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- 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. 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. · II ·
- 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. 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. 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. 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. 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. 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. • 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. 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. 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. 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. 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. 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. ¢¢ 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. 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. 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. 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. Inviscid Fluid Flows(Potential Flows) Viscous Fluid FlowsLaminar Flows Turbulent FlowsInternal Flows External Flows 1.11 Categories of Fluid Dynamics · 13 ·
- 21. · 14 · INTRODUCTION
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. · 40 · INTRODUCTION TO FLUID MOTION I
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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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