Fluid mechanics
Compendium
V. Pavlenko, L. Rosenqvist,
O. Kochukhov
Department of Physics and Astronomy
Uppsala University
Version 11
September, 2016
PART I
Lecture Notes
Contents
1
Fundamental Concepts
1.1 Continuum Hypothesis . . . . . .
1.2 Field Representation . . . . . . .
1.2.1 Density Field . . . . . . .
1.2.2 Velocity Field . . . . . . .
1.2.3 Stress Field . . . . . . . .
1.2.4 Pressure Field . . . . . . .
1.2.5 Viscosity . . . . . . . . .
1.2.6 Surface Tension . . . . . .
1.3 Fluid Motions . . . . . . . . . . .
1.3.1 No-Slip condition . . . . .
1.3.2 Description of Fluid Flows
1.4 Dimensions and Units . . . . . . .
1.5 Summary objectives . . . . . . . .
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9
9
10
10
11
12
16
19
21
24
24
25
27
28
2
Elements of Thermodynamics
29
2.1 First law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.2 Second law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.3 Summary objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3
Kinematics of Fluids
3.1 Eulerian and Lagrangian Representations of Fluid Motion .
3.2 Transition . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Reynolds’ Transport Theorem . . . . . . . . . . . . . . .
3.5 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Summary objectives . . . . . . . . . . . . . . . . . . . . .
4
Basic Laws of Ideal Fluids
4.1 System of Equations . . . . . . . . . . . . . . . . .
4.1.1 Equation of Continuity . . . . . . . . . . . .
4.1.2 Euler Equation . . . . . . . . . . . . . . . .
4.1.3 Completeness of the System of Equations . .
4.2 The Statics of Fluids . . . . . . . . . . . . . . . . .
4.2.1 Basic Equations . . . . . . . . . . . . . . . .
4.2.2 Hydrostatic Equilibrium. Väisälä Frequency
4.3 Bernoulli’s theorem . . . . . . . . . . . . . . . . . .
4.3.1 Bernoulli’s Theorem . . . . . . . . . . . . .
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41
41
42
43
45
46
47
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49
49
50
50
52
54
54
56
60
60
CONTENTS
4.4
4.5
4.6
5
6
CONTENTS
4.3.2 Some Applications of Bernoulli’s Theorem . . . . . . . . . . . . . . .
4.3.3 Bernoulli Theorem as a Consequence of the Energy Conservation Law .
4.3.4 Energy Conservation Law in the General Case of Unsteady Flow . . . .
Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 The Specific Momentum Flux Tensor . . . . . . . . . . . . . . . . . .
4.4.2 Euler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.3 Application of Euler’s Theorem . . . . . . . . . . . . . . . . . . . . .
Motions of Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.1 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.2 Circulation of Velocity . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.3 Kelvin’s Circulation Theorem . . . . . . . . . . . . . . . . . . . . . .
4.5.4 Discussion of Kelvin’s Theorem and Helmholtz Theorems . . . . . . .
4.5.5 Vorticity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Summary objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Potential Flow
5.1 Equations for a Potential Flow . . . . . . . . . . . .
5.1.1 Velocity Potential . . . . . . . . . . . . . . .
5.1.2 Two-Dimensional Flow. Stream Function . .
5.2 Applications . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Complex Flow Potential . . . . . . . . . . .
5.2.2 Some Examples of Two-Dimensional Flows .
5.2.3 Conformal Mapping . . . . . . . . . . . . .
5.3 Steady Flow Around a Cylinder . . . . . . . . . . .
5.3.1 Application of Conformal Mapping . . . . .
5.3.2 The Pressure Coefficient . . . . . . . . . . .
5.3.3 The Paradox of d’Alembert and Euler . . . .
5.3.4 The Flow Around a Cylinder with Circulation
5.4 Irrotational Flow Around a Sphere . . . . . . . . . .
5.4.1 The Flow Potential and the Particle Velocity .
5.4.2 The Induced Mass . . . . . . . . . . . . . .
5.5 Summary objectives . . . . . . . . . . . . . . . . . .
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Flows of Viscous Fluids
6.1 Basic Equations . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Newtonian Viscosity and Viscous Stresses . . . .
6.1.2 The Navier-Stokes Equation . . . . . . . . . . .
6.1.3 The Viscous Force . . . . . . . . . . . . . . . .
6.2 Examples of viscous flow . . . . . . . . . . . . . . . . .
6.2.1 Couette Flow . . . . . . . . . . . . . . . . . . .
6.2.2 Plane Poiseuille Flow . . . . . . . . . . . . . . .
6.2.3 Poiseuille Flow in a Cylindrical Pipe . . . . . . .
6.2.4 Viscous Fluid Flow Around a Sphere . . . . . .
6.2.5 Stokes Formula for Drag . . . . . . . . . . . . .
6.2.6 Laminar and Turbulent Flows . . . . . . . . . .
6.3 Boundary Layer . . . . . . . . . . . . . . . . . . . . . .
6.3.1 Viscous Waves . . . . . . . . . . . . . . . . . .
6.3.2 The Boundary Layer. Qualitative Considerations
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62
64
64
66
66
67
69
69
71
72
73
75
77
78
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81
81
81
84
88
89
89
92
93
94
96
97
98
102
102
104
105
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107
107
107
110
112
113
113
116
118
121
123
124
125
126
128
CONTENTS
6.4
7
8
9
CONTENTS
Summary objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
Dimensional Analysis
7.1 Dimensionless Groups . . . . . . . . . . . . . . . . . . . .
7.2 Similar Flows . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.1 Simple example . . . . . . . . . . . . . . . . . . . .
7.2.2 Similarity Principle of Reynolds . . . . . . . . . . .
7.2.3 Important Dimensionless Groups in Fluid Mechanics
7.3 Repeating Variables . . . . . . . . . . . . . . . . . . . . . .
7.3.1 Complete Dynamic Similarity. Simple Example . . .
7.3.2 Drag Force on Smooth Sphere . . . . . . . . . . . .
7.3.3 Pressure Drop in Pipe Flow . . . . . . . . . . . . .
7.4 Summary objectives . . . . . . . . . . . . . . . . . . . . . .
Theory of Turbulence
8.1 Qualitative Considerations . . . . . . . . . . . . . .
8.1.1 Transition from a Laminar to Turbulent Flow
8.1.2 Flow Around a Cylinder at Different Re . . .
8.2 Statistical Description . . . . . . . . . . . . . . . . .
8.2.1 Reynolds Equation for Mean Flow . . . . . .
8.2.2 Turbulent Viscosity . . . . . . . . . . . . . .
8.2.3 Turbulent Boundary Layer . . . . . . . . . .
8.3 Locally Isotropic Turbulence . . . . . . . . . . . . .
8.4 Similarity Hypothesis . . . . . . . . . . . . . . . . .
8.5 Summary objectives . . . . . . . . . . . . . . . . . .
Surface and Internal Waves
9.1 Stratified Fluids . . . . . . . . . . . . . . . . . . . .
9.1.1 Linearisation of the Hydrodynamic Equations
9.1.2 Linear Boundary Conditions . . . . . . . . .
9.1.3 Equations for an Incompressible Fluid . . . .
9.2 Surface Gravity Waves . . . . . . . . . . . . . . . .
9.2.1 Basic Equations . . . . . . . . . . . . . . . .
9.2.2 Harmonic Waves . . . . . . . . . . . . . . .
9.2.3 Harmonic Surface Waves . . . . . . . . . . .
9.2.4 Shallow and Deep Water Approximations . .
9.2.5 Wave Energy . . . . . . . . . . . . . . . . .
9.3 Capillary Waves . . . . . . . . . . . . . . . . . . . .
9.3.1 “Pure” Capillary Waves . . . . . . . . . . .
9.3.2 Gravity-Capillary Surface Waves . . . . . . .
9.3.3 Waves on a Liquid-Fluid Interface . . . . . .
9.4 Internal Gravity Waves . . . . . . . . . . . . . . . .
9.4.1 Basic Equations. Boussinesq Approximation
9.4.2 Waves in Unlimited Media . . . . . . . . . .
9.5 Summary objectives . . . . . . . . . . . . . . . . . .
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133
133
134
134
136
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142
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149
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151
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155
157
158
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165
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167
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193
10 Stability of Fluid Flows
10.1 Stability and Instability .
10.2 Stability of Steady Flow
10.3 Kelvin-Helmholtz . . . .
10.4 Rayleigh-Taylor . . . . .
10.5 Summary objectives . . .
11 Appendix I: Vector Calculus
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195
195
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201
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207
Introduction
Mechanics is the oldest physical science that deals with both stationary and moving bodies under
the influence of forces. The branch of mechanics that deals with bodies at rest is called statics,
while the branch that deals with bodies in motion is called dynamics. The study of motion of
matter can be divided into the dynamics of rigid bodies and non-rigid bodies. The latter is usually
further divided into elasticity (solid elastic bodies) and fluid mechanics. The subcategory fluid
mechanics is defined as the science that deals with the behaviour of fluids at rest (fluid statics)
or in motion (fluid dynamics), and the interaction of fluids with solids or fluids at the boundaries.
The dynamics of a fluid differ from that of elastic bodies by some important features:
i) An ideal (inviscid) fluid does not support shearing stresses.
ii) Displacements of particles in a fluid flow may be large even when forces are small.
iii) Viscosity (internal friction) fundamentally influences the fluid flow.
In simple terms this means that a fluid is a substance that can offer no permanent resistance to
any force and thus flows under its own weight and takes the shape of any solid body which it is
in contact with. So the definition of a fluid is that a fluid is a substance that deforms continuously
under the application of a shear stress no matter how small the shear stress may be. Fluids can
be classified as liquids or gases. A liquid is difficult to compress (i.e. density varies little with
temperature and pressure) and will fill a container to which it is poured up to the volume of the
liquid regardless of the container’s shape and a “free surface” is formed as a boundary between
the liquid and the air above it. A gas, on the other hand, is easily compressed (the mass, pressure,
temperature and volume are related by the appropriate equation of state of the gas) and will fill
any container into which it is placed.
There are numerous applications of fluid mechanics making it one of the most fundamental of all
engineering and applied science studies. The flows of fluids in pipes and channels are important
to civil engineers. The study of fluid machinery such as pumps, compressors, heat exchangers,
jet and rocket engines etc. illustrates the importance of fluid mechanics to mechanical engineers.
The flow of air over objects, aerodynamics, is of fundamental interest to aeronautical and space
engineers in the design of aircraft, missiles and rockets. Furthermore, the study of fluids is a
basic fundament in meteorology, hydrology and oceanography since the atmosphere and ocean
are fluids. Numerous natural phenomena such as the rain cycle, weather patterns, the rise of
ground water to the top of trees, winds, ocean waves, and currents in large water bodies are also
governed by the principles of fluid mechanics. In addition, fluid mechanics can be combined
with other classical disciplines. For example, fluid mechanics and electromagnetic theory are
studied together as magnetohydrodynamics as you can learn more about in the course Plasma
Physics.
5
CONTENTS
CONTENTS
This compendium aims at providing a solid physical ground to the introduction into fluid mechanics for advanced undergraduate and beginning graduate students in the physical and engineering sciences. The necessary conditions for accomplishing this goal are: (1) a clear, concise
presentation of the fundamentals that you, the student, can read and understand, and (2) your
willingness to read the text material before going to class. We have assumed responsibility for
meeting the first condition. You must assume responsibility for satisfying the second condition. The philosophy of the compendium is to be as rigorous as possible, but, when necessary,
to make approximations that will retain the essential physics and at the same time will be sufficiently simple that solutions to the problems of interest can be given by closed analytic functions.
Even though these solutions are not exact, this approach can be justified by the fact that exact
numerical solutions are often not useful and are much less convenient than approximate analytic
solutions. For example, to evaluate the effect of varying a parameter, a numerical solution must
be repeated many times, whereas if an analytical expression is available, often the effect can be
ascertained by a simple analysis.
The compendium assumes that the reader has had at least an introductory course in physics,
and an adequate background in engineering mechanics (courses in statics and dynamics) as well
as in thermodynamics, plus calculus through differential equations. It is also assumed that students are slightly familiar with vectors and complex variables. For those whose knowledge of
mathematics is limited, brief review of some aspects of vector calculus that will be needed in
this compendium has been included into Appendix I. No prior study of continuum mechanics is
required.
The compendium is divided into three parts mutually connected to each other. The first part,
consisting of Chapters 1–10, provides an accessible and rather comprehensive account of the
subject, emphasising throughout the fundamental physical principles, and stressing the connections with other branches of physics. It starts with fundamental concepts of fluids and fluid
flows and ends with fluid instabilities. Chapter 1 provides the physical properties of fluids and
classification of fluid flows. Elements of thermodynamics important for the treatment of fluid
mechanics phenomena are presented in Chapter 2. Chapter 3 covers topics related to fluid kinematics. Chapters 4–5 are devoted to the treatment of ideal fluid (no internal friction). Here the
basic equations of fluid dynamics governing the motion of a fluid and related to the fundamental
conservation laws of mass, momentum and energy are derived and a number of different aspects
of vortex and potential flows of ideal fluids are presented. These results lay a foundation for the
physics of viscous fluids and the transition from laminar (straight) to turbulent flows which are
presented in Chapters 6–8. Chapter 9 is devoted to the treatment of an important kind of fluid
motions, waves. Surface and internal gravity waves are considered in the simplest, i.e. linear,
approximation and the emphasis in this treatment is mainly on the fluid dynamical properties of
the wave processes. The last chapter, Chapter 10, aims at to present an introduction to linear
stability theory of fluid flows. The mathematical complexity of even this simplest type of theory
is too great to treat any particular case in full detail. Therefore an emphasis is placed here on
the physical insight on the dynamics of the destabilising process which is very valuable and can
be illustrated by considering highly simplified systems. Then, after a general discussion of how
stability principles are extended for full fluid dynamical situation, we investigate two examples
of an unstable fluid flow. The first one is the shear flow instability known also as the KelvinHelmholtz instability, and the second one is the Rayleigh-Taylor instability.
The second part covers problems to accompany the first part. All problems are keyed directly to
6
CONTENTS
CONTENTS
a particular section or topic of fluid mechanics. At the beginning of each Problems section, one
can find a “Memorandum” which provides main formulas and useful framework within which
later solutions can be set. The third part consists of three labs included in this course. The labs
aim at providing the students with practical knowledge and skills in numerical simulation of
different phenomena observed in fluid flows. It is worthwhile to note here that computational
fluid dynamics, even restricted to particular application, is a vast area and one we cannot hope to
cover in one chapter. Nevertheless, hopefully it gives the reader a springboard for exploring the
subject in the future.
7
CONTENTS
CONTENTS
8
Chapter 1
Fundamental Concepts
In the Introduction we indicated that our study of fluid mechanics will build on earlier studies in
mechanics and thermodynamics. Analysis of any problem in fluid mechanics necessarily begins,
either directly or indirectly, with statements of the basic laws governing the fluid motion. The
basic laws which are applicable to any fluid are:
• The conservation of mass.
• Newton’s laws of motion.
• The principle of angular momentum conservation.
• The first law of thermodynamics.
• The second law of thermodynamics.
Clearly, not all basic laws are required to solve any one problem. On the other hand, in many
problems it is necessary to bring into the analysis additional relations in order to understand
the nature of fluid flows. To develop a unified approach, we review some familiar topics and
introduce some new concepts and definitions. The purpose of this chapter is to describe these
fundamental concepts.
1.1
Continuum Hypothesis
Fluid dynamics concerns itself with the study of the motion of fluids (liquids and gases). In
our definition of a fluid, no mention was made of the molecular structure of matter. All fluids
are composed of molecules in constant motion. However, in most engineering applications we
are interested in the average or macroscopic effects of many molecules. It is these macroscopic
effects that we ordinarily perceive and measure. We thus treat a fluid as an infinitely divisible substance with no holes, a continuum, and do not concern ourselves with the behaviour of
individual molecules. We shall suppose, throughout of this compendium, that the fluid’s macroscopic behaviour is the same as if it was perfectly continuous in structure. Furthermore, we will
also assume that all physical quantities such as e.g. mass and momentum associated with the
matter contained within a given small volume will be regarded as being spread uniformly over
that volume instead of, as in strict reality, being concentrated in a small fraction of it. This means
that any small volume element in the fluid is always supposed so large that it still contains a very
great number of molecules. Accordingly, when we speak of infinitely small elements of volume,
we shall always mean those which are “physically” infinitely small, i.e. very small compared
9
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
with the volume of the body under consideration, but large compared with the distances between
the molecules.
To have a sense of the distances involved at the molecular level, consider a container filled with
oxygen at atmospheric conditions. The diameter of the oxygen molecule is about 3 × 10−10 m
and its mass is 5.3 × 10−26 kg. Also, the mean free path of oxygen at 1 atm pressure and 20◦ C
is 6.3 × 10−8 m. That is an oxygen molecule travels, on average, a distance of 6.3 × 10−8 m
(about 200 times its diameter) before it collides with another molecule. Despite the large gaps
between molecules, a substance can be treated as a continuum because of the very large number
of molecules even in an extremely small volume. The continuum model is thus applicable as
long as the characteristic length of the system (such as its diameter) is much larger than the mean
free path of molecules. The expressions fluid particle and point of a fluid are to be understood
in a similar sense. If, for example, we speak of the displacement of some fluid particle, we mean
not the displacement of an individual molecule, but that of volume element containing many
molecules, though still regarded as a point. Once the continuum hypothesis has been introduced,
we can formulate the equations of motion on a continuum basis, and the molecular structure of
the fluid need not be mentioned any more.
1.2
Field Representation
The continuum hypothesis implies that it is possible to attach a definite meaning to the notion
of value “at a point” of the various fluid properties such as density, velocity and temperature,
and that in general the values of these quantities are continuous functions of position in the fluid
and of time with no discontinuity jumps. On this basis we shall be able to establish equations
governing the motion of the fluid which are independent, so far as their form is concerned, of the
nature of the particle structure — so that gases and liquids are treated together — and indeed,
independent of whether any particle structure exists.
1.2.1
Density Field
To illustrate the concept of a property at a point, consider the manner in which we determine
the density at a point. A fluid volume of mass m is shown in Figure 1.1a. We are interested in
determining the density at the point C, whose coordinates are x0 , y0 , z0 . The density is defined
as mass per unit volume. Thus the mean density within the given volume V would be given by
ρ= m
. In general, this will not be equal to the value of density at point C. To determine the
V
density at point C, we must select a small volume δV surrounding point C and determine the
ratio δm
. The question is how small can we make the volume δV ? Let us answer this question
δV
and allowing the volume to shrink continuously in size. Assuming that
by plotting the ratio δm
δV
the volume δV is initially relatively large (but still small compared with the volume V ) a typical
plot of δm
might appear as in Figure 1.1b. In other words, δV must be sufficiently large to yield
δV
a meaningful, a reproducible value for density at a location and yet small enough to be able to
resolve spatial variations in density. The average density tends to approach an asymptotic value
as the volume is shrunk to enclose only homogeneous fluid in the immediate neighbourhood of
point C. When δV becomes so small that it contains a small number of molecules, it becomes
; the value will vary erratically as molecules cross into
impossible to fix a definite value for δm
δV
and out of the volume. Thus there is a lower limiting value of δV , designated δV1 in Figure 1.1b,
allowed to use in defining the fluid density at a point. The density at a “point” is then defined as
10
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
b)
a)
y
Volume, δV
of mass δm
Volume, V
of mass m
y0
δm
δV
C
ρ lim δm
δV --- δV1 δV
x0
x
δV1
δV
z0
z
Figure 1.1: Definition of a density at a point.
δm
(1.1)
δV →δV1 δV
Since point C was arbitrary, the density at any point in the fluid could be determined in a similar
manner. The reciprocal of density is the specific volume which is defined as volume per unit
V
= ρ1 .
mass. That is ν = m
The density of a substance, in general, depends on temperature and pressure. The density of
most gases is proportional to pressure and inversely proportional to temperature. Liquids and
solids, on the other hand, are essentially incompressible substances, and the variation of their
density with pressure is usually negligible. At 20◦ C, for example, the density of water changes
from 998 kg/m3 at 1 atm to 1033 kg/m3 at 100 atm, a change of just 0.5 percent. The density of
liquids and solids depends more strongly on temperature than it does on pressure. At 1 atm, for
example, the density of water changes from 998 kg/m3 at 20◦ C to 975 kg/m3 at 75◦ C, a change
of 2.3 percent, which can still be neglected in many engineering analyses.
If density determinations were made simultaneously at an infinite number of points in the fluid,
we would obtain an expression for the density distribution as a function of space coordinates,
ρ = ρ (x, y, z), at the given instant in time. Clearly, the density at a point may vary in time
as a result of work done on or by the fluid and/or heat transfer to the fluid. Thus the complete
representation of density (the field representation) is given by
ρ ≡ lim
ρ = ρ (x, y, z, t)
(1.2)
Since density is a scalar quantity requiring only the specification of a magnitude for a complete
description, the field represented by (1.2) is a scalar field.
1.2.2
Velocity Field
The continuum assumption leads directly to the notion that other fluid properties also may be
described by field representation. In dealing with fluids in motion, we shall necessarily be con11
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
cerned with description of a velocity field. Refer again to Figure 1.1a. Define the fluid velocity
at point C as the instantaneous velocity of the center of gravity of volume δV surrounding point
C. If we define a fluid particle as a small mass of fluid of fixed volume δV1 , then the velocity
at point C is defined as the instantaneous velocity of the fluid particle which, at a given instant,
is passing through point C. The velocity at any point in the flow field is defined similarly. At a
given instance the velocity field, v, is a function of space coordinates x, y, z. The velocity at any
point in the flow field might vary from one instant to another. Thus the complete representation
of velocity (the velocity field) is given by
v = v (x, y, z, t)
(1.3)
The velocity vector, v, can be written in terms of its three scalar components. Denoting the
components in the x, y and z directions by u, v, and w, then
v = ui + vj + wk
(1.4)
In general, each of the components, u, v, and w will be a function of x, y, z, and t.
If the properties at every point in a flow field do not change with time, the flow is termed steady.
Stated mathematically, the definition of steady flow is
∂β
=0
∂t
where β represents any fluid property. So, for a steady flow we have
∂ρ
= 0 or ρ = ρ (x, y, z)
∂t
and
∂v
= 0 or v = v (x, y, z)
∂t
Thus, in steady flow, any property may vary from point to point in the field, but all properties
remain constant with time at every point.
1.2.3
Stress Field
It is known from classical mechanics (statics) that stress is defined as force per unit area and is
determined by dividing the force by the area upon which it acts. The normal component of the
force acting on a surface per unit area is called normal stress, and the tangential component of
force acting on a surface per unit area is called shear stress (Figure 1.2) In a fluid at rest, the
normal stress is called pressure. A fluid at rest is at a state of zero shear stress. When walls
are removed or a liquid container is tilted, a shear develops as the liquid moves to re-establish a
horizontal free surface.
Surface and body forces are encountered in the study of continuum fluid mechanics. Surface
forces act on the boundaries of a medium through direct contact. Forces developed without
physical contact, and distributed over the volume of the fluid are termed body forces. These
12
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
b)
a)
n
δFn
δA
C
δF
C
δA
δF
δFt
Figure 1.2: The concept of stress in a continuum.
forces (known also as long-range forces) decrease slowly with increasing distance between interacting elements and are still appreciable for distances characteristic of natural fluid flows.
Such forces are capable of penetrating into the interior of the fluid, and act on all elements of
the fluid. Gravity is the obvious and most important example, but two other kinds of long-range
forces of interest in fluid mechanics are the electromagnetic and fictitious forces. The electromagnetic forces may act when the fluid carries an electric charge or when an electric current
passes through it. The fictitious forces, such as centrifugal force, appear to act on fluid elements
when their motion is referred to an accelerating set of axes. A consequence of a slow variation
of the long-range forces with position of the fluid element on which the force is acting is that the
force acts equally on all matter within a small fluid volume. Thus, the total force is proportional
to the size of the volume of the fluid element.
In contrast, surface forces (known also as short-range forces) have a direct molecular origin
and decrease extremely rapidly with increase of distance between interacting elements. They are
negligible unless there is direct mechanical contact between the interacting elements, as in the
case of the reaction between two rigid bodies, because without that contact none of the molecules
of one of the elements is sufficiently close to a molecule of the other element. If a fluid element
is acted on by short-range forces arising from reactions with matter outside this element, these
short-range forces can act only on a thin layer adjacent to the boundary of fluid element of thickness equal to the “penetration” depth of the forces. The total of the short-range forces acting on
the fluid element is thus determined by the surface area of the element, and the volume of the
element is not directly relevant.
Thus, the concept of stress provides convenient means to describe the manner in which forces
acting on the boundaries of the medium are transmitted through the medium. The way in which
body forces depend on the local properties of the fluid is evident, at any rate in the cases of
gravity and the fictitious forces due to accelerating axes, but the dependence of surface forces on
the local properties and motion of the fluid will require examination.
13
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
Imagine any fluid element within a flowing fluid such as in Figure 1.2. The different parts of
a closed surface bounding a fluid element have different orientations. Thus, it is not useful to
specify the short-range forces by their total effect on a finite volume element of fluid (as in Figure 1.2a). Instead we consider a plane surface element in the fluid and specify the local contact
force δF as the total force exerted on the fluid on one side of the element by the fluid on the
other side (Figure 1.2b). Consider a portion δA of the surface acted upon in the neighbourhood
surrounding point C. Provided the penetration depth of the short-range forces is small compared
with the linear dimensions of plane surface element, this total force exerted across the element
will be proportional to its area δA = δAn, where the orientation of δA is given by the unit
vector n. The vector n is the outwardly drawn unit normal to the plane surface element.
The same results can be obtained on the base of more strong considerations. Indeed, the total force on a given fluid element can be written as volume integral
Z
F dV
where F is the forceRper unit volume and F dV is the force on the volume element dV . Each
of three components Fi dV of the resultant of all the internal stresses can be transformed into
an integral over the surface. As it is known from vector analysis, the integral of a scalar over
an arbitrary volume can be transformed into an integral over the surface if the scalar is the
divergence of a vector. In the present case, we have the integral of a vector, and not of scalar.
Hence the vector Fi must be the divergence of a tensor of rank two, i.e. be of the form
Fi =
∂Tik
∂xk
Then the force on any volume can be written as an integral over the closed surface bounding that
volume
Z
Z
I
∂Tik
Fi dV =
dV = Tik dAk
∂xk
The tensor Tik is called the stress tensor. As it is seen from this expression, Tik dAk is the i-th
component of the force on the surface element dA. By taking elements of area in the planes of
xy, yz, zx, we find that the component Tik of the stress tensor is the i-th component of the force
on unit area perpendicular to the xk -axis. For instance, the force on unit area perpendicular to
the x-axis, normal to the area (i.e. along the x-axis) is Txx , and tangential forces (along the y and
z axes) are Tyx and Tzx .
The force δF acting on δA may be resolved into two components, one normal to and the other
tangent to the area. The normal stress εn and the shear stress σn are then defined as
δFn
δAn →0 δAn
εn = lim
and
(1.5)
δFt
(1.6)
δAn
Subscript n on the stress is included as a reminder that the stresses are associated with the surface
δA through C, having an outer normal in the n direction. For any other surface through C the
values of the stresses could be different.
In dealing with vector quantities such as forces, it is customary to consider components in an
orthogonal coordinate system. In rectangular coordinates we might consider the stresses acting
σn = lim
δAn →0
14
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
on planes whose outwardly drawn normals are in the x, y, and z directions. As an example in
Figure 1.3, we consider the stress on the element δAx , whose outwardly drawn normal is in the
x direction. The force δF has been resolved into components along the coordinate directions.
Dividing the magnitude of each force component by the area δAx , and taking the limit as δAx
approaches zero, we define the three components shown in Figure 1.3b:
δFx
δAx →0 δAx
lim δFy ,
δA →0 δAx
εxx = lim
σyx =
(1.7)
δFz
δAx →0 δAx
σzx = lim
x
We have used double subscript notation (known also as tensor notations) to label the stresses.
b)
a)
y
y
δFy
C
σyx
C
δFx
εxx
σzx
δFz
x
z
x
z
Figure 1.3: (a) Force components and (b) stress components, on the element of area δAx .
The first subscript indicates the direction in which the stress acts. The second subscript (in this
case x) indicates the plane on which the stress acts (in this case, a surface perpendicular to the x
axis, i.e. yz -plane). Similarly, consideration of area element δAy would lead to the definitions
of the stresses εyy , σxy , σzy , and use of the area element δAz would lead to the definitions of the
stresses εzz , σxz , σyz . So, εij (or σij ) is the i-component of the force per unit area exerted across
a plane surface element normal to the j-direction, at position x (position of the point C) in the
fluid and at time t.
An infinite number of planes can be passed through point C, resulting in an infinite number
of stresses associated with that point. Fortunately, the state of stress at a point can be described
completely and uniquely by specifying the stresses acting on three mutually perpendicular planes
through the point. The stress at a point is specified by the nine components


εxx σxy σxz
Tij =  σyx εyy σyz 
(1.8)
σzx σzy εzz
where ε has been used to denote a normal stress, and shear stresses are denoted by σ. Equation
(1.8) is also known as the definition of the stress tensor. The three diagonal components of Tij ,
15
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
i.e. εxx , εyy , εzz are normal stresses in the sense that each of them gives the normal component of
surface force acting across a plane surface element parallel to one of the co-ordinate planes. The
six non-diagonal components of Tij are tangential stresses, sometimes also called shear stresses,
since in fluids (as well as in solids) they arise due to a shearing displacement or motion in which
parallel layers of matter slide relative to each other. Hence, in a fluid at rest the tangential stresses
are zero.
Referring to the infinitesimal fluid element shown in Figure 1.4, we see that there are six planes
(two x planes, two y planes, and two z planes) on which stresses may act. In order to designate
the plane of interest, we could use terms like front and back, top and bottom, or left and right.
However it is more logical to name the planes in terms of coordinate axes. The planes are named
and denoted as positive or negative according to the direction of the outwardly drawn normal to
the plane. Thus the top plane, for example, is a positive y-plane and back plane is a negative
z-plane. It is necessary to adopt a sign convention for the stress. A stress component is positive
when the direction of the stress component and the plane on which it acts are both positive or
both negative. For example, σyx = 30 (N/m2 ) represents a shear stress on a positive x-plane in
the positive y-direction, or a shear stress on a negative x-plane in the negative y-direction.
y
ε yy
σ xy
σzy
ε zz
σxz
σzx
σyz
εxx
∆y
σyz
ε zz
εxx
σzx
σ xz
σyx
σyx
x
σzy
σxy
∆z
ε yy
∆x
z
Figure 1.4: Notation for stress.
1.2.4
Pressure Field
Important information about forces applied by fluids at rest may be deduced from the form of
the stress tensor Tij (r) in a fluid at rest. In this case, as it was mentioned above, the tangential
stresses are zero and only normal stresses are non vanishing. Moreover, it can be shown by
considering the surface forces exerted on the fluid within a small sphere by surrounding fluid
that these normal stresses are all the same at all points in the fluid. That is, the stress tensor in a
fluid at rest is everywhere isotropic. Fluids at rest are normally in a state of compression, and it
16
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
is therefore convenient to write the stress tensor Tij (r) as
Tij (r) = −pδij , δij =
1, if i = j
0, if i 6= j
(1.9)
where p may be termed the static fluid pressure and is in general a function of r. It follows that
in a fluid at rest the contact force per unit area exerted across a plane surface element in the fluid
with unit normal n is (−pn), and is a normal force of the same magnitude for all directions of the
normal n at a given point. This is a well-known property of the static fluid pressure meaning that
it is “acting equally in all directions”. It is often established as a consequence of an assumption
that in a fluid at rest, the tangential stresses are zero. So, we may conclude that the fluid property
responsible for forces applied by fluids at rest is the pressure (or normal stress) which is defined
as a normal force exerted by a fluid per unit area.
Pressure is a normal force exerted by a fluid per unit area.
We speak of pressure only when we deal with a gas or a liquid. The counterpart of pressure
in solids is normal stress. Since pressure is defined as force per unit area, it has the unit of newtons per square meter (N/m2 ), which is called a pascal (Pa).
Pressure is the compressive force per unit area, and it gives the impression of being a vector.
However, the pressure at any point in a fluid is the same in all directions. That is, it has magnitude but not a specific direction, and thus it is a scalar quantity. In other words, the pressure at a
point in a fluid has the same magnitude in all directions. It can be shown in the absence of shear
forces that this result is applicable to fluids in motion as well.
It will come as no surprise that the pressure in a fluid at rest does not change in the horizontal direction. This can be proved easily by considering a thin horizontal layer of the fluid and
doing a force balance in any horizontal direction. However, this is not the case in the vertical
direction in a gravity field. Pressure in a fluid increases with depth because more fluid rests on
deep layers, and the effect of this “extra weight” on a deeper layer is balanced by an increase in
pressure, Figure 1.5a. To obtain a relation for the variation of pressure with depth, consider a
rectangular fluid element of height ∆z, length ∆x, and unit depth (∆y = 1 unit into the page) in
equilibrium, as shown in Figure 1.5b. Assuming the density of the fluid ρ to be constant, a force
balance in the vertical z-direction gives
X
Fz = maz = 0 ⇒ P2 ∆x∆y − P1 ∆x∆y − ρg∆x∆y∆z = 0
where W = mg = ρg∆x∆y∆z is the weight of the fluid element. Dividing by ∆x∆y and
rearranging gives
∆P = P2 − P1 = ρg∆z = γs ∆z
(1.10)
where γs = ρg is the specific weight of the fluid. Thus, we conclude that the pressure difference
between two points in a constant density fluid is proportional to the vertical distance ∆z between
the points and the density ρ of the fluid. In other words,
pressure in a static fluid increases linearly with depth.
For a given fluid, the vertical distance ∆z is sometimes used as a measure of pressure, and it
is called the pressure head.
17
1.2. FIELD REPRESENTATION
(a)
CHAPTER 1. FUNDAMENTAL CONCEPTS
(c)
(b)
Figure 1.5: (a) The pressure of a fluid at rest increases with depth. (b) Free-body diagram of a rectangular
fluid element in equilibrium. (c) Pressure in a liquid at rest increases linearly with distance
from the free surface.
One can also conclude from (1.10) that for small to moderate distances, the variation of pressure
with height is negligible for gases because of their low density. If we take point 1 to be at the
free surface of a liquid open to the atmosphere (Figure 1.5c), where the pressure is atmospheric
pressurePatm , then the pressure at a depth h from the free surface becomes
P = Patm + ρgh or Pgage = ρgh
(1.11)
Here Pgage is the difference between the absolute pressure P and the local atmospheric pressure
Patm . This difference is called the gage pressure.
Liquids are essentially incompressible substances, and thus the variation of density with depth is
negligible. This is also the case for gases when the elevation change is not very large. For fluids
whose density changes significantly with elevation, a relation for the variation of pressure can be
obtained by dividing (1.10) by ∆x∆y∆z, and taking the limit as ∆z → 0. It gives
dP
= −ρg
dz
(1.12)
The negative sign is due to our taking the positive z direction to be upward so that dP is negative
when dz is positive since the pressure decreases in the upward direction. When the variation
of the density with elevation is known, the pressure difference between points 1 and 2 can be
determined by integration to be
∆P = P2 − P1 = −
Z2
ρgdz
1
For constant density and constant gravitational acceleration, this relation reduces to (1.10) as
expected.
Pressure in a fluid at rest is independent of the shape or cross section of the container. It changes
18
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
with the vertical distance, but remains constant in other directions. Also, the pressure force exerted by the fluid is always normal to the surface at any special point. A consequence of the
pressure in a fluid remaining constant in the horizontal direction is that
the pressure applied to a confined incompressible fluid increases the pressure throughout the
fluid by the same amount.
This statement is known as Pascal’s law. Pascal also knew that the force applied by a fluid is
proportional to the surface area. He realised that two hydraulic cylinders of different areas could
be connected, and the larger could be used to exert a proportionally greater force than that applied to the smaller. Such “Pascal’s machine” has been the source of many inventions that are a
part of daily lives such as hydraulic brakes and lifts. This is what enables us to lift a car easily
by one arm, as shown in Figure 1.6. Noting that P1 = P2 and both pistons are at the same level
Figure 1.6: Lifting of a large weight by a small force by application of Pascal’s law.
(the effect of small height difference is negligible, especially at high pressure), the ratio of output
force to apply force is determined to be
P1 = P2 ⇒
The area ratio
1.2.5
A2
A1
F1
F2
F2
A2
=
⇒
=
A1
A2
F1
A1
is called the ideal mechanical advantage of the hydrostatic lift.
Viscosity
When two solid bodies in contact move relative to each other, a frictional force develops at the
contact surface in the direction opposite to motion. The situation is similar when a fluid moves
relative to a solid or when two fluids move relative to each other. We move with relative ease in
air, but not so in water. Moving in oil would be even more difficult, as can be observed by the
slower downward motion of a glass ball dropped in a tube filled with oil as compared to water.
It appears that there is property that represents the internal resistance of a fluid to motion or the
“fluidity”, and this property is called viscosity. The force a flowing fluid exerts on a body in
the flow direction is called the drag force, and the magnitude of this force depends, in part, on
19
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
viscosity.
To obtain a relation for viscosity, consider the behaviour of a fluid element between the two
infinite plates shown in Figure 1.7. Let a constant parallel force Fx be applied to the upper plate
δl
M
y
δα
M’
P
Fluid element
at time, t
P’
Force, F
Velocity, δu
Fluid element
at time, t + δt
δy
x
N
O
δx
Figure 1.7: Deformation of a fluid element.
while the lower plate is held fixed. After the initial transients, it is observed that the upper plate
moves continuously under the influence of this force at a constant velocity δu. The fluid in contact with the upper plate sticks to the plate surface and moves with it at the same velocity. The
shear stress σxy applied to the fluid element is given by
dFx
δFx
=
δAy →0 δAy
dAy
σxy = lim
Here δAy is the contact area of a fluid element with the plate, and δFx is the force exerted by
the plate on that element. During time interval δt, the fluid element is deformed from position
MNOP to position M’NOP’. The rate of deformation of the fluid is given by
deformation
δα
dα
=
δt→0 δt
dt
rate = lim
in terms of readily measured quanTo calculate the shear stress σxy , it is desirable to express dα
dt
tities. This can be done easily. The distance δl between the points M and M’ is given by
δl = δuδt
or alternatively, for small angles,
δl = δyδα
Equating these two expressions for δl results in
δu
δα
=
δt
δy
Taking the limits of both sides of the equality yields
dα
du
=
dt
dy
20
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
Thus, the fluid element of Figure 1.7, when subjected to shear stress σxy , experiences a rate
of deformation (shear rate) which is equivalent to the velocity gradient du
. Further, it can be
dy
verified experimentally that for the most fluids the rate of deformation (and thus the velocity
.
gradient) is directly proportional to the shear stress σxy , i.e. σxy ∝ du
dy
Definition:
Fluids in which shear stress is directly proportional to the rate of deformation are Newtonian
fluids.
The term non-Newtonian is used to classify all fluids in which shear stress is not directly proportional to shear rate. Most common fluids such as water, air, gasoline, and oils are Newtonian
fluids. Blood and liquid plastics are examples of non-Newtonian fluids.
If one considers the deformation of two different Newtonian fluids, say glycerin and water, one
recognises that they will deform at different rates under the action of the same applied shear
stress. As glycerin deforms at a slower rate we say that it is much more viscous. The constant
of proportionality in the relation between the shear stress and the rate of deformation is the absolute or dynamic viscosity, η. Thus in terms of the coordinates of Figure 1.7, Newton’s law of
viscosity is given for one dimensional flow by
σxy = η
du
dy
(1.13)
Note that, since the dimensions of σ are LF2 and the dimensions of du
are T1 , η has dimensions
dy
FT . Since the dimensions of force F , mass M , length L, and time
L2
MT are related by Newton’s
second law of motion, the dimensions of η can also be expressed as LT . In fluid mechanics the
ration of absolute viscosity η to density ρ often arises. This ratio is given the name kinematic
M
viscosityh and
i is represented by the symbol ν. Since density has dimensions L3 , the dimensions
2
of ν are LT . Note also that for gases, viscosity increases with temperature, whereas for liquids,
viscosity decreases with increasing temperature.
1.2.6
Surface Tension
A liquid being unable to expand freely, will form an interface with a second liquid or gas. Strictly
speaking, any liquid is in an external medium, such as the atmosphere, and not in a vacuum. Thus
we should speak not merely of surfaces of liquids but of interfaces between two continuous media, for example between fluid and its vapour. In this case, one can say that the interface is
separating two phases of the matter. The sharp boundary between these two media is the most
pronounced characteristic of the liquid state of a matter. In reality, of course, the media are separated by a narrow transition layer, but this is so thin that it may be regarded as a surface.
The existence of such surface layer of liquids brings about the existence of a separate class of
surface phenomena. The physical chemistry of such interfacial surfaces is quite complex and
beyond the scope of this Compendium. Therefore we present here only qualitative picture of
those surface phenomena which are primary related to the interface between fluid and gas and
will be important for the physics of waves excited on fluid surfaces.
21
1.2. FIELD REPRESENTATION
CHAPTER 1. FUNDAMENTAL CONCEPTS
A property that we will discuss is surface tension at the interface of a fluid and a gas. This
phenomenon which is a tensile force distributed along the surface is due primarily to molecular
attraction between like molecules (cohesion) and molecular attraction between unlike molecules
(adhesion). Indeed, in surface phenomena, only those molecules which are actually at the surface
of fluids are involved. The molecules which are in a thin layer adjoin the surface are in conditions
different from those within the fluid. The latter are surrounded by similar molecules on all sides
and therefore the cohesive forces cancel. But at the free surface the fluid cohesive forces from
below exceed the adhesive forces from the gas above and therefore the resulting force felt by
molecules of a surface layer is directed into the fluid. This is due to the fact that the density of a
fluid is much larger than the density of a gas and therefore the attraction force from the molecules
of a fluid is lager with respect to that of gas molecules. Thus to transport a molecule from inside
of the fluid to its surface layer one has to perform a work against this resulting force. Therefore,
each molecule in the surface layer of a fluid has an extra potential energy with respect to that of
the molecules within the fluid. The larger surface of the fluid the more molecules which have this
extra potential energy. The difference between the energy of all the molecules (in both media)
which are near the surface and the energy they would have within the fluid is called the surface
energy. It is evident now that the surface energy is proportional to the area S of the interface
Usurf = σS
The coefficient σ (σ > 0) is the important characteristic of the surface properties of a fluid and
is called the surface tension. Thus,
the surface tension represents here the stretching work that needs to be done to increase the
surface area of the liquid by a unit amount.
Due to the tendency of molecules to migrate from the surface of a fluid to its interior, the fluid
takes the form when its free surface has a minimum value. Indeed, as it is known from the mechanics, forces always act so as to bring a body to a state of minimum energy. In particular, the
surface energy tends to take its least possible value and thus the interface between two media
must always tend to contract. This is the reason why droplets of fluid (and gas bubbles) tend to
be spherical: for a given volume, the sphere is the figure with the least area.
The surface tension force, FS , is shown by the following simple example. Let us imagine a film
of a liquid supported on a wire frame of which one side (of length l) is movable (see Figure
1.8). Because the surface tends to contract, the wire is subject to a force which can be directly
Figure 1.8: Film on a wire plane
measured on the movable part of the frame. The pulling force that causes this tension acts parallel
to the surface and is due to attractive forces between the molecules of the liquid. By general laws
22
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.2. FIELD REPRESENTATION
of mechanics, this force is equal to the derivative of the energy (in this case, the surface energy)
with respect to coordinate x in the direction in which the force acts
FS = −
dUsurf
dS
= −σ
dx
dx
But the area of the film is S = lx, and therefore
FS = −σl
This is the force on a segment l of the frame due to the surface tension on one side of the film;
since the film has two sides, the force on the segment l is twice this value. The minus sign shows
that this force is directed into surface of the film. So, the line bounding the surface of the liquid
(or any part of this surface) is subject to forces, directed into the surface, perpendicular to this
line and tangential to the surface. The force per unit length is equal to the surface tension σ.
Thus it is evident now that the surface tension can also be expressed as
the force applied to the unit length of the line bounding the surface of the liquid perpendicular
to the line and tangential to the surface.
The value of σ depends on the pair of fluids in contact and the temperature. The dimensions of
σ follow from its definition, and may be put in various forms: energy per unit area, or force per
unit length, i.e.
N ·m
N
J
=
[σ] = 2 ≡
2
m
m
m
An important consequence of surface tension is that it gives rise to a pressure jump across the
interface whenever it is curved. Indeed, let us consider, for example, a drop of fluid in air.
The tendency of the drop surface to decrease causes a contraction of the drop and therefore an
increase in its internal pressure. Thus the pressure of the fluid in the drop (p1 ) exceeds the
pressure of the surrounding air (p0 ). The difference between them is called the surface pressure
and will be denoted by psurf , psurf = p1 −p0 . To calculate this quantity, we note that the work done
by the surface forces in reducing the surface area of the drop by dS is equal to the corresponding
decrease σdS in the surface energy. This work can also be written as psurf dV , where dV is the
change in the volume of the drop due its contraction, and therefore
σdS = psurf dV
For a spherical drop of radius r, we have S = 4πr2 and V = 4πr3 /3. Substituting these relations
into the above equation yields the following expression for the surface pressure
psurf ≡ p1 − p0 =
2σ
r
This means that the pressure is greater in the medium whose surface is convex. When r → ∞,
the surface pressure tends to zero. This is in accordance with the fact for a plane interface the
pressures in adjoin media must be the same.
The obtained relation for the surface pressure holds only if the surface is spherical. The curvature
of a general surface can be specified by the radii of curvature along two orthogonal directions,
say R1 and R2 , so called principal curvatures of the surface. A similar analysis based on the
23
1.3. FLUID MOTIONS
CHAPTER 1. FUNDAMENTAL CONCEPTS
condition that the two media be in thermodynamic equilibrium together shows that the pressure
jump across the interface is then given by
1
1
psurf ≡ p1 − p0 = σ
+
R1 R2
that agrees with the previous result if R1 = R2 . This is the Laplace’s formula. The additional
pressure, defined by this formula, is directed to the centre of the curvature. Therefore, if the
surface is convex, the surface pressure is positive, and adds to the normal fluid pressure (p1 ). If,
on the other hand, the surface is concave, the surface pressure is negative and the fluid pressure
will be less than the pressure of the surrounding air (p0 ). Mathematically it corresponds to the
fact that the radius of curvature for the concave surface, i.e. the centre of the curvature is outside
of the fluid, is assumed to be negative, and for the convex surface it is positive.
In general, in order to find the shape of the surface, we need formulae which determine the radii
of curvature. These formulae are obtained in differential geometry and are rather complicated.
They are considerably simplified when the surface deviates only slightly from a plane. Indeed,
let z = ζ (x, y) be the equation of the interface surface. We suppose that the surface deviates only
slightly from the plane z = 0, i.e. ζ is everywhere small. Then, the sum of principal curvatures,
1/R = 1/R1 + 1/R2 , of a slightly curved surface is defined as
2
∂ 2ζ
∂ ζ
1
≡ −∇2 ζ
+
=−
R
∂x2 ∂y 2
Therefore the Laplace’s formula takes the form
psurf ≡ p1 − p0 =
σ
= −σ∇2 ζ
R
It is worthwhile to note also that the surface pressure due to the curvilinear interface is not
identical to the surface tension forces. The latter are tangential to the surface whereas the surface
pressure defined by the Laplace’s formula is perpendicular to the surface. One can say that the
surface tension forces cause the pressure jump through the curvilinear interface. This statement
is illustrated by the Figure 1.9. It is seen that the surface tension forces, FS , which are tangential
to the curvilinear surface, result in the force, F , which is perpendicular to the surface and always
directed to the medium towards which the interface is concave.
1.3
Classification of Fluid Motions
We define fluid mechanics as the science that deals with the behaviour of fluids at rest or in motion, and the interaction of fluids with solids or other fluids at the boundaries. There is a wide
variety of fluid flow problems encountered in practice, and it is usually convenient to classify
them on the basis of some common characteristics to make it feasible to study them in groups.
There are many ways to classify fluid flow problems, and here we present some general categories.
1.3.1
No-Slip condition
Fluid flow is often confined by solid surfaces, and it is important to understand how the presence
of solid surfaces affects fluid flow. We know that water in a river can not flow through large
24
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.3. FLUID MOTIONS
Figure 1.9: Surface forces with curvilinear interface; a) fluid surface is convex; b) fluid surface is concave.
rocks, and must go around them. That is, the water’s velocity normal to the rock surface must
be zero, and water approaching with a direction normal to a surface comes to a complete stop at
the surface. What is not as obvious is that water approaching the rock at any angle also comes
to a complete stop at the rock surface, and thus the tangential velocity of water at the surface
is also zero. Consider, for example, the flow of a fluid over a solid surface. All experimental
observations indicate that a fluid in motion comes to a complete stop at the surface and assumes
a zero velocity relative to the surface. That is, a fluid in direct contact with a solid “sticks” to
the surface, and there is no slip. This is known as the no-slip condition. The fluid property
responsible for the no-slip condition is viscosity. Fluid sticking results in the development of a
velocity gradient in the vicinity of the solid surface. Indeed, the layer that sticks to the surface
slows the adjacent fluid layer because of viscous forces between the fluid layers, which slows
the next layer, and so on. Therefore, the no-slip condition is responsible for the development of
the velocity profile, as illustrated in Figure 1.10. The flow region adjacent to the wall in which
the viscous effects (and thus the velocity gradients) are significant is called the boundary layer.
1.3.2
Description of Fluid Flows
Viscous versus Inviscid Regions of Flow. Consider two fluid layers that move relative to each
other. This results in a friction force which develops between them and the slower layer tries
to slow down the faster layer. This internal resistance to flow is quantified by the fluid property
viscosity, which is the measure of internal stickiness of the fluid. Viscosity is caused by cohesive
forces between the molecules in liquids and by molecular collisions in gases. There is no fluid
with zero viscosity, and thus all fluid flows involve viscous effects to some degree. Flows in
which the frictional effects are significant are called viscous flows. In many flows of practical
interest, there are regions (typically regions not close to solid surfaces) where viscous forces are
negligibly small compared to internal or pressure forces. Neglecting the viscous terms in such
inviscid flow regions greatly simplifies analysis without loss in accuracy. The development of
viscous and inviscid regions of flow as a result of inserting a flat parallel plate into a fluid stream
25
1.3. FLUID MOTIONS
CHAPTER 1. FUNDAMENTAL CONCEPTS
z
v0
x
h(x)
Boundary layer (non-negligible viscous forces)
Outer flow (inviscid region of flow)
Uniform flow
Figure 1.10: Viscous and inviscid regions as a result of inserting a parallel flat plate into a fluid (not to
scale).
of uniform velocity is shown in Figure 1.10. The fluid sticks to the plate because of the non-slip
condition, and the thin boundary layer in which the viscous effects are significant near the plate
surface is the viscous flow region. The region of flow far away from the plate which is unaffected
by the presence of the plate is the inviscid flow region.
Compressible versus Incompressible Flow. A flow is classified as being compressible or incompressible, depending on the level of variation of density during the flow. Incompressibility is
an approximation in which a flow is said to be incompressible if the density remains nearly
constant throughout the entire fluid. Therefore, the volume of every portion of the fluid remains
unchanged over the course of its motion when the flow is incompressible. The density of liquids
is essentially constant, and thus the flow of liquids is typically incompressible. Therefore, liquids
are usually referred to as incompressible substances. For example, a pressure of 210 atmospheres
(atm) causes the density of liquid water to change by just 1 percent. Gases, on the other hand,
are highly compressible. A pressure change of just 0.01 atm, for example, causes a change of 1
percent in the density of atmospheric air. When analysing rockets, spacecraft, and other systems
that involve high speed gas flows, the flow speed is often expressed in terms of the dimensionless
Mach number defined as
Speed of flow
V
=
Ma =
c
Speed of sound
where c is the speed of sound whose value is 346 m/s in air at room temperature at sea level.
A flow is called sonic when Ma = 1, subsonic when Ma < 1, supersonic when Ma > 1, and
hypersonic when Ma ≫ 1. Liquid flows are incompressible to a high level of accuracy, but the
level of variation in density in gas flows and the consequent level of approximation made when
modelling gas flows as incompressible depends on the Mach number. Gas flows can often be approximated as incompressible if the density changes are under about 5 percent, which is usually
the case when Ma < 0.3. Therefore, the compressibility effects of air at room temperature can
be neglected at speeds under about 100 m/s.
Laminar versus Turbulent Flow. Some flows are smooth and orderly while others are rather
chaotic. The highly ordered fluid motion characterised by smooth layers of fluids is called
laminar. The word laminar comes from the movement of adjacent fluid particles together in
“laminae”. The flow of highly viscous fluids such as oils at low velocities is typically laminar.
26
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.4. DIMENSIONS AND UNITS
The highly disordered fluid motion that typically occurs at high velocities and is characterised
by random, three dimensional motions of fluid particles in addition to the mean motion is called
turbulent. A flow that alternates between being laminar and turbulent is called transitional.
The experiments conducted by Osborne Reynolds in the 1880s resulted in the establishment of
the dimensionless Reynolds number, Re, as a key parameter for the determination of the flow
regime in pipes.
In laminar flow there is no macroscopic mixing of adjacent fluid layers. A thin filament of
dye injected into a laminar flow appears as a single line; there is no dispersion of dye in the flow,
except the slow diffusion due to molecular motion. On the other hand, a dye filament injected
into a turbulent flow disperses quickly throughout the flow field; the line of dye breaks up into
myriad entangled threads of dye. This behaviour of turbulent flow is due to the velocity fluctuations present; the macroscopic mixing of fluid particles from adjacent layers of the fluid results
in rapid dispersion of dye.
Steady versus Unsteady Flow. The terms steady and uniform are used frequently in engineering, and thus it is important to have a clear understanding of their meaning. The term steady
implies no change at a point with time. The opposite of steady is unsteady. The term uniform
implies no change with location over a specified region. The terms unsteady and transient are
often used interchangeably, but these terms are not synonyms. In fluid mechanics, unsteady is
the most general term that implies to any flow that is not steady, but transient is typically used
for developing flows. The term periodic refers to the kind of unsteady flow in which the flow
oscillates about a steady mean. Many devices such as turbines, compressors, boilers, condensers,
etc. operate for long periods of time under the same conditions, and they are classified as steadyflow devices. During the steady flow, the fluid properties can change from point to point within
a device, but at any fixed point they remain constant. Therefore, the volume, the mass, and the
total energy content of a steady-flow device or flow section remain constant in steady operation.
One of the most important tasks of an engineer is to determine whether it is sufficient to study
only the time-average steady flow features of a problem, or whether a more detailed study of the
unsteady features is required. If the engineer was interested only in the overall properties of the
flow field (such as the time-averaged drag coefficient, the mean velocity, and pressure fields), a
time averaged description (analytical or numerical) would be sufficient. However, if the engineer
was interested in details about the unsteady-flow field, such as flow-induced vibrations, unsteady
pressure fluctuations, or sound waves emitted from the turbulent eddies or the shock waves, a
time-averaged description of the flow field would be insufficient.
1.4
Dimensions and Units
To study fluid mechanics, we must establish some abstractions to describe those properties of
the fluid flows that interest us. These abstractions are called dimensions. For example, we refer
to physical quantities such as length, time, mass, and temperature as dimensions. In terms of
particular system of dimensions, all measurable quantities can be subdivided into two groups —
primary or fundamental dimensions, and secondary or derived dimensions. Primary dimensions
are those for which we set up arbitrary scales of measure; secondary quantities are those whose
dimensions are expressible in terms of the dimensions of the primary quantities.
27
1.5. SUMMARY OBJECTIVES
CHAPTER 1. FUNDAMENTAL CONCEPTS
A unit is a particular way of attaching a number to the quantitative dimension. Thus, length is
a dimension associated with such variables as distance, displacement, width, deflection, height,
etc., while centimetres and meters are both numerical units for expressing length.
In fluid mechanics there are only four primary dimensions from which all other dimensions
can be derived: mass {M }, length {L}, time {T }, and temperature {Θ}. Units in the SI, International System of Units, corresponding to these dimensions are given by Kilogram (kg), Meter
(m), Second (s), and Kelvin (K). The braces around a symbol like {M } means “the dimension”
of mass. All other variables in fluid mechanics can be expressed in terms of {M }, {L}, {T },
and {Θ}. For example, the acceleration is the secondary dimension quantity and it has the dimensions {LT −2 }, the dimensions of the force is {M LT −2 }, etc.
In order to determine the dimensions of some properties established by laws, we must first
discuss the law of dimensional homogeneity. This states that all theoretical equations in fluid
mechanics (and in other physical sciences) are dimensionally homogeneous, i.e. each additive
term in the equation has the same dimension. In other words,
an analytically derived equation representing a physical phenomenon must be valid for all
systems of units
For this reason, the fundamental equations of physics are dimensionally homogeneous, so all
relations derived from these equations must also be dimensionally homogeneous. Note, that the
presentation of the experimental data in homogeneous form is the subject of the powerful tool
called the dimensional analysis.
1.5
Summary objectives
When you finish studying the Chapter 1, you should be able to do the following:
1. Understand the continuity approximation and define the basic fluid parameters in the field
representation (scalar and vector fields).
2. Define the body and surface forces, state the convention for designating the nine components of the stress field and represent the normal and shear stresses in the tensor notations.
3. Define the pressure field and describe its properties, such as the dependence on the depth
of the fluid and the Pascal’s law.
4. Describe the viscosity as the internal resistance of a fluid to motion, write Newton’s law
of viscosity and determine the shear stress and and shear force that correspond to a given
one dimensional velocity profile.
5. Define the surface tension as the property of the interface between two different continuous
media (work and force representation).
6. Determine the surface tension pressure and write the Laplace’s formula for curvilinear
interface.
7. Formulate the boundary layer concept and define the related fluid flows, to determine the
conditions for the fluid flow to be compressible, or incompressible.
8. List the primary dimensions of the Fluid Mechanics.
28
Chapter 2
Elements of Thermodynamics
In our subsequent discussion of the dynamics of fluids we shall need to make use of some of
the concepts of classical thermodynamics and of the relations between various thermodynamic
quantities, such as temperature and internal energy. Classical thermodynamics is the study of the
equilibrium states of matter, in which the properties are assumed uniform in space and time,
that is with states in which all local mechanical, physical and thermal quantities are virtually
independent of both position and time.
A thermodynamic system is a mass of fluid separated from the surroundings by a flexible
boundary through which the system exchanges heat and work but not mass. This means that
if the external conditions change the state of the thermodynamic system will also change. A
system in an equilibrium state is free of currents, such as those generated by stirring a fluid or
by sudden heating. This definition, however, is not possible in fluid flows, as non-equilibrium
states are common in practical fluid dynamics, and the question arises as to whether the relations
derived in classical thermodynamics are applicable to fluids in constant motion. Experiments
show that the results of classical thermodynamics do hold in most fluid flows if the changes
along the motion are sufficiently slow compared to a relaxation time. Here “sufficiently slow”
means, therefore, so slowly that the fluid is able to establish the equilibrium state corresponding to every instantaneous values of the external parameters and the finite change of the state
can be described as a set of successive equilibrium states. The relaxation time is defined as the
time taken by the thermodynamic system to adjust to a new equilibrium state, and the material
undergoes this adjustment through molecular collisions. This time is very small under ordinary
conditions, since only a few molecular collisions are needed for the adjustment. The relations of
classical thermodynamics are therefore applicable to most fluid flows.
The basic laws of classical thermodynamics are empirical, and can not be proved. Another
way of viewing this is to say that these principles are so basic that they cannot be derived from
anything more basic. They essentially establish certain basic definitions, upon which the subject
is built. The first law of thermodynamics can be regarded as a principle that defines the internal
energy of a system, and the second law can be regarded as the principle that defines the entropy
of a system. The concepts of thermodynamics are helpful to students of fluid mechanics for
additional reason that in both subjects the objective is a set of results which apply to matter as
generally as possible, without regard for the different molecular properties and mechanisms at
work.
The purpose of this chapter is to recapitulate briefly the laws and results of equilibrium ther29
2.1. FIRST LAW
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
modynamics and to set down for future references the relations that will be needed later. For
a proper account of the fundamentals of the subject the reader should refer to one of the many
text-books available.
2.1
First Law of Thermodynamics
It is taken as a fact of experience that the state of a given mass of fluid in equilibrium (the word
being used here and later to imply spatial as well as temporal uniformity) under the simplest possible conditions is specified
by two parameters, which for convenience may be chosen
uniquely
1
as the specific volume, V = ρ , where ρ is the density, and the pressure p as defined above. All
other quantities describing the state of the fluid are thus functions of these two parameters of
state. One of the most important of these quantities is the temperature. A mass of fluid in equilibrium has the same temperature as a test mass of fluid also in equilibrium if the two masses
remain in equilibrium when placed in thermal contact, that is, when separated only by a wall
allowing transmission of heat. This fact is also known as “the zeroth law of thermodynamics”
which states that,
there exist a variable of state, the temperature T , and two systems which are in thermal contact
are in equilibrium only if their temperatures are equal.
The relation between the temperature T and the two parameters of state, which we may write as
f (p, V, T ) = 0
(2.1)
thereby exhibiting formally the arbitrary choice of the two parameters of state, is called an equation of state. For example, if parameters of states of fluid in equilibrium are pressure p and
temperature T , then equation of state (2.1) is reduced to
V = V (p, T ) .
(2.2)
and for sufficiently small variations of the system under these conditions, we have
∂V
∂V
dT +
dp
dV =
∂T p
∂p T
with the partial derivatives
∂V (p,T )
∂T
=
∂V (p,T )
∂p
=
∂V
∂T
∂V
∂p
p
,
T
,
where a suffix indicates the quantity held constant. If, on the other hand, the fluid is involved
in the process in which the pressure, p, is a function of the temperature, T , then the quantities
dT and dp are not independent any more and the process, for example with V = const, is now
described by the equation
∂V
∂V
dT +
dp = 0
∂T p
∂p T
∂p
dp
since dp and dT
, we get the partial derivative ∂T
Solving this equation with respect to dT
V
describes now the changes of pressure and temperature under the condition V = const. So that,
∂V /
∂T p
∂p
= −
∂T V
∂V /
∂p T
30
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
Since
∂V
∂p
2.1. FIRST LAW
1
=
∂p
∂V T
T
the previous relation can evidently be presented by
∂p
∂V
∂p
=−
∂T V
∂V T ∂T p
or in more symmetric form
∂p
∂V
T
∂V
∂T
p
∂T
∂p
V
= −1
It is convenient now to introduce two relative coefficients: thermal expansion coefficient
1 ∂V
,
(2.3)
β=
V ∂T p
and compression coefficient
1
α=−
V
∂V
∂p
.
(2.4)
T
The thermal expansion coefficient shows the relative increasing of fluid volume per unit temperature increasing under constant pressure. The compression coefficient shows the relative
decreasing of fluid volume per unit pressure increasing under constant temperature.
Another important quantity describing the state of the fluid is the internal energy per unit mass,
ǫ. The internal energy (also called “thermal energy”) is a manifestation of the random molecular
motion of the constituents. Since work and heat are regarded as equivalent forms of energy, the
change in the internal energy of a mass of fluid at rest is defined by account of both heat given to
the fluid and work done on the fluid. Thus if the state of a given uniform mass of fluid is changed
by a gain of heat of amount dq per unit mass and by the performance of work on the fluid of
amount of dw per unit mass, the consequential increase in the internal energy per unit mass is
dǫ = dw + dq.
(2.5)
That is,
The change in the internal energy equals the work done on the system plus the heat added to it
during the process.
This important relation expresses the law of conservation of energy for thermal processes and
is called in this connection the first law of thermodynamics. Hence, the first law accounts for
energy entering, leaving and accumulating in a thermodynamic system. If the fluid is thermally
isolated from its surroundings, i.e. does not gain or lose heat so that no exchange of heat can
occur, dq = 0, the change of state of the fluid is said to be adiabatic.
It is important to realise the difference between heat and internal energy. Heat and work are
forms of energy in transition, which appear at the boundary of the system and are not contained within the fluid. Heat is the energy in transition from one mass of fluid to another as a
31
2.1. FIRST LAW
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
result of a temperature difference. On the other hand, work, as we learned in mechanics, is the
energy in transition to or from a mass of fluid which occurs when external forces, acting on the
fluid, move through a distance. In contrast, the internal energy resides within the matter and is
primarily associated with a given mass of a fluid. In fluid mechanics, the internal energy together
with the kinetic energy associated with the motion of the mass of the fluid and the potential energy associated with the position of the mass of fluid in conservative external field are related to
types of stored energy.
There are many ways of performing work on the system, although compression of the fluid by
inward movement of the bounding walls is of specific relevance in fluid mechanics. An analytical expression for the work done by compression is available in the important case in which the
change of the fluid state occurs sufficiently slowly. Such slow processes are called quasistatic.
The quasistatic process is always reversible. This word implies that the change is carried out
so slowly that the fluid passes through a succession of equilibrium states and the direction of
the change being without effect. At each stage of a reversible change the pressure in the fluid
is uniform, and equal to p, so that the work done per unit mass of the fluid as a consequence
compression leading to a infinitesimal decrease in volume given by −pdV . Thus for a reversible
transition from one state to another, neighbouring, state we have
dǫ = dq − pdV.
(2.6)
A finite reversible change of this kind can be described by summing (2.6) over the succession of
infinitesimal steps making up the finite change.
A practical quantity of some importance is the specific heat of the fluid, that is, the amount
of heat given to a unit mass of the fluid per unit rise in temperature in a small reversible change.
The specific heat may be written as
dq
c=
(2.7)
dT
and is not determined uniquely until we specify further the conditions under which the reversible
change occurs. It is worthwhile to note here that the specific heat, defined by (2.7), is always a
positive quantity. It follows from the assumption that the temperature increases when the fluid
is heated. The most usual in physics are the specific heat at constant volume cV and the specific heat at constant pressure cp which gives the quantities of heat when the fluid is heated
under conditions such that its volume and pressure respectively remain constant. To determine
the relations between specific heats and energy we assume that the fluid heating is a quasistatic
process so that the specific heats will characterise the equilibrium state of the fluid.
Let the equilibrium state of the fluid be defined by its volume V and temperature T . The fluid
energy in this state is a function of these two parameters of state ǫ = ǫ(V, T ). So, the heat
obtained by the fluid due to the quasistatic heating equals according to Equation (2.6)
∂ǫ
∂ǫ
dT + ∂V
dV + pdV =
dq = dǫ + pdV = ∂T
V
T
(2.8)
∂ǫ ∂ǫ
dT + ∂V T + p dV.
∂T V
Here the suffix V or T to the derivative indicates that the differentiation is taken under constant
value of V or T . If the volume remains constant in the course of the heating, then dV = 0 and
∂ǫ
dq =
dT.
∂T V
32
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
2.1. FIRST LAW
Taking into account Equation (2.7) yields the specific heat at constant volume
∂ǫ
cV =
.
∂T V
(2.9)
If the quasistatic heating occurs under constant pressure, one obtains from (2.7) and (2.8) the
specific heat at constant pressure
∂ǫ
dq
∂ǫ
∂V
=
cp =
+
+p
.
dT
∂T V
∂V T
∂T p
So that,
cp = cV +
∂ǫ
∂V
+p
T
∂V
∂T
.
(2.10)
p
Here the second term is the heat used for the work done on the fluid, p ∂V
, and change of the
∂T p
∂V ∂ǫ
internal energy of the fluid ∂V T ∂T p per unit temperature in the course of the fluid heating
under constant pressure.
It is convenient in some practical problems to use the pressure p and temperature T as parameters
describing the fluid state in equilibrium instead of the specific volume V and temperature T . The
first law of thermodynamics (Equation (2.6)) can be easily written in these variables. Indeed,
adding and subtracting V dp on the right hand side of the equation dq = dǫ + pdV yields
d(ǫ + pV ) − V dp = dq.
(2.11)
We introduce now a new parameter describing the equilibrium state of the fluid instead of the
internal energy, the enthalpy per unit mass, w, (sometimes it is also called the heat function or
heat content) given by
w = ǫ + pV.
(2.12)
The heat function replaces the internal energy if the equilibrium state of the fluid defined by the
pressure and temperature. Equation (2.11) takes now the form
dq = dw − V dp.
(2.13)
Then the specific heat at the constant pressure can be expressed through the enthalpy w. Indeed,
taking into account that w = w(p, T ) yields
∂w
∂q
∂w
∂w
dT +
− V dp ⇒ cp =
=
.
(2.14)
dq =
∂T p
∂p T
∂T p
∂T p
Thus, it is clear now that
cp 6=
∂ǫ
∂T
.
(2.15)
p
We will see later that in many thermal processes an important property of the fluid is the ratio of
the specific heats cp and cV , usually denoted by γ:
γ=
cp
.
cV
(2.16)
We turn our attention now to adiabatic processes. Let the equilibrium state of the fluid be defined
by two variables V and T (two parameters of state). The equation describing the adiabatic
33
2.1. FIRST LAW
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
compression or expansion of the fluid in the quasistatic approximation is then obtained from
(2.8) by putting dq = 0. This yields:
∂ǫ
∂ǫ
dT +
+ p dV = 0
∂T V
∂V T
We simplify this equation by introducing specific heats cp and cV . Taking into account (2.9) and
(2.10) results in
cp − cV
γ−1
cV dT + ∂V dV = 0 ⇒ dT + ∂V dV = 0.
(2.17)
∂T
∂T
p
p
where γ is defined by (2.16). Equation (2.17) is valid for any fluid or gas. Some important
results can be deduced from (2.17) in the case of an ideal gas. The equation of state of an ideal
gas per one mole is defined by pV = RT . Since the molecules of an ideal gas are assumed not
to interact with one another, the change in their mean distance apart when the volume of the gas
varies cannot affect its internal energy. In other words, the internal energy of an ideal gas is a
∂ǫ
function only of its temperature, and not of its volume or pressure, i.e. ǫ = ǫ(T ) and ∂V
= 0.
T
∂w
∂ǫ
Hence, the specific heats cV = ∂T V and cp = ∂T p of an ideal gas also depend only on the
temperature and ∂V
= Rp . Then Equation (2.10) is reduced to
∂T p
cp = cV + R.
(2.18)
Moreover, Equation (2.17) can be rewritten now in the form
dT
γ−1
=
dV.
T
V
(2.19)
To integrate this differential equation one needs to know the dependence of γ on the temperature
T , i.e. γ = γ(T ) since the specific heats of an ideal gas are functions on the temperature only.
However, experiments show that if the temperature change in the thermal processes is sufficiently
small and slow, i.e. the thermal process is considered in the quasistatic approximation, we can
neglect this dependence and treat γ ∼
= const. Then the solution of (2.19) is given by
T V γ−1 = const.
(2.20)
We see that in the adiabatic process the temperature and volume of an ideal gas vary in such a
way that the product T V γ−1 remains constant. Since γ is always greater than unity, γ − 1 > 0,
and therefore the adiabatic expansion is accompanied by a cooling of the gas, and the adiabatic
compression by heating. Combining the above equation with the formula pV = RT we can
derive a similar relation between the temperature and pressure in the adiabatic process
T γ p1−γ = const,
(2.21)
and the relation
p
= const,
(2.22)
ργ
between the pressure and volume (or density). This last relation is called the equation of Poisson’s adiabatic. It is worthwhile to note here that these relations are valid for great majority of
bodies provided that thermal processes are adiabatic and quasistatic.
pV γ = const ⇔
An interesting application of the adiabatic law (Equation (2.22)) is related to the calculation
of the speed of sound. It is known from mechanics that in a compressible fluid an infinitesimal
34
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
2.1. FIRST LAW
periodic change in density and pressure propagate through the medium at a finite speed in the
form of sound waves. This speed is defined by
s
dp
c=
.
(2.23)
dρ
On the other hand, according to the equation of state (Equation (2.1)), the pressure depends not
only on the density but on the temperature as well. So, one has to indicate in what sense the
dp
has to be calculated. Newton, for example, has assumed that the pressure and the
derivative dρ
density are related by the Bole’s law pV = f (T ), and temperature differences between compressed and decompressed part in the sound wave are relaxed almost instantaneously and thus
the sound propagation is a isothermal process. If this assumption is correct, then the derivative
dp
∂p
should be understood as the partial derivative ∂ρ
. It means that f (T ) = const, i.e. the
dρ
Bole’s law is now pV =
p
ρ
T
= const, and relation 2.23 takes the form
s
r
p
RT
=
cN =
ρ
µ
(2.24)
where R is the universal gas constant, µ is the molecular weight, and suffix N indicates that
the sound speed is calculated using the Newton’s formula. Setting for the dry air (R = 8314,
T = 273, µ = 28.9) we obtain cN = 280 m/s, whereas according to experiments the sound
speed is 330 m/s. This contradiction has been settled by Laplace. He indicated that the density
oscillations and related temperature oscillations in sound waves are taking place so quickly and
the thermal conductivity of the air is so small that the heat exchange between compressed and
decompressed parts in the sound wave is negligible for such processes. So, the temperature
differences are not relaxed and thus the sound propagation should be considered as an adiabatic
process. In this case, taking in Equation (2.8) dq = 0 (adiabatic process) and remembering that
for an ideal fluid ǫ = f (T ) yields
dǫ + pdV = 0 ⇒ cv dT + pdV = 0.
(2.25)
In this equation dT can be expressed through the pressure and specific volume by means of
equation of state, pV = RT ,
dT =
pdV + V dp
d(pV )
⇒ dT =
.
R
cp − cV
Then Equation (2.25) is reduced to
cV V dp + cp pdV = 0 ⇒ V dp + γpdV = 0.
Taking now into account that dV = − ρ12 dρ yields finally
dp
p
∂p
∂p
=
=γ =γ
.
dρ
∂ρ adiabatic
ρ
∂ρ T
So, instead of the Newton’s formula (Equation (2.24)) we obtain the Laplace’ formula
r
p
√
cL = γ = cN γ.
ρ
(2.26)
(2.27)
The√experimentally obtained value of γ for air is 1.4, so that the sound speed in the air is c =
280 1.4 = 330 m/s which is in the perfect agreement with experiments. Therefore, the sound
propagation in the ideal compressible fluid is an adiabatic process.
35
2.2. SECOND LAW
2.2
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
Second Law of Thermodynamics
The first law of thermodynamics does not provide any information concerning the direction in
which a thermal process takes place in nature. If, for instance, an exchange of heat by conduction
takes place between fluids of different temperature, the first law, or the principle of conservation
of energy, merely demands that the quantity of heat given out by one fluid shall be equal to that
taken up by the other. Whether the flow of heat, however, takes place from the colder to the hotter fluid, or vice versa, cannot be answered by the energy principle alone. In contrast, the second
law of thermodynamics can serve to determine the direction of a thermal process. In order to
treat this question in a proper way we turn first to the question of the reversibility of thermal
processes.
The reversibility of mechanical motions may be formulated by saying that they are symmetric as
regards interchanging the future and the past, i.e. with respect to time reversal. The symmetry
of mechanical motions follows at once from the equations of motion themselves, since when the
sign of the time is reversed so is that of the velocity, but the acceleration is left unchanged. The
situation is quite different as regards thermal phenomena. If a thermal process takes place, then
the reverse process (i.e. the process in which the same thermal states are traversed in the opposite order) is in general impossible. Thus thermal processes are as a rule irreversible. Indeed,
if two fluids at different temperatures are brought into contact, the hotter fluid will transmit heat
to the colder fluid, but the reverse process (a spontaneous direct transfer of heat from the colder
to the hotter fluid) never occurs. In general, any system of bodies left to itself tends to reach a
state of thermal equilibrium, in which the bodies are at relative rest, with equal temperatures and
pressures. Having reached such a state, the system will not of its own accord leave that state.
In other words, all thermal phenomena accompanied by processes of approach to the state of
thermal equilibrium are irreversible.
In some cases, however, the degree of irreversibility may be so slight that the process may be
regarded as reversible with sufficient accuracy. It is clear from the foregoing that, in order to
achieve reversibility, it is necessary to eliminate from the system as far as possible all processes
which constitute an approach to thermal equilibrium. Thus, there must be no direct transfer of
heat from a hotter to a colder fluid and no friction in the motion of fluids. An example of a thermal process which is reversible is the adiabatic expansion or compression of an ideal fluid in the
quasistatic approximation. Then numerous reversible thermal processes can at least be imagined
as, for instance, those consisting of a succession of states of equilibrium and therefore directly
reversible in all their parts. Complete reversibility could be achieved only in the ideal case of
an infinitely slow process, and for this reason alone a process occurring at a finite rate cannot be
completely reversible.
The decision as to whether a particular thermal process is reversible or irreversible (i.e. the
process can in any manner be completely reversed or not) depends only on the nature of the
initial and final states, and not the intermediate steps of the process. So, there must exist a new
variable of the fluid in the equilibrium state, termed the entropy, so that, in a reversible transition
from an equilibrium state to another, neighbouring, equilibrium state, the increase in entropy is
proportional to the heat given to the fluid, and the constant of proportionality, itself is a function
of state, depends only on the temperature and can be chosen as the reciprocal of the temperature.
Thus, with entropy per unit mass of a fluid denoted by S, we have
T dS = dq,
36
(2.28)
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
2.2. SECOND LAW
where dq is the infinitesimal amount of heat given reversibly to unit mass of the fluid. If the heat
dq is added to a fluid, the distinction between the final and initial equilibrium states can then be
determined by a change in entropy between these states, which is
Sfinal − Sinitial =
Z
final
initial
dq
.
T
(2.29)
Now we can formulate the second law of thermodynamics. It states that
For any spontaneous process the entropy change is positive or zero. That is
Z B
dq
SB − SA ≥
.
A T
So that, the second law of thermodynamics states that the entropy of the thermodynamic system
either remains constant (in reversible processes) or increases in value (in irreversible processes)
by all changes of the system. It also shows, when two states are given, whether a transformation
is possible in nature from the first to the second, or from the second to the first. In these transformations, the final state of an irreversible process is in some way discriminate from the initial
state, while in reversible processes the two states are in certain respects equivalent. This conclusion permits to predict from the most general points of view the stability of a thermodynamic
system. Indeed, consider a thermodynamic system which is adiabatically isolated, i.e. there is
no heat exchange between the system and its surroundings, and is in the equilibrium state. Let
the entropy of the system be maximal, i.e. the entropy of the equilibrium state is larger with
respect to all possible values of the entropy of all states infinitely close to the given one in which
system can adiabatically be transformed. Then one may conclude that the adiabatic transition of
the system itself to all of these states is impossible, i.e. system is in a stable thermodynamic
equilibrium state. Indeed, if such transition will be possible, then the entropy of the initial (1)
and final (2) states are related by S1 > S2 . But this relation contradicts to the principle of the
entropy growing in adiabatic processes according to which there must be S1 ≤ S2 and therefore
such adiabatic transitions are prohibited. Thus, we are coming to the condition of thermodynamic stability:
If adiabatically isolated system is in some equilibrium state and the entropy of the system in this
state is maximal, then this state is thermodynamically stable.
This condition may be considered (see text books on thermodynamics) as a consequence of a
more general theorem of thermodynamics: an external action which removes a system from a
state of its thermal equilibrium brings about processes in it which try to reduce the effect of this
action. This theorem is also called Le Chatelier’s principle and it permits to predict the direction of the relaxation process if the system was removed by an external action from its stable
equilibrium state.
A process in which the entropy does not change (dS = 0) is termed isentropic. Constant entropy
requires dq = 0, a condition achieved if no heat passes between the fluid and its surroundings
and no mechanical energy is converted to the thermal energy by friction. A frictionless, adiabatic
process is therefore isentropic. An important relation may be deduced for a frictionless adiabatic
flow of a fluid. Referring throughout to unit mass of fluid, we may write for a frictionless (i.e.
37
2.3. SUMMARY OBJECTIVES
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
dS
. Hence
reversible) process: dq = T dS and so that the specific heat c = T dT
∂S
∂T p
cp = T
cV = T
,
(2.30)
∂S
∂T V
.
Now let us assume that the equilibrium state of the fluid is determined by the pressure p and the
specific volume V . Since the entropy is the function of the state, i.e. its infinitesimal increment
is the total differential, we may write
∂S
∂S
dV +
dp,
(2.31)
dS =
∂V p
∂p V
where as usual a suffix indicates the quantity held constant. Temperature does not appear as
a separate variable since it is uniquely related to p and V by the equation of state (2.1). The
considered process is isentropic, dS = 0, and thus, as dp and dV both tend to zero, we obtain
from (2.31)
∂S
( ∂T
)p
∂S
∂p
∂V
( ∂T )p
∂p
∂V p
V ,
= −γ ∂T
(2.32)
= − = − ∂S
∂V
( ∂T )V
∂S
∂V S
∂T
p
∂p
∂p
V
( ∂T
)V
where we use the relations (2.16) and (2.30). If the fluid obeys Bole’s law pV = f (T ), where
f (T ) represents any function of T , then
df
∂p
= V1 dT
,
∂T V
∂V
∂T
p
=
1 df
.
p dT
These expressions substituted in (2.32) give
∂p
p
= −γ .
∂V S
V
(2.33)
For a constant value of γ, integration then yields ln(p) = −γln(V ) + const. Thus
pV γ = const =
p
.
ργ
(2.34)
For an ideal fluid, the equation of state is pV = RT , one easily obtain from (2.34) the relations
(2.20) and (2.21). So, the isentropic flow of an ideal fluid can be described by the Poisson’s
adiabatic law. It was shown in the previous Section that the sound velocity is calculated under the
adiabatic condition, i.e. sound propagation in an ideal fluid is an isentropic process. Therefore
dp
we can conclude that the derivative dρ
in (2.23) should be understood as the partial derivative
∂p
, i.e. the derivative is calculated under constant entropy.
∂ρ
S
2.3
Summary objectives
When you finish studying the Chapter 2, you should be able to do the following:
38
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
2.3. SUMMARY OBJECTIVES
1. Define the equilibrium state of the matter and formulate the condition when the relations
obtained in the classical thermodynamics are applicable to fluids in motion.
2. Define parameters of state and write the equation of state in different forms.
3. Formulate the first law of thermodynamics and point how the assumption that work and
heat are regarded as equivalent forms of energy is used in this law.
4. Derive the equation of state in the form of Poisson’s adiabatic law.
5. Describe the sound propagation as a isothermal process (Newton’s assumption) and adiabatic process (Poisson’s assumption). Which assumption is indeed correct and why?
6. Formulate the second law of thermodynamics and define conditions for the thermal process
to be reversible or irreversible.
7. Formulate the condition for the system to be in a stable thermodynamic equilibrium state
and relate this condition to the Le Chatelier’s principle.
39
2.3. SUMMARY OBJECTIVES
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
40
Chapter 3
Kinematics of Fluids
Fluid kinematics is the study of how fluids flow and how to describe their motion. The subject deals with describing the motion of fluids without necessarily considering the forces and
momentum that cause the motion.
3.1
Eulerian and Lagrangian Representations of Fluid Motion
Fluid motion is determined completely if its velocity v, pressure p, and density ρ are specified
in one or another way, as functions of space and time, namely, (v, p, ρ) = F (r, t). In this
description, we evidently need to decide whether we should use coordinates that move with the
fluid particle or coordinates fixed in space. These two procedures are known respectively as the
Lagrangian and the Eulerian specifications. So, there are two known ways of representation of
the fluid motion:
1. The Lagrangian way of representation
This is the most familiar method which is used for example with Newton’s laws, i. e.
to follow the path of individual objects, for example billiard balls, and keeping track of
the position and velocity vector of each object. When this method is applied to a flowing
fluid particle it is called the Lagrangian description of fluid motion, after the Italian
mathematician Joseph L.Lagrange (1736–1813). Lagrangian analysis is analogous to the
(closed) system analysis, namely, we follow a mass of fixed identity.
The fluid parameters (v, p, ρ) including the coordinates of the given fluid particle
{xi (i = 1, 2, 3)} ≡ (x1 , x2 , x3 ) are specified in terms of time t, and some vector ξ ≡
(ξ1 , ξ2 , ξ3 ) which identifies a given particle of the fluid. For example, p = p(t, ξ1 , ξ2 , ξ3 ) ⇔
p = p(t, ξk ), or


 x1 = x1 (t, ξ1 , ξ2 , ξ3 ) 
x2 = x2 (t, ξ1 , ξ2 , ξ3 )
xi = xi (t, ξk ) ⇔
.


x3 = x3 (t, ξ1 , ξ2 , ξ3 )
In this representation the variables ξ are usually assumed to be coordinates of a given fluid
particle at some time t0 (initial coordinates which identified the given fluid particle), i.e.
xi (t, ξk )|t=t0 = ξi ⇔ ξi = xi (t0 , ξk ), or


 ξ1 = x1 (t0 , ξ1 , ξ2 , ξ3 ) 
ξ2 = x2 (t0 , ξ1 , ξ2 , ξ3 )
⇔ ξi = xi (t0 , ξk ) .


ξ3 = x3 (t0 , ξ1 , ξ2 , ξ3 )
41
3.2. TRANSITION
CHAPTER 3. KINEMATICS OF FLUIDS
Thus, in the Lagrangian representation, our attention is fixed on a chosen fluid particle
and we trace this particle in its motion specifying the variation of (v, p, ρ, T ) etc., in it’s
vicinity (Figure 3.1a). However, in fluids it is not easy to define and identify fluid particles
as they move around. Therefore, it is often more convenient to use the so-called Eulerian
description for fluids.
2. The Eulerian way of representation
A more common methods of describing fluid flow is the Eulerian description of fluid
motion, named after the Swiss mathematician Leonhard Euler (1707–1783). In this description, a finite volume of the fluid, so called control volume, is defined, through which
the fluid flows in and out. Instead of keeping track of the position and velocity of individual field particles we define functions of space and time within the defined volume, such as
the pressure and velocity field (Figure 3.1b). So, the fluid parameters (v, p, ρ) are specified
as functions of coordinates and time of a point in space {xi } ≡ (x1 , x2 , x3 ) and we follow
the time variation of the fluid parameters at this point. Alternatively, if we fix time we can
determine the spatial variation (spatial fields) of the same parameters. Thus, there is no
direct information about the motion of a given fluid particle in the Eulerian representation.,
rather we are concerned with the pressure, velocity, acceleration, etc., of whichever fluid
particle that happened to be at the location of interest at the time of interest.
Henceforth we will mostly use the Eulerian representation.
a)
b)
Figure 3.1: a) Lagrangian viewpoint: sit on a fluid element and move with it as the fluid moves. b)
Eulerian viewpoint: sit at a fixed point in space and watch the fluid move through your volume
element, the “identity” of the fluid in the volume is continually changing.
3.2
Transition from One Representation to Another
While the equations of motions in the Lagrangian description following individual fluid particles
are well known (e.g., Newton’s second law), the equations of motion of fluid flow are not so
readily apparent in the Eulerian description but must be carefully derived. The transition from
one representation to another can be rather complex. Indeed, let the pressure be known in the
Lagrangian representation,
pL = pL (t, ξ1 , ξ2 ξ3 ) ⇔ pL = pL (t, ξk )
42
CHAPTER 3. KINEMATICS OF FLUIDS
3.3. DERIVATIVES
and let us determine this quantity as a function of time at the point {xi }, i.e.
pE = pE (t, x1 , x2 , x3 ) ⇔ pE = pE (t, xk ),
in the Eulerian representation. To make this transformation we have to know the initial coordinates ξ ≡ ξk of the fluid particle which will appear at this point, {xi } , at the time t. Returning
back to the Lagrangian representation of coordinates xi = xi (t, ξk ) ≡ xi (t, ξ1 , ξ2 , ξ3 ) , we conclude that we need to solve the system of equations
xi = xi (t, ξk ) ≡ xi (t, ξ1 , ξ2 , ξ3 ),
for ξk where i = 1, 2, 3 and k = 1, 2, 3, i.e. we need to express


 ξ1 = ξ1 (t, x1 , x2 , x3 ) 
ξ2 = ξ2 (t, x1 , x2 , x3 )
ξi = ξi (t, xk ) ⇔


ξ3 = ξ3 (t, x1 , x2 , x3 )
leading to
pE (t, xk ) = pL [ξi (t, xk ), t] .
It is well known that the solution ξi = ξi (t, xk ) exists if the Jacobian of the transformation is not
equal to zero, i.e.
∂(x1 , x2 , x3 )
∂xi
J=
6= 0
≡ det
∂(ξ1 , ξ2 , ξ3 )
∂ξk
or
J≡
∂x1
∂ξ1
∂x2
∂ξ1
∂x3
∂ξ1
∂x1
∂ξ2
∂x2
∂ξ3
∂x3
∂ξ2
∂x1
∂ξ3
∂x2
∂ξ3
∂x3
∂ξ3
6= 0
(3.1)
In some cases, the time rate of the Jacobian J is of interest (see Exercise 3.7),
1 dJ
∂vk
=
≡ divv ≡ ∇ · v.
J dt
∂xk
3.3
(3.2)
Convective and Local Time Derivatives
In the Lagrangian representation the partial time derivative of some function is its rate of change
i
i
and ai = ∂v
are the velocity and acceleration of
for a fixed fluid particle. In particular, vi = ∂x
∂t
∂t
a given fluid particle respectively. These quantities are defined for a fixed (given) fluid particle
and determined by the vector ξk . On the other hand, in the Eulerian representation the partial
derivatives with respect to xi and t are the component of the gradient along the xi -axis and the
time rate of the corresponding field quantity, respectively. Indeed,
v(t + ∆t, xk ) − v(t, xk )
∂v(t, xk )
= lim
∆t→0
∂t
∆t
can not be treated as the acceleration of the fluid particle which at time t is at the point {xk },
because another particle will be at this very point at time t + ∆t. So, the terms v(t + ∆t, xk )
k)
and v(t, xk ) are related to two different fluid particles and therefore the derivative ∂v(t,x
is not
∂t
the acceleration for the given fluid particle in the Eulerian
representation.
In a waterfall, for
∂v(t,xk )
= 0 , but particles are accelerated
example, the acceleration at a fixed point is constant
∂t
43
3.3. DERIVATIVES
CHAPTER 3. KINEMATICS OF FLUIDS
by the action of the gravitational force.
We now obtain an expression for the acceleration in the Eulerian representation. Consider
the fluid particle which is at the point {xk } at time t. The same particle will be at the point
{xk + ∆xk } at time t + ∆t. Hence, the ith component of acceleration for this fluid particle, ai ,
is given by (using Taylor expansion)
vi (t, xk ) +
vi (t + ∆t, xk + ∆xk ) − vi (t, xk )
= lim
ai = lim
∆t→0
∆t→0
∆t
∂vi
∆xk
∂xk
+
∆t
∂vi
∆t
∂t
− vi (t, xk )
∆xk
∆t→0 ∆t
Taking now into account that lim
or, in vector form,
= vk , yields
∂vi
∂
∂
∂vi
vi
+ vk
=
+ vk
ai =
∂t
∂xk
∂t
∂xk
∂v
+ (v · ∇)v =
a=
∂t
∂
+ v · ∇ v.
∂t
(3.3)
, is called the local acceleration and is nonzero only
Here the first term on the right hand side, ∂v
∂t
for unsteady flows. The second term, (v · ∇)v, is called the advective acceleration (sometimes
the convective acceleration). This term can be nonzero even for steady flows. It accounts for the
effect of the fluid particle moving (convecting) to a new location in the flow, where the velocity
field is different. For example, consider steady flow of water through a garden hose nozzle. We
define steady in the Eulerian frame of reference to be when properties at any point in the flow
field do not change with respect to time. Since the velocity at the exit of the nozzle is larger
than that at the nozzle entrance, fluid particles clearly accelerate, even though the flow is steady.
The acceleration is nonzero because of the advective acceleration term in the relation (3.3). Note
that while the flow is steady from the point of view of a fixed observer in the Eulerian reference
frame, is not steady from the Lagrangian reference frame moving with a fluid particle that enters
the nozzle and accelerates as it passes through the nozzle.
Analogously one can find the time rate of any other vector or scalar field quantity f connected
with the given particle. Such a derivative is called the total or the material derivative; to emphasise that the material derivative is formed by following a fluid particle as it moves through the
flow field it is also called the derivative along the trajectory (the names particle or substantial
(sometimes the notation Df /Dt is
derivatives are also used in the literature). It is denoted by df
dt
used too). In our notation, the relationship between convective and local derivatives is
d
∂
=
+ v
· ∇} .
| {z
dt
∂t
|{z}
|{z}
Advective
Material
(3.4)
Local
Equation (3.4) can also be applied to other fluid properties besides velocity, both scalars and
vectors. For example, the material derivative of pressure is written as
∂p
dp
=
+ (v · ∇)p.
dt
∂t
This equation represents the time rate of change of pressure following a fluid particle as it moves
through the flow and contains both local (unsteady) and advective components.
44
CHAPTER 3. KINEMATICS OF FLUIDS
3.4
3.4. REYNOLDS’ TRANSPORT THEOREM
Reynolds’ Transport Theorem
Let F (xi , t) be an arbitrary vector (scalar) function (any property of the fluid such as its mass,
momentum, or its energy) and let V be a volume that is arbitrary in shape and moving with the
fluid and hence containing the same particle all the time (one can say the volume is “frozen” into
the fluid). Then we can conclude that each conservation principle is applied to an integral over
this volume
Z
I=
F (xi , t)dV.
V
R
For example, the mass of fluid within the volume is V ρdV , where ρ is the fluid density and
the integration is carried out over the entire volume V of the fluid contained within it. So, in
the conservation laws, the combination of the arbitrary volume and the Lagrangian coordinate
system means that the material derivatives of the volume integrals will be encountered. We thus
need to calculate the material derivative of an integral extended over a volume V moving in the
Lagrangian frame of reference
Z
dI
d
=
F (xi , t)dV .
dt
dt
V
As mentioned in the previous section, it is convenient to transform the integral in the right hand
side into an equivalent expression involving volume integrals of the Eulerian derivatives. We
perform the integration using the Lagrangian variables, xi . In these variables, the integration
will be extended over a fixed volume
V0 (position of V at t = 0) at any time. Since the volume is
R
d
fixed, the two operators dt and commute (transpose and remain equal in value) and we have
V0
dI
d
=
dt
dt
Z
d
F (xi , t)dV =
dt
V
Z
F Jdξ1 dξ2 dξ3 =
V0
Z
d
(F J)dξ1 dξ2 dξ3
dt
(3.5)
V0
where J is the Jacobian of the transformation from the Eulerian to Lagrangian variables represented by (3.1). Returning back to the Eulerian variables (inverse transformation) we have
1
1
dx1 dx2 dx3 ≡ dV
J
J
dξ1 dξ2 dξ3 =
dI
=
dt
Z
d
(F J)dξ1 dξ2 dξ3 =
dt
V0
Z
1 d
(F J)dV =
J dt
V
Z V
F dJ
dF
dV
+
dt
J dt
Taking now into account (3.2) and (3.4) yields
dI
d
=
dt
dt
Z
V
F (xi , t)dV =
Z V
∂F
+ (v · ∇)F + F ∇ · v dV.
∂t
(3.6)
Equation (3.6) relates the Lagrangian derivative of a volume integral of a given function F (xi , t)
to a volume integral in which the integrand has Eulerian derivatives only. This result is known
as Reynolds’ transport theorem. It will be used in Chapter 4 to derive the basic conservation
equations based on the various conservation principles.
45
3.5. STREAMLINES
3.5
CHAPTER 3. KINEMATICS OF FLUIDS
Streamlines
A streamline is a curve that is everywhere tangent to the instantaneous local velocity vector.
Streamlines are useful as indicators of the instantaneous direction of fluid motion throughout
the flow field. Streamlines cannot be directly observed experimentally except in steady flow
fields. Mathematically, however, we can write a simple expression for a streamline based on its
definition.
Consider an infinitesimal arc length dr = dxi + dyj + dzk along a streamline, where i, j, k
are unit vectors along x, y and z directions; dr must be parallel to the local velocity vector
v = ui + vj + wk by definition of the streamline (u, v and w are components of the velocity
vector). By simple geometric arguments using similar triangles, we know that the components
of dr must be proportional to those of v. Hence, the equation for a streamline can be written as
|dr|
dx
dy
dz
=
=
=
.
|v|
u
v
w
(3.7)
Here |dr| is the magnitude of dr and |v| is the speed, the magnitude of v. Equation (3.7)
is illustrated in two dimensions for simplicity in Figure 3.2. For a known velocity field, we
integrate Equation (3.7) to obtain an equation for the streamlines. In two dimensions, (x, y) and
(u, v), the following differential equation is obtained
v
dy
= .
(3.8)
dx
u
| {z }
along a streamline
In some simple cases, Equation (3.8) may be solvable analytically; in the general case, it must
be solved numerically. In either case, an arbitrary constant of integration appears. Each chosen
value of the constant represents a different streamline. Therefore the family of curves, that satisfy
(3.8), represents streamlines of the flow field.
Point (x+dx,y+dy)
Streamline
dr
v
x
v
dy
u
dx
y
Point (x,y)
x
Figure 3.2: For two-dimensional flow in the x, y-plane, arc length dr = (dx, dy) along a streamline is
everywhere tangent to the local instantaneous velocity vector v = (u, v).
46
CHAPTER 3. KINEMATICS OF FLUIDS
3.6
3.6. SUMMARY OBJECTIVES
Summary objectives
When you finish studying the Chapter 3, you should be able to do the following:
1. Formulate the key points in the Lagrangian and the Eulerian description of fluid flows.
2. Transform the acceleration of fluid particle from the Lagrangian representation to the Eulerian one and define the local and convective acceleration.
3. Derive the Reynolds transport theorem and relate the result to the conservation principle
of any property of the fluid.
4. Define the streamline and illustrate its definition for two-dimensional flow.
47
3.6. SUMMARY OBJECTIVES
CHAPTER 3. KINEMATICS OF FLUIDS
48
Chapter 4
Basic Laws of Ideal Fluid Dynamics
We shall begin with ideal fluids or so-called inviscid fluids. An ideal fluid refers to a fluid without
internal friction, hence without the transformation of mechanical energy into heat. Naturally, all
real flows have viscosity but it is sometimes useful to consider a fluid as ideal in order to simplify
the problem at hand. Thus, what is meant by an ideal fluid flow is actually a region in the viscous
fluid flow where net viscous forces are negligible compared to pressure and/or internal forces.
We also assume that heat transfer between volumes of the fluid can be neglected, and stresses in
the fluid can be specified in terms only one scalar quantity, namely, the pressure p.
4.1
System of Equations of Fluid Dynamics
We turn now to the derivation of the fundamental equations of fluid dynamics — the starting
point of the theory that will lead to an understanding of different fluid phenomena. These equations are formulations of the usual conservation laws of mechanics — conservation of mass and
Newton’s laws of motion (conservation of momentum and energy) appropriate to a fluid in motion. The conservation laws can be stated in the differential form, applicable at a point, or in the
integral form applicable to an arbitrary volume following the fluid. The latter is the so called
material volume which consists of the same fluid particles and thus its bounding surface moves
with the fluid. As one might suspect, the equations will look different in the two cases. In the
first case the resulting equations are differential equations. Solution of the differential equations
of motion provides a means of determining the detailed (point by point) behaviour of the flow. If,
on the other hand, in the problem under study, the information sought does not require a detailed
knowledge of the flow then it is more appropriate to use the integral formulation of the basic
laws. Before we can proceed with this formulation we need certain preliminary notes.
The mathematical description of the state of a moving fluid is effected by means of functions
which give the distribution of the fluid velocity v = v (x, y, z, t) and of any two thermodynamic quantities pertaining to the fluid, for instance the pressure p (x, y, z, t) and the density
ρ (x, y, z, t). All the thermodynamic quantities are determined by the values of any two of them,
together with the equation of state; hence, if we are given five quantities, namely, the three
components of the velocity v, the pressure p and the density ρ, the state of the moving fluid is
completely determined.
All these quantities are, in general, functions of the coordinates x, y, z and of the time t. We
emphasise that v (x, y, z, t) is the velocity of the fluid at a given point (x, y, z) in space and at
a given time t, i.e. it refers to fixed points in space (Eulerian description) and not to specific
49
4.1. SYSTEM OF EQUATIONS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
particles of the fluid moving about in space in the course of time. The same remarks apply to ρ
and p.
4.1.1
Equation of Continuity
One of the fundamental equations of fluid dynamics is the equation of continuity, or the conservation law of matter. It follows from the evident fact that the mass of a liquid in a volume
containing the same particles all the time is constant. This implies that
Z
d
ρdV = 0
(4.1)
dt
V
with
R
ρdV defined as mass of a liquid. Using Equation (3.6), we rewrite (4.1) as
V
Z V
∂ρ
+ v · ∇ρ + ρ∇ · v dV = 0
∂t
(4.2)
Since the result (4.2) must be true for an arbitrary volume V , the integrand must vanish everywhere. As a result, we obtain the continuity equation in the Eulerian variables, which can be
written in one of three forms:
∂ρ
∂ρ
dρ
+ v · ∇ρ + ρ∇ · v = 0 or
+ ∇ · (ρv) = 0 or
+ ρ∇ · v = 0
∂t
∂t
dt
(4.3)
The integral form of the continuity equation (4.2) can be transformed to show its physical meaning more clearly. For this purpose we use the Gauss divergence theorem to transform the volume
integral of v · ∇ρ + ρ∇ · v = ∇ · (ρv) into an integral over the bounding surface S:
Z
Z
∇ · (ρv)dV = ρv · ndS
V
S
where n is outward normal on S. Then we obtain from (4.2),
Z
Z
∂
ρdV = − ρv · ndS
∂t
V
(4.4)
S
Thus, the time rate of change of the fluid mass inside a fixed volume V is equal (with opposite
sign) to the mass of the fluid flowing out of the volume across the bounding surface.
The vector
j = ρv
is called the mass flux density. Its direction is that of the motion of the fluid, while its magnitude
equals the mass of the fluid flowing per unit time through unit area perpendicular to the velocity.
4.1.2
Euler Equation
Let us consider a particular volume in a fluid. The equation of motion of a volume element in
the fluid – or the Euler equation – can be obtained from Newton’s second law which states that
50
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.1. SYSTEM OF EQUATIONS
the time rate of momentum change of a volume of fluid is equal to the resultant force acting on
this volume, or in mathematical form
Z
d
ρvdV = F + F S
(4.5)
dt
V
where F =
R
ρf dV is the external body force (f being the force per unit mass), F S is the force
V
exerted by the surrounding fluid on the volume V . This force is equal (with opposite sign) to the
force with whichRthe volume acts on the surrounding medium, or for an ideal fluid (no viscous
forces), F S = − pndS, where p is the pressure, S is the surface bounding V , and n is outward
S
normal to S. According to Gauss’ divergence theorem, we have
Z
Z
FS = − pndS = − ∇pdV
S
(4.6)
V
Then Equation (4.5) may be written as
Z Z
d
(ρv) + ρv(∇ · v) dV = (−∇p + ρf ) dV
dt
V
(4.7)
V
if we take into account (3.4) and (3.6). Using the continuity equation (4.3), the integrand on the
left-hand side may be transformed into
dv
dρ
∂v
d
dv
(ρv) + ρv(∇ · v) = ρ
+ v ( + ρ∇ · v) = ρ
= ρ
+ ρ(v · ∇)v
|{z}
dt
dt
dt
∂t
| dt {z
}
(3.4)
=0 (4.3)
Since the volume V is arbitrary, the latter must be equal to the integrand on the right-hand side
of (4.7). This yields the required equation of motion of the fluid volume
∂v
∇p
dv
=
+ (v · ∇)v = −
+f
dt
∂t
ρ
(4.8)
with f being the force per unit mass. Some examples of this force are:
1. The gravitational force, i.e. f = g (g being the gravity acceleration directed to the
earth’s center, g = 9.81 m/s2 ) is a typical example of a body force in Euler’s equation
(4.8), fg = −∇U .
In the dynamics of the ocean and the atmosphere, the non inertial character of the reference
frame caused by Earth’s rotation is very important. Two inertial forces must be included
into (4.8) in this case:
2. The centrifugal force,
2
f cf = Ω r = ∇
Ω2 r 2
2
51
⇔ f cf = −∇U
4.1. SYSTEM OF EQUATIONS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
3. The Coriolis force,
f c = −2Ω × v
with Ω being the vector of the Earth’s rotation frequency.
Note that all these forces mentioned above, except the Coriolis one, are potential, i.e. they can be
represented as a gradient of some scalar function U , f = −∇U , known as a potential function.
Equation (4.8) with f = 0 was first obtained by L. Euler in 1755. It is called the Euler equation
and is one of the fundamental equations of fluid dynamics.
The equations of motion have to be supplemented by the boundary conditions that must be satisfied at the surfaces bounding the fluid. For an ideal fluid, the boundary condition is simply that
the fluid cannot penetrate a solid surface. This means that the component of the fluid velocity
normal to the bounding surface must vanish if that surface is at rest, i.e.
vn = 0
In the general case of a moving surface, vn must be equal to the corresponding component of
the velocity of the surface. At a boundary of two non-penetrating fluids, the condition is that
the pressure and the velocity component normal to the surface of separation must be the same
for the two fluids, and each of these velocity components must be equal to the corresponding
component of the velocity of the surface. Thus,
the general form of boundary conditions: At a boundary of a rigid body at rest, the normal
component of velocity vn must be zero. The tangential velocity component in an ideal fluid can
be arbitrary since shear stresses are absent.
4.1.3
Completeness of the System of Equations
The Euler equation (4.8) together with the continuity equation (4.3) comprises four scalar quantities for five scalar variables (density ρ, pressure p, and the three velocity components vi ) which
determine the fluid motion. Hence, the system is not yet complete. The equation of state
p = p(ρ, S)
(4.9)
as well as the equation for the entropy1 S must be added to it. In deriving the equations of motion
we have taken no account of processes of energy dissipation, which may occur in a moving fluid
in consequence of viscosity (internal friction) of the fluid and heat exchange between different
parts of it. Therefore, the results obtained holds only for motions of fluids in which thermal
conductivity and viscosity are unimportant; such fluids are said to be ideal. The absence of
heat exchange between different parts of the fluid (and also, of course, between the fluid and
bodies adjoining) means that the motion is adiabatic throughout the fluid. Thus the motion of
an ideal fluid must necessarily be adiabatic. In adiabatic motion the entropy of any particle of
fluid remains constant as that particle moves about in space. Denoting the entropy per unit mass
by S, we can express the condition for adiabatic motion as,
∂S
dS
≡
+ v · ∇S = 0,
dt
∂t
1
(4.10)
Entropy-a measure of the energy that disperses or spreads out in a process (at a specific temperature)-must be
added to complete the system. Remember that the 2nd law of thermodynamics tells us that energy spreads out if not
hindered by doing so, e.g., ice melting, frying pan cooling, rock falling etc.
52
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.1. SYSTEM OF EQUATIONS
the so-called adiabatic equation. It does not mean that entropy must remain constant at a fixed
point in space because different fluid elements with different entropy may occur at this point at
different times. Therefore the convective derivative with respect to time is used to express the
rate of change of entropy for a given fluid particle as it moves about. Equation (4.10) is the
general equation describing adiabatic motion of an ideal fluid. Using (4.3), we can rewrite it as
an “equation of continuity” for entropy
∂ (ρS)
+ ∇ · (ρSv) = 0
∂t
The product ρSv is called the entropy flux density.
The adiabatic equation usually takes a much simpler form. If, as usually happens, the entropy is
constant throughout the volume of the fluid (ideal fluid) at some initial instant, it retains everywhere the same constant value at all times and for any subsequent motion of the fluid. In this
case we can write the adiabatic equation simply as
S = const
and we shall usually do so in the following. Such motion is said to be isentropic.
The entropy S can be excluded from our system of equations by applying the operator of a
d
to the equation of state (4.9). Indeed,
convective derivative dt
dp
dρ
dS
∂p
∂p
=
+
dt
∂ρ S=const dt
∂S ρ=const dt
Taking now into account Equation (4.10) and denoting
∂p
2
c =
∂ρ S=const
(4.11)
we obtain
dp
dρ
= c2
dt
dt
(4.12)
where the velocity of sound, c = c(p, ρ), is a function of pressure and density. In the simplest
case, S = const throughout space, Equation (4.9) is reduced to
p = p(ρ)
(4.13)
which makes the system of Equations (4.3, 4.8) complete. The function p(ρ) depends on the
properties of the fluid under consideration. In particular, for an ideal gas, the equation of state is
given by pV γ = const, where γ = cp /cV is the ratio of the specific heat at constant pressure to
that at constant volume. Since V = 1/ρ is the unit mass volume, we have p = p0 (ρ/ρ0 )γ . As
it has been mentioned in the beginning of this Chapter, the state of a moving fluid is determined
by five quantities: the three components of the velocity, v, the pressure p, and the density ρ.
Accordingly, a complete system of equations of fluid dynamics should be five in number. For an
ideal fluid (S = const), these are the Euler equation, the equation of continuity, and the adiabatic
equation.
53
4.2. THE STATICS OF FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Thus, the system of equations
dρ
dt
+ ρ∇ · v = 0
dv
dt
= − ∇p
+f
ρ
dp
dt
= c2 dρ
dt
or
(4.3),
(4.8),
(4.12),
p = p(ρ)
(4.13),
together with the appropriate boundary conditions determines the motion of an ideal fluid
completely.
The case of an incompressible fluid has very important applications. In this case the density
of each fluid particle must remain constant along the trajectory, i.e.,
∂ρ
dρ
≡
+ v · ∇ρ = 0,
dt
∂t
both the local and convective derivatives of the density must be zero. We note that dρ
= 0 does
dt
dp
not imply that dt = 0 which could be deduced from (4.12), since the velocity of sound c is
infinite in an incompressible fluid. The continuity equation (4.3),
dρ
+ ρ∇ · v = 0,
dt
using
dρ
dt
= 0 can now be written as
∇ · v = 0.
(4.14)
Hence, to describe the motion of an incompressible fluid we use (4.14) in addition to Euler’s
equation (4.8), and the system of equations of hydrodynamics remains complete and closed.
4.2
The Statics of Fluids
Fluid statics deals with problems associated with fluids at rest. In fluid statics, there is no
relative motion between adjacent fluid layers, and thus there are no shear (tangential) stresses in
the fluid deforming it. The only stress we deal with in fluid statics is the normal stress, which
is the pressure, and the variation of pressure is due only to the weight of the fluid. Therefore, the
topic of fluid statics has mainly significance in gravity fields, and the force relations developed
naturally involve the gravitational acceleration g.
4.2.1
Basic Equations
Consider a fluid at rest, i.e. v = 0. Then the Euler equation (4.8) is reduced to the so-called
hydrostatic equation
∇p = ρf
(4.15)
This equation describes the mechanical equilibrium of the fluid. If there is no external force, the
equation of equilibrium is simply ∇p = 0, i.e. p = const; the pressure is the same at every point
in the fluid.
54
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.2. THE STATICS OF FLUIDS
The case where the force f is potential, i.e. f = −∇U , is important since we are mainly
concerned with gravity fields which are potential. Equation (4.15) now becomes
∇p = −ρ∇U
(4.16)
A solution to Equation (4.16) does not always exist. Indeed, we have a gradient on the left-hand
side. The right-hand side, however, can be represented as a gradient of some function only in the
case of a specific dependence of the density ρ on the coordinates. In fact, by applying the curl
(rot) operation to (4.16) we obtain
∇ × ∇p = −∇ × (ρ∇U ) = −(ρ∇ × ∇U + ∇ρ × ∇U ) = 0
So, the right-hand side of Equation (4.16) can be represented as a gradient of some function only
if ∇ρ × ∇U = 0 , i.e. the vector ∇ρ must be parallel with ∇U ( ∇ρ||∇U ).
Consider a fluid in a gravitational field when U = gz ; the z-axis is assumed to be directed
upward. Then Equation (4.16) becomes
∇p = −ρ∇U ⇒



∂p
∂y
∂p
∂x
=
∂p
∂z
= −ρg
=0
(4.17)
This implies that the pressure depends only on z: p = p(z). Some applications and examples:
1. The simplest case where the density is constant, ρ = const. Then
p = p0 , z = z 0
p = −ρgz + C ⇒
⇒ p = p0 − ρg(z − z0 )
C = p0 + gρz0
(4.18)
2. The atmosphere ρ = ρ(z). In the atmosphere the change of density ρ with height must be
taken into account. We consider the case of an isothermal (T = const) atmosphere where
the equation of state is that of an ideal gas, i.e.
p=
R
ρT
µ
(4.19)
where R is the universal gas constant and µ is the molecular weight of the gas. Substituting
p from (4.19) into (4.17), we obtain
ρ
µg
RT ∂ρ
= −gρ ⇒ ln
=−
z,
µ ∂z
ρ0
RT
where ρ0 ≡ ρ(z)z=0 ≡ ρ(0). The last equation yields
ρ = ρ0 exp(−µgz/RT )
(4.20)
This is the well-known barometric formula. The density decreases exponentially with
height.
55
4.2. THE STATICS OF FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
3. The barotropic case where pressure depends only on the density and vice versa, thus the
equation of state is givenbyp = p(ρ). Then, analogously, one can examine an isentropic2
γ
atmosphere with p = p0 ρρ0 . In general, we will call the case barotropic if ∇p × ∇ρ =
0 in contrast to the baroclinic case with ∇p × ∇ρ 6= 0.
We may mention here another simple consequence of (4.15). If a fluid (such as the atmosphere)
is in mechanical equilibrium in a gravitational field, the pressure in it can be a function only
of the altitude z (since, if the pressure were different at different points with the same altitude,
dp
motion would result). It then follows from (4.15) and (4.17) that the density ρ = − g1 dz
is also
a function of z only. The pressure and density together determine the temperature, which is
therefore again a function of z only. Thus, in mechanical equilibrium in a gravitational field,
the pressure, density and temperature distributions depend only on the altitude. If, for example,
the temperature is different at different points with the same altitude, mechanical equilibrium is
impossible.
4.2.2
Hydrostatic Equilibrium. Väisälä Frequency and the Condition for
the Absence of Convection
A fluid can be in mechanical equilibrium (i.e. exhibit no macroscopic motion) without being in
thermal equilibrium. Equation (4.15), the condition for mechanical equilibrium, can be satisfied
even if the temperature is not constant throughout the fluid. However, the question then arises of
the stability of such equilibrium. It is found that the equilibrium is stable only when a certain
condition is fulfilled. Otherwise, the equilibrium is unstable, and this leads to the appearance of
flows in the fluid which tend to mix the fluid in such a way as to equalise the temperature. This
motion is called convection. Thus the condition for a mechanical equilibrium to be stable is the
condition that convection must be absent. We will consider this question in detail.
Statement of the problem: Consider a fluid in a gravitational field with the general equation of
state p = p(ρ, S). Write down the equilibrium condition for this fluid and investigate the
stability of this equilibrium.
Using Equation (4.17) for the general equation of state yields
dS
∂p
∂p
dp
dρ
=
+
.
−ρg =
dz p=p(ρ,S)
∂ρ S dz
∂S ρ dz
| {z }
c2
∂p
, we can finally
Assuming the fluid to be inhomogeneous, i.e. ρ = ρ(z), and denoting Y ≡ ∂S
ρ
get
dρ
dS
−ρg = c2
+Y
.
(4.21)
dz
dz
Equation (4.21) describes the equilibrium condition for an inhomogeneous fluid with the general
equation of state in a gravitational field. The next step is to analyse when (at least qualitatively)
this equilibrium will be stable. To answer this question we have to analyse the forces acting on a
fluid element in the equilibrium and the perturbed state.
2
Isentropic process is a process during which the entropy remains constant. An isentropic flow is a flow that is
both adiabatic and reversible, that is no energy is added to the flow, and no energy losses occur due to friction or
dissipative effects.
56
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.2. THE STATICS OF FLUIDS
z
V=V0 + ∆V
F2
ξ
V0
z
F1
0
Figure 4.1: Hydrostatic equilibrium and its stability.
Consider an element of fluid in equilibrium at level z and having a specific volume V0 (p, S)
(see Figure 4.1), where p and S are the equilibrium pressure and entropy at height z. Since the
element is in equilibrium, the forces (body forces) acting on this element balance each other.
Two forces act on this element:
1. The force of gravity, F1 = −m(z)g ≡ −ρ(z)gV0 .
2. Archimedes force F2 = −F1 , Archimedes’ force is directed upward and is the same as F1
in magnitude.
So, the resulting force is given by;
F = F1 + F2 = 0.
Let’s now perturb this equilibrium by an adiabatic vertical displacement ξ acting on a given
fluid element and consider the resulting force F acting on the displaced element. It is easy to
understand that if this force will be in the opposite direction with respect to the displacement,
i.e. the force is restoring and tends to return the element to its original position, the equilibrium
will be stable. If, on the other hand, the direction of the resulting force will coincide with the
displacement, the equilibrium will be unstable and that will result in convective motion. The
specific volume of the displaced element (but not its mass) changes due to the fluid compressibility and becomes V (p′ , S) = V0 + ∆V where p′ is the pressure at height z + ξ while the entropy
of element remains constant (no heat exchange). The two forces now acting on the displaced
element are altered in the following way,
1. The force of gravity is unchanged, F1 = −ρ(z)gV0 since the mass of element remains
constant and changing of the volume is balanced by changing of the density.
2. The Archimedes force is altered according to F2 = g ρ(z +ξ) (V0 +∆V ) due to the vertical
displacement and the changes in the specific volume of the displaced element.
The resulting force will now be
F = −ρ(z)gV0 + ρ(z + ξ)g(V0 + ∆V ).
57
4.2. THE STATICS OF FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
+ . . . and
Since |ξ| ≪ |z|, we can expand ρ(z + ξ) in a Taylor series ρ(z + ξ) = ρ(z) + ξ ∂ρ
∂z
retaining only the terms linear in ξ and ∆V (since they are small) the resulting force becomes
1 ∂ρ
∆V
F = ρ(z)gV0
.
ξ+
ρ(z) ∂z
V0
The volume change ∆V can be found by applying the equation of state to a fluid element and
taking into account that change of entropy, ∆S = 0 (ideal fluid approximation). Indeed,
p = p(ρ) ⇒
dp
dp dρ
dρ
=
= c2 ,
dz
dρ dz
dz
here we assume that the fluid is inhomogeneous only in the z direction, therefore we can replace
, we have
the partial derivatives by ordinarily ones. Taking now into account that ρ = m
V
dp
dρ
= c2
⇔ ∆p = c2 ∆ρ = c2 m∆
dz
dz
∆p ≈ −c2 ρ(z)
m
∆V
1
= c2 2 ∆V ≈ −c2 ρ(z)V0 2 ⇒
V
V
V0
∆V
∆V
∆p
⇔
=− 2
.
V0
V0
c ρ(z)
On the other hand, using Equation (4.17) yields
∆p = ξ
dp
= −gρ(z)ξ.
dz
Then the volume change ∆V is given by
∆V
g
= 2 ξ.
V0
c
We finally find for the force acting on the displaced fluid element
g
1 dρ
F = gρ(z)V0
ξ.
+
ρ(z) dz c2
It is important to note that the resulting force in the frame of the linear analysis is proportional
to the displacement. Introducing now the Väisälä frequency
2
N = −g
g
1 dρ
+ 2
ρ(z) dz c
(4.22)
we can rewrite the force acting on the displaced fluid element in the form
F = −mN 2 ξ.
Conclusion: The fluid element is in a stable equilibrium only if the force is opposite to the
displacement, i.e. when
1 dρ
g
N 2 ≥ 0 or
(4.23)
+ 2 ≤0
ρ(z) dz c
58
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.2. THE STATICS OF FLUIDS
Hence, for the equilibrium to be stable, the density gradient must be negative (ρ decreases with
height) and sufficiently large in magnitude. This means that the considered fluid element which
undergoes an adiabatic displacement through a small interval must be heavier than the fluid
which it “displaces” in its new position. Otherwise the equilibrium is unstable and convective
currents will arise. Indeed, the specific volume of the fluid which was displaced is V (p′ , S ′ ),
where S ′ is the equilibrium entropy at height z + ξ. Thus we have the stability condition
V (p′ , S ′ ) > V (p′ , S) .
Expanding the left hand-side of this equation in powers of S ′ − S = ξ dS
results in the stability
dz
condition in the form
dS
∂V
> 0.
∂S p dz
The formulas of thermodynamics give
∂V
T ∂V
=
,
∂S p cp ∂T p
where cp is the specific heat at constant pressure. Both cp and T are positive, so that we can write
the stability condition as
dS
∂V
> 0.
∂T p dz
The majority of all substances expand while heated, i.e. ∂V
> 0. Then the stability condition,
∂T p
i.e. the absence of convection, takes the form
dS
>0
dz
i.e. the entropy must increase with height. The condition (4.23) can easily be rewritten in the
for the convection to be absent. To
form that must be satisfied by the temperature gradient dT
dz
1 dρ
this end the term ρ dz in the expression for the Väisälä frequency can be excluded by means of
(4.21). Indeed, we have from (4.21)
Y dS
g
1 dρ
=− 2 − 2
⇒
ρ dz
c
ρc dz
2
N = −g
g
1 dρ
+ 2
ρ dz c
g
Y dS
g
= −g − 2 − 2
+ 2
c
ρc dz c
N2 =
gY dS
≥0
ρc2 dz
⇒
(4.24)
The majority of fluids exhibit anincrease in pressure (dp > 0) on heating (dS > 0) under the
∂p
constant density ρ, i.e. Y ≡ ∂S
> 0. Then the condition (4.24) is reduced to the one derived
ρ
dS
above, dz > 0, i.e. the entropy must increase with height. To expand the derivative dS
we use
dz
the thermodynamic relation
dp
cp dT
cp
∂V
∂V
dS
=
−
>0
dS = dT −
dp ⇒
T
∂T p
dz
T dz
∂T p dz
59
4.3. BERNOULLI’S THEOREM
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Taking into account in this relation that
dp
dz
−
= − Vg finally yields
dT
gβT
<
dz
cp
where β = V1 ∂V
is the thermal expansion coefficient, β > 0 since the majority of fluids
∂T p
expand on heating. Convection occurs if this condition is not satisfied, i.e. if the temperature
. If, on the other hand,
decreases upwards with a gradient whose magnitude exceeds the value gβT
cp
the entropy is constant dS
= 0 the equilibrium becomes neutral.
dz
4.3
Bernoulli’s Theorem and the Energy Conservation Law
The Bernoulli equation is an approximate relation between pressure, velocity and elevation,
and is valid in regions of steady, incompressible, ideal fluid flow where net frictional forces are
negligible.
4.3.1
Bernoulli’s Theorem
The equations of fluid dynamics are much simplified in the case of steady flow. By steady flow
we mean a flow in which the velocity is constant in time at any point occupied by the fluid.
= 0. The famous theorem
In other words, v is a function of the coordinates only, so that ∂v
∂t
concerning steady flow of an ideal barotropic fluid (equation of state is given by p = p (ρ)) is
attributed to Bernoulli. To deduce this theorem we start with the Euler equation with ∂v
= 0 and
∂t
the convective term (v · ∇)v replaced by,
2
v
− v × (∇ × v).
(4.25)
(v · ∇)v = ∇
2
Further we introduce a new function w called enthalpy:
p
w = ε + pV = ε + ,
ρ
(4.26)
defined as the sum of the internal energy per unit mass and ρp being the flow energy (flow work)
which is the energy per unit mass needed to move the fluid and maintain the flow. By the internal energy of the system we mean the sum of all microscopic forms of energy, i.e. the forms
of energy related to the molecular structure of the system. In contrast, macroscopic energy of
a system is related to motion and the influence of external effects such as gravity, magnetism,
electricity, and surface tension.
We consider an ideal barotropic fluid, i.e. p = p(ρ) with ε being the internal energy per unit
mass, and V is the volume of a fluid element, thus
m = ρV ⇒ 1 = ρV ⇒ V =
1
dρ
⇒ dV = − 2
ρ
ρ
In differential form, Equation (4.26) can be written as
dw = dε +
dp
dρ
−p 2.
ρ
ρ
60
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.3. BERNOULLI’S THEOREM
For an ideal barotropic fluid the entropy S = const, and thus the entropy differential dS = 0.
Then the internal energy differential is given by (known from thermodynamics)
dε = (T dS − pdV ) |dS=0 = −pdV = p
dρ
.
ρ2
(4.27)
Hence, we have for the enthalpy differential
dw = dε +
dp
dρ
dρ dp
dρ
dp
−p 2 =p 2 +
−p 2 =
.
ρ
ρ
ρ
ρ
ρ
ρ
(4.28)
Now, assuming also that the external force is a potential one (f = −∇u) we may write the Euler
equation for a steady flow as
2
v
∇
+ w + u = v × (∇ × v) .
(4.29)
2
In the following it is convenient to define the concept of streamlines.
Definition: The line with a tangent collinear with the velocity vector v at each point is called a
streamline.
In the case of a steady flow the streamlines are fixed and coincide with the paths of fluid particles.
We choose now some streamline and consider the projection of Equation (4.29) on the tangent
to this line at an arbitrary point. The vector product v × (∇ × v) is orthogonal to v, and hence
the projection of this vector product on a streamline is equal zero. The projection of the gradient
∇ on the tangent to the given streamline
is thederivative dld with dl being the element of the
streamline. As a result, we have
d
dl
v2
2
+ w + u = 0, or
v2
+ w + u = C1
2
(4.30)
2
where C1 is constant and w = ε + ρp . Hence the quantity v2 + w + u is invariant along the
streamline. This statement is known as the Bernoulli theorem. Note that this quantity is also
invariant along the line whose tangent is parallel to the vector ∇×v (vortex line). The expression
(4.30) is called the Bernoulli integral. The constant C1 is different for different streamlines. If
the flow takes place in a gravitational field, u = gz, where g is the acceleration due to gravity,
the Bernoulli theorem states that along streamline
v2
+ w + gz = C1 .
2
Some particular cases:
1. Potential flow, vector ∇ × v = 0. We have then from (4.29)
2
2
v
v
+w+u =0⇒
+ w + u = const
∇
2
2
(4.31)
where the constant is the same for any point.
2. Incompressible fluid flow, ρ = const. In this case w = ρp up to within an arbitrary additive
constant. Then, from (4.30) we obtain the familiar form of the Bernoulli theorem
v2 p
+ + u = C1 .
2
ρ
61
(4.32)
4.3. BERNOULLI’S THEOREM
4.3.2
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Some Applications of Bernoulli’s Theorem
We begin with the flow of incompressible fluid in a tube with a constriction (Figure 4.2). Suppose
we are interested in the question: where is the pressure higher, at the wide section A or at the
narrow section B? The answer follows immediately from the Bernoulli theorem. Indeed, since
the velocity in the narrow section is larger (the mass rate at the both sections is the same), the
pressure at B is less, according to (4.32), if the potential u can be considered constant. If one
makes holes at A and B and connect them by means of a thin glass tube partly filled with
mercury, the difference in levels of the latter in the left and right parts of the connecting tube will
be proportional to the pressure difference. Such a device is called a Venturi velocity meter (see
Figure 4.2).
A
B
Figure 4.2: A venturi velocity meter.
Let us assign the subscripts 1 and 2 to the quantities p and v in the sections A and B respectively.
Then, in accordance with the Bernoulli theorem,
p1 v12
p2 v22
+
=
+ .
ρ
2
ρ
2
In addition, the relation v1 S1 = v2 S2 holds, S1 and S2 being the cross-sectional areas at A and
B, respectively. From the last two equations, we obtain
v
u 2 (p − p )
u
1
2
.
v1 = t 2
ρ S1/S 2 − 1
2
Hence, if the pressure difference (p1 − p2 ) is known, the velocity v1 can be determined.
An interesting example is a two-dimensional flow around two closely placed infinite cylinders
with axes perpendicular to the flow direction (Figure 4.3). An increase in velocity takes place
between the cylinders (streamlines here are situated more densely). It causes a pressure drop,
according to the Bernoulli theorem, and hence an attraction force occurs between the cylinders.
This phenomenon is well known in ship handling because two ships moving closely in the same
direction attract each other. Sometimes it could be the cause of a collision.
In some applications it is important to investigate the fluid outflow through a hole. Consider
some streamline connecting the free surface of the liquid in a vessel and the hole (see dashed
line in Figure 4.4). Let the levels of the surface and the hole be z1 and z2 respectively. At z = z1 ,
62
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.3. BERNOULLI’S THEOREM
Figure 4.3: An attraction force occurs between two cylinders closely placed in a flow.
z1
z2
Figure 4.4: Fluid outflow through a hole.
the pressure is atmospheric p = p0 and the velocity v ∼ 0 (the surface area is assumed large
compared with the area of the hole). At the hole, z = z2 , we have v = v2 and also p = p0 (we
neglect the atmospheric pressure difference between z1 and z2 ). Equating the left-hand sides of
Equation (4.32) for the two cases z = z1 and z = z2 and setting u = gz, yields
gz2 +
v22
= gz1 .
2
This results in the well-know formula
v2 =
p
2g(z1 − z2 ).
63
(4.33)
4.3. BERNOULLI’S THEOREM
4.3.3
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Bernoulli Theorem as a Consequence of the Energy Conservation
Law
Let’s now show that the Bernoulli theorem is just a consequence of the energy conservation law
by analysing the flow mass and energy balance of a flux tube element bounded by two cross
sections whose areas are S1 and S2 . For this purpose we define the concept of fluxtubes.
Definition: A tube formed by all streamlines passing through a small surface element normal to
them is called a flux tube.
Let the flow be directed from S1 towards S2 . At each cross section the flow is defined by the
set of parameters: p, ρ, v, u corresponding to the pressure, density, velocity, and potential of an
external force respectively.
a) Flow mass analysis:
Inflow of fluid mass across S1 during the time ∆t is ρ1 v1 S1 ∆t.
Outflow of fluid mass across S2 during the time ∆t is ρ2 v2 S2 ∆t.
In steady flow, these masses are equal, i.e.
∆M = ρ1 v1 S1 ∆t = ρ2 v2 S2 ∆t.
(4.34)
b) Energy balance analysis:
The pressure p1 does the work p1 v1 S1 ∆t on the inflowing fluid at S1 and the outflowing
fluid does the work p2 v2 S2 ∆t on the external medium at S2 . The difference between these
gives the energy increase, according to the energy conservation law:
p1 v1 S1 ∆t − p2 v2 S2 ∆t = ∆M (E2 − E1 )
(4.35)
where E1 and E2 are the energy per unit mass at S1 and S2 respectively. This energy is the
2
sum of kinetic v2 , potential u, and internal ε energies at each cross section
E=
v2
+ u + ε.
2
(4.36)
Substituting (4.36) into (4.35) and taking into account (4.34), we find
p2 v22
p1 v12
+
+ u 1 + ε1 =
+
+ u 2 + ε2
ρ1
2
ρ2
2
Introducing the corresponding enthalpies wi = εi + ρpii , i = 1, 2, we obtain the Bernoulli theorem
(4.30) which is now just a consequence of the energy conservation law.
4.3.4
Energy Conservation Law in the General Case of Unsteady Flow
In the previous section we considered steady flow. Let’s now generalise the energy conservation
law taking unsteady flow into account. Consider the flow of an ideal fluid under the action of an
external potential force. In this case, the energy of a fluid element per unit volume (the energy
2
2
density) is given by E = ρE = ρ v2 + ρu + ρε with E = v2 + u + ε being the energy per unit
64
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.3. BERNOULLI’S THEOREM
= 0, we obtain for the
mass. Assuming that u (the potential) does not depend on the time, ∂u
∂t
time rate of the energy density:
2
∂E
∂v ∂
v
∂ρ
+ ρv ·
+ (ρε) .
=
+u
∂t
2
∂t | {z∂t} ∂t
{z
}
|
2
1
The first term (1) on the right hand-side of this equation can be rewritten using the continuity
equation (4.3),
2
2
v
∂ρ
v
+u
=−
+ u ∇ · (ρv) .
2
∂t
2
The second term (2) can be simplified using the Euler equation (4.8). Indeed, taking into account
that f = −∇u, we have
ρv ·
Here,
∇p
ρ
∇p
∂v
= −ρv ·
− ρv · ∇u − ρv · (v · ∇) v.
∂t
ρ
2
= ∇w in accordance with (4.28). Using the identity (v · ∇) v = ∇ v2 − v × (∇ × v)
2
with v · [v × (∇ × v)] = 0 we obtain for the last term ρv · (v · ∇) v = ρv · ∇ v2 . Now,
2
∂E
v2
∂
v
=−
+ u ∇ · (ρv) − ρv · ∇w − ρv · ∇u − ρv · ∇ +
(ρε)
∂t
2
2
∂t
∂E
⇒
=−
∂t
2
∂
v2
v
+ u ∇ · (ρv) − ρv · ∇
+u+w +
(ρε) .
2
2
∂t
The time rate of the internal energy density can be determined using (4.26), (4.27) and (4.3).
Indeed,
p
d (ρε) = ρdε + εdρ|dε=p dρ = dρ + εdρ|w= pρ +ε = wdρ
2
ρ
ρ
Then,
∂
∂ρ
(ρε) = w
= −w∇ · (ρv)
∂t
∂t
(4.37)
Finally, we have
∂
∂E
≡
∂t
∂t
v2
ρ + ρu + ρε
2
= −∇ · ρv
v2
+u+w
2
(4.38)
We now integrate the relation (4.38) over an arbitrary fixed volume V . Then, the integral on the
right-hand side can be transformed into the integral
2 over the bounding surface S if one applies
the Gauss divergence theorem to the vector ρv v2 + u + w . Thus, we obtain the integral form
of the energy conservation law for a moving fluid:
∂
∂t
Z V
Z 2
v2
v
ρ + ρu + ρw vn dS
ρ + ρu + ρε dV = −
2
2
(4.39)
S
The left-hand side represents the rate of change of the energy in volume V whereas the the righthand side is the energy flux across the bounding surface S. So, in accordance with the energy
65
4.4. CONSERVATION OF MOMENTUM CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
conservation law, we have
the rate of change of energy in volume V is determined by the flux of energy across the bounding surface S.
Hence, the vector
2
v2
p
v
N ≡ ρv
+ u + w = ρv
+u+ε+
(4.40)
2
2
ρ
2
is the specific energy flux, and the sum v2 + u + w is the energy transported per unit mass
of fluid. Note, that on the left-hand side in Equation (4.39) we have ρεwhereas
on the right-hand
2
side the corresponding term which is added to the mechanical energy v2 + u is ρw 6= ρε. This
is because the fluid is not simply transporting the energy but also does some work on the volume
under consideration, for example by compressing it. It provides the additional energy flux pv.
Indeed, putting w = ε + ρp , we can write the flux of energy through a closed surface in the form
−
Z
S
ρ
Z
v2
+ u + ε vn dS − pv · dS
2
S
The first term is the energy transported through the surface in unit time by the mass of fluid. The
second term is the work done by pressure forces on the fluid within the surface. Note that in
according to (4.28). Since ǫ and w
the case of an incompressible fluid, dε = 0 but dw = dp
ρ
are determined to within additional constants, we can set ǫ = 0 and w = ρp in this case. Using
Equation (4.39), we can easily obtain the Bernoulli theorem. Simply choose the element of the
flux tube between cross sections S1 and S2 as volume V and consider the case of a steady flow
∂
= 0 . Then it follows from (4.39) since vn = 0 at the lateral walls of the tube, that
∂t
ρ1 v 1 S 1
v12
+ u1 + w 1
2
= ρ2 v2 S2
v22
+ u2 + w 2
2
This immediately yields the Bernoulli theorem (4.30), since ρ1 v1 S1 = ρ2 v2 S2 according to the
mass conservation law.
4.4
Conservation of Momentum
4.4.1
The Specific Momentum Flux Tensor
We shall now give a similar series of arguments for the momentum of the fluid. By definition ρv
is the momentum per unit volume. Let us calculate the time rate of this momentum. It is more
convenient in these calculations to use tensor notations. Thus, we have
∂
(ρvi ) =
∂t
∂vi
ρ
∂t}
| {z
∂v
∂p
− ∂x
−ρvk ∂x i +fi ρ
i
⇒
∂
(ρvi ) = −
∂t
k
∂ρ
+ vi
∂t}
| {z
−vi
∂ (ρvk )
∂xk
∂vi
∂p
∂ (ρvk )
+fi ρ − ρvk
− vi
∂x
∂xk
∂xk
|{z}i
|
{z
}
∂
∂xk
∂
∂xk
(δik p)
66
(ρvi vk )
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS 4.4. CONSERVATION OF MOMENTUM
⇒
∂
∂
(ρvi ) = −
(pδik + ρvi vk ) + fi ρ.
∂t
∂xk
(4.41)
Denoting
Πik = (pδik + ρvi vk )
(4.42)
we rewrite (4.41) as
∂
∂
(ρvi ) = −
Πik + fi ρ.
∂t
∂xk
Now we integrate the last equation over an arbitrary fixed volume V :
Z
Z
Z
∂
∂
Πik dV + fi ρdV .
(ρvi ) dV = −
∂t
∂xk
V
(4.43)
V
V
The left-hand side is the rate of change of the ith component of the momentum contained in the
volume considered. The first integral on the right-hand side can be transformed into an integral
over the bounding surface S with the help of the Gauss divergence theorem. Then,
Z
Z
Z
∂
(ρvi ) dV = − Πik nk dS + fi ρdV
(4.44)
∂t
V
S
V
where {nk } are the unit vector components of the outward normal to S. We thus conclude that
the change of the momentum in volume V is related to the flux of the momentum across the
boundary surface and to the action of an external force.
If the later is absent (fi = 0), Equation (4.44) yields the law of momentum conservation
Z
Z
∂
(4.45)
(ρvi ) dV = − Πik nk dS
∂t
V
S
i.e.
the time rate of the momentum in V is equal to the flux of this momentum across the confining
surface.
Consequently, Πik nk is the flux of the ith component of momentum through the unit surface
area. Taking into account (4.42) we may write that Πik nk = pni + ρvi vk nk , or in vector form
pn + ρv (v · n), which determines the momentum flux in the direction of n. Thus Πik is the ith
component of the amount of momentum flowing in unit time through unit area perpendicular to
the xk -axis. Therefore the tensor Πik is called the tensor of the specific momentum flux. We
note that the energy flux is determined by a vector (see 4.40), on the other hand the energy is a
scalar; the momentum flux, however, is determined by a tensor of rank two and the momentum
itself is a vector.
4.4.2
Euler’s Theorem
In applications it is often important to know the force exerted by the moving fluid on the surrounding medium. Consider this force for a steady flow with no external force. Equation (4.45)
then assumes the following form:
Z
Z
Πik nk dS = (pδik + ρvi vk )nk dS.
(4.46)
S
S
67
4.4. CONSERVATION OF MOMENTUM CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
We choose S for the bounding surface of some small part of a flux tube. A sectional view of this
flux tube part can be seen in Figure 4.5. The flow velocity is v = −vn, v = |v| at the inlet face
S2
v
n
S1
n
v
Figure 4.5: A segment of a flux tube.
(S1 ) of the part under consideration and v = vn at the outlet (S2 ). The integral
Z
ρvi vk nk dS
SL
over the lateral walls of the tube vanishes since vk nk = v · n = 0 (the velocity is tangent to the
walls). Thus Equation (4.46) takes the form
Z
Z
pni dS = −
ρvi vk nk dS.
(4.47)
S1 +S2
S
The left-hand side here is the ith component of the required force F acting on the surrounding
medium. Since vk nk = v · n = −v at S1 and vk nk = v · n = v at S2 , Equation (4.47) becomes
Z
Z
2
Fi = − ρv ni dS − ρv 2 ni dS.
(4.48)
S1
S2
Thus, we have deduced the Euler theorem, i.e.
theRforce exerted by the flow in the flux tube upon the surrounding medium is the sum of the forces
− ρv 2 ndS, j = 1, 2, acting on the inlet and outlet faces of the tube.
Sj
Note that the vector
P1 = −
Z
ρv 2 ndS
(4.49)
S1
is the momentum flowing into the chosen part of the flux tube across S1 per unit time. The
minus sign is because v and the outward normal n have opposite directions at S1 . Analogously
Z
P 2 = ρv 2 ndS.
(4.50)
S2
68
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.5. MOTIONS OF IDEAL FLUIDS
is the momentum outflow across S2 per unit time. Hence, Equation (4.48) can be written as
F = P1 − P2
(4.51)
i.e. the force F is equal to the momentum loss by the flux tube per unit time. Multiplying
(4.51) by dt and changing the signs, we obtain the well-known classical form of the momentum
theorem:
−F dt = (P 1 − P 2 )dt
i.e.
the impulse of the force equals the momentum increment in the time interval dt.
Here, −F is the force exerted by the surrounding fluid on the fluid in the flux tube, in accordance with Newton’s third law.
4.4.3
Application of Euler’s Theorem
Consider the force exerted by the flowing fluid on the tube walls. If the tube is cylindrical with
a straight generator, we have the trivial case where P 2 = P 1 and the force is zero. The case is
nontrivial when the flux tube is curved and possibly of variable cross section as for the part of
the flux tube shown in Figure 4.5. Here, the subscripts 1 and 2 refer to the inlet and outlet ends
of the tube, respectively. The total force acting on the boundaries of the part under investigation
is given by
Z
Z
Z
F =
pndS = F w +
S
pndS +
S1
pndS.
S2
The first term on the right-hand side is the required force on the lateral walls of the tube, while
the second and the third are forces on the cross sections S1 and S2 , respectively. We have for the
force on the lateral walls, according to Euler’s theorem (4.48),
Z
Z
2
Fw = −
p + ρv ndS −
(4.52)
p + ρv 2 ndS.
S1
S2
Hence, this force is determined in terms of the flow parameters at the ends of the tube. In the
case where the end faces are the planes of areas S1 and S2 , and the flow parameters p, ρ and v
are constant across these planes, Equation (4.52) becomes
F w = −S1 (p1 + ρ1 v12 )n1 − S2 (p2 + ρ2 v22 )n2 .
4.5
Motions of Ideal Fluids
In fluid mechanics, as in solid mechanics, an element may undergo four fundamental types of
motion or deformation, as illustrated in Figure 4.6:
(a) translation
(b) rotation
(c) linear (extensional) strain
69
4.5. MOTIONS OF IDEAL FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
(d) shear strain
a)
b)
c)
d)
Figure 4.6: Fundamental types of fluid element motion or deformation: (a) translation, (b) rotation, (c)
linear strain, and (d) shear strain.
The study of fluid dynamics is further complicated by the fact that all four types of motion or
deformations occur simultaneously. Because fluid elements may be in constant motion, it is
preferable to describe the motion and deformation of fluid elements in terms of rates. Thus,
the velocity (v) gives the rate of translation and angular velocity (Ω) gives the rate of rotation.
Furthermore we discuss linear shear strain (rate of linear strain) and shear strain rate (rate of
shear strain). In order to describe such decomposition of a local motion mathematically the
standard approach is to consider a relative motion of a fluid in a neighbourhood of an arbitrary
chosen point x, where the velocity is v (x). Writing then the fluid velocity of a neighbouring
point x + r as v + δv at the same instant t, we obtain for the velocity of the relative motion
δv = v (x + r, t) − v (x, t) = (r · ∇) v + O r2
by the Taylor expansion with respect to the separation vector r. To the first order in the linear
dimensions of a small region around point x, the vector δv can be presented as a superposition
of symmetric and anti-symmetric components, δv = δv (s) + δv (a) , as shown in Batchelor3 .
Batchelor shows that the symmetric part of the relative velocity, δv (s) , is proportional to the
displacement, and thus describes translation, linear and shear straining motion, whereas the antisymmetric part describes the rotation at the angular velocity 12 ∇ × v. Namely, every point r in
the neighbourhood of x rotates with the same angular velocity. So, the anti-symmetric part of
the velocity, δv (a) , represents the local rigid-body rotation. In analytical terms, these results
are content of the statement that the velocity field in a small region surrounding the position x
consists of, in general, the superposition of:
a) a uniform translation with velocity v (x) (Figure 4.6a);
b) a pure straining motion characterised by the rate-of strain tensor which itself may be decomposed into an isotropic expansion (Figure 4.6c) and a straining motion without change
of volume (Figure 4.6d);
3
G. K. Batchelor, An Introduction to Fluid Mechanics, Cambridge Mathematical Library, 1967.
70
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.5. MOTIONS OF IDEAL FLUIDS
c) rigid body rotation (Figure 4.6b) with angular velocity given by
1
Ω = ∇ × v.
2
4.5.1
(4.53)
Vorticity
The rate of the rotation vector of a fluid element has been defined by (4.53). A closely related
kinematic property of great importance in the analysis of fluid flows is the vorticity vector or
vortex vector defined as
ω = ∇ × v ≡ curl(v).
(4.54)
Thus, the rate or rotation (Equation 4.53) is equal to half of the vorticity vector which in other
words mean that vorticity equals twice the angular velocity of a fluid particle. From this it
follows that vorticity is a measure of the rotation of a fluid particle. The components of the
vorticity vector are
ω1 =
∂v3
∂v2
∂v1
∂v3
∂v2
∂v1
−
, ω2 =
−
, ω3 =
−
∂x2 ∂x3
∂x3 ∂x1
∂x1 ∂x2
It is thus the vector quantity defined at every point within the fluid. The definition of curl by
I
1
n · ∇ × v = lim
v · dl
S→0 S
l
(where n is the unit vector normal to the surface S, round the edge of which the line integral is
taken) indicates that vorticity corresponds to rotation of the fluid; the line integral is non-zero
only if the fluid is going round the point under consideration. Likewise, if the vorticity in a flow
is zero then fluid particles there are not rotating and the flow is called irrotational. So, a fluid
motion will be irrotational if ω = 0 which would require
∂vj
∂vi
=
, i 6= j
∂xj
∂xi
In irrotational (potential) flow, the velocity vector can be written as the gradient of a scalar func∂ϕ
tion ϕ (xi , t). This is because the assumption vi ≡ ∂x
satisfies the condition of irrotationality.
i
Definition: The fluid flow is called vortex flow if ∇ × v 6= 0 in some volume of the fluid and
the flow is called potential or irrotational if ∇ × v = 0.
Not all flows with circular streamlines are rotational. To illustrate this point, we consider two
incompressible, two-dimensional flows, both of which have circular streamlines in the rθ plane.
The velocity profiles of the two flows, A and B, along with their streamlines are sketched in
Figure 4.7.
Flow A Consider a fluid undergoing rigid-body rotation with angular velocity Ω. The velocity field
is v = Ω × r, giving
vθ = Ωr, vr = vz = 0.
The vorticity field for this flow in polar coordinates is:
1 ∂Ωr2
− 0 = 2Ω, ωr = ωθ = 0
ωz =
r
∂r
The vorticity for rigid-body rotation is thus nonzero and a constant in magnitude of twice
the angular velocity and pointing in the same direction. Flow A is rotational.
71
4.5. MOTIONS OF IDEAL FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Flow B The second example again has all fluid particles moving on a circular path around z-axis,
but with different radial distribution of azimuthal velocity:
vθ =
m
, vr = v z = 0
r
This gives
1 ∂
(rvθ ) = 0 for r 6= 0
r ∂r
but leaves ωz non-determined on the axis. In fact, ωz becomes infinite on the axis, as it can
be seen from the Stoke’s theorem. Indeed,
Z
Z
I
v · dl = curlv·dS = ω · dS
ωr = ωθ = 0, ωz =
l
Carrying out the line integration around a circular path at constant r makes the left-hand
side equal to 2πm independently of r. Hence, there must be a singularity in ωz such that
the contribution ωz dSz from the element at r = 0 to the surface integral is 2πm. Hence
Flow B is irrotational everywhere except on the axis. Obviously exactly this flow can not
occur in practice; however, there are situations in which the vorticity is very high over a
narrow linear region and practically zero elsewhere.
These examples illustrate an important distinction. Rotation, as specified by vorticity, corresponds to a changing orientation in space of the fluid particle and not to motion of the particle on
a closed path. In Flow B each fluid particle moves around a circular path, but its vorticity is zero.
The concept of an isolated rigid particle moving on a circular path with and without rotating is
illustrated on the Figure 4.7. A fluid particle in Flow A behaves like the particle in Figure 4.7
(a), in which the particle both moves on a circular path and rotates. Its vorticity corresponds to
its rotation. The behaviour of a fluid particle in Flow B is analogous to that of the particle in
Figure 4.7 (b), i.e. circulation without rotation.
We will in the following discuss the basic properties of vortex flow (i.e. Flow A).
4.5.2
Circulation of Velocity
The integral
Γ=
I
v · dr
(4.55)
l
extended over the closed contour l is called the circulation of velocity with respect to this contour, where dr is an element of contour. So, the circulation is defined as the line integral of the
tangential component of velocity. By means of the well known Stokes’ theorem we can transform the integral (4.55) into a surface integral over an arbitrary open surface S confined by l.
It is important to note that S must be a simply connected surface, i.e. it can be contracted to a
point. Now we obtain for the circulation of velocity
I
Z
Γ = v · dr = n·(∇ × v)dS
(4.56)
l
S
where n is normal to the surface S. The positive direction of n is related to the positive sense
of traveling l by the right-hand rule. Thus, the circulation around a closed curve is equal to the
72
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Flow A
vϕ
4.5. MOTIONS OF IDEAL FLUIDS
Flow B
vθ =Ωr
vϕ
vθ =m/r
r
r
Figure 4.7: Successive position of fluid particle (smiley) for Flow A: rigid-body rotation and Flow B:
circulation without rotation. Flow A is rotational but Flow B is irrotational everywhere except
the origin.
surface integral of the vorticity, which we can call the flux of vorticity. Equivalently,
vorticity at a point equals the circulation per unit area.
It follows from (4.56) that for a potential flow (∇ × v = 0):
a) The circulation of the velocity over an arbitrary simply connected contour is zero.
b) There are no closed streamlines in a potential flow.
The requirement for the contour to be simply connected is important here. If we choose l as a
contour around a cylinder (such a contour can not be contracted to a point while l is in the fluid
all the time) the circulation can be different from zero even in a potential flow around a cylinder.
4.5.3
Kelvin’s Circulation Theorem
Let us consider a closed contour drawn in the fluid at some instant. We suppose it to be a “fluid
contour”, i.e. composed of the fluid particles that lie on it. In the course of time these particles
move about, and the contour moves with them. Let us investigate what happens to the velocity
circulation. In other words, let us calculate the time derivative
I
d
v · dr
dt
l
We have written here the total derivative with respect to time, since we are seeking the change in
the circulation round a “fluid contour” as it moves about, and not round a contour fixed in space.
The answer to this question is given by the following theorem by Lord Kelvin:
73
4.5. MOTIONS OF IDEAL FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Kelvin’s Circulation Theorem: the circulation of the velocity with respect to any closed “fluid”
contour (i.e. contour consisting of the same fluid particles all the time) remains unchanged
(Lord Kelvin, 1869).
This statement also known as the law of conservation of circulation is an important theorem in
fluid dynamics. To prove it we obtain from (4.55)
I
dΓ
d
=
v · dr
dt
dt
l
where l is a closed fluid contour. To calculate the convective derivative we have to take the
change of the contour’s position in space into account , hence


I
dΓ
dr 
 dv
(4.57)
= 
· dr + v · d
.
dt
dt{z }
dt
|
{z
}
|
l
1
2
In the second term (2)
dr
v2
= v ⇒ vdv = d .
dt
2
To transform the first term (1) we use the Euler equation (4.8) assuming f = −∇u (potential
(isentropic change, see Equation (4.28)). Hence,
force) and taking into account that ∇w = ∇p
ρ
dv
= −∇w − ∇u = −∇(w + u).
dt
Now, Equation (4.57) assumes the form
I 2
dΓ
v
= d
− u − w = 0.
dt
2
l
It vanishes since it is an integral over a closed contour of an exact differential. Thus in an ideal
fluid, the circulation of velocity round a closed “fluid contour” is constant with respect to time
and the theorem is proved.
It should be emphasised that this result has been obtained by using Euler equation and involves
the assumption that the flow is isentropic. The theorem does not hold for flows which are not
isentropic. Indeed, mathematically it is necessary that there should be a one-to-one relation between p and ρ which for isentropic flow is s (p, ρ) = const; then the term, − ρ1 ∇p, can be written
as the gradient of some function, a result which is needed in deriving Kelvin’s theorem.
By applying Kelvin’s theorem to an infinitesimal closed contour δl and transforming the integral according to Stokes’ theorem, we get
I
Z
v · dr = curlv · dS ∼
= δS · ω = const
where dS is a fluid surface element spanning the contour δl. The constancy of the product above
can be intuitively interpreted as meaning that vorticity moves with the fluid. Below this result
will be proved exactly as the statement of one of the Helmholtz theorems.
74
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.5.4
4.5. MOTIONS OF IDEAL FLUIDS
Discussion of Kelvin’s Theorem and Helmholtz Theorems
Since circulation is the surface integral of vorticity (see equation 4.56), Kelvin’s theorem essentially shows that irrotational flows remain irrotational if the following restrictions are satisfied:
1. Ideal fluid approximation. Indeed, in deriving the theorem, the inviscid Euler equation
has been used, but only along the contour l itself. This means that circulation is preserved if
there are no net viscous forces along the path followed by l. If l moves into viscous region
such as for example boundary region along solid surface then the circulation changes. The
presence of viscous effects causes a diffusion of vorticity into or out of a “fluid contour”
and consequently changes the circulation.
2. Potential body forces. Potential body forces such as gravity act through the centre of
mass of a fluid particle and therefore do not tend to rotate it.
3. Barotropic flow. This means that the density ρ must be a function of the pressure p only.
As examples of barotropic flows, we can mention a flow of an incompressible fluid for
which ρ is constant everywhere, and an isentropic (or adiabatic) flow of ideal fluid for
which ρpγ = const.
The Helmholtz theorems can be deduced from Kelvin’s theorem. First we introduce some new
definitions.
Definition: A vortex line is a curve whose tangent at each point has the direction of ω.
Definition: A vortex tube consists of all vortex lines passing through a surface element normal
to them.
Now the Helmholtz theorems can be formulated as:
Theorem 1: The element of an ideal fluid free of vorticity at some initial time remains so indefinitely.
Proof:
To prove it we note that, in accordance to Kelvin’s theorem, the circulation
I
Γ = v · dr
l
with respect to a closed fluid contour remains unchanged. This integral can be transformed
into the integral over the “fluid” open surface S bounded by l
Z
Γ = n · (∇ × v)dS
S
which also must remain unchanged. If at a certain instant ∇ × v = 0 everywhere on S,
then Γ = 0 for any l placed on S and it persists in being zero indefinitely.
Thus, vortices can not be created in an ideal liquid moving in a potential field.
75
4.5. MOTIONS OF IDEAL FLUIDS
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Obviously, the inverse statement also holds, i.e. if ∇ × v 6= 0 at some time, the vorticity cannot be destroyed since Γ remains unchanged. In practice however, it is a common
fact that vortices can appear and disappear (smoke rings, whirlpools, etc.) It is due to the
viscosity of the fluid or its baroclinity (the thermal equation of state does not simply reduce
to p = p(ρ)), or else because the external force is non potential.
Theorem 2: A vortex line consists of the same fluid particles, i.e. it moves together with the
fluid. One can say that vortex line is frozen into the fluid motion.
Proof:
Let us take a vortex tube and consider the circulation Γ over an arbitrary simply connected
closed contour l situated completely on the tube surface
I
Z
Γ = v · dr = n·∇ × vdS.
S
l
So, we have Γ = 0 initially since n · (∇ × v) = 0 due to the definition of a vortex
tube. It follows from the Kelvin’s theorem that Γ = 0 permanently for the contour l, i.e.
n · (∇ × v) = 0 permanently for a surface of a chosen contour l. Since this surface can be
chosen arbitrarily on the tube’s surface the same fluid particles will remain on the tube’s
wall permanently. If we now let the cross sectional area of the vortex tube tend to zero, we
obtain the vortex line.
Consequently, the vortex line will permanently consists of the same fluid particles. This
can also be formulated as the vorticity (ω = ∇ × v) being frozen into the motion of an
ideal fluid.
Theorem 3: The flux of the vortex vector ω = ∇ × v through the vortex tube cross section, i.e.
the quantity
Z
n · ωdS
S
remains unchanged along this tube.
Proof:
To prove this statement we note first that the flux of vector ω through any closed surface
S0 is zero. Indeed, we have according to Gauss’ divergence theorem
Z
Z
Z
n · ωdS = ∇ · ωdV ≡ ∇ · (∇ × v)dV = 0.
V
S0
V
Let now S0 be the bounding surface of the part of a vortex tube:
S 0 = S1 + S2 + S ′
where S ′ is the surface of the walls and S1 and S2 are the end faces of the chosen part of
the vortex tube. We then have
Z
ω · ndS = 0, since n⊥ω,
S′
76
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
0=
Z
n · ωdS =
S0
⇒
Z
n · ωdS+
S1
n · ωdS+
Z
{z
Z
n · ωdS +
S2
n · ωdS = 0 ⇒
S2
S1
|
Z
4.5. MOTIONS OF IDEAL FLUIDS
Z
n · ωdS
S′
Z
|
n · ωdS =
S1
{z
=0
Z
}
n · ωdS.
S2
}
negative since
n is directed
opposite to ω at S 1
So, the influx and outflux of the vector ω are balanced, so their sum is equal zero. Hence,
the theorem is proved.
When S1 and S2 is sufficiently small and can be assumed constant across the vortex tube,
then the quantity ωS, called the vortex strength, remains constant (unchanged) along the
tube. The vortex tube consists of many vortex lines passing through a surface element
normal to the tube. If the vortex strength ωS is unchanged, this means that there are no
vortex lines which begin or end inside the tube.
Thus, the vortex tube has neither a beginning nor an end in the fluid. It is closed, since
otherwise, its ends would be at infinity or at the walls of the vessel.
4.5.5
Vorticity Equation
Flows that are not barotropic are called baroclinic. Consider fluid elements in barotropic and
baroclinic flows (see Figure 4.8). For the barotropic element, lines of constant p are parallel to
lines of constant ρ, which implies that the resultant pressure forces pass through the centre of
mass of the element. For the baroclinic element, the lines of constant p and ρ are not parallel. The
net pressure force does not pass through the centre of mass, and the resulting torque changes the
circulation and thus creates the vorticity. The baroclinic mechanism of vorticity generation can
be described analytically as well. Indeed, when the flow is not isentropic, i.e. flow is baroclinic,
the right-hand side of the Euler equation 4.8 can not be replaced by (−∇w) and to describe this
flow one has to use an equation governing the vorticity. To deduce this equation we start with
the Euler equation 4.8 assuming the body forces are potential, i.e. f = −∇u. Then an equation
for rate of change of vorticity is obtained by taking the curl of the Euler equation. In symbolic
form, we have to perform the operation
∇p
∂v
∇×
+ (v · ∇) v = −
− ∇u
∂t
ρ
Using the vector identity 4.25 and noting that the curl of a gradient vanishes results in
∂ω
1
= ∇ × (v × ω) + 2 (∇ρ × ∇p)
∂t
ρ
This equation is known as the vorticity equation for an ideal fluid and represents the rate of
change of vorticity in an inhomogeneous fluid. Using the vector identity (11.16) and assuming
motion to be incompressible we can rewrite the first term on the right-hand side as
∇ × (ω × v) = (v · ∇) ω − (ω · ∇) v
The vorticity equation then becomes
∂ω
dω
1
+ (v · ∇) ω ≡
= (ω · ∇) v + 2 (∇ρ × ∇p)
∂t
dt
ρ
77
4.6. SUMMARY OBJECTIVES
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
Here the term (ω · ∇) v represents the action of velocity variations on the vorticity and corresponds to rate of change of vorticity due to stretching and twisting of vortex lines. This term is
absent in two-dimensional flows, in which ω is perpendicular to the plane of flow. The second
term on the right-hand side is called the baroclinic vector and describes the rate of vorticity
generation due to baroclinicity of the flow.
net pressure force
net pressure force
G
Baroclinic
Barotropic
ρi
nc
ris
ing
G
p = constant lines
ρ = constant lines
G=center of mass
Figure 4.8: Mechanism of vorticity generation in baroclinic flow, showing that the net pressure force
does not pass through the centre of mass G. The radially inward arrows indicate pressure on
an element.
4.6
Summary objectives
When you finish studying the Chapter 4 , you should be able to do the following:
1. Derive the continuity equation as the conservation law of mass, write different forms of
this equation and consider the case of an incompressible fluid.
2. Derive the Euler equation using the Newton’s second law of mechanics, indicate examples
of external forces and boundary condition that must be satisfied at the surface of the body.
3. Consider the completeness of the continuity equation and Euler equation and prove the
importance of the equation of the state. Write different forms of this equation.
4. Formulate the hydrostatic equation as the equation describing the hydrostatic equilibrium
in the fluid. Describe different forms of this equilibrium and determine the barotropic and
baroclinic fluids.
5. Write the equilibrium condition for the fluid in the gravitational field and investigate the
stability of this equilibrium. Explain the result from the physical points of view, namely,
directions of the resulting force and the displacement of the fluid element.
78
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
4.6. SUMMARY OBJECTIVES
6. Derive the Bernoulli’s theorem, discuss its physical sense and consider different particular
cases with practical applications.
7. Derive and formulate the momentum conservation law in the differential and integral forms
via tensor of the specific momentum flux.
8. Describe the different types of fluid motions and introduce the vorticity vector. Discuss
the physical sense of this quantity. Give examples of rotational and irrotational flows.
9. Introduce the circulation of the velocity, formulate and prove the Kelvin’s circulation theorem and explain the requirements for the ”fluid contour”.
10. Formulate the Helmholtz theorems and discuss the related properties of the ideal fluid
flows.
11. Derive the vorticity equation and discuss the role of the baroclinic vector.
79
4.6. SUMMARY OBJECTIVES
CHAPTER 4. BASIC LAWS OF IDEAL FLUIDS
80
Chapter 5
Potential Flow
In the last chapter we discussed ideal or inviscid fluids which is an approximation that can be
used in regions of the fluid flow where net viscous forces are negligible compared to pressure
and/or inertial forces. In this chapter we will discuss another approximation which can be applied to regions in which fluid particles have no net rotation. As defined in the last chapter a
flow for which ∇ × v = 0 (no rotation) in all space is called a potential flow or irrotational flow.
From this follows that for potential flow, the velocity circulation along any closed contour is zero
and therefore closed streamlines cannot exist in this flow. In general, regions of ideal flow is also
potential, although as illustrated with the two examples in section 4.5.1, there are situations in
which ideal regions of the flow may not be potential (e.g. solid body rotation).
It was shown above that vortices cannot originate or disappear in an ideal (barotropic) fluid
moving in the field of a potential force (work independent on path taken between two points).
Consequently, from the law of conservation of circulation, we might argue as follows. Let us
suppose that at some instant we have potential flow throughout the volume of the fluid. Then the
velocity circulation round any closed contour in the fluid is zero. By Kelvin’s theorem we could
then conclude that, if the flow is potential initially, i.e. ∇ × v = 0 everywhere, it remains so
indefinitely. However, we know that is not true for a real fluid as one can constantly observe the
creation and disappearance of vortices. Nevertheless, we will consider some general properties
of a potential flow since this very often appears to be a good approximation to real fluid flow.
We first recall that the proof of the Kelvin’s theorem, and therefore all its consequences, were
based on the assumption that the flow is isentropic (entropy remains constant). If the flow is
not isentropic, the theorem does not hold, and therefore, even if we have potential flow at some
instant, the vorticity will in general be non-zero at subsequent instants. Thus only isentropic flow
can in fact be potential flow.
5.1
Equations for a Potential Flow
5.1.1
Velocity Potential
The velocity of a potential flow, like any vector field having zero curl, can be expressed in terms
of one scalar function, the velocity potential, ϕ (xi , t), assuming v = ∇ϕ. The condition for
potential flow ∇ × v ≡ ∇ × (∇ϕ) = 0 is satisfied automatically by this representation. We can
now rewrite the Euler equation (4.8) with f = −∇u and the convective term (v · ∇)v replaced
81
5.1. EQUATIONS FOR A POTENTIAL FLOW
by
(v · ∇)v = ∇
v2
2
CHAPTER 5. POTENTIAL FLOW
− v × (∇ × v).
Now recall the definition of w, the enthalpy (Equation 4.26):
p
w = ε + pV = ε + .
ρ
and use the Euler equation for steady flow (Equation 4.29 in section 4.3.1) then the Euler equation
takes the form
2
v
∂v
+∇
+ w + u = v × (∇ × v).
∂t
2
For potential flow we have, v = ∇ϕ ⇒ v × [∇ × (∇ϕ)] = 0 and thus the Euler equation is
reduced to
∂ϕ v 2
∇
+
+ w + u = 0.
∂t
2
Let’s now choose a streamline (a line with a tangent collinear with the velocity vector at each
point) and consider the projection of this expression on the tangent to this line at an arbitrary
point. The projection of the gradient is the derivative dld , dl being the line element of the streamline. Hence we obtain
∂ ∂ϕ v 2
∂ϕ v 2
∇
+
+w+u
+
+w+u =0
=
∂t
2
∂l ∂t
2
l
∂ϕ v 2
+
+ w + u = F (t)
⇒
∂t
2
This equation can be considered as a form of the Bernoulli principle with an arbitrary function
of time F (t). This function can be reduced to zero by the transform
Z
′
ϕ = ϕ + F (t)dt
By omitting the prime at ϕ′ we can rewrite this equation as
∂ϕ v 2
+
+w+u
∂t
2
=0
(5.1)
Thus, we obtain the Euler equation for the potential flow of an ideal fluid.
For steady flow, taking the potential ϕ to be independent of time,
Equation 5.1 becomes Bernoulli’s equation
∂ϕ
∂t
= 0, and F (t) = const,
v2
+ w + u = const
2
It must be emphasised here that there is an important difference between the Bernoulli’s theorem
for potential flow and that for other flows. In the general case, the “constant” on the right-hand
side is a constant along any given streamline, but is different for different streamlines. In potential flow, however, it is constant throughout the fluid.
82
CHAPTER 5. POTENTIAL FLOW
5.1. EQUATIONS FOR A POTENTIAL FLOW
The simplest form of the equations describing the potential flow corresponds to the case of an
incompressible fluid, where dρ
= 0. Moreover we assume also that ρ = const in space. In
dt
⇒ (ρ = const) ⇒ w = ρp up to an additional
this case we obtain from the relation ∇w = ∇p
ρ
constant which can be included in ϕ. The continuity equation is then reduced to
∂ρ
v = 0} ⇒ (v = ∇ϕ)
+ v · ∇ρ + ρ∇ · v = 0 ⇒ (ρ = const) ⇒ |∇ · {z
{z
}
|∂t
ρ=const
continuity equation
⇒ ∇ · ∇ϕ ≡
∆ϕ = 0
| {z }
(5.2)
continuity equation
potential flow
where ∆ ≡ ∇ · ∇ ≡ div grad is the Laplacian operator. Equation 5.2 is known as the Laplace
equation. The Laplace equation is encountered not only in potential flows, but also in heat conduction, elasticity, magnetism, and electricity. Therefore, solutions in one field of study can be
found from a known analogous solution in another field. The Laplace equation is of a type that is
called elliptic. It is known that the solutions of elliptic equations are smooth and do not have discontinuities, except for certain singular points on the boundary of the region. This equation must
be supplemented by boundary conditions at the surfaces where the fluid meets solid bodies. At
fixed solid surfaces, the fluid velocity component vn normal to the surface must be zero, whilst
for moving surfaces it must be equal to the normal component of the velocity of the surface (a
given function of time).
Thus, the problem of a potential flow of inviscid incompressible fluid reduces to the solution of
one scalar equation (5.2) with the appropriate boundary conditions.
As soon as the potential ϕ (xi , t) is known (as a solution of Equation (5.2)), the velocity is
obtained from the relation v = ∇ϕ and the pressure in accordance with (5.1) is given by
2
∂ϕ
v
+u+
.
(5.3)
p = −ρ
2
∂t
We note that one must be careful while applying this formula because the potential in it is defined to within an arbitrary function of time F (t) which can influence p. This function can be
determined if the pressure is prescribed in advance at some point.
We may note here the following important properties of potential flow of an incompressible
fluid. Suppose that some solid body is moving through the fluid. If the result is potential flow,
it depends at any time only on the velocity of the moving body at that time, and not on its acceleration. Indeed, equation (5.2) does not explicitly contain the time, it enters the solution only
through the boundary conditions, and these contain only the velocity of the moving body.
From Bernoulli’s equation, 21 v 2 + ρp = const, we see that in a steady flow of an incompressible fluid (without gravitational field), the largest pressure occurs at points where the velocity is
zero. Such point usually occurs on the surface of a body past by the moving fluid (at the point O
in Figure 5.1) and is called a stagnation point. If v0 is the velocity of the incident flow, i.e. fluid
velocity at infinity, and p0 is the pressure at infinity, the pressure at the stagnation point is given
by
1
pmax = p0 + ρv02 .
2
83
5.1. EQUATIONS FOR A POTENTIAL FLOW
CHAPTER 5. POTENTIAL FLOW
(p0 ,v0 )
0
Figure 5.1: Definition of a stagnation point.
Finally, we consider the conditions under which the fluid may be regarded as incompressible.
When the pressure changes adiabatically by ∆p, the density changes by ∆ρ, where ∆p =
∂p
∆ρ. According to the Bernoulli’s equation, however, ∆p is of the order of ρv 2 in steady
∂ρ
S
∂p
is the square of the velocity c of sound in the fluid
flow. Taking now into account that ∂ρ
2
ρ vc2 .
S
The fluid may be regarded as incompressible if
(see (4.11)) yields ∆ρ ∼
necessary condition for this is that
∆ρ
ρ
≪ 1. We see a
the fluid velocity must be small as compared to the sound velocity, i.e.v ≪ c.
However, this condition is sufficient only in steady flow. In non-steady flow, a further condition
must be fulfilled. Let τ and l be a time and length of the order of times and distances over which
and ρ1 ∇p in Euler equation are
the fluid velocity undergoes significant changes. If the terms ∂v
∂t
comparable, we find in order of magnitude τv ∼ ∆p
or ∆p ∼ lvρ
, and the corresponding change
lρ
τ
∂ρ
∆ρ
lv
in ρ is ρ ∼ τ c2 . On the other hand, comparing the terms ∂t and ρ (∇ · v) in the equation of
may be neglected, i.e. we may suppose ρ to be constant,
continuity, we find that the derivative ∂ρ
∂t
ρv
∆ρ
∆ρ
τv
if τ ≪ l , or ρ ≪ l . Comparing these two estimations for the density changes yields
τ ≫ cl . This condition has an obvious meaning: the time cl taken by sound signal to traverse the
distance l must be small compared with the time τ during which the flow changes appreciably, so
that the propagation of interactions (sound signal) in the fluid may be regarded as instantaneous.
In conclusion, if conditions v ≪ c and τ ≫ cl are both fulfilled, the fluid may be regarded as
incompressible.
5.1.2
Two-Dimensional Flow. Stream Function
In potential regions of flow, one can solve the Laplace equation (5.2) for the velocity potential
function ϕ for both two- and three-dimensional flow fields. For two-dimensional flows, or plane
flows the velocity distribution in a moving fluid depends on only two coordinates (x and y, say)
and the velocity is everywhere parallel to the xy-plane, e.g.,
84
CHAPTER 5. POTENTIAL FLOW
2-D flow:
5.1. EQUATIONS FOR A POTENTIAL FLOW






 x1 = x
x2 = y
{xi } ⇔ (x1 , x2 , x3 ) ⇔ (x, y, z) ⇒


x3 = z



 | {z }
space coordinates





 v1 = vx
v2 = vy
⇔ v (vx , vy , 0)
{vi } ⇔ (v1 , v2 , v3 ) ⇔ (vx , vy , vz ) ⇒


v
=
v
=
0

3
z

|
{z
}
2D flow
However, two-dimensional flow is not limited to flow in the xy-plane but, in fact, we can assume
two-dimensionality in any region of the flow where only two directions of motions are important
and where there is no significant variation in the third direction.
To solve problems of two-dimensional flow of an ideal incompressible fluid, it is convenient
to express the velocity in terms of what is called the stream function. Indeed, if vx and vy are not
dependent of z, then the continuity equation for the flow of incompressible fluid (∇ · v) takes the
form
∂vx ∂vy
+
= 0.
∂x
∂y
It is easy to see that this equation will be satisfied if we introduce the function ψ(x, y, t) in such
a way that

∂ψ
 vx = ∂y ,
(5.4)

∂ψ
vy = − ∂x .
In this case, the 2-D continuity equation will be satisfied automatically:
∇·v =
∂ 2ψ
∂ 2ψ
∂vx ∂vy
+
=
−
= 0.
∂x
∂y
∂x∂y ∂x∂y
The function ψ(x, y, t) is the stream function. If we know the stream function we can immediately determine the form of the streamlines for steady flow. Indeed, the streamline is defined as a
line with a tangent collinear with the velocity vector v at each point. According to this definition
such a line y = y(x) is illustrated in Figure 5.2. The tangent to this line at an arbitrary point is
dy
dy
given by dx
, or dx
= vvxy ⇒ vx dy − vy dx = 0. Taking into account the relations (5.4) in this
equality yields
∂ψ
∂ψ
dx +
dy = dψ = 0 ⇒ ψ = const .
| {z }
∂x
∂y
along the streamline
Thus the streamlines are the family of curves obtained by putting the stream function ψ (x, y)
equal to an arbitrary constant. We therefore conclude that
the lines ψ(x, y, t) = const are the flow streamlines.
If we draw a curve between two points A and B in the xy-plane as in Figure 5.2, the mass flux
85
5.1. EQUATIONS FOR A POTENTIAL FLOW
CHAPTER 5. POTENTIAL FLOW
ψB constant
B
vx
y=y(x)
dx
dy
e
w
Flo
itiv
pos
A
vy
ψA constant
Figure 5.2: The streamlines and the stream function ψ.
Q across this curve is given by the difference in values of the stream function at these points,
regardless of the shape of the curve. Indeed, by definition, this flux is given by
Q=ρ
ZB
v · ndl
(5.5)
A
where n is the unit normal to the curve and dl is an element of its length. The product under the
integral in the definition (5.5) can be determined as follows:
v · ndl = vx dl cos (nx) + vy dl cos (ny)
with
dl cos (nx) ≡ dl cos θ = dy
dl cos (ny) = dl cos π2 + θ = −dl sin θ = −dx.
Taking now relations (5.4) into account we can write
Q=ρ
ZB
A
v · ndl = ρ
ZB ∂ψ
∂ψ
dx +
dy
∂x
∂y
A
=ρ
ZB
dψ = ρ(ψB − ψA )
(5.6)
A
Hence, the flux across a part of a curve is independent of the path taken as long as it connects
the two streamlines ψA and ψB .
In the definition of the stream function we did not use the condition that the flow is potential. So, the stream function can also be introduced for a vortex flow. Then the vorticity vector,
ω = ∇ × v, can also be related to the stream function as
2
∂ ψ ∂ 2ψ
∂vy ∂vx
ẑ = −
+ 2 ẑ
−
ω =∇×v =
∂x
∂y
∂x2
∂y
86
CHAPTER 5. POTENTIAL FLOW
5.1. EQUATIONS FOR A POTENTIAL FLOW
⇒ ω = −∆ψẑ
(5.7)
For the potential flow (v = ∇ϕ and ω = ∇ × ∇ϕ = 0), ψ obeys the equation
∆ψ ≡
∂ 2ψ ∂ 2ψ
+ 2 =0
∂x2
∂y
(5.8)
So, in the case of a potential two-dimensional flow of an ideal incompressible fluid both the
velocity potential ϕ(x, y, t) and the stream function ψ(x, y, t) satisfy the Laplace equation.
Equation (5.8) holds for two-dimensional flows only, because a stream function ψ cannot be
defined in a completely three-dimensional flow. However, a velocity potential ϕ can be defined
in three-dimensional irrotational flows as stated above, because v = ∇ϕ identically satisfies the
irrotationality condition ∇ × v = 0.
Since both the velocity potential ϕ(x, y, t) and the stream function ψ(x, y, t) satisfy the Laplace
equation, they have to relate to each other in one or another way. To establish this relationship
we use the definitions (5.4) and v = ∇ϕ. We then have
(
vx = ∂ϕ
= ∂ψ
∂x
∂y
(5.9)
∂ϕ
vy = ∂y = − ∂ψ
∂x
It is easy to see that if we multiply the first relation by ∂ψ
and the second by
∂x
we obtain
∂ϕ ∂ψ ∂ϕ ∂ψ
+
≡ ∇ϕ · ∇ψ = 0.
∂x ∂x
∂y ∂y
∂ψ
∂y
and add results,
(5.10)
So, the families of curves ϕ = const and ψ = const are mutually orthogonal. The functions ϕ
and ψ are, in a sense, equivalent. One can, for example, imagine the flow where ψ is the velocity
potential and ϕ is the stream function. This flow can be called the “conjugate” to the flow with
ϕ as velocity potential and ψ the stream function.
We note that this demonstration fails at so called stagnation points where the velocity is zero.
The boundary conditions normally encountered in potential flows are of the following types:
a) Condition on solid surface: Component of fluid velocity normal to a solid surface must
equal the velocity of the boundary normal to itself, ensuring that fluid does not penetrate a
solid boundary. For a stationary body, the condition is
∂ψ
∂ϕ
= 0 or
=0
∂n
∂s
where s is the direction along surface, and n is normal to surface.
b) Condition at infinity: For the typical case of a body immersed in a uniform stream flowing
in the x-direction with speed U , the condition is
∂ϕ
∂ψ
= U or
=U
∂x
∂y
87
5.2. APPLICATIONS
CHAPTER 5. POTENTIAL FLOW
We have to note that solving the Laplace equation subject to boundary conditions of the type
of mentioned above is not easy. Historically, potential flow theory was developed by finding
a function that satisfies the Laplace equation and then determining what boundary conditions
are satisfied by the functions, the so-called “inverse approach” to study potential flows. After a
solution of the Laplace equation has been obtained, the velocity components are then determined
by taking derivatives of ϕ or ψ. Lastly, pressure distribution is determined by applying the
Bernoulli equation, p + 12 ρv 2 , between any two points in the flow field. Thus, a solution of the
non-linear equation of motion (the Euler equation) is avoided in potential flows.
5.2
Applications of Analytical Functions to Hydrodynamical
Problems
In this section z will denote the complex variable
z ≡ x + iy = reiθ
√
where i = −1, (x, y) are the Cartesian coordinates, and (r, θ) are the polar coordinates. In the
Cartesian form the complex number z represents a point in the xy-plane whose real axis is x and
imaginary
p axis is y. In the polar form, z represents the position
vector Oz, whose magnitude is
−1 y
2
2
. When x and y are regarded as
r = (x + y ) and whose angle with the x-axis is tan
x
variables, the complex quantity z = x + iy is called a complex variable.
Suppose we define a complex function F whose real and imaginary parts are α and β:
F (z) = α(x, y) + iβ(x, y),
(5.11)
It is shown in theory of complex variables that if α and β are functions of x and y, so is F and
moreover it is a function of the combination z = x + iy. This function has a finite and unique
, i.e. the limit
derivative, F ′ (z) ≡ dF
dz
F ′ (z) = |{z}
lim
∆z→0
F (z + ∆z) − F (z)
∆z
(5.12)
exists, when its real part [Re {F (z)} = α(x, y)] and imaginary part [Im {F (z)} = β(x, y)] satisfy the Cauchy-Riemann conditions :



∂α
∂x
=
∂β
,
∂y
=
− ∂α
.
∂y
(5.13)
∂β
∂x
Then, a single-valued function F (z) is called an analytical function of a complex variable z in
a region if a finite F ′ (z) exists everywhere within the region. Points where F (z) or F ′ (z) is
zero or infinite are called singularities. For example, F (z) = ln z and F (z) = z1 are analytic
everywhere except at the singular point z = 0, where the Cauchy-Riemann conditions are not
satisfied.
88
CHAPTER 5. POTENTIAL FLOW
5.2.1
5.2. APPLICATIONS
Complex Flow Potential
Let us consider the 2-D potential flow of an ideal incompressible fluid. This flow is determined
by the two functions ϕ (velocity potential) and ψ (stream function). The properties of ϕ and
ψ allow us to apply the theory of analytical functions of complex variables to the flow under
consideration. Indeed, if we compare relations (5.13) and (5.9) we conclude that the velocity
potential ϕ and stream function ψ of a 2-D flow of an ideal incompressible fluid satisfy the
Cauchy-Riemann equations exactly. Since the velocity potential and stream function satisfy
(5.13), and the real and imaginary parts of any analytical function
F (z) = ϕ + iψ
of complex variable z also satisfy (5.13), it follows that any analytic function of z represents the
complex potential of some two-dimensional flow.
Hence, ϕ and ψ can be considered as the real and imaginary parts of some analytical function
F (z) called the complex flow potential, with ϕ = Re {F (z)}, ψ = Im {F (z)}.
Since both functions ϕ and ψ satisfy the Laplace equation one can conclude that for any analytical function F (z) (the complex flow potential) we have two mutually conjugate flows. The
velocity potentials and corresponding stream functions of these flows are given by
1. ϕ1 = Re {F (z)} ; ψ1 = Im {F (z)}
2. ϕ2 = Im {F (z)} ; ψ2 = Re {F (z)}.
5.2.2
Some Examples of Two-Dimensional Flows
Let us consider some simple functions F (z) and the respective flows.
a) Uniform flow:
F (z) = az
F (z) = ϕ + iψ = az = {z = x + iy} = ax + iay
⇒
⇒ v = ∇ϕ ⇒
ϕ = ax,
ψ = ay
vx = a = const,
vy = 0.
Thus, it is a uniform flow with constant velocity v = (vx , vy ) = (a, 0) and the streamlines
(ψ = const) are parallel to the x-axis.
The conjugate flow is obtained by multiplying the complex flow potential with −i:
F ∗ (z) = (−i)F (z) = −iaz
F ∗ (z) = ϕ + iψ = ay − iax ⇒
ϕ = ay,
⇒
ψ = −ax
89
5.2. APPLICATIONS
CHAPTER 5. POTENTIAL FLOW
⇒ v = ∇ϕ ⇒
vx = 0,
vy = a = const.
Thus, this is also a uniform flow but with constant velocity v = (vx , vy ) = (0, a) and the
streamlines are parallel to the y-axis.
b) Source and sink:
F (z) = m ln z ,
z 6= 0,
m∈R
(5.14)
Introducing the polar coordinates r and θ so that

 x = r cos θ,
y = r sin θ,

z = x + iy = r exp(iθ)
we can transform equation (5.14) into
⇒ F (z) = m (ln r + iθ) ⇒ ϕ + iψ = m (ln r + iθ) ⇒
ϕ = m ln r,
ψ = mθ.
The flow pattern is shown in Figure 5.3. The streamlines radiate from origin at θ = const
and the equipotential lines (ϕ = const) form concentric circles at r = const. The particles
flow along the streamlines at the velocity

∂ϕ
 vr = ∂r = mr ,
v = ∇ϕ ⇒

vθ = 1r ∂ϕ
= 0.
∂θ
For a point source at r = 0 the particles flow outwards and thus m > 0. For a sink at the
point r = 0 the flow should be inwards and thus m < 0. Let us consider an arbitrary circle
with r = const. The total flow of mass across of this circle is given by
Q = 2πrvr = 2πr
m
= 2πm = const
r
So, the total flow of mass across the circle with r = const is constant and does not depend
on r. The quantity Q = 2πm is called the strength of the source (sink).
The conjugate flow describe the so-called potential vortex flow. We obtain the complex
flow potential for the conjugate flow by multiplying (5.14) by −i:
ϕ = mθ,
∗
F (z) = −im (ln r + iθ) ⇒ ϕ + iψ = −im (ln r + iθ) ⇒
(5.15)
ψ = −m ln r.
Thus, in this case the flow pattern is interchanged from the source and sink. The streamlines in this flow are concentric circles r = const (dotted lines in Figure 5.3) whereas the
equipotential lines radiate from origin at θ = const (full lines in Figure 5.3). The velocity
is now

∂ϕ
 vr = ∂r = 0,
v = ∇ϕ ⇒

vθ = 1r ∂ϕ
= mr .
∂θ
90
CHAPTER 5. POTENTIAL FLOW
Sink
Flow in
5.2. APPLICATIONS
Source
Flow out
y
x
Figure 5.3: Streamlines (full) and equipotential lines (dotted) for the source and sink flow. For the conjugate flow, the potential vortex, the flow pattern is opposite. The concentric circles (dotted)
are the streamlines and the radial lines (full) correspond to the equipotential lines.
Hence, the potential flow defined by the complex flow potential (5.15) corresponds to
revolution of fluid particles about the point r = 0 at a speed inversely proportional to the
distance r from this point. The revolution is counterclockwise, if m > 0 and clockwise if
m < 0.
Here we have a so-called vortex source. Let us calculate the circulation Γ of the velocity
around a closed contour consisting of a vortex source at r = 0 (attention, we consider
potential flow). We then have
Γ=
I
v · dr =
Z2π
vθ rdθ = 2πrvθ = 2πm
0
Γ = 2πrvθ = 2πm 6= 0
(5.16)
Despite the fact that we consider potential flow, the circulation of the velocity Γ 6= 0. The
existence of a non-vanishing circulation does however not contradict Kelvin’s theorem and
the assumption about the potential flow. As a matter of a fact, the point r = 0 is singular in
this case and must be excluded from the region under consideration. Hence, the contour is
not simply connected, i.e. it can not be contracted to a point, and thus the Kelvin’s theorem
is not applicable. The flow is potential everywhere except at the point r = 0. Note, that
this flow corresponds exactly to Flow B presented in Section 4.5.1.
91
5.2. APPLICATIONS
5.2.3
CHAPTER 5. POTENTIAL FLOW
Conformal Mapping
The application of conformal mapping to the theory of two-dimensional flow of an ideal incompressible fluid turns out to be very effective. Let us suppose that we have found a complex flow
potential F (z) for some domain satisfying the appropriate boundary conditions. By conformal
mapping we can transform the complex plane {z} introducing the new variable ξ which is related
to the variable z by z = f (ξ). Then the complex flow potential transforms as
F (z) = F [f (ξ)] = Φ(ξ)
This is the new complex flow potential which in the ξ-plane describes a new flow in a new domain. In such a way, using a suitable conformal mapping, i.e. transformation z → ξ , one can
obtain the solution of rather complicated problem from a simple one.
We will demonstrate this method for wedge-like domains. Namely, let us assume that we need
to describe the flow in the wedge domain ξ(ζ, η) , i.e. we have to first find the complex flow
potential and then having the potential find the velocity and the streamlines of this flow. The
main steps of the problem solution:
1. Find the proper z-plane flow for which the complex flow potential is known.
2. Find the conformal mapping of the z-plane flow in the corresponding domain to the ξ-plane
into the wedge-like domain.
3. Find the new complex flow potential.
y
a)
η
b)
η
c)
ζ
π/α
x
ζ
Figure 5.4: Conformal mapping of the half plane y > 0 into a wedge domain.
Let the flow, which is assumed to be known, be the flow with constant velocity in the half space
y > 0 with a rigid wall at y = 0 (see Figure 5.4a). The flow velocity is v = (vx , vy ) = (v0 , 0)
which is described by the complex flow potential, F (z) = v0 z. Using the relation
α
ξ
z=R
(a, R) > 0
R
we transform the half plane y > 0 into the wedge in the ξ-plane (see Figure 5.4b). The complex
flow potential becomes
α
ξ
.
Φ (ξ) = v0 R
R
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CHAPTER 5. POTENTIAL FLOW
5.3. STEADY FLOW AROUND A CYLINDER
In polar coordinates r and θ we get ξ = r exp(iθ) ≡ r(cos θ + i sin θ), so that the complex flow
potential in the ξ-plane is given by
r α
Φ (ξ) = v0 R
(cos αθ + i sin αθ)
(5.17)
R
This yields for the velocity potential and the stream function

α
 ϕ = v0 R Rr cos αθ,

ψ = v0 R
Then the velocity components are given by

 vr =

vθ =
∂ϕ
∂r
= αv0
1 ∂ϕ
r ∂θ
r α
R
r α−1
R
= −αv0
(5.18)
sin αθ.
cos αθ,
r α−1
R
(5.19)
sin αθ.
We observe that vθ = 0 at radii θ = 0 and θ = απ , where rigid walls can be placed without
influencing the flow. Hence, we have obtained the solution of the problem of flow in a wedge
with vertex angle απ . Figure 5.4b shows the case when α = 4. It follows from (5.19) that
vr , vθ → 0 if r → 0 and a > 1, i.e. the velocity is zero at the vertex of the edge in this case. If
a < 1 we have flow around the wedge. The case α = 32 is shown in Figure 5.4c. It follows from
(5.19) that vr , vθ → ∞ if r → 0 in this case.
5.3
Steady Flow Around a Cylinder
Fluid flows over stationary bodies occur frequently in practice (such as wind blowing over a
building), and other times a body moves through a quiescent fluid (such as a car moving through
air). These two seemingly different processes are indeed equivalent to each other. For convenience such motions are analysed by fixing the coordinate system on the body and are referred
to as flow over bodies or external flow. In general the flow fields and geometries for most external flow problems are too complicated to be solved analytically and therefore in this section
we consider some particular cases where the analytical solution is still possible. In this approach
the velocity of the fluid far away from a body is called free-stream or unperturbed velocity.
The free-stream velocity may vary with location and time. But in the design and analysis, the
free-stream velocity is usually assumed to be uniform and steady for convenience, and this is
what we will do in this section.
The shape of a body has a profound influence on the flow over the body and resulting velocity field. The flow over a body is said to be two-dimensional when the body is very long and
of constant cross section and the flow is normal to the body. The fluid flowing over a long pipe
perpendicular to its axis is an example of two-dimensional external flow. Note that the velocity
component in the axial direction is zero in this case and thus the velocity is two-dimensional.
The two-dimensional idealisation is appropriate when the body is sufficiently long so that the
end effects are negligible and the approach flow is uniform. Flow over a body that can not be
modelled as two-dimensional is three-dimensional.
A fluid may exert forces on a body in various directions. The force a flowing fluid exerts on
93
5.3. STEADY FLOW AROUND A CYLINDER
CHAPTER 5. POTENTIAL FLOW
a body in the flow direction is called the drag force (or just drag). The force in the direction
normal to the flow tends to move the body in that direction and is called the lift force (or just
lift). For external flow of an ideal fluid, these forces can be determined entirely by the pressure
distribution on the surface of the body. For example in the design of airfoils, the primary consideration is then minimising the average pressure at the upper surface while maximising it at
the lower surface. The Bernoulli equation can be used as a guide in identifying the high- and
low-pressure regions: Pressure is low at locations where the flow velocity is high, and pressure is
high at locations where the flow velocity is low. It is worthwhile to remember in this connection
that a point in the flow field where the velocity is identically zero is referred to as a stagnation
point (see Figure 5.1).
5.3.1
Application of Conformal Mapping
Let us consider a 2D steady flow around a circular cylinder with radius r = R in the plane
z = x + iy = r exp(iθ) (see Figure 5.5a). We seek the solution of the Laplace equation ∆ϕ = 0
in the domain outside of the circle r = R under the following conditions:
1. The normal velocity component at the boundary r = R is zero.
2. The unperturbed flow has the velocity component (vx , vy ) = (−v0 , 0) at infinity (r → ∞).
In terms of the velocity potential these conditions are













∂ϕ
∂r r=R
= 0,
∂ϕ
∂x r→∞
= −v0 ,
∂ϕ
∂y
= 0.
r→∞
(5.20)
The potential ϕ and the stream function ψ together give the complex flow potential F (z) =
y
η
a)
b)
h
R
x
-1
1
ζ
Figure 5.5: Conformal mapping of the circular cylinder (a) onto the ξ-plane (b) with the use of the
Joukowski transformation.
ϕ(x, y) + iψ(x, y). Let’s now apply the conformal mapping method in order to transform the
z-plane flow domain which is the exterior of the circle r = R onto the whole ξ-plane except the
94
CHAPTER 5. POTENTIAL FLOW
5.3. STEADY FLOW AROUND A CYLINDER
cut between -1 and +1 as shown in Figure 5.5b. For this purpose the Joukowski transformation
can be used
1 z
R
ξ = ζ + iη =
(5.21)
+
2 R
z
By using the Joukowski transformation (5.21) the circle z = Rexp(iθ) is transformed into the
straight cut between ζ = −1 and ζ = +1 traversed twice. Instead of F (z) we have now
1 z
R
F (z)z→ξ ⇒ Φ(ξ) = Φ
≡ ϕ1 (ζ, η) + iψ(ζ, η).
+
2 R
z
{z
}
|
the new complex flow potential
The boundary conditions now becomes
1. At r = R:
∂ϕ
∂r
2. At r → ∞:
r=R
=0⇒
ξ = ζ + iη ⇒
which yields
∂ϕ1
∂η
=0
η=0
z
x
y
=
+i
2R
2R
2R
x y .
,
ϕ(x, y) |{z}
⇒ ϕ1
2R 2R
r→∞
So, we have
where κ =





p
ζ 2 + η2,
∂ϕ
∂x r→∞
= −v0 =
1 ∂ϕ1
2R ∂ζ κ→∞
⇒
∂ϕ1
∂ζ κ→∞
= −2v0 R,
1 ∂ϕ1
1
= 0 = 2R
⇒ ∂ϕ
=0
∂η κ→∞
∂η κ→∞
p
r = x2 + y 2 , z = reiθ , θ = tan−1 xy .
∂ϕ
∂y r→∞
The result of conformal mapping: The problem of a steady flow around a cylinder is reduced
by the Joukowski transformation to the problem of flow past an infinitely thin strip under
the condition that the fluid velocity is (−2Rv0 ) in front of the strip.
However, the solution of this problem is the flow with the velocity independent of the space
variables and is known from Section 5.2 (the flow with complex flow potential F (z) = az):
v = (vζ , vη ) = (−2v0 R, 0).
The complex flow potential for this case is
Φ (ξ) = −2v0 Rξ.
We can return back to the z variable using the transformation (5.21) and thus obtain the complex
flow potential of the initial problem
z
R
+
⇒
F (z) = −2v0 Rξ(z) = −v0 R
R
z
F (z) = ϕ + iψ = −v0
95
R2 −iθ
re +
.
e
r
iθ
(5.22)
5.3. STEADY FLOW AROUND A CYLINDER
CHAPTER 5. POTENTIAL FLOW
Hence, the velocity potential is
ϕ = Re {F (z)} = −v0
R2
r+
r
and we can calculate the velocity components

∂ϕ

v
=
1−
=
−v

r
0
∂r



 vθ =
1 ∂ϕ
r ∂θ
= v0 1 +
Let us check the boundary conditions (5.20).
1. At r = R:
2. At r → ∞:
∂ϕ
∂r
R2
r2
R2
r2
cos θ.
(5.23)
cos θ,
sin θ.
=0
r=R
⇒ ϕr→∞ = −v0 r cos θ ≡ −v0 x,
hence





∂ϕ
∂x r→∞
= −v0 ,
∂ϕ
∂y
= 0.
r→∞
Finally, the velocity components at the surface of the cylinder (r = R) are given by
vr = 0 (normal component),
v(0, vθ )|r=R ⇒
vθ = 2v0 sin θ.
(5.24)
The maximum velocity is at the midsection θ = ± π2 , where v = vθ = 2v0 , and two stagnation
points are defined by θ = 0 and θ = π (see Figure 5.5).
5.3.2
The Pressure Coefficient
The pressure distribution at the cylindrical surface (r = R) can be determined by the Bernoulli
theorem
v2
v2
p0 + ρ 0 = p + ρ
(5.25)
2
2
where p0 and v0 are the pressure and velocity at infinity and p and v are the same quantities at
the surface. Equation (5.25) is the partial form of the Bernoulli theorem. Indeed, the Bernoulli
theorem for the stationary flow of an ideal fluid is given by
v2
+ w + u = const
2
If now we neglect the gravity effect (u = 0) and take the limit of an incompressible fluid ρ =
const ⇒ w = p/ρ we see that the Bernoulli theorem is reduced to (5.25). Taking into account
that the velocity distribution at the surface of the cylinder is determined by vθ |r=R = 2v0 sin θ
96
CHAPTER 5. POTENTIAL FLOW
5.3. STEADY FLOW AROUND A CYLINDER
(which follows from (5.24)), the pressure distribution at the cylinder surface is obtained from
(5.25)
v2
(5.26)
p = p0 + ρ 0 1 − 4 sin2 θ
2
We now introduce the coefficient
K=
p − p0
= 1 − 4 sin2 θ.
2
ρv0/2
(5.27)
In fluid dynamics this coefficient is known as the pressure coefficient. We note that the pressure coefficient does not depend on the radius of the cylinder, the fluid density and the flow
velocity, which are parameters of the problem. This fact is the manifestation of the principle of
the hydrodynamic similarity. Owing to this principle, the corresponding theory could be experimentally proved by determination of the pressure coefficient dependence on θ for a circular
cylinder with arbitrary fixed R, v0 , ρ. It should be noted that this applies to ideal fluids, however
this theory can be rather close to reality when considering good streamline profiles, such as for
example an airplane wing.
Let us consider the pressure distribution at some particular points on the surface of a cylinder.
1. θ = 0 ⇔ the stagnation point where the flow is divided:
pθ=0
v02
= p 0 + ρ > p0 ,
2
thus the pressure at the stagnation point is larger than the pressure at infinity.
2. θ = π ⇔ the symmetric point at the rear of the cylinder:
pθ=π = pθ=0 > p0 ,
the same pressure as at the stagnation point.
3. As we approach the midsection θ = ± π2 , the pressure drops monotonically to the value
3
pθ=± π2 = p0 − ρv02 < p0 ,
2
thus at the midsection the pressure is less than at the infinity.
5.3.3
The Paradox of d’Alembert and Euler
We have seen that the pressure distribution (5.26) in the flow around the circular cylinder is symmetric with respect to the midsection θ = π2 . Hence the total force on the cylinder is zero. This
is the particular case of more general statement known as d’Alembert-Euler paradox:
There is no drag force on a body of arbitrary but smooth shape in ideal fluid flow.
We can superimpose a homogeneous flow with velocity v0 on the flow considered above. Then
the flow velocity will be zero at the infinity and we obtain the case of a body moving in a fluid
at a rest. It follows from the above that a body moving in an ideal fluid at constant velocity
97
5.3. STEADY FLOW AROUND A CYLINDER
CHAPTER 5. POTENTIAL FLOW
expresses no drag. The origin of this paradox is most clearly seen by considering the drag. The
presence of a drag force in uniform motion of a body would mean that, to maintain the motion,
work must be continually done by some external force. This work must be either dissipated in
the fluid or converted into kinetic energy of the fluid, and the result being a continual flow of
energy to infinity in the fluid. There is, however, by definition no dissipation of energy in an
ideal fluid, and the velocity of the fluid set in motion by the body diminishes so rapidly with
increased distance from the body that there can be no flow of energy to infinity.
5.3.4
The Flow Around a Cylinder with Circulation
In the case of steady flow around the cylinder considered above, the force perpendicular to the
flow velocity is also zero, since the pressure distribution
p = p0 + ρ
v02
1 − 4 sin2 θ
2
is symmetric with respect to the plane θ = 0. Quite another case can be obtained if a circulation
flow around a cylinder is superimposed on the straight steady flow. Indeed, the complex flow
potential will now be the sum of that for symmetric flow (5.22) and that for circulation (potential
vortex) flow (5.15):
R2
R2 −iθ
Γ
iθ
− im ln z = −v0 z +
F (z) = −v0 re +
e
+
ln z
(5.28)
r
z
2πi
where we used the circulation of velocity, Γ = 2πm. The velocity potential ϕ and stream
function ψ are then
Γ
R2
cos θ +
θ
ϕ = −v0 r +
r
2π
R2
Γ
ψ = −v0 r −
sin θ −
ln r
r
2π
There is an infinite family of different flow fields corresponding to different values of v0ΓR . It is
worthwhile to note here that the pattern of streamlines for the case v0ΓR = 0, which is the only
one where the pattern is symmetrical about the x-axis, is shown in Figure 5.6. The boundary
conditions will also be similar. The normal component of the velocity on the cylinder surface
vanishes. The tangential component vθ will be the sum of two superimposed flows
vθ |r=R = 2v0 sin θ +
Γ
2πR
(5.29)
where the following relations for the velocity circulation have been used,
vθ =
Γ
Γ
m
⇒ Γ = 2πm ⇒ m =
⇒ vθ =
.
r
2π
2πr
We gain one view of the effect of changing the value of v0ΓR by noting that the velocity of the
fluid at the surface of the cylinder (5.29) entirely vanishes at two points at which
sin θ = −
Γ
4πv0 R
For v0ΓR < 4π, two values of θ satisfy this relation, implying that there are two stagnation points,
which are at the foremost and rearmost points of the cylinder surface. Both stagnation points
98
CHAPTER 5. POTENTIAL FLOW
5.3. STEADY FLOW AROUND A CYLINDER
Stagnation
points
Figure 5.6: Potential flow over a stationary cylinder.
progressively move down as v0ΓR increases, and coalescence at θ = − 12 π when v0ΓR = 4π.
Streamlines for a case in which 0 < v0ΓR < 4π are shown in Figure 5.7b, and those for the
particular case v0ΓR = 4π in 5.7c. At values of v0ΓR larger than 4π, the velocity is non-zero at all
points of the surface of the cylinder and increasing in the direction of θ. The stagnation point
has here moved away from the cylinder along the line θ = − π2 and is at radial position. The
streamlines in Figure 5.7d show that some of the fluid simply circulates round the cylinder and
remains near it. The radial distance of the stagnation point in this case can be found from the
tangential component vθ given by
R2
Γ
vθ = v0 1 + 2 sin θ +
r
2πR
where vθ =
This gives
1 ∂ϕ
.
r ∂θ
Then the position of the stagnation point is determined by equality
R2
Γ
vθ |θ= − π = −v0 1 + 2 +
=0
2
r
2πr
q
1
2
2
Γ ± Γ − (4πv0 R)
r=
4πv0
one root of which is r > R; the other root corresponds to a stagnation point inside the cylinder
and thus has no physical meaning.
It is evident that the effect of increasing v0ΓR from zero is to cause an increasingly marked difference between the flow regions on the two sides of the x-axis, and in particular to cause the
velocity on the upper surface of the cylinder to be high and those on the lower surface to be low.
Remembering that p + 21 ρv 2 is constant in a steady potential flow (the Bernoulli theorem), we
see plainly the greater pressure on the lower surface which leads to the non-zero side-force on
99
5.3. STEADY FLOW AROUND A CYLINDER
(a) Γ=0
CHAPTER 5. POTENTIAL FLOW
(b) 0< Γ/av0 < 4π
y
y
Γ
x
Γ
x
x
(d) Γ/av0 > 4π
(c) Γ/av0 = 4π
y
Γ
y
x
Figure 5.7: Streamlines for potential flow due to a circular cylinder held in a stream of uniform velocity
(−v0 , 0) at infinity, with circulation Γ (anti-clockwise positive) round the cylinder.
the cylinder. It is possible to calculate the side-force directly and explicitly in this simple case of
a circular cylinder. Indeed, substituting the velocity v into the Bernoulli theorem, we obtain for
the pressure distribution over the cylinder surface
"
2 #
1
2Γv
1
Γ
0
p = p0 + ρ v02 − v 2 = p0 + ρ v02 1 − 4 sin2 θ −
sin θ −
.
(5.30)
2
2
πR
2πR
The pressure p is symmetric with respect to the midsection θ = π2 , therefore the cylinder,
according to the paradox d’Alembert-Euler, experiences no drag (i.e. no force in the direction of
flow). However,
the pressure
is not symmetric with respect to the plane θ = 0 due the presence of
0
sin
θ
in
(5.30).
Hence, the pressure is different on the upper and lower halves
the term − 2Γv
πR
of the cylinder. To understand the reason for such a pressure difference, let’s study Figure 5.8
with a superimposed counterclockwise circulation (Γ > 0) upon a given symmetrical flow. The
velocities (symmetrical flow + circulation) of the flow is of the same sign above the cylinder and
100
CHAPTER 5. POTENTIAL FLOW
5.3. STEADY FLOW AROUND A CYLINDER
opposite sign below it. Hence, the resultant velocity will be larger above the cylinder. Then the
2
pressure, according to the Bernoulli theorem p + ρv2 = const, will be larger below the cylinder.
Thus, a force in the y-direction known as the hydrodynamic lift force will arise.
y
R
x
Figure 5.8: Superimposed circulation flow (dotted lines) on a given symmetrical flow (full line).
Let’s calculate this force per unit length of the cylinder. The area element at the surface of the
cylinder in this case is equal to dS = 1 · Rdθ = Rdθ per unit length. The projection of this force
on the y-axis is given by −p sin θ. The minus sign is due to the fact that direction of the pressure
force and y-axis is opposite. Then
taking into account the symmetry of the pressure distribution
π
about the midsection θ = 2 we will perform the integration in the limits − π2 , π2 instead of
(0, 2π). The result of integration must therefore be multiplied by 2:
Z2π
π
...Rdθ = 2
Z2
...Rdθ
− π2
0
So, we have for the hydrodynamic lift force
Fy = −2
Zπ/2
pR sin θdθ
−π/2
Substituting now p from (5.30) (we will take into account only the antisymmetric term, p ∝
0
−ρ 2Γv
sin θ) into the integral for Fy , we obtain
2πR
π
Fy =
2Γv0 ρ
π
Z2
− π2
2
|sin
{z θ} dθ = ρv0 Γ
(5.31)
1
(1−cos 2θ)
2
Thus, the lift is proportional to the velocity of the symmetric flow v0 as well as to the circulation
Γ. If the later changes sign, the lift also changes direction.
We also note that the expression for the lift Fy = ρv0 Γ holds for any cylindrical body, not
only of circular cross-section. The result that the lift force is proportional to the circulation is of
101
5.4. IRROTATIONAL FLOW AROUND A SPHERE
CHAPTER 5. POTENTIAL FLOW
fundamental importance in aerodynamics. Relation (5.31) was independently proved by the German scientist W. Kutta and Russian scientist N. Joukowski and it known as the Kutta-Joukowski
lift theorem. The interesting question of how certain two-dimensional shapes, such as airfoil,
develop circulation when placed in a stream. The answer to this question is beyond the scope of
the course and therefore we do not present it here. For those who are interesting in these details
we recommend to look for more advanced book on fluid dynamics, for example the book “An
introduction to fluid dynamics” by G. K. Batchelor.
5.4
Irrotational Flow Around a Sphere
5.4.1
The Flow Potential and the Particle Velocity
In the general case of three-dimensional flow, the conformal mapping method cannot be used and
we have to solve the Laplace equation directly under the corresponding boundary conditions. We
will demonstrate this approach in the case of flow around a sphere.
Suppose that a fluid is at rest at infinity and a sphere moves with constant velocity v0 . We
introduce also the velocity potential ϕ (v = ∇ϕ) which must satisfy the Laplace equation. The
boundary condition is the equality of the normal velocity of the spherical particle v 0 · r/r to that
of the fluid v · r/r. A spherical system of coordinates is used here with the origin at the center
of the sphere. So, the problem is formulated mathematically as follows:

∆ϕ ≡ ∇2 ϕ = 0,








lim v = 0,


 | {z }
r→∞
(5.32)

p



r = x2 + y 2 + z 2 ,






 v0 ·r
= v·r
r
r=R
r
r=R
The well-known fundamental solution of the Laplace equation satisfying our condition at infinity
is ϕ = A/r, A being a constant. Such a solution can not serve as the solution to our problem
since it is symmetric whereas our problem has no such symmetry due to the existence of flow
with velocity v 0 in the proper direction. So, we therefore
try another simple solution of Laplace
equation, i.e. the potential of the dipole B∇ 1r , where B is a constant vector also called the
dipole moment. Since the velocity of the sphere v 0 enters the boundary condition at r = R
linearly, we can look for a solution of the form
1
v0 · r
ϕ = Av 0 ∇
= −A 3
r
r
which yields for the particle velocity (v = ∇ϕ)
v=−
3A
A
v 0 + 5 r (v 0 · r)
3
r
r
where we have used the relation
∇ (v 0 · r) ⇒
∂
∂xi
(v0i xi ) = v0i
= v0i δij = voj ⇒ ∇ (v 0 · r) = v 0 .
∂xj
∂xj
102
CHAPTER 5. POTENTIAL FLOW
5.4. IRROTATIONAL FLOW AROUND A SPHERE
Substituting this expression into the boundary conditions at the surface of the sphere yields
v0 · r
r
(r=R)
=
v·r
r
(r=R)
⇔−
3A
R3
A
v
+
v
=
v
⇒
A
=
0r
0r
0r
R3
R3
2
Therefore, the required solution for the flow around the sphere is now given by

R3
 ϕ = − 2r
3 (v 0 · r) ,

v=
R3
− 2r
3 v0
+
3R3
2r5
(5.33)
(v 0 · r) r.
If we superimpose the homogeneous flow of the velocity (−v 0 ) upon the flow (5.33), we obtain
the solution of the problem of the flow around a sphere at rest when the flow velocity at infinity
equals −v 0 . Then the solution (5.33) takes the form
v = −v 0 −
or in components

v·r


 vr = r = −v0 cos θ −


 vθ = v0 sin θ 1 +
R3
2r 3
R3
v
2r 3 0
R3
3R3
v
+
(v 0 · r) r
0
2r3
2r5
cos θ +
3R3
v
2r 3 0
(5.34)
cos θ = −v0 cos θ 1 −
R3
r3
,
(5.35)
It is seen from (5.35) that vr = 0 at r = R.
To obtain (5.35) we have used the following relations

r cos θ

z }| {


e
v
·
r
v
 0
0 x·r
=
= v0 cos θ,
r
r




v 0 · eθ = v0 cos θ + π2 = −v0 sin θ.
At the surface of the sphere (r = R) we thus have
(
vr = 0,
3
⇔ v|(r=R) = (0, v, 0)
vθ ≡ v = v0 sin θ
2
(5.36)
So, at both critical points θ = 0 and θ = π (in front of the sphere and behind it), the velocity
is equal zero. The pressure distribution over the sphere can be determined with the help of the
Bernoulli theorem (5.25) with p0 and v0 being the pressure and velocity at infinity. Substituting
(5.36) into (5.25) yields
9 2
ρv02
1 − sin θ
(5.37)
p = p0 +
2
4
Thus, the pressure is symmetric about the midsection θ = π2 and minimizes at it. The pressure
coefficient is then given by
9
p − p0
K = 1 2 = 1 − sin2 θ
(5.38)
4
ρv0
2
103
5.4. IRROTATIONAL FLOW AROUND A SPHERE
5.4.2
CHAPTER 5. POTENTIAL FLOW
The Induced Mass
Since the pressure distribution is symmetric about the plane θ = π2 , the total force exerted by
the flow on a sphere is zero, which is also in accordance with the d’Alembert-Euler paradox.
However it is true, only if the sphere’s velocity is constant (v0 = const). If on the other hand the
sphere’s velocity is not constant, i.e. the sphere is moving with acceleration, additional induced
inertia arises due to the flow’s action. We note that the mass is the measure of the body’s inertia,
i.e. the larger the body’s mass, the larger force one has to apply in order to change the velocity
of the body.
To determine the additional induced mass of the sphere, we suppose that there is a force F
providing it with constant acceleration a. If the sphere was initially at rest, then under the action
of the force F, it will start to move and reach the velocity v0 at the time T = v0 /a. The distance
covered due to this motion is given by S = aT 2 /2 = v02 /2a. Hence, the work done by the force
F is F s = F v02 /2a. It must be equal to the kinetic energy gain by the system “sphere + fluid”.
Indeed,
Z
v02
ρv 2
mv02
F
=
+
dV.
2a
2
2
V
The integration in the right hand side of this expression is over the exterior of the sphere. Dividing this equation by v02 /2a yields
Z
a
F = ma + 2 ρv 2 dV.
v0
V
For the incompressible fluid we can rewrite this equation in the form
F = ma + M a
where
ρ
M= 2
v0
Z
v 2 dV
(5.39)
V
is called the induced mass. For our case (fluid at rest at infinity and sphere is moving with
velocity v0 ) we have from Equation (5.33):
v = −v 0
with
3 R3
R3
+
(v 0 · r) r
2r3 2 r5

 vr =

Then,
vθ =
R3
v
r3 0
cos θ,
R3
v
2r3 0
sin θ.
R6 2
v (1 + 3 cos2 θ).
4r6 0
The integral in the definition of the induced mass (5.39) can now be calculated
Z
Z
v02 R6
1 + 3 cos2 θ
2
dV,
v dV =
4
r6
v 2 = vr2 + vθ2 =
r>R
r>R
104
CHAPTER 5. POTENTIAL FLOW
with dV = r2 drdϕ sin θdθ, we obtain
Z
5.5. SUMMARY OBJECTIVES
2
v 2 dV = πR3 v02 .
3
r>R
Finally, the induced mass is determined by
2
14 3
1
M = πR3 ρ =
πR ρ ≡ Vsp ρ
(5.40)
3
23
2
where Vsp is the volume of the sphere. We note also that in (5.40) ρ denotes the fluid density.
Hence, the induced mass is equal to half of the mass of the fluid displaced by the sphere.
5.5
Summary objectives
When you finish studying the Chapter 5 , you should be able to do the following:
1. Define the potential flow and relate this definition to the circulation of the velocity (Kelvin’s
theorem).
2. Write the Euler equation for the potential flow and consider the case of an incompressible
fluid. Define the pressure distribution and stagnation point(s) for the potential flow around
smooth body.
3. Determine the condition for a steady and non-steady flow to be incompressible.
4. Consider a 2-D potential flow and define stream function, derive equation for streamlines.
Using the definition of the vorticity vector, derive the equation for stream functions and
show equivalence of the velocity potential and stream function for 2-D potential flow.
5. Introduce the complex flow potential and describe two particular examples of steady potential flows: a) uniform flow and b) source and sink. Give examples of simply connected
and non-simply connected contours, and calculate the velocity circulation around these
contours.
6. Using the conformal mapping method, describe the flow in the wedge-like domain.
7. Using the conformal mapping method (Joukowski transformation), describe the steady
potential flow of an ideal incompressible fluid around a cylinder and determine the pressure
distribution at the cylinder surface. Formulate the paradox of d’Alamber and Euler.
8. Introduce the pressure coefficient and connect its definition to the principle of hydrodynamic similarity.
9. Show that the violation of the symmetry of the steady flow around a cylinder (for example, due to the superposition of the circular flow) results in the hydrodynamic lift force.
Calculate this force in the simplest case and formulate the Kutta-Joukowski lift theorem.
10. Describe the irrotational flow around a sphere and obtain the pressure coefficient for this
flow.
11. Formulate the concept of induced mass and calculate the induced mass for a spherical body
moving in fluid.
105
5.5. SUMMARY OBJECTIVES
CHAPTER 5. POTENTIAL FLOW
106
Chapter 6
Flows of Viscous Fluids
In the preceding chapters we have considered ideal fluids, i.e. we have neglected friction in
the fluid and thus there was no interaction between the flowing fluid and the boundary (slip).
However, all fluids have viscosity which causes friction. The importance of this friction depends
on the type of fluid and the physical configuration or flow pattern. Viscosity can be thought of
as a measure of the fluid’s resistance to shear when the fluid is in motion. A fluid at rest cannot
resist shearing forces without moving in the way a solid can. Thus if such forces act on a fluid
which is in contact with a solid boundary the fluid will flow over the boundary such that the
particles immediately in contact with the boundary have the same velocity as the boundary itself
(no slip). Thus, all components of the velocity vanish on the surface of a body at rest. This is the
reason why dust accumulates on the surfaces of bodies even when flow exist past these bodies
(i.e. the dust on the blades of a fan).
6.1
Basic Equations
6.1.1
Newtonian Viscosity and Viscous Stresses
Let us consider a flow of liquid (or gas) in which the velocity of flow is different at different
points. This is not an equilibrium state, and processes will occur which tend to equalise the
velocities of flow. Such processes are called internal friction or viscosity. Just as there is a heat
flux from the hotter to the colder parts of a medium in thermal conduction, so in internal friction
the thermal motion of the molecules causes a transfer of momentum from the faster to the slower
regions of the flow.
Thus the three phenomena of diffusion, thermal conduction and viscosity have analogous mechanisms. In all three there is an equalisation of a property of the body (density, temperature or
velocity of flow) if this property is originally not uniform through the body; this brings about
an approach to a state of thermal equilibrium. In all three cases this is achieved by a molecular
transport of some quantity from one part of the body to another. In diffusion there is a transport
of number of particles of the various components of the mixture, in thermal conduction a transport of energy, and in internal friction a transport of momentum. All these effects are therefore
often combined under the general name of transport phenomena.
Let us suppose that a liquid is flowing in the same direction at all points, i.e. that the flow
velocity vector u has the same direction throughout the flow (the x axis, for example), and
suppose also that the magnitude u of the velocity varies in only one direction (the y axis, for
107
6.1. BASIC EQUATIONS
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
example), perpendicular to that of the velocity. By analogy with the diffusion and the heat flux,
we can define the momentum flux Π as being the total momentum transported per unit time in
the positive direction of the y axis across unit area perpendicular to that axis. In exactly the same
way as for the other transport processes, we can say that the momentum flux is proportional to
the gradient of the flow velocity u:
du
Π = −η
dy
The quantity η is called the coefficient of viscosity or simply the viscosity of the medium.
The dimensions of the flux Π are those of momentum divided by area and time, i.e. [Π] =
M
is T1 . Hence, [η] = LT
.
The dimension of du
dy
M
.
LT 2
The viscosity determines the rate of transport of momentum from one point in the flow to another. The velocity is equal to the momentum divided by the mass. The rate of equalisation of
the flow velocity is therefore determined by the quantity ηρ , where ρ is the density, i.e. mass of
the liquid per unit volume. The quantity ν = ηρ is called the kinematic viscosity, whereas η itself
2
is called the dynamic viscosity. It is easily seen that [ν] = LT , i.e. ν has the same dimensions as
the diffusion coefficient and the thermal diffusivity. One can say that the kinematic viscosity is
a kind of diffusion coefficient for velocity.
Let us suppose that a liquid flows in contact with a solid surface, for example, along the walls of
a pipe. Between a surface of a solid and any actual liquid (or gas) there always exist forces of
molecular cohesion which have the result that the layer of liquid immediately adjoining the surface is entirely brought to rest and “adheres” to the surface. Thus the flow velocity is zero at the
wall, and increases away from the wall into the fluid. As a result of viscosity, there then occurs a
flux of momentum from the liquid towards the wall. As it is known from mechanics, the change
in the momentum of a body per unit time is the force acting on the body. The momentum Π
transported through unit area per unit time and ultimately transferred from the liquid to the wall
represents the frictional force exerted on unit area (stress) of the solid wall by the fluid flowing
past it.
The following comments should be added concerning the simple formula for momentum flux
Π, given above. Although the formal analogy already mentioned exists between the phenomena of diffusion, thermal conduction and viscosity, there is also an important difference between
them. This is due to the fact that density and temperature are scalar quantities, whereas velocity
is a vector. We have considered here only the simple case where the velocity is everywhere in
the same direction and therefore the momentum flux is a vector. The formula given above for
Π is valid only in this case. If on the other hand, we take the velocity as a vector with different
components at different points in space, the momentum flux Π should be tensor known also as a
viscous stress tensor. An interesting example when this formula can not be applied is given by
the fluid rotating uniformly as a rigid body together with a cylindrical vessel about the axis of
the vessel. In this case the direction of the fluid velocity is different at different points since the
circular velocity of the fluid particles increases with distance from the axis. There is, nevertheless, no flux of momentum, i.e. no frictional forces, in the liquid. A uniform rigid rotation of the
fluid does not affect thermal equilibrium and could continue indefinitely without the velocity’s
becoming uniform.
To define the momentum flux tensor due to viscosity (viscous stress) in the analytical form let’s
108
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.1. BASIC EQUATIONS
consider the following simple experiment. Imagine two infinite parallel plates at a distance h
with a fluid between them (Figure 6.1). Let the upper plate move relative to the lower plate at the
constant velocity v0 . To get this motion we have to apply a force F to the upper plate. According
to experimental results, this force per unit area of the plate is given by
v0
F
v0
F
∝
⇔
=η
S
h
S
h
(6.1)
where S is the area of the plate. The proportionality factor η is called the viscosity of the fluid
v0
F
h
v=0
Figure 6.1: Geometry for Couette flow.
(sometimes the Newtonian viscosity). The ratio of the viscosity coefficient η to the density of
the fluid ρ is also used in fluid mechanics and is called the kinematic viscosity. Denoting this
ratio as ν, we thus have ν = ηρ .
The relation (6.1) can be written in a more exact form for an arbitrary flow. We begin with
considering a small element of a fluid with dimensions ∆x, ∆y, ∆z (see Figure 6.2) such that
∆y ≪ ∆x and ∆y ≪ ∆z. Let the fluid particle velocity at the lower face of the volume be
vx and let’s assume that the velocity at the upper face is vx + ∆vx . Then, according to the experimental law (6.1) a shear force ∆F must exist supporting this shear flow (the velocity shear
x
⇔ ∂v
6= 0). This force according to Equation (6.1), is proportional to the velocity difference
∂y
∆vx , and inversely proportional to the distance ∆y,
∆vx
∆F
=η
∆S
∆y
Let ∆y → 0 and ∆S → 0. Then, we can define the component of the viscous stress tensor σik ,
∆F
∂vx
σxy = |{z}
lim
=η
∆S
∂y
(6.2)
∆S→0
where σxy is the component of the viscous stress tensor.
Definition: A fluid flow when a non-zero viscous stress tensor (σik 6= 0) is taken into account is
called a flow of viscous fluid.
In order to obtain the equations describing the motion of a viscous fluid, we have to include some
additional terms in the equation of motion of an ideal fluid. The equation of continuity, as we see
from its derivation, is equally valid for any fluid, whether viscous or not. The Euler equation, on
the other hand, requires modification.
109
6.1. BASIC EQUATIONS
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
∆z
v+∆v
x
x
vx
∆y
∆x
Figure 6.2: To determine the tensor of viscous stresses.
6.1.2
The Navier-Stokes Equation
We have seen that in the case of an ideal fluid with no external force the Euler equation can be
written as the momentum conservation law,
∂
∂ (ρvi )
=−
Πik
∂t
∂xk
(6.3)
where Πik = pδik + ρvi vk is the specific momentum flux tensor. This tensor represents a completely reversible transfer of momentum, simply due to mechanical transport of different fluid
particles from place to place and to the pressure forces acting in the fluid. The viscosity (internal
friction) causes another, non-reversible, transfer of momentum from points where the velocity
is large to those where it is small. The equation of motion of a viscous fluid may therefore be
obtained by adding to the “ideal” momentum flux a term (−σik )which gives the non-reversible
“viscous” transfer of momentum in the fluid.
Thus we write the specific momentum flux tensor for a viscous fluid in the form
Πik = pδik + ρvi vk − σik
(6.4)
To determine the dependence of σik on the particle velocity in a general form we have to take
into account the following:
1. The viscous force arises only if different fluid particles move at different velocities (as we
have seen before). Hence σik must depend on the velocity gradient (the velocity shear),
rather than the velocity itself.
2. σik = 0 if the fluid rotates as a rigid body.
It follows from these two assumptions that σik depends only on the following combinations of
velocity gradients
∂vi
∂vk ∂vm
(6.5)
+
,
∂xk ∂xi ∂xm
which are zero in the case of rigid motion. Indeed, if the fluid rotates as a whole (rigid motion)
with angular velocity Ω (Ω1 , Ω2 , Ω3 ), we have v = Ω × r (r is the position of the fluid particle
110
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.1. BASIC EQUATIONS
with origin r = 0 assumed to be on the axis of rotation). It follows from this relation that
 ∂v1
∂v
∂v
v1 = Ω2 x3 − Ω3 x2
 ∂x1 + ∂x22 + ∂x33 ≡ ∇ · v = 0
∂v1
v2 = Ω3 x1 − Ω1 x3 ⇒
+ ∂v2 = 0,
 ∂x2 ∂x1
v3 = Ω1 x2 − Ω2 x1
etc.
The most general expression for the second order tensor satisfying the two conditions stated
above is
∂vi
∂vk
∂vm
+
δik
+b
σik = a
(6.6)
∂xk ∂xi
∂xm
where a and b are constants independent on the velocities or its gradients. Now, using Equation
(6.3), we can obtain the equation of fluid motion with friction force due to viscosity of the fluid:
∂Πik
∂ (ρvi )
=−
⇔
∂t
∂xk
ρ
∂vi
∂ρ
∂p
∂vi
∂
∂ ∂vi
∂ ∂vk
∂ ∂vk
+ vi
=−
− ρvk
− vi
(ρvk ) + a
+a
+b
∂t
∂t
∂xi
∂xk
∂xk
∂xk ∂xk
∂xk ∂xi
∂xi ∂xk
Taking now into account the continuity equation,
ρ
∂vi
∂vi
+ vk
∂t
∂xk
=−
∂ρ
∂t
= − ∂x∂ k (ρvk ), we obtain
∂p
∂ 2 vk
+ a∇2 vi + (a + b)
.
∂xi
∂xi ∂xk
(6.7)
Equation (6.7) written in vector form yields
dv
∂v
ρ
≡ρ
+ (v · ∇) v = −∇p + a∇2 v + (a + b) ∇ (∇ · v)
dt
∂t
(6.8)
This is the most general form of equation of motion of viscous fluid and it is known as the NavierStokes equation. In the case of an incompressible fluid the Navier-Stokes equation assumes the
form
dv
∂v
ρ
(6.9)
≡ρ
+ (v · ∇) v = −∇p + a∇2 v
dt
∂t
In discussing different phenomena related to the flows of viscous fluid, we shall always regard
the fluid as incompressible, and accordingly use the equation of motion in the form (6.9). The
viscous stress tensor in an incompressible fluid takes the simple form
∂vk
∂vi
σik = a
.
+
∂xk ∂xi
We must also write down the boundary conditions on equations of motion of a viscous fluid.
There are always forces of molecular attraction between a viscous fluid and the surface of a solid
body, and these forces have the result that the layer of fluid immediately adjusted to the surface is
brought completely to rest, and “adheres” to the surface. Accordingly, the boundary conditions
on the equations of motion of a viscous fluid require that the fluid velocity must vanish at fixed
solid surfaces, i.e.
v=0
111
6.1. BASIC EQUATIONS
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
It should be emphasised that both the normal and the tangential velocity component must vanish,
whereas for an ideal fluid the boundary conditions require the vanishing of vn only.
The Navier-Stokes equation is the cornerstone of fluid mechanics. It may look harmless enough,
but it is an unsteady, nonlinear, second-order, partial differential equation. Unfortunately, analytical solutions are unobtainable except for very simple flow fields. Moreover, equation (6.9)
has five unknowns (three velocity components, pressure and density), yet it represents only three
equations (three components since it is a vector equation). Obviously we need two other equations – the incompressible continuity equation and equation of state – to make the problem
solvable.
6.1.3
The Viscous Force
The term ∂σik/∂xk on the right hand side of Equation (6.3) or (which is the same) the terms
proportional to a or b in Equation (6.7) may be considered as the ith component of the viscous
force per unit volume
fi = a
∂ 2 vk
∂
∂ 2 vi
+
(a
+
b)
= a∇2 vi + (a + b)
(∇ · v)
2
∂xk
∂xi ∂xk
∂xi
(6.10)
Let us now find the force excerted by the viscous fluid on a body in the case of a steady flow.
Imagine a closed surface S ′ surrounding the body (Figure 6.3). In the case of a steady flow,
n
S'
S
Figure 6.3: To determine the viscous forces exerted by a fluid on a body.
Equation (6.3) becomes ∂Πik/∂xk = 0. Integrating this equation over the volume between S ′
and the surface S of the body and using Gauss’ divergence theorem for the tensor Πik yields
Z
Z
(6.11)
− Πik nk dS + Πik nk dS = 0
S
S′
where n = (n1 , n2 , n3 ) is outward normal to S or S ′ . Taking the relation (6.4) into account and
the fact that on S the flow velocity vanishes (adhesion of the viscous fluid to the solid surface),
we write the first term in Equation (6.11) as
Z
Z
Fi = − pni dS + σik nk dS
(6.12)
S
S
112
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2. EXAMPLES OF VISCOUS FLOW
This is the force exerted by the fluid on the body. The first term here is the resultant pressure
force on the body and the second term
Z
′
Fi = σik nk dS
(6.13)
S
is the force due to viscosity. Its density (i.e., the force per unit area of the surface of the body) is
fi = σik nk
(6.14)
Now, taking Equation (6.11) into account we can finally determine the force acting on the body
from the flow of viscous fluid
Z
(6.15)
Fi = (−pni − ρvi vk nk + σik nk )dS
S′
i.e., this force can be calculated if the characteristics of the flow on an arbitrary surface S are
known. This approach may turn out to be much more convenient when compared with calculations by Equation (6.12) because the surface of the body S as well as the flow near it may be
rather complicated. On the other hand, the structure of the flow at large distances from the body
is much simpler and can be determined, for example, by the linearisation of the equation of motion (the momentum equation) with respect to disturbances from an unperturbed state (without
body).
6.2
Some Examples of Viscous Fluid Flow
In this section we show how to apply the differential equation of motion of viscous fluid to particular flows. There are two types of problems for which these differential equations (continuity
and Navier-Stokes) are useful:
• Calculating the pressure field for a known velocity field
• Calculating both the velocity and pressure fields for a flow of known geometry and known
boundary conditions.
The feature of the Navier-Stokes equation which causes most analytical difficulty is the nonlinearity in the velocity v arising from the acceleration of a fluid element in the Eulerian description of the flow field. The mathematical difficulties presented by the complete equation of
motion are so severe that most of the existing solutions are applicable only to circumstances in
which for some reason the equation reduces to a linear form. Among the simplest of such cases
are those in which the velocity vector has the same direction everywhere, and is independent of
distance in the flow direction. The convective rate of change of velocity then vanishes identically
and thus for a steady flow the acceleration of a fluid element is equal to zero.
6.2.1
Couette Flow
The simple flow considered in Section 6.1.1 is called the plane Couette flow. This is the steady
incompressible flow of a viscous fluid in the narrow gap between two infinite parallel plates (see
Figure 6.1). The top plate is moving at speed v0 , and the bottom plate is stationary. The distance
113
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
between these two plates is h. There is no applied pressure other than hydrostatic pressure due
to gravity. To describe this flow let the x-axis be parallel to the velocity vector v 0 and the y-axis
perpendicular to the plates. The origin is on the lower fixed plane.
As a preliminary, we note the precise conditions under which the fluid velocity in the flow region
is exactly unidirectional. Our interest lies in situations in which a general streaming motion in
the x-direction parallel to the generators has “settled down” and is independent of end effects,
so that all three velocity components (vx , vy , vz ) relative to a Cartesian coordinate system are
independent of x. Indeed, due to the symmetry of the problem we can assume the velocity field
∂
to be purely two-dimensional, i.e. w = 0 and ∂z
of any velocity component is zero. Since the
flow is parallel to the x-axis, y-component of the flow velocity is zero, vy = 0. It follows then
x
= 0. So, vx is not a function of x and thus
from the incompressible continuity equation that ∂v
∂x
flow characteristics are invariant in the x-direction. In other words, it does not matter where
we place our origin — the flow is the same at any x-location. The phrase fully developed is
often used to describe this situation (see Figure 6.4). We note that the same result can also be
obtained from the assumption that plates are infinite of length which tells that there is nothing
special about any x-location. Furthermore, since vx is not a function of time (steady flow), or z
(assumption that the flow is two-dimensional), we conclude that the velocity may depend only
on y: vx = v = v(y). Since there is no applied pressure gradient pushing the flow in the x-
v
v
y
h
x
x=x2
x=x1
Figure 6.4: A fully developed region of a flow field is a region where the velocity profile does not change
with downstream distance. Fully developed flows are encountered in long, straight channels
and pipes. Fully developed Couette flow is shown here, the velocity profile at x2 is identical
to that at x1 .
∂p
direction, the Navier-Stokes equation shows that ∇p also must be independent of x, i.e. ∂x
=0
and p = const with respect to x. In this case the flow establishes itself due to viscous stresses
caused by the moving upper plate. Taking now into account that ∂/∂t = 0 since the flow is
steady, we then obtain from the Navier-Stokes equation (6.9) that

∂p

 ŷ : ∂y = 0,

 x̂ :
∂2v
∂y 2
= 0.
It is worthwhile to note here that in the x-component of the Navier-Stokes equation the material
acceleration is zero, implying that fluid particles are not accelerating at all in this flow field,
neither by local (unsteady) acceleration, nor by convective acceleration. Since the convective
acceleration terms make the Navier-Stokes equation nonlinear, this unidirectional assumption
greatly simplifies the problem. In fact, all other terms in this equation vanish except for a lone
114
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2. EXAMPLES OF VISCOUS FLOW
viscous term, which must then itself equal zero.
Integrating these equations yields
p = const,
v = Ay + B.
The first result shows that p is not a function of y. Since p is not a function of time, or x, we
obtain that p is at most a function of z only, i.e. p = p (z). The z-component of the Navier-Stokes
equation simplifies to
∂p
= −ρg
∂z
which after integration yields p = p0 − ρgz, where we assume p = p0 at z = 0. This result
represents a simple hydrostatic pressure distribution due to gravity. We thus conclude that, at
least for the problem under consideration, hydrostatic pressure acts independently of the flow. In
other words, for incompressible flow fields without free surfaces, hydrostatic pressure does not
contribute to the dynamics of the flow field. So, we can neglect the gravity effect on the flow
dynamics and finally get for the pressure distribution in whole region
p = p0 = const.
The constants A and B are determined by the boundary conditions v = 0 at y = 0, v = v0 at
y = h. We thus obtain B = 0, A = v0/h, hence
v = v0/h y.
The velocity field reveals a simple linear velocity profile from vx = 0 at the bottom plate to
vx = v0 at the top plate, as sketched in Figure 6.5. We now determine the force f per unit
v
y
h
v=(v0 /h)y
x
Figure 6.5: The linear velocity profile of the Couette flow between parallel plates.
area exerted by the fluid on the lower plate. This force will be only due to viscosity since the
pressure gives no tangential component and p = const. To calculate this force we can use
expression (6.14) where σik is given by the first term in (6.6) since the fluid is assumed to be
incompressible. So,
∂vm
∂vk
∂vi
+b
i, k = 1, 2, 3.
+
δik
σik = a
∂xk ∂xi
∂xm
| {z }
=0
Here v1 = vx and v2 = vy and thus
f |y=0 = a
∂vx
∂v
av0
≡a
=
∂y
∂y
h
115
(6.16)
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
This is the force exerted by the fluid on the lower plate. The direction of this force is in the
positive x-direction, namely, the fluid tries to pull the bottom wall to the right, due to
viscous
effects (friction). Obviously, this force has to be equal to the force (per unit area) F/S =
η v0/h , hence a = η. Now the Navier-Stokes equation for an incompressible fluid is written as
∂v
+ (v · ∇) v = −∇p + η∇2 v.
ρ
∂t
(6.17)
The important quantity in hydrodynamics (especially in those parts devoted to the flow of viscous
fluid around bodies) is the drag coefficient CD . The drag coefficient is defined as the ratio of the
2
viscous force f to the so called velocity head ρvm
/2. So, the drag coefficient is defined by
CD =
2f
.
2
ρvm
(6.18)
Here vm is the mean (or some characteristic) flow velocity. The mean fluid velocity in the case
of the Couette flow is
Zh
1
1
vm =
vdy = v0
h
2
0
and thus the drag coefficient is given by
CD =
where Re =
6.2.2
vm h
ν
2 · 2ηvm
4ν
4
4η
≡
≡
=
2
hρvm
hρvm
hvm
Re
is the so-called Reynolds number and ν =
η
ρ
(6.19)
is the kinematic viscosity.
Plane Poiseuille Flow
Let us suppose that we have instead two fixed parallel plates a distance h apart. Consider the
fully developed stage of steady flow of viscous fluid between the plates supported by an external
pressure gradient along the plates, the so-called plane Poiseuille flow (Figure 6.6). Again let the
x and y axes be parallel and perpendicular to the plates, respectively, and two-dimensionality of
∂
= 0. The continuity equation is simplified then in the same way as in
the flow requires that ∂z
the plane Couette flow, which yields that the fluid velocity can depend only on y in this case,
vx = v(y). Then, the Navier-Stokes equation will reduce to the following components


 ŷ :

 x̂ :
∂p
∂y
=0
i.e. the pressure is independent of y and depends only on x,
(6.20)
1 ∂p
η ∂x
=
∂2v
.
∂y 2
The y-component of the momentum equation shows that p is not a function of y. In the xcomponent, the right-hand side depends only on y while its left-hand side depends only on x.
The only way this can be satisfied is for both terms to be constant. We denote this constant by b,
b=−
∂p
.
∂x
116
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2. EXAMPLES OF VISCOUS FLOW
Fixed
Fixed
plate
plate
y
h
p1
p2
x
Fixed
Fixedplate
plate
δp p2 - p1
δx = x2 - x1
x1
x2
Figure 6.6: Geometry of viscous flow between two infinite plates with a constant applied pressure gradient (plane Poiseulle flow).
The pressure gradient is therefore a constant, implying that the pressure varies linearly along the
channel. So, the pressure field for the plane Poiseuille flow (we ignore the effects of gravity) is
given by
∂p
p = p0 +
x ≡ p0 − bx
∂x
Integrating the x-momentum equation twice, we get
∂v
∂ 2v
b
b
b
= − y + A ⇒ v = Ay + B − y 2 .
− = 2 ⇒
η
∂y
∂y
η
2η
(6.21)
Using the boundary conditions v = 0 at y = 0, h we find the integration constants B = 0,
A = bh/2η ; hence,
by (h − y)
∂p y (h − y)
v=
=−
≡ v(y)
(6.22)
2η
∂x
2η
So, the velocity profile across the flow is parabolic (see Figure 6.7) and reaching its maximum
∂v
=0
value in the middle. Indeed, taking the derivative of (6.22) with respect to y, we find that ∂y
at y = h2 . The magnitude of the viscous stress tensor (or shear stress) is
σxy
∂v
dp
=−
=η
∂y
dx
h
−y
2
dp
at
which shows that the stress distribution across the channel is linear with a magnitude of h2 dx
the walls. It is important to note that the constancy of the pressure gradient and the linearity of
the shear stress distribution are general results for a fully developed channel flow and hold for
the flows with different Reynolds number.
We now determine the force per unit area exerted by the fluid on each plate. Similar to the
Couette flow considered above (see Equation (6.16)), we have
1
bh
∂p h
∂v
= ηb (h − y − y)y=0 =
≡−
(6.23)
fx = η
∂y y=0
2η
2
∂x 2
The force is positive since the direction of the force is the same as the flow direction, whereas
∂p
the pressure decreases as x increases, i.e. − ∂x
> 0. To determine the drag coefficient CD , we
117
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
y
h
0
v
Figure 6.7: Particle-velocity curve in Poiseuille flow.
have to find the mean flow velocity. This velocity (vm ) is given by
1
vm =
h
Zh
bh2
h2 ∂p
vdy =
≡−
12η
12η ∂x
(6.24)
0
2
Now the drag coefficient CD = 2f /ρvm
takes the form
CD =
12ν
12
≡
hvm
Re
(6.25)
Again, the drag coefficient depends on the Reynolds number only.
6.2.3
Poiseuille Flow in a Cylindrical Pipe
Let us consider now the Poiseuille flow in a cylindrical pipe with a circular cross section (the
same along the whole length of the pipe however). We take the axis of the pipe as the x-axis.
Since the flow is assumed to be parallel, the fluid velocity is evidently along the x-axis at all
points (flow characteristics are invariant in the x-direction) and is a function of y and z only (we
denote this velocity by vx ≡ v), so v = v (y, z). Then the equation of continuity is satisfied
∂p
∂p
= ∂z
= 0,
identically, while y and z components of the Navier-Stokes equation again give ∂y
i.e. the pressure is constant over the cross-section of the pipe. The x-component the momentum
equation gives
∂ 2v ∂ 2v
1 ∂p
+ 2 =
2
∂y
∂z
η ∂x
118
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2. EXAMPLES OF VISCOUS FLOW
∂p
= const. The pressure gradient may therefore be written as
Here we again conclude that ∂x
∂p
∆p
= − l , where ∆p is the difference between the ends of the pipe and l is the length. The
∂x
velocity distribution for flow in a pipe is thus determined by a two-dimensional equation of
the form ∇2 v = const. This equation has to be solved with the boundary condition v = 0 at
the circumference of the cross-section of the pipe. We shall solve this equation for a pipe with
circular cross-section. Taking the origin at thep
center of the circle and using the polar coordinates,
we have by symmetry v = v (r), where r = y 2 + z 2 . Using the expression for the Laplacian
in polar coordinates, we have
dv
∆p
1 d
r
=−
r dr
dr
ηl
Integration gives
∆p 2
r + A ln r + B
v (r) = −
4ηl
The constant A must be put equal to zero, since the velocity must remain finite at the centre of
the pipe. The constant B is determined from the requirement that v = 0 for r = R, where R is
the radius of the pipe. We then find
v (r) =
∆p 2
1 ∂p
R − r2 ≡ −
R2 − r 2
4ηl
4η ∂x
So, the velocity profile in fully developed flow in a pipe is a parabolic of revolution with a
maximum at the axis of the pipe and minimum (zero) at the pipe wall. Also, the axial velocity v
∂p
is positive for any r, and thus the axial pressure gradient ∂x
must be negative, i.e. pressure must
decrease in the flow direction because of viscous effects.
The mean velocity is
1
vm =
πR2
ZR Z2π
0
0
∆p
vrdrdϕ =
2ηlR2
ZR
∆p 2
R2 ∂p
r R − r dr =
R ≡−
8ηl
8η ∂x
0
2
2
Combining the last two equations, the velocity profile is rewritten as
r2
v (r) = 2vm 1 − 2
R
This is a convenient form for the velocity profile since vm can be determined easily from the
flow rate information. The maximum velocity occurs at the axis of the pipe and is obtained by
substituting r = 0, vmax = 2vm . Therefore, the average velocity in fully developed Poiseulle
flow in a cylindrical pipe is one-half of the maximum velocity, (see Figure 6.8). Now we can
obtain the Poiseuille formula for the volume flow rate of fluids through the whole cross-section
of a cylindrical pipe
∆p 4
πR4 ∂p
Q = πR2 vm = π
R ≡−
.
8ηl
8η ∂x
Then the mass of the fluid passing per unit time through the whole cross-section area of a pipe is
determined by
πR4 ∂p
π∆p 4
R ≡−
.
M = ρQ =
8νl
8ν ∂x
The mass of fluid is thus proportional to the fourth power of the radius of the pipe.
119
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
y
vm
h
0
v
Figure 6.8: Average velocity vm is defined as the average speed through a cross section. For fully
developed laminar pipe flow, vm is half of the maximum velocity.
The force acting on the wall per unit area of the tube boundary is determined by (n is directed
opposite to r)
∂v
1 ∆p
1 ∂p
=
R=−
R
(6.26)
fx = σxr nr = −η
∂r r=R 2 l
2 ∂x
∂p
is negative, so the direction of this force is in the positive
For the flow from left to right, ∂x
∂p
is
x-axis, namely, the fluid tries to pull the bottom wall to the right, due to friction, when ∂x
negative. It is worthwhile to note here that the total force per unit length of the tube 2πRfx is
∂p
balanced by the pressure gradient through the cross-section area −πR2 ∂x
, i.e.
2πRfx = −πR2
∂p
.
∂x
This is due to the assumption that the flow under consideration is the fully developed stage of a
steady flow.
The drag coefficient depends again only on the Reynolds number
CD =
2fx
16
8η
≡
,
=
2
ρvm
ρvm R
Re
(6.27)
Re = vmνD , D = 2R. The flows considered above occur when the flow velocities are sufficiently
small (the Reynolds number is small). Formula (6.26) is in good agreement with experiments in
this case.
120
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2.4
6.2. EXAMPLES OF VISCOUS FLOW
Viscous Fluid Flow Around a Sphere
Let us consider a fixed sphere of radius R at rest and assume that a steady flow of an incompressible fluid around the sphere is homogeneous at infinity. The direction of the x-axis is chosen
along the direction of v0 at infinity (see Figure 6.9). To describe such a flow, i.e. find the velocity
distribution and the pressure distribution, we will use the Navier-Stokes equation:


 ∂v

ρ
+ (v · ∇) v  = −∇p + η∇2 v.
∂t
|{z}
=0
The first term on the left hand side is equal zero since we consider a steady flow. The characteristic spatial scale of the flow is D = 2R. Consider the convective term ρ (v · ∇) v and the
2
viscous term η∇2 v. The order of magnitude of each term is given by ρv0/D and η v0/D2 . The
ratio of these terms determines their relative influence on the flow. Indeed,
inertial term
ρv0 D
v0 D
=
=
≡ Re.
viscous term
η
ν
Hence, the inertial term is small compared with the viscous term if Re ≪ 1, i.e. if the velocity
of the flow is rather small. So, to describe the flow of incompressible viscous fluid with Re ≪ 1
around a sphere we can use the reduced form of the Navier-Stokes equation, since we can neglect
the inertial term, and the equation of continuity,

 ∇p = η∇2 v,
(6.28)

∇ · v = 0.
The boundary conditions

 v|r=R = 0,
vx |r→∞ = v0 ,

vy |r→∞ = vz |r→∞ = 0
(6.29)
p
where r = x2 + y 2 + z 2 . We look for a solution of the problem (6.28) and (6.29) as a sum of
a potential part and an additional term
v = ∇ϕ + w(x, y, z).
(6.30)
Substituting (6.30) into the continuity equation (second equation in (6.28)) yields
∇2 ϕ + ∇ · w = 0.
(6.31)
We look for a function f that satisfy the boundary condition ∇ϕr→∞ = v0 . Then we can assume
∂r
∂ 1
ϕ = v0 x + C 1
+ C2
(6.32)
∂x
∂x r
where C1 and C2 are arbitrary constants. Applying
the Laplace operator to (6.32) and taking
2 1
into account that in spherical coordinates ∇ r = 0, since
2the function 1/r is the fundamental
d
r2 dr
= r we obtain
solution of the Laplace equation, and ∇2 r = r12 dr
dr
∂
∂ 2 1
∂ 1
2
2
2
∇ ϕ = ∇ (v0 x) +C1
∇ r + C2 ∇
= 2C1
.
(6.33)
| {z }
∂x
∂x
r
∂x r
|
{z
}
=0
=0
121
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
v0
R
0
x
Figure 6.9: Illustrating viscous flow around a sphere.
We now note that (6.31) is satisfied with w = −2C1
1
r
∂
∇x, in this case ∇ · w = −2C1 ∂x
1
r
.
Now we can determine the velocity of the flow (6.30):
1
∂ 1
∂r
v = ∇ v0 x + C 1
− 2C1
∇x.
+ C2
∂x
∂x r
r
1
∂
∂r
= − rx3 , we obtain for the flow velocity
= xr and ∂x
Taking into account that ∂x
r
1
C1 3C2
1
v = v0 − C1 − C2 3 ∇x + x − 3 + 5 r
(6.34)
r
r
r
r
This expression satisfies the boundary condition v = v0 ∇x as r → ∞. To satisfy the condition
on the sphere’s surface v r=R = 0, we first assume that the second parentheses in (6.34) is equal
zero, i.e.
C1 3C2
− 5
= 0 ⇒ 3C2 = R2 C1 .
3
r
r
r=R
Assuming now that the first parentheses in (6.34) also equals zero yields
4C1
C1 C2
3
0 = v0 −
⇒ v0 −
− 3
= 0 ⇒ C1 = v0 R.
(6.35)
r
r r=R,C2 =R2 C1 /3
3R
4
Now the velocity distribution becomes
3 R 1 R3
3 v0 Rx
R2 r
v = 1−
v0 −
1− 2
−
.
4r
4 r3
4 r2
r
r
Or for the components

3R
1 R3

v
=
1
−
−
v0 −

x
3
4 r


4 r 2
1 − Rr2 ,
vy = − 34 v0rRxy
3



R2
 vz = − 3 v0 Rxz
1
−
.
3
2
4 r
r
122
3 v0 Rx2
4 r3
1−
R2
r2
(6.36)
,
(6.37)
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.2.5
6.2. EXAMPLES OF VISCOUS FLOW
Stokes Formula for Drag
The pressure distribution can now be obtained from the first equation in (6.28). Substituting
the velocity solution
from (6.30) into (6.28) and taking into account that wy = wz = 0, and
∇2 wx ⇒ ∇2 1/r = 0 we obtain that the equation for the pressure can now be written in the
form
∇p = η∇2 (∇ϕ) = ∇ η∇2 ϕ
∂ 1
3
Rx
⇒ ∇p = ∇ 2ηC1
= ∇ − ηv0 3
∂x r
2
r
where we used results from (6.33) and (6.35). The solution of this equation under the condition
that the pressure vanishes at infinity is given by
3
Rx
p = − ηv0 3 .
2
r
(6.38)
Let us now determine the force F exerted by the flow on the sphere. Obviously, Fy = Fz = 0
due to the symmetry of the flow about the x-axis. The viscous force per unit area of the sphere
surface acting in the x-direction is, according to (6.26), given by
∂v1
∂vk
∂vk xk
∂v1
′
nk = η
+
+
.
f1 = σ1k nk = η
∂xk ∂x1
∂xk ∂x1 r
Here, as always, x1 = x, x2 = y, x3 = z. The first term in parentheses gives rise to the force
∂v1 xk
∂v1
∂v1 ∂r xk
∂v1 x2k
=
=
=
.
∂xk r r=R
∂r ∂xk r r=R
∂r r2 r=R
∂r r=R
In performing the differentiation we have disregarded the dependence of v1 ≡ vx on x1 ≡ x given
by the last term in (6.37) for vx . It is possible because this term vanishes after differentiation with
respect to x for r = R. The second term in f1′ becomes
∂vk xk
∂vk x1
∂vk x1 xk
=
=
=0
∂x1 r r=R
∂r r2 r=R
∂xk r r=R
∂vk
=
0
. Now taking the relations (6.37) into acsince the fluid is assumed incompressible ∂x
k
count, we obtain
3ηv0
x2
∂vx
′
′
f1 = fx = η
(6.39)
=
1− 2 .
∂r r=R
2R
R
Besides this tangential force, there is also a force due to pressure acting on the sphere. The
x-component of this force per unit area is, according to the relation (6.38),
x
3 ηv0 x2
=
.
−p
R r=R 2 R R2 r=R
The resultant force per unit area of the sphere’s surface is the sum of the latter and Equation
(6.39):
3 ηv0
.
(6.40)
fx =
2 R
The force exerted by the flow on the whole sphere is equal to fx multiplied by the sphere’s
surface area 4πR2 ,
Fx = 6πηRv0
(6.41)
123
6.2. EXAMPLES OF VISCOUS FLOW
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
This is well-known Stokes’ formula for drag when the sphere moves slowly relative to the fluid.
We note that the problem when the sphere slowly moves relative to the fluid is quite similar to
the problem on the flow of fluid around the sphere at rest under the condition that the flow has
the given velocity v0 at infinity. Note also that in the case of an ideal fluid (η = 0) the drag force
is zero.
The drag coefficient is given by
2Fx
2 · 6πηRv0
24
.
=
=
πR2 ρv02
πR2 ρv02
Re
CD =
(6.42)
In the case of free fall of a sphere in a gravitational field, the constant velocity v∞ is reached as
t → ∞. This velocity can be determined from the condition that the resistance force Fx is equal
to the weight of the sphere minus the Archimedes force:
The resistance force:
Fx = 6πηRv∞ ,
The weight of the sphere:
4
πR3 ρ1 g,
3
The Archimedes force (weight of the fluid displaced by the body):
4 3
πR ρ
|3 {z }
mass of displaced fluid
⇒ 6πηRv∞
4
= πR3 gρ
3
ρ1
−1 .
ρ
(6.43)
This yields for the limiting velocity
v∞
2 gR2
=
9 ν
ρ1
−1 .
ρ
(6.44)
We note that the developed theory is valid only if the Reynolds number is small, Re ≪ 1.
The theory can be improved however by taking into account the nonlinear (inertial) term in the
Navier-Stokes equation.
6.2.6
Laminar and Turbulent Flows
In conclusion we note that in all examples of flows of viscous fluid investigated in this section
the drag coefficient CD may be expressed in general form as a function of the dimensionless
parameter, Re = v0νL , the Reynolds number. Here L is some characteristic length (which might
be the maximum diameter of the body, or the distance between enclosing boundaries) and v0 is
some characteristic velocity (which might be the speed of the steady flow). In other words, for
given geometry of the boundary and initial conditions, the effect on a flow field of changing v0 ,
L, ρ or η or of changing several of these parameters together, can be described uniquely by the
consequent change of Re alone.
The magnitude of the Reynolds number may also be regarded as providing an estimate of the
relative importance of the non-viscous and viscous forces acting on unit volume
ofthe fluid. The
∂p
and the visNavier-Stokes equation contains on the right-hand side the pressure force − ∂x
i
2
vi
i
. These three
, and the sum of the two equals minus the inertia force −ρ dv
cous force µ ∂x∂i ∂x
dt
j
124
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.3. BOUNDARY LAYER
forces together are in equilibrium and the balance between them can be indicated by the ratio of
any of two. It is customary to characterise the flow by the ratio of magnitudes of the inertia and
viscous forces. Then at any point in the fluid (if L and v0 are truly representative parameters of
the flow) this ratio is equal to Re = v0νL , i.e Reynolds number measures the relative magnitude of
the inertia and viscous forces. Thus we can conclude that for a given geometry of the boundary
and initial conditions, changes in Re correspond to changes in the relative magnitude of inertia
and viscous forces.
In particular, the effect of making Re ≪ 1 is to make the inertia force much smaller than the
viscous force, so that pressure and viscous forces are dominant in the flow field, while the effect
of making Re ≫ 1 is to make the inertia force much greater than the viscous force, so that inertia
and pressure forces are dominant. Therefore at large Reynolds numbers, the viscous forces cannot prevent the random and rapid fluctuations in the fluid and the fluid flow is characterised by
strong velocity fluctuations and is highly disordered motion. Such flow is said to be turbulent. In
contrast, at small or moderate Reynolds numbers, the viscous forces are large enough to suppress
these fluctuations and to keep the fluid “in line”. In this case, the flow regime is characterised by
smooth stream lines and is highly ordered motion. Such flow is said to be laminar. Finally, at
values of Reynolds number of the order of unity, all three forces presumably play an important
part in the equation of motion.
6.3
Boundary Layer
The solution for the potential flow of an ideal incompressible fluid ∇2 ϕ = 0, v = ∇ϕ, also
satisfies the Navier-Stokes equation since the viscous term η∇2 v = 0 in this case. Indeed,
∇2 v = ∇2 (∇ϕ) = ∇ ∇2 ϕ = 0.
Hence, the solution for an ideal fluid can also be used to some extent for a viscous one. A drastic
difference may only occur near some body present in the fluid. The solution for the real (viscous)
fluid and for the ideal fluid must satisfy different boundary conditions on the body’s surface:
The tangential velocity on the body’s surface is prohibited in a real fluid and it is not specified
(i.e.arbitrary) in an ideal fluid.
However, in many important cases this difference is essential only in a thin layer near the surface called the boundary layer. The boundary layer approximation has been introduced in fluid
dynamics by Prandtl (1904). Prandtl’s idea was to divide the flow into two regions: an outer
flow region that is inviscid (ideal), and an inner flow region called a boundary layer — a very
thin region of flow near a solid wall where viscous forces cannot be ignored. To illustrate this,
consider the picture in Figure 6.10. A uniform viscous flow suddenly hits a plate which reduces
the velocity to zero on the body surface due to the no-slip condition. Since the effect of viscosity
is to resist fluid motion, the velocity close to the solid surface continuously decreases towards
downstream. But further away from the flat plate the speed still equals the free stream value.
Consequently a velocity gradient is set up in the fluid in a direction normal close to the flat plate,
this is what we call the boundary layer. In the outer flow region, the ideal or inviscid flow, we
use the continuity and Euler equation to obtain the outer flow velocity field, and the Bernoulli
equation to obtain the pressure field. Within the boundary layer, we must solve the Navier-Stokes
equation instead of the Euler equation. Thus the boundary layer approximation corrects some of
the major deficiencies of the Euler equation by providing a way to enforce the non-slip condition
125
6.3. BOUNDARY LAYER
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
z
v0
x
h(x)
Boundary layer (non-negligible viscous forces)
Outer flow (inviscid region of flow)
Uniform flow
Figure 6.10: Prandtl’s boundary layer concept splits the flow into an outer flow region and a thin boundary layer region (not to scale).
at solid walls. This physical picture can be illustrated by imaging the Euler equation and the
Navier-Stokes equation as two mountains separated by a huge chasm. Then, the boundary layer
approximation bridges the gap between the Euler equation and the Navier-Stokes equation, and
between the slip condition and the non-slip condition at solid walls (see Figure 6.11).
The key to successful application of the boundary layer approximation is the assumption that
the boundary layer is very thin. Two classic examples of a boundary layer to be discussed later
are so-called viscous waves and uniform stream flowing parallel to a long flat plane.
6.3.1
Viscous Waves
As a very simple example of a boundary layer, we consider viscous waves near an infinite plate
oscillating in its own plane. Assume that this plane coincide with the x-y plane and the viscous
homogeneous fluid (incompressible, for simplicity) is in the half space (z > 0). Let the oscillations of the plate be harmonic, the x-axis is taken in the direction of oscillation i.e. the velocity
of the points on the plate are given by
(vx )z=0 = v0 exp(−iωt),
(6.45)
(vy )z=0 = (vz )z=0 = 0.
What happens with a fluid if the plate is oscillating under these conditions? The ideal fluid would
remain at rest due to slip at the plate. The viscous fluid adheres to the plate and conditions (6.45)
must be regarded as the boundary conditions at z = 0 for the flow under consideration.
To describe the reaction of the viscous fluid on the plate oscillations we turn to the Navier-Stokes
equation (6.17). Obviously, vy = 0 due to the symmetry of the problem. Indeed, the x-direction
is the direction of oscillations, z > 0 and z < 0 are corresponding to two different media, so
the only symmetrical direction is the y-direction. For the same reasons vx and vz can depend
only on z. Since the fluid is incompressible we have from the continuity equation ∇ · v = 0,
z
= 0 ⇒ vz = const. However, this constant is equal zero since vz = 0 at z = 0. Since
or ∂v
∂z
all quantities are independent of the coordinates x and y, and since vz = 0, it follows that the
126
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.3. BOUNDARY LAYER
Figure 6.11: (a) A huge gap exists between the Euler equation (which allows slip at walls) and the
Navier-Stokes equation (which supports the no-slip condition). (b) the boundary layer approximation bridges that gap.
nonlinear term ρ (v · ∇) v is also identically zero. Then, the z-component of Equation (6.17)
will just be ∂p/∂z = 0 , or p = const, and its x-component becomes (vx ≡ v)
ρ
∂v
∂v
∂ 2v
∂ 2v
=η 2 ⇒
=ν 2
∂t
∂z
∂t
∂z
(6.46)
This is the linear equation of the same type as the equation of heat conduction. We look for its
solution under the condition (6.45) assuming
v = v0 exp [i (kz − ωt)] .
Substitution of the expression (6.47) into (6.46) yields
r
r
iω
ω
2
−iω = −νk ⇒ k =
= ± (1 + i)
.
ν
2ν
(6.47)
(6.48)
Only the “+” sign is appropriate here, otherwise v in (6.47) grows infinitely as z → ∞. Now we
obtain from the solution (6.47) the expression for the viscous wave
r
r
ω
ω
v = v0 exp −
z exp i
z − ωt .
2ν
2ν
127
(6.49)
6.3. BOUNDARY LAYER
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
q
propThis velocity profile may be described as a damped transverse wave of wavelength 2π 2ν
ω
agating in the z-direction. Thus transverse waves can occur in a viscous fluid with velocity
vx ≡ v perpendicular to the direction of propagation. They are, however, rapidly damped as we
move away from the solid surface whose motion generates the waves.
The amplitude of this wave decreases exponentially at the distance
r
2ν
δ=
ω
(6.50)
from the plate. The lower the viscosity, the smaller is δ, the thickness of the layer in which
significant oscillations of the fluid are observed. At distances z ≫ δ, we have v = 0 as in an
ideal fluid.
6.3.2
The Boundary Layer. Qualitative Considerations
Let us consider a fluid flow around a body. One can ask, does the flow of a viscous fluid approach
that of an ideal fluid if the viscosity η → 0. From the first point of view the answer must be “yes”
since the ratio of the viscous term to the inertial one in the Navier-Stokes equation is of the order
of magnitude
ν
1
v0 L
η∇2 v
∼
≡
, Re =
(6.51)
ρ (v · ∇) v
Lv0
Re
ν
where v0 is the characteristic velocity of the flow and L is characteristic dimension of the body.
One has then, Re → ∞ when ν → 0, and the viscous term becomes infinitely small with respect to the inertial one. This conclusion is correct everywhere except in a thin boundary layer
adjacent to the surface of the body. We have seen from our simple example (viscous waves) that
the flow velocity varies from zero on the surface of the body to those characteristic of an ideal
fluid flow velocity at some distance from the surface. The lesser the viscosity, the smaller this
distance which we can consider as the thickness of the boundary layer. As a result, the viscous
0
, where h is the thickness of the boundary layer,
term ν∇2 v having the order of magnitude νv
h2
remains finite and comparable with other terms when ν → 0 and h → 0.
From the mathematical point of view the reason for introducing a boundary layer when Re ≫ 1
is that we reduce the order of the Navier-Stokes equation by dropping the viscous term. It leads
to a reduction of the number of boundary conditions we need for a solution. As a result, we cannot meet the zero condition for the tangential velocity on the body’s surface and have to allow
slip on this surface. But this contradicts the experiment, since the fluid adheres to the surface.
This contradiction can be removed by introducing the concept of a boundary layer across which
the velocity varies from 0 to that corresponding to the flow of an ideal fluid. The viscous waves
considered before is the simplest example of a boundary layer. The width of the zone of existence of the viscous wave decreases with decreasing viscosity. Outside of this zone, the fluid
remains at rest as in the case of an ideal fluid. However, the example of viscous waves is a special one also because the nonlinear (inertial) term in the Navier-Stokes equation is zero. To get
more general result we consider another case which (though being idealistic, too) yields results
of more general importance.
Statement of the problem: Let us have a two-dimensional steady flow of incompressible fluid
past a semi-infinite thin plate. Let the plate lie in the half-plane z = 0 occupying the region
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CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.3. BOUNDARY LAYER
−∞ < y < ∞ and x ≥ 0. At a large distance from the plate (x → −∞) the flow is in
the positive x-direction with the constant velocity v0 . Find the thickness of the boundary
layer as a function of the distance from the plate, the tangential viscous force exerted by
the fluid on the plate per unit area, and the drag coefficient.
The problem can now be formulated mathematically as follows:
 ∂v
+ (v · ∇) v = − ∇p
+ ν∇2 v, ∇ · v = 0

∂t
ρ




v|z=0,x≥0 = 0,



v |
= v0 ,

 x x→−∞
vz |x→−∞ = vy |x→−∞ = 0.
(6.52)
The only characteristic scale of the flow velocity variation along the plate (x-direction) is the
distance of the reference point x from the edge of the plate. The characteristic scale in the zdirection is obviously the thickness of the boundary layer itself, h (x) . Since the boundary layer
is thin, it is clear that the flow in it takes place mainly parallel to the surface, i.e. velocity vz is
small compared with vx . The velocity varies rapidly along z-axis, and an appreciable change in
it is occurring at distances of the order of the thickness of the boundary layer. Along the x-axis,
on the other hand, the velocity varies slowly, an appreciable change in it occurring only over
∂p
is small in compardistances of the order of a length l ≫ h (x). It follows that the derivative ∂z
∂p
vz
ison with ∂x (the ratio being of the same order as vx ). In this approximation we can put simply
∂p
= 0, i.e. suppose that there is no transverse pressure gradient in the boundary layer. In other
∂z
words, the pressure in the boundary layer is equal to the pressure p (x) in the main stream and is
a given function of x for the purpose of solving the boundary layer problem.
Solution: Let us assume that the inertial (v · ∇) v and viscous ν∇2 v terms in the Navier-Stokes
equation (6.52) have the same order of magnitude in the boundary layer. Then by comparing
these two terms we can estimate the thickness of the boundary layer h(x). So, we have to define
the order of magnitude of each these terms. First we consider the viscous term.
2
∂ 2v
∂ 2v
v
v
∂ v ∂ 2v
2
⇒
,
.
+
∝
∝
ν∇ v ≡ ν
2
2
2
2
∂x2
∂z 2
∂x
x
∂z
h
| {z } | {z }
along x
along z
Since the flow is mainly along the plate, i.e. v ≈ vx ∇x, v ≈ vx . Remembering that x ≫ h
2
2
yields ∂∂xv2 ≪ ∂∂zv2 . Finally we obtain for the viscous term,
ν∇2 v ∝ ν
νv
∂ 2v
∝ 2.
2
∂z
h
Now we will estimate the inertial term in the boundary layer. Again, since the flow is mainly
along the plate, i.e. v ≈ vx ∇x, v ≈ vx . We have then
∂vx
x
v2
∂v
∂v
vx ∂x + vz ∂v
x
∂z
(v · ∇) v ≡
∝ .
= vx
⇒v
z
z
vx ∂v
+ vz ∂v
∂x vx =v
∂x
x
∂x
∂z
vz =0
Now equating these two estimates and noting that v ≈ vx ∼ v0 , the velocity of unperturbed
flow,we obtain
1/2
1/2
νv0
v02
ν
x
νx
(6.53)
=x
≈
⇒ h(x) ≈
≡√
2
h (x)
x
v0
v0 x
Re
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6.4. SUMMARY OBJECTIVES
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
where Re = v0νx ≫ 1 is the Reynolds number assumed large, otherwise the concept of boundary
layer is inconsistent.
Conclusion: The thickness of the boundary layer is proportional to the square root of the
distance from the plate’s edge. At a given x-location, the higher the Reynolds number, the
thinner the boundary layer.
Now we can estimate the transverse velocity vz in the boundary layer. Taking into account
p v0 ν the
vx
considered approximations, we obtain from the continuity equation vz ∼ x h (x) ≈
. It is
x
important for the quantitative theory of a boundary layer that the transverse velocity vz tends, as
z → ∞, not to zero but to non-zero value. We may now also estimate
the ratio of the transverse
q
vz
vz
ν
and longitudinal velocities in the boundary layer, vx ≈ v0 ≈ xv0 ∼ √1Re , i.e. the ratio of the
√
transverse and longitudinal velocities in the boundary layer is inversely proportional to Re.
The tangential viscous force exerted by the fluid on the plate per unit area or the skin-friction
resistance per unit area is given by (see Equation 6.14)
∂vx
|f | = η
+
∂z
∂vz
∂x
|{z}
v0
νv0
∂vx
∝η
=ρ
=ρ
∝η
∂z
h(x)
h(x)
0, since vz =0
νv03
x
1/2
Correspondingly, we have for the drag coefficient,
1/2
ν
1
2f
=√ .
CD = 2 ∝
ρv0
v0 x
Re
.
(6.54)
(6.55)
Thus, we reach the important results in the qualitative theory of the boundary layer that, when
the Reynolds number is changed, the whole flow pattern in the boundary layer simply undergoes
a similarity transformation, according to which longitudinal distances and velocities remaining
unchanged, while transverse distances and velocities vary as √1Re .
It is worthwhile to note here that the results concerning the flow of a viscous fluid are valid
for laminar regime of the flow. As we increase the Reynolds number, the infinitesimal disturbances in the flow at some point begin to grow, and the flow can not remain laminar. It begins a
transition process toward a turbulent flow.
6.4
Summary objectives
When you finish studying the Chapter 6, you should be able to do the following:
1. Give the qualitative definition of the viscosity as a rate of transport of momentum from
one point in the flow to another. Consider different examples of transport phenomena
and relate the total momentum flux and velocity gradients. Explain the appearance of the
velocity gradient for the flow of a viscous fluids.
2. Derive the Navier-Stokes equation and formulate the boundary conditions used to solve
this equation. Discuss the difference between Navier-Stokes and Euler equation for an
ideal fluid and explain why the boundary conditions to describe the flow of an ideal and
viscous fluid are different.
130
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
6.4. SUMMARY OBJECTIVES
3. Calculate the viscous force.
4. Describe and calculate the Couette flow. Introduce the drag coefficient and calculate it
for the Couette flow as a function of the Reynolds number. Discuss the result from the
principle of hydrodynamic similarity points of view.
5. Describe and calculate the Poiseulle flow for a plane and cylindrical geometry and find the
drag coefficients for these flows.
6. Define the dimensionless number (the Reynolds number) as a rate of the inertial and viscous terms in the Navier-Stokes equation. Relate the magnitude of the Reynolds number
to the definition of laminar and turbulent flows.
7. Consider the flow of a viscous fluid around a sphere in the small Reynolds number approximation and find the Stokes formula for the drag and corresponding drag coefficient.
8. Introduce the concept of the boundary layer and discuss the flow description using this
concept.
9. Apply the boundary layer concept to describe the viscous waves and determine the spatial
scale of the damping of these waves as a thickness of a boundary layer.
10. Apply the boundary layer concept to describe the two-dimensional steady flow of an incompressible viscous fluid past a semi-infinite plate. Determine the thickness of the boundary layer for this flow.
131
6.4. SUMMARY OBJECTIVES
CHAPTER 6. FLOWS OF VISCOUS FLUIDS
132
Chapter 7
Dimensional Analysis and Similarity
We have seen in the previous chapters that rather few real flows can be solved exactly by analytical methods alone, therefore the development of fluid mechanics has depended on experimental
results. Since experimental work in the laboratory is both time-consuming and expensive, one
obvious goal is to obtain the most information from the fewest experiments. Moreover, it is often a situation whereby the variables involved in a physical phenomenon are known, while the
relationship between the variables is not known. An important tool that helps to achieve this
goal is the study of dimensional aspects of fluid flow. To this end we introduce here a powerful technique called dimensional analysis, i.e. analysis based on the dimension of the physical
quantities. The principal concept of dimensional analysis is the fundamental mathematical law
for equations, the law of dimensional homogeneity, stated as
every additive term in an equation must have the same dimensions.
In this chapter we will show that the dimensional analysis may be useful as a compact way to
present solutions of real problems involving usually a combination of analytical approach and
experimental information on the fluid mechanics data.
7.1
Dimensionless Groups
As it was already mentioned in the Chapter 1, in fluid mechanics there are only four primary
dimensions from which all other dimensions can be derived. Thus to give the most simple dimensional representation of a product of quantities, one need only carry out ordinary algebraic
operations on the primarydimensions
appearing in the dimensional representation of the quan
L
tity. For instance, [V t] = /T T = L which is indicating that the product of the velocity and
time most simply dimensionally is a distance with primary dimension of length [L]. If a group
of quantities has dimensional representation most simply of unity when multiplied together, the
group is called a dimensionless group. Arranging the variables in terms of dimensionless products is especially useful in presenting experimental data. As an example, the product ρV D/η is
a dimensionless group, since
ρV D
(M L−3 ) (L/T ) L
=
= M 0 L0 T 0 = 1
η
(M/LT )
Many of these dimensionless products have been given names; the above group being the wellknown Reynolds number. In a later section the physical significance of the Reynolds number as
well as other dimensionless groups will be discussed.
133
7.2. SIMILAR FLOWS
CHAPTER 7. DIMENSIONAL ANALYSIS
The primary purpose of the dimensional analysis is to generate the dimensionless parameters (or
groups) and thus to reduce the number of parameters in the problem. Then, the fluid mechanics phenomenon may be formulated as a relation between a set of dimensionless groups of the
variables, the groups numbering less than the variables. The immediate advantage of this procedure is that considerably less experimentation is required to establish a relationship between
the variables over a given range. Furthermore, the nature of these experimentations will often be
considerably simplified.
7.2
Similar Flows
Before discussing the technique of dimensional analysis, we first explain the underlying concept
of dimensional analysis – the principle of similarity. What do we mean by similar flows? When
experimental testing of a full-size prototype is either impossible or prohibitively expensive, the
only feasible way of attacking the problem is through the model testing in the laboratory. If we
are to predict the prototype behavior from the measurements on the model, it is obvious that
we cannot run just any test on any model. The model flow and prototype flow must be related
by known scaling laws. There are three necessary conditions for complete similarity between
a model flow and prototype one. The first condition is geometric similarity – the model must
be the same shape as the prototype, but may be scaled by some constant factor. The second
condition is kinematic similarity, which means that the velocity at any point in the model flow
must be proportional (by a constant scale factor) to the velocity at the corresponding point in
the prototype flow. Specifically, for kinematic similarity the velocity at corresponding points
must scale in magnitude and must point in the same direction. The third and most restrictive
similarity condition is that of dynamic similarity. The dynamic similarity is achieved when all
forces (body and stress tensor) in the model flow scale by a constant factor to corresponding
forces in the prototype flow (force-scale equivalence). All three similarity conditions must exist
for complete (or hydrodynamic) similarity to be ensured. Thus,
in general flow field, complete similarity between a model and prototype is achieved
only when there is geometric, kinematic, and dynamic similarity.
Then the sufficient condition for a complete similarity is the requirement of the equality of
certain relevant dimensionless parameters (groups). What this dimensionless parameter should
be depends on the nature of the problem.
There are several methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. Dimensionless parameters for a given problem can be determined
in two ways. They can be deduced directly from the governing differential equations if these
equations are known; this method is illustrated in the next section. If, on the other hand, the governing equations are unknown, then the dimensionless parameters (groups) can be determined
by performing a simple dimensional analysis on the variable involved. This method is illustrated
in Section 7.3.
7.2.1
Simple example
To illustrate the method of determining dimensionless parameters (groups) from the governing
differential equations, consider first the approach when the differential equation of motion can
be integrated analytically. As a simplest illustration, we describe the motion of a particle with
mass m in a uniform and constant field, where the field forces have the same magnitude and
134
CHAPTER 7. DIMENSIONAL ANALYSIS
7.2. SIMILAR FLOWS
direction everywhere and are independent of time, for example the Earth’s gravitational field in
regions small compared with its radius. From the equation of motion of a particle mdv/dt = F
we have when F is constant
F
v = t + v0
m
where v 0 is the initial velocity of the particle. Thus, in a uniform and constant field the velocity
is a linear function of time. The obtained expression shows that the particle moves in the plane
defined by the force vector F and the initial velocity vector v 0 . Let us take this as the xy plane,
and the y axis in the direction of the force F . Then the equation for the velocity v of the particle
gives two equations for the velocity components vx and vy . Assuming now that the initial value
of the velocity component in the x direction is equal zero, vx0 = 0, yelds
vy =
F
t + v0y
m
Since the velocity components are the time derivatives of the corresponding coordinates of the
particle, we can write
F
dx
dy
= t + voy ,
=0
dt
m
dt
Hence,
1F 2
y = y0 + v0y t +
t , x = x0 = const
2m
where x0 and
y0are the initial values of the coordinates of the particle. Taking now x0 = 0 and
replacing F/m by the gravitational acceleration, g, in the negative y-direction, yields
1
y = y0 + v0y t − gt2
2
The result expresses the elevation y (t) for the particle moving in the gravitational field with
two given initial conditions and is written in the dimensional form. To write this equation in
the dimensionless form we need to select scaling parameters, based on the primary dimensions
contained in the original equation. In the case discussed here, there are only two primary dimensions, length and time, and thus we are limited to select only two scaling parameters. We have
some options in the selection of the scaling parameters since we have three available dimensional
constants g, y0 and v0y . We choose y0 and v0y . With these two chosen scaling parameters we
form the dimensionless variables y ′ and t′ given by
y′ =
y ′ v0y t
, t =
y0
y0
Then the expression for the elevation as a function of the time in the dimensionless form is
transforming to
1 ′2
y ′ = 1 + t′ −
t
2F r2
Here we introduce the dimensionless group (parameter) called the Froude number,
v0y
Fr = √
gy0
The Froude number as a dimensionless group (parameter) that represents the ratio of the inertial
force to gravitational force. It is evident that the solution of the equations in the dimensionless
135
7.2. SIMILAR FLOWS
CHAPTER 7. DIMENSIONAL ANALYSIS
form reduces the number of parameters in the problem. Indeed, the original problem of the motion of the particle in the gravitational field contains one dependent variable, y, one independent
variable, t, and three additional dimensional constants, g, v0y and y0 . The solution of the same
problem in the dimensionless form contains only one dependent parameter, y ′ , one independent
parameter, t′ , and only one additional parameter, namely the dimensionless Froude number, Fr.
So that, the number of additional parameters has been reduced from three to one.
We note also that the advantages to transform the original equation into the dimensionless form
are not so clear in this simple example because we were able to analytically integrate the differential equation of motion. In more complicated problems, the differential equation (or more
generally the coupled set of differential equations) cannot be integrated analytically and one must
either integrate equations numerically, or conduct physical experiments to obtain the needed results. In this case, as we will see later, the transforming of the equations into the dimensionless
form can considerably simplify the analysis.
7.2.2
Similarity Principle of Reynolds
To illustrate the method of determining dimensionless parameters (groups) from the governing
differential equations which cannot be integrated analytically, consider a flow of an incompressible viscous fluid past some body. Mathematically the problem is formulated as follows:

 ∂v
+ ρ (v · ∇) v = −∇p + η∇2 v
ρ
(7.1)
∂t
 ∇·v =0
with boundary conditions:
(
at the body surface S: v|S = 0
at infinity: v|r→∞ = v0 x̂
(7.2)
here r2 = x2 + y 2 + z 2 . The surface of the body S can always be specified by some equation
x y z f
=0
, ,
D D D
where D is the characteristic scale of the body. In the case of circular cylinder, for example, f =
(x2 + y 2 ) /D2 − 1/4. The solution of the problem (7.1) and (7.2) depends besides coordinates
and time on four dimensional parameters η, ρ, v0 and D. It would be impossible to perform an
experiment for each of the possible combinations of these quantities because the number of these
combinations is enormously large. However, for a homogeneous fluid (ρ = const) the number of
dimensional parameters can be reduced to three if we use the equation for vorticity ω = ∇ × v
instead of the Navier-Stokes equation. Indeed, applying the curl-operator to the Navier-Stokes
equation (7.1) and taken into account the vector identity (v · ∇) v = ∇ (v 2 /2) − v × ω, we
obtain
2
v
∂
ρ (∇ × v) + ∇ × ∇ − v × ω = −∇ × ∇p + η∇2 (∇ × v)
∂t
2
∂ω
+ ∇ × (ω × v) = ν∇2 ω
(7.3)
∂t
Now our problem includes only three dimensional parameters D, v0 and ν. To further reduce the
number of parameters, it is convenient to write equation (7.3) in the dimensionless form. To this
⇒
136
CHAPTER 7. DIMENSIONAL ANALYSIS
7.2. SIMILAR FLOWS
end we need to select the scaling parameters, based on the primary dimensions contained the
original equation. In the fluid flow problems we have usually some options for the selection of
these scaling parameters. For example, for a flow of an incompressible viscous fluid past some
body the scaling parameters can be the length (the characteristic scale of the body) and velocity
of the developed flow far away from the body. Then the time scaling (the time needed for the
flow to pass a body) is defined as the ratio of these characteristic parameters. Denoting the length
and velocity scaling parameters by D and v0 , we introduce the dimensionless quantities:
1. Velocity v ′ =
v
v0
(v0 )
2. Coordinates x′ = Dx , y ′ =
v0 t
′
3. Time t = D vD0
y
,
D
z′ =
z
D
(D)
In other words, we measure the length, velocity and time in terms of scaling parameters, namely,
D, v0 and D/v0 respectively. The boundary conditions will then be given by
(
v ′ |f (x′ ,y′ ,z′ ) = 0
(7.4)
v ′ |r′ →∞ = x̂′
here ∇′ ≡ D∇ and x′ ≡ r/D. The boundary conditions (7.4) do not contain any dimensional
parameter. We now rewrite equation (7.3) for the vorticity. Since ∂/∂x = D−1 ∂/∂x′ and analogously for the y and z components, we have ω = ∇ × v = v0 D−1 ∇′ × v ′ = v0 D−1 ω ′ . Further
we have ∂/∂t = v0 D−1 ∂/∂t′ and equation (7.3) becomes
v02 ∂ω ′
1 ′ v0 ′
ν ′2 v0 ′ ′
+
∇
=
∇
×
ω
×
v
v
ω
0
D2 ∂t′
D
D
D2
D
Dividing this equation by v02 /D2 and introducing the Reynolds number Re =
obtain
 ∂ω′
1
∇′2 ω ′
 ∂t′ + ∇′ × (ω ′ × v ′ ) = Re

′
v0 D
,
ν
we finally
(7.5)
′
∇ ·v =0
So, the problem of flow past a body is completely defined by these equations, together with the
boundary conditions (7.4), and includes only one parameter, the Reynolds number, Re.
It means that for a fixed Re, all flows past geometrically similar bodies are
similar, i.e. described by the same function of dimensionless variables.
In other words, the fluid velocity measured in terms of v0 depends only on space coordinates and
time measured in terms of D and D/v0 , respectively, and on the Reynolds number Re is given
by
x y z v0 t
, , ,
, Re
v = v0 G
(7.6)
D D D D
where the function G is the same for all geometrically similar bodies. This important result is
called the similarity principle of Reynolds. This principle has wide practical applications, for
example, in model experiments in wind tunnels. In these experiments a scale model is tested to
simulate a prototype flow. To ensure complete similarity between the model and prototype, the
model and prototype must be geometrically and kinematically similar. The dynamic similarity
137
7.2. SIMILAR FLOWS
CHAPTER 7. DIMENSIONAL ANALYSIS
in this problem is provided by the Reynolds number. Thus, the Reynolds number of the model
flow must be identical to that of the prototype one. Using results of these experiments and the
similarity principle we can obtain parameters of flow past real objects.
It is worthwhile to note here that since the dynamic similarity is achieved when all forces in
the model flow scale by a constant factor to corresponding forces in the prototype flow, we can
try to obtain this scale factor by forming the dimensionless combination (group) of dimensional
parameters characterised these forces. For the flow under consideration, the inertia and viscous
forces depends on the four dimensional quantities η, ρ, D and v0 . We will try to look for the
combination (group) of these parameters in the dimensionless form
Re = Dm v0n η k ρl
(7.7)
We can always set m = 1 without loss of generality. Now substituting the dimensions of all
parameters into Eq.(7.7) and taking into account that summed powers of M, L, and T must be
zero, yields
n
k
l
M
Re = L TLn LM
k T k L3l ⇒



k+l =0
k = −l
k = −1
M
 L  ⇒  1 + n − k − 3l = 0 1 + l + l − 3l = 0 l = 1  ⇒
n+k =0
n = −k
n=1
T

Re =
Dv0 ρ
η
≡
v0 D
υ
Thus, we obtain that the dimensionless combination (group) of 4 dimensional parameters describing the flow under consideration is the Reynolds number found earlier.
7.2.3
Important Dimensionless Groups in Fluid Mechanics
In other more complicated cases, the flow can be characterised by more than one dimensionless
parameter. As an example, consider the unsteady flow of incompressible viscous fluid passed
a body in which both pressure gradient and gravity are important. It means that now the flow
characteristics depend, besides the four dimension parameters η, ρ, v0 mentioned above, also on
pressure gradient, ∇p, and gravitational force, g. Then the statement of the problem is given
mathematically as follows:
∇p
∂v
+ (v · ∇) v = g −
+ ν∇2 v
∂t
ρ
∇·v =0
To reduce a number of dimensional parameters of the problem we write these equations in the dimensionless form assuming, as in the previous section, that the length and velocity are measured
in terms of scaling parameters D, v0 respectively, namely,
x = Dx′ , y = Dy ′ , z = Dz ′ , v = v0 v ′ , p = P p′ ≡ ρv02 p′
where P is the characteristic pressure or pressure difference between two points. We note that no
separate pressure scale has been introduced, because in this problem the scale for the pressure
(namely ρv02 ) is decided by that for velocity, essentially according to Bernoulli’s equation p +
138
CHAPTER 7. DIMENSIONAL ANALYSIS
7.2. SIMILAR FLOWS
ρv02 /2 = const, which says that a velocity change of v0 causes a pressure change of order ρv02 /2.
Then the Navier-Stokes and continuity equation becomes
∂v ′ v02 ′
v2
νv0
2
+ (v · ∇′ ) v ′ = g − 0 ∇′ p′ + 2 (∇′ ) v ′
∂t
D
D
D
v0 ′ ′
∇ ·v =0
D
2
Dividing now the Navier-Stokes equation by v0/D, the continuity equation by v0/D and introducing the dimensionless time through the relation t′ = v0 t/D , we finally obtain
v0
∂v ′
gD
ν
2
+ (v ′ · ∇′ ) v ′ = 2 − ∇′ p′ +
(∇′ ) v ′
′
∂t
v0
v0 D
∇′ · v ′ = 0
where g (0, 0, g). These equations together with boundary conditions completely define the problem. Thus, it is apparent that two flows, a model and a prototype, (having different values of
v0 , D or ν) will obey the same dimensionless differential equations if the values of dimensionless groups gD/v 2 and ν/v0 D are identical. We note that the number of parameters in dimen0
sionless equations is less than the number of parameters in the original equations. Since the
dimensionless boundary conditions are also identical in the two flows, it follows that√they will
have the same dimensionless solutions. The dimensionless parameters v0 D/ν and v0 / gD have
been given special names:
Re ≡
inertia force
v0 D
= Reynolds number ≡
ν
friction force
inertia force
v0
= Froude number ≡
Fr ≡ √
gravity force
gD
The dynamic similarity between a model and a prototype flows requires both Re and Fr numbers
have to be equal for two flows in which both viscous and gravitational effects are important.
In the high-speed flow of a gas there are significant changes in pressure, density and temperature
which must be related by an equation of state such as the perfect gas law. These thermodynamic
changes introduce one more dimensional parameter, the sound velocity c. The new dimensionless group widely used to describe the compressible fluid flow is the Mach number,
v0
= Mach number
c
flow speed
root of inertia force
≡
≡
sound speed
root of force due to compressibility
M (M a) ≡
If the heat transfer in the fluid must be taken into account, then the heat transfer equation with
thermal conductivity κ must be incorporated into the set of equations. The dimension of κ is the
same as that of ν. Hence the Prandtl number can be chosen as a new dimensionless group
Pr ≡
ν
viscous diffusion
= Prandtl number ≡
κ
thermal diffusion
Consider now a fluid in the rotating frame of reference with angular velocity vector Ω. Let a fluid
will be of arbitrary depth and of uniform density. The gravity g = −gz and the rotation Ω = Ωz
139
7.2. SIMILAR FLOWS
CHAPTER 7. DIMENSIONAL ANALYSIS
are constant and antiparallel. Let v0 be a characteristic velocity of the fluid (with respect to the
rotating frame) and D is a length characterizing the fluid motion. Then v0 /D is a characteristic
frequency of this motion. We note that it is also a characteristic of the vorticity of this “relative
motion”, while 2Ω is the vorticity of the solid-body rotation. The dimensionless parameter which
is measuring the relative importance of inertial forces to the forces due to rotation (Coriolis force)
is known as the Rossby number
Ro ≡
v0 /D
v0
inertia force
=
= Rossby number ≡
2Ω
2ΩD
Coriolis force
When Ro ≫ 1, rotation effects are negligible and the fluid flow will be fully turbulent if the
Reynolds number is high enough. On the other hand, when Ro ≪ 1, rotation effects will be
dominant. The Rossby number is particularly important for the atmospheric and the oceanic fluid
motions in so-called quasi two-dimensional turbulence, for example in geostrophic turbulence,
resulting from a strong rotation and stratification. So, for flows in which the compressibility,
thermal diffusion and rotation are important, the prototype and model flows will be dynamically
similar if the Mach, Prandtl, Rossby –numbers are equivalent for both flows.
In the foregoing analysis we have assumed that the imposed boundary conditions are steady.
Consider now the situation when these conditions are unsteady. To be specific, consider an
object having a characteristic length scale D oscillating with a frequency ω = 2πf in a fluid at
rest in infinity. This is the problem having an imposed length scale and imposed time scale, 1/f .
The boundary condition then specifies
v = v0 sin (ω t)
Writing this relation in the dimensionless form results in
ωD ′
v
′
= v = sin
t
v0
v0
The argument of the sinus contains the new dimensionless group ωD
known as the Strouhal
v0
number
fD
characteristic flow time
St (S) ≡
= Strouhal number ≡
v0
period of oscillations
We note that some flows, which one might guess to be steady, actually have an oscillatory pattern
which depends on the Reynolds number. For example, the periodic vortex shedding formed
behind a circular cylinder immersed into a steady flow. This regular, periodic shedding is called
a Karman vortex street. The shedding
occurs in the range 102 < Re < 107 with an average
Strouhal number St = ω D/2πv0 ≈ 0.21. So, the dimensionless forces of such an oscillating
flow will be a function of both Reynolds and Strouhal numbers. Therefore the dynamic similarity
between prototype and model oscillating flows will be achieved if these two numbers will be the
same for both flows.
We have discussed several dimensionless parameters (groups) important in fluid mechanics and
there are others. For example, additional parameters arise from dimensionless representation of
the energy equation and its boundary conditions (Eckert number, Grashof number, etc.) This
compendium is primary concerned with Reynolds-, Euler-, Froude-, and Mach- number effects
which dominate most flows.
Under dynamic similarity between the model and prototype flows, the dimensionless solutions
for the prototype and model flows are identical. Therefore the local pressure at point r ≡
140
CHAPTER 7. DIMENSIONAL ANALYSIS
7.2. SIMILAR FLOWS
(x, y, z) must be of the form
r
∆p (r)
=
f
Fr,
Re;
ρv02 /2
D
Here ∆p (r) = p (r) − p∞ is the pressure difference, the local pressure p (r) minus the free
stream pressure, p∞ ; the dimensionless group E = 2∆p/ρv 2 is the ratio of pressure forces to
0
inertia forces and is known as the Euler number (The factor 21 is introduced into the denominator
to give the dynamic pressure). The Euler number is often called the pressure coefficient, K,
introduced earlier. Similar relations also hold for any other dimensionless flow variable such as
velocity v/v0 and acceleration aD/v02 . It follows that in similar flows the dimensionless local
flow variables are identical at corresponding points, i.e. the equality of dimensionless variables
holds locally for identical values of Dr .
In many flow problems the requirement of the locality is not important, i.e. the relations between
overall quantities (scales of the flow) are of the interest only. Consider, for example the pressure
drop, ∆p, for steady flow of viscous incompressible fluid through a straight horizontal pipe. The
pressure coefficient in this case will depend also on the pipe length, l, the gravity effects are not
important, and we finally have
∆p
K≡ 2 =f
ρv0 /2
l
Re,
D
All dimensionless quantities are identical for similar flows. For example, consider a flow around
an immersed body. We can define the dimensionless drag coefficient as follows
CD =
F
1
ρv02 D2
2
≡
F
1
ρv02 A
2
where F is the drag experienced by the whole body, A is the characteristic area of the body; for
blunt bodies such as spheres and cylinders, A is taken to be a cross section perpendicular to the
flow. Therefore A = πD2 /4 for a sphere of diameter D, and A = bD for a cylinder of diameter
D and length b, the axis of the cylinder being perpendicular to the flow. For flow over the flat
plate, A is taken to be the “wetted area”, that is A = bl, where l is the length of the plate in the
direction of the flow and b is the width perpendicular to the flow.
The values of drag coefficient CD are identical for dynamically similar flows. If the drag is
caused both by gravitational and viscous effects, we must have the functional relation for the
drag coefficient of the form
CD = f (F r, Re)
For many flows the gravitational effects are not important, then the F r number is irrelevant, and
we have
CD = f (Re)
We note that this result coincides with that obtained in the previous chapter when one is considering the flow of viscous fluid around blunt bodies. The form of the function still must be
determined experimentally.
141
7.3. REPEATING VARIABLES
7.3
CHAPTER 7. DIMENSIONAL ANALYSIS
Method of Repeating Variables and Pi-theorem
To learn how to generate dimensionless parameters if the governing differential equations are
unknown we use the method of repeating variables. According to this method, we let uppercase
Greek letter Pi (Π) denote a dimensionless parameter. You are familiar with several examples of
Π′ s introduced above, for example, Reynolds number, Froude number, Mach number, etc. In a
general dimension analysis problem, there is one Π that we call the dependent Π giving it the
notation Π1 . The parameter (dimensionless group) Π1 is in general a function of several other
Π’s, which we call independent Π’s. The functional relationship is then given by
Π1 = f (Π2 , Π3 , Π4 , ..., Πk )
where k is the total number of Π’s.
Arranging the variables in terms of dimensionless groups is especially useful in presenting experimental data. Consider an experiment in which a scale model is tested to simulate a prototype
flow. To ensure complete similarity between the model and the prototype, each independent Π of
the model (subscript m) must be identical to the corresponding independent Π of the prototype
(subscript p), i.e. Π2m = Π2p , Π3m = Π3p , ..., Πkm = Πkp . Under these conditions the dependent Π of the model (Π1m ) is guaranteed also to equal the dependent Π of the prototype (Π1p ).
Mathematically, we write the conditional statement for achieving similarity as follows
if Π2m = Π2p , Π3m = Π3p , . . . , Πkm = Πkp , then Π1m =Π1p
This statement is known as the Π-theorem by Riaboushinsky and Buckingham (1911).
The Pi- theorem is a formal method of determining dimensionless groups. We can think of this
method as a step-by-step procedure or “recipe” for obtaining dimensionless parameters. There
are seven steps listed below.
The method of repeating variables:
• Step 1: Write the relation you are looking for, list the parameters of the problem and count
their total number, n.
• Step 2: List the primary dimensions of each of the n parameters.
• Step 3: List and indicate the number of primary dimensions represented in the problem, j
(j is also called the reduction); calculate k = n − j, the expected number of Π’s.
• Step 4: Choose j repeating variables that will be used to construct each Π.
• Step 5: Construct the dependent Π denoted as Π1 .
• Step 6: Construct the independent Π’s; remember that the total number of Π’s is equal k.
• Step 7: Write the final functional relationship between dimensionless groups.
These steps that comprise the method of repeating variables will be explained in further detail
as we work through a number of example problems. We note that the Step 4 is considered often
as the hardest (or at least most mysterious) part of the method of repeating variables. There are
several recommendations for choosing repeating parameters:
1. The chosen repeating parameters must represent all the primary dimensions in the problem.
142
CHAPTER 7. DIMENSIONAL ANALYSIS
7.3. REPEATING VARIABLES
2. It is better to pick parameters with only one or two primary dimensions instead of parameters that composed of several primary dimensions. Pick simple parameters over complex
parameters whenever possible.
3. In fluid flow problems we generally pick a length, a velocity and a mass or density. It is
unwise to pick less common parameters like viscosity η or surface tension σs , since we
would in general not want η or σs to appear in each of the Π’s. Moreover, experience suggests that viscosity should appear in only one dimensionless parameter (group). Therefore
η should not be chosen as a repeating variable.
A wise choice of repeating parameters for most fluid flow problems
is a length, a velocity, and a mass or density .
7.3.1
Complete Dynamic Similarity. Simple Example
To illustrate the technique of the method of repeating variables, we deduce the similarity principle (7.6) obtained above under the condition that the differential equations describing the flow
are known. To this end we consider the flow of an incompressible fluid past a body of an arbitrary
and smooth shape. The solution of this problem, as it was already mentioned, depends, besides
coordinates and time, on four dimensional parameters:
[η]
[ρ]
[D]
[v0 ]
=
=
=
=
M L−1 T −1
M L−3
L
LT −1
where [D] is the dimension of D, etc.; M, L and T are primary dimensions corresponding to
the dimension of mass, length and time respectively. The surface of the body S at which the
boundary conditions must be satisfied can be completely determined by D and some dimensionless function which specifies the shape of the body. So, we are looking for the expression
of a dimensionless flow velocity in terms of dimensionless coordinates, time and dimensionless
combination of parameters above. It is clear that
1. The velocity must be independent of the units which we are using for measuring L, T, M, ρ, η,
etc.
2. This independence implies that only dimensionless combinations
of those 4 dimensional
r
v
parameters may arise in the required expression v0 = f D , etc .
Thus, the statement of the problem can now be formulated as follows: we have to find the
dimensionless relationship between the flow velocity v and other dimensional parameters in the
problem.
Solution. The step-by-step method of repeating variables is employed.
• Step 1: Write the relation you are looking for in the functional form, with the dependent
variable listed as a function of independent variables and list number of relevant parameters:
v = f (r, t, v0 , ρ, η, D)
143
7.3. REPEATING VARIABLES
CHAPTER 7. DIMENSIONAL ANALYSIS
List of relevant parameters: v is the flow velocity, r is the space coordinate, t is time
variable, v0 is the characteristic flow velocity at infinity, ρ is the fluid density, η is the
fluid viscosity, and D is the characteristic scale of a body; the total number of relevant
parameters is n = 7.
• Step 2: List the primary dimensions of each parameter
[v]
LT −1
[r]
L
[t]
T
[v0 ]
LT −1
[ρ]
M L−3
[η]
M L−1 T −1
[D]
L
• Step 3: List and indicate the number of primary dimensions: the primary dimensions
represented in the problem are L (length), T (time), and M (mass), the number of primary
dimensions is thus j = 3 (j is also called the reduction). Then, the expected number of Π′ s
in the problem is equal to n minus j or, according to the Π- theorem, k = n−j = 7−3 = 4.
So, the number of Π′ s predicted by the Π- theorem is k = 4.
• Step 4: Choose j repeating variables (parameters) that will be used to construct each
Π. Since in the present problem j = 3, we are choosing three repeating parameters
v0 , D, and ρ which will appear in each of the Π′ s.
• Step 5: We combine these repeating parameters into a product with the dependent variable
v to create the dependent Π:
Dependent Π: Π1 = v v0α1 Dβ1 ργ1
| {z }
repeating
variables
We apply now the primary dimensions of step 2 into this equation and force Π1 to be
dimensionless:
[Π1 ] = vv0α1 Dβ1 ργ1 = LT −1 Lα1 T −α1 Lβ1 M γ1 L−3γ1 = L0 T 0 M 0
|
{z
}
equate exponents of each primary
dimension to solve for α1 ,β1 ,γ1
This yields the following expression for the dependent Π:




1 + α1 + β1 − 3γ1 = 0 β1 = 0
L
v
 T  ⇒  1 + α1 = 0
α1 = −1  ⇒ Π1 =
v0
γ1 = 0
γ1 = 0
M
Since the number of the Π′ s in the problem is k = 4, we can write now the functional
relationship between the Π′ s in the form
Π1 = f (Π2 , Π3 , Π4 ) ⇔
v
= f (Π2 , Π3 , Π4 )
v0
• Step 6: To determine the independent Π (Π2 ) we combine the repeating parameters into a
product with one of the independent variables, for example r,
Π2 = r v0α2 Dβ2 ργ2
| {z }
repeating
variables
144
CHAPTER 7. DIMENSIONAL ANALYSIS
7.3. REPEATING VARIABLES
Using the primary dimensions of step 2 into this equation and force Π2 to be dimensionless
yields
[Π2 ] = rv0α2 Dβ2 ργ2 = LLα2 T −α2 Lβ2 M γ2 L−3γ2 = L0 T 0 M 0
|
{z
}
equate exponents of each primary
dimension to solve for α2 ,β2 ,γ2
This results in the expression for Π2



1 + α2 + β2 − 3γ2 = 0
L
 T  ⇒  α2 = 0
γ2 = 0
M

β2 = −1
r
α 2 = 0  ⇒ Π2 =
D
γ2 = 0
Similarly, combining the repeating parameters into a product with independent variables
t and η we obtain the independent Π3 and Π4 respectively. Again, using the primary dimensions of step 2 into these Π′ s and forcing Π3 and Π4 to be dimensionless yields
η/
v0 t
η
ν
1
ρ
, Π4 =
=
=
≡
Π3 =
D
v0 Dρ
v0 D
v0 D
Re
• Step 7: Write the final functional relationship between Π′ s.
So, the functional dimensionless form of the relationship between the flow velocity v and other
dimensional parameters in the problem is given by
r v0 t
v
=f
,
, Re
Π1 = f (Π2 , Π3 , ...Πk ) ⇒
v0
D D
which is the similarity principle of Reynolds defined earlier.
The formation of dimensionless parameters (groups) if the governing differential equations describing the flow are unknown is illustrated in the following two sections.
7.3.2
Drag Force on Smooth Sphere
As with other new procedures, the best way to learn is by example and practice. As a simple first
example consider the drag force on a stationary smooth sphere immersed in a uniform stream.
What experiments must be conducted to determine the drag force on the sphere? To answer
this question, we must specify the parameters that are important in determining the drag force.
Clearly, we would expect the drag force to depend on the size of the sphere (characterised by the
diameter D), the fluid velocity, v0 , and the fluid viscosity η. In addition, the density of the fluid,
ρ, also might be important. So that, the statement of the problem can be formulated as follows:
assuming that the drag force, F , on a smooth sphere depends on the relative velocity, V , the
sphere diameter, D, the fluid density, ρ, and the fluid viscosity, η, obtain a set of dimensionless
groups that can be used to correlate experimental data and determine the drag force.
Solution. The step-by-step method of repeating variables is employed.
• Step 1: Write the relation you are looking for in the functional form
F = (V, D, ρ, η)
List the parameters of the problem and count them: F, V, D, ρ, η; n = 5.
145
7.3. REPEATING VARIABLES
CHAPTER 7. DIMENSIONAL ANALYSIS
• Step 2: List the primary dimensions of each parameter
[F ]
M LT −2
[V ]
LT −1
[ρ]
M L−3
[η]
M L−1 T −1
[D]
L
• Step 3: List and indicate the number of primary dimensions: the primary dimensions
representing in the problem are L (length), T (time), and M (mass), the number of primary
dimensions is thus j = 3; according to the Pi-theorem, k = n − j = 5 − 3 = 2. So, the
number of dimensionless groups predicted by the Pi-theorem is k = 2, and the relationship
between the Π’s is Π1 = f (Π2 )
• Step 4: Choose j repeating variables that will be used to construct each Π. Since in the
present problem j = 3, we are choosing three repeating variables V, D and ρ which will
appear in each of the Π’s.
• Step 5: Construct the dependent Π. We combine these repeating variables into the product
with the dependent variable F to create the dependent Π.
Dependent Π: Π1 = F · V α1 Dβ1 ργ1
| {z }
repeating
variables
[Π1 ] = F V α1 Dβ1 ργ1 =
0 0 0
−2 α1 −α1 β1
γ1 −3γ1
|M LT L T {z L M L } = L T M
equate exponents of each primary
dimension to solve for α1 , β1 , γ1
This yields the following expression for the dependent Π:




1 + α1 + β1 − 3γ1 = 0 β1 = −2
L
 ⇒ Π1 = F
 T  ⇒  2 + α1 = 0 α1 = −2
ρV 2 D2
1 + γ1 = 0 γ1 = −1
M
• Step 6: Construct the independent Π. To this end we combine the repeating variables with
independent ones. The only independent variable which left is the fluid viscosity η.
Dependent Π: Π2 = η V α2 Dβ2 ργ2
| {z }
repeating
variables
−α2 β2
L M γ2 L−3γ}2 = L0 T 0 M 0
[Π2 ] = ηV α2 Dβ2 ργ2 = |M L−1 T −1 Lα2 T{z
equate exponents of each primary
dimension to solve for α2 , β2 , γ2
This results for the expression for Π2




−1 + α2 + β2 − 3γ2 = 0 β2 = −1
L
 ⇒ Π2 = η = 1
 T  ⇒  −1 − α2 = 0 α2 = −1
V Dρ
Re
1 + γ2 = 0 γ2 = 0
M
146
CHAPTER 7. DIMENSIONAL ANALYSIS
7.3. REPEATING VARIABLES
• Step 7: Write the final functional relationship between Π’s: Π1 = f (Π2 ), or
F
η
=f
ρV 2 D2
V Dρ
The form of the function, f , must be determined experimentally. Thus, all data for the problem of
determining the drag force for a stationary sphere in terms of quantities that are controllable and
measurable in the laboratory can be plotted as a functional relation between two dimensionless
groups in the form Π1 = f (Π2 ). No longer must we find fluids with a number (10, for example)
of different values of density and viscosity. Nor must we take 10 spheres with different diameters.
Instead, only the ratio, η/V Dρ must be varied. This can be accomplished simply by changing
the velocity, for example.
7.3.3
Pressure Drop in Pipe Flow
The main purpose in this section is to evaluate the pressure changes that result from incompressible flow in pipes. The pressure changes in a flow system result from changes in elevation or flow
velocity (due to area changes) and from friction. In a frictionless flow, the Bernoulli equation
could be used to account for the effects of changes in elevation and flow velocity. Thus the prime
concern in the analysis of real flow is to account for friction. The effect of friction is to decrease
the pressure, causing the pressure “loss” compared with an ideal flow due to the fluid viscosity
and so-called pipe roughness. To simplify analysis, we assume that this “loss” will be mainly
due to fluid viscosity and neglect other frictional effects such as pipe roughness, pressure drop
that occurs at the entrance of a pipe, etc. The results can be easily extended to take into account
these effects by introducing additional parameters describing the flow. Moreover, we consider a
circular cross-section of a pipe and deal with fully developed flows in which the velocity profile
is unvarying in the direction of the flow. Our attention will focus on turbulent flows as most
common in engineering applications. So, the statement of the problem is as follows:
in fully developed turbulent flow, the pressure drop, ∆p, due to friction in a horizontal constantarea pipe is known to depend on the pipe diameter, D, pipe length, l, average flow velocity, V ,
fluid density, ρ, and fluid viscosity, η. Determine a set of dimensionless groups that can be used
to correlate the experimental data and to determine the pressure drop.
Solution. The step-by-step method of repeating variables is employed.
• Step 1: We are looking for the relation
∆p = f (D, l, V, ρ, η)
List the parameters of the problem and count them: ∆p, D, l, V, ρ, η; n = 6.
• Step 2: List the primary dimensions of all n parameters
[∆p]
M L−1 T −2
[l]
L
[V ]
LT −1
[ρ]
M L−3
[η]
M L−1 T −1
[D]
L
• Step 3: List and indicate the number of primary dimensions: the primary dimensions
representing in the problem are L (length), T (time), and M (mass), the number of primary
dimensions is thus j = 3; according to the Pi-theorem, k = n − j = 6 − 3 = 3. So, the
number of dimensionless groups predicted by the Pi-theorem is k = 3 and therefore the
relationship between Π’s we are looking for is as follows: Π1 = f (Π2 , Π3 ).
147
7.3. REPEATING VARIABLES
CHAPTER 7. DIMENSIONAL ANALYSIS
• Step 4: Choose j repeating variables that will be used to construct each Π. Since in the
present problem j = 3, we are choosing three repeating variables V, D and ρ which will
appear in each of the Π’s.
• Step 5: Construct the dependent Π. We combine these repeating variables into the product
with the dependent variable ∆p to create the dependent Π.
Dependent Π: Π1 = ∆p · V α1 Dβ1 ργ1
| {z }
repeating
variables
−α1 β1
L M γ1 L−3γ}1 = L0 T 0 M 0
[Π1 ] = ∆pV α1 Dβ1 ργ1 = |M L−1 T −2 Lα1 T{z
equate exponents of each primary
dimension to solve for α1 , β1 , γ1
This yields the following expression for the dependent Π:




−1 + α1 + β1 − 3γ1 = 0 β1 = 0
L
 ⇒ Π1 = ∆p
 T  ⇒  2 + α1 = 0 α1 = −2
ρV 2
1 + γ1 = 0 γ1 = −1
M
• Step 6: Construct the independent Π. To this end we combine the repeating variables with
independent ones. We start with the fluid viscosity η.
Dependent Π: Π2 = η V α2 Dβ2 ργ2
| {z }
repeating
variables
−α2 β2
L M γ2 L−3γ}2 = L0 T 0 M 0
[Π2 ] = ηV α2 Dβ2 ργ2 = |M L−1 T −1 Lα2 T{z
equate exponents of each primary
dimension to solve for α2 , β2 , γ2
This results for the expression for Π2




−1 + α2 + β2 − 3γ2 = 0 β2 = −1
L
 T  ⇒  −1 − α2 = 0 α2 = −1
 ⇒ Π2 = η ≡ 1
V Dρ
Re
1 + γ2 = 0 γ2 = 0
M
Combining now the repeating variables into a product with independent variable, l, the
length of a pipe and performing the dimensional analysis, we obtain similarly the independent dimensionless group Π3 = l/D.
• Step 7: Write the final functional relationship between Π’s: Π1 = f (Π2 , Π3 ), or
∆p
l
η
−1 l
≡ f Re ,
=f
,
ρV 2
V Dρ D
D
Laboratory experiments have shown that this relationship correlate the data well. Namely, the
pressure drop is directly proportional to the ratio l/D. The dependence on the Reynolds number
is more complicated and description of these factors is beyond of the scope of this compendium.
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CHAPTER 7. DIMENSIONAL ANALYSIS
7.4
7.4. SUMMARY OBJECTIVES
Summary objectives
When you finish studying the Chapter 7, you should be able to do the following:
1. Formulate the law of the dimensional homogeneity.
2. Define the dimensionless group and give a few examples.
3. Formulate the necessary and sufficient conditions for a model and prototype flows to be
similar.
4. Derive the similarity principle of Reynolds.
5. Show that the unsteady flow in which both the pressure gradient and gravity are important is determined completely by two dimensionless groups (Reynolds number and Froude
number). Formulate the condition for the dynamic similarity between a model and prototype flows in this case.
6. What is the dimensionless group is used to describe the compressible fluid flow.
7. Define the Rossby number and Prandtl number. Formulate the condition for the dynamic
similarity between model and prototype flows in which compressibility, thermal diffusion
and rotation are important.
8. Define the Strouhal number and formulate the condition for the dynamic similarity between prototype and model oscillating flows in which the dimensionless forces are function of both Reynolds and Strouhal numbers.
9. Define the dependent and independent dimensionless parameters and write the functional(in
general) dependence between them.
10. Formulate the Π-theorem by Riaboushinsky and Buckingham.
11. List the main steps used in the method of repeating variables and indicate the choice of
repeating parameters for fluid flows.
12. Derive the similarity principle of Reynolds using the method of repeating variables.
13. Determine the drag force on a smooth sphere immersed in a uniform stream by means of
the method of repeating variables.
14. Estimate the pressure drop in a pipe flow using the method of repeating variables.
149
7.4. SUMMARY OBJECTIVES
CHAPTER 7. DIMENSIONAL ANALYSIS
150
Chapter 8
Elements of Turbulence Theory
All the flows considered above were regular or “laminar” ones. Such a terminology is used to
describe a fluid flow in laminae or layers, as opposed to turbulent flow in which the velocity
components have random turbulent fluctuations imposed upon their mean values (see Figure
8.1). A stream of dye or ink inserted into a laminar flow will streak out into a thin line and one
can thus see laminar currents in such flows. Flows are always laminar, according to experiments,
if their velocities are sufficiently small. On the other hand, a laminar flow always turns into
a turbulent one if the velocity increases. Turbulent flow is essentially irregular. Its velocity,
pressure and other parameters vary in a random way at any point, even when the boundary and
all other external conditions are regular. This chapter contains the basic theory of turbulence.
a) Laminar flow
b) Turbulent flow
Figure 8.1: Laminar and turbulent flow. The lines indicate the paths of particles.
8.1
Qualitative Considerations
8.1.1
Transition from a Laminar to Turbulent Flow
A simple example of a transition from laminar to turbulent flow may be seen by observing the
smoke rising from a cigarette or smoke stack (see Figure 8.2). For some distance, the smoke
rises in a smooth laminar manner. Then, rather abruptly, the smoke begins to mix up, become
turbulent, and the smoke column rapidly widens and diffuses. The turbulence aids the diffusion
of the smoke and causes it to diffuse into a widening chaotic stream.
151
8.1. QUALITATIVE CONSIDERATIONS
CHAPTER 8. THEORY OF TURBULENCE
Figure 8.2: Transition from laminar to turbulent flow in the smoke from a cigarette.
A transition from a laminar to a turbulent flow was first observed in 1839 by Hangen in the
flow through circular pipes. Systematic investigation, however, were carried out by Reynolds in
1843. It was shown by him that it is not the velocity itself which is essential for transition but the
dimensionless quantity Re = vD/ν (Reynolds number), v being the characteristic velocity, D
the characteristic space dimension of the flow (for example, the tube’s diameter) and ν the kinematic viscosity. We have seen above that the Reynolds number is the ratio of the inertial term to
the viscous one in the Navier-Stokes equation. It was established by Reynolds that some critical
Recr exists when a transition of laminar to turbulent flow occurs. The laminar-turbulent transition is easily detected by adding a dye into the fluid. Stream lines in a tube are straight when the
flow is laminar. When the transition occurs, however, the stream lines become distorted and the
dye gradually diffuses all over the cross section of the tube. The Reynolds number can be altered
in an experiment by velocity as well as by viscosity. The latter can be achieved by using another
fluid or by heating the same fluid. In the first of such experiments, Reynolds obtained Recr =
12830 when the entrance into the pipe was smooth. The laminar flow could be maintained up
to Recr = 20000 under special precautions taken against distortion of the inward flow (smooth
entrance and smooth pipe walls, no vibrations, etc). It was shown by numerous experiments that
flows are always laminar for Re ≤ 2000. In this case, any distortions in the inward flow decay
away along the flow.
A theoretical description of the laminar-turbulent transition has not been achieved yet. Evidently,
this problem has to be solved on the basis of the hydrodynamic equations. The laminar-turbulent
transition must then be due to the loss of stability of the laminar flow, i.e. due to a dramatic increase of the initially small disturbances which are always present in a real laminar flow. Hence,
exploring the stability of the laminar solution of the hydrodynamic equations, we can find some
critical value of the Reynolds number Re1 when this stability will be lost. According to Landau’s hypothesis, one will have a new stable flow at Re > Re1. This flow will be a combination
of the original laminar flow and one harmonic mode. If the Reynolds number increases further,
then this flow will become unstable in its turn at some Re = Re2 and the second harmonic mode
will appear, etc. According to Landau, this process will continue incessantly with increasing
152
CHAPTER 8. THEORY OF TURBULENCE
8.1. QUALITATIVE CONSIDERATIONS
Re. As a result, a great number of modes will exist at a sufficiently large Re. An ensemble of
these modes with different frequencies behaves as a very irregular system and very closely resembles so-called developed turbulence. Initially Landau’s hypothesis appeared to be verified by
some experiments. The greater the Reynolds number, the more complicated the flow becomes.
However, only a few initial steps of the process could be made in experiments before the flow
suddenly became turbulent.
Recently, new ideas have appeared concerning the laminar-turbulent transition. For its understanding we note that the state of a hydrodynamic system can mathematically be described by
the expansion of its velocity field in terms of an orthogonal set of functions (modes). Then the
coefficients qi of this expansion can be regarded as generalised coordinates of this system and
i
their derivatives with respect to time pi ≡ ∂q
as generalised velocities. In space limited systems,
∂t
the number of generalised coordinates is countable. Moreover, only a finite number N of modes
is important, since very high modes (i is very large) corresponding to small-scale variations of
velocity disappear quickly due to the viscosity. The 2N dimensional coordinate space (qi , pi ) is
called the phase space of the flow. A line in this space describing the change of the state of the
system is called the phase trajectory.
In the case of steady flow, the latter degenerates into a fixed point describing the equilibrium
state of a focus type. For periodical motion it is closed line also called the limit cycle. Only
these two types of features or so called simple attractors in phase space have recently been considered in the hypothesis by Landau, Hopf, and others. The turbulent (random) character of the
flow was related to the existence of large number of modes (degrees of freedom). Recently, however, a new hypothesis was suggested which states that the so-called strange attractors, which
are different from the simple ones, could exist in phase space of hydrodynamic system. The
characteristic example of strange attractors is the Lorenz attractor appearing in the solution of
a model system which includes 3 equations and has two fixed points in the phase space. The
point of the phase space which represents the state of the system (representative point) moves
for the same time in the vicinity of one of the fixed points while the system changes. Then it
follows untwisted cycles and eventually goes over into the vicinity of the other fixed point at a
distance away from the first one. Then the process is repeated over and over in a non-regular
(pseudo random) way. Such strange attractors were also obtained for other model systems, but
their existence in systems described by hydrodynamic equations has not been proved rigorously
yet.
The hypothesis about strange attractors fundamentally changes the notion about the appearance
and development of turbulence. We observe pseudo random behaviour of the system even with
a small number of degrees of freedom in this case. Of course, describing the flows with a large
number of degrees of freedom is required for the turbulence problem. However, such a description is rather far from being achieved. Meanwhile, the practice needs some guidance in treating
turbulent flow. In these circumstances, an experiment together with the similarity theory appears
to be very useful. By means of this theory one can get results for real cases using those obtained
by model experiments.
8.1.2
Flow Around a Cylinder at Different Re
We have seen that if the gravitational force can be neglected, the incompressible ideal fluid flow
is completely specified by the Reynolds number Re. The character of the flow, however, is quite
153
8.1. QUALITATIVE CONSIDERATIONS
CHAPTER 8. THEORY OF TURBULENCE
different at different Re. We consider, as an example, the flow around a circular cylinder with its
axis perpendicular to the flow:
-2
Re∼10
a)
Re∼20
b)
A
0'
0
Figure 8.3: a) Laminar flow around a cylinder for small Re. b) Steady flow past a cylinder with two
vortices
i) At small Reynolds number (Re ≪ 1), laminar flow takes place around the cylinder, which
is very close to that of an ideal fluid (Figure 8.3a).
ii) 1 < Re < 40. The first critical Reynolds number Re1 occurs in the vicinity of Re = 1.
At Re > Re1 , the flow becomes unsteady but the new type of flow is formed only at
Re > 10 when two eddies are formed behind the cylinder (Figure 8.3b). The flow is still
steady and laminar so that at Re = Re1 , we have only a change of one stable flow for
another. The reason for the formation of vortices becomes clear if we observe the change
of pressure and velocity along the surface of the cylinder outside the boundary layer. These
quantities can be found from the solution to the ideal fluid equations. According to (5.24,
5.27), the velocity is zero at the critical points (θ = 0, π) and reaches a maximum at
the mid plane (θ = ±θ/2). In accordance with Bernoulli’s theorem, the pressure has a
maxima at critical points and reaches a minimum at the mid plane. For this reason the fluid
particles downstream of the mid plane move against the increasing pressure which leads to
their retardation. This retardation will have the strongest effect on the fluid particles close
to the surface possessing the smallest velocity. At a point A (Figure 8.3b) downstream
from the mid plane, these particles come to standstill; beyond A they will be moving
backwards with respect to the fluid particles further from the cylinder which have not yet
been retarded. The existence of counterflows near the surface of the body induces the
formation of vortices and also forces the outer flow away from the surface. The latter will
result in the separation of the boundary layer from the surface at point A. The higher Re,
the closer point A is to the mid plane and the grater the drag (the area of high pressure
behind the mid plane is less.)
iii) When Re increases further (Re > 40), the next critical values of this number will be
attained, and the flow becomes unsteady. One of the vortices elongates, then separates
and goes down the stream. After that the other vortices do the same. New vortices appear
154
CHAPTER 8. THEORY OF TURBULENCE
8.2. STATISTICAL DESCRIPTION
at their place and the process is repeated. As a result, the so-called “Karman path” or
“Karman vortex street” is formed in the wake behind the cylinder (Fig 8.4a). The flow is
unsteady but periodical since the process of a separation of vortices is periodical.
iv) At Re > 1000, there will be no time for full formation of these vortices. Instead one can
observe highly turbulent areas rapidly appearing, separating and going down the stream.
At Re ∼ 103 − 104 , turbulence in the wake becomes greater, the flow is now highly
irregular and three dimensional. At Re ∼ 105 , the wake is fully turbulent, beginning just
behind the cylinder (Figure 8.4b).
a)
Re∼200
5
Re∼10
b)
Figure 8.4: a) Illustrating a Karman street. b) The flow with a fully developed turbulent wake.
8.2
Statistical Description of Turbulent Flows
At first sight it is impossible to reveal any regularity in turbulent flow. The flow velocity and
other quantities vary randomly at each point. Therefore it is called an irregular flow. It means
that if the flow is set up repeatedly under the same conditions, exact values of these quantities
will be different each time. However, Reynolds was the first to show that is possible to obtain
reliable relations for average values in turbulent flow considering the latter as a statistical process
and using the formalism of probability theory.
Suppose we repeat the same experiment under the same external conditions many times and
that we measure the fluid velocity at a fixed point during each experiment. The large number
of results of the fluid velocity from these experiments constitutes what is called a statistical ensemble. The result of each experiment represents a separate sample value or realisation in this
ensemble. We denote by hvi i the ensemble average velocity at a fixed point or the velocity at
this point averaged over all the experiments. The actual velocity for a given realization can be
expressed as the sum
vi = hvi i + δvi , hδvi i = 0
(8.1)
where δvi is a random deviation of the velocity from the ensemble average value. Usually, instead
of repeating many experiments one obtains a long sample in one experiment. Figure 8.5 shows
155
8.2. STATISTICAL DESCRIPTION
CHAPTER 8. THEORY OF TURBULENCE
the variation of the instantaneous velocity component v with time at a specified location in a
turbulent flow. We observe that as stated above the instantaneous values of the velocity fluctuate
about the average value and one can express the velocity according to relation 8.1. This is also
the case for other properties such as pressure p = hpi + δp and temperature T = hT i + δT .
Below we will derive Reynolds equations for average flow using the procedure of averaging over
v
δv
<v>
Figure 8.5: Fluctuations of the velocity with time at a specified location in turbulent flow.
an ensemble. Before that, however, we have to note two important items:
1. We will use some rules for averaging, known in probability theory. Most of them are rather
obvious, for example, hv1 + v2 i = hv1 i + hv2 i where the average of the sum is equal to the
sum of averages. Less trivial is the following rule:
∂ hvi
∂v
=
(8.2)
∂s
∂s
where v is one of the components of velocity or pressure or any other quantity (density,
temperature, etc.), and s denotes x1 , x2 , x3 or t. Equation (8.2) expresses the commutation
of the operations of averaging and differentiating.
2. As a rule, we can not carry out a large number of similar experiments under strictly the
same conditions and thus calculate the ensemble average value. Instead, one usually obtains in an experiment only one rather long sample and calculates the time average value,
for example
T
Z2
1
v (t) =
v(t + τ )dτ
(8.3)
T
− T2
To distinguish the two kinds of averaged values, we denote the time average by a bar over
the quantity involved.
An important question regards the connection between these two methods of averaging. Do time
averages converge to the ensemble averages when the averaging interval T becomes infinitely
long limT →∞ v (t) = hvi? If this requirement is met, such a statistical process is called an
156
CHAPTER 8. THEORY OF TURBULENCE
8.2. STATISTICAL DESCRIPTION
ergodic one. It is clear that one of the necessary conditions of ergodicity of the process must be
the independence v(t) of time t, otherwise limT →∞ v (t) does not exist. This requirement must
also be met by the statistical moments of different orders formed using the function v(t). For
example, the autocorrelation function at the points t1 = t and t2 = t + τ ,
b (t) = h[v(t) − hvi] [v(t + τ ) − hvi]i = hδv (t) δv (t + τ )i
(8.4)
may depend only on τ but not on t.
Such random processes satisfying the condition that all their averaged characteristics do not
change when time changes are called stationary.
For a turbulent flow, the condition of stationarity is satisfied if all averaged characteristics of
the flow (the averaged velocity distribution, the average temperature and so on) remains unchanged when time changes. This also implies that all the external conditions (external forces,
position of bodies, etc) remains unchanged.
When a random function v (r, t) depends on time as well as on the space variables, it is possible to obtain space-averaged values by averaging with respect to the coordinates. Analogous to
stationarity is homogeneity of the function under consideration in this case when all moments
of this function depend only on the vector distance between points but not on r. For example,
the correlation function is
b (r2 − r1 ) = hδv (t, r 1 ) δv (t, r 2 )i .
(8.5)
We will proceed further on the assumption that the turbulent flows under consideration are stationary and homogeneous with respect to spatial coordinates. These two conditions (stationarity
and homogeneity) of the random function is a necessary but not sufficient conditions for the
function to be an ergodic one. It can be shown that ergodicity will be secured if the correlation
function (8.4) obeys the condition
lim b (τ ) = 0.
(8.6)
τ →∞
In other words, v(t) and v(t + τ ) must be statistically independent at large τ .
8.2.1
Reynolds Equation for Mean Flow
Now let’s proceed to the averaging of the hydrodynamic equations for an incompressible fluid.
The ensemble average of Equation (6.3) for momentum balance is
ρ
where ρ = const, and
∂
∂ hvi i
∂
∂vi
=−
Πik ⇒ hi ⇒ ρ
=−
hΠik i
∂t
∂xk
∂t
∂xk
(8.7)


 Πik = pδik + ρvi vk − σik ,
∂vi
k
,
+ ∂v
σik = η ∂x
∂xi
k


hΠik i = hpi δik + ρ hvi vk i − hσik i .
(8.8)
Taking now the relations (8.1) into account we have
hvi vk i = h(hvi i + δvi ) (hvk i + δvk )i = hhvi i hvk ii + hδvi δvk i + hhvi i δvk i + hhvk i δvi i =
| {z } | {z }
=0
157
=0
8.2. STATISTICAL DESCRIPTION
CHAPTER 8. THEORY OF TURBULENCE
= hvi i hvk i + hδvi δvk i
since hhvi i hvk ii = hvi i hvk i and hhvi i δvk i = hvi i hδvk i = 0. Hence,
(
hΠik i = hpi
δik + ρ hvi ihvk i − hσik i + ρ hδvi δvk i ,
ii
ki
hσik i = η ∂hv
.
+ ∂hv
∂xk
∂xi
(8.9)
As a result, we have the same equation (8.7) for the averaged quantities hvi i and hpi, as in the
absence of turbulence, with the only difference that an additional term ρ hδvi δvk i occurs in the
momentum flux tensor. Equation (8.7) can also be written in the same form as the Navier-Stokes
equation. Indeed, averaging the continuity equation (incompressibility condition), ∇ · v = 0, we
find,
∂ hvk i
=0
∂xk
and hence,
 ∂
∂
 ∂xk [hvi i hvk i] = hvk i ∂xk hvi i ,

∂ ∂vk
∂xk ∂xi
=
∂ ∂vk
∂xi ∂xk
= 0.
Using Equation (8.7) and Equation (8.9) we obtain finally
∂ hvi i
∂ hpi
∂
∂ hvi i
+ ρ hvk i
=−
+
ρ
∂t
∂xk
∂xi
∂xk
∂ hvi i
η
− ρ hδvi δvk i .
∂xk
(8.10)
Equation (8.10) is known as Reynolds equation for the turbulent flow of an incompressible fluid.
8.2.2
Turbulent Viscosity
If turbulence is absent, the only term in the parentheses on the right-hand side of Equation (8.10)
ii
is η ∂hv
which represents the viscous stresses. In the presence of turbulence, the new term
∂xk
τik = −ρ hδvi δvk i
(8.11)
called Reynolds stresses must be added to it.
Hence, the turbulence can transfer momentum from one part of the fluid to the other in the
same manner as viscous forces.
We can even formally characterise this effect by introducing the viscosity additional to the molecular one. For example, for the flow past a rigid wall x3 = 0 with the average velocity along x1 .
Suppose that the flow is homogeneous in the (x1 , x2 ) plane, i.e. all its average characteristics
do not depend on x1 or x2 . Then in Equation (8.10), we have to take into account only one
component of the complete stress tensor
∂ hv1 i
′
− hδv1 δv3 i .
σ13 = ρ ν
∂x3
Formally it can also be written as
′
σ13
= ρ (ν + νturb )
158
∂ hv1 i
.
∂x3
CHAPTER 8. THEORY OF TURBULENCE
8.2. STATISTICAL DESCRIPTION
The coefficient νturb is called the coefficient of kinematic turbulent viscosity and is defined by
νturb = −
hδv1 δv3 i
∂hv1 i
∂x3
.
(8.12)
It can be different in different parts of the flow and sometimes assumes a negative value. At the
wall, in particular, νturb = 0 because of δv1 = δv2 = 0. Usually νturb is larger by many orders
of magnitude than the molecular viscosity ν.
In a similar way, turbulence may also transfer heat, or some passive scalar (salt, for example), i.e
we can consider the turbulent heat conductivity, or turbulent diffusion which takes place in turbulent media. The coefficient of turbulent heat conductivity or turbulent diffusion can be defined
in the same way as νturb .
8.2.3
Turbulent Boundary Layer
It is natural to assume that as well as a boundary layer is formed close to a body surface for laminar flow, as discussed in Section 6.3, a turbulent boundary layer is formed also for turbulent flow.
For simplicity, we consider two-dimensional flow along a plane portion of the surface. This
plane is taken as the xz-plane, with the x-axis in the direction of the flow. The averaged velocity
distribution is independent of z, and the velocity has no z-component. So, we adopt a system
of coordinates in which x is everywhere parallel to the surface and y is everywhere normal to
the surface. The averaged characteristics of the flow are assumed to be independent of x (homogeneous flow). Let’s first look at the case of a similar but laminar (2D) flow in the x-direction.
Then we would get from the Navier-Stokes (N-S) equation
∂v
ρ
+ (v · ∇) v = −∇p + η∇2 v ⇒ (vx ≡ v)
∂t
 2
∂ v

 ∂y2 = 0 x-component of N-S equation,
(8.13)
⇒

 ∂p = 0 y-component of N-S equation.
∂y
From Equation (8.13) we find v = Ay + B, p = const. Since vy=0 = 0 (boundary condition for
the viscous fluid flow at the wall) ⇒ B = 0. So, the solutions of (8.13) are given by v = Ay and
p = const. The constant of integration A can be determined from the viscous (tangential) stress
defined at the wall (y = 0). Indeed, the tangential stress is given by
∂vi
∂vk
τik = η
.
+
∂xk ∂xi
Since the flow velocity has only one component, we take into account only one term from this
expression. Then,
∂v
τ0
τxy = η
= ηA ⇒ A = .
∂y y=0
η
Hence, the velocity and pressure distribution of the stationary laminar flow of viscous fluid
passed the infinite 2-D plane is given by
v = τη0 y,
(8.14)
p = const
159
8.2. STATISTICAL DESCRIPTION
CHAPTER 8. THEORY OF TURBULENCE
The laminar flow problem is solved. In the next step we consider the turbulent flow. Taking into
account that hvx i = hvx (y)i ≡ hvi we obtain from the averaged Navier-Stokes equation (8.10)
∂
assuming the flow to be stationary, ∂t
hvi = 0 ,
d2 hvi dτ ′
+
= 0,
η
dy 2
dy
(8.15)
where
τ ′ = −ρ hδvx δvy i
is the Reynolds stress. Since the averaged product hδvx δvy i may depend only on y (we have
assumed that the averaged characteristics of the flow may depend on y only and do not depend
on x), we can integrate (8.15) once which yields
η
d hvi
+ τ ′ = C.
dy
d
′
= 0. Hence, C = η dy
Since δvx = δvy |{z}
= 0 ⇒ τy=0
hviy=0 ≡ τ0 ⇒ C = τ0 . So, finally
y=0
d hvi
+ |{z}
τ ′ = τ0 .
η
dy
| {z } turbulent
viscous
stress
(8.16)
stress
We thus conclude that the sum of the viscous stress and the turbulent stress is constant, independently on the distance from the wall and equal τ0 which is the viscous stress at the wall.
Consider now the following particular cases:
(a) Very close to the wall, i.e. at very small y: In this case τ ′ ≈ 0 and the first term in
Equation (8.16) is most important, i.e. the viscous stress. This region is called the viscous
sublayer. The average velocity is linear in the y-direction as in the case of laminar flow, but
it does not mean that the flow in this viscous sublayer is laminar. Considerable turbulent
fluctuations may occur here.
(b) At large distances from the wall: Here we can approximately assume
τ ′ = −ρ hδvx δvy i ∼
= τ0 .
(8.17)
Thus, the momentum flux along the y-axis and the mean velocity gradient are determined
here only by the turbulent viscosity, whereas the molecular viscosity is negligible.
, can
We now will use the dimensional analysis to determine on what quantities the gradient, dhvi
dy
depend on in the region above the viscous sublayer. Since molecular viscosity η is not important
we have only three dimensional quantities ρ, y and τ0 . Their dimensions are [ρ] ≡ hLM3 ; [y]
i ≡L
dhvi
M
≡ T1 .
and τ0 ∝ ρv 2 ⇒ [τ0 ] ≡ LT
2 . The dimension of the velocity gradient is given by
dy
Using these three dimensional quantities ρ, y and τ0 we can form only one combination with
dimension of the velocity gradient. Indeed,






n = 1/2
2n = 1
[T ]
n
l
M
1
M
d hvi
 ⇒  l = −1/2 
l+n=0
⇒ 3l Lm n 2n = ⇒  [M ]  ⇒ 
ρl y m τ0n =
dy
L
L T
T
m = −1
m − 3l − n = 0
[L]
160
CHAPTER 8. THEORY OF TURBULENCE
8.3. LOCALLY ISOTROPIC TURBULENCE
Thus the combination is given by,
τ0
ρ
21
1
.
y
h i
Note that according to (8.17) and estimations made above we have τρ0 ≡ [v 2 ]. So, we can
21
introduce the quantity v ∗ ≡ τρ0 known as the dynamical or frictional velocity. This quantity
is widely used. Now outside the viscous sublayer we have
v∗
d hvi
=A
dy
y
(8.18)
where A is some dimensionless universal constant. Now, we introduce into (8.18) the dimen∗
sionless length ξ = vν y (dimensionless distance from the sublayer), so that (8.18) becomes
d hvi
v∗
=A
dξ
ξ
with general solution
hvi = v ∗ (A ln ξ + B)
(8.19)
where the constants A and B can be determined from experiments. We note that in the definition
of the dimensionless length ξ we use the viscosity. But since y is present in both sides of (8.18)
in the same power, the viscosity is not appearing in the following equation where we introduced
ξ. The interval of ξ where the solution (8.19) holds is called the logarithmic boundary layer.
It is interesting to note that the velocity distribution for a viscous sublayer, hvi = τη0 y, can be
rewritten in terms of ξ as follows
hvi =
τ0
v ∗2 νξ
⇒ hvi = v ∗ ξ
⇒
hvi
=
y
⇒
νξ
y= v∗
ρ ηρ
ν v∗
(8.20)
According to experiments, the velocity distribution for viscous sublayer given by (8.20) holds
at ξ < 8, whereas the logarithmic law determined by (8.19) holds at ξ > 30 with A ≈ 2.5 and
B ≈ 5.5. In the intermediate region 8 < ξ < 30 neither of equations is good approximation.
8.3
Locally Isotropic Turbulence
Up to now we have dealt with averaged characteristics of turbulent flow. For many purposes,
however, the statistical properties of the velocity fluctuations δv and in particular all their correlation functions Bij (r1 , r2 , t) = hδvi (r1 , t) δvj (r2 , t)i (or correlation tensors) are important.
It is known from the results of many experiments that if the main flow has a sufficiently large
Reynolds number, small scale fluctuations generated in this flow must be homogeneous and
isotropic.
We have seen from the probability theory that if the correlation function will depend only on
r = r 1 − r 2 the corresponding statistical process is called homogeneous. So, the turbulence
will be homogeneous if the correlation function of the velocity fluctuations depends only on the
position of the vector r = r 1 − r 2 .
161
8.3. LOCALLY ISOTROPIC TURBULENCE
CHAPTER 8. THEORY OF TURBULENCE
Suppose we have a fluid flow past a body with the characteristic dimension L0 . Let the mean
velocity of the flow be v0 , and the Reynolds number of the flow,
v 0 L0
≫ 1,
ν
i.e. the Reynolds number of the main flow is sufficiently large, Re ≫ Recr , where Recr corresponds to the laminar-turbulent transition. When this transition occurs, turbulent fluctuations
appear and their scale is L1 ≤ L0 with velocity fluctuations comparable with the velocity of
1
the main flow, (δv)2 2 ≤ v0 . These fluctuations are generated directly by the main flow
and hence they will be inhomogeneous across the space in approximately the same manner
as the main flow. The Reynolds number of these fluctuations will still be sufficiently large,
Re1 = v1νL1 ≫ Recr . Hence, these fluctuations are also unstable in their turn and generate fluctuations with scale L2 < L1 , (L2 < L1 < L0 ). The later are also unstable if Re2 = v2νL2 ≫ Recr
and generate fluctuations with scale L3 < L2 and so on. So, the cascading process will take
place and it results in energy going from large scales to small scales. If the Reynolds number for
all scales is very large the viscosity is still of no importance and the whole process of cascading
is governed by nonlinear phenomena. Hence, the power taken away from the main flow goes
through all of the cascade without loss, i.e. it remains unchanged through the full spectrum of
the scales, or through the full number of scales, LN < Li < L0 . This power is an important
parameter of the developed turbulence. Let ε denote the power dissipated per unit time and per
2
energy
≡ TL3 . The order of magnitude of ε can be determined with
unit mass of fluid, [ε] = mass·time
help of dimension analysis. Since the viscosity is not important there are only two parameters
that can determine the quantity of ε, namely the velocity v0 and the scale L0 of the main flow.
The only combination of these quantities in dimension of ε is given by
Re =
ε∝
v03
L0
(8.21)
The cascade process of consecutive breaking up of larger scales of turbulence with energy transition into smaller ones persists until such small scales LN will be reached for which ReN =
vN LN
≈ Recr . The flows of such scales are stable, so that breaking does not take place. The
ν
energy ε received by these fluctuations (with scale LN ) from larger scales (L > LN ) will now
transfer into heat due to viscosity.
In contrast to the main flow, the velocity in fluctuations of scale L1 is generally directed in
all directions and not only in that of the main flow. Analogously, isotropy of the flow with scale
L2 will be greater than that of the flow with scale L1 and so on. As a result after several acts
of cascading the anisotropic character of the main flow will be erased; the developed turbulence
becomes isotropic and statistically homogeneous in space. The scale L1 ≤ L0 is usually called
the external scale of turbulence and LN the inner, or Kolmogorovs scale . The larger Reynolds
number of the main flow the larger number of cascading acts which result in the wide spectrum
of scales Ln , n = 1, 2, 3, ...n, ..., N ≫ 1. The interval LN ≪ Ln ≪ L1 is called the inertial
interval . Fluctuations in the inertial interval have already forgotten about the structure of the
main flow and the action of the viscosity forces are still negligible. Thus, the turbulence in the
inertial interval can be treated as homogeneous and isotropic turbulence. Since this isotropy is
assumed on some small scales, Ln ≪ L1 , (remember that Ln is, for example the wave length, i.e.
the distance between two points and it is much smaller than the characteristic scale of the body
L0 ) the isotropy can be treated as local. Some important results for the developed turbulence in
the inertial interval can be obtained by dimensional analysis under the assumptions that
162
CHAPTER 8. THEORY OF TURBULENCE
8.4. SIMILARITY HYPOTHESIS
1. Turbulence is locally isotropic.
2. Only two quantities Ln and ε specify the flow in this interval (Ln is the characteristic spatial scale of fluctuations and ε is the energy per unit mass taken away from the main flow
per unit time).
Now, we will determine the order of magnitude of the velocity fluctuations on the scale Ln , i.e.
the velocity difference between two points apart r ∼ Ln . The only combination of the quantities
ε and Ln with the dimension of velocity is
1
vn ∼ (εLn ) 3
(8.22)
v3
Taking into account that ε ∼ L00 (this energy can be taken away by the turbulence from the main
flow) we obtain from the relation (8.22)
vn ∼ v0
Ln
L0
31
Introducing the Reynolds number for the arbitrary scale Ln , Ren =
between this Reynolds number and that of the main flow, Re = v0νL0 ,
4
43
v0 (Ln ) 3
Ln
Ren ∼
=
Re
1
L0
(L0 ) 3 ν
(8.23)
vn L n
,
ν
yields the relation
(8.24)
If now we assume that results given by (8.23) and (8.24) also holds for the inner scale Ln with
ReN ∼ 1, one can estimate the order of magnitude of LN and corresponding fluctuation velocity
vN . Indeed, we obtain from (8.23) and (8.24)
4
3
LN 3
(8.25)
1 ∼ ReN =
Re ⇒ LN ∼ L0 Re− 4
L0
1
31
1
LN 3
1
vN ∼ v0
⇒ vN ∼ v0
⇒ vN ∼ v0 (Re)− 4
(8.26)
3
L0
Re 4
Hence, the Kolmogorov scales LN and vN decreases with increasing Reynolds number of the
main flow.
8.4
Kolmogorov’s Similarity Hypothesis
We have seen that small-scale fluctuations (Ln ≪ L0 ) in the case of developed turbulence can be
treated as locally isotropic. If n is rather close to N (ReN ∼ Recr ∼ 1, the flow with such scales
is stable) the properties of this turbulence at such small scales may depend only on ε (energy
of fluctuations) and ν (viscosity). The scale is already fixed because the flow is stable. Let us
define in the order of magnitude the velocity fluctuations of this turbulence and the scale of these
fluctuations under the assumption that viscous forces can be essential. In other words, we have
to determine the combination of ε and ν with dimension of velocity and length:

 v ν = εα ν β ,
(8.27)

α 1 β1
Lν = ε ν ,
163
8.4. SIMILARITY HYPOTHESIS
CHAPTER 8. THEORY OF TURBULENCE
2
2
with dimension [ε] = TL3 , [ν] = LT . Taking into account the dimension of velocity and length
yields correspondingly
1
α = 1/4
2α + 2β = 1
L
L2α L2β
4
⇒
⇒
⇒
v
∼
(εν)
=
ν
3α
β
T
T
T
3α + β = 1
β = 1/4
L=
L2α1 L2β1
T 3α1 T β1
⇒
2α1 + 2β1 = 1
⇒
3α1 + β1 = 0
1
3 4
α1 = −1/4
⇒ Lν ∼ νε
β1 = 3/4
(8.28)
To estimate these fluctuating quantities we use the expression for ε (remember that for this exv3
ample ε is defined by the main flow) given by ε = L00 . We then obtain from (8.28)

3 14
1
1

v0 ν

4
∼ v0 Re− 4 ≡ vN

 vν ∼ (εν) ∼ L0
(8.29)

1 1
1 

 Lν ∼ ν 3 4 ∼ ν 3 L3 0 4 = ν 3 L3 0 L3 30 4 = L0 Re− 43 ≡ LN
ε
v
v L
0
0
0
So, we get rather natural result, namely, scales of Lν and vν coincide with the minimum turbulence spatial scale LN and corresponding velocity fluctuation vN . From this we can conclude that
the statistical properties of locally isotropic turbulence can be described in a universal way by
measuring the distance, for example the wave length, and the velocity fluctuations in terms of
LN and vN respectively.
This statement is known as the first Kolmogorov’s hypothesis.
−1
If on the other hand, n is not close to N , i.e. LN ≪ Ln ≪ L0 (or on k-space, L−1
0 ≪ k ≪ LN )
this corresponds to the developed turbulence in the inertial interval where the viscosity is no
importance.
So, in this interval the mean characteristics of the turbulence can be specified in terms of only
2
one quantity ε, [ε] = TL3 .
This is the second Kolmogorov’s hypothesis .
Now let’s determine the velocity fluctuations on the scale Ln , by calculating the average square
velocity difference between two points a distance Ln ∼ r apart. Having in the mind the relation
1
vn ∼ (εLn ) 3 , we obtain
2 2
(8.30)
(v1 − v2 )2 ≡ (δv)2 ∼ ε 3 r 3
where r ∼ Ln .
So, the mean-squared velocity difference between two points a distance r ∼ Ln apart is proportional to this distance with power (2/3) if this distance is neither too small nor too large,
i.e. LN ≪ r ≪ L0 .
This law is known as Kolmogorov’s (2/3) law.
The Kolmogorov’s (2/3) law can be presented in a equivalent form using so called spectral representation. To show this we introduce instead of Ln -scales the corresponding wave number
164
CHAPTER 8. THEORY OF TURBULENCE
8.5. SUMMARY OBJECTIVES
scales of fluctuations kn ∼ L1n . Then let E(k)dk be the kinetic energy density per unit mass of
the fluid element with fluctuating scales between k and k + dk, i.e.
Z ∞
E(k)dk
E=
k
where E(k) is called spectral energy density. Since the dimension of E corresponds to the
dimension of the energy per unit mass, i.e. [ E] = [v 2 ] and [k] = L1 we have for the dimension
3
of the spectral energy density [E(k)] = [v 2 L] = TL2 . Two quantities which determine the
2
properties of the turbulence in the inertial interval are ε with [ε] = TL3 and k with [k] = 1/L. To
find the combination of these parameters with dimension of spectral energy density we use the
dimensional analysis
L2α 1
L3
α = 32
[L]
2α − β = 3
α β
⇒
E(k) ∼ ε k ⇒ 2 = 3α β ⇒
⇒
[T ]
3α = 2
β = − 35
T
T L
Thus,
2
5
E(k) ∼ ε 3 k − 3
(8.31)
−1
with L−1
0 ≪ k ≪ LN . The relation (8.31) is known as Kolmogorov-Obukhov’s law, or 5/3
law of the turbulent spectrum in an inertial interval.
8.5
Summary objectives
When you finish studying the Chapter 8, you should be able to do the following:
1. Write the definition of the Reynolds number as the ratio of inertial and viscous terms in
the Navier-Stokes equation.
2. Describe the transition from laminar to turbulent flow depending on the magnitude of the
Reynolds number and connect this transition with the relative effects of inertial and viscous
forces.
3. Describe the flow around a cylinder at different Reynolds numbers.
4. Represent the turbulent flow as a statistical process and define the velocity of a flow as a
sum of the mean value and its random deviation. Provide the physical explanation of these
definitions.
5. Derive the Reynolds equations for a mean flow.
6. Define the Reynolds stress and determine its role in transport of the momentum. Derive
the coefficient of the kinematic turbulent viscosity.
7. Consider the phenomenological theory of a turbulent boundary layer. Describe the viscous
sublayer and compare the obtained results with that known for a laminar boundary layer.
Describe the turbulent sublayer.
8. Describe the development of a locally isotropic turbulence in a flow around a smooth
body as a cascading process assuming that the initial Reynolds number of the flow is large
enough.
165
8.5. SUMMARY OBJECTIVES
CHAPTER 8. THEORY OF TURBULENCE
9. Describe the inertial and inner (Kolmogorov) scales of locally isotropic turbulence and
find dimensional relations for velocity and spatial scale of fluctuations. Show that for
inner interval these scales can be expressed as a function of the initial Reynolds number.
10. Formulate the first Kolmogorov’s hypothesis and define the magnitude of velocity fluctuations and spatial scale of these fluctuations in the inner interval.
11. Formulate the second Kolmogorov’s hypothesis and estimate the magnitude of velocity
fluctuations in the inertial interval (Kolmogorov’s 2/3 law and Kolmogorov-Obukhov’s
law).
166
Chapter 9
Surface and Internal Waves in Fluids
It is well known from the classical mechanics that a single particle will execute a periodic motion,
so that the particle remains in a finite region, if the force acting on it is a restoring force, i.e.
proportional to the displacement of the particle and in the opposite direction to this displacement.
If the motion of the particle is constrained so that it can move only along a certain curve, the
motion is said to have one degree of freedom, or to be one-dimensional. One coordinate is then
sufficient to specify the position of the particle; it may be taken, for example, as the distance
along the curve from a point taken as origin. Let this coordinate be denoted by x. The potential
energy of a particle in one-dimensional motion is a function only of this coordinate, Φ = Φ (x).
According to the law of conservation of energy we have E = 21 mv 2 + Φ(x) = const and since
the kinetic energy cannot take negative values the inequality Φ ≤ E must hold. This implies that
the particle during its motion can occupy only points where the potential energy does not exceed
the total energy. If these energies are equal, we have the equation, Φ (x) = E, which determines
the limiting positions of the particle. Let us take a potential energy which, as a function of the
coordinate x, has a form shown in Figure 9.1.
In order to find the boundaries of the motion of a particle in such a force field, as functions of the
total energy, E, of the particle, we draw a straight line, Φ (x) = E, parallel to the x axis. This
line intersects the curve of potential energy Φ = Φ (x) at two points, whose abscissa are denoted
by x1 and x2 . If the motion is to be possible it is necessary that the potential energy should not
exceed the total energy. This means that the motion of a particle with energy E can occur only
between the points x1 and x2 , and a particle of energy E cannot enter the regions right of x2 and
left of x1 . Since the particle remains in a finite region, such motion is called a finite motion. At
the points x1 and x2 the potential energy is equal to the total energy, and therefore at these points
the kinetic energy and hence the particle velocity are zero. At the point x0 the potential energy
is a minimum, and the kinetic energy and velocity have their maximum values. Since the force
F is related to the potential energy, F = −( dΦ
), it is negative between x0 and x2 , and positive
dx
between x0 and x1 . This means that for any particle displacement from the point x0 , the force
acting on it, is opposite to the particle displacement i.e. it is restoring force. In such a force field,
if the particle begins to move from the point x1 , where its velocity is zero, the force to the right
will gradually accelerate it to a maximum velocity at the point x0 . As the particle continues to
move from x0 to x2 under the force which is now to the left, it will slow down until it comes
to rest at x2 . It will then begin to move back from x2 to x0 . This type of motion will continue
infinitely. Thus the particle executes a periodic motion with a period equal to twice the time for
the particle to go from x1 to x2 .
At the point x0 the potential energy is a minimum and the derivative of Φ with respect to x is
167
CHAPTER 9. SURFACE AND INTERNAL WAVES
Figure 9.1: Finite motion of the particle in a potential well.
zero. Thus, the force at this point is zero, and the point x0 is consequently a position of equilibrium of the particle. This position is evidently one of stable equilibrium, since in this case a
departure of the particle from the equilibrium position causes a force which tends to return the
particle to the equilibrium position. This property exists only for minima and not for maxima
of the potential energy, although at the latter the force is likewise zero. If a particle is moved in
either direction from a point of maximum potential energy, the resulting force in either case acts
away from this point, and points, where the potential energy reaches a maximum are therefore
positions of unstable equilibrium.
If, on the other hand, we have a set of connected (or interacting) particles and assume that one of
them is starting to oscillate, it will result in oscillations of other particles as well. Moreover, these
oscillations will not localise at one particular place of a system but instead will transform from
one point to a neighbouring one. This non-locality of oscillations is typical for all continuous
media (gas, fluid, plasma, etc). One can say that in continuous media, oscillations are propagating from one point to another one. Such propagating oscillations are usually called waves.
The main characteristic feature of waves is the possibility of energy transport over considerable
distances without mass transport. In this chapter the basic equations governing the wave propagation in fluids will be derived and three sort of waves related to the gravitational force will
be considered, namely, surface gravity, surface gravity capillary and internal gravity waves. The
first two kind of waves occur at the fluid surface (for example, the ocean’s surface) or near it,
whereas the internal waves occur in inner layers with pronounced vertical gradients of the fluid
density.
168
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.1. STRATIFIED FLUIDS
9.1
Linear Equations for Waves in Stratified Fluids
9.1.1
Linearisation of the Hydrodynamic Equations
The basic hydrodynamic equations are as follows
∂v
− g∇z − 2Ω × v Euler equation
+ (v · ∇) v = − ∇p
ρ
∂t
∂ρ
+ ∇ · (ρv) = 0
∂t
dp
dρ
= c2 ,
dt
dt
c2 =
Continuity equation
∂p
∂ρ
(9.1)
Equation of state
S=const
In the Euler equation we include the gravitational force, the Coriolis force (−2Ω × v) arising in
a coordinate system rotating at the angular velocity Ω. A centrifugal force is a potential force
and can be formally included into the potential of gravitation. In the following we will consider
ideal fluids.
Below we will present the linear theory of waves. It means that we consider waves in which
the velocity of the moving fluid particles is so small that we may neglect the term(v · ∇) v in
in the Euler equation. The physical significance of this is easily seen. Durcomparison with ∂v
∂t
ing a time interval of the order of the period τ of the oscillations of fluid particles in the wave,
these particles travel a distance (i.e. fluid particles displacement) of the order of the amplitude a
of the wave. Their velocity v is therefore of the order of v ∼ τa . It varies noticeably over time
intervals of the order of τ and distances of the order of λ in the direction of propagation (where
λ is the wavelength). Hence the time derivative of the velocity is of the order of τv , and the space
is equivalent to
derivatives are of the order of λv . Thus the condition (v · ∇) v ≪ ∂v
∂t
or
1 a 2
a1
≪
λ τ
ττ
a≪λ
i.e. the amplitude of the oscillations, and thus fluid particle displacement, in the wave must be
small compared with the wavelength. Then the equations (9.1) can be linearised with respect to
these small displacements. Therefore, in this limit of small perturbations of the quantities we can
use the linear wave theory and thus the principle of superposition holds, i.e. waves propagate
independently of each other. Let us assume that a fluid is in static equilibrium, i.e. the initial
state of the fluid is defined by the following parameters: v 0 = 0, p0 , ρ0 . If this equilibrium state
is perturbed in such a way that the pressure and density will differ slightly from their equilibrium
values,
p = p0 + p′ , |p′ | ≪ p0
ρ = ρ0 + ρ′ , |ρ′ | ≪ ρ0
then the perturbed velocity of the fluid element will also be small. Assuming that the velocity
components are defined by v (u, v, w) and omitting the prime (′ ), we obtain after some algebra
169
9.1. STRATIFIED FLUIDS
CHAPTER 9. SURFACE AND INTERNAL WAVES
the linearised set of hydrodynamic equations:
∂v
∂t
+
∇p
ρ0
+ g ρρ0 ∇z + 2Ω × v = 0
∂ρ
∂t
+
dρ0
w
dz
dp
dt
= c2
(9.2)
+ ρ0 ∇ · v = 0
∂ρ
∂t
0
+ w dρ
dz
The last equation of (9.2) (the equation of state) can be rewritten using the definition of the
convective derivative. Indeed, first we transform the left hand side of this equation
dp
∂p
∂p
dp0
∂p
=
+ (v · ∇) p =
+w
=
+ (−gρ0 ) w
dt
∂t
∂t
dz
∂t
Using this result in the equation of the state yields
∂ρ
1 ∂p
= 2
∂t
c ∂t
ρ0
g
h
dρ0
1
− 2 gρ0 w − w
{zi dz}
| c
1 dρ0
+ g2
ρ0 dz
c
−g
w≡
ρ0
N 2 (z)w
g
Finally, the linearized set of hydrodynamic equations can be written as
∂v
∂t
+
∇p
ρ0
+ g ρρ0 ∇z + 2Ω × v = 0
∂ρ
∂t
+
dρ0
w
dz
+ ρ0 ∇ · v = 0
∂ρ
∂t
=
1 ∂p
c2 ∂t
+
ρ0
N2
g
Here
2
N (z) = −g
(9.3)
(z) w
1 dρ0
g
+ 2
ρ0 dz
c
is the Brunt-Väisälä frequency introduced in Chapter 4.2.2, see Equation (4.22).
9.1.2
Linear Boundary Conditions
If the influence of the physical processes reaches some form of boundary (i.e. the boundary
between water and air of the oceans) we have to consider the interaction of this influence and
the outer environment. The boundary will change according to the results of this interaction. A
simple treatment of this case is to look on the boundary as a linear continuous boundary. For
example, let the surface of a fluid be a plane in the equilibrium state. When this state is violated
two kinds of forces arise in order to return this surface to its initial position:
1. the gravitational force,
2. the surface tension force.
As a result, due to inertia, a disturbance arises at the fluid surface which propagates in all directions in the form of a wave. These forces must be included into the boundary conditions for
Equations (9.3).
170
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.1. STRATIFIED FLUIDS
Let z = 0 be the free plane surface at the equilibrium state, i.e. points at this surface are not
assumed to belong to the fluid. The vertical displacement of the surface particle is denoted by
ζ (t, x, y) ≡ ζ. Then the vertical velocity of the surface particles will be given by dζ
. At the
dt
same time these particles can be considered as belonging to the fluid. Since the vertical velocity
of the fluid element is determined by w (t, x, y, z) we have
wz=ζ =
dζ
.
dt
(9.4)
This is the kinematic boundary condition .
To obtain the second boundary condition we have to take into account the existence of a discontinuity of the pressure on both sides of the boundary due to the surface tension. Let p0 (z) + p
be the pressure in the fluid, and pa the pressure at the surface. By means of the well known
Laplace formula we have
σ
[p0 (z) + p]z=ζ − pa = −
(9.5)
R
Here σ is the surface tension coefficient and R1 is the sum of principal curvatures of surface
ζ = ζ (x, y, t). For this surface, R1 is given by


1
∇ζ

(9.6)
= ∇ · q
R
2
1 + (∇ζ)
The condition (9.5) is known as the dynamic boundary condition .
The boundary conditions (9.4) and (9.5) are nonlinear and to simplify them we linearise them
assuming the displacement of the surface ζ to be small with respect to the equilibrium position
z = 0. Results of the linearisation:
1. The kinematic boundary condition (9.4):
∂ζ
wz=ζ = wz=0 + ζ ∂w
+ ...
∂z z=0
⇒ wz=0 =
dζ
∂ζ
= ∂t + v · ∇ζz=0
∂t
dt
(9.7)
2. The dynamic boundary condition (9.5):
−pa + [p0 (z) + p]z=ζ = p0 (z)z=0 +
dp0
ζz=0 + ... + pz=0 − pa
dz
Assuming that pa = p0 (z)z=0 = p0 (0) and taking into account that
∇ζ
∇2 ζ
1
∇2 ζ (∇ζ)2
2
2
q
∇· p
≈ ∇ ζ 1 − (∇ζ) ≈ ∇2 ζ
−
2 =
3/2
2
2
1 + (∇ζ)
1 + (∇ζ)
1 + (∇ζ)2
{z
}
|
nonlinear term
we finally have the dynamic boundary condition in the linearised form
pz=0 +
dp0
ζ
z=0
|dz{z
}
−gρ0 (z)z=0 =−gρ0 (0)
171
= −σ∇2 ζ.
9.2. SURFACE GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
2
2
∂
∂
Since ζ ≡ ζ(x, y, t) is a two dimensional surface, the operator ∇2 ≡ ∂x
2 + ∂y 2 is the
Laplace operator in the horizontal plane. Now the dynamic boundary condition takes the
form
pz=0 = gρ0 (0) ζ − σ∇2 ζ.
(9.8)
The expressions (9.7) and (9.8) are the linear boundary conditions at the undisturbed free surface
z = 0.
9.1.3
Equations for an Incompressible Fluid
Among the “restoring” forces in a fluid is the elasticity force resulting from compression of fluid
elements. The relationship between compression and pressure is described by the first two terms
in the linearised equations (9.3):
1 ∂p
∂ρ
= 2 .
∂t
c ∂t
If we consider the incompressible limit, then c2 → ∞ (but ∂p
6= 0). Indeed, the density of a
∂t
dρ
given incompressible fluid element must be constant, i.e. dt = 0 ⇒ ∇ · v = 0. On the other
dρ
dp
2 dρ
hand, from the exact equationof state
we have dt = c dt . Since for incompressible fluid dt = 0,
we have to assume that c2 =
∂p
∂ρ
S=const
→ ∞, but
dp
dt
6= 0.
Now the linearised equations of incompressible fluid hydrodynamics can be written as
∂v ∇p
ρ
+
+ g ∇z + 2Ω × v = 0
∂t
ρ0
ρ0
(9.9)
∂ρ
N2
− ρ0
w=0
∂t
g
(9.10)
∇·v =0
(9.11)
Here the Väisälä frequency is given by
N 2 (z) = −
g dρ0
ρ0 dz
(9.12)
The boundary conditions (9.7) and (9.8) do not depend on the sound velocity and thus remain
unchanged.
Moreover, we can conclude that the approximation of incompressible fluid (c2 → ∞) is applicable to those waves whose velocity is much less than the velocity of sound.
9.2
Surface Gravity Waves
9.2.1
Basic Equations
The free surface of a liquid in equilibrium in a gravitational field is a plane. If, under the action
of some perturbation, the surface is moved from its equilibrium position at some point, motion
will occur in the liquid. This motion will propagate over the whole surface in a form of waves,
172
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.2. SURFACE GRAVITY WAVES
which are called gravity waves, since they are due to the action of the gravitational field. Gravity
waves appear mainly on the surface of the liquid; they affect the interior also, but less and less
at greater and greater depth. In the case of gravity waves it is the gravitational force that tends
to restore the equilibrium if the free surface of the fluid is perturbed. The effect of gravity in
the boundary conditions (9.7) and (9.8) is described by the term gρ0 (0)ζ which is equal to the
pressure of a fluid column of height ζ. For the time being and in order to calculate the effect of
gravity itself we exclude the effect of the surface tension (σ = 0), the Coriolis force (Ω = 0),
and the Archimedes force (ρ0 = const, and ρ = 0 ⇒ N 2 = 0).
Let u (u, v) be the horizontal component of the fluid particle velocity, and define the horizontal
∂
∂
differential operator as ∇− = ex ∂x
+ ey ∂y
. The momentum equation (9.9) is now reduced to
two equations for u (the horizontal velocity) and w (the vertical velocity):
 ∂u
∇− p
(a)
 ∂t = − ρ0
(9.13)
 ∂w
1 ∂p
= − ρ0 ∂z (b)
∂t
The continuity equation takes the form
∇− · u +
∂w
=0
∂z
(9.14)
We transform now Equations (9.13) and (9.14) by applying the operator ∇− to equation (9.13a)
∂
and ∂t
to equation (9.14). We obtain then
∂
∇
∂t −
∂
∇
∂t −
·u=−
∇2− p
ρ0
(a)
∂2w
· u = − ∂t∂z (b)
⇒ ∇2− p = ρ0
∂ 2w
∂t∂z
(9.15)
As the next step, we apply the operator ∇− to the equation (9.15b)
∂ 2w
∂
∇− (∇− · u) = −∇−
.
∂t
∂t∂z
(9.16)
Finally, as a result of the transformations we have reduced the Equations (9.13a), (9.14) to the
following two equations
∂ 2w
(9.17)
∇2− p = ρ0
∂t∂z
and
∂
∂ 2w
∇− (∇− · u) = −∇−
(9.18)
∂t
∂t∂z
Now we will transform (9.13b) for the vertical component of the fluid velocity. Applying the
∂
to Equation (9.13b) and adding the obtained equation to Equation (9.15) yields
operator ∂z
2
∇ p = 0. Applying then the Laplace operator ∇2 to the (9.13b) and taking into account that
∇2 p = 0, we obtain the equation for the vertical velocity component of the fluid element
∂ 2
∇ w=0
∂t
(9.19)
We transform now the boundary conditions (9.7) assuming that σ = 0, i.e. pz=0 = gρ0 (0)ζ.
Applying the two-dimensional Laplace operator to this equation and taking into account the first
173
9.2. SURFACE GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
2
∂ w
= g∇2− ζ. Taking now the time derivative of the obtained equation
equation (9.17) yields ∂t∂z
, we obtain the boundary condition at the free surface
and using the condition w(z)z=0 = ∂ζ
∂t
z = 0 for w:
3
∂ w
2
− g∇− w
=0
(9.20)
∂t2 ∂z
z=0
If we have a liquid layer of depth H (Figure 9.2) and the bottom is perfectly rigid, another
boundary condition must be imposed on the solution of (9.19)
wz=−H = 0
(9.21)
i.e. the vertical component of the particle velocity at the bottom is zero. So, the basic equations
for surface gravity waves are reduced to (9.19) with boundary conditions (9.20) and (9.21).
z
!x, y, t"
x
0
H
-
Figure 9.2: Surface waves in the presence of a bottom.
9.2.2
Harmonic Waves
Let us consider a wave in which all quantities depend on only one coordinate, say x. That is, the
flow is completely homogeneous in the yz-plane. Such a wave is also called a plane wave. A
simplest example of a plane wave is monochromatic wave. Here all quantities are just periodic
(harmonic) functions of the time and space. It is usually convenient to write such functions as a
complex quantity (remember that only real part of this quantity has some physical meaning). For
example, any quantity describing a monochromatic wave propagating in the positive direction of
the x-axis, is of the form
ϕ (x, t) = A exp [i (kx − ωt)]
174
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.2. SURFACE GRAVITY WAVES
where A is a constant called the amplitude of the wave, ω is the frequency of the wave. We
denote by n a unit vector in the direction of propagation. The vector
k=
2π
n
λ
is called the wave vector, and its magnitude k the wave number. In terms of this vector the
function ϕ (x, t) can be written
ϕ (r, t) = A exp [i (k · r − ωt)]
The argument of the exponential function is called the phase of the wave, φ (r, t) = k · r − ωt.
At the given time t, all space points with the same phase φ form the surface of constant phase
which is also called the wave front. For a plane wave, the wave front is simply a plane which is
perpendicular to the direction of the wave propagation k. The values of the phase at all points
of this plane will be constant at the given time t, i.e. dφ = 0. This plane is called a plane of
a constant phase. Assuming the wave is propagating in the positive direction of the x-axis, we
obtain from the equation for the plane of a constant phase
φ (x, t) = const ⇒ dφ = 0 ⇒ dφ =
∂φ
∂φ
dx +
dt = 0 ⇒ dφ = kdx − ωdt = 0
∂x
∂t
Now we can define the velocity at which the plane of constant phase is propagating in space
cph =
ω
dx
=
dt
k
This velocity is called the phase velocity. If cph = f (k), one say that the corresponding wave
is dispersive, i.e. waves with different wave length have a different phase velocity and thus the
wave packet as a composition of monochromatic waves whose wave vectors lie in a small range
will spread out in the course of its propagation. If, on the other hand, cph = const, then the
corresponding wave is non-dispersive. One has to take into account that the phase velocity despite of its importance in the wave phenomena does not correspond to any physical propagation.
The transport of the energy in the course of wave propagation is related to another wave velocity
called the group velocity.
To determine the group velocity and clarify its meaning, let us consider a wave packet, which occupies some finite region of space. We assume that its spectral composition includes monochromatic components whose frequencies lie in only small range; the same is true of the components
of their wave vectors. Then, at some initial instant, the wave is described by a function having
the form
ϕ = f (r) exp (ik · r)
The function f (r) is appreciably different from zero only in a region which is small (though it
is large compared with the wavelength k1 ). Its expansion as a Fourier integral contains, by the
above assumptions, components having the form exp (ir · ∆k), where ∆k ≪ k. Thus each
monochromatic component is, at the initial instant, proportional to
ϕk = const × exp [i (k + ∆k) · r]
The corresponding frequency is ω (k + ∆k) (we recall that the frequency is a function of the
wave vector). Hence the same component at time t has the form
ϕk = const × exp [i (k + ∆k) · r − iω (k + ∆k) t]
175
9.2. SURFACE GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
· ∆k. Then ϕk becomes
We use the fact that k ≫ ∆k, and put ω (k + ∆k) ∼
= ω (k) + ∂ω
∂k
∂ω
ϕk = const × exp [i (k · r − ωt)] exp i∆k · r − t
∂k
If we now sum all the monochromatic components, with all ∆k that occur in the wave packet,
we see from these formulas that the result is
∂ω
ϕ = exp [i (k · r − ωt)] f r − t
∂k
where f is the same function as above. A comparison with initial description of the wave shows
.
that after a time t, the amplitude distribution has moved as a whole through a distance t ∂ω
∂k
Exponential coefficient of f in this formula affects only the phase. Consequently, the velocity of
the wave is
∂ω
cg =
∂k
The velocity defined by this relation is called the group velocity. This formula gives the velocity
of propagation for any dependence of ω on k. When ω = ck, with c constant, it is of course
gives the simplest result when the phase velocity coincides with the group velocity and equals
c. For example, this result holds for electromagnetic waves in a vacuum where c is the velocity
of the light. In general, when ω (k) is an arbitrary function, the velocity of wave propagation is
a function of the frequency, and the direction of the propagation may not be the same as that of
the wave vector.
In conclusion, we note that the concept of monochromatic waves is very important, because any
wave whatsoever can be represented as a sum of superposed monochromatic plane waves with
various wave vectors and frequencies. This decomposition of wave into monochromatic waves
is simply an expansion as a Fourier series or integral called also spectral resolution. The terms
of this expansion are called the monochromatic components, or Fourier components of the wave.
9.2.3
Harmonic Surface Waves
In this section we will look for the partial solution of Equation (9.19) in the form of a harmonic
wave propagating in the horizontal direction
w(x, y, z, t) = Φ(z) exp(ikr − iωt)
(9.22)
that satisfies the boundary conditions (9.20) and (9.21), where k = (kx , ky ) is the horizontal
, r = (x, y). Then Equation (9.19) is reduced to
wave vector, |k| = 2π
λ
d2 Φ
− k2Φ = 0
dz 2
with boundary conditions
 
−
 dΦ
dz


gk2 Φ
ω2
z=0
= 0, (a)
Φ (z = −H) = 0.
The general solution of (9.23) is given by
Φ(z) = aekz + b̃e−kz .
176
(9.23)
(9.24)
(b)
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.2. SURFACE GRAVITY WAVES
where a and b̃ are arbitrary constants. Taking into account the boundary condition (9.24b) yields
Φ (z = −H) = ae−kH + b̃ekH = 0 ⇒ a = −b̃e2kH ⇒
Φ(z) = −2b̃e
kH e
k(z+H)
− e−k(z+H)
= −2b̃ekH sinh [k (z + H)] .
2
To provide the symmetry of the solution we choose the normalisation factor in the form −2b̃ekH ≡
b
. So, the solution of (9.19) satisfying to the boundary condition (9.21), or (9.24b) is now
sinh(kH)
given by
sinh [k (z + H)]
Φ (z) = b
.
(9.25)
sinh (kH)
Substituting this solution into the boundary condition (9.24a) we get
b cosh (k (z + H))
cosh (kH)
= kb
⇒
sinh (kH)
sinh (kH)
z=0
cosh (kH) gk 2
gk 2
′
= kb
Φ − 2Φ
− 2 b = 0.
ω
sinh (kH)
ω
z=0
Φ′ (z)z=0 =
The last equation yields the relationship between the frequency ω and its wave number k, or the
so-called dispersion relation for the surface gravitational waves
ω 2 = gk tanh (kH)
Let’s now calculate the phase velocity of the surface gravity waves cph = ω/k ⇒
r
ω
tanh (kH)
cph = = gH
.
k
kH
(9.26)
(9.27)
Hence, the phase velocity of surface gravitational waves depends on the wave number, i.e. dispersion takes place. From the physical points of view it means that the non-harmonic surface
wave changes its shape while propagating. Indeed, the non-harmonic wave can be expanded into
a Fourier integral (series) and each term of this expansion corresponds to a surface harmonics
with a period and wave length. If dispersion takes place, it means that each harmonics has its
own phase velocity and different harmonics propagates with different velocities, i.e. the wave
packet spreads out while propagating and changes its shape.
One can also find the group velocity of a surface gravity wave, i.e the propagating velocity
of the envelope of a spectrally narrow wave train cg = dω/dk. From the dispersion relation
(9.26), we have
2ωdω = g tanh (kH) dk + gkH 1 − tanh2 (kH) dk ⇒
dω
kH
cph
1+
=
− kH tanh (kH) .
(9.28)
dk
2
tanh (kH)
Now, the displacement of the surface ζ = ζ (x, y, t) is determined from the boundary condition
(the velocity of the vertical displacement of the surface is equal to the vertical fluid velocity
wz=0 )
sinh[k (z + H)] ikr−iωt
= beikr−iωt
e
w|z=0 = b
sinh(kH)
z=0
177
9.2. SURFACE GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
ib
∂ζ
= wz=0 ⇒ −iωζ = beikr−iωt ⇒ ζ = eikr−iωt ≡ aeikr−iωt .
(9.29)
∂t
ω
Pressure variation at an arbitrary depth z caused by the wave motion will be found from relation
(9.17) with ∇2− ⇒ −k 2
p = ρ0
iω
b
cosh [k (z + H)] ikr−iωt
e
k tanh (kH) cosh (kH)
(9.30)
or by taking into account the dispersion relation (9.26) and denoting a = ib/ω, we can rewrite
(9.30) in the form
cosh [k (z + H)] ikr−iωt
p = ρ0 ga
e
.
(9.31)
cosh (kH)
The horizontal velocity of the fluid particles can be found from (9.18)
k cosh [k (z + H)] ikr−iωt
u=i b
e
.
k
sinh (kH)
(9.32)
So, the fluid particles are moving in the direction of wave propagation, k.
Let ξ be the displacement of fluid particles in the horizontal direction (plane). Then,
k b cosh [k (z + H)] ikr−iωt
dξ
⇒ −iωξ = u ⇒ ξ = −
e
.
dt
k ω sinh (kH)
To relate the amplitude of the horizontal displacement to the vertical one (9.29) it is convenient
to assume that
π
b = ωAei(α− 2 ) ≡ ωA(sin α − i cos α)
π
where α is some angle. Then by direct calculation we obtain (i = ei 2 ) ib/ω = Aeiα ≡ a. So, the
displacement of fluid particles in the horizontal direction (i.e. in the horizontal plane) is given by
k cosh [k (z + H)] ikr−iωt+i(α− π2 )
e
.
ξ=− A
k
sinh (kH)
Since the displacement is a real quantity, we finally have for the displacement of fluid particles
in the horizontal plane
k cosh [k (z + H)]
ξ=− A
sin (kr − ωt + α) .
k
sinh (kH)
Let now η denote the displacement of fluid particles in the vertical direction such that
−iωη = w ⇒
w
ib sinh [k (z + H)] ikr−iωt
η=
=
e
.
(−iω)
ω sinh (kH)
(9.33)
dη
dt
=
Using relation ib/ω = Aeiα ≡ a we obtain
η=A
sinh [k (z + H)] ikr−iωt+iα
e
.
sinh (kH)
Since the vertical displacement of fluid particles is a real quantity, we finally have
η=A
sinh [k (z + H)]
cos (kr − ωt + α)
sinh (kH)
178
(9.34)
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.2. SURFACE GRAVITY WAVES
Excluding the time from (9.33) and (9.34) we find the equation for a path of fluid particle motion
in a surface gravity wave
 2
2
ξ

+ aη2 = 1,

a2ξ

η




(9.35)
,
aξ = A cosh[k(z+H)]
sinh(kH)






 aη = A sinh[k(z+H)] .
sinh(kH)
These are ellipses with semi-axes ratio aη/aξ = tanh [k (z + H)].
A standing wave (surface gravity wave) can be formed by superposition of two propagating
waves w (x, y, z, t) = Φ (z) exp (ikr − iωt) with the same frequencies ω and amplitudes b but
with opposite wave numbers k
1
w = w(1) + w(2) = Φ (z) e−iωt 2 eikr + e−ikr ⇒
2
sinh [k (z + H)]
cos (kr) e−iωt
(9.36)
wst.w = 2b
sinh (kH)
It is interesting to note that the expressions for the particle velocities (9.22) and (9.32) show
that the motion of fluid particles in a surface wave is potential, i.e. ∇ × v = 0. One can
easily prove this fact by direct calculations of ∇ × v, taking into account that the amplitudes
of the horizontal velocity components (u, v) depend only on z whereas the dependence on x
and y is through the exponential factor exp (ikr − iωt). The vertical velocity component w has
an amplitude dependence on z and the (x, y) dependence is only through the exponential factor
exp (ikr − iωt). On the other hand, one can easily calculate the velocity potential v = ∇ϕ,
where
b cosh [k (z + H)] ikr−iωt
e
.
ϕ=
k sinh (kH)
Indeed,

= b sinh[k(z+H)]
eikr−iωt ,
w = ∂ϕ

∂z
sinh(kH)





u = ∂ϕ
= ikkx b cosh[k(z+H)]
eikr−iωt ,
(9.37)
∂x
sinh(kH)






= ikky b cosh[k(z+H)]
eikr−iωt .
v = ∂ϕ
∂y
sinh(kH)
We see that these expressions coincide with the corresponding equations for the velocity components, Equation (9.22) together with (9.25) and Equation (9.32).
9.2.4
Shallow and Deep Water Approximations
Two extreme cases (two limits) are usually considered in the theory of surface gravity waves.
.
These limits differ in the relation between H and the wave length λ = 2π
k
≪ 1, i.e. the depth is small compared to the wave length. In
1. Shallow water kH ≡ 2πH
λ
this case tanh (kH) ≈ kH and hence

√
 ω = gHk
(9.38)
√

cph = gH = const
179
9.2. SURFACE GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
Dispersion is absent. For the fluid particle’s velocity components, we have

ikr−iωt
sinh[k(z+H)] ikr−iωt
z

w
=
b
e
,
e
≈
b
1
+

sinh(kH)
H
(9.39)

 u = i k b cosh[k(z+H)] eikr−iωt ≈ i k b eikr−iωt .
k
sinh(kH)
k
kH
Since kH ≪ 1, |u| ≫ |w|, i.e. the particle’s velocity is practically constant across the
fluid layer and parallel to the bottom.
2πH
λ
2. Deep water kH ≡
this case
≫ 1, the depth is large compared to the wavelength. We have in
tanh (kH) =
ekH −e−kH
ekH +e−kH kH≫1
⇒1
ω 2 = gk tanh (kH) ⇒
(9.40)
ω 2 = gk.
For the phase and group velocities we obtain

pg
g
 cph = ωk = k = ω ,

cg =
dω
dk
=
1
2
pg
k
=
cph
.
2
Then, the fluid particle’s velocities are given by

sinh[k(z+H)] ikr−iωt

≈ bekz eikr−iωt ,
 w = b sinh(kH) e
(9.41)
(9.42)

 u = i k b cosh[k(z+H)] eikr−iωt ≈ i k bekz eikr−iωt .
k
sinh(kH)
k
So, we see that the particle velocity (as well as the pressure variation) decrease exponentially with depth (z < 0). At the depth |z| ≈ λ (z < 0) the wave motion is practically
absent.
A liquid layer with given depth H can be considered either shallow or deep water depending
on the length of propagating wave. For long wave length waves (i.e. Tsunami waves) any sea
represents a shallow water. These waves propagate without dispersion. It is often convenient
to use a graphical representation (Figure 9.3) of the dispersion law (9.26). Here, the linear non
dispersive part is in the vicinity of the origin. At large frequencies (wave numbers) (kH ≫ 1),
we have the dispersion corresponding to waves in deep water.
9.2.5
Wave Energy
The energy of any kind of wave is usually defined as an excess of energy of a fixed volume over
that at equilibrium. For surface waves it is convenient to take such a volume as a fluid column
bounded by the bottom z = −H and by the free surface z = ζ in a vertical direction, and
by two planes x = x0 and x = x0 + λ in the direction of wave propagation. In the direction
perpendicular to wave propagation (y-direction) the volume is supposed to be of unit width. For
a kinetic energy of such volume we have
ρ0
Ek =
2
xZ0 +λ Z0
x0
−H
180
v 2 dxdz.
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.2. SURFACE GRAVITY WAVES
ω
2
ω=gHk
2
2
ω=gk
0
k
Figure 9.3: Graphical representation of the dispersion relation for surface gravity waves.
To calculate v 2 we consider the real parts of the velocity components w and u. Introducing the
π
amplitude factor b exp (−iα′ ) ≡ B and remembering that (−i) = e−i 2 , we obtain from relations
(9.22) and (9.32)
v 2 = (Re{w})2 + (Re{u})2 =
= B2
2
sinh2 [k (z + H)]
π
2
′
2 cosh [k (z + H)]
2
′
.
cos
(kx
−
ωt
+
α
)
+
B
cos
kx
−
ωt
+
α
+
2
sinh2 (kH)
sinh2 (kH)
Taking into account this expression in the kinetic energy integral and integrating first with respect
to x (the direction of wave propagation) yields
xZ0 +λ
cos
x0
ρ0 λB 2
Ek =
4 sinh2 (kH)
ρ0 λB 2
=
4 sinh2 (kH)
Z0
Z0
−H
2
λ
2π
x − f (t) dx = ⇒
λ
2
sinh2 [k (z + H)] + cosh2 [k (z + H)] dz =
ρ0 λB 2
ρ0 λB 2
g.
=
cosh [2k (z + H)] dz =
4k tanh (kH)
4ω 2
−H
It is convenient to introduce now the amplitude of the free surface displacement A = |a| =
Then the kinetic energy of surface gravity waves is given by
Ek =
ρ0 λg 2
ρ0 λg B 2
≡
A
2
4 ω
4
181
ib
ω
.
(9.43)
9.3. CAPILLARY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
In the same way we can calculate the increment of the potential energy


Zλ Zζ
Z0
Zλ  Zζ
Zλ

1
zdz = ρ0 g ζ 2 dx
ρ0 gzdz −
ρ0 gzdz dx = ρ0 g
Ep =


2
0
−H
0
−H
0
0
We have from (9.29)
ζ = aeikx−iωt = Aeikx−iωt+iα ⇒ ℜζ = A cos (kx − ωt + α) .
Now the potential energy can be presented as
1
E p = ρ0 g
2
Zλ
1
A2 cos2 (kx − ωt + α) dx = ρ0 gA2 λ.
4
(9.44)
0
Thus, the kinetic energy and the potential energy of a progressive surface gravity wave are equal
and do not depend on time.
In the case of a standing surface wave ζ = A cos(kx) cos (ωt − α) similar calculations lead
to
2λ
2λ
Ek = ρ0 gA
sin2 (ωt − α) = ρ0 gA
[1 − cos (2ωt − 2α)] ,
4
8
(9.45)
ρ0 gA2 λ
ρ0 gA2 λ
2
Ep = 4 cos (ωt − α) = 8 [1 + cos (2ωt − 2α)] .
i.e. Ek + Ep = const. The kinetic and potential energies both change in time at twice the
frequency, their value oscillating between 0 and ρ0 gλA2 /4. Their sum (total energy of the wave)
remains constant. When time-averaged over a period both kinds of energy of the standing wave
are equal.
9.3
Capillary Waves
Fluid surfaces tend to assume an equilibrium shape, both under the action of the force of gravity and under that of surface tension forces. In studying waves on the surface of a fluid in the
Section 9.2, we did not take the latter forces into account. We shall see below that capillarity
has an important effect on gravity waves with short wave length. As previously, we suppose the
amplitude of the oscillations small compared with the wave length.
A capillary wave is a wave traveling at the interface between two fluids (e.g. air and water),
whose dynamics are dominated by the effects of surface tension. The wavelength of “pure” capillary waves (Section 9.3.1) are typically quite small (less than a centimetre) and they commonly
occur in nature where they are often referred to as ripples. Waves with larger wavelengths are
generally also affected by gravity and are then called gravity-capillary waves (see Section 9.3.2).
See Figure 9.4 for an example of gravity-capillary waves on a beach front. The short wavelength
capillary waves are in front of the ordinarily gravity waves on the beach. The gravity waves
are the larger waves and are propagating “down” toward a region of shallow water. The capillary waves are the very short waves propagating in front of the larger waves. The presence of
two remarkably different waves is due to the non-monotonic speed vs. wavenumber relation of
gravity-capillary waves.
182
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.3. CAPILLARY WAVES
Figure 9.4: This image shows short wavelength capillary waves in front of ordinarily gravity waves on a
beach. The above photo was taken by Fabrice Neyret who kindly granted permission for its
use here.
9.3.1
“Pure” Capillary Waves
To describe the wave phenomena in an incompressible fluid we have derived the linearized equations (v (u, v, w))
 ∂v ∇p
+ ρ0 + g ρρ0 ∇z + 2Ω × v = 0,


∂t







 ∇ · v = 0,










∂ρ
∂t
2
− ρ0 Ng w = 0,
0
N 2 = − ρg0 dρ
,
dz
with the linear boundary conditions

 wz=0 =

∂ζ
,
∂t
pz=0 = gρ0 (0)ζ − σ∇2− ζ.
To consider a simple case we have assumed Ω = 0 (no rotation), ρ0 = const, ρ = 0 (the wave
motion does not cause any density variation). Under these conditions the linearised system of
183
9.3. CAPILLARY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
equations can be written in the form (see Equations 9.13, 9.14)

∂u

= − ∇ρ−0 p ,

∂t





 ∂w
∂p
+ ρ10 ∂z
= 0,
∂t




 ∇− u + ∂w
= 0.

∂z


∂
By applying to these equations the differential operators ∂t
and ∇− , ∇2− we obtain some intermediate equations (Equations 9.17 and 9.18)
 ∂ 2
∂2w
,
 ∂t ∇− u = −∇− ∂t∂z

2
∂ w
∇2− p = ρ0 ∂t∂z
which can be reduced by means of the continuity equation to one equation for vertical velocity
component w (Equation 9.19)
∂
∂ ∂ 2w
∂ 2
=0
∇ w ≡ ∇2− w +
∂t
∂t
∂t ∂z 2
with boundary conditions (Equations 9.7 and 9.8)

,
 wz=0 = ∂ζ
∂t

pz=0 = gρ0 (0)ζ − σ∇2− ζ.
In the case of gravity (surface) waves, it is the gravitational force that tends to restore the perturbed equilibrium of the free surface. Therefore we have solved the problem defined by Equations (9.19) and (9.20) in the limit σ = 0 resulting in the surface gravity waves, i.e. assuming
that the surface tension coefficient is negligible with respect to the gravitational force.
Now we consider the waves caused by a surface tension force. The term −σ∇2− ζ in the boundary condition (9.8) is now responsible for the restoring force. We again eliminate any other type
of wave motions assuming Ω = 0 (no rotation), ρ0 = const, ρ = 0 (the wave motion does not
cause the density variation) and reduce the linearised equations to the Equation (9.19) but the
boundary condition will be different. Namely, at the free surface (z = 0) we put now g = 0
in order to consider the “pure” capillary waves. So, the boundary condition (9.8) is reduced to
∂
pz=0 = −σ∇2− ζ. Then, we apply the operator ∂t
∇2− to this simplified boundary condition which
results in
∂ 2
∂ζ
∇− pz=0 = −σ∇4− ,
∂t
∂t
here
∂ζ
= wz=0 .
∂t
Using now the Equation (9.18) to express ∇2− p through w, we obtain
∂ 2
∂ 3w
.
∇− pz=0 = ρ0 2
∂t
∂t ∂z z=0
184
CHAPTER 9. SURFACE AND INTERNAL WAVES
Combination of last two expressions yields
 3
∂ w

 ∂t2 ∂z + γ∇4− w


γ≡
z=0
9.3. CAPILLARY WAVES
= 0,
(9.46)
σ
ρ0
So, the description of the “pure” capillary waves is reduced to the solution of Equation (9.19)
with boundary condition (9.46). In the most interesting and practical case, we have for capillary
waves the condition (kH) → ∞, i.e. the deep-water approximation. Therefore, instead of
boundary condition at the bottom, we require that the solution (9.19) vanishes at z → −∞, i.e.
wz→−∞ = 0.
(9.47)
So, the capillary wave problem, as the boundary value problem, is described by Equations (9.19),
(9.46) and (9.47). We will look for the solution of this problem in the form of a harmonic wave
w (x, y, z, t) = Φ (z) eikr−iωt ,
with k (kx , ky ) and r (x, y). The solution for Φ (z) is reduced to the solution of equation
d2 Φ
− k 2 Φ = 0,
dz 2
with Φ (z) = C1 ekz + C2 e−kz . Taking into account the boundary condition (9.47) we see that in
order to satisfy this condition we have to put C2 = 0. Let now C1 ≡ b, then Φ (z) = bekz and
w (x, y, z, t) = bekz eikr−iωt .
Substituting this solution into boundary condition (9.46) yields the dispersion relation for capillary waves
ω 2 = γk 3 .
(9.48)
The phase and group velocities are then given by

√
 cph = γk,

cg =
3
2
√
γk ≡ 32 cph .
To calculate the fluid particle velocities we can use expressions for the surface gravity waves in
the deep water approximation. Indeed, for surface gravity waves we have
w (x, y, z, t) = b
sinh [k (z + H)] ikr−iωt
ek(z+H) − e−k(z+H) ikr−iωt
e
=b
ekH≫1 ≈ bekz eikr−iωt .
sinh (kH)
ekH − e−kH
which coincides with the solution for capillary waves. Therefore we can use expressions for
surface gravity waves in the limit kH ≫ 1 to describe the properties of capillary waves. We thus
have
u = ib
p = ρ0
k
k cosh [k (z + H)] ikr−iωt
ekH≫1 ⇒ u = ib ekz eikr−iωt ,
k
sinh (kH)
k
ibω
cosh [k (z + H)] ikr−iωt
ω
ekH≫1 ⇒ p = iρ0 b ekz eikr−iωt .
k tanh (kH) cosh (kH)
k
185
9.3. CAPILLARY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
For the displacement of liquid particles caused by the surface gravity waves we obtained
k cosh [k (z + H)]
k
ξ=− A
sin (kr − ωt + α) ≡ − aξ sin (kr − ωt + α) ,
k
sinh(kH)
k
η=A
sinh [k (z + H)]
cos (kr − ωt + α) ≡ aη cos (kr − ωt + α) .
sinh(kH)
2
2
Excluding the time by calculating aξ 2 and aη2 and adding these expressions we find the equation
η
ξ
for the path of particles motion in a surface gravity waves
ξ 2 η2
= 1.
+
a2ξ a2η
So, a fluid particle is tracing an ellipse in the course of propagation of a surface gravity wave. In
the deep water approximation,

cosh[k(z+H)]
∼ kz

 aξ = A sinh(kH) kH≫1 = Ae ,

 a = A sinh[k(z+H)]
η
sinh(kH)
kH≫1
∼
= Aekz
we get relations for the path of fluid particles motion caused by capillary waves

 ξ = − kk Aekz sin (kr − ωt + α) ,

which yields
η = Aekz cos (kr − ωt + α)
ξ 2 η2
= 1.
+
a2ξ a2η
Since for capillary waves aξ = aη = Aekz , we obtain
ξ 2 + η 2 = A2 e2kz .
Thus, the paths of fluid particles in capillary waves are circles with a radii decreasing exponentially with depth.
9.3.2
Gravity-Capillary Surface Waves
We now consider the joint action of the gravity and surface tension forces in a homogeneous
(ρ0 = const) incompressible (c2 → ∞) and non-rotating (Ω = 0) fluid. Equations (9.13-9.16),
the boundary condition wz=−H = 0 (9.21), and solutions (9.22), (9.25) for w, and solutions
(9.29-9.32) for ξ, p, u in the form of harmonic waves still hold in this case. The boundary
condition at the free surface, however, will now be combination of conditions with (g 6= 0, σ = 0)
and (g = 0, σ 6= 0). Thus, at the free surface we have the boundary condition
p (z = 0) = gρ0 (0) ζ − σ∇2− ζ.
Applying the operator
∂
∇2
∂t −
to this condition yields
∂ 2
∂ζ
∂ζ
∇− p (z = 0) = gρ0 (0) ∇2−
−σ∇4−
.
∂t
∂t
∂t
|{z}
|{z}
w(z=0)
186
w(z=0)
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.3. CAPILLARY WAVES
2
∂ w
. So,
On the other hand, we have received for ∇2− p = ρ0 ∂t∂z
∂ 2
∂ 3w
∇− p (z = 0) = ρ0 2 (z = 0) .
∂t
∂t ∂z
Combination of the last two expressions results in the boundary condition at the free surface
3
∂ w
2
4
= 0.
− g∇− w + γ∇− w
∂t2 ∂z
z=0
So, the problem of the gravity-capillary surface gravity waves is now reduced to the solution of
the eigenvalue problem:
∂ 2
∇ w = 0,
∂t
with boundary conditions


 wz=−H = 0,

 ∂ 32 w − g∇2− w + γ∇4− w
= 0,
∂t ∂z
z=0
where we denote γ = σ/ρ0 . We look for the solution of these equations in the form of a periodic
wave
w (x, y, z, t) = Φ (z) eikr−iωt ,
with k (kx , ky ) and r (x, y). Then we obtain equation to determine Φ (z)
d2 Φ
− k 2 Φ = 0.
dz 2
The solution of this equation satisfying to the boundary condition at the bottom is given by
Φ (z) = b
sinh [k (z + H)]
sinh [k (z + H)] ikr−iωt
⇒ w (x, y, z, t) = b
e
.
sinh(kH)
sinh(kH)
Substituting this solution into boundary condition at the free surface yields the new dispersion
relation for the gravity-capillary surface waves
2 cosh [k (z + H)]
2 sinh [k (z + H)]
4 sinh [k (z + H)]
b −ω k
eikr−iωt = 0.
+ gk
+ γk
sinh(kH)
sinh(kH)
sinh(kH)
z=0
Since b 6= 0 we finally obtain
with phase velocity
ω 2 = gk + γk 3 tanh(kH),
c2ph =
ω2 g
=
+
γk
tanh(kH).
k2
k
(9.49)
(9.50)
Consider now these waves in the deep water approximation, i.e. kH ≫ 1. We then obtain from
(9.49) and (9.50) (tanh(kH)kH≫1 ≈ 1):
 2
 ω = gk + γk 3 ,
(9.51)
 2
g
cph = k + γk.
187
9.3. CAPILLARY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
For small k (long length limit) ω 2 ≈ gk, and the phase velocity decreasing with increasing k,
dcph
< 0. On contrary, for large k (short wave length limit) the second term is more important
dk
dc
and the phase velocity grows with k, dkph > 0 (see Figure 9.5). To find the minimum phase
dc
velocity, we calculate the derivative dkph :
12
− 21 g
dcph
1 g
cph =
− 2 +γ ,
+ γk ⇒
=
+ γk
k
dk
2 k
k
r
r
g
g
γ
dcph
= 0, if 2 = γ ⇒ k ≡ k0 =
⇒ λ0 = 2π
⇒
dk
k
γ
g
√
1
2 (γg) 4 .
cmin
ph =
g
(9.52)
The wave length λ0 separates the gravity and capillary waves. At λ ≫ λ0 the gravity force is
Figure 9.5: (a) Dispersion and (b) phase-velocity curves for gravity-capillary waves
dominant, whereas at λ ≪ λ0 the capillary force is the principal one.
Thus, we may conclude that the two limits of surface waves, namely the surface gravity waves
and the “purely” capillary waves (ripples) have been considered above by omission of the second term (σ = 0) and the first term (g = 0) respectively on the right hand side of the dynamic
boundary condition (9.8). It turns out that this separation can be treated quite precisely by considering the wave length of the surface perturbations. Indeed, using the harmonic representation
in Equation (9.8) we can introduce some critical wave length determined by the balance of the
gravity and the surface tension forces, i.e.
r
gρ0
2π
σ 3
k0 = gk0 ⇒ k0 =
≡
ρ0
σ
λ0
with
λ0 = 2π
r
188
σ
ρ0 g
(9.53)
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.3. CAPILLARY WAVES
For water, the critical wave length λ0 is about 1–7 cm. If the wave length is large compared
2
with λ0 , i.e. λ ≫ λ0 , then σk
≪ 1 and we can neglect the surface tension term in the dynamic
ρ0 g
boundary condition (9.8). The restoring forces which drive surface waves in this limit are due
to the weight of the fluid displaced and such waves are known as surface gravity waves. On the
other hand, if λ ≪ λ0 , the restoring force due to surface tension outweighs that due to gravity
and thus we can neglect the gravity term on the right hand side of Equation (9.8). This results in
the “purely” capillary waves.
9.3.3
Waves on a Liquid-Fluid Interface
We continue to investigate the joint action of the gravity and surface tension forces on interface
of two homogeneous, incompressible and non-rotating fluids. Suppose that a deep layer of stationary liquid of density ρ1 , with its upper surface in the horizontal plane z = 0, has a deep
layer of stationary fluid on top of it, of density ρ2 (< ρ1 ). When the interface is disturbed by
a monochromatic wave of vertical displacement, both fluids move. Repeating the calculations
for the gravity-capillary surface waves for each fluid (see the previous section), we find that
the problem of the gravity-capillary waves on the interface of two fluids is now reduced to the
solution of two eigenvalue problems for vertical velocities w1 and w2 :
∂ 2
∂ 2
∇ w1 = 0,
∇ w2 = 0
∂t
∂t
(9.54)
with boundary conditions in the deep water approximation
lim w1 = 0 lim w2 = 0
z→ − ∞
z →∞
(9.55)
We look for the solution of these equations in the form of a harmonic wave
w1,2 (x, y, z, t) = Φ1,2 (z) eikr−iωt
(9.56)
where k (kx , ky ) and r (x, y). Taking into account solutions (9.56) in equations (9.54) results in
the equations to determine Φ1,2 (z)
d 2 Φ2
d 2 Φ1
2
−
k
Φ
=
0,
− k 2 Φ2 = 0
1
dz 2
dz 2
The solutions of these equations satisfying boundary conditions (9.55) are given by
Φ1 (z) = b1 ekz ⇒ w1 (x, y, z, t) = b1 ekz eikr−iωt
Φ2 (z) = b2 e−kz ⇒ w2 (x, y, z, t) = b2 e−kz eikr−iωt
It is evident now that the kinematic and dynamic boundary conditions must be satisfied at the
, gives in a linear approximation equality of the
interface. The kinematic one, (w1,2 )z=ζ = dζ
dt
vertical velocities. Substituting (9.56) into linearised kinematic boundary condition we obtain
the relationship between the amplitude of waves at both sides of the boundary, b1 = b2 = b. So
that,
∂ζ
lim− w1 = lim+ w2 = beikr−iωt ≡ wz=0 =
z→0
z→0
∂t
To determine the dispersion relation for the gravity-capillary waves on the interface, we have to
use the dynamic boundary condition which describes the pressure difference across the interface
{(p1 − p0 ) − (p2 − p0 )}z=ζ = −σ∇2− ζ
189
(9.57)
9.4. INTERNAL GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
Here p1 and p2 is the total pressure in the corresponding fluid, p1,2 = p0 (z) + p′1,2 , p0 is the nondisturbed pressure at the interface, σ is now the interfacial surface tension, and ζ ≡ ζ (t, x, y) is
the vertical displacement of the surface particle. The boundary condition (9.57) is nonlinear. We
have to linearise it, as above (see Section 9.1), assuming the displacement of the surface ζ to be
small with respect to the equilibrium position z = 0. Then the condition (9.57) takes the form
{(p′1 − gρ1 ζ1 ) − (p′2 − gρ2 ζ2 )}z=0 = − σ∇2− ζ z=0
(9.58)
∂
∇2− to this equation. The result yields
We apply the operator ∂t

 





 ∂




∂ζ
∂
∂ζ
1
2
 −  ∇2− p′2 − gρ2 ∇2−

 ∇2− p′1 − gρ1 ∇2−
 ∂t

∂t   ∂t
∂t 


|{z}
|{z}


w1
w2

z=0

∂ζ 

= −σ ∇4−

∂t
|{z}
w
z=0
To express the perturbed pressure p through the vertical velocity component w we have received
∂2w
(see relation 9.17). Using this relation for the lower and upper fluids
previously for ∇2− p = ρ ∂t∂z
results in the dynamic boundary condition at the interface of two deep fluids
∂ 3 w1
∂ 3 w2
2
2
ρ1 2 − gρ1 ∇− w1
(9.59)
− ρ2 2 − gρ2 ∇− w2
= −σ ∇4− w z=0
∂t ∂z
∂t ∂z
z→0−
z→0+
Substituting now the solutions (9.56) into the dynamic boundary condition (9.59) yields the new
dispersion relation for the gravity-capillary waves at the interface of two deep fluids
ei(kr−ωt) =
b ρ1 −ω 2 k − gρ1 −k 2 − ρ2 −ω 2 (−k) − gρ2 −k 2
z=0
−σ bk 4 ei(kr−ωt)
Since b 6= 0 we finally obtain
ω2 =
z=0
σk 3
ρ1 − ρ2
gk +
ρ1 + ρ 2
ρ 1 + ρ2
(9.60)
This result will be used later in the discussion related to the stability of the liquid-fluid interface.
It is natural enough that the obtained relation (9.60) is reduced to (9.51) if ρ2 = 0 (the interface
between liquid and vacuum) and to (9.40) if ρ2 = 0 and σ = 0 (surface gravity waves in the deep
water approximation).
9.4
Internal Gravity Waves
A characteristic of internal waves is that the vertical particle displacements reach their maximum
not at the free surface (surface gravity waves) but inside the fluid, for example, at the interface
of two fluids with different densities. In the sea, an analogous case takes place when low salinity
water is above heavier water with greater salinity, so called “dead water”. For example, if a ship
is moving in such water it spends a considerable amount of its power for internal-wave generation and as a result its speed diminishes.
Internal waves, however, occur not only in the case of discontinuous variation of the density
along the vertical, but also when the equilibrium density ρ0 (z) is continuous function of vertical
190
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.4. INTERNAL GRAVITY WAVES
coordinate. Such waves are due to an inhomogeneity of the fluid caused by the gravitational
field. The pressure (and therefore the entropy) necessarily varies with height; hence any displacement of a fluid particle in height destroys the mechanical equilibrium, and consequently
causes an oscillatory motion. For, since the motion is adiabatic, the particle carries with it to its
new position its old entropy, which is not the same as the equilibrium value at the new position.
This is practically always the case in the ocean and atmosphere. See Figure 9.6 for an example
of internal gravity waves generated by a tidal flow at the Straight of Gibraltar. Note, the surface
elevation varies by only about 20cm, but the amplitude of the waves well below the surface can
be many tens of meters!
Figure 9.6: Photo of internal gravity waves generated by a tidal flow at the Straight of Gibraltar taken
from the Space Shuttle. The above image was found on the NASA JSC Digital Image site
and is free for noncommercial use.
9.4.1
Basic Equations. Boussinesq Approximation
We start from the linear hydrodynamic equations (9.9-9.11) of an incompressible and nonrotating fluid (Ω = 0). Note that ρ0 (z) and N 2 are functions only the vertical coordinate z.
191
9.4. INTERNAL GRAVITY WAVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
Let v (u, w), so equations (9.13-9.14) can then be written in the form:

∂u

+ ∇ρ−0 p = 0,

∂t






gρ
1 ∂p
∂w

 ∂t + ρ0 ∂z + ρ0 = 0,
(9.61)



∇− u = − ∂w
,

∂z





 1 ∂ρ = N 2 w.
ρ0 ∂t
g
We now reduce these Equations to one equation for the vertical component of the fluid velocity
w. To this end, we first apply the operator ∇− to the momentum equation for u and the operator
∂/ to the continuity equation:
∂t
∇2 p
∂
∂
∂ 2w
∇− u = − − ⇔ ∇− u = −
.
∂t
ρ0
∂t
∂t∂z
This results in the relation
∂ 2w
.
∂t∂z
∂
∇2− to the momentum equation for w:
Second, we apply the operator ∂t
∇2− p = ρ0
(9.62)
∂2 2
∂ρ
1 ∂2 2
g
∇
w
+
∇− p + ∇2−
= 0.
−
2
∂t
ρ0 ∂z∂t
ρ0
∂t
Taking into account (9.61) and (9.62) in this equation yields
1 ∂2
∂ 2w
∂2 2
∇− w +
ρ0
+ ∇2− N 2 w = 0.
2
∂t
ρ0 ∂z∂t ∂z∂t
It is convenient now to rewrite the equation for the vertical velocity component w in the form


∂2
∂t2




2
1
∂
w
∂ρ
∂w
 2

0
+
∇− w +
 + ∇2− N 2 w = 0
2

∂z
ρ0 ∂z ∂z 
{z
}
|

2
≈∂ w/∂z 2
(9.63)
Very often, for example in the theory of ocean waves, a simplified equation
used;the term
is
2
proportional to ∂w/∂z in Equation (9.63) is neglected as compared with ∂ w/∂z 2 , i.e. the
equation for the vertical component of the fluid velocity (9.63) is reduced to
∂2 2
∇ w + N 2 ∇2− w = 0
2
∂t
(9.64)
with N 2 ≡ N 2 (z). The approximation which corresponds to the assumption that ρ0 (z) is constant in Equation (9.63) everywhere except the expression for the N 2 (z), the Brunt-Väisälä
frequency, is called the Boussinesq approximation. Physically it corresponds to the assumption that the wavelength is small in comparison with distances over which the gravitational field
causes a marked change in density. Indeed, the density and the pressure gradients are related
192
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.5. SUMMARY OBJECTIVES
∂p
∇ρ = c2 ∇ρ, where c is the speed of sound in the fluid. Taking into account the
by ∇p = ∂ρ
hydrostatic equation ∇p = ρg yields ∇ρ = cρ2 g ⇒ ∇ρ
= cg2 . Therefore, the density of the fluid
ρ
2
in the gravitational field varies considerably over distances L ∼ cg . For air and water, L ∼10 km
and 200 km respectively. We thus may conclude that in order the Boussinesq approximation to
be valid in the treatment of internal gravity waves, the wave length of these waves λ must satisfy
2
condition λ ≪ L ∼ cg . It also means that for such wave we can neglect the change in the fluid
density caused by the pressure change in the wave.
9.4.2
Waves in Unlimited Media
The simplest form of internal waves propagating in an unlimited medium can be obtained in the
case N 2 = const. For this case we consider harmonic plane internal waves of the form
w (r, z, t) = beiκR−iωt ≡ bei(kr+kz z−ωt)
(9.65)
where κ = {k, kz } ≡ {kx , ky , kz } is the wave vector, ω is the frequency, b is the amplitude, and
R = {r, z} ≡ {x, y, z}. Substituting (9.65) into (9.64) yields the relationship between κ and ω,
the dispersion relation for internal waves
−ω 2 −κ2 + N 2 −k 2 bei(kr+kz z−ωt) = 0 ⇒
ω2 = N 2
2
k2
2k
≡
N
≡ N 2 sin2 θ
k 2 + kz2
κ2
(9.66)
where k = κ cos (90◦ − θ) = κ sin θ. We thus conclude that
1. only waves with ω < N can occur;
2. if θ is fixed, the wave frequency ω is uniquely defined by (9.66), but the wavelength and
the phase velocity of the wave can be arbitrary.
9.5
Summary objectives
When you finish studying the Chapter 9, you should be able to do the following:
1. Define condition when the moving particle execute a periodic motion and describe waves
as oscillations propagating in a continuous media.
2. Define the linear approximation in the wave theory.
3. Describe the process of linearisation and find the linearised set of hydrodynamic equations.
4. Determined the linearised boundary conditions.
5. Write the set of linearised hydrodynamic equations for an incompressible fluid.
6. Describe the physical picture of excitation and propagation of the surface gravity waves.
7. Repeat all steps of calculations to reduce the set of hydrodynamical equations to one equation for the vertical component of the fluid velocity.
193
9.5. SUMMARY OBJECTIVES
CHAPTER 9. SURFACE AND INTERNAL WAVES
8. Consider the plane harmonic waves and define the main parameters describing these waves.
9. Determine the dispersive properties of harmonic waves.
10. Look for a harmonic solution of surface gravity waves. Explain from the physical points of
view why the dependence on the vertical coordinate in this solution is not specified. Show
how this dependence is determined.
11. Find the dispersion relation for the surface gravity waves.
12. Determine the pressure variation, the horizontal components of the fluid velocity and displacement of fluid particles in the surface gravity wave.
13. Show that the motion of fluid particles in the surface gravity wave is potential.
14. Consider the shallow and deep water approximation, explain the physics of these two
extreme case and find the relations describing the properties of waves in these two approximations.
15. Calculate the kinetic and potential energy of a progressive surface gravity wave.
16. Consider the “pure” capillary waves and explain the physics of excitation of these waves.
What is the difference in the boundary conditions between surface gravity waves and capillary waves?
17. Repeat the whole set of calculations to describe the properties of capillary waves: dispersion relation, velocity components, pressure variation, and the path of fluid particles
motion caused by these waves. Explain the deep water approximation used in these relations.
18. Consider the gravity-capillary surface waves and explain how the physics of excitation of
these waves is reflected in the boundary conditions. Determine the dispersion relation for
these waves in general and in the deep water approximations.
19. Find the minimum phase velocity for these waves and determine the critical wave length
which separates two different limits corresponding to the gravity and capillary surface
waves. Explain the obtained results from the used boundary condition points of view.
20. Consider the waves on a liquid-fluid interface. What is the principal difference between
these waves and gravity-capillary waves?
21. Write the dynamic boundary condition and determine the dispersion relation for waves
propagating along liquid-fluid surface. Perform the analysis of this relation in different
limits.
22. Explain the physics of the internal gravity waves. Derive the basic equation for the vertical
fluid velocity component describing the propagation of these waves.
23. Define the Boussinesq approximation to describe the internal gravity waves and indicate
conditions for this approximation to be valid.
24. Consider the waves in the unlimited media and find the dispersion relation for these waves.
194
Chapter 10
Stability of Fluid Flows
For any problem of viscous flow under given steady conditions there must in principle exist
an exact steady solution of the equations of fluid dynamics. These solutions formally exist for
all Reynolds numbers. Yet not every solution of the equations of motion, even if it is exact, can
actually occur in the nature. Those which do must not only obey the equations of fluid dynamics,
but also be stable. Any small perturbations which arise must decrease in the course of time. If, on
the contrary, the small perturbations which inevitably occur in the flow tend to increase with time,
the flow is unstable and can not actually exist. This chapter aims at to present an introduction to
linear stability theory of fluid flows. The mathematical complexity of even this simplest type of
theory is too great to treat any particular case in full. Therefore an emphasis is placed here on the
physical insight on the dynamics of the destabilising process which is very valuable and can be
demonstrated by considering highly simplified systems. Then, after a general discussion of how
stability principles are extended for full fluid dynamical situation, we consider two examples of
an unstable flow, which illustrate the ideas outlined above in a relatively straightforward way.
The first one concerns the so-called shear flow instability known also as the Kelvin-Helmholtz
instability. The second describes the instability of the interface between two fluids of different
densities in which the surface tension plays a key role. This instability is known as the RayleighTaylor instability.
10.1 Stability and Instability
It is well known from the Mechanics that a dynamical system which is in equilibrium may be
stable or unstable. The simplest case of the distinction is that of a particle of mass m which
can move only in one dimension, in circumstances where the particle’s potential energy Φ varies
with its position x in the manner suggested by Figure 10.1.
The particle experiences no forces when it is situated at the minimum, P , or at maximum, Q,
and in principle it can remain at rest indefinitely in either of these positions. However, if it is
slightly displaced from P it accelerates towards P , whereas if it is slightly displaced from Q it
accelerates away from Q; in the first position the equilibrium is stable and in the second it is
can normally be expanded as a
unstable. Near any minimum such as P the restoring force ∂Φ
∂x
Taylor series in powers of displacement ξP = x − xP . Since it is zero at P itself, an adequate
approximation for small value of ξP is
∂Φ
≈
∂x
∂ 2Φ
∂x2
195
ξP
P
10.1. STABILITY AND INSTABILITY
CHAPTER 10. STABILITY OF FLUID FLOWS
Figure 10.1: Position of stable, P , and unstable, Q, equilibrium for a particle whose potential energy
varies with x in a manner shown.
In this case the equation of motion of the particle is linear in ξP and it is given by
∂ 2 ξP
m 2 =−
∂t
∂ 2Φ
∂x2
ξP
P
The oscillations which it describes are then simple harmonic, with angular frequency ωP such
that
1 ∂ 2Φ
2
ωP =
m ∂x2 P
So that, when a particle leaves the position of a stable equilibrium it results in the harmonic
oscillations of a particle around this equilibrium.
An equation
2 of motion similar to that presented above applies in the neighbourhood of Q, but
since ∂ Φ/∂x2
is negative the roots for ω are necessarily imaginary, ωQ = ±isQ with sQ
Q
real. Hence the displacement ξQ = x − xQ of a particle which starts at rest at t = 0 from a
position such that ξQ = ξ0 is given at later times by
1
ξQ = ξ0 esQ t + e−sQ t
2
as long as it remains small. If ξ0 is infinitesimal, then by the time the displacement becomes very
much greater than unity due to the term exp (sQ t) in which case exp (−sQ t) must be negligible.
Thus, when a particle leaves a position of unstable equilibrium its displacement normally grows
in an exponential fashion.
2
Suppose now that ∂Φ/∂x is necessarily always zero, while ∂ Φ/∂x2
can be reduced in
P
P
magnitude and ultimately reversed in sign by altering the external constraints which determine
Φ. In that case P is always an equilibrium position, but the equilibrium is stable in one range of
196
CHAPTER 10. STABILITY OF FLUID FLOWS
10.2. STABILITY OF STEADY FLOW
the constraints and unstable in an adjacent range. Where the changeover occurs one has
1 ∂ 2Φ
2
ωP =
=0
m ∂x2 P
This is the condition for what is called marginal stability.
10.2 Stability of Steady Flow
To investigate the stability of steady flow, we can use a simple method used in Sec. 4.2.2 to derive the condition for mechanical stability of a fluid at rest in a gravitational field. The principle
of the method is to consider any small element of the fluid and to suppose that this element is
displaced from the path which it follows in the steady flow concerned. As a result of this displacement, forces appear which act on the displaced element. If these forces tend to return the
element to its original position, the flow is stable. If, on the other hand, the displacement of the
fluid element, once having arisen, will increase, the steady flow under investigation is unstable.
The underlying notion is that the transition from one type of flow to another one results from
spontaneous amplification of disturbances present in the original flow. One would then expect
the occurrence of the transition to depend on the intensity and structure of disturbances present,
and this is frequently found to be the case. Clearly, a theory covering all possibilities would
be unmanageably complex; moreover, comparison with experiment would be difficult since the
nature of the disturbances is often not in complete experimental control. Therefore, we confine
our attention to so called linear stability theory.
Linear stability theory adopts a less ambitions objective: to ascertain when a flow is unstable
to infinitesimal disturbances (or perturbations). It thus should indicate conditions in which the
flow can not remain in its postulated form and must undergo transition to another type of motion.
Moreover, infinitesimal disturbances are always present, even in the most carefully controlled experiment, and if these tend to be amplified, the flow will necessarily break down.
The linear stability theory for a particular flow starts with a solution of equations of motion
representing the flow. One then considers this solution with a small perturbation superimposed,
and inquires whether this perturbation grows or decays as time passes. In this analysis, all terms
involving the square of the perturbation amplitude are neglected; it is this linearisation that limits
the theory to infinitesimal disturbances. On the other hand, the linearisation gives a means of
allowing for the many different forms that the disturbances can take. Any pattern may be Fourier
analysed spatially. Because of linearity, there are no interactions between different Fourier harmonics. The equations may thus be broken down into separate sets of equations for each Fourier
component, indicating the stability or instability with respect to that component.
The mathematical investigation of the stability of a given flow with respect infinitely small perturbations will proceed as follows. On the steady solution concerned (whose velocity distribution
is v 0 (r), say) we superpose a non-steady small perturbation v 1 (r, t), which must be such that
the resulting velocity v = v 0 + v 1 satisfies the equations of motion. The equation for v 1 is
obtained by substituting in the Navier-Stokes and continuity equations
∇p
∂v
+ (v · ∇) v = −
+ ν∇2 v, ∇ · v = 0
∂t
ρ
197
(10.1)
10.3. KELVIN-HELMHOLTZ
CHAPTER 10. STABILITY OF FLUID FLOWS
the velocity and the pressure
v = v 0 + v 1 , p = p0 + p1
(10.2)
The known functions v 0 and p0 that describe the steady flow satisfy the unperturbed equations
(v 0 · ∇) v 0 = −
∇p0
+ ν∇2 v 0 , ∇ · v 0 = 0
ρ
(10.3)
Omitting terms above the first order in v 1 , we obtain
∇p1
∂v 1
+ (v 0 · ∇) v 1 + (v 1 · ∇) v 0 = −
+ ν∇2 v 1 , ∇ · v 1 = 0
∂t
ρ
(10.4)
The boundary condition is that v 1 vanishes on fixed solid surfaces.
Thus v 1 satisfies a system of homogeneous linear differential equations with coefficients that
are functions of the coordinates only, and not of the time. The general solution of such equations
can be represented as a sum of particular solutions in which v 1 depends on time as e−iωt . The
frequencies ω of the perturbations are not arbitrary, but are determined by solving the equations
(10.4) with the appropriate boundary conditions. The frequencies are in general complex. If
there are ω whose imaginary parts are positive, e−iωt will increase indefinitely with time. In
other words, such perturbations, once having arisen, will increase, i.e. the flow is unstable with
respect to such perturbations. For the flow to be stable it is necessary that the imaginary part of
any possible frequency ω be negative. The perturbations that arise will then decrease exponentially with time.
Such a mathematical investigation of stability is extremely complicated, however. The theoretical problem of the stability of steady flow past bodies with finite dimensions has not yet been
solved. It is certain that steady flow is stable for sufficiently small Reynolds numbers. The experimental data seem to indicate that, when Re increases, it is eventually reaches a value Recr
(the critical Reynolds number) beyond which the flow is unstable with respect to infinitesimal
disturbances. For sufficiently large Reynolds numbers (Re ≫ Recr ), steady flow past solid bodies is therefore impossible. The critical Reynolds number is not, of course, a universal constant,
but takes a different value for each type of flow.
10.3 Kelvin-Helmholtz Instability
A variety of the most important cases of the instability of fluids in motion fall into the general
category of the instability of shear flows. A shear flow is one in which the velocity varies principally in a direction at right angles to the flow direction. The simplest example of this is a
flow with a finite discontinuity in the velocity. The fact that this is subject to instability — the
Kelvin-Helmholtz instability — may be stated as follows:
Flows in which two layers of incompressible fluid move relative to each other, one “sliding”
on the other, are unstable if the fluid is ideal. The surface of separation between these two fluid
layers would be a surface of tangential discontinuity, on which the fluid velocity tangential to
the surface is discontinuous.
This instability is also known as instability of tangential discontinuities.
198
CHAPTER 10. STABILITY OF FLUID FLOWS
10.3. KELVIN-HELMHOLTZ
To prove this statement, we consider a small portion of the surface of discontinuity and the
flow near it. We may regard this portion as plane, and the fluid velocities v 1 and v 2 on each side
of it as constant. Without loss of generality we can suppose that one of these velocities is zero;
this can always be achieved by a suitable choice of the coordinate system. Let v 2 = 0, and v 1 be
denoted by v simply; we take the direction of v as the x-axis, and the z-axis along the normal
to the surface. Let now the surface of discontinuity receive a slight perturbation, in which all
quantities – the coordinates of points on the surface, the pressure, and the fluid velocity – are
periodic functions, proportional to exp (ikx − iωt). The goal of the analysis presented below is
to find the relation between ω and k. First, we consider the fluid on the side where its velocity
is v, and denote by v ′ the small change in the velocity due to the perturbation. According to the
equations (1.4) (with constant v = v 0 and ν = 0), we obtain the following system of equations
for the perturbation v ′ :
∂v ′
∇p′
∇ · v ′ = 0,
+ (v 0 · ∇) v ′ = −
∂t
ρ
Since v = v 0 is along the x-axis, the second equation can be written as
∂v ′
∇p′
∂v ′
+ v0
=−
∂t
∂x
ρ
(10.5)
If we take the divergence of both sides, then the left-hand side gives zero by virtue of ∇ · v ′ = 0,
so that p′ must satisfy Laplace’s equation:
∇2 p′ = 0
(10.6)
p′ = f (z) ei(kx−ωt)
(10.7)
We seek p′ in the form
Substituting into Equation (10.6) yields for the function f (z) the equation
d2 f
− k2f = 0
dz 2
The partial solution of this equation is given by f = const × exp (±kz). Suppose that the space
on the side under consideration (side 1) corresponds to positive values of z. Then we must take
f = const × exp (−kz), so that
p′1 = const × ei(kx−ωt) e−kz
(10.8)
To determine the more exact form of the solution (10.8), we consider z-component of equation
(10.5). Taking into account that v ′ ≡ (u, v, w) yields
∂w
∂w
1 ∂p′1
+ v0
=−
∂t
∂x
ρ ∂z
(10.9)
The perturbed velocity component w may also be sought in the form proportional to the same
exponential factors as in the solution (10.8), i.e.
w = const × ei(kx−ωt) e−kz
Substituting this expression into Equation (10.9), we find
w=−
kp′1
i (ω − kv0 ) ρ1
199
(10.10)
10.3. KELVIN-HELMHOLTZ
CHAPTER 10. STABILITY OF FLUID FLOWS
Let ζ = ζ (x, t) be the displacement in the z-direction of points on the surface of discontinuity
due to the perturbation. The derivative ∂ζ
is the rate of change of the surface coordinate ζ for a
∂t
given value of x. Since the fluid velocity component normal to surface of discontinuity is equal
to the rate of displacement of the surface itself, we have to the necessary approximation
ζ=ζ (x′ , t) , x′ = x − v0 t ⇒
∂ζ
∂ζ dx′
∂ζ
∂ζ
∂ζ
=
+ ′
⇒
= w − v0
∂t
∂t ∂x dt
∂t
∂x
Since the displacement ζ is assumed to be small (we consider the linear theory), the value of w ,
of course, must be taken on the surface of discontinuity, i.e. at z = 0. This finally yields
∂ζ
∂ζ
= wz=0 − v0
∂t
∂x
(10.11)
The displacement ζ may also be sought in a form proportional to the exponential factor ei(kx−ωt) ,
and we obtain from (10.11)
wz=0 = −iζ (ω − kv0 )
(10.12)
This gives together with (10.10) the perturbed pressure in the space on the side 1
p′1 = −ζρ1
(ω − kv0 )2
k
(10.13)
where the value of p′1 is taken on the surface of discontinuity, i.e. at z = 0.
The perturbed pressure p′2 on the other side of the surface of discontinuity is given by a similar formula, where now v0 = 0 and the sign is changed (since in this region z < 0, and all
quantities must be proportional to ekz , not e−kz ). Thus
p′2 = ζρ2
ω2
k
(10.14)
where the value of p′2 should also be taken at the surface of discontinuity, i.e. at z = 0, but
approaching to the surface from the region z < 0. In both expressions (10.13) and (10.14), we
have written different densities ρ1 and ρ2 in order to include the case where we have a boundary
separating two different immiscible fluids. Finally, from the condition that the pressures p′1 and
p′2 be equal on the surface of discontinuity, we obtain
ρ1 (ω − kv0 )2 = −ρ2 ω 2
This yields the desired relation between ω and k
p
ρ1 ± i (ρ1 ρ2 )
ω = kv0
ρ1 + ρ 2
(10.15)
We see that ω is complex, and there are always ω having a positive imaginary part. Thus,
tangential discontinuities are unstable, even with respect to infinitely small perturbations.
We note also that the frequency given by (10.15) has both real part and the imaginary part.
So that in the case of Kelvin-Helmholtz instability perturbations both propagate and grow in
amplitude with the frequency
ρ1
Ω = Re(ω) = kv0
ρ 1 + ρ2
200
CHAPTER 10. STABILITY OF FLUID FLOWS
and growth rate
10.4. RAYLEIGH-TAYLOR
p
(ρ1 ρ2 )
γ = Im(ω) = kv0
ρ1 + ρ2
When finite viscosity is taken into account, the tangential discontinuity is no longer sharp; the
velocity changes from one value to another one across a layer with finite thickness. The problem
of the stability of such a flow is mathematically entirely similar to that of the stability of flow
in a laminar boundary layer with a point of inflexion in the velocity profile. In this case, as
experimental and numerical results indicate, the instability sets in very soon, and perhaps is
always present.
10.4 Rayleigh-Taylor Instability
The Rayleigh-Taylor instability arises when a vessel which contains two fluids separated by a
horizontal interface – at least one of the fluids must of course be a liquid – is suddenly inverted
so that the heavier fluid lies above the lighter one, as illustrated by Figure 10.2.
Figure 10.2: Development of the Rayleigh-Taylor instability. A layer of heavier fluid, blue, is above the
lighter one, yellow.
The gravitational potential energy of the system, which was at its minimum value before inversion, is now at its maximum, and although the system is still in equilibrium while the interface
remains horizontal, the equilibrium is clearly liable to be unstable. Whether or not it is actually
unstable with respect to any particular perturbations depends upon whether the gravitational energy which this releases is greater or less than the increase in surface free energy. The system is
marginally stable with respect to the perturbations when the two are equal.
In order to demonstrate the main features of the Rayleigh-Taylor instability we perform the
calculations in a simplest case corresponding to the large vessel containing two fluids, so large
that the boundary conditions imposed from the vessel have no limitations on results. We assume
also that this vessel is possible to invert almost instantaneously.
The statement of the problem can be formulated then in the following way:
Suppose that a deep layer of light liquid of constant density ρ1 , with its upper surface in the plane
201
10.4. RAYLEIGH-TAYLOR
CHAPTER 10. STABILITY OF FLUID FLOWS
z = 0, has a deep layer of heavy fluid on top of it, of density ρ2 (> ρ1 ). Assuming the system to
be in the equilibrium state and fluids are in the rest, investigate the stability of this equilibrium.
Solution.
Equilibrium state. Two stationary fluids separated by the interface in a vertically downward
gravitational field of acceleration g will be in an equilibrium state if the gravity force is balanced
by the surface tension force. In this state the fluids remain still and the interface is a curvilinear
surface located at z = ξ (x, y) (with the z axis taken vertically upward) with principal curvatures
given by (9.6). In a view of the Laplace’s formula (see (9.5)), the fluid pressure will be different
on both sides of boundary. The difference between them is called the surface pressure.
Perturbed state. Now disturb the system slightly. To this end we suppose that the interface
receives a small vertical displacement ζ (r, t), with r (x, y), which results in the motion of both
fluids. We assume that ζ is small everywhere, i.e. that the surface deviates only slightly from the
plane z = 0. Similar to the previous example (see Sec. 10.3), we consider a small portion of this
plane and the flow near it. To describe the reaction of each fluid on this small perturbation we
(0)
(0)
use the set of linearized Equations (10.4) that for the given problem with v 1 = v 2 = 0 and
ν1 = ν2 = 0 can be written as
∂vi ∇pi
+
= 0,
∂t
ρi
∇ · v i = 0,
i = 1, 2
(10.16)
Here v i (u, w) is the perturbed velocity of the fluid element, pi is the perturbed pressure and
ρi = const. We note that Equations (10.16) are linear equations with constant coefficients and
are similar to (9.13) and (9.14) used to describe the surface gravity waves. We next derive,
by elimination, an equation for the vertical component of the fluid velocity wi (x, y, z, t) alone.
To this goal we write (10.16) for the horizontal ui (x, y, z, t) and vertical wi (x, y, z, t) velocity
components
∂ui
∇− pi ∂wi
1 ∂pi
=−
,
=−
(10.17)
∂t
ρi
∂t
ρi ∂z
∂wi
∇ − · ui +
=0
(10.18)
∂z
∂
∂
where ∇− is the horizontal differential operator ∇− = ex ∂x
+ ey ∂y
. As the next step, we apply
∂
twice the operator ∇− to the momentum equation and the operator ∂t to the continuity equation.
Rearranging slightly the obtained equations results in the following relations
∇2− pi
∂ 2 wi
∂ 2 wi ∂ 2
,
∇ ui = −∇−
= ρi
∂t∂z ∂t −
∂t∂z
(10.19)
Now, in order to describe totally the reaction of fluids on the small perturbation we need the
equation for wi (x, y, z, t). Applying the Laplace operator ∇2 to the second equation of (1.1)
yields
∂ 2
∇ wi = 0
(10.20)
∂t
Thus, the description of a perturbed motion of each fluid is reduced to the solution of the equation for the vertical component of the velocity, Equation (10.19), with boundary conditions corresponding to the deep water approximation
lim w1 = 0,
z→−∞
202
lim w2 = 0
z→∞
(10.21)
CHAPTER 10. STABILITY OF FLUID FLOWS
10.4. RAYLEIGH-TAYLOR
These results can easily be obtained by repeating all steps of calculations with (10.16) on the
both sides of the interface similar to that performed in Equations (9.15)–(9.19). Following now
the general approach accepted in the linear stability theory, we assume that all quantities in a
perturbed motion of each fluid are periodic functions, proportional to exp (ikr − iωt) with k
and ω being non-defined parameters yet. Then the goals of the analysis below are to find the
relation between these parameters, i.e. ω = ω(k), and to determine conditions when ω 2 < 0, i.e.
the equilibrium under consideration is unstable.
Now, derivatives with respect to x and y enter Equation (1.4) only in the combination
(∂ 2 /∂x2 +∂ 2 /∂y 2 ). Thus, there is no preferred horizontal direction in the problem. This permits
separable solutions of the form
wi (x, y, z, t) = Φi (z) eikr−iωt
(10.22)
where k (kx , ky ) and r (x, y), with the amplitude Φi (z) defined by
d 2 Φ1
− k 2 Φ1 = 0
dz 2
,
d 2 Φ2
− k 2 Φ2 = 0
dz 2
Then in view of relations (10.21) and (10.22) it follows that
Φ1 (z) = b1 ekz ⇒ w1 (x, y, z, t) = b1 ekz eikr−iωt
(10.23)
Φ2 (z) = b2 e−kz ⇒ w2 (x, y, z, t) = b2 e−kz eikr−iωt
(10.24)
The solutions (10.22) together with relations (10.19) formally describe the perturbed motion of
both fluids, but amplitude of waves b1 and b2 , as well as wave vector k and frequency ω are still
not determined.
We have not used yet the fact that both fluids are separated by the interface. On the interface
z = ζ (x, y), the solutions (10.22) must satisfy the kinematic and dynamic boundary conditions.
Since the displacement is small, the amplitude of the wave oscillations is also small. Hence, we
can linearized these conditions, i.e. take into account the terms which are linear with respect to
ζ only, and take the derivative at z = 0 instead of z = ζ. Then the kinematic boundary condition
takes the form
dζ
∂ζ
(w1,2 )z=ζ =
⇒
(10.25)
(w1,2 )z=0 =
|{z}
dt
∂t
linearization
Substituting now solutions (10.22) into the kinematic boundary conditions (1.9) yields the relationship between amplitude of waves at both sides of the interface, b1 = b2 ≡ b. Thus,
lim− w1 = lim+ w2 = beikr−iωt ≡ wz=0 =
z→0
z→0
∂ζ
∂t
(10.26)
The dynamic boundary condition relates the pressures in the two fluids at the surface. Now, in
contrast to the surface gravity waves, the pressure difference between two sides of the interface is
not zero, but is given by Laplace’s formula. Since ζ is small, we can write the Laplace’s formula
as
(p1 − p2 )z=ζ = −σ∇2 ζ
⇒
|{z}
linearization
p1 +
dp1
ζ z=0−
dz
− p2 +
dp2
ζ z=0+
dz
203
(10.27)
= −σ (∇2 ζ)z=0
10.4. RAYLEIGH-TAYLOR
CHAPTER 10. STABILITY OF FLUID FLOWS
For (dp/dz) we substitute, according to (1.12), (dp/dz) = −ρg, and apply the operator
the obtained equation. Then the dynamic boundary condition (10.27) becomes
∂ 2
∂
2 ∂ζ1
2
2 ∂ζ2
∇
p
−
gρ
∇
∇
p
−
gρ
∇
−
1
1
2
2
−
− ∂t z=0
− ∂t z=0+ =
∂t −
∂t −
−σ∇4−
∂
∇2
∂t −
to
(10.28)
∂ζ
∂t z=0
Finally, in view of relations (10.19) and (10.25), the dynamic boundary condition (10.28) takes
the form
∂ 3 w1
∂ 3 w2
2
2
ρ1 2 − gρ1 ∇− w1
= −σ ∇4− w z=0 (10.29)
− ρ2 2 − gρ2 ∇− w2
∂t ∂z
∂t ∂z
z=0−
z→0+
The solutions (10.22) will fulfill the boundary condition (10.29) if
b {[ρ1 (−ω 2 ) k − gρ1 (−k 2 )] − [ρ2 (−ω 2 ) (−k) − gρ2 (−k 2 )]}z=0 ei(kr−ωt) =
−σ bk 4 ei(kr−ωt)
z=0
Since b 6= 0, we finally obtain the relation, so called dispersion relation, between ω and k:
ω2 = −
ρ2 − ρ1
σk 3
gk +
ρ1 + ρ2
ρ1 + ρ2
(10.30)
It is interesting to compare this result with the dispersion relation for the gravity-capillary waves
at the interface of two deep fluids given by (9.60). Since in our case ρ1 < ρ2 (the heavy fluid is
above the light one), the equilibrium is unstable, i.e. ω 2 < 0, if the wave length of perturbations
is such that
r
r
(ρ2 − ρ1 ) g
σ
2
⇒ λ ≫ 2π
≡ λ∗
(10.31)
(ρ2 − ρ1 ) g ≫ σk ⇒ k ≪
σ
(ρ2 − ρ1 ) g
It is seen from (10.30) that the marginal stability (ω 2 = 0) is only possible for one wave vector
k ∗ , such that
r
((ρ2 − ρ1 ) g)
2π
∗ 2
∗
≡ ∗
(10.32)
(ρ2 − ρ1 ) g = σ (k ) ⇒ k =
σ
λ
So, the modes for which k < k ∗ (λ > λ∗ ) grow in amplitude. The value of k, say kmax , which
maximises the rate of growth is clearly such that
r
∗
(ρ2 − ρ1 ) g
k
2
(10.33)
(ρ2 − ρ1 ) g = 3σkmax
⇒ kmax = √ ≡
3σ
3
We may conclude from the above results that if it is possible to invert almost instantaneously a
large vessel containing two fluids, so large that the boundary conditions imposed virtually no
limitations on the allowed values of k, the contents would be inherently unstable. The interface
would inevitably develop
corrugations whose periodicity would be the wavelength associated
p
with kmax , i.e. 2π 3σ/(ρ2 − ρ1 ) g, which amounts to about 3 cm when the heavier fluid is
water and the lighter one is air. On the other hand, the modes with
r
σ
∗
∗
(10.34)
k > k ⇒ λ < λ = 2π
(ρ2 − ρ1 ) g
204
CHAPTER 10. STABILITY OF FLUID FLOWS
10.5. SUMMARY OBJECTIVES
are stable since in this case the surface tension force outweighs that due to the gravity.
The result given by (10.34) can be easily illustrated by the well known fact that liquid inside an
inverted bottle is stabilised by surface tension if the opening of the bottle is small enough. To
estimate the linear scale of the opening, suppose, for simplicity, the vessel to be a rectangular
one, with vertical sides and a cross-section in the z = 0 plane of which the larger dimension
is L, see Figure 10.3. In this case, the smallest non-zero value of k consistent with boundary
conditions is π/L. Indeed, it follows from Figure above that
cos (kmin L/2) = 0 ⇒ (kmin L/2) =π/2 ⇒ kmin = π/L
Then the inverted contents are stable provided that Lπ > k ∗ , i.e. provided that
r
σ
π
L< ∗ ≡π
k
(ρ2 − ρ1 ) g
(10.35)
One can consider this result also as an illustration of the experiment in which the water is supported by the atmospheric pressure in an inverted glass if the glass is covered by the gauze with
small enough sells. The linear scale of the sell must then satisfy the inequality (10.35).
Figure 10.3: A layer of one fluid, yellow, with a denser fluid, blue, above it is stabilised by surface
tension against perturbations provided that inequality (10.35) is satisfied.
10.5 Summary objectives
When you finish studying the Chapter 10, you should be able to do the following:
1. Define the general requirement for any steady solution of the fluid dynamics equations and
describe in general the stable and unstable flows.
205
10.5. SUMMARY OBJECTIVES
CHAPTER 10. STABILITY OF FLUID FLOWS
2. Define the stable and unstable equilibrium of a dynamical system. Show that in the case of
a stable equilibrium the perturbed dynamical system performs oscillations and determine
the angular frequency of these oscillations.
3. Consider the unstable equilibrium and show that the displacement from this equilibrium is
normally grows exponentially.
4. Define the condition of a marginal stability.
5. Describe the general approach to the stability problems, define the goal of the linear stability theory of a steady flow, and determine the corresponding mathematical tool used in
the linear stability theory. Give some examples.
6. Formulate the stability problem concerning the flows of two fluid layers with discontinuous
tangential velocities, so called the Kelvin-Helmgoltz instability.
7. Provide the mathematical description of the Kelvin-Helmholtz instability and find its increment.
8. Describe the Rayleigh-Taylor instability and explain the physical mechanism responsible
for its development.
9. Formulate the statement of the problem concerning the Rayleigh-Taylor instability.
10. Provide the mathematical analysis of the Rayleigh-Taylor instability and find the dispersion relation describing this instability.
11. Perform the analysis of this dispersion relation and determine the expression for the critical
wave length separating the stable and unstable solutions.
12. Describe the experiment in which the water can be supported by the atmospheric pressure
in an inverted glass and define the corresponding condition.
206
Chapter 11
Appendix I: Vector Calculus
Vector Calculus: Definitions and fundamental theorems
Unless otherwise indicated, the meanings of the symbols are:
• θ, ψ = two scalar functions of space coordinates, defined and regular in a three-dimensional
domain D;
• A = a vector function of space coordinates, defined and regular in D;
• S = a closed surface, contained in D and regular;
• V = a three-dimensional region, surrounded by the surface S (sometimes, the volume of
that region);
• n̂ = the unit-vector orthogonal to S, oriented towards the external region (see Figure 11.1).
n
V
S
D
Figure 11.1: Illustration of the symbols used in the definitions of the vector calculus operators.
The gradient of a scalar field is the following vector:
1
∇φ ≡ grad φ = lim
V →∞ V
207
Z
S
φn̂dS
(11.1)
CHAPTER 11. APPENDIX I: VECTOR CALCULUS
The divergence of a vector field is the following scalar:
1
∇ · A ≡ divA ≡ lim
V →∞ V
Z
A · n̂dS
(11.2)
Z
n̂ × AdS
(11.3)
Z
n̂AdS
(11.4)
S
The curl of a vector field is the following vector:
1
∇ × A ≡ curl A = lim
V →∞ V
S
The gradient of a vector field is the following dyadic:
1
∇A ≡ gradA = lim
V →0 V
S
The symbolic vector ∇ can be defined as follows:
1
∇ () = lim
V →0 V
Z
n̂ ()dS
(11.5)
S
where the pair of parentheses stands for any of the quantities on which ∇ operates in the previous
four definitions.
The Laplacian of a scalar field is the following scalar:
∇2 φ = ∇ · ∇φ ≡ divgradφ
(11.6)
The Laplacian of a vector field is the following vector:
∇2 A = ∇ (∇ · A) − ∇ × ∇ × A ≡ grad(divA) − curl(curlA)
Gauss’ divergence theorem: with the previously defined symbols, it can be proved that:
Z
I
∇ · AdV = A · n̂dS
V
S
n
T
l
Figure 11.2: Illustration of the symbols used in Stokes’ curl theorem.
208
(11.7)
(11.8)
CHAPTER 11. APPENDIX I: VECTOR CALCULUS
Stokes’ curl theorem: let T be a two-faced regular surface, l be its contour (a connected and
oriented line), n̂ be the unit vector orthogonal to T whose orientation is related to that of l by the
so-called right-hand rule (see Fig. 11.2). Then it can be shown that:
Z
Z
∇ × A · n̂dT = A · dl
(11.9)
T
l
Vector identities.
List of symbols:
• f, g = scalar functions of space coordinates;
• A, B, C = vector functions of space coordinates;
• ·
= scalar (or internal) product;
• × = vector (or external) product;
• (A · ∇) = direction derivative, along the direction of the vector A, multiplied by |A|;
• û = unit-vector.
∇ (f g) = f ∇g + g∇f
∇ (A · B) = (A · ∇) B + (B · ∇) A + A × (∇ × B) + B × (∇ × A)
(11.10)
(11.11)
∇ · (f A) = f ∇ · A + ∇f · A
(11.12)
∇ · (A × B) = B · ∇ × A − A · ∇ × B
(11.13)
∇ · (∇ × A) = 0
(11.14)
∇ × (f A) = f ∇ × A + ∇f × A
(11.15)
∇ × (A × B) = A (∇ · B) − B (∇ · A) + (B · ∇) A − (A · ∇) B
∇ × (∇f ) = 0
(11.16)
(11.17)
∇2 A = ∇ (∇ · A) − ∇ × ∇ × A
(11.18)
A = (A · û) û + û × (A × û)
(11.19)
A × (B × C) = (A · C) B − (A · B) C
(11.20)
209
Index
“Pure” Capillary Waves, 183
Adiabatic equation, 53
adiabatic process, 31
Baroclinic, 56
Baroclinic vector, 78
Barometric formula, 55
Barotropic, 56
Barytropic flow, 75
Bernoulli’s Theorem, 60
Boundary layer, 125
Boussinesq Approximation, 191
Capillary Waves, 182
Cauchy-Riemann conditions, 88
Centrifugal force, 51
Circulation of velocity, 72
Complex flow potential, 89
compression coefficient, 31
Conformal Mapping, 92
Conjugate flow, 87
Conservation of Momentum, 66
Coordinate transformation, 42
Coriolis force, 52
Couette flow, 113
d’Alembert-Euler paradox, 97
Deep Water Approximation, 179
Drag coefficient, 124
Dynamic boundary condition, 171
Dynamical velocity, 161
Ensemble average, 155
Enthalpy, 60
enthalpy, 33
Entropy, 52
Entropy flux density, 53
Equation of continuity, 50
Equation of state, 52
Equipotential lines, 90
Ergodic process, 157
Euler coordinates, 42
Euler equation, 50
Euler’s theorem, 68
first law of thermodynamics, 31
Frictional velocity, 161
Gravity-Capillary Surface Waves, 186
Harmonic Surface Waves, 176
Harmonic waves, 174
Helmholtz Theorems, 75
Hydrodynamic lift force, 101
Hydrodynamical equations, 49
Hydrostatic equation, 54
Hydrostatic Equilibrium, 56
Ideal fluid, 49, 52
Ideal fluid approximation, 75
Ideal Fluids, 49
Incompressible fluid, 54, 172
Induced inertia, 104
Induced mass, 104
Inertial interval, 162
Internal energy, 60
internal energy, 31
Internal Gravity Waves, 190
Irrotational flow, 71
Isentropic motion, 53
Joukowski transformation, 95
Kelvin’s Circulation Theorem, 74
Kinematic boundary condition, 171
Kinematic viscosity, 109
Kinematics of fluids, 41
Kolmogorov’s (2/3) law, 164
Kolmogorov’s first hypothesis, 164
Kolmogorov’s scale, 162
Kolmogorov’s second hypothesis, 164
Kolmogorov’s similarity hypothesis, 163
Kolmogorov-Obukhov’s law, 165
Kutta-Joukowski theorem, 102
210
INDEX
Lagrange coordinates, 42
Laminar-turbulent transition, 151
Linear strain, 69
Logarithmic boundary layer, 161
macroscopic energy, 60
Mass flux density, 50
microscopic energy, 60
Momentum, 66
Navier-Stokes equation, 111
Poiseuille flow, 116
Poisson’s adiabatic law, 34
Potential body forces, 75
Potential flow, 71, 81
Potential force, 61
Principle of hydrodynamic similarity, 97
quasistatic process, 32
reversible process, 32
Reynolds equation for turbulent flow, 158
Reynolds number, 116
Reynolds stresses, 158
Reynolds’ transport theorem, 45
Rotation, 69
INDEX
Turbulent viscosity, 159
Unstable equilibrium, 57
Väisälä frequency, 58, 170, 172
Velocity potential, 81
Viscous fluid, 107
Viscous force, 112
Viscous stress tensor, 109
Viscous sublayer, 160
Viscous waves, 126
Vortex flow, 71
Vortex line, 75
Vortex source, 91
Vortex tube, 75
Vorticity, 71
Vorticity equation, 77
Wave Energy, 180
Waves in Unlimited Media, 193
Waves on a Liquid-Fluid Interface, 189
zeroth law of thermodynamics, 30
second law of thermodynamics, 37
Shallow Water Approximation, 179
Shear strain, 69
Specific energy flux, 66
specific heat, 32
Specific momentum flux tensor, 67
Spectral energy density, 165
Stable equilibrium, 57
stagnation point, 83
Statistical ensemble, 155
Steady flow, 12, 60
Stokes Formula for Drag, 123
stream function, 85
Streamline, 61
Surface Gravity waves, 172
The pressure coefficient, 97
thermal expansion, 31
Translation, 69
Turbulent boundary layer, 159
Turbulent cascading, 162
Turbulent flow, 151
211
PART II
Problem set
Chapter 1
Fundamental Concepts
1.1
For the following velocity
fields:
−bzt
1. v = ae
k;
2. v = axi + bx2 e−ct j;
3. v = axi + by 3 j + cy −2 tk;
4. v = ay 2 i + bx−2 j + cxyk
where the quantities a, b, and c are constants, determine
1. whether the flow field is one-, two-, or three-dimensional, and why.
2. whether the flow is steady or unsteady, and why.
1.2
A velocity field is given by the expression
v = U cos θ 1 −
a 2 r
er − U sin θ 1 +
a 2 r
eθ
where er and eθ are unit vectors in the rθ-plane. Find all points in the rθ-plane where:
1. vr = 0; b) vθ = 0; c) vr = vθ = 0.
Sketch the field and describe a physical situation where this flow can occur.
1.3
A streamline is a line drawn in the flow field so that at each instant it is tangent to the velocity
vector at each point. Consider a two-dimensional, steady flow in the xy-plane with velocity field
given by v = ui + vj. Use the vector cross product, v × ds, where ds is an element of distance
along a streamline, to show that dx
= dy
for the streamline.
u
v
1
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.4
A tornado can be represented in polar coordinates by the velocity field
a
b
v = − er + eθ
r
r
where er and eθ are unit vectors in the r- and θ-directions, respectively. Use the vector cross
product, v × ds, where ds is an element of distance along a streamline, to show that streamlines
form logarithmic spirals, r = ce−(a/b)θ .
1.5
A viscous liquid is sheared between two parallel disks; the upper disk rotates with angular velocity ω and the lower one is fixed. The velocity field between the disks is given by v = eθ rωz/h.
The origin of coordinates is located at the center of the lower disk; the upper disk is located at
z = h. What are the dimensions of this velocity field? Does this velocity field satisfy appropriate
physical boundary conditions? What are they?
1.6
The velocity distribution for laminar flow between fixed parallel plates is given by
2
u
2y
=1−
umax
h
where h is the distance separating the plates and the origin is placed midway between the plates.
Consider flow of water at 15 ˚ C, with umax = 0.30 m/s and h = 0.50 mm. Calculate the shear
stress on the upper plate and give its direction.
1.7
The Reynolds number, which is an important parameter in viscous flow phenomena, is defined
by the equation
ρV L
Re =
η
where ρ and η are the fluid density and viscosity, and V and L are characteristic velocity and
length, respectively. Express each variable in terms of its basic dimensions. Show that the
Reynolds number is dimensionless.
1.8
The fluid mechanics of the human arterial system is of vital importance to our health and safety.
The specific gravity (SG = ρHρ O ) and viscosity of blood are about 1.06 and 3.3 × 10−3 Ns m−2 ,
2
respectively. The mean flow speed in the aorta (30 mm in diameter) of a large human is about
0.15 m/s. Calculate the flow Reynolds number using these parameters. Would you expect this
flow to be laminar or turbulent?
2
CHAPTER 1. FUNDAMENTAL CONCEPTS
1.9
For laminar steady flow through a circular pipe the velocity u(r) varies with radius and takes the
form
∆p 2
u (r) = B
r0 − r 2
η
where η is the fluid viscosity, ∆p is the pressure drop from entrance to exit, and r0 is the pipe
radius. What are the dimensions of the constant B?
1.10
The speed of propagation V of waves traveling at the interface between two fluids is given by
V =
πY
ρa λ
1/2
Where λ is the wave length and ρa the average density of the fluids. If this formula is dimensionally consistent, what are units of Y ? What might it represent?
1.11
Estimate the surface pressure of a spherical soap bubble assuming that the bubbles are quite
small, D ≈ 0.5 cm, and the surface tension of an air-water interface is 0.073 N/m.
3
CHAPTER 1. FUNDAMENTAL CONCEPTS
4
Chapter 2
Elements of Thermodynamics
2.1
Let a(t) be an instantaneous value of any fluid parameter and a0 its value at the equilibrium state,
i.e. a(t) − a0 = ∆a(t), where ∆a(t) is the deviation from the equilibrium state. Write the condition for sufficiently slow variation of the parameter a(t) if the relaxation time is given by τ . Give
the physical explanation of the result and connect it with so-called quasi-static approximation.
2.2
Let we have three independent variables A, B, C and functional dependence of A from B and C
is given by f (A, B, C) = 0. Assuming that each variable A, B, C is differentiable function of
the two others, prove that
∂B ∂C ∂A
= −1
1. ∂B
C ∂C A 1 ∂A B
∂A
2. ∂C B = (∂C/∂A)
B
3. Write the final results with A = p, B = V, C = T .
2.3
Let α = − V1 ∂V
and β = V1 ∂V
be the compression and thermal expansion coefficients
∂p
∂T p
T
(see relations 2.3 and 2.4 in the first part of compendium) for a given fluid. Show that the
Gruneisen’s coefficient (important parameter which is characterized dynamic properties of liqβV
, satisfies the relation
uids), given by Γ = αC
V
Γ=
V (∂p/∂T )V
CV
2.4
Find the derivative of the sound velocity in the ideal gas with respect to temperature. Would you
expect sound to be distorted more at higher or lower tempratures?
5
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
2.5
1
V
11
=
γV
Show that the adiabatic compression coefficient,
, are connected through the relation
cient, V1 ∂V
∂p
∂V
∂p
ad
, and thermal compression coeffi-
T
1
V
∂V
∂p
ad
∂V
∂p
.
T
Hint: use the relation (2.17) and assume the equation of state of the form T = (V, p).
2.6
Using the relations (2.16) and (2.18), show that for an ideal gas the specific heats are given by
γR
R
, and Cp = γ−1
.
CV = γ−1
2.7
Show that the thermal elastic coefficient, KT = −V
∂p
Kad = −V ∂V
, satisfy the equation Kad = γKT .
ad
∂p
,
∂V T
and adiabatic elastic coefficient,
2.8
The measurement of the sound velocity in a gas under investigation is one of the most exact
tools for experimental determination of the ratio of the specific heats, γ, for a given gas. Find
the relation between the sound
velocity, cS , ration of the specific heats, γ, and thermalpelastic
∂p
coefficient, KT = −V ∂V T , if the sound velocity in an elastic media is given by cS = K/ρ.
dp
Here K = −V dV
is the elastic coefficient, and ρ is the gas density.
2.9
A body is moving in an ideal gas. Determine at what point of the surface of the body the
temperature will be maximal. Find this temperature if the temperature of the surrounding gas
is given by T . (Hint: use the equation describing the conservation of the energy in the form
2
w + v2 = const, where w is the enthalpy of the gas per unit mass.)
2.10
Write the integral equation for the second thermodynamics law related to the equilibrium closed
processes, so called Clausius’ equality.
2.11
Let the equilibrium state of the fluid be characterized by two parameters: the temperature, T ,
and the specific volume, V . Show that the main equation of thermodynamics for equilibrium
6
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
processes, T dS = dε + pdV , obtained by combination of the first and second thermodynamics
laws, is equivalent to the relation
∂p
∂ε
T
=
+p
∂T V
∂V T
2.12
∂p
∂ε
Using the relation, T ∂T
=
+ p (see problem 2.11) show that the internal energy of an
∂V
V
T
∂ε
= 0.
ideal gas is a function of its temperature only, i.e. ∂V
T
2.13
Calculate the difference of the specific heats, cp − cV , and show that the result can be presented
2
in the form, cp − cV = V T βα . Here coefficients α and β are determined by the relations (2.3)
and (2.4) given in the first part of the compendium. (Hint: use the result of the problem 2.10 and
relation (2.10) derived in the first part of the compendium.)
7
CHAPTER 2. ELEMENTS OF THERMODYNAMICS
8
Chapter 3
Kinematics of Fluids
3.1
What is the Lagrangian description of fluid motion?
3.2
What is the Eulerian description of fluid motion? How does it differ from the Lagrangian description?
3.3
A weather balloon is launched into the atmosphere by meteorologists. When the balloon reaches
an altitude where it is naturally buoyant, it transmits information about weather conditions to
monitoring stations on the ground. Is this a Lagrangian or an Eulerian measurement? Explain.
3.4
A stationary probe is placed in a fluid flow and measures temperature and pressure as functions
of time at one location in the flow. Is this a Lagrangian or an Eulerian measurement? Explain.
3.5
A so-called Pitot-static probe can often be seen protruding from the underside of an airplane. As
the airplane flies, the probe measures relative wind speed. Is this a Lagrangian or an Eulerian
measurement? Explain.
3.6
Define a steady flow field in the Eulerian reference frame. In such a steady flow, is it possible for
a fluid particle to experience a non-zero acceleration?
9
CHAPTER 3. KINEMATICS OF FLUIDS
3.7
Prove that J −1 dJ
= ∇ · v with J =
dt
Eulerian to Lagrangian coordinates.
∂(x1 ,x2 ,x3 )
∂(ξ1 ,ξ2 ,ξ3 )
being the Jacobian of the transformation from
10
Chapter 4
Basic Laws of Ideal Fluid Dynamics
4.1
A cylindrical vessel with a vertical axis filled by an ideal fluid rotates in the gravitational field
about this axis at the angular velocity Ω. Determine the shape of the free surface of the fluid and
sketch it.
4.2
Now remove the cylinder from problem 4.1 and find the radial dependence of the angular velocity
Ω that makes the flow potential. Prove that circulation of velocity is equal to zero under the right
choice of contour integration.
4.3
A cylindrical container with radius R and height H is partly filled (80 % of height H) with liquid
with density ρ. Now the cylinder is rotated at constant speed. Determine the rotational speed at
which the liquid will start spilling from the edges of the container.
4.4
Determine the height dependence of the horizontal cross-sectional area of a vessel, i.e. S(z), on
the vertical distance under the condition that the time rate of change of the surface level of the
, is constant.
liquid outflowing through the hole, dz
dt
4.5
A large tank open to the atmosphere is filled with water to a height H above an outlet tap. The
tap is now opened and water flows out from the smooth and rounded outlet. Determine the water
velocity at the outlet.
11
CHAPTER 4. BASIC LAWS OF IDEAL FLUID DYNAMICS
4.6
For the case of an ideal gas in the gravitational field, find the pressure dependence on height if
the temperature dependence is T = T0 (1 − Hz ) with z < H.
4.7
Water is flowing from a hose attached to a water main of 400 kPa gauge. A child places his
thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. If the
hose is held upward, what is the maximum height the jet could achieve?
4.8
A simple two-dimensional velocity field called a line sink is often used to simulate fluid being
sucked into a line along the z-axis. Suppose the volume flow rate per unit length along the z-axis,
dV 1
is known, where dV
is a negative quantity. In two dimensions in the rθ-plane, the velocity
dt L
dt
is:
dV 1
vr =
, vθ = 0.
dt 2πrL
Draw the flow streamlines and calculate the vorticity. Is this flow rotational or irrotational?
4.9
A cylindrical tank of water rotates in solid-body rotation, counterclockwise around its vertical
axis at angular speed vϕ . Determine the vorticity of fluid particles in the tank.
12
CHAPTER 4. BASIC LAWS OF IDEAL FLUID DYNAMICS
MEMORANDUM-1
System of equations of hydrodynamics
∂ρ
∂t
+ ∇ · (ρv) = 0
dv
dt
=
∂v
∂t
Equation of continuity
+ (v · ∇)v = − ∇p
+f
ρ
p = p(ρ, s)
Euler’s equation (momentum equation)
Equation of state
∇p = ρf
ρ = ρ0 e
Hydrostatic equation
−µgz
RT
N 2 = −g
Barometric formula
1 dρ
ρ dz
+
v2
2
+ w + u = C1
v2
2
+
∂
∂t
R v2
∂
∂t
p
ρ
V
Väisälä frequency
Bernoulli
theorem (steady flow)
w = ǫ + ρp , ds = 0, f = −∇u
+ u = C1
2
V
R
g
c2
Bernoulli theorem, incompressible fluid
R
R
Πik nk dS
+ u + ǫ dV = −
ρvi dV = −
S
S
ρ
v2
2
+ u + w vn dS
Energy conservation law in integral form
Momentum conservation law (fi = 0)
Πik = pδik + ρvi vk
Tensor of the specifik momentum flux
∇×v =0
Potential or irrotational flow
∇ × v 6= 0
Vortex flow
ω =∇×v
Vorticity
Γ=
H
l
Γ=
R
S
dΓ
dt
v · dr
Circulation of velocity w.r.t contour l
n · ∇ × vdS
Circulation w.r.t surface S confined by l
=0
1 dJ
J dt
=
Kelvin circulation theorem
(contour l goes along closed fluid contour)
∂vk
∂xk
=∇·v
J - Jacobian of the transformation
from Eulerian to Lagrangian coordinates.
13
CHAPTER 4. BASIC LAWS OF IDEAL FLUID DYNAMICS
Thermodynamical formulas
p = p(ρ, s)
Equation of state
∂p
c2 = ( ∂ρ
)s
Velocity of sound
c = c(p, ρ)
c is infinite in an incompressible fluide
p=
R
ρT
µ
w =ǫ+
Equation of state of an ideal gas
p
ρ
dǫ = T dS − pdV
Enthalpy
Internal energy differential
14
Chapter 5
Potential Flow
5.1
What flow property determines whether a region of flow is rotational or potential (irrotational)?
Discuss.
5.2
A subtle point, often missed by students of fluid mechanics (and even their professors!), is that
potential (irrotational) flow is not the same as ideal (inviscid) flow. Discuss the differences and
similarities between these two approximations. Give an example of each.
5.3
A fluid rotates about a vertical axis, the angular velocity being dependent on the radius Ω(r), in
a gravitational field. Find the shape of the free surface of a liquid assuming that the motion is
potential. Sketch it.
5.4
Consider a planar potential flow in the rθ-plane. Show that the stream function Ψ satisfy the
Laplace equation in cylindrical coordinates.
5.5
A horizontal slice through a tornado (Figure 5.1) is modeled by two distinct regions. The inner
(or core) region (0 < r < R) is modeled by rigid-body rotation (a rotational but ideal region of
flow as discussed in Section 1.6). The outer region (r > R) is modeled as an irrotational region
of flow. The flow is 2D in the rθ-plane, and the components of the velocity field are given by:
vr = 0
ωr
vθ =
ωR2
0 < r < R,
r > R,
r
15
CHAPTER 5. POTENTIAL FLOW
where ω is the magnitude of the angular velocity in the inner region. The ambient pressure (far
away from the tornado) is equal to P∞ . Calculate the pressure field in a horizontal slice of the
tornado for 0 < r < ∞. What is the pressure at r = 0?
Figure 5.1: A horizontal slice through a tornado can be modeled by two regions, an inviscid but
rotational inner region of flow and an irrotational outer region of flow.
5.6
The stream function for steady, incompressible, 2D flow over a circular cylinder of radius a
2
and free-stream velocity V∞ is Ψ = V∞ sin θ(r − ar ) for the case in which the flow field is
approximated as potential. Generate an expression for the velocity potential function ϕ for this
flow as a function of r and θ, and parameters V∞ and a.
5.7
What is D’Alembert’s paradox? Why is it a paradox?
5.8
Find the hydrodynamic lift force created by a flow around a cylinder. The flow consists of the
superposition of straight flow with velocity v0 and vortex flow with circulation Γ around the
cylinder. Assume ρ = const.
16
CHAPTER 5. POTENTIAL FLOW
5.9
Estimate the lift force acting on an airplane wing (consider the wing to have a semi-cylindrical
shape). Aircraft velocity is v0 , atmospheric parameters are ρ0 , p0 . For your favourite airplane,
estimate the critical velocity needed to take off. Is it reasonable?
5.10
Consider an potential flow composed of a line source of strength Q = 2πm at location (−a, 0)
and a line sink of the same strength (but opposite sign) at (a, 0) as sketched in Figure 5.2.
• Generate an expression for the stream function in both Cartesian and cylindrical coordinates.
• Add an uniform flow coming from the left. Find the stagnation points. Sketch the stream
function and velocity vectors numerically. What does it look like (try to change the angle
of the incoming flow)? Compare briefly its properties to the airplane wing in exercise 5.9
Figure 5.2: Superposition of a line source of strength Q at (−a, 0) (in figure V/L) and a line sink
−Q at (a, 0). Problem 5.10
5.11
A sphere of mass m and radius R is dropped into the ocean of depth H. What time is needed for
the sphere to reach the bottom of the ocean? The water is assumed to be an ideal fluid.
5.12
A uniform flow is flowing horizontally along a flat surface that has a slit in it (Fig. 5.3). A pump
is sucking in fluid through the slit, which is directed perpendicularly to the water flow. At which
distance above the plane must the incoming water be so as not to be sucked into the hole?
Flow velocity: v0 = 1.2 m/s
Pump rate: Q = −0.01π m2 /s (per unit length)
17
CHAPTER 5. POTENTIAL FLOW
Figure 5.3: Uniform flow over a horizontal surface with a slit in it.
18
CHAPTER 5. POTENTIAL FLOW
MEMORANDUM-2
Set of equations for a flow in inviscid incompressible fluid
ds = 0,
f = −∇u,
∆ϕ = 0,
=
p
ρ
Velocity potential (ϕ)
∂ψ
,
∂y
vy = − ∂ψ
∂x
Stream function for 2-D potential flow (ψ)
∆ψ = 0
∂ϕ
∂x
w=
p
∂ϕ v 2
+
+u+ =0
∂t
2
ρ
v = ∇ϕ
vx =
ρ = const,
Equation for stream function
∂ψ
,
∂y
∂ϕ
∂y
= − ∂ψ
∂x
Relation between ϕ and ψ
F (z) = ϕ(z) + iψ(z)
F (z) = az,
Complex flow potential
a ∈ R,
F (z) = mlnz,
z = x + iy
z 6= 0,
Complex flow potential for straight flow parallel to OX
m∈R
Complex flow potential for source (sink)
F (z) = −imlnz = m(θ − ilnr),
ϕ = mθ, ψ = −mlnr
Complex flow potential for vortex flow
z = f (ξ)
o
F (z) = F (f (ξ)) = φ(ξ)
ζ = ξ + iη =
1
2
z
R
+
R
z
F (z) = −2Rv0 ζ = −v0 reiθ +
Conformal mapping
Zhukovski (Joukovski) transformation
R2 −iθ
e
r
Complex flow potential for the flow around a cylinder
K=
p−p0
ρvo2 /2
= 1 − 4sin2 θ
Pressure coeff. for potential flow around a cylinder
Fy =
2ρv0 Γ
π
R π2
Hydrodynamic lift force for the flow around a cylinder
− π2
sin2 θdθ = ρv0 Γ
v0
3 (v 0 ·r)r
v = −v 0 − R3 2r
3 + 3R
2r5
K=
p−p0
ρv02 /2
Velocity of a flow around a sphere
= 1 − 94 sin2 θ
Pressure coeff. for potential flow around a sphere
M = 21 VS ρ
Induced mass for a sphere in an ideal fluid
19
CHAPTER 5. POTENTIAL FLOW
20
Chapter 6
Flows of Viscous Fluids
6.1
Using the Navier-Stokes equation, derive an equation for the vorticity ω = ∇ × v.
6.2
Consider fully developed Couette flow, illustrated in Figure 4.5 in the compendium. The flow is
steady, incompressible, and two-dimensional in the xy-plane. The velocity field is given by
y
v = (u, v) = v0 ex + 0ey
h
Is this flow rotational or irrotational? If it is rotational, calculate the vorticity component in the
z-direction. Do fluid particles in this flow rotate clockwise or counterclockwise?
6.3
Determine the velocity distribution and drag for Couette’s flow between two coaxial circular
cylindrical tubes: one of these tubes moves along their common axis relative to the other at the
velocity v0 . The fluid is incompressible.
6.4
Determine the velocity distribution in the Poiseuille flow between coaxial cylinders. Find the
flow rate.
6.5
Determine the velocity and pressure distribution in the Couette flow between two coaxial circular
cylinders, with radii R1 and R2 , rotating about its axis at the angular velocities ω1 and ω2 .
21
CHAPTER 6. FLOWS OF VISCOUS FLUID
6.6
Consider the geometry in Figure 4.5 in the compendium but let there also be an applied pressure
gradient in the x-direction as in Figure 4.6. Specifically, let the pressure gradient in the x∂p
direction, ∂x
, be some constant value given by
∂p
p2 − p1
=
= constant
∂x
x2 − x1
where x1 and x2 are two arbitrary locations along the x-axis, and p1 and p2 are the pressures at
these locations.
a) Calculate the velocity and pressure field.
b) Sketch or plot a family of velocity profiles in dimensionless form.
6.7
Consider steady, incompressible, parallel, laminar flow of a film of oil falling slowly down an infinite wall, inclined at an angle θ (Fig. 6.1). The oil film has thickness h, density ρ, and viscosity
η. The oil falls only by the force of gravity, i.e., there is no applied (forced) pressure driving the
flow. Calculate the velocity and pressure fields in the oil film and sketch the normalized velocity
profile. You may neglect changes in the hydrostatic pressure of the surrounding air.
Figure 6.1: Laminar flow of a film of oil falling down on an inclined wall.
6.8
Repeat example 6.7 but for the case when the wall is inclined at an angle θ = 90 degrees.
Generate expressions for both the pressure and velocity fields.
6.9
Consider steady, incompressible, parallel, laminar flow of a viscous fluid with density ρ and viscosity η falling between two infinite walls, inclined at an angle θ (Fig. 6.2). The distance between
the walls is h. There is no applied (forced) pressure driving the flow-the oil falls by gravity alone.
Calculate the velocity field and sketch the velocity profile using appropriate nondimensionalized
variables.
22
CHAPTER 6. FLOWS OF VISCOUS FLUID
Figure 6.2: Laminar flow of a viscous fluid falling down between two inclined walls.
6.10
Compare and summarize the results for the velocity profiles and volumetric flow rates in problems 6.7 and 6.9. Discuss the differences and provide a physical explanation.
6.11
A moving belt is fixed on a platform to skim oil that is floating on the surface of water. The belt
is inclined at an angle θ = 30◦ . The distance between the belt and the platform is 2 mm. The belt
has width w = 5 m, length l = 6 m, and moves with velocity U = 3 m/s. Find the volumetric
flow rate and the power required to move the belt. Assume density of the oil to be 860 kg/m3 .
The dynamic viscosity of the oil is 10−2 Ns/m2 .
Figure 6.3: Sketch for the oil skimmer problem 6.11.
6.12
Consider steady, incompressible, laminar flow of a Newtonian fluid with density ρ and viscosity
η in an infinitely long round pipe of radius R, inclined at an angle θ (Fig. 6.4). There is no
applied pressure gradient. Instead, the fluid flows down the pipe due to gravity alone. Derive
an expression for the velocity profile as a function of the radius R and other parameters of the
problem. Calculate the volumetric flow rate and average axial velocity through the pipe.
23
CHAPTER 6. FLOWS OF VISCOUS FLUID
Figure 6.4: Laminar flow of a viscous fluid inside an inclined pipe.
6.13
An incompressible, viscous fluid flows between two straight walls at a distance h apart. One wall
is moving at a constant velocity U in the x-direction, while the other is at rest. Te flow in the
x-direction is caused by the movement of the wall. The walls are porous and a steady uniform
flow is imposed accross the walls to create a constant velocity V throught the walls. See Fig. 6.5
for a sketch of the problem.
Show the following:
• the velocity profile is given by
u(y) =
1 − eV y/ν
U.
1 − eV h/ν
• u(y) approaches U y/h for small V
• u(y) approaches U e−V (h−y)/ν for large V h/ν,
where ν is kinenematic viscosity of the fluid.
Figure 6.5: Sketch for problem 6.13
24
CHAPTER 6. FLOWS OF VISCOUS FLUID
MEMORANDUM-3
∂(ρvi )
∂t
ik
= − ∂Π
∂xk
Navier-Stokes equation in tensor form
Πik = pδik + ρvi vk − σik
σik = η
∂vi
∂xk
+
∂vk
∂xi
Generalized specific momentum flux tensor
∂vm
δik
+ b ∂x
m
Viscous stress tensor
∂v
ρ dv
=
ρ
+
(v
·
∇)v
dt
∂t
= −∇p + η∆v + (η + b)∇(∇ · v)
Navier-Stokes equation in vector form
2
2
vk
fi = η ∂∂xv2i + (η + b) ∂x∂ i ∂x
=
k
k
∂
η∆vi + (η + b) ∂xi (∇ · v)
i-th component of viscous force/unit volume
fi′ = σik nk
i-th component of viscous force exerted by
the fluid per unit area of the body
2f
2
ρvm
Drag coefficient: Ratio of force, f , to the
CD =
const
Re
Drag coefficient for the Couette flow, the
plane Poiseuille flow, flow around a sphere
Re =
vm L
ν
Reynolds number
CD =
2
velocity head ( ρv2m )
η
ρ
ν=
Kinematic viscosity
v = 1−
3R
4 r
−
1 R3
4 r3
v0 −
3 v0 Rx
4 r2
1−
R2
r2
r
r
Fx = 6πηRv0
v = v0 exp −
δ=
q
νv0
h2 (x)
Stokes’ formula for drag
(slow motion of sphere in a fluid)
p
ω
2νz
p ω
exp i
−
ωt
2νz
Viscous waves
2ν
ω
∼
v02
,
x
Viscous flow around a sphere of radius
R (Re small)
Thickness of layer where significant fluid
oscillations occur
h(x) ∼
12
νx
v0
=x
ν
v0 x
21
=
√x
Re
25
Thickness of the boundary layer
CHAPTER 6. FLOWS OF VISCOUS FLUID
26
Chapter 7
Dimensional Analysis
7.1
List the primary purpose of dimensional analysis.
7.2
List and describe the three necessary conditions for complete similarity between a model and a
prototype.
7.3
Using the dimensional analysis, estimate the characteristic frequency of the pendulum motion.
7.4
Some children are playing with soap bubbles, and you become curious as to the relationship
between the soap bubble radius and the pressure inside the soap bubble. You reason that the
pressure inside the soap bubble must be greater than the atmospheric pressure, and that the shell
of the soap bubble is under tension, much like the skin of a balloon. You also know that the
property surface tension must be important in this problem. Establish a relationship between
pressure difference ∆p = pinside − poutside , soap bubble radius R, and the surface tension σs of
the soap film using dimensional analysis.
7.5
Consider laminar flow through a long section of a cylindrical pipe. For laminar flow it turns out
that wall roughness (ǫ) is not a relevant parameter unless ǫ is very large. The volume flow rate
Ql through the pipe is in fact a function of pipe diameter D, fluid viscosity η, and axial pressure
dp
gradient dx
. If the pipe diameter is doubled, all else being equal, by what factor will volume flow
rate increase? Use dimensional analysis.
27
CHAPTER 7. DIMENSIONAL ANALYSIS
7.6
At very low speeds, the drag on an object is independent of fluid density. Thus the drag force,
F , on a small sphere is a function only of speed, V , fluid viscosity, η, and sphere diameter, D.
Use dimensional analysis to express the drag force as a function of these variables.
7.7
Consider a liquid in a cylindrical container in which both the container and the liquid are rotating
as a rigid body. The elevation difference h between the center of the liquid surface and the rim of
the liquid surface is a function of angular velocity Ω, fluid density ρ, gravitational acceleration g,
and radius R. Use the method of repeating variables to find a dimensionless relationship between
the parameters.
7.8
Consider the case in which the container and liquid of the previous problem are initially at rest.
At t = 0 the container begins to rotate. It takes some time for the liquid to rotate as a rigid
body, and we expect that the liquid’s viscosity is an additional relevant parameter in the unsteady
problem. Repeat the previous problem but with two additional independent parameters included,
namely, fluid viscosity η and time t. (We are interested in the development of height h as a
function of time and the other parameters.)
7.9
A boundary layer is a thin region (usually along the wall) in which viscous forces are significant
and within which the flow is rotational. Consider a boundary layer growing along a thin flat
plate. The flow is steady. The boundary layer thickness h at any downstream distance x is a
function of x, free-stream velocity v0 , and fluid properties ρ (density) and η (viscosity). Use the
method of repeating variables to generate a dimensionless relationship for h as a function of the
other parameters.
7.10
The tangential viscous force f exerted by the fluid on the smooth flat plate in a boundary layer
depends on distance from the leading edge of the body, x , the density ρ , viscosity η of the
fluid, and the free stream speed of the flow V . Obtain the dimensionless groups and express the
functional relationship among them.
7.11
Experiments show that the pressure drop due to fluid flow through a sudden contraction in a circular duct may be expressed as ∆p = p1 − p2 = f (ρ, η, V, d, D). Here p1 and p2 corresponds to
28
CHAPTER 7. DIMENSIONAL ANALYSIS
the pressure at the circular duct with diameter D and d . You have to organize some experimental data. Obtain the resulting dimensionless parameters and express the functional dependence
between them.
29
CHAPTER 7. DIMENSIONAL ANALYSIS
30
Chapter 8
Theory of Turbulence
8.1
Consider a person walking first in air then in water at the same speed. For which motion will the
Reynolds number be higher?
8.2
Consider the flow of air and water in pipes of the same diameter, at the same temperature, and at
the same mean velocity. Which flow is more likely to be turbulent? Why?
8.3
Consider laminar flow in a circular pipe. Will the wall shear stress σw be higher near the inlet of
the pipe or near the exit? Why? What would your response be if the flow were turbulent?
8.4
Prove that for a flow characterized by quantities λ, v, η, ρ, (λ stands for a characteristic length,
all other notations are as usual) the only dimensionless parameter is the Reynolds number, Re,
in some power. Analyze the result obtained.
8.5
A periodic Karman vortex street is formed when a uniform stream flows over a circular cylinder.
Use the method of repeating variables to generate a dimensionless relationship for Karman vortex
shedding frequency fk as a function of free-stream speed v0 , fluid density ρ, fluid viscosity η,
and cylinder diameter D.
8.6
The mean velocity hvi for turbulent flow in a boundary layer may be correlated using the tangential shear stress (sometimes called the wall shear stress) τw , distance from the wall, y, and
31
CHAPTER 8. THEORY OF TURBULENCE
the fluid properties, density, ρ, and viscosity, η. Use the method of repeating variables to find
one dimensionless parameter containing hvi and one containing y that are suitable for organizing
experimental data. Show that the result may be written as
∗
hvi
yu
=
f
u∗
υ
Here u∗ = (τw /ρ)1/2 is the frictional velocity.
8.7
With the help of dimensional analysis, estimate the turbulent viscosity, νturb , dissipation of energy, ǫ, and change of pressure, ∆p, in turbulent flow characterized by ρ, v0 , L0 .
8.8
Estimate the time interval, τ , during which the original distance, L (measured at t = 0), between
two fluid particles become much larger. Use the dimensional analysis and assume that the motion
takes place in the inertial interval.
8.9
Using dimensional analysis, estimate the diffusion coefficient of a turbulent flow in the intertial
interval.
8.10
Obtain the time rate of change of kinetic energy E =
32
ρv2
2
for a viscous incompressible fluid.
CHAPTER 8. THEORY OF TURBULENCE
MEMORANDUM-4
ii
ik i
ρ ∂hv
= − ∂hΠ
∂t
∂xk
Reynolds equation for mean flow
hΠik i = hpiδik + ρhvi ihvk i − hσik i + ρhδvi δvk i
Momentum flux tensor for mean
flow
Viscous stress tensor for mean flow
hσik i = η
∂hvi i
∂xk
+
∂hvk i
∂xi
ii
ii
ρ ∂hv
+ ρhvk i ∂hv
= − ∂hpi
+
∂t
∂xk
∂xi
∂
∂xk
ii
− ρhδvi δvk i
η ∂hv
∂xk
τik = −ρhδvi δvk i
ReN =
vN L N
ν
∼ Recr ∼ 1
The border of laminar-turbulent
transition
Intertial interval
v03
L
Energy decay rate per unit mass
1
vn ∼ (ǫLn ) 3
Ren ∼
Reynolds equation in explicit form
Reynolds stresses
L 1 ≫ L n ≫ LN
ǫ∼
Ln
L0
Magnitude of velocity fluc. on a
scale Ln
34
Reynolds number for a scale Ln
Re
3
Ln ∼ L0 Re− 4
The smallest scale of turbulence
1
vn ∼ v0 Re− 4
The smallest velocity of turbulent
pulsations
33
CHAPTER 8. THEORY OF TURBULENCE
34
Chapter 9
Surface and Internal Waves
9.1
An observer is on the bank of a river which is flowing with the velocity v0 . Under what conditions
does he observe an immobile standing wave (ω = 0) with crests across the river.
9.2
Obtain the dispersion relation for plane internal waves in a liquid with a constant Väisälä frequency without the Boussinesq approximation. Define the criterion of applicability of the latter
in this case.
9.3
Determine the reflection and transmission coefficients for a plane harmonic internal wave incident on the horizontal interface of two fluids with Väisälä frequencies N1 and N2 . The density
is assumed continuos across the interface.
9.4
Determine the velocity of propagation of gravity waves on an unbounded surface of liquid with
depth h.
9.5
Determine the relation between frequency and wavelength for gravity waves on the surface separating two liquids, the upper liquid being bounded above by a fixed horizontal plane, and the
lower liquid similarly bounded below. The density and depth of the lower liquid are ρ and h,
those of the upper liquid are ρ′ and h′ , and ρ > ρ′ .
35
CHAPTER 9. SURFACE AND INTERNAL WAVES
9.6
Determine the relation between frequency and wavelength for gravity waves propagated simultaneously on the surface of separation and on the upper surface of two liquid layers, the lower
(density ρ) being infinitely deep, and the upper (density ρ′ ) having depth h′ and a free upper
surface.
36
CHAPTER 9. SURFACE AND INTERNAL WAVES
MEMORANDUM-5
Surface and internal waves in fluids
1. Linearized set of hydrodynamic equations
∂v
∂t
+
∇p
ρ0
∂ρ
∂t
+
dρ0
w
dz
+ ρ0 ∇ · v = 0
∂ρ
∂t
=
1 ∂p
c2 ∂t
+ ρ0 N g(z) w
+
gρ
∇z
ρ0
+ 2(Ω × v) = 0
2
v = (u, v, w)
N 2 (z) = −g
w|z=0 =
Velocity vector
∂ζ
,
∂t
1 dρ0
ρ0 dz
+
g
c2
Väisälä frequency
p|z=0 = gρ0 (0)ζ − σ∆− ζ
Linearized boundary conditions at a free surface
∇ · v = 0,
For incompressible fluid:
c2 → ∞
2. Surface Gravity waves
∂
∆w
∂t
=0
− g∆− w
∂3w
∂t2 ∂z
z=0
= 0 and













w|z=−H = 0
w(x, y, z, t) = φ(z) exp(ikr − iωt)
Wave equation and boundary conditions
Harmonic wave propagation
φ(z) = b sinh(k(z+H))
sinh(kH)
ω 2 = gktanh(kH)
Dispersion relation
cPh = ωk ,
Phase velocity and group velocity
cg =
dω
dk
√
kH ≪ 1 ⇒ ω = k gH
Shallow water approximation
kH ≫ 1 ⇒ ω =
Deep water approximation
Ek = 41 ρλgA2
√
gk
Kinetic energy of the wave
37
CHAPTER 9. SURFACE AND INTERNAL WAVES
3. Capillary waves
∂
∆w
∂t
∂3w
∂t2 ∂z
=0
+
σ
∆2 w
ρ0
z=0
=0









w(x, y, z, t) = φ(z) exp(ikr − iωt)
Wave equation and boundary conditions
Harmonic wave propagation
φ(z) = b exp(kz)
ω2 =
σ 3
k
ρ0
= γk 3
Dispersion relation
ω 2 = (gk + γk 3 )tanh(kH)
Dispersion relation for gravity-capillary waves
4. Boussinesq approximation for internal waves
∂2
∂t2
→
∂2
∆w
∂t2
∆w +
1 dρ0 ∂w
ρ0 dz ∂z
+ N 2 ∆− w = 0
+ N 2 ∆− w = 0
2
ω 2 = N 2 k2k+k2 = N 2 sin2 θ,
z
N = const
Dispersion relation
38
PART III
Lab Instructions
Lab 1: Channel flow.
1 Goal
The goal of this assignment is to obtain some familiarity with the COMSOL program, and to study the effect of the Reynolds number on laminar channel flow,
including the concept of boundary layers.
2 The physical problem
The physics to be studied is laminar flows of an incompressible and viscous fluid
in an infinitely deep channel (or between two plates). The velocity of the fluid is
specified at the inlet, and the pressure is specified at the outlet.
3 The fluid and the geometry of the problem
The geometry is a channel with a width of 0.6 m and the length 5.0 m, where the
origin (0, 0) is at the lower left corner of the channel. The boundary conditions
are an inlet velocity of first 2.0 m/s and then 0.1 m/s in the x-direction at the left
boundary (inlet), a zero pressure at the right boundary (outlet) and no-slip conditions on the lower and upper boundaries. The fluid to be studied has a dynamic
viscosity of 1.0 kg/ms and a density of 1000 kg/m3 ; these values are valid for
thick oil.
4 The solution to the problem
We will solve this problem using the numerical solver COMSOL. COMSOL is
a MATLAB-based application for modeling and solving engineering problems.
It has a graphical user interface which can be used to do the modeling of the
problem, but it is also possible to do the modeling from a MATLAB program
by calling COMSOL functions. In this course, only the graphical user interface
will be used. More information on COMSOL can be found on the web page:
http://www.comsol.com.
4.1 Log on to the network
To log on to the network you need to have an account at UpUnet-S. Enter your
account name and your password C to log on.
1
4.2 Starting COMSOL
Start COMSOL from the Start menu, by selecting All Programs>Student>Comsol
4.3>Comsol Multiphysics 4.3. The graphical user interface will appear, including the Model Builder and the Model Wizard.
4.3 Model Wizard
1. Go to the Model Wizard window.
2. Click the 2D button.
3. Click Next.
4. In the Add physics tree, select Fluid Flow>Single-Phase Flow>Laminar
Flow (spf).
5. Click Next.
6. Find the Studies subsection. In the tree, select Preset Studies>Stationary.
7. Click Finish.
4.4 Geometry I
Rectangle I
1. In the Model Builder window, right-click Model I>Geometry I and choose
Rectangle.
2. Go to the Settings window for Rectangle.
3. Locate the Size section. In the Width edit field, type 5.0.
4. In the Height edit field, type 0.6.
5. In the Model Builder window, right-click Model I>Geometry I and choose
Build all.
2
4.5 Materials
Material I
1. In the Model Builder window, right-click Model I>Materials and choose
Material.
2. Go to he Settings window for Material.
3. Locate the Material Contents section. In the Material Contents table, enter
the following settings: density ρ = 1000 kg/m3 and the dynamic viscosity
µ = 1.0 kg/ms.
4.6 Laminar Flow
Inlet I
1. In the Model Builder window, right-click Model I>Laminar Flow and
choose Inlet.
2. Select Boundary 1 only.
3. Go to the Settings window for Inlet.
4. Locate the Velocity section. In the U0 edit field, type 2.0 m/s.
Outlet I
1. In the Model Builder window, right-click Model I>Laminar Flow and
choose Outlet.
2. Select Boundary 4 only.
3. Go to the Settings window for Outlet and make sure the pressure is zero.
Wall I
1. In the Model Builder window, select Laminar Flow>Wall I.
2. Go to the Settings window for Wall. Boundary 1 and 4 should now be
overridden. Set the boundary condition to No-slip for the the remaining
walls (boundary 2 and 3).
3
4.7 Mesh I
1. In the Model Builder window, click Model I>Mesh I.
2. Go to the Settings window for Mesh.
3. Locate he Mesh Settings section. From the Element size list, choose Finer.
4. Click the Build All button.
If you zoom in on the inlet and the cylinder you can see the boundary layer that
the physics-controlled mesh gives.
4.8 Study I
1. In the Model Builder window, right-click Study I and choose Compute.
4.9 Results
When the calculation has finished, a surface plot of the velocity field and a contour plot of the pressure field are shown in Model Builder>Results. By clicking
anywhere in the figure panel, the coordinates and field values can be read at the
bottom of the window.
Velocity and streamlines
1. In the Model Builder window, right-click the Results>Velocity (spf) node
and choose Streamline.
2. Go to the Settings window for Streamline.
3. Locate the Streamline Positioning section.
4. From the Positioning list, choose Uniform density.
5. In the Separating distance edit field, type 0.01.
6. In the Model Builder window, right-click Results>Velocity (spf)>Streamline I
and choose Plot.
4
Pressure and contour
1. In the Model Builder window, right-click the Results>Pressure (spf) node
and choose Surface.
2. Go to the Settings window for Surface.
3. Locate the Expression section.
4. In the Expressions edit field, type p.
5. From the Unit list, choose Pa
6. In the Model Builder window, right-click Results>Pressure (spf)>Surface I
and choose Plot.
7. In the Model Builder window, click on Results>Pressure (spf)>Contour.
8. Go to the Settings window for Contour.
9. Locate the Coloring and Style section.
10. From the Coloring list, choose Uniform.
11. From the Color list, choose Black.
12. In the Model Builder window, right-click on Results>Pressure (spf)>Contour
and choose Plot.
4.10 Horizontal Cuts
To see how the velocity and pressure changes along the channel we plot the horizontal cut in the middle of the channel.
1. In the Model Builder window, right-click Results>Data Sets and select
Cut Line 2D.
2. Go to the Settings window for Cut Line 2D.
3. Set the coordinate for the start and end point of the line, (x1 , y1 ) = (0, 0.3)
and (x2 , y2 ) = (5, 0.3).
4. In the Model Builder window, right-click Results and select 1D Plot Group.
5. In the Model Builder window, right-click Results>1D Plot Group 3 and
select Line Graph.
5
6. Go to the Settings window for Line Graph.
7. Locate the Data section.
8. From the Data set list, choose Cut Line 2D 1.
9. Locate the y-Axis Data section.
10. In the Expression edit field, type spf.U to plot the velocity. If you want to
plot pressure instead, type p in the Expression edit field.
11. In the Model Builder window, right-click on Results>1D Plot Group
3>Line Graph 1 and choose Plot.
4.11 Vertical Cuts
To see how the vertical velocity profile changes along the channel we plot multiple
vertical cuts.
1. In the Model Builder window, right-click Results>Data Sets and select
Cut Line 2D.
2. Go to the Settings window for Cut Line 2D.
3. Set the coordinate for the start and end point of the line, (x1 , y1 ) = (5, 0)
and (x2 , y2 ) = (5, 0.6). Tick the box Additional parallel lines and write
range(0,1,5) in the box next to Distances (this creates 6 vertical lines).
4. In the Model Builder window, right-click Results and select 1D Plot Group.
5. In the Model Builder window, right-click Results>1D Plot Group 4 and
select Line Graph.
6. Go to the Settings window for Line Graph.
7. Locate the Data section.
8. From the Data set list, choose Cut Line 2D 2. Now we are only interested
in the velocity, so make sure it says spf.U in the Expression edit field.
9. In the Model Builder window, right-click on Results>1D Plot Group
4>Line Graph 1 and choose Plot.
6
4.12 Decreasing the flow
Change the velocity at the inlet to 0.1 m/s and re-run the calculation. Before you
do, save some figures and plots. See section 4.13 on how to do that and what
figures and plots to save.
4.13 Save images
The report should include answers to the questions and the following plots:
1. Surface plots of both the pressure and the velocity fields.
2. Plots of multiple vertical cuts of the velocity.
3. Plots of a horizontal cut of the pressure and velocity in the middle of the
channel.
This should be done for both inlet velocities. To save a plot, do the following:
1. Locate the camera symbol near the far upper-right corner of your COMSOL
window and click it.
2. A box titled Image Snapshot shows up and tick the box next to Include.
3. From the Target list, choose File and click Browse. . . to select where you
would like to save your image and a filename. Then click OK.
5 Questions
References in the questions relate to the compendium by Pavlenko and Rosenqvist.
1. Explain and give the definition of the Reynolds number and show the dimension of the Reynolds number (see section 6.2.1 and 6.2.4).
2. Calculate the values of the Reynolds number for the two cases in this assignment. Is it likely for any of the two flows to be turbulent according
to theory? (Read section 8.1.1 in the compendium and make some rough
assumptions and conclusions)
3. For an infinitely long channel it is possible to calculate analytic expressions
for a laminar flow.
7
Derive the analytic expressions for the velocity and pressure in an infinitely
long channel (neglecting the effects of the inlet), in terms of the mean velocity vm , the density ρ, viscosity η, coordinates x and y and the width h of
the channel (see section 6.2.2). Assume that vm = v0 .
4. Calculate the analytical values of the velocities and pressures at the center
of the channel, with the assumption of zero-pressure at the outlet for the two
(high and low velocity) cases. Compare with the values for the pressure and
velocity obtained in the numerical simulation at (x, y) = (2.5, 0.3). For
which case is the agreement between experiment and theory best, for the
high or low velocity case? Explain why. Use the plots of the horizontal cut
to see how velocity and pressure varies with x and the vertical cuts to see
how velocity varies with y. Compare to the analytical solution.
5. What is the idea behind the theory of boundary layers (see section 6.3 in
the compendium)? What happens to the boundary layers in the two velocity
cases, i.e. how do they evolve? Use the surface plots and the the plots of the
vertical cuts. For extra information, see the additional info provided for lab
1 which can be found on Studentportalen > Computer labs.
8
Lab 2: Unstable flow
1 Goal
The goal of this assignment is to study the time dependent, unstable flow in a
channel. The instability is caused by an obstacle in the channel, combined with
an increasing Reynolds number.
2 The physical problem
The physics to be studied is time dependent, incompressible and viscous flow in a
channel. In the channel a small cylindrical obstacle is placed. The fluid, initially
at rest, is accelerated until the flow becomes unstable and vortices are created.
3 The fluid and the geometry of the problem
The geometry is a channel with a width of 1.0 m and a length of 4.0 m. A cylindrical obstacle with radius 0.1 m is centered at the coordinate (x, y) = (0.4, 0.4)
where the origin (0, 0) is at the lower left corner of the channel. The boundary condition on the left (inlet) boundary is a time-dependent inflow velocity
vx = 1.2y(1 − y)t which begins with a zero velocity at time t = 0 and then
accelerates with increasing t. On the right (outlet) boundary the pressure is zero.
On all the other boundaries one has the no slip condition, including the surface of
the cylinder.
4 The solution to the problem
We will solve this problem using the numerical solver COMSOL. COMSOL is
a MATLAB-based application for modeling and solving engineering problems.
It has a graphical user interface which can be used to do the modeling of the
problem, but it is also possible to do the modeling from a MATLAB program
by calling COMSOL functions. In this course, only the graphical user interface
will be used. More information on COMSOL can be found on the web page:
http://www.comsol.com.
1
4.1 Log on to the network
To log on to the network you need to have an account at UpUnet-S. Enter your
account name and your password C to log on.
4.2 Starting COMSOL
Start COMSOL from the Start menu, by selecting All Programs>Student>Comsol
4.3>Comsol Multiphysics 4.3. The graphical user interface will appear, including the Model Builder and the Model Wizard.
4.3 Model Wizard
1. Go to the Model Wizard window.
2. Click the 2D button.
3. Click Next.
4. In the Add physics tree, select Fluid Flow>Single-Phase Flow>Laminar
Flow (spf).
5. Click Next.
6. Find the Studies subsection. In the tree, select Preset Studies>Time Dependent.
7. Click Finish.
4.4 Global Definitions
1. In the Model Builder window, right-click Global Definitions and choose
Functions>Ramp.
2. Go to the Settings windows for Ramp.
3. Locate the Parameters section. The Location should be 0, the Slope 1 and
no cut-off.
2
4.5 Geometry I
Rectangle I
1. In the Model Builder window, right-click Model I>Geometry I and choose
Rectangle.
2. Go to the Settings window for Rectangle.
3. Locate the Size section. In the Width edit field, type 4.0.
4. In the Height edit field, type 1.0.
Circle I
1. In the Model Builder window, right-click Geometry I and choose Circle.
2. Go to the Settings window for Circle.
3. Locate the Position section. In the x edit field, type 0.4.
4. In the y edit field, type 0.4.
5. Locate the Size and Shape section. In the Radius edit field, type 0.1.
6. In the Model Builder window, right-click Geometry I and choose Build
All.
Difference I
1. Right-click Geometry I and go to Boolean Operations and choose Difference.
2. Go to the Settings window for Difference.
3. Locate the Difference section. Under Objects to add, click on the empty
field.
4. Select the rectangle and click the + sign. It should now say rI in the Objects
to add field.
5. Under Objects to subtract, click the empty field.
6. Select the circle and click the + sign. It should now say c1 in the Objects
to subtract field.
7. In the Model Builder window, right-click Geometry I and choose Build
All.
3
4.6 Materials
Material I
1. In the Model Builder window, right-click Model I>Materials and choose
Material.
2. Go to he Settings window for Material.
3. Locate the Material Contents section. In the Material Contents table, enter
the following settings: density ρ = 1.0 kg/m3 and the dynamic viscosity
µ = 10−4 kg/ms (compare with air, which has ρ ≈ 1 kg/m3 and µ ≈ 10−5 kg/ms).
4.7 Laminar Flow
Inlet I
1. In the Model Builder window, right-click Model I>Laminar Flow and
choose Inlet.
2. Select Boundary 1 only.
3. Go to the Settings window for Inlet.
4. Locate the Velocity section. In the U0 edit field, type 1.2 ∗ s ∗ (1 − s) ∗ rm1(t[1/s]).
Outlet I
1. In the Model Builder window, right-click Laminar Flow and choose Outlet.
2. Select Boundary 4 only.
3. Go to the Settings window for Outlet and make sure the pressure is zero.
4
4.8 Mesh I
1. In the Model Builder window, click Model I>Mesh I.
2. Go to the Settings window for Mesh.
3. Locate he Mesh Settings section. From the Element size list, choose Finer.
4. Click the Build All button.
If you zoom in on the inlet and the cylinder you can see the boundary layer that
the physics-controlled mesh gives.
4.9 Study I
Step 1: Time Dependent
1. In the Model Builder window, expand the Study I node, then click Step I:
Time Dependent.
2. Go to the Settings window for Time Dependent.
3. Locate the Study Settings section. In the Times edit field, type range(0,0.1,7).
4. In the Model Builder window, right-click Study I and choose Show Default Solver.
Solver I
1. In the Model Builder window, expand the Study I>Solver Configuration>Solver I
node, then click Time-Dependent Solver I.
2. Go to the Settings window for Time-Dependent Solver.
3. Click to expand the Time Stepping section.
4. From the Steps taken by solver list, choose Intermediate.
5. In the Model Builder window, right-click Study I and choose Compute.
5
4.10 Results
Velocity and streamlines
1. In the Model Builder window, right-click the Results>Velocity (spf) node
and choose Streamline.
2. Go to the Settings window for Streamline.
3. Locate the Streamline Positioning section.
4. From the Positioning list, choose Uniform density.
5. In the Separating distance edit field, type 0.01.
6. In the Model Builder window, right-click Results>Velocity (spf)>Streamline I
and choose Plot.
Pressure and contour
1. In the Model Builder window, right-click the Results>Pressure (spf) node
and choose Surface.
2. Go to the Settings window for Surface.
3. Locate the Expression section.
4. In the Expressions edit field, type p.
5. From the Unit list, choose Pa
6. In the Model Builder window, right-click Results>Pressure (spf)>Surface I
and choose Plot.
7. In the Model Builder window, click on Results>Pressure (spf)>Contour
node.
8. Go to the Settings window for Contour.
9. Locate the Coloring and Style section.
10. From the Coloring list, choose Uniform.
11. From the Color list, choose Black.
12. In the Model Builder window, right-click on Results>Pressure (spf)>Contour
and choose Plot.
6
Play movie
1. In the Model Builder window, click on the Results>Velocity (spf) node.
2. Above the Settings window for Velocity (spf), locate the play-button and
press it. There is a slight delay before the movie starts, but be patient!
3. In the Model Builder window, click on the Results>Export>Player I
node.
4. Go to the Settings window for Player I.
5. Locate the Frames section.
6. In the Number of frames edit field, type 71.
7. Locate the Playing section.
8. In the Display each frame for edit field, type 0.5.
9. In the Model Builder window, right-click on the Results>Export>Player
I and choose Play.
10. To see the pressure instead of the velocity, locate the Scene section and from
the Subject list choose Pressure (spf).
11. To manually choose a time frame to display, type the number of that frame
in the Frame number edit field under the Frames section and press the
Generate frame button located above the Settings window.
5 Analyse the solution
The solution is saved for times from 0 s to 7.0 s, in steps of 0.1 s. In Section 8.1
(see 8.1.1 and 8.1.2) of the compendium, the flow around a cylinder is analysed.
Four types of flow are identified:
1. Laminar and stable flow.
2. Stable flow with vortices.
3. Unsteady, periodic flow with vortices created behind the cylinder.
4. Fully turbulent flow without developed vortices.
7
Try to identify these types of flow in the simulation (all of them may not occur).
Write down the time when transitions between different types of flows occur in the
simulation and make estimates of the mean velocities of the fluid at those times.
The interplay between the pressure and the velocity field close to the surface of the
cylinder is important for the formation of vortices. For an analytical estimate on
how the pressure and velocity varies close the the surface of the cylinder and how
vortices are formed see Section 8.1.2 in the compendium or the extra hand-out.
Can you see this process from the numerical results?
5.1 Save images
The report should include answers to the questions above, plots from the simulation documenting the different types of flows that are identified and close up plots
of the pressure and velocity field at the back of the cylinder just as vortices start
to form. To save a plot, do the following:
1. Locate the camera symbol near the far upper-right corner of your COMSOL
window and click it.
2. A box titled Image Snapshot shows up and tick the box next to Include.
3. From the Target list, choose File and click Browse. . . to select where you
would like to save your image and a filename. Then click OK.
6 Questions
1. Which types of flows did you observe around the cylinder, and for which
range of Reynolds number did they occur? Show your calculations, including your approximation of the Reynolds numbers in the simulation. Your
results can be presented in a table with the following structure: type of flow,
time period, calculated range of the Reynolds number.
2. Why do vortices form behind a cylinder, according to theory? Can you
confirm that phenomena with your simulations? See also the additional
information provided at Studentportalen > Computer labs.
8
Lab 3: Dispersive linear waves
1 Goal
The goal of this assignment is to study the time dependent evolution of dispersive
waves, and to understand the connection between the dispersion relation ω = ω(k)
and the time-dependent solution f = f (x, t). We will study the impact of phase
velocities and group velocities on the time-dependent solution.
2 The physical problem
The physics to be studied is the time dependent evolutions of dispersive waves
with different dispersion laws and initial conditions.
3 The geometry of the problem
We will study one-dimensional waves, i.e., waves which travels only in one direction in space. The geometry is periodic, so that waves which travels out through
a boundary will re-appear in the other end of the domain.
4 Theory
When studying linear waves, it is convenient to study “one wave at the time”,
in our case one sinusoidal plane wave at the time, since all waves in infinite or
periodic geometries can be constructed by summing up sinusoidal plane wave solutions.
A plane wave solution in one dimension is a function on the form
f (x, t) = fˆk exp[i(kx − ωt)]
where ω is given by the dispersion relation ω = ω(k). By the principle of superposition, waves with different k can be summed up to give a total solution to the
physical problem. Assume that we know the values of our physical quantity (e.g.
the density or velocity, etc.) at time zero, i.e. the initial condition is given by
f (x, 0) = f0 (x)
1
(1)
This function can be decomposed into its Fourier components by the Fourier transform
f0 (x) =
Z ∞
−∞
fˆ(k) exp(ikx)dk ≡ F[fb(k)]
(2)
where the inverse Fourier transform gives
fb(k)
1 Z∞
f0 (x) exp(−ikx)dx ≡ F−1 [f0 (x)]
=
2π −∞
(3)
Clearly, the function
f (x, t) =
Z ∞
−∞
fb(k) exp[i(kx − ω(k)t)]dk
(4)
fulfills the initial condition (1) at time t = 0. The integrand of (4) is the simple
plane wave solution to the problem, which we discussed at the beginning of this
section, and the integral just represents a superposition of plane wave solutions.
The function f (x, t) is again a solution of the problem because of the superposition principle for linear problems.
The algorithm used by the Matlab application in this lab is:
1. Calculate the inverse Fourier transform of the initial condition,fb(k) = F−1 [f0 (x)]
2. Multiply by the time-harmonic factor, h(k, t) = fb(k) exp(−iω(k)t).
3. Perform the forward Fourier transform to obtain the time-dependent solution: fe(x, t) = F[h(k, t)].
4. Take the real part to obtain the solution f (x, t) = ℜ(fe(x, t))
For the experiments with water waves
the algorithm will correspond to the
below,
∂f
initial conditions ft=0 = f0 (x) and ∂t
= 0, i.e., waves are initiated at rest.
t=0
In the computer program, Fourier transforms are approximated by its discrete
counterparts, which are calculated by using the fast Fourier transform (ifft,
fft) algorithms.
5 Download the waves application
Download the file waves.m from Studentportalen and save it on the C:\TEMP
directory on your workstation.
2
6 Start the waves application
1. Start Matlab from the Start/All Programs/Matlab & Femlab menu. Change
the directory to C:\Temp by typing in the command cd c:/temp at the
Matlab prompt.
2. Type in waves to start the application. You will see the window Integration
of dispersive waves. As default, the dispersion relation is set to ω(k) = k
and the initial function is f (x, 0) = f0 (x) = exp(−x2 /2). At the bottom
of the window you see the time, which you can change with the slider or by
typing in the time.
Feel free to try the functions, including the Animate buttons. Is the behavior of the
wave/pulse expected? Try some other initial functions and dispersion relations if
you like, for example the dispersion relation ω(k) = abs(k) (the absolute value
of k); this relation is the same as for the one-dimensional wave equation, ω 2 = k 2 .
Observe the periodic behavior of the solution, i.e., that waves which leave the
window to the right (or left) re-appear to the left (or right).
7 Gravity-capillary surface waves
We will study gravity-capillary surface waves, which in the deep water limit (see
Section 9.3.2 in the compendium) have the dispersion law
ω 2 = gk + γk 3
(5)
As an initial condition we will have a Gaussian function
f0 (x) = exp(−10x2 )
(6)
where the unit of x is cm (and the unit of k is cm−1 ). We will study water with
γ = 73 cm3 s−2 and the Earth’s gravity g = 981 cms−2 .
1. In the Integration of dispersive waves window, type the dispersion relation ω(k) = sqrt(abs(981*k+73*k b 3)) where abs() means the
absolute value (do you understand why the absolute value is used?).
2. The initial function is written as f0 (x) = exp(-10*x b 2).
3. A suitable End time for the animation is 0.2 s.
3
4. To get the x and y coordinates of a point on the curve, locate Tools in the
menu bar of the Integration of dispersive waves window and select Data
cursor.
5. Click on a desired point on the curve and the coordinates will appear.
If you animate the solution, you will see that the pulse is transformed into two
wave packets, one moving to the left and one moving to the right. After some
time you will see a calm region near x = 0, where all waves have disappeared.
Make a printout of the solution at time t = 0.2 s, and at some more times if
you like, to make references to when answering the questions. Please feel free to
try dispersion relations for other types of waves if you like, such as “gyroscopic
waves” ω 2 = 1/(1 + k 2 ), etc.
8 Questions
References in the questions relate to the compendium by Pavlenko and Rosenqvist.
1. From the dispersion relation of the gravity-capillary waves derive the expressions for the group and phase velocities.
2. Is the group velocity or the phase velocity largest, a) in the long wave limit,
b) in the short wave limit?
3. How far from the origin have the fastest waves moved at the time t = 0.2 s?
Measure the distance to the wavefront of the fastest wave. Use the distance
and time to estimate the average velocity of the fastest waves. Estimate the
wavelength of these waves by measuring the distance from the wavefront to
the previous point with the same y-value and multiplying this with a factor
2. Use this wavelength to calculate a k value and use this k to calculate
the group and phase velocity. Does the group velocity or the phase velocity
agree better with the estimated velocity?
4. The group velocity cg goes to infinity both in the long wavelength limit
and short wavelength limit, so clearly there must be some minimum for the
group velocity, similar to the case for phase velocity described in the book.
Which k gives the smallest group velocity? Do this analytically (no Maple,
Mathematica, Matlab, etc.). What is the corresponding group velocity?
5. How far from the origin have the slowest waves moved at the time 0.2 s?
Measure the distance to the wavefront of the slowest wave. Use the distance
4
and time to estimate the average velocity of the slowest waves. Does this
value agree with the velocity you derived in Q4? Also estimate the wavelength of these waves by measuring the distance from the wavefront to the
previous point with the same y-value and multiplying this with a factor 2.
Calculate the corresponding k and compare with the k you derived in Q4.
The report should contain numerical results requested in the text, plots to support
your data and answers to the questions.
5