8VB00
transport physics
lecture notes – 2021/2022
Frans van de Vosse
department of biomedical engineering — group cardiovascular biomechanics
2
2
transport physics
preface
Transport of heat, nutrients, drugs, and waste products, so heat and particle transport in general, play an essential role in the maintaining of the human body. After a
short introduction in the relevant physical quantities in single-component en multicomponent systems, particle transport based on the balance of mass will be considered and examples of diffusion processes and mass transfer will be given. Next,
transport phenomena based on the balance equations for mass and momentum for
single-component systems will be reviewed. Particle transport in viscid and inviscid
flow will be discussed and applied to physiological and non-physiological example
problems. Special attention will be payed to phenomena that take place in the
boundary layer between fluid flow and rigid walls. Finally, theory of heat transfer
problems based on the balance of energy will be discussed and illustrated.
Learning objectives:
The student will learn about the mechanisms of particle and heat flow as a result of
convection and diffusion. Specific aims are:
- providing basic insight in transport phenomena in terms of particle paths, streamlines and streak lines;
- the formulation of the equations of motion and energy for specific cases (inviscid
flow, boundary layers, viscosity dominated flow);
- the derivation of (approximate) analytical solutions of these equations;
- the modelling of diffusive transport in 1, 2, or 3 dimensions for steady as well as
unsteady cases.
In these Lecture Notes the theory and examples that are presented in the lectures
are written down so students can either prepare or revisit the topics that will be
discussed during the lectures.
February 2019,
Frans N. van de Vosse
Professor of Cardiovasculaire Biomechanica,
Department of Biomedical Engineering,
Eindhoven University of Technology.
Contents
1 Introduction
1.1 Basic physical quantities and definitions . . . . . . .
1.1.1 State variables . . . . . . . . . . . . . . . . .
1.1.2 Volume and density . . . . . . . . . . . . . .
1.1.3 Temperature . . . . . . . . . . . . . . . . . .
1.1.4 Pressure . . . . . . . . . . . . . . . . . . . .
1.1.5 State equations and continuum hypothesis . .
1.1.6 Viscosity . . . . . . . . . . . . . . . . . . . .
1.2 Fluid Kinematics . . . . . . . . . . . . . . . . . . . .
1.2.1 Velocity field . . . . . . . . . . . . . . . . . .
1.2.2 Eulerian and Lagrangian representations . . .
1.2.3 Pathlines, streamlines, and streaklines . . . .
1.2.4 Flow rate and flux . . . . . . . . . . . . . . .
1.3 Deformation and stress . . . . . . . . . . . . . . . .
1.3.1 Velocity gradients . . . . . . . . . . . . . . .
1.3.2 The divergence and rotation of a vector field
1.3.3 State of deformation . . . . . . . . . . . . .
1.3.4 State of stress . . . . . . . . . . . . . . . . .
1.3.5 Constitutive relations . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
7
8
8
8
11
11
12
13
15
15
16
19
22
24
24
25
27
29
33
.
.
.
.
.
.
.
.
.
.
.
.
37
38
38
39
42
43
43
44
45
46
46
50
55
3 Balance of momentum
3.1 Integral form of the momentum balance . . . . . . . . . . . . . . . . .
3.1.1 Momentum balance for a material domain . . . . . . . . . . . .
57
58
58
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2 Balance of mass
2.1 Integral form of the balance of mass . . . . . . . . . . . . . . . . . .
2.1.1 Mass balance for a material domain . . . . . . . . . . . . . .
2.1.2 Mass balance for a fixed or moving domain . . . . . . . . . .
2.2 Differential form of the balance of mass . . . . . . . . . . . . . . . .
2.3 Mass diffusion and mass transfer . . . . . . . . . . . . . . . . . . . .
2.3.1 Mass balance for a multi-component system . . . . . . . . . .
2.3.2 The first law of Fick . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 The second law of Fick and the convection-diffusion equation
2.4 Illustrative examples of mass diffusion problems . . . . . . . . . . . .
2.4.1 Steady diffusion . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Time dependent diffusion . . . . . . . . . . . . . . . . . . . .
2.5 Convection-diffusion problems . . . . . . . . . . . . . . . . . . . . . .
5
3.2
3.3
3.4
3.1.2 Momentum balance for a fixed or moving domain
3.1.3 Hydrostatics . . . . . . . . . . . . . . . . . . . .
Differential form of the momentum balance . . . . . . .
Fluid flow classification . . . . . . . . . . . . . . . . . .
3.3.1 Hydrostatics . . . . . . . . . . . . . . . . . . . .
3.3.2 Inviscid flow: the Euler Equation . . . . . . . . .
3.3.3 Viscous flow: the Navier-Stokes Equation . . . .
Dimensionless groups . . . . . . . . . . . . . . . . . . .
4 Inviscid flow
4.1 Introduction . . . . . . . . . . . . . . . . . .
4.2 Bernoulli’s equation . . . . . . . . . . . . . .
4.2.1 Derivation of Bernoulli’s equation . .
4.2.2 Applications of the Bernoulli equation
4.3 A few remarks on irrotational flow . . . . . .
4.3.1 Elementary potential flow solutions . .
4.4 Inviscid and turbulent tube or channel flow . .
4.4.1 Inviscid 1D tube or channel flow . . .
4.4.2 Turbulent tube flow . . . . . . . . . .
5 Viscous flow
5.1 Internal viscous flow . . . . . . . . . . . . .
5.1.1 Steady flow between two flat plates
5.1.2 Poiseuille tube flow . . . . . . . . .
5.2 External viscous flow . . . . . . . . . . . .
5.2.1 Flow at an impulsively moving plate
5.2.2 Flow at an oscillatory moving plate .
5.2.3 Stokes flow around a sphere . . . .
5.3 Flow in porous media . . . . . . . . . . . .
6 Boundary layer flow
6.1 Introduction . . . . . . . . . . . .
6.2 Boundary layer equations . . . . .
6.3 Drag force induced by a developing
6.4 Inlet flow in a tube . . . . . . . .
6.5 Boundary layer separation . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . . . . .
. . . . . . . . .
boundary layer
. . . . . . . . .
. . . . . . . . .
7 Balance of energy
7.1 First law of thermodynamics . . . . .
7.1.1 Work done by external forces .
7.1.2 Work done by pressure forces .
7.1.3 Integral energy balance . . . .
7.2 Differential form of the energy balance
7.3 Thermal diffusion and heat transfer . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
59
63
65
66
66
67
68
72
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
77
78
80
80
82
86
87
89
89
90
.
.
.
.
.
.
.
.
93
94
94
99
105
105
107
109
111
.
.
.
.
.
115
116
116
120
123
124
.
.
.
.
.
.
129
130
130
131
132
134
136
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Appendix A Appendices
A.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.1.1 Coordinate systems . . . . . . . . . . . . . . . . . . .
A.1.2 Inner product and vector projection . . . . . . . . . . .
A.1.3 A 2nd -order tensor as a diadic product . . . . . . . . .
A.1.4 The gradient operator ∇ . . . . . . . . . . . . . . . .
A.1.5 The divergence operator ∇· . . . . . . . . . . . . . . .
A.1.6 The Laplacian operator ∇2 . . . . . . . . . . . . . . .
A.1.7 Gauss’s divergence theorem . . . . . . . . . . . . . . .
A.1.8 Vector and tensor form of the Gauss theorem . . . . .
A.1.9 Vector differential identities . . . . . . . . . . . . . . .
A.1.10 The error function . . . . . . . . . . . . . . . . . . . .
A.1.11 The Dirac delta function . . . . . . . . . . . . . . . .
A.2 Transport Physics . . . . . . . . . . . . . . . . . . . . . . . .
A.2.1 Reynolds transport theorem . . . . . . . . . . . . . . .
A.2.2 Mass balance in different coordinate systems . . . . .
A.2.3 Balance of momentum in different coordinate systems
8 Index
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
145
146
146
147
148
149
149
151
151
153
153
153
154
154
154
156
157
159
1
Introduction
In this chapter we will first introduce basic quantities important for physical transport
phenomena. Next we will introduce some concepts regarding the kinematics of fluids and
the representation in different coordinate systems. Finally basic concepts of deformation
of fluids and its relation to fluid stresses are introduced.
9
10
1.1
1.1.1
transport physics
Basic physical quantities and definitions
State variables
State variables are physical properties that describe the state of a dynamical system.
In a thermodynamic system, given by a precisely defined region in space, examples
of these variables are: the volume V , the density ρ, the temperature T , the pressure
p, and the amount of matter N . Note that variables such as work W and heat q are
not state variables. They do not define the state of a system but describe the transfer
of energy out of or into a system. Furthermore, we can distinguish intensive and
extensive physical quantities. Intensive quantities such as temperature T , density ρ,
and viscosity η do not depend on the size of the system. If the system is (virtually)
divided in two or more subsystems, intrinsic quantities do not change. Extensive
quantities such as mass m and volume V do change when the system is divided in
subsystems. The mass of a system is the sum of the masses of the subsystems. The
quantities volume, density, temperature, pressure, and viscosity are variables that
will be discussed in more detail in the next sections.
1.1.2
Volume and density
The amount of matter
The amount of matter is expressed in moles, where 1 mole is de amount of a substance
that contains as many molecules as there are atoms in 12 g of the isotope 12 C. This
number is equal to 6, 02214·1023 . So, one mole of a given substance contains NA
molecules with:
NA = 6.02214·1023 mol−1
the Avogadro’s number. Note that in general we will use the units kmol and kg (so
we use NA = 6.02214 · 1026 kmol−1 ).
Molar density and number density
The molar density or molar concentration c of a substance is defined as the amount
of that substance per unit of volume, i.e.:
c=
N
V
(1.1)
and expressed in kmol/m3 . The number of molecules per unit of volume is defined
as the number density n and expressed in m−3 , so we have:
n = cNA .
(1.2)
Mass and mass density
Assume that the mass of 1 molecule of certain substance S is m, then is the mass
per mole of S, the molar mass M , is equal to:
M = NA m ,
(1.3)
Introduction
11
with units kg/kmol. The molar mass of different substances is an intrisic property
of the substance and can be found in handbooks or on internet pages.
Actually, the molar mass of a substance is a measure of the average molecular mass
of the molecules. The molecular mass can be computed from the atomic mass of the
elements in the molecules. Since different isotopes (number of neutrons) may exist,
different molecules of the same substance can have different molecular masses. For
reasons we will discuss in section 1.1.5 we will use the molar mass rather than the
molecular mass.
The mass per unit of volume is the mass density ρ and is expressed in kg/m3 . For
a pure single-component substance S we can readily derive that:
ρ = cM = nm .
(1.4)
If, in the remainder of this course, we talk about the density of a solid or a fluid,
this refers to the mass density.
Example 1.1: Mass of a glucose molecule
The molecular formula of glucose is C6 H12 O6 . The molar masses of C, H, and O
are MC = 12.011 kg/kmol, MH = 1.008 kg/kmol and MO = 15.995 kg/kmol
respectively. Consequently the molar mass of glucose is
MC6 H12 O6
= 6MC + 12MH + 6MO
= 6 · 12.011 + 12 · 1.008 + 6 · 16.000 kg/kmol
= 180.16 kg/kmol .
The mass of a single glucose molecule then is
m = M/NA =
180.16
kg/kmol
= 2.99 · 10−25 kg .
6.02214 · 1026 kmol−1
Multi-component systems
In the previous sections we assumed that the system that was considered contained
one single component and used the mass of its molecules as intrinsic parameter. In
a multi-component system (see e.g. Figure 1.1), such as air, that contains more then
one component we must define the physical properties with more care. Table 1.1
illustrates this for a two-component (or binary) system.
In general for a multi-component mixture of k components we have the relations as
given in Table 1.2. Note that if we mix two substances A and B with mass mA and
mB that initially are contained in two volumes VA and VB respectively, in general
m = mA + mB (no chemical reactions) but V 6= VA + VB . Note also that the density
ρi = ci Mi depends on the mixture and in general is not equal to the specific density
of a material. The latter is the density of a pure substance under specific conditions
(e.g. at T = 273 K and p = 1 atm).
12
transport physics
Table 1.1: Physical quantities for a two-component system that exists of substance A
and substance B
-
amount of particles [kmol]:
substance A
NA
substance B
NB
mixture
N = NA + NB
-
molar density [kmol/m3 ]:
cA = NA /V
cB = NB /V
c = cA + cB
-
molar fraction [-]:
xA = cA /c
xB = cB /c
xA + xB = 1
-
molecular mass [kg]:
mA
mB
-
-
molar mass [kg/kmol]:
MA = NA mA
MB = NA mB
-
-
mass density [kg/m3 ]:
ρA = cA MA
ρB = cB MB
ρ = ρA + ρB
-
mass fraction [-]:
wA = ρA /ρ
wB = ρB /ρ
wA + wB = 1
Table 1.2: Physical quantities for a multi-component system that exists of k substances Si .
-
amount of particles [kmol]
:
substance i
mixture
Ni
N=
k
X
Ni
i
-
molar density [kmol/m3 ]:
molar fraction [-]:
ci = Ni /V
xi = ci /c
k
X
c=
k
X
ci
i
xi = 1
i
-
molecular mass [kg]:
mi
-
-
molar mass [kg/kmol]:
Mi = NA mi
-
-
[kg/m3 ]:
ρi = ci Mi
ρ=
-
massa density
massa fraction [-]:
wi = ρi /ρ
k
X
i
k
X
ρi
i
wi = 1
Introduction
13
single-component system
multi-component system
Figure 1.1: Examples of a single and a multi-component system.
Example 1.2: Molar concentration and mass fraction of glucose solution
We pore water to 10g glucose until we obtain exactly 250 ml glucose solution.
The molecular weight of glucose is 180,16 kg/kmol. Consequently, the molar
concentration is:
cg = 10 g
1.1.3
1 mol
1
= 0.22 mol/l
180.16 g 0.250 l
Temperature
The particle constituents, such as molecules, atoms, and subatomic particles of a material body are always in motion. Roughly spoken the temperature is a measure for
the average (sub)microscopic kinetic energy associated with this motion. A collision
between a particle with high kinetic energy and a particle with low kinetic energy
will transfer energy to the particle with the lower kinetic energy. Consequently,
two bodies in thermal contact with different temperature will transfer energy (heat)
from the high temperature body to the low temperature body until the temperature
is equalized and the bodies are in, what is called, thermal equilibrium. In thermodynamics, the definition of temperature is more subtle and related to the relation
between entropy and internal energy. This more precise definition is beyond the
scope of this course (more details are given in the course 8VM50 Thermodyamics).
Here the temperature is an intensive scalar quantity that in general depends on
spatial coordinates and time, i.e. T = T (~x, t)
1.1.4
Pressure
Also the pressure has a formal thermodynamic definition which we will not consider
in this course. Here we simply define the pressure as a force acting per unit area A
of a surface Γ, i.e.
p=
Fn
A
(1.5)
14
transport physics
with p the pressure, Fn the magnitude of the total normal force, and A the area the
force is acting on. Since the force F and/or the normal direction to the surface area
A can vary from location to location, it is better to define the pressure as:
F~n = Fn~n = (F~ · ~n)~n =
ZZ
p~ndΓ ,
(1.6)
A
~ the total force acting on surface Γ and ~n the unit normal to Γ. The pressure
with F
is a scalar quantity that can depend on space and time. i.e. p = p(~x, t). We will
also use the pressure for situations in which no actual surface area is present. For
example the pressure at some point within a fluid. In that case the pressure is
related to the dynamic or hydrostatic stresses in the fluid. This will be explained in
detail in section 1.3.4 of this chapter.
1.1.5
State equations and continuum hypothesis
A state equation is an equation that relates different state variables. For instance
for an ideal gas we have the equation of Boyle and Gay-Lussac:
NT
.
(1.7)
V
Here the pressure p is related to the amount of moles N , the absolute temperature
T , and the volume V . The constant R is the gas constant and given by R =
8.3145 J·mol−1 K−1 . From the state equation for an ideal gas we can easily see that
it can be compressed by increasing the pressure at a constant temperature, indeed
pV = const at constant temperature. For solid materials and liquids this situation
is completely different and other equations of state must hold. These, however, are
far more complex, for example, on account to their elastic and viscous properties
respectively.
Most substances can exist in two states: solid and fluid. A solid is characterized
by a more or less preferred shape (reference state). It can be deformed to a certain
amount by external forces but will return to its reference state when the external
forces are removed. A solid behaves purely elastic if it completely returns to its
original shape. Liquids and gasses –which we will refer to as fluids– on the contrary,
will continuously deform as long as external forces persist and will not return to a
preferred shape if the external forces are taken away. A purely viscous fluid will
keep its new shape completely once external forces are released. Many substances,
however, are visco-elastic and show some intermediate behaviour. Of course this
behaviour depends on the mutual interaction between the molecules on microscopic
scale. On macroscopic level we observe a statistical average result of these molecular
interactions. Therefore, we can consider the fluid as a continuum of small fluid
particles (or fluid elements) that consist of many molecules of which the individual
motion can be replaced by a continuous average. The continuum hypothesis states
that:
p=R
- the smallest relevant length and time scales in a deforming continuum are that
much larger than the molecular scales that we can describe the deformation on
the basis of a statistical average of the motion on molecular scale.
Introduction
15
- on the other hand, the smallest relevant length and time scales in a deforming
continuum are that much smaller than the macroscopic scales that they can be
assumed to be infinitesimally small in mathematical sense.
Based on the continuum hypothesis we can assume that a fluid consists of an infinite
number of fluid particles and that we can take the mathematical limits ∆x → 0 and
∆t → 0 without arriving at the molecular level with length scale ∆λ and time scale
∆τ . In Figure (1.2) this is demonstrated.
Ω
∆λ
∆z
∆x
∆y
Figure 1.2: A continuum fluid element ∆x, ∆y, ∆z that is infinitesimally small for
volume Ω but still large as compared to the molecular scale ∆λ.
Physical quantities such as density, temperature, pressure, and viscosity then are
continuous and well defined in time and space, i.e. ρ(~x, t), T (~x, t), p(~x, t), and
η(~x, t), respectively.
1.1.6
Viscosity
Although fluids can be deformed relatively easily as compared to solids, they still
show a resistance to deformation. A measure for this resistance is the viscosity of
the fluid. Consider a simple experiment as schematically depicted in Figure (1.3).
Between the two parallel plates at a distance H apart from each other a viscous fluid
is initially at rest and a color pattern as indicated in the left panel of Figure (1.3)
is present. At a certain moment the upper plate starts moving with a velocity U
in positive x direction. What can be observed is that the color pattern will deform
as indicated in the right panel of the figure. At the walls the fluid moves with the
same speed as the wall. This is the result of what is known as the no-slip condition,
stating that the fluid at the wall does not slip and moves with the same speed as
the wall. So at the bottom vx = 0 and at the top vx = U . Later in this course we
will learn that for U not too high, the velocity between the plates is uni-axial (only
in x direction) and linear increases from 0 to U , so vx = (y/H)U . So, it looks as
16
transport physics
if the fluid is flowing in layers that each move with a different velocity but without
exchange of fluid from one layer to the other. This is why this kind of flow is referred
to as laminar flow as opposed to turbulent flow where a random exchange of fluid
(and kinetic energy) between the layers is found.
In the laminar flow regime, each layer will experience a shear stress τ from the
adjacent layers, which results in the deformation as mentioned. This shear stress
can be compared to the stress you need to smoothly rub some sun tan lotion on your
skin. It is clear that the shear stress needed to deform each layer will depend on the
difference in velocity ∆vx of the adjacent layers. In the extreme case that all layers
move with the same velocity, the shear stress and the deformation will be zero. On
the other hand, the thicker the layer, the higher the shear must be to deform it. So,
τ = τ(
∆vx
),
∆h
(1.8)
with ∆h the thickness of the layer. This must also hold for very thin layers so,
τ = τ(
dvx
).
dy
(1.9)
For a Newtonian fluid the relation between the shear stress τ and the shear rate
γ̇ = dvx /dy is linear and the coefficient of proportionality is defined as the dynamic
fluid viscosity η, i.e.:
τ =η
dvx
.
dy
(1.10)
The dimension of the dynamic viscosity is Pa·s. Many fluids, such as water, gasses,
and glycerine, satisfy this linear relation between shear stress and shear rate. The
viscosity of water at room temperature is ηw ≈ 1 · 10−3 Pa·s. Many other fluids,
though, such as blood, paint, polymer melts, and ketchup, show slightly to completely different and non-linear behavior. In these cases fluids behave non-Newtonian.
Blood, for instance, is slightly shear-thinning which means that its viscosity decreases at higher shear rates. The sub-discipline in which the behaviour of nonNewtonian fluids is studied, is known as rheology. At not-too-high shear rates,
however, blood can be considered as a Newtonian fluid with a viscosity that is 4 to
5 times higher than water (so ηb ≈ 4.5·10−3 Pa·s).
τ
y=H
color
pattern
y
x
y=0
U
y
x
velocity profile vx (y)
Figure 1.3: Viscous flow between two parallel flat plates.
Introduction
17
Example 1.3: Viscosity of a fluid
A fluid is enclosed between two parallel vertical plates with surface area
A = 10−2 m. The distance between the plates amounts h = 1 mm. The upper
plate is moved in horizontal direction with a velocity vx = 0.1 m/s. The force
needed to move the plate amouts F = 0.005 N. The viscosity of the fluid
enclosed between the plates can now be computed as follows.
Assuming that, due to the short distance between the plates, the velocity has a
linear change from the lower plate to the upper plate, the shear rate is:
γ̇ =
vx
dv
=
.
dy
h
The shear stress τ is the force per unit area so:
τ=
F
.
A
Consequently, the viscosity is:
η=
τ
Fh N m
Fh
=
=
Pa · s .
2
γ̇
Avx m m/s
Avx
Substitution of the values of the parameters in SI units gives:
η=
1.2
0.005 · 10−3
= 5 · 10−3 Pa · s .
10−2 · 0.1
Fluid Kinematics
1.2.1
Velocity field
The velocity ~v of a fluid particle, or material point, is a vector quantity that represents the magnitude of the velocity as well as its direction. In a Cartesian coordinate
system with standard basis vectors, ~ex , ~ey , ~ez this gives:
~v = vx~ex + vy~ey + vz ~ez ,
(1.11)
with vi the components of the velocity (i = x, y, z). The magnitude of the velocity
then is defined by:
q
√
(1.12)
k~v k = ~v · ~v = vx2 + vy2 + vz2 .
In appendix A.1.1 Cartesian, cylindrical, and spherical coordinate systems, that can
be used to describe the velocity fields, and their mutual relations are defined.
The velocity direction then is given by the unit vector
~ev =
~v
k~v k
(1.13)
18
transport physics
wind direction
20 April 2010
north-westerley
wind direction
24/25 April 2010
south-westerley
Figure 1.4: Velocity field representing the wind at different locations. As can be seen
the magnitude and direction of the wind dependent on both location and time (source:
http://news.bbc.co.uk).
If we attribute a fluid particle, or material point, to each arbitrary position ~x in a
3-dimensional space we can define the velocity field as:
~v = ~v (~x, t) .
(1.14)
The space and time dependence of the velocity field is obviously present in the
weather forecasting pictures as given in Figure 1.4.
1.2.2
Eulerian and Lagrangian representations
Note that a velocity field such as is shown in the example of the wheater forecast
is a snapshot at a certain moment in time. For each vector ~v (~x, t) it tells us what
the velocity is of a fluid particle that accidentally is at the location ~x = ~r of the
tail of that vector at that instant of time t. At a later moment in time the vector
with its tail at the same position represents the velocity of a different fluid particle.
With ~v (~x, t) the velocity field is described in a what is called Eulerian way. Also
other variables, such as the pressure and the temperature, can be written based on
their value at a fixed location ~x, so p(~x, t) and T (~x, t). To be able to clearly relate
the velocity field to the motion of individual fluid particles through space and time,
these fluid particles must be uniquely identified and followed through time. In a
continuum approach, in all relevant locations in a certain volume Ω fluid particles
are present and fluid particles are non-overlapping. Consequently, the position ~x = ξ~
Introduction
19
z
path of particle ξ~
~ t0 )
ξ~ = ~r(ξ,
~ t)
~x = ~r(ξ,
~ t)
~v (~x, t) = ~v (ξ,
y
x
Figure 1.5: Velocity in Eulerian (~v (~x, t): velocity at time t at location ~x) and Lagrangian
~ t): velocity at time t of particle ξ)
~ representation.
(~v (ξ,
of a (fluid) particle at some initial time t0 can be used to uniquely label this particle.
~ t), with ~r the position vector of
The position of a particle ξ~ then is given by ~x = ~r(ξ,
~ t0 ) = ξ.
~ The
particle ξ~ at time t. Of course the particle position ~r must satisfy ~r(ξ,
~
velocity of particle ξ is determined by the rate of change of its position in time and
must be equal to the velocity in Eulerian description, ~v (~x, t) at the same location.
So,
~ t + ∆t) − ~r(ξ,
~ t)
~ t)
~r(ξ,
d~r(ξ,
=
∆t→0
∆t
dt
~ t) = lim
~v (ξ,
= ~v (~x, t) .
(1.15)
ξ~
Note that in (1.15) the velocity is defined along the particle path and hence the
particle identity ξ~ is kept constant. Also other variables, such as the pressure and
the temperature, can be written in a Lagrangian description,based on their value at
~ so p(ξ,
~ t) and T (ξ,
~ t).
a fixed particle ξ,
If we follow a particle (so ξ~ is fixed), an arbitrary physical quantity F = F (~x, t) =
~ t), t) will change its value in time according to1 :
F (~r(ξ,
~ t), t)
dF (~r(ξ,
dt
=
ξ~
=
1
∂F drx
∂F
+
∂t
∂rx dt
ξ~
+
∂F dry
∂ry dt
+
ξ~
∂F drz
∂rz dt
ξ~
(1.16)
∂F
∂F
∂F
∂F
+
vx +
vy +
vz ,
∂t
∂x
∂y
∂z
More information on derivatives of multi-variable functions can be found in appendix A.1.4.
20
transport physics
and thus, with the commonly used notation
DF
Dt
∂F
∂t
=
dF
dt
ξ~
+ ~v · ∇F
≡
DF
, we get:
Dt
.
(1.17)
material
= local
+ convective
derivative
derivative
derivative
The operator
∂
D
=
+ ~v · ∇
Dt
∂t
(1.18)
is called the material derivative and computes the change of physical quantity (F )
for an observer that travels with the particle. The first term, ∂/∂t, is called the local
derivative and represents the change of F for an observer that is sitting on a fixed
location ~x. The second term, ~v · ∇, is the convective derivative2 and represents the
change of F due to the fact that the fluid particle is travelling with velocity ~v and F
is non-homogeneous, i.e. ∇F 6= 0. The material derivative represents the temporal
as well as spatial variations of F .
Note that the above also holds for each of the components of the velocity. This
means:
D~v
Dt
=
∂~v
∂t
+ (~v · ∇)~v
.
(1.19)
material
= local
+ convective
acceleration
acceleration
acceleration
Example 1.4: Acceleration of a particle
A particle moves in a velocity field given by:
~v = αt~ex + βy~ey + γx~ez .
The acceleration of the particle than is given by:
D~v
∂~v
=
+ (~v · ∇)~v
Dt
∂t
∂
∂
∂
= α~ex + (vx~ex + vy ~ey + vz ~ez ) · (~ex
+ ~ey
+ ~ez ) ~v
∂x
∂y
∂z
∂
∂
∂
= α~ex + αt
+ βy
+ γx
(αt~ex + βy~ey + γx~ez )
∂x
∂y
∂z
= α~ex + β 2 y~ey + αtγ~ez .
2
In some literature the terms material derivative and convective derivative are both used for
D/Dt.
Introduction
1.2.3
21
Pathlines, streamlines, and streaklines
From (1.15) it directly follows that the trajectory or pathline of an individual particle
that is located at ~x = ξ~ at t = t0 follows from integration of the velocity field
according to:
~ t) − ~r(ξ,
~ t0 ) =
~r(ξ,
Zt
~v (~x, t)dt ,
(1.20)
t0
~ t0 ). An example of such a pathline is already given in Figure 1.5. In fact
with ξ~ = ~r(ξ,
relation (1.20) gives the relation between the Lagrangian and Eulerian description
of the flow field. Once a particle is at position ~x = ~r the next point at the pathline
can be constructed from dx = vx dt, dy = vy dt, and vz = vz dt. So we have:
dx
= dt ,
vx (~x, t)
dy
= dt ,
vy (~x, t)
dz
= dt .
vz (~x, t)
(1.21)
Another way to describe the velocity field is to make use of streamlines. A streamline
is a line of which the tangent in each point has the same direction as the velocity
~v (~x, t). Consequently there is no motion of fluid particles normal to a streamline.
Note that the velocity magnitude k~v k along a streamline doesn’t need to be constant. In a steady flow streamlines and pathlines overlap so fluid elements move
along streamlines. Fluid particles between two streamlines will stay between these
streamlines. In an unsteady (time dependent) flow streamlines continuously change
shape (see Figure 1.6).
~v
t = t1
t = t2
~v
Figure 1.6: Streamlines.
The physical meaning of what we call the stream function and its relation to the
streamlines becomes clear from Figure 1.7.
From the geometry we see:
tan α =
vy
dy
=
dx
vx
or
dx
dy
=
.
vx
vy
(1.22)
So generalized to 3D, along a streamline at t = t0 we have:
dx
dy
dz
=
=
.
vx (~x, t0 )
vy (~x, t0 )
vz (~x, t0 )
(1.23)
For the description of the streamline pattern of a steady flow in 2D in a Cartesian coordinate system it is convenient to introduce the stream function ψ(x, y). A
22
transport physics
y
vy ~ey
~v
x
α
segment dl
streamline
vx~ex
dl
dy ∝ vy
α
dx ∝ vx
Figure 1.7: Streamline in x, y-plane.
streamline then is represented by ψ(x, y) = ψ0 with ψ0 a constant. Other streamlines are described with other constants ψi .
The total differential of ψ(x, y) is:
dψ =
∂ψ
∂ψ
dx +
dy .
∂x
∂y
(1.24)
Along a streamline the stream function is constant: ψ = ψ0 , so, dψ = 0. Consequently, along a streamline we have:
dψ =
∂ψ
∂ψ
dx +
dy = 0 .
∂x
∂y
(1.25)
From (1.23) we also find
−vy dx + vx dy = 0 .
(1.26)
Combining (1.25) and (1.26), the velocity can be computed from the stream function
according to:
vx =
∂ψ
,
∂y
vy = −
∂ψ
.
∂x
(1.27)
and thus:
~v =
∂ψ
∂ψ
~ex −
~ey ,
∂y
∂x
(1.28)
The vector field ~v thus can be represented by the stream function given by the scalar
field ψ = ψ(x, y).
Finally, a streakline is defined by the curve that connects the location of particles
that have passed through a particular point in the past. If we inject a dye at a fixed
point in a fluid flow, it will produce a streakline.
The differences between streamlines, pathlines and streaklines are depicted in figure
1.8. This figure must be studied very well and in detail to fully understand these
differences.
Introduction
23
velocity
streamlines
pathlines
streaklines
Figure 1.8: Illustration of streamlines, pathlines, and streaklines in the case of a varying
flow around a chimney.
Example 1.5: Streamlines, pathlines, and streaklines
A velocity field is given by:
~v =
y
x
~ex +
~ey .
t0 + t
t0 + 2t
Along a streamline we have:
−vy dx+vx dy = 0 →
y
x
1
1
dx =
dy → (t0 +t) dx = (t0 +2t) dy .
t0 + 2t
t0 + t
x
y
Integration yields:
(t0 + t)(ln x + C1 ) = (t0 + 2t)(ln y + C2 ) .
with C1 = − ln(x0 ) amd C2 = − ln(y0 ) we get:
(t0 + t) ln
y
x
= (t0 + 2t) ln
.
x0
y0
Consequently we have streamlines given by:
y = y0
x
x0
n
with n =
t0 + t
.
t0 + 2t
24
transport physics
1.2.4
Flow rate and flux
If we consider a known velocity field ~v (~x, t) for ~x ∈ IR3 we could be interested in the
volume of fluid VΓ that passes a certain surface area Γ during a certain timespan
∆t (see Figure 1.9). To compute this volume, we divide surface Γ in a number of
infinitesimally small surface areas dΓ with normal ~n. Since dΓ is that small, we can
assume that the velocity ~v (~x, t) is constant in dΓ.
Γ
(~v · ~n)~n
~n
~v (~x, t)
dΓ
(~v · ~n)~n
~v − (~v · ~n)~n
dΓ
(~v · ~n)∆t
dV = (~v · ~n)∆tdΓ
Figure 1.9: Flow through a surface Γ in a given flow field ~v (~x, t) and the details at
location dΓ.
Next, we must realize that the velocity ~v (~x, t) in general is not normal to surface
area dΓ. As a result, fluid particles pass area dΓ with a velocity equal to ~v · ~n: the
normal component3 of ~v . Therefore, during an infinitesimally small timespan ∆t the
total volume of fluid that passes dΓ is equal to:
dV = (~v · ~n)∆tdΓ .
(1.29)
Finally, to compute the total volume of fluid that passes surface Γ we have to add
the contributions of all (say N ) infinitesimal parts dΓ:
V =
N
X
dVi = ∆t
i
N
X
l
(~v · ~n)dΓi .
(1.30)
In mathematics, this sum is called a Riemann sum. If the surface area’s dΓi approach
zero and surface Γ is smooth such that (~v · ~n) is a bounded piecewise continuous
function, we have:
N
X
i
3
(~v · ~n)dΓi ≡
ZZ
Γ
~v · ~ndΓ .
(1.31)
In appendix A.1.2 more information can be found on the inner product and vector projections.
Introduction
25
The volumetric flow rate qV , defined as the volume of fluid that passes surface Γ per
unit of time is equal to V /∆t and, based on what we learned above, is equal to:
qV =
ZZ
Γ
(~v · ~n)dΓ ,
(1.32)
with ~n the unit vector normal to Γ. Note that the dimension of qV is m3 /s. Taking
in mind that the mass of a certain volume of fluid is equal to the volume multiplied
with the density, i.e. dm = ρdV , the mass flow rate is:
qm =
ZZ
Γ
ρ(~v · ~n)dΓ .
(1.33)
In the sequel of this course we will use this expression to derive the balance of mass
equations. The volume flow rate qV and mass flow rate qm are also referred to as
volume flux and mass flux, respectively.
Example 1.6: Volume flux
A velocity field is given by:
~v =
x
x−y
y
~ex + ~ey +
~ez .
t0
t0
t0
The normal velocity to the surface given by Γ : {0 ≤ x ≤ 1, 0 ≤ y ≤ 2, z = 0} is
given by:
~v · ~n = ~v · ~ez =
x−y
.
t0
The volume flux through Γ then reads:
qV =
ZZ
Γ
(~v · ~n)dΓ
and thus:
qV =
Z2 Z1
y=0 x=0
1
x−y
dydx =
t0
t0
Z2
y=0
1
( 21 x2 − xy) dy .
0
This yields:
1
qV =
t0
Z2
y=0
( 21 − y)dy =
1 1
( y − 21 y 2 )
t0 2
2
0
=−
1
.
t0
Lastly, the average velocity v̄ of fluid that passes through surface Γ can be derived
26
transport physics
from the volume flow rate qV and the surface area AΓ . This yields:
v̄ =
qV
=
AΓ
RR
Γ
(~v · ~n)dΓ
RR
dΓ
.
(1.34)
Γ
1.3
Deformation and stress
1.3.1
Velocity gradients
We now consider a rectangular fluid element (infinitesimally small), with dimensions
(dx, dy, dz) in a velocity field ~v . We are interested in de velocity in the corner points
A and B (see Figure 1.10).
A
~v (~x)
~x
B
~v (~x + d~x)
~x + d~x
O
Figure 1.10: Motion and deformation of a rectangular fluid element. The z-direction
is perpendicular to the plane of view.
Now the velocity at location ~x + d~x can be written as a Taylor expansion around
~v (~x):
~v (~x + d~x) = ~v (~x) +
∂~v (~x)
∂~v (~x)
∂~v (~x)
dx +
dy +
dz + ...
∂x
∂y
∂z
(1.35)
or:
~v (~x + d~x) = ~v (~x) + ∇~v (~x) · d~x + ...
(1.36)
with:
∇~v (~x) = ~ex
∂~v (~x)
∂~v (~x)
∂~v (~x)
+ ~ey
+ ~ez
∂x
∂y
∂z
(1.37)
In (1.36) the ~v (~x) term represents a pure translation while the ∇~v (~x) · d~x term
represents rotation, deformation and dilatation. The term (∇~v )c (~x) is called the
velocity gradient tensor L. Here (∇~v )c denotes the conjugate of (∇~v ). We will use
the conjugate of a tensor as the tensor that results in the transposed matrix in case
the tensor is represented by a matrix in a chosen coordinate system.
Introduction
27
In a Cartesian coordinate system (x, y, z) the velocity gradient tensor can be represented by the following matrix:

∂vx
 ∂x



 ∂v

y
L≡
 ∂x



 ∂vz
∂x
∂vx
∂y
∂vy
∂y
∂vz
∂y

∂vx
∂z 



∂vy 

 .
∂z 



∂vz 
(1.38)
∂z
The flow field IRn has n components of which each can vary in n directions. The
velocity gradient therefor includes n2 components and is called a 2nd order tensor
that can be represented by a n × n matrix.
Example 1.7: Velocity gradient
If the velocity field is given by:
~v = (x2 + z)~ex + (z + y)~ey + (x2 + y)~ez .
The velocity gradient tensor then is:

∂vx
 ∂x



 ∂v

y
L=
 ∂x



 ∂vz
∂x
1.3.2
∂vx
∂y
∂vy
∂y
∂vz
∂y

∂vx
∂z 


2x 0

 

∂vy 
 
= 0 1

∂z 
 


2x 1
∂vz 
∂z
1




1  .


0
The divergence and rotation of a vector field
In addition to the velocity gradient, two other important vector operations, that
have a physical meaning in transport physics, are the divergence ∇ ·~v (or div ~v ) and
the rotation ∇ × ~v (or rot ~v ). In Cartesian coordinates these two vector operators
are defined as:
div ~v = ∇ · ~v =
∂vz
∂vx ∂vy
+
+
∂x
∂y
∂z
(1.39)
28
transport physics
and
rot ~v = ∇ × ~v =
=
~ex
~ey
~ez
∂
∂x
∂
∂y
∂
∂z
vx
vy
vz
∂vz
∂vy
−
~ex −
∂y
∂z
(1.40)
∂vx
∂vz
−
~ey +
∂x
∂z
∂vx
∂vy
−
~ez
∂x
∂y
Divergence
The divergence of a velocity field at point ~r is a measure that indicates the rate at
which the vector field diverges from ~r. This divergence is measured by computing the
volume flux per unit volume out of an infinitesimal volume (sphere Sǫ (~r)) centered
at location ~r:
∇ · ~v (~r) = lim
ZZ
ǫ→0
Sǫ (~
r)
(~v · ~n)dS
(1.41)
The proof of this statement is given in appendix A.1.5. From (1.41) it is clear that
the divergence of ~v at location ~r is a measure for the volume flux in the direction
away from ~r. This means that if |∇ · ~v | > 0 location ~r acts as a volume flow source
or sink.
A velocity field with a divergence equal to zero (∇ · ~v = 0) is called divergence free.
Later we will learn that the velocity field of an incompressible fluid is divergence
free.
Rotation
The rotation of ~v is a vector that is perpendicular to ~v . In the next section we
consider the deformation of fluid elements due to gradients in the velocity field. We
will learn that the rotation ∇ × ~v is associated to the rotational components of the
deformation.
Figure 1.11 illustrates the different values that the divergence and rotation can take
for a given complex velocity field. For reference different vector identities that can
be used to find relations between vector operations are given in appendix A.1.9.
Figure 1.11: Vector field ~v = cos(xy)~ex +sin(x2 −xy)~ey (a), corresponding divergence
∇ · ~v (b) and the z-component of the rotation ωz = (∇ × ~v ) · ~ez (c).
Introduction
1.3.3
29
State of deformation
Due to gradients in the flow field the infinitesimal fluid element of Figure 1.10 will
not only change its position but will also change its orientation and shape (see Figure
1.12). In general these effects are combined and can lead to a complex deformation.
translation
~v = ~ex
rotation
~v = −y~ex + x~ey
~v = y~ex
shear
~v = x~ex − y~ey
stretch
expansion
~v = x~ex + y~ey
Figure 1.12: Change of position, orientation and shape of a fluid element for translation,
rotation, simple shear, pure stretch, and isotropic expansion, respectively.
Rate of deformation
The velocity gradient tensor L = (∇~v )c can be decomposed in a symmetric part D,
the rate of deformation tensor and an anti-symmetric part Ω, the rotation tensor:
L=D+Ω
with
D = 21 (L + Lc ) and
Ω = 21 (L − Lc ) .
(1.42)
In Cartesian coordinates this results in the following matrix representations:







D=





∂vx
∂x
1 ∂vx
2 ( ∂y
1 ∂vy
2 ( ∂x
+
∂vx
)
∂y
1 ∂vz
2 ( ∂x
+
∂vx
)
∂z
∂vy
+
)
∂x
∂vy
∂y
1 ∂vz
2 ( ∂y
+
∂vy
)
∂z
1 ∂vx
2 ( ∂z

∂vz
+
)
∂x 



∂vz 

1 ∂vy
2 ( ∂z + ∂y ) 





∂vz
∂z
(1.43)
30
transport physics
and







Ω=





1 ∂vx
2 ( ∂y
0
1 ∂vy
2 ( ∂x
−
∂vx
)
∂y
1 ∂vz
2 ( ∂x
−
∂vx
)
∂z
∂vy
−
)
∂x
0
1 ∂vz
2 ( ∂y
−
∂vy
)
∂z
1 ∂vx
2 ( ∂z

∂vz
−
)
∂x 



∂vz 

1 ∂vy
.
2 ( ∂z − ∂y ) 





(1.44)
0
A compact notation, often found in literature and therefore just mentioned here, for
D and Ω in (1.43) and (1.44) using the ‘Einstein-summation convention’ yields:
1
Dij =
2
∂vj
∂vi
+
∂xj
∂xi
!
(1.45)
1
Ωij =
2
∂vj
∂vi
−
∂xj
∂xi
!
(1.46)
The rate of deformation D is associated with deforming forces and contains dilatation and stretch (diagonal-elements) as well as shear (off-diagonal elements). The
rotation matrix Ω represents only rotation (vorticity) and, just like translation, is
not associated with deforming forces.
Note that with the definition of the rotation of the velocity field (1.40) and the
definition the vorticity ω
~ = ∇ × ~v we can write:


0
−ωz ωy

1
Ω = 2  ωz
0
−ωx  .
−ωy ωx
0
(1.47)
A velocity field is called rotation free in the case that ω
~ = ~0. Note that in a 2D flow
the vorticity has only one non-zero component: ~ω = ωz ~ez .
During deformation such as shear and stretch deforming shear and elongational
forces are involved (see Figure 1.13). Forces acting on the boundary of fluid elements
will not only induce deformation but will also result in internal stresses. In the next
section the state of stress of fluid elements will be defined more precisely.
shear
stretch
Figure 1.13: Strain and stretch imply deforming forces (stresses).
Introduction
31
Example 1.8: Simple shear flow
Simple shear flow in x direction is defined by:
~v = γ̇y~ex ,
with γ̇ the shear rate. Verify that D and Ω are given by:


0 γ̇ 0

1
γ̇
0 0 
D= 2
0 0 0
1.3.4
and


0 γ̇ 0

1
−
Ω = 2  γ̇ 0 0  .
0 0 0
State of stress
Deforming forces
In the previous section we learned that fluid elements in a flow can undergo a deformation as a result to velocity gradients expressed in the velocity gradient tensor
L = (∇~v )c . These deformations will result in stresses in the fluid and vice versa.
The relation between ∇~v and the stress is called a constitutive relation. In general
it is not easy to derive such a constitutive relation from the physical and chemical
properties of the fluid and empirical relations are used. The most simple relation
between stress and rate of deformation was introduced by Newton who stated that
stress and rate of deformation are linearly dependent with the constant of proportionality equal to the viscosity of the fluid.
Before we mathematically define this relation in more detail we first consider what
kind of forces can act on a fluid element. In general we can distinguish two types of
forces:
Volume or body forces. These forces act on all fluid particles inside the fluid
domain considered and ”work from a distance” without physical contact. Examples are given in Figure 1.14:
~grav = ρ~g δV , with ρ the density of the fluid, ~g the gravi- de gravity force F
tational acceleration, and δV the volume of the fluid element.
- electrostatic, magnetic, and electromagnetic forces
Surface forces. These are traction forces ~t that act on or in surface area elements
by the surrounding of the fluid domain considered (see Figure 1.15). They are
mostly defined per unit of area so as surface stresses. Examples are:
- normal- and shear forces
- surface tension
32
transport physics
~n
Ω
F~grav
stress vector ~t
~g
normal stress σn
dS
y
shear stress (σx , σy )
x
Figure 1.14: Volume forces.
Figure 1.15: Surface forces.
As indicated in Figure 1.15, the surface stress can be decomposed in a normal stress
σn (perpendicular to the surface area) and two shear stress components σx and σy .
Although these forces only act on the surface they will be sensed in the complete
domain. Each (infinitesimal) fluid element will experience normal- and shear forces
of neighbouring fluid elements and transmit it to adjacent fluid elements.
As depicted in Figure 1.16 we can define the stress components on an infinitesimal
cubic fluid element and assign these 9 stress components to the point located in the
center of the element. In Cartesian coordinates we can define the stress matrix:


σxx σxy σxz


σ =  σyx σyy σyz 
σzx σzy σzz
(1.48)
This is a Cartesian representation of what in general is defined as the Cauchy stress
tensor σ. In this representation σ contains 3 × 3 elements. The indices of the
elements of the stress tensor σ are defined such that:
- the direction in which the stress component works corresponds with the 2nd index.
- the orientation of the plane on or in which the stress component works corresponds
with the 1st index
So σxy is a stress in y-direction (2nd index) acting on the plane with normal in the
x-direction (1st index).
z
σzz
σzx
σzy
σyz
σxz
σxx
σxy
σyx
σyy
y
x
Figure 1.16: Stress components on the surface of an infinitesimal cubic fluid element.
Introduction
33
Note that stress components with equal indices are normal stresses while unequal
indices are shear stresses.
With this definition of the Cauchy stress tensor the stress vector ~t can be written
as:
~t = σ · ~n
(1.49)
For example, the stress vector ~t on the frontal plane in figure 1.16 has three 3
components, The outer normal is ~n = ~ex , so:
~t = σ · ~n = σxx~ex + σyx~ey + σzx~ez .
(1.50)
Here, σxx is the normal stress component and σxy and σxz are the shear stress
components.
Cauchy’s law
The stress tensor actually defines the state of stress of a medium. To proof this, it
needs to be introduced more formally. Consider a medium (solid or fluid) inside an
arbitrary closed volume Ω with boundary Γ. Assume a number of external forces
F~i are acting on Γ imposing internal stresses in the medium. Suppose we would
cut the volume Ω in two parts Ω1 and Ω2 . In order to keep the state of stress of
both parts identical to the state of stress before the cutting, at each surface element
∆S of cutting plane S a traction force or stress vector ~t needs to be introduced
compensating for the distributed force part Ω2 was exerting on part Ω1 (see Figure
1.17).
We now consider the tetrahedron ∆Tc that can be constructed by taking the origin
of a Cartesian coordinate system at the location indicated in Figure 1.17. The sides
of ∆Tc are defined as ∆S, ∆Sx , ∆Sy , and ∆Sz . The surface area of ∆S can be
computed using the properties of the cross product:
~n =
sy sz ~ex + sx sz ~ey + sy sx~ez
(sy ~ey − sx~ex ) × (sz ~ez − sx~ex )
=
,
||(sy ~ey − sx~ex ) × (sz ~ez − sx~ex )||
2∆A
(1.51)
with ∆A the surface area of ∆S. From this, taking the dot product with ~ex , we get:
2∆A~n · ~ex = 2∆A nx = sy sz = 2∆Ax .
(1.52)
with ∆Ax the surface area of ∆Sx . Consequently, applying the above in different
directions, we have:
∆A =
∆Ay
∆Az
∆Ax
=
=
.
nx
ny
nz
(1.53)
Since we have three right angles at the origin, the volume ∆V is given by:
1
∆V = sx sy sz .
6
(1.54)
34
transport physics
z
sz
∆Sx
∆Sy
F~n
~t
F~1
Γ
~t
σyx
F~2
~n
sy
σyy
σzy
σyz
Ω2
Ω1
~n
σzx
σxy
∆S
σxx
σxz
∆Sz
y
σzz
S
∆S
sx
F~i
Volume Ω
x
Cauchy tetrahedron ∆Tc
Figure 1.17: Stress vector ~t and normal vector ~n on cutting plane S through volume
Ω (left) and stresses acting on an infinitesimal surface elements ∆Sx , ∆Sy , ∆Sz , and
∆S of the Cauchy tetrahedron ∆Tc (right).
Newton’s third law (action equals minus reaction) tells us that the traction per unit
area exerted by the surrounding on each of the sides of the tetrahedron is opposite
to the surface stress given by the stress components defined by the Cauchy stress
tensor. Newton’s second law in x-direction applied to the tetrahedron assuming a
body force per unit volume ρf~ gives:
tx ∆A − σxx nx ∆A − σyx ny ∆A − σzx nz ∆A + ρfx ∆V = ρax ∆V ,
(1.55)
with ρ the density of the medium and ax the acceleration in x-direction. If we take
the tetrahedron Tc infinitesimally small we have ∆V /∆A = 13 sx nx → 0 and thus:
tx − σxx nx − σyx ny − σzx nz = 0 .
(1.56)
Applying this also to the other directions gives:
~t = σ c · ~n
or (see next section) ~t = σ · ~n.
This relation is known as Cauchy’s law.
(1.57)
Introduction
35
z
σxz
z
a
a
σxz
σxy
σxx
σzx
y
y
a
x
x
Figure 1.18: The components of the stress vector ~n · σ for ~n = ~ex (left) and the force
momentum about the y-axis of a cubic fluid element (right).
Symmetry of the Cauchy stress tensor
It can be shown that the Cauchy tensor σ is symmetric, so
σ = σc
or
σij = σji .
(1.58)
Without a formal proof this becomes clear if we consider a small cubic domain with
dimensions a × a × a (see Figure 1.18).
Here, σxz is the stress component in z-direction, in plane x = a and σzx is the stress
component in x-direction, in plane z = a.
These stress components produce a moment of force in y-direction according to
My = (σzx − σxz )a3 .
(1.59)
This will produce an angular acceleration ω̇ equal to ω̇ = My /J with J = 32 ρa5
the moment of inertia of a cube about one of its edges. This would mean that the
angular acceleration would become infinite for a → 0. As this is physically not
possible we must have σzx = σxz to make My = 0.
A consequence of the symmetry of the Cauchy stress tensor is that:
σ · ~n = ~n · σ .
(1.60)
This property will be used later in the course to simplify certain equations.
1.3.5
Constitutive relations
We come back to Newton’s postulate that the stress is linearly related to the rate
of deformation:
σ = σ(∇~v ) .
(1.61)
In case the fluid is at rest (so ∇~v is zero), the stress is completely determined by
the pressure p. The pressure is isotropic (has no directional dependence) so at rest
σ = −pI with I the unity tensor. To distinguish the pressure contribution to the
stress tensor from the velocity gradient dependent part we write:
σ = −pI + τ .
(1.62)
36
transport physics
Here τ is called the extra-stress tensor and depends on the velocity gradient L. Since
pure rotation does not generate stresses in the fluid, the velocity gradient dependence
of the extra-stress τ can be written as τ (D) with D the rate deformation tensor
defined in (1.42). The rate of deformation tensor contains two contributions. The
first is the divergence of the velocity field ∇ · ~v which is defined by the diagonal
elements of D. The second contribution is associated with the shear forces. If we
assume linear dependence on these two contributions we can write
1
(1.63)
σ = −pI + 2η[D − (∇ · ~v )I] + ζ(∇ · ~v )I
3
Here η is defined as the dynamic viscosity and ζ the bulk viscosity.
In the case the fluid can be considered as incompressible (see next chapter) we have
∇ · ~v = 0 and thus:
σ = −pI + 2ηD
(1.64)
This is the mathematical formulation of Newton’s hypothesis and one of the simplest
constitutive relations (relation between the state of stress and strain(rate)) for fluids.
In Cartesian coordinates we find for incompressible media:
σ = −pI + τ = −pI + 2ηD ,
(1.65)
with:
Pressure:


1 0 0


pI = p  0 1 0 
0 0 1
(1.66)
Normal extra stresses: the diagonal elements of τ describe stretch.
τxx = 2η
∂vx
∂x
τyy = 2η
∂vy
∂y
(1.67)
∂vz
∂z
Shear stresses: the off-diagonal elements of τ describe shear.
τzz = 2η
τxy
τxz
= τyx
= τzx
= τzy
∂vx ∂vy
+
=η
∂y
∂x
∂vx ∂vz
+
=η
∂z
∂x
τyz = η
∂vy
∂vz
+
∂z
∂y
(1.68)
Introduction
37
Example 1.9: Normal and shear forces
In the figure we give two examples to illustrate the relation between normal and
shear forces to stretch and deformation, respectively.
vx (x)
vx (x + dx)
vx (x + dx) = vx (x) +
∂ux
∂x dx
vx (y)
vy (x)
+ ...
strain ∝ τxx
τxx
τxx
(a)
τxy
τyx τxy
τyx
∂vx
∂y
∂vy
∂x
>0
>0
(b)
Stretch of and stress in a rectangular fluid element (a). Deformation of and
stress in a rectangular fluid element (b).
In the sequel of this course we will use the state of deformation and state of stress
and their relation to complete the set of equations that result form balance of mass,
momentum, and energy.
38
transport physics
2
Balance of mass
In this chapter we will first derive the intergral form of the mass balance equations for
material, fixed, and moving domains. With the aid of the Reynolds transport theorem
and the divergence theorem of Gauss we will subsequently derive the local or differential
form of the mass balance equations. This differential form then will be applied to a multicomponent system. For dilute solutions the first and second law of Fick will be used
to derive the convection-diffusion or composition equation for the molar concentration
of a solute in a dilute mixture. Solutions will be given for steady and time dependent
diffusion. Finally some remarks on convection diffusion problems will be made.
39
40
2.1
transport physics
Integral form of the balance of mass
Each system, including physiological systems, is characterized by inflow, outflow,
transfer, and storage of chemical substances. In all these processes mass is conserved,
so total mass will not change. In the next sections we will formalize this balance of
mass for material, fixed and arbitrarily moving and deforming volumes.
2.1.1
Mass balance for a material domain
The amount of fluid is defined by its mass m in a closed domain Ω with volume V and
boundary Ω. Note that surface Γ is defined such that Γ closes the entire domain Ω.
Within the domain, fluid particles can move. If we take the surface Γ of the domain
such that it moves with the fluid particles, in general, the whole domain Ω will move
and deform such that all fluid particles keep enclosed by the surface. In this case, we
talk about a material volume or material domain Ωm (t) with outer surface Γm (t).
Γm (t0 )
Ωm (t0 )
Γm (t)
Ωm (t)
dΩ
dΩ
t = t0
t=t
Figure 2.1: Motion and deformation of a material domain Ωm (t) with volume Vm (t)
enclosed by surface Γm (t).
Figure 2.1 shows an arbitrary material domain at two instants of time. It deforms
and moves but no mass will leave or enter the domain. The mass mm of a material
volume is constant and hereby does not change in time, i.e.:
dmm
=0.
dt
(2.1)
Based on the continuum hypothesis, introduced in the previous chapter, we can
define infinitesimal small volumes dΩ with volume dV and a mass ρdV . Here, ρ is
the local mass density of the fluid, which can depend on the local position in space
~x and on the time t, so ρ = ρ(~x, t). The total mass of the material volume Ωm then
is:
mm (t) =
ZZZ
ρ(~x, t)dΩ .
(2.2)
Ωm (t)
The balance of mass for a material domain then reads:
d
dt
ZZZ
Ωm (t)
ρ(~x, t)dΩ = 0 .
(2.3)
Balance of mass
2.1.2
41
Mass balance for a fixed or moving domain
It is not always practical to work with a material volume Ωm (t) and material surfaces
Γm (t). Therefore, we will transfer the mass balance for material volumes to a mass
balance for a control domain Ω that is either fixed to a chosen coordinate system
with origin O or moving with a certain velocity ~vΓ independent of the velocity ~v of
the particles inside.
For a fixed domain Ω = Ω0 the increase of mass inside the domain must be equal
to the mass influx:
d
dt
ZZZ
Ω0
ρ(~x, t)dΩ = −
ZZ
Γ0
ρ(~x, t)~v · ~ndΓ .
(2.4)
This equation is the integral mass balance for a fixed control volume. Note that we
used the minus sign in the right hand side because ~n is the outer normal of domain
Ω0 , i.e influx is minus outflux.
Occasionally it is convenient to work with a control domain Ω(t) that is not fixed but
moves with a certain velocity that at the boundary Γ(t) corresponds with velocity
~vΓ . In that case the influx is proportional to ~v − ~vΓ . Thus for a moving domain
we have:
d
dt
ZZZ
Ω(t)
ρ(~x, t)dΩ = −
ZZ
Γ(t)
ρ(~x, t)(~v − ~vΓ ) · ~ndΓ .
(2.5)
This equation is the integral mass balance for a moving control volume.
Note that in the case that the control volume moves with the fluid particles, i.e.
~vΓ = ~v , we recover the mass balance for a material domain given in (2.3), and for
~vΓ = ~0 we obtain the mass balance for a fixed domain (2.4).
To obtain a general mass balance formulation for all cases we can make use of the
Reynolds transport theorem (see appendix A.2.1 for a proof):
.
d
dt
ZZZ
φ(~x, t)dΩ =
Ω(t)
ZZZ
Ω(t)
∂φ(~x, t)
dΩ +
∂t
ZZ
Γ(t)
φ(~x, t)~vΓ · ~ndΓ
(2.6)
with ~vΓ the boundary velocity of the arbitrarily moving control domain Ω(t).
Finally, substitution of the Reynolds transport theorem (2.6) for φ = ρ gives the
general form:
ZZZ
Ω(t)
∂ρ(~x, t)
dΩ +
∂t
ZZ
Γ(t)
ρ(~x, t)~v · ~ndΓ = 0 .
(2.7)
This general form of the integral mass balance holds for material domains Ωm (t), for
fixed domains Ω0 , as well as for arbitrary moving domains Ω(t). In the next section
42
transport physics
we will use this form to derive a differential form of the mass balance equations.
Note that, although not indicated, in (2.7) not only ρ, but also ~v and ~n generally
depend on location ~x and time t.
Example 2.1: Integral mass balance
Two tubes with a circular cross-sectional of different diameter, D1 and D2 respectively, are connected with a rigid connector.
Γ1
Γ2
Γ3
~v1 (~x)
D2
D1
~v2 (~x)
The relation between the inflow velocity ~v1 (~x) and outflow velocity ~v2 (~x) follows
from the integral mass balance:
ZZZ
Ω
∂ρ(~x, t)
dΩ +
∂t
ZZ
Γ
ρ(~x, t)~v · ~ndΓ = 0 .
If the density is constant we have
∂ρ
= 0 and thus:
∂t
ZZ
Γ
ρ~v · ~ndΓ = 0.
Γ is the surface that closes the entire connector and can be split in three parts,
the inflow area Γ1 , the wall Γ3 , and the outflow area Γ2 . So we have:
ZZ
Γ1
ρ~v · ~n1 dΓ +
ZZ
Γ2
ρ~v · ~n2 dΓ +
ZZ
Γ3
ρ~v · ~n3 dΓ = 0 ,
with ~n1 , ~n2 , and ~n3 the unit normals to Γ1 , Γ2 , and Γ3 respectively. Since
ρ is constant, it can be moved out of the integrals. Moreover, as the wall of
the connector is rigid and impermeable, the velocity at Γ3 is zero. The relation
between ~v1 (~x) and ~v2 (~x) then is given by:
ZZ
Γ1
~v1 (~x) · ~n1 dΓ +
ZZ
Γ2
~v2 (~x) · ~n2 dΓ = 0 .
In case we have the simple situation that v1 (~x) = −V1~n1 and v2 (~x) = V2~n2 we
have ~v1 (~x) · ~n1 = −V1 and ~v2 (~x) · ~n2 = V2 and thus:
−
ZZ
Γ1
V1 dΓ+
ZZ
Γ2
V2 dΓ = 0
→
πD22
πD12
= V2
V1
4
4
→
V2 =
D1
D2
2
V1 .
Balance of mass
43
Example 2.2: Integral mass balance: axi-symmetric
A circular fluid yet with radius r0 is hitting a flat plate. The fluid forms a liquid
film with thickness h of which the front moves with a radial velocity Vf . The
density ρ of the fluid is constant in time and space. The mean velocity in the
yet is V0 in negative z-direction. The first contact of the yet with the plate is at
t = 0. We are intressted in the location rf (t) and the mean front velocity Vf (rf )
for rf (t) ≫ r0 .
Γ1
r0
−v0 (r)~ez
Γ4
vf (rf , z)~er h
z
vf (rf , z)~er
r
Γ2
rf (t)
Γ3
Given the fact that ρ is constant, the integral mass balance for a domain fixed at
t = t > 0 is:
ZZ
Γ1
~v · ~ndΓ +
ZZ
Γ2
~v · ~ndΓ +
ZZ
Γ3
~v · ~ndΓ +
ZZ
Γ4
~v · ~ndΓ = 0 .
At:
Γ1
Γ2
Γ3
Γ4
:
:
:
:
~v
~v
~v
~v
= −v0 (r)~ez ,
= vf (rf , z)~er ,
= ~0,
= ~vΓ ,
→
→
→
→
~n = ~ez
~n = ~er
~n = −~ez
~n ⊥ ~vΓ
~v · ~n = −v0 (r)
~v · ~n = vf (rf , z)
~v · ~n = 0
~v · ~n = 0
Consequently,
ZZ
Γ1
−v0 (r)dΓ +
ZZ
vf (rf , z)dΓ = 0
Γ2
and thus:
Zh Z2π
vf (rf , z)rf dφdz =
Z2π Zr0
v0 (r)rdrdφ .
φ=0 r=0
z=0 φ=0
This yields:
2πrf hVf = πr02 V0
→
Vf =
r02
V0 .
2rf h
,
,
,
.
44
transport physics
Finally,
drf
r2
= Vf = 0 V0
dt
2rf h
2.2
→
drf
1 d 2
r2
rf
=
(rf ) = 0 V0
dt
2 dt
2h
→
rf = r0
s
V0 t
.
h
Differential form of the balance of mass
To derive a differential form of the balance of mass equation, we need a second
important theorem, the Gauss divergence theorem (see appendix A.1.7):
ZZZ
Ω
∇ · f~dΩ =
ZZ
Γ
f~ · ~ndΓ
(2.8)
In the divergence theorem, the integral of the divergence of a vector f~ over a domain
Ω is expressed in terms of the flux of f~ out of the boundary surface Γ enclosing the
domain Ω.
The divergence theorem (2.8) with f~ = ρ~v , yields:
ZZZ
Ω
∇ · (ρ~v )dΩ =
ZZ
Γ
ρ~v · ~ndΓ .
(2.9)
Application to the general form of the integral mass balance (2.7) results in:
ZZZ
Ω(t)
∂ρ
dΩ +
∂t
ZZZ
Ω(t)
∇ · (ρ~v )dΩ = 0 .
(2.10)
or equivalently:
ZZZ Ω(t)
∂ρ
+ ∇ · (ρ~v ) dΩ = 0 .
∂t
(2.11)
Since this equation must hold for any arbitrary subspace of Ω it implies that the
integrand (also referred to as the integral kernel) must be zero, i.e.:
∂ρ
+ ∇ · (ρ~v ) = 0
∂t
in Ω .
(2.12)
This is the differential, or local, form of the mass balance, a partial differential
equation that is satisfied at any location in the domain Ω, independent whether Ω is
a material domain, a fixed domain, or a deforming domain. In appendix A.2.2 the
differential form of the mass balance is given for different coordinate systems.
It can easily be checked that we can also write:
∂ρ
+ ρ(∇ · ~v ) + ~v · ∇ρ = 0 .
∂t
(2.13)
Balance of mass
45
The first and third term in this expression can be combined using the definition of
the material derivative (1.17):
D
d
=
Dt
dt
ξ~
=
∂
+ ~v · ∇ ,
∂t
(2.14)
this gives a third version of the local mass balance:
Dρ
+ ρ(∇ · ~v ) = 0 .
Dt
(2.15)
From the local mass balance it also becomes clear that for incompressible media
with a homogeneous constant density ρ the divergence of the velocity field must be
zero:
∇ · ~v = 0 .
(2.16)
This is also called the incompressibility constraint and will need to be satisfied in
many of the flow problems we will consider in this course.
2.3
Mass diffusion and mass transfer
In the previous section we derived the mass balance equation for a single component
system or a multi-component system with a density of the bulk equal to ρ. For a
multi-component system, though, it will be clear that the mass balance must hold
for each of the components as well as for the bulk.
2.3.1
Mass balance for a multi-component system
If we consider component i of a multi-component system, the mean velocity of all
molecules of this component in an infinitesimal fluid element in the context of a
continuum approach is ~vi (~x, t). For this component we can write the local mass
balance as:
∂ρi
+ ∇ · (ρi~vi ) = 0
∂t
for component i .
(2.17)
If we sum over all components of the system this yields:
X ∂ρi
i
∂t
X
+∇·(
ρi~vi ) = 0 .
(2.18)
i
In the same time, for the mixture we have:
∂ρ
+ ∇ · (ρ~v ) = 0
∂t
Consequently since ρ =
for the bulk:
ρ~v =
X
i
ρi~vi .
for the mixture.
P
i ρi
(2.19)
(see definitions in Table 1.2) the following must hold
(2.20)
46
transport physics
This relation indicates that (per unit of volume) : the momentum of the mixture=
momentum of the components.
From (2.20) we also derive:
~v =
P
i
ρi~vi
.
ρ
P
(2.21)
So, the velocity of the bulk is the mass-averaged velocity of the mixture. The velocity
~vi of component i therefore is different from the velocity of the mixture ~v . This
difference:
~vid = ~vi − ~v
(2.22)
is called the diffusion velocity. Often we use the diffusion flux:
J~i = ρi (~vi − ~v ) .
(2.23)
In thermodynamics and chemistry it is more convenient to use the molar flux ci~vi
in stead of the mass flux ρi~vi . This is because in terms of free energy and chemical
reactions it is easier to regard the number of molecules than their mass. The molar
diffusion flux is defined as
~ji = ci (~vi − ~v ) .
2.3.2
The first law of Fick
Let us consider a binary mixture of substance A and substance B. Molecules of
substance A have an average velocity ~vA (~x, t), those of substance B have an average
velocity ~vB (~x, t). According to (2.21) the mass averaged velocity of the mixture is:
~v =
ρA~vA + ρB ~vB
ρ
(2.24)
with ρ = ρA + ρB . We have the diffusion fluxes (2.23):
J~A = ρA (~vA − ~v ),
J~B = ρB (~vB − ~v ) .
(2.25)
These diffusion fluxes origin from molecular momentum gradients due to concentration gradients: the molecules of substances A and B will move relative to each
other to obtain an optimal energy level in which these concentration differences are
equalized. The diffusion flux that is associated with these concentration gradients
is described by the phenomenological first law of Fick:


 J~A = −ρDAB ∇wA

 ~
JB = −ρDBA ∇wB
(2.26)
The coefficients DAB and DBA are the binary coefficients of diffusion. These in
general are a function of the composition, expressed in mass fractions (wA , wB ),
Balance of mass
47
temperature (T ) and pressure (p).
Note that for a binary mixture:
DAB = DBA .
(2.27)
This immediately follows from the notice that for a mixture that is macroscopically
at rest ~v = ~0 we have J~A + J~B = ~0.
For mixtures of more than two components a relation such as (2.27) in general does
not exist and the coefficient of diffusion of component i depend on all the other
components.
If, however, we have a mixture in which one of the components S is in surplus, the
situation is less complex. Component S, the solvent, then determines the diffusion
behaviour of each of the other components. We talk about a dilute solution. In that
case we have:
J~i = −ρDiS ∇wi
(2.28)
with DiS de coefficient of diffusion of component i in de solution. For fluid mixtures
with a constant density ρ we can substitute ρi = ρwi and obtain
J~i = −DiS ∇ρi ,
(2.29)
the first law of Fick.
2.3.3
The second law of Fick and the convection-diffusion equation
Earlier we derived the mass balance for component i (see (2.17)):
∂ρi
+ ∇ · (ρi~vi ) = 0 .
∂t
(2.30)
Together with the diffusion flux
J~i = ρi (~vi − ~v )
(2.31)
this yields:
∂ρi
+ ∇ · (J~i + ρi~v ) = 0 .
∂t
(2.32)
Substitution of the first law of Fick in the form (2.28), so for a dilute solution gives:
∂ρi
+ ∇ · (−ρDiS ∇wi + ρi~v ) = 0
∂t
(2.33)
∂ρi
+ ∇ · (ρi~v ) − ∇ · (ρDiS ∇wi ) = 0 .
∂t
(2.34)
or:
48
transport physics
In many cases (especially for dilute solutions) we can assume that the density of the
mixture is constant. Together with wi = ρi /ρ this results in
∂ρi
+ ∇ · (ρi~v ) − ∇ · (DiS ∇ρi ) = 0 .
∂t
(2.35)
In addition we have ∇ · ~v = 0 and thus ∇ · (ρi~v ) = (~v · ∇)ρi . This means that (2.35)
changes to:
∂ρi
+ (~v · ∇)ρi − ∇ · (DiS ∇ρi ) = 0 .
∂t
(2.36)
If, in addition, the diffusivity DiS can be assumed to be constant (so not dependent
on ρi , T and p), (2.36) reads:
∂ρi
∂t
+
(~v · ∇)ρi
convection
= DiS ∇2 ρi .
(2.37)
diffusion
This is the convection/diffusion equation for ρi .
Most often this equation is written in terms of the molar concentration ci = ρi /Mi
and the equation is referred to as the composition equation:
∂ci
+ (~v · ∇)ci = DiS ∇2 ci .
∂t
(2.38)
The equation (2.38) describes the transport of matter (ci ) based on two mechanisms:
• convection (~v · ∇)ci (transport due to flow)
• diffusion DiS ∇2 ci (molecular transport as a result of concentration gradients)
In the case that ~v = ~0, so a quiescent medium and thus no convection, (2.38) turns
into:
∂ci
= DiS ∇2 ci
∂t
(2.39)
This is an unsteady diffusion equation and referred to as the second law of Fick.
The definition of the Laplacian operator ∇2 in different coordinate systems is given
in appendix A.1.6.
In the next sections we will consider a few cases in which the medium is at rest and
we will present some standard solution procedures for (2.39).
2.4
2.4.1
Illustrative examples of mass diffusion problems
Steady diffusion
In this section we will consider problems in which the diffusion process is steady. In
that case (2.39) reduces to:
∇2 ci = 0 .
(2.40)
Balance of mass
49
Membrane transport
The membrane of a cell has a thickness d = 8 nm (see Figure 2.2). On both
sides of the membrane a water-like fluid is present in which hydrophobic particles
of substance A are dissolved. The hydrophilic property of the bipolar lipid-layer
results in a much lower concentration of A within the membrane than in the liquid
on both sides. (see Figure 2.2).
ce
membrane
intern
extern
ci
c′e
c′i
c=0
d
x
Figure 2.2: Diffusion through a membrane
Due to this hydrophilic/hydrophobic transitions we have:
c′
c′e
= i =k
ce
ci
(2.41)
with k the partition coefficient for substance A. We ask ourselves how this the diffusion process through the membrane can be quantified in the case that an equilibrium
is established, i.e. can we give an expression for the concentration c(x) as a function
of the position x. For this 1D steady diffusion process equation (2.40) reads:
d2 c
=0
dx2
(2.42)
and thus:
c(x) = ax + b .
(2.43)
The boundary conditions for this problem are:
BC :
(
c(x = 0) = c′e = kce = b
c(x = d) = a · d + b = c′i = kci
From this we readily can derive:
c(x) = k(ci − ce )
x
+ kce .
d
The course of concentration c(x) is depicted in Figure 2.3.
(2.44)
50
transport physics
ce
linear decay
ci
c′e
c′i
x=0
x
x=d
Figure 2.3: Course of the concentration c(x) over the membrane
From this solution for c(x) we can derive the molar diffusion flux j in case of the
steady state equilibrium. From the definition of the diffusion flux (2.28) we get:
j = −D∇c
(2.45)
and thus:
jx = −D
k
dc
= D(ce − ci ) .
dx
d
(2.46)
The quantity Dk/d is called the membrane permeability.
In dynamic processes it is important to know what is the time needed to establish
such a steady state equilibrium. For instance, if at t = 0 the extracellular concentration ce suddenly increases, we would like to know what is the typical time-scale
in which this increase is noticed within the cell.
We can give a first order approximation of this time-scale by an order of magnitude
approximation. For the unsteady diffusion equation
∂2c
∂c
=D 2
∂t
∂x
(2.47)
we can introduce typical scales c0 , τ , and L to make the equation dimensionless. So
here we can introduce
c∗ = c/c0 ,
t∗ = t/τ,
x∗ = x/L .
(2.48)
Since the scales are constants, we get:
c0 ∂c∗
Dc0 ∂ 2 c∗
=
τ ∂t∗
L2 ∂x∗2
(2.49)
If the scales are defined properly, i.e. with the proper magnitude, the differentials
∂c∗
∂ 2 c∗
and
have a magnitude in the order of 1 (we write O(1)). Consequently the
∂t∗
∂x∗ 2
order of magnitude of the left hand side and the right hand side must be O(c0 /τ )
and O(Dc0 /L2 ) respectively. Since the left and right hand side must be equal, we
find:
c0
O
τ
Dc0
=O
L2
.
(2.50)
Balance of mass
51
This order of magnitude analysis can provide us with valuable information about
the relation between the different scales that define a problem.
Here O(·) denotes the order of magnitude, τ is the time-scale, L the length-scale,
and c0 is a characteristic concentration of the problem. For our problem the exact
choice for c0 is not relevant, L = d, and τ is the time we are looking for. From this
it follows that:
τ=
d2
.
D
(2.51)
For the situation that D = 10−10 m2 /s en d = 8 · 10−9 m we get: τ = O(0.6 µsec).
These values are indeed representative for diffusion through a cell membrane involved
in for instance nerve conduction.
Spherical-symmetric diffusion
Consider a spherical pill of a certain material that dissolves slowly in water (see
Figure 2.4). We will assume the dissolution is that slow that the radius R of the
pill can be assumed to be constant during the time we observe the process. The
medicine dissolves that good, that the concentration c at the surface of the pill (so
at r = R) can be assumed to be constant and equal to the saturation concentration
csat . The radial concentration distribution c(r) in the solvent when a steady state
c(r)
R
Figure 2.4: Spherical-symmetric diffusion
is reached follows from (2.39) and gives ∇2 c = 0. In spherical coordinates (r, θ, φ)
this reads (see appendix A.1.6):
1 ∂
∂
∇ = 2
r2
r ∂r
∂r
2
∂
1
∂
+ 2
sin θ
r sin θ ∂θ
∂θ
+
∂2
1
r 2 sin2 θ ∂φ2
(2.52)
In the case of spherical symmetry (∂/∂θ = 0, ∂/∂φ = 0) this reduces to:
∇2 c =
1 ∂
∂c
r2
2
r ∂r
∂r
=0.
Integration over r gives:
a
dc
= 2
dr
r
(a = constant) ,
(2.53)
52
transport physics
a second integration over r gives:
a
c(r) = − + b
(b = constant) .
r
Application of the boundary conditions will provide a and b:
BC :
(
c(r = R) = csat
c(r → ∞) = 0
(2.54)
(2.55)
and the solution:
R
r
The molar flux then can be derived from the first law of Fick:
DR
dc
= 2 csat
jr = −D
dr
r
c(r) = csat ·
(2.56)
(2.57)
It shows the (not unexpected) result that the flux through an arbitrary spherical
surface 4πr 2 is equal to:
jr 4πr 2 = 4πRD · csat = constant .
2.4.2
(2.58)
Time dependent diffusion
We now consider the unsteady, or time dependent, diffusion process governed by the
equation: (2.39):
∂c
= D∇2 c .
(2.59)
∂t
In addition to boundary conditions we also need an initial condition to solve this
equation. We study illustrative examples in detail.
Diffusion between two compartments
Two fluid compartments are separated by a membrane that is permeable for substance A (see Figure 2.5). The fluid in the left compartment (x ≤ 0) is stirred
continuously. The concentration of A then can be assumed to be constant c = c0 .
The right compartment (x > 0) is in rest and is that big that far away from the
membrane the concentration can be assumed to be constant and equal to c∞ < c0 .
In this problem the concentration depends only on x and t, so the unsteady diffusion
equation (2.59) reads:
∂c
∂2c
=D 2
(2.60)
∂t
∂x
with D the diffusion constant. The corresponding boundary and initial conditions
are:


 c(0, t)
= c0 , t ≥ 0
BC/IC :
c(∞, t) = c∞ , t ≥ 0

 c(x, 0) = c
∞ , x>0
(2.61)
Balance of mass
53
x
c = c0
x=∞
c = c∞ < c0
membrane
Figure 2.5: Unsteady diffusion between two compartments.
Since the time t as well as the spatial coordinate x are not upper bounded (they
both can go to infinite), we can not really define a characteristic length-scale L and
characteristic time-scale τ for this problem. Still at a certain time t = τ we have:
1
O
τ
D
=O
L2
.
(2.62)
√
So a time dependent length scale L = O( Dt) could be introduced. This suggest the
introduction of a new parameter, i.e. a dimensionless spatial coordinate, η according
to:
x
,
(2.63)
η= √
k Dt
with k a positive constant of O(1). This implies:
η
x
∂η
= √ (− 12 )t−3/2 = −
∂t
2t
k D
(2.64)
The different terms in (2.60) then change to:
∂c
∂t
∂2c
∂x2
=
dc ∂η
η dc
=−
dη ∂t
2t dη
=
1 d2 c
k2 Dt dη 2
Wat gebeurd hier allemaal???
Waar is deze afleiding voor??
(2.65)
The time dependent diffusion equation (2.60) transforms to:
−
η dc
D d2 c
= 2
2t dη
k Dt dη 2
or
d2 c k2 dc
+ η
=0
dη 2
2 dη
(2.66)
To obtain a differential equation with a well known solution (see further on) it is
convenient to take k = 2, so we get:
dc
d2 c
+ 2η
=0
2
dη
dη
with
η=√
1
.
4Dt
(2.67)
This is a ordinary differential equation with one single parameter, the so called
similarity parameter η. The corresponding boundary conditions
BC :
(
c(η = 0) = c0
c(η = ∞) = c∞
(2.68)
54
transport physics
can easily be verified from (2.61). Note that η can only be introduced if the boundary condition c(∞, t) and the initial condition c(x, 0) must correspond.
If we define
f (η) =
dc
,
dη
(2.69)
we can write (2.67) as:
df (η)
+ 2ηf (η) = 0
dη
(2.70)
with solution:
2
f (η) = Ae−η .
(2.71)
How , why?!
Integration with respect to η gives:
c(η) =
Zη
f (η)dη = A
0
Zη
2
e−η dη + B ,
(2.72)
0
with A and B two integration constants. These integration constants can be obtained
from the boundary conditions (2.68). This produces:
BC :


c(η = 0) = c0






 c(η = ∞) = c∞
→ B = c0
√
2
π
= c∞ → A = − √ (c0 − c∞ )
→ c0 + A
2
π
(2.73)
Together with the definition of the error function:
2
erf(η) ≡ √
π
Zη
2
e−ζ dζ
(2.74)
0
(see also A.45) the solution (2.72) becomes:
c(η) = −(c0 − c∞ ) erf(η) + c0 = c∞ + (c0 − c∞ ) erfc(η)
√
or with η = x/ 4Dt:
x
c(x, t) = c∞ + (c0 − c∞ ) erfc √
4Dt
(2.75)
(2.76)
More information on the error function can be found in appendix A.1.10. In Figure
2.6 the solution c(x, t) is plotted as a function of x for several values of t. The role
of the similarity parameter η is expressed in the fact that the curves for different
instants of time would collapse
to a single curve if not the spatial coordinate x but
√
the scaled version η = x/ 4Dt would be used as the abscissa (horizontal-axis).
Balance of mass
55
c0
c(x, t)
t
c∞
x
δ(t)
x=0
Figure 2.6: Concentration as a function of x for several instants of time.
We now can approximate the thickness δ(t) of the layer in which the diffusion is
perceptible (say c − c∞ = 0.01(c0 − c∞ )). This approximation then follows from:
√
c(δ(t), t) − c∞ = 0.01(c0 − c∞ ) → erfc(δ(t)/ 4Dt) = 0.01 →
√
→ erf(δ(t)/ 4Dt) = 0.99 →
(2.77)
√
→ δ(t)/ 4Dt = 2.
So, the thickness of the layer in which diffusion is perceptible is:
√
δ(t) = 4 Dt .
(2.78)
Finally, the molar flux j(x, t) in this 1D time dependent diffusion problem is:
∂c
jx = −D
=
∂x
s
D
2
(c0 − c∞ )e−x /4Dt ,
πt
(2.79)
At the location of the membrane (x = 0) this is:
s
jx (x = 0) = (c0 − c∞ )
D
.
πt
(2.80)
From this it can be seen that the molar flux through the membrane decreases in
time due to the ongoing increase in concentration in the right compartment.
Diffusion from a point source
A standard problem is the unsteady diffusion from a point source with initial concentration:
c(~x; t = 0) = M δ(~x) .
(2.81)
56
transport physics
Here δ(~x) is the so called Dirac delta-function (see appendixA.1.11).
In one dimension the point source problem is defined as:

∂c
∂2c


=D 2

∂t



∂x
(2.82)
with c(x, 0) = M · δ(x)
Here M is the total amount of substance.
The initial condition is sketched in Figure 2.7.
M δ(x)
x
x=0
Figure 2.7: Initial concentration distribution (2.81)
Again the simularity parameter
x
η≡√
4Dt
is introduced.
A solution c(x, t) for (2.82) can be derived as
c(x, t) =
x2
M
· exp −
4Dt
2(πDt)1/2
!
(2.83)
1
:
.
(2.84)
This is the 1D point source solution. In Figure 2.8 the concentration as a function
of x is given for a 3 instants of time.
Without proof we give the 2D circle-symmetric solution
"
#

∂c
∂2c
∂2c



=D
+ 2

2
∂t




∂x
∂y
r2
M
exp −
→ c(~r, t) =
4πDt
4Dt
!
.
(2.85)
c(t = 0) = M δ(x)δ(y)
and the 3D sphere-symmetric solution
"
#

2c
2c
2c
∂
∂
∂
∂c



=D
+ 2+ 2

2
∂t
1




∂x
∂y
∂z
M
r2
→
·
exp
−
4Dt
8(πDt)3/2
!
. (2.86)
c(x, y, z, t = 0) = M δ(x)δ(y)δ(z)
We will not give a derivation here. One way to derive this is to take a superposition of two
solutions of the diffusion between two compartments shifted over a distance ∆x around x = 0 with
∆x → 0.
Balance of mass
57
t=1
t=2
t=3
x
x=0
Figure 2.8: Concentration according the 1D point souce solution (2.84)
with r =
2.5
p
x2 + y 2 and r =
p
x2 + y 2 + z 2 respectively.
Convection-diffusion problems
We will end this chapter with a few remarks on convection-diffusion problems. If
the velocity field is known, the concentration of a substance will change in time by
diffusion induced by concentration gradients and convection caused by the motion
of the fluid in which the substance is dissolved. These problems are defined by:
∂c
+ (~v · ∇)c = D∇2 c
∂t
(2.87)
Here it is assumed that the velocity field indeed is known and is not influenced by
the concentration of the substance dissolved. In practice this means that both the
density and the viscosity of the fluid should be independent of the concentration. In
general this assumption holds for dilute solutions.
To determine the importance of the different terms we will scale the equations based
on characteristic length-, velocity-, and time-scales L, V , and τ respectively. We
define:
~x∗ = ~x/L, ~v ∗ = ~v /V, c∗ = c/c0 and t∗ = t/τ
This results in:
Dc0 2
V c0 ∗
c0 ∂c∗
(~v · ∇)c∗ = 2 ∇∗ c∗
+
∗
τ ∂t
L
L
(2.88)
After division by V c0 /L and omission of the asterisks (*) we obtain:
Sr
1 2
∂c
+ (~v · ∇)c =
∇ c
∂t
Pe
(2.89)
In this dimensionless form of the convection diffusion equation we introduced two
dimensionless groups: the Strouhal number (ratio between local time derivative and
58
transport physics
convection)
Sr =
[ ∂c
L
∂t ]
=
Vτ
[~v · ∇c]
(2.90)
and the Péclet number (ratio between convection and diffusion)
Pe =
VL
[~v · ∇c]
=
.
D
[D∇2 c]
(2.91)
Note that also in non-isothermal problems (see Chapter 7), where the convection
diffusion equations also hold for the temperature field, the Péclet number is used.
In that case the diffusion constant D is replaced by the thermal diffusivity α
For large values of the Péclet number we get:
Pe ≫ 1 :
∂c
+ (~v · ∇)c = 0
∂t
(2.92)
Diffusion then can be neglected and the composition equation transforms to a
scalar/passive convection equation. This equation describes the motion of passive
particles in a given velocity field and can be used to construct particle trajectories.
For small values of the Péclet number we get:
Pe ≪ 1 :
∂c
− D∇2 c = 0
∂t
(2.93)
In these cases convection can be neglected and the diffusion equation as dealt with
in the previous sections is left.
3
Balance of momentum
In this chapter we will derive the integral form of the momentum balance based on a
continuum interpretation of the second law of Newton. Subsequently a differential form
will be derived with the aid of extended versions of the divergence theorem and the
Reynolds transport theorem. For newtonian incompressible media this leads to the wellknown Navier-Stokes equation. A classification of fluid flow based on the importance of
the different terms in de Navier-Stokes equation will be made and relevant dimensionless
groups will be introduced.
59
60
3.1
3.1.1
transport physics
Integral form of the momentum balance
Momentum balance for a material domain
We consider a material domain Ωm with surface Γm as sketched in Figure 3.1.
Γm
~n
Ωm
~t
dΩ
dΓ
~g
Figure 3.1: Material domain with surface forces ~t and volume forces ~g.
The second law of Newton states that the rate of change of the momentum of a
system is equal to the sum of all forces that act on that system. These forces consist
of volume forces that act on each volume element dΩ and surface forces that act on
each surface element dΓ.
The total momentum P~ in the material volume Ωm must be computed as the integral
over Ω of the momentum density ρ~v , i.e.,
P~ =
ZZZ
ρ~v dΩ
(3.1)
Ωm
This integral is the sum of all contributions of infinitesimal elements dΩ in Ωm . Note
that, unlike the mass density ρ, the momentum density ρ~v is a vector.
The volume forces ~g can be gravity forces, electromagnetic forces, or other body
forces that act on each volume element in the fluid. If we restrict ourselves to
the gravity force, then this force per unit of mass is equal to ~g with g = k~g k the
gravitational acceleration (on earth, g ≈ 9.8 m2 /s). So, on a volume element dΩ the
gravity force is ρ~gdΩ and the total body force is:
F~Ω =
ZZZ
ρ~gdΩ .
(3.2)
Ωm
The surface force on surface element dΓ is equal to ~tdΓ. Here ~t is the Cauchy stress
vector defined as the surface force per unit of area. The stress vector ~t can have a
shear stress component tangential to the surface as well as normal stress component.
The total surface force that acts on the surface of the domain Γm is:
F~Γ =
ZZ
Γm
~tdΓ .
(3.3)
Balance of momentum
61
In many cases, we will write the stress vector ~t in terms of the Cauchy stress tensor
defined by (1.49), i.e. ~t = σ · ~n. The second law of Newton, or integral momentum
balance , now states
dP~
~Γ ,
= F~Ω + F
dt
(3.4)
or:
d
dt
ZZZ
ρ~v dΩ =
Ωm
ZZZ
ρ~gdΩ +
Ωm
ZZ
~tdΓ .
(3.5)
Γm
Note that we derived (3.5) for a material domain Ωm , implying that no particles
cross the domain boundary Γm .
3.1.2
Momentum balance for a fixed or moving domain
For a domain Ω0 that is fixed or domain Ω(t) that is moving with an arbitrary
velocity equal to ~vΓ at the boundary, the momentum wil not only change as a result
of body forces or surface forces but can also change as a result of momentum influx
or outflux. Fluid elements with momentum density ρ~v will enter or leave the domain.
This flux is proportional to the normal component of the velocity of fluid elements
relative to the velocity of the domain (~v − ~vΓ ) · ~n.
The momentum flux through a surface element dΓ then is a vector defined by
d~
qP = ρ~v ((~v − ~vΓ ) · ~n) dΓ
The total momentum flux through surface Γ then follows from:
~qP =
ZZ
Γ
ρ~v ((~v − ~vΓ ) · ~n) dΓ .
(3.6)
As a result, the momentum balance for a fixed (~vΓ = ~0) or moving (~vΓ 6= ~0) domain
reads:
d
dt
ZZZ
Ω
ρ~v dΩ +
ZZ
Γ
ρ~v ((~v − ~vΓ ) · ~n) dΓ =
ZZZ
Ω
ρ~g dΩ +
ZZ
~tdΓ .
(3.7)
Γ
It is important to realize that the momentum balance is a vector equation. The
following example illustrates this.
62
transport physics
Example 3.1: Another integral momentum balance
We revisit the two connected tubes example for in- and outflow ~v1 = Ve~ey and
~v2 = V2~ex , respectively. The pressure at the entrance is p1 and at the outflow is
p2 . The plane of symmetry of the tube system is on the y − z-plane. The fluid is
assumed to be incompressible and gravity forces are neglected. We already know
from the integral mass balance that A1 V1 = A2 V2 . We are now interested in the
total force F~w acting on the wall of the tube system.
Γ1
Γ3
~v1 (~x)
p1
Γ2
z
y
D1
D2
x
p2
~v2 (~x)
For steady flow, fixed boundaries, and without gravity forces, the integral momentum balance (3.7) reduces to:
ZZ
Γ
ρ~v (~v · ~n)dΓ =
ZZ
~tdΓ .
Γ
At the entrance Γ1 we have:
~v = V1~ey , ~n = −~ey
→
ρ~v (~v · ~n) = ρV1~ey (−V1 ) = −ρV12~ey
~t = σ · ~n = −σxy~ex − σyy ~ey − σzy ~ez = p1~ey
Indeed:
incompressible
vx = 0,
∂vy
=0
∂x
∂vy
=0
∂y
vz = 0,
∂vy
=0
∂z
→
∇ · ~v = 0
→
→
σxy = η(
→
σyy = −p1 + 2η
→
σzy = η(
∂vy
∂vx ∂vz
=−
−
=0
∂y
∂x
∂z
∂vx ∂vy
+
)=0
∂y
∂x
∂vy
= −p1
∂y
∂vz
∂vy
+
)=0
∂y
∂z
At the outflow Γ2 we have:
~v = V2~ex , ~n = ~ex
→
ρ~v (~v · ~n) = ρV2~ex (V2 ) = ρV22~ex
~t = σ · ~n = σxx~ex + σyx~ey + σzx~ez = −p2~ex
Balance of momentum
63
Indeed:
∂vx
∂vy
∂vz
=−
−
=0
∂x
∂y
∂z
incompressible
→
∇ · ~v = 0
→
∂vx
=0
∂x
→
σxx = −p2 + 2η
∂vx
= −p2
∂x
vy = 0,
∂vx
=0
∂y
→
σyx = η(
∂vy
∂vx
+
)=0
∂x
∂y
vz = 0,
∂vx
=0
∂z
→
σzx = η(
∂vz
∂vx
+
)=0
∂x
∂z
At the fixed wall Γ3 we have:
~v = ~0
ρ~v (~v · ~n) = ~0
→
~w = −
the force F~w acting on the boundary is F
get:
ZZ
ρ~v (~v ·~n)dΓ+
Γ1
ZZ
Γ2
ρ~v (~v ·~n)dΓ+
ZZ
Γ3
ZZ
~tdΓ. Taking all together we
Γ
ρ~v (~v ·~n)dΓ =
ZZ
~tdΓ+
Γ1
ZZ
Γ2
~tdΓ+
ZZ
~tdΓ
Γ3
and thus:
ZZ
(−ρV12~ey )dΓ +
Γ1
ZZ
(ρV22~ex )dΓ + ~0 =
ZZ
(p1~ey )dΓ +
Γ1
Γ2
ZZ
Γ2
(−p2~ex )dΓ − F~w
so:
F~w = −(p2 + ρV22 )A2~ex + (p1 + ρV12 )A1~ey ,
with:
A1 =
πD12
πD22
, A2 =
4
4
and
V2 =
A1
V1
A2
In the next example the choice of the volume Ω is less obvious and a clever choice
have to be made.
64
transport physics
Example 3.2: Integral momentum balance
To illustrate the use of the integral form of the momentum balance where we
have to chose the control volume we consider the situation depicted in the left
panel of the figure below, where we attempt to close a hole in the side wall of the
hull of a ship. The hole has a cross-sectional area A and is located at a distance
H below the surface of the water. The water, with density ρ, squirts into the
hold of the ship with velocity V . The air pressure in the ship hold and above
the water surface is equal to pa . We want to know what force F1 we need to
close the hole with a flat plate with surface area Ap . The size of the hole is small
compared to the height of the water level H. We assume that viscous forces can
be neglected.
pa
pa
H
Γ2
pa
A
Γ1
F1
V
A
F1
pa
Γ3
y
Γ4
pa
x
(a)
(b)
Fixing a hole in the hull of a ship.
To solve this problem we define a control volume Ω with boundary (Γ1 , ...Γ4 )
as sketched with the dashed rectangle in the right panel of the figure. For this
control volume the integral form of the momentum balance must hold, so:
d
dt
ZZZ
Ω
ρ~v dΩ +
ZZ
Γ
ρ~v (~v − ~vΓ ) · ~ndΓ =
ZZZ
Ω
ρ~g dΩ +
ZZ
~tdΓ .
Γ
We assume that we keep the plate fixed so we can assume that the we have a
steady state and that ~vΓ = ~0. Moreover, the hole is small, so is the dashed
domain. Consequently the influence of the gravity forces in the dashed domain
can be neglected. This gives:
ZZ
Γ
ρ~v (~v · ~n)dΓ =
ZZ
~tdΓ .
Γ
~1 = F1~ex .
Due to symmetry all forces in y-diection are zero or counterparted, i.e. F
~
Consequently, to compute the force F1 we only need to consider the x-component
of the momentum balance.
At Γ1 : inflow in x-direction, ~v = V ~ex and ~n = −~ex
~v · ~n = V ~ex · (−~ex ) = −V
→
ρ~v (~v · ~n) · ~ex = −ρV 2
Balance of momentum
~t = pa~ex
65
~t · ~ex = pa
→
At Γ2 : outflow in y-direction, ~v = Vo~ey and ~n = ~ey
ρ~v (~v · ~n) = ρV02~ey
~t = −pa~ey
→
ρ~v (~v · ~n) · ~ex = 0
~t · ~ex = 0
→
At Γ3 : no flow in x-direction, ~v = ~0 and ~n = ~ex
ρ~v (~v · ~n) = ~0
→
ρ~v (~v · ~n) · ~ex = 0
~t = (−F1 /Ap − pa )~ex
→
~t · ~ex = −F1 /Ap − pa
At Γ4 : outflow in y-direction, ~v = −Vo~ey and ~n = −~ey
ρ~v (~v · ~n) = ρV02~ey
~t = pa · ~ey
→
→
ρ~v (~v · ~n) · ~ex = 0
~t · ~ex = 0
This results in:
ZZ
Γ1
−ρV 2 dΓ =
ZZ
pa dΓ+
ZZ
Γ3
Γ1
(−F1 /Ap −pa )dΓ
→
−ρV 2 A = −(F1 /Ap )Ap .
and thus for the force we were searching for we have:
F1 = ρV 2 A .
√
Later we will learn that from the Bernoulli equation we can compute V = 2gH.
This implies that F1 = 2ρgHA. This is twice as high as the force you need to
keep the hole closed once the plate is pressed to the wall of the hull.
3.1.3
Hydrostatics
We consider a fluid at rest. In that case in each surface element dΓ only normal
surface forces act. Pascal discovered that the magnitude of this normal force is
independent of the direction of the normal ~n. This force acts from the environment
onto the domain Ω and therefore has a direction in negative ~n direction, i.e. ~t = −p~n
where p is the hydrostatic pressure.
Since ~v = ~0, the momentum balance simplifies to:
ZZZ
ρ~g dΩ +
Ω
ZZ
~tdΓ = 0
(3.8)
Γ
with ~t = σ · ~n and, for hydrostatic problems, ~t = −p~n:
ZZZ
Ω
ρ~gdΩ −
ZZ
Γ
p~ndΓ = 0 .
(3.9)
66
transport physics
F~e
water
Ω
~g
Figure 3.2: Gravity force and buoyancy force according to Archimedes
The momentum balance (3.9) for a hydrostatic equilibrium can be applied to understand the law of Archimedes. Assume an object Ω with volume V is immersed
in water at rest with density ρw . The object is kept in place by an external force F~e
(see Figure 3.2).
Suppose the object is filled with water so the density of the object is equal to
the density of water. In that case there would be equilibrium (Fe = 0) and the
momentum balance in z-direction would read:
Fg + Fp = −
ZZZ
Ω
ρw gdΩ −
ZZ
pnz dΓ = 0 .
(3.10)
Γ
So the pressure force in z-direction exerted by the fluid on the object is:
Fp ≡ −
ZZ
pnz dΓ =
Γ
ZZZ
ρw gdΩ .
(3.11)
Ω
If the water inside the object is replaced by another fluid with density ρ and the
object is kept in place by the external force Fe , the water surrounding the object
would not notice this so Fp will not change. The momentum balance in z-direction
now is:
Fe + Fg + Fp = Fe −
ZZZ
Ω
ρgdΩ +
ZZZ
ρw gdΩ = 0 .
(3.12)
Ω
Consequently:
Fe =
ZZZ
Ω
(ρ − ρw )gdΩ .
(3.13)
So compared with the situation in which there would be no water surrounding the
object an extra upward force Fu is experienced equal to the weight of the volume of
water that is displaced, i.e.:
Fu =
ZZZ
ρw gdΩ .
(3.14)
Ω
This is the principle of Archimedes, derived from the integral momentum balance.
Balance of momentum
3.2
67
Differential form of the momentum balance
In order to derive a differential form of the momentum balance we will need different
forms of the Gauss divergence theorem. Hereto, we first introduce the diadic product
of two vectors to define a 2nd order tensor (see appendix A.1.3) and the vector and
tensor form of the divergence theorem of Gauss (see appendix A.1.8).
Application of the Reynolds transport theorem (2.6) with φ = ρ~v to the first term
of the integral momentum balance (3.7) for the general case that the domain Ω(t)
is moving with an arbitrary velocity ~vΓ gives:
d
dt
ZZZ
ρ~v dΩ =
Ω(t)
ZZZ
Ω(t)
∂
(ρ~v )dΩ +
∂t
ZZ
Γ(t)
ρ~v (~vΓ · ~n)dΓ .
(3.15)
With ~t = σ · ~n, the integral momentum balance (3.7) then reads:
ZZZ
Ω(t)
∂
(ρ~v )dΩ +
∂t
ZZ
Γ(t)
ρ~v (~v · ~n)dΓ =
ZZZ
ρ~g dΩ +
Ω(t)
ZZ
Γ(t)
σ · ~ndΓ .
(3.16)
If we apply the different forms of the divergence theorem given in appendix A.1.8
to the two surface integrals of the integral momentum balance (3.16) we obtain1 :
ZZZ Ω(t)
∂
(ρ~v ) + ~v ∇ · (ρ~v ) + ρ~v · ∇~v dΩ =
∂t
ZZZ
Ω(t)
[ρ~g + ∇ · σ c ] dΩ .
(3.17)
The integrand of the left hand side can be rewritten as:
ρ
∂~v
∂ρ
∂~v
+ ~v
+ ~v ∇ · (ρ~v ) + ρ~v · ∇~v = ρ
+ ρ~v · ∇~v .
∂t
∂t
∂t
(3.18)
Here we used the fact that the second and third term vanish because of the local
mass balance equation (2.12). Since (3.17) must also hold for any subvolume of Ω the
integrands of the left and right hand side must be equal. Together with (3.18) this
gives momentum balance in differential form, mostly referred to as the momentum
equation:
ρ
∂~v
+ ρ(~v · ∇)~v = ρ~g + ∇ · σ
∂t
in Ω .
(3.19)
Note that the two terms in the left hand side add up to the material acceleration
D~v /Dt multilied with the density ρ. So, the mass per unit of volume times the
acceleration (left hand side) are related to forces per unit of volume (right hand
side) in accordance with Newton’s second law.
For most applications it is convenient to split the Cauchy stress tensor in a pressure
term and the extra stress tensor τ as in (1.62). So,
σ = −pI + τ
1
Here we used ρ~v(~v · ~n) = ~v (ρ~v · ~n).
(3.20)
68
transport physics
and thus:
∇ · σ = −∇p + ∇ · τ ,
(3.21)
and thus after division by ρ:
∂~v
1
1
+ (~v · ∇)~v = − ∇p + ∇ · τ + ~g
∂t
ρ
ρ
in Ω .
(3.22)
The first term in this momentum balance is the local acceleration and expresses the
time dependence of the velocity field. In steady flow this term is equal to zero. The
second term in the left hand side is the convective acceleration and is associated
with acceleration of fluid particles as a result of their motion in a velocity field with
spatial gradients. The first term in the right hand side are forces acting on fluid
elements due to pressure gradients, viscous forces, and gravity forces respectively.
All terms are vectors with a magnitude and a direction. Of course the momentum
balance must hold for each direction. In Cartesian coordinates this yields:
∂vx
∂vx
∂vx
∂vx
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
1 ∂p 1
= −
+
ρ ∂x ρ
∂τxx ∂τyx ∂τzx
+
+
∂x
∂y
∂z
+ gx
∂vy
∂vy
∂vy
∂vy
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
= −
1 ∂p 1
+
ρ ∂y ρ
∂τxy
∂τyy
∂τzy
+
+
∂x
∂y
∂z
+ gy
∂vz
∂vz
∂vz
∂vz
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
1 ∂p 1
= −
+
ρ ∂z ρ
∂τyz
∂τzz
∂τxz
+
+
∂x
∂y
∂z
+ gz
(3.23)
In the next sections we will make a classification of fluid flow based on the importance
of the different terms in the momentum equation.
3.3
3.3.1
Fluid flow classification
Hydrostatics
When there is no motion (~v = ~0) the left hand side of the momentum equation
(3.22) as well as the extra stress τ are zero. The momentum equation then reduces
to:
~0 = − 1 ∇p + ~g
(3.24)
ρ
with ~g = −g~ez . So apparently:
∂p
= 0,
∂x
∂p
= 0,
∂y
dp
= −ρg .
dz
(3.25)
In the most simple case we have a uniform constant density ρ and after integration
of z:
p(z) = −ρgz + C .
(3.26)
Balance of momentum
69
Example 3.3: Hydrostatic pressure
If, for instance, we consider a layer of water with at z = h an atmospheric pressure
pa = ph (see Figure).
ph
ρ
z=h
~g
~ez
z=0
Hydrostatic pressure p(z) in a layer of fluid.
We have:
p(z) = pa + ρg(h − z) .
In the case that we have an isothermal (T = T0 ) layer of an ideal gas that satisfies
the ideal gas law p = ρRT , the situation is different. In that case we have:
p
dp
= −ρg = −
g
dz
RT0
and thus:
−
p(z) = ph e
g(z − h)
RT0
With ph the pressure at z = h.
This example shows that in a hydrostatic situation the pressure in general can be
a function of the density and accordingly will depend on the coordinate in which
gravitational forces act.
3.3.2
Inviscid flow: the Euler Equation
Another type of fluid flow is the class of inviscid flows. In that case the viscosity is
assumed to be zero so we have no viscous stresses (τ = 0). The momentum equation
then reduces to the Euler equation:
1
∂~v
+ (~v · ∇)~v = − ∇p + ~g
∂t
ρ
(3.27)
70
transport physics
In Cartesian coordinates these equations (3 components) read:
∂vx
∂vx
∂vx
∂vx
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
= −
1 ∂p
+ gx
ρ ∂x
∂vy
∂vy
∂vy
∂vy
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
= −
1 ∂p
+ gy
ρ ∂y
∂vz
∂vz
∂vz
∂vz
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
= −
1 ∂p
+ gz
ρ ∂z
(3.28)
This is a vector equation of 3 components that forms a set of coupled equations with
in general 5 unknowns: the velocity components vx , vy , and vz , the pressure p, and
the density ρ. To solve this set of equations we need two extra relations. The first
one is the balance of mass:
∂ρ
+ ∇ · (ρ~v ) = 0
∂t
(3.29)
∂
∂
∂
∂ρ
+
(ρvx ) +
(ρvy ) +
(ρvz ) = 0
∂t
∂x
∂y
∂z
(3.30)
or
The second relation we need is a constitutive relation ρ = ρ(p) that relates the
density to the pressure.
In addition to the 5 equations we need initial conditions, so the velocity, pressure,
and density at a certain instant of time (say t = 0), as well as boundary conditions
depending on the configuration in which the equations need to be solved. Here, it
is worthwhile to mention that the boundary condition at a fixed wall for the Euler
equations in general are slip conditions. This means that only the normal component
of the velocity is zero:
~vwall · ~n = 0 .
(3.31)
This is different from viscous flow where, due to viscous forces, a no-slip condition
~vwall = ~0 holds. Hence, in inviscid flow, the tangantal component of the velocity at
the wall in general is unequal to zero (slip velocity).
3.3.3
Viscous flow: the Navier-Stokes Equation
In the most general form we have to incluce all terms in the momentum balance.
So:
1
1
∂~v
+ (~v · ∇)~v = − ∇p + ∇ · τ + ~g
∂t
ρ
ρ
(3.32)
must hold. For a Newtonian fluid with viscosity η (see section 1.1) we can substitute
τ = 2ηD and the momentum equation reads:
1
1
∂~v
+ (~v · ∇)~v = − ∇p + (∇ · 2ηD) + ~g .
∂t
ρ
ρ
(3.33)
Balance of momentum
71
If we exam the viscous term in detail we learn:
∇ · τ = ∇ · 2η 12 (∇~v + (∇~v )c )
(3.34)
For a constant viscosity this changes to:
∇·τ
= η(∇ · ∇~v + ∇ · (∇~v )c )
= η(∇2~v + ∇(∇ · ~v ))
(3.35)
Here we used (without prove) the vector identities ∇ · ∇~v = ∇2~v and ∇ · (∇~v )c =
∇(∇ · ~v ). For an incompressibel fluid (∇ · ~v = 0) we find:
1
∇ · τ = ν∇2~v
ρ
(3.36)
with the kinematic viscosity
η
ν= .
ρ
(3.37)
In conclusion, for a Newtonian incompressible fluid with constant viscosity the momentum equation reads:
1
∂~v
+ (~v · ∇)~v = − ∇p + ν∇2~v + ~g .
∂t
ρ
(3.38)
This is the well know Navier-Stokes equation. It holds under the constraint of
incompressibility so together with the continuity equation:
∇ · ~v = 0 .
(3.39)
In Cartesian coordinates the Navier-Stokes equation are written as:
∂vx
∂vx
∂vx
∂vx
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
1 ∂p
= −
+ν
ρ ∂x
∂ 2 vx ∂ 2 vx ∂ 2 vx
+
+
∂x2
∂y 2
∂z 2
!
+ gx
∂vy
∂vy
∂vy
∂vy
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
1 ∂p
= −
+ν
ρ ∂y
∂ 2 vy
∂ 2 vy
∂ 2 vy
+
+
∂x2
∂y 2
∂z 2
!
+ gy
∂vz
∂vz
∂vz
∂vz
+ vx
+ vy
+ vz
∂t
∂x
∂y
∂z
1 ∂p
= −
+ν
ρ ∂z
∂ 2 vz
∂ 2 vz
∂ 2 vz
+
+
2
2
∂x
∂y
∂z 2
!
+ gz
(3.40)
with
∂vx ∂vy
∂vz
+
+
=0.
∂x
∂y
∂z
(3.41)
Different from the Euler equations, a constitutive equation τ = 2ηD is built in
and the relation between density and pressure is present via the continuity equation
(incompressible fluid). So we have 4 equations and 4 unknowns: vx , vy , vz , and p.
At a fixed wall, due to viscous forces, a no-slip conition ~v = ~0 (or ~v = ~vΓ for a
moving wall) holds.
72
transport physics
Example 3.4: Down the drain
To illustrate how the differential form of the mass balance can be used to compute
local characteristics of a velocity field, we will consider the shape of a free surface
near a sink hole (see the figure below) determined by the pressure distribution
that results from an assumed given velocity field.
z
p0
−a
r
a
vφ~eφ
Fluid rotating near a sink hole.
Gravity is in −z-direction (gz = −g). We will assume that the flow is frictionless,
steady, and given by:
~v (r)~eφ =

ωr~eφ



2


 ωa ~
eφ
r
This implies:
vr = 0, vz = 0
for
r≤a
for
r>a
and gr = 0, gφ = 0
and
∂
=0
∂φ
Note that in this example the velocity field is not computed but given. Still the
balance of mass and momentum must be satisfied. The momentum balance in
cylindrical coordinates (see appendix A.2.3) in which we substitute the velocity
properties given above simplifies to:
vφ2
1 ∂p
=−
r
ρ ∂r
r − direction:
−
φ − direction:
0=
z − direction:
∂p
=0
∂φ
1 ∂p
−g
ρ ∂z
Balance of momentum
73
For r ≤ a we get for the r-direction:
vφ2
∂p
ω2r2
=ρ =ρ
= ρω 2 r
∂r
r
r
→
p(r, z) = 21 ρω 2 r 2 + b(z) .
→
b(z) = −ρgz + C1 .
Substitution in the φ-direction gives:
1 ∂p
=g
ρ ∂z
→
∂b
= −ρg
∂z
In other words:
p(r, z) = 12 ρω 2 r 2 − ρgz + C1 .
At the free surface p(r, z) = p0 and thus:
2 2
1
2 ρω r
− ρgz + C1 = p0
→
z=
ω 2 2 C1 − p0
r −
.
2g
ρg
At r = 0 we have z = 0 which means that C1 = p0 . Consequently
z=
ω2r2
2g
for r ≤ a .
For r > a we can take the same steps, now for vφ = ωa2 /r. This yields:
vφ2
ω 2 a4
∂p
=ρ =ρ 3
∂r
r
r
→
p(r, z) = −
ρ ω 2 a4
+ c(z) .
2 r2
Again:
∂c(z)
= −ρg
∂z
→
c(z) = −ρgz + C2 .
In other words:
p(r, z) = −ρ
ω 2 a4
− ρgz + C2 .
2r 2
Now, at the free surface, again p(r, z) = p0 so:
−ρ
ω 2 a4
− ρgz + C2 = p0
2r 2
→
z=−
ω 2 a4 C2 − p0
+
.
2gr 2
ρg
At r = a this solution must be equal to the one for r < a so:
−
ω 2 a2
ω 2 a2 C2 − p0
+
=
2g
ρg
2g
→
Consequently:
ω 2 a2
z=
2g
a2
2− 2
r
!
for r > a .
C2 = p0 + ρω 2 a2
74
3.4
transport physics
Dimensionless groups
To find out whether or not a fluid flow problem can be classified in one of the
flow regimes given above, it is convenient to scale the problem and write it in a
dimensionless form. To do so, we introduce characteristic length, velocity, and time
scales. As an example we consider the flow around an object, with a typical length L,
that makes an oscillatory motion with a typical angular frequency ω. The magnitude
of the velocity of the surrounding fluid is typically V (see Figure 3.3).
V
ω
L
Figure 3.3: Flow around an oscillating object
The scales involved are:
length scale
:
velocity scale :
time scale
:
L
V
τ = 1/ω
In general the flow is described by the Navier-Stokes equations:
1
∂~v
+ (~v · ∇)~v = − ∇p + ν∇2~v + ~g .
∂t
ρ
(3.42)
With the aid of these scales we non-dimensionalize the Navier-Stokes equations
according to:
~x∗ = ~x/L,
~v ∗ = ~v /V,
t∗ = t/τ .
(3.43)
This scaling also has consequences for the gradient operator ∇ and the Laplacian
∇2 . Indeed,
∇ = ~ex
= ~ex
=
∂
∂
∂
+ ~ey
+ ~ez
∂x
∂y
∂z
∂
∂
∂
+ ~ey
+ ~ez
∗
∗
∂(Lx )
∂(Ly )
∂(Lz ∗ )
∂
∂
∂
1
1
1
~ex ∗ + ~ey ∗ + ~ez ∗
L ∂x
L ∂y
L ∂z
(3.44)
=
1 ∗
∇
L
with
∇∗ = ~ex
∂
∂
∂
+ ~ey ∗ + ~ez ∗ .
∗
∂x
∂y
∂z
(3.45)
Balance of momentum
75
Similarly we have:
∇2 =
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
=
∂2
∂2
∂2
+
+
∂(Lx∗ )2 ∂(Ly ∗ )2 ∂(Lz ∗ )2
=
1 ∂2
1 ∂2
1 ∂2
+
+
L2 ∂x∗2 L2 ∂y ∗2 L2 ∂z ∗2
(3.46)
=
1 ∗2
∇
L2
with
∇∗ 2 =
∂2
∂2
∂2
+
+
.
∂x∗ 2 ∂y ∗ 2 ∂z ∗ 2
(3.47)
Next we scale the pressure and gravity forces according to:
p∗ = p/p0 ,
~g∗ = ~g /g .
(3.48)
The Navier-Stokes equations then transform to:
V ∂~v ∗ V 2 ∗
p0
νV
+
(~v · ∇∗ )~v ∗ = − ∇∗ p∗ + 2 ∇∗2~v ∗ + g~g∗ .
∗
τ ∂t
L
ρL
L
(3.49)
All the terms with asterisks are dimensionless and all dimensions are now in the
pre-terms V /τ , V 2 /L, p0 /ρL, νV /L2 and g. Moreover, if we scaled properly, the
terms with the asterisks are O(1). This means that the order of magnitude of the
different terms is determined by the order of magnitude of the pre-terms. If we want
to compare the order of magnitude of the different terms we just need to compare
the pre-terms. For the Navier-Stokes equation it is common to compare all terms
with the convective acceleration. This is achieved by dividing the equation by the
pre-term of (~v ∗ · ∇∗ )~v ∗ , so by V 2 /L. This yields the dimensionless Navier-Stokes
equation:
p0
ν ∗ 2 ∗ gL ∗
L ∂~v ∗
∇ ~v + 2 ~g .
+ (~v ∗ · ∇∗ )~v ∗ = − 2 ∇∗ p∗ +
∗
τ V ∂t
ρV
VL
V
(3.50)
Now, also the pre-terms are dimensionless and we talk about dimensionless groups.
The order of magnitude of the dimensionless groups gives a measure of the relative
importance of the different terms in the equation compared to the convective acceleration (~v · ∇)~v . The dimensionless groups for the Navier-Stokes equations are
Strouhal number
Sr =
L
τV
=
v
[ ∂~
∂t ]
[(~v · ∇)~v ]
Reynolds number
Re =
VL
ν
=
[(~v · ∇)~v ]
[ν∇2~v ]
Froude number
Fr =
V2
gL
=
[~v · ∇)~v ]
[~g ]
(3.51)
76
transport physics
The dimensionless group p0 /ρV 2 can be omitted by making the clever choice p0 =
ρV 2 . The dimensionless Navier-Stokes equations now read:
Sr
1 ∗
1 ∗2 ∗
∂~v ∗
+ (~v ∗ · ∇∗ )~v ∗ = −∇∗ p∗ +
∇ ~v +
~g .
∗
∂t
Re
Fr
(3.52)
We now can use the dimensionless groups to simplify the Navier-Stokes equations. If,
for instance, in our oscillating object the frequency ω is very low, so the characteristic
time τ is very large say τ ≫ L/V , then the Strouhal number is small, Sr ≪ 1. In
that case we can neglect the influence of the local acceleration term and we end up
with the quasi-static Navier-Stokes equations:
(~v ∗ · ∇∗ )~v ∗ = −∇∗ p∗ +
1 ∗
1 ∗2 ∗
∇ ~v +
~g .
Re
Fr
(3.53)
We also learn that the gravity forces can be neglected if the Froude number is large,
i.e. F r ≫ 1. This is the case if V 2 /L ≫ g so for high velocities of small objects.
Finally, if additionally, for example, the viscosity of the fluid is high, this means
ν ≫ V L then the Reynolds number is small, i.e. Re ≪ 1. In that case the viscous
term ν∇2~v is large compared to the convective term (~v · ∇)~v . We the end up with
the steady Stokes equations:
~0 = −∇∗ p∗ + 1 ∇∗2~v ∗ .
Re
(3.54)
Or in dimensionfull form:
~0 = − 1 ∇p + ν∇2~v .
ρ
(3.55)
In conclusion, the dimensionless groups can be used to determine the order of magnitude of the different terms in the Navier-Stokes equation and can lead to significant
simplifications by neglecting the terms that are small.
Another important application of the dimensionless groups is their use in defining
scaled models. Since the dimensionless groups define the relative magnitude of the
different terms in de Navier-Stokes equations they completely determine the solution
if we would solve the equations. Two problems with the same Srouhal number, the
same Reynolds number and the same Froude number, give the same solution of
the Navier-Stokes equation. This means that for these two equivalent problems the
following must hold:
Sr1 =
L2
L1
=
= Sr2
τ 1 V1
τ2 V2
Re1 =
V2 L 2
V1 L 1
=
= Re2
ν1
ν2
F r1 =
V12
V2
= 2 = F r2
g1 L1
g2 L2
(3.56)
Balance of momentum
77
V1
ν1
ω1
ν2
V2
ω2
L1
L2
Figure 3.4: Flow around an oscillating object and a scaled model
Example 3.5: Hydrodynamic equivalence
As an example we consider a scaled model of the oscillating object. Suppose we
make a scaled model that is twice as small L2 = L1 /2 and put it in a flow field
with a velocity that is half of the original velocity V2 = V1 /2 (see Figure 3.4). To
obtain an equivalent model we must have:
Sr1 =
L2
L1
=
= Sr2
τ1 V1
τ 2 V2
Re1 =
V1 L1
V2 L2
=
= Re2
ν1
ν2
→
τ2 = τ2
→
ω2 = ω1
and
→
ν2 = ν1 /4 .
So the viscosity of the fluid needs to be 4 times lower. Moreover, gravity forces
must be small in both cases since the Froude numbers are different (for equal
gravitational acceleration).
78
transport physics
4
Inviscid flow
The next chapters we will focus on convective transport of matter. With convective
transport we mean transport due to a flowing medium and not due to diffusion that can
as well take place in a medium at rest. In this chapter we will consider inviscid flow or
at least flow in which viscous forces can be neglected, i.e. Re ≫ 1. In the next chapters
we will look at viscous flows and flows in boundary layers near fixed boundaries that
enclose (part of) the fluid domain considered.
79
80
transport physics
4.1
Introduction
In the case that viscous forces can be neglected (Re ≫ 1) the Navier-Stokes equation
(3.38) reduces to the Euler equation (3.27):
∂~v
1
+ (~v · ∇)~v = − ∇p + ~g .
∂t
ρ
(4.1)
Although, strictly spoken, this equation only is valid for ideal fluids with zero viscosity, the equation fairly describes fluid flow in which viscous forces play a minor
role, i.e. if the Reynolds number is high. As we learned in the previous chapter,
the Reynolds number is the order of magnitude ratio between the convective forces
ρV 2 /L and viscous forces ηV /L2 . So for inviscid flow we have:
Re =
VL
ρV 2 /L
≫1,
=
2
ηV /L
ν
(4.2)
with η the dynamic viscosity, ρ the density, V the characteristic velocity, L the
characteristic length, and ν = η/ρ the kinematic viscosity. Note that the dynamic
viscosity of fluids such as water and blood strongly differs from the dynamic viscosity
of air; O(10−3 ) Pa · s and O(10−5 ) Pa · s respectively. Because of the difference in
density (ρ = 1 kg/m3 and ρ = 103 kg/m3 ) respectively, the difference in kinematic
viscosity is relatively low; O(10−6 ) m2 /s to O(10−5 ) m2 /s. Due to the relative low
velocity and length scales, in most biological systems the internal flows (blood flow,
airflow in the lungs, flow in cells) are characterized by a low to moderate Reynolds
number (from 0 to the order 1000). The Reynolds number for aircrafts with relatively
large length scale and high velocity therefor are much higher.
Table 4.1: Viscosity and density of water, blood, dry air at atmospheric pressure
water
water
water
blood
dry air
dry air
dry air
temperature
oC
10
25
37
37
10
25
37
density
kg · m3
1000
997
993
1000
1.25
1.19
1.14
dynamic viscosity
Pa · s
1.306 · 10−3
0.890 · 10−3
0.691 · 10−3
(3 − 5) · 10−3
1.76 · 10−5
1.84 · 10−5
1.89 · 10−5
kinematic viscosity
m2 s−1
1.306 · 10−6
0.892 · 10−6
0.696 · 10−6
(3 − 5) · 10−6
1.41 · 10−5
1.55 · 10−5
1.66 · 10−5
Inviscid flow
81
Example 4.1: Reynolds number aircraft flight
For a aircraft the typical length scale is the
chord length of the wing (say 4.5 m), the
kinematic viscosity of air typically is 1.5 ·
10−5 m2 /s. For a speed of 100 m/s this
yields
Re =
100 · 4.5
VL
≈
= O(3 · 107 )
ν
1.5 · 10−5
This is that large that the Euler equations can be applied to perform hydrodynamic analysis of aircraft flight.
External flows, such as in swimming and flight of large objects, generally are associated with high Reynolds numbers (typically between 5 · 103 and 5 · 107 ).
Example 4.2: Reynolds number tuna fish swimming
For a tuna fish the typical length scale is the
distance between the pelvic and dorsal fins
(say 1 m), the kinematic viscosity of water
typically is 1 · 10−6 m2 /s. For a speed of
10 m/s this yields
Re =
10 · 1
VL
≈
= O(107 )
ν
1 · 10−6
Also this is that large that the Euler equations can be applied to perform
hydrodynamic analysis of fish swimming.
For micrometer sized swimmers such as the paramecium, however, the situation is
completely different.
Example 4.3: Reynolds paramecium swimming
For a paramecium the typical length scale
is its witdh (say 100 µm = 100 · 10−6 m),
the kinematic viscosity of water typically is
1 · 10−6 m2 /s. For a speed of 0.5 mm/s this
yields
Re =
5 · 10−4 · 1 · 10−4
VL
≈
= O(5·10−2 )
ν
1 · 10−6
The Euler equations can not be applied to perform hydrodynamic analysis of
microbe swimming.
82
transport physics
For inviscid (frictionless) flows the driving force is mostly a pressure difference (pressure gradient) or a body force such as the gravity force. As a consequence, but only
under certain conditions, it is possible to rewrite the Euler equations in a simpler
equation; the famous Bernoulli equation. This will be the subject of the next section.
4.2
Bernoulli’s equation
4.2.1
Derivation of Bernoulli’s equation
From the Euler equation the Bernoulli equation can be derived. For now, we restrict
v
ourselves to steady flow so ∂~
∂t = 0. This restriction is not really necessary but
convenient for the time being. In addition we assume that the body forces are
restricted to the gravity force. The gravity force is conservative so can be written
as the gradient of a potential field:
~g = ∇ϕgrav = −∇(gz)
(4.3)
with ~g the acceleration of gravity in z-direction and g = ||~g ||.
From vector calculus we know (or can derive):
(~v · ∇)~v = ∇( 12 ~v · ~v ) + ~v × ∇ × ~v .
(4.4)
1
~v · [~v × ω
~ + ∇p + ∇( 12 V 2 + gz)] = 0 .
ρ
(4.5)
√
We now will use V ≡ ||~v || = ~v · ~v as the magnitude of the velocity and ~ω ≡ ∇ × ~v
as the vorticity. The inner product of (4.1) with ~v then reads:
In the case that the flow is barotropic, this means in the case that ρ only depends
on the pressure p, which is certainly also true for incompressible fluids, we have :
1
p
∇p = ∇( ) .
ρ
ρ
(4.6)
The expression ~v · (~v × ~
ω) seems complex but gives a simple result: for the vector
~ω = ∇ × ~v we have ~
ω ⊥~v . Consequently, also (~v × ~ω )⊥~v (see Figure 4.1), and thus:
~v · (~v × ~
ω) = 0
(4.7)
~v × ω
~
~ω
~v
Figure 4.1: Orientation of the vectors ~v , ~ω , and ~v × ω
~
.
Inviscid flow
83
With (4.6) and (4.7) equation (4.5) changes to:
p
~v · ∇( + 12 V 2 + gz) = 0
ρ
(4.8)
In words this is: the vector ∇(p/ρ+ 12 V 2 + gz) is perpendicular to the velocity vector
~v . This means that p/ρ + 12 V 2 + gz only changes in the direction perpendicular to
~v , so perpendicular to a streamline. Consequently,
p + 12 ρV 2 + ρgz = constant
(4.9)
along a streamline.
This is the well known Bernoulli equation valid along streamlines.
We earlier defined the stream function in a 2D flow as the ψ(x, y) that satisfies:
vx =
∂ψ
,
∂y
vy = −
∂ψ
.
∂x
(4.10)
The vector field ~v then is described by the scalar field ψ, moreover this description
automatically satisfies the incompressibility constraint ∇ · ~v = 0. The streamlines,
defined as the lines that are tangent to the velocity field (see Section 1.2.3), turned
out to be lines for which the stream function is constant. So for inviscid flow both
the value of the stream function ψ as well as Bernoulli’s constant p + 12 ρV 2 + ρgh
are preserved.
B0
B1
v
B2
Figure 4.2: Streamlines and the Bernoulli constants.
In the case that ~
ω = ~0, i.e the flow is irrotational we do not need to project the
Euler equation to a streamline and we directly find from (4.1) assuming the steady,
barotropic situation:
p + 21 ρV 2 + ρgz = constant
everywhere in a rotational free field.
(4.11)
This is the Bernoulli equation for rotation free (irrotational) velocity fields which is
valid everywhere in the domain.
The Bernoulli equation relates pressure p, velocity V and height z above a certain
reference level z = 0 and can be used in many practical situations, but only holds
under the following conditions:

• conservative force field 

• inviscid media
Bernoulli along streamline


• barotropic media
84
transport physics
•
•
•
•

conservative force field 



inviscid media
Bernoulli everywhere in rotational free field

barotropic media



irrotational flow
In most cases we use the Bernoulli equation in the form
p1 + 21 ρV12 + ρgz1 = p2 + 12 ρV22 + ρgz2
(4.12)
where the indices 1 and 2 refer to two positions ~r1 and ~r2 on a streamline, in case the
flow is not irrotational, or anywhere in the domain in case of an irrotational flow.
In the next sections we will consider some applications of the Bernoulli equation.
4.2.2
Applications of the Bernoulli equation
Prandtl/Pitot-tube
We can make use of the relation between pressure and velocity magnitude as expressed in the Bernoulli equation to design instruments that measure fluid flow
velocity or the speed of moving object in a fluid at rest such as the airspeed of an
aircraft.
p = p∞
V∞
p∞
stagnation point
p = pstag , ~v = ~0
∆H
Figure 4.3: The Prandtl/Pitot tube
With a pitot tube (see Figure 4.3) the pressure difference (pstag − p∞ ) is measured
with a simple manometer. The Bernoulli equation, ignoring the influence of gravity,
tells us:
2
2
= pstag + 21 ρVstag
.
p∞ + 12 ρV∞
(4.13)
Since Vstag = 0 we find:
V∞ =
s
2
(pstag − p∞ ) .
ρ
(4.14)
Here V∞ is the velocity we want to measure. The stagnation pressure at the tip of
the pitot tube is referred to as pstag . Note that the velocity at the upper hole in
the pitot tube is V∞ . Consequently, the pressure at the upper hole is equal to the
ambient pressure p∞ .
Inviscid flow
85
pa
pa
h1
A1
h2
p1
p2
V1
V2
A2
Figure 4.4: The Venturi flow rate meter
Venturi flow rate meter
The rate of an invisced flow through a tube can be measured by inducing a narrowing
of the tube and measuring the difference between the pressure in the tube and the
pressure at the location of the narrowing. This situation is depicted in figure 4.4.
If we neglect the gravity forces for the flow in the tube, the equation of Bernoulli
gives:
p1 + 21 ρV12 = p2 + 21 ρV22
(4.15)
Together with the continuity equation (balance of mass):
A1 V1 = A2 V2
(4.16)
this yields:
p1 − p2 = 12 ρV12
!
A21
1
1
− 1 = 12 ρV12 A21
− 2
2
2
A2
A2 A1
= 12 ρV12 A21
A21 − A22
A21 A22
!
(4.17)
Note that since A1 > A2 we have V1 < V2 and thus p1 > p2 . This pressure difference
can be computed from:
p1 = pa + ρgh1 ,
p2 = pa + ρgh2
→
p1 − p2 = ρg(h1 − h2 )
(4.18)
Consequently,
2g(h1 − h2 )A21 A22
(V1 A1 ) =
A21 − A22
2
→
qV = A1 V1 = A2
s
2g(h1 − h2 )
. (4.19)
1 − (A2 /A1 )2
The volume flow qV tus can be measured in a ventury tube by measuring the difference between the levels h1 and h2 in the manometer tubes.
We will illustrate the usage of the Bernouilli equations in the following two examples.
86
transport physics
Example 4.4: Emptying of a vessel with a small hole.
Consider the vessel as depicted in the figure below.
pa
Vtop
A1 (≫ A0 )
~g
ρ
h
A0
z=0
V
pa
Conservation of mass implies:
A1 Vtop = A0 V ,
with
Vtop = −
dh
.
dt
If the cross-sectional area of the hole is small, i.e. A1 ≫ A0 then we can neglect
the velocity of the top surface with respect the outflow velocity, i.e. V ≫ Vtop .
Application of the Bernoulli equation (4.9) gives:
2
+ ρgh = pa + 12 ρV 2 + 0
pa + 12 ρVtop
|
{z
top surface
}
|
{z
outflow
(4.20)
}
2 ≪ V 2 we can assume that the flow is quasi-steady and thus:
Since Vtop
V =
p
2gh
Note that this result does not depend on the density ρ of the fluid.
(4.21)
Inviscid flow
87
Example 4.5: Water clock
The water clock is an analogue of the hourglass and consists of an axi-symmetric
surface of revolution C(r, z) containing water. In the bottom (z0 = 0) the water
flows out with velocity V0 (t) through a small hole with cross-section surface area
A0 (see figure below). We assume that viscous forces can be neglected.
pa
zh = h(t)
Ah (t)
Vh
z
C(r, z)
V0 (t)
r
pa
A0
Source right panel: http://academic.brooklyn.cuny.edu
The cross-sectional area depends on the axial coordinate z, to make sure the free
surface will have a constant downward velocity Vh . To compute the shape of the
curve to achieve such a constant velocity of the free surface, we first use the mass
balance:
→
Ah (t)Vh = A0 V0 (t)
V0 (t) =
Vh
Ah (t) .
A0
(4.22)
We also use Bernoulli’s law:
pa + 21 ρVh2 +ρgzh (t) = pa + 21 ρV02 (t)+ρgz0
→
V02 (t) = Vh2 +2gzh . (4.23)
If we combine (4.22) and (4.23) we get:
Vh2 2
A = Vh2 + 2gzh
A20 h
→
A2h = π 2 rz4 = A20 +
2gA20
zh .
Vh2
As long as Ah ≫ A0 so zh ≫ Vh2 /2g we obtain:
rz4 =
2gA20
zh .
Vh2 π 2
and thus:
C(r, z) : ~r(z) = (kz)1/4~er + z~ez ,
k=
2gA20
>0.
Vh2 π 2
For a given outflow area A0 , the constant k must be chosen such that the desired
Vh is given by:
A0
Vh =
π
r
2g
= a20
k
r
2g
k
with a0 the radius of the outflow tube.
88
4.3
transport physics
A few remarks on irrotational flow
For a 2D flow in the x, y-plane the vorticity ω
~ = ∇ × v = ωz~ez with
ωz =
∂vx
∂vy
−
.
∂x
∂y
(4.24)
Substitution of the stream function:
∂ψ
∂ψ
vx =
,
vy = −
,
∂y
∂x
(4.25)
see also (4.10) gives:
ωz = −
∂2ψ ∂2ψ
−
= −∇2 ψ .
∂x2
∂y 2
(4.26)
This is a so called Poisson-equation. For a irrotational flow we have ωz = 0 and
thus:
∇2 ψ = 0 .
(4.27)
So the stream function satisfies the so called Laplace equation. In that case with
ϕ the velocity potential, using ∇ × ~v = 0 and knowing that the vector identity
∇ × ∇ϕ = 0 must hold, the velocity can be written as:
~v = ∇ϕ ,
From the continuity equation
(4.28)
∂vx ∂vy
+
= 0 we find:
∂x
∂y
∇2 ϕ = 0 .
(4.29)
So also the velocity potential ϕ satisfies the Laplace equation. These Laplace equations form the basis of the so called potential theory for inviscid irrotational flows,
also called potential flows.
Note that
∇ψ = −vy~ex + vx~ey
(4.30)
∇ϕ = vx~ex + vy ~ey
(4.31)
and
and thus:
∇ψ · ∇ϕ = −vy vx + vx vy = 0 .
(4.32)
This implies that streamlines ψ = const are perpendicular to equipotential lines
ϕ = const.
In the next section we will study a number of elementary potential flows and illustrate that more complex flows can be constructed using superposition of elementary
solutions. This indeed is possible sinds the Laplace operator is linear and solutions
of (4.27) and (4.29) can be superimposed (added).
Inviscid flow
89
Figure 4.5: Example of a potential flow soltion of the flow around a flat plate.
The outer flow field far away from the boundary of rigid bodies can often be approximated by irrotational flow and constructed by superposition (see e.g figure 4.5).
4.3.1
Elementary potential flow solutions
Uniform flow
Uniform flow given by
~v = U~ex
(4.33)
is described by:
ϕ = Ux ,
ψ = Uy .
(4.34)
Indeed, in that case:
vx =
∂ϕ
=U,
∂x
vy =
∂ϕ
=0
∂y
and
vx =
∂ψ
=U,
∂y
vy = −
∂ψ
= 0 . (4.35)
∂x
Similarly we get for
~v = V ~ey
(4.36)
the potential and streamfunction:
ϕ=Vy ,
ψ = −V x .
(4.37)
Source and sink
We will use polar coordinates so
vr =
∂ϕ
,
∂r
vφ =
1 ∂ϕ
r ∂φ
and vr =
1 ∂ψ
,
r ∂φ
vφ = −
∂ψ
,
∂r
(4.38)
with
∇2 ϕ =
1 ∂ ∂ϕ
1 ∂2ϕ
(r ) + 2 2 = 0 and
r ∂r ∂r
r ∂φ
∇2 ψ =
1 ∂ ∂ψ
1 ∂2ψ
(r
) + 2 2 = 0 . (4.39)
r ∂r ∂r
r ∂φ
90
transport physics
The velocity of a source (or sink) is:
vr =
Q1
,
2π r
vφ = 0 .
(4.40)
The potential and the stream function are:
ϕ=
Q
ln r ,
2π
ψ=
Q
φ.
2π
(4.41)
Flow over an obstacle
By superposition of a uniform flow and a source we obtain the flow over a semiinfinite so-called Rankine body. The velocity field is:
vr = U cos φ +
Q1
,
2π r
vφ = −U sin φ
(4.42)
with a streamfunction derived from vφ = − ∂ψ
∂r :
ψ=−
Z
vφ dr = U r sin φ + Cφ (φ) .
(4.43)
So :
∂C(φ)
∂ψ
= U r cos φ +
.
∂φ
∂φ
Hence, with vr =
1 ∂ψ
r ∂φ
and thus
(4.44)
∂ψ
∂φ
= rvr we get:
Q
∂C(φ)
= rvr − U r cos φ = U r cos φ +
− U r cos φ .
∂φ
2π
(4.45)
This yields:
C(φ) =
Q
φ
2π
(4.46)
and thus:
ψ = U r sin φ +
Q
φ.
2π
(4.47)
These streamlines are plotted in figure 4.6. The body is determined by the streamfunction that passes the stagnation point at which vr = 0 and vφ = 0. This is only
possible when φ = π and thus:
vr = −U +
Q1
=0
2π r
→
r=
Q
.
2πU
(4.48)
In Cartesian coordinates the stagnation point thus is:
~rS = −
Q
.
2πU
(4.49)
As, unlike to the situation in e.g. aerodynamics, in biomedical research irrotational
flow is less relevant, we will not dig further into this interesting subject in this course.
Inviscid flow
91
O
Q
~ex
~rS = − 2πU
Figure 4.6: Example of a potential flow soltion of the flow over a blunt body.
4.4
Inviscid and turbulent tube or channel flow
4.4.1
Inviscid 1D tube or channel flow
Consider the flow of an inviscid medium in a tube (cylindrical cross-section) or
channel (rectangular cross-section) with a varying radius (tube) or height (channel).
In the case that the variations in cross-sectional area A(x) are not too large we can
assume a uniform velocity profile vx (x) in axial (x-) direction (see Figure 4.7).
From mass conservation (the mass balance) we can derive:
qm = ρvx (x)A(x) = const
(4.50)
and consequently:
d
(ρvx (x)A(x)) = 0 .
dx
(4.51)
For an incompressible fluid (ρ=const) we find :
A
dvx
dA
+ vx
=0.
dx
dx
(4.52)
A(x)
x
vx (x)
Figure 4.7: Inviscid flow in a tube or
channel.
dA
>0
dx
dvx
<0
dx
divergent
dA
<0
dx
dvx
>0
dx
convergent
Figure 4.8: Divergent and convergent
1D channel flow.
92
transport physics
Ψ = Ψ2
y
x
Ψ = Ψ1
B
~v
vx
vy
S
dy
dx
Figure 4.9: Computation of volume flux
ΦV between two streamlines.
y
x
ds
~n
A
~v
vy
vx
~n
Figure 4.10: Enlarged view of integration path A → B.
This implies that changes in cross-section area result in changes in velocity. The
larger the cross-sectional area the lower the flow (see Figure 4.8). Bernoulli’s equation gives the relation between the velocity vx (x) and the pressure p(x):
p(x) + 12 ρvx2 (x) + ρgz = constant.
(4.53)
So, the larger the cross-sectional area, the lower the velocity, and the larger the
pressure.
Since a streamline is always tangent to the velocity ~v , we can consider streamlines
as fixed channel walls in the plane of an inviscid flow. The volume flux between
two streamlines ψ = ψ1 and ψ = ψ2 then can be computed in the same way as the
volume flux in a channel. We can determine the flux by integration of the velocity
along an arbitrary path A → B (see Figures 4.9 and 4.10).
The volume flux is
qV =
ZB
A
~v · ~nds =
ZB
A
(vx dy − vy dx)
(4.54)
where we used that ~nds = dy~ex − dx~ey . With the aid of the definition of the stream
function (4.10) this changes to:
qV =
Z
B
A
∂ψ
∂ψ
dy +
dx =
∂y
∂x
Z
B
A
dψ = ψ2 − ψ1 .
(4.55)
Apparently, the volume flux qV in a 2D flow is equal to the difference in the value
of the streamfunction ∆ψ between the streamlines.
As a consequence, the distance between two streamlines is related to the velocity
between the streamlines: we have increase in velocity when streamlines contract and
decrease of velocity when streamlines diverge (see Figure 4.11).
Remark: the stream function is also defined for viscous incompressible media.
4.4.2
Turbulent tube flow
Until now we only considered laminar flow, which is correct for relatively low values
of the Reynolds number. In many cases where the geometry is simple and we have
fully developed flow with a uniform velocity filed, the non-linear convective term in
Inviscid flow
93
the Navier-Stokes equations can be neglected or even essentially is equal to zero.
If, however, small disturbances due to geometrical imperfections or small temporal
vibrations are present these disturbances can grow to rather large fluctuations that
may spread out over the entire flow field. For low Reynolds numbers, viscous forces
will damp these fluctuations and a laminar flow will recover. At relatively high
Reynolds numbers, though, when the non-linear convection term becomes dominant,
these disturbances can persist and a turbulent flow can develop. This process is
illustrated in Figure 4.12 where in the top-panel the streakline in a laminar tube flow
is depicted and in the lower-panel the same streakline is visualized for a turbulent
flow. This phenomenon has been studied by Osborne Reynolds. In a laminar flow
the streakline is smooth and does not show any fluctuations. If we increase the
Reynolds number ReD = V̄ D/ν to values above ≈ 2000 (this value depends on the
smoothness of the wall of the tube), the streak line shows more and more fluctuations
and changes continuously in time.
There still is a mean flow that carries the dye, but large, irregular fluctuations
occur: the flow has become instable and a transition to turbulent flow is observed
(see Figure 4.14). When in this situation one measures the velocity at a certain fixed
location fluctuations as depicted in Figure 4.15 can be found.
Apparently the velocity vx (t) consists of a time averaged value v̄x with superimposed
irregular noisy fluctuations vx′ (t). Also other flow quantities such as vy and p show
this characteristics.
Ψ2
Ψ1
Qv = ∆Ψ = Ψ2 − Ψ1 = const
Figure 4.11: Streamline pattern: high velocities at higher streamline density.
D
laminar flow
V̄
dye injection
streak line
Figure 4.12: Laminar tube flow.
D
turbulent flow
V̄
dye injection
streak line
Figure 4.13: Turbulent tube flow
94
transport physics
We can write the velocity components vx en vy and the pressure p as:
vx (t) = v̄x + vx′ (t)
vy (t) = v̄y + vy′ (t)
p(t) = p̄ + p′ (t)
This decomposition in time averaged values (v̄x , v̄y , p̄) and fluctuations (vx′ , vy′ , p′ ) is
referred to as the Reynolds-decomposition. The time averaging of for instance vx (t)
is obtained via:
v̄x =
1
T
Z
T
0
vx (t)dt ,
(4.56)
with T not too small.
An adequate description of turbulent flow is still not available and especially extended experimental research has learned us about some general characteristics of
turbulent flow. For example it is well established that turbulent mixing is far more
efficient than distributive mixing in a laminar flow. In this remainder of this course,
however, we will restrict ourselves to the laminar flow regime, so to relatively low
Reynolds numbers.
Figure 4.14: Examples of transition to turbulence taken from www.mecaflux.com. Left
panel: transition to turbulent pipe flow. Right panel: transition to turbulent flow over
a cylinder.
T
1.2
vx /v̄x
1
0.8
0.6
v̄x
0.4
0.2
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
t/T
Figure 4.15: Course of the velocity in a turbulent flow.
5
Viscous flow
In this chapter we will consider viscous flow problems. These problems are descibed by
the Navier-Stokes equations. In Chapter 3 these equations have been derived. In this
chapter we will learn how these equations can be used to find descriptions of internal
flows in a bounded domain and external flows in semi-infinite or infinit domains with an
object immersed. Finally, some remarks on flow in porous media are given.
95
96
transport physics
5.1
Internal viscous flow
In this section we will consider flow between two flat plates and flow in a tube.
5.1.1
Steady flow between two flat plates
Consider the 2D flow, ~v = vx~ex + vy ~ey , of an incompressible fluid with viscosity ν,
between two parallel plates (see Figure 5.1). The lower plate (y = 0) is fixed while
the upper plate (y = h) moves with velocity V in x-direction. We will assume that
the flow is fully developed in x-direction, so:
∂vx
= 0 and
∂x
∂vy
= 0.
∂x
(5.1)
From this it follows that the velocity only depends on y: ~v = ~v (y).
In addition, from the continuity equation (mass balance) for incompresible fluids,
we know:
∂vx ∂vy
+
=0
∂x
∂y
→
∂vy
=0
∂y
→
vy = const
(5.2)
Together with the boundary condition at y = 0 (see Figure 5.1): vy (0) = 0 we find
vy (y) = 0 everywhere. This means that ~v = vx~ex , and thus that the velocity only is
in x-direction.
The only unknown left is the x-component of the velocity as a function of y: vx (y).
This unknown can be derived from the x-component of the Navier-Stokes (momentum) equation:
∂vx
∂vx
1 ∂p
∂vx
+ vx
+ vy
=−
+ν
∂t
∂x
∂y
ρ ∂x
∂ 2 vx ∂ 2 vx
+
∂x2
∂y 2
!
.
(5.3)
∂vx
= 0), in x-direction only (vy = 0), and fully developped
Since the flow is steady (
∂t
∂vx
(
= 0), this equation reduces to:
∂x
0=−
∂ 2 vx
1 ∂p
+ν
.
ρ ∂x
∂y 2
(5.4)
V ~ex
y=h
ρ, ν
y
x
y=0
Figure 5.1: Flow between two parallel flat plates. The upper plate, at distance h parallel
to the lower plate, moves with velocity V ~ex , while the lower plate is kept at its place.
Viscous flow
97
V
vx (y)
Figure 5.2: Couette flow. There is no pressure gradient in x direction.
The y-component of the Navier-Stokes equation simplifies to:
0=−
1 ∂p
.
ρ ∂y
(5.5)
Consequently, the pressure does not depend on y and only on x, so p = p(x). Since,
in addition, vx only depends on y, the partial derivatives in equation (5.4) then can
be replaced by ordenary derivatives:
1 dp
d2 vx
=
.
2
dy
η dx
(5.6)
Here the left hand side only depends on y and the right hand side only depends on
x. In other words, both terms must be equal to the same constant. Dependent on
dp
we can distinguish two cases:
dx
Case 1:
If
dp
= 0, then equation (5.6) reads:
dx
d2 vx
= 0,
dy 2
(5.7)
with general solution:
vx (y) = Ay + B.
(5.8)
We need two boundary conditions to determine the two integation constants A and
B. Since the flow is viscous, no-slip conditions will hold at the walls:
y=0:
y=h:

vx = 0 

no-slip conditions
(5.9)

vx = V 
Thus we find: A = V /h and B = 0, so:
vx (y) = yV /h.
This is a linear velocity profile (see Figure 5.2) known as Couette flow.
(5.10)
98
transport physics
Once the velocity field is known it is possible to compute the shear stress. To this
end, we write τ = 2ηD in matrix form for Cartesian coordinates, so τ = 2ηD, which
for two dimensions yields:

∂vx
 ∂x

so:
τ = 2η 


1
2
1
2
∂vx ∂vy
+
∂y
∂x
∂vy
∂vx
+
∂x
∂y

∂vy
∂y

 0

 
=
 
 
dvx
η
dy
η

dvx
dy 

0
τxx = τyy = 0
(viscous normal stresses are zero)
τxy = τyx
(shear stresses))
ηV
dvx
=
=η
dy
h



(5.11)
This stress distribution will induce deformation of volume elements as depicted in
Figure 5.3.
The stress vector acting on the fluid near the upper plate (~n = ~ey ) is thus:
~t = σ · ~n = −p~ey + ηV (~ex~ey + ~ey ~ex ) · ~ey = ηV ~ex − p~ey
h
h
(5.12)
~u that is needed to move the upper plate with velocity V in positive
The force F
x-direction then equals
ηV
A~ex − pA~ey ,
F~u = ~tA =
h
(5.13)
where A represents the surface area of the upper plate. If the pressure between
the plates is equal to the (reference) pressure above the upper plate, the force in
y direction will be delivered by the pressure above the plate. For the lower plate
(~n = −~ey ) we have:
~t = σ · ~n = −p(−~ey ) + ηV (~ex~ey + ~ey ~ex ) · (−~ey ) = − ηV ~ex + p~ey
h
h
(5.14)
and thus a force F~l given by:
ηV
F~l = ~tA = −
A~ex + pA~ey
h
(5.15)
τyx (y + dy)
τxy (x)
dy τxy (x + dx)
dx
τyx (y)
gives deformation
Figure 5.3: Deformation of a volume element in a flow in which shear stresses are
non-zero.
Viscous flow
99
vx (y)
1
2
x
x
dp
>0
dx
dp
<0
dx
Figure 5.4: Poiseuille flow. The walls are both fixed and the flow is driven by a pressure
gradient in x-direction.
This is the force that is needed to keep the lower plate at a fixed position. It is
negative in x-direction because the fluid tends to drag the lower plate in positive
x-direction. It is positive in y-direction because the pressure between the plates
pushes the wall in negative y-direction.
dp
6= 0 we can find vx (y) by integrating (5.6) twice with respect to y.
dx
As mentioned earlier, the left hand side of (5.6) is at most a function of y, while the
right hand side at most is a function of x. From this we can conclude:
Case 2:
If
1 dp
d2 vx
=
= constant.
2
dy
η dx
(5.16)
Note that, to satisfy (5.16), the pressure must be a linear function of x. Two times
integration with respect to y gives:
vx (y) =
1 dp 2
y + Ay + B.
2η dx
(5.17)
Dependent on the boundary conditions we can have two sub-cases:
Sub-case 2a:
both plates are fixed:

vx (y = 0) = 0 


vx (y = h) = 0 
no-slip conditions → B = 0, A = −
h dp
2η dx
(5.18)
with solution:
vx (y) =
1 dp
y(y − h) .
2η dx
(5.19)
this is a parabolic velocity profile (see Figure 5.4) known as Poiseuille flow.
Sub-case 2b: the upper plate moves according to vx (y = h) = V , so we get a
combination of (5.10) and (5.19):
vx (y) =
Vy
h
Couette
+
1 dp
y(y − h)
2η dx
Poiseuille
(5.20)
100
transport physics
dp
>0
dx
vx > 0
V
=
+
Figure 5.5: Combination of Couette and Poiseuille flow.
Also for this case the viscous stresses are given by:
τxx = τyy = 0
(viscous normal stresses are zero)
τxy = τyx
(shear stresses)
dvx
=η
dy
Making use of (5.20) we find for the shear stress in the fluid:
τyx = η
dvx
dy
ηV
h
|{z}
=
+
constant
1 dp
(2y − h) .
2
}
| dx {z
(5.21)
lineair
The shear stress at the upper plate (y = h) is thus:
τyx (y = h) =
ηV
h dp
+
h
2 dx
(5.22)
The stress vector acting on the fluid near te upper plate (~n = ~ey ) now equals:
~t = σ ·~n = −p~ey + ηV + h dp (~ex~ey +~ey ~ex )·~ey =
h
2 dx
h dp
ηV
+
~ex −p~ey (5.23)
h
2 dx
For the force F~u we find:
F~u = ~tA =
ηV
h dp
+
A~ex − p(x)A~ey
h
2 dx
(5.24)
Agian the first term of the x-component represents the force needed to move the
upper plate with speed V in positive x-direction. To interpret the second term we
consider the case that V = 0 and dp/dx < 0. The resulting force acts in negative
x-direction and represents the force needed to prevent the upper plate from being
dragged in positive x-direction by the fluid flow in positive x-direction. The y
~u is the force that is needed to counteract the pressure between the
component of F
plates. This now, according to (5.16), is not a constant, that can be compensated
by the pressure above the upper plate, but a linear function of x.
At the lower plate (y = 0) we have in a similar way:
F~l =
h dp
−ηV
+
A~ex + p(x)A~ey
h
2 dx
(5.25)
Viscous flow
101
Example 5.1: Piston in a cylinder
A piston with a radius of aP moves in a cylinder of radius aS . Oil with viscosity
of ηs fills the gap with width h ≪ aS between the piston and the cylinder. The
pressure at the left side of the psiton is pa while at the right side of the piston
the pressure is ps . The piston moves with velocity Vp~ez . We want to know how
much force F is required to move the cylinder if its velocity is V ~ex ?
pa
r
ap
as
ps
F
z
L
The total force is the sum of the pressure force due to the pressure difference
pa − ps over the piston and the friction forcea Fw between the piston and the
cylinder wall.
F = Fp + Fw = πa2p (ps − pa ) + 2πap Lτw ,
with τw the wall shear stress at the piston wall. If we neglect boundary effects at
the ends of the gap between the piston and the cylinder, the wall shear stress is
given by equation (5.22), i.e.
τw =
ηV
h ps − pa
+
,
h
2 L
with h = as − ap . This is correct if the curvature of the cylinder wall can be
neglected, or, in other words, if h ≪ as . The total force then is given by
F =
πa2p (ps
ηV
h ps − pa
− pa ) + 2πap L
+
h
2 L
.
Note that, in case we keep the piston fixed (V = 0), there will be a small leakage
flow of oil in the negative z −direction introducing an extra force with magnitude
πap h(ps − pa ).
5.1.2
Poiseuille tube flow
Earlier we analysed inviscid flow in a channel or a tube and did not really think about
the driving force of these flows. Especially when the role of viscosity becomes more
important the need for a driving forces becomes evident. Two possible mechanisms
are depicted in the figure 5.6. In (a) muscular contraction in combination with two
valves takes care of a pumping action. The flow of blood in arteries is based on this
principle. In (b) a net flux is accomplished by moving ciliae at the wall of the tube
102
transport physics
Figure 5.6: Illustration of two driving
forces for tube flow.
Figure 5.7: Transport by ciliae at the wall
of a tube.
(see Figure 5.7) : by collaborate and coordinated motion the ciliae sweep the fluid
in a certain direction. For instance the transport of mucus in the lungs is based on
this principle.
Another, more passive, mechanism is based on the existence of a pressure difference
along the axial direction of a tube (see Figure 5.8). It shows that, a pressure difference p1 − p2 > 0 results in a volume flux qv . We will analyse this situation for
a steady, axi-symmetric, fully developed flow in a cylindrical tube with a circular
cross-section and diameter 2a. To solve this flow problem we will use a cylindrical
coordinate system (r, φ, z) with velocity components:
~v = vr~er + vφ~eφ + vz ~ez .
(5.26)
We will assume axi-symmetric flow without swirl
veloped steady flow
∂
= 0, vφ = 0
∂φ
and fully de-
∂~v
∂~v
= 0,
= 0 . Together with the continuity equation this
∂z
∂t
results in:
1 ∂vφ ∂vz
1 ∂
(rvr ) +
+
=0
r ∂r
r ∂φ
∂z
→
Together with boundary condition vr (a) =
1 ∂
(rvr ) = 0
r ∂r
→
vr =
C
.
r
(5.27)
C
= 0, this means C = 0 and results in
a
axial flow: vr = vφ = 0, vz = vz (r).
We are interested in the radial distribution of the axial velocity vz (r). As the
driving force is a pressure difference in axial direction, we will need to analyse the
z-component of the Navier-Stokes equations. Before doing this, note that together
with the results obtained above we find for the r and φ-components of the NavierStokes equation:
∂p
= 0,
∂r
∂p
=0
∂φ
→ p = p(z) .
qv
p1
(5.28)
2a
p2
r
z
φ
ρ, ν
p1 − p2 > 0 → q v
Figure 5.8: Tube flow driven by a pressure
difference p1 − p2 .
Figure 5.9: Poiseuille tube flow.
Viscous flow
r
103
r
vz (r)
z
if
dp
<0
dz
area: 2πrdr
Figure 5.11: Integration in cylindrical coordinates.
Velocity profile of a
Figure 5.10:
Poiseuille flow.
dr
If we substitute this information in the z-component of the Navier-Stokes equation,
this simplifies to 1 :
0=−
dvz
1 dp ν d
+
r
ρ dz
r dr
dr
(5.29)
or with kinematic viscosity ν = η/ρ, η being the dynamic viscosity:
dvz
d
r
dr
dr
=
1 dp
r.
η dz
(5.30)
since the pressure is not a function of r integration over r gives:
r
dvz
r 2 dp
=
+ C1
dr
2η dz
→
dvz
r dp C1
=
+
.
dr
2η dz
r
(5.31)
A second integration over r yields:
vz (r) =
r 2 dp
+ C1 ln r + C2 .
4η dz
(5.32)
The integration constants C1 and C2 are determined by the boundary conditions:
r = a : vz = 0
r = 0 : vz = finite or :
dvz
= 0,
dr
(symmetry)
This results in:
vz (r) = −

(no-slip condition) 
1 dp 2
(a − r 2 ) .
4η dz
(5.33)

(5.34)
The Poiseuille profile has a parabolic shape, with a velocity maximum vmax at r = 0:
vmax = vz (r = 0) = −
1 dp 2
a .
4η dz
(5.35)
Consequently, (5.34) can also be written as
vz (r) = vmax (1 −
1
dvz
1 d
r
Note that:
r dr
dr
r2
).
a2
=
(5.36)
1 ∂vz
∂ 2 vz
+
∂r 2
r ∂r
.
104
transport physics
The corresponding volume flux qv can be found by integration of vz (r) over the
cross-sectional area of the tube:
qv =
Za
0
vz (r) · 2πrdr .
(5.37)
The result then can be written as
dp
8ηqv
=− 4
dz
πa
(5.38)
and is know as the Hagen-Poiseuille relation. The cross-sectional mean velocity v̄z
is:
v̄z =
a2 dp
qv
.
=
−
πa2
8η dz
(5.39)
Substitution in (5.36) gives:
r2
1− 2
a
vz (r) = 2v̄z
!
.
(5.40)
Figure 5.12 schematically shows the analogy between the pressure difference ∆p and
the volume flux qv in a Poiseuille flow with difference in voltage ∆V and electrical
current I in a resistor.
The relation of Hagen-Poiseuille can be written as (see 5.38):
∆p = p1 − p2 = qv ·
8ηL
,
πa4
R=
8ηL
πa4
is a ’resistance’ .
(5.41)
Note that the law of Ohm reads ∆V = V1 − V2 = IRΩ . From (5.41) we see that the
flow resistance for a Poiseuille tube flow ∼ 1/a4 . This implies that a small decrease
in diameter D = 2a leads to a large increase in flow resistance. This is illustrated
by the situation in the capillary outflow of Figure 5.13.
Water flows from a vessel (water height H above outflow tubes) through 3 capillaries
with length L and diameters: D1 = D , D2 = 2D , D3 = 4D . The pressure
p1
pa
qv
p2
~g
L
ρ
RΩ
V1
I
V2
D1 = D
qv1 p
a
Figure 5.12: Analogy between electrical
and hydrodynamical resistance.
D2 = 2D
H
D3 = 4D
qv3
qv2
Figure 5.13: Outflow through capillaries
with different diameters.
Viscous flow
105
difference over the capillaries is: ∆p = ρgH.
From Hagen-Poiseuille’s relation (5.41) we get:



qv2
= 16qv1
qv3

= 256qv1 
(5.42)
If we would compare this with the situation in which the viscous resistance would
be negligible, then the volume flux is
qv =
π 2
D V ,
4
(5.43)
with V the uniform velocity in the tubes. From Bernoulli’s equation this is: V =
√
2gH, and thus:

√
qv1 = π4 D2 2gH 





qv2 = 4qv1
(5.44)






qv3 = 16qv1
Example 5.2: Force on a syringe shaft
We revisit the piston in a cylinder example and assume it was part of a syringe as
depicted in the figure below. We are interested in the pressure ps in the cylinder.
F
r
pa
ap
pa
as ps
2a
z
Ln
L
The inner radius of the needle is a, its length is Ln . The dynamic viscosity of
the fluid in the syringe is η. If we neglect end effects, so assume fully developed
flow in the needle, and assume quasi-static friction dominated flow, the pressure
difference ps − pa is given by (5.41) and thus:
ps − pa = q v
8ηLn
.
πa4
The flow qV is given by:
qv = πa2s V ,
so:
ps = pa +
8ηa2s V Ln
.
a4
106
transport physics
Figure 5.14: Pressure distribution in the circulatory system.
The relation of Hagen-Poiseuille is also of importance for the flow in the small
blood vessels (arterioles and capillaries) and explains the pressure distribution in
the human circulatory system as depicted in Figure 5.14. The figure clearly shows
the large pressure drop over the arterioles and capillaries where the radius is small.
Viscous flow
5.2
5.2.1
107
External viscous flow
Flow at an impulsively moving plate
We will study the so called Rayleigh-problem (see Figure 5.15): the space defined
by the upper x, y-plane (y > 0) consists of an incompressible newtonian fluid with
kinematic viscosity ν and density ρ. An infinite long flat plate is located at the plane
y = 0. Initially the plate and the fluid are at rest. At t = 0 the plate starts to move
with a constant velocity V in the positive x-direction.
d
= 0 (including pressure gradients
We assume that end-effects can be neglected so
dx
in x-direction) and thus ~v (x, y, t) = vx (y, t)~ex .
The question is how the fluid above the plate will move, i.e. what is vx (y, t) under
the following boundary/initial conditions:
t ≤ 0 : ~v = ~0
t ≥ 0 : ~v (y = 0) = V ~ex

all in rest 

(5.45)

plate moves 
The x-component of the momentum equation (Navier-Stokes) reads:
∂vx
∂vx
1 ∂p
∂vx
+ vx
+ vy
=−
+ν
∂t
∂x
∂y
ρ ∂x
With vy = 0 and
∂ 2 vx ∂ 2 vx
+
∂x2
∂y 2
!
.
(5.46)
d
= 0 this equation reduces to:
dx
∂ 2 vx
∂vx
=ν
∂t
∂y 2
(5.47)
In addition to the conditions at y = 0 (5.45) the solution of (5.47) must also satisfy
t ≥ 0,
y→∞
:
vx → 0 .
(5.48)
Equation (5.47) is a ’diffusion equation’ for momentum and is similar to (2.60) in
Section 2.5. The kinematic viscosity takes the role of the diffusion coefficient. We
therefore, again, introduce the similarity parameter ξ(y, t) according to:
ξ=√
y
4νt
(5.49)
y
fluid ρ, ν
x
V
t>0
Figure 5.15: Schematic of the Rayleigh-problem.
108
transport physics
which leads to the following differential equation (see also (2.70) in Section 2.5):
d2 vx
dvx
+ 2ξ
=0
2
dξ
dξ
(5.50)
with boundary/initial conditions in terms of ξ:
→ ξ=0
y=0

y→∞ 

t=0
: vx = V,
(5.51)
→ ξ → ∞ : vx = 0.


and a solution equivalent to (2.72)
vx (ξ) = A erf(ξ) + B
(5.52)
with A and B two integration constants. Application of the conditions (5.51) gives:
vx (ξ = 0) = V



→ B=V
vx (ξ → ∞) = 0 →
The final result is:
(5.53)

V +A=0 
vx (ξ) = V [1 − erf(ξ)]
(5.54)
or:
vx (y, t) = V 1 − erf
y
√
4νt
(5.55)
This velocity profile is plotted in Figure 5.16 for different instants of time. in figuur
5.16.
The motion of the plate can be sensed within a layer with thickness δ(t). Often this
thickness is defined such that on y = δ the velocity is only 1% of the velocity of the
plate V , so :
vx ≃ 0.01V
→
erf
δ(t)
√
4νt
≃ 0.99 .
(5.56)
3
3
3
3
2
2
2
2
1
1
y
0
0
V
0.5
t=0
1
0
0
δ(t)
δ(t)
0.5
t = ∆t
1
1
0
0
1
0.5
t = 2∆t
1
0
0
0.5
t = 3∆t
Figure 5.16: Velocity profiles according to (5.55).
1
Viscous flow
109
Since erf(2) ≃ 0.99 we find:
δ(t)
√
=2
4νt
(5.57)
√
δ(t) = 4 νt
(5.58)
so:
This is a general result for diffusion problems:
(length scale) ∼
5.2.2
q
(diffusion coefficient) × (time scale) .
Flow at an oscillatory moving plate
We now consider a related problem: the oscillating plate flow (see Figure 5.17). The
plate on y = 0 moves harmonically with velocity vx (y = 0) = V cos ωt, with ω the
d
frequency. Knowing that vx = vx (y, t), vy = 0 and dx
= 0 the x-component of the
Navier-Stokes equation reduces again to:
∂vx
∂ 2 vx
=ν
.
∂t
∂y 2
(5.59)
We assume that after some time the flow is periodic with frequency ω, and thus:
n
vx (y, t) = Re f (y)eiωt
o
(5.60)
Here Re{·} means: ’the real part of’ .
Substitution of (5.60) in (5.59) gives:
iω
d2 f
− f (y) = 0 .
dy 2
ν
(5.61)
This equation has a general solution:
f (y) = C1 eα1 y + C2 eα2 y ,
(5.62)
y
fluid ρ, ν
x
vx (0, t) = V cos ωt
Figure 5.17: Schematic of the oscillating plate flow.
110
transport physics
with C1 and C2 integration constants, and
α1,2 = ±
s
r
iω
= ±(1 + i)
ν
ω
.
2ν
(5.63)
The boundary conditions are:
y → ∞ : vx → 0 y=0
: vx = Re V eiωt
→ C1 = 0
→ C2 = V
)
(5.64)
so the solution is:
n
vx (y, t) = V · Re e−y/∆ · ei(ωt−y/∆)
o
(5.65)
r
2ν
the so called viscous penetration depth.
ω
The factor e−y/∆ describes the damping in y-direction. The term y/∆ in the exponential represents a phase shift. The solution (5.65) describes a damped wave that
travels in y-direction (see Figure 5.18), with
with ∆ =
penetration depth : ∆ =
r
2ν
ω
2π
= 2π∆
: λ= r
ω
2ν
√
: c = ω∆ = 2νω
wave length
wave speed













(5.66)












y
t=
t=
π
ω
3π
2ω
t=
π
2ω
t = 0,
2π
ω
Figure 5.18: Plot of the solution (5.65) for different instants of time.
Viscous flow
5.2.3
111
Stokes flow around a sphere
The flow of a fluid with constant density ρ and kinematic viscosity ν around a sphere
with radius a is characterised by the Reynolds number
Re =
aV
,
ν
(5.67)
with V the uniform main stream velocity. In the case that Re ≪ 1 and a steady
flow, the Navier-Stokes equation (3.38) simplifies to:
1
0 = − ∇p + ν∇2~v .
ρ
(5.68)
This equation describes the so called creeping or Stokes flow.
For the stokes flow around a cylinder an analytical solution can be derived (see e.g.
Kundu, 1980, pp. 322-325). Based on that solution the following expression of the
drag force F on the sphere can be derived:
Fd = 6πηaV
(for Re << 1) .
(5.69)
with η = νρ the dynamic viscosity. This is the so called Stokes’ law of resistance.
The Stokes drag force can be used to analyse sedimentation of particles in a fluid
at rest. Consider a spherical particle with radius a = 10−3 m that descends with
velocity V in water (ν = 10−6 m2 /s). The Stokes-formula (5.69) is only valid for
Re =
Va
<< 1, so V ≤ 1 mm/sec .
ν
(5.70)
If the spherical particle has a density ρb we can compute a steady state velocity from
the equilibrium of gravity and drag forces Fg and Fd :
Fg + Fd = 0
→
4 3
πa (ρb − ρ)g = 6πηaV0
3
(5.71)
So:
V0 =
2 (ρb − ρ)ga2
.
9
η
(5.72)
V
a
ρ, ν
Figure 5.19: Stokes flow around a sphere.
112
transport physics
Example 5.3: Sedimentation of blood cells
This result can be used to calculate the time τ needed for sedimentation of a
suspension as depicted in the figure below.
t=0
0<t<τ
t≥τ
~g
H
h
initial state
transitional state
final state
For 0 < t < τ we observe two sharp interfaces: x = x1 is the interface between
the suspension with particle concentration C0 and the ’cleared’ liquid. The top
layer of the sediment with concentration Cm is indicated by x = x2 . In the case
of a dilute suspension the interface x = x1 descends with a velocity V1 equal
to the steady state velocity of single particles V1 = V0 as given by (5.72). The
sediment layer grows with velocity V2 (see Figure below).
x=H
x
x = x1 : interface between suspension
x1
x2
and cleared liquid
V1
C0
V2
x = x2 : top layer of the sediment
Cm
The positions of both interfaces are:
x1 (t) = H − V0 t
and x2 (t) = V2 t .
(5.73)
If we apply the balance of mass (move with velocity V2 at x = x2 (t)) we get with
V1 = V0 :
Cm V2 = C0 (V0 + V2 )
→
V2 =
C0
V0 .
Cm − C0
(5.74)
At t = τ we have x1 = x2 = h and the sedimentation is finished. Condition
(5.73) then leads to the sedimentation layer thickness:
x
H
h
H −h
=
V0
V2
C=0
so:
C = C0
C0
H
=H
h=
.
V0
C
1 + V2
m
h
0
τ
C = Cm
t→
The positions x1 (t) and x2 (t) of both interfaces change in time as depicted in
the figure above.
Viscous flow
5.3
113
Flow in porous media
Porous media are solid materials with pores completely or partly filled with fluid.
Examples of porous media are: granular or fibre like materials such as sand and
paper, brick like materials, but also biological tissue. The pores are often small
irregular capillaries not necessarily directed in a specific direction, but such that
a fluid can flow through the solid material. Since the pores in general are small,
viscous forces will be dominant and inertia forces can be neglected. i.e. Re ≪ 1.
Due to the irregular structure of the pores it is difficult to model the flow in its
complete detail and a third length scale ∆l must be introduced that is in between
the molecular length scale ∆λ and the macroscopic infinitesimal length scale ∆x, i.e.
∆λ ≪ ∆l ≪ ∆x. At the level of length scale ∆x there are many pores or capillaries
with a typical mean radius a and typical length l. To obtain a rough idea which forces
are needed to drive fluid flow in a porous medium we consider a 1-dimensional porous
medium with surface area A that consists of N parallel capillaries perpendicular to
A, let’s say in x-direction. The capillaries all have a radius a and a length l and
experience a pressure difference ∆p between the entrance and the exit (see Figure
5.20).
For each of the capillaries the relation between the pressure difference and the volume
flow is given by the equation of Hagen-Poiseuille (5.41).
qvi =
πa4
∆p ,
8ηl
(5.75)
with η the viscosity of the fluid. Consequently, the total volume flow through surface
area A is given by
qv = N
πa4
∆p
8ηl
(5.76)
which implies a mean velocity v̄x = qv /A:
v̄x =
N πa4
∆p .
A 8ηl
(5.77)
For porous media this relation is normally written as:
v̄x =
kp ∆p
η l
(5.78)
p1
v̄
p2
v̄
p1 − p2 = ∆p
l
Figure 5.20: Flow in a simplified 1D porous medium.
114
transport physics
with kp the specific hydraulic permeability of the porous medium. For the 1D porous
medium considered here we have:
kp =
N πa4
,
A 8
(5.79)
so kp only depends on the microscopic structure of the porous medium (number of
capillaries per unit area and radius of the capillaries). For more complex porous media with capillary-like pores in irregular directions relation (5.78) can be generalized
to:
k
~v = − ∇p ,
η
(5.80)
with ~v the ’homogenized’ fluid velocity, k the permeability tensor. This generalized
formulation is known as Darcy’s law. The permeability tensor is a property of the
porous medium and depends on the, possibly anisotropic, structural properties of
the pores. Often, the porous medium can be considered to be isotropic, so k = kp I.
In the case that the fluid is incompressible we have ∇ · ~v = 0 and thus:
∇·
k
∇p = 0 .
η
(5.81)
For homogeneous isotropic porous media this simplifies to:
kp 2
∇ p=0
η
→
∇2 p = 0 .
(5.82)
and
~v = −K∇p .
(5.83)
Here, K = kp /η is called the hydraulic conductivity of the porous medium. Equations (5.82) and (5.83) are the governing equations for fluid flow in a homogeneous
isotropic porous medium and have exactly the same form as the equations for steady
diffusion (2.40) with diffusion flux (2.45).
Viscous flow
115
Example 5.4: Darcy’s law
Two containers with cross-sectional surface A1 and A2 are connected with a tube
with radius a. The tube is filled with a porous medium with permeability kp . The
viscosity of the fluid is η the density ρ. We are interested in the difference h(t)
of the fluid levels as a function of time. At t = 0 we assume h(0) = h0 .
A1
h
A2
Ø 2a
L
The pressure difference over the porous medium is:
∆p(t) = ρgh(t) .
If the mean velocity in the porous medium is v̄, we have:
v̄(t) =
k ∆p(t)
kp ρg
=
h(t) .
η L
η L
The mass balance yields:
A1
dh
= −πa2 v̄
dt
→
dh
πa2 kp ρg
=−
h(t) .
dt
ηLA1
Consequently,
h(t) = h0 e−t/τ
with
τ=
ηLA1
.
πa2 kp ρg
For large time t ≫ τ , the level of the left container will lower with amount ∆h1
according to:
∆h1 A1 = (h0 − ∆h1 )A2
→
∆h1 = h0
A2
.
A1 + A2
This immediately follows from the integral mass balance of the system.
116
transport physics
6
Boundary layer flow
In the two previous chapters we introduced inviscid flow and viscous flow configurations.
In many cases, however, the flow is characterized by a viscous boundary layer and a
transition to an inviscid flow far a way from the boundary. In this chapter we will discuss
the development of such boundary layers and the length scales that are involved.
117
118
6.1
transport physics
Introduction
Flow problems with Re ≫ 1 can be considered as approximately frictionless or
inviscid (see Chapter 4), even if the fluid has a viscosity ν 6= 0. At fixed rigid walls
of the domain, however, fluid particles can not pass so
~vn = (~v · ~n)~n = ~0
(6.1)
and more importantly for this chapter, for ν > 0, the no-slip condition will always
hold, even for large Reynolds numbers. Thus
~vt = ~v − ~vn = ~0 .
(6.2)
Here, ~vn and ~vt are the normal and tangential velocity at the wall respectively. This
means that, localy (near the wall) viscous effects will always play an essential role
such that (6.2) will be satisfied. As indicated in Figure 6.1, viscous forces will be
important in a (thin) boundary layer of a two-dimensional flow.
For Re ≫ 1 the boundary layer thickness δ(x) will be small compared to the length
L of the boundary region:
δ≪L
(6.3)
For this situations the Navier-Stokes equations that describe the flow in the boundary layer will transfer to the so called boundary layer equations.
6.2
Boundary layer equations
We will first consider a scaling specific for the velocity in the boundary layer where
we take the x-axis along the bondary and the y-axis perpendicular to the boundary.
vx∗ = vx /V
x∗ = x/L
y ∗ = y/δ
main stream velocity scale
length scale along the wall
length scale perpendicular to the wall
(6.4)
Note that the length scale along the wall (L) has been taken different from the length
scale perpendicular to the wall (δ). The continuity equation then gives:
∂vx
∂x
V
O
L
∂vy
∂y
+
vy
O
δ
=
0
(6.5)
and thus:
vy = O
δ
V
L
≪V .
(6.6)
So, the velocity component normal to the wall, vy , is small but we take it not equal
to zero.
Boundary layer flow
119
V
vx (x, y)
inertia forces
(inviscid flow)
main stream
δ(x)
viscous and
inertia forces
(viscous flow)
boundary layer
y
x
L
Figure 6.1: Flow over a flat plate with an inviscid main stream in which viscous forces
can be neglected and a boundary layer in which viscous forces as well as inertia forces
are of importance.
If we take p∗ = p/ρV 2 , the scaling of the x-component of the Navier-Stokes equation
without gravity forces gives:
∂vx
vx
∂x
V
O V
L
∂vx
vy
∂y
+
δ V
V
O
L δ
=
1 ∂p
−
ρ ∂x
O
V2
L
+ ν
!
∂ 2 vx
∂x2
νV
O
L2
+
≪
↑
(δ ≪ L)
∂ 2 vx
∂y 2
!
νV
O
δ2
(6.7)
At the interface between the boundary layer and the main stream (so at y = δ) the
(largest) viscous term, which is of order O(νV /δ2 ), is assumed to be in of the same
order as the inertia term that is of order O(V 2 /L), so:
νV
O
δ2
=O
V2
L
!
→
s
δ = O

νL 
.
V
(6.8)
VL
,
With the definition of the Reynolds number, based on length scale L, ReL =
ν
this can be written as:
p
→
δ = O(L/ ReL )
So, indeed,
δ
1
=O √
L
ReL
(6.9)
δ
≪ 1 in the case that ReL ≫ 1.
L
If we use the local scale along the boundary x (instead of L) equation (6.8) changes
to:
δ(x) = O
r
νx
V
.
This shows that the boundary layer grows with
(6.10)
√
x.
120
transport physics
We can also scale the y-component of the Navier-Stokes equation. This yields:
∂vy
vx
∂x
O
∂vy
vy
∂y
+
δ V2
L L
!
O
δ V2
L L
1 ∂p
−
ρ ∂y
=
!
O
1 ∂p
ρ ∂y
+
ν
O
∂ 2 vy
∂x2
δ νV
L L2
+
∂ 2 vy
∂y 2
≪ O
!
δ νV
L δ2
(6.11)
We see that with this scaling, except for the pressure gradient, all terms are δ/L
smaller than the corresponding terms in the x-component of the Navier-Stokes equation. This is only possible if also
δ ∂p
∂p
=O
∂y
L ∂x
→
O
∂p
∂y
≪O
∂p
∂x
.
(6.12)
From this it is concluded that the pressure in the boundary layer must be equal to
the pressure in the main stream (see Figure 6.2).
The boundary layer equations then read:

∂vx
∂vx


+ vy
v


 x ∂x
∂y




 0
= −
1 ∂p
∂ 2 vx
+ν
ρ ∂x
∂y 2
= −
1 ∂p
ρ ∂y
(6.13)
with, of course, mass conservation:
∂vx ∂vy
+
= 0.
∂x
∂y
(6.14)
In the main stream (outside the boundary layer) the flow is inviscid and Bernoulli’s
equation can be used (see Figure 6.2):
p(x) + 12 ρV 2 (x) = const
along a streamline
(6.15)
Here we again neglected the gravity forces.
streamline
p(x)
y
δ(x)
boundary layer
x
x
Figure 6.2: The pressure in the boundary layer does not depend on y and is equal to
the pressure in the mean stream at the same x-location.
Boundary layer flow
121
From this, after differentiation with respect to x, it follows that:
dV
dp
= −ρV
dx
dx
(6.16)
and (6.13) and (6.14) reduce to:

∂vx
∂vx


vx
+ vy



∂x
∂y





∂vx ∂vy
+
∂x
∂y
= V
dV
∂ 2 vx
+ν
dx
∂y 2
(6.17)
= 0
For a known main stream velocity V (x) and the boundary conditions

vx = vy = 0 at












y=0
vx = V
at y = δ
vy = 0
at y = δ
(6.18)
the solution vx (x, y) and vy (x, y) can be determined although this may still be a
difficult problem to solve analytically.
dV
= 0) is given by Blasius
dx
(1908). For the development of a boundary layer at a flat plate he found:
One of the approximate solutions for a flat plate and (
r
δ(x) ≈ 5
νx
.
V
(6.19)
Here δ is defined as δ0.99 : the position where the velocity u(x, y) is 99% of the main
stream velocity V (see Figure 6.3). The Blasius solution is quite close to the first
order approximation (6.10) that follows from just scaling.
vx (y = δ0.99 ) = 0.99V
vx (x, y)
y
δ0.99
x
Figure 6.3: The definition of the boundary layer thickness δ0.99 .
122
transport physics
V
y
p~n
Γ2
streamline
δ1 (x)
s0
Γ1
δ(x)
y
Γ3
boundary layer
α
δ(x)
0
ds
α
vx
Γ4
x
p~n
sx
dy
dx
x=x
x
Figure 6.4: The boundary layer above a flat plate and the control volume to compute
the total drag force depicted by the dashed box defined by the entrance and outflow
planes Γ1 and Γ3 , the fixed wall Γ4 and the streamline Γ2 that starts at height δ(x) at
the entance. The right panel illustrates the pressure force on an infinitesimal part ds of
the streamline Γ2 .
6.3
Drag force induced by a developing boundary layer
To compute the drag force on a flat plate we consider the situation as depicted in
figure 6.4. First we need to compute the distance δ1 . To this end we apply the
integral mass balance for steady flows on the fixed control volume defined by the
dashed box in figure 6.4:
ZZ
Γ
ρ~v · ~ndΓ = 0
→
ZZ
(−ρV )dΓ +
ZZ
ρvx dΓ = 0 .
(6.20)
Γ3
Γ1
Here we used that there is no volume flux accross the streamline (Γ2 ) and the fixed
wall (Γ4 ). If we assume that b is the width of the plate in the z-direction, we get:
bρ
Zδ
V dy = bρ
0
Zδ
vx dy + bρ
0
δ+δ
Z 1
V dy
δ
(6.21)
= bρ
Zδ
0
vx dy + bρV (δ + δ1 − δ)
→
δ1 V =
Zδ
0
(V − vx )dy .
The distance δ1 is called the displacement thickness and defined as:
δ1 (x) =
δ(x)
Z 0
vx
1−
dy =
V
Z∞ 0
1−
vx
V
dy .
(6.22)
The displacement thickness measures the distance that a streamline outside the
boundary layer is displaced due to the development of the boundary layer. Note
that the upper boundary of integration can be taken any value larger than δ, so also
∞, since the integrant is zero for y > δ.
Boundary layer flow
123
For a steady boundary layer in which we neglect gravity forces, the integral momentum balance for the control volume defined by the dashed box is:
ZZ
Γ
ρ(~v · ~n)~v dΓ =
ZZ
~tdΓ .
(6.23)
Γ
The Cauchy stress tensor in matrix form

∂vx
 −p + 2η ∂x

σ=


∂vy
∂vx
η
+
∂x
∂y
∂vx ∂vy
η
+
∂y
∂x

∂vy
−p + 2η
∂y





(6.24)
yields an x-component of the momentum balance,
For Γ1 , ~n = −~ex , ~v = V ~ex ,
ρ(~v · (−~ex ))~v · ~ex = −ρV 2
tx = (−p + 2η
(6.25)
∂vx
∂vx
)~ex~ex · (−~ex ) · ~ex = p − 2η
.
∂x
∂x
(6.26)
For Γ2 , ~v · ~n = 0,
ρ(~v · ~n)~v · ~ex = 0
tx = −p~n · ~ex
(6.27)
since vy = 0,
∂vx
∂vx
=
=0.
∂x
∂y
(6.28)
For Γ3 , ~n = ~ex , ~v = vx~ex + vy ~ey ,
ρ(~v · ~ex )~v · ~ex = ρvx2
(ρV 2 for y > δ(x))
(6.29)
∂vx
∂vx
)~ex~ex · ~ex · ~ex = −p + 2η
.
∂x
∂x
For Γ4 , ~n = −~ey , ~v = ~0,
tx = (−p + 2η
(6.30)
ρ(~v · ~ey )~v · ~ex = 0
(6.31)
∂vx
∂vy
∂vx
)~ex~ey · (−~ey ) · ~ex = −η
since
= 0.
(6.32)
∂y
∂y
∂x
Taking the above into account the x-component of the momentum balance reduces
to:
tx = (η
ZZ
Γ1
−ρV 2 dΓ +
+
ZZ
Γ2
ZZ
ρvx2 dΓ =
Γ1
Γ3
−p~n · ~ex dΓ +
ZZ
ZZ
Γ3
(p − 2η
∂vx
)dΓ+
∂x
∂vx
)dΓ −
(−p + 2η
∂x
ZZ
Γ4
(6.33)
∂vx
η
dΓ
∂y
124
transport physics
The pressure contributions on Γ1 ,Γ2 , and Γ3 are
Fp = b
Zδ
pdy + b
Zsx
(−p~n · ~ex )ds + b
s0
0
δ+δ
Z 1
(−p)dy ,
(6.34)
0
b being the size of the plate in z-direction. With (see right panel of figure 6.4)
dy
−p~n · ~ex = p sin α and ds =
this yields:
sin α
Fp = b
Zδ
pdy + b
0
δ+δ
Z 1
δ
pdy − b
δ+δ
Z 1
pdy = 0 .
(6.35)
0
So, the sum of the pressure terms on Γ1 and Γ2 is equal but of opposite sign to the
one on Γ3 and cancel each other.
The normal stress contributions on Γ1 and Γ3 scale with O(η(V /L)bδ and are much
smaller than the shear stress contribution on Γ4 which scales with O(η(V /δ)bL).
Consequently we may write:
ZZ
Γ1
2
−ρV dΓ +
ZZ
ρvx2 dΓ
Γ3
ZZ
=−
Γ4
η
∂vx
dΓ .
∂y
(6.36)
The drag force on the plate then is:
Fd =
ZZ
Γ4
η
∂vx
dΓ =
∂y
ZZ
ρV 2 dΓ −
Γ1
= bρ
Zδ
0
2
0
 δ
Z
= ρV 2 b 
0
= ρV b
Zδ
0
ρvx2 dΓ
Γ3
V dy − bρ
 δ
Z
= ρV 2 b 
2
ZZ
Zδ
vx2 dy
0
v2
1 − x2
V
!
v2
1 − x2
V
!
− bρ
δ+δ
Z 1
V 2 dy
δ

dy − δ1 
dy −
(6.37)
Zδ 0
vx
1−
V

dy 
vx
vx
1−
dy ≡ ρV 2 bδ2
V
V
The term δ2 (x) is called the momentum thickness and a dimensionless measure for
the drag force Fd . Also δ2 , just like δ1 , can be defined with any upper bound of the
integral, so also ∞:
δ2 (x) =
Z∞
0
vx
vx
1−
dy
V
V
(6.38)
Boundary layer flow
125
The dimension of δ2 is one of a length scale, as is the dimension of b. This indeed
results in a force Fd if we take Fd = ρV 2 bδ2 . In fluid dynamics, boundary layer flow
is one of the important topics and many analytical approximations for the boundary
layer profile vx (y) are derived. As a first approximation vx = V y/δ can be taken
and yield insight in several phenomena. Later in this chapter we will see that at
least third order polynomials must be taken to describe boundary layer separation.
6.4
Inlet flow in a tube
We consider a viscous fluid with kinematic viscosity ν that enters a tube with a
uniform velocity V . Due to the viscosity and the no-slip condition at the wall a
boundary layer will develop that will increase in thickness with increasing distance
from the entrance of the tube (see Figure 6.5).
Finally the boundary layer will reach the center of the tube and merge with the
boundary layer that developed from the opposite wall of the tube. From that location
z = Le (Le = ‘entry length’) the tube flow is called fully developed and ∂~v /∂z = ~0.
The flow in the region 0 < z < Le is quite complicated and difficult to determine
analytically. Indeed, in the boundary layer (ii) the velocity will decrease. Mass
conservation (dqv /dz = 0) implies that in the core of the tube (i) the flow will
increase. So convective forces play a role and (~v · ∇)~v is essentially unequal to zero.
An empirical expression for the entrance length Le is:
Le
≈ 0.056ReD .
D
(6.39)
with ReD = V D/ν and D the diameter of the tube. For example, if ReD = 2000
we have Le ≈ 100D.
We can understand this empirical relation by assuming the boundary layer will grow
according to (6.10). At x = Le the boundary layer thickness will be D/2 so:
1
2D
s
= O

νLe 
V
(6.40)
uniform velocity V
fully developed flow
’Poiseuille’
developing boundary layer
z
i
vz (r)
ii
z=0
z = Le
Figure 6.5: Entrance flow in a tube
126
transport physics
Figure 6.6: Visualisations of the inlet flow in a tube and between 2 parallel plates (From
Y.Nakayama, Introduction to Fluid Mechanics, Butterworth-Heinemann, 1998) .
and thus:
Le
VD
= O 0.25
D
ν
≈ 0.25ReD .
(6.41)
If we take the Blasius solution (6.19) instead of the first order approximation (6.10)
we even find
Le
≈ 0.05ReD .
D
(6.42)
which is quite close to the empirical expression (6.39). In Figure 6.6 flow visualisations of the entrance flow in a tube and between two flat plates are given.
6.5
Boundary layer separation
Under certain conditions the boundary layer can separate from the wall. In most
cases this is the result of a flow deceleration in the main stream. We already showed
that in the main stream, which can be considered to be inviscid, Bernoulli’s equation
(6.15) can be applied, i.e.:
p(x) + 12 ρV 2 (x) = const
(6.43)
and thus:
dV
dp
= −ρV
.
dx
dx
(6.44)
Boundary layer flow
127
∂p
>0
∂x
y
x
∂vx
∂vx
∂vx
>0
=0
<0
∂y
∂y
∂y
Figure 6.7: Separation of the boundary layer in a decelerating main flow.
In the case that the main stream depends on x, so V = V (x), this implies that
dp
6= 0. Since the pressure in the boundary layer is determined by the pressure in
dx
the main stream this also holds in the boundary layer and near the wall. Near or
νV
at the wall (y = 0), however, viscous forces ( 2 ) are large compared to convective
δ
V2
forces ( ). From the boundary layer equations (6.13) we then find:
L
vx
|
∂vx
∂x
+vy
y=0
∂vx
∂y
{z
≈ 0
y=0
}
=−
1 dp
ρ dx
+ν
y=0
∂2u
∂y 2
(6.45)
y=0
and thus at the wall:
ν
1 dp
∂ 2 vx
≈
.
2
∂y y=0 ρ dx
(6.46)
dp
From this we see that for
> 0 the shear rate increases with increasing distance
dx
∂ ∂vx
from the wall ( (
) > 0). This is clearly visible in the second and third profile
∂y ∂y
sketched in Figure 6.7.
Also indicated in Figure 6.7 (most right profile) is a situation that can occur in
which the flow in the boundary layer has not enough x-momentum to counterpart
the pressure gradient. In that case a reversed flow will result. This phenomenon
is referred to as boundary layer separation and occurs if a main stream in positive
dp
> 0).
x-direction (V (x) > 0) decelerates (
dx
We will give some illustrations of boundary layer separation in the next examples.
128
transport physics
Example 6.1: Flow in a diverging channel
V2
V1
p1
p2
A1
boundary layers
A2 (> A1 )
Top: Flow in a diverging channel. Bottom: Visualisation (H2 -bubbles) of a flow
in a diverging channel (From Y.Nakayama, Introduction to Fluid Mechanics,
Butterworth-Heinemann, 1998) .
From the mass balance we find:
V1 A1 = V2 A2
→
V2 =
A1
< V1 .
A2
(6.47)
If we apply Bernoulli in the main stream:
p1 + 21 ρV12 = p2 + 12 ρV22 .
(6.48)
This gives for a channel of length L:
p2 − p1
ρ(V12 − V22 )
dp
=
=
>0.
dx
L
2L
(6.49)
The flow therefore is in the direction of the higher pressure and is decelerated
which may result in boundary layer separation (see figure ).
Boundary layer flow
129
Example 6.2: Flow around a blunt body
x
C
A
B
C
x
A
C
separation point B
Flow around a blunt body. Top: Streamlines. Middle: Boundary layer
separation. Bottom: Vortices in the wake.
We consider the flow around a blunt body (see figure above). From Bernoulli
along the stream lines we find:
A and B : stagnation points (~v = 0)
C
: maximal velocity
→
→
p = pmax
p = pmin
)
(6.50)
In the thin boundary layer at the surface of the blunt body: Going from A to C
the pressure will decrease from pmax to pmin . As a result the boundary layer will
stay attached. Going from C to B the pressure will increase form pmin to pmax
dp
> 0. as a result, the boundary layer may separate. Using small
so we have
dx
H2 bubbles the boundary layer separation at a blunt body can be visualized in a
clear way (see figure).rom the wall.
Visualisation with H2 -bubbles of boundary layer separation at a blunt body
(From Y.Nakayama, Introduction to Fluid Mechanics, Butterworth-Heinemann,
1998) .
130
transport physics
Example 6.3: Flow at sharp edges.
Two examples of flow separation at sharp edges are given in the figure below.
Here, due to the sudden change in geometry flow separation occurs even when
the boundary layer has not clearly been developed.
Boundary layer or flow separation at sharp edges
7
Balance of energy
In this chapter we will derive the integral and differential form of the balance of energy.
The first law of thermodynamics will be taken as a point of departure. Concepts of
internal energy, heat, and work will be explained and their relation to body and surface
forces that act on a material domain will be elucidated. Using the strategy also used in
chapter 2 and 3 integral and differential forms of the energy balance will be derived and
their application will be illustrated. A link will be made between mass and heat transfer.
131
132
7.1
transport physics
First law of thermodynamics
In the continuum approach we describe the behaviour of material fluid elements on
the basis of a statistical average of the processes on molecular microscopic scale.
On the macroscopic continuum level the internal motion of the individual atoms is
not visible. This motion, however, is observed as thermal energy or heat Q. Since
this motion is associated to momentum transfer between adjacent atoms, heat can
be transferred between fluid elements. Another form of microscopic energy is the
potential or cohesive energy of the chemical bonds between the atoms. This hidden
form of energy is named internal energy U and is determined by the thermodynamic
state in which the system exists. The internal energy is a so called extensive property
which means that it, unlike for instance the density, depends on the extend of the
system and is the sum of the internal energy of subsystems. The relation between
heat and internal energy is, among other things, studied in thermodynamics.
The first law of thermodynamics states that total energy of a closed system in rest
is constant. Energy can be transformed from one form to another but can not be
created nor destroyed. This means that the change of internal energy ∆U of a closed
system, i.e. a material domain Ωm , is equal to the amount of heat Q supplied to the
system minus the amount of work W done by the system on its surroundings:
∆U = Q − W
(7.1)
The thermodynamic state of a closed system is mostly determined by two state
variables, i.e. macroscopically measurable quantities. For a material domain Ωm ,
or subdomain dΩm with a uniform chemical composition, the thermodynamic state
can be determined by the temperature T and the pressure p. Each of these, but not
both at the same time, can be replaced by another state variable, the volume V of
the system. Before we will derive relations between these state variables and go into
phenomena such as heat transfer, we will have a look at different types of work that
play a role in these relations.
7.1.1
Work done by external forces
To introduce the principles of work and its relation to energy, we will consider an
infinitesimal material domain, fluid particle, dΩ, which we can characterise by the
position ~x of its centre of mass and by its mass m. Newton’s laws state:
- If no force acts on the particle, the velocity ~v is constant.
- An action force F~a on the particle will induce an acceleration:
d~v
F~a = m
dt
- An action force F~a acting on the particle will induce a reaction force F~r :
~a
F~r = −F
of the particle on its environment.
Balance of energy
133
If in the timespan t1 ≤ t ≤ t2 a particle follows a trajectory ~x(t) from ~x1 to ~x2 as a
result of force F~a , then the work W done by the reaction force F~r is given by:
W =
Z~x2
~
x1
~r · d~x ,
F
(7.2)
or with d~x = ~v dt:
W =
Zt2
~r · ~v dt .
F
t1
(7.3)
With the second law of Newton this gives:
W =−
Zt2
t1
d~v
m · ~v dt = −m
dt
Z~v2
~v1
~v · d~v = 12 mv12 − 21 mv22 ,
(7.4)
v2
with
= ~v · ~v . This shows that the work done by the particle is counter parted by
a loss of kinetic energy.
7.1.2
Work done by pressure forces
To investigate the work that can be done by pressure forces, we consider a cylinder
with volume V closed by a (frictionless) piston with surface area A. (see Figure 7.1).
A displacement dx of the piston in the direction of the outer normal ~n will cause a
volume increase equal to dV = Adx. The work W done by the system then is:
dW = Fx dx = pAdx = pdV .
(7.5)
For a total volume change from V1 to V2 the work done by the pressure forces then
are:
W =
ZV2
pdV .
(7.6)
V1
This relation can be generalized and holds for any volume change dV of any volume
of arbitrary shape.
dx
p
A
V
dV
Figure 7.1: Work performed by pressure forces of a system
134
7.1.3
transport physics
Integral energy balance
Until here we restricted the first law of thermodynamics to material fluid elements
in rest and stated the following energy balance:
∆U = U2 − U1 = Q − W .
(7.7)
Here ∆U is the change of internal energy between state 2 and state 1. In the case this
fluid element with mass m has a uniform velocity v, the macroscopic kinetic energy
of the system is 12 mv 2 . We already learned that kinetic energy can be transformed
to work done by the system. Hence, the energy balance needs to be extended to
E2 − E1 = Q − W,
(7.8)
with E the total energy defined by:
E = U + 12 mv 2 ,
(7.9)
In general the changes between state 1 and state 2 are changes that take place in
time so we can also write:
dE
dQ dW
=
−
,
dt
dt
dt
(7.10)
of in words:
Per unit of time, the increase in total energy of a material domain is equal to the
heat that is added to the system minus the work the system has performed.
We will specify each of the terms in more detail below.
Total energy
If we now consider a system that is not uniform, the system can be subdivided in
subsystems that are locally in thermodynamic equilibrium and described by their
local state variables T and p. This is named the principle of local thermodynamic
equilibrium and is based on the fact that the internal energy and kinetic energy
are extensive quantities. The total material domain is the sum of material elements
with volume dΩ. With the definition of the mass density ρ we can introduce the
specific internal energy according to:
dU = ρudΩ ,
(7.11)
and the specific kinetic energy 12 ρv 2 . This yields for the internal energy of the total
system:
U=
ZZZ
ρudΩ
(7.12)
Ωm
and for the total energy:
E=
ZZZ
Ωm
ρ(u + 21 v 2 )dΩ .
(7.13)
Balance of energy
135
Heat
Heat Q can be added to the system, either by dissipation in the system itself or by
heat transfer on the surface Γ. This can be written as:
ZZZ
ZZ
dQ
=
Φh dΩ −
(~qh · ~n)dΓ ,
(7.14)
dt
Ωm
Γm
here Φh is the amount of energy that is dissipated per unit of time and volume, and
(~
qh · ~ndΓ) the heat flux that leaves the domain through surface element dΓ. The
function Φh can be heat that is added to the system by electrical dissipation or
chemical reactions.
Work
Both surface forces and body forces perform work on the system. If each infinitesimal
volume element dΩ experiences a body force equal to ρf~dΩ, then the work performed
by this volume element per unit of time is equal to −ρf~ · ~v (see equation 7.3). The
work performed by the complete system per unit of time then follows from:
ZZZ
dWΩ
=−
dt
Ωm
ρf~ · ~v dΩ
(7.15)
The force that acts from the environment on surface element dΓ is equal to ~tdΓ. The
work per unit of time performed by this surface element on its environment then is
−~t · ~v dΓ. The total work resulting from surface forces then is:
dWΓ
=−
dt
ZZ
Γm
~t · ~v dΓ .
(7.16)
Energy balance
We can now write the first law of thermodynamics for a material control volume as:
d
dt
ZZZ
ρ(u+ 12 v 2 )dΩ =
ZZZ
Φh dΩ−
Ωm
Ωm
ZZ
Γm
~qh ·~ndΓ+
ZZZ
Ωm
ρf~·~v dΩ+
ZZ
Γm
~t·~v dΓ . (7.17)
Note that all terms in the right hand side of (7.17) are not specifically only valid for
a material domain Ωm but also hold for a fixed domain Ω0 or even a moving domain
Ω(t). The left hand side, however, must be treated with more care since convective
transport is involved. For a fixed domain Ω0 we get:
d
dE
=
dt
dt
ZZZ
ρ(u +
1 2
2 v )dΩ
+
ZZ
Γ0
Ω0
ρ(u + 12 v 2 )~v · ~ndΓ
(7.18)
The rate of change of internal energy and kinetic energy due to convection (outflow
at the boundary) is added. The integral energy balance for a fixed domain Ω0 with
boundary Γ0 follows from (7.17) with Ωm = Ω0 and (7.18):
d
dt
ZZZ ρ(u +
Ω0
ZZZ
Ω0
1 2
2v )
dΩ +
Φh dΩ −
ZZ
Γ0
ZZ
Γ0
ρ(u + 21 v 2 )~v · ~ndΓ =
~qh · ~ndΓ +
ZZZ
Ω0
ρf~ · ~v dΩ +
ZZ
Γ0
(7.19)
~t · ~v dΓ .
136
transport physics
For a arbitrary moving domain Ω(t) we get:
d
dE
=
dt
dt
ZZZ
ρ(u + 21 v 2 )dΩ +
ZZ
ρ(u + 21 v 2 )(~v − ~vΓ ) · ~ndΓ
Γ(t)
Ω(t)
(7.20)
Again the rate of change of internal energy and kinetic energy due to convection
(outflow at the boundary) is added but now the relative velocity ~v − ~vΓ is involved.
Application of the Reynolds transport equation (2.6) then yields
dE
=
dt
ZZZ
Ω(t)
∂ ρ(u + 12 v 2 ) dΩ +
∂t
ZZ
Γ(t)
ρ(u + 21 v 2 )~v · ~ndΓ
(7.21)
The final integral energy balance for an arbitrary domain Ω(t) with boundary Γ(t)
follows from (7.17) with Ωm = Ω(t) and (7.21):
ZZZ
Ω(t)
|
∂ ρ(u + 21 v 2 ) dΩ +
∂t
Γ(t)
{z
}
change in total energy
ZZZ
−
Φh dΩ
ZZ
Γ(t)
Ω(t)
{z
}
|
heat production
7.2
ZZ
|
|
ρ(u + 12 v 2 )~v · ~ndΓ =
{z
}
flux of total energy
~qh · ~ndΓ +
{z
}
heat flux
ZZZ
Ω(t)
ρf~ · ~v dΩ
+
|
{z
}
work by body forces
ZZ
Γ(t)
~t · ~v dΓ .
{z
}
|
work by surface forces
(7.22)
Differential form of the energy balance
Similar to what have been done for the mass and momentum balance, a differential
form of the energy balance can be derived by applying the divergence theorem of
Gauss to all the surface integrals in (7.22). This yields:
ZZ
Γ(t)
ρ(u + 21 v 2 )~v · ~ndΓ =
ZZ
~qh · ~ndΓ =
ZZ
~t · ~v dΓ =
Γ(t)
ZZZ
Ω(t)
ZZZ
Ω(t)
∇ · (ρu~v + 12 ρv 2~v )dΩ ,
∇·~
qh dΩ ,
(7.23)
(7.24)
and
Γ(t)
ZZ
Γ(t)
~n · σ · ~v dΓ =
ZZZ
Ω(t)
∇ · (σ · ~v )dΩ
(7.25)
Substitution of (7.23), (7.24), and (7.25) in (7.22) yields a relation between volume
integrals. Since this must hold for arbitrary volumes Ω this yields:
∂
∂
(ρu)+∇·(ρu~v )+ ( 12 ρ~v ·~v )+∇· 21 ρ(~v · ~v )~v = Φh −∇·~qh +ρf~·~v +∇·(σ·~v ) (7.26)
∂t
∂t
Balance of energy
137
With the use of several vector identities and application of the mass and momentum
balances the complexity of this equation can be reduced significantly. The first two
terms in (7.26) represent the transport of internal energy and simplify to:
∂u
∂ρ
∂
(ρu) + ∇ · (ρu~v ) = ρ
+u
+ ρ~v · ∇u + u∇ · (ρ~v )
∂t
∂t
∂t
(7.27)
∂u
=ρ
+ ρ~v · ∇u .
∂t
Indeed here the second and last term vanish by virtue of the local mass balance.
The third term in (7.26) is the local time derivative of the kinetic energy and can
be evaluated as:
∂ 1
∂
∂~v
( 2 ρ~v · ~v ) = ~v · ( 21 ρ~v ) + 12 ρ~v ·
∂t
∂t
∂t
= 21 ρ~v ·
= ~v · ρ
∂~v 1
∂ρ 1
∂~v
+ 2 ~v · ~v
+ 2 ρ~v ·
∂t
∂t
∂t
(7.28)
∂~v 1
∂ρ
+ 2 ~v · ~v
.
∂t
∂t
Finally the last term in the left hand side of (7.26), representing the convection of
kimetic energy, gives:
∇·
1
v
2 ρ(~
· ~v )~v
= 12 ρ~v · ~v (∇ · ~v ) + ~v · ∇( 21 ρ~v · ~v )
= 12 ρ~v · ~v (∇ · ~v ) + 21 ρ~v · ∇(~v · ~v ) + 12 (~v · ~v )~v · ∇ρ
(7.29)
= 21 ρ~v · ~v (∇ · ~v ) + ρ~v · (~v · ∇)~v + 12 (~v · ~v )~v · ∇ρ
Combination of the second term of the final right hand side of (7.28) with the first
amd last term of the final right hand side of (7.29) gives:
1
v
2~
· ~v
∂ρ
+ ρ∇ · ~v + ~v · ∇ρ = 0 .
∂t
(7.30)
This again by virtue of the local mass balance.
Making use of this result, the third and forth term of the left hand side of (7.26)
read:
∂~v
∂ 1
( 2 ρ~v · ~v ) + ∇ · 21 ρ(~v · ~v )~v = ~v · ρ
+ ρ~v · (~v · ∇)~v
∂t
∂t
= ~v · ρ
∂~v
+ ρ(~v · ∇)~v
∂t
(7.31)
= ~v · ∇ · σ + ρf~
Substitution of (7.27) and (7.31) in (7.26) gives
ρ
∂u
+ ρ~v · ∇u = −∇ · ~qh + Φh + ∇ · (σ · ~v ) − ~v · (∇ · σ)
∂t
(7.32)
138
transport physics
The last two terms can be combined using a standard tensor identity
∇ · (σ · ~v ) − ~v · (∇ · σ) = σ : (∇~v )c = σ : D .
(7.33)
With the double dot product defined as (ab : cd) = (a · c)(b · d) it can be shown
that σ : D = tr(σ · D). We finally get:
ρ
∂u
+ ρ~v · ∇u = −∇ · ~
q h + Φh + σ : D
∂t
(7.34)
The first two terms describe the material derivative of the internal energy, the terms
in the right hand side are heat flux, heat production, and viscous dissipation respectively.
7.3
Thermal diffusion and heat transfer
The internal energy u in (7.34) is not a directly measurable quantity but, as mentioned earlier, related to state variables such as the temperature T and the density
ρ. For incompressible media the density is constant and the internal energy only
depends on the local temperature:
du = Cp dT
(7.35)
with Cp the specific heat capacity and is a measure for the energy needed to increase
the temperature of one unit of mass by one unit of temperature. So for incompressible media with constant and homogeneous heat capacity, the energy equation reads:
∂T
+ ρCp~v · ∇T = −∇ · ~qh + Φh + σ : D .
(7.36)
∂t
Similar to Fick’s law in mass flux is related to concentration gradients, the heat flux
~qh is reated to temperature gradients according to Fourier’s law:
ρCp
~qh = −k∇T
(7.37)
with k the thermal conductivity of the medium. Substituton of Fourier’s law in the
energy (or heat) equation yields:
ρCp
∂T
+ ρCp~v · ∇T = k∇2 T + Φh + σ : D .
∂t
(7.38)
In the absence of heat production and viscous dissipation, this equation transforms
into a convection-diffusion equation for heat transfer that is completely analog to
the convection diffusion equation (2.38) for mass transfer
∂T
∂t
Here κ =
+
(~v · ∇)T
convection
=
κ∇2 T .
diffusion
k
is the thermal diffusivity.
ρCp
(7.39)
Balance of energy
139
Example 7.1: Heat conduction
T = T0
r
ak
am
z
L
Heat conduction through a wall.
An important form of heat transfer is heat conduction. To illustrate this process
we consider fluid filled container where the temperature must be kept at T0 (see
Figure). The bath is heated via a tube with inner radius ak that carries fluid with
velocity V at a fixed temperature Tk = αT0 . The tube has a wall with thickness
am − ak .
The mean velocity of the fluid in the tube is V [m/s]. The density is ρ and the
heat capacity Cp . The heat is transferred through the wall by diffusion. The
thermal diffusivity is κ. The bath looses heat with an amount Φ [J/s]. We are
interested in the maximum amount of Φ. In the wall of the tube the velocity is
zero so the heat equation in cylindrical coordinates is:
1 ∂
∂T
∂T
−κ
r
∂t
r ∂r
∂r
=0.
(7.40)
The boundary conditions are:


 T (ak , t) = Tk
T (a , t) = T0
m

 T (r, 0) = T
0
(7.41)
After some time τ a steady state will be reached. Before that the order of
magnitude of both terms or non-zero but equal. So O((Tk − T0 )/τ ) = κ(Tk −
T0 )/(am − ak )2 . So we have a steady state for τ ≫ (am − ak )2 /κ. Integration
of the steady state equations yields:
∂T
c1
=
∂r
r
(7.42)
T = c1 ln r + c2
(7.43)
so:
substitution of the boundary conditions gives:
c1 =
Tk − T0
ln(ak /am )
(7.44)
and
c2 = Tk −
Tk − T0
ln(ak ) .
ln(ak /am )
(7.45)
140
transport physics
Consequently the temperature is
ln(r/ak )
.
ln(am /ak )
(7.46)
∂T
κρCp (Tk − T0 )
=
∂r
r ln(am /ak )
(7.47)
T (r) = Tk − (Tk − T0 )
The heat flux is:
qh,r = −κρCp
This means that for the steady state we have:
Φh = 2πam Lqh,r |r=am =
2πLκρCp (Tk − T0 )
.
ln(am /ak )
(7.48)
formuleblad Transportfysica 8VB00
Transport
Kinematica en deformatie
• beschrijving van een deeltjesbaan wordt verkregen door integratie van de volgende uitdrukking(en) naar tijd:
dx = vx (~x, t)dt,
dy = vy (~x, t)dt,
dz = vz (~x, t)dt
• beschrijving van een stroomlijn wordt verkregen door de uitdrukking
dy
vy (t0 )
=
dx
vx (t0 )
in de x, y, z-ruimte te integreren, voor vaste tijd t = t0 .
• voor een snelheidsveld ~v = (vx , vy , 0) kan een stroomfunctie ψ worden gedefinieerd:
~v = vx~ex + vy ~ey
met
vx =
∂ψ
∂ψ
, vy = −
∂y
∂x
• definities van snelheids-gradiënt-tensor L, rotatiesnelheids-tensor Ω, en deformatiesnelheidstensor D:
L = (∇~v )c = D + Ω
D = 21 (L + Lc )
met
en Ω = 21 (L − Lc )
Behoudswetten in integraalvorm
voor een volume Ω, begrensd door oppervlak Γ met snelheid ~vΓ
massa :
impuls :
energie :
d
dt
d
dt
ZZZ
ρdΩ +
Ω
ZZZ
d
dt
Ω
ZZZ
ZZ
Γ
ρ~v dΩ +
ρ(~v − ~vΓ ) · ~ndΓ = 0
ZZ
Γ
ρ~v (~v − ~vΓ ) · ~ndΓ =
ρ(u + 21 v 2 )dΩ +
ZZ
Γ
Ω
ZZZ
Ω
Φh dΩ −
ZZZ
ρ~g dΩ +
Ω
ZZ
~tdΓ
Γ
ρ(u + 12 v 2 )(~v · ~n)dΓ =
ZZ
Γ
~qh · ~ndΓ +
ZZZ
Ω
ρf~ · ~v dΩ +
ZZ
Γ
~t · ~v dΓ
Behoudswetten in differentiaalvorm
massa :
∂ρ
+ ∇ · (ρ~v ) = 0
∂t
impuls :
1
∂~v
+ (~v · ∇)~v = ∇ · σ + ~g
∂t
ρ
energie : ρ
(met ~t = σ · ~n)
∂u
+ ρ~v · ∇u = −∇ · q~h + Φh + σ : D
∂t
Behoudswetten in differentiaalvorm voor een incompressibel, viskeus medium
massa : ∇ · ~v = 0
impuls :
1
∂~v
+ (~v · ∇)~v = − ∇p + ν∇2~v + ~g
∂t
ρ
(met σ = −pI + 2ηD)
Dimensieloze groepen
Reynolds :
Re =
VL
ν
Peclet :
Pe =
VL
D
or
Pe =
Froude :
Fr =
V2
gL
Strouhal : Sr =
ωL
V
or
Sr =
VL
κ
L
Vτ
Bernoulli-vergelijking
p
1 2
V + gz + = constant
2
ρ
(langs een stroomlijn.)
Indien ∇ ×~v = 0 heeft de constante overal in het stromingsveld dezelfde waarde.
Wetten van Fick, Fourier, Darcy
J~ = −D∇c ,
qh = −k∇T ,
~
~v = −
kp
∇p
η
Convectie-diffusie-vergelijkingen
∂c
+ ~v · ∇c = D∇2 c ,
∂t
∂T
+ ~v · ∇T = κ∇2 T
∂t
met
κ = k/ρCp
Coördinaatsystemen
Cartesisch
bol
cilinder
z
z
z
~
ez
~
ez
P
P
~
ex
z
r
z
x
y
Cartesisch
(x, y, z)
cilinder
(r, φ, z)
x=x
y=y
z=z
r = x2 + y 2 ,
φ = arctan(y/x),
z = z,
p
r
φ
~
eθ
z
θ
y
x
~
eφ
P
~
er
y
x
~
er
~
eφ
~
ey
φ
x
x=r cos φ
y=r sin φ
y
x=ρ cos φ
ρ=r sin θ
y=ρ sin φ
bol
(r, θ, φ)
x = r cos φ
y = r sin φ
z=z
p
r = x2 + p
y2 + z2 ,
θ = arctan( x2 + y 2 /z),
φ = arctan(y/x),
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Cartesische coördinaten
coördinaten x, y, x, snelheid ~v = vx~ex + vy ~vy + vz ~ez en scalaire functie ψ = ψ(x, y, z)
grad ψ = ∇ψ =
∂ψ
∂ψ
∂ψ
~ex +
~ey +
~ez
∂x
∂y
∂z
div ~v = ∇ · ~v =
∂vx ∂vy
∂vz
+
+
∂x
∂y
∂z
rot ~v = ∇ × ~v =
∇2 ψ =
∂vy
∂vx
∂vx ∂vz
∂vy
∂vz
−
−
−
~ex +
~ey +
~ez
∂y
∂z
∂z
∂x
∂x
∂y
∂2ψ ∂2ψ ∂2ψ
+
+
∂x2
∂y 2
∂z 2
~v · ∇ψ = vx
∂ψ
∂ψ
∂ψ
+ vy
+ vx
∂x
∂y
∂z
∂vx
∂vx
∂vx
(~v · ∇)~v = vx
+ vy
+ vz
~ex +
∂x
∂y
∂z
+ vx
∂vy
∂vy
∂vy
∂vz
∂vz
∂vz
+ vy
+ vz
+ vy
+ vz
~ey + vx
~ez
∂x
∂y
∂z
∂x
∂y
∂z
#
"
∂ 2 vx ∂ 2 vx ∂ 2 vx
+
+
~ex +
∇ ~v =
∂x2
∂y 2
∂z 2
2
"
#
"
#
∂ 2 vy
∂ 2 vy
∂ 2 vz
∂ 2 vz
∂ 2 vz
∂ 2 vy
+
+
+
+
~
e
+
~ez
+
y
∂x2
∂y 2
∂z 2
∂x2
∂y 2
∂z 2
∂vx
~ex~ex +
∂x
∂vy
~ex~ey +
∂x
∂vz
~ex~ez +
∂x
+
∂vx
~ey ~ex +
∂y
∂vy
~ey ~ey
∂y
+
∂vz
~ey ~ez +
∂y
+
∂vx
~ez ~ex
∂z
∂vy
~ez ~ey
∂z
+
∂vz
~ez ~ez
∂z
∇~v =
+
Cilindercoördinaten
coördinaten r, φ, z, snelheid ~v = vr~er + vφ~eφ + vz ~ez en scalaire functie ψ = ψ(r, φ, z)
grad ψ = ∇ψ =
1 ∂ψ
∂ψ
∂ψ
~er +
~eφ +
~ez
∂r
r ∂φ
∂z
div ~v = ∇ · ~v =
1 ∂vφ ∂vz
1 ∂
(rvr ) +
+
r ∂r
r ∂φ
∂z
∂vφ
1 ∂vz
∂vz
∂vr
1 ∂(rvφ ) 1 ∂vr
rot ~v = (∇ × ~v ) =
−
−
−
~er +
~eφ +
~ez
r ∂φ
∂z
∂z
∂r
r ∂r
r ∂φ
1 ∂
∂ψ
1 ∂2ψ ∂2ψ
r
+ 2 2 +
r ∂r
∂r
r ∂φ
∂z 2
∂ψ
∂ψ vφ ∂ψ
+
+ vz
~v · ∇ψ = vr
∂r
r ∂φ
∂z
∇2 ψ =
(≡ ∇2c ψ)
#
"
vφ2
vφ ∂vr
∂vr
∂vr
+
+ vz
−
~er +
(~v · ∇)~v = vr
∂r
r ∂φ
∂z
r
∂vφ vφ ∂vφ
∂vφ vr vφ
vφ ∂vz
∂vz
∂vz
+ vr
+
+ vz
+
+
+ vz
~eφ + vr
~ez
∂r
r ∂φ
∂z
r
∂r
r ∂φ
∂z
i
h
vφ
2 ∂vφ vr
2 ∂vr
2
2
2
− 2 ~er + ∇c vφ + 2
− 2 ~eφ + ∇2c vz ~ez
∇ ~v = ∇c vr − 2
r ∂φ
r
r ∂φ
r
∂vφ
∂vz
∂vr
~er ~er
+
~er~eφ
+
~er ~ez +
∇~v =
∂r
∂r
∂r
+
+
∂vr
~ez ~er
∂z
1 ∂vr
vφ
−
~eφ~er +
r ∂φ
r
+
1 ∂vφ vr
+
~eφ~eφ +
r ∂φ
r
∂vφ
~ez ~eφ
∂z
+
1 ∂vz
~eφ~ez +
r ∂φ
∂vz
~ez ~ez
∂z
Bolcoördinaten
coördinaten r, θ, φ, snelheid ~v = vr ~er + vθ~eθ + vφ~eφ en scalaire functie ψ = ψ(r, θ, φ)
grad ψ = ∇ψ =
1 ∂ψ
1 ∂ψ
∂ψ
~er +
~eθ +
~eφ
∂r
r ∂θ
r sin θ ∂φ
div ~v = ∇ · ~v =
1 ∂
1 ∂vφ
1 ∂ 2
(r vr ) +
(sin θvθ ) +
2
r ∂r
r sin θ ∂θ
r sin θ ∂φ
rot ~v = ∇ × ~v =
1 ∂
1 ∂vθ
(vφ sin θ) −
~er +
r sin θ ∂θ
r sin θ ∂φ
1 ∂
1 ∂
1 ∂vr
1 ∂vr
−
(rvφ ) ~eθ +
(rvθ ) −
+
~eφ
r sin θ ∂φ
r ∂r
r ∂r
r ∂θ
∂ψ
1 ∂
r2
∇ ψ= 2
r ∂r
∂r
2
~v · ∇ψ = vr
∂
1
∂ψ
+ 2
sin θ
r sin θ ∂θ
∂θ
+
1
∂2ψ
r 2 sin2 θ ∂φ2
(≡ ∇2b ψ)
∂ψ vφ ∂ψ
vφ ∂ψ
+
+
∂r
r ∂θ
r sin θ ∂φ
#
"
vθ2 + vφ2
vθ ∂vr
vφ ∂vr
∂vr
+
+
−
~er +
~v · ∇~v = vr
∂r
r ∂θ
r sin θ ∂φ
r
"
#
vφ2 cot θ
vφ ∂vθ
∂vθ
vθ ∂vθ
vr vθ
+ vr
+
+
+
−
~eθ +
∂r
r ∂θ
r sin θ ∂φ
r
r
vφ ∂vφ vφ vr
vθ vφ cot θ
∂vφ vθ ∂vφ
+
+
+
+
~eφ
+ vr
∂r
r ∂θ
r sin θ ∂φ
r
r
2
∇ ~v =
∇2b vr
+
+
2 ∂vθ
2vθ cot θ
2 ∂vφ
2vr
−
− 2
~er +
− 2 − 2
2
r
r ∂θ
r
r sin θ ∂φ
∇2b vθ
2 ∂vr
vθ
2 cos θ ∂vφ
+ 2
− 2 2 − 2 2
~eθ +
r ∂θ
r sin θ r sin θ ∂φ
∇2b vφ
vφ
2 ∂vr
2 cos θ ∂vθ
+ 2 2
~eφ
− 2 2 + 2
r sin θ r sin θ ∂φ
r sin θ ∂φ
∂vr
∂vθ
∂vφ
~er ~er +
~er ~eθ +
~er ~eφ +
∂r
∂r
∂r
∇~v =
+
1 ∂vr
vθ
−
~eθ ~er +
r ∂θ
r
+
vφ
1 ∂vr
−
~eφ~er +
r sin θ ∂φ
r
+
vθ
1 ∂vφ vr
+
+
cot θ ~eφ~eφ
r sin θ ∂φ
r
r
vr
1 ∂vθ
1 ∂vφ
+
~eθ ~eφ +
~eθ ~eθ +
r ∂θ
r
r ∂θ
vφ
1 ∂vθ
−
cot θ ~eφ~eθ
r sin θ ∂φ
r
Speciale functies
Dirac’s delta-functie
δ(x − x0 ) =
(
∞ voor x = x0
0 voor x 6= x0
en
+∞
Z
−∞
δ(x − x0 )f (x)dx = f (x0 )
Error functie
2
erf(ξ) = √
π
Zξ
2
e−ζ dζ
0
Taylor-reeksontwikkeling
f (x + dx) = f (x) +
1 d2 f 2
df
dx +
dx + . . . O(dx3 ).
dx
2! dx2
Vectoridentiteiten
∇(φψ)
∇ · (φf~)
= φ∇ψ + ψ∇φ
= (∇φ) · f~ + φ(∇ · f~)
∇ × (f~ × ~g )
∇(f~ · ~g)
= (∇ · ~g )f~ + (~g · ∇)f~ − (∇ · f~)~g − (f~ · ∇)~g
= f~ × (∇ × ~g ) + ~g × (∇ × f~) + (f~ · ∇)~g + (~g · ∇)f~
∇ × (φf~)
∇ · (f~ × ~g )
∇ · (∇ × f~)
∇ × (∇φ)
= (∇φ) × f~ + φ(∇ × f~)
= (∇ × f~) · ~g − f~ · (∇ × ~g )
=0
= ~0
∇ × (∇ × f~) = ∇(∇ · f~) − (∇ · ∇)f~
A
Appendices
147
148
A.1
transport physics
General
A.1.1
Coordinate systems
Depending on the geometrical configuration problems in transport physics are defined in rectangular, cylindrical, spherical or even more sophisticated coordinate
systems. Here we will give a short overview of the rectangulari, cylindrical, and
spherical oordinate systems.
rectangular (x, y, z)
cylindrical (r, φ, z)
spherical (r, θ, φ)
(−∞ < x < ∞)
(−∞ < y < ∞)
(−∞ < z < ∞)
(0 ≤ r < ∞)
(0 ≤ φ < 2π)
(−∞ < z < ∞)
(0 ≤ r < ∞)
(0 ≤ θ ≤ π)
(0 ≤ φ < 2π)
(A.1)
The geometrical definitions are depicted in Figure (A.1).
spherical
cylindrical
rectangular
z
z
z
~
ez
~
ez
P
~
ey
P
~
ex
z
y
r
x
x
rectangular
(x, y, z)
cylindrical
(r, φ, z)
x=x
y=y
z=z
r = x2 + y 2 ,
φ = arctan(y/x),
z = z,
p
r
x=r cos φ
y=r sin φ
~
eθ
z
θ
y
φ
~
eφ
P
~
er
z
y
x
~
er
~
eφ
x
φ
y
x=ρ cos φ
ρ=r sin θ
y=ρ sin φ
spherical
(r, θ, φ)
x = r cos φ
y = r sin φ
z=z
p
r = x2 + p
y2 + z2 ,
θ = arctan( x2 + y 2 /z),
φ = arctan(y/x),
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Figure A.1: Rectangular, cylindrical, and spherical coordinate systems and the corresponding relations.
To transform a vector ~v from rectangular to cylindrical coordinates (vx , vy , vz ) →
(vr , vφ , vz ) we write:
vr = ~v · ~er = vx~ex · ~er + vy ~ey · ~er + vz ~ez · ~er = vx~ex~·er + vy ~ey · ~er
vφ = ~v · ~eφ = vx~ex · ~eφ + vy ~ey · ~eφ + vz ~ez · ~eφ = vx~ex~·eφ + vy ~ey · ~eφ
vz = ~v · ~ez = vx~ex · ~ez + vy ~ey · ~ez + vz ~ez · ~ez = vz
Here we used ~ez · ~er = 0, ~ez · ~eφ = 0, ~ex · ~ez = 0, ~ey · ~ez = 0, and ~ez · ~ez = 1.
From figure A.2 we can readily derive that:
~er
= cos φ ~ex + sin φ ~ey
(A.2)
~eφ = − sin φ ~ex + cos φ ~ey
Appendices
149
~er
~eφ
φ
cos φ
φ
sin φ
cos φ
sin φ
ρ
y
φ
z
x
Figure A.2: Cylindrical coordinate system.
and thus:





cos φ sin φ 0
vx
vr


 

 vφ  =  − sin φ cos φ 0   vy 
vz
0
0
1
vz
(A.3)
In most cases in this course we will assume a rectangular coordinate system unless
mentioned otherwise.
A.1.2
Inner product and vector projection
To compute the mass or volume flux through a surface element dΓ we need the
normal component of the velocity, i.e. the component in the direction of the unit
normal to that surface element. From the figure we can derive that the length of
the normal component is:
150
transport physics
vn = k~v k cos θ
(A.4)
With the definition of the inner product:
~v · ~n ≡ k~v kk~nk cos θ
(~v · ~n)~n
~v
(A.5)
this yields:
~n
vn =
θ
~v · ~n
.
k~nk
(A.6)
since, by definition, k~nk = 1, we get:
vn = ~v · ~n .
dΓ
(A.7)
In Cartesian coordinates:
vn = vx nx + vy ny + vz nz .
A.1.3
(A.8)
A 2nd -order tensor as a diadic product
If we define a dyadic or tensor product such that the dyad ~a~b operates on a vector
~c following:
~a~b · ~c ≡ ~a(~b · ~c) ,
(A.9)
then this dyad can be associated with a 2nd order tensor A that transforms ~c into
d~ = ~a(~b · ~c), i.e.:
d~ = A · ~c .
(A.10)
In a Cartesian coordinate system this reads:
~a~b = (ax~ex + ay ~ey + ax~ez )(bx~ex + by ~ey + bx~ez )
= ax bx~ex~ex + ax by ~ex~ey + ax bz ~ex~ez +
ay bx~ey ~ex + ay by ~ey ~ey + ay bz ~ey ~ez +
az bx~ez ~ex + az by ~ez ~ey + az bz ~ez ~ez
(A.11)
So:


0 0 0


~ey ~ex =  1 0 0  ,
0 0 0


0 0 0


~ez ~ey =  0 0 0  ,
0 1 0
etc.
(A.12)
The conjugated tensor Ac is defined according to: Ac ·~c = ~c · A. From this it readily
follows that (~a~b)c = ~b~a.
Appendices
151
Also the vector gradient can be interpreted as a tensor product:
∇~a = (~ex
=
∂
∂
∂
+ ~ey
+ ~ez )(ax~ex + ay ~ey + az ~ez )
∂x
∂y
∂z
∂ax
∂ay
∂az
~ex~ex +
~ex~ey +
~ex~ez +
∂x
∂x
∂x
(A.13)
∂ay
∂az
∂ax
~ey ~ex +
~ey ~ey +
~ey ~ez +
∂y
∂y
∂y
∂ay
∂az
∂ax
~ez ~ex +
~ez ~ey +
~ex~ez
∂z
∂z
∂z
Note that, since ∇ is a differential operator, in this case ∇~a 6= (~a∇)c .
A.1.4
The gradient operator ∇
The partial derivative with respect to variable xm of a function f that depends on
variables x1 , x2 , ..., xn is defined as the derivative in which all variables but xm are
kept fixed, i.e.:
f (x1 , ..., xm + h, ..., xn ) − f (x1 , ..., xm , ..., xn )
∂f
≡ lim
.
∂xm h→0
h
(A.14)
The differential change df of a function f that depends on variables x1 , x2 , ..., xn is:
df =
∂f
∂f
∂f
dx1 +
dx2 + ... +
dxn .
∂x1
∂x2
∂xn
(A.15)
From this the total derivative with respect to variable x1 can be defined as:
df
∂f
∂f dx2
∂f dxn
=
+
+ ... +
.
dx1
∂x1 ∂x2 dx1
∂xn dx1
(A.16)
If the variables x1 , x2 , ... represent space variables such as the Cartesian coordinates
x, y, z in IR3 , it is useful to combine the partial derivatives of a function f into a
vector function called the gradient operator ∇:
∇ = ~ex
∂
∂
∂
+ ~ey
+ ~ez
∂x
∂y
∂z
(A.17)
The gradient operator applied to a function f helps us to find the rate of change of
function f with respect to the directions defined by the space variables.
A.1.5
The divergence operator ∇·
If we take ~r as the origin of a Cartesian coordinate system, we can write the Taylor
series:
~v (~r) = ~v (~0) +
∼
~v 0
+
∂~
v
∂y ~0 y
+
0y
+ ~v,y
0z
+ ~v,z
+ ...
∂~
v
∂x ~0 x
0x
+ ~v,x
∂~
v
∂z ~0 z
+ ...
(A.18)
152
transport physics
The unit normal ~n on sphere Sǫ is ~x/ǫ = 1ǫ (x~ex + y~ey + z~ez ), so
~v · ~n =
1 0
v
ǫ (~
0
+~v,x
0
+~v,y
0
+~v,z
· ~ex x + ~v 0~ey y + ~v 0 · ~ez z+
0 ·~
0 ·~
ez xz
ey xy + ~v,x
· ~ex x2 + ~v,x
0
2
0
· ~ex xy + ~v,y · ~ey y + ~v,y · ~ez yz
0 ·~
0 ·~
ez z 2 + ...)
ey yz + ~v,z
· ~ex xz + ~v,z
(A.19)
In (1.41) all integrals over odd functions (x, y, z, xy, xz, yx) will vanish due to symmetry. Moreover, also because of symmetry,
ZZ
2
x dS =
Sǫ
ZZ
2
y dS =
Sǫ
1
=
3
ZZ
z 2 dS
Sǫ
ZZ
(A.20)
4
1
(x + y + z )dS = (ǫ2 )(4πǫ2 ) = πǫ4
3
3
2
Sǫ
2
2
So we obtain:
ZZ
Sǫ
(~v · ~n)dS =

1

ǫ
ZZ
Sǫ
0
~v,x
· ~ex x2 dx +
ZZ
Sǫ
0
~v,y
· ~ey y 2 dx +
ZZ
Sǫ

0
~v,z
· ~ez z 2 dx (A.21)

Since ~vx0~ex , ~vy0~ey , and ~vz0~ez are evaluations the velocity gradient in ~x = ~0 (see 1.36),
they do not depend on ~x, we get:
ZZ
Sǫ
(~v · ~n)dS =
=

1 0
ex
~v,x · ~
ǫ
ZZ
Sǫ
0
x2 dx + ~v,y
· ~ey
ZZ
Sǫ
0
y 2 dx + ~v,z
· ~ez
4 3 0
0
0
πǫ (~v,x · ~ex + ~v,y
· ~ey + ~v,z
· ~ez )
3
ZZ
Sǫ

z 2 dx

(A.22)
and thus together with (A.20):
lim
ǫ→0
ZZ
Sǫ
(~v · ~n)dS = ∇ · ~v
(A.23)
Divergence in different coordinate systems
Cartesian (x, y, z):
∇ · ~v = (~ex
∂
∂
∂
+ ~ey
+ ~ez ) · (vx~ex + vy ~ey + vz ~ez )
∂x
∂y
∂z
(A.24)
∂vz
∂vx ∂vy
+
+
=
∂x
∂y
∂z
Cilindrical (r, φ, z):
∇ · ~v =
1 ∂
1 ∂vφ ∂vz
(rvr ) +
+
r ∂r
r ∂φ
∂z
(A.25)
Appendices
153
Spherical (r, θ, φ):
∇ · ~v =
1 ∂ 2
1 ∂
1 ∂vφ
(r vr ) +
(vθ sin θ) +
2
r ∂r
r sin θ ∂θ
r sin θ ∂φ
(A.26)
The Laplacian operator ∇2
A.1.6
The Laplacian operator ∇2 is defined as:
∇2 = ∇ · ∇ .
(A.27)
In Cartesian coordinates (x, y, z) this results in:
∇2 = (~ex
=
∂
∂
∂
∂
∂
∂
+ ~ey
+ ~ez ) · (~ex
+ ~ey
+ ~ez )
∂x
∂y
∂z
∂x
∂y
∂z
(A.28)
∂2
∂2
∂2
+
+
.
∂x2 ∂y 2 ∂z 2
In cylindrical coordinates (r, φ, z) this is:
1 ∂
∂
∇ =
r
r ∂r
∂r
2
+
∂2
1 ∂2
+
.
r 2 ∂φ2 ∂z 2
(A.29)
In spherical coordinates (r, θ, φ) we have:
∂
1 ∂
r2
∇ = 2
r ∂r
∂r
2
A.1.7
∂
1
∂
+ 2
sin θ
r sin θ ∂θ
∂θ
+
∂2
1
r 2 sin2 θ ∂φ2
(A.30)
Gauss’s divergence theorem
A rigorous proof of the Gauss divergence theorem is beyond the scope of this lecture.
Instead, we will consider a simplified version that at least give us some rough understanding. Let ∆V be a square box in domain Ω, with dimensions (∆x × ∆y × ∆z)
P
and boundary ∆S = ∆Si for i = 1, ...6 and let us evaluate the integration of the
partial derivative of the z-component of f~ with respect to z:
ZZZ
∆V
∂fz
dV
∂z
=
x
=

 z+∆z
Z
∂f
z

dz  dydx
y+∆y
x+∆x
Z
Z
y
z
∂z
(A.31)
y+∆y
x+∆x
Z
Z
x
y
[fz (z + ∆z) − fz (z)]dydx .
z
Ω
Γ
fz ~n
∆S1
∆V
∆z
∆V
y
∆S2
x
∆x
∆y
fz ~n
154
transport physics
In the same time, if we take
(upper and lower surface):
ZZ
∆S
RR
fz ~ez · ~ndS, we have only two non-zero contributions
S
y+∆y
x+∆x
Z
Z
fz ~ez · ~ndS =
x
y
(fz (z + ∆z)dydx −
y+∆y
x+∆x
Z
Z
fz (z)dydx
x
y
(A.32)
y+∆y
x+∆x
Z
Z
=
x
y
[fz (z + ∆z) − fz (z)]dydx
Consequently:
ZZZ
∆V
∂fz
dV =
∂z
ZZ
fz ~ez · ~ndS .
(A.33)
ZZ
fy~ey · ~ndS .
(A.34)
ZZ
fx~ex · ~ndS .
(A.35)
∆S
and similarly:
ZZZ
∂fy
dV =
∂y
ZZZ
∂fx
dV =
∂x
∆V
∆S
and:
∆V
∆S
Summation of (A.33), (A.34), and (A.35) yields:
ZZZ
∆V
∇ · f~dV =
ZZ
∆S
f~ · ndS .
(A.36)
This holds for any infinitesimal volume ∆Vi in Ω, so:
∇ · f~∆Vi =
6
X
j=1
f~ · ~nij ∆Sij
(A.37)
with Sij the j-the surface j = {1, ..., 6} of volume Vi . Taking in mind that the flux
through two common surfaces of two adjacent volumes is equal but opposite in sign
we have only contributions from the non-overlapping surfaces k if we sum over all
volumes ∆Vi :
X
i
∇ · f~ ∆Vi =
ki
XX
i k=0
f~ · nik dSik
(A.38)
and finally, by interpreting the obove summation as a Riemann sum:
ZZZ
Ω
∇ · f~dΩ =
ZZ
Γ
f~ · ~ndΓ .
(A.39)
Appendices
A.1.8
155
Vector and tensor form of the Gauss theorem
We already introduced the scalar version for a scalar function a and vector function
~b as a direct result of (2.8):
ZZZ h
Ω
i
a∇ · ~b + ~b · ∇a dΩ =
ZZ
Γ
a~b · ~ndΓ .
(A.40)
A vector version for vector functions ~a and ~b can be written as:
ZZZ h
i
~a∇ · ~b + ~b · ∇~a dΩ =
Ω
ZZ
Γ
~a(~b · ~n)dΓ
(A.41)
This can easily be derived from (A.40) by applying it for each of the components
of ~a. Finally by assuming that the tensor A can be written as A = ~a~b we obtain a
tensor version:
ZZZ
Ω
∇ · Ac dΩ =
ZZ
Γ
A · n dΓ
(A.42)
Here the vector identity
∇ · (~a~b) = (∇ · ~a)~b + ~a · ∇~b
(A.43)
is used together with ~b~a = (~a~b)c .
A.1.9
Vector differential identities
Based on the definitions of the divergence and rotation, the following relations can
be derived:
∇(φψ) = φ∇ψ + ψ∇φ
∇ · (φf~) = (∇φ) · f~ + φ(∇ · f~)
∇ × (φf~) = (∇φ) × f~ + φ(∇ × f~)
∇ · (f~ × ~g ) = (∇ × f~) · ~g − f~ · (∇ × ~g )
∇ × (f~ × ~g) = (∇ · ~g)f~ + (~g · ∇)f~ − (∇ · f~)~g − (f~ · ∇)~g
∇(f~ · ~g ) = f~ × (∇ × ~g ) + ~g × (∇ × f~) + (f~ · ∇)~g + (~g · ∇)f~
∇ · (∇ × f~) = 0
∇ × (∇φ) = 0
∇ × (∇ × f~) = ∇(∇ · f~) − (∇ · ∇)f~
A.1.10
(A.44)
The error function
The error function erf(η) originates from probability theory and is proportional to
the integral of a standard normal distribution. It is defined as:
2
erf(η) ≡ √
π
Zη
0
2
e−ζ dζ .
(A.45)
156
transport physics
and is also used as a special function in partial differential equations. The course of
this function is given in the Figure A.3. The function is zero for η = 0 and rapidly
increases to an asymptotic value of 1. For η = 2 the value of the error function is
already 99% of its asymptotic value. The complementary error function erfc(η) is
defined as:
erfc(η) = 1 − erf(η) .
(A.46)
Note that:
d
d
2
2
erf(η) = − erfc(η) = √ e−η .
dη
dη
π
(A.47)
1
erf(η = 2) ≈ 0.99
erf(η)
0.5
0
0
0.5
1
1.5
2
2.5
3
→η
Figure A.3: Plot of the error function erf(η).
A.1.11
The Dirac delta function
The Dirac delta function in more dimensions is defined as:
Z∞
−∞
δ(x − x0 )f (x)dx = f (x0 )
(A.48)
for every smooth function f . For instance the distribution function that satisfies:
dn (x) =


 n/2

 0
if |x − x0 | ≤ 1/n
(A.49)
if |x − x0 | > 1/n
can be interpreted as a Dirac delta function for the case n → ∞.
A.2
Transport Physics
A.2.1
Reynolds transport theorem
Let Φ a property of the fluid and let φ the amount of Φ per unit of volume, so:
Φ=
ZZZ
Ωm
φdΩ
(A.50)
Appendices
157
Γ(t)
Ω(t)
Ω(t + ∆t)
Γ(t + ∆t)
φ(~x, t)
influx
outflux
Figure A.4: Motion and deformation of a control domain Ω(t) in which a fluid property
φ(~x, t) is defined. The control volume is enclosed by surface Γ(t) that moves with
velocity ~vΓ .
Consider an arbitrary control domain Ω(t) and with boundary surface Γ(t) (see
Figure A.4). In general Ω(t) is neither fixed or moving with the fluid and moving
with velocity ~vΓ . The rate of change of Φ in Ω(t) is given by:
dΦ
dt
=
=
d
dt
ZZZ
φ(~x, t)dΩ
Ω(t)

1 

∆t→0 ∆t
lim
ZZZ
Ω(t+∆t)
φ(~x, t + ∆t)dΩ −
ZZZ
Ω(t)


φ(~x, t)dΩ .
Since the boundary moves with velocity ~vΓ this will introduce a flux of φ proportional
to ~vΓ · ~n through the boundary. As a consequence we can write:
ZZZ
Ω(t+∆t)
φ(~x, t + ∆t)dΩ =
ZZZ
Ω(t)
φ(~x, t + ∆t)dΩ + ∆t
ZZ
Γ(t)
φ(~x, t + ∆t)~vΓ · ~ndΓ (A.51)
158
transport physics
The rate of change of Φ then can be written as:
d
dt
ZZZ

1 

∆t→0 ∆t
φ(~x, t)dΩ =
lim
Ω(t)
+ ∆t
ZZ
Γ(t)
ZZZ
=
ZZZ
φ(~x, t + ∆t)dΩ +
Ω(t)
φ(~x, t + ∆t)~vΓ · ~ndΓ −
ZZZ
Ω(t)


φ(~x, t)dΩ
1
[φ(~x, t + ∆t) − φ(~x, t)] dΩ+
∆t→0 ∆t
lim
Ω(t)
+
ZZ
Γ(t)
lim φ(~x, t + ∆t)~vΓ · ~ndΓ
∆t→0
or equivalently, the Reynolds transport theorem:
d
dt
ZZZ
φ(~x, t)dΩ =
Ω(t)
ZZZ
Ω(t)
∂φ(~x, t)
dΩ +
∂t
ZZ
Γ(t)
φ(~x, t)~vΓ · ~ndΓ
(A.52)
with ~vΓ the boundary velocity of the arbitrarily moving control domain Ω(t).
A.2.2
Mass balance in different coordinate systems
rectangular
spherical
cylindrical
z
z
z
~
ez
~
ez
P
~
er
~
eφ
~
ey
P
~
ex
z
r
z
x
x
x
y
rectangular
(x, y, z)
cylindrical
(r, φ, z)
x=x
y=y
z=z
r = x2 + y 2 ,
φ = arctan(y/x),
z = z,
p
θ
y
y
φ
~
eφ
P
~
er
r
x=r cos φ
y=r sin φ
x
φ
~
eθ
z
y
x=ρ cos φ
ρ=r sin θ
y=ρ sin φ
spherical
(r, θ, φ)
x = r cos φ
y = r sin φ
z=z
p
r = x2 + p
y2 + z2 ,
θ = arctan( x2 + y 2 /z),
φ = arctan(y/x),
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Cartesian (x, y, z):
∂
∂
∂
∂ρ
+
(ρvx ) +
(ρvy ) +
(ρvz ) = 0
∂t
∂x
∂y
∂z
(A.53)
Cylindrical (r, φ, z):
∂ρ 1 ∂
1 ∂
∂
+
(ρrvr ) +
(ρvφ ) +
(ρvz ) = 0
∂t
r ∂r
r ∂φ
∂z
(A.54)
Appendices
159
Spherical (r, θ, φ):
∂ρ
∂
1 ∂
1 ∂
1
+ 2 (ρr 2 vr ) +
(ρvθ sin θ) +
(ρvφ ) = 0
∂t
r ∂r
r sin θ ∂θ
r sin θ ∂φ
A.2.3
(A.55)
Balance of momentum in different coordinate systems
We will consider incompressible flow of a Newtonian fluid with a constant viscosity
case (Navier-Stokes).
rectangular
spherical
cylindrical
z
z
z
~
ez
~
ez
P
~
er
~
eφ
~
ey
P
~
ex
z
r
z
x
x
x
y
rectangular
(x, y, z)
cylindrical
(r, φ, z)
x=x
y=y
z=z
r = x2 + y 2 ,
φ = arctan(y/x),
z = z,
p
φ
~
eθ
z
θ
y
y
~
eφ
P
~
er
r
x=r cos φ
y=r sin φ
x
φ
y
x=ρ cos φ
ρ=r sin θ
y=ρ sin φ
spherical
(r, θ, φ)
x = r cos φ
y = r sin φ
z=z
p
r = x2 + p
y2 + z2,
θ = arctan( x2 + y 2 /z),
φ = arctan(y/x),
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Cartesian (x, y, z):
x-component:
"
#
"
#
"
#
∂vx
∂vx
∂vx
∂ 2 vx ∂ 2 vx ∂ 2 vx
∂vx
1 ∂p
+ vx
+ vy
+ vx
=−
+ν
+
+
+ gx
∂t
∂x
∂y
∂z
ρ ∂x
∂x2
∂y 2
∂z 2
y-component:
∂ 2 vy
∂vy
∂vy
∂vy
1 ∂p
∂ 2 vy
∂ 2 vy
∂vy
+ vx
+ vy
+ vx
=−
+ν
+
+
+ gy
∂t
∂x
∂y
∂z
ρ ∂y
∂x2
∂y 2
∂z 2
z-component:
∂vz
∂ 2 vz
∂vz
∂vz
∂vz
1 ∂p
∂ 2 vz
∂ 2 vz
+ vx
+ vz
+ vx
=−
+ν
+
+
+ gz
∂t
∂x
∂y
∂z
ρ ∂z
∂x2
∂y 2
∂z 2
Cylindrical (r, φ, z):
r-component:
∂vr
∂t
vφ2
∂vr
vφ ∂vr
∂vr
+
+ vz
−
=
∂r
∂z
r
"r ∂φ #
1 ∂
∂ 2 vr
2 ∂vφ vr
1 ∂p
∂vr
1 ∂ 2 vr
+ν
− 2 + gr
+
− 2
−
r
+ 2
ρ ∂r
r ∂r
∂r
r ∂φ2
∂z 2
r ∂φ
r
+vr
160
transport physics
φ-component:
∂vφ
∂t
∂vφ vφ ∂vφ
∂vφ vr vφ
+
+ vz
+
=
∂r
r " ∂φ
∂z
r
#
1 1 ∂p
1 ∂
2 ∂vr
vφ
∂vφ
1 ∂ 2 vφ ∂ 2 vφ
−
+ν
+
+ 2
− 2 + gφ
r
+ 2
ρ r ∂φ
r ∂r
∂r
r ∂φ2
∂z 2
r ∂φ
r
+vr
z-component:
∂vz
∂t
vφ ∂vz
∂vz
∂vz
+
+ vz
=
∂r
∂z
"r ∂φ #
1 ∂
∂ 2 vz
∂vz
1 ∂ 2 vz
1 ∂p
+ν
+
+ gz
r
+ 2
−
ρ ∂z
r ∂r
∂r
r ∂φ2
∂z 2
+vr
8
Index
Avogadro’s number, 10
Bernouilli’s equation, 82
boundary layer
equations, 120
thickness, 118
boundary layer equations, 118
boundary layer separation, 126
Cauchy stress, 33
Cauchy stress tensor, 35
Cauchy’s law, 33
composition equation, 48
concentration
molar, 10
conjugate, 26
constitutive relations, 35
continuum approach, 132
continuum hypothesis, 14
convection, 48
convection diffusion equation, 48
convection-diffusion, 57
coordinate systems, 148
cylindrical, 148
rectangular, 148
spherical, 148
Couette flow, 97
Darcy’s law, 114
deformation, 26
density, 10
mass, 11
molar, 10
number, 10
specific, 11
derivative
convective, 20
differential change, 151
local, 20
material, 20
partial derivative, 151
total derivative, 151
derivatives of multi-variable functions, 151
diadic product, 150
diffusion, 48
coefficient of, 46
flux, 46
molar, 46
velocity, 46
dilatation, 30
dimensionless groups, 74
Dirac delta function, 156
Dirac delta-function, 56
displacement thickness, 122
divergence, 27, 28, 151
drag force, 122
Einstein-summation, 30
energy balance
differential form, 136
integral form, 134
error function, 155, 156
Eulerian description, 18
Eulerian representation, 18
extensive property, 10, 132
external flow, 107
extra-stress, 36
Fick
first law, 46
Fick, second law, 47
first law of thermodynamics, 132
flow classification, 68
flow rate, 25
flux, 24, 25
diffusion, 46
molar diffusion, 46
forces
body, 31
traction, 31
Gauss theorem, 155
Gauss’s divergence theorem, 153
gradient operator, 151
heat, 132
heat flux, 136
heat production, 136
heat transfer, 138
hydrostatics, 65, 68
inlet flow, 125
inner product, 149
intensive property, 10
internal energy, 132, 134
internal flow, 96
inviscid flow, 69
irrotational flow, 88
kinetic energy, 134
Lagrangian description, 19
Lagrangian representation, 18
laminar flow, 16
Laplacian operator, 153
162
mass
molar, 10
molecular, 11
mass averaged velocity, 46
mass balance
differential form, 44
fixed domain, 41
integral form, 40
material domain, 40
moving domain, 41
mass density, 10
mass diffusion, 45
mass transfer, 45
membrane transport, 49
molar density, 10
molar mass, 10
momentum, 60
momentum balance, 60
differential form, 67
integral form, 60
integral,fixed domain, 61
integral,material domain, 61
integral,moving domain, 61
momentum density, 60
Newton
second law of, 60
Newton’s laws, 132
Newtonian fluid, 16
number density, 10
Pćlet number, 58
partition coefficient, 49
pathlines, 21
permeability, 50
physical quantities
extensive, 10
intensive, 10
point source, 55
Poiseuille flow, 99, 101
porous media, 113
potential flow, 89
pressure, 13
rate of deformation, 29
Reynolds transport theorem, 41, 156
rotation, 27, 28
rotation tensor, 29
shear, 30
state of deformation, 29
state of stress, 31
state variables, 10
Stokes flow, 111
streaklines, 21
streamlines, 21
stress, 26
stresses
normal, 32
shear, 32
surface, 31
stretch, 30
Strouhal number, 58
systems
multi-component, 11
temperature, 13
tensor
second order, 150
thermal diffusion, 138
thermal diffusivity, 138
thermodynamics
first law, 132
total energy, 134
turbulent flow, 92
vector identities, 155
vector projection, 149
velocity gradient, 26
velocity gradient tensor, 27
viscosity, 15
bulk, 36
dynamic, 16, 36
viscous flow, 70, 96
vorticity, 30
work, 132