Concise Fluid Mechanics - WVU Mechanical and Aerospace ...

Concise Fluid Mechanics
A.V.Smirnov
c Draft date September 12, 2004

Contents
Contents
Preface
Nomenclature
i
v
vii
1 Properties and Variables 1
1.1 Kinematic variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.2 Substantial derivative . . . . . . . . . . . . . . . . . . . . . 3
1.1.3 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.4 Strain rate and vorticity . . . . . . . . . . . . . . . . . . . . 4
1.2 Thermodynamic Variables . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Equations of state . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3.1 Thermodynamic properties . . . . . . . . . . . . . . . . . . 8
1.3.2 Transport Properties . . . . . . . . . . . . . . . . . . . . . . 9
1.3.3 Other properties . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Fundamental Laws 15
2.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . 15
i

ii
CONTENTS
2.1.1 General formulation . . . . . . . . . . . . . . . . . . . . . . 15
2.1.2 Constant density flow . . . . . . . . . . . . . . . . . . . . . 16
2.1.3 Stream function . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . 18
2.2.1 General formulation . . . . . . . . . . . . . . . . . . . . . . 18
2.2.2 Constant density flow . . . . . . . . . . . . . . . . . . . . . 21
2.2.3 Vorticity formulation . . . . . . . . . . . . . . . . . . . . . . 21
2.2.4 Potential flow . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2.5 2D limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.2.6 Viscous limit . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.2.7 Inviscid limit . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.2.8 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . 29
2.3 Pressure Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.3.1 General formulation . . . . . . . . . . . . . . . . . . . . . . 32
2.3.2 Constant density flow . . . . . . . . . . . . . . . . . . . . . 32
2.3.3 Viscous limit . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.3.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . 33
2.4 Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.4.1 General formulation . . . . . . . . . . . . . . . . . . . . . . 34
2.4.2 Constant density flow . . . . . . . . . . . . . . . . . . . . . 37
2.4.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . 39
2.5 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 39
2.5.1 Invariant forms . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.5.2 Non-inertial coordinate systems . . . . . . . . . . . . . . . 40
2.6 The Law of Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.6.1 PI-Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.6.2 Non-dimensional formulations . . . . . . . . . . . . . . . . . 46
2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3 Laminar flows 53

CONTENTS
iii
3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.2 Confined flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.2.1 Flow between parallel plates . . . . . . . . . . . . . . . . . 54
3.2.2 Axially moving concentric cylinders . . . . . . . . . . . . . . 56
3.2.3 Rotating concentric cylinders . . . . . . . . . . . . . . . . . 57
3.2.4 Poiseuille flow through ducts . . . . . . . . . . . . . . . . . 59
3.2.5 Combined Couette-Poiseuille flows . . . . . . . . . . . . . 63
3.2.6 Non-circular ducts . . . . . . . . . . . . . . . . . . . . . . . 64
3.3 Unsteady flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.3.1 Fluid oscillation above infinite plate . . . . . . . . . . . . . . 66
3.3.2 Unsteady flow between infinite plates . . . . . . . . . . . . 68
3.4 Creeping flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.4.1 Stokes flow around a sphere . . . . . . . . . . . . . . . . . 72
3.4.2 2D Creeping flows . . . . . . . . . . . . . . . . . . . . . . . 76
3.4.3 Lubrication theory . . . . . . . . . . . . . . . . . . . . . . . 76
3.5 Boundary layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
3.5.1 Flat plate integral analysis . . . . . . . . . . . . . . . . . . 81
3.5.2 Laminar boundary layer equations . . . . . . . . . . . . . . 85
3.5.3 Blasius solution . . . . . . . . . . . . . . . . . . . . . . . . . 87
3.5.4 Reynolds analogy . . . . . . . . . . . . . . . . . . . . . . . 92
3.5.5 Free shear flows . . . . . . . . . . . . . . . . . . . . . . . . 94
3.6 Integral methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4 Turbulent flows 105
4.1 Transition to turbulence . . . . . . . . . . . . . . . . . . . . . . . . 105
4.2 Turbulence Modeling . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.2.1 LES models . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.2.2 RANS models . . . . . . . . . . . . . . . . . . . . . . . . . 107

iv
CONTENTS
Bibliography 113
A Introduction to Tensor Calculus 115
A.1 Coordinates and Tensors . . . . . . . . . . . . . . . . . . . . . . . 116
A.2 Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
A.2.1 Tensor Notation . . . . . . . . . . . . . . . . . . . . . . . . 120
A.2.2 Tensor Derivatives . . . . . . . . . . . . . . . . . . . . . . . 126
A.3 Curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . 128
A.3.1 Tensor invariance . . . . . . . . . . . . . . . . . . . . . . . 128
A.3.2 Covariant differentiation . . . . . . . . . . . . . . . . . . . . 132
A.3.3 Orthogonal coordinates . . . . . . . . . . . . . . . . . . . . 134
A.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
B Curvilinear coordinate systems 143
C Solutions to problems 147
D Midterm Exam Topics: Laminar Flow Solutions 193
E Final Exam Topics 197
E.1 Fundamental Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
E.2 Analytical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 198
E.3 Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
E.4 Turbulence Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . 202
Index 203

Preface
The idea behind this book is to provide a formal but concise introduction to theoretical
fluid mechanics.
The book covers the traditional topics of fluid mechanics, such as the fundamental
equations of motion, compressible and incompressible forms, and invariant
formulations, special analytical solutions, laminar flows, elements of boundary
layer theory, main aspects of turbulence modeling and numerical methods. The
emphasis is on viscous flow phenomena.
The author tried to put more emphasis on mathematical rigor rather than
on lengthy narrative. Tensor notation is used extensively throughout the book.
However, the knowledge of tensor calculus is not required of the reader, since
enough introductory material is provided in the appendix.
v

!
"
)
¦
'
is defined as ¤ , or ¡ is equivalent to ¤
¦
¦
Nomenclature
. Note: ¡§¦¨¤©¦¡¤
partial derivative over time:
¡£¢¥¤ ¡
¡§¦¨¤©¦
¦ ¡©¦¨¤§¦ ¢
¡
¡ ¦
partial derivative over
control volume
time
:
¡ ¢¡
¦/ .
021¨3
6 ¦7
"54
-th component of a coordinate ( # =0,1,2), or
fluid velocity:
strain tensor
any variable of coordinates and time
stress tensor
viscosity
8
¢ 8(:

¦
¦ "
'
Chapter 1
Properties and Variables
Consider =?>@ a dimensional space of real ACBEDGF numbers representing physical
space of = (= dimension =2,3) and time. The state of the fluid will be represented
, @JIKIL= ), continuous with their
#H
by real functions of coordinates and ( time
derivatives up to the second order. This is an Eulerian description in which both
and represent a set of independent variables. In an alternative Lagrangian
description the fluid is specified by a set of moving particles. These fluid particles
form a continuum and their coordinates are themselves functions of time. Consequently,
time becomes the only independent variable in this case. Our objective
"
will be to formulate the laws of fluid motion in Eulerian, fixed-space coordinates.
The set of independent variables can be extended beyond space coordinates
and time by introducing properties of the fluid. These properties describe
different physical processes and are classified accordingly as thermodynamic
properties, transport properties, etc.
Dependent variables are functions of the independent variables implicitly
expressed in a physical law. Dependent variables can also be classified according
to the physical process they describe, and we shall consider only two types:
kinematic variables and thermodynamic variables.
1.1 Kinematic variables
1.1.1 Velocity
Let’s define a fluid particle as an infinitely small element of the fluid.
1

(1.1) )
¦
'
¦
R
"
F
'
B
)
R
"
R
Y
R
¢
¦
"
'
)
¦
'
'
2 CHAPTER 1. PROPERTIES AND VARIABLES
Definition 1.1.1 Fluid velocity
Fluid velocity, )
¦ 1M3
, 3 ¢N% "54
at a given point in space and time, 1¨3 '
IOIL
is equal to the velocity of a fluid particle
-
:
QP
¦
"54
1¨3
"54
¦ 1M3
"54 ¢SR QP
The definition above can be inverted to define a particle trajectory.
Definition 1.1.2 Particle trajectory
For a given vector field of fluid ) velocities,
a solution to the following problem:
¦ 1¨3
"54
, particle P
¦ 1 "54
trajectory, , is
R QP
¦ 1¨3
"54
¦ 1 "54
P '
QP
¦ 1UT 4
WV
¦
From these definitions it follows that the vector of velocity is tangential to
particle trajectory at each point in the fluid. Particle trajectory is also called a
streamline.
In what follows we shall drop the super-index X , and also adopt the following
dot-notation for particle velocity:
(1.2)
¦¢
meaning that a time derivative of fluid particle position is taken along the particle’s
trajectory at a space point . This notation should not create a confusion since
space
coordinates do not depend on time, while the fluid-particle coordinates
do. Hence when using time derivatives of coordinates, as in (1.2), we will mean
fluid particle coordinates.
We shall use the same dot notation for partial time derivatives of other fluid
variables at a fixed point in space. For example, for any variable, 021¨3 '
"54
(1.3)
0 ¢SY
021¨3
"54
"

¦
¦
¦ "
'
[
¦ "
'
V
"
[
[
¦
0
"
¦
'
¦
'
0
¦
'
)
¦
[
¦ "
'
"
[
R
0
"
>
>
>
¦
"
Y
¦
0
¦
'
¦
'
¦
'
¦
'
¦ "
'
)
¦
'
¦
[
¦
¦
"
>
>
Y
T
Y
¦
'
¦
'
[
"
"
1.1. KINEMATIC VARIABLES 3
1.1.2 Substantial derivative
Definition 1.1.3 Fluid element
By fluid element we shall understand a finite volume of a fluid, which is small
enough that the velocity of all it’s points can be approximated by the velocity of a
single fluid particle inside the element.
Partial derivatives (1.3) describe changes in a variable at a fixed space point
attributed to the explicit time dependence of the variable. There are also changes
brought about solely by the motion of the fluid, i.e. due to the fact that different
fluid elements cross the given space point, changing the fluid variable at that
point. To account for all the changes we introduce a substantial derivative. It is
equal to the rate of change of fluid variable inside a fluid element moving with the
velocity of the fluid.
Consider a change of a fluid variable
fluid element moved to a nearby position 1 0Z1
"54
inside a fluid element as the
:
"54
>\[]
>\[
(1.4)
021
"54
021
"54
021
"54 0Z1
"54
>\[]
>\[
¦ []
" [
Y
where we used a Taylor expansion up to the first order. We can rewrite it in a
more compact tensor notation (Sec.A):
(1.5)
021
021
"54 021
"54
0 ¦ 1
"54
"54
[^ ¦_
[]
[
Considering that the displacement follows the fluid element, it should be
a product of velocity and time: . Then we can rewrite the expression
above in terms of variable change, , as follows:
>^[^
>\[
[^
0 ¢ 021
"54a`
0Z1
"54
(1.6)
>\[]
>\[
021
"54
0 ¦ 1
"54
And dividing both sides by [
, and taking the limit of [
we have:
"cb
(1.7)
dfefg Oi h
¢ R
¦ 0 ¦
>^)

. ¦/k @
l
(1.9)
(1.10) n
)
1
)
R
¦
"
¦
)
1
)
1
)
>
1
)
)
)
4 CHAPTER 1. PROPERTIES AND VARIABLES
this expression represents a substantial derivative of a fluid variable, which describes
the change of that variable in the coordinate system moving with the fluid
, in (1.7) is also called a convective derivative.
element. The last term 1 )
¦ 0 ¦ 4
1.1.3 Acceleration
Applying (1.7) to velocity itself, we have the relation for flow acceleration:
(1.8)
R )
¦f
j
>^)Wj)
where we used the dummy index rule (A.2.16).
1.1.4 Strain rate and vorticity
We can formally represent a velocity ) derivative
an asymmetric parts:
¦f
as a sum of a symmetric and
@
l
¦f `
m ¦ 4
¦fk
@
l
¦f
m ¦ 4
>^)
¦7
where these parts becomes the newly introduced strain rate ( . ¦/
¦/
(n
) tensors.
) and vorticity
>on
. ¦/
Definition 1.1.4 Strain rate tensor
The strain rate tensor, . ¦/
, is defined as:
m ¦ 4
¦f
Definition 1.1.5 Vorticity tensor
The vorticity tensor, n
¦/
is defined as
>^)
¦7k
¦f `
m ¦ 4
@
l

(1.11) n
(1.12) n
(1.13) X
¦
n
n
F
1
1
)
1
)
F
j
rq
F
X
1
)
`
`
1 ; '5s
)
)
j
`
F
4
F
4
1.2. THERMODYNAMIC VARIABLES 5
Definition 1.1.6 Vorticity vector
The vorticity vector, n
, is defined as
¦Gp5¦/
j n
From this definition, and the definition of vorticity tensor (1.10), we have:
¦$
p5¦/
4
@
l
m
j
)Wj
Using the definition (A.23) of the permutation tensor pq¦/ j we can write the
components of (1.11) explicitly as:
r `
@
l
rq
)
r
@
l
)
r 4
@
l
n
1.2 Thermodynamic Variables
Classical thermodynamics was formulated for equilibrium states. Even though
fluid flow is not generally in equilibrium, we can apply thermodynamical concepts
using a quasi-equilibrium approximation, which assumes that the flow changes
slowly enough, so that at each point a local thermodynamic equilibrium is reached.
1.2.1 Equations of state
The equation of state relates important thermodynamic variables, such as pressure,
X , temperature, s , and density, ; :
4

(1.14) X
u
†
l
P
(1.15) R „
(1.16) R ˆ
; Ats
`
P
`
R
u
†
>
l
P
@
6 CHAPTER 1. PROPERTIES AND VARIABLES
such as an ideal gas law:
A where is the gas constant. Depending on the number of parameters in the
system, there can be several equations of state so as to keep the number of
independent uv5wyx variables, to comply to the Gibbs rule [1]:
uzU{}| where is the u
P ~
number of components and is the number of
1}
phases
r,‚
in the
system. For example, for the
4
uz¨{€|
water-vapor mixture we have and
. Thus the state of water-vapor mixture @ can be completely
described by the pressure or temperature only.
, then uƒv5wyx
P ~
uvmwmx
uzU{}|
P ~
1.2.2 Energy
The first law of thermodynamics expresses the principle of energy conservation
related to the thermodynamic variables: internal energy 1 , „ , heat, … , and work,
:
that is, the change of energy of the fluid element is equal to the heat inflow plus
the work done on that element. Rewriting this in terms of specific values, i.e.
values related to the unit of mass (ˆ‰'qŠ'*‹ ), we have:
R …‡>
R ŠŒ> R ‹
Considering only the mechanical work, we have
! :<
@ : ;
where is the specific volume:
, and the minus sign signifies
that the work done on the system is positive when it is compressed and negative
when it is inflated. From the definition of entropy [1] we have:
R ‹
X RŽ
1 In thermodynamics books it is usually denoted by ‘ , but we reserve this symbol for fluid velocity

(1.17) R ˆ
(1.18) R ˆ
(1.19) ˆ
(1.20) ’
(1.21) R ’
`
¢
ˆ
`
1.2. THERMODYNAMIC VARIABLES 7
Thus (1.16) becomes:
R Š
s R .
s R . `
RQ X
which can be expressed in terms of density:
s R . >
X
;
r R ;
This relation implies that ˆ is a function of . and ; only:
1 . ' ; 4
Relation (1.19) constitutes the equation of state as dictated by the energy conservation
law.
as:
Another form of energy identified in thermodynamics is enthalpy. It is defined
X
;
ˆ“>”X
ˆ“>
and analogously to (1.17), we obtain:
X R
;
s R . >
“R X
s R . >
so-called s
Expressing the entropy, , from relations (1.17) and (1.21) we obtain the
.
relations, well known in thermodynamics:
.
s R .
R ˆ•>”X RŽ
R ’
“R X
s R .

(1.22) – v
(1.23) R ˆ
(1.24) –
(1.25) R ’
P
Y ˆ ¢˜—
sZ v Y
’ ¢˜—§Y
sZ Y
–
P
P
8 CHAPTER 1. PROPERTIES AND VARIABLES
1.3 Fluid Properties
Fluid properties are given as free parameters in a physical law. Along with the
space coordinates and time they form the set of independent variables of the
system.
1.3.1 Thermodynamic properties
Specific heat at constant volume – v determines the amount of thermal energy
that is needed to be transfered to the substance to heat it up by one degree,
while keeping it at a constant volume:
For ideal gases the internal energy, ˆ depends only on temperature, and the
relation above can be rewritten as:
– v R s
Specific heat at constant – pressure, , is defined as the amount of energy
needed to be supplied to the substance to heat it up by one degree at constant
pressure:
P
where the enthalpy, ’ , is used instead of internal energy, ˆ , since it accounts for
the work done by the pressure forces to extend/compress the substance. For an
ideal gas this translates to:
R s

(1.26) š$›M)œ
(1.27)
6 ¦/
¢
¥
R
¥
1.3. FLUID PROPERTIES 9
1.3.2 Transport Properties
A general form for the transport law is presumed to obey the gradient approximation
in a form:
1 ! 4 £Ÿž¡ ¢J£
1 ! 4
where is the kinematic or thermodynamic property, which represents a dependent
variable of the problem, flux is the amount of property passed through a
unit area per unit time and gradient is the direction of maximum change in property.
Both flux and gradient are a vectors. The coefficient of proportionality,
!
represents the transport property controlling the transport process (1.26).
The coefficient of viscosity
The coefficient of viscosity is introduced to quantify the process of momentum
transport.
In elasticity theory the resistance force is proportional to defor-
Newtonian fluid:
mation:
¤¥
¢ £ " #
R ˆ§š
=
Similarly in fluid mechanics the resistance to fluid motion is proportional to
the velocity change in the direction normal to the fluid motion (strain).
–ˆt¦
¨ " ¢
. " ¢J£Ž#
=
Since generally the stresses and strains are tensors (Sec.1.1.4, 2.2.1), the
relation above is usually written as:
ˆ . . ¦
¦ . ¦7
Definition 1.3.1 Newtonian fluid
A fluid with the linear relationship between stresses and strains, like (1.27)
is called a Newtonian fluid.

%
6
r 8 )
F
F
¢S©ª
The kinematic viscosity is defined as 9
(1.28)
6
¦
…
1
…
¦
8 R )
® R
R
s
ˆ
…
)
¢
ˆ
4
r
10 CHAPTER 1. PROPERTIES AND VARIABLES
Definition 1.3.2 Coefficient of viscosity
The coefficient of viscosity is introduced as a proportionality constant between
the shear stress and strain. Considering one component of stress tensor
at: i=1, j=2, we have:
r
¦_«%
¦­
Using a more common notation for vector components: )
('q®G',+- we have:
'*‹¬-Q' )('
,¯
Viscosity usually decreases with temperature for liquids and increases for
rarefied gases [2].
Non-newtonian fluid: For non-Newtonian fluids the relation between the stress
and the strain is non-linear, for example
6 ¦/
¦ . B
¦/
Thermal Conductivity
Following the gradient approximation (1.26), we presume the heat transport to
obey the relation:
£ "54
¢J£
1 "
¢

(1.30) º
P
j ¦
¡
r
ˆ
‹
¥
£ . .
"
’
`§½¿¾ ¦ ; j `]½Å¾ ¦ 1 ; Ä
ˆ
4
"
’
4
±
°
P
R
"
ˆ
±
1
"
’
.§. £
"
Ä
®
4
)
¢
ˆ
Ä
j
1.3. FLUID PROPERTIES 11
° ²
conductivity. Dimensionality of , which we shall ± denote as °
by the dimensions of the units in the equation above:
, will be determined
¢ °
¢J£ "
° ²
XWˆ
" #
ž 1
›¨ˆ¡=
›¨ˆ¡=
r
° ²³§·k †
± : 1 ža¸ 4 ° ¹: 1 . ¸ 4
then in SI units: .
Prandtl number: momentum transport / heat transport
¢» 8 –
where –
is specific heat at constant pressure (Sec.1.3.1).
Mass Diffusivity
The gradient approximation (1.26) for the mass transport can be written as:
1 £ . . 4 ¼ ¢

where –
È : –,v
P
–
Ç
ˆ
¨
–
9
½
– ;
P
X —ÃÈ“Y
; Y
°
½
Ê
r
Ä
12 CHAPTER 1. PROPERTIES AND VARIABLES
(1.32)
j ¦
¡
j 4
`§½ ; j ¾ ¦ dfÆ 1
and relating the mass flux to the fluid velocity across the boundary:
we have:
j ¦ ; j ! ¦§¡
,
j ¦ `§½¿¾ ¦ dfÆ 1
!
Ä
j 4
There are two non-dimensional numbers relating the momentum-to-mass
and mass-to-heat transport processes:
Schmidt number:
momentum transport / mass transort
Lewis number:
heat transport / mass transport
1.3.3 Other properties
Speed of sound
The speed of sound for compressible flow is defined as the rate of propagation of
small pressure perturbations, and it is found to be equal to [3]:
F}É

(1.33) Ë
ˆ
À
— Y X ;
; Ê Y
Y ; —
s Y
È
r
À
À
T
T
1.4. PROBLEMS 13
Bulk modulus
The bulk modulus expresses the change of density with increasing pressure at a
constant temperature:
¸¢
; –
and is used in acoustic problems.
Coefficient of thermal expansion
The Coefficient of thermal expansion relates density to temperature changes:
@ `
;
P
and is used in the problems of natural convection.
1.4 Problems
Problem 1.4.1 Mass diffusivity in terms of concentration
Show how to obtain (1.31) from (1.32).
Problem 1.4.2 Lubrication
A plate of mass
incline at Î angle
with viscosity 8
l TÅÍ
with an area
°
slides down a long
–
, on which there is a film of oil of ¡Ì
IO@ ¿ ’ thickness ,
. Assuming the plate does not deform the oil
{ T
: 1 ž . 4 `^Ñ•°
ÐÏ
film estimate (1) the terminal sliding velocity , and (2) the time required for the
plate to accelerate from rest to Ó T !GÒ of the terminal velocity.
IÕÔ‰Ô

14 CHAPTER 1. PROPERTIES AND VARIABLES

¦
R
"
R
;
R
; R
R
"
R
!
>
T
Ö ; )
–
¥
¦
=
¦
R
£
T
Chapter 2
Fundamental Laws
2.1 Conservation of Mass
2.1.1 General formulation
The conservation of mass dictates that:
Ö
!
= . "
which also means that
The total change of mass inside the control volume will consist of changes
of mass inside the volume because of density changes that may occur at each
point of the flow, and the influx or out-flux of mass through the boundary. This can
be expressed as:
Ö
£
= where is the unit normal vector to the R boundary, is the surface area element,
and the last integral spans all the boundary of the control volume.
15

(2.1) R
R
"
R
;
R
!
>
Ö ; )
Ö 1 ; )
(2.4) )
=
¦ £
R
¦ £Ž¦$ Ö×1 ; ) R
; R
" > ; )
R
; R
"
R
R
¥Ö 1 ; !
>
T
T
R
!
1 ; )
R
16 CHAPTER 2. FUNDAMENTAL LAWS
Let’s define the surface area vector, R
£Ž¦
as:
£‰¦$¢
By Gauss theorem we can convert the last integral to the volume integral:
¦ 4 ¦
and finally:
Ö
¦ 4 ¦
¦ 4 ¦ 4
! T
Considering the arbitrary nature of the control volume selection, we conclude:
(2.2)
;
>
1 ; )
¦ 4 ¦$ T
This is a general relation of mass conservation valid for both compressible
and incompressible flows. Differentiating the second term by parts, and using the
relation of substantial differentiation (1.7) the latter can be rewritten as:
(2.3)
¦f ¦G
2.1.2 Constant density flow
For a constant density flow
; T
, and from (2.2) it follows:
¦f ¦G
which is also called the continuity equation or incompressibility condition. Vector
field satisfying (2.4) is also called solenoidal or divergence-free.
Another form of this relation can be obtained by combining (2.3) and (2.4):
(2.5)
T

(2.6)
; )
(2.7)
; )
Ø
F
¦
; )
F
Ø
Ø
Ø
F
F
`
F
)
`
`
'
Ø
Ø
r
j
T
Ø
Ø
F
F
4
4
F
F
F
4 ; r
)
R ;
¦
R
R
¦
2.1. CONSERVATION OF MASS 17
2.1.3 Stream function
Let’s introduce the stream function, which is closely related to the mass flow rate.
The stream Ø function , which is a vector in 3D, also called the streamlinevorticity
function is defined such as to satisfy the relation
or, using the nabla operator (A.32):
¦Gp5¦/
jÙØ­j
¦G£p5¦/
¾
Øaj
which analogously to (1.13) is
1
r `
rq
Ø
(2.8)
) ;
) ;
r• 1
Ø
1
rq
r 4
In two dimensions the streamline function is defined as
¦$Ú% T
'*ØŒi.e.
a vector normal to the plane, and therefore (2.8) are reduced to:
(2.9)
; )
; )
r
r
The following relation between the stream function and the mass flow rate
can be shown for a two dimensional case:
£Ž¦
(2.10)
R
F
r
r“
R Ø
>^Ø
R
; 1 `
>])
R
R
£‰¦
where is the element of the surface normal to the ) velocity R
also be proved more rigorously for a 3D space.
. This relation can
!

(2.11) )
)
)
F
Ö
)
F
R
R
R
R
"
"
R
Ø
¦
¤
F
¦
`
¤
T
Ø
r
F
T
18 CHAPTER 2. FUNDAMENTAL LAWS
Remark 2.1.1 Existence of stream-function
It should be noted that sometimes, instead of definition (2.6), the streamfunction
is defined from a simpler relation:
¦Gp5¦7
jØaj
where the density is omitted. In this case the stream function can only be used to
describe incompressible flow. This can be shown by computing the divergence of
velocity ) vector :
¦K ¦
¦f ¦Gp5¦7
y¦G
¦7
which is true due to the symmetric identity (A.27) and ØÛj
the symmetry of with
respect to the order of differentiation. This becomes especially obvious in a 2D
case:
jØaj
¦f ¦G
rq r•
r
Thus, in terms of definition (2.11) the stream function can only exist for incompressible
flows.
>^)
2.2 Conservation of Momentum
2.2.1 General formulation
According to Newton’s law a particle of mass
force as:
is accelerated by the action of a
1 ¦ 4
which will apply to a particle of both constant and a variable mass. Applying this
in a small control volume, we have
to a fluid particle of density ; and velocity )
(2.12)
1 ; )
¦ 4
!
¦

¦
¤
Ö
Y
"
Y
¤
R —
R
¦
"
R
R
>
"
Ö
¦ !
R š
j
>
¦ !
R š
j
`
š
¦
>
j
j
¦
R
!
¦
£
j
2.2. CONSERVATION OF MOMENTUM 19
The forces acting on a fluid element come from the possible external forces,
like gravity, electromagnetic fields, etc. (body forces), and forces caused by the
interaction of this fluid element with neighboring fluid elements or boundaries
(surface forces). Body forces relate to the unit of volume and surface forces relate
to the unit of area.
¤ƒÜ
¤¬Ý
(2.13)
¤ Ü
¦$
¦ Ö
Ö 6 ¦
j R
š where is the volumetric density of the body force . It corresponds to a force
field like electromagnetic, gravity, etc. Generally it can serve as a source term
connecting this equation to other equations.
Definition 2.2.1 Stress tensor
The surface force term 6 ¦/
in (2.13) is called the stress tensor.
Using this definition and applying Gauss theorem to the last term in (2.13),
we have:
(2.14)
R j
¦
Ö 6 ¦
Now, comparing (2.14) with (2.12), we have
¦ 4a` 6 ¦
! T
1 ; )
j
Using the fact the control volume was chosen arbitrarily, the integral sign
could be dropped, and we have:
(2.15)
1 ; )
jÛ>Þš
¦ 4 6 ¦
Using the definition of substantial derivative (1.7), we have:
(2.16)
1 ; )
1 ; )
4 ¦
j
jc>\š
¦ 4
6 ¦
¦
>])Wj

(2.17)
6 ¦/k `§ß ¦/
(2.18)
6 ¦7Œ `§ß ¦/
(2.19) ã 6 ¦/Œ¢ l 8¬. ¦/ `]ß ¦/Ùá
ã 6 ¦7k l 8¬. ¦/ 8 1
)
(2.20)
¦
Y
"
Y
Y
"
Y
¦
`
X
`
X
¦
20 CHAPTER 2. FUNDAMENTAL LAWS
Using the hypothesis of Newtonian fluid (1.27), and general considerations
of symmetry for the case of isotropic and homogeneous fluid, a general relation
can be written as [2, p.66]:
between the stress tensor 6 ¦7
and the strain tensor . ¦/
j
where X is the pressure, á is the coefficient of bulk viscosity, which is only important
for compressible flows.
Sometimes it is convenient to separate the stress tensor (2.17) into the
pressure-related and viscous parts:
X>
)Wj
l 8¬. ¦/ `àß ¦/Ùá
Xâ>Úã 6 ¦7
where ã 6 ¦/
is the viscous stress tensor defined as
j
Parameter á is the coefficient of bulk viscosity, which can only be important for
variable density flows [2]. Thus, for incompressible flows (2.19) becomes:
)Wj
¦f
m ¦ 4
where we used the definition of strain rate tensor (1.9). Using the definition of the
viscous stress (2.19) we can write (2.16) as:
>^)
(2.21)
1 ; )
1 ; )
4 ¦
j
j
š When
(2.21) as:
represents the gravity forces: š
¦ 4
¦
>])Wj
>Þš
¦H ; ¦
we can rewrite equation
6 ¦
j >äã
(2.22)
¦
¦ 4
> ; ])Wj
1 ; )
1 ; )
4 ¦
j
6 ¦
j ã
j

)
¦
)
ã
j
j
j
j
p
j
j
)
9 )
j
j
p
j
j
j
1
)
P
1
)
P
1
)
P
æ
1
`
`
`
æ
`
æ
X
T
)
¦
> ;
)
æ
)
)
P
æ
æ
4
P
4
P
¦
j
2.2. CONSERVATION OF MOMENTUM 21
2.2.2 Constant density flow
The viscous term in equation (2.22) can be further simplified for constant density
flows. Using (2.20) we can rewrite it as:
l 8(. ¦
8 1
¦K
¦ 4
(2.23)
j
jc>^)Wj
6 ¦
¦f
j*jÛ>^)Wj
j
4 ¦ 4 8 )
8 1
¦
4 8 1
¦f
¦f
j*jÛ>
)Wj
j*j
where we used the continuity )åj relation:
Substituting this into (2.22), we have:
j
for incompressible flow (2.4).
(2.24)
¦f
j
¦f
j*j
>^)WjÙ)
which is an incompressible form of momentum equation, also referred to as the
Navier-Stokes equation (NS).
2.2.3 Vorticity formulation
Our objective will be to replace the velocity vector in a constant density NS equation
(2.24) with a vorticity vector (1.11). For this purpose consider a cross product
between nabla operator (A.32) and vorticity vector:
(2.25)
p5¦/
¾
¼p5¦/
n­j
j n­j
Using (1.12) the cross product (2.25) can be rewritten as:
p5¦/
“
pm¦/
1 p
454
@
l
j n­j
P æ
p5¦/
@
l
(2.26)
Pæ
¦/ p
@
l
Using the tensor identity (A.29):
Pæ

p
j
j
1
)
P
ß ¦ 1
P
`
æ
@
l
æ
)
ß ¦ 1
P
P
)
p
j
æ
j
p5¦
Pæ
4
P
1
)
P æ
`]ß ¦
P æ
(2.29) )
T
j
`
æ
)
æ
ß ¦
P
æ
ß ¦ 1
P
)
æ
P
ß
j
æ
æ
æ
4a`]ß ¦
æ P
`]ß ¦
æ P
@
l
1
)
)
)
)
P
)
)
)
æ
)
æ
æ
ß
P
j
P
P
1
)
P P
æ
1
) )
>
)
)
)
P
`
æ
`
æ
æ
P
)
)
æ
)
æ
4
P
454
P
)
æ
4
P
)
T
22 CHAPTER 2. FUNDAMENTAL LAWS
p5¦/
ß
`]ß
which can be rewritten as
¦/ p
ß
`àß ¦
ß
P
to match the indexes, we can simplify the cross product (2.25):
P æ
@
l
@
l
¦/p
ß
`]ß ¦
ß
4 1
P æ
ß
ß
ß
ß
ß
ß ¦
ß
@
l
¦f} `
m ¦/ `
m ¦/
¦f} 4
(2.27)
>^)
¦f€ `
m ¦/
And finally (2.26) becomes:
(2.28)
p5¦/
“
¦f€ `
m ¦/
jn­j
For constant density flows it follows from (2.4) that )
, and
m ¦7 1
m 4 ¦G
j n­j
¦f}¼p5¦/
Thus we can replace the diffusive ) term
above.
¦f}
in (2.24) with the cross-product
¦f
Now let’s consider the )
convective term in (2.24). Using the constant
()
density assumption ) and now considering the cross product of the type
¦f ¦c
n­j , we can repeate the steps as in (2.26) and obtain 1 :
pm¦/
jÙ)
(2.30)
p5¦/
çžEžEžQ
¦f `
4 ¦
@
l
jÙ)
n­j
1 See Problem 2.7.1

j
ë
¦
> )
¦
> )
¦
> )
1
) )
1
) )
1
) )
)
)
>
)
>
ë
1
) )
j
>
)
4
T
2.2. CONSERVATION OF MOMENTUM 23
Thus
(2.31)
@
l
¦Kk£p5¦7
4 ¦
jè)
n­jc>
And substituting (2.29) and (2.31) into the momentum equation (2.24) we
have:
(2.32)
4 ¦
p5¦7
¦
@
l
¦ ` ;Ãé F X
j)
n­j
j naj
9 pm¦/
Rearranging the terms, and using the relation
¦f
(A.34), we have:
ß ¦/
(2.33)
9 p5¦/
@ 1
l
4
` ¡
4 ¦$ ` p5¦/
> ;Ãé F X
jÙ)
n­jê>
jn­j
This equation can also be rewritten as:
(2.34)
@
X ;
4a` è
4 ¦$¼p5¦/
`
@ 1
l
1 9 n­j
naj
This is a NS equation in vorticity formulation for the incompressible flow 2 .
2.2.4 Potential flow
¦
Let’s consider the irrotational flow where the vorticity vector (n is zero ). This
flow is also called potential flow, since the velocity vector can be replaced by a
gradient of a ë scalar function, , also called a velocity potential function:
¦G
¦
This is possible, because the gradient of a scalar function also satisfies the condition
of zero vorticity, which follows from the definition of the vorticity vector (1.12)
and the symmetry of the second derivative ë of with respect to order of differentiation,
ë
j
:
2 Note that we achieved only a partial success in our objective to replacing the velocity vector with the
vorticity vector, but that’s the best we can do.

(2.35) )
p
Ý
n
¦
>
¦
> )
p
j
Ý
1
)
1
) )
`
s
Ý
¦
ë
>
Ý
s
T
j
1
ë
–
¥
j
`
ë
T
24 CHAPTER 2. FUNDAMENTAL LAWS
@
l
m
j
@
l
4
j
¦G
pm¦/
4
p5¦/
)åj
In the case of steady state incompressible potential flow the continuity condition
(2.4) translates into the Laplace equation for the velocity potential:
¦f ¦G
¦ì¦G
Thus, the solution of the problem in this case is reduced to finding a single scalar
function ë from equation (2.35). Another important relation in this case can be obtained
from equation (2.34), which, after eliminating time derivatives and vorticity
terms reduces to:
(2.36)
X
;
4í` è
@
l
= . "
which is a weak formulation of the Bernoulli’s Equation 3 .
2.2.5 2D limit
Let’s rewrite (2.33) as
(2.37)
¡ ¦G
` p5¦/
jè)
n­jc>
j naj
Where we denoted the term in parentheses by . The equality above is a first
. Let’s now form a cross product between
this equality and ¡ Ý
¦
of the type p î¦ ¾ ¦$p î¦ ¦f
(see also (A.32)):
j p5¦/
rank tensor equality with terms of type s
9 pm¦/
(2.38)
¾ î¦
)
î¦ ¾ ¨¡ ¦$£p
î¦ ¾ 1 ` p5¦/
4
Using the definition of ¾ ¦
(A.32), we get:
jÙ)
n­jê>
jn­j
9 p5¦/
3 See also (2.78)

p
Ý
Ý ¤
ß
Then the term ¤
Ý
j
)
>
p
Ý
j
>
Ý
)
n
j
`
j
Ý
Ý
Ý
Ý
p
Ý
1 Ý
)
)
n
Applying the same manipulations to the Ä
Ä
Ý
¤
j
Ý
'
T
)
Ý
`
n
)
Ý
Ý
ß
Ý
n
n
n
j
Ý
Ý
)
`
)
n
Ý
n
)
Ý
` ³
)
)
j
Ý
Ý
F
n
p
Ý
Ý
Ý
n
n
Ý
j
>
Ý
Ý
r
'
T
)
1
)
T
T
Ý
Ý
Ý
n
n
n
)
Ý
Ý
Ý
2.2. CONSERVATION OF MOMENTUM 25
9 pm¦/
¦f
î¦M¡ ¦ì£p
î¦ 1 ` p5¦7
4
î¦
)
(2.39)
n­j
> 9 p5¦7
n­j
jè)
n­jê>
j naj
` p5¦7
î¦ 1
4
î¦
y
£¤
Ý
Ý
where we used abbreviations
is a symmetric tensor then by (A.2.22), we have p
¤
Noticing that by the definition of vorticity n vector:
the equation (2.39) to the form:
'*Ä
for the last two terms on the RHS. Since
.
î¦M¡ƒ ¦ì
£p
î¦
¦K
¡¬ ¦/
, we can reduce
>\Ä
(2.40)
Using the permutation property (A.24): p Ý
, we have:
£¤
>\Ä
î¦$
` p
¦ì(pm¦
4
` pm¦/
p5¦
1
naj
4 $ ` 1 ß
`]ß y ß
4 1
naj
4
`tß
1
naj
4
ß m ß
naj
4 $ ` 1
4
³
>])
³
>^)
where we used the incompressibility condition (2.4): )
ï% T
In a 2D n
limit we have
the 3-rd component of the equation above, i.e.
'mnŒ- and )
¦aï%
¦f ¦G
'q)
.
- . Let’s consider only
`
³
>^)
(2.41)
n
¼¤
>^Ä
can be simplified as:
term, we have:
T ` T
>^)
(2.42)
yÚžEžEžQ 9 1
î 4
9 p5¦/
pm¦
n­j
And in the 2D limit:

(2.44) )
@ ß ¦
±
P
l
ß ¦ 1
P
Ø
æ
æ
æ
`]ß ¦
æ P
(2.46) n
F
r
n
æ
n
FðF
n
P
n
P
Ø
4
Ø
æ
@
l
)
æ
¦
n
`
j
n
n
`
P
òß
P
1
Ø
`
)
Ø
j
j
@
l
j
P
F
`
`
Ø
n
n
` 9 n
æ
Ø
r
j
æ
j
P
Ø
Ø
`
P
æ
P
æ
æ
`
P
Ø
p
j
j
æ
ß
Ø
j
æ
¦ ß
P
T
T '
¦ ß
P
Ø
Ø
4
Ø
æ
Ø
Ø
æ
æ
P
P
æ
P
P
j
j
4
4
j
²
j
4
j
'
T
26 CHAPTER 2. FUNDAMENTAL LAWS
9 1
rðr 4
Ä
ð
9 1}T `
rðr 4
FðF
FðF
` 9 1
rðr 4
¦Õ¦
After substituting ¡ , Ä into (2.41), it becomes:
>on
(2.43)
9 n
¦ `
¦ì¦
which is the 2D limit of (2.37).
Stream-function formulation
By the definition of the stream-function (2.6) we have:
jØaj
¦Gp5¦7
Substituting (2.44) into (1.12) we can find relation between the vorticity vector
and the stream-function:
@
l
¦$
p5¦/
1 p
pñ
P æ
Pæ
1 p
¦/ p
¦/pñ
@
l
Pæ
Pæ
ß
`]ß ¦
ß
1 ß
ß ¦
`àß
@
l
ß
ß
ß ¦
jc>
m ¦/ `
¦f} `
¦f
¦
(2.45)
j*jê>]Øaj
m ¦/ `
¦f}
In the 2D limit:
the first of the last two terms in (2.45) will Ø vanish:
¦ò«%
'q
- , )
¦ò«%
'*)
- , n
¦òÐ% T
y k
, and we have:
'5nŒ- , and Ø
«% T
'*،- ,
¦ì¦

1
¦
'
(2.48) Ø
Ø
)
n
)
¦
Ø
T
`
T
1
¦
'
2.2. CONSERVATION OF MOMENTUM 27
Now (2.43) can be rewritten in terms of the stream-function only:
(2.47)
jqj
9 Ø
Thus, the flow is now completely defined by a
"54
Ø
scalar field , which
is obtained as a solution of (2.47). Note that since (2.47) contains 4th order
"54
Ø
derivatives of , it involves more complex boundary conditions, and poses
higher differentiability ë requirements on .
In the case of irrotational (n flow
Laplace equation for the stream-function:
), equation (2.46) reduces to the
¦
¦ì¦
j*j
j*j
¦Õ¦
with the boundary conditions derived from the relation between Ø and the velocity
field (2.9). Solving the equation for stream function is usually preferred over solving
an equation for the vorticity, since the velocity field can be obtained from the
stream function by a simple differentiation of type (2.6) or (2.9), whereas obtaining
the velocity field from the vorticity as in (1.12) would require a more laborious
integration.
2.2.6 Viscous limit
Consider the incompressible NS equation (2.24). In the viscous limit we shall
assume the viscous term to be much larger than the convective term. Thus in the
viscous formulation we shall simply neglect the convective terms:
(2.49)
¦
¦G 9 )
¦f
j*j
` º
;
Using the expression of vorticity vector (1.12), we can express the above
equation in terms of vorticity vector only:
(2.50)
9 n
m
j*j
k
which is an incompressible viscous limit of the NS equation in vorticity formulation
(see Problem.2.7.3).

j
)
p
)
j
Y
"
Y
¾ R m¦
R
j
¦
)
)
R
n
R
R
¦
"
>
"
p
p
j
j
)
)
T
;
j
j
`
º
º
)
T
)
28 CHAPTER 2. FUNDAMENTAL LAWS
In the case of steady state flow this equation simplifies to a Laplace equation
for the vorticity vector:
m
j*j
2.2.7 Inviscid limit
The fluid with a zero viscosity is called an ideal fluid, and the flow of such a fluid
is called inviscid. Consider equation (2.22) in the limit of inviscid flow, when the
viscous tensor, 6 ¦
j , vanishes:
(2.51)
1 ; )
1 ; )
4 ¦
j
¦ 4
¦
>])Wj
In the incompressible limit it will simplify to:
(2.52)
¦
` º
¦f
j
>])åj)
;
This is Euler equation for inviscid incompressible flow. It can also be rewritten
in terms of substantial derivative (1.8):
(2.53)
¦
R )
` º
Conservation of vorticity
It can be shown that the inviscid flow preserves vorticity. For this purpose let’s
form a vector product between the nabla operator and equation (2.53):
m¦
¦/k
¦G
m¦
¦Kp
where the last equality is due to the symmetry º of
according to:
y¦
transform the j term p
¦f
¦/
and identity (A.27). We can
" )
(2.54)
¦/
m ¦åp
m¦ 1
¦f `
y ¦ 4
l p
m¦
¦fk¼p
m¦
¦f

T
Y
"
Y
ó ¦
(2.56)
j
Y
"
Y
>
R
R
T
¦
T
1 ; )
ó¦
j
¦
–
`
4
¥
º
2.2. CONSERVATION OF MOMENTUM 29
where we used the skew-symmetric property of pq¦/ j (A.24).
definition of vorticity vector (1.12), we obtain:
Finally, using the
(2.55)
¦
¦$
which means that the vorticity (n = . "
is conserved ). In particular, this
¦G
means
that if the flow (n
was irrotational ), it will remain so 4 . In this case the problem
of inviscid flow can be solved using velocity potential function (Sec.2.2.4).
" n
The momentum flux
Since )åj
for incompressible flow (2.4), we can rewrite (2.51) as:
j
1 ; )
1 ; )
4
j
¦ 4
¦
)Wj
And introducing the momentum flux as
¢ ß ¦
j躇>
)åj
we have the Euler equation in momentum-flux formulation:
(2.57)
1 ; )
¦ 4 `
j
2.2.8 Boundary conditions
Equation system (2.2), (2.21), (2.19) and (1.13) may not have a unique solution
for any boundary conditions. Generally, the character of the equation system,
i.e. hyperbolic, parabolic or eliptic [4, 5], may change depending on the boundary
conditions and the region of space and time. However, there are several types
of boundaries that are typically considered, and that usually lead to well posed
problems.
4 See Problem 2.7.4

j
)
F
)
T
F
r
30 CHAPTER 2. FUNDAMENTAL LAWS
Inlet
At the inlet boundary the value of the velocity is usually specified . This boundary
condition is known as Dirichlet boundary condition. Note, that this is not always
the case, since a pressure can be prescribed as the inlet condition instead, when
the Poisson equation for pressure (2.60) is used.
Outlet
Depending on the character of the equation system the boundary conditions may
or may not need be specified at the outlet boundary. The most common outlet
boundary condition is the condition of the zero boundary-normal velocity derivative
(Neuman boundary).
In more complex flow situations there may not be a clear distinction between
the inlet and the outlet, since the flow may reverse. These types of situations are
hard to solve in a consistent manner and should be avoided by repositioning the
inlet/outlet of the domain so as to comply to either Dirichlet or Neuman boundary
conditions.
Fluid-solid interface (Wall)
In the case of a fluid-solid boundary the flow velocity is set equal to the velocity of
the wall, which covers the cases of both stationary and moving boundaries. This
boundary condition is called a no-slip boundary condition.
In some cases a finite velocity jump may be imposed at the boundary, in
which case this is called a slip boundary condition.
The specification of the velocity alone at the boundary may not be enough,
since the momentum equation (2.21) contains second order velocity derivatives
in the viscous (ã 6 ¦
j term ), which means that the first order derivatives should be
given at the boundary. However, with the no-slip condition at the wall, the velocity
derivatives at the wall can be considered zero. This follows from the continuity
relationship (2.4). For example, if we consider velocity )
components ) and as
being parallel to the wall, and ) normal to the wall, then from the no-slip condition
we have:
r
and consequently:

(2.58) X
)
F
F
`
)
F
)
F
1 @ £ö
A
`
)
>
T
A
@
¯
4
T
2.2. CONSERVATION OF MOMENTUM 31
rq r“
Thus from continuity (2.4) we must have:
rq r“
)
In cases when the forces on the wall need to be estimated they can be
related to the boundary normal forces due to pressures, and shear forces that
can be related to the stress tensor via (2.14).
It should be noted that if the boundary is moving with acceleration additional
non-inertial terms should be introduced into the boundary conditions (Sec.2.5.2).
Fluid-fluid interface (Free surface)
The gas-liquid or liquid-liquid boundary conditions are also called free-boundary
conditions. They consist of the requirements that the pressure, velocities and
the fluxes of mass and momentum be continuous functions across the interface.
This means that these quantities should have the same values on both sides of
the interface. The position of the interface surface will then be determined as a
solution to the flow equations subjected to the free surface boundary conditions.
In cases where surface tension effects are important, they should enter into
the pressure boundary condition, namely the extra boundary pressure should be
added on both sides of the interface. This pressure should be inversely proportional
to the local surface curvature. Since the surface curvature generally
depends on the direction selected on the surface to measure the curvature, one
form of its estimate may be to set it proportional to the sum of inverse curvature
radii in two orthogonal directions:
ÝUô
xyõ
where ö is the coefficient of surface tension. Note, that the forces resulting from
the pressure terms should always act normal to the surface. Shear forces at the
boundary are usually considered to be zero.

¦ Y ¾
Y
(2.60) X
"
>
Y
Y
"
¦ì¦G ` Y
1
Y
>
1
X
>
X
`
>
¦
32 CHAPTER 2. FUNDAMENTAL LAWS
2.3 Pressure Equation
2.3.1 General formulation
In the solution of the equation system (2.2), (2.21), (2.19), and (1.13), is complicated
by the fact that the equation of mass conservation (2.2) does not contain
pressure, and the equation of momentum conservation (2.21) contains both velocity
and pressure [6]. In the case of compressible flow the equation of state
(1.13) can be used as an additional relation between pressure and density. However,
for the incompressible flow the continuity equation (2.4) has velocity only,
and can not be effectively used in combination with the momentum equation.
To make the equation system better conditioned, the continuity equation (2.4) is
usually replaced by the Poisson equation for pressure. To obtain the
¦÷¢
equation
for pressure, let’s apply the divergence operator, , (A.2.32) to the
momentum equation (2.22):
¾
Y : Y
1 ; )
1 ; )
4 ¦
j
6 ¦
j ã
j
4 ¾ ¦ 1 ¦
4 `C¾ ¦
¦
¾ ¦ 1 ; ¦ 4
¾ ¦
)Wj
(2.59)
¦ 4 ¦
4 ¦ `
¦ì¦
1 ; )
1 ; )
4 ¦
j
> ; ¦f ¦
6 ¦
j >äã
¦
j
)Wj
After substituting the first term on the LHS from (2.2), considering that J¦ is
a constant, and rearranging terms, we have:
1 ; )
1 ; )
4 ¦
j
6 ¦
j ã
j
¦ 4 ¦ ` 1
4 ¦
"54
)Wj
which is a Poisson equation for pressure. It has to be solved together with the
momentum equation (2.21) and the relation of the state law (1.13).
2.3.2 Constant density flow
As it was pointed out the pressure equation is mainly used for incompressible
flows where it replaces the continuity equation (2.4). In this case we can simplify
the pressure equation (2.60) by applying the continuity ) condition to (2.60):
¦K ¦

(2.61) X
(2.62) X
T
X
)
)
¦
; `
)
` ;
)
j
T
`
)
j
j
j
j
¦
j
¦
j
1
2.3. PRESSURE EQUATION 33
¦ 4
4 ¦
jÙ)Wj
¦ì¦$ ` 1q1 ; )
ã 6 ¦
` ; 1
¦f ¦ 4
¦f
¦
jÙ)Wj
j )Wj
>Úã
6 ¦
¦K
¦
jÙ)Wj
>Úã
Using the incompressible form of the viscous stress tensor ã 6
, (2.20), and the
continuity relation, )åj
j
, it can be shown that the last term will be zero:
6 ¦
6 ¦
j ã
¦G l 8¬. ¦
j j
j
¦K
j*j
¦ 4 ¦f
j*jc>
j
and finally we obtain:
¦W 8 1
¦
¦ 4 8 151
4 ¦Õ¦ 4 T
>])åj
)Wj
¦
¦ì¦G ` ; )
¦f
j)Wj
This is the incompressible form of the pressure equation, also called the Poisson
equation for pressure. It should be considered together with the momentum
equation (2.24).
2.3.3 Viscous limit
Considering the viscous limit (2.49), and taking divergence of this equation, we
have:
¦ì¦$
which is the Laplace equation for pressure.
2.3.4 Boundary conditions
Pressure equation is an elliptic second order PDE. As such it requires the specification
of two sets of boundary conditions, which usually are the values of the
pressure and it’s boundary normal derivatives.
At the solid walls the boundary-normal derivative of pressure is usually set
to zero, which is a Neuman boundary condition. The value of pressure at the wall

"
R
R
ˆ
R
R
Ö R
R
"
R
R
R
R
"
¦
†
¦
)
>
)
¦
)
l
¦
¦
; R
"Qù R
R
R
!
!
34 CHAPTER 2. FUNDAMENTAL LAWS
comes out as the solution which satisfies this condition. However, because the
Poisson equation is a second order equation, the Neuman boundary condition
alone will result in an indeterminate solution, when adding any constant to the
pressure will still satisfy the equation. Fixing the value for the pressure in at least
one point will remove this uncertainty. Thus, other conditions at the inlet/outlet
boundaries are usually applied. At the inlet/outlet pressure values are given and
the boundary normal derivatives are usually set to zero.
In the case when the velocity is also specified at the inlet/outlet, both pressure
and velocity specifications should be consistent so as not to create a overdefined
problem. Specifying either pressure or velocity alone will be enough in
many cases. However, this will depend on the character of the flow and the discretization
scheme used to solve the equations [4].
2.4 Energy Equation
2.4.1 General formulation
Usually the flow field is a carrier for the transport of other variables of the continuum
media. One important variable is energy.
The balance of energy in a control volume can be written as
(2.63)
" „
" …‡>
where the LHS is the rate of energy change inside the control volume, and the
two terms on the RHS represent total heat inflow into the control volume and work
done on it. This relation is also known as the second law of thermodynamics.
Note that both heat and work are transported into the control volume through its
boundary, so they can be represented by flux-vectors.
Energy:
¦ `

Š
¦
R
R
¦
Š
R
R
R
R
R
R
"
s
s
¦
R
R
!
s
!
2.4. ENERGY EQUATION 35
where the minus sign in front of the gravity term means the potential energy increases
as we move against the gravity force.
Heat: The total heat change inside the volume can be related to the heat flux
through the boundary of the volume:
(2.65)
" …
` Ö ¦
R Š
£Ž¦
where is the heat flux, which is the rate of heat inflow through a unit area per
unit of time, and
£‰¦
the vector-element of the boundary introduced in (2.1). Minus
R
sign occurs because of the convention of surface normal vectors to point outside
of the volume, meaning Š that is actually the ”out-flux” of heat … whereas was
defined as an incoming heat by the virtue of (2.63). Applying the Gauss theorem
to (2.65) we have:
(2.66)
` Ö ¦
R Š
` Ö £‰¦$ ¦K ¦
R Š
" …
It is postulated that the heat flux is proportional to temperature gradient:
¦$ `t°ú¾ ¦
`C°
¦
where the minus sign signifies that the heat flows from higher to lower temperatures.
This relation is known as the Fourier’s law. Substituting it into (2.66), we
have:
(2.67)
Ö×1 °
¦ 4 ¦
" …
Work: In analogy to (2.65) we introduce the flux of external work through the
fluid element:
(2.68)
£‰¦
¥Ö †
‹
The work flux vector ‹
through the area can be computed as

š
¦
R
R
"
)
‹
R
R
"
š
)
¥
¢
‹
R ž
R
¦/m ¦G 6 y¦f ¦G R 6
R
)
— R
R
)
1òR
R
"
"
)
"
. # £
XW›
)
1 ; )
)
R
!
)
)
"
)
!
¦
36 CHAPTER 2. FUNDAMENTAL LAWS
–ˆ
" R
–ˆ
ˆ¡=
We saw in (2.2.1) that the surface forces are described by the stress tensor
¦/
. The time derivative of the displacement of fluid element is given by velocity )
6
(1.1). Thus 5 :
(2.69)
¦G
Applying Gauss theorem to (2.68) and combining it with (2.69) we obtain:
6 ¦/
(2.70)
† Ö1 6 ¦/
4 ¦
Differentiating by parts, we have:
(2.71)
1 6 ¦/
4 ¦G
6 ¦/y ¦ 6 ¦/
>
We can eliminate the derivative of : 6 ¦7m ¦
replaced by the gravity force ` ; J¦
:
6
m ¦
, by expressing it from (2.15) with
; è
where we used the symmetry of 6 : 6 ¦/k
. Now we can rewrite (2.71) as:
4a`
6 m¦
1 6 ¦/
4 ¦G
4a` ; ¡ 4
m ¦
Substituting the above into (2.70), we have:
1 ; )
> 6 ¦/
(2.72)
1 ; )
† Ö ø
m ¦
4a`
; è
> 6 ¦/
ù R
Combining (2.64), (2.67) and (2.72), we have
5 Note that the positive signs in (2.68) and (2.69) follow the convention that the work done on the system
is positive when the direction of external force coincides with the direction of displacement.

"
1 ; ˆ
4a` ¦ ; )
(2.74) – v
; R ’
(2.76) "
R
"
"
Ö
¦
1 ; ˆ“>ûX
X üR
" >
R
s
"
"
"
R
1 ; ˆ
¦
ˆ
"
1 ; ’
s
)
R
R
¦
)
l
"
s
¦
1 ; )
; R
"
R
s
)
s
R
)
s
)
¦
>
>
)
)
¦
)
l
¦
)
l
)
¦
)
l
¦
¦
)
"
)
; R
"ù R
R
; R
"
R
¦
1 ; )
; R
"
R
>
!
!
)
¦
)
l
)
¦
–
; R
"
R
¥
)
2.4. ENERGY EQUATION 37
øR
R
1 ; 1
` ; )
" >
4a`
; ¡
y ¦
¥Ö ø 1 °
¦ 4 ¦
> 6 ¦/
>\)
ù R
which after rearranging terms and dropping the integration sign due to the arbitrariness
of our choice of the control volume becomes:
¦,R )
¦ 4 ¦
¦ 4a`
¦ ; ¦
m ¦
R
R
¦$R
R
1 °
> ; )
" >
And after canceling the same terms on both sides, we have:
> 6 ¦/
>])
(2.73)
R
R
> 6 ¦/
m ¦
4 1 °
¦ 4 ¦
which is the equation for the rate of change of energy density valid for both compressible
and incompressible fluid.
In an ideal gas approximation we can express the LHS in terms of temperature
using the thermodynamic relation (1.23):
y ¦
R
R
4 1 °
¦ 4 ¦
It is useful to express (2.73) in terms of enthalpy (1.20). For this purpose we
to both sides of (2.73) and obtain:
can add R X : R
1 ; s
> 6 ¦/
(2.75)
R
R
R 4
R
ïR X 4
" >
R
> 6 ¦/
m ¦
1 °
¦ 4 ¦
which is the equation for the rate of change of enthalpy valid for both compressible
and incompressible fluid.
2.4.2 Constant density flow
Let’s substitute 6 ¦
:
j from (2.17) into (2.73), and consider that ;
= . "
1 °
¦ 4 ¦
m ¦ 1 `tß ¦/
l 8¬. ¦/ `]ß ¦/á
4
j
>])
XH>
)Wj

R ’ ; "
R
; R ’
(2.77) "
R
X üR
"
R
(2.78) ’>
; –
P
(2.79)
s R
"
R
X R
" >
R
`
X R
" >
R
; –
(2.80)
P
¦
)
°
¦
s
s R
"
R
>
s
–
s
¥
s
–
¥
)
38 CHAPTER 2. FUNDAMENTAL LAWS
Using the definition of viscous stress (2.19), we can rewrite this as:
ã 6 ¦/
which after applying the continuity relation (2.4) this reduces to:
¦f ¦
1 °
¦ 4 ¦
m ¦
X©)
>^)
ã 6 ¦/
1 °
¦ 4 ¦
m ¦
>^)
where we should use the incompressible form of viscous stress (2.20):
¦f
m ¦ 4
Equation (2.77) can be used in combination with the momentum equation
(2.21) to derive a strong form of the Bernoulli’s equation 6 :
6 ¦7k l 8¬. ¦/ 8 1
)
ã
>^)
@
) l
= . "
> +
We can also rewrite (2.77) in terms of temperature change, and velocity. Using
the definition of specific heat (1.25), substituting the velocity from the relation
, we have:
of viscous stress (2.20), and assuming °
= . "
¦ì¦
m ¦ 1
¦K
y ¦ 4
This is an incompressible heat conduction equation.
> 8 )
>])
Heat dominated flow
In the cases with small pressure and velocity gradients, or when fluid viscosity
is small compared to the heat conductivity, which corresponds to small Prandtl
number (1.30), equation (2.79) simplifies to:
°
¦ì¦
6 See Problem 2.7.5

; –
(2.81)
P
¦
°
s
=
¦
s
Š
s
2.5. CURVILINEAR COORDINATES 39
where we also presumed that the heat conduction coefficient, ° , is a constant.
This is a limit case of (2.79) for the case of heat dominated flow. Note that since
the substantial derivative is used for T, we still have the convective terms present:
¦ 4 °
¦ì¦
1
s]>])
which is also called a heat convection equation.
2.4.3 Boundary conditions
Generally temperature may experience a jump at the boundary [2]. Usually this
jump is small and the temperature of the fluid at the wall is considered to be equal
to the temperature of the wall.
Since the equation (2.81) is a second order differential equation, we would
need to specify temperature derivatives in addition to specifying temperature values
at the boundary. These conditions can be obtained from the consideration
of energy conservation. In particular, heat flux across the boundary should be
conserved. Thus from (2.81) we obtain
(2.82)
¦
¦ò
is a boundary-normal unit vector, and Š is the heat flux across the bound-
= where
ary.
2.5 Curvilinear Coordinates
Physical laws should not depend on the choice of a coordinate system. This is
expressed in the terminology of tensor calculus as coordinate invariance. Tensors
are designed to be invariant under coordinate transformations (Remark A.3.3).
Therefore, tensor relations provide a consistent way of writing physical laws.
There are two aspects of expressing physical laws in tensor forms: identifying
, physical components, and forming invariant expressions.

Y
"
Y
R
R
0
"
) ¦
6
0
¦
¦ )
j 6 j
¦
`
º
40 CHAPTER 2. FUNDAMENTAL LAWS
2.5.1 Invariant forms
The scalar product (Definition A.3.4) was constructed to be invariant. By virtue
of its invariance it represents a physical entity. Using the invariant forms of the
scalar product (Corollary A.3.5), we can rewrite the expression for the substantial
derivative (1.7) in invariant form:
(2.83)
0 ¦
>^)
Correspondingly, the mass conservation equation (2.2) will be expressed as
(2.84)
;
>
1 ; )
4 ¦$ T
and the momentum equation (2.22) becomes:
(2.85)
1 ; )
j 1 ; )
4 ¦
j
6 j ¦f
j >äã
¦ 4
¦
>])
where the covariant and contravariant velocities and stress tensors are linked by
the conjugate tensor relations (A.42), (A.43):
¦‡

ý
þ
¡
is
F
r
>
¡
¡
can
2.5. CURVILINEAR COORDINATES 41
non-inertial coordinate systems. This treatment is used in relativistic fluid dynamics
[7]. This approach would also make sense in the treatment of general
moving and deforming coordinate systems. However, in a variety of applications
it makes sense to treat space and time as separate variables. In this case one
has to distinguish between inertial and non-inertial coordinate systems. For any
moving coordinate system one has to formulate an explicit dependence of coordinates
on time. Generally, this time dependence can reveal itself in motions and
deformations. In this section we shall only consider the case of a non-deforming
and moving coordinate system. And in particular, we shall focus attention on an
important case of rotating coordinate systems.
Rotating coordinate systems
Consider an inertial coordinate ý system and a non-inertial coordinate system
, which is moving with respect ý to . Let the coordinates of a particular point in
be described by a vector 7 3 Ú%
'q - :
'q
(2.86)
3 ÿ
> ¡
where is the position vector the origin of system with respect to þ ý
ÿ ¡1 ÿ 1 "54
is the coordinate vector of the point in "54 system .
þ and¡
Differentiating (2.86) over time, we get
(2.87)
3 ÿ
where the time derivative with respect to the inertial frame of ý reference .
We shall consider the motion of as composed of displacement determined by
1 þ
ÿ "54
and rotation, given by angular n velocity vector . Then be represented
as
(2.88)
n
Í ¡
>
is the velocity of the point as measured in the non-inertial coordi-
£¢ ¡
¤
nate system . Since the rotation does not affect the motion of the origin, , we
ÿ þ
7 Here we are using vector notation, since only the vector quantities are involved
where¢ï¢¥¤§¦

3
(2.90)
T
R
R
R©
R
ÿ
¢ ©
R
R
"
¡
n
R
R©
Í ¢
l
R
R
"
Í ¡
>
n
>
¢
"
–
Í ¢
¥
R
R
n
>
r
ÿ
r
"
Í
n
¡
n
–
¥
42 CHAPTER 2. FUNDAMENTAL LAWS
ÿ
¤
have
¢© ¤§¨
and (2.87) becomes
(2.89)
" >
¢
>Ìn
,
3 R
Í ¡
R
" >Ìn
R¡
>
Assuming that the rotation is constant:
©
1 "54
= . "
we can differentiate (2.89) further to obtain
where
©
>
n
(2.91)
and using 2.88 we have
(2.92)
1¢
>
¢ç
Í ¡4
n
Substituting (2.88), (2.91) and (2.92) to (2.90) we get
3 R©
Í ¡4
(2.93)
R
R¢
" >
" > n
Í]1¢
>àn
Í]1
Í ¡4
R
" >
R¢
" >
The extra acceleration terms involving n arise due to rotation and are interpreted
as being the result of the Coriolis force.
An important special case of a non-inertial coordinate systems is a coordinate
system undergoing a pure rotation with a constant angular velocity. Assume
that the moving coordinate system is undergoing a rotation with n = . "
and
¤§¨
þ
. Then we can align the origins of the two coordinate system to ¤
¤ ¤¨

F
r
(2.94)
(2.95) )
'
T
4
R
R
)
r
"
B
'
Ç
B
R
"
F
B
2.6. THE LAW OF SIMILARITY 43
eliminate
plane þ
altogether. Let’s also assume that the axis of rotation is normal to the
, that n is . With these assumptions (2.93) becomes
1UT
'q
'mn
3 R
>à>
are the additional acceleration vectors (Problem 2.7.7). The
corresponding Coriolis forces are introduced into the equations of motion in a rotating
coordinate
r¡
Whereand
Ü
w÷>
system:
with being the displaced mass. In computations of continuum media dynamics
is replaced with mass R element :
à>
; ¡
where is the density of the fluid and is the face-normal velocity across the
face of area of a control volume (see Problem 2.7.8).
¡ ;
2.6 The Law of Similarity
The law of similarity [7, 8] enables in some situations to use a single solution to the
equations of fluid motion to represent a whole family of different cases. Consider
an example of a steady flow past a solid body, where the flow velocity upstream
of the body is. Consider also several cases of such flows when the body has
the same shape but different sizes, . Now, if the only fluid property affecting
this process is the kinematic viscosity, Ç , then in all these cases the distribution
9
, and of at least
-
of velocity should be a function of space coordinates Ú%
three additional parameters, ( Ç 3 9
): '“'
IKI/
¦$
¦ 1¨3
'“' 9 4
The number of parameters 8 can be reduced by considering the dimensions
of physical units in which they are measured:
8 A parameter can be looked at as just another independent variable, like space coordinate or time.
However, we treat them separately, since parameters are specific for each physical law, whereasare
not.

±
"
’
ã
(2.96)
(2.97) &1
(2.98) ± ý
"
’
" ²
’
ý
Ç
A!
¢
¦
F
š
1
ˆ
ˆ
T
" ²
r
’
F
ˆ
44 CHAPTER 2. FUNDAMENTAL LAWS
² é F
The only non-dimensional combination of these parameters is provided by the
Reynolds number:
±/›¨ˆ¡=
±/›¨ˆ¡=
±/›Uˆ¡=
Ç ²$
²
±²$
² é F
±
" #
9 ²G
±
" #
±
9
We can non-dimensionalize other variables by scaling them with the appropriate
length and velocity scales:
Ç
¦$¢
¦G¢ )
Since units dimensions should be preserved in an expression of a physical
law, a law formulated in dimensionless variables can only contain dimensionless
parameters. Hence the new dimensionless variables (2.96) should enter into a
only,
relation with since it’s the only dimensionless parameter derived from
the properties of the system. Following the convention that the velocity is the
dependent variable (2.95), we can write this relation as:
A!
) ã
¦G
Thus, using simple considerations of physical dimensions, we reduced the
number of parameters from three, '“' 9
( ) to one, Similar considerations
allow to reduce the number of parameters in a more general case, which is proved
in a so-called PI-theorem.
(A# Ç
) ã
ã
3
',A"
4
2.6.1 PI-Theorem
).
Let’s consider a physical law formulated for a set of u
variables, ý
IKI/ý%$:
Suppose that the law requires that each variable can be expressed in units
of length, time an mass, which we call the primary dimensions:
IOILý%$4
¦f²å
" # ²('*),+±
².-/),+±
£ .§. ² ©),+
±/›Uˆ¡=

²
ã ý
(2.99)
(2.100) ý
(2.101) &1
ý
F
Y
Y
Y
Y
Y
F
ý
Ç')5+ã 1
Y
Ç
ý
ý
ý
F
r
ý
á0;2Ç')5+é F
á

60=2ý
V
ý
ý
ý
F
F
F
T
T
T
=
=
`
`
46 CHAPTER 2. FUNDAMENTAL LAWS
We can multiply the equality (2.102) by Ç and get:
(2.103)
Y&1
IOILý%$4
á0;2ý
This is an extra relation imposed in addition to our physical law (2.97) by virtue
of scale invariance or homogeneity of our law with respect to length-scaling [9].
In a similar manner we can arrive at two more relations imposed because of
homogeneity with respect to other two primary dimensions: time and mass:
Y ý
(2.104)
Y&1
IOI/ý:$4
Y ý
(2.105)
Y&1
IOI/ý:$4
80=2ý
Thus we have three more relations in addition to our physical law (2.97),
Y ý
which means that the number of variables can be reduced u from u À to .
If we use a more complex law that involves an additional primary dimension,
such as temperature, then we can reduce the number of variables of the problem
by 4. Generally, if we have R primary dimensions and = independent variables in
the problem, then the independent variables can be reduced ã R to nondimensional
parameters. These parameters can be different depending on the
choice of scaling factors used in transformation (2.100). Generally, normalization
(2.100) does not have to be done by primary dimensions, but can be used with
respect to any group of variables that do not form a so called PI-group, i.e. their
products of the type (2.98) can not be reduced to a non-dimensional number, no
matter what powers are used [10, 11]. This constitutes the essence of the PItheorem
[11]. It lays a more rigorous foundation for the law of similarity [7, 8],
which means that the same solution can be reused by rescaling the variables.
2.6.2 Non-dimensional formulations
To formulate a physical law in dimensionless variables we should introduce dimensional
scales for each variable. The scale, representing the variable, will be
denoted with the same symbol, but with the subscript 0. Scales can be introduced
for both scalar, vector, and general tensor variables. Thus, the scale for a vector
variable will be denoted as 10 . In some situations there can be different scales
¦
10 Subscript 0 shouldn’t be confused with the vector component, since vector components are numbered
with one

"
"
V
V
V
¦
V
¾
X
X
¦ "
'
"
; V
V
)
)
¦
"
V
)
`
¦
)
V
2.6. THE LAW OF SIMILARITY 47
for different components, in which case we shall use a different notation.
Space and time derivatives should also be scaled. Thus, if we consider
space and time as independent variables: , and as dependent variables velocity,
density, and pressure we can introduce the following non-dimensional variables:
(2.106)
¦$
¦G
ã "
V ã ¾
ã
V
ã ; ;
;
V
ã ;
V
ã )
V
¦G )
where we use the Nabla operator to denote the space derivative.
After the dimensional variables are replaced withe the dimensionless ones
by means of (2.106), one should look into the physics of the problem and see if
some extra relations between the scales can be applied. For example, in some
problems the characteristic velocity scale can be related to length and time scales
) : "
as: . This can be the case in the problem of a steady flow around a fixed
object. However, a steady flow around a rotating object will have an independent
time scale related to the period of rotation.
After all possible eliminations of scales were done, one should try to construct
dimensionless combinations of scales, or non-dimensional parameters.
There can be several different ways in which these parameters can be selected.
This process can be formalized somewhat [10], but there is still a room for subjective
judgment on which dimensionless combinations of scales are most appropriate
as parameters for the problem at hand. No matter how these parameters are
selected the PI-theorem states that their minimum number can be as low as =
where = is the number of dimensional scales and R is the number of primary dimensions
of the problem. If all the dimensionless parameters have been correctly
identified, it should be possible to replace all the dimensional scales with these
parameters, thereby rendering the physical law in a dimensionless formulation
with the minimum set of independent parameters.
Let’s consider several cases of non-dimensional formulations and of application
of the PI-theorem.
R ,
X ã
;
ã
Mass conservation law
Let’s write a non-dimensional formulation of the mass conservation law (2.2):
(2.107)
;
>
¾ ¦ 1 ; )
¦ 4 T

X
(2.108) )åj
V
V
V
r
V
)
r
V
)
V
V
V
r
¾
9
)
V
V
rV
V
j
¾
V
V
l
X
ã
;
V
>
V
V
V V r
)
V
V
ã
¦
V
V
V
48 CHAPTER 2. FUNDAMENTAL LAWS
where we used the nabla operator (A.32) to simplify further analysis. There are
three primary dimensions in this (R À case ): [length], [time] and [mass]. Using
the scaling transformations (2.106), the non-dimensional form of (2.107) is:
; > ã ¾ ¦ 1
ã ; ã ) ã
¦ 4 T
As can be seen, all the dimensional parameters canceled out from the equation.
Momentum equation
Consider the steady-state limit of the incompressible momentum equation given
by the Navier-Stokes equation (2.24):
¾ ¦
¦G
¦ `
Since this is a constant density formulation, density becomes a
parameter
of the
;
problem: . Considering this, and transforming to the non-dimensional
variables according to (2.106) we obtain:
;
(2.109)
ã ¾ ¦
ã ¾
¾ ¦
9 )
¦ ` X
X ã
) ã
)Wj jÃã
) jã
j ã ¾
To simplify things, let’s select for the pressure X scale the dynamic
pressure:
. By doing this we state that pressure scale, X , is not an independent
;
parameter of our problem, but is related to the density and velocity scales. Now,
let’s make each term of (2.109) dimensionless, by multiplying the whole equation
by :
: )
) ã
)Wj jÃã
) júã
X> ã
ã ¾ ¦
ã ¾
j ã ¾
¦ ` ã ¾ ¦
ã

ã )
(2.112)
¤
ã s
(2.113)
ô
(2.114) u
°
@
¤
¾
ô
x
=
°
¦
9 V
V
V
r
r
Ç
V
¦
¤
@
¦
x
¦
ã
`
R
2.6. THE LAW OF SIMILARITY 49
the number of independent parameters can be as low = as
introduce two non-dimensional numbers: Reynolds number:
Ñ»` l l
. If we
(2.110) A!
)
and Froude number, relating the forces of inertia to gravity:
¢
(2.111)
¢ )
V
then we obtain the non-dimensional form of the momentum equation:
)Wj jÃã
) júã
X> ã
ã ¾ ¦
A!
ã
j ã ¾
¦ ` ã ¾ ¦
with the four non-dimensional ã ' ã
variables:
parameters: x (see also Problem 2.7.9).
A!
'
ºâ' ã
ã ) and
, and two non-dimensional
Boundary conditions
Some non-dimensional parameters arise from the boundary conditions. For example,
non-dimensionalizing the boundary condition of the energy equation (2.82)
leads to:
¦
¦$
u )
where u
is the Nusselt number:
¢ …
[^s
²“ ´
(±L… : 1
…
. 4 [^s
where is the wall heat flux ), the characteristic length-scale,
the heat conduction coefficient, (1.29), and the characteristic temperature
difference between the wall and the fluid.

†
Ü
Ç
)
ö V V
T
¦
r
¦
–
¥
50 CHAPTER 2. FUNDAMENTAL LAWS
Boundary conditions at the free surface give rise to additional parameters,
such as Froude number, relating inertia forces to gravity, (2.111), Weber number,
relating inertia to surface tension:
(2.115)
¢ ;
where
(2.58).
ö
is the coefficient of surface tension entering the boundary condition
Other non-dimensional parameters may appear as new phenomena are
added into the physical law [2].
2.7 Problems
Problem 2.7.1 Derivation of the vorticity equation
Obtain the result outlined in (2.30).
Problem 2.7.2 2D vorticity limit
Perform the missing steps in (2.42).
Problem 2.7.3 Incompressible viscous limit
Derive (2.50) from (2.49).
Problem 2.7.4 Conservation of circulation
The velocity circulation is defined as
(2.116) >
@?
)
R
where the integration is over any closed loop inside the fluid.
Show that for irrotational flow (n
¦G
.
Problem 2.7.5 Bernoulli’s equation
Using the energy equation (2.77):
):>
= . "

¦
Y
"
Y
in (2.94) in terms of n©'q®
F
r
R ’ ;
"
R
R
R
!
"
F
'
X R
" >
R
R
R
ÿ
"
¦
)
r
¦
¦
Ä
Ö
n
s
)
`
¦
=
X
–
¦ £
R
'
T
¥
¦
£
'
'
£
'BAF
'BAr
'CA
2.7. PROBLEMS 51
ã 6 ¦/
1 °
¦ 4 ¦
m ¦
and momentum equation (2.21):
>])
1 ; )
1 ; )
4 ¦
j
6 ¦
j >äã
j
¦ 4
¦
derive the strong formulation of the Bernoulli’s equation:
>^)Wj
>\š
’>
> +
@
) l
= . "
and formulate it’s applicability limits.
Problem 2.7.6 Volume change inside a moving boundary
Suppose that a region of space is enclosed by a moving boundary. The
velocity of motion of the ) boundary, , is given at each point on the boundary.
Show that the rate of change of the volume, , of that region will be equal to:
!
(2.117)
= where is the unit normal vector to the boundary R and
and find the Ä coefficient .
surface area element,
Problem 2.7.7 Rotating coordinates
£Žr
Obtain explicit relations for the components of acceleration vectors in
F
£ . '
'*®
'*® '
Problem 2.7.8 Rotation with separated coordinate origins
Consider a simple rotation with n
T
˜%
n
ý
origin of the rotating coordinate system rotate with the same around the
origin of :
'mnŒ- as in (2.94), but now let the
Í ÿ
Derive the expression for
3
in this case.

V
V
)
– ;
P
s R
"
R
X üR
" >
R
(2.118) „©z
r
V
°
s
–
P
V
s
r
)
52 CHAPTER 2. FUNDAMENTAL LAWS
Problem 2.7.9 Nondimesionalizing energy equation
Write a non-dimensional form of the heat convection equation (2.79):
¦ì¦
m ¦ 1
¦K
y ¦ 4
;
selecting for the pressure scale. Determine the minimum number of
X
dimensionless parameters. Write the equation using the Eckert number (3.114)
as one of the parameters:
> 8 )
>])
¢ )
V

where º
¦
)
(3.4) º
; é F ã 6 ¦
j
(3.5)
P
¦
s
j
s
j
)
¦
T
Chapter 3
Laminar flows
3.1 Assumptions
Flow equations discussed in Chapter 2 provide analytical solutions only in some
special cases. In this chapter we shall consider the equations for incompressible
flow: (2.4), (2.24) and (2.74), assuming that all the coefficients are constant:
(3.1)
(3.2)
(3.3)
; –
>])WjÙ)
1
s]>])
¦f
j
F ã 6 ¦ ;Ãé
¦ 4 °
¦G ¦f
;Ãé F
`
º
¦
¦ì¦
>])
y ¦ 6 ¦/
is the Hydrostatic pressure:
;
For Newtonian incompressible fluids the viscous stress ã 6 ¦
j term, , has the form
(2.23):
j
9 )
¦f
j*j
Definition 3.1.1 Laminar flow
Let’s make an assumption of laminar flow which states that the time scale
of changes in the flow can not be lower than the time-scale of the motion of the
53

F
r
)
54 CHAPTER 3. LAMINAR FLOWS
boundary or any external sources. In other words, if there is any repeatability in
the motion of the boundary or in the external forces then the frequencies associated
with either factors can not be lower than the frequencies of the flow motion.
It means that neither the boundaries not external forces can induce any additional
frequencies in the flow. In the limit case of non-moving boundaries and
non-changing forces the flow should not depend on time, which means that all
the dependent variables should become functions of spatial coordinates only.
The conditions of laminar flow defined by (3.1.1) are realized when the contribution
of the non-linear term in the momentum equation (3.2) is small, or
¦f
when
dominates. This is usually the case
j*j
when non-dimensional Reynolds number (2.110):
the contribution of the viscous term 9 )
(3.6) A!
9
is small. In practical situations the ”smallness” of
Ç
A#
to A!
corresponds
Navier-Stokes equation, (3.2) is known to have very few analytical
¦f
solutions.
This is mainly due to the non-linear )
convective term , which is the main
cause for the rich dynamical features of fluid flow. For this reason, most of the
cases that provide analytical solution do not include the convective term. Below
we shall consider several such cases.
ED@
T‰T
.
3.2 Confined flows
Probably the simplest of confined flows are the flows between moving surfaces,
which belong to the category of Couette flows [2].
3.2.1 Flow between parallel plates
Let’s consider a flow between two parallel plates, one of which is moving relative
to the other with a constant velocity¦
(Fig. 3.1).
We are looking for a two-dimensional solution, since by the assumption of
laminar flow (3.1.1) and from the symmetry of the problem we do not expect
any changes in the transverse direction. We are also looking for a steady-state
solution, thus all the variables will be the functions of axial and vertical coordinates
- only:
, and only two velocity components need to be considered
3 %
'q

)
)
F
r
F
(3.9) )
r
s
r
> 8 —íR ) r
®í R
6 8 R )
(3.10)
® R
(3.11) Äkõ
¢
s
V
1
®
6 l
;r
r
)
r
l 8
T
T
s
F
l
F
3.2. CONFINED FLOWS 55
Figure 3.1: Flow between parallel plates: the lower plate is at rest, the upper plate is
moving with velocity.
¦§%
- . Since the plates are considered to be infinite no variable should
'*)
change in direction either. Thus, the only independent variable of the problem
becomes the vertical
coordinate , which we shall denote ® as . Likewise, from
the symmetry of the problem the only non-zero component of velocity ) is , which
we shall denote ) by . In addition to this we can also assume the pressure to be
constant. This can be explained by the absence of normal stresses in this flow.
With these assumptions the momentum and energy equations (3.2), (3.3) reduce
to:
(3.7)
8 R
r
(3.8)
R ®
° R
Equations (3.7) and (3.8) can be
1UT
solved
4
with
the
1U
boundary
4
conditions
u(0)
s
= s
0 and u(H) = U, and and , with the solution (See Problem
3.7.2):
R ®
4
®
and the shear stress:
The non-dimensional friction coefficient, ÄŒõ , becomes inversely proportional
to the Reynolds number:
8
A!
;

(3.12) º­{
'
T
s
1
®
¤
®
Y
¢
Y
x
r
¢
Y —W¢
¢
Y
)GH
Ä
F
s
F
F
`
`
s
s
V
l
4
T
56 CHAPTER 3. LAMINAR FLOWS
and the Poiseuille number:
Solution to the temperature equation (3.8) produces a quadratic dependence
on ® (See Problem 3.7.2):
— s
ÄkõEA!
` 8 4
°
8
° l
l >
V >
®»>às
V
The dimensionless Brinkman number is introduced as a relative measure of viscous
forces to thermal fluxes:
(3.13)
8r
° 1
In the momentum equation (3.7) the effect gravity was neglected under the
assumption that the gravity force acts normal to the direction of the flow. Generally
it may not be the case, but the solution procedure remains essentially the same
(see Problem 3.7.3).
3.2.2 Axially moving concentric cylinders
¦GÚ%
In this case only the axial component of ) )WxÙ'*)GFè'q)IHÙ-
% T
velocity vector is
¢
non-zero:
, and it only depends on
4
: )IH1 ¢ 4
. The appearance of any
)IH1
',Î'*+
'*)GHÙother
velocity component, or a dependence on other coordinates will lead to the
violation of the assumption of a laminar flow (Definition 3.1.1). ¢ Substituting this
form of the solution into the momentum equations in cylindrical coordinates, we
find that only the an axial momentum equation takes a non-trivial form:
(3.14)
@
¢
A solution that satisfies this equation is:
(3.15) )GH1
r
4
dfÆ 1 ¢ 4
¢
>\Ä

¢
F
r
Ä
F
F
A
4
V
F
A
A
`
V
4
V
V : A
V F
Ä
Ä
V
F
A
A
X R
¢
R
T
F
F
Y
Î Y
'
' )GH1K
V
)GH1
A
V
4 >JF
T
Fr
T
F
4
F
T
`
Ä
1 ¢ : dfÆ
A
A
F
F
V
V
A
4
4
- , and )GF
V
4
)GF1 ¢
3.2. CONFINED FLOWS 57
',Ä
Substituting this into (3.15) we have:
)GH1
r“
dKÆ 1
and the solution becomes:
dfÆ 1
4 ' Ä
(3.16) )IH1
: ¢ 4
dKÆ 1
¢
dKÆ 1
dKÆ 1
Remark 3.2.1 Pulling an infinite rod
Consider the problem above with the boundary conditions:
: A
: A
T
Then, applying this conditions to (3.15), we have
)GH1
4
V
4
dKÆ 1K
4
r“
rk
from which it Ä
follows that . Thus the problem of pulling an infinite rod
does not have a steady-state solution.
>\Ä
3.2.3 Rotating concentric cylinders
In this case we ) have:
Continuity:
¦$Ú%
are the constants, which can be determined from the boundary con-
Ä where
ditions:
)œx¡'*)GFÙ'*)GHè-
Ú% T
'*)GFè'
4
.
-momentum:
(3.17)
; )
¢

R
¢
R
A
A
R —
R
(3.19) )IF
(3.20) )IF
V
T
A
n
V
V
A
A
F
V
F
¢
R
R
r
A
A
s
4
V
n
F
n
V
F
R
R
'
'
: ¢ ` ¢ : A
F : F `
A : A A
V
V
Ä
¢
F
F
n
s
s
>
¢
F
1
A
1
A
V
F
r
Ä
¢
n
F
`
F
T
A
¢ Mr
)GF
s
s
V
F
: ¢
A
A :
F
V
V
`
`
T
A
A
V : A
V
F
F
r
58 CHAPTER 3. LAMINAR FLOWS
Î -momentum (Problem 3.7.5):
(3.18)
)GF
¢NM ¢EL)IF
¢ r >
1 ¢
°
¢
8 L)IF
x >
Boundary conditions:
)IF1
4
4
)IF1
4
4
The equation that determines the velocity
satisfy this equation has the form:
)IF1
4 ¢
is (3.18). A solution that will
Substituting it into the boundary conditions, we find the Ä constants
and the final solution becomes:
, Ä
,
: ¢
>\A
Remark 3.2.2 Flow inside a rotating cylinder
If we set A
the solution above will become:
)GF
which is a solid body rotation. Thus the steady-state solution for the flow inside a
rotating cylinder is a solid body rotation.
Remark 3.2.3 Rotating an infinite rod
If we solve the problem above with the boundary conditions:

(3.23) X
F
T
Ä
X
A
F
X R
¢
R
X
V
V
Ä
>
V
>
Ä
A
¢
V
r
r
V
V
Fr
r
' )GFK
A
Ä
K
r
r
r
V
r
; n
1
@
r
`
V
T
r
A
r
r
4
F
r
1
A
V
X
V
3.2. CONFINED FLOWS 59
T
we can obtain the following system for the coefficients Ä
:
)GFqA
nÅA
',Ä
(3.21)
FK
>
n?A
from which we have: Ä
, and the solution becomes:
r
',Ä
n?A
(3.22) )IF
n?A
¢
Substituting it into (3.17) and integrating it, we can obtain the pressure distribution:
 A
V
; )
¢
and the pressure distribution is
1 ¢ 4 ` ; n
 A
V
where the Ä constant is found from the boundary X condition:
the pressure distribution is:
4
. Thus,
>^Ä
l ¢ r
@ ; n l
It is useful to compute the rotational momentum (torque) that arises in a
system of two rotating cylinders (Problem 3.7.6).
: ¢
3.2.4 Poiseuille flow through ducts
Let’s consider a case of a straight duct with a constant cross-sectional area, .
Since the area does not change its form, the length-scale of the problem will be
the characteristic size of the duct 1 : ¡ .
½
1 See Sec.3.2.6 for another measure of duct diameter
¼¡
F}É

(3.24) )
)
F
¢
)
F
F
)
1
r
º
1
4
F
4
º
`
F
º
F
F
Y º
Y
)
F
)
1
F
F
1
)
r
r
R º
R
4
F
4
'
T
'
4
T
º
)
T
F
)
r
F
F
r
4
60 CHAPTER 3. LAMINAR FLOWS
Compared to the case of infinite plates, a duct has an entrance and it has a
finite cross-sectional area. The existence of an entrance causes entrance effects,
such that the flow is three dimensional over some distance from the
¦íÌ%
entrance,
that is all three components of velocity ) '*) - are non-zero: , and each
component also depends on all three )
spatial coordinates: .
However, we assume that this transition region will end at some distance
after the entrance and be replaced by a fully developed flow region, where axial
velocity does no longer depend on the axial coordinate, i.e.
¦G
¦ 1 '*)
'5
'q
'5
It should be noted that the existence of a fully developed laminar regime
is only an assumption, but it happens so, that there is a solution satisfying this
assumption. However, it also happens that there are other solutions, which do not
satisfy this assumption, i.e. unsteady turbulent regime. Which solution is realized
in reality depends on the magnitude of the Reynolds number. At the low Reynolds
numbers the fully developed regime occurs in circular ducts at distances between
30 and 100 duct diameters from the entrance.
In a fully developed flow in addition to (3.24), the velocity components
¦t%
normal
to the axis should ) 'q - be zero: . This is because the appearance
of any velocity component normal to the axis can not be sustained for a
long period of time since there is no pressure gradient imposed in that direction.
On the other hand, any short time appearance of such components will violate
the assumption of a steady-state laminar nature of the flow in a fully developed
region. Thus, we can use only the axial component, which we shall denote as
. With these assumptions, in a fully developed flow the axial momentum
equation (3.2) can be simplified to:
(3.25)
T
rðr
> 8 1
ð
The other two momentum equations simplify º to
only on º : , and we can write:
r»
. Thus, º depends
>^)
Since )
'q
the second term on the RHS of (3.25) does not depend on
, and then it follows from (3.25) º that should not depend on either. But
F

ã
F
¦
F
r
ã
@
¢
ã
R
¢
ã R
ã
) ã
—
ã
'
@
T
)
¢
1
@ ) ã
@
½
F
r
F
)
3.2. CONFINED FLOWS 61
since º
depends only on is a constant. Thus, dividing
, and introducing dimensionless variables:
equation (3.25) by R º : R
, it follows that R º : R
¾ ¦G ½Å¾ ¦
¦$
8 )
where ½
is the appropriate measure of the duct cross-sectional size, we obtain a
the following boundary value problem for the dimensionless ã ) velocity :
½ ' ã
) ã
R º : R
(3.26)
¾ ¦ ã
¾ ¦
)POQ
where ‹ subscript stands for ”wall” and index spans only the variables that )
#
. This is a Poisson equation in a confined 2D domain with the
'q
depends on:
no-slip velocity at the boundary (Dirichlet boundary conditions).
Remark 3.2.4 Reynolds number independence
Note that the Reynolds number does not enter the equation (3.26), and
therefore, should not affect the solution. This is because we excluded non-steady
solutions and turbulence from our consideration.
The circular pipe
Consider a fully developed flow region in a circular pipe of radius A . The natural
coordinate system for this case is cylindrical: ¢ ',Î',+ , with + being the axial coordinate.
Following the discussion of the previous section, only the axial velocity
ã
component, )IHwill be non zero, which we shall denote for simplicity ) as . As it
was shown, it can only depend on ¢ and Î . However, because of the symmetry of
the duct, Î the dependence on will inevitably lead to
1
time-dependent
¢
solution and
violate the assumption of fully developed flow.
4
)
Thus, we should have ,
and using the Laplacian operator in cylindrical coordinates, (3.26) reduces to:
(3.27)
ã ¾
¢ R ã
) ã
R ã
4 T
which gives the parabolic solution:

(3.28) )
(3.29) …
8 —ÛR ) `
¢
R
(3.31) ÄŒõ
(3.32) º­{
xZYV
l 6Q
º R
R¢
R
r
¢
º R
R
)
r
á
Ñ
r
)
`
r
¢
R
`
A
@
)œ|­w
l
º R
R
UAÂ `
W8
º R
R
62 CHAPTER 3. LAMINAR FLOWS
¢ 4 @ 1
ÑSRã
and in dimensional units:
) ã
@T
¢ 4 @ 1
8 Ñ
This solution is called the Poiseuille flow.
The Volumetric flow rate through the pipe can be computed as
rT
Ö Ö
¡¼
¢»
The mean duct velocity is defined as:
V
l§UÖJV
) R
` A
… ¢
¡
)
X
The wall shear stress:
W8
(3.30)
6Q
Ñ
)
A
8X
A `
l
Darcy friction factor:
áÅ
Skin-friction coefficient
;X W6Q
r
)
where the Reynolds number is based on duct diameter:
;X A! @*[
A\
Poiseuille number
)
X
½ :‰9
.
@[
Äkõ§A!

(3.33)
8 R
ã )
(3.35)
V
ã )
(3.36)
T
F
@
`
Ä
r
R
r
) ã
®
Ä
r
1
r
1
@
r
)
R º r
R
r
F
T
@
º
R
`
Ä
V
Ä
r
º
1
4
3.2. CONFINED FLOWS 63
3.2.5 Combined Couette-Poiseuille flows
Couette flows are driven by shear boundary motion, and Poiseuille flows - by
the axial pressure gradient. The equations describing both types of flows do not
have a nonlinear convective terms, which makes these equations linear, and thus
enables linear superpositions of solutions.
Consider the Couette flow between parallel plates (Sec.3.2.1) but with a
constant pressure gradient applied in the axial direction. The momentum equation
in this case will be a combination of (3.7) and (3.25) in the form:
R ®
where we are using for the axial ® and for the vertical coordinates. Since the
LHS does not depend on and the pressure is a function of º only: , we
conclude R º : R that must be a constant. Introducing dimensionless coordinates:
)
:, we have:
® ã
: '^Ä
®»>^Ä ã
Applying the boundary conditions:
1UT 4
) ã
4
we Ä have:
form:
'*Ä
, and after renaming Ä
, the solution has the
) ã
4
® ã
®»>¼@ Ä¥ã

(3.37) Ä
®
T
½
(3.40)
X
X R
R
½
~
@
l 8
¢
r
º
R
›
64 CHAPTER 3. LAMINAR FLOWS
We can find the constant Ä after substituting this solution into (3.34):
From (3.36) we can see, that Ä @ when the velocity changes sign at the lower
(ã wall ). This is called flow separation point, and according to (3.37) it corresponds
to the pressure gradient:
R
(3.38)
l
r
8
A pressure gradient greater than this will cause the flow at the lower wall to
reverse 2 .
3.2.6 Non-circular ducts
Because the problem of the fully developed duct flow was reduced to a well posed
and well studied boundary value problem based on the Poisson equation there
are variety of analytical solutions obtained for ducts of various shapes [2].
There are several convenient measures that are introduced for ducts of arbitrary
shapes.
The cross-sectional length-scale of the duct is called the Hydraulic diameter,
which is introduced as a generalization of relation for a diameter of a circle.
For a circle of diameter the relation between it’s area and a perimeter]
¡ Ñ ¡ ½
is:
. Thus, for any non-circular duct of º perimeter and area the
¡ :]
hydraulic diameter is defined as:
(3.39)
Ñ ¡
The mean wall shear stress is defined as:
]
6Q
] ? 6QR
2 See Problem 3.7.7

)
1
r
? 6QR
)
¢
› R
X
4
`
¡]
R
º
R
Ö ;Ãé F º
F
R
F
"
3.3. UNSTEADY FLOWS 65
where the integration is done over the perimeter of the duct.
In a fully developed flow each element of the fluid between cross-sectional
planes at and >
that element should be zero. This means that the friction force at the wall should
exactly balance the axial force pushing the fluid element due to the pressure drop
in the duct. Thus, we have the following relation between the mean wall shear
stress and the pressure gradient:
R moves with a constant velocity. Thus the sum of forces on
` ¡
R º
and using the definition of the wall mean shear stress (3.40), we have:
3.3 Unsteady flows
6Q
Some unsteady flows in the ducts can still be solved analytically. To introduce
unsteadiness we formally use the same assumptions that lead to (3.26), but now
we shall reinstall the time derivative from (3.2), which with these assumptions
becomes:
(3.41)
9 )
¦ì¦ ` ; é F º
where as in ) '5
Sec.3.2.4, is the axial component of velocity in the duct,
which is the only non-zero velocity component. Following the same reasoning
as in Sec.3.2.4, we conclude that the last term can not depend on any spatial
coordinate. Since we consider unsteady flows, this term can still depend on time.
However, in this problem we can combine velocity and pressure into a single joint
variable:
)
^
)â>
And the final equation becomes:
(3.42)
9
¦ì¦
)
^
)
^

(3.43) P`_w
1 4
®
1
Y
Y
)
^
"
š
1 4
ˆ ®
r
n
r
)
^
)
^
1
>
r
)
^
4
66 CHAPTER 3. LAMINAR FLOWS
This equation has a parabolic character, which means that one independent
variable - time - is asymmetric with respect to direction. Specifically, at any point
in time the solution will depend only on the previous points on the time axis, but
not on the subsequent points. This difference in the directions of time: the future
and the past, is the result of the first order time derivative in the equation (3.42).
In contrast, the second order Laplacian space derivative,
directions equivalent. It should be noted that since
operator in (3.42) includes only two components:
)
^
¦ì¦G
)
^
rðr
)
^
¦ì¦
makes all space
, the Laplacian
'q
. ð
Equation (3.42) is still simple enough to provide analytical solutions in several
cases. The important cases include:
1. Starting flow in a duct.
2. Pipe flow due to oscillating pressure gradient.
3. Fluid oscillating above an infinite plate.
4. Unsteady flow between infinite plates.
3.3.1 Fluid oscillation above infinite plate
Suppose that the plate is oscillating in direction
, and the velocity of oscillation is:
®
, which we shall denote as
r
ba/ced1
"54
In this case the solution for the fluid velocity will depends on only one ® direction :
. Then equation (3.42) can be rewritten as:
)
^
)
^
(3.44)
9 Y
)
r^
Y ®
We can note that in this case the motion in time is periodic, while the changes
in space are aperiodic. Correspondingly, we may look for a solution form which
is a periodic function in time and a decaying function of space. One form of the
solution that will satisfy this equation is:
(3.45)
¦,f$
"54
)
^
®å'
>\Ä

where Ë
1 4
®
1
T
r
1 4
®
1
'
1
1
1
#
š
1
®
n
9
ˆ énp¯
n
ˆ
r
#
¥
n
¥
n
n
r
á
®
n
4
r
4
4 :nm
3.3. UNSTEADY FLOWS 67
where š
is an unknown function ® of . If we substitute the latter into (3.44), we’ll
get the equation š for :
(3.46)
n
9
¢×á
šg¢
where
has a solution:
š
šGgg
R š : R ® , and the parameter á was introduced for brevity. This equation
4 ¼¡
where ¡
is a constant. Thus, the solution for
) , (3.45), is:
^
ˆ é'¯
"54 ¡ih`jlk1
#
"­`
Using the definition of á , (3.46), and the identity: # F}É
, we can write:
)
^
®å'
>\Ä
1 #
l
>¼@
áÅ —
F}É
1 #
n
l 9
4o
and for
we obtain:
>@
"54
®å'
)
^
¦0f$
)
^
®å'
"54 £¡
>\Ä
. Considering only the real solution, we have:
énp¯2énp¯
1
F}É
n : l 9 4
Ë®
)
^
®å'
"54 ¡
>\Ä
Ä ¡ Constant and can be found from the initial conditions. In case of a moving
plate and a stagnant flow-field at infinity (3.43), we have:
. 1
"a`
–
Thus ¡£
, Ä
, and the final solution is:
)
^
>\Ä
1}T
"54
qarced1
"54 ¼¡
a/ced1
"54
(3.47)
"54
. 1
"a`
4
)
^
®å'
Cˆ énp¯
–
Ë®

1
¥
'
® ts¬
"
V
n
r
ˆ énp¯
Y 9r
Y
Y
Y
)
)
¥
n
)
¥
n
n
68 CHAPTER 3. LAMINAR FLOWS
In case of a stationary plate and a flow-field oscillating asÒ 1 "54
we can obtain the solution by subtracting the equation above fromÒ
(t):
ba/ced1
"54
,
(3.48)
)
^
®å'
"54
H±/–
. 1
"54a`
–
. 1
"a`
Ë®
4 ²
It can be checked by a direct substitution that the above equation satisfies (3.44)
and the boundary conditions: , and
.
1}T
"54 T
"54
"54
)
^
)
^
1K
'
t–
. 1
3.3.2 Unsteady flow between infinite plates
Consider two parallel plates separated by a distance , and a fluid with viscosity
initially at rest is filling up the space between the plates. Suppose that the
upper plate starts moving with velocity. We are looking for the solution for the
unsteady flow-field between the plates. This case is similar to the one described
9
above, but now the solution should be aperiodic in time, and in fact it can be periodic
in space, since any periodic function with a spatial period equal to the plate
separation, , will be suitable. The equation (3.44) is still valid in this case. However,
now we have an explicit length-scale, , and using dimensionless variables
becomes more attractive. Let’s define the non-dimensional variables as:
(3.49)
"
"
)
^
È
ã )
V
Then equation (3.44) can be rewritten in the non-dimensional variables as:
ã ) Y
È r
r^
Ys
And selecting the time scale as " V
:J9
, we have:
ã ) Y
È r
This equation should be solved for the unknown function ã
Ys
conditions:
r^
1MÈ
's4
with the boundary

)
)
1
'
) ã
1
) ã
) ã
@
`
'
) ã
1
) ã
) ã
š
šg
'
)
T
@
T
È
š
@
`
@
Y
Y
r
T
3.3. UNSTEADY FLOWS 69
(3.50)
(3.51)
(3.52)
(3.53)
1¨È
1UT
1MÈ
's4
@‰'s4
'K 4
T 4
where (3.50) describes the fixed lower plate, (3.51) describes the moving upper
plate, (3.52) is the initial and (3.52) the final velocity distributions. Note, that the
final velocity distribution was obtained before as the steady state solution for this
case (Sec.3.2.1). As can be seen, boundary condition (3.51) is non-zero. To
simplify our search for the right solution, it would be nice if we could look for a
function which is zero at the boundaries. To make the boundary conditions both
zero let’s look for a solution that is represented by the difference between the
steady-state ã solution
at both plates:
1MÈ
4
, (3.53), and ã
'K
1¨È
's4
, since this function will be zero
) ã
1¨È
Now, if ã 's4
we replace with the new unknown function:
following boundary value problem:
) ã
1MÈ
's4
, we obtain the
1MÈ
È `
1¨È
's4
's4
(3.54)
Y©
(3.55)
(3.56)
(3.57)
1}T
's4
@‰'s4
4 T
T `
1UT
È r
T ` T T
's4
Ys
@‰'s4
T 4 È ` T È
1¨È
1¨È
È `
4 4
where we have all the boundary values zero. We can look for a solution in form
of separated variables:
'K 'K
1¨È
È `
1¨È
È ` È T
) ã
(3.58)
's4 ‡ 1¨È 4
1¨È
1s4
Substituting this into (3.54) we obtain relationship between š and :
gg
r
¢ ` á

È
¡
B
1
'
¡
B
@
š
4
š
Ö F
F é
4
1
r
š
š
š
Ò
wBYV
>
¡
4
¤
a/ced1
1UT
1MÈ
BdefÆ 1
ù
B/z
é
B/z
R
È
@
`
4
T
@
4
B
70 CHAPTER 3. LAMINAR FLOWS
where as in (3.46) we introduced a constant á . This time, however, á is not known
in advance, and should be determined from the boundary conditions. The relation
above is identically satisfied by:
(3.59)
(3.60)
1MÈ 4 ¼¡
defÆ 1 á È 4
1s4
á
È 4
Ä ˆ é'u
with , Ä and unknown constants in addition to . Out of these the constant
can be absorbed into and , since ¡ ¤
¤ and enter as a product in (3.58). So,
¡ á š Ä
without loss of generality we can set Ä
(3.55) - (3.58) can be translated to 1¨È 4 and š
. The boundary conditions on
@
as:
1MÈ 's4
1s4
(3.61)
(3.62)
(3.63)
(3.64)
1¨È
1MÈ
1}T
T ‡ 1UT 4
's4
T ‡ 1
@‰'s4
'K 4
4 È 1¨È 4 T
È ¼ 1MÈ 4
1s4
1s4 v v
1}T 4 v
1K 4
T 4
4 T 1
4 È
1¨È 4 T
where the last equality (3.64) is satisfied identically by virtue of (3.60). From (3.61)
and (3.59) if follows that ¤ç T . Similarly, from (3.62) it follows that á¿ =U, where
is an integer number. The only way to reconcile boundary condition
1MÈ 4 = (3.63):
, with boundary conditions (3.61) and (3.62) and the analytical form of
given by (3.59), is to express as a Fourier series:
4 1MÈ
1MÈ 4
1¨È 4
È
=UÈ4
Coefficients ¡ B
can be found as:
ÈdefÆ 1
Computing the integral:x\"defÆ 1
"a/cyd1
, we obtain:
=UÈ4
deKÆ 1
4í`
R
4í`
"a/ced1
l 1 `
ødeKÆ 1
=U
Thus for
we have:
=U4
1MÈ 4

wBYV
) R
"
R
(3.66) º
º
¦ì¦G 8 )
U
Ò
(3.67) º
p
¦
æ
j
wBYV
@ h`j|k1
=
9 )
4
B
¦G 8 )
@
`
T
)
4
¦
æ
;
=
=
rUrs4defÆ 1
T
3.4. CREEPING FLOWS 71
` l
1 `
1
=UÈ4
B{deKÆ
1MÈ 4
And the final function becomes:
(3.65)
U
Ò
1 `
1MÈ
` l
=UÈ4
's4
3.4 Creeping flows
If we combine equations (3.2) and (3.5), and use the expression for substantial
derivative (1.7), we obtain yet another form of Navier-Stokes equation (2.24):
¦
¦f} ` º
The LHS of this equation represents the inertial forces. The assumption of creeping
flow or Stokes flow states that the inertial forces are negligible. With this
assumption the last equation becomes:
¦f}
Differentiating over # , we get:
from which we obtain a Laplace equation for pressure:
¦f ¦/}“ 8 1
¦f ¦ 4 }
¦ì¦G
Forming a product with p5¦/ j and using the symmetric identity (A.27), we obtain:
T p
¦f }
j
¦
j¡º
jè)

(3.68) n
(3.69) ë
n
æ
ë
`
ë
x
T
T
Î
Î
72 CHAPTER 3. LAMINAR FLOWS
Swapping indexes # and ° in this equation and subtracting one from another we
can express it in terms of vorticity vector (1.12):
}
Thus both the vorticity and pressure satisfy Laplace equation in a creeping flow.
Important cases of creeping flow include:
1. Fully developed duct flow. Re-number independent.
2. Flow about immersed bodies (Stokes solution or the sphere).
3. Flow in narrow but variable passages. (Lubrication theory).
4. Flow through porous media.
3.4.1 Stokes flow around a sphere
Consider a laminar viscous flow around a sphere of radius A , with the velocity at
infinity. The solution to this problem will be two-dimensional, since by symmetry
nothing should depend on the azimuthal direction. It was shown in Sec.2.2.5 that
in a 2D limit the vorticity vector has only one component and it is related to the
Laplacian of the stream function (2.46):
¦ì¦
With these assumptions (3.68) becomes:
¦ì¦/}
In spherical coordinates the relation between the velocity and stream-function
(2.6) will become:
(3.70)
(3.71)
)œx
`
)GF
¢ rdefÆ
F
¢defÆ
ë

(3.73) ë
(3.76) X
Y
ë
r
ë
@ r Y
¢
Î Y
1
A
Î
X
@
rdeKÆ
—
r
ë
rdefÆ
¥
A
r
` À 8 Ò
l ¢ r A}
Î
A
A
Î
4
`
`
À
A
¢
r
>
ë
A
r
r
T
3.4. CREEPING FLOWS 73
and the Laplacian operator in spherical coordinates is:
(3.72)
" 1
—\Y
¢ r >
r ` –
¢ r Y
where we neglected the azimuthal direction angle because of the symmetry of
the problem. The boundary conditions for this problem are:
Y Îò
x
4
F1
4 T
l
¢
ÎŒ>\Ä
Ä where is any constant.
4
The solution satisfying (3.72) is:
1K
l ¢
1 ¢
',Î
@ 4
Ñ£
A —
¢
which can be checked by direct substitution 3 into (3.72). The velocity components
can be found from (3.70) and (3.71) to be:
(3.74)
À‰A
¢ l
)œx
ba/cedåÎ
@“>
l ¢
(3.75)
qdefÆ
À‰A
¢ Ñ
)GF
— `
The important quantity is the fluid drag on the sphere. It consists of the
contribution of the shear stress (tangential friction at the surface), and the pressure
forces normal to the surface (Fig.3.2). Pressure distribution on the surface
of the sphere is computed from the momentum equation (3.66), and results in the
following expression:
@“>
Ñ ¢ >
arcedWÎ
3 See Problem 3.7.8

½ ¡
=
1~
(3.77)
6 x„F
¤
=
¤
¤
õ
¤
Î
T
4
õ
¤
=
Ö
x
`
¡
=
>
Î
4
Ö
R
¡
>
¡
Î
1~
Î
4
R
¡
¤
=
4
a/ced1
Î
4
74 CHAPTER 3. LAMINAR FLOWS
žI~
=^)GF
L)œx
Â
The total force on the sphere can be obtained by integrating both shear (surface
friction) and the pressure components over the surface of the sphere:
F
¢ M )GF
l ¢
(3.78)
~
¤
=

¤
¤
¤
r
Ö 1
Vz
(3.81) Ĥ
Ö
Ö
z
¤
r
4
4
V
z
1
4
>
l
A"
¤
l
@
a/ced `
r
Ö
Ö `
Vz
À
V
z
4
4
6 x„FGdeKÆ
OzV
¡ l§UA
À
l
Ñ
À
r
l
4
1
Î
Ò
4
3.4. CREEPING FLOWS 75
and substituting the expression for the area R element:
have:
A‹deKÆ
R Î , we
(3.79)
¤xñw‡
X‰defÆ
l§UA
R Î ÎŒa/cydWÎ
Î R Î
where the negative sign in the second term is due to the fact the direction of
increase Î of is opposite to the selected of~
= direction . Substituting x„Ffrom
6
(3.77) and X from (3.76), and setting ¢ A , we obtain:
—WÖ
ÎŒarcedr
where we omitted the term involving the constant pressure X
component , since
its contribution will become zero after the integration. Computing the integrals:
ÀU8CA
ΠR Ω>
Î R Î
¤xðwZ‡
deKÆ
defÆ
a/ced1
À<
Ô{a/cyd1
4a`
R
OzV
V
defÆ 1
4arcedr
we finally obtain:
1
deKÆ
Vz
R
1
(3.80)
1 Ñ
l 4
¤xñw‡
U8tA
[U8
where the plus sign signifies that the force is directed toward the flow velocity.
It is interesting to note, that the contribution of the viscous wall friction due to the
A}
shear stress 6 xFis twice as big as that one from the normal pressure term.
A useful engineering formula for a drag coefficient is obtained by relating the
drag net pressure, to flow inertia (kinetic ;r
energy), :
: 1UA
: l
Ĥ
which in terms of non-dimensional Reynolds number, A\
;rUA
A}:J9
becomes:
l Ñ
which is valid for A#
. Ž@

(3.82)
;t)
(3.83) Ä
which is valid for T Ž
@
A!
—
`
@•> mA! [
X
À
T
Ñ
I
76 CHAPTER 3. LAMINAR FLOWS
3.4.2 2D Creeping flows
Let’s consider the limit of 2D creeping flows over plane surfaces. These flows can
be described by equation (3.69). It can be rigorously shown that this equation
can not have a non-zero steady-state solution with the steady-state boundary
conditions at infinity [12]. Intuitively, it is clear that if an infinitely large plate starts
moving with a constant velocity it will tend to impose this velocity on the rest of
the flow-field, but for an infinite flow field it will take infinite time to accomplish.
Therefore, there will be no steady state solution to this problem. This situation
became known as the Stokes paradox. To remove this paradox it was proposed
to add a convective derivative to a momentum equation [12, 2]:
¦G
¦
¦f€
is the known free-stream velocity.
coefficient becomes:
where
With these assumptions the drag
> 8 )
l Ñ
x which is called the Oseen approximation, and is valid I for .
In addition to this there are various engineering approximations obtained for
T Ž‡À
the drag coefficients of a sphere and a cylinder for various Reynolds numbers [2].
A reasonably good curve-fit approximation for the sphere is:
@“>
@[A" (>¼I¡IèI
(3.84) Ä‘
l Ñ
ÏŽ A\
A"
>
>
Í
l
3.4.3 Lubrication theory
’Ž A!
T”“
Lubrication theory focuses on the study of 2D creeping flows between the contracting
or expanding surfaces, when the surfaces are also moving with respect
to each other.

T
Ç
T
Ç
Ç
Ç
T
T
T
4
3.4. CREEPING FLOWS 77
Figure 3.3: Flow between contracting plates
Pressure inside a non-uniform gap
Let’s consider a flow between two plates one of which is at an angle to the other
(Fig.3.3). We shall use the coordinates in horizontal, and ® in the vertical directions,
and denote the corresponding velocity components ) as and•. Without
loss of generality we may presume the lower plate ® at to be horizontal and
the upper plate is given by a known profile of its height at each -position: .
Also without loss of generality we can assume that the lower plate is moving at
a constant velocity. Let’s consider a flow in a limited section stretching from
à1
coordinate to .
From Sec.3.2.1 we know that the steady flow between parallel plates has
a linear profile given by (3.9). In the case of tilted plates we can not expect this
profile to hold, because it would to the loss of mass-flow conservation, i.e. more
flow will enter the section at than exit at . Indeed, since the flow
velocity at the lower moving plate should be always and at the upper fixed
plate it should always be zero, there will always be more flow entering at ,
where the plates separation is wider than at where it is narrower. This is
because a linear velocity profile takes shape of a triangle, and the mass flow rate
is proportional to the area of this triangle. The area of the triangle at will
always be larger than the one at . From this we conclude that the linear
velocity profile can not be a solution to our problem.
In this situation a pressure distribution arises in the flow that leads to nonlinear
flow profiles at the inlet and the outlet, such that the mass conservation is
satisfied. Thus, the nature of this flow will be that of combined Couette-Poiseuille
flow discussed in Sec.3.2.5.
If we assume that the solution is of combined Couette-Poiseuille type, we

1
)
>•
¯
ã )
(3.87)
4
)
)
)
Ä
l
F
Ä
F
r
R
r
) ã
®
Ä
r
1
@
r
F
Ä
¯
•
V
r
`
r
º
R
F
@
)
º
R
@
@
)
T
)
T
78 CHAPTER 3. LAMINAR FLOWS
can find the equation for the pressure distribution in the section between the converging
plates, that will lead to mass conservation. For this purpose, let’s use the
incompressibility condition (2.4), and apply it to our case:
¦f ¦G
rq r“
If we integrate the last equality over the cros-section, we obtain:
>–•
¯
>])
(3.85)
V
Ö˜—
V
ÖJ—
V
Öš—
V
¥Öš—
R ®»>–•1U 4­`
•1}T 4
R ®
R ®»>
R ®
4 4 T
where we walls:•1UT
•1}
used the no-slip condition at the . The solution
for the combined Couette-Poiseuille flow should be obtained from equation (3.34):
V
ÖJ—
R ®
(3.86)
r
R ã
and boundary conditions:
8
R
1}T 4
) ã
4 T
) ã
Looking for a solution in the same form as (3.35):
® ã
>\Ä
®ƒ>\Ä ã
V
we can obtain the following values for the constants:
>\Ä
r“
8
R

Ä
(3.88) Ä
ã )
(3.89)
Ö
V—
Ç
)
T
ø
ø
Ä
@
À
º
V
V
) R ã ® ù ã
4
ø
Ç
º
r
¢
1
l 8
r
Ä
º
R
Ö
V
T
4
F
r
4
º
º
4
V
V
ù
T
T
3.4. CREEPING FLOWS 79
where the first two equations follow from the boundary conditions, and the last
equality follows from the substitution of (3.87) into (3.86). If we define a constant
as:
we finally obtain:
R
` 1
ã ® Ä
@•>^Ä
Since is now a function of , is also a function of , given by (3.88). Transferring
(3.85) to dimensionless variables, and substituting (3.89), we obtain:
F
—
` 1
4 Ö F
R ®
® ã
@“>^Ä
®»>¼@ ã
(3.90)
` o1
Ö
@•>^Ä
[
>
@ 4
>¼@ l
Substituting Ä
from (3.88), we obtain:
(3.91) R
—
` 1
—
À©>\Ä
T
à1
(
(
º
This is the equation for pressure distribution inside the section between the two
converging plates. It should be solved with a known profile of the gap width
as a function of :
, and with the given pressure distribution at the
inlet ) and the outlet ) of the section. Constant pressure boundary
conditions can be used in a simplified case: .
ù
[8
1}T 4
1 Ç 4
Remark 3.4.1 Contracting vs expanding gap
When we solve equation (3.91) with the constant pressure
4
boundary
4
conditions:
, for a linear profile of we should obtain a parabolic
ºg1}T
ºg1Ç
solution with an à1 extremal point, or
maximum)
4
between
and
. The condition for this º point is . Opening the parentheses on the
LHS of (3.91), we can obtain for the point following relation:
I G I
,(minimum
the

À
r
º
>
)
º
)
,
T
>
)
,
1•
,
º
,
)
¯m¯
4
4
[8
Y
80 CHAPTER 3. LAMINAR FLOWS
T
from which we see that if the slope of the upper wall is negative:
T
(contracting
gap), º
,
then , and the extremal point is a maximum. This means that
the pressure will be everywhere higher than the ambient inside the contracting
gap. The opposite conclusion follows for the expanding gap.
Ž Ž
In reality the case of the expanding gap may lead to higher flow instabilities
and cavitation. This is the result of the counter-flow pressure gradients arising
in the case of expanding gap that may lead to the possibility of flow reversal and
separation as determined by relation (3.38).
Validity of the pressure equation
In arriving at solution (3.91) we assumed that we can use the solution of combined
Couette-Poiseuille flow given by (3.36). But that solution was obtained under the
assumption that flow velocity has ) only component which depends only on one
¦f
® coordinate - . This assumption made the )Gj) j convective term in the Navier-
Stokes equation (3.2) equal to zero. However, this assumption is not strictly valid
in our case since we have a non-zero vertical velocity component,•, and both
horizontal and vertical velocity components are functions ® of and . In this situation
we can still justify dropping the convective term if we use the assumption of
Stokes flow, that is, consider the inertial forces to be negligible as compared to
viscous forces. Mathematically this means that:
¦f}
or expressed in terms of , ® , ) ,•
have two relations:
¦f› 9 )
1
we
)W)
>•§)
>^)
From the form of these equations we can deduce more specific relations between
the parameters of our problem (, , , ). In particular, if we impose
Ç
condition
of a narrow gap: , then it would lead Ç and
› 9 : Y : Y Y ® . With
these conditions we can neglect the equation for•and simplify the ) -equation by
neglecting a smaller order term on the RHS:
› ›
to•
)G•
>•I• 9
¯›
>•
¯5¯
¯› 9

@
)
¯m¯
3.5. BOUNDARY LAYERS 81
Figure 3.4: Integral analysis of a boundary layer
9
Both terms on the LHS are of the same order and can be approximated by
¯›
the order of the first one, which isr
, we finally have:
: Ç
. Approximating the term on the RHS as
)W)
>•§)
9:
(3.92)
r
Ç 9
›
This condition together with justify the Stokes flow assumption and constitute
the validity limits for equation (3.91).
› Ç
3.5 Boundary layers
3.5.1 Flat plate integral analysis
We shall consider a two-dimensional flow above a semi-infinite plate (Fig.3.4).
Our objective is to introduce a quantitative measure of the thickness of the
boundary layer, and to estimate it’s growth with the distance from the edge of the
plate, .

¤
`
Ö
¤
V—
¤
V
Öšœ
1 ;4R ®
¤
`
Ö `
Vœ
V
Ö˜œ
V
1 ; )
Ö
V
Ò
œ
[
1
@
4
)
1
@
)
1 4
®
MR
>
V
֜
` Ö —œ
Vœ
)
)
r
4
82 CHAPTER 3. LAMINAR FLOWS
The displacement thickness
The balance of mass dictates:
(3.93)
R ®
from which we get
1U ` þ 4
1
`
1\>])
4
R ®
R ®
` )
Let’s define the displacement thickness as
R ®
(3.94)
4
` Ö þ
Vœ
4
ß”¢×Ö
` )
R ®
Momentum thickness
General conservation of momentum dictates: ¤
1 4
Applied to the case of the boundary layer (Fig.3.4) it will provide the expression
for a drag force, ¤, on the plate:
) R ®
;r
expressing
from (3.93), and substituting into the above, we get:
r R ®
` )
¤ ;r
®
L@

(3.95) Î
(3.96) Äkõ
(3.97) Ĥ¢
Considering that ¤
(3.98) ÄŒõ
¤
l
¤
¤
@
Ç Ä¤
(3.99)
V
Ò
l
¤
l
)
¢
¤
¤
V
l
;r
6Q
1
4
l
åĤ4
l
Î
Ç 4 1
Ç
R Î l
R
3.5. BOUNDARY LAYERS 83
We shall define the momentum thickness as:
¢ÚÖ
` )
Using this definition we can also define the skin-friction coefficient as:
L@
® MƒR
and the drag coefficient over the length Ç of the plate as:
Î : Ç
;r Ç
¤¥Ö]
6QR
where is the distance from the edge of the plate in the downstream direction,
and consequently, , we can rewrite (3.96) as:
6Q
¤
¤
1
;r —
;r
And the inverse relation:
Äkõ
R
Remark 3.5.1 Displacement vs Momentum thickness Displacement thickness is
a more universal notion than a momentum thickness, since the former is obtained
from the mass conservation, which is a always true in incompressible flows,
whereas the momentum conservation may be violated due to the dissipative effects.
Relation (3.97) will only hold for the flat-plate boundary layers.
V
Ö˜ž

(3.100) )
(3.103) Äkõ
)
)
1
®
1
®
¦
Î
Ä
r
®
r
@
À
Ï
8 R ) 6
® R
ß
8 Ñ
ß ;
F
4
84 CHAPTER 3. LAMINAR FLOWS
Guessed solution
Let’s use a general parabolic velocity profile:
4
>^Ä
®»>^Ä
V
Imposing the boundary conditions appropriate for the boundary layer flow:
1UT 4 T
1 ßJ4
1 ßJ4 T
we can determine the constants Ä
)
, and obtain:
)
¯
®
ß
l ` ® 1
ß
4
Substituting this solution into (3.94) and (3.95), we obtain:
(3.101)
ߧ
(3.102)
l ß
Using the definition of 6 (1.28):
we can rewrite (3.96) as:
Äkõ
8 l
R )
R ® ;r
and using the parabolic velocity profile (3.100), we obtain:
On the other hand, according to (3.98), and (3.102), we can write:

–
P
¦
s
¦
)
ß
ß
ß
R
ß
r
)
¦
À
@
Ñ
Ï
@
Ï 8
¦f“ 9 )
T
R
ß
r
)
) ¦
X
T
¦
3.5. BOUNDARY LAYERS 85
Äkõ
R
Equating this to (3.103), we get:
ß
Integrating, we obtain:
;
R
(3.104)
8 T
;
Introducing the Reynolds number as: A#
:])
; >
¦ 4
¦ì¦
m ¦ 1
¦f
m ¦ 4
1
sà>])
°
s ;
> 9 )
>^)

)
)
F
(3.107) )
F
ã
F
)
F
ã
r
F
r
)
F
F
F
ã s
F
F
)
r
F
)
)
ß
Y
Y F
r
F
F
ã r
s
¦
V
¾
r
r
Y
Y
F›
F
)
)
s ã
r
>
) ã
F
X ã
1
r
>
F
¤
ã
X
V
`
V r X
) ;
F
T x
r
T
V
V
¦ V
V
"
F
86 CHAPTER 3. LAMINAR FLOWS
¦( 1 r 4
where we shall assume a 2D 'q
approximation
¦G 1 r 4
for the boundary layer: ,
, that is, we consider that there is no variation in the span-wise direction.
Consider also, that the layer is thin:
'*)
rŸ›
)
which in our particular case is reflected in relation (3.105):
r ¦
@ r
F}É A
Therefore,
Using this assumption we can introduce the following scales for dimensionless
variables.
@ r
F}É A
(3.108)
V ã ¾
r“
" )
A F}É
ã
r ¾ r*
F ã ¾ r
ã
V
(3.109)
ã

¤
(3.114) „©z
If we apply the momentum equation (3.111) at the wall (®
)
)
¯
'
¢
F
'
)
F
r
T
¢
)
F
)
F
¯
>
¢
–
P
¯
V
s
r
F
F
T
¢
F
r
¢
)
T
F
), where )
¢
)
r
F
1
¢
F
'
)
F T F
F
3.5. BOUNDARY LAYERS 87
x where is the Froude number (2.111), and we introduced the dimensionless
Eckert number:
¢ )
[2].
V
Equations (3.110) - (3.113) represent the Prandtl boundary layer equations
Remark 3.5.2 Parabolic character Note, that all the second derivatives over
have disappeared from the momentum equations. This means the that equations
of motion are parabolic with respect to direction . This in turn enables to use
simpler solution procedures.
Flow separation
, and consider the gravity forces acting normal to the wall ( J¦
r 1
T 4
- ), we obtain:
T 4
% T
)
¯5¯

(3.117)
¢
4
)
1
)
¡
1
Ø
1
1
1
)
1
ß
)
1
1
¦
o
9
l 9
®o
Y Ø 4
® Y
` Y Ø 4
Y
4
¯
šg
4
š
1
T
T
4
4
l 9
4o šg1K
1
)
1
1
4
88 CHAPTER 3. LAMINAR FLOWS
which should be solved with boundary conditions:
(3.116)
T 4 1
T 4
('
('
Relation (3.104) indicates that the boundary layer growth with as:
('K 4
('K 4
Considering this, let’s look for a similarity solution )
4
® : ß

`m
@ 91
lm
(3.123) š
)
`
`
Ø
š
`
Cšgg
l
Ø
`ml 91 @
lm
@
lm
)
¯
1
Cšgg—
¯
Cšgg
)
¯
`
l
)
l
l
9 o
@
Ø
¯
4
mGšg š2>
) 9
4
l
`"
9 l
r
`
l
š
4
4
š
šgg
šgg
šggg
4
3.5. BOUNDARY LAYERS 89
Since
set šg1K šg1K
4
is a constant, we can incorporate it into ¡ 1
, which is to say, that we
:
@ without loss of generality, and obtain for ¡ 1
¡ 1
4 m
9©
Now the expressions of other terms in (3.115):
4
1 4
(3.119)
(3.120)
9§$š
('*®
4 m
Cšg
1
kšg`
l
šg
4
o
)W)
1
kšg`
r
l
Œšgg šgg šg
(3.121)
r
l
¯m¯
And substituting them into (3.115), we obtain:
ššgg
and finally:
(3.122) šggg>Þš$šgg T
»>Þšggšg
`
šggšg
which should be solved with the boundary conditions:
šggg
1}T 4
T šg1K 4
@
šg1}T 4

V
Ò
o
l
1
@
9
(3.124) Î
Ö
Ö
V
V
Ò
Ò
@
@
`
Î
`
o
l
9
Y ) 8
® Y
V
Ò
V
o
l
1
@
Ò
š
`
9
@
9
`
V
Ò
V
l
l
9 o
Ò
`
T
R
Ö
V
Ò
@
`
9
T
I
R
90 CHAPTER 3. LAMINAR FLOWS
where the last condition also implies that . This is a Blasius equation.
It has no analytical solutions in a general case, but can be solved numerically.
šgg1K
Knowing š function
the boundary layer.
1 4
Displacement thickness (3.94):
, we can obtain important integral characteristics of
4
ß ¢ Ö
` )
R ®
4
Ö
šg1
šg4
ߧ
4
@‰I
Momentum thickness (3.95):
¡i Ò 1 `
@[£¢Wo
l
dfeKg
where we can estimate the integral as:
Ö
šg1
šg4
±Áš
ššggR
šg1
R šg4
šg4 ²
Using equation (3.122) and the boundary conditions (3.123), we obtain:
šgggR
šg1
¥Ö
R šg4
šgg1UT 4
Thus:
Ñ[‰Ôy[
šgg1}T 4o
For the wall shear stress we have:
(3.125)
6Q
; šgg1UT 4
l
The friction coefficient:

l
;r 6Q
(3.126) Äkõ
@
Ç Ä¤
(3.127)
4
)
V
1
Ö
V
ž
®o 1
1
V
)
4
Î
¯
4
l
9 l
4
9
3.5. BOUNDARY LAYERS 91
šgg1}T 4o
and comparing the latter to (3.124), we obtain:
Äkõ
©
l
And the total drag on the plate:
1 Ç 4
Äkõ
R
Äkõ
Wedge flows
A similarity solution for the wedge flows was found as an extension of the Blasius
solution. This solution, called after the authors . This solution is obtained by
eliminating -velocity from the continuity equation:
` —åÖ ¯
) R ®
and substituting it into the momentum equation:
` —åÖ ¯
)W)
) R ®
> 9 ) ¤
¯m¯
is a free-stream velocity profile. Then introducing the dimension-
less variable, :
where
1
and looking for the solution in the form:
>¼@
41
('*®
šg1 4
4
1

š
4
Ø
1
4
¦
º
`
1
@
š
`
1
±
P
|
šgr
4 ²
Î
92 CHAPTER 3. LAMINAR FLOWS
4
1 4
where is the first derivative of the so-called Blasius š
stream function, šg1
.
With these assumptions one can obtain the Falkner-Skan equation in terms of
:
1 4
l ¹: 1
@‰>
where , and the boundary conditions are the same as (3.123). The
Ë
Falkner-Skan formulation is consistent with the free-stream velocity distribution of
the type:
šggg>Þš$šgg>^Ë
4 T
4 £¸
I happens so that this equation provides similarity solutions that represent
wedge flows, with the stream function of the type:
1
',Î
>¼@
1 ¢
¢ |
DGFdefÆ
Wall suction or blowing
The case of wall suction or blowing can be modeled in the Blasius solution by a
non-zero vertical velocity at the plate surface. It can be seen from (3.120) that this
be accomplished by specifying a non-zero value of the Blasius stream function at
the wall:
T 4
1}T 4o
('
l
An interesting feature of the solution is the effect of blow-off of the boundary
9
layer, where ) the velocity becomes identically zero. This occurs at the blow-off
limit š of :
1}T 4 ` T
I4[Ã@¡Ôm l
3.5.4 Reynolds analogy
An important empirical relation for the flat plate flows with heat transfer is called
the Reynolds analogy. To introduce it, let’s recall that the Prandtl number was
defined as a relation between momentum transport to the heat transport (1.30):
¢C¢ 8 –
°

Ä
~
P
(3.128) Ä
Ä
u
6
…
~
ô
~
¢
8 R )
® R
…
°
6 l
;r
s
ô
Ç
…
r
3.5. BOUNDARY LAYERS 93
Ä where is the specific heat at constant pressure, (1.24), is the coefficient
of viscosity, defined as a proportionality constant between the shear stress and
velocity gradient (1.28):
8
and ° is the heat conduction coefficient defined through the relation between the
heat flux and temperature gradient (1.29):
,¯
¦$
`t°
¦
The Nusselt number was introduced to quantify the heat conduction at the
walls with the temperature difference [^s (2.114):
where …
is the absolute value of the heat flux:
Reynolds analogy is the statement of proportionality between the heat flux
and shear stress at the wall:
… O
¦O(±/…
²GÚ´ : 1
. 4 ° :‰.
).
[^s
…Ú¦
which is usually formulated as relation between dimensionless Stanton number,
6Q
:
¢ u
and friction coefficient, Äkõ (3.11):
A! 5º_x
such that
Äkõ
£¡
x ÄkõEº9¥
where and are empirical constants depending on geometry. The Reynolds
analogy is approximately valid for shear layers, boundary layers, and pipe flows.

X
(3.130)
¦$
)
)
)
9 ¦K
)
l
)
F
>
¯
¯
4
T
l
l
4
T
94 CHAPTER 3. LAMINAR FLOWS
3.5.5 Free shear flows
In this section we shall be looking for similarity solutions of free shear flows. Such
solutions can only be accurate in the regions of the flow far from the disturbances
that generate the characteristic flow patterns.
In the case of free shear flows we can still assume that the boundary layer
approximations (3.107) are valid when the Reynolds number is
sufficiently r“
large.
Consequently, we could neglect the X
vertical pressure gradient, , (3.112). In
addition to this we shall assume that the axial pressure gradient is small
T
as well:
, which reflects the fact that free-flows by their nature are not pressuredriven
flows. Thus, the equations of motion are:
F§¦
¦f ¦G T #­
@‰'
'„¨
rðr #
¦í 1
which in terms
of explicit )
variables , and
to the Blasius equations for the boundary layer (3.115):
('q®
1 ¦a
)_'
becomes identical
@‰'
(3.129)
9 )
)W)
> )
¯m¯
Shear layer
Consider a shear layer created by two streams with velocitiesF
andr
uniform
initially separated by a flat plate. After passing the plate the streams mix forming a
shear layer. The treatment of this case is done similarly to the Blasius approach,
by selecting the non-dimensional variables as in (3.117), and looking for a solution
in the form (3.119), but applied separately for each of the streams:
®o F
)
which leads to the Blasius equation for each layer:
šI©
9 ¦
F
š©©©
0=2
T
>\š0=2š©©

F
8
F
4
F
š
F
F
š
é
Ò
r
r
–
¥
)
F
F
T
T
)
F
)
T
3.5. BOUNDARY LAYERS 95
Boundary conditions at far ends:
1K 4
r 1 `K
4
F
r
F
)
r 1 `K šI© @
)
F
šI©
, which is the imaginary
continuation of the plate. For this purpose we need to consider boundary conditions
on that plane 5 r 1}T 1}T 4 4
)
. One condition is derived from continuity: ,
:
4
We can sew both solutions at a horizontal plane at
1K
4
1UT 4 r 1}T 4
1}T 4
r 1}T 4 T
1UT 4
and another from the equality of shear
8
stresses:
translates into:
F
š©
š©
¯
1UT 4
8 r
rq
¯
1}T 4
, which
r}ª
1}T 4 °
r 1UT 4
šI©©
; 8 r : 1 ; r
where . It is an interesting fact, that in the case of the same
fluid
in both jets ° ( ) and zero free stream velocity in the one of the (r
jets @ ),
the vertical component of the second jet at infinity is a finite constant, ° r 1 `
determined
T
I,[Ã@EÔml
by , which is the same as the blow-off limit in the case of š
the
plate with suction or blowing considered in Sec.3.5.3.
`K
F
šI©©
F}É
4
Jet
A free jet is characterized by the conservation of momentum in each plane:
(3.131)
´Ÿ
; Ö
= . "
Ò )
R ®
The similarity solution for this case as obtained by Schlichting [2] is:
5 This plane does not represent any physical interface between the two streams. The latter should be
defined as a surface formed by flow streamlines.

Ø
9 `
r
r F}É
r
F}É 9
4
with the equation for the Blasius stream function:
šg
1
T
(3.132) š
r
š
1
É
¯
É
)
1
š
`
r
šggg>\ššgg>\šgr
1
1 4 l £%«B¬Æ®­1
)
1
´
l
š
šg1K 4
r
É
r
T
T
T
1
96 CHAPTER 3. LAMINAR FLOWS
®
À 9
F}É
1 4
F}É
4
šg1
F}É À<
À<
and the boundary ) conditions (axial symmetry), ) and
(quiescent ambient fluid), which translate into:
('K 4
('
('
('K 4
T 4 1
T 4
The solution is
1UT 4
šgg1}T 4
£
4
with the constant £ determined from the conservation of the total jet’s momentum.
Thus, substituting (3.132) into (3.131) we obtain:
£â — Ô
F}É
The maximum centraline velocity drops off as
@[m;Ž8
l £
4
T 4
)œ|aw
('
¦£ é F}É
and the jet spreads as ¦
.
À< F}É
Wake
Wakes are flow patterns generated by bodies moving in a stagnant fluid. We need
to consider these flow patterns far enough from the body for similarity solutions

(3.134) ‹
T
`
r
@
é
É l ¡1
`
r
@
é
É l 1
‹
1
r
É
1
‹
1
ø
@
r
r
É
¡
®
r
r
ø
‹
®
we
r
r 4h`j|k
r
¯
1
)
1
É
4
r
T
@
r
r
ù
r ¯
r
3.5. BOUNDARY LAYERS 97
to be valid. Since it is more convenient to select a coordinate system moving
with the body, we will be considering a non-moving body immersed in a fluid with
the free stream velocity. It is also convenient to describe the wake in terms of
deviation of velocity from the free-stream velocity:
‹ since usually constitutes a small value compared ) to. ‹ Replacing with in
(3.129), ‹ and considering obtain:
› ›
and
('*®
('*®
‹
4
`
(3.133) ‹
9
¯m¯
with the boundary conditions:
4
('`¯K
T 4
('
Equation (3.133) is of a heat-conduction type, with the solution:
4
` ®
('*®
—
Substituting it into (3.133) we can find :
rhjlk
F}É
` ®
>à é F}É
—
£¡
` ®
` l 9
¿
` l 9
®h`j|k —
¡
` l ®
ùhjlkû— `
®
` l 9
` l ®
>à é F}É
and at ®
we obtain:
r 4
r —
(3.135)
Ó
Ñ 9

where ¤
é
Ò
Ò
T
¤
@
l Ĥ;r
¤
¦
R
ß
é
Ò
Ò
T
Ç
ß
ß
U9
T
98 CHAPTER 3. LAMINAR FLOWS
The constant is evaluated from the second Newton law, that the drag force
should equal to the momentum deficit in the wake:
¡
¤
and
[J°
[J°
¥Ö
;
Ö
l ;¡ o
with the final result:
(3.136)
; )W‹ R ®¦
¡¼
ĤÇ
Ñ
U9
o
; ‹ R ®
When substituted into (3.134) it shows that the wake defect is proportional to body
drag coefficient.
3.6 Integral methods
To derive the integral formulations for the boundary layer equations, discussed in
Sec.3.5.2, let’s consider the conservation of mass and momentum in a controlvolume
formed by four ((' points: ), as depicted
in Figure 3.5.
), (('
), (t> R ('
From the conservation of mass we have:
> R
), (t> R ('
£‰¦$
?
)
£‰¦
where is the vector area element. Summing up over all the four faces, and
R
considering that the flow is uniform in the transverse direction (2D flow), we can
expand the equation above as:

Q is
`
V
Ö˜±
; ) R ®
`
;R
; ) R ®
w¦
ß
¤
ß
>
R
R
; ) R ®C> R
¦$ w¦ š$›M)œ
; ) R ®»> ; QR
)
4
; ) R ®
T
3.6. INTEGRAL METHODS 99
Figure 3.5: Control volume for an integral formulation
where the normal velocity at the wall, which is non-zero in the case of wall
suction or effusion. Rewriting the second integral as:
V
Öš±D¤±
and considering constant density, we obtain:
V
Öš±D¤±
V
Öš±
V
Ö±
(3.137)
) R ®t> Q
R
In a similar manner, considering the momentum conservation, we obtain:
V
Ö˜±
R
¦ 1
where the summation is done over all faces, and we are considering projection
of forces and fluxes on the horizontal direction. Writing the terms explicitly, we
obtain:

Now we eliminate R
X
ß
>
R X `tß
R
)
r
1
R X `§ß
R
> ;
R
R
X R
R
R
` ;
>
Ö
V±
`
V r
) ;
r
R
ß
Î
` ;
Ö
V±
`
; )
R
R
r
r
R
`
Ö
V±
; )
; )
R
R
r
r
R
Î
ß
; )
4
r
100 CHAPTER 3. LAMINAR FLOWS
ߧ` 1
4 1 ß
ß R X : l 4
6QR
X>
R X
> R
;r
R ®
` Ö ±
R ®C>
R ®
where X is the pressure. Canceling terms and neglecting higher order terms, we
obtain:
(3.138)
6Q
> ;r
R ®
R
: R
ß
by mens of (3.137), so that (3.138) becomes:
V
Ö˜±
(3.139)
6Q
R ®
) R ®
One can relate the pressure gradient to the free-stream velocity gradient:
V
Öš±
Q
`
V
Ö˜±
` ;9
It is also convenient to express (3.139) in terms of displacement thickness, (3.94)
and momentum thickness (3.95):
R
R
1 ßt`]ߧ,4
) R ®
1 ßt`]ß `
R ®
) R ®
with these transformations (3.139) can be written as:
V
Ö˜±
Ö˜±
V

À
l ß
1 4
®
ß
¥
s
V
@
`
Î
l
1
Î
>
@
>
s
F
Î
`
`
1
@ R
R r
ߧ>
Q
À
l
Î
V
4
)
Î
r
Î
`
`
`
Q
QQ
Q
`
3.7. PROBLEMS 101
>
) R ®
R ®
;r
R
R 1 ßC`]ß”*4a`
@ R
R r
r
ß
R
6Q
Ö±
1 ߧ`àß”c`
*4a`
V
Ö˜±
>
1 ßt`]ߧ,4
ßt`àߧê`
,4a`
1 ßt`]ߧÛ`
,4
ß
R
`
>
>^Î
which finally becomes:
1 l
Î >
For steady flow with an impermeable wall we obtain:
;r 6Q
ߧ,4
(3.140)
Äkõ
l
ߧ,4
1 l
which is called a von Karman integral momentum relation.
Î >
3.7 Problems
Problem 3.7.1 Couette flow equations
Show how to obtain equation (3.7) and (3.8).
Problem 3.7.2 Couette plates solutions
Solve equations (3.7) and (3.8) with the boundary conditions u(0) = 0 and
u(H) = U, s and s and .
1}T 4
1} 4
Problem 3.7.3 Flow of a liquid film
Ï ¿ , flowing steadily
Consider a wide fluid film of ’ @‰I
constant
T
thickness,
due to the gravity down the inclined plate Î
{
at angle . Find an analytical
expression of a fluid velocity distribution as a function of a distance from the plate
surface: . Assuming the viscosity and the density of the fluid are @‰I4[Jˆ
°
) 1 ¹. 4 WT‰T ° :<
respectively, find the volumetric flow rate, 8 , per … :
1m of the plate. Atmospheric pressure can be considered constant.
, and ¢ ’

I
º
¾
r
ë
X
R
R
r
ë
R
R
r
Ñ
ˆ
l
T
@
Ï
F
T
102 CHAPTER 3. LAMINAR FLOWS
Problem 3.7.4 Couette solution for non-Newtonian fluids
How will the solution (3.9) change for a non-Newtonian fluid?
Problem 3.7.5 Momentum equation for Couette flow between concentric cylinders
Using the assumptions on the Couette velocity profile between the rotating
concentric cylinders (Sec.3.2.3) and the expression for the momentum equation
and the Laplacian operator in cylindrical coordinates:
)IF­>])Wx5)GF
x_>
¢ >^)GH,)GFH­>
¢
)œx5)IF
¼
F`
)GFœ)GFF
)œx
; ¢ > 9 1 ¾
)GF­>
r `
¢ ¢ r 4 )GF
@
¢
x
4
x_>
@ r ë ¢
F„F­>Þë
F
F
HH
1 ¢
Derive equation (3.18):
¢ r >
Problem 3.7.6 Rotation torque and power
¢NM ¢EL)IF )GF
In the system of two rotating cylinders (Sec.3.2.3) consider the torque applied
to the inner rotating cylinder when the outer cylinder is (n fixed ). What
is the power required to rotate the inner cylinder?
Problem 3.7.7 Flow between parallel plates under pressure
A viscous fluid with viscosity (
parallel plates
8
` l
ωÏ
) is driven between two
`ÞÑ•° : 1 ¹. 4
Ï @‰I
¿ apart by an imposed pressure gradient of º : R R
volume flow rate per 1m of the plates’ width. What pressure gradient will cause
the flow to reverse?
£ :< . The upper plate is moving with velocity
– ¹:‰.
. Find the
Problem 3.7.8 Verifying the Stokes solution
Verify that the solution (3.73) satisfies the equation (3.72).
Problem 3.7.9 Stokes velocity and shear stress
Using Stokes solution for the stream function (3.73), obtain velocity components
(3.74), (3.75) and the shear stress component 6 x„F(3.77).

Hint: Assume )
@
T
Ï
( 1 ß :
º
)
( 1 ß :
@
(Äkõm A"
@
@
T
3.7. PROBLEMS 103
Problem 3.7.10 Falling sphere in oil
Ý
A sphere of density dropped into oil of density {
and viscosity 8 T £ .
. Estimate the terminal velocity of the sphere if its
; ;
IK@
diameter is (a) R
and (c) R
T ¿ .
¢IW;—rZ²
is
IWeW;—rZ²
@ ¿
Problem 3.7.11 Boundary layer analysis for cubic velocity profile
Repeat the boundary layer analysis of Sec.3.5.1 with assumed velocity profile:
IO@ ¿ , (b) R
À
L®
l
@
L®
ßM
l
ßM
`
4m A! ),
4m A! ),
4m A" ),
Compute ( Î :
1 :‰9
. ©
(Ĥm
where A!
.
1}T 4 T
¯m¯
),
A" ),
Problem 3.7.12 Drag force on a triangle
A thin equilateral triangle plate is immersed parallel to a stream of air with
velocity
l
the
£ "
. Assuming
{
laminar flow, estimate the drag on this plate.
l ¹:‰.
at temperature s
Ä and pressure º
Problem 3.7.13 Boundary layer equations
Derive Prandtl boundary layer equations (3.110) - (3.113):

104 CHAPTER 3. LAMINAR FLOWS

A
V
1
A
F
`
9 ¦
A
V
1
n
F
`
9 ¦
n
V
¦
l
Chapter 4
Turbulent flows
4.1 Transition to turbulence
In the case of free flows transition to turbulence occurs much earlier than in confined
flows. In terms of Reynolds number, it is a matter of several hundred for
the unbounded flows around objects, and a matter of several thousand for the
confined flows.
Thus, the transition to turbulence for the case of parallel plates usually occurs
at:
Ï TJT
The transition to turbulence for the case of rotating cylinders is usually measured
in terms of Taylor number and occurs at the critical value of:
A!
@
4
s$w
4
@³¢TJT
For the flow in ducts the transition to turbulence occurs at A#
.
T‰T‰T
4.2 Turbulence Modeling
Let’s consider the incompressible forms of the mass and momentum equations,
(2.4), (2.24):
105

¦
> )
1
) )
(4.3)
9 9
9 Ê 1
Ä
(4.4)
Ý
[
V
r
4
)
r
T
)
X
Ý
¢
)
)
106 CHAPTER 4. TURBULENT FLOWS
(4.1)
(4.2)
¦f ¦G
where we introduced an abbreviation for pressure-to-density ã ratio:
also use the conservative expression for the pressure ) term:
which is true due to continuity (4.1).
: ;
. We
¦K] 1 ¦ 4 X
The solution to this equation system for high Reynolds numbers will result in
a turbulent flow field. This field is highly unsteady with a broad spectrum of eddies,
which makes it difficult, if not impossible to resolve it on even the most powerful
computers. However, for some limited range of Reynolds numbers the solution
can be obtained numerically. The technique that uses this direct approach of
computing turbulence is called direct numerical simulation (DNS). The difficulty
with this approach is that in order to reproduce all the turbulent eddies from the
largest to the smallest, the simulation has to resolve the smallest space and time
scales of turbulence. This in turn may require a very fine grid and time-resolution.
,
¦ 4 k `
¦
¦f}
X ã
> 9 )
4.2.1 LES models
In order to go beyond the Reynolds number limits of DNS another technique
is employed. In this approach the grid cell sizes used are usually greater than
the smallest turbulent eddies. This amounts to a space-averaging of the Navier-
Stokes equation. In this case only the largest turbulent eddies are resolved, and
the unresolved eddies are modeled as an addition to viscosity:
> 9 Ê
where 9 V
is the molecular viscosity, and 9 Ê is the eddy viscosity, associated with
the cumulative action of unresolved eddies on the resolved large eddies. This
approach to turbulence modeling is called Large eddy simulation. In the first
proposed LES model, Smagorinsky model, the turbulent viscosity 9 Ê is set proportional
to the strain-rate tensor (1.9), and the computational grid size::
1 . ¦7 . ¦/ 4
[ where is the grid cell size, and the coefficient of Ä
proportionality,
Smagorinsky constant.
, called the
F}É

(4.5) Š
)
1
xÊ "54
V
1 It is also common to denote the fluctuating velocity asńµ
)
1
Ý
1
)
R
"
Ö
V
Ê
…
)
Ý
1
>
4
¦
R
4.2. TURBULENCE MODELING 107
In the Smagorinsky model Ä the constant is considered fixed, and is selected
by matching the experimental data or those produced by DNS. In more
sophisticated LES Ä models is no longer considered a constant, and its value
is determined in a more complex way, for example by comparing some integral
measures produced from solutions obtained from space-averaged equations using
different averaging filters [13]. The common feature of LES models is that they
are based on solving for a time-dependent flow field using space-averaged form
of the Navier-Stokes equation.
4.2.2 RANS models
Historically, another turbulence modeling approach was used first. This approach
is based on time averaging, rather than space averaging of the NS equation.
Let’s use the Reynolds assumption that any turbulent quantity, … , can be
decomposed it into the time average and fluctuating components:
"54
"54
('
('
In particular, for the velocity components, we have:
>^Šg1
(4.6)
(4.7)
¦$
¦
X ã
º >”X
, and time averaging of the fluctuating compo-
where1
nent, , is zero 1 :
4 ¢
('
('
"54
(4.8)
¦$
¦ 1
"54
" T
('
The averaging time is usually much greater than the time scale of turbulent fluctuations.
Since the time averaging interval does not stretch to infinity, the resultant
average quantities can still be slowly varying functions of time. But the time scale
of these variations will be larger than largest turbulence time scales. Under this
assumptions decomposition (4.8) is called the Reynolds decomposition.

>
>
>
>
À
l
º
º
¦
)
¦
T
T
>
108 CHAPTER 4. TURBULENT FLOWS
Applying (4.6) to (4.1), (4.2), we have:
(4.9)
¦
¦
1
4 1¦
¦ 4 ²¨k ` 1
¦
¦ 4
¦ 4 €
and after and time averaging the latter and the continuity relation (4.1), and using
(4.8), we obtain:
>±
>ûX
> 9 1¦
¦f ¦G
¦f ¦G
(4.10)
¦
1¦ 4
1 ¦ 4 `
¦
As can be seen, both the mean and fluctuating fields satisfy the continuity. The
last equation is called the Reynolds averaged Navier-Stokes equation (RANS),
thus the name of the approach. As can be seen the equation contains and extra
unknown term composed of derivatives of the so-called the Reynolds
(¢
stress
¦
tensor: .
6
To close the new equation system one needs to formulate extra equations
for the components of Reynolds stress tensor (closure). One of the simplest
and first closure was suggested by Boussinesq, and is called the Boussinesq
approximation. In this approximation the Reynolds stress tensor is considered
proportional to the mean velocity gradient:
> 9¦f€
(4.11)
¦ k
ß ¦7· `
l 9 Ê ¨ ¦/
(4.12)
@ ä¢
) l
¨ ¦/k¢
1¦f
@
l
>¸m ¦ 4
(4.13)
where we introduced the mean strain rate tensor, analogous to (1.9). Quantity
defined by (4.12) is called the Turbulent kinetic energy, and ¨ ¦7 is called Ê
the eddy Both
viscosity,. and 9 represent two new unknown variables in the
Ê
model. Empirical algebraic relations can be devised for 9 and
connecting
Ê
to¦f
them and length-scales of the problem, as is done in the mixing-length
9
theory [2], or in a more sophisticated RANS models discussed below.
On the other end of the turbulence closure spectrum is the Reynolds stress
model (RSM). In this model each component of the Reynolds stress tensor is

)
¦
>
j
>
`
>
>
¦
j
)
>
¦
X
>
j
j
`
j
X
>
¦ ­j
j
¦
¦
X
j
`
)
j
¦
X
¦
j
4.2. TURBULENCE MODELING 109
obtained from a separate equation. Since there are six independent components,
there should be six equations. In the full Reynolds stress model these equations
are represented by the PDEs of the transport type. In an approximate version of
RSM - algebraic Reynolds stress model these equations are given by algebraic
relations.
To obtain the equations for Reynolds stress tensor, let’s first subtract (4.10)
from (4.9):
¦
1 ¦ 4
1 ¦ 4
1 ¦ 4 ` 1 ¦ 4
¦
> 9 ¦f€
which, using continuity (4.1), can be rewritten as:
(4.14)
¦
¦f
¦f
1 ¦ ` ¦ 4 k `
¦
Let’s multiply this equation by )Gj :
> 9 ¦K€
¦
¦K
¦f
1 ¦ ` ¦ 4 k `
¦
¦K€
)Wj
>^)Wj
>])åj
>^)Wj
)WjmX
> 9 )Wj
If we swap the indexes # and ° of this equation, we get:
¦
¦ 1
`
4 k
€
¦ aj
9 ) jc>
jê>])
>^)
>^)
Now add the two last equations:
¦
¦
)åj
>])
¦f
¦f
>C)Wj
>^)
>^)Wj
>])
1 ¦ ` ¦ 4
¦ 1
`
4
>C)Wj
>^)
¦ `
¦f€
} 4
Now apply Reynolds decomposition ) (4.6):
equation:
¦G
¦
, and time average the last
)Wj5X
)Wj
>^)
jc> 9 1

1ajê>
j
` 1aj•> j
` 1aj•> j
¦
)
`
4
X
>
4
X
¦
¦
)
>
`
>
1­jê>
¦
X
>
>
j
j
j
X
X
j
X
>
j
j
1­jê>
j
j
>
j
>
j
>
>
>
j
1ajê>
j
¦
j
>
>
>
j
>
j
>
¦
)
>
¦
1
j
>
j
j
>
>
>
¦
)
>
j
j >
4 ¦ ­j
j >
¦ ­j
j
j
j
j
4
110 CHAPTER 4. TURBULENT FLOWS
4 ¦
1¦
¦ 4
4 ¦f
1¦
¦ 4
4 ¦f
1¦
4 1 ¦ ` ¦ 4
1¦
¦ 4 1
`
4
¦ ` 1¦
¦ 4
4 ¦f}
1¦
¦ 4
} 4
jÛ> 9 151­j•> j
which after taking into account (4.8) simplifies to:
¦
¦
¦
4
¦f
>} 1
> j
1 ¦ ` ¦ 4
1 ¦ 4 `
1 ¦ 4
>}­j
> j
>}¦ 1
`
4
¦ 1
4 ` ¦ 1
4
¦ ` 1¦
¦ 4
4 ¦f}
1¦
¦ 4
} 4
jÛ> 9 151­jê> j
changing the order of time differentiation and time averaging, and canceling terms
we obtain:
4
¦f
jê>¸ 1 ¦
¦ ­j
> j
(4.15)
jÛ> 9 1
1 ¦ 4
¦ 1
4
> j
¦ ` ¦
¦f}
} 4
jqX
and finally:
4
¦f
(4.16)
jê>¸ 1 ¦
j >
9 1
>
¦ ­j
¦ ` ¦
1 ¦
4
¦f}
} 4
j5X
jê>
This is the equation for the components of the Reynolds stress tensor. Since the
Reynolds stress tensor is symmetric, it has only six independent components,
and the number of equations is 6. As can be seen the equation contains 3-rd
order ) )Wj correlations: . One can write an equation ) )Wj for as well, but it
will depend on 4-th order correlations, and so on. In practice, an empirical relationship
is proposed, that links the third order ) )Wj tensor to the second order
) tensor . This relationship is called closure. In fact all the terms on the RHS

is
r
r
¹
r
r
4.2. TURBULENCE MODELING 111
of (4.16) require some kind of closure to make the problem complete. If such
closures are established, the obtained equation system will constitute a turbulence
model. In this particular case, when the equation system provides PDEs
for Reynolds stress tensor components, the corresponding turbulence model is
called Reynolds stress model (RSM). Considering that the Reynolds stress tensor
is symmetric, this model may include as many as 12 PDEs: 3 - velocity, 1
- pressure, 6 - Reynolds stress tensor. The 12-th equation is the one for the
turbulent dissipation rate, which will be discussed in the next section.
Two equation turbulence models
The RSM model described above may be prohibitively expensive in terms of the
number of equations and the complexity of implementation. In this case simpler
models can be used, which are based on smaller number of equations. The
first step to reduce the number of equations is to relate the components of the
Reynolds tensor to the mean velocity gradients following the Boussinesq approximation
(4.11). Then one can formulate a separate transport equation for
and
an algebraic relation for 9 Ê as a function of the latter and the mean velocity gradients.
This approach will constitute a one equation turbulence model.
A more popular approach is to formulate two transport equations: one for
turbulent kinetic energy,
and another for it’s dissipation rate¹. The eddy viscosity,
9 Ê is then related to
and¹as
(4.17)
9 Ê
Ä ©r
±º¹²»
±ƒ² :‰. :J.
±ƒ²? :‰. ± 9 Ê ²? :‰.
the `
which can be shown from dimensional reasoning. Indeed, if¹represents the rate
of change of: then its physical dimensions should be ,
Considering that , and , we can obtain (4.17). This
approach is called ¹turbulence model (KE) [14, 15], which is the most
popular turbulence model for engineering computations today. The equation for
formulated as a transport equation of the form:
(4.18)
1 9§»
¦ 4 ¦ 2
R
"
R
where the effective eddy viscosity of. This equation is solved with the
boundary conditions of zero
at the walls. Usually a non-zero
is set at the
9§»
> 9 ʦK 1¦K
>¸m ¦ 4a`
¹
is

T
T
I
F
F
¦f
F
r
Ä
112 CHAPTER 4. TURBULENT FLOWS
open boundaries, where its value is related to level of turbulent fluctuations expected
in each particular flow case. The equation for the turbulence dissipation
rate,¹, is written in analogy to the one for. Namely, we multiply the-equation
by¹:, and introduce the effective eddy viscosity of¹: 9”¼to obtain:
(4.19)
r¹r
where we introduced two new Ä ',Ä
constants . Boundary conditions on¹can be
obtained from those on, relating¹to
by through equation (4.18) where the
.
R
"
R¹
non-steady term should be set to zero at the R: R
" T
boundary:
Now equations (4.18) and (4.19) have three effective viscosities: Ê (diffusiv-
9
ity of momentum), (diffusivity of turbulent kinetic energy,), and 9§¼(diffusivity
of turbulence dissipation rate,¹). It should be noted that unlike the molecular
viscosity,
V
, which is a property of the fluid and is usually independent on coordinates
or time 2 , all turbulent viscosities are functions of space, and are therefore
represent the new dependent variables of the problem along 9 and¹.
with¦
' 9”»
Another assumption of the model is that and 9”»
9 Ê with different proportionality constants:
9¼are both proportional to
1 9¼¹
4 ¦ ¦
>\Ä
9 Ê ¹
1¦f
>Jm ¦ 4a`
(4.20)
9 Ê
(4.21)
9¼
9 Ê
where j , ö¼
model.
ö
the effective ”Prandtl numbers”, which are constants of the
ö¼
are
The system of equations (4.18), (4.19), (4.20), (4.21), (4.17) is closed and constitutes
¹turbulence model. It has 5 empirical the `
constants:
9”»
ö»
r
l ö»
T ö¼
@‰IÕÀ
the values of which are determined by comparison of computations with experimental
data.
KE model belongs to the class of two-equation turbulence models. Another
important two equation turbulence model is the °Z` n model [16], where instead
of the turbulence dissipation rate¹a turbulence frequency scale, n , is used as an
extra variable.
2 In flows with heat transport this may not be the case
Ä ©
Ô
Ä
щÑ
Ä @‰I
@‰IÕÔ
@‰I

Bibliography
[1] Y.A. Cengel and M.A. Boles. Thermodynamics. McGraw-Hill, Inc., 2002.
[2] Frank White. Viscous Fluid Flow. Second Edition‘. WCB/McGraw-Hill, 1991.
[3] K.Jr. Wark. Advanced Thermodynamics for Engineers. McGraw-Hill, Inc.,
1995.
[4] C.A.J Fletcher. Computational techniques for fluid dynamics. Springer-
Verlag, 1991.
[5] C.R. Chester. Techniques in partial differential equations. McGraw-Hill, NY,
1971.
[6] J.H. Ferziger and M. Peric. Computational Methods for Fluid Dynamics.
Springer Verlag, 1997.
[7] L.D. Landau and E.M. Lifshitz. Fluid Mechanics. Course of Theoretical
Physics, volume 6. Butterworth-Heinemann; 2nd edition, 1987.
[8] O. Reynolds. An experimental investigation of the circumstances which determine
whether the motion of water shall be direct or sinuous, and of the
law of resistance in parallel channels. Royal Society, Phil. Trans., 1883.
[9] E. Buckingham. Model experiments and the form of empirical equations.
Trans. ASME, 37, 1915.
[10] F. White. Fluid Mechanics. Fifth Edition‘. WCB/McGraw-Hill, 2002.
[11] E. Buckingham. On physically similar systems: Illustrations of the use of
dimensional equations. Phys. Rev., 4(4):345–376, 1914.
[12] C.W. Pseen. Ueber die stokes’sche formel und ueber eine verwandte aufgabe
in der hydrodynamik. Ark. f. Math. Astron. och Fys., 6(29), 1910.
113

114 BIBLIOGRAPHY
[13] M. Germano. Turbulence: the filtering approach. Journal of Fluid Mechanics,
238:325–336, 1992.
[14] B.E. Launder and D.B. Spalding. The numerical computation of turbulent
flows. Computer Methods in Applied Mech. and Eng., 3:269–289, 1974.
[15] W.P. Jones and B.E. Launder. The prediction of laminarization with a twoequation
model of turbulence. Int. J. Heat Mass Transfer, 15:301–314, 1972.
[16] D.C. Wilcox. Turbulence modeling for CFD. DCW Industries, Inc., 1993.
[17] Barry Spain. Tensor Calculus. Oliver and Boyd, 1965.
[18] J.L. Synge and A. Schild. Tensor Calculus. Dover Publications, 1969.
[19] P. Morse and H. Feshbach. Methods of Theoretical Physics. McGraw-Hill,
New York, 1953.

Appendix A
Introduction to Tensor Calculus
115

"
¦
-
¦
¦
¡
¦
ã
¦
T
1
¦
¦
¦
¦
4
¦
l
116 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
There are two aspects of tensors that are of practical and fundamental importance:
tensor notation and tensor invariance. Tensor notation is of great practical
importance, since it simplifies handling of complex equation systems. The
idea of tensor invariance is of both practical and fundamental importance, since it
provides a powerful apparatus to describe non-Euclidean spaces in general and
curvilinear coordinate systems in particular.
A definition of a tensor is given in Section A.1. Section A.2 deals with an
important class of Cartesian tensors, and describes the rules of tensor notation.
Section A.3 provides a brief introduction to general curvilinear coordinates, invariant
forms and the rules of covariant differentiation.
A.1 Coordinates and Tensors
Consider a space of real numbers of dimension = , A B , and a single real time,
. Continuum properties in this space can be described by arrays of different
dimensions, , such as scalars @ ( ), vectors ( ), matrices ( ), and
general multi-dimensional arrays. In this space we shall introduce a coordinate
system, , as a way of assigning real numbers 1 for every point of space
¦YF„½º½B
=
There can be a variety of possible coordinate systems. A general transformation
rule between the coordinate systems is
%
(A.1)
Consider a small R
displacement . Then it can be transformed from coordinate
system to a new coordinate ã system using the partial differentiation
rules applied to (A.1):
ã
ã
F IOI/ B
(A.2)
ã R
R
Y ã
This transformation rule 2 can be generalized to a set of vectors that we shall call
contravariant vectors:
Y
(A.3)
ã Y ¡
1
Y
Super-indexes denote components of a vector (¾†¿¸ÀÁ4ÁÂ) and
not the power exponent, for the reason
explained later (Definition A.1.1)
2 The repeated indexes imply summation (See. Proposition A.21)

Y
ã Y
Y ¦
ã Y
¦
¡
£
|
¦
¦
Y ¦
Y
A.1. COORDINATES AND TENSORS 117
That is, a contravariant vector is defined as a vector which transforms to a new
coordinate system according to (A.3). We can also introduce the transformation
matrix as:
(A.4)
¦
¢
Y ã
With which (A.3) can be rewritten as:
Y
(A.5)
¦
¡
£
Transformation rule (A.3) will not apply to all the vectors in our space. For
will transform as:
example, a partial derivative Y : Y
(A.6)
¦ Y
Y
Y
that is, the transformation coefficients are the other way up compared to (A.2).
Now we can generalize this transformation rule, so that each vector that transforms
according to (A.6) will be called a Covariant vector:
Y ã
(A.7)
Y
¦ ¡
ã ¡©¦
This provides the reason for using lower and upper indexes in a general
tensor notation.
Y ã
Definition A.1.1 Tensor
Tensor of order is a set of =
numbers identified by
integer indexes.
For example, a 3rd order tensor ¡ can be denoted as ¡§¦/ j and an -order tensor
can be denoted as ¡§¦6½º½¦ÄÃ.Each index of a tensor changes between 1 and
n. For example, in a 3-dimensional space (n=3) a second order tensor will be
represented À Ô by components.
Each index of a tensor should comply to one of the two transformation rules:
(A.3) or (A.7). An index that complies to the rule (A.7) is called a covariant index
and is denoted as a sub-index, and an index complying to the transformation rule
(A.3) is called a contravariant index and is denoted as a super-index.

¡©¦6½º½¦Ã ¡§¦6½º½¦Ã“1
¨
¦
¦
@
¦
¦
B
118 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
Each index of a tensor can be covariant or a contravariant, thus tensor
j ¦/
is a 2-covariant, 1-contravariant tensor of third order.
¡
Tensors are usually functions of space and time:
which defines a tensor field, i.e. for every point
and time " there are a set of
F IOI/ B '
"54
nubers ¡©¦6½º½¦Ã.
Remark A.1.2 Tensor character of coordinate vectors
Note, that the
coordinates are not tensors, since generally, they are not
transformed as (A.5). Transformation law for the coordinates is actually given
by (A.1). Nevertheless, we shall use the upper (contravariant) indexes for the
coordinates.
Definition A.1.3 Kronecker delta tensor
Second order delta tensor, ß ¦/
is defined as
(A.8)
#­
¦7Œ
¦7Œ T
ß
From this definition and since coordinates
v ¨
it follows that:
are independent of each other
#ª
v ß
(A.9)
Y
ß ¦/
Y
Corollary A.1.4 Delta product
that
From the definition (A.1.3) and the summation convention (A.21), follows
(A.10)
ß ¦/ ¡k¡§¦

£
¦
A
j
¦
¦
¦
j
¦
¦
j
¦
j
¦
j
j
A.2. CARTESIAN TENSORS 119
A¦
:
Assume that there exists the transformation inverse to (A.5), which we call
(A.11)
A¦
Then by analogy (A.4)A¦
to
can be defined as:
ã R
R
A¦
(A.12)
Y
£
A¦
j
j , namely:
Y ã
¦
From this relation and the independence of coordinates (A.9) it follows
A
that £
ß ¦
Y
(A.13)
Y ã
Y
ã j Y
Y
ß ¦
Y ã
Y ã
Y
Y ã
j
j
Y ã
A.2 Cartesian Tensors
Cartesian tensors are a sub-set of general tensors for which the transformation
matrix (A.4) satisfies the following relation:
(A.14)
¦ Y ã
j ¦ £ j £
Y ã
ß ¦/
For Cartesian tensors we have
Y
Y
(A.15)
ïY
Y ã
Y ã
(see Problem A.4.3), which means that both (A.5) and (A.6) are transformed ¦
with
the same j matrix . This in turn means that the difference between the covariant
and contravariant indexes vanishes for the Cartesian tensors. Considering this
we shall only use the sub-indexes whenever we deal with Cartesian tensors.
£
Y j

j
j
„
æ
¤
„
P
æ
¤
120 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
A.2.1
Tensor Notation
Tensor notation simplifies writing complex equations involving multi-dimensional
objects. This notation is based on a set of tensor rules. The rules introduced
in this section represent a complete set of rules for Cartesian tensors and will
be extended in the case of general tensors (Sec.A.3). The importance of tensor
rules is given by the following general remark:
Remark A.2.1 Tensor rules Tensor rules guarantee that if an expression follows
these rules it represents a tensor according to Definition A.1.1.
Thus, following tensor rules, one can build tensor expressions that will preserve
tensor properties of coordinate transformations (Definition A.1.1) and coordinate
invariance (Section A.3).
Tensor rules are based on the following definitions and propositions.
Definition A.2.2 Tensor terms
A tensor term is a product of tensors.
For example:
(A.16)
¡§¦/
¤“
jÙÄ
Pæ
Definition A.2.3 Tensor expression
Tensor expression is a sum of tensor terms. For example:
(A.17)
P
Generally the terms in the expression may come with plus or minus sign.
¡§¦/
¤“
¦ ½
jê>\Ä
Pæ
Proposition A.2.4 Allowed operations
The only allowed algebraic operations in tensor expressions are the addition,
subtraction and multiplication. Divisions are only allowed for constants, like
: Ä . If a tensor index appears in a denominator, such term should be redefined,
@
so as not to have tensor indexes in a denominator. For @ : ¡t¦
example, should be
redefined as: @ : ¡©¦
.
¤§¦¢

Ä
¦ ½
j
P
Ä
P
Ä
j
j
A.2. CARTESIAN TENSORS 121
Definition A.2.5 Tensor equality
Tensor equality is an equality of two tensor expressions.
For example:
(A.18)
¡§¦/Ù¤“k
¦f¤
jÙ„
>^„
Definition A.2.6 Free indexes
A free index is any index that occurs only once in a tensor term. For example,
index # is a free index in the term (A.16).
Proposition A.2.7 Free index restriction
Every term in a tensor equality should have the same set of free indexes.
For example, if index # is a free index in any term of tensor equality, such as
(A.18), it should be the free index in all other terms. For example
¡©¦/¤k
½
is not a valid tensor equality since index # is a free index in the term on the
RHS but not in the LHS.
Definition A.2.8 Rank of a term
A rank of a tensor term is equal to the number of its free indexes.
¤“
For example, the rank of the j Äj term is equal to 1.
It follows from (A.2.7) that ranks of all the terms in a valid tensor ¡§¦/ expression
should be the same. Note, that the difference between the order and the rank is
that the order is equal to the number of indexes of a tensor, and the rank is equal
to the number of free indexes in a tensor term.
Proposition A.2.9 Renaming of free indexes
Any free index in a tensor expression can be named by any symbol as long
as this symbol does not already occur in the tensor expression.

°
¡
j
Ä
„
„
122 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
For example, the equality
(A.19)
is equivalent to
¡§¦7Ù¤
¦ ½
(A.20)
Ù¤ ½
Äj
Here we replaced the free index # with ° .
Definition A.2.10 Dummy indexes
A dummy index is any index that occurs twice in a tensor term.
For example, indexes¨‰'
'€Xò'qŠ in (A.16) are dummy indexes.
Proposition A.2.11 Summation rule
Any dummy index implies summation, i.e.
(A.21)
B w¦
¡§¦M¤§¦$
¡§¦³¤§¦
Proposition A.2.12 Summation rule exception If there should be no summation
over the repeated indices, it can be indicated by enclosing such indices in parentheses.
For example, expression:
½ ¦/
does not imply summation over # .
Ä0¦32¡0¦32¤“k
Corollary A.2.13 Scalar product
A scalar product notation from vector algebra: 1 ¡ûž¤ 4
notation as ¡§¦M¤§¦ .
is expressed in tensor

¡§¦
æ P
j
¤
P
Ä
æ
¡
¤
j
P
P
Ä
„
„
A.2. CARTESIAN TENSORS 123
The scalar product operation is also called a contraction of indexes.
Proposition A.2.14 Dummy index restriction
No index can occur more than twice in any tensor term.
Remark A.2.15 Repeated indexes
In case if an index occurs more than twice in a term this term should be
redefined so as not to contain more than two occurrences of the same index. For
is defined as
j
¤
½
example, term should be rewritten j j as , where jèÄj j
½ ¡t¦ ¡§¦
½
¢¤ 0j2Ä0j2with no summation over ° in the last term.
Proposition A.2.16 Renaming of dummy indexes
Any dummy index in a tensor term can be renamed to any symbol as long
as this symbol does not already occur in this term.
For example, term is equivalent to , and so are terms ¡§¦/ j
.
¡Ù¤“ ¡§¦M¤§¦
¤
Äj and
Remark A.2.17 Renaming rules
Note that while the dummy index renaming rule (A.2.16) is applied to each
tensor term separately, the free index naming rule (A.2.9) should apply to the
whole tensor expression. For example, the equality (A.19) above
¡©¦/¤k
¦ ½
can also be rewritten as
(A.22)
½
without changing its meaning.
Äj
(See Problem A.4.1).
Definition A.2.18 Permutation tensor
The components of a third order permutation tensor pq¦/ j are defined to be
equal to 0 when any index is equal to any other index; equal to 1 when the set of

% £A
£
£A
4
(A.25)
~
j
l
À @
l 4Pv
j
¡¼ ¤ Í ~ ~
j
j
j
j
`
T
@
@
£A
124 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
indexes can be obtained by cyclic permutation of 123; and -1 when the indexes
can be obtained by cyclic permutation from 132. In a mathematical language it
can be expressed as:
#_
¨#Å
°Å%¨
°Æv p5¦7
(A.23)
#¨°ËÇ º9É
1
p5¦/
#­
#¨°ÈÇ º9É
1
4Êv p5¦7
where º9É
–
is a permutation group of a triple of indexes abc, i.e. º9É
£AE- . For example, the permutation group of 123 will consist of three
–
4
@EÀ
combinations: 123, 231 and 312, and the permutation group of 123 consists of
132, 321 and 213.
–§'BA
–
',–
1
1
Corollary A.2.19 Permutation of the permutation tensor indexes
From the definition of the permutation tensor it follows that the permutation
of any of its two indexes changes its sign:
(A.24)
p5¦/
` p5¦
A tensor with this property is called skew-symmetric.
Corollary A.2.20 Vector product
A vector product (cross-product) of two vectors in vector notation is expressed
as
Ä
which in tensor notation can be expressed as
(A.26)
¡§¦$£p5¦/
¤
Äj
Remark A.2.21 Cross product
Tensor expression (A.26) is more accurate than its vector counterpart (A.25),
since it explicitly shows how to compute each component of a vector product.

j
j
p5¦
æ P
j
j
j
¡
j
j
j
j
j
P
j
ß
j
æ
j
j
T
j
j
¡
j
j
æ
j
ß
j
P
A.2. CARTESIAN TENSORS 125
Theorem A.2.22 Symmetric identity
If ¡§¦/ is a symmetric tensor, then the following identity is true:
(A.27)
p5¦/
¡“
Proof:
From the symmetry of ¡§¦/ we have:
(A.28)
p5¦/
¡“
£pm¦/
Let’s rename index¨into ° and ° into¨in the RHS of this expression, according
to rule (A.2.16):
p5¦/
k£pm¦
¡
Using (A.24) we finally obtain:
p5¦
,¡
` p5¦/
¡
j
Comparing the RHS of this expression to the LHS of (A.28) we have:
p5¦/
¡
` p5¦/
¡
from which we conclude that (A.27) is true.
Theorem A.2.23 Tensor identity
The following tensor identity is true:
(A.29)
pq¦7
ß
`]ß
Proof
This identity can be proved by examining the components of equality (A.29)
component-by-component.

(A.30)
~
Í ¤ Í ¡
Ä~
1~
¤ ¡‡ž 4
Ä~
1~ ~
Y
¡
"
¡
¦
4Û`
Ä~
¡‡ž ¤ 4 ~ 1~
126 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
Corollary A.2.24 Vector identity
Using the tensor identity (A.29) it is possible to prove the following important
vector identity:
See Problem A.4.4.
A.2.2
Tensor Derivatives
For Cartesian tensors derivatives introduce the following notation.
Definition A.2.25 Time derivative of a tensor
A partial derivative of a tensor over time is designated as
¡£¢ Y
Definition A.2.26 Spatial derivative of a tensor
A partial derivative of a tensor ¡
over one or its spacial components is denoted
as ¡ ¦ :
(A.31)
¡ ¦¢Y
that is, the index of the spatial component that the derivation is done over is
delimited by a comma (’,’) from other indexes. For example, ¡t¦/y j is a derivative of
a second order tensor ¡§¦7 .
Y
Definition A.2.27 Nabla
Nabla operator acting on a tensor ¡
is defined as
(A.32)
¾ ¦f¡¢¡ƒ ¦

F
F
À
A.2. CARTESIAN TENSORS 127
Even though the notation in (A.31) is sufficient to define the derivative, In
some instances it is convenient to introduce the nabla operator as defined above.
Remark A.2.28 Tensor derivative
In a more general context of non-Cartesian tensors the coordinate independent
derivative will have a different form from (A.31). See the chapter on covariant
differentiation in [17].
Remark A.2.29 Rank of a tensor derivative
The derivative of a zero order tensor (scalar) as given by (A.31) forms a
first order tensor (vector). Generally, the derivative of an -order tensor forms an
order tensor. However, if the derivation index is a dummy index, then the
>×@
rank of the derivative will be lower than that of the original tensor. For example,
the rank of the derivative is one, since there is only one free index in this
term.
¡§¦/m
Remark A.2.30 Gradient
Expression (A.31) represents a gradient, which in a vector notation is ¾ ¡
:
¾ ¡ `åb ¡ƒ ¦
Corollary A.2.31 Derivative of a coordinate
From (A.9) it follows that:
(A.33)
¦Kk
ß ¦/
In particular, the following identity is true:
(A.34)
¦f ¦$
rq r
>]
>]
@•>¼@•>@
Remark A.2.32 Divergence operator
A divergence operator in a vector notation is represented in a tensor notation
as ¡©¦f ¦ :
¾ 1 ¡ 4]`$b ¡©¦f ¦ ž~

r
› R (A.35)
¦
¦
[
¦
128 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
Remark A.2.33 Laplace operator
¡ƒ ¦ì¦
The Laplace operator in vector notation is represented in tensor notation as
:
¡ `åb ¡ƒ ¦ì¦
Remark A.2.34 Tensor notation
Examples (A.2.30), (A.2.32) and (A.2.33) clearly show that tensor notation
is more concise and accurate than vector notation, since it explicitly shows how
each component should be computed. It is also more general since it covers
.
cases that don’t have representation in vector notation, for example: ¡C¦ j
j
A.3 Curvilinear coordinates
In this section 3 we introduce the idea of tensor invariance and introduce the rules
for constructing invariant forms.
A.3.1
Tensor invariance
The distance between the material points in a Cartesian coordinate system is
. The metric tensor, is introduced to generalize the
notion of distance (A.39) to curvilinear coordinates.

› R (A.36)
r
› R (A.39)
¦
¦
¦
¦
¦
ã
¦
¦
¦
j
R ã
¦
¦
j
j
¦
¦
ã
j
¦
¦
¦
¦
¦
¦
T
j
A.3. CURVILINEAR COORDINATES 129
systems, that is, the distance should be independent of the coordinate system,
thus:
¦/
¦/
The metric tensor is symmetric, which can be shown by rewriting (A.35) as
follows:
R
R
ã R
ã R
¦/
¼

¡
ã
¡ X
¦
¦
¦
130 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
Using (A.38) we can also find its inverse as:
(A.41)
¦7
Y ã
Y ã
in various curvilinear coordi-
Using these expression one can compute and
nate systems (see Problem A.4.6).
J¦7
¦7
Y j
Y j
Definition A.3.2 Conjugate tensors
For each index of a tensor we introduce the conjugate tensor where this
index is transfered to its counterpart (covariant/contravariant) using the relations:
(A.42)
(A.43)
¼
¡“
¦/
¡©¦‡

j
¦
)
¦
)
j
)
¤
¦
> 9 6 ¦
j ;
¦
> 9 6 j ¦f
j ;
)
A.3. CURVILINEAR COORDINATES 131
Definition A.3.4 Invariant Scalar Product
The invariant form of the scalar product between ¦7
two
¡©¦M¤
covariant vectors
and is . Similarly, the invariant form of a scalar product between ¡t¦
¦ ¦ ¦
two
contravariant vectors and is , where is the metric tensor (A.40)
and is its conjugate (A.38).
\)
where the rising of indexes was done using relation ) (A.42):
.
6 ¦/
, and j ¦ 6
j ¥ j

¡
¦
Y
¡
¦
Y
¡
¦
>
¡
y͡ j
Y
|
¨
¦
j
|
132 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
A.3.2
Covariant differentiation
A simple scalar value, , is invariant under coordinate transformations. A partial
derivative of an invariant is a first order covariant tensor (vector):
¨
¨ ¦G˜Y
However, a partial derivative of a tensor of the order one and greater is not
generally an invariant under coordinate transformations of type (A.7) and (A.3).
In curvilinear coordinate system we should use more complex differentiation
rules to preserve the invariance of the derivative. These rules are called the rules
of covariant differentiation and they guarantee that the derivative itself is a tensor.
According to these rules the derivatives for covariant and contravariant indices
will be slightly different. They are expressed as follows:
Y
(A.46)
¡§¦
(A.47)
>
̦
j
¡§¦f¢
Y
`Ìj ¦/yÍ¡
¢SY
Y
where contstructÌj
the
¦/£Íis defined as

>
¡
%
%
æ
¦6P
¡
æ
¡
-
-
¦
)
¦
¡
j
)
¡
¦
R
%
¦6½º½¦Î§
P
`
>
>
%
%
¦
º
j
æ
Y
P Y
-
-
>
¡
¡
%
¦6½º½¦Î ¡
6½º½ÃˆÏ6æ
j
¦
j
j
¡
¦
|
A.3. CURVILINEAR COORDINATES 133
the contravariant second order tensor ¡ ¦/
we have:
(A.48)
¦/
j
¦
- |aj
- |­j
¦/
üY
¡ |
Y j >
And for a general = -covariant,
-contravariant tensor we have:
žEžEž
(A.49)
¦6½º½¦Î
>
Despite their seeming complexity, the relations of covariant differentiation
can be easily implemented algorithmically and used in numerical solutions on
arbitrary curved computational grids (Problem A.4.8).
æ
6½º½Ã
¦½º½¦ÄÎ>
¦ÎP
¦6½º½¦ÎÏ6æ
6
½º½Ã
Ã
æfP
žEžEž
6½º½Ã
æfP
6½º½Ã
6½º½Ã ¦6½º½¦Î
Remark A.3.7 Rules of invariant expressions
As was pointed out in Corollary A.3.6, the rules to build invariant expressions
involve raising or lowering indexes (A.42), (A.43). However, since we did not
introduce the notation for contravariant derivative, the only way to raise the index
.
of a covariant derivative, say ¡ ¦ , it to use the relation (A.42) directly, that is:
¦7
¡ƒ
For example, we can re-formulate the momentum equation (A.45) in terms
of contravariant free index # as:
(A.50)
¦
j
j
>^)
; > 9 6
where the index of the pressure term was raised by means of (A.42).
Using the invariance of the scalar product one can
#
construct
¡¼¢¥¡
two ¦ important
differential operators in curvilinear coordinates: R
¦
divergence of a vector
(A.51) [
and Laplacian, (A.55).
¡£¢£
¡
j
j
Definition A.3.8 Divergence
¦
Divergence of a vector is defined
¦
as :
¡
(A.51)
¦
¦
# ¡£¢¡

(A.55) [
¡
Y
¡
¡
¦
> ¦
¡
Y
¡
@
m
m
¦
> ¦
Y
Y
¦
%
-
¦
¡
¦
134 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
From this definition and the rule of covariant differentiation (A.47) we have:
(A.52)
¦
¦
¦
j
¡ j
this can be shown [18] to be equal to:
Y
¦
¦
— @
Y
(A.53)
Q¡
where is the determinant of the metric tensor .

(A.61) ’
F
r
F
4
>
£
–
F
¦
¦
£ T
> F
£
>
rAF
F
F
F
¦
T
¦
T T
-
'BAr '
T
'
T
'
-
-
>
F
T
4
A.3. CURVILINEAR COORDINATES 135
¦
Consider three unit vectors, , each directed along one of the coordinate
axis (tangential unit vectors), that is:
'BA¦
'*–
£
(A.56)
(A.57) A¦
(A.58)
T ×%
T Ú%
ä% £
F '
The condition of orthogonality means that the scalar product between any
two of these unit vectors should be zero. According to the definition of a scalar
product (Definition A.3.4) it should be written in form (A.44), that is, a scalar product
between vectors andA¦
can be written as: £ ¦A¦
or £‰¦A¦
. Let’s use the first
form for definiteness. Then, applying the operation of rising indexes (A.42), we
can express the scalar product in contravariant components only:
£Ž¦
',–
¦A
T ¼£
r £
T‰T
FðF

› R (A.66)
¦
¦
£
£
¦
¦
£ £
F F
£
–
¦
¦
£
@ Ú%
’
¦
F
¦
F
'
'
’
@
’
T
'
F
r
4
@
’
¯
'
T
'
@
’
–
T
T
¦
–
-
-
-
’
F
r
4
–
¦
–
r
1
R ã
¦
@
¦ r
4
@
136 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
£‰¦$
¦
@
Or, expressed in contravariant components only the condition of unity is:
A¦A¦$

F
¦
¡
F
’
¡
F
ˆ
F
'
ˆ
@
ˆ
¦
r
r
¦
F
'
F
¡
ˆ
’0¦32’0¦32
¦
¦
ˆ
¦
¦
T
'
T
A.3. CURVILINEAR COORDINATES 137
Combining the latter with (A.38), we obtain: ß ¦/
that
ß ¦/
, from which it follows
(A.68)
’0¦32
@ : ’0¦32
Physical components of tensors
Consider a direction in space determined by a unit ˆ vector . Then the physical
is given by a scalar product between
ˆ and (Definition A.3.4), namely:
component of a vector ¡§¦ in the direction ˆ
¡§¦
¦/
¡ 1
4 ¼
¡©¦
According to Corollary A.3.5 the above can also be rewritten as:
(A.69)
¡ 1
4 ¼¡§¦
¡
Suppose the unit vector is directed along one of the ˆ axis:
From (A.63) it follows that:
%
ˆ

F
r
[
¡
@ Y ¡£
Y
F
¦
—
¦
@
—
¦
’0¦32¡§¦
¡
Y ’0¦32
Î
›
B ¢
ó¦YF
¦
’
¦
138 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
General rules of covariant differentiation introduced in (Sec.A.3.2) simplify
considerably in orthogonal coordinate systems. In particular, we can define the
nabla operator by the physical components of a covariant vector composed of
partial differentials:
(A.71)
’0¦32
Y
¾ ¦$
Y
where the parentheses indicate that there’s no summation with respect to index # .
In orthogonal coordinate system the general expressions for divergence
(A.53) and Laplacian (A.55)) operators can be expressed in terms of stretching
factors only [19]:
(A.72)
¦
¦ @ Y
Y
Important examples of orthogonal coordinate systems are spherical and cylindrical
coordinate systems. Consider the example of a cylindrical coordinate system:
',Î'W›ñ- :
¦$ä%
¦$ä%§¢
'5
- and ã 'q
¼¢arcedWÎ
rk¢deKÆ
According to (A.40) only few components of the metric tensor will survive
(Problem A.4.5). Then we can compute nabla, divergence and Laplacian operators
according to (A.71), (A.52) and (A.55), or using simplified relations (A.72)-
(A.73):

[
R
1
Y
r
¡
F
¡
Y
Y
r > 4
¡
r
Y
1
Y
> F
¡
@
ã
F
> r
x
r
— Y ¾
Y
@
F ã
1
Y
@
r
Y
Y
¡
1
r >
> r
¡
r
r > 4
>
1
Y
Y
Y
¡
r
Y
1
Y
Î Y
¡
r > 4
¡
r
>
'
Y
+ Y
@
F ã
r > 4
@ ¡
¢
@
ã
¡
F
@
¢
x
F
Y
¡
F
¡
Y
Y
¢
A.3. CURVILINEAR COORDINATES 139
@
¢
¢ '
¡tr
# ¡£ Y
Y ã
>
F
Y ã
Y ã
@
¢
¢ >
Note, that instead of using the contravariant components as implied by the general
definition of the divergence operator (A.51) we are using the covariant components
as dictated by relation (A.70). The expression of the Laplacian becomes:
Y Î
¡F
Y +
¡H
¡
Y ã
r 4
Y ã
Y ã
ã Y
¢ 4
¢ r Y
Y Î
Y +
(see Problems A.4.9,A.4.10).
The advantages of the tensor approach are that it can be used for any type
of curvilinear coordinate transformations, not necessarily analytically defined, like
cylindrical (C.64) or spherical. Another advantage is that the equations above can
be easily produced automatically using symbolic manipulation packages, such
as Mathematica (wolfram.com) (Problems A.4.6,A.4.7,A.4.9). For further reading
see [17, 18].

¤
~
„
„
j
¦
¤
Ä j
j
j
p5¦
Pæ
Ä
P
¤
¤
P
P
ß
½
j
Ä
æ
æ
½
Ä
j
¤
æ
ß
T
j
P
4Û`
Ä~
¤
j
¦
'
j
P
j
½
'
¡
140 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
A.4 Problems
Problem A.4.1 Check tensor expressions for consistency
Check if the following Cartesian tensor expressions violate tensor rules:
¡©¦/ ¤
jê>
P æ
½
P æ
æÉæ
jÛ>
¦/
¡“Ù¤
¦/¡©¦M¤Œ ¦/É
`à½
Problem A.4.2 Construct tensor expression
¡t¦/ ¤§¦/ ½ ¦7
j

^
®
®
+
+
Î
›
ë
A.4. PROBLEMS 141
£¢a/cedWÎ
£¢defÆ
Obtain the components of the metric tensor (A.40) and its inverse
(A.38) in cylindrical coordinates.
J¦7
¦/
Problem A.4.6 Metric tensor in curvilinear coordinates
Using Mathematica Compute the metric tensor,
, (A.38) in spherical coordinate system ( ¢ ' ë_',Î ):
, (A.40) and its conjugate,
¼¢defÆ
ÎŒarcedåë
(A.73)
ÎŒdefÆ
¼¢a/cedåÎ
¼¢defÆ
Problem A.4.7 Christoffel’s symbols with Mathematica
Using the Mathematica package, write the routines for computing Christoffel’s
symbols.
Problem A.4.8 Covariant differentiation with Mathematica
Using the Mathematica package, and the routines developed in Problem A.4.7
write the routines for covariant differentiation of tensors up to second order.
Problem A.4.9 Divergence of a vector in curvilinear coordinates
Using the Mathematica package and the solution of Problem A.4.8, write the
routines for computing divergence of a vector in curvilinear coordinates.
Problem A.4.10 Laplacian in curvilinear coordinates
Using the Mathematica package and the solution of Problem A.4.8, write the
routines for computing the Laplacian in curvilinear coordinates.
Problem A.4.11 Invariant expressions

¡
¤§¦
æ P
(A.76) „
Ä
j
¤
¦
æ
¦
j
`
P
>
¤
j
½
Éj
P
Pj
æ
Ä
æ
½
j
Ä j æ
¦ÉP
¤
P
¦ì³
æ
142 APPENDIX A. INTRODUCTION TO TENSOR CALCULUS
not:
Check if any of these tensor expressions are invariant, and correct them if
(A.74)
¦
Ä j
³
j
½
¡§¦³¤
(A.75)
¦/
j
î¦
j ¡
¦
Problem A.4.12 Contraction invariance
¦
¤©¦
Prove that ¡
is an invariant and ¡§¦M¤©¦ is not.

¾
r
)
¾
r
r
)
Y
®
Y
r
> r
4
Y ) ¢
¢
Y
Y
r
> r
Y
4
Y
r
r
r
)
r
4
Appendix B
Curvilinear coordinate systems
Here we will learn how to express the Laplacian
in polar, cylindrical or spherical coordinates 1
Y
Y ®
Y +
Laplace equation in different coordinate systems. When solving boundary value
problems in more than one dimension it is often necessary to use other coordinate
systems than the Cartesian. It is then important to be able to express the
Laplacian operator in these coordinate systems. We first consider
POLAR COORDINATES
¼¢a/cydWÎ
¼¢defÆ
Î
Laplace’s equation in this coordinate system can be shown to be:
1 ¢
1 ¢
1 ¢
1 ¢
4 Y
',Î
Y )
',Î
',Î
@
¢
@ r Y
¢
',Î
¢ r >
¢ >
Y Î
PROOF; Let us for now apply the convention that subscript implies partial differentiation
e.g.
)œx
1 This material is boroowed from http://www.physics.ubc.ca/ birger/n312l8/
143

)
4
4 ¢
1
1
)GF4
and
¢
¾
r
)
)
)
)
r
r
Î
Î
¢
)
¢
Î
r
¢ÑÐ
Ò«B¬Æ é Î
`
¢
¢
> r
4
¢
Î
r
`
@
Ð
1
¯
4
®
¢ )IF„F`
Â
r
¢ )IF„F­>
Â
Y
r
)
Y
®
l
¢
4
l
l
r
F
`
1
¢
>^)GF„F1
r
r
®
®
r
Y
r
l
Î
`
®
Î
r
r
®
4
4
`
l
l
W®
¢
Â
W®
¢
Â
r
)
)GF
r
4
144 APPENDIX B. CURVILINEAR COORDINATE SYSTEMS
Applying the chain rule we find
)Wx
>^)GF*Î
1
)GF4
)œx
>])œx
>])GF,Î
,
,
,
1 ¢
)œx
)œxñx
>^)œx„F*Î
We have
)IFñx
>]®
r
¢
>^®
@
¢
¢
,
¢ r ®
r 1
r 4
¢ r
@•>
1 `
Collecting terms
W®
¢
Â
4 ¢
,
¢ )Wx­>
W®
)œx„F­> ¢
Similarly we can show that
,
¢ r )œxñx_>
®
W®
)WxFa> ¢
Again, collecting terms
¢ )œx
¯m¯
¢ r )œxñx_>
1 ¢
)GF
1 ¢
1 ¢
Y )
@
¢
@ r Y
¢
',Î
',Î
>])
¢ r >
which is the desired result!
'*Î
> ¢
,
¯m¯
Y Î
CYLINDRICAL COORDINATES
It is easy to generalize the result for polar coordinates to cylindrical coordinates
;arcedåë

¾
r
Î
)
¾
1 ; ' ë_'*+
>
r
)
@
;
Î
r
)
®
+ +
1
r
4
' ë(',+ ; ) r >
; 1 4
r ',ë_',+
>
Y
¢
Y
4
4
>
Y
ë
r
r
)
; 4 1
r ë_'*+ '
r @
Î
Y
¢
Y
r
1 ; ',ë_',+
)
4
4
4
4
145
;deKÆ
@
;
Y )
4 üY
Y ;
Y ;
r Y
Y ë
Y +
SPHERICAL COORDINATES
Finally we give without proof the result for Laplace’s equation in spherical coordinates:
1 ¢
1 ¢
Y )
Î',ë
1 ¢
'*Î',+
@
¢ r 4
±
1 ¢
1 ¢
@
defÆ
Y
Î Y
r ²
Y ë
Y Î deKÆ
For proof see e.g. Chapter 8 of Riley et al. or Chapter 2 of Arfken and Weber.
1deKÆ
Y )
',Î',ë
',Î',ë

146 APPENDIX B. CURVILINEAR COORDINATE SYSTEMS

ˆ
`
À
@
¦ 1 ; ¾
Ä
j Ä
Ä
Ä
j
j
À
T
Appendix C
Solutions to problems
Chapter 1
Problem 1.4.1: Mass diffusivity in terms of concentration
Show how to obtain (1.31) from (1.32).
Solution:
Consider the product ¾ ¦ ; j
and the definition of Ä
j ¢ ; j :

as:
! ` ¢
©
|‚‡~nÔÖÕº×F
Define
R
R
!
"
R
R
R
R
!
"
!
"
R
¡
’
Î
Î
8
Î
’
R
’
Î
’
!
’
Î
148 APPENDIX C. SOLUTIONS TO PROBLEMS
Figure C.1: Sliding plate
!WÒ
defÆ
from which
!åÒ ’‰defÆ
8 ¡
To find the time required to accelerate to the velocity
equation of motion of the accelerated plate:
!
¼ !GÒ , consider the
` 8 ¡
dividing by
, we get:
defÆ
` ! 8 ¡
after rearranging:
¼
defÆ
— ! ` ’‰defÆ
` 8 ¡
. Then:
8 ¡
` 8 ¡
"

whereV
corresponds to the initial velocity ! V
Dividing the above by !åÒ 1 ’‘deKÆ
’
which can be solved for time as a function of
: !GÒ !
!
!åÒ
`
Î
Vhjlkû—
@
` hjlk
8 ¡
hjlk
Î
’
Î
@
"
`
T
"
!
!WÒ
’
"
149
With the solution:
8 ¡
`
. Thus
` 8 ¡
! ` ’‰deKÆ
` ’‘deKÆ
8 ¡
, we obtain:
—
4 1 8 ¡ 4
` 8 ¡
:
—
1 ! : !åÒ 4 ` ’
"
¡ 8
dKÆ —
Matlab solution:
% Lubrication
g=9.8
% m/sˆ2
m=2 % kg
tet=10/180*pi % RAD: angle
mu=5e-3 % kg/(m s)
h=3e-4 % m
alp=0.99
A=0.3*0.4 % mˆ2: area of the plate
Vinf=m*g*sin(tet)*h/(mu*A) % =1.702 m/s
t=-m*h/(mu*A)*log(1-alp) % =4.605 s

j
)
n
Ý
Ä
)
ß
j
Ý
Ý
ß
n
j
j
1
)
n
j
`
Ý
Ý
`
j
j
Ý
Ý 4
;Ãé F
n
Ý
y
PmP
j
j
150 APPENDIX C. SOLUTIONS TO PROBLEMS
Chapter 2
Problem 2.7.2: 2D vorticity limit
Perform the missing steps in (2.42).
Solution:
Using the tensor identity (A.29), we have:
p5¦
m$
9 p5¦/
naj
9 1 ß
`]ß m ß
y
n­j
y `]ß m ß
m 4
9 1 ß
(C.1)
n­j
n­j
9 1
³
`
î 4
Problem 2.7.3: Incompressible viscous limit
Derive (2.50):
(C.2)
k
9 n
m
j*j
from (2.49)
(C.3)
¦ 9 )
¦f
j*j
` ;Ãé F º
¦
Solution:
Using the expression of vorticity vector (1.12):
(C.4)
¦$
pm¦/
4
@
l
m
j
we can form the following equation from (C.3):
)åj
(C.5)
m
j
æ€æ
pm¦/
p5¦7
m
j
9 p5¦7
jè)
jèº
9 pm¦/
jÙ)

¦
R
j
R
¦
"
`
j
?
R
n
¦
?
R
¦
¦
>
?
T
R
¦
j
¦
1
)
¦
æñæ
¦
R
¦
–
`
¥
R
¦
"
l
4
æñæ
T
where the pressure term on the RHS became zero by symmetric identity (A.27).
Now we can form another equation similar to (C.5):
151
(C.6)
9 p5¦/
)Wj
p5¦/
jè)Wj
Subtracting (C.6) from (C.5), we have:
p5¦/
4
9 p5¦7
which after comparison with (C.4) is reduced to:
1
)
m
j
)Wj
m
j
)Wj
Problem 2.7.4: Conservation of circulation
The velocity circulation is defined as
9 n
æñæ
(C.7) >
)
where the integration is over any closed loop inside the fluid.
Ø?
Show that for irrotational (n flow
time, we have:
¦­
R
):>
= . "
. Differentiating (C.7) over
)
— R
" R
)
Using (1.2), we can rewrite the last term on the RHS as:
R
"
R>
¦
r
Ù? R
— R
¦
— )
?
)
)
R )
?
R
the last equality stems from the fact that the integral of a total differential over a
closed loop is zero. Thus we have for the rate of change of circulation:
Substituting the velocity derivative from the Euler equation in form (2.53), we
have:
)
R
"
R>
" R

; R ’
(C.8) "
R
Y
"
Y
Ö
j
º
X R
" >
R
¦
)
¦
s
T
`
X
–
¥
Ö
¦
j
¦
R
T
152 APPENDIX C. SOLUTIONS TO PROBLEMS
` @
¦
;?
º
R
According to the Stokes theorem the contour integral on the RHS can be
rewritten as the integral over the surface encircled by the contour:
R
"
R>
R j
j
p5¦/
¾
£‰¦$
pm¦/
£‰¦$
R
"
R>
jÙº
@ `
;
@ `
;
where the last equality is due to the symmetry º of and the symmetric identity
(A.27). Thus, the time derivative of circulation is zero:
This is the law of circulation or Kelvin’s theorem, which states that in an ideal
fluid the velocity circulation round a closed contour is constant in time.
Problem 2.7.5: Bernoulli’s equation.
R
"
R>
Using the energy equation (2.77):
ã 6 ¦/
1 °
¦ 4 ¦
m ¦
>^)
and momentum equation (2.21):
(C.9)
1 ; )
1 ; )
4 ¦
j
6 ¦
j >äã
j
¦ 4
¦
>])Wj
>Þš
derive the strong formulation of the Bernoulli’s equation:
’¬>
> +
@
) l
= . "
and formulate it’s applicability limits.
Solution:
Method 1
Assuming steady state inviscid fluid at constant temperature, we have according
to (C.8):

(C.12)
; )
and the fact that X
T
¦åR
R
`
X
R ;
R
¦
)
X R
"
R
¦
R ’ ;
"
R
’ R
"
R
`
)
X
X R
"
R
X R
"
R
¦
X
R X `
" > ; R
R
"
R
for a steady-state solution. Solving for¤P ¤
(C.12)
X R
"
R
’ R
"
R
R
R
"
—
R ;
R
R
R
"
"
¦
)
R
R
R
R
"
¦
"
1
)
1
)
¦
)
¦
)
153
which can be rewritten as
(C.10)
@
;
On the other hand with the same assumptions (C.9) can be rewritten as
(C.11)
¦$
¦
" )
> ; ;

`
1
)
¦
¦
¦
¦
R
R
X
¦
)
)
`
¦
’
¦
’
T
’
)
)
–
¥
T
T
`
154 APPENDIX C. SOLUTIONS TO PROBLEMS
which means:
(C.13)
¦ ` ¦
¦$
@
) l
= . "
This is the strong form of a Bernoulli’s equation valid for steady-state inviscid flow.
When gravity is directed opposite to z-axis the last term on the RHS will be equal
+ to .
¦
Method 2
¦$
From (C.10) above it follows that
’>
Ä where is the integration constant. Substituting it into (C.11) above and dividing
by we obtain:
;
; ’>\Ä
¦G
¦
Expansing the expression for substantial derivative (1.7), rearranging terms, and
using the steady state assumption , we have:
¦G
" )
> ; Þ’
¦ `
¦K ^)
)Wj)
¦
¦ `
¦K ^)
¦
¦ ` ¦
The first term can be rewritten as 1 )åj)Wj
, and the last as 1 j Wj
:
)Wjè)Wj
>Þ’
4 ¦ : l
4 ¦
4 ¦
¦ ` 1
T
)Wj)åj
>\’
jWj
>Þ’
j WjM ¦
@
l
¦G
L)WjÙ)åj 4
l

¦
R
R
!
"
in (2.94) in terms of n©'q®
@
À
R
R
3 Ð R©
R
F
@ !
À
"
r
Ö
@ !
À
¦
R
F
Ö
'
r
¦
R
@
À
Ö
@ !
À
l
Ö
n
R
Í ¢
!
R
R
>
!
¦
Ö
¦
R
n
@
À
Ö
n
)
¦
R
'
'
£
'BAF
'BAr
'BA
155
From which we obtain (C.13).
Problem 2.7.6: Volume change inside a moving boundary
Suppose that a region of space is enclosed by a moving boundary. The
velocity of motion of the ) boundary, , is given at each point on the boundary.
Show that the rate of change of the volume, , of that region will be equal to:
!
Solution
Consider any volume of a moving fluid
! Ö
where the integration is done over an arbitrary control volume inside the fluid.
Using identity (A.34), we can rewrite the latter as:
¦f ¦
Then by Gauss theorem:
¦f ¦
£‰¦
£Ž¦
where is a coordinate vector at the boundary R and is an element of the
boundary surface area (2.1). Using the definition of velocity (1.1), we can then
obtain the relation for a volume change:
(C.14)
£‰¦$
£‰¦$
£‰¦
R
" R
Problem 2.7.7: Rotating coordinates
Solution
Writing out (2.93):
£Žr
Obtain explicit relations for the components of acceleration vectors
F
£ . '
'*®
'*® '
Í]1
Í ¡4
" >
by components yields the needed terms in (2.94):
R
" >
R¢

R
R
r
ÿ
r
"
3 R¢
R
" >
; –
(C.18)
P
s R
"
R
3 £¢
l
n
£
F
AF
R
R
X üR
" >
R
n
Í ¢
ÿ
"
>
°
£
`
R
s
`
ÿ
"
l
n
n
n
n
n
r
n
TF
F r ®
r
®
T
r
n
'
n
T
)
n
156 APPENDIX C. SOLUTIONS TO PROBLEMS
` l
£Žr
(C.15)
Ar
Problem 2.7.8: Rotation with separated coordinate origins
Consider a simple rotation with n
A
T
˜%
n
ý
origin of the rotating coordinate system rotate with the same around the
origin of :
'5nŒ- as in (2.94), but now let the
Derive the expression for
3
Solution
in this case.
With these assumptions (2.89) becomes
Í ÿ
(C.16)
> ¡4
Í]1 ÿ
After second differentiation
>Ìn
͇R
Í]1
Í ÿ 4
the realtion (2.93) becomes
(C.17)
Í]1
Í]1 ÿ
> ¡4q4
Problem 2.7.9: Nondimesionalizing energy equation
Write a non-dimensional form of the heat convection equation (2.79):
¦ì¦
m ¦ 1
¦K
y ¦ 4
> 8 )
>])

V
V
)
r
V
Since the density is constant we can replace it with ; ;
)
V
V
V
s
"
;
– s
" P V
s
"
V
–
P
V
–
P
s
"
r
V
V
s
r
V
V
s
X
X
¦
r
" V
V
)
;
V
V
r
V
X
; ã s V
–
P V
°
;
V
V
–
P
°
)
V
s r
s ã
¦
V
–
P
>
–
P
V
r
V
V
s
s ã
)
>
r
V
V
s
V
>
) r
8
;
V
V
V
s
s
) 8
V
r
"
r
V
V
s –
P V
;
V
V
– s
P
V
r
V
V
V
V
V
V
;
V
X
)
V
P
r
V
'
°
157
;
Select for the pressure scale. Determine the minimum number of dimensionless
parameters. Write the equation using the Eckert number as one of
X
the parameters:
¢ )
„Œz
Solution
Introducing the non-dimensional variables:
(C.19)
"
"
¦$ )
ã "
¦$
ã X
V
ã s
V
ã
V
ã )
V
, and obtain:
¦ì¦
m ¦ 4
m ¦ 1
R ã
)
¦f
R ã
R ã
ã " >
R
) ã
) >«ã
) ã
Rearranging:
¦ì¦
y ¦ 4
m ¦ 1
R ã
)
¦f
R ã
°•"
" >
) ã
) ã
) >çã
considering that the velocity scale should relate to length and time scales as
, we have:
R ã
R ã
: "
(C.20)
R ã
8 )
)
R ã
¦ì¦
m ¦ 1
¦f
m ¦ 4
" >
) ã
) ã
) >çã
R ã
R ã
This equation contains 7 dimensional 'q) 'qs ' ; ',– ' 8
parameters: . According
to the PI-theorem, the number of dimensionless parameters can be as low as
three. A conventional choice of these parameters is:
¢ )
„Œz
A!
¢
;
8

s
"
X
@
s ã
>
°
P
)
158 APPENDIX C. SOLUTIONS TO PROBLEMS
¢ 8 –
and (C.20) becomes:
º(x
R ã
R ã
R ã
A! 5º_x
„Œz
" >
A!
ã
) ã
) >çã
R ã
¦ì¦
„Œz
m ¦ 1
¦f
m ¦ 4

)
r
s
)
V
¦
)
– ;
P
Ä
r
s
¦
s
s
> 8 — R ) r
® R
F
)
1
®
)
1
®
F
Ä
)
F
R
s
r
)
r
)
F
V
T
T
T
Ä
:
T
V
F
159
Chapter 3
Problem 3.7.1: Couette flow equations
Show how to obtain equation (3.7) and (3.8).
Solution
Consider (3.2) and (3.3):
(C.21)
(C.22)
(C.23)
>^)Wj)
¦f
j
9 )
1
s]>])
¦f
j*j
¦ì¦
¦G ¦f
;Ãé F
`
º
¦
¦ 4 °
>])
y ¦ 6 ¦/
Using the definition (2.17), and since the only non-zero velocity
¦§%
components
¦Cµ%
- are , and considering
)
continuity (2.4) and constancy º
of pressure , we arrive at (3.7) and (3.8).
and the only independent variable is ¦$
-
Problem 3.7.2: Couette plates solutions
Solve equations (3.7) and (3.8) with the boundary conditions u(0) = 0 and
u(H) = U, s and s and .
Solution
1}T 4
1} 4
(C.24)
r
(C.25)
R ®
° R
Aligning the coordinate origin with the lower plate (y=0) and solving the
equation (C.24) with the boundary conditions: u(0) = 0 and u(H) = U, we have
R ®
4
®C>\Ä
1UT 4
T
1U 4
v
Ä
4
®

@
1 4
®
s
1
®
¥
s
1
®
` 8 4
°
Ä
F
s
®
r
s
F
`
®
l
s
r
F
V
s
`
s
Ä
F
V
V
À
V
160 APPENDIX C. SOLUTIONS TO PROBLEMS
Using this solution we can solve equation (C.25):
>\Ä
®C>\Ä
1UT 4
8
° l
V >
(C.26)
` 8 4
°
8
° l
— s
®»>às
V >
l >
Problem 3.7.3: Flow of a liquid film
Ï ¿ , flowing steadily
Consider a wide fluid film of ’ @‰I
constant
T
thickness,
due to the gravity down the inclined plate Î
{
at angle . Find an analytical
expression of a fluid velocity distribution as a function of a distance from the plate
) surface:
. Assuming the viscosity and the density of the fluid are 8
é ° 1 ¹. 4 T WT‰T °
, and ’ :<
respectively, find the maximum flow velocity
:
and the volumetric ¢ flow rate, , per 1m of the plate. Atmospheric … pressure can
be considered constant.
@‰I,[
ž
Figure C.2: Flow of a liquid film.
Solution
Selecting the coordinate parallel to the plate and in the direction of the
steepest decent, and ® normal to the plate, we can conclude that the steady

(C.27) )
6
1
®
T
’
…
)
r
)
T 8 Y ) 4
® Y
)
V
|
V
Î
Î
¤«
4
4
r
>
Î
’
Î
4
¡
4
`
Î
’
4
r
l
®
4
¤
=
Î
4
1 4
Î
’
4
)
1 4
®
‹
T
state solution should satisfy the following )
constraints: ,
. Then the momentum equation (3.2) reduces to:
161
, and
Y X : Y
Y X : Y ®
8 R
r ` ; defÆ 1
with the solution:
R ®
` ; defÆ 1
l 8 ®
®ƒ>
where the constants ¡ and ¤ can be determined from the boundary conditions:
1}T 4 T v
T
The second condition is the negligible shear stress at the free surface:
` ; defÆ 1
1 l
,¯
> 8 ¡
thus:
O¯Y~
’‰defÆ 1
¡£ ;
Substituting and into (C.27) we obtain the final solution for the axial flow
velocity distribution with :
1 l
; defÆ 1
l 8 ®
The maximum velocity will correspond to ®
’ :
; deKÆ 1
The volumetric flow rate is obtained by integrating the velocity along the
thickness and the width of the film:
)œ|aw
l 8 ’
. #
Ö F
Ö ~
;
) R ® R ›
À 8

¾
r
ë
X
ë
r
r )
r
)
l
T
4
162 APPENDIX C. SOLUTIONS TO PROBLEMS
A cubic dependence of the flow-rate on ’ means that the draining of the film is
strongly dependent on the film thickness.
The values of )W|­w
and …
can be obtained from the following Matlab solution:
h=1.5e-3 % m: film thickness
theta=pi/6 % RAD: inclination of the plate
mu=1.6e-3 % kg/(m s): viscosity
rho=8e2 % kg/mˆ3: fluid density
g=9.8 % m/sˆ2: gravity acceleration
umax=(rho*g*hˆ2*sin(theta))/(2*mu) %=2.756 m/s
Q=(rho*g*hˆ3*sin(theta))/(3*mu) %=0.002756 mˆ3/s
Problem 3.7.4: Couette solution for non-Newtonian fluids
How will the solution (3.9) change for a non-Newtonian fluid?
Solution:
For non-Newtonian fluids the relation between stress and strain is nonlinear,
thus, instead of (3.5) we have:
B
which after the assumptions of Couette flow between parallel plates (Sec.3.2.1)
leads to a modified equation (3.7):
F ã 6 ¦
j ;Ãé
j
¦f
j*j
9 1
B
8 — R
which for any non-zero 8 and = has the same solution as (3.7). So, the solution
for a non-Newtonian fluid will be the same.
Problem 3.7.5: Momentum equation for Couette flow between concentric cylinders
Using the assumptions on the Couette velocity profile between the rotating
concentric cylinders (Sec.3.2.3) and the expression for the momentum equation
and the Laplacian operator in cylindrical coordinates:
R ®
x_>
¢ >^)GH,)GFH­>
¢
)œx5)IF
)IF­>])Wx5)GF
¼
F`
)GFœ)GFF
)œx
; ¢ > 9 1 ¾
¢ r 4
)GF­>
¢ r ` )GF
@
¢
x
4
x_>
@ r ë ¢
F„F­>Þë
F
F
HH
1 ¢

that )
)
¦
)
¢
A
R
R
V
¤
r
¢
V
x
6F
x
A
R
R
V
)
6
V
¼ F¥6¥x
)
¢
)
¢
L)
¢M
x
xñx_>
A
V
6
V
V
Ç
'
T
F
T
163
Derive equation (3.18):
T
¢ r >
¢ ¢ M L)GF
Solution
)IF, and
, the equation (C.28) simplifies to:
)IF
Renaming for simplicity: )
using the assumptions of (Sec.3.2.3)
1 ¢ 4
1 ¢
x
4
x
x
x€x_>
)
¢ r `
)
¢ r `
(C.28)
where we used the identity:
x
)
¢ r `
)
L)
¢‚M
Problem 3.7.6: Rotation torque and power
In the system of two rotating cylinders (Sec.3.2.3) consider the torque applied
to the inner rotating cylinder when the outer cylinder is (n fixed ). What
is the power required to rotate the inner cylinder?
Solution:
In this case we need to multiply the force applied at the surface of the cylinder
by its radius:
(C.29)
l§UA
¡£
where is the length of the cylinder. Since we are using a curvilinear coordinate
system, the expression for the shear stress tensor used in the momentum
(A.45), which in cylindrical coordinate system has a form:
Ç
equation is 6
°
V
and evaluating the expression above at ¢ƒ
¦Gä% T
Considering the specific form of velocity )
dependence:
, we have A
'*)GF1 ¢ 4
- (Sec.3.2.3),

I
6
V
º
R
¢ OxZYVÚ`
R
)GF
A
V
)GF
6
V
ø
¢ M
x
¢L)GF
n
`
V
A
n
V
V
¤
A
…
V
r A
…
F
>
r
A
` V
F
A
ÑU8n
n
V
r
V
A
)
Ç
Ñ
rF
V
A
ˆ
>
r A
1 4
®
`
V
r
@
A
A
` V
ÑU8n
Ä
ø
¢ M
x
ù
¢L)IF
F
A
r
F
V
r
r
F
Ç
A
`
r A
V
n
n
r
V
A
` V
V
A
A
F
A
r
: A
: F
rA
F
A
r
F
rV
V
V
V
`
@
Ï
`
`
A
A
rA
rF
F
V
V
`
A :
A :
@
F
F
164 APPENDIX C. SOLUTIONS TO PROBLEMS
¢ ù
)GFF
xZYVÚ
Computing the derivative from (3.20), we have:
xZYVlÚ
— A
— A
>\A
from which we obtain M:
The power needed rotate the cylinder is:
°
V
†
V°
V
Problem 3.7.7 Flow between parallel plates under pressure
A viscous fluid with viscosity (
parallel plates
8
` l
ωÏ
) is driven between two
`ÞÑ•° : 1 ¹. 4
Ï @‰I
¿ apart by an imposed pressure gradient of º : R R
volume flow rate per 1m of the plates’ width. What pressure gradient will cause
the flow to reverse?
£ :< . The upper plate is moving with velocity
Solution
– ¹:‰.
. Find the
Adapting the general expression for the volumetric flow rate (3.29) to the
case of the flow in a square duct of unit width and height , we have:
which is a volumetric flux per 1m width of the duct. Substituting the solution
(3.36), (3.37) into the above, and performing the integration, we obtain:
V
ÖJ—
R ®
à1
À
4 :[

ë
Y
r
Ä
r
Î
¥
r
º
Î
4
`
À
A
¢
r
>
ë
A
r
r
T
165
where
l
The pressure gradient causing the flow reversal at the lower plate can be
computed from (3.38).
8
Below is the Octave (Matlab) solution to this problem.
% Q=Integrate[u(y),y]
% where u(y) is given by (CFM:Sec.3.2.5):
% u = U/H*y*(C*y/H+1-C)
% where C=Hˆ2/(2*mu*U)*dpdx
% Integrating, we obtain:
% Q=U*H*(3-C)/6
% Express all values in SI units:
U=0.15 % m/s
H=0.005 % m
mu=1.4e-4 % kg/(m s)
dpdx=-2.55 % kg/(mˆ2 sˆ2)
C=Hˆ2/(2*mu*U)*dpdx
Q=U*H*(3-C)/6 %=0.0005647 mˆ2/s =(mˆ3/s)/m
% The pressure gradient that will case the
% reverse flow:
dpdx=2*mu*U/Hˆ2 %=1.68 kg/(mˆ2 sˆ2)
Problem 3.7.8 Verifying the Stokes solution
Verify that the solution (3.73):
(C.30)
rdeKÆ
l ¢
',Î
@ 4
Ñ£
A —
¢
1 ¢
satisfies the equation (3.72):
(C.31)
1 "
r Y
¢
—\Y
¢ r >
r ` –
Solution:
@ r Y
¢
Î Y
Y Îò

166 APPENDIX C. SOLUTIONS TO PROBLEMS
Using the symbolic manipulation library GiNaC (www.ginac.org), we can
easily check the consistency of the solution. Below is the example of a C++
code using the GiNaC library.
/********************************************
This function computes the Laplacian
of the Stokes solution in spherical
coordinates.
It should be compiled as
c++ diff.cc -o diff -lcln -lginac
Using GiNaC system (www.ginac.org)
Running the executable produces
the following output:
Laplace(F)=3/2*sin(t)ˆ2*a*U*rˆ(-1)
Laplace2(F)=0
********************************************/
#include
#include
using namespace std;
using namespace GiNaC;
ex Laplace
(
const ex & F,
const symbol & r,
const symbol & t
)
{ return normal
( F.diff(r,2)
+ F.diff(t,2)/pow(r,2)
- cos(t)*F.diff(t)/(sin(t)*pow(r,2))
);
}
int main()
{

(C.34)
6 x„F
ë
Î
rdeKÆ
x
—
`
r
Î
A
A
`
`
À
A
¢
>
A
Â
r
r
Î
167
symbol r("r"), t("t"), U("U"), a("a");
ex
F=normal
(
U/4*pow(a*sin(t),2)*(a/r-3*r/a+2*pow(r/a,2))
),
LF=Laplace(F,r,t),
LF2=Laplace(LF,r,t);
cout

dropped into oil of density ; {
T
IWeW;—rZ²
168 APPENDIX C. SOLUTIONS TO PROBLEMS
This function computes velocity components
and viscous stress of the Stokes solution.
It should be compiles as:
c++ stokes.cc -o stokes -lcln -lginac
Using GiNaC system (www.ginac.org)
Running the executable produces
the following output:
Ur=1/2*(2*rˆ3+aˆ3-3*a*rˆ2)*cos(t)*U*rˆ(-3)
Ut=3/4*sin(t)*a*U*rˆ(-1)-sin(t)*U+1/4*sin(t)*aˆ3*U*rˆ(-3)
tau=-3/2*mu*sin(t)*aˆ3*U*rˆ(-4)
********************************************/
#include
#include
using namespace std;
using namespace GiNaC;
int main()
{ symbol r("r"), t("t"), U("U"), a("a"), mu("mu");
ex F=normal
(
U/4*pow(a*sin(t),2)
*(a/r-3*r/a+2*pow(r/a,2))
),
Ur= F.diff(t)/(pow(r,2)*sin(t)),
Ut=-F.diff(r)/(r*sin(t)),
dU= Ur.diff(t)/r + Ut.diff(r) - Ut/r;
cout

¤
Ü
F
T
Ï
º
¤
¤
¤
R
R
U[
R
Ü
UÑ R
r
r
¤
F
¤
l V
Ñ : l
A»ˆ
!
@•>
Ð[
V
F
1
R
r
V
r
!
!
V
4
4
T
@
Ñ
I
169
Solution
The balance of forces on the sphere when it reached the terminal velocity
is: ¤
and viscosity 8 T
diameter is (a) R
. £
. Estimate the terminal velocity of the sphere
if
T
its
and (c) R ¿ .
IK@
IO@ ¿ , (b) R
@ ¿
where
is the buoyancy and
¤is a drag force:
A
1 ;
` ;
Ĥ;
;
are densities of the oil and sphere respectively, and the drag coefficient
Ĥcan be first assumed for a laminar
'
case:
where ; V
Ä R
;
8
Or, alternatively the expression for Stokes drag force can be used:
¤
A»ˆ
These equations can be solved for velocity:
ÀU8 ½
` ;
! R
1 ;
@W8
If the computed velocity leads to
Ĥshould be used:
A#
then the turbulent approximation for Û¥@
l Ñ
Ä R
A»ˆ
4 >
>
Below is the Octave solution:
A!

170 APPENDIX C. SOLUTIONS TO PROBLEMS
g=9.8 % gravity
denw=1000 % density of water
den0=0.88*denw % density of oil
den1=7.8*denw % density of the sphere
mu=0.15
% (a):
d=0.1e-3
V=dˆ2*(den1 - den0)*g/(18*mu) % =0.00025
Re=den0*V*d/mu
% =0.00014735: laminar
% (b):
d=1e-3
V=dˆ2*(den1 - den0)*g/(18*mu) % =0.025
Re=den0*V*d/mu
% =0.147: laminar
% (c):
d=1e-2
V=dˆ2*(den1 - den0)*g/(18*mu) % =2.5117
V1=0.0
% It’s a turbulent case so we loop until convergence:
while (abs(V1-V)>0.1*abs(V))
V1=V
Re=den0*V*d/mu
Cd=24/Re + 6/(1+sqrt(Re)) + 0.4
Fb=pi*dˆ3/6*(den1 - den0)*g
Fd=Fb
V=sqrt(Fd/(pi/4*dˆ2*Cd*den0/2))
end
which results in
Re = 48.569
V = 0.78984
% m/s

ß
V
Ö˜±
( 1 ß :
1
T
I
)
R Î l
R
( 1 ß :
@
R ) 8
® R
@
`
`
R l
R
Î
@
L®
ßM
l
V
T
I
(Äkõm A"
ß
)
À
ß
171
Problem 3.7.11: Boundary layer analysis for cubic velocity profile
Repeat the boundary layer analysis of Sec.3.5.1 with assumed velocity profile:
À
L®
l
4m A! ),
4m A! ),
4m A" ),
Compute ( Î :
1 :‰9
. ©
(Ĥm
where A!
Hint: Assume )
Solution
¯m¯
1}T 4 T
.
ßM
`
Using the definition of momentum thickness and introducing dimensionless
¢
coordinate , we obtain:
),
A" ),
® : ß
` )
(C.35)
Ï `
Ï 4 1
Ï 4
À‰Ô
£Ö˜±
L@
® MƒR
@‰I
@‰I
lyWT
Similarly, for displacement thickness:
Ï
»>
R
(C.36)
ß” ß Ö 1
ß :W
R ®
) :4
Using the definition of the wall shear stress:
(C.37)
6Q
@JI
Ï 8: ß
Thus, for the friction coefficient:
Äkõ
l
;r 6Q
8: ß
À
;r
(C.38)
lyWT
— ÀJÔ

s
@
4
@
1
@
ß
ß
Ð
`
Î
ß
R
@
T 8 R
Ñ
@EÀ ;
l Ö 6QR
4
T
T
T
Ñ
l
¡
4
@
r
172 APPENDIX C. SOLUTIONS TO PROBLEMS
and
ß
and solving this for ß , we have:
(C.39)
I4[Ñ
Using (C.35), and (C.36), we obtain:
A! Ð
Substituting ß 1
from (C.39) into (C.38), we obtain:
Ð
A! A!
@JI¢ÑT
I4[Ñ[
Problem 3.7.12: Drag force on a triangle
@‰I Ô‰À A!
ÄkõÐ
A thin equilateral triangle plate (Fig.C.3) with the edge length , is
l £Þ ¹:‰. l
at
ĤРI4[Ñ[
temperature
immersed parallel to a stream of air with the velocity
this plate.
l {
Ä and pressure º
Solution
Using (3.125), we have:
£ "
. Assuming laminar flow, estimate the drag on
F}É
— ;Ž8
l
; šgg1}T 4
IÁÀ‰À
The drag force on one side of the triangle is: ¤
l
9 o
6Q
The area element R is , ’ and
. For two sides of the triangle, we have for the total drag force:
R
: ’
, where
Ç
',’
УdeKÆ 1U: À
¡ÐУ
1}T
4
£mÀ : l

¤
l Ö 6QR
V
1
@
l 4WÀ
Üm
`
4
r
F}É Ý
¤
r
lV
l 4WÀ
r
r
`
r
r
`
V
@
’
:
r
F}É
l
ÀŽ’
T
1
@
V
`
r
Ñ
F}É : ÀŽ’
À
l
r§Üm
r
4
r
F}É £Ý
r
r
173
Figure C.3: Triangular plate pulled in a viscous fluid
£Ö ~
¡¼ l 1}T
IÁÀ‰À
l 4 1 ;Ž8 4
é F}É
F}É
R
Computing the integral:
: ’
Ö ~
Ö ~
: ’
Ö ~
é F}É
R
R
é F}É
R
F}É
’ F}É
£ 1 ;Ž8 4
F}É
1UT
IÕÀJÀ
Tœ1 ;Ž8 £ 4
1UT
1 ;Ž8 £ 4
À
l
IW‰l
IÁÀ‰À
F}É
À‰Ô
F}É
a=2 % m
U=12 % m/s
rho=1.2 % kg/mˆ3
mu=1.8e-5 % kg/(m s)
F=0.332*8/3*(sqrt(3)/2)ˆ(1/2)*(rho*mu*(a*U)ˆ3)ˆ(1/2) %=0.45 N
Problem 3.7.13: Boundary layer equations
Derive Prandtl boundary layer equations (3.110) - (3.113):

F
ã
F
)
r
)
¾
F
¾
F
–
P
)
F
)
F
F
r
F
r
)
F
ã
F
¦
s
r
F
)
)
F
ã s
F
¦
)
)
F
F
F
r
F
V
)
)
F
s
ã r
s
¦f“ 9 )
¾
F
¾
F
V
¾
)
F
)
F
F
s ã
ã
)
F
F
r
T
r
)
r
)
>
) ã
F
X ã
1
)
X
>
¦
> ;
F
)
)
¤
ã
F
F
T x
r
T
V
T
¦
`
r ¾ r*
X
) ;
¾
ã
r
X
F
X
F
V
V
"
¦
V
174 APPENDIX C. SOLUTIONS TO PROBLEMS
(C.40)
(C.41)
) ã
¦f ¦G
r“
rðr `
(C.42)
X ã
) >äã
) ã
) ã
) >äã
) ã
r•
(C.43)
sà>«ã ) ã
r 4
rðr
) ã
r
using equations (3.107) - (3.107):
) >çã
„©z
º_x
¦f ¦G
¦K€ `
>])
°
;
¦ 4
1
s]>])
¦ì¦
m ¦ 1
¦f
m ¦ 4
>^)
> 9 )
and scaling transformations (3.108) - (3.109):
(C.44)
¾ r
" )
r“
V ã ¾
ã
V
A F}É
(C.45)
A F}É
9 ¾ r ¾ r
r ¾ r
9 ¾
>\)
>^)
; >
(C.47)
¾ r
r ¾ r
r
r `
Jr
9 ¾
>\)
>^)
> 9 ¾ r ¾ r
; >

)
r
F
F
F
r V
F
V
@
É
r
and A F}É
dividing by
@ r
) ã
A!
>
@
)
F
A
F
T
F
r
F
ã ¾
F
r ã ¾
>
F
F
@
F
K
F
F
r
) ã
F
)
,
>
r ã ¾ r
F
>
)
V
V
r
V
F
1
r ã ¾ r
r
@
>
r
V
>
¾
F
)
r
F
r
V
4
r V
@ r ã ¾ r ã ¾ r
F}É A
@
A
r ã ¾
F
F
ã ¾
F
F
F
F
>
F
F
r
F
F
r
>
F
>
`
>
r
)
r
V
ã
V
V
F
r
@ ã ¾ r ã
A!
r
r
V
ã
F
F
`
¤
@
¤
@
x
x
ã
F
V
F
r
ã
F
F
V V r
)
V
ã
F
¤
x
175
substituting dimensionless variables from (C.44), (C.45), we obtain:
) ã
ã ¾
r ã ¾ r
A F}É
) ã
) ã
) ã
) ã
(C.48)
9 )
A F}É
9 )
A F}É
` )
) ã
) ã
XH>
ã ¾
ã ¾
¾ r ã ¾ r ã
ã ¾
Using the definition of the Reynolds number (2.110), we have:
) ã
r ã ¾ r
A!
ã
ã ¾ r ã
A"
¾ r
ã ¾
) >äã
ã ¾
X> ã
) ã
) >çã
) ã
ã ¾
) ã
Considering the limit
we have:
A!
and using the definition of Froude number (2.111),
) ã
(C.49)
¾ r ã ¾ r ã
ã ¾
A"
b
) ã
) >çã
X> ã
) ã
) >çã
) ã
) ã
Similarly, introduce the dimensionless variables into the momentum equation for
, (C.47):
ã ) r
ã ¾
r ã
@
F}É A
@ r ã )
F}É A
@ r ã )
F}É A
¾ r
) ã
) ã
r `
¾ r
ã
ã
:
¾ r
ã ¾ r
r `
@ r
F}É A
ã
ã
) ã
In the limit A!
, we have:
(C.50)
b
T
¾ r
X ã

T
V
s )
>
V
>
V
P
–
9 P
9 l
–
–
P
ã 1
) s]>äã
) 9
s –
P
>
) 9
r
V
–
P
ã s
F
V
V V
)
V
F F
F
ã s
r F
V
F
s
F
1
V
¦
s
)
)
s
r
F
F
)
F
>
F
F F
V
F
F
ã r
s
F
l
)
ã r
s
rq
F
F
rq
F
r
s
4
F F
r
s
F
)
– ;
P
>
°
– ;
1
)
°
s
– ;
°
F
V
)
P
F
F
V
r V
F
1
s
1
)
P
r
)
r V
ã 1
s
>
1
s
FðF
F
r
ã 1
s
>
FðF
)
>
FðF
FðF
1
)
@ 1
A"
@ 1
A!
4
F
rq
r
4
F
rq
r
4
F
F
176 APPENDIX C. SOLUTIONS TO PROBLEMS
This means that the pressure only changes in the direction of
Now consider the energy equation (C.44):
.
1
sà>])
°
s ;
¦ 4
¦ì¦
m ¦ 1
¦f
m ¦ 4
> 9 )
>^)
Writing it out explicitly, we have:
1
sà>^)
°
;9
r 4
rðr 4
>])
>]s
rq
r
rq
(C.51)
>C)
>])
>^)
>])
r 1
rq
r 4
rq r 1
rq r
rq r 4
>C)
>])
>^)
>])
Multiplying it by 9ú: –
, and rearranging the RHS, we obtain:
1
s]>])
r 4
rðr 4
>]s
(C.52)
P
R1
¦f ¦ 4
r
rq
F
r 4
rq 4
F
>])
¦K ¦G
where )
(C.44) and (C.45), we obtain:
is zero by continuity (2.4). Transferring, to dimensionless variables
rT
ã s >^A!
r 4
rðr 4
(C.53)
) >çã
— l
r
r 4
) ã
) ã
) ã
) ã
Multiplying the latter by
(2.110), we have:
: s
and using the definition of the Reynolds number
>^A!
1
sà>çã ) ã
r“
rðr 4
) >äã
(C.54)
ã s >\A"
— l
r
r 4
) ã
) ã
) ã
) ã
>^A!
1

@
>
ã s
F
F
ã s
F
l
F
)
F
ã
F
ã s
F
T
) ã
F
rq
F
ã r
s
rq
F
)
ã r
s
>
ã r
s
1
@
qº(x A!
F
F
s ã
r
s ã
r
FðF
>
ã 1
s
>
FðF
>
1
@ 1
A!
) ã r
A
>
s ã
1
4
r
F
F
rq
r
4
F
r
¢
)
r
V
P
s
V
177
using the definition of Prandtl (1.30) and Eckert numbers „§z (3.114),
we obtain:
: –
,
s]>çã ) ã
r“
rðr 4
ã s >\A"
(C.55)
) >çã
r
r 4
— l
) ã
) ã
Collecting the terms with the same powers of A#
A! q„Œz
we have:
) ã
) ã
>\A"
1
r“
rðr
,
s]>çã ) ã
>çã
r 4
A! qº_x
rq
º(x
r
) ã
and considering the limit A#
A! 5„©z
, we obtain:
„Œz
„Œz
r 4
rðr
) ã
b
s]>çã ) ã
r
) >«ã
„©z
º_x

¤
¡©¦
P
j
„
„
j
P
¦
¤
>
Ä j
j
P
æ
Ä
P
Ä
¤
¤
P
½
Ä
æ
½
j
¤
T
¤
P
j
¦
'
j
P
¡
j
T
½
'
¡
178 APPENDIX C. SOLUTIONS TO PROBLEMS
Chapter A
Problem A.4.1: Check tensor expressions
Check if the following Cartesian tensor expressions violate tensor rules:
¡©¦/ ¤
Answer: term (1): ik = free, term (2): pk=free
P æ
jê>
½
Answer: (1): ijq=free (2): p=free (3): kp=free
P æ
æÉæ
jÛ>
¦/
¡“Ù¤
¦/¡©¦M¤Œ ¦/É
`à½
Answer: (1): i=free (2): none, (3): i=free, j = tripple occurrence
Problem A.4.2: Construct tensor expression
¡t¦/ ¤§¦/ ½ ¦7
j

£
¦
¦
j
j
¦
>
£
¦
£
£
£
¦
j
¦
j
j
¦
j
¦
¦
j
j
¦
>
£
>
£
Problem A.4.3: Cartesian identity
Prove identity (A.15).
Proof
Integrating (A.5) in the case of constant transformation marix coefficients,
we have:
179
(C.56)
¦
j
£
>A¦
where the transformation matrix is given by (A.4):
ã
(C.57)
¦
j
¢ Y ã
Y j
By the definition of the Cartesian coordinates (C.58) we have:
(C.58)
ß ¦/
j ¦ £ j £
Y ã
¦ Y ã
¦
Let’s multiply the transformation rule (C.56)
by . Then we get:
£
Y
Y
£
ß
¦
¦
£
¦
A¦
¦
A¦
¦
A¦
ã
j
Differentiation this ã over
, we have:
¦
üY
Now rename index¨into ° :
Y ã
Comparing this with (C.57), we have
üY
Y ã
which proves (A.15).
Y ã
Y
Y
Y ã

(C.60)
~
j
p
j
P
¤
P
ß
j
j
p5¦
Pæ
Ä
P
j
~
ß
j
æ ¡ P
æ
Ä
j
p
æ
j
æ
æ
Ä
j
æ
P
`
j
ß
j
j
¤
j
æ
æ
Ä
Ä
P
Ä
P
æ
ß
Ä
j
j
P
Ä
Ä
æ
æ
æ
4
æ
ß
j
P
4Û`
Ä~
'
j
j
j
¤
¤
ß ¡
j
P
P
¤
P
Ä
Ä
P
¤ P
j
Ä
æ
æ
P
æ
180 APPENDIX C. SOLUTIONS TO PROBLEMS
Problem A.4.4: Tensor identity
Using the tensor identity:
(C.59)
p5¦/
ß
`]ß
prove the vector identity (A.30):
1~
Í ¤ 4 ¡ ¤ ¡‡ž Í 1~ ~ ~ ~ 1~
¡‡ž ¤ 4 ~
Proof
Applying (A.26) twice to the RHS of (C.60), we have:
Í ¡ ¤ Í ~ 1~
pm¦/
¡p
pm¦/
æ ¡“ ¤ P
From (A.24) it follows that p5¦/ j
. Then we have:
P æ
` p5¦
kp
¦7
(C.61)
¡¤
æ P
Now rename the dummy indexes: b # °
, so that the expression
looks like one in (A.29):
p5¦/
¡¤
¼p
¦/p
P æ
# b
¨‰'¨
b °
p5¦
4 ¡
1 p5¦/
4 ¡
Pæ
(C.62)
1 ß
`àß
ß
`àß
Using (A.10), and since ¡â邏 is the same as ¡§¦“朗¦ the latter can be
rewritten as:
j¡Ä
(C.63)
¼¤k ¡
,¡
P

¤ 1~ ~
F Y
ã Y
r
r
¡ž
Ä~
®
Y
Î Y
Y
¢
Y
Y ®
¢
Y
üY ® r
Î Y
¢
¦
4a`
Ä~
¦
+
¢
¢
Î
›
a/cedWÎ
Î
GF
¢
®yF£¢a/cedWÎ
Y +
+ Y
¢
Î
@
¦
¦
181
which is the same as
¡‡ž ¤ 4 ~
1~
Problem A.4.5: Metric tensor in cylindrical coordinates.
%§¢
ã ',Î'å›ð- Cylindrical coordinate system (C.64) is given by the following
transformation rules to a ('*®å',+- Cartesian coordinate system, :
ä%
£¢a/cedWÎ
£¢defÆ
Obtain the components of the metric tensor (A.40) and its inverse
(A.38) in cylindrical coordinates.
J¦7
Solution:
First compute the derivatives of
'*Î'W›ñ- :
¦/
ä%
Ú%§¢
with respect to ã
('*®å',+-
œx
F
Y
defÆ
Y ã
®

^
®
+
ë
r @
@
@
@
182 APPENDIX C. SOLUTIONS TO PROBLEMS
xñx
œxqœxa>^® xq® x
F³F
GFWGFa>^®”Fœ®yF£¢
HZH
xñx
F³F
@ r ¢
HZH
Problem A.4.6: Metric tensor in curvilinear coordinates
Using Mathematica, write a procedure to compute metric tensor in curvilinear
coordinate system, and use it to obtain the components of metric tensor, ,
(A.40) and its conjugate, , (A.38) in spherical coordinate system ( ' ë_',Î ):
¢
¼¢defÆ
ÎŒa/cydåë
(C.65)
ÎŒdeKÆ
¢arcedåÎ
¢defÆ
Solution with Mathematica
NX = 3
(* Curvilinear cooridnate system *)
Y = Array[,NX] (* Spherical coordinate system *)
Y[[1]] = r; (* radius *)
Y[[2]] = th; (* angle theta *)
Y[[3]] = phi; (* angle phi *)
(* Cartesian coordinate system *)
X = Array[,NX]
X[[1]] = r Sin[th] Cos[phi];
X[[2]] = r Sin[th] Sin[phi];
X[[3]] = r Cos[th];
(* Compute the Jacobian: dXi/dYj *)

^2×%Ž%
T
'
T
'
T
T
'
'
¢
r
'
r
é ¢
T
'
T
'
T
'
T
'
r
1 ¢ 4
rdefÆ
r Î
Î
4
183
J = Array[,{NX,NX}]
Do[
J [[i,j]] = D[X[[i]],Y[[j]]],
{j,1,NX},{i,1,NX}
]
(* Covariant Metric tensor *)
g = Array[,{NX,NX}] (* covariant *)
Do[
g [[i,j]] = Sum[J[[k,i]] J[[k,j]],{k,NX}],
{j,1,NX},{i,1,NX}
];
g=Simplify[g]
(* Contravariant metric tensor *)
g1 =Array[,{NX,NX}]
g1=Inverse[g]
With the result:
âä%Ž%
% T
% T
@‰'
-Q'
-Q'
-Ž-
% T
% T
'ard=a1
@‰'
-Q'
-Q'
¢ r -Ž-
Problem A.4.7: Christoffel’s symbols with Mathematica
Using the Mathematica package, write the routines to compute Christoffel’s
symbols
Solution
(************* File g.m *************
The metric tensor
and Christoffel symbols
*************************************)

184 APPENDIX C. SOLUTIONS TO PROBLEMS
DIM = 3
(*
The metric tensor
*)
g = Array[,{DIM,DIM}] (* covariant *)
g1 =Array[,{DIM,DIM}] (* contravariant *)
Do[
g [[i,j]] = 0;
g1[[i,j]] = 0
,
{j,1,DIM},{i,1,DIM}
]
(*
Cylindrical coordinates
*)
Z=Array[,DIM]
Z[[1]] = r
Z[[2]] = th
Z[[3]] = z
g [[1,1]] = 1
g [[2,2]] = rˆ2
g [[3,3]] = 1
g1[[1,1]] = 1
g1[[2,2]] = 1/rˆ2
g1[[3,3]] = 1
(*
Christoffel symbols of the first and second type
*)
Cr1 = Array[,{DIM,DIM,DIM}]
Cr2 = Array[,{DIM,DIM,DIM}]
Do[
Cr1[[i,j,k]] = 1/2
(
D[ g [[i,k]], Z[[j]] ]
+ D[ g [[j,k]], Z[[i]] ]
- D[ g [[i,j]], Z[[k]] ]
),
{k,DIM},{j,DIM},{i,DIM}
]
Do[
Cr2[[l,i,j]] =

185
]
Sum[
g1[[l,k]] Cr1[[i,j,k]],
{k,DIM}
],
{j,DIM},{i,DIM},{l,DIM}
Problem A.4.8: Covariant differentiation with Mathematica
Using the Mathematica package, write the routines for covariant differentiation
of tensors up to second order.
solution
(************** File D.m *******************
Rules of covariant differentiation
********************************************)
(*
B.Spain
Tensor Calculus, 1965
Eq.(22.2)
*)
D1[N_,A_,k_,X_,j_]:=
(*
Computes covariant derivative
of a mixed tensor of second order
with index k - covariant (upper)
*)
Module[
{i,s},
s = Sum[Cr2[[k,i,j]] A[[i]],{i,N}];
D[A[[k]],X[[j]]] + s
]
Dl1[N_,A_,l_,X_,t_]:=
(*
Computes covariant derivative
of a mixed tensor of second order

186 APPENDIX C. SOLUTIONS TO PROBLEMS
with index l - covariant (lower)
*)
Module[
{s,r},
s =Sum[Cr2[[r,l,t]] A[[r]],{r,N}];
D[A[[l]],X[[t]]] - s
]
D1l1[N_,A_,m_,l_,X_,t_]:=
(*
Computes covariant derivative
of a mixed tensor of second order
with index m - contravariant (upper) and
index l - covariant (lower)
*)
Module[
{s1,s2,r},
s1 =Sum[Cr2[[m,r,t]] A[[r,l]],{r,N}];
s2 =Sum[Cr2[[r,l,t]] A[[m,r]],{r,N}];
D[A[[m,l]],X[[t]]] + s1 - s2
]
D2[N_,A_,i_,j_,X_,n_]:=
(*
Computes covariant derivative
of second order tensor with
both m and l contravariant (upper)
indexes
B.Spain
Tensor Calculus, 1965
Eq.(23.3)
*)
Module[
{s1,s2,k},
s1 =Sum[Cr2[[i,k,n]] A[[k,j]],{k,N}];
s2 =Sum[Cr2[[j,k,n]] A[[i,k]],{k,N}];
D[A[[i,j]],X[[n]]] + s1 + s2
]
D2l1[N_,A_,i_,j_,k_,X_,n_]:=
(*
Computes covariant derivative
of third order tensor with
i and j contravariant (upper)

187
and k contravariant (lower)
indexes
B.Spain
Tensor Calculus, 1965
Eq.(23.3)
*)
Module[
{s1,s2,s3,m},
s1 =Sum[Cr2[[i,m,n]] A[[m,j,k]],{m,N}];
s2 =Sum[Cr2[[j,m,n]] A[[i,m,k]],{m,N}];
s3 =Sum[Cr2[[m,k,n]] A[[i,j,m]],{m,N}];
D[A[[i,j,k]],X[[n]]] + s1 + s2 - s3
]
D4l1[N_,A_,i1_,i2_,i3_,i4_,i5,X_,i6_]:=
(*
Computes covariant derivative
of 5 order tensor with
4 first indexes contravariant (upper)
and the last one contravariant (lower)
B.Spain
Tensor Calculus, 1965
Eq.(23.3)
*)
Module[
{k,s1,s2,s3,s4,s5},
s1= Sum[Cr2[[i1,k,n]] A[[k,i2,i3,i4,i5]],{k,N}];
s2= Sum[Cr2[[i2,k,n]] A[[i1,k,i3,i4,i5]],{k,N}];
s3= Sum[Cr2[[i3,k,n]] A[[i1,i2,k,i4,i5]],{k,N}];
s4= Sum[Cr2[[i4,k,n]] A[[i1,i2,i3,k,i5]],{k,N}];
s5=-Sum[Cr2[[k,i5,n]] A[[i1,i2,i3,i4,k]],{k,N}];
D[A[[i1,i2,i3,i4,i5]],X[[i6]]]+s1+s2+s3+s4+s5
]
Problem A.4.9: Divergence of a vector in curvilinear coordinates
Using the Mathematica package and the solution of Problem A.4.8, write the
routines for computing divergence of a vector in curvilinear coordinates.
Solution

188 APPENDIX C. SOLUTIONS TO PROBLEMS
Using the algorithms of covariant differentiation developed in Problem A.4.8
we have:

189
(\cite[5.102-5.110]{SyScTC69})
*)
V = Array[,NX]
Do[
V[[i]] = PowerExpand[V0[[i]]/g[[i,i]]ˆ(1/2)],
{i,1,NX}
]
(*
Transform vectors
as first order contravariant tensors
*)
U = Array[,NX]
SetAttributes[RV1,HoldAll]
RV1[NX,V,U]
(*
Compute first covariant derivatives
of vectors
*)
DV = Array[,{NX,NX}];
Do[
DV[[i,j]] = D1[NX,V,i,Y,j],
{j,1,NX},{i,1,NX}
]
(* Divergence *)
div=0
Do[
div=div+DV[[i,i]],
{i,NX}
]
div0 = div/.th->0
Problem A.4.10: Laplacian in curvilinear coordinates
Using the Mathematica package, write the routines for computing Laplacian
in curvilinear coordinates.
solution
Using the algorithms of covariant differentiation developed in Problem A.4.8

190 APPENDIX C. SOLUTIONS TO PROBLEMS
we have:

¡
¤©¦
æ P
(C.68) „
„
Ä
j
¤
¦
æ
¦
j
¦
j
`
P
P
>
>
¤
j
½
½
Éj
P
¦
Ä
j
P
Pj
Ä
æ
j
³
j
Ä
æ
æ
æ
½
j
Ä j æ
¦ÉP
¦
jÉP
¤
P
¦ì³
æ
]
DDQ = Array[,{NX,NX}];
Do[
DDQ[[i,j]] = Sum[DDP[[k,l]] J1[[k,i]] J1[[l,j]],{k,NX},{l,NX}],
{i,1,NX},{j,1,NX}
]
(* Laplacian *)
(*** lap=lap+Sum[g[[i,j]]*Dl1[NX,DS,j,Y,i],{i,1,NX},{j,1,NX}],*)
lap=Sum[DDQ[[i,i]],{i,NX}]
lap0=lap/.th->0
191
Problem A.4.11: Invariant expressions
not:
Check if any of these tensor expressions are invariant, and correct them if
(C.66)
¦
Ä j
³
j
¡§¦M¤
½
(C.67)
¦7
j
î¦
j ¡
¦
Answers:
A corrected form of (C.66) is:
¡§¦³¤
½
Equality (C.68) requires no corrections. A corrected form of (C.68) is:
¦M¤
½
¦
Since there are two combinations for an invariant combination of dummy
indexes (Corollary A.3.5), there can be several different invariant expressions.

¡
¦
¡ X
¦
¡
¦
j
X
¦
¤
j
¦
j
¦
j
¡
¤
¦
j
¤
j
¦
192 APPENDIX C. SOLUTIONS TO PROBLEMS
Problem A.4.12: Contraction invariance
¦
¤©¦¼¡§¦M¤
Prove that ¡
Proof
, and both are invariant, while ¡§¦M¤©¦ is not.
Using the operation of rising/lowering indexes (A.42), (A.43), we have
¤§¦$‡
¡“ ¦
¤ j
j ß ¡“Ù¤
¡“Ù¤ j ¼¡¤
¦7
¦/
which proves that both forms have the same values. If we now consider the first
form then:
¤§¦$
¦X
Y Wj
ß
£¡
¤Œ¡
¤©¦
Y
which proves the point.
YX
Consider now :
¡§¦³¤§¦ YX
¤§¦$ïY
¡QY Wj
j
which can not be reduced further and, therefore is not invariant, since it has a
different form from the LHS.
YX YX
¡§¦X

Appendix D
Midterm Exam Topics: Laminar Flow
Solutions
Items surrounded in brakets: [. . . ] will not be available during the exam. Other
items will be available.
1. Formulate the equations for incompressible flow [(3.1)], [(3.2)], [(3.3)].
2. Give the definition of hydrostatic pressure [(3.4)].
3. Write the expression for the viscous stress tensor for a Newtonian incompressible
fluid [(3.5)].
4. Give the definition of a laminar flow [(Definition 3.1.1)], and specify the conditions
when it may occur.
5. Formulate the equations for momentum and energy for the incompressible
flow between parallel plates: [(3.7)], [(3.8)].
6. Obtain the solution for velocity and shear stress for the flow between parallel
plates: [(3.9)], [(3.10)].
7. Solve Problem 3.7.3.
8. Give the definition of a friction coefficient, ÄŒõ , and express it in terms of
Reynolds number for the steady flow between parallel plates: [(3.11)].
9. Give the definition of a Poiseuille number, and obtain its value for the flow
between parallel plates: [(3.12)].
10. Give definition of a Brickman number [(3.13)].
193

194 APPENDIX D. MIDTERM EXAM TOPICS: LAMINAR FLOW SOLUTIONS
11. Obtain equation of motion for axially moving concentric cylinders: [(3.14)]
on the basis of general NS equation in cylindrical coordinates.
12. Obtain the solution [(3.16)] to equation (3.14).
13. Show that the problem of pulling an infinite rod does not have a steady-state
solution [Remark 3.2.1].
14. Obtain equations of motion for axially moving concentric cylinders: [(3.17)],
[(3.18)] on the basis of general NS equation in cylindrical coordinates (Problem
3.7.5).
15. Obtain the solution for the flow between axially moving concentric cylinders,
[(3.20)], on the basis of equation (3.18) and the appropriate boundary conditions.
16. Obtain solutions for the flow inside and outside of a rotating cylinder: Remarks
[3.2.2] and [3.2.3].
17. Obtain the pressure distribution in a flow outside a rotating cylinder: [(3.23)]
on the basis of the solution (3.22) and equation and the appropriate momentum
equation [(3.17)].
18. Solve problem 3.7.6.
19. Derive the equation for axial velocity in a fully developed flow region [(3.25)]
using the appropriate assumptions and equation (3.2). Write a non-dimensional
formulation of the boundary value problem [(3.26)].
20. Formulate a boundary value problem for a Poiseuille flow in a circular duct
[(3.27)] on the basis of a NS equation in cylindrical coordinates, and obtain
the Poiseuille solution [(3.28)].
21. For a Poiseuille flow through a duct (3.28) compute the volumetric flow rate
[(3.28)], the wall shear stress [(3.30)], the skin friction coefficient [(3.31)]
and Poiseuille number [(3.32)].
22. For combined Couette-Poiseuille flows formulate the equation of motion
[(3.33)], obtain the solution [(3.36)], and formulate the criterion of separation
[(3.38)].
23. Solve problem 3.7.7.
24. Give a definition of a hydraulic diameter [(3.39)].

4
4
r
l
Ç
4 :nm
4
195
25. Formulate the equation for the fluid oscillating above an infinite plate [(3.44)],
and obtain the solution for the case of oscillating plate [(3.47)] and an oscil-
. @
lating fluid [(3.48)]. Hint: Use the identity: # F}É
, where #­¢ m
26. Obtain the solution for an unsteady flow between two infinite plates
1
[(3.65)].
Hint: relation:x\"defÆ
Use integral . "a/cyd1
deKÆ 1
1 #
l
`
> @
4a`
R
27. Formulate the assumptions of creeping flow, and show how to obtain the
Laplace equations for pressure [(3.67)] and vorticity [(3.68)].
28. Using the expressions for pressure (3.76) and shear stress (3.77) of a Stokes
flow around a sphere, compute the total drag force [(3.80)], and the drag coefficient
as a function of the Reynolds number [(3.81)]. Hint: Use the integral
.
1arced1
relation:xNdefÆ 1
454 : @
4a`
R
À<
Ô{a/ced1
29. Using the assumptions of the lubrication theory, derive the equation for pressure
distribution in a flow between moving plates with a non-uniform gap
[(3.91)].
30. Derive criteria of validity of equation (3.91): [
›
], [(3.92)].

196 APPENDIX D. MIDTERM EXAM TOPICS: LAMINAR FLOW SOLUTIONS

Appendix E
Final Exam Topics
Items surrounded in brakets: [. . . ] will not be available during the exam. Other
items will be available. Abbreviations used: CFM=”Concise Fluid Mechanics”,
A.Smirnov (http://www.mae.wvu.edu/cfm), VFF=”Viscous Fluid Flow”, F.White, 2nd
Edition, McGraw-Hill, 1991.
E.1 Fundamental Laws
1. Definition of substantial derivative and derive its expression [(1.7)].
2. Give definitions of strain tensor [(1.9)], vorticity tensor [(1.10)] and vorticity
vector [(1.11)].
3. Derive the conservation of mass in explicit form [(2.2)] and in substantial
derivative form [(2.3)], and for incompressible flow [(2.4)].
4. Define the stream-function in 3D [(2.6)], [2.8)], and 2D case [(2.9)]. Describe
its realtion to the mass-flow rate [(2.10)].
5. Derive general equation of momentum [(2.22)] based on the definition of
viscous stress tensor [(2.19)], and it’s incompressible limit Navier-Stokes
equation [(2.24)].
6. Derive a vorticity formulation of the incompressible Navier-Stokes equation
[(2.34)].
7. Derive the Poisson equation for pressure for constant density flows [(2.61],
and describe the boundary conditions.
197

; R ’
(E.1) "
R
Y
"
Y
~
j
p5¦
æ P
Ä
X R
" >
R
P
ß
s
j
æ
`
Ä
X
¦
æ
ß
j
P
Ä
¦
198 APPENDIX E. FINAL EXAM TOPICS
8. Formulate the energy equation in terms of temperature [(2.74)] and enthalpy
[(2.75)]. Explain the meaning of each term.
9. Problem CFM.2.7.5: Using the energy equation (2.77):
ã 6 ¦/
1 °
¦ 4 ¦
m ¦
>^)
and momentum equation (2.21):
(E.2)
1 ; )
1 ; )
4 ¦
j
6 ¦
j >Úã
j
¦ 4
¦
>^)Wj
>\š
derive the strong formulation of the Bernoulli’s equation.
10. Derive the expression for Coriolis forces [(2.93)].
11. Problem CFM.A.4.4: Using tensor identity:
p5¦/
ß
`]ß
prove the vector identity (A.30):
Í ¤ 4 ¡ ¤ ¡‡ž Í 4a` 1~ ~ ~ ~ ~ 1~
¡‡ž ¤ 4 ~
¦
¤§¦$¼¡§¦M¤
12. Problem CFM.A.4.12: Prove that ¡
¡§¦M¤©¦
is not.
, and both are invariant, while
1~
E.2 Analytical Solutions
13. Formulate the equations for incompressible flow [(3.1)], [(3.2)], [(3.3)]. Write
the expression for the viscous stress tensor for a Newtonian incompressible
fluid [(3.5)].
14. Formulate the equations for momentum and energy for the incompressible
flow between parallel plates: [(3.7)], [(3.8)]. Obtain the solution for velocity
and shear stress for the flow between parallel plates: [(3.9)], [(3.10)].

F
T
¾
¦
)
1 4
®
r
ë
X
ë
r
l
E.2. ANALYTICAL SOLUTIONS 199
15. Problem CFM.3.7.3: Consider a wide fluid film of constant ’ thickness, ,
flowing steadily due to the gravity down the inclined plate at Î angle . Find
the velocity )
distribution, , and the volumetric flow … rate, . Atmospheric
pressure can be considered constant.
16. Obtain equation of motion for axially moving concentric cylinders: [(3.14)]
on the basis of general NS equation in cylindrical coordinates.
17. Obtain the solution to equation (3.14) for the case of axially moving concentric
cylinders [(3.16)]. Show that the problem of pulling an infinite rod does
not have a steady-state solution [Remark 3.2.1].
18. Problem CFM.3.7.5: Using the assumptions on the Couette velocity profile
between the rotating concentric cylinders (Sec.3.2.3) and the expression for
the momentum equation and the Laplacian operator in cylindrical coordinates:
x_>
¢ >^)GH,)GFH­>
¢
)œx5)IF
)IF­>])Wx5)GF
¼
F`
)GFœ)GFF
)œx
; ¢ > 9 1 ¾
¢ r 4
)GF­>
¢ r ` )GF
@
¢
x
4
x_>
@ r ë ¢
F„F_>Þë
1 ¢
Derive equation for )IF, [(3.18)],
rotating concentric cylinders, [(3.20)], on the basis of the appropriate boundary
conditions.
F
F
and obtain the solution for the flow between
HH
19. Problem CFM.3.7.6: In the system of two rotating cylinders (Sec.3.2.3) consider
the torque applied to the inner rotating cylinder when the outer cylinder
is (n fixed ). What is the power required to rotate the inner cylinder?
20. Derive the equation for axial velocity in a fully developed flow region [(3.25)]
using the appropriate assumptions and the momentum equation (3.2):
¦
Write a non-dimensional formulation of the boundary value problem [(3.26)].
¦f
j
F ã 6 ¦
j ;Ãé
j
` ;Ãé F º
>])åj)
21. Formulate a boundary value problem for a Poiseuille flow in a circular duct
[(3.27)] on the basis of a NS equation in cylindrical coordinates, and obtain
the Poiseuille solution [(3.28)]. Compute the volumetric flow rate [(3.28)], the
wall shear stress [(3.30)], the skin friction coefficient [(3.31)] and Poiseuille
number [(3.32)].

Ñ
ˆ
4
)
1
®
Ä
r
®
4
r
F
r
l
@
Ï
I
º
4 :m
4
200 APPENDIX E. FINAL EXAM TOPICS
22. For combined Couette-Poiseuille flows formulate the equation of motion
[(3.33)], obtain the solution [(3.36)], and formulate the criterion of separation
[(3.38)]. Using the obtained solution, consider a viscous fluid
with
Ï
viscosity
) driven between
¿
two parallel plates
apart
`ÞÑ•° : 1 ¹. 4
(
by an imposed pressure gradient 8 º : R of R
£ :
(Problem 3.7.7).
@‰I
The upper plate is moving velocity
– ¹:‰.
with . Find the volume flow
rate per 1m of the plates’ width. What pressure gradient will cause the flow
to reverse?
` l
ωÏ
23. Formulate the equation for the fluid oscillating above an infinite plate [(3.44)],
and obtain the solution for the case of oscillating plate [(3.47)] and an oscil-
. @
lating fluid [(3.48)]. Hint: Use the identity: # F}É
1 #
l
, where #a¢ m
`
24. Obtain the solution for an unsteady flow between two infinite plates
1
[(3.65)].
Hint: relation:x\!defÆ
Use integral . "a/ced1
defÆ 1
> @
4a`
R
25. Formulate the assumptions of creeping flow, and show how to obtain the
Laplace equations for pressure [(3.67)] and vorticity [(3.68)].
26. Using the expressions for pressure (3.76) and shear stress (3.77) of a Stokes
flow around a sphere, compute the total drag force [(3.80)], and the drag coefficient
as a function of the Reynolds number [(3.81)]. Hint: Use the integral
.
1arced1
relation:xÞdefÆ 1
4q4 : @
4a`
R
À
Ôßarced1
27. Using the assumptions of the lubrication theory, derive the equation for pressure
distribution in a flow between moving plates with a non-uniform gap
[(3.91)], and derive criteria of its validity: [ ], [(3.92)].
›
Ç
E.3 Boundary Layers
28. Using integral analysis derive the expressions for displacement thickness,
[(3.94)] and momentum thickness [(3.95)]. Define the skin friction [(3.96)]
and drag, [(3.97)] coefficients, and derive their relations to the momentum
thickness [(3.98), (3.99)].
29. Using the parabolic velocity profile:
V
obtain the boundary layer growth rate as a function of [(3.104)], [(3.105)].
>\Ä
®C>\Ä
4

F
ã
–
P
)
F
ã
r
‹
¦
s
r
1
)
¦
)
F
F
)
V
)
s
¡
V
¾
r
)
)
r
)
)
X
¦
> ;
r
V
T
¦
X
) ;
r
V
V
"
¦
E.3. BOUNDARY LAYERS 201
30. Problem CFM.3.7.13: Using equations (3.107) - (3.107):
¦f ¦G
¦f“ 9 )
¦K€ `
>])
°
;
¦ 4
¦ì¦
m ¦ 1
¦f
m ¦ 4
1
s]>])
> 9 )
>^)
and scaling transformations (3.108) - (3.109):
r
¾ r“
r ¾ r/
" )
A F}É
F ã
ã
ã
V
V ã ¾
A F}É
)
r“

p
Ä
202 APPENDIX E. FINAL EXAM TOPICS
35. Problem CFM.3.7.12: A thin equilateral triangle plate with the edge length
of is immersed parallel to a 12m/s stream of air at l T {
and 1atm,
as in Fig.C.3. Assuming laminar flow estimate drag of this plate (in N).
l Ç
First give an answer in symbolic form in terms of 8 , ; ,, and Ç . And then
compute it to a number.
E.4 Turbulence Modeling
36. What’s the difference between RANS and LES turbulence modeling. Apply
Reynolds averaging to the Navier-Stokes equation to obtain the RANS
equation [(4.10)]. Formulate the Boussinesq approximation [(4.11)].
37. What assumptions are used in LES turbulence models. Formulate the governing
equations for the Smagorinsky LES model [(4.2)], [(4.3)], [(4.4)].
38. What assumptions are used in RANS turbulence models. Formulate governing
equations for the turbulent model [(4.17)], [(4.18)], [(4.19)],
[(4.20)], [(4.21)].
°Ó`

Index
Acoustic problems, 13
Algebraic Reynolds stress model, 109
Associate tensor, 130
Bernoulli’s Equation, 24
Bernoulli’s equation, 38, 51, 152, 154,
198
Blasius equation, 90
Blasius stream function, 92, 96
Boundary conditions, 29
Boundary layer, 85, 87
Boundary layer blow-off, 92
Boussinesq approximation, 108
Brinkman number, 56
Bulk modulus, 13
Cartesian Tensors, 119
Christoffel’s symbol, 132
Coefficient of bulk viscosity, 20
Coefficient of thermal expansion, 13
Coefficient of viscosity, 9, 10, 93
Conjugate metric tensor, 129
Conjugate tensor, 40, 130, 134
Continuity equation, 16, 32
Contraction of indexes, 123
Contraction operation, 130
Contravariant index, 117
Contravariant tensor, 133
Contravariant vectors, 116
Convective derivative, 4
Coordinate system, 116
Coriolis force, 42, 43
Couette flows, 54
Covariant differentiation, 132
Covariant index, 117
Covariant vectors, 117
Creeping flow, 71, 76
Cross product, 124
Cylindrical coordinates, 138
Darcy friction factor, 62
Dependent variable, 9
Dependent variables, 1
Direct numerical simulation, 106
Dirichlet boundary, 30, 61
Displacement thickness, 82, 90, 100
Divergence operator, 127, 133, 138,
139
Divergence-free vector field, 16
Dot-notation, 2
Drag coefficient, 83, 98
Drag force, 74
Dummy index restriction, 123
Dummy indexes, 122
Eckert number, 52, 87, 157, 177
Eddy viscosity, 106, 108
Enthalpy, 7
Enthapy, 8
Entrance effect, 60
Entropy, 6
Equation of state, 5, 7, 32
Euler equation, 28, 29
Eulerian description, 1
Falkner-Skan equation, 92
Falkner-Skan wedge flow, 91
First law of thermodynamics, 6
Flow separation, 64
203

204 INDEX
Fluid element, 3
Fluid particle, 1
Fluid particles, 1
Fluid properties, 8
Fluid velocity, 2
Flux, 9
Fourier series, 70
Fourier’s law, 10, 35
Free boundary, 31
Free indexes, 121
free jet, 95
Friction coefficient, 55, 90, 93
Froude number, 49, 50, 87, 175
Fully developed flow, 60, 61, 65
Fundamental tensor, 129
Gas constant, 6
Gauss theorem, 16, 19, 35, 36
Gibbs rule, 6
Gradient, 9, 127
Gradient approximation, 9–11
Heat conduction coefficient, 10, 39,
49, 93
Heat conduction equation, 38
Heat conductivity, 10
Heat convection equation, 39
Heat dominated flow, 39
Hydraulic diameter, 64
Hydrostatic pressure, 53
Ideal fluid, 28, 152
Ideal gas law, 6
Incompressibility condition, 16
Independent variables, 1, 6, 8
Inlet boundary, 30, 34
Integral momentum relation, 101
Internal energy, 6, 8
Invariance, 39, 116, 128
Invariant, 40, 130, 132
invariant expressions, 39
Invariant forms, 40, 128, 131
Inviscid flow, 28
Irrotational flow, 23, 27, 29
KE turbulence model, 111
Kelvin’s theorem, 152
Kinematic variables, 1
Kinematic viscosity, 10, 43
Kronecker delta tensor, 118
Lagrangian description, 1
Laminar flow, 56
laminar flow, 53
Laplace equation, 24, 27, 28, 71
Laplace equation for pressure, 33
Laplacian, 61, 66, 72, 73, 102, 128,
134, 138, 139, 162, 199
Large eddy simulation, 106
Law of circulation, 152
Law of similarity, 43, 46
Lewis number, 12
Lift force, 74
Lowering indices, 130
Lubrication theory, 76
Mean wall shear stress, 64
Metric tensor, 40, 128, 131
Momentum equation, 131
Momentum flux, 29
Momentum thickness, 83, 90, 100
Momentum transport, 9
Moving boundaries, 30
Nabla, 17, 28, 47, 48, 86, 126, 138
Natural convection, 13
Neuman boundary, 30, 33
No-slip boundary, 30, 61
Non-dimensional parameters, 46, 47,
49
Non-dimensional variables, 48, 49
Non-inertial coordinate systems, 41
Non-Newtonian, 10
Nusselt number, 49, 93

INDEX 205
One equation turbulence model, 111
Order of a tensor, 117, 121
Orthogonal coordinate system, 137
Orthogonal coordinates, 40, 134
Oseen approximation, 76
Outlet boundary, 30, 34
Parabolic equation, 66
Particle trajectory, 2
Permutation tensor, 123
Physical component, 137
Physical components of tensors, 39
PI-group, 46
PI-theorem, 46, 47, 157
Poiseuille flow, 62
Poiseuille number, 56, 62
Poisson equation for pressure, 33
Potential flow, 23
Prandtl number, 11, 38, 92, 177
Primary dimensions, 44, 46, 48
Properties of the fluid, 1
Quasi-equilibrium approximation, 5
Raising indices, 130
Rank of a tensor derivative, 127
Rank of a term, 121
RANS models, 108
Renaming indexes, 123
Renaming of dummy indexes, 123
Reynolds analogy, 92, 93
Reynolds averaged Navier-Stokes equation,
108
Reynolds averaging, 108
Reynolds decomposition, 107
Reynolds number, 44, 49, 54, 55, 60,
61, 75, 85, 94, 105, 175
Reynolds stress model, 108, 111
Reynolds stress tensor, 108
Scalar product, 40, 122, 123, 130, 131
Schmidt number, 12
Second law of thermodynamics, 34
Shear layer, 94
Similarity solution, 88
Skew-symmetric tensor, 124
Skin-friction coefficient, 62, 83
Slip boundary, 30
Smagorinsky constant, 106
Smagorinsky model, 106
Solenoidal vector field, 16
Solid body rotation, 58
Spatial derivative of a tensor, 126
Specific heat, 8, 11, 38, 93
Speed of sound, 12
Stanton number, 93
Stokes flow, 71, 80
Stokes paradox, 76
Stokes theorem, 152
Strain rate tensor, 4, 20
Stream function, 17, 72
Streamline, 2
Stress tensor, 19
Stretching factors, 135, 138
Substantial derivative, 3, 40
Surface area vector, 16
Surface tension coefficient, 31, 50
T-s relations, 7
Taylor number, 105
Tensor, 117
Tensor derivative, 127
Tensor equality, 121
Tensor expression, 120
Tensor identity, 125
Tensor notation, 116, 120
Tensor rules, 120
Tensor terms, 120
Thermodynamic properties, 1
Thermodynamic variables, 1
Time derivative of a tensor, 126
Transformation matrix, 117, 119
Transformation rule, 116

206 INDEX
Transport properties, 1
Transport property, 9
Turbulence dissipation rate, 112
Turbulence model, 111
Turbulent closure, 110
Turbulent kinetic energy, 108
Two equation turbulence models, 111
Two-equation turbulence models, 112
Vector product, 124
Velocity circulation, 50, 151
Velocity potential function, 23, 29
Viscous flow, 27
Viscous stress tensor, 20, 33
Volumetric flow rate, 62
von Karman, 101
Vorticity tensor, 4
Vorticity vector, 5
Wake, 96
Wall shear stress, 62, 90
Weber number, 50
Wedge flows, 91