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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
- 70. As a result DBsys D = ραdV (5.16) Dt Dt c.v.and ∂Bc.v.II ∂ = ραdV (5.17) ∂t ∂t c.v.5.3 Continuity EquationIf we consider mass variation in a control volume, then we will have α = 1.In terms of Reynold’s transport theorem, the conservation of mass can berevealed as D ∂ ρdV = ρ(u · n)dA + ρdV = 0 (5.18) Dt c.v. c.s. ∂t c.v.This is the continuity equation in integral form.5.4 Momentum EquationSubsequntly, we consider momentum in a control volume, then α will beu. The momentum equation in integral form then is denoted as D ∂ ρudV = ρu(u · n)dA + ρudV . (5.19) Dt c.v. c.s. ∂t c.v.Moreover, the rate of momentum is equal to the resultant force acting onthe control volume, i.e. D ρudV = F = Fbody + Fsurface + Fext . (5.20) Dt c.v.If we consider gravity in body force, then we will have Fbody = ρgdV . (5.21) c.v. 5.3 Continuity Equation · 63 ·
- 71. The surface can be divided into pressure and shear stress, i.e. Fsurface = (−p + τij )ndA . (5.22) c.s.Hence, D ρudV = ρgdV + (−p + τij )ndA + Fext . Dt c.v. c.v. c.s. (5.23)5.5 Moment-of-Momentum EquationNow we consider moment-of-momentum, then α will r×u. Using Reynold’stransport theorem, we obtainD ∂ ρr×udV = ρ(r×u)(u·n)dA+ ρ(r×u)dVDt c.v. c.s. ∂t c.v. (5.24)or ∂Σ(r × F)c.v. = ρ(r × u)(u · n)dA + ρ(r × u)dV = Tshaft , c.s. ∂t c.v. (5.25)where Tshaft is the resultant torque applied to ﬂuids in the control volume.· 64 · EQUATIONS OF MOTION IN INTEGRAL FORM
- 72. Chapter 6DIFFERENTIAL EQUATIONS OFMOTIONSWe obtain governing equations of ﬂuid ﬂows in integral form in Chapter 5.As mentioned before, governing equations in integral form cannot providedetails throughout a control volume. If we would like to know more about aﬂow ﬁeld, such as velocity, pressure and so on, then governing equations ofﬂuid ﬂows in diﬀerential form are necessary. Solving diﬀerential governingequation can get the whole information of a ﬂow ﬁeld.6.1 Lagrangian and Eulerian systemsThese systems are used very often in dynamics. An observer follows aspeciﬁed particle if Lagrangian system is employed. In other words, acoordinate system is ﬁxed at a particular ﬂuid particle when Lagrangiansystem is utilized to describe a ﬂow ﬁeld. In addition, physical variablesdescribed by Lagrangian system are functions of time only. All of spatialcoordinates are functions of time too (x = ut). We can say α = α(t). dα ∂α ∂α ∂α ∂α = +u +v +w (6.1) dt ∂t ∂x ∂y ∂z 65
- 73. The term at the left hand side of the equation is called the total derivativeor material derivative. It means the rate of change of α in a ﬂuid particle,i.e., Lagrangian point of view. It is often to use Dα instead of dα to Dt dtindicate a material derivative, i.e. Dα ∂α ∂α ∂α ∂α = +u +v +w . (6.2) Dt ∂t ∂x ∂y ∂zThe trems at the right hand side of the equation are, in fact, describedby Eulerian system. The ﬁrst term is called a local derivative or unsteadyterm. The rest are called convective terms because they are caused by ﬂowmotions. This equation can shown in vector form, i.e. Dα ∂α = + (u · ∇)α . (6.3) Dt ∂tIt should be noted that u and ∇ cannot be communtive. In other words, (u · ∇) = (∇ · u) . (6.4)Eulerian system is an alternative way to describe a ﬂow ﬁeld. An observeris ﬁxed at a selected point of space if Eulerian system is adopted. Hence,the coordinate system does not move as ﬂuids ﬂow. In addition, spatialcoordinates are required to indicate the point which is observed. Hence,physical variables described by Eulerian system are functions of time andspace, i.e. α = (x, t) (6.5)· 66 · DIFFERENTIAL EQUATIONS OF MOTIONS
- 74. 6.2 Rate of Change Following a Fluid Particleα is a continuous function of space and time. Its inﬁnite samll change canbe described using chain rule of diﬀerentiation, i.e. ∂α ∂α ∂α ∂α dα = dt + dx + dy + dz , (6.6) ∂t ∂x ∂y ∂zwhere Cartesian coordinate system is used. Moreover, the equation isdivided by dt and we get dα ∂α ∂α dx ∂α dy ∂α dz = + + + . (6.7) dt ∂t ∂x dt ∂y dt ∂z dtIn Chapter 5, the change rate of a physical variable in a control volumecan be shown in D ∂ ραdV = ρα(u · n)dA + ραdV (6.8) Dt c.v. c.s. ∂t c.v.The ﬂux across the control surface can be convered to a term in volumeintrgral using Gauss’ theorem, i.e. ρα(u · n)dA = ∇ · (ραu)dV (6.9) c.s. c.v.Subsequently, we can rewrite the control volume formula and obtain D ∂ ραdV = ∇ · (ραu)dV + ραdV (6.10) Dt c.v. c.v. ∂t c.v.If the volume does not change, then it will be D ∂(ρα) ραdV = + ∇ · (ραu) dV (6.11) Dt c.v. c.v. ∂t6.3 Continuity EquationMass conservation is obeyed as ﬂuid particles move. Then it can shown as Dm D = ρdV = 0 . (6.12) Dt Dt c.v. 6.2 Rate of Change Following a Fluid Particle · 67 ·
- 75. It means that mass of a ﬂuid particle does not change with time. Thisis the continuity equation in integral form. The Eulerian system can beapplied to describe the continuity equation, i.e. D ∂ρ ρdV = + ∇ · (ρu) dV = 0 (6.13) Dt c.v. c.v. ∂tIn addition, ∇ · (ρu) = ρ(∇ · u) + (u · ∇)ρ . (6.14)When dV does not change with time, the continuity equation becomes ∂ρ Dρ + ρ(∇ · u) + (u · ∇)ρ = + ρ(∇ · u) = 0 . (6.15) ∂t DtFor an incompressible ﬂuid, Dρ = 0. The continuity equation becomes Dt ∇·u= 0 . (6.16)In a Cartesian coordinate system, it becomes ∂u ∂v ∂w + + =0 . (6.17) ∂x ∂y ∂zIn a cylindrical coordinate system, it becomes ∂ur ur 1 ∂uθ ∂uz + + + =0 . (6.18) ∂r r r ∂θ ∂z6.4 Momentum EquationConsider momentum conservation in ﬂuid particles. It can be shown as D D (mu) = ρudV = F = f dV . (6.19) Dt Dt c.v. c.v.or D (ρu) = f , (6.20) Dt· 68 · DIFFERENTIAL EQUATIONS OF MOTIONS
- 76. where f is the resultant force per unit volume acting on the ﬂuid particle.When dV does not change with time. The total derivative can be revealedas D ∂ (ρu) = (ρu) + (u · ∇)ρu, (6.21) Dt ∂tThere are two kinds of forces, body force and surface force, so it becomes f = fbody + fsurface (6.22)Body force is caused by gravity or electromagnetic forces. Surface forcesare caused by pressure and shear stress. If viscosity of ﬂuids is ignored,then shear stress can be removed from the equation. Now, we have the momentum equation D ∂ (ρu) = (ρu) + (u · ∇)ρu = fbody + fsurface (6.23) Dt ∂tThis is the so-called Navier-Stokes equation. If only gravity appears in thebody force, then fbody = ρg . (6.24)Surface force can be shown in fsurface = −∇p + µ∇2 u . (6.25)Substituting these terms into the N-S equation, we obtain D ∂ (ρu) = (ρu) + (u · ∇)(ρu) = ρg − ∇p + µ∇2 u . (6.26) Dt ∂tIf the viscosity of ﬂuids is ignored, then it will become D (ρu) = ρg − ∇p (6.27) Dtwhich is called the Euler equation. 6.4 Momentum Equation · 69 ·
- 77. 6.5 Boundary ConditionsThe continuity equation and the N-S equation constitute an initial-boundaryvalue problem. Hence approiate boundary and initial conditions are re-quired to get a particular solution. Velocity and pressure are main variablesin the continuity euation and N-S equation. They have to be prescribedbefore solving these equations. It is, however, not easy to obtain condi-tions for pressure. For velocity, non-slip boundary condition is imposed inthe solid boundary, i.e. u=0. (6.28)If viscosity of ﬂuids is ignored, then the non-slip condition cannot be used.Therefore impereable condition is utilized, i.e. u·n=0, (6.29)which means no ﬂuids pass through a solid boundary.· 70 · DIFFERENTIAL EQUATIONS OF MOTIONS
- 78. Chapter 7DIMENSIONAL ANALYSIS7.1 Why dimension analysis? 1. Some important dimensionless parameters are obtained as a dimen- sional analysis is conducted in a ﬂuid mechanic problem. In terms of those dimensionless parameters, we can understand features of a ﬂow problem. 2. Governing equations based on physical laws for ﬂow problems are re- vealed in nondimensional form. This avoids eﬀects of system unit in a ﬂuid ﬂow problem.7.2 Fundamental dimensions 1. MLT: As we consider physical variables, three fundamental dimensions are often involved. They are mass(M), length(L), and time(T). 2. FLT: FLT is an alternative view to adopt fundamental dimensions. They are force(F), Length(L), and time(T). In fact, F = MLT −2 . (7.1) 71
- 79. MLT is more often used than FLT, so it will be considered in the followingdimensional analysis.Examples:velocity: LT −1acceleration: LT −2force: MLT −2density: ML−3volume: L3pressure: ML−1 T −2power: ML2 T −3work: ML2 T −2dynamic viscosity: ML−1T −1kinematic viscosity: L2T −17.3 How to carry out a dimensional analysis? 1. Find out all of physical variables in a ﬂuid ﬂow problem as possible as you can. It depends on your experiences and your fundamental knowledge in ﬂuid dynamics. Assume we have n variables in a ﬂuid ﬂow problem. Find out their dimensions. 2. Check out how many fundamental dimensions are involved in these variables. In most of time, three fundamental dimensions are all in- volved. Assume ℓ fundamental dimensions are involved. 3. The relationships between these physical variables are not obtained.· 72 · DIMENSIONAL ANALYSIS
- 80. Pick up one of them as the dependent variable. For example: A1 = f (A2, · · · , An ) , (7.2) where A1, · · · , An are chosen independent variables and A1 is the de- pendent variable.4. Now we utilize dimensional analysis to non-dimensionalized physical variables. π theory is used to reach the goal. For n physical variables and ℓ fundamental dimensions, (n − ℓ) π products will be obtained. Those π products are dimensionless.5. To ﬁnd out π products, ℓ physical variables have to be chosen ﬁrst. The main principle to choose these physical variables depends on their dimensions. Basically physical variables with less dimensions are cho- sen. The dependent variable cannot be chosen as one of them. For example, we choose A2 , · · · , A2+ℓ−1.6. Now the ﬁrst π1 product will be π1 = A1 (AaAb Ac ) 2 3 4 (7.3) if ℓ = 3. To make π1 dimensionless, check its dimensions in turn, i.e. A1 = M y1 L y2 T y3 (7.4) A2 = M y4 L y5 T y6 (7.5) A3 = M y7 L y8 T y9 (7.6) A4 = M y10 Ly11 T y12 (7.7) M : y1 + ay4 + by7 + cy10 = 0 (7.8) 7.3 How to carry out a dimensional analysis? · 73 ·
- 81. L : y2 + ay5 + by8 + cy11 = 0 (7.9) T : y3 + ay6 + by9 + cy12 = 0 , (7.10) where y1 , ..., y12 are known. a, b, and c are obtained by solving the simultaneous equations. 7. Each π product can be found in turn using the step 6 where the rest of physical variables are used to replace A1 . At last, n − ℓ π products are obtained and Eq. (7.2) becomes π1 = F (π2 , · · · , πn−ℓ) . (7.11)It is the result of a dimensional analysis.Ex: Drag on a sphere u O D Figure 7.1: Flow past a sphere. 1. geometric variable: D(diameter of the sphere), material variables: ρ(density), µ(viscosity),· 74 · DIMENSIONAL ANALYSIS
- 82. kinematic variable: u(velocity), dynamic variable: FD (drag on the sphere)2. D : L ρ : ML−3 µ : ML−1 T −1 u : LT −1 FD : MLT −2 Three fundamental dimensions are involved in physical variables.3. Since the drag on the sphere is the main variable which we would like to know, we choose FD as the dependent variable, i.e. FD = f (D, ρ, u, µ) (7.12)4. 5-3=2. It means that 2 nondimensional parameters should be found in the analysis.5. Choose D, u, and ρ, to nondimensionalize other variables.6. π1 = FD Da ubρc M :1+0+0+c= 0 L : 1 + a + b − 3c = 0 (7.13) T : −2 + 0 − b + 0 = 0 a = −2 → b = −2 (7.14) c = −1 FD ∴ π1 = FD D−2u−2ρ−1 = (7.15) ρD2 u2 7.3 How to carry out a dimensional analysis? · 75 ·
- 83. 7. π2 = µDa ubρc M :1+0+0+c= 0 L : −1 + a + b − 3c = 0 (7.16) T : −1 + 0 − b + 0 = 0 a = −1 → b = −1 (7.17) c = −1 µ ∴ π2 = µD−1 u−1ρ−1 = (7.18) ρuD 8. As a result, FD µ = F( ) (7.19) ρu2D2 ρuD FD ρuD → = F( ) (7.20) ρu2 D2 µ FD π CD = 1 2 : drag coeﬃcient, A = D2 (7.21) 2 ρu A 4 ρuD Re = : Reynolds number (7.22) µEx: Pipe ﬂows 1. geometric variable: D(diameter of a pipe), L(length),ǫ(ruoghness) material variables: ρ(density), µ kinematic variable: uavg (average velocity) dynamic variable: ∆P (pressure drop)· 76 · DIMENSIONAL ANALYSIS
- 84. ua D Figure 7.2: Flow in a straight pipe.2. D:L L:L ǫ:L ρ : ML−3 µ : ML−1T −1 uavg : LT −1 ∆P : ML−1T −2 Three fundamental dimensions are involved in physical variables.3. Choose ∆P as the dependent variable and then ∆P = f (D, L, ǫ, ρ, µ, uavg ) . (7.23)4. Choose D, ρ, uavg to nondimensionalize other variables. There will be 7.3 How to carry out a dimensional analysis? · 77 ·
- 85. 7-3=4 π products. 5. π1 = ∆P Da ρb uc avg M :1+0+b+0=0 L : −1 + a − 3b + c = 0 (7.24) T : −2 + 0 + 0 − c = 0 a=0 → b = −1 (7.25) c = −2 ∆P ∴ π1 = ∆P ρ−1 u−2 = avg (7.26) ρu2 avg 6. π2 = LDa ρb uc avg M :0+0+b+c=0 L : 1 + a − 3b − c = 0 (7.27) T :0+0+0−c=0 a = −1 → b=0 (7.28) c=0 L ∴ π2 = (7.29) D 7. π3 = ǫDa ρb uc avg ǫ π3 = (7.30) D 8. π4 = µDa ρb uc avg µ π4 = (7.31) ρuavg D· 78 · DIMENSIONAL ANALYSIS
- 86. 9. ∆P L ǫ µ L ǫ ρuavg D = F( , , ) = F( , , ) (7.32) ρu2 avg D D ρuavg D D D µ ∆P Cp = 1 , (pressure coeﬃcient) (7.33) 2 ρu2 avg ρuavg D Re = , (Reynolds number) (7.34) µ7.4 Common nondimensional parameters Names of Parameters Formula Physical meanings Reynolds number(Re) ρuL inertia force µ viscous force Froude number(Fr) √u inertia force gL gravitational force P pressure force Euler number(Eu) 1 2 ρu 2 inertia force Mach number(Ma) u inertia force C compressibility force fL Strouhal number(St) u 1 Weber number(We) 2 2 ρu L inertia force σ surface tension force7.5 Nondimensional EquationsEx: P 1 + u2 + gh = const (7.35) ρ 2 characteristic length: L (7.36) characteristic velocity: u0 (7.37) P 1 u2 gh → 2+ + = const (7.38) ρu0 2 u2 u2 0 0 P u2 gh → 1 2 + 2 + 1 2 = const (7.39) 2 ρu0 u0 2 u0 7.4 Common nondimensional parameters · 79 ·
- 87. Ex: ∂u ∂v + =0 (7.40) ∂x ∂y L ∂u L ∂v → + (7.41) u0 ∂x u0 ∂y ∂u∗ ∂v ∗ → + =0 (7.42) ∂x∗ ∂y ∗ u v x y u∗ = , v ∗ = , x∗ = , y ∗ = (7.43) u0 u0 L L7.6 Scale model testsTo establish similitude between a protype system and a model system,geometric, kinematic, and dynamic similarities have to be considered. 1. Geometric Similarity Lm = const , (7.44) Lp where subscripts m and p refer to model and protype, respectively. 2. Kinematic Similarity u1m u2m u3m = = = const (7.45) u1p u2p u3p 3. Dynamic Similarity - Dynamic parameters between a model and a protype must be equal as possible. For example, it viscous force is im- portant in the ﬂow ﬁeld, then Reynolds number will be the important dynamic parameter and Rem = Rep (7.46) and um Lm up Lp = νm νp L p νm um = up . (7.47) L m νp· 80 · DIMENSIONAL ANALYSIS
- 88. Hence, the adjustment of characteristic velocity in the model ﬂow relies on the geometric similarity. If the model is tiny, then the characteristic velocity in the model ﬂow will become very fast. Sometimes, it is impossible to generate very fast characteristic velocity in the model ﬂow. As a result, one has to enlarge the model size. When geometric, kinematic, and dynamic similarities are reached be- tween a model and a protype, a scaled model can be conducted and summarized using the result of a dimensional analysis. Consequently, relationships among various π products can be obtained. In addition, the variation of the dependent valiable for the protype can be pre- dicted. For example, in the previous example concerning a uniform ﬂow past a sphere, we have CD = F (Re) . (7.48) When Rem is equal to Rep , the dynamic similarity is reached. Subse- quently, the drag exerted on the sphere can be determined using the drag coeﬁcient, i.e. 1 (FD )p = ρp upAp CD 2 1 = ρp upAp F (Re) 2 1 ρp up Dp = ρp upAp F ( ) , (7.49) 2 µp where F (Re) is obtained from the scale model test. Distorted model: It is not always possible that all dynamic parametersbetween a model and a protype are qual to others. For example, one cannotpromise dynamic similarities of Reynolds number and Frounde number at 7.6 Scale model tests · 81 ·
- 89. the same time, becausefor Reynolds number L p νm um = up (7.50) L m νpand for Froude number Lm um = up . (7.51) LpOne has to chose one of them to perform dynamic similarity and ignore theother one. When not all dynamic parameters are equal between a modeland a protype, it is the so-called distorted model.· 82 · DIMENSIONAL ANALYSIS
- 90. Chapter 8Viscous Internal FlowWe have investigated governing equations in diﬀerential forms for ﬂuidﬂow problems in Chapter 6. In addition, speciﬁc ﬂuid ﬂow problems suchas pipe ﬂows, ﬂows past obstacles and etc. will be studied. First of all,viscous internal ﬂows are discussed in this chapter. Internal ﬂuid ﬂowsrefer to ﬂows which are bounded by solid walls. For example, a pipe ﬂowis a typical internal ﬂow. It is bounded by pipe walls.8.1 Fully developed ﬂowConsider ﬂuids ﬂow into a pipe from a tank shown in Fig. 8.1. The ﬂow isuniform (U0) at the entrance of the pipe. The uniform veloctiy proﬁle doesnot retain as soon as ﬂuids enter the pipe. Owing to viscosity, ﬂuid arestill on the pipenwall and then velocity of ﬂuids increases along the radialdirection. The viscous eﬀect gradually aﬀects velocity of ﬂuids as ﬂuidsmove downstream. In the region near the entrance, the viscous eﬀect isnot full of the pipe but appears near the pipe wall. The velocity in theaﬀected region is slower than the unaﬀected region. The aﬀected region iscalled a boundary layer. Mathematically, u = 0.99U0 is the artiﬁcial edge 83
- 91. Boundary Layer δ U0 entrance u=0.99U0 fully developed region Figure 8.1: Schematic of development of pipe ﬂow from the inlet.of a boundary layer. The viscous eﬀect disperses from the pipe wall to thecentre as ﬂuids move downstream. Finally, the viscous eﬀect is full of thepipe and the ﬂow in the region is called a fully developed ﬂow.8.2 Laminar, transition and turbulent ﬂowWhen ﬂuid ﬂows in a pipe, various patterns are found according to diﬀerentphysical parameters, such as velocity, viscosity and pipe diameter. If wedye a point in a pipe ﬂow, then we will ﬁnd that the streakline from thepoint may be a straight line or a distorted line. If it is a straight line,it means all ﬂuid particles move along the same straight line as ﬂuidstravel downstream. This is the so-called laminar ﬂow. If it is a distortedline, it means ﬂuid particles do mot move along the same line as they· 84 · Viscous Internal Flow
- 92. Figure 8.2: Experimental appratus of Reynolds’ pipe ﬂow.Source: http : //www.eng.man.ac.uk/historic/reynolds/oreynB.htmtravel downstream but are disturbed. This is the so-called turbulent ﬂow.In 1833 A.D., O. Reynolds explained physical phenomena in pipe ﬂowsusing pipes of various diameters and a control value (see Fig. 8.2 and8.2). He controled the ﬂow rate across a pipe using the vale and dyed theﬂow to visualize the pipe ﬂow. He had same conclusions as the presviousdescription.8.3 2-D Poiseuille ﬂowConsider a fully developed laminar ﬂow between two inﬁnite parallel plates.To analyze the laminar ﬂow, assumptions are made to simplify the wholeproblems. They are 8.3 2-D Poiseuille ﬂow · 85 ·
- 93. Figure 8.3: Flow patterns of laminar, transition, and turbulent pipe ﬂows.Source: http : //www.eng.man.ac.uk/historic/reynolds/oreynB.htm1. 2-D2. steady ﬂow3. incompressible4. v = 05. ignore gravityThe analytical solution can be obtained by solving the continuity equationand the Navier-Stokes equations which are revealed as ∂u ∂v + =0 , (8.1) ∂x ∂y ∂u ∂u ∂u 1 ∂P ∂ 2u ∂ 2u +u +v =− +ν + (8.2) ∂t ∂x ∂y ρ ∂x ∂x2 ∂y 2and ∂v ∂v ∂v 1 ∂P ∂ 2v ∂ 2v +u +v =− +ν + . (8.3) ∂t ∂x ∂y ρ ∂y ∂x2 ∂y 2In additoin, he found that a very important nondimensional parameter,ρuD µ , which is highly relevant to the ﬂow patterns. It is the well-knownReynolds number. Pipe ﬂows are classiﬁed as three kinds of patternsaccording to Reynolds unmber, i.e.,· 86 · Viscous Internal Flow
- 94. y D x C L Figure 8.4: 2-D Poiseuille ﬂow.laminar ﬂow: Re 2,300transition ﬂow: 2,300 Re 4,000turbulent ﬂow: Re 4,000.In laminar ﬂow regime, the pipe ﬂow can be examined using analyticalmethods. The laminar ﬂow solution is called Poiseuille ﬂow which is namedafter J. L. M. Poiseuille. Transition and turbulent ﬂows, however, cannotbe studied using analytical methods. Therefore, most of konwledge oftransition and turbulent ﬂows come from experimental data. First of all, we look into the continuity equation and get ∂u ∂v + =0 (8.4) ∂x ∂yand u = u(y) . (8.5)The continuity equation is satisiﬁed as u is a function of y. Subsequently, 8.3 2-D Poiseuille ﬂow · 87 ·
- 95. the N-S equation in the y-direction is simpliﬁed as ∂v ∂v ∂v 1 ∂P ∂ 2v ∂ 2v +u +v =− +ν + . (8.6) ∂t ∂x ∂y ρ ∂y ∂x2 ∂y 2 (8.7) ∂vSince we assume that the ﬂow is in a steady state, so ∂t = 0, v = 0 andwe get ∂P =0 (8.8) ∂ywhich leads to P = P (x) . (8.9) Therefore, the N-S equation is satiﬁed as pressure is a function of x.Furthermore, the N-S equation in the x-direction is reconsidered as ∂u ∂u ∂u 1 ∂P ∂ 2u ∂ 2 u +u +v =− +ν + . (8.10) ∂t ∂x ∂y ρ ∂x ∂x2 ∂y 2 ∂uAgain, we assume that the ﬂow is in a steady state, so ∂t = 0 , u = u(y) ,v = 0 and we get ∂ 2u 1 dP = . (8.11) ∂y 2 µ dxIn addition, ∂u 1 dP dP = y + C1 , = const. (8.12) ∂y µ dx dxand consequently 1 dP y 2 u= + C1 y + C2 . (8.13) µ dx 2Appropriate boundary conditions are required to obtain arbitary con-stants, C1 and C2 , in the general solution. Since ﬂuids are viscous, non-slipboundary condition can be imposed on solid walls, i.e. D u y= =0 (8.14) 2· 88 · Viscous Internal Flow
- 96. and D u y=− =0 . (8.15) 2Using non-slip conditions, we ﬁnd D 1 dP D2 C1D u y= = + + C2 = 0 (8.16) 2 µ dx 8 2and D 1 dP D2 C1D u y=− = − + C2 = 0 , (8.17) 2 µ dx 8 2where C2 is found by adding these equations and we get d dP D2 + 2C2 = 0 (8.18) µ dx 8 1 dP D2 −→ C2 = − . (8.19) µ dx 8Substitution of C2 into Eq. (8.13) gives 1 dP D2 C1 D 1 dP D2 + − =0 (8.20) µ dx 8 2 µ dx 8which leads to C1 = 0 . (8.21)The solution for velocity in laminar ﬂow between two inﬁnite parallel platesis shown as 1 dP y 2 1 dP D2 u(y) = − µ dx 2 µ dx 8 1 dP 1 2 D2 = · y − µ dx 2 4 2 1 dP D2 2y = (− ) · − +1 2µ dx 4 D 2 D2 dP 2y = (− ) 1 − . (8.22) 8µ dx D 8.3 2-D Poiseuille ﬂow · 89 ·
- 97. It is obvious that velocity in a fully developed laminar ﬂow betweentwo inﬁnite parallel plates is parabolic. The maximum values of velocityproﬁle can be obtained by its ﬁrst derivative, i.e. du D2 dP 2 = (− ) −2 · y = 0 . (8.23) dy 8µ dx DIt happens at y = 0 and the maximum value is D2 dP umax = − . (8.24) 8µ dxMoreover, the ﬂow rate per unit width in a cross area can be obtained byintegrating velocity, i.e. D 2 Q = 2 udy 0 D D2 dP 4 y3 2 = 2 − y− 2 8µ dx D 3 0 D2 dP D 4 1 D3 = 2 − − 2 · 8µ dx 2 D 3 8 2 D dP D D = − − 4µ dx 2 6 2 D dP D = − 4µ dx 3 D3 dP = − . (8.25) 12µ dx The averaged velocity is then found by Q uavg = A 1 D3 dP = · − D 12µ dx D2 dP = − . (8.26) 12µ dxWe ﬁnd 2 uavg = umax . (8.27) 3· 90 · Viscous Internal Flow
- 98. In addition, 2 u 3 2y = 1− . (8.28) uavg 2 DViscous stress on the wall can be found by du τw = µ dy y=± D 2 D2 dP 4 = − − 2y 8µ dx D2 y=± D 2 D2 dP d = µ − − ·D 8µ dx D2 D dP = − 2 dx 6µuavg = . (8.29) DThe friction coeﬃcient Cf will be 6 τw µ D uavg 12µ 12 Cf = 1 = 1 2 = = . (8.30) 2 ρ(uavg )2 2 ρuavg ρDuavg ReThe friction factor will be − dP D f = 1 dx2 2 ρuavg (− dP )D = dx 1 ρ · uavg · 12µ (− dP ) D2 2 dx 24 = . (8.31) Re8.4 Hagen-Poiseuille ﬂowPoiseuille utilized analytical methods to get the solution of a laminar pipeﬂow. He solved the continuity equation and N-S equations based on thefollowing assumptions: 8.4 Hagen-Poiseuille ﬂow · 91 ·
- 99. 1. steady 2. incompressible ∂ 3. ∂θ =0 4. ur , uθ =0 5. ignor the gravitational acceleration.The governing equation for a pipe ﬂow can be revealed as: ∂ur ur 1 ∂uθ ∂uz + + + =0 (8.32) ∂r r r ∂θ ∂z ∂uz = 0, uz = uz (r) ∂zThe 2-D N-S equations in cylindrical coordinate system are denoted asr-direction ∂ur ∂ur uθ ∂ur ∂ur u2 + ur + + uz − θ ∂t ∂r r ∂θ ∂z r 1 ∂P 1 ∂ ∂ur 1 ∂ 2 ur ∂ 2 ur ur 2 ∂u0 =− + ν r + 2 2 + − 2− 2 ρ ∂r r ∂r ∂r r ∂θ ∂z 2 r r ∂θ (8.33) ∂ = 0, ur = 0, uθ = 0 ∂t 1 ∂P − = 0 (8.34) ρ ∂r P = P (z)only (8.35)z-direction ∂uz ∂uz uθ ∂uz ∂uz + ur + + uz ∂t ∂r r ∂θ ∂z 1 ∂P 1∂ ∂uz 1 ∂ 2 uz ∂ 2 uz =− + ν r + 2 2 + (8.36) ρ ∂z r ∂r ∂r r ∂θ ∂z 2· 92 · Viscous Internal Flow
- 100. ∂ = 0, ur = 0, uθ = 0, uz = uz (r) ∂t 1∂ ∂uz 1 dP r = (8.37) r ∂r ∂r µ dz ∂ ∂uz 1 dP r = − r (8.38) ∂r ∂r µ dz ∂uz 1 dP r2 r = − + C1 (8.39) ∂r µ dz 2 ∂uz 1 dP r C1 = − + (8.40) ∂r µ dz 2 r 2 1 dP r uz = − + C1 ln r + C2 (8.41) µ dz 4 (8.42)B.C. D uz (r = ) = 0 2 1 dP D2 D 0 = − + C1 ln + C2 (8.43) µ dz 16 2C1 must be 0, because r cannot be 0 in ln r. As a result, 1 dP D2 ∴ C2 = . (8.44) µ dz 16Substitution of C2 into uz gives 2 1 dP r2 1 dP 1 D uz = − + µ dz 4 µ dz 4 2 2 1 dP 1 2 D = − r − µ dz 4 2 2 1 dP 1 D2 2r = − −1 µ dz 4 4 D 2 1 dP D2 2r = − −1 µ dz 16 D 2 1 dP D2 2r = − −1 . (8.45) µ dz 16 D 8.4 Hagen-Poiseuille ﬂow · 93 ·
- 101. Shear stress on the pipe wall is determined by ∂uz τw = −µ ∂r r= D 2 1 dP D2 4 = −µ · − · 2r µ dz 16 D2 dP D = . (8.46) dz 4In addition, duz 1 dP D2 4 r = (− ) · (8.47) dr µ dz 16 D2 2The maximum velocity appears at duz =0 dr −→ r = 0 .It can be determined by 1 dP D2 (uz )max = uz (r = 0) = . (8.48) µ dz 16The volumetric ﬂow rate per unit width is given by r Q = uz 2πrdr 0 D 2 1 dP D2 4 = (− ) 2 · 2πr3 − 2πr dr 0 µ dz 16 D D 1 dP D2 8π r4 r2 2 = (− ) − 2π µ dz 16 D2 4 2 0 1 dP D2 8π r4 D2 = (− ) −π µ dz 16 D2 4 · 16 4 1 dP D2 π 2 π 2 = (− ) D − D µ dz 16 8 4 4 1 dP πD = (− ) . (8.49) µ dz 128· 94 · Viscous Internal Flow
- 102. ua r z Figure 8.5: Parabolic proﬁle of velocity component in the z-direction.The averaged velocity is Q (uz )avg = A 4 1 dP D2 πD2 πD2 = · (− ) − πD2 µ dz 16 8 4 2 1 dP D = (− ) . (8.50) µ dz 32Furthermore, 1 (uz )avg 32 1 = 1 = . (8.51) (uz )max 16 2We consider uz and uavg and get 2 uz 1 2r = −1 . (8.52) (uz )avg 2 DIt is a parabolic proﬁle for the velocity component in the z-direction. Mean-while, 4µ¯ 8µ¯ u u τw = = (8.53) r0 D 8.4 Hagen-Poiseuille ﬂow · 95 ·
- 103. 4τw 64 f= 1 = (8.54) 2 ρ(uavg )2 Re τw 16 Cf = 1 2 = (8.55) 2 ρ(uavg ) Rewhere f is the Darcy fricition factor.8.5 Transition and turbulent pipe ﬂowsTransition and turbulent pipe ﬂows cannot be solved using analyticalmethods due to extremely complicated physical phenomena. The onlyapproach to investigate transition and turbulent pipe ﬂows is to conductexperiments. L.F. Moody conducted pipe ﬂow experiments and obtainedthe well-known Moody diagram. The Moody diagram explains the rela-tionship between Reynolds unmber, friction factor and relative roughness.The Moody diagram can be used in the following steps: f ε D Re Figure 8.6: Schematic of Moody diagram. 1. Evaluate Reynolds number.· 96 · Viscous Internal Flow
- 104. 64 2. If Re 2,300, then the formula f = Re can be used for the laminar ε pipee ﬂow. If it is not, then evaluate relative roughness, D . ε 3. Find the resultant relative riughness D in the right-hand side of the Moody diagram. 4. Follow the line starting from the resultant relative riughness. Find the point in the line at the resultant Reynolds number. Starting from this point, go to the left hand side and ﬁnd out the friction factor, f. 64 The Moody diagram also proves that f = Re is correct in a laminar pipeﬂow. In addition, some dashed lines are found between Re=2,000-5,000. Itis because those lines are in transition pipe ﬂow. The details in transitionpipe ﬂow are still not very clear. In the region at high Reynolds number,i.e. turbulent ﬂows, it is observed that all lines are parallel to each others.It seems that the friction factor is independent of Re in turbulent ﬂow butonly depends on the relative roughness.8.6 Darcy equationThe major loss comes from friction losses caused by pipe walls. Darcyequation explains how to evaluate the major head loss, i.e. L u2 avg hL = f , (8.56) D 2g 64where f must be found using the Moody diagram or the formula f = Refor a laminar ﬂow. The minor loss comes from ﬁttings such as valves,elbows, expansions and so on. The minor loss is evaluated by u2 avg hL = K , (8.57) 2g 8.6 Darcy equation · 97 ·
- 105. where K is called the K-factor and depends on various ﬁttings. The minorloss can be combined with the major loss using the concept of equivalentlength. The equivalent length, le , is deﬁned as le u 2 avg u2avg f = K , (8.58) D 2g 2g K·D le = . (8.59) fThen the sum of major and minor losses will be (L + le) u2 avg hL = f . (8.60) D 2gHence the Bernoulli equation can be modiﬁed as P1 V12 P2 V22 + + z1 = + + z2 + hL . (8.61) r 2g r 2g valve 1 2 Figure 8.7: Concept of equivalent length.· 98 · Viscous Internal Flow
- 106. 8.7 Hydraulic diameterIf the duct considered is not circular, then its hydraulic diameter can beused. A hydraulic diameter is deﬁned as 4 × cross-sectional area DH = . (8.62) perimeter8.8 Brief Introduction to Turbulence u = u +u′ (8.63) 1 t+T u = lim udt (8.64) T →∞ T t u t Figure 8.8: Fluctuations in turbulent ﬂow. 8.7 Hydraulic diameter · 99 ·
- 107. D laminar flow turbulent flow Figure 8.9: Diﬀerence between laminar and turbulent pipe ﬂows.· 100 · Viscous Internal Flow
- 108. Chapter 9Viscous External FlowsExternal ﬂow are not bounded by solid walls. In external ﬂows, interactionof ﬂuids with solid structures is usually considered. In the past, viscosity ofa ﬂuid was not considered in the potential ﬂow theory. The drag predictedby the potential ﬂow theory for symmetrical bodies in a uniform ﬂow is zerobut it is impossible. This is the so-called d’Alembert’s parabox. Hence, itis obvious that viscosity plays a vital role in drag prediction.9.1 Boundary Layer TheoryPrandtl, a German professor, provided the boundary layer concept. Hethought that viscosity aﬀects ﬂuid ﬂows within a very thin region attachedthe solid body. This region aﬀected by viscosity is the well- known bound-ary layer. Furthermore, he provided the non-slip condition to describleﬂuid kinematic condition on solid walls. The boundary layer concept isshown in Fig. 9.1. The artiﬁcial boundary layer starts from the frontof the solid body. Its thickness (u = 0.99Ue at y = δ) grows along thedownstream direction. In the begin, the ﬂow in the boundary layer islaminar. Traveling downstream, the boundary layer ﬂow becomes transi- 101
- 109. ue ue wake ue y x transtion turbulent laminar sepration stagnation boundary du point =0 dy (inflection point) ue u=0.99ue δ Figure 9.1: Schematic of boundary layer concept.tion and then turbulent (can be characterized by local Reynolds unmber, ρue xRex = µ ) due to disturbances from the surface of the body. If the ad-verse pressure gradient happens in the boundary layer, then a separationappears in the boundary layer. Subsequently, a wake is observed behindthe separation point.9.2 Uniform ﬂow past a ﬂat platConsider a uniform ﬂow past over a ﬂat plate. The control volume conceptcan be used to analyze the boundary layer ﬂow. This idea was providedby von K´rm´n in 1921. For the mass conservation, a a δ δ L − Uedy + udy + vdx = 0 (9.1) 0 0 0 δ L − Ue · δ + udy + vdx = 0 . (9.2) 0 0· 102 · Viscous External Flows
- 110. For the momentum conservation, δ L δ Fx = − ρUe Uedy + ρUevdx + ρu2dy (9.3) 0 0 0 L L δ Fx 2 2 = −Ue δ + Ue δ + Uevdx + u2dy (9.4) ρ 0 0 0From the mass conservation. Ue y δ x L Figure 9.2: Schematic of boundary layer due to a uniform ﬂow past a ﬂat plate L δ vdx = Ueδ − udy (9.5) 0 0 9.2 Uniform ﬂow past a ﬂat plat · 103 ·
- 111. Substitution of this equation into momentum equation gives δ δ Fx 2 δ = −Ue δ + Ue δ − Ue udy + u2dy (9.6) ρUe 0 0 δ u2 = Ue − u dy (9.7) 0 Ue δ u = Ue u − 1 dy (9.8) 0 Ue δ Fx u u 2 = − 1 dy 0 (9.9) ρUe 0 Ue Ue = −θ (9.10) D D = −Fx −→ 2 =θ ρUe The frictional force exerted by the boundary layer ﬂow is obtained, butit depends on the velocity proﬁle within the boundary layer. The rest ofquestion is how to determine the velocity proﬁle.9.3 Boundary Layer Thickness, δThe edge of the boundary layer is deﬁned at the line of u = 0.99Ue. u − 0.99Ue, y = δ (9.11)9.4 Displacement Boundary Layer Thickness, δd Figure 9.3: Concept of displacement boundary layer thickness.· 104 · Viscous External Flows
- 112. ∞ ρδd · Ue = ρ(Ue − u)dy (9.12) 0 ∞ u δd = (1 − )dy (9.13) 0 UeIt means that mass ﬂux the within δd is equal to the absence of mass ﬂuxdue to the presence of the boundary layer.9.5 Momentum Boundary Layer Thickness, θ ∞ 2 ρθUe = ρ [(Ue − u) u] dy (9.14) 0 ∞ u u θ = 1− dy (9.15) 0 Ue UeIt means that the momentum ﬂux within θ is equal to the absence ofmomentum ﬂux due to the presence of the boundary layer. von K´rm´n made a guess of the velocity proﬁle according to the bound- a aary conditions: u(0) = 0, (9.16) u(δ) = Ue (9.17)and ∂u =0 . (9.18) ∂y y=δHe found that a second-order polynominal ﬁts the conditions, i.e. 2y y 2 u ≈ Ue − 2 . (9.19) δ δ 9.5 Momentum Boundary Layer Thickness, θ · 105 ·
- 113. In terms of the velocity proﬁle, the displacement thickness and the mo-mentum thickness are δ 2 δd ≈ and θ≈ δ . (9.20) 3 15In addition, the wall shear stress can be obtained by ∂u 2µU τw = µ ≈ . (9.21) ∂y y=0 δSubsequently, the friction coeﬃcient, Cf , can be found: τw 2µUe 4µ dθ Cf = 1 2 = 1 2 = =2 (9.22) 2 ρUe 2 ρUe δ ρUeδ dx 4µ d 2 4 dδ = 2 δ = (9.23) ρUeδ dx 15 15 dx 15µdx δdδ = (9.24) ρUe 30µx δ2 = (9.25) ρUe δ 5.5 ≈ √ (9.26) x Rex ∗ δ 1δ 1.83 ≈ =√ (9.27) x 3x Rex θ 2 δ 0.73 ≈ =√ = Cf (9.28) x 15 x Rex 2θ CD = (9.29) L 1.46 2Cf (L) = √ (9.30) ReLAlthought the velocity proﬂie is guessed, the results are very close to an-other contributor’s, Blasuis. Blasuis’s solution will be discussed in the next section.· 106 · Viscous External Flows
- 114. 9.6 Boundary Layer EquationPrandtl provided the boundary layer equation which comes from the N-Sequation and on the following assumptions: 1. 2-D 2. steady 3. incompressible 4. ∂P =0 −→ − dP = U dU ∂y dx dxConsequently, N-S equations can be simpliﬁed as: ∂u ∂v + =0 (9.31) ∂x ∂y ∂u ∂u dU ∂ 2u u +v = U +ν 2 (9.32) ∂x ∂y dx ∂y B.C. u(x, 0) = v(x, 0) = 0, u(x, ∞) = U (x) Blasius, one of Pranstl’s students, tried to get the solution using thesimilarity solution approach which is a common approach to transform aP.D.E. to an O.D.E.. However, the transformed O.D.E. is nonlinear and impossible to obtainan analytical solution. Hence a numerical method is reguired to obtain thesolution for the O.D.E. C. T¨epfer (1912) used the Runge-Kutta method oto solve the O.D.E. and obtain the numerical solution. In terms of the numerical solution for velocity, the boundary layer thick-ness, δ, is found according to its deﬁnition and revealed as 2νx δ ≈ 3.5 (u = 0.99Ue) (9.33) Ue 9.6 Boundary Layer Equation · 107 ·
- 115. or δ 5.0 5.5 ≈√ (K´rm´n’s answer √ a a ) (9.34) x Rex RexThe displacement thickness, δd , can be obtained as well and shown as δd 1.7208 5.5 = √ (K’s answer √ ) (9.35) x Rex RexIn addition, the momentum thickness, θ, is θ 0.604 =√ (9.36) x RexThe wall shear stress can be evaluated using the Newton’s viscosity law,i.e. ∂u τw = µ (9.37) ∂y y=0and the friction coeﬃcient, Cf , will be τw 0.664 θ Cf (x) = 1 2 =√ = (9.38) 2 ρUe Rex xFinally, the drag coeﬃcent, Cd , will be L D 0 τw dx 1.328 1.46 Cd = 1 2 = 1 2 =√ (K’s answer: √ ) (9.39) 2 ρUe L 2 ρUe L ReL ReL9.7 Friction coeﬃcient, Cf τw Cf = 1 2 (9.40) 2 ρUeIn boundary layer ﬂow, the wall shear stress, Cf , is a function of its localcoordinates.· 108 · Viscous External Flows
- 116. 9.8 Drag coeﬃcient, CD D CD = 1 2 (9.41) 2 ρUe L 2 D = τw (x)dx = ρUe θ (9.42) 0 2 dθ τw (x) = ρUe (9.43) dx dθ Cf = 2 (9.44) dx 2θ CD = (9.45) L9.9 DragWe mentioned drag calculation in a boundary layer ﬂow. In fact, dragcalculation should consider friction drag caused by the boundary layerand form drag caused by a wake. A low pressure region is generated owingto a wake, so the pressure diﬀerence between the front part and the rearpart of an obstacle is formed. Hence a form drag is produced. frictional drag −→ boundary layer form drag −→ pressure diﬀerence −→ wake9.10 Lift force and attack angle9.11 Streamline bodyA streamline body can reduce the wake region. In other words, most ofﬂuids are attached on the surface of the body. The resultant wake is verysmall. Hence drag is mainly from friction within its boundary layer. Incontrust, if a body’s drag comes mainly from form drag. That means its 9.8 Drag coeﬃcient, CD · 109 ·
- 117. L stall α Figure 9.4: Stallseparation happens very near the front part and its wake, therefore, is verylarge. Such a body is bluﬀ. Some designs to let turbulence happens earlierare made for bluﬀ bodies. For example, a golf ball is given more roughnesson its surface to trigger turbulence in its boundary layer.9.12 SeparationThe separation phenomenon happens in a boundary layer as ﬂows pastan obstacle. Before the separation point, the pressure gradient is nega-tive. Hence ﬂuids before the separation point are accelerated. However,momentum in the boundary layer is lost because of viscous energy dis-sipation. Therefore the pressure gradient gradually increases and ﬁnallybecomes positive. It is called an adverse pressure gradient. When an· 110 · Viscous External Flows
- 118. dP 0 dP dP dx dx =0 dx 0 adverse separation favorable point Figure 9.5: Separation due to an adverse pressure gradient.adverse pressure gradient appears in the boundary layer, ﬂuids are decel-erated. Finally velocity in the boundary layer becomes negative, i.e. areverse ﬂow appears. Then the separation happens.9.13 Separation and TurbulenceAs mentioned, separation is caused by an adverse pressure gradient dueto viscous energy dissipation. On the other hand, turbulence is able toimprove mixing in ﬂows. The momentum mixing in a turbulent boundarylayer is better than in a laminar boundary layer. Therefore, momentumnear the wall within a turbulent boundary layer is higher than a lami-nar boundary layer. This means that the net momentum in turbulentboundary layer is higher than a laminar boundary layer. Hence an ad- 9.13 Separation and Turbulence · 111 ·
- 119. verse pressure gradient does not easily happens and the separation in aturbulent boundary layer. This will make a wake smaller than withoutturbulence. Figure 9.6: K´rm´n vortex street. a a Owing to the feature of a turbulent boundary layer, the form drag canbe reduced. This is useful for a bulﬀ body because its drag is mainlydominated by form drag. Therefore another reason of the separation isbecause of a sharp corner. Hence a sudden expansion or contraction shapegenerates separation. When the separation happens, a wake is generated behind the separa-tion. Eddies are produced in a wake and cause a low pressure region. Itwas found by von K´rm´n and name after him, K´rm´n Vortex Street. a a a aEddy motion in a wake may be periodic, so a frequency may be found.Strouhal number is the nondimensional unmber for the frequency. f ·D St = , (9.46) u· 112 · Viscous External Flows
- 120. where f is the main frequency, D is the characteristic length of the obstacle,and u is the magnitude of characteristic velocity. The eddy motion causes vibration of the obstacle. If the frequency ofthe eddy motion is very close to or even equal to the natural frequencyof the obstacle, then the resonance will happen and cause a very seriousdamage. 9.13 Separation and Turbulence · 113 ·

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