of F in ∆is a simplicial 2-sphere, and hence has a generically 3-rigid graph. This observation, along with the gluing and cone lemmas, allowed Kalai to prove by induction on d ≥4 that the graph of any simplicial (d−1)-manifold is generically d-rigid, and thus to establish the inequality part of Theorem 8.1 for all simplicial manifolds. The relationship of inﬁnitesimal rigidity to the Stanley–Reisner ring was worked out in . Speciﬁcally, let ∆be a (d −1)-dimensional simplicial complex with vertex set V and let ρ : V →Rd be a d-embedding of the graph of ∆, G(∆). Deﬁne d linear forms in R[xv : v ∈V ] by θj := P v∈V ρ(v)jxv for j = 1, . . . , d, where ρ(v)j denotes the j-th coordinate of ρ(v) ∈Rd, and let θd+1 := P v∈V xv (cf. Section 7, and especially the sketch of the proof of Theorem 7.2). Theorem 10 of implies that if (G(∆), ρ) is inﬁnitesimally rigid, then 19 dimR(R[∆]/(θ1, . . . , θd, θd+1))2 = dimR S(G(∆), ρ) = g2(∆). Equivalently, if (G(∆), ρ) is in-ﬁnitesimally rigid, then θi+1 : (R[∆]/(θ1, . . . , θi))1 →(R[∆]/(θ1, . . . , θi))2 is an injection for all 0 ≤i ≤d. Assume now that ∆is centrally symmetric. If there is a d-embedding ρ : V (∆) →Rd that respects symmetry and such that (G(∆), ρ) is inﬁnitesimally rigid, then the previous paragraph along with the N × Z/2Z-grading of R[∆] and relevant computations of Section 7 (e.g., eq. (7.1)) shows that g2(∆) ≥ d 2 −d. (This also follows from the argument in [68, Section 2]). In particular, the following conjecture, if true, would imply the inequality part of Conjecture 8.2. Conjecture 8.4. Let ∆be a (d −1)-dimensional cs simplicial complex with an involution φ. Assume further that ∆is a simplicial sphere (or a connected simplicial manifold or even a normal pseudomanifold) and that d−1 ≥3. Then there exists a d-embedding ρ : V (∆) →Rd that respects symmetry, i.e., ρ(φ(v)) = −ρ(v) for each vertex v of ∆, and such that (G(∆), ρ) is inﬁnitesimally rigid. For instance, if ∆is the boundary complex of a cs simplicial d-polytope P ⊂Rd, then by White-ley’s theorem , the natural d-embedding of the graph of P qualiﬁes for φ as in Conjecture 8.4. One of the reasons Conjecture 8.4 appears to be hard in the general case is that the links of cs complexes are usually not centrally symmetric, and so starting with cs 2-spheres as the base case (and proceeding by induction via the cone and gluing lemmas) does not work. Conjecture 8.4 of proposes a statement about rigidity of graphs of (non cs) simplicial 2-spheres that, if true, will provide an appropriate base case, and imply Conjecture 8.4. In any case, as a step in our quest for a cs analog of the g-conjecture, it would be extremely interesting to shed any light on Conjectures 8.2 and 8.4 as well as to attempt to strengthen the inequality of Conjecture 8.2 in the spirit of results from [54, Theorem 5.3(i)] and . Such strengthened inequalities would provide lower bounds on g2 of a cs simplicial manifold (or even a normal pseudomanifold) ∆in terms of the ﬁrst homology or perhaps even in terms of the fundamental group of ∆and/or of the Z/2Z-quotient of ∆. 9 Concluding remarks The recent decades made the theory of face numbers into a very active and a rather large ﬁeld. Consequently, there are quite a few topics we glossed over or omitted in this paper. Among them are the face numbers of general (not necessarily simplicial) cs polytopes. However, one of the conjectures about general cs polytopes we cannot avoid mentioning is Kalai’s “3d-conjecture” [39, Conjecture A]: it posits that the total number of faces in a cs d-polytope (including the empty face but excluding the polytope itself) is at least 3d. If the conjecture is true, there are multiple minimizers: the class of polytopes with exactly 3d faces includes at least all the Hanner polytopes. (If Mahler’s conjecture holds, then Hanner polytopes are the minimizers of Mahler volume among all the cs convex bodies.) At present the 3d-conjecture is known to hold for all cs simplicial polytopes (this is an immediate corollary of Theorem 7.4) and, by duality, all simple polytopes, as well as for all at most 4-dimensional cs polytopes . The conjecture is wide open in all other cases. We have also only barely touched on the face numbers of cs simplicial manifolds and instead concentrated on the face numbers of cs simplicial polytopes and spheres. Papers , [61, Sec-tion 4], and contain some results pertaining to the Upper Bound Theorem for cs simplicial 20 manifolds as well as to Sparla’s conjecture [71, Conjecture 4.12], on the Euler characteristic of even-dimensional cs simplicial manifolds. While we had to skip quite a few of the topics and could not possibly do justice to all of the existing methods, we hope we have conveyed at least some of the essence and beauty of this fascinating subject! We are very much looking forward to progress on the many existing as well as yet unstated problems about cs polytopes and simplicial complexes, and to new interactions between combinatorics, discrete geometry, commutative algebra, and geometric analysis that will lead to this progress! Acknowledgments I am grateful to Steve Klee, Connor Sawaske, Hailun Zheng, G¨ unter Ziegler, and the anonymous referee for numerous comments on the preliminary version of this paper. References A. A’Campo-Neuen. On toric h-vectors of centrally symmetric polytopes. Arch. Math. (Basel), 87(3):217–226, 2006. R. M. Adin. Combinatorial structure of simplicial complexes with symmetry. PhD thesis, Hebrew University, Jerusalem, 1991. R. M. Adin. On h-vectors and symmetry. 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E. Sparla. An upper and a lower bound theorem for combinatorial 4-manifolds. Discrete Com-put. Geom., 19:575–593, 1998. R. P. Stanley. The upper bound conjecture and Cohen-Macaulay rings. Studies in Applied Math., 54:135–142, 1975. R. P. Stanley. The number of faces of a simplicial convex polytope. Adv. Math., 35:236–238, 1980. R. P. Stanley. On the number of faces of centrally-symmetric simplicial polytopes. Graphs and Combinatorics, 3:55–66, 1987. R. P. Stanley. A monotonicity property of h-vectors and h∗-vectors. European J. Combin., 14(3):251– 258, 1993. R. P. Stanley. Combinatorics and Commutative Algebra. Progress in Mathematics. Birkh¨ auser, Boston, Inc., Boston, MA, 1996. Second edition. T.-S. Tay. Lower-bound theorems for pseudomanifolds. Discrete Comput. Geom., 13(2):203–216, 1995. C. Vinzant. Edges of the Barvinok-Novik orbitope. Discrete Comput. Geom., 46(3):479–487, 2011. D. W. Walkup. The lower bound conjecture for 3- and 4-manifolds. Acta Math., 125:75–107, 1970. W. Whiteley. Cones, inﬁnity and 1-story buildings. Struct. Topology, 8:53–70, 1983. W. Whiteley. Inﬁnitesimally rigid polyhedra. I. Statics of frameworks. Trans. Amer. Math. Soc., 285(2):431–465, 1984. W. Whiteley. Some matroids from discrete applied geometry. In Matroid theory (Seattle, WA, 1995), volume 197 of Contemp. Math., pages 171–311. Amer. Math. Soc., Providence, RI, 1996. G. M. Ziegler. Lectures on polytopes, volume 152 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1995. 24
6154	https://jonochshorn.com/structuralelements/book/4.05-tension-elements.html	Steel: Jonathan Ochshorn's Structural Elements text, Third Edition jonochshorn.com Jonathan Ochshorn's Structural Elements for Architects and Builders, Third Edition Follow @jonochshorn contact \| structural element calculators \| paperback and pdf \|« previous section \| next section » \| contents Tweet Contents \| 1. Introduction to structural design \| 2. Loads \| 3. Wood \| 4. Steel \| 5. Reinforced concrete Introduction to steel \| Material properties \| Sectional properties \| Design approaches \| Construction systems \| Tension elements \| Columns \| Beams \| Connections \| Ch. 4 Appendix Chapter 4 — Steel: Tension elements Unlike tension elements designed in timber, two modes of failure are considered when designing tension members in steel. First, the element might become functionally useless if yielding occurs across its gross area, at the yield stress, F y . Since internal tensile forces are generally uniform throughout the entire length of the element, yielding would result in extremely large deformations. On the other hand, if yielding commenced on the net area (where bolt holes reduce the gross area), the part of the element subjected to yield strains would be limited to the local area around the bolts, and excessive deformations would not occur. However, a second mode of failure might occur at these bolt holes: rupture of the element could occur if, after yielding, the stresses across the net area reached the ultimate stress, F u. As in wood design, typical bolt hole diameters are 1/16 in. larger than the actual bolt diameter (except that for bolts with diameters greater or equal to 1 in., the bolt hole diameter is made 1/8 in. larger). However, because a small amount of material surrounding the bolt hole is damaged as the hole is punched, an additional 1/16 in. is added to the hole diameter for the purpose of calculating net area, resulting in a bolt hole diameter taken as 1/8 in. larger than the nominal bolt diameter for steel elements (or 3/16 in. larger for bolts with diameters of 1 in. or larger). Another difference in the design of wood and steel tension elements occurs because nonrectangular cross sections are often used in steel. If connections are made through only certain parts of the cross section, as illustrated in Figure 4.11, the net area in the vicinity of the connection will be effectively reduced, depending on the geometry of the elements being joined, and the number of bolts being used. This effective net area, A e, is obtained by multiplying the net area, A n, by a coefficient, U, defined in Appendix Table A-4.9. Figure 4.11: Shear lag in steel tension element showing unstressed or under-stressed areas Where all parts (i.e., flanges, webs, etc.) of a cross section are connected, and the so-called shear lag effect described above cannot occur, the coefficient U is taken as 1.0, and the effective net area equals the net area, just as in timber design. For short connection fittings like splice plates and gusset plates, U is also taken as 1.0, but A e = A n cannot exceed 0.85 times the gross area. These short connecting elements may have an effective width less than their actual width to account for the shear lag effect, based on what is known as the "Whitmore section," shown in Figure 4.12. For a length, L, of the fastener group measured in the direction of load, and a distance, W, between the outer rows of bolts or welds, the effective width is computed by extending a 30° line out from both sides of the fastener group; it can be seen that the effective width, l w, is equal to 2 L tan30° + W. Figure 4.12: The Whitmore section for connecting plates limits the effective width of the plate to 2 L tan 30° + W for both (a) welded connections; and (b) bolted connections Finally, the lengths of tension members, other than rods and cables, are recommended, but not required to have, a slenderness ratio — defined as the ratio of effective length to least radius of gyration — of 300, to prevent excessive vibrations and protect against damage during transportation and erection. The radius of gyration, a property of the cross section, is equal to , where I is the moment of inertia and A is the cross-sectional area of the element. From the preceding discussion, it can be seen that two values for available strength, or allowable stress, in tension need to be determined: one for yielding of the gross area and one for failure (rupture) of the effective net area. These two values are: F t gross = 0.6 F y (4.1) and F t net = 0.5 F u (4.2) where F t gross and F t net are the allowable tensile stresses for steel corresponding to the two modes of failure, or limit states: F y is the yield stress and F u is the ultimate stress for steel (Appendix Table A-4.1). The tensile stress is computed on the gross area in the same manner as for wood (see Equation 3.2). Rupture on a failure surface through bolted or welded connections, however, is determined using the effective net area rather than the net area, so Equation 3.3 (for wood) must be modified for steel connections as follows: F t = P/A e (4.3) When computing the capacity based on yielding, the full gross area is available to resist the internal forces: P allow = F t gross × A g (4.4) When computing the capacity on the effective net area: P allow = F t net × A e (4.5) The "available strength" limit states listed in Appendix Table A-4.2 are equivalent to these formulations based on allowable stress. The following example illustrates the application of these principles to a steel tension problem. Different procedures are used for cables, eyebars, threaded rods, and pin-connected plates. Example 4.1 Analyze steel tension element Calculator Problem definition. Find the maximum tension load, P, that can be applied to a W8 × 24 element connected to gusset plates within a truss with 3/4-in.-diameter bolts, as shown in Figure 4.13. Use A36 steel. Find the required thickness of the gusset plates so that their capacity is no smaller than that of the W8 × 24 tension element. The bolt hole diameter = bolt diameter + 1/8 in. = 7/8 in. = 0.875 in. Figure 4.13: Connection detail at gusset plate, with (a) section at gross area, (b) section at net area, and (c) truss elevation with 3-dimensional view for Example 4.1 Solution overview. Find cross-sectional dimensions and material properties; find gross area capacity; find effective net area capacity; the governing capacity is the lower of these two values. For gusset plate thickness, find effective width based on Whitmore section; apply equations for gross and net area capacity to determine required plate thickness. Problem solution 1. From Appendix Table A-4.3, find cross-sectional dimensions (Figure 4.14): A g = 7.08 in 2 d = 7.93 in. b f = 6.50 in. t f = 0.400 in. 2. From Appendix Table A-4.1, find F y = 36 ksi and F u = 58 ksi. Figure 4.14: Cross-sectional dimensions of W8 × 24 for Example 4.1 3. Gross area: find capacity, P: a. Using Equation 4.1 (or Appendix Table A-4.2) find F t gross = 0.6 F y = 0.6(36) = 22 ksi. b. Using Equation 4.4, P = F t gross × A g = 22(7.08) = 156 kips. 4. Effective net area: find capacity, P: a. From Appendix Table A-4.9, find the shear lag coefficient, U: U = 0.90 since the following criteria are met: Bolts connect wide-flange (W) shape? Yes. Flange width, b f is no less than 0.67d? In other words, 6.5 ≥ 0.67(7.93) = 5.3? Yes. Flange is connected with at least 3 bolts per line? Yes. b. Find the net area, A n (same as Equation 3.1 for wood). As shown in Figure 4.15: A n = A g – (number of holes)(d h × t) = 7.08 – 4(0.875 × 0.400) = 5.68 in 2. Figure 4.15: Net area diagram for Example 4.1 c. A e = U(A n) = 0.9(5.68) = 5.11 in 2. d. Using Equation 4.2, find F t net = 0.5 F u = 0.5(58) = 29 ksi. e. Using Equation 4.5, find P = F t net × A e = 29(5.11) = 148 kips. 5. Conclusion: failure on the effective net area governs since 148 kips < 156 kips. The capacity (allowable load) is 148 kips. 6. We now can determine the thickness of the gusset plate, stressed in tension, with two lines of bolt holes per plate, using the Whitmore section to determine the effective width of the plate. As can be seen in Figure 4.12, the effective width, l w = 2(6)(tan 30°) + 3 = 9.9 in. The tensile capacity of the gusset plates may be based on either yielding of the gross area or rupture of the net area. First, the capacity based on yielding of the gross area of both plates is F t A g = 0.6(36)(2)(9.9 t p) = 428 t p kips. Next, the effective net area A e = (2)(9.9 – 2 × 7/8)t p = 16.3 t p in 2, which cannot exceed 85% of the gross area for small gusset plates; i.e., it must be no larger than 0.85(2)(9.9 t p) = 16.8 t p in 2. Therefore, the capacity based on rupture is 0.5(58)(16.3 t p) = 473 t p. Yielding governs, so the required thickness of the plate can be found by setting the required tensile capacity, 428 t p equal to the governing load of 148 kips, from which t p = 0.35 in. Rounding up, we select a 3/8-in. thick gusset plate with t p = 0.375 in. Example 4.2 Design steel tension element Problem definition. Select a W section bolted as shown in Figure 4.16 with 5/8 in. diameter bolts, and 3 bolts per line, to resist a tension force of 100 kips. Assume A36 steel. The effective bolt hole diameter = bolt diameter + 1/8 in. = 5/8 + 1/8 = 3/4 in. = 0.75 in. Figure 4.16: Net area diagram for Example 4.2 Solution overview. Find the required area based on net area capacity, assuming values for shear lag coefficient, U, and flange thickness, t f ; find required area based on gross area capacity; use the larger of the two area values to provisionally select a W section; check using "analysis" method if either U is smaller or t f is larger than assumed values. The area of the selected W section can be somewhat smaller than the "required" area if either U is larger or t f is smaller than assumed values — check using "analysis" method. Problem solution 1.Gross area: find required gross area based on yielding. From Equation 4.4, the required gross area, A g = P/F t gross = 100/(0.6 × 36) = 4.63 in 2. 2.Effective net area: find required gross area after determining effective net area based on rupture through failure surface (assume U = 0.9 and t f = 0.4 in.): a. From Equation 4.5, the required effective net area, A e = P/F t net = 100/(0.5 × 58) = 3.45 in 2. b. Working backwards, the required net area, A n = A e/U = 3.45/0.9 = 3.83 in 2. c. Finally, the required gross area can be computed: A g = A n + (bolt hole area) = 3.83 + 4(0.75 × 0.4) = 5.03 in 2. 3. Since 5.03 in 2> 4.63 in 2, the calculation based on effective net area governs, and a W section must be selected with A g ≥ 5.03 in 2. Many wide-flange shapes could be selected. FromAppendix Table A-4.3, the following candidates are among those that could be considered: a. Check a W8 × 18 with A g = 5.26 in 2. Two assumptions need to be tested: that U = 0.9, and that t f ≤ 0.4 in. From Appendix Table A-4.3, b f = 5.25 in., d = 8.14 in. and t f = 0.330 in. From Appendix Table A-4.9, the criteria for U = 0.9 requires that b f = 5.25 ≥ 0.67 d = 0.67(8.14) = 5.45 in. Since this condition is not met, we must use U = 0.85. Additionally, the flange thickness is different from our assumed value of 0.40 in., so that the calculation of net and effective area will change: A n = A g – (bolt hole area) = 5.26 – 4(0.75 × 0.330) = 4.27 in 2 and A e = U × A n = 0.85(4.27) = 3.63 in 2. The capacity based on rupture through the effective net area is P = F t net × A e = (0.5 × 58)(3.63) = 105 k. The capacity based on yielding on the gross area has already been found satisfactory (since the gross area of the W8 × 18 is greater or equal to the required gross area computed above). Therefore, the W8 × 18 is acceptable. b. Check a W6 × 20 with A g = 5.87 in 2. The same two assumptions need to be tested: that U = 0.9, and that t f ≤ 0.4. From Appendix Table A-4.3, b f = 6.02 in., d = 6.20 in. and t f = 0.365 in. From Appendix Table A-4.9, the criteria for U = 0.9 requires that b f = 6.02 ≥ 0.67 d = 0.67(6.20) = 4.15 in. Since this condition is met, and since its net area is greater than assumed (this is so because its flange thickness, t f , is less that the value assumed, so that the bolt hole area is less than assumed, and therefore the net area is greater than assumed), the W6 × 20 is acceptable. Both the W8 ×18 and the W6 × 20 would work, as would many other wide-flange shapes. Of the two sections considered, the W8 × 18 is lighter (based on the second number in the W-designation that refers to beam weight in pounds per linear foot), and therefore would be less expensive. Steel threaded rods Threaded rods are designed using an allowable tensile stress, F t = 0.375 F u , which is assumed to be resisted by the gross area of the unthreaded part of the rod. This value for the allowable stress is found by dividing the nominal rod tensile strength of 0.75 F u by a safety factor, Ω = 2.00. While there are no limits on slenderness, diameters are normally at least 1/500 of the length, and the minimum diameter rod for structural applications is usually set at 5/8 in. Assuming A36 steel, with F u = 58 ksi (Appendix Table A-4.1), the smallest acceptable rod with area, A = π(5/16)2 can support a tensile load, P = F t × A = 0.375 F u × π(5/16)2 = 21.75 × 0.3068 = 6.67 kips. Pin-connected plates Where plates are connected with a single pin, as shown in Figure 4.17, the net area, A n, is defined, not by the length, b, on either side of the pin hole, but rather by an effective length, b e = 2 t + 0.63 ≤ b, where t is the thickness of the plate (Figure 4.17 b): A n = 2 t × b e (4.6) The plate capacity in tension is governed by either yielding on the gross area or rupture on the net area, whichever is smaller (there is no effective net area in this case), with P gross = 0.6 F y × A g and P net = 0.5 F u × A n as before. It is possible to cut the plate at a 45° angle as shown in Figure 4.17, as long as length c is greater or equal to length a, which in turn must be greater or equal to 1.33 b e. Figure 4.17: (a) Overview of pin-connected plates; (a) rupture on net area and (a) shear failure Aside from failure in tensile rupture or yielding, a third limit state for pin-connected plates is shear failure, or the relative sliding of areas A sf as illustrated in Figure 4.17 c. In this case, the allowable shear stress is taken as 0.3 F u so that P shear = 0.3 F u × A sf. A fourth limit state for pin-connected plates is bearing, or compressive stress caused by the pin itself in direct contact with the adjacent plate. Here, the allowable stress on the projected area (A pb) of the pin that bears on the plate is 0.9 F y so that the allowable bearing strength is 0.9 F y A pb. All four limit states must be checked, with the capacity of the pin-connected plate determined by the lowest of the four limits. Values for yield and ultimate stress used in these calculations, F y and F u , are listed in Appendix Table A-4.1. For such pin-connected plates, as well as for all other bolted connections, the fasteners themselves, and not only the stresses they produce on the elements being joined, must also be checked. This aspect of structural element design is discussed more thoroughly in the section of this chapter dealing with steel connections. « previous section \| next section » \| contents unit abbreviations and conversion calculator \| references \| glossary © 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.
6155	https://peditools.org/growthwho/	PediTools PediTools Bilirubin 2022 Growth Fenton 2013 Growth Fenton 2025 Fenton 2013 Growth Chart Fenton 2025 Growth Chart Growth WHO 0 - 24m Growth CDC 0 - 36m Growth CDC 2 - 20y GA Calc Growth Olsen 2010 BMI Olsen 2015 Down Syndrome 0 - 36m Down Syndrome 2 - 20y CDC arm circ 2m - 18y Mramba arm circ 5y - 19y WHO arm+skinfold 3 - 60m CDC skinfold 1.5 - 20y PN Calc Bilirubin 2004 (old) Growth Fenton 2003 (old) Off site linksNICHD Outcomes NCBI PubMed Infection risk What's new About PediTools Contact us Sitemap iOS Fenton 2013 PediTools Clinical tools for pediatric providers Menu PediTools Bilirubin 2022 Growth Fenton 2013 Growth Fenton 2025 Fenton 2013 Growth Chart Fenton 2025 Growth Chart Growth WHO 0 - 24m Growth CDC 0 - 36m Growth CDC 2 - 20y GA Calc Growth Olsen 2010 BMI Olsen 2015 Down Syndrome 0 - 36m Down Syndrome 2 - 20y CDC arm circ 2m - 18y Mramba arm circ 5y - 19y WHO arm+skinfold 3 - 60m CDC skinfold 1.5 - 20y PN Calc Bilirubin 2004 (old) Growth Fenton 2003 (old) Off site links Off site linksNICHD Outcomes NCBI PubMed Infection risk What's new About PediTools Contact us Sitemap iOS Fenton 2013 Tips and Tricks Valid ages from 0 to 24 months Weight: grams assumed. To enter pounds and ounces, enter #-#; e.g., 8-4 for 8 lb and 4 oz Head circumference: cm assumed. To enter inches, enter #"; e.g., 19" for 19 inches Height: cm assumed. To enter feet and inches, enter #' #"; e.g., 4' 5" for 4 feet 5 inches Gestational age, enter in the form of ## #/7; e.g., 30 3/7 for 30 weeks and 3 days gestation Based on Source: WHO Growth Standard - Data Tables WHO Growth Standard for 0 to 24 months Uses the 2006 WHO growth standard charts to report percentiles and Z-scores on infants from 0 to 24 months of age. Per the CDC, WHO Growth Standards are recommended for use in the U.S. for infants and children 0 to 2 years of age In 2006, the World Health Organization (WHO) released an international growth standard which describes the growth of children living in environments believed to support optimal growth. Citing: If you use PediTools for a publication or clinical guideline, please consider citing: Chou JH et al., J Med Internet Res 2020;22(1):e16204 (available open access [PDF])
6156	https://stackoverflow.com/questions/69511305/how-to-simply-compute-the-travel-time-from-one-point-to-an-other-without-a-plo	python - How to simply compute the travel time from one point to an other? (Without a plot) - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to simply compute the travel time from one point to an other? (Without a plot) Ask Question Asked 3 years, 11 months ago Modified3 years, 11 months ago Viewed 4k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I spent a lot of time reading and testing the example notebooks of OSMnx but I couldn't figure out a way to simply calculate the travel time from a given point (GPS coordonates) to an other one. I would like to estimate, for each point from my list, how long it takes to go to a specific point (sometimes 100km away). I don't need to generate a graph/map/plot, as I only need the duration of each trip (and I think that OSMnx maps render better at a city-scale). I am pretty desperate as I could not find a simple way to do this across different Python libraries... If doing this calculation for +-10k points within a country-scale map is asking too much from OSMnx, could a locally stored pbf file of the country be helpful for another solution? python osmnx osrm Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Oct 10, 2021 at 0:23 Trag0Trag0 asked Oct 9, 2021 at 23:55 Trag0Trag0 35 1 1 silver badge 7 7 bronze badges 10 Try this? movable-type.co.uk/scripts/latlong.htmlJames –James 2021-10-10 00:09:29 +00:00 Commented Oct 10, 2021 at 0:09 Thank you, but I am looking for the travel time (by walking, cycling or driving) and this seems to only measure distance?Trag0 –Trag0 2021-10-10 00:26:41 +00:00 Commented Oct 10, 2021 at 0:26 If you only have GPS points then you can't directly calculate these details. You need to know the allowed speeds. Walking can be estimated by using the average walk speed. It depends on what other information you can get about the location point. Is the information in JSON with other details or is it just a pure GPS location?James –James 2021-10-10 09:07:41 +00:00 Commented Oct 10, 2021 at 9:07 I know you don't want to hear it but have you considered using Google Maps API instead? Google have invested huge amounts of money in mapping all street details for most of the world. Your query against the data is simple in python using Google Maps API. These maps are kept up to date. medium.com/future-vision/…James –James 2021-10-10 09:35:26 +00:00 Commented Oct 10, 2021 at 9:35 @James What I have is pure GPS location of my points. I looked at Google Maps API but there is a ratelimit, that is why I wanted to use OSRM or Graph Hopper (the APIs used by Opens Street Map), as they also seem pretty accurate (and open!)Trag0 –Trag0 2021-10-10 12:49:26 +00:00 Commented Oct 10, 2021 at 12:49 \|Show 5 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. There are inherent trade-offs when you want to model a large study area such as an entire region or an entire country: 1) model precision vs 2) area size vs 3) memory/speed. You need to trade off one of these three. For the first, you can model a coarser-grained network, such as only major roads in the region/country, rather than millions of fine-grained residential streets and paths. For the second, you can study a smaller area. For the third, you can provision a machine with lots of memory and then let the script run for a while to complete the process. What you trade off will be up to your own needs for this analysis. In the example code below, I chose to trade off #1: I've modeled this region (West Midlands) by its motorways and trunk roads. Given a different analytical goal, you may trade off other things instead. After creating the model, I randomly sample 1000 origin and destination lat-long points, snap them to the nearest nodes in the graph, and solve the shortest paths by travel time (accounting for speed limits) with multiprocessing. ```python import osmnx as ox get boundaries of West Midlands region by its OSM ID gdf = ox.geocode_to_gdf('R151283', by_osmid=True) polygon = gdf.iloc['geometry'] get network of motorways and trunk roads, with speed and travel time cf = '["highway"~"motorway\|motorway_link\|trunk\|trunk_link"]' G = ox.graph_from_polygon(polygon, network_type='drive', custom_filter=cf) G = ox.add_edge_speeds(G) G = ox.add_edge_travel_times(G) randomly sample lat-lng points across the graph origin_points = ox.utils_geo.sample_points(ox.get_undirected(G), 1000) origin_nodes = ox.nearest_nodes(G, origin_points.x, origin_points.y) dest_points = ox.utils_geo.sample_points(ox.get_undirected(G), 1000) dest_nodes = ox.nearest_nodes(G, dest_points.x, dest_points.y) %%time solve 1000 shortest paths between origins and destinations minimizing travel time, using all available CPUs paths = ox.shortest_path(G, origin_nodes, dest_nodes, weight='travel_time', cpus=None) elapsed time: 9.8 seconds ``` For faster modeling, you can load the network data from a .osm XML file instead of having to make numerous calls to the Overpass API. OSMnx by default divides your query area into 50km x 50km pieces, then queries Overpass for each piece one a time to not exceed the server's per-query memory limits. You can configure this max_query_area_size parameter, as well as the server memory allocation, if you prefer to use OSMnx's API querying functions rather than its from-file functionality. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 14, 2021 at 19:23 gboeinggboeing 6,570 2 2 gold badges 17 17 silver badges 45 45 bronze badges 2 Comments Add a comment Trag0 Trag0Over a year ago Thank you for your answer! Is there a way, within a region loaded as you did with West Midlands, to get the shortest_path between two precise latlon points (and not randomly sampled ones)? All the best 2021-10-20T06:43:50.467Z+00:00 0 Reply Copy link gboeing gboeingOver a year ago Yes. Just pass whatever coordinates you want into the nearest_nodes function. 2021-10-20T14:58:31.673Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python osmnx osrm See similar questions with these tags. 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6157	https://math.stackexchange.com/questions/706446/solve-the-puzzle-how-many-liars	combinatorics - solve the puzzle how many liars? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more solve the puzzle how many liars? Ask Question Asked 11 years, 6 months ago Modified11 years, 6 months ago Viewed 3k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Each boy in a group of 20 20 boys either always tells thuth or always tells a lie. These boys are sitting around a table. Each boy says that his neighbours are liars. Prove that at least 7 7 out of 20 20 must be truth tellers. combinatorics puzzle Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 10, 2014 at 8:25 Marc van Leeuwen 120k 8 8 gold badges 183 183 silver badges 373 373 bronze badges asked Mar 10, 2014 at 7:56 Murtuza VadhariaMurtuza Vadharia 1,654 4 4 gold badges 18 18 silver badges 33 33 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. If there are ≤6≤6 truth tellers, then there exist 3 3 liars sitting together by pigeon-hole principle. Now the one in the middle of these three is telling the truth when he/she says "my neighbours are liars", a contradiction! Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 10, 2014 at 8:25 Fei GaoFei Gao 406 3 3 silver badges 4 4 bronze badges 5 Would you mind explaining exactly what are the pigeons and what are the pigeonholes?Marc van Leeuwen –Marc van Leeuwen 2014-03-10 08:26:31 +00:00 Commented Mar 10, 2014 at 8:26 @MarcvanLeeuwen Indeed L L T L L T L L T L L T L L T L L T L L L L T L L T L L T L L T L L T L L T L L ...Mark Bennet –Mark Bennet 2014-03-10 08:31:16 +00:00 Commented Mar 10, 2014 at 8:31 @MarcvanLeeuwen pigeon: liars, which is 14; hole: sits between two consecutive truth tellers, which is 6; So there exist two consecutive truth tellers which have at least ⌈14/6⌉=3⌈14/6⌉=3 liars sitting between them.Fei Gao –Fei Gao 2014-03-10 08:44:01 +00:00 Commented Mar 10, 2014 at 8:44 @FeiGao: Thanks, I understand now. The pigeonholes are the gaps between two truth tellers, of which there are at most 6 6 (since the table is round)Marc van Leeuwen –Marc van Leeuwen 2014-03-12 14:45:14 +00:00 Commented Mar 12, 2014 at 14:45 it is possible that k in proof given by me at last is equal to 7. Placing 7 truthtellers around the table, there would be 7 gaps between them. We put one liar in one gap and two liars in each other gaps.Murtuza Vadharia –Murtuza Vadharia 2014-03-13 09:41:44 +00:00 Commented Mar 13, 2014 at 9:41 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. A liar tells that both his neighbours are liars. Therefore at least 1 neighbour of each liar is truthful. Therefore no three successive boys can be liars. Each truth teller's both neighbour are liars, so no two successive boys are truthful. Suppose there are k k truth tellers. Around the table, between pairs of successive truth tellers there are k k gaps. In each gap, there must be at least one liar (otherwise, the successive boys would be truthful) and at most two liars (otherwise, there would be 3 3 successive liars). Therefore the number of liars is at least k k and at most 2 k 2 k. So total number of boys is at least k+k=2 k k+k=2 k and at most k+2 k=3 k k+2 k=3 k. If k≤6 k≤6, then 3 k≤18 3 k≤18 but we have 20 20 boys. Therefore k k is greater than or equal to 7 7. Hence proved. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 14, 2014 at 8:30 community wiki 2 revs, 2 users 62%Murtuza Vadharia Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. No three consecutive boys can be liars, or else the middle one would not say both his neighbours are liars. So every boy either tells the truth or has a neighbour that tells the truth (since the table is round, everybody has two neighbours). If there were at most 6 6 truth tellers, that would account at most for those 6 6, plus their 12 12 neighbours; their might be overlap, but in any case this cannot cover all 20 20 boys, contradiction Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 12, 2014 at 14:43 answered Mar 10, 2014 at 8:24 Marc van LeeuwenMarc van Leeuwen 120k 8 8 gold badges 183 183 silver badges 373 373 bronze badges 3 1 its not a contradiction Murtuza Vadharia –Murtuza Vadharia 2014-03-12 14:21:02 +00:00 Commented Mar 12, 2014 at 14:21 @MurtuzaVadharia: If there are at most 6 6 truth tellers, at most 18 18 people are either a truth teller or next to a truth teller. This contradicts the conclusion above there every one of the 20 20 boys tells the truth or has a neighbour that tells the truth.Marc van Leeuwen –Marc van Leeuwen 2014-03-12 14:43:00 +00:00 Commented Mar 12, 2014 at 14:43 1 it is possible that k in proof given by me at last is equal to 7. Placing 7 truthtellers around the table, there would be 7 gaps between them. We put one liar in one gap and two liars in each other gaps.Murtuza Vadharia –Murtuza Vadharia 2014-03-13 09:43:39 +00:00 Commented Mar 13, 2014 at 9:43 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics puzzle See similar questions with these tags. 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6158	https://www3.risc.jku.at/publications/download/risc_4367/modlifts.pdf	A METHOD FOR DETERMINING THE MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES, WITH APPLICATIONS TO SUBGROUP COUNTING M. KAUERS†, C. KRATTENTHALER‡, AND T. W. M¨ ULLER∗ Abstract. We present a method to obtain congruences modulo powers of 2 for se-quences given by recurrences of ﬁnite depth with polynomial coeﬃcients. We apply this method to Catalan numbers, Fuß–Catalan numbers, and to subgroup counting functions associated with Hecke groups and their lifts. This leads to numerous new results, including many extensions of known results to higher powers of 2. 1. Introduction Ever since the work of Sylow , Frobenius [11, 12], and P. Hall , the study of congruences for subgroup numbers and related numerical quantities of groups has played an important role in group theory. Divisibility properties of subgroup numbers of (ﬁnitely generated) inﬁnite groups may to some extent be viewed as some kind of analogue to these classical results for ﬁnite groups. To the best of our knowledge, the ﬁrst signiﬁcant result in this direction was obtained by Stothers : the number of index-n-subgroups in the inhomogeneous modular group PSL2(Z) is odd if, and only if, n is of the form 2k −3 or 2k+1 −6, for some positive integer k ≥2. A diﬀerent proof of this result was given by Godsil, Imrich, and Razen . The systematic study of divisibility properties of subgroup counting functions for inﬁnite groups begins with . There, the parity of subgroup numbers and the number of free subgroups of given ﬁnite index are determined for arbitrary Hecke groups H(q) = C2 ∗Cq with q ≥3. Subsequently, the results of were generalised to larger classes of groups and arbitrary prime modulus in [3, 18, 23, 24, 26]. A ﬁrst attempt at obtaining congruences modulo higher prime powers was made in , where the behaviour of subgroup numbers in PSL2(Z) ∼ = H(3) is investigated modulo 8 and a congruence modulo 16 is derived for the number of free subgroups of given index in PSL2(Z). A common feature of all the above listed sequences of subgroup numbers is that they obey recurrences of ﬁnite depth with polynomial coeﬃcients. The purpose of this paper is to present a new method for determining congruences modulo arbitrarily large 2000 Mathematics Subject Classiﬁcation. Primary 20E06; Secondary 05A15 05E99 11A07 20E07 33F10 68W30. Key words and phrases. Polynomial recurrences, symbolic summation, subgroup numbers, free sub-group numbers, Catalan numbers, Fuß–Catalan numbers. †Research supported by the Austrian Science Foundation FWF, grant Y464-N18 ‡Research partially supported by the Austrian Science Foundation FWF, grants Z130-N13 and S9607-N13, the latter in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory” ∗Research supported by Lise Meitner Fellowship M1201-N13 of the Austrian Science Foundation FWF. 1 2 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER powers of 2 for sequences described by such recurrences. Our method is inspired by the observation that many of the afore-mentioned results say in essence that the generating function for the subgroup numbers under consideration, when reduced modulo a 2-power, can be expressed as a polynomial in the basic series Φ(z) = X n≥0 z2n (1.1) with coeﬃcients that are Laurent polynomials in z. What our method aﬀords is an algorithmic procedure to ﬁnd such polynomial expressions, provided they exist. By applying our method to Catalan numbers, to (certain) Fuß–Catalan numbers, and to various subgroup counting problems in Hecke groups and their lifts, we obtain far-reaching generalisations and extensions of the previously mentioned results. In order to give a concrete illustration, the recent result [20, Theorems 6.1–6.6] of Liu and Yeh determining the behaviour of Catalan numbers Catn modulo 64 can be compactly written in the form ∞ X n=0 Catn zn = 32z5 + 16z4 + 6z2 + 13z + 1 + 32z4 + 32z3 + 20z2 + 44z + 40 Φ(z) + 16z3 + 56z2 + 30z + 52 + 12 z Φ2(z) + 32z3 + 60z + 60 + 28 z Φ3(z) + 32z3 + 16z2 + 48z + 18 + 35 z Φ4(z) + 32z2 + 44 Φ5(z) + 48z + 8 + 50 z Φ6(z) + 32z + 32 + 4 z Φ7(z) modulo 64, (1.2) as may be seen by a straightforward (but rather tedious) computation. Our method can not only ﬁnd this result, but it produces as well corresponding formulae modulo any given power of 2 in a completely automatic fashion, see Theorems 11 and 12 in Section 5. In a sense, which is made precise in Section 4, our method is very much in the spirit of Doron Zeilberger’s philosophy that mathematicians should train computers to automat-ically produce theorems. Indeed, Theorems 11, 17, 19, 30, 33 imply that our algorithm is able to produce a theorem on the behaviour modulo any given 2-power of the subgroup counting functions featuring in these theorems, and, if fed with a concrete 2-power, our implementation will diligently output the corresponding result (provided the input does not cause the available computer resources to be exceeded . . . ). Moreover, when discussing subgroup numbers of lifts of PSL2(Z), (such as the homogeneous modular group SL2(Z)), a crucial role is also played by an application of the holonomic func-tions approach to ﬁnding recurrences for multi-variate hypergeometric sums, pioneered by Wilf and Zeilberger [35, 36], and further developed in [4, 5, 17]. The rest of this introduction is devoted to a more detailed description of the contents of this paper. In Section 2 we discuss our main character, the formal power series Φ(z) deﬁned in (1.1). While Φ(z) is transcendental over Q, it is easy to see that it is algebraic modulo powers of 2. The focus in that section is on polynomial identities for Φ(z) of minimal degree. Then, in Section 4, we describe our method of expressing the generating function of recursive sequences, when reduced modulo a given 2-power, as MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 3 a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. The method relies in an essential way on the polynomial identities from Section 2. The problem how to extract the explicit value of a concrete coeﬃcient in a polynomial expression in Φ(z) (such as (1.2)) modulo a given 2-power is discussed in Section 3, where we present an eﬃcient algorithm performing this task. This algorithm is of theoretical value (minimal length relations between powers of Φ(z) such as the ones in Proposition 2 are established by applying this algorithm to the powers of Φ(z); see also Appendix A) as well as of practical signiﬁcance, as is demonstrated by the derivations of Theorems 25 and 28. As a ﬁrst illustration of our method, we apply it to Catalan numbers, thereby sig-niﬁcantly improving numerous earlier results in the literature; see Section 5. This is contrasted in Section 6 with an example (concerning particular Fuß–Catalan numbers) where our method is bound to fail. The reason is spelled out in Theorem 13, which, at the same time, also gives a new description for the parity pattern of the numbers of free subgroups of index n in the Hecke group H(7). The subsequent sections contain several applications of our method to the problem of determining congruences modulo a given 2-power for numbers of subgroups of Hecke groups H(q) and their lifts Γm(q) = C2m ∗ Cm Cqm = x, y x2m = yqm = 1, x2 = yq , m ≥1. (1.3) Ubiquitous in these applications is — explicitly or implicitly — the intimate relation between subgroup numbers of a group Γ and numbers of permutation representations of Γ, in the form of identities between the corresponding generating functions. This is directly visible in the folklore result (11.1) (which is not only used in Sections 11, 12, and 14, but is also behind the crucial diﬀerential equation (9.1) in Section 9; cf. its derivation in ), and also indirectly in Lemma 15 via the A-invariants Aµ(H(q)), see [25, Sec. 2.2]. In the cases relevant here, the numbers of permutation representations of Γ satisfy linear recurrences with polynomial coeﬃcients — to make these explicit may require the algorithmic machinery around the “holonomic paradigm” (cf. [4, 5, 17, 21, 29, 35, 36]), see Sections 11 and 12 for corresponding examples. Via the afore-mentioned generating function relation, such a recurrence can be translated into a Riccati-type diﬀerential equation for the generating function of the subgroup numbers that we are interested in. It is here, where our method comes in: it is tailor-made for being applied to formal power series F(z) satisfying this type of diﬀerential equation, and it aﬀords an algorithmic procedure to ﬁnd a polynomial in Φ(z) which agrees, after reduction of the coeﬃcients of F(z) modulo a given power of 2, with the power series F(z). We start in Section 7 with free subgroup numbers of lifts Γm(q), for primes q ≥3, where we prepare the ground for application of our method. More speciﬁcally, in Proposition 16 we present a lower bound for the 2-adic valuation of the number of free subgroups of index n in Γm(q), where q is a Fermat prime. In particular, this result implies that the sequence of free subgroup numbers under consideration is essentially zero modulo a given 2-power in the case when m is even. In Section 8, we show that our method provides an algorithm for determining these numbers of free subgroups of Γm(3) modulo any given 2-power in the case when m is odd. The corresponding results (see Theorems 17 and 18) go far beyond the previous result [27, Theorem 1] on the behaviour of the number of free subgroups of PSL2(Z) modulo 16. Our method provides as well an algorithm for determining the number of all subgroups of index n in PSL2(Z) modulo 4 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER powers of 2, as we demonstrate in Section 9. Not only are we able to provide a new proof of Stothers’ result (which was stated in the second paragraph above), but our method leads as well to reﬁnements modulo arbitrary powers of 2 of Stothers’ result and of the mod-8 results in [27, Theorem 2] mentioned earlier; see Theorems 19 and 22. For the homogeneous modular group SL2(Z) (being isomorphic to the lift Γ2(3)) and for the lift Γ3(3), however, our method works up to modulus 8 = 23, but fails for the modulus 16 = 24. Still, the corresponding results (see Theorems 24 and 27) go signiﬁcantly beyond the earlier parity results [18, Eq. (6.3) with \|H\| = 1] for these groups. This is explained in Sections 11 and 12, with Section 10 preparing the ground by providing formulae for the number of permutation representations of SL2(Z) as well as other lifts of PSL2(Z). A further example where our method works for any 2-power is the subject of Section 13: there we apply the method to a functional equation (see (13.1)) extending the functional equation for Catalan numbers (producing Fuß–Catalan numbers) and show that it works for any given 2-power; see Theorem 30. In fact, we apply a variation of the method here, in that the basic series Φ(z) gets replaced by a slightly diﬀerent series, which we denote by Φh(z) (see (13.3)). If Theorem 30 is combined with results from , then it turns out that our method provides as well an algorithm for determining the number of free subgroups of index n in a Hecke group H(q) and its lifts, where q is a Fermat prime, modulo any given 2-power; see Corollary 31. The same assertion holds as well for the problem of determining the number of subgroups of index n in the Hecke group H(5), again modulo any given 2-power (see Theorems 33 and 34 in Section 14). We conjecture that the same is true for Hecke groups H(q), with q a Fermat prime (see Conjecture 35). The above discussed results in Sections 13 and 14 largely generalise the parity results [25, Cor. A’, respectively Theorem B] for subgroup numbers of Hecke groups, although our results are not independent, in the sense that we base our analyses on prior results from . Concluding the introduction, we remark that there is no principal obstacle to gen-eralising our method to other basic series and moduli. For example, one may think of analysing the behaviour of recursive sequences modulo powers of any prime p in terms of the obvious generalisation of Φ(z), i.e., the series P n≥0 zpn. It is in fact not diﬃcult to see that our results from Sections 5 and 13 for Fuß–Catalan numbers characterised by the functional equation (13.1) for their generating function have rather straightforward analogues for Fuß–Catalan numbers whose generating function satisﬁes the functional equation zf ph(z) −f(z) + 1 = 0. (1.4) However we are not aware of any applications of this (or of variants) to congruence properties of subgroup numbers modulo powers of primes p diﬀerent from 2. In fact, the known results (cf. or [27, Theorem 3]) point strongly to the fact that the phenomena that appear modulo primes diﬀerent from 2 cannot be captured by series of the type P n≥0 zpn. So, currently, we do not know of interesting applications in this direction, but we hope to be able to return to this circle of ideas in future publications. 2. The 2-power series Φ(z) Here we consider the formal power series Φ(z) deﬁned in (1.1). This series is the principal character in the method for determining congruences of recursive sequences which we describe in the next section. It is well known that this series is transcendental MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 5 over Q (this follows for instance from the density argument used in the proof of Lemma 1 below). However, if the coeﬃcients of Φ(z) are considered modulo a 2-power 2γ, then Φ(z) obeys a polynomial relation with coeﬃcients that are polynomials in z. The focus in this section is on what can be said about such polynomial relations, and, in particular, about those of minimal length. Here and in the sequel, given power series f(z) and g(z), we write f(z) = g(z) modulo 2γ to mean that the coeﬃcients of zi in f(z) and g(z) agree modulo 2γ for all i. We say that a polynomial A(z, t) in z and t is minimal for the modulus 2γ, if it is monic (as a polynomial in t), has integral coeﬃcients, satisﬁes A(z, Φ(z)) = 0 modulo 2γ, and there is no monic polynomial B(z, t) of t-degree less than that of A(z, t) with B(z, Φ(z)) = 0 modulo 2γ. Furthermore, we let v2(α) denote the 2-adic valuation of the integer α, that is, the maximal exponent e such that 2e divides α. The lemma below provides a lower bound for the degree of a polynomial that is minimal for the modulus 2γ. Lemma 1. If A(z, t) minimal for the modulus 2γ, then the degree d of A(z, t) in t satisﬁes v2(d!) ≥γ. In particular, the series Φ(z) is transcendental over Q. Proof. We introduce the following density function with respect to a given modulus 2γ for a power series f(z) in z: D(f, 2γ; n) := {e : 2n−1 ≤e < 2n and ⟨ze⟩f(z) ̸≡0 modulo 2γ} , n = 1, 2, . . . , (2.1) where ⟨ze⟩f(z) denotes the coeﬃcient of ze in f(z). Setting Em(z) := X n1>···>nm≥0 z2n1+2n2+···+2nm, (2.2) simple counting yields that D(Em, 2γ; n) = n −1 m −1 ∼ 1 (m −1)!nm−1, as n →∞. (2.3) (The modulus 2γ does not play a role here.) Furthermore, by considering the binary representations of possible exponents e such that the coeﬃcient of ze in Φm(z) does not vanish, we have D(Φm, 2γ; n) = O nm−1 , as n →∞. (2.4) Indeed, by direct expansion, we see that Φm(z) = m! Em(z) + Rm(z), (2.5) where D(Rm, 2γ; n) = O(nm−2) as n →∞. Let us, by way of contradiction, suppose that the degree d of A(z, t) satisﬁes v2(d!) < γ. Considering (2.5) with m = d, we see that Ed(z) appears in Φd(z) with a non-zero coeﬃcient modulo 2γ. Furthermore, the other terms in Φd(z) (denoted by Rd(z) in (2.5)) have a density function modulo 2γ which is asymptotically strictly smaller than the density function of Ed(z). Consequently, if we remember (2.3), we have D(Φd, 2γ; n) ∼ d! (d −1)!nd−1 = dnd−1, as n →∞. 6 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Moreover, by (2.4), all powers Φm(z) with m < d have density function modulo 2γ which is asymptotically strictly smaller than nd−1. Altogether, it is impossible that a linear combination of powers Φm(d), m = 0, 1, . . . , d, with coeﬃcients that are polynomials in z sums up to zero modulo 2γ, a contradiction to our assumption that d is the degree of a minimal polynomial A(z, t). The particular statement is an immediate consequence of the inequality just proven. □ Proposition 2. Minimal polynomials for the moduli 2, 4, 8, 16, 32, 64, 128 are t2 + t + z modulo 2, (t2 + t + z)2 modulo 4, t4 + 6t3 + (2z + 3)t2 + (2z + 6)t + 2z + 5z2 modulo 8, (t2 + t + z)(t4 + 6t3 + (2z + 3)t2 + (2z + 6)t + 2z + 5z2) modulo 16, (t4 + 6t3 + (2z + 3)t2 + (2z + 6)t + 2z + 5z2)2 modulo 32, (t4 + 6t3 + (2z + 3)t2 + (2z + 6)t + 2z + 5z2)2 modulo 64, t8 + 124t7 + t6(68z + 18) + t5(124z + 24) + t4 62z2 + 64z + 81 + t3 20z2 + 76z + 28 + t2 116z3 + 114z2 + 12z + 92 + t 116z3 + 28z2 + 8z + 16 + 9z4 + 124z3 + 12z2 + 112z modulo 128. Proof. In order to be consistent with Section 3, let us change notation and write H1,1,...,1(z) := X n1>···>nm≥0 z2n1+2n2+···+2nm (with m occurrences of 1 in H1,1,...,1(z)). Note that the above series is identical with the series which we earlier denoted by Em(z). Straightforward calculations yield that Φ2(z) = Φ(z) + 2H1,1(z) −z, (2.6) Φ3(z) = −2 X n≥0 z3·2n + 3(1 −z)Φ(z) + 6H1,1(z) + 6H1,1,1(z) −3z, (2.7) Φ4(z) = −12 X n≥0 z3·2n −8 X n1>n2≥0 z3·2n1+2n2 −8 X n1>n2≥0 z2n1+3·2n2 + (13 −18z)Φ(z) + (30 −12z)H1,1(z) + 36H1,1,1(z) + 24H1,1,1,1(z) + 5z2 −13z. (2.8) In particular, relation (2.6), together with Lemma 1, immediately implies the claims about minimal polynomials for the moduli 2 and 4. Moreover, a simple computation using (2.6)–(2.8) shows that Φ4(z) + 6Φ3(z) + (2z + 3)Φ2(z) + (2z + 6)Φ(z) + 2z + 5z2 = 0 modulo 8. (2.9) Together with Lemma 1, this establishes the claims about minimal polynomials for the moduli 8, 16, 32, and 64. In order to prove the claim for the modulus 128, one uses the expressions Φi(z), i = 2, 3, . . . , 8 given above and in Appendix A. □ Based on these observations, we propose the following conjecture. Conjecture 3. The degree of a minimal polynomial for the modulus 2γ, γ ≥1, is the least d such that v2(d!) ≥γ. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 7 Remark 4. (1) Given the binary expansion of d, say d = d0 + d1 · 2 + d2 · 4 + · · · + dr · 2r, 0 ≤γi ≤1, by the well-known formula of Legendre [19, p. 10], we have v2(d!) = ∞ X ℓ=1 d 2ℓ = ∞ X ℓ=1 $ r X i=0 di2i−ℓ % = ∞ X ℓ=1 r X i=ℓ di2i−ℓ = r X i=1 i X ℓ=1 di2i−ℓ= r X i=1 di 2i −1 = d −s(d), (2.10) where s(d) denotes the sum of digits of d in its binary expansion. Consequently, an equivalent way of phrasing Conjecture 3 is to say that the degree of a minimal polyno-mial for the modulus 2γ is the least d with d −s(d) ≥γ. (2) We claim that, in order to establish Conjecture 3, it suﬃces to prove that, for each δ ≥1, there is a polynomial Aδ(z, t) of degree 2δ such that Aδ(z, Φ(z)) = 0 modulo 22δ−1. (2.11) For, arguing by induction, let us suppose that we have already constructed A1(z, t), A2(z, t), . . . , Am(z, t) satisfying (2.11). Let α = α1 · 2 + α2 · 4 + · · · + αm · 2m, 0 ≤αi ≤1, be the binary expansion of the even positive integer α. In this situation, we have m Y δ=1 Aαδ δ (z, Φ(z)) = 0 modulo m Y δ=1 2αδ(2δ−1) = m Y δ=1 2αδ(2δ−1) = 2α−s(α). (2.12) On the other hand, the degree of the left-hand side of (2.12) as a polynomial in Φ(z) is Pm δ=1 αδ2δ = α. Let us put these observations together. In view of (2.10), Lemma 1 says that the degree of a minimal polynomial for the modulus 2γ cannot be smaller than the least integer, d(γ) say, for which d(γ)−s(d(γ)) ≥γ. (We remark that d(γ) must be automatically even.) If we take into account that the quantity α −s(α), as a function in α, is weakly monotone increasing in α, then (2.12) tells us that, as long d(γ) ≤2 + 4 + · · · + 2m = 2m+1 −2, we have found a monic polynomial of degree d(γ), Bγ(z, t) say, for which Bγ(z, Φ(z)) = 0 modulo 2γ, namely the left-hand side of (2.12) with α replaced by d(γ), to wit Bγ(z, t) = m Y δ=1 A d(γ) δ δ (z, t), where d(γ) = d(γ) 1 · 2 + d(γ) 2 · 4 + · · · + d(γ) m · 2m is the binary expansion of d(γ). Hence, it must necessarily be a minimal polynomial for the modulus 2γ. Since (2m+1−2)−s(2m+1−2) = 2m+1−2−m, we have thus found minimal polynomials for all moduli 2γ with γ ≤2m+1−m−2. Now we should note that the quantity α−s(α) makes a jump from 2m+1 −m −2 to 2m+1 −1 when we move from α = 2m+1 −2 to α = 2m+1 (the reader should recall that it suﬃces to consider even α). If we take A2 m(z, t), which has degree 2 · 2m = 2m+1, then, by (2.11), we also have a minimal polynomial for the modulus 22m−12 = 22m+1−2 and, in view of the preceding remark, as well for all moduli 2γ with γ between 2m+1 −m −1 and 2m+1 −2. 8 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER So, indeed, the ﬁrst modulus for which we do not have a minimal polynomial is the modulus 22m+1−1. This is the role which Am+1(z, t) (see (2.11) with m + 1 in place of δ) would have to play. The arguments above show at the same time that, supposing that we have already constructed A1(z, t), A2(z, t), . . . , Am(z, t), the polynomial A2 m(z, t) is a very close “ap-proximation” to the polynomial Am+1(z, t) that we are actually looking for next, which is only “oﬀ” by a factor of 2. In practice, one can recursively compute polynomials Aδ(z, t) satisfying (2.11) by following the procedure outlined in the next-to-last para-graph before Lemma 5 in the next section. It is these computations (part of which are reported in Proposition 2) which have led us to believe in the truth of Conjecture 3. 3. Coefficient extraction from powers of Φ(z) In the next section we are going to describe a method for expressing formal power series satisfying certain diﬀerential equations, after the coeﬃcients of the series have been reduced modulo 2k, as polynomials in the 2-power series Φ(z) (which has been discussed in the previous section; for the deﬁnition see (1.1)), with coeﬃcients being Laurent polynomials in z. Such a method would be without value if we could not, at the same time, provide a procedure for extracting coeﬃcients from powers of Φ(z). The description of such a procedure is the topic of this section. Clearly, a brute force expansion of a power ΦK(z), where K is a given positive integer, yields ΦK(z) = K X r=1 X a1,...,ar≥1 a1+···+ar=K K! a1! a2! · · ·ar!Ha1,a2,...,ar(z), (3.1) where Ha1,a2,...,ar(z) := X n1>n2>···>nr≥0 za12n1+a22n2+···+ar2nr. The expansion (3.1) is not (yet) suited for our purpose, since, when a1, a2, . . . , ar vary over all possible choices such that their sum is K, the series Ha1,a2,...,ar(z) are not linearly independent over the ring Q[z, z−1] of Laurent polynomials in z over the rational numbers1, and, second, coeﬃcient extraction from a series Ha1,a2,...,ar(z) can be a hairy task if some of the ai’s should have non-vanishing 2-adic valuation. However, we shall show (see Corollary 6) that, if we restrict to odd ai’s, then the corresponding series Ha1,a2,...,ar(z), together with the (trivial) series 1, are linearly in-dependent over Q[z, z−1], and there is an eﬃcient algorithm to express all other series Hb1,b2,...,bs(z), where we do not make any restriction on the bi’s, as a linear combination over Q[z, z−1] of 1 and the former series (see Lemma 8). Since coeﬃcient extraction from a series Ha1,a2,...,ar(z) with all ai’s odd is straightforward (see Remark 7), this solves the problem of coeﬃcient extraction from powers of Φ(z). As a side result, the procedure which we described in the previous paragraph, and which will be substantiated below, provides all the means for determining minimal polynomials in the sense of Section 2: as we explained in Item (2) of Remark 4 at the end of that section, it suﬃces to ﬁnd a minimal polynomial for the modulus 22δ−1, δ = 1, 2, . . . . For doing this, we would take the minimal polynomial Aδ−1(z, t) for 1The same is true for an arbitrary ﬁeld in place of the ﬁeld Q of rational numbers. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 9 the modulus 22δ−1−1, expand the square A2 δ−1(z, t), and replace each coeﬃcient cα,β of a monomial zαtβ in A2 δ−1(z, t) by cα,β + 22δ−2xα,β, where xα,β is a variable, thereby obtaining a modiﬁed polynomial, Bδ−1(z, t) say. Now we would substitute Φ(z) for t, so that we obtain Bδ−1(z, Φ(z)). Here, we express powers of Φ(z) in terms of the series Ha1,a2,...,ar(z) with all ai’s being odd, and collect terms. By reading the coeﬃcients of zγHa1,a2,...,ar(z) in this expansion of Bδ−1(z, Φ(z)) and equating them to zero modulo 22δ−1, we produce a system of linear equations modulo 22δ−1 in the unknowns xα,β. By the deﬁnition of Aδ−1(z, t), after division by 22δ−2, this system reduces to a system modulo 2, that is, to a linear system of equations over the ﬁeld with two elements. A priori, this system need not have a solution, but experience seems to indicate that it always does; see Conjecture 3. We start with an auxiliary result pertaining to the uniqueness of representations of integers as sums of powers of 2 with multiplicities, tailor-made for being applied to the series Ha1,a2,...,ar(z). Lemma 5. Let d, r, s be positive integers with r ≥s, c an integer with \|c\| ≤d, and let a1, a2, . . . , ar respectively b1, b2, . . . , bs be two sequences of odd integers, with 1 ≤ai ≤d for 1 ≤i ≤r, and 1 ≤bi ≤d for 1 ≤i ≤s. If a122rd + a222(r−1)d + · · · + ar22d = b12n1 + b22n2 + · · · + bs2ns + c (3.2) for integers n1, n2, . . . , ns with n1 > n2 > · · · > ns ≥0, then r = s, c = 0, ai = bi, and ni = 2d(r + 1 −i) for i = 1, 2, . . . , r. Proof. We use induction on r. First, let r = 1. Then s = 1 as well, and (3.2) becomes a122d = b12n1 + c. (3.3) If n1 > 2d, then the above equation, together with the assumption that a1 is odd, implies 22d ≡c modulo 22d+1. However, by assumption, we have \|c\| ≤d < 22d, which is absurd. If d < n1 < 2d, then it follows from (3.3) that c must be divisible by 2n1. Again by assumption, we have \|c\| ≤d < 2d < 2n1, so that c = 0. But then (3.3) cannot be satisﬁed since b1 is assumed to be odd. If 0 ≤n1 ≤d, then we estimate b12n1 + c ≤d 2d + 1 ≤(2d −1)(2d + 1) < 22d, which is again a contradiction to (3.3). The only remaining possibility is n1 = 2d. If this is substituted in (3.3) and the resulting equation is combined with \|c\| ≤d < 22d, then the conclusion is that the equation can only be satisﬁed if c = 0 and a1 = b1, in accordance with the assertion of the lemma. We now perform the induction step. We assume that the assertion of the lemma is established for all r < R, and we want to show that this implies its validity for r = R. Let t be maximal such that nt ≥2d. Then reduction of (3.2) modulo 22d yields b12nt+1 + b22nt+2 + · · · + bs2ns + c ≡0 modulo 22d. (3.4) 10 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Let us write b · 22d for the left-hand side in (3.4). Then, by dividing (3.2) (with R instead of r) by 22d, we obtain a122(R−1)d+a222(R−2)d+· · ·+aR−122d = b12n1−2d+b22n2−2d+· · ·+bt2nt−2d+b−aR. (3.5) We have 0 ≤b ≤2−2dd 22d−1 + 22d−2 + · · · + 22d−s+t + 1 ≤2−2dd 22d −22d−s+t + 1 ≤d. Consequently, we also have \|b −aR\| ≤d. This means that we are in a position to apply the induction hypothesis to (3.5). The conclusion is that t = R −1, b−aR = 0, ai = bi, and ni = 2d(R + 1 −i) for i = 1, 2, . . . , R −1. If this is used in (3.2) with r = R, then we obtain aR22d = c or aR22d = bR2nR + c, depending on whether s = R −1 or s = R. The ﬁrst case is absurd since c ≤d < 22d ≤ aR22d. On the other hand, the second case has already been considered in (3.3), and we have seen there that it follows that c = 0, aR = bR, and nR = 2d. This completes the proof of the lemma. □ The announced independence of the series Ha1,a2,...,ar(z) with all ai’s odd is now an easy consequence. Corollary 6. The series Ha1,a2,...,ar(z), with all ai’s odd, together with the series 1 are linearly independent over Q[z, z−1]. Proof. Let us suppose that p0(z) + N X i=1 pi(z)Ha(i) 1 ,a(i) 2 ,...,a(i) ri (z) = 0, (3.6) where the pi(z)’s are non-zero Laurent polynomials in z, the ri’s are positive integers, and a(i) j , j = 1, 2, . . . , ri, i = 1, 2, . . . , N, are odd integers. We may also assume that the tuples (a(i) 1 , a(i) 2 , . . . , a(i) ri ), i = 1, 2, . . . , N, are pairwise distinct. Choose i0 such that ri0 is maximal among the ri’s. Without loss of generality, we may assume that the coeﬃcient of z0 in pi0(z) is non-zero (otherwise we could multiply both sides of (3.6) by an appropriate power of z). Let d be the maximum of all a(i) j ’s and the absolute values of exponents of z appearing in monomials with non-zero coeﬃcient in the Laurent polynomials pi(z), i = 0, 1, . . . , N. Then, according to Lemma 5 with r = ri0, aj = a(i0) j , j = 1, 2, . . . , ri0, the coeﬃcient of za(i0) 1 22rd+a(i0) 2 22(r−1)d+···+a(i0) r 22d is 1 in Ha(i0) 1 ,a(i0) 2 ,...,a(i0) ri0 (z), while it is zero in series zeHa(i0) 1 ,a(i0) 2 ,...,a(i0) ri0 (z), where e is a non-zero integer with \|e\| ≤d, and in all other series zeHa(i) 1 ,a(i) 2 ,...,a(i) ri (z), i = 1, . . . , i0 − 1, i0 + 1, . . . , N, where e is a (not necessarily non-zero) integer with \|e\| ≤d. This contradiction to (3.6) establishes the assertion of the lemma. □ MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 11 Remark 7. Coeﬃcient extraction from a series Ha1,a2,...,ar(z) with all ai’s odd is straight-forward: if we want to know whether zM appears in Ha1,a2,...,ar(z), that is, whether we can represent M as M = a12n1 + a22n2 + · · · + ar2nr for some n1, n2, . . . , nr with n1 > n2 > · · · > nr ≥0, then necessarily nr = v2(M), nr−1 = v2(M −ar2nr), etc. The term zM appears in Ha1,a2,...,ar(z) if, and only if, the above process terminates after exactly r steps. This means, that, with nr, nr−1, . . . , n1 constructed as above, we have M −(as2ns + · · · + ar−12nr−1 + ar2nr) > 0 for s > 1, and M −(a12n1 + · · · + ar−12nr−1 + ar2nr) = 0. It should be noted that, given a1, a2, . . . , ar, this procedure of coeﬃcient extraction needs at most O(log M) operations, that is, its computational complexity is linear. Our next goal is to show that a series Hb1,b2,...,bs(z) can be expressed as a linear combination over Q[z, z−1] of the series Ha1,a2,...,ar(z), where all ai’s are odd. In doing this, we are forced to consider the more general series Hβ1,β2,...,βs b1,b2,...,bs (z) := X n1+β1>n2+β2>···>ns+βs≥0 zb12n1+b22n2+···+bs2ns, where, as before, b1, b2, . . . , bs are positive integers, and β1, β2, . . . , βs are integers. Lemma 8. For any positive integers b1, b2, . . . , bs and integers β1, β2, . . . , βs, the series Hβ1,β2,...,βs b1,b2,...,bs (z) can be expressed as a linear combination over Q[z, z−1] of the series 1 and series of the form Ha1,a2,...,ar(z), where all ai’s are odd. Proof. We describe an algorithmic procedure for expressing Hβ1,β2,...,βs b1,b2,...,bs (z) in terms of series Hγ1,γ2,...,γr a1,a2,...,ar(z), where either r < s, or r = s and max{i : ai is even or γi ̸= 0} < max{i : bi is even or βi ̸= 0}. In words, in the second case the length of the string of consecutive 0’s at the tail of the upper parameters respectively the length of the string of consecutive odd numbers at the tail of the lower parameters has been increased. Our algorithmic procedure consists of four recurrence relations, (3.7)–(3.10) below. For the ﬁrst two of these, let bs = b′ s2es, where es = v2(bs). By deﬁnition, the number b′ s is odd. Then we have Hβ1,β2,...,βs b1,b2,...,bs (z) = X n1+β1−βs+es>···>ns−1+βs−1−βs+es>ns+es≥−βs+es zb12n1+b22n2+···+bs−12ns−1+b′ s2ns+es. 12 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER In the above sum on the right-hand side, let n′ s = ns + es be a new summation index. Then, for es ≤βs, one sees that Hβ1,β2,...,βs b1,b2,...,bs (z) = Hβ1−βs+es,β2−βs+es,...,βs−1−βs+es,0 b1,b2,...,bs−1,b′ s (z) + βs−es X k=1 zb′ s2−kHβ1−βs+es+k−1,β2−βs+es+k−1,...,βs−1−βs+es+k−1 b1,b2,...,bs−1 (z). (3.7) On the other hand, for es ≥βs, one has Hβ1,β2,...,βs b1,b2,...,bs (z) = Hβ1−βs+es,β2−βs+es,...,βs−1−βs+es,0 b1,b2,...,bs−1,b′ s (z) − es−βs−1 X k=0 zb′ s2kHβ1−βs+es−k−1,β2−βs+es−k−1,...,βs−1−βs+es−k−1 b1,b2,...,bs−1 (z). (3.8) Now consider Hβ1,...,βh,0,...,0 b1,...,bh,bh+1,...,bs(z), where all of bh+1, . . . , bs are odd. Similar to the proceedings above, let bh = b′ h2eh, where eh = v2(bh). Again, by deﬁnition, the number b′ h is odd. Then we have Hβ1,...,βh,0,...,0 b1,...,bh,bh+1,...,bs(z) = X n1+β1−βh+eh>···>nh−1+βh−1−βh+eh>nh+eh nh+βh>nh+1>···>ns≥0 zb12n1+···+bh−12nh−1+b′ h2nh+eh+bh+12nh+1+···+bs2ns. In the above sum on the right-hand side, let n′ h = nh + eh be a new summation index. Then, for eh ≤βh, one sees that Hβ1,...,βh,0,...,0 b1,...,bh,bh+1,...,bs(z) = Hβ1−βh+eh,...,βh−1−βh+eh,0,0,...,0 b1,...,bh−1,b′ h,bh+1,...,bs (z) + βh−eh−1 X k=0 Hβ1−βh+eh+k,...,βh−1−βh+eh+k,0,0,...,0 b1,...,bh−1,b′ h+bh+12k,bh+2,...,bs (z). (3.9) On the other hand, for eh ≥βh, one has Hβ1,...,βh,0,...,0 b1,...,bh,bh+1,...,bs(z) = Hβ1−βh+eh,...,βh−1−βh+eh,0,0,...,0 b1,...,bh−1,b′ h,bh+1,...,bs (z) − eh−βh X k=1 Hβ1−βh+eh−k,...,βh−1−βh+eh−k,0,0,...,0 b1,...,bh−1,b′ h2k+bh+1,bh+2,...,bs (z). (3.10) It is clear that, if we recursively apply (3.7)–(3.10) to a given series Hβ1,β2,...,βs b1,b2,...,bs (z), and use H∅ ∅(z) = 1 as an initial condition, we will eventually arrive at a linear combination of 1 and series H0,0,...,0 a1,a2,...,ar(z) = Ha1,a2,...,ar(z) with all ai’s being odd, where the coeﬃcients are Laurent polynomials in z. This is exactly the assertion of the lemma. □ Computer computations suggest that, if we restrict our attention to the series Hb1,b2,...,bs(z), which are the ones that we are actually interested in, there is a strength-ening of Lemma 8 (see also Appendix A). MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 13 Conjecture 9. For any positive integers b1, b2, . . . , bs, the series Hb1,b2,...,bs(z) can be expressed as a linear combination over Q[z, z−1] of the series 1 and series of the form Ha1,a2,...,ar(z), where all ai’s are odd, r ≤s, and a1 + a2 + · · · + ar ≤b1 + b2 + · · · + bs. To conclude this section, let us provide an illustration of the above discussion. We set ourselves the task of determining the coeﬃcient of z1099511640192 in Φ5(z). In order to accomplish this task, we ﬁrst express Φ5(z) in terms of series Ha1,...,ar(z) with all ai’s being odd. This is done by means of the expansion (3.1) and the algorithm described in the proof of Lemma 8. The resulting expansion is displayed in Appendix A. Now we have to answer the question, in which of the series Ha1,...,ar(z) that appear in this expansion of Φ5(z) do we ﬁnd the monomial z1099511640192. Using the algorithm described in Remark 7, we see that 1099511640192 = 5 · 27 + 1099511639552, = 3 · 213 + 212 + 27 + 1099511611392, = 240 + 3 · 212 + 27, = 29 + 28 + 3 · 27 + 1099511639040, = 3 · 212 + 27 + 1099511627776, = 28 + 3 · 27 + 1099511639552, = 240 + 213 + 212 + 27, = 1 + 3 · 20 + 1099511640188, = 3 · 27 + 1099511639808, = 1 + 22 + 21 + 20 + 1099511640184, = 213 + 212 + 27 + 1099511627776, = 1 + 21 + 20 + 1099511640188, = 212 + 27 + 1099511635968, = 2 + 21 + 1099511640188, = 1 + 20 + 1099511640190, = 27 + 1099511640064. Here, the third line shows that z1099511640192 appears in H1,3,1(z), and the seventh line shows that it appears in H1,1,1,1(z) (thereby making it impossible to appear in H1,1,1,1,1(z)), while the remaining lines show that it does not appear in any other term in the expansion of Φ5(z) displayed in Appendix A. Hence, by taking into account the coeﬃcients with which the series H1,3,1(z) and H1,1,1,1(z) appear in this expansion, the coeﬃcient of z1099511640192 in Φ5(z) equals −40 + 240 = 200. 4. The method We consider a (formal) diﬀerential equation P(z; F(z), F ′(z), F ′′(z), . . . , F (s)(z)) = 0, (4.1) where P is a polynomial with integer coeﬃcients, which has a power series solution F(z) with integer coeﬃcients. In this situation, we propose the following algorithmic 14 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER approach to determining the series F(z) modulo a 2-power 23·2α, for some positive integer α. We make the Ansatz F(z) = 2α+2−1 X i=0 ai(z)Φi(z) modulo 23·2α, (4.2) with Φ(z) as given in (1.1), and where the ai(z)’s are (at this point) undetermined Laurent polynomials in z. Now we substitute (4.2) into (4.1), and we shall gradually determine approximations ai,β(z) to ai(z) such that (4.1) holds modulo 2β, for β = 1, 2, . . . , 3·2α. To start the procedure, we consider the diﬀerential equation (4.1) modulo 2, with F(z) = 2α+2−1 X i=0 ai,1(z)Φi(z) modulo 2. (4.3) Using the elementary fact that Φ′(z) = 1 modulo 2, we see that the left-hand side of (4.1) is a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. We reduce powers Φk(z) with k ≥2α+2 using the relation (which is implied by the minimal polynomial for the modulus 8 given in Proposition 2) 2 Φ4(z) + 6Φ3(z) + (2z + 3)Φ2(z) + (2z + 6)Φ(z) + 2z + 5z22α = 0 modulo 23·2α. (4.4) Since, at this point, we are only interested in ﬁnding a solution to (4.1) modulo 2, the above relation simpliﬁes to Φ2α+2(z) + Φ2α+1(z) + z2α+1 = 0. (4.5) Now we compare coeﬃcients of powers Φk(z), k = 0, 1, . . . , 2α+2 −1. This yields a system of 2α+2 (diﬀerential) equations (modulo 2) for the unknown Laurent polynomials ai,1(z), i = 0, 1, . . . , 2α+2 −1, which may or may not have a solution. Provided we have already found Laurent polynomials ai,β(z), i = 0, 1, . . . , 2α+2 −1, for some β with 1 ≤β ≤3 · 2α −1, such that F(z) = 2α+2−1 X i=0 ai,β(z)Φi(z) (4.6) solves (4.1) modulo 2β, we put ai,β+1(z) := ai,β(z) + 2βbi,β+1(z), i = 0, 1, . . . , 2α+2 −1, (4.7) 2Actually, if we would like to obtain an optimal result, we should use the relation implied by the minimal polynomial for the modulus 23·2α in the sense of Section 2. But since we have no general formula available for such a minimal polynomial (cf. Item (2) of Remark 4 in that section), and since we wish to prove results for arbitrary moduli, to choose instead powers of the minimal polynomial for the modulus 8 is the best compromise. In principle, it may happen that there exists a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z, which is identical with F(z) after reduction of its coeﬃcients modulo 23·2α, but the Ansatz (4.2) combined with the reduction (4.4) fails because it is too restrictive. However, it turns out that — at least in all the cases that we treat — this obstruction does not occur. Moreover, once we are successful using this (potentially problematic) Ansatz, then the result can be easily converted into an optimal one by reducing the obtained polynomial further using the relation implied by the actual minimal polynomial for the modulus 23·2α. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 15 where the bi,β+1(z)’s are (at this point) undetermined Laurent polynomials in z. Next we substitute F(z) = 2α+2−1 X i=0 ai,β+1(z)Φi(z) in (4.1). Using the fact that Φ′(z) = Pβ n=0 2nz2n−1 modulo 2β+1, we expand the left-hand side as a polynomial in Φ(z) (with coeﬃcients being Laurent polynomials in z), we apply again the reduction using relation (4.4), we compare coeﬃcients of powers Φk(z), k = 0, 1, . . . , 2α+2−1, and, as a result, we obtain a system of 2α+2 (diﬀerential) equations (modulo 2β+1) for the unknown Laurent polynomials bi,β+1(z), i = 0, 1, . . . , 2α+2 −1, which may or may not have a solution. If we manage to push this procedure through until β = 3·2α −1, then, setting ai(z) = ai,3·2α(z), i = 0, 1, . . . , 2α+2 −1, the series F(z) as given in (4.2) is a solution to (4.1) modulo 23·2α, as required. It is not diﬃcult to see that performing the iterative step (4.7) amounts to solving a system of linear diﬀerential equations in the unknown functions bi,β+1(z) modulo 2, where all of them are Laurent polynomials in z, and where only ﬁrst derivatives of the bi,β+1(z)’s occur. Solving such a system is equivalent to solving an ordinary system of linear equations, as is shown by the lemma below. Given a Laurent polynomial p(z), we write po(z) for the odd part 1 2(p(z) −p(−z)) and pe(z) for the even part 1 2(p(z) + p(−z)) of p(z), respectively. Lemma 10. Let ci,j(z) and di,j(z), 1 ≤i, j ≤N, and ri(z), 1 ≤i ≤N, be given Laurent polynomials in z with integer coeﬃcients. Then the system of diﬀerential equations N X j=1 ci,j(z)fj(z) + N X j=1 di,j(z)f ′ j(z) = ri(z) modulo 2, 1 ≤i ≤N, (4.8) has solutions fj(z), 1 ≤j ≤N, that are Laurent polynomials if, and only if, the system of linear equations N X j=1 ce i,j(z)f e j (z) + N X j=1 co i,j(z)f o j (z) + N X j=1 z−1de i,j(z)f o j (z) = re i (z) modulo 2, N X j=1 co i,j(z)f e j (z) + N X j=1 ce i,j(z)f o j (z) + N X j=1 z−1do i,j(z)f o j (z) = ro i (z) modulo 2, 1 ≤i ≤N, (4.9) has a solution in Laurent polynomials f e j (z), f o j (z) for 1 ≤j ≤N. Proof. We write fj(z) = f e j (z) + f o j (z), and observe that f ′ j(z) = z−1f o j (z) modulo 2. If this is used in (4.8), and if we separate the even and odd parts on both sides of the equations, then (4.9) results after little manipulation. □ In general, it is diﬃcult to characterise when the system (4.9) has a solution. One simple case, where a characterisation is possible, is given in Lemma 20, which is crucial for the proof that the generating function for the number of subgroups of index n of 16 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER PSL2(Z), when these are reduced modulo a given power of 2, can always be expressed as a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. We remark that the idea of the method that we have described in this section has certainly further potential. For example, the fact that the series Φ(z) remains invariant under the substitution z →z2 (or, more generally, under the substitution z →z2h, where h is some positive integer) — up to a simple additive correction — can be exploited by extending the range of applicability of our method to equations where we not only allow diﬀerentiation but also this kind of substitution. This is actually already used in a very hidden way in Section 14 (cf. [25, Theorem 12], setting in relation subgroup numbers of the Hecke group H(q) with subgroup numbers of Cq∗Cq modulo 2; in terms of generating functions, the meaning of this theorem is that the generating function for the former numbers can be expressed in terms of the generating function for the latter numbers by a relation which involves a substitution z →z2). Furthermore, as we already mentioned in the Introduction, there is no obstacle to modifying the method presented here to work for recursive sequences which are reduced modulo powers of p, in connection with the series P n≥0 zpn, although at present we are not able to oﬀer any interesting applications in this direction. 5. A sample application: Catalan numbers The Catalan numbers, deﬁned by Catn = 1 n+1 2n n , n = 0, 1, . . . , are ubiquitous in enumerative combinatorics. (Stanley provides a list of 66 sequences of sets enumerated by Catalan numbers in [30, Ex. 6.19], with many more in the addendum .) Recently, there have been several papers on the congruence properties of Catalan numbers modulo powers of 2, see [10, 20, 28]. In particular, in the Catalan numbers are determined modulo 64. As we already mentioned in the Introduction, the corresponding result (cf. [20, Theorems 6.1–6.6]) can be compactly written in the form (1.2). Clearly, once we know the right-hand side of (1.2), the validity of the congruence (1.2) can be routinely veriﬁed by substituting the right-hand side into the well-known functional equation (cf. [34, (2.3.8)]) zC2(z) −C(z) + 1 = 0, (5.1) where C(z) = P∞ n=0 Catn zn denotes the generating function for the Catalan numbers, and reducing powers of Φ(z) whose exponent exceeds 7 by means of the relation (4.4) with α = 1. We shall now demonstrate that the method from Section 4 allows one not only to ﬁnd the congruence (1.2) algorithmically, but also to ﬁnd analogous congruences modulo any power of 2. Theorem 11. Let Φ(z) = P n≥0 z2n, and let α be some positive integer. Then the generating function C(z) for Catalan numbers, reduced modulo 23·2α, can be expressed as a polynomial in Φ(z) of degree at most 2α+2 −1 with coeﬃcients that are Laurent polynomials in z. Proof. We apply the method from Section 4. We start by substituting the Ansatz (4.3) in (5.1) and reducing the result modulo 2. In this way, we obtain z 2α+2−1 X i=0 a2 i,1(z)Φ2i(z) + 2α+2−1 X i=0 ai,1(z)Φi(z) + 1 = 0 modulo 2. (5.2) MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 17 We may reduce Φ2i(z) further using the relation (4.5). This leads to z 2α+1−1 X i=0 a2 i,1(z) + z2α+1a2 i+2α+1,1(z) Φ2i(z) + z 2α−1 X i=0 z2α+1a2 i+3·2α,1(z)Φ2i(z) +z 2α−1 X i=0 a2 i+2α+1,1(z) + a2 i+3·2α,1(z) Φ2i+2α+1(z)+ 2α+2−1 X i=0 ai,1(z)Φi(z)+1 = 0 modulo 2. (5.3) Now we compare coeﬃcients of Φi(z), for i = 0, 1, . . . , 2α+2 −1. For i odd, we see immediately that this implies that ai,1(z) = 0 modulo 2. Proceeding inductively, we now suppose that a2βu,1(z) = 0 modulo 2 for odd u and some positive integer β, β < α. Reading oﬀcoeﬃcients of Φ2β+1i, where i is odd, we then obtain za2 2βi,1(z) + z2α+1+1a2 2βi+2α+1,1(z) + z2α+1+1a2 2βi+3·2α,1(z) + za2 2βi+2α,1(z) + za2 2βi+2α+1,1(z) + a2β+1i,1(z) = 0 modulo 2. However, due to our inductive assumption, all squared terms on the left-hand side of this congruence vanish, and we conclude that a2β+1i,1(z) = 0 modulo 2. So far, we have found that all coeﬃcient Laurent polynomials ai,1(z) vanish modulo 2 except possibly a0,1(z) and a2α+1,1(z). The corresponding congruences that we obtain from extracting coeﬃcients of Φ0(z) and Φ2α+1(z), respectively, in (5.3), are za2 0,1(z) + z2α+1+1a2 2α+1,1(z) + a0,1(z) + 1 = 0 modulo 2, (5.4) za2 2α+1,1(z) + a2α+1,1(z) = 0 modulo 2. (5.5) The only solutions to (5.5) are a2α+1,1(z) = 0 modulo 2, respectively a2α+1,1(z) = z−1 modulo 2. The ﬁrst option is impossible since it would imply that, modulo 2, the series C(z) reduces to a polynomial, a contradiction to the well-known fact (easily derivable from Legendre’s formula [19, p. 10] for the p-adic valuation of factorials; cf. (2.10)) that the Catalan number Catn is odd if, and only if, n = 2k −1 for some k. Thus, a2α+1,1(z) = z−1 modulo 2. Use of this result in (5.4) yields the congruence za2 0,1(z) + a0,1(z) + z2α+1−1 + 1 = 0 modulo 2 (5.6) for a0,1(z). We let a0,1(z) = e a0,1(z) + α X k=0 z2k−1 and substitute this in (5.6). Thereby, we get ze a2 0,1(z) + α X k=0 z2k+1−1 + e a0,1(z) + α X k=0 z2k−1 + z2α+1−1 + 1 = 0 modulo 2, or, after simpliﬁcation, ze a2 0,1(z) + e a0,1(z) = 0 modulo 2. 18 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Again, either e a0,1(z) = 0 modulo 2 or e a0,1(z) = z−1 modulo 2. Here, the second option is impossible, since it would imply that C(z) contains a negative z-power, which is absurd. In summary, we have found that a0,1(z) = α X k=0 z2k−1 modulo 2, a2α+1,1(z) = z−1 modulo 2, with all other ai,1(z) vanishing, forms the unique solution modulo 2 to the system of congruences resulting from (5.3) in Laurent polynomials ai,1(z). After we have completed the “base step,” we now proceed with the iterative steps described in Section 4. We consider the Ansatz (4.6)–(4.7), where the coeﬃcients ai,β(z) are supposed to provide a solution Cβ(z) = P2α+2−1 i=0 ai,β(z)Φi(z) to (5.1) modulo 2β. This Ansatz, substituted in (5.1), produces the congruence zC2 β(z) −Cβ(z) + 2β 2α+2−1 X i=0 bi,β+1(z)Φi(z) + 1 = 0 modulo 2β+1. By our assumption on Cβ(z), we may divide by 2β. Comparison of powers of Φ(z) then yields a system of congruences of the form bi,β+1(z) + Poli(z) = 0 modulo 2, i = 0, 1, . . . , 2α+2 −1, where Poli(z), i = 0, 1, . . . , 2α+2 −1, are certain Laurent polynomials with integer coeﬃcients. This system being trivially uniquely solvable, we have proved that, for an arbitrary positive integer α, the algorithm of Section 4 will produce a solution C23·2α(z) to (5.1) modulo 23·2α which is a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. □ For example, our computer program needs only about 30 seconds to come up with the corresponding congruence modulo 23·22 = 4096. Theorem 12. Let Φ(z) = P n≥0 z2n. Then we have ∞ X n=0 Catn zn = 2048z14 + 3072z13 + 2048z12 + 3584z11 + 640z10 + 2240z9 + 32z8 + 832z7 + 2412z6 + 1042z5 + 2702z4 + 53z3 + 2z2 + z + 1 + 2048z12 + 3840z10 + 2112z8 + 2112z7 + 552z6 +3128z5 + 2512z4 + 4000z3 + 3904z2 Φ(z) + 2048z13 + 3072z11 + 1536z10 + 1152z9 + 1024z8 + 4000z7 + 3440z6 +3788z5 + 3096z4 + 3416z3 + 2368z2 + 288z Φ2(z) + 2048z11 + 2048z10 + 2304z9 + 512z8 + 2752z7 + 3072z6 + 728z5 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 19 +3528z4 + 1032z3 + 3168z2 + 3456z + 3904 Φ3(z) + 2048z12 + 3072z11 + 1024z10 + 2048z9 + 1152z8 + 1728z7 + 2272z6 + 2464z5 +3452z4 + 3154z3 + 2136z2 + 3896z + 1600 + 48 z Φ4(z) + 2048z10 + 2048z9 + 1792z8 + 1792z7 + 1088z6 + 1536z5 +1704z4 + 3648z3 + 3288z2 + 200z + 3728 + 2272 z Φ5(z) + 2048z111024z9 + 1536z8 + 3200z7 + 2816z6 + 1312z5 + 3824z4 +140z3 + 592z2 + 3692z + 488 + 2760 z Φ6(z) + 2048z9 + 2304z7 + 2304z6 + 3520z5 + 960z4 + 2456z3 +2128z2 + 2936z + 1784 + 4024 z Φ7(z) + 2048z10 + 1024z9 + 2048z8 + 512z7 + 3968z6 + 1088z5 + 1888z4 +832z3 + 1444z2 + 2646z + 3258 + 339 z Φ8(z) + 2048z8 + 3328z6 + 1536z5 + 3008z4 +320z3 + 2168z2 + 1144z + 3992 + 3152 z Φ9(z) + 2048z9 + 3072z7 + 512z6 + 1408z5 + 2560z4 +3424z3 + 3408z2 + 1316z + 3608 + 2380 z Φ10(z) + 2048z7 + 2048z6 + 2816z5 + 3072z4 + 1856z3 +2688z2 + 1288z + 3880 + 3904 z Φ11(z) + 2048z8 + 1024z7 + 3072z6 + 2048z5 + 1408z4 +2624z3 + 1440z2 + 224z + 948 + 358 z Φ12(z) + 2048z6 + 2048z5 + 3328z4 + 2816z3 + 1984z2 + 384z + 2488 + 2384 z Φ13(z) 20 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER + 2048z7 + 1024z5 + 512z4 + 2432z3 + 1792z2 + 3040z + 336 + 260 z Φ14(z) + 2048z5 + 768z3 + 256z2 + 64z + 2752 + 2696 z Φ15(z) modulo 4096. (5.7) The reader is reminded that coeﬃcient extraction from an expression such as the one on the right-hand side of (5.7) is straightforward, via the algorithm described in Section 3 (see (3.1) and the proof of Lemma 6). 6. A non-example Consider the formal power series F(z) deﬁned by the equation zF 6(z) −F(z) + 1 = 0. (6.1) This equation has a unique formal power series solution. We note that the coeﬃcients in the series are special instances of numbers that are now commonly known as Fuß– Catalan numbers, which have numerous combinatorial interpretations; cf. [2, pp. 59–60]. It was shown in [25, Eq. (36)] that the coeﬃcient of zλ in the series F(z) has the same parity as the number of free subgroups of index 14λ in the Hecke group H(7) = C2 ∗C7. If we try our method from Section 4, then already at the mod-2 level we fail: let F(z) = a1(z)Φ(z) + a0(z) modulo 2, for some Laurent polynomials a0(z) and a1(z). Upon substitution in (6.1) and simpliﬁcation using (cf. Proposition 2) Φ2(z) + Φ(z) + 1 = 0 modulo 2, we obtain za6 0(z) + za4 0(z)a2 1(z) + a0(z) + za6 1(z) + 1 + Φ(z) za4 0(z)a2 1(z) + za2 0(z)a4 1(z) + a1(z) = 0 modulo 2. or, equivalently, za6 0(z) + za4 0(z)a2 1(z) + za6 1(z) + a0(z) + 1 = 0 modulo 2 (6.2) za4 0(z)a2 1(z) + za2 0(z)a4 1(z) + a1(z) = 0 modulo 2. (6.3) However, this congruence has no solution in Laurent polynomials a0(z) and a1(z). For, the Laurent polynomials a6 0(z), a4 0(z)a2 1(z), a6 1(z), all of them being squares, contain only even powers of z when the coeﬃcients are reduced modulo 2. Consequently, the term a0(z) + 1 on the left-hand side of (6.2) can only contain even powers (modulo 2). In particular, a0(z) must contain the term 1. If we now suppose that a0(z) and/or a1(z) contain negative powers of z, then we obtain a contradiction regardless whether the orders (the minimal e such that ze appears in a Laurent polynomial) of a0(z) and a1(z) are the same or not. This implies that both a0(z) and a1(z) are actually polynomials in z, with a0(z) being of the form a0(z) = 1+e a0(z), where e a0(z) is a polynomial without constant term. Hence, the left-hand side of (6.2) equals z + ze a6 0(z) + za2 1(z) + ze a4 0(z)a2 1(z) + za6 1(z) + e a0(z). (6.4) MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 21 If we now multiply both sides of (6.2) by a2 1(z) and both sides of (6.3) by a2 0(z), and subsequently add the two congruences, then we obtain za8 1(z) + a0(z)a2 1(z) + a2 0(z)a1(z) + a2 1(z) = 0 modulo 2. Equivalently, after replacing a0(z) by 1 + e a0(z), this is za7 1(z) + e a0(z)a1(z) + e a2 0(z) + 1 = 0 modulo 2. This congruence has no solution since e a0(z) has no constant term modulo 2. In the next theorem, we reveal the deeper reason why our method must fail for F(z). Namely, it shows that exponents e of terms ze which survive in F(z) after reduction of its coeﬃcients modulo 2 may have arbitrarily large binary digit sum. In contrast, a polynomial in Φ(z) of degree d with coeﬃcients that are polynomials in z can only have terms ze after reduction of the coeﬃcients of the series modulo 2, where e has binary digit sum at most d. Theorem 13. Let F(z) be the unique formal power series solution to the functional equation (6.1). Then the coeﬃcient of zn in F(z) is odd if, and only if, the sequence of binary digits of n is built by concatenating (in any order) blocks of 0011 and 01. In particular, the number of free subgroups of index n in H(7) = C2 ∗C7 is odd if, and only if, the above condition holds. Proof. By replacing F(z) by 1 + G(z) in (6.1), we obtain z (1 + G(z))6 −G(z) = 0, or, equivalently, z = G(z) (1 + G(z))6, so that G(z) is the compositional inverse of the series z/(1 + z)6. By the Lagrange inversion formula (cf. [30, Theorem 5.4.2 with k = 1]), we obtain for n ≥1 that ⟨zn⟩F(z) = ⟨zn⟩G(z) = 1 n z−1 (1 + z)6n zn = 1 n zn−1 (1 + z)6n = 1 n 6n n −1 = 1 6n + 1 6n + 1 n . By the well-known theorem of Legendre [19, p. 10] (cf. (2.10)), we see that the coeﬃcient of zn in F(z) is odd if, and only if, s(6n + 1) −s(5n + 1) −s(n) = 0, (6.5) where, as in Section 2, s(m) denotes the binary digit sum of m. Another way to phrase (6.5) is to say that, whenever we ﬁnd a 1 in the binary expansion of n, then there must also be a 1 in the binary expansion of 6n + 1 at the same digit place. We are now ready to establish the claim of the theorem. In view of the above considerations, it suﬃces to show that the condition on n in the statement of the theorem is equivalent to (6.5). Let n be a positive integer with the property that its binary expansion is formed by concatenating blocks of the form 0011 and 01. We prove that (6.5) holds in this case by induction on n. It is routine to check that our assertion holds true for n = 1, 2, . . . , 15. Now, let n = 4n1 + 1, with some positive integer n1. In other words, the right-most 22 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER digits in the binary expansion of n are 01 and the binary expansion of n1 is formed by concatenating blocks of the form 0011 and 01. In that case, we have s(6n + 1) −s(5n + 1) −s(n) = s (4(6n1 + 1) + 2 + 1) −s (4(5n1 + 1) + 2) −s(4n1 + 1) = s(6n1 + 1) + 2 −s(5n1 + 1) −1 −s(n1) −1 = 0, (6.6) by the induction hypothesis applied to n1. On the other hand, if n = 16n1 + 3 for some positive integer n1, that is, if the right-most digits in the binary expansion of n are 0011 and the binary expansion of n1 is formed by concatenating blocks of the form 0011 and 01, then s(6n + 1) −s(5n + 1) −s(n) = s (16(6n1 + 1) + 2 + 1) −s (16(5n1 + 1)) −s(16n1 + 2 + 1) = s(6n1 + 1) + 2 −s(5n1 + 1) −s(n1) −2 = 0, (6.7) establishing again the truth of (6.5). In order to prove the converse, let us suppose that n satisﬁes (6.5). We start by showing that the binary expansion of n cannot contain any of the substrings 111, 000, 1011. We call an occurrence of any of these substrings a “violation.” Assuming that the right-most violation is a substring of the form 111, we have n = . . . 1110 . . . , 2n = . . . 1110 . . . 0, 4n = . . . 1110 . . .00, 6n + 1 = . . . 101 . . . 1, since to the right of the substring 111 in n there are only blocks of the form 0011 and 01 according to our assumption, which implies that there cannot be any carries “destroying” the substring 101 in 6n+1. However, this means that at the place where we ﬁnd the bold-face 1 in (the binary expansion of) n we ﬁnd a 0 in (the binary expansion of) 6n + 1, a contradiction to (6.5). Now we assume that the right-most violation is a substring of the form 000. In that case, we have n = . . . 1000 . . . , 2n = . . . 1000 . . . 0, 4n = . . . 1000 . . .00, 6n + 1 = . . . 10 . . . 1, a contradiction to (6.5) for the same reason. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 23 Finally we assume that the right-most violation is a substring of the form 1011. Then we have n = . . . 10110 . . . , 2n = . . . 10110 . . . 0, 4n = . . . 10110 . . . 00, 6n + 1 = . . . 0001 . . . 1, since to the right of the substring 1011 in n there are only blocks of the form 0011 and 01 according to our assumption, which implies that there cannot be any carries “destroying” the substring 0001 in 6n+1. However, this means that at the place where we ﬁnd the bold-face 1 in n we ﬁnd a 0 in 6n + 1, again a contradiction to (6.5). Now let us suppose that n = 2αn1+n0, where n0 is an α-digit (binary) number formed by concatenating blocks of the form 0011 and 01. By applying the computations (6.6) and (6.7) (possibly several times), one sees that n satisﬁes (6.5) if, and only if, n1 does. We claim that n1 cannot be even. For, if it were, say n1 = 2βn2, then n1 has a 1 at the digit place β while 6n1 + 1 has not, a contradiction to (6.5). But then the already established fact that n, and hence n1, cannot contain any of the substrings 111, 000, 1011 implies that the right-most digits in the binary expansion of n1 form either a block 01 or a block 0011. This provides an inductive argument that the binary expansion of n is formed by concatenating blocks of the form 0011 and 01, and thus completes the proof of the theorem. □ 7. Free subgroups in lifts of Hecke groups For integers m, q with m ≥1, q ≥3, and q prime, we consider the group Γm(q) as deﬁned in (1.3). Denote by f (q) λ (m) the number of free subgroups of index 2qmλ in Γm(q). The purpose of this section is to estimate the 2-adic valuation of f (q) λ (m) in the case when q is a Fermat prime. This estimate is based on a recurrence relation for these numbers, which, in turn, results from a specialisation of a diﬀerential equation in [25, Sec. 2]. Moreover, this diﬀerential equation for the generating function of free subgroup numbers in Γm(q) will become of crucial importance in Sections 8 and 13. In order to present the afore-mentioned diﬀerential equation, we ﬁrst need to compute several important invariants of Γm(q). Using notation and deﬁnitions from [25, Sec. 2], we have mΓm(q) = 2qm, χ(Γm(q)) = −q−2 2qm, and thus, µ(Γm(q)) = 1 −mΓm(q)χ(Γm(q)) = q −1 = µ(H(q)). Moreover, for the family of zeta-invariants {ζκ(Γm(q)) : κ \| 2qm} of Γm(q), we ﬁnd that ζκ(Γm(q)) =        1, κ = m, −1, κ = 2qm, 0, otherwise. (7.1) Our ﬁrst result in this section compares the A-invariants of Γm(q), as deﬁned in [25, Eq. (14)], to those of the underlying Hecke group H(q), and provides an estimate for their 2-adic valuation. 24 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Lemma 14. (i) For an integer m ≥1 and a prime q ≥3, we have Aµ(Γm(q)) = mq−1Aµ(H(q)), 0 ≤µ ≤q −1. (ii) We have v2(Aµ(H(q)) ≥µ, 0 ≤µ ≤q −1. Proof. (i) In view of (7.1), we have, for 0 ≤µ ≤q −1, that Aµ(Γm(q)) = 1 µ! µ X j=0 (−1)µ−j µ j 2qm(j + 1)   Y 1≤k≤2qm (k,2qm)=m (2qmj + k)  2qm(j + 1) −1 = 1 µ! µ X j=0 (−1)µ−j µ j Y 1≤k≤2qm (k,2qm)=m (2qmj + k) = 1 µ! µ X j=0 (−1)µ−j µ j Y 1≤k′≤2q (k′,2q)=1 m(2qj + k′) = mϕ(2q) 1 µ! µ X j=0 (−1)µ−j µ j Y 1≤k′≤2q (k′,2q)=1 (2qj + k′) = mq−1Aµ(H(q)), as claimed. (ii) This follows from [25, Lemma 1]. □ Our next result is a recurrence relation for the subgroup numbers f (q) λ (m). Lemma 15. For m, q as in Lemma 14 and λ ≥1, we have f (q) λ+1(m) = q−1 X µ=1 µ X ν=1 X µ1,...,µν>0 µ1+···+µν=µ X λ1,...,λν≥0 λ1+···+λν=λ−µ µ µ1, . . . , µν ν! (2q)ν−1mq−ν−1Aµ(H(q)) × ν Y j=1 h (µj −1)! λj + µj −1 µj −1 i ν Y j=1 f (q) λj+µj(m), (7.2) with initial value f (q) 1 (m) = mq−1A0(H(q)). MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 25 Proof. Setting G = Γm(q) in [25, Eq. (18)], and using Part (i) of Lemma 14 to compute Aµ(Γm(q)) in terms of Aµ(H(q)), leads to the diﬀerential equation Gm(q; z) = mq−1A0(H(q)) + q−1 X µ=1 µ X ν=1 X µ1,...,µν>0 µ1+···+µν=µ µ µ1, . . . , µν ν! (2q)ν−1mq−ν−1Aµ(H(q))zµ ν Y j=1 Gm(q; z) (µj−1) (7.3) for the generating function Gm(q; z) := P λ≥0 f (q) λ+1(m)zλ. Comparing the coeﬃcient of zλ in (7.3) for λ ≥1 yields (7.2); while, for λ = 0, we obtain the required initial value f (q) 1 (m). □ Given these preparations, we can now show the following estimate for the 2-adic valuation of f (q) λ (m) in the case when q is a Fermat prime. Proposition 16. (i) Let m ≥1 be an integer, and let q ≥3 be a Fermat prime. Then we have v2(f (q) λ (m)) ≥v2(m)(λ + q −2), λ ≥1. (7.4) In particular, if m is even, then f (q) λ (m) is zero modulo any given 2-power for all suﬃ-ciently large values of λ. (ii) For q = 3, equality occurs in Inequality (7.4) if, and only if, λ + 1 is a 2-power. Proof. (i) Since (7.4) is trivially true for m odd, we may suppose that v2(m) > 0. We use induction on λ. For λ = 1, we have v2(f (q) 1 (m)) = v2(mq−1A0(H(q))) ≥(q −1)v2(m) = (λ + q −2)v2(m), as desired. Now suppose that our claim (7.4) holds for all f (q) γ (m) such that γ < L, with some integer L ≥2, and consider an arbitrary summand S = S(µ, ν, µ1, . . . , µν, λ1, . . . , λν) 26 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER in the recurrence relation (7.2) with λ = L −1. We ﬁnd that v2(S) ≥v2 µ µ1, . . . , µν −v2(ν!) −ν + (q −ν −1)v2(m) + v2(Aµ(H(q))) + ν X j=1 v2(f (q) λj+µj(m)) ≥v2 µ µ1, . . . , µν −v2(ν!) −ν + (q −ν −1)v2(m) + µ + ν X j=1 (λj + µj + q −2)v2(m) ≥v2 µ µ1, . . . , µν −v2(ν!) −ν + (q −ν −1)v2(m) + µ + (L −1)v2(m) + (q −2)νv2(m) ≥(L + q −2)v2(m) + (q −3)νv2(m) + v2 µ µ1, . . . , µν + µ −ν −v2(ν!), where we have used Part (ii) of Lemma 14 plus the induction hypothesis in the second step. Since ν ≤µ, the desired inequality for the 2-adic valuation of f (q) L (m) will follow, if we can show that v2(ν!) ≤(q −3)νv2(m) + v2 µ µ1, . . . , µν . (7.5) Since q is a Fermat prime, we have q −1 = 2α for some α ≥1. Thus, if ν < q −1, then, by Legendre’s formula for the p-adic valuation of factorials (cf. [19, p. 10]), we get v2(ν!) ≤ X i≥0 j2α −1 2i k < X i≥1 2α−i = q −2, and (7.5) holds, the left-hand side already being compensated by the term (q −3)ν v2(m) ≥q −3. On the other hand, for ν = q −1, we have µ = ν = q −1, µ1 = · · · = µq−1 = 1, v2(ν!) = q −2, and v2 µ µ1, . . . , µν = v2((q −1)!) = q −2, and the desired conclusion holds again. We have thus shown that every summand S on the right-hand side of (7.2) satisﬁes v2(S) ≥(L + q −2)v2(m), which implies that v2(f (q) L (m)) ≥(L + q −2)v2(m), completing the induction. (ii) For q = 3, the recurrence relation (7.2), with λ replaced by λ −1, takes the form f (3) λ (m) = 6mλf (3) λ−1(m) + X µ,ν≥1 µ+ν=λ−1 f (3) µ (m)f (3) ν (m), λ ≥2, (7.6) MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 27 with initial value f (3) 1 (m) = 5m2. In order to establish our second claim, we rewrite Equation (7.6) as f (3) λ (m) = 6mλf (3) λ−1(m) +      2 P λ−2 2 µ=1 f (3) µ (m), f (3) λ−µ−1(m), λ ≡0 (2), 2 P λ−3 2 µ=1 f (3) µ (m)f (3) λ−µ−1(m) + f (3) λ−1 2 (m) 2, λ ≡1 (2), for λ ≥2, (7.7) and argue again by induction on λ. For λ = 1, Inequality (7.4) is sharp, as required. Now suppose that, for λ < L with some L ≥2, that (7.4) with q = 3 is sharp if, and only if, λ + 1 is a 2-power, and consider f (3) L (m) as given by (7.7). Setting m = 2am′ with m′ odd, we have v2(6mLf (3) L−1(m)) = 1 + a + v2(L) + v2(f (3) L−1(m)) ≥a(L + 1) + 1. Consequently, if L is even then, by what we have already shown, v2(f (3) L (m)) ≥a(L + 1) + 1. For L odd, all terms except possibly f (3) L−1 2 (m) 2 are divisible by 2a(L+1)+1 or a higher 2-power, while this exceptional term satisﬁes v2 f (3) L−1 2 (m) 2 ≥a(L + 1), with equality occurring (according to our induction hypothesis) if, and only if, L−1 2 +1 = 2γ for some γ ≥1; that is, if, and only if, L + 1 = 2γ+1 is a 2-power. This completes the induction, and the proof. □ 8. Free subgroup numbers for lifts of the inhomogeneous modular group In this section, we investigate the behaviour of the numbers f (3) λ (m) of free subgroups in lifts of the inhomogeneous modular group PSL2(Z) ∼ = H(3) modulo powers of 2. As we mentioned in the introduction, the best previous result available in the literature is [27, Theorem 1], which determines the behaviour of f (3) λ (1) modulo 16. The results in this section solve the problem of determining f (3) λ (m) modulo powers of 2 not only for m = 1 and the 2-power 24 = 16, but for all m and modulo any power of 2. Let Fm(z) := 1 + P λ≥1 f (3) λ (m) zλ be the generating function for these numbers. (In the notation of the previous section, Fm(z) = 1 + zGm(3; z).) By specialising q = 3 in (7.3), one obtains the diﬀerential equation (1 −(6m −2)z)Fm(z) −6mz2F ′ m(z) −zF 2 m(z) −1 −(1 −6m + 5m2)z = 0. (8.1) Theorem 17. Let Φ(z) = P n≥0 z2n, and let α be some positive integer. Then, for every positive integer m, the generating function Fm(z), when reduced modulo 23·2α, can be expressed as a polynomial in Φ(z) of degree at most 2α+2 −1, with coeﬃcients that are Laurent polynomials in z. 28 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Proof. In view of Proposition 16, the assertion is trivially true for even m, the polyno-mial in Φ(z) being a polynomial of degree zero in this case. We may thus assume from now on that m is odd. We apply the method from Section 4. We start by substituting the Ansatz (4.3) in (8.1) and reducing the result modulo 2. In this way, we obtain 2α+2−1 X i=0 ai,1(z)Φi(z) + z 2α+2−1 X i=0 a2 i,1(z)Φ2i(z) + 1 = 0 modulo 2. This congruence is identical with the congruence (5.2). Hence, we can copy the resulting solution from there. Namely, the unique solution to (5.2) (and, hence, to the above congruence) is given by a0,1(z) = α X k=0 z2k−1 modulo 2, a2α+1,1(z) = z−1 modulo 2, with all other ai,1(z) vanishing. After we have completed the “base step,” we now proceed with the iterative steps described in Section 4. We consider the Ansatz (4.6)–(4.7), where the coeﬃcients ai,β(z) are supposed to provide a solution Fm,β(z) = P2α+2−1 i=0 ai,β(z)Φi(z) to (8.1) modulo 2β. This Ansatz, substituted in (8.1), produces the congruence 2β 2α+2−1 X i=0 bi,β+1(z)Φi(z) + (1 −(6m −2)z)Fm,β(z) −6mz2F ′ m,β(z) −zF 2 m,β(z) −1 −(1 −6m + 5m2)z = 0 modulo 2β+1. By our assumption on Fm,β(z), we may divide by 2β. Comparison of powers of Φ(z) then yields a system of congruences of the form bi,β+1(z) + Poli(z) = 0 modulo 2, i = 0, 1, . . . , 2α+2 −1, where Poli(z), i = 0, 1, . . . , 2α+2 −1, are certain Laurent polynomials with integer coeﬃcients. This system being trivially (uniquely) solvable, we have proved that, for an arbitrary positive integer α, the algorithm of Section 4 will produce a solution Fm,23·2α(z) to (8.1) modulo 23·2α which is a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. □ We have implemented this algorithm. As an illustration, the next theorem contains the result for the modulus 64. Theorem 18. Let Φ(z) = P n≥0 z2n. Then, for all positive odd integers m, we have 1 + X λ≥1 f (3) λ (m) zλ = 32z9 + 48z7 + 32z6 + (16m + 8)z5 + (16m + 8)z4 + 2m2 + 34 z3 + 4m2 −4m + 24 z2 + 5m2 + 12 z + 1 + 48z4 + 24z3 + 12z2 + 60z + 40 Φ(z) MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 29 + 16z5 + (16m + 32)z4 + (4m2 −32m + 68)z3 + 36z2 + 22z + 12 + 12 z Φ2(z) + 32z5 + 32z4 + (16m −16)z3 + 40z2 + 4z + 52 + 28 z Φ3(z) + 32z7 + 32z5 + 32z4 + (16m + 24)z3 + (16m + 40)z2 +(2m2 + 16m + 38 z + 24 + 35 z Φ4(z) + 32z3 + 16z2 + (16m −8)z + 44 Φ5(z) + 16z3 + 16mz2 + 4m2 −16m + 20 z + 44 + 50 z Φ6(z) + 32z3 + 32z2 + (16m + 16)z + 40 + 4 z Φ7(z) modulo 64. (8.2) 9. Subgroup numbers for the inhomogeneous modular group For a ﬁnitely generated group Γ, let sn(Γ) denote the number of subgroups of index n in Γ, and write SΓ(z) for the (shifted) generating function P n≥0 sn+1(Γ) zn. In this section, we focus on the sequence sn(PSL2(Z)) n≥1 and its generating func-tion S(z) := SP SL2(Z)(z). We shall show that our method solves the problem of deter-mining these subgroup numbers modulo any given power of 2, thus reﬁning the parity result of Stothers and the mod-8 result from [27, Theorem 2] mentioned in the Introduction. By the ﬁrst displayed equation on top of p. 276 in (cf. also [18, Eq. (5.29)] with H = {1} and a = b = h = 1), the series S(z) obeys the diﬀerential equation (−1 + 4z3 + 2z4 + 4z6 −2z7 −4z9)S(z) + (z7 −z10)(S′(z) + S2(z)) + 1 + z + 4z2 + 4z3 −z4 + 4z5 −2z6 −2z8 = 0. (9.1) The diﬀerential equation (9.1) has a unique solution since comparison of coeﬃcients of zN ﬁxes the initial values, and yields a recurrence for the sequence sn(PSL2(Z)) n≥1 which computes sn+1(PSL2(Z)) from terms involving only si(PSL2(Z)) with i ≤n. Theorem 19. Let Φ(z) = P n≥0 z2n, and let α be some positive integer. Then the generating function S(z) = SP SL2(Z)(z), when reduced modulo 23·2α, can be expressed as a polynomial in Φ(z) of degree at most 2α+2 −1 with coeﬃcients that are Laurent polynomials in z. 30 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Proof. We apply the method from Section 4. We start by substituting the Ansatz (4.3) in (9.1) and reducing the result modulo 2. In this way, we obtain 2α+2−1 X i=0 ai,1(z)Φi(z) + (z7 + z10) 2α+2−1 X i=0 iai,1(z)Φi−1(z)Φ′(z) + 2α+2−1 X i=0 a′ i,1(z)Φi(z) + 2α+2−1 X i=0 a2 i,1(z)Φ2i(z) ! + 1 + z + z4 = 0 modulo 2. We may reduce Φ2i(z) further using Relation (4.5). This leads to 2α+2−1 X i=0 ai,1(z)Φi(z) + (z7 + z10) 2α+2−1 X i=0 iai,1(z)Φi−1(z)Φ′(z) + 2α+2−1 X i=0 a′ i,1(z)Φi(z) + 2α+1−1 X i=0 a2 i,1(z) + z2α+1a2 i+2α+1,1(z) Φ2i(z) + 2α−1 X i=0 z2α+1a2 i+3·2α,1(z)Φ2i(z) + 2α−1 X i=0 a2 i+2α+1,1(z) + a2 i+3·2α,1(z) Φ2i+2α+1(z) ! + 1 + z + z4 = 0 modulo 2. (9.2) In the same way as in the proof of Theorem 17, one sees that all coeﬃcients ai,1(z) vanish modulo 2, except possibly a0,1(z) and a2α+1,1(z). The corresponding congruences obtained by extracting coeﬃcients of Φ0(z) and Φ2α+1(z), respectively, in (9.2), are a0,1(z)+(z7+z10) a′ 0,1(z) + a2 0,1(z) + z2α+1a2 2α+1,1(z) +1+z+z4 = 0 modulo 2 (9.3) and a2α+1,1(z) + (z7 + z10) a′ 2α+1,1(z) + a2 2α+1,1(z) = 0 modulo 2. (9.4) The only solutions to (9.4) are a2α+1,1(z) = 0 modulo 2, respectively a2α+1,1(z) = z−7 + z−4 modulo 2. The ﬁrst option is impossible, since there is no Laurent polynomial a0,1(z) solving the equation resulting from (9.3). Thus, we have a2α+1,1(z) = z−7 + z−4 modulo 2. (9.5) Use of this result in (9.3) yields the congruence a0,1(z)+z7(1+z3) a′ 0,1(z) + a2 0,1(z) +z2α+1−7(1+z3)3+1+z+z4 = 0 modulo 2. (9.6) for a0,1(z). We let a0,1(z) = e a0,1(z) + z−7 + z−4 + z−3 + (1 + z3) α X k=2 z2k−7 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 31 and substitute this in (9.6). Thereby, we get e a0,1(z) + z7(1 + z3) e a′ 0,1(z) + e a2 0,1(z) + z−7 + z−4 + z−3 + (1 + z3) α X k=2 z2k−7 + z7(1 + z3) z−8 + z−4 + α X k=2 z2k−8 + z−14 + z−8 + z−6 + (1 + z3)2 α X k=2 z2k+1−14 ! + z2α+1−7(1 + z3)3 + 1 + z + z4 = 0 modulo 2, or, after simpliﬁcation, e a0,1(z) + z7(1 + z3) e a′ 0,1(z) + e a2 0,1(z) = 0 modulo 2. Again, either e a0,1(z) = 0 modulo 2 or e a0,1(z) = z−7 + z−4 modulo 2. Here, the second option is impossible, since it would imply that S(z) contains a negative z-power, which is absurd. In summary, we have found that a0,1(z) = z−7 + z−4 + z−3 + (1 + z3) α X k=2 z2k−7 modulo 2, a2α+1,1(z) = z−7 + z−4 modulo 2, with all other ai,1(z) vanishing, forms the unique solution modulo 2 to the system of congruences resulting from (9.2) in Laurent polynomials ai,1(z). It should be noted that all ai,1(z)’s, 1 ≤i ≤22α+2 −1, are divisible by 1 −z3 modulo 2, as is a0,1(z) −1. After we have completed the “base step,” we now proceed with the iterative steps described in Section 4. The arguments turn out to be slightly more delicate here than in the proof of Theorem 17. To be more precise, when considering the Ansatz (4.6)– (4.7), where, inductively, the coeﬃcients ai,β(z) are supposed to provide a solution Sβ(z) = P2α+2−1 i=0 ai,β(z)Φi(z) to (9.1) modulo 2β, we must also assume that ai,β(z), 1 ≤i ≤22α+2 −1, and a0,β(z) −1 −2z2, are all divisible by 1 −z3. The reader should note that the divisibility assumptions do indeed hold for β = 1, the term −2z2 being negligible, since for β = 1 we are computing modulo 2β = 2. 32 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER The above Ansatz, substituted in (9.1), produces the congruence 2β 2α+2−1 X i=0 bi,β+1(z)Φi(z) + 2β(z7 −z10) 2α+2−1 X i=0 ibi,β+1(z)Φi−1(z)Φ′(z) + 2α+2−1 X i=0 b′ i,β+1(z)Φi(z) ! + (−1 + 4z3 + 2z4 + 4z6 −2z7 −4z9) 2α+2−1 X i=0 ai,β(z)Φi(z) + (z7 −z10) 2α+2−1 X i=0 iai,β(z)Φi−1(z)Φ′(z) + 2α+2−1 X i=0 a′ i,β(z)Φi(z) + 2α+2−1 X i=0 ai,β(z)Φi(z) 2! + 1 + z + 4z2 + 4z3 −z4 + 4z5 −2z6 −2z8 = 0 modulo 2β+1. (9.7) By our inductive construction, we know that the terms contained in lines 3–7 of (9.7) are divisible by 2β. Hence, if we were to divide by 2β and compare coeﬃcients of Φi(z), for i = 0, 1, . . . , 2α+2 −1, we would obtain the modular diﬀerential equations bi,β+1(z) + (z7 −z10) b′ i,β+1(z) + (i + 1)bi+1,β+1Φ′(z) + Poli(z) = 0 modulo 2, i = 0, 1, . . . , 2α+2 −1, (9.8) where Poli(z), i = 0, 1, . . . , 2α+2 −1, are certain Laurent polynomials with integer coeﬃcients. If i is odd, then the term (i+ 1)bi+1,β+1 in (9.8) vanishes modulo 2. Hence, in this case, the diﬀerential equation (9.8) is of the form appearing in Lemma 20. The lemma then says that such a diﬀerential equation has a solution if, and only if, the Laurent polynomial Poli(z) satisﬁes the condition given there. We must therefore verify this condition for our Laurent polynomials Poli(z), arising through division of lines 3–7 of (9.7) by 2β. We shall actually prove (see the following paragraph) that Poli(z) is divisible by (1 −z3)2, for all i with 0 ≤i ≤2α+2 −1. For odd i, Corollary 21 thus implies not only unique existence of solutions bi,β+1(z) but also divisibility of these solutions by 1 −z3. If we now consider Equation (9.8) for even i, that is, bi,β+1(z) + (z7 −z10)b′ i,β+1(z) + z7(1 −z3)bi+1,β+1Φ′(z) + Poli(z) = 0 modulo 2, i = 2, 4, . . . , 2α+2 −2, then we see that divisibility of bi+1,β+1(z) by 1−z3 guarantees that we may again apply Corollary 21 to obtain that there is also a unique solution bi,β+1(z) for even i, and that this solution is divisible by 1 −z3. In summary, we would have proved that, for an arbitrary positive integer α, the algorithm of Section 4 produces a solution S23·2α(z) to (9.1) modulo 23·2α, which is a polynomial in Φ(z) with coeﬃcients that are Laurent polynomials in z. This would establish the claim of the theorem. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 33 It remains to prove that lines 3–7 of (9.7) are indeed divisible by (1 −z3)2. In order to see this conveniently, we write ai,β(z) = di,β(z)(1 −z3) modulo (1 −z3)2, i = 1, 2, . . . , 2α+2 −1, and a0,β(z) = −1 −2z2 + d0,β(z)(1 −z3) modulo (1 −z3)2, where the di,β(z)’s are polynomials of the form p0+p1z+p2z2, for some integers p0, p1, p2. (It should be noted that it is at this point where we use our inductive hypothesis on divisibility of the coeﬃcients ai,β(z).) Then, reduction of lines 3–7 of (9.7) modulo (1 −z3)2 leads to the remainder (3 + 2z(1 −z3)) −1 −2z2 + (1 −z3) 2α+2−1 X i=0 di,β(z)Φi(z) ! + z7(1 −z3) −4z −3z2 2α+2−1 X i=0 di,β(z)Φi(z) + 1 + 4z2 + 4z4 ! + 3 + 6z2 + z(1 −z3) modulo (1 −z3)2. (9.9) Using z7(1 −z3) = z(1 −z3) modulo (1 −z3)2 and similar reductions modulo (1 −z3)2, one sees that the expression in (9.9) is in fact divisible by (1 −z3)2, as we claimed. This completes the proof of the theorem. □ Recall that, given a Laurent polynomial p(z), we write po(z) for the odd part 1 2(p(z) −p(−z)) and pe(z) for the even part 1 2(p(z) + p(−z)) of p(z), respectively. Lemma 20. The diﬀerential equation a(z) + (z7 −z10)a′(z) + Pol(z) = 0 modulo 2, (9.10) where Pol(z) is a given Laurent polynomial in z with integer coeﬃcients, has a solution that is a Laurent polynomial if, and only if, Polo(z) is divisible by 1 + z6 (modulo 2). In the latter case, the unique solution is given by a(z) = Pole(z) + 1 + z9 1 + z6Polo(z) = Pole(z) + 1 + z3 + z6 1 −z3 Polo(z) modulo 2. Proof. Let a0(z) = zm. Then it is obvious that a0(z) + (z7 −z10)a′ 0(z) = a0(z) modulo 2 if m is even, and that a0(z) + (z7 −z10)a′ 0(z) = (1 + z6)a0(z) + z9a0(z) modulo 2 if m is odd. The assertion of the lemma follows now immediately. □ Corollary 21. If, in the diﬀerential equation (9.10), the Laurent polynomial Pol(z) is divisible by (1 −z3)2 (modulo 2), then the uniquely determined solution a(z) is divisible by 1 −z3 (modulo 2). We have implemented the algorithm described in the proof of Theorem 19. As an illustration, we present the result for the modulus 64. 34 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Theorem 22. Let Φ(z) = P n≥0 z2n. Then we have X n≥0 sn+1(PSL2(Z)) zn = z57 + 32z50 + 48z44 + 48z41 + 32z36 + 32z35 + 32z33 + 48z32 + 16z28 + 40z26 + 16z25 + 32z24 + 32z23 + 16z22 + 16z21 + 52z20 + 32z19 + 40z18 + 60z17 + 48z16 + 4z14 + 32z13 + 4z12 + 36z11 + 16z10 + 60z9 + 2z8 + 16z7 + 4z6 + 60z5 + 44z4 + 16z3 + 54z2 + 60z + 32 + 56 z + 36 z2 + 51 z3 + 33 z4 + 52 z5 + 32z34 + 32z26 + 32z25 + 32z24 + 16z22 + 32z21 + 32z20 + 32z17 + 32z16 + 48z14 + 16z13 + 16z12 + 16z11 + 32z10 + 32z8 + 48z7 + 8z5 + 8z4 + 48z3 + 24z + 32 + 20 z + 12 z2 + 8 z3 + 36 z4 + 4 z5 + 24 z6 Φ(z) + 32z34 + 32z29 + 32z28 + 32z26 + 32z24 + 32z21 + 48z19 + 32z18 + 48z17 + 32z14 + 48z13 + 32z12 + 56z10 + 8z9 + 16z8 + 48z7 + 24z6 + 56z5 + 44z4 + 16z3 + 48z2 + 40z + 44 + 60 z + 50 z2 + 48 z3 + 8 z4 + 50 z5 + 52 z6 + 52 z7 Φ2(z) + 32z28 + 32z24 + 32z21 + 32z20 + 32z19 + 48z16 + 32z14 + 32z13 + 32z12 + 32z11 + 16z10 + 48z9 + 8z8 + 48z6 + 56z4 + 8z3 + 16z2 + 48z + 56 + 32 z + 20 z2 + 52 z3 + 4 z4 + 36 z5 + 12 z6 + 36 z7 Φ3(z) + 32z44 + 32z41 + 32z33 + 32z32 + 32z31 + 32z30 + 32z28 + 32z27 + 16z26 + 32z24 + 32z23 + 48z22 + 16z21 + 40z20 + 32z19 + 32z18 + 24z17 + 16z16 + 48z15 + 32z14 + 16z13 + 8z12 + 32z11 + 56z10 + 56z9 + 44z8 + 40z7 + 48z6 + 16z5 + 20z4 + 56z3 + 30z2 + 32z + 28 + 40 z + 34 z2 + 52 z3 + 17 z4 + 26 z5 + 40 z6 + 29 z7 Φ4(z) + 32z32 + 32z30 + 32z26 + 32z24 + 32z23 + 32z22 + 32z21 + 48z20 + 48z18 + 32z16 + 48z14 + 32z13 + 48z12 + 48z11 + 32z8 + 16z7 + 56z6 + 48z5 + 48z4 + 40z3 + 16z2 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 35 + 32z + 56 + 24 z + 24 z2 + 20 z3 + 24 z4 + 40 z5 + 20 z6 Φ5(z) + 32z32 + 32z31 + 32z30 + 32z27 + 32z24 + 32z23 + 48z19 + 16z18 + 48z17 + 16z15 + 48z14 + 32z12 + 32z11 + 56z8 + 40z7 + 56z6 + 16z5 + 8z4 + 56z3 + 4z2 + 56z + 32 + 8 z + 52 z2 + 60 z3 + 30 z4 + 20 z5 + 20 z6 + 14 z7 Φ6(z) + 32z30 + 32z26 + 32z21 + 32z20 + 48z18 + 32z16 + 48z14 + 32z13 + 48z10 + 16z9 + 8z6 + 32z5 + 16z4 + 16z3 + 8z2 + 48z + 40 + 48 z + 8 z2 + 40 z3 + 60 z4 + 8 z5 + 24 z6 + 60 z7 Φ7(z) modulo 64. (9.11) 10. Counting permutation representations of Γm(3) for m prime Let m be a prime and, for a ﬁnitely generated group Γ, let hΓ(n) := \|Hom(Γ, Sn)\|. We want to determine the function hΓm(3)(n) counting the permutation representations of the lift Γm(3) of the inhomogeneous modular group PSL2(Z) ∼ = H(3) of degree n. Suppose ﬁrst that m ≥5, and classify the representations Γm(3) →Sn by the image ρ ∈Sn of the central element x2 = y3. The permutation ρ must be of the form ρ = r Y i=1 σi, 0 ≤r ≤⌊n/m⌋, with pairwise disjoint m-cycles σi. For ﬁxed r, the symmetric group Sn contains exactly 1 r! n m, . . . , m, n −mr (m −1)!r = n! r! (n −mr)! mr (10.1) such elements ρ. Next, given such ρ, the image of the generator x will contain a certain number s of 2m-cycles in its disjoint cycle decomposition, 0 ≤s ≤⌊r/2⌋, each of which breaks into two m-cycles when squared. Thus, in order to construct a square root of ρ (i.e., a possible image of x), we need to (i) ﬁx s in the range 0 ≤s ≤⌊r/2⌋, (ii) select s 2-element subsets from the r m-cycles of ρ, which can be done in 1 s! r 2s 2s 2, . . . , 2 = r! 2ss! (r −2s)! diﬀerent ways, (iii) lift each of these s pairs of m-cycles to a 2m-cycle (whose square is the product of the two given m-cycles), which can be done in m ways, 36 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER (iv) compute σω for each of the r −2s remaining m-cycles σ, where ω is the multi-plicative inverse of 2 modulo m, and (v) select a permutation π subject only to the condition that π2 = 1 on the n−mr letters not involved in any of the r m-cycles of ρ. Hence, there are precisely r! hC2(n −mr) ⌊r/2⌋ X s=0 ms 2ss! (r −2s)! (10.2) distinct square roots for each ρ involving r m-cycles. Similarly, the number of cubic roots of such ρ is given by r! hC3(n −mr) ⌊r/3⌋ X t=0 m2t 3tt! (r −3t)! (10.3) (classify the cubic roots by the number t of 3m-cycles, and use the fact that each product of three disjoint m-cycles is the cube of precisely 2m2 3m-cycles). Multiplying (10.1)–(10.3) and summing over r = 0, 1, . . . , ⌊n/m⌋, we ﬁnd that hΓm(3)(n) = n! ⌊n/m⌋ X r=0 ⌊r/2⌋ X s=0 ⌊r/3⌋ X t=0 r! hC2(n −mr)hC3(n −mr) 2s3tmr−s−2ts! t! (n −mr)! (r −2s)! (r −3t)!, m ≥5. (10.4) The cases where m = 2, 3 have to be treated separately. By arguments similar to the ones above, one ﬁnds that hΓ2(3)(n) = n! ⌊n/4⌋ X r=0 ⌊2r/3⌋ X s=0 (2r)! hC2(n −4r)hC3(n −4r) 22(r−s)3sr! s! (n −4r)! (2r −3s)!, (10.5) and that hΓ3(3)(n) = n! ⌊n/9⌋ X r=0 ⌊3r/2⌋ X s=0 (3r)! hC2(n −9r)hC3(n −9r) 2s32r−sr! s! (n −9r)! (3r −2s)!. (10.6) Furthermore, it is well-known that, for a prime p, hCp(n) = ⌊n/p⌋ X k=0 n! pk k! (n −pk)!, which allows us to make Formulae (10.4)–(10.6) completely explicit. 11. Subgroup numbers for the homogeneous modular group In this section we consider the problem of determining the behaviour of the number of index-n-subgroups in the homogeneous modular group SL2(Z) modulo powers of 2. By a folklore result that goes back at least to Dey , these subgroup numbers are in a direct relation to numbers of permutation representations of SL2(Z). Our start-ing point is a recurrence with polynomial coeﬃcients for the latter numbers, which is then translated into a Riccati-type diﬀerential equation for the generating function P n≥0 sn+1(SL2(Z)) zn of the subgroup numbers. (Equation (11.5) displays this equa-tion when reduced modulo 16.) Our method from Section 4 is then applied to this diﬀerential equation. Interestingly, it turns out that it succeeds only up to modulus 8 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 37 (see Theorems 24 and 25), while it fails for the modulus 16 (see Remark 26). This means that a new phenomenon, not covered by our Ansatz, arises in the behaviour of the subgroups numbers at the level of modulus 16. It would be of great interest to ﬁnd an explicit description of the hidden scheme behind the mod-16 behaviour of the number of subgroups of index n in SL2(Z), and, more generally, of the behaviour modulo any power of 2. Let us start with the afore-mentioned result (cf. [8, Theorem 6.10], see [9, Prop. 1] for a conceptual proof, plus generalisations) relating the numbers of subgroups of a ﬁnitely generated group to the numbers of its permutation representations. Proposition 23. Let Γ be a ﬁnitely generated group. Then we have ∞ X n=0 \| Hom(Γ, Sn)\|zn n! = exp ∞ X n=1 sn(Γ)zn n ! . (11.1) We take Γ = Γ2(3) = SL2(Z) and combine (11.1) with (10.5), the latter giving an explicit formula for the homomorphism numbers hn := \| Hom(SL2(Z), Sn)\|. Us-ing the Guessing package , we found a recurrence of order 30 for the sequence (hn/n!)n≥0, with coeﬃcients that are polynomials in n over Z. The validity of the recur-rence was veriﬁed by computing a certiﬁcate using Koutschan’s Mathematica package HolonomicFunctions .3 However, this recurrence is not suitable for our purpose, for which we require a recurrence with coeﬃcients that are polynomials in n over Z, and with leading coeﬃcient n. A recurrence of this form, if it exists, must be a left multiple of the recurrence operator corresponding to the minimal order recurrence. The con-struction of such left multiples is known as desingularisation, and algorithms are known for this purpose . This technique can be used to eliminate factors from the leading coeﬃcient of the recurrence (whenever possible), but it cannot be used to ensure that the leading coeﬃcient be a monic polynomial. The recurrence of order 32 mentioned in Footnote 3, with leading coeﬃcient 1, could indeed be used for our purpose, by simply multiplying it by n. However, since this recurrence has high-degree polynomials as coef-ﬁcients, we preferred to work with a diﬀerent recurrence with lower degree polynomials as coeﬃcients. The price to pay is that the order of such a recurrence will be higher. So, by an indeterminate Ansatz, we computed a candidate for a recurrence of the desired form of order 50, with polynomial coeﬃcients of degree at most 5.4 To be precise, it is the uniquely determined recurrence of the form 50 X k=0 5 X i=0 a(k, i)ni \| Hom(SL2(Z), Sn−k)\| (n −k)! = 0, n ≥50, 3The certiﬁcate has 4 megabytes, and, to obtain it, required about 30 hours of computation time. The coeﬃcients of this recurrence are polynomials in n of degree 34. Interestingly, there is a recurrence of order 32 with leading coeﬃcient 1. Although we did not try to prove it, it is likely that the recurrence of order 30 is the recurrence of minimal order. 4We used the function LinSolveQ of the Guessing package , which uses modular arithmetic, in order to solve the arising system of linear equations. Mathematica’s built-in linear system solver is not capable of solving it on current hardware due to the huge numerators and denominators of rational numbers which arise during the computation. 38 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER where a(0, 0) = a(0, 2) = a(0, 3) = a(0, 4) = a(0, 5) = 0, a(0, 1) = 1, a(50, 5) = 47323476536606893277939021129424044201294092725261226600745838\ 993897087202045010603943040012232525, a(50, 4) = −853333370519051585059335896571817612918194491041969759097679\ 3078743106989966250706985019403282594096, a(49, 5) = 2507660784286104701612089471873568042396155618028516886767837\ 559764248217845308468763736164634176, a(49, 4) = a(48, 5) = a(48, 4) = a(47, 5) = a(47, 4) = a(46, 5) = a(46, 4) = a(45, 5) = a(45, 4) = 0. Subsequently, we checked that this recurrence is a left-multiple of the certiﬁed recurrence of order 30, thereby establishing validity of this candidate recurrence of order 50. This last recurrence was then converted into a linear diﬀerential equation with polynomial coeﬃcients for the series H(z) := ∞ X n=0 hn zn n! = ∞ X n=0 \| Hom(SL2(Z), Sn)\|zn n! . Finally, this last mentioned diﬀerential equation can be translated into a Riccati-type diﬀerential equation for the generating function S(z) = X n≥0 sn+1(SL2(Z)) zn (11.2) for the subgroup numbers of SL2(Z). This is done by diﬀerentiating the relation (11.1), with Γ = SL2(Z), several times and by dividing by H(z). This leads to relations of the form H(k)(z) H(z) = Pk(S(z), S′(z), . . . ), k = 1, 2, . . . , (11.3) where Pk(S(z), S′(z), . . . ) is a polynomial in S(z) and its derivatives, which can be determined explicitly using the Faa di Bruno formula for derivatives of composite func-tions (cf. [6, Sec. 3.4]; but see also [7, 16]). Substituting these relations in the linear diﬀerential equation for H(z), one obtains the announced Riccati-type diﬀerential equa-tion for S(z). It turns out that this diﬀerential equation has integral coeﬃcients, so that it is amenable to our method from Section 4. The diﬀerential equation cannot be displayed here since this would require about ten pages. Its reduction modulo 16 is written out in (11.5). By applying our method from Section 4 with α = 0 to (11.5), we obtain the following theorem. It reﬁnes the parity result [18, Eq. (6.3) with \|H\| = 1, q = 3, m = 2]. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 39 Theorem 24. Let Φ(z) = P n≥0 z2n. Then we have X n≥0 sn+1(SL2(Z)) zn = 4z20 + 4z17 + 4z14 + 4z12 + 4z10 + 4z9 + 6z8 + 4z5 + 6z4 + 4z2 + 4z + 6 + 7 z2 + 3 z3 + 6 z6 + 6 z7 + 4 z8 + 1 z9 + 3 z11 + 6 z12 + 2 z13 + 6 z12 + 4 z10 + 4 z9 + 4 z8 + 2 z7 + 6 z6 + 6 z4 + 4z3 + 6 z3 + 4z2 + 4 z Φ(z) + 4z8 + 4z4 + 4z3 + 6z2 + 4 + 4 z + 6 z2 + 2 z3 + 5 z4 + 2 z5 + 6 z6 + 1 z7 + 4 z8 + 6 z9 + 4 z10 + 6 z11 + 6 z12 + 5 z13 Φ2(z) + 4z2 + 4 z2 + 4 z3 + 2 z4 + 4 z5 + 4 z6 + 2 z7 + 4 z9 + 4 z11 + 4 z12 + 2 z13 Φ3(z) modulo 8. (11.4) Proof. The Riccati-type diﬀerential equation for S(z) (as deﬁned in (11.2)) modulo 16 is5 p0(z) + p1(z)S(z) + p2(z)S(z)2 + p3(z)S(z)3 + p4(z)S(z)4 + p5(z)S(z)5 + p6(z)S′(z) + p7(z)S′(z)2 + p8(z)S(z)S′(z) + p9(z)S(z)2S′(z) + p10(z)S(z)3S′(z) + p11(z)S(z)S′(z)2 + p12(z)S′′(z) + p13(z)S(z)S′′(z) + p14(z)S(z)2S′′(z) + p15(z)S′(z)S′′(z) + p16(z)S′′′(z) + p17(z)S(z)S′′′(z) + p18(z)S′′′′(z) = 0 modulo 16, (11.5) with coeﬃcients pj(z), j = 0, 1, . . . , 18 as displayed in Appendix B. The diﬀerential equation (11.5) has a unique solution since comparison of coeﬃcients of zN ﬁxes the initial values, and yields a recurrence for the sequence sn(SL2(Z)) n≥1 which computes sn+1(SL2(Z)) from terms involving only si(SL2(Z)) with i ≤n. Now we apply the method from Section 4 with α = 0 to the diﬀerential equation (11.5). This yields the above result by means of a straightforward computer calculation. □ If we want to know explicitly for which n the subgroup number sn(SL2(Z)) is con-gruent to a particular value modulo 8, then we should ﬁrst apply the algorithm from Section 3 (see (3.1) and the proof of Lemma 6) in order to express powers of Φ(z) on the right-hand side of (11.4) in terms of the series Ha1,...,ar(z). (The corresponding expansions are in fact listed in (2.6) and (2.7).) The result is 5We display the diﬀerential equation modulo 16 in order to prepare for Remark 26. 40 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER X n≥0 sn+1(SL2(Z)) zn = 4 z4 + 4 z7 + 4 z13 H3(z) + 4 z4 + 4 z7 + 4 z13 H1,1,1(z) + 4z2 + 4 z2 + 4 z3 + 6 z4 + 4 z5 + 4 z6 + 6 z7 + 4 z9 + 4 z11 + 4 z12 + 6 z13 H1,1(z) + 4z4 + 4z3 + 6z2 + 4 + 4z8 + 4 z + 6 z2 + 6 z3 + 5 z4 + 2 z5 + 2 z6 + 1 z7 + 4 z8 + 6 z9 + 4 z10 + 6 z11 + 2 z12 + 5 z13 H1(z) + 4z20 + 4z17 + 4z14 + 4z12 + 4z10 + 6z8 + 2z4 + 6z3 + 4z2 + 2 + 6 z + 1 z2 + 2 z4 + 6 z5 + 7 z6 + 2 z7 + 2 z8 + 5 z9 + 6 z10 + 1 z11 + 3 z12 modulo 8. (11.6) From this expression, it is a routine (albeit tedious) task to extract an explicit de-scription of the behaviour of the subgroup numbers of SL2(Z) modulo 8. Since the corresponding result can be stated within moderate amount of space, we present it in the next theorem. Theorem 25. The subgroup numbers sn(SL2(Z)) obey the following congruences mod-ulo 8 : (i) sn(SL2(Z)) ≡1 (mod 8) if, and only if, n = 1, 2, 4, 10, or if n is of the form 2σ −3 for some σ ≥4; (ii) sn(SL2(Z)) ≡2 (mod 8) if, and only if, n = 7, 12, 17, or if n is of one of the forms 3 · 2σ −3, 3 · 2σ −6, 3 · 2σ −12, for some σ ≥4; (iii) sn(SL2(Z)) ≡4 (mod 8) if, and only if, n = 3, 22, 23, 27, 46, 47, 51, or if n is of one of the forms 2σ + 6, 2σ + 7, 2σ + 11, 2σ + 12, 2σ + 18, 2σ + 21, for some σ ≥5, (11.7) 2σ + 2τ −2, 2σ + 2τ + 1, 2σ + 2τ + 3, for some σ, τ with σ ≥6 and 4 ≤τ ≤σ −1, (11.8) 2σ + 2τ + 2ν −12, 2σ + 2τ + 2ν −6, 2σ + 2τ + 2ν −3, for some σ, τ, ν with σ ≥6, 5 ≤ν ≤σ −1, and 3 ≤τ ≤ν −1; (11.9) (iv) sn(SL2(Z)) ≡5 (mod 8) if, and only if, n = 5, or if n is of one of the forms 2σ −6, 2σ −12, for some σ ≥5; MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 41 (v) sn(SL2(Z)) ≡6 (mod 8) if, and only if, n = 6, 11, 14, 18, 19, 21, 33, 34, 35, 37, or if n is of one of the forms 2σ −2, 2σ −4, for some σ ≥5, (11.10) 2σ + 1, 2σ + 2, 2σ + 3, 2σ + 4, 2σ + 5, 2σ + 10, 2σ + 13, for some σ ≥6, (11.11) 2σ + 2τ −3, 2σ + 2τ −6, 2σ + 2τ −12, for some σ, τ with σ ≥7 and 5 ≤τ ≤σ −2; (11.12) (vi) in the cases not covered by items (i)–(v), sn(SL2(Z)) is divisible by 8; in par-ticular, sn(SL2(Z)) ̸≡3, 7 (mod 8) for all n. Remark 26. If we apply the method from Section 4 with the polynomial (Φ2(z) + Φ(z) + z)(Φ4(z) + 6Φ3(z) + (2z + 3)Φ2(z) + (2z + 6)Φ(z) + 2z + 5z2) in place of the polynomial on the left-hand side of (4.4) with α = 0 (that is, in view of Proposition 2 we are aiming at determining the subgroup numbers sn(SL2(Z)) modulo 16), then, when we arrive at the mod-8-level, we obtain X n≥0 sn+1(SL2(Z)) zn = 4z20 + 4z17 + 4z14 + 4z12 + 4z9 + 6z8 + 4z6 + 4z4 + 4z3 + 2 z + 4 z2 + 3 z3 + 6 z4 + 1 z5 + 4 z6 + 4 z7 + 4 z8 + 3 z9 + 6 z10 + 4z4 + 4 + 6 z2 + 4 z3 + 6 z5 + 4 z7 + 4 z9 + 6 z11 Φ(z) + 4z3 + 4 z + 2 z3 + 4 z4 + 2 z6 + 4 z8 + 4 z10 + 2 z12 Φ2(z) + 4z2 + 4 z2 + 4 z3 + 2 z4 + 4 z5 + 4 z6 + 2 z7 + 4 z9 + 4 z11 + 4 z12 + 2 z13 Φ3(z) + 4z8 + 4z4 + 4z3 + 6z2 + 4 + 4 z + 2 z2 + 6 z3 + 1 z4 + 6 z5 + 2 z6 + 5 z7 + 4 z8 + 6 z9 + 4 z10 + 2 z11 + 2 z12 + 1 z13 Φ4(z) + 4z2 + 4 z2 + 6 z4 + 4 z5 + 6 z7 + 4 z9 + 4 z11 + 6 z13 Φ5(z) modulo 8. (11.13) However, the system of equations for the next level, the mod-16-level, has no polynomial solutions. 12. Subgroup numbers for the lift Γ3(3) Continuing in the spirit of the previous section, we now consider the number of index-n-subgroups in the lift Γ3(3) (of the Hecke group H(3) ∼ = PSL2(Z)) modulo powers 42 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER of 2. We shall see that, again, our method from Section 4 works up to modulus 8 (see Theorems 27 and 28), while it fails for the modulus 16 (see Remark 29). We take Γ = Γ3(3) in (11.1) and combine the resulting formula with (10.6), the lat-ter giving an explicit formula for the homomorphism numbers hn := \| Hom(Γ3(3), Sn)\|. Using the Guessing package , we found a recurrence of order 42 for the sequence (hn/n!)n≥0, with coeﬃcients that are polynomials in n over Z. The validity of the recur-rence was veriﬁed by computing a certiﬁcate using Koutschan’s Mathematica package HolonomicFunctions .6 However, again, this recurrence is not suitable for our pur-pose, for which we require a recurrence with coeﬃcients that are polynomials in n over Z, and with leading coeﬃcient n. By an indeterminate Ansatz, we computed a can-didate for a recurrence of the desired form of order 60, with polynomial coeﬃcients of degree at most 10.7 To be precise, it is the uniquely determined recurrence of the form 60 X k=0 10 X i=0 b(k, i)ni \| Hom(Γ3(3), Sn−k)\| (n −k)! = 0, n ≥60, where b(0, 0) = b(0, 2) = b(0, 3) = b(0, 4) = b(0, 5) = b(0, 6) = b(0, 7) = b(0, 8) = b(0, 9) = b(0, 10) = 0, b(0, 1) = 1, b(60, 8) = 9649124343496238177846526221678676069879148435557456840677\ 68567400990643919180258204664996863270960793634431477\ 96875828563496094243333614632539311543926582958877938\ 09887854513738722474642524334737161421912431106592005\ 22984304410147101964876864298627928130880022459406799\ 539461032349694733915947489297372243661012, b(60, 10) = b(60, 9) = b(60, 4) = b(59, 10) = b(59, 9) = b(59, 8) = b(59, 7) = b(59, 6) = b(59, 5) = b(59, 4) = b(59, 3) = b(59, 2) = b(59, 1) = b(59, 0) = b(58, 10) = b(58, 9) = b(58, 8) = b(58, 7) = b(58, 6) = b(58, 5) = b(58, 4) = b(57, 10) = b(57, 9) = b(57, 8) = b(57, 7) = b(57, 6) = b(57, 5) = b(57, 4) = b(56, 10) = b(56, 9) = b(56, 8) = b(56, 7) = b(56, 6) = b(56, 5) = b(56, 4) = b(55, 10) = b(55, 9) = b(55, 8) = b(55, 7) = b(55, 6) = b(55, 5) = b(55, 4) = b(54, 10) = b(54, 9) = b(54, 8) = b(54, 7) = b(54, 6) = b(54, 5) = b(54, 4) 6The computation took about one week, producing a certiﬁcate of 28 megabytes. The coeﬃcients of this recurrence are polynomials in n of degree 105. In this case, we were not able to ﬁnd a recurrence with leading coeﬃcient 1. (It may still exist.) The best that we found in this direction was a recurrence with leading coeﬃcient a 1828-digit number. Again, although we did not try to prove it, it is likely that the recurrence of order 42 is the recurrence of minimal order. 7Again, we used the function LinSolveQ of the Guessing package in order to solve the arising system of linear equations. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 43 = b(53, 10) = b(53, 9) = b(53, 8) = b(53, 7) = b(53, 6) = b(53, 5) = b(53, 4) = b(52, 10) = b(52, 9) = b(52, 8) = b(50, 7) = b(50, 6) = b(49, 10) = b(49, 9) = b(49, 8) = b(49, 7) = b(49, 6) = b(49, 5) = b(49, 4) = b(1, 7) = 0. Subsequently, we checked that this recurrence is a left-multiple of the certiﬁed recurrence of order 42, thereby establishing validity of this candidate recurrence of order 60. This last recurrence was then converted into a linear diﬀerential equation with polynomial coeﬃcients for the series H(z) := ∞ X n=0 hn zn n! = ∞ X n=0 \| Hom(Γ3(3), Sn)\|zn n! . Finally, this last mentioned diﬀerential equation can be translated into a Riccati-type diﬀential equation for the generating function S(z) = X n≥0 sn+1(Γ3(3)) zn (12.1) for the subgroup numbers of Γ3(3) in the same way as we obtained (11.5) in the previous section. It turns out that this diﬀerential equation has integral coeﬃcients, so that it is amenable to our method from Section 4. The diﬀerential equation cannot be displayed here since this would require about 100 pages.8 Its reduction modulo 16 is written out in (12.3). By applying our method from Section 4 with α = 0 to (12.3), we obtain the following theorem. It reﬁnes the parity result [18, Eq. (6.3) with \|H\| = 1, q = 3, m = 3]. Theorem 27. Let Φ(z) = P n≥0 z2n. Then we have X n≥0 sn+1(Γ3(3)) zn = 4z62 + 4z53 + 4z44 + 4z35 + 6z26 + 4z20 + 4z14 + 4z12 + 4z11 + 4z10 + 4z9 + 4z5 + 6z4 + 4z3 + 4z2 + 4z + 6 + 7 z2 + 7 z3 + 3 z5 + 6 z6 + 4z3 + 4z2 + 4 z + 6 z3 + 6 z4 + 6 z6 + 2 z7 Φ(z) + 4z8 + 4z4 + 4z3 + 6z2 + 4 + 4 z + 6 z2 + 2 z3 + 5 z4 + 6 z5 + 6 z6 + 5 z7 Φ2(z) + 4z2 + 4 z2 + 4 z3 + 2 z4 + 4 z5 + 4 z6 + 2 z7 Φ3(z) modulo 8. (12.2) 8The integers appearing as coeﬃcients have up to 320 digits. 44 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Proof. The Riccati-type diﬀerential equation for S(z) (as deﬁned in (12.1)) modulo 16 is9 q0(z) + q1(z)S(z) + q2(z)S(z)S′(z) + q3(z)S(z)S′(z)2 + q4(z)S(z)S′(z)3 + q5(z)S(z)S′(z)4 + q6(z)S(z)S′′(z) + q7(z)S(z)S′′(z)2 + q8(z)S(z)S′′′(z) + q9(z)S(z)S′′′(z)2 + q10(z)S(z)S′′′′′(z) + q11(z)S(z)S′(z)S′′(z) + q12(z)S(z)S′(z)S′′′(z) + q13(z)S(z)S′(z)2S′′′(z) + q14(z)S(z)2 + q15(z)S(z)2S′(z) + q16(z)S(z)2S′(z)2 + q17(z)S(z)2S′(z)3 + q18(z)S(z)2S′(z)4 + q19(z)S(z)2S′′(z) + q20(z)S(z)2S′′′(z) + q21(z)S(z)2S′′′(z)2 + q22(z)S(z)2S′′′′(z) + q23(z)S(z)2S′(z)S′′(z) + q24(z)S(z)2S′(z)2S′′(z) + q25(z)S(z)2S′(z)S′′′(z) + q26(z)S(z)2S′(z)2S′′′(z) + q27(z)S(z)3 + q28(z)S(z)3S′(z) + q29(z)S(z)3S′(z)2 + q30(z)S(z)3S′(z)3 + q31(z)S(z)3S′′(z) + q32(z)S(z)3S′′′(z) + q33(z)S(z)3S′(z)S′′′(z) + q34(z)S(z)4 + q35(z)S(z)4S′(z) + q36(z)S(z)4S′(z)2 + q37(z)S(z)4S′(z)3 + q38(z)S(z)4S′′(z) + q39(z)S(z)4S′′′(z) + q40(z)S(z)4S′(z)S′′(z) + q41(z)S(z)4S′(z)S′′′(z) + q42(z)S(z)5 + q43(z)S(z)5S′(z) + q44(z)S(z)5S′(z)2 + q45(z)S(z)5S′′(z) + q46(z)S(z)5S′′′(z) + q47(z)S(z)6 + q48(z)S(z)6S′(z) + q49(z)S(z)6S′(z)2 + q50(z)S(z)6S′′(z) + q51(z)S(z)6S′′′(z) + q52(z)S(z)7 + q53(z)S(z)7S′(z) + q54(z)S(z)8 + q55(z)S(z)8S′(z) + q56(z)S(z)9 + q57(z)S(z)10 + q58(z)S′(z) + q59(z)S′(z)S′′(z) + q60(z)S′(z)S′′′(z) + q61(z)S′(z)S′′′(z)2 + q62(z)S′(z)S′′′′(z) + q63(z)S′(z)2 + q64(z)S′(z)2S′′(z) + q65(z)S′(z)2S′′′(z) + q66(z)S′(z)3 + q67(z)S′(z)3S′′(z) + q68(z)S′(z)3S′′′(z) + q69(z)S′(z)4 + q70(z)S′(z)5 + q71(z)S′′(z) + q72(z)S′′(z)S′′′(z) + q73(z)S′′(z)2 + q74(z)S′′′(z) + q75(z)S′′′(z)2 + q76(z)S′′′′(z) + q77(z)S′′′′′(z) = 0 modulo 16, (12.3) with coeﬃcients qj(z), j = 0, 1, . . . , 77 as displayed in Appendix C. The diﬀerential equation (12.3) has a unique solution since comparison of coeﬃcients of zN ﬁxes the initial values, and yields a recurrence for the sequence sn(Γ3(3)) n≥1 which computes sn+1(Γ3(3)) from terms involving only si(Γ3(3)) with i ≤n. 9We display the diﬀerential equation modulo 16 in order to prepare for Remark 29. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 45 Now we apply the method from Section 4 with α = 0 to the diﬀerential equation (12.3). This yields the above result by means of a straightforward computer calcula-tion.10 □ Also here, if we want to know criteria in terms of n when a subgroup number sn(Γ3(3)) is congruent to a particular value modulo 8, then we must ﬁrst apply the algorithm from Section 3 to the right-hand side of (12.2). This leads to the identity X n≥0 sn+1(Γ3(3)) zn = 4 z4 + 4 z7 H3(z) + 4 z4 + 4 z7 H1,1,1(z) + 4z2 + 4 z2 + 4 z3 + 6 z4 + 4 z5 + 4 z6 + 6 z7 H1,1(z) + 4z8 + 4z4 + 4z3 + 6z2 + 4 + 4 z + 6 z2 + 6 z3 + 5 z4 + 6 z5 + 2 z6 + 5 z7 H1(z) + 4z62 + 4z53 + 4z44 + 4z35 + 6z26 + 4z20 + 4z14 + 4z12 + 4z11 + 4z10 + 2z4 + 2z3 + 4z2 + 2 + 6 z + 1 z2 + 4 z3 + 6 z4 + 1 z5 + 3 z6 modulo 8, (12.4) from which we can extract the following explicit description of the behaviour of the subgroup numbers of Γ3(3) modulo 8. Theorem 28. The subgroup numbers sn(Γ3(3)) obey the following congruences modulo 8 : (i) sn(Γ3(3)) ≡1 (mod 8) if, and only if, n = 1, 2, 10, or if n is of the form 2σ −3 for some σ ≥4; (ii) sn(Γ3(3)) ≡2 (mod 8) if, and only if, n = 7, 9, 17, 18, 27, 42, or if n is of one of the forms 3 · 2σ −3, 3 · 2σ −6, for some σ ≥5; (iii) sn(Γ3(3)) ≡4 (mod 8) if, and only if, n = 3, 12, 22, 23, 36, 38, 39, 43, 46, 49, 50, 51, 53, 54, 63, or if n is of one of the forms 2σ + 6, 2σ + 7, 2σ + 11, 2σ + 14, 2σ + 17, 2σ + 18, 2σ + 19, 2σ + 21, for some σ ≥6, (12.5) 2σ + 2τ −2, 2σ + 2τ + 1, 2σ + 2τ + 2, 2σ + 2τ + 3, 2σ + 2τ + 5, 2σ + 2τ + 10, 2σ + 2τ + 13, for some σ, τ with σ ≥6 and 5 ≤τ ≤σ −1, (12.6) 2σ + 2τ + 2ν −6, 2σ + 2τ + 2ν −3, for some σ, τ, ν with σ ≥7, 6 ≤ν ≤σ −1, and 5 ≤τ ≤ν −1; (12.7) (iv) sn(Γ3(3)) ≡5 (mod 8) if, and only if, n = 5, or if n is of the form 2σ −6 for some σ ≥5; 10The calculation being straightforward, it nevertheless required a machine with substantial amount of memory (we had available 32 gigabytes of memory, of which almost 50% were used). 46 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER (v) sn(Γ3(3)) ≡6 (mod 8) if, and only if, n = 6, 11, or if n is of one of the forms 2σ −2, 2σ + 3, 2σ + 4, for some σ ≥4, (12.8) 2σ + 1, 2σ + 2, 2σ + 13, for some σ ≥5, (12.9) 2σ + 10, for some σ ≥6, (12.10) 2σ + 2τ −6, 2σ + 2τ −3, for some σ, τ with σ ≥7 and 5 ≤τ ≤σ −2; (12.11) (vi) in the cases not covered by items (i)–(v), sn(Γ3(3)) is divisible by 8; in partic-ular, sn(Γ3(3)) ̸≡3, 7 (mod 8) for all n. Remark 29. If we apply the method from Section 4 with the polynomial (Φ2(z) + Φ(z) + z)(Φ4(z) + 6Φ3(z) + (2z + 3)Φ2(z) + (2z + 6)Φ(z) + 2z + 5z2) in place of the polynomial on the left-hand side of (4.4) with α = 0 (that is, in view of Proposition 2 we are aiming at determining the subgroup numbers sn(Γ3(3)) modulo 16), then, when we arrive at the mod-8-level, we obtain X n≥0 sn+1(Γ3(3)) zn = 4z62 + 4z53 + 4z44 + 4z35 + 6z26 + 4z20 + 4z14 + 4z12 + 4z11 + 4z9 + 4z6 + 4z4 + 2 z + 4 z2 + 3 z3 + 6 z4 + 4z4 + 4 + 6 z2 + 4 z3 + 6 z5 Φ(z) + 4z3 + 4 z + 2 z3 + 4 z4 + 2 z6 Φ2(z) + 4z2 + 4 z2 + 4 z3 + 2 z4 + 4 z5 + 4 z6 + 2 z7 Φ3(z) + 4z8 + 4z4 + 4z3 + 6z2 + 12 + 4 z + 2 z2 + 6 z3 + 1 z4 + 2 z5 + 2 z6 + 1 z7 Φ4(z) + 4z2 + 4 z2 + 6 z4 + 4 z5 + 6 z7 Φ5(z) modulo 8. (12.12) However, the system of equations for the next level, the mod-16-level, has no polynomial solutions.11 13. A variation I: free subgroup numbers for lifts of Hecke groups In this section, we consider the functional equation zf 2h(z) −f(z) + 1 = 0, (13.1) which generalises the functional equation (5.1) for the generating function of Catalan numbers. It is easy to see that this equation has a unique formal power series solution. 11The corresponding computation took almost 5 hours, using 94% of the 32 gigabytes of memory of the machine on which the computation was performed. MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 47 The coeﬃcients of this uniquely determined series can be calculated explicitly by means of the Lagrange inversion formula, the result being ⟨zn⟩f(z) = 1 n 2hn n −1 , (13.2) but this will not be relevant here. Again, the numbers in (13.2) are special instances of numbers that are now commonly known as Fuß–Catalan numbers (cf. the paragraph containing (6.1)). Our aim is to determine the coeﬃcients of f(z) modulo powers of 2. Our solution of this problem is that, again, the series f(z) can be expressed as a polynomial in a “basic” series. Here, this basic series is Φh(z) = X n≥0 z2nh/(2h−1). (13.3) It will turn out (see Corollary 31) that an adaptation of the proof of the theorem below will allow us to treat as well the behaviour, modulo powers of 2, of free subgroup numbers of lifts of Hecke groups H(q), with q a Fermat prime. The theorem below, in a certain sense, extends Theorem 11. It does not, however, reduce to it for h = 1, due to the choice that, in the proof below, the reductions in our algorithm are based on the polynomial relation (13.7) for the basic series Φh(z), which, for h = 1, is “weaker” than the relation (4.4) which is used in the proof of Theorem 11. Theorem 30. For a positive integer h, let Φh(z) = P n≥0 z2nh/(2h−1), and let α be a further positive integer. Then the unique solution f(z) to (13.1), reduced modulo 22αh, can be expressed as a polynomial in Φh(z) of degree at most 2(α+1)h −1 with coeﬃcients that are Laurent polynomials in z1/(2h−1). Proof. For ease of notation, we replace z by z2h−1 in (13.1), thereby obtaining the equation z2h−1 ˜ f 2h(z) −˜ f(z) + 1 = 0, (13.4) with ˜ f(z) = f(z2h−1). We now have to prove that, modulo 22αh, the series ˜ f(z) can be expressed as a polynomial in ˜ Φh(z) = ∞ X n=0 z2nh (13.5) of degree at most 2(α+1)h −1 with coeﬃcients that are Laurent polynomials in z. It is readily veriﬁed that ˜ Φ2h h (z) + ˜ Φh(z) + z = 0 modulo 2, (13.6) whence ˜ Φ2h h (z) + ˜ Φh(z) + z 2αh = 0 modulo 22αh. (13.7) We modify our Ansatz (4.2) to ˜ f(z) = 2(α+1)h−1 X i=0 ai(z)˜ Φi h(z) modulo 22αh, (13.8) where the ai(z)’s are (at this point) undetermined Laurent polynomials in z. 48 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Next, we gradually ﬁnd approximations ai,β(z) to ai(z) such that (13.4) holds modulo 2β, for β = 1, 2, . . . , 2αh. To start the procedure, we consider the diﬀerential equation (13.4) modulo 2, with ˜ f(z) = 2(α+1)h−1 X i=0 ai,1(z)˜ Φi h(z) modulo 2. (13.9) We substitute the Ansatz (13.9) in (13.4), reduce high powers of ˜ Φh(z) by using the relation (13.7), reduce the resulting expression modulo 2, thereby taking advantage of the elementary fact that ˜ Φ′ h(z) = 1 modulo 2, and we ﬁnally see that the left-hand side of (13.4) becomes a polynomial in ˜ Φh(z) of degree at most 2(α+1)h −1 with coeﬃcients that are Laurent polynomials in z. Now we compare coeﬃcients of powers ˜ Φk h(z), k = 0, 1, . . . , 2(α+1)h −1. This yields a system of 2(α+1)h equations (modulo 2) for the unknown Laurent polynomials ai,1(z), i = 0, 1, . . . , 2(α+1)h −1. Since we have already done similar computations several times before, we content ourselves with stating the result: all Laurent polynomials ai,1(z) must be zero, except for a0,1(z) and a2αh,1(z), which are given by a0,1(z) = α−1 X k=0 z2kh−1, a2αh,1(z) = z−1. (13.10) After we have completed the “base step,” we now proceed with the iterative steps described in Section 4. Our Ansatz here (replacing the corresponding one in (4.6)–(4.7)) is ˜ f(z) = 2(α+1)h−1 X i=0 ai,β+1(z)˜ Φi h(z) modulo 2β+1, (13.11) with ai,β+1(z) := ai,β(z) + 2βbi,β+1(z), i = 0, 1, . . . , 2(α+1)h −1, (13.12) where the coeﬃcients ai,β(z) are supposed to provide a solution ˜ fβ(z) = 2(α+1)h−1 X i=0 ai,β(z)˜ Φi h(z) to (13.4) modulo 2β. This Ansatz, substituted in (13.4), produces the congruence z2h−1 ˜ f 2h β (z) −˜ fβ(z) + 2β 2(α+1)h−1 X i=0 bi,β+1(z)˜ Φi h(z) + 1 = 0 modulo 2β+1. (13.13) By our assumption on ˜ fβ(z), we may divide by 2β. Comparison of powers of ˜ Φh(z) then yields a system of congruences of the form bi,β+1(z) + Poli(z) = 0 modulo 2, i = 0, 1, . . . , 2(α+1)h −1, (13.14) where Poli(z), i = 0, 1, . . . , 2(α+1)h −1, are certain Laurent polynomials with integer coeﬃcients. This system being trivially uniquely solvable, we have proved that, for an arbitrary positive integer α, the modiﬁed algorithm that we have presented here will MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 49 produce a solution ˜ f2αh(z) to (13.4) modulo 22αh which is a polynomial in ˜ Φh(z) with coeﬃcients that are Laurent polynomials in z. □ It has been shown in that the parity pattern of free subgroup numbers in Hecke groups H(q), q a Fermat prime, coincides with the parity pattern of (special) Fuß– Catalan numbers. More precisely, let f (q) λ denote the number of free subgroups of index 2qλ in the Hecke group H(q). (For indices not divisible by 2q, no free subgroups exist in H(q).) Then (see [25, Eq. (37)]) f (q) λ = 1 λ (q −1)λ λ −1 modulo 2. The reader should keep in mind that q −1 is a 2-power. Theorem 30 says that the generating function for the Fuß–Catalan numbers (13.2), when reduced modulo a given power of 2, can be expressed as a polynomial in Φh(z). We are now going to show that the same is true for the generating function for free subgroup numbers in the Hecke group H(q), although the equation it satisﬁes is diﬀerent from the functional equation (13.1) for the generating function of Fuß–Catalan numbers. In the corollary below, we present actually a more general result: even the generating function for free subgroup numbers of the lift Γm(q), when reduced modulo a given power of 2, can be expressed as a polynomial in Φh(z) in the case where q is a Fermat prime. In a certain sense, this extends Theorem 17, although it does not reduce to it for h = 1. Again, the reason lies in the choice that, in the proof below, the reductions in our algorithm are based on the polynomial relation (13.7) for the basic series Φh(z), which, for h = 1, is “weaker” than the relation (4.4) which is used in the proof of Theorem 17. On the other hand, the corollary does largely extend the parity result [25, Cor. A’]. Corollary 31. Let q = 22f + 1 be a Fermat prime, and let γ be some positive in-teger. Then, for every positive integer m, the generating function Fm(q; z) = 1 + P λ≥1 f (q) λ (m)zλ of free subgroup numbers of Γm(q), when reduced modulo 2γ, can be expressed as a polynomial in Φ2f (z) of degree at most 22fγ −1 with coeﬃcients that are Laurent polynomials in z1/(q−2), where the series Φh(z) is deﬁned as in (13.3). Proof. Equation (7.3) provides a Riccati-type diﬀerential equation for Fm(q; z) = 1 + zGm(q; z). Moreover, this equation, considered modulo 2, is the same for every odd m. Namely, we have 1 z Fm(q; z) −1 = A0(H(q)) + q−1 X µ=1 µ X ν=1 X µ1,...,µν>0 µ1+···+µν=µ µ µ1, . . . , µν ν! (2q)ν−1Aµ(H(q))zµ × ν Y j=1 1 z Fm(q; z) −1 (µj−1) modulo 2. Moreover, it is shown in [25, Prop. 2] that, modulo 2, this diﬀerential equation reduces to zF q−1 m (q; z) −Fm(q; z) + 1 = 0 modulo 2. (13.15) 50 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER The latter statement means that reduction of coeﬃcients modulo 2 and usage of the simple fact that F ′′ m(q; z) = 0 modulo 2 (13.16) leads from the original diﬀerential equation (7.3) for Fm(q; z) = 1 + Gm(q; z) to the congruence (13.15). With q −1 being a power of 2 by assumption, we observe that, disregarding the restriction to modulus 2, Equation (13.15) is the special case of (13.1) where h = 2f. In particular, if, for the moment, we assume that γ = 2α2f , for some positive integer α, then we see that the base step of the Ansatz outlined (and applied) in the proof of Theorem 30 (with h = 2f) can be successfully performed here: it would yield exactly the same result as there, namely (13.9) with the Laurent polynomials ai,1(z) being given in the paragraph containing (13.10). However, also the subsequent iterative steps would just work in the same way as in the preceding proof! Indeed, transform the Riccati-type diﬀerential equation (7.3) for Fm(q; z) by the substitution z 7→zq−2 (in analogy with the substitution leading to (13.4)). This yields a Riccati-type diﬀerential equation for Fm(q; zq−2). A ﬁne point to be observed here is that, in this equation, the coeﬃcients will not necessarily be integral; due to the substitution rule for diﬀerentials, denominators that are powers of (q −2) may occur. As in the proof of Theorem 30, we now continue with the Ansatz Fm(q; zq−2) = 2(α+1)2f −1 X i=0 ai,β+1(z)˜ Φi 2f (z) modulo 2β+1, (13.17) with ai,β+1(z) := ai,β(z) + 2βbi,β+1(z), i = 0, 1, . . . , 2(α+1)2f −1 (13.18) (which is analogous to (13.11)–(13.12)), where the coeﬃcients ai,β(z) are supposed to provide a solution Fm,β(q; z) = 2(α+1)2f −1 X i=0 ai,β(z)˜ Φi 2f (z) to the diﬀerential equation for Fm(q; zq−2) modulo 2β. The fact that reduction modulo 2 and usage of (13.16) leads from the original diﬀerential equation for Fm(q; z) to (13.15) implies that substitution of the Ansatz (13.17)–(13.18) in the diﬀerential equation for Fm(q; zq−2) yields an equation completely analogous to (13.13), namely zq−2F q−1 m,β (q; z) −Fm,β(q; z) + 2β 2α+h−1 X i=0 bi,β+1(z)˜ Φi h(z) + 1 + T(z, Fm,β(q; z)) = 0 modulo 2β+1. Here, T(z, Fm,β(q; z)) consists only of terms that may depend on Fm,β(q; z) but do not depend on the bi,β+1(z)’s. The rest of the procedure is then as in the preceding proof: we divide by 2β, compare powers of ˜ Φh(z), and obtain a system of congruences of the form (13.14), which is trivially solvable. The powers of (q −2) that may appear in the denominators of the coeﬃcients in the polynomials involved here are disposed of by interpreting them appropriately as elements of Z/2γZ. Finally, if we are able to express Fm(q; z) as a polynomial in Φ2f (z) modulo 2γ = 22α2f for all α, then the same assertion must hold for every γ. □ MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 51 In order to illustrate the algorithm described in the last proof, let us consider the case of the Hecke group H(5), that is, the case of Corollary 31 where f = 1. The Riccati-type diﬀerential equation for the series Gm(z) := Gm(5; z) = P∞ λ=0 f (5) λ+1(m)zλ that one obtains from (7.3) in this special case reads Gm(z) = 189m4 + 4600m3zGm(z) + 1430m2z2G2 m(z) + 80mz3G3 m(z) + z4Gm(z)4 + 14300m3z2G′ m(z) + 2400m2z3Gm(z)G′ m(z) + 60mz4G2 m(z)G′ m(z) + 300m2z4G′ m(z) 2 + 8000m3z3G′′ m(z) + 400m2z4Gm(z)G′′ m(z) + 1000m3z4G′′′ m(z). Since Gm(z) = Gm(5; z) = 1 z Fm(5; z) −1 , we obtain the diﬀerential equation 1 + (189m4 −300m3 + 130m2 −20m + 1)z + ((300m3 −260m2 + 60m −4)z −1)Fm(5; z) + (130m2 −60m + 6)zF 2 m(5; z) + (20m −4)zF 3 m(5; z) + zF 4 m(5; z) + (4300m3 −1000m2 + 60m)z2F ′ m(5; z) + (1000m2 −120m)z2Fm(5; z)F ′ m(5; z) + 60mz2Fm(5; z)2F ′ m(5; z) + 300m2z3F ′ m(5; z)2 + (5000m3 −400m2)z3F ′′ m(5; z) + 400m2z3Fm(5; z)F ′′ m(5; z) + 1000m3z4F ′′′ m(5; z) = 0 (13.19) for Fm(5; z). We have implemented the algorithm described in the proof of Theorem 31 for this diﬀerential equation. For the modulus 16, it produces the following result. (It is independent of m because of the high divisibility of the coeﬃcients in the diﬀerential equation (13.19) by powers of 2. The parameter m will show up for 2-powers higher than 24 = 16.) Theorem 32. Let Φ2(z) = P n≥0 z4n/3, as before. Then, for the generating function Fm(5; z) = 1 + P λ≥1 f (5) λ (m)zλ for the free subgroup numbers of Γm(5), we have Fm(5; z) = 4z + 1 + 12z2/3Φ2(z) + 10z1/3Φ2 2(z) + 12Φ3 2(z) + 4z2/3 + 7z−1/3 Φ4 2(z) + 4z1/3Φ5 2(z) + 4Φ6 2(z) + 12z−1/3Φ7 2(z) + 8z1/3Φ8 2(z) + 4Φ9 2(z) + 2z−1/3Φ10 2 (z) + 12Φ12 2 (z) + 12z−1/3Φ13 2 (z) modulo 16. (13.20) Clearly, coeﬃcient extraction from powers of Φ2(z) (and, more generally, from powers of Φh(z)) can be accomplished by appropriately adapting the results in Section 3. 14. A variation II: subgroup numbers for Hecke groups In Section 9, we proved that the generating function for the subgroup numbers of the inhomogeneous modular group PSL2(Z) ∼ = H(3), when reduced modulo a power of 2, can always be expressed as a polynomial in the basic series Φ(z) with coeﬃcients that are Laurent polynomials in z. Here, we discuss possible extensions of this result to Hecke groups H(q), where q is a Fermat prime. Again, we have to modify the original method from Section 4 by using the series Φh(z) deﬁned in (13.3) (for suitable h) instead of Φ(z). We conjecture (see Conjecture 35) that this variation of our method 52 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER will always be successful, for arbitrary Fermat primes q. If q = 5, we are actually able to demonstrate this conjecture, thereby largely reﬁning the q = 5 case of [25, Theorem B]. Theorem 33. With notation from the previous section, let Φ2(z) = P n≥0 z4n/3, and let α be a positive integer. Then the generating function S(z) = SH(5)(z) (see the ﬁrst paragraph of Section 9 for the deﬁnition), reduced modulo 24α, can be expressed as a polynomial in Φ2(z) of degree at most 4α+1 −1 with coeﬃcients that are Laurent polynomials in z1/3. Proof. Let h(n) = 1 n!hC2(n)hC5(n). Using the routine to compute recurrences for the Hadamard product of recursive se-quences, implemented in gfun and GeneratingFunctions (cf. [30, Theo-rem 6.4.12] for the theoretical background), one obtains the recurrence n(16 −72n + 174n2 −155n3 + 65n4 −13n5 + n6)h(n) −(184 −620n + 854n2 −555n3 + 177n4 −25n5 + n6)h(n −1) −(856 −1636n + 1250n2 −479n3 + 101n4 −13n5 + n6)h(n −2) −4(n −6)(n −3)2(−7 + 3n)h(n −3) + 8(n −7)(n −4)(3n −7)h(n −4) −(1136 −856n + 1292n2 −2930n3 + 3115n4 −1718n5 + 516n6 −80n7 + 5n8)h(n −5) −2(1856 −5376n + 6828n2 −4868n3 + 2174n4 −651n5 + 133n6 −17n7 + n8)h(n −6) −4(n−6)(n−5)(n−3)2(n−2)(3n−7)h(n−7)+8(n−7)(n−6)(n−4)(n−3)(3n−7)h(n−8) −16(n −8)(n −7)(n −5)(−7 + 3n)h(n −9) −(n−9)(n−8)(n−6)(n−3)(16 + 12n−16n2 −5n3 + 15n4 −7n5 + n6)h(n−10) = 0 (14.1) for the sequence h(n) n≥0. Since the leading coeﬃcient (i.e., the coeﬃcient of h(n)) is not n, this recurrence is not suitable for being translated into a Riccati-type diﬀerential equation with integral coeﬃcients via (11.1), to which we can apply our method from Section 4. Using Euclidean division of diﬀerence operators, one can see that we also have nh(n) −h(n −1) −h(n −2) − 5n2 −11n −44 h(n −5) −2(n −4)(n −2)h(n −6) + 12h(n −7) −4h(n −8) − n4 −20n3 + 95n2 + 260n −2000 h(n −10) + 4(9n −85)h(n −11) −8 n2 −19n + 89 h(n −12) + 4(n −14)(n −13)(n −11)(n −7)h(n −15) −4(n −15)(n −14)(n −12)(n −9)h(n −16) = 0. (14.2) If we now apply the procedure of converting such a recurrence for homomorphism num-bers (divided by n!) into a Riccati-type diﬀerential equation for the generating function MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 53 of the corresponding subgroup numbers as explained in the paragraph containing (11.2), then we obtain the diﬀerential equation 224z15 −256z14 + 40z11 −56z10 + 100z9 + 4z7 −12z6 + 16z5 + 26z4 + z + 1 + 736z16 −824z15 + 48z12 −36z11 + 276z10 + 14z6 + 44z5 −1 S(z) + 448z17 −488z16 + 8z13 + 162z11 + 2z7 + 5z6 S2(z) + 80z18 −84z17 + 26z12 S3(z) + 4z19 −4z18 + z13 S4(z) + 448z17 −488z16 + 8z13 + 162z11 + 2z7 + 5z6 S′(z) + 12z19 −12z18 + 3z13 (S′)2(z) + 240z18 −252z17 + 78z12 S(z)S′(z) + 24z19 −24z18 + 6z13 S2(z)S′(z) + 80z18 −84z17 + 26z12 S′′(z) + 16z19 −16z18 + 4z13 S(z)S′′(z) + 4z19 −4z18 + z13 S′′′(z) = 0. (14.3) For convenience, we replace z by z3 in (14.3). Writing ˜ S(z) = S(z3), the above diﬀer-ential equation translates into 224z45 −256z42 + 40z33 −56z30 + 100z27 + 4z21 −12z18 + 16z15 + 26z12 + z3 + 1 + 736z48 −824z45 + 48z36 −36z33 + 276z30 + 14z18 + 44z15 −1 ˜ S(z) + 448z51 −488z48 + 8z39 + 162z33 + 2z21 + 5z18 ˜ S2(z) + 80z54 −84z51 + 26z36 ˜ S3(z) + 4z57 −4z54 + z39 ˜ S4(z) −1 27z16 96z36 −3688z33 + 3928z30 −72z21 + 24z18 −1312z15 −18z3 −45 ˜ S′(z) + 1 3z35 4z18 −4z15 + 1 ( ˜ S′)2(z) + 80z52 −84z49 + 26z34 ˜ S(z) ˜ S′(z) + 8z55 −8z52 + 2z37 ˜ S2(z) ˜ S′(z) + 4 9z32 4z21 + 14z18 −19z15 + z3 + 6 ˜ S′′(z) + 1 27z33 4z18 −4z15 + 1 ˜ S′′′(z) = 0. (14.4) We have to prove that, modulo 24α, the series ˜ S(z) can be expressed as a polynomial in ˜ Φ2(z) (as deﬁned in (13.5)) of degree at most 4α+1−1 with coeﬃcients that are Laurent polynomials in z. We make the Ansatz ˜ S(z) = 4α+1−1 X i=0 ai(z)˜ Φi 2(z) modulo 24α, (14.5) where the ai(z)’s are (at this point) undetermined Laurent polynomials in z. Next we gradually ﬁnd approximations ai,β(z) to ai(z) such that (14.4) holds modulo 2β, for β = 1, 2, . . . , 4α. To start the procedure, we consider the diﬀerential equation 54 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER (14.4) modulo 2, with ˜ S(z) = 4α+1−1 X i=0 ai,1(z)˜ Φi 2(z) modulo 2. (14.6) We substitute the Ansatz (14.6) in (14.4), reduce high powers of ˜ Φ2(z) by using the relation (13.7) with h = 2, and reduce the resulting expression modulo 2, thereby taking advantage of the elementary fact that ˜ Φ′ 2(z) = 1 modulo 2. The powers of 3 that appear in the denominators of the coeﬃcients in the polynomials involved here are disposed of by interpreting them appropriately as elements of Z/24αZ. We ﬁnally see that the left-hand side of (14.4) becomes a polynomial in ˜ Φ2(z) of degree at most 4α+1 −1 with coeﬃcients that are Laurent polynomials in z. Now we compare coeﬃcients of powers ˜ Φk 2(z), k = 0, 1, . . . , 4α+1 −1. This yields a system of 4α+1 equations (modulo 2) for the unknown Laurent polynomials ai,1(z), i = 0, 1, . . . , 4α+1 −1. Since we have already done similar computations several times before, we content ourselves with stating the result: all Laurent polynomials ai,1(z) must be zero, except for a0,1(z) = z−9 + z−13 α−1 X k=1 z4k + z−8 α−1 X k=1 z2·4k, a4α,1(z) = z−13, a2·4α,1(z) = z−8. (14.7) After we have completed the “base step,” we now proceed with the iterative steps described in Section 4. Our Ansatz here (replacing the corresponding one in (4.6)–(4.7)) is ˜ S(z) = 4α+1−1 X i=0 ai,β+1(z)˜ Φi 2(z) modulo 2β+1, (14.8) with ai,β+1(z) := ai,β(z) + 2βbi,β+1(z), i = 0, 1, . . . , 4α+1 −1, (14.9) where the coeﬃcients ai,β(z) are supposed to provide a solution ˜ Sβ(z) = 4α+1−1 X i=0 ai,β(z)˜ Φi 2(z) to (14.4) modulo 2β. This Ansatz, substituted in (14.4), produces a congruence of the form T( ˜ Sβ(z)) + 2β 4α+1−1 X i=0 bi,β+1(z) + z16b′ i,β+1(z) + (i + 1)bi+1,β+1(z) ˜ Φi 2(z) + 1 = 0 modulo 2β+1, (14.10) where T( ˜ Sβ(z)) represents terms that only depend on ˜ Sβ(z). Inductively, we have al-ready computed ˜ Sβ(z), and we know that T( ˜ Sβ(z)) must be divisible by 2β. Comparison MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 55 of powers of ˜ Φ2(z) then yields a system of congruences that is equivalent to a system of the form bi,β+1(z) + z16b′ i,β+1(z) + Poli(z) = 0 modulo 2, i = 0, 1, . . . , 4α+1 −1, where Poli(z), i = 0, 1, . . . , 4α+1 −1, are certain Laurent polynomials with integer coeﬃcients. By Lemma 10, these equations are solvable for any polynomials Poli(z). Thus, we have proved that, for an arbitrary positive integer α, the modiﬁed algorithm that we have presented here will produce a solution ˜ S4α(z) to (14.4) modulo 24α which is a polynomial in ˜ Φ2(z) of degree at most 4α+1 −1 with coeﬃcients that are Laurent polynomials in z. □ Again, we have implemented the algorithm described in the above proof. For α = 1, that is, for the modulus 16, we obtain the following result. Theorem 34. Let Φ2(z) = P n≥0 z4n/3, as before. Then, for the generating function SH(5)(z) = P∞ n=0 sn+1(H(5))zn for the subgroup numbers of H(5), we have SH(5)(z) = 8z12 + 4z9 + 8z7 + 8z5 + 2z4 + 8z2 + 4z + 14 + 7 z3 + 8z20/3 + 8z5/3 + 8 z1/3 + 8 z4/3 + 12 z10/3 Φ2(z) + 8z19/3 + 8z4/3 + 12 z2/3 + 8 z5/3 + 10 z11/3 Φ2 2(z) + 8z6 + 8z + 8 z + 12 z4 Φ3 2(z) + 8z32/3 + 8z20/3 + 8z11/3 + 8z8/3 + 8z5/3 + 6 z4/3 + 8 z7/3 + 12 z10/3 + 7 z13/3 Φ4 2(z) + 8 z2/3 + 8 z5/3 + 8 z8/3 + 4 z11/3 Φ5 2(z) + 8z6 + 8z + 8 z + 12 z2 + 4 z4 Φ6 2(z) + 8 z4/3 + 8 z7/3 + 12 z13/3 Φ7 2(z) + 8z22/3 + 8z19/3 + 8z16/3 + 8z10/3 + 12z7/3 + 8z4/3 + 4 z2/3 + 8 z5/3 + 5 z8/3 + 2 z11/3 Φ8 2(z) + 8z6 + 8z + 8 z + 4 z4 + 8 z2 Φ9 2(z) + 12 z4/3 + 2 z13/3 Φ10 2 (z) + 8 z8/3 Φ11 2 (z) + 8z6 + 8z4 + 8z + 8 + 8 z + 12 z2 + 12 z3 + 12 z4 Φ12 2 (z) + 8 z4/3 + 8 z7/3 + 12 z13/3 Φ13 2 (z) + 4 z8/3 + 8 z11/3 Φ14 2 (z) modulo 16. (14.11) We conjecture that Theorems 19 and 33 (note that PSL2(Z) ∼ = C2 ∗C3 = H(3)) extend to any Hecke group H(q), where q is a Fermat prime. 56 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Conjecture 35. For a positive integer h, let Φh(z) = P n≥0 z2nh/(2h−1). Let α be a further positive integer, and let q = 22f + 1 be a Fermat prime. Then the generating function SH(q)(z) (see the ﬁrst paragraph of Section 9 for the deﬁnition), reduced modulo 22α2f , can be expressed as a polynomial in Φ2f(z) of degree at most 2(α+1)2f −1 with coeﬃcients that are Laurent polynomials in z1/(q−2). Note that Theorems 19 and 33 are the special cases corresponding to f = 0 and f = 1, respectively. In particular, we conjecture that the obvious extension of the algorithm described in the proofs of the two theorems would be successful modulo any 2-power. In more detail, given a Fermat prime q, the ﬁrst step consists in deriving a recurrence relation for the Hadamard product of the sequences hC2(n) n≥0 and hCq(n)/n! n≥0. By the procedure explained in the paragraph containing (11.2) (where we now use (11.1) with Γ = H(q)), this leads to a Riccati-type diﬀerential equation for the generating function P n≥0 sn+1(H(q)) zn for the subgroup numbers of H(q). The open questions are whether it will be possible to complete the base step, and whether it will always be possible to carry out the subsequent iterative steps in (the variation of) our method. Given the description of the parity pattern of the subgroup numbers proved in [25, Theorem B], it is “more or less” clear that the ﬁrst question has a positive answer. What the answer to the second question is, is entirely open. (The reader should recall that, in Sections 11 and 12, we met cases where our method worked initially, but then stopped to work for modulus 16). Appendix A. Expansions of powers of the 2-power series Φ(z) Recall the notation Ha1,a2,...,ar(z) = X n1>n2>···>nr≥0 za12n1+a22n2+···+ar2nr from Section 3. In this appendix we list the expansions of ΦK(z) for K = 5, 6, 7, 8 in terms of the series Ha1,a2,...,ar(z) where all ai’s are odd, obtained by using the algorithm described in Section 3 (see (3.1) and the proof of Lemma 6). Namely, we have Φ5(z) = 16H5(z) −40H3,1,1(z) −40H1,3,1(z) −40H1,1,3(z) + 120H1,1,1,1,1(z) −80H3,1(z) −80H1,3(z) + 240H1,1,1,1(z) + (20z −90)H3(z) −(60z −270)H1,1,1(z) −(120z −190)H1,1(z) + (25z2 −125z + 75)H1(z) + 50z2 −75z, Φ6(z) = 96H5,1(z) + 96H1,5(z) + 80H3,3(z) −240H3,1,1,1(z) −240H1,3,1,1(z) −240H1,1,3,1(z) −240H1,1,1,3(z) + 720H1,1,1,1,1,1(z) + 240H5(z) −600H3,1,1(z) −600H1,3,1(z) −600H1,1,3(z) + 1800H1,1,1,1,1(z) + (120z −840)H3,1(z) + (120z −840)H1,3(z) −(360z −2520)H1,1,1,1(z) + (300z −764)H3(z) −(900z −2340)H1,1,1(z) + (150z2 −1200z + 1470)H1,1(z) MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 57 + (375z2 −1020z + 525)H1(z) −61z3 + 495z2 −525z, Φ7(z) = −272H7(z) + 672H5,1,1(z) + 672H1,5,1(z) + 672H1,1,5(z) + 560H3,3,1(z) + 560H3,1,3(z) + 560H1,3,3(z) + 2016H5,1(z) + 2016H1,5(z) + 1680H3,3(z) −1680H3,1,1,1,1(z)−1680H1,3,1,1,1(z)−1680H1,1,3,1,1(z)−1680H1,1,1,3,1(z)−1680H1,1,1,1,3(z) +5040H1,1,1,1,1,1,1(z)−5040H3,1,1,1(z)−5040H1,3,1,1(z)−5040H1,1,3,1(z)−5040H1,1,1,3(z) +15120H1,1,1,1,1,1(z)−(336z−3360)H5(z)+(840z−8400)H3,1,1(z)+(840z−8400)H1,3,1(z) + (840z −8400)H1,1,3(z) −(2520z −25200)H1,1,1,1,1(z) + (2520z −9408)H3,1(z) + (2520z −9408)H1,3(z) −(7560z −28560)H1,1,1,1(z) −(350z2 −4060z + 7434)H3(z) + (1050z2 −12180z + 23310)H1,1,1(z) + (3150z2 −13020z + 13230)H1,1(z) −(427z3 −5040z2 + 9555z −4347)H1(z) −1281z3 + 5208z2 −4347z, Φ8(z) = −2176H7,1(z) −2176H1,7(z) −1792H5,3(z) −1792H3,5(z) + 5376H5,1,1,1(z) + 5376H1,5,1,1(z) + 5376H1,1,5,1(z) + 5376H1,1,1,5(z) + 4480H3,3,1,1(z) + 4480H3,1,3,1(z) + 4480H3,1,1,3(z) + 4480H1,3,3,1(z) + 4480H1,3,1,3(z) + 4480H1,1,3,3(z) −13440H3,1,1,1,1,1(z) −13440H1,3,1,1,1,1(z) −13440H1,1,3,1,1,1(z) −13440H1,1,1,3,1,1(z) −13440H1,1,1,1,3,1(z) −13440H1,1,1,1,1,3(z) + 40320H1,1,1,1,1,1,1,1(z) −7616H7(z) + 18816H5,1,1(z) + 18816H1,5,1(z) + 18816H1,1,5(z) + 15680H3,3,1(z) + 15680H3,1,3(z) + 15680H1,3,3(z) −47040H3,1,1,1,1(z) −47040H1,3,1,1,1(z) −47040H1,1,3,1,1(z) −47040H1,1,1,3,1(z) −47040H1,1,1,1,3(z) + 141120H1,1,1,1,1,1,1(z) −(2688z −36288)H5,1(z) −(2688z −36288)H1,5(z) −(2240z −30240)H3,3(z) + (6720z −90720)H3,1,1,1(z) + (6720z −90720)H1,3,1,1(z) +(6720z−90720)H1,1,3,1(z)+(6720z−90720)H1,1,1,3(z)−(20160z−272160)H1,1,1,1,1,1(z) −(9408z −47264)H5(z) + (23520z −119504)H3,1,1(z) + (23520z −119504)H1,3,1(z) + (23520z −119504)H1,1,3(z) −(70560z −361200)H1,1,1,1,1(z) −(2800z2 −44240z + 115304)H3,1(z) −(2800z2 −44240z + 115304)H1,3(z) + (8400z2 −132720z + 355320)H1,1,1,1(z) −(9800z2 −55832z + 80892)H3(z) +(29400z2−168840z+260820)H1,1,1(z)−(3416z3−55020z2+154980z−135982)H1,1(z) −(11956z3−68474z2+101206z−41245)H1(z)+1385z4−22358z3+59961z2−41245z. 58 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER Appendix B. The coefficients in the differential equation (11.5) Here we provide explicit expressions for the coeﬃcients in the Riccati-type diﬀerential equation (11.5), when reduced modulo 16: p0(z) = 8z47 + 8z46 + 12z45 + 4z43 + 12z41 + 12z40 + 4z39 + 12z38 + 8z37 + 4z36 + 4z34 + 2z33 + 11z31 + 6z30 + 14z29 + 14z28 + 13z27 + 6z26 + 9z25 + 11z24 + 4z23 + 7z22 + 9z21 + z20 + 7z19 + 15z18 + 14z17 + 12z16 + 11z15 + z14 + 10z13 + 5z12 + 2z11 + 8z10 + 9z9 + 15z8 + 5z7 + 13z6 + 4z4 + 14z3 + z2 + 11z, p1(z) = 8z48 + 12z46 + 12z45 + 4z43 + 12z42 + 4z41 + 4z40 + 4z39 + 8z38 + 4z36 + 13z34 + 6z33 + 10z31 + z30 + 4z29 + 11z28 + 8z27 + 13z25 + z24 + 8z23 + z22 + z21 + 4z20 + 14z19 + 9z18 + 6z17 + 14z16 + 8z15 + 13z14 + 6z13 + 2z12 + 9z10 + 11z9 + 6z7 + 8z6 + 6z5 + 5z4 + 3z2 + 4z + 1, p2(z) = 12z51 + 2z49 + 12z48 + 4z47 + 4z46 + 12z44 + 10z43 + 2z42 + 4z41 + 8z40 + 14z39 + 5z37 + 2z36 + 6z35 + 10z34 + 8z33 + 10z32 + 5z31 + 7z30 + 12z29 + 7z28 + 4z27 + 12z26 + 9z25 + 12z24 + 2z23 + 14z22 + 10z21 + 12z20 + 10z19 + 3z18 + 6z16 + 9z15 + 12z14 + 15z13 + 14z12 + 10z11 + 2z10 + 12z9 + 14z8 + 13z7 + 7z6 + 8z5 + z3 + 2z2, p3(z) = 6z52 + 8z50 + 12z48 + 8z47 + 14z46 + 4z45 + 8z44 + 6z43 + 8z42 + 4z41 + 5z40 + 6z39 + 12z38 + 2z37 + 2z36 + 7z34 + 6z33 + 7z31 + 14z30 + 8z29 + 5z28 + 14z27 + 8z26 + 14z25 + 14z24 + 2z22 + 8z21 + 12z20 + 14z19 + 6z18 + 12z17 + 9z16 + 4z15 + 2z13 + 2z12 + 15z10 + 14z9, p4(z) = 4z53 + 15z51 + 4z47 + 2z46 + z45 + 9z44 + 15z42 + 4z41 + 10z40 + 4z39 + 6z38 + 4z37 + 2z36 + 6z35 + 10z34 + 11z33 + 8z32 + 8z31 + 10z29 + 12z26 + 12z25 + 2z23 + 4z22, p5(z) = 13z54 + 15z48 + 6z47 + 12z46 + 3z45 + 10z44 + 4z43 + 12z42 + 10z41 + 2z39 + 2z38 + 15z36 + 14z35, p6(z) = 12z51 + 2z49 + 12z48 + 4z47 + 4z46 + 12z44 + 10z43 + 2z42 + 4z41 + 8z40 + 14z39 + 5z37 + 2z36 + 6z35 + 10z34 + 8z33 + 10z32 + 5z31 + 7z30 + 12z29 + 7z28 + 4z27 + 12z26 + 9z25 + 12z24 + 2z23 + 14z22 + 10z21 + 12z20 + 10z19 + 3z18 + 6z16 + 9z15 + 12z14 + 15z13 + 14z12 + 10z11 + 2z10 + 12z9 + 14z8 + 13z7 + 7z6 + 8z5 + z3 + 2z2, p7(z) = 12z53 + 13z51 + 12z47 + 6z46 + 3z45 + 11z44 + 13z42 + 12z41 + 14z40 + 12z39 + 2z38 + 12z37 + 6z36 + 2z35 + 14z34 + z33 + 8z32 + 8z31 + 14z29 + 4z26 + 4z25 + 6z23 + 12z22, p8(z) = 2z52 + 8z50 + 4z48 + 8z47 + 10z46 + 12z45 + 8z44 + 2z43 + 8z42 + 12z41 + 15z40 + 2z39 + 4z38 + 6z37 + 6z36 + 5z34 + 2z33 + 5z31 + 10z30 + 8z29 + 15z28 + 10z27 + 8z26 + 10z25 + 10z24 + 6z22 + 8z21 + 4z20 + 10z19 + 2z18 + 4z17 + 11z16 + 12z15 + 6z13 + 6z12 + 13z10 + 10z9, p9(z) = 8z53 + 10z51 + 8z47 + 12z46 + 6z45 + 6z44 + 10z42 + 8z41 + 12z40 + 8z39 + 4z38 + 8z37 + 12z36 + 4z35 + 12z34 + 2z33 + 12z29 + 8z26 + 8z25 + 12z23 + 8z22, p10(z) = 2z54 + 6z48 + 12z47 + 8z46 + 14z45 + 4z44 + 8z43 + 8z42 + 4z41 + 4z39 + 4z38 + 6z36 + 12z35, MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 59 p11(z) = 3z54 + z48 + 10z47 + 4z46 + 13z45 + 6z44 + 12z43 + 4z42 + 6z41 + 14z39 + 14z38 + z36 + 2z35, p12(z) = 6z52 + 8z50 + 12z48 + 8z47 + 14z46 + 4z45 + 8z44 + 6z43 + 8z42 + 4z41 + 5z40 + 6z39 + 12z38 + 2z37 + 2z36 + 7z34 + 6z33 + 7z31 + 14z30 + 8z29 + 5z28 + 14z27 + 8z26 + 14z25 + 14z24 + 2z22 + 8z21 + 12z20 + 14z19 + 6z18 + 12z17 + 9z16 + 4z15 + 2z13 + 2z12 + 15z10 + 14z9, p13(z) = 12z51 + 8z46 + 4z45 + 4z44 + 12z42 + 8z40 + 8z38 + 8z36 + 8z35 + 8z34 + 12z33 + 8z29 + 8z23, p14(z) = 2z54 + 6z48 + 12z47 + 8z46 + 14z45 + 4z44 + 8z43 + 8z42 + 4z41 + 4z39 + 4z38 + 6z36 + 12z35, p15(z) = 2z54 + 6z48 + 12z47 + 8z46 + 14z45 + 4z44 + 8z43 + 8z42 + 4z41 + 4z39 + 4z38 + 6z36 + 12z35, p16(z) = 4z53 + 15z51 + 4z47 + 2z46 + z45 + 9z44 + 15z42 + 4z41 + 10z40 + 4z39 + 6z38 + 4z37 + 2z36 + 6z35 + 10z34 + 11z33 + 8z32 + 8z31 + 10z29 + 12z26 + 12z25 + 2z23 + 4z22, p17(z) = z54 + 11z48 + 14z47 + 12z46 + 15z45 + 2z44 + 4z43 + 12z42 + 2z41 + 10z39 + 10z38 + 11z36 + 6z35, p18(z) = 13z54 + 15z48 + 6z47 + 12z46 + 3z45 + 10z44 + 4z43 + 12z42 + 10z41 + 2z39 + 2z38 + 15z36 + 14z35. Appendix C. The coefficients in the differential equation (12.3) Here we provide explicit expressions for the coeﬃcients in the Riccati-type diﬀerential equation (12.3), when reduced modulo 16: q0(z) = 8z50 + 8z49 + 4z47 + 12z46 + 8z44 + 10z43 + 2z42 + 2z41 + 12z40 + 12z38 + 7z37 + 11z36 + 10z35 + 11z34 + 13z33 + 13z32 + 14z31 + 15z30 + 10z29 + 14z28 + 8z27 + 10z26 + 6z25 + z23 + 4z22 + 8z20 + 14z19 + 12z18 + 8z17 + 9z16 + 4z15 + 6z14 + 8z13 + z12 + 14z11 + 9z10 + 2z9 + 14z8 + 10z7 + 4z6 + 11z5 + 4z4 + 4z3 + z2 + 8z + 15, q1(z) = 8z51 + 8z50 + 8z49 + 12z48 + 4z47 + 12z46 + 4z45 + 14z43 + 10z42 + 5z40 + 4z39 + 4z38 + 9z37 + 4z36 + 14z35 + 5z34 + 14z33 + 13z31 + 5z30 + 4z29 + 14z28 + 6z27 + 5z26 + 10z25 + 12z23 + 9z22 + 12z21 + 13z20 + 6z19 + z18 + 7z17 + 15z15 + 11z14 + 7z13 + 5z12 + z11 + 6z10 + 10z9 + 2z8 + 4z7 + 8z6 + 2z5 + 11z4 + 4z3 + 8z2 + 7z + 1, q2(z) = 8z57 + 8z56 + 8z55 + 8z54 + 8z53 + 12z51 + 10z50 + 12z49 + 8z48 + 10z47 + 15z46 + 4z45 + 10z43 + z42 + 11z41 + z40 + 2z39 + z38 + 3z37 + 6z36 + 7z35 + 8z34 + z33 + 6z31 + 9z29 + 14z28 + 11z27 + 5z26 + 2z24 + 6z23 + 4z22 + 6z21 + 13z20 + z19 + 10z18 + z17 + 10z15 + 12z14 + z13 + 9z12 + 9z11 + 7z10 + 4z9 + 12z8 + 3z7 + 2z6 + 9z5 + 3z4 + 9z3, q3(z) = 14z56 + 2z55 + 14z54 + 11z52 + 6z51 + 10z50 + 11z49 + 2z48 + 15z47 + 6z46 + 10z45 + 9z44 + 4z43 + 5z42 + 11z41 + 5z40 + 10z39 + 4z38 + 12z37 + 2z36 + 5z35 + z34 + 6z33 + 5z32 + 13z31 + 12z30 + 3z29 + 10z28 + 13z27 + 14z26 + 4z25 + 12z24 + 8z23 + 9z22 + 6z21 + 14z20 + 7z19 + 5z18 + 4z17 + 4z16 + 8z14 + 15z13 + 13z11 + 9z9 + 7z8 + z6 + 4z5, q4(z) = 8z66 + 15z58 + 14z57 + 12z56 + 9z54 + 7z53 + 9z52 + 10z51 + 12z50 + z49 + 10z48 + 14z46 + 6z45 + 6z44 + 6z43 + 14z42 + 11z41 + 13z40 + 10z39 + z38 + 13z37 + 7z36 + 2z34 + 12z33 + 4z32 + 13z31 + 13z30 + 4z29 + 11z28 + 7z27 + 8z26 + 2z25 + 6z24 + 13z23 60 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER + 11z21 + 2z20 + 9z19 + 6z18 + 2z17 + 15z16 + 5z15 + 3z14 + 14z13 + 2z12 + 10z11 + 9z10 + 6z9 + 12z8, q5(z) = 5z59 + 13z58 + 12z56 + z55 + 8z54 + 5z53 + 12z52 + 15z51 + 8z50 + 14z49 + 2z48 + 4z47 + 4z46 + 10z45 + 12z44 + 8z43 + 4z41 + 2z40 + 8z38 + 12z37 + 2z36 + 12z35 + 8z34, q6(z) = 12z49 + 4z48 + 12z46 + 12z45 + 12z44 + 4z43 + 12z42 + 12z41 + 12z39 + 4z38 + 4z37 + 4z36 + 4z32 + 4z31 + 12z27 + 12z24 + 12z22 + 12z21 + 12z20 + 12z19 + 12z18 + 4z15 + 4z13 + 12z12 + 12z11 + 12z10 + 4z9 + 4z8 + 4z7 + 12z6 + 4z5, q7(z) = 10z58 + 6z54 + 10z53 + 6z52 + 6z49 + 2z41 + 14z40 + 6z38 + 14z37 + 10z36 + 14z31 + 14z30 + 2z28 + 10z27 + 14z23 + 2z21 + 6z19 + 10z16 + 14z15 + 2z14 + 6z10, q8(z) = 10z56 + 6z55 + 10z54 + 9z52 + 2z51 + 14z50 + 9z49 + 6z48 + 5z47 + 2z46 + 14z45 + 3z44 + 12z43 + 7z42 + 9z41 + 7z40 + 14z39 + 12z38 + 4z37 + 6z36 + 7z35 + 11z34 + 2z33 + 7z32 + 15z31 + 4z30 + z29 + 14z28 + 15z27 + 10z26 + 12z25 + 4z24 + 3z22 + 2z21 + 10z20 + 13z19 + 7z18 + 12z17 + 12z16 + 5z13 + 15z11 + 3z9 + 13z8 + 11z6 + 12z5, q9(z) = 7z59 + 15z58 + 11z55 + 7z53 + 5z51 + 10z49 + 6z48 + 14z45 + 6z40 + 6z36, q10(z) = z58 + 7z54 + 9z53 + 7z52 + 15z49 + 5z41 + 3z40 + 15z38 + 3z37 + 9z36 + 3z31 + 3z30 + 5z28 + 9z27 + 3z23 + 5z21 + 7z19 + z16 + 11z15 + 13z14 + 7z10, q11(z) = 12z55 + 4z51 + 12z50 + 4z47 + 12z46 + 4z45 + 12z43 + 4z42 + 4z41 + 4z38 + 12z34 + 4z33 + 4z32 + 12z29 + 4z26 + 12z25 + 12z22 + 12z16 + 12z15 + 4z14 + 12z13 + 12z9 + 4z8 + 4z6, q12(z) = 15z58 + 14z57 + 12z56 + 9z54 + 7z53 + 9z52 + 10z51 + 12z50 + z49 + 10z48 + 14z46 + 6z45 + 6z44 + 6z43 + 14z42 + 11z41 + 13z40 + 10z39 + z38 + 13z37 + 7z36 + 2z34 + 12z33 + 4z32 + 13z31 + 13z30 + 4z29 + 11z28 + 7z27 + 2z25 + 6z24 + 13z23 + 11z21 + 2z20 + 9z19 + 6z18 + 2z17 + 15z16 + 5z15 + 3z14 + 14z13 + 2z12 + 10z11 + 9z10 + 6z9 + 12z8, q13(z) = 10z59 + 10z58 + 2z55 + 10z53 + 14z51 + 12z49 + 4z48 + 4z45 + 4z40 + 4z36, q14(z) = 8z57 + 8z56 + 8z55 + 12z53 + 8z51 + 8z50 + 8z49 + 8z48 + 10z47 + 14z46 + 10z45 + 8z44 + 9z43 + 12z42 + 2z41 + 6z40 + 11z39 + 3z38 + 12z37 + 8z36 + 7z35 + 2z34 + z33 + z32 + 7z31 + 12z30 + 12z29 + z28 + z26 + 15z25 + 9z24 + 14z23 + z22 + 3z21 + 3z20 + 4z19 + 7z17 + 7z16 + 6z15 + 3z14 + z13 + 3z12 + 2z11 + 14z9 + 2z8 + 6z7 + 2z5 + 14z4 + 6z3 + 14z2, q15(z) = 12z55 + 12z54 + 8z50 + 2z49 + 14z48 + 8z47 + 10z46 + 10z45 + 10z44 + 14z43 + 10z42 + 10z41 + 2z39 + 14z38 + 6z37 + 14z36 + 12z34 + 14z32 + 14z31 + 4z30 + 8z28 + 10z27 + 12z25 + 10z24 + 12z23 + 2z22 + 2z21 + 2z20 + 2z19 + 10z18 + 8z17 + 4z16 + 6z15 + 8z14 + 14z13 + 2z12 + 2z11 + 2z10 + 6z9 + 14z8 + 14z7 + 2z6 + 14z5 + 12z4, q16(z) = 4z65 + 10z57 + 6z56 + 9z55 + 8z53 + 6z52 + 11z51 + 13z50 + 2z49 + 12z48 + 3z47 + 13z46 + 3z45 + 10z44 + 13z43 + 3z42 + 3z41 + 2z40 + 4z39 + 7z38 + 4z37 + 6z36 + 6z35 + 5z34 + 15z33 + 3z32 + 10z31 + 14z30 + z29 + 2z28 + 14z27 + 11z26 + 9z25 + 2z24 + 9z22 + 8z21 + 2z20 + 8z19 + 10z18 + 12z17 + 9z16 + 5z15 + 11z14 + 5z13 + 8z12 + 10z10 + z9 + 7z8 + 6z7 + 7z6, q17(z) = 8z58 + 12z55 + 8z54 + 8z53 + 8z52 + 4z51 + 12z50 + 8z49 + 8z48 + 4z47 + 12z45 + 8z43 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 61 + 8z42 + 12z41 + 8z40 + 12z39 + 8z38 + 4z36 + 4z35 + 8z34 + 4z33 + 12z32 + 8z31 + 4z29 + 8z28 + 8z26 + 8z22 + 12z21 + 8z18 + 12z17 + 8z16, q18(z) = 6z60 + 11z59 + 4z57 + 10z56 + 11z55 + 2z54 + 4z53, q19(z) = 4z56 + 12z55 + 4z54 + 2z52 + 4z51 + 12z50 + 2z49 + 12z48 + 10z47 + 4z46 + 12z45 + 6z44 + 14z42 + 2z41 + 14z40 + 12z39 + 12z36 + 14z35 + 6z34 + 4z33 + 14z32 + 14z31 + 2z29 + 12z28 + 14z27 + 4z26 + 6z22 + 4z21 + 4z20 + 10z19 + 14z18 + 10z13 + 14z11 + 6z9 + 10z8 + 6z6, q20(z) = 12z65 + 14z57 + 2z56 + 3z55 + 2z52 + 9z51 + 15z50 + 6z49 + 4z48 + z47 + 15z46 + z45 + 14z44 + 15z43 + z42 + z41 + 6z40 + 12z39 + 13z38 + 12z37 + 2z36 + 2z35 + 7z34 + 5z33 + z32 + 14z31 + 10z30 + 11z29 + 6z28 + 10z27 + 9z26 + 3z25 + 6z24 + 3z22 + 6z20 + 14z18 + 4z17 + 3z16 + 7z15 + 9z14 + 7z13 + 14z10 + 11z9 + 13z8 + 2z7 + 13z6, q21(z) = 2z60 + 9z59 + 14z56 + 9z55 + 6z54, q22(z) = 3z58 + 5z54 + 11z53 + 5z52 + 13z49 + 15z41 + 9z40 + 13z38 + 9z37 + 11z36 + 9z31 + 9z30 + 15z28 + 11z27 + 9z23 + 15z21 + 5z19 + 3z16 + z15 + 7z14 + 5z10, q23(z) = 14z58 + 12z57 + 2z54 + 14z53 + 2z52 + 4z51 + 2z49 + 4z48 + 12z46 + 12z45 + 12z44 + 12z43 + 12z42 + 6z41 + 10z40 + 4z39 + 2z38 + 10z37 + 14z36 + 4z34 + 10z31 + 10z30 + 6z28 + 14z27 + 4z25 + 12z24 + 10z23 + 6z21 + 4z20 + 2z19 + 12z18 + 4z17 + 14z16 + 10z15 + 6z14 + 12z13 + 4z12 + 4z11 + 2z10 + 12z9, q24(z) = 4z59 + 4z58 + 4z55 + 4z53 + 12z51, q25(z) = 12z55 + 4z51 + 12z50 + 4z47 + 12z45 + 12z41 + 12z39 + 4z36 + 4z35 + 4z33 + 12z32 + 4z29 + 12z21 + 12z17, q26(z) = 12z60 + 6z59 + 4z56 + 6z55 + 4z54, q27(z) = 8z57 + 8z56 + 8z55 + 8z54 + 8z53 + 4z51 + 14z50 + 4z49 + 8z48 + 14z47 + 5z46 + 12z45 + 14z43 + 11z42 + 9z41 + 11z40 + 6z39 + 11z38 + z37 + 2z36 + 13z35 + 8z34 + 11z33 + 2z31 + 3z29 + 10z28 + 9z27 + 7z26 + 6z24 + 2z23 + 12z22 + 2z21 + 15z20 + 11z19 + 14z18 + 11z17 + 14z15 + 4z14 + 11z13 + 3z12 + 3z11 + 13z10 + 12z9 + 4z8 + z7 + 6z6 + 3z5 + z4 + 3z3, q28(z) = 4z56 + 12z55 + 4z54 + 2z52 + 4z51 + 12z50 + 2z49 + 12z48 + 10z47 + 4z46 + 12z45 + 6z44 + 8z43 + 14z42 + 2z41 + 14z40 + 12z39 + 8z38 + 8z37 + 12z36 + 14z35 + 6z34 + 4z33 + 14z32 + 14z31 + 8z30 + 2z29 + 12z28 + 14z27 + 4z26 + 8z25 + 8z24 + 6z22 + 4z21 + 4z20 + 10z19 + 14z18 + 8z17 + 8z16 + 10z13 + 14z11 + 6z9 + 10z8 + 6z6 + 8z5, q29(z) = 8z66 + 15z58 + 14z57 + 12z56 + 9z54 + 7z53 + 9z52 + 10z51 + 12z50 + z49 + 10z48 + 14z46 + 6z45 + 6z44 + 6z43 + 14z42 + 11z41 + 13z40 + 10z39 + z38 + 13z37 + 7z36 + 2z34 + 12z33 + 4z32 + 13z31 + 13z30 + 4z29 + 11z28 + 7z27 + 8z26 + 2z25 + 6z24 + 13z23 + 11z21 + 2z20 + 9z19 + 6z18 + 2z17 + 15z16 + 5z15 + 3z14 + 14z13 + 2z12 + 10z11 + 9z10 + 6z9 + 12z8, q30(z) = 12z59 + 12z58 + 12z55 + 12z53 + 4z51 + 8z49 + 8z48 + 8z45 + 8z40 + 8z36, q31(z) = 4z55 + 12z51 + 4z50 + 12z47 + 4z46 + 12z45 + 4z43 + 12z42 + 12z41 + 12z38 + 4z34 + 12z33 + 12z32 + 4z29 + 12z26 + 4z25 + 4z22 + 4z16 + 4z15 + 12z14 + 4z13 + 4z9 + 12z8 + 12z6, 62 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER q32(z) = 5z58 + 10z57 + 4z56 + 3z54 + 13z53 + 3z52 + 14z51 + 4z50 + 11z49 + 14z48 + 10z46 + 2z45 + 2z44 + 2z43 + 10z42 + 9z41 + 15z40 + 14z39 + 11z38 + 15z37 + 13z36 + 6z34 + 4z33 + 12z32 + 15z31 + 15z30 + 12z29 + 9z28 + 13z27 + 6z25 + 2z24 + 15z23 + 9z21 + 6z20 + 3z19 + 2z18 + 6z17 + 5z16 + 7z15 + z14 + 10z13 + 6z12 + 14z11 + 3z10 + 2z9 + 4z8, q33(z) = 12z59 + 12z58 + 12z55 + 12z53 + 4z51, q34(z) = 2z55 + 10z54 + 4z50 + 3z49 + 13z48 + 4z47 + 7z46 + 7z45 + 15z44 + 5z43 + 15z42 + 15z41 + 8z40 + 11z39 + 13z38 + z37 + 5z36 + 8z35 + 2z34 + 5z32 + 5z31 + 14z30 + 8z29 + 4z28 + 15z27 + 10z25 + 15z24 + 10z23 + 11z22 + 11z21 + 11z20 + 3z19 + 15z18 + 4z17 + 14z16 + z15 + 12z14 + 13z13 + 11z12 + 11z11 + 11z10 + z9 + 5z8 + 13z7 + 11z6 + 5z5 + 2z4, q35(z) = 12z65 + 14z57 + 2z56 + 3z55 + 8z53 + 2z52 + 9z51 + 15z50 + 6z49 + 4z48 + z47 + 15z46 + z45 + 14z44 + 15z43 + z42 + z41 + 6z40 + 12z39 + 13z38 + 12z37 + 2z36 + 2z35 + 7z34 + 5z33 + z32 + 14z31 + 10z30 + 11z29 + 6z28 + 10z27 + 9z26 + 3z25 + 6z24 + 3z22 + 8z21 + 6z20 + 8z19 + 14z18 + 4z17 + 3z16 + 7z15 + 9z14 + 7z13 + 8z12 + 14z10 + 11z9 + 13z8 + 2z7 + 13z6, q36(z) = 8z67 + 12z58 + 14z55 + 4z54 + 4z53 + 4z52 + 2z51 + 6z50 + 12z49 + 12z48 + 10z47 + 6z45 + 8z44 + 12z43 + 12z42 + 6z41 + 12z40 + 6z39 + 4z38 + 2z36 + 10z35 + 12z34 + 2z33 + 6z32 + 12z31 + 10z29 + 12z28 + 12z26 + 8z25 + 12z22 + 14z21 + 8z19 + 4z18 + 14z17 + 4z16 + 8z15, q37(z) = 4z60 + 2z59 + 8z57 + 12z56 + 2z55 + 12z54 + 8z53, q38(z) = 5z58 + 10z57 + 4z56 + 3z54 + 13z53 + 3z52 + 14z51 + 4z50 + 11z49 + 14z48 + 10z46 + 2z45 + 2z44 + 2z43 + 10z42 + 9z41 + 15z40 + 14z39 + 11z38 + 15z37 + 13z36 + 6z34 + 4z33 + 12z32 + 15z31 + 15z30 + 12z29 + 9z28 + 13z27 + 6z25 + 2z24 + 15z23 + 9z21 + 6z20 + 3z19 + 2z18 + 6z17 + 5z16 + 7z15 + z14 + 10z13 + 6z12 + 14z11 + 3z10 + 2z9 + 4z8, q39(z) = 4z58 + 10z55 + 12z54 + 12z53 + 12z52 + 6z51 + 2z50 + 4z49 + 4z48 + 14z47 + 2z45 + 4z43 + 4z42 + 2z41 + 4z40 + 2z39 + 12z38 + 6z36 + 14z35 + 4z34 + 6z33 + 2z32 + 4z31 + 14z29 + 4z28 + 4z26 + 4z22 + 10z21 + 12z18 + 10z17 + 12z16, q40(z) = 12z59 + 12z58 + 12z55 + 12z53 + 4z51, q41(z) = 4z60 + 2z59 + 12z56 + 2z55 + 12z54, q42(z) = 2z56 + 14z55 + 2z54 + 5z52 + 10z51 + 6z50 + 5z49 + 14z48 + z47 + 10z46 + 6z45 + 7z44 + 12z43 + 11z42 + 5z41 + 11z40 + 6z39 + 12z38 + 4z37 + 14z36 + 11z35 + 15z34 + 10z33 + 11z32 + 3z31 + 4z30 + 13z29 + 6z28 + 3z27 + 2z26 + 12z25 + 4z24 + 8z23 + 7z22 + 10z21 + 2z20 + 9z19 + 11z18 + 12z17 + 12z16 + 8z14 + z13 + 3z11 + 7z9 + 9z8 + 15z6 + 12z5, q43(z) = 8z66 + 3z58 + 6z57 + 12z56 + 5z54 + 11z53 + 5z52 + 2z51 + 12z50 + 13z49 + 2z48 + 6z46 + 14z45 + 14z44 + 14z43 + 6z42 + 15z41 + 9z40 + 2z39 + 13z38 + 9z37 + 11z36 + 10z34 + 12z33 + 4z32 + 9z31 + 9z30 + 4z29 + 15z28 + 11z27 + 8z26 + 10z25 + 14z24 + 9z23 + 15z21 + 10z20 + 5z19 + 14z18 + 10z17 + 3z16 + z15 + 7z14 + 6z13 + 10z12 + 2z11 + 5z10 + 14z9 + 12z8, q44(z) = 2z59 + 2z58 + 8z56 + 10z55 + 2z53 + 8z52 + 6z51 + 12z49 + 4z48 + 8z47 + 8z46 + 4z45 + 8z44 + 8z41 + 4z40 + 8z37 + 4z36 + 8z35, MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 63 q45(z) = 8z21, q46(z) = 6z59 + 6z58 + 14z55 + 6z53 + 2z51 + 4z49 + 12z48 + 12z45 + 12z40 + 12z36, q47(z) = 4z65 + 2z57 + 14z56 + 13z55 + 8z53 + 14z52 + 7z51 + z50 + 10z49 + 12z48 + 15z47 + z46 + 15z45 + 2z44 + z43 + 15z42 + 15z41 + 10z40 + 4z39 + 3z38 + 4z37 + 14z36 + 14z35 + 9z34 + 11z33 + 15z32 + 2z31 + 6z30 + 5z29 + 10z28 + 6z27 + 7z26 + 13z25 + 10z24 + 13z22 + 8z21 + 10z20 + 8z19 + 2z18 + 12z17 + 13z16 + 9z15 + 7z14 + 9z13 + 8z12 + 2z10 + 5z9 + 3z8 + 14z7 + 3z6, q48(z) = 8z58 + 4z55 + 8z54 + 8z53 + 8z52 + 12z51 + 4z50 + 8z49 + 8z48 + 12z47 + 4z45 + 8z43 + 8z42 + 4z41 + 8z40 + 4z39 + 8z38 + 12z36 + 12z35 + 8z34 + 12z33 + 4z32 + 8z31 + 12z29 + 8z28 + 8z26 + 8z22 + 4z21 + 8z18 + 4z17 + 8z16, q49(z) = 4z60 + 10z59 + 8z57 + 12z56 + 10z55 + 12z54 + 8z53, q50(z) = 4z59 + 4z58 + 4z55 + 4z53 + 12z51, q51(z) = 12z60 + 14z59 + 4z56 + 14z55 + 4z54, q52(z) = 8z66 + 7z58 + 14z57 + 12z56 + z54 + 15z53 + z52 + 10z51 + 12z50 + 9z49 + 10z48 + 14z46 + 6z45 + 6z44 + 6z43 + 14z42 + 3z41 + 5z40 + 10z39 + 9z38 + 5z37 + 15z36 + 2z34 + 12z33 + 4z32 + 5z31 + 5z30 + 4z29 + 3z28 + 15z27 + 8z26 + 2z25 + 6z24 + 5z23 + 3z21 + 2z20 + z19 + 6z18 + 2z17 + 7z16 + 13z15 + 11z14 + 14z13 + 2z12 + 10z11 + z10 + 6z9 + 12z8, q53(z) = 4z59 + 4z58 + 4z55 + 4z53 + 12z51 + 8z49 + 8z48 + 8z45 + 8z40 + 8z36, q54(z) = 4z67 + 14z58 + 8z57 + 15z55 + 10z54 + 10z53 + 2z52 + z51 + 11z50 + 14z49 + 6z48 + 5z47 + 8z46 + 3z45 + 12z44 + 6z43 + 6z42 + 3z41 + 6z40 + 3z39 + 2z38 + 9z36 + 5z35 + 6z34 + 9z33 + 3z32 + 6z31 + 8z30 + 13z29 + 6z28 + 6z26 + 4z25 + 8z24 + 14z22 + 15z21 + 4z19 + 2z18 + 15z17 + 10z16 + 4z15, q55(z) = 6z60 + 3z59 + 4z57 + 10z56 + 3z55 + 2z54 + 4z53, q56(z) = 5z59 + 13z58 + 12z56 + z55 + 8z54 + 5z53 + 12z52 + 15z51 + 8z50 + 14z49 + 2z48 + 4z47 + 4z46 + 10z45 + 12z44 + 8z43 + 4z41 + 2z40 + 8z38 + 12z37 + 2z36 + 12z35 + 8z34, q57(z) = 14z60 + 15z59 + 4z57 + 2z56 + 15z55 + 10z54 + 4z53, q58(z) = 8z57 + 8z56 + 8z55 + 12z53 + 8z51 + 8z50 + 8z49 + 8z48 + 10z47 + 14z46 + 10z45 + 8z44 + 9z43 + 12z42 + 2z41 + 6z40 + 11z39 + 3z38 + 12z37 + 8z36 + 7z35 + 2z34 + z33 + z32 + 7z31 + 12z30 + 12z29 + z28 + z26 + 15z25 + 9z24 + 14z23 + z22 + 3z21 + 3z20 + 4z19 + 7z17 + 7z16 + 6z15 + 3z14 + z13 + 3z12 + 2z11 + 14z9 + 2z8 + 6z7 + 2z5 + 14z4 + 6z3 + 14z2, q59(z) = 4z56 + 12z55 + 4z54 + 2z52 + 4z51 + 12z50 + 2z49 + 12z48 + 10z47 + 4z46 + 12z45 + 6z44 + 14z42 + 2z41 + 14z40 + 12z39 + 12z36 + 14z35 + 6z34 + 4z33 + 14z32 + 14z31 + 2z29 + 12z28 + 14z27 + 4z26 + 6z22 + 4z21 + 4z20 + 10z19 + 14z18 + 10z13 + 14z11 + 6z9 + 10z8 + 6z6, q60(z) = 12z65 + 14z57 + 2z56 + 3z55 + 2z52 + 9z51 + 15z50 + 6z49 + 4z48 + z47 + 15z46 + z45 + 14z44 + 15z43 + z42 + z41 + 6z40 + 12z39 + 13z38 + 12z37 + 2z36 + 2z35 + 7z34 + 5z33 + z32 + 14z31 + 10z30 + 11z29 + 6z28 + 10z27 + 9z26 + 3z25 + 6z24 + 3z22 + 6z20 + 14z18 + 4z17 + 3z16 + 7z15 + 9z14 + 7z13 + 14z10 + 11z9 + 13z8 + 2z7 + 13z6, 64 M. KAUERS, C. KRATTENTHALER, AND T. W. M¨ ULLER q61(z) = 2z60 + 9z59 + 14z56 + 9z55 + 6z54, q62(z) = 3z58 + 5z54 + 11z53 + 5z52 + 13z49 + 15z41 + 9z40 + 13z38 + 9z37 + 11z36 + 9z31 + 9z30 + 15z28 + 11z27 + 9z23 + 15z21 + 5z19 + 3z16 + z15 + 7z14 + 5z10, q63(z) = 6z55 + 14z54 + 12z50 + 9z49 + 7z48 + 12z47 + 5z46 + 5z45 + 13z44 + 15z43 + 13z42 + 13z41 + 8z40 + z39 + 7z38 + 3z37 + 15z36 + 8z35 + 6z34 + 15z32 + 15z31 + 10z30 + 8z29 + 12z28 + 13z27 + 14z25 + 13z24 + 14z23 + z22 + z21 + z20 + 9z19 + 13z18 + 12z17 + 10z16 + 3z15 + 4z14 + 7z13 + z12 + z11 + z10 + 3z9 + 15z8 + 7z7 + z6 + 15z5 + 6z4, q64(z) = 15z58 + 14z57 + 12z56 + 9z54 + 7z53 + 9z52 + 10z51 + 12z50 + z49 + 10z48 + 14z46 + 6z45 + 6z44 + 6z43 + 14z42 + 11z41 + 13z40 + 10z39 + z38 + 13z37 + 7z36 + 2z34 + 12z33 + 4z32 + 13z31 + 13z30 + 4z29 + 11z28 + 7z27 + 2z25 + 6z24 + 13z23 + 11z21 + 2z20 + 9z19 + 6z18 + 2z17 + 15z16 + 5z15 + 3z14 + 14z13 + 2z12 + 10z11 + 9z10 + 6z9 + 12z8, q65(z) = 12z58 + 14z55 + 4z54 + 4z53 + 4z52 + 2z51 + 6z50 + 12z49 + 12z48 + 10z47 + 6z45 + 12z43 + 12z42 + 6z41 + 12z40 + 6z39 + 4z38 + 2z36 + 10z35 + 12z34 + 2z33 + 6z32 + 12z31 + 10z29 + 12z28 + 12z26 + 12z22 + 14z21 + 4z18 + 14z17 + 4z16, q66(z) = 12z65 + 14z57 + 2z56 + 3z55 + 8z53 + 2z52 + 9z51 + 15z50 + 6z49 + 4z48 + z47 + 15z46 + z45 + 14z44 + 15z43 + z42 + z41 + 6z40 + 12z39 + 13z38 + 12z37 + 2z36 + 2z35 + 7z34 + 5z33 + z32 + 14z31 + 10z30 + 11z29 + 6z28 + 10z27 + 9z26 + 3z25 + 6z24 + 3z22 + 8z21 + 6z20 + 8z19 + 14z18 + 4z17 + 3z16 + 7z15 + 9z14 + 7z13 + 8z12 + 14z10 + 11z9 + 13z8 + 2z7 + 13z6, q67(z) = 12z59 + 12z58 + 12z55 + 12z53 + 4z51, q68(z) = 4z60 + 2z59 + 12z56 + 2z55 + 12z54, q69(z) = 4z67 + 14z58 + 8z57 + 7z55 + 10z54 + 10z53 + 2z52 + 9z51 + 3z50 + 14z49 + 6z48 + 13z47 + 8z46 + 11z45 + 12z44 + 6z43 + 6z42 + 11z41 + 6z40 + 11z39 + 2z38 + z36 + 13z35 + 6z34 + z33 + 11z32 + 6z31 + 8z30 + 5z29 + 6z28 + 6z26 + 4z25 + 8z24 + 14z22 + 7z21 + 4z19 + 2z18 + 7z17 + 10z16 + 4z15, q70(z) = 14z60 + 15z59 + 4z57 + 2z56 + 15z55 + 10z54 + 4z53, q71(z) = 4z51 + 14z50 + 4z49 + 14z47 + 5z46 + 12z45 + 14z43 + 11z42 + 9z41 + 11z40 + 6z39 + 11z38 + z37 + 2z36 + 13z35 + 11z33 + 2z31 + 3z29 + 10z28 + 9z27 + 7z26 + 6z24 + 2z23 + 12z22 + 2z21 + 15z20 + 11z19 + 14z18 + 11z17 + 14z15 + 4z14 + 11z13 + 3z12 + 3z11 + 13z10 + 12z9 + 4z8 + z7 + 6z6 + 3z5 + z4 + 3z3, q72(z) = 5z58 + 10z57 + 4z56 + 3z54 + 13z53 + 3z52 + 14z51 + 11z49 + 14z48 + 10z46 + 2z45 + 2z44 + 2z43 + 10z42 + 9z41 + 15z40 + 14z39 + 11z38 + 15z37 + 13z36 + 6z34 + 15z31 + 15z30 + 9z28 + 13z27 + 6z25 + 2z24 + 15z23 + 9z21 + 6z20 + 3z19 + 2z18 + 6z17 + 5z16 + 7z15 + z14 + 10z13 + 6z12 + 14z11 + 3z10 + 2z9 + 4z8, q73(z) = 2z55 + 6z51 + 10z50 + 6z47 + 10z46 + 6z45 + 10z43 + 6z42 + 6z41 + 14z38 + 10z34 + 14z33 + 6z32 + 2z29 + 6z26 + 2z25 + 2z22 + 2z16 + 10z15 + 6z14 + 10z13 + 2z9 + 14z8 + 14z6, q74(z) = 2z55 + 10z54 + 4z50 + 3z49 + 13z48 + 4z47 + 7z46 + 7z45 + 15z44 + 5z43 + 15z42 + 15z41 + 11z39 + 13z38 + z37 + 5z36 + 2z34 + 5z32 + 5z31 + 14z30 + 4z28 + 15z27 + 10z25 MOD-2k BEHAVIOUR OF RECURSIVE SEQUENCES 65 + 15z24 + 10z23 + 11z22 + 11z21 + 11z20 + 3z19 + 15z18 + 4z17 + 14z16 + z15 + 12z14 + 13z13 + 11z12 + 11z11 + 11z10 + z9 + 5z8 + 13z7 + 11z6 + 5z5 + 2z4, q75(z) = 10z58 + 13z55 + 14z54 + 14z53 + 6z52 + 3z51 + z50 + 10z49 + 2z48 + 15z47 + 9z45 + 2z43 + 2z42 + 9z41 + 2z40 + 9z39 + 6z38 + 11z36 + 15z35 + 2z34 + 11z33 + 9z32 + 2z31 + 7z29 + 2z28 + 2z26 + 10z22 + 13z21 + 6z18 + 13z17 + 14z16, q76(z) = 5z52 + 5z49 + z47 + 7z44 + 11z42 + 5z41 + 11z40 + 11z35 + 15z34 + 11z32 + 3z31 + 13z29 + 3z27 + 7z22 + 9z19 + 11z18 + z13 + 3z11 + 7z9 + 9z8 + 15z6, q77(z) = 13z55 + 7z51 + z50 + 15z47 + z46 + 15z45 + z43 + 15z42 + 15z41 + 3z38 + 9z34 + 11z33 + 15z32 + 5z29 + 7z26 + 13z25 + 13z22 + 13z16 + 9z15 + 7z14 + 9z13 + 5z9 + 3z8 + 3z6. 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Stanley, Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge, 1999. R. P. Stanley, Catalan Addendum, continuation of Exercise 6.19 from ; available at W. Stothers, The number of subgroups of given index in the modular group, Proc. Royal Soc. Edinburgh, Sec. A 78 (1977), 105–112. L. Sylow, Th´ eor emes sur les groupes de substitutions, Math. Ann. 5 (1872), 584–594. H. S. Wilf, generatingfunctionology, 2nd edition, Academic Press, San Diego, 1994. H. S. Wilf and D. Zeilberger, An algorithmic proof theory for hypergeometric (ordinary and “q”) multisum/integral identities, Invent. Math. 108 (1992), 575–633. D. Zeilberger, A holonomic systems approach to special functions identities, J. Comput. Appl. Math. 32 (1990), 321–368. †Research Institute for Symbolic Computation, Johannes Kepler Universit¨ at, Al-tenbergerstrasze 69, A-4040 Linz, Austria ‡∗Fakult¨ at f¨ ur Mathematik, Universit¨ at Wien, Nordbergstrasze 15, A-1090 Vienna, Austria. WWW: ∗School of Mathematical Sciences, Queen Mary & Westfield College, University of London, Mile End Road, London E1 4NS, United Kingdom.
6159	https://www.sciencedirect.com/science/article/abs/pii/S0363502323006391	Skip to article My account Sign in Access through your organization Purchase PDF Patient Access Article preview Abstract Section snippets References (21) The Journal of Hand Surgery Volume 49, Issue 3, March 2024, Pages 230-236 Editor's Choice A Fresh Cadaver Study on the Innervation of Brachioradialis and Extensor Carpi Radialis Longus Muscles Author links open overlay panel, , , rights and content Purpose Distal nerve transfers have revolutionized peripheral nerve surgery by allowing the transfer of healthy motor nerves to paralyzed ones without causing additional morbidity. Radial nerve branches to the brachialis (Ba), brachioradialis (Br), and extensor carpi radialis longus (ECRL) muscles have not been investigated in fresh cadavers. Methods The radial nerve and its branches were dissected in 34 upper limbs from 17 fresh cadavers. Measurements were taken to determine the number, origin, length, and diameter of the branches. Myelinated fiber counts were obtained through histological analysis. Results The first branch of the radial nerve at the elbow was to the Ba muscle, followed by the branches to the Br and ECRL muscles. The Ba and Br muscles consistently received single innervation. The ECRL muscle showed varying innervation patterns, with one, two, or three branches. The branches to the Br muscles originated from the anterior side of the radial nerve, whereas the branches to the Ba and ECRL muscles originated from the posterior side. The average myelinated fiber counts favored the nerve to Br muscle over that to the ECRL muscle, with counts of 542 versus 350 and 568 versus 302 observed in hematoxylin and eosin and neurofilament staining, respectively. Conclusions This study provides detailed anatomical insights into the motor branches of the radial nerve to the Ba, Br, and ECRL muscles. Clinical relevance Understanding the anatomy of the radial nerve branches at the elbow is of utmost importance when devising a reconstructive strategy for upper limb paralysis. These findings can guide surgeons in selecting appropriate donor or recipient nerves for nerve transfer in cases of high tetraplegia and lower-type brachial plexus injuries. Section snippets Material and Methods In advance of any data collection, the local ethics committee approved the protocol of the present study. We carefully dissected the radial nerve and its branches around the elbow in 34 upper limbs from 17 fresh cadavers. To help differentiate small vascular twigs from nerve branches, in seven cadavers, before dissection, the brachial artery was cannulated and injected with a mixture of green-colored prevulcanized latex plus a catalyst (Dumello) in a ratio of 20:2 mL. Dissections were performed Results The first branch of the radial nerve at the elbow was to Ba, followed by a branch to Br, and more distally, that to the ECRL (Fig. 1). In two dissections, the branch to the ECRL origin was proximal to the Br motor branch. In four dissections, the branch to Ba originated distally to the Br motor branch. We found a motor branch originating from the posteromedial side of the radial nerve that innervated the lateral side of the Ba muscle in 14 dissections. In two specimens, a double branch to the Ba Discussion Radial nerve branches to the Ba muscle were observed in 41% of our 34 dissections, whereas other studies reported higher rates ranging from 50% to 65%.11, 12, 13 However, the higher rates may be attributed to their dissections on embalmed cadavers where vessels can be misinterpreted as nerve branches.14 The motor branch of the radial nerve to Ba could be used to reconstruct wrist extension in high-level spinal cord injury, but its variable frequency and small diameter make it unsuitable for use Acknowledgments We are indebted to Dr. Jhony Chaverra from the Instituto de Medicina Legal y Ciencias Forences Valle Del Cauca, Cali, Colombia, for his endless help with the cadaver dissections. This study was supported by the Iberolatin America foundation for obstetric brachial plexus paralysis, www.fundacionpbo.com. References (21) J.A. Bertelli et al. ### Transfer of supinator motor branches to the posterior interosseous nerve to reconstruct thumb and finger extension in tetraplegia: case report ### J Hand Surg Am (2010) R.A. Abrams et al. ### Anatomy of the radial nerve motor branches in the forearm ### J Hand Surg Am (1997) G. Branovacki et al. ### The innervation pattern of the radial nerve at the elbow and in the forearm ### J Hand Surg Br (1998) A. GarcÃa-LÃ³pez et al. ### Transfer of the nerve to the brachioradialis muscle to the anterior interosseous nerve for treatment for lower brachial plexus lesions: case report ### J Hand Surg Am (2011) F.K. Fuss et al. ### Radial nerve entrapment at the elbow: surgical anatomy ### J Hand Surg Am (1991) J. FridÃ©n et al. ### Brachialis-to-extensor carpi radialis longus selective nerve transfer to restore wrist extension in tetraplegia: case report ### J Hand Surg (2012) E.M. Krauss et al. ### Outcome analysis of medial triceps motor nerve transfer to axillary nerve in isolated and brachial plexus-associated axillary nerve palsy ### Plast Reconstr Surg (2022) J.A. Bertelli ### Transfer of the radial nerve branch to the extensor carpi radialis brevis to the anterior interosseous nerve to reconstruct thumb and finger flexion ### J Hand Surg Am (2015) J.A. Bertelli et al. ### Transfer of supinator motor branches to the posterior interosseous nerve in C7-T1 brachial plexus palsy ### J Neurosurg (2010) J.A. Bertelli et al. ### Transfer of the distal anterior interosseous nerve for thumb motion reconstruction in radial nerve paralysis ### J Hand Surg Am (2020) There are more references available in the full text version of this article. Cited by (0) : No benefits in any form have been received or will be received related directly to this article. View full text © 2024 by the American Society for Surgery of the Hand. All rights reserved.
6160	https://pmc.ncbi.nlm.nih.gov/articles/PMC6698679/	Comparative effectiveness of antiepileptic drugs in juvenile myoclonic epilepsy - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Epilepsia Open . 2019 Jul 4;4(3):420–430. doi: 10.1002/epi4.12349 Search in PMC Search in PubMed View in NLM Catalog Add to search Comparative effectiveness of antiepileptic drugs in juvenile myoclonic epilepsy Katri Silvennoinen Katri Silvennoinen 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK Find articles by Katri Silvennoinen 1,2, Nikola de Lange Nikola de Lange 3 Luxembourg Centre for Systems Biomedicine, University of Luxembourg, Belvaux, Luxembourg Find articles by Nikola de Lange 3, Sara Zagaglia Sara Zagaglia 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK 4 Department of Experimental and Clinical Medicine, Polytechnic University of Marche, Ancona, Italy Find articles by Sara Zagaglia 1,2,4, Simona Balestrini Simona Balestrini 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK 4 Department of Experimental and Clinical Medicine, Polytechnic University of Marche, Ancona, Italy Find articles by Simona Balestrini 1,2,4, Ganna Androsova Ganna Androsova 3 Luxembourg Centre for Systems Biomedicine, University of Luxembourg, Belvaux, Luxembourg Find articles by Ganna Androsova 3, Merel Wassenaar Merel Wassenaar 5 Stichting Epilepsie Instellingen Nederland (SEIN), Heemstede, The Netherlands Find articles by Merel Wassenaar 5, Pauls Auce Pauls Auce 6 Department of Molecular and Clinical Pharmacology, Institute of Translational Medicine, University of Liverpool, Liverpool, UK 7 The Walton Centre NHS Foundation Trust, Liverpool, UK Find articles by Pauls Auce 6,7, Andreja Avbersek Andreja Avbersek 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK Find articles by Andreja Avbersek 1, Felicitas Becker Felicitas Becker 8 Hertie Institute for Clinical Brain Research, University of Tübingen, Tübingen, Germany Find articles by Felicitas Becker 8, Bianca Berghuis Bianca Berghuis 5 Stichting Epilepsie Instellingen Nederland (SEIN), Heemstede, The Netherlands Find articles by Bianca Berghuis 5, Ellen Campbell Ellen Campbell 9 Belfast Health and Social Care Trust, Belfast, UK Find articles by Ellen Campbell 9, Antonietta Coppola Antonietta Coppola 10 Pediatric Neurology and Muscular Diseases Unit, Department of Neurosciences, Rehabilitation, Ophthalmology, Genetics, Maternal and Child Health, University of Genoa, Genoa, Italy 11 Department of Neuroscience, Reproductive and Odontostomatological Sciences, Federico II University, Naples, Italy Find articles by Antonietta Coppola 10,11, Ben Francis Ben Francis 12 Department of Biostatistics, University of Liverpool, Liverpool, UK Find articles by Ben Francis 12, Stefan Wolking Stefan Wolking 8 Hertie Institute for Clinical Brain Research, University of Tübingen, Tübingen, Germany Find articles by Stefan Wolking 8, Gianpiero L Cavalleri Gianpiero L Cavalleri 13 Molecular and Cellular Therapeutics, Royal College of Surgeons in Ireland, Dublin, Ireland Find articles by Gianpiero L Cavalleri 13, John Craig John Craig 9 Belfast Health and Social Care Trust, Belfast, UK Find articles by John Craig 9, Norman Delanty Norman Delanty 13 Molecular and Cellular Therapeutics, Royal College of Surgeons in Ireland, Dublin, Ireland 14 Department of Neurology, Beaumont Hospital, Dublin, Ireland Find articles by Norman Delanty 13,14, Michael R Johnson Michael R Johnson 15 Faculty of Medicine, Division of Brain Sciences, Imperial College, London, UK Find articles by Michael R Johnson 15, Bobby P C Koeleman Bobby P C Koeleman 16 Department of Genetics, University Medical Center Utrecht, Utrecht, The Netherlands Find articles by Bobby P C Koeleman 16, Wolfram S Kunz Wolfram S Kunz 17 Department of Epileptology, University of Bonn, Bonn, Germany Find articles by Wolfram S Kunz 17, Holger Lerche Holger Lerche 8 Hertie Institute for Clinical Brain Research, University of Tübingen, Tübingen, Germany Find articles by Holger Lerche 8, Anthony G Marson Anthony G Marson 6 Department of Molecular and Clinical Pharmacology, Institute of Translational Medicine, University of Liverpool, Liverpool, UK 7 The Walton Centre NHS Foundation Trust, Liverpool, UK Find articles by Anthony G Marson 6,7, Terence J O’Brien Terence J O’Brien 18 Departments of Neuroscience and Neurology, Central Clinical School, Monash University, The Alfred Hospital, Melbourne, Vic. Australia Find articles by Terence J O’Brien 18, Josemir W Sander Josemir W Sander 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK 5 Stichting Epilepsie Instellingen Nederland (SEIN), Heemstede, The Netherlands Find articles by Josemir W Sander 1,2,5, Graeme J Sills Graeme J Sills 6 Department of Molecular and Clinical Pharmacology, Institute of Translational Medicine, University of Liverpool, Liverpool, UK Find articles by Graeme J Sills 6, Pasquale Striano Pasquale Striano 19 Pediatric Neurology and Muscular Diseases Unit, IRCCS Istituto G. Gaslini, Genova, Italy 10 Pediatric Neurology and Muscular Diseases Unit, Department of Neurosciences, Rehabilitation, Ophthalmology, Genetics, Maternal and Child Health, University of Genoa, Genoa, Italy Find articles by Pasquale Striano 19,10, Federico Zara Federico Zara 20 Laboratory of Neurogenetics and Neuroscience, IRCCS Istituto G. Gaslini, Genova, Italy Find articles by Federico Zara 20, Job van der Palen Job van der Palen 21 University of Twente, Enschede, The Netherlands Find articles by Job van der Palen 21, Roland Krause Roland Krause 3 Luxembourg Centre for Systems Biomedicine, University of Luxembourg, Belvaux, Luxembourg Find articles by Roland Krause 3, Chantal Depondt Chantal Depondt 22 Department of Neurology, Hôpital Erasme, Université Libre de Bruxelles, Brussels, Belgium Find articles by Chantal Depondt 22, Sanjay M Sisodiya Sanjay M Sisodiya 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK Find articles by Sanjay M Sisodiya 1,2,✉; the EpiPGX Consortium† Author information Article notes Copyright and License information 1 Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, London, UK 2 Chalfont Centre for Epilepsy, Chalfont St. Peter, UK 3 Luxembourg Centre for Systems Biomedicine, University of Luxembourg, Belvaux, Luxembourg 4 Department of Experimental and Clinical Medicine, Polytechnic University of Marche, Ancona, Italy 5 Stichting Epilepsie Instellingen Nederland (SEIN), Heemstede, The Netherlands 6 Department of Molecular and Clinical Pharmacology, Institute of Translational Medicine, University of Liverpool, Liverpool, UK 7 The Walton Centre NHS Foundation Trust, Liverpool, UK 8 Hertie Institute for Clinical Brain Research, University of Tübingen, Tübingen, Germany 9 Belfast Health and Social Care Trust, Belfast, UK 10 Pediatric Neurology and Muscular Diseases Unit, Department of Neurosciences, Rehabilitation, Ophthalmology, Genetics, Maternal and Child Health, University of Genoa, Genoa, Italy 11 Department of Neuroscience, Reproductive and Odontostomatological Sciences, Federico II University, Naples, Italy 12 Department of Biostatistics, University of Liverpool, Liverpool, UK 13 Molecular and Cellular Therapeutics, Royal College of Surgeons in Ireland, Dublin, Ireland 14 Department of Neurology, Beaumont Hospital, Dublin, Ireland 15 Faculty of Medicine, Division of Brain Sciences, Imperial College, London, UK 16 Department of Genetics, University Medical Center Utrecht, Utrecht, The Netherlands 17 Department of Epileptology, University of Bonn, Bonn, Germany 18 Departments of Neuroscience and Neurology, Central Clinical School, Monash University, The Alfred Hospital, Melbourne, Vic. Australia 19 Pediatric Neurology and Muscular Diseases Unit, IRCCS Istituto G. Gaslini, Genova, Italy 20 Laboratory of Neurogenetics and Neuroscience, IRCCS Istituto G. Gaslini, Genova, Italy 21 University of Twente, Enschede, The Netherlands 22 Department of Neurology, Hôpital Erasme, Université Libre de Bruxelles, Brussels, Belgium Correspondence, Sanjay M. Sisodiya, Department of Clinical and Experimental Epilepsy, UCL Queen Square Institute of Neurology, Box 29, Queen Square, London WC1N 3BG, UK. Email: s.sisodiya@ucl.ac.uk † EpiPGX Consortium contributors are listed in Appendix 1 ✉ Corresponding author. Received 2019 May 8; Revised 2019 Jun 11; Accepted 2019 Jun 22; Collection date 2019 Sep. © 2019 The Authors. Epilepsia Open published by Wiley Periodicals Inc. on behalf of International League Against Epilepsy. This is an open access article under the terms of the License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC6698679 PMID: 31440723 Abstract Objective To study the effectiveness and tolerability of antiepileptic drugs (AEDs) commonly used in juvenile myoclonic epilepsy (JME). Methods People with JME were identified from a large database of individuals with epilepsy, which includes detailed retrospective information on AED use. We assessed secular changes in AED use and calculated rates of response (12‐month seizure freedom) and adverse drug reactions (ADRs) for the five most common AEDs. Retention was modeled with a Cox proportional hazards model. We compared valproate use between males and females. Results We included 305 people with 688 AED trials of valproate, lamotrigine, levetiracetam, carbamazepine, and topiramate. Valproate and carbamazepine were most often prescribed as the first AED. The response rate to valproate was highest among the five AEDs (42.7%), and significantly higher than response rates for lamotrigine, carbamazepine, and topiramate; the difference to the response rate to levetiracetam (37.1%) was not significant. The rates of ADRs were highest for topiramate (45.5%) and valproate (37.5%). Commonest ADRs included weight change, lethargy, and tremor. In the Cox proportional hazards model, later start year (1.10 [1.08‐1.13], P<0.001) and female sex (1.41 [1.07‐1.85], P=0.02) were associated with shorter trial duration. Valproate was associated with the longest treatment duration; trials with carbamazepine and topiramate were significantly shorter (HR [CI]: 3.29 [2.15‐5.02], P<0.001 and 1.93 [1.31‐2.86], P<0.001). The relative frequency of valproate trials shows a decreasing trend since 2003 while there is an increasing trend for levetiracetam. Fewer females than males received valproate (76.2% vs 92.6%, P=0.001). Significance In people with JME, valproate is an effective AED; levetiracetam emerged as an alternative. Valproate is now contraindicated in women of childbearing potential without special precautions. With appropriate selection and safeguards in place, valproate should remain available as a therapy, including as an alternative for women of childbearing potential whose seizures are resistant to other treatments. Keywords: seizures, tolerability, adverse drug reactions, valproate Key Points. We conducted a retrospective study of comparative effectiveness of five commonly used antiepileptic drugs in 305 individuals with JME Valproate was associated with the highest response rate; levetiracetam ranked second Topiramate and valproate were associated with highest rates of ADRs Controlling for start year and sex, valproate was associated with the longest treatment duration Valproate should remain available as a treatment option for people with refractory JME irrespective of sex 1. INTRODUCTION Juvenile myoclonic epilepsy (JME) is a common epilepsy syndrome, comprising 5%‐10% of all epilepsies.1 As JME tends to start during adolescence and lifestyle issues are known to increase the likelihood of seizures, particular attention and care are often required.2, 3, 4 Sodium valproate has long been the antiepileptic drug (AED) of choice for treatment of people with JME, with reported seizure freedom attained in up to 80%.5 Despite its effectiveness, valproate use is limited by adverse drug reactions (ADRs), teratogenicity, and neurotoxicity,6, 7 with important recent restrictions on its use in women of childbearing potential.8 Newer AEDs are taking an increasing role in the management of JME, but there are few data on comparative effectiveness to guide treatment choices.3, 9, 10, 11, 12 People with JME comprised a quarter of those with idiopathic generalized epilepsy (IGE) in the SANAD study, which demonstrated the effectiveness of valproate in IGE overall.13 In a retrospective study of 962 individuals with IGE, of whom 357 had JME, valproate monotherapy was associated with a higher response rate compared to lamotrigine or topiramate, but no statistical comparison was undertaken.9 Three previous studies have addressed AED comparative effectiveness in JME specifically. These include a prospective study of 156 people10 and a retrospective study of 186 individuals11; statistical testing of differences between AED effectiveness was not reported in either study.10, 11 In another retrospective study of 72 individuals, trials with valproate, lamotrigine, or topiramate were associated with better control of myoclonic seizures compared to trials involving phenytoin or carbamazepine.12 The prognosis of JME is relatively good with a reported remission rate of approximately 60%,14 but a subset of individuals is refractory to appropriate medical treatment.4, 5, 14, 15 While AED withdrawal without seizure recurrence may be successful in some,14 the majority of people in remission remain on AEDs,14, 15 raising concerns about long‐term side effects. More information regarding the effectiveness and tolerability of AEDs in JME is required. Our aim was to evaluate AED frequency of use, effectiveness, retention, and tolerability in a real‐world setting. Despite limitations imposed by its observational and retrospective nature, our study provides some much‐needed data on this important topic. 2. METHODS 2.1. Participants People with JME were identified from a large clinical database from the EpiPGX consortium, an international multicenter research project on epilepsy pharmacogenetics (www.epipgx.eu). This database contains demographic and clinical details of nearly 10,000 people with a confirmed diagnosis of epilepsy and detailed information on more than 39,000 treatment regimens (hereafter referred to as trials), all collected retrospectively from medical records. Participants were recruited mainly from tertiary referral centers. Data collection was started in 2012 and completed in 2016. Data collection and use was approved by research ethical committees/institutional review boards of each center, and all participants provided written informed consent for appropriately coded use of their clinical data. The ascertainment of JME cases was based on the criteria of the International League Against Epilepsy (ILAE) and an international consensus statement2, 16: (a) occurrence of myoclonic seizures, (b) onset between age 8‐25 years, (c) no evidence of progressive disease or intellectual disability, (d) EEG showing generalized epileptiform activity (cases with normal EEG during appropriate AED treatment were included, if deemed to otherwise fulfill criteria for JME by the treating specialist), (e) no clinically significant abnormality on neuroimaging, where available. We identified 321 individuals recruited from specialized epilepsy clinics in Belgium, Germany, Ireland, Italy, the Netherlands, the United Kingdom, and Australia (see Table S1 for details). Among them, 12 (3.7%) had already been included in the SANAD study.13 All AED trials from the time of epilepsy diagnosis were considered, including those introduced as add‐on therapy, with some exclusions: (a) AED trials started less than one year before the last clinic visit and (b) AEDs used in fewer than 50 trials. Prescription order, however, was determined considering all AED trials of the individual (see Figure 1). Both regular and extended release AED formulations were included in the analyses. Figure 1. Open in a new tab Flowchart for inclusion of people and AED trials. The number of trials for which defined data were available is indicated for each parameter 2.2. Outcome measure definition The classification of AED trial outcomes was modified from the ILAE consensus.17 Response was determined clinically as seizure freedom, lasting for ≥12 months, attributable to the AED according to either the treating clinician or the person undertaking phenotyping, or both, and occurring prior to initiation of another treatment for epilepsy. Failure of a trial of treatment was defined as persistent seizures at >50% of the pretreatment seizure frequency despite use of an appropriate AED with an adequate trial. We applied the ILAE criteria for assessing adequacy of an intervention.17 To aid outcome assessment, researchers performing the phenotyping used the World Health Organization (WHO) defined daily doses for each AED, as well as the summary table from the ILAE consensus paper defining the minimum dataset to determine whether an intervention is informative.17 If data were available but neither criteria for response nor failure were met, the outcome was considered unclassified. If the data required for assessing outcome were missing, the response was categorized as “unknown” (this included trials which were stopped before the outcome was known). Population percentage response was calculated as the number of responses divided by the total number of known outcomes (response, failure, and unclassified). For trials ongoing at last follow‐up, treatment duration was calculated based on the date of last visit, if available. In total, treatment duration was defined for 527 AED trials. Twelve‐month retention rate was defined as the proportion of trials with minimum duration of 12 months. The reason for discontinuation was recorded as due to ADR, lack of effectiveness, ADR and lack of effectiveness, other reason, remission, or unknown. ADRs were classified into nine categories, and their incidence was calculated for each AED as percentage of all trials. Only ADRs considered attributable to the specific AED either by the treating clinician and/or person undertaking phenotyping, or both, were included. The maximum daily dose was recorded for AED trials. Valproate trials were stratified by maximum daily dosage ≤1 g and >1 g, in keeping with classification used in previous literature.18 2.3. Statistical analyses Median values were used to express central tendency for durations of trials as data were not normally distributed. Pairwise comparisons with χ 2 analyses were performed to compare the AEDs with respect to population percentage response rates and rates of ADRs. Similarly, we compared the proportions of females and males receiving valproate, the number of valproate trials between females and males, and the response rates among first, second, or third or later order AED trials. Bonferroni correction for multiple comparisons was applied. We report corrected P‐values for simplicity; values ≤0.05 were considered significant. We modeled discontinuation patterns and compared retention of different AEDs with a Cox proportional hazards model. The outcome measure was trial duration, with hazard ratios presenting hazard of shorter duration. The cofactors included were AED (five levels, one for each AED, with valproate considered the reference level), trial start year (as a continuous variable), and sex (with male sex considered the default). Trial start year and sex were included as we hypothesized that these could influence retention. The global test for nonproportionality was not significant; thus, the Cox model could be applied. The Mann‐Whitney U test was used to compare maximum daily dose distributions of valproate between trials associated with response and trials associated with failure. All analyses were performed using R.19 3. RESULTS Three hundred and five individuals were included in the final analyses (1). The most commonly used AEDs were valproate, lamotrigine, levetiracetam, carbamazepine, and topiramate, constituting 688 trials (Table 2). Other AEDs were each used in fewer than 50 trials and were excluded from further analyses (see Table S2 and Figure S1 for further demographic and clinical details). Valproate and carbamazepine most often constituted first‐order AED trials, whereas lamotrigine was most commonly started as the second AED trial, levetiracetam as the third trial, and topiramate as the fourth trial (see Tables S3‐S4). Table 1. Demographic and clinical details \| Category \| \| \| :--- \| Age (y) \| Mean \| Range \| \| At last visit \| 31 \| 13‐78 \| \| At epilepsy diagnosis \| 16 \| 8‐43 \| \| At onset \| 15 \| 8‐25 \| \| Epilepsy duration (y) \| 15 \| 1‐65 \| \| AED trials per patient \| 2 \| 1‐8 \| \| \| Total number \| Percentage (%) \| \| Sex \| \| Male \| 95 \| 31.1 \| \| Female \| 210 \| 68.9 \| \| Seizure type \| \| GTCS \| 267 \| 87.5 \| \| Absence \| 92 \| 30.2 \| \| Myoclonic \| 305 \| 100 \| Open in a new tab Abbreviations: AED, antiepileptic drug; GTCS, generalized tonic‐clonic seizures. Table 2. Details for trials of 22 AEDs in the 306 individuals with treatment trials over 12 mo long \| AED name \| AED \| No. of trials \| No. of patients \| 12‐mo retention rate (%) \| Treatment duration (median months±MAD) \| Maximum dose (median mg/d) \| AED 1 (%) \| AED 2 (%) \| AED 3 (%) \| :--- :--- :--- :--- :--- \| \| Valproate \| VPA \| 279 \| 248 \| 86.1 \| 68±71.5 \| 1200 \| 57.0 \| 20.5 \| 22.5 \| \| Lamotrigine \| LTG \| 161 \| 153 \| 83.5 \| 35±35.6 \| 300 \| 26.9 \| 35.2 \| 37.9 \| \| Levetiracetam \| LEV \| 124 \| 122 \| 79 \| 31±33.4 \| 2000 \| 10.8 \| 26.4 \| 62.8 \| \| Carbamazepine \| CBZ \| 64 \| 62 \| 77.1 \| 30±34.1 \| 800 \| 50.0 \| 22.7 \| 27.3 \| \| Topiramate \| TPM \| 60 \| 55 \| 62.7 \| 23±31.1 \| 200 \| 19.1 \| 13.2 \| 67.6 \| \| Clobazam \| CLB \| 33 \| 30 \| 47.1 \| 11±13.5 \| 20 \| 2.8 \| 11.1 \| 86.1 \| \| Phenobarbital \| PB \| 31 \| 29 \| 66.7 \| 39±106.7 \| 100 \| 37.5 \| 15.6 \| 46.9 \| \| Ethosuximide \| ESM \| 27 \| 25 \| 69.2 \| 60±83.7 \| 750 \| 21.4 \| 21.4 \| 57.1 \| \| Phenytoin \| PHT \| 27 \| 24 \| 69.2 \| 48±66.7 \| 300 \| 18.5 \| 25.9 \| 55.6 \| \| Clonazepam \| CNZ \| 25 \| 22 \| 62.5 \| 21±22.6 \| 4 \| 7.7 \| 26.9 \| 65.4 \| \| Zonisamide \| ZNS \| 15 \| 14 \| 86.7 \| 27±22.2 \| 300 \| – \| – \| 100 \| \| Primidone \| PRM \| 12 \| 12 \| 83.3 \| 60±29.3 \| 750 \| 14.3 \| 21.4 \| 64.3 \| \| Oxcarbazepine \| OXC \| 11 \| 11 \| 83.3 \| 44±40.4 \| 1350 \| 25.0 \| 41.7 \| 33.3 \| \| Acetazolamide \| AZM \| 7 \| 7 \| 0 \| 6±8.6 \| 500 \| 28.6 \| – \| 71.4 \| \| Gabapentin \| GBP \| 5 \| 5 \| 0 \| 3±3.7 \| 1200 \| – \| 40.0 \| 60.0 \| \| Diazepam \| DZP \| 3 \| 3 \| – \| – \| 10 \| 66.7 \| 33.3 \| – \| \| Vigabatrin \| VGB \| 3 \| 3 \| 50 \| 34±42.9 \| 2000 \| – \| 66.7 \| 33.3 \| \| Lacosamide \| LCM \| 2 \| 2 \| 100 \| 34±26.4 \| 350 \| – \| – \| 100 \| \| Piracetam \| PIR \| 2 \| 2 \| 0 \| 1±0 \| 2400 \| – \| – \| 100 \| \| Tiagabine \| TGB \| 2 \| 2 \| – \| – \| – \| – \| – \| 100 \| \| Bromide \| BRM \| 1 \| 1 \| – \| – \| – \| 100 \| – \| – \| \| Felbamate \| FBM \| 1 \| 1 \| 100 \| 112±0 \| 1800 \| – \| – \| 100 \| Open in a new tab The three rightmost columns present the proportions of trials started as the individual's first, second, or third or later AED. Missing data are denoted by (–). Abbreviations: AED, antiepileptic drug; MAD, median absolute deviation. 3.1. Secular patterns of AED trials The first AED trial in the study was started in June 1968 and the last in March 2014. The highest number of trials was recorded in 2003 and 2005, with 206 trials in each. Secular changes in the prevalence of AED trials are presented in Figure 2. Since 2003, there has been a decline in the absolute and relative frequency of valproate trials. During this time, the relative frequency of lamotrigine trials remained stable while the relative frequency of levetiracetam trials increased. The majority of carbamazepine trials took place in the 1990s, with only individual trials recorded from 2005 onwards. See Figure S2 for sex‐specific secular changes in the relative frequencies of AED trials. Figure 2. Open in a new tab The secular prevalence of AED trials between 1968 and 2014. The extreme right vertical line indicates the 2013 recommendation by the UK Medicines and Healthcare products Regulatory Agency to restrict valproate use, and referral of valproate to the European Medicines Agency Pharmacovigilance Risk Assessment Committee36 3.2. Effectiveness Response rates to AEDs ranged from 14.1% (carbamazepine) to 42.7% (valproate); see Figure 3 and Table S5 for further details. The population percentage response rate to valproate was significantly higher than the population percentage response rate to lamotrigine (P<0.001), carbamazepine (P=0.03), and topiramate (P=0.02). The differences in population percentage response rates between other AED pairs were not significant (Table S6). Figure 3. Open in a new tab Relative frequencies of trial outcomes for each AED For individual AEDs, correlation between response rate and prescription order could not be performed due to the relatively low response rates observed. Considering all five AEDs together, the response rate of first‐order AED trials was highest (first: 45.6%, second: 38.8% and third or later: 30.0%), but the differences were not statistically significant. 3.3. Drug retention Twelve‐month retention rates ranged from 62.7% for topiramate to 86.1% for valproate; valproate also had the highest median treatment duration (Table 2, Figure S3). In the Cox model, a significant effect of start year on trial duration was observed; later start year was associated with shorter trial duration (HR [CI]: 1.10 [1.08‐1.13], P<0.001). Female sex was associated with shorter trial duration (1.41 [1.07‐1.85], P=0.02). The hazard ratios comparing the duration of the other AEDs to valproate, after adjusting for effect of start year and sex, are presented in Figure S4. Compared to valproate, carbamazepine and topiramate were associated with significantly shorter trial durations. 3.4. AED discontinuation A reason for discontinuation was noted for 69.0% of trials. Lamotrigine had the highest rate of discontinuation due to lack of effectiveness (40.8% of trials), but a low rate of discontinuation due to ADRs (10.2%). The respective figures for levetiracetam were 25.4% and 14.9%. Topiramate was associated with the highest rate of discontinuation due to ADRs (24.5%) (Table S7). 3.5. Adverse drug reactions The frequency of ADRs ranged from 14.5% for carbamazepine to 45.5% for topiramate. The rate of ADRs for carbamazepine was significantly lower than the rate of ADRs for topiramate (P=0.005) and valproate (P=0.010). The rate of ADRs for lamotrigine was also significantly lower than that for topiramate (P<0.001) and valproate (P<0.001). Overall, the three most common ADRs were weight change (reported in 64 trials), lethargy (40 trials), and tremor (37 trials). The incidence of specific ADRs, however, varied for each AED (see Table 3). Additional information on ADRs is presented in Appendix S1. Table 3. Incidence of the nine most frequent adverse drug reactions (ADRs) for each antiepileptic drug (AED), expressed as absolute number of trials with ADR, and as percentage of all trials \| AED \| Patients with ADR (%) \| Weight change \| Lethargy \| Tremor \| Cognitive impairment \| Behavioral disorder \| Depression \| Gastrointestinal ADRs \| Adverse cutaneous reaction \| Speech disorder \| :--- :--- :--- :--- :--- \| VPA \| 93 (37.5) \| 58 (20.8) \| 19 (6.8) \| 30 (10.8) \| 12 (4.3) \| 6 (2.2) \| 2 (0.7) \| 3 (1.1) \| 0 (0.0) \| 1 (0.4) \| \| LTG \| 25 (16.3) \| 2 (1.2) \| 4 (2.5) \| 4 (2.5) \| 3 (1.9) \| 0 (0.0) \| 2 (1.2) \| 2 (1.2) \| 4 (2.5) \| 0 (0.0) \| \| LEV \| 30 (24.6) \| 0 (0.0) \| 11 (8.9) \| 1 (0.8) \| 3 (2.4) \| 16 (12.9) \| 4 (3.2) \| 0 (0.0) \| 0 (0.0) \| 1 (0.8) \| \| CBZ \| 9 (14.5) \| 0 (0.0) \| 1 (1.6) \| 0 (0.0) \| 0 (0.0) \| 0 (0.0) \| 0 (0.0) \| 4 (6.2) \| 4 (6.2) \| 0 (0.0) \| \| TPM \| 25 (45.5) \| 4 (6.7) \| 5 (8.3) \| 2 (3.3) \| 10 (16.7) \| 3 (5.0) \| 3 (5.0) \| 0 (0.0) \| 0 (0.0) \| 3 (5.0) \| Open in a new tab 3.6. Valproate dosage and use in males and females Of the 119 valproate trials associated with response, 61 (51.3%) involved a maximum daily dose of ≤1 g. In trials associated with response, the median maximum daily dose was 1000 mg/d, whereas for failed trials, it was 1500 mg/d. The difference in the maximum daily dose distributions was statistically significant (Mann‐Whitney P=0.004; see Figure S5). Valproate was trialed in 92.6% of males and 76.2% of females; the difference was significant (P=0.001). Females more frequently had interruption of valproate treatment, with more than one valproate trial in 13.8% of females versus 10.2% of males. 4. DISCUSSION We explored the use of AEDs in 305 people with JME over a long period, providing observational real‐world insight into the tolerability and effectiveness of AEDs in this syndrome. The topic is timely as means of managing risks of valproate have recently undergone review by the European Medicines Agency,8 with important implications for management of JME. Prospective JME‐specific trials involving the current array of AEDs, including valproate, are now very unlikely, and observational studies such as ours provide important information for clinical practice. Compared to previous JME‐specific studies,10, 11, 12 a strength of our study is a relatively large number of trials with AEDs other than valproate, formal comparative testing, and stringent criteria for JME diagnostic ascertainment. We applied a uniform method of data collection in all centers; data were stored in a single database. Recently, we successfully applied the same strategy for investigating AED use in mesial temporal lobe epilepsy.20 An obvious weakness of our study is its retrospective design: We could not account for possible effects of the natural history of JME on outcomes. Patients were gathered mainly from tertiary centers, which should increase confidence in the validity of syndromic diagnosis.21 Individuals with refractory epilepsy may, however, be overrepresented in our sample. Other JME studies report a female preponderance.15 In our study, the more pronounced gender difference may further reflect referral bias to our tertiary centers,related, for example, to management issues around pregnancy or family planning. Observed patterns of AED use may be biased by local practices at the participating centers. EpiPGX was not designed to look prospectively at neurodevelopmental outcomes in children exposed to AEDs in utero, and we cannot comment on this very important topic. Lastly, classifying drug response is challenging, with various schemes: We chose a modification of the ILAE scheme.17 The ILAE definition of seizure freedom requires that duration of seizure freedom is three times the previous interseizure interval (“Rule of Three”) or at least 12 months, whichever is longer.17 The limitations of the “Rule of Three” for prediction of ongoing seizure freedom are, however, recognized.22 The conventional definition based only on at least 12 months' seizure freedom is still commonly used.23 The ILAE definition of treatment failure is lack of seizure freedom after an informative trial of an intervention17; for clinical utility, we defined failure as less than 50% reduction in seizure frequency. The frequencies of trials for individual AEDs are influenced by changes in AED availability over time. The predominance of valproate was expected considering its long‐standing availability and typical practice having been to use it as first‐line treatment for JME.24 Carbamazepine, which is not recommended for treatment of JME,3 emerged as the fourth most commonly tried AED. In over half of these trials, carbamazepine was the first AED tried. Most carbamazepine trials in our sample date to the 1990s, when fewer treatment options were available. In some individuals, carbamazepine trials may also have predated their JME diagnosis. The response rate to valproate was the highest, and the population percentage response rate differed significantly from that of lamotrigine, carbamazepine, and topiramate. Levetiracetam had the second highest response rate with no significant difference compared to valproate. Effectiveness measures may be affected by prescription order25; most trials with levetiracetam were the individual's third or later order AED trial. The comparison of response rates is limited by lack of data on whether an AED was introduced as monotherapy or as add‐on. This would have been particularly interesting for valproate and lamotrigine, given the evidence for synergism between these AEDs.26 Further limitations are the high frequencies of unclassified and unknown outcomes. We lacked effectiveness information for specific seizure types; we expect that this may impact upon rating for absence and myoclonic seizures especially. According to previous reports, carbamazepine and lamotrigine are associated with risk of more frequent myoclonic seizures,3, 27 but we were unable to assess whether this could have affected our outcomes. Nevertheless, our results are in keeping with prospective data on the high effectiveness of valproate in the management of IGE.13 They also indicate levetiracetam as an effective alternative, in keeping with previous reports of effectiveness of levetiracetam as add‐on treatment28 or monotherapy29 in JME. Pseudoresistance, that is treatment failure caused by lifestyle factors such as alcohol consumption and sleep deprivation, is a recognized concept in JME.4 As a further limitation, we were unable to assess whether such factors contributed to our treatment outcomes. Valproate was associated with the highest 12‐month retention rate and median treatment duration. In our survival analysis, valproate was associated with a significantly longer trial duration compared to carbamazepine or topiramate, but not lamotrigine or levetiracetam. Retention parameters may be skewed by older trials started when fewer alternative AEDs were available. This was confirmed in the survival analysis. Another possible source of bias is the effect of prescription order, as an individual's first AED trial is more likely to be successful,25 and therefore have longer duration, compared to subsequent trials. In our sample, valproate most commonly constituted an individual's first trial. The highest rate of ADRs was observed for topiramate, with valproate ranking second. For both AEDs, rates of ADRs were similar to those observed in SANAD.13 While there is some evidence that topiramate may be better tolerated than valproate as monotherapy for JME,30 our contrasting findings may reflect effects of polytherapy.31 Due to the retrospective nature of our study, some types of ADRs may have been more frequently recorded than others. The commonest ADR to topiramate was cognitive impairment, whereas for valproate, weight change was most common. These findings are generally in keeping with prospectively collected data.13, 32 The commonest ADR to levetiracetam was behavioral disorder, and indeed concerns have been previously raised about the neuropsychiatric side effects of levetiracetam and topiramate in people with JME,33 in whom psychiatric comorbidities and impulsive personality traits appear overrepresented.34 No valproate dose is considered safe in pregnancy, but some risks associated with fetal exposure to valproate are known to be dose‐dependent,6 as are some of the valproate‐associated ADRs.7 Recommended maintenance doses for valproate are 1‐2 g/d.35 There are suggestions that daily monotherapy doses of ≤1 g are sufficient for maintaining seizure freedom for a significant proportion of people with JME.10, 18 In our group, over half of successful valproate trials involved maximum doses of 1 g/d. Together with previous reports, our findings suggest that in people with JME for whom valproate is a necessary and appropriate treatment choice, it is reasonable to aim initially for lower doses. Valproate is being superseded by other AEDs (see Figure 2). The onset of this change in 2003 precedes the more recent regulatory and pharmacovigilance measures,8, 36 and likely reflects the combination of accumulating evidence of adverse outcomes in valproate‐associated pregnancies6 and increasing availability of alternative AEDs. We also observed sex differences in the patterns of valproate use: Males were significantly more likely to receive valproate than females, and females' valproate trials were more likely to be subject to interruptions. These findings are in keeping with population‐based reports of decline in valproate use in treatment of epilepsy in women.37 While these observations reflect serious concern over adverse effects, teratogenicity, and risk of neurodevelopmental disorders in exposed offspring,6 they also suggest that people with JME, especially female, may be deprived of the most effective treatment for their condition, a concern already voiced by others.11 Poor seizure control carries well‐recognized risks,38 and it is important to consider the possible effects of reduced valproate use on seizure control and other outcomes in JME. Current restrictions on valproate use8 warrant a re‐assessment of therapeutic options for JME. Valproate was associated with a considerable rate of ADRs, and its potential for teratogenicity and inducing neurodevelopmental disorders has significant implications for patient choice, counseling and treatment monitoring. Data on pregnancy‐related risks associated with valproate use have been widely disseminated again recently,8 and new measures for their management are in place. In girls and women, valproate use must take these new measures into account.8 Among other AEDs, our results corroborate the role of levetiracetam in the management of JME. Based on its high retention and response rates, however, valproate should remain available, with the necessary counseling and safeguards, as an alternative for people not responsive to other treatments, irrespective of sex. CONFLICTS OF INTEREST AA is employed by UCB Pharma, Belgium as Associate Director. AC reports grants from Eisai outside the submitted work. JC reports personal fees from UCB Pharma, personal fees from Sanofi‐Synthelabo, personal fees from Glaxo Smith Kline, personal fees from Janssen‐Cilag, personal fees from Pfizer and personal fees from Eisai to undertake lectures, participate in advisory boards and to undertake research. HL has received honoraria for consulting or speaking or travel support from Bial, BioMarine, Desitin, Eisai, and UCB, and research support from Bial, all outside the submitted work. JWS has received research funding from Eisai, and UCB, personal fees from Eisai, Bial, Janssen and UCB outside the submitted work. In the past 36 months, GJS has received personal fees for consulting and/or speaking from UCB Pharma and Eisai, all outside the submitted work. CD has received honoraria and grant funding from UCB, unrelated to the current study. SMS reports representing the Association of British Neurologists and The Royal College of Physicians (London) at the MHRA Valproate Stakeholders Network, is a member of the scientific advisory board of Dravet Syndrome UK, patron of AHC UK. SMS has received honoraria or grant funding from UCB, Eisai, Vitaflo and Nutricia, all outside the submitted work. The remaining authors have no conflicts of interests. We confirm that we have read the Journal's position on issues involved in ethical publication and affirm that this report is consistent with those guidelines. Supporting information Click here for additional data file. (898.3KB, docx) ACKNOWLEDGMENTS This study was supported by EC grant 279062, EpiPGX. This work was partly carried out at NIHR University College London Hospitals Biomedical Research Centre, which receives a proportion of funding from the UK Department of Health's NIHR Biomedical Research Centres funding scheme. Additional funding was provided by the Muir Maxwell Trust and Epilepsy Society. KS is supported by a Wellcome Trust Strategic Award (WT104033AIA). SZ was supported by the Polytechnic University of Marche, Italy, with a 1‐year research fellowship. GA has received funding from the framework of the EU‐funded FP7 research program BioCog (Biomarker Development for Postoperative Cognitive Impairment in the Elderly). The Australian cohort was funded by a grant from the Royal Melbourne Hospital Foundation. We thank Mojgansadat Borghei, Larus J. Gudmundsson, Andres Ingason, Clare Kennedy, Martin Krenn, Benjamin Legros, Ekatarina Pataraia, Sarah Rau, Kari Stefansson, William Stern, Anna Tostevin, Patrick Tugendhaft, Wim Van Paesschen, and Fritz Zimprich for help with data collection. APPENDIX 1. EpiPGX Contributors Martin J. Brodie, Krishna Chinthapalli, Gerrit‐Jan de Haan, Colin P. Doherty, Sinead Heavin, Mark McCormack, Slavé Petrovski, Narek Sargsyan, Lisa Slattery, Joseph Willis Silvennoinen K, de Lange N, Zagaglia S, et al; the EpiPGX Consortium . Comparative effectiveness of antiepileptic drugs in juvenile myoclonic epilepsy. Epilepsia Open. 2019;4:420–430. 10.1002/epi4.12349 Krause, Depondt and Sisodiya contributed equally to the manuscript Data Availability Statement: Requests for data may be addressed to the EpiPGX steering committee via the corresponding author. 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6161	https://brainly.com/question/12145557	[FREE] Help please What is the area of the frame in square units around the picture? - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +26,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +49,6k Ace exams faster, with practice that adapts to you Practice Worksheets +5,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Help please What is the area of the frame in square units around the picture? 2 See answers Explain with Learning Companion NEW Asked by Taehyun • 02/15/2019 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 45476909 people 45M 4.0 2 Upload your school material for a more relevant answer The area of the frame in square units around the picture will be 40s square units. What is the area of the rectangle? Let L be the length and W be the width of the rectangle. Then the area of the rectangle will be Area of the rectangle = L×W square units The area of outer rectangle will be A₁ = 12 x 6s A₁ = 72s square units The area of inner rectangle will be A₂ = 8 x 4s A₂ = 32s square units Then the area of the frame in square units around the picture will be A = A₁ - A₂ A = 72s - 32s A = 40s square units More about the area of the rectangle link is given below. brainly.com/question/20693059 SPJ2 Answered by jainveenamrata •10.5K answers•45.5M people helped Thanks 2 4.0 (3 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 45476909 people 45M 4.0 2 Kinematics fundamentals - Sunil Singh Physics for K-12 - Sunil Singh Energy and Human Ambitions on a Finite Planet - Murphy, Thomas W, Jr Upload your school material for a more relevant answer To find the area of a frame around a picture, calculate the area of the outer rectangle and subtract the area of the inner rectangle. For instance, with outer dimensions of 12x8 units and inner dimensions of 10x6 units, the area of the frame would be 36 square units. This process involves knowing the lengths and widths of both rectangles involved. Explanation To find the area of the frame around a picture, we can follow these steps: Identify Dimensions: First, you need to know the dimensions of the outer rectangle (the frame) and the inner rectangle (the picture). Let's denote the outer rectangle's length as L 1 and width as W 1, and the inner rectangle's length as L 2 and width as W 2. Calculate Areas: The area of the outer rectangle is given by: A 1=L 1×W 1 The area of the inner rectangle is: A 2=L 2×W 2 Find the Area of the Frame: The area of the frame can be found by subtracting the area of the inner rectangle from the area of the outer rectangle: A=A 1−A 2 Substitute the areas you calculated earlier: A=(L 1×W 1)−(L 2×W 2) Example: If the dimensions of the outer rectangle are 12 units by 8 units, and the dimensions of the inner rectangle are 10 units by 6 units: Calculate the area of the outer rectangle: A 1=12×8=96 square units Calculate the area of the inner rectangle: A 2=10×6=60 square units Therefore, the area of the frame is: A=96−60=36 square units Thus, the area of the frame around the picture is 36 square units. Examples & Evidence As an illustration, if the outer frame measures 15 units by 10 units and the picture measures 12 units by 7 units, you can use the same process: 1) Calculate the outer area as 150 square units and the inner area as 84 square units; 2) Subtract to find the frame's area to be 66 square units. The formula for the area of a rectangle (A=L×W) is mathematically proven and universally accepted in geometry. Thanks 2 4.0 (3 votes) Advertisement Community Answer This answer helped 4730 people 4K 3.5 7 Find the area of both squared then subtract 28 units squared Answered by Q00 •10 answers•4.7K people helped Thanks 7 3.5 (8 votes) Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1 Is this equation correct? 6 3⋅7 3=4 2 3 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
6162	https://pmc.ncbi.nlm.nih.gov/articles/PMC9434525/	Digital Eye Strain- A Comprehensive Review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Ophthalmol Ther . 2022 Jul 9;11(5):1655–1680. doi: 10.1007/s40123-022-00540-9 Search in PMC Search in PubMed View in NLM Catalog Add to search Digital Eye Strain- A Comprehensive Review Kirandeep Kaur Kirandeep Kaur 1 Aravind Eye Hospital and Post Graduate Institute of Ophthalmology, Pondicherry, 605007 India Find articles by Kirandeep Kaur 1,✉, Bharat Gurnani Bharat Gurnani 1 Aravind Eye Hospital and Post Graduate Institute of Ophthalmology, Pondicherry, 605007 India Find articles by Bharat Gurnani 1, Swatishree Nayak Swatishree Nayak 2 Department of Ophthalmology, AIIMS, Raipur, Chhattisgarh 492001 India Find articles by Swatishree Nayak 2, Nilutparna Deori Nilutparna Deori 3 Sri Sankaradeva Nethralaya, Guwahati, Assam 781028 India Find articles by Nilutparna Deori 3, Savleen Kaur Savleen Kaur 4 Advanced Eye Center, Post Graduate Institute of Ophthalmology, Chandigarh, 160012 India Find articles by Savleen Kaur 4, Jitendra Jethani Jitendra Jethani 5 Baroda Children Eye Care and Squint Clinic, Vadodara, Gujarat 390007 India Find articles by Jitendra Jethani 5, Digvijay Singh Digvijay Singh 6 Noble Eye Care, Gurugram, 70058 India Find articles by Digvijay Singh 6, Sumita Agarkar Sumita Agarkar 7 Department of Pediatric Ophthalmology and Adult Strabismus, Sankara Nethralaya, 18 College Road, Chennai, Tamil Nadu 600006 India Find articles by Sumita Agarkar 7, Jameel Rizwana Hussaindeen Jameel Rizwana Hussaindeen 8 Sankara Nethralaya, 18 College Road, Chennai, Tamil Nadu 600006 India Find articles by Jameel Rizwana Hussaindeen 8, Jaspreet Sukhija Jaspreet Sukhija 4 Advanced Eye Center, Post Graduate Institute of Ophthalmology, Chandigarh, 160012 India Find articles by Jaspreet Sukhija 4, Deepak Mishra Deepak Mishra 9 Department of Ophthalmology, Regional Institute of Ophthalmology, Institute of Medical Sciences, Banaras Hindu University, Varanasi, Uttar Pradesh 221005 India Find articles by Deepak Mishra 9 Author information Article notes Copyright and License information 1 Aravind Eye Hospital and Post Graduate Institute of Ophthalmology, Pondicherry, 605007 India 2 Department of Ophthalmology, AIIMS, Raipur, Chhattisgarh 492001 India 3 Sri Sankaradeva Nethralaya, Guwahati, Assam 781028 India 4 Advanced Eye Center, Post Graduate Institute of Ophthalmology, Chandigarh, 160012 India 5 Baroda Children Eye Care and Squint Clinic, Vadodara, Gujarat 390007 India 6 Noble Eye Care, Gurugram, 70058 India 7 Department of Pediatric Ophthalmology and Adult Strabismus, Sankara Nethralaya, 18 College Road, Chennai, Tamil Nadu 600006 India 8 Sankara Nethralaya, 18 College Road, Chennai, Tamil Nadu 600006 India 9 Department of Ophthalmology, Regional Institute of Ophthalmology, Institute of Medical Sciences, Banaras Hindu University, Varanasi, Uttar Pradesh 221005 India ✉ Corresponding author. Received 2022 May 8; Accepted 2022 Jun 14; Issue date 2022 Oct. © The Author(s) 2022 Open AccessThis article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, which permits any non-commercial use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit PMC Copyright notice PMCID: PMC9434525 PMID: 35809192 Abstract Digital eye strain (DES) is an entity encompassing visual and ocular symptoms arising due to the prolonged use of digital electronic devices. It is characterized by dry eyes, itching, foreign body sensation, watering, blurring of vision, and headache. Non-ocular symptoms associated with eye strain include stiff neck, general fatigue, headache, and backache. A variable prevalence ranging from 5 to 65% has been reported in the pre-COVID-19 era. With lockdown restrictions during the pandemic, outdoor activities were restricted for all age groups, and digital learning became the norm for almost 2 years. While the DES prevalence amongst children alone rose to 50–60%, the symptoms expanded to include recent onset esotropia and vergence abnormalities as part of the DES spectrum. New-onset myopia and increased progression of existing myopia became one of the most significant ocular health complications. Management options for DES include following correct ergonomics like reducing average daily screen time, frequent blinking, improving lighting, minimizing glare, taking regular breaks from the screen, changing focus to distance object intermittently, and following the 20-20-20 rule to reduce eye strain. Innovations in this field include high-resolution screens, inbuilt antireflective coating, matte-finished glass, edge-to-edge displays, and image smoothening graphic effects. Further explorations should focus on recommendations for digital screen optimization, novel spectacle lens technologies, and inbuilt filters to optimize visual comfort. A paradigm shift is required in our understanding of looking at DES from an etiological perspective, so that customized solutions can be explored accordingly. The aim of this review article is to understand the pathophysiology of varied manifestations, predisposing risk factors, varied management options, along with changing patterns of DES prevalence post COVID-19. Keywords: Accommodation, Convergence, COVID-19, Digital eye strain, Digital revolution, Online classes, Pre-COVID-19 era, Smartphone Key Summary Points Digital eye strain has been an emerging health care problem in recent times. Online education and work from home have become the new norms since the beginning of the COVID-19 pandemic. DES symptoms can be broadly divided into ocular surface-related symptoms like irritation/burning eyes, dry eyes, eyestrain, headache, tired eyes, sensitivity to bright lights, and eye discomfort. Accommodation-related symptoms include blurred near or distance vision after computer use and difficulty refocusing from one distance to another. Recommendations to alleviate DES include the correct ergonomic use of digital devices, limiting daily screen time to ≤ 4 h, frequent breaks, screen time tracking, blue-light filtering glasses with antireflective coating, and an inclination towards outdoor recreational activities. Open in a new tab Introduction Technology has transformed every realm of our lives in the information age, from healthcare to education. The digital revolution, or the third industrial revolution, commenced in the 1980s, showing no signs of deceleration. Prompt communication, extensive availability of information, and most imperative, going paperless or GO GREEN are various advantages. However, every action comes with its opposite reaction, and the digital revolution is no deviation from this rule. As outlined by the American Optometric association, digital eye strain encompasses a cluster of ocular and vision-related problems attributed to prolonged usage of desktops, laptops, mobile phones, tablets, e-readers, and storage devices . The Digital Eye Strain Report of 2016, which included survey responses from over 10,000 adults from the USA, identified an overall self-reported prevalence of 65%, with females more commonly affected than males (69% vs. 60% prevalence) . Its pathophysiology is multifactorial, with several contributing factors being reduced contrast level of letters compared to the background of digital screens, screen glare and reflections, wrong distance and angle of viewing digital screens, poor lighting conditions, improper posture during usage, and infrequent blinking of eyes . The eye focusing and ocular movements required for better visibility of digital screen place additional demand on an intricate balance between accommodation and convergence mechanisms, thus making people with uncorrected or under-corrected refractive errors even more susceptible . The condition can cause an array of symptoms, including eyestrain, watering of eyes, headache, tired eyes, burning sensation, red eyes, irritation, dry eye, foreign body sensation, blurred vision at near, and double vision . According to the American Optometric Association, the usage of digital devices continuously for two hours is adequate to bring about digital eye strain [1, 6]. However, during the recent outbreak of novel Coronavirus disease-19 (COVID-19) declared by World Health Organization, there has been an upsurge in the usage of digital devices. Several countries worldwide declared a nationwide lockdown to shut down activities that necessitate human assembly and interactions, including educational institutions, malls, religious places, offices, airports, and railway stations, to contain the spread of the virus . A major part of the world was compelled to be confined indoors due to the dreaded consequences of this global pandemic, and its effects could be visualized in various sectors. Due to the lockdown, most people resorted to the internet and internet-based services to communicate, interact, and continue with their job responsibilities from home. Working from home became the new norm of working for millions of employees worldwide. Video-conferencing became the new mode of holding meetings and conferences. Education is yet another domain that witnessed a change in paradigm to the online way of implementation. Online learning services served as a panacea during the pandemic. Video conferencing platforms like Zoom and Google Meet have been used by schools, colleges, and universities worldwide since the beginning of the lockdown. There was a rise in usage of internet services from 40 to 100%, compared to pre-lockdown levels . Digital payments and digital currencies played a vital role in the pandemic. Being restricted indoors, digital devices became the only source of entertainment. Thus, the COVID-19 pandemic has added even more fuel to the already existing fire of the digital revolution. Ultimately this has resulted in an upsurge in the symptoms of digital eye strain amongst most individuals irrespective of age, sex, race, or region. The objective of this review article is to comprehensively present an overview of digital eye strain, its pathophysiology, management strategies, role of ophthalmologists and visual health specialists in educating parents or patients and also to understand the impact of COVID-19 on DES prevalence. We have also briefly highlighted the future research prospects in the field of DES. This review article is based on previously conducted studies. The article does not contain any studies with human participants or animals performed by any of the authors. Literature Search A comprehensive systematic literature search was done using PubMed, Google Scholar, and Cochrane database. The search was done using the terms “digital eye strain” or “computer vision syndrome” or “ocular asthenopia secondary to digital devices” or “eye strain”, or “visual fatigue” or “blue-blocking glasses” on May 15, 2022. All the articles with available abstracts along with the referenced articles until the date of search were evaluated. Original research work in the English language on DES and those mentioning prevalence, risk factors, clinical features were considered for inclusion into the present review article. The final reference list was generated on the basis of original work considered for inclusion relevant to the broad scope of this review article. Studies done before and after the COVID-19 pandemic have been summarized in tabular formats. Symptoms of DES Digital eye strain has been used synonymously with ocular asthenopia secondary to digital devices, computer vision syndrome, eye strain post computer or mobile usage, or even visual fatigue [3, 9–12]. The most common symptom is a sense of eye discomfort. This may be in the form of watering, redness, and itching in the eyes. The patients may complain of dryness in the eyes. Apart from this, a frequent complaint is blurry vision. The patient typically complains of blur and clear vision episodes, and eye strain. This usually reduces their concentration but improves after rest. Another set of symptoms is that the patient complains of glare, excessive sensitivity to light, and inability to keep the eyes open. All of these may be associated with headaches and occasionally sore neck or back [3, 10–13]. Broadly, these symptoms can be classified into three categories: Ocular surface-related symptoms are secondary to reduced blink and related to dry eye. These symptoms typically include irritation/burning eyes, dry eyes, eye strain, headache, tired eyes, sensitivity to bright light, and eye discomfort [13, 14]. Accommodation or vergence-related symptoms are secondary to excessive work and related to anomalies of accommodation or binocular visual system. These symptoms include blurred near or distance vision after computer use, difficulty refocusing from one distance to another, or diplopia [13–15]. Extraocular symptoms include musculoskeletal symptoms which can result in inconvenience in daily routine activities. These may include body discomfort like headache, neck or shoulder pain, and back pain . Now, even myopia progression has been linked to the digital eye strain in children. This would remain unique to the pediatric population only . At this point, there is sufficient evidence to suggest that this may be linked, but it would need further work to cement its place in the syndrome complex of digital eye strain. Pathophysiology of DES The symptoms experienced in computer vision syndrome are caused by three potential mechanisms: (i) Extraocular mechanism, (ii) accommodative mechanism, (iii) ocular surface mechanism . Extraocular mechanisms not specifically linked with ocular usage may cause musculoskeletal symptoms such as neck stiffness, neck pain, headache, backache, and shoulder pain . These symptoms are associated with postural problems secondary to improper placement of computer screens, unsuitable table or chair height, or incorrect distance between the eye and screen resulting in unnecessary stretching or forward bending often resulting in a muscular sprain [19, 20]. Accommodative mechanisms cause blurred vision, double vision, presbyopia, myopia, and slowness of focus change [13, 14, 18]. Changes in accommodation lag have been noted secondary to digital device usage over prolonged periods . However, the effects on accommodation, convergence, and pupillary size are mainly due to the demanding near work and not per se due to the screen . The effect of blue light on visual health has also been studied in detail. However, at this point, there is a lack of consensus in the findings of these studies to address the health effects of blue-blocking spectacle lenses . Table 1 summarizes results from research work done on blue-blocking glasses. Table 1. Published studies that explored role of blue-blocking filters on digital eye strain \| Authors, Journal, Year, and Place \| Study participants \| Methodology \| Results \| Conclusion \| :--- :--- \| Vera J et al. Clin Exp Optom. 2022 Jan 20:1–6 Spain \| Twenty-three healthy young adults, mean age 22.9 ± 3.2 years \| Two reading tasks from computer screen with or without blue-blocking filter on two different days. Orbicularis oculi (OO) muscle activity recorded by surface electromyography and DES symptoms noted during 30-min reading task \| No change in orbicularis oculi muscle activity with or without using blue-blocking filter. Reading increased visual fatigue and discomfort but reduced activation levels \| Neither the orbicularis oculi muscle activity nor the visual symptoms altered significantly during 30-min reading task with blue-blocking filters \| \| Rosenfield M et al. Work. 2020;65(2):343–348 United States \| Twenty-four subjects \| 20-min reading task from a tablet computer after wearing either blue-blocking filter lens (TheraBlue 1.67 or TheraBlue polycarbonate) or a CR-39 control lens \| An increase in symptoms was observed immediately after near vision task (p = 0.00001), no significant difference in symptoms was found between the lenses (p = 0.74) \| Use of blue-blocking filters as a treatment for DES is not well proven. Optimal environment for screen viewing, are more likely to benefit in minimizing symptoms \| \| Redondo B et al. Ophthalmic Physiol Opt. 2020 Nov;40(6):790–800 Spain \| Nineteen healthy young adults, mean age 22.0 ± 2.7 years \| 30-min two reading tasks on computer screen placed at 50 cm, with either commercially available blue-blocking filter or without any filter on two different days \| Blue light levels had no effect on lag and variability of accommodation (p = 0.34 and 0.62, respectively) Blue-blocking filter was associated with improved reading speed of 16.5 words per minute (p = 0.02). There was no significant change in pupil dynamics or perceived levels of visual discomfort \| Blue-blocking filter had no effect on accommodation dynamics or visual symptoms related to DES \| \| Palavets T et al. Optom Vis Sci. 2019 Jan;96(1):48–54 United States \| Twenty-three young, visually normal subjects \| 30-min reading task from tablet, with either blue-blocking (BB) or neutral-density (ND) filter producing equal screen luminance. Questionnaire to quantify DES symptoms \| Mean total DES symptom scores for BB and ND filters were 42.83 and 42.61, respectively. Between two filters, no significant differences were found between accommodation and vertical palpebral aperture \| Use of blue-blocking filters to minimize near work-induced asthenopia has limited proven evidence \| Open in a new tab An ocular surface mechanism causes symptoms such as dryness of the eyes, redness, gritty sensation, and burning after an extended period of computer usage. Eyeblink helps maintain a normal ocular surface through a whole cycle of secretion of tears, wetting of ocular surface, evaporation, and finally, drainage of tears . It is now well known that the blink rate reduces significantly during computer usage from 18.4 to 3.6/min in one of the studies and from 22 to 7 blinks/min in another study [28, 29]. The pathophysiology of reduced blink and squinting is bimodal; one, it increases the visual acuity in the presence of a refractive error and decreases the retinal illumination when using a source with glare in the superior visual field, as reported by Sheedy et al. . Rather than the reduced blink rate, an incomplete blink, where the upper eyelid does not cover the entire corneal surface, may be more relevant to dry eye as the tear film stability can be maintained with a reduced blink rate, provided that most blinks are complete . Apart from this, increased surface of cornea exposure caused by horizontal gaze at the computer screen and reduction of tear production due to the aging process and contact lens usage may also increase the digital eye strain. Burden of DES and Associated Conditions There has been massive growth in digital device usage in the past decade, hence increasing the risk of DES. There has been a surge of mobile devices in individuals across all age groups, with more elderly populations also reported to be engaged with digital media . A report by the Vision Council in 2016 noted that in the USA, approximately two-thirds of adults aged 30–49 years spend five or more hours on digital devices . The rampant use of social media is particularly pronounced among younger adults, with reportedly 87% of individuals between 20 and 29 years of age reporting the use of two or more digital devices simultaneously . The burden of DES is challenging to measure because of the variability in symptoms reported across the literature. The computer-related symptoms could be due to accommodation anomalies (such as blurred near vision, blurred distance vision, and difficulty refocusing after prolonged computer work) and those that seemed linked to dry eye (dry eyes, eyestrain, headache, burning eyes, sensitivity to bright lights, and ocular discomfort). Before the COVID-19 pandemic, a highly variable prevalence of DES symptoms ranging from 5 to 65% have been reported [2, 14, 32, 33]. Most of the studies reported dry eyes and accommodation anomalies as the presentation of DES, with refractive error, squinting, and blinking being studied less commonly. The data inconsistency was because these studies were either done through self-reported questionnaires, with variable definitions of DES being used and very little literature reporting the objectively determined DES . Another shortcoming of the older (before COVID-19) studies is that the occurrence of DES amongst children was understudied [34–37]. In children, the prevalence of asthenopia due to presumed DES was about 20% before the pandemic . The COVID-19 pandemic has increased our awareness of the DES and shed more light on the actual disease burden of DES, more so in the younger population. Table 2 summarizes the research work conducted prior to the COVID-19 pandemic. Table 2. Summary of research work published on digital eye strain prior to the COVID-19 pandemic \| S. no. \| Authors and Country \| Demographics \| Risk factors \| Clinical features \| Investigations \| Outcome \| Conclusion \| :--- :--- :--- :--- \| \| 1 \| Sancho et al. Int J Environ Res Public Health, 2022 Apr 8;19(8):4506 Spain \| 241 subjects, 64.3% women, Mean age, 45.49 ± 10.96 years (18–65 years age group) \| Amblyopia, dry eyes, retinal pathologies, occupational use of digital devices (DD), number of hours and years of DD use, scheduled break \| Burning, itching, foreign body sensation, eye blinking, redness, pain, tearing, heaviness, dryness, blurred vision, double vision, etc. \| TBUT, Schirmer’s test, Rasch–Andrich Rating Scale Model Analysis, Italian Version of the Computer Vision Syndrome Questionnaire (CVS-Q IT©) \| Prevalence of CVS was 67.2%, blurred vison-63.5%, worsening of sight, 62.3%, headache 56%. Least prevalent eye strain-11.2%, colored halos-16.2% and double vision-17.4% \| CVS-Q IT© is a simple, reliable, and valuable tool for assessing CVS in adults \| \| 2 \| Auffret et al. J Fr Ophtalmol 2022 Apr;45(4):438–445 France \| 52 participants \| Chronic exposure to digital devices \| Ocular discomfort, blurred vision, photophobia \| Short-term screen exposure, chronic screen exposure, ocular discomfort questionnaire refraction, phoria, near point of accommodation and convergence, fusional vergence and binocular amplitude facility \| No significant difference between control group and exposed group in any objective parameters Exposed group have high discomfort score for near (p-0.04), intermediate (p-0.02) blurred vision and light sensitivity (p-0.04) \| Binocular balance is affected by chronic and intensive screen use \| \| 3 \| Moore et al. Ophthalmic Physiol Opt, 2021 Nov;41(6):1165–1175 United Kingdom \| 406 respondents \| Digital device use \| \| Anonymous online questionnaire, covering attitude and understanding of DES \| Estimations of the proportion of patients affected by DES were lower than reports in the literature (median 25%, IQR 10-50%). Most respondents always (60.6%) or frequently (21.9%) inquired about device usage in routine case history taking, and also asked follow-up questions, although 29.3% only asked about the presence of symptoms half the time or less \| DES causes frequent and persistent symptoms, and practitioners reported high levels of confidence in discussing DES, patients can expect to receive advice on symptoms and management from their optometrist \| \| 4 \| Zayed HAM et al. Environ Sci Pollut Res Int. 2021 May;28(20):25187–25195 Egypt \| 108 IT professionals \| Female gender, age ≥ 35 years, computer use > 6 h/day, refractive error, not adjusting workstation ergonomics, no breaks during computer work, dry environment \| Headache (81.5%), burning of eyes (75.9%), and blurred vision (70.4%) \| Computer vision syndrome questionnaire (CVS-Q) \| Prevalence of DES was found to be 82.41% \| DES can be prevented by increasing knowledge and awareness about eye health, proper ergonomic computer training, and suitable comfortable workplace environment \| \| 5 \| Meyer D et al. Cont Lens Anterior Eye. 2021 Feb;44(1):42–50 United States \| Six hundred and two soft contact lens (SCL) wearers and 127 non-contact lens (non-CL) wearers using digital devices at least 4 h per day \| \| Primary sensations eye strain/pain, soreness, tired eyes, and headaches Secondary or surface sensations burning, eye irritation, tearing and dryness Visual sensations blurred/double vision and words move/float \| Questionnaire assessing frequency and severity of 10 common symptoms associated with eye fatigue related to DES \| 89% of SCL wearers reported eye fatigue more than once per month, and > 60% reported more than once per week Dryness and irritation were more common among SCL wearers \| Eye fatigue is highly common among both soft contact lens and non-contact lens wearers. The frequency or severity is same among SCL users and other group \| \| 6 \| Al Dandan O etal, Acad Radiol. 2021 Aug;28(8):1142–1148 Saudi Arabia \| 198 radiologists (111 males and 87 females), including 40.9% residents, 27.3% senior registrars, and 27.3% consultants \| Female sex and taking breaks once or twice a day only \| \| Online survey \| 26.8% underwent an eye examination within past one year and 50.5% experienced DES \| DES is common among radiologists. It is more common among radiology residents, females, and those not taking frequent breaks \| \| 7 \| Ichhpujani P et al. BMC Ophthalmol. 2019 Mar 12;19(1):76 India \| 576 adolescents attending urban schools \| Preference to lie down \| \| Surveyed regarding their electronic device usage \| 18% (103) experienced eyestrain at the end of the day 18% experienced symptoms related to DES. 20% students aged 11 years use digital devices on daily basis, in comparison with 50% aged 17. In addition to homework aids, one-third of the participants reported using digital devices for reading instead of conventional textbooks. 77% students prefer sitting on a chair while reading, 21% prefer to lie on bed and 2% students alternating between chair and bed \| The increased use of digital devices by adolescents brings a new challenge of digital eyestrain at an early age \| Open in a new tab With the lockdown restrictions during the COVID-19 pandemic, outdoor activities were restricted for all age groups, and digital learning became the norm for almost 2 years. Hence, digital device usage increased throughout the world, exacerbating DES symptoms. DES prevalence amongst children alone rose to 50–60% in the COVID-19 era [44–46]. In children, the symptoms expanded to include recent onset esotropia and vergence abnormalities as part of the DES spectrum [47, 48]. Overall, the incidence of DES was 78%, with participants reporting one or more DES-related symptoms . This was primarily due to the overall time spent on digital devices (7–10 h/day) during the lockdown period, significantly greater than during the pre-curfew period (3–5 h) in all studies [49, 50]. The virtual classes for children and “work from home policy” in office-going adults necessitated additional usage of digital devices. One of the most significant ocular health complications of the COVID-19 pandemic has been new-onset myopia and the increased progression of existing myopia due to excessive near work [17, 50, 51]. The prevalence of myopia has been nearly 50% in the COVID era, with accelerated progression from 0.3D in pre-COVID to 1D in the COVID era . This influence on myopic progression has been maximum in the age group of 6–8 years . Table 3 summarizes the findings from research work done during COVID-19 era. With time, we might have further studies detailing the increase in DES burden due to home confinement in COVID. Table 3. Review of literature of digital eye strain during the COVID-19 pandemic \| S. no. \| Authors and country \| Demographics \| Risk factors \| Clinical features \| Investigations \| Outcome \| Conclusion \| :--- :--- :--- :--- \| \| 1 \| Wangsan et al. Int J Environ Res Public Health. 2022 Apr; 19(7): 3996 Thailand \| 527 students, 70.40% females, mean age 20.04 ± 2.17 years \| Female gender, atopic eye disease, dry eyes, itching, red eye, eye pain, astigmatism, previous refractive surgery, tear substitute use, contact lens use, mobile and tablet use \| Eye pain-96.5%, burning sensation-92.5%, headache 90.08%, defective vision-15.95 \| CVS-Questionnaire (CVS-Q), CVS was diagnosed with a score of CVS-Q ≥ 6 \| Prevalence of CVS was 81%, distance less than 20 cm (52.7 vs. 40%), less brightness less 14.8 vs. 7.0%) and glare or reflection on display (47.8 vs. 29.0%) were associated with CVS \| Social distancing is mandatory, online classes are unavoidable, increased screen was associated with increases prevalence of CVS. Laptop/desktop should be preferred over mobile phone \| \| 2 \| Cai et al. Front Med (Lausanne), 2022 Mar 21;9:853293 China \| 115 children with myopia \| Strict home confinement, hereditary, closed indoor work time, excess exposure to electronic gadgets Protective factors- age, rest time, sleep time, and distance from the device while usage \| Asthenopic symptoms \| Axial length assessment (IOL Master 700) and refractive errors (without cycloplegia), visual function, convergence insufficiency symptom survey (CISS) and eye care habits questionnaire \| Axial length elongation was 35% higher than normal, positively correlated with severe asthenopia (r = 0.711), negative with age (r = − 0.442), distance from eyes (− 0.238) \| Decreased outdoor activities and increased screen time accelerated myopia progression by 1/3 \| \| 3 \| Demirayak et al. Indian J Ophthalmol 2022 Mar;70(3):988–992 Turkey \| 692 children under the age of 18 years, mean age 9.72 ± 3.02 years, 360 (52%) were girls, 62.57% were students in primary school \| Computer use (61.7%), smartphones (57.8%), mean duration of display device use 71.1 ± 36.02 min \| Headache (52.2%), eye fatigue (49.3%), and eye redness (49.3%) and double vision-8.8% \| Online electronic survey using Google Forms \| 48.2% experienced 3 or more symptom, male gender and age were independent risk factors for 3 or more symptoms \| Digital device use during the pandemic exacerbated the DES among children \| \| 4 \| Basnet et al. JNMA J Nepal Med Assoc, 2022 Jan 15;60(245):22–25 Nepal \| 318 subjects \| Digital device use, tablet use, computer, and smartphone \| Eye strain-199 (62.6%), tiredness of eyes-162 (50.9%) \| \| Prevalence of DES was found to be 94.3% \| Prevalence of DES has increased during COVID-19 pandemic \| \| 5 \| Regmi A et al. Clin Exp Optom. 2022 Feb 14:1–7 India \| 1302 participants \| Females spending more than 6 h on digital devices, taking breaks from digital devices after 2 h, inability to maintain a fair sleep schedule, and inability to make ergonomic modifications at home \| \| Electronic communication sources using Google Forms \| 94.5% had one or more visual and ocular symptoms associated with digital devices usage. 43.1% reported that these symptoms began post-lockdown \| A high prevalence of visual/ocular symptoms (43.1%) and work-related musculoskeletal disorder (45%) were reported during COVID-19 lockdown \| \| 6 \| Mohan A et al. Indian J Ophthalmol. 2022 Jan;70(1):241–245 India \| 133 children (266 eyes) \| History of rapid progression in pre-COVID-19 era (p = 0.002) and sun exposure < 1 h/day (p< 0.00001) \| \| \| Annual myopia progression was found to be statistically significant during COVID-19 as compared with pre-COVID-19 (0.90 vs. 0.25 D, p< 0.00001). A total of 45.9% of children showed an annual progression of ≥ 1 D during the pandemic as compared with 10.5% before the COVID-19 (p< 0.00001) \| Rapid myopia progression in children during current pandemic and children should be provided with socially distant outdoor activities to increase their sun exposure and diminish the rate of myopia progression \| \| 7 \| Kaur K, J Pediatr Ophthalmol Strabismus. 2021 Dec 20:1–12 India \| 305 responses \| Digital device use \| Headache was the most common complaint in 100 children (51% of total symptomatic), followed by ocular pain in 19 children (9.64% of total symptomatic) \| Online questionnaire using Google Forms \| Prevalence of DES was found to be 64.6% \| There is a strong need to bridge this knowledge gap and prevent the increased prevalence of myopia and digital eye strain in the future \| \| 8 \| Gupta R et al. J Curr Ophthalmol. 2021 Jul 5;33(2):158–164 India \| 654 students, mean age: 12.02 ± 3.9 years, 332 (58%) females \| Spectacle users, age, and duration of digital device \| Redness (69.1%), heaviness of eyelids (79.7%), blinking (57.8%), blurred vision (56.9%), light sensitivity (56%) \| Rasch-based Computer-Vision Symptom Scale was deployed to measure the DES \| Mean CVS score of. class 1–5 was 26.1 ± 7.8, class 6–9 was 24.8 ± 6.6, class 10–12 was 29.1 ± 7.1. Mean CVS score was lowest in < 4 h group followed by 4–6 h and then > 6 h \| The majority experienced at least one symptom of DES. There is a need to educate the masses about measures to prevent DES \| \| 9 \| Mohan A, et al. J Pediatr Ophthalmol Strabismus. 2021 Jul-Aug;58(4):224–231 India \| 46 children; mean age of 14.47 ± 1.95 years \| Digital devices for 4 h/day or more \| \| Convergence Insufficiency Symptom Survey (CISS) questionnaire \| Mean CISS scores were 21.73 ± 12.81 for digital device use < 4 h/day and 30.34 ± 13.0 for ≥ 4 h/day (p = 0.019). Mean near exophoria (p = 0.03), negative fusional vergence (p = 0.02), negative relative accommodation (p = 0.057), and accommodation amplitude (p = 0.002) were different between the two groups \| Online classes for more than 4 h resulted in abnormal binocular vergence and accommodation \| \| 10 \| Mohan A et al. Strabismus. 2021 Sep;29(3):163–167 India \| 8 children, mean age 12.5 ± 4.2 years, all 8 males \| Emmetropia (5), myopia (1), pseudomyopia (1), hyperopia (1) \| \| Diplopia, Hess chart, visual acuity by Snellen chart, alternate prism cover test, cycloplegic retinoscopy, neurological examination \| Mean duration of smartphone use 4.6 ± 0.7 h, children attending classes for > 4 h/day. The angle of deviation for near and distance were 48.1 ± 16.4 PD and 49.3 ± 15.9 PD, respectively, with normal ocular motility \| Prolonged near work especially using smart phone for e-learning might lead to AACE in children \| \| 11 \| Salinas-Toro D et al. Int J Occup Saf Ergon. 2021 Jul 7:1–6 United States \| 1797 respondents; mean age of respondents 40.5 ± 11.1 years, and 69.9% were female \| Female gender, refractive surgery, rosacea, depression, previous dry eye disease, keratoconus, blepharitis, occupation, contact lens use \| Soreness, pain, foreign body sensation, redness, visual fatigue, redness and blurred vision \| Ocular symptom index, DED (dry eye questionnaire 5 [DEQ-5] questionnaire \| The mean number of teleworking weeks was 10.2 ± 3.0. All DES symptoms presented a significant increase (p< 0.001). The mean DEQ-5 score was 8.3 (SD 4.9). Women had a higher score (p< 0.001) \| Visual display terminal hours are related to increase in DES symptoms and high prevalence of DED \| \| 12 \| Zheng et al. J Med Internet Res, 2021 Apr 30; 23 (4): e 24316 China, Singapore, Ireland, and Australia \| 1009 children, 2 groups – interventional group (485)—exercises and ocular relaxation, and access to a digital behavior change intervention, or control group (469)—health education information only Mean age 13.5 ± 0.5 years, 499 males \| Smartphone use, gender, use of glasses, parental education, smoking and family history \| Eye strain, anxiety, sleep disturbance, \| Health education information promoting exercise and ocular relaxation, and access to a digital behavior change intervention, with live streaming and peer sharing of promoted activities \| Mean anxiety score in the intervention group was greater (− 0.23) as compared to the control group (0.12). A significant reduction in eye strain was observed in the intervention group (− 0.08) as compared to controls (0.07) \| Digital behavior change reduced anxiety and eye strain among children \| \| 13 \| Gammoh Y. Cureus. 2021 Feb 26;13(2):e13575 Jordan \| 382 students, mean age of participants was 21.5 years (± 1.834), male:female ratio was 1:1.56 \| Digital device use for > 6 h per day \| Tearing (59%), headache (53%), and increase sensitivity to light (51%) \| Computer Vision Syndrome Questionnaire (CVS-Q) \| The prevalence of CVS was found to be 94.5%. Tearing was most common-(59%), double vision was least common among students-18.3%. DD use for > 6 h/ day was present in 55.5% patients, and 30.7% of reported pain in joints of fingers and wrists after using a mobile phone \| CVS is highly prevalent among Jordan university students. Safe habits in digital device use are recommended to prevent DES \| \| 14 \| Alabdulkader B. Clin Exp Optom. 2021 Aug;104(6):698–704 Saudi Arabia \| 1939 participants, mean age was 33 ± 12.2 years and 72% were women \| Digital device use duration, use of multiple devices, age, optical correction, and status of employment \| \| Self-reported questionnaire \| Incidence of digital eye strain was 78% \| Importance of regular eye examination, limiting screen time, the 20–20-20 rule, and the use of lubricating drops to help reduce the symptoms of DES should be emphasized \| \| 15 \| Ganne P et al. Ophthalmic Epidemiol. 2021 Aug;28(4):285–292 India \| 941 responses from online classes students (688), online classes teachers (45), and the general population (208) \| Students attending online classes, those with eye diseases, greater screen time, screen distance < 20 cm, using gadgets in dark and infrequent/no breaks \| \| Pre-validated questionnaire \| DES prevalence was higher among students taking online classes (50.6%) compared to the general public (33.2%). An increase in screen time has been observed during the pandemic compared to pre-pandemic time \| There is a need to educate about ergonomics of screen usage. There is need to reduce the online classes duration and working hours for professionals to control the epidemic of DES \| \| 16 \| Mohan A et al. Indian J Ophthalmol. 2021 Jan;69(1):140–144 India \| 217 parents, mean age 13 ± 2.45 years \| Age > 14 years, male gender, smartphone use, > 5 h of digital device use and > 1 h/day of mobile games \| \| Online electronic survey—Computer Vision Syndrome Questionnaire \| Mean digital device use duration during COVID era (3.9 ± 1.9 h) is more than pre COVID era (1.9 ± 1.1 h). 36.9% used digital devices > 5 h in COVID era as compared to 1.8% pre COVID era. Smartphones were most common digital device used (61.7%). 49.8% attended online classes for > 2 h per day \| DES prevalence increased among children in COVID era. Duration, type, and digital device distance ergonomics can avoid DES in children \| \| 17 \| Bahkir FA et al. Indian J Ophthalmol. 2020 Nov;68(11):2378–2383 India \| 407 responses, mean age was 27.4 years, 55.5% were males and 44.5% were female \| Female gender, student population \| Headache, eye pain, heaviness of eyelids, redness, watering, burning sensation, dryness, increased light sensitivity, itching, excessive blinking, difficulty in focusing printed text, blurred vision, foreign body sensation, double vision \| Open online survey through social media platforms \| 93.6% respondents reported increased screen time after lockdown. An average increase of 4.8 ± 2.8 h per day was reported. Total daily usage was found to be 8.65 ± 3.74 h. 62.4% reported sleep disturbances. 95.8% experienced at least one symptom related to DES, and 56.5% agreed to increased frequency and intensity of symptoms post lockdown \| Awareness should be created about prevention of DES, and additional measures should be explored to control the adverse effects related to digital devices \| Open in a new tab Management Strategies Digital screen-time refers to time spent in front of a screen, such as watching television, working on a computer, laptop, or tablet, using a smartphone, and playing video games. It is a sedentary lifestyle habit with excessive visual activity, which has implications both on ocular and general health hygiene . Owing to home confinement during the COVID-19 pandemic, there has been a substantial rise in usage of the digital platform for work and education. As a result of the lack of outdoor activities and social interaction, people have resorted to television and social media for entertainment with an unintentionally increased dependence on these devices [64, 65]. Wong et al. have rightly pointed out that the behavioral changes arising from this growing dependence may persist even after the COVID-19 pandemic . The American Optometric Association has defined digital eye strain (DES) as an entity encompassing visual and ocular symptoms arising from the prolonged use of digital electronic devices [1, 67]. It is characterized by symptoms such as dry eyes, itching, foreign body sensation, watering, blurring of vision, and headaches . The prevalence of DES reported in the literature ranges from 25 to 93% [68–71] and a recent meta-analysis of available data linked to asthenopia associated with DES reported a pooled prevalence of 19.7% in the pediatric population . Continuous staring at the screen leads to a decrease in the blink rate, causing dry eye-related problems. Smartphone use is more commonly associated with dry eye disease than other digital devices . In a case–control study among school-going children, Moon et al. reported an association of 71% among smartphone users . They also documented that symptoms of dry eye diseases were higher in the children above the age of 14 years than in the younger age group. This could be due to older children spending more hours on smartphones . Visual work on a digital screen demands continuous focusing and refocusing in an attempt to see the pixelated characters clearly. Frequent eye movements to maintain focus lead to fatigue and eye strain. Shorter digital screen distance, a constant convergence, and accommodative demand further aggravate the asthenopic symptoms associated with DES [38, 74–76]. Prolonged duration (> 4 h), improper posture, and inadequate lighting conditions are directly proportional to the DES symptoms [38, 77]. Non-ocular symptoms associated with eye strain include stiff neck, general fatigue, headache, and backache [78, 79]. Digital screen-time has also been considered as a potential modifiable environmental risk factor that can increase the risk of myopia progression. Prevention of myopia progression has been prioritized due to the associated risks of myopic macular degeneration, retinal detachment, glaucoma, and cataract . Recommendations to alleviate DES include the ergonomic use of digital devices [74, 81, 82]. Average daily screen time should be reduced to a reasonable limit (≤ 4 h daily). Digital device practices: proper ambient lighting, digital device positioning, adjusting image parameters (resolution, text size, contrast, luminance), and taking frequent breaks (20/20/20 strategy). It is recommended to sit upright at a desk or table with screens approximately 20 inches from the eyes . The height of the screen should be positioned lower than the height of the eyes, such that the viewing distance is 15–20° below the eye level. Frequent blinking of eyes minimizes the chances of developing dry eyes. The refence materials should be placed above the level of keyboard and below the level of monitor. Environments with an illumination of over 1000 lx are known to decline user performance . A contrast setting around 60–70% is considered comfortable by most people. The brightness should be adjusted such that the light coming from monitors matches the light in the surrounding workspace. Anti-glare screens can also help in reducing the amount of light reflected from the screens . A clearly legible font of at least size 12 preferably in a dark color over light background should be chosen. Screen time tracking allows to control excessive screen usage. It encourages to spend less time on digital devices. Refractive error correction and use of glasses with antireflective coating [1, 6]. Public education about the lasting effects of excessive screen time and encouraging healthier lifestyle practices. Parents should be counseled to monitor their child's screen usage and incorporate family time. Encourage children towards outdoor recreational activities. There is strong evidence that increased screen time is associated with higher risks of an unhealthy diet, cognitive outcome, interpersonal relationships, and quality of life among children and young adults . With the recent explosion of digital electronic device usage among children and young adults, there is an urgent need to educate the parents, caregivers, and youth about limiting digital screen time and implementing ergonomic practices of screen exposure. Role of Ophthalmologists and Visual Health Specialists There is a need to increase awareness about digital eye strain since digital screen devices have become an inseparable part of the lifestyle. Recently, the impact of digital eye strain (DES) has been felt across the population with the lockdowns and curfews imposed by the pandemic [53, 62]. In the urban locales, there is some awareness about DES, but this is lacking in the rural and lower socio-economic groups, both of whom have seen an increasing screen exposure in recent years. Eye-health strategies and awareness campaigns need to target the at-risk population. Awareness amongst digital device users can be channelized through doctors (physicians and ophthalmologists), health care workers (optometrists, vision technicians, and nursing staff), and non-medical professionals (wellness professionals, health and fitness experts, and information technology team leaders). A special emphasis should be made to raise awareness among teachers, since they are the ones who can offer early detection of DES symptoms at school, which is more important in the present times considering the increased dependency of education on digital devices. Screen users need to be told to recognize symptoms of digital eye strain such as asthenopia, headache, neckache, red eyes, watery eyes, or burning sensation in the eyes. They need to be encouraged to make specific changes such as improving lighting, minimizing glare, taking regular breaks from the screen, changing focus to a distance object intermittently, following the 20-20-20 rule (taking a 20-s break every 20 min to look at an object 20 feet away) and using ergonomic chairs to reduce eye strain . Frequent blinking needs to be emphasized too. Typically, we blink 14–16 times a minute, but this reduces to 4–6 times a minute when using screens . Persistent symptoms despite these changes mark the need for an ophthalmic exam. Parents and caregivers need to be sensitized to digital eye strain in children. There is a significant gap in the knowledge concerning DES and its potential harm, indicating a need to increase awareness in this group . Parents and caregivers need to pick up on early signs that a child may be straining the eyes. Children often do not express ocular discomfort but may manifest certain mannerisms such as forced blinking or avoidance of screens or complain of transient episodic eye pain, rubbing, or epiphora, which may indicate eye strain . Pediatricians and visual health specialists need to brief parents and teachers to recognize these signs and take remedial measures such as reducing screen time, using larger high-resolution displays, adjusting the lighting, and increasing outdoor activity. Over-the-counter lubricant drops can be considered in case of persistent symptoms, but an ophthalmology consult should be scheduled. Innovation in screen technology has reduced the incidence of digital eye strain. These include high-resolution screens with inbuilt antireflective coating, matte-finished glass, edge-to-edge displays, and image smoothening graphic effects. Specific applications which remind screen users to take regular breaks also help inculcate screen-friendly habits. Innovations in the optical segment such as antireflective coating, blue-light blocking glasses, and polaroid lenses are other recommended measures to reduce eye strain. Research and Knowledge Gaps As it is pretty clear on the date that DES is not going to go away, it is essential from a public health perspective to focus on practical protective and preventive approaches concentrating on improving the vision-related quality of life of individuals affected with DES . Despite the significant strides made concerning the understanding of DES, there are considerable gaps in research and knowledge pertinent to: The symptomology of DES Effective treatment strategies Optimizing and customizing treatment options for different age groups based on the visual demands and symptoms Preventative approaches to ameliorate the onset and severity of DES The current assessment protocols for DES include aspects of understanding the visual symptoms in detail using a structured inventory, understanding task-specific visual demands, ergonomic concerns and considerations, comprehensive eye examination, refractive correction, binocular vision assessment, ocular surface assessment for dry eyes, and management based on the outcomes of the assessment [86–88]. Yet, in the symptomatology of DES, there is a considerable gap in understanding the association between the onset of visual symptoms and pre-existing visual dysfunctions. It has been shown that extensive use of digital devices can induce or exacerbate visual fatigue [3, 82, 83]. It is not clear if individuals who have a pre-existing binocular vision dysfunction, dry eyes, and related anomalies are at an increased risk for DES. Also, there is a considerable gap concerning the context of the type of digital device and the dynamic visual demands imposed by the same. Studies that aim at categorizing the visual symptoms based on the pre-existing visual dysfunction, visual needs, and visual profile can aid in a better understanding of the DES and can also provide insights into preventative approaches to mitigate the visual symptoms . Management options for DES are symptoms-based and include a holistic and comprehensive approach, from the management of refractive errors, binocular vision anomalies, and ocular surface dryness to providing workplace recommendations to improve visual comfort. The global lifestyle disruptions due to COVID-19 resulted in a rapid rise in DES prevalence across all age groups [86, 89]. The impact of DES on children was highlighted by various researchers that pointed out the need for visual protection measures to be followed during online learning. This included using appropriate screen settings, illumination and earning environment settings, posture requirements, adopting a healthy lifestyle, and regular eye examination . Nonetheless, there are barely any studies exploring the optimal environmental conditions and efficacy of visual hygiene measures in ameliorating DES onset and prevalence [90, 91]. Most of these guidelines are primarily expertise and consensus-based and need to be backed up by evidence. There is a clear need for further exploration to understand the cause-and-effect relationship between blue light and DES; when it comes to the effect of blue light illuminance and its association with visual fatigue, dry eyes, and retina damage , there is a clear need for further exploration to understand the cause-and-effect relationship between blue light and DES. Similarly, there is a considerable lacuna in understanding mechanisms based on which anti-fatigue lenses work to reduce visual fatigue. Novel spectacle lens designs are being explored in this context. Hence, further explorations in this field should focus on recommendations for digital screens optimized to improve visual comfort novel spectacle lens technologies to reduce visual fatigue associated with long hours of screen viewing, and inbuilt filters to optimize visual comfort . A paradigm shift is required in our understanding of looking at DES as a man/instrument-made entity to explore customized solutions accordingly . Overall, future research should focus on enhancing our understanding of DES from an etiological perspective, leading to evidence-based management options. Conclusions Digital eye strain has been on the rise since the beginning of the COVID-19 pandemic. An augmented growth pattern has been experienced with prevalence ranging from 5 to 65% in pre-COVID-19 studies to 80–94% in the COVID-19 era. The sudden steep increase in screen and chair time has led way to other silent pandemics like digital eye strain, myopia, musculoskeletal problems, obesity, diabetes etc. Digital device usage of more than 4 h/day, underlying refractive errors, female gender, and prior dry eyes are the most significant risk factors predisposing to DES. There is an urgent need for eye care professionals and vision health specialists to be well informed about DES. Awareness related to effects of excess screen time, ergonomic practices, and preventive measures needs to be spread especially among teachers, youngsters, and professionals exposed to excessive or prolonged screen time. The role of anti-glare screens, anti-fatigue lenses, and blue-blocking filters is still controversial and needs to be further explored. Future studies should focus on understanding the risk factors among different groups and the association between accommodative or binocular vision anomalies and DES. Acknowledgements Funding No funding or sponsorship was received for this article. Authorship All named authors meet the International Committee of Medical Journal Editors (ICMJE) criteria for authorship for this article, take responsibility for the integrity of the work as a whole, and have given their approval for this version to be published. Author Contributions Concept and design: KK, BG; drafting manuscript: KK, BG, SN, SA, ND, JJ, JR, DS, SK, JS, DM; critical revision of manuscript: KK, BG, SK, ND, JJ; supervision: KK, BG, SA, SK, ND. All authors read and approved the final manuscript. Disclosures Bharat Gurnani, Kirandeep Kaur, Swatishree Nayak, Nilutparna Deori, Savleen Kaur, Jitendra Jethani, Digvijay Singh, Sumita Agarkar, Jameel Rizwana Hussaindeen, Jaspreet Sukhija, and Deepak Mishra have nothing to disclose. Compliance with Ethics Guidelines This review article is based on previously conducted studies. The article does not contain any studies with human participants or animals performed by any of the authors. Contributor Information Kirandeep Kaur, Email: beingkirandeep@gmail.com. Bharat Gurnani, Email: drgurnanibharat25@gmail.com. Swatishree Nayak, Email: nswatishree@yahoo.com. Nilutparna Deori, Email: nilutparnadeori@gmail.com. Savleen Kaur, Email: mailsavleen@gmail.com. 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6163	https://study.com/learn/lesson/is-0-natural-number-type.html	What is 0? \| Definition & Types - Lesson \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses / Math for Kids Course What is 0? \| Definition & Types Lesson Transcript Melissa Bialowas, Rayna Cummings, Kathryn Boddie Author Melissa Bialowas Melissa Bialowas has taught preschool through high school for over 20 years. She specializes in math, science, gifted and talented, and special education. She has a Master's Degree in Education from Western Governor's University and a Bachelor's Degree in Sociology from Southern Methodist University. She is a certified teacher in Texas as well as a trainer and mentor throughout the United States. View bio Instructor Rayna Cummings Rayna has taught Elementary Education for 12 years (in both 1st, 2nd, and 3rd grades) and holds a M.Ed in Early Childhood Education from The Ohio State University View bio Expert Contributor Kathryn Boddie Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison. View bio What is 0? Find answers to some of the controversial questions such as "is zero a number" or "is 0 a natural number" or "is 0 a counting number" and many more. Updated: 11/21/2023 Table of Contents What is 0? Natural Number Whole Number Lesson Summary Show FAQs Activities Additional Practice with Natural Numbers and Zero Zero is considered a natural number on a number line and when identifying numbers in a set. Zero is not considered a natural number when counting. In the following problems, students will decide whether zero is considered a natural number in various situations. Practice Problems You are counting the number of rocks in your backyard. Is zero considered a natural number in this situation? Why or why not? You are listing all of the natural numbers in the set {0, 0.5, 3, 4.3, 7, 20}. What are the natural numbers in this set? Is zero considered a natural number in this situation? Why or why not? What are all the natural numbers on the section of the number line shaded below? Is zero considered a natural number in this situation? Why or why not? Solutions Zero is not considered a natural number in this situation, since we are counting. We cannot count something that is not there. The natural numbers in this set are 0, 3, 7, 20. Zero is a natural number in this situation since we are identifying numbers in a set. The number line is shaded from -1 to some number between 2 and 3. The natural numbers shaded in this section are 0, 1, and 2. Zero is considered a natural number in this situation since we are on a number line. Who found the origin of zero in India? The famous astronomer and mathematician, Aryabhatta, is credited with finding zero in India. He is also credited with finding an approximation of pi and recognizing the moon and planets were reflecting the sun's light, not producing their own. What does zero mean in maths? Zero can mean the absence of something, such as I own zero dolphins. It can also be a placeholder in numbers like 101 or 1001. What is the factorial value of 0? The factorial value of 0 equals 1. 0!=1 because if you arrange all of the numbers you need to multiply to get 0 there is an empty set. This means there is exactly 1 arrangement of numbers needed. Create an account Table of Contents What is 0? Natural Number Whole Number Lesson Summary Show What is 0? ---------- There has been much debate in history about the number zero. Many number systems developed throughout the world without a number zero. Some had a placeholder to show the absence of a number, for example, to tell the difference between a 101 and 1001, but is that really the number zero? We know Fibonacci brought Arabic numerals (1, 2, 3...) to the West around 1200, and zero was already a part of the number system. Zero It is believed this system developed the idea for zero from India and the Sanskrit word "shunya" meaning void or emptiness. This is credited to a famous Indian astronomer and mathematician, Aryabhatta. There is also evidence of a zero, or possibly a place holder, in ancient Babylon, the Mayan culture, and one used by the Sumerians. When considering if zero is positive or negative, consider how to count. If there are five sheep most people would count 1, 2, 3, 4, 5. They would likely not count 0, 1, 2, 3, 4, 5. 5 Sheep Zero is not considered a positive number because it is not a counting number, or a number typically used when counting. However, it is also not a negative number. It does not require a negative sign. Zero is considered neither positive nor negative, yet it is considered a number. Many consider zero to be a neutral number. Do we really need the number zero? We certainly need it as a placeholder, but do we need the zero representing nothing? Yes! This would make many everyday math problems difficult, if not impossible, without it. For example, try solving 5 - 5. What if it is currently 5 degrees outside and a cold front comes through and drops it 8 degrees? If one tried this on a number line without the zero represented, they would get the wrong answer. The zero provides us with symmetry and the ability to do higher-level math. Number Line Factorial of 0 When we find a factorial of a number we multiply that number by all of the numbers smaller than it. For example, the factorial of 5, written "5!" is 5 x 4 x 3 x 2 x 1 = 120. So why is the factorial of zero equal to 1? Because the definition of multiplying the smaller integers gets the right answer but doesn't actually explain what a factorial represents. A factorial is the number of ways these numbers can be rearranged (permutation). So there are 120 different ways the numbers 1, 2, 3, 4, 5. can be arranged. How many different ways can the number zero be rearranged? There is exactly one way, which is why 0! = 1 and 1! = 1. Click for sound 2:38 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? 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Your next lesson will play in 10 seconds 0:04 Let's Count 0:26 Is Zero a Natural Number? 2:11 Lesson Summary View Video Only Save Timeline 91K views Video Quiz Course Video Only 91K views Natural Number -------------- A natural number can be defined as any positive integer or as any non-negative integer. These two definitions are very close but actually contradict each other. Some sources present the natural numbers as N: {1, 2, 3...}, thus all positive integers and zero is not included. These same sources then teach about whole numbers which are W: {0, 1, 2, 3...}. Other sources present the natural numbers as N: {0, 1, 2, 3...} thus including zero and not introducing the concept of whole numbers because it has already been combined. Is 0 a Natural Number? When counting, people typically start counting with the number one, thus everyone agrees the counting numbers do not include zero. However, when using a scale we start at 0 and go up, thus allowing numbers smaller than 1 to be counted. So is 0 a counting number, a natural number, a whole number, or something else altogether? Many people say that natural numbers are positive whole integers. Historically, zero was taught as a neutral number, neither positive nor negative, a counting number, and a whole number, but not a natural number. The shift of starting to include zero as a natural number appears to have started with set theory. Set Theory In set theory, the empty, or null, set is crucial to advanced mathematics. Thus the number zero was seen as a natural number in this context. As set theory became better understood, many sources started updating the definition of a natural number throughout all levels of mathematics, and not just in set theory. We are currently at a crossroads with much debate occurring about the definition of natural numbers, the application, and how zero fits into our understanding of mathematics. Currently, we know zero is not a counting number, zero is a whole number, zero is an integer, and zero is sometimes also a natural number. The higher levels of mathematics and the newest published research typically do include zero, while many traditionalists in the world of mathematics do not. Real Number Sets Whole Number ------------ Whole numbers are the set of integers including zero and all positive numbers. This is notated as W: {0, 1, 2, 3...}. Integers themselves include all positive and negative numbers, I: {...-2, -1, 0, 1, 2...}. Integers do not include fractions or decimals. Currently, there is no debate about these definitions. So while all natural numbers are whole numbers, many whole numbers are negative and therefore not natural. Is 0 a Whole Number? Zero is included in every definition of a whole number. Scholars can be confident in always saying 0 is a whole number. Remember decimals and fractions are not considered whole numbers since they are partial numbers (half, quarter, etc.) Lesson Summary -------------- Let's review: the history of zero is long and comes from around the world. Zero could be considered a placeholder or a number. Zero is neither positive nor negative and thus it is considered a neutral number. 0! = 1 Mathematicians agree zero is a counting number, a whole number, and an integer. The addition of set theory in mathematics changed the view of natural numbers and the shift of if it includes the number zero or not. Video Transcript Let's Count Quickly count the numbers one to twenty. No problem right? Well, you just counted natural numbers. A natural number is any positive, whole number such as the numbers you just counted. Positive numbers are also called positive integers. Numbers are not natural numbers if they are negative numbers or fractions, such as 1/3 or 4.20. Is Zero a Natural Number? Now that you know what natural numbers are, is zero considered a natural number? Remember natural numbers are any positive whole number and they count up from zero. We cannot start counting without saying the most important number that starts it all - zero, but do we count something if there is zero of it? Probably not. Think for a minute of a time when you were counting all your cool toys that you got for Christmas. Would you count them by starting with zero, one, two, three, and so on? No... you would say one, two, three, and so on. Usually, when we count items like you did your cool Christmas toys, we do not start with the number zero. Not starting with zero when we count objects makes zero being a natural number or not a debate in the world of mathematics. Most people count starting with the number one not zero. Just like you did with counting your toys. This is the argument some mathematicians make to support zero is not a natural number. Thus, zero is known as the neutral integer, or the whole number that comes in the middle of the positive and negative numbers on a number line. Zero does not have a positive or negative value. However, zero is considered a whole number, which in turn makes it an integer, but not necessarily a natural number. Remember the definition of natural numbers? They have to be positive, whole numbers. Zero is not positive or negative. Even though zero is not a positive number, it's still considered a whole number. Zero's status as a whole number and the fact that it is not a negative number makes it considered a natural number by some mathematicians. So, to answer the question is zero a natural number - yes it is on a number line and when identifying numbers in a set; but also no, because it's not used to count objects. You cannot count something that's not there! Lesson Summary Counting whole numbers, or natural numbers greater than zero, is a fun task to do. These numbers are always positive integers. You have to start with the neutral integer known as zero when using a number line, but you cannot use zero to count different objects if there aren't any objects to count. So get going and count those natural numbers. But, on a number line, always remember to begin with the natural number zero. Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? Log In Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. 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6164	https://medlineplus.gov/druginfo/meds/a624076.html	Gadolinium-based Contrast Agents: MedlinePlus Drug Information Skip navigation An official website of the United States government Here’s how you know Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. National Library of Medicine Menu Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia About MedlinePlus Show Search Search MedlinePlus GO About MedlinePlus What's New Site Map Customer Support Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia Español You Are Here: Home → Drugs, Herbs and Supplements → Gadolinium-based Contrast Agents URL of this page: Gadolinium-based Contrast Agents pronounced as (ga duh ’ nee uhm) Why is this medication prescribed? How should this medicine be used? Other uses for this medicine What special precautions should I follow? What side effects can this medication cause? What other information should I know? Other names IMPORTANT WARNING: Expand Section Gadolinium-based contrast agents (GBCAs) may cause problems with your skin, muscles, and other organs in your body. Tell your doctor if you have or have ever had kidney problems or have previously had a magnetic resonance imaging (MRI) scan. If you experience any of the following symptoms, call your doctor immediately: burning, itching, swelling, scaling, hardening and tightening of the skin; red or dark patches on the skin; stiffness in joints with trouble moving, bending or straightening the arms, hands, legs or feet; pain in the hip bones or ribs; or muscle weakness. GBCAs must not be given by intrathecal (into the space around the spinal cord) injection. Keep all appointments with your doctor and the laboratory. Your healthcare provider will give you the manufacturer's patient information sheet (Medication Guide) before you begin the MRI test with a GBCA. You can also visit the Food and Drug Administration (FDA) website ( or the manufacturer's website to obtain the Medication Guide. Talk to your doctor about the risks of receiving GBCAs. Why is this medication prescribed? Expand Section GBCAs are used at the time of a magnetic resonance imaging (MRI; a test that creates clear images of the structures inside your body using a large magnet, radio waves and a computer) scan to allow your doctor to see images of your brain, spine, heart, liver, breast(s), certain blood vessels, and other areas inside your body. Gadolinium-based contrast agents are in a class of medications called paramagnetic contrast agents. It works to improve the quality of the images seen by a MRI scan. How should this medicine be used? Expand Section GBCAs come as a solution (liquid) to be injected into a vein at the time of your MRI scan by a healthcare provider. Your doctor will give a dose of the medication that is best for you. Other uses for this medicine Expand Section This medication may be prescribed for other uses; ask your doctor or pharmacist for more information. What special precautions should I follow? Expand Section Before receiving a GBCA tell your doctor and pharmacist if you are allergic to dyes or any GBCA, any other medications, or any of the ingredients in the injection. Ask your pharmacist or check the Medication Guide for a list of the ingredients. tell your doctor and pharmacist what prescription and nonprescription medications, vitamins, nutritional supplements, and herbal products you are taking. tell your doctor if you are over 60 years of age or if you have or have ever had high blood pressure, diabetes, or kidney disease. tell your doctor if you are pregnant, plan to become pregnant, or are breast-feeding. What side effects can this medication cause? Expand Section Gadolinium-based contrast agents may cause side effects. Tell your doctor if any of these symptoms are severe or do not go away: burning, pain, or feeling of warmth or coldness at or around the injection site headache nausea Some side effects can be serious. If you experience any of these symptoms or those listed in the IMPORTANT WARNING section, call your doctor immediately or get emergency medical treatment: rash, itching, hives, dizziness, or difficulty breathing or swallowing GBCAs may stay in certain areas of your body such as the brain, bones, and skin in small amounts for months or years, especially if you have kidney problems. Talk to your doctor about the risks of taking this medication. GBCAs may cause other side effects. Call your doctor if you have any unusual problems while taking this medication. If you experience a serious side effect, you or your doctor may send a report to the Food and Drug Administration's (FDA) MedWatch Adverse Event Reporting program online ( or by phone (1-800-332-1088). What other information should I know? Expand Section Keep all appointments with your doctor. 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6165	https://blog.csdn.net/eydwyz/article/details/88557351	算法-有n步台阶，一次只能上1步或2步，共有多少种走法_有n阶楼梯，每步可走1或2或3阶，如果前三步加起来>=6阶则下一步可以后退1阶，请输出所有走法-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作算法-有n步台阶，一次只能上1步或2步，共有多少种走法最新推荐文章于 2022-02-21 20:28:31 发布 eydwyz于 2019-03-14 17:27:33 发布阅读量2.7w收藏 28 点赞数 13 分类专栏：算法算法专栏收录该内容 7 篇文章订阅专栏本文深入探讨了经典的爬楼梯问题，通过递归、迭代和动态规划三种算法解决不同条件下的楼梯攀登方案数量计算，对比了各种算法的优劣，并提供了详细的代码实现。分析 1、n=0 和 n=1 的时候并没有其他可选择的，所以可以得出f(0)=0;f(1)=1; 2、n>=2时情况就变复杂起来，但是这个时候可以操作的步骤也就2种也就是走1步(n-1)与走2步(n-2)。所以可以得到f(n)=f(n-1)+f(n-2); 从当前状态转为下一状态的通用算法既可。 3、验证，使用2以上的数字验证几次。实现实现算法往往是简单的，及时是复杂算法也花费不了太多精力，所以将问题转换为数学问题是一种很好的选择。当前这种简单算法实现方式更为简单，而且往往不止一种方式。递归 csharp public static int f(int n){ if(n<=2) return n; int x = f(n-1)+f(n-2); return x;} AI运行代码优点：可能是最好理解的算法了把。代码简单，好理解。缺点：计算次数颇多，有很多冗余计算。迭代 cobol public static int f(int n){ if(n<=2) return n; if first=1,second=2; int third=0; for(int i=3;i<=n;i++){ third = first+second; first = second; second = third; } return third;} AI运行代码优点: 基本没有冗余计算，效率高缺点: 谁能一次读完就理解的？动态规划原文：动态规划是解决下面这些性质类问题的技术：一个问题可以通过更小子问题的解决方法来解决（译者注：即问题的最优解包含了其子问题的最优解，也就是最优子结构性质）。有些子问题的解可能需要计算多次（译者注：也就是子问题重叠性质）。子问题的解存储在一张表格里，这样每个子问题只用计算一次。需要额外的空间以节省时间。 cobol public static int[] A = new int;public static int f(int n){ if(n<=2){ A[n] = n; } if(A[n]>0){ return A[n]; } else { A[n] = f(n-1)+f(n-2); return A[n]; }} AI运行代码虽然没弄懂为啥叫动态规划，但是代码还是很清晰的。优点：已经计算过的结果就不需要再次计算了。空间换时间缺点：需要额外的开销。楼梯有n阶台阶，上楼可以一步上1阶,2阶，3阶，编程序计算共有多少种不同的走法？ 2017年09月23日 23:35:09圆觉_阅读数：2980 标签：C++递归动态规划编程更多个人分类：C/C++ ```html ``` AI运行代码 html 题目：楼梯有n阶台阶，上楼可以一步上1阶,2阶，3阶，编程序计算共有多少种不同的走法？对于这样一个问题，思路：设n阶台阶的走法数为f(n)。如果只有1个台阶，走法有1种（一步上1个台阶），即f(1)=1；如果有2个台阶，走法有2种（一种是上1阶，再上1阶，另一种是一步上2阶），即f(2)=2；如果有3个台阶，走法有4种（一种每次1阶，共一种；另一种是2+1，共两种；第三种是3，共1种），即f(3)=4；当有n个台阶（n>3）时，我们缩小问题规模，可以这样想：最后是一步上1个台阶的话，之前上了n-1个台阶，走法为f(n-1)种，而最后是一步上2个台阶的话，之前上了n-2个台阶，走法为f(n-2)种，故而f(n)=f(n-1)+f(n-2)。列出的递归方程为：f(1)=1;f(2)=2; f(3)=4; if(n==1) return 1; else if(n==2) return 2; else if(n==3) return 4; else return f(n-1)+f(n-2)+f(n-3),n>3 最后一步可能是从第n-1阶往上走1阶、从n-2阶往上走2阶，或从第n-3阶往上走3阶。因此，抵达最后一阶的走法，其实就是抵达这最后三阶的方式的总和。解决方法1：按照递归的思想；但运算时间很长 ```html ``` AI运行代码 html int countWays (int n) {if (n<0) return 0; if (n==0) return 1; else { return countWays(n-1)+countWays(n-2)+countWays(n-3); } } 解决方法2：采用动态规划的思想优化，当一个问题可以分解成若干重复的子问题时，运用动态规划的思想： html 只需要将子问题求解一次，以后再遇到，直接调用，所以我们新建一个数组用于存储子问题的结果： AI运行代码 html html 将数组元素初始为零，若为新的子问题，我们求解，并把结果赋给对应的数组元素；这样当我们再次遇到相同的子问题，就可以直接调用了。 AI运行代码 html ```html ``` AI运行代码 html ```html ``` AI运行代码 html int countWaysDP(int n, dp[]) { if (n<0) return 0; if (n==0) return dp[n]; if (dp[n]>0) return dp[n]; //如果大于0 说明这个子问题已经计算过，直接调用数组 else { dp[n]=countWays[n-1,dp]+countWays[n-2,dp]+countWays[n-3,dp]; //否则还需计算该数组 return dp[n]; } } ```html ``` AI运行代码 html 接下来贴上实际运行代码吧； ```html ``` AI运行代码 html ```html ``` AI运行代码 html #include<iostream> using namespace std; const int MAX=1000; int countWays(int n) {if (n<0) return 0; if (n==0) return 1; else { return countWays(n-1)+countWays(n-2)+countWays(n-3); } } int countWaysDP(int n, int dp[]) { if (n<0) return 0; if (n==0) return 1; if (dp[n]>0) return dp[n]; //如果大于0 说明这个子问题已经计算过，直接调用数组 else { dp[n]=countWaysDP(n-1,dp)+countWaysDP(n-2,dp)+countWaysDP(n-3,dp); //否则还需计算该数组 return dp[n]; } } int main() { int m[MAX]={0}; // int m[MAX]; for(int i=1;i<10;i++) cout<<countWaysDP(i,m)<<endl; } 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 eydwyz 关注关注 13点赞踩 28 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录【面试题】N级台阶（比如 1 00级），每次可走 1 步或者 2 步，求总共有多少种走法？热门推荐 Mlib 01-05 5万+ 走台阶算法（本质上是斐波那契数列）在面试中常会遇到，描述就如题目那样：总共 1 00级台阶（任意级都行），小明每次可选择走 1 步、2 步或者 3 步，问走完这 1 00级台阶总共有多少种走法？一、题目分析这个问题本质上是斐波那契数列，假设只有一个台阶，那么只有一种跳法，那就是一次跳一级，f(1)= 1；如果有两个台阶，那么有两种跳法，第一种跳法是一次跳一级，第二种跳法是一次跳两级,f(2)= 2。如果有大于 2 级的n级参与评论您还未登录，请先登录后发表或查看评论算法很美:小孩上楼梯有n _阶楼梯,每步可走 1 或 2 或 3 阶,如果前三步加起来>... 9-25 三步问题。有个小孩正在上楼梯,楼梯有n 阶台阶,小孩一次可以上 1 阶、2 阶或 3 阶。实现一种方法,计算小孩有多少种上楼梯的方式。结果可能很大,你需要对结果模 1 000000007。 Code publicstaticvoidmain(String[]args){Scanner scanner = newScanner(System.in);intn = scanner.nextInt();longsolve = solve 2(n);System.out.pri... 有n个台阶的楼梯,有人可以 1 次走 1 步,也可以走两步,也可以走三步,问有... 9-23 交流群中小姐姐提出有n个台阶的楼梯,人一次可走 1 步、2 步或 3 步,求走法数量。这是斐波那契数列问题,表达式为f(n)= f(n - 1)+f(n - 2)+f(n - 3),博主用Java写了demo分享。交流群里有个小姐姐问了个问题: 有n个台阶的楼梯,有人可以 1 次走 1 步,也可以走两步,也可以走三步,问有多少种走法? 显然... 字节算法题 - - N 阶台阶，每次走一步或两步，计算共有多少种走法，并将每种走法打印出来。 weixin_41563161的博客 04-27 2872 N 阶台阶，每次走一步或两步，计算共有多少种走法，打印出每种走法。 / @Auther: liuhaidong Data: 2 0 2 0/4/2 6 00 2 6、1 4:50 Description: @version: 1.0 / public class Test7 { static final int s = 4; //自定义的台阶数 stat... 有N 步台阶，每次可以走 1 步或者 2 步，计算有多少种走法 m0_62962289的博客 10-31 4751 假如台阶为 1 有一种走法 f(1)= 1 假如台阶有 2 有 1+1，2 这两种走法 1 步+1 步 = 2 种 f(2)= 2 假如台阶有 3 有 1+1+1 ，1+2 ，2+1 这三种走法就是f(3)= f(1) + f(2)= 1 + 2 = 3 种走法假如台阶有4 有 1+1+1+1 ，1+2+1，1+1+2，2+1+1，2+2 这五种走法，就是 f(4)= f(2) + f(3)= 2+3 = 5 种走法假如台阶有5 有 1+1+1+1+1，1+2+2，1+2+1+1，1+1+1.. n级阶梯,每次走一步和两步或三步,有多少种走法有n _阶楼梯,每步可走 1 或... 9-13 //主要通过公式,然后让公式一步步的分解成已知条件 //f(n)= f(n - 1)+f(n - 2)+f(n - 3) function step($n){ $res = 0; $a = 1; $b = 2; $c = 4; for($i = 4;$i<=$n;++$i){ $res =$a+$b+$c; $a =$b; $b =$c; $c =$res; if($i = =$n){ return $res; } } }... java 阶梯问题 c语言 c++javac++有n _阶楼梯,每步可走 1 或 2 或 3 阶,如果... 8-31 本文通过递归算法解决了一个经典的数学问题:计算到达第n级阶梯的不同方式数量。当允许每次移动 1 阶或 2 阶时,该问题可以通过递归地求解前两个较小区间来找到解决方案。摘要生成于C知道,由 DeepSeek - R 1 满血版支持,前往体验 > 题目要求:n个台阶,上楼可以一步上 1 阶,也可以一步上 2 阶,一共有多少种上楼的方法 ... 算法! 有n 步台阶,一次只能上 1 步或 2 步,共有多少种走法 qq_40306697的博客 02-21 5095 有 2 种算法, 递归和循环迭代, 依次介绍并比较 1.递归 1 步台阶, 一种走法, 即f(1)= 1; 2 步台阶, 2 种走法, 一步加一步或是直接跨两步, 即f(2)= 2 3 步台阶, 最后一次要么跨 1 步,要么跨 2 步, 一共的走法为最后一次跨 1 步的走法 f(3 - 1)加上最后一次跨 2 步的走法 f(3 - 2), 即f(3)= f(2)+f(1) … n 步台阶, 最后一次跨 1 步的走法 f(n - 1)加上最后一次跨 2 步... 递归求解——n个台阶，每次只能上 1 或 2 个台阶，请问有多少种方法走完这n个台阶黄笳倞的博客 04-28 1万+ 算是一道比较简单的小学奥数题了，主要是为了展示使用递归来解决问题的方法目录一、思想二、代码(Python、C++和C#，含计时）一、思想可以使用逆向思维思考这个问题，即求出上(n - 1)个台阶和(n - 2)个台阶的方法总和为上n个台阶的方法数。二、代码(Python，C++和C#，含计时） 1.Python import time def footstep(n): ... ...楼梯给定一个共有 n 阶的楼梯你每步可以上 1 阶或者 2 阶但不能连续两轮... 9-8 如图 1 4 - 1 所示,对于一个 3 阶楼梯,共有 3 种方案可以爬到楼顶。本题的目标是求解方案数量,我们可以考虑通过回溯来穷举所有可能性。具体来说,将爬楼梯想象为一个多轮选择的过程:从地面出发,每轮选择上 1 阶或 2 阶,每当到达楼梯顶部时就将方案数量加 1 ,当越过楼梯顶部时就将其剪枝。代码如下所示: ... ...规定每步可以迈 1 阶、2 阶或 3 阶。从 1 阶到第n 阶,一共可以有多少种不同... 9-24 @param n 台阶数 @return 迈法总数 / public static long climbingStairs 1(int n) { //递归终止条件 if(n = = 1){ return 1; } if(n = = 2){ return 2; } if(n = = 3){ return 4; } //递归调用 return climbingStairs 1(n - 1)+climbingStairs 1(n - 2)+climbingStairs 1(n - 3); } / 动态... 有一个n 阶的楼梯,一次只能跳 1 阶或 2 阶,程序计算其有多少种跳法? qq_23035703的博客 04-20 2858 假设到n 阶楼梯有f(n)种跳法由于跳到n 阶台阶上只能由n - 1 或者n - 2 两种跳法那么f(n)就有f(n)= f(n - 1)+f(n - 2)种跳法就有 f(n)= 1 (n = 1) f(n)= 2 (n = 2) f(n)= f(n - 1)+f(n - 2) (n>2) 根据其规律可以由递归实... 问题描述：有n级台阶，一个人每次上一级或者两级，问有多少种走完n级台阶的方法。 06-21 问题描述：有n级台阶，一个人每次上一级或者两级，问有多少种走完n级台阶的方法。实际情况：给定一个矩阵m，从左上角开始每次只能向右走或者向下走，最后达到右下角的位置，路径中所有数字累加起来就是路径和，返回... 爬楼梯(python实现)python假设到楼顶的 _台阶共有 n 阶。小明每步可以爬... 9-13 解释: 有三种方法可以爬到楼顶。 4. 1 阶 + 1 阶 + 1 阶 5. 1 阶 + 2 阶 6. 2 阶 + 1 阶来源:力扣(LeetCode) 链接: - cn.com/problems/climbing - stairs 思路:斐波那契数列代码: class Solution: def init(self,n): self.result = self.climbStairs(n) def climbStairs(... 爬楼梯有n级 _台阶(n>0),从下面开始走要走到所有台阶上面,每步可以... 8-15 输入:n = 3 输出:3 解释:有三种方法可以爬到楼顶。 1. 1 阶 + 1 阶 + 1 阶 2. 1 阶 + 2 阶 3. 2 阶 + 1 阶提示: 1<= n <= 45 解题思路: ①如果最后一步是 1 个台阶,那么求 n - 1 台阶需要的步数就可以了 ②如果最后一步是 2 个台阶,就算n - 2 台阶需要的步数,因为只有 2 个可能,所以只需要... n 阶台阶 java_上N 阶楼梯,一次走 1 个台阶或者 2 个台阶,共有多少种走法? weixin_29664739的博客 03-02 3497 假设你需要走 n 阶楼梯才能到达楼顶,走楼梯的方式有两种,一次走 1 个台阶或者一次走 2 个台阶,问有多少种不同的方法可以走完这n 阶楼梯？先穷举几个n值分析下:n = 1,共 1 种;{1}n = 2,共 2 种;{1,1},{2}n = 3,共 3 种{1,2},{1,1,1},{2,1}n = 4,共5种{1,1,2},{2,2},{1,2,1},{1,1,1,1},{2,1,1}n = 5,共8种{1,2,2},{1,1,1,2}... 详细解答【算法题】9个台阶，每次只能上 1 个或者 2 个，一共有多少种走法？阳光大男孩的博客 04-26 2695 思路看到这道题感觉还挺有意思的，自己第一时间也没有想出来，是看了类似博文后才豁然开朗，但是我看到的博文中都没有将下面这个数列的来龙去脉说明清楚， f(n)= f(n−1)+f(n−2)(n>2). f(n) = f(n - 1) + f(n - 2) (n>2). f(n)= f(n−1)+f(n−2)(n>2). 这可是此题的关键，我来详细说明一下。 ... 剑指Offer：一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级。求该青蛙跳上一个n级的台阶总共有多少种跳法 12-20 首先，题目描述了一只青蛙要跳上n级台阶，每次它可以跳 1 级或 2 级。我们要找出青蛙跳上n级台阶的所有可能方法数。这是一个组合问题，因为每一步都有两种选择（跳 1 级或 2 级），而这些选择是相互独立的。 ### 暴力法 ... [C题目]n个台阶，一步只能走 1 个台阶或者 2 个台阶，有几种走法？(非递归)(待改正) weixin_46603347的博客 01-01 1652 #include int com(int a, int b)//高中学的排列组合中的组合C(a b)，下面有一张图。 { int mother = 1, son = 1;//mother是图片中分母的值。son是图片中分子的值。 int p = a;//因为a存的值，既要用来决定循环次数，又要参与分母的运算，所以用另一个变量p保存a的值来参与mother的运算。 for (int i = 0; i < a; i++)//求分母的值，分母中有几个数相乘就循环几次。 . 有n 步台阶一次只能上 1 步或 2 步共有多少种走法？ qq_33732195的博客 01-16 595 public class test { //递归算法 public static long f(int n){ if(n = = 0) return 0; if(n = = 1 \|\| n = = 2) return n; return f(n - 1) + f(n - 2); } //循环遍历 public static int loop(int n){ if(n<1){... N 阶楼梯上楼问题：一次可以走两阶或一阶，问有多少种上楼方式 yan 05-27 7036 N 阶楼梯上楼问题：一次可以走两阶或一阶，问有多少种上楼方式？思路一：设有x次走一阶，y次走两阶，则一定满足x+2y = n，且x、y均为整数，那么对于任何一个满足的x的可能走法共有 C（x+(n - x)/2,x）种走法，即从数x+(n - x)/2 中取x种组合，值为（x+(n - x)/2）的阶乘除以x的阶乘与(n - x)/2 的阶乘的乘积。依次取可能的x值，然后相加每一种的可能情况就可以了。代码... n个台阶,每次只能走一步或者两步,求多少种走法 wxyBlog 07-17 3970 最重要的就是最后一步:如果走一步就需要f(n - 1)种,如果走两步就需要f(n - 2)种走一个台阶:1 种 - f(1) 走两个台阶:2 种 - f(2) 三个台阶:先走一个台阶 f 1 和最后一次走两个台阶:f(1) 先走二个台阶 f 2 和最后一次走一个台阶:f(2) f(3)= f(1)+f(2) 四个台阶:先走三个台阶和最后一次走一个:f(3) 先走两个台阶和最后一次两个台阶:f(2) f(4)= f(2)+f(3) n个台阶:f(n)= f(n - 1)+f(n - 2) package myjava.递归和迭代; 1 2 05 shengdai20的专栏 02-20 561 题目描述： N 阶楼梯上楼问题：一次可以走两阶或一阶，问有多少种上楼方式。（要求采用非递归）输入：输入包括一个整数N,(1 输出：可能有多组测试数据，对于每组数据，输出当楼梯阶数是N时的上楼方式个数。样例输入： 4 样例输出： 5 注意用long long #include #include int main() 有n 阶楼梯，一次上一阶或两阶，共有多少种方法？(c) weixin_54013177的博客 02-21 925 有n 阶楼梯，一次上一阶或两阶，共有多少种方法？(c) n级阶梯，每次走一步或两步，问最多有多少种走法 redstraw的专栏 04-09 7548 遇到这道题两次了，现在来总结下。方法一：运用组合数学的思想。假设在整个过程中，我有k次走了两步，那么剩下的n - 2 k次我走的都是一步，那么我总的走的步数是k+n - 2 k。则结果为从n - k里面选了k次走两步，结果为，代码如下： public BigDecimal jiecheng(int n){ if(n = = 0\|\|n = = 1){ return new BigDecimal 有n个台阶，每次只能走一步或两步，用递归求有多少种走法用c语言最新发布 03-03 好的，我现在要解决这个问题：有n个台阶，每次可以走 1 步或 2 步，用递归的方法求有多少种走法，并且用C语言实现。让我仔细想想该怎么处理。首先，我得理解问题。题目是说，当有n级台阶时，每次可以跨 1 步或者 2 步，问... 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 eydwyz 博客等级码龄15年 24 原创626 点赞 3415 收藏 525 粉丝关注私信热门文章 NAT的四种类型 137817 EtherType ：以太网类型字段及值 87680 MCS简介-编码方式 41119 GRE、PPTP、L2TP隧道协议 28084 算法-有n步台阶，一次只能上1步或2步，共有多少种走法 27058 分类专栏 PCIE Linux调试8篇 Linux环境搭建5篇 GIT1篇 weston5篇电脑 pjsip1篇 gstreamer8篇 openmax3篇 wayland4篇 OpenCL 海思1篇 LINUX Makefile1篇 Linux105篇 Linux iptables34篇 Linux TC29篇路由技术3篇 WIFI知识10篇 wifidog18篇 LSDK1篇 proc5篇 p2p17篇 C20篇 dhcp12篇 uboot5篇 openwrt23篇硬件6篇 80220篇 linux驱动13篇 TC--higher9篇 iptables - L77篇社会3篇 PPPOE12篇 PPTP4篇蓝牙4篇 STM327篇 android50篇 hostapd1篇 WPA10篇 L2TP1篇 PPP5篇 LORA8篇 C++2篇 WPA supplicant8篇 D-BUS8篇音频30篇 USB2篇 PCI10篇设备树3篇 interview3篇 QT13篇 ROS1篇 ffmpeg35篇 openCV2篇 V4L2-UVC9篇 JNI2篇 java8篇 Vulkan1篇 openGL7篇算法7篇 CAN3篇 H2646篇 MP44篇 GPU6篇展开全部收起上一篇： Linux V4L2驱动架构解析与开发导引下一篇：八大排序算法最新评论 NAT的四种类型 rainbow_947:通俗易懂关于fread函数读取到的数据和实际统计的和数据不一样石下听海:感谢博主！！我还一直在想这个为什么会不一样，问ai也问不出所以然，还是网上能人多啊 QT5的软键盘输入法实现 simen-wang:我重载了QTextEdit为自己的QMyTextEdit后却不能触发软键盘了这是为什么呢？ Qt/C++ 模仿酷狗音乐播放器Qt/C++ 模仿酷狗音乐播放器乔慧理子:源码链接好像失效了 dlmalloc, ptmalloc,tcmalloc和jemalloc内存分配策略研究 nanfengnan?:dlmalloc呢？你在搞笑吗大家在看当用户同时打开多个标签页时，如何同步它们之间的状态和数据呢？ TikTok新店冷启动避坑指南：如何避免0曝光和封号？ uni-app 模板语法修复说明实时交易平台性能优化：Go+React全栈改造 153 【C++】类和对象（下） 627 最新文章 Linux设备驱动模块自加载示例与原理解析关于Wifi WDS的两种模式浅谈 linux驱动加载动态加载静态加载自动加载 2022年 18篇 2021年 43篇 2020年 93篇 2019年 48篇 2018年 24篇 2017年 259篇 2016年 195篇目录分析实现递归迭代动态规划楼梯有n阶台阶，上楼可以一步上1阶,2阶，3阶，编程序计算共有多少种不同的走法？展开全部收起目录分析实现递归迭代动态规划楼梯有n阶台阶，上楼可以一步上1阶,2阶，3阶，编程序计算共有多少种不同的走法？展开全部收起上一篇： Linux V4L2驱动架构解析与开发导引下一篇：八大排序算法分类专栏 PCIE Linux调试8篇 Linux环境搭建5篇 GIT1篇 weston5篇电脑 pjsip1篇 gstreamer8篇 openmax3篇 wayland4篇 OpenCL 海思1篇 LINUX Makefile1篇 Linux105篇 Linux iptables34篇 Linux TC29篇路由技术3篇 WIFI知识10篇 wifidog18篇 LSDK1篇 proc5篇 p2p17篇 C20篇 dhcp12篇 uboot5篇 openwrt23篇硬件6篇 80220篇 linux驱动13篇 TC--higher9篇 iptables - L77篇社会3篇 PPPOE12篇 PPTP4篇蓝牙4篇 STM327篇 android50篇 hostapd1篇 WPA10篇 L2TP1篇 PPP5篇 LORA8篇 C++2篇 WPA supplicant8篇 D-BUS8篇音频30篇 USB2篇 PCI10篇设备树3篇 interview3篇 QT13篇 ROS1篇 ffmpeg35篇 openCV2篇 V4L2-UVC9篇 JNI2篇 java8篇 Vulkan1篇 openGL7篇算法7篇 CAN3篇 H2646篇 MP44篇 GPU6篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
6166	https://artofproblemsolving.com/wiki/index.php/Diophantine_equation?srsltid=AfmBOopVax6SPeLPPdOJySXCvVvgEODxzlvZMB-p_aB347feymqFFh-L	Art of Problem Solving Diophantine equation - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Diophantine equation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Diophantine equation A Diophantine equation is an equation relating integer (or sometimes natural number or whole number) quanitites. Finding the solution or solutions to a Diophantine equation is closely tied to modular arithmetic and number theory. Often, when a Diophantine equation has infinitely many solutions, parametric form is used to express the relation between the variables of the equation. Diophantine equations are named for the ancient Greek/Alexandrian mathematician Diophantus. Contents [hide] 1 Linear Combination 2 Pythagorean Triples 2.1 Method of Pythagoras 2.2 Method of Plato 2.3 Babylonian Method 3 Sum of Fourth Powers 4 Pell Equations 5 Methods of Solving 5.1 Coordinate Plane 5.2 Modular Arithmetic 5.3 Induction 5.4 General Solutions 6 Fermat's Last Theorem 7 Problems 7.1 Introductory 7.2 Intermediate 7.3 Olympiad 8 References 9 See also Linear Combination A Diophantine equation in the form is known as a linear combination. If two relatively prime integers and are written in this form with , the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when . If , then there are no solutions to the equation. To see why, consider the equation . is a divisor of the LHS (also notice that must always be an integer). However, will never be a multiple of , hence, no solutions exist. Now consider the case where . Thus, . If and are relatively prime, then all solutions are obviously in the form for all integers . If they are not, we simply divide them by their greatest common divisor. See also: Bézout's identity. Pythagorean Triples Main article: Pythagorean triple A Pythagorean triple is a set of three integers that satisfy the Pythagorean Theorem, . There are three main methods of finding Pythagorean triples: Method of Pythagoras If is an odd number, then is a Pythagorean triple. Method of Plato If , is a Pythagorean triple. Babylonian Method For any (), we have is a Pythagorean triple. Sum of Fourth Powers An equation of form has no integer solutions, as follows: We assume that the equation does have integer solutions, and consider the solution which minimizes . Let this solution be . If then their GCD must satsify . The solution would then be a solution less than , which contradicts our assumption. Thus, this equation has no integer solutions. If , we then proceed with casework, in . Note that every square, and therefore every fourth power, is either or . The proof of this is fairly simple, and you can show it yourself. Case 1: This would imply , a contradiction. Case 2: This would imply , a contradiction since we assumed . Case 3: , and We also know that squares are either or . Thus, all fourth powers are either or . By similar approach, we show that: , so . This is a contradiction, as implies is odd, and implies is even. QED [Oops, this doesn't work. 21 (or ) are equal to and not even...] Pell Equations Main article: Pell equation A Pell equation is a type of Diophantine equation in the form for natural number. The solutions to the Pell equation when is not a perfect square are connected to the continued fraction expansion of . If is the period of the continued fraction and is the th convergent, all solutions to the Pell equation are in the form for positive integer . Methods of Solving Coordinate Plane Note that any linear combination can be transformed into the linear equation , which is just the slope-intercept equation for a line. The solutions to the diophantine equation correspond to lattice points that lie on the line. For example, consider the equation or . One solution is (0,1). If you graph the line, it's easy to see that the line intersects a lattice point as x and y increase or decrease by the same multiple of and , respectively (wording?). Hence, the solutions to the equation may be written parametrically (if we think of as a "starting point"). Modular Arithmetic Sometimes, modular arithmetic can be used to prove that no solutions to a given Diophantine equation exist. Specifically, if we show that the equation in question is never true mod , for some integer , then we have shown that the equation is false. However, this technique cannot be used to show that solutions to a Diophantine equation do exist. Induction Sometimes, when a few solutions have been found, induction can be used to find a family of solutions. Techniques such as infinite Descent can also show that no solutions to a particular equation exist, or that no solutions outside of a particular family exist. General Solutions It is natural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equations. This is known as Hilbert's tenth problem. The answer, however, is no. Fermat's Last Theorem Main article: Fermat's Last Theorem is known as Fermat's Last Theorem for the condition . In the 1600s, Fermat, as he was working through a book on Diophantine Equations, wrote a comment in the margins to the effect of "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." Fermat actually made many conjectures and proposed plenty of "theorems," but wasn't one to write down the proofs or much other than scribbled comments. After he died, all his conjectures were re-proven (either false or true) except for Fermat's Last Theorem. After over 350 years of failing to be proven, the theorem was finally proven by Andrew Wiles after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof. Problems Introductory Two farmers agree that pigs are worth dollars and that goats are worth dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way? (Source) Intermediate Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product . (Source) Olympiad Determine the maximum value of , where and are integers satisfying and . (Source) Solve in integers the equation . References Proof of Fermat's Last Theorem See also Number Theory Pell equation Retrieved from " Category: Number theory Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. 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6167	https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Oxides/Physical_Properties_of_Period_3_Oxides	Skip to main content Physical Properties of Period 3 Oxides Last updated : Jun 30, 2023 Save as PDF Oxides Hard Water Page ID : 545 Jim Clark Truro School in Cornwall ( \newcommand{\kernel}{\mathrm{null}\,}) This page explains the relationship between the physical properties of the oxides of Period 3 elements (sodium to chlorine) and their structures. Argon is obviously omitted because it does not form an oxide. The oxides The oxides we'll be looking at are: \| \| \| \| \| \| \| \| --- --- --- \| Na2O \| MgO \| Al2O3 \| SiO2 \| P4O10 \| SO3 \| Cl2O7 \| \| \| \| \| \| P4O6 \| SO2 \| Cl2O \| Those oxides in the top row are known as the highest oxides of the various elements. These are the oxides where the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 element are being involved in the bonding - from just the one with sodium, to all seven of chlorine's outer electrons. The structures The trend in structure is from the metallic oxides containing giant structures of ions on the left of the period via a giant covalent oxide (silicon dioxide) in the middle to molecular oxides on the right. Melting and boiling points The giant structures (the metal oxides and silicon dioxide) will have high melting and boiling points because a lot of energy is needed to break the strong bonds (ionic or covalent) operating in three dimensions. The oxides of phosphorus, sulfur and chlorine consist of individual molecules; some are small and simple and others are polymeric. The attractive forces between these molecules will be van der Waals dispersion and dipole-dipole interactions. These vary in size depending on the size, shape and polarity of the various molecules - but will always be much weaker than the ionic or covalent bonds you need to break in a giant structure. These oxides tend to be gases, liquids or low melting point solids. Electrical conductivity None of these oxides has any free or mobile electrons. That means that none of them will conduct electricity when they are solid. The ionic oxides can, however, undergo electrolysis when they are molten. They can conduct electricity because of the movement of the ions towards the electrodes and the discharge of the ions when they get there. The metallic oxides The structures Sodium, magnesium and aluminium oxides consist of giant structures containing metal ions and oxide ions. Magnesium oxide has a structure just like sodium chloride. The other two have more complicated arrangements. Melting and boiling points There are strong attractions between the ions in each of these oxides and these attractions need a lot of heat energy to break. These oxides therefore have high melting and boiling points. Electrical conductivity None of these conducts electricity in the solid state, but electrolysis is possible if they are molten. They conduct electricity because of the movement and discharge of the ions present. The only important example of this is in the electrolysis of aluminium oxide in the manufacture of aluminium . Whether you can electrolyse molten sodium oxide depends, of course, on whether it actually melts instead of subliming or decomposing under ordinary circumstances. If it sublimes, you will not get any liquid to electrolyse! Magnesium and aluminium oxides have melting points far too high to be able to electrolyse them in a simple lab. Silicon dioxide (silicon(IV) oxide) The structure The electronegativity of the elements increases as you go across the period, and by the time you get to silicon, there is not enough electronegativity difference between the silicon and the oxygen to form an ionic bond. Silicon dioxide is a giant covalent structure. There are three different crystal forms of silicon dioxide. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms. Notice that each silicon atom is bridged to its neighbours by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending in all 3 dimensions. Note If you want to be fussy, the Si-O-Si bond angles are wrong in this diagram. In reality the "bridge" from one silicon atom to its neighbour is not in a straight line, but via a "V" shape (similar to the shape around the oxygen atom in a water molecule). It's extremely difficult to draw that convincingly and tidily in a diagram involving this number of atoms. The simplification is perfectly acceptable. Melting and boiling points: Silicon dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but they are all around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Silicon dioxide boils at 2230°C. Because you are talking about a different form of bonding, it doesn't make sense to try to compare these values directly with the metallic oxides. What you can safely say is that because the metallic oxides and silicon dioxide have giant structures, the melting and boiling points are all high. Electrical conductivity: Silicon dioxide does not have any mobile electrons or ions and hence does not conduct electricity either as a solid or a liquid. The molecular oxides Phosphorus, sulfur and chlorine all form oxides which consist of molecules. Some of these molecules are fairly simple - others are polymeric. We are just going to look at some of the simple ones. Melting and boiling points of these oxides will be much lower than those of the metal oxides or silicon dioxide. The intermolecular forces holding one molecule to its neighbors will be van der Waals dispersion forces or dipole-dipole interactions. The strength of these will vary depending on the size of the molecules. None of these oxides conducts electricity either as solids or as liquids. None of them contains ions or free electrons. The Phosphorus Oxides Phosphorus has two common oxides, phosphorus(III) oxide, P4O6, and phosphorus(V) oxide, P4O10. Phosphorus(III) oxide Phosphorus(III) oxide is a white solid, melting at 24°C and boiling at 173°C. The structure of its molecule is best worked out starting from a P4 molecule which is a little tetrahedron. Pull this apart so that you can see the bonds . . . . . . and then replace the bonds by new bonds linking the phosphorus atoms via oxygen atoms. These will be in a V-shape (rather like in water), but you probably wouldn't be penalised if you drew them on a straight line between the phosphorus atoms in an exam. The phosphorus is using only three of its outer electrons (the 3 unpaired p electrons) to form bonds with the oxygens. Phosphorus(V) oxide Phosphorus(V) oxide is also a white solid, subliming (turning straight from solid to vapour) at 300°C. In this case, the phosphorus uses all five of its outer electrons in the bonding. Solid phosphorus(V) oxide exists in several different forms - some of them polymeric. We are going to concentrate on a simple molecular form, and this is also present in the vapor. This is most easily drawn starting from P4O6. The other four oxygens are attached to the four phosphorus atoms via double bonds. If you look carefully, the shape of this molecule looks very much like the way we usually draw the repeating unit in the diamond giant structure. Don't confuse the two, though! The molecule stops here. This is not a little bit of a giant structure - it's all there is. In diamond, of course, the structure just continues almost endlessly in three dimensions. The Sulfur Oxides Sulfur has two common oxides, sulfur dioxide (sulfur(IV) oxide), SO2, and sulfur trioxide (sulfur(VI) oxide), SO3. sulfur dioxide Sulfur dioxide is a colourless gas at room temperature with an easily recognised choking smell. It consists of simple SO2 molecules. The sulfur uses 4 of its outer electrons to form the double bonds with the oxygen, leaving the other two as a lone pair on the sulfur. The bent shape of SO2 is due to this lone pair. sulfur trioxide Pure sulfur trioxide is a white solid with a low melting and boiling point. It reacts very rapidly with water vapour in the air to form sulfuric acid. That means that if you make some in the lab, you tend to see it as a white sludge which fumes dramatically in moist air (forming a fog of sulfuric acid droplets). Gaseous sulfur trioxide consists of simple SO3 molecules in which all six of the sulfur's outer electrons are involved in the bonding. There are various forms of solid sulfur trioxide. The simplest one is a trimer, S3O9, where three SO3 molecules are joined up and arranged in a ring. There are also other polymeric forms in which the SO3 molecules join together in long chains. For example: It is difficult to draw this convincingly. In fact, on each sulfur atom, one of the double bonded oxygens is coming out of the diagram towards you, and the other one is going back in away from you. The fact that the simple molecules join up in this way to make bigger structures is what makes the sulfur trioxide a solid rather than a gas. The chlorine oxides Chlorine forms several oxides. Here we are just looking at two of them : chlorine(I) oxide (Cl2O) and chlorine(VII) oxide (Cl2O7). Chlorine(I) oxide Chlorine(I) oxide is a yellowish-red gas at room temperature. It consists of simple small molecules. There's nothing in the least surprising about this molecule and it's physical properties are just what you would expect for a molecule this size. Chlorine(VII) oxide In chlorine(VII) oxide, the chlorine uses all of its seven outer electrons in bonds with oxygen. This produces a much bigger molecule, and so you would expect its melting point and boiling point to be higher than chlorine(I) oxide. Chlorine(VII) oxide is a colourless oily liquid at room temperature. In the diagram, for simplicity I have drawn a standard structural formula. In fact, the shape is tetrahedral around both chlorines, and V-shaped around the central oxygen. Problems I intended at this point to quote values for each of the oxides, hoping to show that the melting and boiling points increase as the charges on the positive ion increase from 1+ in sodium to 3+ in aluminium. You would expect that the greater the charge, the greater the attractions. Unfortunately, the oxide with the highest melting and boiling point is magnesium oxide, not aluminium oxide! So that theory bit the dust! The reason for this probably lies in the increase in electronegativity as you go from sodium to magnesium to aluminium. That would mean that the electronegativity difference between the metal and the oxygen is decreasing. The smaller difference means that the bond won't be so purely ionic. It is also likely that molten aluminium oxide contains complex ions containing both aluminium and oxygen rather than simple aluminium and oxide ions. All this means, of course, that you aren't really comparing like with like - so wouldn't necessarily expect a neat trend. The other problems I came across lie with sodium oxide. Most sources say that this sublimes (turns straight from solid to vapour) at 1275°C. However, the usually reliable Webelements gives a melting point of 1132°C followed by a decomposition temperature (before boiling) of 1950°C. Other sources talk about it decomposing (to sodium and sodium peroxide) above 400°C. I have no idea what the truth of this is - although I suspect that the Webelements melting point value is probably for a pressure above atmospheric pressure (although it doesn't say so). Oxides Hard Water
6168	https://www.cracksat.net/digital/math/test9.html	Home Digital SAT SAT Tests SAT Downloads PSAT AP Prep SAT Test SAT Prep SAT Skills Digital SAT Reading and Writing Math SAT Grammar SAT Downloads SAT Books PSAT College Admission Digital SAT Math Practice Test 9: Averages Home Digital SAT Test Digital SAT Math Practice Tests Test Information 12 questions 19 minutes See All test questions More Tests Digital SAT Math Practice Test 1 Digital SAT Math Practice Test 2 Digital SAT Math Practice Test 3 Digital SAT Math Practice Test 4 Digital SAT Math Practice Test 5 Digital SAT Math Practice Test 6 Digital SAT Math Practice Test 7 Digital SAT Math Practice Test 8 Digital SAT Math Practice Test 9 Digital SAT Math Practice Test 10 Digital SAT Math Practice Test 11 Digital SAT Math Practice Test 12 Digital SAT Math Practice Test 13 Digital SAT Math Practice Test 14 Digital SAT Math Practice Test 15 Digital SAT Math Practice Test 16 Digital SAT Math Practice Test 17 Digital SAT Math Practice Test 18 Digital SAT Math Practice Test 19 Digital SAT Math Practice Test 20 Digital SAT Math Practice Test 21 Digital SAT Math Practice Test 22 Digital SAT Math Practice Test 23 Digital SAT Math Practice Test 24 Digital SAT Math Practice Test 25 Digital SAT Math Practice Test 26 Quick View Digital SAT Tests Reading and Writing Math Switch to Mobile Version? Yes
6169	https://en.wikipedia.org/wiki/Linear_space_(geometry)	Jump to content Linear space (geometry) Deutsch Русский Українська Edit links From Wikipedia, the free encyclopedia A linear space is a basic structure in incidence geometry. A linear space consists of a set of elements called points, and a set of elements called lines. Each line is a distinct subset of the points. The points in a line are said to be incident with the line. Each two points are in a line, and any two lines may have no more than one point in common. Intuitively, this rule can be visualized as the property that two straight lines never intersect more than once. Linear spaces can be seen as a generalization of projective and affine planes, and more broadly, of 2- block designs, where the requirement that every block contains the same number of points is dropped and the essential structural characteristic is that 2 points are incident with exactly 1 line. The term linear space was coined by Paul Libois in 1964, though many results about linear spaces are much older. Definition [edit] Let L = (P, G, I) be an incidence structure, for which the elements of P are called points and the elements of G are called lines. L is a linear space if the following three axioms hold: (L1) two distinct points are incident with exactly one line. (L2) every line is incident to at least two distinct points. (L3) L contains at least two distinct lines. Some authors drop (L3) when defining linear spaces. In such a situation the linear spaces complying to (L3) are considered as nontrivial and those that do not are trivial. Examples [edit] The regular Euclidean plane with its points and lines constitutes a linear space, moreover all affine and projective spaces are linear spaces as well. The table below shows all possible nontrivial linear spaces of five points. Because any two points are always incident with one line, the lines being incident with only two points are not drawn, by convention. The trivial case is simply a line through five points. In the first illustration, the ten lines connecting the ten pairs of points are not drawn. In the second illustration, seven lines connecting seven pairs of points are not drawn. \| \| \| \| \| --- --- \| \| \| \| \| \| \| 10 lines \| 8 lines \| 6 lines \| 5 lines \| A linear space of n points containing a line being incident with n − 1 points is called a near pencil. (See pencil) \| \| \| \| \| near pencil with 10 points \| Properties [edit] The De Bruijn–Erdős theorem shows that in any finite linear space which is not a single point or a single line, we have . See also [edit] Block design Fano plane Projective space Affine space Molecular geometry Partial linear space References [edit] Shult, Ernest E. (2011), Points and Lines, Universitext, Springer, doi:10.1007/978-3-642-15627-4, ISBN 978-3-642-15626-7. Albrecht Beutelspacher: Einführung in die endliche Geometrie II. Bibliographisches Institut, 1983, ISBN 3-411-01648-5, p. 159 (German) J. H. van Lint, R. M. Wilson: A Course in Combinatorics. Cambridge University Press, 1992, ISBN 0-521-42260-4. p. 188 L. M. Batten, Albrecht Beutelspacher: The Theory of Finite Linear Spaces. Cambridge University Press, Cambridge, 1992. Retrieved from " Category: Incidence geometry
6170	https://matheducators.stackexchange.com/questions/28148/what-is-the-simplest-formula-for-calculating-the-circumference-of-a-circle	Skip to main content Mathematics Educators Asked Modified 11 months ago Viewed 5k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. I am studying for a test to become a "paraeducator/instructional aid." One of the pages in the study guide has this information on computing area and perimeter and circumference and such: Wouldn't it be simpler to say "multiply the diameter by pi"? geometry math-circle Share CC BY-SA 4.0 Follow this question to receive notifications asked Sep 24, 2024 at 17:28 B. Clay Shannon-B. Crow RavenB. Clay Shannon-B. Crow Raven 20911 silver badge55 bronze badges 9 9 It depends on what value you have, or which you want to parameterize by. Using the diameter of an existing circle is probably easier when measuring with a ruler, but when creating the circle with a compass, the radius is more natural. Adam – Adam 2024-09-24 21:59:02 +00:00 Commented Sep 24, 2024 at 21:59 15 Using the diameter d=2r makes the circumference formula exactly one symbol shorter, while it turns the area formula into e.g. A=πd24. Torsten Schoeneberg – Torsten Schoeneberg 2024-09-25 03:43:36 +00:00 Commented Sep 25, 2024 at 3:43 3 2 pi or not 2 pi, that is the question. chux – chux 2024-09-25 08:50:49 +00:00 Commented Sep 25, 2024 at 8:50 5 Are you interested in the simplest mathematical method or are you interested at the simplest teaching method? These two things are not always the same. David S – David S 2024-09-25 15:22:08 +00:00 Commented Sep 25, 2024 at 15:22 3 As an aside, "there are 360 degrees in a circle" is one of those imprecise statements which people know the meaning of without considering what it actually says. An angle is between two lines; what does it mean for an angle to be "in a circle"? There are other things on this sheet to take issue with: bisect means to find the midpoint of a line or half of an angle; pi is not equal to 3.14; the three angles of a triangle add to 180 degrees, but do not each add to 180 degrees. Clear thinking is more important than facial simplicity. kaya3 – kaya3 2024-09-26 17:50:34 +00:00 Commented Sep 26, 2024 at 17:50 \| Show 4 more comments 6 Answers 6 Reset to default This answer is useful 25 Save this answer. Show activity on this post. Conservation of Difficulty You should not be focused on what formula is "simpler". I have often joked with colleagues that mathematics has a kind of Law of Conservation of Difficulty: any particular problem has a certain difficulty associated to it, and there is not really any way of getting around that difficulty. My usual go-to example of this is a proof of the Fundamental Theorem of Algebra: there are very elementary proofs which basically boil down to gradient descent, and fancier proofs which rely on Liouville's theorem. The elementary proofs are typically very long and tedious—because an elementary proof uses elementary tools, there are no big theorems which can be used to make the proof shorter or "simpler"—one must simply brute force the problem. On the other hand, a proof using Liouville is a two liner—but one has to spend a lot of time building up the theory of complex analysis before getting to a proof of Liouville so that one can then proceed to prove the Fundamental Theorem of Algebra. The difficulty of the proof can be moved to a different place, but it can't be entirely removed. As an instructor, it is your job to take complicated ideas and find a path through those ideas which will develop the theory in a coherent manner, so that the steps from one idea to the next will each appear simple, until you have a chance to zoom out or look back and see the material taught in total. An Ahistorical History In geometry, a circle is defined by two things: a point C (the center of the circle) and a distance r (the radius of the circle). By definition, the circle with center C and radius r is the set of points which are exactly r units from C. In classical Euclidean geometry, this makes sense, since one of the basic tools is the compass, which can be used to construct a circle in exactly the manner described above, the center and radius are fundamental. This implies that any formula relating to a circle should start from these two data: the center and the radius. After some investigation, we learn that the circumference of a circle is proportional to the radius. That is, there exists some constant k such that the circumference P of a circle with radius r is P=kr. A bit more work shows that k≈6.28. This number shows up in a lot of places, so one might want to go ahead and give it a name: τ≈6.28 is, by definition, the ratio of the circumference of a circle to its radius. However, in the real world, measuring the radius of a circle is a bit of a pain—one has to find the center of the circle, and then measure the radius. It turns out that it is easier to measure the diameter (e.g. with calipers; or by constructing a chord and considering a perpendicular bisector of that chord). So, from a practical point of view, it is often easier to get one's hands on the diameter, D, of a circle. The diameter is twice the radius, so another formula for the circumference of a circle is P=τD2=τ2D. Because the term τ/2 appears repeatedly, it might be wise to give this a name, too. So define π=τ2=circumferencediameter≈3.14. At the end of the day, there are two (equivalent) formulae for the circumference of a circle: P=πDandP=2πr. The former is, perhaps, more practical (because it is, maybe, easier to measure a diameter); but the latter is more fundamental, as it relates directly to the construction of the circle in terms of a center and radius. The Language of Mathematics Another job of an instructor is to teach students the language used in their field. In mathematics, one "multiplies two numbers" (one does not "times two numbers); one "evaluates an integral" (one does not "solve an integral); one "solves a polynomial equation" and "determines [or finds] the zeros of a polynomial function" (one does not "solve a function"); and so on. The use of the radius in the formula for the circumference of a circle is a matter of correct mathematical "spelling" or "grammar"—it is the correct way to use language as mathematicians use that language. Whether or not it is "more simple" (which, as I have argued above, it really isn't), it is how the language is used, and students need to get used to that use of language. This is particularly important, as there are many, many other places in mathematics where the radius plays a role. In three-dimensional geometry, it shows up in the formulae for the volume and surface area of a sphere; it is fundamental to the definitions of the cosine and sine functions as functions on R; and so on. The radius shows up everywhere in the language of mathematics, so one might as well get students used to this as early as possible. Tangential Note: In the formula sheet quoted in the question, it is asserted that π=3.14. This is wrong—don't say this to students; π is not equal to 3.14. This is a reasonable approximation for π in most circumstances, but there is no equality here. The constant π is irrational—it does not have a terminating or repeating decimal expansion. It is totally reasonable to tell students that π≈3.14, but be clear that this is an approximation, not an equality. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 25, 2024 at 13:46 Xander Henderson♦Xander Henderson 9,32211 gold badge2525 silver badges6363 bronze badges 2 2 "in the real world, measuring the radius of a circle is a bit of a pain" ─ true, but also constructing a circle of a given radius is easier than constructing a circle of a given diameter. So which formula for computing the circumference is more practical, depends on if it is a circle you found (and must measure), vs. a circle you constructed. kaya3 – kaya3 2024-09-26 17:42:34 +00:00 Commented Sep 26, 2024 at 17:42 @kaya3 Sure... but I don't think that I made a definitive statement. I think what I said is that measuring the diameter is often more practical (not that it is always more practical). In the "real world", we really want both formulae, which is why they both exist. Xander Henderson – Xander Henderson ♦ 2024-09-26 18:08:43 +00:00 Commented Sep 26, 2024 at 18:08 Add a comment \| This answer is useful 18 Save this answer. Show activity on this post. You could say: τr. Circles are defined geometrically by their radius (note it has precedence before diameter in the study guide), so a lot of formulas would be conceptually more meaningful, and often simpler, if the radius was in the picture. On the other hand, real-world circles are arguably easier to measure via the rim-to-rim diameter. As a result, we've gotten in the historically unfortunate habit of shifting a factor of 2 from the circle constant to the circle measure. A lot of people now note that many formulas would be simplified if the circle constant used was τ=2π. E.g., that's the number of radians in one full circle (whereas π only gets you halfway around). And then like many other formulas, here you're looking at a factor of r, which is more fundamental than the diameter in the definition of a circle (e.g., as in the area formula immediately prior in the given study guide). This larger story is probably why the study guide prefers to see r in the formula. It's part of other circle-based formulas (as opposed to diameter), so students should be most familiar with it. See more at: The Tau Manifesto. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Sep 25, 2024 at 2:52 answered Sep 24, 2024 at 18:08 Daniel R. CollinsDaniel R. Collins 28.3k8181 silver badges132132 bronze badges 2 Seems unfortunate that the constant with a larger value is denoted by a letter lacking a leg. I think ᲅ, or even better ന്ന or 𦉫 to properly reflect the doubling, would be more suitable. Ruslan – Ruslan 2024-09-27 19:00:43 +00:00 Commented Sep 27, 2024 at 19:00 1 @Ruslan: Helps to think of it as a denominator. Tau is 1 circle, pi is 12 circle. Daniel R. Collins – Daniel R. Collins 2024-09-27 19:18:30 +00:00 Commented Sep 27, 2024 at 19:18 Add a comment \| This answer is useful 7 Save this answer. Show activity on this post. It is a matter of taste. But there are some reasons for this preference. It is easiest to use either the radius or the diameter and to stick with it. There is less conversion, so less mental load and less room for error. So given that we absolutely want this to be the case and we also only consider the formulas for area and circumference, we would have to choose between A=πr2 and C=2πr or A=14πD2 and C=πD. Just from these formulas I would not say there is a clear winner. How about other dimensions? We can do the same comparison for higher dimensions and compare which expressions look nicer. The formulas using r behave better when the number of dimensions get larger. When using the diameter in 6D, one of the coefficients is 384, yuck! When you start doing calculus, at some point you will likely calculate the area of a circle using integration. For that you will use some form of the formula x2+y2≤r2. Not only is the radius a clear winner here in terms of computation, conceptually it is also easier to define it in terms of radius. The formula x2+y2≤r2 is also used a lot in programming. So to summarise, the decision is kind of arbitrary. But some considerations that come from more "advanced" uses, combined with a fair bit of historical bagage, manage to tip the scale towards the radius in favor of the diameter. There are some things to be said though for the diameter. First of all, π is often defined as the ratio between the circumference of the circle and its diameter, i.e. π=C/D. Also, in some areas in physics (for example soft matter) the size of spheres is defined using the diameter instead of the radius. For those interested, I used the following Mathematica code to create the tables: ``` TableForm[ Table[{ToString[d] <> "D", (2Pi^(d/2)r^(-1 + d))/ Gamma[d/2], (Pi^(d/2)r^d)/Gamma[1 + d/2]}, {d, 1, 10}], TableHeadings -> {None, {"Dim.", "Area", "Volume"}}] ``` Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 26, 2024 at 9:09 AccidentalTaylorExpansionAccidentalTaylorExpansion 34611 silver badge44 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Apologies for the commenty-answer, but have not seen this aspect raised yet: In no way, wanting to cut you off from wondering about math or the like. But you should concentrate on teaching the standard methods, content, notation, etc. Especially as an "aide". But even if you were a head of the NYC public schools or the like. There are plenty of more important issues like how to teach fractions or keep control in the classroom. (And these are non-trivial!) Of course, every question can't (and shouldn't) be on the highest priority topic. But I just want to warn you. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 26, 2024 at 19:38 guest trollguest troll 3111 bronze badge Add a comment \| This answer is useful -1 Save this answer. Show activity on this post. Given an arc-length of l which represents a proportion, p of the circle, the circumference is lp. For example, if an arc-length is 1 and that represents 13 of the circle, then the circumference is 3. Some may scoff at division being "simpler," but with just 2 terms, no constants to remember, and no π:τ debates, lp is hard to ignore. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 25, 2024 at 6:17 user121330user121330 10922 bronze badges 2 1 But if you have the means to directly measure an arc-length and know exactly what fraction of the circle it is, then you have the means to directly measure the circumference and not need this formula at all. :P Idran – Idran 2024-09-25 15:06:41 +00:00 Commented Sep 25, 2024 at 15:06 @Idran I must have misread the title of question! user121330 – user121330 2024-09-25 16:44:02 +00:00 Commented Sep 25, 2024 at 16:44 Add a comment \| This answer is useful -2 Save this answer. Show activity on this post. Well, the fun fact is, diameter does not "exist" in math. All formulas are defined in terms of radius r. The only time you will see diameter d mentioned is in the formula introducing concept of Pi (ratio of circumference to diameter) Another fun fact is that you will see that "2 Pi" pop up in majority of formulas. Truly, it would be more convenient to have a constant of value 2 Pi (). ()Which we actually have, look for Tau Manifesto. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Sep 26, 2024 at 13:57 ThomasThomas 1 8 4 I very much disagree. The diameter of a set exists. For a circle, the diameter is exactly twice the radius. Otherwise, the diameter of A (where A is some subset of a metric space) is sup{d(x,y):x,y∈A} (i.e. it is the greatest distance between any two points in the space). This notion of diameter is extremely important in some contexts, e.g. in the definition of the Hausdorff measure. There are also interesting questions about the diameter, e.g. "are there any non-circular shapes with constant diameter?" en.wikipedia.org/wiki/Curve_of_constant_width Xander Henderson – Xander Henderson ♦ 2024-09-26 18:12:09 +00:00 Commented Sep 26, 2024 at 18:12 1 You are moving the goalposts. In your answer, you assert that "diameter does not 'exist' in math". When I demonstrated that the diameter manifestly does "exist" in mathematics, you change the argument. Now you are asserting that (1) it only "exists" once in pre-calculus, and that (2) the majority of people will won't need to know anything about the diameter. Xander Henderson – Xander Henderson ♦ 2024-09-30 13:39:50 +00:00 Commented Sep 30, 2024 at 13:39 1 Regarding (1), the diameter shows up a lot in high-school geometry. There are some interesting theorems related to the diameter, e.g. if the endpoints of a diameter are joined to any other point on the circle, the triangle produced is a right triangle (Thales' Theorem). Note that this is one example---I am not giving me because space in comments is limited. Xander Henderson – Xander Henderson ♦ 2024-09-30 13:43:10 +00:00 Commented Sep 30, 2024 at 13:43 1 (2) From a practical point of view, the diameter is actually more commonly used outside of mathematics. Formulae related to the diameter are more helpful, as the tool used to measure circles in the real world is typically a set of calipers (e.g. for measuring pipe). Xander Henderson – Xander Henderson ♦ 2024-09-30 13:45:16 +00:00 Commented Sep 30, 2024 at 13:45 1 Finally, you seem to argue that we should use some other word, instead, e.g. "width". The problem with that is that "diameter" (like "circumference", which is really just the perimeter) is already used all over the place, and part of teaching students is teaching them to recognize the language being used. "Diameter" is used enough in the "real world" that it does a disservice to students to pretend that it doesn't exist. Xander Henderson – Xander Henderson ♦ 2024-09-30 13:46:42 +00:00 Commented Sep 30, 2024 at 13:46 \| Show 3 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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6171	https://www.geeksforgeeks.org/c/modulus-on-negative-numbers/	Modulus on Negative Numbers Last Updated : 23 Jul, 2025 Suggest changes 78 Likes The modulus operator, denoted as %, returns the remainder when one number (the dividend) is divided by another number (the divisor). Modulus of Positive Numbers Problem: What is 7 mod 5? Solution: From Quotient Remainder Theorem, Dividend=DivisorQuotient + Remainder 7 = 51 + 2, which gives 2 as the remainder. Thus, to find 7 mod 5, find the largest number that is less than or equal to 7 and divisible by 5. We get 5 as that number and in order to get the remainder we subtract 5 from 7, which gives 2 that is the remainder. This gives the definition of the remainder, A remainder isleast positive integer that should be subtracted from a to make it divisible by b (mathematically if, a = qb + r then 0 ? r < \|b\|), where a is dividend, b is divisor, q is quotient and r is remainder. Modulus of Negative Numbers Problem: What is -7 mod 5? Solution: To find -7 mod 5, we firstly find the largest number that is less than or equal to -7 and divisible by 5. That number is -10. Now, in order to get remainder subtract -10 from -7, which gives 3 that is the remainder. Remainder can also be computed from the above definition. Since, remainder isleast positive integer that should be subtracted from dividend to make it divisible by divisor. Thus the least positive integer that should be subtracted from -7 to make it divisible by 5 is 3. Hence the remainder comes out to be 3. Modulus Operation in Different Programming Languages: In programming languages, the sign of remainder depends upon sign of divisor and dividend. Let's consider the following cases: 1. When Dividend is negative and Divisor is positive Let's see the result of -7 mod 5 in different programming languages: C++ ```` include using namespace std; int main() { int a = -7, b = 5; cout << a % b; return 0; } ```` ``` include #include ``` using namespace std; using namespace std int main() int main { int a = -7, b = 5; int a = - 7 b = 5 cout << a % b; cout<< a% b return 0; return 0 } C ```` include int main() { int a = -7, b = 5; printf("%d", a % b); return 0; } ```` Java ```` /package whatever //do not write package name here / import java.io.; class GFG { public static void main(String[] args) { int a = -7; int b = 5; System.out.println(a % b ); } } ```` Python ```` a = -7 b = 5 print(a % b) ```` C# ```` using System; public class GFG { static public void Main() { int a = -7; int b = 5; Console.WriteLine(a % b); } } ```` JavaScript ```` var a = -7; var b = 5; document.write(a % b); ```` Output -2 Note: The python program gives 3 as the remainder, meanwhile the other programming languages (C/C++) gives -2 as the remainder of -7 mod 5. The reason behind this is Python uses floored division to find modulus. As we know that Remainder = Dividend – (Divisor Quotient) and Quotient can be computed from Dividend and Divisor. To find the quotient there are two methods, which determine the sign of the remainder. Floored division: In this method, Quotient is determined by the floor division of Dividend and Divisor. It rounds off the quotient to the nearest smallest integer. So remainder can be computed from the following expression: r = a - b floor(a/b), where r, a, b are Remainder, Dividend and Divisor respectively. Example: => -7 % 5 = -7 - 5 ( floor(-7/5) ) = -7 - 5( floor(-1.4) ) = -7 - 5( -2) = -7+10 = 3 In Python, floor division is used to determine the remainder. Truncated division: In this method, Quotient is determined by the truncate divison of Dividend and Divisor. It rounds off the quotient towards zero. So remainder can be computed from the following expression: r = a - b trunc(a/b), where r, a, b are Remainder, Dividend and Divisor respectively. Example: => -7 % 5 = -7 - 5 ( trunc(-7/5) ) = -7 - 5( trunc(-1.4) ) = -7 - 5( -1) = -7+5 = -2 In C/C++, truncate division is used to determine the remainder Thus, In programming languages which uses truncate divison to find the remainder, we always find remainder as (a%b + b)%b (add quotient to remainder and again take remainder) to avoid negative remainder and get the correct result. Below is the correct implementation to find remainder of negative number: C++ ```` include using namespace std; int main() { int a = -7, b = 5; cout << ((a % b) + b) % b; return 0; } ```` ``` include #include ``` using namespace std; using namespace std int main() int main { int a = -7, b = 5; int a = - 7 b = 5 cout << ((a % b) + b) % b; cout<< a% b + b% b return 0; return 0 } C ```` include int main() { int a = -7, b = 5; printf("%d", ((a % b) + b) % b); return 0; } ```` Output 3 2. When Divisor is negative and Dividend is positive: Let's see the result of 7 mod -5 in different programming languages: C++ ```` include using namespace std; int main() { int a = 7, b = -5; cout << a % b; return 0; } ```` ``` include #include ``` using namespace std; using namespace std int main() int main { int a = 7, b = -5; int a = 7 b = - 5 cout << a % b; cout<< a% b return 0; return 0 } C ```` include int main() { // a positive and b negative. int a = 7, b = -5; printf("%d", a % b); return 0; } ```` Java ```` /package whatever //do not write package name here / import java.io.; class GFG { public static void main(String[] args) { int a = 7; int b = -5; System.out.println(a % b); } } // This code is contributed by ksrikanth0498. ```` Python ```` a = 7 b = -5 print(a % b) ```` C# ```` using System; public class GFG { static public void Main() { int a = 7; int b = -5; Console.WriteLine(a % b); } } // This code is contributed by sarajadhav12052009 ```` JavaScript ```` var a = 7; var b = -5; document.write(a % b ); // This code is contributed by ksrikanth0498. ```` Output 2 Floored division: The sign of remainder is negative using floor division to determine remainder when only divisor is negative. => 7 % -5 = 7 - (-5) ( floor(-7/5) ) = 7 + 5( floor(-1.4) ) = 7 +5( -2) = 7-10 = -3 Thus, In above code python code gives -3 as remainder because it uses floored division to find modulus. Truncated division: The sign of remainder is positive using truncate division to determine remainder when only divisor is negative. => 7 % -5 = 7 - (-5) ( trunc(-7/5) ) = 7 + 5( trunc(-1.4) ) = 7 + 5( -1) = 7 - 5 = 2 Thus, the C/C++ code gives 2 as remainder because it uses truncate divison to find modulus. 3. When Divisor is negative and Dividend is negative: Let's see the result of -7 mod -5 in different programming languages: C++ ```` include using namespace std; int main() { // a and b both negative int a = -7, b = -5; cout << a % b; return 0; } ```` ``` include #include ``` using namespace std; using namespace std int main() {int main // a and b both negative // a and b both negative int a = -7, b = -5; int a = - 7 b = - 5 cout << a % b; cout<< a% b return 0; return 0 } C ```` include int main() { // a and b both negative int a = -7, b = -5; printf("%d", a % b); return 0; } ```` Java ```` /package whatever //do not write package name here / import java.io.; class GFG { public static void main (String[] args) { int a=-7,b=-5; System.out.println(a%b); } } // This code is contributed by ksrikanth0498. ```` Python ```` if name == 'main': a = -7 b = -5 print(a % b) ```` C# ```` using System; public class GFG{ static public void Main () { int a = -7; int b = -5; Console.WriteLine(a % b); } } // This code is contributed by sarajadhav12052009 ```` JavaScript ```` var a = -7; var b = -5; document.write(a % b ); // This code is contributed by laxmigangarajula03. ```` Output -2 Floored division: The sign of remainder is negative using floor division to determine remainder when both divisor and dividend is negative. => -7 % -5 = -7 - (-5) ( floor(-7/-5) ) = -7 + 5( floor(1.4) ) = -7 +5( 1) = -7+5 = -2 Truncated division: The sign of remainder is negative using truncate division to determine remainder when both divisor and dividend is negative. => 7 % -5 = 7 - (-5) ( trunc(-7/-5) ) = 7 + 5( trunc(1.4) ) = 7 + 5( 1) = -7 + 5 = -2 Related Articles Modulo Operator (%) in C/C++ with Examples K kartik Improve Article Tags : C Language C-Operators Explore C Programming Language Tutorial 4 min read C Basics C Language Introduction 6 min readFeatures of C Programming Language 3 min readC Programming Language Standard 6 min readC Hello World Program 1 min readCompiling a C Program: Behind the Scenes 4 min readC Comments 3 min readTokens in C 4 min readKeywords in C 2 min read C Variables and Constants C Variables 4 min readConstants in C 4 min readConst Qualifier in C 6 min readDifferent ways to declare variable as constant in C 2 min readScope rules in C 5 min readInternal Linkage and External Linkage in C 4 min readGlobal Variables in C 3 min read C Data Types Data Types in C 5 min readLiterals in C 4 min readEscape Sequence in C 5 min readbool in C 5 min readInteger Promotions in C 2 min readCharacter Arithmetic in C 2 min readType Conversion in C 4 min read C Input/Output Basic Input and Output in C 4 min readFormat Specifiers in C 5 min readprintf in C 5 min readscanf in C 3 min readScansets in C 2 min readFormatted and Unformatted Input/Output in C 6 min read C Operators Operators in C 11 min readArithmetic Operators in C 5 min readUnary Operators in C 5 min readRelational Operators in C 4 min readBitwise Operators in C 6 min readC Logical Operators 4 min readAssignment Operators in C 4 min readIncrement and Decrement Operators in C 4 min readConditional or Ternary Operator (?:) in C 3 min readsizeof operator in C 3 min readOperator Precedence and Associativity in C 7 min read C Control Statements Decision-Making Decision Making in C (if , if..else, Nested if, if-else-if ) 7 min readC - if Statement 4 min readC if else Statement 3 min readC if , else if ladder 4 min readSwitch Statement in C 5 min readUsing Range in switch Case in C 2 min readC - Loops 6 min readC for Loop 4 min readwhile Loop in C 5 min readdo...while Loop in C 4 min readFor vs. While 4 min readContinue Statement in C 4 min readBreak Statement in C 5 min readgoto Statement in C 4 min read C Functions C Functions 6 min readUser-Defined Function in C 6 min readParameter Passing Techniques in C 3 min readFunction Prototype in C 4 min readHow can I return multiple values from a function? 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6172	https://www.flinnsci.com/api/library/Download/7ee0bddcb4334e6aa1c61cd0f4b91de7?srsltid=AfmBOop2-2KyZZqSCp9nCVYQp1yXqvrLQ5h3R-2yYlZo_oE4rkI5Q7_A	Pressure vs. Volume and Boyle’s Law Boyle’s Law Introduction In 1642 Evangelista Torricelli, who had worked as an assistant to Galileo, conducted a famous experiment demonstrating that the weight of air would support a column of mercury about 30 inches high in an inverted tube. Torricelli’s experiment provided the first measurement of the invisible pressure of air. Robert Boyle, a “skeptical chemist” working in England, was inspired by Torricelli’s experiment to measure the pressure of air when it was compressed or expanded. The results of Boyle’s experiments were published in 1662 and became essentially the first gas law—a mathematical equation describing the relationship between the volume and pressure of air. What is Boyle’s law and how can it be demonstrated? Concepts • Gas properties • Pressure • Boyle’s law • Kinetic-molecular theory Background Robert Boyle built a simple apparatus to measure the relationship between the pressure and volume of air. The apparatus consisted of a J-shaped glass tube that was sealed at one end and open to the atmosphere at the other end. A sample of air was trapped in the sealed end by pouring mercury into the tube (see Figure 1). In the beginning of the experiment, the height of the mercury column was equal in the two sides of the tube. The pressure of the air trapped in the sealed end was equal to that of the surrounding air and equivalent to 29.9 inches (760 mm) of mercury. When Boyle added more mercury to the open end of the tube, the air trapped in the sealed end was com - pressed into a smaller volume (see Figure 2). The difference in height of the two columns of mercury ( ∆h in Figure 2) was due to the additional pressure exerted by the compressed air compared to the surrounding air. Boyle found that when the volume of trapped air was reduced to one-half its original volume, the additional height of the column of mercury in the open end of the tube measured 29.9 inches. The pressure exerted by the compressed air was twice as great as atmospheric pressure. The mathematical relationship between the volume of the air and the pressure it exerts was confirmed through a series of measurements. Experiment Overview The purpose of this experiment is to perform a modern version of Boyle’s classic experiment without the hazards associated with the use of mercury. The experiment will be carried out using air trapped inside a sealed syringe within a “pressure bottle.” The bottle will be pressurized by pumping in air to obtain a pressure several times greater than that of the sur - rounding air. As some of the excess pressure within the bottle is then released, the volume of the trapped air inside the syringe will change. Volume measurements will be made at several different pressures and the results will be analyzed by graphing to derive the mathematical relationship between pressure and volume. © 2016 Flinn Scientific, Inc. All Rights Reserved. 1Publication No. 91651 061616 S C I E N T I F I C Sealed end V2=1/2V1Open end Mercury (Hg) Trapped air (V 1) ∆h∆h = 29.9 in. Hg Figure 1. Figure 2. Pressure vs. Volume and Boyle’s Law continued 2© 2016 Flinn Scientific, Inc. All Rights Reserved. Pre-Lab Questions According to our modern understanding of the gas laws, there are four measurable properties (variables) of a gas. These variables are P (pressure), V (volume), T (temperature), and n (number of moles). In Boyle’s experiment, which two vari - ables were held constant? Fill in the blanks to summarize the relationship among the gas properties in Boyle’s experiment: For a fixed _____ of gas at constant _____, the _____ of a gas increases as the ________ of its container decreases. Pressure is defined in physics as force divided by area (P = force/area). According to the kinetic-molecular theory, the par - ticles in a gas are constantly moving and colliding with the walls of their container. The pressure of the gas is related to the total force exerted by the individual collisions. Use the kinetic theory to explain the results of Boyle’s experiment. The pressure scale on a tire gauge is marked in units of pounds per square inch (psi). The scale starts at zero when the gauge is exposed to the surrounding air. This means that the total pressure is equal to the gauge pressure plus the pressure of the surrounding air. Standard atmospheric pressure (1 atm) is equal to 14.7 psi. Assume that you have just inflated the tire on your bicycle to 82 psi using a bicycle pump. What is the total pressure of air in the tire in psi? In atmospheres? Materials Bicycle pump with pressure gauge, or electric air pump Syringe, 10-mL, with syringe tip cap Graph paper, 2 sheets Barometer (optional) Petroleum jelly, small bead Tire gauge (optional) Pressure bottle, 1-L, with tire valve Design of the Pressure Bottle The “pressure bottle” is a 1-L PETE (polyethylene terephthalate) soda bottle. The bottle cap has been fitted with a tire valve to give an airtight seal (see Figure 3). Pumping air into the bottle using an ordinary bicycle pump makes it possible to pressurize the bottle above atmospheric pressure. The bottle retains its volume when it is pressurized—any expansion is negligible. The plastic used to make these bottles will withstand pressures up to about 100 psi. Safety Precautions The pressure bottle is safe if used properly. The bottle should not be inflated above 100 psi. Even if the bottle should “pop ,” the plastic construction will only result in a quick release of air, an accompanying loud noise, and a hole in the bottle. The bottle will split but will not shatter. Wear eye protection (safety glasses or chemical splash goggles) when working with the pressure bottle. Procedure Using a barometer, measure the value of the local air pressure. Note: If a barometer is not available, consult an Internet site such as the national weather service site ( to obtain a current pressure reading for your area. Record the barometric pressure in the Data Table. Obtain a 1-L pressure bottle and a 10-mL syringe with a rubber tip cap. Remove the tip cap from the syringe and pull on the plunger to draw about 9 mL of air into the syringe. Replace the tip cap to seal the air inside the syringe. Place the sealed syringe inside the 1-L pressure bottle. Run a small bead of petroleum jelly around the rim of the bottle. Close the bottle with the special cap fitted with a tire valve. Tighten the cap securely. Connect the tire valve to a bicycle pump or an electric air pump. Note: Exercise caution if using an electric air pump. Do Figure 3. Pressure vs. Volume and Boyle’s Law continued 3© 2016 Flinn Scientific, Inc. All Rights Reserved. not exceed the maximum suggested pressure of 50–60 psi. 8. Pump air into the pressure bottle to obtain a pressure read - ing of 50–60 psi on the tire gauge. Do NOT exceed 100 psi. Note: Using a manual pump provides its own safety feature— it is very difficult to pump more than about 70 psi into the pressure bottle by hand. Loosen the connection between the pressure bottle–tire valve and the pump to release a small amount of pressure. As soon as you see the syringe plunger start to move, immediately retighten the tire valve to the pump. Using the pump gauge, measure and record the pressure to within ±1 psi. Measure and record the volume of air trapped in the syringe at this bottle pressure. Note: Measure the volume at the black rubber seal, not at the inverted V-shaped projection (see Figure 4). The syringe barrel has major scale divisions marked every milliliter, and minor scale divisions every 0.2 mL. The volume should be estimated to within ±0.1 mL. Loosen the connection between the pressure bottle-tire valve and the bicycle pump to release a small amount of pressure from the pressure bot - tle. Try to reduce the pressure by no more than about 10 psi. Immediately retighten the tire valve to the pump. Measure both the new pressure on the pump gauge and the new volume of the air trapped inside the syringe. Record all data in the Data Table. Note: If you are using a tire gauge to measure pressure, press lightly on the brass pin in the tire valve to release some air pressure. It may be necessary to bleed off enough air initially to get the first pressure reading below 50 psi, which is the scale maximum on many tire gauges. Repeat steps 10 and 11 to measure the volume of gas trapped in the syringe at several different pressures down to about 5 psi. It should be possible to obtain at least 5–6 pressure and volume measurements in this range. When the pressure on the tire gauge measures close to zero, remove the tire valve from the pump. Press down on the brass pin inside the tire valve to release all of the excess pressure within the pressure bottle. Record the final volume of air in the syringe at atmospheric pressure. If time permits, repeat steps 6–13 to obtain a second, independent set of pressure–volume data. Record this data as Trial 2 in the Data Table. Estimated volume = 2.3 mL Tip cap Syringe barrel Plunger Figure 4. Pressure vs. Volume and Boyle’s Law continued 4© 2016 Flinn Scientific, Inc. All Rights Reserved. Pressure vs. Volume and Boyle’s Law Data and Results Table See Post-Lab Question #2. †See Post-Lab Question #5. Post-Lab Questions Convert the local barometric pressure to psi units and enter the value to the nearest psi in the Data and Results Table. Some appropriate conversion factors are shown below. 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi The pressure gauge measures the relative pressure in psi above atmospheric pressure. For each pressure reading in the Data and Results Table, add the local barometric pressure to the gauge pressure to determine the total pressure of air inside the pressure bottle. Enter the total pressure to the nearest psi in the table. Plot a graph of volume on the y-axis versus total pressure on the x-axis. Note: The origin of the graph should be (0,0). Choose a suitable scale for each axis so that the data points fill the graph as completely as possible. Remember to label each axis and give the graph a title. Barometric Pressure Trial 1 Trial 2 Gauge Volume of Air Total Gauge Volume of Air Total Pressure in Syringe Pressure 1/V† Pressure in Syringe Pressure 1/V† Pressure vs. Volume and Boyle’s Law continued 5© 2016 Flinn Scientific, Inc. All Rights Reserved. Describe the shape of the graph. Draw a best-fit straight or curved line, whichever seems appropriate, to illustrate how the volume of a gas changes as the pressure changes. The relationship between pressure and volume is called an “inverse” relationship—as the pressure increases the volume of air trapped in the syringe decreases. This inverse relationship may be expressed mathematically as P ∝ 1/V. Calculate the value of 1/V for each volume measurement and enter the results in the Data and Results Table. Plot a graph of pressure on the y-axis versus 1/V on the x-axis and draw a best-fit straight line through the data. Note: The origin of the graph should be (0,0). Choose a suitable scale for each axis so that the data points fill the graph as completely as possible. Another way of expressing an inverse relationship between two variables (P ∝ 1/V) is to say that the product of the two variables is a constant (P × V = constant). Multiply the total pressure (P) times the volume (V) for each set of data points. Construct a Results Table to summarize the P × V values. Calculate the average value of the P × V “constant” and the average deviation. What is the relative percent error (uncer - tainty) in this constant? Relative percent error = (Average deviation/Average value) × 100% At constant temperature, the pressure of a gas is proportional to the concentration of gas particles in the container. When some of the pressure was released from the bottle, the syringe plunger moved up. Why did this happen? Use diagrams and explain in words what happens to the gas particles moving around both inside and outside the syringe before and after the pressure is released. (Optional) Research the properties of PETE on the Internet. What characteristics of PETE make it an ideal plastic for use in soda bottles? Pressure vs. Volume and Boyle’s Law continued 6© 2016 Flinn Scientific, Inc. All Rights Reserved. Teacher’s Notes Pressure vs. Volume and Boyle’s Law Disposal The pressure bottles and syringes are reusable—no disposal required. Connecting to the National Standards This laboratory activity relates to the following National Science Education Standards (1996): Unifying Concepts and Processes: Grades K–12 Constancy, change, and measurement Content Standards: Grades 9–12 Content Standard A: Science as Inquiry Content Standard B: Physical Science, structure and properties of matter, motions and forces Lab Hints • The laboratory work for this experiment can reasonably be completed in about 20 minutes. This should allow ample time in a normal 50-minute class period for students to share equipment, if student groups of two are preferred. • Pressure bottles may be prepared using readily available 2-L soda bottles. See the Supplementary Information section for instructions on how to prepare home-made pressure bottles. Cut or saw off the flanges on the top of the syringes. Otherwise, the syringes will not fit inside the pressure bottles. Flinn Scientific sells a pre-made pressure bottle (Catalog No. AP5930) equipped with the attached tire valve and syringe with tip cap. • If the bottle leaks air around the cap when pressurized, remove the cap assembly and put additional petroleum jelly around the inside seal of the cap. Vaseline ® may also be used in place of the petroleum jelly. • Best results are obtained using a bicycle pump with an attached pressure gauge. If enough bicycle pumps cannot be obtained with student help, the experiment may be done using common automobile tire gauges. The teacher should prepare the syringe/bottle assemblies and pressurize the bottles ahead of time. Since the scale maximum on many tire gauges is 50 psi, students may have to bleed off enough air initially to get the pressure below 50 psi. The units shown on some pressure gauges may be psig (gauge pressure per square inch. The correct units for the “total pressures” calculated in the Data and Results Table are psia (pounds per square inch absolute) rather than psi. • For best results, use a barometer to measure the local barometric pressure. The national weather service site reports corrected, sea-level air pressures. Note that these are not actual barometric pressure readings. Meteorologists convert station pressure values to what they would be if they had been taken at sea level. The following equation can be used to recalculate the barometric pressure (in inches Hg) from the reported sea-level pressure (in inches Hg). Elevation must be in meters. barometric pressure = sea-level pressure – (elevation/312 m) On many syringes, the black rubber seals have two “seal lines.” Make sure students are consistent in where they measure the volume! A small amount of pressure is released in step 7 before volume measurements are made—this is to overcome the friction between the rubber seal and the syringe barrel. • Tire gauges come in many shapes and sizes. The best gauges for this experiment are those with an attached pressure dial or digital readout rather than a “pop-out” sliding scale. • Students may need to be instructed in the use of a tire gauge. Students usually are too cautious—they tend to press the tire gauge softly against the valve, which allows air to escape and does not give an accurate pressure reading. The best advice is to work quickly and deliberately. There is an obvious value for students in learning to use a tire gauge so that they can inflate their automobile tires to the proper pressure. Maintaining proper tire pressure improves safety, tire wear, and gas mileage. Pressure vs. Volume and Boyle’s Law continued 7© 2016 Flinn Scientific, Inc. All Rights Reserved. • Try not to exceed the maximum pressure recommended in the Procedure section. It was found that the graph of P versus 1/V became nonlinear as the pressure bottle was pressurized above about 70 psi (total pressure 85 psi). This can be used as a teaching point—deviations from ideal gas behavior are more important at higher pressures. Many textbooks show graphs of real versus ideal gas behavior as a function of pressure. For many gases, deviations from ideal behavior become significant at pressures greater than about 200 atm. Even at modest pressures, however, small deviations are common in the P × V “constant.” Some of this deviation may also be due to a change in temperature. Compressing the gas will increase the temperature of the gas. • This lab provides excellent data and is a great way to introduce the use of computer spreadsheet or graphical analysis programs. Using these programs, it is possible with just a “click of the mouse” to draw best-fit, straight, or curved lines (trendlines) through data and obtain regression equations. High school students are not expected to do the math involved in generating a best-fit straight line (linear regression). It is worthwhile, however, for students to learn how this important statistical tool is used to evaluate the reliability of results. • See the Supplementary Information section for an example of how the data from this experiment can be analyzed to extrapolate the value of the atmospheric pressure in units of psi. Teaching Tips • What does a pressure of 14.7 psi feel like? The “Atmosphere Bar” available from Flinn Scientific (Catalog No. AP5882) is a 52-inch steel bar that weighs 14.7 lbs. The base of the bar is one inch square. The resulting pressure of 14.7 pounds per square inch certainly feels impressive! • If necessary, review the definition of the average deviation with your students. The average deviation is obtained by finding the difference between each individual value (X i) and the average value (X –), taking the sum ( Σ) of their absolute values, and dividing by the number of measurements. n Σ \| x i – x– \| i = 1 average deviation = ————— n • Does Boyle’s law depend on the nature of the gas? The experiment can be extended to test the behavior of gases other than air by filling the syringes with pure gases such as helium, hydrogen, and nitrogen. Any source of gas may be used. Contact Flinn Scientific to obtain a complimentary copy of our ChemFax No. 10527, “Construction of a Simple Gas Delivery Apparatus.” • The approaches shown here may be conceptually challenging for students but they lead to some interesting discussions. Are negative pressures possible? Why or why not? • Boyle’s law can also be demonstrated with every breath you take! When you inhale, your diaphragm moves down, and the volume of the body cavity surrounding the lungs expands. As the volume increases, the pressure decreases, and the outside air, which is at a higher pressure, rushes into the lungs. The opposite occurs when you exhale—the diaphragm moves upward, the lung cavity contracts, and the pressure increases relative to the outside air, pushing air out of the lungs. See the “Functioning Lung Model” (Catalog No. FB1110) available from Flinn Scientific for a large, demonstration-size model of the pressure–volume relationship involved in breathing. • The great 17th century physicist Isaac Newton was a contemporary of Robert Boyle, and Boyle’s results prompted Newton to propose an explanation for the pressure of a gas. Newton assumed that the particles in a gas were motionless (static) and that the pressure of a gas was due to the mutual repulsion of gas particles. That no less a person than the “father of physics” and the discoverer of the calculus proposed this model might serve as a cautionary tale for all teachers—our modern, obvious explanations for gas behavior may not always be obvious to students. • The foundation of the modern kinetic-molecular theory of gases is usually attributed to Daniel Bernoulli. In 1734 Bernoulli proposed that a gas was composed of “hard” (inelastic) particles that were in constant and random motion. The force exerted by these particles as they collided with the walls of their container was the source of the gas pressure. Using this model, Bernoulli derived a mathematical equation for the pressure of a gas. His mathematical treatment correctly “predicted” the inverse pressure–volume relationship that had been observed empirically by Robert Boyle. The development of the kinetic theory culminated in the mid-1800s with the definition of temperature and the relationship between temperature and kinetic energy. Pressure vs. Volume and Boyle’s Law continued 8© 2016 Flinn Scientific, Inc. All Rights Reserved. Answers to Pre-Lab Questions (Student answers will vary.) According to our modern understanding of the gas laws, there are four measurable properties (variables) of a gas. These variables are P (pressure), V (volume), T (temperature), and n (number of moles). In Boyle’s experiment, which two vari - ables were held constant? Both the temperature (T) and the number of moles of gas (n) were held constant in Boyle’s J-tube experiment. Fill in the blanks to summarize the relationship among the gas properties in Boyle’s experiment: For a fixed number of moles of gas at constant temperature , the pressure of a gas increases as the volume of its container decreases. Pressure is defined in physics as force divided by area (P = force/area). According to the kinetic-molecular theory, the par - ticles in a gas are constantly moving and colliding with the walls of their container. The pressure of the gas is related to the total force exerted by the individual collisions. Use the kinetic theory to explain the results of Boyle’s experiment. In Boyle’s experiment, the pressure of air increased when it was compressed into a smaller volume container. According to the kinetic theory, confining the gas particles in a smaller volume will increase the number of collisions and hence the total force of the collisions with the container walls. (The distance the particles must travel between collisions decreases as the volume is reduced.) The pressure scale on a tire gauge is marked in units of pounds per square inch (psi). The scale starts at zero when the gauge is exposed to the surrounding air. This means that the total pressure is equal to the gauge pressure plus the pressure of the surrounding air. Standard atmospheric pressure (1 atm) is equal to 14.7 psi. Assume that you have just inflated the tire on your bicycle to 82 psi using a bicycle pump. What is the total pressure of air in the tire in psi? In atmospheres? Relative (gauge) pressure = Total pressure – Atmospheric pressure 82 psi = Total pressure – 14.7 psi Total pressure = (82 + 14.7) psi = 97 psi (two significant figures) 1 atm Convert to atm: 97 psi × ———— = 6.6 atm 14.7 psi Sample Data and Results Table (Student data will vary.) See Post-Lab Question #2. †See Post-Lab Question #5. Barometric Pressure 29.9 inches Hg = 15 psi Trial 1 Trial 2 Gauge Volume of Air Total Gauge Volume of Air Total Pressure in Syringe Pressure 1/V† Pressure in Syringe Pressure 1/V† 42 psi 2.0 mL 57 psi 0.50 mL –1 54 psi 1.6 mL 69 psi 0.63 mL –1 35 psi 2.4 mL 50 psi 0.42 mL –1 35 psi 2.4 mL 50 psi 0.42 mL –1 22 psi 3.3 mL 37 psi 0.30 mL –1 23 psi 3.2 mL 38 psi 0.31 mL –1 17 psi 3.8 mL 32 psi 0.26 mL –1 14 psi 4.4 mL 29 psi 0.23 mL –1 11 psi 5.0 mL 26 psi 0.20 mL –1 6 psi 6.2 mL 21 psi 0.16 mL –1 6 psi 6.2 mL 21 psi 0.16 mL –1 3 psi 7.3 mL 18 psi 0.14 mL –1 1 psi 8.4 mL 16 psi 0.12 mL –1 0 psi 9.0 mL 15 psi 0.11 mL –1 Pressure vs. Volume and Boyle’s Law continued 9© 2016 Flinn Scientific, Inc. All Rights Reserved. Answers to Post-Lab Questions (Student answers will vary.) Convert the local barometric pressure to psi units and enter the value to the nearest psi in the Data and Results Table. Some appropriate conversion factors are shown below. 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi 14.7 psi 29.9 in Hg × —————– = 15 psi (rounded to the nearest psi) 29.92 in Hg The pressure gauge measures the relative pressure in psi above atmospheric pressure. For each pressure reading in the Data and Results Table, add the local barometric pressure to the gauge pressure to determine the total pressure of air inside the pressure bottle. Enter the total pressure to the nearest psi in the table. Sample calculation for Trial 1: Total pressure = 42 psi + 15 psi = 57 psi Refer to the Data and Results Table for the results of the other calculations. Plot a graph of volume on the y-axis versus total pressure on the x-axis. Note: The origin of the graph should be (0,0). Choose a suitable scale for each axis so that the data points fill the graph as completely as possible. Remember to label each axis and give the graph a title. 12.0 10.0 8.0 6.0 4.0 2.0 0 Volume of Air, mL Total Pressure, psi Pressure Bottle Experiment: Pressure vs. Volume 20 040 60 80 Describe the shape of the graph. Draw a best-fit straight or curved line, whichever seems appropriate, to illustrate how the volume of a gas changes as the pressure changes. The graph is curved. The volume decreases as the pressure increases. At first, there is a sharp reduction in the volume as the pressure increases. The decrease in volume then becomes more gradual and the volume appears to level off as the pressure increases further. Mathematically, the shape of the curve is described as hyperbolic. A hyperbolic curve of this type is obtained when there is an inverse relationship between two variables (y ∝ 1/x). See the graph for the best-fit curved line through the data. The relationship between pressure and volume is called an “inverse” relationship—as the pressure increases the volume of air trapped in the syringe decreases. This inverse relationship may be expressed mathematically as P ∝ 1/V. Calculate the value of 1/V for each volume measurement and enter the results in the Data and Results Table. Sample calculation for Trial 1: V = 2.0 mL 1/V = 0.50 mL –1 Refer to the Sample Data and Results Table for the results of the other calculations. Pressure vs. Volume and Boyle’s Law continued 10 © 2016 Flinn Scientific, Inc. All Rights Reserved. Plot a graph of pressure on the y-axis versus 1/V on the x-axis and draw a best-fit straight line through the data. Note: The origin of the graph should be (0,0). Choose a suitable scale for each axis so that the data points fill the graph as completely as possible. 100 80 60 40 20 0 Total Pressure, psi Pressure Bottle Experiment: Pressure vs. 1/Volume 0.20 0.00 0.40 0.60 0.80 Another way of expressing an inverse relationship between two variables (P ∝ 1/V) is to say that the product of the two variables is a constant (P × onstant). Multiply the total pressure (P) times the volume (V) for each set of data points. Construct a Results Table to summarize the P × values. Sample Results Table Calculate the average value of the P × V “constant” and the average deviation. What is the relative percent error (uncer - tainty) in this constant? Relative percent error = (Average deviation/Average value) × 100% Average value of the (P × V) constant = 120 psi  mL Average deviation = 7 psi  mL Relative percent error = (7 psi  mL/120 psi  mL) × 100% = 6% It appears that the mathematical product (P × V) is a constant within plus-or-minus 6%. Trial 1 Trial 2 P × V P × V P (psi) V (mL) (psi  mL) P (psi) V (mL) (psi  mL) 57 2.0 110 69 1.6 110 50 2.4 120 50 2.4 120 37 3.3 120 38 3.2 120 32 3.8 120 29 4.4 130 26 5.0 130 21 6.2 130 21 6.2 130 18 7.3 130 16 8.4 130 15 9.0 140 Average Value 120 Average Deviation 7Pressure vs. Volume and Boyle’s Law continued 11 © 2016 Flinn Scientific, Inc. All Rights Reserved. At constant temperature, the pressure of a gas is proportional to the concentration of gas particles in the container. When some of the pressure was released from the pressure bottle, the syringe plunger moved up. Why did this happen? Use dia - grams and explain in words what happens to the gas particles moving around both inside and outside the syringe before and after the pressure is released. Initially, the pressure was the same both inside and outside the syringe. This means there were equal concentrations of gas particles colliding with the inside and outside of the plunger, so the plunger stayed in place (A). When some of the pressure from the bottle was released, some air particles escaped, leaving fewer air particles to collide on the outside of the plunger (B). The concentration of par - ticles inside the syringe overpowered those outside the syringe and pushed the plunger outward (the volume increased). As the volume increased, the particle concentration inside the syringe decreased. When the concentration of particles (pressure) inside the syringe was the same as that outside the syringe, the plunger stopped moving (C). Fout Fin Fout Fin Fout Fin A B C Equal Forces Force in > Force out Equal Forces Initial Conditions Unstable Conditions Final Conditions (Optional) Research the properties of high density PETE on the Internet. What characteristics of PETE make it an ideal plastic for use in soda bottles? PETE is an ideal plastic for soda bottles for several reasons. It is transparent, crystal clear, pure, tough, and unbreakable. PETE also has very low permeability to oxygen, carbon dioxide, and water, and excellent chemical resistance to acids and bases. If the PETE does not contain additives, it is relatively easy to recycle. Supplementary Information Construction of a Pressure Bottle Cut off the top portion of a 2-L soda bottle. This will be used as a base or cap handle to hold the bottle caps that must be drilled. Screw one bottle cap securely onto the bottle top. Using a 1⁄2-inch drill and drill press, drill a hole through the center of the cap, as shown in Figure 5. (Notice you are drilling through the bottom of the cap.) Remove the cap from the bottle top, add rubber cement to the lip of the valve stem, and use pliers to pull the valve stem through the hole from the inside, as shown below. The tire valve will stick out of the top of the bottle cap when it is placed on the soda bottle. Clear away any burrs that may remain in the drilled hole. They will cause the seal to break. Valve stem Pull valve stem with pliers. Cut-off pop bottle holds cap in place. Bottle cap Add rubber cement Figure 5. Pressure vs. Volume and Boyle’s Law continued 12 © 2016 Flinn Scientific, Inc. All Rights Reserved. Graphical Analysis and the Value of the Atmospheric Pressure As observed in Post-Lab Question #6, the graph of total pressure versus 1/V is a straight line that passes through the origin (y-intercept equal to zero). This graph corresponds to an equation of the form P total = m × 1/V, where m is the slope of the line. If instead of the total pressure, students plot the gauge pressure (P gauge = Ptotal – Pair ) versus 1/V, they will obtain a straight line of the form P gauge = (m × 1/V) – Pair (where the y-intercept is equal to –P air ). The value of P air can be estimated by extending the best-fit straight line backwards to “negative” gauge pressures. The best estimate we were able to obtain using this method was about 12 psi (18% error). 60 50 40 30 20 10 0–10 –20 Gauge Pressure, psi Gauge Pressure vs. 1/Volume 00.1 0.2 0.3 0.4 0.5 0.6 0.7 Algebraic Determination of the Value of the Atmospheric Pressure It is also possibe to use the gauge pressure–volume measurements to estimate a value for the atmospheric pressure in psi. This is a nice algebraic challenge for students. According to Boyle’s law, multiplying any two sets of P–V data should give the same result (P 1V1 = P2V2). This is only true if the pressures are total pressures rather than relative gauge pressures (P total = Pgauge + x, where x = value of atmospheric pressure in psi). The following equation is obtained by combining the equations for Boyle’s law and the total pressure: (P gauge, 1 + x)(V 1) = (P gauge, 2 + x)(V 2) Substituting the raw data for 17 psi and 1 psi (see Sample Data on page 11) into this equation gives: (17 psi + x)(3.8 mL) = (1 psi + x)(8.4 mL) 64.6 + 3.8 x = 8.4 + 8.4 x 56.2 = 4.6 x x = 12 psi Reference This experiment has been adapted from Flinn ChemTopic ™ Labs, Volume 9, The Gas Laws, Cesa, I., ed., Flinn Scientific, Batavia, IL, 2003. Pressure vs. Volume and Boyle’s Law continued 13 © 2016 Flinn Scientific, Inc. All Rights Reserved. Flinn Scientific—Teaching Chemistry ™ eLearning Video Series A video of the Pressure vs. Volume and Boyle’s Law activity, presented by Bob Becker, is available in Boyle’s Law, part of the Flinn Scientific—Teaching Chemistry eLearning Video Series. Materials for Pressure vs. Volume and Boyle’s Law are available from Flinn Scientific, Inc. Materials required to perform this activity are available in the Boyle’s Law in a Bottle—Student Laboratory Kit available from Flinn Scientific. Materials may also be purchased separately. Catalog No. Description AP6855 Boyle’s Law in a Bottle—Student Laboratory Kit AP5930 Pressure Bottle AP6884 Bicycle Pump, with Gauge and Release Valve AP1730 Syringe, without Needle, 10 mL AP8958 Syringe Tip Cap, Pkg/10
6173	https://www.shaalaa.com/question-bank-solutions/give-a-balanced-equation-for-thermal-decomposition-of-mercury-ii-oxide_136145	Give a Balanced Equation for Thermal Decomposition of Mercury [Ii] Oxide - Chemistry Advertisements Advertisements Question Give a balanced equation for thermal decomposition of mercury [II] oxide Advertisements SolutionShow Solution Decomposition of Mercury [II] oxide ["HgO"]: [\ce{2HgO->[\Delta]2Hg + O2}] APPEARS IN RELATED QUESTIONS Explain, giving one example of the following chemical changes: Double decomposition (a) Define the neutralization reaction with an example. (b) Give a balanced equation for this reaction. (c) Give three applications of neutralization reactions. Decomposition reactions can occur by sunlight. Give two balanced reactions. State which of the substance given below evolve oxygen gas on thermal decomposition. Lead nitrate Give a balanced equation for a direct combination reaction involving: Two elements – one of which is a neutral gas and the other a yellow non-metal Give a reason why zinc displaces hydrogen from dilute sulphuric acid but copper does not. Complete the statement by filling in the blank with the correct word: Direct combination reaction of sulphur dioxide with water gives ________. Give a balanced equation for the following type of reaction: A soluble salt of lead formed from an insoluble base by double decomposition – neutralization. Give a balanced equation for the following type of reaction: A decomposition reaction of a salt which leaves behind a silvery metal. Chemical reaction in which reactants react to form products and the products formed react with each other directly to form the original reactants back is known as ______ reaction. Select a course
6174	https://www.youtube.com/watch?v=hlN9ChM6w_k	Introducing Chemical Equilibrium \| OpenStax Chemistry 2e 13.1 Michael Evans 29900 subscribers 14 likes Description 1458 views Posted: 22 Jun 2022 00:00 Introduction 01:57 Chapter Outline 04:19 Ocean and Beach Metaphor 05:28 Reversible Reactions and Equilibrium Transcript: Introduction so far in thinking about chemical reactions we've focused on situations where only the forward reaction matters so reactant a for example going to a product b we didn't worry at all about the conversion of b back to the reactant a and this is a very common situation in chemical reactions but it might give us some pause right because if we can just as well write the reaction a going to b well we can write the reaction b going to a and here b is now the reactant and a is the product and it seems like if nature wants to do the conversion of a to b then the conversion of b to a is something that nature quote unquote does not want to do on the other hand there are reactions and reaction systems where the conversion of a to b and the conversion of b to a are both important and these are so-called reversible reactions or reversible reaction systems a big focus of this chapter we're really going to zero in on this situation in this chapter where a reaction doesn't go all the way to products at the state of what's called chemical equilibrium when macroscopic change appears to stop there's a little bit of a and a little bit of b left a little bit of reactant and a little bit of product this is a situation where chemical equilibrium is most important and although it appears that things are static from the macroscopic scale on the molecular level a lot is still happening in a system in chemical equilibrium and we can take advantage of that and harness that and use that idea to make predictions about what's going to happen to a system in equilibrium when we apply a stress to it or when we change conditions somehow so this chapter is all about fundamental concepts of chemical equilibrium and the fundamental conceptual tools and problem solving strategies we use to tackle equilibrium problems Chapter Outline we've got four sections in the first section we're going to introduce the concepts of chemical equilibrium really the underlying conceptual model and this is really really important actually having that right mental picture of a system in chemical equilibrium is going to give you good intuition when you're solving problems about whether for example your answer makes sense numerically does it make sense for a reaction to go forward or backward given conditions for example keep that conceptual foundation from section one in your mind throughout this chapter in section two we're going to introduce a key quantity for the study of chemical equilibrium known as the equilibrium constant this was represented with a capital k in contrast to the lowercase k that we used in the kinetics chapter and it's a measure of how much a reaction wants to go forward quote unquote based on the natural and really the thermodynamic situation associated with the reactants and products in section three we're going to see what happens when we take a system that begins in chemical equilibrium and apply some kind of change to the conditions the so-called stress to the system this is going to cause the reaction system to change in response to the change in conditions using an idea known as le chatelier's principle and le chatelier's principle is really based on the dynamic nature of chemical equilibrium a very important point that we'll return to later finally in section four we'll dig into the math and this is one of the most common areas of struggle and something almost universally taught in introductory chemistry courses how to do calculations related to systems in equilibrium and an idea here in section four that we're going to see very early on is that where we would apply basic stoichiometry in problems where for example we wanted to know how much product we would end up with given an amount of reactant that we start with that becomes impossible in reactions that don't go to completion for reversible reactions to apply directly stoichiometry because the reaction doesn't go all the way so we can't assume that all of the reactants are converted to products we're actually going to loop back to section two and the equilibrium constant and see how that plays into these equilibrium calculations in section four Ocean and Beach Metaphor let's start with the conceptual underpinnings of chemical equilibrium and the text introduces this sort of interesting metaphor of bathers on a beach the idea that as people for example are relaxing on the beach and they get hot they're going to go into the water let's say we've got some water over here at the same time as bathers in the ocean get tired they are going to leave the water and return to the beach so there's some rate of sunbathers going into the water and some rate of ocean goers ocean people coming out of the water onto the beach an equilibrium situation is one in which the two rates of people entering and leaving the water are equal and the key conceptual point here is that there are still people going in and out of the water but if we look at the numbers of people in and out of the water they appear to be unchanging because the rates of going in and out are equal a dynamic situation we'll have another metaphor for this here in a second Reversible Reactions and Equilibrium the key to chemical equilibrium really and why we care about it is the idea of a reversible reaction a reversible chemical reaction proceeds in both the forward and reverse directions the products are not so much more stable than the reactants that the reaction goes all the way to products with no reactant left or vice versa the reactants in fact may be more stable than the product so the reaction hardly goes at all for the most important reversible reactions the rate constants for the forward and reverse reactions are both non-negligible negligible they're on a similar order of magnitude for example and the slide gives two examples of irreversible and reversible reactions the precipitation of silver chloride which is the first reaction you see is an example of an irreversible reaction we can actually tell that from the reaction scheme based on this forward only arrow right here that i've highlighted in blue this indicates that for all practical purposes all of the silver cations and chloride anions in aqueous solution get together and form silver chloride and assuming we've got the stoichiometric ratio right no reactant remains at the end of the reaction a very different situation is shown in the second case hcn in aqueous solution is reacting with water to form cn minus that's the cyanide anion and h3o plus now here the reaction arrow indicates both the forward and reverse directions are taking place and this arrow highlighted in orange is a reversible reaction arrow just like the bathers on the beach reactants are going to products and products are going to reactants in this reaction system specifically in a state of chemical equilibrium the rates of the forward and reverse reactions are equal and this is key to the definition of chemical equilibrium so in general for a reversible reaction system the rates need not be equal right if we start only with hcn and water well clearly only the forward reaction can take place likewise if we start with only cn minus and h3o plus only the reverse reaction can happen key to the state of equilibrium is that the rates of the forward and reverse reactions are equal to each other this leads to a situation where the concentrations of reactants and products appear to be unchanging with time so macroscopically it looks like nothing is happening however on the microscopic level these forward and reverse reactions are always occurring just like bathers going in and out of the ocean
6175	https://opentextbc.ca/introductorybusinessstatistics/chapter/regression-basics-2/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Main Body Chapter 8. Regression Basics Regression analysis, like most multivariate statistics, allows you to infer that there is a relationship between two or more variables. These relationships are seldom exact because there is variation caused by many variables, not just the variables being studied. If you say that students who study more make better grades, you are really hypothesizing that there is a positive relationship between one variable, studying, and another variable, grades. You could then complete your inference and test your hypothesis by gathering a sample of (amount studied, grades) data from some students and use regression to see if the relationship in the sample is strong enough to safely infer that there is a relationship in the population. Notice that even if students who study more make better grades, the relationship in the population would not be perfect; the same amount of studying will not result in the same grades for every student (or for one student every time). Some students are taking harder courses, like chemistry or statistics; some are smarter; some study effectively; and some get lucky and find that the professor has asked them exactly what they understood best. For each level of amount studied, there will be a distribution of grades. If there is a relationship between studying and grades, the location of that distribution of grades will change in an orderly manner as you move from lower to higher levels of studying. Regression analysis is one of the most used and most powerful multivariate statistical techniques for it infers the existence and form of a functional relationship in a population. Once you learn how to use regression, you will be able to estimate the parameters — the slope and intercept — of the function that links two or more variables. With that estimated function, you will be able to infer or forecast things like unit costs, interest rates, or sales over a wide range of conditions. Though the simplest regression techniques seem limited in their applications, statisticians have developed a number of variations on regression that greatly expand the usefulness of the technique. In this chapter, the basics will be discussed. Once again, the t-distribution and F-distribution will be used to test hypotheses. What is regression? Before starting to learn about regression, go back to algebra and review what a function is. The definition of a function can be formal, like the one in my freshman calculus text: “A function is a set of ordered pairs of numbers (x,y) such that to each value of the first variable (x) there corresponds a unique value of the second variable (y)” (Thomas, 1960).. More intuitively, if there is a regular relationship between two variables, there is usually a function that describes the relationship. Functions are written in a number of forms. The most general is y = f(x), which simply says that the value of y depends on the value of x in some regular fashion, though the form of the relationship is not specified. The simplest functional form is the linear function where: α and β are parameters, remaining constant as x and y change. α is the intercept and β is the slope. If the values of α and β are known, you can find the y that goes with any x by putting the x into the equation and solving. There can be functions where one variable depends on the values values of two or more other variables where x1 and x2 together determine the value of y. There can also be non-linear functions, where the value of the dependent variable (y in all of the examples we have used so far) depends on the values of one or more other variables, but the values of the other variables are squared, or taken to some other power or root or multiplied together, before the value of the dependent variable is determined. Regression allows you to estimate directly the parameters in linear functions only, though there are tricks that allow many non-linear functional forms to be estimated indirectly. Regression also allows you to test to see if there is a functional relationship between the variables, by testing the hypothesis that each of the slopes has a value of zero. First, let us consider the simple case of a two-variable function. You believe that y, the dependent variable, is a linear function of x, the independent variable — y depends on x. Collect a sample of (x, y) pairs, and plot them on a set of x, y axes. The basic idea behind regression is to find the equation of the straight line that comes as close as possible to as many of the points as possible. The parameters of the line drawn through the sample are unbiased estimators of the parameters of the line that would come as close as possible to as many of the points as possible in the population, if the population had been gathered and plotted. In keeping with the convention of using Greek letters for population values and Roman letters for sample values, the line drawn through a population is: while the line drawn through a sample is: y = a + bx In most cases, even if the whole population had been gathered, the regression line would not go through every point. Most of the phenomena that business researchers deal with are not perfectly deterministic, so no function will perfectly predict or explain every observation. Imagine that you wanted to study the estimated price for a one-bedroom apartment in Nelson, BC. You decide to estimate the price as a function of its location in relation to downtown. If you collected 12 sample pairs, you would find different apartments located within the same distance from downtown. In other words, you might draw a distribution of prices for apartments located at the same distance from downtown or away from downtown. When you use regression to estimate the parameters of price = f(distance), you are estimating the parameters of the line that connects the mean price at each location. Because the best that can be expected is to predict the mean price for a certain location, researchers often write their regression models with an extra term, the error term, which notes that many of the members of the population of (location, price of apartment) pairs will not have exactly the predicted price because many of the points do not lie directly on the regression line. The error term is usually denoted as ε, or epsilon, and you often see regression equations written: Strictly, the distribution of ε at each location must be normal, and the distributions of ε for all the locations must have the same variance (this is known as homoscedasticity to statisticians). Simple regression and least squares method In estimating the unknown parameters of the population for the regression line, we need to apply a method by which the vertical distances between the yet-to-be estimated regression line and the observed values in our sample are minimized. This minimized distance is called sample error, though it is more commonly referred to as residual and denoted by e.In more mathematical form, the difference between the y and its predicted valueis the residual in each pair of observations for x and y. Obviously, some of these residuals will be positive (above the estimated line) and others will be negative (below the line). If we add all these residuals over the sample size and raise them to the power 2 in order to prevent the chance those positive and negative signs are cancelling each other out, we can write the following criterion for our minimization problem: S is the sum of squares of the residuals. By minimizing S over any given set of observations for x and y, we will get the following useful formula: After computing the value of b from the above formula out of our sample data, and the means of the two series of data on xand y, one can simply recover the intercept of the estimated line using the following equation: For the sample data, and given the estimated intercept and slope, for each observation we can define a residual as: Depending on the estimated values for intercept and slope, we can draw the estimated line along with all sample data in a y–x panel. Such graphs are known as scatter diagrams. Consider our analysis of the price of one-bedroom apartments in Nelson, BC. We would collect data for y=price of one bedroom apartment, x1=its associated distance from downtown, and x2=the size of the apartment, as shown in Table 8.1. Table 8.1 Data for Price, Size, and Distance of Apartments in Nelson, BC \| y = price of apartments in $1000 x1 = distance of each apartment from downtown in kilometres x2 = size of the apartment in square feet \| \| \| \| y \| x1 \| x2 \| \| 55 \| 1.5 \| 350 \| \| 51 \| 3 \| 450 \| \| 60 \| 1.75 \| 300 \| \| 75 \| 1 \| 450 \| \| 55.5 \| 3.1 \| 385 \| \| 49 \| 1.6 \| 210 \| \| 65 \| 2.3 \| 380 \| \| 61.5 \| 2 \| 600 \| \| 55 \| 4 \| 450 \| \| 45 \| 5 \| 325 \| \| 75 \| 0.65 \| 424 \| \| 65 \| 2 \| 285 \| The graph (shown in Figure 8.1) is a scatter plot of the prices of the apartments and their distances from downtown, along with a proposed regression line. In order to plot such a scatter diagram, you can use many available statistical software packages including Excel, SAS, and Minitab. In this scatter diagram, a negative simple regression line has been shown. The estimated equation for this scatter diagram from Excel is: Where a=71.84 and b=-5.38. In other words, for every additional kilometre from downtown an apartment is located, the price of the apartment is estimated to be $5380 cheaper, i.e. 5.38$1000=$5380. One might also be curious about the fitted values out of this estimated model. You can simply plug the actual value for x into the estimated line, and find the fitted values for the prices of the apartments. The residuals for all 12 observations are shown in Figure 8.2. You should also notice that by minimizing errors, you have not eliminated them; rather, this method of least squares only guarantees the best fitted estimated regression line out of the sample data. In the presence of the remaining errors, one should be aware of the fact that there are still other factors that might not have been included in our regression model and are responsible for the fluctuations in the remaining errors. By adding these excluded but relevant factors to the model, we probably expect the remaining error will show less meaningful fluctuations. In determining the price of these apartments, the missing factors may include age of the apartment, size, etc. Because this type of regression model does not include many relevant factors and assumes only a linear relationship, it is known as a simple linear regression model. Testing your regression: does y really depend on x? Understanding that there is a distribution of y (apartment price) values at each x (distance) is the key for understanding how regression results from a sample can be used to test the hypothesis that there is (or is not) a relationship between x and y. When you hypothesize that y = f(x), you hypothesize that the slope of the line (β in y = α + βx + ε) is not equal to zero. If β was equal to zero, changes in x would not cause any change in y. Choosing a sample of apartments, and finding each apartment’s distance to downtown, gives you a sample of (x, y). Finding the equation of the line that best fits the sample will give you a sample intercept, α, and a sample slope, β. These sample statistics are unbiased estimators of the population intercept, α, and slope, β. If another sample of the same size is taken, another sample equation could be generated. If many samples are taken, a sampling distribution of sample β’s, the slopes of the sample lines, will be generated. Statisticians know that this sampling distribution of b’s will be normal with a mean equal to β, the population slope. Because the standard deviation of this sampling distribution is seldom known, statisticians developed a method to estimate it from a single sample. With this estimated sb, a t-statistic for each sample can be computed: where n = sample size m = number of explanatory (x) variables b = sample slope β= population slope sb = estimated standard deviation of b’s, often called the standard error These t’s follow the t-distribution in the tables with n–m-1 df. Computing sb is tedious, and is almost always left to a computer, especially when there is more than one explanatory variable. The estimate is based on how much the sample points vary from the regression line. If the points in the sample are not very close to the sample regression line, it seems reasonable that the population points are also widely scattered around the population regression line and different samples could easily produce lines with quite varied slopes. Though there are other factors involved, in general when the points in the sample are farther from the regression line, sb is greater. Rather than learn how to compute sb, it is more useful for you to learn how to find it on the regression results that you get from statistical software. It is often called the standard error and there is one for each independent variable. The printout in Figure 8.3 is typical. You will need these standard errors in order to test to see if y depends on x or not. You want to test to see if the slope of the line in the population, β, is equal to zero or not. If the slope equals zero, then changes in x do not result in any change in y. Formally, for each independent variable, you will have a test of the hypotheses: If the t-score is large (either negative or positive), then the sample b is far from zero (the hypothesized β), and Ha should be accepted. Substitute zero for b into the t-score equation, and if the t-score is small, b is close enough to zero to accept Ha. To find out what t-value separates “close to zero” from “far from zero”, choose an alpha, find the degrees of freedom, and use a t-table from any textbook, or simply use the interactive Excel template from Chapter 3, which is shown again in Figure 8.4. Figure 8.4 Interactive Excel Template for Determining t-Value from the t-Table – see Appendix 8. Remember to halve alpha when conducting a two-tail test like this. The degrees of freedom equal n – m -1, where n is the size of the sample and m is the number of independent x variables. There is a separate hypothesis test for each independent variable. This means you test to see if y is a function of each x separately. You can also test to see if β > 0 (or β < 0) rather than β ≠ 0 by using a one-tail test, or test to see if β equals a particular value by substituting that value for β when computing the sample t-score. Testing your regression: does this equation really help predict? To test to see if the regression equation really helps, see how much of the error that would be made using the mean of all of the y’s to predict is eliminated by using the regression equation to predict. By testing to see if the regression helps predict, you are testing to see if there is a functional relationship in the population. Imagine that you have found the mean price of the apartments in our sample, and for each apartment, you have made the simple prediction that price of apartment will be equal to the sample mean, y. This is not a very sophisticated prediction technique, but remember that the sample mean is an unbiased estimator of population mean, so on average you will be right. For each apartment, you could compute your error by finding the difference between your prediction (the sample mean, y) and the actual price of an apartment. As an alternative way to predict the price, you can have a computer find the intercept, α, and slope, β, of the sample regression line. Now, you can make another prediction of how much each apartment in the sample may be worth by computing: Once again, you can find the error made for each apartment by finding the difference between the price of apartments predicted using the regression equation ŷ, and the observed price, y. Finally, find how much using the regression improves your prediction by finding the difference between the price predicted using the mean, y, and the price predicted using regression, ŷ. Notice that the measures of these differences could be positive or negative numbers, but that error or improvement implies a positive distance. Coefficient of Determination If you use the sample mean to predict the amount of the price of each apartment, your error is (y–y) for each apartment. Squaring each error so that worries about signs are overcome, and then adding the squared errors together, gives you a measure of the total mistake you make if you want to predict y. Your total mistake is Σ(y–y)2. The total mistake you make using the regression model would be Σ(y-ŷ)2. The difference between the mistakes, a raw measure of how much your prediction has improved, is Σ(ŷ–y)2. To make this raw measure of the improvement meaningful, you need to compare it to one of the two measures of the total mistake. This means that there are two measures of “how good” your regression equation is. One compares the improvement to the mistakes still made with regression. The other compares the improvement to the mistakes that would be made if the mean was used to predict. The first is called an F-score because the sampling distribution of these measures follows the F-distribution seen in Chapter 6, “F-test and One-Way ANOVA”. The second is called R2, or the coefficient of determination. All of these mistakes and improvements have names, and talking about them will be easier once you know those names. The total mistake made using the sample mean to predict, Σ(y–y)2, is called the sum of squares, total. The total mistake made using the regression, Σ(y-ŷ)2, is called the sum of squares, error (residual). The general improvement made by using regression, Σ(ŷ–y)2 is called the sum of squares, regression or sum of squares, model. You should be able to see that: sum of squares, total = sum of squares, regression + sum of squares, error (residual) ŷŷ In other words, the total variations in y can be partitioned into two sources: the explained variations and the unexplained variations. Further, we can rewrite the above equation as: where SST stands for sum of squares due to total variations, SSR measures the sum of squares due to the estimated regression model that is explained by variable x, and SSE measures all the variations due to other factors excluded from the estimated model. Going back to the idea of goodness of fit, one should be able to easily calculate the percentage of each variation with respect to the total variations. In particular, the strength of the estimated regression model can now be measured. Since we are interested in the explained part of the variations by the estimated model, we simply divide both sides of the above equation by SST, and we get: We then isolate this equation for the explained proportion, also known as R-square: Only in cases where an intercept is included in a simple regression model will the value of R2 be bounded between zero and one. The closer R2 is to one, the stronger the model is. Alternatively, R2 is also found by: This is the ratio of the improvement made using the regression to the mistakes made using the mean. The numerator is the improvement regression makes over using the mean to predict; the denominator is the mistakes (errors) made using the mean. Thus R2 simply shows what proportion of the mistakes made using the mean are eliminated by using regression. In the case of the market for one-bedroom apartments in Nelson, BC, the percentage of the variations in price for the apartments is estimated to be around 50%. This indicates that only half of the fluctuations in apartment prices with respect to the average price can be explained by the apartments’ distance from downtown. The other 50% are not controlled (that is, they are unexplained) and are subject to further research. One typical approach is to add more relevant factors to the simple regression model. In this case, the estimated model is referred to as a multiple regression model. While R2 is not used to test hypotheses, it has a more intuitive meaning than the F-score. The F-score is the measure usually used in a hypothesis test to see if the regression made a significant improvement over using the mean. It is used because the sampling distribution of F-scores that it follows is printed in the tables at the back of most statistics books, so that it can be used for hypothesis testing. It works no matter how many explanatory variables are used. More formally, consider a population of multivariate observations, (y, x1, x2, …, xm), where there is no linear relationship between y and the x’s, so that y ≠ f(y, x1, x2, …, xm). If samples of n observations are taken, a regression equation estimated for each sample, and a statistic, F, found for each sample regression, then those F’s will be distributed like those shown in Figure 8.5, the F-table with (m, n–m-1) df. Figure 8.5 Interactive Excel Template of an F-Table – see Appendix 8. The value of F can be calculated as: where n is the size of the sample, and m is the number of explanatory variables (how many x’s there are in the regression equation). If Σ(ŷ–y)2 the sum of squares regression (the improvement), is large relative to Σ(ŷ–y)3, the sum of squares residual (the mistakes still made), then the F-score will be large. In a population where there is no functional relationship between y and the x’s, the regression line will have a slope of zero (it will be flat), and the ŷ will be close to y. As a result very few samples from such populations will have a large sum of squares regression and large F-scores. Because this F-score is distributed like the one in the F-tables, the tables can tell you whether the F-score a sample regression equation produces is large enough to be judged unlikely to occur if y ≠ f(y, x1, x2, …, xm). The sum of squares regression is divided by the number of explanatory variables to account for the fact that it always decreases when more variables are added. You can also look at this as finding the improvement per explanatory variable. The sum of squares residual is divided by a number very close to the number of observations because it always increases if more observations are added. You can also look at this as the approximate mistake per observation. To test to see if a regression equation was worth estimating, test to see if there seems to be a functional relationship: This might look like a two-tailed test since Ho has an equal sign. But, by looking at the equation for the F-score you should be able to see that the data support Ha only if the F-score is large. This is because the data support the existence of a functional relationship if the sum of squares regression is large relative to the sum of squares residual. Since F-tables are usually one-tail tables, choose an α, go to the F-tables for that α and (m, n–m-1) df, and find the table F. If the computed F is greater than the table F, then the computed F is unlikely to have occurred if Ho is true, and you can safely decide that the data support Ha. There is a functional relationship in the population. Now that you have learned all the necessary steps in estimating a simple regression model, you may take some time to re-estimate the Nelson apartment model or any other simple regression model, using the interactive Excel template shown in Figure 8.6. Like all other interactive templates in this textbook, you can change the values in the yellow cells only. The result will be shown automatically within this template. For this template, you can only estimate simple regression models with 30 observations. You use special paste/values when you paste your data from other spreadsheets. The first step is to enter your data under independent and dependent variables. Next, select your alpha level. Check your results in terms of both individual and overall significance. Once the model has passed all these requirements, you can select an appropriate value for the independent variable, which in this example is the distance to downtown, to estimate both the confidence intervals for the average price of such an apartment, and the prediction intervals for the selected distance. Both these intervals are discussed later in this chapter. Remember that by changing any of the values in the yellow areas in this template, all calculations will be updated, including the tests of significance and the values for both confidence and prediction intervals. Figure 8.6 Interactive Excel Template for Simple Regression – see Appendix 8. Multiple Regression Analysis When we add more explanatory variables to our simple regression model to strengthen its ability to explain real-world data, we in fact convert a simple regression model into a multiple regression model. The least squares approach we used in the case of simple regression can still be used for multiple regression analysis. As per our discussion in the simple regression model section, our low estimated R2 indicated that only 50% of the variations in the price of apartments in Nelson, BC, was explained by their distance from downtown. Obviously, there should be more relevant factors that can be added into this model to make it stronger. Let’s add the second explanatory factor to this model. We collected data for the area of each apartment in square feet (i.e., x2). If we go back to Excel and estimate our model including the new added variable, we will see the printout shown in Figure 8.7. The estimates equation of the regression model is: predicted price of apartments= 60.041 – 5.393distance + .03area This is the equation for a plane, the three-dimensional equivalent of a straight line. It is still a linear function because neither of the x’s nor y is raised to a power nor taken to some root nor are the x’s multiplied together. You can have even more independent variables, and as long as the function is linear, you can estimate the slope, β, for each independent variable. Before using this estimated model for prediction and decision-making purposes, we should test three hypotheses. First, we can use the F-score to test to see if the regression model improves our ability to predict price of apartments. In other words, we test the overall significance of the estimated model. Second and third, we can use the t-scores to test to see if the slopes of distance and area are different from zero. These two t-tests are also known as individual tests of significance. To conduct the first test, we choose an α = .05. The F-score is the regression or model mean square over the residual or error mean square, so the df for the F-statistic are first the df for the regression model and, second, the df for the error. There are 2 and 9 df for the F-test. According to this F-table, with 2 and 9 df, the critical F-score for α = .05 is 4.26. The hypotheses are: H0: price ≠ f(distance, area) Ha: price = f(distance, area) Because the F-score from the regression, 6.812, is greater than the critical F-score, 4.26, we decide that the data support Ho and conclude that the model helps us predict price of apartments. Alternatively, we say there is such a functional relationship in the population. Now, we move to the individual test of significance. We can test to see if price depends on distance and area. There are (n-m-1)=(12-2-1)=9 df. There are two sets of hypotheses, one set for β1, the slope for distance, and one set for β2, the slope for area. For a small town, one may expect that β1, the slope for distance, will be negative, and expect that β2 will be positive. Therefore, we will use a one-tail test on β1, as well as for β2: Since we have two one-tail tests, the t-values we choose from the t-table will be the same for the two tests. Using α = .05 and 9 df, we choose .05/2=.025 for the t-score for β1 with a one-tail test, and come up with 2.262. Looking back at our Excel printout and checking the t-scores, we decide that distance does affect price of apartments, but area is not a significant factor in explaining the price of apartments. Notice that the printout also gives a t-score for the intercept, so we could test to see if the intercept equals zero or not. Alternatively, one may go ahead and compare directly the p-values out of the Excel printout against the assumed level of significance (i.e., α = .05). We can easily see that the p-values associated with the intercept and price are both less than alpha, and as a result we reject the hypothesis that the associated coefficients are zero (i.e., both are significant). However, area is not a significant factor since its associated p-value is greater than alpha. While there are other required assumptions and conditions in both simple and multiple regression models (we encourage students to consult an intermediate business statistics open textbook for more detailed discussions), here we only focus on two relevant points about the use and applications of multiple regression. The first point is related to the interpretation of the estimated coefficients in a multiple regression model. You should be careful to note that in a simple regression model, the estimated coefficient of our independent variable is simply the slope of the line and can be interpreted. It refers to the response of the dependent variable to a one-unit change in the independent variable. However, this interpretation in a multiple regression model should be adjusted slightly. The estimated coefficients under multiple regression analysis are the response of the dependent variable to a one-unit change in one of the independent variables when the levels of all other independent variables are kept constant. In our example, the estimated coefficient of price of an apartment in Nelson, BC, indicates that — for a given size of apartment— it will drop by 5.2481000=$5248 for every one kilometre that the apartment is away from downtown. The second point is about the use of R2 in multiple regression analysis. Technically, adding more independent variables to the model will increase the value of R2, regardless of whether the added variables are relevant or irrelevant in explaining the variation in the dependent variable. In order to adjust the inflated R2 due to the irrelevant variables added to the model, the following formula is recommended in the case of multiple regression: where n is the sample size, and k is number of the estimated parameters in our model. Back to our earlier Excel results for the multiple regression model estimated for the apartment example, we can see that while the R2 has been inflated from .504 to .612 due to the new added factor, apartment size, the adjusted R2 has dropped the inflated value to .526. To understand it better, you should pay attention to the associated p-value for the newly added factor. Since this value is more than .05, we cannot reject the hypothesis that the true coefficient of apartment size (area) is significantly different from zero. In other words, in its current situation, apartment size is not a significant factor, yet the value of R2 has been inflated! Furthermore, the adjusted R2 indicates that only 61.2% of variations in price of one-bedroom apartments in Nelson, BC, can be explained by their locations and sizes. Almost 40% of the variations of the price still cannot be explained by these two factors. One may seek to improve this model, by searching for more relevant factors such as style of the apartment, year built, etc. and add them in to this model. Using the interactive Excel template shown in Figure 8.8, you can estimate a multiple regression model. Again, enter your data into the yellow cells only. For this template you are allowed to use up to 50 observations for each column. Like all other interactive templates in this textbook, you use special paste/values when you paste your data from other spreadsheets. Specifically, if you have fewer than 50 data entries, you must also fill out the rest of the empty yellow cells under X1, X2, and Y with zeros. Now, select your alpha level. By clicking enter, you will not only have all your estimated coefficients along with their t-values, etc., you will also be guided as to whether the model is significant both overall and individually. If your p-value associated with F-value within the ANOVA table is not less than the selected alpha level, you will see a message indicating that your estimated model is not overall significant, and as a result, no values for C.I. and P.I. will be shown. By either changing the alpha level and/or adding more accurate data, it is possible to estimate a more significant multiple regression model. Figure 8.8 Interactive Excel Template for Multiple Regression Model - see Appendix 8. One more point is about the format of your assumed multiple regression model. You can see that the nature of the associations between the dependent variable and all the independent variables may not always be linear. In reality, you will face cases where such relationships may be better formed by a nonlinear model. Without going into the details of such a non-linear model, just to give you an idea, you should be able to transform your selected data for X1, X2, and Y before estimating your model. For instance, one possible multiple regression non-linear model may be a model in which both the dependent and independent variables have been transformed to a natural logarithm rather than a level. In order to estimate such a model within Figure 8.5, all you need to do is transform the data in all three columns in a separate sheet from level to logarithm. In doing this, simply use =log(say A1) where in cell A1 you have the first observation of X1, and =log(say B1),.... Finally, simply cut and special paste/value into the yellow columns within the template. Now you have estimated a multiple regression model with both sides in a non-linear form (i.e., log form). Predictions using the estimated simple regression If the estimated regression line fits well into the data, the model can then be used for predictions. Using the above estimated simple regression model, we can predict the price of an apartment a given distance to downtown. This is known as the prediction interval or P.I. Alternatively, we may predict the mean price of the apartment, also known as the confidence interval or C.I., for the mean value. In predicting intervals for the price of an apartment that is six kilometres away from downtown, we simply set x=6 , and substitute it back into the estimated equation: You should pay attention to the scale of data. In this case, the dependent variable is measured in $1000s. Therefore, the predicted value for an apartment six kilometres from downtown is 39.561000=$39,560. This value is known as the point estimate of the prediction and is not reliable, as we are not clear how close this value is to the true value of the population. A more reliable estimate can be constructed by setting up an interval around the point estimate. This can be done in two ways. We can predict the particular value of y for a given value of x,or we can estimate the expected value (mean) of y,for a given value of x. For the particular value of y, we use the following formula for the interval: where the standard error, S.E., of the prediction is calculated based on the following formula: In this equation, x is the particular value of the independent variable, which in our case is 6, and sis the standard error of the regression, calculated as: From the Excel printout for the simple regression model, this standard error is estimated as 7.02. The sum of squares of the independent variable, can also be calculated as shown in Figure 8.9. All these calculated values can be substituted back into the formula for the S.E. of the prediction: Now that the S.E. of the confidence interval has been calculated, you can pick up the cut-off point from the t-table. Given the degrees of freedom 12-2=10, the appropriate value from the t-table is 2.23. You use this information to calculate the margin of error as 6.522.23=14.54. Finally, construct the prediction interval for the particular value of the price of an apartment located six kilometres away from downtown as: This is a compact version of the prediction interval. For a more general version of any confidence interval for any given confidence level of alpha, we can write: Intuitively, for say a .05 level of confidence, we are 95% confident that the true parameter of the population will be within these two lower and upper limits: Based on our simple regression model that only includes distance as a significant factor in predicting the price of an apartment, and for a particular apartment six kilometres away from downtown, we are 95% confident that the true price of an apartments in Nelson, BC, is between $25,037 and $54,096, with a width of $29,059. One should not be surprised there is such a wide width, given the fact that the coefficient of determination of this model was only 50%, and the fact that we have selected a distance far away from the mean distance from downtown. We can always improve these numbers by adding more explanatory variables to our simple regression model. Alternatively, we can predict only for the numbers as much as possible close to the downtown area. Now we estimate the expected value (mean) of y for a given value of x, the so-called prediction interval. The process of constructing intervals is very similar to the previous case, except we use a new formula for S.E. and of course we set up the intervals for the mean value of the apartment price (i.e., =59.33). You should be very careful to note the difference between this formula and the one introduced earlier for S.E. for predicting the particular value of y for a given value of x.They look very similar but this formula comes with an extra 1 inside the radical! The margin of error is then calculated as 2.1793.82=8.32. We use this to set up directly the lower and upper limits of the estimates: Thus, for the average price of apartments located in Nelson, BC, six kilometres away from downtown, we are 95% confident that this average price will be between $18,200 and $60,920, with a width of $47,720. Compared with the earlier width for C.I., it is obvious that we are less confident in predicting the average price. The reason is that the S.E. for the prediction is always larger than the S.E. for the confidence interval. This process can be repeated for all different levels of x, to calculate the associated confidence and prediction intervals. By doing this, we will have a range of lower and upper levels for both P.I.s and C.I.s. All these numbers can be reproduced within the interactive Excel template shown in Figure 8.8. If you use a statistical software such as Minitab, you will directly plot a scatter diagram with all P.I.s and C.I.s as well as the estimated linear regression line all in one diagram. Figure 8.10 shows such a diagram from Minitab for our example. Figure 8.10 indicates that a more reliable prediction should be made as close as possible to the mean of our observations for x. In this graph, the widths of both intervals are at the lowest levels closer to the means of x and y. You should be careful to note that Figure 8.10 provides the predicted intervals only for the case of a simple regression model. For the multiple regression model, you may use other statistical software packages, such as SAS, SPSS, etc., to estimate both P.I. and C.I. For instance, by selecting x1=3, and x2=300, and coding these figures into Minitab, you will see the results as shown in Figure 8.11. Alternatively, you may use the interactive Excel template provided in Figure 8.8 to estimate your multiple regression model, and to check for the significance of the estimated parameters. This template can also be used to construct both the P.I. and C.I. for the given values of x1=3, and x2=300 or any other values of your choice. Furthermore, this template enables you to test if the estimated multiple regression model is overall significant. When the estimated multiple regression model is not overall significant, this template will not provide the P.I. and C.I. To practice this case, you may want to change the yellow columns of x1 and x2 with different random numbers that are not correlated with the dependent variable. Once the estimated model is not overall significant, no prediction values will be provided. The 95% C.I., and P.I. figures in the brackets are the lower and upper limits of the intervals given the specific values for distance and size of apartments. The fitted value of the price of apartment, as well as the standard error of this value, are also estimated. We have just given you some rough ideas about how the basic regression calculations are done. We left out other steps needed to calculate more detailed results of regression without a computer on purpose, for you will never compute a regression without a computer (or a high-end calculator) in all of your working years. However, by working with these interactive templates, you will have a much better chance to play around with any data to see how the outcomes can be altered, and to observe their implications for the real-world business decision-making process. Correlation and covariance The correlation between two variables is important in statistics, and it is commonly reported. What is correlation? The meaning of correlation can be discovered by looking closely at the word—it is almost co-relation, and that is what it means: how two variables are co-related. Correlation is also closely related to regression. The covariance between two variables is also important in statistics, but it is seldom reported. Its meaning can also be discovered by looking closely at the word—it is co-variance, how two variables vary together. Covariance plays a behind-the-scenes role in multivariate statistics. Though you will not see covariance reported very often, understanding it will help you understand multivariate statistics like understanding variance helps you understand univariate statistics. There are two ways to look at correlation. The first flows directly from regression and the second from covariance. Since you just learned about regression, it makes sense to start with that approach. Correlation is measured with a number between -1 and +1 called the correlation coefficient. The population correlation coefficient is usually written as the Greek rho, ρ, and the sample correlation coefficient as r. If you have a linear regression equation with only one explanatory variable, the sign of the correlation coefficient shows whether the slope of the regression line is positive or negative, while the absolute value of the coefficient shows how close to the regression line the points lie. If ρ is +.95, then the regression line has a positive slope and the points in the population are very close to the regression line. If r is -.13 then the regression line has a negative slope and the points in the sample are scattered far from the regression line. If you square r, you will get R2, which is higher if the points in the sample lie very close to the regression line so that the sum of squares regression is close to the sum of squares total. The other approach to explaining correlation requires understanding covariance, how two variables vary together. Because covariance is a multivariate statistic, it measures something about a sample or population of observations where each observation has two or more variables. Think of a population of (x,y) pairs. First find the mean of the x’s and the mean of the y’s, μx and μy. Then for each observation, find (x - μx)(y - μy). If the x and the y in this observation are both far above their means, then this number will be large and positive. If both are far below their means, it will also be large and positive. If you found Σ(x - μx)(y - μy), it would be large and positive if x and y move up and down together, so that large x’s go with large y’s, small x’s go with small y’s, and medium x’s go with medium y’s. However, if some of the large x’s go with medium y’s, etc. then the sum will be smaller, though probably still positive. A Σ(x - μx)(y - μy) implies that x’s above μx are generally paired with y’s above μy, and those x’s below their mean are generally paired with y’s below their mean. As you can see, the sum is a measure of how x and y vary together. The more often similar x’s are paired with similar y’s, the more x and y vary together and the larger the sum and the covariance. The term for a single observation, (x - μx)(y - μy), will be negative when the x and y are on opposite sides of their means. If large x’s are usually paired with small y’s, and vice versa, most of the terms will be negative and the sum will be negative. If the largest x’s are paired with the smallest y’s and the smallest x’s with the largest y’s, then many of the (x - μx)(y - μy) will be large and negative and so will the sum. A population with more members will have a larger sum simply because there are more terms to be added together, so you divide the sum by the number of observations to get the final measure, the covariance, or cov: The maximum for the covariance is the product of the standard deviations of the x values and the y values, σxσy. While proving that the maximum is exactly equal to the product of the standard deviations is complicated, you should be able to see that the more spread out the points are, the greater the covariance can be. By now you should understand that a larger standard deviation means that the points are more spread out, so you should understand that a larger σx or a larger σy will allow for a greater covariance. Sample covariance is measured similarly, except the sum is divided by n-1 so that sample covariance is an unbiased estimator of population covariance: Correlation simply compares the covariance to the standard deviations of the two variables. Using the formula for population correlation: At its maximum, the absolute value of the covariance equals the product of the standard deviations, so at its maximum, the absolute value of r will be 1. Since the covariance can be negative or positive while standard deviations are always positive, r can be either negative or positive. Putting these two facts together, you can see that r will be between -1 and +1. The sign depends on the sign of the covariance and the absolute value depends on how close the covariance is to its maximum. The covariance rises as the relationship between x and y grows stronger, so a strong relationship between x and y will result in r having a value close to -1 or +1. Covariance, correlation, and regression Now it is time to think about how all of this fits together and to see how the two approaches to correlation are related. Start by assuming that you have a population of (x, y) which covers a wide range of y-values, but only a narrow range of x-values. This means that σy is large while σx is small. Assume that you graph the (x, y) points and find that they all lie in a narrow band stretched linearly from bottom left to top right, so that the largest y’s are paired with the largest x’s and the smallest y’s with the smallest x’s. This means both that the covariance is large and a good regression line that comes very close to almost all the points is easily drawn. The correlation coefficient will also be very high (close to +1). An example will show why all these happen together. Imagine that the equation for the regression line is y=3+4x, μy = 31, and μx = 7, and the two points farthest to the top right, (10, 43) and (12, 51), lie exactly on the regression line. These two points together contribute ∑(x-μx)(y-μy) =(10-7)(43-31)+(12-7)(51-31)= 136 to the numerator of the covariance. If we switched the x’s and y’s of these two points, moving them off the regression line, so that they became (10, 51) and (12, 43), μx, μy, σx, and σy would remain the same, but these points would only contribute (10-7)(51-31)+(12-7)(43-31)= 120 to the numerator. As you can see, covariance is at its greatest, given the distributions of the x’s and y’s, when the (x, y) points lie on a straight line. Given that correlation, r, equals 1 when the covariance is maximized, you can see that r=+1 when the points lie exactly on a straight line (with a positive slope). The closer the points lie to a straight line, the closer the covariance is to its maximum, and the greater the correlation. As the example in Figure 8.12 shows, the closer the points lie to a straight line, the higher the correlation. Regression finds the straight line that comes as close to the points as possible, so it should not be surprising that correlation and regression are related. One of the ways the goodness of fit of a regression line can be measured is by R2. For the simple two-variable case, R2 is simply the correlation coefficient r, squared. Correlation does not tell us anything about how steep or flat the regression line is, though it does tell us if the slope is positive or negative. If we took the initial population shown in Figure 8.12, and stretched it both left and right horizontally so that each point’s x-value changed, but its y-value stayed the same, σx would grow while σy stayed the same. If you pulled equally to the right and to the left, both μx and μy would stay the same. The covariance would certainly grow since the (x-μx) that goes with each point would be larger absolutely while the (y-μy)’s would stay the same. The equation of the regression line would change, with the slope b becoming smaller, but the correlation coefficient would be the same because the points would be just as close to the regression line as before. Once again, notice that correlation tells you how well the line fits the points, but it does not tell you anything about the slope other than if it is positive or negative. If the points are stretched out horizontally, the slope changes but correlation does not. Also notice that though the covariance increases, correlation does not because σx increases, causing the denominator in the equation for finding r to increase as much as covariance, the numerator. The regression line and covariance approaches to understanding correlation are obviously related. If the points in the population lie very close to the regression line, the covariance will be large in absolute value since the x’s that are far from their mean will be paired with y’s that are far from theirs. A positive regression slope means that x and y rise and fall together, which also means that the covariance will be positive. A negative regression slope means that x and y move in opposite directions, which means a negative covariance. Summary Simple linear regression allows researchers to estimate the parameters — the intercept and slopes — of linear equations connecting two or more variables. Knowing that a dependent variable is functionally related to one or more independent or explanatory variables, and having an estimate of the parameters of that function, greatly improves the ability of a researcher to predict the values the dependent variable will take under many conditions. Being able to estimate the effect that one independent variable has on the value of the dependent variable in isolation from changes in other independent variables can be a powerful aid in decision-making and policy design. Being able to test the existence of individual effects of a number of independent variables helps decision-makers, researchers, and policy-makers identify what variables are most important. Regression is a very powerful statistical tool in many ways. The idea behind regression is simple: it is simply the equation of the line that comes as close as possible to as many of the points as possible. The mathematics of regression are not so simple, however. Instead of trying to learn the math, most researchers use computers to find regression equations, so this chapter stressed reading computer printouts rather than the mathematics of regression. Two other topics, which are related to each other and to regression, were also covered: correlation and covariance. Something as powerful as linear regression must have limitations and problems. There is a whole subject, econometrics, which deals with identifying and overcoming the limitations and problems of regression. Thomas, G.B. (1960). Calculus and analytical geometry (3rd ed.). Boston, MA: Addison-Wesley. ↵
6176	https://www.nceyes.org/digital-eye-strain	Digital Eye Strain Menu Home Eye Health Importance of Regular Eye Exams Maintain Healthy Eyes for Life Contact Lenses Diet and Nutrition Protecting Your Eyes at Work Digital Eye Strain UV Protection Sports and Vision Low Vision Your Vision Throughout Life Eye and Vision Problems Eye Exams for Low-Income Uninsured Individuals Helpful Resources and Links FAQs NC Eye Doctors About North Carolina Optometrists NC Optometrists in Your Community Find an Optometrist Diabetic Eye Care Collaborative Optometrists Membership Join NCOS Benefits & Services Member Benefit Partners Students & New Licensees Events Virtual Seminar Legislative Day Spring Congress Exhibitor Information Fall Congress Partnership & Ads Our Partners Become a Partner Sponsorship Opportunities Advertising Opportunities About About NCOS Leadership and Staff Past Award Recipents Contact Us Member Login Home Eye Health Importance of Regular Eye Exams Maintain Healthy Eyes for Life Contact Lenses Diet and Nutrition Protecting Your Eyes at Work Digital Eye Strain UV Protection Sports and Vision Low Vision Your Vision Throughout Life Eye and Vision Problems Eye Exams for Low-Income Uninsured Individuals Helpful Resources and Links FAQs NC Eye Doctors About North Carolina Optometrists NC Optometrists in Your Community Find an Optometrist Diabetic Eye Care Collaborative Optometrists Membership Join NCOS Benefits & Services Member Benefit Partners Students & New Licensees Events Virtual Seminar Legislative Day Spring Congress Exhibitor Information Fall Congress Partnership & Ads Our Partners Become a Partner Sponsorship Opportunities Advertising Opportunities About About NCOS Leadership and Staff Past Award Recipents Contact Us Digital Eye Strain Digital Eye Strain describes a group of eye and vision-related problems that result from prolonged computer, tablet, e-reader and cell phone use. Many individuals experience eye discomfort and vision problems when viewing digital screens for extended periods. The level of discomfort appears to increase with the amount of digital screen use. The average American worker spends seven hours a day on the computer either in the office or working from home. To help alleviate digital eye strain, follow the 20-20-20 rule. The most common symptoms associated with Digital Eye Strain are: eyestrain headaches blurred vision dry eyes neck and shoulder pain These symptoms may be caused by: poor lighting glare on a digital screen improper viewing distances poor seating posture uncorrected vision problems a combination of these factors The extent to which individuals experience visual symptoms often depends on the level of their visual abilities and the amount of time spent looking at a digital screen. Uncorrected vision problems like farsightedness andastigmatism, inadequate eye focusing or eye coordination abilities, and aging changes of the eyes, such aspresbyopia, can all contribute to the development of visual symptoms when using a computer or digital screen device. Many of the visual symptoms experienced by users are only temporary and will decline after stopping computer work or use of the digital device. However, some individuals may experience continued reduced visual abilities, such as blurred distance vision, even after stopping work at a computer. If nothing is done to address the cause of the problem, the symptoms will continue to recur and perhaps worsen with future digital screen use. Prevention or reduction of the vision problems associated with Digital Eye Strain involves taking steps to control lighting and glare on the device screen, establishing proper working distances and posture for screen viewing, and assuring that even minor vision problems are properly corrected. What causes Digital Eye Strain? Viewing a computer or digital screen often makes the eyes work harder. As a result, the unique characteristics and high visual demands of computer and digital screen device viewing make many individuals susceptible to the development of vision-related symptoms. Uncorrected vision problems can increase the severity of Digital Eye Strain symptoms. Viewing a computer or digital screen is different than reading a printed page. Often the letters on the computer or handheld device are not as precise or sharply defined, the level of contrast of the letters to the background is reduced, and the presence of glare and reflections on the screen may make viewing difficult. Viewing distances and angles used for this type of work are also often different from those commonly used for other reading or writing tasks. As a result, the eye focusing and eye movement requirements for digital screen viewing can place additional demands on the visual system. In addition, the presence of even minor vision problems can often significantly affect comfort and performance at a computer or while using other digital screen devices. Uncorrected or under corrected vision problems can be major contributing factors to computer-related eyestrain. Even people who have an eyeglass or contact lens prescription may find it's not suitable for the specific viewing distances of their computer screen. Some people tilt their heads at odd angles because their glasses aren't designed for looking at a computer. Or they bend toward the screen in order to see it clearly. Their postures can result in muscle spasms or pain in the neck, shoulder or back. In most cases, symptoms of Digital Eye Strain occur because the visual demands of the task exceed the visual abilities of the individual to comfortably perform them. At greatest risk for developing Digital Eye Strain are those persons who spend two or more continuous hours at a computer or using a digital screen device every day. How is Digital Eye Strain diagnosed? Digital Eye Strain can be diagnosed through a comprehensive eye examination. Testing, with special emphasis on visual requirements at the computer or digital device working distance, may include: Patient historyto determine any symptoms the patient is experiencing and the presence of any general health problems, medications taken, or environmental factors that may be contributing to the symptoms related to computer use. Visual acuity measurementsto assess the extent to which vision may be affected. Arefractionto determine the appropriate lens power needed to compensate for any refractive errors (nearsightedness, farsightedness or astigmatism). Testing how the eyes focus, move and work together. In order to obtain a clear, single image of what is being viewed, the eyes must effectively change focus, move and work in unison. This testing will look for problems that keep your eyes from focusing effectively or make it difficult to use both eyes together. This testing may be done without the use of eye drops to determine how the eyes respond under normal seeing conditions. In some cases, such as when some of the eyes' focusing power may be hidden, eye drops may be used. They temporarily keep the eyes from changing focus while testing is done. Using the information obtained from these tests, along with results of other tests, your optometrist can determine if you have Digital Eye Strain and advise you on treatment options. How is Digital Eye Strain treated? Solutions to digital screen-related vision problems are varied. However, they can usually be alleviated by obtaining regular eye care and making changes in how you view the screen. Eye Care In some cases, individuals who do not require the use of eyeglasses for other daily activities may benefit from glasses prescribed specifically for computer use. In addition, persons already wearing glasses may find their current prescription does not provide optimal vision for viewing a computer. Eyeglasses or contact lenses prescribed for general use may not be adequate for computer work. Lenses prescribed to meet the unique visual demands of computer viewing may be needed. Special lens designs, lens powers or lens tints or coatings may help to maximize visual abilities and comfort. Some computer users experience problems with eye focusing or eye coordination that can't be adequately corrected with eyeglasses or contact lenses. A program of vision therapy may be needed to treat these specific problems. Vision therapy, also called visual training, is a structured program of visual activities prescribed to improve visual abilities. It trains the eyes and brain to work together more effectively. These eye exercises help remediate deficiencies in eye movement, eye focusing and eye teaming and reinforce the eye-brain connection. Treatment may include office-based as well as home training procedures. Viewing the Computer Some important factors in preventing or reducing the symptoms of Digital Eye Strain have to do with the computer and how it is used. This includes lighting conditions, chair comfort, location of reference materials, position of the monitor, and the use of rest breaks. Location of computer screen- Most people find it more comfortable to view a computer when the eyes are looking downward. Optimally, the computer screen should be 15 to 20 degrees below eye level (about 4 or 5 inches) as measured from the center of the screen and 20 to 28 inches from the eyes. Reference materials- These materials should be located above the keyboard and below the monitor. If this is not possible, a document holder can be used beside the monitor. The goal is to position the documents so you do not need to move your head to look from the document to the screen. Lighting- Position the computer screen to avoid glare, particularly from overhead lighting or windows. Use blinds or drapes on windows and replace the light bulbs in desk lamps with bulbs of lower wattage. Anti-glare screens- If there is no way to minimize glare from light sources, consider using a screen glare filter. These filters decrease the amount of light reflected from the screen. Seating position- Chairs should be comfortably padded and conform to the body. Chair height should be adjusted so your feet rest flat on the floor. If your chair has arms, they should be adjusted to provide arm support while you are typing. Your wrists shouldn't rest on the keyboard when typing. Rest breaks- To prevent eyestrain, try to rest your eyes when using the computer for long periods. Rest your eyes for 15 minutes after two hours of continuous computer use. Also, for every 20 minutes of computer viewing, look into the distance for 20 seconds to allow your eyes a chance to refocus. Blinking- To minimize your chances of developingdry eyewhen using a computer, make an effort to blink frequently. Blinking keeps the front surface of your eye moist. Regular eye examinations and proper viewing habits can help to prevent or reduce the development of the symptoms associated with Digital Eye Strain. 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6177	http://www.wallace.ccfaculty.org/book/10.3%20Inverse%20Functions%20Practice.pdf	10.3 Practice - Inverse Functions State if the given functions are inverses. 1) g(x) = −x5 −3 f(x) = −x −3 5 √ 3) f(x) = −x −1 x −2 g(x) = −2x + 1 −x −1 5) g(x) = −10x + 5 f(x) = x −5 10 7) f(x) = − 2 x + 3 g(x) = 3x + 2 x + 2 9) g(x) = x −1 2 5 q f(x) = 2x5 + 1 2) g(x) = 4 −x x f(x) = 4 x 4) h(x) = −2 −2x x f(x) = −2 x + 2 6) f(x) = x −5 10 h(x) = 10x + 5 8) f(x) = x + 1 2 5 q g(x) = 2x5 −1 10) g(x) = 8 + 9x 2 f(x) = 5x −9 2 Find the inverse of each functions. 11) f(x) = (x −2)5 + 3 13) g(x) = 4 x + 2 15) f(x) = −2x −2 x + 2 17) f(x) = 10−x 5 19) g(x) = −(x −1)3 21) f(x) = (x −3)3 23) g(x) = x x −1 25) f(x) = x −1 x + 1 27) g(x) = 8 −5x 4 29) g(x) = −5x + 1 31) g(x) = −1 + x3 33) h(x) = 4 − 4x 3 √ 2 35) f(x) = x + 1 x + 2 37) f(x) = 7 −3x x −2 39) g(x) = −x 12) g(x) = x + 1 3 √ + 2 14) f(x) = −3 x −3 16) g(x) = 9 + x 3 18) f(x) = 5x −15 2 20) f(x) = 12−3x 4 22) g(x) = −x + 2 2 5 q 1 24) f(x) = −3 −2x x + 3 26) h(x) = x x + 2 28) g(x) = −x + 2 3 30) f(x) = 5x −5 4 32) f(x) = 3 −2x5 34) g(x) = (x −1)3 + 2 36) f(x) = −1 x + 1 38) f(x) = −3x 4 40) g(x) = −2x + 1 3 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. ( 2 10.3 Answers - Inverse Functions 1) Yes 2) No 3) Yes 4) Yes 5) No 6) Yes 7) No 8) Yes 9) Yes 10) No 11) f −1(x) = x −3 5 √ + 2 12) g−1(x) = (x −2)3 −1 13) g−1(x) = 4 −2x x 14) f −1(x) = −3 + 3x x 15) f −1(x) = −2x −2 x + 2 16) g−1(x) = 3x −9 17) f −1(x) = −5x + 10 18) f −1(x) = 15 + 2x 5 19) g−1(x) = − x 3 √+ 1 20) f −1(x) = −4x + 12 3 21) f −1(x) = x 3 √+ 3 22) g−1(x) = −2x5 + 2 23) g−1(x) = x x −1 24) f −1(x) = −3x −3 x + 2 25) f −1(x) = −x −1 x −1 26) h−1(x) = −2x x −1 27) g−1(x) = −4x + 8 5 28) g−1(x) = −3x + 2 29) g−1(x) = −x + 1 5 30) f −1(x) = 5 + 4x 5 31) g−1(x) = x + 1 3 √ 32) f −1(x) = −x + 3 2 5 q 33) h−1(x) = ( −2x + 4)3 4 34) g−1(x) = x −2 3 √ + 1 35) f −1(x) = −2x + 1 x −1 36) f −1(x) = −1 −x x 37) f −1(x) = 2x + 7 x + 3 38) f −1(x) = −4x 3 39) g−1(x) = −x 40) g−1(x) = −3x + 1 2 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. ( 3
6178	https://dokumen.pub/fundamentals-of-fluid-mechanics-7nbsped-1118116135.html	Fundamentals of Fluid Mechanics [7 ed.] 1118116135 - DOKUMEN.PUB Anmelden Registrierung Deutsch English Español Português Français Dom Najlepsze kategorie CAREER & MONEY PERSONAL GROWTH POLITICS & CURRENT AFFAIRS SCIENCE & TECH HEALTH & FITNESS LIFESTYLE ENTERTAINMENT BIOGRAPHIES & HISTORY FICTION Najlepsze historie Najlepsze historie Dodaj historię Moje historie Home Fundamentals of Fluid Mechanics [7 ed.] 1118116135 Fundamentals of Fluid Mechanics [7 ed.] 1118116135 This book is intended for junior and senior engineering students who are interested in learning some fundamental aspects 18,257 1,139 25MB English Pages 747 Year 2013 Report DMCA / Copyright DOWNLOAD FILE Polecaj historie ###### Fundamentals of Fluid Mechanics [second ed.] 9789380578859 7,089 536 123MB Read more ###### Fundamentals of Fluid Mechanics [3 ed.] 9789390455201, 9789390455256 1,038 139 7MB Read more ###### Fluid Mechanics: Fundamentals and Applications: 2024 Release ISE 1266968369, 9781266968365 ISBN: 9781266968365 Fluid Mechanics: Fundamentals and Applications: 2024 Release by Yunus A. 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Rothmayer Categories Physics Mechanics: Fluid Mechanics _Table of contents : Front Cover Title Page Copyright Page About the Authors Preface Featured in this Book CONTENTS 1 Introduction Learning Objectives 1.1 Some Characteristics of Fluids 1.2 Dimensions, Dimensional Homogeneity, and Units 1.2.1 Systems of Units 1.3 Analysis of Fluid Behavior 1.4 Measures of Fluid Mass and Weight 1.4.1 Density 1.4.2 Specific Weight 1.4.3 Specific Gravity 1.5 Ideal Gas Law 1.6 Viscosity 1.7 Compressibility of Fluids 1.7.1 Bulk Modulus 1.7.2 Compression and Expansion of Gases 1.7.3 Speed of Sound 1.8 Vapor Pressure 1.9 Surface Tension 1.10 A Brief Look Back in History 1.11 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 2 Fluid Statics Learning Objectives 2.1 Pressure at a Point 2.2 Basic Equation for Pressure Field 2.3 Pressure Variation in a Fluid at Rest 2.3.1 Incompressible Fluid 2.3.2 Compressible Fluid 2.4 Standard Atmosphere 2.5 Measurement of Pressure 2.6 Manometry 2.6.1 Piezometer Tube 2.6.2 U-Tube Manometer 2.6.3 Inclined-Tube Manometer 2.7 Mechanical and Electronic Pressure-Measuring Devices 2.8 Hydrostatic Force on a Plane Surface 2.9 Pressure Prism 2.10 Hydrostatic Force on a Curved Surface 2.11 Buoyancy, Flotation, and Stability 2.11.1 Archimedes’ Principle 2.11.2 Stability 2.12 Pressure Variation in a Fluid with Rigid-Body Motion 2.12.1 Linear Motion 2.12.2 Rigid-Body Rotation 2.13 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 3 Elementary Fluid Dynamics—The Bernoulli Equation Learning Objectives 3.1 Newton’s Second Law 3.2 F = ma along a Streamline 3.3 F = ma Normal to a Streamline 3.4 Physical Interpretation 3.5 Static, Stagnation, Dynamic, and Total Pressure 3.6 Examples of Use of the Bernoulli Equation 3.6.1 Free Jets 3.6.2 Confined Flows 3.6.3 Flowrate Measurement 3.7 The Energy Line and the Hydraulic Grade Line 3.8 Restrictions on Use of the Bernoulli Equation 3.8.1 Compressibility Effects 3.8.2 Unsteady Effects 3.8.3 Rotational Effects 3.8.4 Other Restrictions 3.9 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 4 Fluid Kinematics Learning Objectives 4.1 The Velocity Field 4.1.1 Eulerian and Lagrangian Flow Descriptions 4.1.2 One-, Two-, and Three-Dimensional Flows 4.1.3 Steady and Unsteady Flows 4.1.4 Streamlines, Streaklines, and Pathlines 4.2 The Acceleration Field 4.2.1 The Material Derivative 4.2.2 Unsteady Effects 4.2.3 Convective Effects 4.2.4 Streamline Coordinates 4.3 Control Volume and System Representations 4.4 The Reynolds Transport Theorem 4.4.1 Derivation of the Reynolds Transport Theorem 4.4.2 Physical Interpretation 4.4.3 Relationship to Material Derivative 4.4.4 Steady Effects 4.4.5 Unsteady Effects 4.4.6 Moving Control Volumes 4.4.7 Selection of a Control Volume 4.5 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 5 Finite Control Volume Analysis Learning Objectives 5.1 Conservation of Mass—The Continuity Equation 5.1.1 Derivation of the Continuity Equation 5.1.2 Fixed, Nondeforming Control Volume 5.1.3 Moving, Nondeforming Control Volume 5.1.4 Deforming Control Volume 5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 5.2.1 Derivation of the Linear Momentum Equation 5.2.2 Application of the Linear Momentum Equation 5.2.3 Derivation of the Moment-of-Momentum Equation 5.2.4 Application of the Moment-of-Momentum Equation 5.3 First Law of Thermodynamics—The Energy Equation 5.3.1 Derivation of the Energy Equation 5.3.2 Application of the Energy Equation 5.3.3 Comparison of the Energy Equation with the Bernoulli Equation 5.3.4 Application of the Energy Equation to Nonuniform Flows 5.3.5 Combination of the Energy Equation and the Moment-of-Momentum Equation 5.4 Second Law of Thermodynamics—Irreversible Flow 5.5 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 6 Differential Analysis of Fluid Flow Learning Objectives 6.1 Fluid Element Kinematics 6.1.1 Velocity and Acceleration Fields Revisited 6.1.2 Linear Motion and Deformation 6.1.3 Angular Motion and Deformation 6.2 Conservation of Mass 6.2.1 Differential Form of Continuity Equation 6.2.2 Cylindrical Polar Coordinates 6.2.3 The Stream Function 6.3 Conservation of Linear Momentum 6.3.1 Description of Forces Acting on the Differential Element 6.3.2 Equations of Motion 6.4 Inviscid Flow 6.4.1 Euler’s Equations of Motion 6.4.2 The Bernoulli Equation 6.4.3 Irrotational Flow 6.4.4 The Bernoulli Equation for Irrotational Flow 6.4.5 The Velocity Potential 6.5 Some Basic, Plane Potential Flows 6.5.1 Uniform Flow 6.5.2 Source and Sink 6.5.3 Vortex 6.5.4 Doublet 6.6 Superposition of Basic, Plane Potential Flows 6.6.1 Source in a Uniform Stream—Half-Body 6.6.2 Rankine Ovals 6.6.3 Flow around a Circular Cylinder 6.7 Other Aspects of Potential Flow Analysis 6.8 Viscous Flow 6.8.1 Stress-Deformation Relationships 6.8.2 The Navier–Stokes Equations 6.9 Some Simple Solutions for Viscous, Incompressible Fluids 6.9.1 Steady, Laminar Flow between Fixed Parallel Plates 6.9.2 Couette Flow 6.9.3 Steady, Laminar Flow in Circular Tubes 6.9.4 Steady, Axial, Laminar Flow in an Annulus 6.10 Other Aspects of Differential Analysis 6.10.1 Numerical Methods 6.11 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 7 Dimensional Analysis, Similitude, and Modeling Learning Objectives 7.1 Dimensional Analysis 7.2 Buckingham Pi Theorem 7.3 Determination of Pi Terms 7.4 Some Additional Comments about Dimensional Analysis 7.4.1 Selection of Variables 7.4.2 Determination of Reference Dimensions 7.4.3 Uniqueness of Pi Terms 7.5 Determination of Pi Terms by Inspection 7.6 Common Dimensionless Groups in Fluid Mechanics 7.7 Correlation of Experimental Data 7.7.1 Problems with One Pi Term 7.7.2 Problems with Two or More Pi Terms 7.8 Modeling and Similitude 7.8.1 Theory of Models 7.8.2 Model Scales 7.8.3 Practical Aspects of Using Models 7.9 Some Typical Model Studies 7.9.1 Flow through Closed Conduits 7.9.2 Flow around Immersed Bodies 7.9.3 Flow with a Free Surface 7.10 Similitude Based on Governing Differential Equations 7.11 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 8 Viscous Flow in Pipes Learning Objectives 8.1 General Characteristics of Pipe Flow 8.1.1 Laminar or Turbulent Flow 8.1.2 Entrance Region and Fully Developed Flow 8.1.3 Pressure and Shear Stress 8.2 Fully Developed Laminar Flow 8.2.1 From F = ma Applied Directly to a Fluid Element 8.2.2 From the Navier–Stokes Equations 8.2.3 From Dimensional Analysis 8.2.4 Energy Considerations 8.3 Fully Developed Turbulent Flow 8.3.1 Transition from Laminar to Turbulent Flow 8.3.2 Turbulent Shear Stress 8.3.3 Turbulent Velocity Profile 8.3.4 Turbulence Modeling 8.3.5 Chaos and Turbulence 8.4 Dimensional Analysis of Pipe Flow 8.4.1 Major Losses 8.4.2 Minor Losses 8.4.3 Noncircular Conduits 8.5 Pipe Flow Examples 8.5.1 Single Pipes 8.5.2 Multiple Pipe Systems 8.6 Pipe Flowrate Measurement 8.6.1 Pipe Flowrate Meters 8.6.2 Volume Flowmeters 8.7 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 9 Flow Over Immersed Bodies Learning Objectives 9.1 General External Flow Characteristics 9.1.1 Lift and Drag Concepts 9.1.2 Characteristics of Flow Past an Object 9.2 Boundary Layer Characteristics 9.2.1 Boundary Layer Structure and Thickness on a Flat Plate 9.2.2 Prandtl/Blasius Boundary Layer Solution 9.2.3 Momentum Integral Boundary Layer Equation for a Flat Plate 9.2.4 Transition from Laminar to Turbulent Flow 9.2.5 Turbulent Boundary Layer Flow 9.2.6 Effects of Pressure Gradient 9.2.7 Momentum Integral Boundary Layer Equation with Nonzero Pressure Gradient 9.3 Drag 9.3.1 Friction Drag 9.3.2 Pressure Drag 9.3.3 Drag Coefficient Data and Examples 9.4 Lift 9.4.1 Surface Pressure Distribution 9.4.2 Circulation 9.5 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 10 Open-Channel Flow Learning Objectives 10.1 General Characteristics of Open-Channel Flow 10.2 Surface Waves 10.2.1 Wave Speed 10.2.2 Froude Number Effects 10.3 Energy Considerations 10.3.1 Specific Energy 10.3.2 Channel Depth Variations 10.4 Uniform Depth Channel Flow 10.4.1 Uniform Flow Approximations 10.4.2 The Chezy and Manning Equations 10.4.3 Uniform Depth Examples 10.5 Gradually Varied Flow 10.6 Rapidly Varied Flow 10.6.1 The Hydraulic Jump 10.6.2 Sharp-Crested Weirs 10.6.3 Broad-Crested Weirs 10.6.4 Underflow Gates 10.7 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 11 Compressible Flow Learning Objectives 11.1 Ideal Gas Relationships 11.2 Mach Number and Speed of Sound 11.3 Categories of Compressible Flow 11.4 Isentropic Flow of an Ideal Gas 11.4.1 Effect of Variations in Flow Cross-Sectional Area 11.4.2 Converging–Diverging Duct Flow 11.4.3 Constant Area Duct Flow 11.5 Nonisentropic Flow of an Ideal Gas 11.5.1 Adiabatic Constant Area Duct Flow with Friction (Fanno Flow) 11.5.2 Frictionless Constant Area Duct Flow with Heat Transfer (Rayleigh Flow) 11.5.3 Normal Shock Waves 11.6 Analogy between Compressible and Open-Channel Flows 11.7 Two-Dimensional Compressible Flow 11.8 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems 12 Turbomachines Learning Objectives 12.1 Introduction 12.2 Basic Energy Considerations 12.3 Basic Angular Momentum Considerations 12.4 The Centrifugal Pump 12.4.1 Theoretical Considerations 12.4.2 Pump Performance Characteristics 12.4.3 Net Positive Suction Head (NPSH) 12.4.4 System Characteristics and Pump Selection 12.5 Dimensionless Parameters and Similarity Laws 12.5.1 Special Pump Scaling Laws 12.5.2 Specific Speed 12.5.3 Suction Specific Speed 12.6 Axial-Flow and Mixed-Flow Pumps 12.7 Fans 12.8 Turbines 12.8.1 Impulse Turbines 12.8.2 Reaction Turbines 12.9 Compressible Flow Turbomachines 12.9.1 Compressors 12.9.2 Compressible Flow Turbines 12.10 Chapter Summary and Study Guide References Review Problems Conceptual Questions Problems APPENDICES A: Computational Fluid Dynamics B: Physical Properties of Fluids C: Properties of the U.S. Standard Atmosphere D: Compressible Flow Graphs for an Ideal Gas (k =1.4) ANSWERS to Selected Even-Numbered Homework Problems ch01 ch02 ch03-04 ch05 ch06 ch07-08 ch09-10 ch11-12 INDEX A B C D E F G-H I J-K-L-M N O P Q-R-S T U-V W-Z Video Index (with WEB link) Useful Reference Pages Physical Properties of Common Liquids Physical Properties of Common Gases at Standard Atmospheric Pressure Conversion Factors Conversion Factors_ Citation preview Fundamentals of seventh edition Fluid Mechanics Munson Okiishi Huebsch Rothmayer ■ TA B L E 1 . 5 Approximate Physical Properties of Some Common Liquids (BG Units) Liquid Temperature (ⴗF) Density, R (slugsⲐ ft3) Carbon tetrachloride Ethyl alcohol Gasolinec Glycerin Mercury SAE 30 oilc Seawater Water 68 68 60 68 68 60 60 60 3.09 1.53 1.32 2.44 26.3 1.77 1.99 1.94 Specific Weight, G (lbⲐft3) Dynamic Viscosity, M (lb ⴢ sⲐft2) Kinematic Viscosity, N (ft2Ⲑs) Surface Tension,a S (lbⲐ ft) 99.5 49.3 42.5 78.6 847 57.0 64.0 62.4 2.00 E ⫺ 5 2.49 E ⫺ 5 6.5 E ⫺ 6 3.13 E ⫺ 2 3.28 E ⫺ 5 8.0 E ⫺ 3 2.51 E ⫺ 5 2.34 E ⫺ 5 6.47 E ⫺ 6 1.63 E ⫺ 5 4.9 E ⫺ 6 1.28 E ⫺ 2 1.25 E ⫺ 6 4.5 E ⫺ 3 1.26 E ⫺ 5 1.21 E ⫺ 5 1.84 E ⫺ 3 1.56 E ⫺ 3 1.5 E ⫺ 3 4.34 E ⫺ 3 3.19 E ⫺ 2 2.5 E ⫺ 3 5.03 E ⫺ 3 5.03 E ⫺ 3 Vapor Pressure, pv [lbⲐ in.2 (abs)] Bulk Modulus,b Ev (lbⲐ in.2) E⫹0 E⫺1 E⫹0 E⫺6 E⫺5 — 2.56 E ⫺ 1 2.56 E ⫺ 1 1.91 E ⫹ 5 1.54 E ⫹ 5 1.9 E ⫹ 5 6.56 E ⫹ 5 4.14 E ⫹ 6 2.2 E ⫹ 5 3.39 E ⫹ 5 3.12 E ⫹ 5 Vapor Pressure, pv [NⲐm2 (abs)] Bulk Modulus,b Ev (NⲐ m2) E⫹4 E⫹3 E⫹4 E⫺2 E⫺1 — 1.77 E ⫹ 3 1.77 E ⫹ 3 1.31 E ⫹ 9 1.06 E ⫹ 9 1.3 E ⫹ 9 4.52 E ⫹ 9 2.85 E ⫹ 10 1.5 E ⫹ 9 2.34 E ⫹ 9 2.15 E ⫹ 9 1.9 8.5 8.0 2.0 2.3 a In contact with air. Isentropic bulk modulus calculated from speed of sound. c Typical values. Properties of petroleum products vary. b ■ TA B L E 1 . 6 Approximate Physical Properties of Some Common Liquids (SI Units) Liquid Temperature (ⴗC) Density, R (kgⲐ m3) Carbon tetrachloride Ethyl alcohol Gasolinec Glycerin Mercury SAE 30 oilc Seawater Water 20 20 15.6 20 20 15.6 15.6 15.6 1,590 789 680 1,260 13,600 912 1,030 999 a In contact with air. Isentropic bulk modulus calculated from speed of sound. c Typical values. Properties of petroleum products vary. b Specific Weight, G (kNⲐm3) Dynamic Viscosity, M (N ⴢ sⲐm2) Kinematic Viscosity, N (m2 Ⲑ s) Surface Tension,a S (NⲐm) 15.6 7.74 6.67 12.4 133 8.95 10.1 9.80 9.58 E ⫺ 4 1.19 E ⫺ 3 3.1 E ⫺ 4 1.50 E ⫹ 0 1.57 E ⫺ 3 3.8 E ⫺ 1 1.20 E ⫺ 3 1.12 E ⫺ 3 6.03 E ⫺ 7 1.51 E ⫺ 6 4.6 E ⫺ 7 1.19 E ⫺ 3 1.15 E ⫺ 7 4.2 E ⫺ 4 1.17 E ⫺ 6 1.12 E ⫺ 6 2.69 E ⫺ 2 2.28 E ⫺ 2 2.2 E ⫺ 2 6.33 E ⫺ 2 4.66 E ⫺ 1 3.6 E ⫺ 2 7.34 E ⫺ 2 7.34 E ⫺ 2 1.3 5.9 5.5 1.4 1.6 ■ TA B L E 1 . 7 Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (BG Units) Gas Temperature (ⴗF) Air (standard) Carbon dioxide Helium Hydrogen Methane (natural gas) Nitrogen Oxygen 59 68 68 68 68 68 68 Density, R (slugsⲐ ft3) 2.38 E 3.55 E 3.23 E 1.63 E 1.29 E 2.26 E 2.58 E ⫺3 ⫺3 ⫺4 ⫺4 ⫺3 ⫺3 ⫺3 Specific Weight, G (lbⲐ ft3) 7.65 E 1.14 E 1.04 E 5.25 E 4.15 E 7.28 E 8.31 E ⫺ ⫺ ⫺ ⫺ ⫺ ⫺ ⫺ Dynamic Viscosity, M (lb ⴢ sⲐft2) 2 1 2 3 2 2 2 3.74 E 3.07 E 4.09 E 1.85 E 2.29 E 3.68 E 4.25 E ⫺ ⫺ ⫺ ⫺ ⫺ ⫺ ⫺ 7 7 7 7 7 7 7 Kinematic Viscosity, N (ft2Ⲑs) 1.57 E ⫺ 8.65 E ⫺ 1.27 E ⫺ 1.13 E ⫺ 1.78 E ⫺ 1.63 E ⫺ 1.65 E ⫺ 4 5 3 3 4 4 4 Gas Constant,a R (ft ⴢ lbⲐ slug ⴢ ⴗR) Specific Heat Ratio,b k 1.716 E ⫹ 3 1.130 E ⫹ 3 1.242 E ⫹ 4 2.466 E ⫹ 4 3.099 E ⫹ 3 1.775 E ⫹ 3 1.554 E ⫹ 3 1.40 1.30 1.66 1.41 1.31 1.40 1.40 a Values of the gas constant are independent of temperature. Values of the specific heat ratio depend only slightly on temperature. b ■ TA B L E 1 . 8 Approximate Physical Properties of Some Common Gases at Standard Atmospheric Pressure (SI Units) Gas Temperature (ⴗC) Density, R (kgⲐ m3) Specific Weight, G (N Ⲑm3) Dynamic Viscosity, M (N ⴢ sⲐ m2) Kinematic Viscosity, N (m2 Ⲑs) Gas Constant,a R (JⲐ kg ⴢ K) Specific Heat Ratio,b k Air (standard) Carbon dioxide Helium Hydrogen Methane (natural gas) Nitrogen Oxygen 15 20 20 20 20 20 20 1.23 E ⫹ 0 1.83 E ⫹ 0 1.66 E ⫺ 1 8.38 E ⫺ 2 6.67 E ⫺ 1 1.16 E ⫹ 0 1.33 E ⫹ 0 1.20 E ⫹ 1 1.80 E ⫹ 1 1.63 E ⫹ 0 8.22 E ⫺ 1 6.54 E ⫹ 0 1.14 E ⫹ 1 1.30 E ⫹ 1 1.79 E ⫺ 5 1.47 E ⫺ 5 1.94 E ⫺ 5 8.84 E ⫺ 6 1.10 E ⫺ 5 1.76 E ⫺ 5 2.04 E ⫺ 5 1.46 E ⫺ 5 8.03 E ⫺ 6 1.15 E ⫺ 4 1.05 E ⫺ 4 1.65 E ⫺ 5 1.52 E ⫺ 5 1.53 E ⫺ 5 2.869 E ⫹ 2 1.889 E ⫹ 2 2.077 E ⫹ 3 4.124 E ⫹ 3 5.183 E ⫹ 2 2.968 E ⫹ 2 2.598 E ⫹ 2 1.40 1.30 1.66 1.41 1.31 1.40 1.40 a Values of the gas constant are independent of temperature. Values of the specific heat ratio depend only slightly on temperature. b This page is intentionally left blank WileyPLUS is a research-based online environment for effective teaching and learning. WileyPLUS builds students’ confidence because it takes the guesswork out of studying by providing students with a clear roadmap: • • • what to do how to do it if they did it right It offers interactive resources along with a complete digital textbook that help students learn more. With WileyPLUS, students take more initiative so you’ll have greater impact on their achievement in the classroom and beyond. For more information, visit www.wileyplus.com ALL THE HELP, RESOURCES, AND PERSONAL SUPPORT YOU AND YOUR STUDENTS NEED! www.wileyplus.com/resources 2-Minute Tutorials and all of the resources you and your students need to get started Student support from an experienced student user Collaborate with your colleagues, find a mentor, attend virtual and live events, and view resources www.WhereFacultyConnect.com Pre-loaded, ready-to-use assignments and presentations created by subject matter experts Technical Support 24/7 FAQs, online chat, and phone support www.wileyplus.com/support Your WileyPLUS Account Manager, providing personal training and support 7 th edition Fundamentals of Fluid Mechanics Bruce R. Munson Department of Aerospace Engineering Iowa State University Ames, Iowa Theodore H. Okiishi Department of Mechanical Engineering Iowa State University Ames, Iowa Wade W. Huebsch Department of Mechanical and Aerospace Engineering West Virginia University Morgantown, West Virginia Alric P. Rothmayer Department of Aerospace Engineering Iowa State University Ames, Iowa John Wiley & Sons, Inc. Executive Publisher: Don Fowley Senior Editor and Product Designer: Jennifer Welter Content Manager: Kevin Holm Senior Content Editor: Wendy Ashenberg Creative Director: Harry Nolan Senior Designer: Madelyn Lesure Executive Marketing Manager: Christopher Ruel Editorial Operations Manager: Melissa Edwards Photo Researcher: Sheena Goldstein Assistant Editor: Samantha Mandel Senior Production Editor: John Curley Media Specialist: Lisa Sabatini Production Management Services: Ingrao Associates/Suzanne Ingrao Cover Design: Madelyn Lesure Cover Photo: Graham Jeffery/Sensitive Light This book was set in 10/12 Times Roman by Aptara®, Inc., and printed and bound by R.R. Donnelley/Jefferson City. The cover was printed by R.R. Donnelley/Jefferson City. 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Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. Library of Congress Cataloging-in-Publication Data Munson, Bruce Roy, 1940Fundamentals of fluid mechanics / Bruce R. Munson, Theodore H. Okiishi, Wade W. Huebsch, Alric P. Rothmayer—7th edition. pages cm Includes indexes. ISBN 978-1-118-11613-5 1. Fluid mechanics—Textbooks. I. Okiishi, T. H. (Theodore Hisao), 1939- II. Huebsch, Wade W. III. Rothmayer, Alric P., 1959- IV. Title. TA357.M86 2013 532–dc23 2012011618 ISBN 978-1-118-11613-5 (Main Book) ISBN 978-1-118-39971-2 (Binder-Ready Version) Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 About the Authors Bruce R. Munson, Professor Emeritus of Engineering Mechanics at Iowa State University, received his B.S. and M.S. degrees from Purdue University and his Ph.D. degree from the Aerospace Engineering and Mechanics Department of the University of Minnesota in 1970. Prior to joining the Iowa State University faculty in 1974, Dr. Munson was on the mechanical engineering faculty of Duke University from 1970 to 1974. From 1964 to 1966, he worked as an engineer in the jet engine fuel control department of Bendix Aerospace Corporation, South Bend, Indiana. Dr. Munson’s main professional activity has been in the area of fluid mechanics education and research. He has been responsible for the development of many fluid mechanics courses for studies in civil engineering, mechanical engineering, engineering science, and agricultural engineering and is the recipient of an Iowa State University Superior Engineering Teacher Award and the Iowa State University Alumni Association Faculty Citation. He has authored and coauthored many theoretical and experimental technical papers on hydrodynamic stability, low Reynolds number flow, secondary flow, and the applications of viscous incompressible flow. He is a member of The American Society of Mechanical Engineers. Ted H. Okiishi, Professor Emeritus of Mechanical Engineering at Iowa State University, joined the faculty there in 1967 after receiving his undergraduate and graduate degrees from that institution. From 1965 to 1967, Dr. Okiishi served as a U.S. Army officer with duty assignments at the National Aeronautics and Space Administration Lewis Research Center, Cleveland, Ohio, where he participated in rocket nozzle heat transfer research, and at the Combined Intelligence Center, Saigon, Republic of South Vietnam, where he studied seasonal river flooding problems. Professor Okiishi and his students have been active in research on turbomachinery fluid dynamics. Some of these projects have involved significant collaboration with government and industrial laboratory researchers, with two of their papers winning the ASME Melville Medal (in 1989 and 1998). Dr. Okiishi has received several awards for teaching. He has developed undergraduate and graduate courses in classical fluid dynamics as well as the fluid dynamics of turbomachines. He is a licensed professional engineer. His professional society activities include having been a vice president of The American Society of Mechanical Engineers (ASME) and of the American Society for Engineering Education. He is a Life Fellow of The American Society of Mechanical Engineers and past editor of its Journal of Turbomachinery. He was recently honored with the ASME R. Tom Sawyer Award. Wade W. Huebsch, Associate Professor in the Department of Mechanical and Aerospace Engineering at West Virginia University, received his B.S. degree in aerospace engineering from San Jose State University where he played college baseball. He received his M.S. degree in mechanical engineering and his Ph.D. in aerospace engineering from Iowa State University in 2000. Dr. Huebsch specializes in computational fluid dynamics research and has authored multiple journal articles in the areas of aircraft icing, roughness-induced flow phenomena, and boundary layer flow control. He has taught both undergraduate and graduate courses in fluid mechanics and has developed a new undergraduate course in computational fluid dynamics. He has received multiple teaching awards such as Outstanding Teacher and Teacher of the Year from the College of Engineering and Mineral Resources at WVU as well as the Ralph R. v vi About the Authors Teetor Educational Award from SAE. He was also named as the Young Researcher of the Year from WVU. He is a member of the American Institute of Aeronautics and Astronautics, the Sigma Xi research society, the Society of Automotive Engineers, and the American Society of Engineering Education. Alric P. Rothmayer, Professor of Aerospace Engineering at Iowa State University, received his undergraduate and graduate degrees from the Aerospace Engineering Department at the University of Cincinnati, during which time he also worked at NASA Langley Research Center and was a visiting graduate research student at the Imperial College of Science and Technology in London. He joined the faculty at Iowa State University (ISU) in 1985 after a research fellowship sponsored by the Office of Naval Research at University College in London. Dr. Rothmayer has taught a wide variety of undergraduate fluid mechanics and propulsion courses for over 25 years, ranging from classical low and high speed flows to propulsion cycle analysis. Dr. Rothmayer was awarded an ISU Engineering Student Council Leadership Award, an ISU Foundation Award for Early Achievement in Research, an ISU Young Engineering Faculty Research Award, and a National Science Foundation Presidential Young Investigator Award. He is an Associate Fellow of the American Institute of Aeronautics and Astronautics (AIAA), and was chair of the 3rd AIAA Theoretical Fluid Mechanics Conference. Dr. Rothmayer specializes in the integration of Computational Fluid Dynamics with asymptotic methods and low order modeling for viscous flows. His research has been applied to diverse areas ranging from internal flows through compliant tubes to flow control and aircraft icing. In 2001, Dr. Rothmayer won a NASA Turning Goals into Reality (TGIR) Award as a member of the Aircraft Icing Project Team, and also won a NASA Group Achievement Award in 2009 as a member of the LEWICE Ice Accretion Software Development Team. He was also a member of the SAE AC-9C Aircraft Icing Technology Subcommittee of the Aircraft Environmental Systems Committee of SAE and the Fluid Dynamics Technical Committee of AIAA. Preface This book is intended for junior and senior engineering students who are interested in learning some fundamental aspects of fluid mechanics. We developed this text to be used as a first course. The principles considered are classical and have been well-established for many years. However, fluid mechanics education has improved with experience in the classroom, and we have brought to bear in this book our own ideas about the teaching of this interesting and important subject. This seventh edition has been prepared after several years of experience by the authors using the previous editions for introductory courses in fluid mechanics. On the basis of this experience, along with suggestions from reviewers, colleagues, and students, we have made a number of changes in this edition. The changes (listed below, and indicated by the word New in descriptions in this preface) are made to clarify, update, and expand certain ideas and concepts. New to This Edition In addition to the continual effort of updating the scope of the material presented and improving the presentation of all of the material, the following items are new to this edition. With the widespread use of new technologies involving the web, DVDs, digital cameras and the like, there is an increasing use and appreciation of the variety of visual tools available for learning. As in recent editions, this fact has been addressed in the new edition by continuing to include additional new illustrations, graphs, photographs, and videos. Illustrations: New illustrations and graphs have been added to this edition, as well as updates to past ones. The book now contains nearly 1600 illustrations. These illustrations range from simple ones that help illustrate a basic concept or equation to more complex ones that illustrate practical applications of fluid mechanics in our everyday lives. Photographs: This edition has also added new photographs throughout the book to enhance the text. The total number of photographs now exceeds 300. Some photos involve situations that are so common to us that we probably never stop to realize how fluids are involved in them. Others involve new and novel situations that are still baffling to us. The photos are also used to help the reader better understand the basic concepts and examples discussed. Combining the illustrations, graphs and photographs, the book has approximately 1900 visual aids. Videos: The video library has been enhanced by the addition of 19 new video segments directly related to the text material, as well as multiple updates to previous videos (i.e. same topic with an updated video clip). In addition to being strategically located at the appropriate places within the text, they are all listed, each with an appropriate thumbnail photo, in the video index. They illustrate many of the interesting and practical applications of real-world fluid phenomena. There are now 175 videos. Examples: The book contains 5 new example problems that involve various fluid flow fundamentals. Some of these examples also incorporate new PtD (Prevention through Design) discussion material. The PtD project, under the direction of the National Institute for Occupational Safety and Health, involves, in part, the use of textbooks to encourage the proper design and use of workday equipment and material so as to reduce accidents and injuries in the workplace. Problems and Problem Types: Approximately 30% new homework problems have been added for this edition, with a total number of 1484 problems in the text (additional problems in WileyPLUS ). Also, new multiple-choice concept questions (developed by Jay Martin and John Mitchell of the University of Wisconsin-Madison) have been added at the beginning of each Problems section. These questions test the students’ knowledge of basic chapter concepts. This edition has also significantly improved the homework problem integration with the vii viii Preface WileyPLUS course management system. New icons have been introduced in the Problems section to help instructors and students identify which problems are available to be assigned within WileyPLUS for automatic grading, and which problems have tutorial help available. Author: A new co-author was brought on board for this edition. We are happy to welcome Dr. Alric P. Rothmayer. Within WileyPLUS: New What an Engineer Sees animations demonstrate an engineer’s perspective of everyday objects, and relates the transfer of theory to real life through the solution of a problem involving that everyday object. New Office-Hours Videos demonstrate the solution of selected problems, focusing specifically on those areas in which students typically run into difficulty, with video and voiceover. Over 700 homework problems from the text that can be assigned for automatic feedback and grading (34 new for the 7th edition). Including 65 GO (Guided Online) Tutorial problems (26 new for this edition). Key Features Illustrations, Photographs, and Videos y Fr < 1 Fr = 1 Fr > 1 E (© Photograph courtesy of Pend Oreille Public Utility District.) Fluid mechanics has always been a “visual” subject—much can be learned by viewing various aspects of fluid flow. In this new edition we have made several changes to reflect the fact that with new advances in technology, this visual component is becoming easier to incorporate into the learning environment, for both access and delivery, and is an important component to the learning of fluid mechanics. Thus, new photographs and illustrations have been added to the book. Some of these are within the text material; some are used to enhance the example problems; and some are included as margin figures of the type shown in the left margin to more clearly illustrate various points discussed in the text. In addition, new video segments have been added, bringing the total number of video segments to 175. These video segments illustrate many interesting and practical applications of real-world fluid phenomena. Each video segment is identified at the appropriate location in the text material by a video icon and thumbnail photograph of the type shown in the left margin. The full video library is shown in the video index at the back of the book. Each video segment has a separate associated text description of what is shown in the video. There are many homework problems that are directly related to the topics in the videos. Examples V1.9 Floating razor blade One of our aims is to represent fluid mechanics as it really is—an exciting and useful discipline. To this end, we include analyses of numerous everyday examples of fluid-flow phenomena to which students and faculty can easily relate. In the seventh edition there are 5 new examples and a total of 164 examples that provide detailed solutions to a variety of problems. Some of the new examples incorporate Prevention through Design (PtD) material. Many of the examples illustrate what happens if one or more of the parameters is changed. This gives the user a better feel for some of the basic principles involved. In addition, many of the examples contain new photographs of the actual device or item involved in the example. Also, all of the examples are outlined and carried out with the problem solving methodology of “Given, Find, Solution, and Comment” as discussed on page 5 in the “Note to User” before Example 1.1. Fluids in the News The set of approximately 60 short “Fluids in the News” stories reflect some of the latest important, and novel, ways that fluid mechanics affects our lives. Many of these problems have homework problems associated with them. Preface ix Homework Problems A set of more than 1480 homework problems (approximately 30% new to this edition) stresses the practical application of principles. The problems are grouped and identified according to topic. An effort has been made to include several new, easier problems at the start of each group. The following types of problems are included: 1) new conceptual multiple-choice problems 9) Excel-based lab problems, 2) “standard” problems, 10) “Lifelong learning” problems, 3) computer problems, 11) problems that require the user to obtain a 4) discussion problems, photograph/image of a given flow situation and 5) supply-your-own-data problems, write a brief paragraph to describe it, 6) review problems with solutions, 12) simple CFD problems to be solved using 7) problems based on the “Fluids in the ANSYS Academic CFD Software, News” topics, 13) Fundamental of Engineering (FE) exam 8) problems based on the fluid videos, questions available on book website. Lab Problems—There are 30 extended, laboratory-type problems that involve actual experimental data for simple experiments of the type that are often found in the laboratory portion of many introductory fluid mechanics courses. The data for these problems are provided in Excel format. Lifelong Learning Problems—Each chapter has lifelong learning problems that involve obtaining additional information about various new state-of-the-art fluid mechanics topics and writing a brief report about this material. Review Problems—There is a set of 186 review problems covering most of the main topics in the book. Complete, detailed solutions to these problems can be found in the Student Solutions Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Well-Paced Concept and Problem-Solving Development Since this is an introductory text, we have designed the presentation of material to allow for the gradual development of student confidence in fluid problem solving. Each important concept or notion is considered in terms of simple and easy-to-understand circumstances before more complicated features are introduced. Many pages contain a brief summary (a highlight) sentence that serves to prepare or remind the reader about an important concept discussed on that page. Several brief components have been added to each chapter to help the user obtain the “big picture” idea of what key knowledge is to be gained from the chapter. A brief Learning Objectives section is provided at the beginning of each chapter. It is helpful to read through this list prior to reading the chapter to gain a preview of the main concepts presented. Upon completion of the chapter, it is beneficial to look back at the original learning objectives to ensure that a satisfactory level of understanding has been acquired for each item. Additional reinforcement of these learning objectives is provided in the form of a Chapter Summary and Study Guide at the end of each chapter. In this section a brief summary of the key concepts and principles introduced in the chapter is included along with a listing of important terms with which the student should be familiar. These terms are highlighted in the text. A list of the main equations in the chapter is included in the chapter summary. System of Units Two systems of units continue to be used throughout most of the text: the International System of Units (newtons, kilograms, meters, and seconds) and the British Gravitational System (pounds, slugs, feet, and seconds). About one-half of the examples and homework problems are in each set of units. The English Engineering System (pounds, pounds mass, feet, and seconds) is used in the discussion of compressible flow in Chapter 11. This usage is standard practice for the topic. x Preface Topical Organization In the first four chapters the student is made aware of some fundamental aspects of fluid motion, including important fluid properties, regimes of flow, pressure variations in fluids at rest and in motion, fluid kinematics, and methods of flow description and analysis. The Bernoulli equation is introduced in Chapter 3 to draw attention, early on, to some of the interesting effects of fluid motion on the distribution of pressure in a flow field. We believe that this timely consideration of elementary fluid dynamics increases student enthusiasm for the more complicated material that follows. In Chapter 4 we convey the essential elements of kinematics, including Eulerian and Lagrangian mathematical descriptions of flow phenomena, and indicate the vital relationship between the two views. For teachers who wish to consider kinematics in detail before the material on elementary fluid dynamics, Chapters 3 and 4 can be interchanged without loss of continuity. Chapters 5, 6, and 7 expand on the basic analysis methods generally used to solve or to begin solving fluid mechanics problems. Emphasis is placed on understanding how flow phenomena are described mathematically and on when and how to use infinitesimal and finite control volumes. The effects of fluid friction on pressure and velocity distributions are also considered in some detail. A formal course in thermodynamics is not required to understand the various portions of the text that consider some elementary aspects of the thermodynamics of fluid flow. Chapter 7 features the advantages of using dimensional analysis and similitude for organizing test data and for planning experiments and the basic techniques involved. Owing to the growing importance of computational fluid dynamics (CFD) in engineering design and analysis, material on this subject is included in Appendix A. This material may be omitted without any loss of continuity to the rest of the text. This introductory CFD overview includes examples and problems of various interesting flow situations that are to be solved using ANSYS Academic CFD software. Chapters 8 through 12 offer students opportunities for the further application of the principles learned early in the text. Also, where appropriate, additional important notions such as boundary layers, transition from laminar to turbulent flow, turbulence modeling, and flow separation are introduced. Practical concerns such as pipe flow, open-channel flow, flow measurement, drag and lift, the effects of compressibility, and the fluid mechanics fundamentals associated with turbomachines are included. Students who study this text and who solve a representative set of the exercises provided should acquire a useful knowledge of the fundamentals of fluid mechanics. Faculty who use this text are provided with numerous topics to select from in order to meet the objectives of their own courses. More material is included than can be reasonably covered in one term. All are reminded of the fine collection of supplementary material. We have cited throughout the text various articles and books that are available for enrichment. Student and Instructor Resources Student Solutions Manual and Study Guide, by Munson et al. (© 2013 John Wiley and Sons, Inc.)—This short paperback book is available as a supplement for the text. It provides detailed solutions to the Review Problems and a concise overview of the essential points of most of the main sections of the text, along with appropriate equations, illustrations, and worked examples. This supplement is available through WileyPLUS, your local bookstore, or you may purchase it on the Wiley website at www.wiley.com/college/munson. Student Companion Site—The student section of the book website at www.wiley.com/ college/munson contains the assets listed below. Access is free-of-charge. Video Library Comprehensive Table of Conversion Factors Review Problems with Answers CFD Driven Cavity Example Lab Problems Instructor Companion Site—The instructor section of the book website at www.wiley.com/ college/munson contains the assets in the Student Companion Site, as well as the following, which are available only to professors who adopt this book for classroom use: Preface xi Instructor Solutions Manual, containing complete, detailed solutions to all of the problems in the text. Figures from the text, appropriate for use in lecture slides. These instructor materials are password-protected. Visit the Instructor Companion Site to register for a password. WileyPLUS. WileyPLUS combines the complete, dynamic online text with all of the teaching and learning resources you need, in one easy-to-use system. This edition offers a much tighter integration between the book and WileyPLUS. The instructor assigns WileyPLUS, but students decide how to buy it: they can buy the new, printed text packaged with a WileyPLUS registration code at no additional cost or choose digital delivery of WileyPLUS, use the online text and integrated read, study, and practice tools, and save off the cost of the new book. WileyPLUS offers today’s engineering students the interactive and visual learning materials they need to help them grasp difficult concepts—and apply what they’ve learned to solve problems in a dynamic environment. A robust variety of examples and exercises enable students to work problems, see their results, and obtain instant feedback including hints and reading references linked directly to the online text. Contact your local Wiley representative, or visit www.wileyplus.com for more information about using WileyPLUS in your course. Acknowledgments We wish to express our gratitude to the many persons who provided suggestions for this and previous editions through reviews and surveys. In addition, we wish to express our appreciation to the many persons who supplied photographs and videos used throughout the text. Finally, we thank our families for their continued encouragement during the writing of this seventh edition. Working with students over the years has taught us much about fluid mechanics education. We have tried in earnest to draw from this experience for the benefit of users of this book. Obviously we are still learning, and we welcome any suggestions and comments from you. BRUCE R. MUNSON THEODORE H. OKIISHI WADE W. HUEBSCH ALRIC P. ROTHMAYER Featured in this Book 5 Finite Control Volume Analysis CHAPTER OPENING PHOTO: Wind turbine farms (this is the Middelgrunden Offshore Wind Farm in Denmark) are becoming more common. Finite control volume analysis can be used to estimate the amount of energy transferred between the moving air and each turbine rotor. (Photograph courtesy of Siemens Wind Power.) Learning Objectives Learning Objectives at the beginning of each chapter focus students’ attention as they read the chapter. After completing this chapter, you should be able to: I select an appropriate finite control volume to solve a fluid mechanics problem. I apply conservation of mass and energy and Newton’s second law of motion to the contents of a finite control volume to get important answers. I know how velocity changes and energy transfers in fluid flows are related to forces and torques. I understand why designing for minimum loss of energy in fluid flows is so important. Viscosity not important Photographs, Illustrations, and Videos Boundary layer (a) low Reynolds number flow, (b) moderate Reynolds number flow, (c) large Reynolds number flow. Most familiar flows involve large Reynolds numbers. E XAMPLE s 9.2 Characteristics of Flow Past Objects FIND Would the flow characteristics for these three situations be similar? Explain. i n Spreading of oil spills With the large traffic in oil tankers there is great interest in the prevention of and response to oil spills. As evidenced by the famous Exxon Valdez oil spill in Prince William Sound in 1989, oil spills can create disastrous environmental problems. A more recent example of this type of catastrophe is the oil spill that occurred in the Gulf of Mexico in 2010. It is not surprising that much attention is given to the rate at which an oil spill spreads. When spilled, most oils tend to spread horizontally into a smooth and slippery surface, called a xii field. Since the shear stress 1i.e., viscous effect2 is the product of the fluid viscosity and the velocity gradient, it follows that viscous effects are confined to the boundary layer and wake regions. The characteristics described in Figs. 9.5 and 9.6 for flow past a flat plate and a circular cylinder are typical of flows past streamlined and blunt bodies, respectively. The nature of the flow depends strongly on the Reynolds number. (See Ref. 31 for many examples illustrating this behavior.) Most familiar flows are similar to the large Reynolds number flows depicted in Figs. 9.5c and 9.6c, rather than the low Reynolds number flow situations. (See the photograph at the beginning of Chapters 7 and 11.) In the remainder of this chapter we will investigate more thoroughly these ideas and determine how to calculate the forces on immersed bodies. GIVEN It is desired to experimentally determine the various characteristics of flow past a car as shown in Fig E9.2. The following tests could be carried out: 1a2 U ⫽ 20 mmⲐs flow of glycerin past a scale model that is 34-mm tall, 100-mm long, and 40-mm wide, 1b2 U ⫽ 20 mmⲐs air flow past the same scale model, or 1c2 U ⫽ 25 mⲐs airflow past the actual car, which is 1.7-m tall, 5-m long, and 2-m wide. relate brief stories involving current, sometimes novel, applications of fluids phenomena. Many are associated with homework problems. d (c) I Figure 9.6 Character of the steady, viscous flow past a circular cylinder: Fluids in the News i x Separated region Re = 10 V9.3 Human aerodynamic wake emphasize important points for the reader. u Wake region δ tyield V1.6 NonNewtonian behavior 17 For shear thickening fluids the apparent viscosity increases with increasing shear rate—the harder the fluid is sheared, the more viscous it becomes. Common examples of this type of fluid include water–corn starch mixture and water–sand mixture 1“quicksand”2. Thus, the difficulty in removing an object from quicksand increases dramatically as the speed of removal increases. The other type of behavior indicated in Fig. 1.7 is that of a Bingham plastic, which is neither a fluid nor a solid. Such material can withstand a finite, nonzero shear stress, ␶yield , the yield stress, without motion 1therefore, it is not a fluid2, but once the yield stress is exceeded it flows like a fluid 1hence, it is not a solid2. Toothpaste and mayonnaise are common examples of Bingham plastic materials. As indicated in the figure in the margin, mayonnaise can sit in a pile on a slice of bread 1the shear stress less than the yield stress2,but it flows smoothly into a thin layer when the knife increases the stress above the yield stress. From Eq. 1.9 it can be readily deduced that the dimensions of viscosity are FTL2. Thus, in BG units viscosity is given as lb # sft2 and in SI units as N # s m2. Values of viscosity for several common liquids and gases are listed in Tables 1.5 through 1.8. A quick glance at these tables reveals the wide variation in viscosity among fluids. Viscosity is only mildly dependent on pressure and the effect of pressure is usually neglected. However, as previously mentioned, and as illustrated in Fig. 1.8, viscosity is very sensitive to temperature. For example, as the temperature of water changes from 60 to 100 °F the density decreases by less than 1%, but the viscosity decreases by about 40%. It is thus clear that particular attention must be given to temperature when determining viscosity. Figure 1.8 shows in more detail how the viscosity varies from fluid to fluid and how for a given fluid it varies with temperature. It is to be noted from this figure that the viscosity of liquids decreases with an increase in temperature, whereas for gases an increase in temperature causes an increase in viscosity. This difference in the effect of temperature on the viscosity of liquids and gases can again be traced back to the difference in molecular structure. The liquid molecules are closely spaced, with strong cohesive forces between molecules, and the resistance to relative motion between adjacent layers of fluid is related to these intermolecular forces. As 4.0 2.0 1.0 8 6 4 Gl yc 2 SA E -1 1 × 10 Dynamic viscosity, m, N • s/m2 The various types of non-Newtonian fluids are distinguished by how their apparent viscosity changes with shear rate. Viscosity 8 6 4 10 W er in oi l 2 -2 1 × 10 8 6 4 2 1 × 10-3 8 6 4 Water 2 -4 1 × 10 8 6 4 Air 2 1 × 10-5 8 6 -20 Hydrogen 0 20 40 60 Temperature, °C 80 100 120 ■ Figure 1.8 Dynamic (absolute) viscosity of some common fluids as a function of temperature. 18 Chapter 1 ■ Introduction the temperature increases, these cohesive forces are reduced with a corresponding reduction in resistance to motion. Since viscosity is an index of this resistance, it follows that the viscosity is reduced by an increase in temperature. In gases, however, the molecules are widely spaced and intermolecular forces negligible. In this case, resistance to relative motion arises due to the exchange of momentum of gas molecules between adjacent layers. As molecules are transported by random motion from a region of low bulk velocity to mix with molecules in a region of higher bulk velocity 1and vice versa2, there is an effective momentum exchange that resists the relative motion between the layers. As the temperature of the gas increases, the random molecular activity increases with a corresponding increase in viscosity. The effect of temperature on viscosity can be closely approximated using two empirical formulas. For gases the Sutherland equation can be expressed as Viscosity is very sensitive to temperature. m CT 32 TS (1.10) where C and S are empirical constants, and T is absolute temperature. Thus, if the viscosity is known at two temperatures, C and S can be determined. Or, if more than two viscosities are known, the data can be correlated with Eq. 1.10 by using some type of curve-fitting scheme. For liquids an empirical equation that has been used is m DeBT (1.11) where D and B are constants and T is absolute temperature. This equation is often referred to as Andrade’s equation. As was the case for gases, the viscosity must be known at least for two temperatures so the two constants can be determined. A more detailed discussion of the effect of temperature on fluids can be found in Ref. 1. E XAMPLE 1.4 Viscosity and Dimensionless Quantities GIVEN A dimensionless combination of variables that is important in the study of viscous flow through pipes is called the Reynolds number, Re, defined as rVDm where, as indicated in Fig. E1.4, r is the fluid density, V the mean fluid velocity, D the pipe diameter, and m the fluid viscosity. A Newtonian fluid having a viscosity of 0.38 N # sm2 and a specific gravity of 0.91 flows through a 25-mmdiameter pipe with a velocity of 2.6 ms. V FIND Determine the value of the Reynolds number using 1a2 SI units and 1b2 BG units. r, m D SOLUTION (a) The fluid density is calculated from the specific gravity as r SG rH2O@4 °C 0.91 11000 kgm3 2 910 kgm3 ■ Figure E1.4 and from the definition of the Reynolds number 1910 kgm3 212.6 ms2125 mm21103 mmm2 rVD m 0.38 N # sm2 156 1kg # ms2 2 N Re However, since 1 N 1 kg # ms2 it follows that the Reynolds number is unitless—that is, Re 156 (Ans) The value of any dimensionless quantity does not depend on the system of units used if all variables that make up the quantity are expressed in a consistent set of units. To check this, we will calculate the Reynolds number using BG units. (b) We first convert all the SI values of the variables appearing in the Reynolds number to BG values by using the conversion factors from Table 1.4. Thus, r 1910 kgm3 211.940 103 2 1.77 slugsft3 V 12.6 ms213.2812 8.53 fts D 10.025 m213.2812 8.20 102 ft m 10.38 N # sm2 212.089 102 2 7.94 103 lb # sft2 1.6 11.77 slugsft 2 18.53 ft s218.20 10 Re 7.94 103 lb # sft2 2 ft2 156 1slug # fts2 2 lb 156 (Ans) since 1 lb 1 slug # fts . 2 E XAMPLE same, as expected. Dimensionless quantities play an important role in fluid mechanics, and the significance of the Reynolds number as well as other important dimensionless combinations will be discussed in detail in Chapter 7. It should be noted that in the Reynolds number it is actually the ratio mr that is important, and this is the property that is defined as the kinematic viscosity. Newtonian Fluid Shear Stress 1.5 GIVEN The velocity distribution for the flow of a Newtonian fluid between two fixed wide, parallel plates (see Fig. E1.5a) is given by the equation u 19 COMMENTS The values from part 1a2 and part 1b2 are the and the value of the Reynolds number is 3 Viscosity y 2 3V c1 a b d 2 h where V is the mean velocity. The fluid has a viscosity of 0.04 lb # s ft2. Also, V 2 fts and h 0.2 in. FIND Determine: (a) the shearing stress acting on the bottom wall, and (b) the shearing stress acting on a plane parallel to the walls and passing through the centerline (midplane). SOLUTION For this type of parallel flow the shearing stress is obtained from Eq. 1.9, tm du dy 3Vy du 2 dy h (2) (a) Along the bottom wall y h so that (from Eq. 2) du 3V dy h u h (1) Thus, if the velocity distribution u u1y2 is known, the shearing stress can be determined at all points by evaluating the velocity gradient, du dy. For the distribution given y h ■ Figure E1.5a COMMENT From Eq. 2 we see that the velocity gradient (and therefore the shearing stress) varies linearly with y and in this particular example varies from 0 at the center of the channel to 14.4 lb ft2 at the walls. This is shown in Fig. E1.5b. For the more general case the actual variation will, of course, depend on the nature of the velocity distribution. and therefore the shearing stress is wall 10.04 lb # sft2 213212 fts2 3V b h 10.2 in.211 ft12 in.2 14.4 lbft2 1in direction of flow2 15 (Ans) This stress creates a drag on the wall. Since the velocity distribution is symmetrical, the shearing stress along the upper wall would have the same magnitude and direction. (b) Along the midplane where y 0 it follows from Eq. 2 that du 0 dy 10 5 midplane = 0 0 0.2 and thus the shearing stress is tmidplane 0 bottom wall = 14.4 lb/ft2 = top wall , lb/ft 2 tbottom m a (Ans) 0.1 0 y, in. 0.1 0.2 ■ Figure E1.5b Quite often viscosity appears in fluid flow problems combined with the density in the form n m r 20 Chapter 1 ■ Introduction This ratio is called the kinematic viscosity and is denoted with the Greek symbol n 1nu2. The dimensions of kinematic viscosity are L2T, and the BG units are ft2 s and SI units are m2s. Values of kinematic viscosity for some common liquids and gases are given in Tables 1.5 through 1.8. More extensive tables giving both the dynamic and kinematic viscosities for water and air can be found in Appendix B 1Tables B.1 through B.42, and graphs showing the variation in both dynamic and kinematic viscosity with temperature for a variety of fluids are also provided in Appendix B 1Figs. B.1 and B.22. Although in this text we are primarily using BG and SI units, dynamic viscosity is often expressed in the metric CGS 1centimeter-gram-second2 system with units of dyne # scm2. This combination is called a poise, abbreviated P. In the CGS system, kinematic viscosity has units of cm2s, and this combination is called a stoke, abbreviated St. Kinematic viscosity is defined as the ratio of the absolute viscosity to the fluid density. 1.7 Compressibility of Fluids 1.7.1 Bulk Modulus p An important question to answer when considering the behavior of a particular fluid is how easily can the volume 1and thus the density2 of a given mass of the fluid be changed when there is a change in pressure? That is, how compressible is the fluid? A property that is commonly used to characterize compressibility is the bulk modulus, Ev, defined as V Ev p + dp dp dV V (1.12) where dp is the differential change in pressure needed to create a differential change in volume, dV , of a volume V . This is illustrated by the figure in the margin. The negative sign is included since an increase in pressure will cause a decrease in volume. Since a decrease in volume of a given mass, m rV , will result in an increase in density, Eq. 1.12 can also be expressed as V – dV Ev (1.13) The bulk modulus 1also referred to as the bulk modulus of elasticity2 has dimensions of pressure, FL2. In BG units, values for Ev are usually given as lbin.2 1psi2 and in SI units as Nm2 1Pa2. Large values for the bulk modulus indicate that the fluid is relatively incompressible—that is, it takes a large pressure change to create a small change in volume. As expected, values of Ev for common liquids are large 1see Tables 1.5 and 1.62. For example, at atmospheric pressure and a temperature of 60 °F it would require a pressure of 3120 psi to compress a unit volume of water 1%. This result is representative of the compressibility of liquids. Since such large pressures are required to effect a change in volume, we conclude that liquids can be considered as incompressible for most practical engineering applications. As liquids are compressed the bulk modulus increases, but the bulk modulus near atmospheric pressure is usually the one of interest. The use of bulk modulus as a property describing compressibility is most prevalent when dealing with liquids, although the bulk modulus can also be determined for gases. V1.7 Water balloon F dp drr l u i d s i n This water jet is a blast Usually liquids can be treated as incompressible fluids. However, in some applications the compressibility of a liquid can play a key role in the operation of a device. For example, a water pulse generator using compressed water has been developed for use in mining operations. It can fracture rock by producing an effect comparable to a conventional explosive such as gunpowder. The device uses the energy stored in a water-filled accumulator to generate an ultrahigh-pressure water pulse ejected through a 10- to 25-mm-diameter discharge valve. At the ultrahigh pressures used (300 to 400 MPa, or 3000 to 4000 atmos- t h e N e w s pheres), the water is compressed (i.e., the volume reduced) by about 10 to 15%. When a fast-opening valve within the pressure vessel is opened, the water expands and produces a jet of water that upon impact with the target material produces an effect similar to the explosive force from conventional explosives. Mining with the water jet can eliminate various hazards that arise with the use of conventional chemical explosives, such as those associated with the storage and use of explosives and the generation of toxic gas by-products that require extensive ventilation. (See Problem 1.110.) 1.7 Compressibility of Fluids 21 1.7.2 Compression and Expansion of Gases When gases are compressed 1or expanded2, the relationship between pressure and density depends on the nature of the process. If the compression or expansion takes place under constant temperature conditions 1isothermal process2, then from Eq. 1.8 Isentropic (k = 1.4) p Isothermal ρ The value of the bulk modulus depends on the type of process involved. p (1.14) constant r If the compression or expansion is frictionless and no heat is exchanged with the surroundings 1isentropic process2, then p (1.15) constant rk where k is the ratio of the specific heat at constant pressure, cp, to the specific heat at constant volume, cv 1i.e., k cp cv 2. The two specific heats are related to the gas constant, R, through the equation R cp cv. As was the case for the ideal gas law, the pressure in both Eqs. 1.14 and 1.15 must be expressed as an absolute pressure. Values of k for some common gases are given in Tables 1.7 and 1.8 and for air over a range of temperatures, in Appendix B 1Tables B.3 and B.42. The pressure–density variations for isothermal and isentropic conditions are illustrated in the margin figure. With explicit equations relating pressure and density, the bulk modulus for gases can be determined by obtaining the derivative dpdr from Eq. 1.14 or 1.15 and substituting the results into Eq. 1.13. It follows that for an isothermal process Ev p (1.16) Ev kp (1.17) and for an isentropic process, Note that in both cases the bulk modulus varies directly with pressure. For air under standard atmospheric conditions with p 14.7 psi 1abs2 and k 1.40, the isentropic bulk modulus is 20.6 psi. A comparison of this figure with that for water under the same conditions 1Ev 312,000 psi2 shows that air is approximately 15,000 times as compressible as water. It is thus clear that in dealing with gases, greater attention will need to be given to the effect of compressibility on fluid behavior. However, as will be discussed further in later sections, gases can often be treated as incompressible fluids if the changes in pressure are small. E XAMPLE Isentropic Compression of a Gas 1.6 GIVEN A cubic foot of air at an absolute pressure of 14.7 psi is compressed isentropically to 12 ft3 by the tire pump shown in Fig. E1.6a. FIND What is the final pressure? SOLUTION For an isentropic compression pi rki pf rkf where the subscripts i and f refer to initial and final states, respectively. Since we are interested in the final pressure, pf, it follows that rf k pf a b pi ri ■ Figure E1.6a 22 Chapter 1 ■ Introduction As the volume, V . is reduced by one-half, the density must double, since the mass, m r V . of the gas remains constant. Thus, with k 1.40 for air pf 122 1.40 114.7 psi2 38.8 psi 1abs2 400 350 300 (Ans) of the ratio of the final volume to the initial volume, Vf Vi , the results shown in Fig. E1.6b are obtained. Note that even though air is often considered to be easily compressed (at least compared to liquids), it takes considerable pressure to significantly reduce a given volume of air as is done in an automobile engine where the compression ratio is on the order of V f V i 1/8 0.125. pf, psi 250 COMMENT By repeating the calculations for various values 200 150 100 (0.5, 38.8 psi) 50 0 0 0.2 0.4 0.6 0.8 1 Vf /Vi ■ Figure E1.6b 1.7.3 Speed of Sound The velocity at which small disturbances propagate in a fluid is called the speed of sound. Another important consequence of the compressibility of fluids is that disturbances introduced at some point in the fluid propagate at a finite velocity. For example, if a fluid is flowing in a pipe and a valve at the outlet is suddenly closed 1thereby creating a localized disturbance2, the effect of the valve closure is not felt instantaneously upstream. It takes a finite time for the increased pressure created by the valve closure to propagate to an upstream location. Similarly, a loudspeaker diaphragm causes a localized disturbance as it vibrates, and the small change in pressure created by the motion of the diaphragm is propagated through the air with a finite velocity. The velocity at which these small disturbances propagate is called the acoustic velocity or the speed of sound, c. It will be shown in Chapter 11 that the speed of sound is related to changes in pressure and density of the fluid medium through the equation 6000 c water 4000 (1.18) c, ft/s or in terms of the bulk modulus defined by Eq. 1.13 c 2000 air 0 dp B dr 0 100 T, F 200 Ev Br (1.19) Since the disturbance is small, there is negligible heat transfer and the process is assumed to be isentropic. Thus, the pressure–density relationship used in Eq. 1.18 is that for an isentropic process. For gases undergoing an isentropic process, Ev kp 1Eq. 1.172 so that c kp Br and making use of the ideal gas law, it follows that c 1kRT V1.8 As fast as a speeding bullet (1.20) Thus, for ideal gases the speed of sound is proportional to the square root of the absolute temperature. For example, for air at 60 °F with k 1.40 and R 1716 ft # lbslug # °R, it follows that c 1117 fts. The speed of sound in air at various temperatures can be found in Appendix B 1Tables B.3 and B.42. Equation 1.19 is also valid for liquids, and values of Ev can be used to determine the speed of sound in liquids. For water at 20 °C, Ev 2.19 GNm2 and r 998.2 kgm3 so that c 1481 ms or 4860 ft s. As shown by the figure in the margin, the speed of sound is much higher in water than in air. If a fluid were truly incompressible 1Ev q 2 1.8 Vapor Pressure 23 the speed of sound would be infinite. The speed of sound in water for various temperatures can be found in Appendix B 1Tables B.1 and B.22. E XAMPLE 1.7 Speed of Sound and Mach Number GIVEN A jet aircraft flies at a speed of 550 mph at an altitude FIND Determine the ratio of the speed of the aircraft, V, to that of 35,000 ft, where the temperature is 66 F and the specific heat ratio is k 1.4. of the speed of sound, c, at the specified altitude. SOLUTION 0.9 From Eq. 1.20 the speed of sound can be calculated as c 2kRT 211.402 11716 ft lbslug °R2166 4602 °R 973 ft /s V 1550 mi/hr215280 ft /mi2 13600 s/hr2 Ma = V/c Since the air speed is 807 ft /s (–66 F, 0.829) 0.7 0.6 the ratio is V 807 ft /s 0.829 c 973 ft /s (Ans) COMMENT This ratio is called the Mach number, Ma. If Ma 1.0 the aircraft is flying at subsonic speeds, whereas for Ma 1.0 it is flying at supersonic speeds. The Mach number is an important dimensionless parameter used in the study of the flow of gases at high speeds and will be further discussed in Chapters 7 and 11. By repeating the calculations for different temperatures, the results shown in Fig. E1.7 are obtained. Because the speed of 1.8 0.8 0.5 –100 –50 0 T, F 50 100 ■ Figure E1.7 sound increases with increasing temperature, for a constant airplane speed, the Mach number decreases as the temperature increases. Vapor Pressure Liquid Vapor, pv Liquid A liquid boils when the pressure is reduced to the vapor pressure. It is a common observation that liquids such as water and gasoline will evaporate if they are simply placed in a container open to the atmosphere. Evaporation takes place because some liquid molecules at the surface have sufficient momentum to overcome the intermolecular cohesive forces and escape into the atmosphere. If the container is closed with a small air space left above the surface, and this space evacuated to form a vacuum, a pressure will develop in the space as a result of the vapor that is formed by the escaping molecules. When an equilibrium condition is reached so that the number of molecules leaving the surface is equal to the number entering, the vapor is said to be saturated and the pressure that the vapor exerts on the liquid surface is termed the vapor pressure, pv. Similarly, if the end of a completely liquid-filled container is moved as shown in the figure in the margin without letting any air into the container, the space between the liquid and the end becomes filled with vapor at a pressure equal to the vapor pressure. Since the development of a vapor pressure is closely associated with molecular activity, the value of vapor pressure for a particular liquid depends on temperature. Values of vapor pressure for water at various temperatures can be found in Appendix B 1Tables B.1 and B.22, and the values of vapor pressure for several common liquids at room temperatures are given in Tables 1.5 and 1.6. Boiling, which is the formation of vapor bubbles within a fluid mass, is initiated when the absolute pressure in the fluid reaches the vapor pressure. As commonly observed in the kitchen, water Boiling temperature, F 24 Chapter 1 ■ Introduction 250 150 50 0 0 20 40 60 Altitude, thousands of feet In flowing liquids it is possible for the pressure in localized regions to reach vapor pressure, thereby causing cavitation. 1.9 Surface Tension Surface tension, lb/ft V1.9 Floating razor blade 6 × 10−3 4 Water 2 0 at standard atmospheric pressure will boil when the temperature reaches 212 °F 1100 °C2 —that is, the vapor pressure of water at 212 °F is 14.7 psi 1abs2. However, if we attempt to boil water at a higher elevation, say 30,000 ft above sea level 1the approximate elevation of Mt. Everest2, where the atmospheric pressure is 4.37 psi 1abs2, we find that boiling will start when the temperature is about 157 °F. At this temperature the vapor pressure of water is 4.37 psi 1abs2. For the U.S. Standard Atmosphere 1see Section 2.42, the boiling temperature is a function of altitude as shown in the figure in the margin. Thus, boiling can be induced at a given pressure acting on the fluid by raising the temperature, or at a given fluid temperature by lowering the pressure. An important reason for our interest in vapor pressure and boiling lies in the common observation that in flowing fluids it is possible to develop very low pressure due to the fluid motion, and if the pressure is lowered to the vapor pressure, boiling will occur. For example, this phenomenon may occur in flow through the irregular, narrowed passages of a valve or pump. When vapor bubbles are formed in a flowing fluid, they are swept along into regions of higher pressure where they suddenly collapse with sufficient intensity to actually cause structural damage. The formation and subsequent collapse of vapor bubbles in a flowing fluid, called cavitation, is an important fluid flow phenomenon to be given further attention in Chapters 3 and 7. 0 50 100 150 200 Temperature, F F l u At the interface between a liquid and a gas, or between two immiscible liquids, forces develop in the liquid surface that cause the surface to behave as if it were a “skin” or “membrane” stretched over the fluid mass. Although such a skin is not actually present, this conceptual analogy allows us to explain several commonly observed phenomena. For example, a steel needle or a razor blade will float on water if placed gently on the surface because the tension developed in the hypothetical skin supports it. Small droplets of mercury will form into spheres when placed on a smooth surface because the cohesive forces in the surface tend to hold all the molecules together in a compact shape. Similarly, discrete bubbles will form in a liquid. (See the photograph at the beginning of Chapter 1.) These various types of surface phenomena are due to the unbalanced cohesive forces acting on the liquid molecules at the fluid surface. Molecules in the interior of the fluid mass are surrounded by molecules that are attracted to each other equally. However, molecules along the surface are subjected to a net force toward the interior. The apparent physical consequence of this unbalanced force along the surface is to create the hypothetical skin or membrane. A tensile force may be considered to be acting in the plane of the surface along any line in the surface. The intensity of the molecular attraction per unit length along any line in the surface is called the surface tension and is designated by the Greek symbol s 1sigma2. For a given liquid the surface tension depends on temperature as well as the other fluid it is in contact with at the interface. The dimensions of surface tension are FL1 with BG units of lb ft and SI units of N m. Values of surface tension for some common liquids 1in contact with air2 are given in Tables 1.5 and 1.6 and in Appendix B 1Tables B.1 and B.22 for water at various temperatures. As indicated by the figure in the margin, the value of the surface tension decreases as the temperature increases. i d s i n Walking on water Water striders are insects commonly found on ponds, rivers, and lakes that appear to “walk” on water. A typical length of a water strider is about 0.4 in., and they can cover 100 body lengths in one second. It has long been recognized that it is surface tension that keeps the water strider from sinking below the surface. What has been puzzling is how they propel themselves at such a high speed. They can’t pierce the water surface or they would sink. A team of mathematicians and engineers from the Massachusetts Institute of Technology (MIT) applied conventional flow visualization techniques and high-speed video to t h e N e w s examine in detail the movement of the water striders. They found that each stroke of the insect’s legs creates dimples on the surface with underwater swirling vortices sufficient to propel it forward. It is the rearward motion of the vortices that propels the water strider forward. To further substantiate their explanation, the MIT team built a working model of a water strider, called Robostrider, which creates surface ripples and underwater vortices as it moves across a water surface. Waterborne creatures, such as the water strider, provide an interesting world dominated by surface tension. (See Problem 1.131.) 1.9 σ Surface Tension 25 R σ Δ p π R2 ■ Figure 1.9 Forces acting on one-half of a liquid drop. The pressure inside a drop of fluid can be calculated using the free-body diagram in Fig. 1.9. If the spherical drop is cut in half 1as shown2, the force developed around the edge due to surface tension is 2pRs. This force must be balanced by the pressure difference, ¢p, between the internal pressure, pi, and the external pressure, pe, acting over the circular area, pR2. Thus, 2pRs ¢p pR2 V1.10 Capillary rise or ¢p pi pe h 2s R (1.21) It is apparent from this result that the pressure inside the drop is greater than the pressure surrounding the drop. 1Would the pressure on the inside of a bubble of water be the same as that on the inside of a drop of water of the same diameter and at the same temperature?2 Among common phenomena associated with surface tension is the rise 1or fall2 of a liquid in a capillary tube. If a small open tube is inserted into water, the water level in the tube will rise above the water level outside the tube, as is illustrated in Fig. 1.10a. In this situation we have a liquid–gas–solid interface. For the case illustrated there is an attraction 1adhesion2 between the wall of the tube and liquid molecules which is strong enough to overcome the mutual attraction 1cohesion2 of the molecules and pull them up the wall. Hence, the liquid is said to wet the solid surface. The height, h, is governed by the value of the surface tension, s, the tube radius, R, the specific weight of the liquid, g, and the angle of contact, u, between the fluid and tube. From the freebody diagram of Fig. 1.10b we see that the vertical force due to the surface tension is equal to 2pRs cos u and the weight is gpR2h, and these two forces must balance for equilibrium. Thus, h ~ 1 R R gpR2h 2pRs cos u V1.11 Contact angle so that the height is given by the relationship h Capillary action in small tubes, which involves a liquid– gas–solid interface, is caused by surface tension. 2s cos u gR (1.22) The angle of contact is a function of both the liquid and the surface. For water in contact with clean glass u ⬇ 0°. It is clear from Eq. 1.22 that the height is inversely proportional to the tube radius, and therefore, as indicated by the figure in the margin, the rise of a liquid in a tube as a result of capillary action becomes increasingly pronounced as the tube radius is decreased. If adhesion of molecules to the solid surface is weak compared to the cohesion between molecules, the liquid will not wet the surface and the level in a tube placed in a nonwetting liquid will actually be depressed, as shown in Fig. 1.10c. Mercury is a good example of a nonwetting liquid when it is in contact with a glass tube. For nonwetting liquids the angle of contact is greater than 90°, and for mercury in contact with clean glass u ⬇ 130°. θ 2π Rσ θ h γ π R2h h ■ Figure 1.10 Effect of capillary action in small 2R (a) (b) (c) tubes. (a) Rise of column for a liquid that wets the tube. (b) Free-body diagram for calculating column height. (c) Depression of column for a nonwetting liquid. 26 Chapter 1 ■ Introduction E XAMPLE Capillary Rise in a Tube 1.8 GIVEN Pressures are sometimes determined by measuring the height of a column of liquid in a vertical tube. FIND What diameter of clean glass tubing is required so that the rise of water at 20 °C in a tube due to capillary action 1as opposed to pressure in the tube2 is less than h 1.0 mm? SOLUTION From Eq. 1.22 h 2s cos u gR R 2s cos u gh Note that as the allowable capillary rise is decreased, the diameter of the tube must be significantly increased. There is always some capillarity effect, but it can be minimized by using a large enough diameter tube. so that 100 For water at 20 °C 1from Table B.22, s 0.0728 Nm and g 9.789 kNm3. Since u ⬇ 0° it follows that for h 1.0 mm, 19.789 10 N m 211.0 mm2110 0.0149 m 3 3 3 D, mm 210.0728 N m2112 R 80 mmm2 60 40 (1 mm, 29.8 mm) 20 and the minimum required tube diameter, D, is D 2R 0.0298 m 29.8 mm (Ans) 0 0 0.5 of the capillary rise, h, the results shown in Fig. E1.8 are obtained. l 1.5 2 ■ Figure E1.8 Surface tension effects play a role in many fluid mechanics problems, including the movement of liquids through soil and other porous media, flow of thin films, formation of drops and bubbles, and the breakup of liquid jets. For example, surface tension is a main factor in the formation of drops from a leaking faucet, as shown in the photograph in the margin. Surface phenomena associated with liquid–gas, liquid–liquid, and liquid–gas–solid interfaces are exceedingly complex, and a more detailed and rigorous discussion of them is beyond the scope of this text. Fortunately, in many fluid mechanics problems, surface phenomena, as characterized by surface tension, are not important, since inertial, gravitational, and viscous forces are much more dominant. (Photograph copyright 2007 by Andrew Davidhazy, Rochester Institute of Technology.) F 1 h, mm COMMENT By repeating the calculations for various values u i d s i n Spreading of oil spills With the large traffic in oil tankers there is great interest in the prevention of and response to oil spills. As evidenced by the famous Exxon Valdez oil spill in Prince William Sound in 1989, oil spills can create disastrous environmental problems. A more recent example of this type of catastrophe is the oil spill that occurred in the Gulf of Mexico in 2010. It is not surprising that much attention is given to the rate at which an oil spill spreads. When spilled, most oils tend to spread horizontally into a smooth and slippery surface, called a t h e N e w s slick. There are many factors that influence the ability of an oil slick to spread, including the size of the spill, wind speed and direction, and the physical properties of the oil. These properties include surface tension, specific gravity, and viscosity. The higher the surface tension the more likely a spill will remain in place. Since the specific gravity of oil is less than one, it floats on top of the water, but the specific gravity of an oil can increase if the lighter substances within the oil evaporate. The higher the viscosity of the oil, the greater the tendency to stay in one place. 1.10 1.10 A Brief Look Back in History 27 A Brief Look Back in History Some of the earliest writings that pertain to modern fluid mechanics can be traced back to the ancient Greek civilization and subsequent Roman Empire. Before proceeding with our study of fluid mechanics, we should pause for a moment to consider the history of this important engineering science. As is true of all basic scientific and engineering disciplines, their actual beginnings are only faintly visible through the haze of early antiquity. But we know that interest in fluid behavior dates back to the ancient civilizations. Through necessity there was a practical concern about the manner in which spears and arrows could be propelled through the air, in the development of water supply and irrigation systems, and in the design of boats and ships. These developments were, of course, based on trial-and-error procedures without any knowledge of mathematics or mechanics. However, it was the accumulation of such empirical knowledge that formed the basis for further development during the emergence of the ancient Greek civilization and the subsequent rise of the Roman Empire. Some of the earliest writings that pertain to modern fluid mechanics are those of Archimedes 1287–212 B.C.2, a Greek mathematician and inventor who first expressed the principles of hydrostatics and flotation. Elaborate water supply systems were built by the Romans during the period from the fourth century B.C. through the early Christian period, and Sextus Julius Frontinus 1A.D. 40–1032, a Roman engineer, described these systems in detail. However, for the next 1000 years during the Middle Ages 1also referred to as the Dark Ages2, there appears to have been little added to further understanding of fluid behavior. As shown in Fig. 1.11, beginning with the Renaissance period 1about the fifteenth century2 a rather continuous series of contributions began that forms the basis of what we consider to be the science of fluid mechanics. Leonardo da Vinci 11452–15192 described through sketches and writings many different types of flow phenomena. The work of Galileo Galilei 11564–16422 marked the beginning of experimental mechanics. Following the early Renaissance period and during the seventeenth and eighteenth centuries, numerous significant contributions were made. These include theoretical and mathematical advances associated with the famous names of Newton, Bernoulli, Euler, and d’Alembert. Experimental aspects of fluid mechanics were also advanced during this period, but unfortunately the two different approaches, theoretical and experimental, developed along separate paths. Hydrodynamics was the term associated with the theoretical or mathematical study of idealized, frictionless fluid behavior, with the term hydraulics being used to describe the applied or experimental aspects of real fluid behavior, particularly the behavior of water. Further contributions and refinements were made to both theoretical hydrodynamics and experimental hydraulics during the nineteenth century, with the general differential equations describing fluid motions that are used in modern fluid mechanics being developed in this period. Experimental hydraulics became more of a science, and many of the results of experiments performed during the nineteenth century are still used today. At the beginning of the twentieth century, both the fields of theoretical hydrodynamics and experimental hydraulics were highly developed, and attempts were being made to unify the two. In 1904 a classic paper was presented by a German professor, Ludwig Prandtl 11875–19532, who introduced the concept of a “fluid boundary layer,” which laid the foundation for the unification of the theoretical and experimental aspects of fluid mechanics. Prandtl’s idea was that for flow next to Geoffrey Taylor Theodor von Karman Ludwig Prandtl Osborne Reynolds Ernst Mach George Stokes Jean Poiseuille Louis Navier Leonhard Euler Daniel Bernoulli Isaac Newton Galileo Galilei Leonardo da Vinci 1200 1300 1400 1500 1600 1700 1800 1900 Year ■ Figure 1.11 Time line of some contributors to the science of fluid mechanics. 2000 28 Chapter 1 ■ Introduction The rich history of fluid mechanics is fascinating, and many of the contributions of the pioneers in the field are noted in the succeeding chapters. a solid boundary a thin fluid layer 1boundary layer2 develops in which friction is very important, but outside this layer the fluid behaves very much like a frictionless fluid. This relatively simple concept provided the necessary impetus for the resolution of the conflict between the hydrodynamicists and the hydraulicists. Prandtl is generally accepted as the founder of modern fluid mechanics. Also, during the first decade of the twentieth century, powered flight was first successfully demonstrated with the subsequent vastly increased interest in aerodynamics. Because the design of aircraft required a degree of understanding of fluid flow and an ability to make accurate predictions of the effect of airflow on bodies, the field of aerodynamics provided a great stimulus for the many rapid developments in fluid mechanics that took place during the twentieth century. As we proceed with our study of the fundamentals of fluid mechanics, we will continue to note the contributions of many of the pioneers in the field. Table 1.9 provides a chronological Table 1.9 Chronological Listing of Some Contributors to the Science of Fluid Mechanics Noted in the Texta ARCHIMEDES 1287– 212 B.C.2 Established elementary principles of buoyancy and flotation. SEXTUS JULIUS FRONTINUS 1A.D. 40–1032 Wrote treatise on Roman methods of water distribution. Leonardo da Vinci LEONARDO da VINCI 11452–15192 Expressed elementary principle of continuity; observed and sketched many basic flow phenomena; suggested designs for hydraulic machinery. GALILEO GALILEI 11564–16422 Indirectly stimulated experimental hydraulics; revised Aristotelian concept of vacuum. EVANGELISTA TORRICELLI 11608–16472 Related barometric height to weight of atmosphere, and form of liquid jet to trajectory of free fall. BLAISE PASCAL 11623–16622 Finally clarified principles of barometer, hydraulic press, and pressure transmissibility. Isaac Newton LOUIS MARIE HENRI NAVIER 11785–18362 Extended equations of motion to include “molecular” forces. AUGUSTIN LOUIS de CAUCHY 11789–18572 Contributed to the general field of theoretical hydrodynamics and to the study of wave motion. GOTTHILF HEINRICH LUDWIG HAGEN 11797–18842 Conducted original studies of resistance in and transition between laminar and turbulent flow. JEAN LOUIS POISEUILLE 11799–18692 Performed meticulous tests on resistance of flow through capillary tubes. HENRI PHILIBERT GASPARD DARCY 11803–18582 HENRI de PITOT 11695–17712 JULIUS WEISBACH 11806–18712 DANIEL BERNOULLI 11700–17822 Experimented and wrote on many phases of fluid motion, coining name “hydrodynamics”; devised manometry technique and adapted primitive energy principle to explain velocity-head indication; proposed jet propulsion. WILLIAM FROUDE 11810–18792 Developed many towing-tank techniques, in particular the conversion of wave and boundary layer resistance from model to prototype scale. Explored various aspects of fluid resistance— inertial, viscous, and wave; discovered jet contraction. Performed extensive tests on filtration and pipe resistance; initiated open-channel studies carried out by Bazin. Incorporated hydraulics in treatise on engineering mechanics, based on original experiments; noteworthy for flow patterns, nondimensional coefficients, weir, and resistance equations. LEONHARD EULER 11707–17832 First explained role of pressure in fluid flow; formulated basic equations of motion and so-called Bernoulli theorem; introduced concept of cavitation and principle of centrifugal machinery. ROBERT MANNING 11816–18972 Proposed several formulas for open-channel resistance. Originated notion of velocity and acceleration components, differential expression of continuity, and paradox of zero resistance to steady nonuniform motion. ERNST MACH 11838–19162 One of the pioneers in the field of supersonic aerodynamics. JEAN le ROND d’ALEMBERT 11717–17832 Ernst Mach GIOVANNI BATTISTA VENTURI 11746–18222 Performed tests on various forms of mouthpieces— in particular, conical contractions and expansions. ISAAC NEWTON 11642–17272 Constructed double-tube device to indicate water velocity through differential head. Daniel Bernoulli ANTOINE CHEZY 11718–17982 Formulated similarity parameter for predicting flow characteristics of one channel from measurements on another. GEORGE GABRIEL STOKES 11819–19032 Derived analytically various flow relationships ranging from wave mechanics to viscous resistance— particularly that for the settling of spheres. 1.11 Chapter Summary and Study Guide 29 Table 1.9 (continued) OSBORNE REYNOLDS 11842–19122 Described original experiments in many fields— cavitation, river model similarity, pipe resistance— and devised two parameters for viscous flow; adapted equations of motion of a viscous fluid to mean conditions of turbulent flow. JOHN WILLIAM STRUTT, LORD RAYLEIGH Osborne Reynolds 11842–19192 Investigated hydrodynamics of bubble collapse, wave motion, jet instability, laminar flow analogies, and dynamic similarity. VINCENZ STROUHAL 11850–19222 Investigated the phenomenon of “singing wires.” EDGAR BUCKINGHAM 11867–19402 Stimulated interest in the United States in the use of dimensional analysis. Ludwig Prandtl MORITZ WEBER 11871–19512 Emphasized the use of the principles of similitude in fluid flow studies and formulated a capillarity similarity parameter. LUDWIG PRANDTL 11875–19532 Introduced concept of the boundary layer and is generally considered to be the father of present-day fluid mechanics. LEWIS FERRY MOODY 11880–19532 Provided many innovations in the field of hydraulic machinery. Proposed a method of correlating pipe resistance data that is widely used. THEODOR VON KÁRMÁN 11881–19632 One of the recognized leaders of twentieth century fluid mechanics. Provided major contributions to our understanding of surface resistance, turbulence, and wake phenomena. PAUL RICHARD HEINRICH BLASIUS 11883–19702 One of Prandtl’s students who provided an analytical solution to the boundary layer equations. Also demonstrated that pipe resistance was related to the Reynolds number. a Used by permission of IIHR—Hydroscience & Engineering, The University of Iowa. listing of some of these contributors and reveals the long journey that makes up the history of fluid mechanics. This list is certainly not comprehensive with regard to all past contributors but includes those who are mentioned in this text. As mention is made in succeeding chapters of the various individuals listed in Table 1.9, a quick glance at this table will reveal where they fit into the historical chain. It is, of course, impossible to summarize the rich history of fluid mechanics in a few paragraphs. Only a brief glimpse is provided, and we hope it will stir your interest. References 2 to 5 are good starting points for further study, and in particular Ref. 2 provides an excellent, broad, easily read history. Try it—you might even enjoy it! 1.11 Chapter Summary and Study Guide This introductory chapter discussed several fundamental aspects of fluid mechanics. Methods for describing fluid characteristics both quantitatively and qualitatively are considered. For a quantitative description, units are required, and in this text, two systems of units are used: the British Gravitational (BG) system (pounds, slugs, feet, and seconds) and the International (SI) System (newtons, kilograms, meters, and seconds). For the qualitative description the concept of dimensions is introduced in which basic dimensions such as length, L, time, T, and mass, M, are used to provide a description of various quantities of interest. The use of dimensions is helpful in checking the generality of equations, as well as serving as the basis for the powerful tool of dimensional analysis discussed in detail in Chapter 7. Various important fluid properties are defined, including fluid density, specific weight, specific gravity, viscosity, bulk modulus, speed of sound, vapor pressure, and surface tension. The ideal gas law is introduced to relate pressure, temperature, and density in common gases, along with a brief discussion of the compression and expansion of gases. The distinction between absolute and gage pressure is introduced and this important idea is explored more fully in Chapter 2. 30 Chapter 1 ■ Introduction fluid units basic dimensions dimensionally homogeneous density specific weight specific gravity ideal gas law absolute pressure gage pressure no-slip condition rate of shearing strain absolute viscosity Newtonian fluid non-Newtonian fluid kinematic viscosity bulk modulus speed of sound vapor pressure surface tension The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. determine the dimensions of common physical quantities. determine whether an equation is a general or restricted homogeneous equation. use both BG and SI systems of units. calculate the density, specific weight, or specific gravity of a fluid from a knowledge of any two of the three. calculate the density, pressure, or temperature of an ideal gas (with a given gas constant) from a knowledge of any two of the three. relate the pressure and density of a gas as it is compressed or expanded using Eqs. 1.14 and 1.15. use the concept of viscosity to calculate the shearing stress in simple fluid flows. calculate the speed of sound in fluids using Eq. 1.19 for liquids and Eq. 1.20 for gases. determine whether boiling or cavitation will occur in a liquid using the concept of vapor pressure. use the concept of surface tension to solve simple problems involving liquid–gas or liquid– solid–gas interfaces. Some of the important equations in this chapter are: Specific weight Specific gravity Ideal gas law Newtonian fluid shear stress Bulk modulus Speed of sound in an ideal gas Capillary rise in a tube g rg r rH2O@4 °C p r RT du tm dy dp Ev dV V c 1kRT 2s cos u h gR SG (1.6) (1.7) (1.8) (1.9) (1.12) (1.20) (1.22) References 1. Reid, R. C., Prausnitz, J. M., and Sherwood, T. K., The Properties of Gases and Liquids, 3rd Ed., McGraw-Hill, New York, 1977. 2. Rouse, H. and Ince, S., History of Hydraulics, Iowa Institute of Hydraulic Research, Iowa City, 1957, Dover, New York, 1963. 3. Tokaty, G. A., A History and Philosophy of Fluid Mechanics, G. T. Foulis and Co., Ltd., Oxfordshire, Great Britain, 1971. 4. Rouse, H., Hydraulics in the United States 1776–1976, Iowa Institute of Hydraulic Research, Iowa City, Iowa, 1976. 5. Garbrecht, G., ed., Hydraulics and Hydraulic Research—A Historical Review, A. A. Balkema, Rotterdam, Netherlands, 1987. 6. Brenner, M. P., Shi, X. D., Eggens, J., and Nagel, S. R., Physics of Fluids, Vol. 7, No. 9, 1995. 7. Shi, X. D., Brenner, M. P., and Nagel, S. R., Science, Vol. 265, 1994. 31 Problems Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. © 2013 John Wiley and Sons, Inc.). Conceptual Questions 1.1C The correct statement for the definition of density is a) Density is the mass per unit volume. b) Density is the volume per unit mass. c) Density is the weight per unit volume. d) Density is the weight divided by gravity. e) Density is the mass divided by the weight. 1.2C Given the following equation where p is pressure in lb/ft2, is the specific weight in lb/ft3, V is the magnitude of velocity in ft/s, g is in ft/s2, and z is height in feet. If values are substituted into the equation, will the correct value of C be determined? p V2 zC g 2g a) Yes if the constant C has units of ft. b) Yes if the constant C is dimensionless. c) No, the equation cannot produce the correct value of C. d) Yes if the constant C has units of ft and the specific weight is multiplied by the conversion factor from lbm to lbf. 1.3C The no-slip condition is: a) An experimental observation that the velocity of a fluid in contact with a solid surface is equal to the velocity of the surface. b) Valid only for liquids. c) Useful only for very low density gases. d) Indicates that two solids in contact will not slip if the joining force is large. du 1.4C In fluids, the shearing strain rate for a Newtonian fluid dy has dimensions of: a) L/T2. b) 1/T. c) L2/T. d) L2/T2. 1.5C The laminar velocity profile for a Newtonian fluid is shown below. Height y Velocity u 0 Shear Stress Shear Stress τ Shear Stress τ 0 y Shear Stress τ 0 y Shear Stress τ 0 y τ 0 y 0 y Which figure best describes the variation of shear stress with distance from the plate? Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the even-numbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. product of force times volume, and (c) kinetic energy divided by area. Section 1.2 Dimensions, Dimensional Homogeneity, and Units 1.4 Determine the dimensions, in both the FLT system and the MLT system, for (a) the product of force times acceleration, (b) the product of force times velocity divided by area, and (c) momentum divided by volume. 1.1 The force, F, of the wind blowing against a building is given by F CD rV 2 A2, where V is the wind speed, r the density of the air, A the cross-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient. 1.2 Determine the dimensions, in both the FLT system and the MLT system, for (a) the product of mass times velocity, (b) the 1.3 Verify the dimensions, in both the FLT and MLT systems, of the following quantities which appear in Table 1.1: (a) volume, (b) acceleration, (c) mass, (d) moment of inertia (area), and (e) work. 1.5 Verify the dimensions, in both the FLT and MLT systems, of the following quantities which appear in Table 1.1: (a) angular velocity, (b) energy, (c) moment of inertia (area), (d) power, and (e) pressure. 32 Chapter 1 ■ Introduction 1.6 Verify the dimensions, in both the FLT system and the MLT system, of the following quantities which appear in Table 1.1: (a) frequency, (b) stress, (c) strain, (d) torque, and (e) work. 1.7 If u is a velocity, x a length, and t a time, what are the dimensions 1in the MLT system2 of (a) 0u 0t, (b) 0 2u 0x0t, and (c) 兰 1 0u 0t2 dx? 1.8 Verify the dimensions, in both the FLT system and the MLT system, of the following quantities which appear in Table 1.1: (a) acceleration, (b) stress, (c) moment of a force, (d) volume, and (e) work. 1.9 If p is a pressure, V a velocity, and a fluid density, what are the dimensions (in the MLT system) of (a) p/, (b) pV, and (c) p rV 2? 1.10 If P is a force and x a length, what are the dimensions (in the FLT system) of (a) dP/dx, (b) d 3P/dx3, and (c) 兰 P dx? 1.11 If V is a velocity, / a length, and n a fluid property (the kinematic viscosity) having dimensions of L2T 1, which of the following combinations are dimensionless: (a) V/n, (b) V/ n, (c) V 2n, (d) V/n? 1.12 If V is a velocity, determine the dimensions of Z, a, and G, which appear in the dimensionally homogeneous equation V Z1a 12 G 1.13 The volume rate of flow, Q, through a pipe containing a slowly moving liquid is given by the equation pR4 ¢p Q 8m/ where R is the pipe radius, ¢p the pressure drop along the pipe, m a fluid property called viscosity 1FL2T2 , and / the length of pipe. What are the dimensions of the constant p 8? Would you classify this equation as a general homogeneous equation? Explain. 1.14 According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula h 10.04 to 0.092 1Dd2 4V 22g where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units? Explain. 1.15 The pressure difference, ¢p, across a partial blockage in an artery 1called a stenosis2 is approximated by the equation ¢p Kv 2 mV A0 Ku a 1b rV 2 D A1 where V is the blood velocity, m the blood viscosity 1FL2T 2, r the blood density 1ML3 2, D the artery diameter, A0 the area of the unobstructed artery, and A1 the area of the stenosis. Determine the dimensions of the constants Kv and Ku. Would this equation be valid in any system of units? 1.16 Assume that the speed of sound, c, in a fluid depends on an elastic modulus, Ev, with dimensions FL2, and the fluid density, r, in the form c 1Ev 2 a 1r2 b. If this is to be a dimensionally homogeneous equation, what are the values for a and b? Is your result consistent with the standard formula for the speed of sound? 1See Eq. 1.19.2 1.17 A formula to estimate the volume rate of flow, Q, flowing over a dam of length, B, is given by the equation Q 3.09 BH 32 where H is the depth of the water above the top of the dam 1called the head2. This formula gives Q in ft3/s when B and H are in feet. Is the constant, 3.09, dimensionless? Would this equation be valid if units other than feet and seconds were used? 1.18 The force, P, that is exerted on a spherical particle moving slowly through a liquid is given by the equation P 3pmDV where m is a fluid property (viscosity) having dimensions of FL2T, D is the particle diameter, and V is the particle velocity. What are the dimensions of the constant, 3p? Would you classify this equation as a general homogeneous equation? †1.19 Cite an example of a restricted homogeneous equation contained in a technical article found in an engineering journal in your field of interest. Define all terms in the equation, explain why it is a restricted equation, and provide a complete journal citation 1title, date, etc.2. 1.20 Make use of Table 1.3 to express the following quantities in SI units: (a) 10.2 in.min, (b) 4.81 slugs, (c) 3.02 lb, (d) 73.1 fts2, (e) 0.0234 lb # sft2. 1.21 Make use of Table 1.4 to express the following quantities in BG units: (a) 14.2 km, (b) 8.14 Nm3, (c) 1.61 kgm3, (d) 0.0320 N # ms, (e) 5.67 mmhr. 1.22 Express the following quantities in SI units: (a) 160 acres, (b) 15 gallons (U.S.), (c) 240 miles, (d) 79.1 hp, (e) 60.3 °F. 1.23 For Table 1.3 verify the conversion relationships for: (a) area, (b) density, (c) velocity, and (d) specific weight. Use the basic conversion relationships: 1 ft 0.3048 m; 1 lb 4.4482 N; and 1 slug 14.594 kg. 1.24 For Table 1.4 verify the conversion relationships for: (a) acceleration, (b) density, (c) pressure, and (d) volume flowrate. Use the basic conversion relationships: 1 m 3.2808 ft; 1N 0.22481 lb; and 1 kg 0.068521 slug. 1.25 Water flows from a large drainage pipe at a rate of 1200 galmin. What is this volume rate of flow in (a) m3s, (b) liters min, and (c) ft3s? 1.26 Dimensionless combinations of quantities (commonly called dimensionless parameters) play an important role in fluid mechanics. Make up five possible dimensionless parameters by using combinations of some of the quantities listed in Table 1.1. 1.27 GO An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V 1g/, where V is a velocity, g the acceleration of gravity, and ᐉ a length. Determine the value of the Froude number for V 10 fts, g 32.2 fts2, and / 2 ft. Recalculate the Froude number using SI units for V, g, and /. Explain the significance of the results of these calculations. Section 1.4 Measures of Fluid Mass and Weight 1.28 Obtain a photograph/image of a situation in which the density or specific weight of a fluid is important. Print this photo and write a brief paragraph that describes the situation involved. 1.29 A tank contains 500 kg of a liquid whose specific gravity is 2. Determine the volume of the liquid in the tank. 1.30 Clouds can weigh thousands of pounds due to their liquid water content. Often this content is measured in grams per cubic meter (g/m3). Assume that a cumulus cloud occupies a volume of one cubic kilometer, and its liquid water content is 0.2 g/m3. (a) What is the volume of this cloud in cubic miles? (b) How much does the water in the cloud weigh in pounds? Problems 1.31 A tank of oil has a mass of 25 slugs. (a) Determine its weight in pounds and in newtons at the Earth’s surface. (b) What would be its mass 1in slugs2 and its weight 1in pounds2 if located on the moon’s surface where the gravitational attraction is approximately one-sixth that at the Earth’s surface? 1.32 A certain object weighs 300 N at the Earth’s surface. Determine the mass of the object 1in kilograms2 and its weight 1in newtons2 when located on a planet with an acceleration of gravity equal to 4.0 fts2. 3 1.33 The density of a certain type of jet fuel is 775 kg/m . Determine its specific gravity and specific weight. 1.34 A hydrometer is used to measure the specific gravity of liquids. (See Video V2.8.) For a certain liquid, a hydrometer reading indicates a specific gravity of 1.15. What is the liquid’s density and specific weight? Express your answer in SI units. 1.35 The specific weight of a certain liquid is 85.3 lb/ft3. Determine its density and specific gravity. 1.36 An open, rigid-walled, cylindrical tank contains 4 ft3 of water at 40 °F. Over a 24-hour period of time the water temperature varies from 40 to 90 °F. Make use of the data in Appendix B to determine how much the volume of water will change. For a tank diameter of 2 ft, would the corresponding change in water depth be very noticeable? Explain. †1.37 Estimate the number of pounds of mercury it would take to fill your bathtub. List all assumptions and show all calculations. 1.38 A mountain climber’s oxygen tank contains 1 lb of oxygen when he begins his trip at sea level where the acceleration of gravity is 32.174 ft/s2. What is the weight of the oxygen in the tank when he reaches the top of Mt. Everest where the acceleration of gravity is 32.082 ft/s2? Assume that no oxygen has been removed from the tank; it will be used on the descent portion of the climb. 1.39 GO The information on a can of pop indicates that the can contains 355 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at 20 °C. Express your results in SI units. 33 Section 1.5 Ideal Gas Law 1.44 Determine the mass of air in a 2 m3 tank if the air is at room temperature, 20 °C, and the absolute pressure within the tank is 200 kPa (abs). 1.45 Nitrogen is compressed to a density of 4 kg/m3 under an absolute pressure of 400 kPa. Determine the temperature in degrees Celsius. 1.46 The temperature and pressure at the surface of Mars during a Martian spring day were determined to be 50 °C and 900 Pa, respectively. (a) Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the Earth’s atmosphere during a spring day when the temperature is 18 °C and the pressure 101.6 kPa (abs). 1.47 A closed tank having a volume of 2 ft3 is filled with 0.30 lb of a gas. A pressure gage attached to the tank reads 12 psi when the gas temperature is 80 °F. There is some question as to whether the gas in the tank is oxygen or helium. Which do you think it is? Explain how you arrived at your answer. 1.48 A tire having a volume of 3 ft3 contains air at a gage pressure of 26 psi and a temperature of 70 F. Determine the density of the air and the weight of the air contained in the tire. 1.49 A compressed air tank contains 5 kg of air at a temperature of 80 °C. A gage on the tank reads 300 kPa. Determine the volume of the tank. 1.50 A rigid tank contains air at a pressure of 90 psia and a temperature of 60 F. By how much will the pressure increase as the temperature is increased to 110 F? 1.51 The density of oxygen contained in a tank is 2.0 kg/m3 when the temperature is 25 C. Determine the gage pressure of the gas if the atmospheric pressure is 97 kPa. 1.52 The helium-filled blimp shown in Fig. P1.52 is used at various athletic events. Determine the number of pounds of helium within it if its volume is 68,000 ft3 and the temperature and pressure are 80 °F and 14.2 psia, respectively. 1.40 The variation in the density of water, r, with temperature, T, in the range 20 °C T 50 °C, is given in the following table. Density 1kg m32 Temperature 1°C2 998.2 20 997.1 995.7 25 30 994.1 992.2 35 40 990.2 988.1 45 50 Use these data to determine an empirical equation of the form r c1 c2T c3T 2 which can be used to predict the density over the range indicated. Compare the predicted values with the data given. What is the density of water at 42.1 °C? 3 1.41 If 1 cup of cream having a density of 1005 kg/m is turned into 3 cups of whipped cream, determine the specific gravity and specific weight of the whipped cream. 1.42 A liquid when poured into a graduated cylinder is found to weigh 8 N when occupying a volume of 500 ml (milliliters). Determine its specific weight, density, and specific gravity. †1.43 The presence of raindrops in the air during a heavy rainstorm increases the average density of the air–water mixture. Estimate by what percent the average air–water density is greater than that of just still air. State all assumptions and show calculations. ■ Figure P1.52 1.53 Develop a computer program for calculating the density of an ideal gas when the gas pressure in pascals 1abs2, the temperature in degrees Celsius, and the gas constant in J kg # K are specified. Plot the density of helium as a function of temperature from 0 °C to 200 °C and pressures of 50, 100, 150, and 200 kPa (abs). Section 1.6 Viscosity (also see Lab Problems 1.1LP and 1.2LP) 1.54 Obtain a photograph/image of a situation in which the viscosity of a fluid is important. Print this photo and write a brief paragraph that describes the situation involved. 34 Chapter 1 ■ Introduction 1.55 For flowing water, what is the magnitude of the velocity gradient needed to produce a shear stress of 1.0 N/m2? 1.63 A liquid has a specific weight of 59 lb/ft3 and a dynamic viscosity of 2.75 lb # s/ft2. Determine its kinematic viscosity. 1.56 Make use of the data in Appendix B to determine the dynamic viscosity of glycerin at 85 °F. Express your answer in both SI and BG units. 1.64 The kinematic viscosity of oxygen at 20 °C and a pressure of 150 kPa 1abs2 is 0.104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure. 1.57 Make use of the data in Appendix B to determine the dynamic viscosity of mercury at 75 F. Express your answer in BG units. 1.65 1.58 One type of capillary-tube viscometer is shown in Video V1.5 and in Fig. P1.58. For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line. The kinematic viscosity, , in m2/s is then obtained from the equation n KR 4t where K is a constant, R is the radius of the capillary tube in mm, and t is the drain time in seconds. When glycerin at 20 C is used as a calibration fluid in a particular viscometer, the drain time is 1430 s. When a liquid having a density of 970 kg/m3 is tested in the same viscometer the drain time is 900 s. What is the dynamic viscosity of this liquid? Fluids for which the shearing stress, , is not linearly related to the rate of shearing strain, ᝽, are designated as nonNewtonian fluids. Such fluids are commonplace and can exhibit unusual behavior, as shown in Video V1.6. Some experimental data obtained for a particular non-Newtonian fluid at 80 F are shown below. (lb/ft2) 1 ᝽ (s ) 0 2.11 7.82 18.5 31.7 0 50 100 150 200 Plot these data and fit a second-order polynomial to the data using a suitable graphing program. What is the apparent viscosity of this fluid when the rate of shearing strain is 70 s 1? Is this apparent viscosity larger or smaller than that for water at the same temperature? 1.66 Water flows near a flat surface and some measurements of the water velocity, u, parallel to the surface, at different heights, y, above the surface are obtained. At the surface y 0. After an analysis of the data, the lab technician reports that the velocity distribution in the range 0 6 y 6 0.1 ft is given by the equation Glass strengthening bridge u 0.81 9.2y 4.1 103y3 Etched lines with u in ft/s when y is in ft. (a) Do you think that this equation would be valid in any system of units? Explain. (b) Do you think this equation is correct? Explain. You may want to look at Video 1.4 to help you arrive at your answer. 1.67 Calculate the Reynolds numbers for the flow of water and for air through a 4-mm-diameter tube, if the mean velocity is 3 ms and the temperature is 30 °C in both cases 1see Example 1.42. Assume the air is at standard atmospheric pressure. Capillary tube 1.68 SAE 30 oil at 60 F flows through a 2-in.-diameter pipe with a mean velocity of 5 ft/s. Determine the value of the Reynolds number (see Example 1.4). ■ Figure P1.58 1.59 The viscosity of a soft drink was determined by using a capillary tube viscometer similar to that shown in Fig. P1.58 and Video V1.5. For this device the kinematic viscosity, , is directly proportional to the time, t, that it takes for a given amount of liquid to flow through a small capillary tube. That is, n Kt. The following data were obtained from regular pop and diet pop. The corresponding measured specific gravities are also given. Based on these data, by what percent is the absolute viscosity, , of regular pop greater than that of diet pop? Regular pop Diet pop t(s) 377.8 300.3 SG 1.044 1.003 1.60 Determine the ratio of the dynamic viscosity of water to air at a temperature of 60 °C. Compare this value with the corresponding ratio of kinematic viscosities. Assume the air is at standard atmospheric pressure. 1.61 The viscosity of a certain fluid is 5 104 poise. Determine its viscosity in both SI and BG units. 1.62 The kinematic viscosity and specific gravity of a liquid are 3.5 104 m2/s and 0.79, respectively. What is the dynamic viscosity of the liquid in SI units? 1.69 For air at standard atmospheric pressure the values of the constants that appear in the Sutherland equation 1Eq. 1.102 are C 1.458 106 kg 1m # s # K12 2 and S 110.4 K. Use these values to predict the viscosity of air at 10 °C and 90 °C and compare with values given in Table B.4 in Appendix B. 1.70 Use the values of viscosity of air given in Table B.4 at temperatures of 0, 20, 40, 60, 80, and 100 °C to determine the constants C and S which appear in the Sutherland equation 1Eq. 1.102. Compare your results with the values given in Problem 1.69. 1Hint: Rewrite the equation in the form 1 S T 3 2 a bT m C C and plot T 32m versus T. From the slope and intercept of this curve, C and S can be obtained.2 1.71 The viscosity of a fluid plays a very important role in determining how a fluid flows. (See Video V1.3.) The value of the viscosity depends not only on the specific fluid but also on the fluid temperature. Some experiments show that when a liquid, under the action of a constant driving pressure, is forced with a low velocity, V, through a small horizontal tube, the velocity is given by the equation V Km. In this equation K is a constant for a given tube and pressure, and is the dynamic viscosity. For a particular Problems liquid of interest, the viscosity is given by Andrade’s equation (Eq. 1.11) with D 5 10 7 lb # sft2 and B 4000 °R. By what percentage will the velocity increase as the liquid temperature is increased from 40 F to 100 F? Assume all other factors remain constant. 35 3.5 105 lb # sft2. Determine the thickness of the water layer under the runners. Assume a linear velocity distribution in the water layer. 1.72 Use the value of the viscosity of water given in Table B.2 at temperatures of 0, 20, 40, 60, 80, and 100 °C to determine the constants D and B which appear in Andrade’s equation 1Eq. 1.112. Calculate the value of the viscosity at 50 °C and compare with the value given in Table B.2. 1Hint: Rewrite the equation in the form ln m 1B2 1 ln D T and plot ln m versus 1 T. From the slope and intercept of this curve, B and D can be obtained. If a nonlinear curve-fitting program is available, the constants can be obtained directly from Eq. 1.11 without rewriting the equation.2 1.73 For a certain liquid m 7.1 105 lb # s/ft2 at 40 F and m 1.9 105 lb # s/ft2 at 150 F. Make use of these data to determine the constants D and B which appear in Andrade’s equation (Eq. 1.11). What would be the viscosity at 80 F? 1.74 GO For a parallel plate arrangement of the type shown in Fig. 1.5 it is found that when the distance between plates is 2 mm, a shearing stress of 150 Pa develops at the upper plate when it is pulled at a velocity of 1 m/s. Determine the viscosity of the fluid between the plates. Express your answer in SI units. 1.75 Two flat plates are oriented parallel above a fixed lower plate as shown in Fig. P1.75. The top plate, located a distance b above the fixed plate, is pulled along with speed V. The other thin plate is located a distance cb, where 0 c 1, above the fixed plate. This plate moves with speed V1, which is determined by the viscous shear forces imposed on it by the fluids on its top and bottom. The fluid on the top is twice as viscous as that on the bottom. Plot the ratio V1/V as a function of c for 0 c 1. V 2μ b V1 cb μ ■ Figure P1.75 1.76 There are many fluids that exhibit non-Newtonian behavior (see, for example, Video V1.6). For a given fluid the distinction between Newtonian and non-Newtonian behavior is usually based on measurements of shear stress and rate of shearing strain. Assume that the viscosity of blood is to be determined by measurements of shear stress, , and rate of shearing strain, du/dy, obtained from a small blood sample tested in a suitable viscometer. Based on the data given below, determine if the blood is a Newtonian or non-Newtonian fluid. Explain how you arrived at your answer. (N/m2) 1 du/dy (s ) 0.04 0.06 0.12 0.18 0.30 0.52 1.12 2.10 2.25 4.50 11.25 22.5 45.0 90.0 225 450 1.77 The sled shown in Fig. P1.77 slides along on a thin horizontal layer of water between the ice and the runners. The horizontal force that the water puts on the runners is equal to 1.2 lb when the sled’s speed is 50 ft/s. The total area of both runners in contact with the water is 0.08 ft2, and the viscosity of the water is ■ Figure P1.77 1.78 A 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in Fig. P1.78. The lubricant that fills the 0.3-mm gap between the shaft and bearing is an oil having a kinematic viscosity of 8.0 104 m2s and a specific gravity of 0.91. Determine the force P required to pull the shaft at a velocity of 3 m/s. Assume the velocity distribution in the gap is linear. Bearing Shaft Lubricant P 0.5 m ■ Figure P1.78 1.79 A piston having a diameter of 5.48 in. and a length of 9.50 in. slides downward with a velocity V through a vertical pipe. The downward motion is resisted by an oil film between the piston and the pipe wall. The film thickness is 0.002 in., and the cylinder weighs 0.5 lb. Estimate V if the oil viscosity is 0.016 lb # s/ft2. Assume the velocity distribution in the gap is linear. 1.80 A 10-kg block slides down a smooth inclined surface as shown in Fig. P1.80. Determine the terminal velocity of the block if the 0.1-mm gap between the block and the surface contains SAE 30 oil at 60 °F. Assume the velocity distribution in the gap is linear, and the area of the block in contact with the oil is 0.1 m2. V 0.1 mm gap 20° ■ Figure P1.80 1.81 A layer of water flows down an inclined fixed surface with the velocity profile shown in Fig. P1.81. Determine the magnitude and direction of the shearing stress that the water exerts on the fixed surface for U 2 ms and h 0.1 m. 36 Chapter 1 ■ Introduction h 1.85 The space between two 6-in.-long concentric cylinders is filled with glycerin 1viscosity ⫽ 8.5 ⫻ 10⫺3 lb # sⲐft2 2. The inner cylinder has a radius of 3 in. and the gap width between cylinders is 0.1 in. Determine the torque and the power required to rotate the inner cylinder at 180 revⲐmin. The outer cylinder is fixed. Assume the velocity distribution in the gap to be linear. U u y u y y2 __ = 2 __ – __2 U h h ■ Figure P1.81 1.82 A thin layer of glycerin flows down an inclined, wide plate with the velocity distribution shown in Fig. P1.82. For h ⫽ 0.3 in. and a ⫽ 20⬚, determine the surface velocity, U. Note that for equilibrium, the component of weight acting parallel to the plate surface must be balanced by the shearing force developed along the plate surface. In your analysis assume a unit plate width. 1.86 A pivot bearing used on the shaft of an electrical instrument is shown in Fig. P1.86. An oil with a viscosity of ␮ ⫽ 0.010 lb . s/ft2 fills the 0.001-in. gap between the rotating shaft and the stationary base. Determine the frictional torque on the shaft when it rotates at 5000 rpm. 5000 rpm 0.2 in. U α y u y y2 u = __ __ 2 – __2 h U h h 30° μ = 0.010 lb • s/ft2 0.001 in. ■ Figure P1.82 1.83 Standard air flows past a flat surface, and velocity measurements near the surface indicate the following distribution: y 1ft2 u 1ftⲐs2 0.005 0.01 0.02 0.04 0.06 0.08 0.74 1.51 3.03 6.37 10.21 14.43 The coordinate y is measured normal to the surface and u is the velocity parallel to the surface. (a) Assume the velocity distribution is of the form u ⫽ C1y ⫹ C2 y3 and use a standard curve-fitting technique to determine the constants C1 and C2. (b) Make use of the results of part 1a2 to determine the magnitude of the shearing stress at the wall 1y ⫽ 02 and at y ⫽ 0.05 ft. 1.84 A new computer drive is proposed to have a disc, as shown in Fig. P1.84. The disc is to rotate at 10,000 rpm, and the reader head is to be positioned 0.0005 in. above the surface of the disc. Estimate the shearing force on the reader head as a result of the air between the disc and the head. ■ Figure P1.86 1.87 The viscosity of liquids can be measured through the use of a rotating cylinder viscometer of the type illustrated in Fig. P1.87. In this device the outer cylinder is fixed and the inner cylinder is rotated with an angular velocity, v. The torque t required to develop ␻ is measured and the viscosity is calculated from these two measurements. (a) Develop an equation relating m, v, t, /, Ro, and Ri. Neglect end effects and assume the velocity distribution in the gap is linear. (b) The following torque-angular velocity data were obtained with a rotating cylinder viscometer of the type discussed in part (a). Torque 1ft # lb2 13.1 26.0 39.5 52.7 64.9 78.6 Angular velocity 1radⲐs2 1.0 2.0 3.0 4.0 5.0 6.0 For this viscometer Ro ⫽ 2.50 in., Ri ⫽ 2.45 in., and / ⫽ 5.00 in. Make use of these data and a standard curve-fitting program to determine the viscosity of the liquid contained in the viscometer. ᐀ Liquid Stationary reader head ω 0.2-in.dia. 10,000 rpm 0.0005 in. 2 in. Rotating disc ■ Figure P1.84 Fixed outer cylinder Ri Ro Rotating inner cylinder ᐉ ■ Figure P1.87 Problems 1.88 One type of rotating cylinder viscometer, called a Stormer viscometer, uses a falling weight, ᐃ, to cause the cylinder to rotate with an angular velocity, v, as illustrated in Fig. P1.88. For this device the viscosity, m, of the liquid is related to ᐃ and v through the equation ᐃ Kmv, where K is a constant that depends only on the geometry (including the liquid depth) of the viscometer. The value of K is usually determined by using a calibration liquid (a liquid of known viscosity). (a) Some data for a particular Stormer viscometer, obtained using glycerin at 20 C as a calibration liquid, are given below. Plot values of the weight as ordinates and values of the angular velocity as abscissae. Draw the best curve through the plotted points and determine K for the viscometer. ᐃ(lb) 0.22 0.66 1.10 1.54 2.20 v (rev/s) 0.53 1.59 2.79 3.83 5.49 (b) A liquid of unknown viscosity is placed in the same viscometer used in part (a), and the data given below are obtained. Determine the viscosity of this liquid. ᐃ(lb) 0.04 0.11 0.22 0.33 0.44 v (rev/s) 0.72 1.89 3.73 5.44 7.42 37 Section 1.7 Compressibility of Fluids 1.92 Obtain a photograph/image of a situation in which the compressibility of a fluid is important. Print this photo and write a brief paragraph that describes the situation involved. 1.93 A sound wave is observed to travel through a liquid with a speed of 1500 m/s. The specific gravity of the liquid is 1.5. Determine the bulk modulus for this fluid. 1.94 A rigid-walled cubical container is completely filled with water at 40 F and sealed. The water is then heated to 100 F. Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi. 1.95 In a test to determine the bulk modulus of a liquid it was found that as the absolute pressure was changed from 15 to 3000 psi the volume decreased from 10.240 to 10.138 in.3 Determine the bulk modulus for this liquid. 1.96 Estimate the increase in pressure (in psi) required to decrease a unit volume of mercury by 0.1%. 1.97 A 1-m3 volume of water is contained in a rigid container. Estimate the change in the volume of the water when a piston applies a pressure of 35 MPa. 1.98 Determine the speed of sound at 20 °C in (a) air, (b) helium, and (c) natural gas (methane). Express your answer in m/s. 1.99 Calculate the speed of sound in m/s for (a) gasoline, (b) mercury, and (c) seawater. ω ᐃ Weight Liquid Rotating inner cylinder 1.100 Air is enclosed by a rigid cylinder containing a piston. A pressure gage attached to the cylinder indicates an initial reading of 25 psi. Determine the reading on the gage when the piston has compressed the air to one-third its original volume. Assume the compression process to be isothermal and the local atmospheric pressure to be 14.7 psi. Fixed outer cylinder 1.101 Repeat Problem 1.100 if the compression process takes place without friction and without heat transfer (isentropic process). ■ Figure P1.88 1.89 A 12-in.-diameter circular plate is placed over a fixed bottom plate with a 0.1-in. gap between the two plates filled with glycerin as shown in Fig. P1.89. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible. Torque Rotating plate 0.1-in. gap 1.102 Carbon dioxide at 30 °C and 300 kPa absolute pressure expands isothermally to an absolute pressure of 165 kPa. Determine the final density of the gas. 1.103 Oxygen at 30 C and 300 kPa absolute pressure expands isothermally to an absolute pressure of 120 kPa. Determine the final density of the gas. 1.104 Natural gas at 70 °F and standard atmospheric pressure of 14.7 psi (abs) is compressed isentropically to a new absolute pressure of 70 psi. Determine the final density and temperature of the gas. 1.105 Compare the isentropic bulk modulus of air at 101 kPa 1abs2 with that of water at the same pressure. 1.106 Develop a computer program for calculating the final gage ■ Figure P1.89 †1.90 Vehicle shock absorbers damp out oscillations caused by road roughness. Describe how a temperature change may affect the operation of a shock absorber. 1.91 Some measurements on a blood sample at 37 °C 198.6 °F2 indicate a shearing stress of 0.52 Nm2 for a corresponding rate of shearing strain of 200 s1. Determine the apparent viscosity of the blood and compare it with the viscosity of water at the same temperature. pressure of gas when the initial gage pressure, initial and final volumes, atmospheric pressure, and the type of process 1isothermal or isentropic2 are specified. Use BG units. Check your program against the results obtained for Problem 1.100. 1.107 Often the assumption is made that the flow of a certain fluid can be considered as incompressible flow if the density of the fluid changes by less than 2%. If air is flowing through a tube such that the air pressure at one section is 9.0 psi and at a downstream section it is 8.6 psi at the same temperature, do you think that this flow could be considered an incompressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressure. 38 Chapter 1 ■ Introduction 1.108 An important dimensionless parameter concerned with very high-speed flow is the Mach number, defined as V/c, where V is the speed of the object such as an airplane or projectile, and c is the speed of sound in the fluid surrounding the object. For a projectile traveling at 800 mph through air at 50 F and standard atmospheric pressure, what is the value of the Mach number? 1.109 Jet airliners typically fly at altitudes between approximately 0 to 40,000 ft. Make use of the data in Appendix C to show on a graph how the speed of sound varies over this range. 1.110 (See Fluids in the News article titled “This water jet is a blast,” Section 1.7.1.) By what percent is the volume of water decreased if its pressure is increased to an equivalent to 3000 atmospheres (44,100 psi)? Section 1.8 Vapor Pressure 1.111 During a mountain climbing trip it is observed that the water used to cook a meal boils at 90 °C rather than the standard 100 °C at sea level. At what altitude are the climbers preparing their meal? (See Tables B.2 and C.2 for data needed to solve this problem.) 1.112 When a fluid flows through a sharp bend, low pressures may develop in localized regions of the bend. Estimate the minimum absolute pressure 1in psi2 that can develop without causing cavitation if the fluid is water at 160 °F. 1.113 A partially filled closed tank contains ethyl alcohol at 68 F. If the air above the alcohol is evacuated, what is the minimum absolute pressure that develops in the evacuated space? 1.114 Estimate the minimum absolute pressure 1in pascals2 that can be developed at the inlet of a pump to avoid cavitation if the fluid is carbon tetrachloride at 20 °C. 1.115 When water at 70 °C flows through a converging section of pipe, the pressure decreases in the direction of flow. Estimate the minimum absolute pressure that can develop without causing cavitation. Express your answer in both BG and SI units. 1.116 At what atmospheric pressure will water boil at 35 °C? Express your answer in both SI and BG units. Section 1.9 Surface Tension 1.117 Obtain a photograph/image of a situation in which the surface tension of a fluid is important. Print this photo and write a brief paragraph that describes the situation involved. 1.118 When a 2-mm-diameter tube is inserted into a liquid in an open tank, the liquid is observed to rise 10 mm above the free surface of the liquid (see Video V1.10). The contact angle between the liquid and the tube is zero, and the specific weight of the liquid is 1.2 104 N/m3. Determine the value of the surface tension for this liquid. 1.119 An open 2-mm-diameter tube is inserted into a pan of ethyl alcohol, and a similar 4-mm-diameter tube is inserted into a pan of water. In which tube will the height of the rise of the fluid column due to capillary action be the greatest? Assume the angle of contact is the same for both tubes. 1.120 Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 mm, what is the difference in pressure between the inside and outside of the droplets? 1.121 A 12-mm-diameter jet of water discharges vertically into the atmosphere. Due to surface tension the pressure inside the jet will be slightly higher than the surrounding atmospheric pressure. Determine this difference in pressure. 1.122 Estimate the excess pressure inside a raindrop having a diameter of 3 mm. 1.123 What is the difference between the pressure inside a soap bubble and atmospheric pressure for a 3-in.-diameter bubble? Assume the surface tension of the soap film to be 70% of that of water at 70 F. 1.124 As shown in Video V1.9, surface tension forces can be strong enough to allow a double-edge steel razor blade to “float” on water, but a single-edge blade will sink. Assume that the surface tension forces act at an angle relative to the water surface as shown in Fig. P1.124. (a) The mass of the double-edge blade is 0.64 10 3 kg, and the total length of its sides is 206 mm. Determine the value of required to maintain equilibrium between the blade weight and the resultant surface tension force. (b) The mass of the single-edge blade is 2.61 10 3 kg, and the total length of its sides is 154 mm. Explain why this blade sinks. Support your answer with the necessary calculations. Surface tension force Blade θ ■ Figure P1.24 1.125 To measure the water depth in a large open tank with opaque walls, an open vertical glass tube is attached to the side of the tank. The height of the water column in the tube is then used as a measure of the depth of water in the tank. (a) For a true water depth in the tank of 3 ft, make use of Eq. 1.22 (with u ⯝ 0°) to determine the percent error due to capillarity as the diameter of the glass tube is changed. Assume a water temperature of 80 F. Show your results on a graph of percent error versus tube diameter, D, in the range 0.1 in. 6 D 6 1.0 in. (b) If you want the error to be less than 1%, what is the smallest tube diameter allowed? 1.126 Under the right conditions, it is possible, due to GO surface tension, to have metal objects float on water. (See Video V1.9.) Consider placing a short length of a small diameter steel (g 490 lb/ft3) rod on a surface of water. What is the maximum diameter that the rod can have before it will sink? Assume that the surface tension forces act vertically upward. Note: A standard paper clip has a diameter of 0.036 in. Partially unfold a paper clip and see if you can get it to float on water. Do the results of this experiment support your analysis? 1.127 An open, clean glass tube, having a diameter of 3 mm, is inserted vertically into a dish of mercury at 20 °C (see Video V1.10). How far will the column of mercury in the tube be depressed? 1.128 An open, clean glass tube 1u 0°2 is inserted vertically into a pan of water (see Video V1.10). What tube diameter is needed if the water level in the tube is to rise one tube diameter (due to surface tension)? 1.129 Determine the height that water at 60 °F will rise due to capillary action in a clean, 14-in.-diameter tube (see Video V1.10). What will be the height if the diameter is reduced to 0.01 in.? 1.130 Two vertical, parallel, clean glass plates are spaced a distance of 2 mm apart. If the plates are placed in water, how high will the water rise between the plates due to capillary action? Problems 1.131 (See Fluids in the News article titled “Walking on water,” Section 1.9.) (a) The water strider bug shown in Fig. P1.131 is supported on the surface of a pond by surface tension acting along the interface between the water and the bug’s legs. Determine the minimum length of this interface needed to support the bug. Assume the bug weighs 104 N and the surface tension force acts vertically upwards. (b) Repeat part (a) if surface tension were to support a person weighing 750 N. 39 ■ Lifelong Learning Problems 1.1 LL Although there are numerous non-Newtonian fluids that occur naturally (quicksand and blood among them), with the advent of modern chemistry and chemical processing, many new manufactured non-Newtonian fluids are now available for a variety of novel applications. Obtain information about the discovery and use of newly developed non-Newtonian fluids. Summarize your findings in a brief report. 1.2 LL For years, lubricating oils and greases obtained by refining crude oil have been used to lubricate moving parts in a wide variety of machines, motors, and engines. With the increasing cost of crude oil and the potential for the reduced availability of it, the need for non-petroleum-based lubricants has increased considerably. Obtain information about non-petroleum-based lubricants. Summarize your findings in a brief report. ■ Figure P1.131 ■ Lab Problems 1.1LP This problem involves the use of a Stormer viscometer to determine whether a fluid is a Newtonian or a non-Newtonian fluid. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www. wiley.com/college/munson. 1.2LP This problem involves the use of a capillary tube viscometer to determine the kinematic viscosity of water as a function of temperature. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 1.3 LL It is predicted that nano technology and the use of nano sized objects will allow many processes, procedures, and products that, as of now, are difficult for us to comprehend. Among new nano technology areas is that of nano scale fluid mechanics. Fluid behavior at the nano scale can be entirely different than that for the usual everyday flows with which we are familiar. Obtain information about various aspects of nano fluid mechanics. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam question for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 2 Fluid Statics CHAPTER OPENING PHOTO: Floating iceberg: An iceberg is a large piece of fresh water ice that originated as snow in a glacier or ice shelf and then broke off to float in the ocean. Although the fresh water ice is lighter than the salt water in the ocean, the difference in densities is relatively small. Hence, only about one ninth of the volume of an iceberg protrudes above the ocean’s surface, so that what we see floating is literally “just the tip of the iceberg.” (© oversnap/iStockphoto.) Learning Objectives After completing this chapter, you should be able to: ■ determine the pressure at various locations in a fluid at rest. ■ explain the concept of manometers and apply appropriate equations to determine pressures. ■ calculate the hydrostatic pressure force on a plane or curved submerged surface. ■ calculate the buoyant force and discuss the stability of floating or submerged objects. In this chapter we will consider an important class of problems in which the fluid is either at rest or moving in such a manner that there is no relative motion between adjacent particles. In both instances there will be no shearing stresses in the fluid, and the only forces that develop on the surfaces of the particles will be due to the pressure. Thus, our principal concern is to investigate pressure and its variation throughout a fluid and the effect of pressure on submerged surfaces. The absence of shearing stresses greatly simplifies the analysis and, as we will see, allows us to obtain relatively simple solutions to many important practical problems. 2.1 Pressure at a Point As we briefly discussed in Chapter 1, the term pressure is used to indicate the normal force per unit area at a given point acting on a given plane within the fluid mass of interest. A question that immediately arises is how the pressure at a point varies with the orientation of the plane passing through the point. To answer this question, consider the free-body diagram, illustrated in Fig. 2.1, 40 2.1 Pressure at a Point 41 z ps δ x δ s θ δs py δ x δ z y δz δx θ δy x δx δyδz γ __ 2 pz δ x δ y ■ Figure 2.1 Forces on an arbitrary wedge-shaped element of fluid. that was obtained by removing a small triangular wedge of fluid from some arbitrary location within a fluid mass. Since we are considering the situation in which there are no shearing stresses, the only external forces acting on the wedge are due to the pressure and the weight. For simplicity the forces in the x direction are not shown, and the z axis is taken as the vertical axis so the weight acts in the negative z direction. Although we are primarily interested in fluids at rest, to make the analysis as general as possible, we will allow the fluid element to have accelerated motion. The assumption of zero shearing stresses will still be valid as long as the fluid element moves as a rigid body; that is, there is no relative motion between adjacent elements. The equations of motion 1Newton’s second law, F ma2 in the y and z directions are, respectively, a Fy py dx dz ps dx ds sin u r dx dy dz ay 2 a Fz pz dx dy ps dx ds cos u g dx dy dz dx dy dz r az 2 2 where ps, py, and pz are the average pressures on the faces, g and r are the fluid specific weight and density, respectively, and ay, az the accelerations. Note that a pressure must be multiplied by an appropriate area to obtain the force generated by the pressure. It follows from the geometry that dy ds cos u dz ds sin u so that the equations of motion can be rewritten as The pressure at a point in a fluid at rest is independent of direction. py ps ray dy 2 pz ps 1raz g2 dz 2 Since we are really interested in what is happening at a point, we take the limit as dx, dy, and dz approach zero 1while maintaining the angle u2, and it follows that py ps py pz py pz pz ps or ps py pz. The angle u was arbitrarily chosen so we can conclude that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present. This important result is known as Pascal’s law, named in honor of Blaise Pascal 11623– 16622, a French mathematician who made important contributions in the field of hydrostatics. Thus, as shown by the photograph in the margin, at the junction of the side and bottom of the beaker, the pressure is the same on the side as it is on the bottom. In Chapter 6 it will be shown that for moving fluids in which there is relative motion between particles 1so that shearing stresses develop2, the normal stress at a point, which corresponds to pressure in fluids at rest, is not necessarily the same 42 Chapter 2 ■ Fluid Statics in all directions. In such cases the pressure is defined as the average of any three mutually perpendicular normal stresses at the point. 2.2 Basic Equation for Pressure Field The pressure may vary across a fluid particle. ∂p δy ––– ––– ∂y 2 Although we have answered the question of how the pressure at a point varies with direction, we are now faced with an equally important question—how does the pressure in a fluid in which there are no shearing stresses vary from point to point? To answer this question, consider a small rectangular element of fluid removed from some arbitrary position within the mass of fluid of interest as illustrated in Fig. 2.2. There are two types of forces acting on this element: surface forces due to the pressure and a body force equal to the weight of the element. Other possible types of body forces, such as those due to magnetic fields, will not be considered in this text. If we let the pressure at the center of the element be designated as p, then the average pressure on the various faces can be expressed in terms of p and its derivatives, as shown in Fig. 2.2. We are actually using a Taylor series expansion of the pressure at the element center to approximate the pressures a short distance away and neglecting higher order terms that will vanish as we let dx, dy, and dz approach zero. This is illustrated by the figure in the margin. For simplicity the surface forces in the x direction are not shown. The resultant surface force in the y direction is dFy ap p 0p dy 0p dy b dx dz ap b dx dz 0y 2 0y 2 or δy ––– 2 y dFy 0p dx dy dz 0y Similarly, for the x and z directions the resultant surface forces are dFx 0p dx dy dz 0x dFz 0p dx dy dz 0z The resultant surface force acting on the element can be expressed in vector form as dFs dFxˆi dFy jˆ dFzkˆ ( ) ∂p δ z p + ––– ––– δ x δ y ∂z 2 z δz ( ) ∂p δy p – ––– ––– δ x δ z ∂y 2 δx ( γ δx δyδz δy ( ^ k ^ ^ j ) ∂p δy p + ––– ––– δ x δ z ∂y 2 ) ∂p δ z p – ––– ––– δ x δ y ∂z 2 y i x ■ Figure 2.2 Surface and body forces acting on small fluid element. 2.3 The resultant surface force acting on a small fluid element depends only on the pressure gradient if there are no shearing stresses present. Pressure Variation in a Fluid at Rest 43 or dFs a 0p ˆ 0p ˆ 0p ˆ i j kb dx dy dz 0x 0y 0z (2.1) where ˆi , jˆ, and kˆ are the unit vectors along the coordinate axes shown in Fig. 2.2. The group of terms in parentheses in Eq. 2.1 represents in vector form the pressure gradient and can be written as 0p ˆ 0p ˆ 0p ˆ i j k §p 0x 0y 0z where §1 2 01 2 0x ˆi 01 2 0y jˆ 01 2 0z kˆ and the symbol § is the gradient or “del” vector operator. Thus, the resultant surface force per unit volume can be expressed as dFs §p dx dy dz Since the z axis is vertical, the weight of the element is dwkˆ g dx dy dz kˆ where the negative sign indicates that the force due to the weight is downward 1in the negative z direction2. Newton’s second law, applied to the fluid element, can be expressed as a dF dm a where dF represents the resultant force acting on the element, a is the acceleration of the element, and dm is the element mass, which can be written as r dx dy dz. It follows that ˆ a dF dFs dwk dm a or §p dx dy dz g dx dy dz kˆ r dx dy dz a and, therefore, §p gkˆ ra (2.2) Equation 2.2 is the general equation of motion for a fluid in which there are no shearing stresses. We will use this equation in Section 2.12 when we consider the pressure distribution in a moving fluid. For the present, however, we will restrict our attention to the special case of a fluid at rest. 2.3 Pressure Variation in a Fluid at Rest For a fluid at rest a 0 and Eq. 2.2 reduces to §p gkˆ 0 or in component form 0p 0 0x 0p 0 0y 0p g 0z (2.3) These equations show that the pressure does not depend on x or y. Thus, as we move from point to point in a horizontal plane 1any plane parallel to the x–y plane2, the pressure does not 44 Chapter 2 ■ Fluid Statics For liquids or gases at rest, the pressure gradient in the vertical direction at any point in a fluid depends only on the specific weight of the fluid at that point. p dz g dp 1 dp ––– = −g dz z change. Since p depends only on z, the last of Eqs. 2.3 can be written as the ordinary differential equation dp g dz (2.4) Equation 2.4 is the fundamental equation for fluids at rest and can be used to determine how pressure changes with elevation. This equation and the figure in the margin indicate that the pressure gradient in the vertical direction is negative; that is, the pressure decreases as we move upward in a fluid at rest. There is no requirement that g be a constant. Thus, it is valid for fluids with constant specific weight, such as liquids, as well as fluids whose specific weight may vary with elevation, such as air or other gases. However, to proceed with the integration of Eq. 2.4 it is necessary to stipulate how the specific weight varies with z. If the fluid is flowing (i.e., not at rest with a 0), then the pressure variation is usually much more complex than that given by Eq. 2.4. For example, the pressure distribution on your car as it is driven along the road varies in a complex manner with x, y, and z. This idea is covered in detail in Chapters 3, 6, and 9. 2.3.1 Incompressible Fluid Since the specific weight is equal to the product of fluid density and acceleration of gravity 1g rg2, changes in g are caused by a change in either r or g. For most engineering applications the variation in g is negligible, so our main concern is with the possible variation in the fluid density. In general, a fluid with constant density is called an incompressible fluid. For liquids the variation in density is usually negligible, even over large vertical distances, so that the assumption of constant specific weight when dealing with liquids is a good one. For this instance, Eq. 2.4 can be directly integrated 冮 p2 dp g p1 to yield dz z1 p2 p1 g1z2 z1 2 or V2.1 Pressure on a car 冮 z2 p1 p2 g1z2 z1 2 (2.5) where p1 and p2 are pressures at the vertical elevations z1 and z2, as is illustrated in Fig. 2.3. Equation 2.5 can be written in the compact form p1 p2 gh (2.6) p1 gh p2 (2.7) or where h is the distance, z2 z1, which is the depth of fluid measured downward from the location of p2. This type of pressure distribution is commonly called a hydrostatic distribution, and Eq. 2.7 Free surface (pressure = p0) V2.2 Demonstration of atmospheric pressure. p2 z h = z2 – z1 z2 p1 z1 y ■ Figure 2.3 Notation for pressure variax tion in a fluid at rest with a free surface. 2.3 pA = 0 A = 1 in.2 Pressure Variation in a Fluid at Rest 45 shows that in an incompressible fluid at rest the pressure varies linearly with depth. The pressure must increase with depth to “hold up” the fluid above it. It can also be observed from Eq. 2.6 that the pressure difference between two points can be specified by the distance h since 23.1 ft h ᐃ = 10 lb p1 p2 g In this case h is called the pressure head and is interpreted as the height of a column of fluid of specific weight g required to give a pressure difference p1 p2. For example, a pressure difference of 10 psi can be specified in terms of pressure head as 23.1 ft of water 1g 62.4 lbft3 2, or 518 mm of Hg 1g 133 kNm3 2. As illustrated by the figure in the margin, a 23.1-ft-tall column of water with a cross-sectional area of 1 in.2 weighs 10 lb. Water pA = 10 lb F l u i d s i n Giraffe’s blood pressure A giraffe’s long neck allows it to graze up to 6 m above the ground. It can also lower its head to drink at ground level. Thus, in the circulatory system there is a significant hydrostatic pressure effect due to this elevation change. To maintain blood to its head throughout this change in elevation, the giraffe must maintain a relatively high blood pressure at heart level—approximately two and a half times that in humans. To prevent rupture of blood vessels in the high-pressure lower leg t h e N e w s regions, giraffes have a tight sheath of thick skin over their lower limbs that acts like an elastic bandage in exactly the same way as do the g-suits of fighter pilots. In addition, valves in the upper neck prevent backflow into the head when the giraffe lowers its head to ground level. It is also thought that blood vessels in the giraffe’s kidney have a special mechanism to prevent large changes in filtration rate when blood pressure increases or decreases with its head movement. (See Problem 2.17.) When one works with liquids there is often a free surface, as is illustrated in Fig. 2.3, and it is convenient to use this surface as a reference plane. The reference pressure p0 would correspond to the pressure acting on the free surface 1which would frequently be atmospheric pressure2, and thus if we let p2 p0 in Eq. 2.7 it follows that the pressure p at any depth h below the free surface is given by the equation: p gh p0 (2.8) As is demonstrated by Eq. 2.7 or 2.8, the pressure in a homogeneous, incompressible fluid at rest depends on the depth of the fluid relative to some reference plane, and it is not influenced by the size or shape of the tank or container in which the fluid is held. Thus, in Fig. 2.4 Liquid surface (p = p0) h B A ■ Figure 2.4 Fluid pressure in Specific weight γ containers of arbitrary shape. 46 Chapter 2 ■ Fluid Statics the pressure is the same at all points along the line AB, even though the containers have very irregular shapes. The actual value of the pressure along AB depends only on the depth, h, the surface pressure, p0, and the specific weight, g, of the liquid in the container. E XAMPLE 2.1 Pressure–Depth Relationship GIVEN Because of a leak in a buried gasoline storage tank, Open water has seeped in to the depth shown in Fig. E2.1. The specific gravity of the gasoline is SG 0.68. FIND Determine the pressure at the gasoline–water interface and at the bottom of the tank. Express the pressure in units of lbft2, lb in.2, and as a pressure head in feet of water. 17 ft Gasoline (1) Water (2) SOLUTION 3 ft ■ Figure E2.1 Since we are dealing with liquids at rest, the pressure distribution will be hydrostatic, and therefore the pressure variation can be found from the equation: p gh p0 With p0 corresponding to the pressure at the free surface of the gasoline, then the pressure at the interface is p2 gH2O hH2O p1 p1 SGgH2O h p0 10.682162.4 lbft3 2117 ft2 p0 721 p0 1lbft2 2 162.4 lb ft3 213 ft2 721 lbft2 908 lbft2 If we measure the pressure relative to atmospheric pressure 1gage pressure2, it follows that p0 0, and therefore p1 721 lbft2 p1 721 lb ft2 5.01 lb in.2 144 in.2ft2 p1 721 lbft 11.6 ft gH2O 62.4 lb ft3 (Ans) p2 908 lb ft2 6.31 lbin.2 144 in.2 ft2 p2 908 lbft2 14.6 ft gH2O 62.4 lbft3 (Ans) (Ans) (Ans) (Ans) COMMENT Observe that if we wish to express these pres- (Ans) sures in terms of absolute pressure, we would have to add the local atmospheric pressure 1in appropriate units2 to the previous results. A further discussion of gage and absolute pressure is given in Section 2.5. 2 The transmission of pressure throughout a stationary fluid is the principle upon which many hydraulic devices are based. It is noted that a rectangular column of water 11.6 ft tall and 1 ft2 in cross section weighs 721 lb. A similar column with a 1-in.2 cross section weighs 5.01 lb. We can now apply the same relationship to determine the pressure at the tank bottom; that is, The required equality of pressures at equal elevations throughout a system is important for the operation of hydraulic jacks (see Fig. 2.5a), lifts, and presses, as well as hydraulic controls on aircraft and other types of heavy machinery. The fundamental idea behind such devices and systems is demonstrated in Fig. 2.5b. A piston located at one end of a closed system filled with a liquid, such as oil, can be used to change the pressure throughout the system, and thus transmit an applied force F1 to a second piston where the resulting force is F2. Since the pressure p acting on the faces of both pistons is the same 1the effect of elevation changes is usually negligible for this type of hydraulic device2, it follows that F2 1A2 A1 2F1. The piston area A2 can be made much larger than A1 and therefore a large mechanical advantage can be developed; that is, a small force applied at the smaller piston can be used to develop a large force at the larger piston. The applied force could be created manually through some type of mechanical device, such as a hydraulic jack, or through compressed air acting directly on the surface of the liquid, as is done in hydraulic lifts commonly found in service stations. 2.3 Pressure Variation in a Fluid at Rest 47 A2 F2 = pA2 F1 = pA1 A2 A1 A1 (b) (a) ■ Figure 2.5 (a) Hydraulic jack, (b) Transmission of fluid pressure. 2.3.2 Compressible Fluid We normally think of gases such as air, oxygen, and nitrogen as being compressible fluids because the density of the gas can change significantly with changes in pressure and temperature. Thus, although Eq. 2.4 applies at a point in a gas, it is necessary to consider the possible variation in g before the equation can be integrated. However, as was discussed in Chapter 1, the specific weights of common gases are small when compared with those of liquids. For example, the specific weight of air at sea level and 60 °F is 0.0763 lb ft3, whereas the specific weight of water under the same conditions is 62.4 lb ft3. Since the specific weights of gases are comparatively small, it follows from Eq. 2.4 that the pressure gradient in the vertical direction is correspondingly small, and even over distances of several hundred feet the pressure will remain essentially constant for a gas. This means we can neglect the effect of elevation changes on the pressure for stationary gases in tanks, pipes, and so forth in which the distances involved are small. For those situations in which the variations in heights are large, on the order of thousands of feet, attention must be given to the variation in the specific weight. As is described in Chapter 1, the equation of state for an ideal 1or perfect2 gas is r If the specific weight of a fluid varies significantly as we move from point to point, the pressure will no longer vary linearly with depth. p RT where p is the absolute pressure, R is the gas constant, and T is the absolute temperature. This relationship can be combined with Eq. 2.4 to give gp dp dz RT and by separating variables 冮 p2 p1 dp p2 g ln p p1 R 冮 z2 z1 dz T (2.9) where g and R are assumed to be constant over the elevation change from z1 to z2. Although the acceleration of gravity, g, does vary with elevation, the variation is very small 1see Tables C.1 and C.2 in Appendix C2, and g is usually assumed constant at some average value for the range of elevation involved. 48 Chapter 2 ■ Fluid Statics 1 Before completing the integration, one must specify the nature of the variation of temperature with elevation. For example, if we assume that the temperature has a constant value T0 over the range z1 to z2 1isothermal conditions2, it then follows from Eq. 2.9 that p2 / p1 Isothermal 0.8 0.6 p2 p1 exp c Incompressible 0 5000 10,000 z2 – z1, ft E XAMPLE g1z2 z1 2 RT0 d (2.10) This equation provides the desired pressure–elevation relationship for an isothermal layer. As shown in the margin figure, even for a 10,000-ft altitude change, the difference between the constant temperature 1isothermal2 and the constant density 1incompressible2 results is relatively minor. For nonisothermal conditions a similar procedure can be followed if the temperature–elevation relationship is known, as is discussed in the following section. 2.2 Incompressible and Isothermal Pressure–Depth Variations GIVEN In 2010, the world’s tallest building, the Burj Khalifa skyscraper, was completed and opened in the United Arab Emirates. The final height of the building, which had remained a secret until completion, is 2717 ft (828 m). FIND (a) Estimate the ratio of the pressure at the 2717-ft top of the building to the pressure at its base, assuming the air to be at a common temperature of 59 °F. (b) Compare the pressure calculated in part (a) with that obtained by assuming the air to be incompressible with g 0.0765 lbft3 at 14.7 psi 1abs2 1values for air at standard sea level conditions2. SOLUTION (a) For the assumed isothermal conditions, and treating air as a compressible fluid, Eq. 2.10 can be applied to yield g1z2 z1 2 p2 exp c d p1 RT0 exp e 0.906 132.2 fts2 212717 ft2 f 11716 ft # lbslug # °R2 3 159 4602°R 4 (Ans) (b) If the air is treated as an incompressible fluid we can apply Eq. 2.5. In this case ■ Figure E2.2 (Figure courtesy of Emaar Properties, Dubai, UAE.) p2 p1 g1z2 z1 2 or g1z2 z1 2 p2 1 p1 p1 10.0765 lbft3 2 12717 ft2 1 0.902 114.7 lbin.2 21144 in.2 ft2 2 (Ans) COMMENTS Note that there is little difference between the two results. Since the pressure difference between the bottom and top of the building is small, it follows that the variation in fluid density is small and, therefore, the compressible fluid and incompressible fluid analyses yield essentially the same result. We see that for both calculations the pressure decreases by approximately 10% as we go from ground level to the top of this tallest building. It does not require a very large pressure difference to support a 2717-ft-tall column of fluid as light as air. This result supports the earlier statement that the changes in pressures in air and other gases due to elevation changes are very small, even for distances of hundreds of feet. Thus, the pressure differences between the top and bottom of a horizontal pipe carrying a gas, or in a gas storage tank, are negligible since the distances involved are very small. 2.4 2.4 49 Standard Atmosphere The standard atmosphere is an idealized representation of mean conditions in the earth’s atmosphere. 300 Space shuttle 150 An important application of Eq. 2.9 relates to the variation in pressure in the earth’s atmosphere. Ideally, we would like to have measurements of pressure versus altitude over the specific range for the specific conditions 1temperature, reference pressure2 for which the pressure is to be determined. However, this type of information is usually not available. Thus, a “standard atmosphere” has been determined that can be used in the design of aircraft, missiles, and spacecraft and in comparing their performance under standard conditions. The concept of a standard atmosphere was first developed in the 1920s, and since that time many national and international committees and organizations have pursued the development of such a standard. The currently accepted standard atmosphere is based on a report published in 1962 and updated in 1976 1see Refs. 1 and 22, defining the so-called U.S. standard atmosphere, which is an idealized representation of middle-latitude, year-round mean conditions of the earth’s atmosphere. Several important properties for standard atmospheric conditions at sea level are listed in Table 2.1, and Fig. 2.6 shows the temperature profile for the U.S. standard atmosphere. As is shown in this figure, the temperature decreases with altitude in the region nearest the earth’s surface 1troposphere2, then becomes essentially constant in the next layer 1stratosphere2, and subsequently starts to increase in the next layer. Typical events that occur in the atmosphere are shown in the figure in the margin. Since the temperature variation is represented by a series of linear segments, it is possible to integrate Eq. 2.9 to obtain the corresponding pressure variation. For example, in the troposphere, which extends to an altitude of about 11 km 1苲36,000 ft2, the temperature variation is of the form T Ta bz Aurora Meteor 50 Property SI Units BG Units Temperature, T Pressure, p 288.15 K 115 °C2 101.33 kPa 1abs2 Density, r Specific weight, g Viscosity, m 1.225 kgm3 12.014 Nm3 1.789 105 N # sm2 518.67 °R 159.00 °F2 2116.2 lbft2 1abs2 314.696 lbin.2 1abs2 4 0.002377 slugsft3 0.07647 lb ft3 3.737 107 lb # sft2 Ozone layer Mt. Everest Acceleration of gravity at sea level 9.807 ms2 32.174 ft s2. -2.5 °C 0 50 47.3 km (p = 0.1 kPa) -44.5 °C Thunder storm Commercial jet (2.11) Table 2.1 Properties of U.S. Standard Atmosphere at Sea Level 40 32.2 km (p = 0.9 kPa) 30 20 Stratosphere -56.5 °C 100 Altitude z, km Altitude z, km Standard Atmosphere 10 20.1 km (p = 5.5 kPa) 11.0 km (p = 22.6 kPa) p = 101.3 kPa (abs) 15 °C Troposphere 0 -100 -80 -60 -40 -20 Temperature T, °C 0 +20 ■ Figure 2.6 Variation of temperature with altitude in the U.S. standard atmosphere. 50 Chapter 2 ■ Fluid Statics where Ta is the temperature at sea level 1z ⫽ 02 and b is the lapse rate 1the rate of change of temperature with elevation2. For the standard atmosphere in the troposphere, b ⫽ 0.00650 KⲐm or 0.00357 °RⲐft. Equation 2.11 used with Eq. 2.9 yields p ⫽ pa a1 ⫺ bz gⲐRb b Ta (2.12) where pa is the absolute pressure at z ⫽ 0. With pa, Ta, and g obtained from Table 2.1, and with the gas constant R ⫽ 286.9 JⲐkg # K or 1716 ft # lbⲐslug # °R, the pressure variation throughout the troposphere can be determined from Eq. 2.12. This calculation shows that at the outer edge of the troposphere, where the temperature is ⫺56.5 °C 1⫺69.7 °F2, the absolute pressure is about 23 kPa 13.3 psia2. It is to be noted that modern jetliners cruise at approximately this altitude. Pressures at other altitudes are shown in Fig. 2.6, and tabulated values for temperature, acceleration of gravity, pressure, density, and viscosity for the U.S. standard atmosphere are given in Tables C.1 and C.2 in Appendix C. Measurement of Pressure Pressure is designated as either absolute pressure or gage pressure. Since pressure is a very important characteristic of a fluid field, it is not surprising that numerous devices and techniques are used in its measurement. As is noted briefly in Chapter 1, the pressure at a point within a fluid mass will be designated as either an absolute pressure or a gage pressure. Absolute pressure is measured relative to a perfect vacuum 1absolute zero pressure2, whereas gage pressure is measured relative to the local atmospheric pressure. Thus, a gage pressure of zero corresponds to a pressure that is equal to the local atmospheric pressure. Absolute pressures are always positive, but gage pressures can be either positive or negative depending on whether the pressure is above atmospheric pressure 1a positive value2 or below atmospheric pressure 1a negative value2. A negative gage pressure is also referred to as a suction or vacuum pressure. For example, 10 psi 1abs2 could be expressed as ⫺4.7 psi 1gage2, if the local atmospheric pressure is 14.7 psi, or alternatively 4.7 psi suction or 4.7 psi vacuum. The concept of gage and absolute pressure is illustrated graphically in Fig. 2.7 for two typical pressures located at points 1 and 2. In addition to the reference used for the pressure measurement, the units used to express the value are obviously of importance. As is described in Section 1.5, pressure is a force per unit area, and the units in the BG system are lbⲐft2 or lbⲐin.2, commonly abbreviated psf or psi, respectively. In the SI system the units are NⲐm2; this combination is called the pascal and written as Pa 11 NⲐm2 ⫽ 1 Pa2. As noted earlier, pressure can also be expressed as the height of a column of liquid. Then the units will refer to the height of the column 1in., ft, mm, m, etc.2, and in addition, the liquid in the column must be specified 1H2O, Hg, etc.2. For example, standard atmospheric pressure can be expressed as 760 mm Hg 1abs2. In this text, pressures will be assumed to be gage pressures unless specifically designated absolute. For example, 10 psi or 100 kPa would be gage pressures, whereas 10 psia or 100 kPa 1abs2 would refer to absolute pressures. It is to be 1 Gage pressure @ 1 Pressure 2.5 Local atmospheric pressure reference 2 Absolute pressure @1 Gage pressure @ 2 (suction or vacuum) Absolute pressure @2 ■ Figure 2.7 Graphical representation Absolute zero reference of gage and absolute pressure. 2.5 51 Measurement of Pressure pvapor A h patm B Mercury Water ■ Figure 2.8 Mercury barometer. noted that pressure differences are independent of the reference, so that no special notation is required in this case. The measurement of atmospheric pressure is usually accomplished with a mercury barometer, which in its simplest form consists of a glass tube closed at one end with the open end immersed in a container of mercury as shown in Fig. 2.8. The tube is initially filled with mercury 1inverted with its open end up2 and then turned upside down 1open end down2, with the open end in the container of mercury. The column of mercury will come to an equilibrium position where its weight plus the force due to the vapor pressure 1which develops in the space above the column2 balances the force due to the atmospheric pressure. Thus, patm gh pvapor (2.13) where g is the specific weight of mercury. For most practical purposes the contribution of the vapor pressure can be neglected since it is very small [for mercury, the fluid most commonly used in barometers, pvapor 0.000023 lbin.2 1abs2 at a temperature of 68 °F], so that patm ⬇ gh. It is conventional to specify atmospheric pressure in terms of the height, h, in millimeters or inches of mercury. Note that if water were used instead of mercury, the height of the column would have to be approximately 34 ft rather than 29.9 in. of mercury for an atmospheric pressure of 14.7 psia! This is shown to scale in the figure in the margin. The concept of the mercury barometer is an old one, with the invention of this device attributed to Evangelista Torricelli in about 1644. Mercury E XAMPLE 2.3 Barometric Pressure GIVEN A mountain lake has an average temperature of 10 °C and a maximum depth of 40 m. The barometric pressure is 598 mm Hg. FIND Determine the absolute pressure 1in pascals2 at the deepest part of the lake. SOLUTION The pressure in the lake at any depth, h, is given by the equation p gh p0 where p0 is the pressure at the surface. Since we want the absolute pressure, p0 will be the local barometric pressure expressed in a consistent system of units; that is pbarometric 598 mm 0.598 m gHg and for gHg 133 kNm3 p0 10.598 m2 1133 kNm3 2 79.5 kNm2 From Table B.2, gH2 O 9.804 kNm3 at 10 °C and therefore p 19.804 kNm3 2140 m2 79.5 kNm2 392 kNm2 79.5 kNm2 472 kPa 1abs2 (Ans) COMMENT This simple example illustrates the need for close attention to the units used in the calculation of pressure; that is, be sure to use a consistent unit system, and be careful not to add a pressure head 1m2 to a pressure 1Pa2. 52 Chapter 2 ■ Fluid Statics F l u i d s i n Weather, barometers, and bars One of the most important indicators of weather conditions is atmospheric pressure. In general, a falling or low pressure indicates bad weather; rising or high pressure, good weather. During the evening TV weather report in the United States, atmospheric pressure is given as so many inches (commonly around 30 in.). This value is actually the height of the mercury column in a mercury barometer adjusted to sea level. To determine the true atmospheric pressure at a particular location, the elevation relative to sea level must be known. Another unit used by meteorologists to indicate atmospheric pressure is the bar, first used in 2.6 t h e N e w s weather reporting in 1914 and defined as 105 N m2. The definition of a bar is probably related to the fact that standard sealevel pressure is 1.0133 105 Nm2, that is, only slightly larger than one bar. For typical weather patterns, “sea-level equivalent” atmospheric pressure remains close to one bar. However, for extreme weather conditions associated with tornadoes, hurricanes, or typhoons, dramatic changes can occur. The lowest atmospheric sea-level pressure ever recorded was associated with a typhoon, Typhoon Tip, in the Pacific Ocean on October 12, 1979. The value was 0.870 bar (25.8 in. Hg). (See Problem 2.24.) Manometry Manometers use vertical or inclined liquid columns to measure pressure. A standard technique for measuring pressure involves the use of liquid columns in vertical or inclined tubes. Pressure-measuring devices based on this technique are called manometers. The mercury barometer is an example of one type of manometer, but there are many other configurations possible depending on the particular application. Three common types of manometers include the piezometer tube, the U-tube manometer, and the inclined-tube manometer. 2.6.1 Piezometer Tube Tube open at top Column of mercury The simplest type of manometer consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired, as illustrated in Fig. 2.9. The figure in the margin shows an important device whose operation is based on this principle. It is a sphygmomanometer, the traditional instrument used to measure blood pressure. Since manometers involve columns of fluids at rest, the fundamental equation describing their use is Eq. 2.8 Container of mercury p gh p0 Arm cuff which gives the pressure at any elevation within a homogeneous fluid in terms of a reference pressure p0 and the vertical distance h between p and p0. Remember that in a fluid at rest pressure will increase as we move downward and will decrease as we move upward. Application of this equation to the piezometer tube of Fig. 2.9 indicates that the pressure pA can be determined by a measurement of h1 through the relationship pA g1h1 where g1 is the specific weight of the liquid in the container. Note that since the tube is open at the top, the pressure p0 can be set equal to zero 1we are now using gage pressure2, with the height Open γ1 A h1 (1) ■ Figure 2.9 Piezometer tube. 2.6 Manometry 53 Open γ1 A (1) h2 h1 (2) (3) γ2 (gage fluid) ■ Figure 2.10 Simple U-tube manometer. h1 measured from the meniscus at the upper surface to point 112. Since point 112 and point A within the container are at the same elevation, pA p1. Although the piezometer tube is a very simple and accurate pressure-measuring device, it has several disadvantages. It is suitable only if the pressure in the container is greater than atmospheric pressure 1otherwise air would be sucked into the system2, and the pressure to be measured must be relatively small so the required height of the column is reasonable. Also the fluid in the container in which the pressure is to be measured must be a liquid rather than a gas. 2.6.2 U-Tube Manometer The contribution of gas columns in manometers is usually negligible since the weight of the gas is so small. V2.3 Blood pressure measurement To overcome the difficulties noted previously, another type of manometer which is widely used consists of a tube formed into the shape of a U, as is shown in Fig. 2.10. The fluid in the manometer is called the gage fluid. To find the pressure pA in terms of the various column heights, we start at one end of the system and work our way around to the other end, simply utilizing Eq. 2.8. Thus, for the U-tube manometer shown in Fig. 2.10, we will start at point A and work around to the open end. The pressure at points A and 112 are the same, and as we move from point 112 to 122 the pressure will increase by g1h1. The pressure at point 122 is equal to the pressure at point 132, since the pressures at equal elevations in a continuous mass of fluid at rest must be the same. Note that we could not simply “jump across” from point 112 to a point at the same elevation in the right-hand tube since these would not be points within the same continuous mass of fluid. With the pressure at point 132 specified, we now move to the open end where the pressure is zero. As we move vertically upward the pressure decreases by an amount g2h2. In equation form these various steps can be expressed as pA g1h1 g2h2 0 and, therefore, the pressure pA can be written in terms of the column heights as pA g2h2 g1h1 (2.14) A major advantage of the U-tube manometer lies in the fact that the gage fluid can be different from the fluid in the container in which the pressure is to be determined. For example, the fluid in A in Fig. 2.10 can be either a liquid or a gas. If A does contain a gas, the contribution of the gas column, g1h1, is almost always negligible so that pA ⬇ p2 , and in this instance Eq. 2.14 becomes pA g2h2 Thus, for a given pressure the height, h2, is governed by the specific weight, g2, of the gage fluid used in the manometer. If the pressure pA is large, then a heavy gage fluid, such as mercury, can be used and a reasonable column height 1not too long2 can still be maintained. Alternatively, if the pressure pA is small, a lighter gage fluid, such as water, can be used so that a relatively large column height 1which is easily read2 can be achieved. 54 Chapter 2 ■ Fluid Statics E XAMPLE 2.4 Simple U-Tube Manometer GIVEN A closed tank contains compressed air and oil Pressure gage 1SGoil 0.902 as is shown in Fig. E2.4. A U-tube manometer using mercury 1SGHg 13.62 is connected to the tank as shown. The column heights are h1 36 in., h2 6 in., and h3 9 in. Air FIND Determine the pressure reading 1in psi2 of the gage. Open h1 Oil SOLUTION h3 Following the general procedure of starting at one end of the manometer system and working around to the other, we will start at the air–oil interface in the tank and proceed to the open end where the pressure is zero. The pressure at level 112 is h2 (1) Hg p1 pair goil 1h1 h2 2 ■ Figure E2.4 This pressure is equal to the pressure at level 122, since these two points are at the same elevation in a homogeneous fluid at rest. As we move from level 122 to the open end, the pressure must decrease by gHgh3, and at the open end the pressure is zero. Thus, the manometer equation can be expressed as Since the specific weight of the air above the oil is much smaller than the specific weight of the oil, the gage should read the pressure we have calculated; that is, pair goil 1h1 h2 2 gHgh3 0 pgage or pair 1SGoil 21gH2O 2 1h1 h2 2 1SGHg 21gH2O 2 h3 0 440 lb ft2 3.06 psi 144 in.2ft2 (Ans) COMMENTS Note that the air pressure is a function of the height of the mercury in the manometer and the depth of the oil (both in the tank and in the tube). It is not just the mercury in the manometer that is important. Assume that the gage pressure remains at 3.06 psi, but the manometer is altered so that it contains only oil. That is, the mercury is replaced by oil. A simple calculation shows that in this case the vertical oil-filled tube would need to be h3 11.3 ft tall, rather than the original h3 9 in. There is an obvious advantage of using a heavy fluid such as mercury in manometers. For the values given 36 6 ftb 12 9 113.62 162.4 lbft3 2 a ftb 12 pair 10.92 162.4 lbft3 2 a so that pair 440 lbft2 Manometers are often used to measure the difference in pressure between two points. (2) The U-tube manometer is also widely used to measure the difference in pressure between two containers or two points in a given system. Consider a manometer connected between containers A and B as is shown in Fig. 2.11. The difference in pressure between A and B can be found B (5) h3 γ3 (4) γ1 γ2 A (1) h2 h1 (2) (3) ■ Figure 2.11 Differential U-tube manometer. 2.6 Manometry 55 by again starting at one end of the system and working around to the other end. For example, at A the pressure is pA, which is equal to p1, and as we move to point 122 the pressure increases by g1h1. The pressure at p2 is equal to p3, and as we move upward to point 142 the pressure decreases by g2h2. Similarly, as we continue to move upward from point 142 to 152 the pressure decreases by g3h3. Finally, p5 pB, since they are at equal elevations. Thus, γ1h1 γ2h2 pA g1h1 g2h2 g3h3 pB pA Or, as indicated in the figure in the margin, we could start at B and work our way around to A to obtain the same result. In either case, the pressure difference is pA pB g2h2 g3h3 g1h1 When the time comes to substitute in numbers, be sure to use a consistent system of units! Capillarity due to surface tension at the various fluid interfaces in the manometer is usually not considered, since for a simple U-tube with a meniscus in each leg, the capillary effects cancel 1assuming the surface tensions and tube diameters are the same at each meniscus2, or we can make the capillary rise negligible by using relatively large bore tubes 1with diameters of about 0.5 in. or larger; see Section 1.92. Two common gage fluids are water and mercury. Both give a well-defined meniscus 1a very important characteristic for a gage fluid2 and have wellknown properties. Of course, the gage fluid must be immiscible with respect to the other fluids in contact with it. For highly accurate measurements, special attention should be given to temperature since the various specific weights of the fluids in the manometer will vary with temperature. pA − pB γ3h3 pB E XAMPLE 2.5 U-Tube Manometer GIVEN As will be discussed in Chapter 3, the volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated in Fig. E2.5a. The nozzle creates a pressure drop, pA pB, along the pipe that is related to the flow through the equation Q K 1pA pB, where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer of the type illustrated. FIND 1a2 Determine an equation for pA pB in terms of the specific weight of the flowing fluid, g1, the specific weight of the gage fluid, g2, and the various heights indicated. 1b2 For g1 9.80 kN m3, g2 15.6 kN m3, h1 1.0 m, and h2 0.5 m, what is the value of the pressure drop, pA pB? γ1 (4) (5) (3) γ1 h2 (1) SOLUTION (2) h1 (a) Although the fluid in the pipe is moving, the fluids in the columns of the manometer are at rest so that the pressure variation in the manometer tubes is hydrostatic. If we start at point A and move vertically upward to level 112, the pressure will decrease by g1h1 and will be equal to the pressure at 122 and at 132. We can now move from 132 to 142 where the pressure has been further reduced by g2h2. The pressures at levels 142 and 152 are equal, and as we move from 152 to B the pressure will increase by g1 1h1 h2 2. Thus, in equation form pA g1h1 g2h2 g1 1h1 h2 2 pB (Ans) COMMENT It is to be noted that the only column height of importance is the differential reading, h2. The differential A B Flow nozzle ■ Figure E2.5a manometer could be placed 0.5 or 5.0 m above the pipe (h1 0.5 m or h1 5.0 m), and the value of h2 would remain the same. (b) or pA pB h2 1g2 g1 2 Flow γ2 The specific value of the pressure drop for the data given is pA pB 10.5 m2115.6 kNm3 9.80 kNm3 2 (Ans) 2.90 kPa COMMENT By repeating the calculations for manometer fluids with different specific weights, ␥2, the results shown in Fig. E2.5b are obtained. Note that relatively small pressure 56 Chapter 2 ■ Fluid Statics 3 differences can be measured if the manometer fluid has nearly the same specific weight as the flowing fluid. It is the difference in the specific weights, ␥2 ␥1, that is important. Hence, by rewriting the answer as h2 1pA pB 2 1g2 g1 2 it is seen that even if the value of pA pB is small, the value of h2 can be large enough to provide an accurate reading provided the value of g2 g1 is also small. pA – pB, kPa (15.6 kN/m3, 2.90 kPa) 2 1 ␥2 = ␥1 0 8 10 12 ␥2, kN/m3 14 16 ■ Figure E2.5b 2.6.3 Inclined-Tube Manometer To measure small pressure changes, a manometer of the type shown in Fig. 2.12 is frequently used. One leg of the manometer is inclined at an angle u, and the differential reading /2 is measured along the inclined tube. The difference in pressure pA pB can be expressed as pA g1h1 g2/2 sin u g3 h3 pB or Inclined-tube manometers can be used to measure small pressure differences accurately. pA pB g2/2 sin u g3 h3 g1h1 (2.15) where it is to be noted the pressure difference between points 112 and 122 is due to the vertical distance between the points, which can be expressed as /2 sin u. Thus, for relatively small angles the differential reading along the inclined tube can be made large even for small pressure differences. The inclined-tube manometer is often used to measure small differences in gas pressures so that if pipes A and B contain a gas, then pA pB g2/2 sin u or ᐉ2 ~ 1 sinθ ᐉ2 0 30 60 θ, deg /2 90 pA pB g2 sin u (2.16) where the contributions of the gas columns h1 and h3 have been neglected. Equation 2.16 and the figure in the margin show that the differential reading /2 1for a given pressure difference2 of the inclined-tube manometer can be increased over that obtained with a conventional U-tube manometer by the factor 1sin u. Recall that sin u S 0 as u S 0. γ3 B γ1 h3 A (2) γ2 h1 ᐉ2 (1) θ ■ Figure 2.12 Inclined-tube manometer. 2.7 2.7 Mechanical and Electronic Pressure-Measuring Devices 57 Mechanical and Electronic Pressure-Measuring Devices A Bourdon tube pressure gage uses a hollow, elastic, and curved tube to measure pressure. V2.4 Bourdon gage Although manometers are widely used, they are not well suited for measuring very high pressures or pressures that are changing rapidly with time. In addition, they require the measurement of one or more column heights, which, although not particularly difficult, can be time consuming. To overcome some of these problems numerous other types of pressure-measuring instruments have been developed. Most of these make use of the idea that when a pressure acts on an elastic structure the structure will deform, and this deformation can be related to the magnitude of the pressure. Probably the most familiar device of this kind is the Bourdon pressure gage, which is shown in Fig. 2.13a. The essential mechanical element in this gage is the hollow, elastic curved tube 1Bourdon tube2 which is connected to the pressure source as shown in Fig. 2.13b. As the pressure within the tube increases the tube tends to straighten, and although the deformation is small, it can be translated into the motion of a pointer on a dial as illustrated. Since it is the difference in pressure between the outside of the tube 1atmospheric pressure2 and the inside of the tube that causes the movement of the tube, the indicated pressure is gage pressure. The Bourdon gage must be calibrated so that the dial reading can directly indicate the pressure in suitable units such as psi, psf, or pascals. A zero reading on the gage indicates that the measured pressure is equal to the local atmospheric pressure. This type of gage can be used to measure a negative gage pressure 1vacuum2 as well as positive pressures. The aneroid barometer is another type of mechanical gage that is used for measuring atmospheric pressure. Since atmospheric pressure is specified as an absolute pressure, the conventional Bourdon gage is not suitable for this measurement. The common aneroid barometer contains a hollow, closed, elastic element that is evacuated so that the pressure inside the element is near absolute zero. As the external atmospheric pressure changes, the element deflects, and this motion can be translated into the movement of an attached dial. As with the Bourdon gage, the dial can be calibrated to give atmospheric pressure directly, with the usual units being millimeters or inches of mercury. For many applications in which pressure measurements are required, the pressure must be measured with a device that converts the pressure into an electrical output. For example, it may be desirable to continuously monitor a pressure that is changing with time. This type of pressuremeasuring device is called a pressure transducer, and many different designs are used. One possible type of transducer is one in which a Bourdon tube is connected to a linear variable differential transformer 1LVDT2, as is illustrated in Fig. 2.14. The core of the LVDT is connected to the free end of the Bourdon tube so that as a pressure is applied the resulting motion of the end of the tube moves the core through the coil and an output voltage develops. This voltage is a linear function of the pressure and could be recorded on an oscillograph or digitized for storage or processing on a computer. (a) (b) ■ Figure 2.13 (a) Liquid-filled Bourdon pressure gages for various pressure ranges. (b) Internal elements of Bourdon gages. The “C-shaped” Bourdon tube is shown on the left, and the “coiled spring” Bourdon tube for high pressures of 1000 psi and above is shown on the right. (Photographs courtesy of Weiss Instruments, Inc.) 58 Chapter 2 ■ Fluid Statics Bourdon C-tube Output LVDT Core Mounting block Pressure line ■ Figure 2.14 Pressure transducer that combines a linear variable differential transformer (LVDT) with a Bourdon gage. (From Ref. 4, used by permission.) Input Spring F l u i d s i n Tire pressure warning Proper tire inflation on vehicles is important for more than ensuring long tread life. It is critical in preventing accidents such as rollover accidents caused by underinflation of tires. The National Highway Traffic Safety Administration is developing a regulation regarding four-tire tire-pressure monitoring systems that can warn a driver when a tire is more than 25% underinflated. Some of these devices are currently in operation on select vehicles; it is expected that they will soon be required on all vehicles. A typical tire-pressure monitoring It is relatively complicated to make accurate pressure transducers for the measurement of pressures that vary rapidly with time. t h e N e w s system fits within the tire and contains a pressure transducer (usually either a piezo-resistive or a capacitive-type transducer) and a transmitter that sends the information to an electronic control unit within the vehicle. Information about tire pressure and a warning when the tire is underinflated is displayed on the instrument panel. The environment (hot, cold, vibration) in which these devices must operate, their small size, and required low cost provide challenging constraints for the design engineer. One disadvantage of a pressure transducer using a Bourdon tube as the elastic sensing element is that it is limited to the measurement of pressures that are static or only changing slowly 1quasistatic2. Because of the relatively large mass of the Bourdon tube, it cannot respond to rapid changes in pressure. To overcome this difficulty, a different type of transducer is used in which the sensing element is a thin, elastic diaphragm that is in contact with the fluid. As the pressure changes, the diaphragm deflects, and this deflection can be sensed and converted into an electrical voltage. One way to accomplish this is to locate strain gages either on the surface of the diaphragm not in contact with the fluid, or on an element attached to the diaphragm. These gages can accurately sense the small strains induced in the diaphragm and provide an output voltage proportional to pressure. This type of transducer is capable of measuring accurately both small and large pressures, as well as both static and dynamic pressures. For example, strain-gage pressure transducers of the type shown in Fig. 2.15 are used to measure arterial blood pressure, which is a relatively small pressure that varies periodically with a fundamental frequency of about 1 Hz. The transducer is usually connected to the blood vessel by means of a liquid-filled, small-diameter tube called a pressure catheter. Although the strain-gage type of transducer can be designed to have very good frequency response 1up to approximately 10 kHz2, they become less sensitive at the higher frequencies since the diaphragm must be made stiffer to achieve the higher frequency response. As an alternative, the diaphragm can be constructed of a piezoelectric crystal to be used as both the elastic element and the sensor. When a pressure is applied to the crystal, a voltage develops because of the deformation of the crystal. This voltage is directly related to the applied pressure. Depending on the design, this type of transducer can be used to measure both very low and high pressures 1up to approximately 100,000 psi2 at high frequencies. Additional information on pressure transducers can be found in Refs. 3, 4, and 5. 2.8 Hydrostatic Force on a Plane Surface 59 (a) Case Diaphragm stop Electrical connections Armature Diaphragm Link pin Beam (strain gages deposited on beam) (b) ■ Figure 2.15 (a) Photograph of a typical pressure transducer with a male thread fitting in front of the diaphragm for system connection and an electrical connector in the rear of the device. (b) Schematic diagram of a typical pressure transducer device (male thread connector not shown). Deflection of the diaphragm due to pressure is measured with a silicon beam on which strain gages and an associated bridge circuit have been deposited. 2.8 Hydrostatic Force on a Plane Surface V2.5 Hoover dam When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. We also know that the pressure will vary linearly with depth as shown in Fig. 2.16 if the fluid is incompressible. For a horizontal surface, such as the bottom of a liquidfilled tank 1Fig. 2.16a2, the magnitude of the resultant force is simply FR ⫽ pA, where p is the uniform pressure on the bottom and A is the area of the bottom. For the open tank shown, p ⫽ gh. Note that if atmospheric pressure acts on both sides of the bottom, as is illustrated, the resultant force on the bottom is simply due to the liquid in the tank. Since the pressure is constant and uniformly distributed over the bottom, the resultant force acts through the centroid of the area as shown in Fig. 2.16a. As shown in Fig. 2.16b, the pressure on the ends of the tank is not uniformly distributed. Determination of the resultant force for situations such as this is presented as follows. 60 Chapter 2 ■ Fluid Statics Free surface p=0 Free surface p=0 Specific weight = γ h Specific weight = γ p = γh FR p = γh p=0 p=0 (a) Pressure on tank bottom (b) Pressure on tank ends ■ Figure 2.16 (a) Pressure distribution and resultant hydrostatic force on the bottom of an open tank. (b) Pressure distribution on the ends of an open tank. The resultant force of a static fluid on a plane surface is due to the hydrostatic pressure distribution on the surface. For the more general case in which a submerged plane surface is inclined, as is illustrated in Fig. 2.17, the determination of the resultant force acting on the surface is more involved. For the present we will assume that the fluid surface is open to the atmosphere. Let the plane in which the surface lies intersect the free surface at 0 and make an angle u with this surface as in Fig. 2.17. The x–y coordinate system is defined so that 0 is the origin and y 0 (i.e., the x axis) is directed along the surface as shown. The area can have an arbitrary shape as shown. We wish to determine the direction, location, and magnitude of the resultant force acting on one side of this area due to the liquid in contact with the area. At any given depth, h, the force acting on dA 1the differential area of Fig. 2.172 is dF gh dA and is perpendicular to the surface. Thus, the magnitude of the resultant force can be found by summing these differential forces over the entire surface. In equation form FR 冮 gh dA 冮 gy sin u dA A A Free surface 0 θ h hc y yc yR dF FR x x A c y xc xR CP dA Centroid, c Location of resultant force (center of pressure, CP) ■ Figure 2.17 Notation for hydrostatic force on an inclined plane surface of arbitrary shape. 2.8 Hydrostatic Force on a Plane Surface 61 where h y sin u. For constant g and u FR g sin u 冮 y dA (2.17) A The integral appearing in Eq. 2.17 is the first moment of the area with respect to the x axis, so we can write 冮 y dA y A c A where yc is the y coordinate of the centroid of area A measured from the x axis which passes through 0. Equation 2.17 can thus be written as FR gAyc sin u The magnitude of the resultant fluid force is equal to the pressure acting at the centroid of the area multiplied by the total area. γ hc or more simply as FR ghc A (2.18) where, as shown by the figure in the margin, hc is the vertical distance from the fluid surface to the centroid of the area. Note that the magnitude of the force is independent of the angle u. As indicated by the figure in the margin, it depends only on the specific weight of the fluid, the total area, and the depth of the centroid of the area below the surface. In effect, Eq. 2.18 indicates that the magnitude of the resultant force is equal to the pressure at the centroid of the area multiplied by the total area. Since all the differential forces that were summed to obtain FR are perpendicular to the surface, the resultant FR must also be perpendicular to the surface. Although our intuition might suggest that the resultant force should pass through the centroid of the area, this is not actually the case. The y coordinate, yR, of the resultant force can be determined by summation of moments around the x axis. That is, the moment of the resultant force must equal the moment of the distributed pressure force, or FRyR 冮 y dF 冮 g sin u y dA 2 A A and, therefore, since FR gAyc sin u FR = γhc A c 冮 y dA 2 A yR A yc A The integral in the numerator is the second moment of the area (moment of inertia), Ix, with respect to an axis formed by the intersection of the plane containing the surface and the free surface 1x axis2. Thus, we can write Ix yR yc A Use can now be made of the parallel axis theorem to express Ix as Ix Ixc Ay2c where Ixc is the second moment of the area with respect to an axis passing through its centroid and parallel to the x axis. Thus, yc FR c Ixc yc A yR Ixc yc yc A (2.19) As shown by Eq. 2.19 and the figure in the margin, the resultant force does not pass through the centroid but for nonhorizontal surfaces is always below it, since Ixcyc A 7 0. The x coordinate, xR, for the resultant force can be determined in a similar manner by summing moments about the y axis. Thus, FR xR 冮 g sin u xy dA A 62 Chapter 2 ■ Fluid Statics The resultant fluid force does not pass through the centroid of the area. and, therefore, xR 冮 xy dA A yc A Ixy yc A where Ixy is the product of inertia with respect to the x and y axes. Again, using the parallel axis theorem,1 we can write xR Gate c FR FR right left Ixyc yc A xc (2.20) where Ixyc is the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area and formed by a translation of the x–y coordinate system. If the submerged area is symmetrical with respect to an axis passing through the centroid and parallel to either the x or y axis, the resultant force must lie along the line x xc, since Ixyc is identically zero in this case. The point through which the resultant force acts is called the center of pressure. It is to be noted from Eqs. 2.19 and 2.20 that as yc increases the center of pressure moves closer to the centroid of the area. Since yc hc sin u, the distance yc will increase if the depth of submergence, hc, increases, or, for a given depth, the area is rotated so that the angle, u, decreases. Thus, the hydrostatic force on the right-hand side of the gate shown in the margin figure acts closer to the centroid of the gate than the force on the left-hand side. Centroidal coordinates and moments of inertia for some common areas are given in Fig. 2.18. c a –– 2 x a –– 2 y b –– 2 b –– 2 A = ba A = π R2 1 ba3 Ixc = ––– R 12 x c 1 ab3 Iyc = ––– 12 Ixyc = 0 (b) Circle π R2 A = ––––– ab A = ––– d 2 2 Ixc = 0.1098R R x 4R ––– 3π R Iyc = 0.3927R4 ba3 Ixc = –––-– 36 2 Ixyc = –ba –––– (b – 2d) 4 c y 4 Ixyc = 0 y (a) Rectangle π R4 Ixc = Iyc = ––––– 72 a c x y Ixyc = 0 a –– 3 b+d ––––––– 3 b (c) Semicircle (d) Triangle π R2 A = ––––– 4R ––– 3π 4R ––– 3π c 4 Ixc = Iyc = 0.05488R4 x R Ixyc = –0.01647R4 y (e) Quarter circle ■ Figure 2.18 Geometric properties of some common shapes. 1 Recall that the parallel axis theorem for the product of inertia of an area states that the product of inertia with respect to an orthogonal set of axes 1x–y coordinate system2 is equal to the product of inertia with respect to an orthogonal set of axes parallel to the original set and passing through the centroid of the area, plus the product of the area and the x and y coordinates of the centroid of the area. Thus, Ixy Ixyc Axcyc. 2.8 u i d s i n The Three Gorges Dam The Three Gorges Dam being constructed on China’s Yangtze River will contain the world’s largest hydroelectric power plant when in full operation. The dam is of the concrete gravity type, having a length of 2309 m with a height of 185 m. The main elements of the project include the dam, two power plants, and navigation facilities consisting of a ship lock and lift. The power plants will contain 26 Francis-type turbines, each with a capacity of 700 megawatts. The spillway section, which is the center section of the dam, is 483 m long with 23 bottom outlets and 22 surface sluice gates. E XAMPLE t h e N e w s The maximum discharge capacity is 102,500 cu m per second. After more than 10 years of construction, the dam gates were finally closed, and on June 10, 2003, the reservoir had been filled to its interim level of 135 m. Due to the large depth of water at the dam and the huge extent of the storage pool, hydrostatic pressure forces have been a major factor considered by engineers. When filled to its normal pool level of 175 m, the total reservoir storage capacity is 39.3 billion cu m. All of the originally planned components of the project (except for the ship lift) were completed in 2008. (See Problem 2.111.) 2.6 Hydrostatic Force on a Plane Circular Surface GIVEN The 4-m-diameter circular gate of Fig. E2.6a is located in the inclined wall of a large reservoir containing water 1g 9.80 kNm3 2. The gate is mounted on a shaft along its horizontal diameter, and the water depth is 10 m above the shaft. 0 60° (a) the magnitude and location of the resultant force exerted on the gate by the water and (b) the moment that would have to be applied to the shaft to open the gate. x –––10 y Stop yc 10 m FIND Determine 0 si –––m yR n 60–°–– l 63 = F Hydrostatic Force on a Plane Surface Shaft FR A 4m c c (b) A SOLUTION (a) To find the magnitude of the force of the water we can apply Eq. 2.18, Center of pressure (a) Oy FR ᐃ and since the vertical distance from the fluid surface to the centroid of the area is 10 m, it follows that FR 19.80 103 Nm3 2 110 m214p m2 2 1230 103 N 1.23 MN (Ans) To locate the point 1center of pressure2 through which FR acts, we use Eqs. 2.19 and 2.20, Ixyc yc A xc yR Ixc yc yc A For the coordinate system shown, xR 0 since the area is symmetrical, and the center of pressure must lie along the diameter A-A. To obtain yR, we have from Fig. 2.18 Ixc pR 4 4 and yc is shown in Fig. E2.6b. Thus, 1p4212 m2 4 10 m sin 60° 110 msin 60°2 14p m 2 0.0866 m 11.55 m 11.6 m yR Ox M FR ghc A xR c 2 (c) ■ Figure E2.6a–c and the distance 1along the gate2 below the shaft to the center of pressure is yR yc 0.0866 m (Ans) We can conclude from this analysis that the force on the gate due to the water has a magnitude of 1.23 MN and acts through a point along its diameter A-A at a distance of 0.0866 m 1along the gate2 below the shaft. The force is perpendicular to the gate surface as shown in Fig. E2.6b. COMMENT By repeating the calculations for various values of the depth to the centroid, hc, the results shown in Fig. E2.6d are obtained. Note that as the depth increases, the distance between the center of pressure and the centroid decreases. (b) The moment required to open the gate can be obtained with the aid of the free-body diagram of Fig. E2.6c. In this diagram w 64 Chapter 2 ■ Fluid Statics is the weight of the gate and Ox and Oy are the horizontal and vertical reactions of the shaft on the gate. We can now sum moments about the shaft 0.5 0.4 yR – yc, m a Mc 0 and, therefore, M FR 1 yR yc 2 11230 103 N2 10.0866 m2 1.07 105 N # m 0.3 0.2 (10m, 0.0886 m) 0.1 (Ans) 0 0 5 10 15 20 25 30 hc, m ■ Figure E2.6d E XAMPLE 2.7 Hydrostatic Pressure Force on a Plane Triangular Surface GIVEN An aquarium contains seawater 1g 64.0 lbft3 2 to a depth of 1 ft as shown in Fig. E2.7a. To repair some damage to one corner of the tank, a triangular section is replaced with a new section as illustrated in Fig. E2.7b. FIND Determine (a) the magnitude of the force of the seawater on this triangular area, and (b) the location of this force. SOLUTION (a) The various distances needed to solve this problem are shown in Fig. E2.7c. Since the surface of interest lies in a vertical plane, yc hc 0.9 ft, and from Eq. 2.18 the magnitude of the force is FR ghc A 164.0 lb ft3 2 10.9 ft2 3 10.3 ft2 2 2 4 2.59 lb (Ans) COMMENT Note that this force is independent of the tank length. The result is the same if the tank is 0.25 ft, 25 ft, or 25 miles long. 0.3 ft 1 ft 0.3 ft 0.9 ft 2.5 ft (b) The y coordinate of the center of pressure 1CP2 is found from Eq. 2.19, yR (b) Ixc yc yc A x y and from Fig. 2.18 yc 1 ft c δA c CP 0.1 ft (c) ■ Figure E2.7b–d ■ Figure E2.7a (Photograph courtesy of Tenecor Median line 0.2 ft 0.1 ft Tanks, Inc.) yR xR CP 0.15 ft 0.15 ft (d ) 2.9 Ixc 10.3 ft210.3 ft2 3 0.0081 4 ft 36 36 xR 0.0081 36 ft4 0.9 ft 10.9 ft210.092 ft2 2 0.00556 ft 0.9 ft 0.906 ft yR (Ans) Similarly, from Eq. 2.20 Ixyc yc A xc and from Fig. 2.18 Ixyc 2.9 65 so that so that xR Pressure Prism 10.3 ft210.3 ft2 2 0.0081 4 10.3 ft2 ft 72 72 0.008172 ft4 0 0.00278 ft 10.9 ft210.092 ft2 2 (Ans) COMMENT Thus, we conclude that the center of pressure is 0.00278 ft to the right of and 0.00556 ft below the centroid of the area. If this point is plotted, we find that it lies on the median line for the area as illustrated in Fig. E2.7d. Since we can think of the total area as consisting of a number of small rectangular strips of area dA 1and the fluid force on each of these small areas acts through its center2, it follows that the resultant of all these parallel forces must lie along the median. Pressure Prism An informative and useful graphical interpretation can be made for the force developed by a fluid acting on a plane rectangular area. Consider the pressure distribution along a vertical wall of a tank of constant width b, which contains a liquid having a specific weight g. Since the pressure must vary linearly with depth, we can represent the variation as is shown in Fig. 2.19a, where the pressure is equal to zero at the upper surface and equal to gh at the bottom. It is apparent from this diagram that the average pressure occurs at the depth h2 and, therefore, the resultant force acting on the rectangular area A bh is h FR pav A g a b A 2 which is the same result as obtained from Eq. 2.18. The pressure distribution shown in Fig. 2.19a applies across the vertical surface, so we can draw the three-dimensional representation of the pressure distribution as shown in Fig. 2.19b. The base of this “volume” in pressure-area space is the plane surface of interest, and its altitude at each point is the pressure. This volume is called the pressure prism, and it is clear that the magnitude of the resultant force acting on the rectangular surface is equal to the volume of the pressure prism. Thus, for the prism of Fig. 2.19b the fluid force is The magnitude of the resultant fluid force is equal to the volume of the pressure prism and passes through its centroid. FR volume 1 h 1gh21bh2 g a b A 2 2 where bh is the area of the rectangular surface, A. The resultant force must pass through the centroid of the pressure prism. For the volume under consideration the centroid is located along the vertical axis of symmetry of the surface and at a distance of h3 above the base 1since the centroid of a triangle is located at h 3 above its base2. This result can readily be shown to be consistent with that obtained from Eqs. 2.19 and 2.20. h p h CP FR h– 3 h– 3 FR b γh ( a) γh (b) ■ Figure 2.19 Pressure prism for vertical rectangular area. 66 Chapter 2 ■ Fluid Statics h1 γ h1 B A y1 h2 F1 FR p yA y2 F2 C D E γ (h2 - h1) (a) (b) ■ Figure 2.20 Graphical representation of hydrostatic forces on a vertical rectangular surface. This same graphical approach can be used for plane rectangular surfaces that do not extend up to the fluid surface, as illustrated in Fig. 2.20a. In this instance, the cross section of the pressure prism is trapezoidal. However, the resultant force is still equal in magnitude to the volume of the pressure prism, and it passes through the centroid of the volume. Specific values can be obtained by decomposing the pressure prism into two parts, ABDE and BCD, as shown in Fig. 2.20b. Thus, FR F1 F2 where the components can readily be determined by inspection for rectangular surfaces. The location of FR can be determined by summing moments about some convenient axis, such as one passing through A. In this instance FRyA F1y1 F2 y2 The use of the pressure prism concept to determine the force on a submerged area is best suited for plane rectangular surfaces. and y1 and y2 can be determined by inspection. For inclined plane rectangular surfaces the pressure prism can still be developed, and the cross section of the prism will generally be trapezoidal, as is shown in Fig. 2.21. Although it is usually convenient to measure distances along the inclined surface, the pressures developed depend on the vertical distances as illustrated. The use of pressure prisms for determining the force on submerged plane areas is convenient if the area is rectangular so the volume and centroid can be easily determined. However, for other nonrectangular shapes, integration would generally be needed to determine the volume and centroid. In these circumstances it is more convenient to use the equations developed in the previous section, in which the necessary integrations have been made and the results presented in a convenient and compact form that is applicable to submerged plane areas of any shape. The effect of atmospheric pressure on a submerged area has not yet been considered, and we may ask how this pressure will influence the resultant force. If we again consider the pressure distribution on a plane vertical wall, as is shown in Fig. 2.22a, the pressure varies from zero at the surface to gh at the bottom. Since we are setting the surface pressure equal to zero, we are using γ h1 h1 h2 γ h2 ■ Figure 2.21 Pressure variation along an inclined plane area. 2.9 patm patm h p FR Pressure Prism 67 patm patm A patm A FR γh (a) (b) ■ Figure 2.22 Effect of atmospheric pressure on the resultant force acting on a plane vertical wall. The resultant fluid force acting on a submerged area is affected by the pressure at the free surface. E XAMPLE atmospheric pressure as our datum, and thus the pressure used in the determination of the fluid force is gage pressure. If we wish to include atmospheric pressure, the pressure distribution will be as is shown in Fig. 2.22b. We note that in this case the force on one side of the wall now consists of FR as a result of the hydrostatic pressure distribution, plus the contribution of the atmospheric pressure, patm A, where A is the area of the surface. However, if we are going to include the effect of atmospheric pressure on one side of the wall, we must realize that this same pressure acts on the outside surface 1assuming it is exposed to the atmosphere2, so that an equal and opposite force will be developed as illustrated in the figure. Thus, we conclude that the resultant fluid force on the surface is that due only to the gage pressure contribution of the liquid in contact with the surface— the atmospheric pressure does not contribute to this resultant. Of course, if the surface pressure of the liquid is different from atmospheric pressure 1such as might occur in a closed tank2, the resultant force acting on a submerged area, A, will be changed in magnitude from that caused simply by hydrostatic pressure by an amount ps A, where ps is the gage pressure at the liquid surface 1the outside surface is assumed to be exposed to atmospheric pressure2. 2.8 Use of the Pressure Prism Concept GIVEN A pressurized tank contains oil 1SG 0.902 and has a square, 0.6-m by 0.6-m plate bolted to its side, as is illustrated in Fig. E2.8a. The pressure gage on the top of the tank reads 50 kPa, and the outside of the tank is at atmospheric pressure. FIND What is the magnitude and location of the resultant force on the attached plate? p = 50 kPa γ h1 ps Oil surface Air h1 = 2 m Oil 2m 0.6 m F1 F2 0.6 m FR 0.2 m yO O γ (h2 – h1) (a) ■ Figure E2.8 h2 = 2.6 m (b) Plate 0.3 m 68 Chapter 2 ■ Fluid Statics SOLUTION The pressure distribution acting on the inside surface of the plate is shown in Fig. E2.8b. The pressure at a given point on the plate is due to the air pressure, ps, at the oil surface and the pressure due to the oil, which varies linearly with depth as is shown in the figure. The resultant force on the plate 1having an area A2 is due to the components, F1 and F2, where F1 and F2 are due to the rectangular and triangular portions of the pressure distribution, respectively. Thus, F1 1 ps gh1 2 A 3 50 103 Nm2 10.902 19.81 103 Nm3 212 m2 4 10.36 m2 2 24.4 103 N and 0.954 103 N 2.10 FR F1 F2 25.4 103 N 25.4 kN (Ans) The vertical location of FR can be obtained by summing moments around an axis through point O so that FR yO F1 10.3 m2 F2 10.2 m2 or yO 124.4 103 N210.3 m2 10.954 103 N210.2 m2 25.4 103 N 0.296 m (Ans) Thus, the force acts at a distance of 0.296 m above the bottom of the plate along the vertical axis of symmetry. h2 h1 F2 g a bA 2 10.902 19.81 103 Nm3 2 a The magnitude of the resultant force, FR, is therefore 0.6 m b 10.36 m2 2 2 COMMENT Note that the air pressure used in the calculation of the force was gage pressure. Atmospheric pressure does not affect the resultant force 1magnitude or location2, since it acts on both sides of the plate, thereby canceling its effect. Hydrostatic Force on a Curved Surface V2.6 Pop bottle The equations developed in Section 2.8 for the magnitude and location of the resultant force acting on a submerged surface only apply to plane surfaces. However, many surfaces of interest 1such as those associated with dams, pipes, and tanks2 are nonplanar. The domed bottom of the beverage bottle shown in the figure in the margin shows a typical curved surface example. Although the resultant fluid force can be determined by integration, as was done for the plane surfaces, this is generally a rather tedious process and no simple, general formulas can be developed. As an alternative approach, we will consider the equilibrium of the fluid volume enclosed by the curved surface of interest and the horizontal and vertical projections of this surface. For example, consider a curved portion of the swimming pool shown in Fig. 2.23a. We wish to find the resultant fluid force acting on section BC (which has a unit length perpendicular to the plane of the paper) shown in Fig. 2.23b. We first isolate a volume of fluid that is bounded by the surface of interest, in this instance section BC, the horizontal plane surface AB, and the vertical plane surface AC. The free-body diagram for this volume is shown in Fig. 2.23c. The magnitude and location of forces F1 and F2 can be determined from the relationships for planar surfaces. The weight, w, is simply the specific weight of the fluid times the enclosed volume and acts through the center of gravity 1CG2 of the mass of fluid contained within the volume. The forces FH and FV represent the components of the force that the tank exerts on the fluid. In order for this force system to be in equilibrium, the horizontal component FH must be equal in magnitude and collinear with F2, and the vertical component FV equal in magnitude and collinear with the resultant of the vertical forces F1 and w. This follows since the three forces acting on the fluid mass 1F2, the resultant of F1 and w, and the resultant force that the tank exerts on the mass2 must form a concurrent force system. That is, from the principles of statics, it is known that when a body is held in equilibrium by three nonparallel forces, they must be concurrent 1their lines of action intersect at a common point2 and coplanar. Thus, FH F2 FV F1 w and the magnitude of the resultant is obtained from the equation FR 21FH 2 2 1FV 2 2 2.10 69 Hydrostatic Force on a Curved Surface F1 ᐃ A B A FR = √(FH)2 + (FV)2 B B CG F2 FH O O C C C FV (b) (a) (c) (d) ■ Figure 2.23 Hydrostatic force on a curved surface. (Photograph courtesy of Intex Marketing, Ltd.) The resultant FR passes through the point O, which can be located by summing moments about an appropriate axis. The resultant force of the fluid acting on the curved surface BC is equal and opposite in direction to that obtained from the free-body diagram of Fig. 2.23c. The desired fluid force is shown in Fig. 2.23d. E XAMPLE 2.9 Hydrostatic Pressure Force on a Curved Surface GIVEN A 6-ft-diameter drainage conduit of the type shown in Fig. E2.9a is half full of water at rest, as shown in Fig. E2.9b. FIND Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall. 1.27 ft 3 ft A A B A FR = 523 lb B CG 32.5° ᐃ F1 FH 1 ft O 1 ft C C (a) FV (b) (c) (d) ■ Figure E2.9 (Photograph courtesy of CONTECH Construction Products, Inc.) SOLUTION We first isolate a volume of fluid bounded by the curved section BC, the horizontal surface AB, and the vertical surface AC, as shown in Fig. E2.9c. The volume has a length of 1 ft. The forces acting on the volume are the horizontal force, F1, which acts on the vertical surface AC, the weight, w, of the fluid contained within the volume, and the horizontal and vertical components of the force of the conduit wall on the fluid, FH and FV, respectively. The magnitude of F1 is found from the equation F1 ⫽ ghc A ⫽ 162.4 lbⲐft3 2 1 32 ft2 13 ft2 2 ⫽ 281 lb and this force acts 1 ft above C as shown. The weight w ⫽ gV ⫺, where ⫺ V is the fluid volume, is w ⫽ g⫺ V ⫽ 162.4 lbⲐft3 2 19pⲐ4 ft2 2 11 ft2 ⫽ 441 lb and acts through the center of gravity of the mass of fluid, which according to Fig. 2.18 is located 1.27 ft to the right of AC as shown. Therefore, to satisfy equilibrium FH ⫽ F1 ⫽ 281 lb FV ⫽ w ⫽ 441 lb and the magnitude of the resultant force is FR ⫽ 21FH 2 2 ⫹ 1FV 2 2 ⫽ 21281 lb2 2 ⫹ 1441 lb2 2 ⫽ 523 lb (Ans) The force the water exerts on the conduit wall is equal, but opposite in direction, to the forces FH and FV shown in Fig. E2.9c. Thus, the resultant force on the conduit wall is shown in Fig. E2.9d. This force acts through the point O at the angle shown. 70 Chapter 2 ■ Fluid Statics COMMENT An inspection of this result will show that the line of action of the resultant force passes through the center of the conduit. In retrospect, this is not a surprising result since at each point on the curved surface of the conduit the elemental force due to the pressure is normal to the surface, and each line of action must pass through the center of the conduit. It therefore follows that the resultant of this concurrent force system must also pass through the center of concurrence of the elemental forces that make up the system. This same general approach can also be used for determining the force on curved surfaces of pressurized, closed tanks. If these tanks contain a gas, the weight of the gas is usually negligible in comparison with the forces developed by the pressure. Thus, the forces 1such as F1 and F2 in Fig. 2.23c2 on horizontal and vertical projections of the curved surface of interest can simply be expressed as the internal pressure times the appropriate projected area. F l u i d s i n Miniature, exploding pressure vessels Our daily lives are safer because of the effort put forth by engineers to design safe, lightweight pressure vessels such as boilers, propane tanks, and pop bottles. Without proper design, the large hydrostatic pressure forces on the curved surfaces of such containers could cause the vessel to explode with disastrous consequences. On the other hand, the world is a friendlier place because of miniature pressure vessels that are designed to explode under the proper conditions— popcorn kernels. Each grain of popcorn contains a small amount 2.11 t h e N e w s of water within the special, impervious hull (pressure vessel) which, when heated to a proper temperature, turns to steam, causing the kernel to explode and turn itself inside out. Not all popcorn kernels have the proper properties to make them pop well. First, the kernel must be quite close to 13.5% water. With too little moisture, not enough steam will build up to pop the kernel; too much moisture causes the kernel to pop into a dense sphere rather than the light fluffy delicacy expected. Second, to allow the pressure to build up, the kernels must not be cracked or damaged. Buoyancy, Flotation, and Stability 2.11.1 Archimedes’ Principle (Photograph courtesy of Cameron Balloons.) When a stationary body is completely submerged in a fluid 1such as the hot air balloon shown in the figure in the margin2, or floating so that it is only partially submerged, the resultant fluid force acting on the body is called the buoyant force. A net upward vertical force results because pressure increases with depth and the pressure forces acting from below are larger than the pressure forces acting from above, as shown by the figure in the margin. This force can be determined through an approach similar to that used in the previous section for forces on curved surfaces. V , that is immersed in a fluid as illustrated Consider a body of arbitrary shape, having a volume in Fig. 2.24a. We enclose the body in a parallelepiped and draw a free-body diagram of the parallelepiped with the body removed as shown in Fig. 2.24b. Note that the forces F1, F2, F3, and F4 are simply the forces exerted on the plane surfaces of the parallelepiped 1for simplicity the forces in the x direction are not shown2, w is the weight of the shaded fluid volume 1parallelepiped minus body2, and FB is the force the body is exerting on the fluid. The forces on the vertical surfaces, such as F3 and F4, are all equal and cancel, so the equilibrium equation of interest is in the z direction and can be expressed as FB F2 F1 w (2.21) If the specific weight of the fluid is constant, then F2 F1 g1h2 h1 2A where A is the horizontal area of the upper 1or lower2 surface of the parallelepiped, and Eq. 2.21 can be written as FB g1h2 h1 2A g3 1h2 h1 2A V4 Simplifying, we arrive at the desired expression for the buoyant force FB gV (2.22) 2.11 Buoyancy, Flotation, and Stability 71 c h1 h2 A FB z B Centroid of displaced volume (d) y x D C Centroid (a) c F1 y1 y2 A V2.7 Cartesian Diver FB B ᐃ F3 yc F4 FB D (c) C F2 (b) ■ Figure 2.24 Buoyant force on submerged and floating bodies. V is the volume of the body. The effects of the spewhere g is the specific weight of the fluid and cific weight (or density) of the body as compared to that of the surrounding fluid are illustrated by the figure in the margin. The direction of the buoyant force, which is the force of the fluid on the body, is opposite to that shown on the free-body diagram. Therefore, the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward. This result is commonly referred to as Archimedes’ principle in honor of Archimedes 1287–212 B.C.2, a Greek mechanician and mathematician who first enunciated the basic ideas associated with hydrostatics. The location of the line of action of the buoyant force can be determined by summing moments of the forces shown on the free-body diagram in Fig. 2.24b with respect to some convenient axis. For example, summing moments about an axis perpendicular to the paper through point D we have γ1 < γ γ2 = γ γ γ3 > γ FByc F2 y1 F1y1 wy2 and on substitution for the various forces Archimedes’ principle states that the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward. F l V yc V Ty1 1V T V 2 y2 (2.23) V T is the total volume 1h2 h1 2A. The right-hand side of Eq. 2.23 is the first moment of where V with respect to the x–z plane so that yc is equal to the y coordinate of the the displaced volume centroid of the volume V . In a similar fashion, it can be shown that the x coordinate of the buoyant force coincides with the x coordinate of the centroid. Thus, we conclude that the buoyant force passes through the centroid of the displaced volume as shown in Fig. 2.24c. The point through which the buoyant force acts is called the center of buoyancy. u i d s i n Concrete canoes A solid block of concrete thrown into a pond or lake will obviously sink. But if the concrete is formed into the shape of a canoe it can be made to float. Of, course the reason the canoe floats is the development of the buoyant force due to the displaced volume of water. With the proper design, this vertical force can be made to balance the weight of the canoe plus passengers—the canoe floats. Each year since 1988 a National Concrete Canoe Competition for university teams is jointly spon- t h e N e w s sored by the American Society of Civil Engineers and Master Builders Inc. The canoes must be 90% concrete and are typically designed with the aid of a computer by civil engineering students. Final scoring depends on four components: a design report, an oral presentation, the final product, and racing. In 2011 California Polytechnic State University, San Luis Obispo, won the national championship with its 208-pound, 20-foot-long canoe. (See Problem 2.147.) 72 Chapter 2 ■ Fluid Statics These same results apply to floating bodies that are only partially submerged, as illustrated in Fig. 2.24d, if the specific weight of the fluid above the liquid surface is very small compared with the liquid in which the body floats. Since the fluid above the surface is usually air, for practical purposes this condition is satisfied. In the derivations presented above, the fluid is assumed to have a constant specific weight, g. If a body is immersed in a fluid in which g varies with depth, such as in a layered fluid, the magnitude of the buoyant force remains equal to the weight of the displaced fluid. However, the buoyant force does not pass through the centroid of the displaced volume, but rather, it passes through the center of gravity of the displaced volume. V2.8 Hydrometer E XAMPLE 2.10 Buoyant Force on a Submerged Object GIVEN A Type I offshore life jacket (personal flotation device) of the type worn by commercial fishermen is shown in Fig. E2.10a. It is designed for extended survival in rough, open water. According to U.S. Coast Guard regulations, the life jacket must provide a minimum 22-lb net upward force on the user. Consider such a life jacket that uses a foam material with a specific weight of 2.0 lb/ft3 for the main flotation material. The remaining material (cloth, straps, fasteners, etc.) weighs 1.3 lb and is of negligible volume. FIND Determine the minimum volume of foam needed for this life jacket. SOLUTION A free-body diagram of the life jacket is shown in Fig. E2.10b, where FB is the buoyant force acting on the life jacket, ᐃF is the weight of the foam, ᐃS 1.3 lb is the weight of the remaining material, and FU 22 lb is the required force on the user. For equilibrium it follows that FB ᐃF ᐃS FU (1) where from Eq. 2.22 FB gwater V Here gwater 64.0 lbft3 is the specific weight of seawater and V is the volume of the foam. Also ᐃfoam gfoam V , where gfoam 2.0 lbft3 is the specific weight of the foam. Thus, from Eq. 1 gwater V gfoam V ᐃS FU FB ᐃF ■ Figure E2.10b or V 1ᐃS FU 2 1gwater gfoam 2 11.3 lb 22 lb2 164.0 lbft3 2.0 lbft3 2 0.376 ft3 (Ans) COMMENTS In this example, rather than using difficult-to- ᐃS FU ■ Figure E2.10a calculate hydrostatic pressure force on the irregularly shaped life jacket, we have used the buoyant force. The net effect of the pressure forces on the surface of the life jacket is equal to the upward buoyant force. Do not include both the buoyant force and the hydrostatic pressure effects in your calculations—use one or the other. There is more to the proper design of a life jacket than just the volume needed for the required buoyancy. According to regulations, a Type I life jacket must also be designed so that it provides proper protection to the user by turning an unconscious person in the water to a face-up position as shown in Fig. E2.10a. This involves the concept of the stability of a floating object (see Section 2.11.2). The life jacket should also provide minimum interference under ordinary working conditions so as to encourage its use by commercial fishermen. 2.11 l 73 The effects of buoyancy are not limited to the interaction between a solid body and a fluid. Buoyancy effects can also be seen within fluids alone, as long as a density difference exists. Consider the shaded portion of Fig. 2.24c to be a volume of fluid instead of a solid. This volume of fluid is submerged in the surrounding fluid and therefore has a buoyant force due to the fluid it displaces (like the solid). If this volume contains fluid with a density of r1, then the downward force due to weight is w r1gV. In addition, if the surrounding fluid has the same density, then the buoyant . As expected, in this case force on the volume due to the displaced fluid will be FB gV r1gV the weight of the volume is exactly balanced by the buoyant force acting on the volume so there is no net force. However, if the density of fluid in the volume is r2, then w and FB will not balance and there will be a net force in the upward or downward direction depending on whether the density in the volume (r2) is less than or greater than, respectively, the density of the surrounding fluid. Note that this difference can develop from two different fluids with different densities or from temperature differences within the same fluid causing density variations in space. For example, smoke from a fire rises because it is lighter (due to its higher temperature) than the surrounding air. V2.9 Atmospheric buoyancy F Buoyancy, Flotation, and Stability u i d s i n Explosive lake In 1986 a tremendous explosion of carbon dioxide (CO2) from Lake Nyos, west of Cameroon, killed more than 1700 people and livestock. The explosion resulted from a buildup of CO2 that seeped into the high-pressure water at the bottom of the lake from warm springs of CO2-bearing water. The CO2-rich water is heavier than pure water and can hold a volume of CO2 more than five times the water volume. As long as the gas remains dissolved in the water, the stratified lake (i.e., pure water on top, CO2 water on the bottom) is stable. But if some mechanism causes the gas t h e N e w s bubbles to nucleate, they rise, grow, and cause other bubbles to form, feeding a chain reaction. A related phenomenon often occurs when a pop bottle is shaken and then opened. The pop shoots from the container rather violently. When this set of events occurred in Lake Nyos, the entire lake overturned through a column of rising and expanding buoyant bubbles. The heavier-than-air CO2 then flowed through the long, deep valleys surrounding the lake and asphyxiated human and animal life caught in the gas cloud. One victim was 27 km downstream from the lake. 2.11.2 Stability V2.10 Density differences in fluids Another interesting and important problem associated with submerged or floating bodies is concerned with the stability of the bodies. As illustrated by the figure in the margin, a body is said to be in a stable equilibrium position if, when displaced, it returns to its equilibrium position. Conversely, it is in an unstable equilibrium position if, when displaced 1even slightly2, it moves to a new equilibrium position. Stability considerations are particularly important for submerged or floating bodies since the centers of buoyancy and gravity do not necessarily coincide. A small rotation can result in either a restoring or overturning couple. For example, for the completely submerged body shown in Fig. 2.25, which has a center of gravity below the center of buoyancy, a rotation from its equilibrium position will create a restoring couple formed by the weight, w, and the buoyant force, FB, which causes the body to rotate back to its original position. Thus, for this configuration the body is stable. It is to be noted that as long as the center of gravity falls below the center of buoyancy, this will always be true; that is, the body is in a stable equilibrium position with respect to small rotations. However, as is illustrated in Fig. 2.26, if the center of gravity of the Stable Unstable c The stability of a body can be determined by considering what happens when it is displaced from its equilibrium position. FB CG FB c CG CG ᐃ FB ᐃ Stable ᐃ CG c Restoring couple ■ Figure 2.25 Stability of a completely immersed body—center of gravity below centroid. ᐃ c FB Unstable Overturning couple ■ Figure 2.26 Stability of a completely immersed body—center of gravity above centroid. 74 Chapter 2 ■ Fluid Statics ᐃ V2.11 Stability of a floating cube ᐃ CG CG c c' FB FB c = centroid of original c' = centroid of new displaced volume Restoring couple displaced volume Stable ■ Figure 2.27 Stability of a floating body—stable configuration. ᐃ ᐃ CG CG c FB Marginally stable c = centroid of original displaced volume c' FB c' = centroid of new displaced volume Overturning couple Unstable Very stable © RiverNorthPhotography/ iStockphoto V2.12 Stability of a model barge 2.12 ■ Figure 2.28 Stability of a floating body—unstable configuration. completely submerged body is above the center of buoyancy, the resulting couple formed by the weight and the buoyant force will cause the body to overturn and move to a new equilibrium position. Thus, a completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position. For floating bodies the stability problem is more complicated, since as the body rotates the location of the center of buoyancy 1which passes through the centroid of the displaced volume2 may change. As is shown in Fig. 2.27, a floating body such as a barge that rides low in the water can be stable even though the center of gravity lies above the center of buoyancy. This is true since as the body rotates the buoyant force, FB, shifts to pass through the centroid of the newly formed displaced volume and, as illustrated, combines with the weight, w, to form a couple that will cause the body to return to its original equilibrium position. However, for the relatively tall, slender body shown in Fig. 2.28, a small rotational displacement can cause the buoyant force and the weight to form an overturning couple as illustrated. It is clear from these simple examples that determining the stability of submerged or floating bodies can be difficult since the analysis depends in a complicated fashion on the particular geometry and weight distribution of the body. Thus, although both the relatively narrow kayak and the wide houseboat shown in the figures in the margin are stable, the kayak will overturn much more easily than the houseboat. The problem can be further complicated by the necessary inclusion of other types of external forces such as those induced by wind gusts or currents. Stability considerations are obviously of great importance in the design of ships, submarines, bathyscaphes, and so forth; such considerations play a significant role in the work of naval architects 1see, for example, Ref. 62. Pressure Variation in a Fluid with Rigid-Body Motion Although in this chapter we have been primarily concerned with fluids at rest, the general equation of motion 1Eq. 2.22 ⫺§p ⫺ gkˆ ⫽ ra Even though a fluid may be in motion, if it moves as a rigid body there will be no shearing stresses present. was developed for both fluids at rest and fluids in motion, with the only stipulation being that there were no shearing stresses present. Equation 2.2 in component form, based on rectangular coordinates with the positive z axis being vertically upward, can be expressed as ⫺ 0p ⫽ rax 0x ⫺ 0p ⫽ ray 0y ⫺ 0p ⫽ g ⫹ raz 0z (2.24) A general class of problems involving fluid motion in which there are no shearing stresses occurs when a mass of fluid undergoes rigid-body motion. For example, if a container of fluid 2.12 75 Pressure Variation in a Fluid with Rigid-Body Motion accelerates along a straight path, the fluid will move as a rigid mass 1after the initial sloshing motion has died out2 with each particle having the same acceleration. Since there is no deformation, there will be no shearing stresses and, therefore, Eq. 2.2 applies. Similarly, if a fluid is contained in a tank that rotates about a fixed axis, the fluid will simply rotate with the tank as a rigid body, and again Eq. 2.2 can be applied to obtain the pressure distribution throughout the moving fluid. Specific results for these two cases 1rigid-body uniform motion and rigid-body rotation2 are developed in the following two sections. Although problems relating to fluids having rigid-body motion are not, strictly speaking, “fluid statics” problems, they are included in this chapter because, as we will see, the analysis and resulting pressure relationships are similar to those for fluids at rest. There is no shear stress in fluids that move with rigidbody motion or with rigid-body rotation. 2.12.1 Linear Motion We first consider an open container of a liquid that is translating along a straight path with a constant acceleration a as illustrated in Fig. 2.29. Since ax 0, it follows from the first of Eqs. 2.24 that the pressure gradient in the x direction is zero 10p 0x 02. In the y and z directions 0p ray 0y (2.25) 0p r1g az 2 0z (2.26) The change in pressure between two closely spaced points located at y, z, and y dy, z dz can be expressed as z dp dy or in terms of the results from Eqs. 2.25 and 2.26 dz ay 0p 0p dy dz 0y 0z dp ray dy r1g az 2 dz g + az y (2.27) Along a line of constant pressure, dp 0, and therefore from Eq. 2.27 it follows that the slope of this line is given by the relationship ay dz (2.28) dy g az This relationship is illustrated by the figure in the margin. Along a free surface the pressure is constant, so that for the accelerating mass shown in Fig. 2.29 the free surface will be inclined if ay 0. In addition, all lines of constant pressure will be parallel to the free surface as illustrated. Free surface slope = dz/dy g + az ay p1 Constant p2 pressure p3 lines z x (a) az y (b) ■ Figure 2.29 Linear acceleration of a liquid with a free surface. a ay (c) 76 Chapter 2 ■ Fluid Statics The pressure distribution in a fluid mass that is accelerating along a straight path is not hydrostatic. For the special circumstance in which ay 0, az 0, which corresponds to the mass of fluid accelerating in the vertical direction, Eq. 2.28 indicates that the fluid surface will be horizontal. However, from Eq. 2.26 we see that the pressure distribution is not hydrostatic, but is given by the equation dp r 1g az 2 dz For fluids of constant density this equation shows that the pressure will vary linearly with depth, but the variation is due to the combined effects of gravity and the externally induced acceleration, r1g az 2, rather than simply the specific weight rg. Thus, for example, the pressure along the bottom of a liquid-filled tank which is resting on the floor of an elevator that is accelerating upward will be increased over that which exists when the tank is at rest 1or moving with a constant velocity2. It is to be noted that for a freely falling fluid mass 1az g2, the pressure gradients in all three coordinate directions are zero, which means that if the pressure surrounding the mass is zero, the pressure throughout will be zero. The pressure throughout a “blob” of orange juice floating in an orbiting space shuttle 1a form of free fall2 is zero. The only force holding the liquid together is surface tension 1see Section 1.92. E XAMPLE 2.11 Pressure Variation in an Accelerating Tank ay GIVEN The cross section for the fuel tank of an experimental vehicle is shown in Fig. E2.11. The rectangular tank is vented to the atmosphere, and the specific gravity of the fuel is SG 0.65. A pressure transducer is located in its side as illustrated. During testing of the vehicle, the tank is subjected to a constant linear acceleration, ay. FIND (a) Determine an expression that relates ay and the pressure 1in lb ft2 2 at the transducer. (b) What is the maximum acceleration that can occur before the fuel level drops below the transducer? Vent z Air y z1 Fuel (2) (1) 0.75 ft 0.5 ft Transducer 0.75 ft ■ Figure E2.11 SOLUTION (a) For a constant horizontal acceleration the fuel will move as a rigid body, and from Eq. 2.28 the slope of the fuel surface can be expressed as ay dz g dy since az 0. Thus, for some arbitrary ay, the change in depth, z1, of liquid on the right side of the tank can be found from the equation where h is the depth of fuel above the transducer, and therefore p 10.652162.4 lb ft3 2 30.5 ft 10.75 ft21ayg2 4 ay (Ans) 20.3 30.4 g for z1 0.5 ft. As written, p would be given in lbft2. (b) The limiting value for 1ay 2 max 1when the fuel level reaches the transducer2 can be found from the equation ay z1 g 0.75 ft 0.5 ft 10.75 ft2 c 1ay 2 max g d or or ay z1 10.75 ft2 a b g Since there is no acceleration in the vertical, z, direction, the pressure along the wall varies hydrostatically as shown by Eq. 2.26. Thus, the pressure at the transducer is given by the relationship p gh 1a y 2 max 2g 3 and for standard acceleration of gravity 1ay 2 max 23 132.2 ft s2 2 21.5 fts2 (Ans) COMMENT Note that the pressure in horizontal layers is not constant in this example since 0p 0y ray 0. Thus, for example, p1 p2. 2.12 Pressure Variation in a Fluid with Rigid-Body Motion 77 2.12.2 Rigid-Body Rotation After an initial “start-up” transient, a fluid contained in a tank that rotates with a constant angular velocity v about an axis as is shown in Fig. 2.30 will rotate with the tank as a rigid body. It is known from elementary particle dynamics that the acceleration of a fluid particle located at a distance r from the axis of rotation is equal in magnitude to rv2, and the direction of the acceleration is toward the axis of rotation, as is illustrated in the figure. Since the paths of the fluid particles are circular, it is convenient to use cylindrical polar coordinates r, u, and z, defined in the insert in Fig. 2.30. It will be shown in Chapter 6 that in terms of cylindrical coordinates the pressure gradient § p can be expressed as A fluid contained in a tank that is rotating with a constant angular velocity about an axis will rotate as a rigid body. §p 0p 0p 1 0p ê ê ê r 0u u 0r r 0z z (2.29) Thus, in terms of this coordinate system ar rv2 êr au 0 az 0 and from Eq. 2.2 0p rrv2 0r 0p 0 0u 0p g 0z (2.30) These results show that for this type of rigid-body rotation, the pressure is a function of two variables r and z, and therefore the differential pressure is dp 0p 0p dr dz 0r 0z or dp rrv2 dr g dz p (2.31) On a horizontal plane (dz 0), it follows from Eq. 2.31 that dpdr ␳␻2r, which is greater than zero. Hence, as illustrated in the figure in the margin, because of centrifugal acceleration, the pressure increases in the radial direction. Along a surface of constant pressure, such as the free surface, dp 0, so that from Eq. 2.31 1using g rg2 z = constant dp dr dp ––– = rw2r dr dz rv2 g dr r Integration of this result gives the equation for surfaces of constant pressure as z v2r 2 constant 2g Axis of rotation (2.32) z ω r ar = rω 2 r θ ω y ez eθ x er (a) (b) ■ Figure 2.30 Rigid-body rotation of a liquid in a tank. (Photograph courtesy of Geno Pawlak.) (c) 78 Chapter 2 ■ Fluid Statics z p1 p1 Constant pressure lines p2 p2 p3 p3 p4 p4 2 2 ω r ____ 2g r y ■ Figure 2.31 Pressure distribution in a rotating liquid. x The free surface in a rotating liquid is curved rather than flat. This equation reveals that these surfaces of constant pressure are parabolic, as illustrated in Fig. 2.31. Integration of Eq. 2.31 yields 冮 dp rv 冮 r dr g 冮 dz 2 or p rv2r 2 gz constant 2 (2.33) where the constant of integration can be expressed in terms of a specified pressure at some arbitrary point r0, z0. This result shows that the pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies hydrostatically in the vertical direction as shown in Fig. 2.31. E XAMPLE 2.12 Free Surface Shape of Liquid in a Rotating Tank r GIVEN It has been suggested that the angular velocity, v, of a rotating body or shaft can be measured by attaching an open cylinder of liquid, as shown in Fig. E2.12a, and measuring with some type of depth gage the change in the fluid level, H h0, caused by the rotation of the fluid. R Depth gage Initial depth r FIND Determine the relationship between this change in fluid level and the angular velocity. h h0 z SOLUTION H dr h 0 The height, h, of the free surface above the tank bottom can be determined from Eq. 2.32, and it follows that h v2r 2 h0 2g ω (a) (b) ■ Figure E2.12 The initial volume of fluid in the tank, V i, is equal to V i pR2H The volume of the fluid with the rotating tank can be found with the aid of the differential element shown in Fig. E2.12b. This cylindrical shell is taken at some arbitrary radius, r, and its volume is dV 2prh dr 2.13 冮 0 R ra v2r 2 pv2R 4 h0 b dr pR2h0 2g 4g H h0 Since the volume of the fluid in the tank must remain constant 1assuming that none spills over the top2, it follows that pR 2H F l u i 79 or The total volume is, therefore, V 2p Chapter Summary and Study Guide pv2R 4 pR2h0 4g d s i n Rotating mercury mirror telescope A telescope mirror has the same shape as the parabolic free surface of a liquid in a rotating tank. The liquid mirror telescope (LMT) consists of a pan of liquid (normally mercury because of its excellent reflectivity) rotating to produce the required parabolic shape of the free surface mirror. With recent technological advances, it is possible to obtain the vibrationfree rotation and the constant angular velocity necessary to produce a liquid mirror surface precise enough for astronomical use. Construction of the largest LMT, located at the University of British v2R2 4g (Ans) COMMENT This is the relationship we were looking for. It shows that the change in depth could indeed be used to determine the rotational speed, although the relationship between the change in depth and speed is not a linear one. t h e N e w s Columbia, has recently been completed. With a diameter of 6 ft and a rotation rate of 7 rpm, this mirror uses 30 liters of mercury for its 1-mm thick, parabolic-shaped mirror. One of the major benefits of a LMT (compared to a normal glass mirror telescope) is its low cost. Perhaps the main disadvantage is that a LMT can look only straight up, although there are many galaxies, supernova explosions, and pieces of space junk to view in any part of the sky. The nextgeneration LMTs may have movable secondary mirrors to allow a larger portion of the sky to be viewed. (See Problem 2.163.) 2.13 Chapter Summary and Study Guide Pascal’s law surface force body force incompressible fluid hydrostatic pressure distribution pressure head compressible fluid U.S. standard atmosphere absolute pressure gage pressure vacuum pressure barometer manometer Bourdon pressure gage center of pressure buoyant force Archimedes’ principle center of buoyancy In this chapter the pressure variation in a fluid at rest is considered, along with some important consequences of this type of pressure variation. It is shown that for incompressible fluids at rest the pressure varies linearly with depth. This type of variation is commonly referred to as hydrostatic pressure distribution. For compressible fluids at rest the pressure distribution will not generally be hydrostatic, but Eq. 2.4 remains valid and can be used to determine the pressure distribution if additional information about the variation of the specific weight is specified. The distinction between absolute and gage pressure is discussed along with a consideration of barometers for the measurement of atmospheric pressure. Pressure-measuring devices called manometers, which utilize static liquid columns, are analyzed in detail. A brief discussion of mechanical and electronic pressure gages is also included. Equations for determining the magnitude and location of the resultant fluid force acting on a plane surface in contact with a static fluid are developed. A general approach for determining the magnitude and location of the resultant fluid force acting on a curved surface in contact with a static fluid is described. For submerged or floating bodies the concept of the buoyant force and the use of Archimedes’ principle are reviewed. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. calculate the pressure at various locations within an incompressible fluid at rest. calculate the pressure at various locations within a compressible fluid at rest using Eq. 2.4 if the variation in the specific weight is specified. use the concept of a hydrostatic pressure distribution to determine pressures from measurements using various types of manometers. determine the magnitude, direction, and location of the resultant hydrostatic force acting on a plane surface. 80 Chapter 2 ■ Fluid Statics determine the magnitude, direction, and location of the resultant hydrostatic force acting on a curved surface. use Archimedes’ principle to calculate the resultant hydrostatic force acting on floating or submerged bodies. analyze, based on Eq. 2.2, the motion of fluids moving with simple rigid-body linear motion or simple rigid-body rotation. Some of the important equations in this chapter are: Pressure variation in a stationary incompressible fluid dp g dz p1 gh p2 Hydrostatic force on a plane surface FR ghc A Pressure gradient in a stationary fluid Location of hydrostatic force on a plane surface Buoyant force Pressure gradient in rigid-body motion 0p rax, 0x Pressure gradient in rigid-body rotation yR 0p 0, 0u (2.7) (2.18) Ixc yc yc A Ixyc xR xc ycA FB g V 0p 0p ray, g raz 0y 0z 0p rrv2, 0r (2.4) 0p g 0z (2.19) (2.20) (2.22) (2.24) (2.30) References 1. The U.S. Standard Atmosphere, 1962, U.S. Government Printing Office, Washington, DC, 1962. 2. The U.S. Standard Atmosphere, 1976, U.S. Government Printing Office, Washington, DC, 1976. 3. Benedict, R. P., Fundamentals of Temperature, Pressure, and Flow Measurements, 3rd Ed., Wiley, New York, 1984. 4. Dally, J. W., Riley, W. F., and McConnell, K. G., Instrumentation for Engineering Measurements, 2nd Ed., Wiley, New York, 1993. 5. Holman, J. P., Experimental Methods for Engineers, 4th Ed., McGraw-Hill, New York, 1983. 6. Comstock, J. P., ed., Principles of Naval Architecture, Society of Naval Architects and Marine Engineers, New York, 1967. 7. Hasler, A. F., Pierce, H., Morris, K. R., and Dodge, J., “Meteorological Data Fields ‘In Perspective’,” Bulletin of the American Meteorological Society, Vol. 66, No. 7, July 1985. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). 81 Problems Conceptual Questions 2.1C Two tubes connected to two water reservoirs are as shown below. The tube on the left is straight, and that on the right is a cone with the top area four times that of the base. The bottom area of the cone equals that of the straight tube. The height of the water is the same for both cases. Atmosphere 4A Water H Water p1 p2 A A The relation between the pressures p1 and p2 at the base of the tubes is: 1 a) p2 4 p1. b) p2 2 p1. c) p2 p1. 2 1 d) p2 p1. e) p2 p1. 3 2.3C For a fluid element at rest, the forces acting on the element are: a) gravity, shear, and normal forces. b) gravity and normal forces. c) gravity and shear forces. d) normal and shear forces. 2.4C A tank is filled with a liquid, and the surface is exposed to the atmosphere. Which of the following accurately represents the absolute pressure distribution on the right-hand side of the tank? (a) (b) Force Final position wire F2 A1 (d) 2.5C A wire is attached to a block of metal that is submerged in a tank of water as shown below. The graph that most correctly describes the relation between the force in the wire and time as the block is pulled slowly out of the water is 2.2C A system filled with a liquid is shown below. On the left there is a piston in a tube of cross-sectional area A1 with a force F1 applied to it, and on the right there is a piston in a tube of crosssectional area A2 that is twice that of A1 and a force F2. The pistons are weightless. The two liquid levels are the same. F1 (c) Water Block initial position A2 = 2 A1 Force The relation between the force F2 and F1 is a) F2 F1 b) F2 4 F1 c) F2 2 F1 d) F2 F1/2 e) F2 F1/4 Time Time Time Time Time (a) (b) (c) (d) (e) Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the evennumbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 2.3 Pressure Variation in a Fluid at Rest 2.1 Obtain a photograph/image of a situation in which the fact that in a static fluid the pressure increases with depth is important. Print this photo and write a brief paragraph that describes the situation involved. 2.2 A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicated by a pressure gage at the top of the tank, is at a pressure of 20 kPa. Determine the pressure that the water exerts on the bottom of the tank. 2.3 GO A closed tank is partially filled with glycerin. If the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft, what is the pressure in lb/ft2 at the bottom of the tank? 2.4 Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure). As shown in Video V2.3, such pressures are commonly measured with a mercury manometer. A typical value for this ratio for 82 Chapter 2 ■ Fluid Statics a human would be 120/70, where the pressures are in mm Hg. (a) What would these pressures be in pascals? (b) If your car tire was inflated to 120 mm Hg, would it be sufficient for normal driving? very tall buildings so that the hydrostatic pressure difference is within acceptable limits. 2.5 An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown liquid is 1.5 m and the depth of the oil (specific weight 8.5 kN/m3) floating on top is 5.0 m. A pressure gage connected to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid? decreases with increasing elevation. In some situations, however, a temperature inversion may exist so that the air temperature increases with elevation. A series of temperature probes on a mountain give the elevation–temperature data shown in the table below. If the barometric pressure at the base of the mountain is 12.1 psia, determine by means of numerical integration the pressure at the top of the mountain. 2.6 The water level in an open standpipe is 80 ft above the ground. What is the static pressure at a fire hydrant that is connected to the standpipe and located at ground level? Express your answer in psi. 2.7 How high a column of SAE 30 oil would be required to give the same pressure as 700 mm Hg? 2.8 What pressure, expressed in pascals, will a skin diver be subjected to at a depth of 40 m in seawater? 2.9 Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure at a depth of 5 km, assuming that seawater has a constant specific weight of 10.1 kN/m3? Express your answer in pascals and psi. 2.10 For the great depths that may be encountered in the ocean the compressibility of seawater may become an important consideration. (a) Assume that the bulk modulus for seawater is constant and derive a relationship between pressure and depth which takes into account the change in fluid density with depth. (b) Make use of part (a) to determine the pressure at a depth of 6 km assuming seawater has a bulk modulus of 2.3 109 Pa and a density of 1030 kg/m3 at the surface. Compare this result with that obtained by assuming a constant density of 1030 kg/m3. 2.11 Sometimes when riding an elevator or driving up or down a hilly road a person’s ears “pop” as the pressure difference between the inside and outside of the ear is equalized. Determine the pressure difference (in psi) associated with this phenomenon if it occurs during a 150-ft elevation change. 2.15 Under normal conditions the temperature of the atmosphere Temperature (ⴗF ) Elevation (ft) 50.1 1base2 55.2 60.3 62.6 67.0 68.4 70.0 69.5 68.0 67.1 1top2 5000 5500 6000 6400 7100 7400 8200 8600 9200 9900 †2.16 Although it is difficult to compress water, the density of water at the bottom of the ocean is greater than that at the surface because of the higher pressure at depth. Estimate how much higher the ocean’s surface would be if the density of seawater were instantly changed to a uniform density equal to that at the surface. 2.17 (See Fluids in the News article titled “Giraffe’s blood pressure,” Section 2.3.1.) (a) Determine the change in hydrostatic pressure in a giraffe’s head as it lowers its head from eating leaves 6 m above the ground to getting a drink of water at ground level as shown in Fig. P2.17. Assume the specific gravity of blood is SG 1. (b) Compare the pressure change calculated in part (a) to the normal 120 mm of mercury pressure in a human’s heart. 2.12 Develop an expression for the pressure variation in a liquid in which the specific weight increases with depth, h, as g Kh g0, where K is a constant and g0 is the specific weight at the free surface. 2.13 In a certain liquid at rest, measurements of the specific weight at various depths show the following variation: h (ft) G (lbft3) 0 10 20 30 40 50 60 70 80 90 100 70 76 84 91 97 102 107 110 112 114 115 The depth h 0 corresponds to a free surface at atmospheric pressure. Determine, through numerical integration of Eq. 2.4, the corresponding variation in pressure and show the results on a plot of pressure (in psf) versus depth (in feet). †2.14 Because of elevation differences, the water pressure in the second floor of your house is lower than it is in the first floor. For tall buildings this pressure difference can become unacceptable. Discuss possible ways to design the water distribution system in 6m Water ■ Figure P2.17 Section 2.4 Standard Atmosphere 2.18 What would be the barometric pressure reading, in mm Hg, at an elevation of 4 km in the U.S. standard atmosphere? (Refer to Table C.2 in Appendix C.) 2.19 An absolute pressure of 7 psia corresponds to what gage pressure for standard atmospheric pressure of 14.7 psia? 2.20 Assume that a person skiing high in the mountains at an altitude of 15,000 ft takes in the same volume of air with each Problems breath as she does while walking at sea level. Determine the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level. 2.21 Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. (a) Determine the pressure at this elevation, based on Eq. 2.12. (b) If the air is assumed to have a constant specific weight of 0.07647 lb/ft3, what would the pressure be at this altitude? (c) If the air is assumed to have a constant temperature of 59 F, what would the pressure be at this elevation? For all three cases assume standard atmospheric conditions at sea level (see Table 2.1). 2.22 Equation 2.12 provides the relationship between pressure and elevation in the atmosphere for those regions in which the temperature varies linearly with elevation. Derive this equation and verify the value of the pressure given in Table C.2 in Appendix C for an elevation of 5 km. 2.23 As shown in Fig. 2.6 for the U.S. standard atmosphere, the troposphere extends to an altitude of 11 km where the pressure is 22.6 kPa (abs). In the next layer, called the stratosphere, the temperature remains constant at 56.5 C. Determine the pressure and density in this layer at an altitude of 15 km. Assume g 9.77 m/s2 in your calculations. Compare your results with those given in Table C.2 in Appendix C. 2.24 (See Fluids in the News article titled “Weather, barometers, and bars,” Section 2.5.) The record low sea-level barometric pressure ever recorded is 25.8 in. of mercury. At what altitude in the standard atmosphere is the pressure equal to this value? Section 2.5 Measurement of Pressure 2.25 On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument? 2.26 Aneroid barometers can be used to measure changes in altitude. If a barometer reads 30.1 in. Hg at one elevation, what has been the change in altitude in meters when the barometer reading is 28.3 in. Hg? Assume a standard atmosphere and that Eq. 2.12 is applicable over the range of altitudes of interest. 2.27 Bourdon gages (see Video V2.4 and Fig. 2.13) are commonly used to measure pressure. When such a gage is attached to the closed water tank of Fig. P2.27 the gage reads 5 psi. What is the absolute air pressure in the tank? Assume standard atmospheric pressure of 14.7 psi. 2.28 On the suction side of a pump, a Bourdon pressure gage reads 40 kPa vacuum. What is the corresponding absolute pressure if the local atmospheric pressure is 100 kPa (abs)? 2.29 A Bourdon pressure gage attached to the outside of a tank containing air reads 77.0 psi when the local atmospheric pressure is 760 mm Hg. What will be the gage reading if the atmospheric pressure increases to 773 mm Hg? Section 2.6 Manometry 2.30 Obtain a photograph/image of a situation in which the use of a manometer is important. Print this photo and write a brief paragraph that describes the situation involved. 2.31 A water-filled U-tube manometer is used to measure the pressure inside a tank that contains air. The water level in the U-tube on the side that connects to the tank is 5 ft above the base of the tank. The water level in the other side of the U-tube (which is open to the atmosphere) is 2 ft above the base. Determine the pressure within the tank. 2.32 A barometric pressure of 29.4 in. Hg corresponds to what value of atmospheric pressure in psia, and in pascals? 2.33 For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in barometers containing one of the following liquids: (a) mercury, (b) water, and (c) ethyl alcohol. Calculate the heights including the effect of vapor pressure and compare the results with those obtained neglecting vapor pressure. Do these results support the widespread use of mercury for barometers? Why? 2.34 The closed tank of Fig. P.2.34 is filled with water and is 5 ft long. The pressure gage on the tank reads 7 psi. Determine: (a) the height, h, in the open water column, (b) the gage pressure acting on the bottom tank surface AB, and (c) the absolute pressure of the air in the top of the tank if the local atmospheric pressure is 14.7 psia. Open 7 psi Air h 2 ft Water 2 ft A Air 83 B ■ Figure P2.34 2.35 A mercury manometer is connected to a large reservoir of water as shown in Fig. P2.35. Determine the ratio, hw/hm, of the distances hw and hm indicated in the figure. 12 in. Bourdon gage Water 15 20 10 hw Water 25 30 5 0 hm 35 hm 6 in. Mercury ■ Figure P2.27 ■ Figure P2.35 84 Chapter 2 ■ Fluid Statics 2.36 A U-tube mercury manometer is connected to a closed pressurized tank as illustrated in Fig. P2.36. If the air pressure is 2 psi, determine the differential reading, h. The specific weight of the air is negligible. pair = 2 psi Air 2 ft to 0.6. Determine the manometer reading, h, if the barometric pressure is 14.7 psia and the pressure gage reads 0.5 psi. The effect of the weight of the air is negligible. 2.39 GO A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system as shown in Fig. P2.39. The liquid in the top part of the piping system has a specific gravity of 0.8, and the remaining parts of the system are filled with water. If the pressure gage reading at A is 60 kPa, determine (a) the pressure in pipe B, and (b) the pressure head, in millimeters of mercury, at the top of the dome (point C). Hemispherical dome Water 2 ft 2 ft SG = 0.8 C pA = 60 kPa 4m 3m A 3m h Water 2m Mercury (SG = 13.6) B ■ Figure P2.36 Water 2.37 A U-tube manometer is connected to a closed tank containing air and water as shown in Fig. P2.37. At the closed end of the manometer the air pressure is 16 psia. Determine the reading on the pressure gage for a differential reading of 4 ft on the manometer. Express your answer in psi (gage). Assume standard atmospheric pressure and neglect the weight of the air columns in the manometer. ■ Figure P2.39 2.40 Two pipes are connected by a manometer as shown in Fig. P2.40. Determine the pressure difference, pA pB, between the pipes. A Closed valve 0.5 m Water Air pressure = 16 psia 0.6 m 1.3 m 4 ft Air Gage fluid (SG = 2.6) Water 2 ft B Water ■ Figure P2.40 Gage fluid ( γ = 90 lb / ft 3 ) Pressure gage ■ Figure P2.37 2.41 An inverted open tank is held in place by a force R as shown in Fig. P2.41. If the specific gravity of the manometer fluid is 2.5, determine the value of h. 2.38 Compartments A and B of the tank shown in Fig. P2.38 are closed and filled with air and a liquid with a specific gravity equal 1-in.-diameter tube 0.5 psi Open R 2-ftdiameter tank Air 3 ft h A 2 ft 0.1 ft Water B ■ Figure P2.38 Liquid (SG = 0.6) h Air 1 ft Mercury (SG = 13.6) ■ Figure P2.41 Water 85 Problems 2.42 A U-tube manometer is connected to a closed tank as shown in Fig. P2.42. The air pressure in the tank is 0.50 psi and the liquid in the tank is oil ( 54.0 lb/ft3). The pressure at point A is 2.00 psi. Determine: (a) the depth of oil, z, and (b) the differential reading, h, on the manometer. z 2.46 The cylindrical tank with hemispherical ends shown in Fig. P2.46 contains a volatile liquid and its vapor. The liquid density is 800 kg/m3, and its vapor density is negligible. The pressure in the vapor is 120 kPa (abs) and the atmospheric pressure is 101 kPa (abs). Determine: (a) the gage pressure reading on the pressure gage, and (b) the height, h, of the mercury, manometer. Open Air pressure p1 p2, is applied, a differential reading, h, develops. It is desired to have this reading sufficiently large (so that it can be easily read) for small pressure differentials. Determine the relationship between h and p1 p2 when the area ratio At /Ar is small, and show that the differential reading, h, can be magnified by making the difference in specific weights, 2 1, small. Assume that initially (with p1 p2) the fluid levels in the two reservoirs are equal. Oil h A Vapor 2 ft SG = 3.05 1m Water B Water A 3 in. 8 in. 3 in. h Liquid ■ Figure P2.42 2.43 For the inclined-tube manometer of Fig. P2.43, the pressure in pipe A is 0.6 psi. The fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific gravity of 2.6. What is the pressure in pipe B corresponding to the differential reading shown? Open 1m 1m Mercury ■ Figure P2.46 2.47 Determine the elevation difference, h, between the water levels in the two open tanks shown in Fig. P2.47. SG = 0.90 SG = 2.6 30° 0.4 m ■ Figure P2.43 2.44 A flowrate measuring device is installed in a horizontal pipe through which water is flowing. A U-tube manometer is connected to the pipe through pressure taps located 3 in. on either side of the device. The gage fluid in the manometer has a specific weight of 112 lb/ft3. Determine the differential reading of the manometer corresponding to a pressure drop between the taps of 0.5 lb/in.2 2.45 Small differences in gas pressures are commonly measured with a micromanometer of the type illustrated in Fig. P2.45. This device consists of two large reservoirs each having a crosssectional area Ar which are filled with a liquid having a specific weight 1 and connected by a U-tube of cross-sectional area At containing a liquid of specific weight 2. When a differential gas p1 p2 γ1 γ1 Δh 1m Water ■ Figure P2.47 2.48 Water, oil, and an unknown fluid are contained in the vertical tubes shown in Fig. P2.48. Determine the density of the unknown fluid. 1-in. diameter 2-in. diameter 1 ft Oil, SG = 0.9 Water 2 ft h γ2 ■ Figure P2.45 1 ft ■ Figure P2.48 Unknown fluid 86 Chapter 2 ■ Fluid Statics 2.49 For the configuration shown in Fig. P2.49 what must be the value of the specific weight of the unknown fluid? Express your answer in lb/ft3. Open Open Water 5.5 in. Valve pg 4.9 in. 3.3 in. Unknown fluid 1.4 in. 2.52 Both ends of the U-tube mercury manometer of Fig. P2.52 are initially open to the atmosphere and under standard atmospheric pressure. When the valve at the top of the right leg is open, the level of mercury below the valve is h1. After the valve is closed, air pressure is applied to the left leg. Determine the relationship between the differential reading on the manometer and the applied gage pressure, pg. Show on a plot how the differential reading varies with pg for h1 25, 50, 75, and 100 mm over the range 0 pg 300 kPa. Assume that the temperature of the trapped air remains constant. hi ■ Figure P2.49 Mercury 2.50 An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 in. as shown in Fig. P2.50. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the outside water pressure indicates differential reading of 735 mm Hg as illustrated. Based on these data, what is the atmospheric pressure at the ocean surface? ■ Figure P2.52 2.53 The inverted U-tube manometer of Fig. P2.53 contains oil (SG 0.9) and water as shown. The pressure differential between pipes A and B, pA pB, is 5 kPa. Determine the differential reading h. Ocean surface Oil Shell wall Seawater h 0.2 m 735 mm 360 mm 10 m A 0.3 m Water Mercury B Shell ■ Figure P2.53 ■ Figure P2.50 2.51 Water, oil, and salt water fill a tube as shown in Fig. P2.51. Determine the pressure at point 1 (inside the closed tube). 2.54 An inverted U-tube manometer containing oil (SG 0.8) is located between two reservoirs as shown in Fig. P2.54. The reservoir on the left, which contains carbon tetrachloride, is closed and pressurized to 8 psi. The reservoir on the right contains water and is open to the atmosphere. With the given data, determine the depth of water, h, in the right reservoir. Oil density Carbon tetrachloride = 1.20 slugs/ft3 1-in. diameter 2-in. diameter 3 ft (1) 2 ft Water ■ Figure P2.51 8 psi Salt water, SG = 1.20 0.7 ft 3 ft 4 ft 1 ft 1 ft ■ Figure P2.54 Oil Water h Problems 2.55 Three different liquids with properties as indicated fill the tank and manometer tubes as shown in Fig. P2.55. Determine the specific gravity of Fluid 3. B 0.4 m Fluid 1: specific weight = 62.4 lb/ft3 Fluid 2: density = 1.60 slugs/ft3 Fluid 3: specific gravity = ? Fluid 2 87 Oil A 7 ft Gasoline 0.3 m Mercury A Fluid 1 Fluid 3 6 ft 5 ft ■ Figure P2.58 3 ft 2.59 The mercury manometer of Fig. P2.59 indicates a differential reading of 0.30 m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure in pipe B. ■ Figure P2.55 2.56 Determine the pressure of the water in pipe A shown in Fig. P2.56 if the gage pressure of the air in the tank is 2 psi. Water p = 2 psi 0.50 m Oil A SG = 0.9 Air Mercury 1 ft B 0.15 m 0.30 m 4 ft ■ Figure P2.59 2 ft A Water ■ Figure P2.56 2.57 In Fig. P2.57 pipe A contains carbon tetrachloride (SG 1.60) and the closed storage tank B contains a salt brine (SG 1.15). Determine the air pressure in tank B if the pressure in pipe A is 25 psi. 2.60 The inclined differential manometer of Fig. P2.60 contains carbon tetrachloride. Initially the pressure differential between pipes A and B, which contain a brine (SG 1.1), is zero as illustrated in the figure. It is desired that the manometer give a differential reading of 12 in. (measured along the inclined tube) for a pressure differential of 0.1 psi. Determine the required angle of inclination, u. A B Brine Air B A 4 ft 4 ft 3 ft Brine Brine 12 in. Carbon tetrachloride θ 3 ft Carbon tetrachloride ■ Figure P2.57 ■ Figure P2.60 2.58 In Fig. P2.58 pipe A contains gasoline (SG 0.7), pipe B contains oil (SG 0.9), and the manometer fluid is mercury. Determine the new differential reading if the pressure in pipe A is decreased 25 kPa, and the pressure in pipe B remains constant. The initial differential reading is 0.30 m as shown. 2.61 The manometer fluid in the manometer of Fig. P2.61 has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer. 88 Chapter 2 ■ Fluid Statics crease in pressure of 5 psi in pipe A while the pressure in pipe B remains constant. A 2.65 The U-shaped tube shown in Fig. 2.65 initially contains water only. A second liquid with specific weight, , less than water is placed on top of the water with no mixing occurring. Can the height, h, of the second liquid be adjusted so that the left and right levels are at the same height? Provide proof of your answer. 2 ft Water Water 1 ft B 1 ft Gage fluid (SG = 3.46) γ h D1 = 1.5 D2 ■ Figure P2.61 D2 Water 2.62 Determine the new differential reading along the inclined leg of the mercury manometer of Fig. P2.62, if the pressure in pipe A is decreased 10 kPa and the pressure in pipe B remains unchanged. The fluid in A has a specific gravity of 0.9 and the fluid in B is water. A SG = 0.9 Water 100 mm 30° B 80 mm ■ Figure P2.65 2.66 An inverted hollow cylinder is pushed into the water as is shown in Fig. P2.66. Determine the distance, /, that the water rises in the cylinder as a function of the depth, d, of the lower edge of the cylinder. Plot the results for 0 d H, when H is equal to 1 m. Assume the temperature of the air within the cylinder remains constant. 50 mm Mercury ■ Figure P2.62 Water H d 2.63 Determine the ratio of areas, A1/A2, of the two manometer legs of Fig. P2.63 if a change in pressure in pipe B of 0.5 psi gives a corresponding change of 1 in. in the level of the mercury in the right leg. The pressure in pipe A does not change. D ᐉ Open end ■ Figure P2.66 A B Section 2.8 Hydrostatic Force on a Plane Surface (also see Lab Problems 2.1LP, 2.2LP, 2.3LP, and 2.4LP) Oil (SG = 0.8) Area = A1 Water 2.67 Obtain a photograph/image of a situation in which the hydrostatic force on a plane surface is important. Print this photo and write a brief paragraph that describes the situation involved. Area = A2 2.68 The basic elements of a hydraulic press are shown in Fig. P2.68. The plunger has an area of 1 in.2, and a force, F1, can be applied to the plunger through a lever mechanism having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in.2, what load, F2, can be raised by a force of 30 lb applied to the lever? Neglect the hydrostatic pressure variation. Mercury ■ Figure P2.63 2.64 Determine the change in the elevation of the mercury in the left leg of the manometer of Fig. P2.64 as a result of an inWater F2 Oil (SG = 0.9) A 18 in. 6 in. B 12 in. Plunger 1 _ -in.2 diameter 30° 1 _ -in.-diameter 4 Hydraulic fluid Mercury ■ Figure P2.64 ■ Figure P2.68 F1 Problems 2.69 A 0.3-m-diameter pipe is connected to a 0.02-m-diameter pipe, and both are rigidly held in place. Both pipes are horizontal with pistons at each end. If the space between the pistons is filled with water, what force will have to be applied to the larger piston to balance a force of 90 N applied to the smaller piston? Neglect friction. 2.70 A Bourdon gage (see Fig. 2.13 and Video V2.4) is often used to measure pressure. One way to calibrate this type of gage is to use the arrangement shown in Fig. P2.70a. The container is filled with a liquid and a weight, w, placed on one side with the gage on the other side. The weight acting on the liquid through a 0.4-in.-diameter opening creates a pressure that is transmitted to the gage. This arrangement, with a series of weights, can be used to determine what a change in the dial movement, u, in Fig. P2.70b, corresponds to in terms of a change in pressure. For a particular gage, some data are given below. Based on a plot of these data, determine the relationship between u and the pressure, p, where p is measured in psi. ᐃ (lb) u (deg.) 0 0 1.04 20 2.00 40 3.23 60 Bourdon gage 4.05 80 5.24 100 6.31 120 P Piston h Water h1 Mercury ■ Figure P2.73 2.74 A 6-in.-diameter piston is located within a cylinder that is connected to a 12-in.-diameter inclined-tube manometer as shown in Fig. P2.74. The fluid in the cylinder and the manometer is oil 1specific weight 59 lb ft3 2. When a weight, w, is placed on the top of the cylinder, the fluid level in the manometer tube rises from point (1) to (2). How heavy is the weight? Assume that the change in position of the piston is negligible. ᐃ ᐃ 6 in. 0.4-in.-diameter Liquid (2) (1) Piston θ (b) 89 30° Oil ( a) ■ Figure P2.70 ■ Figure P2.74 2.71 An inverted 0.1-m-diameter circular cylinder is partially filled with water and held in place as shown in Fig. P2.71. A force of 20 N is needed to pull the flat plate from the cylinder. Determine the air pressure within the cylinder. The plate is not fastened to the cylinder and has negligible mass. 2.76 A large, open tank contains water and is connected to a 6ft-diameter conduit as shown in Fig. P2.76. A circular plug is used to seal the conduit. Determine the magnitude, direction, and location of the force of the water on the plug. 0.1 m Air Water 2.75 A square gate (4 m by 4 m) is located on the 45 face of a dam. The top edge of the gate lies 8 m below the water surface. Determine the force of the water on the gate and the point through which it acts. 0.2 m 9 ft Plate Water Plug Open 6 ft F = 20 N ■ Figure P2.71 ■ Figure P2.76 2.72 You partially fill a glass with water, place an index card on top of the glass, and then turn the glass upside down while holding the card in place. You can then remove your hand from the card and the card remains in place, holding the water in the glass. Explain how this works. 2.77 A circular 2-m-diameter gate is located on the sloping side of a swimming pool. The side of the pool is oriented 60 relative to the horizontal bottom, and the center of the gate is located 3 m below the water surface. Determine the magnitude of the water force acting on the gate and the point through which it acts. 2.73 A piston having a cross-sectional area of 0.07 m2 is located in a cylinder containing water as shown in Fig. P2.73. An open U-tube manometer is connected to the cylinder as shown. For h1 60 mm and h 100 mm, what is the value of the applied force, P, acting on the piston? The weight of the piston is negligible. 2.78 A vertical rectangular gate is 8 ft wide and 10 ft long and weighs 6000 lb. The gate slides in vertical slots in the side of a reservoir containing water. The coefficient of friction between the slots and the gate is 0.03. Determine the minimum vertical force required to lift the gate when the water level is 4 ft above the top edge of the gate. 90 Chapter 2 ■ Fluid Statics 2.79 A horizontal 2-m-diameter conduit is half filled with a liquid (SG 1.6) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is 200 kPa. Determine the resultant force of the fluid acting on one of the end caps, and locate this force relative to the bottom of the conduit. 2.80 Concrete is poured into the forms as shown in Fig. P2.80 to produce a set of steps. Determine the weight of the sandbag needed to keep the bottomless forms from lifting off the ground. The weight of the forms is 85 lb, and the specific weight of the concrete is 150 lb/ft3. Open top 3 ft 8 in. risers Sand 10 in. tread Open bottom ■ Figure P2.80 2.81 Forms used to make a concrete basement wall are shown in Fig. P2.81. Each 4-ft-long form is held together by four ties—two at the top and two at the bottom as indicated. Determine the tension in the upper and lower ties. Assume concrete acts as a fluid with a weight of 150 lbft3. hinged on one edge. Determine the minimum air pressure, p1, within the container that will open the hatch. Neglect the weight of the hatch and friction in the hinge. 2.83 An open rectangular container contains a liquid that has a specific weight that varies according to the equation c1 c2h, where c1 and c2 are constants and h is a vertical coordinate measured downward from the free surface. Derive an equation for the magnitude of the liquid force exerted on one wall of the container having a width b and height H and an equation that gives the vertical coordinate of this force. 2.84 A long, vertical wall separates seawater from fresh water. If the seawater stands at a depth of 7 m, what depth of freshwater is required to give a zero resultant force on the wall? When the resultant force is zero, will the moment due to the fluid forces be zero? Explain. 2.85 A gate having the shape shown in Fig. P2.85 is located in the vertical side of an open tank containing water. The gate is mounted on a horizontal shaft. (a) When the water level is at the top of the gate, determine the magnitude of the fluid force on the rectangular portion of the gate above the shaft and the magnitude of the fluid force on the semicircular portion of the gate below the shaft. (b) For this same fluid depth determine the moment of the force acting on the semicircular portion of the gate with respect to an axis that coincides with the shaft. Water 6m 10 in. Shaft 1 ft 3m Tie Side view of gate Concrete ■ Figure P2.85 10 ft Form 2.86 A pump supplies water under pressure to a large tank as shown in Fig. P2.86. The circular-plate valve fitted in the short discharge pipe on the tank pivots about its diameter A–A and is held shut against the water pressure by a latch at B. Show that the force on the latch is independent of the supply pressure, p, and the height of the tank, h. 1 ft ■ Figure P2.81 Pressure p 2.82 A structure is attached to the ocean floor as shown in Fig. P2.82. A 2-m-diameter hatch is located in an inclined wall and Free surface Water h Seawater 10 m 30° Hatch Hinge A A Air pressure, p1 B ■ Figure P2.82 ■ Figure P2.86 Supply 91 Problems 2.87 GO A homogeneous, 4-ft-wide, 8-ft-long rectangular gate weighing 800 lb is held in place by a horizontal flexible cable as shown in Fig. P2.87. Water acts against the gate, which is hinged at point A. Friction in the hinge is negligible. Determine the tension in the cable. 2.92 A vertical plane area having the shape shown in Fig. P2.92 is immersed in an oil bath (specific weight 8.75 kN/m3). Determine the magnitude of the resultant force acting on one side of the area as a result of the oil. 4m Cable 4m 45° 60° Oil bath ■ Figure P2.92 Water Gate 6 ft 8 ft Hinge A ■ Figure P2.87 2.93 GO A 3-m-wide, 8-m-high rectangular gate is located at the end of a rectangular passage that is connected to a large open tank filled with water as shown in Fig. P2.93. The gate is hinged at its bottom and held closed by a horizontal force, FH, located at the center of the gate. The maximum value for FH is 3500 kN. (a) Determine the maximum water depth, h, above the center of the gate that can exist without the gate opening. (b) Is the answer the same if the gate is hinged at the top? Explain your answer. 2.88 A rectangular gate 6 ft tall and 5 ft wide in the side of an open tank is held in place by the force F as indicated in Fig. P.2.88. The weight of the gate is negligible, and the hinge at O is frictionless. (a) Determine the water depth, h, if the resultant hydrostatic force of the water acts 2.5 ft above the bottom of the gate, i.e., it is collinear with the applied force F. (b) For the depth of part (a), determine the magnitude of the resultant hydrostatic force. (c) Determine the force that the hinge puts on the gate under the above conditions. h 4m FH 4m Hinge ■ Figure P2.93 Water Hinge 0 h Gate F 6 ft 2.5 ft ■ Figure P2.88 2.94 Two square gates close two openings in a conduit connected to an open tank of water as shown in Fig. P2.94. When the water depth, h, reaches 5 m it is desired that both gates open at the same time. Determine the weight of the homogeneous horizontal gate and the horizontal force, R, acting on the vertical gate that is required to keep the gates closed until this depth is reached. The weight of the vertical gate is negligible, and both gates are hinged at one end as shown. Friction in the hinges is negligible. †2.89 Sometimes it is difficult to open an exterior door of a building because the air distribution system maintains a pressure difference between the inside and outside of the building. Estimate how big this pressure difference can be if it is “not too difficult” for an average person to open the door. 2.90 An area in the form of an isosceles triangle with a base width of 6 ft and an attitude of 8 ft lies in the plane forming one wall of a tank that contains a liquid having a specific weight of 79.8 lb/ft3. The side slopes upward, making an angle of 60 with the horizontal. The base of the triangle is horizontal, and the vertex is above the base. Determine the resultant force the fluid exerts on the area when the fluid depth is 20 ft above the base of the triangular area. Show, with the aid of a sketch, where the center of pressure is located. 2.91 Solve Problem 2.90 if the isosceles triangle is replaced with a right triangle having the same base width and altitude as the isosceles triangle. Horizontal gate, 4m × 4m Hinge h 3m Water R Vertical gate, 4m × 4m Hinge ■ Figure P2.94 2.95 A gate having the cross section shown in Fig. 2.95 closes an opening 5 ft wide and 4 ft high in a water reservoir. The gate weighs 500 lb, and its center of gravity is 1 ft to the left of AC and 2 ft above BC. Determine the horizontal reaction that is developed on the gate at C. 92 Chapter 2 ■ Fluid Statics Water 5 ft Water 8 ft A Hinge A 4 ft Moveable stop 12 ft Gate Brace θ ■ Figure P2.98 Gate C B force that the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted on the gate by the brace as a function of the angle of the gate, u, for 0 u 90 . (b) Repeat the calculations for the case in which the weight of the gate is negligible. Common on the result as u → 0. 3 ft ■ Figure P2.95 2.96 A gate having the cross section shown in Fig. P2.96 is 4 ft wide and is hinged at C. The gate weighs 18,000 lb, and its mass center is 1.67 ft to the right of the plane BC. Determine the vertical reaction at A on the gate when the water level is 3 ft above the base. All contact surfaces are smooth. 2.99 An open tank has a vertical partition and on one side contains gasoline with a density r 700 kg/m3 at a depth of 4 m, as shown in Fig. P2.99. A rectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h, will the gate start to open? 5 ft Partition Hinge Stop C 9 ft Water Water surface 4m h Gasoline Hinge ■ Figure P2.99 3 ft B A ■ Figure P2.96 2.97 The massless, 4-ft-wide gate shown in Fig. P2.97 pivots about the frictionless hinge O. It is held in place by the 2000 lb counterweight, w. Determine the water depth, h. 2.100 A 4-ft by 3-ft massless rectangular gate is used to close the end of the water tank shown in Fig. P2.100. A 200-lb weight attached to the arm of the gate at a distance / from the frictionless hinge is just sufficient to keep the gate closed when the water depth is 2 ft, that is, when the water fills the semicircular lower portion of the tank. If the water were deeper, the gate would open. Determine the distance /. ᐉ 1 ft Water Gate Hinge Hinge 200 lb h Gate Water 2 ft radius 3 ft 4 ft ■ Figure P2.100 Pivot O Width = 4 ft 2 ft 3 ft ■ Figure P2.97 2.101 The rigid gate, OAB, of Fig. P2.101 is hinged at O and rests against a rigid support at B. What minimum horizontal force, P, is required to hold the gate closed if its width is 3 m? Neglect the weight of the gate and friction in the hinge. The back of the gate is exposed to the atmosphere. 2.98 A 200-lb homogeneous gate 10 ft wide and 5 ft long is hinged at point A and held in place by a 12-ft-long brace as shown in Fig. P2.98. As the bottom of the brace is moved to the right, the water level remains at the top of the gate. The line of action of the 2.102 A rectangular gate that is 2 m wide is located in the vertical wall of a tank containing water as shown in Fig. P2.102. It is desired to have the gate open automatically when the depth of water above the top of the gate reaches 10 m. (a) At what distance, d, should the frictionless horizontal shaft be located? (b) What is the magnitude of the force on the gate when it opens? ᐃ Problems 93 Open to atmosphere D Hinge B, C Water 3m A O Water 30° 4 ft 2 ft C y′ Stop Hinge B 4m y′ = 4(x′)2 A B A P Plan of gate x′ ■ Figure P2.104 2m ■ Figure P2.101 2.105 An open rectangular tank is 2 m wide and 4 m long. The tank contains water to a depth of 2 m and oil (SG 0.8) on top of the water to a depth of 1 m. Determine the magnitude and location of the resultant fluid force acting on one end of the tank. 2.106 An open rectangular settling tank contains a liquid suspension that at a given time has a specific weight that varies approximately with depth according to the following data: Water 10 m d h (m) G (N m3) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 10.0 10.1 10.2 10.6 11.3 12.3 12.7 12.9 13.0 13.1 4m ■ Figure P2.102 2.103 GO A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point O, as shown in Fig. P2.103. The horizontal portion of the gate covers a 1-ft-diameter drain pipe that contains air at atmospheric pressure. Determine the minimum water depth, h, at which the gate will pivot to allow water to flow into the pipe. Width = 4 ft Right-angle gate h Water Hinge The depth h 0 corresponds to the free surface. Determine, by means of numerical integration, the magnitude and location of the resultant force that the liquid suspension exerts on a vertical wall of the tank that is 6 m wide. The depth of fluid in the tank is 3.6 m. 2.107 The closed vessel of Fig. P2.107 contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout that is closed by a 6-in.-diameter circular gate that is hinged along one side as illustrated. The horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that must be applied at the hinge to hold the gate shut. Neglect the weight of the gate and friction at the hinge. O 1-ft-diameter pipe 10 psi 3 ft Air ■ Figure P2.103 2.104 The inclined face AD of the tank of Fig. P2.104 is a plane surface containing a gate ABC, which is hinged along line BC. The shape of the gate is shown in the plan view. If the tank contains water, determine the magnitude of the force that the water exerts on the gate. 10 ft Water Axis 6-in.-diameter gate 3 ■ Figure P2.107 4 94 Chapter 2 ■ Fluid Statics 2.108 A 4-ft-tall, 8-in.-wide concrete (150 lb/ft3) retaining wall is built as shown in Fig. P2.108. During a heavy rain, water fills the space between the wall and the earth behind it to a depth h. Determine the maximum depth of water possible without the wall tipping over. The wall simply rests on the ground without being anchored to it. 2.111 (See Fluids in the News article titled “The Three Gorges Dam,” Section 2.8.) (a) Determine the horizontal hydrostatic force on the 2309-m-long Three Gorges Dam when the average depth of the water against it is 175 m. (b) If all of the 6.4 billion people on Earth were to push horizontally against the Three Gorges Dam, could they generate enough force to hold it in place? Support your answer with appropriate calculations. 8 in. Section 2.10 Hydrostatic Force on a Curved Surface 2.112 Obtain a photograph/image of a situation in which the hydrostatic force on a curved surface is important. Print this photo and write a brief paragraph that describes the situation involved. 2.113 A 2-ft-diameter hemispherical plexiglass “bubble” is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2-ft-diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere. 4 ft h ■ Figure P2.108 2.109 Water backs up behind a concrete dam as shown in Fig. 2.109. Leakage under the foundation gives a pressure distribution under the dam as indicated. If the water depth, h, is too great, the dam will topple over about its toe (point A). For the dimensions given, determine the maximum water depth for the following widths of the dam, / 20, 30, 40, 50, and 60 ft. Base your analysis on a unit length of the dam. The specific weight of the concrete is 150 lb/ft3 2.114 Two round, open tanks containing the same type of fluid rest on a table top as shown in Fig. P2.114. They have the same bottom area, A, but different shapes. When the depth, h, of the liquid in the two tanks is the same, the pressure force of the liquids on the bottom of the two tanks is the same. However, the force that the table exerts on the two tanks is different because the weight in each of the tanks is different. How do you account for this apparent paradox? 80 ft Water h h B A Water hT = 10 ft pA = γ hT pB = γ h Area = A Area = A ■ Figure P2.114 ᐉ ■ Figure P2.109 2.110 The concrete dam of Fig. P2.110 weighs 23.6 kN/m3 and rests on a solid foundation. Determine the minimum coefficient of friction between the dam and the foundation required to keep the dam from sliding at the water depth shown. Assume no fluid uplift pressure along the base. Base your analysis on a unit length of the dam. 2.115 Two hemispherical shells are bolted together as shown in Fig. P2.115. The resulting spherical container, which weighs 300 lb, is filled with mercury and supported by a cable as shown. The container is vented at the top. If eight bolts are symmetrically located around the circumference, what is the vertical force that each bolt must carry? 2m Cable Water 5m Vent Sphere diameter = 3 ft 4m 6m ■ Figure P2.110 ■ Figure P2.115 95 Problems 2.116 A 4-m-long curved gate is located in the side of a reservoir containing water as shown in Fig. P2.116. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. Will this force pass through point A? Explain. Water Tank wall 18 ft A 6 ft Water B 6m ■ Figure P2.119 A 2m Gate ■ Figure P2.116 2.117 The 18-ft-long lightweight gate of Fig. P2.117 is a quarter circle and is hinged at H. Determine the horizontal force, P, required to hold the gate in place. Neglect friction at the hinge and the weight of the gate. 2.120 Hoover Dam (see Video 2.5) is the highest arch-gravity type of dam in the United States. A cross section of the dam is shown in Fig. P2.120(a). The walls of the canyon in which the dam is located are sloped, and just upstream of the dam the vertical plane shown in Figure P2.120(b) approximately represents the cross section of the water acting on the dam. Use this vertical cross section to estimate the resultant horizontal force of the water on the dam, and show where this force acts. 45 ft 880 ft P 6 ft Water 727 ft 715 ft. Gate H Hinge ■ Figure P2.117 660 ft 290 ft (a) 2.118 The air pressure in the top of the 2-liter pop bottle shown in Video V2.6 and Fig. P2.118 is 40 psi, and the pop depth is 10 in. The bottom of the bottle has an irregular shape with a diameter of 4.3 in. (a) If the bottle cap has a diameter of 1 in. what is the magnitude of the axial force required to hold the cap in place? (b) Determine the force needed to secure the bottom 2 in. of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c) By how much does the weight of the pop increase the pressure 2 in. above the bottom? Assume the pop has the same specific weight as that of water. (b) ■ Figure P2.120 2.121 A plug in the bottom of a pressurized tank is conical in shape, as shown in Fig. P2.121. The air pressure is 40 kPa, and the liquid in the tank has a specific weight of 27 kN/m3. Determine the magnitude, direction, and line of action of the force exerted on the curved surface of the cone within the tank due to the 40-kPa pressure and the liquid. 40 kPa 1-in. diameter Air pair = 40 psi Liquid 12 in. 10 in. 3m 4.3-in. diameter 1m 60° ■ Figure P2.118 2.119 A tank wall has the shape shown in Fig. P2.119. Determine the horizontal and vertical components of the force of the water on a 4-ft length of the curved section AB. ■ Figure P2.121 2.122 The homogeneous gate shown in Fig. P2.122 consists of one quarter of a circular cylinder and is used to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, the gate 96 Chapter 2 ■ Fluid Statics 6 ft Pivot Bulge Water 4m 3 ft 1m ■ Figure P2.128 ■ Figure P2.122 pA opens slightly and lets the water flow under it. Determine the weight of the gate per meter of length. 2.123 The concrete (specific weight 150 lb/ft3) seawall of Fig. P2.123 has a curved surface and restrains seawater at a depth of 24 ft. The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A). Air 4-ft diameter 20 psi 5 ft 2 ft Water 2 ft Seawater y = 0.2x2 Gage fluid (SG = 3.0) ■ Figure P2.129 24 ft y A x 15 ft 2.130 A 3-m-diameter open cylindrical tank contains water and has a hemispherical bottom as shown in Fig. P2.130. Determine the magnitude, line of action, and direction of the force of the water on the curved bottom. ■ Figure P2.123 2.124 A 12-in.-diameter pipe contains a gas under a pressure of 140 psi. If the pipe wall thickness is 14 in., what is the average circumferential stress developed in the pipe wall? 2.125 A cylindrical tank with its axis horizontal has a diameter of 2.0 m and a length of 4.0 m. The ends of the tank are vertical planes. A vertical 0.1-m-diameter pipe is connected to the top of the tank. The tank and the pipe are filled with ethyl alcohol to a level of 1.5 m above the top of the tank. Determine the resultant force of the alcohol on one end of the tank and show where it acts. 2.126 Imagine the tank of Problem 2.125 split by a horizontal plane. Determine the magnitude of the resultant force of the alcohol on the bottom half of the tank. 2.127 If the tank ends in Problem 2.125 are hemispherical, what is the magnitude of the resultant horizontal force of the alcohol on one of the curved ends? 2.128 GO An open tank containing water has a bulge in its vertical side that is semicircular in shape as shown in Fig. P2.128. Determine the horizontal and vertical components of the force that the water exerts on the bulge. Base your analysis on a 1-ft length of the bulge. 2.129 A closed tank is filled with water and has a 4-ftdiameter hemispherical dome as shown in Fig. 2.129. A U-tube manometer is connected to the tank. Determine the vertical force of the water on the dome if the differential manometer reading is 7 ft and the air pressure at the upper end of the manometer is 12.6 psi. 8m Water 3m ■ Figure P2.130 2.131 Three gates of negligible weight are used to hold back water in a channel of width b as shown in Fig. P2.131. The force of the gate against the block for gate (b) is R. Determine (in terms of R) the force against the blocks for the other two gates. Section 2.11 Buoyancy, Flotation, and Stability 2.132 Obtain a photograph/image of a situation in which Archimede’s principle is important. Print this photo and write a brief paragraph that describes the situation involved. 2.133 A freshly cut log floats with one fourth of its volume protruding above the water surface. Determine the specific weight of the log. Problems h 2 97 V is floated in the liqcylinder of specific weight 2 and volume uid (see Fig. P2.137b), the liquid level rises by an amount h ( 2/ 1) V /A. Hinge h h Block (a) (b) h 2 h 2.138 When the Tucurui Dam was constructed in northern GO Brazil, the lake that was created covered a large forest of valuable hardwood trees. It was found that even after 15 years underwater the trees were perfectly preserved and underwater logging was started. During the logging process a tree is selected, trimmed, and anchored with ropes to prevent it from shooting to the surface like a missile when cut. Assume that a typical large tree can be approximated as a truncated cone with a base diameter of 8 ft, a top diameter of 2 ft, and a height of 100 ft. Determine the resultant vertical force that the ropes must resist when the completely submerged tree is cut. The specific gravity of the wood is approximately 0.6. †2.139 Estimate the minimum water depth needed to float a canoe carrying two people and their camping gear. List all assumptions and show all calculations. (c) ■ Figure P2.131 2.134 A 3 ft 3 ft 3 ft wooden cube (specific weight 37 lb/ft3) floats in a tank of water. How much of the cube extends above the water surface? If the tank were pressurized so that the air pressure at the water surface was increased by 1.0 psi (i.e., 1 psig), how much of the cube would extend above the water surface? Explain how you arrived at your answer. 2.135 The homogeneous timber AB of Fig. P2.135 is 0.15 m by 0.35 m in cross section. Determine the specific weight of the timber and the tension in the rope. 2.140 An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle as shown in Video V2.7 and Fig. P2.140. The amount of air in the tube has been adjusted so that it just floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon. Air Test tube 0.15 m 2m B 8m Water A Rope Water Plastic bottle ■ Figure P2.140 ■ Figure P2.135 2.136 A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 28 ft wide and 90 ft long. When unloaded, its draft (depth of submergence) is 5 ft. and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain. 2.141 An irregularly shaped piece of a solid material weighs 8.05 lb in air and 5.26 lb when completely submerged in water. Determine the density of the material. 2.142 A 1-ft-diameter, 2-ft-long cylinder floats in an open tank containing a liquid having a specific weight . A U-tube manometer is connected to the tank as shown in Fig. P2.142. When the 2.137 A tank of cross-sectional area A is filled with a liquid of specific weight 1 as shown in Fig. P2.137a. Show that when a V pa = – 0.1 psi 1 ft A Δh γ2 2 ft γ1 Cylinder Sp. wt. = g (a) ■ Figure P2.137 (b) ■ Figure P2.142 1 ft 2 ft 0.5 ft Water Gage fluid SG = 1.5 98 Chapter 2 ■ Fluid Statics pressure in pipe A is 0.1 psi below atmospheric pressure, the various fluid levels are as shown. Determine the weight of the cylinder. Note that the top of the cylinder is flush with the fluid surface. replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface? The hydrometer weighs 0.042 lb. 2.143 A 1-m-diameter cylindrical mass, M, is connected to a 2-mwide rectangular gate as shown in Fig. P2.143. The gate is to open when the water level, h, drops below 2.5 m. Determine the required value for M. Neglect friction at the gate hinge and the pulley. 2.146 A 2-ft-thick block constructed of wood (SG 0.6) is submerged in oil (SG 0.8) and has a 2-ft-thick aluminum (specific weight 168 lb/ft3) plate attached to the bottom as indicated in Fig. P2.146. Determine completely the force required to hold the block in the position shown. Locate the force with respect to point A. 4m 1-mdiameter Water h 6 ft Oil 1m ■ Figure P2.143 4 ft 2.144 The thin-walled, 1-m-diameter tank of Fig. P2.144 is closed at one end and has a mass of 90 kg. The open end of the tank is lowered into the water and held in the position shown by a steel block having a density of 7840 kg/m3. Assume that the air that is trapped in the tank is compressed at a constant temperature. Determine: (a) the reading on the pressure gage at the top of the tank, and (b) the volume of the steel block. Tank 0.6 m Air Water 3.0 m Open end Cable Aluminum 0.5 ft A 10 ft ■ Figure P2.146 2.147 (See Fluids in the News article titled “Concrete canoes,” Section 2.11.1.) How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb Kevlar canoe of the same size carrying the same load? 2.148 An iceberg (specific gravity 0.917) floats in the ocean (specific gravity 1.025). What percent of the volume of the iceberg is under water? Section 2.12 Pressure Variation in a Fluid with Rigid-Body Motion 2.149 Obtain a photograph/image of a situation in which the pressure variation in a fluid with rigid-body motion is involved. Print this photo and write a brief paragraph that describes the situation involved. ■ Figure P2.144 2.150 It is noted that while stopping, the water surface in a glass of water sitting in the cup holder of a car is slanted at an angle of 15 relative to the horizontal street. Determine the rate at which the car is decelerating. 2.145 When a hydrometer (see Fig. P2.145 and Video V2.8) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is 2.151 An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi/hr. As the truck slows uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration? Steel block Hydrometer Fluid surface 2.152 A 5-gal, cylindrical open container with a bottom area of 120 in.2 is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of 3 ft/s2. (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal 231 in.3) 2.153 An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, what is the maximum horizontal acceleration (along the long axis of the tank) that can develop before the gasoline would begin to spill? ■ Figure P2.145 2.154 If the tank of Problem 2.153 slides down a frictionless plane that is inclined at 30 with the horizontal, determine the angle the free surface makes with the horizontal. Problems 99 2.155 A closed cylindrical tank that is 8 ft in diameter and 24 ft long is completely filled with gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal surface. Determine the pressure difference between the ends (along the long axis of the tank) when the truck undergoes an acceleration of 5 ft/s2. 2.159 An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank. 2.156 The open U-tube of Fig. P2.156 is partially filled with a liquid. When this device is accelerated with a horizontal acceleration a, a differential reading h develops between the manometer legs which are spaced a distance / apart. Determine the relationship between a, /, and h. 2.160 An open, 2-ft-diameter tank contains water to a depth of 3 ft when at rest. If the tank is rotated about its vertical axis with an angular velocity of 180 rev/min, what is the minimum height of the tank walls to prevent water from spilling over the sides? a h 2.161 A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the balloon tilts to the right. On the other hand, the child tends to be forced forward as the car decelerates and to the left as the car makes a righthand turn. Explain these observed effects on the balloon and child. 2.162 A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9) and rotates about its vertical longitudinal axis with an angular velocity of 40 rad/s. Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis. ᐉ ■ Figure P2.156 2.157 The U-tube of Fig. P2.157 is partially filled with water and rotates around the axis a–a. Determine the angular velocity that will cause the water to start to vaporize at the bottom of the tube (point A). 2.163 (See Fluids in the News article titled “Rotating mercury mirror telescope,” Section 2.12.2.) The largest liquid mirror telescope uses a 6-ft-diameter tank of mercury rotating at 7 rpm to produce its parabolic-shaped mirror as shown in Fig. P2.163. Determine the difference in elevation of the mercury, h, between the edge and the center of the mirror. Receiver a 4 in. Light rays 4 in. 6 ft Δh Mercury ω ω = 7 rpm ■ Figure P2.163 12 in. ■ Lab Problems A 2.1 LP This problem involves the force needed to open a gate that covers an opening in the side of a water-filled tank. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. a ■ Figure P2.157 2.158 The U-tube of Fig. P2.158 contains mercury and rotates about the off-center axis a–a. At rest, the depth of mercury in each leg is 150 mm as illustrated. Determine the angular velocity for which the difference in heights between the two legs is 75 mm. 2.3 LP This problem involves determining the weight needed to hold down an open-bottom box that has slanted sides when the box is filled with water. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley. com/college/munson. a ω 150 mm Mercury 220 mm 90 mm a ■ Figure P2.158 2.2 LP This problem involves the use of a cleverly designed apparatus to investigate the hydrostatic, pressure force on a submerged rectangle. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/ college/munson. 2.4 LP This problem involves the use of a pressurized air pad to provide the vertical force to support a given load. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Lifelong Learning Problems 2.1 LL Although it is relatively easy to calculate the net hydrostatic pressure force on a dam, it is not necessarily easy to design and 100 Chapter 2 ■ Fluid Statics construct an appropriate, long-lasting, inexpensive dam. In fact, inspection of older dams has revealed that many of them are in peril of collapse unless corrective action is soon taken. Obtain information about the severity of the poor conditions of older dams throughout the country. Summarize your findings in a brief report. 2.2 LL Over the years the demand for high-quality, first-growth timber has increased dramatically. Unfortunately, most of the trees that supply such lumber have already been harvested. Recently, however, several companies have started to reclaim the numerous high-quality logs that sank in lakes and oceans during the logging boom times many years ago. Many of these logs are still in excellent condition. Obtain information about the use of fluid mechanics concepts in harvesting sunken logs. Summarize your findings in a brief report. 2.3 LL Liquid-filled manometers and Bourdon tube pressure gages have been the mainstay for measuring pressure for many, many years. However, for many modern applications these triedand-true devices are not sufficient. For example, various new uses need small, accurate, inexpensive pressure transducers with digital outputs. Obtain information about some of the new concepts used for pressure measurement. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 3 Elementary Fluid Dynamics— The Bernoulli Equation CHAPTER OPENING PHOTO: Flow past a blunt body: On any object placed in a moving fluid there is a stagnation point on the front of the object where the velocity is zero. This location has a relatively large pressure and divides the flow field into two portions—one flowing to the left, and one flowing to the right of the body. 1Dye in water.2 Learning Objectives After completing this chapter, you should be able to: ■ discuss the application of Newton’s second law to fluid flows. ■ explain the development, uses, and limitations of the Bernoulli equation. ■ use the Bernoulli equation (stand-alone or in combination with the continuity equation) to solve simple flow problems. ■ apply the concepts of static, stagnation, dynamic, and total pressures. ■ calculate various flow properties using the energy and hydraulic grade lines. The Bernoulli equation may be the most used and abused equation in fluid mechanics. 3.1 In this chapter we investigate some typical fluid motions (fluid dynamics) in an elementary way. We will discuss in some detail the use of Newton’s second law (F ma) as it is applied to fluid particle motion that is “ideal” in some sense. We will obtain the celebrated Bernoulli equation and apply it to various flows. Although this equation is one of the oldest in fluid mechanics and the assumptions involved in its derivation are numerous, it can be used effectively to predict and analyze a variety of flow situations. However, if the equation is applied without proper respect for its restrictions, serious errors can arise. Indeed, the Bernoulli equation is appropriately called the most used and the most abused equation in fluid mechanics. A thorough understanding of the elementary approach to fluid dynamics involved in this chapter will be useful on its own. It also provides a good foundation for the material in the following chapters where some of the present restrictions are removed and “more nearly exact” results are presented. Newton’s Second Law As a fluid particle moves from one location to another, it usually experiences an acceleration or deceleration. According to Newton’s second law of motion, the net force acting on the fluid particle under consideration must equal its mass times its acceleration, F ma 101 102 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation In this chapter we consider the motion of inviscid fluids. That is, the fluid is assumed to have zero viscosity. If the viscosity is zero, then the thermal conductivity of the fluid is also zero and there can be no heat transfer 1except by radiation2. In practice there are no inviscid fluids, since every fluid supports shear stresses when it is subjected to a rate of strain displacement. For many flow situations the viscous effects are relatively small compared with other effects. As a first approximation for such cases it is often possible to ignore viscous effects. For example, often the viscous forces developed in flowing water may be several orders of magnitude smaller than forces due to other influences, such as gravity or pressure differences. For other water flow situations, however, the viscous effects may be the dominant ones. Similarly, the viscous effects associated with the flow of a gas are often negligible, although in some circumstances they are very important. We assume that the fluid motion is governed by pressure and gravity forces only and examine Newton’s second law as it applies to a fluid particle in the form: 1Net pressure force on particle2 1net gravity force on particle2 1particle mass2 1particle acceleration2 Inviscid fluid flow is governed by pressure and gravity forces. z x y Rectangular z θ r Cylindrical The results of the interaction between the pressure, gravity, and acceleration provide numerous useful applications in fluid mechanics. To apply Newton’s second law to a fluid 1or any other object2, we must define an appropriate coordinate system in which to describe the motion. In general the motion will be threedimensional and unsteady so that three space coordinates and time are needed to describe it. There are numerous coordinate systems available, including the most often used rectangular 1x, y, z2 and cylindrical 1r, u, z2 systems shown by the figures in the margin. Usually the specific flow geometry dictates which system would be most appropriate. In this chapter we will be concerned with two-dimensional motion like that confined to the x–z plane as is shown in Fig. 3.1a. Clearly we could choose to describe the flow in terms of the components of acceleration and forces in the x and z coordinate directions. The resulting equations are frequently referred to as a two-dimensional form of the Euler equations of motion in rectangular Cartesian coordinates. This approach will be discussed in Chapter 6. As is done in the study of dynamics 1Ref. 12, the motion of each fluid particle is described in terms of its velocity vector, V, which is defined as the time rate of change of the position of the particle. The particle’s velocity is a vector quantity with a magnitude 1the speed, V 0 V 0 2 and direction. As the particle moves about, it follows a particular path, the shape of which is governed by the velocity of the particle. The location of the particle along the path is a function of where the particle started at the initial time and its velocity along the path. If it is steady flow 1i.e., nothing changes with time at a given location in the flow field2, each successive particle that passes through a given point [such as point 112 in Fig. 3.1a] will follow the same path. For such cases the path is a fixed line in the x–z plane. Neighboring particles that pass on either side of point 112 follow their own paths, which may be of a different shape than the one passing through 112. The entire x–z plane is filled with such paths. For steady flows each particle slides along its path, and its velocity vector is everywhere tangent to the path. The lines that are tangent to the velocity vectors throughout the flow field are called streamlines. For many situations it is easiest to describe the flow in terms of the z z Fluid particle V (2) s (1) ᏾ = ᏾(s) n = n1 n n=0 V Streamlines (a) x (b) ■ Figure 3.1 (a) Flow in the x–z plane. (b) Flow in terms of streamline and normal coordinates. x 3.1 Fluid particles accelerate normal to and along streamlines. V3.1 Streamlines past an airfoil as = an = 0 103 “streamline” coordinates based on the streamlines as are illustrated in Fig. 3.1b. The particle motion is described in terms of its distance, s s1t2, along the streamline from some convenient origin and the local radius of curvature of the streamline, r r1s2. The distance along the streamline is related to the particle’s speed by V dsdt, and the radius of curvature is related to the shape of the streamline. In addition to the coordinate along the streamline, s, the coordinate normal to the streamline, n, as is shown in Fig. 3.1b, will be of use. To apply Newton’s second law to a particle flowing along its streamline, we must write the particle acceleration in terms of the streamline coordinates. By definition, the acceleration is the time rate of change of the velocity of the particle, a dVdt. For two-dimensional flow in the x–z plane, the acceleration has two components—one along the streamline, as, the streamwise acceleration, and one normal to the streamline, an, the normal acceleration. The streamwise acceleration results from the fact that the speed of the particle generally varies along the streamline, V V1s2. For example, in Fig. 3.1a the speed may be 50 ft s at point 112 and 100 ft s at point 122. Thus, by use of the chain rule of differentiation, the s component of the acceleration is given by as dVdt 10V0s21dsdt2 10V0s2V. We have used the fact that speed is the time rate of change of distance, V dsdt. Note that the streamwise acceleration is the product of the rate of change of speed with distance along the streamline, 0V0s, and the speed, V. Since 0V0s can be positive, negative, or zero, the streamwise acceleration can, therefore, be positive (acceleration), negative (deceleration), or zero (constant speed). The normal component of acceleration, the centrifugal acceleration, is given in terms of the particle speed and the radius of curvature of its path. Thus, an V 2 r, where both V and r may vary along the streamline. These equations for the acceleration should be familiar from the study of particle motion in physics 1Ref. 22 or dynamics 1Ref. 12. A more complete derivation and discussion of these topics can be found in Chapter 4. Thus, the components of acceleration in the s and n directions, as and an, are given by as V as > 0 as < 0 Newton’s Second Law 0V , 0s an V2 r (3.1) where r is the local radius of curvature of the streamline, and s is the distance measured along the streamline from some arbitrary initial point. In general there is acceleration along the streamline 1because the particle speed changes along its path, 0V0s 02 and acceleration normal to the streamline 1because the particle does not flow in a straight line, r q 2. Various flows and the accelerations associated with them are shown in the figure in the margin. As discussed in Section 3.6.2, for incompressible flow the velocity is inversely proportional to the streamline spacing. Hence, converging streamlines produce positive streamwise acceleration. To produce this acceleration there must be a net, nonzero force on the fluid particle. To determine the forces necessary to produce a given flow 1or conversely, what flow results from a given set of forces2, we consider the free-body diagram of a small fluid particle as is shown in Fig. 3.2. The particle of interest is removed from its surroundings, and the reactions of the surroundings on the particle are indicated by the appropriate forces present, F1, F2, and so forth. For the present case, the important forces are assumed to be gravity and pressure. Other forces, an > 0 z Fluid particle F5 F4 θ as > 0, an > 0 Streamline F1 F3 F2 x ■ Figure 3.2 Isolation of a small fluid particle in a flow field. (Photo courtesy of Diana Sailplanes.) g 104 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation such as viscous forces and surface tension effects, are assumed negligible. The acceleration of gravity, g, is assumed to be constant and acts vertically, in the negative z direction, at an angle u relative to the normal to the streamline. 3.2 F ⴝ ma along a Streamline Consider the small fluid particle of size ds by dn in the plane of the figure and dy normal to the figure as shown in the free-body diagram of Fig. 3.3. Unit vectors along and normal to the streamline are denoted by ˆs and nˆ, respectively. For steady flow, the component of Newton’s second law along the streamline direction, s, can be written as 0V 0V V a dFs dm as dm V 0s r dV 0s (3.2) where g dFs represents the sum of the s components of all the forces acting on the particle, which has mass dm r dV ds dn dy is , and V 0V0s is the acceleration in the s direction. Here, dV the particle volume. Equation 3.2 is valid for both compressible and incompressible fluids. That is, the density need not be constant throughout the flow field. The gravity force 1weight2 on the particle can be written as dw g dV , where g rg is the specific weight of the fluid 1lbft3 or Nm3 2. Hence, the component of the weight force in the direction of the streamline is dws dw sin u g dV sin u In a flowing fluid the pressure varies from one location to another. If the streamline is horizontal at the point of interest, then u 0, and there is no component of particle weight along the streamline to contribute to its acceleration in that direction. As is indicated in Chapter 2, the pressure is not constant throughout a stationary fluid 1§p 02 because of the fluid weight. Likewise, in a flowing fluid the pressure is usually not constant. In general, for steady flow, p p1s, n2. If the pressure at the center of the particle shown in Fig. 3.3 is denoted as p, then its average value on the two end faces that are perpendicular to the streamline are p dps and p dps. Since the particle is “small,” we can use a one-term Taylor series expansion for the pressure field 1as was done in Chapter 2 for the pressure forces in static fluids2 to obtain dps 0p ds 0s 2 g (p + δ pn) δ s δ y τ δs δ y = 0 Particle thickness = δ y θ n (p + δ ps) δ n δ y δs s δn δ n δ θ δ s (p – δ ps) δ n δ y δs δz θ Along streamline τ δs δ y = 0 θ δz δn (p – δ pn) δ s δ y Normal to streamline ■ Figure 3.3 Free-body diagram of a fluid particle for which the important forces are those due to pressure and gravity. F ⴝ ma along a Streamline 3.2 105 Thus, if dFps is the net pressure force on the particle in the streamline direction, it follows that dFps 1 p dps 2 dn dy 1 p dps 2 dn dy 2 dps dn dy 0p 0p ds dn dy dV 0s 0s The net pressure force on a particle is determined by the pressure gradient. Note that the actual level of the pressure, p, is not important. What produces a net pressure force is the fact that the pressure is not constant throughout the fluid. The nonzero pressure gradient, §p 0p 0s ˆs 0p 0n nˆ, is what provides a net pressure force on the particle. Viscous forces, represented by t ds dy, are zero, since the fluid is inviscid. Thus, the net force acting in the streamline direction on the particle shown in Fig. 3.3 is given by 0p a dFs dws dFps ag sin u 0s b dV (3.3) By combining Eqs. 3.2 and 3.3, we obtain the following equation of motion along the streamline direction: g sin u 0p 0V rV ras 0s 0s (3.4) V, that appears in both the force and We have divided out the common particle volume factor, d the acceleration portions of the equation. This is a representation of the fact that it is the fluid density 1mass per unit volume2, not the mass, per se, of the fluid particle that is important. The physical interpretation of Eq. 3.4 is that a change in fluid particle speed is accomplished by the appropriate combination of pressure gradient and particle weight along the streamline. For fluid static situations this balance between pressure and gravity forces is such that no change in particle speed is produced—the right-hand side of Eq. 3.4 is zero, and the particle remains stationary. In a flowing fluid the pressure and weight forces do not necessarily balance—the force unbalance provides the appropriate acceleration and, hence, particle motion. E XAMPLE 3.1 Pressure Variation along a Streamline GIVEN Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a, as shown in Fig. E3.1a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is V V0 a1 a3 b x3 FIND Determine the pressure variation along the streamline from point A far in front of the sphere 1xA and VA V0 2 to point B on the sphere 1xB a and VB 02. 1 Vo as shown in Fig. E3.1b. 0.75 Vo z V VA = VO ˆi V = V ˆi VB = 0 B A x a 0.5 Vo 0.25 Vo 0 –3a –2a –1a 0 x (b) (a) ∂__ p ∂x –3a –2a –a (c) p 0.610 ρV02/a 0 x 0.5 ρV02 –3a –2a –a (d) 0 x ■ Figure E3.1 106 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation SOLUTION Since the flow is steady and inviscid, Eq. 3.4 is valid. In addition, since the streamline is horizontal, sin u sin 0° 0 and the equation of motion along the streamline reduces to 0p 0V rV 0s 0s (1) With the given velocity variation along the streamline, the acceleration term is V 3V0 a3 0V 0V a3 V V0 a1 3 b a 4 b 0s 0x x x 3 3 a a 3V 20 a1 3 b 4 x x 1ax2 6 a 3 d p rV 20 c a b x 2 3ra3V 20 11 a3x3 2 0p 0x x4 l u i d s n = constant n s p = p(s) VB 0, is the highest pressure along the streamline 1 pB rV 20 22. As shown in Chapter 9, this excess pressure on the front of the sphere 1i.e., pB 7 02 contributes to the net drag force on the sphere. Note that the pressure gradient and pressure are directly proportional to the density of the fluid, a representation of the fact that the fluid inertia is proportional to its mass. (2) i n Incorrect raindrop shape The incorrect representation that raindrops are teardrop shaped is found nearly everywhere— from children’s books to weather maps on the Weather Channel. About the only time raindrops possess the typical teardrop shape is when they run down a windowpane. The actual shape of a falling raindrop is a function of the size of the drop and results from a balance between surface tension forces and the air pressure exerted on the falling drop. Small drops with a radius less than about 0.5 mm have a spherical shape because the surface tension effect (which is inversely proportional to drop Streamline (Ans) COMMENT The pressure at B, a stagnation point since where we have replaced s by x since the two coordinates are identical 1within an additive constant2 along streamline A–B. It follows that V 0V0s 6 0 along the streamline. The fluid slows down from V0 far ahead of the sphere to zero velocity on the “nose” of the sphere 1x a2. Thus, according to Eq. 1, to produce the given motion the pressure gradient along the streamline is F This variation is indicated in Fig. E3.1c. It is seen that the pressure increases in the direction of flow 1 0p0x 7 02 from point A to point B. The maximum pressure gradient 10.610 rV 20 a2 occurs just slightly ahead of the sphere 1x 1.205a2. It is the pressure gradient that slows the fluid down from VA V0 to VB 0 as shown in Fig. E3.1b. The pressure distribution along the streamline can be obtained by integrating Eq. 2 from p 0 1gage2 at x to pressure p at location x. The result, plotted in Fig. E3.1d, is t h e N e w s size) wins over the increased pressure, rV 202, caused by the motion of the drop and exerted on its bottom. With increasing size, the drops fall faster and the increased pressure causes the drops to flatten. A 2-mm drop, for example, is flattened into a hamburger bun shape. Slightly larger drops are actually concave on the bottom. When the radius is greater than about 4 mm, the depression of the bottom increases and the drop takes on the form of an inverted bag with an annular ring of water around its base. This ring finally breaks up into smaller drops. Equation 3.4 can be rearranged and integrated as follows. First, we note from Fig. 3.3 that along the streamline sin u dz ds. Also we can write V dVds 12d1V 2 2 ds. Finally, along the streamline the value of n is constant 1dn 02 so that dp 10p0s2 ds 10p0n2 dn 10p0s2 ds. Hence, as indicated by the figure in the margin, along a given streamline p(s, n) p(s) and 0p0s dpds. These ideas combined with Eq. 3.4 give the following result valid along a streamline g 2 dp dz 1 d1V 2 r ds ds 2 ds This simplifies to For steady, inviscid flow the sum of certain pressure, velocity, and elevation effects is constant along a streamline. dp 1 rd1V 2 2 g dz 0 2 1along a streamline2 (3.5) which, for constant acceleration of gravity, can be integrated to give dp 1 V 2 gz C r 2 1along a streamline2 (3.6) where C is a constant of integration to be determined by the conditions at some point on the streamline. 3.2 F ⴝ ma along a Streamline 107 In general it is not possible to integrate the pressure term because the density may not be constant and, therefore, cannot be removed from under the integral sign. To carry out this integration we must know specifically how the density varies with pressure. This is not always easily determined. For example, for a perfect gas the density, pressure, and temperature are related according to r pRT, where R is the gas constant. To know how the density varies with pressure, we must also know the temperature variation. For now we will assume that the density and specific weight are constant 1incompressible flow2. The justification for this assumption and the consequences of compressibility will be considered further in Section 3.8.1 and more fully in Chapter 11. With the additional assumption that the density remains constant 1a very good assumption for liquids and also for gases if the speed is “not too high”2, Eq. 3.6 assumes the following simple representation for steady, inviscid, incompressible flow. V3.2 Balancing ball p 12 rV 2 gz constant along streamline (3.7) This is the celebrated Bernoulli equation—a very powerful tool in fluid mechanics. In 1738 Daniel Bernoulli 11700–17822 published his Hydrodynamics in which an equivalent of this famous equation first appeared. To use it correctly we must constantly remember the basic assumptions used in its derivation: 112 viscous effects are assumed negligible, 122 the flow is assumed to be steady, 132 the flow is assumed to be incompressible, and 142 the equation is applicable along a streamline. In the derivation of Eq. 3.7, we assume that the flow takes place in a plane 1the x–z plane2. In general, this equation is valid for both planar and nonplanar 1three-dimensional2 flows, provided it is applied along the streamline. We will provide many examples to illustrate the correct use of the Bernoulli equation and will show how a violation of the basic assumptions used in the derivation of this equation can lead to erroneous conclusions. The constant of integration in the Bernoulli equation can be evaluated if sufficient information about the flow is known at one location along the streamline. V3.3 Flow past a biker E XAMPLE 3.2 The Bernoulli Equation GIVEN Consider the flow of air around a bicyclist moving through still air with velocity V0, as is shown in Fig. E3.2. V2 = 0 FIND Determine the difference in the pressure between points 112 and 122. (2) V1 = V0 (1) SOLUTION In a coordinate fixed to the ground, the flow is unsteady as the bicyclist rides by. However, in a coordinate system fixed to the bike, it appears as though the air is flowing steadily toward the bicyclist with speed V0. Since use of the Bernoulli equation is restricted to steady flows, we select the coordinate system fixed to the bike. If the assumptions of Bernoulli’s equation are valid 1steady, incompressible, inviscid flow2, Eq. 3.7 can be applied as follows along the streamline that passes through 112 and 122 p1 12 rV 21 gz1 p2 12 rV 22 gz2 We consider 112 to be in the free stream so that V1 V0 and 122 to be at the tip of the bicyclist’s nose and assume that z1 z2 and V2 0 1both of which, as is discussed in Section 3.4, are reasonable assumptions2. It follows that the pressure at 122 is greater than that at 112 by an amount p2 p1 12 rV 21 12 rV 20 (Ans) COMMENTS A similar result was obtained in Example 3.1 by integrating the pressure gradient, which was known because ■ Figure E3.2 the velocity distribution along the streamline, V1s2, was known. The Bernoulli equation is a general integration of F ma. To determine p2 p1, knowledge of the detailed velocity distribution is not needed—only the “boundary conditions” at 112 and 122 are required. Of course, knowledge of the value of V along the streamline is needed to determine the pressure at points between 112 and 122. Note that if we measure p2 p1 we can determine the speed, V0. As discussed in Section 3.5, this is the principle on which many velocity-measuring devices are based. If the bicyclist were accelerating or decelerating, the flow would be unsteady 1i.e., V0 constant2 and the above analysis would be incorrect since Eq. 3.7 is restricted to steady flow. 108 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation The difference in fluid velocity between two points in a flow field, V1 and V2, can often be controlled by appropriate geometric constraints of the fluid. For example, a garden hose nozzle is designed to give a much higher velocity at the exit of the nozzle than at its entrance where it is attached to the hose. As is shown by the Bernoulli equation, the pressure within the hose must be larger than that at the exit 1for constant elevation, an increase in velocity requires a decrease in pressure if Eq. 3.7 is valid2. It is this pressure drop that accelerates the water through the nozzle. Similarly, an airfoil is designed so that the fluid velocity over its upper surface is greater 1on the average2 than that along its lower surface. From the Bernoulli equation, therefore, the average pressure on the lower surface is greater than that on the upper surface. A net upward force, the lift, results. F ⴝ ma Normal to a Streamline 3.3 V3.4 Hydrocyclone separator In this section we will consider application of Newton’s second law in a direction normal to the streamline. In many flows the streamlines are relatively straight, the flow is essentially one-dimensional, and variations in parameters across streamlines 1in the normal direction2 can often be neglected when compared to the variations along the streamline. However, in numerous other situations valuable information can be obtained from considering F ma normal to the streamlines. For example, the devastating low-pressure region at the center of a tornado can be explained by applying Newton’s second law across the nearly circular streamlines of the tornado. We again consider the force balance on the fluid particle shown in Fig. 3.3 and the figure in the margin. This time, however, we consider components in the normal direction, n ˆ , and write Newton’s second law in this direction as a dFn n V δm r d V V2 dm V 2 r r (3.8) where g dFn represents the sum of n components of all the forces acting on the particle and dm is particle mass. We assume the flow is steady with a normal acceleration an V 2r, where r is the local radius of curvature of the streamlines. This acceleration is produced by the change in direction of the particle’s velocity as it moves along a curved path. We again assume that the only forces of importance are pressure and gravity. The component of the weight 1gravity force2 in the normal direction is dwn dw cos u g dV cos u To apply F ma normal to streamlines, the normal components of force are needed. If the streamline is vertical at the point of interest, u 90°, and there is no component of the particle weight normal to the direction of flow to contribute to its acceleration in that direction. If the pressure at the center of the particle is p, then its values on the top and bottom of the particle are p dpn and p dpn, where dpn 10p0n21dn22. Thus, if dFpn is the net pressure force on the particle in the normal direction, it follows that dFpn 1 p dpn 2 ds dy 1 p dpn 2 ds dy 2 dpn ds dy V3.5 Aircraft wing tip vortex 0p 0p ds dn dy dV 0n 0n Hence, the net force acting in the normal direction on the particle shown in Fig 3.3 is given by 0p a dFn dwn dFpn ag cos u 0n b dV (3.9) By combining Eqs. 3.8 and 3.9 and using the fact that along a line normal to the streamline cos u dzdn 1see Fig. 3.32, we obtain the following equation of motion along the normal direction: g 0p rV 2 dz dn 0n r (3.10a) Weight and/or pressure can produce curved streamlines. 109 F ⴝ ma Normal to a Streamline 3.3 The physical interpretation of Eq. 3.10 is that a change in the direction of flow of a fluid particle 1i.e., a curved path, r 6 q 2 is accomplished by the appropriate combination of pressure gradient and particle weight normal to the streamline. A larger speed or density or a smaller radius of curvature of the motion requires a larger force unbalance to produce the motion. For example, if gravity is neglected 1as is commonly done for gas flows2 or if the flow is in a horizontal 1dz dn 02 plane, Eq. 3.10 becomes 0p rV 2 0n r (3.10b) This indicates that the pressure increases with distance away from the center of curvature 1 0p0n is negative since rV 2r is positive—the positive n direction points toward the “inside” of the curved streamline2. Thus, the pressure outside a tornado 1typical atmospheric pressure2 is larger than it is near the center of the tornado 1where an often dangerously low partial vacuum may occur2. This pressure difference is needed to balance the centrifugal acceleration associated with the curved streamlines of the fluid motion. (See Fig. E6.6a in Section 6.5.3.) V3.6 Free vortex E XAMPLE 3.3 Pressure Variation Normal to a Streamline y y GIVEN Shown in Figs. E3.3a,b are two flow fields with circular streamlines. The velocity distributions are V1r2 1V0 /r0 2r and V1r2 1V0 r0 2 r V = (V0 /r0)r for case (a) r= for case (b) V = (V0 r0)/r x n x (a) where V0 is the velocity at r r0. (b) 6 FIND Determine the pressure distributions, p p(r), for each, (a) given that p p0 at r r0. 4 2 SOLUTION p – p0 We assume the flows are steady, inviscid, and incompressible with streamlines in the horizontal plane (dz/dn 0). Because the streamlines are circles, the coordinate n points in a direction opposite that of the radial coordinate, ∂/∂n ∂/∂r, and the radius of curvature is given by r r. Hence, Eq. 3.10b becomes 0p rV 2 r 0r 0 (b) 2 4 6 0 0.5 1 1.5 2 2.5 r/r0 (c) For case (a) this gives ■ Figure E3.3 0p r1V0 /r0 2 2r 0r whereas for case (b) it gives for case (a) and r1V0 r0 2 2 0p 0r r3 For either case the pressure increases as r increases since ∂p/∂r 0. Integration of these equations with respect to r, starting with a known pressure p p0 at r r0, gives p p0 1rV0222 3 1r/r0 2 2 14 ρV 02/2 (Ans) p p0 1rV0222 3 1 1r0 / r2 2 4 (Ans) for case (b). These pressure distributions are shown in Fig. E3.3c. COMMENT The pressure distributions needed to balance the centrifugal accelerations in cases (a) and (b) are not the same because the velocity distributions are different. In fact, for case (a) the 110 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation pressure increases without bound as r → q , whereas for case (b) the pressure approaches a finite value as r → q . The streamline patterns are the same for each case, however. Physically, case (a) represents rigid-body rotation (as obtained in a can of water on a turntable after it has been “spun up”) and The sum of pressure, elevation, and velocity effects is constant across streamlines. case (b) represents a free vortex (an approximation to a tornado, a hurricane, or the swirl of water in a drain, the “bathtub vortex”). See Fig. E6.6 for an approximation of this type of flow. If we multiply Eq. 3.10 by dn, use the fact that 0p0n dpdn if s is constant, and integrate across the streamline 1in the n direction2 we obtain dp r r dn gz constant across the streamline V2 (3.11) To complete the indicated integrations, we must know how the density varies with pressure and how the fluid speed and radius of curvature vary with n. For incompressible flow the density is constant and the integration involving the pressure term gives simply pr. We are still left, however, with the integration of the second term in Eq. 3.11. Without knowing the n dependence in V V1s, n2 and r r1s, n2 this integration cannot be completed. Thus, the final form of Newton’s second law applied across the streamlines for steady, inviscid, incompressible flow is pr r dn gz constant across the streamline V2 (3.12) As with the Bernoulli equation, we must be careful that the assumptions involved in the derivation of this equation are not violated when it is used. 3.4 Physical Interpretation In the previous two sections, we developed the basic equations governing fluid motion under a fairly stringent set of restrictions. In spite of the numerous assumptions imposed on these flows, a variety of flows can be readily analyzed with them. A physical interpretation of the equations will be of help in understanding the processes involved. To this end, we rewrite Eqs. 3.7 and 3.12 here and interpret them physically. Application of F ma along and normal to the streamline results in p 12 rV 2 gz constant along the streamline (3.13) and pr z p + 1 rV2 + gz 2 = constant p + r V dn + gz 2 = constant V2 dn gz constant across the streamline r (3.14) as indicated by the figure in the margin. The following basic assumptions were made to obtain these equations: The flow is steady, and the fluid is inviscid and incompressible. In practice none of these assumptions is exactly true. A violation of one or more of the above assumptions is a common cause for obtaining an incorrect match between the “real world” and solutions obtained by use of the Bernoulli equation. Fortunately, many “real-world” situations are adequately modeled by the use of Eqs. 3.13 and 3.14 because the flow is nearly steady and incompressible and the fluid behaves as if it were nearly inviscid. The Bernoulli equation was obtained by integration of the equation of motion along the “natural” coordinate direction of the streamline. To produce an acceleration, there must be an unbalance of the resultant forces, of which only pressure and gravity were considered to be important. Thus, 3.4 Physical Interpretation 111 there are three processes involved in the flow—mass times acceleration 1the rV 22 term2, pressure 1the p term2, and weight 1the gz term2. Integration of the equation of motion to give Eq. 3.13 actually corresponds to the work– energy principle often used in the study of dynamics [see any standard dynamics text 1Ref. 12]. This principle results from a general integration of the equations of motion for an object in a way very similar to that done for the fluid particle in Section 3.2. With certain assumptions, a statement of the work–energy principle may be written as follows: The work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle. The Bernoulli equation can be written in terms of heights called heads. E XAMPLE The Bernoulli equation is a mathematical statement of this principle. As the fluid particle moves, both gravity and pressure forces do work on the particle. Recall that the work done by a force is equal to the product of the distance the particle travels times the component of force in the direction of travel 1i.e., work F ⴢ d2. The terms gz and p in Eq. 3.13 are related to the work done by the weight and pressure forces, respectively. The remaining term, rV 22, is obviously related to the kinetic energy of the particle. In fact, an alternate method of deriving the Bernoulli equation is to use the first and second laws of thermodynamics 1the energy and entropy equations2, rather than Newton’s second law. With the appropriate restrictions, the general energy equation reduces to the Bernoulli equation. This approach is discussed in Section 5.4. An alternate but equivalent form of the Bernoulli equation is obtained by dividing each term of Eq. 3.7 by the specific weight, g, to obtain p V2 z constant on a streamline g 2g Each of the terms in this equation has the units of energy per weight 1LFF L2 or length 1feet, meters2 and represents a certain type of head. The elevation term, z, is related to the potential energy of the particle and is called the elevation head. The pressure term, p g, is called the pressure head and represents the height of a column of the fluid that is needed to produce the pressure p. The velocity term, V 22g, is the velocity head and represents the vertical distance needed for the fluid to fall freely 1neglecting friction2 if it is to reach velocity V from rest. The Bernoulli equation states that the sum of the pressure head, the velocity head, and the elevation head is constant along a streamline. 3.4 Kinetic, Potential, and Pressure Energy GIVEN Consider the flow of water from the syringe shown in Fig. E3.4a. As indicated in Fig. E3.4b, a force, F, applied to the (3) g plunger will produce a pressure greater than atmospheric at point 112 within the syringe. The water flows from the needle, point 122, with relatively high velocity and coasts up to point 132 at the top of its trajectory. FIND Discuss the energy of the fluid at points 112, 122, and 132 by using the Bernoulli equation. (2) Energy Type (1) Point 1 2 3 F (a) ■ Figure E3.4 (b) Kinetic RV 22 Potential Gz Pressure p Small Large Zero Zero Small Large Large Zero Zero 112 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation SOLUTION If the assumptions 1steady, inviscid, incompressible flow2 of the Bernoulli equation are approximately valid, it then follows that the flow can be explained in terms of the partition of the total energy of the water. According to Eq. 3.13, the sum of the three types of energy 1kinetic, potential, and pressure2 or heads 1velocity, elevation, and pressure2 must remain constant. The table above indicates the relative magnitude of each of these energies at the three points shown in the figure. The motion results in 1or is due to2 a change in the magnitude of each type of energy as the fluid flows from one location to another. An alternate way to consider this flow is as follows. The F l u i d s i E XAMPLE COMMENT If friction 1viscous2 effects were important, there would be an energy loss between 112 and 132 and for the given p1 the water would not be able to reach the height indicated in the figure. Such friction may arise in the needle 1see Chapter 8 on pipe flow2 or between the water stream and the surrounding air 1see Chapter 9 on external flow2. n Armed with a water jet for hunting Archerfish, known for their ability to shoot down insects resting on foliage, are like submarine water pistols. With their snout sticking out of the water, they eject a high-speed water jet at their prey, knocking it onto the water surface where they snare it for their meal. The barrel of their water pistol is formed by placing their tongue against a groove in the roof of their mouth to form a tube. By snapping shut their gills, water is forced through the tube and directed with the tip of The pressure variation across straight streamlines is hydrostatic. pressure gradient between 112 and 122 produces an acceleration to eject the water from the needle. Gravity acting on the particle between 122 and 132 produces a deceleration to cause the water to come to a momentary stop at the top of its flight. t h e N e w s their tongue. The archerfish can produce a pressure head within their gills large enough so that the jet can reach 2 to 3 m. However, it is accurate to only about 1 m. Recent research has shown that archerfish are very adept at calculating where their prey will fall. Within 100 milliseconds (a reaction time twice as fast as a human’s), the fish has extracted all the information needed to predict the point where the prey will hit the water. Without further visual cues it charges directly to that point. A net force is required to accelerate any mass. For steady flow the acceleration can be interpreted as arising from two distinct occurrences—a change in speed along the streamline and a change in direction if the streamline is not straight. Integration of the equation of motion along the streamline accounts for the change in speed 1kinetic energy change2 and results in the Bernoulli equation. Integration of the equation of motion normal to the streamline accounts for the centrifugal acceleration 1V 2 r2 and results in Eq. 3.14. When a fluid particle travels along a curved path, a net force directed toward the center of curvature is required. Under the assumptions valid for Eq. 3.14, this force may be either gravity or pressure, or a combination of both. In many instances the streamlines are nearly straight 1r q 2 so that centrifugal effects are negligible and the pressure variation across the streamlines is merely hydrostatic 1because of gravity alone2, even though the fluid is in motion. 3.5 Pressure Variation in a Flowing Stream GIVEN Water flows in a curved, undulating waterslide as shown in Fig. E3.5a. As an approximation to this flow, consider z (4) Free surface (p = 0) g (2) (3) h4-3 ^ n h2-1 C (1) A D B ■ Figure E3.5b the inviscid, incompressible, steady flow shown in Fig. E3.5b. From section A to B the streamlines are straight, while from C to D they follow circular paths. ■ Figure E3.5a (Photo courtesy of Schlitterbahn® Waterparks.) FIND Describe the pressure variation between points 112 and 122 and points 132 and 142. 3.5 Static, Stagnation, Dynamic, and Total Pressure 113 SOLUTION With the above assumptions and the fact that r for the portion from A to B, Eq. 3.14 becomes With p4 0 and z4 z3 h4–3 this becomes p3 gh4–3 r p gz constant z3 The constant can be determined by evaluating the known variables at the two locations using p2 0 1gage2, z1 0, and z2 h2–1 to give p1 p2 g1z2 z1 2 p2 gh2–1 (Ans) Note that since the radius of curvature of the streamline is infinite, the pressure variation in the vertical direction is the same as if the fluid were stationary. However, if we apply Eq. 3.14, between points 132 and 142, we obtain 1using dn dz2 p4 r z4 z3 3.5 z4 V2 1dz2 gz4 p3 gz3 r V2 dz r (Ans) To evaluate the integral, we must know the variation of V and r with z. Even without this detailed information we note that the integral has a positive value. Thus, the pressure at 132 is less than the hydrostatic value, gh4–3, by an amount equal to r zz34 1V 2 r2 dz. This lower pressure, caused by the curved streamline, is necessary to accelerate the fluid around the curved path. COMMENT Note that we did not apply the Bernoulli equation 1Eq. 3.132 across the streamlines from 112 to 122 or 132 to 142. Rather we used Eq. 3.14. As is discussed in Section 3.8, application of the Bernoulli equation across streamlines 1rather than along them2 may lead to serious errors. Static, Stagnation, Dynamic, and Total Pressure Each term in the Bernoulli equation can be interpreted as a form of pressure. A useful concept associated with the Bernoulli equation deals with the stagnation and dynamic pressures. These pressures arise from the conversion of kinetic energy in a flowing fluid into a “pressure rise” as the fluid is brought to rest 1as in Example 3.22. In this section we explore various results of this process. Each term of the Bernoulli equation, Eq. 3.13, has the dimensions of force per unit area—psi, lb ft2, Nm2. The first term, p, is the actual thermodynamic pressure of the fluid as it flows. To measure its value, one could move along with the fluid, thus being “static” relative to the moving fluid. Hence, it is normally termed the static pressure. Another way to measure the static pressure would be to drill a hole in a flat surface and fasten a piezometer tube as indicated by the location of point 132 in Fig. 3.4. As we saw in Example 3.5, the pressure in the flowing fluid at 112 is p1 gh3–1 p3, the same as if the fluid were static. From the manometer considerations of Chapter 2, we know that p3 gh4–3. Thus, since h3–1 h4–3 h it follows that p1 gh. The third term in Eq. 3.13, gz, is termed the hydrostatic pressure, in obvious regard to the hydrostatic pressure variation discussed in Chapter 2. It is not actually a pressure but does represent the change in pressure possible due to potential energy variations of the fluid as a result of elevation changes. The second term in the Bernoulli equation, rV 2 2, is termed the dynamic pressure. Its interpretation can be seen in Fig. 3.4 by considering the pressure at the end of a small tube inserted into the flow and pointing upstream. After the initial transient motion has died out, the liquid will fill the tube to a height of H as shown. The fluid in the tube, including that at its tip, 122, will be stationary. That is, V2 0, or point 122 is a stagnation point. If we apply the Bernoulli equation between points 112 and 122, using V2 0 and assuming that z1 z2, we find that p2 p1 12 rV 21 Open (4) H h h4-3 V h3-1 (3) ρ (1) (2) V1 = V V2 = 0 ■ Figure 3.4 Measurement of static and stagnation pressures. 114 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation Stagnation streamline Stagnation point Stagnation point (b) (a) ■ Figure 3.5 Stagnation points. Hence, the pressure at the stagnation point is greater than the static pressure, p1, by an amount rV 21 2, the dynamic pressure. It can be shown that there is a stagnation point on any stationary body that is placed into a flowing fluid. Some of the fluid flows “over” and some “under” the object. The dividing line 1or surface for three-dimensional flows2 is termed the stagnation streamline and terminates at the stagnation point on the body. 1See the photograph at the beginning of the chapter.2 For symmetrical objects 1such as a baseball2 the stagnation point is clearly at the tip or front of the object as shown in Fig. 3.5a. For other flows such as a water jet against a car as shown in Fig. 3.5b, there is also a stagnation point on the car. If elevation effects are neglected, the stagnation pressure, p rV 2 2, is the largest pressure obtainable along a given streamline. It represents the conversion of all of the kinetic energy into a pressure rise. The sum of the static pressure, hydrostatic pressure, and dynamic pressure is termed the total pressure, pT. The Bernoulli equation is a statement that the total pressure remains constant along a streamline. That is, V3.7 Stagnation point flow p 12 rV 2 gz pT constant along a streamline (3.15) Again, we must be careful that the assumptions used in the derivation of this equation are appropriate for the flow being considered. F l u i d s i n Pressurized eyes Our eyes need a certain amount of internal pressure in order to work properly, with the normal range being between 10 and 20 mm of mercury. The pressure is determined by a balance between the fluid entering and leaving the eye. If the pressure is above the normal level, damage may occur to the optic nerve where it leaves the eye, leading to a loss of the visual field termed glaucoma. Measurement of the pressure within the eye can be done by several different noninvasive types of instru- t h e N e w s ments, all of which measure the slight deformation of the eyeball when a force is put on it. Some methods use a physical probe that makes contact with the front of the eye, applies a known force, and measures the deformation. One noncontact method uses a calibrated “puff” of air that is blown against the eye. The stagnation pressure resulting from the air blowing against the eyeball causes a slight deformation, the magnitude of which is correlated with the pressure within the eyeball. (See Problem 3.28.) Knowledge of the values of the static and stagnation pressures in a fluid implies that the fluid speed can be calculated. This is the principle on which the Pitot-static tube is based [H. de Pitot (1695–1771)]. As shown in Fig. 3.6, two concentric tubes are attached to two pressure gages 1or a differential gage2 so that the values of p3 and p4 1or the difference p3 p42 can be determined. The center tube measures the stagnation pressure at its open tip. If elevation changes are negligible, p3 p 12 rV 2 3.5 115 Static, Stagnation, Dynamic, and Total Pressure (3) (2) (1) (4) (1) V p (2) (a) (b) ■ Figure 3.6 The Pitot-static tube. where p and V are the pressure and velocity of the fluid upstream of point 122. The outer tube is made with several small holes at an appropriate distance from the tip so that they measure the static pressure. If the effect of the elevation difference between 112 and 142 is negligible, then p4 p1 p By combining these two equations we see that Pitot-static tubes measure fluid velocity by converting velocity into pressure. p3 p4 12 rV 2 which can be rearranged to give V 221 p3 p4 2 r (3.16) The actual shape and size of Pitot-static tubes vary considerably. A typical Pitot-static probe used to determine aircraft airspeed is shown in Fig. 3.7. (See Fig. E3.6a also.) F l u i d s i n t Bugged and plugged Pitot tubes Although a Pitot tube is a simple device for measuring aircraft airspeed, many airplane accidents have been caused by inaccurate Pitot tube readings. Most of these accidents are the result of having one or more of the holes blocked and, therefore, not indicating the correct pressure (speed). Usually this is discovered during takeoff when time to resolve the issue is short. The two most common causes for such a blockage are either that the pilot (or ground crew) has forgotten to remove the protective Pitot tube cover or that insects have built h e N e w s their nest within the tube where the standard visual check cannot detect it. The most serious accident (in terms of number of fatalities) caused by a blocked Pitot tube involved a Boeing 757 and occurred shortly after takeoff from Puerto Plata in the Dominican Republic. Incorrect airspeed data were automatically fed to the computer, causing the autopilot to change the angle of attack and the engine power. The flight crew became confused by the false indications; the aircraft stalled and then plunged into the Caribbean Sea killing all aboard. Four static pressure ports Heated outer case Stagnation pressure port V3.8 Airspeed indicator Mounting flange 2 in. Stagnation pressure fitting Static pressure fitting Heater leads (a) ■ Figure 3.7 Airplane Pitot-static probe. (a) Schematic, (b) Photograph. (Photograph courtesy of Aero-Instruments Co., LLC.) (b) 116 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation E XAMPLE 3.6 Pitot-Static Tube GIVEN An airplane flies 200 mph at an elevation of 10,000 ft in a standard atmosphere as shown in Fig. E3.6a. FIND Determine the pressure at point 112 far ahead of the air- plane, the pressure at the stagnation point on the nose of the airplane, point 122, and the pressure difference indicated by a Pitotstatic probe attached to the fuselage. SOLUTION (2) (1) V1 = 200 mph Pitot-static tube ■ Figure E3.6a (Photo courtesy of Hawker Beechcraft Corporation.) From Table C.1 we find that the static pressure at the altitude given is p1 1456 lbft2 1abs2 10.11 psia 1 Also the density is r 0.001756 slugft3. If the flow is steady, inviscid, and incompressible and elevation changes are neglected, Eq. 3.13 becomes 0.8 p1/p2 (Ans) rV 21 p2 p1 2 p2 1456 lbft2 10.001756 slugsft3 212932 ft2s2 2 2 11456 75.42 lbft2 1abs2 Hence, in terms of gage pressure p2 75.4 lbft2 0.524 psi (Ans) Thus, the pressure difference indicated by the Pitot-static tube is rV 21 0.524 psi 2 (Ans) COMMENTS Note that it is very easy to obtain incorrect results by using improper units. Do not add lb in.2 and lb ft2. Recall that 1slugft3 21ft2 s2 2 1slug # fts2 2 1ft2 2 lbft2. Accurate measurement of static pressure requires great care. 0.6 0.4 0.2 With V1 200 mph 293 fts and V2 0 1since the coordinate system is fixed to the airplane2, we obtain p2 p1 (200 mph, 0.951) 0 0 100 200 300 400 500 600 V1, mph ■ Figure E3.6b It was assumed that the flow is incompressible—the density remains constant from 112 to 122. However, since r pRT, a change in pressure 1or temperature2 will cause a change in density. For this relatively low speed, the ratio of the absolute pressures is nearly unity 3i.e., p1p2 110.11 psia2 110.11 0.524 psia2 0.951 4, so that the density change is negligible. However, by repeating the calculations for various values of the speed, V1, the results shown in Fig. E3.6b are obtained. Clearly at the 500- to 600-mph speeds normally flown by commercial airliners, the pressure ratio is such that density changes are important. In such situations it is necessary to use compressible flow concepts to obtain accurate results. 1See Section 3.8.1 and Chapter 11.2 The Pitot-static tube provides a simple, relatively inexpensive way to measure fluid speed. Its use depends on the ability to measure the static and stagnation pressures. Care is needed to obtain these values accurately. For example, an accurate measurement of static pressure requires that none of the fluid’s kinetic energy be converted into a pressure rise at the point of measurement. This requires a smooth hole with no burrs or imperfections. As indicated in Fig. 3.8, such imperfections can cause the measured pressure to be greater or less than the actual static pressure. Also, the pressure along the surface of an object varies from the stagnation pressure at its stagnation point to values that may be less than the free stream static pressure. A typical pressure variation for a Pitot-static tube is indicated in Fig. 3.9. Clearly it is important that V p V p V p (1) (1) (1) p1 > p p1 < p p1 = p ■ Figure 3.8 Incorrect and correct design of static pressure taps. 3.6 Examples of Use of the Bernoulli Equation 117 p (2) V (1) Stagnation pressure at tip Tube Stagnation pressure on stem Stem (1) (2) 0 Static pressure ■ Figure 3.9 Typical pressure distribution along a Pitotstatic tube. β V p θ (3) (2) β (1) If θ = 0 p1 = p3 = p _ ρ V2 p2 = p + 1 2 ■ Figure 3.10 Cross section of a directional-finding Pitot-static tube. the pressure taps be properly located to ensure that the pressure measured is actually the static pressure. In practice it is often difficult to align the Pitot-static tube directly into the flow direction. Any misalignment will produce a nonsymmetrical flow field that may introduce errors. Typically, yaw angles up to 12 to 20° 1depending on the particular probe design2 give results that are less than 1% in error from the perfectly aligned results. Generally it is more difficult to measure static pressure than stagnation pressure. One method of determining the flow direction and its speed 1thus the velocity2 is to use a directional-finding Pitot tube as is illustrated in Fig. 3.10. Three pressure taps are drilled into a small circular cylinder, fitted with small tubes, and connected to three pressure transducers. The cylinder is rotated until the pressures in the two side holes are equal, thus indicating that the center hole points directly upstream. The center tap then measures the stagnation pressure. The two side holes are located at a specific angle 1b 29.5°2 so that they measure the static pressure. The speed is then obtained from V 321 p2 p1 2 r4 12. The above discussion is valid for incompressible flows. At high speeds, compressibility becomes important 1the density is not constant2 and other phenomena occur. Some of these ideas are discussed in Section 3.8, while others 1such as shockwaves for supersonic Pitot-tube applications2 are discussed in Chapter 11. The concepts of static, dynamic, stagnation, and total pressure are useful in a variety of flow problems. These ideas are used more fully in the remainder of the book. 3.6 Examples of Use of the Bernoulli Equation In this section we illustrate various additional applications of the Bernoulli equation. Between any two points, 112 and 122, on a streamline in steady, inviscid, incompressible flow the Bernoulli equation can be applied in the form p1 12 rV 21 gz1 p2 12 rV 22 gz2 (3.17) Obviously, if five of the six variables are known, the remaining one can be determined. In many instances it is necessary to introduce other equations, such as the conservation of mass. Such considerations will be discussed briefly in this section and in more detail in Chapter 5. 118 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation (1) V3.9 Flow from a tank h z (3) (2) (2) d H (4) V ■ Figure 3.11 Vertical (5) flow from a tank. 3.6.1 Free Jets One of the oldest equations in fluid mechanics deals with the flow of a liquid from a large reservoir. A modern version of this type of flow involves the flow of coffee from a coffee urn as indicated by the figure in the margin. The basic principles of this type of flow are shown in Fig. 3.11 where a jet of liquid of diameter d flows from the nozzle with velocity V. 1A nozzle is a device shaped to accelerate a fluid.2 Application of Eq. 3.17 between points 112 and 122 on the streamline shown gives gh 12 rV 2 We have used the facts that z1 h, z2 0, the reservoir is large 1V1 02 and open to the atmosphere 1 p1 0 gage2, and the fluid leaves as a “free jet” 1 p2 02. Thus, we obtain V V The exit pressure for an incompressible fluid jet is equal to the surrounding pressure. B 2 gh 12gh r (3.18) which is the modern version of a result obtained in 1643 by Torricelli 11608–16472, an Italian physicist. The fact that the exit pressure equals the surrounding pressure 1 p2 02 can be seen by applying F ma, as given by Eq. 3.14, across the streamlines between 122 and 142. If the streamlines at the tip of the nozzle are straight 1r q 2, it follows that p2 p4. Since 142 is on the surface of the jet, in contact with the atmosphere, we have p4 0. Thus, p2 0 also. Since 122 is an arbitrary point in the exit plane of the nozzle, it follows that the pressure is atmospheric across this plane. Physically, since there is no component of the weight force or acceleration in the normal 1horizontal2 direction, the pressure is constant in that direction. Once outside the nozzle, the stream continues to fall as a free jet with zero pressure throughout 1 p5 02 and as seen by applying Eq. 3.17 between points 112 and 152, the speed increases according to V 12g 1h H2 V=0 h V = 2gh where, as shown in Fig. 3.11, H is the distance the fluid has fallen outside the nozzle. Equation 3.18 could also be obtained by writing the Bernoulli equation between points 132 and 142 using the fact that z4 0, z3 /. Also, V3 0 since it is far from the nozzle, and from hydrostatics, p3 g1h /2. As learned in physics or dynamics and illustrated in the figure in the margin, any object dropped from rest that falls through a distance h in a vacuum will obtain the speed V 12gh, the same as the water leaving the spout of the watering can shown in the figure in the margin on the next page. This is consistent with the fact that all of the particle’s potential energy is converted to kinetic energy, provided viscous 1friction2 effects are negligible. In terms of heads, the elevation head at point 112 is converted into the velocity head at point 122. Recall that for the case shown in Fig. 3.11 the pressure is the same 1atmospheric2 at points 112 and 122. For the horizontal nozzle of Fig. 3.12a, the velocity of the fluid at the centerline, V2, will be slightly greater than that at the top, V1, and slightly less than that at the bottom, V3, due to the differences in elevation. In general, d h as shown in Fig. 3.12b and we can safely use the centerline velocity as a reasonable “average velocity.” 3.6 h Examples of Use of the Bernoulli Equation d 119 h (1) (2) d (3) (a) (b) ■ Figure 3.12 Horizontal flow from a tank. (1) dh dj a (2) (3) a ■ Figure 3.13 Vena contracta effect for a sharp-edged orifice. (1) h V = √2gh (2) The diameter of a fluid jet is often smaller than that of the hole from which it flows. F l u If the exit is not a smooth, well-contoured nozzle, but rather a flat plate as shown in Fig. 3.13, the diameter of the jet, dj, will be less than the diameter of the hole, dh. This phenomenon, called a vena contracta effect, is a result of the inability of the fluid to turn the sharp 90° corner indicated by the dotted lines in the figure. Since the streamlines in the exit plane are curved 1r 6 q 2, the pressure across them is not constant. It would take an infinite pressure gradient across the streamlines to cause the fluid to turn a “sharp” corner 1r 02. The highest pressure occurs along the centerline at 122 and the lowest pressure, p1 p3 0, is at the edge of the jet. Thus, the assumption of uniform velocity with straight streamlines and constant pressure is not valid at the exit plane. It is valid, however, in the plane of the vena contracta, section a–a. The uniform velocity assumption is valid at this section provided dj h, as is discussed for the flow from the nozzle shown in Fig. 3.12. The vena contracta effect is a function of the geometry of the outlet. Some typical configurations are shown in Fig. 3.14 along with typical values of the experimentally obtained contraction coefficient, Cc AjAh, where Aj and Ah are the areas of the jet at the vena contracta and the area of the hole, respectively. i d s i n Cotton candy, glass wool, and steel wool Although cotton candy and glass wool insulation are made of entirely different materials and have entirely different uses, they are made by similar processes. Cotton candy, invented in 1897, consists of sugar fibers. Glass wool, invented in 1938, consists of glass fibers. In a cotton candy machine, sugar is melted and then forced by centrifugal action to flow through numerous tiny orifices in a spinning “bowl.” Upon emerging, the thin streams of liquid sugar cool very quickly and become solid threads that are collected on a stick or cone. Making glass wool in- t h e N e w s sulation is somewhat more complex, but the basic process is similar. Liquid glass is forced through tiny orifices and emerges as very fine glass streams that quickly solidify. The resulting intertwined flexible fibers, glass wool, form an effective insulation material because the tiny air “cavities” between the fibers inhibit air motion. Although steel wool looks similar to cotton candy or glass wool, it is made by an entirely different process. Solid steel wires are drawn over special cutting blades that have grooves cut into them so that long, thin threads of steel are peeled off to form the matted steel wool. 120 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation dj dh (a) Knife edge CC = 0.61 CC = 1.0 CC = A j /A h = (dj /dh)2 (b) Well rounded CC = 0.50 CC = 0.61 (c) Sharp edge (d) Re-entrant ■ Figure 3.14 Typical flow patterns and contraction coefficients for various round exit configurations. (a) Knife edge, (b) Well rounded, (c) Sharp edge, (d) Re-entrant. 3.6.2 Confined Flows The continuity equation states that mass cannot be created or destroyed. V2 = 2V1 A2 (2) r1A1V1 r2A2V2 Q A1 = 2A2 V1 In many cases the fluid is physically constrained within a device so that its pressure cannot be prescribed a priori as was done for the free jet examples above. Such cases include nozzles and pipes of variable diameter for which the fluid velocity changes because the flow area is different from one section to another. For these situations it is necessary to use the concept of conservation of mass 1the continuity equation2 along with the Bernoulli equation. The derivation and use of this equation are discussed in detail in Chapters 4 and 5. For the needs of this chapter we can use a simplified form of the continuity equation obtained from the following intuitive arguments. Consider a fluid flowing through a fixed volume 1such as a syringe2 that has one inlet and one outlet as shown in Fig. 3.15a. If the flow is steady so that there is no additional accumulation of fluid within the volume, the rate at which the fluid flows into the volume must equal the rate at which it flows out of the volume 1otherwise, mass would not be conserved2. # # The mass flowrate from an outlet, m 1slugss or kgs2, is given by m rQ, where Q 1ft3s or m3s2 is the volume flowrate. If the outlet area is A and the fluid flows across this area 1normal to the area2 with an average velocity V, then the volume of the fluid crossing this area in a time interval dt is VA dt, equal to that in a volume of length V dt and cross-sectional area A 1see Fig. 3.15b2. Hence, the vol# ume flowrate 1volume per unit time2 is Q VA. Thus, m rVA. To conserve mass, the inflow rate # # must equal the outflow rate. If the inlet is designated as 112 and the outlet as 122, it follows that m1 m2. Thus, conservation of mass requires If the density remains constant, then r1 r2, and the above becomes the continuity equation for incompressible flow A1V1 A2V2, or Q1 Q2 (3.19) (1) For example, if as shown by the figure in the margin the outlet flow area is one-half the size of the inlet flow area, it follows that the outlet velocity is twice that of the inlet velocity, since 3.6 Examples of Use of the Bernoulli Equation 121 V1 V2 (2) (1) V1 δ t V2 δ t Volume = V1 δ t A1 V2 V1 (2) Volume = V2 δ t A2 (1) Same parcel at t = δ t Fluid parcel at t = 0 ■ Figure 3.15 (a) Flow through a syringe. (b) Steady flow into and out of a volume. V2 A1V1A2 2V1. Use of the Bernoulli equation and the flowrate equation 1continuity equation2 is demonstrated by Example 3.7. E XAMPLE 3.7 Flow from a Tank—Gravity Driven GIVEN A stream of refreshing beverage of diameter d 0.01 m flows steadily from the cooler of diameter D 0.20 m as shown in Figs. E3.7a and b. FIND Determine the flowrate, Q, from the bottle into the cooler if the depth of beverage in the cooler is to remain constant at h 0.20 m. 1.10 Q (1) Q/Q0 1.05 D = 0.20 m h = 0.20 m (2) (3) d = 0.01 m (a) ■ Figure E3.7 (b) (0.05, 1.000003) 1.00 0 0.2 0.4 d/ D (c) 0.6 0.8 122 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation SOLUTION For steady, inviscid, incompressible flow, the Bernoulli equation applied between points 112 and 122 is p1 1 2 2 rV 1 gz1 p2 1 2 2 rV 2 gz2 (1) With the assumptions that p1 p2 0, z1 h, and z2 0, Eq. 1 becomes 1 2 2V1 gh 12 V 22 (2) Although the liquid level remains constant 1h constant2, there is an average velocity, V1, across section 112 because of the flow from the tank. From Eq. 3.19 for steady incompressible flow, conservation of mass requires Q1 Q2, where Q AV. Thus, A1V1 A2V2, or p 2 p D V1 d 2V2 4 4 Hence, d 2 V1 a b V2 D (3) Equations 1 and 3 can be combined to give V2 2gh B 1 1dD2 4 219.81 ms2 2 10.20 m2 B 1 10.01 m0.20 m2 4 1.98 ms Thus, Q A1V1 A2V2 p 10.01 m2 2 11.98 ms2 4 1.56 10 4 m3s (Ans) COMMENTS In this example we have not neglected the kinetic energy of the water in the tank 1V1 02. If the tank diameter is large compared to the jet diameter 1D d2, Eq. 3 indicates that V1 V2 and the assumption that V1 0 would be reasonable. The error associated with this assumption can be seen by calculating the ratio of the flowrate assuming V1 0, denoted Q, to that assuming V1 0, denoted Q0. This ratio, written as 22gh 31 1dD2 4 4 Q V2 1 Q0 V2 0 D 22gh 21 1dD2 4 is plotted in Fig. E3.7c. With 0 6 dD 6 0.4 it follows that 1 6 QQ0 1.01, and the error in assuming V1 0 is less than 1%. For this example with d/D 0.01 m/0.20 m 0.05, it follows that Q/Q0 1.000003. Thus, it is often reasonable to assume V1 0. Note that this problem was solved using points (1) and (2) located at the free surface and the exit of the pipe, respectively. Although this was convenient (because most of the variables are known at those points), other points could be selected and the same result would be obtained. For example, consider points (1) and (3) as indicated in Fig. E3.7b. At (3), located sufficiently far from the tank exit, V3 0 and z3 z2 0. Also, p3 h since the pressure is hydrostatic sufficiently far from the exit. Use of this information in the Bernoulli equation applied between (3) and (2) gives the exact same result as obtained using it between (1) and (2). The only difference is that the elevation head, z1 h, has been interchanged with the pressure head at (3), p3/ h. The fact that a kinetic energy change is often accompanied by a change in pressure is shown by Example 3.8. E XAMPLE 3.8 Flow from a Tank—Pressure Driven GIVEN Air flows steadily from a tank, through a hose of diameter D 0.03 m, and exits to the atmosphere from a nozzle of diameter d 0.01 m as shown in Fig. E3.8. The pressure in the tank remains constant at 3.0 kPa 1gage2 and the atmospheric conditions are standard temperature and pressure. FIND Determine p1 = 3.0 kPa (1) Air D = 0.03 m d = 0.01 m Q (2) (3) ■ Figure E3.8a (a) the flowrate and (b) the pressure in the hose. SOLUTION (a) If the flow is assumed steady, inviscid, and incompressible, we can apply the Bernoulli equation along the streamline from (1) to (2) to (3) as p1 12 rV 21 gz1 p2 12 rV 22 gz2 p3 12 rV 23 gz3 With the assumption that z1 z2 z3 1horizontal hose2, V1 0 1large tank2, and p3 0 1free jet2, this becomes V3 2p1 B r and p2 p1 12 rV 22 (1) The density of the air in the tank is obtained from the perfect gas law, using standard absolute pressure and temperature, as 3.6 r p1 RT1 13.0 1012 kN m2 103 NkN 3000 1286.9 N ᝽ mkg ᝽ K2115 2732K 1.26 kgm3 V3 B 1.26 kgm3 (0.01 m, 2963 N/m2) p2, N/m2 2000 Thus, we find that 213.0 103 Nm2 2 123 Examples of Use of the Bernoulli Equation 1000 69.0 ms 0 or p p Q A3 V3 d 2 V3 10.01 m2 2 169.0 ms2 4 4 0.00542 m3 s 0 0.01 0.02 0.03 d, m ■ Figure E3.8b (Ans) 0.05 COMMENT Note that the value of V3 is determined strictly by 0.04 Q, m3/s the value of p1 1and the assumptions involved in the Bernoulli equation2, independent of the “shape” of the nozzle. The pressure head within the tank, p1 g 13.0 kPa2 19.81 ms2 211.26 kgm3 2 243 m, is converted to the velocity head at the exit, V 222g 169.0 ms2 2 12 9.81 ms2 2 243 m. Although we used gage pressure in the Bernoulli equation 1 p3 02, we had to use absolute pressure in the perfect gas law when calculating the density. 0.03 0.02 0.01 (b) The pressure within the hose can be obtained from Eq. 1 and the continuity equation 1Eq. 3.192 0 (0.01 m, 0.00542 m3/s) 0 0.01 0.02 0.03 d, m A2V2 A3V3 ■ Figure E3.8c Hence, d V2 A3V3 A2 a b V3 D 0.01 m 2 a b 169.0 ms2 7.67 ms 0.03 m 2 and from Eq. 1 p2 3.0 103 Nm2 12 11.26 kgm3 217.67 ms2 2 13000 37.12Nm2 2963 Nm2 (Ans) COMMENTS In the absence of viscous effects, the pressure throughout the hose is constant and equal to p2. Physically, the decreases in pressure from p1 to p2 to p3 accelerate the air and increase its kinetic energy from zero in the tank to an intermediate value in the hose and finally to its maximum value at the nozzle exit. Since the air velocity in the nozzle exit is nine F l u i d s i n Hi-tech inhaler The term inhaler often brings to mind a treatment for asthma or bronchitis. Work is underway to develop a family of inhalation devices that can do more than treat respiratory ailments. They will be able to deliver medication for diabetes and other conditions by spraying it to reach the bloodstream through the lungs. The concept is to make the spray droplets fine enough to penetrate to the lungs’ tiny sacs, the alveoli, where exchanges between blood and the outside world take place. This is accomplished by use of a laser-machined nozzle containing an array of very fine holes that cause the liquid to divide into a mist of times that in the hose, most of the pressure drop occurs across the nozzle 1 p1 3000 Nm2, p2 2963 N m2, and p3 02. Since the pressure change from 112 to 132 is not too great 3 i.e., in terms of absolute pressure 1p1 p3 2 p1 3.0101 0.034, it follows from the perfect gas law that the density change is also not significant. Hence, the incompressibility assumption is reasonable for this problem. If the tank pressure were considerably larger or if viscous effects were important, application of the Bernoulli equation to this situation would be incorrect. By repeating the calculations for various nozzle diameters, d, the results shown in Figs. E3.8b,c are obtained. The flowrate increases as the nozzle is opened (i.e., larger d). Note that if the nozzle diameter is the same as that of the hose (d 0.03 m), the pressure throughout the hose is atmospheric (zero gage). t h e N e w s micron-scale droplets. The device fits the hand and accepts a disposable strip that contains the medicine solution sealed inside a blister of laminated plastic and the nozzle. An electrically actuated piston drives the liquid from its reservoir through the nozzle array and into the respiratory system. To take the medicine, the patient breathes through the device and a differential pressure transducer in the inhaler senses when the patient’s breathing has reached the best condition for receiving the medication. At that point, the piston is automatically triggered. 124 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation In many situations the combined effects of kinetic energy, pressure, and gravity are important. Example 3.9 illustrates this. E XAMPLE 3.9 Flow in a Variable Area Pipe GIVEN Water flows through a pipe reducer as is shown in Fig. E3.9. The static pressures at 112 and 122 are measured by the inverted U-tube manometer containing oil of specific gravity, SG, less than one. SG h FIND Determine the manometer reading, h. D2 SOLUTION (2) With the assumptions of steady, inviscid, incompressible flow, the Bernoulli equation can be written as p1 12 rV 21 gz1 p2 12 rV 22 gz2 The continuity equation 1Eq. 3.192 provides a second relationship between V1 and V2 if we assume the velocity profiles are uniform at those two locations and the fluid incompressible: Q A1V1 A2V2 p1 p2 g1z2 z1 2 2 (1) This pressure difference is measured by the manometer and can be determined by using the pressure–depth ideas developed in Chapter 2. Thus, p1 g 1z2 z1 2 g/ gh SG gh g/ p2 (2) As discussed in Chapter 2, this pressure difference is neither merely gh nor g1h z1 z2 2. Equations 1 and 2 can be combined to give the desired result as follows: V3.10 Venturi channel Water θ or since V2 QA2 h 1QA2 2 2 1 1A2A1 2 2 2g11 SG2 (Ans) COMMENT The difference in elevation, z1 z2, was not or 11 SG2gh (1) ■ Figure E3.9 31 1A2A1 2 4 p1 p2 g1z2 z1 2 11 SG2gh γ D1 By combining these two equations we obtain 1 2 2 rV 2 z2 – z1 needed because the change in elevation term in the Bernoulli equation exactly cancels the elevation term in the manometer equation. However, the pressure difference, p1 p2, depends on the angle u, because of the elevation, z1 z2, in Eq. 1. Thus, for a given flowrate, the pressure difference, p1 p2, as measured by a pressure gage would vary with u, but the manometer reading, h, would be independent of u. A2 2 1 rV22 c 1 a b d 2 A1 In general, an increase in velocity is accompanied by a decrease in pressure. For example, the velocity of the air flowing over the top surface of an airplane wing is, on the average, faster than that flowing under the bottom surface. Thus, the net pressure force is greater on the bottom than on the top—the wing generates a lift. If the differences in velocity are considerable, the differences in pressure can also be considerable. For flows of gases, this may introduce compressibility effects as discussed in Section 3.8 and Chapter 11. For flows of liquids, this may result in cavitation, a potentially dangerous situation that results when the liquid pressure is reduced to the vapor pressure and the liquid “boils.” 3.6 Examples of Use of the Bernoulli Equation 125 Q (1) (2) (3) p (Absolute pressure) Small Q Moderate Q pv Large Q Incipient cavitation 0 x ■ Figure 3.16 Pressure variation and cavitation in a variable area pipe. Cavitation occurs when the pressure is reduced to the vapor pressure. Cavitation can cause damage to equipment. E XAMPLE ■ Figure 3.17 Tip cavitation from a propeller. (Photograph courtesy of The Pennsylvania State University, Applied Research Laboratory, Garfield Thomas Water Tunnel.) As discussed in Chapter 1, the vapor pressure, pv, is the pressure at which vapor bubbles form in a liquid. It is the pressure at which the liquid starts to boil. Obviously this pressure depends on the type of liquid and its temperature. For example, water, which boils at 212 °F at standard atmospheric pressure, 14.7 psia, boils at 80 °F if the pressure is 0.507 psia. That is, pv 0.507 psia at 80 °F and pv 14.7 psia at 212 °F. 1See Tables B.1 and B.2.2 One way to produce cavitation in a flowing liquid is noted from the Bernoulli equation. If the fluid velocity is increased 1for example, by a reduction in flow area as shown in Fig. 3.162, the pressure will decrease. This pressure decrease 1needed to accelerate the fluid through the constriction2 can be large enough so that the pressure in the liquid is reduced to its vapor pressure. A simple example of cavitation can be demonstrated with an ordinary garden hose. If the hose is “kinked,” a restriction in the flow area in some ways analogous to that shown in Fig. 3.16 will result. The water velocity through this restriction will be relatively large. With a sufficient amount of restriction the sound of the flowing water will change—a definite “hissing” sound is produced. This sound is a result of cavitation. In such situations boiling occurs 1though the temperature need not be high2, vapor bubbles form, and then they collapse as the fluid moves into a region of higher pressure 1lower velocity2. This process can produce dynamic effects 1imploding2 that cause very large pressure transients in the vicinity of the bubbles. Pressures as large as 100,000 psi 1690 MPa2 are believed to occur. If the bubbles collapse close to a physical boundary they can, over a period of time, cause damage to the surface in the cavitation area. Tip cavitation from a propeller is shown in Fig. 3.17. In this case the high-speed rotation of the propeller produced a corresponding low pressure on the propeller. Obviously, proper design and use of equipment are needed to eliminate cavitation damage. 3.10 Siphon and Cavitation GIVEN A liquid can be siphoned from a container as shown in Fig. E3.10a, provided the end of the tube, point (3), is below the free surface in the container, point (1), and the maximum elevation of the tube, point (2), is “not too great.” Consider water at 60 °F being siphoned from a large tank through a constant-diameter hose as shown in Fig. E3.10b. The end of the siphon is 5 ft below the bottom of the tank, and the atmospheric pressure is 14.7 psia. FIND Determine the maximum height of the hill, H, over which the water can be siphoned without cavitation occurring. 126 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation SOLUTION (2) If the flow is steady, inviscid, and incompressible we can apply the Bernoulli equation along the streamline from 112 to 122 to 132 as follows: p1 ⫹ 12 rV 21 ⫹ gz1 ⫽ p2 ⫹ 12 rV 22 ⫹ gz2 ⫽ p3 ⫹ 12 rV 23 ⫹ gz3 (1) (1) With the tank bottom as the datum, we have z1 ⫽ 15 ft, z2 ⫽ H, and z3 ⫽ ⫺5 ft. Also, V1 ⫽ 0 1large tank2, p1 ⫽ 0 1open tank2, p3 ⫽ 0 1free jet2, and from the continuity equation A2V2 ⫽ A3V3, or because the hose is constant diameter, V2 ⫽ V3. Thus, the speed of the fluid in the hose is determined from Eq. 1 to be (3) V3 ⫽ 22g1z1 ⫺ z3 2 ⫽ 22132.2 ftⲐs2 2 315 ⫺ 1⫺52 4 ft ⫽ 35.9 ftⲐs ⫽ V2 Use of Eq. 1 between points 112 and 122 then gives the pressure p2 at the top of the hill as ■ Figure E3.10a p2 ⫽ p1 ⫹ 12 rV 21 ⫹ gz1 ⫺ 12 rV 22 ⫺ gz2 ⫽ g1z1 ⫺ z2 2 ⫺ 12 rV 22 (2) From Table B.1, the vapor pressure of water at 60 °F is 0.256 psia. Hence, for incipient cavitation the lowest pressure in the system will be p ⫽ 0.256 psia. Careful consideration of Eq. 2 and Fig. E3.10b will show that this lowest pressure will occur at the top of the hill. Since we have used gage pressure at point 112 1 p1 ⫽ 02, we must use gage pressure at point 122 also. Thus, p2 ⫽ 0.256 ⫺ 14.7 ⫽ ⫺14.4 psi and Eq. 2 gives (2) (1) H 15 ft Water 1⫺14.4 lbⲐin.2 21144 in.2 Ⲑft2 2 ⫽ 162.4 lbⲐft3 2115 ⫺ H2ft ⫺ 12 11.94 slugsⲐft3 2135.9 ftⲐs2 2 or 5 ft (3) ■ Figure E3.10b (Ans) 40 35 30 25 20 10 5 0 Carbon tet 15 Gasoline sure throughout 1p2 ⫽ 0.256 psia and p1 ⫽ 14.7 psia2 and obtained the same result. The lower the elevation of point 132, the larger the flowrate and, therefore, the smaller the value of H allowed. We could also have used the Bernoulli equation between 122 and 132, with V2 ⫽ V3, to obtain the same value of H. In this case it would not have been necessary to determine V2 by use of the Bernoulli equation between 112 and 132. The above results are independent of the diameter and length of the hose 1provided viscous effects are not important2. Proper design of the hose 1or pipe2 is needed to ensure that it will not collapse due to the large pressure difference 1vacuum2 between the inside and outside of the hose. Water COMMENTS Note that we could have used absolute pres- By using the fluid properties listed in Table 1.5 and repeating the calculations for various fluids, the results shown in Fig. E3.10c are obtained. The value of H is a function of both the specific weight of the fluid, g, and its vapor pressure, pv. Alcohol For larger values of H, vapor bubbles will form at point 122 and the siphon action may stop. H, ft H ⫽ 28.2 ft Fluid ■ Figure E3.10c 3.6.3 Flowrate Measurement Many types of devices using principles involved in the Bernoulli equation have been developed to measure fluid velocities and flowrates. The Pitot-static tube discussed in Section 3.5 is an example. Other examples discussed below include devices to measure flowrates in pipes and 3.6 (1) Examples of Use of the Bernoulli Equation 127 (2) Orifice Nozzle Venturi (1) ■ Figure 3.18 Typical devices for measuring (2) flowrate in pipes. conduits and devices to measure flowrates in open channels. In this chapter we will consider “ideal” flowmeters—those devoid of viscous, compressibility, and other “real-world” effects. Corrections for these effects are discussed in Chapters 8 and 10. Our goal here is to understand the basic operating principles of these simple flowmeters. An effective way to measure the flowrate through a pipe is to place some type of restriction within the pipe as shown in Fig. 3.18 and to measure the pressure difference between the low-velocity, high-pressure upstream section 112 and the high-velocity, low-pressure downstream section 122. Three commonly used types of flowmeters are illustrated: the orifice meter, the nozzle meter, and the Venturi meter. The operation of each is based on the same physical principles— an increase in velocity causes a decrease in pressure. The difference between them is a matter of cost, accuracy, and how closely their actual operation obeys the idealized flow assumptions. We assume the flow is horizontal 1z1 z2 2, steady, inviscid, and incompressible between points 112 and 122. The Bernoulli equation becomes p1 12 rV 21 p2 12 rV 22 1The effect of nonhorizontal flow can be incorporated easily by including the change in elevation, z1 z2, in the Bernoulli equation.2 If we assume the velocity profiles are uniform at sections 112 and 122, the continuity equation 1Eq. 3.192 can be written as The flowrate varies as the square root of the pressure difference across the flowmeter. Q Q ~ Δp Δ p = p1 – p2 Q A1V1 A2V2 where A2 is the small 1A2 6 A1 2 flow area at section 122. Combination of these two equations results in the following theoretical flowrate Q A2 21p1 p2 2 B r31 1A2 A1 2 2 4 (3.20) Thus, as shown by the figure in the margin, for a given flow geometry 1A1 and A2 2 the flowrate can be determined if the pressure difference, p1 p2, is measured. The actual measured flowrate, Qactual, will be smaller than this theoretical result because of various differences between the “real world” and the assumptions used in the derivation of Eq. 3.20. These differences 1which are quite consistent and may be as small as 1 to 2% or as large as 40%, depending on the geometry used2 can be accounted for by using an empirically obtained discharge coefficient as discussed in Section 8.6.1. 128 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation E XAMPLE 3.11 Venturi Meter GIVEN Kerosene 1SG 0.852 flows through the Venturi meter shown in Fig. E3.11a with flowrates between 0.005 and 0.050 m3s. Kerosene, SG = 0.85 D1 = 0.1 m (1) FIND Determine the range in pressure difference, p1 p2, needed to measure these flowrates. (2) D2 = 0.06 m Q 0.005 m3/s < Q < 0.050 m3/s ■ Figure E3.11a SOLUTION If the flow is assumed to be steady, inviscid, and incompressible, the relationship between flowrate and pressure is given by Eq. 3.20. This can be rearranged to give p1 p2 Q2r3 1 1A2A1 2 2 4 2 A22 With the density of the flowing fluid r SG rH2O 0.8511000 kgm3 2 850 kgm3 and the area ratio A2 A1 1D2 D1 2 2 10.06 m0.10 m2 2 0.36 the pressure difference for the smallest flowrate is p1 p2 10.005 m3s2 2 1850 kgm3 2 1160 Nm 1.16 kPa 2 results presented here are independent of the particular flowmeter geometry—an orifice, nozzle, or Venturi meter 1see Fig. 3.182. It is seen from Eq. 3.20 that the flowrate varies as the square root of the pressure difference. Hence, as indicated by the numerical results and shown in Fig. E3.11b, a 10-fold increase in flowrate requires a 100-fold increase in pressure difference. This nonlinear relationship can cause difficulties when measuring flowrates over a wide range of values. Such measurements would require pressure transducers with a wide range of operation. An alternative is to use two flowmeters in parallel—one for the larger and one for the smaller flowrate ranges. 11 0.362 2 2 3 1p4210.06 m2 2 4 2 120 Likewise, the pressure difference for the largest flowrate is 11 0.362 2 p1–p2, kPa p1 p2 10.052 18502 2 2 3 1p42 10.062 2 4 2 1.16 105 Nm2 116 kPa (0.05 m3/s, 116 kPa) 100 80 60 40 (0.005 m3/s, 1.16 kPa) 20 Thus, 1.16 kPa p1 p2 116 kPa (Ans) 0 0 0.01 0.02 0.03 0.04 0.05 Q, m3/s COMMENTS These values represent the pressure differ- ■ Figure E3.11b ences for inviscid, steady, incompressible conditions. The ideal Other flowmeters based on the Bernoulli equation are used to measure flowrates in open channels such as flumes and irrigation ditches. Two of these devices, the sluice gate and the sharp-crested weir, are discussed below under the assumption of steady, inviscid, incompressible flow. These and other open-channel flow devices are discussed in more detail in Chapter 10. Sluice gates like those shown in Fig. 3.19a are often used to regulate and measure the flowrate in open channels. As indicated in Fig. 3.19b, the flowrate, Q, is a function of the water depth upstream, z1, the width of the gate, b, and the gate opening, a. Application of the Bernoulli equation and continuity equation between points 112 and 122 can provide a good approximation to the actual flowrate obtained. We assume the velocity profiles are uniform sufficiently far upstream and downstream of the gate. 3.6 129 Examples of Use of the Bernoulli Equation V1 (1) Sluice gate width = b Sluice gates b z1 a V2 Q (3) a ( a) z2 (2) (4) (b) ■ Figure 3.19 Sluice gate geometry. (Photograph courtesy of Plasti-Fab, Inc.) give Thus, we apply the Bernoulli equation between points on the free surfaces at 112 and 122 to p1 12 rV 21 gz1 p2 12 rV 22 gz2 Also, if the gate is the same width as the channel so that A1 bz1 and A2 bz2, the continuity equation gives Q A1V1 bV1z1 A2V2 bV2z2 The flowrate under a sluice gate depends on the water depths on either side of the gate. With the fact that p1 p2 0, these equations can be combined and rearranged to give the flowrate as Q z2b 2g1z1 z2 2 B 1 1z2 z1 2 2 (3.21) In the limit of z1 z2 this result simply becomes Q z2b12gz1 This limiting result represents the fact that if the depth ratio, z1z2, is large, the kinetic energy of the fluid upstream of the gate is negligible and the fluid velocity after it has fallen a distance 1z1 z2 2 z1 is approximately V2 12gz1. The results of Eq. 3.21 could also be obtained by using the Bernoulli equation between points 132 and 142 and the fact that p3 gz1 and p4 gz2 since the streamlines at these sections are straight. In this formulation, rather than the potential energies at 112 and 122, we have the pressure contributions at 132 and 142. The downstream depth, z2, not the gate opening, a, was used to obtain the result of Eq. 3.21. As was discussed relative to flow from an orifice 1Fig. 3.142, the fluid cannot turn a sharp 90° corner. A vena contracta results with a contraction coefficient, Cc z2 a, less than 1. Typically Cc is approximately 0.61 over the depth ratio range of 0 6 az1 6 0.2. For larger values of az1 the value of Cc increases rapidly. E XAMPLE 3.12 Sluice Gate GIVEN Water flows under the sluice gate shown in Fig. E3.12a. FIND Determine the approximate flowrate per unit width of the channel. 130 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation SOLUTION 6.0 m Under the assumptions of steady, inviscid, incompressible flow, we can apply Eq. 3.21 to obtain Qb, the flowrate per unit width, as 5.0 m 2g1z1 z2 2 Q z2 b B 1 1z2z1 2 2 Q In this instance z1 5.0 m and a 0.80 m so the ratio az1 0.16 6 0.20, and we can assume that the contraction coefficient is approximately Cc 0.61. Thus, z2 Cc a 0.61 10.80 m2 0.488 m and we obtain the flowrate 219.81 ms 2 15.0 m 0.488 m2 Q 10.488 m2 b B 1 10.488 m5.0 m2 2 2 4.61 m s (Ans) 0.8 m ■ Figure E3.12a 9 8 2 Q/b, m2/s COMMENT If we consider z1 z2 and neglect the kinetic 7 energy of the upstream fluid, we would have 6 5 (5m, 4.61 m2/s) 4 3 2 1 Q z2 12gz1 0.488 m 2219.81 ms2 215.0 m2 b 4.83 m2s 0 0 5 z1, m 10 15 ■ Figure E3.12b In this case the difference in Q with or without including V1 is not too significant because the depth ratio is fairly large 1z1z2 5.00.488 10.22. Thus, it is often reasonable to neglect the kinetic energy upstream from the gate compared to that downstream of it. By repeating the calculations for various flow depths, z1, the results shown in Fig. E3.12b are obtained. Note that the flowrate is not directly proportional to the flow depth. Thus, for example, if during flood conditions the upstream depth doubled from z1 5 m to z1 10 m, the flowrate per unit width of the channel would not double but would increase only from 4.61 m2s to 6.67 m2s. Another device used to measure flow in an open channel is a weir. A typical rectangular, sharp-crested weir is shown in Fig. 3.20. For such devices the flowrate of liquid over the top of the weir plate is dependent on the weir height, Pw, the width of the channel, b, and the head, H, of the water above the top of the weir. Application of the Bernoulli equation can provide a simple approximation of the flowrate expected for these situations, even though the actual flow is quite complex. Between points 112 and 122 the pressure and gravitational fields cause the fluid to accelerate from velocity V1 to velocity V2. At 112 the pressure is p1 gh, while at 122 the pressure is essentially atmospheric, p2 0. Across the curved streamlines directly above the top of the weir plate 1section a–a2, the pressure changes from atmospheric on the top surface to some maximum value within the fluid stream and then to atmospheric again at the bottom surface. This distribution is indicated in Fig. 3.20. Such a pressure distribution, combined with the streamline curvature and gravity, produces a rather nonuniform velocity profile across this section. This velocity distribution can be obtained from experiments or a more advanced theory. b a Pressure distribution Width = b H h Q a V1 (1) Weir plate (2) V2 Pw Weir plate ■ Figure 3.20 Rectangular, sharpcrested weir geometry. 3.7 The Energy Line and the Hydraulic Grade Line 131 For now, we will take a very simple approach and assume that the weir flow is similar in many respects to an orifice-type flow with a free streamline. In this instance we would expect the average velocity across the top of the weir to be proportional to 12gH and the flow area for this rectangular weir to be proportional to Hb. Hence, it follows that Q Q C1Hb 12gH C1b 12g H32 Q ~ H3/2 H E XAMPLE where C1 is a constant to be determined. Simple use of the Bernoulli equation has provided a method to analyze the relatively complex flow over a weir. The correct functional dependence of Q on H has been obtained 1Q H 32, as indicated by the figure in the margin), but the value of the coefficient C1 is unknown. Even a more advanced analysis cannot predict its value accurately. As is discussed in Chapter 10, experiments are used to determine the value of C1. 3.13 Weir GIVEN Water flows over a triangular weir, as is shown in Fig. _θ H tan 2 E3.13. θ FIND Based on a simple analysis using the Bernoulli equation, H determine the dependence of the flowrate on the depth H. If the flowrate is Q0 when H H0, estimate the flowrate when the depth is increased to H 3H0. SOLUTION ■ Figure E3.13 With the assumption that the flow is steady, inviscid, and incompressible, it is reasonable to assume from Eq. 3.18 that the average speed of the fluid over the triangular notch in the weir plate is proportional to 12gH. Also, the flow area for a depth of H is H 3 H tan 1u22 4 . The combination of these two ideas gives Q AV H 2 tan u u 1C 12gH2 C2 tan 12g H 52 2 2 2 Q3H0 QH0 C2 tan1u22 12g 13H0 2 52 C2 tan1u22 12g 1H0 2 52 15.6 (Ans) (Ans) COMMENT Note that for a triangular weir the flowrate is where C2 is an unknown constant to be determined experimentally. Thus, an increase in the depth by a factor of three 1from H0 to 3H02 results in an increase of the flowrate by a factor of 3.7 H proportional to H 52, whereas for the rectangular weir discussed above, it is proportional to H 32. The triangular weir can be accurately used over a wide range of flowrates. The Energy Line and the Hydraulic Grade Line The hydraulic grade line and energy line are graphical forms of the Bernoulli equation. As was discussed in Section 3.4, the Bernoulli equation is actually an energy equation representing the partitioning of energy for an inviscid, incompressible, steady flow. The sum of the various energies of the fluid remains constant as the fluid flows from one section to another. A useful interpretation of the Bernoulli equation can be obtained through use of the concepts of the hydraulic grade line 1HGL2 and the energy line 1EL2. These ideas represent a geometrical interpretation of a flow and can often be effectively used to better grasp the fundamental processes involved. For steady, inviscid, incompressible flow the total energy remains constant along a streamline. The concept of “head” was introduced by dividing each term in Eq. 3.7 by the specific weight, g rg, to give the Bernoulli equation in the following form p V2 z constant on a streamline H g 2g (3.22) 132 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation V12/2g 2 V ___ 2 2g V 2/2g p2/γ Energy line (EL) Hydraulic grade line (HGL) p/γ p1/ γ H Static z2 z Q z1 Datum Stagnation ■ Figure 3.21 Representation of the energy line and the hydraulic grade line. V 2/2g p/␥ H z Under the assumptions of the Bernoulli equation, the energy line is horizontal. Each of the terms in this equation has the units of length 1feet or meters2 and represents a certain type of head. The Bernoulli equation states that the sum of the pressure head, the velocity head, and the elevation head is constant along a streamline. This constant is called the total head, H, and is shown in the figure in the margin. The energy line is a line that represents the total head available to the fluid. As shown in Fig. 3.21, the elevation of the energy line can be obtained by measuring the stagnation pressure with a Pitot tube. 1A Pitot tube is the portion of a Pitot-static tube that measures the stagnation pressure. See Section 3.5.2 The stagnation point at the end of the Pitot tube provides a measurement of the total head 1or energy2 of the flow. The static pressure tap connected to the piezometer tube shown, on the other hand, measures the sum of the pressure head and the elevation head, pg z. This sum is often called the piezometric head. The static pressure tap does not measure the velocity head. According to Eq. 3.22, the total head remains constant along the streamline 1provided the assumptions of the Bernoulli equation are valid2. Thus, a Pitot tube at any other location in the flow will measure the same total head, as is shown in the figure. The elevation head, velocity head, and pressure head may vary along the streamline, however. The locus of elevations provided by a series of Pitot tubes is termed the energy line, EL. The locus provided by a series of piezometer taps is termed the hydraulic grade line, HGL. Under the assumptions of the Bernoulli equation, the energy line is horizontal. If the fluid velocity changes along the streamline, the hydraulic grade line will not be horizontal. If viscous effects are important 1as they often are in pipe flows2, the total head does not remain constant due to a loss in energy as the fluid flows along its streamline. This means that the energy line is no longer horizontal. Such viscous effects are discussed in Chapters 5 and 8. The energy line and hydraulic grade line for flow from a large tank are shown in Fig. 3.22. If the flow is steady, incompressible, and inviscid, the energy line is horizontal and at the elevation of the liquid in the tank 1since the fluid velocity in the tank and the pressure on the surface V1 = p1 = 0 H = z1 EL 2 V ___ 2 2g (1) p__2 γ HGL 2 V ___ 3 2g p3 = 0 (2) z2 z3 (3) ■ Figure 3.22 The energy line and hydraulic grade line for flow from a tank. 3.7 The Energy Line and the Hydraulic Grade Line EL 2 V __ 2g p0 For flow below (above) the hydraulic grade line, the pressure is positive (negative). E XAMPLE 133 z z ■ Figure 3.23 Use of the energy line and the hydraulic grade line. are zero2. The hydraulic grade line lies a distance of one velocity head, V 22g, below the energy line. Thus, a change in fluid velocity due to a change in the pipe diameter results in a change in the elevation of the hydraulic grade line. At the pipe outlet the pressure head is zero 1gage2, so the pipe elevation and the hydraulic grade line coincide. The distance from the pipe to the hydraulic grade line indicates the pressure within the pipe, as is shown in Fig. 3.23. If the pipe lies below the hydraulic grade line, the pressure within the pipe is positive 1above atmospheric2. If the pipe lies above the hydraulic grade line, the pressure is negative 1below atmospheric2. Thus, a scale drawing of a pipeline and the hydraulic grade line can be used to readily indicate regions of positive or negative pressure within a pipe. 3.14 Energy Line and Hydraulic Grade Line GIVEN Water is siphoned from the tank shown in Fig. E3.14 through a hose of constant diameter. A small hole is found in the hose at location 112 as indicated. FIND When the siphon is used, will water leak out of the hose, or will air leak into the hose, thereby possibly causing the siphon to malfunction? (1) HGL with valve closed and EL with valve open or closed 2 V __ 2g z p_ γ HGL with valve open SOLUTION Valve Whether air will leak into or water will leak out of the hose depends on whether the pressure within the hose at 112 is less than or greater than atmospheric. Which happens can be easily determined by using the energy line and hydraulic grade line concepts. With the assumption of steady, incompressible, inviscid flow it follows that the total head is constant—thus, the energy line is horizontal. Since the hose diameter is constant, it follows from the continuity equation 1AV constant2 that the water velocity in the hose is constant throughout. Thus, the hydraulic grade line is a constant distance, V 22g, below the energy line as shown in Fig. E3.14. Since the pressure at the end of the hose is atmospheric, it follows that the hydraulic grade line is at the same elevation as the end of the hose outlet. The fluid within the hose at any point above the hydraulic grade line will be at less than atmospheric pressure. Thus, air will leak into the hose through the hole at point 112. (Ans) ■ Figure E3.14 COMMENT In practice, viscous effects may be quite impor- tant, making this simple analysis 1horizontal energy line2 incorrect. However, if the hose is “not too small diameter,” “not too long,” the fluid “not too viscous,” and the flowrate “not too large,” the above result may be very accurate. If any of these assumptions are relaxed, a more detailed analysis is required 1see Chapter 82. If the end of the hose were closed so that the flowrate were zero, the hydraulic grade line would coincide with the energy line 1V 22g 0 throughout2, the pressure at 112 would be greater than atmospheric, and water would leak through the hole at 112. The above discussion of the hydraulic grade line and the energy line is restricted to ideal situations involving inviscid, incompressible flows. Another restriction is that there are no “sources” or “sinks” of energy within the flow field. That is, there are no pumps or turbines involved. Alterations in the energy line and hydraulic grade line concepts due to these devices are discussed in Chapters 5 and 8. 134 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation 3.8 Restrictions on Use of the Bernoulli Equation Proper use of the Bernoulli equation requires close attention to the assumptions used in its derivation. In this section we review some of these assumptions and consider the consequences of incorrect use of the equation. 3.8.1 Compressibility Effects Δp Δp ~ V 2 V One of the main assumptions is that the fluid is incompressible. Although this is reasonable for most liquid flows, it can, in certain instances, introduce considerable errors for gases. In the previous section, we saw that the stagnation pressure, pstag, is greater than the static pressure, pstatic, by an amount ¢p pstag pstatic rV 22, provided that the density remains constant. If this dynamic pressure is not too large compared with the static pressure, the density change between two points is not very large and the flow can be considered incompressible. However, since the dynamic pressure varies as V 2, the error associated with the assumption that a fluid is incompressible increases with the square of the velocity of the fluid, as indicated by the figure in the margin. To account for compressibility effects, we must return to Eq. 3.6 and properly integrate the term dpr when r is not constant. A simple, although specialized, case of compressible flow occurs when the temperature of a perfect gas remains constant along the streamline—isothermal flow. Thus, we consider p rRT, where T is constant. 1In general, p, r, and T will vary.2 For steady, inviscid, isothermal flow, Eq. 3.6 becomes RT dp 1 V 2 gz constant p 2 where we have used r pRT. The pressure term is easily integrated and the constant of integration evaluated if z1, p1, and V1 are known at some location on the streamline. The result is V 21 p1 V 22 RT z1 ln a b z2 g p2 2g 2g The Bernoulli equation can be modified for compressible flows. (3.23) Equation 3.23 is the inviscid, isothermal analog of the incompressible Bernoulli equation. In the limit of small pressure difference, p1p2 1 1p1 p2 2 p2 1 e, with e 1 and Eq. 3.23 reduces to the standard incompressible Bernoulli equation. This can be shown by use of the approximation ln11 e2 e for small e. The use of Eq. 3.23 in practical applications is restricted by the inviscid flow assumption, since 1as is discussed in Section 11.52 most isothermal flows are accompanied by viscous effects. A much more common compressible flow condition is that of isentropic 1constant entropy2 flow of a perfect gas. Such flows are reversible adiabatic processes—“no friction or heat transfer”— and are closely approximated in many physical situations. As discussed fully in Chapter 11, for isentropic flow of a perfect gas the density and pressure are related by prk C, where k is the specific heat ratio and C is a constant. Hence, the dpr integral of Eq. 3.6 can be evaluated as follows. The density can be written in terms of the pressure as r p1kC1k so that Eq. 3.6 becomes C1k p 1k dp 1 2 V gz constant 2 The pressure term can be integrated between points 112 and 122 on the streamline and the constant C evaluated at either point 1C1k p11kr1 or C1k p12kr2 2 to give the following: C1k p2 p1 p 1k dp C1k a a k b 3p21k12k p11k12k 4 k1 p2 p1 k ba b r1 k 1 r2 3.8 Restrictions on Use of the Bernoulli Equation 135 Thus, the final form of Eq. 3.6 for compressible, isentropic, steady flow of a perfect gas is a p1 V 21 p2 V 22 k k b ⫹ ⫹ gz1 ⫽ a b ⫹ ⫹ gz2 k ⫺ 1 r1 2 k ⫺ 1 r2 2 (3.24) The similarities between the results for compressible isentropic flow 1Eq. 3.242 and incompressible isentropic flow 1the Bernoulli equation, Eq. 3.72 are apparent. The only differences are the factors of 3k Ⲑ 1k ⫺ 12 4 that multiply the pressure terms and the fact that the densities are different 1r1 ⫽ r2 2. In the limit of “low-speed flow” the two results are exactly the same, as is seen by the following. We consider the stagnation point flow of Section 3.5 to illustrate the difference between the incompressible and compressible results. As is shown in Chapter 11, Eq. 3.24 can be written in dimensionless form as kⲐk⫺1 p2 ⫺ p1 k⫺1 ⫽ c a1 ⫹ Ma21 b ⫺ 1d p1 2 1compressible2 (3.25) where 112 denotes the upstream conditions and 122 the stagnation conditions. We have assumed z1 ⫽ z2, V2 ⫽ 0, and have denoted Ma1 ⫽ V1 Ⲑc1 as the upstream Mach number—the ratio of the fluid velocity to the speed of sound, c1 ⫽ 1kRT1. A comparison between this compressible result and the incompressible result is perhaps most easily seen if we write the incompressible flow result in terms of the pressure ratio and the Mach number. Thus, we divide each term in the Bernoulli equation, rV12Ⲑ2 ⫹ p1 ⫽ p2, by p1 and use the perfect gas law, p1 ⫽ rRT1, to obtain p2 ⫺ p1 V 21 ⫽ p1 2RT1 Since Ma1 ⫽ V1Ⲑ 1kRT1 this can be written as p2 ⫺ p1 kMa21 ⫽ p1 2 1incompressible2 (3.26) Equations 3.25 and 3.26 are plotted in Fig. 3.24. In the low-speed limit of Ma1 S 0, both of the results are the same. This can be seen by denoting 1k ⫺ 12Ma21 Ⲑ2 ⫽ e~ and using the binomial expan~ ⫹ n1n ⫺ 12 ~ sion, 11 ⫹ ~ e 2 n ⫽ 1 ⫹ ne e 2Ⲑ2 ⫹ p , where n ⫽ kⲐ 1k ⫺ 12, to write Eq. 3.25 as p2 ⫺ p1 kMa21 1 2⫺k ⫽ a1 ⫹ Ma21 ⫹ Ma41 ⫹ p b p1 2 4 24 1compressible2 For Ma1 Ⰶ 1 this compressible flow result agrees with Eq. 3.26. The incompressible and compressible equations agree to within about 2% up to a Mach number of approximately Ma1 ⫽ 0.3. For larger Mach numbers the disagreement between the two results increases. 1 0.8 Compressible (Eq. 3.25) 0.6 p1 p2 – p1 ______ For small Mach numbers the compressible and incompressible results are nearly the same. k = 1.4 0.4 Incompressible (Eq. 3.26) 0.2 0 0 0.2 0.4 0.6 Ma1 0.8 1 ■ Figure 3.24 Pressure ratio as a function of Mach number for incompressible and compressible (isentropic) flow. 136 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation Thus, a rule of thumb is that the flow of a perfect gas may be considered as incompressible provided the Mach number is less than about 0.3. In standard air 1T1 59 °F, c1 1kRT1 1117 fts2 this corresponds to a speed of V1 Ma1c1 0.311117 fts2 335 fts 228 mihr. At higher speeds, compressibility may become important. E XAMPLE 3.15 Compressible Flow—Mach Number GIVEN The jet shown in Fig. E3.15 flies at Mach 0.82 at an altitude of 10 km in a standard atmosphere. FIND Determine the stagnation pressure on the leading edge of its wing if the flow is incompressible; and if the flow is compressible isentropic. SOLUTION From Tables 1.8 and C.2 we find that p1 26.5 kPa 1abs2, T1 49.9 °C, r 0.414 kgm3, and k 1.4. Thus, if we assume incompressible flow, Eq. 3.26 gives 10.822 2 p2 p1 kMa21 1.4 0.471 p1 2 2 ■ Figure E3.15 (© RobHowarth/iStockphoto) or p2 p1 0.471126.5 kPa2 12.5 kPa (Ans) On the other hand, if we assume isentropic flow, Eq. 3.25 gives 1.411.412 11.4 12 p2 p1 e c1 10.822 2 d 1f p1 2 0.555 or p2 p1 0.555 126.5 kPa2 14.7 kPa (Ans) lift and drag on the airplane; see Chapter 92 is approximately 14.712.5 1.18 times greater according to the compressible flow calculations. This may be very significant. As discussed in Chapter 11, for Mach numbers greater than 1 1supersonic flow2 the differences between incompressible and compressible results are often not only quantitative but also qualitative. Note that if the airplane were flying at Mach 0.30 1rather than 0.822 the corresponding values would be p2 p1 1.670 kPa for incompressible flow and p2 p1 1.707 kPa for compressible flow. The difference between these two results is about 2%. COMMENT We see that at Mach 0.82 compressibility effects are of importance. The pressure 1and, to a first approximation, the 3.8.2 Unsteady Effects The Bernoulli equation can be modified for unsteady flows. Another restriction of the Bernoulli equation 1Eq. 3.72 is the assumption that the flow is steady. For such flows, on a given streamline the velocity is a function of only s, the location along the streamline. That is, along a streamline V V1s2. For unsteady flows the velocity is also a function of time, so that along a streamline V V1s, t2. Thus when taking the time derivative of the velocity to obtain the streamwise acceleration, we obtain as 0V0t V 0V0s rather than just as V 0V0s as is true for steady flow. For steady flows the acceleration is due to the change in velocity resulting from a change in position of the particle 1the V 0V0s term2, whereas for unsteady flow there is an additional contribution to the acceleration resulting from a change in velocity with time at a fixed location 1the 0V0t term2. These effects are discussed in detail in Chapter 4. The net effect is that the inclusion of the unsteady term, 0V0t, does not allow the equation of motion to be easily integrated 1as was done to obtain the Bernoulli equation2 unless additional assumptions are made. The Bernoulli equation was obtained by integrating the component of Newton’s second law 1Eq. 3.52 along the streamline. When integrated, the acceleration contribution to this equation, the 1 2 2 rd1V 2 term, gave rise to the kinetic energy term in the Bernoulli equation. If the steps leading 3.8 Restrictions on Use of the Bernoulli Equation 137 to Eq. 3.5 are repeated with the inclusion of the unsteady effect 10VⲐ0t ⫽ 02, the following is obtained: r 0V 1 ds ⫹ dp ⫹ rd1V 2 2 ⫹ g dz ⫽ 0 0t 2 1along a streamline2 For incompressible flow this can be easily integrated between points 112 and 122 to give p1 ⫹ V3.11 Oscillations in a U-tube 1 rV 21 ⫹ gz1 ⫽ r 2 冮 s2 s1 0V 1 ds ⫹ p2 ⫹ rV 22 ⫹ gz2 0t 2 1along a streamline2 (3.27) Equation 3.27 is an unsteady form of the Bernoulli equation valid for unsteady, incompressible, inviscid flow. Except for the integral involving the local acceleration, 0VⲐ0t, it is identical to the steady Bernoulli equation. In general, it is not easy to evaluate this integral because the variation of 0VⲐ0t along the streamline is not known. In some situations the concepts of “irrotational flow” and the “velocity potential” can be used to simplify this integral. These topics are discussed in Chapter 6. E XAMPLE Unsteady Flow—U-Tube 3.16 Open tube GIVEN An incompressible, inviscid liquid is placed in a vertical, constant diameter U-tube as indicated in Fig. E3.16. When released from the nonequilibrium position shown, the liquid column will oscillate at a specific frequency. V2 (2) g z FIND Determine this frequency. z= 0 Equilibrium position (1) SOLUTION V1 The frequency of oscillation can be calculated by use of Eq. 3.27 as follows. Let points 112 and 122 be at the air–water interfaces of the two columns of the tube and z ⫽ 0 correspond to the equilibrium position of these interfaces. Hence, p1 ⫽ p2 ⫽ 0 and if z2 ⫽ z, then z1 ⫽ ⫺z. In general, z is a function of time, z ⫽ z1t2. For a constant diameter tube, at any instant in time the fluid speed is constant throughout the tube, V1 ⫽ V2 ⫽ V, and the integral representing the unsteady effect in Eq. 3.27 can be written as 冮 s2 s1 0V dV ds ⫽ 0t dt 冮 s2 ds ⫽ / s1 dV dt where / is the total length of the liquid column as shown in the figure. Thus, Eq. 3.27 can be written as g1⫺z2 ⫽ r/ dV ⫹ gz dt Since V ⫽ dzⲐdt and g ⫽ rg, this can be written as the secondorder differential equation describing simple harmonic motion ᐉ ■ Figure E3.16 which has the solution z1t2 ⫽ C1 sin1 12gⲐ/ t2 ⫹ C2 cos 1 12gⲐ/ t2. The values of the constants C1 and C2 depend on the initial state 1velocity and position2 of the liquid at t ⫽ 0. Thus, the liquid oscillates in the tube with a frequency v ⫽ 22g Ⲑ/ (Ans) COMMENT This frequency depends on the length of the col- umn and the acceleration of gravity 1in a manner very similar to the oscillation of a pendulum2. The period of this oscillation 1the time required to complete an oscillation2 is t0 ⫽ 2p 1/Ⲑ2g. 2g d 2z ⫹ z⫽0 2 / dt In a few unsteady flow cases, the flow can be made steady by an appropriate selection of the coordinate system. Example 3.17 illustrates this. 138 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation E XAMPLE 3.17 Unsteady or Steady Flow GIVEN A submarine moves through seawater 1SG 1.032 at a depth of 50 m with velocity V0 5.0 ms as shown in Fig. E3.17. FIND Determine the pressure at the stagnation point 122. γ h = 50 m (1) SOLUTION In a coordinate system fixed to the ground, the flow is unsteady. For example, the water velocity at 112 is zero with the submarine in its initial position, but at the instant when the nose, 122, reaches point 112 the velocity there becomes V1 V0 ˆi . Thus, 0V1 0t 0 and the flow is unsteady. Application of the steady Bernoulli equation between 112 and 122 would give the incorrect result that p1 p2 rV 20 2. According to this result, the static pressure is greater than the stagnation pressure—an incorrect use of the Bernoulli equation. We can either use an unsteady analysis for the flow 1which is outside the scope of this text2 or redefine the coordinate system so that it is fixed on the submarine, giving steady flow with respect to this system. The correct method would be p2 (2) V0 = 5 m/s x ■ Figure E3.17 112,900 505,0002 Nm2 518 kPa (Ans) similar to that discussed in Example 3.2. COMMENT If the submarine were accelerating, 0V0 0t 0, the flow would be unsteady in either of the above coordinate systems and we would be forced to use an unsteady form of the Bernoulli equation. rV 21 gh 3 11.032110002 kgm3 4 15.0 ms2 2 2 2 19.80 103 Nm3 2 11.032 150 m2 Some unsteady flows may be treated as “quasisteady” and solved approximately by using the steady Bernoulli equation. In these cases the unsteadiness is “not too great” 1in some sense2, and the steady flow results can be applied at each instant in time as though the flow were steady. The slow draining of a tank filled with liquid provides an example of this type of flow. 3.8.3 Rotational Effects Care must be used in applying the Bernoulli equation across streamlines. E XAMPLE Another of the restrictions of the Bernoulli equation is that it is applicable along the streamline. Application of the Bernoulli equation across streamlines 1i.e., from a point on one streamline to a point on another streamline2 can lead to considerable errors, depending on the particular flow conditions involved. In general, the Bernoulli constant varies from streamline to streamline. However, under certain restrictions this constant is the same throughout the entire flow field. Example 3.18 illustrates this fact. 3.18 Use of Bernoulli Equation across Streamlines Fluid particles spin GIVEN Consider the uniform flow in the channel shown in Fig. E3.18a. The liquid in the vertical piezometer tube is stationary. p5 = 0 FIND Discuss the use of the Bernoulli equation between points 112 and 122, points 132 and 142, and points 142 and 152. SOLUTION If the flow is steady, inviscid, and incompressible, Eq. 3.7 written between points 112 and 122 gives p1 12 rV 21 gz1 p2 12 rV 22 gz2 constant C12 (5) H V0 (3) h (1) p1 = p0 V0 (a) ■ Figure E3.18 (4) z=h (2) z=0 (b) 3.9 Since V1 V2 V0 and z1 z2 0, it follows that p1 p2 p0 and the Bernoulli constant for this streamline, C12, is given by C12 12 rV 20 p0 Along the streamline from 132 to 142 we note that V3 V4 V0 and z3 z4 h. As was shown in Example 3.5, application of F ma across the streamline 1Eq. 3.122 gives p3 p1 gh because the streamlines are straight and horizontal. The above facts combined with the Bernoulli equation applied between 132 and 142 show that p3 p4 and that the Bernoulli constant along this streamline is the same as that along the streamline between 112 and 122. That is, C34 C12, or p3 12 rV 23 gz3 p4 12 rV 24 gz4 C34 C12 Similar reasoning shows that the Bernoulli constant is the same for any streamline in Fig. E3.18. Hence, p 12 rV 2 gz constant throughout the flow V3.12 Flow over a cavity Chapter Summary and Study Guide 139 Again from Example 3.5 we recall that p4 p5 gH gH If we apply the Bernoulli equation across streamlines from 142 to 152, we obtain the incorrect result H p4g V 24 2g. The correct result is H p4g. From the above we see that we can apply the Bernoulli equation across streamlines 112–122 and 132–142 1i.e., C12 C342 but not across streamlines from 142 to 152. The reason for this is that while the flow in the channel is “irrotational,” it is “rotational” between the flowing fluid in the channel and the stationary fluid in the piezometer tube. Because of the uniform velocity profile across the channel, it is seen that the fluid particles do not rotate or “spin” as they move. The flow is “irrotational.” However, as seen in Fig. E3.18b, there is a very thin shear layer between 142 and 152 in which adjacent fluid particles interact and rotate or “spin.” This produces a “rotational” flow. A more complete analysis would show that the Bernoulli equation cannot be applied across streamlines if the flow is “rotational” 1see Chapter 62. As is suggested by Example 3.18, if the flow is “irrotational” 1i.e., the fluid particles do not “spin” as they move2, it is appropriate to use the Bernoulli equation across streamlines. However, if the flow is “rotational” 1fluid particles “spin”2, use of the Bernoulli equation is restricted to flow along a streamline. The distinction between irrotational and rotational flow is often a very subtle and confusing one. These topics are discussed in more detail in Chapter 6. A thorough discussion can be found in more advanced texts 1Ref. 32. 3.8.4 Other Restrictions The Bernoulli equation is not valid for flows that involve pumps or turbines. 3.9 Another restriction on the Bernoulli equation is that the flow is inviscid. As is discussed in Section 3.4, the Bernoulli equation is actually a first integral of Newton’s second law along a streamline. This general integration was possible because, in the absence of viscous effects, the fluid system considered was a conservative system. The total energy of the system remains constant. If viscous effects are important the system is nonconservative 1dissipative2 and energy losses occur. A more detailed analysis is needed for these cases. Such material is presented in Chapter 5. The final basic restriction on use of the Bernoulli equation is that there are no mechanical devices 1pumps or turbines2 in the system between the two points along the streamline for which the equation is applied. These devices represent sources or sinks of energy. Since the Bernoulli equation is actually one form of the energy equation, it must be altered to include pumps or turbines, if these are present. The inclusion of pumps and turbines is covered in Chapters 5 and 12. In this chapter we have spent considerable time investigating fluid dynamic situations governed by a relatively simple analysis for steady, inviscid, incompressible flows. Many flows can be adequately analyzed by use of these ideas. However, because of the rather severe restrictions imposed, many others cannot. An understanding of these basic ideas will provide a firm foundation for the remainder of the topics in this book. Chapter Summary and Study Guide In this chapter, several aspects of the steady flow of an inviscid, incompressible fluid are discussed. Newton’s second law, F ma, is applied to flows for which the only important forces are those due to pressure and gravity (weight)—viscous effects are assumed negligible. The result is the oftenused Bernoulli equation, which provides a simple relationship among pressure, elevation, and velocity variations along a streamline. A similar but less often used equation is also obtained to describe the variations in these parameters normal to a streamline. The concept of a stagnation point and the corresponding stagnation pressure is introduced, as are the concepts of static, dynamic, and total pressure and their related heads. 140 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation steady flow streamline Bernoulli equation elevation head pressure head velocity head static pressure dynamic pressure stagnation point stagnation pressure total pressure Pitot-static tube free jet volume flowrate continuity equation cavitation flowmeter hydraulic grade line energy line Several applications of the Bernoulli equation are discussed. In some flow situations, such as the use of a Pitot-static tube to measure fluid velocity or the flow of a liquid as a free jet from a tank, a Bernoulli equation alone is sufficient for the analysis. In other instances, such as confined flows in tubes and flowmeters, it is necessary to use both the Bernoulli equation and the continuity equation, which is a statement of the fact that mass is conserved as fluid flows. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. explain the origin of the pressure, elevation, and velocity terms in the Bernoulli equation and how they are related to Newton’s second law of motion. apply the Bernoulli equation to simple flow situations, including Pitot-static tubes, free jet flows, confined flows, and flowmeters. use the concept of conservation of mass (the continuity equation) in conjunction with the Bernoulli equation to solve simple flow problems. apply Newton’s second law across streamlines for appropriate steady, inviscid, incompressible flows. use the concepts of pressure, elevation, velocity, and total heads to solve various flow problems. explain and use the concepts of static, stagnation, dynamic, and total pressures. use the energy line and the hydraulic grade line concepts to solve various flow problems. explain the various restrictions on use of the Bernoulli equation. Some of the important equations in this chapter are: Streamwise and normal acceleration Force balance along a streamline for steady inviscid flow Bernoulli equation Pressure gradient normal to streamline for inviscid flow in absence of gravity Force balance normal to a streamline for steady, inviscid, incompressible flow Velocity measurement for a Pitot-static tube Free jet Continuity equation Flowmeter equation Sluice gate equation Total head as V r 2V dp 1 2 0V , 0s gz C V2 r (3.1) 1along a streamline2 (3.6) an p 12rV2 gz constant along streamline rV2 0p 0n r pr (3.10b) r dn gz constant across the streamline V2 V 22 1p3 p4 2 r gh V 2 12gh B r A1V1 A2V2, or Q1 Q2 Q A2 (3.7) 21p1 p2 2 B r31 1A2 A1 2 2 4 2g1z1 z2 2 Q z2b B 1 1z2 z1 2 2 p V2 z constant on a streamline H g 2g (3.12) (3.16) (3.18) (3.19) (3.20) (3.21) (3.22) Problems 141 References 1. Riley, W. F., and Sturges, L. D., Engineering Mechanics: Dynamics, 2nd Ed., Wiley, New York, 1996. 2. Tipler, P. A., Physics, Worth, New York, 1982. 3. Panton, R. L., Incompressible Flow, Wiley, New York, 1984. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Conceptual Questions 3.1C In low-speed flow, the following device is called a diffuser: Wall Flow direction 3.3C The stagnation pressure for a fluid is a) the pressure due to the velocity of the fluid. b) not defined unless the fluid is moving. c) the total pressure when the fluid is brought to rest. d) the sum of the static and dynamic pressures. 3.4C Pitot tubes are placed in two ducts in which air is flowing and the static pressures of the flows are equal. The stagnation pressure tap and the static pressure tap are connected to manometers as shown. The manometer reading for duct A is height h, and that for duct B is 2h. The relation between the velocity VA in duct A and the velocity VB in duct B is Duct A VA The fluid velocity: a) increases in the direction of flow. b) decreases in the direction of flow. c) stays the same in the direction of flow. d) None of the above. 3.2C Consider constant altitude, steady flow along a streamline with a flow that satisfies the assumptions necessary for Bernoulli’s equation. Which of the following have a constant value along the streamline? a) Internal energy c) Stagnation or total pressure b) Static or local pressure d) Dynamic or velocity pressure Duct B VB h 2h a) VB equals VA/2. d) VB equals 2VA. b) VB equals 22VA. e) VB equals VA. c) VB equals 4VA. Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the even-numbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 3.2 F ⫽ ma along a Streamline 3.1 Obtain a photograph/image of a situation that can be analyzed by use of the Bernoulli equation. Print this photo and write a brief paragraph that describes the situation involved. 142 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation 3.2 Air flows steadily along a streamline from point (1) to point (2) with negligible viscous effects. The following conditions are measured: At point (1) z1 2 m and p1 0 kPa; at point (2) z2 10 m, p2 20 N/m2, and V2 0. Determine the velocity at point (1). 3.3 Water flows steadily through the variable area horizontal pipe shown in Fig. P3.3. The centerline velocity is given by V 1011 x2 ˆi fts, where x is in feet. Viscous effects are neglected. (a) Determine the pressure gradient, 0p0x, 1as a function of x2 needed to produce this flow. (b) If the pressure at section 112 is 50 psi, determine the pressure at 122 by 1i2 integration of the pressure gradient obtained in (a), 1ii2 application of the Bernoulli equation. is important. Print this photo and write a brief paragraph that describes the situation involved. 3.11 Air flows along a horizontal, curved streamline with a 20 ft radius with a speed of 100 ft/s. Determine the pressure gradient normal to the streamline. 3.12 Water flows around the vertical two-dimensional bend with circular streamlines and constant velocity as shown in Fig. P3.12. If the pressure is 40 kPa at point 112, determine the pressures at points 122 and 132. Assume that the velocity profile is uniform as indicated. g V(x) Q 4m = 3 ft (2) (1) (3) x 2m ■ Figure P3.3 1m V = 10m/s (2) 3.4 What pressure gradient along the streamline, dpds, is required to accelerate water in a horizontal pipe at a rate of 30 m s2? 3.5 GO At a given location the airspeed is 20 m/s and the pressure gradient along the streamline is 100 N/m3. Estimate the airspeed at a point 0.5 m farther along the streamline. 3.6 What pressure gradient along the streamline, dp/ds, is required to accelerate water upward in a vertical pipe at a rate of 30 ft/s2? What is the answer if the flow is downward? 3.7 The Bernoulli equation is valid for steady, inviscid, incompressible flows with constant acceleration of gravity. Consider flow on a planet where the acceleration of gravity varies with height so that g g0 cz, where g0 and c are constants. Integrate “F ma” along a streamline to obtain the equivalent of the Bernoulli equation for this flow. (1) ■ Figure P3.12 3.13 Water flows around the vertical two-dimensional bend with circular streamlines as is shown in Fig. P3.13. The pressure at point 112 is measured to be p1 25 psi and the velocity across section a–a is as indicated in the table. Calculate and plot the pressure across section a–a of the channel [p p(z) for 0 z 2 ft]. z (ft) V (fts) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0 8.0 14.3 20.0 19.5 15.6 8.3 6.2 3.7 2.0 0 3.8 An incompressible fluid flows steadily past a circular cylinder as shown in Fig. P3.8. The fluid velocity along the dividing streamline 1 x a2 is found to be V V0 11 a2x2 2, where a is the radius of the cylinder and V0 is the upstream velocity. (a) Determine the pressure gradient along this streamline. (b) If the upstream pressure is p0, integrate the pressure gradient to obtain the pressure p1x2 for x a. (c) Show from the result of part (b) that the pressure at the stagnation point 1x a2 is p0 rV 20 2, as expected from the Bernoulli equation. Dividing streamline x=0 V0 p0 g x Stagnation point 20 ft a P1 = 25 psi ■ Figure P3.8 3.9 Consider a compressible liquid that has a constant bulk modulus. Integrate “F ma” along a streamline to obtain the equivalent of the Bernoulli equation for this flow. Assume steady, inviscid flow. a (1) 2 ft V = V (z) z Section 3.3 F ⫽ ma Normal to a Streamline 3.10 Obtain a photograph/image of a situation in which Newton’s second law applied across the streamlines (as given by Eq. 3.12) a ■ Figure P3.13 Problems 3.14 Water in a container and air in a tornado flow in horizontal circular streamlines of radius r and speed V as shown in Video V3.6 and Fig. P3.14. Determine the radial pressure gradient, 0pⲐ0r, needed for the following situations: (a) The fluid is water with r ⫽ 3 in. and V ⫽ 0.8 ft Ⲑs. (b) The fluid is air with r ⫽ 300 ft and V ⫽ 200 mph. 143 p1 ⫺ p2, are generated to provide a fresh airflow within the burrow? 1.07 V0 V0 (2) (1) y Q V ■ Figure P3.20 r 3.21 A loon is a diving bird equally at home “flying” in the air or water. What swimming velocity under water will produce a dynamic pressure equal to that when it flies in the air at 40 mph? x †3.22 Estimate the pressure on your hand when you hold it in the stream of air coming from the air hose at a filling station. List all assumptions and show calculations. Warning: Do not try this experiment; it can be dangerous! ■ Figure P3.14 3.15 Air flows smoothly over the hood of your car and up past the windshield. However, a bug in the air does not follow the same path; it becomes splattered against the windshield. Explain why this is so. Section 3.5 Static, Stagnation, Dynamic, and Total Pressure 3.16 Obtain a photograph/image of a situation in which the concept of the stagnation pressure is important. Print this photo and write a brief paragraph that describes the situation involved. 3.17 At a given point on a horizontal streamline in flowing air, the static pressure is ⫺2.0 psi (i.e., a vacuum) and the velocity is 150 ft/s. Determine the pressure at a stagnation point on that streamline. †3.18 A drop of water in a zero-g environment (as in the International Space Station) will assume a spherical shape as shown in Fig. P3.18a. A raindrop in the cartoons is typically drawn as in Fig. P3.18b. The shape of an actual raindrop is more nearly like that shown in Fig. 3.18c. Discuss why these shapes are as indicated. (a) (b) (c) ■ Figure P3.18 3.19 GO When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing? 3.20 Some animals have learned to take advantage of the Bernoulli effect without having read a fluid mechanics book. For example, a typical prairie dog burrow contains two entrances—a flat front door and a mounded back door as shown in Fig. P3.20. When the wind blows with velocity V0 across the front door, the average velocity across the back door is greater than V0 because of the mound. Assume the air velocity across the back door is 1.07V0. For a wind velocity of 6 m/s, what pressure differences, 3.23 A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the “car” were an Indy 500 racer traveling 220 mph? 3.24 GO A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 40 °F and 25 psia. A water manometer connected to the Pitot-static tube indicates a reading of 2.3 in. Determine the helium velocity. Is it reasonable to consider the flow as incompressible? Explain. 3.25 A Bourdon-type pressure gage is used to measure the pressure from a Pitot tube attached to the leading edge of an airplane wing. The gage is calibrated to read in miles per hour at standard sea level conditions (rather than psi). If the airspeed meter indicates 150 mph when flying at an altitude of 10,000 ft, what is the true airspeed? †3.26 Estimate the force of a hurricane strength wind against the side of your house. List any assumptions and show all calculations. 3.27 A 40-mph wind blowing past your house speeds up as it flows up and over the roof. If elevation effects are negligible, determine (a) the pressure at the point on the roof where the speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psia. Would this effect tend to push the roof down against the house, or would it tend to lift the roof? (b) Determine the pressure on a window facing the wind if the window is assumed to be a stagnation point. 3.28 (See Fluids in the News article titled “Pressurized eyes,” Section 3.5.) Determine the air velocity needed to produce a stagnation pressure equal to 10 mm of mercury. Section 3.6.1 Free Jets 3.29 Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible. †3.30 GO Estimate the pressure needed at the pumper truck in order to shoot water from the street level onto a fire on the roof of a five-story building. List all assumptions and show all calculations. 3.31 Water flows from the faucet on the first floor of the building shown in Fig. P3.31 with a maximum velocity of 20 ft兾s. For steady inviscid flow, determine the maximum water velocity 144 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation from the basement faucet and from the faucet on the second floor 1assume each floor is 12 ft tall2. 3.35 Several holes are punched into a tin can as shown in Fig. P3.35. Which of the figures represents the variation of the water velocity as it leaves the holes? Justify your choice. 4 ft 8 ft (a) (b) (c) ■ Figure P3.35 V = 20 ft/s 4 ft 3.36 Water flows from a pressurized tank, through a 6-in.diameter pipe, exits from a 2-in.-diameter nozzle, and rises 20 ft above the nozzle as shown in Fig. P3.36. Determine the pressure in the tank if the flow is steady, frictionless, and incompressible. 12 ft 4 ft ■ Figure P3.31 3.32 Laboratories containing dangerous materials are often kept at a pressure slightly less than ambient pressure so that contaminants can be filtered through an exhaust system rather than leaked through cracks around doors, etc. If the pressure in such a room is 0.1 in. of water below that of the surrounding rooms, with what velocity will air enter the room through an opening? Assume viscous effects are negligible. †3.33 The “supersoaker” water gun shown in Fig. P3.33 can shoot more than 30 ft in the horizontal direction. Estimate the minimum pressure, p1, needed in the chamber in order to accomplish this. List all assumptions and show all calculations. 20 ft 2 in. Air 2 ft 6 in. ■ Figure P3.36 3.37 An inviscid, incompressible liquid flows steadily from the large pressurized tank shown in Fig. P.3.37. The velocity at the exit is 40 ft / s. Determine the specific gravity of the liquid in the tank. (1) 10 psi Air 5 ft Liquid ■ Figure P3.33 3.34 Streams of water from two tanks impinge upon each other as shown in Fig. P3.34. If viscous effects are negligible and point A is a stagnation point, determine the height h. 10 ft 40 ft/s ■ Figure P3.37 Section 3.6.2 Confined Flows (also see Lab Problems 3.1LP and 3.3LP) h Free jets A p1 = 25 psi Air 20 ft 8 ft ■ Figure P3.34 3.38 Obtain a photograph/image of a situation that involves a confined flow for which the Bernoulli and continuity equations are important. Print this photo and write a brief paragraph that describes the situation involved. 3.39 Air flows steadily through a horizontal 4-in.-diameter pipe and exits into the atmosphere through a 3-in.-diameter nozzle. The velocity at the nozzle exit is 150 ft/s. Determine the pressure in the pipe if viscous effects are negligible. 145 Problems 3.40 For the pipe enlargement shown in Fig. P3.40, the pressures at sections (1) and (2) are 56.3 and 58.2 psi, respectively. Determine the weight flowrate (lb/s) of the gasoline in the pipe. 3.44 Water flows steadily through the large tanks shown in Fig. P3.44. Determine the water depth, hA. Q Q 2.05 in. Gasoline 0.03–m diameter hA 3.71 in. (1) A (2) ■ Figure P3.40 hB = 2 m 0.05–m diameter B 3.41 GO A fire hose nozzle has a diameter of 118 in. According to some fire codes, the nozzle must be capable of delivering at least 250 galmin. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flowrate? 3.42 Water flowing from the 0.75-in.-diameter outlet shown in Video V8.15 and Fig. P3.42 rises 2.8 in. above the outlet. Determine the flowrate. ■ Figure P3.44 3.45 Water (assumed inviscid and incompressible) flows steadily in the vertical variable-area pipe shown in Fig. P3.45. Determine the flowrate if the pressure in each of the gages reads 50 kPa. 2m 10 m p = 50 kPa 2.8 in. 1m 0.75 in. Q Q ■ Figure P3.45 ■ Figure P3.42 3.43 Pop (with the same properties as water) flows from a 4-in.-diameter pop container that contains three holes as shown in Fig. P3.43 (see Video 3.9). The diameter of each fluid stream is 0.15 in., and the distance between holes is 2 in. If viscous effects are negligible and quasi-steady conditions are assumed, determine the time at which the pop stops draining from the top hole. Assume the pop surface is 2 in. above the top hole when t 0. Compare your results with the time you measure from the video. 3.46 Air is drawn into a wind tunnel used for testing automobiles as shown in Fig. P3.46. (a) Determine the manometer reading, h, when the velocity in the test section is 60 mph. Note that there is a 1-in. column of oil on the water in the manometer. (b) Determine the difference between the stagnation pressure on the front of the automobile and the pressure in the test section. Wind tunnel 60 mph Open h Surface at t = 0 2 in. 2 in. 2 in. 4 in. ■ Figure P3.43 0.15 in. Water Fan 1 in. Oil (SG = 0.9) ■ Figure P3.46 3.47 Natural gas (methane) flows from a 3-in.-diameter gas main, through a 1-in.-diameter pipe, and into the burner of a furnace at a rate of 100 ft3/hour. Determine the pressure in the gas main if the pressure in the 1-in. pipe is to be 6 in. of water greater than atmospheric pressure. Neglect viscous effects. 3.48 Small-diameter, high-pressure liquid jets can be used to cut various materials as shown in Fig. P3.48. If viscous effects are negligible, estimate the pressure needed to produce a 0.10-mm- 146 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation diameter water jet with a speed of 700 ms. Determine the flowrate. 3.51 Water flows through the pipe contraction shown in Fig. P3.51. For the given 0.2-m difference in manometer level, determine the flowrate as a function of the diameter of the small pipe, D. 0.2 m Q D 0.1 m 0.1 mm ■ Figure P3.51 3.52 Water flows through the pipe contraction shown in Fig. P3.52. For the given 0.2-m difference in the manometer level, determine the flowrate as a function of the diameter of the small pipe, D. ■ Figure P3.48 3.49 Water (assumed frictionless and incompressible) flows steadily from a large tank and exits through a vertical, constant diameter pipe as shown in Fig. P3.49. The air in the tank is pressurized to 50 kN/m2. Determine (a) the height h, to which the water rises, (b) the water velocity in the pipe, and (c) the pressure in the horizontal part of the pipe. 0.2 m 0.1 m Q D ■ Figure P3.52 Pipe exit 50 kN/m2 3.53 A 0.15-m-diameter pipe discharges into a 0.10-mdiameter pipe. Determine the velocity head in each pipe if they are carrying 0.12 m3s of kerosene. 3.54 Carbon tetrachloride flows in a pipe of variable diameter with negligible viscous effects. At point A in the pipe the pressure and velocity are 20 psi and 30 ft/s, respectively. At location B the pressure and velocity are 23 psi and 14 ft/s. Which point is at the higher elevation and by how much? h Air 4m 2m Water ■ Figure P3.49 3.50 Water (assumed inviscid and incompressible) flows steadily with a speed of 10 ft/s from the large tank shown in Fig. P3.50. Determine the depth, H, of the layer of light liquid 1specific weight 50 lb ft3 2 that covers the water in the tank. 3.55 Water flows from a 20-mm-diameter pipe with a flowrate Q as shown in Fig. P3.55. Plot the diameter of the water stream, d, as a function of distance below the faucet, h, for values of 0 h 1 m and 0 Q 0.004 m3/s. Discuss the validity of the one-dimensional assumption used to calculate d d(h), noting, in particular, the conditions of small h and small Q. Q 20 mm 3 50 lb/ft 10 ft/s d H 5 ft Water ■ Figure P3.50 h 4 ft ■ Figure P3.55 3.56 Water flows upward through a variable area pipe with a constant flowrate, Q, as shown in Fig. P3.56. If viscous effects are Problems negligible, determine the diameter, D(z), in terms of D1 if the pressure is to remain constant throughout the pipe. That is, p(z) p1. 147 will collapse and siphon will stop. If viscous effects are negligible, determine the minimum value of h allowed without the siphon stopping. 2m 4m h D(z) z ■ Figure P3.60 D1 (1) 3.61 Water is siphoned from the tank shown in Fig. P3.61. Determine the flowrate from the tank and the pressure at points (1), (2), and (3) if viscous effects are negligible. Q ■ Figure P3.56 3.57 The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determine the flowrate. 2-in.-diameter hose 2 ft (3) 3 ft 3.58 GO Water is siphoned from the tank shown in Fig. P3.58. The water barometer indicates a reading of 30.2 ft. Determine the maximum value of h allowed without cavitation occurring. Note that the pressure of the vapor in the closed end of the barometer equals the vapor pressure. 8 ft (2) (1) ■ Figure P3.61 Closed end 3.62 Redo Problem 3.61 if a 1-in-diameter nozzle is placed at the end of the tube. 3-in. diameter 3.63 A smooth plastic, 10-m-long garden hose with an inside diameter of 20 mm is used to drain a wading pool as is shown in Fig. P3.63. If viscous effects are neglected, what is the flowrate from the pool? 30.2 ft 6 ft 0.2 m h 0.23 m 5-in.-diameter ■ Figure P3.63 ■ Figure P3.58 3.59 Water is siphoned from a tank as shown in Fig. P3.59. Determine the flowrate and the pressure at point A, a stagnation point. 3.64 Water exits a pipe as a free jet and flows to a height h above the exit plane as shown in Fig. P3.64. The flow is steady, incompressible, and frictionless. (a) Determine the height h. (b) Determine the velocity and pressure at section (1). 0.04-m diameter h 3m V = 16 ft/s A 6-in. diameter 8 ft ■ Figure P3.59 (1) 3.60 A 50-mm-diameter plastic tube is used to siphon water from the large tank shown in Fig. P3.60. If the pressure on the outside of the tube is more than 30 kPa greater than the pressure within the tube, the tube 4-in. diameter ■ Figure P3.64 148 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation 3.65 Water flows steadily from a large, closed tank as shown in Fig. P3.65. The deflection in the mercury manometer is 1 in. and viscous effects are negligible. (a) Determine the volume flowrate. (b) Determine the air pressure in the space above the surface of the water in the tank. 3.69 Determine the flowrate through the pipe in Fig. P3.69. ρm = 900 kg/m3 2.5 m Water Air 0.08 m Q ■ Figure P3.69 1-ft diameter 3-in. diameter 8 ft 3.70 The specific gravity of the manometer fluid shown in Fig. P3.70 is 1.07. Determine the volume flowrate, Q, if the flow is inviscid and incompressible and the flowing fluid is (a) water, (b) gasoline, or (c) air at standard conditions. 1 in. Mercury Q ■ Figure P3.65 0.05 m 3.66 Carbon dioxide flows at a rate of 1.5 ft3/s from a 3-in. pipe in which the pressure and temperature are 20 psi (gage) and 120 F into a 1.5-in. pipe. If viscous effects are neglected and incompressible conditions are assumed, determine the pressure in the smaller pipe. 3.67 Oil of specific gravity 0.83 flows in the pipe shown in Fig. P3.67. If viscous effects are neglected, what is the flowrate? 0.09-m diameter 20 mm 10 mm ■ Figure P3.70 3.71 Water flows steadily with negligible viscous effects through the pipe shown in Fig. P3.71. It is known that the 4-in.diameter section of thin-walled tubing will collapse if the pressure within it becomes less than 10 psi below atmospheric pressure. Determine the maximum value that h can have without causing collapse of the tubing. 4 in. Water 4 ft 4-in.-diameter thin-walled tubing SG = 0.83 3 in. Q 4 in. h 6 in. ■ Figure P3.67 ■ Figure P3.71 3.68 Water flows steadily through the variable area pipe shown in Fig. P3.68 with negligible viscous effects. Determine the manometer reading, H, if the flowrate is 0.5 m3/s and the density of the manometer fluid is 600 kg/m3. 3.73 Water is pumped from a lake through an 8-in. pipe at a rate of 10 ft3s. If viscous effects are negligible, what is the pressure in the suction pipe 1the pipe between the lake and the pump2 at an elevation 6 ft above the lake? Density = 600 kg/m3 H Area = 0.05 m2 ■ Figure P3.68 3.72 Helium flows through a 0.30-m-diameter horizontal pipe with a temperature of 20 °C and a pressure of 200 kPa (abs) at a rate of 0.30 kg/s. If the pipe reduces to 0.25-m-diameter, determine the pressure difference between these two sections. Assume incompressible, inviscid flow. Area = 0.07 m2 3.74 Air is drawn into a small open-circuit wing tunnel as shown in Fig. P3.74. Atmospheric pressure is 98.7 kPa (abs) and the temperature is 27 C. If viscous effects are negligible, determine the pressure at the stagnation point on the nose of the airplane. Also determine the manometer reading, h, for the manometer attached to the static pressure tap within the test section of the wind tunnel if the air velocity within the test section is 50 m/s. 149 Problems Test section Inlet nozzle x x=L x=0 L Diffuser V0 Q Q Fan h Inlet Air H(x) H0 Water ■ Figure P3.74 dmax 3.75 Air flows through the device shown in Fig. P3.75. If the flowrate is large enough, the pressure within the constriction will be low enough to draw the water up into the tube. Determine the flowrate, Q, and the pressure needed at section (1) to draw the water into section (2). Neglect compressibility and viscous effects. Free jet 25 mm (2) (1) Air Q d Water ■ Figure P3.78 3.79 Water flows from a large tank and through a pipe of variable area as shown in Fig. 3.79. The area of the pipe is given by A A0[1 x(1 x/ᐉ)/2ᐉ], where A0 is the area at the beginning (x 0) and end (x ᐉ) of the pipe. Plot graphs of the pressure within the pipe as a function of distance along the pipe for water depths of h 1, 4, 10, and 25 m. 0.3 m 50 mm 50 mm Water ■ Figure P3.75 h 艎 3.76 Water flows steadily from the large open tank shown in Fig. 3.76. If viscous effects are negligible, determine (a) the flowrate, Q, and (b) the manometer reading, h. Free jet x x=0 ■ Figure P3.79 2m h 4m Q 0.10 m 0.08 m 3.80 If viscous effects are neglected and the tank is large, determine the flowrate from the tank shown in Fig. P3.80. Mercury ■ Figure P3.76 Oil, SG = 0.81 3.77 Water from a faucet fills a 16-oz glass (volume 28.9 in.3) in 20 s. If the diameter of the jet leaving the faucet is 0.60 in., what is the diameter of the jet when it strikes the water surface in the glass which is positioned 14 in. below the faucet? 3.78 Air flows steadily through a converging–diverging rectangular channel of constant width as shown in Fig. 3.78 and Video V3.10. The height of the channel at the exit and the exit velocity are H0 and V0 respectively. The channel is to be shaped so that the distance, d, that water is drawn up into tubes attached to static pressure taps along the channel wall is linear with distance along the channel. That is d (dmax/L) x, where L is the channel length and dmax is the maximum water depth (at the minimum channel height: x L). Determine the height, H(x), as a function of x and the other important parameters. 2m 0.7 m 50-mm diameter Water ■ Figure P3.80 3.81 Water flows steadily downward in the pipe shown in Fig. P.3.81 with negligible losses. Determine the flowrate. 150 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation Open 1 in. Oil SG = 0.7 Q = 2 gal/min 0.4-in.-diameter holes 1.5 m Stopper 1.2 m 2m 1m ■ Figure P3.84 ■ Figure P3.81 3.82 Water flows steadily from a large open tank and discharges into the atmosphere though a 3-in.-diameter pipe as shown in Fig. P3.82. Determine the diameter, d, in the narrowed section of the pipe at A if the pressure gages at A and B indicate the same pressure. 3.85 What pressure, p1, is needed to produce a flowrate of 0.09 ft3/s from the tank shown in Fig. P3.85? p1 Air B 9 ft Gasoline 2.0 ft Salt water SG = 1.1 3.6 ft 3-in. diameter 8 ft A 16 ft diameter = d 0.06-ft diameter ■ Figure P3.82 ■ Figure P3.85 3.83 Water flows from a large tank as shown in Fig. P3.83. Atmospheric pressure is 14.5 psia, and the vapor pressure is 1.60 psia. If viscous effects are neglected, at what height, h, will cavitation begin? To avoid cavitation, should the value of D1 be increased or decreased? To avoid cavitation, should the value of D2 be increased or decreased? Explain. 3.86 The vent on the tank shown in Fig. P3.86 is closed and the tank pressurized to increase the flowrate. What pressure, p1, is needed to produce twice the flowrate of that when the vent is open? p1 Vent 4 ft h Air D3 = 4 in. 10 ft Water D1 = 1 in. D2 = 2 in. Q ■ Figure P3.83 3.84 Water flows into the sink shown in Fig. P3.84 and Video V5.1 at a rate of 2 gal/min. If the drain is closed, the water will eventually flow through the overflow drain holes rather than over the edge of the sink. How many 0.4-in.-diameter drain holes are needed to ensure that the water does not overflow the sink? Neglect viscous effects. ■ Figure P3.86 3.87 Water is siphoned from the tank shown in Fig. P3.87. Determine the flowrate from the tank and the pressures at points (1), (2), and (3) if viscous effects are negligible. Problems 151 A (1) 3 ft (2) (3) 3 ft 4 ft 0.1 ft 5 ft 0.15-ft diameter Q 0.1-ft diameter ■ Figure P3.87 End of pipe ■ Figure P3.90 3.88 Water is siphoned from a large tank and discharges into the atmosphere through a 2-in.-diameter tube as shown in Fig. P3.88. The end of the tube is 3 ft below the tank bottom, and viscous effects are negligible. (a) Determine the volume flowrate from the tank. (b) Determine the maximum height, H, over which the water can be siphoned without cavitation occurring. Atmospheric pressure is 14.7 psia, and the water vapor pressure is 0.26 psia. 3.91 JP-4 fuel (SG 0.77) flows through the Venturi meter shown in Fig. P3.91 with a velocity of 15 ft/s in the 6-in. pipe. If viscous effects are negligible, determine the elevation, h, of the fuel in the open tube connected to the throat of the Venturi meter. H 6 ft 6 in. 2-in. diameter 9 ft JP-4 fuel 8 in. 20° 4 in. h 3 ft 6 in. V = 15 ft/s ■ Figure P3.91 ■ Figure P3.88 3.89 Determine the manometer reading, h, for the flow shown in Fig. P3.89. 3.92 Water, considered an inviscid, incompressible fluid, flows steadily as shown in Fig. P3.92. Determine h. h Air Q = 4 ft3/s h 0.37 m Water 0.5-ft diameter 0.08-m diameter Free jet 1-ft diameter 3 ft 0.05-m diameter ■ Figure P3.89 3.90 Water flows steadily from the pipe shown in Fig. P3.90 with negligible viscous effects. Determine the maximum flowrate if the water is not to flow from the open vertical tube at A. ■ Figure P3.92 3.93 Determine the flowrate through the submerged orifice shown in Fig. P3.93 if the contraction coefficient is Cc 0.63. 152 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation ■ Figure P3.97 6 ft 4 ft 3-in. diameter 3.98 When the drain plug is pulled, water flows from a hole in the bottom of a large, open cylindrical tank. Show that if viscous effects are negligible and if the flow is assumed to be quasisteady, then it takes 3.41 times longer to empty the entire tank than it does to empty the first half of the tank. Explain why this is so. 2 ft ■ Figure P3.93 3.94 An ancient device for measuring time is shown in Fig. P3.94. The axisymmetric vessel is shaped so that the water level falls at a constant rate. Determine the shape of the vessel, R R(z), if the water level is to decrease at a rate of 0.10 m/hr and the drain hole is 5.0 mm in diameter. The device is to operate for 12 hr without needing refilling. Make a scale drawing of the shape of the vessel. 3.99 The surface area, A, of the pond shown in Fig. P3.99 varies with the water depth, h, as shown in the table. At time t 0 a valve is opened and the pond is allowed to drain through a pipe of diameter D. If viscous effects are negligible and quasisteady conditions are assumed, plot the water depth as a function of time from when the valve is opened (t 0) until the pond is drained for pipe diameters of D 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0 ft. Assume h 18 ft at t 0. Area A R(z) h 3 ft z D ■ Figure P3.99 5.0-mm diameter ■ Figure P3.94 3.95 A long water trough of triangular cross section is formed from two planks as is shown in Fig. P3.95. A gap of 0.1 in. remains at the junction of the two planks. If the water depth initially was 2 ft, how long a time does it take for the water depth to reduce to 1 ft? 90° 2 ft h (ft) A [acres (1 acre ⫽ 43,560 ft2)] 0 2 4 6 8 10 12 14 16 18 0 0.3 0.5 0.8 0.9 1.1 1.5 1.8 2.4 2.8 3.100 Water flows through a horizontal branching pipe as shown in Fig. P3.100. Determine the pressure at section (3). 0.1 in. V3 A3 = 0.07 m2 ■ Figure P3.95 (3) 3.96 A spherical tank of diameter D has a drain hole of diameter d at its bottom. A vent at the top of the tank maintains atmospheric pressure at the liquid surface within the tank. The flow is quasisteady and inviscid and the tank is full of water initially. Determine the water depth as a function of time, h h(t), and plot graphs of h(t) for tank diameters of 1, 5, 10, and 20 ft if d 1 in. †3.97 A small hole develops in the bottom of the stationary rowboat shown in Fig. P3.97. Estimate the amount of time it will take for the boat to sink. List all assumptions and show all calculations. (2) (1) V1 = 4 m/s p1 = 400 kPa A1 = 0.1 m2 ■ Figure P3.100 V2 p2 = 350 kPa A2 = 0.02 m2 Problems 3.101 Water flows through the horizontal Y-fitting shown in Fig. P3.101. If the flowrate and pressure in pipe (1) are Q1 2.3 ft3/s and p1 50 lb/ft3, determine the pressures p2 and p3, in pipes (2) and (3) under the assumption that the flowrate divides evenly between pipes (2) and (3). 153 3m 7m 0.03-m diameter 0.05-m diameter Q3 (1) (3) 0.02-m diameter ■ Figure P3.104 0.20 ft 0.3 ft Q1 3.105 Air, assumed incompressible and inviscid, flows into the outdoor cooking grill through nine holes of 0.40-in. diameter as shown in Fig. P3.105. If a flowrate of 40 in.3/s into the grill is required to maintain the correct cooking conditions, determine the pressure within the grill near the holes. (1) (2) 0.25 ft Q2 ■ Figure P3.101 3.102 Water flows through the branching pipe shown in Fig. P3.102. If viscous effects are negligible, determine the pressure at section (2) and the pressure at section (3). Q1 = 1 m3/s A1 = 0.1 m2 p1 = 300 kPa z1 = 0 A3 = 0.035 m2 z3 = 10 m 9 holes, each 0.40-in. diameter (3) V2 = 14 m/s A2 = 0.03 m2 z2 = 0 ■ Figure P3.105 (2) (1) ■ Figure P3.102 3.103 Water flows through the horizontal branching pipe shown in Fig. P3.103 at a rate of 10 ft3/s. If viscous effects are negligible, determine the water speed at section (2), the pressure at section (3), and the flowrate at section (4). A2 = 0.07 ft2 p2 = 5.0 psi 3.106 An air cushion vehicle is supported by forcing air into the chamber created by a skirt around the periphery of the vehicle as shown in Fig. P3.106. The air escapes through the 3-in. clearance between the lower end of the skirt and the ground (or water). Assume the vehicle weighs 10,000 lb and is essentially rectangular in shape, 30 by 65 ft. The volume of the chamber is large enough so that the kinetic energy of the air within the chamber is negligible. Determine the flowrate, Q, needed to support the vehicle. If the ground clearance were reduced to 2 in., what flowrate would be needed? If the vehicle weight were reduced to 5000 lb and the ground clearance maintained at 3 in., what flowrate would be needed? (2) A3 = 0.2 ft2 V3 = 20 ft /s (1) Fan Q Vehicle Skirt A1 = 1 ft2 Q1 = 10 ft3/s p1 = 10 psi (3) (4) 3 in. ■ Figure P3.103 ■ Figure P3.106 3.104 Water flows from a large tank through a large pipe that splits into two smaller pipes as shown in Fig. P3.104. If viscous effects are negligible, determine the flowrate from the tank and the pressure at point (1). 3.107 Water flows from the pipe shown in Fig. P3.107 as a free jet and strikes a circular flat plate. The flow geometry shown is axisymmetrical. Determine the flowrate and the manometer reading, H. 154 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation blows, the harder the card “sticks” to the spool. In fact by blowing hard enough it is possible to keep the card against the spool with the spool turned upside down. (Note: It may be necessary to use a thumb tack to prevent the card from sliding from the spool.) Explain this phenomenon. H 0.1-m diameter V Card 0.4 mm 0.2 m 0.01-m diameter Spool Pipe Q Q ■ Figure P3.107 3.108 A conical plug is used to regulate the airflow from the pipe shown in Fig. P3.108. The air leaves the edge of the cone with a uniform thickness of 0.02 m. If viscous effects are negligible and the flowrate is 0.50 m3/s, determine the pressure within the pipe. V Free jet ■ Figure P3.110 3.111 Observations show that it is not possible to blow the table tennis ball from the funnel shown in Fig. P3.111a. In fact, the ball can be kept in an inverted funnel, Fig P3.111b, by blowing through it. The harder one blows through the funnel, the harder the ball is held within the funnel. Explain this phenomenon. Pipe Q 0.20 m Q = 0.50 m3/s 0.23 m Cone 0.02 m V ■ Figure P3.108 Q 3.109 Water flows steadily from a nozzle into a large tank as shown in Fig. P3.109. The water then flows from the tank as a jet of diameter d. Determine the value of d if the water level in the tank remains constant. Viscous effects are negligible. 4 ft 0.1-ft diameter 0.15-ft diameter (a) (b) ■ Figure P3.111 3.112 Water flows down the sloping ramp shown in Fig. P3.112 with negligible viscous effects. The flow is uniform at sections (1) and (2). For the conditions given show that three solutions for the downstream depth, h2, are obtained by use of the Bernoulli and continuity equations. However, show that only two of these solutions are realistic. Determine these values. V1 = 10 ft/s h1 = 1 ft H = 2 ft h2 V2 d 3 ft 1 ft ■ Figure P3.109 3.110 A small card is placed on top of a spool as shown in Fig. P3.110. It is not possible to blow the card off the spool by blowing air through the hole in the center of the spool. The harder one ■ Figure P3.112 3.113 Water flows in a rectangular channel that is 2.0 m wide as shown in Fig. P3.113. The upstream depth is 70 mm. The water surface rises 40 mm as it passes over a portion where the channel bottom rises 10 mm. If viscous effects are negligible, what is the flowrate? Problems 100 mm 70 mm Q 10 mm 155 3.120 The flowrate in a water channel is sometimes determined by use of a device called a Venturi flume. As shown in Fig. P3.120, this device consists simply of a bump on the bottom of the channel. If the water surface dips a distance of 0.07 m for the conditions shown, what is the flowrate per width of the channel? Assume the velocity is uniform and viscous effects are negligible. ■ Figure P3.113 0.07 m Section 3.6.3 Flowrate Measurement (also see Lab Problems 3.2LP and 3.4LP) 3.114 Obtain a photograph/image of a situation that involves some type of flowmeter. Print this photo and write a brief paragraph that describes the situation involved. 3.115 A Venturi meter with a minimum diameter of 3 in. is to be used to measure the flowrate of water through a 4-in.-diameter pipe. Determine the pressure difference indicated by the pressure gage attached to the flowmeter if the flowrate is 0.5 ft3/s and viscous effects are negligible. 3.116 Determine the flowrate through the Venturi meter shown in Fig. P3.116 if ideal conditions exist. p1 = 735 kPa Q V1 V2 1.2 m 0.2 m ■ Figure P3.120 3.121 Water flows under the inclined sluice gate shown in Fig. P3.121. Determine the flowrate if the gate is 8 ft wide. p2 = 550 kPa 19 mm 31 mm 30° 6 ft γ = 9.1 kN/m3 ■ Figure P3.116 1.6 ft 1 ft 3.117 GO For what flowrate through the Venturi meter of Problem 3.116 will cavitation begin if p1 275 kPa gage, atmospheric pressure is 101 kPa (abs), and the vapor pressure is 3.6 kPa (abs)? ■ Figure P3.121 3.118 What diameter orifice hole, d, is needed if under ideal conditions the flowrate through the orifice meter of Fig. P3.118 is to be 30 gal/min of seawater with p1 p2 2.37 lb/in.2? The contraction coefficient is assumed to be 0.63. Section 3.7 The Energy Line and the Hydraulic Grade Line p1 p2 d Q 2-in. diameter ■ Figure P3.118 3.119 A weir (see Video V10.13) of trapezoidal cross section is used to measure the flowrate in a channel as shown in Fig. P3.119. If the flowrate is Q0 when H /2, what flowrate is expected when H /? 3.122 Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressure of 200 kPa at an elevation of 25 m. Determine the velocity head and pressure head at elevations of 20 and 55 m. 3.123 Draw the energy line and the hydraulic grade line for the flow of Problem 3.83. 3.124 Draw the energy line and hydraulic grade line for the flow shown in Problem 3.71. Section 3.8 Restrictions on Use of the Bernoulli Equation 3.125 Obtain a photograph/image of a flow in which it would not be appropriate to use the Bernoulli equation. Print this photo and write a brief paragraph that describes the situation involved. 3.126 Listed below are typical flight speeds for two aircraft. For which of these conditions would it be reasonable to use the incompressible Bernoulli equation to study the aerodynamics associated with their flight? Explain. H 30° Flight speed, km/hr ᐉ ■ Figure P3.119 Aircraft Cruise Landing approach Boeing 787 F-22 fighter 913 1960 214 250 156 Chapter 3 ■ Elementary Fluid Dynamics—The Bernoulli Equation 3.127 A meteorologist uses a Pitot-static tube to measure the wind speed in a tornado. Based on the damage caused by the storm, the tornado is rated as EF5 on the Enhanced Fujita Scale. This means that the wind speed is estimated to be in the range of 261 to 318 mph. Is it reasonable to use the incompressible Pitot-tube equation (Eq. 3.16) to determine the actual wind speed, or must compressible effects to taken into account? Explain. ■ Lab Problems 3.1 LP This problem involves the pressure distribution between two parallel circular plates. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 3.2 LP This problem involves the calibration of a nozzle-type flowmeter. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley. com/college/munson. 3.3 LP This problem involves the pressure distribution in a twodimensional channel. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 3.4 LP This problem involves the determination of the flowrate under a sluice gate as a function of the water depth. To proceed with this problem, go to Appendix H which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Lifelong Learning Problems 3.1 LL The concept of the use of a Pitot-static tube to measure the airspeed of an airplane is rather straightforward. However, the design and manufacture of reliable, accurate, inexpensive Pitotstatic tube airspeed indicators is not necessarily simple. Obtain information about the design and construction of modern Pitot-static tubes. Summarize your findings in a brief report. 3.2 LL In recent years damage due to hurricanes has been significant, particularly in the southeastern United States. The low barometric pressure, high winds, and high tides generated by hurricanes can combine to cause considerable damage. According to some experts, in the coming years hurricane frequency may increase because of global warming. Obtain information about the fluid mechanics of hurricanes. Summarize your findings in a brief report. 3.3 LL Orifice, nozzle, or Venturi flowmeters have been used for a long time to predict accurately the flowrate in pipes. However, recently there have been several new concepts suggested or used for such flowrate measurements. Obtain information about new methods to obtain pipe flowrate information. Summarize your findings in a brief report. 3.4 LL Ultra-high-pressure, thin jets of liquids can be used to cut various materials ranging from leather to steel and beyond. Obtain information about new methods and techniques proposed for liquid jet cutting and investigate how they may alter various manufacturing processes. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 4 Fluid Kinematics CHAPTER OPENING PHOTO: A vortex ring: The complex, three-dimensional structure of a smoke ring is indicated in this cross-sectional view. 1Smoke in air.2 3Photograph courtesy of R. H. Magarvey and C. S. MacLatchy 1Ref. 42.4 Learning Objectives After completing this chapter, you should be able to: ■ discuss the differences between the Eulerian and Lagrangian descriptions of fluid motion. ■ identify various flow characteristics based on the velocity field. ■ determine the streamline pattern and acceleration field given a velocity field. ■ discuss the differences between a system and control volume. ■ apply the Reynolds transport theorem and the material derivative. V4.1 Streaklines 4.1 In this chapter we will discuss various aspects of fluid motion without being concerned with the actual forces necessary to produce the motion. That is, we will consider the kinematics of the motion—the velocity and acceleration of the fluid, and the description and visualization of its motion. The analysis of the specific forces necessary to produce the motion 1the dynamics of the motion2 will be discussed in detail in the following chapters. A wide variety of useful information can be gained from a thorough understanding of fluid kinematics. Such an understanding of how to describe and observe fluid motion is an essential step to the complete understanding of fluid dynamics. The Velocity Field In general, fluids flow. That is, there is a net motion of molecules from one point in space to another point as a function of time. As is discussed in Chapter 1, a typical portion of fluid contains so many molecules that it becomes totally unrealistic 1except in special cases2 for us to attempt to account for the motion of individual molecules. Rather, we employ the continuum hypothesis and consider fluids to be made up of fluid particles that interact with each other and with their surroundings. Each particle contains numerous molecules. Thus, we can describe the flow of a fluid in terms of the motion of fluid particles rather than individual molecules. This motion can be described in terms of the velocity and acceleration of the fluid particles. The infinitesimal particles of a fluid are tightly packed together 1as is implied by the continuum assumption2. Thus, at a given instant in time, a description of any fluid property 1such as density, 157 158 Chapter 4 ■ Fluid Kinematics z Particle A at time t + δ t Particle path Particle A at time t rA(t) rA(t + δ t) y x V4.2 Velocity field ■ Figure 4.1 Particle location in terms of its position vector. pressure, velocity, and acceleration2 may be given as a function of the fluid’s location. This representation of fluid parameters as functions of the spatial coordinates is termed a field representation of the flow. Of course, the specific field representation may be different at different times, so that to describe a fluid flow we must determine the various parameters not only as a function of the spatial coordinates 1x, y, z, for example2 but also as a function of time, t. Thus, to completely specify the temperature, T, in a room we must specify the temperature field, T ⫽ T 1x, y, z, t2, throughout the room 1from floor to ceiling and wall to wall2 at any time of the day or night. Shown in the margin figure is one of the most important fluid variables, the velocity field, w V u v z ^ j ^ k ^ x i V ⫽ u1x, y, z, t2iˆ ⫹ v1x, y, z, t2jˆ ⫹ w1x, y, z, t2kˆ y where u, v, and w are the x, y, and z components of the velocity vector. By definition, the velocity of a particle is the time rate of change of the position vector for that particle. As is illustrated in Fig. 4.1, the position of particle A relative to the coordinate system is given by its position vector, rA, which 1if the particle is moving2 is a function of time. The time derivative of this position gives the velocity of the particle, drAⲐdt ⫽ VA. By writing the velocity for all of the particles, we can obtain the field description of the velocity vector V ⫽ V1x, y, z, t2. Since the velocity is a vector, it has both a direction and a magnitude. The magnitude of V, denoted V ⫽ 0V 0 ⫽ 1u2 ⫹ v2 ⫹ w2 2 1Ⲑ2, is the speed of the fluid. 1It is very common in practical situations to call V velocity rather than speed, i.e., “the velocity of the fluid is 12 m兾s.”2 As is discussed in the next section, a change in velocity results in an acceleration. This acceleration may be due to a change in speed and/or direction. V4.3 Cylindervelocity vectors F l u i d s i n Follow those particles Superimpose two photographs of a bouncing ball taken a short time apart and draw an arrow between the two images of the ball. This arrow represents an approximation of the velocity (displacement/time) of the ball. The particle image velocimeter (PIV) uses this technique to provide the instantaneous velocity field for a given cross section of a flow. The flow being studied is seeded with numerous micron-sized particles that are small enough to follow the flow yet big enough to reflect enough light to be captured by the camera. The flow is t h e N e w s illuminated with a light sheet from a double-pulsed laser. A digital camera captures both light pulses on the same image frame, allowing the movement of the particles to be tracked. By using appropriate computer software to carry out a pixel-by-pixel interrogation of the double image, it is possible to track the motion of the particles and determine the two components of velocity in the given cross section of the flow. By using two cameras in a stereoscopic arrangement, it is possible to determine all three components of velocity. (See Problem 4.18.) 4.1 E XAMPLE The Velocity Field 159 4.1 Velocity Field Representation GIVEN A velocity field is given by V ⫽ 1V0 Ⲑ/2 1⫺xiˆ ⫹ yjˆ 2 where V0 and / are constants. FIND At what location in the flow field is the speed equal to V0? Make a sketch of the velocity field for x ⱖ 0 by drawing arrows representing the fluid velocity at representative locations. SOLUTION The x, y, and z components of the velocity are given by u ⫽ ⫺V0 xⲐ/, v ⫽ V0 yⲐ/, and w ⫽ 0 so that the fluid speed, V, is V ⫽ 1u2 ⫹ v2 ⫹ w2 2 1Ⲑ2 ⫽ V0 2 1x ⫹ y2 2 1Ⲑ 2 / (1) The speed is V ⫽ V0 at any location on the circle of radius / centered (Ans) at the origin 3 1x2 ⫹ y2 2 1Ⲑ2 ⫽ /4 as shown in Fig. E4.1a. The direction of the fluid velocity relative to the x axis is given in terms of u ⫽ arctan 1vⲐu2 as shown in Fig. E4.1b. For this flow tan u ⫽ V0 yⲐ/ y v ⫽ ⫽ u ⫺x ⫺V0 xⲐ/ Thus, along the x axis 1y ⫽ 02 we see that tan u ⫽ 0, so that u ⫽ 0° or u ⫽ 180°. Similarly, along the y axis 1x ⫽ 02 we obtain tan u ⫽ ⫾q so that u ⫽ 90° or u ⫽ 270°. Also, for y ⫽ 0 we find V ⫽ 1⫺V0 xⲐ/2iˆ, while for x ⫽ 0 we have V ⫽ 1V0yⲐ/2jˆ, indicating 1if V0 7 02 that the flow is directed away from the origin along the y axis and toward the origin along the x axis as shown in Fig. E4.1a. By determining V and u for other locations in the x–y plane, the velocity field can be sketched as shown in the figure. For example, on the line y ⫽ x the velocity is at a 45° angle relative to the x axis 1tan u ⫽ vⲐu ⫽ ⫺yⲐx ⫽ ⫺12. At the origin x ⫽ y ⫽ 0 so that V ⫽ 0. This point is a stagnation point. The farther from the origin the fluid is, the faster it is flowing 1as seen from Eq. 12. By careful consideration of the velocity field it is possible to determine considerable information about the flow. COMMENT The velocity field given in this example approximates the flow in the vicinity of the center of the sign shown in Fig. E4.1c. When wind blows against the sign, some air flows over the sign, some under it, producing a stagnation point as indicated. 2V0 y 2ᐉ V0 V ᐉ θ v u (b) V0/2 2V0 V0 0 x y −ᐉ V0 V=0 −2ᐉ 2V0 (c) (a) ■ Figure E4.1 x 160 Chapter 4 ■ Fluid Kinematics The figure in the margin shows the velocity field (i.e., velocity vectors) at a given instant for flow past two square bars. It is possible to obtain much qualitative and quantitative information for complex flows by using plots such as this. 4.1.1 Eulerian and Lagrangian Flow Descriptions Flow (Photograph courtesy of Alvaro Valencia) Flow (Courtesy of Chris Griffin) Either Eulerian or Lagrangian methods can be used to describe flow fields. There are two general approaches in analyzing fluid mechanics problems 1or problems in other branches of the physical sciences, for that matter2. The first method, called the Eulerian method, uses the field concept introduced above. In this case, the fluid motion is given by completely prescribing the necessary properties 1pressure, density, velocity, etc.2 as functions of space and time. From this method we obtain information about the flow in terms of what happens at fixed points in space as the fluid flows through those points. A typical Eulerian representation of the flow is shown by the figure in the margin which involves flow past an airfoil at angle of attack. The pressure field is indicated by using a contour plot showing lines of constant pressure, with gray shading indicating the intensity of the pressure. The second method, called the Lagrangian method, involves following individual fluid particles as they move about and determining how the fluid properties associated with these particles change as a function of time. That is, the fluid particles are “tagged” or identified, and their properties determined as they move. The difference between the two methods of analyzing fluid flow problems can be seen in the example of smoke discharging from a chimney, as is shown in Fig. 4.2. In the Eulerian method one may attach a temperature-measuring device to the top of the chimney 1point 02 and record the temperature at that point as a function of time. At different times there are different fluid particles passing by the stationary device. Thus, one would obtain the temperature, T, for that location 1x ⫽ x0, y ⫽ y0, and z ⫽ z0 2 as a function of time. That is, T ⫽ T 1x0, y0, z0, t2. The use of numerous temperature-measuring devices fixed at various locations would provide the temperature field, T ⫽ T1x, y, z, t2. The temperature of a particle as a function of time would not be known unless the location of the particle were known as a function of time. In the Lagrangian method, one would attach the temperature-measuring device to a particular fluid particle 1particle A2 and record that particle’s temperature as it moves about. Thus, one would obtain that particle’s temperature as a function of time, TA ⫽ TA 1t2. The use of many such measuring devices moving with various fluid particles would provide the temperature of these fluid particles as a function of time. The temperature would not be known as a function of position unless the location of each particle were known as a function of time. If enough information in Eulerian form is available, Lagrangian information can be derived from the Eulerian data—and vice versa. Example 4.1 provides an Eulerian description of the flow. For a Lagrangian description we would need to determine the velocity as a function of time for each particle as it flows along from one point to another. In fluid mechanics it is usually easier to use the Eulerian method to describe a flow—in either experimental or analytical investigations. There are, however, certain instances in which the Lagrangian method is more convenient. For example, some numerical fluid mechanics calculations are based on determining the motion of individual fluid particles 1based on the appropriate interactions among the particles2, thereby describing the motion in Lagrangian terms. Similarly, in y Location 0: T = T(x0, y0, t) 0 y0 Particle A: TA = TA(t) x x0 ■ Figure 4.2 Eulerian and Lagrangian descriptions of temperature of a flowing fluid. 4.1 V4.4 Follow the particles (experiment) V4.5 Follow the particles (computer) The Velocity Field 161 some experiments individual fluid particles are “tagged” and are followed throughout their motion, providing a Lagrangian description. Oceanographic measurements obtained from devices that flow with the ocean currents provide this information. Similarly, by using X-ray opaque dyes, it is possible to trace blood flow in arteries and to obtain a Lagrangian description of the fluid motion. A Lagrangian description may also be useful in describing fluid machinery 1such as pumps and turbines2 in which fluid particles gain or lose energy as they move along their flow paths. Another illustration of the difference between the Eulerian and Lagrangian descriptions can be seen in the following biological example. Each year thousands of birds migrate between their summer and winter habitats. Ornithologists study these migrations to obtain various types of important information. One set of data obtained is the rate at which birds pass a certain location on their migration route 1birds per hour2. This corresponds to an Eulerian description—“flowrate” at a given location as a function of time. Individual birds need not be followed to obtain this information. Another type of information is obtained by “tagging” certain birds with radio transmitters and following their motion along the migration route. This corresponds to a Lagrangian description— “position” of a given particle as a function of time. 4.1.2 One-, Two-, and Three-Dimensional Flows Most flow fields are actually threedimensional. V4.6 Flow past a wing Generally, a fluid flow is a rather complex three-dimensional, time-dependent phenomenon— V ⫽ V1x, y, z, t2 ⫽ uiˆ ⫹ vjˆ ⫹ wkˆ. In many situations, however, it is possible to make simplifying assumptions that allow a much easier understanding of the problem without sacrificing needed accuracy. One of these simplifications involves approximating a real flow as a simpler one- or twodimensional flow. In almost any flow situation, the velocity field actually contains all three velocity components 1u, v, and w, for example2. In many situations the three-dimensional flow characteristics are important in terms of the physical effects they produce. (See the photograph at the beginning of Chapter 4.) For these situations it is necessary to analyze the flow in its complete three-dimensional character. Neglect of one or two of the velocity components in these cases would lead to considerable misrepresentation of the effects produced by the actual flow. The flow of air past an airplane wing provides an example of a complex three-dimensional flow. A feel for the three-dimensional structure of such flows can be obtained by studying Fig. 4.3, which is a photograph of the flow past a model wing; the flow has been made visible by using a flow visualization technique. In many situations one of the velocity components may be small 1in some sense2 relative to the two other components. In situations of this kind it may be reasonable to neglect the smaller component and assume two-dimensional flow. That is, V ⫽ uiˆ ⫹ vjˆ, where u and v are functions of x and y 1and possibly time, t2. It is sometimes possible to further simplify a flow analysis by assuming that two of the velocity components are negligible, leaving the velocity field to be approximated as a onedimensional flow field. That is, V ⫽ uiˆ. As we will learn from examples throughout the remainder of the book, although there are very few, if any, flows that are truly one-dimensional, there are ■ Figure 4.3 Flow visualization of the complex three-dimensional flow past a delta wing. (Photograph courtesy of T. T. Lim and Y. D. Cui, National University of Singapore.) 162 Chapter 4 ■ Fluid Kinematics many flow fields for which the one-dimensional flow assumption provides a reasonable approximation. There are also many flow situations for which use of a one-dimensional flow field assumption will give completely erroneous results. 4.1.3 Steady and Unsteady Flows Solenoid off,valve closed Spring Solenoid Diaphragm Outlet Inlet Solenoid on,valve open V4.7 Flow types F l u In the previous discussion we have assumed steady flow—the velocity at a given point in space does not vary with time, 0V0t 0. In reality, almost all flows are unsteady in some sense. That is, the velocity does vary with time. It is not difficult to believe that unsteady flows are usually more difficult to analyze 1and to investigate experimentally2 than are steady flows. Hence, considerable simplicity often results if one can make the assumption of steady flow without compromising the usefulness of the results. Among the various types of unsteady flows are nonperiodic flow, periodic flow, and truly random flow. Whether or not unsteadiness of one or more of these types must be included in an analysis is not always immediately obvious. An example of a nonperiodic, unsteady flow is that produced by turning off a faucet to stop the flow of water. Usually this unsteady flow process is quite mundane, and the forces developed as a result of the unsteady effects need not be considered. However, if the water is turned off suddenly 1as with the electrically operated valve in a dishwasher shown in the figure in the margin2, the unsteady effects can become important [as in the “water hammer” effects made apparent by the loud banging of the pipes under such conditions 1Ref. 12]. In other flows the unsteady effects may be periodic, occurring time after time in basically the same manner. The periodic injection of the air–gasoline mixture into the cylinder of an automobile engine is such an example. The unsteady effects are quite regular and repeatable in a regular sequence. They are very important in the operation of the engine. i d s i n New pulsed liquid-jet scalpel High-speed liquid-jet cutters are used for cutting a wide variety of materials such as leather goods, jigsaw puzzles, plastic, ceramic, and metal. Typically, compressed air is used to produce a continuous stream of water that is ejected from a tiny nozzle. As this stream impacts the material to be cut, a high pressure (the stagnation pressure) is produced on the surface of the material, thereby cutting the material. Such liquid-jet cutters work well in air but are difficult to control if the jet must pass through a liquid as often happens V4.8 Jupiter red spot t h e N e w s in surgery. Researchers have developed a new pulsed jet cutting tool that may allow surgeons to perform microsurgery on tissues that are immersed in water. Rather than using a steady water jet, the system uses unsteady flow. A high-energy electrical discharge inside the nozzle momentarily raises the temperature of the microjet to approximately 10,000 °C. This creates a rapidly expanding vapor bubble in the nozzle and expels a tiny fluid jet from the nozzle. Each electrical discharge creates a single, brief jet, which makes a small cut in the material. In many situations the unsteady character of a flow is quite random. That is, there is no repeatable sequence or regular variation to the unsteadiness. This behavior occurs in turbulent flow and is absent from laminar flow. The “smooth” flow of highly viscous syrup onto a pancake represents a “deterministic” laminar flow. It is quite different from the turbulent flow observed in the “irregular” splashing of water from a faucet onto the sink below it. The “irregular” gustiness of the wind represents another random turbulent flow. The differences between these types of flows are discussed in considerable detail in Chapters 8 and 9. It must be understood that the definition of steady or unsteady flow pertains to the behavior of a fluid property as observed at a fixed point in space. For steady flow, the values of all fluid properties 1velocity, temperature, density, etc.2 at any fixed point are independent of time. However, the value of those properties for a given fluid particle may change with time as the particle flows along, even in steady flow. Thus, the temperature of the exhaust at the exit of a car’s exhaust pipe may be constant for several hours, but the temperature of a fluid particle that left the exhaust pipe five minutes ago is lower now than it was when it left the pipe, even though the flow is steady. 4.1.4 Streamlines, Streaklines, and Pathlines Although fluid motion can be quite complicated, there are various concepts that can be used to help in the visualization and analysis of flow fields. To this end we discuss the use of streamlines, 4.1 dx v u x dy v u dx V4.9 Streamlines (4.1) If the velocity field is known as a function of x and y 1and t if the flow is unsteady2, this equation can be integrated to give the equation of the streamlines. For unsteady flow there is no easy way to produce streamlines experimentally in the laboratory. As discussed below, the observation of dye, smoke, or some other tracer injected into a flow can provide useful information, but for unsteady flows it is not necessarily information about the streamlines. E XAMPLE 4.2 Streamlines for a Given Velocity Field GIVEN Consider the two-dimensional steady flow discussed in Example 4.1, V 1V0 /21xiˆ yjˆ 2. FIND Determine the streamlines for this flow. SOLUTION Since y u 1V0 /2x and v 1V0 /2y (1) it follows that streamlines are given by solution of the equation 1V0/2y dy y v u x dx 1V0/2x in which variables can be separated and the equation integrated to give 冮 or dy y 冮 4 2 C=9 dx x C=4 ln y ln x constant 0 Thus, along the streamline xy C, 163 streaklines, and pathlines in flow analysis. The streamline is often used in analytical work, while the streakline and pathline are often used in experimental work. A streamline is a line that is everywhere tangent to the velocity field. If the flow is steady, nothing at a fixed point 1including the velocity direction2 changes with time, so the streamlines are fixed lines in space. (See the photograph at the beginning of Chapter 6.) For unsteady flows the streamlines may change shape with time. Streamlines are obtained analytically by integrating the equations defining lines tangent to the velocity field. As illustrated in the margin figure, for two-dimensional flows the slope of the streamline, dydx, must be equal to the tangent of the angle that the velocity vector makes with the x axis or y dy The Velocity Field where C is a constant C=1 2 4 C = –1 C = –4 (Ans) By using different values of the constant C, we can plot various lines in the x–y plane—the streamlines. The streamlines for x 0 are plotted in Fig. E4.2. A comparison of this figure with Fig. E4.1a illustrates the fact that streamlines are lines tangent to the velocity field. COMMENT Note that a flow is not completely specified by the shape of the streamlines alone. For example, the streamlines for the flow with V0/ 10 have the same shape as those for the flow with V0/ 10. However, the direction of the flow is opposite for these two cases. The arrows in Fig. E4.2 representing the flow direction are correct for V0/ 10 since, from Eq. 1, u 10x and v 10y. That is, the flow is from right to left. For V0/ 10 the arrows are reversed. The flow is from left to right. –2 –4 ■ Figure E4.2 C = –9 x 164 Chapter 4 ■ Fluid Kinematics A streakline consists of all particles in a flow that have previously passed through a common point. Streaklines are more of a laboratory tool than an analytical tool. They can be obtained by taking instantaneous photographs of marked particles that all passed through a given location in the flow field at some earlier time. Such a line can be produced by continuously injecting marked fluid 1neutrally buoyant smoke in air or dye in water2 at a given location 1Ref. 22. (See Fig. 9.1.) If the flow is steady, each successively injected particle follows precisely behind the previous one, forming a steady streakline that is exactly the same as the streamline through the injection point. For unsteady flows, particles injected at the same point at different times need not follow the same path. An instantaneous photograph of the marked fluid would show the streakline at that instant, but it would not necessarily coincide with the streamline through the point of injection at that particular time nor with the streamline through the same injection point at a different time 1see Example 4.32. The third method used for visualizing and describing flows involves the use of pathlines. A pathline is the line traced out by a given particle as it flows from one point to another. The pathline is a Lagrangian concept that can be produced in the laboratory by marking a fluid particle 1dying a small fluid element2 and taking a time exposure photograph of its motion. (See the photograph at the beginning of Chapter 7.) V4.10 Streaklines F l u i d s i n Air bridge spanning the oceans It has long been known that large quantities of material are transported from one location to another by airborne dust particles. It is estimated that 2 billion metric tons of dust are lifted into the atmosphere each year. Most of these particles settle out fairly rapidly, but significant amounts travel large distances. Scientists are beginning to understand the full impact of this phenomenon—it is not only the tonnage transported, but the type of material transported that is significant. In addition to the mundane inert material we all term “dust,” it is now known that a wide variety of hazardous For steady flow, streamlines, streaklines, and pathlines are the same. E XAMPLE t h e N e w s materials and organisms are also carried along these literal particle paths. Satellite images reveal the amazing rate by which desert soils and other materials are transformed into airborne particles as a result of storms that produce strong winds. Once the tiny particles are aloft, they may travel thousands of miles, crossing the oceans and eventually being deposited on other continents. For the health and safety of all, it is important that we obtain a better understanding of the air bridges that span the oceans and also understand the ramification of such material transport. If the flow is steady, the path taken by a marked particle 1a pathline2 will be the same as the line formed by all other particles that previously passed through the point of injection 1a streakline2. For such cases these lines are tangent to the velocity field. Hence, pathlines, streamlines, and streaklines are the same for steady flows. For unsteady flows none of these three types of lines need be the same 1Ref. 32. Often one sees pictures of “streamlines” made visible by the injection of smoke or dye into a flow as is shown in Fig. 4.3. Actually, such pictures show streaklines rather than streamlines. However, for steady flows the two are identical; only the nomenclature is incorrectly used. 4.3 Comparison of Streamlines, Pathlines, and Streaklines GIVEN Water flowing from the oscillating slit shown in Fig. E4.3a produces a velocity field given by V u0 sin 3v1t y v0 2 4 ˆi v0 jˆ, where u0, v0, and v are constants. Thus, the y component of velocity remains constant 1v v0 2, and the x component of velocity at y 0 coincides with the velocity of the oscillating sprinkler head 3 u u0 sin1vt2 at y 04 . FIND 1a2 Determine the streamline that passes through the origin at t 0; at t p2v. 1b2 Determine the pathline of the particle that was at the origin at t 0; at t p2. 1c2 Discuss the shape of the streakline that passes through the origin. SOLUTION (a) Since u u0 sin 3 v1t yv0 2 4 and v v0, it follows from Eq. 4.1 that streamlines are given by the solution of dy v0 v u dx u0 sin 3 v1t yv0 2 4 in which the variables can be separated and the equation integrated 1for any given time t2 to give 冮 u0 sin c v at 冮 y b d dy v0 dx v0 4.1 The Velocity Field 165 This can be integrated to give the x component of the pathline as or u0 1v0v2 cos c v at y b d v0x C v0 x c u0 sin a (1) where C is a constant. For the streamline at t 0 that passes through the origin 1x y 02, the value of C is obtained from Eq. 1 as C u0v0 v. Hence, the equation for this streamline is vy u0 x c cos a b 1 d v0 v x0 x u0 at y vy u0 u0 p p b x cos c v a b d cos a v0 v0 v 2v v 2 p b and 2v y (3) (Ans) (6) (Ans) y v0 at p b 2v (7) (8) (Ans) Fig. E4.3c, are straight lines from the origin 1rays2. The pathlines and streamlines do not coincide because the flow is unsteady. not the same because the flow is unsteady. For example, at the origin 1x y 02 the velocity is V v0 jˆ at t 0 and V u0ˆi v0 jˆ at t p2v. Thus, the angle of the streamline passing through the origin changes with time. Similarly, the shape of the entire streamline is a function of time. (c) The streakline through the origin at time t 0 is the locus of particles at t 0 that previously 1t 6 02 passed through the origin. The general shape of the streaklines can be seen as follows. Each particle that flows through the origin travels in a straight line 1pathlines are rays from the origin2, the slope of which lies between v0u0 as shown in Fig. E4.3d. Particles passing through the origin at different times are located on different rays from the origin and at different distances from the origin. The net result is that a stream of dye continually injected at the origin 1a streakline2 would have the shape shown in Fig. E4.3d. Because of the unsteadiness, the streakline will vary with time, although it will always have the oscillating, sinuous character shown. (b) The pathline of a particle 1the location of the particle as a function of time2 can be obtained from the velocity field and the definition of the velocity. Since u dxdt and v dydt we obtain dy v0 dt The y equation can be integrated 1since v0 constant2 to give the y coordinate of the pathline as y v0 t C1 v0 x u0 COMMENT The pathlines given by Eqs. 6 and 8, shown in COMMENT These two streamlines, plotted in Fig. E4.3b, are and y v0 t The pathline can be drawn by plotting the locus of x1t2, y1t2 values for t 0 or by eliminating the parameter t from Eq. 7 to give or y dx u0 sin c v at b d v0 dt and Similarly, for the particle that was at the origin at t p2v, Eqs. 4 and 5 give C1 pv0 2v and C2 pu0 2v. Thus, the pathline for this particle is (2) (Ans) vy u0 sin a b v0 v (5) where C2 is a constant. For the particle that was at the origin 1x y 02 at time t 0, Eqs. 4 and 5 give C1 C2 0. Thus, the pathline is Similarly, for the streamline at t p2v that passes through the origin, Eq. 1 gives C 0. Thus, the equation for this streamline is x C1v b d t C2 v0 (4) COMMENT Similar streaklines are given by the stream of where C1 is a constant. With this known y y1t2 dependence, the x equation for the pathline becomes water from a garden hose nozzle that oscillates back and forth in a direction normal to the axis of the nozzle. In this example neither the streamlines, pathlines, nor streaklines coincide. If the flow were steady, all of these lines would be the same. v0 t C1 C1 v dx u0 sin c v at b d u0 sin a b v0 v0 dt y y 2π v0/ω Streamlines through origin x 0 t=0 π v0/ω t = π /2ω Oscillating sprinkler head Q (a) –2u0/ω 0 (b) 2u0/ω x ■ Figure E4.3(a), (b) 166 Chapter 4 ■ Fluid Kinematics y y t=0 v0 /u0 Pathlines of particles at origin at time t v0 t = π /2ω –1 0 1 Pathline u0 x x 0 (c) 4.2 Streaklines through origin at time t (d) ■ Figure E4.3(c), (d) The Acceleration Field V4.11 Pathlines Acceleration is the time rate of change of velocity for a given particle. As indicated in the previous section, we can describe fluid motion by either 112 following individual particles 1Lagrangian description2 or 122 remaining fixed in space and observing different particles as they pass by 1Eulerian description2. In either case, to apply Newton’s second law 1F ma2 we must be able to describe the particle acceleration in an appropriate fashion. For the infrequently used Lagrangian method, we describe the fluid acceleration just as is done in solid body dynamics— a a1t2 for each particle. For the Eulerian description we describe the acceleration field as a function of position and time without actually following any particular particle. This is analogous to describing the flow in terms of the velocity field, V V 1x, y, z, t2, rather than the velocity for particular particles. In this section we will discuss how to obtain the acceleration field if the velocity field is known. The acceleration of a particle is the time rate of change of its velocity. For unsteady flows the velocity at a given point in space 1occupied by different particles2 may vary with time, giving rise to a portion of the fluid acceleration. In addition, a fluid particle may experience an acceleration because its velocity changes as it flows from one point to another in space. For example, water flowing through a garden hose nozzle under steady conditions 1constant number of gallons per minute from the hose2 will experience an acceleration as it changes from its relatively low velocity in the hose to its relatively high velocity at the tip of the nozzle. 4.2.1 The Material Derivative Consider a fluid particle moving along its pathline as is shown in Fig. 4.4. In general, the particle’s velocity, denoted VA for particle A, is a function of its location and the time. That is, VA VA 1rA, t2 VA 3xA 1t2, yA 1t2, zA 1t2, t4 z VA(rA, t) Particle A at time t Particle path vA(rA, t) rA wA(rA, t) uA(rA, t) y zA(t) yA(t) x xA(t) ■ Figure 4.4 Velocity and position of particle A at time t. 4.2 The Acceleration Field 167 where xA xA 1t2, yA yA 1t2, and zA zA 1t2 define the location of the moving particle. By definition, the acceleration of a particle is the time rate of change of its velocity. Since the velocity may be a function of both position and time, its value may change because of the change in time as well as a change in the particle’s position. Thus, we use the chain rule of differentiation to obtain the acceleration of particle A, denoted aA, as aA 1t2 dVA 0VA 0VA dxA 0VA dyA 0VA dzA dt 0t 0x dt 0y dt 0z dt (4.2) Using the fact that the particle velocity components are given by uA dxAdt, vA dyAdt, and wA dzAdt, Eq. 4.2 becomes aA 0VA 0VA 0VA 0VA uA vA wA 0t 0x 0y 0z Since the above is valid for any particle, we can drop the reference to particle A and obtain the acceleration field from the velocity field as 0V 0V 0V 0V u v w 0t 0x 0y 0z a (4.3) This is a vector result whose scalar components can be written as ax 0u 0u 0u 0u u v w 0t 0x 0y 0z ay 0v 0v 0v 0v u v w 0t 0x 0y 0z az 0w 0w 0w 0w u v w 0t 0x 0y 0z (4.4) and The material derivative is used to describe time rates of change for a given particle. where ax, ay, and az are the x, y, and z components of the acceleration. The above result is often written in shorthand notation as DV Dt a where the operator D1 2 Dt ⬅ 01 2 0t u 01 2 0x v 01 2 0y w 01 2 0z (4.5) is termed the material derivative or substantial derivative. An often-used shorthand notation for the material derivative operator is T = T (x, y, z, t) D1 2 Dt V Particle A z x y 01 2 0t 1V ⴢ § 21 2 (4.6) The dot product of the velocity vector, V, and the gradient operator, § 1 2 01 2 0x ˆi 0 1 2 0y jˆ 01 2 0z kˆ 1a vector operator2 provides a convenient notation for the spatial derivative terms appearing in the Cartesian coordinate representation of the material derivative. Note that the notation V ⴢ § represents the operator V ⴢ § 1 2 u01 2 0x v01 2 0y w01 2 0z. The material derivative concept is very useful in analysis involving various fluid parameters, not just the acceleration. The material derivative of any variable is the rate at which that variable changes with time for a given particle 1as seen by one moving along with the fluid—the Lagrangian description2. For example, consider a temperature field T T1x, y, z, t2 associated with a given flow, like the flame shown in the figure in the margin. It may be of interest to determine the time rate of change of temperature of a fluid particle 1particle A2 as it moves through this temperature 168 Chapter 4 ■ Fluid Kinematics field. If the velocity, V ⫽ V 1x, y, z, t2, is known, we can apply the chain rule to determine the rate of change of temperature as dTA 0TA 0TA dxA 0TA dyA 0TA dzA ⫽ ⫹ ⫹ ⫹ dt 0t 0x dt 0y dt 0z dt This can be written as DT 0T 0T 0T 0T 0T ⫽ ⫹u ⫹v ⫹w ⫽ ⫹ V ⴢ §T Dt 0t 0x 0y 0z 0t As in the determination of the acceleration, the material derivative operator, D1 2ⲐDt, appears. E XAMPLE 4.4 Acceleration along a Streamline y GIVEN An incompressible, inviscid fluid flows steadily past a tennis ball of radius R, as shown in Fig. E4.4a. According to a more advanced analysis of the flow, the fluid velocity along streamline A–B is given by R A R3 V ⫽ u1x2iˆ ⫽ V0 a1 ⫹ 3 b ˆi x x B V0 where V0 is the upstream velocity far ahead of the sphere. (a) FIND Determine the acceleration experienced by fluid particles as they flow along this streamline. ax _ (V02/R) SOLUTION –3 Along streamline A–B there is only one component of velocity 1v ⫽ w ⫽ 02 so that from Eq. 4.3 a⫽ –2 –1 x/R B A –0.2 0V 0u 0u 0V ⫹u ⫽ a ⫹ u b ˆi 0t 0x 0t 0x –0.4 –0.6 or (b) ax ⫽ 0u 0u ⫹u , 0t 0x ay ⫽ 0, az ⫽ 0 ■ Figure E4.4 Since the flow is steady, the velocity at a given point in space does not change with time. Thus, 0uⲐ 0t ⫽ 0. With the given velocity distribution along the streamline, the acceleration becomes ax ⫽ u 0u R3 ⫽ V0 a1 ⫹ 3 b V0 3R3 1⫺3x⫺4 2 4 0x x or ax ⫽ ⫺31V 20 ⲐR2 1 ⫹ 1RⲐx2 3 1xⲐR2 4 (Ans) COMMENTS Along streamline A–B 1⫺ q ⱕ x ⱕ ⫺R and y ⫽ 02 the acceleration has only an x component, and it is negative 1a deceleration2. Thus, the fluid slows down from its upstream velocity of V ⫽ V0ˆi at x ⫽ ⫺ q to its stagnation point velocity of V ⫽ 0 at x ⫽ ⫺R, the “nose” of the ball. The variation of ax along streamline A–B is shown in Fig. E4.4b. It is the same result as is obtained in Example 3.1 by using the streamwise component of the acceleration, ax ⫽ V 0VⲐ0s. The maximum deceleration occurs at x ⫽ ⫺1.205R and has a value of ax,max ⫽ ⫺0.610 V02ⲐR. Note that this maximum deceleration increases with increasing velocity and decreasing size. As indicated in the following table, typical values of this deceleration can be quite large. For example, the ax,max ⫽ ⫺4.08 ⫻ 10 4 ftⲐs2 value for a pitched baseball is a deceleration approximately 1500 times that of gravity. 4.2 Object Rising weather balloon Soccer ball Baseball Tennis ball Golf ball V0 1ftⲐs2 R 1ft2 ax,max 1ftⲐs2 2 1 20 90 100 200 4.0 0.80 0.121 0.104 0.070 ⫺0.153 ⫺305 ⫺4.08 ⫻ 10 4 ⫺5.87 ⫻ 104 ⫺3.49 ⫻ 10 5 The Acceleration Field 169 In general, for fluid particles on streamlines other than A–B, all three components of the acceleration 1ax, ay, and az 2 will be nonzero. 4.2.2 Unsteady Effects The local derivative is a result of the unsteadiness of the flow. V4.12 Unsteady flow As is seen from Eq. 4.5, the material derivative formula contains two types of terms—those involving the time derivative 3 01 2Ⲑ0t4 and those involving spatial derivatives 3 0 1 2 Ⲑ 0x, 01 2Ⲑ0y, and 01 2Ⲑ0z 4. The time derivative portions are denoted as the local derivative. They represent effects of the unsteadiness of the flow. If the parameter involved is the acceleration, that portion given by 0VⲐ0t is termed the local acceleration. For steady flow the time derivative is zero throughout the flow field 3 01 2Ⲑ0t ⬅ 04, and the local effect vanishes. Physically, there is no change in flow parameters at a fixed point in space if the flow is steady. There may be a change of those parameters for a fluid particle as it moves about, however. If a flow is unsteady, its parameter values 1velocity, temperature, density, etc.2 at any location may change with time. For example, an unstirred 1V ⫽ 02 cup of coffee will cool down in time because of heat transfer to its surroundings. That is, DTⲐDt ⫽ 0TⲐ0t ⫹ V ⴢ §T ⫽ 0TⲐ0t 6 0. Similarly, a fluid particle may have nonzero acceleration as a result of the unsteady effect of the flow. Consider flow in a constant diameter pipe as is shown in Fig. 4.5. The flow is assumed to be spatially uniform throughout the pipe. That is, V ⫽ V0 1t2 ˆi at all points in the pipe. The value of the acceleration depends on whether V0 is being increased, 0V0 Ⲑ0t 7 0, or decreased, 0V0Ⲑ0t 6 0. Unless V0 is independent of time 1V0 ⬅ constant2 there will be an acceleration, the local acceleration term. Thus, the acceleration field, a ⫽ 0V0 Ⲑ0t ˆi, is uniform throughout the entire flow, although it may vary with time 1 0V0 Ⲑ0t need not be constant2. The acceleration due to the spatial variations of velocity 1u 0uⲐ0x, v 0vⲐ0y, etc.2 vanishes automatically for this flow, since 0uⲐ0x ⫽ 0 and v ⫽ w ⫽ 0. That is, a⫽ 0V0 ˆ 0V 0V 0V 0V 0V ⫹u ⫹v ⫹w ⫽ ⫽ i 0t 0x 0y 0z 0t 0t 4.2.3 Convective Effects V2 > V1 V1 The portion of the material derivative 1Eq. 4.52 represented by the spatial derivatives is termed the convective derivative. It represents the fact that a flow property associated with a fluid particle may vary because of the motion of the particle from one point in space where the parameter has one value to another point in space where its value is different. For example, the water velocity at the inlet of the garden hose nozzle shown in the figure in the margin is different (both in direction and speed) than it is at the exit. This contribution to the time rate of change of the parameter for the particle can occur whether the flow is steady or unsteady. V0(t) x V0(t) ■ Figure 4.5 Uniform, unsteady flow in a constant diameter pipe. 170 Chapter 4 ■ Fluid Kinematics Hot Water heater Tout > Tin Pathline ∂T = 0 ___ ∂t DT ≠ 0 ___ Dt Cold Tin The convective derivative is a result of the spatial variation of the flow. u 0 x ax 0 x ■ Figure 4.6 Steady-state operation of a water heater. (Photograph courtesy of American Water Heater Company.) It is due to the convection, or motion, of the particle through space in which there is a gradient 3 § 1 2 ⫽ 01 2 Ⲑ 0x ˆi ⫹ 01 2 Ⲑ 0y jˆ ⫹ 01 2 Ⲑ 0z kˆ4 in the parameter value. That portion of the acceleration given by the term 1V ⴢ § 2V is termed the convective acceleration. As is illustrated in Fig. 4.6, the temperature of a water particle changes as it flows through a water heater. The water entering the heater is always the same cold temperature, and the water leaving the heater is always the same hot temperature. The flow is steady. However, the temperature, T, of each water particle increases as it passes through the heater— Tout 7 Tin. Thus, DTⲐDt ⫽ 0 because of the convective term in the total derivative of the temperature. That is, 0TⲐ0t ⫽ 0, but u 0TⲐ0x ⫽ 0 1where x is directed along the streamline2, since there is a nonzero temperature gradient along the streamline. A fluid particle traveling along this nonconstant temperature path 1 0TⲐ0x ⫽ 02 at a specified speed 1u2 will have its temperature change with time at a rate of DTⲐDt ⫽ u 0TⲐ0x even though the flow is steady 10TⲐ0t ⫽ 02. The same types of processes are involved with fluid accelerations. Consider flow in a variable area pipe as shown in Fig. 4.7. It is assumed that the flow is steady and one-dimensional with velocity that increases and decreases in the flow direction as indicated. As the fluid flows from section 112 to section 122, its velocity increases from V1 to V2. Thus, even though 0VⲐ0t ⫽ 0 1steady flow2, fluid particles experience an acceleration given by ax ⫽ u 0uⲐ 0x 1convective acceleration2. For x1 6 x 6 x2, it is seen that 0uⲐ0x 7 0 so that ax 7 0 —the fluid accelerates. For x2 6 x 6 x3, it is seen that 0uⲐ0x 6 0 so that ax 6 0 —the fluid decelerates. This acceleration and deceleration are shown in the figure in the margin. If V1 ⫽ V3, the amount of acceleration precisely balances the amount of deceleration even though the distances between x2 and x1 and x3 and x2 are not the same. The concept of the material derivative can be used to determine the time rate of change of any parameter associated with a particle as it moves about. Its use is not restricted to fluid mechanics alone. The basic ingredients needed to use the material derivative concept are the field description of the parameter, P ⫽ P1x, y, z, t2, and the rate at which the particle moves through that field, V ⫽ V 1x, y, z, t2. u = V1 u = V2 > V1 u = V3 = V1 < V2 x x1 ■ Figure 4.7 x2 x3 Uniform, steady flow in a variable area pipe. 4.2 E XAMPLE 4.5 The Acceleration Field 171 Acceleration from a Given Velocity Field GIVEN Consider the steady, two-dimensional flow field dis- FIND Determine the acceleration field for this flow. cussed in Example 4.2. SOLUTION In general, the acceleration is given by y V a DV 0V a 1V ⴢ § 2 1V2 Dt 0t Streamline 0V 0V 0V 0V u v w 0t 0x 0y 0z (1) where the velocity is given by V 1V0 /21xiˆ yjˆ 2 so that u 1V0/2 x and v 1V0/2y. For steady 3 01 2 0t 0 4, twodimensional 3w 0 and 0 1 2 0z 04 flow, Eq. l becomes 0V 0V v 0x 0y 0u 0u 0v 0v au v b iˆ au v b jˆ 0x 0y 0x 0y au x ■ Figure E4.5 Hence, for this flow the acceleration is given by a c a V0 V0 V0 b 1x2 a b a b 1y2102 d ˆi / / / c a Also, the acceleration vector is oriented at an angle u from the x axis, where V0 V0 V0 b 1x2102 a b 1y2 a b d jˆ / / / tan u or ax V 20 x /2 , ay V 20 y /2 (Ans) COMMENTS The fluid experiences an acceleration in both the x and y directions. Since the flow is steady, there is no local acceleration—the fluid velocity at any given point is constant in time. However, there is a convective acceleration due to the change in velocity from one point on the particle’s pathline to another. Recall that the velocity is a vector—it has both a magnitude and a direction. In this flow both the fluid speed 1magnitude2 and flow direction change with location 1see Fig. E4.1a2. For this flow the magnitude of the acceleration is constant on circles centered at the origin, as is seen from the fact that 0a 0 1a2x a2y a2z 2 12 a E XAMPLE 4.6 V0 2 2 b 1x y2 2 12 / ay ax y x This is the same angle as that formed by a ray from the origin to point 1x, y2. Thus, the acceleration is directed along rays from the origin and has a magnitude proportional to the distance from the origin. Typical acceleration vectors 1from Eq. 22 and velocity vectors 1from Example 4.12 are shown in Fig. E4.5 for the flow in the first quadrant. Note that a and V are not parallel except along the x and y axes 1a fact that is responsible for the curved pathlines of the flow2 and that both the acceleration and velocity are zero at the origin 1x y 02. An infinitesimal fluid particle placed precisely at the origin will remain there, but its neighbors 1no matter how close they are to the origin2 will drift away. (2) The Material Derivative GIVEN A fluid flows steadily through a two-dimensional nozzle of length / as shown in Fig. E4.6a. The nozzle shape is given by y / ; 0.5 31 1x/2 4 If viscous and gravitational effects are negligible, the velocity field is approximately u V0冤1 x/冥, v V0y/ (1) 172 Chapter 4 ■ Fluid Kinematics and the pressure field is p p0 1rV 20 22 3 1x 2 V0 y 2 / 2x/4 2 2 where V0 and p0 are the velocity and pressure at the origin, x y 0. Note that the fluid speed increases as it flows through the nozzle. For example, along the centerline 1y 02 , V V0 at x 0 and V 2V0 at x /. y 0 0.5 _y = _ (1 + x/) 0.5 _y = – _ (1 + x/) 2V0 FIND Determine, as a function of x and y, the time rate of x change of pressure felt by a fluid particle as it flows through the nozzle. SOLUTION The time rate of change of pressure at any given, fixed point in this steady flow is zero. However, the time rate of change of pressure felt by a particle flowing through the nozzle is given by the material derivative of the pressure and is not zero. Thus, Dp 0p 0p 0p 0p 0p u v u v Dt 0t 0x 0y 0x 0y (2) ■ Figure E4.6a where the x and y components of the pressure gradient can be written as 0p rV20 x a 1b 0x / / (3) rV20 y 0p a b 0y / / (4) and Therefore, by combining Eqs. (1), (2), (3), and (4) we obtain COMMENT Lines of constant pressure within the nozzle are indicated in Fig. E4.6b, along with some representative streamlines of the flow. Note that as a fluid particle flows along its streamline, it moves into areas of lower and lower pressure. Hence, even though the flow is steady, the time rate of change of the pressure for any given particle is negative. This can be verified from Eq. (5) which, when plotted in Fig. E4.6c, shows that for any point within the nozzle Dp Dt 6 0. y rV20 x rV20 y Dp x V0 a1 b a b a 1b aV0 b a ba b Dt / / / / / / Dp/Dt _ (ρV03/) or 2 Dp rV30 x y 2 c a 1b a b d Dt / / / _y –0.5 (5) (Ans) _x = 0 –1 p – p0 ______ /2 rV02 0.5 _x 1 –0.5 _y 0.5 0 –1.0 –1.5 –2.0 –2.5 –2.25 –3.0 _x = 0.5 _x 1 _x = 1 ■ Figure E4.6b ■ Figure E4.6c –4 4.2 The Acceleration Field 173 4.2.4 Streamline Coordinates V4.13 Streamline coordinates In many flow situations, it is convenient to use a coordinate system defined in terms of the streamlines of the flow. An example for steady, two-dimensional flows is illustrated in Fig. 4.8. Such flows can be described either in terms of the usual x, y Cartesian coordinate system 1or some other system such as the r, u polar coordinate system2 or the streamline coordinate system. In the streamline coordinate system the flow is described in terms of one coordinate along the streamlines, denoted s, and the second coordinate normal to the streamlines, denoted n. Unit vectors in these two directions are denoted by ˆs and nˆ, as shown in the figure. Care is needed not to confuse the coordinate distance s 1a scalar2 with the unit vector along the streamline direction, ˆs. The flow plane is therefore covered by an orthogonal curved net of coordinate lines. At any point the s and n directions are perpendicular, but the lines of constant s or constant n are not necessarily straight. Without knowing the actual velocity field 1hence, the streamlines2 it is not possible to construct this flow net. In many situations, appropriate simplifying assumptions can be made so that this lack of information does not present an insurmountable difficulty. One of the major advantages of using the streamline coordinate system is that the velocity is always tangent to the s direction. That is, V V ˆs V a an This allows simplifications in describing the fluid particle acceleration and in solving the equations governing the flow. For steady, two-dimensional flow we can determine the acceleration as as a DV as ˆs annˆ Dt where as and an are the streamline and normal components of acceleration, respectively, as indicated by the figure in the margin. We use the material derivative because by definition the acceleration is the time rate of change of the velocity of a given particle as it moves about. If the streamlines y s = s2 n = n2 n = n1 s = s1 s=0 n=0 Streamlines ^ n ^ s V s x ■ Figure 4.8 Streamline coordinate system for two-dimensional flow. 174 Chapter 4 ■ Fluid Kinematics are curved, both the speed of the particle and its direction of flow may change from one point to another. In general, for steady flow both the speed and the flow direction are a function of location— V V1s, n2 and ˆs ˆs 1s, n2. For a given particle, the value of s changes with time, but the value of n remains fixed because the particle flows along a streamline defined by n constant. 1Recall that streamlines and pathlines coincide in steady flow.2 Thus, application of the chain rule gives a D1V ˆs2 Dt DV Dsˆ ˆs V Dt Dt or aa 0V 0V ds 0V dn 0sˆ 0sˆ ds 0sˆ dn b ˆs V a b 0t 0s dt 0n dt 0t 0s dt 0n dt This can be simplified by using the fact that for steady flow nothing changes with time at a given point so that both 0V0t and 0sˆ0t are zero. Also, the velocity along the streamline is V dsdt, and the particle remains on its streamline 1n constant2 so that dndt 0. Hence, a aV The orientation of the unit vector along the streamline changes with distance along the streamline. 0V 0sˆ b ˆs V aV b 0s 0s The quantity 0sˆ0s represents the limit as ds S 0 of the change in the unit vector along the streamline, dsˆ, per change in distance along the streamline, ds. The magnitude of ˆs is constant 1 0sˆ 0 1; it is a unit vector2, but its direction is variable if the streamlines are curved. From Fig. 4.9 it is seen that the magnitude of 0sˆ0s is equal to the inverse of the radius of curvature of the streamline, r, at the point in question. This follows because the two triangles shown 1AOB and A¿O¿B¿ 2 are similar triangles so that dsr 0dˆs 0 0sˆ 0 0dsˆ 0 , or 0dsˆds 0 1r. Similarly, in the limit ds S 0, the direction of dsˆds is seen to be normal to the streamline. That is, 0sˆ dsˆ nˆ lim 0s dsS0 ds r Hence, the acceleration for steady, two-dimensional flow can be written in terms of its streamwise and normal components in the form aV 0V V2 0V ˆs nˆ or as V , 0s r 0s an V2 r (4.7) The first term, as V 0V0s, represents the convective acceleration along the streamline, and the second term, an V 2 r, represents centrifugal acceleration 1one type of convective acceleration2 normal to the fluid motion. These components can be noted in Fig. E4.5 by resolving the acceleration vector into its components along and normal to the velocity vector. Note that the unit vector nˆ is directed from the streamline toward the center of curvature. These forms of the acceleration were used in Chapter 3 and are probably familiar from previous dynamics or physics considerations. O O δθ δθ ^ n δs s A ^ s (s) B ^s (s + ) δs δs A ^s (s O' B ) δs δθ B' ^ δs ^ s (s) ■ Figure 4.9 Relationship between the unit vector along the streamline, sˆ , and the radius of curvature of the streamline, r. A' 4.3 4.3 Control Volume and System Representations 175 Control Volume and System Representations Both control volume and system concepts can be used to describe fluid flow. (Photograph courtesy of NASA.) As is discussed in Chapter 1, a fluid is a type of matter that is relatively free to move and interact with its surroundings. As with any matter, a fluid’s behavior is governed by fundamental physical laws that are approximated by an appropriate set of equations. The application of laws such as the conservation of mass, Newton’s laws of motion, and the laws of thermodynamics forms the foundation of fluid mechanics analyses. There are various ways that these governing laws can be applied to a fluid, including the system approach and the control volume approach. By definition, a system is a collection of matter of fixed identity 1always the same atoms or fluid particles2, which may move, flow, and interact with its surroundings. A control volume, on the other hand, is a volume in space 1a geometric entity, independent of mass2 through which fluid may flow. A system is a specific, identifiable quantity of matter. It may consist of a relatively large amount of mass 1such as all of the air in the Earth’s atmosphere2, or it may be an infinitesimal size 1such as a single fluid particle2. In any case, the molecules making up the system are “tagged” in some fashion 1dyed red, either actually or only in your mind2 so that they can be continually identified as they move about. The system may interact with its surroundings by various means 1by the transfer of heat or the exertion of a pressure force, for example2. It may continually change size and shape, but it always contains the same mass. A mass of air drawn into an air compressor can be considered as a system. It changes shape and size 1it is compressed2, its temperature may change, and it is eventually expelled through the outlet of the compressor. The matter associated with the original air drawn into the compressor remains as a system, however. The behavior of this material could be investigated by applying the appropriate governing equations to this system. One of the important concepts used in the study of statics and dynamics is that of the free-body diagram. That is, we identify an object, isolate it from its surroundings, replace its surroundings by the equivalent actions that they put on the object, and apply Newton’s laws of motion. The body in such cases is our system—an identified portion of matter that we follow during its interactions with its surroundings. In fluid mechanics, it is often quite difficult to identify and keep track of a specific quantity of matter. A finite portion of a fluid contains an uncountable number of fluid particles that move about quite freely, unlike a solid that may deform but usually remains relatively easy to identify. For example, we cannot as easily follow a specific portion of water flowing in a river as we can follow a branch floating on its surface. We may often be more interested in determining the forces put on a fan, airplane, or automobile by air flowing past the object than we are in the information obtained by following a given portion of the air 1a system2 as it flows along. Similarly, for the Space Shuttle launch vehicle shown in the figure in the margin, we may be more interested in determining the thrust produced than we are in the information obtained by following the highly complex, irregular path of the exhaust plume from the rocket engine nozzle. For these situations we often use the control volume approach. We identify a specific volume in space 1a volume associated with the fan, airplane, or automobile, for example2 and analyze the fluid flow within, through, or around that volume. In general, the control volume can be a moving volume, although for most situations considered in this book we will use only fixed, nondeformable control volumes. The matter within a control volume may change with time as the fluid flows through it. Similarly, the amount of mass within the volume may change with time. The control volume itself is a specific geometric entity, independent of the flowing fluid. Examples of control volumes and control surfaces 1the surface of the control volume2 are shown in Fig. 4.10. For case 1a2, fluid flows through a pipe. The fixed control surface consists of the inside surface of the pipe, the outlet end at section 122, and a section across the pipe at 112. One portion of the control surface is a physical surface 1the pipe2, while the remainder is simply a surface in space 1across the pipe2. Fluid flows across part of the control surface but not across all of it. Another control volume is the rectangular volume surrounding the jet engine shown in Fig. 4.10b. If the airplane to which the engine is attached is sitting still on the runway, air flows through this control volume because of the action of the engine within it. The air that was within the engine itself at time t t1 1a system2 has passed through the engine and is outside of the control volume at a later time t t2 as indicated. At this later time other air 1a different system2 is within the engine. If the airplane is moving, the control volume is fixed relative to an observer on the airplane, but it 176 Chapter 4 ■ Fluid Kinematics Jet engine Pipe Balloon V (1) (2) (a) Control volume surface (b) System at time t1 (c) System at time t2 > t1 ■ Figure 4.10 Typical control volumes: (a) fixed control volume, (b) fixed or moving control volume, (c) deforming control volume. The governing laws of fluid motion are stated in terms of fluid systems, not control volumes. 4.4 is a moving control volume relative to an observer on the ground. In either situation air flows through and around the engine as indicated. The deflating balloon shown in Fig. 4.10c provides an example of a deforming control volume. As time increases, the control volume 1whose surface is the inner surface of the balloon2 decreases in size. If we do not hold onto the balloon, it becomes a moving, deforming control volume as it darts about the room. The majority of the problems we will analyze can be solved by using a fixed, nondeforming control volume. In some instances, however, it will be advantageous, in fact necessary, to use a moving, deforming control volume. In many ways the relationship between a system and a control volume is similar to the relationship between the Lagrangian and Eulerian flow description introduced in Section 4.1.1. In the system or Lagrangian description, we follow the fluid and observe its behavior as it moves about. In the control volume or Eulerian description, we remain stationary and observe the fluid’s behavior at a fixed location. 1If a moving control volume is used, it virtually never moves with the system—the system flows through the control volume.2 These ideas are discussed in more detail in the next section. All of the laws governing the motion of a fluid are stated in their basic form in terms of a system approach. For example, “the mass of a system remains constant,” or “the time rate of change of momentum of a system is equal to the sum of all the forces acting on the system.” Note the word system, not control volume, in these statements. To use the governing equations in a control volume approach to problem solving, we must rephrase the laws in an appropriate manner. To this end we introduce the Reynolds transport theorem in the following section. The Reynolds Transport Theorem V We are sometimes interested in what happens to a particular part of the fluid as it moves about. Other times we may be interested in what effect the fluid has on a particular object or volume in space as fluid interacts with it. Thus, we need to describe the laws governing fluid motion using both system concepts 1consider a given mass of the fluid2 and control volume concepts 1consider a given volume2. To do this we need an analytical tool to shift from one representation to the other. The Reynolds transport theorem provides this tool. All physical laws are stated in terms of various physical parameters. Velocity, acceleration, mass, temperature, and momentum are but a few of the more common parameters. Let B represent any of these 1or other2 fluid parameters and b represent the amount of that parameter per unit mass. That is, B mb m B b = B/m m 1 mV V 1 _ mV 2 2 1 _V 2 2 where m is the mass of the portion of fluid of interest. For example, as shown by the figure in the margin, if B m, the mass, it follows that b 1. The mass per unit mass is unity. If B mV 2 2, the kinetic energy of the mass, then b V 2 2, the kinetic energy per unit mass. The parameters B and b may be scalars or vectors. Thus, if B mV, the momentum of the mass, then b V. 1The momentum per unit mass is the velocity.2 The parameter B is termed an extensive property, and the parameter b is termed an intensive property. The value of B is directly proportional to the amount of the mass being considered, whereas the value of b is independent of the amount of mass. The amount of an extensive property that a system possesses at a given instant, Bsys, can be determined by adding up the amount associated with each fluid particle in the system. For infinitesimal fluid particles of size d V and mass r dV , 4.4 The Reynolds Transport Theorem 177 this summation 1in the limit of dV ⫺ S 02 takes the form of an integration over all the particles in the system and can be written as Bsys ⫽ lim a bi 1ri dV ⫺i 2 ⫽ dV ⫺S0 i 冮 rb dV ⫺ sys The limits of integration cover the entire system—a 1usually2 moving volume. We have used the fact that the amount of B in a fluid particle of mass r d ⫺ V is given in terms of b by dB ⫽ br dV ⫺. Most of the laws governing fluid motion involve the time rate of change of an extensive property of a fluid system—the rate at which the momentum of a system changes with time, the rate at which the mass of a system changes with time, and so on. Thus, we often encounter terms such as dBsys dt da ⫽ 冮 sys rb dV ⫺b (4.8) dt To formulate the laws into a control volume approach, we must obtain an expression for the time rate of change of an extensive property within a control volume, Bcv, not within a system. This can be written as dBcv ⫽ dt 冮 rb dV⫺b cv (4.9) dt where the limits of integration, denoted by cv, cover the control volume of interest. Although Eqs. 4.8 and 4.9 may look very similar, the physical interpretation of each is quite different. Mathematically, the difference is represented by the difference in the limits of integration. Recall that the control volume is a volume in space 1in most cases stationary, although if it moves it need not move with the system2. On the other hand, the system is an identifiable collection of mass that moves with the fluid 1indeed it is a specified portion of the fluid2. We will learn that even for those instances when the control volume and the system momentarily occupy the same volume in space, the two quantities dBsys Ⲑdt and dBcv Ⲑdt need not be the same. The Reynolds transport theorem provides the relationship between the time rate of change of an extensive property for a system and that for a control volume—the relationship between Eqs. 4.8 and 4.9. Differences between control volume and system concepts are subtle but very important. E XAMPLE da 4.7 Time Rate of Change for a System and a Control Volume GIVEN Fluid flows from the fire extinguisher tank shown in Fig. E4.7a. FIND Discuss the differences between dBsys Ⲑdt and dBcvⲐdt if B represents mass. SOLUTION With B ⫽ m, the system mass, it follows that b ⫽ 1 and Eqs. 4.8 and 4.9 can be written as dBsys dt ⬅ dmsys dt da ⫽ 冮 sys r dV ⫺b dt and dBcv dmcv ⬅ ⫽ dt dt da 冮 r dV⫺b cv dt ■ Figure E4.7 (a) (© DNY59/iStockphoto) 178 Chapter 4 ■ Fluid Kinematics Physically, these represent the time rate of change of mass within the system and the time rate of change of mass within the control volume, respectively. We choose our system to be the fluid within the tank at the time the valve was opened 1t 02 and the control volume to be the tank itself as shown in Fig. E4.7b. A short time after the valve is opened, part of the system has moved outside of the control volume as is shown in Fig. E4.7c. The control volume remains fixed. The limits of integration are fixed for the control volume; they are a function of time for the system. Clearly, if mass is to be conserved 1one of the basic laws governing fluid motion2, the mass of the fluid in the system is constant, so that da 冮 sys t=0 t>0 r d Vb dt System 0 On the other hand, it is equally clear that some of the fluid has left the control volume through the nozzle on the tank. Hence, the amount of mass within the tank 1the control volume2 decreases with time, or da The actual numerical value of the rate at which the mass in the control volume decreases will depend on the rate at which the fluid flows through the nozzle 1i.e., the size of the nozzle and the speed and density of the fluid2. Clearly the meanings of dBsys dt and dBcvdt are different. For this example, dBcv dt 6 dBsys dt. Other situations may have dBcvdt dBsys dt. 冮 r dV b cv dt Control surface (b) (c) ■ Figure E4.7 6 0 4.4.1 Derivation of the Reynolds Transport Theorem A simple version of the Reynolds transport theorem relating system concepts to control volume concepts can be obtained easily for the one-dimensional flow through a fixed control volume such as the variable area duct section shown in Fig. 4.11a. We consider the control volume to be that stationary volume within the duct between sections 112 and 122 as indicated in Fig. 4.11b. The system that we consider is that fluid occupying the control volume at some initial time t. A short time later, at time t dt, the system has moved slightly to the right. The fluid particles that coincided with section 122 of the control surface at time t have moved a distance d/2 V2 dt to the right, where V2 is the velocity of the fluid as it passes section 122. Similarly, the fluid initially at section 112 has moved a distance d/1 V1 dt, where V1 is the fluid velocity at section 112. We assume the fluid flows across sections 112 and 122 in a direction normal to these surfaces and that V1 and V2 are constant across sections 112 and 122. As is shown in Fig. 4.11c, the outflow from the control volume from time t to t dt is denoted as volume II, the inflow as volume I, and the control volume itself as CV. Thus, the system at time t consists of the fluid in section CV; that is, “SYS CV” at time t. At time t dt the system consists of the same fluid that now occupies sections 1CV I2 II. That is, “SYS CV I II” at time t dt. The control volume remains as section CV for all time. The moving system flows through the fixed control volume. δ 2 = V2 δ t V1 δ 1 = V1 δ t V2 I CV – I (1) (1) (2) (2) Fixed control surface and system boundary at time t System boundary at time t + δ t (a) II (b) ■ Figure 4.11 Control volume and system for flow through a variable area pipe. (c) 4.4 The Reynolds Transport Theorem 179 If B is an extensive parameter of the system, then the value of it for the system at time t is Bsys 1t2 Bcv 1t2 since the system and the fluid within the control volume coincide at this time. Its value at time t dt is Bsys 1t dt2 Bcv 1t dt2 BI 1t dt2 BII 1t dt2 Thus, the change in the amount of B in the system, dBsys, in the time interval dt divided by this time interval is given by dBsys dt Bsys 1t dt2 Bsys 1t2 dt Bcv 1t dt2 BI 1t dt2 BII 1t dt2 Bsys 1t2 dt By using the fact that at the initial time t we have Bsys 1t2 Bcv 1t2, this ungainly expression may be rearranged as follows. The time rate of change of a system property is a Lagrangian concept. dBsys dt dt BI 1t dt2 dt BII 1t dt2 dt (4.10) In the limit dt S 0, the left-hand side of Eq. 4.10 is equal to the time rate of change of B for the system and is denoted as DBsys Dt. We use the material derivative notation, D1 2 Dt, to denote this time rate of change to emphasize the Lagrangian character of this term. 1Recall from Section 4.2.1 that the material derivative, DPDt, of any quantity P represents the time rate of change of that quantity associated with a given fluid particle as it moves along.2 Similarly, the quantity DBsys Dt represents the time rate of change of property B associated with a system 1a given portion of fluid2 as it moves along. In the limit dt S 0, the first term on the right-hand side of Eq. 4.10 is seen to be the time rate of change of the amount of B within the control volume lim V2 Bcv 1t dt2 Bcv 1t2 dtS0 Bcv 1t dt2 Bcv 1t2 dt 0Bcv 0t 0a 冮 rb dVb cv 0t (4.11) The third term on the right-hand side of Eq. 4.10 represents the rate at which the extensive parameter B flows from the control volume across the control surface. As indicated by the figure in the margin, during the time interval from t 0 to t dt the volume of fluid that flows across section 122 is given by dV II A2 d/2 A2 1V2dt2. Thus, the amount of B within region II, the outflow region, is its amount per unit volume, rb, times the volume BII 1t dt2 1r2b2 21dV II 2 r2b2A2V2 dt (2) t=0 where b2 and r2 are the constant values of# b and r across section 122. Thus, the rate at which this property flows from the control volume, Bout, is given by δ VII V2δ t # BII 1t dt2 Bout lim r2 A2V2 b2 dtS0 dt (4.12) Similarly, the inflow of B into the control volume across section 112 during the time interval dt corresponds to that in region I and is given by the amount per unit volume times the volume, dV I A1 d/1 A1 1V1 dt2. Hence, BI 1t dt2 1r1b1 21dV 1 2 r1b1A1V1 dt where b1 and r1 are the constant values #of b and r across section 112. Thus, the rate of inflow of the property B into the control volume, Bin, is given by (2) t = δt # BI 1t dt2 Bin lim r1A1V1b1 dtS0 dt (4.13) 180 Chapter 4 ■ Fluid Kinematics If we combine Eqs. 4.10, 4.11, 4.12, and 4.13, we see that the relationship between the time rate of change of B for the system and that for the control volume is given by DBsys # # 0Bcv Bout Bin dt (4.14) 0Bcv r2A2V2b2 r1A1V1b1 0t (4.15) Dt or DBsys Dt This is a version of the Reynolds transport theorem valid under the restrictive assumptions associated with the flow shown in Fig. 4.11—fixed control volume with one inlet and one outlet having uniform properties 1density, velocity, and the parameter b2 across the inlet and outlet with the velocity normal to sections 112 and 122. Note that the time rate of change of B for the system 1the left-hand side of Eq. 4.15 or the quantity in Eq. 4.82 is not necessarily the same as the rate of change of B within the control volume 1the first term on the right-hand side of Eq. 4.15 or the quantity in Eq. 4.92. This is true because the inflow rate 1b1r1V1A1 2 and the outflow rate 1b2 r2V2 A2 2 of the property B for the control volume need not be the same. The time derivative associated with a system may be different from that for a control volume. E XAMPLE 4.8 Use of the Reynolds Transport Theorem GIVEN Consider again the flow from the fire extinguisher FIND Write the appropriate form of the Reynolds transport shown in Fig. E4.7. Let the extensive property of interest be the system mass 1B m, the system mass, or b 12. theorem for this flow. SOLUTION Again we take the control volume to be the fire extinguisher and the system to be the fluid within it at time t 0. For this case there is no inlet, section 112, across which the fluid flows into the control volume 1A1 02. There is, however, an outlet, section 122. Thus, the Reynolds transport theorem, Eq. 4.15, along with Eq. 4.9 with b 1 can be written as Dmsys Dt 0a 冮 cv r d Vb 0t r2 A2V2 0a (1) (Ans) law of conservation of mass, we may set the left-hand side of this equation equal to zero 1the amount of mass in a system is constant2 and rewrite Eq. 1 in the form 冮 cv r d Vb 0t Right Atrium Left Atrium Left Ventricle Right Ventricle r2 A2V2 冮 r d Vb cv 0t COMMENT If we proceed one step further and use the basic 0a The physical interpretation of this result is that the rate at which the mass in the tank decreases in time is equal in magnitude but opposite to the rate of flow of mass from the exit, r2 A2V2. Note the units for the two terms of Eq. 2 1kg兾s or slugs兾s2. If there were both an inlet and an outlet to the control volume shown in Fig. E4.7, Eq. 2 would become (2) r1 A1V1 r2 A2V2 (3) In addition, if the flow were steady, the left-hand side of Eq. 3 would be zero 1the amount of mass in the control would be constant in time2 and Eq. 3 would become r1 A1V1 r2 A2V2 This is one form of the conservation of mass principle discussed in Section 3.6.2—the mass flowrates into and out of the control volume are equal. Other more general forms are discussed in Chapter 5. Equation 4.15 is a simplified version of the Reynolds transport theorem. We will now derive it for much more general conditions. A general, fixed control volume with fluid flowing through it is shown in Fig. 4.12. The flow field may be quite simple 1as in the above one-dimensional flow considerations2, or it may involve a quite complex, unsteady, three-dimensional situation such as the flow through a human heart as illustrated by the figure in the margin. In any case we again consider the system to be the fluid within the control volume at the initial time t. A short time later a portion of the fluid 1region II2 has exited from the control volume, and additional fluid 1region I, not part of the original system2 has entered the control volume. 4.4 181 The Reynolds Transport Theorem V2 V3 Ib Inflow IIc II CV–I Ia IId V1 I V5 Outflow II b IIa Fixed control surface and system boundary at time t System boundary at time t + δ t ■ Figure 4.12 Control volume and system for flow through an arbitrary, fixed control volume. The simplified Reynolds transport theorem can be easily generalized. V4 V6 ■ Figure 4.13 Typical control volume with more than one inlet and outlet. We consider an extensive fluid property B and seek to determine how the rate of change of B associated with the system is related to the rate of change of B within the control volume at any instant. By repeating the exact steps that we did for the simplified control volume shown in Fig. 4.11, we see that Eq. 4.14 # is valid # for the general case also, provided that we give the correct interpretation to the terms Bout and Bin. In general, the control volume may contain more 1or less2 than one inlet and one outlet. A typical pipe system may contain several inlets and outlets as are shown in Fig. 4.13. In such instances we think of all inlets grouped together 1I Ia Ib Ic p 2 and all outlets grouped together 1II IIa IIb IIc p 2, at least conceptually. # The term Bout represents the net flowrate of the property B from the control volume. Its value can be thought of as arising from the addition 1integration2 of the contributions through each infinitesimal area element of size dA on the portion of the control surface dividing region II and the control volume. This surface is denoted CSout. As is indicated in Fig. 4.14, in time dt the volume of fluid that passes across each area element is given by dV d/n dA, where d/n d/ cos u is the height 1normal to the base, dA2 of the small volume element, and u is the angle between the velocity vector and the outward pointing normal to the surface, nˆ. Thus, since d/ V dt, the amount of the property B carried across the area element dA in the time interval dt is given by dB br dV br1V cos u dt2 dA The # rate at which B is carried out of the control volume across the small area element dA, denoted dBout, is # 1rbV cos u dt2 dA rb dV dBout lim lim rbV cos u dA dtS0 dt dtS0 dt Outflow portion of control surface CSout δ V = δ n δ A n n δA θ δA V (a) δ n ^ ^ ^ n θ θ δ = Vδt V (b) ■ Figure 4.14 Outflow across a typical portion of the control surface. V δ (c) 182 Chapter 4 ■ Fluid Kinematics Inflow portion of control surface CSin δ n δ = V δt δA ^ n ^ n ^ n θ θ θ V (a) δ V δ V = δ n δ A (b) V (c) ■ Figure 4.15 Inflow across a typical portion of the control surface. By integrating over the entire outflow portion of the control surface, CSout, we obtain 冮 # Bout # dBout csout The flowrate of a parameter across the control surface is written in terms of a surface integral. 冮 rbV cos u dA csout The quantity V cos u is the component of the velocity normal to the area element dA. From the definition of the dot product, this can be written as V cos u V ⴢ nˆ. Hence, an alternate form of the outflow rate is # Bout 冮 rbV ⴢ nˆ dA (4.16) csout In a similar fashion, by considering the inflow portion of the control surface, CSin, as shown in Fig. 4.15, we find that the inflow rate of B into the control volume is 冮 # Bin rbV cos u dA csin 冮 rbV ⴢ nˆ dA (4.17) csin We use the standard notation that the unit normal vector to the control surface, nˆ, points out from the control volume. Thus, as is shown in Fig. 4.16, 90° 6 u 6 90° for outflow regions 1the normal component of V is positive; V ⴢ nˆ 7 02. For inflow regions 90° 6 u 6 270° 1the normal component of V is negative; V ⴢ nˆ 6 02. The value of cos u is, therefore, positive on the CVout portions of the control surface and negative on the CVin portions. Over the remainder of the control surface, there is no inflow or outflow, leading to V ⴢ nˆ V cos u 0 on those portions. On such portions either V 0 1the fluid “sticks” to the surface2 or cos u 0 1the fluid “slides” along the surface without crossing it2 1see Fig. 4.162. Therefore, the net flux 1flowrate2 of parameter B across the entire control surface is # # Bout Bin 冮 csout rbV ⴢ nˆ dA a 冮 rbV ⴢ nˆ dAb cs in 冮 rbV ⴢ nˆ dA (4.18) cs where the integration is over the entire control surface. CS CSin CSout V=0 ^ n ^ n ^ n V •^ n>0 V •^ n0 V •^ n t0 System at time t1 > t0 At t1 VB VCV = Control volume velocity ■ Figure 4.21 Typical moving control volume and system. 4.4 VA = Absolute velocity of A The Reynolds Transport Theorem 187 VB VCV VCV The Reynolds transport theorem for a moving control volume involves the relative velocity. WB = Velocity of B relative WA = Velocity of A relative to control volume to control volume ■ Figure 4.22 Relationship between absolute and relative velocities. velocity, V, and the control volume velocity, Vcv, will not be in the same direction so that the relative and absolute velocities will have different directions 1see Fig. 4.222. The Reynolds transport theorem for a moving, nondeforming control volume can be derived in the same manner that it was obtained for a fixed control volume. As is indicated in Fig. 4.23, the only difference that needs be considered is the fact that relative to the moving control volume the fluid velocity observed is the relative velocity, not the absolute velocity. An observer fixed to the moving control volume may or may not even know that he or she is moving relative to some fixed coordinate system. If we follow the derivation that led to Eq. 4.19 1the Reynolds transport theorem for a fixed control volume2, we note that the corresponding result for a moving control volume can be obtained by simply replacing the absolute velocity, V, in that equation by the relative velocity, W. Thus, the Reynolds transport theorem for a control volume moving with constant velocity is given by DBsys Dt 0 0t 冮 rb dV cv 冮 rb W ⴢ nˆ dA (4.23) cs where the relative velocity is given by Eq. 4.22. 4.4.7 Selection of a Control Volume V4.14 Control volume Any volume in space can be considered as a control volume. It may be of finite size or it may be infinitesimal in size, depending on the type of analysis to be carried out. In most of our cases, the control volume will be a fixed, nondeforming volume. In some situations, we will consider control volumes that move with constant velocity. In either case, it is important that considerable thought go into the selection of the specific control volume to be used. The selection of an appropriate control volume in fluid mechanics is very similar to the selection of an appropriate free-body diagram in dynamics or statics. In dynamics, we select the body in which we are interested, represent the object in a free-body diagram, and then apply the appropriate governing laws to that body. The ease of solving a given dynamics problem is often very dependent on the specific object that we select for use in our free-body diagram. Similarly, the ease of solving a given fluid mechanics problem is often very dependent on the choice of the control volume used. Only by practice can we develop skill at selecting the “best” control volume. None are “wrong,” but some are “much better” than others. Solution of a typical problem will involve determining parameters such as velocity, pressure, and force at some point in the flow field. It is usually best to ensure that this point is located on the control surface, not “buried” within the control volume. The unknown will then appear in the convective term 1the surface integral2 of the Reynolds transport theorem. If possible, the control Control volume and system at time t System at time t + δt Pathlines as seen from the moving control volume W = V – VCV Flow as seen by an observer moving with velocity VCV ■ Figure 4.23 Control volume and system as seen by an observer moving with the control volume. 188 Chapter 4 ■ Fluid Kinematics Control surface V (1) V (a) (1) (b) V (1) (c) ■ Figure 4.24 Various control volumes for flow through a pipe. ^ n θ V Control surface 4.5 surface should be normal to the fluid velocity so that the angle u 1V ⴢ nˆ V cos u as shown by the figure in the margin2 in the flux terms of Eq. 4.19 will be 0 or 180°. This will usually simplify the solution process. Figure 4.24 illustrates three possible control volumes associated with flow through a pipe. If the problem is to determine the pressure at point 112, the selection of the control volume 1a2 is better than that of 1b2 because point 112 lies on the control surface. Similarly, control volume 1a2 is better than 1c2 because the flow is normal to the inlet and exit portions of the control volume. None of these control volumes are wrong—1a2 will be easier to use. Proper control volume selection will become much clearer in Chapter 5 where the Reynolds transport theorem is used to transform the governing equations from the system formulation into the control volume formulation, and numerous examples using control volume ideas are discussed. Chapter Summary and Study Guide field representation velocity field Eulerian method Lagrangian method one-, two-, and threedimensional flow steady and unsteady flow streamline streakline pathline acceleration field material derivative local acceleration convective acceleration system control volume Reynolds transport theorem This chapter considered several fundamental concepts of fluid kinematics. That is, various aspects of fluid motion are discussed without regard to the forces needed to produce this motion. The concepts of a field representation of a flow and the Eulerian and Lagrangian approaches to describing a flow are introduced, as are the concepts of velocity and acceleration fields. The properties of one-, two-, or three-dimensional flows and steady or unsteady flows are introduced along with the concepts of streamlines, streaklines, and pathlines. Streamlines, which are lines tangent to the velocity field, are identical to streaklines and pathlines if the flow is steady. For unsteady flows, they need not be identical. As a fluid particle moves about, its properties (i.e., velocity, density, temperature) may change. The rate of change of these properties can be obtained by using the material derivative, which involves both unsteady effects (time rate of change at a fixed location) and convective effects (time rate of change due to the motion of the particle from one location to another). The concepts of a control volume and a system are introduced, and the Reynolds transport theorem is developed. By using these ideas, the analysis of flows can be carried out using a control volume (a volume, usually fixed, through which the fluid flows), whereas the governing principles are stated in terms of a system (a flowing portion of fluid). The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. understand the concept of the field representation of a flow and the difference between Eulerian and Lagrangian methods of describing a flow. explain the differences among streamlines, streaklines, and pathlines. calculate and plot streamlines for flows with given velocity fields. use the concept of the material derivative, with its unsteady and convective effects, to determine time rate of change of a fluid property. determine the acceleration field for a flow with a given velocity field. understand the properties of and differences between a system and a control volume. interpret, physically and mathematically, the concepts involved in the Reynolds transport theorem. Conceptual Questions 189 Some of the important equations in this chapter are: dy v u dx Equation for streamlines (4.1) 0V 0V 0V 0V (4.3) u v w 0t 0x 0y 0z D1 2 01 2 (4.6) 1V ⴢ § 21 2 Dt 0t 0V V2 (4.7) as V , an 0s r DBsys 0Bcv r2A2V2b2 r1A1V1b1 (4.15) Dt 0t DBsys 0 rb d V rb V ⴢ n ˆ dA (4.19) Dt 0t cv cs V W Vcv (4.22) a Acceleration Material derivative Streamwise and normal components of acceleration Reynolds transport theorem (restricted form) Reynolds transport theorem (general form) Relative and absolute velocities 冮冮 References 1. Streeter, V. L., and Wylie, E. B., Fluid Mechanics, 8th Ed., McGraw-Hill, New York, 1985. 2. Goldstein, R. J., Fluid Mechanics Measurements, Hemisphere, New York, 1983. 3. Homsy, G. M., et al., Multimedia Fluid Mechanics CD-ROM, 2nd Ed., Cambridge University Press, New York, 2007. 4. Magarvey, R. H., and MacLatchy, C. S., The Formation and Structure of Vortex Rings, Canadian Journal of Physics, Vol. 42, 1964. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www.wiley. com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Conceptual Questions 4.1C A velocity field is given by: V 6xjˆ ms. Velocity fields are plotted below, where the arrows indicate the magnitude and direction of the velocity vectors. The picture that best describes the field is: 190 Chapter 4 ■ Fluid Kinematics y y y 0 x (a) 0 x (b) 4.2C Consider the path of a football thrown by Brett Favre. Suppose that you are interested in determining how far the ball travels. Then you are interested in: a) neither Lagrangian nor Eulerian views. b) the Eulerian view of the field. c) the Lagrangian or particle view of the football. y 0 x 0 x (c) (d) a) The equations apply to a fixed quantity of mass. b) The equations are formulated in terms of extensive properties. c) The equations apply to a fixed volume. Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. 4.3C The following is true of the system (versus control volume) form of the basic equations (mass, momentum, angular momentum, and energy): Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the evennumbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley. com/college/munson. vθ vr r θ Section 4.1 The Velocity Field 4.1 Obtain a photograph/image of a situation in which a fluid is flowing. Print this photo and draw in some lines to represent how you think some streamlines may look. Write a brief paragraph to describe the acceleration of a fluid particle as it flows along one of these streamlines. 4.2 The velocity field of a flow is given by GO V 13y 22 ˆi 1x 82 ˆj 5zkˆ ft s , where x, y, and z are in feet. Determine the fluid speed at the origin (x y z 0) and on the y axis (x z 0). 4.3 The velocity field of a flow is given by V 2x2tiˆ [4y(t 1) 2x2t]jˆ m/s, where x and y are in meters and t is in seconds. For fluid particles on the x axis, determine the speed and direction of flow. 4.4 A two-dimensional velocity field is given by u 1 y and v 1. Determine the equation of the streamline that passes through the origin. On a graph, plot this streamline. 4.5 The velocity field of a flow is given by V 15z 32iˆ 1x 42jˆ 4ykˆ fts, where x, y, and z are in feet. Determine the fluid speed at the origin 1x y z 02 and on the x axis 1y z 02. 4.6 A flow can be visualized by plotting the velocity field as velocity vectors at representative locations in the flow as shown in Video V4.2 and Fig. E4.1. Consider the velocity field given in polar coordinates by vr 10r, and vu 10r. This flow approximates a fluid swirling into a sink as shown in Fig. P4.6. Plot the velocity field at locations given by r 1, 2, and 3 with u 0, 30, 60, and 90°. ■ Figure P4.6 4.7 The velocity field of a flow is given by V 20y 1x2 y2 2 12 ˆi 20x 1x2 y2 2 12 jˆ fts, where x and y are in feet. Determine the fluid speed at points along the x axis; along the y axis. What is the angle between the velocity vector and the x axis at points 1x, y2 15, 02, 15, 52, and 10, 52? 4.8 The components of a velocity field are given by u x y, v xy3 16, and w 0. Determine the location of any stagnation points 1V 02 in the flow field. 4.9 The x and y components of velocity for a two-dimensional flow are u 6y fts and v 3 fts, where y is in feet. Determine the equation for the streamlines and sketch representative streamlines in the upper half plane. 4.10 The velocity field of a flow is given by u V0y(x2 2 1/2 y ) and v V0 x(x2 y2)12, where V0 is a constant. Where in the flow field is the speed equal to V0? Determine the equation of the streamlines and discuss the various characteristics of this flow. 4.11 A velocity field is given by V x ˆi x1x 121y 12jˆ, where u and v are in fts and x and y are in feet. Plot the streamline that passes through x 0 and y 0. Compare this streamline with the streakline through the origin. 191 Problems 4.12 From time t 0 to t 5 hr radioactive steam is released from a nuclear power plant accident located at x 1 mile and y 3 miles. The following wind conditions are expected: V 10 ˆi 5 ˆj mph for 0 6 t 6 3 hr, V 15 ˆi 8 ˆj mph for 3 6 t 6 10 hr, and V 5 ˆi mph for t 7 10 hr. Draw to scale the expected streakline of the steam for t 3, 10, and 15 hr. 4.13 The x and y components of a velocity field are given by u x2y and v xy2. Determine the equation for the streamlines of this flow and compare it with those in Example 4.2. Is the flow in this problem the same as that in Example 4.2? Explain. 4.14 In addition to the customary horizontal velocity components of the air in the atmosphere 1the “wind”2, there often are vertical air currents 1thermals2 caused by buoyant effects due to uneven heating of the air as indicated in Fig. P4.14. Assume that the velocity field in a certain region is approximated by u u0, v v0 11 yh2 for 0 6 y 6 h, and u u0, v 0 for y 7 h. Plot the shape of the streamline that passes through the origin for values of u0 v0 0.5, 1, and 2. 4.18 (See Fluids in the News article titled “Follow those particles,” Section 4.1.) Two photographs of four particles in a flow past a sphere are superposed as shown in Fig. P4.18. The time interval between the photos is ¢t 0.002 s. The locations of the particles, as determined from the photos, are shown in the table. (a) Determine the fluid velocity for these particles. (b) Plot a graph to compare the results of part (a) with the theoretical velocity which is given by V V0 11 a3x3 2, where a is the sphere radius and V0 is the fluid speed far from the sphere. Particle x at t ⴝ 0 s 1ft2 x at t ⴝ 0.002 s 1ft2 1 2 3 4 0.500 0.250 0.140 0.120 0.480 0.232 0.128 0.112 y y, ft t=0 t = 0.002 s u0 –0.4 x, ft –0.2 a = 0.1 ft ■ Figure P4.18 0 x ■ Figure P4.14 4.15 As shown in Video V4.6 and Fig. P4.15, a flying airplane produces swirling flow near the end of its wings. In certain circumstances this flow can be approximated by the velocity field u Ky 1x2 y2 2 and v Kx 1x2 y2 2, where K is a constant depending on various parameters associated with the airplane (i.e., its weight, speed) and x and y are measured from the center of the swirl. (a) Show that for this flow the velocity is inversely proportional to the distance from the origin. That is, V K 1x2 y2 2 12. (b) Show that the streamlines are circles. †4.19 Pathlines and streaklines provide ways to visualize flows. Another technique would be to instantly inject a line of dye across streamlines and observe how this line moves as time increases. For example, consider the initially straight dye line injected in front of the circular cylinder shown in Fig. P4.19. Discuss how this dye line would appear at later times. How would you calculate the location of this line as a function of time? Dye at t = 0 V y v u x ■ Figure P4.19 ■ Figure P4.15 Section 4.2 The Acceleration Field †4.16 For any steady flow the streamlines and streaklines are the same. For most unsteady flows this is not true. However, there are unsteady flows for which the streamlines and streaklines are the same. Describe a flow field for which this is true. 4.17 A 10-ft-diameter dust devil that rotates one revolution per second travels across the Martian surface (in the x-direction) with a speed of 5 ft/s. Plot the pathline etched on the surface by a fluid particle 10 ft from the center of the dust devil for time 0 t 3 s. The particle position is given by the sum of that for a stationary swirl [x 10 cos(2 t), y 10 sin(2 t)] and that for a uniform velocity (x 5t, y constant), where x and y are in feet and t is in seconds. 4.20 A velocity field is given by u cx2 and v cy2, where c is a constant. Determine the x and y components of the acceleration. At what point 1points2 in the flow field is the acceleration zero? 4.21 Determine the acceleration field for a three-dimensional flow with velocity components u x, v 4x2y2, and w x y. 4.22 A three-dimensional velocity field is given by u 2x, v y, and w z. Determine the acceleration vector. 4.23 Water flows through a constant diameter pipe with a uniform velocity given by V (8/t 5)jˆ m/s, where t is in seconds. Determine the acceleration at time t 1, 2, and 10 s. 192 Chapter 4 ■ Fluid Kinematics 4.24 The velocity of air in the diverging pipe shown in Fig. P4.24 is given by V1 4t fts and V2 2t fts, where t is in seconds. (a) Determine the local acceleration at points 112 and 122. (b) Is the average convective acceleration between these two points negative, zero, or positive? Explain. V1 = 4t ft/s 4.31 As a valve is opened, water flows through the diffuser shown in Fig. P4.31 at an increasing flowrate so that the velocity along the centerline is given by V uiˆ V0 11 ect 2 11 x/2 ˆi, where u0, c, and / are constants. Determine the acceleration as a function of x and t. If V0 10 fts and / 5 ft, what value of c 1other than c 02 is needed to make the acceleration zero for any x at t 1 s? Explain how the acceleration can be zero if the flowrate is increasing with time. V2 = 2t ft/s y (1) /2 (2) ■ Figure P4.24 4.25 Water flows in a pipe so that its velocity triples every 20 s. At t 0 it has u 5 fts. That is, V u1t2 ˆi 5 13t20 2iˆ ft/s. Determine the acceleration when t 0, 10, and 20 s. u u = 1– V0(1 – e–ct) 2 x –ct u = V0(1 – e ) 4.26 When a valve is opened, the velocity of water in a certain pipe is given by u 1011 et 2, v 0, and w 0, where u is in ft兾s and t is in seconds. Determine the maximum velocity and maximum acceleration of the water. 4.27 The velocity of the water in the pipe shown in Fig. P4.27 is given by V1 0.50t ms and V2 1.0t ms, where t is in seconds. Determine the local acceleration at points (1) and (2). Is the average convective acceleration between these two points negative, zero, or positive? Explain. V1 = V2 = 1.0t m/s 0.50t m/s ■ Figure P4.31 4.32 The fluid velocity along the x axis shown in Fig. P4.32 changes from 6 m/s at point A to 18 m/s at point B. It is also known that the velocity is a linear function of distance along the streamline. Determine the acceleration at points A, B, and C. Assume steady flow. (2) (1) ■ Figure P4.27 4.28 A shock wave is a very thin layer (thickness /) in a high-speed (supersonic) gas flow across which the flow properties (velocity, density, pressure, etc.) change from state (1) to state (2) as shown in Fig. P4.28. If V1 1800 fps, V2 700 fps, and / 104 in., estimate the average deceleration of the gas as it flows across the shock wave. How many g’s deceleration does this represent? V1 V2 V1 VB = 18 m/s A C B x 0.05 m 0.1 m V ■ Figure P4.32 V2 Shock wave VA = 6 m/s x ■ Figure P4.28 †4.29 Estimate the average acceleration of water as it travels through the nozzle on your garden hose. List all assumptions and show all calculations. †4.30 A stream of water from the faucet strikes the bottom of the sink. Estimate the maximum acceleration experienced by the water particles. List all assumptions and show calculations. 4.33 A fluid flows along the x axis with a velocity given by V 1xt2 ˆi, where x is in feet and t in seconds. (a) Plot the speed for 0 x 10 ft and t 3 s. (b) Plot the speed for x 7 ft and 2 t 4 s. (c) Determine the local and convective acceleration. (d) Show that the acceleration of any fluid particle in the flow is zero. (e) Explain physically how the velocity of a particle in this unsteady flow remains constant throughout its motion. 4.34 A hydraulic jump is a rather sudden change in depth of a liquid layer as it flows in an open channel as shown in Fig. P4.34 and Video V10.12. In a relatively short distance Problems 1thickness ⫽ /2 the liquid depth changes from z1 to z2, with a corresponding change in velocity from V1 to V2. If V1 ⫽ 1.20 ftⲐs, V2 ⫽ 0.30 ftⲐs, and / ⫽ 0.02 ft, estimate the average deceleration of the liquid as it flows across the hydraulic jump. How many g’s deceleration does this represent? s B A C D E F V0 1.5 V0 (a) Hydraulic jump V2 ᐉ 193 V1 V z2 1.5 V0 V0 z1 ■ Figure P4.34 A B C D E F (b) ■ Figure P4.38 4.35 A fluid particle flowing along a stagnation streamline, as shown in Video V4.9 and Fig. P4.35, slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance s ⫽ 0.6 ft upstream of the stagnation point at t ⫽ 0 is given approximately by s ⫽ 0.6e⫺0.5t, where t is in seconds and s is in feet. (a) Determine the speed of a fluid particle as a function of time, Vparticle 1t2, as it flows along the streamline. (b) Determine the speed of the fluid as a function of position along the streamline, V ⫽ V1s2. (c) Determine the fluid acceleration along the streamline as a function of position, as ⫽ as 1s2. Stagnation point, s = 0 4.39 Air flows steadily through a variable area pipe with a veloc- ity of V ⫽ u1x2iˆ ftⲐs, where the approximate measured values of u1x2 are given in the table. Plot the acceleration as a function of x for 0 ⱕ x ⱕ 12 in. Plot the acceleration if the flowrate is increased by a factor of N 1i.e., the values of u are increased by a factor of N 2 for N ⫽ 2, 4, 10. x (in.) u (ft兾s) x (in.) u (ft兾s) 0 1 2 3 4 5 6 10.0 10.2 13.0 20.1 28.3 28.4 25.8 7 8 9 10 11 12 13 20.1 17.4 13.5 11.9 10.3 10.0 10.0 Fluid particle s V 4.40 ■ Figure P4.35 4.36 A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is, V ⫽ ax ⫹ b, where a and b are constants. If the flow is constant with V1 ⫽ 10 mⲐs at x1 ⫽ 0 and V2 ⫽ 25 mⲐs at x2 ⫽ 1 m, determine the local acceleration, the convective acceleration, and the acceleration of the fluid at points 112 and 122. As is indicated in Fig. P4.40, the speed of exhaust in a car’s exhaust pipe varies in time and distance because of the periodic nature of the engine’s operation and the damping effect with distance from the engine. Assume that the speed is given by V ⫽ V0 31 ⫹ ae⫺bx sin1vt2 4, where V0 ⫽ 8 fps, a ⫽ 0.05, b ⫽ 0.2 ft⫺1, and v ⫽ 50 radⲐs. Calculate and plot the fluid acceleration at x ⫽ 0, 1, 2, 3, 4, and 5 ft for 0 ⱕ t ⱕ pⲐ25 s. 5 ft V 4.37 Repeat Problem 4.36 with the assumption that the flow is not steady, but at the time when V1 ⫽ 10 mⲐs and V2 ⫽ 25 mⲐs, it is known that 0V1Ⲑ 0t ⫽ 20 mⲐs2 and 0V2Ⲑ 0t ⫽ 60 mⲐs2. ■ Figure P4.40 4.38 An incompressible fluid flows through the converging duct shown in Fig. P4.38a with velocity V0 at the entrance. Measurements indicate that the actual velocity of the fluid near the wall of the duct along streamline A–F is as shown in Fig. P4.38b. Sketch the component of acceleration along this streamline, a, as a function of s. Discuss the important characteristics of your result. 4.41 Water flows down the face of the dam shown in Fig. P4.41. The face of the dam consists of two circular arcs with radii of 10 and 20 ft as shown. If the speed of the water along streamline A–B is approximately V ⫽ (2gh)1/2, where the distance h is as indicated, plot the normal acceleration as a function of distance along the streamline, an ⫽ an(s). x V = V0[1 + ae–bx sin(ω t)] 194 Chapter 4 ■ Fluid Kinematics V 4 ft s h A ᏾1 = 10 ft V0 20° 40° A a θ C ■ Figure P4.46 ᏾2 = 20 ft V B ■ Figure P4.41 4.47 The velocity components for steady flow through the nozzle shown in Fig. P4.47 are u V0 x/ and v V0 3 1 1y/2 4 , where V0 and / are constants. Determine the ratio of the magnitude of the acceleration at point (1) to that at point (2). y 4.42 Water flows over the crest of a dam with speed V as shown in Fig. P4.42. Determine the speed if the magnitude of the normal acceleration at point (1) is to equal the acceleration of gravity, g. (1) V (1) (2) = 2 ft /2 x ■ Figure P4.47 ■ Figure P4.42 4.43 Water flows under the sluice gate shown in Fig. P4.43. If V1 3 m/s, what is the normal acceleration at point (1)? 4.48 Water flows through the curved hose shown in Fig. P4.48 with an increasing speed of V 10t ft/s, where t is in seconds. For t 2 s determine (a) the component of acceleration along the streamline, (b) the component of acceleration normal to the streamline, and (c) the net acceleration (magnitude and direction). Sluice gate V ᏾ = 20 ft ᏾ = 0.12 m (1) ■ Figure P4.48 V1 = 3 m/s ■ Figure P4.43 4.44 For the flow given in Problem 4.41, plot the streamwise acceleration, as, as a function of distance, s, along the surface of the dam from A to C. 4.45 Assume that the streamlines for the wingtip vortices from an airplane (see Fig. P4.15 and Video V4.6) can be approximated by circles of radius r and that the speed is V K/r, where K is a constant. Determine the streamline acceleration, as, and the normal acceleration, an, for this flow. 4.46 A fluid flows past a sphere with an upstream velocity of V0 40 m/s as shown in Fig. P4.46. From a more advanced theory it is found that the speed of the fluid along the front part of the sphere is V 32V0 sin u. Determine the streamwise and normal components of acceleration at point A if the radius of the sphere is a 0.20 m. 4.49 A fluid flows past a circular cylinder of radius a with an upstream speed of V0 as shown in Fig. P4.49. A more advanced theory indicates that if viscous effects are negligible, the velocity of the fluid along the surface of the cylinder is given by V 2V0 sin u. Determine the streamline and normal components of acceleration on the surface of the cylinder as a function of V0, a, and u and plot graphs of as and an for 0 u 90° with V0 10 ms and a 0.01, 0.10, 1.0, and 10.0 m. V V0 θ a ■ Figure P4.49 Problems Disks 4.50 Determine the x and y components of acceleration for the flow u c(x2 y2) and 2cxy, where c is a constant. If c 0, is the particle at point x x0 0 and y 0 accelerating or decelerating? Explain. Repeat if x0 0. 4.51 When the floodgates in a channel are opened, water flows along the channel downstream of the gates with an increasing speed given by V 411 0.1t2 fts, for 0 t 20 s, where t is in seconds. For t 7 20 s the speed is a constant V 12 fts. Consider a location in the curved channel where the radius of curvature of the streamlines is 50 ft. For t 10 s, determine (a) the component of acceleration along the streamline, (b) the component of acceleration normal to the streamline, and (c) the net acceleration (magnitude and direction). Repeat for t 30 s. 4.52 Water flows steadily through the funnel shown in Fig. P4.52. Throughout most of the funnel the flow is approximately radial 1along rays from O2 with a velocity of V cr 2, where r is the radial coordinate and c is a constant. If the velocity is 0.4 m兾s when r 0.1 m, determine the acceleration at points A and B. 0.12 m 195 R r V0 Pipe V ■ Figure P4.54 4.55 Air flows into a pipe from the region between a circular disk and a cone as shown in Fig. P4.55. The fluid velocity in the gap between the disk and the cone is closely approximated by V V0 R2r2, where R is the radius of the disk, r is the radial coordinate, and V0 is the fluid velocity at the edge of the disk. Determine the acceleration for r 0.5 and 2 ft if V0 5 fts and R 2 ft. Cone Pipe V Disk r V R A ■ Figure P4.55 0.2 m B r Section 4.2.1 The Material Derivative 0.1 m 4.56 Air flows steadily through a long pipe with a speed of u 50 0.5x, where x is the distance along the pipe in feet, and u is in ft/s. Due to heat transfer into the pipe, the air temperature, T, within the pipe is T 300 10x °F. Determine the rate of change of the temperature of air particles as they flow past the section at x 5 ft. O ■ Figure P4.52 4.53 Water flows though the slit at the bottom of a twodimensional water trough as shown in Fig. P4.53. Throughout most of the trough the flow is approximately radial 1along rays from O2 with a velocity of V cr, where r is the radial coordinate and c is a constant. If the velocity is 0.04 m兾s when r 0.1 m, determine the acceleration at points A and B. A V r 0.8 m B 0.2 m O ■ Figure P4.53 4.54 Air flows from a pipe into the region between two parallel circular disks as shown in Fig. P4.54. The fluid velocity in the gap between the disks is closely approximated by V V0 Rr, where R is the radius of the disk, r is the radial coordinate, and V0 is the fluid velocity at the edge of the disk. Determine the acceleration for r 1, 2, or 3 ft if V0 5 fts and R 3 ft. 4.57 A gas flows along the x axis with a speed of V 5x m/s and a pressure of p 10x2 N/m2, where x is in meters. (a) Determine the time rate of change of pressure at the fixed location x 1. (b) Determine the time rate of change of pressure for a fluid particle flowing past x 1. (c) Explain without using any equations why the answers to parts (a) and (b) are different. 4.58 Assume the temperature of the exhaust in an exhaust pipe can be approximated by T T0(1 aebx) [1 c cos(vt)], where T0 100 °C, a 3, b 0.03 m1, c 0.05, and v 100 rad/s. If the exhaust speed is a constant 3 m/s, determine the time rate of change of temperature of the fluid particles at x 0 and x 4 m when t 0. 4.59 GO A bicyclist leaves from her home at 9 A.M. and rides to a beach 40 mi away. Because of a breeze off the ocean, the temperature at the beach remains 60 °F throughout the day. At the cyclist’s home the temperature increases linearly with time, going from 60 °F at 9 A.M. to 80 °F by 1 P.M. The temperature is assumed to vary linearly as a function of position between the cyclist’s home and the beach. Determine the rate of change of temperature observed by the cyclist for the following conditions: (a) as she pedals 10 mph through a town 10 mi from her home at 10 A.M.; (b) as she eats lunch at a rest stop 30 mi from her home at noon; (c) as she arrives enthusiastically at the beach at 1 P.M., pedaling 20 mph. 4.60 The temperature distribution in a fluid is given by T 10x 5y, where x and y are the horizontal and vertical coordinates in meters and T is in degrees centigrade. Determine the time rate of change of temperature of a fluid particle traveling (a) horizontally with u 20 m s, v 0 or (b) vertically with u 0, v 20 m s. 196 Chapter 4 ■ Fluid Kinematics Section 4.4 The Reynolds Transport Theorem 4.61 Obtain a photograph/image of a situation in which a fluid is flowing. Print this photo and draw a control volume through which the fluid flows. Write a brief paragraph that describes how the fluid flows into and out of this control volume. 4.62 Water flows through a duct of square cross section as shown in Fig. P4.62 with a constant, uniform velocity of V ⫽ 20 mⲐs. Consider fluid particles that lie along line A–B at time t ⫽ 0. Determine the position of these particles, denoted by line A¿⫺B¿, when t ⫽ 0.20 s. Use the volume of fluid in the region between lines A–B and A¿⫺B¿ to determine the flowrate in the duct. Repeat the problem for fluid particles originally along line C–D; along line E–F. Compare your three answers. B B' D F V = 20 m/s A' d dt 冮 x21t2 x11t2 C f 1x, t2dx ⫽ 冮 x2 x1 0f dx2 dx1 dx ⫹ f 1x2, t2 ⫺ f 1x1, t2 0t dt dt Discuss how this formula is related to the time derivative of the total amount of a property in a system and to the Reynolds transport theorem. 4.67 Air enters an elbow with a uniform speed of 10 m/s as shown in Fig. P4.67. At the exit of the elbow, the velocity profile is not uniform. In fact, there is a region of separation or reverse flow. The fixed control volume ABCD coincides with the system at time t ⫽ 0. Make a sketch to indicate (a) the system at time t ⫽ 0.01 s and (b) the fluid that has entered and exited the control volume in that time period. D 0.5 m 45° A †4.66 From calculus, one obtains the following formula 1Leibnitz rule2 for the time derivative of an integral that contains time in both the integrand and the limits of the integration: E 1m ■ Figure P4.62 V = 10 m/s A Control volume 4.63 Repeat Problem 4.62 if the velocity profile is linear from 0 to 20 m兾s across the duct as shown in Fig. P4.63. 1m 5 m/s Reverse B flow C 15 m/s 20 m/s ■ Figure P4.67 0 m/s ■ Figure P4.63 4.64 In the region just downstream of a sluice gate, the water may develop a reverse flow region as is indicated in Fig. P4.64 and Video V10.9. The velocity profile is assumed to consist of two uniform regions, one with velocity Va ⫽ 10 fps and the other with Vb ⫽ 3 fps. Determine the net flowrate of water across the portion of the control surface at section 122 if the channel is 20 ft wide. 4.68 A layer of oil flows down a vertical plate as shown in Fig. P4.68 with a velocity of V ⫽ 1V0 Ⲑh2 2 12hx ⫺ x2 2 ˆj where V0 and h are constants. (a) Show that the fluid sticks to the plate and that the shear stress at the edge of the layer 1x ⫽ h2 is zero. (b) Determine the flowrate across surface AB. Assume the width of the plate is b. (Note: The velocity profile for laminar flow in a pipe has a similar shape. See Video V6.13.) x v(x) Plate Control surface Sluice gate Vb = 3 ft/s A B 1.8 ft 1.2 ft (1) (2) Va = 10 ft/s Oil y h ■ Figure P4.64 ■ Figure P4.68 4.65 At time t ⫽ 0 the valve on an initially empty 1perfect vacuum, r ⫽ 02 tank is opened and air rushes in. If the tank has a volume of ⫺ V 0 and the density of air within the tank increases as r ⫽ rq 11 ⫺ e⫺bt 2, where b is a constant, determine the time rate of change of mass within the tank. 4.69 Water flows in the branching pipe shown in Fig. P4.69 with uniform velocity at each inlet and outlet. The fixed control volume indicated coincides with the system at time t ⫽ 20 s. Make a sketch to indicate (a) the boundary of the system at time t ⫽ 20.1 s, (b) the fluid that left the control volume during that 0.1-s interval, and (c) the fluid that entered the control volume during that time interval. Problems V1 = 2 m/s A D θ 0.8 m 0.5 m (1) 0.5 m V = 3 m/s (3) V3 = 2.5 m/s 197 C B Control surface ■ Figure P4.72 (2) 0.6 m 4.73 The wind blows across a field with an approximate velocity profile as shown in Fig. P4.73. Use Eq. 4.16 with the parameter b equal to the velocity to determine the momentum flowrate across the vertical surface A–B, which is of unit depth into the paper. V2 = 1 m/s Control volume ■ Figure P4.69 4.70 Two plates are pulled in opposite directions with speeds of 1.0 ft/s as shown in Fig. P4.70. The oil between the plates moves with a velocity given by V 10 y ˆi ft/s, where y is in feet. The fixed control volume ABCD coincides with the system at time t 0. Make a sketch to indicate (a) the system at time t 0.2 s and (b) the fluid that has entered and exited the control volume in that time period. 15 ft/s B 20 ft y 0.2 ft B Control volume 0.2 ft C 10 ft 1 ft/s u(y) = 10y ft/s 0.1 ft A x 0.1 ft ■ Figure P4.73 1 ft/s A D ■ Figure P4.70 4.71 Water enters a 5-ft-wide, 1-ft-deep channel as shown in Fig. P4.71. Across the inlet the water velocity is 6 ft s in the center portion of the channel and 1 ft s in the remainder of it. Farther downstream the water flows at a uniform 2 fts velocity across the entire channel. The fixed control volume ABCD coincides with the system at time t 0. Make a sketch to indicate (a) the system at time t 0.5 s and (b) the fluid that has entered and exited the control volume in that time period. 1 ft/s A 6 ft/s 1 ft Nozzle B 2 ft 2 ft/s 5 ft 4.74 Water flows from a nozzle with a speed of V 10 m/s and is collected in a container that moves toward the nozzle with a speed of Vcv 2 m/s as shown in Fig. P4.74. The moving control surface consists of the inner surface of the container. The system consists of the water in the container at time t 0 and the water between the nozzle and the tank in the constant diameter stream at t 0. At time t 0.1 s what volume of the system remains outside of the control volume? How much water has entered the control volume during this time period? Repeat the problem for t 0.3 s. Stream diameter = 0.1 m Container at t = 0 Stream at t = 0 V = 10 m/s Vcv = 2 m/s 1 ft/s D 2 ft C Control surface 3m Moving control volume ■ Figure P4.74 ■ Figure P4.71 4.72 Water flows through the 2-m-wide rectangular channel shown in Fig. P4.72 with a uniform velocity of 3 m兾s. (a) Directly integrate Eq. 4.16 with b 1 to determine the mass flowrate 1kg兾s2 across section CD of the control volume. (b) Repeat part 1a2 with b 1r, where r is the density. Explain the physical interpretation of the answer to part (b). ■ Lifelong Learning Problems 4.1LL Even for the simplest flows it is often not easy to visually represent various flow field quantities such as velocity, pressure, or temperature. For more complex flows, such as those involving threedimensional or unsteady effects, it is extremely difficult to “show the data.” However, with the use of computers and appropriate software, 198 Chapter 4 ■ Fluid Kinematics novel methods are being devised to more effectively illustrate the structure of a given flow. Obtain information about methods used to present complex flow data. Summarize your findings in a brief report. 4.2LL For centuries people have obtained qualitative and quantitative information about various flow fields by observing the motion of objects or particles in a flow. For example, the speed of the current in a river can be approximated by timing how long it takes a stick to travel a certain distance. The swirling motion of a tornado can be observed by following debris moving within the tornado funnel. Recently, various high-tech methods using lasers and minute particles seeded within the flow have been developed to measure velocity fields. Such techniques include the laser doppler anemometer (LDA), the particle image velocimeter (PIV), and others. Obtain information about new laser-based techniques for measuring velocity fields. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 5 Finite Control Volume Analysis CHAPTER OPENING PHOTO: Wind turbine farms (this is the Middelgrunden Offshore Wind Farm in Denmark) are becoming more common. Finite control volume analysis can be used to estimate the amount of energy transferred between the moving air and each turbine rotor. (Photograph courtesy of Siemens Wind Power.) Learning Objectives After completing this chapter, you should be able to: ■ select an appropriate finite control volume to solve a fluid mechanics problem. ■ apply conservation of mass and energy and Newton’s second law of motion to the contents of a finite control volume to get important answers. ■ know how velocity changes and energy transfers in fluid flows are related to forces and torques. ■ understand why designing for minimum loss of energy in fluid flows is so important. Many fluid mechanics problems can be solved by using control volume analysis. To solve many practical problems in fluid mechanics, questions about the behavior of the contents of a finite region in space 1a finite control volume2 are answered. For example, we may be asked to estimate the maximum anchoring force required to hold a turbojet engine stationary during a test. Or we may be called on to design a propeller to move a boat both forward and backward. Or we may need to determine how much power it would take to move natural gas from one location to another many miles away. The bases of finite control volume analysis are some fundamental laws of physics, namely, conservation of mass, Newton’s second law of motion, and the first and second laws of thermodynamics. While some simplifying approximations are made for practicality, the engineering answers possible with the estimates of this powerful analysis method have proven valuable in numerous instances. Conservation of mass is the key to tracking flowing fluid. How much enters and leaves a control volume can be ascertained. Newton’s second law of motion leads to the conclusion that forces can result from or cause changes in a flowing fluid’s velocity magnitude and/or direction. Moment of force 1torque2 can result from or cause changes in a flowing fluid’s moment of velocity. These forces and torques can be associated with work and power transfer. 199 200 Chapter 5 ■ Finite Control Volume Analysis The first law of thermodynamics is a statement of conservation of energy. The second law of thermodynamics identifies the loss of energy associated with every actual process. The mechanical energy equation based on these two laws can be used to analyze a large variety of steady, incompressible flows in terms of changes in pressure, elevation, speed, and of shaft work and loss. Good judgment is required in defining the finite region in space, the control volume, used in solving a problem. What exactly to leave out of and what to leave in the control volume are important considerations. The formulas resulting from applying the fundamental laws to the contents of the control volume are easy to interpret physically and are not difficult to derive and use. Because a finite region of space, a control volume, contains many fluid particles and even more molecules that make up each particle, the fluid properties and characteristics are often average values. Chapter 6 presents an analysis of fluid flow based on what is happening to the contents of an infinitesimally small region of space or control volume through which numerous molecules simultaneously flow (what we might call a point in space). 5.1 Conservation of Mass—The Continuity Equation 5.1.1 Derivation of the Continuity Equation A system is defined as a collection of unchanging contents, so the conservation of mass principle for a system is simply stated as Time rate of change of the system mass 0 The amount of mass in a system is constant. or DMsys Dt 0 (5.1) where the system mass, Msys, is more generally expressed as Msys 冮 r dV (5.2) sys and the integration is over the volume of the system. In words, Eq. 5.2 states that the system mass is equal to the sum of all the density-volume element products for the contents of the system. For a system and a fixed, nondeforming control volume that are coincident at an instant of time, as illustrated in Fig. 5.1, the Reynolds transport theorem 1Eq. 4.192 with B mass and b 1 allows us to state that D Dt 冮 r dV sys 0 0t 冮 cv r dV 冮 rV ⴢ nˆ dA cs Control volume V2 (2) V1 System (a) (1) (b) (c) ■ Figure 5.1 System and control volume at three different instances of time. (a) System and control volume at time t ⴚ Dt. (b) System and control volume at time t, coincident condition. (c) System and control volume at time t ⴙ Dt. (5.3) 5.1 Conservation of Mass—The Continuity Equation 201 or time rate of change Time rate of change of the mass of the of the mass of the contents of the coincoincident system cident control volume net rate of flow of mass through the control surface In Eq. 5.3, we express the time rate of change of the system mass as the sum of two control volume quantities, the time rate of change of the mass of the contents of the control volume, 0 0t 冮 r dV cv and the net rate of mass flow through the control surface, 冮 rV ⴢ nˆ dA cs Control surface V ^ n ^ Vn > 0 When a flow is steady, all field properties 1i.e., properties at any specified point2 including density remain constant with time and the time rate of change of the mass of the contents of the control volume is zero. That is, 冮 0 0t V ^ n V^ n0 V V ^ ^ n n Control volume y x ^ ^ n n V V VrV ^ n 0 r V W Vr V U (5.50) by applying the same kind of analysis used with the sprinkler of Fig. 5.4. The “” is used with # mass flowrate into the control volume, min, and the “” is used with mass flowrate out of the # control volume, mout, to account for the sign of the dot product, V ⴢ nˆ , involved. Whether “” or “” is used with the rVu product depends on the direction of 1r ⴛ V2 axial. A simple way to determine the sign of the rVu product is to compare the direction of Vu and the blade speed, U. As shown in the margin, if Vu and U are in the same direction, then the rVu product is positive. If Vu and U are in opposite directions, the rVu product is negative. The sign of the shaft torque is “” if Tshaft is in the same direction along the axis of rotation as v, and “” otherwise. # The shaft power, Wshaft, is related to shaft torque, Tshaft, by # W shaft Tshaft v (5.51) Thus, using Eqs. 5.50 and 5.51 with a “” sign for Tshaft in Eq. 5.50, we obtain # # # Wshaft 1min 21 rinvVuin 2 mout 1 routvVuout 2 r rV < 0 V Vr V W U When shaft torque and shaft rotation are in the same (opposite) direction, power is into (out of) the fluid. (5.52) or since rv U # # # Wshaft 1min 21 UinVuin 2 mout 1 UoutVuout 2 (5.53) The “” is used for the UVu product when U and Vu are in the same direction; the “” is used # when U and Vu are in opposite directions. Also, since Tshaft was used to # obtain Eq. 5.53, when Wshaft is positive, power is into the fluid 1for example, a pump2, and when Wshaft, is negative, power is out of the fluid 1for example, a turbine2. # The shaft work per unit mass, wshaft, can be obtained from the shaft power, Wshaft, by divid# ing Eq. 5.53 by the mass flowrate, m. By conservation of mass, # # # m min mout From Eq. 5.53, we obtain wshaft 1 UinVuin 2 1 UoutVuout 2 (5.54) The application of Eqs. 5.50, 5.53, and 5.54 is demonstrated in Example 5.19. More examples of the application of Eqs. 5.50, 5.53, and 5.54 are included in Chapter 12. 5.2 E XAMPLE Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 235 Moment-of-Momentum—Power 5.19 GIVEN An air fan has a bladed rotor of 12-in. outside diameter and 10-in. inside diameter as illustrated in Fig. E5.19a. The height of each rotor blade is constant at 1 in. from blade inlet to outlet. The flowrate is steady, on a timeaverage basis, at 230 ft3/min and the absolute velocity of the air at blade inlet, V1, is radial. The blade discharge angle is 30 from the tangential direction. The rotor rotates at a constant speed of 1725 rpm. FIND Estimate the power required to run the fan. SOLUTION We select a fixed and nondeforming control volume that includes the rotating blades and the fluid within the blade row at an instant, as shown with a dashed line in Fig. E5.19a. The flow within this control volume is cyclical, but steady in the mean. The only torque we consider is the driving shaft torque, Tshaft. This torque is provided by a motor. We assume that the entering and leaving flows are each represented by uniformly distributed velocities and flow properties. Since shaft power is sought, Eq. 5.53 is appropriate. Application of Eq. 5.53 to the contents of the control volume in Fig. E5.19 gives 0 1V1 is radial2 # # # Wshaft m1 1 U1Vu1 2 m2 1 U2Vu2 2 (1) 12.38 103 slugft3 21230 ft3 min2 # m rQ 160 smin2 0.00912 slugs 10.0766 lbmft3 21230 ft3 min2 # 0.294 lbms m 160 s min2 The rotor exit blade speed, U2, is From Eq. 1 we see that to calculate fan power, we need mass # flowrate, m, rotor exit blade velocity, U2, and fluid tangential ve# locity at blade exit, V␪2. The mass flowrate, m, is easily obtained from Eq. 5.6 as U2 r2v 16 in.211725 rpm212p radrev2 90.3 fts 112 in.ft2160 smin2 Section (1) 30° Fixed control volume Tshaft W2 V1 ω Section (2) D2 = 2r2 = 12 in. D1 = 2r1 = 10 in. h= 1 in. ω Tshaft (a) ■ Figure E5.19 Fixed control volume (2) Often, problems involving fans are solved using English Engineering units. Since 1 slug 32.174 lbm, we could have used as the density of air rair 12.38 103 slugft3 2132.174 lbmslug2 0.0766 lbmft3. Then U2 Wr2 W2 Vr2 V2 Vθ 2 30° (b) (3) 236 Chapter 5 ■ Finite Control Volume Analysis To determine the fluid tangential speed at the fan rotor exit, V␪2, we use Eq. 5.43 to get V2 W2 U2 Substituting known values into Eq. 10, we obtain W2 (4) By using this value of W2 in Eq. 5 we get Vu2 U2 W2 cos 30° 90.3 fts 129.3 fts210.8662 64.9 fts (5) To solve Eq. 5 for V␪2 we need a value of W2, in addition to the value of U2 already determined (Eq. 3). To get W2, we recognize that W2 sin 30° Vr 2 Equation 1 can now be used to obtain # # Wshaft m U2Vu2 (6) with BG units. With EE units where Vr2 is the radial component of either W2 or V2. Also, using Eq. 5.6, we obtain # (7) m rA2Vr 2 # Wshaft or since 10.00912 slug/s2190.3 ft / s2164.9 ft / s2 31 1slug ⴢ ft/s2 2 lb 4 3550 1ft ⴢ lb2 / 1hp ⴢ s2 4 10.294 lbm/s2190.3 ft /s2164.9 ft / s2 332.174 1lbm ⴢ ft2 1lbs2 2 4 3 550 1ft ⴢ lb2 / 1hp ⴢ s2 4 In either case A2 2 pr2h (8) where h is the blade height, Eqs. 7 and 8 combine to form # m r2pr2hVr 2 Taking Eqs. 6 and 9 together we get # rQ m W2 r2pr2h sin 30° r2pr2h sin 30° Q 2pr2h sin 30° 5.3 160 smin22p16 in.211 in.2 sin 30° 29.3 fts The vector addition of Eq. 4 is shown in the form of a “velocity triangle” in Fig. E5.19b. From Fig. E5.19b, we can see that Vu2 U2 W2 cos 30° 1230 ft3min2112 in.ft2112 in.ft2 # Wshaft 0.097 hp (Ans) (9) COMMENT Note that the “” was used with the U2V␪2 (10) product because U2 and V␪2 are in the same direction. This result, 0.097 hp, is the power that needs to be delivered through the fan shaft for the given conditions. Ideally, all of this power would go into the flowing air. However, because of fluid friction, only some of this power will produce useful effects (e.g., movement and pressure rise) on the air. How much useful effect depends on the efficiency of the energy transfer between the fan blades and the fluid. First Law of Thermodynamics—The Energy Equation 5.3.1 Derivation of the Energy Equation The first law of thermodynamics for a system is, in words, Time rate of net time rate of net time rate of increase of the energy addition by energy addition by total stored energy heat transfer into work transfer into of the system the system the system In symbolic form, this statement is # # # # D er dV a a Qin a Qout b a a Win a Wout b Dt 冮 sys sys The first law of thermodynamics is a statement of conservation of energy. sys or D Dt 冮 sys # # er dV 1Qnet Wnet 2 sys in (5.55) in Some of these variables deserve a brief explanation before proceeding further. The total stored energy per unit mass for each particle in the system, e, is related to the internal energy per unit mass, uˇ, the kinetic energy per unit mass, V 2 2, and the potential energy per unit mass, gz, by the equation e uˇ V2 gz 2 (5.56) 5.3 First Law of Thermodynamics—The Energy Equation 237 # The net rate of heat transfer into # the system is denoted with Qnet in, and the net rate of work transfer into the system is labeled Wnet in. Heat transfer and work transfer are considered “” going into the system and “” coming out. Equation 5.55 is valid for inertial and noninertial reference systems. We proceed to develop the control volume statement of the first law of thermodynamics. For the control volume that is coincident with the system at an instant of time # # # # 1Qnet Wnet 2 sys 1Qnet Wnet 2 coincident (5.57) in in in in control volume Furthermore, for the system and the contents of the coincident control volume that is fixed and nondeforming, the Reynolds transport theorem 1Eq. 4.19 with the parameter b set equal to e2 allows us to conclude that 冮 D Dt er dV sys 0 0t 冮 er dV cv 冮 erV ⴢ nˆ dA (5.58) cs or in words, Time rate time rate of increase of increase of the total stored of the total energy of the contents stored energy of the control volume of the system net rate of flow of the total stored energy out of the control volume through the control surface Combining Eqs. 5.55, 5.57, and 5.58, we get the control volume formula for the first law of thermodynamics: The energy equation involves stored energy and heat and work transfer. • Q3 • Q2 • Q4 • Q1 • Control Volume • • • • Qnet = Q1 + Q2 – Q3 – Q4 in 0 0t 冮 er dV cv 冮 erV ⴢ nˆ dA 1Q # cs net in # Wnet 2 cv (5.59) in The total stored energy per unit mass, e, in Eq. 5.59 is for fluid particles entering, leaving, and within the control volume. Further explanation of the heat transfer and work transfer involved in this equation follows. # The heat transfer rate, Q, represents all of the ways in which energy is exchanged between the control volume contents and surroundings because of a temperature difference. Thus, radiation, conduction, and/or convection are possible. As shown by the figure in the margin, heat transfer into the control volume is considered positive; heat transfer out # is negative. In many engineering applications, the process is adiabatic; the heat # # # transfer rate, Q, is zero. The net heat transfer rate, Qnet in, can also be zero when g # Qin g Qout 0. The work transfer rate, W, also called power, is positive when work is done on the contents of the control volume by the surroundings. Otherwise, it is considered negative. Work can be transferred across the control surface in several ways. In the following paragraphs, we consider some important forms of work transfer. In many instances, work is transferred across the control surface by a moving shaft. In rotary devices such as turbines, fans, and propellers, a rotating shaft transfers work across that portion of the control surface that slices through the shaft. Even in reciprocating machines like positive displacement internal combustion engines and compressors that utilize piston-in-cylinder arrangements, a rotating crankshaft is used. Since work is the dot product of force and related displacement, rate of work 1or power2 is the dot product of force and related displacement per unit time. For a rotat# ing shaft, the power transfer, Wshaft, is related to the shaft torque that causes the rotation, Tshaft, and the angular velocity of the shaft, v, by the relationship # Wshaft Tshaftv When the control surface cuts through the shaft material, the shaft torque is exerted by shaft material at the control surface. To allow for consideration of problems involving more than one shaft we use the notation # # # Wshaft a Wshaft a Wshaft (5.60) net in in out 238 Chapter 5 ■ Finite Control Volume Analysis Section (1) R Control volume Section (2) umax r u1 = umax 1 - _r R [ ()] 2 Pipe umax u2 = umax 1 - _r R ■ Figure 5.6 Simple, fully developed [ ()] 2 pipe flow. Work transfer can also occur at the control surface when a force associated with fluid normal stress acts over a distance. Consider the simple pipe flow illustrated in Fig. 5.6 and the control volume shown. For this situation, the fluid normal stress, s, is simply equal to the negative of fluid pressure, p, in all directions; that is, s p V F (5.61) This relationship can be used with varying amounts of approximation for many engineering problems 1see Chapter 62. # The power transfer, W, associated with a force F acting on an object moving with velocity V is given by the dot product F ⴢ V. This is illustrated by the figure in the margin. Hence, the power # transfer associated with normal stresses acting on a single fluid particle, dWnormal stress, can be evaluated as the dot product of the normal stress force, dFnormal stress, and the fluid particle velocity, V, as # dWnormal stress dFnormal stress ⴢ V θ • W = F•V = FV cosθ If the normal stress force is expressed as the product of local normal stress, s p, and fluid particle surface area, nˆ dA, the result is # dWnormal stress snˆ dA ⴢ V pnˆ dA ⴢ V pV ⴢ nˆ dA For all fluid particles on the# control surface of Fig. 5.6 at the instant considered, power transfer due to fluid normal stress, Wnormal stress, is # Wnormal stress τ V ^ n • δ Wtangential stress = 0 cs (5.62) cs # Note that the value of Wnormal stress #for particles on the wetted inside surface of the pipe is zero because V ⴢ nˆ is zero there. Thus, Wnormal stress can be nonzero only where fluid enters and leaves the control volume. Although only a simple pipe flow was considered, Eq. 5.62 is quite general, and the control volume used in this example can serve as a general model for other cases. Work transfer can also occur at the control surface because of tangential stress forces. Rotating shaft work is transferred by tangential stresses in the shaft material. For a fluid particle, shear # stress force power, dWtangential stress, can be evaluated as the dot product of tangential stress force, dFtangential stress, and the fluid particle velocity, V. That is, # dWtangential stress dFtangential stress ⴢ V Work is transferred by rotating shafts, normal stresses, and tangential stresses. p 冮 sV ⴢ nˆ dA 冮 pV ⴢ nˆ dA For the control volume of Fig. 5.6, the fluid particle velocity is zero everywhere on the wetted inside surface of the pipe. Thus, no tangential stress work is transferred across that portion of the control surface. Furthermore, if we select the control surface so that it is perpendicular to the fluid particle velocity, then the tangential stress force is also perpendicular to the velocity. Therefore, the tangential stress work transfer is zero on that part of the control surface. This is illustrated in the figure in the margin. Thus, in general, we select control volumes like the one of Fig. 5.6 and consider fluid tangential stress power transfer to be negligibly small. Using the information we have developed about power, we can express the first law of thermodynamics for the contents of a control volume by combining Eqs. 5.59, 5.60, and 5.62 to obtain 0 0t 冮 cv er dV 冮 erV ⴢ nˆ dA Q # cs net in # Wshaft net in 冮 pV ⴢ nˆ dA cs (5.63) 5.3 First Law of Thermodynamics—The Energy Equation 239 When the equation for total stored energy 1Eq. 5.562 is considered with Eq. 5.63, we obtain the energy equation: 0 0t 冮 er dV cv 冮 cs auˇ # # p V2 gzb rV ⴢ nˆ dA Qnet Wshaft r 2 in net in (5.64) 5.3.2 Application of the Energy Equation In Eq. 5.64, the term 0 0t 兰cv er dV represents the time rate of change of the total stored energy, e, of the contents of the control volume. This term is zero when the flow is steady. This term is also zero in the mean when the flow is steady in the mean 1cyclical2. In Eq. 5.64, the integrand of 冮 cs auˇ p V2 gzb rV ⴢ nˆ dA r 2 can be nonzero only where fluid crosses the control surface 1V ⴢ nˆ 02. Otherwise, V ⴢ nˆ is zero and the integrand is zero for that portion of the control surface. If the properties within parentheses, uˇ, pr, V 2 2, and gz, are all assumed to be uniformly distributed over the flow cross-sectional areas involved, the integration becomes simple and gives 冮 auˇ r p cs p V2 V2 # gzb rV ⴢ nˆ dA a auˇ gzb m r 2 2 flow out p V2 # a auˇ gzb m r 2 flow (5.65) in Furthermore, if there is only one stream entering and leaving the control volume, then 冮 cs auˇ p V2 gzb rV ⴢ nˆ dA r 2 auˇ p p V2 V2 # # gzb mout auˇ gzb min r r 2 2 out in (5.66) Uniform flow as described above will occur in an infinitesimally small diameter streamtube as illustrated in Fig. 5.7. This kind of streamtube flow is representative of the steady flow of a particle of fluid along a pathline. We can also idealize actual conditions by disregarding nonuniformities in a finite cross section of flow. We call this one-dimensional flow, and although such uniform flow rarely occurs in reality, the simplicity achieved with the one-dimensional approximation often justifies its use. More details about the effects of nonuniform distributions of velocities and other fluid flow variables are considered in Section 5.3.4 and in Chapters 8, 9, and 10. If shaft work is involved, the flow must be unsteady, at least locally 1see Refs. 1 and 22. The flow in any fluid machine that involves shaft work is unsteady within that machine. For example, the velocity and pressure at a fixed location near the rotating blades of a fan are unsteady. However, upstream and downstream of the machine, the flow may be steady. Most often shaft work is associated with flow that is unsteady in a recurring or cyclical way. On a time-average basis for flow that is one-dimensional, cyclical, and involves only one stream of fluid entering and leaving m• out dA Streamtube • min V ■ Figure 5.7 Streamtube flow. 240 Chapter 5 ■ Finite Control Volume Analysis m• out • min m• in = m• out = m• the control volume, such as the hair dryer shown in the figure in the margin, Eq. 5.64 can be simplified with the help of Eqs. 5.9 and 5.66 to form # # p p V 2out V 2in # m c uˇout uˇin a b a b g1zout zin 2 d Q net W shaft r out r in 2 in net in (5.67) We call Eq. 5.67 the one-dimensional energy equation for steady-in-the-mean flow. Note that Eq. 5.67 is valid for incompressible and compressible flows. Often, the fluid property called enthalpy, hˇ, where hˇ uˇ The energy equation is sometimes written in terms of enthalpy. p r (5.68) is used in Eq. 5.67. With enthalpy, the one-dimensional energy equation for steady-in-the-mean flow 1Eq. 5.672 is # # V 2out V 2in # m c hˇout hˇin g1zout zin 2 d Qnet Wshaft 2 in net in (5.69) Equation 5.69 is often used for solving compressible flow problems. Examples 5.20 and 5.21 illustrate how Eqs. 5.67 and 5.69 can be used. E XAMPLE Energy—Pump Power 5.20 • GIVEN A pump delivers water at a steady rate of 300 gal/min as shown in Fig. E5.20. Just upstream of the pump [section (1)] where the pipe diameter is 3.5 in., the pressure is 18 psi. Just downstream of the pump [section (2)] where the pipe diameter is 1 in., the pressure is 60 psi. The change in water elevation across the pump is zero. The rise in internal energy of water, uˇ 2 uˇ 1, associated with a temperature rise across the pump is 93 ft ⴢ lb/lbm. The pumping process is considered to be adiabatic. Wshaft = ? Control volume D2 = 1 in. Q= 300 gal/min. D1 = Pump 3.5 in. Section (2) Section (1) p1 = 18 psi FIND Determine the power (hp) required by the pump. p2 = 60 psi ^ ^ u2 – u1 = 93 ft • lb/lbm ■ Figure E5.20 SOLUTION We include in our control volume the water contained in the pump between its entrance and exit sections. Application of Eq. 5.67 to the contents of this control volume on a time-average basis yields 0 (no elevation change) p p V 22 V 21 # g1z2 z1 2 d m c uˇ 2 uˇ 1 a b a b r 2 r 1 2 0 (adiabatic flow) # # (1) Qnet Wshaft in out of the pump, V2. All other quantities in Eq. 1 are given in the problem statement. From Eq. 5.6, we get 11.94 slugsft3 21300 galmin2132.174 lbmslug2 # m rQ 17.48 galft3 2160 smin2 41.8 lbms (2) Also from Eq. 5.6, V net in so We # can solve directly for the power required by the pump, Wshaft net in, from Eq. 1, after we first determine the mass flowrate, # m, the speed of flow into the pump, V1, and the speed of the flow V1 Q Q A pD24 1300 galmin24 112 in.ft2 2 Q A1 17.48 galft3 2160 smin2p 13.5 in.2 2 10.0 ft s (3) 5.3 and V2 First Law of Thermodynamics—The Energy Equation 123 ft s 11.94 slugsft3 2132.174 lbmslug2 1123 fts2 2 110.0 fts2 2 d 23 32.174 1lbmⴢft2 1lbⴢs2 2 4 (4) Substituting the values of Eqs. 2, 3, and 4 and values from the problem statement into Eq. 1 we obtain 141.8 lbms2 c 193 ftⴢlblbm2 # Wshaft net in 1 32.2 hp 35501ftⴢlbs2 hp4 (Ans) COMMENT Of the total 32.2 hp, internal energy change accounts for 7.09 hp, the pressure rise accounts for 7.37 hp, and the kinetic energy increase accounts for 17.8 hp. 160 psi2 1144 in.2ft2 2 11.94 slugsft3 2132.174 lbmslug2 E XAMPLE 118 psi21144 in.2ft2 2 1300 galmin24 112 in.ft2 2 Q A2 17.48 galft3 2160 smin2p 11 in.2 2 241 5.21 Energy—Turbine Power per Unit Mass of Flow GIVEN A steam turbine generator unit used to produce electricity is shown in Fig. E5.21a. Assume the steam enters a turbine with a velocity of 30 m/s and enthalpy, hˇ 1, of 3348 kJ/kg (see Fig. E5.21b). The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy of 2550 kJ/kg. The flow through the turbine is adiabatic, and changes in elevation are negligible. FIND Determine the work output involved per unit mass of steam through-flow. SOLUTION We use a control volume that includes the steam in the turbine from the entrance to the exit as shown in Fig. E5.21b. Applying Eq. 5.69 to the steam in this control volume we get ■ Figure E5.21a Control volume 0 (elevation change is negligible) 0 (adiabatic flow) # m c hˇ 2 hˇ 1 V 22 2 V 21 Steam turbine # # g1z2 z1 2 d Qnet Wshaft in (1) net in The work output per unit mass of steam through-flow, wshaft net in, can # be obtained by dividing Eq. 1 by the mass flowrate, m, to obtain # Wshaft V 22 V 21 net in (2) wshaft # hˇ 2 hˇ 1 m 2 net in Since wshaft net out wshaft net in, we obtain V 21 V 22 wshaft hˇ 1 hˇ 2 2 net out or wshaft 3348 kJkg 2550 kJ kg net out 3 130 ms2 2 160 ms2 2 4 31 J 1Nⴢm2 4 2 31 1kgⴢm2 1Nⴢs2 2 4 11000 JkJ2 Section (1) V ^1 = 30 m/s h1 = 3348 kJ/kg wshaft = ? Section (2) V ^2 = 60 m/s h2 = 2550 kJ/kg ■ Figure E5.21b Thus, wshaft 3348 kJkg 2550 kJkg 1.35 kJkg net out 797 kJ /kg (Ans) COMMENT Note that in this particular example, the change in kinetic energy is small in comparison to the difference in enthalpy involved. This is often true in applications involving steam # turbines. To determine the power output, Wshaft , we must know # the mass flowrate, m. 242 Chapter 5 ■ Finite Control Volume Analysis If the flow is steady throughout, one-dimensional, and only one fluid stream is involved, then the shaft work is zero and the energy equation is # p p V 2out V 2in # m c uˇout uˇin a b a b g1zout zin 2 d Qnet r out r in 2 in (5.70) We call Eq. 5.70 the one-dimensional, steady-flow energy equation. This equation is valid for incompressible and compressible flows. For compressible flows, enthalpy is most often used in the one-dimensional, steady-flow energy equation and, thus, we have V5.14 Pelton wheel turbine # V 2out V 2in # m c hˇout hˇin g1zout zin 2 d Qnet 2 in (5.71) An example of the application of Eq. 5.70 follows. E XAMPLE 5.22 Energy—Temperature Change GIVEN The 420-ft waterfall shown in Fig. E5.22a involves steady flow from one large body of water to another. FIND Determine the temperature change associated with this flow. SOLUTION To solve this problem, we consider a control volume consisting of a small cross-sectional streamtube from the nearly motionless surface of the upper body of water to the nearly motionless surface of the lower body of water as is sketched in Fig. E5.22b. We need to determine T2 T1. This temperature change is related to the change of internal energy of the water, uˇ2 uˇ1, by the relationship T2 T1 uˇ2 uˇ1 cˇ (1) Section (1) ■ Figure E5.22a Control volume [Photograph of Akaka Falls (Hawaii) courtesy of Scott and Margaret Jones.] where cˇ 1 Btu1lbm # °R2 is the specific heat of water. The application of Eq. 5.70 to the contents of this control volume leads to 420 ft Section (2) p p V 22 V 21 # m c uˇ2 uˇ1 a b a b g1z2 z1 2 d r 2 r 1 2 # (2) Qnet in # We assume that the flow is adiabatic. Thus Qnet in 0. Also, ■ Figure E5.22b p p a b a b r 1 r 2 (3) 5.3 First Law of Thermodynamics—The Energy Equation because the flow is incompressible and atmospheric pressure prevails at sections 112 and 122. Furthermore, V1 V2 0 (4) because the surface of each large body of water is considered motionless. Thus, Eqs. 1 through 4 combine to yield T2 T1 g1z1 z2 2 cˇ 243 so that with cˇ 3 1 Btu 1lbm # °R2 4 1778 ft # lbBtu2 3 778 ft # lb 1lbm # °R2 4 132.2 fts2 21420 ft2 3778 ft # lb 1lbm # °R2 4 3 32.2 1lbm # ft2 1lb # s2 2 4 0.540 °R (Ans) T2 T1 COMMENT Note that it takes a considerable change of potential energy to produce even a small increase in temperature. A form of the energy equation that is most often used to solve incompressible flow problems is developed in the next section. 5.3.3 Comparison of the Energy Equation with the Bernoulli Equation When the one-dimensional energy equation for steady-in-the-mean flow, Eq. 5.67, is applied to a flow that is steady, Eq. 5.67 becomes the one-dimensional, steady-flow energy # equation, Eq. 5.70. The only difference between Eq. 5.67 and Eq. 5.70 is that shaft power, W shaft net in, is zero if the flow is steady throughout the control volume 1fluid machines involve locally unsteady flow2. If in addition to being steady, the flow is incompressible, we get from Eq. 5.70 # pin V 2out V 2in pout # m c uˇout uˇin g1zout zin 2 d Qnet r r 2 in # Dividing Eq. 5.72 by the mass flowrate, m, and rearranging terms we obtain V 2out V 2in pout pin gzout gzin 1uˇout uˇin qnet 2 r r 2 2 in (5.72) (5.73) where # Qnet in qnet # m in is the heat transfer rate per mass flowrate, or heat transfer per unit mass. Note that Eq. 5.73 involves energy per unit mass and is applicable to one-dimensional flow of a single stream of fluid between two sections or flow along a streamline between two sections. If the steady, incompressible flow we are considering also involves negligible viscous effects 1frictionless flow2, then the Bernoulli equation, Eq. 3.7, can be used to describe what happens between two sections in the flow as pout rV 2out rV 2in gzout pin gzin 2 2 (5.74) where g rg is the specific weight of the fluid. To get Eq. 5.74 in terms of energy per unit mass, so that it can be compared directly with Eq. 5.73, we divide Eq. 5.74 by density, r, and obtain pout V 2out pin V 2in gzout gzin r r 2 2 (5.75) A comparison of Eqs. 5.73 and 5.75 prompts us to conclude that uˇout uˇin qnet 0 (5.76) in when the steady incompressible flow is frictionless. For steady incompressible flow with friction, we learn from experience (second law of thermodynamics) that uˇout uˇin qnet 7 0 in (5.77) 244 Chapter 5 ■ Finite Control Volume Analysis In Eqs. 5.73 and 5.75, we can consider the combination of variables p V2 gz r 2 as equal to useful or available energy. Thus, from inspection of Eqs. 5.73 and 5.75, we can conclude that uˇout uˇin qnet in represents the loss of useful or available energy that occurs in an incompressible fluid flow because of friction. In equation form we have Minimizing loss is the central goal of fluid mechanical design. uˇout uˇin qnet loss (5.78) in For a frictionless flow, Eqs. 5.73 and 5.75 tell us that loss equals zero. It is often convenient to express Eq. 5.73 in terms of loss as pout V 2out V 2in pin gzout gzin loss r r 2 2 (5.79) An example of the application of Eq. 5.79 follows. E XAMPLE 5.23 Energy—Effect of Loss of Available Energy GIVEN As shown in Fig. E5.23a, air flows from a room through two different vent configurations: a cylindrical hole in the wall having a diameter of 120 mm and the same diameter cylindrical hole in the wall but with a well-rounded entrance. The room pressure is held constant at 1.0 kPa above atmospheric pressure. Both vents exhaust into the atmosphere. As discussed in Section 8.4.2, the loss in available energy associated with flow through the cylindrical vent from the room to the vent SOLUTION exit is 0.5V22/2 where V2 is the uniformly distributed exit velocity of air. The loss in available energy associated with flow through the rounded entrance vent from the room to the vent exit is 0.05V22/2, where V2 is the uniformly distributed exit velocity of air. FIND Compare the volume flowrates associated with the two different vent configurations. Control volume We use the control volume for each vent sketched in Fig. E5.23a. What is sought is the flowrate, Q A2V2, where A2 is the vent exit cross-sectional area, and V2 is the uniformly distributed exit velocity. For both vents, application of Eq. 5.79 leads to V 22 (1) D2 = 120 mm where 1loss2 is the loss between sections (1) and (2). Solving Eq. 1 for V2 we get V2 B 2ca p1 p2 b 1loss2 d r (2) V 22 2 (3) V2 Section (2) Section (1) for both vents is in the room and involves V1 = 0 p1 = 1.0 kPa 0 (no elevation change) p2 p1 V 21 gz2 gz1 1loss2 r r 2 2 0 1V1 ⬇ 02 D2 = 120 mm V2 Section (2) Control volume Since 1loss2 KL where KL is the loss coefficient (KL 0.5 and 0.05 for the two vent configurations involved), we can combine Eqs. 2 and 3 to get V2 B 2ca p1 p2 V 22 d b KL r 2 (4) ■ Figure E5.23a Solving Eq. 4 for V2 we obtain V2 p1 p2 B r 3 11 KL 2 24 (5) 5.3 Therefore, for flowrate, Q, we obtain Q A2V2 p1 p2 4 B r3 11 KL 2 24 pD 22 (6) For the rounded entrance cylindrical vent, Eq. 6 gives Q 245 First Law of Thermodynamics — The Energy Equation p1120 mm2 2 411000 mmm2 2 11.0 kPa2 11000 PakPa2 311Nm2 2 1Pa2 4 B 11.23 kg m3 2 3 11 0.052 24 311Nⴢs2 2 1kgⴢm2 4 COMMENT By repeating the calculations for various values of the loss coefficient, KL, the results shown in Fig. E5.23b are obtained. Note that the rounded entrance vent allows the passage of more air than does the cylindrical vent because the loss associated with the rounded entrance vent is less than that for the cylindrical one. For this flow the pressure drop, p1 p2, has two purposes: (1) overcome the loss associated with the flow, and (2) produce the kinetic energy at the exit. Even if there were no loss (i.e., KL 0), a pressure drop would be needed to accelerate the fluid through the vent. 0.5 or Q 0.445 m3 s (Ans) 0.4 (0.05, 0.445 m3/s) (0.5, 0.372 m3/s) Q Q, m3/s For the cylindrical vent, Eq. 6 gives us p1120 mm2 2 411000 mm m2 2 11.0 kPa211000 PakPa2 3 11Nm2 2 1Pa2 4 B 11.23 kgm3 2 3 11 0.52 24 3 11Nⴢs2 2 1kgⴢm2 4 or Q 0.372 m3 s 0.3 0.2 0.1 0 (Ans) 0 0.1 0.2 0.3 0.4 0.5 KL ■ Figure E5.23b An important group of fluid mechanics problems involves one-dimensional, incompressible, steady-in-the-mean flow with friction and shaft work. Included in this category are constant density flows through pumps, blowers, fans, and turbines. For this kind of flow, Eq. 5.67 becomes # # pin V 2out V 2in pout # m c uˇout uˇin g1zout zin 2 d Qnet Wshaft r r 2 in net in (5.80) # # Dividing Eq. 5.80 by mass flowrate and using the work per unit mass, wshaft Wshaft m, we net in net in obtain The mechanical energy equation can be written in terms of energy per unit mass. V 2out V 2in pout pin gzout gzin wshaft 1uˇout uˇin qnet 2 r r 2 2 net in in If the flow is steady throughout, Eq. 5.81 becomes identical to Eq. 5.73, and the previous observation that uˇout uˇin qnet in equals the loss of available energy is valid. Thus, we conclude that Eq. 5.81 can be expressed as pout V 2out pin V 2in gzout gzin wshaft loss r r 2 2 net in V5.15 Energy transfer (5.81) (5.82) This is a form of the energy equation for steady-in-the-mean flow that is often used for incompressible flow problems. It is sometimes called the mechanical energy equation or the extended Bernoulli equation. Note that Eq. 5.82 involves energy per unit mass 1ft # lbslug ft2s2 or N # m m2s2 2. According to Eq. 5.82, when the shaft work is into the control volume, as for example with a pump, a larger amount of loss will result in more shaft work being required for the same rise in available energy. Similarly, when the shaft work is out of the control volume 1for example, a turbine2, a larger loss will result in less shaft work out for the same drop in available energy. Designers spend a great deal of effort on minimizing losses in fluid flow components. The following examples demonstrate why losses should be kept as small as possible in fluid systems. 246 Chapter 5 ■ Finite Control Volume Analysis E XAMPLE 5.24 Energy—Fan Work and Efficiency GIVEN An axial-flow ventilating fan driven by a motor that FIND Determine how much of the work to the air actually pro- delivers 0.4 kW of power to the fan blades produces a 0.6-mdiameter axial stream of air having a speed of 12 m/s. The flow upstream of the fan involves negligible speed. duces useful effects, that is, fluid motion and a rise in available energy. Estimate the fluid mechanical efficiency of this fan. SOLUTION We select a fixed and nondeforming control volume as is illustrated in Fig. E5.24. The application of Eq. 5.82 to the contents of this control volume leads to 0 (atmospheric pressures cancel) Section (1) Stream surface V1 = 0 0 (V1 ⬇ 0) Control volume p2 p1 V 22 V 21 gz2 b a gz1 b (1) r r 2 2 wshaft loss a net in D2 = 0.6 m Fan motor Fan 0 (no elevation change) Section (2) V2 = 12 m/s where wshaft net in loss is the amount of work added to the air that produces a useful effect. Equation 1 leads to wshaft loss net in 112 m s2 2 V 22 2 2 311kgⴢm2 1Nⴢs2 2 4 72.0 Nⴢm/kg (2) (Ans) ■ Figure E5.24 A reasonable estimate of efficiency, ␩, would be the ratio of amount of work that produces a useful effect, Eq. 2, to the amount of work delivered to the fan blades. That is, h wshaft net in # Wshaft wshaft loss net in (3) wshaft net in To calculate the efficiency, we need a value of w # shaft net in, which is related to the power delivered to the blades, Wshaft net in. We note that # Wshaft (4) net in wshaft # m net in or From Eqs. 2, 3, and 6 we obtain h pD 22 # m rAV r V2 4 i d s 72.0 Nⴢm/kg 0.752 95.8 Nⴢm/kg (Ans) (5) For fluid density, ␳, we use 1.23 kg/m3 (standard air) and, thus, from Eqs. 4 and 5 we obtain u (6) net in # l 11.23 kgm3 2 3 1p210.6 m2 244 112 ms2 wshaft 95.8 Nⴢm/kg where the mass flowrate, m, is (from Eq. 5.6) F net in 1rpD 2242V2 10.4 kW2 3 1000 1Nm2 1skW2 4 i n Curtain of air An air curtain is produced by blowing air through a long rectangular nozzle to produce a high-velocity sheet of air, or a “curtain of air.” This air curtain is typically directed over a doorway or opening as a replacement for a conventional door. The air curtain can be used for such things as keeping warm air from infiltrating dedicated cold spaces, preventing dust and other contaminants from entering a clean environment, and even just keeping insects out of the workplace, still allowing people to enter or exit. A disadvantage over conventional doors is the added COMMENT Note that only 75% of the power that was delivered to the air resulted in useful effects and, thus, 25% of the shaft power is lost to air friction. t h e N e w s power requirements to operate the air curtain, although the advantages can outweigh the disadvantage for various industrial applications. New applications for current air curtain designs continue to be developed. For example, the use of air curtains as a means of road tunnel fire security is currently being investigated. In such an application, the air curtain would act to isolate a portion of the tunnel where fire has broken out and not allow smoke and fumes to infiltrate the entire tunnel system. (See Problem 5.127.) 5.3 First Law of Thermodynamics — The Energy Equation 247 If Eq. 5.82, which involves energy per unit mass, is multiplied by fluid density, r, we obtain pout V5.16 Water plant aerator rV 2out rV 2in gzout pin gzin rwshaft r1loss2 2 2 net in (5.83) where g rg is the specific weight of the fluid. Equation 5.83 involves energy per unit volume and the units involved are identical with those used for pressure 1ft # lbft3 lbft2 or N # mm3 Nm2 2. If Eq. 5.82 is divided by the acceleration of gravity, g, we get pout V 2out V 2in pin zout zin hs hL g g 2g 2g (5.84) where # # Wshaft Wshaft net in net in hs wshaft net in g # mg gQ is the shaft work head and hL lossg is the head loss. Equation 5.84 involves energy per unit weight 1ft # lblb ft or N # mN m2. In Section 3.7, we introduced the notion of “head,” which is energy per unit weight. Units of length 1for example, ft, m2 are used to quantify the amount of head involved. If a turbine is in the control volume, hs is negative because it is associated with shaft work out of the control volume. For a pump in the control volume, hs is positive because it is associated with shaft work into the control volume. We can define a total head, H, as follows: H p V2 z g 2g Then Eq. 5.84 can be expressed as Hout Hin hs hL Some important possible values of Hout in comparison to Hin are shown in Fig. 5.8. Note that hL (head loss) always reduces the value of Hout, except in the ideal case when it is zero. Note also that hL lessens the effect of shaft work that can be extracted from a fluid. When hL 0 (ideal condition) the shaft work head, hs, and the change in total head are the same. This head change is sometimes called ideal head change. The corresponding ideal shaft work head is the minimum required to achieve a desired effect. For work out, it is the maximum possible. Designers usually strive to minimize loss. In Chapter 12 we learn of one instance when minimum loss is sacrificed for survivability of fish coursing through a turbine rotor. Hout hs hs – hL Hin hs + hL hL > 0, hs < 0 hL = 0, hs < 0 hL > 0, hs > 0 hs hL = 0, hs > 0 hL > 0, hs = 0 hL hL = 0, hs = 0 The energy equation written in terms of energy per unit weight involves heads. (5.85) ■ Figure 5.8 Total-head change in fluid flows. 248 Chapter 5 ■ Finite Control Volume Analysis E XAMPLE 5.25 Energy—Head Loss and Power Loss GIVEN The pump shown in Fig. E5.25a adds 10 horsepower Section (2) to the water as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the head loss is 15 ft. Control volume FIND Determine (a) 30 ft the flowrate and Flow (b) the power loss associated with this flow. Section (1) SOLUTION (a) Pump The energy equation (Eq. 5.84) for this flow is Flow p2 p1 V22 V12 z2 z1 hs hL g g 2g 2g (1) where points 2 and 1 (corresponding to “out” and “in” in Eq. 5.84) are located on the lake surfaces. Thus, p2 p1 0 and V2 V1 0 so that Eq. 1 becomes hs hL z2 z1 (2) where z 2 30 ft, z1 0, and hL 15 ft. The pump head is obtained from Eq. 5.85 as # hs Wshaft net in g Q 110 hp2 1550 ft # lbshp2162.4 lb ft3 2 Q 88.1Q where hs is in ft when Q is in ft3s. Hence, from Eq. 2, 88.1 Q 15 ft 30 ft or Q 1.96 ft3s (Ans) COMMENT Note that in this example the purpose of the ■ Figure E5.25a (a 15-ft head); it does not, overall, alter the water’s pressure or velocity. (b) The power lost due to friction can be obtained from Eq. 5.85 as # Wloss g QhL 162.4 lb / ft3 211.96 ft3/s2115 ft2 1830 ft # lb /s 11 hp550 ft # lb /s2 (Ans) 3.33 hp COMMENTS The remaining 10 hp 3.33 hp 6.67 hp that the pump adds to the water is used to lift the water from the lower to the upper lake. This energy is not “lost,” but it is stored as potential energy. By repeating the calculations for various head losses, hL, the results shown in Fig. E5.25b are obtained. Note that as the head loss increases, the flowrate decreases because an increasing portion of the 10 hp supplied by the pump is lost and, therefore, not available to lift the fluid to the higher elevation. pump is to lift the water (a 30-ft head) and overcome the head loss 3.5 3 Q, ft3/s 2.5 (15 ft, 1.96 ft3/s) 2 1.5 1 0.5 0 0 5 10 15 hL, ft ■ Figure E5.25b 20 25 5.3 First Law of Thermodynamics—The Energy Equation 249 A comparison of the energy equation and the Bernoulli equation has led to the concept of loss of available energy in incompressible fluid flows with friction. In Chapter 8, we discuss in detail some methods for estimating loss in incompressible flows with friction. In Section 5.4 and Chapter 11, we demonstrate that loss of available energy is also an important factor to consider in compressible flows with friction. F l u i d s i n Smart shocks Vehicle shock absorbers are dampers used to provide a smooth, controllable ride. When going over a bump, the relative motion between the tires and the vehicle body displaces a piston in the shock and forces a viscous fluid through a small orifice or channel. The viscosity of the fluid produces a head loss that dissipates energy to dampen the vertical motion. Current shocks use a fluid with fixed viscosity. However, recent technology has been developed that uses a synthetic oil with millions of tiny iron balls suspended in it. These tiny balls react to a magnetic field t h e N e w s generated by an electric coil on the shock piston in a manner that changes the fluid viscosity, going anywhere from essentially no damping to a solid almost instantly. A computer adjusts the current to the coil to select the proper viscosity for the given conditions (i.e., wheel speed, vehicle speed, steering-wheel angle, lateral acceleration, brake application, and temperature). The goal of these adjustments is an optimally tuned shock that keeps the vehicle on a smooth, even keel while maximizing the contact of the tires with the pavement for any road conditions. (See Problem 5.110.) 5.3.4 Application of the Energy Equation to Nonuniform Flows The forms of the energy equation discussed in Sections 5.3.2 and 5.3.3 are applicable to onedimensional flows, flows that are approximated with uniform velocity distributions where fluid crosses the control surface. If the velocity profile at any section where flow crosses the control surface is not uniform, inspection of the energy equation for a control volume, Eq. 5.64, suggests that the integral V2 rV ⴢ nˆ dA cs 2 冮 The kinetic energy coefficient is used to account for nonuniform flows. will require special attention. The other terms of Eq. 5.64 can be accounted for as already discussed in Sections 5.3.2 and 5.3.3. For one stream of fluid entering and leaving the control volume, we can define the relationship ainV 2in V2 # aoutV 2out rV ⴢ nˆ dA m a b 2 2 cs 2 冮 where a is the kinetic energy coefficient and V is the average velocity defined earlier in Eq. 5.7. From the above we can conclude that # maV 2 2 V2 rV ⴢ nˆ dA A 2 冮 for flow through surface area A of the control surface. Thus, a Parabolic (laminar) ~ ~ 1.08 Turbulent 冮 1V 22rV ⴢ nˆ dA 2 =2 A # mV 22 (5.86) It can be shown that for any velocity profile, a 1, with a 1 only for uniform flow. Some typical velocity profile examples for flow in a conventional pipe are shown in the sketch in the margin. Therefore, for nonuniform velocity profiles, the energy equation on an energy per unit mass basis for the incompressible flow of one stream of fluid through a control volume that is steady in the mean is =1 Uniform pout aoutV 2out pin ainV 2in gzout gzin wshaft loss r r 2 2 net in (5.87) 250 Chapter 5 ■ Finite Control Volume Analysis On an energy per unit volume basis we have pout raoutV 2out rainV 2in gzout pin gzin rwshaft r1loss2 2 2 net in (5.88) and on an energy per unit weight or head basis we have pout aoutV 2out ainV 2in pin zout zin g g 2g 2g wshaft net in g hL (5.89) The following examples illustrate the use of the kinetic energy coefficient. E XAMPLE 5.26 Energy—Effect of Nonuniform Velocity Profile GIVEN The small fan shown in Fig. E5.26 moves air at a mass flowrate of 0.1 kg兾min. Upstream of the fan, the pipe diameter is 60 mm, the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, a1, is equal to 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent, the velocity profile is quite uniform, and the kinetic energy coefficient, a2, is equal to 1.08. The rise in static pressure across the fan is 0.1 kPa, and the fan motor draws 0.14 W. FIND Compare the value of loss calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions. D2 = 30 mm Turbulent flow Section (2) α 2 = 1.08 Control volume D1 = 60 mm SOLUTION Section (1) α 1 = 2.0 Application of Eq. 5.87 to the contents of the control volume shown in Fig. E5.26 leads to 0 1change in gz is negligible2 p2 p1 a2V 22 a1V 21 gz2 gz1 r r 2 2 loss wshaft • ■ Figure E5.26 (1) net in net in p2 p1 a1V 21 a2V 22 b r 2 2 (2) To proceed further, we need values of wshaft net in, V1, and V2. These quantities can be obtained as follows. For shaft work wshaft net in power to fan motor # m wshaft net in 0.1 kgmin 84.0 N # mkg For the average velocity at section 122, V2, V2 10.1 kgmin2 11 min60 s2 11000 mmm2 2 11.23 kgm3 2 3p130 mm2 244 1.92 ms (5) (a) For the assumed uniform velocity profiles 1a1 a2 1.02, Eq. 2 yields or 10.14 W2 3 11 N # ms2 W4 10.1 kgmin2 11 min60 s2 11000 mmm2 2 11.23 kgm3 2 3p160 mm2 244 0.479 ms or solving Eq. 1 for loss we get loss wshaft a Laminar flow m = 0.1 kg/min loss wshaft a 160 s min2 net in (3) For the average velocity at section 112, V1, from Eq. 5.11 we obtain # m V1 rA1 # m (4) r1pD21 42 p2 p1 V 21 V 22 b r 2 2 (6) Using Eqs. 3, 4, and 5 and the pressure rise given in the problem statement, Eq. 6 gives loss 84.0 10.1 kPa211000 PakPa211 Nm2 Pa2 N#m kg 1.23 kgm3 5.3 or 10.479 m s2 2 11.92 ms2 2 2 31 1kg # m2 1N # s2 2 4 23 1 1kg # m2 1N # s2 2 4 loss 84.0 N # mkg 81.3 N # mkg 0.115 N # mkg 1.84 N # mkg 0.975 N # mkg 1.0811.92 m s2 2 210.479 m s2 2 23 1 1kg # m2 1N # s2 2 4 23 1 1kg # m2 1 # s2 2 4 or loss 84.0 N # mkg 81.3 N # mkg 0.230 N # mkg 1.99 N # mkg 0.940 N # mkg (Ans) (b) For the actual velocity profiles 1a1 2, a2 1.082, Eq. 1 gives p2 p1 V 21 V 22 loss wshaft a b a1 a2 r 2 2 net in 251 First Law of Thermodynamics—The Energy Equation (7) (Ans) COMMENT The difference in loss calculated assuming uniform velocity profiles and actual velocity profiles is not large compared to wshaft net in for this fluid flow situation. If we use Eqs. 3, 4, and 5 and the given pressure rise, Eq. 7 yields loss 84 N # mkg E XAMPLE 10.1 kPa211000 PakPa211 Nm2Pa2 1.23 kgm3 5.27 Energy—Effect of Nonuniform Velocity Profile GIVEN Consider the flow situation of Example 5.14. FIND Apply Eq. 5.87 to develop an expression for the fluid pressure drop that occurs between sections 112 and 122. By compar- ing the equation for pressure drop obtained presently with the result of Example 5.14, obtain an expression for loss between sections 112 and 122. SOLUTION Application of Eq. 5.87 to the flow of Example 5.14 1see Fig. E5.142 leads to Then, substituting Eq. 4 into Eq. 3, we obtain 0 1no shaft work2 p2 p1 a2w 22 a1w 21 gz2 gz1 loss wshaft r r 2 2 net in r8w 232p a2 # mw 22 a2 冮 0 R 16 R2 2 a2 (2) (3) Substituting the parabolic velocity profile equation into Eq. 3 we obtain r 12w1 2 3 31 1rR2 2 4 3 2pr dr 1rA2w2 2w 22 冮 R 0 3 1 31rR2 2 31rR2 4 1rR2 6 4 r dr (5) Now we combine Eqs. 2 and 5 to get p1 p2 r c 2.0w 22 1.0w 21 g1z2 z1 2 loss d 2 2 (6) However, from conservation of mass w2 w1 w so that Eq. 6 becomes p1 p2 rw 2 rg1z2 z1 2 r1loss2 2 (7) rg1z2 z1 2 (8) The term associated with change in elevation, rg1z2 z1 2, is equal to the weight per unit cross-sectional area, wA, of the water contained between sections 112 and 122 at any instant, w A Thus, combining Eqs. 7 and 8 we get From conservation of mass, since A1 A2 w1 w2 3 1 1rR2 2 4 3r dr or rw 32 dA2 A2 R rpR2 w 32 (1) Since the fluid velocity at section 112, w1, is uniformly distributed over cross-sectional area A1, the corresponding kinetic energy coefficient, a1, is equal to 1.0. The kinetic energy coefficient at section 122, a2, needs to be determined from the velocity profile distribution given in Example 5.14. Using Eq. 5.86 we get 冮 0 a2 Solving Eq. 1 for the pressure drop, p1 p2, we obtain a2w 22 a1w 21 g1z2 z1 2 loss d p1 p2 r c 2 2 冮 (4) p1 p2 rw 2 w r1loss2 2 A (9) 252 Chapter 5 ■ Finite Control Volume Analysis The pressure drop between sections 112 and 122 is due to: The change in kinetic energy between sections 112 and 122 associated with going from a uniform velocity profile to a parabolic velocity profile. 2. The weight of the water column, that is, hydrostatic pressure effect. 3. Viscous loss. Comparing Eq. 9 for pressure drop with the one obtained in Example 5.14 1i.e., the answer of Example 5.142, we obtain Rz rw 2 rw 2 w w r1loss2 2 A 3 A A (10) or loss Rz rA w2 6 (Ans) COMMENT We conclude that while some of the pipe wall friction force, Rz, resulted in loss of available energy, a portion of this friction, rAw 26, led to the velocity profile change. 5.3.5 Combination of the Energy Equation and the Moment-of-Momentum Equation3 If Eq. 5.82 is used for one-dimensional incompressible flow through a turbomachine, we can use Eq. 5.54, developed in Section 5.2.4 from the moment-of-momentum equation 1Eq. 5.422, to evaluate shaft work. This application of both Eqs. 5.54 and 5.82 allows us to ascertain the amount of loss that occurs in incompressible turbomachine flows as is demonstrated in Example 5.28. E XAMPLE 5.28 Energy—Fan Performance GIVEN Consider the fan of Example 5.19. ciency equation and a practical means for estimating lost shaft energy. FIND Show that only some of the shaft power into the air is converted into useful effects. Develop a meaningful effi- SOLUTION We use the same control volume used in Example 5.19. Application of Eq. 5.82 to the contents of this control volume yields p2 p1 V 22 V 21 gz2 gz1 wshaft loss r r 2 2 net in (1) useful effect wshaft loss net in a V 21 p1 p2 gz2 b a gz1 b r r 2 2 (2) (Ans) In other words, only a portion of the shaft work delivered to the air by the fan blades is used to increase the available energy of the air; the rest is lost because of fluid friction. A meaningful efficiency equation involves the ratio of shaft work converted into a useful effect 1Eq. 22 to shaft work into the air, wshaft net in. Thus, we can express efficiency, h, as wshaft h net in loss wshaft net in 3 wshaft U2Vu2 (4) net in As in Example 5.26, we can see with Eq. 1 that a “useful effect” in this fan can be defined as V 22 However, when Eq. 5.54, which was developed from the momentof-momentum equation 1Eq. 5.422, is applied to the contents of the control volume of Fig. E5.19, we obtain (3) Combining Eqs. 2, 3, and 4, we obtain h 5 3 1p2r2 1V 22 22 gz2 4 3 1 p1r2 1V 2122 gz1 4 6 U2Vu2 (5) (Ans) Equation 5 provides us with a practical means to evaluate the efficiency of the fan of Example 5.19. Combining Eqs. 2 and 4, we obtain loss U2Vu2 c a a p2 V 22 gz2 b r 2 p1 V 21 gz1 b d r 2 (6) (Ans) COMMENT Equation 6 provides us with a useful method of evaluating the loss due to fluid friction in the fan of Example 5.19 in terms of fluid mechanical variables that can be measured. This section may be omitted without loss of continuity in the text material. This section should not be considered without prior study of Sections 5.2.3 and 5.2.4. All of these sections are recommended for those interested in Chapter 12. 5.5 5.4 Chapter Summary and Study Guide 253 Second Law of Thermodynamics—Irreversible Flow This section is available on WileyPLUS. Note that this entire section may be omitted without loss of continuity in the text material. 5.5 Chapter Summary and Study Guide conservation of mass continuity equation mass flowrate linear momentum equation moment-ofmomentum equation shaft power shaft torque first law of thermodynamics heat transfer rate energy equation loss shaft work head head loss kinetic energy coefficient In this chapter the flow of a fluid is analyzed by using important principles including conservation of mass, Newton’s second law of motion, and the first law of thermodynamics as applied to control volumes. The Reynolds transport theorem is used to convert basic system-oriented laws into corresponding control volume formulations. The continuity equation, a statement of the fact that mass is conserved, is obtained in a form that can be applied to any flow—steady or unsteady, incompressible or compressible. Simplified forms of the continuity equation enable tracking of fluid everywhere in a control volume, where it enters, where it leaves, and within. Mass or volume flowrates of fluid entering or leaving a control volume and rate of accumulation or depletion of fluid within a control volume can be estimated. The linear momentum equation, a form of Newton’s second law of motion applicable to flow of fluid through a control volume, is obtained and used to solve flow problems. Net force results from or causes changes in linear momentum (velocity magnitude and/or direction) of fluid flowing through a control volume. Work and power associated with force can be involved. The moment-of-momentum equation, which involves the relationship between torque and changes in angular momentum, is obtained and used to solve flow problems dealing with turbines (energy extracted from a fluid) and pumps (energy supplied to a fluid). The steady-state energy equation, obtained from the first law of thermodynamics (conservation of energy), is written in several forms. The first (Eq. 5.69) involves power terms. The second form (Eq. 5.82 or 5.84) is termed the mechanical energy equation or the extended Bernoulli equation. It consists of the Bernoulli equation with extra terms that account for energy losses due to friction in the flow, as well as terms accounting for the work of pumps or turbines in the flow. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. select an appropriate control volume for a given problem and draw an accurately labeled control volume diagram. use the continuity equation and a control volume to solve problems involving mass or volume flowrate. use the linear momentum equation and a control volume, in conjunction with the continuity equation as necessary, to solve problems involving forces related to linear momentum change. use the moment-of-momentum equation to solve problems involving torque and related work and power due to angular momentum change. use the energy equation, in one of its appropriate forms, to solve problems involving losses due to friction (head loss) and energy input by pumps or extraction by turbines. use the kinetic energy coefficient in the energy equation to account for nonuniform flows. Some of the important equations in this chapter are: Conservation of mass Mass flowrate 0 0t 冮 cv r dV 冮 rV ⴢ nˆ dA 0 (5.5) cs # m rQ rAV (5.6) 254 Chapter 5 ■ Finite Control Volume Analysis V Average velocity Moving control volume mass conservation 冮 0 0t DMsys Dt Force related to change in linear momentum A 0 0t 冮 r dV cv 冮 rW ⴢ nˆ dA 0 0 0t 冮 r dV cv Vr dV 冮 VrV ⴢ nˆ dA a Fcontents of the 冮 WrW ⴢ nˆ dA a F 冮 Conservation of power Conservation of mechanical energy cv contents of the control volume cs VWU er dV Tshaft # # # Wshaft 1min 21 UinVuin 2 mout 1 UoutVuout 2 冮 cs auˇ (5.22) (5.29) (5.43) (5.45) axial # # Tshaft 1min 21 rinVuin 2 mout 1 routVuout 2 Shaft power related to change in moment-of-momentum (angular momentum) (5.17) control volume cs control volume 0 0t 冮 rW ⴢ nˆ dA 0 a B1r ⴛ F2 contents of the R Shaft torque related to change in moment-of-momentum (angular momentum) (5.16) cs Vector addition of absolute and relative velocities Shaft torque from force (5.9) cs cv Moving control volume force related to change in linear momentum First law of thermodynamics (Conservation of energy) (5.7) rA # # a mout a m in 0 Steady-flow mass conservation Deforming control volume mass conservation 冮 rV ⴢ nˆ dA (5.50) (5.53) # # p V2 gzb rV ⴢ nˆ dA Qnet Wshaft (5.64) r 2 in net in # # V 2out V 2in # m c hˇout hˇin g1zout zin 2 d Qnet Wshaft (5.69) 2 in net in pout V 2out pin V 2in gzout gzin wshaft loss r r 2 2 net in (5.82) References 1. Eck, B., Technische Stromungslehre, Springer-Verlag, Berlin, Germany, 1957. 2. Dean, R. C., “On the Necessity of Unsteady Flow in Fluid Machines,” ASME Journal of Basic Engineering 81D; 24–28, March 1959. 3. Moran, M. J., and Shapiro, H. N., Fundamentals of Engineering Thermodynamics, 6th Ed., Wiley, New York, 2008. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Problems 255 Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Conceptual Questions 5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its initial value as it flows through the pipe. The correct statement about the average velocity V is r1 r2 = r1/2 5.3C The nozzle on a fire hose is connected to the hose via a coupling. When the fire hose is in use with water flowing through it and the hose is stationary, the coupling is: a) in equilibrium, so there is no force on the coupling. b) in tension. c) in compression. Flow V2 = ? V1 a) V2 equals 2 V1. d) V2 equals V1/4. b) V2 equals V1/2. e) V2 equals 4 V1. c) V2 equals V1. 5.2C Water flows steadily into and out of the system with four pipes shown below. The mass flowrate through three of the pipes in kg/s is indicated. 3 kg/s 5.4C Two fluid jets are pointed at surfaces as shown in the following figures. The fluids are incompressible, and the effects of gravity can be neglected. The mass flowrates and the velocities of the jets are identical. The cross-sectional areas of the jets do not change significantly as the fluid flows. The correct statement regarding the horizontal forces is a) F1 equals 2 F2. b) F1 is greater than 0 and F2 equals 0. c) F1 equals F2/2. d) F1 equals 0 and F2 is greater than 0. e) F1 equals F2. Surface 6 kg/s System m• Nozzle F1 Fluid stream Surface 4 kg/s Nozzle # The mass flowrate m in the fourth pipe is a) 13 kg/s. b) 9 kg/s. d) 5 kg/s. e) 4kg/s. c) 6 kg/s. F2 Fluid stream Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the even-numbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 5.1.1 Derivation of the Continuity Equation 5.1 Use the Reynolds transport theorem (Eq. 4.19) with B volume and, therefore, b volume/mass 1/density to obtain the continuity equation for steady or unsteady incompressible flow through a fixed control volume: 冮 V ⴢ nˆ dA 0. cv 5.2 An incompressible fluid flows horizontally in the x–y plane with a velocity given by u 301y h2 12 ms, v 0 where y and h are in meters and h is a constant. Determine the average velocity for the portion of the flow between y 0 and y h. 256 Chapter 5 ■ Finite Control Volume Analysis Section 5.1.2 Fixed, Nondeforming Control Volume— Section (1) Uniform Velocity Profile or Average Velocity 5.3 Water flows steadily through the horizontal piping system shown in Fig. P5.3. The velocity is uniform at section (1), the mass flowrate is 10 slugs/s at section (2), and the velocity is nonuniform D at section (3). (a) Determine the value of the quantity rdV , Dt sys where the system is the water contained in the pipe bounded by sections (1), (2), and (3). (b) Determine the mean velocity at section (2). (c) Determine, if possible, the value of the integral Pump V2 Section (2) D2 冮冮 rV ⴢ nˆ dA over section (3). If it is not possible, explain what ■ Figure P5.6 additional information is needed to do so. 5.7 Water flows into a sink as shown in Video V5.1 and Fig. P5.7 at a rate of 2 gal/min. Determine the average velocity through each of the three 0.4-in.-diameter overflow holes if the drain is closed and the water level in the sink remains constant. 132 10 slugs/s Area = 0.3 ft2 (2) 15 ft/s Three 0.4-in.-diameter overflow holes Q = 2 gal/min (1) Area = 0.7 ft2 (3) Area = 0.7 ft2 ■ Figure P5.3 Drain 5.4 Water flows out through a set of thin, closely spaced blades as shown in Fig. P5.4 with a speed of V 10 fts around the entire circumference of the outlet. Determine the mass flowrate through the inlet pipe. ■ Figure P5.7 Inlet 0.08-ft diameter 0.1 ft 5.8 The wind blows through a 7 ft 10 ft garage door opening with a speed of 5 ft兾s as shown in Fig. P5.8. Determine the average speed, V, of the air through the two 3 ft 4 ft openings in the windows. Blades 0.6 ft V V 3 ft 3 ft 60° 16 ft 10 ft 5 ft/s V = 10 ft/s 20° ■ Figure P5.4 22 ft † 5.5 Estimate the rate (in gal/hr) that your car uses gasoline when it is being driven on an interstate highway. Determine how long it would take to empty a 12-oz soft-drink container at this flowrate. List all assumptions and show calculations. 5.6 The pump shown in Fig. P5.6 produces a steady flow of 10 gal/s through the nozzle. Determine the nozzle exit diameter, D2, if the exit velocity is to be V2 100 fts. ■ Figure P5.8 5.9 The human circulatory system consists of a complex branching pipe network ranging in diameter from the aorta (largest) to the capillaries (smallest). The average radii and the number of these vessels are shown in the table. Does the average blood velocity increase, decrease, or remain constant as it travels from the aorta to the capillaries? Problems Vessel Aorta Arteries Arterioles Capillaries Average Radius, mm Number 12.5 2.0 0.03 0.006 1 159 1.4 107 3.9 109 257 5.13 An evaporative cooling tower (see Fig. P5.13) is used to cool water from 110 to 80 °F. Water enters the tower at a rate of 250,000 lbm hr. Dry air (no water vapor) flows into the tower at a rate of 151,000 lbm hr. If the rate of wet airflow out of the tower is 156,900 lbm hr, determine the rate of water evaporation in lbm hr and the rate of cooled water flow in lbm hr. Wet air • m = 156,900 lbm/hr 5.10 Air flows steadily between two cross sections in a long, straight section of 0.1-m-inside-diameter pipe. The static temperature and pressure at each section are indicated in Fig. P5.10. If the average air velocity at section 112 is 205 m兾s, determine the average air velocity at section 122. Warm water • m = 250,000 lbm/hr D = 0.1 m Section (1) Section (2) p1 = 77 kPa (abs) T1 = 268 K V1 = 205 m/s p2 = 45 kPa (abs) T2 = 240 K ■ Figure P5.10 5.11 A hydraulic jump (see Video V10.11) is in place downstream from a spillway as indicated in Fig. P5.11. Upstream of the jump, the depth of the stream is 0.6 ft and the average stream velocity is 18 ft兾s. Just downstream of the jump, the average stream velocity is 3.4 ft兾s. Calculate the depth of the stream, h, just downstream of the jump. Cooled water Dry air • m = 151,000 lbm/hr ■ Figure P5.13 5.14 At cruise conditions, air flows into a jet engine at a steady rate of 65 lbm/s. Fuel enters the engine at a steady rate of 0.60 lbm/s. The average velocity of the exhaust gases is 1500 ft/s relative to the engine. If the engine exhaust effective cross-sectional area is 3.5 ft2, estimate the density of the exhaust gases in lbm/ft3. 5.15 Water at 0.1 m3/s and alcohol (SG 0.8) at 0.3 m3/s are mixed in a y-duct as shown in Fig. 5.15. What is the average density of the mixture of alcohol and water? Water and alcohol mix Water Q = 0.1 m3/s 0.6 ft h 3.4 ft/s 18 ft/s Alcohol (SG = 0.8) Q = 0.3 m3/s ■ Figure P5.11 ■ Figure P5.15 5.12 Water enters a rigid, sealed, cylindrical tank at a steady rate of 100 liters/hr and forces gasoline (SG 0.68) out as is indicated in Fig. P5.12. What is the time rate of change of mass of gasoline contained in the tank? 5.16 Oil having a specific gravity of 0.9 is pumped as illustrated in Fig. P5.16 with a water jet pump. The water volume flowrate is 1 m3/s. The water and oil mixture has an average specific gravity of 0.95. Calculate the rate, in m3/s, at which the pump moves oil. Section (1) Section (3) Gasoline (SG = 0.68) Water Q1 = 1 m3/s Water and oil mix (SG = 0.95) Section (2) Water Oil (SG = 0.9) ■ Figure P5.12 ■ Figure P5.16 258 Chapter 5 ■ Finite Control Volume Analysis 5.17 Fresh water flows steadily into an open 55-gal drum initially filled with seawater. The fresh water mixes thoroughly with the seawater, and the mixture overflows out of the drum. If the fresh water flowrate is 10 gal/min, estimate the time in seconds required to decrease the difference between the density of the mixture and the density of fresh water by 50%. 0.8 V 30 ft 70 ft 3 ft/s V 50 ft Section 5.1.2 Fixed, Nondeforming Control Volume— Nonuniform Velocity Profile 5.18 A water jet pump 1see Fig. P5.182 involves a jet crosssectional area of 0.01 m2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m2. Determine the pumping rate 1i.e., the entrained fluid flowrate2 involved in liters/s. 6 m/s Entrained water Depth = 3 ft 80 ft Depth = 5 ft 4 ft/s ■ Figure P5.20 5.21 Various types of attachments can be used with the shop vac shown in Video V5.2. Two such attachments are shown in Fig. P5.21—a nozzle and a brush. The flowrate is 1 ft3s. (a) Determine the average velocity through the nozzle entrance, Vn. (b) Assume the air enters the brush attachment in a radial direction all around the brush with a velocity profile that varies linearly from 0 to Vb along the length of the bristles as shown in the figure. Determine the value of Vb. 30 m/s jet Q = 1 ft3/s Entrained water Q = 1 ft3/s ■ Figure P5.18 5.19 To measure the mass flowrate of air through a 6-in.-insidediameter pipe, local velocity data are collected at different radii from the pipe axis (see Table). Determine the mass flowrate corresponding to the data listed in the following table. 1.5 in. 2-in. dia. Vn r (in.) Axial Velocity (ft/s) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 2.9 2.95 2.98 3.00 30 29.71 29.39 29.06 28.70 28.31 27.89 27.42 26.90 26.32 25.64 24.84 23.84 22.50 20.38 18.45 16.71 14.66 0 Vb 3-in. dia. ■ Figure P5.21 5.22 An appropriate turbulent pipe flow velocity profile is V uc a R r 1n ˆ b i R where uc centerline velocity, r local radius, R pipe radius, and ˆi unit vector along pipe centerline. Determine the ratio of average velocity, u, to centerline velocity, uc, for (a) n 4, (b) n 6, (c) n 8, (d) n 10. Compare the different velocity profiles. 5.23 GO As shown in Fig. P5.23, at the entrance to a 3-ft-wide channel the velocity distribution is uniform with a velocity V. Further downstream the velocity profile is given by u 4y 2y2, where u is in ft兾s and y is in ft. Determine the value of V. V 0.75 ft y 1 ft u = 4y – 2y2 x ■ Figure P5.23 5.20 Two rivers merge to form a larger river as shown in Fig. P5.20. At a location downstream from the junction 1before the two streams completely merge2, the nonuniform velocity profile is as shown and the depth is 6 ft. Determine the value of V. 5.24 An incompressible flow velocity field (water) is given as 1 1 V eˆ r eˆu ms r r Problems where r is in meters. (a) Calculate the mass flowrate through the cylindrical surface at r 1 m from z 0 to z 1 m as shown in Fig. P5.24a. (b) Show that mass is conserved in the annular control volume from r 1 m to r 2 m and z 0 to z 1 m as shown in Fig. P5.24b. z 259 5.27 Estimate the time required to fill with water a coneshaped container (see Fig. P5.27) 5 ft high and 5 ft across at the top if the filling rate is 20 galmin. 5 ft z 2m 1m 1m 1m 5 ft 1m ■ Figure P5.27 5.28 How long would it take to fill a cylindrical-shaped swimming pool having a diameter of 8 m to a depth of 1.5 m with water from a garden hose if the flowrate is 1.0 liter/s? (a) (b) Section 5.1.3 Moving, Nondeforming Control Volume ■ Figure P5.24 5.25 Flow of a viscous fluid over a flat plate surface results in the development of a region of reduced velocity adjacent to the wetted surface as depicted in Fig. P5.25. This region of reduced flow is called a boundary layer. At the leading edge of the plate, the velocity profile may be considered uniformly distributed with a value U. All along the outer edge of the boundary layer, the fluid velocity component parallel to the plate surface is also U. If the x-direction velocity profile at section 122 is y 1 7 u a b U d develop an expression for the volume flowrate through the edge of the boundary layer from the leading edge to a location downstream at x where the boundary layer thickness is d. †5.29 For an automobile moving along a highway, describe the control volume you would use to estimate the flowrate of air across the radiator. Explain how you would estimate the velocity of that air. Section 5.1.4 Deforming Control Volume 5.30 A hypodermic syringe (see Fig. P5.30) is used to apply a vaccine. If the plunger is moved forward at the steady rate of 20 mm/s and if vaccine leaks past the plunger at 0.1 of the volume flowrate out the needle opening, calculate the average velocity of the needle exit flow. The inside diameters of the syringe and the needle are 20 mm and 0.7 mm. Qleak Qout ■ Figure P5.30 Section (2) U Section (1) Outer edge of boundary layer U δ x ■ Figure P5.25 Section 5.1.2 Fixed, Nondeforming Control Volume— Unsteady Flow 5.26 Air at standard conditions enters the compressor shown in Fig. P5.26 at a rate of 10 ft3s. It leaves the tank through a 1.2-in.diameter pipe with a density of 0.0035 slugsft3 and a uniform speed of 700 fts. (a) Determine the rate 1slugss2 at which the mass of air in the tank is increasing or decreasing. (b) Determine the average time rate of change of air density within the tank. Compressor Tank volume = 20 ft3 10 ft3/s 0.00238 slugs/ft3 ■ Figure P5.26 1.2 in. 700 ft/s 0.0035 slugs/ft3 5.31 The Hoover Dam (see Video V2.4) backs up the Colorado River and creates Lake Mead, which is approximately 115 miles long and has a surface area of approximately 225 square miles. If during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and the outflow from the dam is 8000 cfs, how many feet per 24-hour day will the lake level rise? 5.32 Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 1500 ft2. What capacity 1gal兾min2 pump would you rent to (a) keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, and (b) reduce the water accumulation in your basement at a rate of 3 in.兾hr even while the backup problem exists? 5.33 (See Fluids in the News article “New 1.6-gpf standards,” Section 5.1.2.) When a toilet is flushed, the water depth, h, in the tank as a function of time, t, is as given in the table. The size of the rectangular tank is 19 in. by 7.5 in. (a) Determine the volume of water used per flush, gpf. (b) Plot the flowrate for 0 t 6 s. t (s) h (in.) 0 0.5 1.0 2.0 3.0 4.0 5.0 6.0 5.70 5.33 4.80 3.45 2.40 1.50 0.75 0 260 Chapter 5 ■ Finite Control Volume Analysis Section 5.2.1 Derivation of the Linear Momentum Equation 5.34 A fluid flows steadily in the x direction through a control volume. Measurements indicate that to cause this flow the force acting on the contents of the control volume is 120 N in the negative x direction. Determine the net rate of flow of linear momentum through the control surface. 5.35 Consider the unsteady flow of a fluid in the x direction through a control volume. The linear momentum of the fluid within the control volume is a function of time given by 200 t ˆi slug ⴢ fts, where t is in seconds and ˆi is a unit vector in the x direction. Measurements indicate that to cause this flow the force acting on the contents of the control volume is 40 ˆi lb. Determine the net rate of flow of linear momentum through the control surface. D1 = 0.6 in. Section (1) 1.2 in. Section (2) D2 = 0.2 in. Q = 10 gal/min ■ Figure P5.38 Section 5.2.2 Application of the Linear Momentum Equation (also see Lab Problems 5.1LP, 5.2LP, 5.3LP, and 5.4LP) 5.36 A 10-mm-diameter jet of water is deflected by a homogeneous rectangular block (15 mm by 200 mm by 100 mm) that weighs 6 N as shown in Video V5.6 and Fig. P5.36. Determine the minimum volume flowrate needed to tip the block. 5.39 Water flows through a horizontal, 180 pipe bend as is illustrated in Fig. P5.39. The flow cross-sectional area is constant at a value of 9000 mm2. The flow velocity everywhere in the bend is 15 m/s. The pressures at the entrance and exit of the bend are 210 and 165 kPa, respectively. Calculate the horizontal (x and y) components of the anchoring force needed to hold the bend in place. z 0.015 m 0.010 m Q y 0.10 m 0.050 m x ■ Figure P5.39 ■ Figure P5.36 †5.37 When a baseball player catches a ball, the force of the ball on her glove is as shown as a function of time in Fig. P5.37. Describe how this situation is similar to the force generated by the deflection of a jet of water by a vane. Note: Consider many baseballs being caught in quick succession. 5.40 Water flows through a horizontal bend and discharges into the atmosphere as shown in Fig. P5.40. When the pressure gage reads 10 psi, the resultant x-direction anchoring force, FAx, in the horizontal plane required to hold the bend in place is shown on the figure. Determine the flowrate through the bend and the y-direction anchoring force, FAy, required to hold the bend in place. The flow is not frictionless. F 10 psi Bend Q=? Water FAx = 1440 lb Area = 0.2 ft2 t FAy = ? Area = 0.1 ft2 ■ Figure P5.37 45° 5.38 Determine the anchoring force required to hold in place the conical nozzle attached to the end of the laboratory sink faucet shown in Fig. P5.38 when the water flowrate is 10 gal/min. The nozzle weight is 0.2 lb. The nozzle inlet and exit inside diameters are 0.6 and 0.2 in., respectively. The nozzle axis is vertical, and the axial distance between sections (1) and (2) is 1.2 in. The pressure at section (1) is 68 psi. ■ Figure P5.40 5.41 A free jet of fluid strikes a wedge as shown in Fig. P5.41. Of the total flow, a portion is deflected 30; the remainder is not deflected. The horizontal and vertical components of force needed Problems to hold the wedge stationary are FH and FV, respectively. Gravity is negligible, and the fluid speed remains constant. Determine the force ratio, FH/FV. D = 12 in. Section (1) FH Free jet FV V ■ Figure P5.41 5.42 Water enters the horizontal, circular cross-sectional, sudden contraction nozzle sketched in Fig. P5.42 at section (1) with a uniformly distributed velocity of 25 ft/s and a pressure of 75 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determine the axial component of the anchoring force required to hold the contraction in place. Section (2) p1 = 690 kPa (abs) T1 = 300 K V q = 30° V 261 p2 = 127 kPa (abs) T2 = 252 K V2 = 320 m/s ■ Figure P5.45 5.46 Water flows steadily from a tank mounted on a cart as shown in Fig. 5.46. After the water jet leaves the nozzle of the tank, it falls and strikes a vane attached to another cart. The cart’s wheels are frictionless, and the fluid is inviscid. (a) Determine the speed of the water leaving the tank, V1, and the water speed leaving the cart, V2. (b) Determine the tension in rope A. (c) Determine the tension in rope B. Nozzle area = 0.01 m2 2m Horizontal free jets V1 4m V2 Section (2) p2 = D1 = 3 in. p1 = 75 psi V1 = 25 ft/s 0 psi V2 = 100 ft/s Rope A Rope B ■ Figure P5.46 Section (1) ■ Figure P5.42 †5.43 A truck carrying chickens is too heavy for a bridge that it needs to cross. The empty truck is within the weight limits; with the chickens it is overweight. It is suggested that if one could get the chickens to fly around the truck (i.e., by banging on the truck side) it would be safe to cross the bridge. Do you agree? Explain. 5.44 Exhaust (assumed to have the properties of standard air) leaves the 4-ft-diameter chimney shown in Video V5.4 and Fig. P5.44 with a speed of 6 ft/s. Because of the wind, after a few diameters downstream the exhaust flows in a horizontal direction with the speed of the wind, 15 ft/s. Determine the horizontal component of the force that the blowing wind exerts on the exhaust gases. 5.47 Determine the magnitude and direction of the anchoring force needed to hold the horizontal elbow and nozzle combination shown in Fig. P5.47 in place. Atmospheric pressure is 100 kPa(abs). The gage pressure at section (1) is 100 kPa. At section (2), the water exits to the atmosphere. 160 mm V2 y Section (2) x 300 mm V1 Water Section (1) 15 ft/s p1 = 100 kPa V1 = 2 m/s 15 ft/s ■ Figure P5.47 6 ft/s 4 ft 5.48 Water is added to the tank shown in Fig. P5.48 through a vertical pipe to maintain a constant (water) level. The tank is placed Constant water level ■ Figure P5.44 5.45 Air flows steadily between two cross sections in a long, straight section on 12-in.-inside-diameter pipe. The static temperature and pressure at each section are indicated in Fig P5.45. If the average air velocity at section (2) is 320 m/s, determine the average air velocity at section (1). Determine the frictional force exerted by the pipe wall on the air flowing between sections (1) and (2). Assume uniform velocity distributions at each section. Jet area = 1250 mm2 1m 1m F Frictionless surface ■ Figure P5.48 Jet area = 625 mm2 262 Chapter 5 ■ Finite Control Volume Analysis intake air velocity 700 fts exhaust gas velocity 1640 ft s intake cross section area 10 ft2 intake static pressure 11.4 psia intake static temperature 480 °R exhaust gas pressure 0 psi on a horizontal plane which has a frictionless surface. Determine the horizontal force, F, required to hold the tank stationary. Neglect all losses. 5.49 GO Water flows as two free jets from the tee attached to the pipe shown in Fig. P5.49. The exit speed is 15 m/s. If viscous effects and gravity are negligible, determine the x and y components of the force that the pipe exerts on the tee. estimate a nominal thrust to design for. V = 15 m/s Area = 0.3 m2 Area = 1 m2 y x V = 15 m/s 5.53 A vertical jet of water leaves a nozzle at a speed of 10 m/s and a diameter of 20 mm. It suspends a plate having a mass of 1.5 kg as indicated in Fig. P5.53. What is the vertical distance h? Area = 0.5 m2 Pipe Tee ■ Figure P5.49 h 5.50 A nozzle is attached to a vertical pipe and discharges water into the atmosphere as shown in Fig. P5.50. When the discharge is 0.1 m3/s, the gage pressure at the flange is 40 kPa. Determine the vertical component of the anchoring force required to hold the nozzle in place. The nozzle has a weight of 200 N, and the volume of water in the nozzle is 0.012 m3. Is the anchoring force directed upward or downward? 30° Area = 0.01 m2 ■ Figure P5.53 5.54 A horizontal, circular cross-sectional jet of air having a diameter of 6 in. strikes a conical deflector as shown in Fig. P5.54. A horizontal anchoring force of 5 lb is required to hold the cone in place. Estimate the nozzle flowrate in ft3/s. The magnitude of the velocity of the air remains constant. Nozzle g 60° 6 in. Area = 0.02 m2 p = 40 kPa FA = 5 lb ■ Figure P5.54 0.10 m3/s ■ Figure P5.50 5.51 The hydraulic dredge shown in Fig. P5.51 is used to dredge sand from a river bottom. Estimate the thrust needed from the propeller to hold the boat stationary. Assume the specific gravity of the sand/water mixture is SG 1.4. 2-ft diameter 5.55 A vertical, circular cross-sectional jet of air strikes a conical deflector as indicated in Fig. P5.55. A vertical anchoring force of 0.1 N is required to hold the deflector in place. Determine the mass (kg) of the deflector. The magnitude of velocity of the air remains constant. FA = 0.1 N 30 ft/s 30° 60° 9 ft 7 ft Prop 0.1 m ■ Figure P5.51 5.52 A static thrust stand is to be designed for testing a specific jet engine, knowing the following conditions for a typical test. V = 30 m/s ■ Figure P5.55 263 Problems 5.56 A vertical jet of water having a nozzle exit velocity of 15 ft/s with a diameter of 1 in. suspends a hollow hemisphere as indicated in Fig. P5.56. If the hemisphere is stationary at an elevation of 12 in., determine its weight. 5.59 Two water jets of equal size and speed strike each other as shown in Fig. P5.59. Determine the speed, V, and direction, u, of the resulting combined jet. Gravity is negligible. V θ 12 in. V2 = 10 ft /s 0.1 ft 90° 0.1 ft ■ Figure P5.56 5.57 Air flows into the atmosphere from a nozzle and strikes a vertical plate as shown in Fig. P5.57. A horizontal force of 12 N is required to hold the plate in place. Determine the reading on the pressure gage. Assume the flow to be incompressible and frictionless. V1 =10 ft /s ■ Figure P5.59 5.60 GO Assuming frictionless, incompressible, one-dimensional flow of water through the horizontal tee connection sketched in Fig. P5.60, estimate values of the x and y components of the force exerted by the tee on the water. Each pipe has an inside diameter of 1 m. p=? z Section (3) Section (2) 12N Q3 = 10 m3/s y 2 Area = 0.003 m x Area = 0.01 m2 Section (1) V1 = 6 m/s p1 = 200 kPa ■ Figure P5.57 ■ Figure P5.60 5.58 Water flows from a large tank into a dish as shown in Fig. P5.58. (a) If at the instant shown the tank and the water in it weigh W1 lb, what is the tension, T1, in the cable supporting the tank? (b) If at the instant shown the dish and the water in it weigh W2 lb, what is the force, F2, needed to support the dish? 5.61 Water discharges into the atmosphere through the device shown in Fig. P5.61. Determine the x component of force at the flange required to hold the device in place. Neglect the effect of gravity and friction. A = 0.4 ft2 T1 30 ft/s 10 psi Flange Tank 20 ft/s 10 ft A = 0.8 ft2 43° A = 0.8 ft2 V ■ Figure P5.61 0.1-ft diameter 12 ft 2 ft F2 ■ Figure P5.58 Dish 5.62 Determine the magnitude of the horizontal component of the anchoring force required to hold in place the sluice gate shown in Fig. 5.62. Compare this result with the size of the horizontal component of the anchoring force required to hold in place the sluice gate when it is closed and the depth of water upstream is 10 ft. 264 Chapter 5 ■ Finite Control Volume Analysis hold the plate stationary; (b) the fraction of mass flow along the plate surface in each of the two directions shown; (c) the magnitude of FA, the anchoring force required to allow the plate to move to the right at a constant speed of 10 m/s. 10 ft 1.5 ft 5.66 Air discharges from a 2-in.-diameter nozzle and strikes a curved vane, which is in a vertical plane as shown in Fig. P5.66. A stagnation tube connected to a water U-tube manometer is located in the free air jet. Determine the horizontal component of the force that the air jet exerts on the vane. Neglect the weight of the air and all friction. 4 ft/s ■ Figure P5.62 5.63 Water flows steadily into and out of a tank that sits on frictionless wheels as shown in Fig. P5.63. Determine the diameter D so that the tank remains motionless if F 0. Stagnation tube Air 2-in. dia. Open d Fixed vane Water Free air jet 7 in. 30° F ■ Figure P5.66 D 5.67 Water is sprayed radially outward over 180 as indicated in Fig. P5.67. The jet sheet is in the horizontal plane. If the jet velocity at the nozzle exit is 20 ft/s, determine the direction and magnitude of the resultant horizontal anchoring force required to hold the nozzle in place. d ■ Figure P5.63 5.64 The rocket shown in Fig. P5.64, is held stationary by the horizontal force, Fx, and the vertical force, Fz. The velocity and pressure of the exhaust gas are 5000 ft/s and 20 psia at the nozzle exit, which has a cross section area of 60 in.2. The exhaust mass flowrate is constant at 21 lbm/s. Determine the value of the restraining force Fx. Assume the exhaust flow is essentially horizontal. 8 in. 0.5 in. V= 20 ft/s ■ Figure P5.67 Fx 5.68 A sheet of water of uniform thickness (h 0.01 m) flows from the device shown in Fig. P5.68. The water enters vertically through the inlet pipe and exits horizontally with a speed that varies linearly from 0 to 10 m/s along the 0.2-m length of the slit. Determine the y component of anchoring force necessary to hold this device stationary. Fz ■ Figure P5.64 Q 5.65 GO A horizontal circular jet of air strikes a stationary flat plate as indicated in Fig. P5.65. The jet velocity is 40 m/s and the jet diameter is 30 mm. If the air velocity magnitude remains constant as the air flows over the plate surface in the directions shown, determine: (a) the magnitude of FA, the anchoring force required to V2 Dj = 30 mm x h = 0.01 m 0.2 m 90° Vj = 40 m/s 10 m/s 30° y 0 m/s V3 ■ Figure P5.65 FA ■ Figure P5.68 Problems 5.69 The results of a wind tunnel test to determine the drag on a body (see Fig. P5.69) are summarized below. The upstream [section (1)] velocity is uniform at 100 ft/s. The static pressures are given by p1 p2 14.7 psia. The downstream velocity distribution, which is symmetrical about the centerline, is given by 冟 y冟冟 y冟 3 ft u 100 30 a1 b 3 u 100 冟y冟 7 3 ft where u is the velocity in ft/s and y is the distance on either side of the centerline in feet (see Fig. P5.69). Assume that the body shape does not change in the direction normal to the paper. Calculate the drag force (reaction force in x direction) exerted on the air by the body per unit length normal to the plane of the sketch. V2 = 100 ft/s u 3 ft r 2 u uc c 1 a b d R as is illustrated in Fig. P5.72. Compare the axial direction momentum flowrate calculated with the average velocity, u, with the axial direction momentum flowrate calculated with the nonuniform velocity distribution taken into account. r R u uc ■ Figure P5.72 5.74 A Pelton wheel vane directs a horizontal, circular cross-sectional jet of water symmetrically as indicated in Fig. P5.74 and Video V5.6. The jet leaves the nozzle with a velocity of 100 ft/s. Determine the x-direction component of anchoring force required to (a) hold the vane stationary, (b) confine the speed of the vane to a value of 10 ft/s to the right. The fluid speed magnitude remains constant along the vane surface. 3 ft Section (1) 5.72 In a laminar pipe flow that is fully developed, the axial velocity profile is parabolic. That is, †5.73 Water from a garden hose is sprayed against your car to rinse dirt from it. Estimate the force that the water exerts on the car. List all assumptions and show calculations. V1 = 100 ft/s Body 265 Section (2) ■ Figure P5.69 45° 5.70 A variable mesh screen produces a linear and axisymmetric velocity profile as indicated in Fig. P5.70 in the airflow through a 2-ft-diameter circular cross-sectional duct. The static pressures upstream and downstream of the screen are 0.2 and 0.15 psi and are uniformly distributed over the flow cross-sectional area. Neglecting the force exerted by the duct wall on the flowing air, calculate the screen drag force. 45° 100 ft/s y 100 ft/s 10 ft/s x D = 1 in. D = 1 in. 45° 45° Variable mesh screen (a) (b) ■ Figure P5.74 D = 2 ft Section (1) Section (2) p1 = 0.2 psi V1 = 100 ft/s p2 = 0.15 psi 5.75 The thrust developed to propel the jet ski shown in Video V9.18 and Fig. P5.75 is a result of water pumped through the vehicle and exiting as a high-speed water jet. For the conditions shown in the figure, what flowrate is needed to produce a 300-lb thrust? Assume the inlet and outlet jets of water are free jets. ■ Figure P5.70 5.71 Consider unsteady flow in the constant diameter, horizontal pipe shown in Fig. P5.71. The velocity is uniform throughout the entire pipe, but it is a function of time: V u1t2 ˆi. Use the x component of the unsteady momentum equation to determine the pressure difference p1 p2. Discuss how this result is related to Fx max. 3.5-in.-diameter outlet jet 30° 25-in.2 inlet area u (t) ᐉ ■ Figure P5.75 D x ρ = density (1) ■ Figure P5.71 (2) 5.76 Thrust vector control is a technique that can be used to greatly improve the maneuverability of military fighter aircraft. It consists of using a set of vanes in the exit of a jet engine to deflect the exhaust gases as shown in Fig. P5.76. (a) Determine the pitching moment (the moment tending to rotate the nose of the aircraft 266 Chapter 5 ■ Finite Control Volume Analysis pout = 0 Vout = 1500 ft/s Vin = 300 ft/s pin = 0 Vane θ = 8 deg the flowrate and the velocity of the exiting jet if the thrust is to be 300 lb. Neglect the momentum of the water entering the pump. cg 20 ft m• in = m• out = 16 slugs/s ■ Figure P5.76 up) about the aircraft’s mass center (cg) for the conditions indicated in the figure. (b) By how much is the thrust (force along the centerline of the aircraft) reduced for the case indicated compared to normal flight when the exhaust is parallel to the centerline? 5.77 The exhaust gas from the rocket shown in Fig. P5.77a leaves the nozzle with a uniform velocity parallel to the x axis. The gas is assumed to be discharged from the nozzle as a free jet. (a) Show that the thrust that is developed is equal to rAV 2, where A ⫽ pD2/4. (b) The exhaust gas from the rocket nozzle shown in Fig. P5.77b is also uniform, but rather than being directed along the x axis, it is directed along rays from point 0 as indicated. Determine the thrust for this rocket. m• ■ Figure P5.79 5.80 (See Fluids in the News article titled “Bow Thrusters,” Section 5.2.2.) The bow thruster on the boat shown in Fig. P5.80 is used to turn the boat. The thruster produces a 1-m-diameter jet of water with a velocity of 10 m/s. Determine the force produced by the thruster. Assume that the inlet and outlet pressures are zero and that the momentum of the water entering the thruster is negligible. D V V = 10 m/s (a) D=1m O m• q D ■ Figure P5.80 V (b) ■ Figure P5.77 5.78 (See Fluids in the News article titled “Where the Plume goes,” Section 5.2.2.) Air flows into the jet engine shown in Fig. P5.78 at a rate of 9 slugs/s and a speed of 300 ft/s. Upon landing, the engine exhaust exits through the reverse thrust mechanism with a speed of 900 ft/s in the direction indicated. Determine the reverse thrust applied by the engine to the airplane. Assume the inlet and exit pressures are atmospheric and that the mass flowrate of fuel is negligible compared to the air flowrate through the engine. 5.81 Water flows from a two-dimensional open channel and is diverted by an inclined plate as illustrated in Fig. P5.81. When the velocity at section (1) is 10 ft/s, what horizontal force (per unit width) is required to hold the plate in position? At section (1) the pressure distribution is hydrostatic, and the fluid acts as a free jet at section (2). Neglect friction. 20° Section (1) Plate V3 = 900 ft/s 10 ft/s 4 ft (3) 4-ft diameter (1) V1 = 300 ft/s Section (2) 30° 1.0 ft (2) V2 = 900 ft/s ■ Figure P5.78 5.79 (See Fluids in the News article titled “Motorized Surfboard,” Section 5.2.2.) The thrust to propel the powered surfboard shown in Fig. P5.79 is a result of water pumped through the board that exits as a high-speed 2.75-in.-diameter jet. Determine ■ Figure P5.81 †5.82 If a valve in a pipe is suddenly closed, a large pressure surge may develop. For example, when the electrically operated shutoff valve in a dishwasher closes quickly, the pipes supplying the dishwasher may rattle or “bang” because of this large pressure pulse. Explain the physical mechanism for this “water hammer” phenomenon. How could this phenomenon be analyzed? Problems 5.83 A snowplow mounted on a truck clears a path 12 ft through heavy wet snow, as shown in Figure P5.83. The snow is 8 in. deep and its density is 10 lbm/ft3. The truck travels at 30 mph. The snow is discharged from the plow at an angle of 45⬚ from the direction of travel and 45⬚ above the horizontal, as shown in Figure P5.83. Estimate the force required to push the plow. 267 Nozzle exit area = 0.04 in.2 r = 8 in. U = 30 mph θ = 45° (in plane of blade) Q = 16 gal/min d = 8 in. ■ Figure P5.87 ■ Figure P5.83 Section 5.2.3 Derivation of the Moment-of-Momentum Equation 5.84 Describe a few examples (include photographs/images) of turbines where the force/torque of a flowing fluid leads to rotation of a shaft. 5.88 GO Five liters/s of water enter the rotor shown in Video V5.10 and Fig. P5.88 along the axis of rotation. The cross-sectional area of each of the three nozzle exits normal to the relative velocity is 18 mm2. How large is the resisting torque required to hold the rotor stationary? How fast will the rotor spin steadily if the resisting torque is reduced to zero and (a) u ⫽ 0⬚, (b) u ⫽ 30⬚, (c) u ⫽ 60⬚? 5.85 Describe a few examples (include photographs/images) of pumps where a fluid is forced to move by “blades” mounted on a rotating shaft. 5.86 An incompressible fluid flows outward through a blower as indicated in Fig. P5.86. The shaft torque involved, Tshaft, is estimated with the following relationship: # Tshaft ⫽ mr2Vu2 # where m ⫽ mass flowrate through the blower, r2 ⫽ outer radius of blower, and Vu2 ⫽ tangential component of absolute fluid velocity leaving the blower. State the flow conditions that make this formula valid. Nozzle exit area normal to relative velocity = 18 mm2 r = 0.5m θ Q = 5 liters/s ■ Figure P5.88 5.89 (See Fluids in the News article titled “Tailless Helicopters,” Section 5.2.4.) Shown in Fig. P5.89 is a toy “helicopter” powered V2 Vθ2 r2 r1 ω ω ■ Figure P5.86 Section 5.2.4 Application of the Moment-of-Momentum Equation 5.87 Water enters a rotating lawn sprinkler through its base at the steady rate of 16 gal/min as shown in Fig. P5.87. The exit cross-sectional area of each of the two nozzles is 0.04 in.2, and the flow leaving each nozzle is tangential. The radius from the axis of rotation to the centerline of each nozzle is 8 in. (a) Determine the resisting torque required to hold the sprinkler head stationary. (b) Determine the resisting torque associated with the sprinkler rotating with a constant speed of 500 rev/min. (c) Determine the angular velocity of the sprinkler if no resisting torque is applied. Balloon ■ Figure P5.89 268 Chapter 5 ■ Finite Control Volume Analysis Q = 230 ft3/min by air escaping from a balloon. The air from the balloon flows radially through each of the three propeller blades and out through small nozzles at the tips of the blades. Explain physically how this flow can cause the rotation necessary to rotate the blades to produce the needed lifting force. 30° 5.90 A water turbine wheel rotates at the rate of 50 rpm in the direction shown in Fig. P5.90. The inner radius, r2, of the blade row is 2 ft, and the outer radius, r1, is 4 ft. The absolute velocity vector at the turbine rotor entrance makes an angle of 20 with the tangential direction. The inlet blade angle is 60 relative to the tangential direction. The blade outlet angle is 120. The flowrate is 20 ft3/s. For the flow tangent to the rotor blade surface at inlet and outlet, determine an appropriate constant blade height, b, and the corresponding power available at the rotor shaft. V1 12 in. 5 in. 1725 rpm b 1 in. 20° ■ Figure P5.92 V1 W1 60° 50 rpm 120° measured with respect to the tangential direction at the outside diameter of the rotor. (a) What would be a reasonable blade inlet angle (measured with respect to the tangential direction at the inside diameter of the rotor)? (b) Find the power required to run the fan. r1 = 4 ft r2 = 5.93 The radial component of velocity of water leaving the centrifugal pump sketched in Fig. P5.93 is 30 ft/s. The magnitude of the absolute velocity at the pump exit is 60 ft/s. The fluid enters the pump rotor radially. Calculate the shaft work required per unit mass flowing through the pump. 2 ft W2 Section (1) Section (2) VR2 = 30 ft/s Q = 20 ft3/s ■ Figure P5.90 0.5 ft V2 = 60 ft/s V1 5.91 A water turbine with radial flow has the dimensions shown in Fig. P5.91. The absolute entering velocity is 50 ft/s, and it makes an angle of 30 with the tangent to the rotor. The absolute exit velocity is directed radially inward. The angular speed of the rotor is 120 rpm. Find the power delivered to the shaft of the turbine. 0.2 ft 2000 rpm V1 = 50 ft/s 30° 1 ft ■ Figure P5.93 r1 = 2 ft 5.94 An axial flow turbomachine rotor involves the upstream (1) and downstream (2) velocity triangles shown in Fig. P5.94. Is this turbomachine a turbine or a fan? Sketch an appropriate blade section and determine energy transferred per unit mass of fluid. V2 r2 = 1 ft 120 rpm W1 = W2 W1 Section (1) W2 U1 = 30 ft/s U2 = 30 ft/s Section (2) ■ Figure P5.91 1 V1 = 20 ft/s 60° ■ Figure P5.94 5.92 A fan (see Fig. P5.92) has a bladed rotor of 12-in. outside diameter and 5-in. inside diameter and runs at 1725 rpm. The width of each rotor blade is 1 in. from blade inlet to outlet. The volume flowrate is steady at 230 ft3/min, and the absolute velocity of the air at blade inlet, V1, is purely radial. The blade discharge angle is 30 5.95 An inward flow radial turbine (see Fig. P5.95) involves a nozzle angle, 1, of 60 and an inlet rotor tip speed, U1, of 6 m/s. The ratio of rotor inlet to outlet diameters is 1.8. The absolute velocity leaving the rotor at section (2) is radial with a magnitude of 269 Problems Section 5.3.2 Application of the Energy Equation—No Shaft Work and Section 5.3.3 Comparison of the Energy Equation with the Bernoulli Equation Section (1) α1 5.100 A 100-ft-wide river with a flowrate of 2400 ft3/s flows over a rock pile as shown in Fig. P5.100. Determine the direction of flow and the head loss associated with the flow across the rock pile. r1 r2 V2 = 12 m/s (1) Section (2) (2) 4 ft 2 ft Rock pile ■ Figure P5.100 ■ Figure P5.95 12 m/s. Determine the energy transfer per unit mass of fluid flowing through this turbine if the fluid is (a) air, (b) water. 5.96 A sketch of the arithmetic mean radius blade sections of an axial-flow water turbine stage is shown in Fig. P5.96. The rotor speed is 1000 rpm. (a) Sketch and label velocity triangles for the flow entering and leaving the rotor row. Use V for absolute velocity, W for relative velocity, and U for blade velocity. Assume flow enters and leaves each blade row at the blade angles shown. (b) Calculate the work per unit mass delivered at the shaft. 5.101 A horizontal Venturi flow meter consists of a converging–diverging conduit as indicated in Fig. P5.101. The diameters of cross sections (1) and (2) are 6 and 4 in. The velocity and static pressure are uniformly distributed at cross sections (1) and (2). Determine the volume flowrate (ft3/s) through the meter if p1 p2 3 psi, the flowing fluid is oil (p 56 lbm/ft3), and the loss per unit mass from (1) to (2) is negligibly small. D1 = 6 in. Section (1) D2 = 4 in. rm = 6 in. Section (2) 1000 rpm Rotor Stator U 70° U 45° Blade sections at the arithmetic mean radius ■ Figure P5.101 5.102 Oil (SG 0.9) flows downward through a vertical pipe contraction as shown in Fig. P5.102. If the mercury manometer reading, h, is 100 mm, determine the volume flowrate for frictionless flow. Is the actual flowrate more or less than the frictionless value? Explain. 45° 300 mm ■ Figure P5.96 5.97 By using velocity triangles for flow upstream (1) and downstream (2) of a turbomachine rotor, prove that the shaft work in per unit mass flowing through the rotor is wshaft net in V22 V21 U22 U21 W21 W22 2 0.6 m where V absolute flow velocity magnitude, W relative flow velocity magnitude, and U blade speed. h Section 5.3.1 Derivation of the Energy Equation 5.98 Distinguish between shaft work and other kinds of work associated with a flowing fluid. ■ Figure P5.102 5.99 An incompressible fluid flows along a 0.20-m-diameter pipe with a uniform velocity of 3 m/s. If the pressure drop between the upstream and downstream sections of the pipe is 20 kPa, determine the power transferred to the fluid due to fluid normal stresses. 5.103 An incompressible liquid flows steadily along the pipe shown in Fig. P5.103. Determine the direction of flow and the head loss over the 6-m length of pipe. 100 mm 270 Chapter 5 ■ Finite Control Volume Analysis 0.75 m 1.0 m 3m 12 in. 6m 1.5 m ■ Figure P5.103 24 in. 5.104 A siphon is used to draw water at 70 F from a large container as indicated in Fig. P5.104. The inside diameter of the siphon line is 1 in. and the pipe centerline rises 3 ft above the essentially constant water level in the tank. Show that by varying the length of the siphon below the water level, h, the rate of flow through the siphon can be changed. Assuming frictionless flow, determine the maximum flowrate possible through the siphon. The limiting condition is the occurrence of cavitation in the siphon. Will the actual maximum flow be more or less than the frictionless value? Explain. ■ Figure P5.106 5.107 A gas expands through a nozzle from a pressure of 300 psia to a pressure of 5 psia. The enthalpy change involved, hˇ1 hˇ2, is 150 Btu/lbm. If the expansion is adiabatic but with frictional effects and the inlet gas speed is negligibly small, determine the exit gas velocity. 5.108 For the 180 elbow and nozzle flow shown in Fig. P5.108, determine the loss in available energy from section (1) to section (2). How much additional available energy is lost from section (2) to where the water comes to rest? 3 ft 6 in. y Section (2) h x 12 in. p1 = 15 psi V1 = 5 ft/s 1 in. Section (1) ■ Figure P5.108 ■ Figure P5.104 5.105 A water siphon having a constant inside diameter of 3 in. is arranged as shown in Fig. P5.105. If the friction loss between A and B is 0.8V 2/2, where V is the velocity of flow in the siphon, determine the flowrate involved. 4 ft A 12 ft 4 ft 5.109 An automobile engine will work best when the back pressure at the interface of the exhaust manifold and the engine block is minimized. Show how reduction of losses in the exhaust manifold, piping, and muffler will also reduce the back pressure. How could losses in the exhaust system be reduced? What primarily limits the minimization of exhaust system losses? 5.110 (See Fluids in the News article titled “Smart Shocks,” Section 5.3.3.) A 200-lb force applied to the end of the piston of the shock absorber shown in Fig. P5.110 causes the two ends of the shock absorber to move toward each other with a speed of 5 ft/s. Determine the head loss associated with the flow of the oil through the channel. Neglect gravity and any friction force between the piston and cylinder walls. 3 in. B Gas Oil ■ Figure P5.105 5.106 Water flows through a valve (see Fig. P5.106) at the rate of 1000 lbm/s. The pressure just upstream of the valve is 90 psi and the pressure drop across the valve is 50 psi. The inside diameters of the valve inlet and exit pipes are 12 and 24 in. If the flow through the valve occurs in a horizontal plane, determine the loss in available energy across the valve. p=0 Piston Channel 1-in. diameter 200 lb ■ Figure P5.110 Problems Section 5.3.2 Application of the Energy Equation— With Shaft Work †5.111 Based on flowrate and pressure rise information, estimate the power output of a human heart. 5.112 What is the maximum possible power output of the hydroelectric turbine shown in Fig. P5.112? 271 pressure at the 4-in.-diameter outlet of the hydrant is 10 psi. If head losses are negligibly small, determine the power that the pump must add to the water. 5.115 The hydroelectric turbine shown in Fig. P5.115 passes 8 million gal/min across a head of 600 ft. What is the maximum amount of power output possible? Why will the actual amount be less? 50 m 600 ft 1m 6 m/s ■ Figure P5.112 Turbine 5.113 GO Oil (SG 0.88) flows in an inclined pipe at a rate of 5 ft3/s as shown in Fig. P5.113. If the differential reading in the mercury manometer is 3 ft, calculate the power that the pump supplies to the oil if head losses are negligible. Turbine ■ Figure P5.115 5.116 A pump is to move water from a lake into a large, pressurized tank as shown in Fig. P5.116 at a rate of 1000 gal in 10 min or less. Will a pump that adds 3 hp to the water work for this purpose? Support your answer with appropriate calculations. Repeat the problem if the tank were pressurized to 3, rather than 2, atmospheres. 6 in. p = 2 atm h P Air Oil H 20 ft Pump 12 in. 3 ft ■ Figure P5.116 ■ Figure P5.113 5.114 The pumper truck shown in Fig. P5.114 is to deliver 1.5 ft3/s to a maximum elevation of 60 ft above the hydrant. The 5.117 Water is supplied at 150 ft3/s and 60 psi to a hydraulic turbine through a 3-ft inside-diameter inlet pipe as indicated in Fig. P5.117. The turbine discharge pipe has a 4-ft inside diameter. The static pressure at section (2), 10 ft below the turbine inlet, is 10-in. Hg vacuum. If the turbine develops 2500 hp, determine the power lost between sections (1) and (2). Section (1) p1 = 60 psi Q = 150 ft3/s D1 = 3 ft 60 ft Turbine 10 psi 4-in. diameter 10 ft p2 = 10-in. Hg vacuum D2 = 4 ft Hydrant ■ Figure P5.114 Section (2) ■ Figure P5.117 272 Chapter 5 ■ Finite Control Volume Analysis 5.118 Water is pumped from the tank shown in Fig. P5.118a. The head loss is known to be 1.2 V 2/2g, where V is the average velocity in the pipe. According to the pump manufacturer, the relationship between the pump head and the flowrate is as shown in Fig. P5.118b: hp 20 2000 Q2, where hp is in meters and Q is in m3/s. Determine the flowrate, Q. p3 p4 h 12 in. T 12 in. 20 hp, m Free jet 6m ■ Figure P5.121 10 hp = 20–2000Q2 0 0 Pump 0.05 0.10 Q, m3/s 0.07 m ( b) (a) ■ Figure P5.118 5.122 Water is pumped from a tank, point (1), to the top of a water plant aerator, point (2), as shown in Video V5.16 and Fig. P5.122 at a rate of 3.0 ft3/s. (a) Determine the power that the pump adds to the water if the head loss from (1) to (2) where V2 0 is 4 ft. (b) Determine the head loss from (2) to the bottom of the aerator column, point (3), if the average velocity at (3) is V3 2 ft/s. Aerator column (2) 5.119 Water is pumped from the large tank shown in Fig. P5.119. The head loss is known to be equal to 4V2/2g and the pump head is hp 20 4Q2, where hp is in ft when Q is in ft3/s. Determine the flowrate. 10 ft (1) (3) 5 ft 13 ft Q V 3 ft Pump Pump Pipe area = 0.10 ft2 ■ Figure P5.122 ■ Figure P5.119 5.120 Water flows by gravity from one lake to another as sketched in Fig. P5.120 at the steady rate of 80 gpm. What is the loss in available energy associated with this flow? If this same amount of loss is associated with pumping the fluid from the lower lake to the higher one at the same flowrate, estimate the amount of pumping power required. 5.123 Water is to be moved from one large reservoir to another at a higher elevation as indicated in Fig. P5.123. The loss of available energy associated with 2.5 ft3/s being pumped from sections (1) to (2) is loss 61V 2 2 ft2 s2, where V is the average velocity of water in the 8-in. inside-diameter piping involved. Determine the amount of shaft power required. Section (2) 8-in. insidediameter pipe 50 ft 50 ft Section (1) Pump ■ Figure P5.123 ■ Figure P5.120 5.121 The turbine shown in Fig. P5.121 develops 100 hp when the flowrate of water is 20 ft3/s. If all losses are negligible, determine (a) the elevation h, (b) the pressure difference across the turbine, and (c) the flowrate expected if the turbine were removed. 5.124 A 34-hp motor is required by an air ventilating fan to produce a 24-in.-diameter stream of air having a uniform speed of 40 ft/s. Determine the aerodynamic efficiency of the fan. 5.125 A pump moves water horizontally at a rate of 0.02 m3/s. Upstream of the pump where the pipe diameter is 90 mm, the pressure is 120 kPa. Downstream of the pump where the pipe diameter Problems is 30 mm, the pressure is 400 kPa. If the loss in energy across the pump due to fluid friction effects is 170 N ⴢ m/kg, determine the hydraulic efficiency of the pump. 5.126 Water is to be pumped from the large tank shown in Fig. P5.126 with an exit velocity of 6 m/s. It was determined that the original pump (pump 1) that supplies 1 kW of power to the water did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How much power must pump 2 add to the water? The head loss for this flow is hL ⫽ 250 Q2, where hL is in m when Q is in m3/s. 273 of the object to a location in the wake at the exit plane of the pipe. (b) Determine the force that the air exerts on the object. Air 4 m/s Wake 1-m dia. 2-m-dia. 12 m/s Exit p = 50 N/m2 V = 10 m/s ■ Figure P5.129 2 V = 6 m/s Nozzle area = 0.01 m Pipe area = 0.02 m2 Pump #2 Pump #1 2m 5.130 Near the downstream end of a river spillway, a hydraulic jump often forms, as illustrated in Fig. P5.130 and Video V10.11. The velocity of the channel flow is reduced abruptly across the jump. Using the conservation of mass and linear momentum principles, derive the following expression for h2, h2 ⫽ ⫺ ■ Figure P5.126 5.127 (See Fluids in the News article titled “Curtain of Air,” Section 5.3.3.) The fan shown in Fig. P5.127 produces an air curtain to separate a loading dock from a cold storage room. The air curtain is a jet of air 10 ft wide, 0.5 ft thick moving with speed V ⫽ 30 ft/s. The loss associated with this flow is loss ⫽ KLV2/2, where KL ⫽ 5. How much power must the fan supply to the air to produce this flow? h1 2 2V 21h1 h1 ⫹ a b ⫹ g 2 B 2 The loss of available energy across the jump can also be determined if energy conservation is considered. Derive the loss expression g1h2 ⫺ h1 2 3 jump loss ⫽ 4h1h2 Fan h1 V = 30 ft/s Air curtain (0.5-ft thickness) Open door 10 ft ■ Figure P5.127 Section 5.3.3 Application of the Energy Equation— Combined with Linear Momentum 5.128 If a 34-hp motor is required by a ventilating fan to produce a 24-in. stream of air having a velocity of 40 ft/s as shown in Fig. P5.128, estimate (a) the efficiency of the fan and (b) the thrust of the supporting member on the conduit enclosing the fan. V1 ■ Figure P5.130 5.131 Water flows steadily down the inclined pipe as indicated in Fig P5.131. Determine the following: (a) the difference in pressure p1 ⫺ p2, (b) the loss between sections (1) and (2), (c) the net axial force exerted by the pipe wall on the flowing water between sections (1) and (2). Flo w 5 ft Section (1) 24 in. h2 6 in. 40 ft/s 30° Section (2) ■ Figure P5.128 5.129 Air flows past an object in a pipe of 2-m diameter and exits as a free jet as shown in Fig. P5.129. The velocity and pressure upstream are uniform at 10 m/s and 50 N/m2, respectively. At the pipe exit the velocity is nonuniform as indicated. The shear stress along the pipe wall is negligible. (a) Determine the head loss associated with a particle as it flows from the uniform velocity upstream 6 in. Mercury ■ Figure P5.131 274 Chapter 5 ■ Finite Control Volume Analysis 5.132 Water flows steadily in a pipe and exits as a free jet through an end cap that contains a filter as shown in Fig. P5.132. The flow is in a horizontal plane. The axial component, Ry, of the anchoring force needed to keep the end cap stationary is 60 lb. Determine the head loss for the flow through the end cap. Ry = 60 lb Area = 0.10 ft2 Rx Filter 30° Pipe Area = 0.12 ft2 V = 10 ft/s area is uniform. At section (2), the velocity profile is V wc a R r 1 7 ˆ b k R where V local velocity vector, wc centerline velocity in the axial direction, R pipe inside radius, and r radius from pipe axis. Develop an expression for the loss in available energy between sections (1) and (2). 5.136 A small fan moves air at a mass flowrate of 0.004 lbm/s. Upstream of the fan, the pipe diameter is 2.5 in., the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, a1, is equal to 2.0. Downstream of the fan, the pipe diameter is 1 in., the flow is turbulent, the velocity profile is quite flat, and the kinetic energy coefficient, a2, is equal to 1.08. If the rise in static pressure across the fan is 0.015 psi and the fan shaft draws 0.00024 hp, compare the value of loss calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions. ■ Figure P5.132 5.133 When fluid flows through an abrupt expansion as indicated in Fig. P5.133, the loss in available energy across the expansion, lossex, is often expressed as lossex a1 A1 2 V 12 b A2 2 where A1 cross-sectional area upstream of expansion, A2 cross-sectional area downstream of expansion, and V1 velocity of flow upstream of expansion. Derive this relationship. Section (1) Section (2) ■ Figure P5.133 5.134 Two water jets collide and form one homogeneous jet as shown in Fig. P5.134. (a) Determine the speed, V, and direction, u, of the combined jet. (b) Determine the loss for a fluid particle flowing from (1) to (3), from (2) to (3). Gravity is negligible. Section 5.3.5 Combination of the Energy Equation and the Moment-of-Momentum Equation 5.137 Air enters a radial blower with zero angular momentum. It leaves with an absolute tangential velocity, Vu, of 200 ft/s. The rotor blade speed at rotor exit is 170 ft/s. If the stagnation pressure rise across the rotor is 0.4 psi, calculate the loss of available energy across the rotor and the rotor efficiency. 5.138 Water enters a pump impeller radially. It leaves the impeller with a tangential component of absolute velocity of 10 m/s. The impeller exit diameter is 60 mm, and the impeller speed is 1800 rpm. If the stagnation pressure rise across the impeller is 45 kPa, determine the loss of available energy across the impeller and the hydraulic efficiency of the pump. 5.139 Water enters an axial-flow turbine rotor with an absolute velocity tangential component, Vu, of 15 ft/s. The corresponding blade velocity, U, is 50 ft/s. The water leaves the rotor blade row with no angular momentum. If the stagnation pressure drop across the turbine is 12 psi, determine the hydraulic efficiency of the turbine. 5.140 An inward flow radial turbine (see Fig. P5.140) involves a nozzle angle, a1, of 60 and an inlet rotor tip speed, U1, of 30 ft/s. The ratio of rotor inlet to outlet diameters is 2.0. The radial component of velocity remains constant at 20 ft/s through the rotor, and the flow leaving the rotor at section (2) is without angular momentum. If the flowing fluid is water and the stagnation pressure drop across the rotor is 16 psi, determine the loss of available energy across the rotor and the hydraulic efficiency involved. V 0.12 m (3) θ V2 = 6 m/s 60° (2) 90° 0.10 m Vr1 = 20 ft/s (1) V1 = 4 m/s U1 = 30 ft/s ■ Figure P5.134 Section 5.3.4 Application of the Energy Equation to Nonuniform Flows 5.135 Water flows vertically upward in a circular cross-sectional pipe. At section (1), the velocity profile over the cross-sectional ■ Figure P5.140 r1 r2 2 1 Problems Section 5.3.4 Application of the Energy Equation to Nonuniform Flows 5.141 The distribution of axial direction velocity, u, in a pipe flow is linear from zero at the wall to maximum of uc at the centerline. Determine the average velocity, u, and the kinetic energy coefficient, a. ■ Lab Problems 5.1LP This problem involves the force that a jet of air exerts on a flat plate as the air is deflected by the plate. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 5.2LP This problem involves the pressure distribution produced on a flat plate that deflects a jet of air. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 5.3LP This problem involves the force that a jet of water exerts on a vane when the vane turns the jet through a given angle. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 275 5.4LP This problem involves the force needed to hold a pipe elbow stationary. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www. wiley.com/college/munson. ■ Lifelong Learning Problems 5.1LL What are typical efficiencies associated with swimming and how can they be improved? 5.2LL Explain how local ionization of flowing air can accelerate it. How can this be useful? 5.3LL Discuss the main causes of loss of available energy in a turbo-pump and how they can be minimized. What are typical turbo-pump efficiencies? 5.4LL Discuss the main causes of loss of available energy in a turbine and how they can be minimized. What are typical turbine efficiencies? ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 6 Differential Analysis of Fluid Flow CHAPTER OPENING PHOTO: Flow past an airfoil at angle of attack: The flow of air around a model airfoil is made visible by smoke line flow visualization. Note the flow separation on the upper surface—see Ch. 9 for more details. (Photograph courtesy of NASA.2 Learning Objectives After completing this chapter, you should be able to: ■ determine various kinematic elements of the flow given the velocity field. ■ explain the conditions necessary for a velocity field to satisfy the continuity equation. ■ apply the concepts of stream function and velocity potential. ■ characterize simple potential flow fields. ■ analyze certain types of flows using the Navier–Stokes equations. In the previous chapter attention is focused on the use of finite control volumes for the solution of a variety of fluid mechanics problems. This approach is very practical and useful, since it does not generally require detailed knowledge of the pressure and velocity variations within the control volume. Typically, we found that only conditions on the surface of the control volume were needed, and thus problems could be solved without detailed knowledge of the flow field. Unfortunately, many situations arise in which the details of the flow are important and the finite control volume approach will not yield the desired information. For example, we may need to know how the velocity varies over the cross section of a pipe, or how the pressure and shear stress vary along the surface of an airplane wing. In these circumstances we need to develop relationships that apply at a point, or at least in a very small infinitesimal region within a given flow field. This approach, which involves an infinitesimal control volume, as distinguished from a finite control volume, is commonly referred to as differential analysis, since 1as we will soon discover2 the governing equations are differential equations. In this chapter we will provide an introduction to the differential equations that describe 1in detail2 the motion of fluids. Unfortunately, we will also find that these equations are rather complicated, nonlinear partial differential equations that cannot be solved exactly except in a few cases, where 276 6.1 V6.1 Spinning football-pressure contours V6.2 Spinning football-velocity vectors 6.1 Fluid Element Kinematics 277 simplifying assumptions are made. Thus, although differential analysis has the potential for supplying very detailed information about flow fields, this information is not easily extracted. Nevertheless, this approach provides a fundamental basis for the study of fluid mechanics. We do not want to be too discouraging at this point, since there are some exact solutions for laminar flow that can be obtained, and these have proved to be very useful. A few of these are included in this chapter. In addition, by making some simplifying assumptions, many other analytical solutions can be obtained. For example, in some circumstances it may be reasonable to assume that the effect of viscosity is small and can be neglected. This rather drastic assumption greatly simplifies the analysis and provides the opportunity to obtain detailed solutions to a variety of complex flow problems. Some examples of these so-called inviscid flow solutions are also described in this chapter. It is known that for certain types of flows the flow field can be conceptually divided into two regions—a very thin region near the boundaries of the system in which viscous effects are important, and a region away from the boundaries in which the flow is essentially inviscid. By making certain assumptions about the behavior of the fluid in the thin layer near the boundaries, and using the assumption of inviscid flow outside this layer, a large class of problems can be solved using differential analysis. These boundary layer problems are discussed in Chapter 9. Finally, it is to be noted that with the availability of powerful computers it is feasible to attempt to solve the differential equations using the techniques of numerical analysis. Although it is beyond the scope of this book to delve extensively into this approach, which is generally referred to as computational fluid dynamics 1CFD2, the reader should be aware of this approach to complex flow problems. CFD has become a common engineering tool, and a brief introduction can be found in Appendix A. To introduce the power of CFD, two animations based on the numerical computations are provided as shown in the margin. We begin our introduction to differential analysis by reviewing and extending some of the ideas associated with fluid kinematics that were introduced in Chapter 4. With this background the remainder of the chapter will be devoted to the derivation of the basic differential equations 1which will be based on the principle of conservation of mass and Newton’s second law of motion2 and to some applications. Fluid Element Kinematics Fluid element motion consists of translation, linear deformation, rotation, and angular deformation. In this section we will be concerned with the mathematical description of the motion of fluid elements moving in a flow field. A small fluid element in the shape of a cube that is initially in one position will move to another position during a short time interval dt as illustrated in Fig. 6.1. Because of the generally complex velocity variation within the field, we expect the element not only to translate from one position but also to have its volume changed 1linear deformation2, to rotate, and to undergo a change in shape 1angular deformation2. Although these movements and deformations occur simultaneously, we can consider each one separately as illustrated in Fig. 6.1. Since element motion and deformation are intimately related to the velocity and variation of velocity throughout the flow field, we will briefly review the manner in which velocity and acceleration fields can be described. Element at t0 Element at t0 + δ t = General motion + Translation + Linear deformation ■ Figure 6.1 Types of motion and deformation for a fluid element. + Rotation Angular deformation 278 Chapter 6 ■ Differential Analysis of Fluid Flow 6.1.1 Velocity and Acceleration Fields Revisited As discussed in detail in Section 4.1, the velocity field can be described by specifying the velocity V at all points, and at all times, within the flow field of interest. Thus, in terms of rectangular coordinates, the notation V 1x, y, z, t2 means that the velocity of a fluid particle depends on where it is located within the flow field 1as determined by its coordinates, x, y, and z2 and when it occupies the particular point 1as determined by the time, t2. As is pointed out in Section 4.1.1, this method of describing the fluid motion is called the Eulerian method. It is also convenient to express the velocity in terms of three rectangular components so that (6.1) V uiˆ vjˆ wkˆ z V w ^ k ^ i x ^ j v y u where u, v, and w are the velocity components in the x, y, and z directions, respectively, and ˆi , jˆ, and kˆ are the corresponding unit vectors, as shown by the figure in the margin. Of course, each of these components will, in general, be a function of x, y, z, and t. One of the goals of differential analysis is to determine how these velocity components specifically depend on x, y, z, and t for a particular problem. With this description of the velocity field it was also shown in Section 4.2.1 that the acceleration of a fluid particle can be expressed as 0V 0V 0V 0V u v w 0t 0x 0y 0z a (6.2) and in component form: 0u 0u 0u 0u u v w 0t 0x 0y 0z 0v 0v 0v 0v ay u v w 0t 0x 0y 0z 0w 0w 0w 0w az u v w 0t 0x 0y 0z ax The acceleration of a fluid particle is described using the concept of the material derivative. (6.3a) (6.3b) (6.3c) The acceleration is also concisely expressed as a where the operator D1 2 Dt 01 2 0t u DV Dt 01 2 0x (6.4) v 01 2 0y w 01 2 0z (6.5) is termed the material derivative, or substantial derivative. In vector notation D1 2 where the gradient operator, ⵱1 2, is Dt ⵱1 2 01 2 01 2 0x 0t ˆi 1V ⴢ ⵱21 2 01 2 0y jˆ 01 2 0z (6.6) kˆ (6.7) which was introduced in Chapter 2. As we will see in the following sections, the motion and deformation of a fluid element depend on the velocity field. The relationship between the motion and the forces causing the motion depends on the acceleration field. 6.1.2 Linear Motion and Deformation The simplest type of motion that a fluid element can undergo is translation, as illustrated in Fig. 6.2. In a small time interval dt a particle located at point O will move to point O¿ as is illustrated in the figure. If all points in the element have the same velocity 1which is only true if there are no velocity gradients2, then the element will simply translate from one position to another. However, 6.1 v Fluid Element Kinematics 279 O' vδt u O u δt ■ Figure 6.2 Translation of a fluid element. because of the presence of velocity gradients, the element will generally be deformed and rotated as it moves. For example, consider the effect of a single velocity gradient, 0u 0x, on a small cube having sides dx, dy, and dz. As is shown in Fig. 6.3a, if the x component of velocity of O and B is u, then at nearby points A and C the x component of the velocity can be expressed as u 1 0u 0x2 dx. This difference in velocity causes a “stretching” of the volume element by an amount 10u 0x2 1dx21dt2 during the short time interval dt in which line OA stretches to OA¿ and BC to BC¿ 1Fig. 6.3b2. The corresponding change in the original volume, d V dx dy dz, would be Change in d V a 0u dxb 1dy dz21dt2 0x and the rate at which the volume d V is changing per unit volume due to the gradient 0u 0x is 2 10u 0x2 dt 1 d1dV 0u lim c d dV dt dt 0x dtS0 The rate of volume change per unit volume is related to the velocity gradients. (6.8) If velocity gradients 0v 0y and 0w 0z are also present, then using a similar analysis it follows that, in the general case, 2 1 d1dV 0u 0v 0w ⵱ⴢV dV dt 0x 0y 0z (6.9) This rate of change of the volume per unit volume is called the volumetric dilatation rate. Thus, we see that the volume of a fluid may change as the element moves from one location to another in the flow field. However, for an incompressible fluid the volumetric dilatation rate is zero, since the element volume cannot change without a change in fluid density 1the element mass must be conserved2. Variations in the velocity in the direction of the velocity, as represented by the derivatives 0u 0x, 0v 0y, and 0w 0z, simply cause a linear deformation of the element in the sense that the shape of the element does not change. Cross derivatives, such as 0u 0y and 0v 0x, will cause the element to rotate and generally to undergo an angular deformation, which changes the shape of the element. 6.1.3 Angular Motion and Deformation For simplicity we will consider motion in the x–y plane, but the results can be readily extended to the more general three-dimensional case. The velocity variation that causes rotation and angular deformation is illustrated in Fig. 6.4a. In a short time interval dt the line segments OA and OB will B u C ∂u u + __ δ x ∂x δy C C' A A' δy ∂u u + __ δ x ∂x u O B δx A O δx ) ∂∂__ux δx ) δt (a) (b) ■ Figure 6.3 Linear deformation of a fluid element. 280 Chapter 6 ■ Differential Analysis of Fluid Flow ) ∂∂__uy δ y) δt ∂u u + ___ δ y ∂y B B B' δβ δy δy v O A' ∂v v + ___ δ x ∂x u δx δα O A (a) δx ) ∂∂__vx δ x) δ t A ■ Figure 6.4 Angular motion and deformation of a fluid element. (b) rotate through the angles da and db to the new positions OA¿ and OB¿ , as is shown in Fig. 6.4b. The angular velocity of line OA, vOA, is vOA lim V6.3 Shear deformation dtS0 For small angles tan da ⬇ da so that vOA lim c dtS0 da dt 10v 0x2 dx dt dx 10v 0x2 dt dt d 0v dt 0x (6.10) 0v 0x Note that if 0v 0x is positive, vOA will be counterclockwise. Similarly, the angular velocity of the line OB is db dtS0 dt vOB lim and tan db ⬇ db so that vOB lim c dtS0 Rotation of fluid particles is related to certain velocity gradients in the flow field. 10u 0y2 dy dt dy 10u 0y2 dt dt d 0u dt 0y (6.11) 0u 0y In this instance if 0u 0y is positive, vOB will be clockwise. The rotation, vz, of the element about the z axis is defined as the average of the angular velocities vOA and vOB of the two mutually perpendicular lines OA and OB.1 Thus, if counterclockwise rotation is considered to be positive, it follows that vz 1 0v 0u a b 2 0x 0y (6.12) Rotation of the field element about the other two coordinate axes can be obtained in a similar manner with the result that for rotation about the x axis vx 1 0w 0v a b 2 0y 0z (6.13) vy 1 0u 0w a b 2 0z 0x (6.14) and for rotation about the y axis 1 With this definition vz can also be interpreted to be the angular velocity of the bisector of the angle between the lines OA and OB. 6.1 Fluid Element Kinematics 281 The three components, vx, vy, and vz can be combined to give the rotation vector, ␻, in the form (6.15) ␻ vxˆi vy ˆj vzkˆ An examination of this result reveals that ␻ is equal to one-half the curl of the velocity vector. That is, ␻ 12 curl V 12 ⵱ ⴛ V (6.16) since by definition of the vector operator ⵱ ⴛ V ˆi 1 1 0 ⵱ⴛV ∞ 2 2 0x u Vorticity in a flow field is related to fluid particle rotation. ˆj 0 0y v kˆ 0 ∞ 0z w 0v 1 0u 0w ˆ 1 a 0v 0u b kˆ 1 0w a b ˆi a b j 2 0x 0y 2 0y 0z 2 0z 0x The vorticity, Z, is defined as a vector that is twice the rotation vector; that is, Z2␻⵱ⴛV (6.17) The use of the vorticity to describe the rotational characteristics of the fluid simply eliminates the 1 12 2 factor associated with the rotation vector. The figure in the margin shows vorticity contours of the wing tip vortex flow shortly after an aircraft has passed. The lighter colors indicate stronger vorticity. (See also Fig. 4.3.) We observe from Eq. 6.12 that the fluid element will rotate about the z axis as an undeformed block 1i.e., vOA vOB 2 only when 0u 0y 0v 0x. Otherwise the rotation will be associated with an angular deformation. We also note from Eq. 6.12 that when 0u 0y 0v 0x the rotation around the z axis is zero. More generally if ⵱ ⴛ V 0, then the rotation 1and the vorticity2 are zero, and flow fields for which this condition applies are termed irrotational. We will find in Section 6.4 that the condition of irrotationality often greatly simplifies the analysis of complex flow fields. However, it is probably not immediately obvious why some flow fields would be irrotational, and we will need to examine this concept more fully in Section 6.4. Wing Engine E XAMPLE 6.1 Vorticity GIVEN For a certain two-dimensional flow field the velocity FIND Is this flow irrotational? is given by the equation V 1x2 y2 2iˆ 2xyjˆ SOLUTION For an irrotational flow the rotation vector, ␻, having the components given by Eqs. 6.12, 6.13, and 6.14 must be zero. For the prescribed velocity field u x2 y2 v 2xy w0 zero, since by definition of two-dimensional flow u and v are not functions of z, and w is zero. In this instance the condition for irrotationality simply becomes vz 0 or 0v 0x 0u 0y. The streamlines for the steady, two-dimensional flow of this example are shown in Fig. E6.1. (Information about how to calculate and therefore vx 1 0w 0v a b0 2 0y 0z vy 0w 1 0u a b0 2 0z 0x y 0u 1 1 0v vz a b 3 12y2 12y2 4 0 2 0x 0y 2 Thus, the flow is irrotational. (Ans) COMMENTS It is to be noted that for a two-dimensional flow field 1where the flow is in the x–y plane2 vx and vy will always be x ■ Figure E6.1 282 Chapter 6 ■ Differential Analysis of Fluid Flow OA and OB of Fig. 6.4 rotate with the same speed but in opposite directions. As shown by Eq. 6.17, the condition of irrotationality is equivalent to the fact that the vorticity, Z, is zero or the curl of the velocity is zero. streamlines for a given velocity field is given in Sections 4.1.4 and 6.2.3.) It is noted that all of the streamlines (except for the one through the origin) are curved. However, because the flow is irrotational, there is no rotation of the fluid elements. That is, lines In addition to the rotation associated with the derivatives 0u 0y and 0v 0x, it is observed from Fig. 6.4b that these derivatives can cause the fluid element to undergo an angular deformation, which results in a change in shape of the element. The change in the original right angle formed by the lines OA and OB is termed the shearing strain, dg, and from Fig. 6.4b dg da db where dg is considered to be positive if the original right angle is decreasing. The rate of change of dg is called the rate of shearing strain or the rate of angular deformation and is commonly denoted with the symbol g᝽. The angles da and db are related to the velocity gradients through Eqs. 6.10 and 6.11 so that 10v 0x2 dt 10u 0y2 dt dg lim c d dt dtS0 dt dtS0 g᝽ lim and, therefore, g᝽ 0v 0u 0x 0y (6.18) As we will learn in Section 6.8, the rate of angular deformation is related to a corresponding shearing stress that causes the fluid element to change in shape. From Eq. 6.18 we note that if 0u 0y 0v 0x, the rate of angular deformation is zero, and this condition corresponds to the case in which the element is simply rotating as an undeformed block 1Eq. 6.122. In the remainder of this chapter we will see how the various kinematical relationships developed in this section play an important role in the development and subsequent analysis of the differential equations that govern fluid motion. 6.2 Conservation of Mass Conservation of mass requires that the mass of a system remain constant. As is discussed in Section 5.1, conservation of mass requires that the mass, M, of a system remain constant as the system moves through the flow field. In equation form this principle is expressed as DMsys 0 Dt We found it convenient to use the control volume approach for fluid flow problems, with the control volume representation of the conservation of mass written as 0 0t 冮 r dV cv 冮 r V ⴢ nˆ dA 0 (6.19) cs where the equation 1commonly called the continuity equation2 can be applied to a finite control volume 1cv2, which is bounded by a control surface 1cs2. The first integral on the left side of Eq. 6.19 represents the rate at which the mass within the control volume is changing, and the second integral represents the net rate at which mass is flowing out through the control surface 1rate of mass outflow rate of mass inflow2. To obtain the differential form of the continuity equation, Eq. 6.19 is applied to an infinitesimal control volume. 6.2.1 Differential Form of Continuity Equation We will take as our control volume the small, stationary cubical element shown in Fig. 6.5a. At the center of the element the fluid density is r, and the velocity has components u, v, and w. Since the element is small, the volume integral in Eq. 6.19 can be expressed as 0 0t 冮 cv r dV ⬇ 0r dx dy dz 0t (6.20) 6.2 v ρ y (ρu) δ__x δ y δ z ρu – ∂_____ ∂x 2 δy δy ρu y δz δz x δ y δz (ρu) δ__ ρ u + ∂_____ ∂x 2 δx x z 283 u w δx Conservation of Mass x z (a) (b) ■ Figure 6.5 A differential element for the development of conservation of mass equation. The rate of mass flow through the surfaces of the element can be obtained by considering the flow in each of the coordinate directions separately. For example, in Fig. 6.5b flow in the x direction is depicted. If we let ru represent the x component of the mass rate of flow per unit area at the center of the element, then on the right face ru 0 x 1dx22 ru 01ru2 dx 0x 2 (6.21) ru 0 x 1dx22 ru 01ru2 dx 0x 2 (6.22) and on the left face ∂ (ρ u) δx ∂x 2 (ρ u) Slope = ∂ ∂x ρu δx 2 δx 2 Note that we are really using a Taylor series expansion of ru and neglecting higher order terms such as 1dx2 2, 1dx2 3, and so on. This is indicated by the figure in the margin. When the right-hand sides of Eqs. 6.21 and 6.22 are multiplied by the area dy dz, the rate at which mass is crossing the right and left sides of the element is obtained as is illustrated in Fig. 6.5b. When these two expressions are combined, the net rate of mass flowing from the element through the two surfaces can be expressed as 01ru2 dx Net rate of mass c ru d dy dz outflow in x direction 0x 2 c ru x 01ru2 dx 01ru2 d dy dz dx dy dz (6.23) 0x 2 0x For simplicity, only flow in the x direction has been considered in Fig. 6.5b, but, in general, there will also be flow in the y and z directions. An analysis similar to the one used for flow in the x direction shows that 01rv2 Net rate of mass dx dy dz outflow in y direction 0y (6.24) 01rw2 Net rate of mass dx dy dz outflow in z direction 0z (6.25) 01ru2 01rv2 01rw2 Net rate of c d dx dy dz mass outflow 0x 0y 0z (6.26) and Thus, From Eqs. 6.19, 6.20, and 6.26 it now follows that the differential equation for conservation of mass is The continuity equation is one of the fundamental equations of fluid mechanics. 01ru2 01rv2 01rw2 0r 0 0t 0x 0y 0z (6.27) As previously mentioned, this equation is also commonly referred to as the continuity equation. 284 Chapter 6 ■ Differential Analysis of Fluid Flow The continuity equation is one of the fundamental equations of fluid mechanics and, as expressed in Eq. 6.27, is valid for steady or unsteady flow, and compressible or incompressible fluids. In vector notation, Eq. 6.27 can be written as 0r ⵱ ⴢ rV 0 0t (6.28) Two special cases are of particular interest. For steady flow of compressible fluids ⵱ ⴢ rV 0 or 01ru2 0x 01rv2 0y 01rw2 0z 0 (6.29) This follows since by definition r is not a function of time for steady flow, but could be a function of position. For incompressible fluids the fluid density, r, is a constant throughout the flow field so that Eq. 6.28 becomes For incompressible fluids the continuity equation reduces to a simple relationship involving certain velocity gradients. ⵱ⴢV0 (6.30) 0u 0v 0w 0 0x 0y 0z (6.31) or Equation 6.31 applies to both steady and unsteady flow of incompressible fluids. Note that Eq. 6.31 is the same as that obtained by setting the volumetric dilatation rate 1Eq. 6.92 equal to zero. This result should not be surprising since both relationships are based on conservation of mass for incompressible fluids. However, the expression for the volumetric dilatation rate was developed from a system approach, whereas Eq. 6.31 was developed from a control volume approach. In the former case the deformation of a particular differential mass of fluid was studied, and in the latter case mass flow through a fixed differential volume was studied. E XAMPLE 6.2 Continuity Equation GIVEN The velocity components for a certain incompress- FIND Determine the form of the z component, w, required to ible, steady-flow field are satisfy the continuity equation. ux y z v xy yz z w? 2 2 2 so that the required expression for 0w 0z is 0w 2x 1x z2 3x z 0z SOLUTION Integration with respect to z yields Any physically possible velocity distribution must for an incompressible fluid satisfy conservation of mass as expressed by the continuity equation 0v 0w 0u 0 0x 0y 0z For the given velocity distribution 0u 2x 0x and 0v xz 0y w 3xz z2 f 1x, y2 2 (Ans) COMMENT The third velocity component cannot be explic- itly determined since the function f 1x, y2 can have any form and conservation of mass will still be satisfied. The specific form of this function will be governed by the flow field described by these velocity components—that is, some additional information is needed to completely determine w. 6.2 y Conservation of Mass 285 ^ eθ P vθ P r vr ^ e r ^ e z vz x θ z ■ Figure 6.6 The representation of velocity components in cylindrical polar coordinates. 6.2.2 Cylindrical Polar Coordinates For some problems, velocity components expressed in cylindrical polar coordinates will be convenient. For some problems it is more convenient to express the various differential relationships in cylindrical polar coordinates rather than Cartesian coordinates. As is shown in Fig. 6.6, with cylindrical coordinates a point is located by specifying the coordinates r, u, and z. The coordinate r is the radial distance from the z axis, u is the angle measured from a line parallel to the x axis 1with counterclockwise taken as positive2, and z is the coordinate along the z axis. The velocity components, as sketched in Fig. 6.6, are the radial velocity, vr, the tangential velocity, vu, and the axial velocity, vz. Thus, the velocity at some arbitrary point P can be expressed as V vrêr vuêu vzêz (6.32) where êr, êu, and êz are the unit vectors in the r, u, and z directions, respectively, as are illustrated in Fig. 6.6. The use of cylindrical coordinates is particularly convenient when the boundaries of the flow system are cylindrical. Several examples illustrating the use of cylindrical coordinates will be given in succeeding sections in this chapter. The differential form of the continuity equation in cylindrical coordinates is 01rvz 2 0r 1 01rrvr 2 1 01rvu 2 0 r 0r r 0u 0t 0z (6.33) This equation can be derived by following the same procedure used in the preceding section. For steady, compressible flow 01rvz 2 1 01rrvr 2 1 01rvu 2 0 r 0r r 0u 0z (6.34) For incompressible fluids 1for steady or unsteady flow2 0vz 1 0vu 1 01rvr 2 0 r 0r r 0u 0z (6.35) 6.2.3 The Stream Function Steady, incompressible, plane, two-dimensional flow represents one of the simplest types of flow of practical importance. By plane, two-dimensional flow we mean that there are only two velocity components, such as u and v, when the flow is considered to be in the x–y plane. For this flow the continuity equation, Eq. 6.31, reduces to 0u 0v 0 0x 0y (6.36) 286 Chapter 6 ■ Differential Analysis of Fluid Flow We still have two variables, u and v, to deal with, but they must be related in a special way as indicated by Eq. 6.36. This equation suggests that if we define a function c1x, y2, called the stream function, which relates the velocities shown by the figure in the margin as Velocity components in a twodimensional flow field can be expressed in terms of a stream function. u v 0c 0x (6.37) then the continuity equation is identically satisfied. This conclusion can be verified by simply substituting the expressions for u and v into Eq. 6.36 so that y V u= v=– ∂c ∂y x V v u Streamlines 0c 0y y x 0c 0 2c 0 2c 0 0c 0 a b a b 0 0x 0y 0y 0x 0x 0y 0y 0x ∂c ∂x Thus, whenever the velocity components are defined in terms of the stream function we know that conservation of mass will be satisfied. Of course, we still do not know what c1x, y2 is for a particular problem, but at least we have simplified the analysis by having to determine only one unknown function, c1x, y2, rather than the two functions, u1x, y2 and v1x, y2. Another particular advantage of using the stream function is related to the fact that lines along which c is constant are streamlines. Recall from Section 4.1.4 that streamlines are lines in the flow field that are everywhere tangent to the velocities, as is illustrated in Fig. 6.7 and the figure in the margin. It follows from the definition of the streamline that the slope at any point along a streamline is given by dy v u dx The change in the value of c as we move from one point 1x, y2 to a nearby point 1x dx, y dy2 is given by the relationship: dc 0c 0c dx dy v dx u dy 0x 0y Along a line of constant c we have dc 0 so that v dx u dy 0 and, therefore, along a line of constant c dy v u dx which is the defining equation for a streamline. Thus, if we know the function c1x, y2, we can plot lines of constant c to provide the family of streamlines that are helpful in visualizing the pattern V v Streamlines u y x ■ Figure 6.7 Velocity and velocity components along a streamline. 6.2 Conservation of Mass 287 c + dc C dq c2 c q u dy A y B c1 – v dx x (a) (b) ■ Figure 6.8 The flow between two streamlines. The change in the value of the stream function is related to the volume rate of flow. of flow. An infinite number of streamlines make up a particular flow field, since for each constant value assigned to c a streamline can be drawn. The actual numerical value associated with a particular streamline is not of particular significance, but the change in the value of c is related to the volume rate of flow. Consider two closely spaced streamlines, shown in Fig. 6.8a. The lower streamline is designated c and the upper one c dc. Let dq represent the volume rate of flow 1per unit width perpendicular to the x–y plane2 passing between the two streamlines. Note that flow never crosses streamlines, since by definition the velocity is tangent to the streamline. From conservation of mass we know that the inflow, dq, crossing the arbitrary surface AC of Fig. 6.8a must equal the net outflow through surfaces AB and BC. Thus, dq u dy v dx or in terms of the stream function dq 0c 0c dy dx 0y 0x (6.38) The right-hand side of Eq. 6.38 is equal to dc so that dq dc c2 > c1 (6.39) Thus, the volume rate of flow, q, between two streamlines such as c1 and c2 of Fig. 6.8b can be determined by integrating Eq. 6.39 to yield q q c1 冮 c2 dc c2 c1 (6.40) c1 c2 < c1 q The relative value of c2 with respect to c1 determines the direction of flow, as shown by the figure in the margin. In cylindrical coordinates the continuity equation 1Eq. 6.352 for incompressible, plane, twodimensional flow reduces to 1 01rvr 2 1 0vu 0 r 0r r 0u c1 (6.41) and the velocity components, vr and vu, can be related to the stream function, c1r, u2, through the equations vr ∂c vr = 1r ∂θ vθ = – r V ∂c ∂r θ 1 0c r 0u vu 0c 0r (6.42) as shown by the figure in the margin. Substitution of these expressions for the velocity components into Eq. 6.41 shows that the continuity equation is identically satisfied. The stream function concept can be extended to axisymmetric flows, such as flow in pipes or flow around bodies of revolution, and to twodimensional compressible flows. However, the concept is not applicable to general three-dimensional flows using a single stream function. 288 Chapter 6 ■ Differential Analysis of Fluid Flow E XAMPLE Stream Function 6.3 GIVEN The velocity components in a steady, incompressible, FIND two-dimensional flow field are (a) u 2y v 4x Determine the corresponding stream function and (b) Show on a sketch several streamlines. Indicate the direction of flow along the streamlines. SOLUTION (a) From the definition of the stream function 1Eqs. 6.372 u y ψ=0 ψ=0 0c 2y 0y and v 0c 4x 0x x The first of these equations can be integrated to give c y2 f1 1x2 where f1 1x2 is an arbitrary function of x. Similarly from the second equation c 2x2 f2 1y2 where f2 1y2 is an arbitrary function of y. It now follows that in order to satisfy both expressions for the stream function c 2x2 y2 C (Ans) or where C is an arbitrary constant. y 12x COMMENT Since the velocities are related to the derivatives of the stream function, an arbitrary constant can always be added to the function, and the value of the constant is actually of no consequence. Usually, for simplicity, we set C 0 so that for this particular example the simplest form for the stream function is c 2x2 y2 (1) Other streamlines can be obtained by setting c equal to various constants. It follows from Eq. 1 that the equations of these streamlines 1for c 02 can be expressed in the form y2 x2 1 c c2 (Ans) Either answer indicated would be acceptable. (b) Streamlines can now be determined by setting c constant and plotting the resulting curve. With the preceding expression for c 1with C 02 the value of c at the origin is zero, so that the equation of the streamline passing through the origin 1the c 0 streamline2 is 0 2x2 y2 6.3 ■ Figure E6.3 which we recognize as the equation of a hyperbola. Thus, the streamlines are a family of hyperbolas with the c 0 streamlines as asymptotes. Several of the streamlines are plotted in Fig. E6.3. Since the velocities can be calculated at any point, the direction of flow along a given streamline can be easily deduced. For example, v 0c 0x 4x so that v 7 0 if x 7 0 and v 6 0 if x 6 0. The direction of flow is indicated on the figure. Conservation of Linear Momentum To develop the differential momentum equations we can start with the linear momentum equation F DP ` Dt sys where F is the resultant force acting on a fluid mass, P is the linear momentum defined as P 冮 sys V dm (6.43) 6.3 Conservation of Linear Momentum 289 and the operator D1 2Dt is the material derivative 1see Section 4.2.12. In the last chapter it was demonstrated how Eq. 6.43 in the form 0 a Fcontents of the 0t control volume 冮 Vr dV cv 冮 VrV ⴢ nˆ dA (6.44) cs could be applied to a finite control volume to solve a variety of flow problems. To obtain the differential form of the linear momentum equation, we can either apply Eq. 6.43 to a differential system, consisting of a mass, dm, or apply Eq. 6.44 to an infinitesimal control volume, dV , which initially bounds the mass dm. It is probably simpler to use the system approach since application of Eq. 6.43 to the differential mass, dm, yields dF D1V dm2 Dt where dF is the resultant force acting on dm. Using this system approach dm can be treated as a constant so that dF dm DV Dt But DVDt is the acceleration, a, of the element. Thus, dF dm a (6.45) which is simply Newton’s second law applied to the mass dm. This is the same result that would be obtained by applying Eq. 6.44 to an infinitesimal control volume 1see Ref. 12. Before we can proceed, it is necessary to examine how the force dF can be most conveniently expressed. 6.3.1 Description of Forces Acting on the Differential Element Both surface forces and body forces generally act on fluid particles. In general, two types of forces need to be considered: surface forces, which act on the surface of the differential element, and body forces, which are distributed throughout the element. For our purpose, the only body force, dFb, of interest is the weight of the element, which can be expressed as dFb dm g (6.46) where g is the vector representation of the acceleration of gravity. In component form dFbx dm gx (6.47a) dFby dm gy (6.47b) dFbz dm gz (6.47c) where gx, gy, and gz are the components of the acceleration of gravity vector in the x, y, and z directions, respectively. Surface forces act on the element as a result of its interaction with its surroundings. At any arbitrary location within a fluid mass, the force acting on a small area, dA, that lies in an arbitrary surface, can be represented by dFs, as is shown in Fig. 6.9. In general, dFs will be inclined with respect to the surface. The force dFs can be resolved into three components, dFn, dF1, and dF2, where dFn is normal to the area, dA, and dF1 and dF2 are parallel to the area and orthogonal to each other. The normal stress, sn, is defined as sn lim dAS0 dFn dA δ Fn δ Fs δA δ F2 δ F1 Arbitrary surface ■ Figure 6.9 Components of force acting on an arbitrary differential area. 290 Chapter 6 ■ Differential Analysis of Fluid Flow C C' τ xz τ xz σxx τ xy y τ xy D D' σxx B B' A A' (b) (a) x z ■ Figure 6.10 Double subscript notation for stresses. and the shearing stresses are defined as t1 lim dAS0 dF1 dA and dF2 dAS0 dA t2 lim Surface forces can be expressed in terms of the shear and normal stresses. We will use s for normal stresses and t for shearing stresses. The intensity of the force per unit area at a point in a body can thus be characterized by a normal stress and two shearing stresses, if the orientation of the area is specified. For purposes of analysis it is usually convenient to reference the area to the coordinate system. For example, for the rectangular coordinate system shown in Fig. 6.10 we choose to consider the stresses acting on planes parallel to the coordinate planes. On the plane ABCD of Fig. 6.10a, which is parallel to the y–z plane, the normal stress is denoted sxx and the shearing stresses are denoted as txy and txz. To easily identify the particular stress component we use a double subscript notation. The first subscript indicates the direction of the normal to the plane on which the stress acts, and the second subscript indicates the direction of the stress. Thus, normal stresses have repeated subscripts, whereas the subscripts for the shearing stresses are always different. It is also necessary to establish a sign convention for the stresses. We define the positive direction for the stress as the positive coordinate direction on the surfaces for which the outward normal is in the positive coordinate direction. This is the case illustrated in Fig. 6.10a where the outward normal to the area ABCD is in the positive x direction. The positive directions for sxx, txy, and txz are as shown in Fig. 6.10a. If the outward normal points in the negative coordinate direction, as in Fig. 6.10b for the area A¿B¿C¿D¿, then the stresses are considered positive if directed in the negative coordinate directions. Thus, the stresses shown in Fig. 6.10b are considered to be positive when directed as shown. Note that positive normal stresses are tensile stresses; that is, they tend to “stretch” the material. It should be emphasized that the state of stress at a point in a material is not completely defined by simply three components of a “stress vector.” This follows, since any particular stress vector depends on the orientation of the plane passing through the point. However, it can be shown that the normal and shearing stresses acting on any plane passing through a point can be expressed in terms of the stresses acting on three orthogonal planes passing through the point 1Ref. 22. We now can express the surface forces acting on a small cubical element of fluid in terms of the stresses acting on the faces of the element as shown in Fig. 6.11. It is expected that in general the stresses will vary from point to point within the flow field. Thus, through the use of Taylor series expansions we will express the stresses on the various faces in terms of the corresponding stresses at the center of the element of Fig. 6.11 and their gradients in the coordinate directions. For simplicity only the forces in the x direction are shown. Note that the stresses must be multiplied by the area on which they act to obtain the force. Summing all these forces in the x direction yields dFsx a 0tyx 0tzx 0sxx b dx dy dz 0x 0y 0z (6.48a) 6.3 (τ yx Conservation of Linear Momentum 291 ∂τ yx __ δy δ x δ z + ____ ∂y 2 ( (τ zx ∂τ z x __ δ z δx δy – ____ ∂z 2 ( δy (σ xx (τ ∂σ xx δ__x δ yδ z – ____ ∂x 2 (σ ( zx + ∂τz x __ δz δx δy ____ ∂z 2 xx + ∂σxx δ__x ____ δ y δz ∂x 2 ( δz ( δx y (τ ∂τ yx __ δy δ x δ z – ____ ∂y 2 ( yx x z ■ Figure 6.11 Surface forces in the x direction acting on a fluid element. for the resultant surface force in the x direction. In a similar manner the resultant surface forces in the y and z directions can be obtained and expressed as dFsy a 0txy dFsz a 0x 0txz 0x 0syy 0y 0tyz 0y 0tzy 0z 0szz 0z b dx dy dz (6.48b) b dx dy dz (6.48c) The resultant surface force can now be expressed as dFs dFsxˆi dFsy jˆ dFszkˆ (6.49) and this force combined with the body force, dFb, yields the resultant force, dF, acting on the differential mass, dm. That is, dF dFs dFb. 6.3.2 Equations of Motion The expressions for the body and surface forces can now be used in conjunction with Eq. 6.45 to develop the equations of motion. In component form Eq. 6.45 can be written as dFx dm ax dFy dm ay dFz dm az where dm r dx dy dz, and the acceleration components are given by Eq. 6.3. It now follows 1using Eqs. 6.47 and 6.48 for the forces on the element2 that The motion of a fluid is governed by a set of nonlinear differential equations. rgx rgy rgz 0tyx 0tzx 0sxx 0u 0u 0u 0u ra u v w b 0x 0y 0z 0t 0x 0y 0z 0txy 0x 0txz 0x 0syy 0y 0tyz 0y 0tzy 0z 0szz 0z (6.50a) ra 0v 0v 0y 0v u v w b 0t 0x 0y 0z (6.50b) ra 0w 0w 0w 0w u v w b 0t 0x 0y 0z (6.50c) where the element volume dx dy dz cancels out. Equations 6.50 are the general differential equations of motion for a fluid. In fact, they are applicable to any continuum 1solid or fluid2 in motion or at rest. However, before we can use the equations to solve specific problems, some additional information about the stresses must be obtained. 292 Chapter 6 ■ Differential Analysis of Fluid Flow Otherwise, we will have more unknowns 1all of the stresses and velocities and the density2 than equations. It should not be too surprising that the differential analysis of fluid motion is complicated. We are attempting to describe, in detail, complex fluid motion. 6.4 Inviscid Flow z p = –σzz y p = – σxx p = – σyy As is discussed in Section 1.6, shearing stresses develop in a moving fluid because of the viscosity of the fluid. We know that for some common fluids, such as air and water, the viscosity is small, and therefore it seems reasonable to assume that under some circumstances we may be able to simply neglect the effect of viscosity 1and thus shearing stresses2. Flow fields in which the shearing stresses are assumed to be negligible are said to be inviscid, nonviscous, or frictionless. These terms are used interchangeably. As is discussed in Section 2.1, for fluids in which there are no shearing stresses the normal stress at a point is independent of direction—that is, sxx syy szz. In this instance we define the pressure, p, as the negative of the normal stress so that, as indicated by the figure in the margin, p sxx syy szz x The negative sign is used so that a compressive normal stress 1which is what we expect in a fluid2 will give a positive value for p. In Chapter 3 the inviscid flow concept was used in the development of the Bernoulli equation, and numerous applications of this important equation were considered. In this section we will again consider the Bernoulli equation and will show how it can be derived from the general equations of motion for inviscid flow. 6.4.1 Euler’s Equations of Motion For an inviscid flow in which all the shearing stresses are zero, and the normal stresses are replaced by p, the general equations of motion 1Eqs. 6.502 reduce to Euler’s equations of motion apply to an inviscid flow field. rgx 0p 0u 0u 0u 0u ra u v w b 0x 0t 0x 0y 0z (6.51a) rgy 0p 0v 0v 0v 0v ra u v w b 0y 0t 0x 0y 0z (6.51b) rgz 0p 0w 0w 0w 0w ra u v w b 0z 0t 0x 0y 0z (6.51c) These equations are commonly referred to as Euler’s equations of motion, named in honor of Leonhard Euler 11707–17832, a famous Swiss mathematician who pioneered work on the relationship between pressure and flow. In vector notation Euler’s equations can be expressed as 0V rg ⵱p r c 1V ⴢ ⵱2V d (6.52) 0t Although Eqs. 6.51 are considerably simpler than the general equations of motion, Eqs. 6.50, they are still not amenable to a general analytical solution that would allow us to determine the pressure and velocity at all points within an inviscid flow field. The main difficulty arises from the nonlinear velocity terms 1u 0u 0x, v 0u 0y, etc.2, which appear in the convective acceleration, or the (V ⴢ ⵱)V term. Because of these terms, Euler’s equations are nonlinear partial differential equations for which we do not have a general method of solving. However, under some circumstances we can use them to obtain useful information about inviscid flow fields. For example, as shown in the following section we can integrate Eq. 6.52 to obtain a relationship 1the Bernoulli equation2 between elevation, pressure, and velocity along a streamline. 6.4.2 The Bernoulli Equation In Section 3.2 the Bernoulli equation was derived by a direct application of Newton’s second law to a fluid particle moving along a streamline. In this section we will again derive this important 6.4 Inviscid Flow 293 Streamline ds Streamline z z y y x x ■ Figure 6.12 The notation for differential length along a streamline. equation, starting from Euler’s equations. Of course, we should obtain the same result since Euler’s equations simply represent a statement of Newton’s second law expressed in a general form that is useful for flow problems and maintains the restriction of zero viscosity. We will restrict our attention to steady flow so Euler’s equation in vector form becomes rg ⵱p r1V ⴢ ⵱2V (6.53) z g y g gz ^ ^ ^ 0i 0j gk We wish to integrate this differential equation along some arbitrary streamline 1Fig. 6.122 and select the coordinate system with the z axis vertical 1with “up” being positive2 so that, as shown by the figure in the margin, the acceleration of gravity vector can be expressed as g g⵱z where g is the magnitude of the acceleration of gravity vector. Also, it will be convenient to use the vector identity 1V ⴢ ⵱2V 12⵱1V ⴢ V2 V ⴛ 1⵱ ⴛ V2 Equation 6.53 can now be written in the form rg⵱z ⵱p r ⵱1V ⴢ V2 rV ⴛ 1⵱ ⴛ V2 2 and this equation can be rearranged to yield ⵱p 1 ⵱1V 2 2 g⵱z V ⴛ 1⵱ ⴛ V2 r 2 We next take the dot product of each term with a differential length ds along a streamline 1Fig. 6.122. Thus, ⵱p 1 ⴢ ds ⵱1V 2 2 ⴢ ds g⵱z ⴢ ds 3V ⴛ 1⵱ ⴛ V2 4 ⴢ ds r 2 V ( Δ V) Δ x ds V V Euler’s equations can be arranged to give the relationship among pressure, velocity, and elevation for inviscid fluids. (6.54) Since ds has a direction along the streamline, the vectors ds and V are parallel. However, as shown by the figure in the margin, the vector V ⴛ 1⵱ ⴛ V2 is perpendicular to V 1why?2, so it follows that 3V ⴛ 1⵱ ⴛ V2 4 ⴢ ds 0 Recall also that the dot product of the gradient of a scalar and a differential length gives the differential change in the scalar in the direction of the differential length. That is, with ds dx ˆi dy jˆ dz kˆ we can write ⵱p ⴢ ds 10p 0x2 dx 10p 0y2dy 10p 0z2dz dp. Thus, Eq. 6.54 becomes dp 1 d1V 2 2 g dz 0 r 2 (6.55) where the change in p, V, and z is along the streamline. Equation 6.55 can now be integrated to give 冮 dp V2 gz constant r 2 (6.56) which indicates that the sum of the three terms on the left side of the equation must remain a constant along a given streamline. Equation 6.56 is valid for both compressible and incompressible 294 Chapter 6 ■ Differential Analysis of Fluid Flow inviscid flows, but for compressible fluids the variation in r with p must be specified before the first term in Eq. 6.56 can be evaluated. For inviscid, incompressible fluids 1commonly called ideal fluids2 Eq. 6.56 can be written as p V2 gz constant along a streamline r 2 (6.57) and this equation is the Bernoulli equation used extensively in Chapter 3. It is often convenient to write Eq. 6.57 between two points 112 and 122 along a streamline and to express the equation in the “head” form by dividing each term by g so that V 21 V 22 p1 p2 z1 z2 g g 2g 2g (6.58) It should be again emphasized that the Bernoulli equation, as expressed by Eqs. 6.57 and 6.58, is restricted to the following: inviscid flow steady flow incompressible flow flow along a streamline You may want to go back and review some of the examples in Chapter 3 that illustrate the use of the Bernoulli equation. 6.4.3 Irrotational Flow The vorticity is zero in an irrotational flow field. If we make one additional assumption—that the flow is irrotational—the analysis of inviscid flow problems is further simplified. Recall from Section 6.1.3 that the rotation of a fluid element is equal to 12 1⵱ ⴛ V2, and an irrotational flow field is one for which ⵱ ⴛ V 0 1i.e., the curl of velocity is zero2. Since the vorticity, Z, is defined as ⵱ ⴛ V, it also follows that in an irrotational flow field the vorticity is zero. The concept of irrotationality may seem to be a rather strange condition for a flow field. Why would a flow field be irrotational? To answer this question we note that if 12 1⵱ ⴛ V2 0, then each of the components of this vector, as are given by Eqs. 6.12, 6.13, and 6.14, must be equal to zero. Since these components include the various velocity gradients in the flow field, the condition of irrotationality imposes specific relationships among these velocity gradients. For example, for rotation about the z axis to be zero, it follows from Eq. 6.12 that vz 1 0v 0u a b0 2 0x 0y and, therefore, 0v 0u 0x 0y (6.59) 0w 0v 0y 0z (6.60) 0u 0w 0z 0x (6.61) Similarly from Eqs. 6.13 and 6.14 A general flow field would not satisfy these three equations. However, a uniform flow as is illustrated in Fig. 6.13 does. Since u U 1a constant2, v 0, and w 0, it follows that Eqs. 6.59, 6.60, and 6.61 are all satisfied. Therefore, a uniform flow field 1in which there are no velocity gradients2 is certainly an example of an irrotational flow. Uniform flows by themselves are not very interesting. However, many interesting and important flow problems include uniform flow in some part of the flow field. Two examples are 6.4 Inviscid Flow 295 u = U (constant) v=0 y w=0 x ■ Figure 6.13 Uniform flow in the x direction. z Flow fields involving real fluids often include both regions of negligible shearing stresses and regions of significant shearing stresses. shown in Fig. 6.14. In Fig. 6.14a a solid body is placed in a uniform stream of fluid. Far away from the body the flow remains uniform, and in this far region the flow is irrotational. In Fig. 6.14b, flow from a large reservoir enters a pipe through a streamlined entrance where the velocity distribution is essentially uniform. Thus, at the entrance the flow is irrotational. For an inviscid fluid there are no shearing stresses—the only forces acting on a fluid element are its weight and pressure forces. Since the weight acts through the element center of gravity, and the pressure acts in a direction normal to the element surface, neither of these forces can cause the element to rotate. Therefore, for an inviscid fluid, if some part of the flow field is irrotational, the fluid elements emanating from this region will not take on any rotation as they progress through the flow field. This phenomenon is illustrated in Fig. 6.14a in which fluid elements flowing far away from the body have irrotational motion, and as they flow around the body the motion remains irrotational except very near the boundary. Near the boundary the velocity changes rapidly from zero at the boundary 1no-slip condition2 to some relatively large value in a short distance from the boundary. This rapid change in velocity gives rise to a large velocity gradient normal to the boundary and produces significant shearing stresses, even though the viscosity is small. Of course if we had a truly inviscid fluid, the fluid would simply “slide” past the boundary and the flow would be irrotational everywhere. But this is not the case for real fluids, so we will typically have a layer 1usually relatively thin2 near any fixed surface in a moving stream in which shearing stresses are not negligible. This layer is called the boundary layer. Outside the boundary layer the flow can be treated as an irrotational flow. Another possible consequence of the boundary layer is that the main stream may “separate” from the surface and form a wake downstream from the body. (See the Boundary layer Uniform approach velocity Separation Wake Inviscid irrotational flow region (a) Entrance region Uniform entrance velocity Fully developed region Inviscid, irrotational core Boundary layer (b) ■ Figure 6.14 Various regions of flow: (a) around bodies; (b) through channels. 296 Chapter 6 ■ Differential Analysis of Fluid Flow photographs at the beginning of Chapters 7, 9, and 11.) The wake would include a region of slow, perhaps randomly moving fluid. To completely analyze this type of problem it is necessary to consider both the inviscid, irrotational flow outside the boundary layer, and the viscous, rotational flow within the boundary layer and to somehow “match” these two regions. This type of analysis is considered in Chapter 9. As is illustrated in Fig. 6.14b, the flow in the entrance to a pipe may be uniform 1if the entrance is streamlined2 and thus will be irrotational. In the central core of the pipe the flow remains irrotational for some distance. However, a boundary layer will develop along the wall and grow in thickness until it fills the pipe. Thus, for this type of internal flow there will be an entrance region in which there is a central irrotational core, followed by a so-called fully developed region in which viscous forces are dominant. The concept of irrotationality is completely invalid in the fully developed region. This type of internal flow problem is considered in detail in Chapter 8. The two preceding examples are intended to illustrate the possible applicability of irrotational flow to some “real fluid” flow problems and to indicate some limitations of the irrotationality concept. We proceed to develop some useful equations based on the assumptions of inviscid, incompressible, irrotational flow, with the admonition to use caution when applying the equations. 6.4.4 The Bernoulli Equation for Irrotational Flow In the development of the Bernoulli equation in Section 6.4.2, Eq. 6.54 was integrated along a streamline. This restriction was imposed so the right side of the equation could be set equal to zero; that is, 3V ⴛ 1⵱ ⴛ V2 4 ⴢ ds 0 1since ds is parallel to V2. However, for irrotational flow, ⵱ ⴛ V 0 , so the right side of Eq. 6.54 is zero regardless of the direction of ds. We can now follow the same procedure used to obtain Eq. 6.55, where the differential changes dp, d1V 2 2, and dz can be taken in any direction. Integration of Eq. 6.55 again yields 冮 dp V2 gz constant r 2 (6.62) where for irrotational flow the constant is the same throughout the flow field. Thus, for incompressible, irrotational flow the Bernoulli equation can be written as p1 p2 V 21 V 22 z1 z2 g g 2g 2g The Bernoulli equation can be applied between any two points in an irrotational flow field. (6.63) between any two points in the flow field. Equation 6.63 is exactly the same form as Eq. 6.58 but is not limited to application along a streamline. However, Eq. 6.63 is restricted to inviscid flow steady flow incompressible flow irrotational flow It may be worthwhile to review the use and misuse of the Bernoulli equation for rotational flow as is illustrated in Example 3.18. 6.4.5 The Velocity Potential For an irrotational flow the velocity gradients are related through Eqs. 6.59, 6.60, and 6.61. It follows that in this case the velocity components can be expressed in terms of a scalar function f1x, y, z, t2 as u 0f 0x v 0f 0y w 0f 0z (6.64) 6.4 Inviscid Flow 297 where f is called the velocity potential. Direct substitution of these expressions for the velocity components into Eqs. 6.59, 6.60, and 6.61 will verify that a velocity field defined by Eqs. 6.64 is indeed irrotational. In vector form, Eqs. 6.64 can be written as V ⵱f (6.65) so that for an irrotational flow the velocity is expressible as the gradient of a scalar function f. The velocity potential is a consequence of the irrotationality of the flow field, whereas the stream function is a consequence of conservation of mass (see Section 6.2.3). It is to be noted, however, that the velocity potential can be defined for a general three-dimensional flow, whereas a single stream function is restricted to two-dimensional flows. For an incompressible fluid we know from conservation of mass that ⵱ⴢV0 Inviscid, incompressible, irrotational flow fields are governed by Laplace’s equation and are called potential flows. 2 0 Streamlines and therefore for incompressible, irrotational flow 1with V ⵱f2 it follows that ⵱2f 0 where ⵱ 1 2 ⵱ ⴢ ⵱ 1 2 is the Laplacian operator. In Cartesian coordinates (6.66) 2 0 2f 0x2 0 2f 0y2 0 2f 0z2 0 This differential equation arises in many different areas of engineering and physics and is called Laplace’s equation. Thus, inviscid, incompressible, irrotational flow fields are governed by Laplace’s equation. This type of flow is commonly called a potential flow. To complete the mathematical formulation of a given problem, boundary conditions have to be specified. These are usually velocities specified on the boundaries of the flow field of interest. It follows that if the potential function can be determined, then the velocity at all points in the flow field can be determined from Eq. 6.64, and the pressure at all points can be determined from the Bernoulli equation 1Eq. 6.632. Although the concept of the velocity potential is applicable to both steady and unsteady flow, we will confine our attention to steady flow. Potential flows, governed by Eqs. 6.64 and 6.66, are irrotational flows. That is, the vorticity is zero throughout. If vorticity is present (e.g., boundary layer, wake), then the flow cannot be described by Laplace’s equation. The figure in the margin illustrates a flow in which the vorticity is not zero in two regions—the separated region behind the bump and the boundary layer next to the solid surface. This is discussed in detail in Chapter 9. For some problems it will be convenient to use cylindrical coordinates, r, u, and z. In this coordinate system the gradient operator is 2 0 ⵱1 2 01 2 0r êr 01 2 1 01 2 êu ê r 0u 0z z (6.67) so that Vorticity contours ⵱f 0f 0f 1 0f ê ê ê r 0u u 0r r 0z z (6.68) where f f1r, u, z2. Since V vrêr vuêu vzêz (6.69) it follows for an irrotational flow 1with V ⵱f2 vr 0f 0r vu 1 0f r 0u vz 0f 0z (6.70) Also, Laplace’s equation in cylindrical coordinates is 0f 0 2f 1 0 1 0 2f ar b 2 2 2 0 r 0r 0r r 0u 0z (6.71) 298 Chapter 6 ■ Differential Analysis of Fluid Flow E XAMPLE 6.4 Velocity Potential and Inviscid Flow Pressure GIVEN The two-dimensional flow of a nonviscous, incompressible fluid in the vicinity of the 90° corner of Fig. E6.4a is described by the stream function that is, there is no difference in elevation between points 112 and 122. FIND c 2r2 sin 2u where c has units of m2 s when r is in meters. Assume the fluid density is 103 kgm3 and the x – y plane is horizontal— (a) Determine, if possible, the corresponding velocity potential. (b) If the pressure at point 112 on the wall is 30 kPa, what is the pressure at point 122? y y Streamline ( ψ = constant) Equipotential line ( φ = constant) (2) r 0.5 m θ (1) x 1m x (a) (b) α α (c) ■ Figure E6.4 SOLUTION (a) The radial and tangential velocity components can be obtained from the stream function as 1see Eq. 6.422 1 0c vr 4r cos 2u r 0u and therefore by integration f 2r2 cos 2u f1 1u2 where f1 1u2 is an arbitrary function of u. Similarly vu and vu 0c 4r sin 2u 0r Since vr 0f 0r it follows that 0f 4r cos 2u 0r (1) 1 0f 4r sin 2u r 0u and integration yields f 2r2 cos 2u f2 1r2 (2) f 2r2 cos 2u C (Ans) where f2 1r2 is an arbitrary function of r. To satisfy both Eqs. 1 and 2, the velocity potential must have the form where C is an arbitrary constant. As is the case for stream functions, the specific value of C is not important, and it is customary to let C 0 so that the velocity potential for this corner flow is f 2r2 cos 2u (Ans) 6.5 COMMENT In the statement of this problem, it was implied by the wording “if possible” that we might not be able to find a corresponding velocity potential. The reason for this concern is that we can always define a stream function for two-dimensional flow, but the flow must be irrotational if there is a corresponding velocity potential. Thus, the fact that we were able to determine a velocity potential means that the flow is irrotational. Several streamlines and lines of constant f are plotted in Fig. E6.4b. These two sets of lines are orthogonal. The reason why streamlines and lines of constant f are always orthogonal is explained in Section 6.5. (b) Since we have an irrotational flow of a nonviscous, incompressible fluid, the Bernoulli equation can be applied between any two points. Thus, between points 112 and 122 with no elevation change This result indicates that the square of the velocity at any point depends only on the radial distance, r, to the point. Note that the constant, 16, has units of s2. Thus, V 21 116 s 2 211 m2 2 16 m2s2 and V 22 116 s 2 210.5 m2 2 4 m2s2 Substitution of these velocities into Eq. 3 gives 103 kg m3 116 m2 s2 4 m2s2 2 p2 30 103 Nm2 2 (Ans) 36 kPa COMMENT The stream function used in this example could also be expressed in Cartesian coordinates as or c 4xy or (3) since x r cos u and y r sin u. However, in the cylindrical polar form the results can be generalized to describe flow in the vicinity of a corner of angle a 1see Fig. E6.4c2 with the equations Since V 2 v 2r v 2u V 2 14r cos 2u2 2 14r sin 2u2 2 16r 2 1cos2 2u sin2 2u2 16r 2 c Arpa sin pu a f Arpa cos pu a and it follows that for any point within the flow field 6.5 299 c 2r 2 sin 2u 4r 2 sin u cos u p1 p2 V 21 V 22 g g 2g 2g r p2 p1 1V 21 V 22 2 2 Some Basic, Plane Potential Flows where A is a constant. Some Basic, Plane Potential Flows For potential flow, basic solutions can be simply added to obtain more complicated solutions. y V u= v= ∂φ ∂x x vr = ∂φ ∂y u V θ 0f 0x v 0f 0y (6.72) 1 0f r 0u (6.73) or by using cylindrical coordinates vr ∂φ ∂r ∂φ vθ = 1r ∂θ r A major advantage of Laplace’s equation is that it is a linear partial differential equation. Since it is linear, various solutions can be added to obtain other solutions—that is, if f1 1x, y, z2 and f2 1x, y, z2 are two solutions to Laplace’s equation, then f3 f1 f2 is also a solution. The practical implication of this result is that if we have certain basic solutions we can combine them to obtain more complicated and interesting solutions. In this section several basic velocity potentials, which describe some relatively simple flows, will be determined. In the next section these basic potentials will be combined to represent complicated flows. For simplicity, only plane 1two-dimensional2 flows will be considered. In this case, by using Cartesian coordinates 0f 0r vu as shown by the figure in the margin. Since we can define a stream function for plane flow, we can also let u 0c 0y v 0c 0x (6.74) 300 Chapter 6 ■ Differential Analysis of Fluid Flow or vr 1 0c r 0u vu 0c 0r (6.75) where the stream function was previously defined in Eqs. 6.37 and 6.42. We know that by defining the velocities in terms of the stream function, conservation of mass is identically satisfied. If we now impose the condition of irrotationality, it follows from Eq. 6.59 that 0u 0v 0y 0x and in terms of the stream function 0c 0 0c 0 a b a b 0y 0y 0x 0x or 0 2c 0x2 0 2c 0y2 0 Thus, for a plane irrotational flow we can use either the velocity potential or the stream function— both must satisfy Laplace’s equation in two dimensions. It is apparent from these results that the velocity potential and the stream function are somehow related. We have previously shown that lines of constant c are streamlines; that is, dy v ` dx along cconstant u (6.76) The change in f as we move from one point 1x , y2 to a nearby point 1x dx, y dy2 is given by the relationship y a b df b a a_ × – b_ = –1 b a ( ) Along a line of constant f we have df 0 so that dy u ` v dx along fconstant x ψ Streamwise acceleration 0f 0f dx dy u dx v dy 0x 0y (6.77) A comparison of Eqs. 6.76 and 6.77 shows that lines of constant f 1called equipotential lines2 are orthogonal to lines of constant c 1streamlines2 at all points where they intersect. 1Recall that two lines are orthogonal if the product of their slopes is 1, as illustrated by the figure in the margin.2 For any potential flow field a “flow net ” can be drawn that consists of a family of streamlines and equipotential lines. The flow net is useful in visualizing flow patterns and can be used to obtain graphical solutions by sketching in streamlines and equipotential lines and adjusting the lines until the lines are approximately orthogonal at all points where they intersect. An example of a flow net is shown in Fig. 6.15. Velocities can be estimated from the flow net, since the velocity is inversely proportional to the streamline spacing, as shown by the figure in the margin. Thus, for example, from Fig. 6.15 we can see that the velocity near the inside corner will be higher than the velocity along the outer part of the bend. (See the photographs at the beginning of Chapters 3 and 6.) 6.5.1 Uniform Flow The simplest plane flow is one for which the streamlines are all straight and parallel, and the magnitude of the velocity is constant. This type of flow is called a uniform flow. For example, consider a uniform flow in the positive x direction as is illustrated in Fig. 6.16a. In this instance, u U and v 0, and in terms of the velocity potential Streamwise deceleration 0f U 0x 0f 0 0y 6.5 Some Basic, Plane Potential Flows 301 Equipotential line ( φ = constant) d1 > d d V V1 < V d V1 V V2 d2 < d V2 > V Streamline (ψ = constant) ■ Figure 6.15 Flow net for a 90ⴗ bend. (Adapted from Ref. 3.) These two equations can be integrated to yield f ⫽ Ux ⫹ C where C is an arbitrary constant, which can be set equal to zero. Thus, for a uniform flow in the positive x direction f ⫽ Ux (6.78) The corresponding stream function can be obtained in a similar manner, since 0c ⫽U 0y 0c ⫽0 0x and, therefore, c ⫽ Uy (6.79) These results can be generalized to provide the velocity potential and stream function for a uniform flow at an angle a with the x axis, as in Fig. 6.16b. For this case f ⫽ U1x cos a ⫹ y sin a2 (6.80) c ⫽ U1y cos a ⫺ x sin a2 (6.81) and 6.5.2 Source and Sink Consider a fluid flowing radially outward from a line through the origin perpendicular to the x–y plane as is shown in Fig. 6.17. Let m be the volume rate of flow emanating from the line 1per unit length2, and therefore to satisfy conservation of mass 12pr2vr ⫽ m or vr ⫽ y = y U = m 2pr 1 = 1 2 = 3 = 4 U x (b) 4 φ = φ2 φ = φ1 φ = φ2 x (a) 3 = α φ = φ1 2 = = ■ Figure 6.16 Uniform flow: (a) in the x direction; (b) in an arbitrary direction, A. 302 Chapter 6 ■ Differential Analysis of Fluid Flow y ψ = constant φ = constant vr r θ x ■ Figure 6.17 The streamline pattern for a source. A source or sink represents a purely radial flow. Also, since the flow is a purely radial flow, vu 0, the corresponding velocity potential can be obtained by integrating the equations 0f m 0r 2pr 1 0f 0 r 0u It follows that f vr ~ 1r vr m ln r 2p (6.82) If m is positive, the flow is radially outward, and the flow is considered to be a source flow. If m is negative, the flow is toward the origin, and the flow is considered to be a sink flow. The flowrate, m, is the strength of the source or sink. As shown by the figure in the margin, at the origin where r 0 the velocity becomes infinite, which is of course physically impossible. Thus, sources and sinks do not really exist in real flow fields, and the line representing the source or sink is a mathematical singularity in the flow field. However, some real flows can be approximated at points away from the origin by using sources or sinks. Also, the velocity potential representing this hypothetical flow can be combined with other basic velocity potentials to approximately describe some real flow fields. This idea is further discussed in Section 6.6. The stream function for the source can be obtained by integrating the relationships r vr 1 0c m r 0u 2pr vu 0c 0 0r to yield c m u 2p (6.83) It is apparent from Eq. 6.83 that the streamlines 1lines of c constant2 are radial lines, and from Eq. 6.82 that the equipotential lines 1lines of f constant2 are concentric circles centered at the origin. E XAMPLE 6.5 Potential Flow—Sink GIVEN A nonviscous, incompressible fluid flows between y wedge-shaped walls into a small opening as shown in Fig. E6.5. The velocity potential 1in ft2 s2, that approximately describes this flow, is f 2 ln r FIND Determine the volume rate of flow 1per unit length2 into the opening. vr π _ 6 R r θ x ■ Figure E6.5 6.5 Some Basic, Plane Potential Flows 303 SOLUTION COMMENT Note that the radius R is arbitrary since the flowrate crossing any curve between the two walls must be the same. The negative sign indicates that the flow is toward the opening, that is, in the negative radial direction. The components of velocity are vr 0f 2 r 0r vu 1 0f 0 r 0u which indicates we have a purely radial flow. The flowrate per unit width, q, crossing the arc of length Rp6 can thus be obtained by integrating the expression q 冮 p6 vr R du 0 冮 p6 0 2 a b R du R p 1.05 ft2s 3 (Ans) 6.5.3 Vortex We next consider a flow field in which the streamlines are concentric circles—that is, we interchange the velocity potential and stream function for the source. Thus, let A vortex represents a flow in which the streamlines are concentric circles. vθ r (6.84) c K ln r (6.85) and where K is a constant. In this case the streamlines are concentric circles as are illustrated in Fig. 6.18, with vr 0 and vu vθ ~ 1 r f Ku 0c 1 0f K r 0u r 0r (6.86) This result indicates that the tangential velocity varies inversely with the distance from the origin, as shown by the figure in the margin, with a singularity occurring at r 0 1where the velocity becomes infinite2. It may seem strange that this vortex motion is irrotational 1and it is since the flow field is described by a velocity potential2. However, it must be recalled that rotation refers to the orientation of a fluid element and not the path followed by the element. Thus, for an irrotational vortex, if a pair of small sticks were placed in the flow field at location A, as indicated in Fig. 6.19a, the sticks would rotate as they move to location B. One of the sticks, the one that is aligned along the streamline, would follow a circular path and rotate in a counterclockwise y ψ = constant vθ r θ x φ = constant ■ Figure 6.18 The streamline pattern for a vortex. 304 Chapter 6 ■ Differential Analysis of Fluid Flow B _ vθ ~ 1 r A vθ ~ r A B r r (a) (b) ■ Figure 6.19 Motion of fluid element from A to B: (a) for irrotational (free) vortex; (b) for rotational (forced) vortex. direction. The other stick would rotate in a clockwise direction due to the nature of the flow field—that is, the part of the stick nearest the origin moves faster than the opposite end. Although both sticks are rotating, the average angular velocity of the two sticks is zero since the flow is irrotational. If the fluid were rotating as a rigid body, such that vu K1r where K1 is a constant, then sticks similarly placed in the flow field would rotate as is illustrated in Fig. 6.19b. This type of vortex motion is rotational and cannot be described with a velocity potential. The rotational vortex is commonly called a forced vortex, whereas the irrotational vortex is usually called a free vortex. The swirling motion of the water as it drains from a bathtub is similar to that of a free vortex, whereas the motion of a liquid contained in a tank that is rotated about its axis with angular velocity v corresponds to a forced vortex. A combined vortex is one with a forced vortex as a central core and a velocity distribution corresponding to that of a free vortex outside the core. Thus, for a combined vortex Vortex motion can be either rotational or irrotational. vu vr r r0 (6.87) K r r 7 r0 (6.88) and vu where K and v are constants and r0 corresponds to the radius of the central core. The pressure distribution in both the free and forced vortex was previously considered in Example 3.3. (See Fig. E6.6a for an approximation of this type of flow.) F l u i d s i n Some hurricane facts One of the most interesting, yet potentially devastating, naturally occurring fluid flow phenomenon is a hurricane. Broadly speaking, a hurricane is a rotating mass of air circulating around a low-pressure central core. In some respects the motion is similar to that of a free vortex. The Caribbean and Gulf of Mexico experience the most hurricanes, with the official hurricane season being from June 1 to November 30. Hurricanes are usually 300 to 400 miles wide and are structured around a central eye in which the air is relatively calm. The eye is surrounded by an eye wall, which is the region of strongest winds and precipitation. As one goes from the eye wall to the eye, the wind speeds decrease sharply and within the eye the air is relatively calm and clear of t h e N e w s clouds. However, in the eye the pressure is at a minimum and may be 10% less than standard atmospheric pressure. This low pressure creates strong downdrafts of dry air from above. Hurricanes are classified into five categories based on their wind speeds: Category Category Category Category Category one—74–95 mph two—96–110 mph three—111–130 mph four—131–155 mph five—greater than 155 mph (See Problem 6.60.) 6.5 ds Arbitrary curve C Some Basic, Plane Potential Flows 305 V ■ Figure 6.20 The notation for determining circulation around closed curve C. A mathematical concept commonly associated with vortex motion is that of circulation. The circulation, , is defined as the line integral of the tangential component of the velocity taken around a closed curve in the flow field. In equation form, can be expressed as 冮 ≠ ˇ䊊 V ⴢ ds (6.89) C where the integral sign means that the integration is taken around a closed curve, C, in the counterclockwise direction, and ds is a differential length along the curve as is illustrated in Fig. 6.20. For an irrotational flow, V ⵱f so that V ⴢ ds ⵱f ⴢ ds df and, therefore, 冮 ≠ ˇ䊊 df 0 C This result indicates that for an irrotational flow the circulation will generally be zero. (Chapter 9 has further discussion of circulation in real flows.) However, if there are singularities enclosed within the curve the circulation may not be zero. For example, for the free vortex with vu Kr the circulation around the circular path of radius r shown in Fig. 6.21 is ≠ 冮 2p 0 The numerical value of the circulation may depend on the particular closed path considered. K 1r du2 2pK r which shows that the circulation is nonzero and the constant K 2p. However, for irrotational flows the circulation around any path that does not include a singular point will be zero. This can be easily confirmed for the closed path ABCD of Fig. 6.21 by evaluating the circulation around that path. The velocity potential and stream function for the free vortex are commonly expressed in terms of the circulation as f u 2p (6.90) ln r 2p (6.91) and V6.4 Vortex in a beaker c The concept of circulation is often useful when evaluating the forces developed on bodies immersed in moving fluids. This application will be considered in Section 6.6.3. vθ ds B C r A D dθ θ ■ Figure 6.21 Circulation around various paths in a free vortex. 306 Chapter 6 ■ Differential Analysis of Fluid Flow E XAMPLE 6.6 Potential Flow—Free Vortex GIVEN A liquid drains from a large tank through a small opening as illustrated in Fig. E6.6a. A vortex forms whose velocity distribution away from the tank opening can be approximated as that of a free vortex having a velocity potential f u 2p FIND Determine an expression relating the surface shape to the strength of the vortex as specified by the circulation . SOLUTION Since the free vortex represents an irrotational flow field, the Bernoulli equation V 21 ■ Figure E6.6a V 22 p1 p2 z1 z2 g g 2g 2g z can be written between any two points. If the points are selected at the free surface, p1 p2 0, so that r p = patm (1) y zs x V 21 V 22 zs 2g 2g (1) (2) where the free surface elevation, zs, is measured relative to a datum passing through point 112 as shown in Fig. E6.6b. The velocity is given by the equation vu 1 0f r 0u 2pr ■ Figure E6.6b We note that far from the origin at point 112, V1 vu ⬇ 0 so that Eq. 1 becomes zs 8p2r2g 2 (Ans) COMMENT The negative sign indicates that the surface falls as the origin is approached as shown in Fig. E6.6. This solution is not valid very near the origin since the predicted velocity becomes excessively large as the origin is approached. which is the desired equation for the surface profile. 6.5.4 Doublet A doublet is formed by an appropriate source–sink pair. The final, basic potential flow to be considered is one that is formed by combining a source and sink in a special way. Consider the equal strength, source–sink pair of Fig. 6.22. The combined stream function for the pair is c y P r2 θ2 m 1u u2 2 2p 1 θ r r1 θ1 x Source Sink a a ■ Figure 6.22 The combination of a source and sink of equal strength located along the x axis. 6.5 Some Basic, Plane Potential Flows 307 which can be rewritten as tan a 2pc tan u1 tan u2 b tan1u1 u2 2 m 1 tan u1 tan u2 (6.92) From Fig. 6.22 it follows that tan u1 r sin u r cos u a tan u2 r sin u r cos u a and These results substituted into Eq. 6.92 give tan a y 2pc 2ar sin u b 2 m r a2 so that Source Sink c x m 2ar sin u tan1 a 2 b 2p r a2 (6.93) The figure in the margin shows typical streamlines for this flow. For small values of the distance a c A doublet is formed by letting a source and sink approach one another. m 2ar sin u mar sin u 2 2 2p r a p1r2 a2 2 (6.94) since the tangent of an angle approaches the value of the angle for small angles. The so-called doublet is formed by letting the source and sink approach one another 1a S 02 while increasing the strength m 1m S q 2 so that the product map remains constant. In this case, since r 1r2 a2 2 S 1r, Eq. 6.94 reduces to c K sin u r (6.95) where K, a constant equal to map, is called the strength of the doublet. The corresponding velocity potential for the doublet is f K cos u r (6.96) Plots of lines of constant c reveal that the streamlines for a doublet are circles through the origin tangent to the x axis as shown in Fig. 6.23. Just as sources and sinks are not physically realistic entities, neither are doublets. However, the doublet when combined with other basic potential flows y x ■ Figure 6.23 Streamlines for a doublet. 308 Chapter 6 ■ Differential Analysis of Fluid Flow Table 6.1 Summary of Basic, Plane Potential Flows Description of Flow Field Uniform flow at angle a with the x axis 1see Fig. 6.16b2 Source or sink 1see Fig. 6.172 m 7 0 source m 6 0 sink Free vortex 1see Fig. 6.182 7 0 counterclockwise motion 6 0 clockwise motion Doublet 1see Fig. 6.232 Velocity Potential Stream Function f U1x cos a y sin a2 c U1 y cos a x sin a2 f m ln r 2p f m u 2p c u 2p c ln r 2p Velocity Componentsa u U cos a v U sin a m 2pr vu 0 vr vr 0 vu f K cos u r c K sin u r 2pr K cos u r2 K sin u vu r2 vr a Velocity components are related to the velocity potential and stream function through the relationships: 0c 0f 0c 0f 0c 0f 1 0c 1 0f v vr vu . u 0x 0y 0y 0x 0r r 0u r 0u 0r provides a useful representation of some flow fields of practical interest. For example, we will determine in Section 6.6.3 that the combination of a uniform flow and a doublet can be used to represent the flow around a circular cylinder. Table 6.1 provides a summary of the pertinent equations for the basic, plane potential flows considered in the preceding sections. 6.6 Superposition of Basic, Plane Potential Flows As was discussed in the previous section, potential flows are governed by Laplace’s equation, which is a linear partial differential equation. It therefore follows that the various basic velocity potentials and stream functions can be combined to form new potentials and stream functions. 1Why is this true?2 Whether such combinations yield useful results remains to be seen. It is to be noted that any streamline in an inviscid flow field can be considered as a solid boundary, since the conditions along a solid boundary and a streamline are the same—that is, there is no flow through the boundary or the streamline. Thus, if we can combine some of the basic velocity potentials or stream functions to yield a streamline that corresponds to a particular body shape of interest, that combination can be used to describe in detail the flow around that body. This method of solving some interesting flow problems, commonly called the method of superposition, is illustrated in the following three sections. 6.6.1 Source in a Uniform Stream—Half-Body Flow around a half-body is obtained by the addition of a source to a uniform flow. Consider the superposition of a source and a uniform flow as shown in Fig. 6.24a. The resulting stream function is c cuniform flow csource Ur sin u m u 2p (6.97) 6.6 y U 309 Superposition of Basic, Plane Potential Flows ψ = π bU Stagnation point Stagnation point r πb θ x πb Source b b (a) (b) ■ Figure 6.24 The flow around a half-body: (a) superposition of a source and a uniform flow; (b) replacement of streamline C ⴝ PbU with solid boundary to form a half-body. and the corresponding velocity potential is f Ur cos u m ln r 2p (6.98) It is clear that at some point along the negative x axis the velocity due to the source will just cancel that due to the uniform flow and a stagnation point will be created. For the source alone V6.5 Half-body vr m 2pr so that the stagnation point will occur at x b where U m 2pb b m 2pU or (6.99) The value of the stream function at the stagnation point can be obtained by evaluating c at r b and u p, which yields from Eq. 6.97 cstagnation For inviscid flow, a streamline can be replaced by a solid boundary. Since m2 pbU 1from Eq. 6.992 it follows that the equation of the streamline passing through the stagnation point is pbU Ur sin u bUu or r Increasing m U = constant m 2 b1p u2 sin u (6.100) where u can vary between 0 and 2p. A plot of this streamline is shown in Fig. 6.24b. If we replace this streamline with a solid boundary, as indicated in the figure, then it is clear that this combination of a uniform flow and a source can be used to describe the flow around a streamlined body placed in a uniform stream. The body is open at the downstream end and thus is called a half-body. Other streamlines in the flow field can be obtained by setting c constant in Eq. 6.97 and plotting the resulting equation. A number of these streamlines are shown in Fig. 6.24b. Although the streamlines inside the body are shown, they are actually of no interest in this case, since we are concerned with the flow field outside the body. It should be noted that the singularity in the flow field 1the source2 occurs inside the body, and there are no singularities in the flow field of interest 1outside the body2. The width of the half-body asymptotically approaches 2pb mU. Thus, as shown by the figure in the margin, for a given free stream velocity, U, the width of the half-body increases as the source strength. This follows from Eq. 6.100, which can be written as y b1p u2 310 Chapter 6 ■ Differential Analysis of Fluid Flow so that as u S 0 or u S 2p the half-width approaches bp. With the stream function 1or velocity potential2 known, the velocity components at any point can be obtained. For the half-body, using the stream function given by Eq. 6.97, vr 1 0c m U cos u r 0u 2pr and vu 0c U sin u 0r Thus, the square of the magnitude of the velocity, V, at any point is V 2 v2r v2u U 2 m 2 Um cos u a b pr 2pr and since b m2pU V 2 U 2 a1 2 b b2 cos u 2 b r r (6.101) With the velocity known, the pressure at any point can be determined from the Bernoulli equation, which can be written between any two points in the flow field since the flow is irrotational. Thus, applying the Bernoulli equation between a point far from the body, where the pressure is p0 and the velocity is U, and some arbitrary point with pressure p and velocity V, it follows that p0 12 rU 2 p 12 rV 2 For a potential flow the fluid is allowed to slip past a fixed solid boundary. E XAMPLE (6.102) where elevation changes have been neglected. Equation 6.101 can now be substituted into Eq. 6.102 to obtain the pressure at any point in terms of the reference pressure, p0, and the upstream velocity, U. This relatively simple potential flow provides some useful information about the flow around the front part of a streamlined body, such as a bridge pier or strut placed in a uniform stream. An important point to be noted is that the velocity tangent to the surface of the body is not zero; that is, the fluid “slips” by the boundary. This result is a consequence of neglecting viscosity, the fluid property that causes real fluids to stick to the boundary, thus creating a “no-slip” condition. All potential flows differ from the flow of real fluids in this respect and do not accurately represent the velocity very near the boundary. However, outside this very thin boundary layer the velocity distribution will generally correspond to that predicted by potential flow theory if flow separation does not occur. 1See Section 9.2.6.2 Also, the pressure distribution along the surface will closely approximate that predicted from the potential flow theory, since the boundary layer is thin and there is little opportunity for the pressure to vary through the thin layer. In fact, as discussed in more detail in Chapter 9, the pressure distribution obtained from potential flow theory is used in conjunction with viscous flow theory to determine the nature of flow within the boundary layer. 6.7 Potential Flow—Half-body GIVEN A 40-mi/hr wind blows toward a hill arising from a 40 mi/hr plain that can be approximated with the top section of a half-body as illustrated in Fig. E6.7a. The height of the hill approaches 200 ft as shown. Assume an air density of 0.00238 slugs/ft3. FIND (a) What is the magnitude of the air velocity at a point on the hill directly above the origin [point (2)]? (b) What is the elevation of point (2) above the plain and what is the difference in pressure between point (1) on the plain far from the hill and point (2)? (2) 200 ft y (1) r (3) b ■ Figure E6.7a θ x 6.6 Superposition of Basic, Plane Potential Flows 311 SOLUTION (a) The velocity is given by Eq. 6.101 as V 2 U 2 a1 2 and V2 147.4 mihr2 a b2 b cos u 2 b r r At point 122, u p2, and since this point is on the surface 1Eq. 6.1002, r b1p u2 sin u pb 2 it follows that Thus, b2 d 1pb22 2 4 U 2 a1 2 b p COMMENTS This result indicates that the pressure on the and the magnitude of the velocity at 122 for a 40 mi兾hr approaching wind is V2 a1 (b) 4 12 b 140 mihr2 47.4 mihr p2 (Ans) The elevation at 122 above the plain is given by Eq. 1 as y2 pb 2 hill at point 122 is slightly lower than the pressure on the plain at some distance from the base of the hill with a 0.0533-psi difference due to the elevation increase and a 0.0114-psi difference due to the velocity increase. By repeating the calculations for various values of the upstream wind speed, U, the results shown in Fig. E6.7b are obtained. Note that as the wind speed increases, the pressure difference increases from the calm conditions of p1 p2 0.0533 psi. The maximum velocity along the hill surface does not occur at point 122 but farther up the hill at u 63°. At this point Vsurface 1.26U. The minimum velocity 1V 02 and maximum pressure occur at point 132, the stagnation point. Since the height of the hill approaches 200 ft and this height is equal to pb, it follows that so that 0.12 (Ans) From the Bernoulli equation 1with the y axis the vertical axis2 p1 p2 V 21 V 22 y1 y2 g g 2g 2g 0.14 0.10 p1 – p2, psi 200 ft 100 ft y2 2 10.00238 slugsft3 2 3 169.5 ft s2 2 158.7 fts2 2 4 2 10.00238 slugsft3 2132.2 fts2 21100 ft 0 ft2 9.31 lbft2 0.0647 psi (Ans) p1 p2 (1) V 22 U 2 c 1 5280 ftmi b 69.5 fts 3600 shr 0.08 0.06 (40 mph, 0.0647 psi) 0.04 p1 p2 r 2 1V V 21 2 g1y2 y1 2 2 2 and with 0.02 0 5280 ftmi b 58.7 fts V1 140 mihr2 a 3600 shr 0 20 60 40 80 100 U, mph ■ Figure E6.7b 6.6.2 Rankine Ovals The half-body described in the previous section is a body that is “open” at one end. To study the flow around a closed body, a source and a sink of equal strength can be combined with a uniform flow as shown in Fig. 6.25a. The stream function for this combination is m 1u u2 2 2p 1 (6.103) m 1ln r1 ln r2 2 2p (6.104) c Ur sin u and the velocity potential is f Ur cos u 312 Chapter 6 ■ Differential Analysis of Fluid Flow y Stagnation point U r2 θ2 θ r r1 θ1 +m –m x Source a Sink a Stagnation point ψ=0 h h a a ᐉ ( a) ᐉ (b) ■ Figure 6.25 The flow around a Rankine oval: (a) superposition of source–sink pair and a uniform flow; (b) replacement of streamline C ⴝ 0 with solid boundary to form Rankine oval. As discussed in Section 6.5.4, the stream function for the source–sink pair can be expressed as in Eq. 6.93, and, therefore, Eq. 6.103 can also be written as c Ur sin u 2ar sin u m tan1 a 2 b 2p r a2 or c Uy Rankine ovals are formed by combining a source and sink with a uniform flow. 2ay m tan1 a 2 b 2p x y2 a2 (6.105) The corresponding streamlines for this flow field are obtained by setting c constant. If several of these streamlines are plotted, it will be discovered that the streamline c 0 forms a closed body as is illustrated in Fig. 6.25b. We can think of this streamline as forming the surface of a body of length 2/ and width 2h placed in a uniform stream. The streamlines inside the body are of no practical interest and are not shown. Note that since the body is closed, all of the flow emanating from the source flows into the sink. These bodies have an oval shape and are termed Rankine ovals. Stagnation points occur at the upstream and downstream ends of the body, as are indicated in Fig. 6.25b. These points can be located by determining where along the x axis the velocity is zero. The stagnation points correspond to the points where the uniform velocity, the source velocity, and the sink velocity all combine to give a zero velocity. The locations of the stagnation points depend on the value of a, m, and U. The body half-length, / 1the value of 0x 0 that gives V 0 when y 02, can be expressed as /a 12 ma a2 b pU (6.106) or 12 m / a 1b a pUa Large Ua/m (6.107) The body half-width, h, can be obtained by determining the value of y where the y axis intersects the c 0 streamline. Thus, from Eq. 6.105 with c 0, x 0, and y h, it follows that h h2 a2 2pUh tan m 2a (6.108) or 1 h 2 h pUa h c a b 1 d tan c 2 a b d a m a 2 a Small Ua/m (6.109) Equations 6.107 and 6.109 show that both /a and ha are functions of the dimensionless parameter, pUam. Although for a given value of Uam the corresponding value of /a can be determined directly from Eq. 6.107, ha must be determined by a trial-and-error solution of Eq. 6.109. A large variety of body shapes with different length to width ratios can be obtained by using different values of Uam, as shown by the figure in the margin. As this parameter becomes large, flow around a long slender body is described, whereas for small values of the parameter, flow around a 6.6 313 Superposition of Basic, Plane Potential Flows more blunt shape is obtained. Downstream from the point of maximum body width, the surface pressure increases with distance along the surface. This condition 1called an adverse pressure gradient2 typically leads to separation of the flow from the surface, resulting in a large low-pressure wake on the downstream side of the body. Separation is not predicted by potential theory 1which simply indicates a symmetrical flow2. This is illustrated by the figure in the margin for an extreme blunt shape, showing the differences between potential and viscous flow. Therefore, the potential solution for the Rankine ovals will give a reasonable approximation of the velocity outside the thin, viscous boundary layer and the pressure distribution on the front part of the body only. Potential Flow © S. T. Thoroddsen & Stanford University 6.6.3 Flow around a Circular Cylinder Viscous Flow © Stanford University, with permission A doublet combined with a uniform flow can be used to represent flow around a circular cylinder. As was noted in the previous section, when the distance between the source–sink pair approaches zero, the shape of the Rankine oval becomes more blunt and in fact approaches a circular shape. Since the doublet described in Section 6.5.4 was developed by letting a source–sink pair approach one another, it might be expected that a uniform flow in the positive x direction combined with a doublet could be used to represent flow around a circular cylinder. This combination gives for the stream function c ⫽ Ur sin u ⫺ K sin u r (6.110) f ⫽ Ur cos u ⫹ K cos u r (6.111) and for the velocity potential In order for the stream function to represent flow around a circular cylinder, it is necessary that c ⫽ constant for r ⫽ a, where a is the radius of the cylinder. Since Eq. 6.110 can be written as c ⫽ aU ⫺ V6.6 Circular cylinder K b r sin u r2 it follows that c ⫽ 0 for r ⫽ a if U⫺ K ⫽0 a2 which indicates that the doublet strength, K, must be equal to Ua2. Thus, the stream function for flow around a circular cylinder can be expressed as c ⫽ Ur a1 ⫺ a2 b sin u r2 (6.112) a2 b cos u r2 (6.113) and the corresponding velocity potential is V6.7 Ellipse f ⫽ Ur a1 ⫹ A sketch of the streamlines for this flow field is shown in Fig. 6.26. The velocity components can be obtained from either Eq. 6.112 or 6.113 as 0f 1 0c a2 ⫽ ⫽ U a1 ⫺ 2 b cos u r 0u 0r r (6.114) 0c 1 0f a2 ⫽⫺ ⫽ ⫺U a1 ⫹ 2 b sin u r 0u 0r r (6.115) vr ⫽ and 2 vu ⫽ υ θs U On the surface of the cylinder 1r ⫽ a2 it follows from Eq. 6.114 and 6.115 that vr ⫽ 0 and 1 vus ⫽ ⫺2U sin u 0 0 ±π 2 θ ±π As shown by the figure in the margin, the maximum velocity occurs at the top and bottom of the cylinder 1u ⫽ ⫾pⲐ22 and has a magnitude of twice the upstream velocity, U. As we move away from the cylinder along the ray u ⫽ p Ⲑ2 the velocity varies, as is illustrated in Fig. 6.26. 314 Chapter 6 ■ Differential Analysis of Fluid Flow U 2U Ψ=0 r θ a ■ Figure 6.26 The flow around a circular cylinder. The pressure distribution on the cylinder surface is obtained from the Bernoulli equation. The pressure distribution on the cylinder surface is obtained from the Bernoulli equation written from a point far from the cylinder where the pressure is p0 and the velocity is U so that p0 12 rU 2 ps 12 rv2us where ps is the surface pressure. Elevation changes are neglected. Since vus 2U sin u, the surface pressure can be expressed as ps p0 12 rU 2 11 4 sin2 u2 V6.8 Circular cylinder with separation (6.116) A comparison of this theoretical, symmetrical pressure distribution expressed in dimensionless form with a typical measured distribution is shown in Fig. 6.27. This figure clearly reveals that only on the upstream part of the cylinder is there approximate agreement between the potential flow and the experimental results. Because of the viscous boundary layer that develops on the cylinder, the main flow separates from the surface of the cylinder, leading to the large difference between the theoretical, frictionless fluid solution and the experimental results on the downstream side of the cylinder 1see Chapter 92. The resultant force 1per unit length2 developed on the cylinder can be determined by integrating the pressure over the surface. From Fig. 6.28 it can be seen that Fx 冮 2p ps cos u a du (6.117) 0 3 U V6.9 Forces on suspended ball β 2 1 ps – p0 _ 1 _ ρU2 2 0 Experimental –1 –2 Theoretical (inviscid) –3 0 30 60 90 β (deg) 120 150 180 ■ Figure 6.27 A comparison of theoretical (inviscid) pressure distribution on the surface of a circular cylinder with typical experimental distribution. 6.6 315 Superposition of Basic, Plane Potential Flows y ps dθ Fy θ x Fx a ■ Figure 6.28 The notation for determining lift and drag on a circular cylinder. and Fy V6.10 Potential and viscous flow ps sin u a du (6.118) 0 where Fx is the drag 1force parallel to direction of the uniform flow2 and Fy is the lift 1force perpendicular to the direction of the uniform flow2. Substitution for ps from Eq. 6.116 into these two equations, and subsequent integration, reveals that Fx 0 and Fy 0. These results indicate that both the drag and lift as predicted by potential theory for a fixed cylinder in a uniform stream are zero. Since the pressure distribution is symmetrical around the cylinder, this is not really a surprising result. However, we know from experience that there is a significant drag developed on a cylinder when it is placed in a moving fluid. This discrepancy is known as d’Alembert’s paradox. The paradox is named after Jean le Rond d’Alembert 11717–17832, a French mathematician and philosopher, who first showed that the drag on bodies immersed in inviscid fluids is zero. It was not until the latter part of the nineteenth century and the early part of the twentieth century that the role viscosity plays in the steady fluid motion was understood and d’Alembert’s paradox explained 1see Section 9.12. Potential theory incorrectly predicts that the drag on a cylinder is zero. E XAMPLE 冮 2p 6.8 Potential Flow—Cylinder GIVEN When a circular cylinder is placed in a uniform FIND stream, a stagnation point is created on the cylinder as is shown in Fig. E6.8a. If a small hole is located at this point, the stagnation pressure, pstag, can be measured and used to determine the approach velocity, U. (a) U U U y p0 r Show how pstag and U are related. (b) If the cylinder is misaligned by an angle a (Fig. E6.8b), but the measured pressure is still interpreted as the stagnation pressure, determine an expression for the ratio of the true velocity, U, to the predicted velocity, U . Plot this ratio as a function of a for the range 20 a 20 . θ x α β β a a Stagnation point (a) (b) (d) 1.5 1.4 U 1.3 __ U' 1.2 1.1 1.0 –20° –10° 0° α (c) 10° 20° ■ Figure E6.8 316 Chapter 6 ■ Differential Analysis of Fluid Flow SOLUTION (a) The velocity at the stagnation point is zero, so the Bernoulli equation written between a point on the stagnation streamline upstream from the cylinder and the stagnation point gives If we now write the Bernoulli equation between a point upstream of the cylinder and the point on the cylinder where r a, u a, it follows that pstag p0 U2 g g 2g p0 12 rU 2 pa 12 r12U sin a2 2 and, therefore, pa p0 12 rU 2 11 4 sin2a2 Thus, 2 U c 1 pstag p0 2 d r 12 (Ans) Since pstag p0 1 2 2 rU it follows from Eqs. 1 and 2 that U1true2 U¿ 1predicted2 COMMENT A measurement of the difference between the pressure at the stagnation point and the upstream pressure can be used to measure the approach velocity. This is, of course, the same result that was obtained in Section 3.5 for Pitot-static tubes. (b) If the direction of the fluid approaching the cylinder is not known precisely, it is possible that the cylinder is misaligned by some angle, a. In this instance the pressure actually measured, pa, will be different from the stagnation pressure, but if the misalignment is not recognized the predicted approach velocity, U¿, would still be calculated as U¿ c 12 2 1 pa p0 2 d r Thus, U1true2 U¿1predicted2 a pstag p0 12 b pa p0 (1) (2) 11 4 sin2a2 12 (Ans) This velocity ratio is plotted as a function of the misalignment angle a in Fig. E6.8c. COMMENT It is clear from these results that significant errors can arise if the stagnation pressure tap is not aligned with the stagnation streamline. As is discussed in Section 3.5, if two additional, symmetrically located holes are drilled on the cylinder, as are illustrated in Fig. E6.8d, the correct orientation of the cylinder can be determined. The cylinder is rotated until the pressures in the two symmetrically placed holes are equal, thus indicating that the center hole coincides with the stagnation streamline. For b 30° the pressure at the two holes theoretically corresponds to the upstream pressure, p0. With this orientation a measurement of the difference in pressure between the center hole and the side holes can be used to determine U. The velocity on the surface of the cylinder, vu, where r a, is obtained from Eq. 6.115 as vu 2U sin u An additional, interesting potential flow can be developed by adding a free vortex to the stream function or velocity potential for the flow around a cylinder. In this case c Ur a1 a2 b sin u ln r 2 2p r (6.119) a2 b cos u u 2 2p r (6.120) and f Ur a1 where is the circulation. We note that the circle r a will still be a streamline 1and thus can be replaced with a solid cylinder2, since the streamlines for the added free vortex are all circular. However, the tangential velocity, vu, on the surface of the cylinder 1r a2 now becomes vus Flow around a rotating cylinder is approximated by the addition of a free vortex. 0c ` 2U sin u 0r ra 2pa (6.121) This type of flow field could be approximately created by placing a rotating cylinder in a uniform stream. Because of the presence of viscosity in any real fluid, the fluid in contact with the rotating cylinder would rotate with the same velocity as the cylinder, and the resulting flow field would resemble that developed by the combination of a uniform flow past a cylinder and a free vortex. 6.6 Superposition of Basic, Plane Potential Flows Γ=0 Γ 1 4π Ua (c) (d) ■ Figure 6.29 The location of stagnation points on a circular cylinder: (a) without circulation; (b, c, d) with circulation. A variety of streamline patterns can be developed, depending on the vortex strength, . For example, from Eq. 6.121 we can determine the location of stagnation points on the surface of the cylinder. These points will occur at u ustag where vu 0 and therefore from Eq. 6.121 sin ustag 4pUa (6.122) If 0, then ustag 0 or p —that is, the stagnation points occur at the front and rear of the cylinder as are shown in Fig. 6.29a. However, for 1 4pUa 1, the stagnation points will occur at some other location on the surface as illustrated in Figs. 6.29b,c. If the absolute value of the parameter 4pUa exceeds 1, Eq. 6.122 cannot be satisfied, and the stagnation point is located away from the cylinder as shown in Fig. 6.29d. The force per unit length developed on the cylinder can again be obtained by integrating the differential pressure forces around the circumference as in Eqs. 6.117 and 6.118. For the cylinder with circulation, the surface pressure, ps, is obtained from the Bernoulli equation 1with the surface velocity given by Eq. 6.1212 p0 1 1 2 rU 2 ps r a2U sin u b 2 2 2pa or ps p0 1 2 sin u 2 rU 2 a1 4 sin2 u b 2 paU 4p2a2U 2 (6.123) Equation 6.123 substituted into Eq. 6.117 for the drag, and integrated, again yields Fx 0 Potential flow past a cylinder with circulation gives zero drag but nonzero lift. That is, even for the rotating cylinder no force in the direction of the uniform flow is developed. However, use of Eq. 6.123 with the equation for the lift, Fy 1Eq. 6.1182, yields Fy rU (6.124) Thus, for the cylinder with circulation, lift is developed equal to the product of the fluid density, the upstream velocity, and the circulation. The negative sign means that if U is positive 1in the positive x direction2 and is positive 1a free vortex with counterclockwise rotation2, the direction of the Fy is downward. Of course, if the cylinder is rotated in the clockwise direction ( 0), the direction of Fy would be upward. This can be seen by studying the surface pressure distribution (Eq. 6.123), which is plotted in Fig. 6.30 for two situations. One has 4pUa 0, which corresponds to no rotation of the cylinder. The other has 4pUa 0.25, which corresponds to clockwise rotation of the cylinder. With no 318 Chapter 6 ■ Differential Analysis of Fluid Flow 1 Γ/4πUa = 0 Γ/4πUa = –0.25 0 –1 ps – p0 _ 1 2 ρU2 –2 –3 θ U –4 –5 Top half –6 0 Bottom half 90 180 θ, deg 270 360 ■ Figure 6.30 Pressure distribution on a circular cylinder with and without rotation. rotation, the flow is symmetrical both top to bottom and front to back on the cylinder. With rotation, the flow is symmetrical front to back but not top to bottom. In this case the two stagnation points [i.e., ( ps p0)(rU22) 1] are located on the bottom of the cylinder, and the average pressure on the top half of the cylinder is less than that on the bottom half. The result is an upward lift force. It is this force acting in a direction perpendicular to the direction of the approach velocity that causes baseballs and golf balls to curve when they spin as they are propelled through the air. The development of this lift on rotating bodies is called the Magnus effect. (See Section 9.4 for further comments.) Although Eq. 6.124 was developed for a cylinder with circulation, it gives the lift per unit length for any two-dimensional object of any cross-sectional shape placed in a uniform, inviscid stream. The circulation is determined around any closed curve containing the body. The generalized equation relating lift to fluid density, velocity, and circulation is called the Kutta–Joukowski law and is commonly used to determine the lift on airfoils (see Section 9.4.2 and Refs. 2–6). F l u i d s i n A sailing ship without sails A sphere or cylinder spinning about its axis when placed in an airstream develops a force at right angles to the direction of the airstream. This phenomenon is commonly referred to as the Magnus effect and is responsible for the curved paths of baseballs and golf balls. Another lesser-known application of the Magnus effect was proposed by a German physicist and engineer, Anton Flettner, in the 1920s. Flettner’s idea was to use the Magnus effect to make a ship move. To demonstrate the practicality of the “rotor-ship” he purchased a sailing schooner and replaced the ship’s masts and rigging with 6.7 t h e N e w s two vertical cylinders that were 50 ft high and 9 ft in diameter. The cylinders looked like smokestacks on the ship. Their spinning motion was developed by 45-hp motors. The combination of a wind and the rotating cylinders created a force 1Magnus effect2 to push the ship forward. The ship, named the Baden Baden, made a successful voyage across the Atlantic, arriving in New York Harbor on May 9, 1926. Although the feasibility of the rotor-ship was clearly demonstrated, it proved to be less efficient and practical than more conventional vessels and the idea was not pursued. 1See Problem 6.79.2 Other Aspects of Potential Flow Analysis In the preceding section the method of superposition of basic potentials has been used to obtain detailed descriptions of irrotational flow around certain body shapes immersed in a uniform stream. For the cases considered, two or more of the basic potentials were combined, and the question is asked: What kind of flow does this combination represent? This approach is relatively simple and does not require the use of advanced mathematical techniques. It is, however, restrictive in its general applicability. It does not allow us to specify a priori the body shape and then determine the velocity potential or stream function that describes the flow around the particular body. 6.8 Potential flow solutions are always approximate because the fluid is assumed to be frictionless. V6.11 Potential flow 6.8 Viscous Flow 319 Determining the velocity potential or stream function for a given body shape is a much more complicated problem. It is possible to extend the idea of superposition by considering a distribution of sources and sinks, or doublets, which when combined with a uniform flow can describe the flow around bodies of arbitrary shape. Techniques are available to determine the required distribution to give a prescribed body shape. Also, for plane potential flow problems it can be shown that complex variable theory 1the use of real and imaginary numbers2 can be effectively used to obtain solutions to a great variety of important flow problems. There are, of course, numerical techniques that can be used to solve not only plane two-dimensional problems but also the more general threedimensional problems. Since potential flow is governed by Laplace’s equation, any procedure that is available for solving this equation can be applied to the analysis of irrotational flow of frictionless fluids. Potential flow theory is an old and well-established discipline within the general field of fluid mechanics. The interested reader can find many detailed references on this subject, including Refs. 2–6 given at the end of this chapter. An important point to remember is that regardless of the particular technique used to obtain a solution to a potential flow problem, the solution remains approximate because of the fundamental assumption of a frictionless fluid. Thus, “exact” solutions based on potential flow theory represent, at best, only approximate solutions to real fluid problems. The applicability of potential flow theory to real fluid problems has been alluded to in a number of examples considered in the previous section. As a rule of thumb, potential flow theory will usually provide a reasonable approximation in those circumstances when we are dealing with a low viscosity fluid moving at a relatively high velocity, in regions of the flow field in which the flow is accelerating. Under these circumstances we generally find that the effect of viscosity is confined to the thin boundary layer that develops at a solid boundary. Outside the boundary layer, the velocity distribution and the pressure distribution are closely approximated by the potential flow solution. However, in those regions of the flow field in which the flow is decelerating 1for example, in the rearward portion of a bluff body or in the expanding region of a conduit2, the pressure near a solid boundary will increase in the direction of flow. This so-called adverse pressure gradient can lead to flow separation, a phenomenon that causes dramatic changes in the flow field which are generally not accounted for by potential theory. However, as discussed in Chapter 9, in which boundary layer theory is developed, it is found that potential flow theory is used to obtain the appropriate pressure distribution that can then be combined with the viscous flow equations to obtain solutions near the boundary 1and also to predict separation2. The general differential equations that describe viscous fluid behavior and some simple solutions to these equations are considered in the remaining sections of this chapter. Viscous Flow To incorporate viscous effects into the differential analysis of fluid motion we must return to the previously derived general equations of motion, Eqs. 6.50. Since these equations include both stresses and velocities, there are more unknowns than equations and, therefore, before proceeding it is necessary to establish a relationship between the stresses and velocities. 6.8.1 Stress–Deformation Relationships For incompressible Newtonian fluids it is known that the stresses are linearly related to the rates of deformation and can be expressed in Cartesian coordinates as 1for normal stresses2 sxx p 2m 0u 0x (6.125a) syy p 2m 0v 0y (6.125b) szz p 2m 0w 0z (6.125c) 320 Chapter 6 ■ Differential Analysis of Fluid Flow 1for shearing stresses2 z szz syy sxx x p=– 1 (s + syy + szz) 3 xx For Newtonian fluids, stresses are linearly related to the rate of strain. y txy tyx m a 0u 0v b 0y 0x (6.125d) tyz tzy m a 0v 0w b 0z 0y (6.125e) tzx txz m a 0w 0u b 0x 0z (6.125f) where p is the pressure, the negative of the average of the three normal stresses; that is, as indicated by the figure in the margin, p 1 13 21sxx syy szz 2. For viscous fluids in motion the normal stresses are not necessarily the same in different directions, thus, the need to define the pressure as the average of the three normal stresses. For fluids at rest, or frictionless fluids, the normal stresses are equal in all directions. 1We have made use of this fact in the chapter on fluid statics and in developing the equations for inviscid flow.2 Detailed discussions of the development of these stress– velocity gradient relationships can be found in Refs. 3, 7, and 8. An important point to note is that whereas for elastic solids the stresses are linearly related to the deformation 1or strain2, for Newtonian fluids the stresses are linearly related to the rate of deformation 1or rate of strain2. In cylindrical polar coordinates the stresses for incompressible Newtonian fluids are expressed as 1for normal stresses2 srr p 2m 0vr 0r vr 1 0vu b r 0u r 0vz (6.126a) suu p 2m a (6.126b) szz p 2m (6.126c) 0z 1for shearing stresses2 tru tur m c r 0 vu 1 0vr a b d r 0u 0r r 0vu 1 0vz tuz tzu m a b r 0u 0z 0vz 0vr tzr trz m a b 0z 0r (6.126d) (6.126e) (6.126f) The double subscript has a meaning similar to that of stresses expressed in Cartesian coordinates— that is, the first subscript indicates the plane on which the stress acts and the second subscript the direction. Thus, for example, srr refers to a stress acting on a plane perpendicular to the radial direction and in the radial direction 1thus a normal stress2. Similarly, tru refers to a stress acting on a plane perpendicular to the radial direction but in the tangential 1u direction2 and is therefore a shearing stress. 6.8.2 The Navier–Stokes Equations The stresses as defined in the preceding section can be substituted into the differential equations of motion 1Eqs. 6.502 and simplified by using the continuity equation for incompressible flow 1Eq. 6.312. For rectangular coordinates (see the figure in the margin) the results are 1x direction2 w z u v y x ra 0p 0u 0u 0u 0u 0 2u 0 2u 0 2u u v w b rgx m a 2 2 2 b 0t 0x 0y 0z 0x 0x 0y 0z (6.127a) ra 0p 0v 0v 0v 0v 0 2v 0 2v 0 2v u v w b rgy m a 2 2 2 b 0t 0x 0y 0z 0y 0x 0y 0z (6.127b) 1y direction2 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids 321 1z direction2 ra The Navier–Stokes equations are the basic differential equations describing the flow of Newtonian fluids. 0p 0w 0w 0w 0w 0 2w 0 2w 0 2w u v w b rgz m a 2 2 2 b 0t 0x 0y 0z 0z 0x 0y 0z (6.127c) where u, v, and w are the x, y, and z components of velocity as shown in the figure in the margin of the previous page. We have rearranged the equations so that the acceleration terms are on the left side and the force terms are on the right. These equations are commonly called the Navier– Stokes equations, named in honor of the French mathematician L. M. H. Navier 11785–18362 and the English mechanician Sir G. G. Stokes 11819–19032, who were responsible for their formulation. These three equations of motion, when combined with the conservation of mass equation 1Eq. 6.312, provide a complete mathematical description of the flow of incompressible Newtonian fluids. We have four equations and four unknowns 1u, v, w, and p2, and therefore the problem is “well-posed” in mathematical terms. Unfortunately, because of the general complexity of the Navier–Stokes equations 1they are nonlinear, second-order, partial differential equations2, they are not amenable to exact mathematical solutions except in a few instances. However, in those few instances in which solutions have been obtained and compared with experimental results, the results have been in close agreement. Thus, the Navier–Stokes equations are considered to be the governing differential equations of motion for incompressible Newtonian fluids. In terms of cylindrical polar coordinates 1see the figure in the margin2, the Navier–Stokes equations can be written as 1r direction2 ra z ra vθ vr y x (6.128a) 1u direction2 vz r θ 0vr 0vr vu 0vr v2u 0vr vr vz b r 0u r 0t 0r 0z 0p 0vr vr 0 2vr 1 0 1 0 2vr 2 0vu rgr m c ar b 2 2 2 2 2d r 0r 0r 0r r r 0u r 0u 0z 0vu 0vu vu 0vu vrvu vr vz r 0u r 0t 0r 1 1 0p rgu m c r 0u r 1z direction2 ra 0vz 0t vr 0vz 0r 0vu b 0z 0vu vu 0 2vu 0 1 0 2vu 2 0vr ar b 2 2 2 2 2 d 0r 0r r r 0u r 0u 0z (6.128b) 0vz vu 0vz vz b r 0u 0z 2 0vz 0 2vz 0p 1 0 1 0 vz rgz m c ar b 2 2 2 d r 0r 0z 0r r 0u 0z (6.128c) To provide a brief introduction to the use of the Navier–Stokes equations, a few of the simplest exact solutions are developed in the next section. Although these solutions will prove to be relatively simple, this is not the case in general. In fact, only a few other exact solutions have been obtained. 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids A principal difficulty in solving the Navier–Stokes equations is a result of their nonlinearity arising from the convective acceleration terms 1i.e., u 0u 0x, w 0v 0z, etc.2. There are no general analytical schemes for solving nonlinear partial differential equations 1e.g., superposition of solutions cannot be used2, and each problem must be considered individually. For most practical flow problems, fluid particles do have accelerated motion as they move from one location to another in the flow field. Thus, the convective acceleration terms are usually important. However, there are a few special cases for which the convective acceleration vanishes because of the nature of the geometry of the flow 322 Chapter 6 ■ Differential Analysis of Fluid Flow system. In these cases exact solutions are often possible. The Navier–Stokes equations apply to both laminar and turbulent flow, but for turbulent flow each velocity component fluctuates randomly with respect to time and this added complication makes an analytical solution intractable. Thus, the exact solutions referred to are for laminar flows in which the velocity is either independent of time 1steady flow2 or dependent on time 1unsteady flow2 in a well-defined manner. 6.9.1 Steady, Laminar Flow between Fixed Parallel Plates An exact solution can be obtained for steady laminar flow between fixed parallel plates. We first consider flow between the two horizontal, infinite parallel plates of Fig. 6.31a. For this geometry the fluid particles move in the x direction parallel to the plates, and there is no velocity in the y or z direction—that is, v 0 and w 0. In this case it follows from the continuity equation 1Eq. 6.312 that 0u 0x 0. Furthermore, there would be no variation of u in the z direction for infinite plates, and for steady flow 0u 0t 0 so that u u1y2. If these conditions are used in the Navier–Stokes equations 1Eqs. 6.1272, they reduce to 0 0p 0 2u m a 2b 0x 0y (6.129) 0 0p rg 0y (6.130) 0 0p 0z (6.131) where we have set gx 0, gy g, and gz 0. That is, the y axis points up. We see that for this particular problem the Navier–Stokes equations reduce to some rather simple equations. Equations 6.130 and 6.131 can be integrated to yield p rgy f1 1x2 (6.132) which shows that the pressure varies hydrostatically in the y direction. Equation 6.129, rewritten as d 2u 1 0p 2 m 0x dy can be integrated to give du 1 0p a b y c1 m 0x dy V6.12 No-slip boundary condition and integrated again to yield u 1 0p 2 a b y c1y c2 2m 0x (6.133) Note that for this simple flow the pressure gradient, 0p 0x, is treated as constant as far as the integration is concerned, since 1as shown in Eq. 6.1322 it is not a function of y. The two constants c1 and c2 must be determined from the boundary conditions. For example, if the two plates are y u u h umax x h g z (a) (b) ■ Figure 6.31 The viscous flow between parallel plates: (a) coordinate system and notation used in analysis; (b) parabolic velocity distribution for flow between parallel fixed plates. 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids 323 fixed, then u 0 for y h 1because of the no-slip condition for viscous fluids2. To satisfy this condition c1 0 and c2 1 0p 2 a bh 2m 0x Thus, the velocity distribution becomes u 1 0p a b 1y2 h2 2 2m 0x (6.134) Equation 6.134 shows that the velocity profile between the two fixed plates is parabolic as illustrated in Fig. 6.31b. The volume rate of flow, q, passing between the plates 1for a unit width in the z direction2 is obtained from the relationship q V6.13 Liquid– liquid no-slip 冮 h h u dy 冮 h 1 0p a b 1y2 h2 2 dy h 2m 0x or q 2h3 0p a b 3m 0x (6.135) The pressure gradient 0p 0x is negative, since the pressure decreases in the direction of flow. If we let ¢p represent the pressure drop between two points a distance / apart, then ¢p 0p / 0x and Eq. 6.135 can be expressed as q 2h3 ¢p 3m/ (6.136) The flow is proportional to the pressure gradient, inversely proportional to the viscosity, and strongly dependent 1⬃h3 2 on the gap width. In terms of the mean velocity, V, where V q2h, Eq. 6.136 becomes V h2 ¢p 3m/ (6.137) Equations 6.136 and 6.137 provide convenient relationships for relating the pressure drop along a parallel-plate channel and the rate of flow or mean velocity. The maximum velocity, umax, occurs midway 1y 02 between the two plates, as shown in Fig. 6.31b, so that from Eq. 6.134 umax h2 0p a b 2m 0x or umax 32V The Navier–Stokes equations provide detailed flow characteristics for laminar flow between fixed parallel plates. (6.138) The details of the steady laminar flow between infinite parallel plates are completely predicted by this solution to the Navier–Stokes equations. For example, if the pressure gradient, viscosity, and plate spacing are specified, then from Eq. 6.134 the velocity profile can be determined, and from Eqs. 6.136 and 6.137 the corresponding flowrate and mean velocity determined. In addition, since the pressure gradient in the x direction is constant, from Eq. 6.132 it follows that f1 1x2 a 0p b x p0 0x where p0 is a reference pressure at x y 0, and the pressure variation throughout the fluid can be obtained from p rgy a 0p b x p0 0x (6.139) 324 Chapter 6 ■ Differential Analysis of Fluid Flow For a given fluid and reference pressure, p0, the pressure at any point can be predicted. This relatively simple example of an exact solution illustrates the detailed information about the flow field which can be obtained. The flow will be laminar if the Reynolds number, Re ⫽ rV12h2 Ⲑm, remains below about 1400. For flow with larger Reynolds numbers the flow becomes turbulent and the preceding analysis is not valid since the flow field is complex, three-dimensional, and unsteady. F l u i d s i n t 10 tons on 8 psi Place a golf ball on the end of a garden hose and then slowly turn the water on a small amount until the ball just barely lifts off the end of the hose, leaving a small gap between the ball and the hose. The ball is free to rotate. This is the idea behind the new “floating ball water fountains” developed in Finland. Massive, 10-ton, 6-ft-diameter stone spheres are supported by the pressure force of the water on the curved surface within a pedestal and rotate so easily that even a small child can change their direction of rotation. The key to the h e N e w s fountain design is the ability to grind and polish stone to an accuracy of a few thousandths of an inch. This allows the gap between the ball and its pedestal to be very small (on the order of 5/1000 in.) and the water flowrate correspondingly small (on the order of 5 gal/min). Due to the small gap, the flow in the gap is essentially that of flow between parallel plates. Although the sphere is very heavy, the pressure under the sphere within the pedestal needs to be only about 8 psi. (See Problem 6.91.) 6.9.2 Couette Flow For a given flow geometry, the character and details of the flow are strongly dependent on the boundary conditions. Another simple parallel-plate flow can be developed by fixing one plate and letting the other plate move with a constant velocity, U, as is illustrated in Fig. 6.32a. The Navier–Stokes equations reduce to the same form as those in the preceding section, and the solution for the pressure and velocity distribution are still given by Eqs. 6.132 and 6.133, respectively. However, for the moving plate problem the boundary conditions for the velocity are different. For this case we locate the origin of the coordinate system at the bottom plate and designate the distance between the two plates as b 1see Fig. 6.32a2. The two constants c1 and c2 in Eq. 6.133 can be determined from the no-slip boundary conditions, u ⫽ 0 at y ⫽ 0 and u ⫽ U at y ⫽ b. It follows that u⫽U y 1 0p ⫹ a b 1y2 ⫺ by2 b 2m 0x (6.140) or, in dimensionless form, y y u b2 0p y ⫽ ⫺ a b a b a1 ⫺ b U b 2mU 0x b b (6.141) U Moving plate 1.0 0.8 Backflow U P = –3 –2 –1 0 1 2 3 0.6 _y b b b 0.4 y x 0.2 0 z Fixed plate -0.4 -0.2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 u __ U (a) (b) ■ Figure 6.32 The viscous flow between parallel plates with bottom plate fixed and upper plate moving (Couette flow): (a) coordinate system and notation used in analysis; (b) velocity distribution as a function of parameter, P, where P ⴝ ⴚ(b 2兾2MU) ⵲p兾⵲x. (Data from Ref. 8.) 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids 325 Lubricating oil ω ro Rotating shaft ri Housing ■ Figure 6.33 Flow in the narrow gap of a journal bearing. The actual velocity profile will depend on the dimensionless parameter P Flow between parallel plates with one plate fixed and the other moving is called Couette flow. b2 0p a b 2mU 0x Several profiles are shown in Fig. 6.32b. This type of flow is called Couette flow. The simplest type of Couette flow is one for which the pressure gradient is zero; that is, the fluid motion is caused by the fluid being dragged along by the moving boundary. In this case, with 0p 0x 0, Eq. 6.140 simply reduces to uU y b (6.142) which indicates that the velocity varies linearly between the two plates as shown in Fig. 6.31b for P 0. This situation would be approximated by the flow between closely spaced concentric cylinders in which one cylinder is fixed and the other cylinder rotates with a constant angular velocity, v. As illustrated in Fig. 6.33, the flow in an unloaded journal bearing might be approximated by this simple Couette flow if the gap width is very small 1i.e., ro ri ri 2. In this case U ri v, b ro ri, and the shearing stress resisting the rotation of the shaft can be simply calculated as t mri v 1ro ri 2. When the bearing is loaded 1i.e., a force applied normal to the axis of rotation2, the shaft will no longer remain concentric with the housing and the flow cannot be treated as flow between parallel boundaries. Such problems are dealt with in lubrication theory 1see, for example, Ref. 92. E XAMPLE 6.9 Plane Couette Flow GIVEN A wide moving belt passes through a container of a viscous liquid. The belt moves vertically upward with a constant velocity, V0, as illustrated in Fig. E6.9a. Because of viscous forces the belt picks up a film of fluid of thickness h. Gravity tends to make the fluid drain down the belt. Assume that the flow is laminar, steady, and fully developed. h Fluid layer V0 y FIND Use the Navier–Stokes equations to determine an expression for the average velocity of the fluid film as it is dragged up the belt. g x SOLUTION Since the flow is assumed to be fully developed, the only velocity component is in the y direction 1the v component2 so that u w 0. It follows from the continuity equation that 0v 0y 0, and for steady flow 0v 0t 0, so that v v1x2. Under these conditions the Navier–Stokes equations for the x direction 1Eq. 6.127a2 and the z direction 1perpendicular to the paper2 1Eq. 6.127c2 simply reduce to 0p 0p 0 0 0x 0z ■ Figure E6.9a This result indicates that the pressure does not vary over a horizontal plane, and since the pressure on the surface of the film 1x h2 is atmospheric, the pressure throughout the film must be 326 Chapter 6 ■ Differential Analysis of Fluid Flow atmospheric 1or zero gage pressure2. The equation of motion in the y direction 1Eq. 6.127b2 thus reduces to d 2v 0 rg m 2 dx q V0 h V V0 (1) Integration of Eq. 1 yields (2) On the film surface 1x h2 we assume the shearing stress is zero—that is, the drag of the air on the film is negligible. The shearing stress at the free surface 1or any interior parallel surface2 is designated as txy, where from Eq. 6.125d txy m a dv b dx Thus, if txy 0 at x h, it follows from Eq. 2 that c1 (Ans) gh m form as v x 2 x c a b 2c a b 1 V0 h h where c gh2/2mV0. This velocity profile is shown in Fig. E6.9b. Note that even though the belt is moving upward, for c 1 (e.g., for fluids with small enough viscosity or with a small enough belt speed) there are portions of the fluid that flow downward (as indicated by v/V0 0). It is interesting to note from this result that there will be a net upward flow of liquid (positive V) only if V0 gh2/3m. It takes a relatively large belt speed to lift a small viscosity fluid. A second integration of Eq. 2 gives the velocity distribution in the film as g 2 gh x x c2 2m m 0.8 0.6 and the velocity distribution is therefore c = 0.5 0.4 0.2 v/V0 c2 V0 g 2 gh x x V0 2m m c=0 1 At the belt 1x 02 the fluid velocity must match the belt velocity, V0, so that v gh2 3m COMMENT Equation (3) can be written in dimensionless g dv x c1 dx m v gh3 3m The average film velocity, V 1where q Vh2, is therefore or g d 2v m dx2 and thus c = 1.0 0 0 0.2 0.4 0.6 0.8 1 –0.2 (3) c = 1.5 –0.4 –0.6 With the velocity distribution known we can determine the flowrate per unit width, q, from the relationship q 冮 h v dx 0 冮 0 h a g 2 gh x x V0 b dx 2m m –0.8 c = 2.0 –1 x/h ■ Figure E6.9b 6.9.3 Steady, Laminar Flow in Circular Tubes An exact solution can be obtained for steady, incompressible, laminar flow in circular tubes. Probably the best known exact solution to the Navier–Stokes equations is for steady, incompressible, laminar flow through a straight circular tube of constant cross section. This type of flow is commonly called Hagen–Poiseuille flow, or simply Poiseuille flow. It is named in honor of J. L. Poiseuille 11799– 18692, a French physician, and G. H. L. Hagen 11797–18842, a German hydraulic engineer. Poiseuille was interested in blood flow through capillaries and deduced experimentally the resistance laws for laminar flow through circular tubes. Hagen’s investigation of flow in tubes was also experimental. It was actually after the work of Hagen and Poiseuille that the theoretical results presented in this section were determined, but their names are commonly associated with the solution of this problem. Consider the flow through a horizontal circular tube of radius R as is shown in Fig. 6.34a. Because of the cylindrical geometry it is convenient to use cylindrical coordinates. We assume that the flow is parallel to the walls so that vr 0 and vu 0, and from the continuity equation 16.342 0vz 0z 0. Also, for steady, axisymmetric flow, vz is not a function of t or u, so the velocity, vz, 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids y 327 g vz r R vz q r z z dr (a) (b) ■ Figure 6.34 The viscous flow in a horizontal, circular tube: (a) coordinate system and notation used in analysis; (b) flow through differential annular ring. is only a function of the radial position within the tube—that is, vz vz 1r2. Under these conditions the Navier–Stokes equations 1Eqs. 6.1282 reduce to 0 rg sin u 0p 0r (6.143) 0 rg cos u 1 0p r 0u (6.144) 0 0vz 0p 1 0 mc ar bd r 0r 0z 0r (6.145) where we have used the relationships gr g sin u and gu g cos u 1with u measured from the horizontal plane2. Equations 6.143 and 6.144 can be integrated to give p rg1r sin u2 f1 1z2 or p rgy f1 1z2 (6.146) Equation 6.146 indicates that the pressure is hydrostatically distributed at any particular cross section, and the z component of the pressure gradient, 0p 0z, is not a function of r or u. The equation of motion in the z direction 1Eq. 6.1452 can be written in the form V6.14 Laminar flow 0vz 1 0 1 0p ar b r 0r m 0z 0r and integrated 1using the fact that 0p 0z constant2 to give r 0vz 0r 1 0p 2 a b r c1 2m 0z Integrating again we obtain vz vz 0 c1 = 0, c2>0 0.5 R R c1>0 r 1 0p 2 a b r c1 ln r c2 4m 0z As indicated by the figure in the margin, the shape of the velocity profile depends on the values of the two constants c1 and c2. Since we wish vz to be finite at the center of the tube 1r 02, it follows that c1 0 3since ln 102 4. At the wall 1r R2 the velocity must be zero so that c2 The velocity distribution is parabolic for steady, laminar flow in circular tubes. (6.147) 1 0p 2 a bR 4m 0z and the velocity distribution becomes vz 1 0p a b 1r2 R2 2 4m 0z (6.148) Thus, at any cross section the velocity distribution is parabolic. To obtain a relationship between the volume rate of flow, Q, passing through the tube and the pressure gradient, we consider the flow through the differential, washer-shaped ring of 328 Chapter 6 ■ Differential Analysis of Fluid Flow Fig. 6.34b. Since vz is constant on this ring, the volume rate of flow through the differential area dA ⫽ 12pr2 dr is dQ ⫽ vz 12pr2 dr and therefore Q ⫽ 2p 冮 R vzr dr (6.149) 0 Equation 6.148 for vz can be substituted into Eq. 6.149, and the resulting equation integrated to yield pR 4 0p (6.150) Q⫽⫺ a b 8m 0z V6.15 Complex pipe flow This relationship can be expressed in terms of the pressure drop, ¢p, which occurs over a length, /, along the tube, since ¢p 0p ⫽⫺ / 0z and therefore Q⫽ pR 4 ¢p 8m/ (6.151) For a given pressure drop per unit length, the volume rate of flow is inversely proportional to the viscosity and proportional to the tube radius to the fourth power. A doubling of the tube radius produces a 16-fold increase in flow! Equation 6.151 is commonly called Poiseuille’s law. In terms of the mean velocity, V, where V ⫽ QⲐpR2, Eq. 6.151 becomes Poiseuille’s law relates pressure drop and flowrate for steady, laminar flow in circular tubes. V⫽ R2 ¢p 8m/ (6.152) The maximum velocity vmax occurs at the center of the tube, where from Eq. 6.148 vmax ⫽ ⫺ R2 ¢p R2 0p a b⫽ 4m 0z 4m/ (6.153) so that vmax ⫽ 2V vz _r vmax = 1 – R The velocity distribution, as shown by the figure in the margin, can be written in terms of vmax as ( ) 2 vz R r vmax (6.154) As was true for the similar case of flow between parallel plates 1sometimes referred to as plane Poiseuille flow2, a very detailed description of the pressure and velocity distribution in tube flow results from this solution to the Navier–Stokes equations. Numerous experiments performed to substantiate the theoretical results show that the theory and experiment are in agreement for the laminar flow of Newtonian fluids in circular tubes or pipes. In general, the flow remains laminar for Reynolds numbers, Re ⫽ rV12R2 Ⲑm, below 2100. Turbulent flow in tubes is considered in Chapter 8. vmax F r 2 ⫽1⫺a b R l u i d s i n Poiseuille’s law revisited Poiseuille’s law governing laminar flow of fluids in tubes has an unusual history. It was developed in 1842 by a French physician, J. L. M. Poiseuille, who was interested in the flow of blood in capillaries. Poiseuille, through a series of carefully conducted experiments using water flowing through very small tubes, arrived at the formula, Q ⫽ K¢p D4Ⲑ/. In this formula Q is the flowrate, K an empirical constant, ¢p the pressure drop over the length /, and D the tube diameter. Another t h e N e w s formula was given for the value of K as a function of the water temperature. It was not until the concept of viscosity was introduced at a later date that Poiseuille’s law was derived mathematically and the constant K found to be equal to pⲐ8m, where m is the fluid viscosity. The experiments by Poiseuille have long been admired for their accuracy and completeness considering the laboratory instrumentation available in the mid-nineteenth century. 6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids vz 329 ro r ri z ■ Figure 6.35 The viscous flow through an annulus. 6.9.4 Steady, Axial, Laminar Flow in an Annulus An exact solution can be obtained for axial flow in the annular space between two fixed, concentric cylinders. The differential equations 1Eqs. 6.143, 6.144, 6.1452 used in the preceding section for flow in a tube also apply to the axial flow in the annular space between two fixed, concentric cylinders 1Fig. 6.352. Equation 6.147 for the velocity distribution still applies, but for the stationary annulus the boundary conditions become vz 0 at r ro and vz 0 for r ri. With these two conditions the constants c1 and c2 in Eq. 6.147 can be determined and the velocity distribution becomes vz r 2i r 2o r 1 0p a b c r 2 r 2o ln d 4m 0z ln1ro ri 2 ro The corresponding volume rate of flow is Q 冮 ro ri vz 12pr2 dr (6.155) 1r 2o r 2i 2 2 p 0p a b c r 4o r 4i d 8m 0z ln1ro ri 2 or in terms of the pressure drop, ¢p, in length / of the annulus Q 1r 2o r 2i 2 2 p¢p 4 c r o r 4i d 8m/ ln1ro ri 2 (6.156) The velocity at any radial location within the annular space can be obtained from Eq. 6.155. The maximum velocity occurs at the radius r rm where 0vz 0r 0. Thus, rm c r 2o r 2i 12 d 2 ln1ro ri 2 (6.157) An inspection of this result shows that the maximum velocity does not occur at the midpoint of the annular space, but rather it occurs nearer the inner cylinder. The specific location depends on ro and ri. These results for flow through an annulus are valid only if the flow is laminar. A criterion based on the conventional Reynolds number 1which is defined in terms of the tube diameter2 cannot be directly applied to the annulus, since there are really “two” diameters involved. For tube cross sections other than simple circular tubes it is common practice to use an “effective” diameter, termed the hydraulic diameter, Dh, which is defined as Dh 4 cross-sectional area wetted perimeter The wetted perimeter is the perimeter in contact with the fluid. For an annulus Dh 4p1ro2 ri2 2 2p1ro ri 2 21ro ri 2 In terms of the hydraulic diameter, the Reynolds number is Re rDhVm 1where V Q cross-sectional area2, and it is commonly assumed that if this Reynolds number remains below 2100 the flow will be laminar. A further discussion of the concept of the hydraulic diameter as it applies to other noncircular cross sections is given in Section 8.4.3. 330 Chapter 6 ■ Differential Analysis of Fluid Flow E XAMPLE 6.10 Laminar Flow in an Annulus 1r 1.18 103 kg m3; m 2 # 0.0045 N s m 2 flows at a rate of 12 ml兾s through a horizontal, 4-mm-diameter tube. GIVEN A viscous liquid FIND 1a2 Determine the pressure drop along a l-m length of the of velocity is parallel to the tube axis. 1b2 If a 2-mm-diameter rod is placed in the 4-mm-diameter tube to form a symmetric annulus, what is the pressure drop along a l-m length if the flowrate remains the same as in part 1a2? tube which is far from the tube entrance so that the only component SOLUTION (a) We first calculate the Reynolds number, Re, to determine whether or not the flow is laminar. With the diameter D 4 mm 0.004 m, the mean velocity is Q V 1p42D2 0.955 ms so that ¢p 112 mls2 1106 m3ml2 11.18 103 kg m3 210.955 ms210.004 m2 rVD Re m 0.0045 N # sm2 1000 Since the Reynolds number is well below the critical value of 2100, we can safely assume that the flow is laminar. Thus, we can apply Eq. 6.151, which gives for the pressure drop 8m/Q pR4 810.0045 N # sm2 211 m2 112 106 m3s2 p10.002 m2 4 8.59 kPa (Ans) (b) For flow in the annulus with an outer radius ro 0.002 m and an inner radius ri 0.001 m, the mean velocity is V Q ri2 2 1.27 ms p1ro2 p e 10.002 m2 4 10.001 m2 4 1p42 10.004 m2 2 3 10.002 m2 2 10.001 m2 2 4 2 ln10.002 m0.001 m2 68.2 kPa and, therefore, ¢p 810.0045 N # sm2 211 m2112 106 m3s2 f 1 (Ans) COMMENTS The pressure drop in the annulus is much larger than that of the tube. This is not a surprising result, since to maintain the same flow in the annulus as that in the open tube, the average velocity must be larger (the cross-sectional area is smaller) and the pressure difference along the annulus must overcome the shearing stresses that develop along both an inner and an outer wall. By repeating the calculations for various radius ratios, ri ro, the results shown in Fig. E6.10 are obtained. It is seen that the pressure drop ratio, ¢pannulus ¢ptube (i.e., the pressure drop in the annulus compared to that in a tube with a radius equal to the outer radius of the annulus, ro), is a strong function of the radius ratio. Even an annulus with a very small inner radius will have a pressure drop significantly larger than that of a tube. For example, if the inner radius is only 1100 of the outer radius, ¢pannulus¢ptube 1.28. As shown in the figure, for larger inner radii, the pressure drop ratio is much larger [i.e., pannulus ptube 7.94 for riro 0.50 as in part (b) of this example]. 12 106 m3 s 1p2 3 10.002 m2 2 10.001 m2 2 4 8 (0.50, 7.94) 7 and the Reynolds number [based on the hydraulic diameter, Dh 21ro ri 2 210.002 m 0.001 m2 0.002 m] is rDhV m 11.18 103 kg m3 2 10.002 m2 11.27 ms2 0.0045 N # sm2 666 This value is also well below 2100, so the flow in the annulus should also be laminar. From Eq. 6.156, 1r2o r2i 2 2 8m/Q 4 ¢p d c ro r4i p ln1ro ri 2 1 1.2 6 Re 1.1 5 Δpannulus __ Δptube (0.01, 1.28) 1.3 1 4 0 0.005 0.01 3 2 1 0 0.1 0.2 0.3 ri /r0 ■ Figure E6.10 0.4 0.5 6.10 6.10 Other Aspects of Differential Analysis 331 Other Aspects of Differential Analysis In this chapter the basic differential equations that govern the flow of fluids have been developed. The Navier–Stokes equations, which can be compactly expressed in vector notation as ra 0V V ⴢ ⵱Vb ⵱p rg m§ 2V 0t (6.158) along with the continuity equation § ⴢV0 (6.159) are the general equations of motion for incompressible Newtonian fluids. Although we have restricted our attention to incompressible fluids, these equations can be readily extended to include compressible fluids. It is well beyond the scope of this introductory text to consider in depth the variety of analytical and numerical techniques that can be used to obtain both exact and approximate solutions to the Navier–Stokes equations. Students, however, should be aware of the existence of these very general equations, which are frequently used as the basis for many advanced analyses of fluid motion. A few relatively simple solutions have been obtained and discussed in this chapter to indicate the type of detailed flow information that can be obtained by using differential analysis. However, it is hoped that the relative ease with which these solutions were obtained does not give the false impression that solutions to the Navier–Stokes equations are readily available. This is certainly not true, and as previously mentioned there are actually very few practical fluid flow problems that can be solved by using an exact analytical approach. In fact, there are no known analytical solutions to Eq. 6.158 for flow past any object such as a sphere, cube, or airplane. Because of the difficulty in solving the Navier–Stokes equations, much attention has been given to various types of approximate solutions. For example, if the viscosity is set equal to zero, the Navier–Stokes equations reduce to Euler’s equations. Thus, the frictionless fluid solutions discussed previously are actually approximate solutions to the Navier–Stokes equations. At the other extreme, for problems involving slowly moving fluids, viscous effects may be dominant and the nonlinear 1convective2 acceleration terms can be neglected. This assumption greatly simplifies the analysis, since the equations now become linear. There are numerous analytical solutions to these “slow flow” or “creeping flow” problems. Another broad class of approximate solutions is concerned with flow in the very thin boundary layer. L. Prandtl showed in 1904 how the Navier–Stokes equations could be simplified to study flow in boundary layers. Such “boundary layer solutions” play a very important role in the study of fluid mechanics. A further discussion of boundary layers is given in Chapter 9. Very few practical fluid flow problems can be solved using an exact analytical approach. 6.10.1 Numerical Methods Numerical methods using digital computers are, of course, commonly utilized to solve a wide variety of flow problems. As discussed previously, although the differential equations that govern the flow of Newtonian fluids [the Navier–Stokes equations 16.1582] were derived many years ago, there are few known analytical solutions to them. With the advent of high-speed digital computers it has become possible to obtain numerical solutions to these 1and other fluid mechanics2 equations for many different types of problems. A brief introduction to computational fluid dynamics (CFD) is given in Appendix A. V6.16 CFD example F l u i d s i n Fluids in the Academy Awards A computer science professor at Stanford University and his colleagues were awarded a Scientific and Technical Academy Award for applying the Navier–Stokes equations for use in Hollywood movies. These researchers make use of computational algorithms to numerically solve the Navier–Stokes equations (also termed computational fluid dynam- t h e N e w s ics, or CFD) and simulate complex liquid flows. The realism of the simulations has found application in the entertainment industry. Movie producers have used the power of these numerical tools to simulate flows from ocean waves in Pirates of the Caribbean to lava flows in the final duel in Star Wars: Revenge of the Sith. Therefore, even Hollywood has recognized the usefulness of CFD. 332 6.11 Chapter 6 ■ Differential Analysis of Fluid Flow Chapter Summary and Study Guide volumetric dilatation rate vorticity irrotational flow continuity equation stream function Euler’s equations of motion ideal fluid Bernoulli equation velocity potential potential flow equipotential lines flow net uniform flow source and sink vortex circulation doublet method of superposition half-body Rankine oval Navier–Stokes equations Couette flow Poiseuille’s law Differential analysis of fluid flow is concerned with the development of concepts and techniques that can be used to provide a detailed, point-by-point, description of a flow field. Concepts related to the motion and deformation of a fluid element are introduced, including the Eulerian method for describing the velocity and acceleration of fluid particles. Linear deformation and angular deformation of a fluid element are described through the use of flow characteristics such as the volumetric dilatation rate, rate of angular deformation, and vorticity. The differential form of the conservation of mass equation (continuity equation) is derived in both rectangular and cylindrical polar coordinates. Use of the stream function for the study of steady, incompressible, plane, two-dimensional flow is introduced. The general equations of motion are developed, and for inviscid flow these equations are reduced to the simpler Euler equations of motion. The Euler equations are integrated to give the Bernoulli equation, and the concept of irrotational flow is introduced. Use of the velocity potential for describing irrotational flow is considered in detail, and several basic velocity potentials are described, including those for a uniform flow, source or sink, vortex, and doublet. The technique of using various combinations of these basic velocity potentials, by superposition, to form new potentials is described. Flows around a half-body, a Rankine oval, and around a circular cylinder are obtained using this superposition technique. Basic differential equations describing incompressible, viscous flow (the Navier–Stokes equations) are introduced. Several relatively simple solutions for steady, viscous, laminar flow between parallel plates and through circular tubes are included. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. determine the acceleration of a fluid particle, given the equation for the velocity field. determine the volumetric dilatation rate, vorticity, and rate of angular deformation for a fluid element, given the equation for the velocity field. show that a given velocity field satisfies the continuity equation. use the concept of the stream function to describe a flow field. use the concept of the velocity potential to describe a flow field. use superposition of basic velocity potentials to describe simple potential flow fields. use the Navier – Stokes equations to determine the detailed flow characteristics of incompressible, steady, laminar, viscous flow between parallel plates and through circular tubes. Some of the important equations in this chapter are: Acceleration of fluid particle Vorticity Conservation of mass Stream function Euler’s equations of motion 0V 0V 0V 0V u v w 0t 0x 0y 0z z2␻ § ⴛV 01ru2 01rv2 01rw2 0r 0 0t 0x 0y 0z 0c 0c u v 0y 0x 0p 0u 0u 0u 0u rgx ra u v w b 0x 0t 0x 0y 0z 0p 0v 0v 0v 0v rgy ra u v w b 0y 0t 0x 0y 0z 0p 0w 0w 0w 0w ra u v w b rgz 0z 0t 0x 0y 0z a (6.2) (6.17) (6.27) (6.37) (6.51a) (6.51b) (6.51c) References Velocity potential V ⵱f Laplace’s equation ⵱2f 0 Uniform potential flow f U1x cos a y sin a2 c U1y cos a x sin a2 Source and sink f f Vortex m ln r 2p ≠ u 2p c c m u 2p ≠ ln r 2p f K cos u r c K sin u r (6.65) (6.66) u U cos a v U sin a m 2pr vu 0 vr vr 0 vu Doublet 333 2pr K cos u r2 K cos u vu r2 vr The Navier–Stokes equations 1x direction2 ra 0p 0u 0u 0u 0u 0 2u 0 2u 0 2u u v w b rgx m a 2 2 2 b (6.127a) 0t 0x 0y 0z 0x 0x 0y 0z 1y direction2 ra 0p 0v 0v 0v 0v 0 2v 0 2v 0 2v u v w b rgy m a 2 2 2 b (6.127b) 0t 0x 0y 0z 0y 0x 0y 0z 1z direction2 ra 0p 0w 0w 0w 0w 0 2w 0 2w 0 2w u v w b rgz m a 2 2 2 b (6.127c) 0t 0x 0y 0z 0z 0x 0y 0z References 1. 2. 3. 4. 5. 6. 7. 8. 9. White, F. M., Fluid Mechanics, 5th Ed., McGraw-Hill, New York, 2003. Streeter, V. L., Fluid Dynamics, McGraw-Hill, New York, 1948. Rouse, H., Advanced Mechanics of Fluids, Wiley, New York, 1959. Milne-Thomson, L. M., Theoretical Hydrodynamics, 4th Ed., Macmillan, New York, 1960. Robertson, J. M., Hydrodynamics in Theory and Application, Prentice-Hall, Englewood Cliffs, N.J., 1965. Panton, R. L., Incompressible Flow, 3rd Ed., Wiley, New York, 2005. Li, W. H., and Lam, S. H., Principles of Fluid Mechanics, Addison-Wesley, Reading, Mass., 1964. Schlichting, H., Boundary-Layer Theory, 8th Ed., McGraw-Hill, New York, 2000. Fuller, D. D., Theory and Practice of Lubrication for Engineers, Wiley, New York, 1984. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. 334 Chapter 6 ■ Differential Analysis of Fluid Flow Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www.wiley. com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Conceptual Questions 6.1C Looking at Eq. 6.37, what are the units of stream function? a) LT1 b) L2 T1 1 c) MLT d) T1 a) the diameter of pipe 2 is less than the diameter of pipe 1. b) the diameter of pipe 2 is greater than the diameter of pipe 1. c) the diameter of pipe 2 is equal to the diameter of pipe 1. 6.2C Two tanks filled with water are connected by two straight circular pipes that have diameters D1 and D2, as shown in the figure. 6.3C A uniform flow moving to the left (in the negative x-axis direction) is imposed on the doublet shown in Fig. 6.23. For the resulting flow field, a) there are two stagnation points, one above and one below the doublet. b) there is one stagnation point to the left of the doublet. c) there is one stagnation point to the right of the doublet. d) there are no stagnation points anywhere in the flow. D2 2L L D1 The water level in the left tank is twice that of the right tank. If the flow through the connection pipes is laminar and can be approximated by the fully developed Poiseuille solution, then the flow through pipe 2 will have the same velocity as the flow through pipe 1 when: 6.4C For steady, laminar flow between fixed parallel plates, the velocity profile is parabolic in shape, as shown in Fig. 6.31. If the shear stress is given by t m0u 0y, what is the shear stress at the midpoint between the two plates? a) maximum shear b) negative shear c) zero shear d) equal to the wall shear Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the even-numbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 6.1 Fluid Element Kinematics 6.1 The velocity in a certain flow field is given by the equation V yziˆ x2zjˆ xkˆ Determine the expressions for the three rectangular components of acceleration. 6.4 given by The three components of velocity in a flow field are u x2 y2 z2 v xy yz z2 w 3xz z22 4 (a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field? 6.5 vector: Determine the vorticity field for the following velocity V 1x 2 y 2 2 ˆi 2xyjˆ 6.2 The velocity in a certain two-dimensional flow field is given by the equation V 2xtiˆ 2ytˆj 6.6 Determine an expression for the vorticity of the flow field described by V xy3 ˆi y4ˆj where the velocity is in ft/s when x, y, and t are in feet and seconds, respectively. Determine expressions for the local and convective components of acceleration in the x and y directions. What is the magnitude and direction of the velocity and the acceleration at the point x y 2 ft at the time t 0? Is the flow irrotational? 6.3 tion The velocity in a certain flow field is given by the equaV xiˆ x2zjˆ yzkˆ Determine the expressions for the three rectangular components of acceleration. 6.7 GO A two-dimensional flow field described by V 12x2y x2iˆ 12xy2 y 12jˆ where the velocity is in m/s when x and y are in meters. Determine the angular rotation of a fluid element located at x 0.5 m, y 1.0 m. 6.8 For a certain incompressible, two-dimensional flow field the velocity component in the y direction is given by the equation v 3xy x2y Problems Determine the velocity component in the x direction so that the volumetric dilatation rate is zero. 6.9 An incompressible viscous fluid is placed between two large parallel plates as shown in Fig. P6.9. The bottom plate is fixed and the upper plate moves with a constant velocity, U. For these conditions the velocity distribution between the plates is linear and can be expressed as y uU b 335 6.14 For each of the following stream functions, with units of m2/s, determine the magnitude and the angle the velocity vector makes with the x axis at x 1 m, y 2 m. Locate any stagnation points in the flow field. (a) c xy (b) c 2x2 y 6.15 The stream function for an incompressible, twodimensional flow field is c ay by3 Determine: (a) the volumetric dilatation rate, (b) the rotation vector, (c) the vorticity, and (d) the rate of angular deformation. U where a and b are constants. Is this an irrotational flow? Explain. 6.16 The stream function for an incompressible, two-dimensional flow field is c ay2 bx Moving plate u where a and b are constants. Is this an irrotational flow? Explain. b 6.17 The velocity components in an incompressible, twodimensional flow field are given by the equations y u x2 y 2xy x Fixed plate ■ Figure P6.9 Determine, if possible, the corresponding stream function. 6.10 A viscous fluid is contained in the space between concentric cylinders. The inner wall is fixed, and the outer wall rotates with an angular velocity v. (See Fig. P6.10a and Video V6.3.) Assume that the velocity distribution in the gap is linear as illustrated in Fig. P6.10b. For the small rectangular element shown in Fig. P6.10b, determine the rate of change of the right angle g due to the fluid motion. Express your answer in terms of ro, ri, and v. ω 6.18 The velocity components of an incompressible, twodimensional velocity field are given by the equations u 2xy y x2 y2 Show that the flow is irrotational and satisfies conservation of mass. 6.19 For a certain two-dimensional flow field u0 yV ro ω u ri ro γ y ro – ri x (a) (b) ■ Figure P6.10 Section 6.2 Conservation of Mass 6.11 For incompressible fluids the volumetric dilatation rate must be zero; that is, § V 0. For what combination of constants, a, b, c, and e can the velocity components u ax by y cx ey w0 be used to describe an incompressible flow field? 6.12 For a certain incompressible flow field it is suggested that the velocity components are given by the equations u 2xy v x2y w 0 Is this a physically possible flow field? Explain. (a) What are the corresponding radial and tangential velocity components? (b) Determine the corresponding stream function expressed in Cartesian coordinates and in cylindrical polar coordinates. 6.20 Some velocity measurements in a three-dimensional incompressible flow field indicate that u 6xy2 and v 4y2z. There is some conflicting data for the velocity component in the z direction. One set of data indicates that w 4yz2 and the other set indicates that w 4yz2 6y2z. Which set do you think is correct? Explain. 6.21 A two-dimensional, incompressible flow is given by u y and v x. Show that the streamline passing through the point x 10 and y 0 is a circle centered at the origin. 6.22 In a certain steady, two-dimensional flow field the fluid density varies linearly with respect to the coordinate x: that is, r Ax where A is a constant. If the x component of velocity u is given by the equation u y, determine an expression for v. 6.23 In a two-dimensional, incompressible flow field, the x component of velocity is given by the equation u 2x. (a) Determine the corresponding equation for the y component of velocity if v 0 along the x axis. (b) For this flow field, what is the magnitude y, ft 1.0 A 6.13 The velocity components of an incompressible, twodimensional velocity field are given by the equations u y2 x11 x2 v y12x 12 Show that the flow is irrotational and satisfies conservation of mass. O 1.0 x, ft ■ Figure P6.23 336 Chapter 6 ■ Differential Analysis of Fluid Flow of the average velocity of the fluid crossing the surface OA of Fig. P6.23? Assume that the velocities are in feet per second when x and y are in feet. 6.24 The radial velocity component in an incompressible, two-dimensional flow field (vz 0) is yr 2r 3r 2 sin u Determine the corresponding tangential velocity component, vu, required to satisfy conservation of mass. 6.25 The stream function for an incompressible flow field is given by the equation c 3x2y y3 where the stream function has the units of m2/s with x and y in meters. (a) Sketch the streamline(s) passing through the origin. (b) Determine the rate of flow across the straight path AB shown in Fig. P6.25. (a) Determine the corresponding stream function. (b) What is the relationship between the discharge, q (per unit width normal to plane of paper) passing between the walls and the coordinates xi, yi of any point on the curved wall? Neglect body forces. Section 6.3 Conservation of Linear Momentum 6.30 A fluid with a density of 2000 kg/m3 flows steadily between two flat plates as shown in Fig. P6.30. The bottom plate is fixed and the top one moves at a constant speed in the x direction. The velocity is V 0.20 y ˆi ms where y is in meters. The acceleration of gravity is g 9.8 ˆj ms2. The only nonzero shear stresses, tyx txy, are constant throughout the flow with a value of 5 N/m2. The normal stress at the origin (x y 0) is sxx 100 kPa. Use the x and y components of the equations of motion (Eqs. 6.50a and b) to determine the normal stress throughout the fluid. Assume that sxx syy. y, m 1.0 Moving plate B U b 0 1.0 g y A x, m ■ Figure P6.25 6.26 The streamlines in a certain incompressible, two-dimensional flow field are all concentric circles so that vr 0. Determine the stream function for (a) vu Ar and for (b) vu Ar1, where A is a constant. 6.27 It is proposed that a two-dimensional, incompressible flow field be described by the velocity components u Ay v Bx where A and B are both positive constants. (a) Will the continuity equation be satisfied? (b) Is the flow irrotational? (c) Determine the equation for the streamlines and show a sketch of the streamline that passes through the origin. Indicate the direction of flow along this streamline. 6.28 The stream function for an incompressible, twodimensional flow field is x z Fixed plate ■ Figure P6.30 6.31 A fluid with a density of 2 slug/ft3 flows steadily between two stationary flat plates as shown in Fig. P6.31. The velocity is V 0.5 3 1 1yh2 2 4 ˆi fts where y and h are in feet. The only nonzero shear stresses, tyx txy, are given by tyx 4.0 y lb/ft2 and the acceleration of gravity is negligible. The normal stress at the origin (x y 0) is sxx 10 lb/ft2. Use the x and y components of the equations of motion (Eqs. 6.50a and b) to determine the normal stress throughout the fluid. Assume that sxx syy. c 3x2y y For this flow field, plot several streamlines. 6.29 GO Consider the incompressible, two-dimensional flow of a nonviscous fluid between the boundaries shown in Fig. P6.29. The velocity potential for this flow field is u y h x h umax z f x2 y2 y ■ Figure P6.31 Section 6.4 Inviscid Flow q 6.32 Given the stream function for a flow as c 4x2 4y2, show that the Bernoulli equation can be applied between any two points in the flow field. (xi, yi) B q 6.33 A two-dimensional flow field for a nonviscous, incompressible fluid is described by the velocity components ψ =0 A x ■ Figure P6.29 u U0 2y y0 Problems y 337 6.40 The stream function for a two-dimensional, nonviscous, incompressible flow field is given by the expression c 21x y2 B(0, 1) where the stream function has the units of ft2/s with x and y in feet. (a) Is the continuity equation satisfied? (b) Is the flow field irrotational? If, so, determine the corresponding velocity potential. (c) Determine the pressure gradient in the horizontal x direction at the point x 2 ft, y 2 ft. A(1, 0) p0 x ■ Figure P6.33 6.41 The velocity potential for a certain inviscid, incompressible flow field is given by the equation where U0 is a constant. If the pressure at the origin (Fig. P6.33) is p0, determine an expression for the pressure at (a) point A, and (b) point B. Explain clearly how you obtained your answer. Assume that the units are consistent and body forces may be neglected. where f has the units of m2/s when x and y are in meters. Determine the pressure at the point x 2 m, y 2 m if the pressure at x 1 m, y 1 m is 200 kPa. Elevation changes can be neglected, and the fluid is water. 6.34 In a certain two-dimensional flow field, the velocity is constant with components u 4 ft/s and v 2 ft/s. Determine the corresponding stream function and velocity potential for this flow field. Sketch the equipotential line f 0 which passes through the origin of the coordinate system. 6.42 A steady, uniform, incompressible, inviscid, two-dimensional flow makes an angle of 30 with the horizontal x axis. (a) Determine the velocity potential and the stream function for this flow. (b) Determine an expression for the pressure gradient in the vertical y direction. What is the physical interpretation of this result? 6.35 field is 6.43 The streamlines for an incompressible, inviscid, twodimensional flow field are all concentric circles, and the velocity varies directly with the distance from the common center of the streamlines; that is The stream function for a given two-dimensional flow c 5x2y 1532y3 Determine the corresponding velocity potential. 6.36 yu Kr A certain flow field is described by the stream function c A u B r sin u where A and B are positive constants. Determine the corresponding velocity potential and locate any stagnation points in this flow field. 6.37 It is known that the velocity distribution for twodimensional flow of a viscous fluid between wide parallel plates (Fig. P6.37) is parabolic; that is, y 2 u Uc c 1 a b d h with v 0. Determine, if possible, the corresponding stream function and velocity potential. h f 2x2y 1 23 2y3 y u x where K is a constant. (a) For this rotational flow, determine, if possible, the stream function. (b) Can the pressure difference between the origin and any other point be determined from the Bernoulli equation? Explain. 6.44 The velocity potential f k1x2 y2 2 1k constant2 may be used to represent the flow against an infinite plane boundary, as illustrated in Fig. P6.44. For flow in the vicinity of a stagnation point, it is frequently assumed that the pressure gradient along the surface is of the form 0p Ax 0x where A is a constant. Use the given velocity potential to show that this is true. Uc y h ■ Figure P6.37 6.38 The velocity potential for a certain inviscid flow field is f 13x2y y3 2 where f has the units of ft2/s when x and y are in feet. Determine the pressure difference (in psi) between the points (1, 2) and (4, 4), where the coordinates are in feet, if the fluid is water and elevation changes are negligible. 6.39 The velocity potential for a flow is given by a f 1x2 y2 2 2 where a is a constant. Determine the corresponding stream function and sketch the flow pattern. x ■ Figure P6.44 6.45 In a certain steady, two-dimensional flow field the fluid may be assumed to be ideal and the weight of the fluid (specific weight 50 lb/ft3) is the only body force. The x component of velocity is known to be u 6x, which gives the velocity in ft/s when x is measured in feet, and the y component of velocity is known to be a function of only y. The y axis is vertical, and at the origin the 338 Chapter 6 ■ Differential Analysis of Fluid Flow velocity is zero. (a) Determine the y component of velocity so that the continuity equation is satisfied. (b) Can the difference in pressures between the points x 1 ft, y 1 ft and x 1 ft, y 4 ft be determined from the Bernoulli equation? If so, determine the value in lb/ft2. If not, explain why not. radius of the core. Velocity measurements at points A and B indicate that VA 125 ft/s and VB 60 ft/s. Determine the distance from point A to the center of the tornado. Why can the free vortex model not be used to approximate the tornado throughout the flow field (r 0)? 6.46 Water is flowing between wedge-shaped walls into a small opening as shown in Fig. P6.46. The velocity potential with units m2/s for this flow is f 2 ln r with r in meters. Determine the pressure differential between points A and B. 6.51 The streamlines in a particular two-dimensional flow field are all concentric circles, as shown in Fig. P6.51. The velocity is given by the equation vu vr where v is the angular velocity of the rotating mass of fluid. Determine the circulation around the path ABCD. B π 6 r C A θ D ω 0.5 m A Δθ B ■ Figure P6.46 1.0 m a r b 6.47 A certain flow field is described by the velocity potential f A ln r Br cos u where A and B are positive constants. Determine the corresponding stream function and locate any stagnation points in this flow field. 6.48 GO The velocity potential for a given two-dimensional flow field is f 1 53 2x3 5xy2 Show that the continuity equation is satisfied and determine the corresponding stream function. 6.49 It is suggested that the velocity potential for the incompressible, nonviscous, two-dimensional flow along the wall shown in Fig. P6.49 is ■ Figure P6.51 6.52 The motion of a liquid in an open tank is that of a combined vortex consisting of a forced vortex for 0 r 2 ft and a free vortex for r 2 ft. The velocity profile and the corresponding shape of the free surface are shown in Fig. P6.52. The free surface at the center of the tank is a depth h below the free surface at r q. Determine the value of h. Note that h hforced hfree, where hforced and hfree are the corresponding depths for the forced vortex and the free vortex, respectively. (See Section 2.12.2 for further discussion of the forced vortex.) f r43 cos 43 u Is this a suitable velocity potential for flow along the wall? Explain. 10 vθ , ft/s r 3π /4 0 θ 2 z 2 ■ Figure P6.49 Section 6.5 Some Basic, Plane Potential Flows r, ft r, ft h 6.50 As illustrated in Fig. P6.50, a tornado can be approximated by a free vortex of strength for r Rc, where Rc is the ■ Figure P6.52 y 6.53 A source of strength m is located a distance / from a vertical solid wall as shown in Fig. P6.53. The velocity potential for this incompressible, irrotational flow is given by r Rc A B x f 100 ft ■ Figure P6.50 m 5ln 3 1x /2 2 y2 4 ln 3 1x /2 2 y2 4 6 4p (a) Show that there is no flow through the wall. (b) Determine the velocity distribution along the wall. (c) Determine the pressure distribution along the wall, assuming p p0 far from the source. Neglect the effect of the fluid weight on the pressure. Problems y Source 339 6.56 Water flows over a flat surface at 4 ft/s, as shown in Fig. P6.56. A pump draws off water through a narrow slit at a volume rate of 0.1 ft3/s per foot length of the slit. Assume that the fluid is incompressible and inviscid and can be represented by the combination of a uniform flow and a sink. Locate the stagnation point on the wall (point A) and determine the equation for the stagnation streamline. How far above the surface, H, must the fluid be so that it does not get sucked into the slit? x 4 ft/s ᐉ H A ■ Figure P6.53 6.54 GO Water flows through a two-dimensional diffuser having a 20 expansion angle as shown in Fig. P6.54. Assume that the flow in the diffuser can be treated as a radial flow emanating from a source at the origin O. (a) If the velocity at the entrance is 20 m/s, determine an expression for the pressure gradient along the diffuser walls. (b) What is the pressure rise between the entrance and exit? 0.1 ft3/s (per foot of length of slit) ■ Figure P6.56 6.57 Two sources, one of strength m and the other with strength 3m, are located on the x axis as shown in Fig. P6.57. Determine the location of the stagnation point in the flow produced by these sources. Diffuser wall y 2ᐉ 2m r x 20° O +m Flow +3m ■ Figure P6.57 Entrance 7m Exit 6.58 The velocity potential for a spiral vortex flow is given by f ( /2p) u (m/2p) ln r, where and m are constants. Show that the angle, a, between the velocity vector and the radial direction is constant throughout the flow field (see Fig. P6.58). ■ Figure P6.54 6.55 When water discharges from a tank through an opening in its bottom, a vortex may form with a curved surface profile, as shown in Fig. P6.55 and Video V6.4. Assume that the velocity distribution in the vortex is the same as that for a free vortex. At the same time the water is being discharged from the tank at point A, it is desired to discharge a small quantity of water through the pipe B. As the discharge through A is increased, the strength of the vortex, as indicated by its circulation, is increased. Determine the maximum strength that the vortex can have in order that no air is sucked in at B. Express your answer in terms of the circulation. Assume that the fluid level in the tank at a large distance from the opening at A remains constant and viscous effects are negligible. B 2 ft A ■ Figure P6.55 3ᐉ 1 ft y V α r θ x ■ Figure P6.58 6.59 For a free vortex (see Video V6.4) determine an expression for the pressure gradient (a) along a streamline, and (b) normal to a streamline. Assume that the streamline is in a horizontal plane, and express your answer in terms of the circulation. 6.60 (See Fluids in the News article titled “Some Hurricane Facts,” Section 6.5.3.) Consider a category five hurricane that has a maximum wind speed of 160 mph at the eye wall, 10 mi from the center of the hurricane. If the flow in the hurricane outside of the hurricane’s eye is approximated as a free vortex, determine the wind speeds at locations 20 mi, 30 mi, and 40 mi from the center of the storm. 340 Chapter 6 ■ Differential Analysis of Fluid Flow Section 6.6 Superposition of Basic, Plane Potential Flows 6.61 A plane flow field is developed by the addition of a free vortex and a uniform stream in the positive x direction. If the vortex is located at the origin, determine the pressure variation in this flow field. Neglect the effect of the fluid weight. Express your answer in terms of the uniform velocity, U, the strength of the vortex, , and the pressure, p0, far from the origin. 6.62 Potential flow against a flat plate (Fig. P6.62a) can be described with the stream function c Axy where A is a constant. This type of flow is commonly called a stagnation point flow since it can be used to describe the flow in the vicinity of the stagnation point at O. By adding a source of strength m at O, stagnation point flow against a flat plate with a “bump” is obtained as illustrated in Fig. P6.62b. Determine the relationship between the bump height, h, the constant, A, and the source strength, m. B y Windshield r U = 55 mph 1.5 ft θ A x 2.0 ft ■ Figure P6.66 6.67 A body having the general shape of a half-body is placed in a stream of fluid. At a great distance upstream the velocity is U as shown in Fig. P6.67. Show how a measurement of the differential pressure between the stagnation point and point A can be used to predict the free-stream velocity, U. Express the pressure differential in terms of U and fluid density. Neglect body forces and assume that the fluid is nonviscous and incompressible. y U A y x ■ Figure P6.67 O (a) 6.68 One end of a pond has a shoreline that resembles a half-body as shown in Fig. P6.68. A vertical porous pipe is located near the end of the pond so that water can be pumped out. When water is pumped at the rate of 0.08 m3/s through a 3-m-long pipe, what will be the velocity at point A? Hint: Consider the flow inside a half-body. (See Video V6.5.) x y h Pipe x A Source (b) ■ Figure P6.62 6.63 Show on a plot several “bumps” that can be generated by combining stagnation point flow against a flat plate and a source as described in Problem 6.62. 6.64 Consider a uniform flow in the positive x direction combined with a free vortex located at the origin of the coordinate system. The streamline c 0 passes through the point x 4, y 0. Determine the equation of this streamline. 6.65 The combination of a uniform flow and a source can be used to describe flow around a streamlined body called a half-body. (See Video V6.5.) Assume that a certain body has the shape of a half-body with a thickness of 0.5 m. If this body is placed in an airstream moving at 15 m/s, what source strength is required to simulate flow around the body? 6.66 A vehicle windshield is to be shaped as a portion of a half-body with the dimensions shown in Fig. P6.66. (a) Make a scale drawing of the windshield shape. (b) For a free-stream velocity of 55 mph, determine the velocity of the air at points A and B. 5m 15 m ■ Figure P6.68 6.69 Consider two sources having equal strengths located along the x axis at x 0 and x 2 m, and a sink located on the y axis at y 2 m. Determine the magnitude and direction of the fluid velocity at x 5 m and y 0 due to this combination if the flowrate from each of the sources is 0.5 m3/s per m and the flowrate into the sink is 1.0 m3/s per m. 6.70 Consider a uniform flow with velocity U in the positive x direction combined with two free vortices of equal strength located along the y-axis. Let one vortex located at y a be a clockwise vortex 1c K ln r2 and the other at y a be a counterclockwise vortex, where K is a positive constant. It can be shown by plotting streamlines that for Ua/K 2 the streamline c 0 forms a closed contour, as shown in Fig. P6.70. Thus, this combination Problems can be used to represent flow around a family of bodies (called Kelvin ovals). Show, with the aid of a graph, how the dimensionless height, H/a, varies with the parameter Ua/K in the range 0.3 Ua/K 1.75. U 341 y p2 p1 a 60° x y ■ Figure P6.74 U Ua < 2 K H a Ua = 2 K x a 6.75 Water flows around a 6-ft-diameter bridge pier GO with a velocity of 12 ft/s. Estimate the force (per unit length) that the water exerts on the pier. Assume that the flow can be approximated as an ideal fluid flow around the front half of the cylinder, but due to flow separation (see Video V6.8), the average pressure on the rear half is constant and approximately equal to 1⁄2 the pressure at point A (see Fig. P6.75). A ■ Figure P6.70 U = 12 ft/s 6.71 A Rankine oval is formed by combining a source–sink pair, each having a strength of 36 ft2/s and separated by a distance of 12 ft along the x axis, with a uniform velocity of 10 ft/s (in the positive x direction). Determine the length and thickness of the oval. 6.72 Make use of Eqs. 6.107 and 6.109 to construct a table showing how /a, ha, and /h for Rankine ovals depend on the parameter pUa m. Plot /h versus pUa m and describe how this plot could be used to obtain the required values of m and a for a Rankine oval having a specific value of / and h when placed in a uniform fluid stream of velocity, U. 6.73 An ideal fluid flows past an infinitely long, semicircular “hump” located along a plane boundary, as shown in Fig. P6.73. Far from the hump the velocity field is uniform, and the pressure is p0. (a) Determine expressions for the maximum and minimum values of the pressure along the hump, and indicate where these points are located. Express your answer in terms of r, U, and p0. (b) If the solid surface is the c 0 streamline, determine the equation of the streamline passing through the point u p2, r 2a. 6 ft ■ Figure P6.75 6.76 Consider the steady potential flow around the circular cylinder shown in Fig. 6.26. On a plot show the variation of the magnitude of the dimensionless fluid velocity, V/U, along the positive y axis. At what distance, y/a (along the y axis), is the velocity within 1% of the free-stream velocity? 6.77 The velocity potential for a cylinder (Fig. P6.77) rotating in a uniform stream of fluid is f Ur a1 where is the circulation. For what value of the circulation will the stagnation point be located at: (a) point A; (b) point B? U y r U, p0 θ r a a2 u b cos u 2p r2 a θ A x B ■ Figure P6.73 ■ Figure P6.77 6.74 Assume that the flow around the long circular cylinder of Fig. P6.74 is nonviscous and incompressible. Two pressures, p1 and p2, are measured on the surface of the cylinder, as illustrated. It is proposed that the free-stream velocity, U, can be related to the pressure difference p p1 p2 by the equation 6.78 Show that for a rotating cylinder in a uniform flow, the following pressure ratio equation is true. UC ¢p B r where r is the fluid density. Determine the value of the constant C. Neglect body forces. ptop pbottom pstagnation 8q U Here U is the velocity of the uniform flow and q is the surface speed of the rotating cylinder. 6.79 (See Fluids in the News article titled “A Sailing Ship without Sails,” Section 6.6.3.) Determine the magnitude of the total 342 Chapter 6 ■ Differential Analysis of Fluid Flow force developed by the two rotating cylinders on the Flettner “rotorship” due to the Magnus effect. Assume a wind-speed relative to the ship of (a) 10 mph and (b) 30 mph. Each cylinder has a diameter of 9 ft, a length of 50 ft, and rotates at 750 rev/min. Use Eq. 6.124 and calculate the circulation by assuming the air sticks to the rotating cylinders. Note: This calculated force is at right angles to the direction of the wind and it is the component of this force in the direction of motion of the ship that gives the propulsive thrust. Also, due to viscous effects, the actual propulsive thrust will be smaller than that calculated from Eq. 6.124 which is based on inviscid flow theory. 6.80 A long porous pipe runs parallel to a horizontal plane surface as shown in Fig. P6.80. The longitudinal axis of the pipe is perpendicular to the plane of the paper. Water flows radially from the pipe at a rate of 0.5 p ft3/s per foot of pipe. Determine the difference in pressure (in lb/ft2) between point B and point A. The flow from the pipe may be approximated by a two-dimensional source. Hint: To develop the stream function or velocity potential for this type of flow, place (symmetrically) another equal source on the other side of the wall. With this combination there is no flow across the x axis, and this axis can be replaced with a solid boundary. This technique is called the method of images. Pipe 3 ft y A B x Show that this flow field is also a solution of the Navier–Stokes equations of motion for viscous flow. 6.83 The stream function for a certain incompressible, twodimensional flow field is c 3r3 sin 2u 2u 2 where c is in ft /s when r is in feet and u in radians. Determine the shearing stress, tru , at the point r 2 ft, u p/3 radians if the fluid is water. 6.84 Determine the shearing stress for an incompressible Newtonian fluid with a velocity distribution of V 13xy2 4x3 2iˆ 112x2y y3 2jˆ. 6.85 The two-dimensional velocity field for an incompressible Newtonian fluid is described by the relationship V 112xy2 6x3 2iˆ 118x2y 4y3 2jˆ where the velocity has units of m/s when x and y are in meters. Determine the stresses sxx, syy, and txy at the point x 0.5 m, y 1.0 m if pressure at this point is 6 kPa and the fluid is glycerin at 20 C. Show these stresses on a sketch. 6.86 The velocity of a fluid particle moving along a horizontal streamline that coincides with the x axis in a plane, twodimensional, incompressible flow field was experimentally found to be described by the equation u x2. Along this streamline determine an expression for (a) the rate of change of the v component of velocity with respect to y, (b) the acceleration of the particle, and (c) the pressure gradient in the x direction. The fluid is Newtonian. Section 6.9.1 Steady, Laminar Flow between Fixed Parallel Plates 4 ft ■ Figure P6.80 6.81 Typical inviscid flow solutions for flow around bodies indicate that the fluid flows smoothly around the body, even for blunt bodies as shown in Video V6.11. However, experience reveals that due to the presence of viscosity, the main flow may actually separate from the body, creating a wake behind the body. As discussed in a later section (Section 9.2.6), whether or not separation takes place depends on the pressure gradient along the surface of the body, as calculated by inviscid flow theory. If the pressure decreases in the direction of flow (a favorable pressure gradient), no separation will occur. However, if the pressure increases in the direction of flow (an adverse pressure gradient), separation may occur. For the circular cylinder of Fig. P6.81 placed in a uniform stream with velocity, U, determine an expression for the pressure gradient in the direction of flow on the surface of the cylinder. For what range of values for the angle u will an adverse pressure gradient occur? 6.87 GO Oil (SAE 30) at 15.6 C flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the channel is 30 kPa/m, and the distance between the plates is 4 mm. The flow is laminar. Determine: (a) the volume rate of flow (per meter of width), (b) the magnitude and direction of the shearing stress acting on the bottom plate, and (c) the velocity along the centerline of the channel. 6.88 Two fixed, horizontal, parallel plates are spaced 0.4 in. apart. A viscous liquid (m 8 103 lb s/ft2, SG 0.9) flows between the plates with a mean velocity of 0.5 ft/s. The flow is laminar. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel? 6.89 A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates of Fig. P6.89. Determine, by use of the Navier–Stokes equations, an expression for the pressure gradient in the direction of flow. Express your answer in terms of the mean velocity. Assume that the flow is laminar, steady, and uniform. U Direction of flow θ a y ■ Figure P6.81 z x Section 6.8 Viscous Flow 6.82 The velocity field for a two-dimensional source or sink flow is a solution of the Euler equations of motion for inviscid flow. h h ■ Figure P6.89 Problems 6.90 A fluid of density r flows steadily downward between the two vertical, infinite, parallel plates shown in the figure for Problem 6.89. The flow is fully developed and laminar. Make use of the Navier–Stokes equation to determine the relationship between the discharge and the other parameters involved, for the case in which the change in pressure along the channel is zero. 6.91 (See Fluids in the News article titled “10 Tons on 8 psi,” Section 6.9.1.) A massive, precisely machined, 6-ft-diameter granite sphere rests on a 4-ft-diameter cylindrical pedestal as shown in Fig. P6.91. When the pump is turned on and the water pressure within the pedestal reaches 8 psi, the sphere rises off the pedestal, creating a 0.005-in. gap through which the water flows. The sphere can then be rotated about any axis with minimal friction. (a) Estimate the pump flowrate, Q0, required to accomplish this. Assume the flow in the gap between the sphere and the pedestal is essentially viscous flow between fixed, parallel plates. (b) Describe what would happen if the pump flowrate were increased to 2Q0. 343 plates move in opposite directions with constant velocities, U1 and U2, as shown. The pressure gradient in the x direction is zero, and the only body force is due to the fluid weight. Use the Navier–Stokes equations to derive an expression for the velocity distribution between the plates. Assume laminar flow. U1 b y x U2 ■ Figure P6.94 6.95 Two immiscible, incompressible, viscous fluids having the same densities but different viscosities are contained between two infinite, horizontal, parallel plates (Fig. P6.95). The bottom plate is fixed, and the upper plate moves with a constant velocity U. Determine the velocity at the interface. Express your answer in terms of U, m1, and m2. The motion of the fluid is caused entirely by the movement of the upper plate; that is, there is no pressure gradient in the x direction. The fluid velocity and shearing stress are continuous across the interface between the two fluids. Assume laminar flow. 6 ft U 0.005 in. ρ , μ1 h 4 in. 4 ft y h p 8 psi ρ, μ 2 x Fixed plate ■ Figure P6.95 Pump ■ Figure P6.91 Section 6.9.2 Couette Flow 6.92 Two horizontal, infinite, parallel plates are spaced a distance b apart. A viscous liquid is contained between the plates. The bottom plate is fixed, and the upper plate moves parallel to the bottom plate with a velocity U. Because of the no-slip boundary condition (see Video V6.12), the liquid motion is caused by the liquid being dragged along by the moving boundary. There is no pressure gradient in the direction of flow. Note that this is a socalled simple Couette flow discussed in Section 6.9.2. (a) Start with the Navier–Stokes equations and determine the velocity distribution between the plates. (b) Determine an expression for the flowrate passing between the plates (for a unit width). Express your answer in terms of b and U. 6.93 A layer of viscous liquid of constant thickness (no velocity perpendicular to plate) flows steadily down an infinite, inclined plane. Determine, by means of the Navier–Stokes equations, the relationship between the thickness of the layer and the discharge per unit width. The flow is laminar, and assume air resistance is negligible so that the shearing stress at the free surface is zero. 6.94 An incompressible, viscous fluid is placed between horizontal, infinite, parallel plates as is shown in Fig. P6.94. The two 6.96 The viscous, incompressible flow between the parallel plates shown in Fig. P6.96 is caused by both the motion of the bottom plate and a pressure gradient, 0p 0x. As noted in Section 6.9.2, an important dimensionless parameter for this type of problem is P (b2/2 mU) 1 0p 0x2 where m is the fluid viscosity. Make a plot of the dimensionless velocity distribution (similar to that shown in Fig. 6.32b) for P 3. For this case where does the maximum velocity occur? Fixed plate b y x U ■ Figure P6.96 6.97 A viscous fluid (specific weight 80 lb/ft3; viscosity 0.03 lb s/ft2) is contained between two infinite, horizontal parallel plates as shown in Fig. P6.97. The fluid moves between the plates under the action of a pressure gradient, and the upper plate moves with a velocity U while the bottom plate is fixed. A U-tube 344 Chapter 6 ■ Differential Analysis of Fluid Flow manometer connected between two points along the bottom indicates a differential reading of 0.1 in. If the upper plate moves with a velocity of 0.02 ft/s, at what distance from the bottom plate does the maximum velocity in the gap between the two plates occur? Assume laminar flow. U = 0.02 ft/s 1.0 in. 6.103 A simple flow system to be used for steady-flow tests consists of a constant head tank connected to a length of 4-mmdiameter tubing as shown in Fig. P6.103. The liquid has a viscosity of 0.015 N ⭈ s/m2, a density of 1200 kg/m3, and discharges into the atmosphere with a mean velocity of 2 m/s. (a) Verify that the flow will be laminar. (b) The flow is fully developed in the last 3 m of the tube. What is the pressure at the pressure gage? (c) What is the magnitude of the wall shearing stress, trz, in the fully developed region? y 6 in. x Fixed plate 0.1 in. γ = 100 lb/ft3 Pressure gage ■ Figure P6.97 6.98 An infinitely long, solid, vertical cylinder of radius R is located in an infinite mass of an incompressible fluid. Start with the Navier–Stokes equation in the u direction and derive an expression for the velocity distribution for the steady flow case in which the cylinder is rotating about a fixed axis with a constant angular velocity v. You need not consider body forces. Assume that the flow is axisymmetric and the fluid is at rest at infinity. 6.99 A vertical shaft passes through a bearing and is lubricated with an oil having a viscosity of 0.2 N ⭈ s/m2 as shown in Fig. P6.99. Assume that the flow characteristics in the gap between the shaft and bearing are the same as those for laminar flow between infinite parallel plates with zero pressure gradient in the direction of flow. Estimate the torque required to overcome viscous resistance when the shaft is turning at 80 rev/min. Shaft Bearing 160 mm 0.25 mm ■ Figure P6.103 6.104 (a) Show that for Poiseuille flow in a tube of radius R the magnitude of the wall shearing stress, trz, can be obtained from the relationship 4mQ 0 1trz 2 wall 0 ⫽ pR3 for a Newtonian fluid of viscosity m. The volume rate of flow is Q. (b) Determine the magnitude of the wall shearing stress for a fluid having a viscosity of 0.004 N ⭈ s/m2 flowing with an average velocity of 130 mm/s in a 2-mm-diameter tube. 6.105 An infinitely long, solid, vertical cylinder of radius R is located in an infinite mass of an incompressible fluid. Start with the Navier–Stokes equation in the u direction and derive an expression for the velocity distribution for the steady-flow case in which the cylinder is rotating about a fixed axis with a constant angular velocity v. You need not consider body forces. Assume that the flow is axisymmetric and the fluid is at rest at infinity. 75 mm Oil 3m Diameter = 4 mm ■ Figure P6.99 6.100 A viscous fluid is contained between two long concentric cylinders. The geometry of the system is such that the flow between the cylinders is approximately the same as the laminar flow between two infinite parallel plates. (a) Determine an expression for the torque required to rotate the outer cylinder with an angular velocity v. The inner cylinder is fixed. Express your answer in terms of the geometry of the system, the viscosity of the fluid, and the angular velocity. (b) For a small, rectangular element located at the fixed wall, determine an expression for the rate of angular deformation of this element. (See Video V6.3 and Fig. P6.10.) 6.101 Oil (SAE 30) flows between parallel plates spaced 5 mm apart. The bottom plate is fixed, but the upper plate moves with a velocity of 0.2 m/s in the positive x direction. The pressure gradient is 60 kPa/m, and it is negative. Compute the velocity at various points across the channel and show the results on a plot. Assume laminar flow. Section 6.9.3 Steady, Laminar Flow in Circular Tubes 6.102 Ethyl alcohol flows through a horizontal tube having a diameter of 10 mm. If the mean velocity is 0.15 m/s, what is the pressure drop per unit length along the tube? What is the velocity at a distance of 2 mm from the tube axis? 6.106 As is shown by Eq. 6.150, the pressure gradient for laminar flow through a tube of constant radius is given by the expression 0p 8mQ ⫽⫺ 4 0z pR For a tube whose radius is changing very gradually, such as the one illustrated in Fig. P6.106, it is expected that this equation can be used to approximate the pressure change along the tube if the actual radius, R(z), is used at each cross section. The following measurements were obtained along a particular tube. zⲐ/ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 R1z2 ⲐRo 1.00 0.73 0.67 0.65 0.67 0.80 0.80 0.71 0.73 0.77 1.00 Compare the pressure drop over the length / for this nonuniform tube with one having the constant radius Ro. Hint: To solve this problem you will need to numerically integrate the equation for the pressure gradient given previously. Ro R(z) z ᐉ ■ Figure P6.106 Problems 6.107 GO A liquid (viscosity 0.002 N s/m2; density 1000 kg/m3) is forced through the circular tube shown in Fig. P6.107. A differential manometer is connected to the tube as shown to measure the pressure drop along the tube. When the differential reading, h, is 9 mm, what is the mean velocity in the tube? 4 mm 2m 345 that there are no velocity components other than the tangential component. The only body force is the weight. 6.110 For flow between concentric cylinders, with the outer cylinder rotating at an angular velocity v and the inner cylinder fixed, it is commonly assumed that the tangential velocity (vu) distribution in the gap between the cylinders is linear. Based on the exact solution to this problem (see Problem 6.109) the velocity distribution in the gap is not linear. For an outer cylinder with radius ro 2.00 in. and an inner cylinder with radius ri 1.80 in., show, with the aid of a plot, how the dimensionless velocity distribution, vuro v, varies with the dimensionless radial position, r兾ro, for the exact and approximate solutions. 6.111 A viscous liquid (m 0.012 lb s/ft2, r 1.79 slugs/ft3) flows through the annular space between two horizontal, fixed, concentric cylinders. If the radius of the inner cylinder is 1.5 in. and the radius of the outer cylinder is 2.5 in., what is the pressure drop along the axis of the annulus per foot when the volume flowrate is 0.14 ft3/s? Δh Density of gage fluid = 2000 kg/m3 ■ Figure P6.107 6.112 Section 6.9.4 Steady, Axial, Laminar Flow in an Annulus 6.108 An incompressible Newtonian fluid flows steadily between two infinitely long, concentric cylinders as shown in Fig. P6.108. The outer cylinder is fixed, but the inner cylinder moves with a longitudinal velocity V0 as shown. The pressure gradient in the axial direction is ¢p/. For what value of V0 will the drag on the inner cylinder be zero? Assume that the flow is laminar, axisymmetric, and fully developed. 6.113 A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pipe, by how much does the presence of the wire reduce the flowrate if (a) d兾D 0.1; (b) d兾D 0.01? Section 6.10 Other Aspects of Differential Analysis 6.114 Obtain a photograph/image of a situation in which CFD has been used to solve a fluid flow problem. Print this photo and write a brief paragraph that describes the situation involved. ■ Lifelong Learning Problems Fixed wall ro V0 Show how Eq. 6.155 is obtained. ri ■ Figure P6.108 6.109 A viscous fluid is contained between two infinitely long, vertical, concentric cylinders. The outer cylinder has a radius ro and rotates with an angular velocity v. The inner cylinder is fixed and has a radius ri. Make use of the Navier–Stokes equations to obtain an exact solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position u within the gap) and 6.1LL What sometimes appear at first glance to be simple fluid flows can contain subtle, complex fluid mechanics. One such example is the stirring of tea leaves in a teacup. Obtain information about “Einstein’s tea leaves” and investigate some of the complex fluid motions interacting with the leaves. Summarize your findings in a brief report. 6.2LL Computational fluid dynamics (CFD) has moved from a research tool to a design tool for engineering. Initially, much of the work in CFD was focused in the aerospace industry, but now has expanded into other areas. Obtain information on what other industries (e.g., automotive) make use of CFD in their engineering design. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 7 Dimensional Analysis, Similitude, and Modeling CHAPTER OPENING PHOTO: Flow past a circular cylinder with Re ⫽ rVDⲐm ⫽ 2000: The pathlines of flow past any circular cylinder 1regardless of size, velocity, or fluid2 are as shown provided that the dimensionless parameter called the Reynolds number, Re, is equal to 2000. For other values of Re, the flow pattern will be different 1air bubbles in water2. (Photograph courtesy of ONERA, The French Aerospace Lab.) Learning Objectives After completing this chapter, you should be able to: ■ apply the Buckingham pi theorem. ■ develop a set of dimensionless variables for a given flow situation. ■ discuss the use of dimensionless variables in data analysis. ■ apply the concepts of modeling and similitude to develop prediction equations. Experimentation and modeling are widely used techniques in fluid mechanics. V7.1 Real and model flies 346 Although many practical engineering problems involving fluid mechanics can be solved by using the equations and analytical procedures described in the preceding chapters, there remain a large number of problems that rely on experimentally obtained data for their solution. In fact, it is probably fair to say that very few problems involving real fluids can be solved by analysis alone. The solution to many problems is achieved through the use of a combination of theoretical and numerical analysis and experimental data. Thus, engineers working on fluid mechanics problems should be familiar with the experimental approach to these problems so that they can interpret and make use of data obtained by others, such as might appear in handbooks, or be able to plan and execute the necessary experiments in their own laboratories. In this chapter we consider some techniques and ideas that are important in the planning and execution of experiments, as well as in understanding and correlating data that may have been obtained by other experimenters. An obvious goal of any experiment is to make the results as widely applicable as possible. To achieve this end, the concept of similitude is often used so that measurements made on one system 1for example, in the laboratory2 can be used to describe the behavior of other similar systems 1outside the laboratory2. The laboratory systems are usually thought of as models and are used to study the phenomenon of interest under carefully controlled conditions. From these model studies, empirical formulations can be developed, or specific predictions of one or more characteristics of some other similar system can be made. To do this, it is necessary to establish the relationship between the laboratory model and the “other” system. In the following sections, we find out how this can be accomplished in a systematic manner. 7.1 F l u i d s i n Model study of New Orleans levee breach caused by Hurricane Katrina Much of the devastation to New Orleans from Hurricane Katrina in 2005 was a result of flood waters that surged through a breach of the 17th Street Outfall Canal. To better understand why this occurred and to determine what can be done to prevent future occurrences, the U.S. Army Engineer Research and Development Center Coastal and Hydraulics Laboratory is conducting tests on a large (1:50 length scale) 15,000-square-foot hydraulic model that replicates 0.5 mile of the canal surrounding the 7.1 t h e Dimensional Analysis N e w 347 s breach and more than a mile of the adjacent Lake Pontchartrain front. The objective of the study is to obtain information regarding the effect that waves had on the breaching of the canal and to investigate the surging water currents within the canals. The waves are generated by computer-controlled wave generators that can produce waves of varying heights, periods, and directions similar to the storm conditions that occurred during the hurricane. Data from the study will be used to calibrate and validate information that will be fed into various numerical model studies of the disaster. Dimensional Analysis To illustrate a typical fluid mechanics problem in which experimentation is required, consider the steady flow of an incompressible Newtonian fluid through a long, smooth-walled, horizontal, circular pipe. An important characteristic of this system, which would be of interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe as a result of friction. Although this would appear to be a relatively simple flow problem, it cannot generally be solved analytically 1even with the aid of large computers2 without the use of experimental data. The first step in the planning of an experiment to study this problem would be to decide on the factors, or variables, that will have an effect on the pressure drop per unit length, ¢p/ 3 1lbft2 2 ft lbft3 or Nm3 4. We expect the list to include the pipe diameter, D, the fluid density, r, fluid viscosity, m, and the mean velocity, V, at which the fluid is flowing through the pipe. Thus, we can express this relationship as ¢p/ f 1D, r, m, V2 It is important to develop a meaningful and systematic way to perform an experiment. (7.1) which simply indicates mathematically that we expect the pressure drop per unit length to be some function of the factors contained within the parentheses. At this point the nature of the function is unknown, and the objective of the experiments to be performed is to determine the nature of this function. To perform the experiments in a meaningful and systematic manner, it would be necessary to change one of the variables, such as the velocity, while holding all others constant, and measure the corresponding pressure drop. This series of tests would yield data that could be represented graphically as is illustrated in Fig. 7.1a. It is to be noted that this plot would only be valid for the specific pipe and for the specific fluid used in the tests; this certainly does not give us the general formulation we are looking for. We could repeat the process by varying each of the other variables in turn, as is illustrated in Figs. 7.1b, 7.1c, and 7.1d. This approach to determining the functional relationship between the pressure drop and the various factors that influence it, although logical in concept, is fraught with difficulties. Some of the experiments would be hard to carry out—for example, to obtain the data illustrated in Fig. 7.1c it would be necessary to vary fluid density while holding viscosity constant. How would you do this? Finally, once we obtained the various curves shown in Figs. 7.1a, 7.1b, 7.1c, and 7.1d, how could we combine these data to obtain the desired general functional relationship between ¢p/, D, r, m, and V which would be valid for any similar pipe system? Fortunately, there is a much simpler approach to this problem that will eliminate such difficulties. In the following sections we will show that rather than working with the original list of variables, as described in Eq. 7.1, we can collect these into two nondimensional combinations of variables 1called dimensionless products or dimensionless groups2 so that D ¢p/ rV 2 fa rVD b m (7.2) 348 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Δ pᐉ Δ pᐉ D, ρ , μ – constant V, ρ, μ– constant V D (a) (b) Δ pᐉ Δ pᐉ D, ρ , V– constant D, V, μ – constant ρ μ (c) (d) ■ Figure 7.1 Illustrative plots showing how the pressure drop in a pipe may be affected by several different factors. Dimensionless products are important and useful in the planning, execution, and interpretation of experiments. Thus, instead of having to work with five variables, we now have only two. The necessary experiment would simply consist of varying the dimensionless product rVDm and determining the corresponding value of D ¢p/rV 2. The results of the experiment could then be represented by a single, universal curve as is illustrated in Fig. 7.2. This curve would be valid for any combination of smooth-walled pipe and incompressible Newtonian fluid. To obtain this curve we could choose a pipe of convenient size and a fluid that is easy to work with. Note that we wouldn’t have to use different pipe sizes or even different fluids. It is clear that the experiment would be much simpler, easier to do, and less expensive 1which would certainly make an impression on your boss2. The basis for this simplification lies in a consideration of the dimensions of the variables involved. As was discussed in Chapter 1, a qualitative description of physical quantities can be given in terms of basic dimensions such as mass, M, length, L, and time, T.1 Alternatively, we could use force, F, L, and T as basic dimensions, since from Newton’s second law F ⬟ MLT 2 D Δpᐉ _____ ρV 2 ρ VD ____ μ 1 ■ Figure 7.2 An illustrative plot of pressure drop data using dimensionless parameters. As noted in Chapter 1, we will use T to represent the basic dimension of time, although T is also used for temperature in thermodynamic relationships 1such as the ideal gas law2. 7.2 Buckingham Pi Theorem 349 1Recall from Chapter 1 that the notation ⬟ is used to indicate dimensional equality.2 The dimensions of the variables in the pipe flow example are ¢p/ ⬟ FL3, D ⬟ L, r ⬟ FL4T 2, m ⬟ FL 2T, and V ⬟ LT 1. 3Note that the pressure drop per unit length has the dimensions of 1FL2 2 L FL3.4 A quick check of the dimensions of the two groups that appear in Eq. 7.2 shows that they are in fact dimensionless products; that is, D ¢p/ rV 2 and ⬟ L1FL3 2 1FL4T 2 21LT 1 2 2 ⬟ F 0L0T 0 1FL 4T 2 21LT 1 21L2 rVD ⬟ ⬟ F 0L0T 0 m 1FL2T2 Not only have we reduced the number of variables from five to two, but also the new groups are dimensionless combinations of variables, which means that the results presented in the form of Fig. 7.2 will be independent of the system of units we choose to use. This type of analysis is called dimensional analysis, and the basis for its application to a wide variety of problems is found in the Buckingham pi theorem described in the following section. 7.2 Buckingham Pi Theorem A fundamental question we must answer is how many dimensionless products are required to replace the original list of variables? The answer to this question is supplied by the basic theorem of dimensional analysis that states the following: If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k ⴚ r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables. Dimensional analysis is based on the Buckingham pi theorem. The dimensionless products are frequently referred to as pi terms, and the theorem is called the Buckingham pi theorem.2 Edgar Buckingham used the symbol ß to represent a dimensionless product, and this notation is commonly used. Although the pi theorem is a simple one, its proof is not so simple and we will not include it here. Many entire books have been devoted to the subject of similitude and dimensional analysis, and a number of these are listed at the end of this chapter 1Refs. 1–152. Students interested in pursuing the subject in more depth 1including the proof of the pi theorem2 can refer to one of these books. The pi theorem is based on the idea of dimensional homogeneity which was introduced in Chapter 1. Essentially we assume that for any physically meaningful equation involving k variables, such as u1 f 1u2, u3, . . . , uk 2 the dimensions of the variable on the left side of the equal sign must be equal to the dimensions of any term that stands by itself on the right side of the equal sign. It then follows that we can rearrange the equation into a set of dimensionless products 1pi terms2 so that ß 1 f1ß 2, ß 3, . . . , ß kr 2 where f1ß 2, ß 3, . . . , ß kr 2 is a function of ß 2 through ß kr. The required number of pi terms is fewer than the number of original variables by r, where r is determined by the minimum number of reference dimensions required to describe the original list of variables. Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T. However, in some instances perhaps only two dimensions, such as L and T, are required, or maybe just one, such as L. Also, in a few rare cases Although several early investigators, including Lord Rayleigh 11842–19192 in the nineteenth century, contributed to the development of dimensional analysis, Edgar Buckingham’s 11867–19402 name is usually associated with the basic theorem. He stimulated interest in the subject in the United States through his publications during the early part of the twentieth century. See, for example, E. Buckingham, On Physically Similar Systems: Illustrations of the Use of Dimensional Equations, Phys. Rev., 4 119142, 345–376. 2 350 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling the variables may be described by some combination of basic dimensions, such as MT 2 and L, and in this case r would be equal to two rather than three. Although the use of the pi theorem may appear to be a little mysterious and complicated, we will actually develop a simple, systematic procedure for developing the pi terms for a given problem. 7.3 Determination of Pi Terms A dimensional analysis can be performed using a series of distinct steps. Several methods can be used to form the dimensionless products, or pi terms, that arise in a dimensional analysis. Essentially we are looking for a method that will allow us to systematically form the pi terms so that we are sure that they are dimensionless and independent and that we have the right number. The method we will describe in detail in this section is called the method of repeating variables. It will be helpful to break the repeating variable method down into a series of distinct steps that can be followed for any given problem. With a little practice you will be able to readily complete a dimensional analysis for your problem. Step 1 List all the variables that are involved in the problem. This step is the most difficult one, and it is, of course, vitally important that all pertinent variables be included. Otherwise the dimensional analysis will not be correct! We are using the term variable to include any quantity, including dimensional and nondimensional constants, which play a role in the phenomenon under investigation. All such quantities should be included in the list of variables to be considered for the dimensional analysis. The determination of the variables must be accomplished by the experimenter’s knowledge of the problem and the physical laws that govern the phenomenon. Typically the variables will include those that are necessary to describe the geometry of the system 1such as a pipe diameter2, to define any fluid properties 1such as a fluid viscosity2, and to indicate external effects that influence the system 1such as a driving pressure drop per unit length2. These general classes of variables are intended as broad categories that should be helpful in identifying variables. It is likely, however, that there will be variables that do not fit easily into one of these categories, and each problem needs to be carefully analyzed. Since we wish to keep the number of variables to a minimum, so that we can minimize the amount of laboratory work, it is important that all variables be independent. For example, if in a certain problem the cross-sectional area of a pipe is an important variable, either the area or the pipe diameter could be used, but not both, since they are obviously not independent. Similarly, if both fluid density, r, and specific weight, g, are important variables, we could list r and g, or r and g 1acceleration of gravity2, or g and g. However, it would be incorrect to use all three since g rg; that is, r, g, and g are not independent. Note that although g would normally be constant in a given experiment, that fact is irrelevant as far as a dimensional analysis is concerned. Step 2 Express each of the variables in terms of basic dimensions. For the typical fluid mechanics problem the basic dimensions will be either M, L, and T or F, L, and T. Dimensionally these two sets are related through Newton’s second law 1F ma2 so that F ⬟ MLT 2. For example, r ⬟ ML3 or r ⬟ FL4T 2. Thus, either set can be used. The basic dimensions for typical variables found in fluid mechanics problems are listed in Table 1.1 in Chapter 1. Step 3 Determine the required number of pi terms. This can be accomplished by means of the Buckingham pi theorem, which indicates that the number of pi terms is equal to k r, where k is the number of variables in the problem 1which is determined from Step 12 and r is the number of reference dimensions required to describe these variables 1which is determined from Step 22. The reference dimensions usually correspond to the basic dimensions and can be determined by an inspection of the dimensions of the variables obtained in Step 2. As previously noted, there may be occasions 1usually rare2 in which the basic dimensions appear in combinations so that the number of reference dimensions is less than the number of basic dimensions. This possibility is illustrated in Example 7.2. Step 4 Select a number of repeating variables, where the number required is equal to the number of reference dimensions. Essentially what we are doing here is selecting from the original list of variables several of which can be combined with each of the remaining 7.3 Step 5 Step 6 Step 7 Step 8 Determination of Pi Terms 351 variables to form a pi term. All of the required reference dimensions must be included within the group of repeating variables, and each repeating variable must be dimensionally independent of the others 1i.e., the dimensions of one repeating variable cannot be reproduced by some combination of products of powers of the remaining repeating variables2. This means that the repeating variables cannot themselves be combined to form a dimensionless product. For any given problem we usually are interested in determining how one particular variable is influenced by the other variables. We would consider this variable to be the dependent variable, and we would want this to appear in only one pi term. Thus, do not choose the dependent variable as one of the repeating variables, since the repeating variables will generally appear in more than one pi term. Form a pi term by multiplying one of the nonrepeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. Essentially each pi term will be of the form uiua1iub2iuc3i where ui is one of the nonrepeating variables; u1, u2, and u3 are the repeating variables; and the exponents ai, bi, and ci are determined so that the combination is dimensionless. Repeat Step 5 for each of the remaining nonrepeating variables. The resulting set of pi terms will correspond to the required number obtained from Step 3. If not, check your work—you have made a mistake! Check all the resulting pi terms to make sure they are dimensionless. It is easy to make a mistake in forming the pi terms. However, this can be checked by simply substituting the dimensions of the variables into the pi terms to confirm that they are all dimensionless. One good way to do this is to express the variables in terms of M, L, and T if the basic dimensions F, L, and T were used initially, or vice versa, and then check to make sure the pi terms are dimensionless. Express the final form as a relationship among the pi terms, and think about what it means. Typically the final form can be written as ß 1 f1ß 2, ß 3, . . . , ß kr 2 By using dimensional analysis, the original problem is simplified and defined with pi terms. (1) (2) D V ρ, μ ᐉ Δpᐉ = (p1 – p2)/ᐉ where ß 1 would contain the dependent variable in the numerator. It should be emphasized that if you started out with the correct list of variables 1and the other steps were completed correctly2, then the relationship in terms of the pi terms can be used to describe the problem. You need only work with the pi terms—not with the individual variables. However, it should be clearly noted that this is as far as we can go with the dimensional analysis; that is, the actual functional relationship among the pi terms must be determined by experiment. To illustrate these various steps we will again consider the problem discussed earlier in this chapter which was concerned with the steady flow of an incompressible Newtonian fluid through a long, smooth-walled, horizontal circular pipe. We are interested in the pressure drop per unit length, ¢p/, along the pipe as illustrated by the figure in the margin. First (Step 1) we must list all of the pertinent variables that are involved based on the experimenter’s knowledge of the problem. In this problem we assume that ¢p/ f 1D, r, m, V2 where D is the pipe diameter, r and m are the fluid density and viscosity, respectively, and V is the mean velocity. Next 1Step 22 we express all the variables in terms of basic dimensions. Using F, L, and T as basic dimensions it follows that ¢p/ ⬟ FL3 D⬟L r ⬟ FL4T 2 m ⬟ FL2T V ⬟ LT 1 352 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Special attention should be given to the selection of repeating variables as detailed in Step 4. We could also use M, L, and T as basic dimensions if desired—the final result will be the same. Note that for density, which is a mass per unit volume 1ML3 2, we have used the relationship F ⬟ MLT 2 to express the density in terms of F, L, and T. Do not mix the basic dimensions; that is, use either F, L, and T or M, L, and T. We can now apply the pi theorem to determine the required number of pi terms 1Step 32. An inspection of the dimensions of the variables from Step 2 reveals that all three basic dimensions are required to describe the variables. Since there are five 1k 52 variables 1do not forget to count the dependent variable, ¢p/2 and three required reference dimensions 1r 32, then according to the pi theorem there will be 15 32, or two pi terms required. The repeating variables to be used to form the pi terms 1Step 42 need to be selected from the list D, r, m, and V. Remember, we do not want to use the dependent variable as one of the repeating variables. Since three reference dimensions are required, we will need to select three repeating variables. Generally, we would try to select as repeating variables those that are the simplest, dimensionally. For example, if one of the variables has the dimension of a length, choose it as one of the repeating variables. In this example we will use D, V, and r as repeating variables. Note that these are dimensionally independent, since D is a length, V involves both length and time, and r involves force, length, and time. This means that we cannot form a dimensionless product from this set. We are now ready to form the two pi terms 1Step 52. Typically, we would start with the dependent variable and combine it with the repeating variables to form the first pi term; that is, ß 1 ¢p/DaV b rc Since this combination is to be dimensionless, it follows that 1FL3 21L2 a 1LT 1 2 b 1FL4T 2 2 c ⬟ F 0L0T 0 The exponents, a, b, and c must be determined such that the resulting exponent for each of the basic dimensions—F, L, and T—must be zero 1so that the resulting combination is dimensionless2. Thus, we can write 1c0 3 a b 4c 0 b 2c 0 1for F2 1for L2 1for T 2 The solution of this system of algebraic equations gives the desired values for a, b, and c. It follows that a 1, b 2, c 1 and, therefore, ß1 ¢p/D rV 2 The process is now repeated for the remaining nonrepeating variables 1Step 62. In this example there is only one additional variable 1m2 so that ß 2 mDaV brc or 1FL2T 21L2 a 1LT 1 2 b 1FL4T 2 2 c ⬟ F 0L0T 0 and, therefore, 1c0 2 a b 4c 0 1 b 2c 0 1for F2 1for L2 1for T 2 Solving these equations simultaneously it follows that a 1, b 1, c 1 so that ß2 m DVr 7.3 Determination of Pi Terms 353 Note that we end up with the correct number of pi terms as determined from Step 3. At this point stop and check to make sure the pi terms are actually dimensionless 1Step 72. We will check using both FLT and MLT dimensions. Thus, ß1 Step 1 ß2 Δpᐉ = f (D, r, m, V ) Step 2 Δpᐉ = FL3, ... ¢p/D ⬟ rV 2 1FL3 21L2 1FL4T 2 21LT 1 2 2 1FL2T 2 m ⬟ ⬟ F 0L0T 0 DVr 1L21LT 1 21FL4T 2 2 or alternatively, ß1 Step 3 ¢p/D rV 2 1ML2T 2 21L2 ⬟ 1ML3 21LT 1 2 2 ⬟ M 0L0T 0 1ML1T 1 2 m ⬟ ß2 ⬟ M 0L0T 0 DVr 1L21LT 1 21ML3 2 k–r=3 Step 4 Finally 1Step 82, we can express the result of the dimensional analysis as D, V, r ¢p/D Step 5 a b c 1 = ΔpᐉD V r Step 6 a b c 2 = mD V r Step 7 ΔpᐉD 0 0 0 _____ =F L T ρV2 Step 8 m ΔpᐉD ∼ _ __ =␾ DVρ ρV2 ⬟ F 0L0T 0 2 rV m f˜ a DVr b This result indicates that this problem can be studied in terms of these two pi terms, rather than the original five variables we started with. The eight steps carried out to obtain this result are summarized by the figure in the margin. Dimensional analysis will not provide the form of the function f˜ . This can only be obtained from a suitable set of experiments. If desired, the pi terms can be rearranged; that is, the reciprocal of mDVr could be used and, of course, the order in which we write the variables can be changed. Thus, for example, ß 2 could be expressed as ß2 rVD m and the relationship between ß 1 and ß 2 as D ¢p/ rV DΔpᐉ ______ 2 fa rVD b m ρV2 ρVD _____ μ The method of repeating variables can be most easily carried out by following a step-bystep procedure. as shown by the figure in the margin. This is the form we previously used in our initial discussion of this problem 1Eq. 7.22. The dimensionless product rVDm is a very famous one in fluid mechanics—the Reynolds number. This number has been briefly alluded to in Chapters 1 and 6 and will be further discussed in Section 7.6. To summarize, the steps to be followed in performing a dimensional analysis using the method of repeating variables are as follows: Step Step Step Step 1 2 3 4 Step 5 Step 6 Step 7 Step 8 List all the variables that are involved in the problem. Express each of the variables in terms of basic dimensions. Determine the required number of pi terms. Select a number of repeating variables, where the number required is equal to the number of reference dimensions 1usually the same as the number of basic dimensions2. Form a pi term by multiplying one of the nonrepeating variables by the product of repeating variables each raised to an exponent that will make the combination dimensionless. Repeat Step 5 for each of the remaining nonrepeating variables. Check all the resulting pi terms to make sure they are dimensionless and independent. Express the final form as a relationship among the pi terms and think about what it means. 354 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling E XAMPLE 7.1 Method of Repeating Variables GIVEN A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid as shown in Fig. E7.1. Assume the drag, d, that the fluid exerts on the plate is a function of w and h, the fluid viscosity and density, ␮ and ␳, respectively, and the velocity V of the fluid approaching the plate. FIND Determine a suitable set of pi terms to study this problem experimentally. V7.2 Flow past a flat plate w V ρ, μ SOLUTION h From the statement of the problem we can write d f 1w, h, m, r, V 2 where this equation expresses the general functional relationship between the drag and the several variables that will affect it. The dimensions of the variables 1using the MLT system2 are ■ Figure E7.1 It follows that 1L21L2 a 1LT 1 2 b 1ML3 2 c ⬟ M 0L0T 0 d ⬟ MLT2 and w⬟L c0 h⬟L 1 a b 3c 0 m ⬟ ML1T1 b0 r ⬟ ML3 We see that all three basic dimensions are required to define the six variables so that the Buckingham pi theorem tells us that three pi terms will be needed 1six variables minus three reference dimensions, k r 6 32. We will next select three repeating variables such as w, V, and r. A quick inspection of these three reveals that they are dimensionally independent, since each one contains a basic dimension not included in the others. Note that it would be incorrect to use both w and h as repeating variables since they have the same dimensions. Starting with the dependent variable, d, the first pi term can be formed by combining d with the repeating variables such that ß2 1for T 2 ß 3 mwaV brc with 1ML1T 1 21L2 a 1LT 1 2 b 1ML3 2 c ⬟ M 0L0T 0 and, therefore, 1c0 1 a b 3c 0 1 b 0 and in terms of dimensions 1MLT 2 21L2 a 1LT 1 2 b 1ML3 2 c ⬟ M 0L0T 0 1for M2 1for L2 1 a b 3c 0 1for T 2 2 b 0 and, therefore, a 2, b 2, and c 1. The pi term then becomes d w V 2r ß3 ß 2 hw V r b c 1for L2 1for T2 m wVr Now that we have the three required pi terms we should check to make sure they are dimensionless. To make this check we use F, L, and T, which will also verify the correctness of the original dimensions used for the variables. Thus, ß1 2 Next the procedure is repeated with the second nonrepeating variable, h, so that 1for M2 Solving for the exponents, we obtain a 1, b 1, c 1 so that Thus, for ß 1 to be dimensionless it follows that 1c0 h w The remaining nonrepeating variable is m so that ß 1 dwaV brc a 1for L2 so that a 1, b 0, c 0, and therefore V ⬟ LT1 ß1 1for M2 ß2 ß3 1F2 d ⬟ ⬟ F 0L0T 0 2 2 1 2 wV r 1L2 1LT 2 1FL4T 2 2 2 1L2 h ⬟ F 0L0T 0 ⬟ w 1L2 1FL2T 2 m ⬟ F 0L0T 0 ⬟ wVr 1L21LT 1 21FL4T 2 2 7.4 Some Additional Comments about Dimensional Analysis If these do not check, go back to the original list of variables and make sure you have the correct dimensions for each of the variables and then check the algebra you used to obtain the exponents a, b, and c. Finally, we can express the results of the dimensional analysis in the form h m d b f˜ a , 2 w wVr wV r 2 example, we could express the final result in the form d w rVw fa , b 2 h m w rV 2 (Ans) which would be more conventional, since the ratio of the plate width to height, wh, is called the aspect ratio, and rVw m is the Reynolds number. (Ans) Since at this stage in the analysis the nature of the function f˜ is unknown, we could rearrange the pi terms if we so desire. For 7.4 355 COMMENT To proceed, it would be necessary to perform a set of experiments to determine the nature of the function f, as discussed in Section 7.7. Some Additional Comments about Dimensional Analysis The preceding section provides a systematic approach for performing a dimensional analysis. Other methods could be used, although we think the method of repeating variables is the easiest for the beginning student to use. Pi terms can also be formed by inspection, as is discussed in Section 7.5. Regardless of the specific method used for the dimensional analysis, there are certain aspects of this important engineering tool that must seem a little baffling and mysterious to the student 1and sometimes to the experienced investigator as well2. In this section we will attempt to elaborate on some of the more subtle points that, based on our experience, can prove to be puzzling to students. 7.4.1 Selection of Variables It is often helpful to classify variables into three groups— geometry, material properties, and external effects. One of the most important and difficult steps in applying dimensional analysis to any given problem is the selection of the variables that are involved. As noted previously, for convenience we will use the term variable to indicate any quantity involved, including dimensional and nondimensional constants. There is no simple procedure whereby the variables can be easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the governing physical laws. If extraneous variables are included, then too many pi terms appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these experimentally. If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly and difficult to ascertain. It is, therefore, imperative that sufficient time and attention be given to this first step in which the variables are determined. Most engineering problems involve certain simplifying assumptions that have an influence on the variables to be considered. Usually we wish to keep the problem as simple as possible, perhaps even if some accuracy is sacrificed. A suitable balance between simplicity and accuracy is a desirable goal. How “accurate” the solution must be depends on the objective of the study; that is, we may be only concerned with general trends and, therefore, some variables that are thought to have only a minor influence in the problem may be neglected for simplicity. For most engineering problems 1including areas outside of fluid mechanics2, pertinent variables can be classified into three general groups—geometry, material properties, and external effects. Geometry. The geometric characteristics can usually be described by a series of lengths and angles. In most problems the geometry of the system plays an important role, and a sufficient number of geometric variables must be included to describe the system. These variables can usually be readily identified. Material Properties. Since the response of a system to applied external effects such as forces, pressures, and changes in temperature is dependent on the nature of the materials involved in the system, the material properties that relate the external effects and the responses must be included as variables. For example, for Newtonian fluids the viscosity of the fluid is the property that relates the applied forces to the rates of deformation of the fluid. As the material behavior becomes more complex, such as would be true for non-Newtonian fluids, the determination of material properties becomes difficult, and this class of variables can be troublesome to identify. 356 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling External Effects. This terminology is used to denote any variable that produces, or tends to produce, a change in the system. For example, in structural mechanics, forces 1either concentrated or distributed2 applied to a system tend to change its geometry, and such forces would need to be considered as pertinent variables. For fluid mechanics, variables in this class would be related to pressures, velocities, or gravity. The above general classes of variables are intended as broad categories that should be helpful in identifying variables. It is likely, however, that there will be important variables that do not fit easily into one of the above categories and each problem needs to be carefully analyzed. Since we wish to keep the number of variables to a minimum, it is important that all variables are independent. For example, if in a given problem we know that the moment of inertia of the area of a circular plate is an important variable, we could list either the moment of inertia or the plate diameter as the pertinent variable. However, it would be unnecessary to include both moment of inertia and diameter, assuming that the diameter enters the problem only through the moment of inertia. In more general terms, if we have a problem in which the variables are f 1 p, q, r, . . . , u, v, w, . . .2 0 (7.3) and it is known that there is an additional relationship among some of the variables, for example, q f1 1u, v, w, . . .2 (7.4) then q is not required and can be omitted. Conversely, if it is known that the only way the variables u, v, w, . . . enter the problem is through the relationship expressed by Eq. 7.4, then the variables u, v, w, . . . can be replaced by the single variable q, therefore reducing the number of variables. In summary, the following points should be considered in the selection of variables: 1. Clearly define the problem. What is the main variable of interest 1the dependent variable2? 2. Consider the basic laws that govern the phenomenon. Even a crude theory that describes the essential aspects of the system may be helpful. 3. Start the variable selection process by grouping the variables into three broad classes: geometry, material properties, and external effects. 4. Consider other variables that may not fall into one of the above categories. For example, time will be an important variable if any of the variables are time dependent. 5. Be sure to include all quantities that enter the problem even though some of them may be held constant 1e.g., the acceleration of gravity, g2. For a dimensional analysis it is the dimensions of the quantities that are important—not specific values! 6. Make sure that all variables are independent. Look for relationships among subsets of the variables. 7.4.2 Determination of Reference Dimensions Typically, in fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two are required. For any given problem it is obviously desirable to reduce the number of pi terms to a minimum; therefore, we wish to reduce the number of variables to a minimum—that is, we certainly do not want to include extraneous variables. It is also important to know how many reference dimensions are required to describe the variables. As we have seen in the preceding examples, F, L, and T appear to be a convenient set of basic dimensions for characterizing fluid-mechanical quantities. There is, however, really nothing “fundamental” about this set and, as previously noted, M, L, and T would also be suitable. Actually any set of measurable quantities could be used as basic dimensions provided that the selected combination can be used to describe all secondary quantities. However, the use of FLT or MLT as basic dimensions is the simplest, and these dimensions can be used to describe fluid-mechanical phenomena. Of course, in some problems only one or two of these are required. In addition, we occasionally find that the number of reference dimensions needed to describe all variables is smaller than the number of basic dimensions. This point is illustrated in Example 7.2. Interesting discussions, both practical and philosophical, relative to the concept of basic dimensions can be found in the books by Huntley 1Ref. 42 and by Isaacson and Isaacson 1Ref. 122. 7.4 E XAMPLE 357 Some Additional Comments about Dimensional Analysis Determination of Pi Terms 7.2 GIVEN An open, cylindrical paint can having a diameter D is filled to a depth h with paint having a specific weight g. The vertical deflection, d, of the center of the bottom is a function of D, h, d, g, and E, where d is the thickness of the bottom and E is the modulus of elasticity of the bottom material. FIND Determine the functional relationship between the vertical deflection, d, and the independent variables using dimensional analysis. g SOLUTION h From the statement of the problem d f 1D, h, d, g, E2 E, d and the dimensions of the variables are d d⬟L D D⬟L ■ Figure E7.2 h⬟L d⬟L g ⬟ FL3 ⬟ ML2T 2 E ⬟ FL2 ⬟ ML1T 2 where the dimensions have been expressed in terms of both the FLT and MLT systems. We now apply the pi theorem to determine the required number of pi terms. First, let us use F, L, and T as our system of basic dimensions. There are six variables and two reference dimensions 1F and L2 required so that four pi terms are needed. For repeating variables, we can select D and g so that ß 1 d Dagb 1L21L2 a 1FL3 2 b ⬟ F 0L0 and 1 a 3b 0 b0 1for L2 1for F 2 Therefore, a 1, b 0, and ß1 d D Similarly, ß 2 h Dagb and following the same procedure as above, a 1, b 0 so that ß2 h D The remaining two pi terms can be found using the same procedure, with the result d ß3 D E ß4 Dg Thus, this problem can be studied by using the relationship h d E d fa , , b D D D Dg (Ans) COMMENTS Let us now solve the same problem using the MLT system. Although the number of variables is obviously the same, it would seem that there are three reference dimensions required, rather than two. If this were indeed true, it would certainly be fortuitous, since we would reduce the number of required pi terms from four to three. Does this seem right? How can we reduce the number of required pi terms by simply using the MLT system of basic dimensions? The answer is that we cannot, and a closer look at the dimensions of the variables listed above reveals that actually only two reference dimensions, MT 2 and L, are required. This is an example of the situation in which the number of reference dimensions differs from the number of basic dimensions. It does not happen very often and can be detected by looking at the dimensions of the variables 1regardless of the systems used2 and making sure how many reference dimensions are actually required to describe the variables. Once the number of reference dimensions has been determined, we can proceed as before. Since the number of repeating variables must equal the number of reference dimensions, it follows that two reference dimensions are still required and we could again use D and g as repeating variables. The pi terms would be determined in the same manner. For example, the pi term containing E would be developed as ß 4 EDagb 1ML1T 2 21L2 a 1ML2T 2 2 b ⬟ 1MT 2 2 0L0 1b0 1 a 2b 0 1for MT 2 2 1for L2 358 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling and, therefore, a 1, b 1 so that ß4 This will always be true—you cannot affect the required number of pi terms by using M, L, and T instead of F, L, and T, or vice versa. E Dg which is the same as ß 4 obtained using the FLT system. The other pi terms would be the same, and the final result is the same; that is, d h d E fa , , b D D D Dg (Ans) 7.4.3 Uniqueness of Pi Terms A little reflection on the process used to determine pi terms by the method of repeating variables reveals that the specific pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe, we selected D, V, and r as repeating variables. This led to the formulation of the problem in terms of pi terms as ¢p/D rV 2 fa rVD b m (7.5) What if we had selected D, V, and m as repeating variables? A quick check will reveal that the pi term involving ¢p/ becomes ¢p/D2 Vm D Δpᐉ ρV 2 and the second pi term remains the same. Thus, we can express the final result as ¢p/D2 rVD b f1 a m Vm ρ VD ____ μ D2Δpᐉ Vm ρ____ VD μ Once a correct set of pi terms is obtained, any other set can be obtained by manipulation of the original set. (7.6) Both results are correct, and both would lead to the same final equation for ¢p/. Note, however, that the functions f and f1 in Eqs. 7.5 and 7.6 will be different because the dependent pi terms are different for the two relationships. As shown by the figure in the margin, the resulting graph of dimensionless data will be different for the two formulations. However, when extracting the physical variable, ¢p/, from the two results, the values will be the same. We can conclude from this illustration that there is not a unique set of pi terms which arises from a dimensional analysis. However, the required number of pi terms is fixed, and once a correct set is determined, all other possible sets can be developed from this set by combinations of products of powers of the original set. Thus, if we have a problem involving, say, three pi terms, ß 1 f1ß 2, ß 3 2 we could always form a new set from this one by combining the pi terms. For example, we could form a new pi term, ß¿2, by letting ß¿2 ß a2 ß b3 where a and b are arbitrary exponents. Then the relationship could be expressed as ß 1 f1 1ß¿2, ß 3 2 or ß 1 f2 1ß 2, ß¿2 2 All of these would be correct. It should be emphasized, however, that the required number of pi terms cannot be reduced by this manipulation; only the form of the pi terms is altered. By using 7.5 Determination of Pi Terms by Inspection 359 this technique, we see that the pi terms in Eq. 7.6 could be obtained from those in Eq. 7.5; that is, we multiply ß 1 in Eq. 7.5 by ß 2 so that a ¢p/D rV 2 ba rVD ¢p/D2 b m Vm which is the ß 1 of Eq. 7.6. There is no simple answer to the question: Which form for the pi terms is best? Usually our only guideline is to keep the pi terms as simple as possible. Also, it may be that certain pi terms will be easier to work with in actually performing experiments. The final choice remains an arbitrary one and generally will depend on the background and experience of the investigator. It should again be emphasized, however, that although there is no unique set of pi terms for a given problem, the number required is fixed in accordance with the pi theorem. 7.5 Determination of Pi Terms by Inspection The method of repeating variables for forming pi terms has been presented in Section 7.3. This method provides a step-by-step procedure that if executed properly will provide a correct and complete set of pi terms. Although this method is simple and straightforward, it is rather tedious, particularly for problems in which large numbers of variables are involved. Since the only restrictions placed on the pi terms are that they be 112 correct in number, 122 dimensionless, and 132 independent, it is possible to simply form the pi terms by inspection, without resorting to the more formal procedure. To illustrate this approach, we again consider the pressure drop per unit length along a smooth pipe. Regardless of the technique to be used, the starting point remains the same—determine the variables, which in this case are ¢p/ f 1D, r, m, V2 Next, the dimensions of the variables are listed: ¢p/ ⬟ FL3 D⬟L r ⬟ FL4T 2 m ⬟ FL2T V ⬟ LT 1 Pi terms can be formed by inspection by simply making use of the fact that each pi term must be dimensionless. and subsequently the number of reference dimensions determined. The application of the pi theorem then tells us how many pi terms are required. In this problem, since there are five variables and three reference dimensions, two pi terms are needed. Thus, the required number of pi terms can be easily obtained. The determination of this number should always be done at the beginning of the analysis. Once the number of pi terms is known, we can form each pi term by inspection, simply making use of the fact that each pi term must be dimensionless. We will always let ß 1 contain the dependent variable, which in this example is ¢p/. Since this variable has the dimensions FL3, we need to combine it with other variables so that a nondimensional product will result. One possibility is to first divide ¢p/ by r so that 1FL3 2 ¢p/ L ⬟ ⬟ 2 4 2 r 1FL T 2 T 1cancels F2 The dependence on F has been eliminated, but ¢p/r is obviously not dimensionless. To eliminate the dependence on T, we can divide by V 2 so that a ¢p/ 1 L 1 1 b ⬟ a 2b ⬟ r V2 L T 1LT 1 2 2 1cancels T2 360 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Finally, to make the combination dimensionless we multiply by D so that a 1 b D ⬟ a b 1L2 ⬟ L0 L rV ¢p/ 2 1cancels L2 Thus, ß1 ¢p/D rV 2 Next, we will form the second pi term by selecting the variable that was not used in ß 1, which in this case is m. We simply combine m with the other variables to make the combination dimensionless 1but do not use ¢p/ in ß 2, since we want the dependent variable to appear only in ß 12. For example, divide m by r 1to eliminate F2, then by V 1to eliminate T2, and finally by D 1to eliminate L2. Thus, 1FL2T 2 m ß2 ⬟ ⬟ F 0L0T 0 rVD 1FL4T 2 21LT 1 21L2 and, therefore, ¢p/D rV 2 fa m b rVD which is, of course, the same result we obtained by using the method of repeating variables. An additional concern, when one is forming pi terms by inspection, is to make certain that they are all independent. In the pipe flow example, ß 2 contains m, which does not appear in ß 1, and therefore these two pi terms are obviously independent. In a more general case a pi term would not be independent of the others in a given problem if it can be formed by some combination of the others. For example, if ß 2 can be formed by a combination of say ß 3, ß 4, and ß 5 such as ß2 ß 23 ß 4 ß5 then ß 2 is not an independent pi term. We can ensure that each pi term is independent of those preceding it by incorporating a new variable in each pi term. Although forming pi terms by inspection is essentially equivalent to the repeating variable method, it is less structured. With a little practice the pi terms can be readily formed by inspection, and this method offers an alternative to more formal procedures. 7.6 Common Dimensionless Groups in Fluid Mechanics A useful physical interpretation can often be given to dimensionless groups. At the top of Table 7.1 is a list of variables that commonly arise in fluid mechanics problems. The list is obviously not exhaustive but does indicate a broad range of variables likely to be found in a typical problem. Fortunately, not all of these variables would be encountered in all problems. However, when combinations of these variables are present, it is standard practice to combine them into some of the common dimensionless groups 1pi terms2 given in Table 7.1. These combinations appear so frequently that special names are associated with them, as indicated in the table. It is also often possible to provide a physical interpretation to the dimensionless groups which can be helpful in assessing their influence in a particular application. For example, the Froude number is an index of the ratio of the force due to the acceleration of a fluid particle to the force due to gravity 1weight2. This can be demonstrated by considering a fluid particle moving along a streamline 1Fig. 7.32. The magnitude of the component of inertia force FI along the streamline can be expressed as FI asm, where as is the magnitude of the acceleration along the streamline for a particle having a mass m. From our study of particle motion along a curved path 1see Section 3.12 we know that as dVs dVs Vs dt ds 7.6 Common Dimensionless Groups in Fluid Mechanics 361 Table 7.1 Some Common Variables and Dimensionless Groups in Fluid Mechanics Variables: Acceleration of gravity, g; Bulk modulus, Ev; Characteristic length, ; Density, r; Frequency of oscillating flow, v; Pressure, p (or ¢ p); Speed of sound, c; Surface tension, s; Velocity, V; Viscosity, m Dimensionless Groups Special names along with physical interpretations are given to the most common dimensionless groups. Name Interpretation (Index of Force Ratio Indicated) Types of Applications rV/ m Reynolds number, Re inertia force viscous force Generally of importance in all types of fluid dynamics problems V 1g/ Froude number, Fr inertia force gravitational force Flow with a free surface p Euler number, Eu pressure force inertia force Problems in which pressure, or pressure differences, are of interest rV 2 Ev Cauchy number,a Ca inertia force compressibility force Flows in which the compressibility of the fluid is important V c Mach number,a Ma inertia force compressibility force Flows in which the compressibility of the fluid is important v/ V Strouhal number, St rV 2/ s Weber number, We rV 2 inertia 1local2 force inertia 1convective2 force Unsteady flow with a characteristic frequency of oscillation inertia force surface tension force Problems in which surface tension is important a The Cauchy number and the Mach number are related, and either can be used as an index of the relative effects of inertia and compressibility. See accompanying discussion. where s is measured along the streamline. If we write the velocity, Vs, and length, s, in dimensionless form, that is, V s Vs V s s / where V and / represent some characteristic velocity and length, respectively, then as dV s V2 V s / ds and FI dV s V2 V s m / ds Vs e lin am S tre gm ■ Figure 7.3 The force of gravity acting on a fluid particle moving along a streamline. 362 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling The magnitude of the weight of the particle, FG, is FG gm, so the ratio of the inertia to the gravitational force is FI dV s V2 V s FG g/ ds Thus, the force ratio FIFG is proportional to V 2g/, and the square root of this ratio, V1g/, is called the Froude number. We see that a physical interpretation of the Froude number is that it is a measure of, or an index of, the relative importance of inertial forces acting on fluid particles to the weight of the particle. Note that the Froude number is not really equal to this force ratio, but is simply some type of average measure of the influence of these two forces. In a problem in which gravity 1or weight2 is not important, the Froude number would not appear as an important pi term. A similar interpretation in terms of indices of force ratios can be given to the other dimensionless groups, as indicated in Table 7.1, and a further discussion of the basis for this type of interpretation is given in the last section in this chapter. Some additional details about these important dimensionless groups are given below, and the types of application or problem in which they arise are briefly noted in the last column of Table 7.1. V7.3 Reynolds number Reynolds Number. The Reynolds number is undoubtedly the most famous dimensionless parameter in fluid mechanics. It is named in honor of Osborne Reynolds 11842–19122, a British engineer who first demonstrated that this combination of variables could be used as a criterion to distinguish between laminar and turbulent flow. In most fluid flow problems there will be a characteristic length, /, and a velocity, V, as well as the fluid properties of density, r, and viscosity, m, which are relevant variables in the problem. Thus, with these variables the Reynolds number Re No separation Re ≈ 0.2 Laminar boundary layer, wide turbulent wake Re ≈ 20,000 V7.4 Froude number rV/ m arises naturally from the dimensional analysis. The Reynolds number is a measure of the ratio of the inertia force on an element of fluid to the viscous force on an element. When these two types of forces are important in a given problem, the Reynolds number will play an important role. However, if the Reynolds number is very small 1Re 12, this is an indication that the viscous forces are dominant in the problem, and it may be possible to neglect the inertial effects; that is, the density of the fluid will not be an important variable. Flows at very small Reynolds numbers are commonly referred to as “creeping flows” as discussed in Section 6.10. Conversely, for large Reynolds number flows, viscous effects are small relative to inertial effects and for these cases it may be possible to neglect the effect of viscosity and consider the problem as one involving a “nonviscous” fluid. This type of problem is considered in detail in Sections 6.4 through 6.7. An example of the importance of the Reynolds number in determining the flow physics is shown in the figure in the margin for flow past a circular cylinder at two different Re values. This flow is discussed further in Chapter 9. Froude Number. The Froude number Fr V 1g/ is distinguished from the other dimensionless groups in Table 7.1 in that it contains the acceleration of gravity, g. The acceleration of gravity becomes an important variable in a fluid dynamics problem in which the fluid weight is an important force. As discussed, the Froude number is a measure of the ratio of the inertia force on an element of fluid to the weight of the element. It will generally be important in problems involving flows with free surfaces since gravity principally affects this type of flow. Typical problems would include the study of the flow of water around ships 1with the resulting wave action2 or flow through rivers or open conduits. The Froude number is named in honor of William Froude 11810–18792, a British civil engineer, mathematician, and naval architect who pioneered the use of towing tanks for the study of ship design. It is to be noted that the Froude number is also commonly defined as the square of the Froude number listed in Table 7.1. 7.6 Common Dimensionless Groups in Fluid Mechanics 363 Euler Number. The Euler number Eu p rV 2 can be interpreted as a measure of the ratio of pressure forces to inertial forces, where p is some characteristic pressure in the flow field. Very often the Euler number is written in terms of a pressure difference, ¢p, so that Eu ¢prV 2. Also, this combination expressed as ¢p12 rV2 is called the pressure coefficient. Some form of the Euler number would normally be used in problems in which pressure or the pressure difference between two points is an important variable. The Euler number is named in honor of Leonhard Euler 11707–17832, a famous Swiss mathematician who pioneered work on the relationship between pressure and flow. For problems in which cavitation is of concern, the dimensionless group 1 pr pv 2 12 rV 2 is commonly used, where pv is the vapor pressure and pr is some reference pressure. Although this dimensionless group has the same form as the Euler number, it is generally referred to as the cavitation number. Cauchy Number and Mach Number. The Cauchy number Ca The Mach number is a commonly used dimensionless parameter in compressible flow problems. rV 2 Ev and the Mach number Ma V c are important dimensionless groups in problems in which fluid compressibility is a significant factor. Since the speed of sound, c, in a fluid is equal to c 1Ev r 1see Section 1.7.32, it follows that Ma V r A Ev and the square of the Mach number Ma2 rV 2 Ca Ev is equal to the Cauchy number. Thus, either number 1but not both2 may be used in problems in which fluid compressibility is important. Both numbers can be interpreted as representing an index of the ratio of inertial forces to compressibility forces. When the Mach number is relatively small 1say, less than 0.32, the inertial forces induced by the fluid motion are not sufficiently large to cause a significant change in the fluid density, and in this case the compressibility of the fluid can be neglected. The Mach number is the more commonly used parameter in compressible flow problems, particularly in the fields of gas dynamics and aerodynamics. The Cauchy number is named in honor of Augustin Louis de Cauchy 11789–18572, a French engineer, mathematician, and hydrodynamicist. The Mach number is named in honor of Ernst Mach 11838–19162, an Austrian physicist and philosopher. Strouhal Number. The Strouhal number St V7.5 Strouhal number v/ V is a dimensionless parameter that is likely to be important in unsteady, oscillating flow problems in which the frequency of the oscillation is v. It represents a measure of the ratio of inertial forces due to the unsteadiness of the flow 1local acceleration2 to the inertial forces due to changes in velocity from point to point in the flow field 1convective acceleration2. This type of unsteady flow may develop when a fluid flows past a solid body 1such as a wire or cable2 placed in the moving stream. For example, in a certain Reynolds number range, a periodic flow will develop downstream from a cylinder placed in a moving fluid due to a regular pattern of vortices that are shed from the body. 1See the photograph at the beginning of this chapter and Fig. 9.21.2 This system of vortices, called a Kármán vortex trail [named after Theodor von Kármán 11881–19632, a famous fluid 364 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling mechanician], creates an oscillating flow at a discrete frequency, v, such that the Strouhal number can be closely correlated with the Reynolds number. When the frequency is in the audible range, a sound can be heard and the bodies appear to “sing.” In fact, the Strouhal number is named in honor of Vincenz Strouhal 11850–19222, who used this parameter in his study of “singing wires.” The most dramatic evidence of this phenomenon occurred in 1940 with the collapse of the Tacoma Narrows Bridge. The shedding frequency of the vortices coincided with the natural frequency of the bridge, thereby setting up a resonant condition that eventually led to the collapse of the bridge. There are, of course, other types of oscillating flows. For example, blood flow in arteries is periodic and can be analyzed by breaking up the periodic motion into a series of harmonic components 1Fourier series analysis2, with each component having a frequency that is a multiple of the fundamental frequency, v 1the pulse rate2. Rather than use the Strouhal number in this type of problem, a dimensionless group formed by the product of St and Re is used; that is St Re rv/2 m The square root of this dimensionless group is often referred to as the frequency parameter. Weber Number. The Weber number We may be important in problems in which there is an interface between two fluids. In this situation the surface tension may play an important role in the phenomenon of interest. The Weber number can be thought of as an index of the inertial force to the surface tension force acting on a fluid element. Common examples of problems in which this parameter may be important include the flow of thin films of liquid, or the formation of droplets or bubbles. Clearly, not all problems involving flows with an interface will require the inclusion of surface tension. The flow of water in a river is not affected significantly by surface tension, since inertial and gravitational effects are dominant 1We 12. However, as discussed in a later section, for river models 1which may have small depths2 caution is required so that surface tension does not become important in the model, whereas it is not important in the actual river. The Weber number is named after Moritz Weber 11871–19512, a German professor of naval mechanics who was instrumental in formalizing the general use of common dimensionless groups as a basis for similitude studies. V7.6 Weber number F l u i d s i n Slip at the micro scale A goal in chemical and biological analyses is to miniaturize the experiment, which has many advantages including reduction in sample size. In recent years, there has been significant work on integrating these tests on a single microchip to form the “lab-on-a-chip” system. These devices are on the millimeter scale with complex passages for fluid flow on the micron scale (or smaller). While there are advantages to miniaturization, care must be taken in moving to smaller and smaller flow regimes, as you will eventually bump into the continuum assumption. To 7.7 rV 2/ s t h e N e w s characterize this situation, a dimensionless number termed the Knudsen number, Kn l /, is commonly employed. Here l is the mean free path and / is the characteristic length of the system. If Kn is smaller than 0.01, then the flow can be described by the Navier–Stokes equations with no-slip at the walls. For 0.01 6 Kn 6 0.3, the same equations can be used, but there can be slip between the fluid and the wall so the boundary conditions need to be adjusted. For Kn 7 10, the continuum assumption breaks down and the Navier–Stokes equations are no longer valid. Correlation of Experimental Data One of the most important uses of dimensional analysis is as an aid in the efficient handling, interpretation, and correlation of experimental data. Since the field of fluid mechanics relies heavily on empirical data, it is not surprising that dimensional analysis is such an important tool in this field. As noted previously, a dimensional analysis cannot provide a complete answer to any given problem, since the analysis only provides the dimensionless groups describing the phenomenon, and not the specific relationship among the groups. To determine this relationship, suitable experimental data must be obtained. The degree of difficulty involved in this process depends on the number of pi terms, and the nature of the experiments 1How hard is it to obtain 7.7 Correlation of Experimental Data 365 the measurements?2. The simplest problems are obviously those involving the fewest pi terms, and the following sections indicate how the complexity of the analysis increases with the increasing number of pi terms. 7.7.1 Problems with One Pi Term Application of the pi theorem indicates that if the number of variables minus the number of reference dimensions is equal to unity, then only one pi term is required to describe the phenomenon. The functional relationship that must exist for one pi term is If only one pi term is involved in a problem, it must be equal to a constant. E XAMPLE ß1 C where C is a constant. This is one situation in which a dimensional analysis reveals the specific form of the relationship and, as is illustrated by the following example, shows how the individual variables are related. The value of the constant, however, must still be determined by experiment. 7.3 Flow with Only One Pi Term = f(D,V, μ) GIVEN As shown in Fig. E7.3, assume that the drag, d, acting on a spherical particle that falls very slowly through a viscous fluid, is a function of the particle diameter, D, the particle velocity, V, and the fluid viscosity, ␮. V7.7 Stokes flow D FIND Determine, with the aid of dimensional analysis, how the drag depends on the particle velocity. SOLUTION μ From the information given, it follows that d f (D, V, ␮) V and the dimensions of the variables are ■ Figure E7.3 d⬟F D⬟L V ⬟ LT 1 m ⬟ FL2T COMMENTS Actually, the dimensional analysis reveals that We see that there are four variables and three reference dimensions (F, L, and T) required to describe the variables. Thus, according to the pi theorem, one pi term is required. This pi term can be easily formed by inspection and can be expressed as ß1 d mVD Because there is only one pi term, it follows that d C mVD where C is a constant. Thus, d 3pmVD d CmVD Thus, for a given particle and fluid, the drag varies directly with the velocity so that d r V the drag not only varies directly with the velocity, but it also varies directly with the particle diameter and the fluid viscosity. We could not, however, predict the value of the drag, since the constant, C, is unknown. An experiment would have to be performed in which the drag and the corresponding velocity are measured for a given particle and fluid. Although in principle we would only have to run a single test, we would certainly want to repeat it several times to obtain a reliable value for C. It should be emphasized that once the value of C is determined it is not necessary to run similar tests by using different spherical particles and fluids; that is, C is a universal constant as long as the drag is a function only of particle diameter, velocity, and fluid viscosity. An approximate solution to this problem can also be obtained theoretically, from which it is found that C 3␲ so that (Ans) This equation is commonly called Stokes law and is used in the study of the settling of particles. Our experiments would reveal that this result is only valid for small Reynolds numbers (␳VD/␮ 1). This follows, since in the original list of variables, we have 366 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling neglected inertial effects (fluid density is not included as a variable). The inclusion of an additional variable would lead to another pi term so that there would be two pi terms rather than one. Consider a free-body diagram of a sphere in Stokes flow; there would be a buoyant force in the same direction as the drag in Fig. E7.3, as well as a weight force in the opposite direction. As shown above, the drag force is proportional to the product of the diameter and fall velocity, d r VD. The weight and buoyant force are proportional to the diameter cubed, W and FB r D3. Given equilibrium conditions, the force balance can be written as d W FB Based on the scaling laws for these terms, it follows that VD r D3. Hence, the fall velocity will be proportional to the square of the diameter, V r D2. Therefore, for two spheres, one having twice the diameter of the other, and falling through the same fluid, the sphere with the larger diameter will fall four times faster (see Video V7.7). Π1 k–r=2 Valid range ■ Figure 7.4 The graphical presentation of data Π2 for problems involving two pi terms, with an illustration of the potential danger of extrapolation of data. 7.7.2 Problems with Two or More Pi Terms If a given phenomenon can be described with two pi terms such that ß 1 f1ß 2 2 For problems involving only two pi terms, results of an experiment can be conveniently presented in a simple graph. E XAMPLE the functional relationship among the variables can then be determined by varying ß 2 and measuring the corresponding values of ß 1. For this case the results can be conveniently presented in graphical form by plotting ß 1 versus ß 2 as is illustrated in Fig. 7.4. It should be emphasized that the curve shown in Fig. 7.4 would be a “universal” one for the particular phenomenon studied. This means that if the variables and the resulting dimensional analysis are correct, then there is only a single relationship between ß 1 and ß 2, as illustrated in Fig. 7.4. However, since this is an empirical relationship, we can only say that it is valid over the range of ß 2 covered by the experiments. It would be unwise to extrapolate beyond this range, since as illustrated with the dashed lines in the figure, the nature of the phenomenon could dramatically change as the range of ß 2 is extended. In addition to presenting the data graphically, it may be possible 1and desirable2 to obtain an empirical equation relating ß 1 and ß 2 by using a standard curve-fitting technique. 7.4 Dimensionless Correlation of Experimental Data GIVEN The relationship between the pressure drop per unit length along a smooth-walled, horizontal pipe and the variables that affect the pressure drop is to be determined experimentally. In the laboratory the pressure drop was measured over a 5-ft length of smooth-walled pipe having an inside diameter of 0.496 in. The fluid used was water at 60 °F 1m 2.34 10 5 lb # sft2, r 1.94 slugsft3 2. Tests were run in which the velocity was varied and the corresponding pressure drop measured. The results of these tests are shown below: SOLUTION The first step is to perform a dimensional analysis during the planning stage before the experiments are actually run. As was discussed in Section 7.3, we will assume that the pressure drop per unit length, ¢p/, is a function of the pipe diameter, D, fluid Velocity (ft/s) Pressure drop for 5-ft length (lb/ft2) 1.17 1.95 2.91 5.84 11.13 16.92 23.34 28.73 6.26 15.6 30.9 106 329 681 1200 1730 FIND Make use of these data to obtain a general relationship between the pressure drop per unit length and the other variables. 7.7 density, r, fluid viscosity, m, and the velocity, V. Thus, 0.020 0.018 rV 2 ß2 and rVD m D Δp ______ ρV2 and application of the pi theorem yields two pi terms D ¢p/ 367 0.022 ¢p/ f 1D, r, m, V 2 ß1 Correlation of Experimental Data 0.016 0.014 0.012 Hence, rV 2 fa To determine the form of the relationship, we need to vary the Reynolds number, Re rVDm, and to measure the corresponding values of D ¢p/rV 2. The Reynolds number could be varied by changing any one of the variables, r, V, D, or m, or any combination of them. However, the simplest way to do this is to vary the velocity, since this will allow us to use the same fluid and pipe. Based on the data given, values for the two pi terms can be computed, with the result: D¢p RV2 0.0195 0.0175 0.0155 0.0132 0.0113 0.0101 0.00939 0.00893 0.010 rVD b m RVDM 4.01 6.68 9.97 2.00 3.81 5.80 8.00 9.85 103 103 103 104 104 104 104 104 These are dimensionless groups so that their values are independent of the system of units used so long as a consistent system is used. For example, if the velocity is in fts, then the diameter should be in feet, not inches or meters. Note that since the Reynolds numbers are all greater than 2100, the flow in the pipe is turbulent 1see Section 8.1.12. A plot of these two pi terms can now be made with the results shown in Fig. E7.4a. The correlation appears to be quite good, and if it was not, this would suggest that either we had large experimental measurement errors or that we had perhaps omitted an important variable. The curve shown in Fig. E7.4a represents the general relationship between the pressure drop and the other factors in the range of Reynolds numbers between 4.01 103 and 9.85 104. Thus, for this range of Reynolds numbers it is not necessary to repeat the tests for other pipe sizes or other fluids provided the assumed independent variables 1D, r, m, V 2 are the only important ones. Since the relationship between ß 1 and ß 2 is nonlinear, it is not immediately obvious what form of empirical equation might be used to describe the relationship. If, however, the same data are 0.008 0 20,000 40,000 60,000 80,000 100,000 ρ VD Re = _____ μ (a) 4 2 D Δp ______ ρV2 D ¢p/ 10–2 8 6 4 × 10–3 103 2 4 6 8 104 2 4 6 8 105 ρ VD Re = _____ μ (b) ■ Figure E7.4 plotted on a logarithmic scale, as is shown in Fig. E7.4b, the data form a straight line, suggesting that a suitable equation is of the form ß 1 Aß n2 where A and n are empirical constants to be determined from the data by using a suitable curve-fitting technique, such as a nonlinear regression program. For the data given in this example, a good fit of the data is obtained with the equation ß 1 0.150 ß 20.25 (Ans) COMMENT In 1911, H. Blasius 11883 – 19702, a German fluid mechanician, established a similar empirical equation that is used widely for predicting the pressure drop in smooth pipes in the range 4 103 6 Re 6 105 1Ref. 162. This equation can be expressed in the form D ¢p/ rV 2 0.1582 a rVD 14 b m The so-called Blasius formula is based on numerous experimental results of the type used in this example. Flow in pipes is discussed in more detail in the next chapter, where it is shown how pipe roughness 1which introduces another variable2 may affect the results given in this example 1which is for smooth-walled pipes2. As the number of required pi terms increases, it becomes more difficult to display the results in a convenient graphical form and to determine a specific empirical equation that describes the phenomenon. For problems involving three pi terms ß 1 f1ß 2, ß 3 2 it is still possible to show data correlations on simple graphs by plotting families of curves as illustrated in Fig. 7.5. This is an informative and useful way of representing the data in a general 368 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Π1 k–r=3 Π3 = C1 (constant) Π 3 = C2 Π3 = C3 Π3 = C4 ■ Figure 7.5 The graphical presentaΠ2 For problems involving more than two or three pi terms, it is often necessary to use a model to predict specific characteristics. 7.8 tion of data for problems involving three pi terms. way. It may also be possible to determine a suitable empirical equation relating the three pi terms. However, as the number of pi terms continues to increase, corresponding to an increase in the general complexity of the problem of interest, both the graphical presentation and the determination of a suitable empirical equation become intractable. For these more complicated problems, it is often more feasible to use models to predict specific characteristics of the system rather than to try to develop general correlations. Modeling and Similitude V Prototype Vm m Model V7.8 Model airplane Models are widely used in fluid mechanics. Major engineering projects involving structures, aircraft, ships, rivers, harbors, dams, air and water pollution, and so on, frequently involve the use of models. Although the term model is used in many different contexts, the “engineering model” generally conforms to the following definition. A model is a representation of a physical system that may be used to predict the behavior of the system in some desired respect. The physical system for which the predictions are to be made is called the prototype. Although mathematical or computer models may also conform to this definition, our interest will be in physical models, that is, models that resemble the prototype but are generally of a different size, may involve different fluids, and often operate under different conditions 1pressures, velocities, etc.2. As shown by the figure in the margin, usually a model is smaller than the prototype. Therefore, it is more easily handled in the laboratory and less expensive to construct and operate than a large prototype (it should be noted that variables or pi terms without a subscript will refer to the prototype, whereas the subscript m will be used to designate the model variables or pi terms). Occasionally, if the prototype is very small, it may be advantageous to have a model that is larger than the prototype so that it can be more easily studied. For example, large models have been used to study the motion of red blood cells, which are approximately 8 mm in diameter. With the successful development of a valid model, it is possible to predict the behavior of the prototype under a certain set of conditions. We may also wish to examine a priori the effect of possible design changes that are proposed for a hydraulic structure or fluid flow system. There is, of course, an inherent danger in the use of models in that predictions can be made that are in error and the error not detected until the prototype is found not to perform as predicted. It is, therefore, imperative that the model be properly designed and tested and that the results be interpreted correctly. In the following sections we will develop the procedures for designing models so that the model and prototype will behave in a similar fashion. 7.8.1 Theory of Models The theory of models can be readily developed by using the principles of dimensional analysis. It has been shown that any given problem can be described in terms of a set of pi terms as ß 1 f1ß 2, ß 3, . . . , ß n 2 (7.7) In formulating this relationship, only a knowledge of the general nature of the physical phenomenon, and the variables involved, is required. Specific values for variables 1size of components, fluid properties, and so on2 are not needed to perform the dimensional analysis. Thus, Eq. 7.7 applies 7.8 Modeling and Similitude 369 to any system that is governed by the same variables. If Eq. 7.7 describes the behavior of a particular prototype, a similar relationship can be written for a model of this prototype; that is, ß 1m f1ß 2m, ß 3m, . . . , ß nm 2 (7.8) where the form of the function will be the same as long as the same phenomenon is involved in both the prototype and the model. Variables, or pi terms, without a subscript will refer to the prototype, whereas the subscript m will be used to designate the model variables or pi terms. The pi terms can be developed so that ß 1 contains the variable that is to be predicted from observations made on the model. Therefore, if the model is designed and operated under the following conditions, ß 2m ß 2 ß 3m ß 3 (7.9) o ß nm ß n then with the presumption that the form of f is the same for model and prototype, it follows that ß 1 ß 1m The similarity requirements for a model can be readily obtained with the aid of dimensional analysis. (7.10) Equation 7.10 is the desired prediction equation and indicates that the measured value of ß 1m obtained with the model will be equal to the corresponding ß 1 for the prototype as long as the other pi terms are equal. The conditions specified by Eqs. 7.9 provide the model design conditions, also called similarity requirements or modeling laws. As an example of the procedure, consider the problem of determining the drag, d, on a thin rectangular plate 1w h in size2 placed normal to a fluid with velocity, V, as shown by the figure in the margin. The dimensional analysis of this problem was performed in Example 7.1, where it was assumed that d f 1w, h, m, r, V2 Application of the pi theorem yielded w rVw d fa , b 2 h m w rV 2 w V ρ, μ h (7.11) We are now concerned with designing a model that could be used to predict the drag on a certain prototype 1which presumably has a different size than the model2. Since the relationship expressed by Eq. 7.11 applies to both prototype and model, Eq. 7.11 is assumed to govern the prototype, with a similar relationship dm 2 wmrmV 2m fa wm rmVmwm , b mm hm (7.12) for the model. The model design conditions, or similarity requirements, are therefore wm w hm h and rmVmwm rVw mm m The size of the model is obtained from the first requirement which indicates that wm hm w h (7.13) We are free to establish the height ratio between the model and prototype, hmh, but then the model plate width, wm, is fixed in accordance with Eq. 7.13. The second similarity requirement indicates that the model and prototype must be operated at the same Reynolds number. Thus, the required velocity for the model is obtained from the relationship Vm mm r w V m rm wm (7.14) 370 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Note that this model design requires not only geometric scaling, as specified by Eq. 7.13, but also the correct scaling of the velocity in accordance with Eq. 7.14. This result is typical of most model designs—there is more to the design than simply scaling the geometry! With the foregoing similarity requirements satisfied, the prediction equation for the drag is dm d ⫽ 2 2 w rV wmrmVm2 2 or d⫽a Similarity between a model and a prototype is achieved by equating pi terms. V7.9 Environmental models © Stanford University (with permission) F l u w 2 r V 2 b a b a b dm wm rm Vm Thus, a measured drag on the model, dm, must be multiplied by the ratio of the square of the plate widths, the ratio of the fluid densities, and the ratio of the square of the velocities to obtain the predicted value of the prototype drag, d. Generally, as is illustrated in this example, to achieve similarity between model and prototype behavior, all the corresponding pi terms must be equated between model and prototype. Usually, one or more of these pi terms will involve ratios of important lengths 1such as w Ⲑh in the foregoing example2; that is, they are purely geometrical. Thus, when we equate the pi terms involving length ratios, we are requiring that there be complete geometric similarity between the model and prototype. This means that the model must be a scaled version of the prototype. Geometric scaling may extend to the finest features of the system, such as surface roughness, or small protuberances on a structure, since these kinds of geometric features may significantly influence the flow. Any deviation from complete geometric similarity for a model must be carefully considered. Sometimes complete geometric scaling may be difficult to achieve, particularly when dealing with surface roughness, since roughness is difficult to characterize and control. Another group of typical pi terms 1such as the Reynolds number in the foregoing example2 involves force ratios as noted in Table 7.1. The equality of these pi terms requires the ratio of like forces in model and prototype to be the same. Thus, for flows in which the Reynolds numbers are equal, the ratio of viscous forces in model and prototype is equal to the ratio of inertia forces. This is further illustrated by the photographs in the margin. The top photograph is of flow past a cylinder with diameter, D, and velocity, Vo, while the velocity is doubled and the diameter halved in the bottom photograph (image shown in 2x enlargement). Since the Reynolds numbers match, the flows look and behave the same. If other pi terms are involved, such as the Froude number or Weber number, a similar conclusion can be drawn; that is, the equality of these pi terms requires the ratio of like forces in model and prototype to be the same. Thus, when these types of pi terms are equal in model and prototype, we have dynamic similarity between model and prototype. It follows that with both geometric and dynamic similarity the streamline patterns will be the same and corresponding velocity ratios 1Vm ⲐV2 and acceleration ratios 1amⲐa2 are constant throughout the flow field. Thus, kinematic similarity exists between model and prototype. To have complete similarity between model and prototype, we must maintain geometric, kinematic, and dynamic similarity between the two systems. This will automatically follow if all the important variables are included in the dimensional analysis, and if all the similarity requirements based on the resulting pi terms are satisfied. i d s i n Modeling parachutes in a water tunnel The first use of a parachute with a free-fall jump from an aircraft occurred in 1914, although parachute jumps from hot-air balloons had occurred since the late 1700s. In more modern times parachutes are commonly used by the military and for safety and sport. It is not surprising that there remains interest in the design and characteristics of parachutes, and researchers at the Worcester Polytechnic Institute have been studying various aspects of the aerodynamics associated with parachutes. An unusual part of their study is that they are using small-scale para- t h e N e w s chutes tested in a water tunnel. The model parachutes are reduced in size by a factor of 30 to 60 times. Various types of tests can be performed, ranging from the study of the velocity fields in the wake of the canopy with a steady free-stream velocity to the study of conditions during rapid deployment of the canopy. According to the researchers, the advantage of using water as the working fluid, rather than air, is that the velocities and deployment dynamics are slower than in the atmosphere, thus providing more time to collect detailed experimental data. (See Problem 7.57.) 7.8 E XAMPLE 371 Modeling and Similitude Prediction of Prototype Performance from Model Data 7.5 GIVEN A long structural component of a bridge has an elliptical cross section shown in Fig. E7.5. It is known that when a steady wind blows past this type of bluff body, vortices may develop on the downwind side that are shed in a regular fashion at some definite frequency. Since these vortices can create harmful periodic forces acting on the structure, it is important to determine the shedding frequency. For the specific structure of interest, D 0.1 m, H 0.3 m, and a representative wind velocity is 50 kmhr. Standard air can be assumed. The shedding frequency is to be determined through the use of a small-scale model that is to be tested in a water tunnel. For the model Dm 20 mm and the water temperature is 20 °C. D V H ■ Figure E7.5 V7.10 Flow past an ellipse FIND Determine the model dimension, Hm, and the velocity at which the test should be performed. If the shedding frequency for the model is found to be 49.9 Hz, what is the corresponding frequency for the prototype? SOLUTION We expect the shedding frequency, v, to depend on the lengths D and H, the approach velocity, V, and the fluid density, r, and viscosity, m. Thus, v f 1D, H, V, r, m2 The second similarity requirement indicates that the Reynolds number must be the same for model and prototype so that the model velocity must satisfy the condition Vm where D⬟L H⬟L V ⬟ LT 1 r ⬟ ML3 m ⬟ ML T V 1 Since there are six variables and three reference dimensions 1MLT2, three pi terms are required. Application of the pi theorem yields vD D rVD fa , b V H m We recognize the pi term on the left as the Strouhal number, and the dimensional analysis indicates that the Strouhal number is a function of the geometric parameter, D H, and the Reynolds number. Thus, to maintain similarity between model and prototype Dm D Hm H 150 103 mhr2 13600 shr2 31.00 103 kg 1m # s2 4 11.23 kgm3 2 31.79 105 kg 1m # s2 4 1998 kg m3 2 10.1 m2 113.9 ms2 120 103 m2 Vm 4.79 ms Vm vmDm vD V Vm and the predicted prototype vortex shedding frequency is From the first similarity requirement (Ans) This is a reasonable velocity that could be readily achieved in a water tunnel. With the two similarity requirements satisfied, it follows that the Strouhal numbers for prototype and model will be the same so that rmVmDm rVD mm m Dm H D 120 103 m2 13.9 ms The required velocity can now be calculated from Eq. 1 as and Hm (1) For air at standard conditions, m 1.79 10 5 kgm # s, r 1.23 kgm3, and for water at 20 °C, m 1.00 10 3 kgm # s, r 998 kgm3. The fluid velocity for the prototype is v ⬟ T 1 1 mm r D V m rm Dm v 10.3 m2 10.1 m2 Hm 60 103 m 60 mm (Ans) V Dm v Vm D m 113.9 ms2 120 103 m2 14.79 ms2 v 29.0 Hz 10.1 m2 149.9 Hz2 (Ans) 372 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling COMMENT This same model could also be used to predict the drag per unit length, d/ 1lbft or Nm2, on the prototype, since the drag would depend on the same variables as those used for the frequency. Thus, the similarity requirements would be the same and with these requirements satisfied it follows that the drag per unit length expressed in dimensionless form, such as d/DrV 2, would be equal in model and prototype. The measured drag per unit length on the model could then be related to the corresponding drag per unit length on the prototype through the relationship d/ a r D V 2 b a b a b d/m Dm rm Vm 7.8.2 Model Scales It is clear from the preceding section that the ratio of like quantities for the model and prototype naturally arises from the similarity requirements. For example, if in a given problem there are two length variables /1 and /2, the resulting similarity requirement based on a pi term obtained from these two variables is /1 /1m /2 /2m so that /2m /1m /1 /2 We define the ratio /1m/1 or /2m/2 as the length scale. For true models there will be only one length scale, and all lengths are fixed in accordance with this scale. There are, however, other scales such as the velocity scale, VmV, density scale, rmr, viscosity scale, mmm, and so on. In fact, we can define a scale for each of the variables in the problem. Thus, it is actually meaningless to talk about a “scale” of a model without specifying which scale. We will designate the length scale as l/, and other scales as lV, lr, lm, and so on, where the subscript indicates the particular scale. Also, we will take the ratio of the model value to the prototype value as the scale 1rather than the inverse2. Length scales are often specified, for example, as 1 : 10 or as a 101 scale model. The meaning of this specification is that the model is one-tenth the size of the prototype, and the tacit assumption is that all relevant lengths are scaled accordingly so the model is geometrically similar to the prototype. The ratio of a model variable to the corresponding prototype variable is called the scale for that variable. F l u i d s i n “Galloping Gertie” One of the most dramatic bridge collapses occurred in 1940 when the Tacoma Narrows Bridge, located near Tacoma, Washington, failed due to aerodynamic instability. The bridge had been nicknamed “Galloping Gertie” due to its tendency to sway and move in high winds. On the fateful day of the collapse the wind speed was 65 km hr. This particular combination of a high wind and the aeroelastic properties of the bridge created large oscillations leading to its failure. The bridge was replaced in 1950, and a second bridge parallel to the existing structure was opened in t h e N e w s To determine possible wind interference effects due to two bridges in close proximity, wind tunnel tests were run in a 9 m 9 m wind tunnel operated by the National Research Council of Canada. Models of the two side-by-side bridges, each having a length scale of 1:211, were tested under various wind conditions. Since the failure of the original Tacoma Narrows Bridge, it is now common practice to use wind tunnel model studies during the design process to evaluate any bridge that is to be subjected to windinduced vibrations. (See Problem 7.84.) 7.8.3 Practical Aspects of Using Models Validation of Model Design. Most model studies involve simplifying assumptions with regard to the variables to be considered. Although the number of assumptions is frequently less stringent than that required for mathematical models, they nevertheless introduce some uncertainty in the model design. It is, therefore, desirable to check the design experimentally whenever possible. In some situations the purpose of the model is to predict the effects of certain proposed changes in a given prototype, and in this instance some actual prototype data may be available. The model can be designed, constructed, and tested, and the model prediction can be compared 7.8 V7.11 Model of fish hatchery pond Models for which one or more similarity requirements are not satisfied are called distorted models. Modeling and Similitude 373 with these data. If the agreement is satisfactory, then the model can be changed in the desired manner, and the corresponding effect on the prototype can be predicted with increased confidence. Another useful and informative procedure is to run tests with a series of models of different sizes, where one of the models can be thought of as the prototype and the others as “models” of this prototype. With the models designed and operated on the basis of the proposed design, a necessary condition for the validity of the model design is that an accurate prediction be made between any pair of models, since one can always be considered as a model of the other. Although suitable agreement in validation tests of this type does not unequivocally indicate a correct model design 1e.g., the length scales between laboratory models may be significantly different than required for actual prototype prediction2, it is certainly true that if agreement between models cannot be achieved in these tests, there is no reason to expect that the same model design can be used to predict prototype behavior correctly. Distorted Models. Although the general idea behind establishing similarity requirements for models is straightforward 1we simply equate pi terms2, it is not always possible to satisfy all the known requirements. If one or more of the similarity requirements are not met, for example, if ß 2m Z ß 2, then it follows that the prediction equation ß 1 ß 1m is not true; that is, ß 1 Z ß 1m. Models for which one or more of the similarity requirements are not satisfied are called distorted models. Distorted models are rather commonplace, and they can arise for a variety of reasons. For example, perhaps a suitable fluid cannot be found for the model. The classic example of a distorted model occurs in the study of open channel or free-surface flows. Typically, in these problems both the Reynolds number, rV/ m, and the Froude number, V1g/, are involved. Froude number similarity requires Vm V 1gm/m 1g/ If the model and prototype are operated in the same gravitational field, then the required velocity scale is Vm /m 1l/ V B/ Reynolds number similarity requires rmVm/m rV/ mm m and the velocity scale is V7.12 Distorted river model mm r / Vm m rm /m V Since the velocity scale must be equal to the square root of the length scale, it follows that nm m m rm 1l/ 2 32 n mr (7.15) where the ratio mr is the kinematic viscosity, n. Although in principle it may be possible to satisfy this design condition, it may be quite difficult, if not impossible, to find a suitable model fluid, particularly for small length scales. For problems involving rivers, spillways, and harbors, for which the prototype fluid is water, the models are also relatively large so that the only practical model fluid is water. However, in this case 1with the kinematic viscosity scale equal to unity2 Eq. 7.15 will not be satisfied, and a distorted model will result. Generally, hydraulic models of this type are distorted and are designed on the basis of the Froude number, with the Reynolds number different in model and prototype. Distorted models can be successfully used, but the interpretation of results obtained with this type of model is obviously more difficult than the interpretation of results obtained with true models for which all similarity requirements are met. There are no general rules for handling distorted 374 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling models, and essentially each problem must be considered on its own merits. The success of using distorted models depends to a large extent on the skill and experience of the investigator responsible for the design of the model and in the interpretation of experimental data obtained from the model. Distorted models are widely used, and additional information can be found in the references at the end of the chapter. References 14 and 15 contain detailed discussions of several practical examples of distorted fluid flow and hydraulic models. F l u i d s i n Old Man River in (large) miniature One of the world’s largest scale models, a Mississippi River model, resides near Jackson, Mississippi. It is a detailed, complex model that covers many acres and replicates the 1,250,000-acre Mississippi River basin. Built by the Army Corps of Engineers and used from 1943 to 1973, today it has mostly gone to ruin. As with many hydraulic models, this is a distorted model, with a horizontal scale of 1:2000 and a vertical scale of 1:100. One step along the model river corresponds to one mile along the river. All essential river basin elements such as geological features, 7.9 t h e N e w s levees, and railroad embankments were sculpted by hand to match the actual contours. The main purpose of the model was to predict floods. This was done by supplying specific amounts of water at prescribed locations along the model and then measuring the water depths up and down the model river. Because of the length scale, there is a difference in the time taken by the corresponding model and prototype events. Although it takes days for the actual floodwaters to travel from Sioux City, Iowa, to Omaha, Nebraska, it would take only minutes for the simulated flow in the model. Some Typical Model Studies Models are used to investigate many different types of fluid mechanics problems, and it is difficult to characterize in a general way all necessary similarity requirements, since each problem is unique. We can, however, broadly classify many of the problems on the basis of the general nature of the flow and subsequently develop some general characteristics of model designs in each of these classifications. In the following sections we will consider models for the study of 112 flow through closed conduits, 122 flow around immersed bodies, and 132 flow with a free surface. Turbomachine models are considered in Chapter 12. 7.9.1 Flow through Closed Conduits Geometric and Reynolds number similarity is usually required for models involving flow through closed conduits. Common examples of this type of flow include pipe flow and flow through valves, fittings, and metering devices. Although the conduit cross sections are often circular, they could have other shapes as well and may contain expansions or contractions. Since there are no fluid interfaces or free surfaces, the dominant forces are inertial and viscous so that the Reynolds number is an important similarity parameter. For low Mach numbers 1Ma 6 0.32, compressibility effects are usually negligible for both the flow of liquids and gases. For this class of problems, geometric similarity between model and prototype must be maintained. Generally the geometric characteristics can be described by a series of length terms, /1, /2, /3, . . . , /i, and /, where / is some particular length dimension for the system. Such a series of length terms leads to a set of pi terms of the form ßi /i / where i 1, 2, . . . , and so on. In addition to the basic geometry of the system, the roughness of the internal surface in contact with the fluid may be important. If the average height of surface roughness elements is defined as e, then the pi term representing roughness will be e /. This parameter indicates that for complete geometric similarity, surface roughness would also have to be scaled. Note that this implies that for length scales less than 1, the model surfaces should be smoother than those in the prototype since em l/e. To further complicate matters, the pattern of roughness elements in model and prototype would have to be similar. These are conditions that are virtually impossible to satisfy exactly. Fortunately, in some problems the surface roughness plays 7.9 Some Typical Model Studies 375 a minor role and can be neglected. However, in other problems 1such as turbulent flow through pipes2 roughness can be very important. It follows from this discussion that for flow in closed conduits at low Mach numbers, any dependent pi term 1the one that contains the particular variable of interest, such as pressure drop2 can be expressed as /i e rV/ b Dependent pi term f a , , / / m (7.16) This is a general formulation for this type of problem. The first two pi terms of the right side of Eq. 7.16 lead to the requirement of geometric similarity so that /im /i /m / em e /m / or /m /im em l/ e /i / This result indicates that the investigator is free to choose a length scale, l/, but once this scale is selected, all other pertinent lengths must be scaled in the same ratio. The additional similarity requirement arises from the equality of Reynolds numbers rmVm/m rV/ mm m Accurate predictions of flow behavior require the correct scaling of velocities. From this condition the velocity scale is established so that Vm mm r / m rm /m V (7.17) and the actual value of the velocity scale depends on the viscosity and density scales, as well as the length scale. Different fluids can be used in model and prototype. However, if the same fluid is used 1with mm m and rm r2, then Vm / V /m Thus, Vm Vl/, which indicates that the fluid velocity in the model will be larger than that in the prototype for any length scale less than 1. Since length scales are typically much less than unity, Reynolds number similarity may be difficult to achieve because of the large model velocities required. With these similarity requirements satisfied, it follows that the dependent pi term will be equal in model and prototype. For example, if the dependent variable of interest is the pressure differential,3 ¢p, between two points along a closed conduit, then the dependent pi term could be expressed as ß1 ¢p rV 2 The prototype pressure drop would then be obtained from the relationship ¢p r V 2 a b ¢pm rm Vm so that from a measured pressure differential in the model, ¢pm, the corresponding pressure differential for the prototype could be predicted. Note that in general ¢p ¢pm. 3 In some previous examples the pressure differential per unit length, ¢p/, was used. This is appropriate for flow in long pipes or conduits in which the pressure would vary linearly with distance. However, in the more general situation the pressure may not vary linearly with position so that it is necessary to consider the pressure differential, ¢p, as the dependent variable. In this case the distance between pressure taps is an additional variable 1as well as the distance of one of the taps measured from some reference point within the flow system2. 376 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling E XAMPLE 7.6 Reynolds Number Similarity GIVEN Model tests are to be performed to study the flow through a large check valve having a 2-ft-diameter inlet and carrying water at a flowrate of 30 cfs as shown in Fig. E7.6a. The working fluid in the model is water at the same temperature as that in the prototype. Complete geometric similarity exists between model and prototype, and the model inlet diameter is 3 in. Q = 30 cfs FIND Determine the required flowrate in the model. (Qm = ?) SOLUTION D = 2 ft (Dm = 3 in.) To ensure dynamic similarity, the model tests should be run so that ■ Figure E7.6a Rem Re or VmDm VD nm n where V and D correspond to the inlet velocity and diameter, respectively. Since the same fluid is to be used in model and prototype, n nm, and therefore Vm D V Dm The discharge, Q, is equal to VA, where A is the inlet area, so Fig. E7.6b. For this particular example, DmD 0.125, and the corresponding velocity scale is 8 (see Fig. E7.6b). Thus, with the prototype velocity equal to V 130 ft3s2 1p4212 ft2 2 9.50 fts, the required model velocity is Vm 76.4 ft s. Although this is a relatively large velocity, it could be attained in a laboratory facility. It is to be noted that if we tried to use a smaller model, say one with D 1 in., the required model velocity is 229 fts, a very high velocity that would be difficult to achieve. These results are indicative of one of the difficulties encountered in maintaining Reynolds number similarity—the required model velocities may be impractical to obtain. 2 Qm Vm Am D 3 1p42Dm 4 a b Q VA Dm 3 1p42D2 4 Dm D and for the data given 1312 ft2 12 ft2 Qm 3.75 cfs 20 15 Vm /V Qm 25 130 ft3 s2 10 (0.125, 8) (Ans) 5 COMMENT As indicated by the foregoing analysis, to maintain Reynolds number similarity using the same fluid in model and prototype, the required velocity scale is inversely proportional to the length scale, that is, VmV 1DmD2 1. This strong influence of the length scale on the velocity scale is shown in In some problems Reynolds number similarity may be relaxed. 0 0 0.2 0.4 0.6 0.8 1 Dm /D ■ Figure E7.6b Two additional points should be made with regard to modeling flows in closed conduits. First, for large Reynolds numbers, inertial forces are much larger than viscous forces, and in this case it may be possible to neglect viscous effects. The important practical consequence of this is that it would not be necessary to maintain Reynolds number similarity between model and prototype. However, both model and prototype would have to operate at large Reynolds numbers. Since we do not know, a priori, what a “large Reynolds number” is, the effect of Reynolds numbers would 7.9 377 Some Typical Model Studies have to be determined from the model. This could be accomplished by varying the model Reynolds number to determine the range 1if any2 over which the dependent pi term ceases to be affected by changes in Reynolds number. The second point relates to the possibility of cavitation in flow through closed conduits. For example, flow through the complex passages that may exist in valves may lead to local regions of high velocity 1and thus low pressure2, which can cause the fluid to cavitate. If the model is to be used to study cavitation phenomena, then the vapor pressure, pv, becomes an important variable and an additional similarity requirement such as equality of the cavitation number 1pr ⫺ pv 2 Ⲑ12rV 2 is required, where pr is some reference pressure. The use of models to study cavitation is complicated, since it is not fully understood how vapor bubbles form and grow. The initiation of bubbles seems to be influenced by the microscopic particles that exist in most liquids, and how this aspect of the problem influences model studies is not clear. Additional details can be found in Ref. 17. 7.9.2 Flow around Immersed Bodies Geometric and Reynolds number similarity is usually required for models involving flow around bodies. Models have been widely used to study the flow characteristics associated with bodies that are completely immersed in a moving fluid. Examples include flow around aircraft, automobiles, golf balls, and buildings. 1These types of models are usually tested in wind tunnels as is illustrated in Fig. 7.6.2 Modeling laws for these problems are similar to those described in the preceding section; that is, geometric and Reynolds number similarity is required. Since there are no fluid interfaces, surface tension 1and therefore the Weber number2 is not important. Also, gravity will not affect the flow patterns, so the Froude number need not be considered. The Mach number will be important for high-speed flows in which compressibility becomes an important factor, but for incompressible fluids 1such as liquids or for gases at relatively low speeds2 the Mach number can be omitted as a similarity requirement. In this case, a general formulation for these problems is /i e rV/ b Dependent pi term ⫽ f a , , / / m V7.13 Wind engineering models (7.18) where / is some characteristic length of the system and /i represents other pertinent lengths, eⲐ/ is the relative roughness of the surface 1or surfaces2, and rV/Ⲑm is the Reynolds number. Frequently, the dependent variable of interest for this type of problem is the drag, d, developed on the body, and in this situation the dependent pi term would usually be expressed in the form of a drag coefficient, CD, where CD ⫽ d 1 2 2 2 rV / The numerical factor, 12, is arbitrary but commonly included, and /2 is usually taken as some representative area of the object. Thus, drag studies can be undertaken with the formulation d 1 2 2 2 rV / /i e rV/ ⫽ CD ⫽ f a , , b / / m (7.19) ■ Figure 7.6 Model of the National Bank of Commerce, San Antonio, Texas, for measurement of peak, rms, and mean pressure distributions. The model is located in a long-test-section, meteorological wind tunnel. (Photograph courtesy of CPP Wind Engineering; www.cppwind.com.) 378 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling It is clear from Eq. 7.19 that geometric similarity em e /m / /im /i /m / as well as Reynolds number similarity rmVm/m rV/ mm m must be maintained. If these conditions are met, then For flow around bodies, drag is often the dependent variable of interest. d 1 2 2 2 rV / dm 1 2 2 2 rmVm/m or d r V 2 / 2 a b a b dm rm Vm /m Measurements of model drag, dm, can then be used to predict the corresponding drag, d, on the prototype from this relationship. As was discussed in the previous section, one of the common difficulties with models is related to the Reynolds number similarity requirement which establishes the model velocity as Vm mm r / V m rm /m (7.20) nm / V n /m (7.21) or Vm where nm n is the ratio of kinematic viscosities. If the same fluid is used for model and prototype so that nm n, then Vm V7.14 Model airplane test in water V7.15 Large scale wind tunnel / V /m and, therefore, the required model velocity will be higher than the prototype velocity for //m greater than 1. Since this ratio is often relatively large, the required value of Vm may be large. For example, for a 101 length scale, and a prototype velocity of 50 mph, the required model velocity is 500 mph. This is a value that is unreasonably high to achieve with liquids, and for gas flows this would be in the range where compressibility would be important in the model 1but not in the prototype2. As an alternative, we see from Eq. 7.21 that Vm could be reduced by using a different fluid in the model such that nm n 6 1. For example, the ratio of the kinematic viscosity of water to that of air is approximately 101 , so that if the prototype fluid were air, tests might be run on the model using water. This would reduce the required model velocity, but it still may be difficult to achieve the necessary velocity in a suitable test facility, such as a water tunnel. Another possibility for wind tunnel tests would be to increase the air pressure in the tunnel so that rm 7 r, thus reducing the required model velocity as specified by Eq. 7.20. Fluid viscosity is not strongly influenced by pressure. Although pressurized tunnels have been used, they are obviously more complicated and expensive. The required model velocity can also be reduced if the length scale is modest; that is, the model is relatively large. For wind tunnel testing, this requires a large test section which greatly increases the cost of the facility. However, large wind tunnels suitable for testing very large models 1or prototypes2 are in use. One such tunnel, located at the NASA Ames Research Center, Moffett Field, California, has a test section that is 40 ft by 80 ft and can accommodate test speeds to 345 mph. Such a large and expensive test facility is obviously not feasible for university or industrial laboratories, so most model testing has to be accomplished with relatively small models. 7.9 E XAMPLE 7.7 379 Some Typical Model Studies Model Design Conditions and Predicted Prototype Performance GIVEN The drag on the airplane shown in Fig. E7.7 cruising at 240 mph in standard air is to be determined from tests on a 1:10 scale model placed in a pressurized wind tunnel. To minimize compressibility effects, the airspeed in the wind tunnel is also to be 240 mph. V = 240 mph FIND Determine (a) the required air pressure in the tunnel (assuming the same air temperature for model and prototype) and (b) the drag on the prototype corresponding to a measured force of 1 lb on the model. ■ Figure E7.7 (Photograph courtesy of NASA) SOLUTION (a) From Eq. 7.19 it follows that drag can be predicted from a geometrically similar model if the Reynolds numbers in model and prototype are the same. Thus, rV/ rm Vm/m ⫽ mm m For this example, Vm ⫽ V and /m Ⲑ/ ⫽ 101 so that for constant temperature 1T ⫽ Tm 2. Therefore, the wind tunnel would need to be pressurized so that pm ⫽ 10 p Since the prototype operates at standard atmospheric pressure, the required pressure in the wind tunnel is 10 atmospheres or pm ⫽ 10 114.7 psia2 ⫽ 147 psia mm V / rm ⫽ r m Vm /m ⫽ mm 112 1102 m and therefore rm mm ⫽ 10 r m This result shows that the same fluid with rm ⫽ r and mm ⫽ m cannot be used if Reynolds number similarity is to be maintained. One possibility is to pressurize the wind tunnel to increase the density of the air. We assume that an increase in pressure does not significantly change the viscosity so that the required increase in density is given by the relationship rm ⫽ 10 r For an ideal gas, p ⫽ rRT so that pm rm ⫽ p r V7.16 Full Scale Race Car Model (Ans) COMMENT Thus, we see that a high pressure would be required and this could not be achieved easily or inexpensively. However, under these conditions, Reynolds similarity would be attained. (b) The drag could be obtained from Eq. 7.19 so that d 1 2 rV 2/2 ⫽ dm 1 2 rmVm2 /2m or r V 2 / 2 a b a b dm rm Vm /m 1 ⫽ a b 112 2 1102 2dm 10 ⫽ 10dm d⫽ Thus, for a drag of 1 lb on the model the corresponding drag on the prototype is d ⫽ 10 lb (Ans) Fortunately, in many situations the flow characteristics are not strongly influenced by the Reynolds number over the operating range of interest. In these cases we can avoid the rather stringent similarity requirement of matching Reynolds numbers. To illustrate this point, consider the variation in the drag coefficient with the Reynolds number for a smooth sphere of diameter d placed in a uniform stream with approach velocity, V. Some typical data are shown in Fig. 7.7. We observe that for Reynolds numbers between approximately 103 and 2 ⫻ 105 the drag coefficient is relatively constant and does not strongly depend on the specific value of the Reynolds number. Thus, exact Reynolds number similarity is not required in this range. For other geometric shapes we would typically find that for high Reynolds numbers, inertial forces are dominant 1rather than viscous forces2, and the drag is essentially independent of the Reynolds number. 380 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Drag coefficient, CD 100 10 1 0.1 10–1 100 101 102 103 104 ρVd Reynolds number, Re = ____ μ 105 106 107 ■ Figure 7.7 The effect of Reynolds number on the drag coefficient, CD, for a smooth sphere with CD ⫽ dⲐ12 ARV 2, where A is the projected area of sphere, Pd 2Ⲑ4. (Adapted from Kurt Gramoll, www.eCourses.ou.edu. Data from Ref. 16.) At high Reynolds numbers the drag is often essentially independent of the Reynolds number. ᑞ V Another interesting point to note from Fig. 7.7 is the rather abrupt drop in the drag coefficient near a Reynolds number of 3 ⫻ 105. As is discussed in Section 9.3.3, this is due to a change in the flow conditions near the surface of the sphere. These changes are influenced by the surface roughness and, in fact, the drag coefficient for a sphere with a “rougher” surface will generally be less than that of the smooth sphere for high Reynolds number. For example, the dimples on a golf ball are used to reduce the drag over that which would occur for a smooth golf ball. Although this is undoubtedly of great interest to the avid golfer, it is also important to engineers responsible for fluid-flow models, since it does emphasize the potential importance of the surface roughness. However, for bodies that are sufficiently angular with sharp corners, the actual surface roughness is likely to play a secondary role compared with the main geometric features of the body. One final note with regard to Fig. 7.7 concerns the interpretation of experimental data when plotting pi terms. For example, if r, m, and d remain constant, then an increase in Re comes from an increase in V. Intuitively, it would seem in general that if V increases, the drag would increase. However, as shown in the figure, the drag coefficient generally decreases with increasing Re. When interpreting data, one needs to be aware if the variables are nondimensional. In this case, the physical drag force is proportional to the drag coefficient times the velocity squared. Thus, as shown by the figure in the margin, the drag force does, as expected, increase with increasing velocity. The exception occurs in the Reynolds number range 2 ⫻ 105 6 Re 6 4 ⫻ 105 where the drag coefficient decreases dramatically with increasing Reynolds number (see Fig. 7.7). This phenomena is discussed in Section 9.3. For problems involving high velocities in which the Mach number is greater than about 0.3, the influence of compressibility, and therefore the Mach number 1or Cauchy number2, becomes significant. In this case complete similarity requires not only geometric and Reynolds number similarity but also Mach number similarity so that Vm V ⫽ cm c (7.22) n /m c ⫽ cm nm / (7.23) This similarity requirement, when combined with that for Reynolds number similarity 1Eq. 7.212, yields 7.9 Some Typical Model Studies 381 Clearly the same fluid with c cm and n nm cannot be used in model and prototype unless the length scale is unity 1which means that we are running tests on the prototype2. In high-speed aerodynamics the prototype fluid is usually air, and it is difficult to satisfy Eq. 7.23 for reasonable length scales. Thus, models involving high-speed flows are often distorted with respect to Reynolds number similarity, but Mach number similarity is maintained. 7.9.3 Flow with a Free Surface Froude number similarity is usually required for models involving freesurface flows. Flows in canals, rivers, spillways, and stilling basins, as well as flow around ships, are all examples of flow phenomena involving a free surface. For this class of problems, both gravitational and inertial forces are important and, therefore, the Froude number becomes an important similarity parameter. Also, since there is a free surface with a liquid–air interface, forces due to surface tension may be significant, and the Weber number becomes another similarity parameter that needs to be considered along with the Reynolds number. Geometric variables will obviously still be important. Thus, a general formulation for problems involving flow with a free surface can be expressed as /i e rV/ V rV 2/ (7.24) , b Dependent pi term f a , , , / / m 2g/ s As discussed previously, / is some characteristic length of the system, /i represents other pertinent lengths, and e/ is the relative roughness of the various surfaces. Since gravity is the driving force in these problems, Froude number similarity is definitely required so that Vm 2gm/m V7.17 River flow model V 2g/ The model and prototype are expected to operate in the same gravitational field 1gm g2, and therefore it follows that Vm /m 1l/ V B/ (7.25) Thus, when models are designed on the basis of Froude number similarity, the velocity scale is determined by the square root of the length scale. As is discussed in Section 7.8.3, to simultaneously have Reynolds and Froude number similarity it is necessary that the kinematic viscosity scale be related to the length scale as nm 1l/ 2 32 n (7.26) The working fluid for the prototype is normally either freshwater or seawater and the length scale is small. Under these circumstances it is virtually impossible to satisfy Eq. 7.26, so models involving free-surface flows are usually distorted. The problem is further complicated if an attempt is made to model surface tension effects, since this requires the equality of Weber numbers, which leads to the condition sm rm 1l/ 2 2 sr V7.18 Boat model (7.27) for the kinematic surface tension 1sr2. It is again evident that the same fluid cannot be used in model and prototype if we are to have similitude with respect to surface tension effects for l/ 1. Fortunately, in many problems involving free-surface flows, both surface tension and viscous effects are small and consequently strict adherence to Weber and Reynolds number similarity is not required. Certainly, surface tension is not important in large hydraulic structures and rivers. Our only concern would be if in a model the depths were reduced to the point where surface tension becomes an important factor, whereas it is not in the prototype. This is of particular importance in the design of river models, since the length scales are typically small 1so that the width of the model is reasonable2, but with a small length scale the required model depth may be very small. To overcome this problem, different horizontal and vertical length scales are often used for river 382 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling ■ Figure 7.8 A scale hydraulic model (1:197) of the Guri Dam in Venezuela which is used to simulate the characteristics of the flow over and below the spillway and the erosion below the spillway. (Photograph courtesy of St. Anthony Falls Hydraulic Laboratory.) models. Although this approach eliminates surface tension effects in the model, it introduces geometric distortion that must be accounted for empirically, usually by increasing the model surface roughness. It is important in these circumstances that verification tests with the model be performed 1if possible2 in which model data are compared with available prototype river flow data. Model roughness can be adjusted to give satisfactory agreement between model and prototype, and then the model subsequently used to predict the effect of proposed changes on river characteristics 1such as velocity patterns or surface elevations2. For large hydraulic structures, such as dam spillways, the Reynolds numbers are large so that viscous forces are small in comparison to the forces due to gravity and inertia. In this case, Reynolds number similarity is not maintained and models are designed on the basis of Froude number similarity. Care must be taken to ensure that the model Reynolds numbers are also large, but they are not required to be equal to those of the prototype. This type of hydraulic model is usually made as large as possible so that the Reynolds number will be large. A spillway model is shown in Fig. 7.8. Also, for relatively large models the geometric features of the prototype can be accurately scaled, as well as surface roughness. Note that em ⫽ l/e, which indicates that the model surfaces must be smoother than the corresponding prototype surfaces for l/ 6 1. V7.19 Dam model F l u i d s i n Ice engineering Various types of models have been studied in wind tunnels, water tunnels, and towing tanks for many years. But another type of facility is needed to study ice and ice-related problems. The U.S. Army Cold Regions Research and Engineering Laboratory has developed a unique complex that houses research facilities for studies related to the mechanical behavior of ice and ice–structure interactions. The laboratory contains three separate cold-rooms—a test basin, a flume, and a general research area. In the test basin, large-scale model studies of ice forces on structures E XAMPLE 7.8 t h e N e w s such as dams, piers, ships, and offshore platforms can be performed. Ambient temperatures can be controlled as low as ⫺20 °F, and at this temperature a 2-mm-per-hour ice growth rate can be achieved. It is also possible to control the mechanical properties of the ice to properly match the physical scale of the model. Tests run in the recirculating flume can simulate river processes during ice formation. And in the large research area, scale models of lakes and rivers can be built and operated to model ice interactions with various types of engineering projects. (See Problem 7.85.) Froude Number Similarity GIVEN The spillway for the dam shown in Fig. E7.8a is 20 m wide and is designed to carry 125 m3/s at flood stage. A 1:15 model is constructed to study the flow characteristics through the spillway. The effects of surface tension and viscosity are to be neglected. FIND (a) Determine the required model width and flowrate. (b) What operating time for the model corresponds to a 24-hr period in the prototype? ■ Figure E7.8a (Photo by Marty Melchior.) 7.9 Some Typical Model Studies 383 SOLUTION The width, wm, of the model spillway is obtained from the length scale, l/, so that or /m tm V /m 2l/ t Vm / B/ wm 1 l/ w 15 This result indicates that time intervals in the model will be smaller than the corresponding intervals in the prototype if l/ 6 1. For l/ 151 and a prototype time interval of 24 hr Therefore, wm 20 m 1.33 m 15 Of course, all other geometric features 1including surface roughness2 of the spillway must be scaled in accordance with the same length scale. With the neglect of surface tension and viscosity, Eq. 7.24 indicates that dynamic similarity will be achieved if the Froude numbers are equal between model and prototype. Thus, Vm 2gm/m tm 2151 124 hr2 6.20 hr (Ans) V 2g/ and for gm g Vm /m V B/ COMMENT As indicated by the foregoing analysis, the time scale varies directly as the square root of the length scale. Thus, as shown in Fig. E7.8b, the model time interval, tm, corresponding to a 24-hr prototype time interval can be varied by changing the length scale, l/. The ability to scale times may be very useful, since it is possible to “speed up” events in the model that may occur over a relatively long time in the prototype. There is, of course, a practical limit to how small the length scale (and the corresponding time scale) can become. For example, if the length scale is too small then surface tension effects may become important in the model whereas they are not in the prototype. In such a case the present model design, based simply on Froude number similarity, would not be adequate. Since the flowrate is given by Q VA, where A is an appropriate cross-sectional area, it follows that 20 Qm Vm Am /m /m 2 a b Q VA B/ / 16 1l/ 2 52 tm , hr 12 where we have made use of the relationship AmA 1/m/2 2. For l/ 151 and Q 125 m3s Qm 1 151 2 52 1125 m3 s2 0.143 m3s 8 (1/15, 6.20 hr) (Ans) 4 The time scale can be obtained from the velocity scale, since the velocity is distance divided by time 1V /t2, and therefore 0 / tm V Vm t /m V7.20 Testing of large yacht model (Ans) 0 0.1 0.2 0.3 0.4 0.5 m ___ ■ Figure E7.8b There are, unfortunately, problems involving flow with a free surface in which viscous, inertial, and gravitational forces are all important. The drag on a ship as it moves through water is due to the viscous shearing stresses that develop along its hull, as well as a pressure-induced component of drag caused by both the shape of the hull and wave action. The shear drag is a function of the Reynolds number, whereas the pressure drag is a function of the Froude number. Since both Reynolds number and Froude number similarity cannot be simultaneously achieved by using water as the model fluid 1which is the only practical fluid for ship models2, some technique other than a straightforward model test must be employed. One common approach is to measure the total drag on a small, geometrically similar model as it is towed through a model basin at Froude numbers matching those of the prototype. The shear drag on the model is calculated using analytical techniques of the type described in Chapter 9. This calculated value is then subtracted from the total drag to obtain pressure drag, and using Froude number scaling the pressure drag on the prototype can then be predicted. The experimentally determined value can then be combined with a calculated value of the shear drag 1again using analytical techniques2 to provide the desired total drag 384 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling ■ Figure 7.9 Instrumented, smallwaterplane-area, twin hull (SWATH) model suspended from a towing carriage. (Photograph courtesy of the U.S. Navy’s David W. Taylor Research Center.) on the ship. Ship models are widely used to study new designs, but the tests require extensive facilities 1see Fig. 7.92. It is clear from this brief discussion of various types of models involving free-surface flows that the design and use of such models requires considerable ingenuity, as well as a good understanding of the physical phenomena involved. This is generally true for most model studies. Modeling is both an art and a science. Motion picture producers make extensive use of model ships, fires, explosions, and the like. It is interesting to attempt to observe the flow differences between these distorted model flows and the real thing. 7.10 Similitude Based on Governing Differential Equations Similarity laws can be directly developed from the equations governing the phenomenon of interest. In the preceding sections of this chapter, dimensional analysis has been used to obtain similarity laws. This is a simple, straightforward approach to modeling, which is widely used. The use of dimensional analysis requires only a knowledge of the variables that influence the phenomenon of interest. Although the simplicity of this approach is attractive, it must be recognized that omission of one or more important variables may lead to serious errors in the model design. An alternative approach is available if the equations 1usually differential equations2 governing the phenomenon are known. In this situation similarity laws can be developed from the governing equations, even though it may not be possible to obtain analytic solutions to the equations. To illustrate the procedure, consider the flow of an incompressible Newtonian fluid. For simplicity we will restrict our attention to two-dimensional flow, although the results are applicable to the general three-dimensional case. From Chapter 6 we know that the governing equations are the continuity equation 0u 0v 0 0x 0y (7.28) and the Navier–Stokes equations ra 0p 0u 0u 0u 0 2u 0 2u u v b ma 2 2 b 0t 0x 0y 0x 0x 0y (7.29) ra 0p 0v 0v 0v 0 2v 0 2v u v b rg m a 2 2 b 0t 0x 0y 0y 0x 0y (7.30) where the y axis is vertical, so that the gravitational body force, rg, only appears in the “y equation.” To continue the mathematical description of the problem, boundary conditions are required. For example, velocities on all boundaries may be specified; that is, u uB and v vB at all boundary points x xB and y yB. In some types of problems it may be necessary to specify the pressure over some part of the boundary. For time-dependent problems, initial conditions would also have to be provided, which means that the values of all dependent variables would be given at some time 1usually taken at t 02. Once the governing equations, including boundary and initial conditions, are known, we are ready to proceed to develop similarity requirements. The next step is to define a new set of 7.10 Similitude Based on Governing Differential Equations 385 variables that are dimensionless. To do this we select a reference quantity for each type of variable. In this problem the variables are u, v, p, x, y, and t so we will need a reference velocity, V, a reference pressure, p0, a reference length, /, and a reference time, t. These reference quantities should be parameters that appear in the problem. For example, / may be a characteristic length of a body immersed in a fluid or the width of a channel through which a fluid is flowing. The velocity, V, may be the free-stream velocity or the inlet velocity. The new dimensionless 1starred2 variables can be expressed as Each variable is made dimensionless by dividing by an appropriate reference quantity. r, m v=0 p = p0 x= x v V x x / y y / p t p p0 t t and V 0 0u 0x V 0 2u 0 2u a b / 0x 0x 0x 0x2 /2 0x2 y v = 0 p = 1 v 0u 0Vu 0x V 0u 0x 0x 0x / 0x Actual u = 1 u V as shown in the figure in the margin. The governing equations can now be rewritten in terms of these new variables. For example, y u=V u Re x = 1 x The other terms that appear in the equations can be expressed in a similar fashion. Thus, in terms of the new variables the governing equations become Dimensionless 0u 0v 0 0x 0y (7.31) and c rV 2 p0 0p mV rV 0u 0u 0u 0 2u 0 2u d c d au v b c d c 2 d a b t 0t / 0x 0y / 0x / 0x2 0y2 c rV 0v rV 2 0v 0v d c d au v b t 0t / 0x 0y FI/ FIc c p0 0p mV 0 2v 0 2v b d 3rg 4 c 2 d a / 0y / 0x2 0y2 FP FG FV (7.32) (7.33) The terms appearing in brackets contain the reference quantities and can be interpreted as indices of the various forces 1per unit volume2 that are involved. Thus, as is indicated in Eq. 7.33, FI/ inertia 1local2 force, FIc inertia 1convective2 force, Fp pressure force, FG gravitational force, and FV viscous force. As the final step in the nondimensionalization process, we will divide each term in Eqs. 7.32 and 7.33 by one of the bracketed quantities. Although any one of these quantities could be used, it is conventional to divide by the bracketed quantity rV 2/ which is the index of the convective inertia force. The final nondimensional form then becomes c p0 0p m / 0u 0u 0u 0 2u 0 2u d u v c 2 d c d a b tV 0t 0x 0y rV/ rV 0x 0x2 0y2 (7.34) c g/ m p0 0p / 0v 0v 0v 0 2v 0 2v d u v c 2d c d a b c 2 d tV 0t 0x 0y rV/ rV 0y V 0x2 0y2 (7.35) We see that bracketed terms are the standard dimensionless groups 1or their reciprocals2 which were developed from dimensional analysis; that is, /tV is a form of the Strouhal number, p0 rV 2 386 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Governing equations expressed in terms of dimensionless variables lead to the appropriate dimensionless groups. the Euler number, g/V 2 the reciprocal of the square of the Froude number, and m rV/ the reciprocal of the Reynolds number. From this analysis it is now clear how each of the dimensionless groups can be interpreted as the ratio of two forces, and how these groups arise naturally from the governing equations. Although we really have not helped ourselves with regard to obtaining an analytical solution to these equations 1they are still complicated and not amenable to an analytical solution2, the dimensionless forms of the equations, Eqs. 7.31, 7.34, and 7.35, can be used to establish similarity requirements. From these equations it follows that if two systems are governed by these equations, then the solutions 1in terms of u, v, p, x, y, and t2 will be the same if the four parameters /tV, p0 rV 2, V 2g/, and rV/ m are equal for the two systems. The two systems will be dynamically similar. Of course, boundary and initial conditions expressed in dimensionless form must also be equal for the two systems, and this will require complete geometric similarity. These are the same similarity requirements that would be determined by a dimensional analysis if the same variables were considered. However, the advantage of working with the governing equations is that the variables appear naturally in the equations, and we do not have to worry about omitting an important one, provided the governing equations are correctly specified. We can thus use this method to deduce the conditions under which two solutions will be similar even though one of the solutions will most likely be obtained experimentally. In the foregoing analysis we have considered a general case in which the flow may be unsteady, and both the actual pressure level, p0, and the effect of gravity are important. A reduction in the number of similarity requirements can be achieved if one or more of these conditions is removed. For example, if the flow is steady, the dimensionless group, /tV, can be eliminated. The actual pressure level will only be of importance if we are concerned with cavitation. If not, the flow patterns and the pressure differences will not depend on the pressure level. In this case, p0 can be taken as rV 2 1or 12rV 2 2, and the Euler number can be eliminated as a similarity requirement. However, if we are concerned about cavitation 1which will occur in the flow field if the pressure at certain points reaches the vapor pressure, pv2, then the actual pressure level is important. Usually, in this case, the characteristic pressure, p0, is defined relative to the vapor pressure such that p0 pr pv where pr is some reference pressure within the flow field. With p0 defined in this manner, the similarity parameter p0 rV 2 becomes 1p r pv 2 rV 2. This parameter is frequently written as 1 pr pv 2 12rV 2, and in this form, as was noted previously in Section 7.6, is called the cavitation number. Thus we can conclude that if cavitation is not of concern we do not need a similarity parameter involving p0, but if cavitation is to be modeled, then the cavitation number becomes an important similarity parameter. The Froude number, which arises because of the inclusion of gravity, is important for problems in which there is a free surface. Examples of these types of problems include the study of rivers, flow through hydraulic structures such as spillways, and the drag on ships. In these situations the shape of the free surface is influenced by gravity, and therefore the Froude number becomes an important similarity parameter. However, if there are no free surfaces, the only effect of gravity is to superimpose a hydrostatic pressure distribution on the pressure distribution created by the fluid motion. The hydrostatic distribution can be eliminated from the governing equation 1Eq. 7.302 by defining a new pressure, p¿ p rgy, and with this change the Froude number does not appear in the nondimensional governing equations. We conclude from this discussion that for the steady flow of an incompressible fluid without free surfaces, dynamic and kinematic similarity will be achieved if 1for geometrically similar systems2 Reynolds number similarity exists. If free surfaces are involved, Froude number similarity must also be maintained. For free-surface flows we have tacitly assumed that surface tension is not important. We would find, however, that if surface tension is included, its effect would appear in the free-surface boundary condition, and the Weber number, rV 2/s, would become an additional similarity parameter. In addition, if the governing equations for compressible fluids are considered, the Mach number, Vc, would appear as an additional similarity parameter. It is clear that all the common dimensionless groups that we previously developed by using dimensional analysis appear in the governing equations that describe fluid motion when these equations are expressed in terms of dimensionless variables. Thus, use of the governing equations to obtain similarity laws provides an alternative to dimensional analysis. This approach has the 7.11 Chapter Summary and Study Guide 387 advantage that the variables are known and the assumptions involved are clearly identified. In addition, a physical interpretation of the various dimensionless groups can often be obtained. 7.11 Chapter Summary and Study Guide similitude dimensionless product basic dimensions pi term Buckingham pi theorem method of repeating variables model modeling laws prototype prediction equation model design conditions similarity requirements modeling laws length scale distorted model true model Many practical engineering problems involving fluid mechanics require experimental data for their solution. Thus, laboratory studies and experimentation play a significant role in this field. It is important to develop good procedures for the design of experiments so they can be efficiently completed with as broad applicability as possible. To achieve this end the concept of similitude is often used in which measurements made in the laboratory can be utilized for predicting the behavior of other similar systems. In this chapter, dimensional analysis is used for designing such experiments, as an aid for correlating experimental data, and as the basis for the design of physical models. As the name implies, dimensional analysis is based on a consideration of the dimensions required to describe the variables in a given problem. A discussion of the use of dimensions and the concept of dimensional homogeneity (which forms the basis for dimensional analysis) was included in Chapter 1. Essentially, dimensional analysis simplifies a given problem described by a certain set of variables by reducing the number of variables that need to be considered. In addition to being fewer in number, the new variables are dimensionless products of the original variables. Typically these new dimensionless variables are much simpler to work with in performing the desired experiments. The Buckingham pi theorem, which forms the theoretical basis for dimensional analysis, is introduced. This theorem establishes the framework for reducing a given problem described in terms of a set of variables to a new set of fewer dimensionless variables. A simple method, called the repeating variable method, is described for actually forming the dimensionless variables (often called pi terms). Forming dimensionless variables by inspection is also considered. It is shown how the use of dimensionless variables can be of assistance in planning experiments and as an aid in correlating experimental data. For problems in which there are a large number of variables, the use of physical models is described. Models are used to make specific predictions from laboratory tests rather than formulating a general relationship for the phenomenon of interest. The correct design of a model is obviously imperative for the accurate predictions of other similar, but usually larger, systems. It is shown how dimensional analysis can be used to establish a valid model design. An alternative approach for establishing similarity requirements using governing equations (usually differential equations) is presented. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. use the Buckingham pi theorem to determine the number of independent dimensionless variables needed for a given flow problem. form a set of dimensionless variables using the method of repeating variables. form a set of dimensionless variables by inspection. use dimensionless variables as an aid in interpreting and correlating experimental data. use dimensional analysis to establish a set of similarity requirements (and prediction equation) for a model to be used to predict the behavior of another similar system (the prototype). rewrite a given governing equation in a suitable nondimensional form and deduce similarity requirements from the nondimensional form of the equation. Some of the important equations in this chapter are: Reynolds number Re rV/ m Froude number Fr V 1g/ 388 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling Euler number Eu Cauchy number Ca p rV 2 rV 2 Ev V Ma c Mach number v/ V rV 2/ We s St Strouhal number Weber number References 1. Bridgman, P. W., Dimensional Analysis, Yale University Press, New Haven, Conn., 1922. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. Murphy, G., Similitude in Engineering, Ronald Press, New York, 1950. Langhaar, H. L., Dimensional Analysis and Theory of Models, Wiley, New York, 1951. Huntley, H. E., Dimensional Analysis, Macdonald, London, 1952. Duncan, W. J., Physical Similarity and Dimensional Analysis: An Elementary Treatise, Edward Arnold, London, 1953. Sedov, K. I., Similarity and Dimensional Methods in Mechanics, Academic Press, New York, 1959. Ipsen, D. C., Units, Dimensions, and Dimensionless Numbers, McGraw-Hill, New York, 1960. Kline, S. J., Similitude and Approximation Theory, McGraw-Hill, New York, 1965. Skoglund, V. J., Similitude—Theory and Applications, International Textbook, Scranton, Pa., 1967. Baker, W. E., Westline, P. S., and Dodge, F. T., Similarity Methods in Engineering Dynamics—Theory and Practice of Scale Modeling, Hayden 1Spartan Books2, Rochelle Park, N.J., 1973. Taylor, E. S., Dimensional Analysis for Engineers, Clarendon Press, Oxford, UK, 1974. Isaacson, E. de St. Q., and Isaacson, M. de St. Q., Dimensional Methods in Engineering and Physics, Wiley, New York, 1975. Schuring, D. J., Scale Models in Engineering, Pergamon Press, New York, 1977. Yalin, M. S., Theory of Hydraulic Models, Macmillan, London, 1971. Sharp, J. J., Hydraulic Modeling, Butterworth, London, 1981. Schlichting, H., Boundary-Layer Theory, 7th Ed., McGraw-Hill, New York, 1979. Knapp, R. T., Daily, J. W., and Hammitt, F. G., Cavitation, McGraw-Hill, New York, 1970. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.) Problems 389 Conceptual Questions 7.1C For flow in a pipe, the Reynolds number is defined as Re VrD m where V is velocity, r is density, m is viscosity, and D is diameter. Expressed this way, the Reynolds number can be interpreted as a) the ratio of viscous to inertial forces. b) the ratio of inertial to viscous forces. c) the ratio of momentum flux to viscous forces. d) the ratio of kinetic energy to viscous forces. 7.2C A 1/20th scale model of an airplane is used to determine forces on the actual airplane. The 1/20th scale refers to the a) lengths. b) velocity. c) forces. d) Reynolds number. e) all of the above. 7.3C A 1/4th scale model of a new automobile design is tested in a wind tunnel. The Reynolds number of the model is the same as that of the full-scale prototype. Assuming the model and prototype are exposed to the same air conditions, the velocity in the wind tunnel is then a) b) c) d) 1/4th that of the full-scale vehicle. The same as that of the full-scale vehicle. 4 times that of the full-scale vehicle. 40 times that of the full-scale vehicle. 7.4C A 1/10th scale model of a new automobile design is tested in a wind tunnel at the same Reynolds number as that of the full-scale prototype. The force coefficient 1FA2 1 12rV 2 2 of the model is the same as that of the prototype. Assuming the model and prototype are both tested in air, the force on the scale model, Fm, is then a) 1/10,000th that of the full-scale vehicle. b) 1/1000th that of the full-scale vehicle. c) 1/100th that of the full-scale vehicle. d) 1/10th that of the full-scale vehicle. e) The same as that of the full-scale vehicle. Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the evennumbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/ college/munson. Section 7.1 Dimensional Analysis 7.1 What are the dimensions of density, pressure, specific weight, surface tension, and dynamic viscosity in (a) the FLT system, and (b) the MLT system? Compare your results with those given in Table 1.1 in Chapter 1. 7.2 Verify the left-hand side of Eq. 7.2 is dimensionless using the MLT system. 7.3 The Reynolds number, rVD/m, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity of 3 m/s through a 2-in.-diameter pipe. 7.4 What are the dimensions of acceleration of gravity, density, dynamic viscosity, kinematic viscosity, specific weight, and speed of sound in (a) the FLT system, and (b) the MLT system? Compare your results with those given in Table 1.1 in Chapter 1. 7.5 For the flow of a thin film of a liquid with a depth h and a free surface, two important dimensionless parameters are the Froude number, V 1gh, and the Weber number, rV2h/␴. Determine the value of these two parameters for glycerin (at 20 C) flowing with a velocity of 0.7 m/s at a depth of 3 mm. 7.6 The Mach number for a body moving through a fluid with velocity V is defined as V/c, where c is the speed of sound in the fluid. This dimensionless parameter is usually considered to be important in fluid dynamics problems when its value exceeds 0.3. What would be the velocity of a body at a Mach number of 0.3 if the fluid is (a) air at standard atmospheric pressure and 20 C, and (b) water at the same temperature and pressure? Section 7.3 Determination of Pi Terms 7.7 It is desired to determine the wave height when wind blows across a lake. The wave height, H, is assumed to be a function of the wind speed, V, the water density, r, the air density, ra, the water depth, d, the distance from the shore, /, and the acceleration of gravity, g, as shown in Fig. P7.7. Use d, V, and r as repeating variables to determine a suitable set of pi terms that could be used to describe this problem. V H d ■ Figure P7.7 7.8 GO Water flows over a dam as illustrated in Fig. P7.8. Assume the flowrate, q, per unit length along the dam depends on the head, H, width, b, acceleration of gravity, g, fluid density, r, and H q b ■ Figure P7.8 390 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling fluid viscosity, m. Develop a suitable set of dimensionless parameters for this problem using b, g, and r as repeating variables. of elasticity of the wire material. Develop a suitable set of pi terms for this problem. 7.9 The excess pressure inside a bubble (discussed in Chapter 1) is known to be dependent on bubble radius and surface tension. After finding the pi terms, determine the variation in excess pressure if we (a) double the radius and (b) double the surface tension. 7.17 The water flowrate, Q, in an open rectangular channel can be measured by placing a plate across the channel as shown in Fig. P7.17. This type of a device is called a weir. The height of the water, H, above the weir crest is referred to as the head and can be used to determine the flowrate through the channel. Assume that Q is a function of the head, H, the channel width, b, and the acceleration of gravity, g. Determine a suitable set of dimensionless variables for this problem. 7.10 The flowrate, Q, of water in an open channel is assumed to be a function of the cross-sectional area of the channel, A, the height of the roughness of the channel surface, e, the acceleration of gravity, g, and the slope, So, of the hill on which the channel sits. Put this relationship into dimensionless form. 7.11 At a sudden contraction in a pipe the diameter changes from D1 to D2. The pressure drop, ¢p, which develops across the contraction is a function of D1 and D2, as well as the velocity, V, in the larger pipe, and the fluid density, r, and viscosity, m. Use D1, V, and m as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable? 7.12 Water sloshes back and forth in a tank as shown in Fig. P7.12. The frequency of sloshing, v, is assumed to be a function of the acceleration of gravity, g, the average depth of the water, h, and the length of the tank, /. Develop a suitable set of dimensionless parameters for this problem using g and / as repeating variables. ω h b H Weir plate ■ Figure P7.17 7.18 Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface as shown in Fig. P7.18. (See Video V1.9.) The maximum thickness, h, of a square of material that can be supported is assumed to be a function of the length of the side of the square, /, the density of the material, r, the acceleration of gravity, g, and the surface tension of the liquid, s. Develop a suitable set of dimensionless parameters for this problem. ■ Figure P7.12 7.13 The drag, d, on a washer-shaped plate placed normal to a stream of fluid can be expressed as d f 1d1, d2, V, m, r2 where d1 is the outer diameter, d2 the inner diameter, V the fluid velocity, m the fluid viscosity, and r the fluid density. Some experiments are to be performed in a wind tunnel to determine the drag. What dimensionless parameters would you use to organize these data? 7.14 Assume that the flowrate, Q, of a gas from a smokestack is a function of the density of the ambient air, ra, the density of the gas, rg, within the stack, the acceleration of gravity, g, and the height and diameter of the stack, h and d, respectively. Use ra, d, and g as repeating variables to develop a set of pi terms that could be used to describe this problem. 7.15 as The pressure rise, ¢p, across a pump can be expressed ¢p f 1D, r, v, Q2 where D is the impeller diameter, r the fluid density, v the rotational speed, and Q the flowrate. Determine a suitable set of dimensionless parameters. ■ Figure P7.18 7.19 Under certain conditions, wind blowing past a rectangular speed limit sign can cause the sign to oscillate with a frequency v. (See Fig. P7.19 and Video V9.9.) Assume that v is a function of the sign width, b, sign height, h, wind velocity, V, air density, r, and an elastic constant, k, for the supporting pole. The constant, k, has dimensions of FL. Develop a suitable set of pi terms for this problem. SPEED LIMIT 40 ω 7.16 A thin elastic wire is placed between rigid supports. A fluid flows past the wire, and it is desired to study the static deflection, d, at the center of the wire due to the fluid drag. Assume that d f 1/, d, r, m, V, E2 where / is the wire length, d the wire diameter, r the fluid density, m the fluid viscosity, V the fluid velocity, and E the modulus h ■ Figure P7.19 Problems 7.20 The height, h, that a liquid will rise in a capillary tube is a function of the tube diameter, D, the specific weight of the liquid, g, and the surface tension, s. (See Video V1.10.) Perform a dimensional analysis using both the FLT and MLT systems for basic dimensions. Note: The results should obviously be the same regardless of the system of dimensions used. If your analysis indicates otherwise, go back and check your work, giving particular attention to the required number of reference dimensions. 391 Section 7.5 Determination of Pi Terms by Inspection 7.26 A viscous fluid is poured onto a horizontal plate as shown in Fig. P7.26. Assume that the time, t, required for the fluid to flow a certain distance, d, along the plate is a function of the volume of fluid poured, V , acceleration of gravity, g, fluid density, r, and fluid viscosity, m. Determine an appropriate set of pi terms to describe this process. Form the pi terms by inspection. 7.21 A cone and plate viscometer consists of a cone with a very small angle a that rotates above a flat surface as shown in Fig. P7.21. The torque, t, required to rotate the cone at an angular velocity v is a function of the radius, R, the cone angle, a, and the fluid viscosity, m, in addition to v. With the aid of dimensional analysis, determine how the torque will change if both the viscosity and angular velocity are doubled. Volume V R d ω ■ Figure P7.26 Fluid α ■ Figure P7.21 7.22 The pressure drop, ¢p, along a straight pipe of diameter D has been experimentally studied, and it is observed that for laminar flow of a given fluid and pipe, the pressure drop varies directly with the distance, /, between pressure taps. Assume that ¢p is a function of D and /, the velocity, V, and the fluid viscosity, m. Use dimensional analysis to deduce how the pressure drop varies with pipe diameter. 7.23 A cylinder with a diameter D floats upright in a liquid as shown in Fig. P7.23. When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequency, v. Assume that this frequency is a function of the diameter, D, the mass of the cylinder, m, and the specific weight, g, of the liquid. Determine, with the aid of dimensional analysis, how the frequency is related to these variables. If the mass of the cylinder were increased, would the frequency increase or decrease? 7.27 The velocity, c, at which pressure pulses travel through arteries (pulse-wave velocity) is a function of the artery diameter, D, and wall thickness, h, the density of blood, r, and the modulus of elasticity, E, of the arterial wall. Determine a set of nondimensional parameters that can be used to study experimentally the relationship between the pulse-wave velocity and the variables listed. Form the nondimensional parameters by inspection. 7.28 As shown in Fig. P7.28 and Video V5.6, a jet of liquid directed against a block can tip over the block. Assume that the velocity, V, needed to tip over the block is a function of the fluid density, r, the diameter of the jet, D, the weight of the block, w, the width of the block, b, and the distance, d, between the jet and the bottom of the block. (a) Determine a set of dimensionless parameters for this problem. Form the dimensionless parameters by inspection. (b) Use the momentum equation to determine an equation for V in terms of the other variables. (c) Compare the results of parts (a) and (b). b Cylinder diameter = D V ρ D d ■ Figure P7.28 Liquid ■ Figure P7.23 †7.24 Consider a typical situation involving the flow of a fluid that you encounter almost every day. List what you think are the important physical variables involved in this flow and determine an appropriate set of pi terms for this situation. 7.25 The speed of sound in a gas, c, is a function of the gas pressure, p, and density, r. Determine, with the aid of dimensional analysis, how the velocity is related to the pressure and density. Be careful when you decide on how many reference dimensions are required. 7.29 Assume that the drag, d, on an aircraft flying at supersonic speeds is a function of its velocity, V, fluid density, r, speed of sound, c, and a series of lengths, /1, . . . ., /i, which describe the geometry of the aircraft. Develop a set of pi terms that could be used to investigate experimentally how the drag is affected by the various factors listed. Form the pi terms by inspection. Section 7.6 Common Dimensionless Groups in Fluid Mechanics 7.30 Shown in the following table are several flow situations and the associated characteristic velocity, size, and fluid kinematic viscosity. Determine the Reynolds number for each of the 392 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling flows and indicate for which ones the inertial effects are small relative to viscous effects. Flow Velocity ft/s Size ft Kinematic viscosity ft2/s 400 0.02 0.03 60 0.015 0.12 4.3 104 1.6 104 2.5 103 0.001 5 105 1.2 105 Airplane Mosquito Syrup on pancake Blood in capillary Section 7.7 Correlation of Experimental Data (also see Lab Problems 7.1LP, 7.2LP, 7.3LP, 7.4LP) 7.32 The pressure rise, ¢p p2 p1, across the abrupt expansion of Fig. P7.32 through which a liquid is flowing can be expressed as ¢p f 1A1, A2, r, V1 2 where A1 and A2 are the upstream and downstream cross-sectional areas, respectively, r is the fluid density, and V1 is the upstream velocity. Some experimental data obtained with A2 1.25 ft2, V1 5.00 ft/s, and using water with r 1.94 slugs/ft3 are given in the following table: 0.10 3.25 0.25 7.85 0.37 10.3 0.52 11.6 V, fts ¢p, lbft2 3 11 17 20 7.31 A liquid spray nozzle is designed to produce a specific size droplet with diameter, d. The droplet size depends on the nozzle diameter, D, nozzle velocity, V, and the liquid properties r, m, s. Using the common dimensionless terms found in Table 7.1, determine the functional relationship for the dependent diameter ratio of d D. A1 (ft2) ¢p (lb/ft2) functional relationship between the pressure coefficient and the Reynolds number. (c) What are the limitations on the applicability of your equation obtained in part (b)? 0.61 12.3 Plot the results of these tests using suitable dimensionless parameters. With the aid of a standard curve fitting program determine a general equation for ¢p and use this equation to predict ¢p for water flowing through an abrupt expansion with an area ratio A1/A2 0.35 at a velocity V1 3.75 ft/s. p2 192 704 1088 1280 7.34 GO The pressure drop across a short hollowed plug placed in a circular tube through which a liquid is flowing (see Fig. P7.34) can be expressed as ¢p f 1r, V, D, d2 where r is the fluid density, and V is the mean velocity in the tube. Some experimental data obtained with D 0.2 ft, r 2.0 slugs/ft3, and V 2 ft/s are given in the following table: d 1ft2 0.06 ¢p 1lbft2 2 0.08 493.8 0.10 156.2 64.0 0.15 12.6 Plot the results of these tests, using suitable dimensionless parameters, on a log–log scale. Use a standard curve-fitting technique to determine a general equation for ¢p. What are the limits of applicability of the equation? Δp V D d ■ Figure P7.34 7.35 The pressure drop per unit length, ¢p/, for the flow of blood through a horizontal small-diameter tube is a function of the volume rate of flow, Q, the diameter, D, and the blood viscosity, m. For a series of tests in which D 2 mm and m 0.004 N s/m2, the following data were obtained, where the ¢p listed was measured over the length, / 300 mm. p1 Q (m3/s) V1 Area = A1 Area = A2 ■ Figure P7.32 7.33 The pressure drop, ¢p, over a certain length of horizontal pipe is assumed to be a function of the velocity, V, of the fluid in the pipe, the pipe diameter, D, and the fluid density and viscosity, r and m. (a) Show that this flow can be described in dimensionless form as a “pressure coefficient,” Cp ¢p/(0.5 rV2) that depends on the Reynolds number, Re rVD/m. (b) The following data were obtained in an experiment involving a fluid with r 2 slugs/ft3, m 2 103 lb s/ft2, and D 0.1 ft. Plot a dimensionless graph and use a power law equation to determine the 3.6 4.9 6.3 7.9 9.8 106 106 106 106 106 ⌬p (N/m2) 1.1 1.5 1.9 2.4 3.0 104 104 104 104 104 Perform a dimensional analysis for this problem, and make use of the data given to determine a general relationship between ¢p/ and Q (one that is valid for other values of D, /, and m). 7.36 As shown in Fig. 2.26. Fig. P7.36, and Video V2.12, a rectangular barge floats in a stable configuration provided the distance between the center of gravity, CG, of the object (boat and load) and the center of buoyancy, C, is less than a certain amount, H. If this distance is greater than H, the boat will tip over. Assume H is a function of the boat’s width, b, length, /, and draft, h. (a) Put this relationship into dimensionless form. (b) The results of a set Problems of experiments with a model barge with a width of 1.0 m are shown in the table. Plot these data in dimensionless form and determine a power-law equation relating the dimensionless parameters. /, m h, m H, m 2.0 4.0 2.0 4.0 2.0 4.0 0.10 0.10 0.20 0.20 0.35 0.35 0.833 0.833 0.417 0.417 0.238 0.238 393 Plot these data by using appropriate dimensionless variables. Could any of the original variables have been omitted? 7.40 In order to maintain uniform flight, smaller birds must beat their wings faster than larger birds. It is suggested that the relationship between the wingbeat frequency, v, beats per second, and the bird’s wingspan, /, is given by a power law relationship, v /n. (a) Use dimensional analysis with the assumption that the wingbeat frequency is a function of the wingspan, the specific weight of the bird, g, the acceleration of gravity, g, and the density of the air, ra, to determine the value of the exponent n. (b) Some typical data for various birds are given in the following table. Do these data support your result obtained in part (a)? Provide appropriate analysis to show how you arrived at your conclusion. CG Bird H Wingspan, m Wingbeat frequency, beats/s 0.28 0.36 0.46 1.00 1.50 1.80 5.3 4.3 3.2 2.2 2.6 2.0 h Purple martin Robin Mourning dove Crow Canada goose Great blue heron C b ■ Figure P7.36 7.37 When a sphere of diameter d falls slowly in a highly viscous fluid, the settling velocity, V, is known to be a function of d, the fluid viscosity, m, and the difference, ¢g, between the specific weight of the sphere and the specific weight of the fluid. Due to a tight budget situation, only one experiment can be performed, and the following data were obtained: V 0.42 ft/s for d 0.1 in., m 0.03 lb s/ft2, and ¢g 10 lb/ft3. If possible, based on this limited amount of data, determine the general equation for the settling velocity. If you do not think it is possible, indicate what additional data would be required. 7.38 The time, t, it takes to pour a certain volume of liquid from a cylindrical container depends on several factors, including the viscosity of the liquid. (See Video V1.3.) Assume that for very viscous liquids the time it takes to pour out two-thirds of the initial volume depends on the initial liquid depth, /, the cylinder diameter, D, the liquid viscosity, m, and the liquid specific weight, g. The data shown in the following table were obtained in the laboratory. For these tests / 45 mm, D 67 mm, and g 9.60 kN/m3. (a) Perform a dimensional analysis, and based on the data given, determine if variables used for this problem appear to be correct. Explain how you arrived at your answer. (b) If possible, determine an equation relating the pouring time and viscosity for the cylinder and liquids used in these tests. If it is not possible, indicate what additional information is needed. m (N s/m2) t(s) 11 15 17 23 39 53 61 83 107 145 7.39 A liquid flows with a velocity V through a hole in the side of a large tank. Assume that V f 1h, g, r, s2 where h is the depth of fluid above the hole, g is the acceleration of gravity, r the fluid density, and s the surface tension. The following data were obtained by changing h and measuring V, with a fluid having a density 103 kg/m3 and surface tension 0.074 N/m. V (m/s) h (m) 3.13 0.50 4.43 1.00 5.42 1.50 6.25 2.00 7.00 2.50 7.41 A fluid flows through the horizontal curved pipe of Fig. P7.41 with a velocity V. The pressure drop, ¢p, between the entrance and the exit to the bend is thought to be a function of the velocity, bend radius, R, pipe diameter, D, and fluid density, r. The data shown in the following table were obtained in the laboratory. For these tests r 2.0 slugs/ft3, R 0.5 ft, and D 0.1 ft. Perform a dimensional analysis and based on the data given, determine if the variables used for this problem appear to be correct. Explain how you arrived at your answer. V (ft/s) 2 ¢p (lb/ft ) 2.1 3.0 3.9 5.1 1.2 1.8 6.0 6.5 V R D ■ Figure P7.41 7.42 GO The concentric cylinder device of the type shown in Fig. P7.42 is commonly used to measure the viscosity, m, of liquids by relating the angle of twist, u, of the inner cylinder to the angular velocity, v, of the outer cylinder. Assume that u f 1v, m, K, D1, D2, /2 where K depends on the suspending wire properties and has the dimensions FL. The following data were obtained in a series of tests for which m 0.01 lb s/ft2, K 10 lb ft, / 1 ft, and D1 and D2 were constant. 394 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling U (rad) V (rad/s) 0.89 1.50 2.51 3.05 4.28 5.52 6.40 0.30 0.50 0.82 1.05 1.43 1.86 2.14 Determine from these data, with the aid of dimensional analysis, the relationship between u, v, and m for this particular apparatus. Hint: Plot the data using appropriate dimensionless parameters, and determine the equation of the resulting curve using a standard curve-fitting technique. The equation should satisfy the condition that u 0 for v 0. Fixed support Wire Liquid maintain Re and St similarity for the testing. Based on standard college quarterbacks, the prototype parameters are set at V 40 mph and v 300 rpm, where V and v are the speed and angular velocity of the football. The prototype football has a 7-in. diameter. Due to instrumentation required to measure pressure and shear stress on the surface of the football, the model will require a length scale of 2:1 (the model will be larger than the prototype). Determine the required model freestream velocity and model angular velocity. 7.48 When a fluid flows slowly past a vertical plate of height h and width b (see Fig. P7.48), pressure develops on the face of the plate. Assume that the pressure, p, at the midpoint of the plate is a function of plate height and width, the approach velocity, V, and the fluid viscosity, m. Make use of dimensional analysis to determine how the pressure, p, will change when the fluid velocity, V, is doubled. V p Rotating outer cylinder h Plate width = b θ ■ Figure P7.48 Inner cylinder D1 D2 ■ Figure P7.42 Section 7.8 Modeling and Similitude 7.43 Air at 80 F is to flow through a 2-ft pipe at an average velocity of 6 ft/s. What size pipe should be used to move water at 60 F and average velocity of 3 ft/s if Reynolds number similarity is enforced? 7.44 The design of a river model is to be based on Froude number similarity, and a river depth of 3 m is to correspond to a model depth of 100 mm. Under these conditions what is the prototype velocity corresponding to a model velocity of 2 m/s? 7.45 Glycerin at 20 C flows with a velocity of 4 m/s through a 30-mm-diameter tube. A model of this system is to be developed using standard air as the model fluid. The air velocity is to be 2 m/s. What tube diameter is required for the model if dynamic similarity is to be maintained between model and prototype? 7.46 To test the aerodynamics of a new prototype automobile, a scale model will be tested in a wind tunnel. For dynamic similarity, it will be required to match Reynolds number between model and prototype. Assuming that you will be testing a one-tenth-scale model and both model and prototype will be exposed to standard air pressure, will it be better for the wind tunnel air to be colder or hotter than standard sea-level air temperature of 15 C? Why? 7.47 You are to conduct wind tunnel testing of a new football design that has a smaller lace height than previous designs (see Videos V6.1 and V6.2). It is known that you will need to 7.49 A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while the prototype will be operated in seawater. Determine the speed of the prototype to ensure Reynolds number similarity. 7.50 SAE 30 oil at 60 F is pumped through a 3-ft-diameter pipeline at a rate of 6400 gal/min. A model of this pipeline is to be designed using a 3-in.-diameter pipe and water at 60 F as the working fluid. To maintain Reynolds number similarity between these two systems, what fluid velocity will be required in the model? 7.51 The water velocity at a certain point along a 1:10 scale model of a dam spillway is 3 m/s. What is the corresponding prototype velocity if the model and prototype operate in accordance with Froude number similarity? 7.52 The drag characteristics of a torpedo are to be studied in a water tunnel using a 1:5 scale model. The tunnel operates with fresh water at 20 C, whereas the prototype torpedo is to be used in seawater at 15.6 C. To correctly simulate the behavior of the prototype moving with a velocity of 30 m/s, what velocity is required in the water tunnel? 7.53 For a certain fluid flow problem it is known that both the Froude number and the Weber number are important dimensionless parameters. If the problem is to be studied by using a 1:15 scale model, determine the required surface tension scale if the density scale is equal to 1. The model and prototype operate in the same gravitational field. 7.54 The fluid dynamic characteristics of an airplane flying 240 mph at 10,000 ft are to be investigated with the aid of a 1:20 scale model. If the model tests are to be performed in a wind tunnel using standard air, what is the required air velocity in the wind tunnel? Is this a realistic velocity? 7.55 The pressure drop between the entrance and exit of a 150mm-diameter 90 elbow, through which ethyl alcohol at 20 C is flowing, is to be determined with a geometrically similar model. The velocity of the alcohol is 5 m/s. The model fluid is to be water at 20 C, and the model velocity is limited to 10 m/s. (a) What is the required diameter of the model elbow to maintain dynamic similarity? (b) A measured pressure drop of 20 kPa in the model will correspond to what prototype value? Problems 7.56 If an airplane travels at a speed of 1120 km/hr at an altitude of 15 km, what is the required speed at an altitude of 8 km to satisfy Mach number similarity? Assume the air properties correspond to those for the U.S. standard atmosphere. 7.57 (See Fluid in the News article “Modeling Parachutes in a Water Tunnel,” Section 7.8.1.) Flow characteristics for a 30-ftdiameter prototype parachute are to be determined by tests of a 1-ft-diameter model parachute in a water tunnel. Some data collected with the model parachute indicate a drag of 17 lb when the water velocity is 4 ft/s. Use the model data to predict the drag on the prototype parachute falling through air at 10 ft/s. Assume the drag to be a function of the velocity, V, the fluid density, r, and the parachute diameter, D. 7.58 The lift and drag developed on a hydrofoil are to be determined through wind tunnel tests using standard air. If fullscale tests are to be run, what is the required wind tunnel velocity corresponding to a hydrofoil velocity in seawater at 15 mph? Assume Reynolds number similarity is required. 7.59 A 1/50 scale model is to be used in a towing tank to study the water motion near the bottom of a shallow channel as a large barge passes over. (See Video V7.16.) Assume that the model is operated in accordance with the Froude number criteria for dynamic similitude. The prototype barge moves at a typical speed of 15 knots. (a) At what speed (in ft/s) should the model be towed? (b) Near the bottom of the model channel a small particle is found to move 0.15 ft in one second so that the fluid velocity at that point is approximately 0.15 ft/s. Determine the velocity at the corresponding point in the prototype channel. 7.60 GO When small particles of diameter d are transported by a moving fluid having a velocity V, they settle to the ground at some distance / after starting from a height h as shown in Fig. P7.60. The variation in / with various factors is to be studied with a model having a length scale of 101 . Assume that / f1h, d, V, g, m2 where g is the particle specific weight and m is the fluid viscosity. The same fluid is to be used in both the model and the prototype, but g (model) 9 g (prototype). (a) If V 50 mph, at what velocity should the model tests be run? (b) During a certain model test it was found that / (model) 0.8 ft. What would be the predicted / for this test? 395 7.62 A thin layer of an incompressible fluid flows steadily over a horizontal smooth plate as shown in Fig. P7.62. The fluid surface is open to the atmosphere, and an obstruction having a square cross section is placed on the plate as shown. A model with a length scale of 14 and a fluid density scale of 1.0 is to be designed to predict the depth of fluid, y, along the plate. Assume that inertial, gravitational, surface tension, and viscous effects are all important. What are the required viscosity and surface tension scales? Free surface y V ■ Figure P7.62 7.63 The drag on a 2-m-diameter satellite dish due to an 80km/hr wind is to be determined through a wind tunnel test using a geometrically similar 0.4-m-diameter model dish. Assume standard air for both model and prototype. (a) At what air speed should the model test be run? (b) With all similarity conditions satisfied, the measured drag on the model was determined to be 170 N. What is the predicted drag on the prototype dish? 7.64 During a storm, a snow drift is formed behind a snow fence as shown in Fig. P7.64. (See Video V9.9.) Assume that the height of the drift, h, is a function of the number of inches of snow deposited by the storm, d, height of the fence, H, width of slats in the fence, b, wind speed, V, acceleration of gravity, g, air density, r, and specific weight of snow, gs. (a) If this problem is to be studied with a model, determine the similarity requirements for the model and the relationship between the drift depth for model and prototype (prediction equation). (b) A storm with winds of 30 mph deposits 16 in. of snow having a specific weight of 5.0 lb/ft3. A 12sized scale model is to be used to investigate the effectiveness of a proposed snow fence. If the air density is the same for the model and the storm, determine the required specific weight for the model snow and required wind speed for the model. Fence b b Drift Particle V H h H V h ■ Figure P7.64 ■ Figure P7.60 7.61 A solid sphere having a diameter d and specific weight gs is immersed in a liquid having a specific weight gf (gf gs) and then released. It is desired to use a model system to determine the maximum height, h, above the liquid surface that the sphere will rise upon release from a depth H. It can be assumed that the important liquid properties are the density, gf /g, specific weight, gf, and viscosity, mf. Establish the model design conditions and the prediction equation, and determine whether the same liquid can be used in both the model and prototype systems. 7.65 A large, rigid, rectangular billboard is supported by an elastic column as shown in Fig. P7.65. There is concern about the deflection, d, of the top of the structure during a high wind of velocity V. A wind tunnel test is to be conducted with a 1:15 scale model. Assume the pertinent column variables are its length and cross-sectional dimensions, and the modulus of elasticity of the material used for the column. The only important “wind” variables are the air density and velocity. (a) Determine the model design conditions and the prediction equation for the deflection. (b) If the same structural materials are used for the model and prototype, and the wind tunnel operates under standard atmospheric conditions, what is the required wind tunnel velocity to match an 80 km/hr wind? 396 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling V δ what will be the critical velocity scale (assuming all similarity requirements are satisfied)? Billboard Vc ■ Figure P7.69 Front View Side View ■ Figure P7.65 7.66 A thin flat plate having a diameter of 0.3 ft is towed through a tank of oil (g 53 lb/ft3) at a velocity of 5 ft/s. The plane of the plate is perpendicular to the direction of motion, and the plate is submerged so that wave action is negligible. Under these conditions the drag on the plate is 1.4 lb. If viscous effects are neglected, predict the drag on a geometrically similar, 2-ftdiameter plate that is towed with a velocity of 3 ft/s through water at 60 F under conditions similar to those for the smaller plate. 7.67 Air bubbles discharge from the end of a submerged tube as shown in Fig. P7.67. The bubble diameter, D, is assumed to be a function of the air flowrate, Q, the tube diameter, d, the acceleration of gravity, g, the density of the liquid, r, and the surface tension of the liquid, s. (a) Determine a suitable set of dimensionless variables for this problem. (b) Model tests are to be run on the Earth for a prototype that is to be operated on a planet where the acceleration of gravity is 10 times greater than that on Earth. The model and prototype are to use the same fluid, and the prototype tube diameter is 0.25 in. Determine the tube diameter for the model and the required model flowrate if the prototype flowrate is to be 0.001 ft3/s. Q 7.70 The pressure rise, ¢p, across a blast wave, as shown in Fig. P7.70, and Video V11.8, is assumed to be a function of the amount of energy released in the explosion, E, the air density, r, the speed of sound, c, and the distance from the blast, d. (a) Put this relationship in dimensionless form. (b) Consider two blasts: the prototype blast with energy release E and a model blast with 1 1000th the energy release (Em 0.001 E). At what distance from the model blast will the pressure rise be the same as that at a distance of 1 mile from the prototype blast? Air (ρ , c) Δp = p2 – p1 (2) (1) d ■ Figure P7.70 7.71 The drag, d, on a sphere located in a pipe through which a fluid is flowing is to be determined experimentally (see Fig. P7.71). Assume that the drag is a function of the sphere diameter, d, the pipe diameter, D, the fluid velocity, V, and the fluid density, r. (a) What dimensionless parameters would you use for this problem? (b) Some experiments using water indicate that for d 0.2 in., D 0.5 in., and V 2 ft/s, the drag is 1.5 103 lb. If possible, estimate the drag on a sphere located in a 2-ft-diameter pipe through which water is flowing with a velocity of 6 ft/s. The sphere diameter is such that geometric similarity is maintained. If it is not possible, explain why not. V d Sphere D d D ■ Figure P7.67 ■ Figure P7.71 7.68 For a certain model study involving a 1:5 scale model it is known that Froude number similarity must be maintained. The possibility of cavitation is also to be investigated, and it is assumed that the cavitation number must be the same for model and prototype. The prototype fluid is water at 30 C, and the model fluid is water at 70 C. If the prototype operates at an ambient pressure of 101 kPa (abs), what is the required ambient pressure for the model system? 7.72 An incompressible fluid oscillates harmonically (V V0 sin vt, where V is the velocity) with a frequency of 10 rad/s in a 4-in.-diameter pipe. A 14 scale model is to be used to determine the pressure difference per unit length, ¢p/ (at any instant) along the pipe. Assume that 7.69 A thin layer of particles rests on the bottom of a horizontal tube as shown in Fig. P7.69. When an incompressible fluid flows through the tube, it is observed that at some critical velocity the particles will rise and be transported along the tube. A model is to be used to determine this critical velocity. Assume the critical velocity, Vc, to be a function of the pipe diameter, D, particle diameter, d, the fluid density, r, and viscosity, m, the density of the particles, rp, and the acceleration of gravity, g. (a) Determine the similarity requirements for the model, and the relationship between the critical velocity for model and prototype (the prediction equation). (b) For a length scale of 12 and a fluid density scale of 1.0, ¢p/ f 1D, V0, v, t, m, r2 where D is the pipe diameter, v the frequency, t the time, m the fluid viscosity, and r the fluid density. (a) Determine the similarity requirements for the model and the prediction equation for ¢p/. (b) If the same fluid is used in the model and the prototype, at what frequency should the model operate? 7.73 As shown in Fig. P7.73, a “noisemaker” B is towed behind a minesweeper A to set off enemy acoustic mines such as at C. The drag force of the noisemaker is to be studied in a water tunnel at a 14 scale model (model 1⁄4 the size of the prototype). The drag force is assumed to be a function of the speed of the ship, the density and viscosity of the fluid, and the diameter of the noisemaker. (a) If the Problems prototype towing speed is 3 m/s, determine the water velocity in the tunnel for the model tests. (b) If the model tests of part (a) produced a model drag of 900 N, determine the drag expected on the prototype. 397 Fig. P7.77. For these tests the viscosity of the water was 2.3 105 lb # sft2 and the water density was 1.94 slugsft3. Estimate the drag on an 8-ft-diameter balloon moving in air at a velocity of 3 fts. Assume the air to have a viscosity of 3.7 107 lb s/ft2 and a density of 2.38 103 slugs/ft3. Section 7.9 Some Typical Model Studies A B C 7.78 Drag measurements were taken for a sphere with a diameter of 5 cm, moving at 4 m/s in water at 20 C. The resulting drag on the sphere was 10 N. For a balloon with 1-m diameter rising in air with standard temperature and pressure, determine (a) the velocity if Reynolds number similarity is enforced and (b) the drag force if the drag coefficient (Eq. 7.19) is the dependent pi term. 7.79 The pressure rise, ¢p, across a centrifugal pump of a given shape can be expressed as ¢p f 1D, v, r, Q2 ■ Figure P7.73 7.75 The drag characteristics of an airplane are to be determined by model tests in a wind tunnel operated at an absolute pressure of 1300 kPa. If the prototype is to cruise in standard air at 385 km/hr, and the corresponding speed of the model is not to differ by more than 20% from this (so that compressibility effects may be ignored), what range of length scales may be used if Reynolds number similarity is to be maintained? Assume the viscosity of air is unaffected by pressure, and the temperature of air in the tunnel is equal to the temperature of the air in which the airplane will fly. 7.76 Wind blowing past a flag causes it to “flutter in the breeze.” The frequency of this fluttering, v, is assumed to be a function of the wind speed, V, the air density, r, the acceleration of gravity, g, the length of the flag, /, and the “area density,” rA 1with dimensions of ML22 of the flag material. It is desired to predict the flutter frequency of a large / 40 ft flag in a V 30 fts wind. To do this a model flag with / 4 ft is to be tested in a wind tunnel. (a) Determine the required area density of the model flag material if the large flag has rA 0.006 slugsft2. (b) What wind tunnel velocity is required for testing the model? (c) If the model flag flutters at 6 Hz, predict the frequency for the large flag. 7.77 The drag on a sphere moving in a fluid is known to be a function of the sphere diameter, the velocity, and the fluid viscosity and density. Laboratory tests on a 4-in.-diameter sphere were performed in a water tunnel and some model data are plotted in Model drag,lb 6 Model data ω m = 40π rad/s Dm = 8 in. 6 4 2 0 0 0.5 1.0 1.5 2.0 Qm (ft3/s) ■ Figure P7.79 7.80 A prototype automobile is designed to travel at 65 km/hr. A model of this design is tested in a wind tunnel with identical standard sea-level air properties at a 1:5 scale. The measured model drag is 400 N, enforcing dynamic similarity. Determine (a) the drag force on the prototype and (b) the power required to overcome this drag. See Eq. 7.19. 7.81 A new blimp will move at 6 m/s in 20 C air, and we want to predict the drag force. Using a 1:13-scale model in water at 20 C and measuring a 2500-N drag force on the model, determine (a) the required water velocity, (b) the drag on the prototype blimp and, (c) the power that will be required to propel it through the air. 4 2 0 8 Δ pm (psi) 7.74 The drag characteristics for a newly designed automobile having a maximum characteristic length of 20 ft are to be determined through a model study. The characteristics at both low speed (approximately 20 mph) and high speed (90 mph) are of interest. For a series of projected model tests, an unpressurized wind tunnel that will accommodate a model with a maximum characteristic length of 4 ft is to be used. Determine the range of air velocities that would be required for the wind tunnel if Reynolds number similarity is desired. Are the velocities suitable? Explain. where D is the impeller diameter, v the angular velocity of the impeller, r the fluid density, and Q the volume rate of flow through the pump. A model pump having a diameter of 8 in. is tested in the laboratory using water. When operated at an angular velocity of 40p rad/s the model pressure rise as a function of Q is shown in Fig. P7.79. Use this curve to predict the pressure rise across a geometrically similar pump (prototype) for a prototype flowrate of 6 ft3/s. The prototype has a diameter of 12 in. and operates at an angular velocity of 60p rad/s. The prototype fluid is also water. 0 2 4 6 8 Model velocity, ft/s ■ Figure P7.77 10 12 7.82 GO At a large fish hatchery the fish are reared in open, water-filled tanks. Each tank is approximately square in shape with curved corners, and the walls are smooth. To create motion in the tanks, water is supplied through a pipe at the edge of the tank. The water is drained from the tank through an opening at the center. (See Video V7.9.) A model with a length scale of 1⬊13 is to be used 398 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling to determine the velocity, V, at various locations within the tank. Assume that V f (/, /i, r, m, g, Q) where / is some characteristic length such as the tank width, /i represents a series of other pertinent lengths, such as inlet pipe diameter, fluid depth, etc., r is the fluid density, m is the fluid viscosity, g is the acceleration of gravity, and Q is the discharge through the tank. (a) Determine a suitable set of dimensionless parameters for this problem and the prediction equation for the velocity. If water is to be used for the model, can all of the similarity requirements be satisfied? Explain and support your answer with the necessary calculations. (b) If the flowrate into the fullsized tank is 250 gpm, determine the required value for the model discharge assuming Froude number similarity. What model depth will correspond to a depth of 32 in. in the full-sized tank? 7.83 Flow patterns that develop as winds blow past a vehicle, such as a train, are often studied in low-speed environmental (meteorological) wind tunnels. (See Video V7.16.) Typically, the air velocities in these tunnels are in the range of 0.1 ms to 30 m s. Consider a cross wind blowing past a train locomotive. Assume that the local wind velocity, V, is a function of the approaching wind velocity (at some distance from the locomotive), U, the locomotive length, /, height, h, and width, b, the air density, r, and the air viscosity, m. (a) Establish the similarity requirements and prediction equation for a model to be used in the wind tunnel to study the air velocity, V, around the locomotive. (b) If the model is to be used for cross winds gusting to U 25 m/s, explain why it is not practical to maintain Reynolds number similarity for a typical length scale 1:50. 7.84 (See Fluids in the News article titled “Galloping Gertie,” Section 7.8.2.) The Tacoma Narrows Bridge failure is a dramatic example of the possible serious effects of wind-induced vibrations. As a fluid flows around a body, vortices may be created that are shed periodically, creating an oscillating force on the body. If the frequency of the shedding vortices coincides with the natural frequency of the body, large displacements of the body can be induced as was the case with the Tacoma Narrows Bridge. To illustrate this type of phenomenon, consider fluid flow past a circular cylinder. (See Video V7.5.) Assume the frequency, n, of the shedding vortices behind the cylinder is a function of the cylinder diameter, D, the fluid velocity, V, and the fluid kinematic viscosity, v. (a) Determine a suitable set of dimensionless variables for this problem. One of the dimensionless variables should be the Strouhal number, nD/V. (b) Some results of experiments in which the shedding frequency of the vortices (in Hz) was measured, using a particular cylinder and Newtonian, incompressible fluid, are shown in Fig. P7.84. Is this a “universal curve” that can be used to predict the shedding frequency for any cylinder placed in any fluid? Explain. (c) A certain structural component in the form of a 1-in.-diameter, 12-ft-long rod acts as a cantilever beam 0.22 St=nD/ V 0.20 0.18 0.16 with a natural frequency of 19 Hz. Based on the data in Fig. P7.84, estimate the wind speed that may cause the rod to oscillate at its natural frequency. Hint: Use a trial-and-error solution. 7.85 (See Fluids in the News article titled “Ice Engineering,” Section 7.9.3.) A model study is to be developed to determine the force exerted on bridge piers due to floating chunks of ice in a river. The piers of interest have square cross sections. Assume that the force, R, is a function of the pier width, b, the thickness of the ice, d, the velocity of the ice, V, the acceleration of gravity, g, the density of the ice, ri, and a measure of the strength of the ice, Ei, where Ei has the dimensions FL2. (a) Based on these variables determine a suitable set of dimensionless variables for this problem. (b) The prototype conditions of interest include an ice thickness of 12 in. and an ice velocity of 6 fts. What model ice thickness and velocity would be required if the length scale is to be 110? (c) If the model and prototype ice have the same density, can the model ice have the same strength properties as that of the prototype ice? Explain. 7.86 As illustrated in Video V7.9, models are commonly used to study the dispersion of a gaseous pollutant from an exhaust stack located near a building complex. Similarity requirements for the pollutant source involve the following independent variables: the stack gas speed, V, the wind speed, U, the density of the atmospheric air, r, the difference in densities between the air and the stack gas, r rs, the acceleration of gravity, g, the kinematic viscosity of the stack gas, vs, and the stack diameter, D. (a) Based on these variables, determine a suitable set of similarity requirements for modeling the pollutant source. (b) For this type of model a typical length scale might be 1:200. If the same fluids were used in model and prototype, would the similarity requirements be satisfied? Explain and support your answer with the necessary calculations. 7.87 A 1:40 scale model of a ship is to be tested in a towing tank. (See Video V7.20.) Determine the required kinematic viscosity of the model fluid so that both the Reynolds number and the Froude number are the same for model and prototype. Assume the prototype fluid to be seawater at 60 F. Could any of the liquids with viscosities given in Fig. B.2 in Appendix B be used as the model fluid? 7.88 River models are used to study many different types of flow situations. (See, for example, Video V7.12.) A certain small river has an average width and depth of 60 ft and 4 ft, respectively, and carries water at a flowrate of 700 ft3/s. A model is to be designed based on Froude number similarity so that the discharge scale is 1/250. At what depth and flowrate would the model operate? 7.89 As winds blow past buildings, complex flow patterns can develop due to various factors such as flow separation and interactions between adjacent buildings. (See Video V7.13.) Assume that the local gage pressure, p, at a particular location on a building is a function of the air density, r, the wind speed, V, some characteristic length, /, and all other pertinent lengths, /i, needed to characterize the geometry of the building or building complex. (a) Determine a suitable set of dimensionless parameters that can be used to study the pressure distribution. (b) An eight-story building that is 100 ft tall is to be modeled in a wind tunnel. If a length scale of 1:300 is to be used, how tall should the model building be? (c) How will a measured pressure in the model be related to the corresponding prototype pressure? Assume the same air density in model and prototype. Based on the assumed variables, does the model wind speed have to be equal to the prototype wind speed? Explain. 0.14 Section 7.10 Similitude Based on Governing Differential 0.12 10 Equations 100 1,000 Re =VD/v ■ Figure P7.84 10,000 7.90 Start with the two-dimensional continuity equation and the Navier–Stokes equations (Eqs. 7.28, 7.29, and 7.30) and verify the nondimensional forms of these equations (Eqs. 7.31, 7.34, and 7.35). Problems 7.91 A viscous fluid is contained between wide, parallel plates spaced a distance h apart as shown in Fig. P7.91. The upper plate is fixed, and the bottom plate oscillates harmonically with a velocity amplitude U and frequency v. The differential equation for the velocity distribution between the plates is h 399 u y x h ■ Figure P7.94 2 r 0u 0u m 2 0t 0y fluid oscillates harmonically with a frequency v. The differential equation describing the fluid motion is where u is the velocity, t is time, and r and m are fluid density and viscosity, respectively. Rewrite this equation in a suitable nondimensional form using h, U, and v as reference parameters. ■ Lab Problems u h 0u 0 2u X cos vt m 2 0t 0y where X is the amplitude of the pressure gradient. Express this equation in nondimensional form using h and v as reference parameters. Fixed plate y r x ■ Figure P7.91 u = Ucos ω t 7.92 The deflection of the cantilever beam of Fig. P7.92 is governed by the differential equation. EI d2y dx2 P1x /2 where E is the modulus of elasticity and I is the moment of inertia of the beam cross section. The boundary conditions are y 0 at x 0 and dydx 0 at x 0. (a) Rewrite the equation and boundary conditions in dimensionless form using the beam length, /, as the reference length. (b) Based on the results of part (a), what are the similarity requirements and the prediction equation for a model to predict deflections? y P x ■ Figure P7.92 7.1LP This problem involves the time that it takes water to drain from two geometrically similar tanks. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/ college/munson. 7.2LP This problem involves determining the frequency of vortex shedding from a circular cylinder as water flows past it. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/munson. 7.3LP This problem involves the determination of the head loss for flow through a valve. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/munson. 7.4LP This problem involves the calibration of a rotameter. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/munson. ■ Lifelong Learning Problems 7.1LL Microfluidics is the study of fluid flow in fabricated devices at the micro scale. Advances in microfluidics have enhanced the ability of scientists and engineers to perform laboratory experiments using miniaturized devices known as a “lab-on-a-chip.” Obtain information about a lab-on-a-chip device that is available commercially and investigate its capabilities. Summarize your findings in a brief report. 7.93 A liquid is contained in a pipe that is closed at one end as shown in Fig. P7.93. Initially the liquid is at rest, but if the end is suddenly opened the liquid starts to move. Assume the pressure p1 remains constant. The differential equation that describes the resulting motion of the liquid is 0yz 0 2yz l 0yz p1 r ma 2 b 0t / r 0r 0r 7.2LL For some types of aerodynamic wind tunnel testing, it is difficult to simultaneously match both the Reynolds number and Mach number between model and prototype. Engineers have developed several potential solutions to the problem including pressurized wind tunnels and lowering the temperature of the flow. Obtain information about cryogenic wind tunnels and explain the advantages and disadvantages. Summarize your findings in a brief report. where vz is the velocity at any radial location, r, and t is time. Rewrite this equation in dimensionless form using the liquid density, r, the viscosity, m, and the pipe radius, R, as reference parameters. ■ FE Exam Problems p1 End initially closed vz r z R ■ Figure P7.93 7.94 An incompressible fluid is contained between two infinite parallel plates as illustrated in Fig. P7.94. Under the influence of a harmonically varying pressure gradient in the x direction, the Sample FE (Fundamental of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Computational Fluid Dynamics (CFD) The CFD problems associated with this chapter have been developed for use with the ANSYS Academic CFD software package that is associated with this text. See WileyPLUS or the book’s web site (www.wiley.com/college/munson) for additional details. 7.1CFD This CFD problem involves investigation of the Reynolds number significance in fluid dynamics through the simulation of flow past a cylinder. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/munson. There are additional CFD problems located in WileyPLUS. 8 Viscous Flow in Pipes CHAPTER OPENING PHOTO: Turbulent jet: The jet of water from the pipe is turbulent. The complex, irregular, unsteady structure typical of turbulent flows is apparent. (Laser-induced fluorescence of dye in water.) (© P. Dimotakis. Dimotakis, P.E., R.C. Miake-Lye, and D.A. Papantoniou 1983 Structure and Dynamics of Round Turbulent Jets. Phys. Fluids 26(11):3185–3192.) Learning Objectives V8.1 Turbulent jet Pipe flow is very important in our daily operations. 400 After completing this chapter, you should be able to: ■ identify and understand various characteristics of the flow in pipes. ■ discuss the main properties of laminar and turbulent pipe flow and appreciate their differences. ■ calculate losses in straight portions of pipes as well as those in various pipe system components. ■ apply appropriate equations and principles to analyze a variety of pipe flow situations. ■ predict the flowrate in a pipe by use of common flowmeters. In the previous chapters we have considered a variety of topics concerning the motion of fluids. The basic governing principles concerning mass, momentum, and energy were developed and applied, in conjunction with rather severe assumptions, to numerous flow situations. In this chapter we will apply the basic principles to a specific, important topic—the incompressible flow of viscous fluids in pipes and ducts. The transport of a fluid 1liquid or gas2 in a closed conduit 1commonly called a pipe if it is of round cross section or a duct if it is not round2 is extremely important in our daily operations. A brief consideration of the world around us will indicate that there is a wide variety of applications of pipe flow. Such applications range from the large, man-made Alaskan pipeline that carries crude oil almost 800 miles across Alaska, to the more complex 1and certainly not less useful2 natural systems of “pipes” that carry blood throughout our body and air into and out of our lungs. Other examples include the water pipes in our homes and the distribution system that delivers the water from the city well to the house. Numerous hoses and pipes carry hydraulic fluid or other fluids to various components of vehicles and machines. The air quality within our buildings is maintained at comfortable levels by the distribution of conditioned 1heated, cooled, humidified兾dehumidified2 air through a maze of pipes and ducts. Although all of these systems are different, the fluid mechanics 8.1 General Characteristics of Pipe Flow 401 Outlet Pipe Pump Elbow Tee Valve Inlet ■ Figure 8.1 Typical pipe system components. principles governing the fluid motions are common. The purpose of this chapter is to understand the basic processes involved in such flows. Some of the basic components of a typical pipe system are shown in Fig. 8.1. They include the pipes themselves 1perhaps of more than one diameter2, the various fittings used to connect the individual pipes to form the desired system, the flowrate control devices 1valves2, and the pumps or turbines that add energy to or remove energy from the fluid. Even the most simple pipe systems are actually quite complex when they are viewed in terms of rigorous analytical considerations. We will use an “exact” analysis of the simplest pipe flow topics 1such as laminar flow in long, straight, constant diameter pipes2 and dimensional analysis considerations combined with experimental results for the other pipe flow topics. Such an approach is not unusual in fluid mechanics investigations. When “real-world” effects are important 1such as viscous effects in pipe flows2, it is often difficult or “impossible” to use only theoretical methods to obtain the desired results. A judicious combination of experimental data with theoretical considerations and dimensional analysis often provides the desired results. The flow in pipes discussed in this chapter is an example of such an analysis. 8.1 General Characteristics of Pipe Flow The pipe is assumed to be completely full of the flowing fluid. Before we apply the various governing equations to pipe flow examples, we will discuss some of the basic concepts of pipe flow. With these ground rules established, we can then proceed to formulate and solve various important flow problems. Although not all conduits used to transport fluid from one location to another are round in cross section, most of the common ones are. These include typical water pipes, hydraulic hoses, and other conduits that are designed to withstand a considerable pressure difference across their walls without undue distortion of their shape. Typical conduits of noncircular cross section include heating and air conditioning ducts that are often of rectangular cross section. Normally the pressure difference between the inside and outside of these ducts is relatively small. Most of the basic principles involved are independent of the cross-sectional shape, although the details of the flow may be dependent on it. Unless otherwise specified, we will assume that the conduit is round, although we will show how to account for other shapes. For all flows involved in this chapter, we assume that the pipe is completely filled with the fluid being transported as is shown in Fig. 8.2a. Thus, we will not consider a concrete pipe through 402 Chapter 8 ■ Viscous Flow in Pipes p2 ≠ p1 (1) (1) Q p1 = p2 Q (2) (2) (a) (b) ■ Figure 8.2 (a) Pipe flow. (b) Open-channel flow. which rainwater flows without completely filling the pipe, as is shown in Fig. 8.2b. Such flows, called open-channel flow, are treated in Chapter 10. The difference between open-channel flow and the pipe flow of this chapter is in the fundamental mechanism that drives the flow. For openchannel flow, gravity alone is the driving force—the water flows down a hill. For pipe flow, gravity may be important 1the pipe need not be horizontal2, but the main driving force is likely to be a pressure gradient along the pipe. If the pipe is not full, it is not possible to maintain this pressure difference, p1 p2. 8.1.1 Laminar or Turbulent Flow V8.2 Laminar/ turbulent pipe flow A flow may be laminar, transitional, or turbulent. The flow of a fluid in a pipe may be laminar flow or it may be turbulent flow. Osborne Reynolds 11842 –19122, a British scientist and mathematician, was the first to distinguish the difference between these two classifications of flow by using a simple apparatus as shown by the figure in the margin, which is a sketch of Reynolds’ dye experiment. Reynolds injected dye into a pipe in which water flowed due to gravity. The entrance region of the pipe is depicted in Fig. 8.3a. If water runs through a pipe of diameter D with an average velocity V, the following characteristics are observed by injecting neutrally buoyant dye as shown. For “small enough flowrates” the dye streak 1a streakline2 will remain as a well-defined line as it flows along, with only slight blurring due to molecular diffusion of the dye into the surrounding water. For a somewhat larger “intermediate flowrate” the dye streak fluctuates in time and space, and intermittent bursts of irregular behavior appear along the streak. On the other hand, for “large enough flowrates” the dye streak almost immediately becomes blurred and spreads across the entire pipe in a random fashion. These three characteristics, denoted as laminar, transitional, and turbulent flow, respectively, are illustrated in Fig. 8.3b. The curves shown in Fig. 8.4 represent the x component of the velocity as a function of time at a point A in the flow. The random fluctuations of the turbulent flow 1with the associated particle mixing2 are what disperse the dye throughout the pipe and cause the blurred appearance illustrated in Fig. 8.3b. For laminar flow in a pipe there is only one component of velocity, Turbulent Dye Pipe D Q = VA Dye streak Transitional Smooth, well-rounded entrance Laminar (a) ■ Figure 8.3 (a) Experiment to illustrate type of flow. (b) Typical dye streaks. (b) 8.1 General Characteristics of Pipe Flow 403 uA Turbulent Q A x Transitional Laminar t ■ Figure 8.4 Time dependence of fluid velocity at a point. V uiˆ. For turbulent flow the predominant component of velocity is also along the pipe, but it is unsteady 1random2 and accompanied by random components normal to the pipe axis, V uiˆ vjˆ wkˆ . Such motion in a typical flow occurs too fast for our eyes to follow. Slowmotion pictures of the flow can more clearly reveal the irregular, random, turbulent nature of the flow. As was discussed in Chapter 7, we should not label dimensional quantities as being “large” or “small,” such as “small enough flowrates” in the preceding paragraphs. Rather, the appropriate dimensionless quantity should be identified and the “small” or “large” character attached to it. A quantity is “large” or “small” only relative to a reference quantity. The ratio of those quantities results in a dimensionless quantity. For pipe flow the most important dimensionless parameter is the Reynolds number, Re—the ratio of the inertia to viscous effects in the flow. Hence, in the previous paragraph the term flowrate should be replaced by Reynolds number, Re rVDm, where V is the average velocity in the pipe. That is, the flow in a pipe is laminar, transitional, or turbulent provided the Reynolds number is “small enough,” “intermediate,” or “large enough.” It is not only the fluid velocity that determines the character of the flow—its density, viscosity, and the pipe size are of equal importance. These parameters combine to produce the Reynolds number. The distinction between laminar and turbulent pipe flow and its dependence on an appropriate dimensionless quantity was first pointed out by Osborne Reynolds in 1883. The Reynolds number ranges for which laminar, transitional, or turbulent pipe flows are obtained cannot be precisely given. The actual transition from laminar to turbulent flow may take place at various Reynolds numbers, depending on how much the flow is disturbed by vibrations of the pipe, roughness of the entrance region, and the like. For general engineering purposes 1i.e., without undue precautions to eliminate such disturbances2, the following values are appropriate: The flow in a round pipe is laminar if the Reynolds number is less than approximately 2100. The flow in a round pipe is turbulent if the Reynolds number is greater than approximately 4000. For Reynolds numbers between these two limits, the flow may switch between laminar and turbulent conditions in an apparently random fashion 1transitional flow2. Pipe flow characteristics are dependent on the value of the Reynolds number. V8.3 Intermittent turbulent burst in pipe flow F l u i d s i n Nanoscale flows The term nanoscale generally refers to objects with characteristic lengths from atomic dimensions up to a few hundred nanometers (nm). (Recall that 1 nm 109 m.) Nanoscale fluid mechanics research has recently uncovered many surprising and useful phenomena. No doubt many more remain to be discovered. For example, in the future researchers envision using nanoscale tubes to push tiny amounts of water-soluble drugs to exactly where they are needed in the human body. Because of the tiny diameters involved, the Reynolds numbers for such flows are extremely small and the flow is definitely laminar. In addition, some t h e N e w s standard properties of everyday flows (for example, the fact that a fluid sticks to a solid boundary) may not be valid for nanoscale flows. Also, ultratiny mechanical pumps and valves are difficult to manufacture and may become clogged by tiny particles such as biological molecules. As a possible solution to such problems, researchers have investigated the possibility of using a system that does not rely on mechanical parts. It involves using light-sensitive molecules attached to the surface of the tubes. By shining light onto the molecules, the light-responsive molecules attract water and cause motion of water through the tube. (See Problem 8.9.) 404 Chapter 8 ■ Viscous Flow in Pipes E XAMPLE 8.1 Laminar or Turbulent Flow GIVEN Water at a temperature of 50 °F flows through a pipe of diameter D 0.73 in. and into a glass as shown in Fig. E8.1a. V FIND Determine (a) the minimum time taken to fill a 12-oz glass (volume 0.0125 ft3) with water if the flow in the pipe is to be laminar. Repeat the calculations if the water temperature is 140 °F. D Q ␮, ␳ (b) the maximum time taken to fill the glass if the flow is to be turbulent. Repeat the calculations if the water temperature is 140 °F. SOLUTION (a) If the flow in the pipe is to remain laminar, the minimum time to fill the glass will occur if the Reynolds number is the maximum allowed for laminar flow, typically Re rVDm 2100. Thus, V 2100 mrD, where from Table B.1, r 1.94 slugsft3 and m 2.73 105 lb # sft2 at 50 °F, while r 1.91 slugsft3 and m 0.974 105 lb # sft2 at 140 °F. Thus, the maximum average velocity for laminar flow in the pipe is 210012.73 105 lb # sft2 2 2100m rD 11.94 slugsft3 210.7312 ft2 0.486 lb # sslug 0.486 fts V ■ Figure E8.1a Similarly, V 0.176 fts at 140 °F. With V volume of glass and V Qt we obtain t 410.0125 ft3 2 V V Q 1p42D2V 1p 30.7312 4 2ft2 210.486 ft s2 8.85 s at T 50 °F (Ans) Similarly, t 24.4 s at 140 °F. To maintain laminar flow, the less viscous hot water requires a lower flowrate than the cold water. (b) If the flow in the pipe is to be turbulent, the maximum time to fill the glass will occur if the Reynolds number is the minimum allowed for turbulent flow, Re 4000. Thus, V 4000m rD 0.925 fts and t 4.65 s at 50 °F If the flowing fluid had been honey with a kinematic viscosity (␯ ␮/␳) 3000 times greater than that of water, the velocities given earlier would be increased by a factor of 3000 and the times reduced by the same factor. As shown in the following sections, the pressure needed to force a very viscous fluid through a pipe at such a high velocity may be unreasonably large. (Ans) Similarly, V 0.335 fts and t 12.8 s at 140 °F. 40 laminar flow COMMENTS Note that because water is “not very viscous,” 30 t, s the velocity must be “fairly small” to maintain laminar flow. In general, turbulent flows are encountered more often than laminar flows because of the relatively small viscosity of most common fluids (water, gasoline, air). By repeating the calculations at various water temperatures, T (i.e., with different densities and viscosities), the results shown in Fig. E8.1b are obtained. As the water temperature increases, the kinematic viscosity, ␯ ␮/␳, decreases and the corresponding times to fill the glass increase as indicated. (Temperature effects on the viscosity of gases are the opposite; increase in temperature causes an increase in viscosity.) 20 turbulent flow (50°F, 8.85 s) 10 0 (50°F, 4.65 s) 0 50 100 150 T, °F ■ Figure E8.1b 200 250 8.1 General Characteristics of Pipe Flow 405 8.1.2 Entrance Region and Fully Developed Flow Any fluid flowing in a pipe had to enter the pipe at some location. The region of flow near where the fluid enters the pipe is termed the entrance region and is illustrated in Fig. 8.5. It may be the first few feet of a pipe connected to a tank or the initial portion of a long run of a hot air duct coming from a furnace. As is shown in Fig. 8.5, the fluid typically enters the pipe with a nearly uniform velocity profile at section 112. As the fluid moves through the pipe, viscous effects cause it to stick to the pipe wall 1the no-slip boundary condition2. This is true whether the fluid is relatively inviscid air or a very viscous oil. Thus, a boundary layer in which viscous effects are important is produced along the pipe wall such that the initial velocity profile changes with distance along the pipe, x, until the fluid reaches the end of the entrance length, section 122, beyond which the velocity profile does not vary with x. The boundary layer has grown in thickness to completely fill the pipe. Viscous effects are of considerable importance within the boundary layer. For fluid outside the boundary layer [within the inviscid core surrounding the centerline from 112 to 122], viscous effects are negligible. The shape of the velocity profile in the pipe depends on whether the flow is laminar or turbulent, as does the length of the entrance region, /e. As with many other properties of pipe flow, the dimensionless entrance length, /e D, correlates quite well with the Reynolds number. Typical entrance lengths are given by The entrance length is a function of the Reynolds number. 100 e D 10 1 0.1 100 102 104 Re 106 /e 0.06 Re for laminar flow D (8.1) /e 4.4 1Re2 16 for turbulent flow D (8.2) and For very low Reynolds number flows the entrance length can be quite short 1/e 0.6D if Re 102, whereas for large Reynolds number flows it may take a length equal to many pipe diameters before the end of the entrance region is reached 1/e 120D for Re 20002. For many practical engineering problems, 104 6 Re 6 105 so that as shown by the figure in the margin, 20D 6 /e 6 30D. Calculation of the velocity profile and pressure distribution within the entrance region is quite complex. However, once the fluid reaches the end of the entrance region, section 122 of Fig. 8.5, the flow is simpler to describe because the velocity is a function of only the distance from the pipe centerline, r, and independent of x. This is true until the character of the pipe changes in some way, such as a change in diameter, or the fluid flows through a bend, valve, or some other component at section 132. The flow between 122 and 132 is termed fully developed flow. Beyond the interruption of the fully developed flow [at section 142], the flow gradually begins its Fully developed flow Entrance region flow D Boundary layer Inviscid core r (1) x (2) (3) e (6) (5) x6 – x5 Fully developed flow (4) x5 – x4 Developing flow ■ Figure 8.5 Entrance region, developing flow, and fully developed flow in a pipe system. 406 Chapter 8 ■ Viscous Flow in Pipes return to its fully developed character [section 152] and continues with this profile until the next pipe system component is reached [section 162]. In many cases the pipe is long enough so that there is a considerable length of fully developed flow compared with the developing flow length 3 1x3 x2 2 /e and 1x6 x5 2 1x5 x4 2 4 . In other cases the distances between one component 1bend, tee, valve, etc.2 of the pipe system and the next component is so short that fully developed flow is never achieved. 8.1.3 Pressure and Shear Stress Laminar flow characteristics are different than those for turbulent flow. Fully developed steady flow in a constant diameter pipe may be driven by gravity and兾or pressure forces. For horizontal pipe flow, gravity has no effect except for a hydrostatic pressure variation across the pipe, gD, that is usually negligible. It is the pressure difference, ¢p p1 p2, between one section of the horizontal pipe and another which forces the fluid through the pipe. Viscous effects provide the restraining force that exactly balances the pressure force, thereby allowing the fluid to flow through the pipe with no acceleration. If viscous effects were absent in such flows, the pressure would be constant throughout the pipe, except for the hydrostatic variation. In non-fully developed flow regions, such as the entrance region of a pipe, the fluid accelerates or decelerates as it flows 1the velocity profile changes from a uniform profile at the entrance of the pipe to its fully developed profile at the end of the entrance region2. Thus, in the entrance region there is a balance between pressure, viscous, and inertia 1acceleration2 forces. The result is a pressure distribution along the horizontal pipe as shown in Fig. 8.6. The magnitude of the pressure gradient, 0p 0x, is larger in the entrance region than in the fully developed region, where it is a constant, 0p 0x ¢p/ 6 0. The fact that there is a nonzero pressure gradient along the horizontal pipe is a result of viscous effects. As is discussed in Chapter 3, if the viscosity were zero, the pressure would not vary with x. The need for the pressure drop can be viewed from two different standpoints. In terms of a force balance, the pressure force is needed to overcome the viscous forces generated. In terms of an energy balance, the work done by the pressure force is needed to overcome the viscous dissipation of energy throughout the fluid. If the pipe is not horizontal, the pressure gradient along it is due in part to the component of weight in that direction. As is discussed in Section 8.2.1, this contribution due to the weight either enhances or retards the flow, depending on whether the flow is downhill or uphill. The nature of the pipe flow is strongly dependent on whether the flow is laminar or turbulent. This is a direct consequence of the differences in the nature of the shear stress in laminar and turbulent flows. As is discussed in some detail in Section 8.3.3, the shear stress in laminar flow is a direct result of momentum transfer among the randomly moving molecules 1a microscopic phenomenon2. The shear stress in turbulent flow is largely a result of momentum transfer among the randomly moving, finite-sized fluid particles 1a macroscopic phenomenon2. The net result is that the physical properties of the shear stress are quite different for laminar flow than for turbulent flow. p Fully developed flow: ∂ p/∂ x = constant Entrance flow Entrance pressure drop Δp x3 – x2 = x1 = 0 x2 = e ■ Figure 8.6 Pressure distribution along a horizontal pipe. x3 x 8.2 8.2 Fully Developed Laminar Flow 407 Fully Developed Laminar Flow As is indicated in the previous section, the flow in long, straight, constant diameter sections of a pipe becomes fully developed. That is, the velocity profile is the same at any cross section of the pipe. Although this is true whether the flow is laminar or turbulent, the details of the velocity profile 1and other flow properties2 are quite different for these two types of flow. As will be seen in the remainder of this chapter, knowledge of the velocity profile can lead directly to other useful information such as pressure drop, head loss, flowrate, and the like. Thus, we begin by developing the equation for the velocity profile in fully developed laminar flow. If the flow is not fully developed, a theoretical analysis becomes much more complex and is outside the scope of this text. If the flow is turbulent, a rigorous theoretical analysis is as yet not possible. Although most flows are turbulent rather than laminar, and many pipes are not long enough to allow the attainment of fully developed flow, a theoretical treatment and full understanding of fully developed laminar flow is of considerable importance. First, it represents one of the few theoretical viscous analyses that can be carried out “exactly” 1within the framework of quite general assumptions2 without using other ad hoc assumptions or approximations. An understanding of the method of analysis and the results obtained provides a foundation from which to carry out more complicated analyses. Second, there are many practical situations involving the use of fully developed laminar pipe flow. There are numerous ways to derive important results pertaining to fully developed laminar flow. Three alternatives include: 112 from F ma applied directly to a fluid element, 122 from the Navier –Stokes equations of motion, and 132 from dimensional analysis methods. 8.2.1 From F ⴝ ma Applied Directly to a Fluid Element Steady, fully developed pipe flow experiences no acceleration. Velocity profiles Streamlines We consider the fluid element at time t as is shown in Fig. 8.7. It is a circular cylinder of fluid of length / and radius r centered on the axis of a horizontal pipe of diameter D. Because the velocity is not uniform across the pipe, the initially flat ends of the cylinder of fluid at time t become distorted at time t dt when the fluid element has moved to its new location along the pipe as shown in the figure. If the flow is fully developed and steady, the distortion on each end of the fluid element is the same, and no part of the fluid experiences any acceleration as it flows, as shown by the figure in the margin. The local acceleration is zero 10V 0t 02 because the flow is steady, and the convective acceleration is zero 1V ⴢ ⵱ V u 0u 0x ˆi 02 because the flow is fully developed. Thus, every part of the fluid merely flows along its streamline parallel to the pipe walls with constant velocity, although neighboring particles have slightly different velocities. The velocity varies from one pathline to the next. This velocity variation, combined with the fluid viscosity, produces the shear stress. If gravitational effects are neglected, the pressure is constant across any vertical cross section of the pipe, although it varies along the pipe from one section to the next. Thus, if the pressure is p p1 at section 112, it is p2 p1 ¢p at section 122 where ¢p is the pressure drop between sections (1) and (2). We anticipate the fact that the pressure decreases in the direction of flow so that ¢p 7 0. A shear stress, t, acts on the surface of the cylinder of fluid. This viscous stress is a function of the radius of the cylinder, t t1r2. As was done in fluid statics analysis 1Chapter 22, we isolate the cylinder of fluid as is shown in Fig. 8.8 and apply Newton’s second law, Fx max. In this case, even though the fluid is moving, it is not accelerating, so that ax 0. Thus, fully developed horizontal pipe flow is merely a Element at time t + δ t Fluid element at time t Velocity profile r D ^ V = u(r)i (1) x ■ Figure 8.7 Motion of a cylindri(2) cal fluid element within a pipe. 408 Chapter 8 ■ Viscous Flow in Pipes τ 2 π r r p1 π r2 (p1 – Δp) π r 2 x ■ Figure 8.8 Free-body diagram of a cylinder of fluid. balance between pressure and viscous forces—the pressure difference acting on the end of the cylinder of area pr 2, and the shear stress acting on the lateral surface of the cylinder of area 2pr/. This force balance can be written as 1 p1 2pr 2 1 p1 ¢p2pr 2 1t22pr/ 0 which can be simplified to give ¢p 2t r / (8.3) Equation 8.3 represents the basic balance in forces needed to drive each fluid particle along the pipe with constant velocity. Since neither ¢p nor / are functions of the radial coordinate, r, it follows that 2tr must also be independent of r. That is, t Cr, where C is a constant. At r 0 1the centerline of the pipe2 there is no shear stress 1t 02. At r D2 1the pipe wall2 the shear stress is a maximum, denoted tw, the wall shear stress. Hence, C 2tw D and the shear stress distribution throughout the pipe is a linear function of the radial coordinate t 2twr D (8.4) as is indicated in Fig. 8.9. The linear dependence of t on r is a result of the pressure force being proportional to r 2 1the pressure acts on the end of the fluid cylinder; area pr 22 and the shear force being proportional to r 1the shear stress acts on the lateral sides of the cylinder; area 2pr/2. If the viscosity were zero there would be no shear stress, and the pressure would be constant throughout the horizontal pipe 1¢p 02. As is seen from Eqs. 8.3 and 8.4, the pressure drop and wall shear stress are related by Basic horizontal pipe flow is governed by a balance between viscous and pressure forces. ¢p 4/tw D (8.5) A small shear stress can produce a large pressure difference if the pipe is relatively long 1/D 12. Although we are discussing laminar flow, a closer consideration of the assumptions involved in the derivation of Eqs. 8.3, 8.4, and 8.5 reveals that these equations are valid for both laminar and turbulent flow. To carry the analysis further we must prescribe how the shear stress is related to the velocity. This is the critical step that separates the analysis of laminar from that of turbulent flow—from being able to solve for the laminar flow properties and not being able to solve for the turbulent flow properties without additional ad hoc assumptions. As is discussed in Section 8.3, the shear stress dependence for turbulent flow is very complex. However, for laminar flow of a Ideal (inviscid) profile Laminar profile τ (D/2) = τ w τ (r) u(r) r x τ (0) = 0 ■ Figure 8.9 Shear stress τw Vc V = Vc /2 distribution within the fluid in a pipe (laminar or turbulent flow) and typical velocity profiles. 8.2 Fully Developed Laminar Flow 409 Newtonian fluid, the shear stress is simply proportional to the velocity gradient, “t m du dy” 1see Section 1.62. In the notation associated with our pipe flow, this becomes t m du dr (8.6) The negative sign is included to give t 7 0 with du dr 6 0 1the velocity decreases from the pipe centerline to the pipe wall2. Equations 8.3 and 8.6 represent the two governing laws for fully developed laminar flow of a Newtonian fluid within a horizontal pipe. The one is Newton’s second law of motion, and the other is the definition of a Newtonian fluid. By combining these two equations, we obtain ¢p du a br dr 2m/ which can be integrated to give the velocity profile as follows: 冮 du 2m/ 冮 r dr ¢p or u a ¢p b r 2 C1 4m/ where C1 is a constant. Because the fluid is viscous it sticks to the pipe wall so that u 0 at r D2. Thus, C1 1¢p16m/2D2. Hence, the velocity profile can be written as u1r2 a Under certain restrictions the velocity profile in a pipe is parabolic. R ¢pD2 2r 2 2r 2 b c 1 a b d Vc c 1 a b d 16m/ D D (8.7) where Vc ¢pD2 116m/2 is the centerline velocity. An alternative expression can be written by using the relationship between the wall shear stress and the pressure gradient 1Eqs. 8.5 and 8.72 to give twD r 2 c1 a b d 4m R u1r2 where R D2 is the pipe radius. This velocity profile, plotted in Fig. 8.9, is parabolic in the radial coordinate, r, has a maximum velocity, Vc , at the pipe centerline, and a minimum velocity 1zero2 at the pipe wall. The volume flowrate through the pipe can be obtained by integrating the velocity profile across the pipe. Since the flow is axisymmetric about the centerline, the velocity is constant on small area elements consisting of rings of radius r and thickness dr, as shown in the figure in the margin. Thus, r Q dr 冮 u dA 冮 rR u1r22pr dr 2p Vc r0 冮 0 R r 2 c 1 a b d r dr R or dA = 2π r dr Q pR2Vc 2 By definition, the average velocity is the flowrate divided by the cross-sectional area, V QA QpR2, so that for this flow V pR2Vc 2pR2 Vc ¢pD2 2 32m/ (8.8) and Q pD4 ¢p 128m/ (8.9) 410 Chapter 8 ■ Viscous Flow in Pipes Poiseuille’s law is valid for laminar flow only. 16Q0 Q ~ D4 Q Q0 D0 D 2D0 As is indicated in Eq. 8.8, the average velocity is one-half of the maximum velocity. In general, for velocity profiles of other shapes 1such as for turbulent pipe flow2, the average velocity is not merely the average of the maximum 1Vc 2 and minimum 102 velocities as it is for the laminar parabolic profile. The two velocity profiles indicated in Fig. 8.9 provide the same flowrate—one is the fictitious ideal 1m 02 profile; the other is the actual laminar flow profile. The above results confirm the following properties of laminar pipe flow. For a horizontal pipe the flowrate is 1a2 directly proportional to the pressure drop, 1b2 inversely proportional to the viscosity, 1c2 inversely proportional to the pipe length, and 1d2 proportional to the pipe diameter to the fourth power. With all other parameters fixed, an increase in diameter by a factor of 2 will increase the flowrate by a factor of 24 16—the flowrate is very strongly dependent on pipe size. This dependence is shown by the figure in the margin. Likewise, a small error in pipe diameter can cause a relatively large error in flowrate. For example, a 2% error in diameter gives an 8% error in flowrate 1Q ⬃ D4 or dQ ⬃ 4D3 dD, so that dQQ 4 dDD2. This flow, the properties of which were first established experimentally by two independent workers, G. Hagen 11797–18842 in 1839 and J. Poiseuille 11799–18692 in 1840, is termed Hagen – Poiseuille flow. Equation 8.9 is commonly referred to as Poiseuille’s law. Recall that all of these results are restricted to laminar flow 1those with Reynolds numbers less than approximately 21002 in a horizontal pipe. The adjustment necessary to account for nonhorizontal pipes, as shown in Fig. 8.10, can be easily included by replacing the pressure drop, ¢p, by the combined effect of pressure and gravity, ¢p g/ sin u, where u is the angle between the pipe and the horizontal. 1Note that u 7 0 if the flow is uphill, while u 6 0 if the flow is downhill.2 This can be seen from the force balance in the x direction 1along the pipe axis2 on the cylinder of fluid shown in Fig. 8.10b. The method is exactly analogous to that used to obtain the Bernoulli equation 1Eq. 3.62 when the streamline is not horizontal. The net force in the x direction is a combination of the pressure force in that direction, ¢ppr 2, and the component of weight in that direction, gpr 2/ sin u. The result is a slightly modified form of Eq. 8.3 given by ¢p g/ sin u 2t r / (8.10) Thus, all of the results for the horizontal pipe are valid provided the pressure gradient is adjusted for the elevation term, that is, ¢p is replaced by ¢p g/ sin u so that V 1¢p g/ sin u2D2 (8.11) 32m/ and Q p1¢p g/ sin u2D4 (8.12) 128m/ It is seen that the driving force for pipe flow can be either a pressure drop in the flow direction, ¢p, or the component of weight in the flow direction, g/ sin u. If the flow is downhill, gravity helps the flow 1a smaller pressure drop is required; sin u 6 02. If the flow is uphill, gravity works against the flow 1a larger pressure drop is required; sin u 7 02. Note that g/ sin u g¢z 1where Q pπ r 2 τ 2π r Fluid cylinder θ r (p + Δ p) π r 2 θ sin θ = γ π r2 sin θ (a) (b) ■ Figure 8.10 Free-body diagram of a fluid cylinder for flow in a nonhorizontal pipe. x 8.2 Fully Developed Laminar Flow 411 ¢z is the change in elevation2 is a hydrostatic type pressure term. If there is no flow, V 0 and ¢p g/ sin u g¢z, as expected for fluid statics. E XAMPLE 8.2 Laminar Pipe Flow GIVEN An oil with a viscosity of m 0.40 N # sm2 and density r 900 kgm3 flows in a pipe of diameter D 0.020 m. FIND (a) What pressure drop, p1 p2, is needed to produce a flowrate of Q 2.0 105 m3s if the pipe is horizontal with x1 0 and x2 10 m? SOLUTION (a) If the Reynolds number is less than 2100 the flow is laminar and the equations derived in this section are valid. Since the average velocity is V QA 12.0 105 m3 s2 3 p10.0202 2m2 4 4 0.0637 ms, the Reynolds number is Re rVDm 2.87 6 2100. Hence, the flow is laminar and from Eq. 8.9 with / x2 x1 10 m, the pressure drop is ¢p p1 p2 128m/Q pD4 12810.40 N # sm2 2 110.0 m212.0 105 m3s2 p10.020 m2 4 or ¢p 20,400 Nm 20.4 kPa 2 (Ans) (b) If the pipe is on a hill of angle u such that ¢p p1 p2 0, Eq. 8.12 gives sin u 128mQ (1) prgD4 12810.40 N # sm2 212.0 105 m3s2 p1900 kgm 219.81 ms 210.020 m2 3 (c) For the conditions of part 1b2, if p1 200 kPa, what is the pressure at section x3 5 m, where x is measured along the pipe? 12.31 m2 20,400 N m2, which is equivalent to that needed for the horizontal pipe. For the horizontal pipe it is the work done by the pressure forces that overcomes the viscous dissipation. For the zero-pressure-drop pipe on the hill, it is the change in potential energy of the fluid “falling” down the hill that is converted to the energy lost by viscous dissipation. Note that if it is desired to increase the flowrate to Q 1.0 104 m3s with p1 p2, the value of u given by Eq. 1 is sin u 1.15. Since the sine of an angle cannot be greater than 1, this flow would not be possible. The weight of the fluid would not be large enough to offset the viscous force generated for the flowrate desired. A larger-diameter pipe would be needed. (c) With p1 p2 the length of the pipe, /, does not appear in the flowrate equation 1Eq. 12. This is a statement of the fact that for such cases the pressure is constant all along the pipe 1provided the pipe lies on a hill of constant slope2. This can be seen by substituting the values of Q and u from case 1b2 into Eq. 8.12 and noting that ¢p 0 for any /. For example, ¢p p1 p3 0 if / x3 x1 5 m. Thus, p1 p2 p3 so that p3 200 kPa or sin u (b) How steep a hill, u, must the pipe be on if the oil is to flow through the pipe at the same rate as in part 1a2, but with p1 p2? 2 4 (Ans) Thus, u 13.34°. COMMENT This checks with the previous horizontal result as is seen from the fact that a change in elevation of ¢z / sin u 110 m2 sin113.34°2 2.31 m is equivalent to a pressure change of ¢p rg ¢z 1900 kgm3 219.81 ms2 2 (Ans) COMMENT Note that if the fluid were gasoline 1m 3.1 104 N # sm2 and r 680 kg m3 2, the Reynolds number would be Re 2790, the flow would probably not be laminar, and use of Eqs. 8.9 and 8.12 would give incorrect results. Also note from Eq. 1 that the kinematic viscosity, n mr, is the important viscous parameter. This is a statement of the fact that with constant pressure along the pipe, it is the ratio of the viscous force 1⬃m2 to the weight force 1⬃g rg2 that determines the value of u. 8.2.2 From the Navier–Stokes Equations Poiseuille’s law can be obtained from the Navier–Stokes equations. In the previous section we obtained results for fully developed laminar pipe flow by applying Newton’s second law and the assumption of a Newtonian fluid to a specific portion of the fluid— a cylinder of fluid centered on the axis of a long, round pipe. When this governing law and assumptions are applied to a general fluid flow 1not restricted to pipe flow2, the result is the Navier –Stokes equations as discussed in Chapter 6. In Section 6.9.3 these equations were solved for the specific geometry of fully developed laminar flow in a round pipe. The results are the same as those given in Eq. 8.7. 412 Chapter 8 ■ Viscous Flow in Pipes We will not repeat the detailed steps used to obtain the laminar pipe flow from the Navier – Stokes equations 1see Section 6.9.32 but will indicate how the various assumptions used and steps applied in the derivation correlate with the analysis used in the previous section. General motion of an incompressible Newtonian fluid is governed by the continuity equation 1conservation of mass, Eq. 6.312 and the momentum equation 1Eq. 6.1272, which are rewritten here for convenience: ⵱ⴢV0 ⵱p 0V g n⵱2V 1V ⴢ ⵱2V r 0t (8.13) (8.14) For steady, fully developed flow in a pipe, the velocity contains only an axial component, which is a function of only the radial coordinate 3V u1r2iˆ 4. For such conditions, the left-hand side of the Eq. 8.14 is zero. This is equivalent to saying that the fluid experiences no acceleration as it flows along. The same constraint was used in the previous section when considering F ma for the fluid cylinder. Thus, with g gkˆ the Navier –Stokes equations become ⵱ⴢV0 ⵱p rgkˆ m⵱2 V (8.15) The flow is governed by a balance of pressure, weight, and viscous forces in the flow direction, similar to that shown in Fig. 8.10 and Eq. 8.10. If the flow were not fully developed 1as in an entrance region, for example2, it would not be possible to simplify the Navier –Stokes equations to that form given in Eq. 8.15 1the nonlinear term 1V ⴢ ⵱2V would not be zero2, and the solution would be very difficult to obtain. Because of the assumption that V u1r2iˆ, the continuity equation, Eq. 8.13, is automatically satisfied. This conservation of mass condition was also automatically satisfied by the incompressible flow assumption in the derivation in the previous section. The fluid flows across one section of the pipe at the same rate that it flows across any other section 1see Fig. 8.82. When it is written in terms of polar coordinates 1as was done in Section 6.9.32, the component of Eq. 8.15 along the pipe becomes The governing differential equations can be simplified by appropriate assumptions. 0p 1 0 0u rg sin u m ar b r 0r 0x 0r (8.16) Since the flow is fully developed, u u1r2 and the right-hand side is a function of, at most, only r. The left-hand side is a function of, at most, only x. It was shown that this leads to the condition that the pressure gradient in the x direction is a constant— 0p 0x ¢p/. The same condition was used in the derivation of the previous section 1Eq. 8.32. It is seen from Eq. 8.16 that the effect of a nonhorizontal pipe enters into the Navier –Stokes equations in the same manner as was discussed in the previous section. The pressure gradient in the flow direction is coupled with the effect of the weight in that direction to produce an effective pressure gradient of ¢p/ rg sin u. The velocity profile is obtained by integration of Eq. 8.16. Since it is a second-order equation, two boundary conditions are needed—112 the fluid sticks to the pipe wall 1as was also done in Eq. 8.72 and 122 either of the equivalent forms that the velocity remains finite throughout the flow 1in particular u 6 q at r 02 or, because of symmetry, that 0u 0r 0 at r 0. In the derivation of the previous section, only one boundary condition 1the no-slip condition at the wall2 was needed because the equation integrated was a first-order equation. The other condition 10u 0r 0 at r 02 was automatically built into the analysis because of the fact that t m dudr and t 2twrD 0 at r 0. The results obtained by either applying F ma to a fluid cylinder 1Section 8.2.12 or solving the Navier –Stokes equations 1Section 6.9.32 are exactly the same. Similarly, the basic assumptions regarding the flow structure are the same. This should not be surprising because the two methods are based on the same principle—Newton’s second law. One is restricted to fully developed laminar pipe flow from the beginning 1the drawing of the free-body diagram2, and the other starts with the general governing equations 1the Navier –Stokes equations2 with the appropriate restrictions concerning fully developed laminar flow applied as the solution process progresses. 8.2 Fully Developed Laminar Flow 413 8.2.3 From Dimensional Analysis (1) (2) D V μ Δp = p1 – p2 = F(V, , D, ␮) Although fully developed laminar pipe flow is simple enough to allow the rather straightforward solutions discussed in the previous two sections, it may be worthwhile to consider this flow from a dimensional analysis standpoint. Thus, we assume that the pressure drop in the horizontal pipe, ¢p, is a function of the average velocity of the fluid in the pipe, V, the length of the pipe, /, the pipe diameter, D, and the viscosity of the fluid, m, as shown by the figure in the margin. We have not included the density or the specific weight of the fluid as parameters because for such flows they are not important parameters. There is neither mass 1density2 times acceleration nor a component of weight 1specific weight times volume2 in the flow direction involved. Thus, ¢p F1V, /, D, m2 Five variables can be described in terms of three reference dimensions 1M, L, T2. According to the results of dimensional analysis 1Chapter 72, this flow can be described in terms of k r 5 3 2 dimensionless groups. One such representation is D ¢p / fa b mV D (8.17) where f1/ D2 is an unknown function of the length to diameter ratio of the pipe. Although this is as far as dimensional analysis can take us, it seems reasonable to impose a further assumption that the pressure drop is directly proportional to the pipe length. That is, it takes twice the pressure drop to force fluid through a pipe if its length is doubled. The only way that this can be true is if f1/D2 C/D, where C is a constant. Thus, Eq. 8.17 becomes D ¢p C/ mV D which can be rewritten as ¢p Cm V / D2 or Q AV Dimensional analysis can be used to put pipe flow parameters into dimensionless form. 1p4C2 ¢pD4 m/ (8.18) The basic functional dependence for laminar pipe flow given by Eq. 8.18 is the same as that obtained by the analysis of the two previous sections. The value of C must be determined by theory 1as done in the previous two sections2 or experiment. For a round pipe, C 32. For ducts of other cross-sectional shapes, the value of C is different 1see Section 8.4.32. It is usually advantageous to describe a process in terms of dimensionless quantities. To this end we rewrite the pressure drop equation for laminar horizontal pipe flow, Eq. 8.8, as ¢p 32m/VD2 and divide both sides by the dynamic pressure, rV 2 2, to obtain the dimensionless form as ¢p 1 2 rV 2 132m/VD2 2 1 2 rV 2 64 a m / 64 / ba b a b rVD D Re D This is often written as ¢p f / rV 2 D 2 where the dimensionless quantity f ¢p1D/2 1rV 2 22 is termed the friction factor, or sometimes the Darcy friction factor [H. P. G. Darcy (1803–1858)]. 1This parameter should not be confused with the less-used Fanning friction 414 Chapter 8 ■ Viscous Flow in Pipes 10 Laminar flow 1 factor, which is defined to be f4. In this text we will use only the Darcy friction factor.2 Thus, the friction factor for laminar fully developed pipe flow is simply f f 0.1 0.01 10 100 1000 Re 64 Re (8.19) as shown by the figure in the margin. By substituting the pressure drop in terms of the wall shear stress 1Eq. 8.52, we obtain an alternate expression for the friction factor as a dimensionless wall shear stress f 8tw rV 2 (8.20) Knowledge of the friction factor will allow us to obtain a variety of information regarding pipe flow. For turbulent flow the dependence of the friction factor on the Reynolds number is much more complex than that given by Eq. 8.19 for laminar flow. This is discussed in detail in Section 8.4. 8.2.4 Energy Considerations In the previous three sections we derived the basic laminar flow results from application of F ma or dimensional analysis considerations. It is equally important to understand the implications of energy considerations of such flows. To this end we consider the energy equation for incompressible, steady flow between two locations as is given in Eq. 5.89 p1 V 21 p2 V 22 a1 z1 a2 z2 hL g g 2g 2g Recall that the kinetic energy coefficients, a1 and a2, compensate for the fact that the velocity profile across the pipe is not uniform. For uniform velocity profiles, a 1, whereas for any nonuniform profile, a 7 1. The head loss term, hL, accounts for any energy loss associated with the flow. This loss is a direct consequence of the viscous dissipation that occurs throughout the fluid in the pipe. For the ideal 1inviscid2 cases discussed in previous chapters, a1 a2 1, hL 0, and the energy equation reduces to the familiar Bernoulli equation discussed in Chapter 3 1Eq. 3.72. Even though the velocity profile in viscous pipe flow is not uniform, for fully developed flow it does not change from section 112 to section 122 so that a1 a2. Thus, the kinetic energy is the same at any section 1a1 V 21 2 a2 V 2222 and the energy equation becomes hL p1 p2 (2) (1) z1 (8.21) a p1 p2 z1 b a z2 b hL g g (8.22) z2 The energy dissipated by the viscous forces within the fluid is supplied by the excess work done by the pressure and gravity forces as shown by the figure in the margin. A comparison of Eqs. 8.22 and 8.10 shows that the head loss is given by hL 2t/ gr 1recall p1 p2 ¢p and z2 z1 / sin u2, which, by use of Eq. 8.4, can be rewritten in the form The head loss in a pipe is a result of the viscous shear stress on the wall. hL 4/tw gD (8.23) It is the shear stress at the wall 1which is directly related to the viscosity and the shear stress throughout the fluid2 that is responsible for the head loss. A closer consideration of the assumptions involved in the derivation of Eq. 8.23 will show that it is valid for both laminar and turbulent flow. 8.2 E XAMPLE 8.3 Fully Developed Laminar Flow Laminar Pipe Flow Properties GIVEN The flowrate, Q, of corn syrup through the horizontal pipe shown in Fig. E8.3a is to be monitored by measuring the pressure difference between sections 112 and 122. It is proposed that Q K ¢p, where the calibration constant, K, is a function of temperature, T, because of the variation of the syrup’s viscosity and density with temperature. These variations are given in Table E8.3. 6 ft Q (2) (1) 3-in. diameter (a) FIND (a) Plot K1T 2 versus T for 60 °F T 160 °F. (b) De- termine the wall shear stress and the pressure drop, ¢p p1 p2, for Q 0.5 ft3 s and T 100 °F. (c) For the conditions of part 1b2, determine the net pressure force, 1pD242 ¢p, and the net shear force, pD/tw , on the fluid within the pipe between the sections 112 and 122. 100 K, ft5/(lb•s) 10–1 SOLUTION (a) 10–2 10–3 If the flow is laminar, it follows from Eq. 8.9 that Q p1 123 ft2 4 ¢p pD4 ¢p 128m/ 128m16 ft2 10–4 60 100 (1) 1.60 105 (Ans) m where the units of K are ft5lb # s. By using values of the viscosity from Table E8.3, the calibration curve shown in Fig. E8.3b is obtained. This result is valid only if the flow is laminar. K COMMENT As shown in Section 8.5, for turbulent flow the flowrate is not linearly related to the pressure drop so it would not be possible to have Q K ¢p. Note also that the value of K is independent of the syrup density 1r was not used in the calculations2 since laminar pipe flow is governed by pressure and viscous effects; inertia is not important. (b) For T 100 °F, the viscosity is m 3.8 103 lb # sft2 so that with a flowrate of Q 0.5 ft3s the pressure drop 1according to Eq. 8.92 is ■ Figure E8.3 pD4 12813.8 103 lb # sft2 216 ft210.5 ft3s2 119 lbft (Ans) provided the flow is laminar. For this case so that Q 0.5 ft3s 10.2 fts p 3 2 A 1 12 ft2 4 12.05 slugsft 2110.2 ft ft2 rVD 3 2 # m 13.8 10 lb sft 2 1380 6 2100 3 Re T (ⴗF) R (slugs兾ft3) M (lb ⴢ sft2) 60 80 100 120 140 160 2.07 2.06 2.05 2.04 2.03 2.02 4.0 102 1.9 102 3.8 103 4.4 104 9.2 105 2.3 105 Hence, the flow is laminar. From Eq. 8.5 the wall shear stress is tw Fp p1 123 ft2 4 2 Table E8.3 1119 lbft2 21 123 ft2 ¢pD 1.24 lbft2 4/ 416 ft2 (Ans) (c) For the conditions of part 1b2, the net pressure force, Fp, on the fluid within the pipe between sections 112 and 122 is 128m/Q V 180 (b) 1.60 105 ¢p m where the units on Q, ¢p, and m are ft3s, lbft2, and lb # sft2, respectively. Thus ¢p 140 T, °F or Q K ¢p 415 2 p 2 p 3 D ¢p a ftb 1119 lb ft2 2 5.84 lb 4 4 12 (Ans) Similarly, the net viscous force, Fv, on that portion of the fluid is D Fv 2p a b /tw 2 2p c 3 ft d 16 ft211.24 lbft2 2 5.84 lb 21122 (Ans) s21 123 COMMENT Note that the values of these two forces are the same. The net force is zero; there is no acceleration. 416 Chapter 8 ■ Viscous Flow in Pipes 8.3 Fully Developed Turbulent Flow In the previous section various properties of fully developed laminar pipe flow were discussed. Since turbulent pipe flow is actually more likely to occur than laminar flow in practical situations, it is necessary to obtain similar information for turbulent pipe flow. However, turbulent flow is a very complex process. Numerous persons have devoted considerable effort in attempting to understand the variety of baffling aspects of turbulence. Although a considerable amount of knowledge about the topic has been developed, the field of turbulent flow still remains the least understood area of fluid mechanics. In this book we can provide only some of the very basic ideas concerning turbulence. The interested reader should consult some of the many books available for further reading 1Refs. 1–32. 8.3.1 Transition from Laminar to Turbulent Flow Random, turbulent fluctuations Turbulent 3 Turbulent bursts 2 4000 Transitional 2000 1 Laminar 0 t, sec ■ Figure 8.11 Transition from laminar to turbulent flow in a pipe. 0 Re = VD/v u, ft/s Turbulent flows involve randomly fluctuating parameters. Flows are classified as laminar or turbulent. For any flow geometry, there is one 1or more2 dimensionless parameter such that with this parameter value below a particular value the flow is laminar, whereas with the parameter value larger than a certain value the flow is turbulent. The important parameters involved 1i.e., Reynolds number, Mach number2 and their critical values depend on the specific flow situation involved. For example, flow in a pipe and flow along a flat plate 1boundary layer flow, as is discussed in Section 9.2.42 can be laminar or turbulent, depending on the value of the Reynolds number involved. As a general rule for pipe flow, the value of the Reynolds number must be less than approximately 2100 for laminar flow and greater than approximately 4000 for turbulent flow. For flow along a flat plate the transition between laminar and turbulent flow occurs at a Reynolds number of approximately 500,000 1see Section 9.2.42, where the length term in the Reynolds number is the distance measured from the leading edge of the plate. Consider a long section of pipe that is initially filled with a fluid at rest. As the valve is opened to start the flow, the flow velocity and, hence, the Reynolds number increase from zero 1no flow2 to their maximum steady-state flow values, as is shown in Fig. 8.11. Assume this transient process is slow enough so that unsteady effects are negligible 1quasi-steady flow2. For an initial time period the Reynolds number is small enough for laminar flow to occur. At some time the Reynolds number reaches 2100, and the flow begins its transition to turbulent conditions. Intermittent spots or bursts of turbulence appear. As the Reynolds number is increased, the entire flow field becomes turbulent. The flow remains turbulent as long as the Reynolds number exceeds approximately 4000. A typical trace of the axial component of velocity measured at a given location in the flow, u u1t2, is shown in Fig. 8.12. Its irregular, random nature is the distinguishing feature of turbulent flow. The character of many of the important properties of the flow 1pressure drop, heat transfer, etc.2 depends strongly on the existence and nature of the turbulent fluctuations or randomness 8.3 u Fully Developed Turbulent Flow 417 u' u(t) _ u = time-averaged (or mean) value T tO tO + T t ■ Figure 8.12 The time-averaged, u, and fluctuating, uⴕ, description of a parameter for turbulent flow. V8.4 Stirring color into paint V8.5 Laminar and turbulent mixing indicated. In previous considerations involving inviscid flow, the Reynolds number is 1strictly speaking2 infinite 1because the viscosity is zero2, and the flow most surely would be turbulent. However, reasonable results were obtained by using the inviscid Bernoulli equation as the governing equation. The reason that such simplified inviscid analyses gave reasonable results is that viscous effects were not very important and the velocity used in the calculations was actually the timeaveraged velocity, u, indicated in Fig. 8.12. Calculation of the heat transfer, pressure drop, and many other parameters would not be possible without inclusion of the seemingly small, but very important, effects associated with the randomness of the flow. Consider flow in a pan of water placed on a stove. With the stove turned off, the fluid is stationary. The initial sloshing has died out because of viscous dissipation within the water. With the stove turned on, a temperature gradient in the vertical direction, 0T 0z, is produced. The water temperature is greatest near the pan bottom and decreases toward the top of the fluid layer. If the temperature difference is very small, the water will remain stationary, even though the water density is smallest near the bottom of the pan because of the decrease in density with an increase in temperature. A further increase in the temperature gradient will cause a buoyancy-driven instability that results in fluid motion—the light, warm water rises to the top, and the heavy, cold water sinks to the bottom. This slow, regular “turning over” increases the heat transfer from the pan to the water and promotes mixing within the pan. As the temperature gradient increases still further, the fluid motion becomes more vigorous and eventually turns into a chaotic, random, turbulent flow with considerable mixing, vaporization (boiling) and greatly increased heat transfer rate. The flow has progressed from a stationary fluid, to laminar flow, and finally to turbulent, multiphase (liquid and vapor) flow. Mixing processes and heat and mass transfer processes are considerably enhanced in turbulent flow compared to laminar flow. This is due to the macroscopic scale of the randomness in turbulent flow. We are all familiar with the “rolling,” vigorous eddy-type motion of the water in a pan being heated on the stove 1even if it is not heated to boiling2. Such finite-sized random mixing is very effective in transporting energy and mass throughout the flow field, thereby increasing the various rate processes involved. Laminar flow, on the other hand, can be thought of as very small but finite-sized fluid particles flowing smoothly in layers, one over another. The only randomness and mixing take place on the molecular scale and result in relatively small heat, mass, and momentum transfer rates. Without turbulence it would be virtually impossible to carry out life as we now know it. Mixing is one positive application of turbulence, as discussed above, but there are other situations where turbulent flow is desirable. To transfer the required heat between a solid and an adjacent fluid 1such as in the cooling coils of an air conditioner or a boiler of a power plant2 would require an enormously large heat exchanger if the flow were laminar. Similarly, the required mass transfer of a liquid state to a vapor state 1such as is needed in the evaporated cooling system associated with sweating2 would require very large surfaces if the fluid flowing past the surface were 418 Chapter 8 ■ Viscous Flow in Pipes laminar rather than turbulent. As shown in Chapter 9, turbulence can also aid in delaying flow separation. F l u i d s i n t Smaller heat exchangers Automobile radiators, air conditioners, and refrigerators contain heat exchangers that transfer energy from (to) the hot (cold) fluid within the heat exchanger tubes to (from) the colder (hotter) surrounding fluid. These units can be made smaller and more efficient by increasing the heat transfer rate across the tubes’ surfaces. If the flow through the tubes is laminar, the heat transfer rate is relatively small. Significantly larger heat transfer rates are obtained if the flow within the tubes is turbulent. Even greater heat transfer rates can be obtained by the use of turbulence promoters, sometimes V8.6 Stirring cream into coffee h e N e w s termed “turbulators,” which provide additional turbulent mixing motion than would normally occur. Such enhancement mechanisms include internal fins, spiral wire or ribbon inserts, and ribs or grooves on the inner surface of the tube. While these mechanisms can increase the heat transfer rate by 1.5 to 3 times over that for a bare tube at the same flowrate, they also increase the pressure drop (and therefore the power) needed to produce the flow within the tube. Thus, a compromise involving increased heat transfer rate and increased power consumption is often needed. Turbulence is also of importance in the mixing of fluids. Smoke from a stack would continue for miles as a ribbon of pollutant without rapid dispersion within the surrounding air if the flow were laminar rather than turbulent. Under certain atmospheric conditions this is observed to occur. Although there is mixing on a molecular scale 1laminar flow2, it is several orders of magnitude slower and less effective than the mixing on a macroscopic scale 1turbulent flow2. It is considerably easier to mix cream into a cup of coffee 1turbulent flow2 than to thoroughly mix two colors of a viscous paint 1laminar flow2. In other situations laminar 1rather than turbulent2 flow is desirable. The pressure drop in pipes 1hence, the power requirements for pumping2 can be considerably lower if the flow is laminar rather than turbulent. Fortunately, the blood flow through a person’s arteries is normally laminar, except in the largest arteries with high blood flowrates. The aerodynamic drag on an airplane wing can be considerably smaller with laminar flow past it than with turbulent flow. 8.3.2 Turbulent Shear Stress The fundamental difference between laminar and turbulent flow lies in the chaotic, random behavior of the various fluid parameters. Such variations occur in the three components of velocity, the pressure, the shear stress, the temperature, and any other variable that has a field description. Turbulent flow is characterized by random, three-dimensional vorticity 1i.e., fluid particle rotation or spin; see Section 6.1.32. As is indicated in Fig. 8.12, such flows can be described in terms of their mean values 1denoted with an overbar2 on which are superimposed the fluctuations 1denoted with a prime2. Thus, if u u1x, y, z, t2 is the x component of instantaneous velocity, then its time mean 1or time-average2 value, u, is Turbulent flow parameters can be described in terms of mean and fluctuating portions. 1 u T 冮 t0 T u1x, y, z, t2 dt (8.24) t0 where the time interval, T, is considerably longer than the period of the longest fluctuations, but considerably shorter than any unsteadiness of the average velocity. This is illustrated in Fig. 8.12. The fluctuating part of the velocity, u¿, is that time-varying portion that differs from the average value u u u¿ or u¿ u u (8.25) Clearly, the time average of the fluctuations is zero, since u¿ 1 T 冮 t0 T t0 1u u 2 dt 1 1T u T u2 0 T 1 a T 冮 t0 T t0 u dt u 冮 t0 T t0 dtb 8.3 Fully Developed Turbulent Flow 419 u' (u')2 u' or (u')2 (u')2 > 0 0 t u' = 0 ■ Figure 8.13 Average of the fluctuations and average of the square of the fluctuations. The fluctuations are equally distributed on either side of the average. It is also clear, as is indicated in Fig. 8.13, that since the square of a fluctuation quantity cannot be negative 3 1u¿2 2 04, its average value is positive. Thus, 1u¿2 2 V, ft/s 冮 t0 T t0 1u¿2 2 dt 7 0 On the other hand, it may be that the average of products of the fluctuations, such as u¿v¿, are zero or nonzero 1either positive or negative2. The structure and characteristics of turbulence may vary from one flow situation to another. For example, the turbulence intensity 1or the level of the turbulence2 may be larger in a very gusty wind than it is in a relatively steady 1although turbulent2 wind. The turbulence intensity, i, is often defined as the square root of the mean square of the fluctuating velocity divided by the timeaveraged velocity, or 15 10 5 0 1 T 0 20 40 t, min i The relationship between fluid motion and shear stress is very complex for turbulent flow. 2 1u¿2 u 2 c 1 T 冮 t0 T t0 1u¿2 2 dt d u 12 The larger the turbulence intensity, the larger the fluctuations of the velocity 1and other flow parameters2. Well-designed wind tunnels have typical values of i ⬇ 0.01, although with extreme care, values as low as i 0.0002 have been obtained. On the other hand, values of i 0.1 are found for the flow in the atmosphere and rivers. A typical atmospheric wind-speed graph is shown in the figure in the margin. Another turbulence parameter that is different from one flow situation to another is the period of the fluctuations—the time scale of the fluctuations shown in Fig. 8.12. In many flows, such as the flow of water from a faucet, typical frequencies are on the order of 10, 100, or 1000 cycles per second 1cps2. For other flows, such as the Gulf Stream current in the Atlantic Ocean or flow of the atmosphere of Jupiter, characteristic random oscillations may have a period on the order of hours, days, or more. It is tempting to extend the concept of viscous shear stress for laminar flow 1t m dudy2 to that of turbulent flow by replacing u, the instantaneous velocity, by u, the time-averaged velocity. However, numerous experimental and theoretical studies have shown that such an approach leads to completely incorrect results. That is, t m d udy. A physical explanation for this behavior can be found in the concept of what produces a shear stress. Laminar flow is modeled as fluid particles that flow smoothly along in layers, gliding past the slightly slower or faster ones on either side. As is discussed in Chapter 1, the fluid actually consists of numerous molecules darting about in an almost random fashion as is indicated in Fig. 8.14a. The motion is not entirely random—a slight bias in one direction produces the flowrate we associate 420 Chapter 8 ■ Viscous Flow in Pipes y y Average velocity profile, u = u(y) Velocity profile, u = u(y) (2) A A A A u u u1 < u2 (1) Turbulent eddies (a) (b) ■ Figure 8.14 (a) Laminar flow shear stress caused by random motion of molecules. (b) Turbulent flow as a series of random, three-dimensional eddies. with the motion of fluid particles, u. As the molecules dart across a given plane 1plane A– A, for example2, the ones moving upward have come from an area of smaller average x component of velocity than the ones moving downward, which have come from an area of larger velocity. The momentum flux in the x direction across plane A– A gives rise to a drag 1to the left2 of the lower fluid on the upper fluid and an equal but opposite effect of the upper fluid on the lower fluid. The sluggish molecules moving upward across plane A– A must be accelerated by the fluid above this plane. The rate of change of momentum in this process produces 1on the macroscopic scale2 a shear force. Similarly, the more energetic molecules moving down across plane A– A must be slowed down by the fluid below that plane. This shear force is present only if there is a gradient in u u1y2, otherwise the average x component of velocity 1and momentum2 of the upward and downward molecules is exactly the same. In addition, there are attractive forces between molecules. By combining these effects we obtain the well-known Newton viscosity law: t m dudy, where on a molecular basis m is related to the mass and speed 1temperature2 of the random motion of the molecules. Although the above random motion of the molecules is also present in turbulent flow, there is another factor that is generally more important. A simplistic way of thinking about turbulent flow is to consider it as consisting of a series of random, three-dimensional eddy type motions as is depicted 1in one dimension only2 in Fig. 8.14b. (See the photograph at the beginning of this chapter.) These eddies range in size from very small diameter 1on the order of the size of a fluid particle2 to fairly large diameter 1on the order of the size of the object or flow geometry considered2. They move about randomly, conveying mass with an average velocity u u1y2. This eddy structure greatly promotes mixing within the fluid. It also greatly increases the transport of x momentum across plane A– A. That is, finite particles of fluid 1not merely individual molecules as in laminar flow2 are randomly transported across this plane, resulting in a relatively large 1when compared with laminar flow2 shear force. These particles vary in size but are much larger than molecules. Turbulent flow shear stress is larger than laminar flow shear stress because of the irregular, random motion. F l u i d s i n Listen to the flowrate Sonar systems are designed to listen to transmitted and reflected sound waves in order to locate submerged objects. They have been used successfully for many years to detect and track underwater objects such as submarines and aquatic animals. Recently, sonar techniques have been refined so that they can be used to determine the flowrate in pipes. These new flowmeters work for turbulent, not laminar, pipe flows because their operation depends strictly on the existence of turbu- t h e N e w s lent eddies within the flow. The flowmeters contain a sonar-based array that listens to and interprets pressure fields generated by the turbulent motion in pipes. By listening to the pressure fields associated with the movement of the turbulent eddies, the device can determine the speed at which the eddies travel past an array of sensors. The flowrate is determined by using a calibration procedure that links the speed of the turbulent structures to the volumetric flowrate. 8.3 Fully Developed Turbulent Flow 421 The random velocity components that account for this momentum transfer 1hence, the shear force2 are u¿ 1for the x component of velocity2 and v¿ 1for the rate of mass transfer crossing the plane2. A more detailed consideration of the processes involved will show that the apparent shear stress on plane A– A is given by the following 1Ref. 22: tm V8.7 Turbulence in a bowl The shear stress is the sum of a laminar portion and a turbulent portion. du ru¿v¿ tlam tturb dy (8.26) Note that if the flow is laminar, u¿ v¿ 0, so that u¿v¿ 0 and Eq. 8.26 reduces to the customary random molecule-motion-induced laminar shear stress, tlam m d udy. For turbulent flow it is found that the turbulent shear stress, tturb ru¿v¿, is positive. Hence, the shear stress is greater in turbulent flow than in laminar flow. Note the units on tturb are 1density21velocity2 2 1slugsft3 21fts2 2 1slugs # fts2 2 ft2 lbft2, or Nm2, as expected. Terms of the form ru¿v¿ 1or rv¿w¿, etc.2 are called Reynolds stresses in honor of Osborne Reynolds who first discussed them in 1895. It is seen from Eq. 8.26 that the shear stress in turbulent flow is not merely proportional to the gradient of the time-averaged velocity, u1y2. It also contains a contribution due to the random fluctuations of the x and y components of velocity. The density is involved because of the momentum transfer of the fluid within the random eddies. Although the relative magnitude of tlam compared to tturb is a complex function dependent on the specific flow involved, typical measurements indicate the structure shown in Fig. 8.15a. 1Recall from Eq. 8.4 that the shear stress is proportional to the distance from the centerline of the pipe.2 In a very narrow region near the wall 1the viscous sublayer2, the laminar shear stress is dominant. Away from the wall 1in the outer layer2 the turbulent portion of the shear stress is dominant. The transition between these two regions occurs in the overlap layer. The corresponding typical velocity profile is shown in Fig. 8.15b. The scale of the sketches shown in Fig. 8.15 is not necessarily correct. Typically the value of tturb is 100 to 1000 times greater than tlam in the outer region, while the converse is true in the viscous sublayer. A correct modeling of turbulent flow is strongly dependent on an accurate knowledge of tturb. This, in turn, requires an accurate knowledge of the fluctuations u¿ and v¿, or ru¿v¿. As yet it is not possible to solve the governing equations 1the Navier –Stokes equations2 for these details of the flow, although numerical techniques 1see Appendix A2 using the largest and fastest computers available have produced important information about some of the characteristics of turbulence. Considerable effort has gone into the study of turbulence. Much remains to be learned. Perhaps studies in the new areas of chaos and fractal geometry will provide the tools for a better understanding of turbulence 1see Section 8.3.52. The vertical scale of Fig. 8.15 is also distorted. The viscous sublayer is usually a very thin layer adjacent to the wall. For example, for water flow in a 3-in.-diameter pipe with an average velocity of 10 ft s, the viscous sublayer is approximately 0.002 in. thick. Since the fluid motion within this thin layer is critical in terms of the overall flow 1the no-slip condition and the wall shear stress occur in this layer2, it is not surprising to find that turbulent pipe flow properties can be quite Viscous sublayer Pipe wall R R τ lam τ turb Overlap layer τ r r Outer layer Pipe centerline 0 0 τw 0 0 Vc τ (r) u(r) (a) (b) ■ Figure 8.15 Structure of turbulent flow in a pipe. (a) Shear stress. (b) Average velocity. 422 Chapter 8 ■ Viscous Flow in Pipes dependent on the roughness of the pipe wall, unlike laminar pipe flow which is independent of roughness. Small roughness elements 1scratches, rust, sand or dirt particles, etc.2 can easily disturb this viscous sublayer 1see Section 8.42, thereby affecting the entire flow. An alternate form for the shear stress for turbulent flow is given in terms of the eddy viscosity, h, where tturb h Various ad hoc assumptions have been used to approximate turbulent shear stresses. du dy (8.27) This extension of laminar flow terminology was introduced by J. Boussinesq, a French scientist, in 1877. Although the concept of an eddy viscosity is intriguing, in practice it is not an easy parameter to use. Unlike the absolute viscosity, m, which is a known value for a given fluid, the eddy viscosity is a function of both the fluid and the flow conditions. That is, the eddy viscosity of water cannot be looked up in handbooks—its value changes from one turbulent flow condition to another and from one point in a turbulent flow to another. The inability to accurately determine the Reynolds stress, ru¿v¿, is equivalent to not knowing the eddy viscosity. Several semiempirical theories have been proposed 1Ref. 32 to determine approximate values of h. L. Prandtl 11875–19532, a German physicist and aerodynamicist, proposed that the turbulent process could be viewed as the random transport of bundles of fluid particles over a certain distance, /m, the mixing length, from a region of one velocity to another region of a different velocity. By the use of some ad hoc assumptions and physical reasoning, it was concluded that the eddy viscosity was given by h r/2m ` du ` dy Thus, the turbulent shear stress is tturb r/2m a du 2 b dy (8.28) The problem is thus shifted to that of determining the mixing length, /m. Further considerations indicate that /m is not a constant throughout the flow field. Near a solid surface the turbulence is dependent on the distance from the surface. Thus, additional assumptions are made regarding how the mixing length varies throughout the flow. The net result is that as yet there is no general, all-encompassing, useful model that can accurately predict the shear stress throughout a general incompressible, viscous turbulent flow. Without such information it is impossible to integrate the force balance equation to obtain the turbulent velocity profile and other useful information, as was done for laminar flow. 8.3.3 Turbulent Velocity Profile V8.8 Turbulent channel flow Considerable information concerning turbulent velocity profiles has been obtained through the use of dimensional analysis, experimentation, numerical simulations, and semiempirical theoretical efforts. As is indicated in Fig. 8.15, fully developed turbulent flow in a pipe can be broken into three regions which are characterized by their distances from the wall: the viscous sublayer very near the pipe wall, the overlap region, and the outer turbulent layer throughout the center portion of the flow. Within the viscous sublayer the viscous shear stress is dominant compared with the turbulent 1or Reynolds2 stress, and the random, eddying nature of the flow is essentially absent. In the outer turbulent layer the Reynolds stress is dominant, and there is considerable mixing and randomness to the flow. The character of the flow within these two regions is entirely different. For example, within the viscous sublayer the fluid viscosity is an important parameter; the density is unimportant. In the outer layer the opposite is true. By a careful use of dimensional analysis arguments for the flow in each layer and by a matching of the results in the common overlap layer, it has been possible to obtain the following conclusions about the turbulent velocity profile in a smooth pipe 1Ref. 52. In the viscous sublayer the velocity profile can be written in dimensionless form as yu u n u (8.29) 8.3 Fully Developed Turbulent Flow 423 25 Experimental data 20 Eq. 8.29 Pipe centerline Eq. 8.30 15 u ___ u Turbulent effects 10 Viscous and turbulent effects 5 Viscous sublayer 0 1 10 102 103 104 yu ____ v ■ Figure 8.16 Typical structure of the turbulent velocity profile in a pipe. where y R r is the distance measured from the wall, u is the time-averaged x component of velocity, and u 1tw r2 12 is termed the friction velocity. Note that u is not an actual velocity of the fluid—it is merely a quantity that has dimensions of velocity. As is indicated in Fig. 8.16, Eq. 8.29 1commonly called the law of the wall2 is valid very near the smooth wall, for 0 yun f 5. Dimensional analysis arguments indicate that in the overlap region the velocity should vary as the logarithm of y. Thus, the following expression has been proposed: A turbulent flow velocity profile can be divided into various regions. V8.9 Laminar to turbulent flow from a pipe yu u b 5.0 2.5 ln a n u (8.30) where the constants 2.5 and 5.0 have been determined experimentally. As is indicated in Fig. 8.16, for regions not too close to the smooth wall, but not all the way out to the pipe center, Eq. 8.30 gives a reasonable correlation with the experimental data. Note that the horizontal scale is a logarithmic scale. This tends to exaggerate the size of the viscous sublayer relative to the remainder of the flow. As is shown in Example 8.4, the viscous sublayer is usually quite thin. Similar results can be obtained for turbulent flow past rough walls 1Ref. 172. A number of other correlations exist for the velocity profile in turbulent pipe flow. In the central region 1the outer turbulent layer2 the expression 1Vc u2 u 2.5 ln1Ry2, where Vc is the centerline velocity, is often suggested as a good correlation with experimental data. Another often-used 1and relatively easy to use2 correlation is the empirical power-law velocity profile u r 1n a1 b Vc R (8.31) In this representation, the value of n is a function of the Reynolds number, as is indicated in Fig. 8.17. The one-seventh power-law velocity profile 1n 72 is often used as a reasonable approximation for many practical flows. Typical turbulent velocity profiles based on this power-law representation are shown in Fig. 8.18. A closer examination of Eq. 8.31 shows that the power-law profile cannot be valid near the wall, since according to this equation the velocity gradient is infinite there. In addition, Eq. 8.31 cannot be precisely valid near the centerline because it does not give du dr 0 at r 0. However, it does provide a reasonable approximation to the measured velocity profiles across most of the pipe. Note from Fig. 8.18 that the turbulent profiles are much “flatter” than the laminar profile and that this flatness increases with Reynolds number 1i.e., with n2. Recall from Chapter 3 that 424 Chapter 8 ■ Viscous Flow in Pipes 11 10 9 n 8 7 6 5 104 105 106 ρVD Re = ____ μ ■ Figure 8.17 Exponent, n, for power-law velocity profiles. (Adapted from Ref. 1.) 1.0 n = 10 n=6 Laminar r __ R n=8 0.5 V8.10 Laminar/ turbulent velocity profiles Turbulent 0 ■ Figure 8.18 0 0.5 _ u __ 1.0 Vc Typical laminar flow and turbulent flow velocity profiles. reasonable approximate results are often obtained by using the inviscid Bernoulli equation and by assuming a fictitious uniform velocity profile. Since most flows are turbulent and turbulent flows tend to have nearly uniform velocity profiles, the usefulness of the Bernoulli equation and the uniform profile assumption is not unexpected. Of course, many properties of the flow cannot be accounted for without including viscous effects. E XAMPLE 8.4 Turbulent Pipe Flow Properties GIVEN Water at 20 °C 1r 998 kgm3 and n 1.004 10 m s2 flows through a horizontal pipe of 0.1-m diameter with a flowrate of Q 4 102 m3s and a pressure gradient of 2.59 kPam. 6 2 FIND (a) Determine the approximate thickness of the viscous sublayer. (b) Determine the approximate centerline velocity, Vc. (c) Determine the ratio of the turbulent to laminar shear stress, tturb tlam, at a point midway between the centerline and the pipe wall 1i.e., at r 0.025 m2. 8.3 425 Fully Developed Turbulent Flow SOLUTION (a) According to Fig. 8.16, the thickness of the viscous sublayer, ds, is approximately obtained by integration of the power-law velocity profile as follows. Since the flow is axisymmetric, dsu 5 n Q AV 冮 u dA Vc 冮 rR r0 a1 r 1n b 12pr2 dr R which can be integrated to give Therefore, ds 5 n u Q 2pR2Vc n2 1n 1212n 12 Thus, since Q pR2V, we obtain where tw 12 u a b r (1) The wall shear stress can be obtained from the pressure drop data and Eq. 8.5, which is valid for either laminar or turbulent flow. Thus, 10.1 m2 12.59 10 Nm 2 D ¢p 64.8 Nm2 4/ 411 m2 3 tw 2n2 V Vc 1n 1212n 12 With n 8.4 in the present case, this gives Vc 2n2 6.04 ms 2 Hence, from Eq. 1 we obtain 1n 1212n 12 Recall that Vc 2V for laminar pipe flow. t so that 511.004 106 m2s2 0.255 ms 1.97 105 m ⯝ 0.02 mm t tlam tturb 32.4 Nm2 (Ans) very thin. Minute imperfections on the pipe wall will protrude into this sublayer and affect some of the characteristics of the flow 1i.e., wall shear stress and pressure drop2. where tlam m dudr. From the power-law velocity profile 1Eq. 8.312 we obtain the gradient of the average velocity as Vc r 11n2n du a1 b dr nR R which gives 16.04 m s2 du 0.025 m 118.428.4 a1 b dr 8.410.05 m2 0.05 m 26.5s (b) The centerline velocity can be obtained from the average velocity and the assumption of a power-law velocity profile as follows. For this flow with Q 0.04 m3 s 5.09 ms A p10.1 m2 2 4 the Reynolds number is Re 2164.8 Nm2 210.025 m2 2twr D 10.1 m2 or COMMENT As stated previously, the viscous sublayer is V (Ans) (c) From Eq. 8.4, which is valid for laminar or turbulent flow, the shear stress at r 0.025 m is 64.8 Nm2 12 u a b 0.255 ms 998 kg m3 ds V 1.186V 1.186 15.09 ms2 15.09 ms210.1 m2 VD 5.07 105 n 11.004 106 m2s2 Thus, from Fig. 8.17, n 8.4 so that r 18.4 u ⬇ a1 b Vc R To determine the centerline velocity, Vc, we must know the relationship between V 1the average velocity2 and Vc. This can be Thus, du du 1nr2 dr dr 11.004 106 m2s21998 kgm3 21–26.5s2 0.0266 Nm2 tlam m Thus, the ratio of turbulent to laminar shear stress is given by tturb t tlam 32.4 0.0266 1220 tlam tlam 0.0266 (Ans) COMMENT As expected, most of the shear stress at this location in the turbulent flow is due to the turbulent shear stress. The turbulent flow characteristics discussed in this section are not unique to turbulent flow in round pipes. Many of the characteristics introduced 1i.e., the Reynolds stress, the viscous sublayer, the overlap layer, the outer layer, the general characteristics of the velocity profile, etc.2 are found in other turbulent flows. In particular, turbulent pipe flow and turbulent flow past a solid wall 1boundary layer flow2 share many of these common traits. Such ideas are discussed more fully in Chapter 9. 426 Chapter 8 ■ Viscous Flow in Pipes 8.3.4 Turbulence Modeling Although it is not yet possible to theoretically predict the random, irregular details of turbulent flows, it would be useful to be able to predict the time-averaged flow fields 1pressure, velocity, etc.2 directly from the basic governing equations. To this end one can time average the governing Navier– Stokes equations 1Eqs. 6.31 and 6.1272 to obtain equations for the average velocity and pressure. However, because the Navier –Stokes equations are nonlinear, the resulting time-averaged differential equations contain not only the desired average pressure and velocity as variables, but also averages of products of the fluctuations—terms of the type that one tried to eliminate by averaging the equations! For example, the Reynolds stress ru¿v¿ 1see Eq. 8.262 occurs in the timeaveraged momentum equation. Thus, it is not possible to merely average the basic differential equations and obtain governing equations involving only the desired averaged quantities. This is the reason for the variety of ad hoc assumptions that have been proposed to provide “closure” to the equations governing the average flow. That is, the set of governing equations must be a complete or closed set of equations—the same number of equation as unknowns. Various attempts have been made to solve this closure problem 1Refs. 1, 322. Such schemes involving the introduction of an eddy viscosity or the mixing length 1as introduced in Section 8.3.22 are termed algebraic or zero-equation models. Other methods, which are beyond the scope of this book, include the one-equation model and the two-equation model. These turbulence models are based on the equation for the turbulence kinetic energy and require significant computer usage. Turbulence modeling is an important and extremely difficult topic. Although considerable progress has been made, much remains to be done in this area. 8.3.5 Chaos and Turbulence Chaos theory may eventually provide a deeper understanding of turbulence. V8.11 2D grid turbulence in soap film. 8.4 Chaos theory is a relatively new branch of mathematical physics that may provide insight into the complex nature of turbulence. This method combines mathematics and numerical 1computer2 techniques to provide a new way to analyze certain problems. Chaos theory, which is quite complex and is currently under development, involves the behavior of nonlinear dynamical systems and their response to initial and boundary conditions. The flow of a viscous fluid, which is governed by the nonlinear Navier– Stokes equations 1Eq. 6.1272, may be such a system. To solve the Navier–Stokes equations for the velocity and pressure fields in a viscous flow, one must specify the particular flow geometry being considered 1the boundary conditions2 and the condition of the flow at some particular time 1the initial conditions2. If, as some researchers predict, the Navier – Stokes equations allow chaotic behavior, then the state of the flow at times after the initial time may be very, very sensitive to the initial conditions. A slight variation to the initial flow conditions may cause the flow at later times to be quite different than it would have been with the original, only slightly different initial conditions. When carried to the extreme, the flow may be “chaotic,” “random,” or perhaps 1in current terminology2, “turbulent.” The occurrence of such behavior would depend on the value of the Reynolds number. For example, it may be found that for sufficiently small Reynolds numbers the flow is not chaotic 1i.e., it is laminar2, while for large Reynolds numbers it is chaotic with turbulent characteristics. Thus, with the advancement of chaos theory it may be found that the numerous ad hoc turbulence ideas mentioned in previous sections 1i.e., eddy viscosity, mixing length, law of the wall, etc.2 may not be needed. It may be that chaos theory can provide the turbulence properties and structure directly from the governing equations. As of now we must wait until this exciting topic is developed further. The interested reader is encouraged to consult Ref. 4 for a general introduction to chaos or Ref. 33 for additional material. Dimensional Analysis of Pipe Flow As noted previously, turbulent flow can be a very complex, difficult topic—one that as yet has defied a rigorous theoretical treatment. Thus, most turbulent pipe flow analyses are based on experimental data and semi-empirical formulas. These data are expressed conveniently in dimensionless form. 8.4 Dimensional Analysis of Pipe Flow 427 It is often necessary to determine the head loss, hL, that occurs in a pipe flow so that the energy equation, Eq. 5.84, can be used in the analysis of pipe flow problems. As shown in Fig. 8.1, a typical pipe system usually consists of various lengths of straight pipe interspersed with various types of components (valves, elbows, etc.). The overall head loss for the pipe system consists of the head loss due to viscous effects in the straight pipes, termed the major loss and denoted hL major, and the head loss in the various pipe components, termed the minor loss and denoted hL minor. That is, hL hL major hL minor The head loss designations of “major” and “minor” do not necessarily reflect the relative importance of each type of loss. For a pipe system that contains many components and a relatively short length of pipe, the minor loss may actually be larger than the major loss. 8.4.1 Major Losses Δp = p1 – p2 (1) (2) ρ, μ D V ε Turbulent pipe flow properties depend on the fluid density and the pipe roughness. A dimensional analysis treatment of pipe flow provides the most convenient base from which to consider turbulent, fully developed pipe flow. An introduction to this topic was given in Section 8.3. As is discussed in Sections 8.2.1 and 8.2.4, the pressure drop and head loss in a pipe are dependent on the wall shear stress, tw, between the fluid and pipe surface. A fundamental difference between laminar and turbulent flow is that the shear stress for turbulent flow is a function of the density of the fluid, r. For laminar flow, the shear stress is independent of the density, leaving the viscosity, m, as the only important fluid property. Thus, as indicated by the figure in the margin, the pressure drop, ¢p, for steady, incompressible turbulent flow in a horizontal round pipe of diameter D can be written in functional form as ¢p F1V, D, /, e, m, r2 (8.32) where V is the average velocity, / is the pipe length, and e is a measure of the roughness of the pipe wall. It is clear that ¢p should be a function of V, D, and /. The dependence of ¢p on the fluid properties m and r is expected because of the dependence of t on these parameters. Although the pressure drop for laminar pipe flow is found to be independent of the roughness of the pipe, it is necessary to include this parameter when considering turbulent flow. As is discussed in Section 8.3.3 and illustrated in Fig. 8.19, for turbulent flow there is a relatively thin viscous sublayer formed in the fluid near the pipe wall. In many instances this layer is very thin; dsD 1, where ds is the sublayer thickness. If a typical wall roughness element protrudes sufficiently far into 1or even through2 this layer, the structure and properties of the viscous sublayer 1along with ¢p and tw2 will be different than if the wall were smooth. Thus, for turbulent flow the pressure drop is expected to be a function of the wall roughness. For laminar flow there is no thin viscous layer—viscous effects are important across the entire pipe. Thus, relatively small roughness elements have completely negligible effects on laminar pipe flow. Of course, for pipes with very large wall “roughness” 1eD g 0.12, such as that in corrugated pipes, the flowrate may be a function of the “roughness.” We will consider only typical constant diameter pipes with relative roughnesses in the range 0 eD f 0.05. Analysis of flow in corrugated pipes does not fit into the standard constant diameter pipe category, although experimental results for such pipes are available 1Ref. 302. The list of parameters given in Eq. 8.32 is apparently a complete one. That is, experiments have shown that other parameters 1such as surface tension, vapor pressure, etc.2 do not affect the pressure drop for the conditions stated 1steady, incompressible flow; round, horizontal pipe2. Since there are seven variables 1k 72 that can be written in terms of the three reference dimensions MLT 1r 32, Eq. 8.32 can be written in dimensionless form in terms of k r 4 dimensionless groups. As was discussed in Section 7.9.1, one such representation is ¢p 1 2 2 rV ~ rVD / e fa , , b m D D This result differs from that used for laminar flow 1see Eq. 8.172 in two ways. First, we have chosen to make the pressure dimensionless by dividing by the dynamic pressure, rV 2 2, rather than a characteristic viscous shear stress, mVD. This convention was chosen in recognition of the fact that the shear stress for turbulent flow is normally dominated by tturb, which is a stronger function 428 Chapter 8 ■ Viscous Flow in Pipes Velocity profile, u = u(y) R = D/2 y δs x Viscous sublayer or e δs e Rough wall ■ Figure 8.19 Flow in the viscous sublayer near rough and smooth walls. Smooth wall of the density than it is of viscosity. Second, we have introduced two additional dimensionless parameters, the Reynolds number, Re rVDm, and the relative roughness, eD, which are not present in the laminar formulation because the two parameters r and e are not important in fully developed laminar pipe flow. As was done for laminar flow, the functional representation can be simplified by imposing the reasonable assumption that the pressure drop should be proportional to the pipe length. 1Such a step is not within the realm of dimensional analysis. It is merely a logical assumption supported by experiments.2 The only way that this can be true is if the /D dependence is factored out as ¢p 1 2 2 rV / e f aRe, b D D As was discussed in Section 8.2.3, the quantity ¢pD 1/rV 2 22 is termed the friction factor, f. Thus, for a horizontal pipe ¢p f / rV 2 D 2 (8.33) where f f aRe, e b D For laminar fully developed flow, the value of f is simply f 64Re, independent of e D. For turbulent flow, the functional dependence of the friction factor on the Reynolds number and the relative roughness, f f1Re, eD2, is a rather complex one that cannot, as yet, be obtained from a theoretical analysis. The results are obtained from an exhaustive set of experiments and usually presented in terms of a curve-fitting formula or the equivalent graphical form. From Eq. 5.89 the energy equation for steady incompressible flow is p1 V 21 p2 V 22 z1 z2 hL a1 a2 g g 2g 2g The major head loss in pipe flow is given in terms of the friction factor. where hL is the head loss between sections 112 and 122. With the assumption of a constant diameter 1D1 D2 so that V1 V2 2, horizontal 1z1 z2 2 pipe with fully developed flow 1a1 a2 2, this becomes ¢p p1 p2 ghL, which can be combined with Eq. 8.33 to give hL major f / V2 D 2g (8.34) 8.4 Dimensional Analysis of Pipe Flow 429 Equation 8.34, called the Darcy–Weisbach equation, is valid for any fully developed, steady, incompressible pipe flow—whether the pipe is horizontal or on a hill. On the other hand, Eq. 8.33 is valid only for horizontal pipes. In general, with V1 V2 the energy equation gives p1 p2 g1z2 z1 2 ghL g1z2 z1 2 f (Photograph courtesy of CorrView) 0.08 f Completely turbulent flow 0.06 0.04 e D 0.1 0.01 0.001 0.0001 0 0.00001 0.02 / rV 2 D 2 Part of the pressure change is due to the elevation change and part is due to the head loss associated with frictional effects, which are given in terms of the friction factor, f. It is not easy to determine the functional dependence of the friction factor on the Reynolds number and relative roughness. Much of this information is a result of experiments conducted by J. Nikuradse in 1933 1Ref. 62 and amplified by many others since then. One difficulty lies in the determination of the roughness of the pipe. Nikuradse used artificially roughened pipes produced by gluing sand grains of known size onto pipe walls to produce pipes with sandpaper-type surfaces. The pressure drop needed to produce a desired flowrate was measured, and the data were converted into the friction factor for the corresponding Reynolds number and relative roughness. The tests were repeated numerous times for a wide range of Re and eD to determine the f f1Re, eD2 dependence. In commercially available pipes the roughness is not as uniform and well defined as in the artificially roughened pipes used by Nikuradse. However, it is possible to obtain a measure of the effective relative roughness of typical pipes and thus to obtain the friction factor. Typical roughness values for various pipe surfaces are given in Table 8.1. Figure 8.20 shows the functional dependence of f on Re and eD and is called the Moody chart in honor of L. F. Moody, who, along with C. F. Colebrook, correlated the original data of Nikuradse in terms of the relative roughness of commercially available pipe materials. It should be noted that the values of e D do not necessarily correspond to the actual values obtained by a microscopic determination of the average height of the roughness of the surface. They do, however, provide the correct correlation for f f1Re, eD2. It is important to observe that the values of relative roughness given pertain to new, clean pipes. After considerable use, most pipes 1because of a buildup of corrosion or scale2 may have a relative roughness that is considerably larger 1perhaps by an order of magnitude2 than that given. As shown by the upper figure in the margin, very old pipes may have enough scale buildup to not only alter the value of e but also to change their effective diameter by a considerable amount. The following characteristics are observed from the data of Fig. 8.20. For laminar flow, f 64 Re, which is independent of relative roughness. For turbulent flows with very large Reynolds numbers, f f1eD2, which, as shown by the figure in the margin, is independent of the Reynolds number. For such flows, commonly termed completely turbulent flow 1or wholly turbulent flow2, the laminar sublayer is so thin 1its thickness decreases with increasing Re2 that the surface roughness completely dominates the character of the flow near the wall. Hence, the pressure drop required is a Table 8.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7) and Colebrook (Ref. 8)] Equivalent Roughness, E Pipe Feet Millimeters Riveted steel Concrete Wood stave Cast iron Galvanized iron Commercial steel or wrought iron Drawn tubing Plastic, glass 0.003–0.03 0.001–0.01 0.0006–0.003 0.00085 0.0005 0.9–9.0 0.3–3.0 0.18–0.9 0.26 0.15 0.00015 0.000005 0.0 1smooth2 0.045 0.0015 0.0 1smooth2 430 Chapter 8 ■ Viscous Flow in Pipes 0.1 0.09 Wholly turbulent flow 0.08 0.05 0.04 0.07 0.06 0.03 0.05 0.02 0.015 0.04 0.01 0.008 0.006 f 0.03 0.004 e __ D 0.025 0.002 0.02 0.001 0.0008 0.0006 Laminar flow 0.0004 0.015 0.0002 Smooth Transition range 0.0001 0.00005 0.01 0.009 0.008 2(103) 103 4 2(104) 6 8 104 4 2(105) 6 8 4 105 6 2(106) 8 106 4 2(107) 6 8 4 6 8 0.00001 107 ρ VD Re = _____ μ ■ Figure 8.20 Friction factor as a function of Reynolds number and relative roughness for round pipes—the Moody chart. (Data from Ref. 7 with permission.) For any pipe, even smooth ones, the head loss is not zero. result of an inertia-dominated turbulent shear stress rather than the viscosity-dominated laminar shear stress normally found in the viscous sublayer. For flows with moderate values of Re, the friction factor is indeed dependent on both the Reynolds number and relative roughness— f f1Re, eD2. The gap in the figure for which no values of f are given 1the 2100 6 Re 6 4000 range2 is a result of the fact that the flow in this transition range may be laminar or turbulent 1or an unsteady mix of both2 depending on the specific circumstances involved. Note that even for smooth pipes 1e 02 the friction factor is not zero. That is, there is a head loss in any pipe, no matter how smooth the surface is made. This is a result of the no-slip boundary condition that requires any fluid to stick to any solid surface it flows over. There is always some microscopic surface roughness that produces the no-slip behavior 1and thus f 02 on the molecular level, even when the roughness is considerably less than the viscous sublayer thickness. Such pipes are called hydraulically smooth. Various investigators have attempted to obtain an analytical expression for f f1Re, eD2. Note that the Moody chart covers an extremely wide range in flow parameters. The nonlaminar region covers more than four orders of magnitude in Reynolds number—from Re 4 103 to Re 108. Obviously, for a given pipe and fluid, typical values of the average velocity do not cover this range. However, because of the large variety in pipes 1D2, fluids 1r and m2, and velocities 1V2, such a wide range in Re is needed to accommodate nearly all applications of pipe flow. In many cases the particular pipe flow of interest is confined to a relatively small region of the Moody chart, and simple semiempirical expressions can be developed for those conditions. For example, a company that manufactures cast iron water pipes with diameters between 2 and 12 in. may use a simple equation valid for their conditions only. The Moody chart, on the other hand, is universally valid for all steady, fully developed, incompressible pipe flows. 8.4 431 Dimensional Analysis of Pipe Flow The following equation from Colebrook is valid for the entire nonlaminar range of the Moody chart The turbulent portion of the Moody chart is represented by the Colebrook formula. 1 eD 2.51 2.0 log a b 3.7 1f Re1f (8.35a) In fact, the Moody chart is a graphical representation of this equation, which is an empirical fit of the pipe flow pressure drop data. Equation 8.35 is called the Colebrook formula. A difficulty with its use is that it is implicit in the dependence of f. That is, for given conditions 1Re and e D2, it is not possible to solve for f without some sort of iterative scheme. With the use of modern computers and calculators, such calculations are not difficult. A word of caution is in order concerning the use of the Moody chart or the equivalent Colebrook formula. Because of various inherent inaccuracies involved 1uncertainty in the relative roughness, uncertainty in the experimental data used to produce the Moody chart, etc.2, the use of several place accuracy in pipe flow problems is usually not justified. As a rule of thumb, a 10% accuracy is the best expected. It is possible to obtain an equation that adequately approximates the Colebrook兾Moody chart relationship but does not require an iterative scheme. For example, the Haaland equation (Ref. 34), which is easier to use, is given by 1 2f 1.8 log c a eD 1.11 6.9 b d 3.7 Re (8.35b) where one can solve for f explicitly. E XAMPLE 8.5 Comparison of Laminar or Turbulent Pressure Drop GIVEN Air under standard conditions flows through a 4.0-mm- diameter drawn tubing with an average velocity of V 50 ms. For such conditions the flow would normally be turbulent. However, if precautions are taken to eliminate disturbances to the flow 1the entrance to the tube is very smooth, the air is dust free, the tube does not vibrate, etc.2, it may be possible to maintain laminar flow. FIND (a) Determine the pressure drop in a 0.1-m section of the tube if the flow is laminar. (b) Repeat the calculations if the flow is turbulent. SOLUTION Under standard temperature and pressure conditions the density and viscosity are r 1.23 kgm3 and m 1.79 105 N # sm2. Thus, the Reynolds number is COMMENT Note that the same result is obtained from Eq. 8.8: ¢p 11.23 kgm 2150 ms210.004 m2 rVD 13,700 m 1.79 105 N # sm2 3 Re 179 Nm (a) If the flow were laminar, then f 64Re 6413,700 0.00467, and the pressure drop in a 0.1-m-long horizontal section of the pipe would be or / 1 rV 2 D2 10.1 m2 1 10.004672 11.23 kgm3 2150 ms2 2 10.004 m2 2 ¢p 0.179 kPa V D2 3211.79 105 N # sm2 210.1 m2150 ms2 2 which would normally indicate turbulent flow. ¢p f 32m/ (Ans) 10.004 m2 2 (b) If the flow were turbulent, then f f1Re, eD2, where from Table 8.1, e 0.0015 mm so that eD 0.0015 mm 4.0 mm 0.000375. From the Moody chart with Re 1.37 104 and eD 0.000375 we obtain f 0.028. Thus, the pressure drop in this case would be approximately ¢p f 10.1 m2 1 / 1 rV 2 10.0282 11.23 kgm3 2150 ms2 2 D2 10.004 m2 2 or ¢p 1.076 kPa (Ans) 432 Chapter 8 ■ Viscous Flow in Pipes COMMENT A considerable savings in effort to force the fluid through the pipe could be realized 10.179 kPa rather than 1.076 kPa2 if the flow could be maintained as laminar flow at this Reynolds number. In general this is very difficult to do, although laminar flow in pipes has been maintained up to Re ⬇ 100,000 in rare instances. An alternate method to determine the friction factor for the turbulent flow would be to use the Colebrook formula, Eq. 8.35a. Thus, 1 1f 2.0 log a eD 2.51 0.000375 2.51 b 2.0 log a b 3.7 3.7 Re1f 1.37 104 1f or 1.83 104 1 2.0 log a1.01 104 b 1f 1f (1) By using a root-finding technique on a computer or calculator, the solution to Eq. 1 is determined to be f 0.0291, in agreement 1within the accuracy of reading the graph2 with the Moody chart method of f 0.028. Equation 8.35b provides an alternate form to the Colebrook formula that can be used to solve for the friction factor directly. 1 1f 1.8 log ca 6.9 eD 1.11 6.9 0.000375 1.11 b d 1.8 log ca b d 3.7 Re 3.7 1.37 104 0.0289 This agrees with the Colebrook formula and Moody chart values obtained above. Numerous other empirical formulas can be found in the literature 1Ref. 52 for portions of the Moody chart. For example, an often- used equation, commonly referred to as the Blasius formula, for turbulent flow in smooth pipes 1eD 02 with Re 6 105 is f 0.316 Re14 For our case this gives f 0.316113,7002 0.25 0.0292 which is in agreement with the previous results. Note that the value of f is relatively insensitive to eD for this particular situation. Whether the tube was smooth glass 1eD 02 or the drawn tubing 1eD 0.0003752 would not make much difference in the pressure drop. For this flow, an increase in relative roughness by a factor of 30 to eD 0.0113 1equivalent to a commercial steel surface; see Table 8.12 would give f 0.043. This would represent an increase in pressure drop and head loss by a factor of 0.0430.0291 1.48 compared with that for the original drawn tubing. The pressure drop of 1.076 kPa in a length of 0.1 m of pipe corresponds to a change in absolute pressure [assuming p 101 kPa 1abs2 at x 0] of approximately 1.076101 0.0107, or about 1%. Thus, the incompressible flow assumption on which the above calculations 1and all of the formulas in this chapter2 are based is reasonable. However, if the pipe were 2-m long the pressure drop would be 21.5 kPa, approximately 20% of the original pressure. In this case the density would not be approximately constant along the pipe, and a compressible flow analysis would be needed. Such considerations are discussed in Chapter 11. 8.4.2 Minor Losses Losses due to pipe system components are given in terms of loss coefficients. As discussed in the previous section, the head loss in long, straight sections of pipe, the major losses, can be calculated by use of the friction factor obtained from either the Moody chart or the Colebrook equation. Most pipe systems, however, consist of considerably more than straight pipes. These additional components 1valves, bends, tees, and the like2 add to the overall head loss of the system. Such losses are generally termed minor losses, with the corresponding head loss denoted hL minor. In this section we indicate how to determine the various minor losses that commonly occur in pipe systems. The head loss associated with flow through a valve is a common minor loss. The purpose of a valve is to provide a means to regulate the flowrate. This is accomplished by changing the geometry of the system 1i.e., closing or opening the valve alters the flow pattern through the valve2, which in turn alters the losses associated with the flow through the valve. The flow resistance or head loss through the valve may be a significant portion of the resistance in the system. In fact, with the valve closed, the resistance to the flow is infinite—the fluid cannot flow. Such minor losses may be very important indeed. With the valve wide open the extra resistance due to the presence of the valve may or may not be negligible. The flow pattern through a typical component such as a valve is shown in Fig. 8.21. It is not difficult to realize that a theoretical analysis to predict the details of such flows to obtain the head loss for these components is not, as yet, possible. Thus, the head loss information for essentially all components is given in dimensionless form and based on experimental data. The most common method used to determine these head losses or pressure drops is to specify the loss coefficient, KL, which is defined as KL hL minor 1V 2g2 2 ¢p 1 2 2 rV 8.4 Dimensional Analysis of Pipe Flow Q 433 Q (b) (a) ■ Figure 8.21 Flow through a valve. so that ¢p KL 12rV 2 or hL minor KL hL, minor hL, minor ~ V 2 (8.36) The pressure drop across a component that has a loss coefficient of KL 1 is equal to the dynamic pressure, rV 2 2. As shown by Eq. 8.36 and the figure in the margin, for a given value of KL the head loss is proportional to the square of the velocity. The actual value of KL is strongly dependent on the geometry of the component considered. It may also be dependent on the fluid properties. That is, V For most flows the loss coefficient is independent of the Reynolds number. V2 2g KL f1geometry, Re2 where Re rVDm is the pipe Reynolds number. For many practical applications the Reynolds number is large enough so that the flow through the component is dominated by inertia effects, with viscous effects being of secondary importance. This is true because of the relatively large accelerations and decelerations experienced by the fluid as it flows along a rather curved, variable area 1perhaps even torturous2 path through the component 1see Fig. 8.212. In a flow that is dominated by inertia effects rather than viscous effects, it is usually found that pressure drops and head losses correlate directly with the dynamic pressure. This is why the friction factor for very large Reynolds number, fully developed pipe flow is independent of the Reynolds number. The same condition is found to be true for flow through pipe components. Thus, in most cases of practical interest the loss coefficients for components are a function of geometry only, KL f1geometry2. Minor losses are sometimes given in terms of an equivalent length, /eq. In this terminology, the head loss through a component is given in terms of the equivalent length of pipe that would produce the same head loss as the component. That is, hL minor KL /eq V 2 V2 f 2g D 2g or /eq KLD f 434 Chapter 8 ■ Viscous Flow in Pipes (a) (b) (c) (d) ■ Figure 8.22 Entrance flow conditions and loss coefficient (Data from Refs. 28, 29). (a) Reentrant, KL ⴝ 0.8, (b) sharp-edged, KL ⴝ 0.5, (c) slightly rounded, KL ⴝ 0.2 (see Fig. 8.24), (d) well-rounded, KL ⴝ 0.04 (see Fig. 8.24). Minor head losses are often a result of the dissipation of kinetic energy. where D and f are based on the pipe containing the component. The head loss of the pipe system is the same as that produced in a straight pipe whose length is equal to the pipes of the original system plus the sum of the additional equivalent lengths of all of the components of the system. Most pipe flow analyses, including those in this book, use the loss coefficient method rather than the equivalent length method to determine the minor losses. Many pipe systems contain various transition sections in which the pipe diameter changes from one size to another. Such changes may occur abruptly or rather smoothly through some type of area change section. Any change in flow area contributes losses that are not accounted for in the fully developed head loss calculation 1the friction factor2. The extreme cases involve flow into a pipe from a reservoir 1an entrance2 or out of a pipe into a reservoir 1an exit2. A fluid may flow from a reservoir into a pipe through any number of differently shaped entrance regions as are sketched in Fig. 8.22. Each geometry has an associated loss coefficient. A typical flow pattern for flow entering a pipe through a square-edged entrance is sketched in Fig. 8.23. As was discussed in Chapter 3, a vena contracta region may result because the fluid cannot turn a sharp right-angle corner. The flow is said to separate from the sharp corner. The maximum velocity at section 122 is greater than that in the pipe at section 132, and the pressure there is lower. If this high-speed fluid could slow down efficiently, the kinetic energy could be converted into pressure 1the Bernoulli effect2, and the ideal pressure distribution indicated in Fig. 8.23 would result. The head loss for the entrance would be essentially zero. Such is not the case. Although a fluid may be accelerated very efficiently, it is very difficult to slow down 1decelerate2 a fluid efficiently. Thus, the extra kinetic energy of the fluid at section 122 is partially lost because of viscous dissipation, so that the pressure does not return to the ideal value. An entrance head loss 1pressure drop2 is produced as is indicated in Fig. 8.23. The majority of this loss is due to inertia effects that are eventually dissipated by the shear stresses within the fluid. Only a small portion of the loss is due to the wall shear stress within the entrance region. The net effect is that the loss coefficient for a square-edged entrance is approximately KL 0.50. One-half of a velocity head is lost as the fluid enters the pipe. If the pipe protrudes into the tank 1a reentrant entrance2 as is shown in Fig. 8.22a, the losses are even greater. An obvious way to reduce the entrance loss is to round the entrance region as is shown in Fig. 8.22c, thereby reducing or eliminating the vena contracta effect. Typical values for the loss coefficient for entrances with various amounts of rounding of the lip are shown in Fig. 8.24. A significant reduction in KL can be obtained with only slight rounding. 8.4 Dimensional Analysis of Pipe Flow 435 Separated flow Vena contracta (1) V1 = 0 x (2) V > V 2 3 (3) Flow separation at corner (a) p1 ρV 23 ____ 2 Ideal full recovery of kinetic energy ρV 22 ____ ρV 23 KL ____ 2 2 p Actual x1 x2 p3 x3 x (b) ■ Figure 8.23 Flow pattern and pressure distribution for a sharp-edged entrance. 0.6 Sharp-edged entrance r 0.5 0.4 D KL 0.3 0.2 0.1 0 0 0.05 r __ D V8.12 Entrance/exit flows 0.1 ■ Figure 8.24 Entrance loss coefficient as a function of rounding of the inlet edge (Data from Ref. 9). A head loss 1the exit loss2 is also produced when a fluid flows from a pipe into a tank as is shown in Fig. 8.25. In these cases the entire kinetic energy of the exiting fluid 1velocity V12 is dissipated through viscous effects as the stream of fluid mixes with the fluid in the tank and eventually comes to rest 1V2 02. The exit loss from points 112 and 122 is therefore equivalent to one velocity head, or KL 1. Losses also occur because of a change in pipe diameter as is shown in Figs. 8.26 and 8.27. The sharp-edged entrance and exit flows discussed in the previous paragraphs are limiting cases of this type of flow with either A1 A2 q , or A1A2 0, respectively. The loss coefficient for a sudden contraction, KL hL 1V 22 2g2, is a function of the area ratio, A2 A1, as is shown in Fig. 8.26. The value of KL changes gradually from one extreme of a sharp-edged entrance 1A2 A1 0 with KL 0.502 to the other extreme of no area change 1A2 A1 1 with KL 02. In many ways, the flow in a sudden expansion is similar to exit flow. As is indicated in Fig. 8.28, the fluid leaves the smaller pipe and initially forms a jet-type structure as it enters the larger pipe. Within a few diameters downstream of the expansion, the jet becomes dispersed across the pipe, and fully developed flow becomes established again. In this process [between sections 122 and 132] a portion of the kinetic energy of the fluid is dissipated as a result of viscous effects. A squareedged exit is the limiting case with A1A2 0. A sudden expansion is one of the few components 1perhaps the only one2 for which the loss coefficient can be obtained by means of a simple analysis. To do this we consider the continuity 436 Chapter 8 ■ Viscous Flow in Pipes (2) (1) (a) (b) (c) (d) ■ Figure 8.25 Exit flow conditions and loss coefficient. (a) Reentrant, KL ⴝ 1.0, (b) sharp-edged, KL ⴝ 1.0, (c) slightly rounded, KL ⴝ 1.0, (d ) well-rounded, KL ⴝ 1.0. 0.6 A1 A2 V22 hL = KL ___ 2g 0.4 KL 0.2 0 0 0.2 0.4 0.6 0.8 1.0 A2/A1 ■ Figure 8.26 Loss coefficient for a sudden contraction (Ref. 10). 1.0 A1 0.8 A2 V12 hL = KL ___ 2g 0.6 KL 0.4 0.2 0 0 0.2 0.4 0.6 A1/A2 0.8 1.0 ■ Figure 8.27 Loss coefficient for a sudden expansion (Ref. 10). 8.4 Dimensional Analysis of Pipe Flow 437 Control volume a b V1 (1) V1 V3 c (2) (3) ■ Figure 8.28 Control volume used to calculate the loss coefficient for a sudden expansion. and momentum equations for the control volume shown in Fig. 8.28 and the energy equation applied between 122 and 132. We assume that the flow is uniform at sections 112, 122, and 132 and the pressure is constant across the left-hand side of the control volume 1 pa pb pc p1 2. The resulting three governing equations 1mass, momentum, and energy2 are A1V1 A3V3 p1A3 p3A3 rA3V3 1V3 V1 2 and The loss coefficient for a sudden expansion can be theoretically calculated. p1 V 21 p3 V 23 hL g g 2g 2g These can be rearranged to give the loss coefficient, KL hL 1V 21 2g2, as KL a1 where we have used the fact that A2 A3. This result, plotted in Fig. 8.27, is in good agreement with experimental data. As with so many minor loss situations, it is not the viscous effects directly 1i.e., the wall shear stress2 that cause the loss. Rather, it is the dissipation of kinetic energy 1another type of viscous effect2 as the fluid decelerates inefficiently. The losses may be quite different if the contraction or expansion is gradual. Typical results for a conical diffuser with a given area ratio, A2 A1, are shown in Fig. 8.29. 1A diffuser is a device shaped to decelerate a fluid.2 Clearly the included angle of the diffuser, u, is a very important parameter. For very small angles, the diffuser is excessively long and most of the head loss is due to the wall shear stress as in fully developed flow. For moderate or large angles, the flow separates from the walls and the losses are due mainly to a dissipation of the kinetic energy of the jet leaving the smaller diameter pipe. In fact, for moderate or large values of u 1i.e., u 7 35° for the case 1.4 1.2 1.0 KL ______ (1 – A1/A2)2 V8.13 Separated flow in a diffuser A1 2 b A2 0.8 V21 hL = KL ___ 2g 0.6 V1 0.4 A2 ___ fixed A1 V2 θ 0.2 0 0 30 60 90 θ , degrees 120 150 ■ Figure 8.29 Loss coefficient for a typical conical diffuser (Ref. 5). 180 438 Chapter 8 ■ Viscous Flow in Pipes V8.14 Car exhaust system Extensive tables are available for loss coefficients of standard pipe components. shown in Fig. 8.292, the conical diffuser is, perhaps unexpectedly, less efficient than a sharp-edged expansion which has KL 11 A1 A2 2 2. There is an optimum angle 1u ⬇ 8° for the case illustrated2 for which the loss coefficient is a minimum. The relatively small value of u for the minimum KL results in a long diffuser and is an indication of the fact that it is difficult to efficiently decelerate a fluid. It must be noted that the conditions indicated in Fig. 8.29 represent typical results only. Flow through a diffuser is very complicated and may be strongly dependent on the area ratio A2 A1, specific details of the geometry, and the Reynolds number. The data are often presented in terms of a pressure recovery coefficient, Cp 1 p2 p1 2 1rV 2122, which is the ratio of the static pressure rise across the diffuser to the inlet dynamic pressure. Considerable effort has gone into understanding this important topic 1Refs. 11, 122. Flow in a conical contraction 1a nozzle; reverse the flow direction shown in Fig. 8.292 is less complex than that in a conical expansion. Typical loss coefficients based on the downstream 1high-speed2 velocity can be quite small, ranging from KL 0.02 for u 30°, to KL 0.07 for u 60°, for example. It is relatively easy to accelerate a fluid efficiently. Bends in pipes produce a greater head loss than if the pipe were straight. The losses are due to the separated region of flow near the inside of the bend 1especially if the bend is sharp2 and the swirling secondary flow that occurs because of the imbalance of centripetal forces as a result of the curvature of the pipe centerline. These effects and the associated values of KL for large Reynolds number flows through a 90° bend are shown in Fig. 8.30. The friction loss due to the axial length of the pipe bend must be calculated and added to that given by the loss coefficient of Fig. 8.30. For situations in which space is limited, a flow direction change is often accomplished by use of miter bends, as is shown in Fig. 8.31, rather than smooth bends. The considerable losses in such bends can be reduced by the use of carefully designed guide vanes that help direct the flow with less unwanted swirl and disturbances. Another important category of pipe system components is that of commercially available pipe fittings such as elbows, tees, reducers, valves, and filters. The values of KL for such components depend strongly on the shape of the component and only very weakly on the Reynolds number for typical large Re flows. Thus, the loss coefficient for a 90° elbow depends on whether the pipe joints are threaded or flanged but is, within the accuracy of the data, fairly independent of the pipe diameter, flowrate, or fluid properties 1the Reynolds number effect2. Typical values of KL for such components are given in Table 8.2. These typical components are designed more for ease of manufacturing and costs than for reduction of the head losses that they produce. The flowrate from a faucet in a typical house is sufficient whether the value of KL for an elbow is the typical KL 1.5, or it is reduced to KL 0.2 by use of a more expensive long-radius, gradual bend 1Fig. 8.302. 1.0 b b Separated flow 0.8 a D 0.6 a Secondary flow 90° KL Primary flow 0.4 e __ D = 0.01 0.002 0.2 0.001 0 0 0 2 4 6 8 10 /D ■ Figure 8.30 Character of the flow in a 90ⴗ bend and the associated loss coefficient (Ref. 5). 12 8.4 Dimensional Analysis of Pipe Flow Guide vanes Q Q Separated flow KL ≈ 1.1 KL ≈ 0.2 (a) (b) ■ Figure 8.31 Character of the flow in a 90ⴗ mitered bend and the associated loss coefficient: (a) without guide vanes, (b) with guide vanes. Table 8.2 Loss Coefficients for Pipe Components ahL ⴝ KL Component a. Elbows Regular 90°, flanged Regular 90°, threaded Long radius 90°, flanged Long radius 90°, threaded Long radius 45°, flanged Regular 45°, threaded V2 b (Data from Refs. 5, 10, 27) 2g KL 0.3 1.5 0.2 0.7 0.2 0.4 V V 90° elbow 45° elbow b. 180ⴗ return bends 180° return bend, flanged 180° return bend, threaded 0.2 1.5 c. Tees Line flow, flanged Line flow, threaded Branch flow, flanged Branch flow, threaded 0.2 0.9 1.0 2.0 V d. Union, threaded 0.08 V Tee V Tee e. See Valves Globe, fully open Angle, fully open Gate, fully open Gate, 14 closed Gate, 12 closed Gate, 34 closed Swing check, forward flow Swing check, backward flow Ball valve, fully open Ball valve, 13 closed Ball valve, 23 closed Fig. 8.32 for typical valve geometry. 10 2 0.15 0.26 2.1 17 2 0.05 5.5 210 180° return bend V Union 439 440 Chapter 8 ■ Viscous Flow in Pipes ■ Figure 8.32 Internal structure of various valves: (a) globe valve, (b) gate valve, (c) swing check valve, (d ) stop check valve. (Courtesy of Crane Co., Fluid Handling Division.) F l u i d s i n Safety in fluids The following incident is taken from a NIOSH (National Institute for Occupational Safety and Health) fatality investigation report that emphasizes the dangers of oxygendeficient environments in confined spaces. A 35-year-old male water system operator (victim) was asphyxiated after entering a valve vault at a municipal water system plant. The victim was assigned to turn on a waterline valve serving a nearby tree farm. The valve was located at the water treatment plant inside an underground valve vault that “always had normal air.” The victim entered the valve vault through a ground-level manhole without t h e N e w s testing or ventilating the vault to atmosphere. A coworker, who had last seen the victim one hour earlier, checked the manhole and saw the victim lying on his back at the bottom. The victim did not respond to any calls. Other workers summoned from the plant building and local fire department personnel ventilated the valve vault and removed the victim. The vault atmosphere was subsequently found to be oxygen deficient. There were no witnesses to the incident, but evidence suggests that the victim lost consciousness and fell from the ladder railings to the bottom of the vault. Valves control the flowrate by providing a means to adjust the overall system loss coefficient to the desired value. When the valve is closed, the value of KL is infinite and no fluid flows. Opening of the valve reduces KL, producing the desired flowrate. Typical cross sections of various types of valves are shown in Fig. 8.32. Some valves 1such as the conventional globe valve2 are designed for general use, providing convenient control between the extremes of fully closed and fully open. Others 1such as a needle valve2 are designed to provide very fine control of the flowrate. The check valve provides a diode type operation that allows fluid to flow in one direction only. Loss coefficients for typical valves are given in Table 8.2. As with many system components, the head loss in valves is mainly a result of the dissipation of kinetic energy of a high-speed portion of the flow. This high speed, V3, is illustrated in Fig. 8.33. 8.4 V1 V3 >> V1 (1) (2) Dimensional Analysis of Pipe Flow 441 V2 = V1 ■ Figure 8.33 Head loss in a valve is due to dissipation of the kinetic energy of the large-velocity fluid near the valve seat. E XAMPLE 8.6 Minor Losses GIVEN The closed-circuit wind tunnel shown in Fig. E8.6a is a smaller version of that depicted in Fig. E8.6b in which air at standard conditions is to flow through the test section [between sections (5) and (6)] with a velocity of 200 ft/s. The flow is driven by a fan that essentially increases the static pressure by the amount p1 ⫺ p9 that is needed to overcome the head losses experienced by the fluid as it flows around the circuit. FIND Estimate the value of p1 ⫺ p9 and the horsepower supplied to the fluid by the fan. V5 = 200 ft/s (4) (6) (5) (7) Test section Flow-straightening screens (3) (8) ■ Figure E8.6a Q (1) (© SMART BLADE GmbH.) (9) (2) SOLUTION Fan The maximum velocity within the wind tunnel occurs in the test section 1smallest area; see Table E8.6 on the next page2. Thus, the maximum Mach number of the flow is Ma5 ⫽ V5 Ⲑc5, where V5 ⫽ 200 ftⲐs and from Eq. 1.20 the speed of sound is c5 ⫽1kRT5 2 1Ⲑ 2 ⫽ 51.4 11716 ft # lbⲐslug # °R2 3 1460 ⫹ 592 °R4 61Ⲑ2 ⫽ 1117 ft Ⲑs. Thus, Ma5 ⫽ 200Ⲑ1117 ⫽ 0.179. As was indicated in Chapter 3 and discussed fully in Chapter 11, most flows can be considered as incompressible if the Mach number is less than about 0.3. Hence, we can use the incompressible formulas for this problem. The purpose of the fan in the wind tunnel is to provide the necessary energy to overcome the net head loss experienced by the air as it flows around the circuit. This can be found from the energy equation between points 112 and 192 as p1 p9 V 21 V 29 ⫹ ⫹ z1 ⫽ ⫹ ⫹ z9 ⫹ hL1⫺9 g g 2g 2g where hL1⫺9 is the total head loss from 112 to 192. With z1 ⫽ z9 and V1 ⫽ V9 this gives p1 p9 ⫺ ⫽ hL1⫺9 g g (1) Similarly, by writing the energy equation 1Eq. 5.842 across the fan, from 192 to 112, we obtain p9 p1 V 29 V 21 ⫹ ⫹ ⫹ z9 ⫹ hp ⫽ ⫹ z1 g g 2g 2g ■ Figure E8.6 b where hp is the actual head rise supplied by the pump 1fan2 to the air. Again since z9 ⫽ z1 and V9 ⫽ V1 this, when combined with Eq. 1, becomes hp ⫽ 1 p1 ⫺ p9 2 g ⫽ hL1⫺9 The actual power supplied to the air 1horsepower, pa2 is obtained from the fan head by pa ⫽ gQhp ⫽ gA5V5hp ⫽ gA5V5hL1⫺9 (2) Thus, the power that the fan must supply to the air depends on the head loss associated with the flow through the wind tunnel. To obtain a reasonable, approximate answer we make the following assumptions. We treat each of the four turning corners as a mitered bend with guide vanes so that from Fig. 8.31 KL corner ⫽ 0.2. Thus, for each corner hL corner ⫽ KL V2 V2 ⫽ 0.2 2g 2g where, because the flow is assumed incompressible, V ⫽ V5 A5ⲐA. The values of A and the corresponding velocities throughout the tunnel are given in Table E8.6. We also treat the enlarging sections from the end of the test section 162 to the beginning of the nozzle 142 as a conical diffuser 442 Chapter 8 ■ Viscous Flow in Pipes Table E8.6 250 Area (ft ) Velocity (ft s) 1 2 3 4 5 6 7 8 9 22.0 28.0 35.0 35.0 4.0 4.0 10.0 18.0 22.0 36.4 28.6 22.9 22.9 200.0 200.0 80.0 44.4 36.4 200 a, hp Location 2 150 100 (200 ft/s, 62.3 hp) 50 0 0 50 100 150 200 250 300 V5, ft/s ■ Figure E8.6c with a loss coefficient of KL dif 0.6. This value is larger than that of a well-designed diffuser 1see Fig. 8.29, for example2. Since the wind tunnel diffuser is interrupted by the four turning corners and the fan, it may not be possible to obtain a smaller value of KL dif for this situation. Thus, hL dif KL dif V 26 2g 0.6 V 26 2g The loss coefficients for the conical nozzle between sections 142 and 152 and the flow-straightening screens are assumed to be KL noz 0.2 and KL scr 4.0 1Ref. 132, respectively. We neglect the head loss in the relatively short test section. Thus, the total head loss is hL19 hL corner7 hL corner8 hL corner2 hL corner3 hL dif hL noz hL scr or hL 19 3 0.21V 27 V 28 V 22 V 23 2 0.6V 26 0.2V 25 4.0V 24 4 2g 3 0.2180.02 44.42 28.62 22.92 2 0.612002 2 0.212002 2 4.0122.92 2 4 ft2 s2 32132.2 fts2 2 4 or hL19 560 ft Hence, from Eq. 1 we obtain the pressure rise across the fan as p1 p9 ghL19 10.0765 lbft3 21560 ft2 42.8 lbft2 0.298 psi (Ans) From Eq. 2 we obtain the power added to the fluid as pa 10.0765 lbft3 214.0 ft2 2 1200 fts21560 ft2 34,300 ft # lbs or pa 34,300 ft # lbs 62.3 hp 550 1ft # lbs2 hp (Ans) COMMENTS By repeating the calculations with various test section velocities, V5, the results shown in Fig. E8.6c are obtained. Since the head loss varies as V52 and the power varies as head loss times V5, it follows that the power varies as the cube of the velocity. Thus, doubling the wind tunnel speed requires an eightfold increase in power. With a closed-return wind tunnel of this type, all of the power required to maintain the flow is dissipated through viscous effects, with the energy remaining within the closed tunnel. If heat transfer across the tunnel walls is negligible, the air temperature within the tunnel will increase in time. For steady-state operations of such tunnels, it is often necessary to provide some means of cooling to maintain the temperature at acceptable levels. It should be noted that the actual size of the motor that powers the fan must be greater than the calculated 62.3 hp because the fan is not 100% efficient. The power calculated above is that needed by the fluid to overcome losses in the tunnel, excluding those in the fan. If the fan were 60% efficient, it would require a shaft power of p 62.3 hp 10.602 104 hp to run the fan. Determination of fan 1or pump2 efficiencies can be a complex problem that depends on the specific geometry of the fan. Introductory material about fan performance is presented in Chapter 12; additional material can be found in various references 1Refs. 14, 15, 16, for example2. It should also be noted that the above results are only approximate. Clever, careful design of the various components 1corners, diffuser, etc.2 may lead to improved 1i.e., lower2 values of the various loss coefficients, and hence lower power requirements. Since hL is proportional to V 2, the components with the larger V tend to have the larger head loss. Thus, even though KL 0.2 for each of the four corners, the head loss for corner 172 is 1V7V3 2 2 18022.92 2 12.2 times greater than it is for corner 132. 8.4.3 Noncircular Conduits Many of the conduits that are used for conveying fluids are not circular in cross section. Although the details of the flows in such conduits depend on the exact cross-sectional shape, many round pipe results can be carried over, with slight modification, to flow in conduits of other shapes. Theoretical results can be obtained for fully developed laminar flow in noncircular ducts, although the detailed mathematics often becomes rather cumbersome. For an arbitrary 8.4 z z A = cross-sectional Dimensional Analysis of Pipe Flow 443 V = u(y,z) area x y P = perimeter of pipe Dh = 4A/P = hydraulic diameter (a) (b) ■ Figure 8.34 Noncircular duct. The hydraulic diameter is used for noncircular duct calculations. cross section, as is shown in Fig. 8.34, the velocity profile is a function of both y and z 3V u1y, z2iˆ 4. This means that the governing equation from which the velocity profile is obtained 1either the Navier –Stokes equations of motion or a force balance equation similar to that used for circular pipes, Eq. 8.62 is a partial differential equation rather than an ordinary differential equation. Although the equation is linear 1for fully developed flow the convective acceleration is zero2, its solution is not as straightforward as for round pipes. Typically the velocity profile is given in terms of an infinite series representation 1Ref. 172. Practical, easy-to-use results can be obtained as follows. Regardless of the cross-sectional shape, there are no inertia effects in fully developed laminar pipe flow. Thus, the friction factor can be written as f C Reh, where the constant C depends on the particular shape of the duct, and Reh is the Reynolds number, Reh rVDh m, based on the hydraulic diameter. The hydraulic diameter defined as Dh 4AP is four times the ratio of the cross-sectional flow area divided by the wetted perimeter, P, of the pipe as is illustrated in Fig. 8.34. It represents a characteristic length that defines the size of a cross section of a specified shape. The factor of 4 is included in the definition of Dh so that for round pipes the diameter and hydraulic diameter are equal 3Dh 4AP 41pD242 1pD2 D4. The hydraulic diameter is also used in the definition of the friction factor, hL f 1/Dh 2V 22g, and the relative roughness, eDh. The values of C f Reh for laminar flow have been obtained from theory and or experiment for various shapes. Typical values are given in Table 8.3 along with the hydraulic diameter. Note Table 8.3 Friction Factors for Laminar Flow in Noncircular Ducts (Data from Ref. 18) Shape Parameter I. Concentric Annulus Dh = D2 – D1 C ⴝ f Reh D1D2 0.0001 0.01 0.1 0.6 1.00 71.8 80.1 89.4 95.6 96.0 a兾b 0 0.05 0.10 0.25 0.50 0.75 1.00 96.0 89.9 84.7 72.9 62.2 57.9 56.9 D1 D2 II. Rectangle 2ab Dh = _____ a+b a b 444 Chapter 8 ■ Viscous Flow in Pipes that the value of C is relatively insensitive to the shape of the conduit. Unless the cross section is very “thin” in some sense, the value of C is not too different from its circular pipe value, C 64. Once the friction factor is obtained, the calculations for noncircular conduits are identical to those for round pipes. Calculations for fully developed turbulent flow in ducts of noncircular cross section are usually carried out by using the Moody chart data for round pipes with the diameter replaced by the hydraulic diameter and the Reynolds number based on the hydraulic diameter. Such calculations are usually accurate to within about 15%. If greater accuracy is needed, a more detailed analysis based on the specific geometry of interest is needed. The Moody chart, developed for round pipes, can also be used for noncircular ducts. E XAMPLE 8.7 Noncircular Conduit GIVEN Air at a temperature of 120 ºF and standard pressure flows from a furnace through an 8-in.-diameter pipe with an average velocity of 10 ft/s. It then passes through a transition section similar to the one shown in Fig. E8.7 and into a square duct whose side is of length a. The pipe and duct surfaces are smooth 1e 02 . The head loss per foot is to be the same for the pipe and the duct. FIND Determine the duct size, a. SOLUTION V We first determine the head loss per foot for the pipe, hL / 1 fD2 V 2 2g, and then size the square duct to give the same value. For the given pressure and temperature we obtain 1from Table B.32 n 1.89 104 ft2s so that 110 fts21 128 ft2 VD 35,300 Re n 1.89 104 ft2s With this Reynolds number and with eD 0 we obtain the friction factor from Fig. 8.20 as f 0.022 so that 110 fts2 2 Thus, for the square duct we must have (1) where Dh 4AP 4a24a a and 2 p 8 a ftb 110 fts2 Q 4 12 3.49 2 Vs 2 A a a (2) is the velocity in the duct. By combining Eqs. 1 and 2 we obtain 0.0512 2 2 f 13.49a 2 a 2132.22 13.49a2 2a VsDh 1.85 104 4 n a 1.89 10 (4) We have three unknowns 1a, f, and Reh 2 and three equations— Eqs. 3, 4, and either in graphical form the Moody chart 1Fig. 8.202 or the Colebrook equation (Eq. 8.35a). If we use the Moody chart, we can use a trial-and-error solution as follows. As an initial attempt, assume the friction factor for the duct is the same as for the pipe. That is, assume f 0.022. From Eq. 3 we obtain a 0.606 ft, while from Eq. 4 we have Reh 3.05 104. From Fig. 8.20, with this Reynolds number and the given smooth duct we obtain f 0.023, which does not quite agree with the assumed value of f. Hence, we do not have the solution. We try again, using the latest calculated value of f 0.023 as our guess. The calculations are repeated until the guessed value of f agrees with the value obtained from Fig. 8.20. The final result 1after only two iterations2 is f 0.023, Reh 3.03 104, and a 0.611 ft 7.34 in. or a 1.30 f 15 where a is in feet. Similarly, the Reynolds number based on the hydraulic diameter is Reh hL 0.022 0.0512 8 / 1 12 ft2 2132.2 ft s2 2 f V 2s hL 0.0512 / Dh 2g ■ Figure E8.7 (Ans) COMMENTS Alternatively, we can use the Colebrook equa(3) tion or the Haaland equation (rather than the Moody chart) to obtain 8.5 the solution as follows. For a smooth pipe 1eDh 02 the Colebrook equation, Eq. 8.35a, becomes e Dh 2.51 1 2.0 log a b 3.7 1f Reh 1f 2.0 log a 2.51 b Reh 1f (5) where from Eq. 3, f 0.269 a5 (6) If we combine Eqs. 4, 5, and 6 and simplify, Eq. 7 is obtained for a. 1.928 a52 2 log 12.62 104 a32 2 8.5 (7) 445 Pipe Flow Examples By using a root-finding technique on a computer or calculator, the solution to Eq. 7 is determined to be a 0.614 ft, in agreement (given the accuracy of reading the Moody chart) with that obtained by the trial-and-error method given earlier. Note that the length of the side of the equivalent square duct is a D 7.348 0.918, or approximately 92% of the diameter of the equivalent duct. It can be shown that this value, 92%, is a very good approximation for any pipe flow—laminar or turbulent. The cross-sectional area of the duct 1A a2 53.9 in.2 2 is greater than that of the round pipe 1A pD2 4 50.3 in.2 2. Also, it takes less material to form the round pipe 1perimeter pD 25.1 in.2 than the square duct 1perimeter 4a 29.4 in.2. Circles are very efficient shapes. Pipe Flow Examples In the previous sections of this chapter, we discussed concepts concerning flow in pipes and ducts. The purpose of this section is to apply these ideas to the solutions of various practical problems. The application of the pertinent equations is straightforward, with rather simple calculations that give answers to problems of engineering importance. The main idea involved is to apply the energy equation between appropriate locations within the flow system, with the head loss written in terms of the friction factor and the minor loss coefficients. We will consider two classes of pipe systems: those containing a single pipe 1whose length may be interrupted by various components2, and those containing multiple pipes in parallel, series, or network configurations. Pipe systems may contain a single pipe with components or multiple interconnected pipes. F l u i d s i n New hi-tech fountains Ancient Egyptians used fountains in their palaces for decorative and cooling purposes. Current use of fountains continues but with a hi-tech flair. Although the basic fountain still consists of a typical pipe system (i.e., pump, pipe, regulating valve, nozzle, filter, and basin), recent use of computercontrolled devices has led to the design of innovative fountains with special effects. For example, by using several rows of multiple nozzles, it is possible to program and activate control valves to produce water jets that resemble symbols, letters, or the time of day. Other fountains use specially designed nozzles to produce t h e N e w s coherent, laminar streams of water that look like glass rods flying through the air. By using fast-acting control valves in a synchronized manner it is possible to produce mesmerizing three-dimensional patterns of water droplets. The possibilities are nearly limitless. With the initial artistic design of the fountain established, the initial engineering design (i.e., the capacity and pressure requirements of the nozzles and the size of the pipes and pumps) can be carried out. It is often necessary to modify the artistic and/or engineering aspects of the design in order to obtain a functional, pleasing fountain. (See Problem 8.60.) 8.5.1 Single Pipes II: Q = ? III: D = ? I: Δ p = ? The nature of the solution process for pipe flow problems can depend strongly on which of the various parameters are independent parameters 1the “given”2 and which is the dependent parameter 1the “determine”2. The three most common types of problems are shown in Table 8.4 in terms of the parameters involved. We assume the pipe system is defined in terms of the length of pipe sections used and the number of elbows, bends, and valves needed to convey the fluid between the desired locations. In all instances we assume the fluid properties are given. In a Type I problem we specify the desired flowrate or average velocity and determine the necessary pressure difference or head loss. For example, if a flowrate of 2.0 gal兾min is required for a dishwasher that is connected to the water heater by a given pipe system as shown by the figure in the margin, what pressure is needed in the water heater? In a Type II problem we specify the applied driving pressure 1or, alternatively, the head loss2 and determine the flowrate. For example, how many gal兾min of hot water are supplied to the dishwasher if the pressure within the water heater is 60 psi and the pipe system details 1length, diameter, roughness of the pipe; number of elbows; etc.2 are specified? 446 Chapter 8 ■ Viscous Flow in Pipes Table 8.4 Pipe Flow Types Pipe flow problems can be categorized by what parameters are given and what is to be calculated. Variable a. Fluid Density Viscosity b. Pipe Diameter Length Roughness c. Flow Flowrate or Average Velocity d. Pressure Pressure Drop or Head Loss Type I Type II Type III Given Given Given Given Given Given Given Given Given Given Given Given Determine Given Given Given Determine Given Determine Given Given In a Type III problem we specify the pressure drop and the flowrate and determine the diameter of the pipe needed. For example, what diameter of pipe is needed between the water heater and dishwasher if the pressure in the water heater is 60 psi 1determined by the city water system2 and the flowrate is to be not less than 2.0 gal兾min 1determined by the manufacturer2? Several examples of these types of problems follow. E XAMPLE 8.8 Type I, Determine Pressure Drop GIVEN Water at 60 ºF flows from the basement to the second KL = 2 based on floor through the 0.75-in. (0.0625-ft)-diameter copper pipe (a drawn tubing) at a rate of Q 12.0 galmin 0.0267 ft3s and exits through a faucet of diameter 0.50 in. as shown in Fig. E8.8a. 10 ft 0.75-in.-diameter copper pipe (a) all losses are neglected, (b) the only losses included are major losses, or (c) all losses are included. Since the fluid velocity in the pipe is given by V1 QA1 Q 1pD2 42 10.0267 ft3s2 3p 10.0625 ft2 24 4 8.70 ft s, and the fluid properties are r 1.94 slugsft3 and m 2.34 105 lb # sft2 1see Table B.12, it follows that Re rVDm 11.94 slugsft3 218.70 fts210.0625 ft2 12.34 105lb # sft2 2 45,000. Thus, the flow is turbulent. The governing equation for either case 1a2, 1b2, or 1c2 is the energy equation given by Eq. 8.21, Wide open globe valve 10 ft 0.50-in. diameter (5) Q= 12.0 gal/min 10 ft g (3) 15 ft Threaded 90° elbows ■ Figure E8.8a p1 gz2 12r1V 22 V 21 2 ghL (1) where the head loss is different for each of the three cases. (a) If all losses are neglected 1hL 02, Eq. 1 gives p1 162.4 lbft3 2120 ft2 p1 p2 V 21 V 22 z1 z2 hL a1 a2 g g 2g 2g where z1 0, z2 20 ft, p2 0 1free jet2, g rg 62.4 lbft3, and the outlet velocity is V2 QA2 10.0267 ft3s2 3 p10.50 122 2ft24 4 19.6 fts. We assume that the kinetic energy coefficients a1 and a2 are unity. This is reasonable because turbulent velocity profiles are nearly uniform across the pipe. Thus, (2) (4) (1) SOLUTION (6) 5 ft FIND Determine the pressure at point (1) if pipe velocity 10 ft (7) (8) 1.94 slugsft3 ft 2 ft 2 c a19.6 b a8.70 b d 2 s s 11248 2992 lbft2 1547 lbft2 or p1 10.7 psi (Ans) 8.5 COMMENT Note that for this pressure drop, the amount due to elevation change 1the hydrostatic effect2 is g1z2 z1 2 8.67 psi and the amount due to the increase in kinetic energy is r1V 22 V 21 2 2 2.07 psi. 27.1 p, psi 1 / V 21 r1V 22 V 21 2 rf 2 D 2 11248 2992 lbft2 18.70 fts2 2 60 ft b 11.94 slugsft3 210.02152 a 0.0625 ft 2 p1 gz2 10.7 Elevation and kinetic energy 0 0 6.37 3.09 6.37 4.84 2.07 2.07 10 20 30 40 p2 = 0 50 60 Distance along pipe from point (1), ft (3) (4) (5) (6) (7) (8) (2) ■ Figure E8.8b taching reservoir where the kinetic energy is zero. Thus, by combining Eqs. 2 and 3 we obtain the entire pressure drop as p1 121.3 9.172 psi 30.5 psi If major and minor losses are included, Eq. 1 becomes 1 / V 12 V2 r1V 22 V 21 2 fg a rKL 2 D 2g 2 or V2 p1 21.3 psi a rKL 2 (2) where the 21.3 psi contribution is due to elevation change, kinetic energy change, and major losses [part 1b2], and the last term represents the sum of all of the minor losses. The loss coefficients of the components 1KL 1.5 for each elbow and KL 10 for the wide-open globe valve2 are given in Table 8.2 1except for the loss coefficient of the faucet, which is given in Fig. E8.8a as KL 22. Thus, 18.70 ft2 2 V2 11.94 slugsft3 2 310 411.52 2 4 2 2 1321 lbft2 or (3) Note that we did not include an entrance or exit loss because points 112 and 122 are located within the fluid streams, not within an at- (Ans) This pressure drop calculated by including all losses should be the most realistic answer of the three cases considered. COMMENTS More detailed calculations will show that the pressure distribution along the pipe is as illustrated in Fig. E8.8b for cases 1a2 and 1c2—neglecting all losses or including all losses. Note that not all of the pressure drop, p1 p2, is a “pressure loss.” The pressure change due to the elevation and velocity changes is completely reversible. The portion due to the major and minor losses is irreversible. This flow can be illustrated in terms of the energy line and hydraulic grade line concepts introduced in Section 3.7. As is shown in Fig. E8.8c, for case 1a2 there are no losses and the energy line 1EL2 is horizontal, one velocity head 1V 22g2 above the hydraulic grade line 1HGL2, which is one pressure head 1gz2 above the pipe itself. For cases 1b2 or 1c2 the energy line is not horizontal. Each bit of friction in the pipe or loss in a component reduces the available 80 H, elevation to energy line, ft (Ans) COMMENT Of this pressure drop, the amount due to pipe friction is approximately 121.3 10.72 psi 10.6 psi. V2 rK 9.17 psi L a 2 9.93 10 or a rKL 11.7 12.4 10.7 Location: (1) 11248 299 15152 lbft2 3062 lbft2 p1 gz2 21.0 20.2 19.3 18.5 20 From Table 8.1 the roughness for a 0.75-in.-diameter copper pipe 1drawn tubing2 is e 0.000005 ft so that e D 8 105. With this eD and the calculated Reynolds number 1Re 45,0002, the value of f is obtained from the Moody chart as f 0.0215. Note that the Colebrook equation 1Eq. 8.35a2 would give the same value of f. Hence, with the total length of the pipe as / 115 5 10 10 202 ft 60 ft and the elevation and kinetic energy portions the same as for part 1a2, Eq. 1 gives (c) (a) No losses (c) Including all losses 27.8 Pressure loss / V 21 D 2g p1 21.3 psi 447 30.5 psi 30 (b) If the only losses included are the major losses, the head loss is hL f Pipe Flow Examples Slope due to pipe friction Sharp drop due to component loss 60 Energy line including all losses, case (c) 40 20 0 Energy line with no losses, case (a) 0 10 20 30 40 50 Distance along pipe from point (1), ft ■ Figure E8.8c 60 448 Chapter 8 ■ Viscous Flow in Pipes energy, thereby lowering the energy line. Thus, for case 1a2 the total head remains constant throughout the flow with a value of 18.70 fts2 2 11547 lbft2 2 p1 V 21 0 z1 g 2g 162.4 lbft3 2 2132.2 ft s2 2 26.0 ft. 2 p2 p3 V2 V 33 z2 z3 p g g 2g 2g The elevation of the energy line can be calculated at any point along the pipe. For example, at point 172, 50 ft from point 112, H H7 For case 1c2 the energy line starts at H1 19.93 1442 lbft2 162.4 lbft3 2 44.1 ft 18.70 ft s2 2 2132.2 fts2 2 20 ft The head loss per foot of pipe is the same all along the pipe. That is, V 21 p1 z1 g 2g 130.5 1442lbft2 162.4 lb ft3 2 18.70 ft s2 2 2132.2 ft s2 2 119.6 fts2 p2 20 ft z2 0 g 2g 2132.2 ft s2 2 26.0 ft 2 V 22 0.021518.70 fts2 2 hL V2 0.404 ftft f / 2gD 2132.2 ft s2 210.0625 ft2 0 71.6 ft and falls to a final value of H2 p7 V 27 z7 g 2g Thus, the energy line is a set of straight-line segments of the same slope separated by steps whose height equals the head loss of the minor component at that location. As is seen from Fig. E8.8c, the globe valve produces the largest of all the minor losses. Although the governing pipe flow equations are quite simple, they can provide very reasonable results for a variety of applications, as is shown in the next example. EXAMPLE 8.9 Type I, Determine Head Loss GIVEN A utility worker is fixing power lines in a manhole space at a busy intersection as shown in Fig. E8.9a. A major portion of car exhaust is carbon dioxide (CO2), which has a density of 3.55 103 slugs/ft3. Since the density of air at standard conditions is 2.38 103 slugs/ft3, the carbon dioxide will tend to settle into the bottom of the manhole space and displace the oxygen. This is a very relevant danger to utility workers and is further discussed in the related “Fluids in the News” article on page 440. To avoid suffocation, a fan is used to supply fresh air from the surface into the manhole space as shown in Fig. E8.9b. The air is routed through an 8-in.-diameter plastic hose that is 30 ft in length. FIND Determine the horsepower needed by the fan to overcome the head loss with a required flowrate of 600 cfm, when (a) neglecting minor losses and (b) including minor losses for one sharp-edged entrance, one exit, and one 90° miter bend (no guide vanes). (1) Fan V (2) ■ Figure E8.9a (© TIM MCCAIG/iStockphoto.) ■ Figure E8.9b 8.5 449 Pipe Flow Examples SOLUTION The purpose of the fan is to deliver air at a certain flowrate, which also requires the fan to overcome the net head loss within the pipe. This can be found from the energy equation between points (1) and (2) as p1 p2 V 21 V 22 z1 hp z2 hL g g 2g 2g where hL is the total head loss from (1) to (2) and hp is the head rise supplied by the fan to the air. Points (1) and (2) are assumed to be sufficiently away from the entrance and exit of the plumbing so that p1 p2 V1 V2 0. Also, since the fluid is air, the change in elevation is negligible. Thus, the energy equation reduces to hp hL. (a) Neglecting minor losses it follows that hP f / V2 D 2g 1.21 105 From Table 8.1 the equivalent roughness, e, for plastic is 0.0. Therefore, eD 0. Using this information with the Moody chart, we find the friction factor, f, to be 0.017. We now have enough information to solve Eq. 1. 128.6 ft s2 2 30 ft d 9.72 ft hP 10.0172 c dc 18122 ft 2132.2 ft s2 2 The actual power the fan must supply to the air (horsepower, pa) can be calculated by pa gQhP pa 7.44 s (2) 1 min b 19.72 ft2 60 s 1 hp a b 1.35 10 2 hp 550 ft # lbs (Ans) COMMENTS Typically, flexible hosing is used to route air into confined spaces. Minor losses were neglected in part (a), but Some pipe flow problems require a trial-and-error solution technique. hL minor 33.0 ft The power required to account for both the major and the minor losses is a 1 hp 1 min b 133 ft2 a b 60 s 550 ft # lbs (Ans) the horsepower required is more than double what is needed for a straight pipe. It is easy to see that in this situation neglecting minor losses would provide an underestimation of the power required to pump 600 cfm of air. The table that follows contains different power requirements for a few different pipe situations using the preceding flow properties. 12.38 10 3 slugs ft3 2128.6 ft s21 128 ft2 rVD m 3.74 10 7 lb # sft2 ft # lb s ft # lb 128.6 ft s2 2 V2 10.5 1.1 1.02 2g 2132.2 ft s2 2 COMMENTS Note that with the inclusion of minor losses, it follows that 7.44 hL minor gKL 5.94 10 2 hp Q 600 ft3min V 1719 ft min 28.6 ft s p A 1812 ft2 2 4 pa 17.65 102 lbft 3 2 1600 ft3min2 a (b) For the additional minor losses, we need to include the entrance, exit, and bend. The loss coefficient for an entrance with sharp edges in 0.5, the loss coefficient for a 90 bend with no guide vanes is 1.1, and the loss coefficient for the exit is 1.0. Therefore, the total minor loss is given as pa 1.35 10 2 hp 17.65 10 2 lbft3 21600 ft3 min2 (1) where V is the velocity within the hose. Therefore, we must use the Moody chart to find the friction factor, f. Since Re we need to be cautious in making this assumption with a relatively short overall pipe length. Pipe Setup Straight pipe Entrance and exit Entrance, exit, 1 miter bend Entrance, exit, 2 miter bends Entrance, exit, 4 miter bends hL minor (ft) hL major (ft) hP (ft) pa (hp) 0 9.72 9.72 0.014 19.1 9.72 28.8 0.040 33.0 9.72 42.7 0.059 47.0 9.72 56.7 0.079 74.9 9.72 84.7 0.118 Caution also needs to be taken when placing the fan outside the confined space so that clean air is supplied and hazardous gases emitted from the manhole are not circulated back into the confined space. There is also the question of whether to have air forced into the confined space or have the contaminated air exhausted from the space and naturally draw in fresh air. Pipe flow problems in which it is desired to determine the flowrate for a given set of conditions 1Type II problems2 often require trial-and-error or numerical root-finding techniques. This is because it is necessary to know the value of the friction factor to carry out the calculations, but the friction factor is a function of the unknown velocity 1flowrate2 in terms of the Reynolds number. The solution procedure is indicated in Example 8.10. 450 Chapter 8 ■ Viscous Flow in Pipes EXAMPLE 8.10 Type II, Determine Flowrate GIVEN The fan shown in Fig. E8.10a is to provide airflow through the spray booth and fume hood so that workers are protected from harmful vapors and aerosols while mixing chemicals within the hood. For proper operation of the hood, the flowrate is to be between 6 ft3/s and 12 ft3/s. With the initial setup the flowrate is 9 ft3/s, the loss coefficient for the system is 5, and the duct is short enough so that major losses are negligible. It is proposed that when the factory is remodeled the 8-in.-diameter galvanized iron duct will be 100 ft long and the total loss coefficient will be 10. 8-in.-diameter galvanized iron duct Fan Q (2) FIND Determine if the flowrate will be within the required 6 ft3/s to 12 ft3/s after the proposed remodeling. Assume that the head added to the air by the fan remains constant. ■ Figure E8.10a (1) SOLUTION We can determine the head that the fan adds to the air by considering the initial situation (i.e., before remodeling). To do this we write the energy equation between section (1) in the room and section (2) at the exit of the duct as shown in Fig. E8.10a. p1 p2 V21 V22 z1 hp z2 hL g g 2g 2g hp 2g hL 2ghp V R (1) Since we are dealing with air, we can assume any change in elevation is negligible. We can also assume the pressure inside the room and at the exit of the duct is equal to atmospheric pressure and the air in the room has zero velocity. Therefore, Eq. 1 reduces to V 22 where V2 V and gKL 10. We can now rearrange and solve for the velocity in ft/s. 1f / gKL D 2132.2 fts2 2161.9 ft2 100 ft R 1 f a 812 ft b 10 3990 B 11 150 f (3) The value of f is dependent on Re, which is dependent on V, an unknown. Re (2) 12.38 10 3 slugsft3 21V21 128 ft2 rVD m 3.74 10 7 lb # sft2 or The diameter of the duct is given as 8 in., so the velocity at the exit can be calculated from the flowrate, where V QA 19 ft3 s2 冤p18122 24冥 25.8 fts. For the original configuration the duct is short enough to neglect major losses and only minor losses contribute to the total head loss. This head loss can be found from hL, minor ©KLV2 12g2 5125.8 ft s2 2 3 2132.2 fts2 24 51.6 ft. With this information the simplified energy equation, Eq. 2, can now be solved to find the head added to the air by the fan. hp 125.8 fts2 2 2132.2 fts2 2 51.6 ft 61.9 ft The energy equation now must be solved with the new configuration after remodeling. Using the same assumptions as before gives the same reduced energy equation as shown in Eq. 2. With the increase in duct length to 100 ft the duct is no longer short enough to neglect major losses. Thus, hp V 22 / V2 V2 f a KL 2g D 2g 2g Re 4240V (4) where again V is in feet per second. Also, since eD 10.0005 ft2 1812 ft2 0.00075 (see Table 8.1 for the value of e2, we know which particular curve of the Moody chart is pertinent to this flow. Thus, we have three relationships (Eqs. 3, 4, and /D 0.00075 curve of Fig. 8.20) from which we can solve for the three unknowns, f, Re, and V. This is done easily by an iterative scheme as follows. It is usually simplest to assume a value of f, calculate V from Eq. 3, calculate Re from Eq. 4, and look up the new value of f in the Moody chart for this value of Re. If the assumed f and the new f do not agree, the assumed answer is not correct—we do not have the solution to the three equations. Although values of f, V, or Re could be assumed as starting values, it is usually simplest to assume a value of f because the correct value often lies on the relatively flat portion of the Moody chart for which f is quite insensitive to Re. 8.5 9.5 Thus, we assume f 0.019, approximately the large Re limit for the given relative roughness. From Eq. 3 we obtain Pipe Flow Examples 451 Original system 9 8.5 8 Q, ft3/s 3990 17.0 fts V B 11 15010.0192 and from Eq. 4 7.5 KL = 5 7 6.5 Re 4240(17.0) 72,100 With this Re and /D, Fig. 8.20 gives f 0.022, which is not equal to the assumed solution of f 0.019 (although it is close!). We try again, this time with the newly obtained value of f 0.022, which gives V 16.7 ft/s and Re 70,800. With these values, Fig. 8.20 gives f 0.022, which agrees with the assumed value. Thus, the solution is V 16.7 ft/s, or 2 8 p Q VA 116.7 fts2 a b a ftb 5.83 ft3s 4 12 (Ans) COMMENT It is seen that operation of the system after the proposed remodeling will not provide enough airflow to protect workers from inhalation hazards while mixing chemicals within the hood. By repeating the calculations for different duct lengths and different total minor loss coefficients, the results shown in Fig. E8.10b are obtained, which give flowrate as a function of E XAMPLE KL = 8 6 Minimum flowrate 5.5 needed 5 Proposed remodeled system 0 25 50 75 ᐉ, ft KL = 10 100 125 ■ Figure E8.10b duct length for various values of the minor loss coefficient. It will be necessary to redesign the remodeled system (e.g., larger fan, shorter ducting, larger-diameter duct) so that the flowrate will be within the acceptable range. In many companies, teams of occupational safety and health experts and engineers work together during the design phase of remodeling (or when a new operation is being planned) to consider and prevent potential negative impacts on workers’ safety and health in an effort called “Prevention through Design.” They also may be required to ensure that exhaust from such a system exits the building away from human activity and into an area where it will not be drawn back inside the facility. 8.11 Type II, Determine Flowrate GIVEN The turbine shown in Fig. E8.11 extracts 50 hp from z1 = 90 ft (1) the water flowing through it. The 1-ft-diameter, 300-ft-long pipe is assumed to have a friction factor of 0.02. Minor losses are negligible. 300-ft-long, 1-ft-diameter pipe Free jet Turbine FIND Determine the flowrate through the pipe and turbine. f = 0.02 (2) z2 = 0 ■ Figure E8.11 SOLUTION The energy equation 1Eq. 8.212 can be applied between the surface of the lake [point 112] and the outlet of the pipe as p1 p2 V 21 V 22 z1 z2 hL hT g g 2g 2g 1300 ft2 / V2 V2 0.0932V 2 ft hL f 0.02 D 2g 11 ft2 2132.2 fts2 2 where V is in ft兾s. Also, the turbine head is hT (1) where hT is the turbine head, p1 V1 p2 z2 0, z1 90 ft, and V2 V, the fluid velocity in the pipe. The head loss is given by 150 pa pa gQ g1p42D2V 150 hp2 3 1550 ft # lbs2 hp 4 162.4 lb ft3 2 3 1p4211 ft2 2V4 561 ft V Thus, Eq. 1 can be written as 90 V2 561 0.0932V 2 2132.22 V 452 Chapter 8 ■ Viscous Flow in Pipes or 0.109V 3 90V 561 0 (2) where V is in ft兾s. The velocity of the water in the pipe is found as the solution of Eq. 2. Surprisingly, there are two real, positive roots: V 6.58 fts or V 24.9 fts. The third root is negative 1V 31.4 fts2 and has no physical meaning for this flow. Thus, the two acceptable flowrates are Q p 2 p D V 11 ft2 2 16.58 ft s2 5.17 ft3s 4 4 (Ans) or Q p 11 ft2 2 124.9 ft s2 19.6 ft3 s 4 (Ans) COMMENTS Either of these two flowrates gives the same power, pa gQhT. The reason for two possible solutions can be seen from the following. With the low flowrate 1Q 5.17 ft3s2, we obtain the head loss and turbine head as hL 4.04 ft and hT 85.3 ft. Because of the relatively low velocity there is a relatively small head loss and, therefore, a large head available for the turbine. With the large flowrate 1Q 19.6 ft3s2, we find hL 57.8 ft and hT 22.5 ft. The high-speed flow in the pipe produces a relatively large loss due to friction, leaving a relatively small head for the turbine. However, in either case the product of the turbine head times the flowrate is the same. That is, the power extracted 1pa gQhT 2 is identical for each case. Although either flowrate will allow the extraction of 50 hp from the water, the details of the design of the turbine itself will depend strongly on which flowrate is to be used. Such information can be found in Chapter 12 and various references about turbomachines 1Refs. 14, 19, 202. If the friction factor were not given, the solution to the problem would be lengthier. A trial-and-error solution similar to that in Example 8.10 would be required, along with the solution of a cubic equation. In pipe flow problems for which the diameter is the unknown 1Type III2, an iterative or numerical root-finding technique is required. This is, again, because the friction factor is a function of the diameter—through both the Reynolds number and the relative roughness. Thus, neither Re rVDm 4rQ pmD nor eD are known unless D is known. Examples 8.12 and 8.13 illustrate this. E XAMPLE 8.12 Type III without Minor Losses, Determine Diameter GIVEN Air at standard temperature and pressure flows through a horizontal, galvanized iron pipe 1e 0.0005 ft2 at a rate of 2.0 ft3s. The pressure drop is to be no more than 0.50 psi per 100 ft of pipe. FIND Determine the minimum pipe diameter. SOLUTION We assume the flow to be incompressible with r 0.00238 slugsft3 and m 3.74 107 lb # sft2. Note that if the pipe were too long, the pressure drop from one end to the other, p1 p2, would not be small relative to the pressure at the beginning, and compressible flow considerations would be required. For example, a pipe length of 200 ft gives 1p1 p2 2 p1 310.50 psi2 1100 ft24 1200 ft2 14.7 psia 0.068 6.8%, which is probably small enough to justify the incompressible assumption. With z1 z2 and V1 V2 the energy equation 1Eq. 8.212 becomes / rV 2 p1 p2 f D 2 (1) where V QA 4Q 1pD 2 412.0 ft s2 pD , or 2 V 3 2.55 D2 2 where D is in feet. Thus, with p1 p2 10.5 lbin.2 2 1144 in.2 ft2 2 and / 100 ft, Eq. 1 becomes p1 p2 10.5211442 lbft2 f 1100 ft2 D 10.00238 slugsft3 2 1 2.55 ft 2 a b 2 D2 s or D 0.404 f 15 (2) where D is in feet. Also Re rVDm 10.00238 slugsft3 2 3 12.55D2 2 fts 4D 13.74 10 7 lb # sft2 2, or Re 1.62 104 D (3) 8.5 and Pipe Flow Examples 453 0.25 (4) Thus, we have four equations 3Eqs. 2, 3, 4, and either the Moody chart, the Colebrook equation (8.35a) or the Haaland equation (8.35b)] and four unknowns 1 f, D, eD, and Re2 from which the solution can be obtained by trial-and-error methods. If we use the Moody chart, it is probably easiest to assume a value of f, use Eqs. 2, 3, and 4 to calculate D, Re, and eD, and then compare the assumed f with that from the Moody chart. If they do not agree, try again. Thus, we assume f 0.02, a typical value, and obtain D 0.40410.022 15 0.185 ft, which gives eD 0.00050.185 0.0027 and Re 1.62 1040.185 8.76 104. From the Moody chart we obtain f 0.027 for these values of eD and Re. Since this is not the same as our assumed value of f, we try again. With f 0.027, we obtain D 0.196 ft, eD 0.0026, and Re 8.27 104, which in turn give f 0.027, in agreement with the assumed value. Thus, the diameter of the pipe should be D 0.196 ft (Ans) COMMENT If we use the Colebrook equation 1Eq. 8.35a2 with eD 0.0005 0.404 f 15 0.00124f 15 and Re 1.62 104 0.404 f 15 4.01 104f 15, we obtain 2.51 eD 1 2.0 log a b 3.7 1f Re 1f 0.20 (2 ft3/s, 0.196 ft) D, ft 0.0005 e D D 0.15 0.10 0.05 0 0 0.5 1 2 1.5 2.5 3 Q, ft3/s ■ Figure E8.12 By using a root-finding technique on a computer or calculator, the solution to this equation is determined to be f 0.027, and hence D 0.196 ft, in agreement with the Moody chart method. By repeating the calculations for various values of the flowrate, Q, the results shown in Fig. E8.12 are obtained. Although an increase in flowrate requires a larger diameter pipe (for the given pressure drop), the increase in diameter is minimal. For example, if the flowrate is doubled from 1 ft3 s to 2 ft3 s, the diameter increases from 0.151 ft to 0.196 ft. or 3.35 104 6.26 105 1 2.0 log a b f 15 f 310 1f In the previous example we only had to consider major losses. In some instances the inclusion of major and minor losses can cause a slightly lengthier solution procedure, even though the governing equations are essentially the same. This is illustrated in Example 8.13. E XAMPLE 8.13 Type III with Minor Losses, Determine Diameter GIVEN Water at 60 °F 1n 1.21 105 ft2s, see Table 1.52 is to flow from reservoir A to reservoir B through a pipe of length 1700 ft and roughness 0.0005 ft at a rate of Q 26 ft3s as shown in Fig. E8.13a. The system contains a sharp-edged entrance and four flanged 45° elbows. FIND Determine the pipe diameter needed. Elevation z1 = 44 ft (1) Total length = 1700 ft A Elevation z2 = 0 (2) D B ■ Figure E8.13a 454 Chapter 8 ■ Viscous Flow in Pipes SOLUTION The energy equation 1Eq. 8.212 can be applied between two points on the surfaces of the reservoirs 1p1 p2 V1 V2 z2 02 as follows: ment. A few rounds of calculation will reveal that the solution is given by p1 p2 V 21 V 22 z1 z2 hL g g 2g 2g COMMENTS Alternatively, we can use the Colebrook equa- z1 V2 / af K b 2g D a L (1) where V QA 4QpD2 4126 ft3 s2 pD2, or V 33.1 D2 (2) is the velocity within the pipe. 1Note that the units on V and D are fts and ft, respectively.2 The loss coefficients are obtained from Table 8.2 and Figs. 8.22 and 8.25 as KL ent 0.5, KL elbow 0.2, and KL exit 1. Thus, Eq. 1 can be written as 44 ft 1700 V2 e f 3410.22 0.5 14 f D 2132.2 ft s2 2 or, when combined with Eq. 2 to eliminate V, f 0.00152 D5 0.00135 D (3) To determine D we must know f, which is a function of Re and e D, where Re 3 133.12 D2 4D VD 2.74 106 5 n D 1.21 10 (4) and 0.0005 e D D (5) where D is in feet. Again, we have four equations 1Eqs. 3, 4, 5, and the Moody chart or the Colebrook equation2 for the four unknowns D, f, Re, and eD. Consider the solution by using the Moody chart. Although it is often easiest to assume a value of f and make calculations to determine if the assumed value is the correct one, with the inclusion of minor losses this may not be the simplest method. For example, if we assume f 0.02 and calculate D from Eq. 3, we would have to solve a fifth-order equation. With only major losses 1see Example 8.122, the term proportional to D in Eq. 3 is absent, and it is easy to solve for D if f is given. With both major and minor losses included, this solution for D 1given f 2 would require a trial-and-error or iterative technique. Thus, for this type of problem it is perhaps easier to assume a value of D, calculate the corresponding f from Eq. 3, and with the values of Re and eD determined from Eqs. 4 and 5, look up the value of f in the Moody chart 1or the Colebrook equation2. The solution is obtained when the two values of f are in agree- (Ans) tion rather than the Moody chart to solve for D. This is easily done by using the Colebrook equation (Eq. 8.35a) with f as a function of D obtained from Eq. 3 and Re and eD as functions of D from Eqs. 4 and 5. The resulting single equation for D can be solved by using a root-finding technique on a computer or calculator to obtain D 1 .63 ft. This agrees with the solution obtained using the Moody chart. By repeating the calculations for various pipe lengths, /, the results shown in Fig. E8.13b are obtained. As the pipe length increases it is necessary, because of the increased friction, to increase the pipe diameter to maintain the same flowrate. It is interesting to attempt to solve this example if all losses are neglected so that Eq. 1 becomes z1 0. Clearly from Fig. E8.13a, z1 44 ft. Obviously something is wrong. A fluid cannot flow from one elevation, beginning with zero pressure and velocity, and end up at a lower elevation with zero pressure and velocity unless energy is removed 1i.e., a head loss or a turbine2 somewhere between the two locations. If the pipe is short 1negligible friction2 and the minor losses are negligible, there is still the kinetic energy of the fluid as it leaves the pipe and enters the reservoir. After the fluid meanders around in the reservoir for some time, this kinetic energy is lost and the fluid is stationary. No matter how small the viscosity is, the exit loss cannot be neglected. The same result can be seen if the energy equation is written from the free surface of the upstream tank to the exit plane of the pipe, at which point the kinetic energy is still available to the fluid. In either case the energy equation becomes z1 V 22g in agreement with the inviscid results of Chapter 3 1the Bernoulli equation2. 1.8 1.6 (1700 ft, 1.63 ft) 1.4 1.2 D, ft or D ⬇ 1.63 ft 1.0 0.8 0.6 0.4 0.2 0.0 0 500 ■ Figure E8.13b 1000 ᐉ, ft 1500 2000 8.5 Pipe Flow Examples 455 8.5.2 Multiple Pipe Systems Trachea Lung Bronchiole F l u In many pipe systems there is more than one pipe involved. The complex system of tubes in our lungs 1beginning as shown by the figure in the margin, with the relatively large-diameter trachea and ending in tens of thousands of minute bronchioles after numerous branchings2 and the maze of pipes in a city’s water distribution system are typical of such systems. The governing mechanisms for the flow in multiple pipe systems are the same as for the single pipe systems discussed in this chapter. However, because of the numerous unknowns involved, additional complexities may arise in solving for the flow in multiple pipe systems. Some of these complexities are discussed in this section. i d s i n t h e N e w s forms and shore. The steel pipe used is 28 in. in diameter with a wall thickness of 1 1兾8 in. The thick-walled pipe is needed to withstand the large external pressure which is about 3250 psi at a depth of 7300 ft. The pipe is installed in 240-ft sections from a vessel the size of a large football stadium. Upon completion, the deepwater pipeline system will have a total length of more than 450 miles and the capability of transporting more than 1 million barrels of oil per day and 1.5 billion cu ft of gas per day. (See Problem 8.115.) Deepwater pipeline Pipelines used to transport oil and gas are commonplace. But south of New Orleans, in deep waters of the Gulf of Mexico, a not-so-common multiple pipe system is being built. The new so-called Mardi Gras system of pipes is being laid in water depths of 4300 to 7300 ft. It will transport oil and gas from five deepwater fields with the interesting names of Holstein, Mad Dog, Thunder Horse, Atlantis, and Na Kika. The deepwater pipelines will connect with lines at intermediate water depths to transport the oil and gas to shallow-water fixed plat- The simplest multiple pipe systems can be classified into series or parallel flows, as are shown in Fig. 8.35. The nomenclature is similar to that used in electrical circuits. Indeed, an analogy between fluid and electrical circuits is often made as follows. In a simple electrical circuit, there is a balance between the voltage 1e2, current 1i2, and resistance 1R2 as given by Ohm’s law: e iR. In a fluid circuit there is a balance between the pressure drop 1¢p2, the flowrate or velocity 1Q or V2, and the flow resistance as given in terms of the friction factor and minor loss coefficients 1 f and KL 2. ~ ~ For a simple flow 3¢p f 1/D21rV 222 4 , it follows that ¢p Q2R, where R, a measure of the resistance to the flow, is proportional to f. The main differences between the solution methods used to solve electrical circuit problems and those for fluid circuit problems lie in the fact that Ohm’s law is a linear equation 1doubling the voltage doubles the current2, while the fluid equations are generally nonlinear 1doubling the pressure drop does not double the flowrate unless the flow is laminar2. Thus, although some of the Q A V1 (1) V2 D1 V3 D2 (2) (3) D3 Q B (a) A D1 B V1 Q1 (1) V2 D2 Q2 (2) V3 D3 (3) Q3 ■ Figure 8.35 (a) Series and (b) parallel pipe (b) systems. 456 Chapter 8 ■ Viscous Flow in Pipes standard electrical engineering methods can be carried over to help solve fluid mechanics problems, others cannot. One of the simplest multiple pipe systems is that containing pipes in series, as is shown in Fig. 8.35a. Every fluid particle that passes through the system passes through each of the pipes. Thus, the flowrate 1but not the velocity2 is the same in each pipe, and the head loss from point A to point B is the sum of the head losses in each of the pipes. The governing equations can be written as follows: Q1 Q2 Q3 and hLA – B hL1 hL2 hL3 Series and parallel pipe systems are often encountered. where the subscripts refer to each of the pipes. In general, the friction factors will be different for each pipe because the Reynolds numbers 1Rei rViDim2 and the relative roughnesses 1eiDi 2 will be different. If the flowrate is given, it is a straightforward calculation to determine the head loss or pressure drop 1Type I problem2. If the pressure drop is given and the flowrate is to be calculated 1Type II problem2, an iteration scheme is needed. In this situation none of the friction factors, fi, are known, so the calculations may involve more trial-and-error attempts than for corresponding single pipe systems. The same is true for problems in which the pipe diameter 1or diameters2 is to be determined 1Type III problems2. Another common multiple pipe system contains pipes in parallel, as is shown in Fig. 8.35b. In this system a fluid particle traveling from A to B may take any of the paths available, with the total flowrate equal to the sum of the flowrates in each pipe. However, by writing the energy equation between points A and B it is found that the head loss experienced by any fluid particle traveling between these locations is the same, independent of the path taken. Thus, the governing equations for parallel pipes are Q Q1 Q2 Q3 and hL1 hL2 hL3 Again, the method of solution of these equations depends on what information is given and what is to be calculated. Another type of multiple pipe system called a loop is shown in Fig. 8.36. In this case the flowrate through pipe 112 equals the sum of the flowrates through pipes 122 and 132, or Q1 Q2 Q3. As can be seen by writing the energy equation between the surfaces of each reservoir, the head loss for pipe 122 must equal that for pipe 132, even though the pipe sizes and flowrates may be different for each. That is, pA pB V 2A V 2B zA zB hL1 hL2 g g 2g 2g for a fluid particle traveling through pipes 112 and 122, while pA V 2A pB V 2B zA zB hL1 hL3 g g 2g 2g A B Node, N D1 Q1 V2 D2 Q2 (2) V1 V3 (1) (3) D3 ■ Figure 8.36 Multiple-pipe loop system. Q3 8.5 457 Pipe Flow Examples A B D1, ᐉ1 (1) D2, ᐉ2 (2) C D3, ᐉ3 ■ Figure 8.37 A three- (3) reservoir system. For some pipe systems, the direction of flow is not known a priori. E XAMPLE for fluid that travels through pipes 112 and 132. These can be combined to give hL2 hL3. This is a statement of the fact that fluid particles that travel through pipe 122 and particles that travel through pipe 132 all originate from common conditions at the junction 1or node, N2 of the pipes and all end up at the same final conditions. The flow in a relatively simple looking multiple pipe system may be more complex than it appears initially. The branching system termed the three-reservoir problem shown in Fig. 8.37 is such a system. Three reservoirs at known elevations are connected together with three pipes of known properties 1length, diameter, and roughness2. The problem is to determine the flowrates into or out of the reservoirs. If valve 112 were closed, the fluid would flow from reservoir B to C, and the flowrate could be easily calculated. Similar calculations could be carried out if valves 122 or 132 were closed with the others open. With all valves open, however, it is not necessarily obvious which direction the fluid flows. For the conditions indicated in Fig. 8.37, it is clear that fluid flows from reservoir A because the other two reservoir levels are lower. Whether the fluid flows into or out of reservoir B depends on the elevation of reservoirs B and C and the properties 1length, diameter, roughness2 of the three pipes. In general, the flow direction is not obvious, and the solution process must include the determination of this direction. This is illustrated in Example 8.14. 8.14 Three-Reservoir, Multiple Pipe System GIVEN Three reservoirs are connected by three pipes as are shown in Fig. E8.14. For simplicity we assume that the diameter of each pipe is 1 ft, the friction factor for each is 0.02, and because of the large length-to-diameter ratio, minor losses are negligible. A Elevation = 100 ft D1 = 1 ft (1) ᐉ = 1000 ft 1 FIND Determine the flowrate into or out of each reservoir. D2 = 1 ft ᐉ2 = 500 ft B Elevation = 20 ft (2) SOLUTION It is not obvious which direction the fluid flows in pipe 122. However, we assume that it flows out of reservoir B, write the governing equations for this case, and check our assumption. The continuity equation requires that Q1 Q2 Q3, which, since the diameters are the same for each pipe, becomes simply V1 V2 V3 ᐉ3 = 400 ft (3) C Elevation = 0 ft ■ Figure E8.14 By using the fact that pA pC VA VC zC 0, this becomes (1) The energy equation for the fluid that flows from A to C in pipes 112 and 132 can be written as pA pC V 2A V 2C /1 V 21 /3 V 23 zA zC f1 f3 g g 2g 2g D1 2g D3 2g D3 = 1 ft zA f1 /1 V 21 /3 V 23 f3 D1 2g D3 2g For the given conditions of this problem we obtain 100 ft 1 0.02 3 11000 ft2V 21 1400 ft2V 23 4 2 2132.2 ft s 2 11 ft2 458 Chapter 8 ■ Viscous Flow in Pipes or which, with the given data, become 322 V 21 0.4V 23 (2) where V1 and V3 are in ft兾s. Similarly the energy equation for fluid flowing from B and C is V 2B V 2C /2 V 22 /3 V 23 pB pC zB zC f2 f3 g g 2g 2g D2 2g D3 2g 258 V 21 0.5 V 22 (5) 322 V 21 0.4 V 23 (6) and Equations 4, 5, and 6 can be solved as follows. By subtracting Eq. 5 from 6 we obtain V3 2160 1.25V 22 or zB f2 /2 V 22 /3 V 23 f3 D2 2g D3 2g Thus, Eq. 5 can be written as 258 1V2 V3 2 2 0.5V 22 1V2 2160 1.25V 22 2 2 0.5V 22 For the given conditions this can be written as 64.4 0.5V 22 0.4V 23 (3) Equations 1, 2, and 3 1in terms of the three unknowns V1, V2, and V32 are the governing equations for this flow, provided the fluid flows from reservoir B. It turns out, however, that there is no solution for these equations with positive, real values of the velocities. Although these equations do not appear to be complicated, there is no simple way to solve them directly. Thus, a trial-and-error solution is suggested. This can be accomplished as follows. Assume a value of V1 7 0, calculate V3 from Eq. 2, and then V2 from Eq. 3. It is found that the resulting V1, V2, V3 trio does not satisfy Eq. 1 for any value of V1 assumed. There is no solution to Eqs. 1, 2, and 3 with real, positive values of V1, V2, and V3. Thus, our original assumption of flow out of reservoir B must be incorrect. To obtain the solution, assume the fluid flows into reservoirs B and C and out of A. For this case the continuity equation becomes or 2V2 2160 1.25V 22 98 2.75V 22 which, upon squaring both sides, can be written as V 42 460 V 22 3748 0 By using the quadratic formula, we can solve for V 22 to obtain either V 22 452 or V 22 8.30. Thus, either V2 21.3 fts or V2 2.88 fts. The value V2 21.3 fts is not a root of the original equations. It is an extra root introduced by squaring Eq. 7, which with V2 21.3 becomes “1140 1140.” Thus, V2 2.88 fts and from Eq. 5, V1 15.9 fts. The corresponding flowrates are Q1 A1V1 or (4) Application of the energy equation between points A and B and A and C gives zA zB f1 /1 V 21 /2 V 22 f2 D1 2g D2 2g and /1 V 21 /3 V 23 zA zC f1 f3 D1 2g D3 2g Pipe network problems can be solved using node and loop concepts. p 2 p D1V1 11 ft2 2 115.9 fts2 4 4 12.5 ft3s from A (Ans) p 2 p D2V2 11 ft2 2 12.88 fts2 4 4 2.26 ft3s into B (Ans) Q3 Q1 Q2 112.5 2.262 ft3 s 10.2 ft3s into C (Ans) Q2 A2V2 Q1 Q2 Q3 V1 V2 V3 (7) and Note the slight differences in the governing equations depending on the direction of the flow in pipe 122—compare Eqs. 1, 2, and 3 with Eqs. 4, 5, and 6. COMMENT If the friction factors were not given, a trial-anderror procedure similar to that needed for Type II problems 1see Section 8.5.12 would be required. The ultimate in multiple pipe systems is a network of pipes such as that shown in Fig. 8.38. Networks like these often occur in city water distribution systems and other systems that may have multiple “inlets” and “outlets.” The direction of flow in the various pipes is by no means obvious—in fact, it may vary in time, depending on how the system is used from time to time. The solution for pipe network problems is often carried out by use of node and loop equations similar in many ways to that done in electrical circuits. For example, the continuity equation requires that for each node 1the junction of two or more pipes2 the net flowrate is zero. What flows into a node must flow out at the same rate. In addition, the net pressure difference completely around a loop 1starting at one location in a pipe and returning to that location2 must be zero. By combining these ideas with the usual head loss and pipe flow equations, the flow throughout the entire network can 8.6 459 Pipe Flowrate Measurement ■ Figure 8.38 A general pipe network. be obtained. Of course, trial-and-error solutions are usually required because the direction of flow and the friction factors may not be known. Such a solution procedure using matrix techniques is ideally suited for computer use 1Refs. 21, 222. 8.6 Pipe Flowrate Measurement It is often necessary to determine experimentally the flowrate in a pipe. In Chapter 3 we introduced various types of flow-measuring devices 1Venturi meter, nozzle meter, orifice meter, etc.2 and discussed their operation under the assumption that viscous effects were not important. In this section we will indicate how to account for the ever-present viscous effects in these flowmeters. We will also indicate other types of commonly used flowmeters. 8.6.1 Pipe Flowrate Meters Orifice, nozzle, and Venturi meters involve the concept “high velocity gives low pressure.” Three of the most common devices used to measure the instantaneous flowrate in pipes are the orifice meter, the nozzle meter, and the Venturi meter. As was discussed in Section 3.6.3, each of these meters operates on the principle that a decrease in flow area in a pipe causes an increase in velocity that is accompanied by a decrease in pressure. Correlation of the pressure difference with the velocity provides a means of measuring the flowrate. In the absence of viscous effects and under the assumption of a horizontal pipe, application of the Bernoulli equation 1Eq. 3.72 between points 112 and 122 shown in Fig. 8.39 gave Qideal A2V2 A2 21 p1 p2 2 B r11 b4 2 (8.37) where b D2 D1. Based on the results of the previous sections of this chapter, we anticipate that there is a head loss between 112 and 122 so that the governing equations become Q A1V1 A2V2 and p1 V 21 p2 V 22 hL g g 2g 2g The ideal situation has hL 0 and results in Eq. 8.37. The difficulty in including the head loss is that there is no accurate expression for it. The net result is that empirical coefficients are used in the flowrate equations to account for the complex real-world effects brought on by the nonzero viscosity. The coefficients are discussed in this section. Q D1 V1 (1) V2 (2) D2 ■ Figure 8.39 Typical pipe flowmeter geometry. 460 Chapter 8 ■ Viscous Flow in Pipes Pressure taps Q D1 = D (1) d A0 A2 A1 (2) D2 ■ Figure 8.40 Typical orifice meter construction. An orifice discharge coefficient is used to account for nonideal effects. A typical orifice meter is constructed by inserting between two flanges of a pipe a flat plate with a hole, as shown in Fig. 8.40. The pressure at point 122 within the vena contracta is less than that at point 112. Nonideal effects occur for two reasons. First, the vena contracta area, A2, is less than the area of the hole, Ao, by an unknown amount. Thus, A2 Cc Ao, where Cc is the contraction coefficient 1Cc 6 12. Second, the swirling flow and turbulent motion near the orifice plate introduce a head loss that cannot be calculated theoretically. Thus, an orifice discharge coefficient, Co, is used to take these effects into account. That is, Q CoQideal Co Ao Q ~ p1 p2 Q p1 p 2 21 p1 p2 2 B r11 b4 2 (8.38) where Ao pd 24 is the area of the hole in the orifice plate. The value of Co is a function of b dD and the Reynolds number, Re rVDm, where V QA1. Typical values of Co are given in Fig. 8.41. As shown by Eq. 8.38 and the figure in the margin, for a given value of Co, the flowrate is proportional to the square root of the pressure difference. Note that the value of Co depends on the specific construction of the orifice meter 1i.e., the placement of the pressure taps, whether the orifice plate edge is square or beveled, etc.2. Very precise conditions governing the construction of standard orifice meters have been established to provide the greatest accuracy possible 1Refs. 23, 242. Another type of pipe flowmeter that is based on the same principles used in the orifice meter is the nozzle meter, three variations of which are shown in Fig. 8.42. This device uses a contoured nozzle 1typically placed between flanges of pipe sections2 rather than a simple 1and less expensive2 plate with a hole as in an orifice meter. The resulting flow pattern for the nozzle meter is closer to ideal than the orifice meter flow. There is only a slight vena contracta and the secondary 0.66 D D 0.64 D __ 2 V d Co 0.62 d = 0.7 β = __ D 0.6 0.5 0.4 0.2 0.60 0.58 4 10 105 106 Re = ρVD/μ 107 108 ■ Figure 8.41 Orifice meter discharge coefficient (Data from Ref. 24). 8.6 Pipe Flowrate Measurement 461 d D Pressure taps (a) (b) (c) ■ Figure 8.42 Typical nozzle meter construction. The nozzle meter is more efficient than the orifice meter. flow separation is less severe, but there still are viscous effects. These are accounted for by use of the nozzle discharge coefficient, Cn, where Q CnQideal Cn An 21 p1 p2 2 B r11 b4 2 (8.39) with An pd 24. As with the orifice meter, the value of Cn is a function of the diameter ratio, b dD, and the Reynolds number, Re rVDm. Typical values obtained from experiments are shown in Fig. 8.43. Again, precise values of Cn depend on the specific details of the nozzle design. Accepted standards have been adopted 1Ref. 242. Note that Cn 7 Co; the nozzle meter is more efficient 1less energy dissipated2 than the orifice meter. The most precise and most expensive of the three obstruction-type flowmeters is the Venturi meter shown in Fig. 8.44 [G. B. Venturi (1746–1822)]. Although the operating principle for this device is the same as for the orifice or nozzle meters, the geometry of the Venturi meter is designed to reduce head losses to a minimum. This is accomplished by providing a relatively streamlined contraction 1which eliminates separation ahead of the throat2 and a very gradual expansion downstream of the throat 1which eliminates separation in this decelerating portion of the device2. Most of the head loss that occurs in a well-designed Venturi meter is due to friction losses along the walls rather than losses associated with separated flows and the inefficient mixing motion that accompanies such flow. 1.00 0.2 0.4 0.6 0.98 d = 0.8 β = __ D Cn 0.96 D 0.94 4 10 105 V 106 d 107 Re = ρVD/μ D d Q ■ Figure 8.44 Typical Venturi meter construction. 108 ■ Figure 8.43 Nozzle meter discharge coefficient (Data from Ref. 24). 462 Chapter 8 ■ Viscous Flow in Pipes 1.00 0.98 Range of values depending on specific geometry Cv 0.96 0.94 104 105 106 107 108 Re = ρ VD/ μ ■ Figure 8.45 Venturi meter discharge coefficient (Ref. 23). Thus, the flowrate through a Venturi meter is given by Q CvQideal CAT The Venturi discharge coefficient is a function of the specific geometry of the meter. E XAMPLE 21p1 p2 2 B r11 b4 2 (8.40) where AT pd 24 is the throat area. The range of values of Cv, the Venturi discharge coefficient, is given in Fig. 8.45. The throat-to-pipe diameter ratio 1b dD2, the Reynolds number, and the shape of the converging and diverging sections of the meter are among the parameters that affect the value of Cv. Again, the precise values of Cn, Co, and Cv depend on the specific geometry of the devices used. Considerable information concerning the design, use, and installation of standard flowmeters can be found in various books 1Refs. 23, 24, 25, 26, 312. 8.15 Nozzle Flowmeter GIVEN Ethyl alcohol flows through a pipe of diameter FIND Determine the diameter, d, of the nozzle. D 60 mm in a refinery. The pressure drop across the nozzle meter used to measure the flowrate is to be ¢p 4.0 kPa when the flowrate is Q 0.003 m3s. SOLUTION From Table 1.6 the properties of ethyl alcohol are r 789 kgm3 and m 1.19 103 N # sm2. Thus, Re rVD 4rQ m pDm Hence, with an initial guess of d 0.0346 m or b dD 0.03460.06 0.577, we obtain from Fig. 8.43 1using Re 42,2002 a value of Cn 0.972. Clearly this does not agree with our initial assumption of Cn 1.0. Thus, we do not have the solution to Eq. 1 and Fig. 8.43. Next we assume b 0.577 and Cn 0.972 and solve for d from Eq. 1 to obtain 214 103 Nm2 2 p 2 d 4 B 789 kgm3 11 b4 2 or C nd 2 21 b4 (2) d 11.20 103 2 12 0.0346 m From Eq. 8.39 the flowrate through the nozzle is 1.20 103 d 11.20 103 21 b4 2 12 In addition, for many cases 1 b4 ⬇ 1, so that an approximate value of d can be obtained from Eq. 2 as 41789 kg m3 2 10.003 m3s2 42,200 p10.06 m2 11.19 103 N # sm2 2 Q 0.003 m3 s Cn As a first approximation we assume that the flow is ideal, or Cn 1.0, so that Eq. 1 becomes (1) where d is in meters. Note that b dD d0.06. Equation 1 and Fig. 8.43 represent two equations for the two unknowns d and Cn that must be solved by trial and error. da 12 1.20 103 21 0.5774 b 0.972 or d 0.0341 m. With the new value of b 0.03410.060 0.568 and Re 42,200, we obtain 1from Fig. 8.432 Cn ⬇ 0.972 in 8.6 agreement with the assumed value. Thus, 60 d=D (Ans) 50 COMMENTS If numerous cases are to be investigated, it may There are many types of flowmeters. D d 40 Mercury Carbon tet 20 Water 30 Alcohol d, mm be much easier to replace the discharge coefficient data of Fig. 8.43 by the equivalent equation, Cn f1b, Re2, and use a computer to iterate for the answer. Such equations are available in the literature 1Ref. 242. This would be similar to using the Colebrook equation rather than the Moody chart for pipe friction problems. By repeating the calculations, the nozzle diameters, d, needed for the same flowrate and pressure drop but with different fluids are shown in Fig. E8.15. The diameter is a function of the fluid viscosity because the nozzle coefficient, Cn, is a function of the Reynolds number (see Fig. 8.43). In addition, the diameter is a function of the density because of this Reynolds number effect and, perhaps more importantly, because the density is involved directly in the flowrate equation, Eq. 8.39. These factors all combine to produce the results shown in the figure. Gasoline d 34.1 mm 463 Pipe Flowrate Measurement 10 0 ■ Figure E8.15 Numerous other devices are used to measure the flowrate in pipes. Many of these devices use principles other than the high-speed/low-pressure concept of the orifice, nozzle, and Venturi meters. A quite common, accurate, and relatively inexpensive flowmeter is the rotameter, or variable area meter as is shown in Fig. 8.46. In this device a float is contained within a tapered, transparent metering tube that is attached vertically to the pipeline. As fluid flows through the meter 1entering at the bottom2, the float will rise within the tapered tube and reach an equilibrium height that is a function of the flowrate. This height corresponds to an equilibrium condition for which the net force on the float 1buoyancy, float weight, fluid drag2 is zero. A calibration scale in the tube provides the relationship between the float position and the flowrate. Q V8.15 Rotameter Float at large end of tube indicates maximum flowrate Position of edge of float against scale gives flowrate reading Tapered metering tube Metering float is freely suspended in process fluid Float at narrow end of tube indicates minimum flowrate Q ■ Figure 8.46 Rotameter-type flowmeter. (Courtesy of ABB.) 464 Chapter 8 ■ Viscous Flow in Pipes Magnetic sensor O T U Flow out IN Flow in ■ Figure 8.47 FLOW Turbine Turbine-type flowmeter. (Courtesy of FTI Flow Technology, Inc.) Another useful pipe flowrate meter is a turbine meter as is shown in Fig. 8.47. A small, freely rotating propeller or turbine within the turbine meter rotates with an angular velocity that is a function of 1nearly proportional to2 the average fluid velocity in the pipe. This angular velocity is picked up magnetically and calibrated to provide a very accurate measure of the flowrate through the meter. 8.6.2 Volume Flowmeters Volume flowmeters measure volume rather than volume flowrate. In many instances it is necessary to know the amount 1volume or mass2 of fluid that has passed through a pipe during a given time period, rather than the instantaneous flowrate. For example, we are interested in how many gallons of gasoline are pumped into the tank in our car rather than the rate at which it flows into the tank. Numerous quantity-measuring devices provide such information. The nutating disk meter shown in Fig. 8.48 is widely used to measure the net amount of water used in domestic and commercial water systems as well as the amount of gasoline delivered to your gas tank. This meter contains only one essential moving part and is relatively inexpensive and accurate. Its operating principle is very simple, but it may be difficult to understand its operation without actually inspecting the device firsthand. The device consists of a metering chamber with spherical sides and conical top and bottom. A disk passes through a central sphere and divides the chamber into two portions. The disk is constrained to be at an angle not normal to the axis of symmetry of the chamber. A radial plate 1diaphragm2 divides the chamber so that the entering fluid causes the disk to wobble 1nutate2, with fluid flowing alternately above or below the disk. The fluid exits the chamber after the disk has completed one wobble, which corresponds to a specific volume of fluid passing through the chamber. During each wobble of the disk, the pin attached to the tip V8.16 Water meter Pin Metering Chamber Flow In Flow Out Casing Diaphram Sphere ■ Figure 8.48 Nutating disk flowmeter. (Courtesy of Badger Meter, Inc. © 2011.) 8.7 Chapter Summary and Study Guide Inlet 465 Outlet Diaphragm-driven slider valves Left case Right case Left diaphragm Right diaphragm (a) (b) (c) (d) Center partition ■ Figure 8.49 Bellows-type flowmeter. (Based on information from BTR—Rockwell Gas Products.) (a) Back case emptying, back diaphragm filling. (b) Front diaphragm filling, front case emptying. (c) Back case filling, back diaphragm emptying. (d) Front diaphragm emptying, front case filling. The nutating disk meter has only one moving part; the bellows meter has a complex set of moving parts. 8.7 of the center sphere, normal to the disk, completes one circle. The volume of fluid that has passed through the meter can be obtained by counting the number of revolutions completed. Another quantity-measuring device that is used for gas flow measurements is the bellows meter as shown in Fig. 8.49. It contains a set of bellows that alternately fill and empty as a result of the pressure of the gas and the motion of a set of inlet and outlet valves. The common household natural gas meter 1see photograph in the margin2 is of this type. For each cycle [1a2 through 1d2] a known volume of gas passes through the meter. The nutating disk meter 1water meter2 is an example of extreme simplicity—one cleverly designed moving part. The bellows meter 1gas meter2, on the other hand, is relatively complex—it contains many moving, interconnected parts. This difference is dictated by the application involved. One measures a common, safe-to-handle, relatively high-pressure liquid, whereas the other measures a relatively dangerous, low-pressure gas. Each device does its intended job very well. There are numerous devices used to measure fluid flow, only a few of which have been discussed here. The reader is encouraged to review the literature to gain familiarity with other useful, clever devices 1Refs. 25, 262. Chapter Summary and Study Guide This chapter discussed the flow of a viscous fluid in a pipe. General characteristics of laminar, turbulent, fully developed, and entrance flows are considered. Poiseuille’s equation is obtained to describe the relationship among the various parameters for fully developed laminar flow. 466 Chapter 8 ■ Viscous Flow in Pipes laminar flow transitional flow turbulent flow entrance length fully developed flow wall shear stress Poiseuille’s law friction factor turbulent shear stress major loss minor loss relative roughness Moody chart Colebrook formula loss coefficient hydraulic diameter multiple pipe systems orifice meter nozzle meter Venturi meter Various characteristics of turbulent pipe flow are introduced and contrasted to laminar flow. It is shown that the head loss for laminar or turbulent pipe flow can be written in terms of the friction factor (for major losses) and the loss coefficients (for minor losses). In general, the friction factor is obtained from the Moody chart or the Colebrook formula and is a function of the Reynolds number and the relative roughness. The minor loss coefficients are a function of the flow geometry for each system component. Analysis of noncircular conduits is carried out by use of the hydraulic diameter concept. Various examples involving flow in single pipe systems and flow in multiple pipe systems are presented. The inclusion of viscous effects and losses in the analysis of orifice, nozzle, and Venturi flowmeters is discussed. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. determine which of the following types of flow will occur: entrance flow or fully developed flow; laminar flow or turbulent flow. use the Poiseuille equation in appropriate situations and understand its limitations. explain the main properties of turbulent pipe flow and how they are different from or similar to laminar pipe flow. use the Moody chart and the Colebrook equation to determine major losses in pipe systems. use minor loss coefficients to determine minor losses in pipe systems. determine the head loss in noncircular conduits. incorporate major and minor losses into the energy equation to solve a variety of pipe flow problems, including Type I problems (determine the pressure drop or head loss), Type II problems (determine the flow rate), and Type III problems (determine the pipe diameter). solve problems involving multiple pipe systems. determine the flowrate through orifice, nozzle, and Venturi flowmeters as a function of the pressure drop across the meter. Some of the important equations in this chapter are: Entrance length /e 0.06 Re for laminar flow D /e 4.4 1Re2 16 for turbulent flow D Pressure drop for fully developed laminar pipe flow ¢p Velocity profile for fully developed laminar pipe flow u1r2 a Volume flowrate for fully developed laminar horizontal pipe flow Q Friction factor for fully developed laminar pipe flow Pressure drop for a horizontal pipe Head loss due to major losses f 4/tw D ¢pD2 2r 2 2r 2 b c 1 a b d Vc c 1 a b d 16m/ D D pD4 ¢p 128m/ 64 Re ¢p f hL major f (8.1) (8.2) (8.5) (8.7) (8.9) (8.19) / rV 2 D 2 (8.33) / V2 D 2g (8.34) References 467 Colebrook formula 1 eD 2.51 2.0 log a b 3.7 1f Re1f (8.35a) Haaland formula 1 eD 1.11 6.9 1.8 log ca b d 3.7 Re 1f (8.35b) Head loss due to minor losses Volume flowrate for orifice, nozzle, or Venturi meter hL minor KL V2 2g Q Ci Ai (8.36) 21 p1 p2 2 B r11 b4 2 (8.38, 8.39, 8.40) References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 34. Hinze, J. O., Turbulence, 2nd Ed., McGraw-Hill, New York, 1975. Panton, R. L., Incompressible Flow, 3rd Ed., Wiley, New York, 2005. Schlichting, H., Boundary Layer Theory, 8th Ed., McGraw-Hill, New York, 2000. Gleick, J., Chaos: Making a New Science, Viking Penguin, New York, 1987. White, F. M., Fluid Mechanics, 6th Ed., McGraw-Hill, New York, 2008. Nikuradse, J., “Stomungsgesetz in Rauhen Rohren,” VDI-Forschungsch, No. 361, 1933; or see NACA Tech Memo 1922. Moody, L. F., “Friction Factors for Pipe Flow,” Transactions of the ASME, Vol. 66, 1944. Colebrook, C. F., “Turbulent Flow in Pipes with Particular Reference to the Transition Between the Smooth and Rough Pipe Laws,” Journal of the Institute of Civil Engineers London, Vol. 11, 1939. ASHRAE Handbook of Fundamentals, ASHRAE, Atlanta, 1981. Streeter, V. L., ed., Handbook of Fluid Dynamics, McGraw-Hill, New York, 1961. Sovran, G., and Klomp, E. D., “Experimentally Determined Optimum Geometries for Rectilinear Diffusers with Rectangular, Conical, or Annular Cross Sections,” in Fluid Mechanics of Internal Flow, Sovran, G., ed., Elsevier, Amsterdam, 1967. Runstadler, P. W., “Diffuser Data Book,” Technical Note 186, Creare, Inc., Hanover, N.H., 1975. Laws, E. M., and Livesey, J. L., “Flow Through Screens,” Annual Review of Fluid Mechanics, Vol. 10, Annual Reviews, Inc., Palo Alto, Calif., 1978. Balje, O. E., Turbomachines: A Guide to Design, Selection and Theory, Wiley, New York, 1981. Wallis, R. A., Axial Flow Fans and Ducts, Wiley, New York, 1983. Karassick, I. J. et al., Pump Handbook, 2nd Ed., McGraw-Hill, New York, 1985. White, F. M., Viscous Fluid Flow, 3rd Ed., McGraw-Hill, New York, 2006. Olson, R. M., Essentials of Engineering Fluid Mechanics, 4th Ed., Harper & Row, New York, 1980. Dixon, S. L., Fluid Mechanics of Turbomachinery, 3rd Ed., Pergamon, Oxford, UK, 1978. Finnemore, E. J., and Franzini, J. R., Fluid Mechanics, 10th Ed., McGraw-Hill, New York, 2002. Streeter, V. L., and Wylie, E. B., Fluid Mechanics, 8th Ed., McGraw-Hill, New York, 1985. Jeppson, R. W., Analysis of Flow in Pipe Networks, Ann Arbor Science Publishers, Ann Arbor, Mich., 1976. Bean, H. S., ed., Fluid Meters: Their Theory and Application, 6th Ed., American Society of Mechanical Engineers, New York, 1971. “Measurement of Fluid Flow by Means of Orifice Plates, Nozzles, and Venturi Tubes Inserted in Circular Cross Section Conduits Running Full,” Int. Organ. Stand. Rep. DIS-5167, Geneva, 1976. Goldstein, R. J., ed., Flow Mechanics Measurements, 2nd Ed., Taylor and Francis, Philadelphia, 1996. Benedict, R. P., Measurement of Temperature, Pressure, and Flow, 2nd Ed., Wiley, New York, 1977. Hydraulic Institute, Engineering Data Book, 1st Ed., Cleveland Hydraulic Institute, 1979. Harris, C. W., University of Washington Engineering Experimental Station Bulletin, 48, 1928. Hamilton, J. B., University of Washington Engineering Experimental Station Bulletin, 51, 1929. Miller, D. S., Internal Flow Systems, 2nd Ed., BHRA, Cranfield, UK, 1990. Spitzer, D. W., ed., Flow Measurement: Practical Guides for Measurement and Control, Instrument Society of America, Research Triangle Park, N.C., 1991. Wilcox, D. C., Turbulence Modeling for CFD, DCW Industries, Inc., La Canada, California, 1994. Mullin, T., ed., The Nature of Chaos, Oxford University Press, Oxford, UK, 1993. Haaland, S.E., “Simple and Explicit Formulas for the Friction-Factor in Turbulent Pipe Flow,” Transactions of the ASME, Journal of Fluids Engineering, Vol. 105, 1983. 468 Chapter 8 ■ Viscous Flow in Pipes Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www.wiley. com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013 John Wiley and Sons, Inc.). Conceptual Questions 8.1C For fully developed viscous flow in a horizontal pipe, which of the following is true? a) Pressure forces are balanced by shear forces. b) Pressure forces result in fluid acceleration. c) Shear forces result in fluid deceleration. d) Pressure forces are balanced by body forces. Flow P1 P2 0 Length = ᐉ A 8.2C There is a steady laminar flow of water in a pipe of length /. B Pressure drop (P2−P1) C Flow D E Length = ᐉ 0 As the volume flowrate increases, the pressure drop over the length /. a) becomes larger. b) becomes smaller. c) stays the same. 8.3C Water flows steadily through a horizontal circular pipe. Which curve most correctly describes the pressure change through the pipe as the length is increased? Length,ᐉ 8.4C Water flows steadily through a smooth pipe. For turbulent flow, if the velocity is increased, in general a) the friction factor increases. b) the friction factor decreases. c) the friction factor stays the same. Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the evennumbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 8.1 General Characteristics of Pipe Flow (also see Lab Problem 8.5LP) †8.1 Under normal circumstances is the airflow though your trachea (your windpipe) laminar or turbulent? List all assumptions and show all calculations. 469 Problems 8.2 Rainwater runoff from a parking lot flows through a 3-ftdiameter pipe, completely filling it. Whether flow in a pipe is laminar or turbulent depends on the value of the Reynolds number. (See Video V8.2.) Would you expect the flow to be laminar or turbulent? Support your answer with appropriate calculations. 8.3 Blue and yellow streams of paint at 60 F (each with a density of 1.6 slugs/ft3 and a viscosity 1000 times greater than water) enter a pipe with an average velocity of 4 ft/s as shown in Fig. P8.3. Would you expect the paint to exit the pipe as green paint or separate streams of blue and yellow paint? Explain. Repeat the problem if the paint were “thinned” so that it is only 10 times more viscous than water. Assume the density remains the same. Yellow Green? 2 in. diameter is 4 cm, find the maximum and average velocities in the pipe as well as the volume flow rate. 8.11 The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 1.85 lb/ft2. Determine the pressure gradient, 0p 0x, where x is in the flow direction, if the pipe is (a) horizontal, (b) vertical with flow up, or (c) vertical with flow down. 8.12 The pressure drop needed to force water through a horizontal 1-in.-diameter pipe is 0.60 psi for every 12-ft length of pipe. Determine the shear stress on the pipe wall. Determine the shear stress at distances 0.3 and 0.5 in. away from the pipe wall. 8.13 Repeat Problem 8.12 if the pipe is on a 20 hill. Is the flow up or down the hill? Explain. 8.14 Water flows in a constant-diameter pipe with the following conditions measured: At section (a) pa 32.4 psi and za 56.8 ft; at section (b) pb 29.7 psi and zb 68.2 ft. Is the flow from (a) to (b) or from (b) to (a)? Explain. Splitter Blue 25 ft ■ Figure P8.3 8.4 Air at 200 F flows at standard atmospheric pressure in a pipe at a rate of 0.08 lb/s. Determine the minimum diameter allowed if the flow is to be laminar. 8.5 To cool a given room it is necessary to supply 4 ft3/s of air through an 8-in.-diameter pipe. Approximately how long is the entrance length in this pipe? 8.6 The flow of water in a 3-mm-diameter pipe is to remain laminar. Plot a graph of the maximum flowrate allowed as a function of temperature for 0 T 100 C. 8.7 Carbon dioxide at 20 C and a pressure of 550 kPa (abs) flows in a pipe at a rate of 0.04 N/s. Determine the maximum diameter allowed if the flow is to be turbulent. 8.8 The pressure distribution measured along a straight, horizontal portion of a 50-mm-diameter pipe attached to a tank is shown in the table below. Approximately how long is the entrance length? In the fully developed portion of the flow, what is the value of the wall shear stress? x (m) (ⴞ0.01 m) p (mm H2O) (ⴞ5 mm) 0 (tank exit) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 (pipe exit) 520 427 351 288 236 188 145 109 73 36 0 8.15 Some fluids behave as a non-Newtonian power-law fluid characterized by ␶ C(du/dr)n, where n 1, 3, 5, and so on, and C is a constant. (If n 1, the fluid is the customary Newtonian fluid.) (a) For flow in a round pipe of a diameter D, integrate the force balance equation (Eq. 8.3) to obtain the velocity profile u1r2 1n12n ¢p 1n 1n12 n n aDb a b cr d 1n 12 2/C 2 (b) Plot the dimensionless velocity profile u/Vc, where Vc is the centerline velocity (at r 0), as a function of the dimensionless radial coordinate r/(D/2), where D is the pipe diameter. Consider values of n 1, 3, 5, and 7. 8.16 For laminar flow in a round pipe of diameter D, at what distance from the centerline is the actual velocity equal to the average velocity? 8.17 Water at 20 C flows through a horizontal 1-mm-diameter tube to which are attached two pressure taps a distance 1 m apart. (a) What is the maximum pressure drop allowed if the flow is to be laminar? (b) Assume the manufacturing tolerance on the tube diameter is D 1.0 0.1 mm. Given this uncertainty in the tube diameter, what is the maximum pressure drop allowed if it must be assured that the flow is laminar? 8.18 GO Glycerin at 20 C flows upward in a vertical 75-mmdiameter pipe with a centerline velocity of 1.0 m/s. Determine the head loss and pressure drop in a 10-m length of the pipe. 8.19 A large artery in a person’s body can be approximated by a tube of diameter 9 mm and length 0.35 m. Also assume that blood has a viscosity of approximately 4 103 N s/m2, a specific gravity of 1.0, and that the pressure at the beginning of the artery is equivalent to 120 mm Hg. If the flow were steady (it is not) with V 0.2 m/s, determine the pressure at the end of the artery if it is oriented (a) vertically up (flow up) or (b) horizontal. 8.20 At time t 0 the level of water in tank A shown in Fig. P8.20 is 2 ft above that in tank B. Plot the elevation of the water in tank A 8.9 (See Fluids in the News article titled “Nanoscale Flows,” Section 8.1.1.) (a) Water flows in a tube that has a diameter of D 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per second. (b) Repeat the calculations if the tube is a nanoscale tube with a diameter of D 100 nm. 3 ft Section 8.2 Fully Developed Laminar Flow B 3 ft 2 ft at t = 0 25 ft 8.10 For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by u(r) 2 (1 r2/R2) in m/s, where R is the inner radius of the pipe. Assuming that the pipe A 0.1-in. diameter, galvanized iron ■ Figure P8.20 470 Chapter 8 ■ Viscous Flow in Pipes as a function of time until the free surfaces in both tanks are at the same elevation. Assume quasisteady conditions—that is, the steady pipe flow equations are assumed valid at any time, even though the flowrate does change (slowly) in time. Neglect minor losses. Note: Verify and use the fact that the flow is laminar. 8.21 A fluid flows through a horizontal 0.1-in.-diameter pipe. When the Reynolds number is 1500, the head loss over a 20-ft length of the pipe is 6.4 ft. Determine the fluid velocity. 8.22 A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If the flow is laminar, determine the centerline velocity and the flowrate. 10-mm-diameter 0.12 m 0.25-mm-diameter 0.10-m-long needle 8.23 Oil flows through the horizontal pipe shown in Fig. P8.23 under laminar conditions. All sections are the same diameter except one. Which section of the pipe (A, B, C, D, or E) is slightly smaller in diameter than the others? Explain. 15 ft 5 ft 10 ft 6 ft 46 in. 60 in. Q patm = 101 kPa (abs) 15 ft 39 in. 26 in. 56 in. A B C D E 20-foot sections ■ Figure P8.23 8.24 Asphalt at 120 ⬚F, considered to be a Newtonian fluid with a viscosity 80,000 times that of water and a specific gravity of 1.09, flows through a pipe of diameter 2.0 in. If the pressure gradient is 1.6 psi/ft determine the flowrate assuming the pipe is (a) horizontal; (b) vertical with flow up. 8.25 Oil of SG ⫽ 0.87 and a kinematic viscosity ␯ ⫽ 2.2 ⫻ 10⫺4 m2/s flows through the vertical pipe shown in Fig. P8.25 at a rate of 4 ⫻ 10⫺4 m3/s. Determine the manometer reading, h. ■ Figure P8.26 Section 8.3 Fully Developed Turbulent Flow 8.27 For oil (SG ⫽ 0.86, ␮ ⫽ 0.025 Ns/m2) flow of 0.3 m3/s through a round pipe with diameter of 500 mm, determine the Reynolds number. Is the flow laminar or turbulent? 8.28 For air at a pressure of 200 kPa (abs) and temperature of 15 ⬚C, determine the maximum laminar volume flowrate for flow through a 2.0-cm-diameter tube. 8.29 If the velocity profile for turbulent flow in a pipe is approximated by the power-law profile (Eq. 8.31), at what radial location should a Pitot tube be placed if it is to measure the average velocity in the pipe? Assume n ⫽ 7, 8, or 9. 8.30 As shown in Video V8.10 and Fig. P8.30, the velocity profile for laminar flow in a pipe is quite different from that for turbulent flow. With laminar flow the velocity profile is parabolic; with turbulent flow at Re ⫽ 10,000 the velocity profile can be approximated by the power-law profile shown in the figure. (a) For laminar flow, determine at what radial location you would place a Pitot tube if it is to measure the average velocity in the pipe. (b) Repeat part (a) for turbulent flow with Re ⫽ 10,000. SG = 0.87 Turbulent with Re = 10,000 r 1/5 u = 1 – __ __ 1.0 Vc [ R ] r __ R 20 mm Laminar with Re < 2100 u = 1 – __ r 2 __ 4m (R) Vc 0.5 u r R h SG = 1.3 Vc 0 Q ■ Figure P8.25 8.26 A liquid with SG ⫽ 0.96, ␮ ⫽ 9.2 ⫻ 10⫺4 N ⭈ s/m2, and vapor pressure p␯ ⫽ 1.2 ⫻ 104 N/m2(abs) is drawn into the syringe as is indicated in Fig P8.26. What is the maximum flowrate if cavitation is not to occur in the syringe? 0.5 u __ 1.0 Vc ■ Figure P8.30 8.31 GO Water at 10 °C flows through a smooth 60-mm-diameter pipe with an average velocity of 8 m/s. Would a layer of rust of height 0.005 mm on the pipe wall protrude through the viscous sublayer? Explain. Problems 8.32 When soup is stirred in a bowl, there is considerable turbulence in the resulting motion (see Video V8.7). From a very simplistic standpoint, this turbulence consists of numerous intertwined swirls, each involving a characteristic diameter and velocity. As time goes by, the smaller swirls (the fine scale structure) die out relatively quickly, leaving the large swirls that continue for quite some time. Explain why this is to be expected. 8.33 Determine the thickness of the viscous sublayer in a smooth 8-in.-diameter pipe if the Reynolds number is 25,000. 8.34 Water at 60 F flows through a 6-in.-diameter pipe with an average velocity of 15 ft/s. Approximately what is the height of the largest roughness element allowed if this pipe is to be classified as smooth? Section 8.4.1 Major Losses (also see Lab Problem 8.1LP) 8.35 Water flows in a constant-diameter pipe with the following conditions measured: At section (a) pa 32.4 psi and za 56.8 ft; at section (b) pb 29.7 psi and zb 68.2 ft. Is the flow from (a) to (b) or from (b) to (a)? Explain. 8.36 Water is pumped between two tanks as shown in Fig. P8.36. The energy line is as indicated. Is the fluid being pumped from A to B or B to A? Explain. Which pipe has the larger diameter: A to the pump or B to the pump? Explain. Energy line A P B ■ Figure P8.36 8.37 A person with no experience in fluid mechanics wants to estimate the friction factor for 1-in.-diameter galvanized iron pipe at a Reynolds number of 8,000. The person stumbles across the simple equation of f 64/Re and uses this to calculate the friction factor. Explain the problem with this approach and estimate the error. 8.38 During a heavy rainstorm, water from a parking lot completely fills an 18-in.-diameter, smooth, concrete storm sewer. If the flowrate is 10 ft3/s, determine the pressure drop in a 100-ft horizontal section of the pipe. Repeat the problem if there is a 2-ft change in elevation of the pipe per 100 ft of its length. 8.39 Water flows through a horizontal plastic pipe with a diameter of 0.2 m at a velocity of 10 cm/s. Determine the pressure drop per meter of pipe using the Moody chart. 8.40 GO For Problem 8.39, calculate the power lost to the friction per meter of pipe. drop data indicated in the following table. Determine the average friction factor over this range of data. V (ftⲐ min) ⌬p (in. water) 3950 3730 3610 3430 3280 3000 2700 0.35 0.32 0.30 0.27 0.24 0.20 0.16 8.45 Water flows through a horizontal 60-mm-diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per 10 m of pipe, do you think this pipe is (a) a new pipe, (b) an old pipe with a somewhat increased roughness due to aging, or (c) a very old pipe that is partially clogged by deposits? Justify your answer. 8.46 Water flows at a rate of 10 gallons per minute in a new horizontal 0.75-in.-diameter galvanized iron pipe. Determine the pressure gradient, ¢p/, along the pipe. 8.47 Two equal length, horizontal pipes, one with a diameter of 1 in., the other with a diameter of 2 in., are made of the same material and carry the same fluid at the same flow rate. Which pipe produces the larger head loss? Justify your answer. 8.48 Carbon dioxide at a temperature of 0 C and a pressure of 600 kPa (abs) flows through a horizontal 40-mm-diameter pipe with an average velocity of 2 m/s. Determine the friction factor if the pressure drop is 235 N/m2 per 10-m length of pipe. 8.49 Blood (assume ␮ 4.5 105 lb s/ft2, SG 1.0) flows through an artery in the neck of a giraffe from its heart to its head at a rate of 2.5 104 ft3/s. Assume the length is 10 ft and the diameter is 0.20 in. If the pressure at the beginning of the artery (outlet of the heart) is equivalent to 0.70 ft Hg, determine the pressure at the end of the artery when the head is (a) 8 ft above the heart, or (b) 6 ft below the heart. Assume steady flow. How much of this pressure difference is due to elevation effects, and how much is due to frictional effects? 8.50 A 40-m-long, 12-mm-diameter pipe with a friction factor of 0.020 is used to siphon 30 C water from a tank as shown in 10 m 7m 3m 8.41 Oil (SG 0.9), with a kinematic viscosity of 0.007 ft2/s, flows in a 3-in.-diameter pipe at 0.01 ft3/s. Determine the head loss per unit length of this flow. 30 m 8.42 Water flows through a 6-in.-diameter horizontal pipe at a rate of 2.0 cfs and a pressure drop of 4.2 psi per 100 ft of pipe. Determine the friction factor. h 8.43 Water flows downward through a vertical 10-mm-diameter galvanized iron pipe with an average velocity of 5.0 m/s and exits as a free jet. There is a small hole in the pipe 4 m above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is 0.5 m/s. 8.44 Air at standard conditions flows through an 8-in.diameter, 14.6-ft-long, straight duct with the velocity versus pressure 471 ■ Figure P8.50 472 Chapter 8 ■ Viscous Flow in Pipes Fig. P8.50. Determine the maximum value of h allowed if there is to be no cavitation within the hose. Neglect minor losses. 8.51 Gasoline flows in a smooth pipe of 40-mm diameter at a rate of 0.001 m3/s. If it were possible to prevent turbulence from occurring, what would be the ratio of the head loss for the actual turbulent flow compared to that if it were laminar flow? Vent 27 in. 8.52 A 3-ft-diameter duct is used to carry ventilating air into a vehicular tunnel at a rate of 9000 ft3/min. Tests show that the pressure drop is 1.5 in. of water per 1500 ft of duct. What is the value of the friction factor for this duct and the approximate size of the equivalent roughness of the surface of the duct? 1 -in.-diameter galvanized iron pipe 2 with threaded fittings 18 in. 3 in. 8.53 A fluid flows in a smooth pipe with a Reynolds number of 6000. By what percent would the head loss be reduced if the flow could be maintained as laminar flow rather than the expected turbulent flow? 2 in. 32 in. ■ Figure P8.59 Section 8.4.2 Minor Losses (also see Lab Problem 8.6LP) 8.54 Air at standard temperature and pressure flows through a 1-in.-diameter galvanized iron pipe with an average velocity of 8 ft/s. What length of pipe produces a head loss equivalent to (a) a flanged 90 elbow, (b) a wide-open angle valve, or (c) a sharpedged entrance? 8.55 Given 90 threaded elbows used in conjunction with copper pipe (drawn tubing) of 0.75-in. diameter, convert the loss for a single elbow to equivalent length of copper pipe for wholly turbulent flow. 8.60 (See Fluids in the News article titled “New Hi-tech Fountains,” Section 8.5.) The fountain shown in Fig. P8.60 is designed to provide a stream of water that rises h 10 ft to h 20 ft above the nozzle exit in a periodic fashion. To do this the water from the pool enters a pump, passes through a pressure regulator that maintains a constant pressure ahead of the flow control valve. The valve is electronically adjusted to provide the desired water height. With h 10 ft the loss coefficient for the valve is KL 50. Determine the valve loss coefficient needed for h 20 ft. All losses except for the flow control valve are negligible. The area of the pipe is 5 times the area of the exit nozzle. 8.56 Air at 80 F and standard atmospheric pressure flows through a furnace filter with an average velocity of 2.4 ft/s. If the pressure drop across the filter is 0.06 in. of water, what is the loss coefficient for the filter? h 8.57 To conserve water and energy, a “flow reducer” is installed in the shower head as shown in Fig. P8.57. If the pressure at point (1) remains constant and all losses except for that in the flow reducer are neglected, determine the value of the loss coefficient (based on the velocity in the pipe) of the flow reducer if its presence is to reduce the flowrate by a factor of 2. Neglect gravity. 1 __ in. 2 4 ft Flow reducer washer Pump (1) Flow control valve Pressure regulator ■ Figure P8.60 Q 50 holes of diameter 0.05 in. ■ Figure P8.57 8.58 Water flows at a rate of 0.040 m3/s in a 0.12-m-diameter pipe that contains a sudden contraction to a 0.06-m-diameter pipe. Determine the pressure drop across the contraction section. How much of this pressure difference is due to losses and how much is due to kinetic energy changes? 8.59 Water flows from the container shown in Fig. P8.59. Determine the loss coefficient needed in the valve if the water is to “bubble up” 3 in. above the outlet pipe. The entrance is slightly rounded. 8.61 Water flows through the screen in the pipe shown in Fig. P8.61 as indicated. Determine the loss coefficient for the screen. Water V = 20 ft/s Screen 6 in. SG = 3.2 ■ Figure P8.61 Problems 8.62 Air flows though the mitered bend shown in Fig. P8.62 at a rate of 5.0 cfs. To help straighten the flow after the bend, a set of 0.25-in.-diameter drinking straws is placed in the pipe as shown. Estimate the extra pressure drop between points (1) and (2) caused by these straws. 6-in. length 6-in. length 90° threaded elbows 0.60-in. dia. Reducer 1-in. length (1) 4-in. length Q = 0.020 cfs Tee Tightly packed 0.25-in.-diameter, 12-in.-long straws (2) 473 Closed ball valve 12 in. ■ Figure P8.62 ■ Figure P8.66 8.63 Repeat Problem 8.62 if the straws are replaced by a piece of porous foam rubber that has a loss coefficient equal to 5.4. Section 8.4.3 Noncircular Conduits 8.64 As shown in Fig. P8.64, water flows from one tank to another through a short pipe whose length is n times the pipe diameter. Head losses occur in the pipe and at the entrance and exit. (See Video V8.10.) Determine the maximum value of n if the major loss is to be no more than 10% of the minor loss and the friction factor is 0.02. 8.67 Given two rectangular ducts with equal cross-sectional area, but different aspect ratios (width/height) of 2 and 4, which will have the greater frictional losses? Explain your answer. 8.68 A viscous oil with a specific gravity SG 0.85 and a viscosity of 0.10 Pas flows from tank A to tank B through the six rectangular slots indicated in Fig. P8.68. If the total flowrate is 30 mm3/s and minor losses are negligible, determine the pressure in tank A. A D Same elevation a 0.6 m B Each slot 3 mm × 1 mm ᐉ = nD a Section a−a ■ Figure P8.68 ■ Figure P8.64 8.65 Air flows through the fine mesh gauze shown in Fig. P8.65 with an average velocity of 1.50 m/s in the pipe. Determine the loss coefficient for the gauze. Gauze over end of pipe V = 1.5 m/s Water 8.69 Air at standard temperature and pressure flows at a rate of 7.0 cfs through a horizontal, galvanized iron duct that has a rectangular cross-sectional shape of 12 in. by 6 in. Estimate the pressure drop per 200 ft of duct. 8.70 Air flows through a rectangular galvanized iron duct of size 0.30 m by 0.15 m at a rate of 0.068 m3/s. Determine the head loss in 12 m of this duct. 8.71 Air at standard conditions flows through a horizontal 1 ft by 1.5 ft rectangular wooden duct at a rate of 5000 ft3/min. Determine the head loss, pressure drop, and power supplied by the fan to overcome the flow resistance in 500 ft of the duct. 8 mm Section 8.5.1 Single Pipes—Determine Pressure Drop ■ Figure P8.65 8.66 Water flows steadily through the 0.75-in.-diameter galvanized iron pipe system shown in Video V8.16 and Fig. P8.66 at a rate of 0.020 cfs. Your boss suggests that friction losses in the straight pipe sections are negligible compared to losses in the threaded elbows and fittings of the system. Do you agree or disagree with your boss? Support your answer with appropriate calculations. 8.72 Assume a car’s exhaust system can be approximated as 14 ft of 0.125-ft-diameter cast-iron pipe with the equivalent of six 90 flanged elbows and a muffler. (See Video V8.14.) The muffler acts as a resistor with a loss coefficient of KL 8.5. Determine the pressure at the beginning of the exhaust system if the flowrate is 0.10 cfs, the temperature is 250 F, and the exhaust has the same properties as air. 8.73 The pressure at section (2) shown in Fig. P8.73 is not to fall below 60 psi when the flowrate from the tank varies from 0 to 1.0 cfs and the branch line is shut off. Determine the minimum 474 Chapter 8 ■ Viscous Flow in Pipes 8.79 Water at 10 C is pumped from a lake as shown in Fig. P8.79. If the flowrate is 0.011 m3/s, what is the maximum length inlet pipe, /, that can be used without cavitation occurring? 10 ft Length ᐉ h 6 ft All pipe is 6-in.-diameter plastic ( e /D = 0), flanged fittings Elevation 653 m Branch line 600 ft with 15 90° elbows Elevation 650 m D = 0.07 m e = 0.08 mm (2) Main line ■ Figure P8.79 900 ft ■ Figure P8.73 height, h, of the water tank under the assumption that (a) minor losses are negligible, (b) minor losses are not negligible. 8.74 Repeat Problem 8.73 with the assumption that the branch line is open so that half of the flow from the tank goes into the branch, and half continues in the main line. 8.75 The 12-in.-diameter hose shown in Fig. P8.75 can withstand a maximum pressure of 200 psi without rupturing. Determine the maximum length, /, allowed if the friction factor is 0.022 and the flowrate is 0.010 cfs. Neglect minor losses. 8.80 At a ski resort, water at 40 F is pumped through a 3-in.diameter, 2000-ft-long steel pipe from a pond at an elevation of 4286 ft to a snow-making machine at an elevation of 4623 ft at a rate of 0.26 ft3/s. If it is necessary to maintain a pressure of 180 psi at the snow-making machine, determine the horsepower added to the water by the pump. Neglect minor losses. 8.81 Water at 40 F flows through the coils of the heat exchanger as shown in Fig. P8.81 at a rate of 0.9 gal/min. Determine the pressure drop between the inlet and outlet of the horizontal device. 18 in. Q Q= 0.010 cfs Nozzle tip diameter = 0.30 in. Threaded 180° return bend D = 0.50 in. Water 10 ft ᐉ L Q= 0.011 m3/s 3 ft Pump 0.5-in. copper pipe (drawn tubing) ■ Figure P8.75 ■ Figure P8.81 8.76 The hose shown in Fig. P8.75 will collapse if the pressure within it is lower than 10 psi below atmospheric pressure. Determine the maximum length, /, allowed if the friction factor is 0.015 and the flowrate is 0.010 cfs. Neglect minor losses. 8.82 A 70-ft-long, 0.5-in.-diameter hose with a roughness of e 0.0009 ft is fastened to a water faucet where the pressure is p1. Determine p1 if there is no nozzle attached and the average velocity in the hose is 6 ft/s. Neglect minor losses and elevation changes. 8.77 GO According to fire regulations in a town, the pressure drop in a commercial steel horizontal pipe must not exceed 1.0 psi per 150 ft of pipe for flowrates up to 500 gal/min. If the water temperature is above 50 F, can a 6-in.-diameter pipe be used? 8.78 GO As shown in Video V8.15 and Fig. P8.78, water “bubbles up” 3 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length of the 0.75-in.diameter galvanized iron pipe between point (1) and the exit is 21 in. Determine the pressure needed at point (1) to produce this flow. 8.83 Repeat Problem 8.82 if there is a nozzle of diameter 0.25 in. attached to the end of the hose. 8.84 Water flows through a 2-in.-diameter pipe with a velocity of 15 ft/s as shown in Fig. P8.84. The relative roughness of the pipe is 0.004, and the loss coefficient for the exit is 1.0. Determine the height, h, to which the water rises in the piezometer tube. Open h (1) 8 ft 2 in. 15 ft/s 3 in. 8 ft 4 in. ■ Figure P8.78 ■ Figure P8.84 8.85 Water is pumped through a 60-m-long, 0.3-m-diameter pipe from a lower reservoir to a higher reservoir whose surface is 10 m Problems above the lower one. The sum of the minor loss coefficient for the system is KL 14.5. When the pump adds 40 kW to the water the flowrate is 0.20 m3/s. Determine the pipe roughness. 8.86 Natural gas (r 0.0044 slugs/ft3 and 5.2 105 ft2/s) is pumped through a horizontal 6-in.-diameter cast-iron pipe at a rate of 800 lb/hr. If the pressure at section (1) is 50 psi (abs), determine the pressure at section (2) 8 mi downstream if the flow is assumed incompressible. Is the incompressible assumption reasonable? Explain. 8.87 As shown in Fig. P8.87, a standard household water meter is incorporated into a lawn irrigation system to measure the volume of water applied to the lawn. Note that these meters measure volume, not volume flowrate. (See Video V8.16.) With an upstream pressure of p1 50 psi the meter registered that 120 ft3 of water was delivered to the lawn during an “on” cycle. Estimate the upstream pressure, p1, needed if it is desired to have 150 ft3 delivered during an “on” cycle. List any assumptions needed to arrive at your answer. (1) 8.90 Water flows from the nozzle attached to the spray tank shown in Fig. P8.90. Determine the flowrate if the loss coefficient for the nozzle (based on upstream conditions) is 0.75 and the friction factor for the rough hose is 0.11. Nozzle diameter = 7.5 mm p = 150 kPa D = 15 mm ᐉ = 1.9 m 0.80 m 40° ■ Figure P8.90 8.91 Water flows through the pipe shown in Fig. P8.91. Determine the net tension in the bolts if minor losses are neglected and the wheels on which the pipe rests are frictionless. Irrigation system: pipes, fittings, nozzles, etc. WATER METER 475 ■ Figure P8.87 3.0 m 80 mm 8.88 A fan is to produce a constant air speed of 40 m/s throughout the pipe loop shown in Fig. P8.88. The 3-m-diameter pipes are smooth, and each of the four 90 elbows has a loss coefficient of 0.30. Determine the power that the fan adds to the air. Bolts Galvanized iron 10 m 20 m ■ Figure P8.91 20 m 8.92 When the pump shown in Fig. P8.92 adds 0.2 horsepower to the flowing water, the pressures indicated by the two gages are equal. Determine the flowrate. V = 40 m/s Length of pipe between gages 60 ft Pipe diameter 0.1 ft Pipe friction factor 0.03 Filter loss coefficient 12 D=3m 10 m Fan Filter Pump ■ Figure P8.88 Section 8.5.1 Single Pipes—Determine Flowrate (also see Lab Problems 8.3LP and 8.4LP) 8.89 The turbine shown in Fig. P8.89 develops 400 kW. Determine the flowrate if (a) head losses are negligible or (b) head loss due to friction in the pipe is considered. Assume f 0.02. Note: There may be more than one solution or there may be no solution to this problem. ■ Figure P8.92 8.93 Water at 40 F is pumped from a lake as shown in Fig. P8.93. What is the maximum flowrate possible without cavitation occurring? Threaded elbow Q 20 m Diffuser 10 ft T 1m 20 ft 120 m of 0.30-m-diameter cast-iron pipe ■ Figure P8.89 30 ft ■ Figure P8.93 3-in. diameter, f = 0.02 Pump 476 Chapter 8 ■ Viscous Flow in Pipes 8.94 The pump shown in Fig. P8.94 adds 25 kW to the water and causes a flowrate of 0.04 m3/s. Determine the flowrate expected if the pump is removed from the system. Assume f 0.016 for either case and neglect minor losses. Free jet F f = 0.020 Pipe weighs 0.20 lb/ft 60-mm-diameter, 30-m-long pipe; 40-mm-diameter f = 0.016 nozzle D = 0.40 ft Bellows Pump V ■ Figure P8.94 ■ Figure P8.97 †8.95 Gasoline is unloaded from the tanker truck shown in Fig. P8.95 through a 4-in.-diameter rough-surfaced hose. This is a “gravity dump” with no pump to enhance the flowrate. It is claimed that the 8800-gallon capacity truck can be unloaded in 28 minutes. Do you agree with this claim? Support your answer with appropriate calculations. 8.98 Water is circulated from a large tank, through a filter, and back to the tank as shown in Fig. P8.98. The power added to the water by the pump is 200 ft lb/s. Determine the flowrate through the filter. KL exit = 1.0 KL elbow = 1.5 KL valve = 6.0 KL ent = 0.8 KL filter = 12.0 Midstate Gasoline Filter Pump 200 ft. of 0.1-ft-diameter pipe with ε/D = 0.01 ■ Figure P8.98 Section 8.5.1 Single Pipes—Determine Diameter ■ Figure P8.95 8.99 A certain process requires 2.3 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90 elbows, determine the pipe diameter. Elevation differences are negligible. 8.96 GO The pump shown in Fig. P8.96 delivers a head of 250 ft to the water. Determine the power that the pump adds to the water. The difference in elevation of the two ponds is 200 ft. 8.100 A flowrate of 3.5 ft3/s is to be maintained in a horizontal aluminum pipe (e 5 106 ft). The inlet and outlet pressures are 65 psi and 30 psi, respectively, and the pipe length is 500 ft. Determine the diameter of the pipe. KL exit = 1.0 Pump KL KL valve elbow = 1.5 = 5.0 KL ent = 0.8 Pipe length = 500 ft Pipe diameter = 0.75 ft Pipe roughness = 0 ■ Figure P8.96 8.97 Water flows through two sections of the vertical pipe shown in Fig. P8.97. The bellows connection cannot support any force in the vertical direction. The 0.4-ft-diameter pipe weights 0.2 lb/ft, and the friction factor is assumed to be 0.02. At what velocity will the force, F, required to hold the pipe be zero? 8.101 Water is pumped between two large open reservoirs through 1.5 km of smooth pipe. The water surfaces in the two reservoirs are at the same elevation. When the pump adds 20 kW to the water, the flowrate is 1 m3/s. If minor losses are negligible, determine the pipe diameter. 8.102 Determine the diameter of a steel pipe that is to carry 2000 gal/min of gasoline with a pressure drop of 5 psi per 100 ft of horizontal pipe. 8.103 Water is to be moved from a large, closed tank in which the air pressure is 20 psi into a large, open tank through 2000 ft of smooth pipe at the rate of 3 ft3/s. The fluid level in the open tank is 150 ft below that in the closed tank. Determine the required diameter of the pipe. Neglect minor losses. 8.104 Water flows downward through a vertical smooth pipe. When the flowrate is 0.5 ft3/s, there is no change in pressure along the pipe. Determine the diameter of the pipe. 8.105 Rainwater flows through the galvanized iron downspout shown in Fig. P8.105 at a rate of 0.006 m3/s. Determine the size of the downspout cross section if it is a rectangle with an aspect ratio Problems 70 mm 477 8.112 With the valve closed, water flows from tank A to tank B as shown in Fig. P8.112. What is the flowrate into tank B when the valve is opened to allow water to flow into tank C also? Neglect all minor losses and assume that the friction factor is 0.02 for all pipes. g Elevation = 15 m 4m Diameter of each pipe = 0.10 m A Elevations = 0 80 m 40 m B 3m C ■ Figure P8.105 75 m of 1.7 to 1 and it is completely filled with water. Neglect the velocity of the water in the gutter at the free surface and the head loss associated with the elbow. 8.106 Repeat Problem 8.105 if the downspout is circular. 8.107 For a given head loss per unit length, what effect on the flowrate does doubling the pipe diameter have if the flow is (a) laminar, or (b) completely turbulent? ■ Figure P8.112 8.113 Repeat Problem 8.112 if the friction factors are not known, but the pipes are steel pipes. 8.114 GO The three water-filled tanks shown in Fig. P8.114 are connected by pipes as indicated. If minor losses are neglected, determine the flowrate in each pipe. Section 8.5.2 Multiple Pipe Systems Elevation = 60 m 8.108 Air, assumed incompressible, flows through the two pipes shown in Fig. P8.108. Determine the flowrate if minor losses are neglected and the friction factor in each pipe is 0.015. Determine the flowrate if the 0.5-in.-diameter pipe were replaced by a 1-in.diameter pipe. Comment on the assumption of incompressibility. p = 0.5 psi T = 150°F 1 in. Elevation = 20 m Elevation = 0 D = 0.10 m ᐉ = 200 m f = 0.015 0.50 in. 20 ft D = 0.08 m ᐉ = 200 m f = 0.020 20 ft D = 0.08 m ᐉ = 400 m f = 0.020 ■ Figure P8.114 ■ Figure P8.108 8.109 Repeat Problem 8.108 if the pipes are galvanized iron and the friction factors are not known a priori. †8.110 Estimate the power that the human heart must impart to the blood to pump it through the two carotid arteries from the heart to the brain. List all assumptions and show all calculations. 8.111 The flowrate between tank A and tank B shown in Fig. P8.111 is to be increased by 30% (i.e., from Q to 1.30Q) by the addition of a second pipe (indicated by the dotted lines) running from node C to tank B. If the elevation of the free surface in tank A is 25 ft above that in tank B, determine the diameter, D, of this new pipe. Neglect minor losses and assume that the friction factor for each pipe is 0.02. 8.115 (See Fluids in the News article titled “Deepwater Pipeline,” Section 8.5.2.) Five oil fields, each producing an output of Q barrels per day, are connected to the 28-in.-diameter “mainline pipe” (A–B–C) by 16-in.-diameter “lateral pipes” as shown in Fig. P8.115. The friction factor is the same for each of the pipes and elevation effects are negligible. (a) For section A–B determine the ratio of the pressure drop per mile in the mainline pipe to that in the lateral pipes. (b) Repeat the calculations for section B–C. Q A Q 6-in. diameter; 600 ft long A 6-in. diameter; 500 ft long Q B Main line C Q ■ Figure P8.115 C B Diameter D, 500 ft long ■ Figure P8.111 Lateral Q †8.116 As shown in Fig. P8.116, cold water (T 50 F) flows from the water meter to either the shower or the hot water heater. In the hot water heater it is heated to a temperature of 150 F. Thus, with equal amounts of hot and cold water, the shower is at a comfortable 100 F. However, when the dishwasher is turned on, the 478 Chapter 8 ■ Viscous Flow in Pipes 6 in. Q 3 in. 2 in. Hot SG = 1.52 Dishwasher ■ Figure P8.125 Shower 8.126 If the fluid flowing in Problem 8.125 were air, what would the flowrate be? Would compressibility effects be important? Explain. Cold Water meter Hot water heater 8.127 Water flows through the orifice meter shown in Fig. P8.127 at a rate of 0.10 cfs. If d 0.1 ft, determine the value of h. ■ Figure P8.116 h shower water becomes too cold. Indicate how you would predict this new shower temperature (assume the shower faucet is not adjusted). State any assumptions needed in your analysis. Section 8.6 Pipe Flowrate Measurement (also see Lab Problem 8.2LP) 8.117 A 2-in.-diameter orifice plate is inserted in a 3-in.diameter pipe. If the water flowrate through the pipe is 0.90 cfs, determine the pressure difference indicated by a manometer attached to the flowmeter. 8.118 Air to ventilate an underground mine flows through a large 2-m-diameter pipe. A crude flowrate meter is constructed by placing a sheet metal “washer” between two sections of the pipe. Estimate the flowrate if the hole in the sheet metal has a diameter of 1.6 m and the pressure difference across the sheet metal is 8.0 mm of water. 8.119 Water flows through a 40-mm-diameter nozzle meter in a 75-mm-diameter pipe at a rate of 0.015 m3/s. Determine the pressure difference across the nozzle if the temperature is (a) 10 C, or (b) 80 C. d Q 2 in. ■ Figure P8.127 8.128 Water flows through the orifice meter shown in Fig P8.127 such that h 1.6 ft with d 1.5 in. Determine the flowrate. 8.129 Water flows through the orifice meter shown in Fig. P8.127 at a rate of 0.10 cfs. If h 3.8 ft, determine the value of d. 8.130 The scale reading on the rotameter shown in Fig. P8.130 and Video V8.15 (also see Fig. 8.46) is directly proportional to the volumetric flowrate. With a scale reading of 2.6 the water bubbles up approximately 3 in. How far will it bubble up if the scale reading is 5.0? 8.120 Gasoline flows through a 35-mm-diameter pipe at a rate of 0.0032 m3/s. Determine the pressure drop across a flow nozzle placed in the line if the nozzle diameter is 20 mm. 8.121 Air at 200 F and 60 psia flows in a 4-in.-diameter pipe at a rate of 0.52 lb/s. Determine the pressure at the 2-in.-diameter throat of a Venturi meter placed in the pipe. 8.122 A 2.5-in.-diameter flow nozzle meter is installed in a 3.8in.-diameter pipe that carries water at 160 F. If the air–water manometer used to measure the pressure difference across the meter indicates a reading of 3.1 ft, determine the flowrate. 8.123 A 0.064-m-diameter nozzle meter is installed in a 0.097-mdiameter pipe that carries water at 60 C. If the inverted air– water U-tube manometer used to measure the pressure difference across the meter indicates a reading of 1 m, determine the flowrate. 8.124 A 50-mm-diameter nozzle meter is installed at the end of a 80-mm-diameter pipe through which air flows. A manometer attached to the static pressure tap just upstream from the nozzle indicates a pressure of 7.3 mm of water. Determine the flowrate. 8.125 Water flows through the Venturi meter shown in Fig. P8.125. The specific gravity of the manometer fluid is 1.52. Determine the flowrate. 6 5 4 3 2 1 0 3 in. Rotameter ■ Figure P8.130 ■ Lab Problems 8.1LP This problem involves the determination of the friction factor in a pipe for laminar and transitional flow conditions. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 8.2LP This problem involves the calibration of an orifice meter and a Venturi meter. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. Problems 8.3LP This problem involves the flow of water from a tank and through a pipe system. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 8.4LP This problem involves the flow of water pumped from a tank and through a pipe system. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 8.5LP This problem involves the pressure distribution in the entrance region of a pipe. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 8.6LP This problem involves the power loss due to friction in a coiled pipe. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Lifelong Learning Problems 8.1LL The field of bioengineering has undergone significant growth in recent years. Some universities have undergraduate and graduate programs in this field. Bioengineering applies engineering principles to help solve problems in the medical field for human health. Obtain information about bioengineering applications in blood flow. Summarize your findings in a brief report. 8.2LL Data used in the Moody diagram were first published in 1944. Since then, there have been many innovations in pipe material, pipe design, and measurement techniques. Investigate whether 479 there have been any improvements or enhancements to the Moody chart. Summarize your findings in a brief report. 8.3LL As discussed in Section 8.4.2, flow separation in pipes can lead to losses (we will also see in Chapter 9 that external flow separation is a significant problem). For external flows, many mechanisms have been devised to help mitigate and control flow separation from the surface, for example, from the wing of an airplane. Investigate either passive or active flow control mechanisms that can reduce or eliminate internal flow separation (e.g., flow separation in a diffuser). Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Computational Fluid Dynamics (CFD) The CFD problems associated with this chapter have been developed for use with the ANSYS Academic CFD software package that is associated with this text. See WileyPLUS or the book’s web site (www.wiley.com/college/munson) for additional details. 8.1CFD This CFD problem involves the simulation of fully developed pipe flow and how the Reynolds number affects the wall friction. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/munson. There are additional CFD problems located in WileyPLUS. 9 Flow over Immersed Bodies CHAPTER OPENING PHOTO: Impulsive start of flow past an array of cylinders: The complex structure of laminar flow past a relatively simple geometric structure illustrates why it is often difficult to obtain exact analytical results for external flows. 1Dye in water.2 (Photograph courtesy of ONERA, The French Aerospace Lab.) Learning Objectives After completing this chapter, you should be able to: ■ identify and discuss the features of external flow. ■ explain the fundamental characteristics of a boundary layer, including laminar, transitional, and turbulent regimes. ■ calculate boundary layer parameters for flow past a flat plate. ■ provide a description of boundary layer separation. ■ calculate the lift and drag forces for various objects. Many practical situations involve flow past objects. 480 In this chapter we consider various aspects of the flow over bodies that are immersed in a fluid. Examples include the flow of air around airplanes, automobiles, and falling snowflakes or the flow of water around submarines and fish. In these situations the object is completely surrounded by the fluid and the flows are termed external flows. External flows involving air are often termed aerodynamics in response to the important external flows produced when an object such as an airplane flies through the atmosphere. Although this field of external flows is extremely important, there are many other examples that are of equal importance. The fluid force 1lift and drag2 on surface vehicles 1cars, trucks, bicycles2 has become a very important topic. By correctly designing cars and trucks, it has become possible to greatly decrease the fuel consumption and improve the handling characteristics of the vehicle. Similar efforts have resulted in improved ships, whether they are surface vessels 1surrounded by two fluids, air and water2 or submersible vessels 1surrounded completely by water2. Other applications of external flows involve objects that are not completely surrounded by fluid, although they are placed in some external-type flow. For example, the proper design of a building 1whether it is your house or a tall skyscraper2 must include consideration of the various wind effects involved. As with other areas of fluid mechanics, various approaches 1theoretical, numerical, and experimental2 are used to obtain information on the fluid forces developed by external flows. 9.1 General External Flow Characteristics 481 (a) ■ Figure 9.1 (a) Flow past a full-sized (b) streamlined vehicle in the GM aerodynamics laboratory wind tunnel, an 18-ft by 34-ft test section facility driven by a 4000-hp, 43-ftdiameter fan. (Photograph courtesy of General Motors Corporation.) (b) Predicted streamlines for flow past a Formula 1 race car as obtained by using computational fluid dynamics techniques. (Courtesy of ANSYS, Inc.) Theoretical 1i.e., analytical2 techniques can provide some of the needed information about such flows. However, because of the complexities of the governing equations and the complexities of the geometry of the objects involved, the amount of information obtained from purely theoretical methods is limited. Much of the information about external flows comes from experiments carried out, for the most part, on scale models of the actual objects. Such testing includes the obvious wind tunnel testing of model airplanes, buildings, and even entire cities. In some instances the actual device, not a model, is tested in wind tunnels. Figure 9.1a shows a test of a vehicle in a wind tunnel. Better performance of cars, bikes, skiers, and numerous other objects has resulted from testing in wind tunnels. The use of water tunnels and towing tanks also provides useful information about the flow around ships and other objects. With advancement in computational fluid dynamics, or CFD, numerical methods are also capable of predicting external flows past objects. Figure 9.1b shows streamlines around a Formula 1 car as predicted by CFD. Appendix A provides an introduction to CFD. In this chapter we consider characteristics of external flow past a variety of objects. We investigate the qualitative aspects of such flows and learn how to determine the various forces on objects surrounded by a moving liquid. 9.1 General External Flow Characteristics For external flows it is usually easiest to use a coordinate system fixed to the object. A body immersed in a moving fluid experiences a resultant force due to the interaction between the body and the fluid surrounding it. In some instances 1such as an airplane flying through still air2 the fluid far from the body is stationary and the body moves through the fluid with velocity U. In other instances 1such as the wind blowing past a building2 the body is stationary and the fluid flows past the body with velocity U. In any case, we can fix the coordinate system in the body and treat the situation as fluid flowing past a stationary body with velocity U, the upstream velocity. For the purposes of this book, we will assume that the upstream velocity is constant in both time and location. That is, there is a uniform, constant velocity fluid flowing past the object. In actual situations this is often not true. For example, the wind blowing past a smokestack is nearly always turbulent and gusty 1unsteady2 and probably not of uniform velocity from the top to the bottom of the stack. Usually the unsteadiness and nonuniformity are of minor importance. 482 Chapter 9 ■ Flow over Immersed Bodies y y U y U U x x x z z ( b) (a) (c) ■ Figure 9.2 Flow classification: (a) two-dimensional, (b) axisymmetric, (c) three-dimensional. V9.1 Space Shuttle landing Even with a steady, uniform upstream flow, the flow in the vicinity of an object may be unsteady. Examples of this type of behavior include the flutter that is sometimes found in the flow past airfoils 1wings2, the regular oscillation of telephone wires that “sing” in a wind, and the irregular turbulent fluctuations in the wake regions behind bodies. The structure of an external flow and the ease with which the flow can be described and analyzed often depend on the nature of the body in the flow. Three general categories of bodies are shown in Fig. 9.2. They include 1a2 two-dimensional objects 1infinitely long and of constant crosssectional size and shape2, 1b2 axisymmetric bodies 1formed by rotating their cross-sectional shape about the axis of symmetry2, and 1c2 three-dimensional bodies that may or may not possess a line or plane of symmetry. In practice there can be no truly two-dimensional bodies—nothing extends to infinity. However, many objects are sufficiently long so that the end effects are negligibly small. Another classification of body shape can be made depending on whether the body is streamlined or blunt. The flow characteristics depend strongly on the amount of streamlining present. In general, streamlined bodies 1i.e., airfoils, racing cars, etc.2 have little effect on the surrounding fluid, compared with the effect that blunt bodies 1i.e., parachutes, buildings, etc.2 have on the fluid. Usually, but not always, it is easier to force a streamlined body through a fluid than it is to force a similar-sized blunt body at the same velocity. There are important exceptions to this basic rule. 9.1.1 Lift and Drag Concepts A body interacts with the surrounding fluid through pressure and shear stresses. When any body moves through a fluid, an interaction between the body and the fluid occurs; this effect can be given in terms of the forces at the fluid–body interface. These forces can be described in terms of the stresses—wall shear stresses on the body, tw, due to viscous effects and normal stresses due to the pressure, p. Typical shear stress and pressure distributions are shown in Figs. 9.3a and 9.3b. Both tw and p vary in magnitude and direction along the surface. It is often useful to know the detailed distribution of shear stress and pressure over the surface of the body, although such information is difficult to obtain. Many times, however, only the integrated or resultant effects of these distributions are needed. The resultant force in the direction of the upstream velocity is termed the drag, d, and the resultant force normal to the upstream p0 τw U (a) Shear stress distribution ( c) U (b) ■ Figure 9.3 Forces from the surrounding fluid on a two-dimensional object: (a) pressure force, (b) viscous force, and (c) resultant force (lift and drag). 9.1 y pdA θ General External Flow Characteristics 483 τ w dA dA U θ x ■ Figure 9.4 Pressure and shear forces on a small element of the surface of a body. velocity is termed the lift, l, as is indicated in Fig. 9.3c. For some three-dimensional bodies there may also be a side force that is perpendicular to the plane containing d and l. The resultant of the shear stress and pressure distributions can be obtained by integrating the effect of these two quantities on the body surface as is indicated in Fig. 9.4. The x and y components of the fluid force on the small area element dA are dFx 1 p dA2 cos u 1tw dA2 sin u and dFy 1 p dA2 sin u 1tw dA2 cos u Thus, the net x and y components of the force on the object are d dF p cos u dA t x w sin u dA (9.1) cos u dA (9.2) and l l y w Of course, to carry out the integrations and determine the lift and drag, we must know the body shape 1i.e., u as a function of location along the body2 and the distribution of tw and p along the surface. These distributions are often extremely difficult to obtain, either experimentally or theoretically. The pressure distribution can be obtained experimentally by use of a series of static pressure taps along the body surface. On the other hand, it is usually quite difficult to measure the wall shear stress distribution. Lift and drag on a section of a body depend on the orientation of the surface. F dF p sin u dA t u i d s i n Pressure-sensitive paint For many years, the conventional method for measuring surface pressure has been to use static pressure taps consisting of small holes on the surface connected by hoses from the holes to a pressure measuring device. Pressuresensitive paint (PSP) is now gaining acceptance as an alternative to the static surface pressure ports. The PSP material is typically a luminescent compound that is sensitive to the pressure on it and can be excited by an appropriate light that is captured by special video imaging equipment. Thus, it provides a quantitative mea- t h e N e w s sure of the surface pressure. One of the biggest advantages of PSP is that it is a global measurement technique, measuring pressure over the entire surface, as opposed to discrete points. PSP also has the advantage of being nonintrusive to the flow field. Although static pressure port holes are small, they do alter the surface and can slightly alter the flow, thus affecting downstream ports. In addition, the use of PSP eliminates the need for a large number of pressure taps and connecting tubes. This allows pressure measurements to be made in less time and at a lower cost. It is seen that both the shear stress and pressure force contribute to the lift and drag, since for an arbitrary body u is neither zero nor 90° along the entire body. The exception is a flat plate aligned either parallel to the upstream flow 1u 90°2 or normal to the upstream flow 1u 02 as is discussed in Example 9.1. 484 Chapter 9 ■ Flow over Immersed Bodies E XAMPLE 9.1 Drag from Pressure and Shear Stress Distributions the surface are as indicated 1obtained either by experiment or theory2. GIVEN Air at standard conditions flows past a flat plate as is indicated in Fig. E9.1. In case 1a2 the plate is parallel to the upstream flow, and in case 1b2 it is perpendicular to the upstream flow. The pressure and shear stress distributions on FIND Determine the lift and drag on the plate. y ( b = width = 10 ft p = p(x) = 0 y p = –0.893 lb/ft2 y2 2 p = 0.744 1 – __ 4 lb/ft where y is in feet ) U = 25 ft/s U = 25 ft/s x p = 0 (gage) τw(y) = –τw(–y) τw x p=0 4 ft τw = τw (x) = (1.24 × 10–3)/ x lb/ft2 where x is in feet (b) (a) ■ Figure E9.1 SOLUTION For either orientation of the plate, the lift and drag are obtained from Eqs. 9.1 and 9.2. With the plate parallel to the upstream flow we have u 90° on the top surface and u 270° on the bottom surface so that the lift and drag are given by l p dA top symmetrical about the center of the plate. With the given relatively large pressure on the front of the plate 1the center of the plate is a stagnation point2 and the negative pressure 1less than the upstream pressure2 on the back of the plate, we obtain the following drag p dA 0 d bottom 2 ft y2 and d tw dA top tw dA 2 bottom tw dA or where we have used the fact that because of symmetry the shear stress distribution is the same on the top and the bottom surfaces, as is the pressure also [whether we use gage 1 p 02 or absolute 1 p patm 2 pressure]. There is no lift generated —the plate does not know up from down. With the given shear stress distribution, Eq. 1 gives d2 4 ft x0 a 1.24 103 lbft2 b 110 ft2 dx x12 or d 0.0992 lb (Ans) With the plate perpendicular to the upstream flow, we have u 0° on the front and u 180° on the back. Thus, from Eqs. 9.1 and 9.2 l tw dA front d 55.6 lb for the drag. On the ultimately streamlined body 1a zero-thickness flat plate parallel to the flow2 the drag is entirely due to the shear stress at the surface and, in this example, is relatively small. For the ultimately blunted body 1a flat plate normal to the upstream flow2 the drag is entirely due to the pressure difference between the front and back portions of the object and, in this example, is relatively large. If the flat plate were oriented at an arbitrary angle relative to the upstream flow as indicated in Fig. E9.1c, there would be both a lift and a drag, each of which would be dependent on both the shear stress and the pressure. Both the pressure and shear stress distributions would be different for the top and bottom surfaces. tw dA 0 Low p τw U front p dA p dA ≠0 ≠0 High p τw back Again there is no lift because the pressure forces act parallel to the upstream flow 1in the direction of d not l2 and the shear stress is (Ans) COMMENTS Clearly there are two mechanisms responsible back and d y2 b lbft2 4 10.8932 lbft2 d 110 ft2 dy (1) top c 0.744 a1 (c) ■ Figure E9.1 (Continued) 9.1 General External Flow Characteristics 485 Although Eqs. 9.1 and 9.2 are valid for any body, the difficulty in their use lies in obtaining the appropriate shear stress and pressure distributions on the body surface. Considerable effort has gone into determining these quantities, but because of the various complexities involved, such information is available only for certain simple situations. Without detailed information concerning the shear stress and pressure distributions on a body, Eqs. 9.1 and 9.2 cannot be used. The widely used alternative is to define dimensionless lift and drag coefficients and determine their approximate values by means of either a simplified analysis, some numerical technique, or an appropriate experiment. The lift coefficient, CL, and drag coefficient, CD, are defined as Lift coefficients and drag coefficients are dimensionless forms of lift and drag. CL l 1 2 2 rU A and CD where A is a characteristic area of the object 1see Chapter 72. Typically, A is taken to be frontal area—the projected area seen by a person looking toward the object from a direction parallel to the upstream velocity, U, as indicated by the figure in the margin. It would be the area of the shadow of the object projected onto a screen normal to the upstream velocity as formed by a light shining along the upstream flow. In other situations A is taken to be the planform area—the projected area seen by an observer looking toward the object from a direction normal to the upstream velocity 1i.e., from “above” it2. Obviously, which characteristic area is used in the definition of the lift and drag coefficients must be clearly stated. D U A = D U d 1 2 2 rU A c 9.1.2 Characteristics of Flow Past an Object A = c 100 Air 10 106 Re 4 10 U, m/s = 1 0.1 102 0.01 1 0.001 0.001 0.01 0.1 , m 1 10 External flows past objects encompass an extremely wide variety of fluid mechanics phenomena. Clearly the character of the flow field is a function of the shape of the body. Flows past relatively simple geometric shapes 1i.e., a sphere or circular cylinder2 are expected to have less complex flow fields than flows past a complex shape such as an airplane or a tree. However, even the simplestshaped objects produce rather complex flows. For a given-shaped object, the characteristics of the flow depend very strongly on various parameters such as size, orientation, speed, and fluid properties. As is discussed in Chapter 7, according to dimensional analysis arguments, the character of the flow should depend on the various dimensionless parameters involved. For typical external flows the most important of these parameters are the Reynolds number, Re rU/m U/n, the Mach number, Ma Uc, and for flows with a free surface 1i.e., flows with an interface between two fluids, such as the flow past a surface ship2, the Froude number, Fr U 1g/. 1Recall that / is some characteristic length of the object and c is the speed of sound.2 For the present, we consider how the external flow and its associated lift and drag vary as a function of the Reynolds number. Recall that the Reynolds number represents the ratio of inertial effects to viscous effects. In the absence of all viscous effects 1m 02, the Reynolds number is infinite. On the other hand, in the absence of all inertial effects 1negligible mass or r 02, the Reynolds number is zero. Clearly, any actual flow will have a Reynolds number between 1but not including2 these two extremes. The nature of the flow past a body depends strongly on whether Re 1 or Re 1. Most external flows with which we are familiar are associated with moderately sized objects with a characteristic length on the order of 0.01 m 6 / 6 10 m. In addition, typical upstream velocities are on the order of 0.01 ms 6 U 6 100 ms, and the fluids involved are typically water or air. The resulting Reynolds number range for such flows is approximately 10 6 Re 6 109. This is shown by the figure in the margin for air. As a rule of thumb, flows with Re 7 100 are dominated by inertial effects, whereas flows with Re 6 1 are dominated by viscous effects. Hence, most familiar external flows are dominated by inertia. On the other hand, there are many external flows in which the Reynolds number is considerably less than 1, indicating in some sense that viscous forces are more important than inertial 486 Chapter 9 ■ Flow over Immersed Bodies For low Reynolds number flows, viscous effects are felt far from the object. forces. The gradual settling of small particles of dirt in a lake or stream is governed by low Reynolds number flow principles because of the small diameter of the particles and their small settling speed. Similarly, the Reynolds number for objects moving through large-viscosity oils is small because m is large. The general differences between small and large Reynolds number flow past streamlined and blunt objects can be illustrated by considering flows past two objects—one a flat plate parallel to the upstream velocity and the other a circular cylinder. Flows past three flat plates of length / with Re rU/m 0.1, 10, and 107 are shown in Fig. 9.5. If the Reynolds number is small, the viscous effects are relatively strong and the plate affects the uniform upstream flow far ahead, above, below, and behind the plate. To reach that portion of the flow field where the velocity has been altered by less than 1% of its undisturbed value 1i.e., U u 6 0.01 U2, we must travel relatively far from the plate. In low Reynolds number flows, the viscous effects are felt far from the object in all directions. As the Reynolds number is increased 1by increasing U, for example2, the region in which viscous effects are important becomes smaller in all directions except downstream, as is shown in Viscous effects important Re = U/v = 0.1 u < 0.99U y x U Streamlines deflected considerably U (a) Viscous effects important Re = 10 Viscosity not important y u < 0.99U x Streamlines deflected somewhat U U (b) y Re = 107 Viscosity not important Boundary layer Wake region Viscous effects important δ 104 D T-beam A = bD 1.80 1.65 Re > 104 D I-beam A = bD 2.05 Re > 104 D Angle A = bD 1.98 1.82 Re > 104 Hexagon A = bD 1.0 Re > 104 D 529 Drag coefficient D Square rod with rounded corners Lift D D Rectangle A = bD /D CD < 0.1 0.5 0.65 1.0 2.0 3.0 1.9 2.5 2.9 2.2 1.6 1.3 Re = 105 ■ Figure 9.28 Typical drag coefficients for regular two-dimensional objects (Refs. 5, 6). which is obtained from experiments, advanced analysis, or numerical considerations. The lift coefficient is a function of the appropriate dimensionless parameters and, as the drag coefficient, can be written as The lift coefficient is a function of other dimensionless parameters. CL f1shape, Re, Ma, Fr, e/2 The Froude number, Fr, is important only if there is a free surface present, as with an underwater “wing” used to support a high-speed hydrofoil surface ship. Often the surface roughness, e, is relatively unimportant in terms of lift—it has more of an effect on the drag. The Mach number, Ma, is of importance for relatively high-speed subsonic and supersonic flows 1i.e., Ma 7 0.82, and the Reynolds number effect is often not great. The most important parameter that affects the lift coefficient is the shape of the object. Considerable effort has gone into designing optimally shaped liftproducing devices. We will emphasize the effect of the shape on lift—the effects of the other dimensionless parameters can be found in the literature 1Refs. 13, 14, and 292. 530 Chapter 9 ■ Flow over Immersed Bodies Shape D Solid hemisphere D Hollow hemisphere D Thin disk Reference area Drag coefficient A CD π 2 A = __ D 4 1.17 0.42 Re > 104 π 2 A = __ D 4 1.42 0.38 Re > 104 π 2 A = __ D 1.1 Re > 103 4 D θ D D Circular rod parallel to flow Reynolds number Re = ρ UD/μ π 2 A = __ D 4 /D CD 0.5 1.0 2.0 4.0 1.1 0.93 0.83 0.85 θ , degrees CD 10 30 60 90 0.30 0.55 0.80 1.15 Re > 105 Cone π 2 A = __ D Cube A = D2 1.05 Re > 104 Cube A = D2 0.80 Re > 104 π 2 A = __ D 0.04 Re > 105 4 Re > 104 D D 4 Streamlined body ■ Figure 9.29 Typical drag coefficients for regular three-dimensional objects (Ref. 5). Usually most lift comes from pressure forces, not viscous forces. Most common lift-generating devices 1i.e., airfoils, fans, spoilers on cars, etc.2 operate in the large Reynolds number range in which the flow has a boundary layer character, with viscous effects confined to the boundary layers and wake regions. For such cases the wall shear stress, tw, contributes little to the lift. Most of the lift comes from the surface pressure distribution. A typical pressure distribution on a moving car is shown in Fig. 9.31. The distribution, for the most part, is consistent with simple Bernoulli equation analysis. Locations with high-speed flow 1i.e., over the roof and hood2 have low pressure, while locations with low-speed flow 1i.e., on the grill and windshield2 have high pressure. It is easy to believe that the integrated effect of this pressure distribution would provide a net upward force. For objects operating in very low Reynolds number regimes 1i.e., Re 6 12, viscous effects are important, and the contribution of the shear stress to the lift may be as important as that of the pressure. Such situations include the flight of minute insects and the swimming of microscopic organisms. The relative importance of tw and p in the generation of lift in a typical large Reynolds number flow is shown in Example 9.14. 9.4 Shape Reference area D Parachute Frontal area π 2 A = __ D 4 Porous parabolic dish Frontal area π 2 A = __ D 4 Drag coefficient CD 1.4 0 0.2 0.5 1.42 1.20 0.82 0.95 0.90 0.80 Porosity D Porosity = open area/total area Average person D Fluttering flag Standing CD A = 9 ft2 Sitting CD A = 6 ft2 Crouching CD A = 2.5 ft2 A = ᐉD l Empire State Building ᐉ/D CD 1 2 3 0.07 0.12 0.15 Frontal area 1.4 Frontal area 1.8 Upright commuter A = 5.5 ft2 1.1 Racing A = 3.9 ft2 0.88 Drafting A = 3.9 ft2 0.50 Streamlined A = 5.0 ft2 0.12 Standard Frontal area 0.96 With fairing Frontal area 0.76 With fairing and gap seal Frontal area 0.70 U = 10 m/s U = 20 m/s U = 30 m/s Frontal area 0.43 0.26 0.20 Dolphin Wetted area 0.0036 at Re = 6 × 106 (flat plate has CDf = 0.0031) Large birds Frontal area 0.40 Six-car passenger train Bikes Tractor-trailer trucks Fairing Gap seal Tree U ■ Figure 9.30 Typical drag coefficients for objects of interest (Refs. 5, 6, 15, 20). Lift 531 532 Chapter 9 ■ Flow over Immersed Bodies Denotes p > p0 Denotes p < p0 U, p0 ■ Figure 9.31 Pressure distribution on the surface of an automobile. E XAMPLE Lift from Pressure and Shear Stress Distributions 9.14 GIVEN When a uniform wind of velocity U blows past the semicircular building shown in Fig. E9.14a,b, the wall shear stress and pressure distributions on the outside of the building are as given previously in Figs. E9.8b and E9.9a, respectively. FIND If the pressure in the building is atmospheric 1i.e., the value, p0, far from the building2, determine the lift coefficient and the lift on the roof. SOLUTION curves of 31p p0 2 1rU222 4 sin u versus u and F1u2 cos u versus u plotted in Figs. E9.14c and E9.14d. The results are From Eq. 9.2 we obtain the lift as l p sin u dA t w cos u dA (1) As is indicated in Fig. E9.14b, we assume that on the inside of the building the pressure is uniform, p p0, and that there is no shear stress. Thus, Eq. 1 can be written as p l 0 0 D 1 p p0 2 sin u b a b du 2 p p 1 p p0 2 1 2 2 rU 0 sin u du 1.76 and p F1u2 cos u du 3.92 0 Thus, the lift is D tw cos u b a b du 2 l 1 1 1 rU 2A c a b 11.762 13.922 d 2 2 21Re or or l bD c 2 0 p 1 p p0 2 sin u du p 0 tw cos u du d where b and D are the length and diameter of the building, respectively, and dA b1D22du. Equation 2 can be put into dimensionless form by using the dynamic pressure, rU 22, planform area, A bD, and dimensionless shear stress F1u2 tw 1Re2 1 2 1rU 222 to give l 1 1 rU 2A c 2 2 1 2 1Re 0 p 0 p 1 p p0 2 1 2 2 rU l a0.88 (2) (Ans) and CL l 1 2 2 rU A 0.88 1.96 1Re (4) (Ans) COMMENTS Consider a typical situation with D 20 ft, U 30 fts, b 50 ft, and standard atmospheric conditions 1r 2.38 103 slugs ft3 and n 1.57 104 ft2s2, which gives a Reynolds number of sin u du F1u2 cos u du d 1.96 1 b a rU 2Ab 1Re 2 Re (3) From the data in Figs. E9.8b and E9.9a, the values of the two integrals in Eq. 3 can be obtained by determining the area under the 130 fts2120 ft2 UD 3.82 106 n 1.57 104 ft2s Hence, the lift coefficient is CL 0.88 1.96 0.88 0.001 0.881 13.82 106 2 1 2 9.4 U p0 Lift 533 U p0 pdA τ w dA p0 dA θ D/2 dθ ( a) (b) 1.0 0 6 –0.5 4 F(θ ) cos θ 2 p – p0 Cp sin θ = _ sin θ 1 ρU2 __ 0.5 –1.0 –1.5 0 π __ 4 π __ 2 θ, rad 3π ___ 4 π (c) 2 0 0 π __ 4 π __ 2 θ, rad 3π ___ 4 π (d ) ■ Figure E9.14 Note that the pressure contribution to the lift coefficient is 0.88, whereas that due to the wall shear stress is only 1.96 1Re1 2 2 0.001. The Reynolds number dependency of CL is quite minor. The lift is pressure dominated. Recall from Example 9.9 that this is also true for the drag on a similar shape. From Eq. 4 with A 20 ft 50 ft 1000 ft2 , we obtain the lift for the assumed conditions as l 12rU 2ACL 12 10.00238 slugsft3 2130 fts2 2 11000 ft2 210.8812 ~ U2 U or l 944 lb There is a considerable tendency for the building to lift off the ground. Clearly this is due to the object being nonsymmetrical. The lift force on a complete circular cylinder is zero, although the fluid forces do tend to pull the upper and lower halves apart. A typical device designed to produce lift does so by generating a pressure distribution that is different on the top and bottom surfaces. For large Reynolds number flows these pressure distributions are usually directly proportional to the dynamic pressure, rU 2 2, with viscous effects being of secondary importance. Hence, as indicated by the figure in the margin, for a given airfoil the lift is proportional to the square of the airspeed. Two airfoils used to produce lift are indicated in Fig. 9.32. Clearly the symmetrical one cannot produce lift unless the angle of attack, a, is nonzero. Because of the asymmetry of the nonsymmetric airfoil, the pressure distributions on the upper and lower surfaces are different, and a lift is produced even with a 0. Of course, there will be a certain value of a 1less than zero for this case2 for which the lift is zero. For this situation, the pressure distributions on the upper and lower surfaces are different, but their resultant 1integrated2 pressure forces will be equal and opposite. Since most airfoils are thin, it is customary to use the planform area, A bc, in the definition of the lift coefficient. Here b is the length of the airfoil and c is the chord length —the length from the leading edge to the trailing edge as indicated in Fig. 9.32. Typical lift coefficients so defined are on the order of unity. That is, the lift force is on the order of the dynamic pressure times the planform area of the wing, l 1rU 2 22A. The wing loading, defined as the average lift per unit area of the wing, lA, therefore, increases with speed. For example, the wing loading of the 534 Chapter 9 ■ Flow over Immersed Bodies U α Symmetrical U c α ■ Figure 9.32 Symmetrical and nonsym- Nonsymmetrical Not stalled Stalled Courtesy of NASA At large angles of attack the boundary layer separates and the wing stalls. metrical airfoils. 1903 Wright Flyer aircraft was 1.5 lb Ⲑft2, while for the present-day Boeing 747 aircraft it is 150 lb Ⲑft2. The wing loading for a bumble bee is approximately 1 lbⲐft2 1Ref. 152. Typical lift and drag coefficient data as a function of angle of attack, a, and aspect ratio, a, are indicated in Figs. 9.33a and 9.33b. The aspect ratio is defined as the ratio of the square of the wing length to the planform area, a ⫽ b2 ⲐA. If the chord length, c, is constant along the length of the wing 1a rectangular planform wing2, this reduces to a ⫽ bⲐc. In general, the lift coefficient increases and the drag coefficient decreases with an increase in aspect ratio. Long wings are more efficient because their wing tip losses are relatively more minor than for short wings. The increase in drag due to the finite length 1a 6 q 2 of the wing is often termed induced drag. It is due to the interaction of the complex swirling flow structure near the wing tips 1see Fig. 9.372 and the free stream 1Ref. 132. High-performance soaring airplanes and highly efficient soaring birds 1i.e., the albatross and sea gull2 have long, narrow wings. Such wings, however, have considerable inertia that inhibits rapid maneuvers. Thus, highly maneuverable fighter or acrobatic airplanes and birds 1i.e., the falcon2 have small-aspect-ratio wings. Although viscous effects and the wall shear stress contribute little to the direct generation of lift, they play an extremely important role in the design and use of lifting devices. This is because of the viscosity-induced boundary layer separation that can occur on nonstreamlined bodies such as airfoils that have too large an angle of attack 1see Fig. 9.182. As is indicated in Fig. 9.33, up to a certain point, the lift coefficient increases rather steadily with the angle of attack. If a is too large, the boundary layer on the upper surface separates, the flow over the wing develops a wide, turbulent wake region, the lift decreases, and the drag increases. This condition, as indicated by the figures in the margin, is termed stall. Such conditions are extremely dangerous if they occur while the airplane is flying at a low altitude where there is not sufficient time and altitude to recover from the stall. 1.4 1.2 Ꮽ=7 1.0 Ꮽ=3 0.03 0.8 0.6 Ꮽ=1 CL Ꮽ=1 0.4 Ꮽ=3 0.02 Ꮽ=7 CD 0.2 0.01 0 –0.2 –0.4 –10 0 10 20 α , degrees (a) 0 –10 0 10 20 α , degrees (b) ■ Figure 9.33 Typical lift and drag coefficient data as a function of angle of attack and the aspect ratio of the airfoil: (a) lift coefficient, (b) drag coefficient. 9.4 535 Lift 120 NACA 64(1) – 412 airfoil Re = 7 × 105 Stall 100 1.5 80 1.0 60 α = 8° α = 6° C __L CD 40 α = 4° α = 2° CL 0.5 20 α = 0° α = –2° 0 0 V9.21 Stalled airfoil α = –4° –20 –0.5 –40 –8 –4 0 4 8 0 0.005 α = –6° 0.015 0.010 0.02 CD α , degrees (a) (b) ■ Figure 9.34 Two representations of the same lift and drag data for a typical airfoil: (a) lift-to-drag ratio as a function of angle of attack, with the onset of boundary layer separation on the upper surface indicated by the occurrence of stall, (b) the lift and drag polar diagram with the angle of attack indicated (Ref. 27). In many lift-generating devices the important quantity is the ratio of the lift to drag developed, ld CL CD. Such information is often presented in terms of CL CD versus a, as is shown in Fig. 9.34a, or in a lift-drag polar of CL versus CD with a as a parameter, as is shown in Fig. 9.34b. The most efficient angle of attack 1i.e., largest CL CD2 can be found by drawing a line tangent to the CL CD curve from the origin, as is shown in Fig. 9.34b. High-performance airfoils generate lift that is perhaps 100 or more times greater than their drag. This translates into the fact that in still air they can glide a horizontal distance of 100 m for each 1 m drop in altitude. V9.22 Bat flying F l u i d s i n Bats feel turbulence Researchers have discovered that at certain locations on the wings of bats, there are special touch-sensing cells with a tiny hair poking out of the center of the cell. These cells, which are very sensitive to air flowing across the wing surface, can apparently detect turbulence in the flow over the wing. If these hairs are removed, the bats fly well in a straight line, but when maneuver- V9.23 Trailing edge flap t h e N e w s ing to avoid obstacles, their elevation control is erratic. When the hairs grow back, the bats regain their complete flying skills. It is proposed that these touch-sensing cells are used to detect turbulence on the wing surface and thereby tell bats when to adjust the angle of attack and curvature of their wings in order to avoid stalling out in midair. As is indicated above, the lift and drag on an airfoil can be altered by changing the angle of attack. This actually represents a change in the shape of the object. Other shape changes can be used to alter the lift and drag when desirable. In modern airplanes it is common to utilize leading edge and trailing edge flaps as is shown in Fig. 9.35. To generate the necessary lift during the relatively low-speed landing and takeoff procedures, the airfoil shape is altered by extending special flaps on the front andor rear portions of the wing. Use of the flaps considerably enhances the lift, although it is at the expense of an increase in the drag 1the airfoil is in a “dirty” configuration2. This increase in drag is not of much concern during landing and takeoff operations—the decrease in landing or takeoff speed is more important than is a temporary increase in drag. During normal flight with the flaps retracted 1the “clean” configuration2, the drag is relatively small, and the needed lift force is achieved with the smaller lift coefficient and the larger dynamic pressure 1higher speed2. 536 Chapter 9 ■ Flow over Immersed Bodies 3.0 2.0 CL No flaps 1.0 Trailing edge slotted flap V9.24 Leading edge flap Double slotted trailing edge flaps (Data not Leading edge flap shown) 0 0 ■ Figure 9.35 Typical lift and 0.1 0.2 0.3 CD F l u i d s i n Learning from nature For hundreds of years humans looked toward nature, particularly birds, for insight about flying. However, all early airplanes that closely mimicked birds proved to be unsuccessful. Only after much experimenting with rigid (or at least nonflapping) wings did human flight become possible. Recently, however, engineers have been turning to living systems—birds, insects, and other biological models—in an attempt to produce breakthroughs in aircraft design. Perhaps it is possible that nature’s basic design concepts can be applied to airplane systems. For example, by morphing and rotating their wings in three dimensions, birds t h e drag alterations possible with the use of various types of flap designs (Data from Ref. 21). N e w s have remarkable maneuverability that to date has no technological parallel. Birds can control the airflow over their wings by moving the feathers on their wingtips and the leading edges of their wings, providing designs that are more efficient than the flaps and rigid, pivoting tail surfaces of current aircraft (Ref. 15). On a smaller scale, understanding the mechanism by which insects dynamically manage unstable flow to generate lift may provide insight into the development of microscale air vehicles. With new hi-tech materials, computers, and automatic controls, aircraft of the future may mimic nature more than was once thought possible. (See Problem 9.122.) A wide variety of lift and drag information for airfoils can be found in standard aerodynamics books 1Ref. 13, 14, 292. E XAMPLE 9.15 Lift and Power for Human-Powered Flight GIVEN In 1977 the Gossamer Condor, shown in Fig. E9.15a, FIND Determine won the Kremer prize by being the first human-powered aircraft to complete a prescribed figure-of-eight course around two turning points 0.5 mi apart 1Ref. 222.The following data pertain to this aircraft: (a) the lift coefficient, CL, and (b) the power, p, required by the pilot. flight speed U 15 fts wing size b 96 ft, c 7.5 ft 1average2 weight 1including pilot2 w 210 lb U drag coefficient CD 0.046 1based on planform area2 power train efficiency h power to overcome drag pilot power 0.8 ■ Figure E9.15a (Photograph © Don Monroe.) 9.4 537 Lift SOLUTION (a) For steady flight conditions the lift must be exactly balanced by the weight, or wl 1 2 2 rU ACL Thus, CL 2w rU 2A where A bc 96 ft 7.5 ft 720 ft2, w 210 lb, and r 2.38 103 slugsft3 for standard air. This gives 21210 lb2 CL 12.38 103 slugs ft3 2115 fts2 2 1720 ft2 2 1.09 COMMENT This power level is obtainable by a well- conditioned athlete 1as is indicated by the fact that the flight was successfully completed2. Note that only 80% of the pilot’s power 1i.e., 0.8 0.302 0.242 hp, which corresponds to a drag of d 8.86 lb2 is needed to force the aircraft through the air. The other 20% is lost because of the power train inefficiency. By repeating the calculations for various flight speeds, the results shown in Fig. E9.15b are obtained. Note from Eq. 1 that for a constant drag coefficient, the power required increases as U 3—a doubling of the speed to 30 ft/s would require an eightfold increase in power (i.e., 2.42 hp, well beyond the range of any human). (Ans) 2.5 a reasonable number. The overall lift-to-drag ratio for the aircraft is CL CD 1.090.046 23.7. hp dU where 2.0 1.5 , hp (b) The product of the power that the pilot supplies and the power train efficiency equals the useful power needed to overcome the drag, d. That is, 1.0 0.5 (15, 0.302) d 12 rU 2ACD 0 Thus, p or p dU h 1 2 2 rU ACDU h rACDU 2h 0 5 10 15 20 25 30 U, ft/s 3 (1) ■ Figure E9.15b 12.38 103 slugs ft3 21720 ft2 210.0462115 fts2 3 210.82 1 hp p 166 ft # lbs a b 0.302 hp 550 ft # lbs (Ans) 9.4.2 Circulation Inviscid flow analysis can be used to obtain ideal flow past airfoils. Since viscous effects are of minor importance in the generation of lift, it should be possible to calculate the lift force on an airfoil by integrating the pressure distribution obtained from the equations governing inviscid flow past the airfoil. That is, the potential flow theory discussed in Chapter 6 should provide a method to determine the lift. Although the details are beyond the scope of this book, the following is found from such calculations 1Ref. 42. The calculation of the inviscid flow past a two-dimensional airfoil gives a flow field as indicated in Fig. 9.36. The predicted flow field past an airfoil with no lift 1i.e., a symmetrical airfoil at zero angle of attack, Fig. 9.36a2 appears to be quite accurate 1except for the absence of thin boundary layer regions2. However, as is indicated in Fig. 9.36b, the calculated flow past the same airfoil at a nonzero angle of attack 1but one small enough so that boundary layer separation would not occur2 is not proper near the trailing edge. In addition, the calculated lift for a nonzero angle of attack is zero—in conflict with the known fact that such airfoils produce lift. In reality, the flow should pass smoothly over the top surface as is indicated in Fig. 9.36c, without the strange behavior indicated near the trailing edge in Fig. 9.36b. As is shown in Fig. 9.36d, the unrealistic flow situation can be corrected by adding an appropriate clockwise swirling flow around the airfoil. The results are twofold: 112 The unrealistic behavior near the trailing edge is eliminated 1i.e., 538 Chapter 9 ■ Flow over Immersed Bodies α=0 ᏸ=0 (a) α>0 ᏸ=0 (b) α>0 ᏸ>0 (c) + = “(b) + circulation = (c)” (d) V9.25 Wing tip vortices (Photograph courtesy of NASA.) ■ Figure 9.36 Inviscid flow past an airfoil: (a) symmetrical flow past the symmetrical airfoil at a zero angle of attack; (b) same airfoil at a nonzero angle of attack—no lift, flow near trailing edge not realistic; (c) same conditions as for (b) except circulation has been added to the flow—nonzero lift, realistic flow; (d) superposition of flows to produce the final flow past the airfoil. the flow pattern of Fig. 9.36b is changed to that of Fig. 9.36c2, and 122 the average velocity on the upper surface of the airfoil is increased while that on the lower surface is decreased. From the Bernoulli equation concepts 1i.e., p Ⲑg ⫹ V 2 Ⲑ2g ⫹ z ⫽ constant2, the average pressure on the upper surface is decreased and that on the lower surface is increased. The net effect is to change the original zero lift condition to that of a lift-producing airfoil. The addition of the clockwise swirl is termed the addition of circulation. The amount of swirl 1circulation2 needed to have the flow leave the trailing edge smoothly is a function of the airfoil size and shape and can be calculated from potential flow 1inviscid2 theory 1see Section 6.6.3 and Ref. 292. Although the addition of circulation to make the flow field physically realistic may seem artificial, it has well-founded mathematical and physical grounds. For example, consider the flow past a finite-length airfoil, as is indicated in Fig. 9.37. For lift-generating conditions the average pressure on the lower surface is greater than that on the upper surface. Near the tips of the wing this pressure difference will cause some of the fluid to attempt to migrate from the lower to the upper surface, as is indicated in Fig. 9.37b. At the same time, this fluid is swept downstream, forming a trailing vortex 1swirl2 from each wing tip 1see Fig. 4.32. It is speculated that the reason some birds migrate in vee-formation is to take advantage of the updraft produced by the trailing vortex of the preceding bird. [It is calculated that for a given expenditure of energy, a flock of 25 birds flying in vee-formation could travel 70% farther than if each bird were to fly separately 1Ref. 152.] The trailing vortices from the right and left wing tips are connected by the bound vortex along the length of the wing. It is this vortex that generates the circulation that produces the lift. The combined vortex system 1the bound vortex and the trailing vortices2 is termed a horseshoe vortex. The strength of the trailing vortices 1which is equal to the strength of the bound vortex2 is proportional to the lift generated. Large aircraft 1for example, a Boeing 7472 can generate very strong trailing vortices that persist for a long time before viscous effects and instability mechanisms finally cause them to die out. Such vortices are strong enough to flip smaller aircraft out of control if they follow too closely behind the large aircraft. The figure in the margin clearly shows a trailing vortex produced during a wake vortex study in which an airplane flew through a column of smoke. 9.4 Lift 539 U B Bound vortex A Trailing vortex (a) Bound vortex Low pressure A B High pressure Trailing vortex (b) ■ Figure 9.37 Flow past a finite-length wing: (a) the horseshoe vortex system produced by the bound vortex and the trailing vortices; (b) the leakage of air around the wing tips produces the trailing vortices. F l u i d s i n Why winglets? Winglets, those upward turning ends of airplane wings, boost the performance by reducing drag. This is accomplished by reducing the strength of the wing tip vortices formed by the difference between the high pressure on the lower surface of the wing and the low pressure on the upper surface of the wing. These vortices represent an energy loss and an increase in drag. In essence, the winglet provides an effective increase in the aspect ratio of the wing without extending the wingspan. Winglets come in a variety of styles—the Airbus A320 has a very small upper and A spinning sphere or cylinder can generate lift. t h e N e w s lower winglet; the Boeing 747-400 has a conventional, vertical upper winglet; and the Boeing Business Jet (a derivative of the Boeing 737) has an 8-ft winglet with a curving transition from wing to winglet. Since the airflow around the winglets is quite complicated, the winglets must be carefully designed and tested for each aircraft. In the past, winglets were more likely to be retrofitted to existing wings, but new airplanes are being designed with winglets from the start. Unlike tailfins on cars, winglets really do work. (See Problem 9.123.) As is indicated above, the generation of lift is directly related to the production of a swirl or vortex flow around the object. A nonsymmetric airfoil, by design, generates its own prescribed amount of swirl and lift. A symmetric object like a circular cylinder or sphere, which normally provides no lift, can generate swirl and lift if it rotates. As is discussed in Section 6.6.3, the inviscid flow past a circular cylinder has the symmetrical flow pattern indicated in Fig. 9.38a. By symmetry the lift and drag are zero. However, if the cylinder is rotated about its axis in a stationary real 1m 02 fluid, the rotation will drag some of the fluid around, producing circulation about the cylinder as in Fig. 9.38b. When this circulation is combined with an ideal, uniform upstream flow, the flow pattern indicated in Fig. 9.38c is obtained. The flow is no longer symmetrical about the horizontal plane through the center of the cylinder; the average pressure is greater on the lower half of the cylinder than on the upper half, and a lift is generated. This effect is called the Magnus effect, after Heinrich Magnus 11802–18702, a German chemist and physicist who first investigated this phenomenon. A similar lift is generated on a rotating sphere. It accounts for the various types of pitches in baseball 1i.e., curve ball, floater, sinker, etc.2, the ability of a soccer player to hook the ball, and the hook or slice of a golf ball. Typical lift and drag coefficients for a smooth, spinning sphere are shown in Fig. 9.39. Although the drag coefficient is fairly independent of the rate of rotation, the lift coefficient is strongly 540 Chapter 9 ■ Flow over Immersed Bodies ω ω S S S S S = stagnation point (highest pressure) “(a) + (b) = (c)” (a) (b) (c) ■ Figure 9.38 Inviscid flow past a circular cylinder: (a) uniform upstream flow without circulation, (b) free vortex at the center of the cylinder, (c) combination of free vortex and uniform flow past a circular cylinder giving nonsymmetric flow and a lift. 0.8 CD = ______ π 2 1 ρU2 __ __ D 2 4 0.6 ω CD, CL U D Smooth sphere 0.4 CL = ______ π 2 1 ρU2 __ __ D 2 4 0.2 ___ = 6 × 104 Re = UD v 0 0 1 2 3 4 ω D/2U A dimpled golf ball has less drag and more lift than a smooth one. E XAMPLE 5 ■ Figure 9.39 Lift and drag coefficients for a spinning smooth sphere (Ref. 23). dependent on it. In addition 1although not indicated in the figure2, both CL and CD are dependent on the roughness of the surface. As was discussed in Section 9.3, in a certain Reynolds number range an increase in surface roughness actually decreases the drag coefficient. Similarly, an increase in surface roughness can increase the lift coefficient because the roughness helps drag more fluid around the sphere, increasing the circulation for a given angular velocity. Thus, a rotating, rough golf ball travels farther than a smooth one because the drag is less and the lift is greater. However, do not expect a severely roughed up 1cut2 ball to work better—extensive testing has gone into obtaining the optimum surface roughness for golf balls. 9.16 Lift on a Rotating Sphere GIVEN A table tennis ball weighing 2.45 102 N with di- ameter D 3.8 102 m is hit at a velocity of U 12 m s with a back spin of angular velocity v as is shown in Fig. E9.16. FIND What is the value of v if the ball is to travel on a horizontal path, not dropping due to the acceleration of gravity? 9.5 Chapter Summary and Study Guide 541 SOLUTION For horizontal flight, the lift generated by the spinning of the ball must exactly balance the weight, w, of the ball so that w l 12 rU 2ACL Horizontal path with backspin or ω 2w CL rU 2 1p42D2 where the lift coefficient, CL, can be obtained from Fig. 9.39. For Path without spin standard atmospheric conditions with r 1.23 kgm3 we obtain CL U 212.45 102 N2 11.23 kg m3 2112 ms2 2 1p4213.8 102 m2 2 ■ Figure E9.16 0.244 which, according to Fig. 9.39, can be achieved if vD 0.9 2U or v 2U10.92 D 2112 ms2 10.92 3.8 102 m 568 rad s Thus, v 1568 rads2 160 smin211 rev2p rad2 5420 rpm 9.5 (Ans) COMMENT Is it possible to impart this angular velocity to the ball? With larger angular velocities the ball will rise and follow an upward curved path. Similar trajectories can be produced by a well-hit golf ball — rather than falling like a rock, the golf ball trajectory is actually curved up and the spinning ball travels a greater distance than one without spin. However, if topspin is imparted to the ball 1as in an improper tee shot2 the ball will curve downward more quickly than under the action of gravity alone — the ball is “topped” and a negative lift is generated. Similarly, rotation about a vertical axis will cause the ball to hook or slice to one side or the other. Chapter Summary and Study Guide drag lift lift coefficient drag coefficient wake region boundary layer laminar boundary layer turbulent boundary layer boundary layer thickness transition free-stream velocity favorable pressure gradient adverse pressure gradient boundary layer separation friction drag pressure drag stall circulation Magnus effect In this chapter the flow past objects is discussed. It is shown how the pressure and shear stress distributions on the surface of an object produce the net lift and drag forces on the object. The character of flow past an object is a function of the Reynolds number. For large Reynolds number flows, a thin boundary layer forms on the surface. Properties of this boundary layer flow are discussed. These include the boundary layer thickness, whether the flow is laminar or turbulent, and the wall shear stress exerted on the object. In addition, boundary layer separation and its relationship to the pressure gradient are considered. The drag, which contains portions due to friction (viscous) effects and pressure effects, is written in terms of the dimensionless drag coefficient. It is shown how the drag coefficient is a function of shape, with objects ranging from very blunt to very streamlined. Other parameters affecting the drag coefficient include the Reynolds number, Froude number, Mach number, and surface roughness. The lift is written in terms of the dimensionless lift coefficient, which is strongly dependent on the shape of the object. Variation of the lift coefficient with shape is illustrated by the variation of an airfoil’s lift coefficient with angle of attack. The following checklist provides a study guide for this chapter. When your study of the entire chapter and end-of-chapter exercises has been completed, you should be able to write out meanings of the terms listed here in the margin and understand each of the related concepts. These terms are particularly important and are set in italic, bold, and color type in the text. 542 Chapter 9 ■ Flow over Immersed Bodies determine the lift and drag on an object from the given pressure and shear stress distributions on the object. for flow past a flat plate, calculate the boundary layer thickness, the wall shear stress, and the friction drag, and determine whether the flow is laminar or turbulent. explain the concept of the pressure gradient and its relationship to boundary layer separation. for a given object, obtain the drag coefficient from appropriate tables, figures, or equations and calculate the drag on the object. explain why golf balls have dimples. for a given object, obtain the lift coefficient from appropriate figures and calculate the lift on the object. Some of the important equations in this chapter are: Lift coefficient and drag coefficient CL Boundary layer displacement thickness d l 1 2 2 rU A q q 0 Boundary layer momentum thickness ™ 0 Blasius boundary layer thickness, displacement thickness, and momentum thickness for flat plate , CD a1 u b dy U u u a1 b dy U U d 5 d 1.721 ™ 0.664 , , x x x 1Rex 1Rex 1Rex Blasius wall shear stress for flat plate tw 0.332U32 Drag on flat plate d rbU 2 ™ Blasius wall friction coefficient and friction drag coefficient for flat plate d 1 2 2 rU A cf 0.664 , 1Rex CDf rm B x 1.328 1Re/ (9.39), (9.36) (9.3) (9.4) (9.15), (9.16), (9.17) (9.18) (9.23) (9.32) References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Schlichting, H., Boundary Layer Theory, 8th Ed., McGraw-Hill, New York, 2000. Rosenhead, L., Laminar Boundary Layers, Oxford University Press, London, 1963. White, F. M., Viscous Fluid Flow, 3rd Ed., McGraw-Hill, New York, 2005. Currie, I. G., Fundamental Mechanics of Fluids, McGraw-Hill, New York, 1974. Blevins, R. D., Applied Fluid Dynamics Handbook, Van Nostrand Reinhold, New York, 1984. Hoerner, S. F., Fluid-Dynamic Drag, published by the author, Library of Congress No. 64,19666, 1965. Happel, J., Low Reynolds Number Hydrodynamics, Prentice-Hall, Englewood Cliffs, N.J., 1965. Van Dyke, M., An Album of Fluid Motion, Parabolic Press, Stanford, Calif., 1982. Thompson, P. A., Compressible-Fluid Dynamics, McGraw-Hill, New York, 1972. Zucrow, M. J., and Hoffman, J. D., Gas Dynamics, Vol. I, Wiley, New York, 1976. Clayton, B. R., and Bishop, R. E. D., Mechanics of Marine Vehicles, Gulf Publishing Co., Houston, Tex., 1982. CRC Handbook of Tables for Applied Engineering Science, 2nd Ed., CRC Press, Boca Raton, Fla., 1973. Shevell, R. S., Fundamentals of Flight, 2nd Ed., Prentice-Hall, Englewood Cliffs, N.J., 1989. Kuethe, A. M., and Chow, C. Y., Foundations of Aerodynamics, Bases of Aerodynamics Design, 4th Ed., Wiley, New York, 1986. Conceptual Questions 543 Vogel, J., Life in Moving Fluids, 2nd Ed., Willard Grant Press, Boston, 1994. 16. Kreider, J. F., Principles of Fluid Mechanics, Allyn and Bacon, Newton, Mass., 1985. 17. Dobrodzicki, G. A., Flow Visualization in the National Aeronautical Establishment’s Water Tunnel, National Research Council of Canada, Aeronautical Report LR-557, 1972. 18. White, F. M., Fluid Mechanics, 6th Ed., McGraw-Hill, New York, 2008. 19. Vennard, J. K., and Street, R. L., Elementary Fluid Mechanics, 7th Ed., Wiley, New York, 1995. 20. Gross, A. C., Kyle, C. R., and Malewicki, D. J., The Aerodynamics of Human Powered Land Vehicles, Scientific American, Vol. 249, No. 6, 1983. 21. Abbott, I. H., and Von Doenhoff, A. E., Theory of Wing Sections, Dover Publications, New York, 1959. 22. MacReady, P. B., “Flight on 0.33 Horsepower: The Gossamer Condor,” Proc. AIAA 14th Annual Meeting 1Paper No. 78-3082, Washington, D.C., 1978. 23. Goldstein, S., Modern Developments in Fluid Dynamics, Oxford Press, London, 1938. 24. Achenbach, E., Distribution of Local Pressure and Skin Friction around a Circular Cylinder in CrossFlow Up to Re 5 106, Journal of Fluid Mechanics, Vol. 34, Pt. 4, 1968. 25. Inui, T., Wave-Making Resistance of Ships, Transactions of the Society of Naval Architects and Marine Engineers, Vol. 70, 1962. 26. Sovran, G., et al., eds., Aerodynamic Drag Mechanisms of Bluff Bodies and Road Vehicles, Plenum Press, New York, 1978. 27. Abbott, I. H., von Doenhoff, A. E., and Stivers, L. S., Summary of Airfoil Data, NACA Report No. 824, Langley Field, Va., 1945. 28. Society of Automotive Engineers Report HSJ1566, “Aerodynamic Flow Visualization Techniques and Procedures,” 1986. 29. Anderson, J. D., Fundamentals of Aerodynamics, 4th Ed., McGraw-Hill, New York, 2007. 30. Hucho, W. H., Aerodynamics of Road Vehicles, Butterworth–Heinemann, 1987. 31. Homsy, G. M., et al., Multimedia Fluid Mechanics, 2nd Ed., CD-ROM, Cambridge University Press, New York, 2008. Problem available in WileyPLUS at instructor’s discretion. GO Tutoring problem available in WileyPLUS at instructor’s discretion. Problem is related to a chapter video available in WileyPLUS. † Problem to be solved with aid of programmable calculator or computer. Open-ended problem that requires critical thinking. These problems require various assumptions to provide the necessary input data. There are not unique answers to these problems. Review Problems Go to Appendix G (WileyPLUS or the book’s web site, www. wiley.com/college/munson) for a set of review problems with answers. Detailed solutions can be found in the Student Solution Manual and Study Guide for Fundamentals of Fluid Mechanics, by Munson et al. (© 2013, John Wiley and Sons, Inc.). Conceptual Questions 9.1C The flow of two different fluids over thin flat plates is shown for two cases, A and B. The velocity profiles and boundary thicknesses are to scale. For these two situations: a) There is not enough information to determine the relation between the Reynolds numbers. b) The Reynolds number for case A is lower than that for case B. c) The Reynolds numbers for cases A and B are the same. d) The Reynolds number for case A is higher than that for case B. Case A UA Case B UB 544 Chapter 9 ■ Flow over Immersed Bodies 9.2C Water flows over two flat plates. The free-stream velocity is the same for both plates, the flow is laminar, and both plates have the same width. The length of the first plate is L1, and that of the second plate is twice as long (L2 2 L1). The relation between the shear force F2 for the plate of length L2 2 L1 relative to the shear force F1 for the plate of length L1 is a) F2 2 F1. b) F2 2 F1. c) F2 F1. d) F2 12 F1. e) F2 F1. Free-stream velocity U 9.3C In two different experiments, air and water flow over an object that has the same shape in both experiments. It is found that the drag coefficient on each of the objects is the same for both the air and water flow. For the same velocity: a) The drag force in the water flow is about equal to that for the airflow. b) The drag force in the water flow is about 10 times that for the airflow. c) The drag force in the water flow is about 100 times that for the airflow. d) The drag force in the water flow is about 1000 times that for the airflow. Additional conceptual questions are available in WileyPLUS at the instructor’s discretion. Drag force F Length L Problems Note: Unless specific values of required fluid properties are given in the problem statement, use the values found in the tables on the inside of the front cover. Answers to the evennumbered problems are listed at the end of the book. The Lab Problems as well as the videos that accompany problems can be accessed in WileyPLUS or the book’s web site, www.wiley.com/college/munson. Section 9.1 General External Flow Characteristics 9.1 GO Assume that water flowing past the equilateral triangular bar shown in Fig. P9.1 produces the pressure distributions indicated. Determine the lift and drag on the bar and the corresponding lift and drag coefficients (based on frontal area). Neglect shear forces. p = –0.25 ρU2 Linear distribution U = 5 ft/s 0.1 ft p = 0.5 ρ U 2 τ avg p = 1 ␳U2 U p = –0.2 1 ␳U2 2 2 h Width = b τ avg 10h ■ Figure P9.2 drag due to viscous effects. (a) Determine tavg in terms of the dynamic pressure, rU2/2. (b) Determine the drag coefficient for this object. 9.3 Repeat Problem 9.1 if the object is a cone (made by rotating the equilateral triangle about the horizontal axis through its tip) rather than a triangular bar. 9.4 A small 15-mm-long fish swims with a speed of 20 mm/s. Would a boundary layer type flow be developed along the sides of the fish? Explain. 9.5 The average pressure and shear stress acting on the surface of the 1-m-square flat plate are as indicated in Fig. P9.5. Determine the lift and drag generated. Determine the lift and drag if the shear stress is neglected. Compare these two sets of results. b = length = 4 ft pave = –1.2 kN/m2 τ ave = 5.8 × 10–2 kN/m2 ■ Figure P9.1 α = 7° U 9.2 Fluid flows past the two-dimensional bar shown in Fig. P9.2. The pressures on the ends of the bar are as shown, and the average shear stress on the top and bottom of the bar is tavg. Assume that the drag due to pressure is equal to the pave = 2.3 kN/m2 τ ave = 7.6 × 10–2 kN/m2 ■ Figure P9.5 Problems 9.6 The pressure distribution on the l-m-diameter circular disk in Fig. P9.6 is given in the table. Determine the drag on the disk. r (m) p (kN m2) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 4.34 4.28 4.06 3.72 3.10 2.78 2.37 1.89 1.41 0.74 0.0 p = –5 kN/m2 545 Section 9.2 Boundary Layer Characteristics (also see Lab Problems 9.1LP and 9.2LP.) 9.11 A 12-ft-long kayak moves with a speed of 5 ft/s. Would a boundary layer type flow be developed along the sides of the boat? Explain. 9.12 Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.5 m/s. Determine the approximate location downstream from the leading edge where the boundary layer becomes turbulent. What is the boundary layer thickness at this location? p = p(r) r U D = 1m 9.13 A viscous fluid flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm. Determine the boundary layer thickness at distances of 0.20, 2.0, and 20 m from the leading edge. Assume laminar flow. 9.14 If the upstream velocity of the flow in Problem 9.13 is U 1.5 m/s, determine the kinematic viscosity of the fluid. ■ Figure P9.6 9.7 When you walk through still air at a rate of 1 m/s, would you expect the character of the airflow around you to be most like that depicted in Fig. 9.6a, b, or c? Explain. (See Video 9.3.) 9.15 Water flows past a flat plate with an upstream velocity of U 0.02 m/s. Determine the water velocity a distance of 10 mm from the plate at distances of x 1.5 m and x 15 m from the leading edge. 9.8 A 0.10-m-diameter circular cylinder moves through air with a speed U. The pressure distribution on the cylinder’s surface is approximated by the three straight-line segments shown in Fig. P9.8. Determine the drag coefficient on the cylinder. Neglect shear forces. 9.16 Approximately how fast can the wind blow past a 0.25in.-diameter twig if viscous effects are to be of importance throughout the entire flow field (i.e., Re 1)? Explain. Repeat for a 0.004in.-diameter hair and a 6-ft-diameter smokestack. 3 9.17 The typical shape of small cumulus clouds is as indicated in Fig. P9.17. Based on boundary layer ideas, explain why it is clear that the wind is blowing from right to left as indicated. 2 1 p, N /m2 0 20 40 60 θ , deg 80 100 120 140 160 180 –1 –2 U –3 –4 ■ Figure P9.17 –5 –6 ■ Figure P9.8 9.9 Typical values of the Reynolds number for various animals moving through air or water are listed below. For which cases is inertia of the fluid important? For which cases do viscous effects dominate? For which cases would the flow be laminar; turbulent? Explain. Animal 1a2 1b2 1c2 1d2 1e2 large whale flying duck large dragonfly invertebrate larva bacterium Speed Re 10 ms 20 ms 7 ms 1 mms 0.01 mm s 300,000,000 300,000 30,000 0.3 0.00003 9.18 Consider flow past the flat plate described in Problem 9.15. Based on the nature of the boundary layer equations that govern this flow, Eqs. 9.8 and 9.9, explain why the answer to Problem 9.15 is independent of the plate length. 9.19 Because of the velocity deficit, U u, in the boundary layer, the streamlines for flow past a flat plate are not exactly parallel to the plate. This deviation can be determined by use of the displacement thickness, d. For air blowing past the flat plate shown in Fig. P9.19, plot the streamline AB that passes through the edge of the boundary layer (y dB at x ) at point B. That is, plot y y(x) for streamline AB. Assume laminar boundary layer flow. U= y Streamline A–B 1 m/s A †9.10 Estimate the Reynolds numbers associated with the following objects moving through water. (a) a kayak, (b) a minnow, (c) a submarine, (d) a grain of sand settling to the bottom, (e) you swimming. Edge of boundary layer B δB x =4m ■ Figure P9.19 546 Chapter 9 ■ Flow over Immersed Bodies 9.20 GO Air enters a square duct through a 1-ft opening as is shown in Fig. P9.20. Because the boundary layer displacement thickness increases in the direction of flow, it is necessary to increase the cross-sectional size of the duct if a constant U 2 ft/s velocity is to be maintained outside the boundary layer. Plot a graph of the duct size, d, as a function of x for 0 x 10 ft if U is to remain constant. Assume laminar flow. U= 2 ft/s d(x) 1 ft 2 ft/s x ■ Figure P9.20 9.21 A smooth, flat plate of length 6 m and width b 4 m is placed in water with an upstream velocity of U 0.5 m/s. Determine the boundary layer thickness and the wall shear stress at the center and the trailing edge of the plate. Assume a laminar boundary layer. 9.22 An atmospheric boundary layer is formed when the wind blows over the Earth’s surface. Typically, such velocity profiles can be written as a power law: u ayn, where the constants a and n depend on the roughness of the terrain. As is indicated in Fig. P9.22, typical values are n 0.40 for urban areas, n 0.28 for woodland or suburban areas, and n 0.16 for flat open country (Ref. 23). (a) If the velocity is 20 ft/s at the bottom of the sail on your boat (y 4 ft), what is the velocity at the top of the mast (y 30 ft)? (b) If the average velocity is 10 mph on the tenth floor of an urban building, what is the average velocity on the sixtieth floor? u ~ y0.40 y, m 450 300 u ~ y0.28 u ~ y0.16 150 0 ■ Figure P9.22 9.23 A 30-story office building (each story is 12 ft tall) is built in a suburban industrial park. Plot the dynamic pressure, ru22, as a function of elevation if the wind blows at hurricane strength (75 mph) at the top of the building. Use the atmospheric boundary layer information of Problem 9.22. 9.24 Show that by writing the velocity in terms of the similarity variable h and the function f(h), the momentum equation for boundary layer flow on a flat plate (Eq. 9.9) can be written as the ordinary differential equation given by Eq. 9.14. 9.25 Integrate the Blasius equation (Eq, 9.14) numerically to determine the boundary layer profile for laminar flow past a flat plate. Compare your results with those of Table 9.1. 9.26 GO An airplane flies at a speed of 400 mph at an altitude of 10,000 ft. If the boundary layers on the wing surfaces behave as those on a flat plate, estimate the extent of laminar boundary layer flow along the wing. Assume a transitional Reynolds number of Rexcr 5 105. If the airplane maintains its 400-mph speed but descends to sea-level elevation, will the portion of the wing covered by a laminar boundary layer increase or decrease compared with its value at 10,000 ft? Explain. †9.27 If the boundary layer on the hood of your car behaves as one on a flat plate, estimate how far from the front edge of the hood the boundary layer becomes turbulent. How thick is the boundary layer at this location? 9.28 A laminar boundary layer velocity profile is approximated by u/U [ 2 (y/d)] (y/d) for y d, and u U for y d. (a) Show that this profile satisfies the appropriate boundary conditions. (b) Use the momentum integral equation to determine the boundary layer thickness, d d(x). 9.29 A laminar boundary layer velocity profile is approximated by the two straight-line segments indicated in Fig. P9.29. Use the momentum integral equation to determine the boundary layer thickness, d d(x), and wall shear stress, tw tw(x). Compare these results with those in Table 9.2. y δ δ /2 u 0 2U ___ 3 U ■ Figure P9.29 9.30 A laminar boundary layer velocity profile is approximated by u/U 2(y/d) 2(y/d)3 (y/d)4 for y d, and u U for y d. (a) Show that this profile satisfies the appropriate boundary conditions. (b) Use the momentum integral equation to determine the boundary layer thickness, d d(x). 9.31 GO For a fluid of specific gravity SG 0.86 flowing past a flat plate with an upstream velocity of U 5 m/s, the wall shear stress on the flat plate was determined to be as indicated in the table below. Use the momentum integral equation to determine the boundary layer momentum thickness, (x). Assume 0 at the leading edge, x 0. x (m) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Tw (Nm2) — 13.4 9.25 7.68 6.51 5.89 6.57 6.75 6.23 5.92 5.26 Problems Section 9.3 Drag 9.32 Should a canoe paddle be made rough to get a “better grip on the water” for paddling purposes? Explain. 9.33 Two different fluids flow over two identical flat plates with the same laminar free-stream velocity. Both fluids have the same viscosity, but one is twice as dense as the other. What is the relationship between the drag forces for these two plates? 9.34 Fluid flows past a flat plate with a drag force d1. If the free-stream velocity is doubled, will the new drag force, d2, be larger or smaller than d1 and by what amount? 9.35 A model is placed in an airflow with a given velocity and then placed in water flow with the same velocity. If the drag coefficients are the same between these two cases, how do the drag forces compare between the two fluids? 9.44 GO Determine the drag on a small circular disk of 0.01-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.87 and a viscosity 10,000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow? 9.45 The square, flat plate shown in Fig. P9.45a is cut into four equal-sized pieces and arranged as shown in Fig. P9.45b. Determine the ratio of the drag on the original plate [case (a)] to the drag on the plates in the configuration shown in (b). Assume laminar boundary flow. Explain your answer physically. U 9.36 The drag coefficient for a newly designed hybrid car is predicted to be 0.21. The cross-sectional area of the car is 30 ft2. Determine the aerodynamic drag on the car when it is driven through still air at 55 mph. 9.37 A 5-m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average vertical speed of 3 m/s. The total weight of the load and parachute is 200 N. Determine the approximate drag coefficient for the parachute. 9.38 A 50-mph wind blows against an outdoor movie screen that is 70 ft wide and 20 ft tall. Estimate the wind force on the screen. 9.39 The aerodynamic drag on a car depends on the “shape” of the car. For example, the car shown in Fig. P9.39 has a drag coefficient of 0.35 with the windows and roof closed. With the windows and roof open, the drag coefficient increases to 0.45. With the windows and roof open, at what speed is the amount of power needed to overcome aerodynamic drag the same as it is at 65 mph with the windows and roof closed? Assume the frontal area remains the same. Recall that power is force times velocity. 547 (a) U /4 4 (b) ■ Figure P9.45 9.46 If the drag on one side of a flat plate parallel to the upstream flow is d when the upstream velocity is U, what will the drag be when the upstream velocity is 2U; or U/2? Assume laminar flow. 9.47 Water flows past a triangular flat plate oriented parallel to the free stream as shown in Fig. P9.47. Integrate the wall shear stress over the plate to determine the friction drag on one side of the plate. Assume laminar boundary layer flow. 45° Windows and roof closed: CD = 0.35 Windows open; roof open: CD = 0.45 U = 0.2 m/s 1.0 m ■ Figure P9.39 45° 9.40 A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9 m2. Speculate whether the rider is in the upright or racing position. 9.41 A baseball is thrown by a pitcher at 95 mph through standard air. The diameter of the baseball is 2.82 in. Estimate the drag force on the baseball. 9.42 A logging boat is towing a log that is 2 m in diameter and 8 m long at 4 m/s through water. Estimate the power required if the axis of the log is parallel to the tow direction. 9.43 A sphere of diameter D and density rs falls at a steady rate through a liquid of density r and viscosity m. If the Reynolds number, Re rDU/m, is less than 1, show that the viscosity can be determined from m gD2(rs r)/18 U. ■ Figure P9.47 9.48 For small Reynolds number flows, the drag coefficient of an object is given by a constant divided by the Reynolds number (see Table 9.4). Thus, as the Reynolds number tends to zero, the drag coefficient becomes infinitely large. Does this mean that for small velocities (hence, small Reynolds numbers) the drag is very large? Explain. 9.49 A rectangular cartop carrier of 1.6-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car. Estimate the additional power required to drive the car with the carrier at 60 mph through still air compared with the power required to drive only the car at 60 mph. 548 Chapter 9 ■ Flow over Immersed Bodies 9.50 As shown in Video V9.2 and Fig. P9.50a, a kayak is a relatively streamlined object. As a first approximation in calculating the drag on a kayak, assume that the kayak acts as if it were a smooth, flat plate 17 ft long and 2 ft wide. Determine the drag as a function of speed and compare your results with the measured values given in Fig. P9.50b. Comment on reasons why the two sets of values may differ. and the temperature within the balloon is 165 F, estimate the rate at which it will rise under steady-state conditions if the atmospheric pressure is 14.7 psi. 9.56 It is often assumed that “sharp objects can cut through the air better than blunt ones.” Based on this assumption, the drag on the object shown in Fig. P9.56 should be less when the wind blows from right to left than when it blows from left to right. Experiments show that the opposite is true. Explain. U? U? ■ Figure P9.56 (a) 9.57 An object falls at a rate of 100 ft/s immediately prior to the time that the parachute attached to it opens. The final descent rate with the chute open is 10 ft/s. Calculate and plot the speed of falling as a function of time from when the chute opens. Assume that the chute opens instantly, that the drag coefficient and air density remain constant, and that the flow is quasisteady. Measured drag , lb 8 6 9.58 Estimate the velocity with which you would contact the ground if you jumped from an airplane at an altitude of 5,000 ft and (a) air resistance is negligible, (b) air resistance is important, but you forgot your parachute, or (c) you use a 25-ft-diameter parachute. 4 2 9.59 As is discussed in Section 9.3, the drag on a rough golf ball is less than that on an equal-sized smooth ball. Does it follow that a 10-m-diameter spherical water tank resting on a 20-m-tall support should have a rough surface so as to reduce the moment needed at the base of the support when a wind blows? Explain. 0 2 6 4 Kayak speed U, ft/s 8 (b) ■ Figure P9.50 9.51 A three-bladed helicopter blade rotates at 200 rpm. If each blade is 12 ft long and 1.5 ft wide, estimate the torque needed to overcome the friction on the blades if they act as flat plates. 9.52 A ceiling fan consists of five blades of 0.80-m length and 0.10-m width which rotate at 100 rpm. Estimate the torque needed to overcome the friction on the blades if they act as flat plates. 9.53 A thin smooth sign is attached to the side of a truck as is indicated in Fig. P9.53. Estimate the friction drag on the sign when the truck is driven at 55 mph. 9.60 A 12-mm-diameter cable is strung between a series of poles that are 50 m apart. Determine the horizontal force this cable puts on each pole if the wind velocity is 30 m/s. 9.61 How fast do small water droplets of 0.06-mm (6 108 m) diameter fall through the air under standard sea-level conditions? Assume the drops do not evaporate. Repeat the problem for standard conditions at 5000-m altitude. 9.62 A strong wind can blow a golf ball off the tee by pivoting it about point 1 as shown in Fig. P9.62. Determine the wind speed necessary to do this. Radius = 0.845 in. 5 ft 20 ft U Weight = 0.0992 lb SIGN 4 ft (1) 0.20 in. ■ Figure P9.62 ■ Figure P9.53 9.54 A 38.1-mm-diameter, 0.0245-N table tennis ball is released from the bottom of a swimming pool. With what velocity does it rise to the surface? Assume it has reached its terminal velocity. 9.55 A hot-air balloon roughly spherical in shape has a volume of 70,000 ft3 and a weight of 500 lb (including passengers, basket, balloon fabric, etc.). If the outside air temperature is 80 F 9.63 A 22 in. by 34 in. speed limit sign is supported on a 3-in.-wide, 5-ft-long pole. Estimate the bending moment in the pole at ground level when a 30-mph wind blows against the sign. (See Video V9.14.) List any assumptions used in your calculations. 9.64 Determine the moment needed at the base of a 20-mtall, 0.12-m-diameter flag pole to keep it in place in a 20-m/s wind. 9.65 Repeat Problem 9.64 if a 2-m by 2.5-m flag is attached to the top of the pole. See Fig. 9.30 for drag coefficient data for flags. Problems †9.66 During a flash flood, water rushes over a road as shown in Fig. P9.66 with a speed of 12 mph. Estimate the maximum water depth, h, that would allow a car to pass without being swept away. List all assumptions and show all calculations. 12 ft 7 V U = 12 mph 100 h 549 Truck width = 10 ft ■ Figure P9.71 9.72 As shown in Video V9.19 and Fig. P9.72, the aerodynamic drag on a truck can be reduced by the use of appropriate air deflectors. A reduction in drag coefficient from CD 0.96 to CD 0.70 corresponds to a reduction of how many horsepower needed at a highway speed of 65 mph? ■ Figure P9.66 9.67 How much more power is required to pedal a bicycle at 15 mph into a 20-mph head-wind than at 15 mph through still air? Assume a frontal area of 3.9 ft2 and a drag coefficient of CD 0.88. †9.68 Estimate the wind velocity necessary to knock over a 20-lb garbage can that is 3 ft tall and 2 ft in diameter. List your assumptions. 9.69 On a day without any wind, your car consumes x gallons of gasoline when you drive at a constant speed, U, from point A to point B and back to point A. Assume that you repeat the journey, driving at the same speed, on another day when there is a steady wind blowing from B to A. Would you expect your fuel consumption to be less than, equal to, or greater than x gallons for this windy round-trip? Support your answer with appropriate analysis. 9.70 The structure shown in Fig. P9.70 consists of three cylindrical support posts to which an elliptical flat plate sign is attached. Estimate the drag on the structure when a 50-mph wind blows against it. b = width = 10 ft 12 ft Lily 2011 (a) CD = 0.70 Lily 2011 (b) CD = 0.96 16 ft LAUREN’S PEANUT SHOP 0.6 ft ■ Figure P9.72 5 ft 15 ft 0.8 ft 15 ft 1 ft 15 ft 9.73 As shown in Video V9.11 and Fig. P9.73, a vertical wind tunnel can be used for skydiving practice. Estimate the vertical wind speed needed if a 150-lb person is to be able to “float” motionless when the person (a) curls up as in a crouching position or (b) lies flat. See Fig. 9.30 for appropriate drag coefficient data. ■ Figure P9.70 U 9.71 A 25-ton (50,000-lb) truck coasts down a steep 7% mountain grade without brakes, as shown in Fig. P9.71. The truck’s ultimate steady-state speed, V, is determined by a balance between weight, rolling resistance, and aerodynamic drag. Determine V if the rolling resistance for a truck on concrete is 1.2% of the weight and the drag coefficient based on frontal area is 0.76. ■ Figure P9.73 9.74 Compare the rise velocity of an 81-in.-diameter air bubble in water to the fall velocity of an 18-in.-diameter water drop in air. Assume each to behave as a solid sphere. 550 Chapter 9 ■ Flow over Immersed Bodies 9.75 A 50-lb box shaped like a 1-ft cube falls from the cargo hold of an airplane at an altitude of 30,000 ft. If the drag coefficient of the falling box is 1.2, determine the time it takes for the box to hit the ocean. Assume that it falls at the terminal velocity corresponding to its current altitude and use a standard atmosphere (see Table C.1). 9.76 GO A 500-N cube of specific gravity SG 1.8 falls through water at a constant speed U. Determine U if the cube falls (a) as oriented in Fig. P9.76a, (b) as oriented in Fig. P9.76b. 10-mm diameter 1 m long 0.6 m 20-mm diameter 1.5 m long 0.5 m 40-mm diameter 5 m long 0.25 m ■ Figure P9.79 g (a) (b) ■ Figure P9.76 9.77 The helium-filled balloon shown in Fig P9.77 is to be used as a wind-speed indicator. The specific weight of the helium is y 0.011 lb/ft3, the weight of the balloon material is 0.20 lb, and the weight of the anchoring cable is negligible. Plot a graph of u as a function of U for 1 U 50 mph. Would this be an effective device over the range of U indicated? Explain. †9.80 Estimate the wind force on your hand when you hold it out of your car window while driving 55 mph. Repeat your calculations if you were to hold your hand out of the window of an airplane flying 550 mph. †9.81 Estimate the energy that a runner expends to overcome aerodynamic drag while running a complete marathon race. This expenditure of energy is equivalent to climbing a hill of what height? List all assumptions and show all calculations. 9.82 A 2-mm-diameter meteor of specific gravity 2.9 has a speed of 6 km/s at an altitude of 50,000 m where the air density is 1.03 103 kg/m3. If the drag coefficient at this large Mach number condition is 1.5, determine the deceleration of the meteor. 9.83 Air flows past two equal sized spheres (one rough, one smooth) that are attached to the arm of a balance as is indicated in Fig. P9.83. With U 0 the beam is balanced. What is the minimum air velocity for which the balance arm will rotate clockwise? 2-ft diameter U D = 0.1 m Rough sphere ∋ /D = 1.25 × 10–2 Smooth sphere 0.3 m θ 0.5 m U ■ Figure P9.77 ■ Figure P9.83 9.78 A 0.30-m-diameter cork ball (SG 0.21) is tied to an object on the bottom of a river as is shown in Fig. P9.78. Estimate the speed of the river current. Neglect the weight of the cable and the drag on it. 9.84 A 2-in.-diameter sphere weighing 0.14 lb is suspended by the jet of air shown in Fig. P9.84 and Video V3.2. The drag coefficient for the sphere is 0.5. Determine the reading on the pressure gage if friction and gravity effects can be neglected for the flow between the pressure gage and the nozzle exit. U Area = 0.3 ft2 30° ■ Figure P9.78 Area = 0.6 ft2 9.79 A shortwave radio antenna is constructed from circular tubing, as is illustrated in Fig. P9.79. Estimate the wind force on the antenna in a 100-km/hr wind. Air Pressure gage ■ Figure P9.84 Problems 9.86 The United Nations Building in New York is approximately 87.5 m wide and 154 m tall. (a) Determine the drag on this building if the drag coefficient is 1.3 and the wind speed is a uniform 20 m/s. (b) Repeat your calculations if the velocity profile against the building is a typical profile for an urban area (see Problem 9.22) and the wind speed halfway up the building is 20 m/s. 9.87 A regulation football is 6.78 in. in diameter and weighs 0.91 lb. If its drag coefficient is CD 0.2, determine its deceleration if it has a speed of 20 ft/s at the top of its trajectory. 9.88 An airplane tows a banner that is b 0.8 m tall and 25 m long at a speed of 150 km/hr. If the drag coefficient based on the area b is CD 0.06, estimate the power required to tow the banner. Compare the drag force on the banner with that on a rigid flat plate of the same size. Which has the larger drag force and why? 9.89 The paint stirrer shown in Fig. P9.89 consists of two circular disks attached to the end of a thin rod that rotates at 80 rpm. The specific gravity of the paint is SG 1.1 and its viscosity is m 2 102 lb s/ft2. Estimate the power required to drive the mixer if the induced motion of the liquid is neglected. 0.6 L 0.5 U Calm wind 0.4 Strong wind CD 9.85 A 60 mph wind blows against a football stadium scoreboard that is 36 ft tall, 80 ft wide, and 8 ft thick (parallel to the wind). Estimate the wind force on the scoreboard. See Fig. 9.28 for drag coefficient data. 551 0.3 0.2 U H W 0.1 0 10,000 100,000 Re = ρUL /μ 1,000,000 ■ Figure P9.96 mated as shown. Consider a tree with leaves of length L 0.3 ft. What wind speed will produce a drag on the tree that is 6 times greater than the drag on the tree in a 15 ft/s wind? 9.97 The blimp shown in Fig. P9.97 is used at various athletic events. It is 128 ft long and has a maximum diameter of 33 ft. If its drag coefficient (based on the frontal area) is 0.060, estimate the power required to propel it (a) at its 35-mph cruising speed, or (b) at its maximum 55-mph speed. 80 rpm 7 in. 8 ■ Figure P9.97 1.5 in. ■ Figure P9.89 †9.90 If the wind becomes strong enough, it is “impossible” to paddle a canoe into the wind. Estimate the wind speed at which this will happen. List all assumptions and show all calculations. 9.91 A fishnet consists of 0.10-in.-diameter strings tied into squares 4 in. per side. Estimate the force needed to tow a 15-ft by 30-ft section of this net through seawater at 5 ft/s. 9.92 It is suggested that the power, p, needed to overcome the aerodynamic drag on a vehicle traveling at a speed U varies as p U n. What is an appropriate value for the constant n? Explain. 9.93 Estimate the power needed to overcome the aerodynamic drag of a person who runs at a rate of 100 yds in 10 s in still air. Repeat the calculations if the race is run into a 20-mph headwind; a 20-mph tailwind. Explain. 9.94 GO By appropriate streamlining, the drag coefficient for an airplane is reduced by 12% while the frontal area remains the same. For the same power output, by what percentage is the flight speed increased? 9.95 Two bicycle racers ride 30 km/hr through still air. By what percentage is the power required to overcome aerodynamic drag for the second cyclist reduced if she drafts closely behind the first cyclist rather than riding alongside her? Neglect any forces other than aerodynamic drag. (See Fig. 9.30.) 9.96 As indicated in Fig. P9.96, the orientation of leaves on a tree is a function of the wind speed, with the tree becoming “more streamlined” as the wind increases. The resulting drag coefficient for the tree (based on the frontal area of the tree, HW) as a function of Reynolds number (based on the leaf length, L) is approxi- 9.98 If for a given vehicle it takes 20 hp to overcome aerodynamic drag while being driven at 65 mph, estimate the horsepower required at 75 mph. 9.99 (See Fluids in the News article “Dimpled Baseball Bats,” Section 9.3.3.) How fast must a 3.5-in.-diameter, dimpled baseball bat move through the air in order to take advantage of drag reduction produced by the dimples on the bat? Although there are differences, assume the bat (a cylinder) acts the same as a golf ball in terms of how the dimples affect the transition from a laminar to a turbulent boundary layer. 9.100 (See Fluids in the News article “At 12,600 mpg It Doesn’t Cost Much to ‘Fill ’er Up,’” Section 9.3.3.) (a) Determine the power it takes to overcome aerodynamic drag on a small (6 ft2 cross section), streamlined (CD 0.12) vehicle traveling 15 mph. (b) Compare the power calculated in part (a) with that for a large (36 ft2 cross-sectional area), nonstreamlined (CD 0.48) SUV traveling 65 mph on the interstate. Section 9.4 Lift 9.101 GO A rectangular wing with an aspect ratio of 6 is to generate 1000 lb of lift when it flies at a speed of 200 ft/s. Determine the length of the wing if its lift coefficient is 1.0. 9.102 A 1.2-lb kite with an area of 6 ft2 flies in a 20-ft/s wind such that the weightless string makes an angle of 55 relative to the horizontal. If the pull on the string is 1.5 lb, determine the lift and drag coefficients based on the kite area. 9.103 A Piper Cub airplane has a gross weight of 1750 lb, a cruising speed of 115 mph, and a wing area of 179 ft2. Determine the lift coefficient of this airplane for these conditions. 9.104 A light aircraft with a wing area of 200 ft2 and a weight of 2000 lb has a lift coefficient of 0.40 and a drag coefficient of 0.05. Determine the power required to maintain level flight. 552 Chapter 9 ■ Flow over Immersed Bodies 9.105 As shown in Video V9.25 and Fig. P9.105, a spoiler is used on race cars to produce a negative lift, thereby giving a better tractive force. The lift coefficient for the airfoil shown is CL 1.1, and the coefficient of friction between the wheels and the pavement is 0.6. At a speed of 200 mph, by how much would use of the spoiler increase the maximum tractive force that could be generated between the wheels and ground? Assume the airspeed past the spoiler equals the car speed and that the airfoil acts directly over the drive wheels. b = spoiler length = 4 ft Spoiler sfglfbkjxfdbaerg 200 mph 33 1.5 ft akkjbgfkgbsgboiabkv GOOD YEAR NOLAN'S Happy Birds ■ Figure P9.105 x(% c) y(% c) u⁄U 0 2.5 5.0 7.5 10 20 30 40 50 60 70 80 90 100 0 3.72 5.30 6.48 7.43 9.92 11.14 11.49 10.45 9.11 6.46 3.62 1.26 0 0 0.971 1.232 1.273 1.271 1.276 1.295 1.307 1.308 1.195 1.065 0.945 0.856 0.807 9.110 When air flows past the airfoil shown in Fig. P9.110, the velocity just outside the boundary layer, u, is as indicated. Estimate the lift coefficient for these conditions. 9.106 The wings of old airplanes are often strengthened by the use of wires that provided cross-bracing as shown in Fig. P9.106. If the drag coefficient for the wings was 0.020 (based on the planform area), determine the ratio of the drag from the wire bracing to that from the wings. 1.6 Upper surface 1.2 Lower surface u U 2 ( ) 0.8 NACA 632−015 u 0.4 U 0 Speed: 70 mph Wing area: 148 ft 2 Wire: length = 160 ft diameter = 0.05 in. 0 0.2 0.4 0.6 0.8 1.0 x c ■ Figure P9.110 ■ Figure P9.106 9.107 For a given airplane, compare the power to maintain level flight at a 5000-ft altitude with that at 30,000 ft at the same velocity. Assume CD remains constant. 9.108 A wing generates a lift l when moving through sealevel air with a velocity U. How fast must the wing move through the air at an altitude of 10,000 m with the same lift coefficient if it is to generate the same lift? 9.109 Air blows over the flat-bottomed, two-dimensional object shown in Fig. P9.109. The shape of the object, y y(x), and the fluid speed along the surface, u u(x), are given in the table. Determine the lift coefficient for this object. u = u(x) U ■ Figure P9.109 9.112 Show that for unpowered flight (for which the lift, drag, and weight forces are in equilibrium) the glide slope angle, u, is given by tan u CD/CL. 9.113 A sail plane with a lift-to-drag ratio of 25 flies with a speed of 50 mph. It maintains or increases its altitude by flying in thermals, columns of vertically rising air produced by buoyancy effects of nonuniformly heated air. What vertical airspeed is needed if the sail plane is to maintain a constant altitude? 9.114 If the lift coefficient for a Boeing 777 aircraft is 15 times greater than its drag coefficient, can it glide from an altitude of 30,000 ft to an airport 80 mi away if it loses power from its engines? Explain. (See Problem 9.112.) y u=U 9.111 A Boeing 747 aircraft weighing 580,000 lb when loaded with fuel and 100 passengers takes off with an airspeed of 140 mph. With the same configuration (i.e., angle of attack, flap settings, etc.), what is its takeoff speed if it is loaded with 372 passengers? Assume each passenger with luggage weighs 200 lb. x c 9.115 On its final approach to the airport, an airplane flies on a flight path that is 3.0 relative to the horizontal. What lift-todrag ratio is needed if the airplane is to land with its engines idled back to zero power? (See Problem 9.112.) Problems 9.116 Over the years there has been a dramatic increase in the flight speed (U), altitude (h), weight (w), and wing loading (w/A weight divided by wing area) of aircraft. Use the data given in the table below to determine the lift coefficient for each of the aircraft listed. 553 9.123 (See Fluids in the News article “Why Winglets?,” Section 9.4.2.) It is estimated that by installing appropriately designed winglets on a certain airplane the drag coefficient will be reduced by 5%. For the same engine thrust, by what percent will the aircraft speed be increased by use of the winglets? ■ Lab Problems Aircraft Year w, lb U, mph w A, lb ft2 h, ft Wright Flyer Douglas DC-3 Douglas DC-6 Boeing 747 1903 1935 1947 1970 750 25,000 105,000 800,000 35 180 315 570 1.5 25.0 72.0 150.0 0 10,000 15,000 30,000 9.117 If the takeoff speed of a particular airplane is 120 mi/hr at sea level, what will it be at Denver (elevation 5000 ft)? Use properties of the U.S. Standard Atmosphere. 9.118 The landing speed of an airplane such as the Space Shuttle is dependent on the air density. (See Video V9.1.) By what percent must the landing speed be increased on a day when the temperature is 110 F compared to a day when it is 50 F? Assume that the atmospheric pressure remains constant. 9.119 Commercial airliners normally cruise at relatively high altitudes (30,000 to 35,000 ft). Discuss how flying at this high altitude (rather than 10,000 ft, for example) can save fuel costs. 9.120 A pitcher can pitch a “curve ball” by putting sufficient spin on the ball when it is thrown. A ball that has absolutely no spin will follow a “straight” path. A ball that is pitched with a very small amount of spin (on the order of one revolution during its flight between the pitcher’s mound and home plate) is termed a knuckle ball. A ball pitched this way tends to “jump around” and “zig-zag” back and forth. Explain this phenomenon. Note: A baseball has seams. 9.121 For many years, hitters have claimed that some baseball pitchers have the ability to actually throw a rising fastball. Assuming that a top major leaguer pitcher can throw a 95-mph pitch and impart an 1800-rpm spin to the ball, is it possible for the ball to actually rise? Assume the baseball diameter is 2.9 in. and its weight is 5.25 oz. 9.122 (See Fluids in the News article “Learning from Nature,” Section 9.4.1.) As indicated in Fig. P9.122, birds can significantly alter their body shape and increase their planform area, A, by spreading their wing and tail feathers, thereby reducing their flight speed. If during landing the planform area is increased by 50% and the lift coefficient increased by 30% while all other parameters are held constant, by what percent is the flight speed reduced? 9.1LP This problem involves measuring the boundary layer profile on a flat plate. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. 9.2LP This problem involves measuring the pressure distribution on a circular cylinder. To proceed with this problem, go to Appendix H, which is located in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Lifelong Learning Problems 9.1LL One of the “Fluids in the News” articles in this chapter discusses pressure-sensitive paint—a new technique of measuring surface pressure. There have been other advances in fluid measurement techniques, particularly in velocity measurements. One such technique is particle image velocimetry, or PIV. Obtain information about PIV and its advantages. Summarize your findings in a brief report. 9.2LL For typical aircraft flying at cruise conditions, it is advantageous to have as much laminar flow over the wing as possible because there is an increase in friction drag once the flow becomes turbulent. Various techniques have been developed to help promote laminar flow over the wing, both in airfoil geometry configurations as well as active flow control mechanisms. Obtain information on one of these techniques. Summarize your findings in a brief report. 9.3LL We have seen in this chapter that streamlining an automobile can help to reduce the drag coefficient. One of the methods of reducing the drag has been to reduce the projected area. However, it is difficult for some road vehicles, such as a tractor-trailer, to reduce this projected area due to the storage volume needed to haul the required load. Over the years, work has been done to help minimize some of the drag on this type of vehicle. Obtain information on a method that has been developed to reduce drag on a tractortrailer. Summarize your findings in a brief report. ■ FE Exam Problems Sample FE (Fundamentals of Engineering) exam questions for fluid mechanics are provided in WileyPLUS or on the book’s web site, www.wiley.com/college/munson. ■ Computational Fluid Dynamics (CFD) The CFD problems associated with this chapter have been developed for use with the ANSYS Academic CFD software package that is associated with this text. See WileyPLUS or the book’s web site (www.wiley.com/college/munson) for additional details. 9.1CFD This CFD problem involves simulating the pressure distribution for flow past a cylinder and investigating the differences between inviscid and viscous flows. To proceed with this problem, go to WileyPLUS or the book’s web site, www.wiley.com/college/ munson. There are additional CFD problems located in WileyPLUS. ■ Figure P9.122 10 Open-Channel Flow CHAPTER OPENING PHOTO: Hydraulic jump: Under certain conditions, when water flows in an open channel, even if it has constant geometry, the depth of the water may increase considerably over a short distance along the channel. This phenomenon is termed a hydraulic jump (water flow from left to right). Learning Objectives After completing this chapter, you should be able to: ■ discuss the general characteristics of open-channel flow. ■ use a specific energy diagram. ■ apply appropriate equations to analyze open-channel flow with uniform depth. ■ calculate key properties of a hydraulic jump. ■ determine flowrates based on open-channel flow-measuring devices. V10.1 Offshore oil drilling platform. 554 Open-channel flow involves the flow of a liquid in a channel or conduit that is not completely filled. A free surface exists between the flowing fluid (usually water) and fluid above it (usually the atmosphere). The main driving force for such flows is the fluid weight—gravity forces the fluid to flow downhill. Most open-channel flow results are based on correlations obtained from model and full-scale experiments. Additional information can be gained from various analytical and numerical efforts. Open-channel flows are essential to the world as we know it. The natural drainage of water through the numerous creek and river systems is a complex example of open-channel flow. Although the flow geometry for these systems is extremely complex, the resulting flow properties are of considerable economic, ecological, and recreational importance. Other examples of open-channel flows include the flow of rainwater in the gutters of our houses; the flow in canals, drainage ditches, sewers, and gutters along roads; the flow of small rivulets and sheets of water on a road during a rain (as shown by the photograph in the margin); and the flow in the chutes of water rides in amusement parks. Open-channel flow involves the existence of a free surface which can distort into various shapes. Thus, a brief introduction into the properties and characteristics of surface waves is included. The purpose of this chapter is to investigate the concepts of open-channel flow. Because of the amount and variety of material available, only a brief introduction to the topic can be presented. Further information can be obtained from the references indicated. 10.1 10.1 General Characteristics of Open-Channel Flow 555 General Characteristics of Open-Channel Flow Open-channel flow can have a variety of characteristics. Uniform flow Rapidly varying flow (Photo by Marty Melchior) In our study of pipe flow 1Chapter 82, we found that there are many ways to classify a flow— developing, fully developed, laminar, turbulent, and so on. For open-channel flow, the existence of a free surface allows additional types of flow. The extra freedom that allows the fluid to select its free-surface location and configuration 1because it does not completely fill a pipe or conduit2 allows important phenomena in open-channel flow that cannot occur in pipe flow. Some of the classifications of the flows are described below. The manner in which the fluid depth, y, varies with time, t, and distance along the channel, x, is used to partially classify a flow. For example, the flow is unsteady or steady depending on whether the depth at a given location does or does not change with time. Some unsteady flows can be viewed as steady flows if the reference frame of the observer is changed. For example, a tidal bore 1difference in water level2 moving up a river is unsteady to an observer standing on the bank, but steady to an observer moving along the bank with the speed of the wave front of the bore. Other flows are unsteady regardless of the reference frame used. The complex, time-dependent, wind-generated waves on a lake are in this category. In this book we will consider only steady open-channel flows. An open-channel flow is classified as uniform flow 1UF2 if the depth of flow does not vary along the channel 1dyⲐdx ⫽ 02. Conversely, it is nonuniform flow or varied flow if the depth varies with distance 1dyⲐdx ⫽ 02. Nonuniform flows are further classified as rapidly varying flow 1RVF2 if the flow depth changes considerably over a relatively short distance; dy Ⲑdx ⬃ 1. Gradually varying flows 1GVF2 are those in which the flow depth changes slowly with distance along the channel; dy Ⲑdx Ⰶ 1. Examples of these types of flow are illustrated in Fig. 10.1 and the photographs in the margin. The relative importance of the various types of forces involved 1pressure, weight, shear, inertia2 is different for the different types of flows. As for any flow geometry, open-channel flow may be laminar, transitional, or turbulent, depending on various conditions involved. Which type of flow occurs depends on the Reynolds number, Re ⫽ rVRh Ⲑm, where V is the average velocity of the fluid and Rh is the hydraulic radius of the channel 1see Section 10.42. A general rule is that open-channel flow is laminar if Re 6 500, turbulent if Re 7 12,500, and transitional otherwise. The values of these dividing Reynolds numbers are only approximate—a precise knowledge of the channel geometry is necessary to obtain specific values. Since most open-channel flows involve water 1which has a fairly small viscosity2 and have relatively large characteristic lengths, it is rare to have laminar open-channel flows. For example, flow of 50 °F water 1n ⫽ 1.41 ⫻ 10⫺5 ft 2 Ⲑs2 with an average velocity of V ⫽ 1 ftⲐs in a river with a hydraulic radius of Rh ⫽ 10 ft has Re ⫽ VRh Ⲑn ⫽ 7.1 ⫻ 105. The flow is turbulent. However, flow of a thin sheet of water down a driveway with an average velocity of V ⫽ 0.25 ftⲐs such that Rh ⫽ 0.02 ft 1in such cases the hydraulic radius is approximately equal to the fluid depth; see Section 10.42 has Re ⫽ 355. The flow is laminar. In some cases stratified flows are important. In such situations layers of two or more fluids of different densities flow in a channel. A layer of oil on water is one example of this type of flow. All of the open-channel flows considered in this book are homogeneous flows. That is, the fluid has uniform properties throughout. Open-channel flows involve a free surface that can deform from its undisturbed, relatively flat configuration to form waves. Such waves move across the surface at speeds that depend on UF uniform flow GVF gradually varying flow RVF rapidly varying flow y RVF UF RVF UF RVF GVF ■ Figure 10.1 Classification of open-channel flow. RVF UF 556 Chapter 10 ■ Open-Channel Flow Fr = V √gy 1 their size 1height, length2 and properties of the channel 1depth, fluid velocity, etc.2. The character of an open-channel flow may depend strongly on how fast the fluid is flowing relative to how fast a typical wave moves relative to the fluid. The dimensionless parameter that describes this behavior is termed the Froude number, Fr ⫽ VⲐ 1g/2 1Ⲑ2, where / is an appropriate characteristic length of the flow. This dimensionless parameter was introduced in Chapter 7 and is discussed more fully in Section 10.2. As shown by the figure in the margin, the special case of a flow with a Froude number of unity, Fr ⫽ 1, is termed a critical flow. If the Froude number is less than 1, the flow is subcritical 1or tranquil2. A flow with the Froude number greater than 1 is termed supercritical 1or rapid2. Supercritical Critical Subcritical 0 10.2 Surface Waves The distinguishing feature of flows involving a free surface 1as in open-channel flows2 is the opportunity for the free surface to distort into various shapes. The surface of a lake or the ocean is seldom “smooth as a mirror.” It is usually distorted into ever-changing patterns associated with surface waves as shown in the photos in the margin. Some of these waves are very high, some barely ripple the surface; some waves are very long 1the distance between wave crests2, some are short; some are breaking waves that form whitecaps, others are quite smooth. Although a general study of this wave motion is beyond the scope of this book, an understanding of certain fundamental properties of simple waves is necessary for open-channel flow considerations. The interested reader is encouraged to use some of the excellent references available for further study about wave motion 1Refs. 1, 2, 32. F l u i d s i n t Rogue waves There is a long history of stories concerning giant rogue ocean waves that come out of nowhere and capsize ships. The movie Poseidon (2006) is based on such an event. Although these giant, freakish waves were long considered fictional, recent satellite observations and computer simulations prove that, although rare, they are real. Such waves are single, sharply peaked mounds of water that travel rapidly across an otherwise relatively calm ocean. Although most ships are designed to withstand waves up to 15 m high, satellite measurements and data from offshore oil plat- h e N e w s forms indicate that such rogue waves can reach a height of 30 m. Although researchers still do not understand the formation of these large rogue waves, there are several suggestions as to how ordinary smaller waves can be focused into one spot to produce a giant wave. Additional theoretical calculations and wave tank experiments are needed to adequately grasp the nature of such waves. Perhaps it will eventually be possible to predict the occurrence of these destructive waves, thereby reducing the loss of ships and life because of them. 10.2.1 Wave Speed V10.2 Filling your car’s gas tank. Consider the situation illustrated in Fig. 10.2a in which a single elementary wave of small height, dy, is produced on the surface of a channel by suddenly moving the initially stationary end wall with speed dV. The water in the channel was stationary at the initial time, t ⫽ 0. A stationary observer will observe a single wave move down the channel with a wave speed c, with no fluid motion ahead of the wave and a fluid velocity of dV behind the wave. The motion is unsteady for such an observer. For an observer moving along the channel with speed c, the flow will appear steady as shown in Fig. 10.2b. To this observer, the fluid velocity will be V ⫽ ⫺ciˆ on the observer’s right and V ⫽ 1⫺c ⫹ dV2iˆ to the left of the observer. The relationship between the various parameters involved for this flow can be obtained by application of the continuity and momentum equations to the control volume shown in Fig. 10.2b as follows. With the assumption of uniform one-dimensional flow, the continuity equation 1Eq. 5.122 becomes ⫺cyb ⫽ 1⫺c ⫹ dV21y ⫹ dy2b where b is the channel width. This simplifies to c⫽ 1y ⫹ dy2dV dy 10.2 δy δV Surface Waves 557 c = wave speed Moving end wall δV Stationary fluid y x (a) Control surface Channel width = b Stationary wave ^ V = (– c + δ V )i ^ y +δy V = – ci y x (1) (b) (2) ■ Figure 10.2 (a) Production of a single elementary wave in a channel as seen by a stationary observer. (b) Wave as seen by an observer moving with a speed equal to the wave speed. or in the limit of small-amplitude waves with dy Ⰶ y c⫽y dV dy (10.1) Similarly, the momentum equation 1Eq. 5.222 is ⫺ 12 g1y ⫹ dy2 2b ⫽ rbcy 3 1c ⫺ dV2 ⫺ c4 # where we have written the mass flowrate as m ⫽ rbcy and have assumed that the pressure variation is hydrostatic within the fluid. That is, the pressure forces on the channel cross sections 112 and 122 are F1 ⫽ gyc1A1 ⫽ g1y ⫹ dy2 2b Ⲑ2 and F2 ⫽ gyc2A2 ⫽ gy2bⲐ2, respectively. If we again impose the assumption of small-amplitude waves [i.e., 1dy2 2 Ⰶ y dy], the momentum equation reduces to 1 2 2 gy b The wave speed can be obtained from the continuity and momentum equations. g dV ⫽ c dy Combination of Eqs. 10.1 and 10.2 gives the wave speed c ⫽ 1gy 10 c, m/s 8 6 c = √gy 4 2 0 0 2 4 6 y, m 8 10 (10.2) (10.3) as indicated by the figure in the margin. The speed of a small-amplitude solitary wave as is indicated in Fig. 10.2 is proportional to the square root of the fluid depth, y, and independent of the wave amplitude, dy. The fluid density is not an important parameter, although the acceleration of gravity is. This is a result of the fact that such wave motion is a balance between inertial effects 1proportional to r2 and weight or hydrostatic pressure effects 1proportional to g ⫽ rg2. A ratio of these forces eliminates the common factor r but retains g. For very small waves (like those produced by insects on water as shown in Video V10.3), Eq. 10.3 is not valid because the effects of surface tension are significant. The wave speed can also be calculated by using the energy and continuity equations rather than the momentum and continuity equations as is done above. A simple wave on the surface is shown in Fig. 10.3. As seen by an observer moving with the wave speed, c, the flow is steady. Since the pressure is constant at any point on the free surface, the Bernoulli equation for this frictionless flow is simply V2 ⫹ y ⫽ constant 2g or by differentiating 558 Chapter 10 ■ Open-Channel Flow δy Stationary wave V=c V = c + δV Moving fluid y ■ Figure 10.3 Stationary simple wave in a flowing fluid. V dV ⫹ dy ⫽ 0 g V10.3 Water strider Also, by differentiating the continuity equation, Vy ⫽ constant, we obtain y dV ⫹ V dy ⫽ 0 We combine these two equations to eliminate dV and dy and use the fact that V ⫽ c for this situation 1the observer moves with speed c2 to obtain the wave speed given by Eq. 10.3. The above results are restricted to waves of small amplitude because we have assumed onedimensional flow. That is, dyⲐy Ⰶ 1. More advanced analysis and experiments show that the wave speed for finite-sized solitary waves exceeds that given by Eq. 10.3. To a first approximation, one obtains 1Ref. 42 10 dy y = 0.4 c, m/s 8 6 dy y = 0.2 4 dy y =0 2 0 dy 1Ⲑ 2 c 1gy a1 ⫹ y b 0 2 4 6 8 10 y, m As indicated by the figure in the margin, the larger the amplitude, the faster the wave travels. A more general description of wave motion can be obtained by considering continuous 1not solitary2 waves of sinusoidal shape as is shown in Fig. 10.4. By combining waves of various wavelengths, l, and amplitudes, dy, it is possible to describe very complex surface patterns found in nature, such as the wind-driven waves on a lake. Mathematically, such a process consists of using a Fourier series 1each term of the series represented by a wave of different wavelength and amplitude2 to represent an arbitrary function 1the free-surface shape2. A more advanced analysis of such sinusoidal surface waves of small amplitude shows that the wave speed varies with both the wavelength and fluid depth as 1Ref. 12 V10.4 Sinusoidal waves c⫽ c gl 2py 1Ⲑ2 tanh a bd 2p l (10.4) where tanh12pyⲐl2 is the hyperbolic tangent of the argument 2py Ⲑl. The result is plotted in Fig. 10.5. For conditions for which the water depth is much greater than the wavelength 1y Ⰷ l, as in the ocean2, the wave speed is independent of y and given by c⫽ gl A 2p Surface at time t δy = amplitude c λ = length y = mean depth ■ Figure 10.4 Sinusoidal surface wave. Surface at time t + δ t c δt 10.2 Surface Waves 559 c2 __ gy c = √gy Shallow layer 1.0 g c = __λ √ 2π y > λ 0 c, m/s 150 50 0 2 4 6 8 10 l, km F l y ■ Figure 10.5 Wave speed as a function of wavelength. This result, shown in the figure in the margin, follows from Eq. 10.4, since tanh12pyⲐl2 S 1 as yⲐl S ⬁. Note that waves with very long wavelengths [e.g., waves created by a tsunami (“tidal wave”) with wavelengths on the order of several kilometers] travel very rapidly. On the other hand, if the fluid layer is shallow 1 y Ⰷ l, as often happens in open channels2, the wave speed is given by c ⫽ 1gy2 1Ⲑ2, as derived for the solitary wave in Fig. 10.2. This result also follows from Eq. 10.4, since tanh12pyⲐl2 S 2pyⲐl as yⲐl S 0. These two limiting cases are shown in Fig. 10.5. For moderate depth layers 1y l2, the results are given by the complete Eq. 10.4. Note that for a given fluid depth, the long wave travels the fastest. Hence, for our purposes we will consider the wave speed to be this limiting situation, c ⫽ 1gy2 1Ⲑ2. y >> l 100 0 __ λ 0 u i d s i n Tsunami, the nonstorm wave A tsunami, often miscalled a “tidal wave,” is a wave produced by a disturbance (for example, an earthquake, volcanic eruption, or meteorite impact) that vertically displaces the water column. Tsunamis are characterized as shallow-water waves, with long periods, very long wavelengths, and extremely large wave speeds. For example, the waves of the great December 2005, Indian Ocean tsunami traveled with speeds of 500–1000 m/s. Typically, these waves were of small amplitude in deep water far from land. Satellite radar measured the wave height as less than 1 m in these areas. However, as the V–c t h e N e w s waves approached shore and moved into shallower water, they slowed down considerably and reached heights up to 30 m. Because the rate at which a wave loses its energy is inversely related to its wavelength, tsunamis, with their wavelengths on the order of 100 km, not only travel at high speeds, but they also travel great distances with minimal energy loss. The furthest reported death from the Indian Ocean tsunami occurred approximately 8000 km from the epicenter of the earthquake that produced it. A more recent example is the tsunami that struck Japan in 2011. (See Problem 10.14.) 10.2.2 Froude Number Effects V>c Stationary V=c c–V V Report "Fundamentals of Fluid Mechanics [7 ed.] 1118116135" × Close Submit Contact information Michael Browner [email protected] Address: 1918 St.Regis, Dorval, Quebec, H9P 1H6, Canada. Support & Legal O nas Skontaktuj się z nami Prawo autorskie Polityka prywatności Warunki FAQs Cookie Policy Subscribe to our newsletter Be the first to receive exclusive offers and the latest news on our products and services directly in your inbox. 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6179	https://ijmtlm.org/index.php/journal/article/view/28	International Journal of Medical Toxicology and Legal Medicine Effectiveness of Using Potassium Levels in Vitreous Humour for Estimating Postmortem Interval - A Systematic Review Article Sidebar PDF Published: Aug 10, 2024 Keywords: Autopsy, Potassium, Postmortem Interval, Time Since Death, Vitreous Humor Main Article Content Jagadish Ramasamy, Benjy Tom Varughese, Malathi Murugesan, Daniel Manoj, Antony L Arakkal, Latif Rajesh Johnson, Ranjit Immanuel James Abstract Background: Estimating the Postmortem Interval (PMI) is a requirement in medico-legal autopsy, and often, it poses challenges due to environmental factors and body conditions. A biochemical approach, especially using potassium levels in vitreous humor (VH), is a widely proposed method to estimate PMI. Hence, this systematic review is done to analyse the effectiveness of using vitreous humor potassium levels in estimating PMI. Methodology: We searched three databases (MEDLINE, Scopus, and ProQuest) to identify studies analysing vitreous potassium concentration for PMI estimation. Data extracted from included studies encompassed analytical methods employed, the range of vitreous potassium, and the impact of temperature and humidity on PMI. Results: The electronic search identified 471 articles that were subsequently screened based on the inclusion criteria, and 53 studies were found eligible for qualitative synthesis. Forty studies have reported the actual PMI for the subjects they studied, and it ranged from 0 hours to 408 hours. 5 studies (12.5%) had PMI < 24 hours and 15 studies (37.5 %) included subjects with PMI > 72 hours. Among the eligible studies, 25 studies proposed regression equations to estimate PMI using vitreous humor potassium levels. The majority of them used only potassium (n=21), and few studies (n=4) have used vitreous levels of chloride, uric acid, hypoxanthine, albumin, sodium along with potassium to derive a regression equation to estimate PMI. Conclusions: Most of the studies have validated their proposed regression equations in the same subjects from which they were derived. Advanced statistical methods like generalized additive modelling and artificial neural networks have been shown to predict PMI much better than simple regression equations. The reporting of the standard error of the regression coefficient is recommended to enable quantitative analysis of the data, i.e., meta-analysis. Article Details How to Cite Jagadish Ramasamy. (2024). Effectiveness of Using Potassium Levels in Vitreous Humour for Estimating Postmortem Interval - A Systematic Review. International Journal of Medical Toxicology and Legal Medicine, 27(1), 143–153. Retrieved from ACM ACS APA ABNT Chicago Harvard IEEE MLA Turabian Vancouver Endnote/Zotero/Mendeley (RIS) BibTeX Issue Vol. 27 No. 1 (2024) Section Articles
6180	https://math.stackexchange.com/questions/4753616/extremal-of-a-functional-on-a-constraint	Skip to main content Extremal of a Functional on a Constraint Ask Question Asked Modified 2 years ago Viewed 86 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am trying to understand how to extremize a function S:C∞(R→Rn)→R, which takes as input a function x(t) and gives as output a scalar, on a constraint g:Rn→Rm regarding x(t), meaning the constraints are constraining the x(t), which thus reduces the domain of S. However, I am not really sure if S is mapping from a submanifold, since I was not able to express the domain as the kernel of a function. I tried setting ddϵS=!0, where we define x=x0+ϵη and x0 is the function over which S is extremal and therefore for ϵ=0 ddϵS indeed becomes 0, but I don't know how to incorporate the constraints on x(t) using this method. Since the constraints are affecting x(t) and not S, I refrained from directly using the method of Lagrange multipliers: ∇S=∑iλ∇g which is I think not possible, since S is depdendant on only one variable, therefore the left hand side is a scalar, whereas the right hand side is a vector. Any help is much appreciated! calculus-of-variations constraints Share CC BY-SA 4.0 Follow this question to receive notifications asked Aug 15, 2023 at 23:52 gluongluon 101 Add a comment \| 1 Answer 1 Reset to default This answer is useful 1 Save this answer. Show activity on this post. In general, Lagrange multipliers should lie in the dual space of whatever the codomain of the constraint is, which leads to the Lagrangian L(x,λ)=S(x)+(λ,g(x)), where ( , ) denotes the duality pairing. Often, this is an inner product, such as in the case where g takes values in Rm. In this case, λ can be interpreted as an m-vector: λ=(λ1,…,λm)⊤ and the objective can be written as L(x)=S(x)+λ⊤g(x). Your formulation of the constraint is a bit odd, as it is unclear how it relates to x. g is a mapping from Rn and x is a mapping to Rn, so does it act pointwise on a single value of x(t)? If we interpret g as a mapping from the function space to Rm, then the gradient of the Lagrangian w.r.t. x is well-posed, as we would have ∇xL=∇xS+λ⊤∇xg. The key here is that ∇xg\|x is a linear map from the function space to Rm, so the operator by λ⊤∇xg is now an operator from the function space to R, just like ∇xS, so we are good. The key points are You need m Lagrange multipliers, one for each degree of freedom in the output of the constraint. You need to formulate the constraint as an operation on the state x so that it has the same domain as S. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Aug 18, 2023 at 18:14 whpowell96whpowell96 7,91911 gold badge1212 silver badges2424 bronze badges Add a comment \| You must log in to answer this question. 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6181	https://pubs.acs.org/doi/10.1021/acs.jpcc.8b03459	S1 Supporting Information Barium in High Oxidation States in Pressure-stabilized Barium Fluorides Dongbao Luo, 1 Yanchao Wang, 1 Guochun Yang 2 and Yanming Ma 1,3 1.State Key Laboratory of Superhard Materials, College of Physics, Jilin University, Changchun 130012, China 2. Centre for Advanced Optoelectronic Functional Materials Research and Laboratory for UV Light-Emitting Materials and Technology of Ministry of Education, Northeast Normal University, Changchun 130024, China 3. International Center of Future Science, Jilin University, Changchun 130012, China Address correspondence to: yanggc468@nenu.edu.cn ;mym@calypso.cn Index page 1. Computational details ·····································································S2 Pressure-composition phase diagram of BaF 2 at low pressure range ···············S3 Phonon dispersion curves and phonon density of states of Ba-F compounds······S3 Calculated electron partial density of states ············································S4 Relativistic corrections on the Ba (Sr, Ca) orbital levels ·····························S4 Band structures of BaF 2 at 200 GPa and BaF 3 at 50 GPa······························S5 Spin DOS and projected spin DOS of BaF 3 at 50 GPa and BaF 5 at 200 GPa ······S5 Bond length of BaF n (n > 2) compounds·················································S6 Details structures used for the calculations of decomposition enthalpy ·············S7 References ··················································································S8 S2 Computational Details The structural prediction approach is based on a global minimization of free energy surfaces merging ab initio total-energy calculations through CALYPSO (Crystal structure AnaLYsis by Particle Swarm Optimization) methodology as implemented in its same-name CALYPSO code. 1,2 Our structure search for BaF n (n >1) compounds is performed with the formula units (f.u.) up to 4 at 0, 50, 100 and 200 GPa, respectively. In the first generation, a population of structures with certain symmetry are constructed randomly. Local optimizations of the generating structures are done with the conjugate gradients method through the VASP code, with an economy set of input parameters and an enthalpy convergence of 1 × 10 -5 eV per cell. 3,4 Step in the second generation, 60% of them with lower enthalpies are selected to produce the next generation structures by PSO , and 40% of the structures in the new generation are randomly generated. A structure fingerprinting technique of bond characterization matrix is applied to the generated structures, so that identical structures are strictly forbidden. These procedures significantly enhance the diversity of the structures, which is crucial for the efficiency of the global search of structures. For most of the cases, the structure searching simulation for each calculation was stopped after we generated 1000 ~ 1200 structures (e.g., about 20 ~ 30 generations). Higher precision of structure optimization by VASP was adopted for a number of structures with low enthalpies. The cut-off energy for the expansion of wavefunctions into plane waves is set to 800 eV in all calculations, and the Monkhorst-Pack k-mesh with a maximum spacing is 0.032, which gives total energy well converged within ~ 1 meV/atom. The electron-ion interaction was described by means of projector augmented wave with 5 s25p66s2 and 2 s22p5 electrons as valence for Ba and F atoms, respectively. Afterwards, the structures with lowest enthalpies were used to investigate the energetic stabilities for BaF x (x = 1 - 6) compounds by the following formula: ∆ Hf = [ H(BaF x) – xH (Ba) – yH (F)]/( x + 1), where H(BaF x) is the enthalpy of S3 the considered compound, H(Ba) is the enthalpy of elemental Ba, and H(F) is the enthalpy of elemental F. Supplementary Figures Figure S1. Pressure-composition phase diagram of BaF 2 at low pressure range. 5,6 Fm-3m-structured BaF 2 ranges from 0 to 3 GPa. Pnma -structured BaF 2 is stable from 3 to 16 GPa. P63/mmc -structured BaF 2 ranges from 16 to 110 GPa. Figure S2. Phonon dispersion curves and phonon density of states (PHDOS) projected on Ba and F atoms of Ba-F compounds. S4 Figure S3. Calculated partial density of states (PDOS) of (a) Pm-3m-structured BaF 3 at 50 GPa. (b) Cmcm -structured BaF 3 at 200 GPa. (c) R-3m-structured BaF 4 at 50 GPa. (d) I4/mmm -structured BaF 4 at 200 GPa. Figure S4. The comparison between influence of relativistic and non-relativistic effect on the atomic orbital levels of Ba, Sr, and Ca. S5 Figure S5. Calculated band structures of (a) P-3m1-structured BaF 2 at 200 GPa. (b) Pm-3n-structured BaF 3 at 50 GPa. Figure S6. Spin DOS and projected spin DOS for (a, b) Pm-3n-structured BaF 3 at 50 GPa and (c, d) F-43 m-structured BaF 5 at 200 GPa. S6 Supplementary Tables Table S1. Bond length for BaF n (n > 2) compound. The Ba-Ba bond length in solid are 4.356 Å ( Im-3m -structured Ba) at 0 GPa, 3.121 Å ( P63/mmc -structured Ba) at 50 GPa, and 2.807 Å ( P6 3/mmc -structured Ba) at 200 GPa, respectively. Phase Pressure (GPa) Ba-F (Å) F-F (Å) Ba-Ba (Å) BaF 2-Fm-3m 0 2.723 3.144 4.449 BaF 2-P63/mmc 50 2.362 2.544 3.471 BaF 2-P-3m1 200 2.167 2.278 2.916 BaF 3-Pm-3n 50 2.494 2.231 3.864 BaF 3-Cmcm 200 2.234 1.957 3.303 BaF 4-R-3m 50 2.417 1.596 4.163 BaF 4- I4/mmm 200 2.319 1.701 3.655 BaF 5-F-43 m 200 2.278 1.962 3.881 S7 Table S2. Details structures used for the calculations of decomposition enthalpy. Phase Pressure (GPa) Lattice parameters (Å, °°°°)Atomic coordinates (fractional) BaF 2 Fm-3m 0 a = b = c = 6.2887 Ba(4a) 0.000 0.000 0.000 α = β = γ = 90.000 F(8c) 0.750 0.750 0.750 BaF 2 P63/mmc 50 a = b = 4.0918 Ba(2a) 0.000 0.000 0.000 c = 5.0873 Ba(2c) 0.333 0.667 0.250 α = β = 90.000 F(2d) 0.333 0.667 0.750 γ= 120.000 BaF 2 P-3m1 200 a = b = 3.6947 Ba(1a) 0.000 0.000 0.000 c = 4.5551 Ba(1b) 0.000 0.000 0.500 α = β = 90.000 Ba(2d) 0.333 0.667 0.302 γ= 120.000 F(6i) 0.333 0.667 0.782 BaF 3 Pm-3n 50 a = b = c = 4.4622 Ba(2a) 0.000 0.000 0.000 α = β = γ = 90.000 F(6c) 0.500 0.250 1.000 BaF 3 Cmcm 200 a = 4.7960 Ba(4c) 0.500 0.285 0.750 b = 5.8889 F(8e) 0.296 0.000 0.500 c = 4.4689 F(4c) 0.500 0.285 0.250 α = β = γ = 90.000 BaF 4 R-3m 50 a = b = c = 4.1805 Ba(3a) 0.000 0.000 0.000 α = β = γ = 59.716 F(18h) 0.333 0.667 0.431 F(18h) 0.333 0.667 0.244 BaF 4 I4/mmm 200 a = b = 4.4418 Ba(2a) 0.500 0.500 0.500 c = 3.7403 F(8j) 0.691 0.000 0.500 α = β = γ = 90.000 BaF 5 F-43 m 200 a = b = c = 5.4890 Ba(4b) 0.000 0.000 0.500 α = β = γ = 90.000 F(16e) 0.124 0.124 0.876 F(4c) 0.250 0.250 0.250 S8 Reference Wang, Y.; Lv, J.; Zhu, L.; Ma, Y., Crystal Structure Prediction Via Particle-Swarm Optimization. Phys. Rev. B 2010 , 82 , 094116. 2. Wang, Y.; Lv, J.; Zhu, L.; Ma, Y., Calypso: A Method for Crystal Structure Prediction. Comput. Phys. Commun. 2012 , 183 , 2063-2070. 3. Kresse, G.; Furthmüller, J., Efficient Iterative Schemes for Ab Initio Total-Energy Calculations Using a Plane-Wave Basis Set. Phys. Rev. B 1996 , 54 , 11169. 4. Perdew, J. P.; Burke, K.; Ernzerhof, M., Generalized Gradient Approximation Made Simple. Phys. Rev. Lett. 1996 , 77 , 3865-3868. 5. Smith, J. S.; Desgreniers, S.; Tse, J. S.; Sun, J.; Klug, D. D.; Ohishi, Y., High-Pressure Structures and Vibrational Spectra of Barium Fluoride: Results Obtained under Nearly Hydrostatic Conditions. Phys. Rev. B 2009 , 79 .6. Ai-Min, H.; Xiao-Cui, Y.; Jie, L.; Wei, X.; Su-Hong, Z.; Xin-Yu, Z.; Ri-Ping, L., First-Principles Study of Structural Stabilities, Electronic and Optical Properties of Srf2 under High Pressure. Chin. Rev. Lett. 2009 , 26 , 077103.
6182	https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/04%3A_Stoichiometry_of_Chemical_Reactions/4.03%3A_Classifying_Chemical_Reactions	4.3: Classifying Chemical Reactions - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode 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Home 2. Bookshelves 3. General Chemistry 4. Chemistry 1e (OpenSTAX) 5. 4: Stoichiometry of Chemical Reactions 6. 4.3: Classifying Chemical Reactions Expand/collapse global location 4.3: Classifying Chemical Reactions Last updated Oct 27, 2022 Save as PDF 4.2: Writing and Balancing Chemical Equations 4.4: Reaction Stoichiometry Page ID 38155 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Precipitation Reactions and Solubility Rules 1. Example 4.3.1: Predicting Precipitation Reactions 1. Solution 2. Exercise 4.3.1/04:_Stoichiometry_of_Chemical_Reactions/4.03:_Classifying_Chemical_Reactions#Exercise_.5C(.5CPageIndex.7B1.7D.5C)) Acid-Base Reactions Example 4.3.2: Writing Equations for Acid-Base Reactions Solution Exercise 4.3.21 Oxidation-Reduction Reactions Example 4.3.3: Assigning Oxidation Numbers 1. Solution Exercise 4.3.3 Example 4.3.4: Describing Redox Reactions Solution Exercise 4.3.4 Balancing Redox Reactions via the Half-Reaction Method Example 4.3.5: Balancing Redox Reactions in Acidic Solution Solution Exercise 4.3.5 Summary Footnotes Glossary Learning Objectives Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction) Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations Identify common acids and bases Predict the solubility of common inorganic compounds by using solubility rules Compute the oxidation states for elements in compounds Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction. Precipitation Reactions and Solubility Rules A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter). The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table 4.3.1). Table 4.3.1: Solubilities of Common Ionic Compounds in WaterSoluble compounds containExceptions to these solubility rules include group 1 metal cations (Li+, Na+, K+, Rb+, and Cs+) and ammonium ion (NH⁢A 4⁢A+) the halide ions (Cl−, Br−, and I−) the acetate (C⁢A 2⁢H⁡A 3⁢O⁢A 2⁢A−), bicarbonate (HCO⁢A 3⁢A−), nitrate (NO⁢A 3⁢A−), and chlorate (ClO⁢A 3⁢A−) ions the sulfate (SO⁢A 4⁢A−) ion halides of Ag+, Hg⁢A 2⁢A 2+, and Pb 2+ sulfates of Ag+, Ba 2+, Ca 2+, Hg⁢A 2⁢A 2+, Pb 2+, and Sr 2+ Insoluble compounds containExceptions to these insolubility rules include carbonate (CO⁢A 3⁢A 2−), chromate (CrO⁢A 4⁢A 2−), phosphate (PO⁢A 4⁢A 3−), and sulfide (S 2−) ions hydroxide ion (OH−) compounds of these anions with group 1 metal cations and ammonium ion hydroxides of group 1 metal cations and Ba 2+ A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide: 2 KI⁢(a⁢q)+Pb⁡(NO⁢A 3)⁢A 2⁢(a⁢q)→PbI⁢A 2⁢(s)+2 KNO⁢A 3⁢(a⁢q) This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts. The net ionic equation representing this reaction is: Pb⁢A 2+⁢(a⁢q)+2 I⁢A−⁢(a⁢q)→PbI⁢A 2⁢(s) Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure 4.3.1). The properties of pure PbI 2 crystals make them useful for fabrication of X-ray and gamma ray detectors. Figure 4.3.1: A precipitate of PbI 2 forms when solutions containing Pb 2+ and I− are mixed. (credit: Der Kreole/Wikimedia Commons) A photograph is shown of a yellow green opaque substance swirled through a clear, colorless liquid in a test tube. The solubility guidelines in Table 4.3.1 may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag+, NO⁢A 3⁢A−, Na+, and F− ions. Aside from the two ionic compounds originally present in the solutions, AgNO 3 and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO 3 and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations: NaF⁡(a⁢q)+AgNO⁢A 3⁢(a⁢q)→AgF⁡(s)+NaNO⁢A 3⁢(a⁢q)⁢(molecular) Ag⁢A+⁢(a⁢q)+F⁡A−⁢(a⁢q)→AgF⁡(s)⁢(net ionic) Example 4.3.1: P redicting Precipitation Reactions Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction. potassium sulfate and barium nitrate lithium chloride and silver acetate lead nitrate and ammonium carbonate Solution (a) The two possible products for this combination are KNO 3 and BaSO 4. The solubility guidelines indicate BaSO 4 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is Ba⁢A 2+⁢(a⁢q)+SO⁢A 4⁢A 2−⁢(a⁢q)→BaSO⁢A 4⁢(s) (b) The two possible products for this combination are LiC 2 H 3 O 2 and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is Ag⁢A+⁢(a⁢q)+Cl⁢A−⁢(a⁢q)→AgCl⁡(s) (c) The two possible products for this combination are PbCO 3 and NH 4 NO 3. The solubility guidelines indicate PbCO 3 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is Pb⁢A 2+⁢(a⁢q)+CO⁢A 3⁢A 2−⁢(a⁢q)→PbCO⁢A 3⁢(s) Exercise 4.3.1 Which solution could be used to precipitate the barium ion, Ba 2+, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate? Answer sodium sulfate, BaSO 4 Acid-Base Reactions An acid-base reaction is one in which a hydrogen ion, H+, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text. For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, H 3 O+. As an example, consider the equation shown here: HCl⁡(a⁢q)+H⁡A 2⁢O⁢(a⁢q)→Cl⁢A−⁢(a⁢q)+H⁡A 3⁢O⁢A+⁢(a⁢q) The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H 3 O+ ions are produced by a chemical reaction in which H+ ions are transferred from HCl molecules to H 2 O molecules (Figure 4.3.2). Figure 4.3.2: When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions) This figure shows two flasks, labeled a and b. The flasks are both sealed with stoppers and are nearly three-quarters full of a liquid. Flask a is labeled H C l followed by g in parentheses. In the liquid there are approximately twenty space-filling molecular models composed of one red sphere and two smaller attached white spheres. The label H subscript 2 O followed by a q in parentheses is connected with a line to one of these models. In the space above the liquid in the flask, four space filling molecular models composed of one larger green sphere to which a smaller white sphere is bonded are shown. To one of these models, the label H C l followed by g in parentheses is attached with a line segment. An arrow is drawn from the space above the liquid pointing down into the liquid below. Flask b is labeled H subscript 3 O superscript positive sign followed by a q in parentheses. This is followed by a plus sign and C l superscript negative sign which is also followed by a q in parentheses. In this flask, no molecules are shown in the open space above the liquid. A label, C l superscript negative sign followed by a q in parentheses, is connected with a line segment to a green sphere. This sphere is surrounded by four molecules composed each of one red sphere and two white smaller spheres. A few of these same molecules appear separate from the green spheres in the liquid. A line segment connects one of them to the label H subscript 2 O which is followed by l in parentheses. There are a few molecules formed from one central larger red sphere to which three smaller white spheres are bonded. A line segment is drawn from one of these to the label H subscript 3 O superscript positive sign, followed by a q in parentheses. The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 4.3.1). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars: CH⁢A 3⁢CO⁢A 2⁢H⁢(a⁢q)+H⁡A 2⁢O⁢(l)⇌CH⁢A 3⁢CO⁢A 2⁢A−⁢(a⁢q)+H⁡A 3⁢O⁢A+⁢(a⁢q) When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, CH⁢A 3⁢CO⁢A 2⁢A− (Figure 4.3.3). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.) Figure 4.3.3: (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons) This figure contains two images, each with an associated structural formula provided in the lower left corner of the image. The first image is a photograph of a variety of thinly sliced, circular cross sections of citrus fruits ranging in color for green to yellow, to orange and reddish-orange. The slices are closely packed on a white background. The structural formula with this picture shows a central chain of five C atoms. The leftmost C atom has an O atom double bonded above and to the left and a singly bonded O atom below and to the left. This single bonded O atom has an H atom indicated in red on its left side which is highlighted in pink. The second C atom moving to the right has H atoms bonded above and below. The third C atom has a single bonded O atom above which has an H atom on its right. This third C atom has a C atom bonded below it which has an O atom double bonded below and to the left and a singly bonded O atom below and to the right. An H atom appears in red and is highlighted in pink to the right of the singly bonded O atom. The fourth C atom has H atoms bonded above and below. The fifth C atom is at the right end of the structure. It has an O atom double bonded above and to the right and a singly bonded O atom below and to the right. This single bonded O atom has a red H atom on its right side which is highlighted in pink. The second image is a photograph of bottles of vinegar. The bottles are labeled, “Balsamic Vinegar,” and appear to be clear and colorless. The liquid in this bottle appears to be brown. The structural formula that appears with this image shows a chain of two C atoms. The leftmost C atom has H atoms bonded above, below, and to the left. The C atom on the right has a doubly bonded O atom above and to the right and a singly bonded O atom below and to the right. This O atom has an H atom bonded to its right which is highlighted in pink. Table 4.3.2: Common Strong Acids\| Compound Formula \| Name in Aqueous Solution \| --- \| \| HBr \| hydrobromic acid \| \| HCl \| hydrochloric acid \| \| HI \| hydroiodic acid \| \| HNO 3 \| nitric acid \| \| HClO 4 \| perchloric acid \| \| H 2 SO 4 \| sulfuric acid \| A base is a substance that will dissolve in water to yield hydroxide ions, OH−. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for example, NaOH and Ca(OH)2. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)2 dissolve in water and dissociate completely to produce cations (K+ and Ba 2+, respectively) and hydroxide ions, OH−. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases. Consider as an example the dissolution of lye (sodium hydroxide) in water: NaOH⁡(s)→Na⁢A+⁢(a⁢q)+OH⁢A−⁢(a⁢q) This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na+ and OH− ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases. Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 4.3.4). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here: NH⁢A 3⁢(a⁢q)+H⁡A 2⁢O⁢(l)⇌NH⁢A 4⁢A+⁢(a⁢q)+OH⁢A−⁢(a⁢q) This is, by definition, an acid-base reaction, in this case involving the transfer of H+ ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as NH⁢A 4⁢A+ ions. Figure 4.3.4: Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139) This photograph shows a large agricultural tractor in a field pulling a field sprayer and a large, white cylindrical tank which is labeled “Caution Ammonia.” The chemical reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base, the products are often a salt and water, and neither reactant is the water itself: acid+base→salt+water To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid Mg(OH)2) is ingested to ease symptoms associated with excess stomach acid (HCl): Mg⁡(OH)⁢A 2⁢(s)+2 HCl⁢(a⁢q)→MgCl⁢A 2⁢(a⁢q)+2⁢H⁡A 2⁢O⁢(l). Note that in addition to water, this reaction produces a salt, magnesium chloride. Example 4.3.2: Writing Equations for Acid-Base Reactions Write balanced chemical equations for the acid-base reactions described here: the weak acid hydrogen hypochlorite reacts with water a solution of barium hydroxide is neutralized with a solution of nitric acid Solution (a) The two reactants are provided, HOCl and H 2 O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H+ from HOCl to H 2 O to generate hydronium ions, H 3 O+ and hypochlorite ions, OCl−. HOCl⁡(a⁢q)+H⁡A 2⁢O⁢(l)⇌OCl⁢A−⁢(a⁢q)+H⁡A 3⁢O⁢A+⁢(a⁢q) A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely. (b) The two reactants are provided, Ba(OH)2 and HNO 3. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba 2+) and the anion generated when the acid transfers its hydrogen ion (NO⁢A 3⁢A−). Ba⁡(OH)⁢A 2⁢(a⁢q)+2 HNO⁢A 3⁢(a⁢q)→Ba⁡(NO⁢A 3)⁢A 2⁢(a⁢q)+2⁢H⁡A 2⁢O⁢(l) Exercise 4.3.21 Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.) Answer H⁡A 3⁢O⁢A+⁢(a⁢q)+OH⁢A−⁢(a⁢q)→2⁢H⁡A 2⁢O⁢(l) Explore the microscopic view of strong and weak acids and bases. Oxidation-Reduction Reactions Earth’s atmosphere contains about 20% molecular oxygen, O 2, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O 2, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification. Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride: 2 Na⁢(s)+Cl⁢A 2⁢(g)→2 NaCl⁢(s) It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction: 2 Na⁢(s)→2 Na⁢A+⁢(s)+2 e⁢A−Cl⁢A 2⁢(g)+2 e⁢A−→2 Cl⁢A−⁢(s) These equations show that Na atoms lose electrons while Cl atoms (in the Cl 2 molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur: (4.3.1)oxidation=loss of electrons(4.3.2)reduction=gain of electrons In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium. (4.3.3)reducing agent=species that is oxidized(4.3.4)oxidizing agent=species that is reduced Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl: H⁡A 2⁢(g)+Cl⁢A 2⁢(g)→2 HCl⁢(g) The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion. The oxidation number of an atom in an elemental substance is zero. The oxidation number of a monatomic ion is equal to the ion’s charge. Oxidation numbers for common nonmetals are usually assigned as follows: Hydrogen: +1 when combined with nonmetals, −1 when combined with metals Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, O⁢A 2⁢A 2−), very rarely −1 2 (so-called superoxides, O⁢A 2⁢A−), positive values when combined with F (values vary) Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values) The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion. Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties. Example 4.3.3: Assigning Oxidation Numbers Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species: H 2 S SO⁢A 3⁢A 2− Na 2 SO 4 Solution (a) According to guideline 1, the oxidation number for H is +1. Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur: charge on H⁡A 2⁢S=0=(2×+1)+(1×x) x=0−(2×+1)=−2 (b) Guideline 3 suggests the oxidation number for oxygen is −2. Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur: charge on SO⁢A 3⁢A 2−=−2=(3×−2)+(1×x) x=−2−(3×−2)=+4 (c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately. According to guideline 2, the oxidation number for sodium is +1. Assuming the usual oxidation number for oxygen (−2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4: charge on SO⁢A 4⁢A 2−=−2=(4×−2)+(1×x) x=−2−(4×−2)=+6 Exercise 4.3.3 Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions: K N O 3 Al H 3 N―⁢H 4+ msubsup Answer a N, +5 Answer b Al, +3 Answer c N, −3 Answer d P, +5 Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist as shown below). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here: (4.3.5)oxidation=increase in oxidation number(4.3.6)reduction=decrease in oxidation number Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H 2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in HCl). Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted below are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation: 10 Al⁢(s)+6 NH⁢A 4⁢ClO⁢A 4⁢(s)→4 Al⁢A 2⁢O⁢A 3⁢(s)+2 AlCl⁢A 3⁢(s)+12⁢H⁡A 2⁢O⁢(g)+3 N⁢A 2⁢(g) Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at 3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture. Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals: Zn⁡(s)+2 HCl⁢(a⁢q)→ZnCl⁢A 2⁢(a⁢q)+H⁡A 2⁢(g) Metallic elements may also be oxidized by solutions of other metal salts; for example: Cu⁡(s)+2 AgNO⁢A 3⁢(a⁢q)→Cu⁡(NO⁢A 3)⁢A 2⁢(a⁢q)+2 Ag⁢(s) This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu 2+ ions dissolve in the solution to yield a characteristic blue color (Figure 4.3.4). Figure 4.3.4: (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott) This figure contains three photographs. In a, a coiled copper wire is shown beside a test tube filled with a clear, colorless liquid. In b, the wire has been inserted into the test tube with the clear, colorless liquid. In c, the test tube contains a light blue liquid and the coiled wire appears to have a fuzzy silver gray coating. Example 4.3.4: Describing Redox Reactions Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant. ZnCO⁢A 3⁢(s)→ZnO⁡(s)+CO⁢A 2⁢(g) 2 Ga⁢(l)+3 Br⁢A 2⁢(l)→2 GaBr⁢A 3⁢(s) 2⁢H⁡A 2⁢O⁢A 2⁢(a⁢q)→2⁢H⁡A 2⁢O⁢(l)+O⁢A 2⁢(g) BaCl⁢A 2⁢(a⁢q)+K⁢A 2⁢SO⁢A 4⁢(a⁢q)→BaSO⁢A 4⁢(s)+2 KCl⁢(a⁢q) C⁢A 2⁢H⁡A 4⁢(g)+3 O⁢A 2⁢(g)→2 CO⁢A 2⁢(g)+2⁢H⁡A 2⁢O⁢(l) S olution Redox reactions are identified per definition if one or more elements undergo a change in oxidation number. This is not a redox reaction, since oxidation numbers remain unchanged for all elements. This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr 3(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br 2(l) to −1 in GaBr 3(s). The oxidizing agent is Br 2(l). This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H 2 O 2(aq) to 0 in O 2(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H 2 O 2(aq) to −2 in H 2 O(l). For disproportionation reactions, the same substance functions as an oxidant and a reductant. This is not a redox reaction, since oxidation numbers remain unchanged for all elements. This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C 2 H 4(g) to +4 in CO 2(g). The reducing agent (fuel) is C 2 H 4(g). Oxygen is reduced, its oxidation number decreasing from 0 in O 2(g) to −2 in H 2 O(l). The oxidizing agent is O 2(g). Exercise 4.3.4 This equation describes the production of tin(II) chloride: Sn⁡(s)+2 HCl⁢(g)→SnCl⁢A 2⁢(s)+H⁡A 2⁢(g) Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant. Answer Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant. Balancing Redox Reactions via the Half-Reaction Method Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps: Write the two half-reactions representing the redox process. Balance all elements except oxygen and hydrogen. Balance oxygen atoms by adding H 2 O molecules. Balance hydrogen atoms by adding H+ ions. Balance charge1 by adding electrons. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps: Add OH− ions to both sides of the equation in numbers equal to the number of H+ ions. On the side of the equation containing both H+ and OH− ions, combine these ions to yield water molecules. Simplify the equation by removing any redundant water molecules. Finally, check to see that both the number of atoms and the total charges2 are balanced. Example 4.3.5: Balancing Redox Reactions in Acidic Solution Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution. Cr⁢A 2⁢O⁢A 7⁢A 2−+Fe⁢A 2+Cr⁢A 3++Fe⁢A 3+ S olution Write the two half-reactions. Each half-reaction will contain one reactant and one product with one element in common. Fe⁢A 2+Fe⁢A 3+ Cr⁢A 2⁢O⁢A 7⁢A 2−Cr⁢A 3+ Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms. Fe⁢A 2+Fe⁢A 3+ Cr⁢A 2⁢O⁢A 7⁢A 2−2 Cr⁢A 3+ Balance oxygen atoms by adding H 2 O molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side. Fe⁢A 2+Fe⁢A 3+ Cr⁢A 2⁢O⁢A 7⁢A 2−2 Cr⁢A 3++7⁢H⁡A 2⁢O Balance hydrogen atoms by adding H+ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side. Fe⁢A 2+Fe⁢A 3+ Cr⁢A 2⁢O⁢A 7⁢A 2−+14⁢H⁡A+2 Cr⁢A 3++7⁢H⁡A 2⁢O Balance charge by adding electrons. The iron half-reaction shows a total charge of 2+ on the left side (1 Fe 2+ ion) and 3+ on the right side (1 Fe 3+ ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved. The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side (1 Cr⁢A 2⁢O⁢A 7⁢A 2− ion and 14 H+ ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr 3+ ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6−) = 6+, and charge balance is achieved. Fe⁢A 2+Fe⁢A 3++e⁢A− Cr⁢A 2⁢O⁢A 7⁢A 2−+14⁢H⁡A++6 e⁢A−2 Cr⁢A 3++7⁢H⁡A 2⁢O Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6. 6 Fe⁢A 2+6 Fe⁢A 3++6 e⁢A− Cr⁢A 2⁢O⁢A 7⁢A 2−+6 e⁢A−+14⁢H⁡A+2 Cr⁢A 3++7⁢H⁡A 2⁢O Add the balanced half-reactions and cancel species that appear on both sides of the equation. 6 Fe⁢A 2++Cr⁢A 2⁢O⁢A 7⁢A 2−+6 e⁢A−+14⁢H⁡A+6 Fe⁢A 3++6 e⁢A−+2 Cr⁢A 3++7⁢H⁡A 2⁢O Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here: 6 Fe⁢A 2++Cr⁢A 2⁢O⁢A 7⁢A 2−+14⁢H⁡A+6 Fe⁢A 3++2 Cr⁢A 3++7⁢H⁡A 2⁢O A final check of atom and charge balance confirms the equation is balanced. Final check of atom and charge balance confirms the equation is balanced.\| \| Reactants \| Products \| --- \| Fe \| 6 \| 6 \| \| Cr \| 2 \| 2 \| \| O \| 7 \| 7 \| \| H \| 14 \| 14 \| \| charge \| 24+ \| 24+ \| Exercise 4.3.5 In acidic solution, hydrogen peroxide reacts with Fe 2+ to produce Fe 3+ and H 2 O. Write a balanced equation for this reaction. Answer H⁡A 2⁢O⁢A 2⁢(a⁢q)+2⁢H⁡A+⁢(a⁢q)+2 Fe⁢A 2+→2⁢H⁡A 2⁢O⁢(l)+2 Fe⁢A 3+ Summary Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method. Footnotes 1 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. 2 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. Glossary acid substance that produces H 3 O+ when dissolved in water acid-base reaction reaction involving the transfer of a hydrogen ion between reactant species base substance that produces OH− when dissolved in water combustion reaction vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light half-reaction an equation that shows whether each reactant loses or gains electrons in a reaction.insoluble of relatively low solubility; dissolving only to a slight extent neutralization reaction reaction between an acid and a base to produce salt and water oxidation process in which an element’s oxidation number is increased by loss of electrons oxidation-reduction reaction(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements oxidation number(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic oxidizing agent(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced precipitate insoluble product that forms from reaction of soluble reactants precipitation reaction reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis reduction process in which an element’s oxidation number is decreased by gain of electrons reducing agent(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized salt ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide single-displacement reaction(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species soluble of relatively high solubility; dissolving to a relatively large extent solubility the extent to which a substance may be dissolved in water, or any solvent strong acid acid that reacts completely when dissolved in water to yield hydronium ions strong base base that reacts completely when dissolved in water to yield hydroxide ions weak acid acid that reacts only to a slight extent when dissolved in water to yield hydronium ions weak base base that reacts only to a slight extent when dissolved in water to yield hydroxide ions 4.3: Classifying Chemical Reactions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by LibreTexts. Back to top 4.2: Writing and Balancing Chemical Equations 4.4: Reaction Stoichiometry Was this article helpful? Yes No Recommended articles 4.2: Classifying Chemical ReactionsChemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: p... 8.6: Classifying Chemical ReactionsChemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: p... 4.2: Classifying Chemical ReactionsChemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: p... 7.3: Classifying Chemical ReactionsChemical reactions are classified according to similar patterns of behavior. This section will help you to differentiate between the different types o... 3.3: Classifying Chemical ReactionsChemical reactions are classified according to similar patterns of behavior. This section will help you to differentiate between the different types o... Article typeSection or PageAutonumber Section Headingstitle with colon delimitersLicenseCC BYLicense Version4.0Show Page TOCno on page Tags combustion reaction oxidation oxidation-reduction reaction oxidizing agent precipitate precipitation reaction reduction single-displacement reaction © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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6183	https://www.cuemath.com/numbers/binary-to-hexadecimal/	LearnPracticeDownload Binary to Hexadecimal Binary to hexadecimal conversion is another type of conversion that occurs in the number system. There are 4 types of the number system in mathematics i.e. binary, octal, decimal, and hexadecimal. Each of these forms can be converted to the other type of number system by using the conversion table or the conversion method. Let us explore the different ways of converting binary numbers to hexadecimal numbers and solve a few examples for a better understanding. \| \| \| --- \| \| 1. \| What is Binary to Hexadecimal Conversion? \| \| 2. \| Steps to Convert Binary to Hexadecimal \| \| 3. \| Convert Binary to Hexadecimal With Decimal Point \| \| 4. \| FAQs on Binary to Hexadecimal \| What is Binary to Hexadecimal Conversion? Binary to hexadecimal conversion is the process of converting binary numbers to hexadecimal numbers. Binary numbers have a base number of 2 while the base number of hexadecimal is 16. The conversion from binary to hexadecimal occurs with the help of the base numbers. There are ways through which the conversion is done, the first is by converting the binary to a decimal number then a hexadecimal number. The second is by using the binary to the hexadecimal conversion table. Before we get to the method of converting, let us see what binary and hexadecimal are. Binary Number System Binary number system is one of the simplest number systems that use the digits 0 and 1 only along with the base number as 2. Binary numbers are mostly used in computers that are very handy for engineers, networking and communication specialist, and in many modern computers. Digits 0 and 1 are called bits and 8 bits together make a byte. The binary number system does not deal with other numbers such as 2,3,4,5 and so on. For example: 101100012,110011012,10110012 are some examples of numbers in the binary number system. Hexadecimal Number System Hexadecimal number system is the positional numeral system in the number system that uses the base number of 16 along with sixteen digits/alphabets: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and A, B, C, D, E, F. Here, A-F of the hexadecimal system means the numbers 10-15 of the decimal number system respectively. Each digit in the hexadecimal number system represents the power of the base (16). For example: 4E716,3F16,6D2C16 are some examples of numbers in the hexadecimal number system. Steps to Convert Binary to Hexadecimal To convert binary to hexadecimal numbers, we need to use both the base numbers i.e 2 for binary and 16 for hexadecimal. The conversion process happens in two methods, the first method is by using the binary to hexadecimal conversion table where 1 hexadecimal number is equivalent to 4 binary numbers. The second method is by converting the hexadecimal number to a decimal number then convert it to a binary. Let us see both the methods in detail. Method 1: Convert Binary to Hexadecimal With Conversion Table One of the simplest and easiest methods to convert from binary to hexadecimal is by using the conversion table. Since binary numbers only have 0 and 1 that are called bits and hexadecimal numbers are also positional number system, every 4 bits or numbers is equivalent to 1 hexadecimal number that includes the alphabets A - F as well. The conversion table is as follows: Let us look at an example for a better understanding. For example: Convert (00110110101)2 to Hexadecimal. We first group the numbers in a set of 4. Since every 4 digit in binary becomes one 1 digit in hexadecimal. Add zeros to the left of the last digit if there aren't enough digits to make a set of four: 0001 1011 0101 By looking at the conversion table, we can find the equivalent hexadecimal number. 0001 = 1 , 1011 = B , 0101 = 5 We arrange the numbers together to get the final number. Therefore, (00110110101)2 = (1B5)16. Method 2: Convert Binary to Hexadecimal Without Conversion Table Binary numbers can be converted to hexadecimal numbers without using the conversion table as well. Binary numbers are first converted to decimal number then to a hexadecimal number. Here, the base number of a decimal number is 10. The binary number can be converted to a decimal number by expressing each digit as a product of the given number 1 or 0 to the respective power of 2. And to convert from decimal to hexadecimal we divide the number 16 until the quotient is zero. Let us look at an example for a better understanding. For example: Convert (0111000101001)2 to Hexadecimal. We first convert the binary number to a decimal number. To do that each digit is multiplied with the corresponding power of two. (0111000101001)2 = 0 × 212 + 1 × 211 + 1 × 210 + 1 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 (0111000101001)2 = 0 × 4096 + 1 × 2048 + 1 × 1024 + 1 × 512 + 0 × 256 + 0 × 128 + 0 × 64 + 1 × 32 + 0 × 16 + 1× 8 + 0 × 4 + 0 × 2 + 1 × 1 (0111000101001)2 = 0 + 2048 + 1024 + 512 + 0 + 0 + 0 + 32 + 0 + 8 + 0 + 0 + 1 (0111000101001)2 = 3625 Therefore, (0111000101001)2 = (3625)10. Once the decimal number is obtained, we convert this decimal number to a hexadecimal number. The number is divided by 16 until the quotient is zero. 3625/16 = 226 is the quotient, the remainder is 9 226/16 = 14 is the quotient, the remainder is 2 14/16 = 0 is the quotient, remainder is 14 The final number is obtained by arranging the numbers from bottom to top i.e. 1429. Since the hexadecimal number system only deals with 0 - 9 in numbers and 10 -15 in alphabets as A - F, so the number is E29. Therefore, (0111000101001)2 = (E29)16. Convert Binary to Hexadecimal With Decimal Point To convert the binary to hexadecimal with a decimal point, we use a similar method as used in the previous section. We use the conversion table to convert binary to hexadecimal numbers. With a decimal point, the binary number will have the fractional part as well which is considered after the decimal point. While conversion the decimal point does not affect the position of the numbers. Let us look at an example to understand this better. For example: (0100110.10110110)2 We first group the numbers in a set of 4. Since every 4 digit in binary becomes one 1 digit in hexadecimal. Add zeros to the left of the last digit if there aren't enough digits to make a set of four: 0010 0110 . 1011 0110 By looking at the conversion table mentioned in the previous section, we can find the equivalent hexadecimal number. 0010 = 2 , 0110 = 6 , 1011 = B , 0110 = 6 We arrange the numbers together to get the final number. The decimal point will be at the same position as it is in the binary number. Therefore, (0100110.10110110)2 = (26.B6)16. Related Topics Listed below are a few interesting topics related to binary to hexadecimal, take a look. Hexadecimal to Binary Decimal to Octal Octal to Decimal Read More Download FREE Study Materials Numbers Worksheets Binary to Hexadecimal Worksheets on Binary to Hexadecimal Numbers Examples on Binary to Hexadecimal Example 1: Find the hexadecimal equivalent of (001101011)2. Solution: We first group the numbers in a set of 4. Since every 4 digit in binary becomes one 1 digit in hexadecimal. Add zeros to the left of the last digit if there aren't enough digits to make a set of four: 0000 0110 1011 By looking at the conversion table, we can find the equivalent hexadecimal number. 0000 = 0 0110 = 6 1011 = B We arrange the numbers together to get the final number. Therefore, (001101011)2 = (6B)16. 2. Example 2: Find the decimal equivalent of (111001001110)2. Solution: Multiply each digit of the binary number by the corresponding power of two: (111001001110)2 = 1 × 211 + 1 × 210 + 1 × 29 + 0 × 28 + 0 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 (111001001110)2 = 1 × 2048 + 1 × 1024 + 1 × 512 + 0 × 256 + 0 × 128 + 1 × 64 + 0 × 32 + 0 × 16 + 1 × 8 + 1 × 4 + 1 × 2 + 0 × 1 (111001001110)2 = 2048 + 1024 + 512 + 0 + 0 + 64 + 0 + 0 + 8 + 4 + 2 + 0 (111001001110)2 = 3662 Therefore, (111001001110)2 = (3662)10. 3. Example 3: Find the hexadecimal equivalent of (11001.001110)2 Solution: We first group the numbers in a set of 4. Since every 4 digit in binary becomes one 1 digit in hexadecimal. Add zeros to the left and right to the last digits if there aren't enough digits to make a set of four: 0001 1001 . 0011 1000 By looking at the conversion table mentioned in the previous section, we can find the equivalent hexadecimal number. 0001 = 1 , 1001 = 9 , 0011 = 3 , 1000 = 8 We arrange the numbers together to get the final number. The decimal point will be at the same position as it is in the binary number. Therefore, (11001.001110)2 = (19.38)16. View Answer > Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class Practice Questions on Binary to Hexadecimal Check Answer > FAQs on Binary to Hexadecimal What is Binary to Hexadecimal? Binary to hexadecimal is a form of conversion where a binary number with the base of 2 is converted to a hexadecimal number with the base of 16. Binary numbers only have 2 numbers or bits i.e. 0 and 1. While hexadecimal numbers deal with numbers and alphabets, 0 - 9 and A - F (10 -15). How Do You Convert Binary to Hexadecimal? The steps to convert binary to hexadecimal are: Break down the binary number into groups with 4 digits in each group. By looking at the conversion table, write the hexadecimal equivalent of each of the groups. Combine all the numbers together to get the hexadecimal number. What is the Binary Number 11000011 in Hexadecimal? Break the binary into groups with 4 digits in each group. 1100 0011. By looking the conversion table, 1100 = C and 0011 = 3. Hence, (11000011)2 = (C3)16. What is FFFF in Binary? (FFFF)16 =((1111111111111111)2. What is the Hexadecimal Number F Equal to in Binary? Looking at the binary to hexadecimal conversion table, we can say F = 1111. How to Convert a Binary Number to a Decimal Number? We can convert a binary number to a decimal number expressing each digit as a product of the given number 1 or 0 to the respective power of 2. If a binary number has n digits, B = (a)n−1.. (a)3 (a)2 (a)1 (a)0, the decimal number for it is given as, D = ( (a)0×20) + ( (a)1×21) + ( (a)2×22) + ... 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6184	https://chem.libretexts.org/Courses/UW-Whitewater/Chem_260%3A_Inorganic_Chemistry_(Girard)/02%3A_Molecular_Orbital_Theory/2.10%3A_Second-Row_Diatomic_Molecules	2.10: Second-Row Diatomic Molecules - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter 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Home 2. Campus Bookshelves 3. UW-Whitewater 4. Chem 260: Inorganic Chemistry (Girard) 5. 2: Molecular Orbital Theory 6. 2.10: Second-Row Diatomic Molecules Expand/collapse global location 2.10: Second-Row Diatomic Molecules Last updated Jul 26, 2020 Save as PDF 2.9: Orbital Filling 2.11: Periodic Trends in π Bonding Page ID 240408 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. four key points to remember when drawing molecular orbital diagrams: 3. Example 2.10.3: Diatomic Sulfur 1. Strategy: 2. Solution: Exercise 2.10.3: The Peroxide Ion Molecular Orbitals for Heteronuclear Diatomic Molecules An Odd Number of Valence Electrons: NO Nonbonding Molecular Orbitals Example 2.10.4: The Cyanide Ion Strategy: Solution: Exercise 2.10.4: The Hypochlorite Ion Summary Contributors and Attributions Learning Objectives To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period. If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N 2, O 2, and F 2. four key points to remember when drawing molecular orbital diagrams: The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the "law of conservation of orbitals"). As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized. The interaction between atomic orbitals is greatest when they have the same energy. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F 2. We use the diagram in part (a) in Figure 2.10.1; the n = 1 orbitals (σ 1 s and σ 1 s ) are located well below those of the _n = 2 level and are not shown. As illustrated in the diagram, the σ 2 s and σ 2 s _ molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 p atomic orbitals because of the large difference in energy between the 2 s and 2 p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 p orbitals on each F is σ 2⁢p z and the next most stable are the two degenerate orbitals, π 2⁢p x and π 2⁢p y. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the σ 2⁢p z⋆ orbital is higher in energy than either of the degenerate π 2⁢p x⋆ and π 2⁢p y⋆ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F 2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ 2 s and σ 2 s __ orbitals, 2 fill the σ 2⁢p z orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F 2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F 2 is predicted to have a stable F–F single bond, in agreement with experimental data. Figure 2.10.1: Molecular Orbital Energy-Level Diagrams for Homonuclear Diatomic Molecules.(a) For F 2, with 14 valence electrons (7 from each F atom), all of the energy levels except the highest, σ 2⁢p z⋆ are filled. This diagram shows 8 electrons in bonding orbitals and 6 in antibonding orbitals, resulting in a bond order of 1. (b) For O 2, with 12 valence electrons (6 from each O atom), there are only 2 electrons to place in the (π n⁢p x⋆,π n⁢p y⋆) pair of orbitals. Hund’s rule dictates that one electron occupies each orbital, and their spins are parallel, giving the O 2 molecule two unpaired electrons. This diagram shows 8 electrons in bonding orbitals and 4 in antibonding orbitals, resulting in a predicted bond order of 2. We now turn to a molecular orbital description of the bonding in O 2. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O 2, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure 2.10.1. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ 2 s and σ 2 s __ orbitals, 2 more to fill the σ 2⁢p z orbital, and 4 to fill the degenerate π 2⁢p x⋆ and π 2⁢p y⋆ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. None of the other bonding models can predict the presence of two unpaired electrons in O 2. Chemists had long wondered why, unlike most other substances, liquid O 2 is attracted into a magnetic field. As shown in Figure 2.10.2, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O 2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. Figure 2.10.2: Liquid O 2 Suspended between the Poles of a Magnet.Because the O 2 molecule has two unpaired electrons, it is paramagnetic. Consequently, it is attracted into a magnetic field, which allows it to remain suspended between the poles of a powerful magnet until it evaporates. Full video can be found at www.youtube.com/watch?featur...&v=Lt4P6ctf06Q. The magnetic properties of O 2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H 2 O, CO 2, and N 2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H 2 O, CO 2, and N 2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O 2 with organic compounds to give H 2 O, CO 2, and N 2 would require that at least one of the electrons on O 2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier. Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N 2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the σ 2⁢p z molecular orbital is slightly higher in energy than the degenerate π n⁢p x⋆ and π n⁢p y⋆ orbitals. The difference in energy between the 2 s and 2 p atomic orbitals increases from Li 2 to F 2 due to increasing nuclear charge and poor screening of the 2 s electrons by electrons in the 2 p subshell. The bonding interaction between the 2 s orbital on one atom and the 2 pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ 2 s orbital and increases the energy of the σ 2⁢p z orbital. Thus for Li 2, Be 2, B 2, C 2, and N 2, the σ 2⁢p z orbital is higher in energy than the σ 3⁢p z orbitals, as shown in Figure 2.10.3 Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure 2.10.3 ). Thus for example, the σ 2⁢p z molecular orbital is at a lower energy than the π 2⁢p x,y pair. Figure 2.10.3: Molecular Orbital Energy-Level Diagrams for the Diatomic Molecules of the Period 2 Elements. Unlike earlier diagrams, only the molecular orbital energy levels for the molecules are shown here. For simplicity, the atomic orbital energy levels for the component atoms have been omitted. For Li 2 through N 2, the σ 2⁢p z orbital is higher in energy than the π 2⁢p x,y orbitals. In contrast, the σ 2⁢p z orbital is lower in energy than the π 2⁢p x,y orbitals for O 2 and F 2 due to the increase in the energy difference between the 2 s and 2 p atomic orbitals as the nuclear charge increases across the row. Completing the diagram for N 2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N 2 versus 141.2 pm in F 2), and the bond energy is much greater for N 2 than for F 2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N 2 bond is much shorter and stronger than the F 2 bond, consistent with what we would expect when comparing a triple bond with a single bond. Example 2.10.3: Diatomic Sulfur Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S 2, a bright blue gas at high temperatures. Given:chemical species Asked for:molecular orbital energy-level diagram, bond order, and number of unpaired electrons Strategy: Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S 2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S 2. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule. Calculate the bond order and describe the bonding. Solution: A Sulfur has a [Ne]3 s 2 3 p 4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure 2.10.1 and Figure 2.10.3, we need to know how close in energy the 3 s and 3 p atomic orbitals are because their energy separation will determine whether the π 3⁢p x,y or the σ 3⁢p z> molecular orbital is higher in energy. Because the ns–np energy gap increases as the nuclear charge increases (Figure 2.10.3), the σ 3⁢p z molecular orbital will be lower in energy than the π 3⁢p x,y pair. B The molecular orbital energy-level diagram is as follows: Two filled sigma orbitals, one filled sigma orbital, two filled pi orbitals, and two pi orbitals with one unpaired electron. Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. C Ten valence electrons are used to fill the orbitals through π 3⁢p x and π 3⁢p y, leaving 2 electrons to occupy the degenerate π 3⁢p x⋆ and π 3⁢p y⋆ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S 2 is (σ 3⁢s)2⁢(σ 3⁢s⋆)2⁢(σ 3⁢p z)2⁢(π 3⁢p x,y)4⁢(π 3⁢p x,y⋆)2 with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Exercise 2.10.3: The Peroxide Ion Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O 2 2−). Answer (σ 2⁢s)2⁢(σ 2⁢s⋆)2⁢(σ 2⁢p z)2⁢(π 2⁢p x,y)4⁢(π 2⁢p x,y⋆)4 bond order of 1; no unpaired electrons Molecular Orbitals for Heteronuclear Diatomic Molecules Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ B> χ A), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure 2.10.4. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. Figure 2.10.4: Molecular Orbital Energy-Level Diagram for a Heteronuclear Diatomic Molecule AB, Where χ B> χ A. The bonding molecular orbitals are closer in energy to the atomic orbitals of the more electronegative B atom. Consequently, the electrons in the bonding orbitals are not shared equally between the two atoms. On average, they are closer to the B atom, resulting in a polar covalent bond. A molecular orbital energy-level diagram is always skewed toward the more electronegative atom. An Odd Number of Valence Electrons: NO Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O 2 with N 2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O 2 to produce NO 2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure 2.10.13) shows that the general pattern is similar to that for the O 2 molecule (Figure 2.10.11). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 p atomic orbitals, the 11th electron must occupy one of the degenerate π orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N 2 and O 2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. Figure 2.10.13: Molecular Orbital Energy-Level Diagram for NO. Because NO has 11 valence electrons, it is paramagnetic, with a single electron occupying the (π 2⁢p x⋆,π 2⁢p y⋆) pair of orbitals. Note that electronic structure studies show the ground state configuration of NO to be (σ 2⁢s)2⁢(σ 2⁢s⋆)2⁢(π 2⁢p x,y)4⁢(σ 2⁢p z)2⁢(π 2⁢p x,y⋆)1 in order of increasing energy. Hence, the π 2⁢p x,y orbitals are lower in energy than the σ 2⁢p z orbital. This is because the NO molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond (Figure 2.10.11). Molecular orbital theory can also tell us something about the chemistry of N⁢O. As indicated in the energy-level diagram in Figure 2.10.13, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, N⁢O is easily oxidized to the N⁢O+ cation, which is isoelectronic with N 2 and has a bond order of 3, corresponding to an N≡O triple bond. Nonbonding Molecular Orbitals Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure 2.10.6 that the 1 s orbital of atomic hydrogen is closest in energy to the 3 p orbitals of chlorine. Consequently, the filled Cl 3 s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 s and Cl 3 p orbitals. Of the three p orbitals, only one, designated as 3 p z, can interact with the H 1 s orbital. The 3 p x and 3 p y atomic orbitals have no net overlap with the 1 s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 s, 3 p x, and 3 p y orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 p z than to the H 1 s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give H δ+−−C⁢l δ−. Figure 2.10.6: Molecular Orbital Energy-Level Diagram for HCl. The hydrogen 1 s atomic orbital interacts most strongly with the 3 p z orbital on chlorine, producing a bonding/antibonding pair of molecular orbitals. The other electrons on Cl are best viewed as nonbonding. As a result, only the bonding σ orbital is occupied by electrons, giving a bond order of 1. Electrons in nonbonding molecular orbitals have no effect on bond order. Example 2.10.4: The Cyanide Ion Use a “skewed” molecular orbital energy-level diagram like the one in Figure 2.10.4 to describe the bonding in the cyanide ion (CN−). What is the bond order? Given:chemical species Asked for:“skewed” molecular orbital energy-level diagram, bonding description, and bond order Strategy: Calculate the total number of valence electrons in CN−. Then place these electrons in a molecular orbital energy-level diagram like Figure 2.10.4 in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and describe the bonding in CN−. Solution: A The CN− ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure 2.10.4 fills the five lowest-energy orbitals, as shown here: Because χ N>χ C, the atomic orbitals of N (on the right) are lower in energy than those of C. B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN− ion has a triple bond, analogous to that in N 2. Exercise 2.10.4: The Hypochlorite Ion Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl−). What is the bond order? Answer All molecular orbitals except the highest-energy σ are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. Summary Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules, molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O 2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules, using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry. Contributors and Attributions Modified by Joshua Halpern (Howard University) 2.10: Second-Row Diatomic Molecules is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. 9.8: Second-Row Diatomic Molecules is licensed CC BY-NC-SA 3.0. Toggle block-level attributions Back to top 2.9: Orbital Filling 2.11: Periodic Trends in π Bonding Was this article helpful? Yes No Recommended articles 9.8: Second-Row Diatomic MoleculesMolecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a ... 10.8: Second-Row Diatomic MoleculesMolecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a ... 9.8: Second-Row Diatomic MoleculesMolecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a ... 9.7: Molecular OrbitalsA molecular orbital is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomi... 10.7: Molecular OrbitalsA molecular orbital is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomi... Article typeSection or PageLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on pageTranscludedyes Tags heteronuclear diatomic molecule homonuclear diatomic molecule nonbonding molecular orbitals source-chem-21758 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 2.9: Orbital Filling 2.11: Periodic Trends in π Bonding
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6186	https://www.scribd.com/document/690767562/Analytical-Reasoning-Puzzles-NMAT-OG	Analytical Reasoning+ Puzzles - NMAT OG \| PDF \| Matrix (Mathematics) Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(1)100% found this document useful (1 vote) 2K views 17 pages Analytical Reasoning+ Puzzles - NMAT OG This document contains 16 logical reasoning questions with information provided about people or animals sitting in a circular or linear arrangement. The questions require determining positio… Full description Uploaded by Kishan kumar AI-enhanced description Go to previous items Go to next items Download Save Save Analytical Reasoning+ Puzzles - NMAT OG For Later Share 100%100% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Analytical Reasoning+ Puzzles - NMAT OG For Later You are on page 1/ 17 Search Fullscreen 7.0 Logical Reasoning Practice the one who owns Lycan. Y owns Lycan and is sitting exactly between I is sitting second to the right of N. The one who owns Ariel is second to the right of the person who owns Garmin. P sits third to the left of the person who owns W. Neither Q nor L is the immediate neighbour of N. Q is fourth to the left of L. N does not own Creta or Ariel. person who owns Lambretta is sitting second to the right of the person who owns Creta. R owns Lambretta and he is not an immediate neighbour of N. 3 Analytical Puzzles and L. The one who owns Punto Directions for Questions 1-4: Read the below information and answer the questions that follow. Six people: C, D, E, F, G and H are standing in a straight line facing North, not necessarly in the same order. F s standing second to the left of D. C s standing fourth to the left of H and H is not standing on the extreme end of the line. D is standing second to the left of E. 1, What is the position of F with respect to E? Who sits fourth to the right of L? (A) Immediate left (A)P (B) Fourth to the left (B) Y (C) Third to the left (C) R (D) Third to the right (D) Q E) None of these E) None of these Which of the following pairs represents the people standing at the extreme ends of the line? Which of the following cars does N own? (A) Punto (A) CE (B) Scorpio (B) FH (C) Garmin (C) FE (D) w (D) CH E) Lycan (E) None of these What is P's position with respect to Q? Who is standing second to the right of F? (A) Third to the left (A) H (B) Fifth to the left (B) D C) Second to the right (C) G (D) Third to the right (E) Fourth to the left D) E E) None of these Which of the following combinations is correct? Four of the following five pairs are alike in a certain way based on their position in the above arrangement and so form a group. Which of the following pairs do not belong to the group? (A) R-Garmin B) Y-Creta (C) N-Scorpio D) T-Ariel A) GC (E) All are true (B) GE Directions for Questions 9-12: Read the following information and answer the questions that follow. C) HG (D) DE Sarah, Caitin, Megan, Amy, Jessica, Emma, Amber and Zoe are sitting around a circular table facing the centre. Each one of them has a favourite fruit o r vegetable-dandelion, eggplant, apple, tomato, cherry, spinach, broccoli and beetroot. Sarah sits third to the right of the person who sits second to the left of Amber. E) FD Directions for Questions 5-8: Read the below information and answer the questions that follow. Eight friends N, L, M, R, P, Y, T and Q are sitting around a circular desk facing away from the centre. Each friend has a different carLambretta, WW, Lycan, Creta, Ariel, Garmin, Punto and Scorpio, but not necessarily in that order. T is sitting third to the right of P. h e one who owns Lambretta is second to the left of kes tomato. Amy Amber is not an immediate neighbour of the person who likes tomato. Only one person sits between Caitin who likes spinach and the person who likes tomato. 497 adDownload to read ad-free The person who likes apple sits third to the right of the person who likes spinach. Zoe sits between the person who likes apple and the person who likes eggplant. Jessica is not an immediate neighbour of Zoe. Who will be sitting second to the right of D? (A) S (B) M (C) R The person who likes eggplant sits third to the right of the person who likes cherry. Only one person sits between the person who likes broccoli and Emma. Jessica likes neither broccoli nor dandelion. (D) P (E) K Four of the following five pairs are alike in a certain way based on their positions in the above arrangement and so form a group. Which of the following pairs do not belong to the group? 9 Who sits third to the right of Mega? (A) Amber (B) Emma (C) The person who likes dandelion (A) SR (D) Zoe (B) DM E) The person who likes beetroot (C) PS Which of the following statements must be true? (D) KM A) Zoe likes dandelion. E) RP (B) The person who likes cherry sits to the immediate left of Sarah. Who will be sitting opposite to $? (A) R (C) Amy sits two places to the right of Amber. (D) Caitlin sits two places to the left of the person who likes tomato. (B) M (C) K (D) D (E) Jessica likes broccoli. (E) P What is the position of the person who likes broccoli with respect to Zoe? Who wll be sitting to the immediate left of K? (A) P (A) 3rd to the left (B) D B) 4th to the ri ght (C) R (C) 5th to the left (D) S (D) 4th to the left (E) None of these (E) 2nd to the right Directions for Questions 17-20: Read the below information and answer the questions that follow. Which of the following combinations is not correct? (A) Amber- cherry Four dogs F, G, H, J and two cats-K and M-will be assigned to exactly six cages numbered 1 to 6. Cage 1 faces Cage 4, Cage 2 faces Cage 5 and Cage 3 faces Cage (B) Emma-apple (C) Amy- dandelion (D) Zoe-beetroot E) Caitlin spinach The following conditions apply: The cats cann ot face each other, else theyll start fighting. A dog must be put in Cage 1. H must be put in Cage 6. J must be put in a cage whose number is 1 more than the number of K's cage K and H cannot be opposite each other. Directions for Questions 13-16: Read the below information and answer the questions that follow. Six friends-S, R, P, D, M and K-are sitting around a circular sofa. S is sitting opposite to R. P is sitting to the right of R but left of D. M s sitting to the left of R. Kis sitting to the right of S and left of M. Now, D and K interchang e their positions and so do M and R. Which one of the following must be true? (A) Fis assigned t o an even numbered cage. (B) Fis assigned to Cage (C) Jis assigned to Cage 2 or Cage (D) Jis assigned to Cage 3 or Cage 4 (E) Kis assigned to Cage 2 or Cage4. 498 adDownload to read ad-free 7.0 Logical Reasoning Practice If Jis assigned to Cage 3, which one of the following could be true? (D) Susmita (E) Gaurav (A) Fis assigned to Cage (B) Fis assigned to Cage 4 (C) Gis assigned to Cage (D) G is assigned to Cage (E) M is assigned to Cage In what capacity is Devanshu working? (A) Assistant Manager in Marketing (B) Assistat Manager in Personnel (C) Supervisor in Marketing (D) Director in Marketing E) Manager in Accounts 19 Which one of the following must be true? (A) A cat is assigned to Cage Which of the following is correct about Kuldeep? (B) A cat is assigned to Cage (C) K's cage is in a different row from M s cage. (D) Each cat is assigned to an even-numbered cage. (E) Each dog is assigned to a cage that faces a cat's cage. A) He is the Assistant Manager in Marketing. (B) He is the Assistant Manager in Personnel. C) He is the Supervisor in Marketing. (D) He is the Director in Marketing. E) He is the Manager in Accounts. If K's cage is in the same row a s H's cage, which one of the following must be true? Which one is the correct combination? (A) Fs cage is in the same roW a s J's cage. (B) Fis assigned to a lower-numbered cage than G. C) Gis assigned to a lower-numbered cage than M. (D) G's cage faces H's cage. (A) Gaurav-Chennai-Accounts (B) Pratima-Chennai-Marketing (C) Pratima-Chennai-Marketing (D) Pratima-Chennai-Marketing E) Kuldeep-Assistant Manager-Personnel E) M's cage is in the same row as Gs cage. Directions for Questions 21-24: Based on the information given beloW, answer the questions that follow. (Real NMAT Question) Directions for Questions 25-28: Read the below information and answer the questions that follow on the basis of this information. Five people, Kuldeep, Gaurav, Susmita, Pratima, and Devanshu are working in three branches of a company based at Bangalore, Chennai, and Hyderabad. Two people work in Bangalore and two people work in Chennai. Of these four employees, one works in Accounts and another works in Personnel. The remaining are in Marketing. There are 2 Assistant Managers, one Manager, one Director and one Nine people, Richard, Emmanuel, Luke, Andre, Patrick, Ethan, Jason, Shane and Joshua, stay on different floors of a 9-storey building. All of them own one car each, and each car is of a different colour: blue, white, grey, black, green, yelow, orange, red and pink, not necessarily in that order. The ground floor is numbered 1 and the topmost floor is numbered Shane owns a black coloured car and stays on a n even numbered floor. Richard stays on any even numbered floor below the floor on which Shane ervisor. Gaurav is the Director in the Marketing division at Chennai. stays. The person who owns the orange coloured car stays on the fourth floor. Pratima is the Manager at neither the Bangalore nor the Chennai branch. She is in the Accounts department. Patrick stays on the second floor and owns the white coloured car. The person who owns a pink coloured car stays on the third floor. Richard does not own a green coloured car. There are two floors between the floors on which the people owning the red and the black coloured cars stay. .Luke owns a grey coloured car. There are three floors between the floors on which Luke and Jason stay. Andre stays on a floor immediately above Joshua's floor. There is one floor between the floors on which Ethan and Jason stay. The person in the Person nel department is an Assistant Manager in Bangalore. Susmita is at the Bangalore branch working as Supervisor and Devanshu is at the Chennai branch. Who is in the Personnel department: A) Devanshu B) Kuldeep C) Pratima 499 adDownload to read ad-free NMAT by GMACTM Official Guide 2020 Ethan does not own the pink coloured car and does not stay on the ground floor. The person who owns the blue car stays on the top-most floor. 29 Who among the following is pursuing B.Com? (A) J, V and W 25 Who stays on floor number 8? (B) V, W and T (C) J, V and T (A) Emmanuel (D) J,P and R (B) Andre (E) None of the above (C) Richard (D) Ethan What is the favourite musical instrument of M? (E) Shane (A) Flute (B) Sitar ow many persons are staying between Jason and Emmanuel? (C) Guitar (A) Three (D) Banjo B) Four (E) Saxophone (C) Two What are the favourite musical instruments of those who (D) One are pursuing B.Sc? (E) None of these (A) Guitar and Violin 27 Who stays o n the floor immediately below Joshua's floor? (B) Sitar and Tabla (C) Tabla and violin (A) Ethan (D) Flute and sitar E) Violin and saxophone (B) Andre 32 (A) J-B.A. - Guitar (B) M-B.Com- Banjo (C) Patrick Which of the following combinations is correct? (D) Richard (E) Emmanuel or Luke C) T-B.A. Tabla Who owns the yellow coloured car? (D) T-B.Sc - Sitar (A) Andre (E) W-B.Com - Saxophone (B) Ethan Directions for Question 33: A word is represented by only one set of numbers a s given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in two matrices given below. The columns and rows of Matrixl are numbered from 0 to 4 and that of matrix ll are numbered from 5 to (C) Emmanuel (D) Richard (E) Luke Directions for Questions 29-32: Read the following information and answer the questions that follow. A letter from these matrices can be represented first by its row and next by its column, e.g., N can be represented by 10, 40 etc. and 'S' can be represented by 14, 44 etc. Similarly, you have to identify the set for the word NOISE. Seven friends-P, T, M, J, V, R and Ware pursuing B.Com, B.A. and B.Sc courses. Three of them are pursuing B.Com, two are pursuing B.A. and two are pursuing B.Sc. Each of them has a favourite musical instrument ranging from banjo, sitar, guitar, flute, violin, saxophone and tabla but not necessarily in the same order. None of those pursuing B.Com like either sitar or violin. M is pursuing B.A. and he likes banjo. R is pursuing B.Sc and likes tabla. J is pursuing B.Com and likes guitar. P, who does not like sitar, is pursuing the same discipline as R. T is pursuing the same discipline as M. V doe s not like saxophone. Matrix1 Matrix l 0 1 2 3 4 0 R E O N G 5 GVE A C \|1\| N P V Es6 R 0 N S S 2 MT\|0 N 3 E A C \| T\|TA 4 N T A R S 9 N S N E P 5 6 7 8 9 7 M N E S (A) 76, 85, 79, 68, 78 500 adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like 1 - PEDDINSERT - Insertion Machines No ratings yet 1 - PEDDINSERT - Insertion Machines 26 pages Matrix Arrangements Assignment: All The Puzzles Are Exactly CAT Level. 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6187	https://codesciencelove.wordpress.com/2013/01/07/given-an-n-x-n-grid-find-the-number-of-ways-of-moving-from-one-corner-to-the-other/	Given an n x n grid, find the number of ways of moving from one corner to the other \| Vishaal's coding shenanigans Vishaal's coding shenanigans About About Coding Experiments and Adventures Given an n x n grid, find the number of ways of moving from one corner to the other January 7, 2013 This was a fairly simple problem to code once I figured out an algorithm. Given an n x n square and moving only two directions – one must travel n places in one direction and n places in the other. For example in a 2 x 2 square traveling from the top left to the bottom right: The key to understanding this problem is to write down a set of paths and reach the simple logical conclusion that: One must travel 2 steps down and 2 steps to the right eventually to reach the bottom right: D D R R However, it doesn’t have to necessarily be in this order – one can take whichever option first as long as they do not exceed the limit n. D R R D D R D R R R D D R D R D R D D R Giving a grand total of 6. Next, I went about trying to find a formula for such a problem. I knew that there were a given number of slots n. At each slot, I had two options, either D or R. However, I had a limited number of each, which limits my total. There are a total of n slots with two options:D or R and each can be repeated only twice. This is simply the permutation with repeats formula: Advertisement where n represents the total number of letters and a,b,c… represent the number of repeats Although this formula was a little difficult to implement for massive numbers (even long double can’t hold values like 40!), I simplified the formula for permutations so that the processor would never have to crunch such massive numbers. Time for an input of a 20 x 20 grid: real 0m0.002s user 0m0.000s sys 0m0.001s Output: 1.37847E11 Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Related Using e^x to approximate piJuly 2, 2014 Solving (a fairly complex) Recurrence RelationAugust 1, 2014 Optimal Trajectory of an Ant – The most difficult problem I’ve ever solved!August 20, 2014 Category: Uncategorized Leave a comment Finding the length of the longest chain in the Collatz Sequence with an input less than n Finding the number of letters to write all the numbers from 1-1000 Leave a comment Cancel reply Δ Search: Go! Recent Posts Finding the three Roots of a Cubic Functions Phigital Number Phase Optimal Trajectory of an Ant – The most difficult problem I’ve ever solved! Expected Value of Empty Chair Fractions – My favorite problem yet! Number of Solutions to the Diophantine Reciprocal Problem Recent Comments Dhruv on Finding the nth lexicographic… Archives September 2014 August 2014 July 2014 April 2014 March 2014 December 2013 August 2013 July 2013 June 2013 May 2013 March 2013 February 2013 January 2013 December 2012 November 2012 September 2012 July 2012 June 2012 Categories Problems Uncategorized Meta Create account Log in Entries feed Comments feed WordPress.com Create a free website or blog at WordPress.com. Comment Reblog SubscribeSubscribed Vishaal's coding shenanigans Sign me up Already have a WordPress.com account? Log in now. Vishaal's coding shenanigans SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... Email (Required) Name (Required) Website Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy %d Design a site like this with WordPress.com Get started Advertisement
6188	https://en.wikipedia.org/wiki/1967_Marburg_virus_disease_outbreak	Jump to content 1967 Marburg virus disease outbreak Add links From Wikipedia, the free encyclopedia Disease outbreak in Germany and Yugoslavia \| 1967 Marburg virus disease outbreak \| \| --- \| \| \| \| \| Disease \| Marburg virus disease \| \| Virus strain \| Marburg virus \| \| Location \| West Germany and Yugoslavia \| \| First outbreak \| Marburg and Frankfurt \| \| Confirmed cases \| 32 \| \| Deaths \| 7 \| \| Territories \| 2 \| The 1967 Marburg virus disease outbreak was the first recorded outbreak of Marburg virus disease. It started in early August 1967 when 30 people became ill in the West German towns of Marburg and Frankfurt and later two in Belgrade, Yugoslavia (now Serbia). The infections were traced back to three laboratories in the separate locations which received a shared shipment of infected African green monkeys. The outbreak involved 25 primary Marburg virus infections and seven deaths, and six non-lethal secondary cases. Overview [edit] In early August 1967, patients with unusual symptoms indicating an infectious disease were admitted to the university hospitals in Marburg and Frankfurt. The first patients were treated in their homes for up to 10 days, even though the illness was described as beginning suddenly with extreme malaise, myalgia, headache, and a rapid increase in body temperature to as high as 39 °C (102.2 °F) or more. Although the clinical symptoms were not very alarming during the first 3–4 days, additional symptoms and signs appeared at the end of the first week. The patients were therefore admitted to a hospital. In some cases, patients died from severe hemorrhagic shock on the day after hospital admission. Severe hemorrhagic shock occurred in about 25% of patients. All patients who died had hemorrhagic shock. The first infections occurred in laboratory workers who were conducting necropsies on imported African green monkeys. The incubation time of Marburg virus disease could only be estimated retrospectively, after the source of infection and the date of exposure were known. Incubation ranged from 5 to 9 days, with an average of 8 days. The ratio of primary to secondary infections was 21:3 in Marburg, 4:2 in Frankfurt, and 1:1 in Belgrade. Three cases of secondary infection resulted from inadvertent needle-stick inoculations; in one case, a pathology technician cut himself on the forearm with a knife during a postmortem examination. Airborne transmission between humans did not occur, as indicated, for example, by the instance of a young man who slept in the same bed with his brother only a couple of days before he died; the brother did not develop disease and was seronegative for Marburg virus disease six months later. Origin [edit] The origin of the outbreaks was investigated at the same time as the microbiological studies. Early on in the investigation, it was realized that the patients in Marburg were employees of Behringwerke, a producer of sera and vaccines. The patients in Frankfurt were employees of Paul Ehrlich Institute, a control institute of sera and vaccines. The primary case in Belgrade was an employee involved in testing of live vaccines. All the patients at the three locations had contact with blood, organs, and cell cultures from African green monkeys (Cercopithecus aethiops). The monkeys' organs were used to make kidney cell culture for the production and safety testing of vaccines. The separate outbreaks were traced back to a shared shipment of infected green monkeys. Generally, shipments of green monkeys went directly from Uganda to Frankfurt. However, because of the Six Day War (5–10 June 1967), this shipment of monkeys was rerouted through London, where they were placed in animal storage because of a strike at the airport. After a two day delay, the monkeys were shipped to Frankfurt, and then to the laboratories in Frankfurt, Marburg, and Belgrade in June and July. The subsequent processing of the monkeys for cell culture at the three locations led to the laboratory-related outbreaks. The monkeys were believed to have been infected in Uganda, although infection from other animals in storage in London was also possible. History [edit] The Marburg virus disease made reappearances in other countries in 1975, 1980, 1987, 1990, 1998–2000, 2004–05, 2007, 2008, 2017 and 2021-24. The seven deaths out of the 31 initially diagnosed infections during the 1967 Marburg virus outbreak represent a case fatality rate of 23%. The 32nd case was diagnosed retroactively via serology. See also [edit] List of epidemics List of Filoviridae outbreaks References [edit] ^ "Marburg Hemorrhagic Fever (Marburg HF) \| CDC". www.cdc.gov. 25 February 2019. Archived from the original on 19 October 2022. Retrieved 16 February 2020. ^ a b "Marburg Virus - an overview \| ScienceDirect Topics". www.sciencedirect.com. Archived from the original on 2021-10-21. Retrieved 2020-08-12. ^ a b c d Slenczka, Werner; Klenk, Hans Dieter (2007-11-15). "Forty Years of Marburg Virus". The Journal of Infectious Diseases. 196 (Supplement_2): S131 – S135. doi:10.1086/520551. ISSN 0022-1899. PMID 17940940. ^ Brauburger, Kristina; Hume, Adam J.; Mühlberger, Elke; Olejnik, Judith (2012-10-01). "Forty-Five Years of Marburg Virus Research". Viruses. 4 (10): 1878–1927. doi:10.3390/v4101878. ISSN 1999-4915. PMC 3497034. PMID 23202446. ^ "Outbreaks Chronology: Marburg Hemorrhagic Fever \| Marburg Hemorrhagic Fever (Marburg HF) \| CDC". www.cdc.gov. 2019-02-25. Archived from the original on 2020-05-06. Retrieved 2020-02-16. Further reading [edit] Bebell, Lisa M.; Riley, Laura E. (June 2015). "Ebola Virus Disease and Marburg Disease in Pregnancy". Obstetrics & Gynecology. 125 (6): 1293–1298. doi:10.1097/AOG.0000000000000853. ISSN 1873-233X. PMC 4443859. PMID 26000499. Murphy, R. Henry and F. A. (2017). "Etymologia: Marburg Virus". Emerging Infectious Diseases. 23 (10): 1689. doi:10.3201/eid2310.ET2310. PMC 5621541. "Ebola Virus Disease & Marburg Virus Disease - Chapter 3 - 2018 Yellow Book \| Travelers' Health \| CDC". wwwnc.cdc.gov. Centers for Disease Control. Retrieved 8 November 2017. Martini, G. A.; Siegert, R. (2013). Marburg Virus Disease. Berlin, Heidelberg: Springer Science & Business Media. ISBN 9783662015933. Retrieved 8 November 2017. "Marburg virus disease: origins, reservoirs, transmission and guidelines – GOV.UK". Government of the United Kingdom. Retrieved 8 November 2017. Scholia has a topic profile for 1967 Marburg virus disease outbreak. \| v t e Filoviridae \| \| --- \| \| Ebolavirus \| \| \| \| --- \| \| Outbreaks \| 1976 Sudan outbreak 1976 Zaire outbreak 2013−2016 West African Ebola virus epidemic + Timeline + Reported cases and deaths + Responses + United Nations Ebola Response Fund + Operation United Assistance + in Guinea + in Liberia + in Mali + in Nigeria + in Sierra Leone + in Spain + in the US + in the UK + Ouse to Ouse Tock + Womey massacre Recent DR Congo outbreaks and epidemics + 2014 + 2017 + 2018 Équateur + Kivu epidemic + 2020 Équateur + 2021 North Kivu \| \| Species \| Bundibugyo ebolavirus + BDBV Reston ebolavirus + RESTV Sudan ebolavirus + SUDV Taï Forest ebolavirus + TAFV Zaire ebolavirus + EBOV \| \| Drug candidates \| BCX4430 Brincidofovir DZNep Favipiravir FGI-103 FGI-104 FGI-106 JK-05 Lamivudine mAb114 TKM-Ebola (failed) Triazavirin ZMapp Vaccines + cAd3-ZEBOV \| \| Drugs \| Vaccines + rVSV-ZEBOV \| \| Notable people \| Ebola 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6189	https://glreview.org/article/the-conspiracy-of-the-homintern/	The ‘Conspiracy’ of the ‘Homintern’ - The Gay & Lesbian Review From the G&LR's Blog Becoming Daddy Avian Art Mysteries for Queer-Coded Histories Hope Can Be Found in the Gayborhood Revisiting 2020: Then and Now My Aunt Alice The Paradox of “Tolerant” LGBT Communities How I Got Out A View of Queer History through Photography We Had to Pretend Joan E. Biren’s Eye to Eye Search FacebookInstagramLinkedInFeed Log In My Account A bimonthly magazine of history, culture, and politics. A bimonthly magazine of history, culture and politics. Navigate Home Current Issue Online Features Blog Here’s My Story Popular Podcast ITBR Podcast Print Magazine Past Issues Online Purchase Past Issues Print Editions: $12.00 Digital Edition Read Past Issues Purchase Past Issues Digital Editions: $2.99 Access the Current Issue About Contact Mission & History Staff/Board of Directors Legal/Privacy Donate Subscribe New Subscription Renew Your Subscription Give a Gift Subscription Give a Gift Subscription Renew Gift Subscription Check Status of Gifts HomeArticles The ‘Conspiracy’ of the ‘Homintern’ This article is only a portion of the full article. If you are already a premium subscriber please login. If you are not a premium subscriber, please subscribe for access to all of our content. 0 By Gregory Woodson November 7, 2017 Published in:May-June 2003issue. THERE WAS NO such thing as “the Homintern.” It existed only as a camp joke or an imagined plot. The Comintern, or Communist International, was a real organization set up by Lenin in 1919 and dissolved in 1943. The camp term “Homintern” is often said to have been coined by Cyril Connolly or W. H. Auden, but Harold Norse claimed it for himself. Most likely, various people invented it at the same time. The humorous implication was that queer people made up a secret, world-wide network of lovers and friends, a “Homosexual International.” The only people who ever took this simple play on words seriously were those who feared the spread of homosexual influence. In mere conversations, they saw plots; in groups of friends, conspiracies. While it’s true that some gay people became central to literature and modern culture in general, the “Homosexual International” was often superficially international and incidentally homosexual. Commented one writer on the persistence of this conspiracy theory through much of the 20th century: “The Homintern theory … is a constant obsession of certain journalists and crops up from time-to-time not only in the popular press but in the pages of otherwise respectable literary journals.” The writer was Gore Vidal, in 1970, but it could just as well have been said decades earlier or later. In a 1936 attack on the poet Stefan George (“Aufzeichnung Stefan George betreffend”), for example, Rudolf Borchardt claimed that the German press and publishing houses were dominated by gay men. Merely a variation on the anti-Semitic myths of Nazism, such paranoia seems unsurprising for the place and time. But one finds the same complaint being voiced in England at the same time, now coming from the Left. The best-known example is George Orwell’s attitude toward the Auden group. In Keep the Aspidistra Flying (1936), Orwell’s alter ego, the failed poet Gordon Comstock, rails against an English cultural review called The Primrose Quarterly, “one of those poisonous literary papers in which the fashionable Nancy Boy and the professional Roman Catholic walk bras dessus, bras dessous.” Comstock, who was, like Orwell, neither gay nor Catholic, regards the quarterly’s contributors as “a coterie of moneyed highbrows” and “that pansy crowd.” (“The sods! The bloody sods!”) In a later rant he comes back to the idea of the excessive cultural influence of “fashionable Nancy boys.” In The Road to Wigan Pier (1937), Orwell, speaking now in his own voice, refers several times to the Auden group as “the Nancy poets,” and he speaks of their tendency to “scratch each other’s backs”—presumably a veiled reference to sodomy as well as to mutual assistance. Fifty years after Orwell, Valentine Cunningham marred his otherwise impressive book, British Writers in the Thirties (1989), with the repeated suggestion that some cultural networks wielded a power that was somehow illegitimate because its members were predominantly gay men. Speaking of the Auden group, he seems both bewildered and disturbed to have to report that: “The shared male bed lay behind many of the coterie’s dedications”—as if heterosexuals never dedicate their books to their lovers! Cunningham describes this group and others in conventionally conspiratorial terms as “coteries bonded by shared private codes, covert languages and publicly inadmissible passions.” One group even becomes “the magic homosexual circle.” Cunningham cannot leave the point alone. He speaks of “the homosexual nature of much 30’s cliquery,” “the period’s crowd of homosexuals,” “the homosexual core of the clique,” and “this homosexual coterie.” One group of friends and collaborators is “a covey of homosexual chums.” (Homosexual men always have “chums” in this book.) Although there is an implied triviality and exoticism to all of these expressions, Cunningham chooses to take them very seriously indeed, as labels that carry a strong suggestion of a subversive departure from the values of family, religion, and state. A coterie or a clique is assumed to be held together by misplaced loyalties. Any group that sees itself as marginal, he implies, is likely to be only weakly committed to the national (or imperial) project. When Wyndham Lewis wrote in the 1930’s of “the intense ‘outcast’ esprit de corps of the pathic,” he was using a French phrase that evokes the militaristic togetherness of an enemy to describe what might, from a different point of view, be interpreted as the solidarity of the oppressed. That Cunningham, writing in the late 1980’s, was still using this kind of language to diminish the Homintern, such as it was, is rather depressing. It was during the Cold War period that the various national security services of the “Free World”—notably the FBI under J. Edgar Hoover—followed the Soviets and the Nazis by taking seriously the possibility that gay people could constitute a potentially subversive conspiracy. In 1953 the Eisenhower administration enacted a purge of homosexual “security risks” in government. The argument was that such people were vulnerable to blackmail by Soviet agents and were themselves, therefore, prone to become Russian spies. Pressure was put on the USA’s allies in NATO to take similar purgative measures. Thus, “sex perverts” came to be closely associated with spies in the public imagination. And yet, the entry under “Homosexuality” in Norman Polmar and Thomas B. Allen’s reliable Encyclopedia of Espionage (1998) names just nine gay men and one bisexual: Alfred Redl, Guy Burgess, the bisexual Donald Maclean, Anthony Blunt, Alan Turing, James A. Mintkenbaugh, William Martin, Bernon Mitchell, John Vassall, and Maurice Oldfield. The latter had to resign his position as co-ordinator of UK security and intelligence in Northern Ireland after he was found to be gay; there was no suggestion that in his previous incarnation as Director General of MI6 he ever spied for anyone but his own Whitehall masters. Similarly, there has never been any suggestion that Alan Turing ever betrayed the Allied war effort; on the contrary, he is known to have saved that effort by cracking the Germans’ secret code for military planning. So, we arrive at a grand total of just seven gay individuals who actually betrayed the interests of their own nations. In the liberal arts, where few great national interests are at stake and matters of life and death tend to be only theoretically engaged, similar prejudices have often prevailed. For over a century, anxieties about homosexual exclusivity have consistently shaped the critical reception of work by artists known by the critics to be gay. Following a review of a book about Benjamin Britten, for example, correspondence in the Times Literary Supplement (February 19, 1949) hinted at Britten’s homosexuality and spoke of “the small but powerful sect that threatens to kill with kindness one of the most naturally gifted of contemporary British composers.” It is not clear whether this was a reference to the group of colleagues and friends with whom Britten surrounded himself professionally at Aldeburgh or, considering the hints these remarks follow, gay friends in particular. The ambiguity seems deliberate. The composer William Walton responded to the feeling that he was being eclipsed by Britten by creating his own conspiracy theory. When Britten was offered the post of musical director at Covent Garden, Walton remarked: “There are enough buggers in the place already, it’s time it was stopped.” According to Michael Tippett, Walton mixed with a group of composers (Constant Lambert, Elizabeth Lutyens, and Alan Rawsthorne) who “all had great chips on their shoulder and entertained absurd fantasies about a homosexual conspiracy in music, led by Britten and Peter Pears.” On one occasion Walton whined, “Everyone is queer and I’m just normal, so my music will never succeed.” The heterosexual coterie of Walton, Lambert, Thomas Beecham and their chums coped with the conspicuous success of Britten and Pears with sodomy jokes based on fractured titles: “Twilight of the Sods,” “The Bugger’s Opera,” “The Stern of the Crew,” and so on. On one occasion, Charles Mackerras, who was working as the musical director on Britten’s Noye’s Fludde, made a disparaging remark to John Cranko about the number of boys in the piece. Cranko, who was gay himself, passed on the remark to Britten. (Cranko explained: “When suddenly you hear something like that, however long you may have worked together, suddenly you hate that person.”) Disgruntled at having thus been betrayed to the maestro, Mackerras conjured up the usual spectre, saying that Cranko “was a homosexual, and I’m not, and there is a sort of Freemasonry among them.” Similarly, the presence of gay men in the vanguard of the visual and performing arts gave rise to talk of a network of “queer” artists, dealers, and curators who were allegedly conspiring to promote their favorites at the expense of other talent. Such a rumor cropped up with reference to the Abstract Expressionists in New York. In 1959, in the magazine Arts, Hilton Kramer expressed his thinly veiled homophobia when he attacked Robert Rauschenberg and Jasper Johns as purveyors of “the window decorator’s æsthetic.” In the early 1960’s, like-minded critics whipped up a flurry of disquiet around the plays of Tennessee Williams, William Inge, and Edward Albee. That the three major American dramatists were known to be gay was bad enough; that they were purveying an unwholesome version of masculinity and femininity—creating weak male and strong female characters—was seen as intolerable. In 1963, Howard Taubman, drama critic for The New York Times, offered a “Primer” of “Helpful hints on how to scan the intimations and symbols of homosexuality in our theater”—in other words, on how to survey and police dramatic works for putatively queer material. Once these rumors got started, they didn’t readily go away. In The Real Life of Laurence Olivier (1996), Roger Lewis, while mistakenly insisting that Olivier was entirely heterosexual, gratuitously holds forth about a gay conspiracy in English public life: [L]et’s face it, in the acting profession, and in the arts in England (and in politics: there are over eighty homosexual Members of Parliament [out of 650]), it is practically impossible to become successful, or to garner honors, if you are too exclusively heterosexual. … It is a conspiracy—as bad as anti-Semitism. Literary editors, television producers, theatre critics, publishers, opera, ballet, museum curating: domains all controlled by [homosexuals].” (How, then, does he account for the “straight” Olivier’s undeniable success?) He also speaks of “the extent of theatre’s homosexual mafia,” and compiles a little list of its alleged membership, including H. M. Tennant, Hugh Beaumont, the Ivy Restaurant, Terence Rattigan, the Royal Court Theatre, Tony Richardson, and John Dexter. All important people and institutions, to be sure, but hardly a who’s who of London’s postwar theatrical establishment. In the mid-90’s the fact that media moguls David Geffen, Barry Diller, and Sandy Gallin, designer Calvin Klein, painter Ross Bleckner, and writer Fran Lebowitz were all close friends gave rise to rumors of a “Gay Mafia” (or “Velvet Mafia”) in charge of Hollywood. Once such clusters were identified as working in each other’s interests, it was a short step to the inference that they were working in the interests of gayness as a whole and excluding heterosexuals. The presence of more than one gay person evokes the possibility of a perverse cultural coup. I’m reminded of how President Robert Mugabe of Zimbabwe and his supporters keep calling the cabinet of British Prime Minister Tony Blair a “gay mafia” merely because it used to include two gay ministers. One viable response to such outspoken attacks on any sign of gay influence would be to embrace this notion of cultural conspiracy, whether ironically or in earnest. Just as W. H. Auden and Harold Norse cheerfully took ownership of the Homintern joke, we might decide that being thought to wield such cultural power isn’t such a bad thing. Indeed some gay people have contributed to the rumor that there’s a secret cabal of some sort. Even before the so-called Homintern, Proust wrote of “a freemasonry far more extensive, more powerful and less suspected than that of the Lodges.” Painter Francis Bacon ranted about “a Jewish, homosexual mafia” working against his interests in Manhattan after he failed to secure a collaboration with the photographer Peter Beard. Proust, a gay man and a Jew himself, recognized that certain persecuted minorities are forced to adopt defensive formations or a “freemasonry,” but saw this tendency as more likely to play a beneficial role in society than a destructive one. (Note: even as late as 1948, Gore Vidal was still using the term “freemasonry” to mean a discreet network. “It was a form of freemasonry,” he said in The City and the Pillar.) Loyal alliances among artists and writers are creative and productive. One has only to think of a few such groups to be persuaded of this: Natalie Barney’s salons at 20, rue Jacob in Paris, the Ballets Russes, the Ballets Suèdois, the Beats, the Bloomsbury Group, the “Nancy poets,” and so on. Where would modern culture be without them? The willingness of gay men and lesbians to associate across national boundaries throughout the last century led to some extraordinary encounters, some fleeting, others more enduring. To begin to understand the full cultural potential of such meetings, just imagine the conversations that took place between these pairs of individuals: Sergei Eisenstein and Noël Coward, Tamara de Lempicka and Adrienne Monnier, Yukio Mishima and James Merrill, Una Troubridge and Vaslav Nijinsky, Angus Wilson and Alberto Arbasino, Yves Saint Laurent and Andy Warhol, Roger Casement and Magnus Hirschfeld, Willa Cather and Stephen Tennant, John Minton and Gerard Reve, Anthony Blunt and Ludwig Wittgenstein, René Crevel and Gertrude Stein, Federico García Lorca and Hart Crane, May Sarton and Virginia Woolf. This is a random list of international encounters, all of them richly suggestive meetings of creative individuals who just happened to be gay. But there are those who think any such encounter must inevitably be perverted or sinister—or both. Where we see gay cultural transactions, they see a subversive plot. Gregory Woods’s latest poetry collection is The District Commissioner’s Dreams (Carcanet Press). Subscribe to The G&LR FacebookTwitterGoogle+LinkedInmore PinterestTumblrEmail Gregory Woods Read More from Gregory Woods Show Comments (0) Share Your ThoughtsCancel Reply Your Name (required) Your Email (required) Your Website (optional) Notify me by email of follow-up comments Δ Previous ArticleA Maori Writer in Two Worlds Next Article Women’s Lib Hits London, 1972 Related Stories The Ascension of the Retail Queen The Attraction of Confederate Gents A Romance Almost Lost to History September-October 2025 Features From Current Issue Cultural History The Ascension of the Retail Queen Feature Essay By William Benemann: By the 1830s, male store clerks had become a recognized subcategory of the labor market: they were the “counter jumpers.” Essays The Attraction of Confederate Gents Feature Essay By Andrew Holleran: Amid the animosity and violence between North and South, homosocial and even homoerotic elements endured. Confederate Sympathies is the study of these elements in American literature and politics before, during, and after the Civil War. Essays A Romance Almost Lost to History A few decades ago, researching something else, I stumbled upon a bit of history that had long been silenced, a tale that may be one of the oddest gay American love stories of the early 20th century. In his day, Ludwig Lewisohn (1882–1955) was hailed as a popular and prolific novelist as well as aMore Essays The French Evolution of Identities As far as I know, “lesbienne” first appears in print with its current denotation in neurologist Jules Cotard’s Études médicales (1870). He was part of a wave of European neuropsychiatrists obsessed with the “sexual perversions,” including “sexual inversion.” Historian Tamara Chaplin opens Becoming Lesbian: A Queer History of Modern France with this 19th-century context forMore ##### Blog 1. September 19, 2025Avian Art Mysteries for Queer-Coded Histories 2. August 29, 2025Revisiting 2020: Then and Now Sign Up for The Gay & Lesbian Review’s Newsletter! Receive news from The G&LR in your inbox once a month. Subscribe Donate Advertise The Gay & Lesbian Review / Worldwide(The G&LR) is a bimonthly magazine targeting an educated readership of LGBT individuals. Under the tagline, “a bimonthly journal of history, culture, and politics,”The G&LR publishes essays in a wide range of disciplines as well as reviews of books, movies, and plays. About GLR Books Outer Appearances Casual Outings In Search of Stonewall The Line of Dissent Writers’ Guidelines Contact Cart Copyright 2024 © The Gay & Lesbian Review. All rights reserved. Site by BWG. ✓ Thanks for sharing! AddToAny More… AddToAny More…
6190	https://brainstellar.com/puzzles/116/	Consecutive Heads \| Brainstellar Puzzles EasyMediumHardDeadly Consecutive Heads \| BRAINSTELLAR mediumConsecutive Headsprobability What is the expected number of coin tosses required to get N N N consecutive heads? Denote the term by E N E_N E N Hint Try to find a recurrence relation between E N E_{N}E N and E N−1 E_{N-1}E N−1. Answer Recurrence form: E N=2 E N−1+2 E_N = 2E_{N-1} + 2 E N=2 E N−1+2; Closed form: E N=2 N+1−2 E_N = 2^{N+1}-2 E N=2 N+1−2 Solution Method 1 We can simply reuse the method from problem "Innocent Monkey" (the law of total probability) Define let X N X_N X N = number of tosses to get N N N consecutive heads E N E_N E N = expected number of coin tosses we require from now on, to get N N N consecutive heads. Note that if we already have N−1 N-1 N−1 Heads, then with probability (1/2), we can get N N N consecutive heads, and with probability (1/2), we get a tail, and the state is reset. E N=1 2(E(N−1)+1)+1 2(E N−1+1+E N)E_N = \frac{1}{2}( E_{(N-1)} + 1) + \frac{1}{2}(E_{N-1} + 1 + E_N)E N=2 1(E(N−1)+1)+2 1(E N−1+1+E N) the extra 1 is because we just used a toss E N=2 E N−1+2 E_N = 2E_{N-1} + 2 E N=2 E N−1+2 This is a recurrence relation and a good enough solution. But we can solve this into its closed form too. The right way to solve this is by using generating functions or characteristic equations. But we will use a more intuitive approach. We can tell it is an exponential function, involving powers of 2 2 2 because the N N N th term is about 2 times N−1 N-1 N−1 th term. Suppose E N=C 1⋅2 N+C 2 E_N = C_1 \cdot 2^{N} + C_2 E N=C 1⋅2 N+C 2 Assume that E 0=0 E_0 = 0 E 0=0 (no tosses needed) ⟹0=C 1+C 2\implies 0 = C_1 + C_2⟹0=C 1+C 2 And E 1=2 E_1 = 2 E 1=2 (refer to method 2, step 1) ⟹2=C 1⋅2+C 2\implies 2 = C_1\cdot 2 + C_2⟹2=C 1⋅2+C 2 Simplify to get C 1=2,C 2=−2 C_1 =2, C_2 = -2 C 1=2,C 2=−2 This simplifies to E N=2(N+1)−2 E_N = 2^{(N+1)}-2 E N=2(N+1)−2 Method 2 This approach is more rigorous and algebraic. Let us solve a simpler question first. Step 1: What's the expected number of coin tosses to get N=1 N=1 N=1 heads? There are two possibilities: We get a head in the first toss itself. This happens with probability 1/2 1/2 1/2. We get a tail in the first toss, and then we start a new game, where number of chance is 1 more than the expected value. E 1=1 2⋅1+1 2⋅(1+E 1)E_1 = \dfrac{1}{2} \cdot 1 + \dfrac{1}{2} \cdot (1 + E_1)E 1=2 1⋅1+2 1⋅(1+E 1) Simplify to get E 1=2 E_1 = 2 E 1=2 Step 2: What's the expected number of coin tosses to get N=2 N=2 N=2 consecutive heads? There are three possibilities: We can get two heads in the first two tosses itself. This happens with probability 1/4 1/4 1/4, and takes 2 tosses. We can get a tail in the first toss, and then we start a new game, where number of attempts is 1 more than the expected value. We can get a heads in the first toss, and then tails in the second. Then we start a new game, where number of attempts is 2 more than the expected value. E 2=1 4⋅2+1 2⋅(1+E 2)+1 4⋅(2+E 2)E_2 = \dfrac{1}{4} \cdot 2 + \dfrac{1}{2} \cdot (1 + E_2) + \dfrac{1}{4} \cdot (2 + E_2)E 2=4 1⋅2+2 1⋅(1+E 2)+4 1⋅(2+E 2) Simplify to get E 2=6 E_2 = 6 E 2=6 Step 3: What's the expected number of coin tosses to get N N N consecutive heads? There are N+1 N+1 N+1 possibilities. We can get N N N heads in the first N N N tosses itself. This happens with probability 1/2 N 1/2^N 1/2 N, and takes N N N tosses. We can get a tail in the first toss, and then we start a new game, where number of attempts is 1 more than the expected value. We can get a heads in the first toss, and then tails in second. Then we start a new game, where number of attempts is 2 more than the expected value. ⋮\vdots⋮ We can get N−1 N-1 N−1 heads in the first N−1 N-1 N−1 tosses, and then a tail in the N t h N^{th}N t h toss. Then we start a new game, where number of attempts is N N N more than the expected value. E N=1 2 N⋅N+1 2⋅(1+E N)+1 2 2⋅(2+E N)+…+1 2 N⋅(N+E N)E_N = \dfrac{1}{2^N} \cdot N + \dfrac{1}{2} \cdot (1 + E_N) + \dfrac{1}{2^2} \cdot (2 + E_N) + \ldots + \dfrac{1}{2^{N}} \cdot (N + E_N)E N=2 N 1⋅N+2 1⋅(1+E N)+2 2 1⋅(2+E N)+…+2 N 1⋅(N+E N) This looks quite complex, but be assured that this can be simplified. I did not spend time on solving this. Simplify to get E N=2 N+1−2 E_N = 2^{N+1}-2 E N=2 N+1−2 Method 3 (for N=1 N=1 N=1 only) We can use linearity of expectation for N=1 N=1 N=1 Let X X X be the number of tosses required to get 1 head. Define indicator variable, X i=1 X_i = 1 X i=1 if no Heads appeared before the i i i th step and 0 0 0 otherwise. Once we get a "heads", we stop tossing, so all the later X i=0 X_i=0 X i=0. Note that X=∑i=1∞X i X = \sum_{i=1}^{\infty} X_i X=∑i=1∞X i E[X]=E[∑i=1∞X i]=∑i=1∞E[X i]E[X] = E[\sum_{i=1}^{\infty} X_i] = \sum_{i=1}^{\infty} E[X_i]E[X]=E[∑i=1∞X i]=∑i=1∞E[X i] (using linearity of expectation) E[X 1]=1⋅1 2+1⋅1 2=1 E[X_1] = 1 \cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{2} = 1 E[X 1]=1⋅2 1+1⋅2 1=1 (if heads, then also count 1, if tails, then also count 1) E[X 2]=P(First toss was Tails)⋅1 E[X_2] = P(\text{First toss was Tails}) \cdot 1 E[X 2]=P(First toss was Tails)⋅1 E[X 3]=P(First two toss were Tails)⋅1=1 4 E[X_3] = P(\text{First two toss were Tails}) \cdot 1 = \dfrac{1}{4}E[X 3]=P(First two toss were Tails)⋅1=4 1 ⋮\vdots⋮ E[X]=∑i=1∞E[X i]=1+1 2+1 4+…=2 E[X] = \sum_{i=1}^{\infty} E[X_i] = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots = 2 E[X]=∑i=1∞E[X i]=1+2 1+4 1+…=2 Okay, same answer as before. Unfortunately, I'm unable to extend this solution to N=2 N=2 N=2 and further because of the presence of the state. Method 4 (for N=1 N=1 N=1 only) This method uses the basic definition of expected value. E[X]=P(X=1)⋅1+P(X=2)⋅2+…E[X] = P(X=1) \cdot 1 + P(X=2) \cdot 2 + \ldots E[X]=P(X=1)⋅1+P(X=2)⋅2+… P(X=1)=1/2 P(X=1) = 1/2 P(X=1)=1/2 (because we can get a head in the first toss itself) P(X=2)=1/4 P(X=2) = 1/4 P(X=2)=1/4 (first tails, and second hands) ⋮\vdots⋮ E[X]=1 2⋅1+1 2 2⋅2+1 2 3⋅3 E[X] = \dfrac{1}{2} \cdot 1 + \dfrac{1}{2^2} \cdot 2 + \dfrac{1}{2^3} \cdot{3}E[X]=2 1⋅1+2 2 1⋅2+2 3 1⋅3 This is an Arithmetic Geometric Progression of type a,(a+d)r,(a+2 d)r 2,…,[a+(n−1)d]r n−1 a, (a+d)r, (a+2d)r^2, \ldots, [a + (n-1)d] r^{n-1}a,(a+d)r,(a+2 d)r 2,…,[a+(n−1)d]r n−1 with a=0,d=1,r=1/2 a=0, d=1, r=1/2 a=0,d=1,r=1/2 The sum of infinite such terms is S=a 1−r+d r(1−r)2=1/2 1/4=2 S = \dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2} = \dfrac{1/2}{1/4} = 2 S=1−r a+(1−r)2 d r=1/4 1/2=2 Comments PreviousNext Puzzle © BRAINSTELLAR
6191	http://www.mhtl.uwaterloo.ca/courses/ece309_mechatronics/lectures/pdffiles/chapter4.pdf	Entropy and the Second Law of Thermodynamics Reading Problems 5-1 →5-3, 5.6, 5.7 5-25, 5-71, 5-74 6-1 →6-11 6-31, 6-33, 6-42, 6-57, 6-59, 6-65, 6-72, 6-77 Introduction Zeroth Law: led to the concept of temperature First Law: led to the thermodynamic property called internal energy Second Law: leads to the thermodynamic property called entropy Why do we need another law in thermodynamics? Answer: While the 1st law allowed us to determine the quantity of energy transfer in a process it does not provide any information about the direction of energy transfer nor the quality of the energy transferred in the process. In addition, we can not determine from the 1st law alone whether the process is possible or not. The second law will provide answers to these unanswered questions. A process will not occur unless it satisﬁes both the ﬁrst and the second laws of thermody-namics. 1. Direction: Consider an isolated system where Q = W = 0 From the 1st law we know (U + KE)1 E1 = U2 E2      same both ways 1 →2 2 →1 1 gas gas possible impossible 1: cold system propeller & gas rotating 2: warm system propeller & gas stationary The physical interpretation of this is: State 1: Most of the energy is in a highly organized form residing in the macroscopic KE of the propeller and the rotating gas. State 2: All of the energy is now in a disorganized form residing in the microscopic E, i.e. U of the propeller and the gas. • The process 1 →2 has resulted in a higher state of molecular chaos. ENT ROP Y is the thermodynamic property that describes the degree of molecular disorder in matter. Hence, S2 > S1. Entropy can be considered a quantitative index that describes the quality of energy. • The process 2 →1 is impossible because it would require disorganized KE to pro-duce macroscopically organized KE. That is S2 < S1 which is impossible for an isolated system. • A thermodynamic process is only possible if it satisﬁes both the 1st and 2nd laws simultaneously. 2. Quality of Energy: A heat engine produces reversible work as it transfers heat from a high temperature reservoir at TH to a low temperature reservoir at TL. If we ﬁx the low temperature reservoir at TL = 300 K, we can determine the relationship between the efﬁciency of the heat engine, η = 1 −QL QH = 1 −TL TH as the temperature of the high temperature reservoir changes. In effect we are determin-ing the quality of the energy transferred at high temperature versus that transferred at low temperature. 2 TH TL reversible heat engine Wrev QH QL TL (K) TH (K) η 300 1000 0.700 300 800 0.625 300 600 0.500 300 400 0.250 Since the purpose of the heat engine is to convert heat energy to work energy, we can clearly see that more of the high temperature thermal energy can be converted to work. Therefore the higher the temperature, the higher the quality of the energy. 3 Second Law of Thermodynamics The second law of thermodynamics states: The entropy of an isolated system can never decrease. When an isolated system reaches equilibrium, its entropy attains the maximum value possible under the constraints of the system Deﬁnition Unlike mass and energy, entropy can be produced but it can never be destroyed. That is, the entropy of a system plus its surroundings (i.e. an isolated system) can never decrease (2nd law). Pm = m2 −m1 = 0 (conservation of mass) PE = E2 −E1 = 0 (conservation of energy) →1st law PS = Sgen = S2 −S1 ≥0 →2nd law • we have conservation of mass and energy, but not entropy. Entropy is not conserved. • the 2nd law dictates why processes occur in a speciﬁc direction i.e., Sgen cannot be −ve • Sgen is created by irreversibilities in a system, and therefore can be used to quantify the efﬁciency of a process. It is a measure of the perfectness of a process. The smaller Sgen is, the more efﬁcient the process. • if heat is leaving, then ∆S can be negative for a system. ∆Stotal cannot be negative. 4 System Surroundings Work Heat Non-Isolated Systems - their entropy may decrease Isolated System - its entropy may never decrease The second law states, for an isolated system: (∆S)system + (∆S)surr. ≥0 where ∆≡final −initial Entropy 1. Like mass and energy, every system has entropy. Entropy is a measure of the degree of microscopic disorder and represents our uncertainty about the microscopic state. 2. Reference: In a prefect crystal of a pure substance at T = 0 K, the molecules are com-pletely motionless and are stacked precisely in accordance with the crystal structure. Since entropy is a measure of microscopic disorder, then in this case S = 0. That is, there is no uncertainty about the microscopic state. Example: A freezing process 3. Relationship to Work: For a given system, an increase in the microscopic disorder (that is an increase in entropy) results in a loss of ability to do useful work. Work can completely (and easily) be transformed into thermal energy and then exchanged as heat. Example: Unrestrained expansion of an ideal gas A 1st law balance gives U1 = U2. For an ideal gas, we know u = u(T ) only and T1 = T2. The internal energy and the temperature of the gas remain unchanged but the gas in state 2 is in a more random or disordered state than 1 because now any gas molecules can move at 5 random within the whole volume V rather than V/2. In other words, uncertainty of a gas molecule has increased. S2 > S1 The reverse process 2 →1 is extremely unlikely to occur because the probability of ﬁnding all of the gas molecules in 1/2V at an instant of time is extremely remote, although not zero. Hence the process is irreversible. 4. Work: Energy transfer by work is microscopically organized and therefore entropy-free. Work can completely (and easily) be transformed into thermal energy and then exchanged as heat. 5. Heat: Energy transfer as heat takes place as work at the microscopic level but in a random, disorganized way. This type of energy transfer carries with it some chaos and thus results 6 in entropy ﬂow in or out of the system. Heat transfer from a heat source cannot be used to exchange the same amount of energy as work. 6. Reversibility: According to the 2nd law, no real process can be reversed without leaving any trace on the surroundings. In fact, the 2nd law says that it is impossible to do things perfectly (it is ideal but impossible). It is our job (as engineers) to make the process as close as possible to ideal, i.e. no loss and perfect recovery. A reversible process is the theoretical limit for an irreversible (real) process. Some factors that cause a process to become irreversible: • friction • unrestrained expansion and compression • mixing • heat transfer (ﬁnite ∆T ) • elastic resistance • inelastic deformation • chemical reactions In a reversible process things happen very slowly, without any resisting force, without any space limitation →Everything happens in a highly organized way (it is not physically pos-sible - it is an idealization). 7. Entropy is an extensive property with the dimensions of energy/temperature S : J K S = SA + SB = mAsA + mBsB 7 The T −s diagram s = (1 −x) sf + x sg where x ⇒quality →fraction of the total mass which is saturated vapor. 8 Gibb’s Equation From a 1st law energy balance when KE and PE are neglected Energy Input = Energy Output + Increase in Energy Storage δQ amount = δW + dU differential (1) We know that the differential form of entropy is dS = δQ T (2) δW = P dV (3) Combining Eqs. 1, 2 and 3 dS = dU T + P dV T ⇒ ds = du T + P dv T per unit mass Alternate Derivation: S = S(U, V ) dS = ∂S ∂U V = 1 T dU + ∂S ∂V U = P T dV T dS = dU + P dV This form of Gibb’s equation is very general and very useful. 9 2nd Law Analysis for a Closed System (Control Mass) MER TER CM dW dQ TTER We can ﬁrst perform a 1st law energy balance on the system shown above. dU = δQ + δW (1) For a simple compressible system δW = −P dV (2) From Gibb’s equation we know TT ER dS = dU + P dV (3) Combining (1), (2) and (3) we get TT ER dS = δQ where net in-ﬂow dS = δQ TT ER net out-ﬂow dS = −δQ TT ER 10 Therefore (dS)CM ≡storage = δQ TT ER ≡entropy flow + dPS ≡production Integrating gives (S2 −S1)CM = Q1−2 TT ER + Sgen ≥0 where Q1−2 TT ER - the entropy associated with heat transfer across a ﬁnite temperature difference, i.e. T > 0 Question? If all heat transfer processes are inherently irreversible →how can we speak of a reversible system involving heat transfer? S2 −S1 = Q1−2 TT ER + Sgen ↗0 (reversible) Possible only if Q occurs across an inﬁnitesimal temperature difference, i.e. quasi-equilibrium T dQ T =T - e TER TER CM 11 let ϵ = TT ER −TCM Sgen = (dS)CM + (dS)T ER = δQ (TT ER −ϵ) + δQ TT ER as ϵ →0 then Sgen →0. Thus (dS)CM ≈ δQ TT ER ≈δQ TCM In summary, δQ = T dS for a slow heat transfer process across a inﬁnitesimal temperature difference, therefore we can assume that it is a reversible process. Why use TT ER as the sink temperature regardless of the direction of heat ﬂow? Lets examine heat transfer from system A to system B, where system A is at TT ER and system B is changing in temperature as heat is transferred. (S2 −S1) = 2 1 Q T A = − 2 1 Q T B If the system is externally reversible →there are no other additions of S other than the heat transfer process. Therefore the magnitude of the entropy transfer is identical regardless of the sink temperature used, only the sign changes. Since TA = TT ER is constant it is convenient to use the form that allows us to use the constant temperature process (avoiding integration where we would have to monitor the temperature change in B), therefore we can say ∆S = Q1−2 TT ER 12 2nd Law Analysis for Open Systems (Control Volume) TER TER MER A A B B FR FR CV S= s m S= -s m B A 1-2 1-2 1-2 1-2 1-2 1-2 1-2 1-2 1-2 B A A B A B B A A B S= - Q d S= Q d T T TER TER dQ dQ m m SCV dW S=0 isolated S 0 gen ≥ where: FR -ﬂuid reservoir TER -thermal energy reservoir MER -mechanical energy reservoir 13 For the isolated system going through a process from 1 →2 δSgen = (∆S)sys + (∆S)sur δSgen = ∆SCV system + −sAmA 1−2 + sBmB 1−2 −δQA 1−2 T A T ER + δQB 1−2 T B T ER surroundings or as a rate equation ˙ Sgen = dS dt CV +  s ˙ m + ˙ Q TT ER   OUT −  s ˙ m + ˙ Q TT ER   IN This can be thought of as generation = accumulation + OUT −IN 14 Reversible Process Example: Slow adiabatic compression of a gas A process 1 →2 is said to be reversible if the reverse process 2 →1 restores the system to its original state without leaving any change in either the system or its surroundings. →idealization where S2 = S1 ⇒ Sgen = 0 T2 > T1 ⇒ increased microscopic disorder V2 < V1 ⇒ reduced uncertainty about the whereabouts of molecules Reversible Sgen=0 + Adiabatic P rocess Q=0 ⇒Isentropic P rocess S1=S2 The 2nd law states: (∆S)system + (∆S)surr = Sgen ≥0 where: > 0 irreversible (real world) = 0 reversible (frictionless, perfectly elastic, inviscid ﬂuid) • for all real processes, the entropy of a closed system (i.e. the universe) is continuously increasing. The best scenario, a reversible world. 15 • the closest man has gotten to a reversible process is with charge ﬂow through superconduc-tors at very low temperatures (approaching 0 K). • bad for engineers because good quality energy is disappearing • physicists use this argument to prove that the universe will eventually die • mathematicians have chaos theory But does: Isentropic P rocess ⇒Reversible + Adiabatic NOT ALWAYS - the entropy increase of a substance during a process as a result of irreversibilities may be offset by a decrease in entropy as a result of heat losses. At steady state: ˙ W = ˙ Q (1st law) SCV = constant ⇒isentropic From the 2nd law d dtSCV =  ˙ ms + ˙ Q TT ER   IN −  ˙ ms + ˙ Q TT ER   OUT + ˙ Sgen 16 but d dtSCV = 0 (steady state) ˙ ms = 0 (no ﬂow) ˙ Q TT ER = 0 (no heat transfer in) Therefore ˙ Sgen = ˙ Q Tsur , and ˙ Sgen > 0 (ﬂuid friction) ˙ Q > 0 (non-adiabatic system) If ˙ Sgen and ˙ Q/TT ER and ∆S = 0, i.e. the system is isentropic, however, the system is not adiabatic nor is it reversible. 17 Reversible Compression and Expansion real or ideal gas dQ dW m From the 1st law increase in energy storage = energy input dU = δW −δQ mdu = δW −δQ (1) From the 2nd law for a control mass dS = −δQ T + Sgen where Sgen = 0 for a reversible process. Therefore mds = −δQ T (2) Combining (1) and (2) through the −δQ term gives mdu = δW + mT ds (3) 18 From Gibb’s Eq. T ds = du + P dv mdu = mT ds −mP dv (4) Combining Eqs. (3) and (4) mT ds −mP dv = δW + mT ds δW = −mP dv Integrating gives W1−2 = −m 2 1 P dv or on a per unit mass basis w1−2 = W1−2 m = − 2 1 P dv Good for: • reversible process • any gas, real or ideal 19 Reversible Isothermal Expansion for an Ideal Gas We know for an ideal gas P v = constant = RT or P = constant v (1) P v = constant = P1v1 = P2v2 (2) The work done at the boundary of a simple, compressible substance (S.C.S.) during a reversible process w1−2 = − 2 1 P dv = −constant 2 1 dv v = −P1v1 ln v2 v1 = P1v1 ln v1 v2 W1−2 = mRT ln v1 v2 = mRT ln V1 V2 = mRT ln P2 P1 20 Calculation of ∆s in Processes For a simple compressible system, Gibb’s equations are given as T ds = du + P dv T ds = dh −vdP • these relations are valid for open and closed, and reversible and irreversible processes • ds can be solved by integrating the equations in combination with other known relations (dh = CpdT, du = CvdT, P v = RT , etc) • an adiabatic and reversible process must also be an isentropic process (∆s = 0) (but not visa versa) • T −s and h −s (Mollier) diagrams are very useful – the area under the curve on a T −s diagram is the heat transfer for internally reversible processes qint,rev = 2 1 T ds and qint,rev,isothermal = T ∆s T s dA = T ds process path 1 2 ds – Notes: ∗it is not good to attribute a value to s at a speciﬁc state ∗as before, we must assign a reference state where s = 0 (i.e. for water, we use the triple point) ∗one can use T ds relations to calculate ds and ∆s ∗we can deal with s exactly as we deal with h and u 21 Tabulated Calculation of ∆s for Pure Substances Depending on the phase of the substance: Calculation of the Properties of Wet Vapor: Use Tables A-4 and A-5 to ﬁnd sf, sg and/or sfg for the following s = (1 −x)sf + xsg s = sf + xsfg Calculation of the Properties of Superheated Vapor: Given two properties or the state, such as temperature and pressure, use Table A-6. Calculation of the Properties of a Compressed Liquid: Use Table A-7. In the absence of compressed liquid data for a property sT,P ≈sf@T Calculation of ∆s for Incompressible Materials T ds = du + P dv (the other T ds equation reduces the same way) - for an incompressible substance, dv = 0, and Cp = Cv = C ds = du T = C dT T s2 −s1 = 2 1 C(T ) dT T ∆s = Cavg ln T2 T1 where Cavg = [C(T1) + C(T2)]/2 - if the process is isentropic, then T2 = T1, and ∆s = 0 22 Calculation of ∆s for Ideal Gases For an ideal gas with constant Cp and Cv Ideal Gas Equation ⇒ P v = RT du = Cv dT ⇒ u2 −u1 = Cv(T2 −T1) dh = Cp dT ⇒ h2 −h1 = Cp(T2 −T1) There are 3 forms of a change in entropy as a function of T & v, T & P , and P & v. s2 −s1 = Cv ln T2 T1 + R ln v2 v1 = Cp ln T2 T1 −R ln P2 P1 = Cp ln v2 v1 + cv ln P2 P1 with the gas constant R given as R = Cp −Cv The equations for ∆s in an ideal gas can be readily attained by combining Gibb’s equation, the Ideal gas law and the differential form of internal energy and entropy as follows: ∆s for T and v given: From Gibb’s Equation ds = du T + P T dv = Cv dT T + Rdv v where from the ideal gas equation P T = R v and du = Cv dT . 23 If we integrate ds, we obtain (s2 −s1) = Cv ln T2 T1 + R ln v2 v1 ∆s for T and P given: Noting that dh = du + P dv + vdP we can substitute into ds to get ds = dh T −P T dv −v T dP + P T dv = dh T −v T dP = Cp dT T −RdP P = Cp ln T2 T1 −R ln P2 P1 ∆s for P and v given: And ﬁnally, s2 −s1 = Cv ln T2 T1 + R ln v2 v1 = Cv ln T2 T1 + (Cp −Cv) ln v2 v1 = Cv ln T2 T1 + Cp ln v2 v1 −Cv ln v2 v1 = Cp ln v2 v1 + Cv ln T2/T1 v2/v1 = Cp ln v2 v1 + Cv ln P2 P1 where T2/v2 T1/v1 = P2/R P1/R = P2/P1 24 System Level Analysis Energy Reservoirs Idealized Mechanical Energy Reservoir (MER) This is a source or sink for organized energy. M dW dE MER frictionless pulleys adiabatic non-isolated The system shown above is open to work transfer only, therefore it is entropy free. dS = 0 for a MER Idealized Thermal Energy Reservoir (TER) This is a source or sink for disorganized energy. dU TTER dQ non-isolated uniformly at T at all times large thermal inertia TER ﬁ TER Examples include; oceans, rivers, lakes, the atmosphere, etc. 25 Reversible Heat Engine A heat engine is a device in which a working substance (control mass) undergoes a cyclic process while operating between two temperature reservoirs (TER). As an example we can simplify the conventional power plant where a fossil fuel or a nuclear pro-cess is used to provide high temperature energy to a boiler. Through a vapor compression/expansion process, mechanical energy is produced in a turbine that in turn is used to produce electrical energy. combustion gases cooling water boiler condenser WP WT QH QH QL QL TER TER TH TL Wnet heat engines isolated systems S 0 gen ≥ Find: 1. Wnet = WT −Wp =? 2. Wnet,max =? 3. ηmax Solution: A 1st law energy balance gives ∆E ↗0= QH −QL −Wnet = 0 (1) The 2nd law gives Sgen ≥0 for an isolated system 26 An entropy balance gives Sgen = (∆S)CM ≡0 (cyclic) + (∆S)T ER−H ≡−QH/TH + (∆S)T ER−L ≡+QL/TL +(∆S)MER ↗0 Therefore QL TL = QH TH + Sgen or QL QH = TL TH + Sgen TL QH (2) Combining Eqs. (1) and (2) Wnet = QH −QL = QH 1 −QL QH = QH 1 −TL TH −Sgen TL QH = QH 1 −TL TH Wmax possible − TLSgen Wlost due to irreversibilties The engine efﬁciency is deﬁned as the beneﬁt over the cost η = benefit cost = Wnet QH = 1 −TL TH −TLSgen QH For a perfect (reversible) system where Sgen = 0 η = 1 −TL TH ⇐ Carnot efﬁciency It is clear that even for a “perfect” the efﬁciency cannot be 100% 27 If we were to assume an ambient temperature of TL = 293 K and TH = 1000 K, which is roughly equivalent to the maximum temperature of high grade metals, then ηmax = 1 −293 1000 ≈71% In reality most heat engines have an efﬁciency of less that 40% due to irreversibilities. The Carnot Cycle • an ideal theoretical cycle that is the most efﬁcient conceivable • based on a fully reversible heat engine - it does not include any of the irreversibilities asso-ciated with friction, viscous ﬂow, etc. • in practice the thermal efﬁciency of real world heat engines are about half that of the ideal, Carnot cycle T T T s s s Q Q Q W P = P P = P H H L L 1 1 2 4 4 3 in out 28 Process State Points Description Pump 1 →2 isentropic compression from TL →TH to return vapor to a liquid state Heat Supply 2 →3 heat is supplied at constant temperature and pressure Work Output 3 →4 the vapor expands isentropically from the high pressure and temperature to the low pressure Condenser 4 →1 the vapor which is wet at 4 has to be cooled to state point 1 The cycle is totally reversible. The reversed Carnot cycle is called the Carnot refrigeration cycle. Cycle Efﬁciency • deﬁned as the net work output divided by the gross heat supplied η = Wnet QH = QH −QL QH = 1 −TL TH From the ﬁgure the gross heat supplied is QH = area(s1 →s4 →3 →2 →s1) = TH(s4 −s1) The net work output is QH −QL = area(1 →4 →3 →2) = (TH −TL)(s4 −s1) 29 Therefore the Carnot efﬁciency is η = (TH −TL)(s4 −s1) TH(s4 −s1) = 1 −TL TH Practical Problems • at state point 1 the steam is wet at TL and it is difﬁcult to pump water/steam (two phase) to state point 2 • the pump can be sized smaller if the ﬂuid is 100% liquid water • the pump is smaller, cheaper and more efﬁcient • can we devise a Carnot cycle to operate outside the wet vapor region T s s s Q Q P = P P P H L 1 1 4 4 3 2 – between state points 2 and 3 the vapor must be isothermal and at different pressures -this is difﬁcult to achieve – the high temperature and pressure at 2 and 3 present metallurgical limitations The net effect is that the Carnot cycle is not feasible for steam power plants. 30 Example: A steady ﬂow process Find: ˙ Wmax =? Solution: Assume S.S.S.F and ∆P E = 0, ∆KE = 0 From the 1st law we know dE ↗0 dt CV = ˙ Q −˙ W + ( ˙ mh)in −( ˙ mh)out From the 2nd law we know dS ↗0 dt CV =  ˙ ms + ˙ Q TR   in −( ˙ ms)out + ˙ Sgen Rearranging the 2nd law to isolate ˙ Q ˙ Q = ( ˙ ms)out TR −( ˙ ms)in TR −˙ SgenTR Then substituting this into the 1st law ˙ W =     ˙ m[h −TRs]in bin −˙ m[h −TRs]out bout     −TR ˙ Sgen Therefore we can write ˙ W ≤˙ m(bin −bout) 31 where for a reversible process with ˙ Sgen = 0 we can write ˙ Wmax = ˙ m(bin −bout) Aside: b = h −TRs is referred to as the availability and is a measure of the “available energy” after the energy destroyed as a result of irreversibilities is removed. 32 Steady State, Steady Flow in a Flow Channel of Arbitrary Cross-section with Work and Heat Transfer TER m m x x x+dx x+dx E E dW dQ d ˙ E = ˙ Efinal −˙ Einitial = ˙ Ex+dx −˙ Ex where ˙ E = ˙ m(e + P v) = ˙ m u + V2 2 + gz + P v From the 1st law rate of energy storage = rate of work + rate of heat transfer + net rate of energy leaving the system dECV dt = δ ˙ W −δ ˙ Q −δ ˙ E (1) 33 where dECV dt = 0 for steady state. Equation (1) becomes 0 = δ ˙ W −δ ˙ Q −˙ m δ u + P v + V2 2 + gz (2) From the 2nd law rate of entropy storage =      rate of entropy inflow − rate of entropy outflow     + rate of entropy production dSCV dt = [ ˙ ms]x −[ ˙ ms]x+dx − δ ˙ Q TT ER + ˙ Sgen where dSCV dt = 0 for steady state. 0 = −˙ mds − δ ˙ Q TT ER + ˙ Sgen or δ ˙ Q = TT ER ˙ Sgen −TT ER ˙ m ds (3) Combining (2) and (3) through δ ˙ Q TT ER ˙ Sgen −TT ER ˙ m ds = δ ˙ W −˙ m δ u + P v + V2 2 + gz (4) Equation (4) can be used for any SS-SF process. 34 Special Cases Reversible, SS-SF Process Reversible implies ⇒˙ Sgen = 0 • frictionless process • heat transfer is allowed but must be across ∆T →0 • which means TT ER ≈TCV = T Equation 4 becomes δ ˙ W ˙ m = −T ds + du + d(P v) =du + P dv =T ds +vdP +δ V2 2 + δ(gz) (5) Therefore d ˙ W ˙ m = vdP + d V2 2 + d(gz) (6) Integrating Eq. (6) between the inlet and the outlet ˙ W ˙ m = out in vdP + V2 2 out in ∆KE + gz out in ∆P E (7) but ∆KE and ∆P E are usually negligible. If ∆KE + ∆P E = 0 ˙ W ˙ m = out in vdP (8) Equation can be used for a reversible, SS-SF ﬂow in a liquid or a gas. 35 If we keep in mind ρliq >> ρgas ⇒ vliq << vgas i.e. water @ 25 ◦C ρ = 997 kg/m3 and air @ 25 ◦C ρ = 1.18 kg/m3 Therefore  ˙ W ˙ m   liq <<  ˙ W ˙ m   gas For example: the work required to operate a pump is much less that that required to operate a compressor. Incompressible Substance This is a special case of Eq. (8) where v = constant = vin −vout. From Equation (8) ˙ W ˙ m = vin(Pout −Pin) (9) The work term represents the minimum work required to pump a liquid from Pin to Pout with negligible ∆KE and ∆P E. Incompressible Substance and δ ˙ W = 0 From Eq. (6) vdP + δ V2 2 + δ(gz) = 0 (10) Therefore d P ρ + d V2 2 + d(gz) = 0 36 d P ρ + V2 2 + gz = 0 (11) Integrating gives P ρ + V2 2 + gz = constant (12) Equation (12) is Bernoulli’s equation for frictionless ﬂow with constant density. The constant is Bernoulli’s constant, which remains constant along a streamline for steady, frictionless, incom-pressible ﬂow. Isothermal Ideal Gas, Compression/Expansion This is a special case of Eq. (8) for an ideal gas where P v = RT P v = constant = (P v)in = (P v)out ˙ W ˙ m = out in vdP = out in (P v)in dP P Therefore ˙ W ˙ m = Pinvin ln Pout Pin (13) Isentropic Ideal Gas, Compression/Expansion Isentropic implies a reversible and adiabatic process where s = constant. With an ideal gas, P vk = constant and (P vk)in = (P vk)out. Equation (8) becomes ˙ W ˙ m = out in vdP = out in (P vk)in P 1/k dP 37 ˙ W ˙ m = k k −1 (P v)in   Pout Pin (k−1)/k −1  = Cp(Tout −Tin) (14) The right side of Eq. (14) is based on the fact that ∆KE + ∆P E = 0 and dh = du + dP v and du = 0. Which leads to h = vdP . Note: for the same inlet state and pressure ratio ⇒  ˙ W ˙ m   rev.,isothermal <  ˙ W ˙ m   rev.,adiabatic 38
6192	https://www.bbc.co.uk/bitesize/topics/zm982hv/watch/zrs8d6f	Positive and negative numbers (signed) - Number and number processes: Video playlist - BBC Bitesize BBC Homepage Skip to content Accessibility Help Your account Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds More menu More menu Search Bitesize Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Close menu Bitesize Menu Home Learn Study support Careers Teachers Parents Trending My Bitesize More England Early years KS1 KS2 KS3 GCSE Functional Skills Northern Ireland Foundation Stage KS1 KS2 KS3 GCSE Scotland Early Level 1st Level 2nd Level 3rd Level 4th Level National 4 National 5 Higher Core Skills An Tràth Ìre A' Chiad Ìre An Dàrna Ìre 3mh ìre 4mh ìre Nàiseanta 4 Nàiseanta 5 Àrd Ìre Wales Foundation Phase KS2 KS3 GCSE WBQ Essential Skills Cyfnod Sylfaen CA2 CA3 CBC TGAU International KS3 IGCSE More from Bitesize About us All subjects All levels Primary games Secondary games 2nd level Number and number processes: Video playlist Part ofMaths and NumeracyNumber and number processes Now playing video 6 of 15 Positive and negative numbers (signed) Description An introduction to positive and negative numbers. Cartoon-style numbers initially form a line from 0 to 9. The negative numbers 1 to -9 then join the line in the correct sequence. The clip continues with 3 added to -7, and then added to -4, and so on, to show how the result of the addition becomes more positive each time. Classroom Ideas Use as an introduction to negative numbers. After watching the clip, create a horizontal number line (or have one already prepared) for the pupils to further practise. Try starting at the number 9 and subtracting 5 until you reach the negative numbers. Ask the children which negative number would come next. Alternatively, start at -9 and add 4 each time. Ask the children to record their ideas on mini whiteboards to ensure the whole class understands. You could progress onto a vertical number line and ask the children for real life reasons for using negative numbers. Number and number processes Now playing video 6 of 15 Measuring length (signed) Video Measuring length (signed) 1 of 15 2:36 Numbers - odd or even? Video Numbers - odd or even? 2 of 15 1:37 Using basic algebra to find the weight of three cats (signed) Video Using basic algebra to find the weight of three cats (signed) 3 of 15 0:40 Dividing by ten (signed) Video Dividing by ten (signed) 4 of 15 1:33 Multiples of three (signed) Video Multiples of three (signed) 5 of 15 2:21 Now playing. Positive and negative numbers (signed) Video Positive and negative numbers (signed) 6 of 15 Now playing 1:30 Up next. Using multiplication to create models (signed) Video Using multiplication to create models (signed) 7 of 15 Up next 1:23 Place value and three-digit numbers. Video Place value and three-digit numbers 8 of 15 1:48 Making totals with a specified number and value of coins (signed) Video Making totals with a specified number and value of coins (signed) 9 of 15 2:18 Measuring temperature - freezing point. Video Measuring temperature - freezing point 10 of 15 2:48 Measurement of length (signed) Video Measurement of length (signed) 11 of 15 4:05 Adding three small numbers by counting on. Video Adding three small numbers by counting on 12 of 15 1:01 Estimating and calculating distance. Video Estimating and calculating distance 13 of 15 5:01 Making weekly weather recordings. Video Making weekly weather recordings 14 of 15 3:22 Making three-digit numbers. Video Making three-digit numbers 15 of 15 0:45 Language: Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Terms of Use About the BBC Privacy Policy Cookies Accessibility Help Parental Guidance Contact the BBC BBC emails for you Advertise with us Do not share or sell my info Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking.
6193	https://www.youtube.com/watch?v=3HCkUrhqKiU	Graphing Functions on a Restricted Domain DJones Mathematics 1050 subscribers 26 likes Description 3096 views Posted: 24 Nov 2019 In this video, we'll teach you how to graph linear functions when the domain is limited to specific values. You'll learn how to identify and apply restrictions to the domain, graph functions accordingly, and interpret the results. The video will also cover examples that restrict that range and force us to work backwards. With clear explanations and step-by-step examples, you'll gain confidence in graphing functions on a restricted domain. Don’t forget to like, comment, and subscribe for more educational math videos! For additional practice, be sure to check out the workbook linked below. Algebra 1: A Problem Set for Perfecting Your Skills 3 comments Transcript: hey everyone in this video we're going to learn how to graph functions on a restricted domain more specifically linear functions as that's the only type of function that we've learned how to graph so far now this picks up from the last video where we talked about the domain and range of a function and the different ways to define them now we're going to be given a domain and a function of graph and we have to combine the ideas together our first example graph the functions restrict the domain and then state the range of the graph so the equation or the function is y equals 3 over 2x minus 6 and the domain is X is greater than 0 so the function they gave us is a line now they didn't write f of X equals they wrote y equals but we've discussed how those are synonymous with one another and can be interchanged if we were to graph this line without a domain restriction we would identify the slope as 3 over 2 a point on the line is the y-intercept which from this equation can be identified as the point 0 negative 6 because this is in slope intercept form and normally we would go plot the y-intercept and then start plotting points using the slope 3 over 2 so up 3 and over 2 boxes and we would go in the other direction as well but I'm not going to go past x equals 0 and the reason for that is the domain restriction X is greater than 0 so we only want to plot points on the line where the x value is greater than 0 so that's only to the right of the y intercept my x equals 1 x equals 2 x equals 3 those are X values that are greater than 0 now we do have to deal with the fact that X is greater than 0 not greater than or equal to 0 and the way we'll deal with that is similar to how we dealt with it when we were graphing a solution set on the number line we're going to make that point at the y-intercept an open Circle now that's going to indicate that we're not equal to zero but only greater than zero and now we're going to connect the points on the line that we've drawn but we're only going to include an arrow on one part of our line so going from the y-intercept through the points that we've plotted there is our line notice there's only an arrow on this side of the line no arrow here because we don't want to go past zero right we want any value that's less than zero now identifying the range remember we work our way from the bottom of the graph up so the lowest y-value on our graph is negative six well not negative six but that's the lowest point on the graph the y-values are all greater than negative six so that's exactly what we're going to state as the range y is greater than negative six now we could choose to write this in interval notation and might as well practice with this so that's an inequality if they asked us for interval notation we would say negative six to infinity we're not equal to negative six so parenthesis on that side we can't be equal to infinity so a parenthesis there and then we had one more way to express the range based off the graph that was set builder notation so curly bracket Y such that Y is greater than negative six now in this particular example since they gave us an inequality as the domain I would have stated the range as an inequality as well but all of these are correct answers so basically what is this question this question is giving us practice graphing a line incorporating the new concepts of domain and range in so it's just a little more challenging than what we were doing before when we're graphing our lines next same directions the function y minus 4 equals negative 2 times the quantity x + 1 and the domain is negative 2 is less than or equal to X which is less than or equal to 4 so the function or the line is in point-slope form so we'll identify our two key pieces of information and that's the fact that the slope is negative 2 and a point on the line is negative 1/4 so again normally we would go negative 1 1 2 3 4 and start plotting our slope of negative 2 or points using our slope of negative 2 but now we have to plot points by using the domain restriction that they gave us so negative 2 is less or equal to X which is less than or equal to 4 well that's negative 1 so let's start by going backwards so instead of down 2 will go up 2 into the left one and we'll stop there because that's x equals negative 2 I don't want to go any further than that and then go back negative 2 slope of negative 2 slope so down two to the right one down two to the right one so right now I'm at x equals one x equals two I'm at x equals three so I should only go one more point so down two to the right one and that's x equals 4 so I've only plotted the points such that X is between negative 2 & 4 inclusive so now we're going to connect our points from x equals negative 2 to x equals 4 I'll fix this just so it looks a little bit better going through a points there we have it now notice there are no arrows we don't put an arrow here we don't put an arrow there because we're only graphing on this restricted domain so our range working from the lowest Y value up so the lowest by Y value negative 1 negative 2 negative 3 and and 85 negative 6 so we have negative 6 is less than or equal to Y which is less than or equal to so we work our way up the graph the highest Y value 1 2 3 4 5 6 is positive 6 so there is our range now we could have done this a different way so let's say we don't understand why I stopped where we plotted our points you could set up a table of values instead you could have said negative 2 negative 1 0 1 2 3 & 4 plug each of these X values into the equation into the function and got the output but now I want you to weigh those two options the way that we had done it previously or filling out this table and then plotting the points I think the way we originally did it by using the fact that it was in point-slope form identifying the slope and the point and then graphing only on these x-values was more efficient than filling out this table would be okay so just something to keep in mind they're both correct but one way is definitely more efficient going to save you some time our next example same directions y equals 3x plus 2y our domain is negative 3 is less than or equal to X which is less than or equal to 1 X is an integer now so we'll deal with that part laughs let's continue the same way that we've been doing so we have our line the slope is 3 a point on the line is the y-intercept at 0 2 so we'll plot 0 2 which is in our domain and we'll graph a slope of 3 so up three to the right one I'll stop going that way because that's already x equals one now working in the other direction so down three to the left one down three to the left one down three to the left one and we'll stop there because that's x equals negative 3 now let's say you just plotted a bunch of points you could always go back with an eraser and erase the ones that are not in this domain that's definitely another option okay so don't think about the domain just plot the points as you normally would and then erase the points that are not in the domain restriction that they gave now we have to deal with the fact that X is an integer well where are the integers in this domain that would be one set up a table of values just in case you wanted to do it this way so the integers negative 3 negative 2 negative 1 0 and 1 and we get our Y values so plug in negative 3 we have negative 9 plus 2 so that's negative 7 and then negative 6 plus 2 is negative 4 negative 3 plus 2 is negative 1 2 & 5 so this is how you complete the table now these are the only X values that are integers on this restricted domain so these are the only points that should be on our graph so now that brings us to our next question do we connect the points with a line these are the only X values that are integers in this domain and they answer that question is no we should not connect our points with a line because let's just focus in on a little piece of this graph if I connected these two points here from x equals 0 to x equals 1 those would be all the points like x equals 0.5 x equals point 9 all x values that aren't integers inputs that are not integers so we don't want to include them in our graph so we don't want to draw a line so now we need to state the range of this particular function so we could do it in two ways we can list the y-values from least to greatest we've already done that in the table so we can just say negative 7 negative 4 negative 1 2 & 5 and that's going to be the simplest way so that's the only way that I'm going to to suggest doing this the other way is a little bit more complex and I don't think I don't think it'd be as easy for us to wrap our head around so let's stick with what simple right list out the outputs in this domain so negative seven negative or negative one two and five are the range perfect graph the function on the given restricted range y equals negative x plus five the range is negative three is less or equal to Y which is less than or equal to one so they've changed it up on us a little bit instead of giving us the domain they now gave us the range huh whoa we're not used to inputting Y values to graph an equation graph a line right if we look at our equation you know the slope is negative one you know the y-intercept is the point 0 five but we're not sure how to incorporate this so my suggestion would be to solve for the x value the input that outputs negative three and one so for instance substituting negative three in for y you have negative x plus five we can subtract five from both sides so we have negative 8 equals negative x so eight equals x so that means on this graph on this function is the point 8 comma negative three and then we can do the similar argument for y equals one so substituting 1 for Y and then negative x plus 5 so we can subtract 5 from both sides negative 4 equals negative X divide both sides by negative 1/4 equals x so the point 4 comma 1 is on our graph now that's enough information think back to graphing lines we've graphed plenty of lines where we found two points on the line and we connected those two points so we have 4 comma 1 1 2 3 4 comma 1 and then 5 6 7 8 negative 3 now our ringing is only from negative 3 to 1 so we're going to connect these two points with a line with no arrow right we don't want an arrow we don't want any y-value outside of the range that they gave us okay now then you tell us to find the domain but as good practice let's do that the domain we're working from left to right the lowest x-value is for the largest x-value is 8 so we'd have 4 is less than or equal to X which is less than or equal to 8 so they change up the problem a little bit on us here they give us the rangers head of the domain but we were still able to work our way through the problem and get our graph our last question the function f of X equals negative 2x plus 7 is defined on the domain negative 1 is less than or equal to X which is less or equal to 6 what is the maximum value of the range so we could of course graph this and then decide based off the graph and that's always an option but let's try to reason out an approach here without a graph so the first thing looking at our function is that it's a line right so it's a line and it's a line with a slope of negative 2 so think if you work from left to right what a line with a negative slope looks like we can even draw one line with the negative slope looks like that right it wouldn't have arrows here because we are restricted domain so we can turn these into points ok so that's what line with a negative slope looks like so where is the maximum or the largest Y value or output going to be well it's going to be at the first input in our domain which in this case it's going to be negative one all right so the maximum value occurs when x equals negative one so we can evaluate the function at negative one so we have negative two times negative one plus seven which is two plus seven which is not so the maximum value in the range of this function on the domain negative 1 is less than or equal to X which is less than or equal to 6 is 9 and we can make a similar argument for a positive slope line and go through the same reasoning and decide where the maximum value if the line had a positive slope okay and again we could always graph this line and come with our answer from there but I think it's nice to reason through where that maximum value would occur then they follow this up with what would the range be if the domain was changed to be the set of all real numbers so if we didn't have points here for a restricted domain if we turn those into arrows so we can draw a picture so if we turn those into arrows how would that change the range well the arrows indicate that we're going to continue infinitely down and infinitely up on our graph so the lowest Y value would be negative infinity the highest Y value would be positive infinity so I'm running it in interval notation right now so that would be the range in interval notation which is just an way another way to write the set of all real numbers so if we took those points and turn them into arrows and change the domain to the set of all real numbers the range would also change to be the set of all real numbers because our function is a line as we study different functions that are going to have different graphs the answers to these questions are going to change but for right now as we're incorporating just linear functions into our discussion of domain and range these are two nice questions to wrap that discussion up so re-watch any parts of the video if you weren't sure about our approach or why the answer turned out that way and then practice go through examples so that you see a bunch of different presentations different changes in the questions and you've wrapped your head truly around the idea of finding the domain and range click the Amazon link down below for my algebra workbook so you can practice on your own give the video a like and before you go click that subscribe button so you can see more videos just like this thanks for watching
6194	https://up.codes/s/hydraulic-design-of-roadside-channels	Topic 866 — Hydraulic Design of Roadside Channels \| \| \| \| --- \| H \| = \| Total head, in feet of water \| \| z \| = \| Distance above some datum, in feet \| \| d \| = \| Depth of flow, in feet \| \| \| = \| Velocity head, in feet \| \| g \| = \| Acceleration of gravity \| \| \| = \| 32.2 feet per second squared \| \| \| \| \| --- \| hL \| = \| Intervening head losses, in feet \| \| \| \| \| --- \| V \| = \| Mean velocity, in feet per second \| \| n \| = \| Manning coefficient of roughness \| \| S \| = \| Channel slope, in foot per feet \| \| R \| = \| Hydraulic Radius, in feet \| \| \| = \| A/WP \| \| \| \| \| --- \| A \| = \| Cross sectional flow area, in square feet \| \| WP \| = \| Wetted perimeter, in feet \| \| \| \| \| --- \| E \| = \| Specific energy, in feet \| \| d \| = \| Depth of flow, in feet \| \| \| = \| Velocity head, in feet \| \| \| \| \| --- \| A \| = \| Cross sectional area, ft2 \| \| T \| = \| Top width of water surface, ft \| \| Q \| = \| Discharge, CFS \| \| g \| = \| Acceleration of gravity, 32.2 ft/s2 \| \| \| \| \| --- \| q \| = \| Flow per unit width, CFS \| \| \| \| \| --- \| D = A/T \| = \| Hydraulic depth, in feet \| \| Fr < 1.0 \| ==> \| Subcritical flow \| \| Fr = 1.0 \| ==> \| Critical flow \| \| Fr > 1.0 \| ==> \| Supercritical flow \| \| Type of Channel \| n value \| --- \| \| Unlined Channels: \| \| \| Clay Loam \| 0.023 \| \| Sand \| 0.020 \| \| Gravel \| 0.030 \| \| Rock \| 0.040 \| \| Lined Channels: \| \| \| Portland Cement Concrete \| 0.014 \| \| Air Blown Mortar (troweled) \| 0.012 \| \| Air Blown Mortar (untroweled) \| 0.016 \| \| Air Blown Mortar (roughened) \| 0.025 \| \| Asphalt Concrete \| 0.016-0.018 \| \| Sacked Concrete \| 0.025 \| \| Pavement and Gutters: \| \| \| Portland Cement Concrete \| 0.013-0.015 \| \| Hot Mix Asphalt Concrete \| 0.016-0.018 \| \| Depressed Medians: \| \| \| Earth (without growth) \| 0.016-0.025 \| \| Earth (with growth) \| 0.050 \| \| Gravel (D50 = 1 in. flow depth ≤ 6 in.) \| 0.040 \| \| Gravel (D50 = 2 in. flow depth ≤ 6 in.) \| 0.056 \| Related Code Sections
6195	http://www.mathisfunforum.com/viewtopic.php?id=11591	mutilated chessboard variations (Page 1) / Euler Avenue / Math Is Fun Forum Math Is Fun Forum Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° Index User list Rules Search Register Login You are not logged in. Topics: Active \| Unanswered Index »Euler Avenue »mutilated chessboard variations Pages: 1 12009-02-18 09:51:40 bossk171MemberRegistered: 2007-07-16 Posts: 305 mutilated chessboard variations The mutilated chessboard problem is a very popular classic math puzzle. If you're not familiar with it, google "mutilated chessboard" and read the proof as to why it's impossible, it's pretty cool. Essentially it boils down to the question, "If two opposing corners of a chessboard are removed, can the mutilated board be completely covered with 31 dominoes in such a way that each domino covers 2 squares?" Let's discuss variations on this problem here! Variation 1: Consider an 8x8 chessboard with one corner missing (so a total of 64-1=63 squares). Can it be covered with 21 triominoes (dominoes that are 31 squares instead of the traditional 21)? What if the triominoes are in an L shape instead of a line? Board with one corner missing: \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|\|\| \|\|\|\|\|\|\|_\| __ _____ \|\| triomino: \|\|\|\| trinomino (L variation): \|\|_\| Based on playing around with pictures my suspicions are no, a board with a missing corner piece cannot be covered, but I lack a proof. So go ahead, post your variations here! (PS: anyone who wants to produce some legitimate pictures as apposed to my sloppily typed one is more than encouraged to). Last edited by bossk171 (2009-02-18 09:52:53) There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction. Offline 22009-02-18 10:53:01 mathsypersonModeratorRegistered: 2005-06-22 Posts: 4,900 Re: mutilated chessboard variations It's impossible for the straight trionimo. The proof is similar to the standard one, just label the board with three colours instead of two. One of my favourite problems is this one: Prove that given a 2[sup]n[/sup] x 2[sup]n[/sup] chessboard with any one of its squares removed, it is possible to completely cover using L-shaped trionimoes. (Your second question is a specific case of this.) Why did the vector cross the road? It wanted to be normal. Offline 32009-02-18 13:19:03 bossk171MemberRegistered: 2007-07-16 Posts: 305 Re: mutilated chessboard variations mathsyperson wrote: It's impossible for the straight trionimo. The proof is similar to the standard one, just label the board with three colours instead of two. I tried that, but I'm pretty sure I missed something: A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C There are 21 each of A,B and C. The original proof relied on there be a different number A squares and B squares. There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction. Offline 42009-02-19 00:00:23 mathsypersonModeratorRegistered: 2005-06-22 Posts: 4,900 Re: mutilated chessboard variations You just got unlucky. There's a difference if you label it using antidiagonal stripes. Alternatively, consider what would happen on your diagram if the top-right corner was missing instead. Why did the vector cross the road? It wanted to be normal. Offline 52009-02-19 04:24:32 bossk171MemberRegistered: 2007-07-16 Posts: 305 Re: mutilated chessboard variations Isn't how I color it arbitrary? I purposely removed that spot because it would give me 21 each of A,B, and C. When yo say antidiagonal, do you mean vertical or horizontal? If so, I would think that wouldn't make sense... I tried to "color" it so that any triomino place would have to cover and A, B, and C (like a domino has to cover a black and a white). If I cover it vertically than it would be possible to lay a triomino that would cover 3 As (or Bs or Cs) which doesn't seem right... I'm not trying to be difficult, I honestly am having trouble getting this. Am I missing something very simple... I think am. There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction. Offline 62009-02-19 04:58:34 TheDudeMemberRegistered: 2007-10-23 Posts: 361 Re: mutilated chessboard variations I'm not sure what he means by antidiagonal stripes, but do consider his second point. If you reinsert the A that you removed and instead remove the B in the top-right corner of the board you'll get a board with 22 A's, 20 B's, and 21 C's. Since it's still setup such that every straight triomino will have to cover exactly 1 A, B, and C, you can once again see that no such covering is possible. Wrap it in bacon Offline 72009-02-19 05:34:03 mathsypersonModeratorRegistered: 2005-06-22 Posts: 4,900 Re: mutilated chessboard variations Sorry for being confusing. All I meant was to draw the lines from bottom-left to top-right, instead of how you had them. Why did the vector cross the road? It wanted to be normal. Offline 82009-02-19 05:53:40 TheDudeMemberRegistered: 2007-10-23 Posts: 361 Re: mutilated chessboard variations That is an interesting problem mathsy. Unfortunately I'm not familiar with giving rigorous proofs of something like this, but I can see it visually. First thing is to show that you can cover any 2^n X 2^n board with a corner square missing. n = 0 is a trivial case, so let's look at n = 1. We have a 2 X 2 board, and since all 4 squares are corners we can assume WLOG that the bottom-right square is missing, so our board looks like X X X _ It's quite obvious how we place our triomino to cover the remaining squares on the board. So now we know how to do a 2 X 2 board. Let's see what happens when we turn that into a 4 X 4 board, again only considering cases where the bottom-left square is removed: X X X X X X X X X X X X X X X _ Let's place our first triomino the same way that we did in the 2 X 2 case (H represents squares covered by a triomino): X X X X X X X X X X H H X X H _ Let's pretend that the board is divided into four 2 X 2 squares. The bottom-right square is either covered by a triomino or has an empty square. That leaves us with three 2 X 2 squares to cover. How can we do that? By placing another triomino such that each of the 3 tiles that it covers are corner squares to one of the 2 X 2 squares: X X X X X H H X X H H H X X H _ Now those other three 2 X 2 squares are identical to our original 2 X 2 board problem, which we already know how to solve. Thus, we now know how to solve a 4 X 4 board. How do we do an 8 X 8 board? We use the same trick as before. We mentally divide the board into four 4 X 4 boards. We can solve the bottom-right board with the method we just used, which will leave us with three other completely uncovered 4 X 4 boards. If we place our next triomino such that each tile covers a corner of one of the remaining 4 X 4 boards they'll each become idential to the original 4 X 4 board problem, which we already know how to solve. Covering those gives us a completely filled 8 X 8 board. The process can continue indefinitely to any 2^n X 2^n board. Now, mathsy's actual question was to show that it could be done with any square missing. This is actually not much of an issue thanks to how triominoes work. Let's say that instead of the lower left corner tile being removed, we actually removed the piece immediately to the left of that piece. Here's what that would look like on a 4 X 4 board: X X X X X X X X X X X X X X _ X To solve this new problem we can simply rotate our first triomino such that the corner tile is in the top-right instead of top-left, so it would look like this: X X X X X X X X X X H H X X _ H Notice that once again we are left with a completed 2 X 2 board in the bottom right, so the rest of our 4 X 4 solution can remain unchanged. The same thing goes for if we move the empty square one tile up: X X X X X X X X X X _ H X X H H Any other tile that is removed from the 4 X 4 board is just a reflection or rotation of what we've just shown, so now we can solve a 4 X 4 board with any square missing. And if we can solve any 4 X 4 board with a missing square we can generalize that to the 8 X 8 case, since we need solve only one 4 X 4 board and then we add a triomino that turns the other three parts of the board into the "corner missing" case. And if we can do any 8 X 8 board we can do any 16 X 16 board, etc. Wrap it in bacon Offline 92009-02-19 06:44:42 mathsypersonModeratorRegistered: 2005-06-22 Posts: 4,900 Re: mutilated chessboard variations Yep, that works. I solved it by observing that 4 L-shaped trionimoes can fit together to make a double-sized block: ``` \| \| \| - - \| \| \| \| - - - \| \| \| - - ``` Fitting 4 of the size-2 blocks together will create a size-4 block, and so iterating this can get any size-2^n block you want. So then given a 2^n x 2^n chessboard, you can place a block of appropriate size on it so that the removed square isn't covered, and you're left with the same problem, only now the chessboard is 2^(n-1) x 2^(n-1). Do this until there's only a 2x2 board left, and then place one final block to finish the covering. (This can be rephrased as a proof by induction if you want to be formal about it) Why did the vector cross the road? It wanted to be normal. Offline 102009-02-19 10:16:00 bossk171MemberRegistered: 2007-07-16 Posts: 305 Re: mutilated chessboard variations Cool proof TheDude. How about if any two (oppositely colored) squares are removed on and 8x8 board (or more generally a 2^n x 2^n board) can it still be covered with dominoes? I haven't put any thought into this yet, I just want to keep up the momentum. I would think that the proof would be a lot the one The Dude wrote for Mathsyperson's variation, but with minor changes... EDIT: I like Mathsyperson's proof too, it just took me a few more minutes to get it. Last edited by bossk171 (2009-02-19 10:22:03) There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction. Offline 112009-02-19 12:33:35 mathsypersonModeratorRegistered: 2005-06-22 Posts: 4,900 Re: mutilated chessboard variations Interesting question! A few minutes of scribbling leads me to think it's probably possible, because there aren't any easy counterexamples I can find. Edit: I haven't figured out exactly how to work it, but I think laying dominoes to make a rectangle such that the two removed squares are opposite corners would be a good start. Also, I'm pretty sure it's possible for all boards of length 2n, not just the 2^n ones. Edit2: I came across a much nicer way today. First draw a snakey path so on the board so that all squares are covered, and the path starts and finishes in the same square. Cover the board with dominoes by following this path, and then once two of the squares have been removed, you can get the new solution by removing a domino and shifting. Why did the vector cross the road? It wanted to be normal. 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6196	https://cran.r-project.org/web/packages/CalibrationCurves/vignettes/CalibrationCurves.html	Introduction to the CalibrationCurves package Bavo De Cock Campo 2025-07-06 1 Assessing the performance of risk prediction models 1.1 Risk prediction models 1.1.1 Mathematical details on existing predictive models 1.2 Different aspects of the predictive performance 1.3 Assessing the calibration performance of a risk prediction model 1.3.1 A mathematical perspective 1.3.2 A practical perspective 1.3.3 How do we construct a calibration curve? 1.3.4 Calibration intercept and slope 1.4 Illustration of the CalibrationCurves package 1.4.1 Training the model 1.4.2 Assessing the calibration performance 1.4.3 ggplot version 2 Assessing the performance of survival models 2.1 Cox Proportional hazards model 2.1.1 Calibration Curve 3 Assessing the performance of other types of prediction models 3.1 Generalized calibration curves 3.2 Illustration of the generalized calibration framework 3.2.1 Training the model 3.2.2 Assessing the calibration performance 4 FAQ 4.1 Why is the calibration intercept different in the rms package? 4.2 I have predicted probabilities of 0 or 1. Why is this not allowed by default and why do I get these annoying warning messages? 5 References In this document, we give you a brief overview of the basic functionality of the CalibrationCurves package. In addition, we present the theoretical framework behind calibration and provide some illustrative examples to give the reader a better insight into the calibration assessment of a predictive model. We advise you to also consult the help-pages of the functions to get an exhaustive overview of the functionality. We tried to tailor the explanation of the concepts to professionals with different backgrounds. Please, do contact me if you feel that something is unclear so that I can adjust (and hopefully improve) it. In addition, don’t hesitate to send any suggestions you might have and bug reports to the package author. 1 Assessing the performance of risk prediction models 1.1 Risk prediction models In this package, we focus on risk prediction models that estimate the probability (\pi_i) of observing an event. We use (y_i \in (0, 1)) to denote the variable that captures this outcome which takes on the value 0 in case of a non-event and 1 in case of an event. Here, (i) serves as an index for the observations (mostly the patient within medical predictive analytics) with (i = (1, \dots, n)) and where (n) denotes the total number of observations. We assume that the response variable (y_i) follows a Bernoulli distribution (y_i \sim \text{Bern}(\pi_i)). For example, we could be interested in estimating the probability (\pi_i) of observing a malignant tumour for patient (i). In this case, the event (y_i = 1) is the tumour being malignant and (y_i = 0) when the tumour is benign. With no available information on the patient characteristics, we might rely on the prevalence in the general population to estimate this probability. Using risk prediction models, we model the outcome as a function of the observed risk/patient characteristics. The risk characteristics are contained in the covariate vector (\boldsymbol{x}_i). This vector contains all observed information for patient (i) (e.g. maximum diameter of the lesion, proportion of solid tissue, …). This allows us to obtain a more accurate prediction that is based on the relation between the patient characteristics and the outcome. To construct a clinical prediction model, we either rely on a statistical models such as logistic regression or machine learning methods. A general expression that encompasses both types of models is [\begin{align} E[y_i \| \boldsymbol{x}_i] = f(\boldsymbol{x}_i). \end{align}] This expression states that we model the response (y_i) as a function of the observed risk characteristics (\boldsymbol{x}_i). 1.1.1 Mathematical details on existing predictive models To construct a risk prediction model, we could rely on a logistic regression model [\begin{align} E[y_i \| \boldsymbol{x}_i] = \pi_i(\boldsymbol{\beta}) = \frac{e^{\boldsymbol{x}_i^\top \boldsymbol{\beta}}}{1 + e^{\boldsymbol{x}_i^\top \boldsymbol{\beta}}} \end{align}] where (\boldsymbol{\beta}) denotes the parameter vector. (\pi_i(\boldsymbol{\beta}) = P(y_i = 1\| \boldsymbol{x}_i)) denotes the probability of observing the event, given the covariate vector (\boldsymbol{x}_i). We can rewrite the equation to its more well-known form [\begin{align} \log\left( \frac{\pi_i(\boldsymbol{\beta})}{1 - \pi_i(\boldsymbol{\beta})} \right) &= \boldsymbol{x}_i^\top \boldsymbol{\beta}\[0.5em] \text{logit}(\pi_i(\boldsymbol{\beta})) &= \eta_i \end{align}] where (\eta_i) denotes the linear predictor. Here, we have the well-known logit function at the left side of the equation. With machine learning methods, (f(\cdot)) depends on the specific algorithm. With tree-based methods, for example, this correspond to the observed proportion in the leaf nodes. For neural networks, (f(\cdot)) is determined by the weights in the layers and the chosen activation functions. 1.2 Different aspects of the predictive performance To assess how well the model is able to predict (the probability of) the outcome, we assess two different aspects of the model (Van Calster et al. 2016, 2019; Alba et al. 2017): discrimination; calibration. With discrimination, we refer to the model’s ability to differentiate between observations that have the event and observations that have not. In this context, this translates to giving higher risk estimates for patients with the event than patients without the event. We commonly assess this using the area under the receiver operating characteristic curve. However, discrimination performance does not tell us how accurate the predictions are. The estimated risk may result in good discrimination and can be inaccurate at the same time. We refer to the accuracy of the predictions as the calibration. Hence, hereby we assess the agreement between the estimated and observed number of events (Van Calster et al. 2016). We say that a prediction model is calibrated if the predicted risks correspond to the observed proportions of the event. 1.3 Assessing the calibration performance of a risk prediction model 1.3.1 A mathematical perspective One way to examine the calibration of risk predictions, is by using calibration curves (Van Calster et al. 2016, 2019; Steyerberg 2019; De Cock Campo 2023). A calibration curve maps the predicted probabilities (f(\boldsymbol{x}_i)) to the actual event probabilities (P(y_i = 1\| f(\boldsymbol{x}_i))) and visualizes the correspondence between the model’s predicted risks and the true probabilities. For perfectly calibrated predictions, the calibration curve equals the diagonal, i.e. (P(y_i = 1 \| f(\boldsymbol{x}_i)) = f(\boldsymbol{x}_i) \ \forall \ i) where (\forall \ i) denotes for all (i). 1.3.2 A practical perspective In practice, we typically assess the model’s calibration on a validation set. In this setting, a calibration curve visualizes the correspondence between the model’s predicted risks and the observed proportion. When we have a perfect agreement between the observed and predicted proportion the calibration curve coincides with the ideal curve (a diagonal line). This scenario is visualized in Figure 1.1. Figure 1.1: Example of a perfectly calibrated model By assessing the calibration performance on a data set other than the training set, we obtain an indication of how well our risk prediction is able to generalize to other data sets and how accurate its out-of-sample predictions are. In general, the prediction model will show some miscalibration and the calibration curve gives us a visual depiction of how badly the model is miscalibrated. The further from the diagonal line, the worse the calibration. Figure 1.2 depicts an example of a model that is miscalibrated and is a typical example of a model that is overfitted to the training data. This particular model has predictions that are too extreme: high risks are overestimated and low risks are underestimated. Figure 1.2: Example of a miscalibrated model due to overfitting Its counterpart, an underfitted model, occurs less frequently. 1.3 shows the calibration curve of an underfitted model. Here, there is an overestimation for the low risks and an underestimation for the high risks. Figure 1.3: Example of a miscalibrated model due to underfitting 1.3.3 How do we construct a calibration curve? Fitting a logistic regression model to the training data results in an estimate for the parameter vector (\boldsymbol{\beta}), which we denote as (\widehat{\boldsymbol{\beta}}). The latter contains the estimated effects of the included covariates (e.g. proportion of solid tissue). To obtain a risk estimate for patient (i), we multiply the covariate vector (\boldsymbol{x}_i) (which contains all the patient-specific characteristics) with the estimated parameter vector (\widehat{\boldsymbol{\beta}}) to obtain the linear predictor (\widehat{\eta}_i) [\begin{align} \widehat{\eta}_i = \boldsymbol{x}_i^\top \widehat{\boldsymbol{\beta}}. \end{align}] To differentiate between the training and test set, we append the subscript () to the quantities of the test set. Hence, ({}_{} y_i) denotes the outcome in the test set. Similarly, we use ({}_{} \boldsymbol{x}_i) to denote the covariate vector for patient (i) in the test set. We then calculate the linear predictor on the test set as [\begin{align} {}_{} \widehat{\eta}_i = {}_{} \boldsymbol{x}_i^\top \widehat{\boldsymbol{\beta}} \tag{1}. \end{align}] Similarly, we can predict the probability (\widehat{f}({}_{} \boldsymbol{x}_i)) for patient (i) in the test set using machine learning methods. We use [\begin{align} {}_{} \widehat{\pi}_i = \widehat{f}({}_{} \boldsymbol{x}_i) \end{align}] as a general notation to denote the predicted probability of the risk prediction model. One way to compute the calibration curve, is by using a logistic regression model [\begin{align} \text{logit}(P({}_{} y_i = 1\| {}_{} \widehat{\pi}_i)) &= \alpha + \zeta \ \text{logit}({}_{} \widehat{\pi}_i) \tag{1.1} \end{align}] where we estimate the observed proportions as a function of the predicted probabilities. This model fit yields a logistic calibration curve. Note that (\text{logit}({}_{} \widehat{\pi}_i) = {}_{} \widehat{\eta}_i) when ({}_{} \widehat{\pi}_i) is estimated using a logistic regression model (see (1.1)). Alternatively, we can obtain flexible, nonlinear calibration curve using a non-parametric smoother such as loess or restricted cubic splines. In our package, we provide both types of calibration curves. 1.3.4 Calibration intercept and slope In addition to the calibration curve, we have two measures that summarize different aspects of the calibration performance: the calibration intercept (\alpha_c) (calibration-in-the-large); the calibration slope (\zeta). We have a perfectly calibrated model when the calibration curve coincides with the diagonal line or when (\alpha =\alpha_c = 0) and (\zeta = 1). To compute the calibration slope (\zeta), we rely on the model used to obtain the logistic calibration curve (see equation (1.1)). The value of the calibration slope (\zeta) tells us whether the model is over- or underfitted. When (\zeta < 1) the model is overfitted. (\zeta < 1) indicates that ({}_{} \eta_i) is too extreme and needs to be lower to ensure that the predicted risks coincide with the observed risks. Conversely, we have a model that is underfitted when (\zeta > 1). To calculate the calibration intercept or calibration-in-the-large, we fix the calibration slope at (1) and denote this as (\alpha\|\zeta = 1) or the short-hand notation (\alpha_c). To estimate (\alpha_c), we fit the model [\begin{align} \text{logit}(P({}_{} y_i = 1\| {}_{} \widehat{\pi}_i)) &= \alpha_c + \text{offset}(\text{logit}({}_{} \widehat{\pi}_i)) \tag{1.2} \end{align}] where we enter (\text{logit}({}_{} \widehat{\pi}_i)) as an offset variable. Hereby, we fix (\zeta = 1). The calibration intercept tells us whether the risks are overestimated ((\alpha_c < 0)) or underestimated ((\alpha_c > 0)) on average. 1.4 Illustration of the CalibrationCurves package 1.4.1 Training the model To illustrate the functionality, the package has two example data sets: traindata and testdata. These are two synthetically generated data sets (using the same underlying process/settings to generate the data) to illustrate the functionality of the CalibrationCurves package. The traindata data frame represents the data that we will use to develop our risk prediction model library(CalibrationCurves) #> Loading required package: rms #> Loading required package: Hmisc #> Warning: package 'Hmisc' was built under R version 4.4.2 #> #> Attaching package: 'Hmisc' #> The following objects are masked from 'package:base': #> #> format.pval, units #> Loading required package: ggplot2 data("traindata") In this data frame, we have four covariates and one response variable y. head(traindata) #> y x1 x2 x3 x4 #> 1 0 -0.19981624 0.2982990 1.0277486 -0.1146414 #> 2 1 -1.37127488 0.5940002 -0.8234645 2.0927676 #> 3 1 1.04050541 0.5440481 -1.3576457 1.3126813 #> 4 0 -1.11652476 -0.5382577 -1.1651439 1.0987873 #> 5 1 1.39659613 1.1325081 0.6053029 -1.0598506 #> 6 0 -0.04645095 -0.8167364 1.0196761 -0.4867560 Next, we fit a logistic regression model to obtain the estimated parameter vector (\widehat{\beta}). glmFit = glm(y ~ . , data = traindata, family = binomial) summary(glmFit) #> #> Call: #> glm(formula = y ~ ., family = binomial, data = traindata) #> #> Coefficients: #> Estimate Std. Error z value Pr(>\|z\|) #> (Intercept) 0.08915 0.08016 1.112 0.266 #> x1 0.60585 0.08475 7.148 8.79e-13 #> x2 1.38035 0.10554 13.079 < 2e-16 #> x3 -0.75109 0.08854 -8.483 < 2e-16 #> x4 0.82757 0.08759 9.448 < 2e-16 #> --- #> Signif. codes: 0 '' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> (Dispersion parameter for binomial family taken to be 1) #> #> Null deviance: 1385.89 on 999 degrees of freedom #> Residual deviance: 950.28 on 995 degrees of freedom #> AIC: 960.28 #> #> Number of Fisher Scoring iterations: 5 1.4.2 Assessing the calibration performance Hereafter, we assess the calibration performance on the testdata set. Hereto, we first have to compute the predicted probabilities on this data set. data("testdata") pHat = predict(glmFit, newdata = testdata, type = "response") We then store the response in the testdata in a separate vector yTest. yTest = testdata$y Now we have everything we need to assess the calibration performance of our prediction model. We can either use val.prob.ci.2 or valProbggplot to visualize the calibration performance and to obtain the statistics. val.prob.ci.2 makes the plot using base R and valProbggplot uses the ggplot2 package. By default, the flexible calibration curve (based on a loess smoother) will be plotted. calPerf = val.prob.ci.2(pHat, yTest) In addition to the plot, the function returns an object of the class CalibrationCurve. calPerf #> Call: #> val.prob.ci.2(p = pHat, y = yTest) #> #> A 95% confidence interval is given for the calibration intercept, calibration slope and c-statistic. #> #> Dxy C (ROC) R2 D D:Chi-sq D:p #> 0.62853462 0.81426731 0.38019823 0.33282644 167.41322219 0.00000000 #> U U:Chi-sq U:p Q Brier Intercept #> 0.01286390 8.43195136 0.01475792 0.31996254 0.17703339 0.22404680 #> Slope Emax Brier scaled Eavg ECI #> 0.82278297 0.08288689 0.28730517 0.04747448 0.32064056 This object contains the calculated statistics as well as the calculated coordinates of the calibration curve. str(calPerf) #> List of 7 #> $ call : language val.prob.ci.2(p = pHat, y = yTest) #> $ stats : Named num [1:17] 0.629 0.814 0.38 0.333 167.413 ... #> ..- attr(, "names")= chr [1:17] "Dxy" "C (ROC)" "R2" "D" ... #> $ cl.level : num 0.95 #> $ Calibration :List of 2 #> ..$ Intercept: Named num [1:3] 0.22405 0.00339 0.4447 #> .. ..- attr(, "names")= chr [1:3] "Point estimate" "Lower confidence limit" "Upper confidence limit" #> ..$ Slope : Named num [1:3] 0.823 0.666 0.98 #> .. ..- attr(, "names")= chr [1:3] "Point estimate" "Lower confidence limit.2.5 %" "Upper confidence limit.97.5 %" #> $ Cindex : Named num [1:3] 0.814 0.774 0.848 #> ..- attr(, "names")= chr [1:3] "Point estimate" "Lower confidence limit" "Upper confidence limit" #> $ warningMessages : NULL #> $ CalibrationCurves:List of 1 #> ..$ FlexibleCalibration:'data.frame': 500 obs. of 4 variables: #> .. ..$ x : num [1:500] 0.00561 0.00651 0.00706 0.01183 0.01235 ... #> .. ..$ y : num [1:500] 0.0504 0.0514 0.052 0.0572 0.0578 ... #> .. ..$ ymin: num [1:500] 0 0 0 0 0 0 0 0 0 0 ... #> .. ..$ ymax: num [1:500] 0.177 0.177 0.177 0.179 0.179 ... #> - attr(, "class")= chr "CalibrationCurve" The coordinates are stored in the CalibrationCurves slot and can be extracted as follows. flexCal = calPerf$CalibrationCurves$FlexibleCalibration plot(flexCal[, 1:2], type = "l", xlab = "Predicted probability", ylab = "Observed proportion", lwd = 2, xlim = 0:1, ylim = 0:1) polygon( x = c(flexCal$x, rev(flexCal$x)), y = c( flexCal$ymax, rev(flexCal$ymin) ), col = rgb(177, 177, 177, 177, maxColorValue = 255), border = NA ) Alternatively, we can use restricted cubic splines to obtain the flexible calibration curve. rcsFit = tryCatch(val.prob.ci.2(pHat, yTest, smooth = "rcs"), error = function(e) TRUE) if(is.logical(rcsFit)) { plot(1, type = "n", xlab = "", ylab = "", xlim = c(0, 10), ylim = c(0, 10)) text(x = 5, y = 5, labels = paste0("There was a problem estimating\n", "the calibration curve using rcs"), cex = 2) } else { rcsFit } #> Call: #> val.prob.ci.2(p = pHat, y = yTest, smooth = "rcs") #> #> A 95% confidence interval is given for the calibration intercept, calibration slope and c-statistic. #> #> Dxy C (ROC) R2 D D:Chi-sq D:p #> 0.62853462 0.81426731 0.38019823 0.33282644 167.41322219 0.00000000 #> U U:Chi-sq U:p Q Brier Intercept #> 0.01286390 8.43195136 0.01475792 0.31996254 0.17703339 0.22404680 #> Slope Emax Brier scaled #> 0.82278297 0.08288689 0.28730517 We obtain the logistic calibration curve using the following code. invisible(val.prob.ci.2(pHat, yTest, logistic.cal = TRUE, smooth = "none")) We can plot both using invisible(val.prob.ci.2(pHat, yTest, logistic.cal = TRUE, col.log = "orange")) The package also allows to change the colors, change the position of the legend and much more. Check out the help-function to see what other arguments the functions have. invisible(val.prob.ci.2(pHat, yTest, col.ideal = "black", col.smooth = "red", CL.smooth = TRUE, legendloc = c(0, 1), statloc = c(0.6, 0.25))) Finally, we can also decide which statistics appear on the plot. invisible(val.prob.ci.2(pHat, yTest, dostats = c("C (ROC)", "Intercept", "Slope", "ECI"))) 1.4.3 ggplot version The ggplot version (i.e.valProbggplot) uses virtually the same arguments. Hence, we can easily obtain a ggplot using the same code. valProbggplot(pHat, yTest) ``` > Call: #> valProbggplot(p = pHat, y = yTest) #> #> A 95% confidence interval is given for the calibration intercept, calibration slope and c-statistic. #> #> Dxy C (ROC) R2 D D:Chi-sq D:p #> 0.62853462 0.81426731 0.38019823 0.33282644 167.41322219 0.00000000 #> U U:Chi-sq U:p Q Brier Intercept #> 0.01286390 8.43195136 0.01475792 0.31996254 0.17703339 0.22404680 #> Slope Emax Brier scaled Eavg ECI #> 0.82278297 0.08288689 0.28730517 0.04747448 0.32064056 ``` 2 Assessing the performance of survival models 2.1 Cox Proportional hazards model The Cox proportional hazards model is a widely used method for analyzing survival data. It estimates the hazard function, which represents the instantaneous risk of an event occurring at time (t), given that the subject has survived up to (t). For patient (i) with covariate vector (\boldsymbol{x}_i), the hazard function is given by: [\begin{align} h(t \| \boldsymbol{x}_i) &= h_0(t) \exp(\boldsymbol{x}_i^\top \beta)\ &= h_0(t) \exp(\eta_i) \end{align}] where (h(t \| \boldsymbol{x}_i)) is the hazard function at time (t) for subject (i), (h_0(t)) is the baseline hazard function at time (t), which is shared by all individuals. In this model, the survival probability for an individual (i) at time (t) is given by [ S_i(t) = \exp\left(-H_i(t)\right) ] where (H_i(t)) is the cumulative hazard function: [ H_i(t) = H_0(t) \exp( \eta_i) ] where (H_0(t)) is the baseline cumulative hazard function. 2.1.1 Calibration Curve We can estimate the calibration curve in this scenario by entering the linear predictor (\eta_i) as a covariate [\begin{align} h(t \| {}_{} \eta_i) &= h_0(t) \exp(\zeta {}_{} \eta_i)\ \end{align}] where (\zeta) is the calibration slope that quantifies the relationship between the predicted and observed hazards. Similarly to (1.1), (\zeta) indicates whether the model is over- ((\zeta < 1)) or underfitted ((\zeta >1)). Note that we do not have a calibration intercept in this model. This is because the baseline hazard function (h_0(t)) is not estimated in the Cox model, and the model is only identified up to a proportionality constant. As a result, any intercept term would be absorbed into the baseline hazard function, and would not be separately identifiable. We do not explicitly specify the time point (t), as calibration can be assessed either across all follow-up time points or at a specific time point of interest. To estimate this model, we can fit a Cox proportional hazards model. Below is a short example illustrating how you can assess the calibration performance of a Cox proportional hazards model using the package. Here, we first fit the model using coxph from the survival package. to assess the calibration performance, we pass the object containing the model fit along with the external validation data set to the valProbSurvival function. library(survival) data(trainDataSurvival) data(testDataSurvival) sFit = coxph(Surv(ryear, rfs) ~ csize + cnode + grade3, data = trainDataSurvival, x = TRUE, y = TRUE) calPerf = valProbSurvival(sFit, testDataSurvival, plotCal = "ggplot", nk = 5) Next to the plot, you also get a range of statistics that assess the calibration performance. calPerf #> Call: #> valProbSurvival(fit = sFit, valdata = testDataSurvival, nk = 5, #> plotCal = "ggplot") #> #> A 95% confidence interval is given for the statistics. #> #> Calibration performance: #> ------------------------ #> #> In the large #> #> OE 2.5 % 97.5 % #> 1.0444489 0.9299645 1.1730270 #> #> Slope #> #> calibration slope 2.5 % 97.5 % #> 1.0703257 0.8202242 1.3204271 #> #> Additional statistics #> #> ICI E50 E90 Emax #> 0.02844123 0.04046305 0.05838470 0.05857903 #> model times Brier se lower upper IPA #> #> 1: Null model 4.99 0.2499302 0.0004004949 0.2491452 0.2507151 0.0000000 #> 2: cox 4.99 0.2245471 0.0077937209 0.2092717 0.2398225 0.1015608 #> #> #> Discrimination performance: #> ------------------------------- #> #> Concordance statistic #> #> Estimate 2.5 % 97.5 % #> Harrell C 0.6517240 0.6193261 0.6841220 #> Uno C 0.6388712 0.6071328 0.6706096 #> #> #> Time-dependent AUC #> #> Uno AUC 2.5 % 97. 5 % #> 0.6856354 0.6305826 0.7406882 3 Assessing the performance of other types of prediction models In my recent paper (De Cock Campo 2023), I propose an extension of the logistic calibration framework to distributions that belong to the exponential family with probability density function (pdf) [\begin{align} f(y_i; \theta_i, \phi, w_i) = \exp\left( \frac{y_i \theta_i - b(\theta_i)}{\phi} w_i + c(y_i, \phi, w_i)\right). \end{align}] Here, (\theta_i) is the natural parameter, (\phi) the dispersion parameter and (w_i) the weight. (b(\cdot)) and (c(\cdot)) are known functions. Similar to before, we assume that there is an unknown regression function (r(\boldsymbol{x}_i) = E[y_i \| \boldsymbol{x}_i]). To approximate this unknown function, we rely on prediction models with the following functional form [\begin{align} E[y_i \| \boldsymbol{x}_i] = \mu_i = f(\boldsymbol{x}_i). \tag{3.1} \end{align}] To estimate (3.1), we can use a generalized linear model [\begin{align} g(E[y_i \| \boldsymbol{x}_i]) = \boldsymbol{x}_i^\top \boldsymbol{\beta} = \eta_i. \tag{3.2} \end{align}] where (g(\cdot)) denotes the link function. Alternatively, we can estimate (3.1) using machine learning methods. Using the model fit, we obtain the predictions (\widehat{\mu}_i = \widehat{f}(\boldsymbol{x}_i)). 3.1 Generalized calibration curves To examine the calibration of prediction models where the outcome is a member of the exponential family, we redefine the framework in more general terms. In this context, a calibration curve maps the predicted values (f(\boldsymbol{x}_i)) to (E[y_i\| f(\boldsymbol{x}_i)]), the actual conditional mean of (y_i) given (f(\boldsymbol{x}_i)). As before, a model is perfectly calibrated if the calibration curve equals the diagonal, i.e. (E[y_i \| f(\boldsymbol{x}_i)] = f(\boldsymbol{x}_i) \ \forall \ i). Hence, in this context, the calibration curve captures the correspondence between the predicted values and the conditional mean. We propose two methods to estimate the calibration curve. Firstly, we can estimate the calibration curve using a generalized linear model [\begin{align} g(E[{}_{} y_i \| {}_{} \widehat{\mu}_i]) = \alpha + \zeta \ g({}_{} \widehat{\mu}_i). \tag{3.3} \end{align}] By transforming ({}_{} \widehat{\mu}_i) using the appropriate (g(\cdot)), we map ({}_{} \widehat{\mu}_i) to the whole real line to better fit the model. If ({}_{} \widehat{\mu}_i) is estimated using a generalized linear model with the same link function (i.e. (g(\cdot)) is identical in (3.2) and (3.3)), it follows that (g({}_{} \widehat{\mu}_i) = {}_{} \widehat{\eta}_i). Using equation (3.3), we estimate the empirical average as a function of the predicted values. Further, similarly to (1.1), (\zeta) tells us whether the model is over- ((\zeta < 1)) or underfitted ((\zeta >1)). We estimate the calibration-in-the-large (\alpha_c) as [\begin{align} g(E[{}_{} y_i \| {}_{} \widehat{\mu}_i]) = \alpha_c + \text{offset}(g({}_{} \widehat{\mu}_i)). \tag{3.4} \end{align}] Hereby, we assess to which extent the observed empirical average equals the average predicted value. Secondly, as with the logistic regression model, we can employ non-parametric smoothers to estimate the calibration curve. 3.2 Illustration of the generalized calibration framework 3.2.1 Training the model To illustrate the functionality, the package has two example data sets with a poisson distributed outcome variable: poissontraindata and poissontestdata. These are two synthetically generated data sets (using the same underlying process/settings to generate the data) to illustrate the functionality of the CalibrationCurves package. The poissontraindata data frame represents the data that we will use to develop our prediction model. data("poissontraindata") In this data frame, we have five covariates and one response variable y. head(traindata) #> y x1 x2 x3 x4 #> 1 0 -0.19981624 0.2982990 1.0277486 -0.1146414 #> 2 1 -1.37127488 0.5940002 -0.8234645 2.0927676 #> 3 1 1.04050541 0.5440481 -1.3576457 1.3126813 #> 4 0 -1.11652476 -0.5382577 -1.1651439 1.0987873 #> 5 1 1.39659613 1.1325081 0.6053029 -1.0598506 #> 6 0 -0.04645095 -0.8167364 1.0196761 -0.4867560 Next, we fit a Poisson GLM with log link to obtain the estimated parameter vector (\widehat{\beta}). glmFit = glm(Y ~ . , data = poissontraindata, family = poisson) summary(glmFit) #> #> Call: #> glm(formula = Y ~ ., family = poisson, data = poissontraindata) #> #> Coefficients: #> Estimate Std. Error z value Pr(>\|z\|) #> (Intercept) -2.33425 0.05301 -44.034 < 2e-16 #> x1 1.28147 0.17645 7.262 3.80e-13 #> x2 2.02783 0.18019 11.254 < 2e-16 #> x3 -1.16815 0.16907 -6.909 4.88e-12 #> x4 -1.88795 0.17840 -10.583 < 2e-16 #> x5 -1.84003 0.17866 -10.299 < 2e-16 #> --- #> Signif. codes: 0 '' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 #> #> (Dispersion parameter for poisson family taken to be 1) #> #> Null deviance: 3087.4 on 4999 degrees of freedom #> Residual deviance: 2660.4 on 4994 degrees of freedom #> AIC: 4053.3 #> #> Number of Fisher Scoring iterations: 6 3.2.2 Assessing the calibration performance Hereafter, we assess the calibration performance on the poissontestdata set. Hereto, we first have to compute the predicted values on this data set. data("poissontestdata") yHat = predict(glmFit, newdata = poissontestdata, type = "response") We then store the response in the poissontestdata in a separate vector yTest. yTest = poissontestdata$Y Now we have everything we need to assess the calibration performance of our prediction model. We can use genCalCurve to visualize the calibration performance and to obtain the statistics. genCalCurve makes the plot using base R and a ggplot version will be included in one of the next updates. By default, the calibration curve as estimated by a GLM will be plotted. Further, in addition to the outcome and the predicted values, we have to specify the distribution of the response variable. calPerf = genCalCurve(yTest, yHat, family = poisson) #> Waiting for profiling to be done... #> Waiting for profiling to be done... In addition to the plot, the function returns an object of the class GeneralizedCalibrationCurve. calPerf #> Call: #> genCalCurve(y = yTest, yHat = yHat, family = poisson) #> #> A 95% confidence interval is given for the calibration intercept and calibration slope. #> #> Calibration intercept Calibration slope #> 0.02710876 0.98320991 This object contains the calculated statistics as well as the calculated coordinates of the calibration curve. str(calPerf) #> List of 6 #> $ call : language genCalCurve(y = yTest, yHat = yHat, family = poisson) #> $ stats : Named num [1:2] 0.0271 0.9832 #> ..- attr(, "names")= chr [1:2] "Calibration intercept" "Calibration slope" #> $ cl.level : num 0.95 #> $ Calibration :List of 2 #> ..$ Intercept: Named num [1:3] 0.0271 -0.1368 0.1825 #> .. ..- attr(, "names")= chr [1:3] "Point estimate.Calibration intercept" "Lower confidence limit.2.5 %" "Upper confidence limit.97.5 %" #> ..$ Slope : Named num [1:3] 0.983 0.768 1.198 #> .. ..- attr(, "names")= chr [1:3] "Point estimate.Calibration slope" "Lower confidence limit.2.5 %" "Upper confidence limit.97.5 %" #> $ warningMessages : NULL #> $ CalibrationCurves:List of 1 #> ..$ GLMCalibration:'data.frame': 1000 obs. of 2 variables: #> .. ..$ x: num [1:1000] 0.0111 0.012 0.0124 0.0142 0.0149 ... #> .. ..$ y: num [1:1000] 0.012 0.013 0.0134 0.0152 0.0159 ... #> - attr(, "class")= chr "GeneralizedCalibrationCurve" The coordinates are stored in the CalibrationCurves slot and can be extracted as follows. GLMCal = calPerf$CalibrationCurves$GLMCalibration plot(GLMCal[, 1:2], type = "l", xlab = "Predicted value", ylab = "Empirical average", lwd = 2, xlim = 0:1, ylim = 0:1, col = "red", lty = 2) abline(0, 1, lty = 1) 4 FAQ 4.1 Why is the calibration intercept different in the rms package? To construct the logistic calibration curve (see How do we construct a calibration curve?), we fit the model [\begin{align} \text{logit}(E[{}_{} y_i \| {}_{} \widehat{\pi}_i]) = \alpha + \zeta \ \text{logit}({}_{} \widehat{\pi}_i) \end{align}] Here, (\zeta) corresponds to the calibration slope. The calibration intercept from the val.prob function from the rms package corresponds to (\alpha \neq \alpha_c). In the CalibrationCurves package, the calibration intercept corresponds to (\alpha_c) which assesses the calibration in the large. Using this formulation, the calibration intercept indicates whether the predicted risks are under- or overestimated on average and this is conform with the definition of the calibration intercept in the article ‘A calibration hierarchy for risk models was defined: from utopia to empirical data’ (and other articles published on this topic) (Van Calster et al. 2016, 2019). We compute (\alpha_c) using [\begin{align} \text{logit}(E[{}_{} y_i \| {}_{} \widehat{\pi}_i]) = \alpha_c + \text{offset}(\text{logit}({}_{} \widehat{\pi}_i)). \end{align}] where we fix (\zeta = 1) by including (\text{logit}({}_{} \widehat{\pi}_i)) as an offset variable. Consequently, both types of calibration intercepts need to be interpreted differently: (\alpha): this corresponds to the constant you have to add after you multiplied the linear predictor with the ‘correction’ factor (i.e. the calibration slope) to get the predicted probabilities to correspond to the observed ones. In essence: once we have multiplied the linear predictor by a correction factor, what is the constant that we still have to add to make the predicted probabilities correspond to the observed ones? (\alpha_c): (> 0): ({}_{} \widehat{\pi}_i) is too low on average and hence, on average the risks are underestimated. You have to increase it to make it correspond to the observed probabilities. (< 0): ({}_{} \widehat{\pi}_i) is too high on average and hence, on average the risks are overestimated. You have to decrease it to make it correspond to the observed probabilities. 4.2 I have predicted probabilities of 0 or 1. Why is this not allowed by default and why do I get these annoying warning messages? Predicted probabilities of 0 or 1 imply that there is no more randomness and that the process is deterministic. If the process was truly deterministic, we would not have to model it. Mostly the presence of perfect predictions signifies that something went wrong when fitting the model or that the model is severely overfitted. We therefore make sure that this is not allowed by default and delete these observations. We observe this behavior in the following cases: - logistic regression: with quasi-complete separation, the coefficients tend to infinity; - tree-based methods: one of the leaf nodes contains only observations with either 0 or 1; - neural networks: the weights tend to infinity and this is known as weight/gradient explosion. If you are confident that nothing is wrong with the model fit, then you can obtain a calibration curve by setting the argument allowPerfectPredictions to TRUE. In this case, predictions of 0 and 1 are replaced by values 1e-8 and 1 - 1e-8, respectively. Do take this into account when interpreting the performance measures, as these are not calculated with the original values. set.seed(1) yTest = testdata$y pHat[sample(1:length(pHat), 5, FALSE)] = sample(0:1, 5, TRUE) x = val.prob.ci.2(pHat, yTest, allowPerfectPredictions = TRUE) #> Warning in val.prob.ci.2(pHat, yTest, allowPerfectPredictions = TRUE): There are predictions with value 0 or 1! These are replaced by values 1e-8 and 1 - 1e-8, respectively. Take this into account when interpreting the performance measures, as these are not calculated with the original values. #> #> Please check your model, as this may be an indication of overfitting. Predictions of 0 or 1 imply that these predicted values are deterministic. #> #> We observe this in the following cases: #> - logistic regression: with quasi-complete separation, the coefficients tend to infinity; #> - tree-based methods: one of the leaf nodes contains only observations with either 0 or 1; #> - neural networks: the weights tend to infinity and this is known as weight/gradient explosion. 5 References Alba, Ana Carolina, Thomas Agoritsas, Michael Walsh, Steven Hanna, Alfonso Iorio, P. J Devereaux, Thomas McGinn, and Gordon Guyatt. 2017. “Discrimination and Calibration of Clinical Prediction Models: Users’ Guides to the Medical Literature.” JAMA : The Journal of the American Medical Association 318 (14): 1377–84. De Cock Campo, Bavo. 2023. “Towards Reliable Predictive Analytics: A Generalized Calibration Framework.” Steyerberg, Ewout W. 2019. Clinical Prediction Models: A Practical Approach to Development, Validation, and Updating. Statistics for Biology and Health. Cham: Springer International Publishing AG. Van Calster, Ben, David J McLernon, Maarten van Smeden, Laure Wynants, Ewout W Steyerberg, Patrick Bossuyt, Gary S Collins, Petra MacAskill, Karel G. M Moons, and Anew J Vickers. 2019. “Calibration: The Achilles Heel of Predictive Analytics.” BMC Medicine 17 (1): 230–30. Van Calster, Ben, Daan Nieboer, Yvonne Vergouwe, Bavo De Cock, Michael J Pencina, and Ewout W Steyerberg. 2016. “A Calibration Hierarchy for Risk Models Was Defined: From Utopia to Empirical Data.” Journal of Clinical Epidemiology 74: 167–76.
6197	https://mathmonks.com/parabola	Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Eulers Number Inequalities Sets De Morgans Laws Transcendental Numbers About Us Table of Contents Last modified on November 25th, 2024 chapter outline Let us imagine a basketball game. As a player shoots the ball, the ball follows a curved path before landing in the hoop. The path that the ball takes is a parabola. A parabola is a U-shaped curve that is also found in the path of water in a fountain or the shape of satellite dishes. Mathematically, it is the graph of a quadratic function in two variables, y = ax2 + bx + c, that can be defined as the set of all points that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix. The vertex is the point where the parabola is closest to the directrix and where it changes direction. The y = ax2 + bx + c is thus a parabolic function. The shape of a parabola is found in the trajectory of rockets and the arc of water from the fountains. It has many applications, such as in headlights, bridge cables, and solar panels. Parts Vertex: It is the turning point of the parabola and also the highest or lowest point, depending on whether it opens upward or downward. The vertex of the above parabola is at (0, 0). Axis of Symmetry: It is the dashed line passing through the vertex of the parabola and dividing it into two symmetric halves. Focus: Has 1 focus, with coordinates (0, a) Directrix: It is the line drawn perpendicular to the axis of the parabola. Here, the line parallel to the x-axis and passing through the point (0, -a) is the directrix. Focal Chord: It is the chord that intersects the parabola at two distinct points and passes through the focus. Latus Rectum: It is the focal chord of the parabola that is perpendicular to its axis and passes through the focus. The endpoints of the latus rectum are (2a, a) and (-2a, a), and its length is 4a. Focal Distance: It is the distance from a point (x1, y1) on the parabola to the focus. The focal distance is also equal to the perpendicular distance from this point to the directrix. Eccentricity: It is the ratio of the distance between any point on the curve and the focus to the distance between that point and the directrix. For a parabola, the eccentricity is always equal to 1. Equations The equation of a parabola depends on its orientation, the position of its vertex, and whether it is centered at the origin or elsewhere. Standard Form A horizontal parabola is a parabola that opens sideways, either to the left or to the right. y2 = 4ax Here, The equation of the axis of symmetry is y = 0 The equation of directrix is x = -a Vertex at (0, 0) Focus at (a, 0) The length of the latus rectum is 4a This parabola opens to the right, forming a ‘’ shaped curve. Derivation: y2 = 4ax Let us derive the above equation. Let us consider a point P on the parabola with the coordinates (x, y) By the definition of a parabola, the distance of point P from the focus F and the distance of point P from the directrix is the same. Now, we will draw a perpendicular line from point P that intersects the directrix at B, as shown. As we know, the eccentricity of a parabola is always 1 We have e = 1 ${\dfrac{PF}{PB}=1}$ PF = PB Since the coordinates of the focus are F(a,0), we can determine its distance from the point P (x, y) Using the coordinate distance formula, we get PF = ${\sqrt{\left( x-a\right) ^{2}+\left( y-0\right) ^{2}}}$ = ${\sqrt{\left( x-a\right) ^{2}+y^{2}}}$ Also, the equation of the directrix is x + a = 0. Thus, by using the perpendicular distance formula, we get PB = ${\dfrac{x+a}{\sqrt{1^{2}+0^{2}}}}$ = ${x+a}$ Now, PF = PB ${\sqrt{\left( x-a\right) ^{2}+y^{2}}=x+a}$ By squaring the equation on both sides, (x – a)2 + y2 = (x + a)2 x2 – 2ax + a2 + y2 = x2 + 2ax + a2 y2 – 2ax = 2ax y2 = 4ax If the parabola is of the form y2 = 4ax, then its parametric coordinates are (at2, 2at). It represents all the points on the parabola. y2 = -4ax Here, The equation of the axis of symmetry is y = 0 The equation of directrix is x = a Vertex at (0, 0) Focus at (-a, 0) The length of the latus rectum is 4a This parabola opens to the left, forming a ‘’ shaped curve. When a parabolaopens either upward or downward, it is called a vertical parabola. x2 = 4ay Here, The equation of the axis of symmetry is x = 0 The equation of directrix is y = -a Vertex at (0, 0) Focus at (0, a) The length of the latus rectum is 4a This parabola opens upward, forming a ‘U’ shaped curve. x2 = -4ay Here, The equation of the axis of symmetry is x = 0 The equation of directrix is y = a Vertex at (0, 0) Focus at (0, -a) The length of the latus rectum is 4a This parabola opens downward, forming a ‘’ shaped curve. As we derived the equation for the parabola y2 = 4ax, we can similarly derive the equations for other types of parabolas. Below is a summary of the formulas to determine the axis, directrix, vertex, focus, and length of the latus rectum for parabolas in different orientations when expressed in standard form: : Find the focus from the given equation of the parabola:a) y2 = 20xb) x2 = -4y Solution: a) Given y2 = 20x ¦..(i)As we know, the standard equation of the parabola is y2 = 4ax ¦..(ii)Comparing the equations (i) and (ii),a = 5Thus, the focus is at (a, 0) = (5, 0)b) Given x2 = -4y ¦..(i)As we know, the standard equation of the parabola is x2 = -4ay ¦..(ii)Comparing the equations (i) and (ii),a = 1Thus, the focus is at (0, -a) = (0, -1) : The equation of a parabola is y2 = 16x. Find the length of the latus rectum, focus, and vertex. Solution: As we know,The equation of a parabola is y2 = 16xNow, 4a = 16a = 4Thus,The length of the latus rectum = 4a = 4 Ã 4 = 16Focus = (a, 0) = (4, 0)Vertex = (0, 0) Vertex Form Another useful form of writing the equation of a parabola is the vertex form, which highlights the vertex of the parabola when graphed. The vertex form of the equation of a vertical parabola is given by: y = a(x – h)2 + k Here, The equation of the axis of symmetry is x = h The equation of directrix is ${y=k-\dfrac{1}{4a}}$ Vertex at (h, k) Focus at ${\left( h,k+\dfrac{1}{4a}\right)}$ The length of the latus rectum is ${\dfrac{1}{a}}$ Derivation of Quadratic Function: y = ax2 + bx + c By rearranging the vertex form of the parabola, we get y = a(x – h)2 + k y = a(x2 – 2hx + h2) + k y = ax2 – 2ahx + ah2 + k y = ax2 + (-2ah)x + (ah2 + k) If we replace (-2ah) with b and (ah2 + k) with c, the standard form of a parabola is also written as: y = ax2 + bx + c, here, a 0, b, and c are constants. The value of a determines whether the parabola opens upward or downward: If a > 0, then the parabola opens upward. If a < 0, then the parabola opens downward. The vertex form of the equation of a horizontal parabola is given by: x = a(y – k)2 + h Here, The equation of the axis of symmetry is y = k The equation of directrix is ${x=h-\dfrac{1}{4a}}$ Vertex at (h, k) Focus at ${\left( h+\dfrac{1}{4a},k\right)}$ The length of the latus rectum is ${\dfrac{1}{a}}$ Here is a table listing all the formulas that are used to find the axis, directrix, vertex, focus, and length of the latus rectum of the parabolas whose vertex is at (h, k): As we obtained the quadratic form for the parabola y = ax2 + bx + c, we can similarly derive the equations for vertical parabolas. The quadratic equation of a vertical parabola is written as x = ay2 + by + c. Here, a 0, b, and c are constants. Graphing To graph a parabola, we find the vertex of the parabola and the axis of symmetry, and then, sketch the curve. For the equation of the parabola y = ax2 + bx + c, the x-coordinate for the vertex is ${h=-\dfrac{b}{2a}}$ By substituting this value in the equation, the y-coordinate for the vertex is: k = a(h)2 + b(h) + c Let us graph the parabola y = x2 Comparing the given equation with the standard form of a parabola y = ax2 + bx + c, we get a = 1, b = 0, and c = 0 Since a is positive, the parabola opens up. The vertex can be calculated as: ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{0}{2\times 1}}$ = ${0}$ Now, substituting h in the original equation y = x2, we get k = f(h) = 02 = 0 The vertex is (0, 0) The length of the latus rectum = ${\dfrac{1}{a}}$ = ${1}$ Focus = ${\left( h,k+\dfrac{1}{4a}\right)}$ = ${\left( 0,0+\dfrac{1}{4\times 1}\right)}$ = ${\left( 0,0.25\right)}$ The axis of symmetry is x = 0 Directrix is ${y=k-\dfrac{1}{4a}}$ ${y=0-\dfrac{1}{4\times 1}}$ ${y=-0.25}$ Now, plotting the given quadratic equation y = x2 on the graph, we get : Graph the parabola x = 3(y – 2)2 + 4. Find the focus, vertex, and length of the latus rectum. Solution: Given x = 3(y – 2)2 + 4 ¦..(i)As we know, the standard form of the parabola is x = a(y – k)2 +h ¦..(ii) Comparing the equations (i) and (ii), we get a = 3, h = 4, and k = 2Since a is positive, the parabola opens to the right.Now,The vertex is at (h, k) = (4, 2)The length of the latus rectum = ${\dfrac{1}{a}}$ = ${\dfrac{1}{3}}$Focus = ${\left( h+\dfrac{1}{4a},k\right)}$ = ${\left( 4+\dfrac{1}{4\times 3},2\right)}$ = ${\left( \dfrac{49}{12},2\right)}$The axis of symmetry is y = k y = 2Directrix is ${x=h-\dfrac{1}{4a}}$ ${x=4-\dfrac{1}{4\times 3}}$ ${x=\dfrac{47}{12}}$Now, plotting the given quadratic equation y = x2 on the graph, we get the required graph. : Find the equation of the parabola whose graph is shown below. Solution: The graph has a vertex at (2, 3). Hence, the equation of the parabola in vertex form is written asy = a(x – 2)2 + 3We now use the y-intercept at (0,1) to find the coefficient a-1 = a(0 – 2) + 3Now, we will solve for aa = 2Thus, the equation of the parabola, whose graph is shown above, isy = 2(x – 2)2 + 3 Properties Tangent A line that touches the parabola at a single point is called its tangent. If the equation of the parabola is y2 = 4ax, the equation of its tangent at the point of contact (x1, y1) is yy1 = 2a(x + x1) Normal The line drawn perpendicular to the tangent that passes through both the point of contact and the focus of the parabola is called the normal. If the parabola is of the form y2 = 4ax and the normal passes through the point (x1, y1) with a slope of ${m=-\dfrac{y_{1}}{2a}}$, then its equation is ${\left( y-y_{1}\right) =-\dfrac{y_{1}}{2a}\left( x-x_{1}\right)}$ Chord of Contact The chord of contact is formed by joining the points of contact of the tangents drawn from an external point to the parabola. If a point (x1, y1) is outside the parabola, the equation of the chord of contact is yy1 = 2x(x + x1) Here, P is any point with coordinates (x1, y1) CD and EF are tangents AB is chord of contact BG is normal Problem: Finding the equation of a parabola when the VERTEX and DIRECTRIX are given : If the vertex of the parabola is at (3, 2) and the directrix of the parabola is y = 5, find the equation of the parabola. Solution: The equation of the parabola in the vertex form is generally written as (x – h)2 = 4p(y – k) ¦..(i)Here, (h, k) = vertex of the parabolap = distance from the vertex to the directrix (or the focus)Given the vertex is (3, 2) and the directrix is y = 5Here, h = 3, k = 2, and p = 2 – 5 = -3Substituting the values in the equation (i), we get(x – 3)2 = 4(-3)(y – 2) x2 – 6x + 9 = -12y + 24 12y = -x2 + 6x – 9 + 24 12y = -x2 + 6x + 15 ${y=-\dfrac{1}{12}x^{2}+\dfrac{6}{12}x+\dfrac{15}{12}}$ ${y=-\dfrac{1}{12}x^{2}+\dfrac{1}{2}x+\dfrac{5}{4}}$Thus, the equation of the parabola is ${y=-\dfrac{1}{12}x^{2}+\dfrac{1}{2}x+\dfrac{5}{4}}$ More Resources Vertex of a Parabola Graphing Parabolas Axis of Symmetry of a Parabola Finding X and Y-Intercepts of a Parabola Domain and Range of a Parabola Transformations of a Parabola Focus and Directrix of a Parabola Standard and Vertex Form of a Parabola Parabola vs Hyperbola Area Under Parabola Last modified on November 25th, 2024 About us Contact us Privacy Policy Categories Algebra Arithmetic Geometry Statistics Trigonometry Grades 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade 9th Grade 10th Grade 11th Grade 12th Grade Join Our Newsletter © 2025 Mathmonks.com. 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6198	https://teachy.ai/en/summaries/high-school-ZA/10th-grade-za/mathematics-za/polygons-diagonal-or-traditional-summary-366a	Log In Summary of Polygons: Diagonal Lara from Teachy SubjectMathematics Mathematics SourceTeachy Original Teachy Original TopicPolygons: Diagonal Polygons: Diagonal Summary Tradisional \| Polygons: Diagonal Contextualization Polygons are flat geometric figures made up of straight lines that connect to form a closed shape. Common examples include triangles, squares, pentagons, and hexagons. Each polygon is defined by the number of sides it has, and these sides meet at points known as vertices. Grasping the properties of polygons is essential for various fields in mathematics and has practical applications in engineering, architecture, and computer graphics. An important concept within the study of polygons is that of a diagonal. A diagonal is a line segment that connects two non-adjacent vertices of a polygon. Being able to identify and calculate the number of diagonals in a polygon is vital, as diagonals play a crucial role in the structural integrity and rendering of three-dimensional shapes. In today’s lesson, we’ll delve into the formula for calculating the number of diagonals in a polygon and tackle practical problems related to this concept. To Remember! Definition of Diagonal A diagonal is a line segment that connects two vertices of a polygon which are not directly next to each other. This concept is crucial for comprehending the internal structure of polygons since diagonals segment the polygon into triangles, which aids in a more detailed analysis of their properties. In a polygon, not all vertices are connected directly by sides; those that are not connected directly are linked through diagonals. This characteristic allows for a complex inner structure, which is essential in practical fields such as engineering and architecture. The definition of a diagonal might be straightforward, but its applications are considerable. It's used to calculate the area of polygons, evaluate the stability of structures, and even in rendering images in computer graphics. For instance, by segmenting a polygon into triangles using its diagonals, we can calculate the total area of the polygon by adding together the areas of the triangles formed. This approach is especially useful for irregular polygons, where simpler area calculation methods may not work. Moreover, diagonals contribute to the structural integrity of buildings and bridges. By incorporating diagonals into a structure, engineers can distribute forces more effectively, enhancing the strength and durability of the construction. Therefore, understanding how to calculate and apply diagonals is an essential skill for those in these professions. A diagonal connects two vertices in a polygon that are not adjacent. Diagonals segment the polygon into triangles, which aids in area calculations. They enhance structural stability in engineering and architecture. Formula for Calculating Diagonals The formula to calculate the number of diagonals in a polygon with 'n' sides is D = n(n - 3) / 2. This formula is derived from the fact that each vertex of the polygon can connect to 'n-3' other vertices, steering clear of the two adjacent vertices and itself. Since each diagonal gets counted twice (once from each end), we divide the result by 2. To illustrate the formula better, let’s look at a pentagon (a polygon with 5 sides). Each vertex of the pentagon can connect to 5 - 3 = 2 other vertices, which leads to a total of 5 2 = 10 connections. Dividing by 2 to avoid double counting gives us 10 / 2 = 5 diagonals. This systematic approach ensures the formula is applicable to any polygon, regardless of the number of sides. The formula D = n(n - 3) / 2 is a robust tool not only for solving academic exercises but also for practical applications. For example, when conceptualising a new architectural design, engineers can use this formula to establish how many diagonals are necessary for stability. Similarly, in the realm of computer graphics, it assists in the efficient rendering of complex shapes. The formula for calculating diagonals is D = n(n - 3) / 2. Each vertex connects to 'n-3' other vertices. The formula is halved to avoid double counting. Practical Examples To put the formula for calculating diagonals into action, let’s go through some practical examples. First, we’ll calculate the number of diagonals in a triangle (n = 3). Using the formula D = 3(3 - 3) / 2, we get D = 3 0 / 2 = 0. This makes sense as a triangle has no diagonals. Next, let’s consider a square (n = 4). Applying the formula gives us D = 4(4 - 3) / 2 = 4 1 / 2 = 2. This confirms that a square has 2 diagonals, which intersect at the centre of the polygon. For a pentagon (n = 5), the formula results in D = 5(5 - 3) / 2 = 5 2 / 2 = 5. Thus, a pentagon has 5 diagonals. These instances illustrate how the formula can be applicable to any regular or irregular polygon. By mastering and applying this formula, learners can tackle complex geometry challenges with more confidence. Additionally, solving various examples reinforces knowledge and hones problem-solving abilities. Triangle (n = 3): 0 diagonals. Square (n = 4): 2 diagonals. Pentagon (n = 5): 5 diagonals. Problem Solving Solving problems that involve counting diagonals in polygons is a practical skill that allows for the application of theory in real-world scenarios. For instance, calculating the number of diagonals in a decagon (n = 10) with the formula D = 10(10 - 3) / 2 = 10 7 / 2 = 35. This kind of problem is common in tests and mathematics competitions. Another example would be calculating the number of diagonals in a polygon with 20 sides. Using the formula, we get D = 20(20 - 3) / 2 = 20 17 / 2 = 170. These problems not only assess understanding of the formula but also evaluate the ability to manage larger numbers and execute exact calculations. Additionally, more intricate challenges may require applying the concept of diagonals in practical contexts, such as deciding the optimal way to segment a structure into smaller parts using diagonals to enhance efficacy and stability. These tasks cultivate critical thinking and sharpen problem-solving skills. Example of a decagon (n = 10): 35 diagonals. Example of a 20-sided polygon: 170 diagonals. Problem-solving that applies diagonals in real scenarios. Key Terms Polygon: A flat geometric figure with straight sides. Diagonal: A line segment connecting two vertices of a polygon that are not adjacent. Vertex: A point where two sides of a polygon converge. Diagonal Formula: D = n(n - 3) / 2, used to calculate the number of diagonals in a polygon. Structural Stability: The role of diagonals in distributing forces and enhancing structural strength. Rendering: The process of creating three-dimensional images in computer graphics where diagonals help define shapes. Important Conclusions In this lesson, we looked at diagonals in polygons, defining them as line segments connecting two non-adjacent vertices. We learned the formula D = n(n - 3) / 2 for calculating the number of diagonals in a polygon and practiced with examples like triangles, squares, and pentagons to understand its relevance in various scenarios. We also highlighted the importance of diagonals in practical domains such as engineering, architecture, and computer graphics, where they ensure structural integrity and assist in rendering three-dimensional forms. Tackling more complex problems, like finding diagonals in polygons with multiple sides, has reinforced learning and developed our calculation and analytical skills. Understanding the diagonals of polygons is crucial not only for solving mathematical problems but also for practical applications that involve structural stability and visual representation of shapes. We encourage students to keep exploring this topic and apply the knowledge gained in their daily experiences and other study areas. Study Tips Practice applying the formula D = n(n - 3) / 2 to various types of polygons to strengthen understanding and calculation skills. Investigate the practical applications of diagonals in sectors like engineering, architecture, and computer graphics to appreciate the concept’s relevance in the real world. Engage with additional exercises and problems that involve counting diagonals in polygons with diverse side counts to build confidence and accuracy in computations. 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6199	https://en.wikipedia.org/wiki/Univariate_distribution	Published Time: 2007-02-01T19:28:35Z Univariate distribution - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Examples 2 See also 3 References 4 Further reading Univariate distribution [x] 3 languages Català Deutsch 粵語 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia In statistics, a univariate distribution is a probability distribution of only one random variable. This is in contrast to a multivariate distribution, the probability distribution of a random vector (consisting of multiple random variables). Examples [edit] Continuous uniform distribution One of the simplest examples of a discrete univariate distribution is the discrete uniform distribution, where all elements of a finite set are equally likely. It is the probability model for the outcomes of tossing a fair coin, rolling a fair die, etc. The univariate continuous uniform distribution on an interval [a, b] has the property that all sub-intervals of the same length are equally likely. Binomial distribution with normal approximation for n=6 and p=0.5 Other examples of discrete univariate distributions include the binomial, geometric, negative binomial, and Poisson distributions. At least 750 univariate discrete distributions have been reported in the literature. Examples of commonly applied continuous univariate distributions include the normal distribution, Student's t distribution, chisquare distribution, F distribution, exponential and gamma distributions. See also [edit] Univariate Bivariate distribution List of probability distributions References [edit] ^Johnson, N.L., Kemp, A.W., and Kotz, S. (2005) Discrete Univariate Distributions, 3rd Edition, Wiley, ISBN978-0-471-27246-5. ^Wimmer G, Altmann G (1999) Thesaurus of univariate discrete probability distributions. STAMM Verlag GmbH Essen, 1st ed XXVII ISBN3-87773-025-6 ^Johnson N.L., Kotz S, Balakrishnan N. (1994) Continuous Univariate Distributions Vol 1. Wiley Series in Probability and Statistics. Further reading [edit] Leemis, L. M.; McQueston, J. T. (2008). "Univariate Distribution Relationships"(PDF). The American Statistician. 62: 45–53. doi:10.1198/000313008X270448. This statistics-related article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Types of probability distributions Statistics stubs Hidden category: All stub articles This page was last edited on 26 April 2025, at 20:32(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Univariate distribution 3 languagesAdd topic

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