Fluid mechanics lectur notes

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

PREFACE

Fluid mechanics is one of important subjects in engineering science. Although it has been developing for

more than one hundred years, the area which fluid mechanics covers is getting wider, e.g. biomechanics

and nanofluids. I started to write up this manuscript when I was assigned to give lectures on fluid

mechanics for senior undergraduate students. The main purpose of this lecture is to bring physics of

fluid 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 when

they were familiar with physics in this subject. As I finished this book, I do hope that readers can get

something from this book. Meanwhile, I wold like to express my graditude to those who helped me finish

this book.

Ming-Jyh Chern

Associate Professor

Department of Mechanical Engineering

National Taiwan University of Science and Technology

[email protected]

May 29, 2007

I

· II ·

Contents

PREFACE 2

1 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 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

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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 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 ·

4.6.6 Venturi tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.6.7 Flowrate pass through a sluice gate . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

7 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

8 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

9 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 ·

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 ·

Chapter 1

INTRODUCTION

1.1 Why study FLUID MECHANICS?

Fluid mechanics is highly relevant to our daily life. We live in the world

full 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 industrial

applications.

1.2 What is a fluid?

The main difference between fluid and solid is their behaviour when shear

forces acting on them. A certain amount of displacement is found when

a shear force is applied to a solid element. The displacement disappears

as the shear force is released from the solid element. A fluid deforms

continuously under the application of a shear force. Liquids and gases are

both regarded as fluids.

1

1.3 Approaches to study Fluid Mechanics

• Analytical Methods

• Experiments

• Computations

1.3.1 Analytical Methods

Using advanced mathematics, we can solve governing equations of fluid

motions and obtain specific solutions for various flow problems. For ex-

ample: pipe flows.

1.3.2 Expenmental Fluid Mechanics

This approach utilities facilities to measure considered flow fields or uses

various 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, costs

of 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

• 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.

Y0

Y

Z

Z0

XX0

Volumeof massm

Volumeof mass, m

C

Vδδ

V

Figure 1.1: Concept of a continuum.

1.4 History of Fluid Mechanics · 3 ·

δ

Vρ= lim

δ

δ

δ

Vδ

V

V

’

’

δ

m

δ δmV

V

Figure 1.2: Variation of a physical property with respect to the size of a continuum. Density is used as

an example.

1.5 Fluid as a continuum

The concept of a continuum is the basis of classic fluid mechanics. The

continuum assumption is valid in treating the behaviour of fluids under

normal conditions. However, it breaks down whenever the mean free path

of 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 into

the upper reaches of the atmosphere), we must abandon the concept of a

continuum in favor of the microscopic and statistical points of view.

As a consequence of the continuum, each fluid property is assumed to

have a definite value at every point in space. Thus fluid properties such as

density, temperature, velocity, and so on, are considered to be continuous

functions of position and time.

There exists a nondimensional number which is utilizd to judge whether


BOLTZMANN EQUATIONCOLLISIONLESS

BOLTZMANNEQUATION

CONSERVATION EQUATIONSDO NOT FROM A

CLOSED SETNAVER-STOKES

EQUATIONSEUL.ER

EQS.

DISCRETEPARTICLE ORMOLECULAR

MODEL

CONTINUUMMODEL

INVISCIDLIMIT LOCAL KNUDSEN NUMBER

FREE-MOLECULELIMIT

0 0.01 0.1 1 10 100 00

Figure 1.3: Knusden number and continuum.

fluids are continuous or not. Its definition is

Kn =ℓ

L, (1.1)

where ℓ is the free mean path of a fluid molecule and L is the smallest

characteristic length of a flow field. Kn is the so-called Knusen number.

1.5 Fluid as a continuum · 5 ·

1.6 Macroscopic physical properties of fluids

1.6.1 density, ρ

kg · m−3

Air 1.204

Water 998.2

Sea Water 1025

Mercury 13550

1.6.2 specific gravity, SG

SG =density of substance

density of water(1.2)

Air 0.001206

Oil 0.79

Ice 0.917

1.6.3 specific volume, ν

ν =1

ρ(1.3)

1.6.4 specific weight, γ

γ = ρg (1.4)

1.6.5 Compressibility of fluids

When fluids are pressurized, the total volume V is changed. The amount

of volume change is the compressibility of fluids. In fluid mechanics, we

use bulk modulus which is denoted as

Ev = −VdP

dV= ρ

dP

dρ, (1.5)


A high bulk modulus means that fluids are not easy to be compressed.

Hence, fluids with a high bulk modulus are incompressible. Units and

dimensions 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 regarded

as compressible fluids due to their small bulk moduluses.

1.7 Ideal gas law

The ideal gas law describes the relationship among pressure, density, and

temperature for an ideal gas. It can be shown that P = ρRT where R is

the gas constant. For air

R = 287.03 m2s−2K−1 = 1716.4 ft2s−2R−2 (1.6)

1.8 Pascal’s law

The Pascal’s law indicates that pressure transmission does not decrease

within a closed container filled with fluids. As shown in Fig. 1.4, pressure

at point A and point B are equal in terms of Pascal law. Therefore, if we

apply a force to the area A, it will produce a force on B and the force is

larger than the force on A.

1.9 Speed of sound

When disturbances are intorduced into fluid, they are propagated at a

finite velocity. The velocity depends on the compressibility of considered

fluids. It is called the acoustic velocity or the speed of sound, C. It is

defind as

1.7 Ideal gas law · 7 ·

A B

Figure 1.4: Concept of Pascal’s law.

C =

√

dP

dρ=

√

Ev

ρ

For ideal gases,

C =

√

d(ρRT )

dρ=

√RT

Example: Determine acoustic velocities of air and water where the tem-

perature is 20oC.

Cwater =

√

Ev

ρ=

√

2.19 × 109 N · m−2

998.2 kg · m−3= 1480 m · s−1 (1.7)

Consider air as an ideal gas

Cair =√

RT = 290 m · s−1 (1.8)

It implies that sound in incompressible fluids propagates faster than in

compressible fluids.


�

x

y � �y

�u�

t

�x

Figure 1.5: Deformation of a fluid experiencing shear stress.

1.9.1 Viscosity, µ & ν

Newtonian fluids

Consider fluids are full of two parallel walls. A shear stress, τ , is applied

to 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 Newtonian

fluid, i.e.

τ ∝ dγ

dt, (1.9)

where γ is shear angle or

τ = µdγ

dt. (1.10)

In addition,dγ

dt=

du

dy. (1.11)

Hence,

τ = µdu

dy. (1.12)

Again, the relationship between shear stress acting on a Newtonian fluid

and rate of strain (or velocity gradient) is linear. If it is not linear, then

1.9 Speed of sound · 9 ·

the fluid will be called a non-Newtonian fluid. µ is the so-called dynamic

viscosity. Its units are dyne · scm2 or Poise (cP). In addition, lb· s

in2 or Ryne

in B.G. 1 microRyne = 0.145 µ (cP)

Another definition of viscosity is the kinematic viscosity which is ν = µρ

Its units are cm2

s or Stoke(cS) in S.I. In addition, in2

s or Newt in B.G. 1

Newt = 0.00155 (cS).

Example: Determine the shear stress exerted on the bottom.

Solution:

U = 10 cm/s

x

y d =5.0 mm oil ( = 0.036 N·s/m2)

u(y)

According to Newton’s viscosity law, we have

τb = µdu

dy

∣

∣

∣

∣

y=0

. (1.13)

The velocity profile is available by a non-slip boundary condition, i.e.

u =U

dy

=0.1 m · s−1

0.005 my

= 20y . (1.14)

In addition, the velocity gradient on the bottom can be obtained by


du

dy

∣

∣

∣

∣

y=0

=U

d= 20 . (1.15)

Therefore, the shear stress is

τb = 0.036× 20 = 0.72 N · m−2. (1.16)

Saybolt viscometer

When we try to measure the viscosity for a fluid, we do not measure the

shear stress, and the volocity gradient but another variable, time.

Saybolt viscometer is designed to measure the viscosity of a fluid in

constant temperature. The principle of a fluids drain from a container in

constant temperature and we measure the total time till it takes for 60 ml

of fluids. Then we use empirical formulae to evaluate kinematic viscosity,

ν. The time, measured in second, is the viscosity of the oil in offficial units

called Saybolt Universal Seconds (SOS).

ν(cS) = 0.226t− 195

t, t ≤ 100 SOS (1.17)

ν(cS) = 0.22t− 135

t, t ≥ 100 SOS (1.18)

(temperature= 1500 F )

1.10 Hooke’s law and Newton’s viscosity law

Hooke’s law for a solid element

σ = Eǫ = Eδ

L, (1.19)

where 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 ·

Sample

temperatureis constant

60ml

Figure 1.6: Saybolt viscosmeter

Newton’s viscosity law

τ = µǫ = µdu

dy(1.20)

solid σ δ E

fluid τ u µ

In solid mechains, we utilize displacement to describe solid motions or

respons. 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 a

fluid motion.

1.11 Categories of Fluid Dynamics

Hydrodynamics & Hydraulics


Inviscid Fluid Flows(Potential Flows) & Viscous Fluid Flows

Laminar Flows & Turbulent Flows

Internal Flows & External Flows

1.11 Categories of Fluid Dynamics · 13 ·

Chapter 2

FLUID STATICS

In fluid statics, fluids at rest are considered. No relative motion between

adjacent 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 to

keep equilibrium.

2.1 Review of Taylor Expansion

For a continuous function, f(x), it can be expanded in a power series in

the neighborhood of x = α . This is the so-called Taylor Expansion given

by

f(x) = f(α)+f ′(α)

1!(x−α)+

f ′′(α)

2!(x−α)2+. . .+

fn(α)

n!(x−α)n+. . . (2.1)

2.2 Pressure

Pressure is continuous throughout a flow field in terms of continuum con-

cept. Pressure is isotropic. In other words, pressure is independent of

15

direction. Positive pressure means compression. On the other hand, neg-

ative pressure means tension. It is opposite to a normal stress. Pressure

can be regarded as a scalar.

y

z

x

g ds dz

x

z P1dA

P2dydz

P3dxdy

dA = ds · dy =dy · dz/sin

ρgdxdydz/2

Figure 2.1: Fluid element in a static fluid domain.

∑

F = 0 (2.2)

Fx = P2dydz − P1dA sin θ = 0 (2.3)

P2dydz = P1dydz

sin θsin θ (2.4)

P2 = P1 (2.5)

Fz = P3dydx =1

2ρgdxdydz + P1dy

dx

cos θcos θ (2.6)

P3 = P1 +1

2ρgdz (2.7)

dz → 0, P3 = P1 (2.8)

∴ P1 = P2 = P3 (2.9)

units of pressure

S.I.

1 N · m−2 = 1 Pascal(Pa) = 0.01 mbar(mb) (2.10)

· 16 · FLUID STATICS

B.G.

1 lb · in−2 = 1 psi = 144 psf(lbf · ft−2) (2.11)

2.3 The Hydrostatic Equation

Consider a fluid particle at rest shown in Figure 2.2. The centroid of the

y

x

z �z �

x �y

�O

Figure 2.2: Concept of a fluid element.

fluid element is at the original point O. The fluid element has a small

volume 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, we

can 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 be

given by

W = ρgδV = ρg∆x∆y∆z . (2.14)

2.3 The Hydrostatic Equation · 17 ·

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. P1

and P2 can be expanded using Taylor Expansion, i.e.

P1 = P (0) +P ′(0)

1!

(

+∆z

2

)

+P ′′(0)

2!

(

+∆z

2

)2

+ . . . (2.16)

and

P2 = P (0) +P ′(0)

1!

(

−∆z

2

)

+P ′′(0)

2!

(

−∆z

2

)2

+ . . . (2.17)

Substituting formulae above into the surface force, the surface force be-

comes

Fs = −2

[

P ′(0)∆z

2+ P ′′′(0)

(

∆z

2

)3

+ . . .

]

∆x∆y . (2.18)

Consider static equilibrium again, then we find

∑

Fz = Fs+W = −2

[

P ′(0)∆z

2+ P ′′′(0)

(

∆z

2

)3

+ . . .

]

∆x∆y−ρg∆x∆y∆z = 0

(2.19)

2

[

P ′(0)∆z

2+ P ′′′(0)

(

∆z

2

)3

+ . . .

]

= −ρg∆z (2.20)

In terms of continuum concept, ∆z should be very small (not zero), so we

can negelect high order terms in the formula, i.e.

P ′(0)∆z = −ρg∆z (2.21)

ordP

dz

∣

∣

∣

∣

z=0

= −ρg . (2.22)

We can use a notation directional gradient to show the equation again, i.e.

∇P = ρg . (2.23)

This is called the hydrostatic equation.


2.4 Pressure variation in incompressible fluids

Density is constant throughout an incompressible fluid domain. Hence, we

can evaluate the pressure difference between two points(z = z1 and z2 ),

i.e.

∆P |21 =

∫ 2

1

dP

dzdz

=

∫ 2

1

−ρgdz

= −ρg

∫ 2

1

dz

= −ρg (z2 − z1) . (2.24)

∆Pρg is called a pressure head and equal to −∆z .

2.5 Pressure variation in compressible fluids

Density is not constant throughout a compressible fluid domain. In other

words, density may be affected by temperature and pressure. If we consider

a 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

dz= −ρg = −Pg

RT⇒ dP

P= − g

RTdz (2.26)

In addition, the pressure difference between two points (z = z1 and z2)

can be evaluated by integrating the hydrostatic equation:∫ 2

1

dP

P=

∫ 2

1

− g

RTdz (2.27)

⇒=lnP|21=- gRT (z2 − z1)

2.4 Pressure variation in incompressible fluids · 19 ·

⇒=lnP2

P1

=- gRT (z2 − z1)

P2=P1exp[- gRT (z2 − z1)]

△P |21=P2-P1=-P1

[

1 − exp[

− gRT (z2 − z1)

]]

Example: Determine the pressure at the gasoline-water interface, and at

the bottom of the tank (see Fig. 2.3). Gasoline and water can be both

open

gasoline

water

S.G.=0.68

17ft

3ft P1

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 Atmosphere

Sea level conditions of the U.S. Standard Atmosphere.


10 20

40

50

-40 -60 -20 0 20 40 40 80 120

z(km)

surface

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 15oC 59oC

Pressure 101.33 kPa 2116.2 psf

Density 1.225 kg/m3 0.002377 slug/ft3

Homework: Derive the formula for the pressure variation within the con-

vection layer. Remember pressure and temperature are both functions of

elevation.

Ans:

P = P0

[

1 − α(z − z0)

T0

]g/αR

(2.31)

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 ·

2.6.1 Absolute pressure

Pressure measured relative to an absolute vacuum.(Pb)

2.6.2 Gauge pressure

Pressure measured relative to atmospheric pressure.(Pg)

d

h

z

Pa

Pressure caused by fluid

weight. 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 pressure

at 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 =

∫

PzdA =

∫

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)


2.7 Facilities for pressure measurement

2.7.1 Manometers

Z1

�h

Z2

A

B

P1 P2

Figure 2.6: Schematic of a manometer.

Manometers are utilized to measure pressure difference between two

points,

∆P = P1 − P2 = ρgδh . (2.39)

2.7.2 Barometers

Barometers 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 ·

2.8 Inclined-tube Manometer

The 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.

A

B

γ1 γ2

γ3

h1

h2

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 = γ3h3 + γ2(l2 sin θ) − γ1h1 (2.43)

If we ignore γ1 and γ3, then

PA − PB = γ2l2 sin θ (2.44)

and

l2 =PA − PB

γ2 sin θ. (2.45)

If 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 = Pbwdz (2.46)


Pa

d F d z

z

h

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

0

[Pa + ρg(h − z)]wdz

=

∫ h

0

Pawdz +

∫ h

0

ρg(h − z)wdz (2.49)

Pawh ρgh2

2w

If we just consider pressure caused by the weight of fluids, then the force

will be

2.9 Hydrostatic force on vertical walls of constant width · 25 ·

Fs =ρgh2

2w . (2.50)

The 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

0

ρg(h − z)wzdz

= ρgw

(

hz2

2− z3

3

)∣

∣

∣

∣

h

0

= ρgw

(

h3

2− h3

3

)

=ρgh3w

6. (2.52)

We can evaluate the moment arm z, i.e.

z =M0

F=

ρgh3w6

ρgh2w2

=h

3. (2.53)

2.10 Hydrostatic force on an inclined surface

Consider an inclined surface shown in Fig. 2.10, then

dF = ρghdA, h = y sin θ

= ρgy sin θdA, dA = wdy (2.54)


h

dF

dA

Y

X

Y

O

w

θ

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 can

say∫

ydA = ycA, (2.56)

where yc is the centroid of the area. Furthermore, the resultant force

becomes

F = ρg sin θycA

= ρghcA (2.57)

We consider the moment caused by the resultant force with respect to

2.10 Hydrostatic force on an inclined surface · 27 ·

the original point O. First of all,we know

dM = ydF (2.58)

and then

M =

∫

dM

=

∫

ydF

=

∫

ρgy2 sin θdA . (2.59)

∫

y2dA is called the second moment of the area with respect to the x-axis,

Ix. We know

M = F · yR (2.60)

and

yR =M

F=

ρg sin θ∫

y2dA

ρg sin θycA=

Ix

ycA, (2.61)

where yR is the acting point of the resultant force or so-called the centre

of pressure.

Example: Consider a dam of width 100 m and depth 6 m. Determine the

resultant hydrostatic force and the moment with respect to A.


h

A

Figure 2.11: Problem of hydrostatic force exerted on a dam.

Sol:

F = γhcA

= γh

2A

= 1000 × 9.8 × 0.5 × 6 × (6 × 100)

= 17660 kN

M = F · hf

= F · 1

3h

= 35320 kN-m (2.62)

2.11 Hydrostatic force on a curved surface

Consider a curved surface shown in Fig. 2.12. The resultant force acting

2.11 Hydrostatic force on a curved surface · 29 ·

h

Z

X

dF

Fx

Fz

dAα

F

Figure 2.12: Hydrostatic force exerted on a curved surface.

on a small element of the curved surface is given by

dF = Pn · 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 the

small area. In addition,

Fx =

∫

dFx

=

∫

ρg(h − z) sin αdA

= ρg

∫

(h − z) sin αdA

= ρg

∫

(h − z)dAv, (2.65)


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 vertical

plane. 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 resultant

force F can be given by

|F| =√

F 2x + F 2

z . (2.68)

2.12 Buoyance

It is well-knoen that Archimede provided the buoyance principle to eval-

uate the buoyant force acting on a submerged solid body. In fact, we can

derive the buoyance principle from the hydrostatic equation. Let us con-

sider a submerged body shown in Fig. 2.13. The resultant force caused by

pressure on the small wetted area is given by

dF = P2dA − P1dA = (−ρgz2 + ρgz1)dA (2.69)

and

F =

∫

dF = ρg

∫

(z1 − z2)dA = −ρgV . (2.70)

2.12 Buoyance · 31 ·

P1

P2

dAZ1

Z2

Z

Figure 2.13: Schematic of buoyance exerted on an immersed body.

Therefore, we know the resultant force caused by static pressure or called

the buoyant force is equal to the weight of liquids of volume equal to the

submerged body. In addition, the point where the buoyant force exerts is

called the centre of buoyance.


Chapter 3

INTRODUCTION TO FLUID

MOTION I

The chapter demonstrates basic concepts of fluid kinematics and funda-

mental laws which fluids conserve.

3.1 Lagrangian and Eulerian Systems

When we describle physical quantities, such as density, pressure, and so

on, of adynamic problem, we usually chose either Lagrangine or Eulerian

system. In terms of Lagrangine system, we move with the considered

system or particles, so physical quantities, say φ, is only a function of

time, 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 often

employed in solid dynamic. On the other hand, we fix a point in space and

observe the variation at this point in terms of Eulerian system. Therefore

physical quantities, φ, are not only functions of time but also functions of

33

space, i.e.

φ = φ(x, y, z, t) , (3.2)

where x, y, z, and t are independent. Eulerian system is commonly used in

fluid dynamics. It may be because lots of fluid particles are involved in a

considered flow. It contains different fluid particles at the observed point

as time goes in Eulerian system. Hence it is hard to describe a system or

its physical quantities in terms of a specified fluid particle. Therefore, we

utilize Eulerian system to describe a system.

3.2 Control Volume

In addition, we utilize a control volume concept to describe a fluid flow

problem. Coupled with Eulerian system, a control volume is a fixed region

with artifical boundaries in a fluid field. A control volume contains lots of

and various fluid particles as time goes. Fluid flows in and out through its

control surface and then physical quantities in a control volume change.

3.3 Steady and Unsteady flow

If physical quantities of a flow field are independent of time, then the flow

will be called steady. Otherwise, it is unsteady.

3.3.1 Streamlines

A steamline is defined as a line that is everywhere tangential to the in-

stantaneous velocity direction, i.e.

dy

dx=

v

u,dy

dz=

v

w, and

dx

dz=

u

w. (3.3)

Streamlines cannot cross.

· 34 · INTRODUCTION TO FLUID MOTION I

3.3.2 Pathlines

A pathline is defined as the path along which a specified fluid particle

flows. It is a Lagrangine concept. Hence, coordinates of a pathline are

functions of time.

3.3.3 Streaklines

A streakline is the line traced out by particles that pass through a partic-

ular point.

3.3.4 Streamtubes

A 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 properties

of flows are only functions of a spatial variable. The spatial variable can

be coordinates of an axis, such as x, or along a streamline. For example,

3.3 Steady and Unsteady flow · 35 ·

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 volume

Consider a control volume in a flow field (see Fig. 3.2). The rate of

variation of a physical property in a control volume shall be equal to the

sum of the flux through its control surface and the surface of the physical

property.

uφ φsource of

Figure 3.2: Control volume

d

dt

∫ ∫ ∫

φdV =

∫ ∫

control surfaceφu · dA +

∫

∂φ

∂tdV (3.6)

φ: physical property in a unit volume. For example, mass in a unit volume

is density. (mV = ρ)


3.5 Mass conservation of 1-D flows

When 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 in

a 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. If

we consider the control volume between point 1 and point 2 and the mass

conservation law should be satisfied in the control volume. If we donot

consider any mass source or sink in the control volume, then the rest will

be 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

ρ1u1A1 = ρ2u2A2 , (3.11)

where u1 and u2 are average velocities at points 1 and 2, respectively. If

density of fluids are the same at surface 1 and 2, i.e.

Q = u1A1 = u2A2 , (3.12)

where Q is the volumetric flow rate. In terms of the mass conservation

law, we find that average velocity on a small area is higher than one on a

large area.

3.5 Mass conservation of 1-D flows · 37 ·

3.6 Momemtum conservation for 1-D flows

According 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 means

the change rate of momentum in the object should be zero. We look into

the 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

dt(mu) = 0 . (3.14)

If we donot consider any force source in a control volume for a 1-D flow

like Fig. 3.2, then only momentum fluxes on surface 1, 2 are considered,

i.e.

F =d

dt(m1u1 + m2u2) = 0 (3.15)

or

d

dt(ρ1A1u1 · u1 + ρ2A2u2 · u2) = 0 (3.16)

If the 1-D flow is steady, then we can remove the total derivative, i.e.

ρ1A1(u1 · u1) + ρ2A2(u2 · u2) = 0 (3.17)

or

ρ1A1u21 = ρ2A2u

22 . (3.18)

If we consider other forces acting on the control volume, then

∑

F = 0 = F0 +d

dt(mu) = 0 (3.19)


F0 +d

dt(ρ1A1u1 · u1 + ρ2A2u2 · u2) = 0 . (3.20)

This is consistent with Newton third law. F can be divided into two parts:

1. body forces such as gravity forces, magnetic forces; 2. surface forces

such as pressure.

3.6 Momemtum conservation for 1-D flows · 39 ·

Chapter 4

INTRODUCATION TO FLUID

MOTION II

4.1 The Bernoulli equation

Consider a steady inviscid flow. If we apply Newton’s second law along a

stream line, we will obtain the Bernoulli equation

P1 +1

2ρu2

1 + ρgz1 = P2 +1

2ρu2

2 + ρgz2 = const . (4.1)

The detailed deviation of the Bernoulli equation will be given later. The

Bernoulli equation above is based on four assumptions:

1. along a same streamline

2. steady flow

3. same density

4. inviscid

41

4.2 Derive the Bernoulli equation

Consider a steady flow shown in Fig. 4.1. For a fluid particle in the

streamline A, the momentum should be conserved. Assume the volume of

the fluid is ∆x∆n∆s. The total force along the streamline should be

β

β

∆

∆

ρ ∆ ∆ ∆

∂

∂

g x n s

Z

Y

n

s

n

n

Adndx( P+ )

ds2

Ps

∆

P( P- )ds2s

∂

dndx∂

g

β

s

Figure 4.1: Force balance for a fluid element in the tangential direction of a streamline.

ΣFs =

[(

P − ∂P

∂s

ds

2

)

−(

P +∂P

∂s

ds

2

)]

dndx − ρg∆x∆n∆s sinβ

= −∂P

∂sdsdndx − ρg∆x∆n∆s sinβ . (4.2)

· 42 · INTRODUCATION TO FLUID MOTION II

The momentum change along the streamline should be

∂

∂t(Σmu) =

1

∆t

[

− (ρ∆x∆n∆s) (u) + (ρ∆x∆n∆s)

(

u +∂u

∂sds

)]

=1

∆t

[

ρ(∆x∆n∆s)∂u

∂sds

]

= ρ (∆x∆n∆s)u∂u

∂s, (4.3)

where u is the tangential velocity component. Let us consider Newton’s

second law, i.e.

ΣFs =∂

∂t(Σmu) (4.4)

Substitution of Eq. (4.2) into (4.3) gives

−∂P

∂s− ρg sin β = ρu

∂u

∂s, sinβ =

∂z

∂s(4.5)

and then

−∂P

∂s− ρg

∂z

∂s= ρu

∂u

∂s. (4.6)

This is the so-called Euler equation along a streamline in a steady flow. If

the Euler equation is multiplied by ds, it will become

−dP − ρgdz = ρudu (4.7)

Futhermore, we integrate the whole equation and obtain the Bernoulli

equation, i.e.

P

ρ+

1

2u2 + gz = constant . (4.8)

The 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 fluid

along the streamline. The integral of the resultant equation is constant

along a streamline. It turns out that the Bernoulli equation refers to en-

4.2 Derive the Bernoulli equation · 43 ·

ergy conservation along a streamline. Pρ + gz can be regarded as potential

energy and u2

2 , of course, is the kinetic energy.

Moreover, we consider force balance across a streamline. The resultant

force should be

β

W

∂

∂n

dsdx( P+ )dn2

Pn

∆

P( P- )dn2n

∂

dsdx∂

Figure 4.2: Force balance of a fluid element in the normal direction of a streamline.

ΣFn =

[(

P − ∂P

∂n

dn

2

)

dsdx −(

P +∂P

∂n

dn

2

)

dsdx

]

−ρg∆x∆s∆n cosρ .

(4.9)

Its momentum change across a streamline should be

∂

∂tΣmun = −ρ

u2

R∆x∆s∆n , (4.10)

where un is the velocity component normal to a streamline and R is the


curvature radius. Let us consider Newton’s second law again.

ΣFn =∂

∂tΣmun (4.11)

Substitution of Eq. (4.9) into (4.10) gives

−∂P

∂ndndsdx − ρg∆x∆s∆n cosβ = −ρ

u2

R∆x∆s∆n (4.12)

cos β =∂z

∂n(4.13)

and then

∂P

∂n+ ρg

∂z

∂n= ρ

u2

R. (4.14)

This is the Euler equation across a streamline. If the Euler equation is

multiplied by dn and integrated along the normal direction, it will become

−∫

dP −∫

ρgdz =

∫

ρu2

Rdn . (4.15)

It is the Bernoulli equation along the normal direction of a stream.

Example: Determine the pressure variation along the streamline from

z

xO BA

u=u0(1+ )a3

x3

a

Figure 4.3: 2-D flow past a circle.

point A to point B.

4.2 Derive the Bernoulli equation · 45 ·

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

A

dP =

∫ O

A

ρudu .

We know that

du = u0a3(−3)x−4dx

= −3u0a3

x4dx . (4.18)

As a result,

PO − PA = ρ

∫ O

A

u0

(

1 +a3

x3

)(

−3u0a3

x4

)

dx

= ρ

∫ O

A

−3u20

(

a3

x4+

a6

x7

)

dx

= ρu20

(

a3

x3+

a6

2x6

)∣

∣

∣

∣

O

A

= ρa3u20

x3

(

1 +1

2x3

)∣

∣

∣

∣

O

A

. (4.19)

The x-coordinate of point B is -a, so

PB − PA = −ρu20

(

1 − 1

2a3

)

− ρa3 u20

x3A

(

1 +1

2x3A

)

. (4.20)

4.3 Stagnation Pressure and Dynamic Pressure

Consider fluids flow toward a horizontal plate far upstream. Fluids moves

at u∞ and pressure is P∞ upstream. Because fluids cannot pass through a


stagnation point

stagnation streamline

P0

∞P∞u

Figure 4.4: Stagnation point

plate, fluids must flow along the plate. Subsequently we can find a point

where fluids are at rest. This is the so-called stagnation point. Further-

more, we can find a stagnation steamline which leads to the stagnation

point. Owing to no variation of altitude in the whole flow, pressure and

velocity are considered in the Bernoulli equation. If we apply the Bernoulli

equation along the stagnation line, we will find

P∞ρ

+u2∞2

=P0

ρ, (4.21)

where P0 is called the stagnation pressure or total pressure, P∞ is called

the static pressure, and ρu2

∞

2 is called dynamic pressure which is distincted

from the pressure due to hydrostatic pressure, P∞ .

4.3 Stagnation Pressure and Dynamic Pressure · 47 ·

Pressure coefficient is defined as

Cp =P − P∞

12ρu2

∞= 1 − (

u

u∞)2 . (4.22)

Its means the ratio of pressure difference to inertia force. At a stagnation

point, Cp = 1, that means all of kinetic energy is transfered to pressure

energy. Cp is zero far upstream. It means no kinetic energy is transfered

to pressure energy.

4.4 Mass conservation in channel flows

Consider fluid flow in a channel with various cross section areas show in

Fig. 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, the

amount of mass passing a cross section per unit time, must be equal at

every cross section, i.e.

m = m1 = m2 , (4.23)


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,

ρ1Q1 = ρ2Q2 (4.25)

or

ρ1u1A1 = ρ2u2A2 , (4.26)

where u1 and u2 are average velocity at cross sections 1 and 2, A1 and

A2 are cross sectional areas. For incompressible fluids, ρ1 = ρ2 and conse-

quently u1A1 = u2A2.

4.5 Relationship between cross area, velocity ana pressure

Consider a steady flow in a channel with varied cross sectional areas. In

terms of the continuity equation, velocity decreases as its cross sectional

area diverages for incompressible fluids. In addition, pressure increases

as velocity decreases in terms of the Bernoulli equation. For a converged

channel, cross sectional area decreases so velocity increases. Subsequently,

pressure decreases owing to increasing velocity.

4.6 Applications of Bernoulli equation

4.6.1 Pitot tube

P∞ +1

2ρau

2∞ + ρagz∞ = PO +

1

2ρau

2O + ρagzO (4.27)

z∞ = zO, uO = 0 (4.28)

∴ P∞ +1

2ρau

2∞ = PO (4.29)

4.5 Relationship between cross area, velocity ana pressure · 49 ·

A

V

P

A

V

P

Figure 4.6: Variations of velocity and pressure in converged and diverged channels.

(PO − P∞) =1

2ρau

2∞ (4.30)

PO − P∞ = ρℓg∆h (4.31)

ρℓg∆h =1

2ρau

2∞ (4.32)

u2∞ = 2

ρℓ

ρag∆h (4.33)

4.6.2 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 is

assumed 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 Bernoulli

equation is applied to analyze the flow in a siphon. Consider conditions

at points 1 and 3 and

Pa

ρ+ 0 + gz1 =

Pa

ρ+

u23

2+ gz3 , (4.34)


z0

Pu

zh∆

ρl

∞

∞

∞

O

∞

Figure 4.7: Schematic of Pitot tube.

where z1 = 0, z3 = −h3. Velocity at point 3 is obtained from the equation

i.e.

u3 =√

2gh3 . (4.35)

Another interesting location is at point 2. In terms of Bernoulli equation,

we findPa

ρ+ 0 + gz1 =

P2

ρ+

u22

2+ gz2 , (4.36)

where z1 = 0, z2 = h2. Then we find pressure at point 2 is

P2

ρ=

Pa

ρ− u2

2

2− gh2 , (4.37)

where u2

2

2 and gh2 must be positive. It turns out that P2 should be less

than the atmospheric pressure. If point 2 is high enough to let pressure at

point 2 less than vapor presure, then gas in fluids will form bubbles. These

bubbles will move with fluids. If pressure around bubbles increases and is

higher than vapor pressure, then bubbles will burst. The phenomenon is

4.6 Applications of Bernoulli equation · 51 ·

1

2

3

z h2

h3

Figure 4.8: Schematic of siphon tube.

called cavitation. Cavitation is often found in flow fields around a inside

propeller or fluid machinery.

4.6.3 Torricelli’s Theorem

H

Pa

Pa2

1

Figure 4.9: Torricelli’s theorem.

Consider a liquid tank of high H. There is a hole, shown in Fig. 4.9, near

the ground. Liquids drain from the hole. It is assumed that the tank is

quite 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


pressure. Now we can apply Bernoulli equation to point 1 and 2, i.e.

P1

ρ+

u21

2+ gz1 =

P2

ρ+

u22

2+ gz2 , (4.38)

where P1 = P2 = Pa, u1 = 0, and (z1 − z2) = H. It then becomes

Pa

ρ+ gH =

Pa

ρ+

u22

2. (4.39)

It turns out that

u2 =√

2gH . (4.40)

This is the Torricelli’s Theorem.

4.6.4 vena contracta effect

dh

dj

Figure 4.10: Vena contracta effect

contraction coefficient

Cc =Aj

Ah=

(dj)2

(dh)2(4.41)


l

h

1

2

Figure 4.11: Free jet

4.6.5 Free jets

Consider fluids in a tank. A nozzle is arranged at the bottom of the

tank. Fluids flow through the nozzle due to the gravitational force and

consequently a jet is observed. Suppose no energy loss in the nozzle.

Bernoulli equation can be utilized to determine the jet condition at the

exit of the nozzle. The free surface of the tank is assumed to be still if

the tank is large enough. Therefore, u1 = 0. According to the Bernoulli

equation, the total energy along a streamline from the free surface to the

exit should be the same, i.e.

P1

ρ+

u21

2+ gz1 =

P2

ρ+

u22

2+ gz2 = constant . (4.42)

We know u1 = 0, P1 = P2 = Pa and (z1 − z2) = h + l. The equation

becomesu2

2

2= g(h + l) (4.43)


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 energy

loss at the nozzle. As a result, Bernoulli equation has to be modified.

4.6.6 Venturi tube

A B

Figure 4.12: Venturi tube.


uAAA = uBAB

uB = uAAA

AB

AA > AB

uA < uB

PA

ρ+

u2A

2=

PB

ρ+

u2B

2

PA − PB

ρ=

u2B − u2

A

2

=u2

A(AA

AB) − u2

A

2

=u2

a

2

[

(

AA

AB

)2

− 1

]

(4.45)

A Venturi tube is a device made up of a contraction followed by a diverging

section. Fluids moving toward the contraction are speeded up according

to the continuity equation. In addition, pressure decreases as velocity

increases in terms of the Bernoulli equation. A famous application of a

Ventui tube is a carburetor. A carburetor is shown in Fig. 4.13. Fuel is

sucked 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 measure

the flow rate in a pipe. Fluids flow a contraction part and then a expansion

part in a Venturi tube.


ButterflyValve

Throat ofVenturi

Air-FuelMixture

Q

FUEL

Q(Air)

Figure 4.13: Schematic of caburetor.

4.6.7 Flowrate pass through a sluice gate

Form the Bernoulli equation,

P1

r+

u21

2g+ z1 =

P2

r+

u22

2g+ z2

P1 = P2 = Pa

u21

2g+ z1 =

u22

2g+ z2 (4.46)

Form mass conservation

u1z1 = u2z2

u1 =z2

z1u2 . (4.47)


Substituting into Bernoulli equation

u22

2g

(

z2

z1

)2

+ z1 =u2

2

2g+ z2

u22

2g

[

(

z2

z1

)2

− 1

]

= z2 − z1

u2 =

√

2gz2 − z1

(z2

z1

)2 − 1. (4.48)

The flowrate pass through the sluice gate must be

Q = u2 · z2

= z2

√

2g(z2 − z1)

(z2

z1

)2 − 1. (4.49)


Chapter 5

EQUATIONS OF MOTION IN

INTEGRAL FORM

We consider one-dimensional flows in Chapter 3 and 4. Conservation laws

of 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 control

volume concept. Equations in integral form do not give any information

throughout a flow field, but they can provide resultant forces acting on a

control volume. On the other hand, equations in differential form provide

details regarding variations in a flow field, so we can get values of physical

variables throughout a flow field.

In this chapter, we consider governing equation of fluid flows in integral

form first.

59

5.1 Flux

We mentioned the control volume concept in Chapter 3. A control volume

is bounded artificially in a flow field. Physical properties in a control

volume may vary in space or in time, because fluids with various physical

properties flow in and out a control volume and it causes variations of

physical properties in a control volume. The amount of a physical property

cross an unit surface per second is called flux.

A flux can be revealed as b(u · A), where b is a physical property per

unit volume, u is the velocity over the area and A is the area vector. We

may use nA instead of A and n is the unit vector in the normal direction

of 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 the

area but negative in the inward direction.

· 60 · EQUATIONS OF MOTION IN INTEGRAL FORM

5.2 Reynolds’ Transport Theorem

I

II

III

1

2

Figure 5.1: Flow through a control volume.

We consider a control volume I+II in a flow field. Fluids contained in

the control volume at t = t will flow, so the control volume containing

same fluids at t = t+δt will be II+III. The rate of change of a physical

property in the control volume can be shown in DDt

∫ ∫ ∫

c.v. ραdV where α

is 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 two

parts. 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 the

region 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 ·

fluxes across two surfaces and get∫ ∫

c.s. ρα(u · n)dA. As δt → 0, we will

have

D

Dt

∫ ∫ ∫

c.v.

ραdV =

∫ ∫

c.s.

ρα(u · n)dA +∂

∂t

∫ ∫ ∫

c.v.

ραdV (5.6)

At t = t0

Bsys = BI(t) + BII(t) . (5.7)

At t = t0 + ∆t

Bsys = BII(t + δt) + BIII(t + δt) (5.8)

lim∆t→0

∆Bsys

∆t=

DBsys

Dt(5.9)

or

∆Bsys

∆t=

BII(t + δt) + BIII(t + δt) − BII(t) − BI(t)

∆t(5.10)

lim∆t→0

BII(t + δt) − BII(t)

∆t=

∂BII

∂t(5.11)

−BI(t)∆t is the flux flow through in C.S.1 and is denote as

∫ ∫

C.S.1

ρα(u · dA) (5.12)

In addition, BIII(t+δt)∆t

is the flux flow out C.S.2 and is denoted as

∫ ∫

C.S.2

ρα(u · dA) (5.13)

∫ ∫

C.S.1

ρα(u · dA) +

∫ ∫

C.S.2

ρα(u · dA) =

∫ ∫

C.S.

ρα(u · dA) (5.14)

Besides,

lim∆t→0

(C.V.I+C.V.II) = lim∆t→0

(C.V.III+C.V.II) = C.V.II = C.V. =

∫ ∫ ∫

C.V.

ραdV

(5.15)


As a result

DBsys

Dt=

D

Dt

∫ ∫ ∫

c.v.

ραdV (5.16)

and

∂Bc.v.II

∂t=

∂

∂t

∫ ∫ ∫

c.v.

ραdV (5.17)

5.3 Continuity Equation

If 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 be

revealed as

D

Dt

∫ ∫ ∫

c.v.

ρdV =

∫ ∫

c.s.

ρ(u · n)dA +∂

∂t

∫ ∫ ∫

c.v.

ρdV = 0 (5.18)

This is the continuity equation in integral form.

5.4 Momentum Equation

Subsequntly, we consider momentum in a control volume, then α will be

u. The momentum equation in integral form then is denoted as

D

Dt

∫ ∫ ∫

c.v.

ρudV =

∫ ∫

c.s.

ρu(u · n)dA +∂

∂t

∫ ∫ ∫

c.v.

ρudV . (5.19)

Moreover, the rate of momentum is equal to the resultant force acting on

the control volume, i.e.

D

Dt

∫ ∫ ∫

c.v.

ρudV = F = Fbody + Fsurface + Fext . (5.20)

If we consider gravity in body force, then we will have

Fbody =

∫ ∫ ∫

c.v.

ρgdV . (5.21)

5.3 Continuity Equation · 63 ·

The surface can be divided into pressure and shear stress, i.e.

Fsurface =

∫ ∫

c.s.

(−p + τij)ndA . (5.22)

Hence,

D

Dt

∫ ∫ ∫

c.v.

ρudV =

∫ ∫ ∫

c.v.

ρgdV +

∫ ∫

c.s.

(−p + τij)ndA + Fext .

(5.23)

5.5 Moment-of-Momentum Equation

Now we consider moment-of-momentum, then α will r×u. Using Reynold’s

transport theorem, we obtain

D

Dt

∫ ∫ ∫

c.v.

ρr×udV =

∫ ∫

c.s.

ρ(r×u)(u·n)dA+∂

∂t

∫ ∫ ∫

c.v.

ρ(r×u)dV

(5.24)

or

Σ(r×F)c.v. =

∫ ∫

c.s.

ρ(r×u)(u ·n)dA+∂

∂t

∫ ∫ ∫

c.v.

ρ(r×u)dV = Tshaft ,

(5.25)

where Tshaft is the resultant torque applied to fluids in the control volume.


Chapter 6

DIFFERENTIAL EQUATIONS OF

MOTIONS

We obtain governing equations of fluid flows in integral form in Chapter 5.

As mentioned before, governing equations in integral form cannot provide

details throughout a control volume. If we would like to know more about a

flow field, such as velocity, pressure and so on, then governing equations of

fluid flows in differential form are necessary. Solving differential governing

equation can get the whole information of a flow field.

6.1 Lagrangian and Eulerian systems

These systems are used very often in dynamics. An observer follows a

specified particle if Lagrangian system is employed. In other words, a

coordinate system is fixed at a particular fluid particle when Lagrangian

system is utilized to describe a flow field. In addition, physical variables

described by Lagrangian system are functions of time only. All of spatial

coordinates are functions of time too (x = ut). We can say α = α(t).

dα

dt=

∂α

∂t+ u

∂α

∂x+ v

∂α

∂y+ w

∂α

∂z(6.1)

65

The term at the left hand side of the equation is called the total derivative

or 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

Dtinstead of dα

dtto

indicate a material derivative, i.e.

Dα

Dt=

∂α

∂t+ u

∂α

∂x+ v

∂α

∂y+ w

∂α

∂z. (6.2)

The trems at the right hand side of the equation are, in fact, described

by Eulerian system. The first term is called a local derivative or unsteady

term. The rest are called convective terms because they are caused by flow

motions. This equation can shown in vector form, i.e.

Dα

Dt=

∂α

∂t+ (u · ∇)α . (6.3)

It should be noted that u and ∇ cannot be communtive. In other words,

(u · ∇) 6= (∇ · u) . (6.4)

Eulerian system is an alternative way to describe a flow field. An observer

is fixed at a selected point of space if Eulerian system is adopted. Hence,

the coordinate system does not move as fluids flow. In addition, spatial

coordinates are required to indicate the point which is observed. Hence,

physical variables described by Eulerian system are functions of time and

space, i.e.

α = (x, t) (6.5)

· 66 · DIFFERENTIAL EQUATIONS OF MOTIONS

6.2 Rate of Change Following a Fluid Particle

α is a continuous function of space and time. Its infinite samll change can

be described using chain rule of differentiation, i.e.

dα =∂α

∂tdt +

∂α

∂xdx +

∂α

∂ydy +

∂α

∂zdz , (6.6)

where Cartesian coordinate system is used. Moreover, the equation is

divided by dt and we get

dα

dt=

∂α

∂t+

∂α

∂x

dx

dt+

∂α

∂y

dy

dt+

∂α

∂z

dz

dt. (6.7)

In Chapter 5, the change rate of a physical variable in a control volume

can be shown in

D

Dt

∫ ∫ ∫

c.v.

ραdV =

∫ ∫

c.s.

ρα(u · n)dA +∂

∂t

∫ ∫ ∫

c.v.

ραdV (6.8)

The flux across the control surface can be convered to a term in volume

intrgral using Gauss’ theorem, i.e.∫ ∫

c.s.

ρα(u · n)dA =

∫ ∫ ∫

c.v.

∇ · (ραu)dV (6.9)

Subsequently, we can rewrite the control volume formula and obtain

D

Dt

∫ ∫ ∫

c.v.

ραdV =

∫ ∫ ∫

c.v.

∇ · (ραu)dV +∂

∂t

∫ ∫ ∫

c.v.

ραdV (6.10)

If the volume does not change, then it will be

D

Dt

∫ ∫ ∫

c.v.

ραdV =

∫ ∫ ∫

c.v.

[

∂(ρα)

∂t+ ∇ · (ραu)

]

dV (6.11)

6.3 Continuity Equation

Mass conservation is obeyed as fluid particles move. Then it can shown as

Dm

Dt=

D

Dt

∫ ∫ ∫

c.v.

ρdV = 0 . (6.12)

6.2 Rate of Change Following a Fluid Particle · 67 ·

It means that mass of a fluid particle does not change with time. This

is the continuity equation in integral form. The Eulerian system can be

applied to describe the continuity equation, i.e.

D

Dt

∫ ∫ ∫

c.v.

ρdV =

∫ ∫ ∫

c.v.

[

∂ρ

∂t+ ∇ · (ρu)

]

dV = 0 (6.13)

In addition,

∇ · (ρu) = ρ(∇ · u) + (u · ∇)ρ . (6.14)

When dV does not change with time, the continuity equation becomes

∂ρ

∂t+ ρ(∇ · u) + (u · ∇)ρ =

Dρ

Dt+ ρ(∇ · u) = 0 . (6.15)

For an incompressible fluid,Dρ

Dt= 0. The continuity equation becomes

∇ · u = 0 . (6.16)

In a Cartesian coordinate system, it becomes

∂u

∂x+

∂v

∂y+

∂w

∂z= 0 . (6.17)

In a cylindrical coordinate system, it becomes

∂ur

∂r+

ur

r+

1

r

∂uθ

∂θ+

∂uz

∂z= 0 . (6.18)

6.4 Momentum Equation

Consider momentum conservation in fluid particles. It can be shown as

D

Dt(mu) =

D

Dt

∫ ∫ ∫

c.v.

ρudV = F =

∫ ∫ ∫

c.v.

fdV . (6.19)

orD

Dt(ρu) = f , (6.20)


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 revealed

asD

Dt(ρu) =

∂

∂t(ρu) + (u · ∇)ρu, (6.21)

There 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 forces

are 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

Dt(ρu) =

∂

∂t(ρu) + (u · ∇)ρu = fbody + fsurface (6.23)

This is the so-called Navier-Stokes equation. If only gravity appears in the

body force, then

fbody = ρg . (6.24)

Surface force can be shown in

fsurface = −∇p + µ∇2u . (6.25)

Substituting these terms into the N-S equation, we obtain

D

Dt(ρu) =

∂

∂t(ρu) + (u · ∇)(ρu) = ρg −∇p + µ∇2u . (6.26)

If the viscosity of fluids is ignored, then it will become

D

Dt(ρu) = ρg −∇p (6.27)

which is called the Euler equation.

6.4 Momentum Equation · 69 ·

6.5 Boundary Conditions

The continuity equation and the N-S equation constitute an initial-boundary

value problem. Hence approiate boundary and initial conditions are re-

quired to get a particular solution. Velocity and pressure are main variables

in the continuity euation and N-S equation. They have to be prescribed

before solving these equations. It is, however, not easy to obtain condi-

tions for pressure. For velocity, non-slip boundary condition is imposed in

the 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.


Chapter 7

DIMENSIONAL ANALYSIS

7.1 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

MLT is more often used than FLT, so it will be considered in the following

dimensional analysis.

Examples:

velocity: LT−1

acceleration: LT−2

force: MLT−2

density: ML−3

volume: L3

pressure: ML−1T−2

power: ML2T−3

work: ML2T−2

dynamic viscosity: ML−1T−1

kinematic viscosity: L2T−1

7.3 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

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(Aa2A

b3A

c4) (7.3)

if ℓ = 3. To make π1 dimensionless, check its dimensions in turn, i.e.

A1 = My1Ly2T y3 (7.4)

A2 = My4Ly5T y6 (7.5)

A3 = My7Ly8T y9 (7.6)

A4 = My10Ly11T y12 (7.7)

M : y1 + ay4 + by7 + cy10 = 0 (7.8)

7.3 How to carry out a dimensional analysis? · 73 ·

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),


kinematic variable: u(velocity),

dynamic variable: FD(drag on the sphere)

2. D : L

ρ : ML−3

µ : ML−1T−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 = FDDaubρc

M : 1 + 0 + 0 + c = 0

L : 1 + a + b − 3c = 0

T : −2 + 0 − b + 0 = 0

(7.13)

→

a = −2

b = −2

c = −1

(7.14)

∴ π1 = FDD−2u−2ρ−1 =FD

ρD2u2(7.15)


7. π2 = µDaubρc

M : 1 + 0 + 0 + c = 0

L : −1 + a + b − 3c = 0

T : −1 + 0 − b + 0 = 0

(7.16)

→

a = −1

b = −1

c = −1

(7.17)

∴ π2 = µD−1u−1ρ−1 =µ

ρuD(7.18)

8. As a result,

FD

ρu2D2= F (

µ

ρuD) (7.19)

→ FD

ρu2D2= F (

ρuD

µ) (7.20)

CD =FD

12ρu2A

: drag coefficient, A =π

4D2 (7.21)

Re =ρuD

µ: Reynolds number (7.22)

Ex: 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)


Dua

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=4 π products.

5. π1 = ∆PDaρbucavg

M : 1 + 0 + b + 0 = 0

L : −1 + a − 3b + c = 0

T : −2 + 0 + 0 − c = 0

(7.24)

→

a = 0

b = −1

c = −2

(7.25)

∴ π1 = ∆Pρ−1u−2avg =

∆P

ρu2avg

(7.26)

6. π2 = LDaρbucavg

M : 0 + 0 + b + c = 0

L : 1 + a − 3b − c = 0

T : 0 + 0 + 0 − c = 0

(7.27)

→

a = −1

b = 0

c = 0

(7.28)

∴ π2 =L

D(7.29)

7. π3 = ǫDaρbucavg

π3 =ǫ

D(7.30)

8. π4 = µDaρbucavg

π4 =µ

ρuavgD(7.31)


9.

∆P

ρu2avg

= F (L

D,

ǫ

D,

µ

ρuavgD) = F (

L

D,

ǫ

D,ρuavgD

µ) (7.32)

Cp =∆P

12ρu2

avg

, (pressure coefficient) (7.33)

Re =ρuavgD

µ, (Reynolds number) (7.34)

7.4 Common nondimensional parameters

Names of Parameters Formula Physical meanings

Reynolds number(Re) ρuLµ

inertia forceviscous force

Froude number(Fr) u√gL

inertia forcegravitational force

Euler number(Eu) P1

2ρu2

pressure forceinertia force

Mach number(Ma) uC

inertia forcecompressibility force

Strouhal number(St) fLu

Weber number(We)1

2ρu2L

σinertia force

surface tension force

7.5 Nondimensional Equations

Ex:

P

ρ+

1

2u2 + gh = const (7.35)

characteristic length: L (7.36)

characteristic velocity: u0 (7.37)

→ P

ρu20

+1

2

u2

u20

+gh

u20

= const (7.38)

→ P12ρu2

0

+u2

u20

+gh12u

20

= const (7.39)

7.4 Common nondimensional parameters · 79 ·

Ex:∂u

∂x+

∂v

∂y= 0 (7.40)

→ L

u0

∂u

∂x+

L

u0

∂v

∂y(7.41)

→ ∂u∗

∂x∗ +∂v∗

∂y∗= 0 (7.42)

u∗ =u

u0, v∗ =

v

u0, x∗ =

x

L, y∗ =

y

L(7.43)

7.6 Scale model tests

To establish similitude between a protype system and a model system,

geometric, kinematic, and dynamic similarities have to be considered.

1. Geometric SimilarityLm

Lp= const , (7.44)

where subscripts m and p refer to model and protype, respectively.

2. Kinematic Similarity

u1m

u1p=

u2m

u2p=

u3m

u3p= const (7.45)

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

umLm

νm=

upLp

νp

um = upLp

Lm

νm

νp. (7.47)


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.

(FD)p =1

2ρpupApCD

=1

2ρpupApF (Re)

=1

2ρpupApF (

ρpupDp

µp) , (7.49)

where F (Re) is obtained from the scale model test.

Distorted model: It is not always possible that all dynamic parameters

between a model and a protype are qual to others. For example, one cannot

promise dynamic similarities of Reynolds number and Frounde number at

7.6 Scale model tests · 81 ·

the same time, because

for Reynolds number

um = upLp

Lm

νm

νp(7.50)

and for Froude number

um = up

√

Lm

Lp. (7.51)

One has to chose one of them to perform dynamic similarity and ignore the

other one. When not all dynamic parameters are equal between a model

and a protype, it is the so-called distorted model.


Chapter 8

Viscous Internal Flow

We have investigated governing equations in differential forms for fluid

flow problems in Chapter 6. In addition, specific fluid flow problems such

as pipe flows, flows past obstacles and etc. will be studied. First of all,

viscous internal flows are discussed in this chapter. Internal fluid flows

refer to flows which are bounded by solid walls. For example, a pipe flow

is a typical internal flow. It is bounded by pipe walls.

8.1 Fully developed flow

Consider fluids flow into a pipe from a tank shown in Fig. 8.1. The flow is

uniform (U0) at the entrance of the pipe. The uniform veloctiy profile does

not retain as soon as fluids enter the pipe. Owing to viscosity, fluid are

still on the pipenwall and then velocity of fluids increases along the radial

direction. The viscous effect gradually affects velocity of fluids as fluids

move downstream. In the region near the entrance, the viscous effect is

not full of the pipe but appears near the pipe wall. The velocity in the

affected region is slower than the unaffected region. The affected region is

called a boundary layer. Mathematically, u = 0.99U0 is the artificial edge

83

U0

Boundary Layer

fully developedregion

u=0.99U0

δ

entrance

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 the

centre as fluids move downstream. Finally, the viscous effect is full of the

pipe and the flow in the region is called a fully developed flow.

8.2 Laminar, transition and turbulent flow

When fluid flows in a pipe, various patterns are found according to different

physical parameters, such as velocity, viscosity and pipe diameter. If we

dye a point in a pipe flow, then we will find that the streakline from the

point 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 fluids

travel downstream. This is the so-called laminar flow. If it is a distorted

line, it means fluid particles do mot move along the same line as they

· 84 · Viscous Internal Flow

Figure 8.2: Experimental appratus of Reynolds’ pipe flow.

Source: http : //www.eng.man.ac.uk/historic/reynolds/oreynB.htm

travel downstream but are disturbed. This is the so-called turbulent flow.

In 1833 A.D., O. Reynolds explained physical phenomena in pipe flows

using pipes of various diameters and a control value (see Fig. 8.2 and

8.2). He controled the flow rate across a pipe using the vale and dyed the

flow to visualize the pipe flow. He had same conclusions as the presvious

description.

8.3 2-D Poiseuille flow

Consider a fully developed laminar flow between two infinite parallel plates.

To analyze the laminar flow, assumptions are made to simplify the whole

problems. They are

8.3 2-D Poiseuille flow · 85 ·

Figure 8.3: Flow patterns of laminar, transition, and turbulent pipe flows.

Source: http : //www.eng.man.ac.uk/historic/reynolds/oreynB.htm

1. 2-D

2. steady flow

3. incompressible

4. v = 0

5. ignore gravity

The analytical solution can be obtained by solving the continuity equation

and the Navier-Stokes equations which are revealed as

∂u

∂x+

∂v

∂y= 0 , (8.1)

∂u

∂t+ u

∂u

∂x+ v

∂u

∂y= −1

ρ

∂P

∂x+ ν

(

∂2u

∂x2+

∂2u

∂y2

)

(8.2)

and∂v

∂t+ u

∂v

∂x+ v

∂v

∂y= −1

ρ

∂P

∂y+ ν

(

∂2v

∂x2+

∂2v

∂y2

)

. (8.3)

In additoin, he found that a very important nondimensional parameter,

ρuDµ

, which is highly relevant to the flow patterns. It is the well-known

Reynolds number. Pipe flows are classified as three kinds of patterns

according to Reynolds unmber, i.e.,


D

y

x CL

Figure 8.4: 2-D Poiseuille flow.

laminar flow: Re < 2,300

transition flow: 2,300 < Re < 4,000

turbulent flow: Re > 4,000.

In laminar flow regime, the pipe flow can be examined using analytical

methods. The laminar flow solution is called Poiseuille flow which is named

after J. L. M. Poiseuille. Transition and turbulent flows, however, cannot

be studied using analytical methods. Therefore, most of konwledge of

transition and turbulent flows come from experimental data.

First of all, we look into the continuity equation and get

∂u

∂x+

∂v

∂y= 0 (8.4)

and

u = u(y) . (8.5)

The continuity equation is satisified as u is a function of y. Subsequently,


the N-S equation in the y-direction is simplified as

∂v

∂t+ u

∂v

∂x+ v

∂v

∂y= −1

ρ

∂P

∂y+ ν

(

∂2v

∂x2+

∂2v

∂y2

)

. (8.6)

(8.7)

Since we assume that the flow is in a steady state, so ∂v∂t = 0, v = 0 and

we get∂P

∂y= 0 (8.8)

which 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

∂t+ u

∂u

∂x+ v

∂u

∂y= −1

ρ

∂P

∂x+ ν

(

∂2u

∂x2+

∂2u

∂y2

)

. (8.10)

Again, we assume that the flow is in a steady state, so ∂u∂t

= 0 , u = u(y) ,

v = 0 and we get

∂2u

∂y2=

1

µ

dP

dx. (8.11)

In addition,∂u

∂y=

1

µ

dP

dxy + C1,

(

dP

dx= const.

)

(8.12)

and consequently

u =1

µ

dP

dx

y2

2+ C1y + C2 . (8.13)

Appropriate boundary conditions are required to obtain arbitary con-

stants, C1 and C2, in the general solution. Since fluids are viscous, non-slip

boundary condition can be imposed on solid walls, i.e.

u

(

y =D

2

)

= 0 (8.14)


and

u

(

y = −D

2

)

= 0 . (8.15)

Using non-slip conditions, we find

u

(

y =D

2

)

=1

µ

dP

dx

D2

8+

C1D

2+ C2 = 0 (8.16)

and

u

(

y = −D

2

)

=1

µ

dP

dx

D2

8− C1D

2+ C2 = 0 , (8.17)

where C2 is found by adding these equations and we get

d

µ

dP

dx

D2

8+ 2C2 = 0 (8.18)

−→ C2 = −1

µ

dP

dx

D2

8. (8.19)

Substitution of C2 into Eq. (8.13) gives

1

µ

dP

dx

D2

8+

C1D

2− 1

µ

dP

dx

D2

8= 0 (8.20)

which leads to

C1 = 0 . (8.21)

The solution for velocity in laminar flow between two infinite parallel plates

is shown as

u(y) =1

µ

dP

dx

y2

2− 1

µ

dP

dx

D2

8

=1

µ

dP

dx· 1

2

[

y2 − D2

4

]

=1

2µ(−dP

dx) · D2

4

[

−(

2y

D

)2

+ 1

]

=D2

8µ(−dP

dx)

[

1 −(

2y

D

)2]

. (8.22)


It is obvious that velocity in a fully developed laminar flow between

two infinite parallel plates is parabolic. The maximum values of velocity

profile can be obtained by its first derivative, i.e.

du

dy=

D2

8µ(−dP

dx)

[

−2 · 2

Dy

]

= 0 . (8.23)

It happens at y = 0 and the maximum value is

umax =D2

8µ

(

−dP

dx

)

. (8.24)

Moreover, the flow rate per unit width in a cross area can be obtained by

integrating velocity, i.e.

Q = 2

∫ D2

0

udy

= 2D2

8µ

(

−dP

dx

)[

y − 4

D2

y3

3

]∣

∣

∣

∣

D2

0

= 2D2

8µ

(

−dP

dx

)[

D

2− 4

D2

1

3· D3

8

]

=D2

4µ

(

−dP

dx

)[

D

2− D

6

]

=D2

4µ

(

−dP

dx

)[

D

3

]

=D3

12µ

(

−dP

dx

)

. (8.25)

The averaged velocity is then found by

uavg =Q

A

=1

D· D3

12µ

(

−dP

dx

)

=D2

12µ

(

−dP

dx

)

. (8.26)

We find

uavg =2

3umax . (8.27)


In addition,

u

uavg=

3

2

[

1 −(

2y

D

)2]

. (8.28)

Viscous stress on the wall can be found by

τw = µdu

dy

∣

∣

∣

∣

y=±D2

=D2

8µ

(

−dP

dx

) [

− 4

D22y

]∣

∣

∣

∣

y=±D2

= µD2

8µ

(

−dP

dx

)(

− d

D2· D)

=D

2

(

−dP

dx

)

=6µuavg

D. (8.29)

The friction coefficient Cf will be

Cf =τw

12ρ(uavg)2

=µ 6

Duavg

12ρu2

avg

=12µ

ρDuavg=

12

Re. (8.30)

The friction factor will be

f =−dP

dxD

12ρu2

avg

=(−dP

dx)D

12ρ · uavg · D2

12µ(−dP

dx)

=24

Re. (8.31)

8.4 Hagen-Poiseuille flow

Poiseuille utilized analytical methods to get the solution of a laminar pipe

flow. He solved the continuity equation and N-S equations based on the

following assumptions:

8.4 Hagen-Poiseuille flow · 91 ·

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

∂r+

ur

r+

1

r

∂uθ

∂θ+

∂uz

∂z= 0 (8.32)

∂uz

∂z= 0, uz = uz(r)

The 2-D N-S equations in cylindrical coordinate system are denoted as

r-direction

∂ur

∂t+ ur

∂ur

∂r+

uθ

r

∂ur

∂θ+ uz

∂ur

∂z− u2

θ

r

= −1

ρ

∂P

∂r+ ν

[

1

r

∂

∂r

(

r∂ur

∂r

)

+1

r2

∂2ur

∂θ2+

∂2ur

∂z2− ur

r2− 2

r2

∂u0

∂θ

]

(8.33)

∂

∂t= 0, ur = 0, uθ = 0

−1

ρ

∂P

∂r= 0 (8.34)

P = P (z)only (8.35)

z-direction

∂uz

∂t+ ur

∂uz

∂r+

uθ

r

∂uz

∂θ+ uz

∂uz

∂z

= −1

ρ

∂P

∂z+ ν

[

1

r

∂

∂r

(

r∂uz

∂r

)

+1

r2

∂2uz

∂θ2+

∂2uz

∂z2

]

(8.36)


∂

∂t= 0, ur = 0, uθ = 0, uz = uz(r)

1

r

∂

∂r

(

r∂uz

∂r

)

=1

µ

dP

dz(8.37)

∂

∂r

(

r∂uz

∂r

)

= −1

µ

dP

dzr (8.38)

r∂uz

∂r= −1

µ

dP

dz

r2

2+ C1 (8.39)

∂uz

∂r= −1

µ

dP

dz

r

2+

C1

r(8.40)

uz = −1

µ

dP

dz

r2

4+ C1 ln r + C2 (8.41)

(8.42)

B.C.

uz(r =D

2) = 0

0 = −1

µ

dP

dz

D2

16+ C1 ln

D

2+ C2 (8.43)

C1 must be 0, because r cannot be 0 in ln r. As a result,

∴ C2 =1

µ

dP

dz

D2

16. (8.44)

Substitution of C2 into uz gives

uz = −1

µ

dP

dz

r2

4+

1

µ

dP

dz

1

4

(

D

2

)2

= −1

µ

dP

dz

1

4

(

r2 −(

D

2

)2)

= −1

µ

dP

dz

1

4

(

D2

4

)

[

(

2r

D

)2

− 1

]

= −1

µ

dP

dz

D2

16

[

(

2r

D

)2

− 1

]

=1

µ

(

−dP

dz

)

D2

16

[

(

2r

D

)2

− 1

]

. (8.45)


Shear stress on the pipe wall is determined by

τw = −µ∂uz

∂r

∣

∣

∣

∣

r=D2

= −µ · 1

µ

(

−dP

dz

)

D2

16

[

4

D2· 2r]

=dP

dz

D

4. (8.46)

In addition,

duz

dr=

1

µ(−dP

dz)D2

16

[

4

D2· r

2

]

(8.47)

The maximum velocity appears at

duz

dr= 0

−→ r = 0 .

It can be determined by

(uz)max = uz(r = 0) =1

µ

dP

dz

D2

16. (8.48)

The volumetric flow rate per unit width is given by

Q =

∫ r

0

uz2πrdr

=

∫ D2

0

1

µ(−dP

dz)D2

16

[

4

D2· 2πr3 − 2πr

]

dr

=1

µ(−dP

dz)D2

16

[

8π

D2

r4

4− 2π

r2

2

]∣

∣

∣

∣

D2

0

=1

µ(−dP

dz)D2

16

[

8π

D2

r4

4 · 16− π

D2

4

]

=1

µ(−dP

dz)D2

16

[π

8D2 − π

4D2]

=1

µ(−dP

dz)πD4

128. (8.49)


rzua

Figure 8.5: Parabolic profile of velocity component in the z-direction.

The averaged velocity is

(uz)avg =Q

A

=4

πD2· 1

µ(−dP

dz)D2

16

[

πD2

8− πD2

4

]

=1

µ(−dP

dz)D2

32. (8.50)

Furthermore,

(uz)avg

(uz)max=

132116

=1

2. (8.51)

We consider uz and uavg and get

uz

(uz)avg=

1

2

[

(

2r

D

)2

− 1

]

. (8.52)

It is a parabolic profile for the velocity component in the z-direction. Mean-

while,

τw =4µu

r0=

8µu

D(8.53)


f =4τw

12ρ(uavg)2

=64

Re(8.54)

Cf =τw

12ρ(uavg)2

=16

Re(8.55)

where f is the Darcy fricition factor.

8.5 Transition and turbulent pipe flows

Transition and turbulent pipe flows cannot be solved using analytical

methods due to extremely complicated physical phenomena. The only

approach to investigate transition and turbulent pipe flows is to conduct

experiments. L.F. Moody conducted pipe flow experiments and obtained

the 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 εD

Re

Figure 8.6: Schematic of Moody diagram.

1. Evaluate Reynolds number.


2. If Re < 2,300, then the formula f = 64

Re can be used for the laminar

pipee flow. If it is not, then evaluate relative roughness, εD .

3. Find the resultant relative riughness εD

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.

The Moody diagram also proves that f = 64

Re is correct in a laminar pipe

flow. In addition, some dashed lines are found between Re=2,000-5,000. It

is because those lines are in transition pipe flow. The details in transition

pipe 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 but

only depends on the relative roughness.

8.6 Darcy equation

The major loss comes from friction losses caused by pipe walls. Darcy

equation explains how to evaluate the major head loss, i.e.

hL = fL

D

u2avg

2g, (8.56)

where f must be found using the Moody diagram or the formula f = 64

Re

for a laminar flow. The minor loss comes from fittings such as valves,

elbows, expansions and so on. The minor loss is evaluated by

hL = Ku2

avg

2g, (8.57)

8.6 Darcy equation · 97 ·

where K is called the K-factor and depends on various fittings. The minor

loss can be combined with the major loss using the concept of equivalent

length. The equivalent length, le, is defined as

fle

D

u2avg

2g= K

u2avg

2g, (8.58)

le =K · D

f. (8.59)

Then the sum of major and minor losses will be

hL = f(L + le)

D

u2avg

2g. (8.60)

Hence the Bernoulli equation can be modified as

P1

r+

V 21

2g+ z1 =

P2

r+

V 22

2g+ z2 + hL . (8.61)

1 2

valve

Figure 8.7: Concept of equivalent length.


8.7 Hydraulic diameter

If the duct considered is not circular, then its hydraulic diameter can be

used. A hydraulic diameter is defined as

DH =4 × cross-sectional area

perimeter. (8.62)

8.8 Brief Introduction to Turbulence

u = < u > +u′ (8.63)

< u > = limT→∞

1

T

∫ t+T

t

udt (8.64)

u

t

Figure 8.8: Fluctuations in turbulent flow.

8.7 Hydraulic diameter · 99 ·

D laminar flow

turbulent flow

Figure 8.9: Difference between laminar and turbulent pipe flows.


Chapter 9

Viscous External Flows

External flow are not bounded by solid walls. In external flows, interaction

of fluids with solid structures is usually considered. In the past, viscosity of

a fluid was not considered in the potential flow theory. The drag predicted

by the potential flow theory for symmetrical bodies in a uniform flow is zero

but it is impossible. This is the so-called d’Alembert’s parabox. Hence, it

is obvious that viscosity plays a vital role in drag prediction.

9.1 Boundary Layer Theory

Prandtl, a German professor, provided the boundary layer concept. He

thought that viscosity affects fluid flows within a very thin region attached

the solid body. This region affected by viscosity is the well- known bound-

ary layer. Furthermore, he provided the non-slip condition to describle

fluid kinematic condition on solid walls. The boundary layer concept is

shown in Fig. 9.1. The artificial boundary layer starts from the front

of the solid body. Its thickness (u = 0.99Ue at y = δ) grows along the

downstream direction. In the begin, the flow in the boundary layer is

laminar. Traveling downstream, the boundary layer flow becomes transi-

101

u=0.99ueue

ue

ue

ue

laminarboundary

transtion turbulent

sepration

(inflectionpoint)

dudy =0

δ

stagnationpoint

xy

wake

Figure 9.1: Schematic of boundary layer concept.

tion and then turbulent (can be characterized by local Reynolds unmber,

Rex = ρuexµ ) due to disturbances from the surface of the body. If the ad-

verse pressure gradient happens in the boundary layer, then a separation

appears in the boundary layer. Subsequently, a wake is observed behind

the separation point.

9.2 Uniform flow past a flat plat

Consider a uniform flow past over a flat plate. The control volume concept

can be used to analyze the boundary layer flow. This idea was provided

by von Karman in 1921. For the mass conservation,

−∫ δ

0

Uedy +

∫ δ

0

udy +

∫ L

0

vdx = 0 (9.1)

− Ue · δ +

∫ δ

0

udy +

∫ L

0

vdx = 0 . (9.2)

· 102 · Viscous External Flows

For the momentum conservation,

Fx = −∫ δ

0

ρUeUedy +

∫ L

0

ρUevdx +

∫ δ

0

ρu2dy (9.3)

Fx

ρ= −U 2

e δ +

∫ L

0

U 2e δ +

∫ L

0

Uevdx +

∫ δ

0

u2dy (9.4)

From the mass conservation.

y

xδ

Ue

L

Figure 9.2: Schematic of boundary layer due to a uniform flow past a flat plate

∫ L

0

vdx = Ueδ −∫ δ

0

udy (9.5)

9.2 Uniform flow past a flat plat · 103 ·

Substitution of this equation into momentum equation gives

Fx

ρUe= −U 2

e δ + U δe δ − Ue

∫ δ

0

udy +

∫ δ

0

u2dy (9.6)

= Ue

∫ δ

0

[

u2

Ue− u

]

dy (9.7)

= Ue

∫ δ

0

u

[

u

Ue− 1

]

dy (9.8)

Fx

ρU 2e

=

∫ δ

0

u

Ue

[

u

Ue− 1

]

dy < 0 (9.9)

= −θ (9.10)

D = −Fx −→ DρU 2

e

= θ

The frictional force exerted by the boundary layer flow is obtained, but

it depends on the velocity profile within the boundary layer. The rest of

question is how to determine the velocity profile.

9.3 Boundary Layer Thickness, δ

The 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.


ρδd · Ue =

∫ ∞

0

ρ(Ue − u)dy (9.12)

δd =

∫ ∞

0

(1 − u

Ue)dy (9.13)

It means that mass flux the within δd is equal to the absence of mass flux

due to the presence of the boundary layer.

9.5 Momentum Boundary Layer Thickness, θ

ρθU 2e =

∫ ∞

0

ρ [(Ue − u)u] dy (9.14)

θ =

∫ ∞

0

[

u

Ue

(

1 − u

Ue

)]

dy (9.15)

It means that the momentum flux within θ is equal to the absence of

momentum flux due to the presence of the boundary layer.

von Karman made a guess of the velocity profile according to the bound-

ary conditions:

u(0) = 0, (9.16)

u(δ) = Ue (9.17)

and

∂u

∂y

∣

∣

∣

∣

y=δ

= 0 . (9.18)

He found that a second-order polynominal fits the conditions, i.e.

u ≈ Ue

(

2y

δ− y2

δ2

)

. (9.19)

9.5 Momentum Boundary Layer Thickness, θ · 105 ·

In terms of the velocity profile, the displacement thickness and the mo-

mentum thickness are

δd ≈δ

3and θ ≈ 2

15δ . (9.20)

In addition, the wall shear stress can be obtained by

τw = µ∂u

∂y

∣

∣

∣

∣

y=0

≈ 2µU

δ. (9.21)

Subsequently, the friction coefficient, Cf , can be found:

Cf =τw

12ρU 2

e

=2µUe

12ρU 2

e δ=

4µ

ρUeδ= 2

dθ

dx(9.22)

4µ

ρUeδ= 2

d

dx

(

2

15δ

)

=4

15

dδ

dx(9.23)

δdδ =15µdx

ρUe(9.24)

δ2 =30µx

ρUe(9.25)

δ

x≈ 5.5√

Rex

(9.26)

δ∗

x≈ 1

3

δ

x=

1.83√Rex

(9.27)

θ

x≈ 2

15

δ

x=

0.73√Rex

= Cf (9.28)

CD =2θ

L(9.29)

2Cf(L) =1.46√ReL

(9.30)

Althought 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.


9.6 Boundary Layer Equation

Prandtl provided the boundary layer equation which comes from the N-S

equation and on the following assumptions:

1. 2-D

2. steady

3. incompressible

4. ∂P∂y

= 0 −→ −dP

dx= U dU

dx

Consequently, N-S equations can be simplified as:

∂u

∂x+

∂v

∂y= 0 (9.31)

u∂u

∂x+ v

∂u

∂y= U

dU

dx+ ν

∂2u

∂y2(9.32)

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 the

similarity solution approach which is a common approach to transform a

P.D.E. to an O.D.E..

However, the transformed O.D.E. is nonlinear and impossible to obtain

an analytical solution. Hence a numerical method is reguired to obtain the

solution for the O.D.E. C. Toepfer (1912) used the Runge-Kutta method

to 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

δ ≈ 3.5

√

2νx

Ue(u = 0.99Ue) (9.33)

9.6 Boundary Layer Equation · 107 ·

or

δ

x≈ 5.0√

Rex

(Karman’s answer5.5√Rex

) (9.34)

The displacement thickness, δd, can be obtained as well and shown as

δd

x=

1.7208√Rex

(K’s answer5.5√Rex

) (9.35)

In addition, the momentum thickness, θ, is

θ

x=

0.604√Rex

(9.36)

The wall shear stress can be evaluated using the Newton’s viscosity law,

i.e.

τw = µ∂u

∂y

∣

∣

∣

∣

y=0

(9.37)

and the friction coefficient, Cf , will be

Cf(x) =τw

12ρU 2

e

=0.664√

Rex

=θ

x(9.38)

Finally, the drag coefficent, Cd, will be

Cd =D

12ρU 2

e L=

∫ L

0 τwdx12ρU 2

e L=

1.328√ReL

(K’s answer:1.46√ReL

) (9.39)

9.7 Friction coefficient, Cf

Cf =τw

12ρU 2

e

(9.40)

In boundary layer flow, the wall shear stress, Cf , is a function of its local

coordinates.


9.8 Drag coefficient, CD

CD =D

12ρU 2

e

(9.41)

D =

∫ L

0

τw(x)dx = ρU 2e θ (9.42)

τw(x) = ρU 2e

dθ

dx(9.43)

Cf = 2dθ

dx(9.44)

CD =2θ

L(9.45)

9.9 Drag

We mentioned drag calculation in a boundary layer flow. In fact, drag

calculation should consider friction drag caused by the boundary layer

and form drag caused by a wake. A low pressure region is generated owing

to a wake, so the pressure difference between the front part and the rear

part of an obstacle is formed. Hence a form drag is produced.

frictional drag −→ boundary layer

form drag −→ pressure difference −→ wake

9.10 Lift force and attack angle

9.11 Streamline body

A streamline body can reduce the wake region. In other words, most of

fluids are attached on the surface of the body. The resultant wake is very

small. Hence drag is mainly from friction within its boundary layer. In

contrust, if a body’s drag comes mainly from form drag. That means its

9.8 Drag coefficient, CD · 109 ·

L

α

stall

Figure 9.4: Stall

separation happens very near the front part and its wake, therefore, is very

large. Such a body is bluff. Some designs to let turbulence happens earlier

are made for bluff bodies. For example, a golf ball is given more roughness

on its surface to trigger turbulence in its boundary layer.

9.12 Separation

The separation phenomenon happens in a boundary layer as flows past

an 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 finally

becomes positive. It is called an adverse pressure gradient. When an


dPdPdx

dx dxdP>0

<0=0

favorableadverse separation

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. a

reverse flow appears. Then the separation happens.

9.13 Separation and Turbulence

As mentioned, separation is caused by an adverse pressure gradient due

to viscous energy dissipation. On the other hand, turbulence is able to

improve mixing in flows. The momentum mixing in a turbulent boundary

layer is better than in a laminar boundary layer. Therefore, momentum

near the wall within a turbulent boundary layer is higher than a lami-

nar boundary layer. This means that the net momentum in turbulent

boundary layer is higher than a laminar boundary layer. Hence an ad-

9.13 Separation and Turbulence · 111 ·

verse pressure gradient does not easily happens and the separation in a

turbulent boundary layer. This will make a wake smaller than without

turbulence.

Figure 9.6: Karman vortex street.

Owing to the feature of a turbulent boundary layer, the form drag can

be reduced. This is useful for a bulff body because its drag is mainly

dominated by form drag. Therefore another reason of the separation is

because of a sharp corner. Hence a sudden expansion or contraction shape

generates 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. It

was found by von Karman and name after him, Karman Vortex Street.

Eddy motion in a wake may be periodic, so a frequency may be found.

Strouhal number is the nondimensional unmber for the frequency.

St =f · D

u, (9.46)


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 of

the eddy motion is very close to or even equal to the natural frequency

of the obstacle, then the resonance will happen and cause a very serious

damage.

9.13 Separation and Turbulence · 113 ·

Fluid mechanics lectur notes

Documents

Transcript of Fluid mechanics lectur notes