Fluid mechanics lectur notes

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Fluid mechanics lectur notes

  1. 1. LECTURE NOTES ON FLUID MECHANICS Version 1.1 Ming-Jyh Chern, D.Phil. Oxon Department of Mechanical Engineering National Taiwan University of Science and Technology 43 Sec. 4 Keelung Road Taipei 10607 Taiwan
  2. 2. PREFACEFluid mechanics is one of important subjects in engineering science. Although it has been developing formore than one hundred years, the area which fluid mechanics covers is getting wider, e.g. biomechanicsand nanofluids. I started to write up this manuscript when I was assigned to give lectures on fluidmechanics for senior undergraduate students. The main purpose of this lecture is to bring physics offluid motion to students during a semester. Mathematics was not addressed in the lecture. However,students were also required to learn use mathematics to describe phenomena of fluid dynamics whenthey were familiar with physics in this subject. As I finished this book, I do hope that readers can getsomething from this book. Meanwhile, I wold like to express my graditude to those who helped me finishthis book. Ming-Jyh Chern Associate Professor Department of Mechanical Engineering National Taiwan University of Science and Technology mjchern@mail.ntust.edu.tw May 29, 2007 I
  3. 3. · II ·
  4. 4. ContentsPREFACE 21 INTRODUCTION 1 1.1 Why study FLUID MECHANICS? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 What is a fluid? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.3 Approaches to study Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3.1 Analytical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3.2 Expenmental Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3.3 Computation Fluid Dynamics (CFD) . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.4 History of Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.5 Fluid as a continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.6 Macroscopic physical properties of fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6.1 density, ρ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6.2 specific gravity, SG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6.3 specific volume, ν . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6.4 specific weight, γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6.5 Compressibility of fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.7 Ideal gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.8 Pascal’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.9 Speed of sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.9.1 Viscosity, µ & ν . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.10 Hooke’s law and Newton’s viscosity law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.11 Categories of Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 FLUID STATICS 15 2.1 Review of Taylor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.3 The Hydrostatic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.4 Pressure variation in incompressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 III
  5. 5. 2.5 Pressure variation in compressible fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.6 Standard Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.6.1 Absolute pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.6.2 Gauge pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.7 Facilities for pressure measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.7.1 Manometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.7.2 Barometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.8 Inclined-tube Manometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.9 Hydrostatic force on vertical walls of constant width . . . . . . . . . . . . . . . . . . . . . 24 2.10 Hydrostatic force on an inclined surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.11 Hydrostatic force on a curved surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.12 Buoyance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 INTRODUCTION TO FLUID MOTION I 33 3.1 Lagrangian and Eulerian Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 Control Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.3 Steady and Unsteady flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.3.1 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.3.2 Pathlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3.3 Streaklines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3.4 Streamtubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.3.5 Definition of 1-D flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.4 Variation of physical properties in a control volume . . . . . . . . . . . . . . . . . . . . . . 36 3.5 Mass conservation of 1-D flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.6 Momemtum conservation for 1-D flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 INTRODUCATION TO FLUID MOTION II 41 4.1 The Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.2 Derive the Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.3 Stagnation Pressure and Dynamic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . 46 4.4 Mass conservation in channel flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.5 Relationship between cross area, velocity ana pressure . . . . . . . . . . . . . . . . . . . . 49 4.6 Applications of Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.6.1 Pitot tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.6.2 Siphon(ÞÜ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.6.3 Torricelli’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 4.6.4 vena contracta effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.6.5 Free jets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54· IV ·
  6. 6. 4.6.6 Venturi tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.6.7 Flowrate pass through a sluice gate . . . . . . . . . . . . . . . . . . . . . . . . . . . 575 EQUATIONS OF MOTION IN INTEGRAL FORM 59 5.1 Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 5.2 Reynolds’ Transport Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 5.3 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5.4 Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5.5 Moment-of-Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646 DIFFERENTIAL EQUATIONS OF MOTIONS 65 6.1 Lagrangian and Eulerian systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.2 Rate of Change Following a Fluid Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 6.3 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 6.4 Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 6.5 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 DIMENSIONAL ANALYSIS 71 7.1 Why dimension analysis? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 7.2 Fundamental dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 7.3 How to carry out a dimensional analysis? . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 7.4 Common nondimensional parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 7.5 Nondimensional Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 7.6 Scale model tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808 Viscous Internal Flow 83 8.1 Fully developed flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 8.2 Laminar, transition and turbulent flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 8.3 2-D Poiseuille flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 8.4 Hagen-Poiseuille flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 8.5 Transition and turbulent pipe flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 8.6 Darcy equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 8.7 Hydraulic diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 8.8 Brief Introduction to Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999 Viscous External Flows 101 9.1 Boundary Layer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 9.2 Uniform flow past a flat plat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 9.3 Boundary Layer Thickness, δ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 ·V·
  7. 7. 9.4 Displacement Boundary Layer Thickness, δd . . . . . . . . . . . . . . . . . . . . . . . . . . 104 9.5 Momentum Boundary Layer Thickness, θ . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.6 Boundary Layer Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 9.7 Friction coefficient, Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 9.8 Drag coefficient, CD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.9 Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.10 Lift force and attack angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.11 Streamline body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 9.12 Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 9.13 Separation and Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111· VI ·
  8. 8. Chapter 1INTRODUCTION1.1 Why study FLUID MECHANICS?Fluid mechanics is highly relevant to our daily life. We live in the worldfull of fluids! Fluid mechanics covers many areas such as meteorology, oceanography,aerodynamics, biomechanics, hydraulics, mechanical engineering, civil en-gineering, naval architecture engineering, and etc. It does not only explain scientific phenomena but also leads industrialapplications.1.2 What is a fluid?The main difference between fluid and solid is their behaviour when shearforces acting on them. A certain amount of displacement is found whena shear force is applied to a solid element. The displacement disappearsas the shear force is released from the solid element. A fluid deformscontinuously under the application of a shear force. Liquids and gases areboth regarded as fluids. 1
  9. 9. 1.3 Approaches to study Fluid Mechanics • Analytical Methods • Experiments • Computations1.3.1 Analytical MethodsUsing advanced mathematics, we can solve governing equations of fluidmotions and obtain specific solutions for various flow problems. For ex-ample: pipe flows.1.3.2 Expenmental Fluid MechanicsThis approach utilities facilities to measure considered flow fields or usesvarious visualization methods to visualize flow pattern. For example: LDA(Laser Doppler Anemometer), hot wire, wind-tunnel test.1.3.3 Computation Fluid Dynamics (CFD)For most of flow problems, we cannnot obtain an analytical solution.Hence, we can adopt numerical methods to solve governing equations.The results are so-called numerical solutions. On the other hands, costsof experiments become very expensive. Numerical solutions proides an al-ternative approach to observe flow fields without built-up a real flow field.For example: finite volume method, finite element method.1.4 History of Fluid Mechanics • Archmides (207-212 B.C.): buoyance theory.·2· INTRODUCTION
  10. 10. • Leodnado da Vinci (1452-1519): He described wave motions, hydraulic jump, jet and vortex motion.• Torricelli (1608-1647): He is well known for measuring atmospheric pressure.• Newton (1643-1727): He explained his famous second law in ” Philosophiae Naturalis Principia Mathematica”. This is one of main laws governing fluid motions. He also provided the idea of linear viscosity describing the relationship between fluid deformation and shearing forces.• Bernoulli (1700-1782): Bernoulli equation.• Euler (1707-1783): Euler equation.• Reynolds (1842-1919): Pipe flows, Reynolds stress, turbulence theory.• Prandtl (1875-1953), Boundary layer theory. Y Volume V of mass m Y0 Volume δ V of mass, δ m C X0 X Z0 Z Figure 1.1: Concept of a continuum. 1.4 History of Fluid Mechanics ·3·
  11. 11. δm δV ρ =δlimδ V ’ δ m V δV δV δV ’Figure 1.2: Variation of a physical property with respect to the size of a continuum. Density is used asan example.1.5 Fluid as a continuumThe concept of a continuum is the basis of classic fluid mechanics. Thecontinuum assumption is valid in treating the behaviour of fluids undernormal conditions. However, it breaks down whenever the mean free pathof the magnitude as the smallest characteristic dimension of the problem.In a problem such as rare fied gas flow (e.g. as encountered in flights intothe upper reaches of the atmosphere), we must abandon the concept of acontinuum in favor of the microscopic and statistical points of view. As a consequence of the continuum, each fluid property is assumed tohave a definite value at every point in space. Thus fluid properties such asdensity, temperature, velocity, and so on, are considered to be continuousfunctions of position and time. There exists a nondimensional number which is utilizd to judge whether·4· INTRODUCTION
  12. 12. DISCRETE COLLISIONLESS PARTICLE OR BOLTZMANN EQUATION BOLTZMANN MOLECULAR EQUATION MODEL CONSERVATION EQUATIONS CONTINUUM EUL.ER NAVER-STOKES DO NOT FROM A MODEL EQS. EQUATIONS CLOSED SET 0 0.01 0.1 1 10 100 00 INVISCID FREE-MOLECULE LIMIT LOCAL KNUDSEN NUMBER LIMIT Figure 1.3: Knusden number and continuum.fluids are continuous or not. Its definition is ℓ Kn = , (1.1) Lwhere ℓ is the free mean path of a fluid molecule and L is the smallestcharacteristic length of a flow field. Kn is the so-called Knusen number. 1.5 Fluid as a continuum ·5·
  13. 13. 1.6 Macroscopic physical properties of fluids1.6.1 density, ρ kg · m−3 Air 1.204 Water 998.2 Sea Water 1025 Mercury 135501.6.2 specific gravity, SG density of substance SG = (1.2) density of water Air 0.001206 Oil 0.79 Ice 0.9171.6.3 specific volume, ν 1 ν= (1.3) ρ1.6.4 specific weight, γ γ = ρg (1.4)1.6.5 Compressibility of fluidsWhen fluids are pressurized, the total volume V is changed. The amountof volume change is the compressibility of fluids. In fluid mechanics, weuse bulk modulus which is denoted as dP dP Ev = −V =ρ , (1.5) dV dρ·6· INTRODUCTION
  14. 14. A high bulk modulus means that fluids are not easy to be compressed.Hence, fluids with a high bulk modulus are incompressible. Units anddimensions of bulk modulus are as same as pressure. For most of liquids, they have very large bulk moduluses (109 in S.I.).It means liquids are incompressible. For most of gases, they are regardedas compressible fluids due to their small bulk moduluses.1.7 Ideal gas lawThe ideal gas law describes the relationship among pressure, density, andtemperature for an ideal gas. It can be shown that P = ρRT where R isthe gas constant. For air R = 287.03 m2s−2 K−1 = 1716.4 ft2 s−2R−2 (1.6)1.8 Pascal’s lawThe Pascal’s law indicates that pressure transmission does not decreasewithin a closed container filled with fluids. As shown in Fig. 1.4, pressureat point A and point B are equal in terms of Pascal law. Therefore, if weapply a force to the area A, it will produce a force on B and the force islarger than the force on A.1.9 Speed of soundWhen disturbances are intorduced into fluid, they are propagated at afinite velocity. The velocity depends on the compressibility of consideredfluids. It is called the acoustic velocity or the speed of sound, C. It isdefind as 1.7 Ideal gas law ·7·
  15. 15. A B Figure 1.4: Concept of Pascal’s law. dP Ev C= = dρ ρFor ideal gases, d(ρRT ) √ C= = RT dρExample: Determine acoustic velocities of air and water where the tem-perature is 20o C. Ev 2.19 × 109 N · m−2 Cwater = = −3 = 1480 m · s−1 (1.7) ρ 998.2 kg · mConsider air as an ideal gas √ Cair = RT = 290 m · s−1 (1.8)It implies that sound in incompressible fluids propagates faster than incompressible fluids.·8· INTRODUCTION
  16. 16.   ¢¢ u t y ¢ y ¡ ¢ x x Figure 1.5: Deformation of a fluid experiencing shear stress.1.9.1 Viscosity, µ νNewtonian fluidsConsider fluids are full of two parallel walls. A shear stress, τ , is appliedto the upper wall. Fluids are deformed continuously because fluids can-not support shear stresses. The deformation rate, however, is constant.Furthermore, if the deformation rate or the so-called rate of strain is pro-portional to the shear stress, then the fluid will be classified as a Newtonianfluid, i.e. dγ τ∝ , (1.9) dtwhere γ is shear angle or dγ τ =µ . (1.10) dtIn addition, dγ du = . (1.11) dt dyHence, du τ =µ . (1.12) dyAgain, the relationship between shear stress acting on a Newtonian fluidand rate of strain (or velocity gradient) is linear. If it is not linear, then 1.9 Speed of sound ·9·
  17. 17. the fluid will be called a non-Newtonian fluid. µ is the so-called dynamicviscosity. Its units are dyne · cm2 or Poise (cP). In addition, lb· s 2 or Ryne s inin B.G. 1 microRyne = 0.145 µ (cP) Another definition of viscosity is the kinematic viscosity which is ν = µ ρ 2Its units are cm or Stoke(cS) in S.I. In addition, in or Newt in B.G. 1 2 s sNewt = 0.00155 (cS).Example: Determine the shear stress exerted on the bottom. Solution: U = 10 cm/s oil ( = 0.036 N·s/m2) y d =5.0 mm u(y) xAccording to Newton’s viscosity law, we have du τb = µ . (1.13) dy y=0The velocity profile is available by a non-slip boundary condition, i.e. U u = y d 0.1 m · s−1 = y 0.005 m = 20y . (1.14)In addition, the velocity gradient on the bottom can be obtained by· 10 · INTRODUCTION
  18. 18. du U = = 20 . (1.15) dy y=0 dTherefore, the shear stress is τb = 0.036 × 20 = 0.72 N · m−2. (1.16)Saybolt viscometerWhen we try to measure the viscosity for a fluid, we do not measure theshear stress, and the volocity gradient but another variable, time. Saybolt viscometer is designed to measure the viscosity of a fluid inconstant temperature. The principle of a fluids drain from a container inconstant temperature and we measure the total time till it takes for 60 mlof fluids. Then we use empirical formulae to evaluate kinematic viscosity,ν. The time, measured in second, is the viscosity of the oil in offficial unitscalled Saybolt Universal Seconds (SOS). 195 ν(cS) = 0.226t − , t ≤ 100 SOS (1.17) t 135 ν(cS) = 0.22t − , t ≥ 100 SOS (1.18) t(temperature= 1500 F )1.10 Hooke’s law and Newton’s viscosity lawHooke’s law for a solid element δ σ = Eǫ = E , (1.19) Lwhere σ is stress, ǫ is strain and E is the so-called Young’s modulus. 1.10 Hooke’s law and Newton’s viscosity law · 11 ·
  19. 19. Sample temperature is constant 60ml Figure 1.6: Saybolt viscosmeterNewton’s viscosity law du τ = µǫ = µ ˙ (1.20) dy solid σ δ E fluid τ u µ In solid mechains, we utilize displacement to describe solid motions orrespons. Velocity, however, is employed in fluid motions instead of dis-placement. It is because fluid deformation under shear stress is continu-ous, so it is hard to find a displacement to indicate the magnitude of afluid motion.1.11 Categories of Fluid DynamicsHydrodynamics Hydraulics· 12 · INTRODUCTION
  20. 20. Inviscid Fluid Flows(Potential Flows) Viscous Fluid FlowsLaminar Flows Turbulent FlowsInternal Flows External Flows 1.11 Categories of Fluid Dynamics · 13 ·
  21. 21. · 14 · INTRODUCTION
  22. 22. Chapter 2FLUID STATICSIn fluid statics, fluids at rest are considered. No relative motion betweenadjacent fluid particles. Since there is no relative motion between fluids,viscous stress shoud not exist. Otherwise, fluids would not be at rest.Weight of fluids is the only force in fluid statics. To keep static equilibrium,resultant forces must be zero. Therefore pressure should be included tokeep equilibrium.2.1 Review of Taylor ExpansionFor a continuous function, f (x), it can be expanded in a power series inthe neighborhood of x = α . This is the so-called Taylor Expansion givenby f ′(α) f ′′ (α) 2 f n (α) f (x) = f (α)+ (x−α)+ (x−α) +. . .+ (x−α)n +. . . (2.1) 1! 2! n!2.2 PressurePressure is continuous throughout a flow field in terms of continuum con-cept. Pressure is isotropic. In other words, pressure is independent of 15
  23. 23. direction. Positive pressure means compression. On the other hand, neg-ative pressure means tension. It is opposite to a normal stress. Pressurecan be regarded as a scalar. z z P1dA g dz ds y P2dydz ρgdxdydz/2 P3dxdy x x dA = ds · dy =dy · dz/sin Figure 2.1: Fluid element in a static fluid domain. F=0 (2.2) Fx = P2 dydz − P1 dA sin θ = 0 (2.3) dz P2 dydz = P1 dy sin θ (2.4) sin θ P2 = P1 (2.5) 1 dx Fz = P3 dydx = ρgdxdydz + P1 dy cos θ (2.6) 2 cos θ 1 P3 = P1 + ρgdz (2.7) 2 dz → 0, P3 = P1 (2.8) ∴ P1 = P2 = P3 (2.9)units of pressureS.I. 1 N · m−2 = 1 Pascal(Pa) = 0.01 mbar(mb) (2.10)· 16 · FLUID STATICS
  24. 24. B.G. 1 lb · in−2 = 1 psi = 144 psf(lbf · ft−2) (2.11)2.3 The Hydrostatic EquationConsider a fluid particle at rest shown in Figure 2.2. The centroid of the z   z ¡ O     x y y x Figure 2.2: Concept of a fluid element.fluid element is at the original point O. The fluid element has a smallvolume δV = ∆x∆y∆z . Furthermore, the fluid is at static equilibrium,so resultant forces acting on the fluid element should be zero, i.e. F=0 . (2.12)No shear stresses should exist owing to static equilibrium. Therefore, wecan just consider resultant forces in the z-direction, i.e. Fz = 0 . (2.13)Resultant forces in the z-direction include the weight of the fluid and sur-face forces caused by pressure. The weight of the fluid particle can begiven by W = ρgδV = ρg∆x∆y∆z . (2.14) 2.3 The Hydrostatic Equation · 17 ·
  25. 25. Subsequently, surface forces acting on the fluid element can be given by Fs = (P2 − P1 )∆x∆y , (2.15)where P1 and P2 are pressures on the top and the bottom respectively. P1and P2 can be expanded using Taylor Expansion, i.e. 2 P ′ (0) ∆z P ′′ (0) ∆z P1 = P (0) + + + + + ... (2.16) 1! 2 2! 2and 2 P ′ (0) ∆z P ′′ (0) ∆z P2 = P (0) + − + − + ... (2.17) 1! 2 2! 2Substituting formulae above into the surface force, the surface force be-comes 3 ∆z ′ ∆z Fs = −2 P (0) + P ′′′ (0) + . . . ∆x∆y . (2.18) 2 2Consider static equilibrium again, then we find 3 ∆z′ ∆z Fz = Fs +W = −2 P (0) + P ′′′ (0) + . . . ∆x∆y−ρg∆x∆y∆z = 0 2 2 (2.19) 3 ∆z ∆z 2 P ′ (0) + P ′′′ (0) + . . . = −ρg∆z (2.20) 2 2In terms of continuum concept, ∆z should be very small (not zero), so wecan negelect high order terms in the formula, i.e. P ′ (0)∆z = −ρg∆z (2.21)or dP = −ρg . (2.22) dz z=0We can use a notation directional gradient to show the equation again, i.e. ∇P = ρg . (2.23)This is called the hydrostatic equation.· 18 · FLUID STATICS
  26. 26. 2.4 Pressure variation in incompressible fluidsDensity is constant throughout an incompressible fluid domain. Hence, wecan evaluate the pressure difference between two points(z = z1 and z2 ),i.e. 2 dP ∆P |2 1 = dz 1 dz 2 = −ρgdz 1 2 = −ρg dz 1 = −ρg (z2 − z1 ) . (2.24)∆Pρg is called a pressure head and equal to −∆z .2.5 Pressure variation in compressible fluidsDensity is not constant throughout a compressible fluid domain. In otherwords, density may be affected by temperature and pressure. If we considera perfect gas, then the equation of state for a perfect gas can be used: P = ρRT (2.25)Substituting the perfect gas law to the hydrostatic equation, we obtain dP Pg dP g = −ρg = − ⇒ =− dz (2.26) dz RT P RTIn addition, the pressure difference between two points (z = z1 and z2 )can be evaluated by integrating the hydrostatic equation: 2 2 dP g = − dz (2.27) 1 P 1 RT g ⇒=lnP|2=- RT (z2 − z1 ) 1 2.4 Pressure variation in incompressible fluids · 19 ·
  27. 27. P2 g ⇒=ln P1 =- RT (z2 − z1 ) g P2 =P1 exp[- RT (z2 − z1 )] g △P |2 =P2 -P1 =-P1 1 − exp − RT (z2 − z1 ) 1Example: Determine the pressure at the gasoline-water interface, and atthe bottom of the tank (see Fig. 2.3). Gasoline and water can be both open 17ft gasoline S.G.=0.68 P1 water 3ft P2 Figure 2.3: Problem of hydrostatic force on bottom of a tank.regarded as incompressible fluids. Hence, P1 = γgasoline · h + P0 (2.28)If we assume P0 =0, then P1 = 0.68 · 62.4 lb/ft3 · 17 = 721 psf (2.29)In addition, the pressure at the bottom is determined by P2 = γwater · 3 + P1 = 62.4 · 3 + 721 = 908 psf . (2.30)2.6 Standard AtmosphereSea level conditions of the U.S. Standard Atmosphere.· 20 · FLUID STATICS
  28. 28. 50 z(km) 40 20 10 surface -60 -40 -20 0 20 40 40 80 120 Temperature Pressure T = T0- (z-zo)     = 6.5Kkm-1 Figure 2.4: Variation of atmospheric pressure. Table 2.1: sea level condition S.I. B.G. Temperature 15o C 59oC Pressure 101.33 kPa 2116.2 psf Density 1.225 kg/m3 0.002377 slug/ft3Homework: Derive the formula for the pressure variation within the con-vection layer. Remember pressure and temperature are both functions ofelevation.Ans: g/αR α(z − z0 ) P = P0 1− (2.31) T0 α = 6.5 Kkm−1 (2.32) R = 287 Jkg−1K−1 (2.33) g = 9.8 ms−2 (2.34) 2.6 Standard Atmosphere · 21 ·
  29. 29. 2.6.1 Absolute pressurePressure measured relative to an absolute vacuum.(Pb)2.6.2 Gauge pressurePressure measured relative to atmospheric pressure.(Pg ) Pa d h . Pressure caused by fluid weight. z Pressure caused by atmospheric. Figure 2.5: Variation of static pressure. Pb = Pg + Pa , (2.35)(Pa : atmospheric pressure)Consider fluids shown in Fig. 2.5. Its depth is h. If we evaluate pressureat z = h − d, pressure at z = h − d should include two components,atmospheric pressure and static pressure, i.e. Pz = Pa + ρgd = Pa + ρg(h − z) . (2.36)The resultant force acting on a small area dA at z can be given by dF = Pz dA = Pa + ρg(h − z)dA . (2.37)If we evaluate the resultant force on the bottom, then we obtain F = (Pa + ρgh)dA . (2.38)· 22 · FLUID STATICS
  30. 30. 2.7 Facilities for pressure measurement2.7.1 Manometers P1 P2 B   h A Z2 Z1 Figure 2.6: Schematic of a manometer. Manometers are utilized to measure pressure difference between twopoints, ∆P = P1 − P2 = ρgδh . (2.39)2.7.2 BarometersBarometers are devices designed to measure absolute pressure, ¦¥¥¤£¢ ¡ h Figure 2.7: Schematic of a barometer. Pb = ρg∆h . (2.40) 2.7 Facilities for pressure measurement · 23 ·
  31. 31. 2.8 Inclined-tube ManometerThe main purpose of an inclined-tube manometer is to improve its resolu-tion. Therefore, if a small pressure change is expected in an experiment,then an inclined-tube manometer should be considered. γ 3 h2 B γ 1 γ 2 A h1 l2   Figure 2.8: Inclined manometer. P1 = P2 + γ2(l2 sin θ) (2.41) PA + γ1h1 = PB + γ3h2 + γ2 (l2 sin θ) (2.42) PA − PB = γ3 h3 + γ2(l2 sin θ) − γ1 h1 (2.43)If we ignore γ1 and γ3, then PA − PB = γ2 l2 sin θ (2.44)and PA − PB l2 = . (2.45) γ2 sin θIf PA -PB and γ2 are constant, l2 is quite large as θ is small.2.9 Hydrostatic force on vertical walls of constant width dF = Pb wdz (2.46)· 24 · FLUID STATICS
  32. 32. Pa dF dz h z Figure 2.9: Hydrostatic force exerted on a vertical gate. Pb = Pa + ρg(h − z) (2.47) dF = [Pa + ρg(h − z)]wdz (2.48)For the whole vertical wall, the resultant force is F = dF h = [Pa + ρg(h − z)]wdz 0 h h = Pa wdz + ρg(h − z)wdz (2.49) 0 0 Pawh ρgh2 w 2If we just consider pressure caused by the weight of fluids, then the forcewill be 2.9 Hydrostatic force on vertical walls of constant width · 25 ·
  33. 33. ρgh2 Fs = w . (2.50) 2The force exerts a moment at point z = 0 and the moment is given by dM0 = zdFs = z · ρg(h − z)wdz (2.51)and then M0 = dM0 h = ρg(h − z)wzdz 0 h hz 2 z 3 = ρgw − 2 3 0 3 3 h h = ρgw − 2 3 ρgh3 w = . (2.52) 6We can evaluate the moment arm z , i.e. ¯ ρgh3 w M0 6 h z= ¯ = ρgh2 w = . (2.53) F 3 22.10 Hydrostatic force on an inclined surfaceConsider an inclined surface shown in Fig. 2.10, then dF = ρghdA, h = y sin θ = ρgy sin θdA, dA = wdy (2.54)· 26 · FLUID STATICS
  34. 34. O θ Y h dF w X dA Y Figure 2.10: Hydrostatic force exerted on an inclined gate.and F = dF = ρgy sin θdA = ρg sin θ ydA . (2.55) ydA is the first moment of the area with respect to the x-axis, so we cansay ydA = yc A, (2.56)where yc is the centroid of the area. Furthermore, the resultant forcebecomes F = ρg sin θyc A = ρghc A (2.57) We consider the moment caused by the resultant force with respect to 2.10 Hydrostatic force on an inclined surface · 27 ·
  35. 35. the original point O. First of all,we know dM = ydF (2.58)and then M = dM = ydF = ρgy 2 sin θdA . (2.59) y 2 dA is called the second moment of the area with respect to the x-axis,Ix. We know M = F · yR (2.60)and M ρg sin θ y 2 dA Ix yR = = = , (2.61) F ρg sin θyc A yc Awhere yR is the acting point of the resultant force or so-called the centreof pressure.Example: Consider a dam of width 100 m and depth 6 m. Determine theresultant hydrostatic force and the moment with respect to A.· 28 · FLUID STATICS
  36. 36. h A Figure 2.11: Problem of hydrostatic force exerted on a dam.Sol: F = γhc A h = γ A 2 = 1000 × 9.8 × 0.5 × 6 × (6 × 100) = 17660 kN M = F · hf 1 = F· h 3 = 35320 kN-m (2.62)2.11 Hydrostatic force on a curved surfaceConsider a curved surface shown in Fig. 2.12. The resultant force acting 2.11 Hydrostatic force on a curved surface · 29 ·
  37. 37. F Fx h dF Fz α Z dA X Figure 2.12: Hydrostatic force exerted on a curved surface.on a small element of the curved surface is given by dF = P n · dA = ρg(h − z)n · dA (2.63)The resultant force in the x-direction, Fx , can be denoted as dFx = ρg(h − z) sin αdA, (2.64)where α is the angle between the z-axis and the normal direction of thesmall area. In addition, Fx = dFx = ρg(h − z) sin αdA = ρg (h − z) sin αdA = ρg (h − z)dAv , (2.65)· 30 · FLUID STATICS
  38. 38. where dAv is the project area of dA on the z-axis. In terms of the formula,the resultant force in the x-axisis equal to the force acting on a verticalplane. On the other hand, the resulatant force in the z-axis is given by dFz = −ρg(h − z) cos αdA (2.66)In addition, Fz = dFz = −ρg(h − z) cos αdA = −ρg (h − z)dAh , (2.67)where dAh is the project area of dA on the x-axis. In terms of this formula,Fz is equal to the weight of liquids above the curved surface. The resultantforce F can be given by |F| = Fx + Fz2 . 2 (2.68)2.12 BuoyanceIt is well-knoen that Archimede provided the buoyance principle to eval-uate the buoyant force acting on a submerged solid body. In fact, we canderive the buoyance principle from the hydrostatic equation. Let us con-sider a submerged body shown in Fig. 2.13. The resultant force caused bypressure on the small wetted area is given by dF = P2 dA − P1 dA = (−ρgz2 + ρgz1 )dA (2.69)and F = dF = ρg (z1 − z2 )dA = −ρgV . (2.70) 2.12 Buoyance · 31 ·
  39. 39. P1 Z Z1 dA Z2 P2 Figure 2.13: Schematic of buoyance exerted on an immersed body.Therefore, we know the resultant force caused by static pressure or calledthe buoyant force is equal to the weight of liquids of volume equal to thesubmerged body. In addition, the point where the buoyant force exerts iscalled the centre of buoyance.· 32 · FLUID STATICS
  40. 40. Chapter 3INTRODUCTION TO FLUIDMOTION IThe chapter demonstrates basic concepts of fluid kinematics and funda-mental laws which fluids conserve.3.1 Lagrangian and Eulerian SystemsWhen we describle physical quantities, such as density, pressure, and soon, of adynamic problem, we usually chose either Lagrangine or Euleriansystem. In terms of Lagrangine system, we move with the consideredsystem or particles, so physical quantities, say φ, is only a function oftime, i.e. φ = φ(t) = φ(x(t), y(t), z(t), t) . (3.1)Its coordinates are also functions of time. Lagrangian system is oftenemployed in solid dynamic. On the other hand, we fix a point in space andobserve the variation at this point in terms of Eulerian system. Thereforephysical quantities, φ, are not only functions of time but also functions of 33
  41. 41. space, i.e. φ = φ(x, y, z, t) , (3.2)where x, y, z, and t are independent. Eulerian system is commonly used influid dynamics. It may be because lots of fluid particles are involved in aconsidered flow. It contains different fluid particles at the observed pointas time goes in Eulerian system. Hence it is hard to describe a system orits physical quantities in terms of a specified fluid particle. Therefore, weutilize Eulerian system to describe a system.3.2 Control VolumeIn addition, we utilize a control volume concept to describe a fluid flowproblem. Coupled with Eulerian system, a control volume is a fixed regionwith artifical boundaries in a fluid field. A control volume contains lots ofand various fluid particles as time goes. Fluid flows in and out through itscontrol surface and then physical quantities in a control volume change.3.3 Steady and Unsteady flowIf physical quantities of a flow field are independent of time, then the flowwill be called steady. Otherwise, it is unsteady.3.3.1 StreamlinesA steamline is defined as a line that is everywhere tangential to the in-stantaneous velocity direction, i.e. dy v dy v dx u = , = , and = . (3.3) dx u dz w dz wStreamlines cannot cross.· 34 · INTRODUCTION TO FLUID MOTION I
  42. 42. 3.3.2 PathlinesA pathline is defined as the path along which a specified fluid particleflows. It is a Lagrangine concept. Hence, coordinates of a pathline arefunctions of time.3.3.3 StreaklinesA streakline is the line traced out by particles that pass through a partic-ular point.3.3.4 StreamtubesA streamtube is formed by steamlines. Since streamlines cannot cross,they are parallel in a streamtube.3.3.5 Definition of 1-D flows 1 2 Figure 3.1: 1-D flow 1-D flows are idealizd flows (see Fig. 3.1). It means physical propertiesof flows are only functions of a spatial variable. The spatial variable canbe coordinates of an axis, such as x, or along a streamline. For example, 3.3 Steady and Unsteady flow · 35 ·
  43. 43. density ρ, for 1-D flows can be given: ρ = ρ(x) . (3.4)In addition, 1-D flows can be steady or unsteady, so it may be ρ = ρ(x, t) . (3.5)3.4 Variation of physical properties in a control volumeConsider a control volume in a flow field (see Fig. 3.2). The rate ofvariation of a physical property in a control volume shall be equal to thesum of the flux through its control surface and the surface of the physicalproperty. uφ source of φ Figure 3.2: Control volume d ∂φ φdV = φu · dA + dV (3.6) dt control surface ∂tφ: physical property in a unit volume. For example, mass in a unit volumeis density. ( m = ρ) V· 36 · INTRODUCTION TO FLUID MOTION I
  44. 44. 3.5 Mass conservation of 1-D flowsWhen fluids move, the mass conservation law should be satisfied through-out a flow field. In terms of a control volume, the change rate of mass ina control volume should be zero, i.e. ˙ m=0 . (3.7) Consider a 1-D flow like the figure and fluids move along a streamline. Ifwe consider the control volume between point 1 and point 2 and the massconservation law should be satisfied in the control volume. If we donotconsider any mass source or sink in the control volume, then the rest willbe mass flux on the surface 1 and 2, i.e. mc = m1 + m2 = 0 . ˙ ˙ ˙ (3.8) m1 = −m2 ˙ ˙ (3.9)In addition, m = ρu · A ˙ (3.10)and then ρ1 u1 A1 = ρ2 u2A2 , (3.11)where u1 and u2 are average velocities at points 1 and 2, respectively. Ifdensity of fluids are the same at surface 1 and 2, i.e. Q = u 1 A1 = u 2 A2 , (3.12)where Q is the volumetric flow rate. In terms of the mass conservationlaw, we find that average velocity on a small area is higher than one on alarge area. 3.5 Mass conservation of 1-D flows · 37 ·
  45. 45. 3.6 Momemtum conservation for 1-D flowsAccording to Newton’s second law, an object should retain the same ve-locity or be at rest if the resultant force exerted on it is zero. That meansthe change rate of momentum in the object should be zero. We look intothe control volume concept again. If a control volume is not accelerated,then the resultant force should be zero in the control volume. i.e. F=0 , (3.13)or d (mu) = 0 . (3.14) dt If we donot consider any force source in a control volume for a 1-D flowlike Fig. 3.2, then only momentum fluxes on surface 1, 2 are considered,i.e. d F= (m1u1 + m2 u2) = 0 (3.15) dtor d (ρ1A1 u1 · u1 + ρ2 A2u2 · u2) = 0 (3.16) dtIf the 1-D flow is steady, then we can remove the total derivative, i.e. ρ1 A1(u1 · u1 ) + ρ2 A2 (u2 · u2) = 0 (3.17)or ρ1 A1 u2 = ρ2 A2u2 . 1 2 (3.18)If we consider other forces acting on the control volume, then d F = 0 = F0 + (mu) = 0 (3.19) dt· 38 · INTRODUCTION TO FLUID MOTION I
  46. 46. d F0 + (ρ1A1u1 · u1 + ρ2 A2u2 · u2) = 0 . (3.20) dtThis is consistent with Newton third law. F can be divided into two parts:1. body forces such as gravity forces, magnetic forces; 2. surface forcessuch as pressure. 3.6 Momemtum conservation for 1-D flows · 39 ·
  47. 47. · 40 · INTRODUCTION TO FLUID MOTION I
  48. 48. Chapter 4INTRODUCATION TO FLUIDMOTION II4.1 The Bernoulli equationConsider a steady inviscid flow. If we apply Newton’s second law along astream line, we will obtain the Bernoulli equation 1 1 P1 + ρu2 + ρgz1 = P2 + ρu2 + ρgz2 = const . (4.1) 2 1 2 2The detailed deviation of the Bernoulli equation will be given later. TheBernoulli equation above is based on four assumptions: 1. along a same streamline 2. steady flow 3. same density 4. inviscid 41
  49. 49. 4.2 Derive the Bernoulli equationConsider a steady flow shown in Fig. 4.1. For a fluid particle in thestreamline A, the momentum should be conserved. Assume the volume ofthe fluid is ∆x∆n∆s. The total force along the streamline should be Z g A ∂ P ds dndx n ( P+ ) ∂s 2 s ∆s β ∆n ∆n β β( P- ∂P ds ) dndx ∂s 2 ρg∆x∆n∆s Y Figure 4.1: Force balance for a fluid element in the tangential direction of a streamline. ∂P ds ∂P ds ΣFs = P− − P+ dndx − ρg∆x∆n∆s sin β ∂s 2 ∂s 2 ∂P = − dsdndx − ρg∆x∆n∆s sin β . (4.2) ∂s· 42 · INTRODUCATION TO FLUID MOTION II
  50. 50. The momentum change along the streamline should be ∂ 1 ∂u (Σmu) = − (ρ∆x∆n∆s) (u) + (ρ∆x∆n∆s) u + ds ∂t ∆t ∂s 1 ∂u = ρ(∆x∆n∆s) ds ∆t ∂s ∂u = ρ (∆x∆n∆s) u , (4.3) ∂swhere u is the tangential velocity component. Let us consider Newton’ssecond law, i.e. ∂ ΣFs = (Σmu) (4.4) ∂tSubstitution of Eq. (4.2) into (4.3) gives ∂P ∂u ∂z − − ρg sin β = ρu , sin β = (4.5) ∂s ∂s ∂sand then ∂P ∂z ∂u − − ρg = ρu . (4.6) ∂s ∂s ∂sThis is the so-called Euler equation along a streamline in a steady flow. Ifthe Euler equation is multiplied by ds, it will become −dP − ρgdz = ρudu (4.7)Futhermore, we integrate the whole equation and obtain the Bernoulliequation, i.e. P 1 + u2 + gz = constant . (4.8) ρ 2The Euler equation refers to force balance along a streamline, so the prod-uct of the Euler equation and ds can be regarded as work done by a fluidalong the streamline. The integral of the resultant equation is constantalong a streamline. It turns out that the Bernoulli equation refers to en- 4.2 Derive the Bernoulli equation · 43 ·
  51. 51. Pergy conservation along a streamline. ρ + gz can be regarded as potential u2energy and 2, of course, is the kinetic energy. Moreover, we consider force balance across a streamline. The resultantforce should be ∂ P dn dsdx ( P+ ) ∂n 2 ∆n β ( P-∂P dn ) dsdx ∂n 2 W Figure 4.2: Force balance of a fluid element in the normal direction of a streamline. ∂P dn ∂P dnΣFn = P− dsdx − P + dsdx − ρg∆x∆s∆n cos ρ . ∂n 2 ∂n 2 (4.9)Its momentum change across a streamline should be ∂ u2 Σmun = −ρ ∆x∆s∆n , (4.10) ∂t Rwhere un is the velocity component normal to a streamline and R is the· 44 · INTRODUCATION TO FLUID MOTION II
  52. 52. curvature radius. Let us consider Newton’s second law again. ∂ ΣFn = Σmun (4.11) ∂tSubstitution of Eq. (4.9) into (4.10) gives ∂P u2 − dndsdx − ρg∆x∆s∆n cos β = −ρ ∆x∆s∆n (4.12) ∂n R ∂z cos β = (4.13) ∂nand then ∂P ∂z u2 + ρg =ρ . (4.14) ∂n ∂n RThis is the Euler equation across a streamline. If the Euler equation ismultiplied by dn and integrated along the normal direction, it will become u2 − dP − ρgdz = ρ dn . (4.15) RIt is the Bernoulli equation along the normal direction of a stream.Example: Determine the pressure variation along the streamline from z A a x O B 3 u=u0(1+a3 ) x Figure 4.3: 2-D flow past a circle.point A to point B. 4.2 Derive the Bernoulli equation · 45 ·
  53. 53. Solution:From the Bernoulli equation along a streamline, −dP − ρgdz = ρudu (4.16)Since point A and B are at the horizontal streamline, dz = 0 Hence −dP = ρudu . (4.17)In additions, O O dP = ρudu . A AWe know that du = u0a3 (−3)x−4dx a3 = −3u0 4 dx . (4.18) xAs a result, O a3 a3 PO − PA = ρ u0 1 + −3u0 4 dx A x3 x O 2 a3 a6 = ρ −3u0 + dx A x4 x7 O a3 a6 = ρu2 0 + x3 2x6 A O u2 0 1 = ρa3 1+ . (4.19) x3 2x3 AThe x-coordinate of point B is -a, so 1 u2 0 1 PB − PA = −ρu2 0 1− 3 − ρa 3 3 1+ . (4.20) 2a xA 2x3A4.3 Stagnation Pressure and Dynamic PressureConsider fluids flow toward a horizontal plate far upstream. Fluids movesat u∞ and pressure is P∞ upstream. Because fluids cannot pass through a· 46 · INTRODUCATION TO FLUID MOTION II
  54. 54. P ∞ u ∞ P0 stagnation point stagnation streamline Figure 4.4: Stagnation pointplate, fluids must flow along the plate. Subsequently we can find a pointwhere fluids are at rest. This is the so-called stagnation point. Further-more, we can find a stagnation steamline which leads to the stagnationpoint. Owing to no variation of altitude in the whole flow, pressure andvelocity are considered in the Bernoulli equation. If we apply the Bernoulliequation along the stagnation line, we will find P∞ u2 P0 + ∞= , (4.21) ρ 2 ρwhere P0 is called the stagnation pressure or total pressure, P∞ is called 2 ρu∞the static pressure, and 2 is called dynamic pressure which is distinctedfrom the pressure due to hydrostatic pressure, P∞ . 4.3 Stagnation Pressure and Dynamic Pressure · 47 ·
  55. 55. Pressure coefficient is defined as P − P∞ u Cp = 1 2 = 1 − ( )2 . (4.22) 2 ρu∞ u∞Its means the ratio of pressure difference to inertia force. At a stagnationpoint, Cp = 1, that means all of kinetic energy is transfered to pressureenergy. Cp is zero far upstream. It means no kinetic energy is transferedto pressure energy.4.4 Mass conservation in channel flowsConsider fluid flow in a channel with various cross section areas show inFig. 4.5. Fluids connot accumulate at any cross sections. In other words, 1 2 Figure 4.5: Mass conservation in 1-D flow.mass must be conserved at any cross section. Hence mass flowrates, theamount of mass passing a cross section per unit time, must be equal atevery cross section, i.e. m = m1 = m2 , ˙ ˙ ˙ (4.23)· 48 · INTRODUCATION TO FLUID MOTION II
  56. 56. where m is the mass flow rate in the channel. In addition, ˙ m = ρQ , ˙ (4.24)where ρ is fluid density and Q is volumeric flowrate. Then, ρ1 Q1 = ρ2 Q2 (4.25)or ρ1 u1 A1 = ρ2 u2A2 , (4.26)where u1 and u2 are average velocity at cross sections 1 and 2, A1 andA2 are cross sectional areas. For incompressible fluids, ρ1 = ρ2 and conse-quently u1 A1 = u2A2.4.5 Relationship between cross area, velocity ana pressureConsider a steady flow in a channel with varied cross sectional areas. Interms of the continuity equation, velocity decreases as its cross sectionalarea diverages for incompressible fluids. In addition, pressure increasesas velocity decreases in terms of the Bernoulli equation. For a convergedchannel, cross sectional area decreases so velocity increases. Subsequently,pressure decreases owing to increasing velocity.4.6 Applications of Bernoulli equation4.6.1 Pitot tube 1 1 P∞ + ρa u2 + ρa gz∞ = PO + ρa u2 + ρa gzO ∞ O (4.27) 2 2 z∞ = zO , uO = 0 (4.28) 1 ∴ P∞ + ρa u2 = PO ∞ (4.29) 2 4.5 Relationship between cross area, velocity ana pressure · 49 ·
  57. 57. A A V V P P Figure 4.6: Variations of velocity and pressure in converged and diverged channels. 1 (PO − P∞ ) = ρa u2 ∞ (4.30) 2 PO − P∞ = ρℓ g∆h (4.31) 1 ρℓ g∆h = ρa u2 ∞ (4.32) 2 ρℓ u2 = 2 g∆h ∞ (4.33) ρa4.6.2 Siphon(ÞÜ)A siphon is a device transfering fluids from a lower level to a higher level.Consider a siphon shown in Fig. 4.8. The free surface in the tank isassumed to be still owing to the flow rate to the siphon is very slow.Hence the velocity is zero at the free surface. Furthermore, the Bernoulliequation is applied to analyze the flow in a siphon. Consider conditionsat points 1 and 3 and Pa Pa u2 + 0 + gz1 = + 3 + gz3 , (4.34) ρ ρ 2· 50 · INTRODUCATION TO FLUID MOTION II
  58. 58. O P∞ u∞ z0 ∞ z∞ ∆h ρl Figure 4.7: Schematic of Pitot tube.where z1 = 0, z3 = −h3 . Velocity at point 3 is obtained from the equationi.e. u3 = 2gh3 . (4.35)Another interesting location is at point 2. In terms of Bernoulli equation,we find Pa P2 u2 + 0 + gz1 = + 2 + gz2 , (4.36) ρ ρ 2where z1 = 0, z2 = h2 . Then we find pressure at point 2 is P2 Pa u2 = − 2 − gh2 , (4.37) ρ ρ 2 u2where 2 2 and gh2 must be positive. It turns out that P2 should be lessthan the atmospheric pressure. If point 2 is high enough to let pressure atpoint 2 less than vapor presure, then gas in fluids will form bubbles. Thesebubbles will move with fluids. If pressure around bubbles increases and ishigher than vapor pressure, then bubbles will burst. The phenomenon is 4.6 Applications of Bernoulli equation · 51 ·
  59. 59. 2 z h2 1 h3 3 Figure 4.8: Schematic of siphon tube.called cavitation. Cavitation is often found in flow fields around a insidepropeller or fluid machinery.4.6.3 Torricelli’s Theorem 1 Pa H Pa 2 Figure 4.9: Torricelli’s theorem. Consider a liquid tank of high H. There is a hole, shown in Fig. 4.9, nearthe ground. Liquids drain from the hole. It is assumed that the tank isquite large, so the location of the free surface is almost still. Hence, u1 = 0.Moreover, pressure at the hole is assumed to be equal to the atmospheric· 52 · INTRODUCATION TO FLUID MOTION II
  60. 60. pressure. Now we can apply Bernoulli equation to point 1 and 2, i.e. P1 u21 P2 u2 + + gz1 = + 2 + gz2 , (4.38) ρ 2 ρ 2where P1 = P2 = Pa , u1 = 0, and (z1 − z2 ) = H. It then becomes Pa Pa u2 + gH = + 2 . (4.39) ρ ρ 2It turns out that u2 = 2gH . (4.40)This is the Torricelli’s Theorem.4.6.4 vena contracta effect dj dh Figure 4.10: Vena contracta effect contraction coefficient Aj (dj )2 Cc = = (4.41) Ah (dh)2 4.6 Applications of Bernoulli equation · 53 ·
  61. 61. 1 h l 2 Figure 4.11: Free jet4.6.5 Free jetsConsider fluids in a tank. A nozzle is arranged at the bottom of thetank. Fluids flow through the nozzle due to the gravitational force andconsequently a jet is observed. Suppose no energy loss in the nozzle.Bernoulli equation can be utilized to determine the jet condition at theexit of the nozzle. The free surface of the tank is assumed to be still ifthe tank is large enough. Therefore, u1 = 0. According to the Bernoulliequation, the total energy along a streamline from the free surface to theexit should be the same, i.e. P1 u21 P2 u2 + + gz1 = + 2 + gz2 = constant . (4.42) ρ 2 ρ 2We know u1 = 0, P1 = P2 = Pa and (z1 − z2 ) = h + l. The equationbecomes u2 2 = g(h + l) (4.43) 2· 54 · INTRODUCATION TO FLUID MOTION II
  62. 62. or u2 = 2g(h + l) . (4.44)The result is as same as Torricelli’s Theorem. However, if the nozzle is not designed well, then there will be energyloss at the nozzle. As a result, Bernoulli equation has to be modified.4.6.6 Venturi tube A B Figure 4.12: Venturi tube. 4.6 Applications of Bernoulli equation · 55 ·
  63. 63. u A AA = u B AB AA uB = uA AB AA AB uA uB 2 PA uA PB u2 + = + B ρ 2 ρ 2 PA − PB uB − u2 2 A = ρ 2 AA u2 ( AB ) − u2 A A = 2 2 2 ua AA = −1 (4.45) 2 ABA Venturi tube is a device made up of a contraction followed by a divergingsection. Fluids moving toward the contraction are speeded up accordingto the continuity equation. In addition, pressure decreases as velocityincreases in terms of the Bernoulli equation. A famous application of aVentui tube is a carburetor. A carburetor is shown in Fig. 4.13. Fuel issucked into the throat due to the low pressure at the throat. Subsequently,fuel is mixed with air at the throat. Venturi tube is a facility to measurethe flow rate in a pipe. Fluids flow a contraction part and then a expansionpart in a Venturi tube.· 56 · INTRODUCATION TO FLUID MOTION II
  64. 64. Q(Air) Butterfly Valve Throat of Venturi FUEL Air-Fuel Mixture Q Figure 4.13: Schematic of caburetor.4.6.7 Flowrate pass through a sluice gateForm the Bernoulli equation, P1 u21 P2 u2 + + z1 = + 2 + z2 r 2g r 2g P1 = P2 = Pa u2 1 u2 + z1 = 2 + z2 (4.46) 2g 2gForm mass conservation u1z1 = u2 z2 z2 u1 = u2 . (4.47) z1 4.6 Applications of Bernoulli equation · 57 ·
  65. 65. Substituting into Bernoulli equation 2 u2 2 z2 u2 + z1 = 2 + z2 2g z1 2g 2 u2 2 z2 −1 = z2 − z1 2g z1 z2 − z1 u2 = 2g z2 . (4.48) ( z1 )2 − 1The flowrate pass through the sluice gate must be Q = u2 · z2 2g(z2 − z1 ) = z2 z2 . (4.49) ( z1 )2 − 1· 58 · INTRODUCATION TO FLUID MOTION II
  66. 66. Chapter 5EQUATIONS OF MOTION ININTEGRAL FORMWe consider one-dimensional flows in Chapter 3 and 4. Conservation lawsof mass, momentum and energy are obtained for one-dimensional flows.Most of fluid flows, however, cannot be simplified as one-dimensional flows.Therefore, we have to look into conservation laws again and derive gov-erning equations for general fluid flows. These equations for fluid flows can be either in integral form or in differ-ential form. Equations in integral form are derived in terms of the controlvolume concept. Equations in integral form do not give any informationthroughout a flow field, but they can provide resultant forces acting on acontrol volume. On the other hand, equations in differential form providedetails regarding variations in a flow field, so we can get values of physicalvariables throughout a flow field. In this chapter, we consider governing equation of fluid flows in integralform first. 59
  67. 67. 5.1 FluxWe mentioned the control volume concept in Chapter 3. A control volumeis bounded artificially in a flow field. Physical properties in a controlvolume may vary in space or in time, because fluids with various physicalproperties flow in and out a control volume and it causes variations ofphysical properties in a control volume. The amount of a physical propertycross an unit surface per second is called flux. A flux can be revealed as b(u · A), where b is a physical property perunit volume, u is the velocity over the area and A is the area vector. Wemay use nA instead of A and n is the unit vector in the normal directionof the area. Physical properties considered in this chapter can be mass,momentum or energy, so we have different fluxes: mass flux : ρ(u · n)A (5.1) momentum flux : ρu(u · n)A (5.2) energy flux : e(u · n)A (5.3) e : energy contained in a unit volume, (5.4) i.e., specific energy (5.5) It should be noticed that n is positive in the outward direction of thearea but negative in the inward direction.· 60 · EQUATIONS OF MOTION IN INTEGRAL FORM
  68. 68. 5.2 Reynolds’ Transport Theorem 2 1 III II I Figure 5.1: Flow through a control volume. We consider a control volume I+II in a flow field. Fluids contained inthe control volume at t = t will flow, so the control volume containingsame fluids at t = t+δt will be II+III. The rate of change of a physicalproperty in the control volume can be shown in DtD c.v. ραdV where αis the amount of the physical property per unit mass. In terms of Fig. 5.1,we know the rate of change in the control volume can be divided into twoparts. The first is the local chang at the region II, which can be shown ∂in ∂t II ραdV . The second is the net flux including the flux from theregion I to the region II and the flux from the region II to the region III,so we have c.s.1 ρα(u · n)dA and c.s.2 ρα(u · n)dA. We can combine 5.2 Reynolds’ Transport Theorem · 61 ·
  69. 69. fluxes across two surfaces and get c.s. ρα(u · n)dA. As δt → 0, we willhave D ∂ ραdV = ρα(u · n)dA + ραdV (5.6) Dt c.v. c.s. ∂t c.v.At t = t0 Bsys = BI (t) + BII (t) . (5.7)At t = t0 + ∆t Bsys = BII (t + δt) + BIII (t + δt) (5.8) ∆Bsys DBsys lim = (5.9) ∆t→0 ∆t Dtor ∆Bsys BII (t + δt) + BIII (t + δt) − BII (t) − BI (t) = (5.10) ∆t ∆t BII (t + δt) − BII (t) ∂BII lim = (5.11) ∆t→0 ∆t ∂t −BI (t) ∆t is the flux flow through in C.S.1 and is denote as ρα(u · dA) (5.12) C.S.1 BIII (t+δt)In addition, ∆t is the flux flow out C.S.2 and is denoted as ρα(u · dA) (5.13) C.S.2 ρα(u · dA) + ρα(u · dA) = ρα(u · dA) (5.14) C.S.1 C.S.2 C.S.Besides, lim (C.V.I +C.V.II ) = lim (C.V.III +C.V.II ) = C.V.II = C.V. = ραdV∆t→0 ∆t→0 C.V. (5.15)· 62 · EQUATIONS OF MOTION IN INTEGRAL FORM

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