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HYDRODYNAMICS
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HYDRODYNAMICS
Jørgen Fredsøe
MEK-DTU 2008
jf@mek.dtu.dk
2
3
List of Content
1. The kinematics and forces of flows
7
1.1 Velocities and accelerations
8
1.2 The stresses in a fluid
15
2. Laminar flows
25
2.1 The constitutive equations
26
2.2 The deformation of fluid elements in laminar flows
31
2.3 The equation of continuity
34
2.4 The flow equations (the Navier-Stokes equations)
36
2.5 Other examples of laminar flows
38
3. Potential flow
44
3.1 Definition of the potential
44
3.2 The Bernoulli equation
45
3.3 The Laplace equation
47
3.4 An example of a potential flow: flow over a wavy bed
47
4
4.0 Progressive sinusoidal waves
60
4.1 Shallow water theory
60
4.2 Sinusoidal wave at an arbitrary water depth
65
4.3 Wave energy
75
4.4 Waves over variable water depth
80
4.5 The wave reaction force
86
4.6 Wave induced currents
96
5. Laminar Boundary Layers
102
5.1 The boundary layer equations
103
5.2 The general momentum equation
104
5.3 The momentum equation for boundary layers
105
5.4 Boundary layers with outer pressure gradients
119
6. Turbulent flows
129
6.1 The stability of a laminar flow
131
6.2 Turbulent fluctuations
138
6.3 Reynolds stresses
139
6.4 The mixing length theory
143
6.5 Velocity profiles in uniform channel flows
146
6.6 The structure of turbulence in a pipe
161
5
7. Turbulent boundary layers
165
7.1 The turbulent boundary layer equation
166
7.2 Stationary boundary layer along a plate
168
7.3 Examples of non-stationary boundary layers
175
7.4 Free turbulence: the turbulent jet
179
7.5 Free turbulence: the plane wake
186
8. Turbulence modelling
193
8.1 Mixing length
193
8.2 The 1-equation model (k-model)
198
8.3 2-equation models (k-epsilon model)
207
Appendix
References
Index
6
1. The kinematics and forces of flows
The solution of a flow problem does in principle imply solving Newtons
second law
G
G
du
∑ F = ∫ ρ dt dX
(1.1)
G
G
where ΣF is the sum of all forces acting on the volume X with the density ρ . u is the
velocity vector for the small elements dX of this volume. In generalthe velocity and
the force vectors have three components.
In hydrodynamics the starting point will often be to consider the forces acting
G
on a small element with the velocity u , and thereafter by integration obtain (1.1). For
a small element dX Newtons second law reads
G
G
du
dF = ρ dX
dt
(1.2)
In the following the velocity term and force term for the forces acting on the small
volume.
To simplify we will normally consider the plane situation only. This is defined
G
as a flow where all the velocity vectors of the fluid particles u are parallel to a fixed
plane, and do not change by a shift in the direction normal to this plane.
7
1.1 Velocities and accelerations
G
In equation (1.1) we see the velocity vector u which is the velocity of a fluid
particle, and the total derivative
G
G du
a=
dt
(1.3)
which describes the acceleration of a fluid particle. In the following we consider a
flow, where a fluid particle at the time t0 is in the point A, cf. figure 1.1. The particle
moves along a curve called the particle path.
If the flow is stationary, the flow pattern is the same at all times, which means
that at any point (x,y) in the flow there is no change in the velocity with time, so that
G
∂u
=0
∂t
(1.4)
In this case a new particle that passes point A at a new time t2 also pass through point
B at a later time: t = t2 + (t1 – t0).
Fig. 1.1
Definition of particle path and velocity vector.
8
Example 1.1: Example of non-stationary flow.
Figure 1.2 shows a flow through a narrow duct where a block moves up and
down between the walls of the duct. The discharge (coming from left towards right) is
constant, but it is easy to see that a particle passing point A while the block is at the
top will have a particle path different from that of the particle passing A while the
block is at the low position.
Fig. 1.2
Example of a non-stationary flow: the block is moving up and
down in a narrow duct.
Streamlines
A streamline is a curve which has the velocity vector as tangent. In a nonstationary, or unsteady, flow the streamlines are in general different from the particle
paths, cf. example 1.1, while the streamlines and the particle paths are identical in a
stationary flow, cf. figure 1.1. Consider figure 1.1; at time t the particle has the
position
G
x = ( x, y )
(1.5)
G
G
and at time t + dt it will have the position x + dx = ( x + dx, y + dy ) . In the flow
situation the streamline is therefore defined by
G
G
dx = const × u
or
9
dx dy
=
u
v
(1.6)
where u is the velocity component in the x-direction and v is the velocity component
in the y-direction,
G
u = (u , v)
(1.7)
Let us then calculate the velocity of a fluid particle. Consider once more figure
1.1, where the particle starting from A at t=t0 later passes through B. In the general
non-stationary case we have seen, that a particle passing A at a later time t2 will not
necessarily pass through B, cf. figure 1.3. This means that the particle paths given by
(1.5) in general will be functions of time
G G
x = x (t ) = ( x(t ), y (t ))
(1.8)
The velocity is now found as the change in the position of a fluid particle per unit
time, or
Fig. 1.3
Non-stationary flow: particle paths through A at different
times.
10
G
G dx ⎛ d x(t ) d y (t ) ⎞
=⎜
u=
,
⎟
dt ⎝ dt
dt ⎠
(1.9)
In all of the x-y-plane a velocity vector is therefore defined for every point:
G
u = (u ( x, y, t ), v( x, y, t ))
(1.10)
Example 1.2: The velocity field in a plane drain.
Fig. 1.4
A plane drain.
Figure 1.4 shows the flow which is formed when water is sucked out in point 0
with a discharge Q = Q(t), where the discharge may vary with time. All velocity
vectors are directed towards 0. This particular flow is characterised as being axisymmetric, because the velocity field is not changed by an arbitrary rotation about an
axis through O and normal to the plane of the paper.
The velocity at a distance r from 0 can be calculated by using the continuity
equation, which in this case expresses that the amount of water flowing in through the
circle with radius r, shown in figure 1.4, must also be sucked out in point 0.
11
Because of the axi-symmetry it is much simpler to to work with polar
coordinates rather than a cartesian coordinate system.
Fig. 1.5
The velocity components in a polar coordinate system.
As shown in figure 1.5 a point A can be characterised by either (x,y) or (r , θ) where
x = r cos(θ)
and
(1.11)
y = sin(θ)
The velocity in the direction of the r-axis is called v r and the velocity normal to it is
called v θ (see figure 1.5).
In the flow of the plane drain the component v θ is seen to be zero due to
symmetry, and v r is independent of θ due to symmetry. The continuity equation
reads
− v r 2πr = Q
(1.12)
or
vr = −
Q
2πr
(1.13)
which describes the velocity field. The velocity field can also be specified in a
cartesian coordinate system with point 0 as origo:
12
u = v r cos(θ) − vθ sin(θ) = −
Qx
Q x
=−
2π r
2π( x 2 + y 2 )
v = v r sin(θ) + vθ cos(θ) = −
Qy
Q y
=−
2π r
2π( x 2 + y 2 )
and
From the velocity field, equation (1.9), the acceleration field, defined by (1.3)
can be found. As described a fluid particle follows the particle path given by (1.8).
The acceleration of the particle is found as
G
G du ⎛ d (u ( x, y, t )) d (v( x, y, t )) ⎞
a=
,
=⎜
⎟
dt ⎝
dt
dt
⎠
(1.14)
where the coordinates for the particle are given by (1.8), so that the acceleration in the
direction of the x-axis is found as
d
d
(u ( x, y, t )) = (u ( x(t ), y (t ), t ))
dt
dt
∂u dx(t ) ∂u dy (t ) ∂u
=
+
+
∂x dt
∂y dt
∂t
ax =
(1.15)
where the last part of the equation has been obtained by making a total differentiation.
The term ∂u / ∂x × dx(t ) / dt has the following physical interpretation:
dx(t ) / dt (= u ) describes how far the fluid particle has moved along its path in the xdirection per unit time. The quantity ∂u / ∂x describes how much the u-velocity varies
in the x-direction at the location of the particle per unit length. The product of the two
terms gives the change in u for the particle, because it is moved from the position x
where the velocity is u to another position x + dx / dt with a new velocity u + du / dx .
13
Example 1.5: Flow through a constriction.
Figure 1.6 shows a plane flow in a narrow conduit with a constriction. For
simplicity it is assumed that the velocity u in the direction of the x-axis is constant
across the conduit. Due to continuity (mass/volume conservation) it is easy to see that
the velocity u2 is larger at the constriction than the velocity u1 at the original crosssection of the conduit. This means that, even if the fluid discharge Q is constant in
time, a fluid particle must be accelerated from u1 to u2 as it moves into the
constriction. This acceleration represents the term (∂u / ∂x)(dx / dt ) = (∂u / ∂t ) u , which
is also know as an advective- or convective term, because it represents the
acceleration due to changes in the flow direction.
Fig. 1.6
Flow in a narrow, non-uniform conduit.
Equation (1.15) can thus be written
ax = u
∂u
∂u ∂u
+v
+
∂x
∂y ∂t
(1.16)
Similarly the acceleration in the y-direction is written
ay = u
∂v
∂v ∂v
+v +
∂x
∂y ∂t
(1.17)
The two first terms on the right hand side represents the convective terms. The last
term ∂u / ∂t expresses that the flow in general is not stationary, so that in a fixed point
14
(x,y) there may also be changes with time so that fluid particle in a fixed point are
also changing their velocity.
Fig. 1.7
Flow forced by a moving piston.
Example 1.6: Unsteady flow.
Figure 1.7 shows the flow in a narrow conduit between two parallel walls. The
flow is forced from left to right by a moving piston. The velocity profile shown to the
right of the piston is taken to be constant over the depth.
Due to the continuity (the fluid is incompressible) the fluid velocity must
everywhere be identical to the velocity of the piston. This means that the convective
terms are identical to zero. This always the case in a uniform flow, which is defined as
a flow where the conditions are the same from one cross section to the next, i.e. there
is no variation in the flow direction.
The acceleration of the fluid in this case is
du ∂u ∂ ds d 2 s
=
=
=
dt ∂t ∂t dt dt 2
(1.18)
where s it the position of the piston at the time t.
1.2 The stresses in a fluid
Let us consider a small unit cube of the fluid, cf. figure 1.8. At the left part of
the figure the forces acting on the four sides of the cube are shown. These forces, F1
to F4, can be resolved into two components: normal forces acting in the direction
normal area and tangential forces acting in the direction of the plane of the area.
15
The tangential force per unit area is known as shear stress. The definition of
the direction for a positive shear stress is shown in figure 1.9.
Fig. 1.8
Forces on a unit cube, resolved into normal and shear forces at
the cube to the right.
Fig. 1.9
Definition of shear stresses.
Fig. 1.10
Normal stresses acting on a fluid element.
16
The notation in figure 1.9 is as follows: τ xy is the shear stress in the ydirection acting on a surface with the area vector (cf. example 1.7) in the direction of
the x-axis.
This can easily be extended to the more general case of a three-dimensional
stress condition.
Analogously the normal stress σ xx and σ yy are defined as the normal force
per unit area, cf. figure 1.10.
Example 1.7: Area vector
Let us consider a volume X as shown in figure 1.11. This volume is limited by
a surface with the area A. On this surface we can cut out an infinitesimal part of the
area: dA.
G
This part area is characterised by the area vector dA , which has a magnitude
equal to the part area (dA) and a direction normal to the surface at the part area and
taken to be positive in the direction out of the volume X.
Fig. 1.11
Definition of the area vector.
17
If consider the specific case of a plane infinitesimal cube with sides parallel to
the coordinate axes, the area vectors per unit width (the width is in the
G direction
G
normal to the plane of the drawing) are as shown in figure 1.12, where i and j are
the unit vectors in the x- and y-direction respective
Fig. 1.12
The area vectors of an infinitesimal cube.
18
Example 1.9: Pressure in a fluid at rest.
In a fluid at rest the shear stresses are zero. This is because if τ ≠ 0 , it will in
one way of the other set the fluid in motion, which intuitively can be seen from figure
1.9.
In a fluid at rest we therefore have only normal stresses. Let us consider a
triangular body, figure 1.13, where dx and dy are of equal length, forming an isosceles
triangle.
Figure 1.13
Forces acting on a triangle in a fluid at rest.
The normal stress at the hypotenuse is simply called σ . The y-axis is pointing
directly up, and the force of gravity has the magnitude ρ g × volume , where ρ is the
density of the fluid and g is the acceleration of gravity. The vertical force balance
reads
− σ yy dx − 12 ρ g dx dy + σ 2dx ×
2
=0
2
If we make the triangle shrink to become infinitely small (dx → 0, and dx = dy ) we
get
σ yy = σ
(1.19)
The horizontal force balance reads
19
− σ xx dy + σ 2dx ×
2
=0
2
or, using dx = dy, we find
σ xx = σ
(1.20)
It is thus seen that σ xx = σ yy . Following this principle it is easy to
demonstrate that the normal stress is the same in all directions.
In a fluid at motion σ xx is in general not equal to σ yy (but in practise often
very close to being equal). A measure for the strength of the normal stresses is called
pressure, which is defined as
p = − 13 (σ xx + σ yy + σ zz )
(1.21)
where the z-axis is normal to the plane of the drawing. The pressure is thus the mean
value of the normal stresses in the three directions of the coordinate axes, but with
opposite sign, so that the p is positive when the normal forces are directed into the
fluid volume considered.
Example 1.10: Proof that
τ xy = τ yx
(1.22)
Determine the moment of the forces (torque) acting on the cube shown in
figure 1.9 and neglect higher order terms:
Moment = surface area × stress × arm =
⎡
⎡
∂τ yx ⎞ dx ⎤
∂τ xy ⎞ dy ⎤
dy ⎛
dx ⎛
+ ⎜⎜ τ yx +
+ ⎜⎜ τ xy +
dx ⎟⎟ ⎥
dx ⎢τ xy
dy ⎟⎟ ⎥ − dy ⎢τ yx
∂
2
y
2
2
x
∂
⎠ 2 ⎥⎦
⎝
⎠ ⎥⎦
⎝
⎣⎢
⎣⎢
20
After division by the factor dydx we have terms with and with out the factors
dx or dy. The terms with the factors dx or dy are infinitely small and can be neglected
and we end with 1.22. The moment must be zero because the moment of inertia of an
infinitesimal cube around its axis normal to the x-y plane is smaller (of a higher order
in dx and dy) than the moment of the forces.
Example 1.11: The shear stress distribution in infinitely long narrow plane gap.
Let us consider two horizontal plates of infinite extension. The lower is at rest
while the upper is moving with the constant velocity v0 in its own plane. There is a
fluid between the two plates. A condition for the fluid is that it adheres to the solid
surfaces, so that the fluid particles adjacent to the fixed plate at the bottom has the
velocity 0, while the fluid particle adjacent to the upper plate has the velocity vo. This
means that there will be a flow in the gap between the two plates: the fluid will move
with a certain velocity u parallel to direction of motion of the upper plate, which is
taken to be the x-axis, cf. figure 1.14.
Fig. 1.14
The velocity profile between two plates.
The upper plate moves with the constant velocity vo for a very long time, and
the induced flow is therefore steady or stationary, which means that
21
∂u
= 0 , u = u ( x, y )
∂t
(1.23)
Furthermore the flow is uniform because the conditions do not vary from one
cross section to the next. Mathematically this is written
∂u
=0
∂x
(1.24)
which means that u is only a function of y. Due to symmetry the vertical velocities are
identical to zero everywhere. By applying eqs. (1.16) and (1.17) we find that the
acceleration of any fluid particle in the flow is identical to zero. If we consider a unit
cube in the flow the resulting force acting on it must be equal to zero, cf (1.1) and
(1.2).
The plates are placed horizontally, and the force of gravity acts in the negative
y-direction. In figure 1.15 the force balance in the direction of the x-axis is
considered.
The conditions are fully uniform and the flow is driven only by the moving
lid. The horizontal forces in the fluid are therefore not varying in the flow direction,
and we can take
∂p
=0
∂x
Fig. 1.15.
The forces acting in the direction of the x-axis (horizontal).
22
The horizontal force balance the reads
⎡
⎤
∂
⎢τ xy + (τ xy ) dy ⎥ dx − τ xy dx = 0
∂y
⎣
⎦
(1.25)
or
∂τ xy
∂y
=0
τ xy is often simply written as τ
τ xy ≡ τ
(1.26)
Like the pressure p, τ cannot vary with x, and we get:
∂τ dτ
=
∂y dy
(1.27)
τ = const
(1.28)
or
The variation in the pressure in the vertical can be found by a projection of the
forces on the direction of the y-axis. The forces are shown in figure 1.16. The two
shear stresses are of the same magnitude and cancel each other out. The can be sen
directly from the conditions of symmetry or mathematically from (1.28) together with
(1.22).
The vertical force balance therefore reads:
−
∂p
dy dx − ρg dx dy = 0
∂y
or
p = −ρ g y + const
(1.29)
which is the hydrostatic pressure distribution.
23
Fig. 1.16
The forces acting in the direction of the y-axis (vertical).
24
2. Laminar flows
In general a distinction is made between laminar and turbulent flows. If colour
is injected into a flow by a thin tube it can be seen, that at small flow velocities the
colour forms a thin thread, which keeps its identity and can be followed over a long
distance in the flow and is only dispersed very slowly. This type of flow is known as a
laminar flow and is found to occur for small values of the Reynolds number (cf.
chapter 6) or at small flow velocities. The very slow dispersion of the thread is
because the molecules in the fluid - in addition to the motion of the flow – are moving
among each other, the so-called Brownian motions.
Fig. 2.1
Laminar (a) and turbulent (b) flow.
At larger flow velocities the nature of the flow changes. A large number of
irregular eddies are superposed and added to the mean flow. The eddies are of very
different magnitude and character. The thread of colour will be spread much faster
and cannot be observed as an entity over a long stretch of the flow. This type of flow
is known as turbulent flow and will be treated in chapter 6.
2.1 The constitutive equations
By a constitutive equation we mean an equation giving a relation between the
forces and deformations of the fluid. Let us continue from example 1.11, where we
25
considered the flow in a closed channel between two plates and the upper plate moved
forward in its own plane while the lower plate was stationary. As described in the
example there is a variation of the velocity u across the channel because of the no-slip
boundary conditions due to the adhesion
u=0
u = v0
for
for
y=0
y=a
(2.1)
where a is the width of the channel.
This is approximately the you find (disregarding the small effects of the
curvature) between two concentric cylinders with slightly different cylinders, cf.
figure 2.2.
Fig. 2.2
Experimental set-up for determination of the viscosity of fluids.
The void between the two cylinders is filled with fluid and the outer cylinder
is made to rotate with the angular velocity ω , so that the velocity along the outer
cylinder becomes
v0 = ω (r + a)
(2.2)
The inner cylinder may rotate as well, but is restricted by a spring. The tension
in the spring increases as the outer cylinder rotates because the shear stress τ is
transferred through the fluid to the inner cylinder. For given values of a and r ordinary
fluids will show a linear relation between the speed of rotation of the outer cylinder ω
(or vo) and the deformation of the spring, which is a measure of the force τ 2πr acting
on the inner cylinder, cf. figure 2.3. for different fluids, for example air and water, the
deformation will be quite different for the same vo (air will give much less
deformation of the spring).
26
Fig. 2.3
The relation between the velocity of the outer cylinder and the
shear stress acting on the inner cylinder for given a and r.
If in the other hand a is changed for the same fluid while maintaining r, the
deformation of the spring will turn out to be inversely proportional to the width of the
gap for the same velocity vo.
From these two experiments we can conclude that
τ=
v0
μ
a
(2.3)
where μ is found to be different for different fluids. The ratio vo/a is a measure of the
velocity gradient of the flow
du v0
~
dy
a
(2.4)
where the y-axis is in the radial direction (normal to the gap). Normally (2.2) and
(2.4) are written as
du
τ=μ
(2.5)
dy
which is known as Newton's formula. The quantity μ is the dynamic viscosity of the
fluid and is larger the more thick and viscous the fluid is. In addition to μ it is useful
to introduce the kinematic viscosity ν defined by
μ = ρν
(2.6)
where ρ is the density of the fluid.
27
The relation between vo and τ for different widths of the gap.
Fig. 2.4
In a general two-dimensional flow (2.5) reads
⎛ ∂u ∂v ⎞
τ xy = μ⎜⎜ + ⎟⎟
⎝ ∂y ∂x ⎠
(2.7)
Example 2.1 The velocity distribution in a closed channel between two plane
parallel walls.
In (2.4) it was assumed that the velocity gradient was constant over the gap.
This is in fact the case as described in example 1.11
du
τ = τ0 = μ
(2.8)
dy
which implies
u=
τ0
y+k
μ
(2.9)
By use of the boundary conditions (2.1) we get
τ 0 v0
=
a
μ
(2.10)
and k = 0, i.e.
u = v0
y
a
(2.11)
28
Example 2.2 The power exerted by the outer cylinder.
We consider again the experiment illustrated in figure 2.2 where the outer
cylinder is rotated, while the inner one I kept stationary by the tension of the spring.
The outer cylinder may be driven by a motor, and the question addressed is how much
power the motor must supply to rotate it. There are 2 different types of energy loss, a)
loss due friction in bearings and transmission and b) loss due to friction I the fluid.
Only b) is considered here. The force acting on the outer cylinder from the fluid is
F = τ 0 2π ( r + a )
(2.12)
or, from (2.10)
F=
v0
μ 2π ( r + a )
a
(2.13)
The power is defined as the work performed per time unit, where the work is force
time path. The total power exerted E is therefore force times velocity
(r + a)
(2.14)
E = Fv0 = v02 μ 2π
a
Calculation example: Consider a flow with vo = 0.10 m/s, a = 2 mm and water as the
fluid. For water the viscosity is ν = 10-6 m2/s at the temperature 10O C. The power is
found to be
⎛ 0.102 m ⎞
⎟
E = (0.1 m / s ) 2 ⋅ 10 3 kg / m 3 ⋅ 10 −6 m 2 / s ⋅ 2π ⋅ ⎜⎜
⎟
0
.
002
m ⎠
⎝
or
E = 0.0032 kg m s −3 = 0.0032Watt / m width
The 'm width' is included in the unit for power because the power exerted is
proportional to the width of the gap (or height of the cylinder) taken normal to the
plane of the drawing in figure 2.2.
This transfer of energy to the fluid between the two cylinders must be spent in
the fluid. Since the flow is stationary it does not gain any kinetic or potential energy
and the power supplied must be dissipated to heat.
29
Fig. 2.5
The distribution of τ , u and the energy dissipation in the gap
between the two cylinders.
From figure 2.5 it is seen that the term
∂
∂u
(τ u ) = τ
∂y
∂y
(2.15)
is constant across the gap. This must represent the power per unit volume, which
therefore is equal to the loss of energy to heat (the dissipation ε ). In this example the
dissipation is constant over the depth and is given as
⎛ ∂u ⎞
∂u
ε=τ
= μ⎜⎜ ⎟⎟
∂y
⎝ ∂y ⎠
2
(2.16)
2.2 The deformation of fluid elements in laminar flows
In section 1.1 we considered the paths of fluid particle in a flow. I this section
we shall, investigate the deformation of a small volume of fluid. The starting point is
the flow between two parallel walls introduced in example 1.1.
The velocity profile is linear (cf. figure 2.5) and is given by
u = v0
y
a
(2.17)
where vo is the velocity of the upper wall and a is the width of the gap. The vertical
velocities are identical equal to zero. It is thus seen that the streamlines are straight
lines parallel to the walls.
In figure 2.6 an element ABCD is shown. At time t = 0 the element is
quadratic with sides parallel to the x- and y-axis.
30
Fig. 2.6
The deformation of a quadrate in a flow with a linear velocity
profile.
First of all it is noted that the centre of gravity of the element has moved in the
direction of the flow. This part of the motion is called translation. Next it can be seen
that point A has moved closer to point C while the distance between points B and D is
increased. This is called the deformation, cf. figure 2.7a. Finally the deformed
quadrate as a whole has been given a small rotation in the clockwise direction, cf.
figure 2.7b
Fig. 2.7
a: The deformation of a quadrate.
b: The rotation of the deformed quadrate.
In a more general flow situation the deformation can be described as follows:
In figure 2.8 it is shown how a quadrate ABCD is deformed to A'B'C'D' during the
short time interval dt. The latter is also shown in figure 2.8 with point D' placed over
point D, so that the translation (u dt , v dt ) has been compensated for.
The rotation, which is taken to be positive in the counterclockwise direction,
is now determined as a sort of average of the change in angle per unit time
G (α + α 2 )
rot v = 1
dt
(2.18)
31
The angle α1 is determined as
∂v
dt dx
∂
x
α1 =
∂u
dt dx
dx +
∂x
or, using that (1 + α) −1 ≈ 1 − α
α1 =
Fig. 2.8
(2.19)
for
∂v ⎛ ∂u ⎞
dt ⎜1 − dt ⎟
∂t ⎝ ∂x ⎠
α << 1
(2.19)
The general deformation of a unite quadrate in a plane flow.
Since the term dt is very small, the term with dt2 is one order smaller and can be
neglected. Equation (2.19) then becomes
α1 =
∂v
dt
∂t
(2.20)
32
The angle α 2 is determined similarly, so the result of (2.18) is
G ⎛ ∂v ∂u ⎞
rot v = ⎜⎜ − ⎟⎟ = 2 ω3
⎝ ∂x ∂y ⎠
(2.21)
The rotation or vorticity is commonly written as 2ω and in a three-dimensional
flow it will become a vector, where (2.21) designates the component of this vector
2ω3 in the direction normal to the x-y-plane.
The deformation or strain is determined in an analogous way as
G
∂v ∂u
def v = (α1 − α 2 ) / dt =
+
∂x ∂y
(2.22)
The deformation of a fluid element is primarily the result of the shear stresses.
Newton's formula (2.7) indicates the relation between the shear stresses and the
deformation.
2.3 The equation of continuity
The continuity equation for a fluid expresses the conservation of mass. For a
non-compressible fluid the density ρ is constant and the conservation of mass will in
this case be the same as conservation of volume. For a plane flow the continuity
equation will in this way express that the areas of ABCD and A'B'C'D' in figure 2.8
are identical. The simplest way to derive the continuity equation mathematically is by
considering the flow of water in and out of a small unit cube, which is placed in a
fixed position. Figure 2.9 shows the plane case. The volume flux per time through the
side AD is u dy, because the flow component v is parallel with AD and does not
contribute to the transport across AD. The remaining contributions for the three other
sides are given in the figure.
The continuity equation now expresses that the total flow into the area through AD
and DC must correspond to the flux out through AB and BC. Hereby we find
∂u ∂v
+
=0
∂x ∂y
(2.23)
33
Fig. 2.9
The flux of fluid through a unit element in a fixed position.
In the general three-dimensional case the continuity equation is written
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
(2.24)
In mathematical terms the continuity equation expresses that the divergence of
the velocity field is 0.
Example 2.3 The continuity equation for a plane flow in polar coordinates
In many cases flow is described most easily by use of polar coordinates, cf.
figure 1.4 illustrating a plane drain. The polar velocity components are shown in
figure 1.5. To derive the continuity equation in polar coordinates it is convenient to
consider the element shown in figure 2.10. AB and DC are here sections of a circle
with O as its centre, while AD and BC are parts of straight lines passing through O.
The continuity equation now reads, cf. figure 2.10
∂v
∂
(v r r ) + θ = 0
(2.25)
∂r
∂θ
or
vr + r
∂v r ∂vθ
+
=0
∂r
∂θ
(2.26)
34
Fig. 2.10
The flux of fluid in an element in polar coordinates.
The first term, which makes the equation differ from the Cartesian version in (2.23),
is included because the length of the circular sections increases in the r-direction.
2.4 The flow equations (the Navier-Stokes equations)
We can now formulate Newton's second law (1.2) for a fluid particle by use of
the forces due to the pressure and the shear stresses. Figure shows a unit volume with
sides parallel to the x- and y- axes. The x-axis forms the angle β with horizontal.
Newton's second law applied in the x-direction now reads
ρ ax =
∂τ ∂p
−
+ ρ g sin β
∂y ∂x
(2.27)
Similarly we find for the y-direction
ρay =
∂τ ∂p
−
+ ρ g cos β
∂x ∂y
(2.28)
35
Fig. 2.11
Forces acting in the x-direction on a unit quadrate.
The two components of the force of gravity in the x- and y-directions can be written
g x = g sin β
and
g y = g cos β
(2.29)
By use of (2.29), Newton's formula (2.7) and the kinematic equations (1.16) and
(1.17) equations (2.23) and (2.24) can formulated as the general flow equations, know
as the Navier-Stokes equations. The derivation of the completely correct equations
requires a more accurate description of the normal stresses in the fluid than the simple
application of the averaged pressure term p, cf. (1.21). The formal derivation is given
in Appendix 1. The complete Navier-Stokes equations read
ρ
⎛ ∂ 2 u ∂ 2 u ⎞ ∂p
⎛ ∂u
du
∂u
∂u ⎞
+ ρ gx
= ρ⎜⎜ + u
+ v ⎟⎟ = μ⎜⎜ 2 + 2 ⎟⎟ −
dt
∂x
∂y ⎠
∂y ⎠ ∂x
⎝ ∂t
⎝ ∂x
(2.30)
ρ
⎛ ∂ 2 v ∂ 2 v ⎞ ∂p
⎛ ∂v
dv
∂v
∂v ⎞
+ρgy
= ρ⎜⎜ + u + v ⎟⎟ = μ⎜⎜ 2 + 2 ⎟⎟ −
dt
∂
y
∂x
∂y ⎠
x
y
∂
∂
⎝ ∂t
⎠
⎝
(2.31)
2.5 Other examples of laminar flows
Poiseuille flow: The plane Poiseuille flow is the flow between two parallel
plates, where the plates are fixed and the flow is driven by a pressure gradient (for
example when a fluid flows from one reservoir to another).
Due to symmetry it is seen that the flow in the y-direction is 0. As the flow is
uniform and stationary the velocity u is only a function of y
36
u = u( y)
v=0
(2.32)
The Navier-Stokes equations (2.30) and (2.31) now are reduced to
μ
d 2 u ∂p
−
+ ρ g sin β
dy 2 ∂x
(2.33)
and
∂p
+ ρ g cos β = 0
∂y
Fig. 2.12
(2.34)
The flow between two parallel stationary walls.
It is often found to be convenient to introduce the excess pressure p+, defined as
p + = p + ρg z
(2.35)
where z is a vertical coordinate relative to a horizontal base level, cf. figure 2.11. We
have therefore
∂p + ∂p
=
+ ρg cos β
∂y
∂y
and
(2.36)
+
∂p
∂p
=
− ρg sin β
∂x
∂x
Using this (2.33) and (2.34) become
∂ 2 u ∂p
μ 2 =
∂x
∂y
(2.37)
37
and
∂p +
=0
∂y
(2.38)
Physically the pressure p+ is seen from (2.35) to be the pressure in excess of
the hydrostatic pressure. The hydrostatic pressure is not of interest for the description
of the flow
because it does not contribute to the driving forces (the hydrostatic pressure
corresponds to the conditions in stagnant water). It is thus the pressure in excess of the
static, which can put a fluid in motion. Equation (2.38) expresses that the pressure p+
is constant across the gap
p + = p x (x)
(2.39)
∂p +
= const
∂x
(2.40)
so that
since ∂p + / ∂x must be the same for all values of x (uniform flow conditions). The
constant in (2.40) may be written − γI , where I is called the hydraulic gradient.
Equation (2.37) then can be written
μ
∂ 2u
= − γI
∂y 2
or
∂ 2u
Ig
=−
2
ν
∂y
(2.41)
which can be integrated twice to give
u=−
Ig y 2
+ C1 y + C 2
ν 2
(2.42)
Where the constants C1 and C2 can be found from the two boundary conditions
u = 0
for
y= 0
(2.43)
u = 0
for
y = a (the width of the gap)
from which we get
38
u=
Ig
(a − y ) y
2ν
(2.44)
giving a parabolic velocity profile as shown in figure 2.12.
Unsteady flow, the wave boundary layer: In this example we consider the flow which
occurs when a plate is given a harmonic oscillation in its own plane. The fluid above
the plate is taken to be of infinite extent.
The velocity of the plate is given as
t⎞
⎛
u = u 0 cos⎜ 2π ⎟
⎝ T⎠
(2.45)
u = u 0 cos(ωt ) , ω = 2π / T
(2.46)
or
Here T is known as the wave period since the motion of the plate is seen (cf.
(2.45)) to be periodic, repeating itself after the time T, 2T, 3T etc. The quantity ω is
known as the cyclic frequency or the angular velocity.
The motion of the plate in its own plane cannot cause any pressure gradients
so ∂p / ∂x = 0 . Further it is seen that ∂u / ∂x = 0 and v = 0. The flow equation (2.30)
is therefore reduced to
+
ρ
∂ 2u
∂u
=μ 2
∂t
∂y
(2.47)
where the hydrostatic pressure has been neglected as in the previous example.
Fig. 2.13
The forces acting on a unit quadrate above an oscillating bed
and the resulting velocity profile.
39
Equation (2.47) expresses that the fluid is accelerated by the shear stresses, as
indicated in figure 2.13.
When the plate oscillates with the periodically (with the period T) a natural
solution would be that the velocity u at a given distance y from the bed also varies
periodically in time. The period must be the same everywhere, but the fluid can be out
of phase with the motion of the plate. The solution can therefore be written
u = f ( y ) cos(ωt ) + g ( y ) sin(ωt )
(2.48)
A simpler way of treating the problem is by the use of complex numbers in
which case (2.48) can be written
u = h( y ) exp(i ωt )
(2.49)
which written in full reads
u = [hr cos(ωt ) − hi sin(ωt )]
+ i[hr sin(ωt ) + hi cos(ωt )]
(2.50)
where hr is the real part of h and hi is the imaginary part of h. When using complex
number two equations emerge, one from the real and one from the imaginary.
Normally only the real part of the solution is taken to have physical significance.
Insertion of (2.49) into (2.47) gives
d 2h
ρ iω h = μ 2
dy
(2.51)
having the solution
⎛ iω ⎞
⎛
iω ⎞
h = C1 exp⎜⎜
y ⎟⎟ + C 2 exp⎜⎜ −
y ⎟⎟
ν
ν
⎝
⎠
⎝
⎠
(2.52)
Using that i = (1 + i ) / 2 it is seen that the first term (with C1) will grow to infinity
away from the wall. As the boundary condition must be that the motion is limited
infinitely far away from the bed, we must have C1 = 0.
Equation (2.49) then becomes
⎛⎛
⎛
ω ⎞
ω ⎞ ⎞⎟
u = C 2 exp⎜⎜ −
y ⎟⎟ exp⎜ i⎜⎜ ωt −
y ⎟⎟
(2.53)
⎜
⎟
ν
ν
2
2
⎝
⎠
⎠⎠
⎝⎝
C2 is found from the condition expressing that the fluid must follow the plate, given in
(2.46) for y = 0. This gives C2 = uo. The term
δ=
2ν
ω
(2.54)
40
is known as the Stokes length and is a measure of how far from the plate its motion
can be felt. For ν = 10-6 m2/s (i.e. water at 200) and T = 5 s we find δ = 1.26 mm, a
quite small quantity. Equation (2.53) describes how u decreases exponentially from
the wall (while it oscillates) with the length scale δ , cf. figure 2.13.
The shear stress can be found by application of Newton's formula
⎛⎛
∂u νu 0 ⎧
τ
y ⎞⎞
⎛ y⎞
=
=ν
⎨− exp⎜ − ⎟ exp⎜⎜ i⎜ ωt − ⎟ ⎟⎟
δ ⎠⎠
δ ⎩
∂y
ρ
⎝ δ⎠
⎝⎝
(2.55)
⎛⎛
y ⎞ ⎞⎫
⎛ y⎞
− i exp⎜ − ⎟ exp⎜⎜ i⎜ ωt − ⎟ ⎟⎟⎬
δ ⎠ ⎠⎭
⎝ δ⎠
⎝⎝
or
τ0
νu
= − 0 [cos(ωt ) − sin (ωt )]
ρ
δ
(2.56)
at the bed, where y = 0. It is seen that τ 0 has a phase shift of 450 in the dimensionless
time ωt relative to the velocity of the plate. The wall shear stress does not have its
maximum at the same time as the velocity because we have an unsteady flow and
fluid has to be accelerated to follow the oscillatory motion.
41
3. Potential flow
Potential flow is a particular type of flow, which can be described in
mathematical terms by application of the so-called potential φ . Physically this means
that the shear stresses can be neglected, i.e. the flow can be described as an
equilibrium between inertial forces (mass times acceleration), pressure forces and
gravity. This is also known as an ideal flow or an ideal fluid.
One consequence of neglecting the shear stresses is that the no-slip boundary
condition at a solid surface cannot be fulfilled. If we consider the wave boundary
layer, cf. chapter 2, equation 2.45 will degenerate to
ρ
du
=0
dt
(3.1)
which with the boundary condition u = 0 for y → ∞ has the solution u = 0. This
describes actually the flow over the entire area except for the very thin layer near the
plate, where the flow occurs because the water adheres to the oscillating plate.
This illustrates the general strength of the theory for ideal fluids: in a relatively
simple way it can describe the flow in the main part of the domain, so that you only
needs to take the effect of the shear stresses into account in thin layers along the solid
boundaries. These thin layers are known as boundary layers and are treated in chapter
5 and 7.
3.1 Definition of the potential
The potential is a scalar, called φ . It is related to the velocity field through the
relation
∂ϕ
(3.2)
u=−
∂x
and
v=−
∂φ
∂y
(3.3)
There are a number of mathematical condition associated to the existence of a
potential: it can be shown that it is a necessary and sufficiently condition that the
rotation of the velocity field is zero in all points of the flow domain, i.e. the flow must
be irrotational.
44
Example 3.1
It is seen from the definition of rotation, equation (2.21), that this is a
necessary condition
G ∂v ∂u
∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞
−
= − ⎜⎜ ⎟⎟ + ⎜ ⎟ = 0
2 rot v =
∂x ∂y
∂x ⎝ ∂y ⎠ ∂y ⎝ ∂x ⎠
(3.4)
G
That rot v = 0 is also a sufficient condition, is for example shown in [1].
3.2 The Bernoulli equation
As mentioned shear streses are neglected in ideal fluids, which means that the
viscosity is taken to be zero. I this way the Navier-Stokes equation (2.30) and (2.31)
turns into
⎛ ∂u
∂u
∂p
∂u ⎞
ρ⎜⎜ + u
+ v ⎟⎟ = − + ρg x
∂x
∂x
∂y ⎠
⎝ ∂t
(3.5)
⎛ ∂v
∂v
∂v ⎞
∂p
ρ⎜⎜ + u + v ⎟⎟ = −
+ ρg y
∂x
∂y ⎠
∂y
⎝ ∂t
(3.6)
and
These two equations are known as the Euler equations. For a potential flow the
rotation is also known to be zero, so that
∂v ∂u
=
∂x ∂y
(3.7)
cf. equation (3.4). Inserting this into (3.5) and (3.6) gives
∂p
⎛ ∂u ⎞ ∂ ⎛ 1 ⎞
ρ⎜ ⎟ + ⎜ V 2 ⎟ = − + ρg x
∂x
⎝ ∂t ⎠ ∂x ⎝ 2 ⎠
(3.8)
∂p
⎛ ∂v ⎞ ∂ ⎛ 1
⎞
ρ⎜ ⎟ + ⎜ V 2 ⎟ = −
+ ρg y
∂y
⎝ ∂t ⎠ ∂y ⎝ 2 ⎠
(3.9)
and
where
V 2 = u2 + v2
(3.10)
45
If we apply (3.2) and (3.3) in (3.8) and (3.9) and integrates with respect to x and y
respectively, both equation give
z+
p V 2 1 ∂φ
= C (t )
+
−
γ 2 g g ∂t
(3.11)
where z is a vertical coordinate (cf. figure 2.11 and equation (2.27)). Equation (3.11)
is the general version of Bernoulli’s equation. For a stationary flow ( ∂ / ∂t = 0 ) it
turns into
z+
p V2
+
= const
γ 2g
(3.12)
3.3 The Laplace equation
While laminar (and turbulent) flows are solved by use of the dynamic flow
equation and the boundary conditions, a potential flow is determined by solving the
continuity equation with the corresponding boundary conditions. The continuity
equation (2.21) can by applying (3.2) and (3.3) be written
∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂ϕ ⎞
⎜− ⎟ + ⎜− ⎟ = 0
∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠
(3.13)
Δφ = 0
(3.14)
or
Where Δ is the Laplace operator
∂2
∂2
Δ= 2 + 2
∂x
∂y
(3.15)
and (3.14) is know as the Laplace equation, which is seen simply to be the continuity
equation.
3.4 An example of a potential flow: flow over a wavy bed
In this (and the following) chapter we consider potential flows, where the flow
is composed of a simple basic flow, on which we superpose a small additional flow
field, also known as a perturbation. In this example the flow over a sinusoidal bed
(or a ripple bed) is considered. The technique applied is as follows:
a:
First the basic flow is determined. In this case it is the flow over a horizontal
bed. The Laplace equation with boundary conditions are applied.
46
b:
The disturbance to the basic flow due to the bed undulations is then
determined. This is also done by application of the Laplace equation and
boundary conditions. The boundary conditions expanded in Taylor series and
it is used that the perturbation is small, so that only the most significant terms
are maintained.
Fig. 3.1 The basic flow and the perturbed flow.
Figure 3.1a shows the basic flow, which is simply a uniform flow over a
horizontal bed. The Laplace equation reads
∂ 2φ ∂ 2φ
+
=0
∂x 2 ∂y 2
(3.16)
Since the flow is uniform we have that ∂u / ∂x = 0 , or by use of (3.2)
∂ 2φ
=0
∂x 2
(3.17)
then (3.16) turns into
∂ 2φ
=0
∂y 2
(3.18)
Integration of (3.17) gives
φ = f1 ( y ) x + f 2 ( y )
(3.19)
while the integration of (3.18) similarly gives
47
φ = g1 ( x ) y + g 2 ( x )
(3.20)
Combination of (3.19) and (3.20) gives
φ = k1 xy + k 2 x + k 3 y + k 4
(3.21)
The boundary condition is: the vertical velocity is zero at y = 0 (no flux through the
bed), i.e.
∂φ
=0
∂y
for
y=0
(3.22)
k1 x + k 3 = 0
for
y=0
(3.23)
v=−
or
which implies that both k1 and k3 must be 0, since (3.23) must hold for all values of x.
Adding a constant to potential gives no change in the velocity field, cf. (3.2) and (3.3).
The constant k4 can therefore have any value and is arbitrarily set to zero. The
potential is then
φ = k2 x
(3.24)
From (3.24) we find
u=−
∂φ
= −k 2
∂x
(3.25)
v=−
∂φ
=0
∂y
(3.26)
and
it is seen that –k2 is equal to the undisturbed velocity V in the flow direction, and the
potential is therefore given as
φ 0 = −Vx
(3.27)
where the index 0 shows that this is the basic (plane bed) flow. The velocity is
constant and equal to V in the x-direction over the entire water depth D, which
isbecause the shear stresses have been neglected so that the velocity cannot become 0
at the bed.
48
We are now at point b in the procedure, and the bed is no longer plane, but has now
superposed a small sinusoidal perturbation. The bed level is given by
h = h0 sin( kx)
(3.28)
where ho is the amplitude of the bed undulation and k is the so-called wave number.
It is seen from (3.28) that h jhas the same value for kx and for kx + 2π , we therefore
find that
kL = 2π
or
k=
2π
L
(3.29)
where L is the wave length of the bed undulation.
This perturbation of the initially plane bed introduces a change in the potential
~
from φ 0 to φ . The change is called φ so that
~
φ = φ0 + φ
(3.30)
~
~
For h0 → 0 we see that φ → 0 , and therefore the order of magnitude of φ is
~
φ = O (h0 )
Normally the amplitude is described by a dimensional term like ho/D or ho/L. In the
following we choose
~
φ = O(h0 / D)
(3.31)
where the quantity
ε=
h0
D
(3.32)
is a small quantity, ε << 1 . This will be used in the following when determining the
~
boundary conditions. First φ is found by solving the Laplace equation, which gives
~
~
~
Δφ = 0 = Δ ( φ 0 + φ ) = Δφ 0 + Δ φ = Δ φ
(3.33)
or
~
Δφ = 0
Since the bed oscillation oscillates periodically with the wave number k in the
~
x-direction it is reasonable to assume that φ behaves similarly. Therefore a qualified
guess for the solution to (3.33) could be
49
~
φ = f ( y )[cos(kx) + a sin(kx)]
(3.34)
which, when inserted into (3.33) gives
f ′′ − k 2 f = 0
(3.35)
f = c1 exp( ky ) + c 2 exp( −ky )
(3.36)
or
The two constants c1 and c2 are found from the two boundary conditions, one
at the water surface and one at the bed.
1) Boundary condition at the bed:
As we use potential theory we cannot require that u = 0 (the water adheres to
the solid boundary) at the bed. But no water can pass the bed, which is impermeable.
This can be expressed by saying that the velocity
Fig. 3.2 Kinematic boundary condition at the bed.
vector must be parallel to the bed. As seen from figure 3.2 this means that
v ∂h
=
u ∂x
for
y=h
(3.37)
The vertical velocity v is found as
~
⎛ ∂φ ⎞
⎛ ∂φ ⎞
v = ⎜⎜ − ⎟⎟
= ⎜⎜ − ⎟⎟
⎝ ∂y ⎠ y =h ⎝ ∂y ⎠ y = h
~
⎛ ∂ ⎛ ∂~
⎛ ∂φ ⎞
φ ⎞⎞
⎜
⎟
≈ ⎜− ⎟
+ h⎜ ⎜⎜ − ⎟⎟ ⎟
+ O (ε 3 )
⎜
⎟
⎝ ∂y ⎠ y =0
⎝ ∂y ⎝ ∂y ⎠ ⎠ y =0
(3.38)
In (3.38) expansion in a Taylor series has been used to express the boundary condition
at y = h by quantities given at y = 0. The two first terms at the right hand side of
(3.38) has the following orders of magnitude
50
~
⎛ ∂φ ⎞
⎜− ⎟
= O(h0 / D)
⎜ ∂y ⎟
⎠ y =0
⎝
(3.39)
cf. (3.31) and
~
h ∂ ⎛ ∂φ ⎞
⎜− ⎟
= O((h0 / D) 2 )
⎜
⎟
D ∂y ⎝ ∂y ⎠ y =0
(3.40)
which shows that the last term in (3.38) consists of terms with ho/D in the power 3 or
higher.
The horizontal velocity can in a similar way be written
~
~
⎛ ∂φ ⎞
⎛ ∂φ ⎞
⎛ ∂φ ⎞
= V − ⎜⎜ ⎟⎟
= V − ⎜⎜ ⎟⎟
u = ⎜⎜ − ⎟⎟
+ O((h0 / D) 2 )
⎝ ∂y ⎠ y =h
⎝ ∂y ⎠ y =h
⎝ ∂y ⎠ y =0
(3.41)
Equation (3.37) can now be written
~
⎛ ∂φ ⎞
∂h
⎜− ⎟
+ O((h0 / D) 2 ) = [V + O(h0 / D)]
⎜ ∂y ⎟
∂x
⎝
⎠ y =0
where
∂h
= O(h0 / D)
∂x
(3.42)
(3.43)
When it is assumed that ho/D << 1 we are allowed only to keep terms which
are proportional to ho/D and neglect all terms of higher order, which means terms that
contain ho/D raised to a power of 2 or higher. In principle (3.37) can be written as
2
3
h
⎛h ⎞
⎛h ⎞
A 0 + B⎜ 0 ⎟ + C ⎜ 0 ⎟ + ..... = 0
D
⎝D⎠
⎝D⎠
(3.44)
In a linear theory it is only required that A = 0, which in our case gives
~
⎡ ∂φ ⎤
∂h
=V
⎢− ⎥
∂x
⎣ ∂y ⎦ y =0
(4.45)
[In a higher order theory additional requirements are to be met: 2. order theory: B = 0,
3. order theory: C = 0 etc.].
By application of (3.28) and (3.36) equation (3.45) can now be written
(−c1 + c 2 )k [cos(kx) + a sin(kx)] = V h0 k cos(kx)
(3.46)
from which we get
51
a=0
and
c 2 − c1 = h0 V
(3.47)
2) Boundary condition at the surface
Flow with a free surface are complicated to describe because the shape of the
surface (and therefore the geometry of the flow) is not know in advance (in contrast to
for example flow in a pipeline). In general the deviation of the surface from the mean
water surface (cf. figure 3.1) can be written as a Fourier expansion
∞
η = ∑ [a n cos(nkx) + bn sin(nkx)]
(3.48)
n =0
The kinematic boundary condition at the surface can be expressed as, cf.
(3.37)
v ∂η
=
u ∂x
for
y = D+η
(3.49)
This can (similar to (3.37)) be linearised to give
v y=D = V
∂η
∂x
or
− k (c1 exp(kD) − c 2 exp(−kD)) cos(kx) =
∞
V ∑ [a n cos(nkx) + bn sin( nkx)]
(3.50)
n =0
where it has been used that a = 0 in (3.34). It follows therefore that only b1 ≠ 0 , while
all the other coefficients (an and bn) are identically equal to 0. Equation (3.50)
therefore is reduced to
− V b1 = c1 exp(kD) − c 2 exp(− kD)
(3.51)
One additional boundary condition is needed for the surface, because we have
only 2 equations to determine the three unknowns c1, c2 and b1. The last boundary
condition expresses that the pressure at the surface is constant and equal to the
atmospheric pressure.
p = const
for
y = D+η
(3.52)
This is know as a dynamic boundary condition, because it is using information
on the forces – in contrast to a kinematic boundary condition, which is related to the
direction of motion only.
52
The relevant equation for the dynamics of the potential flow is the Bernoulli
equation (3.12), which in combination with ((3.52) gives
1 2
(u + v 2 ) + gη = const for y = D + η
(3.53)
2
As before the boundary condition are to be linearised. We shall identify and
maintain only the most significant terms of the square velocities u2 and v2. First we
consider u2
~ 2
~
⎡⎛ h ⎞ 2 ⎤
⎛ ∂φ ⎞
⎛
∂φ ⎞
2
⎟
⎟
⎜
⎜
u = ⎜V − ⎟ = V − 2V ⎜ ⎟
+ O ⎢⎜ 0 ⎟ ⎥
∂x ⎠
⎢⎣⎝ D ⎠ ⎦⎥
⎝ ∂x ⎠ y = D
⎝
2
(3.54)
where the term V2is constant and may be included in the constant in (3.53). Similarly
we have
~ 2
⎡⎛ h ⎞ 2 ⎤
⎛ ∂φ ⎞
v = ⎜⎜ − ⎟⎟
= O ⎢⎜ 0 ⎟ ⎥
⎢⎣⎝ D ⎠ ⎥⎦
⎝ ∂x ⎠ y = D + η
2
(3.55)
Hereby (3.53) turns into
~
⎛ ∂φ ⎞
+ gη = 0
− V ⎜⎜ ⎟⎟
⎝ ∂x ⎠ y = D
(3.56)
or
V k [c1 exp(kD) + c 2 exp(−kD)]sin( kx) + gb1 sin( kx) = 0
from which we get
c1 exp(kD) + c 2 exp(−kD) = −
g
b1
Vk
(3.57)
From the equations (3.47), (3.51) and (3.57) the coefficients c1, c2 and b1 can
be determined. The coefficient b1 is thus found to be described by
b1 ⎡
1
⎤
sinh(kD)⎥
= ⎢cosh(kD) −
2
h0 ⎣
kDF
⎦
−1
(3.58)
where
F=
V
gD
(3.59)
is the Froude number.
53
The variation in the surface elevation with the flow velocity can now be found
from the expression
η=
b1
h
h0
(3.60)
where b1 is given by (3.59). It is seen that the surface is either in phase or 180 degree
out of phase with the bed oscillation. For small values of F (or for small flow
velocities at a given water depth) we find
b1
kD
→ −F 2
sinh(kD)
h0
(3.61)
which means that the water surface is plane (almost as for a non-moving fluid). For
F2 <
tanh(kD)
= Fc2
kD
(3.62)
b1/ho will be negative as shown in figure 3.3b, which means that the surface is in
counter-phase (180 degree out of phase) with the bed oscillation until the critical
Froude number Fc is exceeded. For larger values of the Froude number the surface
will be in phase with with the bed as shown in figure 3.3c. Theoretically
b1 → ∞ for F → Fc , which is because the present theory is linear and terms like
b12 and higher powers have been neglected. These terms become important as F
approaches Fc.
Shallow water theory
For small values of kD (that is for very long oscillations of the bed) the previous
calculations can be simplified considerably. In this case the pressure can locally be
taken to vary hydrostatically over the vertical, increasing from zero as you move
down from the water surface
p = ρg ( D − y + η)
(3.63)
54
Fig. 3.3 The variation of the water surface with the flow velocity.
This different from the general case considered until now, where the pressure
cannot be taken to be hydrostatic because of the curvature of the streamlines, cf.
figure 3.4. For very weak curvature the pressure variations indicated in figure 3.4 can
be neglected.
Equation (3.63) implies that the horizontal velocity u is constant over a given
vertical, because the horizontal pressure gradient, which accelerates the flow, is the
same at all levels for a given x
u = U = U (x)
(3.64)
Fig. 3.4 The pressure variation associated with curvature of the streamlines.
55
The flow equation reads
ρ
dU
∂p
=−
dt
∂x
(3.65)
cf. (2.28). Since the flow is stationary and the pressure is given by (3.63) equation
(3.65) can be written
U
dU
dη
+g
dx
dx
(3.66)
The continuity equation can be formulated for the entire discharge across a
vertical
U ( D + η − h) = Q
(3.67)
which simply expresses that the discharge Q is constant in the flow direction.
As done for the general case it is assumed that the bed undulation is small and
causes only a small perturbation to flow. The velocity can then be written
U ( x) = U 0 + u~ ( x)
(3.68)
where u~ is the perturbation caused by the bed oscillation. When linearised (3.66) and
(3.67) now reads
du~
dη
(3.69)
U0
+g
=0
dx
dx
and
U 0 (η − h) + u~D = const
(3.70)
By differentiating (3.70) with respect to x the quantity u~ can be eliminated from
(3.69) which gives
U 02 d (η − h) dη
−
=0
gD
dx
dx
(3.71)
⎛ F2 ⎞
⎟
η = h⎜⎜ 2
⎟
⎝ F − 1⎠
(3.72)
or
equation (3.72) is the shallow water version of (3.58). As seen the change in the phase
of the water surface now occurs for a critical Froude number of Fc = 1 instead of the
general expression (3.62), where (cf. figure 3.5) Fc is always less than 1.
56
Fig. 3.5 The variation in Fc with kD.
57
4. Progessive sinusoidal waves
Small progressive water waves is an example, where potential flow theory can
give a very good representation of nature. As it is also an important subject in coastal
and offshore engineering a whole chapter has been devoted to these waves.
4.1 Shallow water theory
Contrary to chapter 3.4 we start with the simple case where the pressure
distribution can be taken to be hydrostatic and proceeds thewn to the derivation of the
general expressions for an arbitrary water depth.
Consider a wave with the height H propagating with the celerity c towards
right, cf. figure 4.1. The water surface elevation η can then be written as
η=
H
sin(ωt − kx)
2
(4.1)
ω=
2π
T
(4.2)
where
Is called the cyclic or angular frequency and T is the wave period. If you are at a fixed
place, for example x = 0, (4.1) is reduced to:
H
t⎞
⎛
sin ⎜ 2π ⎟
(4.3)
2
⎝ T⎠
From this it is seen that two successive wave crests will arrive with a time interval of
T (= the wave period). If you consider the wave train at a given time, for example t =
0, (4.1) gives
η=
η=
H
H
⎛ 2π ⎞
x⎟
sin(kx) = sin ⎜
2
2
⎝ L ⎠
(4.4)
cf. (3.39), it is seen that the distance between two successive wave crests is L (= the
wave length).
60
Fig. 4.1 Definition of progressive shallow water waves and the pressure
distribution.
The wave described by (4.1) is propagating with the velocity
c=
ω L
=
k T
(4.5)
Which can be seen by considering (4.1) at two closely spaced points in time, t and
t + Δt . During this time you have to move the distance Δx forward to maintain the
same value of η and stay at the same place in the wave, so that
H
H
sin(ωt − kx) = sin(ωt − kx + ωΔt − kΔx)
(4.6)
2
2
where
ωΔt − kΔx = 0
(4.7)
Δx
ω
=c=
Δt
k
(4.8)
giving
As an alternative (4.5) can be obtained directly by realizing tha a wave will
move one wave length forward during one wave period. The propagation velocity c is
also know as the phase velocity.
The deformation of the water surface as the wave moves causes velocities in
the fluid below. These velocities are periodic in time and space and are normally
much smaller than the propagation velocity c.
Consider the point A at location between the wave crest and the wave trough,
figure 4.1. As mentioned we can for shallow water waves take the pressure to be
hydrostatic – as indicated under the crest and trough in figure 4.1. This means that
there is a negative pressure gradient from left to right in point A, because the pressure
at a given distance from the bottom is large at the left side than at the right side of A
due to the difference in the water surface elevation. The hydrostatic pressure is
described as
61
p = ρg ( D − y + η)
(cf. (3.63)). The pressure gradient in the x-direction at all water depths is therefore
found as
∂p
∂η
= ρg
∂x
∂x
(4.9)
This pressure gradient is giving an acceleration of the fluid
⎡ ∂u
∂p
du
∂u
∂u ⎤
= −ρ
= −ρ ⎢ + u
+v ⎥
∂x
dt
∂x
∂y ⎦
⎣ ∂t
(4.10)
Here the term u is small (being of the order H/L, as u → 0 for H → 0). The term v is
also small, and the two convective terms (the two last at the right hand side) in 4.10
are therefore
Small of second order (~(H/L)2), while the term ∂u / ∂t is proportional to H/L and will
therefore be the dominant term on the right hand side of (4.10). When linearized
(4.10) becomes
ρg
∂u
∂η
= −ρ
∂t
∂x
(4.11)
since ∂η / ∂x is not a function of y we have the same pressure gradient the entire water
depth. (4.11) therefore shows that we can take u to be constant over the depth. The
continuity equation now reads
∂
[( D + η)u ]dx dt = − ∂η dx dt
∂t
∂x
or
u
∂u ∂η
∂η
∂u
+η +D
=
∂x
∂x
∂x ∂t
The first two terms on the right hand side are small of 2. order (~(H/D)2) and
can be neglected reducing the equation to
D
∂u
∂η
=−
∂x
∂t
(4.12)
62
Fig. 4.2 The continuity equation for a water column
From (4.11) and (4.12) second order partial differential equations can be formed for
either u or η . For η the equation reads:
1 ∂ 2η
∂ 2η
=
∂x 2 gD ∂t 2
(4.13)
By inserting (4.1) we get
−k2 = −
1 2
ω
gD
or
c=
ω
= gD
k
(4.14)
which expresses that (small) shallow water waves propagates with a velocity
depending only on the water depth and not on for example the wave length or the
wave height.
The fluid velocity u may for example be derived from (4.12)
∂u
1 ∂η
1 H
=−
=−
ω cos(ωt − kx)
D ∂t
D 2
∂x
(4.15)
giving
u=
1 H ω
sin(ωt − kx)
D 2 k
or simply
63
u=
c
η
D
(4.16)
It is thus seen that the water particles under the wave crest have velocities in the
direction of the wave propagation and the particle under the troughs have velocities in
the opposite direction, see figure 4.3.
Fig. 4.3
The wave induced velocities in shallow water waves.
4.2 Sinusoidal wave at an arbitrary water depth
The expression (4.16) is of course not general: at large water depths surface
water waves cannot be felt at the bottom, so the velocity field and the propagation
velocity must depend on other parameters than the water depth (in this case the wave
length). At large water depth the pressure variation under the wave is spread out and
the assumption of hydrostatic pressure is no longer valid. In this case vertical
accelerations are of importance and a more detailed description of the flow field is
necessary. This can be obtained by use of the potential flow theory.
We consider once more a wave given by
η=
H
sin(ωt − kx)
2
(4.17)
propagating with the velocity c over a fluid with the water depth D. The bed is
horizontal at y=0. By expressing the balance between pressure forces and the
acceleration, i.e. neglecting the effect of viscosity, potential flow theory can be
applied.
The Laplace equation reads
Δφ = 0
(4.18)
At the bottom the kinematic condition is
v=0
for
y=0
or
−
∂φ
=0
∂y
for
y=0
(4.19)
64
At the surface y = D + η the pressure (in excess of the atmospheric pressure) must be
zero, in which case the generakised Bernoulli equation (3.11) reads
D +η+
u 2 + v 2 1 ∂φ
= C (t )
−
2g
g ∂t
for
y = D+η
(4.20)
Finally there is the kinematic condition expressing that a water particle at the water
surface stays at the water follows the water surface. Mathematically this can be
expressed as
v=
dη
dt
for
y = d +η
(4.21)
In (4.21) the total derivative related to a water particle has been applied, where
∂η ∂η
dη ∂η
=
+u
≈
dt
∂t
∂x ∂t
(4.22)
using that the convective terms can be neglected in a linearized analysis. (4.21) now
reads
v=
∂η
∂t
for
y = D+η
(4.23)
The vertical velocity v at the surface can be written as an expansion
⎛ ∂v ⎞
v(d + η) = v( D) + η⎜⎜ ⎟⎟
+ ......
y
∂
⎝ ⎠ y =D
(4.24)
In a linear theory the last terms on the right hand side can be neglected as thy are of
the order (H/D)2 or smaller, so (4.23) becomes:
⎛ ∂φ ⎞
∂η
=
v( D) = ⎜⎜ − ⎟⎟
⎝ ∂y ⎠ y = D ∂t
Analogous (4.20) can be linearized to give
η−
1 ⎛ ∂φ ⎞
=0
⎜ ⎟
g ⎝ ∂t ⎠ y =D
(4.25)
(4.26)
where C(t) has been included in φ . Changing φ with a constant does not affect the
velocity field.
The water surface has been taken to vary periodically in time and x (4.17). It is
therefore natural to assume the induced velocities – and therefore also φ - to vary
periodically in the same way, possibly with a phase shift, introducing also a sine term
65
φ = f ( y )[a sin(ωt − kx) + cos(ωt − kx)]
(4.27)
where f(y) is an unknown function and a is a constant. By inserting (4.27) into the
Laplace equation (4.18) we get
∂2 f
−k2 f = 0
2
∂y
(4.28)
f = c1 exp(ky ) + c 2 exp(−ky )
(4.29)
having the solution
The three constant, c1, c2 and a are found from the three boundary condition (4.19),
(4.25) and (4.26). Equation (4.19) gives
c1k − c 2 k = 0
or
c1 = c 2
(4.30)
so that f can be written
f = c1 [exp( ky ) + exp( −ky )] = 2c1 cosh( ky )
(4.31)
(4.25) now gives
− 2c1k sinh( kD)[a sin(ωt − kx) + cos(ωt − kx)] =
(4.32)
H
ω cos(ωt − kx)
2
from which we find
a = 0
(4.33)
and
2c1k sinh( kD) = −
H
ω
2
(4.34)
Finally the boundary condition (4.26) gives
H
1
sin(ωt − kx) + ω sin(ωt − kx)2c1 cosh(kD) = 0
2
g
(4.35)
or
66
2c1
ω
H
cosh(kD) = −
2
g
(4.36)
From (4.34) and (4.36) we find the relation
kg tanh(kD) = ω 2
(4.37)
Noting that the equation (4.8) for the wave celerity is general, (4.37) can be written
c=
ω
=
k
g
tanh(kD)
k
(4.38)
which is the general equation describing the celerity of small sinusoidal waves at an
arbitrary water depth.
For shallow water kD will tend to zero, so (4.38) approaches
c = gD
tanh(kD)
→ gD
kD
for
kD → 0
(4.39)
which is the expression for shallow water waves, (4.14).
For deep water we have kD → ∞ so that tanh(kD) → 1 which gives
c→
g
k
(4.40)
In deep water the wave celerity for a linear theory is thus only a function of the wave
length.
The particle velocities can be found from the potential φ , from (4.27), (4.31),
(4.33) and (4.36) is found as
φ=−
H g cosh(ky)
cos(ωt − kx)
2 ω cosh(kD)
(4.41)
From this the velocities can be found as
u=−
∂φ kH g cosh(ky)
=
sin(ωt − kx)
∂x
2 ω cosh(kD)
(4.42)
v=−
∂φ kH g sinh(ky )
=
cos(ωt − kx)
∂y
2 ω cosh(kD)
(4.43)
and
It can be noticed that in a linear theory the velocities u and v are proportional to the
wave height and for moderate values of H they will be much smaller than the
propagation velocity c. The velocity profiles under the wave crest and trough are
shown in figure 4.4.
67
Fig. 4.4 Wave induced velocities under a sinusoidal wave. Compare with
figure 4.3.
Example 4.1
Refraction
As waves propagates towards a coastline with an angle different from 0o, the
wave fronts will be deflected (a wave front is a curve indicating the position of the
wave crest of a wave at a given time). This is because the wave celerity c decreases
with decreasing water depth. In figure 4.5 the wave fronts at time t and t + dt are
shown for a wave approaching a straight coast with depth contour lines parallel with
the coastline. During this time interval the wave has moved from AC to BD, where
AB and CD are known as wave orthogonals, defined as curves that are orthogonal to
the wave fronts. The distance between the two wave orthogonals is called Δb .
Fig. 4.5 Deflection of wave fronts approaching a coast.
68
The distances A-B and C-D are given as
A − B = c dt
(4.44)
∂c
⎛
⎞
C − D = ⎜ c − Δb sin β ⎟ dt
∂x
⎝
⎠
(4.45)
Where c is the wave celerity at x = 0, cf. figure 4.5.
From this it is seen that CD is slightly longer than AB, namely by
ds = −
∂c
Δb sin β dt
∂x
(4.46)
ds dc
= sin β dt
Δb dx
(4.47)
By defining dβ as
dβ = −
it is found that the deflexion dβ that takes place as the wave moves towards the
coastline by the distance dx = (c dt cos β ) is given as
∂β 1 ∂c
=
tan β
∂x c ∂x
(4.48)
This equation can be integrated to give
c
= const
sin β
(4.49)
which is know as Snell’s law
Example 4.2
Particle paths in waves.
From the particle velocities given by (4.42) and (4.43) it is possible to
calculate they position at different times. Consider a particle with a mean position in
(xo,yo). The particles position in the wave motion can then be found from
dx
dy
(4.49)
=u ,
=v
dt
dt
By nitration of (4.42) and (4.43) the particle coordinates are found as
69
x = x0 −
H cosh(ky )
cos(ωt − kx)
2 sinh(kD)
(4.50)
y = y0 +
H sinh(ky )
sin(ωt − kx)
2 sinh(kD)
applying also (4.37). The particle paths are shown in figure 4.6 for deep, intermediate
and shallow water. In deep water the particle paths are circular (in the linear theory),
while the particles in shallow water move back and forth in an almost horizontal
oscillation. It may be noted that if higher order terms in H/D are included in the
analysis the particle paths are no longer closed.
Fig. 4.6 Particle paths in deep, intermediate and shallow water waves.
Example 4.3
Wave groups
In nature waves are generated by the blowing over the water surface. The water
surface can be shown to be unstable under the action from the wind develops
infinitely small undulations which grow to become waves.
These waves are not all identical but vary in length and height. Wave with different
periods and heights form a quite complicated and time-varying ocean surface. Here
we consider a simple example, the pattern formed by superposition of two sinusoidal
waves with the same wave height but with slightly different wave lengths, having the
wave numbers k and k + Δk respectively.
The first wave is described as
η1 =
H
H
sin(ωt − kx) = sin( k (ct − x))
2
2
(4.51)
while the second wave is given by
70
η2 =
H
sin[(k + Δk )((c + Δc)t − x)]
2
(4.52)
By superposition and application of formulas for addition of sines we get
η = η1 + η 2 =
(4.53)
⎡ Δk ⎧⎛
⎫⎤
Δc ⎞
H cos ⎢ ⎨⎜ c + k
⎟t − x ⎬⎥ sin[k (ct − x)]
Δk ⎠
⎭⎦
⎣ 2 ⎩⎝
The surface described by (4.53) is illustrated in figure 4.7
Fig. 4.7
Wave group.
The last tewrm on the right hand side of (4.53) represents the normal
propagating wave, while the cosine factor makes the amplitude of the wave varies in
the x direction so that the wave amplitude is modulated by the broken curve shown in
figure 4.7. This curve has the wave length Lg, given as
Lg =
4π
Δk
(4.54)
A difference in the wave number by Δk corresponds to a difference in the wave length
of ΔL , which is seen from
Δk = ( k + Δk ) − k =
2π
2π 2π ⎛ ΔL ⎞
ΔL
−
≈
− 1⎟ = −k
⎜1 −
L + ΔL L
L ⎝
L
L
⎠
(4.55)
so that
Lg
L
= −2
L
ΔL
(4.56)
where ΔL < 0 for Δk > 0. It is seen from (4.56) that Lg >> L for small relative
differences in the wave length of the two waves. The two points A and B, where the
wave length is 0, moves forward with the velocity cg (the group velocity), determined
as
71
cg = c + k
Δc
∂c
≈c+k
∂k
Δk
(4.57)
cf. (4.53). From (4.38) we have
c2 =
g
tanh( kD)
k
giving
2c
⎡
2kD ⎤
dc
g
g
D
g
= − 2 tanh(kd ) +
= − 2 tanh(kD) ⎢1 −
⎥
2
dk
k cosh (kD)
k
k
⎣ sinh( 2kD) ⎦
(4.58)
From which (4.57) gives
cg = c −
c⎡
2kD ⎤
⎢1 −
⎥
2 ⎣ sinh( 2kD) ⎦
cg =
c⎡
2kD ⎤
⎢1 +
⎥
2 ⎣ sinh( 2kD) ⎦
or
(4.59)
The wave group is seen to propagate with a velocity varying between c/2 in
deep water (kD → ∞) and c in shallow water (kD → 0) . The individual waves (drawn
with a full cure in figure 4.7) propagate of course with the celerity c.
It may be noted that large natural waves are often observed to form groups of
3-8 waves, and that the groups are important for the design of offshore as well as
coastal structures.
The explanation for cg being smaller than c in deep water and equal to c in
shallow water is that all waves regardless of the period or wave length have the same
celerity in shallow water (no frequency dispersion), while short wave propagate at a
slower rate than long waves in deep water.
4.3 Wave energy
When compared to an undisturbed water surface a wave possesses 2 types of
energy, namely potential energy and kinetic energy. The potential energy is present
because water so to speak has been moved up from the wave trough to the wave crest,
see figure 4.8.
The total potential energy can thus be found by calculating the potential
energy gained by a single strip with the length dx, the height ηand the centre of
gravity η /2 beneath the still water level when moved to a position with the centre of
gravity placed η/2 above the still water level, giving
72
E pot
1
=
L
L/2
∫
ρgη(η dx) =
0
1
ρgH 2
16
(4.60)
The kinetic energy is found by integrating over the entire volume of water
(using that the upper limit of integration D + η can be approximated by D)
E kin
L D
⎤
1 ⎡ 1 2
= ρ ∫ ⎢ ∫ (u + v 2 )dy ⎥ dx
L 0 ⎣⎢ 0 2
⎦⎥
(4.61)
Considering first the depth integrated kinetic energy we get
D
2
1 2
1 ⎛ kHg ⎞
1
2
∫ 2 (u + v )dy = 2 ⎜⎝ 2ω ⎟⎠ cosh 2 (kD) ×
0
(4.62)
D
D
⎤
⎡ 2
2
2
sin
(
t
kx
)
cosh
(
ky
)
dy
cos
(
t
kx
)
sinh 2 (ky ) dy ⎥
ω
−
+
ω
−
⎢
∫
∫
⎥⎦
⎢⎣
0
0
by applying (4.42) and (4.43). Using the rule cosh2(ky) = 1 + sinh2(ky) (4.62) is
reduced to
D
⎞
⎛
1 2
1 ⎛ kHg ⎞
1
2
2
2
⎟ (4.63)
⎜
(
u
v
)
dy
D
sin
(
t
kx
)
sinh
(
ky
)
dy
+
=
ω
−
+
⎜
⎟
∫2
∫
2
⎟
⎜
2
2
ω
cosh
(
kD
)
⎝
⎠
0
0
⎠
⎝
D
Using that sinh2(ky) = ½(cosh(2ky)-1) (4.63) can be further modified to
Fig. 4.8 Determining the potential energy
2
(
)
⎞ ⎡
1 2
kHg
1⎛
D
⎤
2
2
∫ 2 (u + v )dy = 2 ⎜⎜⎝ 2ω cosh(kD) ⎟⎟⎠ ⎢⎣ D sin (ωt − kx) − 12 + 4kD sinh(2kD)⎥⎦ (4.64)
0
D
Averaging (in time or space) make the term ( sin 2 (ωt − kx) − 1 / 2 ) = 0, and the
average kinetic energy per unit bed area is found as (cf. (4.61))
73
2
E kin
⎞ D
kHg
1 ⎛
⎟
sinh( 2kD)
= ρ⎜⎜
2 ⎝ 2ω cosh(kD) ⎟⎠ 4kD
(4.65)
The relation sinh(2kD) = 2sinh(kD)cosh(kD) inserted in (4.65) reduced this to
E kin = ρ
1
gH 2
16
(4.66)
Exactly identical to the result for the potential energy.
Example 4.4
Transport of wave energy, wave energy flux
As the wave moves forward with the celerity c it also transports energy in the
same direction, but the transport velocity ce of the energy is generally smaller than the
velocity of propagation.
The energy transport velocity ce can be determined in two ways, by
considering the kinematics or the dynamics. The kinematic analysis has in fact already
been made in example 4.3 considering the wave pattern resulting from the
superposition of two waves of the same amplitude, but with slightly different wave
numbers. There it was found that the modulation curve shown as broken lines in
figure 4.7 moved forward with the velocity cg. In the points A and B the wave energy
is zero, as the amplitude here is zero. A ‘parcel’ of energy is contained between A and
B. This ‘parcel’ is transported forward with the same velocity as point A and B; that is
cg. The amount of energy contained between A and B is constant as long as energy
dissipation is not considered.
The dynamic analysis of the transport and transport velocity of energy in waves is
more fundamental and can be made as follows:
The significant part of the energy transport can be expressed through the work
per time unit (power) made by the wave motion at the vertical line A-A, i.e. the work
made by the fluid at the left hand side of A-A, cf. figure 4.9. This work can be
expressed as
W =
D+η
D
0
0
∫ pu dy ≈ ∫ pu dy
(4.67)
using that η<< D.
This work results in a transfer of energy to the fluid at the right hand side of
A-A, i.e. a flux of energy through A-A from left to right. The energy flux Ef through
A-A can thus be written
Ef =W
(4.68)
It may be noted that there is also a flux of kinetic energy through A-A. This
contribution is of higher order (larger than (H/D)2) than the work, and can therefore
be neglected in this analysis.
74
Fig. 4.9 The rate of work (power) made on A-A by the fluid at the left hand
side of A-A.
The pressure in (4.69) can (as described for (2.35)) be decomposed as
p + = p − ρg ( D − y )
(4.69)
so that p+ describes the excess pressure above the mean hydrostatic pressur. This
pressure can be found from the generilized Bernoulli equation (3.11), where the
potential is known from (4.41). After some calculations the pressure is found as
p + = ρg
H cosh(ky )
sin(ωt − kx)
2 cosh(kD)
(4.70)
(4.67) now gives
D
W =
D
+
∫ p u dy − ∫ ρgyu dy
0
(4.71)
0
where u can be obtained from (4.42). Only the first term on the right hand side of
(4.71) gives a contribution with a time average different from zero, expressing the
transport of energy through the line A-A. Physically the energy transport can be
explained by the relation between the excess pressure, the instantaneous water surface
elevation and the horizontal water velocity. When the surface elevation is high (wave
crest) the pressure is high and the water moves forward, and correspondingly when
the elevation is low the pressure is also low (p+ is negative) and the water moves
backwards. The product of p+ and u is therefore always positive.
The term can be written
D
+
∫ p u dy =
0
2kD ⎞ 2
ρgH 2 c ⎛
⎟ sin (ωt )
⎜⎜1 +
8 ⎝ sinh(2kD) ⎟⎠
(4.72)
Which gives the time averaged rate of work
75
T D
⎤
⎛
1 ⎡ +
1
2kD ⎞
⎟⎟
W = ∫ ⎢ ∫ p u dy ⎥ dt = ρgH 2 c⎜⎜1 +
T 0 ⎣⎢ 0
16
sinh(
2
kD
)
⎝
⎠
⎦⎥
(4.73)
Introducing the group velocity from (4.59) and that
1
E = E kin + E pot = ρgH 2
8
into (4.73) we get
W = E f = cg E
(4.74)
(4.75)
Which expresses directly that the energy flux corresponds to the wave energy being
moved with the group velocity cg.
4.4 Waves over variable water depth
When waves approach a coast their dimensions (height and length will change
due to the decreasing water depth. For example will the wave height at sufficiently
shallow water increase with a decreasing water depth until the wave breaks. The
breaking begins when the wave height is comparable to the water depth (often the
criterion H/D = 0.8) is used). Simultaneously the wave length will be smaller as the
water depth decreases.
Change in the wave length
Equation (4.39) gave the relation for the variation in the wave celerity at an
arbitrary water depth
c = gD
tanh(kD)
kD
(4.76)
76
Fig. 4.10
Change in the propagation velocity, the wave length and wave height
with the water depth. Subscript o indicates infinitely deep water.
As shown in figure 4.10 the wave celerity decreases as the wave propagates
into more shallow water.
At the same time we have from (4.5)
T=
L
c
(4.77)
where T is the wave period. This we assume to be constant, which just means that the
same number of waves pass a given cross-section. From (4.77) together with (4.76)
we get
tanh(kD)
L
= gD
T
kD
(4.78)
For deep water we specifically get
Lo
=
T
g
=
k0
gLo
2π
(4.79)
or
Lo =
gT 2
2π
(4.80)
which inserted into (4.78) gives
L = Lo tanh(kD)
(4.81)
77
Fig. 4.11 Variation in wave height over variable water depth.
Analogously (4.80) and (4.77) gives
c = co tanh(kD)
(4.81)
This relation is shown in figure 4.10. Equation (4.80) the general expression for the
variation in the wave length with the water depth, provided that the slope of the bed is
not too large. If the bed slope is large other phenomena, such as reflection, may be of
importance.
Change in the wave height
The variation in the wave height over a gently sloping bed is determined by
requiring that there is no energy loss in the wave motion. Normally there is a small
energy loss in the near-bed wave boundary layer, which is negligible over a length of
some wave lengths. breaking waves there is a considerable energy loss, and the
considerations made in the following cannot be applied in the surf zone.
Let us consider a two-dimensional plane situation where the waves propagate
over a bed with depth contours which are normal to the plane of the paper, see figure
4.11.
When there is is no energy loss the energy flux in A and B must be the same
and (4.75) then gives
( c g E ) A = (c g E ) B
(4.82)
E is found from (4.60) and (4.66)
1
E = E pot + E kin = ρgH 2
8
(4.83)
The group velocity cg may be found from (4.59) and (4.81)
cg =
⎡
1
2kD ⎤
co tanh(kD) ⎢1 +
⎥
2
⎣ sinh( kD) ⎦
(4.84)
78
Using (4.84) and (4.83) in (4.82) and that in deep water cgo = ½co we get
⎡
2kD ⎤
2
H 2 tanh(kD) ⎢1 +
⎥ = const = H o
sinh(
2
)
kD
⎣
⎦
(4.85)
where Ho is the wave height in (infinitely) deep water. The variation in the wave
height can therefore be written
⎡
H
2kD ⎤
= coth(kD) ⎢1 +
⎥
Ho
⎣ sinh(2kD) ⎦
(4.86)
which expresses the relative change in wave height as function of the dimensionless
parameter kD. This parameter can for a given wave period be found from (4.80).
Example 4.5:
Wave table.
The variation in wave height and wave length as a wave propagates from deep
to more shallow water can be found from (4.79), (4.80) and (4.86). These equations
may be solved use of a hand calculator or by use of a wave table as given in figure
4.12.
Consider for example a 2 m high wave in 40 m of water and with a period of
11 s. We wish to determine the height and length at 10 m water depth. The table can
be used as follows.
1) The wave length at infinitely deep water is calculated:
Lo = 1.56T2 = 189 m
2) D/Lo is calculated: 40m/189m = 0.212
Now the value for column 1 is known, and H/Ho can be read from column 8
H
= 0.921
Ho
giving a wave height Ho in infinitely deep water of
Ho = 2m/0.921 = 2.17 m
The deep water wave height is seen to be larger than at 40 m water depth. This is
because cg and thus the energy flux has a maximum for D/L = 0.19, cf. (4.84) and
figure 4.10.
79
Wave table:
Formulae and relations
General:
ω = 2π / T
Abbreviation:
G = 2kD/sinh(2kD), θ = ωt − kx
Wave surface:
η = H / 2 cos θ = h / 2 cos(ωt − kx)
, k = 2π / L , c = L / T
, Lo = 1.56T 2
Formula No.
Wave propagation
velocity, celerity:
c=
g
tanh(kD)
k
(4.38)
Group velocity:
cg =
c
(1 + G )
2
(4.59)
Horizontal orbital velocity:
u=
Vertical orbital velocity:
v=−
Horizontal particle
motion:
x = xo +
H cosh(ky )
sin(θ)
2 sinh(kD)
(4.50) & (4.80)
Vertical particle
motion:
y = yo +
H sinh(ky )
sin(θ)
2 sinh(kD)
(4.50) & (4.80)
Excess pressure:
p + = ρg
H cosh(ky)
cos(θ)
2 cosh(kD)
(4.70)
Reaction force, pressure:
ρgH 2 G
Fp =
16
Reaction force, momentum:
Fm =
ρgH 2 (1 + G )
16
(4.97)
Energy flux:
Ef =
ρgH 2 (1 + G )c
16
(4.59) & (4.75)
πH cosh( ky )
cos(θ)
T sinh( kD )
(4.42) & (4.80)
πH sinh( ky )
cos(θ)
T sinh(kD)
(4.43) & (4.80)
(4.92)
Fig. 4.12 Wave table (from [3]. Note that the water surface in these formulae is
shifted by 900 relative to the formulae in the text, cos(θ) instead of
sin θ ).
80
81
At 10 m water depth the parameter for column 1 is found as
10 m
D
=
= 0.0529
Lo 189 m
And column 3 and 8 then gives
D
= 0.0971
L
⇒
L = 10 m / 0.0971 = 103 m
H
= 1.015
Ho
⇒
H = 2.17 ⋅ 1.015 m = 2.20 m
and
Finally it should be noted that the limit for using shallow water wave theory is D/L <
1/20, while the limit for using deep water wave theory is D/L > ½.
4.5 The wave reaction force
In the last part of this chapter more time-averaged quantities will be
considered. Time averaged terms are of higher order than one (H/L)2, (H/L)3 etc. In
the following only terms of second order will be retained. In section 4.3 the energy in
the wave motion was considered, now we will analyse the momentum in water waves.
Consider a shallow water wave. The pressure distribution is hydrostatic as
shown in figure 4.13.
Now we will show that the presence of the waves makes the fluid at the left
hand side of the vertical section A-A exert a force F on the fluid to the right, which is
larger than the hydrostatic pressure force po, we would find if there were no wave
motion
po =
1
ρgD 2
2
(4.87)
The force F consists of two contributions, a pressure force Fp and a momentum force
Fm.
82
Fig. 4.13 The pressure distribution under a shallow water wave.
A: The pressure force Fp
The instantaneous pressure force acting on A-A is
D +η
Fp =
∫ p dy
(4.88)
0
For a shallow water wave the pressure p is given as
p = ρg (d + η − y )
(4.89)
by which (4.88) is found to give
D +η
1 ⎤
⎡
F p = ρg ⎢( D + η) y − y 2 ⎥
2 ⎦0
⎣
(4.90)
= ρg
( D + η)
1
= po + ρgDη + ρgη 2
2
2
2
The water surface elevation is
H
η = sin(ωt − kx)
2
and we find therefore the following time-averaged value of Fp (averaging over a wave
period)
T
1
1
F p = ∫ F p dt = po + ρgH 2
T0
16
(4.91)
83
As seen in (4.91) the pressure force is larger than po. Physically the force is
centred around the level of the mean water surface. At all levels below the wave
trough there is fluid during the entire wave period and the time averaged pressure
force corresponds to the hydrostatic pressure for quiescent water. This is seen from
p = ρg ( D + η − y )
so that
p = ρg ( D − y )
At the levels D – H/2 < y < D+H/2 there is only water during a part of the wave
period (at the trough level practically all the time, at the crest level practically none of
the time). If we consider the level above the mean water level there is no water under
quiescent conditions (pressure = 0) and therefore no force. With the wave ther is
water part of the time and therefore a positive pressure force.
Similarly for the levels between the mean water surface and the wave trough
level: Here we have a higher pressure when the wave crest passes, but not a pressure
lower than zero when the trough passes, this gives also a positive pressure force.
In case of intermediate or deep water waves the analysis of the pressure force is
considerably more complicated because the mean pressure distribution deviates from
the hydrostatic. The general first order (sinusoidal) wave theory then gives
F p = po +
1
2kD
ρgH 2
16
sinh(2kD)
(4.92)
Equation (4.92) is seen to approach (4.91) for kD → 0 , and for deep water waves:
Fp → 0
for
kD → ∞ , so there is no additional pressure force for deep water
waves.
B: The momentum force
In the case of shallow water waves the horizontal particle velocity is
u=
c
η
D
(4.93)
(cf. (4.16)). This horizontal velocity gives rise to a flux of momentum which acts as a
force. From basic hydraulics the momentum force is know, and can be written as
G G G
Fm = −ρ ∫ v (v ⋅ dA)
(4.94)
G
where dA is an infinitesimal area vector on the control surface, for the vertical surface
A-A the area vector is horizontal, pointing towards left. The horizontal component of
(4.94) for A-A is
84
D +η
Fm = ρ
∫u
2
dy
(4.94a)
0
(4.93) and (4.95) gives
Fm = ρ
c2 2
η ( D + η)
D
(4.95)
Time-averaging of (4.96) then gives
Fm = ρ
c2 H 2 1
= ρgH 2
D 8
8
(4.96)
using that c = gD .
In contrast to F p
Fm is evenly distributed over the vertical from the bed to
the mean water surface.
In the general case of arbitrary water depth the horizontal wave orbital
velocity (4.42) inserted into (4.94a) gives (after time-averaging)
Fm =
⎛
1
2kD ⎞
⎟⎟
ρgH 2 ⎜⎜1 +
16
⎝ sinh( 2kD) ⎠
(4.97)In total the
resulting time-averaged force F acting on section A-A in figure 4.13 is
F = F p + Fm
= po +
1 ⎛
2kD ⎞
⎟
E ⎜⎜1 +
2 ⎝ sinh(2kD) ⎟⎠
(4.98)
= po + Fr
where E is the wave energy given by (4.74). By deducting po (the hydrostatic pressure
under quiescent conditions) from F we get Fr which is normally known as the
reaction force of the wave.
Example 4.6:
The change in the mean water surface for waves approaching a
coast.
85
Let us consider waves approaching a long straight coast in the direction
normal to the coastline, cf. figure 4.14. As described in section 4.4 the wave height
will increase towards the shoreline until the waves start to break. The breaking
criterion being expressed by the wave height and depth at the breaker point
H b = Khb ≈ KDb
(4.99)
where K is approximately 0.8.
Fig. 4.14 The variation in the wave height as the waves propagate towards the
coast.
Outside the breaker point the loss of wave energy is very small and may for
practical purposes be neglected (the energy loss only takes place in the bed boundary
layer). When the energy loss is neglected the Bernoulli equation (4.20) may be
applied. It reads
u2 + v2
1 ∂φ
+ ( D + η) −
= c(t )
2g
g ∂t
for
y = d +η
(4.100)
In this equation we include terms of second order (rather than linearizing as done in
(4.26)) and get
u 2 ( D) + v 2 ( D)
1 ⎛ ∂φ ⎞
1 ⎛ ∂ 2φ ⎞
⎟
+η− ⎜ ⎟
− η⎜⎜
= c(t )
g ⎝ ∂t ⎠ y = D g ⎝ ∂t∂y ⎟⎠ y = D
2g
(4.101)
Time-averaging of (4.101) over a wave period gives
η=−
u 2 ( D) + v 2 ( D) η ∂ 2 φ
+
+ c(t )
2g
g ∂t∂y
(4.102)
Inserting φ, η, u and v from the solution found earlier (4.41), (4.42) and
(4.43) gives
86
η=
⎡⎛ kH g ⎞ 2 1 ⎛ kH g ⎞ 2
1⎤
2
− ⎢⎜
+⎜
⎟
⎟ tanh (kD) ⎥
2 ⎦⎥
⎣⎢⎝ 2 ω ⎠ 2 ⎝ 2 ω ⎠
2g
(4.103)
+
H H g
1
k tanh(kD)ω
2 2 ω
2
g
+ c(t )
By use of (4.78) k 2 g / ω 2 can be written as
k 2g T 2
k
= 2 g=
2
tanh(kD)
ω
L
(4.104)
by which (4.103) may be reduced to
2
⎞
k
1
⎛H ⎞ ⎛1
η = −⎜ ⎟ ⎜⎜
− k tanh( kD) ⎟⎟ + c(t )
⎝ 2 ⎠ ⎝ 4 tanh(kD) 4
⎠
2
2
⎛ H ⎞ ⎛ cosh (kD) − sinh (kD) ⎞⎟
+ c(t )
= −⎜ ⎟ k ⎜⎜
⎝ 2 ⎠ ⎝ 4 cosh(kD) sinh( kD) ⎟⎠
2
(4.105)
2
k
⎛H⎞
= −⎜ ⎟
+ c(t )
⎝ 2 ⎠ 2 sinh( 2kD)
If we put η = 0 in deep water we get c(t ) = 0 giving
2
k
⎛H⎞
η = −⎜ ⎟
⎝ 2 ⎠ 2 sinh( 2kD)
(4.106)
From this it is seen that there is a small lowering of the mean water surface as
the wave propagate into shallow water, provided we are outside the surf zone with
breaking waves. This phenomenon is called set-down. That it is in fact a small
variation can be seen from (4.106) and (4.99). Just before the point of wave breaking
we find
1 ⎛H
= − ⎜⎜ b
Db
16 ⎝ Db
η
2
⎞
2kD
⎟⎟
⎠ sinh( 2kD)
(4.107)
1 ⎛H
≈ − ⎜⎜ b
16 ⎝ Db
2
⎞
K2
⎟⎟ = −
≈ −0.04
16
⎠
that is about 4% of the water depth at the point of wave breaking.
87
Inside of the point of wave breaking there is a considerable loss of energy
(associated with a large amount of turbulence in the surface rollers). Here ther is a a
strong decrease in the wave height as the waves approach the shoreline. A some
distance inshore of the point of wave breaking experiments show that the local wave
height is approximately equal to the local water depth h times a constant. For the sake
of simplicity we take this constant to be equal to K
H = Kh , h = D + η
(4.108)
In reality the coefficient K decreases from the value 0.8 just after the point of
wave breaking to approach 0.5 in zone of similarity. In the following we will use K =
0.8 to make the calculations as simple as possible. Because of the loss of energy the
Bernoulli equation cannot be used in the surf zone. In stead a force balance may be
used. The idea is that due to the decrease in the wave height towards the shoreline, the
wave reaction force will also decrease towards the shore. To have a balance of the
forces (averaged over time) for a water column there must therefore also be a force
directed offshore. This force can be obtained from a slope in the mean water surface.
First consider the forces on a column with height D (equal to the still water
depth) and of width dx due to the pressure in quiescent water, figure 4.15a. Here a
horizontal projections of the hydrostatic pressure forces acting on the column are in
equilibrium: the total horizontal component of the hydrostatic pressure is 0. If we give
a small perturbation η to this horizontal water surface, figure 4.15b, a resulting
horizontal force on the water column emerges. This force can be found a the resulting
force of the three hatched areas in figure 4.15b. The force can be found by assuming
hydrostatic pressure and calculate the areas in figure 4.15b, or it may be found from
the gradient theorem, which is derived from Gauss’s theorem, saying that the integral
of a quantity of the area of a closed surface is identical to the integral of its gradient
over the volume enclosed by the surface
Fig. 4.15.
Pressure acting on a water column with horizontal water
surface (a), and sloping water surface (b).
G
∫ φ dA = ∫ grad (φ) dX
A
(4.109)
X
If we apply (4.109) on the pressure (i.e. p = φ ) we get that the resulting pressure
force on a surface (in our case the water column) is identical to the the pressure
88
gradient over the volume enclosed by the surface. If we consider the horizontal force
component, the pressure gradient over the entire water column is given by
∂p
∂η
= ρg
∂x
∂x
(4.110)
given a total horizontal pressure force of
∂p
∂η
dX = −ρg h dx
∂x
∂x
X
P = −∫
(4.111)
Let us consider a water column, figure 4.16, in a situation where sinusoidal
waves propagate towards the shore over decreasing water depth.
On the water column there is a resulting reaction force from the wave with
varying wave height. Taken to be positive towards the right hand side, the reaction
force is
−
∂Fr
dx
∂x
Fig. 4.16
Forces acting on a water column. The magnitude of the set-up
shown is exaggerated.
Since the resulting reaction force is different from 0, a compensating force
must be introduced to obtain a time-averaged total force on the water column of 0.
The compensating force is obtained by having a pressure gradient from the sloping
mean water surface. This gives a resulting mean pressure force (4.111) and we will
have a force equilibrium if the slope of the mean water surface fulfils
ρgh
∂F
∂η
=− r
∂x
∂x
(4.112)
Inserting Fr from (4.98) and using the shallow water approximation
2kD
≈1
sinh(2kD)
we now get
89
ρgh
∂η
∂ ⎛31
∂ ⎛31
⎞
⎞
=− ⎜
ρg H 2 ⎟ = − ⎜
ρg (kh) 2 ⎟
∂x
∂x ⎝ 8 8
∂x ⎝ 8 8
⎠
⎠
(4.113)
inserting (4.108) to describe the variation in the wave height in the surf zone (known
as the wave set-up). Equation (4.113) can be reduced
∂η
3
1 ∂(h 2 )
3
∂h
= − K2
= − K2
∂x
16
h ∂x
8
∂x
(4.114)
from which
3
η = − K 2 h + const
8
(4.115)
The constant may be determined from the mean water level at the point of
wave breaking (4.107), giving
η=−
K 2 Db 3 2
+ K ( Db − h)
16
8
At the shoreline h = 0, and here we find
5
η(0) = K 2 Db ≈ 0.20 Db
16
(4.116)
(4.117)
So the maximum in the wave set-up is 20% of the water depth at the point of wave
breaking, a quantity which often has to be taken into consideration.
4.6 Wave induced currents
In the previous sections we have considered the forces acting on a section
normal to the direction of wave propagation, considering a plane situation with depth
contours normal to the direction of propagation. This corresponds to forces Fm and Fp
on section A-A in figure 4.17. If we in stead consider the forces on section B-B
which is normal to the wave fronts it is easy to realize that the momentum force now
is 0. Because all velocities are parallel to B-B there cannot be any flux of momentum
through this section. On the other hand pressure is acting in all directions, and the
pressure force Fp is the same on section B-B as on A-A. Therefore the reaction force
of the wave on a given section will depend on the angle α between the section and the
wave fronts. For α = 0 0 and 90 0
90
Fig. 4.17. Variation in the reaction force with the angle between the section
considered and the waves (waves seen from above)
we have only a normal force, but since these two normal forces are not identical
(being Fm+Fp and Fp respectively) ther will for other directions be normal forces Sxx
as well as shear forces Ssz, these forces are defined as stresses, integrated over the
water depth, having the units of force per length of the section considered. For an
arbitrary angle α (see section C-C in figure 4.17) the variation in Sxx and Ssz can be
found from a force balance for triangular body, cf. figure 4.18.
By expressing force equilibrium in the x-direction we find
S xx dz − S11 cos α dz cos α − S 22 sin α dz sin α = 0
or , using
S11 = ( Fm + F p )
and
S 22 = − F p
91
Fig. 4.18 Forces acting on a triangular body.
we get
S xx = −( Fm + F p ) cos 2 α − F p sin 2 α
or
S xx = − F p − Fm cos 2 α
(4.118)
Similarly a force balance for the triangular body in figure 4.18 in the z-direction gives
S xz dz − S11 cos α dz sin α + S 22 sin α dz cos α = 0
or
S xz = − Fm sin α cos α
(4.119)
We consider now wave approaching a coastline at an angle. The resulting
force on the longshore direction (the z-direction, cf. figure 4.19) on a water column
with a height equal to the water depth D and covering an area of dxdz is
F=
∂S xz
∂S
dx dz + zz dx dz
∂x
∂z
(4.120)
The coast is assumed to be infinitely long with constant wave conditions along
the coast. For such uniform conditions the term ∂S zz / ∂z become 0, giving
92
F=
∂S xz
dx dz
∂x
(4.121)
As long as there is no energy dissipation (i.e. when the potential flow theory is
valid) it can be shown that Sxz is constant, even when the waves are shoaling and
refracting as they propagate towards the coast. This can also be realised just by
considering that we cannot have shear stresses in a potential flow. Outside the surf
zone the energy dissipation is very small and can be disregarded. After the waves
have broken there is a considerable energy loss. Here Sxz changes rapidly
because the decreasing wave height (and wave energy flux) makes the magnitude of
Sxz also decease towards the shoreline. Since Sxz < 0 (cf. (4.119)) this implies that
∂S xz
>0
∂x
(4.122)
Therefore a resulting force is acting in the alongshore direction. The force in the
direction wave propagation, projected on the coastline, i.e. in the positive z-direction
in figure 4.19.
The force generates a longshore wave-driven current also know as the littoral
current. The strength of the current is determined by a force balance: the bed friction
τ b must of the same magnitude (and in the opposite direction) as the gradient in Sxz. If
we use a relation as
τ b = ρK1U c
(4.123)
where Uc is the longshore current velocity, figure 4.20. The force balance for the unit
water column, see figure 4.21, can be written
Fig. 4.19 The wave forces acting on a unit water column in the alongshore
direction.
93
∂S xz
dx dz = ρK1U c dx dz
∂x
(4.124)
In general this equation have to be solved numerically, using (4.119).
A simple assessment of the variation of Uc across the surf zone may be made
by using the assumptions that the angle between the waves and the coastline is small:
cos β ≈ 1
(4.125)
By application of shallow water theory Sxz can be written
S xz = −
ρg 2
ρg 2 sin β
H sin β = −
H c
8
8
c
(4.126)
=−
ρg 2
sin β
H gD
8
c
using (4.119) and (4.96). Due to the effect of refraction, described by Snell’s law
(4.48a) the quantity sin β / c is constant through the surf zone.
Fig. 4.20 Schematic illustration of the distribution of the wave-driven
longshore current.
If we use the approximation H =0.8D in the surf zone, where D is the local
water depth (4.126) gives, for a coast with constant beach slope:
∂S xz
= K2 D3/ 2
∂x
(4.127)
where K2 is a positive constant.
From the force balance (4.124) we now find
94
U c = K3D3/ 2
(4.128)
where K3 is a positive constant. It is seen that the wave driven current is strongest just
after the point of wave breaking, where the water depth is largest. The velocity Uc
decreases towards the coast as indicated in figure 4.20. In field conditions Uc can
easily be 1 or 2 m/s, and this longshore current is obviously very important for the
sediment transport along the coast, also know as the littoral drift.
It may be noted that the wave induced stresses Sxx and Sxz here have been
defined as positive when expressing a tension (cf. 4.118 and figure 4.18), this is
similar to the conventions for stresses in hydrodynamics or structural engineering. In
many books on coastal engineering these stresses, known as the radiation stresses
associated with the wave motion, are defined with an opposite sign, being positive
when expressing a normal force equivalent to a positive pressure.
Fig. 4.21 Forces acting on a unit water column in the alongshore direction.
95
5. Laminar Boundary Layers
Let us consider a thin plate moving in its own plane in a quiescent fluid with a
constant velocity uo. Due to the adhesion of the fluid to the plate some of the
surrounding fluid will be carried along with the plate, see figure 5.1. The flow is only
occurring in a thin layer along the boundary to the plate. This thin layer is known as a
boundary layer. If we in stead consider the completely analogous case where the fluid
flows along a fixed plate, figure 5.2, we see a reduction in the flow velocity near the
plate due to the adhesion. The zone where the velocity decreases from the free outer
velocity uo down to 0 at the wall is also known as a boundary layer.
Fig. 5.1
The boundary layer formed along a plate moving through a
quiescent fluid.
A boundary layer can in this way be described as a domain where the flow
velocity has large variations when you move across the main flow direction. Inside
this layer the viscous forces must be included in the flow description while the shear
stresses can be neglected in the flow outside. In the outer domain the flow may
therefore be described by potential flow theory. This is of course a simplification of
the real flow conditions, because there must be a gradual transition from the domain
where the viscosity is of importance to the domain where it may be totally neglected.
102
Fig. 5.2
The boundary layer along a fixed thin plate.
5.1 The boundary layer equations
The assumption that the thickness of the boundary layer is very small implies
that some of the terms in the Navier-Stokes equations are small and may be neglected.
If we disregard the gravity, the Navier-Stokes equation in the flow direction reads (cf.
figure 5.2)
⎛ ∂ 2u ∂ 2u ⎞
1 ∂p
∂u
∂u
∂u
+u
+v
=−
+ ν⎜⎜ 2 + 2 ⎟⎟
∂t
∂x
∂y
ρ ∂x
∂y ⎠
⎝ ∂x
(5.1)
Colse to the wall the v is skmall, because the stream lines in the boundary layer must
be approximately parallel to the vall. In contrast the term ∂u / ∂y must be large
because the velocity grows from 0 to the outer velocity over a very short distance,
namely the thin boundary layer. The only term which a priori may be neglected is
therefore ν ∂ 2 u / ∂x 2 , which must be very small compared to ν ∂ 2 u / ∂y 2 . In this way
(5.1) is reduced to
∂u
∂u
∂u
1 ∂p
∂ 2u
(5.2)
+u
+v
=−
+ν 2
∂t
∂x
∂y
ρ ∂x
∂y
If we consider the Navier-Stokes equation in the direction normal to the wall,
⎛ ∂ 2v ∂ 2v ⎞
∂v
∂v
∂v
1 ∂p
+u +v
=−
+ ν⎜⎜ 2 + 2 ⎟⎟
∂t
∂x
∂y
ρ ∂y
∂y ⎠
⎝ ∂x
(5.3)
we can from the same argumentation conclude that all terms except ∂p / ∂y are small,
so that (5.3) can be approximated by
∂p
=0
∂y
(5.4)
103
This a result of great importance because it shows that the pressure inside the
boundary layer is determined by the pressure outside it (i.e. in the potential flow).
This means that the pressure gradient in the flow direction in (5.2) is determined from
the outer potential flow.
Equations (5.2) and (5.4) are known as the boundary layer equations for
laminar boundary layers.
5.2 The general momentum equation
The momentum equation is one of the most frequently used equations in
hydraulics (cf. [2]). It is derived by expressing Newtons 2. law for a finite volume of
fluid X, which is enclosed by the control surface A.
Since Newtons 2. law expresses that mass times acceleration inside the control
surface is identical to the resulting force acting on X , we find
dvi
∫ ρ dt
dX = − ∫ p dAi + ∫ ρg i dX + si
X
A
(5.5)
X
Here the index i can have the values i = 1,2 and 3, so that v1 is the velocity in the
direction of the x1-axis (similar to u and x), v2 is the velocity in the x2 direction
(similar to v and y) etc. Equation (5.5) is in this way actually three equations,
corresponding to a force balance in each of the three directions of the axes in the
coordinate system.
In (5.5) the resulting force on the right hand side is composed of 3
contributions: the pressure force (which acts as positive into the volume - that is in the
direction opposite the area vector dAi), the force of gravity (which is a volume force,
other volume forces such as the Coriolis force could have been included as well) and
finaly si, which is the vector sum of the shear stresses acting on the control surface
plus the stresses in any solid bodies, which are cut by the control surface.
Equation (5.5) can be modified, using that the left hand side can be written
dvi
∫ ρ dt
X
⎛ ∂v
⎞
∂v
dX = ∫ ρ⎜⎜ i + i v k ⎟⎟ dX
∂t ∂x k ⎠
X ⎝
(5.6)
cf. (1.16) and (1.17). It should be noted that when the same index is used twice in the
same term (as does index k in (5.6)), a summation is to be made over this index.
The last term in (5.6) can be modified by application of Gauss’s gradient
theorem (cf. (4.109)), which applied on the term vivk gives
∂
∫ vi vk dAk = ∫ ∂xk (vi vk ) dX
A
X
⎡ ∂v
⎤
∂v
= ∫ ⎢ i v k + k vi ⎥ dX
∂x k
∂x k ⎦
X⎣
(5.7)
Here it follows from the continuity equation that the last term in the integral equal to
zero.
104
Equation (5.5) can thus be written
∂vi
∫ ρ ∂t
dX + ∫ ρvi v k dAk = − ∫ p dAi + ∫ g i dX + si
X
A
A
(5.8)
X
Equation (5.8) can be written in vector form
G
G
G G G
G
G
∂v
ρ
dX
+
ρ
v
(
v
⋅
d
A
)
=
−
p
d
A
∫ ∂t
∫
∫ + ∫ ρg dX + s
X
A
A
X
(5.9)
5.3 The momentum equation for boundary layers
We will now apply the general momentum equation (5.9) specifically on
boundary layers. This is because the momentum equation has proven to be a very
efficient tool for calculating
the wall shear stress in boundary layers and the thickness of boundary layers. In the
following the term with the acceleration of gravity will be neglected, so p represents
the pressure in excess of the hydrostatic pressure.
As an approximation the boundary layer is taken to have a well-defined finite
thickness δ and the control surface is taken as the closed surface A-A’-B’-B-A,
shown in figure 5.3. The two sides A-A’ and B-B’ are normal to the wall, A-B
follows the wall, and A’-B’ follows the outer boundary of the boundary layer:
y = δ(x ) .
As mentioned (5.9) is a vector equation and for this application we use the
projection along the x-axis. The contribution from each term in the momentum
equation is found individually.
Fig. 5.3
The control surface for the momentum equation, and the
definition of the displacement thickness.
1. The first term on the left hand side of (5.9) represents the change in
momentum due to the flow being non-steady. The contribution in the direction
of the x-axis is
δ
dx ∫ ρ
0
∂u
dy
∂t
(5.10)
105
2.
The second term on the left hand side of (5.9) is known as the convective
term, which is found in non-uniform flows where the fluid particles change
G G
their velocity as thay move along the stream lines. The term ρv ⋅ dA is the
mass flux through a surface element.
This can be found as follows: first the mass flux through the surface A-A’ is given as
δ
M 1 = − ∫ ρu dy
(5.11)
0
G
the minus sign is because dA is directed out of the control surface and
therefore against the positive x-direction. Through B-B’ we have
δ
M 2 = ∫ ρu dy +
0
δ
∂
ρu dy dx
∂x ∫0
(5.12)
From this the mass flux down through A’-B’ can be found from the
requirement that the total flux mass flux must be zero.
δ
∂
M 3 = −( M 2 + M 1 ) = − ∫ ρu dy
∂x 0
(5.13)
The contributions to the term
∫A
G G G
ρv (v ⋅ dA)
can now be found for the four sides of the control surface.
Along A-B we have no flux and U = 0 so the contribution is zero.
Along A’-B’ at the top of the boundary layer the velocity is equal to the outer
velocity: u = uo. When making the projection on the x-axis the momentum
flux is found to be
δ
G G
∂
⋅
=
−
u
(
v
d
A
)
u
ρu dy dx
0
∫ 0
∂x ∫0
(5.14)
From A-A’ the contribution is
δ
∂
2
∫ ρu (−u ) dy = −∫ ρu dy
0
(5.15)
0
And finally the contribution from B-B’ we get
106
δ
2
∫ ρu dy +
0
∂
∂
ρu 2 dy dx
∫
∂x 0
(5.16)
giving a total momentum flux of
∂
δ
∂
∂
ρu 2 dy dx − u 0
ρu dy dx
∫
∂x 0
∂x ∫0
3.
(5.17)
The pressure term in (5.9) can by use of the gradient theorem be written
G
− ∫ p dA = − ∫ grad p dX
A
X
As the pressure does not vary with y in the boundary layer (cf. (5.4)) the f
contribution in the x-direction is found to be
−
4.
∂p
δ dx
∂x
(5.18)
G
The term s represents the shear stresses along the control surface and stresses
in any solid bodies piercing the control surface. In the present case there are
no shear stress at the top of the boundary layer, forming the boundary to the
outer frictionless flow. Along the wall the shear stress contribution is found to
be
− τ 0 dx
(5.19)
From the contributions (5.10), (5.17), (5.18) and (5.19) the momentum
equation (5.9) is found to give
δ
∂
δ
∂u
∂
∂
∂p
2
∫ ∂t dy + ∂x ∫ ρu dy − u 0 ∂x ∫ ρu dy = −δ ∂x − τ 0
0
0
0
(5.20)
which is known as the momentum equation for boundary layers.
_____________________________________________________________________
Example 5.1 Flow along a plate with a sudden movement.
Let us consider a plate which is suddenly accelerated from the velocity 0 to the
velocity uo. The plate moves in its own plane and is taken to be of infinite extension.
107
Fig. 5.4
The development of the velocity profile over a plate with a
sudden movement.
This plate will entrain more and more and fluid as time passes, as shown in figure 5.4.
This flow will be solved in two basically different ways: by an exact solution and by a
boundary layer approximation.
A. The exact solution: This is done by use of the principles given in chapter 2
by formulating the flow equation with the appropriate boundary conditions. The flow
equation in the flow direction reads (cf. (2.30))
∂u
∂ 2u
=ν 2
∂t
∂x
(5.21)
with the boundary conditions
u = u0
u=0
for
for
y=0
t≤0
and
t>0
(5.22)
(5.23)
and
u→0
for
y→∞
(5.24)
(5.21) is the so-called equation for heat conduction. It can be reformulated from being
a partial differential equation to an ordinary differential equation by introduction of
dimensionless parameters.
The velocity u is made dimensionless by division by uo
U=
u
u0
(5.25)
Another dimensionless parameter can be found as a combination of y, t and ν
108
Y=
y
(5.26)
νt
( ν has the dimension m2/s). Insertion of (5.25) and (5.26) into (5.21) gives
d 2U 1 dU
+ Y
=0
dY 2 2 dY
(5.27)
where the following derivations have been applied
∂U dU ∂Y dU ⎡ 1 Y ⎤
=
=
−
dY ∂t
dY ⎢⎣ 2 t ⎥⎦
∂t
∂U dU ∂Y dU Y
=
=
∂y
dY ∂y dY y
∂ 2U
Y dU 1 d ⎛ dU ⎞ ∂Y Y 2 d 2U
=
−
+
=
⎜Y
⎟
∂y 2
y 2 dy y dY ⎝ dY ⎠ ∂y y 2 dY 2
Equation (5.27) can be integrated once to give
⎡ Y2⎤
dU
= c exp ⎢−
⎥
dY
⎣ 4 ⎦
(5.28)
which, by a second integration yields
Y
⎛ s2 ⎞
u
= c ∫ exp⎜⎜ − ⎟⎟ ds + c1
u0
⎝ 4⎠
0
(5.29)
In equation (5.29) we recognise the error function
F (Y ) =
2
π
Y
−x
∫ e dx
2
(5.30)
0
which as shown in figure 5.5 asymptotically approaches ± 1
By insertion of (5.30) into (5.29) we obtain
for
Y → ±∞ .
109
Fig. 5.5
The error function F(Y):
u
π ⎛Y ⎞
π ⎛ y ⎞
F ⎜ ⎟ + c1 = 2c
F⎜
⎟ + c1
= 2c
u0
2 ⎝ 2⎠
2 ⎜⎝ 2 νt ⎟⎠
(5.31)
where the constants c and c1 is determined from the boundary conditions.
Equation (5.22) gives
c1 = 1
and (5.24) gives
0 = c π +1
from which (5.31) is found to give
⎛ y ⎞
u
= 1 − F ⎜⎜
⎟⎟
u0
⎝ 2 νt ⎠
(5.32)
We see from figure 5.5 that this expression describes a velocity profile of the
shape indicated in figure 5.4.
The error function F(Y) approaches 1 asymptotically for Y → ∞ and the
movement of the wall can therefore in principle be felt infinitely far from it at any
positive time. The main part of the flow is, however, restricted to a zone near the wall
(the boundary layer!). The definition of this boundary layer can be made somewhat
arbitrarily. As an example the thickness may be defined as the distance where the
velocity is reduced to one percent of the velocity of the wall, or
u
= 0.01
u0
⇒
F (Y ) = 0.99
(5.33)
From tables we find that (5.33) means Y = 1.82. (If the choice had been one
per mille we had found Y = 2.31).
110
The boundary layer thickness is by this definition found to be
δ = 1.82 × 2 νt = 3.64 νt
(5.34)
From which we see that the boundary layer thickness starts from zero at t = 0 when
the plate is set into motion, then δ grows as the square root of the time.
The shear stress at the wall τ 0 can be found as
⎛ dU ∂Y ⎞
⎛ ∂u ⎞
ν u0c
τ0
1 u0 ν
⎟⎟
=
= ν u 0 ⎜⎜
=−
= ν⎜⎜ ⎟⎟
ρ
νt
π νt
⎝ dY ∂y ⎠ y ,Y =0
⎝ ∂y ⎠ y =0
(5.35)
It is seen that τ 0 is infinitely large at t = 0 (where the plate is given an infinite
acceleration) and thereafter decreases with time while δ increases.
B. Approximate solution by use of the momentum equation: The advantage of
using the momentum equation for boundary layer calculations is that the calculations
are made considerably simpler. This is obtained at the expense of knowledge of the
details in the velocity profile. The momentum equation (5.20) is integrated over the
boundary layer, and the exact shape of the velocity profile is therefore not crucial for
the assessment of the variation in the boundary layer thickness and the bed shear
stress.
In this example we can, for example, make the guess that the velocity profile
is given by a polynomium, which fulfils the two boundary conditions
1:
u = u0
2:
u=0
for
y=0
(5.36)
and
for
y=δ
(5.37)
These conditions are met by the first order polynomium
f1 (η) =
u
y
= 1− = 1− η
u0
δ
(5.38)
where the dimensionless coordinate η = y / δ has been introduced. This profile is of
course a very crude approximation to the actual velocity distribution. It has not been
used that
the boundary layer flow has a smooth transition to the outer flow (where the velocity
in this example is 0), so that the gradient in the velocity is zero at the outer boundary
3:
∂u
=0
∂y
for
y=δ
(5.39)
This boundary condition can in addition to the two first be satisfied by the
polynomium of second order
111
f 2 (η) = 1 − 2η + η 2
(5.40)
but also, for example, by the function
⎛π ⎞
f 3 (η) = 1 − sin ⎜ η ⎟
⎝2 ⎠
(5.41)
In the following all three assumed profiles (5.38), (5.40) and (5.41) are used
with the momentum equation.
Because of the uniformity in the x-direction the momentum equation (5.20)
can be reduced to
δ
∂u
∫ ρ ∂t dy = −τ 0
(5.42)
0
where the wall shear stress τ 0 in this case will be negative. As uo is constant the timederivative of u is found as
df ⎛ y ⎞ dδ
df ∂η
∂f
∂u
= u0
= u0
= u0
⎜− ⎟
dη ⎝ δ 2 ⎠ dt
dη ∂t
∂t
∂t
(5.43)
In a similar way τ 0 is found as
τ0 = ρ ν
∂u
df ∂η
= ρ ν u0
, η=0
∂y
dη ∂y
(5.44)
or
τ0 = ρ ν u0
df 1
dη δ
, η=0
(5.45)
Insertion of (5.43) and (5.45) into (5.42) gives
1
dδ df
f ′(0)
η dη = ρνu 0
ρu 0
∫
dt 0 dη
δ
(5.46)
or
d
dt
( δ )= ν
1
2
f ′(0)
2
1
(5.47)
df
∫ dη η dη
0
which by integration gives
112
δ 2 = 2ν t
f ′(0)
(5.48)
1
∫ η f ′ dη
0
By inserting the three estimated velocity profiles f1, f2 and f3 into (5.48) the
sensitivity of the momentum equation for different choices of the profile can be
assessed.
All the profiles give the correct variation of δ and τ 0 with respect to time and
viscosity, but as seen in table 5.1 the constants involved differ. In the column next to
δ the variation in so-called displacement thickness δ * is given. This is introduced to
have a more well-defined measure of the boundary layer thickness, applicable for the
exact solution as well as the approximation obtained from the momentum equation. In
connection with the exact solution we found that the thickness of the boundary layer
depended on the choice of definition, based on u = 0.01uo or u = 0.001uo. This way of
defining the boundary layer thickness is also very difficult to apply in connection with
analysis of measurements. To overcome this ambiguity δ * is defined as the distance
from the wall, corresponding to the 'displacement' of the outer flow due to the deficit
in the velocity inside the boundary layer, giving
∞
u 0 δ = ∫ u dy
*
(5.49)
0
cf. figure 5.6.
δ/ νt
δ* / ν t
f1 (η) = 1 − η
2.00
1.000
0.500
f 2 (η) = 1 − 2η + η 2
3.46
1.155
0.578
f 3 (η) = 1 − sin(ηπ / 2)
2.94
1.068
0.534
exact (u δ / u 0 = 0.01)
3.64
1.129
0.564
exact (u δ / u 0 = 0.001)
4.62
1.129
0.564
Velocity distribution
τ 0 ν t / ρν u 0
Table 5.1
It is seen from table 5.1 that the variation in δ * is quite moderate when
changing from one velocity profile to the other.
113
Definition of the displacement thickness δ * : the two hatched
areas shall be of the same size.
Fig. 5.6
Example 5.2 Stationary, non-uniform flow along a plate.
In the previous example the flow was identical for all values of x because the
plate was taken to be of infinite extension. In the following we will instead consider
the boundary layer formed along a plate when it is met by a flow of constant velocity
uo parallel to the plane of the plate, cf. figure 5.2.
Since uo is constant the pressure p outside the boundary layer is also constant
giving ∂p / ∂x = 0 outside the boundary layer. According to the boundary layer
equation (5.4) it follows immediately
that the pressure gradient in the boundary layer also is zero and the momentum
equation (5.20) can be reduced to
∂
δ
∂
∂
ρu 2 dy − u 0
ρu dy = −τ 0
∫
∂x 0
∂x ∫0
(5.50)
We try as in the previous example to make a guess for the velocity profile. We
assumed a profile described by
u
⎛ y⎞
= f⎜ ⎟
u0
⎝δ⎠
(5.51)
in which the time-variation is included only through the variation of δ with t. The
profiles are in this way said to be similar or affine. As we shall see when considering
flow separation this assumption is not valid in general. For the present situation we
can, however, use the assumption (5.51) because the outer pressure gradient is 0.
The procedure is exactly as in the previous example. Possible velocity profiles
to use can be
⎛π ⎞
f1 (η) = sin ⎜ η ⎟
⎝2 ⎠
(5.52)
similar to (5.41), or
114
f 2 (η) = 3η − 3η 2 + η3
(5.53)
Equations (5.52) and (5.53) both satisfy the boundary conditions
u=0
for
η=0
u = u0
for
η =1
(5.54)
and
du
=0
dx
for
η =1
Equation (5.53) is an extension of (5.40) from a polynomium of sendond to third
order. In this way the additional boundary condition
d 2u
(5.55)
= 0 for η = 1
dy 2
is also satisfied.
Only the calculations for the first approximation, the sinus-shaped profile f1,
will be carried out here. The two first terms in (5.50) gives
δ
π/2
⎤
∂ ⎡ 2 2δ
∂
u
(
u
u
)
dy
u
sin ξ (1 − sin ξ) dξ⎥
ρ
−
=
−
ρ
⎢ 0
0
∫
∫
∂x ⎢⎣
π 0
∂x 0
⎥⎦
=−
∂ ⎡ 2 2δ ⎛ π ⎞⎤
⎜1 − ⎟ ⎥
⎢ρu 0
∂x ⎣
π ⎝ 4 ⎠⎦
= −ρu 02
(5.56)
2 ⎛ π ⎞ dδ
⎜1 − ⎟
π ⎝ 4 ⎠ dx
while τ 0 is given as
⎛ ∂u ⎞
πu
= ρν 0
τ 0 = ρ ν⎜⎜ ⎟⎟
2δ
⎝ ∂y ⎠ y =0
(5.57)
the momentum equation (5.50) is thus reduced to
ρu 0
πu
2 ⎛ π ⎞ dδ
= ρν 0
⎜1 − ⎟
π ⎝ 4 ⎠ dx
2δ
(5.58)
which has the solution
115
δ=
π
2 − π/ 2
νx
νx
= 4.8
u0
u0
(5.59)
taking x = 0 at the leading edge of the plate, cf. figure 5.2, where the boundary layer
thickness starts from 0. As done in the previous example the displacement thickness
can be determined. The definition is slightly different when the wall is fixed, compare
figures 5.6 and 5.7
∞
u 0 δ * = ∫ (u 0 − u ) dy
(5.60)
0
Fro the sinus profile we get
νx
π−2
δ* =
δ = 1.74
π
u0
(5.61)
Table 5.2 gives, similarly to table 5.1, the variation in the displacement thickness and
wall shear stress for the two assumed profiles and for the exact solution, which for
this case can only be obtained numerically.
Velocity distribution
δ* / ν x / u 0
τ 0 u 0 x / ν / ρu 02
f1 (η) = sin(ηπ / 2)
1.741
0.328
f 2 (η) = 3η − 3η 2 + η3
1.740
0.323
1.721
0.332
exact
Table 5.2
Fig. 5.7
Definition of the displacement thickness: the two hatched areas
must be of equal size.
116
5.4 Boundary layers with outer pressure gradients
As mentioned in example 5.2 the assumption of similarity is not valid in
general. The calculational work increases dramatically when the pressure gradient is
different from zero, and the problem will only be treated cursory in the following. Let
us consider a plane flow condition as sketched in figure 5.8.
The water flows from left to right, first through a channel with converging
walls. Here the flow velocity increases in the flow direction, which means that the
pressure decreases in the flow direction, cf. the Bernoulli equation. At the right side
the water flows through diverging walls so that the pressure increases in the flow
direction. In other words: in the converging part we have ∂p / ∂x < 0 and in the
diverging part ∂p / ∂x > 0 .
It is only in the converging section that the flow has the character of a
boundary layer flow. The flow in the diverging section will tend to separate with a
return flow close to the wall. The velocity profiles are thus completely different in the
two sections and may not be approximated by similar profiles.
A more general way of describing the velocities in stationary flows with
pressure gradients is the Pohlhausen method. The main ideas of this method are
outlines in the following.
It is assumed that the velocity profiles can be described by a polynomium such
as
u
= a η + b η 2 + c η3 + d η 4
u0
,
η=
y
δ
(5.62)
which automatically fulfils the condition u = 0 for η = 0.
To determine the coefficients in (5.62) we apply as before the following
boundary conditions
u = u0
for
η =1
∂u
=0
∂y
for
η =1
∂ 2u
=0
∂y 2
Fig. 5.8
for
(5.63)
η =1
The velocity profiles in converging and diverging flow.
117
To introduce the pressure gradient in the description we now revert to the boundary
layer equation (5.2), which at the wall degenerates to
−
1 ∂p
∂ 2u
+ν 2 =0
ρ ∂x
∂x
(5.64)
using that u and v are exactly 0 at the wall. Using (5.64) the last boundary condition
may be formulated as
∂ 2 (u / u 0 )
∂η 2
=
δ 2 ∂p
= −Λ ,
ρν u 0 ∂x
η=0
(5.65)
where Λ is a measure of the pressure gradient, which we know is related to the outer
flow velocity uo by
∂u
1 ∂p
= −u 0 0
ρ ∂x
∂x
(5.66)
Using the boundary conditions (5.63) and (5.65) the four coefficients in (5.62)
are determined, giving
u
= F (η) + Λ G (η)
u0
(5.67)
where
F = 2η − 2η3 + η 4
and
(5.68)
1
G = (η − 3η 2 + 3η3 − η 4 )
6
It is thus seen that the velocity profiles in plane stationary boundary layers can
be described as a one-parameter family with Λ as the shape-parameter.
Examples of possible shapes of profiles are shown in figure 5.9. In general the
velocity profile will vary along the wall corresponding to the variation in Λ . The
parameter Λ is constant only for affine boundary layers, as for example in the flow
along a plane plate parallel to the flow direction (example 5.2) where Λ = 0,
By use of (5.67) and the momentum equation for the boundary layer any plane
stationary boundary layer flow may in principle be calculated.
Figure 5.9 illustrates the very important difference between flow with a
positive (opposing) and a negative (favourable) pressure gradient:
118
Fig. 5.9
Velocity profiles in boundary layers with positive and negative
pressure gradients.
In a boundary layer flow with a flow in the direction of a pressure drop Λ will
be positive and the velocity profiles will show very little variation from one crosssection to the other.
Conversely, if the flow is in the direction of a pressure increase Λ will be
negative, which drastically changes the shape of the velocity profile. For Λ = -12 the
velocity gradient is zero at the wall corresponding to the wall shear stress being zero.
For even larger negative values of Λ , there is a return flow near the wall, a situation
described as the boundary layer separating from the wall. The point where Λ = -12 is
known as the separation point. In a separating flow the assumptions behind the
boundary layer theory normally breaks down: the thickness of the boundary layer
increases drastically and the assumption that velocity component normal to the wall is
always small compared to the longitudinal velocity is no longer valid.
119
Example 5.3 Flow around a circular cylinder
Fig. 5.10
Flow around a circular cylinder. Top: potential flow. Bottom: laminar
flow, photo: Re (= 2u 0 R / ν) = 26 (from [4]).
The potential flow around a circular cylinder may be calculated based on the
principles given in chapter 3. The solution for the potential becomes
⎡
R2 ⎤
φ = −u 0 x ⎢1 + 2
2⎥
⎣ x +y ⎦
(5.69)
The stream lines are shown in figure 5.10 (top). This flow pattern is, however,
only a good approximation to the real flow, figure 5.10 (bottom), at the upstream part
of the cylinder – and then outside the boundary layer. At the downstream part of the
cylinder the flow is diverging, and separation will take place at the two points S,
figure 5.10 (bottom).
The flow separation is crucial for the force from the current – the drag force –
acting on a cylinder, which is fixed in flowing water.
If there are no separation, there is a thin boundary layer along the entire
perimeter of the cylinder, and the only force acting on the cylinder is due to the wall
friction. The pressure in the boundary layer flow is symmetrical around the y-axis as
shown in figure 5.11, because the pressure outside the boundary layer can be
determined from Bernoulli's equation. The flow separation will change the pressure
distribution drastically. The velocities in the separation zone are much smaller than
120
outside it. The separation zone thus forms a wake with a rather small variation in the
pressure. The transition between the outer flow and the wake is very sharp as
Fig. 5.11
Sketch of the pressure distribution around a circular cylinder.
indicated in the photo of figure 5.12. As an approximation the pressure in the wake
may be taken to be equal to the pressure at the separation points. The actual pressure
distribution is therefore as sketched with the dashed line in figure 5.11, which is seen
to result in pressure force acting on the cylinder in the direction of the flow. Figure
5.13 shows an example of the measured pressure distribution around the perimeter of
a cylinder submerged in flowing water. The measurement indicates the so-called
pressure coefficient, which is defined as
( p − p0 )
Cp =
(5.70)
2
1
U
ρ
0
2
where po is the undisturbed pressure far upstream. In this case the measurements are
made with a high value of the Reynolds number (The Reynolds number is defined in
chapter 6, cf. (6.19)), and the boundary layer at the upstream side is turbulent (cf.
Fig. 5.12
Photo of laminar and turbulent separation around a cylinder,
from [4].
121
chapter 6 and 7), but qualitatively we find the same pressure distribution at high and
low Re numbers, with the exception of very small Re numbers (<5), where separation
does not occur. This is because the velocities in that case are so small that the square
of the velocity (non-linear terms) can be neglected when compared to the shear
stresses. This means that the inertia force (the convective term u ∂u / ∂x ) is neglected.
The flow in this regime is known as creeping, cf. figures 5.14 and 5.15.
When the Reynolds number is larger than approximately 40 the wake
downstream of the cylinder becomes unstable. Small disturbances will be amplified
and one of the two downstream circulation vortices will be shed from the cylinder and
carried with the flow. While a new vortex is being formed, the remaining vortex at the
other side of the cylinder will be shed, and the process continues. This is known as
rhythmically vortex shedding, cf. figure 5.14 and the photography in figure 5.16.
Fig. 5.13
Measured pressure distribution around a circular cylinder, Re =
3.6 106.
The vortex shedding occurs with a frequency f determined by the
relation
fd
=S
U0
where d is the diameter, Uo the undisturbed velocity and S is a dimensionless number
known as the Strouhals number, which for a wide range of Re has the value 0.2.
122
Fig. 5.14
The flow pattern around a cylinder at different values of
Reynolds number.
Fig. 5.15
Photo of creeping flow. The undisturbed velocity is 1 mm/s.
The flow takes place between two parallel glass plates with a
mutual distance of 1 mm. (Hele-Shaw apparatus).
123
Fig. 5.16
Rhythmic vortex shedding forming a vortex street.
Fig. 5.17
The variation in Strouhals number with Reynolds number,
based on physical experiments.
124
125
6. Turbulent flows
If we consider the flow through a pipe, two different flow situations can occur
for the same fluid depending on the flow velocity. For low flow velocities the flow is
laminar (cf. chapter 2), but when the velocity exceeds a certain critical value,
Fig. 6.1
Reynolds' experiment, showing the transition from laminar to
turbulent flow in a pipe (from [4]).
The flow changes to become turbulent. This is illustrated in figure 6 by O. Reynolds'
famous experiment from the 1860-ties, where the transition from the regular
organized laminar flow to the more irregular and stochastic turbulent flow is clearly
visible. If the pipe diameter or the viscosity of the fluid is changed it will be found
that the critical flow velocity is also changed. It was shown that the point of transition
from laminar to turbulent flow can be characterized by the dimensionless number
129
Re =
UD
ν
(6.1)
where U is a characteristic velocity (for example the mean velocity in the pipe), D a
typical dimension (for example the diameter of the pipe) and ν is the kinematic
viscosity of the fluid. For flow in a pipe the critical Reynolds number is approx. 2000.
If we consider a pipe with a diameter of 0.1 m we find for water ( ν = 10-6 m2/s) the
flow will be turbulent for flow velocities larger than about 0.02 m/s. In the field of
hydraulic engineering the flows to be dealt with are therefore in most cases turbulent.
An important example of laminar pipe flow is found in the human blood
vessels where the Reynolds number is moderate due to the small dimensions (the flow
in aorta can, however, be turbulent).
Fig. 6.2
A turbulent spot (from[4]).
The transition from laminar to turbulent flow is gradual. If the velocity is is
only slightly above the critical the turbulence may emerge locally as shown in figure
6.2, which is a picture of a 'turbulent spot'.
The Reynolds number has to be considerably larger (by a factor 10-50) for the
flow to evolve to become fully turbulent.
6.1 The stability of a laminar flow
The emergence of turbulence can be explained as the result of an instability in
the laminar flow: basically any flow could be laminar since the laminar flow fulfils
the flow equations and the proper boundary conditions. It can, however, be shown that
under some circumstances (for large values of the Reynolds number) any small
perturbation to the laminar flow will grow in time so that the nice regular laminar
flow breaks up to become turbulent. This section treats the mathematical formulation
of such a stability analysis.
Normally the starting point of a hydrodynamic stability analysis is the socalled vorticity transport equation, which can be derived from the Navier-Stokes
equation (2.30) and (2.31), cf. also (A8), appendix I:
130
⎛ ∂ 2u ∂ 2u ⎞
du
1 ∂p
=−
+ g x + ν⎜ 2 + 2 ⎟
⎜ ∂x
ρ ∂x
dt
∂y ⎟⎠
⎝
and
(6.2)
⎛∂ v ∂ v⎞
dv
1 ∂p
=−
+ g y + ν⎜ 2 + 2 ⎟
⎜ ∂x
ρ ∂y
dt
∂y ⎟⎠
⎝
2
2
By making a differentiation of the first equation with respect to y and similarly a
differentiation of the second with respect to x the pressure p can be eliminated to give
⎛ ∂u ⎞
∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u
∂u ⎞
⎜ ⎟ + ⎜u
+ v ⎟⎟ − νΔ⎜⎜ ⎟⎟ =
⎜
⎟
⎜
∂t ⎝ ∂y ⎠ ∂y ⎝ ∂x
∂y ⎠
⎝ ∂y ⎠
(6.3)
∂v ⎞
∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v
⎛ ∂v ⎞
+ v ⎟⎟ − νΔ⎜ ⎟
⎜ ⎟ + ⎜⎜ u
∂y ⎠
∂t ⎝ ∂x ⎠ ∂x ⎝ ∂x
⎝ ∂x ⎠
By application of the equation of continuity
∂u ∂v
=0
+
∂x ∂y
equation (6.3) can be modified to
∂ω
∂ω
∂ω
= νΔω
+v
+u
∂y
∂x
∂t
(6.4)
where
2ω =
∂v ∂u
−
∂x ∂y
(6.5)
cf. equation (2.21). Equation (6.4) can be written
dω
= νΔω
dt
(6.6)
which is known as the vorticity transport equation.
In a general three-dimensional flow we find similarly
131
dω1
∂u
= νΔω1 + ω k
dt
∂x k
dω 2
∂v
= νΔω 2 + ω k
dt
∂x k
(6.7)
dω3
∂w
= νΔω3 + ω k
dt
∂x k
which is quite similar to (6.6) except for the last terms, expressing the effect of
stretching of vortices, which can not occur in a purely two-dimensional flow.
We now consider a laminar flow, for example the laminar boundary layer
forming along a plane plate as described in section 5.2.
The streamlines in this laminar flow are parallel to the plate. These streamlines
are given a small perturbation, see figure 6.3, so that the original velocity field
u = U ( y)
v=0
(6.8)
u = U ( y ) + u~
v = v~
(6.9)
is changed to
where u~ and v~ are the small velocity perturbations introduced to the flow.
Fig. 6.3
A perturbation of a boundary layer flow.
The basic flow has a vorticity Ω , determined from (6.8) to be
Ω=−
1 ∂U
2 ∂y
(6.10)
while the vorticity of the perturbation is called ω . For the combined flow the vorticity
transport equation now reads, cf. (6.4)
132
∂
∂ω
∂
+ (U + u~ ) (Ω + ω) + v~ (Ω + ω) = ν(ΔΩ + Δω)
∂t
∂x
∂y
(6.11)
For the unperturbed basic flow the vorticity transport equation is all valid and
it reads
U
∂Ω
= νΔΩ
∂x
(6.12)
By subtracting (6.12) from (6.11) and using that the basic flow is a boundary layer
where
∂Ω
∂Ω
<<
∂x
∂y
we finally get
dω ∂Ω ~
+
v = νΔω
dt ∂y
(6.13)
By multiplying (6.13) by ω and using (6.10) equation (6.13) can be reformulated as
d ⎛1 2⎞
⎜ ω ⎟=
dt ⎝ 2 ⎠
I
1
U ′′ωv~ + νωΔω
2
II
III
(6.14)
The physical significance of the three terms is as follows
I:
Represents the time variation of the rotational energy of a fluid particle, given
as 1/2 ω 2 (equivalent to the kinetic energy).
II:
Is the production of rotational energy. This can be seen from figure 6.4:
We consider a fluid particle at the distance a from the bed, which at the
time t = 0 has the rotation Ω + ω . After the time interval dt the particle is now
at the level (a+vdt) above the bed, but the vorticity will be practically the same
because the effect of the viscosity almost negligible for so short a time.
This means that because of the introduced velocity perturbation v the
fluid particle will now have a perturbed vorticity of
2
dΩ ~ ⎞
1⎛
1
dΩ ~
⎜⎜ ω −
v dt ⎟⎟ = ω 2 −
ωv dt + ....
dy
2⎝
2
dy
⎠
(6.15)
From this it is seen that the term
−
dΩ
ω ν = U ′′ω ν
dy
(6.16)
133
which is representing II, signifies the increase in rotational energy ω 2 / 2 due
to transfer from the basic flow per unit time.
III:
is the work made by the viscous forces, whereby kinetic energy is transformed
to heat, cf. (2.16). This term (dissipation) is always negative because any
motion will dye out with time if no energy is transferred from outside.
Fig. 6.4
Production of rotation
Equation (6.14) is now integrated of the thickness of the boundary layer, and
all quantities are made dimensionless as follows
ω
Ω
, Ω′ =
ω′ =
Ω0
Ω0
v
U
x
x′ =
δ
v′ =
,
,
Ut
δ
y
y′ =
δ
t′ =
(6.17)
where Ω 0 is a typical value of the undisturbed vorticity in the boundary layer and U
is a characteristic velocity, for example the velocity immediately outside the boundary
layer.
In this way (6.14) turns into
1
1
⎛
⎞
′
d 1
2
′
′
(ω ) dy = ∫ ⎜⎜ − dΩ ω′v ′ ⎟⎟dy ′
∫
dt 0 2
dy ′
⎠
0⎝
(6.18)
ν
ω′Δω′ dy ′
Uδ ∫0
1
+
134
It is seen from (6.18) that for identical boundary conditions all types of flows
will dynamically similar if the number
Re =
Uδ
ν
(6.19)
is the same. This is Reynolds number, cf. (6.1). If two flows are dynamically similar
(that is have the same Reynolds number) then both flows will be either stable (that is
they will remain laminar) or unstable (they will break up and become turbulent). The
Re number can therefore be used as the criterion for a flow being laminar or turbulent.
As mentions term III (the heat loss) will always be negative. From (6.18) it is
seen that for small values of the Re number I will also be negative because term II
(the production of energy) is negligible. When I is negative the vorticity of the
perturbation will decrease with time meaning that the flow is stable and will remain
laminar.
For larger Re numbers the term I will become positive, provided that the term
1
dΩ ′
∫ − dy ′ ω′v′ dy ′
0
is positive. Normally ther will always exist a perturbation with a combination of ω′
and
v ′ for which this integral becomes positive. As I in that case will be positive for
large Re numbers the perturbation will grow with time and the flow will become
turbulent.
Example 6.1: Boundary layer along a flat plate
If we consider a flow without any pressure gradients along a plate (cf. example
5.2) the Reynolds number will grow along the plate because the boundary layer
thickness δ increases in the flow direction. When the Reynolds number reaches a
critical value the flow will break up and become turbulent as show at section A-A in
figure 6.5.
Refined stability analyses has shown the critical Reynolds number for this
flow to be about 400.
Inserting
δ = 4.8
νx
U
from (5.55) the position of the transition can be estimated because
135
Fig. 6.5
Transition from laminar to turbulent boundary layer flow along
a plate.
Re = 400 =
Uδ 4.8U
=
ν
ν
νx
U
or
x=
160000 ν
4.8 2 U
For ν = 10-6 the transition to turbulent flow will take place 0.7 cm from the leading
edge for U = 1 m/s and 7 cm from the edge for U = 0.1 m/s.
6.2 Turbulent fluctuations
A fully developed turbulent flow may be interpreted as a regular mean motion
superposed by irregular turbulent fluctuations associated with eddies or vortices. The
flow defined as stationary when the mean motion does not change with time. If we as
an example consider a stationary pipe flow, the flow in the direction of the axis will
vary with time as sketched in figure 6.6A. The instantaneous velocity u can be written
as
u = U + u′
(6.20)
wher U is the mean value and u' is the turbulent fluctuation. The mean flow normal to
the axis of the pipe is zero, and the other two velocity components are given as (cf.
figure 6.6B)
v = v′
(6.21)
w = w′
(6.22)
and
In a turbulent flow the turbulent fluctuations will always be three-dimensional.
136
Fig. 6.6
A registration of the velocity components as function of time in
a pipe flow.
A deterministic description of the turbulent fluctuations u', v' and w' is not
possible, but by use of turbulence models it is today possible to describe a many of
the mean values characterising the turbulence, such as mean flow velocities and mean
shear stresses.
6.3 Reynolds stresses
Consider a uniform, stationary open channel flow over a plane bed, cf. figure
6.7.
In this flow the shear stress τ is determined by a static force balance and
given by
τ = ρg ( D − y ) sin β
(6.23)
cf. [2], p. 40. At small velocities the flow is laminar, and vi have
τ
du
=ν
ρ
dy
(6.24)
From (6.23) and (6.24) we find
137
U=
Fig. 6.7
g sin β ⎛
y2 ⎞
⎜ Dy −
⎟
ν ⎜⎝
2 ⎟⎠
(6.25)
The distribution of the shear stress and velocity profiles in
uniform channel flow.
When the flow becomes turbulent equation (6.23) is still valid because τ as
mentioned is determined by a static force balance. Equation (6.24) is, however,
changed drastically due to the turbulent velocity fluctuations. The eddies cause an
exchange of momentum in the direction normal to the flow direction, quite similar to
the momentum exchange due to the heat motion of the molecules which is the
mechanism behind the viscosity. In a turbulent flow the exchange due to the eddies is
normally several orders of magnitude larger than the molecular exchange, and the
viscosity can often be neglected.
To analyse the effect of turbulent fluctuations on the equation of motion for
the mean flow in a turbulent flow situation we take the Navier-Stokes eqautions as
starting point. The Navier-Stokes equations are valid for the instantaneous flow
conditions in turbulent as well as laminar flows.
The turbulent velocity fluctuations are three-dimensional and it is therefore
convenient to derive the flow equations for the mean flow by application of the index
(or tensor) notation, cf. [2] p. 29. In index notation the Navier-Stokes equation reads
dvi
∂ 2 vi
∂p
(6.26)
+μ
ρ
=−
dt
∂x j ∂x j
∂xi
Consider the left hand side, which can be written
∂v j
∂v
∂v
dvi ∂vi
∂
=
+vj i = i +
(v i v j ) − vi
∂t
∂x j
∂t ∂x j
dt
∂x j
(6.27)
Here the last term on the right hand side is zero, because the continuity equation must
be satisfied. By introduction of (6.27) into (2.26) we get
vi = U i + u i′
(6.28)
where U i represents the mean motion and u i′ the turbulent velocity components we
get
138
ρ
∂ (U i + u i′ )
∂
((U i + u ′)(U i + u ′) =
+ρ
∂t
∂x j
∂ ( P + p ′)
∂2
(U i + u ′)
−
+μ
∂xi
∂x j ∂xi
(6.29)
Each term in (6.29) is time averaged by taking
T
φ=
1
φ dt
T ∫0
(6.30)
where T is a sufficiently large time interval and φ can be taken as any of the variables
in (6.29). The second term in (6.29) can now be written
∂
(U iU j + U i u ′j + u i′U j + u i′u ′j ) =
∂x j
∂
(U iU j + U i u ′j + u i′U j + u i′u ′j ) =
∂x j
∂
∂
∂
∂
(U iU j ) +
(U i u ′j ) +
(u i′U j ) +
(u i′u ′j ) =
∂x j
∂x j
∂x j
∂x j
(6.31)
∂
∂
(U iU j ) +
(u i′u ′j )
∂x j
∂x j
because
Ui = Ui
and
u i′ = 0
In this way equation (6.29) can - when time averaged - be written
ρ
∂U i
∂
+ρ
(U i U j )
∂t
∂x j
=−
∂ 2U i
∂
∂P
+μ
−ρ
( u i′u ′j )
∂x i
∂x j ∂x j
∂x j
(6.32)
or
dU i
∂ 2U i
∂P
∂
ρ
=−
+μ
−ρ
(u i′u ′j )
dt
∂xi
∂x j ∂x j
∂x j
(6.33)
where the continuity equation once more has been applied on the left hand side of
(6.33). Equation (6.33) may also be written
139
ρ
dU i
∂
∂P
σ ij
+
=−
dt
∂xi ∂x j
(6.34)
where
σ ij = ρν
∂U i
∂x j
− ρu i′u ′j
(6.35)
When we compare (6.34) to (6.26) we see that the equations for the mean motion are
identical to the Navier-Stokes equations except for some additional stresses in the
form of the term u i′u ′j . These additional stresses are know as the Reynolds stresses,
which emerge because the turbulent velocity fluctuations causes an exchange of
momentum between the different parts of the flow. This is easily seen when only two
dimensions are considered:
If we consider a section parallel to the x-axis in the direction of the mean flow,
the flow through this section per time will be ρv ′ . This corresponds to a transport of
momentum of ρv ′(U + u ′) or in average ρv ′u ′ . This exchange of momentum must in
the equation for the mean motion be represented as an apparent stress, which on
average is given as − ρv ′u ′ .
Fig. 6.8
Transverse exchange of momentum.
We return to the flow shown in figure 6.7 where the shear stress is determined
from a static force balance. In a laminar flow the shear stress is due to the molecular
heat motion, but in the case of a turbulent flow it can also be carried by the exchange
of momentum of the eddies, equation (6.24) is then changed to
τ
dU
=ν
− u ′v ′
ρ
dy
(6.36)
Since τ is unchanged, the mean velocity profile must be modified when the flow
becomes turbulent. This change can be calculated if the term u ′v ′ is known. At
present only semi-empirical theories are available for making simple estimates of this
term. Some of the most simple theories will be described in the following.
140
6.4 The mixing length theory
Let us again consider the uniform open channel flow in figure 6.7. The flow is
assumed to be turbulent, so that the mean flow is superposed by eddies of different
size and rotational speed.
The section A-A parallel to the bed (cf. figure 6.9) is considered. Across this
section there is an exchange of fluid due to the eddies. These eddies are assumed to
have
Fig. 6.9
The exchange of momentum normal to the mean flow direction.
a characteristic size, described by the length scale A . An eddy with this length scale
will be able to transport fluid from one level I to another higher level II with a mutual
distance of the order A . At the same time there will due to continuity be transported
the same amount of water down from II to I.
Through section A-A there will be a mass flux per unit area of the order of
magnitude ρq / 2 - going up as well as down. Her q may be interpreted as a
characteristic value of the turbulent fluctuations normal to the mean flow direction.
The fluid particles transported from level I are assumed to maintain their
longitudinal flow velocity unchanged until they reach level II, where they are mixed
with the surrounding fluid and obtain the mean flow velocity at this level (this is the
reason for the name 'mixing length').
As seen from figure 6.9 these particles will increase their velocity by
U II − U I ≈ A
dU
dy
(6.37)
Since there is a flux of q up per unit time (and similarly a flux q down) the total
exchange of momentum per unit area through A-A is found to be
dU
ΔM = ρqA
dy
When formulating an equation of motion for the mean flow, in which the turbulent
fluctuations are averaged out, you must as seen in (6.36) include the effect of the
turbulent fluctuations as a shear stress given by
141
τ = ρ qA
dU
dy
(6.38)
q is as mentioned of the same order of magnitude as the vertical velocity fluctuations.
For reasons of continuity these must be of the same order of magnitude as the
horizontal velocity fluctuations, which are assumed to be of the same order of
magnitude as the difference in the mean flow velocity over the distance of the mixing
length, giving
q≈A
dU
dy
whereby (6.38) becomes
τ = ρA 2
dU dU
dy dy
(6.39)
To get further it is, however, necessary with knowledge of the mixing length A . This
may be expected to be a function of the distance from the wall A = A (y) and must
decrease towards 0 at the wall where the turbulence ceases. Since A is a length scale
the most simple assumption is to take it to vary linearly with the distance y from the
wall,
A = κy
(6.40)
where κ is a constant (know as von Kármáns constant; κ is close to 0.40.
The transfer of momentum due to the eddies in a turbulent flow is as described
analogous to the transfer by the motions of the molecules. The latter gave rise to
Newton's formula (2.5), and because of the analogy you operate with technical
turbulence theories where an expression similar to (2.5) is introduced to describe the
shear stress, namely
dU
τ = ρν t
(6.41)
dy
where ν t is called the eddy viscosity. When comparing (6.41) with (6.39) we find
νt = A2
dU
dy
(6.42)
In the derivation of (6.42) the moleculous viscosity has been neglected. It is
easily seen that if the kinematic viscosity is to be included (6.41) is modified to
τ = ρ(ν t + ν)
dU
dy
(6.43)
142
where ν t is still found from (6.42). normally ν t >> ν , so that ν can be neglected,
except in situations, as will be described in the following.
6.5 Velocity profiles in uniform channel flows
The velocity distribution in an infinitely wide channel can easily be calculated
by use of the mixing length theory, i.e. from (6.40), (6.42) and (6.43).
The shear stress is given by
τ τ0 ⎛ Y ⎞
= ⎜1 − ⎟
ρ ρ ⎝ D⎠
(6.44)
where τ 0 is the shear stress at the bed. This must balance the longitudinal component
of the gravity force of the water and is therefore given by
τ 0 = ρgDI
(6.45)
where I is the slope of the water surface. The friction velocity (or shear velocity) Uf is
by definition given as
τ0
Uf =
(6.46)
ρ
From (6.44), (6.45) in combination with (6.40), (6.41) and (6.42) we get'
y⎞
dU
⎛
U 2f ⎜1 − ⎟ = (ν t + ν)
dy
⎝ D⎠
2
⎞
dU
⎟⎟ + ν
= A ⎜⎜
dy
⎝ dy ⎠
2 ⎛ dU
(6.47)
2
⎞
dU
⎟⎟ + ν
= κ y ⎜⎜
dy
⎝ dy ⎠
2
A.
2 ⎛ dU
A smooth bed
Let us consider the flow over a completely smooth bed. It is now seen that no matter
how small ν is, the viscous term (2. term on the right hand side of (6.47)) will be
larger than the turbulent contribution (1. term on the right hand side of (6.47)) when
we are sufficiently close to the bed and y tends towards zero. If the turbulent
contribution is neglected, equation (6.47) is reduced to (valid for small values of y)
143
U 2f = ν
dU
dy
(6.48)
which with the boundary condition U = 0 for y = 0 gives
Uf
U
=
y
ν
Uf
(6.49)
However, we only have to move a small distance from the wall to have ν t >> ν .
Away from the wall (6.47) reads
dU ⎞
⎟⎟
U = κ y ⎜⎜
⎝ dy ⎠
2
f
2
2⎛
2
(6.50)
where the weak variation in τ with y has been neglected on the left hand side of
(6.50). Equation (6.50) can then be written
dU 1 dy
(6.51)
=
Uf
κ y
which can be integrated to give
1
U
= ln( y ) + const
Uf
κ
(6.52)
We have thus a layer very close to the wall where the velocity varies linearly
(6.49), while the velocity a little further from the wall varies logarithmicaly as
illustrated in figure 6.10, where the dimensionless velocity U/Uf is show as function
of the dimensionless vertical coordinate y+, given by
y+ = y
Fig. 6.10
Uf
ν
(6.53)
The velocity profile close to the bed for a smooth bed.
144
Figure 6.11 shows similar measurements made near the wall in a pipe flow. Away
from the wall the measurements follow a straight line with a logarithmically scaled
y+-axis, cf. figure 6.11B.
Fig. 6.11
Measurements of the mean velocity profile over a smooth wall.
If equation (6.52) is rewritten as
1
U
= ln y + + K1
Uf
κ
(6.54)
we see from figure 6.11 that all measurements are following the same straight line,
and that K1 in (6.54) shall be chosen to be
K1 = 5.7
(6.55)
This value is therefore determined experimentally, but today it may also be
derived theoretically by application of more sophisticated models. Very close to the
wall the points deviate from the straight line given by (6.54), which is because the
viscous forces become dominant. The distance from the wall to the point of
intersection between the two velocity profiles (6.54) and (6.53) is know as the
thickness of the viscous sub-layer δ v .
The thickness of the viscous sub-layer is thus found from
U f δv
ν
⎛ U f δv ⎞
⎟
= 5.7 + 2.5 ln⎜⎜
⎟
ν
⎠
⎝
(6.56)
145
Fig. 6.12
Photography of the viscous syb-layer. a: seen fromt the side, b:
seen from above, c: detail from b. From [4].
or
U f δv
ν
= 11.7
(6.57)
From this the thickness of the viscous sub-layer is found to be
δ v = 11.7
ν
Uf
(6.58)
This description of the near-wall conditions is naturally based on
simplifications. In reality the nature of viscous sub-layer is not like a laminar flow, as
indicated in the photograph in figure 6.12, where a a complex pattern with a large
number of 'streaks' in this layer are clearly visible.
_____________________________________________________________________
Example 6.2 The flow resistance and the thickness of the viscous sub-layer in a
smooth channel flow.
Let us consider a channel flow with a discharge of Q = 1 m3/m/s and a slope I
of 10 . We want then to determine the thickness of the viscous sub-layer. For this we
need to know the shear velocity Uf, cf. (6.58). If we assume the thickness of the
viscous sub-layer is very small compared to the total water depth, the total discharge
can be related to Uf and D by integration of (6.54) over the depth
-4
146
D
Q = ∫ Udy ≈
0
D
Uf
⎛ yU f ⎞
⎟ dy + 5.7 DU f
ν ⎟⎠
∫ ln⎜⎜⎝
κ
0
(6.59)
or
DU f
Q=
Uf ν
κ Uf
ν
∫ ln ς dς + 5.7 DU f
0
DU f
ν
= [ς ln ς − ς]0 ν
κ
DU f ⎡ ⎛ DU f
=
⎢ln⎜
κ ⎣⎢ ⎜⎝ ν
= 3.2 DU f +
DU f
κ
+ 5.7 DU f
(6.60)
⎞ ⎤
⎟ − 1⎥ + 5.7 DU f
⎟
⎠ ⎦⎥
⎛ DU f
ln⎜⎜
⎝ ν
⎞
⎟
⎟
⎠
This equation corresponds to the formula for flow resistance know from hydraulics
V
1 ⎛ DU f
= 3.2 + ln⎜⎜
κ ⎝ ν
Uf
⎞
⎟
⎟
⎠
(6.61)
where V is the mean flow velocity. Equation (6.60) has two unknown, Uf and D.
These are, however, correlated by the relation
U 2f = gDI
(6.62)
2
cf. (6.45) and (6.46). By using the given discharge Q = 1 m /s and a viscosity of
ν = 10 −6 m 2 / s iteration of equations (6.62) and (6.61) gives the following quantity
U 3f
gI
= 0.034 m 2 / s
From this we find Uf = 0.032 m/s, while the depth is found to be 1.06 m. From (6.58)
we find
δ v = 11.7 ×
10 −6
m = 0.37 mm
0.032
(6.63)
which I very small compared to the water depth. For a larger discharge the viscous
sub-layer becomes even thinner.
B.
A rough bed
147
If the bed is not completely smooth the irregularities in the bed may cause
an important change in the velocity distribution over the entire water depth. As an
example let us consider the flow over a sand bed where the sand is assumed not to
move. If the sand grains lie evenly distributed along the bed as sketched in figure
6.13, the bed has a roughness k, which is the same as the grain diameter. When
δ v >> k the roughness elements of the wall are completely embedded in the
viscous sub-layer and the turbulent flow is not affected by the presence of the
roughness of the wall. This situation therefore corresponds to smooth bed
conditions. When δ v << k (as indicated by the dotted curve in figure 6.13) the
viscous sub-layer loses its significance because the shear stress is no longer
transferred to the bed through viscous forces. In stead the shear stress is
transferred as forces acting on the roughness elements though a higher pressure on
the front of the elements and a lower pressure on the back quite similar to the drag
force on a cylinder, cf. example 5.3. In a region just above the rough bed the mean
flow in non-uniform and can in this way transfer a part of the stress, while the
Reynolds' stresses transfer the rest. As an approximation it can be assumed that
the turbulent Reynolds' stress carries the entire shear stress at a distance of the
order k from the top of the roughness elements.
Fig. 6.13
Flow over a rough bed.
When the bed shear stress is transferred as drag forces the viscosity is of very
small significance, cf. example 5.3, and over a rough bed the flow resistance and
velocity distribution is practically independent of the viscosity of the fluid. As
mentioned the shear stress is transferred as Reynolds' stresses just a small distance
above the roughness elements and by similar arguments as for a smooth bed (outside
the sub-layer) the velocity profile can be described by (6.52), that is
1
U
= ln y + const
Uf
κ
(6.64)
In the case of a rough bed it is of course more difficult to make a definition of
the position where y is 0. It is close to the top if the sand grains and the exact position
is of minor practical importance. The vertical dimensionless coordinate may be taken
as y/k, in which case experiments have shown that the constant in (6.64) shall be 8.6.
Giving
148
1 ⎛ y⎞
1 ⎛ y ⎞
U
= ln⎜ ⎟ + 8.6 ≈ ln⎜
⎟
Uf
κ ⎝ k / 30 ⎠
κ ⎝k⎠
(6.65)
This expression may as an approximation be used all the way from the bed to the
surface and gives rise to formula for the flow resistance quite analogous to (6.61)
V
⎛ D⎞
= 6.1 + 2.5 ln⎜ ⎟
Uf
⎝k⎠
(6.66)
which as described is valid for δ v << k .
There is a complex area of transition between the conditions of smooth and
rough bed, but the expression
⎛k
3.2ν ⎞⎟
V
= 6.1 − 2.5 ln⎜ +
⎜ D DU f ⎟
Uf
⎝
⎠
(6.67)
gives an adequate description of the flow resistance and includes (6.66) and (6.61) as
asymptotic cases.
Example 6.3 Suspended transport of sand in water
Let us consider sand of uniform grain size, which in quiescent water has the
settling velocity (i.e. fall velocity of the individual sand grains) w. Even though each
gains settles due to the effect of gravity experience shows that if the current (or wind
velocity) is strong enough considerable amounts of sand may be carried in
suspension. This can be explained by the presence of turbulent eddies. Following the
basic idea of the mixing length theory there is an upward flux of fluid through section
A-A per unit area of q., cf. figure 6.14. This fluid will carry an amount of sand taken
on average from the level ½ A below the section, where the average concentration is
c − ½ A dc / dy , where c(y) is the mean concentration of sand by volume. The upward
flux of sand is then
⎛
dc ⎞
(v − w)⎜⎜ c − 12 A ⎟⎟
dy ⎠
⎝
(6.68)
where v is the vertical velocity of the fluid. In (6.69) it is expressed that the sand is
settling with the velocity w relative to the motion of the water. The upward flux of
sand is countered by a downward flux due to the downward fluid velocity, which is
also q. The downward flux is analogous to (6.68) and given as
149
⎛
dc ⎞
(v + w)⎜⎜ c + 12 A ⎟⎟
dy ⎠
⎝
Fig. 6.14
(6.69)
Suspension of sediment in a turbulent flow.
In the case of a stationary distribution of sediment the flux up and down
through section must be equal, so that (6.68) and (6.69) are of the same magnitude,
giving
cw + 12 vA
dc
=0
dy
(6.70)
From (6.38) we have (using that ½v = q) that
qA = 12 vA =
τ/ρ
⎛ dU ⎞
⎟⎟
⎜⎜
⎝ dy ⎠
(6.71)
The velocity gradient is here found as
dU U f
=
dy
κy
(6.72)
cf. (6.65), and τ / ρ is taken from (6.44) by which (6.70) can be written
y ⎞ dc
⎛
cw + κU f ⎜1 − ⎟ y
=0
⎝ D ⎠ dy
The variables can be separated to give
dc
w
D dy
=−
c
κU f D − y y
(6.73)
(6.74)
which can be integrated to give
150
⎛D− y⎞
⎟⎟
c = const × ⎜⎜
⎝ y ⎠
Z
(6.75)
where the coefficient Z has been introduced as
Z=
w
κU f
(6.76)
It is seen from (6.75) that the heavier the sand (i.e. larger w) the closer the
sand will be to the bed. Very fine sand (small w) will be more unifromly distributed
over the vertical. Similarly strong turbulence (large Uf) will tend to carry more
sediment away from the bed.
The eddy viscosity, defined by (6.41) is as seen from (6.71) given as
y⎞
⎛
ν t = qA = κU f y⎜1 − ⎟
⎝ D⎠
(6.77)
Insertion of ν t in (6.70) makes this read
cw + ν t
dc
=0
dy
(6.78)
A simple physical interpretation can be made of each term in (6.78). The first term cw
is the flux of sand through a unit area due to the settling of the grains. The second
term expresses the concept known as 'turbulent diffusion' with the eddy viscosity as
the diffusion coefficient. Typically diffusion is a process where a scalar
(concentration, temperature etc.) is spread when its magnitude varies from one place
to another, and the flux of the scalar is proportional to its gradient ( in the present case
the gradient is dc/dy).
In the case of quiescent water or a laminar flow we have molecular diffusion,
because the spreading is cause by the heat motion (Brownian motion) of the
molecules (in addition to any advection). In general the molecular diffusivity is of the
same order as the kinematic viscosity of the fluid.
Turbulence is a much stronger mechanism for mixing than the molecular
motion and ν t is therefore most often many orders of magnitude larger than ν . The
quantity ν t can be interpreted as the diffusivity for the momentum. The diffusivity
for suspended or dissolved matter, temperature etc. is normally assumed to be of the
same order of magnitude as ν t .
151
Example 6.4 The energy balance in a turbulent flow.
If we consider the flow in a pipe or a uniform channel flow the turbulence in a
given point will be maintained, which means that the intensity and the statistical
properties of the turbulent fluctuations are not changing with time in a given point, or
in the downstream direction. This is in contrast to turbulence generated by a grid,
which is moved swiftly through a fluid at rest or alternatively by letting a fluid flow
through a fixed grid. In these cases the turbulence will decay with time or space
respectively.
The reason for the turbulence being maintained at the same level is that energy
is constantly being transferred from the mean motion to the turbulence due to the
gradient in the mean flow velocity profile. In section 6.1 it was explained how a small
perturbation for sufficiently large values of the Reynolds' number received energy
from the basic flow and therefore grew in strength to form turbulence. It is an
analogous mechanism, which causes a continuous production of turbulent energy and
maintains the turbulence.
Let us consider a fluid particle with the components u' and v' of the turbulent
velocity fluctuations. During a small time increment Δt this particle has moved the
distance v ′Δt away from the wall. If we disregard the effect of outer forces on the
particle it will keep the same absolute velocity after the time Δt . But since it is now at
a level with a different mean velocity the relative velocity fluctuation will be changed
to
dU
v ′Δt
(6.79)
u′ −
dy
as indicated in figure 6.15.
Fig. 6.15
The production of turbulent kinetic energy.
The other two velocity fluctuations v' and w' are not affected. The turbulent
kinetic energy k per unit volume is now defined as
k = 12 (u ′ 2 + v ′ 2 + w′ 2 )
(6.80)
At the end position k is found as
152
k=
⎛
2
1⎜ ′
u −
2⎜
⎝
⎞
dU
v ′Δt ⎟⎟ + 12 (v ′ 2 + w′ 2 )
dy
⎠
(6.81)
dU
= k 0 − u ′v ′
Δt
dy
where ko is the initial kinetic energy. When time averaged it is seen that the term
− u ′v ′
dU
dy
(6.82)
describes the increase in the turbulent kinetic energy due to the gradient in the mean
flow velocity. The factor − ρu ′v ′ is equal to the Reynolds stress which for fully
developed turbulent flow is equal to the shear stress. The term (6.82) may therefore be
written
− ρu ′v ′
dU
dU
=τ
dy
dy
(6.83)
which thus described the rate of production of turbulent kinetic energy.
The term at the right hand side expresses the the work made by the shear stress
on a unit cube. The dissipation (heat loss9 is according to (2.16) given as
⎛ ∂u ⎞
ε = ρν⎜⎜ ⎟⎟
⎝ ∂y ⎠
2
(6.84)
(a more general proof can be found in appendix I). For a laminar flow (6.84) can also
be written
ε=τ
∂u
∂y
(6.85)
so that the energy dissipation in a laminar is directly found as the work by the shear
stress.
In a turbulent flow the picture is more complicated because the work by the
shear stress (6.83) first is transformed to turbulent energy, that is stronger motion of
the eddies. Experiments show that the eddies containing most of the turbulent energy
are the larger ones corresponding to a smaller wave number. The energy is, however,
dissipated to heat in connection with the small eddies because the velocity gradients
are largest for these (cf. the expression for the dissipation (6.84)). From theoretical
considerations as well as by experiments it has been found that the the energy
dissipated per time and per volume can be approximated by
k 3/ 2
ε=A
Ad
(6.86)
153
where A is a dimensionless factor of the order 0.08 and A d is the dissipation length –
also know as the length scale of the turbulence. This is a measure of the size of the
large energy containing eddies and is of the same order of magnitude as the mixing
length A , cf. section 8.2.
It may seem to be a paradox that the expression (6.86) for the total dissipation
does not depend on the viscosity, since we know that in the end it is the viscous forces
which cause the dissipation. The explanation can found in a model proposed by
Kolmogorov:
It must be expected that eddies of widely different orders of magnitude cannot
interact directly. A very large eddy can transport very small eddies by advection
without any other direct influence. Eddies can only exchange substantial amounts of
energy between them if they are of comparable size.
In broad terms the pattern is so that the large eddies, which contain most
energy, transfer some of this to eddies which is one order of magnitude smaller. These
the pass the energy to new eddies yet one order smaller. This process continues to
smaller and smaller eddies until the energy has been passed to eddies so small that the
energy can be dissipated through the action of viscous forces. If the amount of energy
passing through this chain of processes ('the energy cascade') is determined by the
transfer between the larger eddies, then only the last part of the process is affected by
the viscosity. The model can be summarized: The energy containing large eddies is a
reservoir of energy, which is drained gradually by transfer of energy from the larger
to the smaller eddies (or from the smaller to the larger wave numbers) through a sort
of cascade process. The largest wave numbers (the smallest eddies) represent an
energy drain, where the viscous forces transform the turbulent energy to heat.
It is thus quite natural to call the expression (6.83) the 'production of turbulent
energy'.
From these considerations it is seen that contribution to the turbulent energy
immediately is given so that only the longitudinal fluctuations u' are increased.
Thereafter the 'surplus' is gradually distributed to the other components through a
process involving the pressure fluctuations.
Figure 6.16 shows a series of measurements of the energy balance in smooth
pipe flow. The balance of energy is quite similar in open channel flow, showing very
small deviations from figure 6.16.
The energy balance can be expressed through an equation, which mainly
consists of the terms
production + resulting diffusion = dissipation
Fig. 6.16
Measurements of the energy balance in pipe flow.
154
The dissipation is largest near the pipe wall and smallest – but not zero – at the
axis of the pipe. Production and dissipation are dominant in the main part of the crosssection of the pipe, but there exist a contribution of some significance which is due to
the diffusion of turbulent energy toward the axis across the direction of the mean
flow. The diffusion is of relative importance in the central area away from the wall.
As seen from the figure the production is very large close to the wall and
decreases to zero at the axis of the pipe.
6.6 The structure of turbulence in a pipe
In the following the most important characteristics of a turbulent pipe flow
will be briefly described. As mentioned the characteristics of turbulence in a pipe and
in an open channel flow are almost identical.
Figure 6.17 shows five distribution functions for turbulent quantities in pipe
flow. These distributions are almost universal and resemble also open channel flow
conditions.
Figure 6.17a shows the distribution of the mean flow velocity, which as
described in section 6.5 is logarithmic. In the form given the logarithmic distribution
is, however, not universal, but it is easy to make a universal formulation in the form
of the so-called deficit law
U0 −U
Uf
= 2.45 ln
r
y′
(r = the radiusof the pipe)
which is valid for smooth as well as for rough pipes, when Uo is taken as the velocity
at the pipe axis.
Fig. 6.17
Measurements of the characteristics of turbulence in a pipe.
155
Figure 6.17b shows the distribution of the shear stress τ . τ is zero at the axis
and attains the value τ 0 = ρU 2f at the wall. In a turbulent flow most of τ is
transferred through the Reynolds stresses, that is τ = −ρu ′v ′ . However, for a smooth
pipe this is not the case in the viscous wall layer, where the viscosity is dominant,
especially very close to the wall. Figure 6.17c shows the distribution of the RMS (root
mean square) of the turbulent fluctuations in the three principal directions. The RMS
value (or standard fluctuations) is a measure of the variation around the mean value
and is defined as
u = u′2
(6.87)
In formula 6.17c u is in the direction of the axis and v is in the direction
normal to the wall. Close to the wall u is largest and v smallest, but close to the axis
the three standard fluctuations are practically of the same magnitude which means that
the turbulence is almost isotropic (i.e. has the same characteristics in all three
directions). Figure 6.17d shows the distribution of the turbulent kinetic energy k. The
variation is approximately linear except in a region close to the axis. This has been
used as basis for some turbulence models especially for boundary layers where
proportionality between τ and k has been assumed, cf. chapter 8.
Finally the correlation coefficient between u' and v' is shown in figure 6.17e.
This is defined as
R=
u ′v ′
uv
(6.88)
R lies between –1 and +1. In turbulent pipe flows R is constant (= -0.4) over most of
the cross-section, increases towards the axis to zero.
By full correlation we mean in practice that there is a linear relation between
the two variables, for example u' = -v', which is seen to give R = -1. If all the
measured values are plotted together in a diagram as in figure 6.18a, the points will on
a straight line passing through origo.
Korrelation zero is found when the two velocity flutuations are independent
variables
u ′v ′ = 0
Fig. 6.18
and
R=0
Examples of different degrees of correlation.
156
If simultaneous measurements are plotted, the point will be positioned at random
locations around origo, and no orientation can be distinguished.
The concept of correlation is of most interest when two fluctuations are partly
correlated as is the case is a point in the pipe flow. Simultaneous values of u' and v'
will then mainly fall within an ellipse-shaped area as indicated in figure 6.18c.
A simple illustration of this relationship in turbulent pipe flow can be made by
plotting the end points of the velocity vectors at different times as shown in figure
6.19. Statistically the points will then lie in areas of variation, which close to the wall
are ellipse-shaped (as in figure 6.18c) and near the axisis is circular (cf. figure 6.18b).
It is the (partial) correlation between u' and v' which makes the transfer of
shear stresses as Reynolds stresses in a turbulent flow possible.
157
7. Turbulent boundary layers
As mentioned in example 6.1 a laminar boundary layer will be unstable at
sufficiently large Reynolds numbers and go through a transition to become turbulent.
Such a turbulent boundary formes along a plane plate is shown in the photograph in
figure 7.1.
Fig. 7.1
Photograph of a turbulent boundary layer, Re ≈ 3000.
The transition between the turbulence in the boundary layer and the outer
potential flow is very irregular, as seen in figure 7.1. In a point places near the edge of
the boundary layer there will at some instances be a fully turbulent flow and at other
instances practically no turbulence at all.
This is called intermittent turbulence. At the same time the transition between areas
with turbulence and areas with potential flow is remarkable sharp. In these transitions
the viscosity is very significant, while the outer flow (potential flow) and the fully
developed turbulent flow are independent of the viscosity (except for the small
turbulent structures).
The outer potential flow in the vicinity of a turbulent boundary layer is more
irregular because of these fluctuations of boundary between the turbulence and the
potential flow.
7.1 The turbulent boundary layer equation
The boundary layer equation for a laminar boundary layer was described in
section 5.1. The main idea was that the variation in the different terms (velocity,
165
pressure) is much larger across the flow direction (the y-direction) than in the flow
direction (the x-direction). In the turbulent case the x-component of the Navier-Stokes
equation (6.33) reads
dU 1
∂ 2U 1
∂
∂
∂P
− ρ (u ′) 2 − ρ (u ′v ′)
+ ρν
ρ
=−
2
dt
∂x
∂y
∂x
∂y
(7.1)
where the term ρν ∂ 2U 1 / ∂x 2 has been neglected (just as in the laminar case). The ycomponent of the flow equation reads similarly
0=−
∂
∂P
− ρ (v ′) 2
∂y
∂y
(7.2)
Here the term ρ ∂ (u ′v ′) / ∂x has been neglected, as is involves a differentiation in the
x-direction.
Integration of (7.2) gives
P = −ρ(v ′) 2 + const
(7.3)
Outside the boundary layer in the potential flow (v ′) 2 = 0 , and here the pressure is
called po. Equation (7.3) can therefore be written
(7.4)
P = p0 − ρ(v ′) 2
In most cases the last term in (7.4) can be neglected, so that we in turbulents as well
as in laminar boundary layers can assume that the pressure inside the boundary layer
is determined by the pressure outside in the potential flow, i.e.
P = p0
(7.5)
Insertion of (7.4) into (7.1) gives
∂p0
dU 1
∂ 2U 1
∂
ρ
=−
+ ρν
− ρ (u ′v ′)
2
dt
∂x
∂y
∂y
[
∂
−ρ
(u ′) 2 − (v ′) 2
∂x
]
(7.6)
In practice the last term in the right hand side can also be neglected, so that (7.6) can
be reduced to
ρ
⎞
∂p
dU 1
∂ ⎛ ∂U
= − 0 + ρ ⎜⎜ ν 1 − (u ′v ′) ⎟⎟
∂x
∂y ⎝ ∂y
dt
⎠
(7.7)
Using (6.35) this gives
166
ρ
∂p
dU 1
∂τ
=− 0 +
dt
∂x ∂y
(7.8)
which can be seen as the equation of motion for a unit cube. On the left hand side we
have mass times acceleration and on the right hand side we have the resulting forces
from the pressure and the shear stress in the flow direction.
In addition to the boundary layer equations (7.5) and (7.8) the integrated
momentum equation (5.20) can also be applied. The derivation of (5.20) is not based
on the assumption of the flow being laminar. But it should be kept in mind that
Newton's formula (2.5) for τ is normally not applicable for turbulent flows. In stead
one should apply the more general expression for the shear stress at the wall
τ 0 = ρU 2f
(7.9)
The actual application of (7.9) illustrated by examples in the following.
7.2 Stationary boundary layer along a plate
In the following we shall consider the variation in the boundary layer thickness for a
turbulent flow along a plate. The plate is rough, and there is no outer pressure gradient
so the case is analogous to the laminar flow treated in example 5.2. Corresponding to
(5.50) the integrated momentum equation now reads
δ
δ
∂
∂
ρU 2 dy − u 0 ∫ ρUdy = − ρU 2f
∫
∂x 0
∂x 0
(7.10)
Just as in the laminar case we can now assume a velocity profile for the mean flow
velocity U.
In order to make a reasonable guess for the velocity profile in the boundary
layer we can compare the conditions in the boundary layer to the conditions in pipe
and channel flow as described in chapter 6.
The conditions in the boundary layer are illustrated in figure 7.2.
Fig. 7.2
Typical velocity profile and shear stress distribution in a
boundary layer.
167
Apart from the fluctuating boundary, the main difference from the pipe flow is
that the shear stress is not determined from a static balance. Further the boundary
layer flow is non-uniform, because the boundary layer is growing in the flow
direction.
Close to the wall experiments do, however, show that an equilibrium layer
exists similar to the channel flow, where the production and the dissipation of
turbulent energy practically balance each other. Mathematically this can be expressed
as
τ dU
k 3/ 2
=A
ρ dy
A
(7.11)
where the production is given by (6.83) and the dissipation of turbulent energy is
given by (6.86).
Measurements has further confirmed that the turbulent fluctuations shown in
figure 6.17 and the distinction between smooth and rough walls in all essentials can
be used in boundary layer flows as well as channel- and pipe flows.
This is, however, only valid for the affine or self-similar boundary layers (cf.
example 5.2 for the laminar case), where the velocity profiles can be formulated as
⎛ y⎞
u0 − U = U f f ⎜ ⎟
⎝δ ⎠
(7.12)
uo is the velocity outside the boundary layer, see figure 7.2. For such an affine
boundary layer measurements as shown in figure 6.17 can be used to estimated the
near-bed turbulent kinetic energy to be
k = 12 ρ (u ′ 2 + v ′ 2 + w′ 2 ) ≈ 3.5U 2f
(7.13)
Equation (7.11) can now be used to assess the velocity profile in the equilibrium
layer, using that
τ τ0
≈
= U 2f
ρ ρ
(7.14)
where τ 0 is the bed shear stress, cf. figure 7.2. By inserting (7.13) and (7.14) together
with
A = κy
(7.15)
Into equation (7.11) we get
Uf
dU
= const1 ×
y
dy
(7.16)
or
168
U
= cons1 × ln( y ) + const 2
Uf
(7.17)
In other words the velocity profile in an affine boundary layer is logarithmic, as it is
in channel- and pipe flows.
Because the conditions in a boundary layer are quite similar to a channel flow,
we can in the case of a hydraulically rough bed write (7.17) as
1 ⎛ y ⎞
U
= ln⎜
⎟
U f κ ⎝ k / 30 ⎠
(7.18)
(7.18) is completely analogous to (6.65). This expression is now used in the
momentum equation (7.10). First we determine the term
δ
∫ U dy =
⎡ ⎛ δ ⎞ ⎤
⎟ − 1⎥
⎣ ⎝ k / 30 ⎠ ⎦
Uf
δ ⎢ln⎜
κ
0
(7.19)
Using that U = uo for y = δ , we find
u0
1 ⎛ δ ⎞
= ln⎜
⎟
U f κ ⎝ k / 30 ⎠
(7.20)
which inserted into (7.19) gives
δ
∫ U dy = δ u 0 −
δUf
0
(7.21)
κ
Similarly we find
δ
⎛U f
∫ U dy = ⎜⎜⎝ κ
0
2
⎛U f
= ⎜⎜
⎝ κ
=
u 02δ
⎞ k 30δ / k 2
⎟
⎟ 30 ∫ ln (ς ) dς
⎠
0
2
2
[
⎞ k
2
⎟
⎟ 30 ς (ln ς ) − 2ς ln ς + 2ς
⎠
− 2u 0
δUf
κ
+ 2δ
]
30δ / k
0
(7.22)
U 2f
κ2
Insertion of (7.21) and (7.22) into (7.10) gives
169
Uf
δU f
∂ ⎛⎜ 2
+ 2δ 2
u 0 δ − 2u 0
∂x ⎜⎝
κ
κ
2
⎞
⎟
⎟
⎠
(7.23)
− u0
U
∂ ⎛
⎜δ u0 − δ f
∂x ⎜⎝
κ
⎞
⎟ = −U 2f
⎟
⎠
In the present case uo is constant and (7.23) can be reduced to
∂
(2Z 2δ − Zδ ) = −κ 2 Z 2
∂x
(7.24)
Where Z is a new dimensionless variable, defined as
Z=
Uf
κ u0
(7.25)
equation (7.20) can be written
⎛ δ ⎞
Z ln⎜
⎟ =1
⎝ k / 30 ⎠
(7.26)
together equation (7.24) and (7.26) form a system of equations in Z, from which δ
can be determined. Differentiation of equation (7.26) gives
⎛ δ ⎞ Z ′
Z ′ ln⎜
⎟+ δ =0
⎝ k / 30 ⎠ δ
or
Z′ δ ′
+ =0
Z2 δ
(7.27)
Differentiation of (7.24) gives
(4Z − 1) δ Z ′ + (2 Z 2 − Z )δ ′ = −κ 2 Z 2
(7.28)
Which together with (7.27) gives
κ 2Z 2
dδ
=
dx 4 Z 2 − 3Z 2 + Z
(7.29)
170
Fig. 7.3
The variation in the boundary layer thickness along a rough
plate.
this equation must (together with (7.26)) be solved numerically.
Figure 7.3 shows such a solution, where the boundary condition has been
taken as: δ = δ 0 for x = 0 , where δ 0 is the thickness of the boundary layer at its
transition from being laminar to being turbulent, cf. figure 6.5. It is seen from figure
7.3 that δ grows approximately linearly with x (in contrast to the laminar case, where
δ ∝ x ). This can be realized by the following considerations: the quantity Z defined
in (7.25) is normally small because the friction velocity is only a few percent of the
outer flow velocity. Further the variation in Z along the x-axis is weak as can be seen
from (7.26) where the ln-function varies very slowly. In this way (7.29) can as an
approximation be written as
dδ
≈ κ 2 Z ≈ const
dx
(7.30)
which gives δ ≈ const × x
Example 7.2: Wind-driven flow.
As an even more simple example of the use of the momentum equation we can
calculate how the wind creates a flow in a body of water with land (in the form of a
vertical wall) to the left and extending infinitely to the right. On the surface (along the
x-axis) a wind is blowing, which creates a constant shear stress τ 0 , figure 7.4. This
shear stress creates a boundary layer flow near the surface. For simplicity the flow is
taken to be turbulent everywhere, why it is natural to assume the velocity profile to be
logarithmic. Since the velocity must be zero for y = δ , this can be written
171
U
1 ⎛δ⎞
= ln⎜⎜ ⎟⎟
Uf
κ ⎝ y⎠
Fig.
7.4
(7.31)
Stationary wind-driven flow behind an edge.
The flow is stationary, and the outer flow velocity is in this case equal zero. The
integrated momentum equation (5.20) now reads:
δ
∂
ρU 2 dy = S
∫
∂x 0
(7.32)
where S is the resulting outer force in the x-direction. Along a plate we have S = −τ 0 ,
where τ 0 is the bed shear stress acting against the flow direction, cf. (7.10). In the
present case S is an external force, which is actually acting in the x-direction, see
figure 7.4. Therefore S will now be given as
S = τ 0 = ρU 2f
(7.33)
By inserting (7.33) into (7.32) an integration and a few calculations gives
∂ ⎛ ρ 2 ⎞
2
⎜ 2 U f 2 δ ⎟ = ρU f
∂x ⎝ κ
⎠
(7.34)
Since the τ 0 and therefore U 2f is constant in this example, (7.34) is esily integrated to
give
δ=
κ2
x = 0.08 x
2
(7.35)
using that δ = 0 for x = 0. We thus see that once again δ grows linearly with x, which
is a characteristic for turbulent flows.
Until now we have assumed that there is no flow outside the boundary layer.
Because of the mass flux in the boundary layer, which grows in the wind direction,
this does not hold exactly. Due to continuity there must therefore be a flow of water
172
from the stagnant water and up into the boundary layer. Figure 7.5 shows this balance
of continuity.
As seen in the figure the continuity equation reads
dq
= v0
dx
(7.36)
Fig. 7.5
The continuity for the area surrounding the boundary layer.
Here the discharge q is given as
δ
q = ∫ u dy =
0
1
U f δ = 0.2U f x
κ
(7.37)
from which we get
v0 = 0.2U f
(7.38)
It is thus seen that more and more water is flowing up to be part of the boundary layer
flow as illustrated in figure 7.6. This process is called entrainment.
Fig. 7.6
Entrainment.
7.3 Examples of non-stationary boundary layers
In this section we shall continue to work on the two cases considered in
section 7.2. In that section the flow was non-uniform in the x-direction, but stationary.
Now the flow is taken to be uniform but unsteady.
173
Example 7.3: Wind-driven flow (continued from example 7.2)
We now consider a body of water with a free surface of infinite extent. From
time t = 0 a wind is blowing over the surface, causing a constant surface shear stress.
This shear stress generates a boundary layer flow which is assumed to be turbulent
with a velocity profile described by
U
1 ⎛δ⎞
= ln⎜⎜ ⎟⎟
Uf
κ ⎝ y⎠
(7.39)
Due to the uniformity the boundary layer thickness δ is constant for all x, but varies
with time, as more and more of the fluid takes part in the boundary layer flow. The
momentum equation (5.20) now reads (taking uo = 0)
δ
∫ρ
0
Fig. 7.7
dU
= S = ρ U 2f
dt
(7.40)
Wind-driven current.
Where the sign of the outer force is discussed in example 7.2. From (7.39) we find
dU U f 1 dδ
=
κ δ dt
dt
(7.41)
this is inserted into (7.40) which after integration gives
U f dδ
= U 2f
κ dt
(7.42)
From (7.42) we get (with initial condition δ = 0 for t = 0)
δ = κU f t
(7.43)
174
The boundary layer thickness δ is seen to grow linearly in time. Since there is no
variation in the x-direction there is no vertical velocity, but more and water is still
entrained to take part in the boundary layer flow.
Example 7.4: The boundary layer along a plate suddenly put in motion
This example is the turbulent version of example 5.1 where the laminar
boundary layer along a plate suddenly put into motion was treated. So we consider a
plate which at time t = 0 is given a sudden acceleration from velocity 0 to velocity uo.
The plate moves in its own plane and of infinite extent. The boundary layer flow over
the plate is assumed to be turbulent for all times t > 0, and the velocity profile is
assumed to be described by
U − u0
1 ⎛ y ⎞
= − ln⎜
⎟
κ ⎝ k / 30 ⎠
Uf
(7.44)
which fulfils the boundary condition U = uo at the bed. At the top of the boundary
layer we have
u0
Uf
=
1 ⎛ δ ⎞
ln⎜
⎟
κ ⎝ k / 30 ⎠
(7.45)
in accordance with the boundary condition U = 0 for y = δ .
Fig. 7.8
The boundary layer along a plate suddenly given the velocity
uo.
By combining (7.449 and (7.45) U can as an alternative be written as
1 ⎛δ⎞
U
= ln⎜⎜ ⎟⎟
Uf
κ ⎝ y⎠
(7.46)
The expression for U is now inserted into the momentum equation (5.20) (see also
(5.42)), which in this case reads
175
δ
∂U
∫ ρ ∂t
dy = τ 0 = ρU 2f
(7.47)
0
the shear stress from the plate is in the positive direction and τ 0 is positive in (7.47).
By differentiation of (7.46) we get
∂U U f dδ dU f U
+
=
∂t
κ δ dt
dt U f
(7.48)
This can be inserted into (7.47), which by integration gives
U f dδ 1 dU f
+
δ = U 2f
κ dt κ dt
(7.49)
An additional relation between Uf and δ can be obtained by differentiation of (7.45)
with respect to time, which gives
⎛ δ ⎞ U f dδ
ln⎜
=0
⎟+
dt
⎝ k / 30 ⎠ δ dt
dU f
(7.50)
If we introduce
Z=
Uf
κ u0
similarly to (7.25) in section 7.1, we can write (7.50) as
dZ Z 2 dδ
=0
+
dt
δ dt
Similarly equation (7.49) can be written
dδ
dZ
Z
+δ
= u0 κ 2 Z 2
dt
dt
(7.51)
(7.52)
176
Fig. 7.9
Variation in δ as function of time, rough bed case.
From (7.51) and (7.52) we may eliminate the term dZ/dt and get
dδ
[1 − Z ] = u 0 κ 2 Z
dt
(7.53)
which together with (7.26) yields a first order differential equation in δ that has to be
solved numerically to describe the growth in δ with time. Figure 7.9 shows results
from such a numerical solution. It is seen that δ grows faster with time the larger the
roughness of the bed.
7.4 Free turbulence: the turbulent jet
In this and the next section we shall look at flows characterized by being far
away from any solid boundaries. Yet the flow is still assumed to have boundary layer
characteristics, which means that the variation across the mean flow direction is much
stronger than the longitudinal variation.
Let us consider a fluid flowing through a circular hole into a space filled with
a stagnant fluid with similar properties (density, viscosity) as the fluid coming from
the hole. For sufficiently large discharges the fluid coming through the hole will form
a submerged turbulent jet. In the following we shall consider the spreading of this jet
at some distance from the orifice, where we use some assumptions regarding the
turbulence. An example of a turbulent jet can be seen in the photograph in figure 7.11.
Just as for the boundary layer formed along a wall there is also an irregular fluctuating
boundary between the turbulence n the jet and the quiescent fluid outside the jet,
which will cause intermittency in the measured turbulence.
177
Fig. 7.10
The spreading of a jet.
When we are at a distance from the orifice (typically 50-100 times its
diameter) the flow in the jet has reached a stage where the velocity profiles can be
taken to be similar, which means the velocity can be described as
⎛r⎞
u = U 0 ( x) f ⎜ ⎟
⎝δ⎠
(7.54)
where Uo is the velocity at the centreline and δ is half the width of the jet. Because of
rotational symmetry it is convenient to use polar coordinates normal to the x-axis
(cylindrical coordinates) and r is the coordinate normal to the x-axis.
To calculate the behaviour of the jet we shall first apply the momentum
equation for boundary layers (5.20), which is written
δ
∂
ρ2πru 2 dr = 0
∂x ∫0
(7.55)
where it has been used that there are no outer forces (shear stresses), the flow is
steady, the outer velocity is zero and there are no pressure gradients.
Fig. 7.11
Photograph of the spreading of a jet.
178
By inserting (7.54) into (7.55) we get
1
∂ ⎡ 2 2 r 2 ⎛ r ⎞ ⎛ r ⎞⎤
δ
U
⎢ 0 ∫ f ⎜ ⎟d ⎜ ⎟⎥
∂x ⎢⎣
δ ⎝ δ ⎠ ⎝ δ ⎠⎥⎦
0
(7.56)
or
U 0 δ = const
(7.57)
Additional assumptions have to be introduced to proceed further in the
analysis of the free jet. This is in contrast to a boundary layer along a wall because
there is no outer force in the form of the wall shear stress to control the development
in the boundary layer.
Let us consider the flow equation for the boundary layer. In equation (7.8) it
was interpreted as the equation of motion for a unit cube, where mass times
acceleration was given at the resulting pressure- and shear stresses in the flow
direction. In the present case the pressure gradient is zero, and the flow equation can in polar coordinates – be written
ρ2πr dr
du ∂
= (τ2πr ) dr
dt ∂r
(7.58)
which is the force balance for a ring-shaped element (torus) with the cross-section of a
unit cube, cf. figure 7.10. Equation (7.58) can be written
u
∂u
∂u 1 ∂ ⎛ r τ ⎞
⎜ ⎟
=
+v
∂x
∂r r ∂r ⎜⎝ ρ ⎟⎠
(7.59)
where v is the velocity on the r-direction. In the boundary layer flow τ is not known,
and we have to introduce assumption no. 2 to utilize equation (8.59). This assumption
is that the distribution of τ across the boundary layer has a given shape which is
similar in the x-direction. As τ / ρ has the dimension (m/s)2 we can now write τ as
τ
⎛r⎞
= U 02 g ⎜ ⎟
ρ
⎝δ⎠
(7.60)
Since τ in a turbulent flow is due to the Reynolds stresses, which are formed
by the eddies, equation (7.60) says that the pattern of the energy containing eddies is
self-similar along the jet axis. For large Reynolds numbers detailed theoretical studies
and experimental investigations has shown that this is in fact the case.
Equation (7.60) can now be inserted into (7.59) expressing v through the
continuity equation for polar coordinates
∂u 1 ∂ (rv)
+
∂x r ∂r
(7.61)
cf. example 2.3. From (7.61) and (7.54) we find that the velocity v at the distance r
from the x-axis is given as
179
r
⎛r⎞
⎛ r ⎞ δ′( x) r
′
r v = − ∫ rU 0 ( x) f ⎜ ⎟ dr + ∫ rU 0 ( x) f ′⎜ ⎟ 2
dr
⎝δ⎠
⎝ δ ⎠ δ ( x)
0
0
r
ζ
ζ
= −U 0′ ( x) δ 2 ( x) ∫ s f ( s ) ds + U 0 ( x) δ( x) ∫ s 2 f ′( s ) δ( x) ds
0
(7.62)
0
= −U 0′ ( x) δ 2 ( x) F (ζ ) + U 0 ( x) δ( x) G (ζ )
where ζ = r / δ has been introduced as a new dimensionless coordinate and F and G
are unknown functions in ζ . Now (7.59) reads
r ⎤ ⎡U 0′ δ 2 F δ′U 0 δ G ⎤
f′
⎡ ′
′
−
U 0 f ⎢U 0 f − U 0 f 2 δ′⎥ − ⎢
⎥U 0
δ
r
δ
⎣
⎦ ⎢⎣ r
⎥⎦
g′
1
= U 02 g + U 02
r
δ
(7.63)
Another relation between U 0′ and δ′ is obtained by differentiation of (7.57), which
gives
U 0′ δ + U 0 δ′ = 0
(7.64)
We now have two rather complicated expressions with U 0′ and δ′ , namely
(7.63) and (7.64). Equation can (by multiplication with δ /U 02 ) be written
U 0′
U0
δ f 2 − δ′f f ′ζ −
U 0′ δ 2
δ
g
F f ′ + δ′G f ′ = + g ′
U0 r
r
ζ
or by use of (7.64)
⎡
F f ′ G f ′⎤ g
δ′⎢− f 2 − f f ′ζ +
+
⎥ = + g′
ζ
ζ ⎦ ζ
⎣
(7.65)
This relation is in the form
dδ
G1 (ζ ) = G2 (ζ )
dx
(7.66)
from which it is easily seen that
180
dδ
= const
dx
(7.67)
because G1 and G2 are functions of ζ only. The width of the jet is thus proportional to
x
δ∝ x
(7.68)
The assumption of self similarity is not valid immediately after the orifice, and a
suitable choice of the origin for the x-axis has to be made for (7.68) to be valid. From
(7.57) and (7.68) we find the corresponding variation of the velocity at the centreline
to be
U 0 ( x) ∝
1
x
(7.69)
More detailed information on the characteristics of the jet, such as velocity
profiles and the distribution of the shear stress, can be obtained by use of the technical
turbulence theory if we introduce the eddy viscosity ν t . For the flow along a wall
(section 6.4) we found according to (6.41) that
2
τ/ρ U f
y⎞
⎛
νt =
=
= κU f y⎜1 − ⎟
du
Uf
⎝ D⎠
dy
κy
(7.70)
It is thus seen that the eddy viscosity close to the wall grows linearly with the distance
y from it. In a free jet there is no solid boundary damping the free turbulent motions
and measurements does show that it is a good approximation to take ν t to be constant
across the entire width of the jet. Generally ν t is proportional to a typical velocity
scale (here: Uo) and to a typical length scale (here: δ ), so ν t may be written
νt ∝ U 0δ
(7.71)
According to (7.57) this product is constant, and we can therefore also expect ν t to be
constant and independent of x.
The most convenient way to proceed is by application of (7.68) and (7.69) so
that u can be written
⎛r⎞ U δ ⎛r⎞ k ν
⎛ r⎞
u = U 0 f ⎜ ⎟ = 0 f ⎜ ⎟ = 1 t f1 ⎜ k ⎟
δ ⎝δ⎠
x
⎝δ⎠
⎝ x⎠
(7.72)
=
ν t k1r ⎛ kr ⎞ ν t
f1 ⎜ ⎟ =
H ′(ζ )k
r x ⎝ x⎠ r
181
where the new function
H ′(ζ ) =
k1
k
2
ζ f 1 (ζ )
where
ζ=
kr
x
has been introduced. By inserting into the continuity equation (7.61) it is seen that
v=
νt
(ζH ′ − H )
r
(7.73)
is the solution for the radial velocity v.
By inserting (7.72) and (7.73) together with
ν t2
∂u ν t2 k 2
τ
H ′′ − 2 H ′k
=νt
=
ρ
r x
∂r
r
(7.74)
into the flow equation (7.59) we get the following ordinary differential equation in H
HH ′ ( H ′) 2 HH ′′ d ⎡
H ′⎤
−
−
=
⎢ H ′′ −
⎥
2
ζ
ζ
dζ ⎣
ζ ⎦
ζ
(7.75)
This can be integrated to give
HH ′ = H ′ − ζH ′′
(7.76)
which has the solution
H=
ζ2
1+ ζ2 / 4
(7.77)
The velocity profile can the be found by differentiation of (7.77), which gives
1
ζ
u∝
r (1 + ζ 2 / 4) 2
or
u∝
1
1
x (1 + ζ 2 / 4) 2
(7.78)
Since Uo is the velocity at the centreline (7.78) can be written
u=
U0
(1 + ζ 2 / 4) 2
(7.79)
Which give a bell-shaped profile as shown in figure 7.12.
182
Fig. 7.12
Measurements of the velocity profile, compared to equation
(7.79).
As seen from figure 7.12, the measurements are in good agreement with (7.79)
except at the outer edges of the jet where there is intermittence and the assumption of
ν t being constant is a crude assumption.
7.5 Free turbulence: the plane wake
In this section we consider the flow field downstream of a cylinder which is
kept in a fixed position. Far upstream of the cylinder the flow is a parallel flow with
the constant velocity uo, cf. figure 7.13. Close to the cylinder the flow is strongly
disturbed as described in example 5.3.
For example in most cases there will be vortex shedding downstream of the
cylinder. These vortices will die out when we are sufficiently far downstream from
the cylinder (approximately 100 timed the diameter or more), but around the x-axis
the flow will still be affected by the cylinder, and the velocity at the centreline will be
smaller than uo. The cylinder thus creates a lee zone for the current, and this is the
flow field we shall consider in the following.
Fig. 7.13
The flow around a circular cylinder: definition of control
surfaces for the momentum equation.
As we have done before, we can start with formulating the momentum
equation. This is used for the control surface A'BCD', which is quite analogous to the
control surface shown in figure 5.3 used for the derivation o fthe integrated
momentum equation for boundary layers, equation (5.20).
In our case there are no shear stresses, and (5.20) can therefore be written
183
δ
δ
∂
∂
ρu 2 dy − u 0
ρu dy = 0
∫
∂x −δ
∂x −∫δ
(7.80)
or, using that uo is constant
δ
∫ u (u − u 0 ) dy = const
(7.81)
−δ
At this stage we must (as in the previous example) introduce additional
assumptions to get further with the analysis. The first assumption is that the velocity
profiles are affine or self similar, which can be written as
⎛ y⎞
u0 − u = U 0 f ⎜ ⎟
⎝δ⎠
(8.82)
where Uo is the maximum velocity deficit at the give position at the x-axis, cf. figure
7.13. Inserting (7.82) into (7.81) gives
δ
∫U 0
−δ
⎛ y⎞
f ⎜ ⎟ dy = const
⎝δ⎠
(8.83)
where it has been assumed that Uo/uo is small (i.e. the velocity reduction in the wake
this far downstream is small), so that the term (Uo/uo)2 may be neglected. Equation
(7.83) can be written
1
U 0 δ ∫ f (ζ ) dζ = const
−1
which gives that
U 0 δ = const
(7.84)
To determine the variation both in Uo and in δ with x we must (as also done
in section 7.4) introduce the assumption of affinity (self similarity) in the distribution
of the shear stress distribution
τ
⎛ y⎞
= −u ′v ′ = U 02 g ⎜ ⎟
ρ
⎝δ⎠
(7.85)
This expression is now inserted into the flow equation for boundary layers,
which in this case can be written
184
u
∂u
∂u ∂ ⎛ τ ⎞
+v
= ⎜ ⎟
∂x
∂y ∂y ⎜⎝ ρ ⎟⎠
(7.86)
Her v is found from the equation of continuity
y
∂u
dy
∂
y
0
v = −∫
(7.87)
From (7.82) it is seen that v must be proportional to Uo, and similarly the term
∂u / ∂y . By neglecting higher order terms in Uo/uo equation (7.87) may be
approximated as
u
∂u ∂ ⎛ τ ⎞
= ⎜ ⎟
∂x ∂y ⎜⎝ ρ ⎟⎠
(7.88)
which by inserting (7.82) and (7.85) can be written
U0 y
U 02
⎛
⎞
′
′
′
U 0 ⎜U 0 f −
f δ ⎟=−
g′
δ δ
δ
⎝
⎠
(7.89)
From (7.84) we find by differentiation
U 0′ δ + U 0 δ′ = 0
(7.90)
From this δ′ can be expressed and then eliminated from (7.89), which now can be
written
U 0′ [ f + ζ f ′] = −
U 03
U 0δ
g′
(7.91)
or
U 0′
U 03
=−
g′
1
U 0δ f + ζ f ′
(7.92)
Since U 0 δ is constant the right hand side of (7.92) is only a function of the
dimensionless parameter ζ = y / δ , while the left hand side is only a function of x.
This means that both must be constant, which gives
U0 ∝
1
x
(7.93)
and
δ∝ x
(7.94)
185
If we finally want to know the shape of the velocity profile, this can be
obtained by introducing the eddy viscosity ν t , which may be taken to be constant.
Now the flow equation (7.88) reads
∂u ∂ ⎛ ∂u ⎞
∂ 2u
⎜
⎟
u
=
νt
= νt 2
∂y ∂y ⎜⎝ ∂y ⎟⎠
∂y
A solution to (7.95) fulfilling the boundary conditions u → u 0
(7.95)
for
⎛ u0 y 2 ⎞
⎟
exp⎜⎜ −
u0 − u = K
⎟
4
νt x
ν
x
t ⎠
⎝
u0
x, y → ∞ is
(7.96)
where K is a constant. This velocity profile is shown in figure 7.14. The eddy
viscosity has the dimension m2/s, and experiments have shown that the velocity
profile (7.96) shows the best agreement if ν t is put to
ν t = 0.016 u 0 d
(7.97)
where d is the diameter of the cylinder. The integrated velocity deficit
δ
∫ (u 0 − u ) dy
(7.98)
−δ
has also the dimension m2/s and is constant. It can be shown that a more general
expression for ν t may be based on the velocity deficit, writing ν t as
δ
ν t = β ∫ (u 0 − u ) dy
(7.99)
−δ
where β is a dimensionless constant.
186
Fig. 7.14
The distribution of velocity and shear stress across a plane
wake.
Example 7.3: The force on a cylinder in a turbulent flow.
The force acting on the cylinder from the flow can be determined by applying
the general momentum equation (5.9) on the control surface ABCD shown in figure
7.13. Here AD is placed so far upstream of the cylinder that the velocity over the
entire section can be taken to be the undisturbed velocity uo. The momentum equation
(5.9) reads
G
G G G
ρ
(
⋅
)
=
−
v
v
d
A
F
∫
(7.100)
A
G
where F is the force acting on the cylinder from the fluid. We now look at the xcomponent of the momentum equation and calculate each term on the left hand side of
(7.100). From section AD we get
δ
− ∫ ρ u 02 dy
(7.101)
−δ
From BC we get
δ
∫ρu
2
dy
(7.102)
−δ
From the two sections AB and DC the sum of the contributions is
ρ Q u0
(7.103)
because the velocity here is undisturbed uo because we are outside the boundary layer.
Q is the total volume flux through the two sides, which is there because of weak
187
vertical velocities in the outer flow due to entrainment, cf. section 7.4. Q is found as
the difference between the volume flux through AD and BC
δ
δ
−δ
−δ
Q = ∫ u 0 dy − ∫ u dy
(7.104)
By inserting (7.101)-(7.104) into (7.100) we get
δ
−F =
∫ [− ρ u0 + ρ u
2
2
]
+ ρ u 0 (u 0 − u ) dy
(7.105)
−δ
or
δ
f = ρ ∫ u (u − u 0 ) dy
(7.106)
−δ
This expression is according to (7.81) a constant, which it should be because
the force F cannot be dependent on the position of the sections of the control surface.
Equation (7.106) can be linearized to give
δ
f = ρ u 0 ∫ (u − u 0 ) dy
(7.107)
−δ
which by using (7.97) and (7.99) can be written
F ∝ ρ u 02 d
(7.108)
It is thus seen that the force is proportional to the square of the undisturbed
velocity and proportional to the diameter of the cylinder.
188
8. Turbulence modelling
In the two previous chapters we have seen how turbulent flows may be described
in more or less detail, depending on how much information we require about the flow.
Since it is not feasible to solve the complete Navier-Stokes equations numerically
in the turbulent case because of the presence of the very small eddies (of the scale where
dissipation to heat takes place), it is necessary in stead to introduce physically plausible
assumptions to construct an approximate model of the flow. This is actually what has
been done in the last two chapters, where the concepts of mixing length and eddy
viscosity have been introduced.
8.1 Mixing length
The concept of the mixing length is explained in section 6.4. The basic idea is that
eddies transfer momentum across the plane parallel to the direction of the mean flow.
This exchange of momentum is equivalent to a shear force, which as shown in section 6.4
can be written as
τ = ρ A2
∂U ∂U
∂y ∂y
(8.1)
The eddy viscosity may be defined through the expression
τ = ρν t
∂U
∂y
(8.2)
From which we seen that in a turbulent flow the eddy viscosity and the mixing length is
related as
∂U
ν t = A2
(8.3)
∂y
By combining (8.1) and (8.3) we get alternatively
νt = A
τ
ρ
(8.4)
This expression will later be used in section 8.2.
Turbulence modelling is now aiming at making as good a representation of the
length scale A as possible. The models are to a large degree based on dimensional
193
considerations and interpretation of experimental data. In the following some simple
ways of representing A in different flow situations are described.
Boundary layer along a wall: Disregarding a possible viscous sublayer at the
wall, A will grow linearly with the distance from the wall:
A =κ y
for
y <δ
(8.5)
which was also applied in chapter 6. For some cases rthe description can be improved by
restricting the linear growth of A to a certain fraction λ of the boundary layer thickness,
outside which A is taken to be constant. This can be formulated
A =κ y
A = κλδ
for
for
y < λδ
y > λδ
(8.6)
where we may take λ ≈ 0.25 .
Turbulent pipe flow: Here A can be take to be parabolically distributed, described
as
2
A
y⎞
y⎞
⎛
⎛
= 0.14 − 0.08⎜1 − ⎟ − 0.06⎜1 − ⎟
D
⎝ D⎠
⎝ D⎠
4
(8.7)
Where D is the pipe diameter. In equation (8.7) A varies as A = 0.4 y (= κy ) for small
values of y.
Plane jet in a waterbody at rest: In this flow the mixing length varies only weakly
across the jet. Measurements has given the value
A = 0.2δ
where δ is half the width of the jet. The width and therefore also the mixing length
increases in the flow direction.
Example 8.1: The turbulent wave boundary layer.
The turbulent wave boundary layer is the flow found immediately above the sea
bed, because the wave-induced oscillatory flow has to obey the no-slip boundary
condition at the bed.
194
Fig. 8.1
The turbulent wave boundary layer.
We treated the laminar case in example 2.5, where we considered the flow generated by
oscillating a plate in its own plane. In that case a small part of the fluid (with a thickness
of 2ν / ω , cf equation 2.54) was taking part in the oscillatory motion. Under waves it is
the fluid that oscillates back and forth
over the immobile plate (the sea bed), cf. figure 8.1. For a laminar flow the solution to
this case completely similar to flow treated in section 2.5.
We now consider the turbulent wave boundary layer. Outside the boundary layer,
away from the bed the shear stress can be neglected, so the flow equation can be written
ρ
dU 0
∂U 0 ∂U 0 ⎞
∂U 0
∂p
⎛
=−
= ρ ⎜U 0
+
⎟≈ρ
dt
∂x
∂t ⎠
∂t
∂x
⎝
(8.8)
Here U0 is the horizontal velocity outside the boundary layer. The boundary layer is thin
compared to the depth, and we are so close to the bed that vertical velocities can be
neglected even outside the boundary layer.
Inside the boundary layer the horizontal velocity is U, and here shear stresses are
present so that the flow equation reads
ρ
∂U
∂p ∂τ
=− +
∂t
∂x ∂y
(8.9)
The pressure p is transferred unchanged from the outer flow to the boundary layer, and
the pressure gradient in (8.9) can therefore be taken from (8.8) giving
ρ
∂ (U 0 − U )
∂τ
=−
∂t
∂y
(8.10)
Now it is convenient to introduce the concept of velocity deficit, defined as
Ud = U −U0
(8.11)
195
As shown in figure 8.1. By use of Ud (8.10) can be written
ρ
∂U d ∂τ
=
∂t
∂y
(8.12)
The distribution of the shear stress τ over the boundary layer is not known, and (8.12)
cannot be solved directly. By introducing the mixing length concept from (8.1) a relation
between τ and U can be established, so that τ can be eliminated.
Equation (8.1) can be written
∂U d ∂U d
τ = ρA 2
(8.13)
∂y ∂y
because there are no shear stresses in the outer potential flow. Now the combination of
(8.12) and (8.13) gives
∂U d
∂ ⎛ ∂U d ∂U d ⎞
⎟
= ⎜⎜ A 2
∂t
∂y ⎝
∂y ∂y ⎟⎠
(8.14)
This partial equation in Ud can be solved numerically if A is given. The simplest
formulation to use is (8.5). The boundary conditions are as follows: at the top of the
boundary layer the velocity deficit Ud must tend to zero
Ud → 0
for
y→∞
(8.15)
If the bed is hydraulically rough, the velocity must be zero for y = kN/30, or
U d = −U 0
for
y = k N / 30
(8.16)
where kN is the equivalent bed roughness.
196
Fig. 8.2
The instantaneous velocity profiles in a turbulent boundary layer.
Finally the flow must be periodic with the period T, i.e.
U d (t ) = U d (t + T )
(8.17)
Figure 8.2 shows the variation of U over half a wave period.
The variation of the eddy viscosity in time and space may be found from (8.3),
which together with (8.5) gives
νt = κ 2 y2
∂U d
∂y
(8.18)
The time variation in ν t at two different levels is shown in figure 8.3. It is seen
that ν t increases with the distance from the bed, but that ν t twice every period becomes
zero. This is because the velocity gradient due to the oscillatory nature of the flow has to
cross zero twice every wave period.
8.2 The 1-equation model (k-model)
The mixing length model described ion section 8.1 has two obvious shortcomings.
a) Equation (8.18) means that ν t is zero whenever the velocity gradient is zero.
This for example not correct at the surface in an open channel flow,
whereν t at the surface is about 80% of the maximum value. In the oscillatory
boundary layer we also saw that ν t was zero twice during a wave cycle. Since
197
new turbulent eddies are produced all the time and it takes some time for the
eddies to dissipate, it should be expected that ν t in reality always is positive
in this boundary layer.
b) No advection or diffusion processes of the turbulent energy are included in the
description: the local level of the turbulence is normally not determined only
by what is happening at the place in question, but depends also on what has
happened some distance upsteam. In a recirculating flow processes occurring
downstream may also be of significance
c)
Fig. 8.3
The variation in ν t for a / k N = U 1 / ω k N = 10 2 and 10 3 . Broken
line: 1-equation model. Fully drawn curve: mixing length model.
A description of these processes requires the representation of memory effects in
the turbulence. The strength of the turbulence can be characterized by the turbulent
kinetic energy, defined as
k = 12 (u ′ 2 + v ′ 2 + w′ 2 )
(8.19)
198
cf. equation (6.80). The turbulence may now be described in more detail by adding an
equation for the variation in the turbulent kinetic energy in the turbulent flow. The
variation in k in the flow is described
by the so-called k-equation, which is a transport equation for k, keeping track of the
production of k (i.e. how many eddies are formed), the dissipation of k, and finally the
transport of k to or away from the actual location. Such a transport equation may be
formulated by taking the Navier-Stokes equation (A8) for each coordinate direction,
multiply each equation with the corresponding turbulent velocity component ui and
finally adding all three equations. This will give
d (U i + u i )
∂ 2 (U i + u i )
∂p
ρ ui
= −u i
+ ui ρ g i + ui μ
dt
∂xi
∂x j ∂x j
(8.20)
Equation (8.20) is time-averaged. The term on the left hand side is time-averaged as
follows
ui
d (U i + u i )
∂ (U i + u i )
∂ (U i + u i )
= ui
+ u i (U j + u j )
dt
∂t
∂x j
= ui
∂U i
∂u
∂U i
+ ui i + ui U j
∂t
∂t
∂x j
∂U i
∂u
∂u
+ ui u j
+ U j ui i + ui u j i
∂x j
∂x j
∂x j
= 0+
(8.21)
∂U i
∂k
∂k
∂k
+ 0 + ui u j
+U j
+uj
∂t
∂x j
∂x j
∂x j
When calculating the last term it has been used that the continuity equation must also
hold for the turbulent velocity fluctuations:
ui, j = 0
(8.22)
The whole time-averaged equation (8.20) now becomes
ρ
∂U i
∂k
∂k
∂
+ ρU j
+ρ
(u j k ) + ρu i u j
∂t
∂x j
∂x j
∂x j
∂ 2ui
∂p
= −u i
+ μ ui
∂xi
∂x j ∂x j
or
199
∂U i
∂ 2 ui
∂
dk
=
−
+
−
+
u
p
u
k
u
u
u
ρ
( j
ρ j ) ρ i j
μ i
∂x j
∂x j
dt
∂x j ∂x j
(8.23)
For boundary layer flow equation (8.23) can be reduced to
ρ
∂ 2ui
∂U
dk
∂
+ μ ui
= − (v ′p ′ + ρ v ′k ′) − ρ u ′v ′
∂x j ∂x j
∂y
dt
∂y
I
II
III
IV
(8.24)
V
Where u’ and v’ are the longitudinal and transverse velocity fluctuations respectively.
The parameter p’ is the fluctuating part of the pressure and k’ is the fluctuating part of the
turbulent kinetic energy.
The five terms in (8.24) can be interpreted as follows:
I is the total rate of change of the turbulent energy for a fluid particle.
II & III can be interpreted as net diffusion of energy. II is expressing the change
in energy due to the work done by the fluctuating pressure, while III represents the net
transport of energy to the particle due to the turbulent fluctuating velocity field. The
diffusive character of these terms can be illustrated by integration of (8.24) over the
boundary layer thickness. If we consider a flow along a wall with a free stream flow with
no turbulence above it the integrals of II and III will be zero, so these terms do not
contribute to a change in the total energy level, but redistribute the energy across the
boundary layer.
IV describes the production of turbulent energy due to the transfer of energy from
the mean flow, cf. example 6.4, page 142.
Finally V represents a drain of the turbulent energy due to dissipation to heat.
Application of (8.24) requires that the terms on the right hand side are
approximated by use of parameters which are known or can be determined.
The diffusive terms II-III are often assumed to be modelled analogously to a
normal gradient diffusion process by
− (v' p' + ρ v' k ') = ρ
ν t ∂k
σ k ∂y
(8.25)
whereν t is the eddy viscosity. This expression is analogous to the second term in (6.78)
which describe the distribution of suspended sediment. There the term represented the
redistribution of sand caused by the turbulent eddies.
The term IV in (8.24) is well known because − ρ u 'v' is the shear Reynolds stress,
giving
− ρ u ′v ′
∂U
∂U
=τ
∂y
∂y
(8.26)
200
here τ may be modelled in several ways. We can use the concept of eddy viscosity
which gives
τ = ρν t
∂U
∂y
(8.27a)
But there are also other possibilities. As mentioned in chapter 6 it is for open channel
flow found that k is proportional toτ over a large part of the cross-section. This is used in
some turbulence models, because it has turned out that the relation
τ = 0.3ρ k
(8.27b)
fits well with experiments. Equation (8.27b) has the advantage relative to (8.27a) that the
local shear stress is not necessarily zero where the velocity gradient is zero.
Finally the dissipation of energy (term V in (8.24)) is found from (6.86) as
μ ui
∂ 2 ui
k 3/ 2
= − Aρ
Ad
∂x j ∂x j
(8.28)
where A from measurements are found to be about 0.08. If the eddy viscosity is used to
model τ , equation (8.24) now reads
2
⎛ ∂U ⎞
∂ ⎛ ν ∂k ⎞
k 3/ 2
dk
⎟⎟ + ρν t ⎜⎜
⎟⎟ − Aρ
= ρ ⎜⎜ t
ρ
∂y ⎝ σ k ∂y ⎠
dt
Ad
⎝ ∂y ⎠
(8.29)
In this equation there are three unknows, namely k, ν t , and U, and we therefore need two
additional equations to describe the flow. One of these is the flow equation
⎛ ∂U ⎞
⎜ν t
⎟
(8.30)
⎜ ∂x j ⎟
⎝
⎠
Here the pressure in the boundary layer is as usual determined from the potential flow
outside the boundary layer.
The other equation correlates the eddy viscosity to the turbulent kinetic energy k.
The eddy viscosity has the units of m2/s and can be formulated as the product of a length
scale and a velocity scale. A typical velocity scale in a turbulent flow is k , which is the
square root of the mean square of the velocity fluctuations. k thus represents a typical
velocity in the most energy containing eddies. A measure for the size of these eddies is
the length scale A d , cf. equation (6.86). The eddy viscosity can now be written
∂
1 ∂p
dU
+
=−
dt
ρ ∂x ∂x j
νt = Ad k
(8.31)
201
Combined, equations (8.29), (8.30) and (8.31) are sufficient to determine k and U.
Example 8.2: The k-equation in the equilibrium layer.
In example 6.4 we considered the energy balance in a channel- or pipe flow. In
general this balance expressed that the formation of eddies (the production of turbulent
kinetic energy) corresponds to the decay of eddies due to the heat loss (dissipation) and
dispersion of eddies (diffusion). Finally eddies can be transport by advection. In the
equilibrium layer, which is found in the lower 10-20% of the flow (cf. figure 6.17), the
dominating terms are the production and the dissipation, which approximately balance
each other out. From this we get that the two terms IV (given by (8.26)) and V (given by
(8.28)) must balance each other, giving
⎛ ∂U
ν t ⎜⎜
⎝ ∂y
2
⎞
k 3/ 2
⎟⎟ = A
Ad
⎠
(8.32)
Application of the eddy viscosity (8.27a) in (8.32) gives
2
⎛τ ⎞
k 3/ 2
⎜⎜ ⎟⎟ = ν t A
= Ak 2
Ad
⎝ρ⎠
(8.33)
τ = ρ Ak
(8.34)
or
which is seen to be identical to (8.27b). In the equilibrium layer there is therefore no
contradiction between the two different ways of modelling τ (by (8.27a) and (8.27b)).
The distribution of τ can be related to A d and the velocity gradient as follows:
From the eddy viscosity we have
2
2
⎛ ∂U ⎞
⎛ ∂U ⎞
⎛τ ⎞
⎟⎟ = A 2d k ⎜⎜
⎟⎟
⎜⎜ ⎟⎟ = ν t2 ⎜⎜
⎝ρ⎠
⎝ ∂y ⎠
⎝ ∂y ⎠
2
(8.35)
where (8.31) has been applied. By then inserting
k=
τ 1
ρ A
(8.36)
from (8.34) into (8.35) we get
202
τ
1 2 ⎛ ∂U ⎞
⎟⎟
=
A d ⎜⎜
ρ
A ⎝ ∂y ⎠
2
(8.37)
As seen this expression is equivalent to the mixing length model (6.39)
τ
∂U ∂U
= A2
ρ
∂y ∂y
(8.38)
The mixing length theory can therefore be taken as a turbulence model for domains with
local equilibrium. Further it is seen that the mixing length A and the length scale A d are
closely related, since (8.37) and (8.38) gives
A d = A1 / 4 A ≈ 0.53 A
Example 8.3: The turbulent wave boundary layer modelled by a one-equation
model.
This example is a continuation of example 8.1, where the turbulent wave
boundary layer was modelled by use of the mixing length theory. We thus once more
consider the flow shown in figure 8.1, now with the following system of equations at our
disposal
A:
The flow equation, cf. (8.12)
ρ
∂U d ∂τ
∂ ⎛ ∂U d ⎞
⎟
= ρ ⎜⎜ν t
=
∂y ⎝
∂y ⎟⎠
∂y
∂t
(8.39)
where Ud is the deficit velocity, cf. figure 8.1.
B:
The transport equation for turbulent kinetic energy
⎛ ∂U d
∂ ⎛ ∂k ⎞
∂k
⎟⎟ + ν t ⎜⎜
= ⎜⎜ν t
∂t ∂y ⎝ ∂y ⎠
⎝ ∂y
2
⎞
k 3/ 2
⎟⎟ − A
Ad
⎠
(8.40)
where A d in the boundary layer is given by
A d = 0.53A = 0.53κy = 0.21y
C:
(8.41)
The eddy viscosity is given by
203
νt = Ad k
(8.42)
These three equations form a complete system for the three unknown: Ud,ν t ,
and k.
The boundary conditions are that U close to the bed is described by the familiar
logarithmic distribution
U→
⎛ y ⎞
⎟
ln⎜⎜
κ ⎝ k n / 30 ⎟⎠
Uf
for
y→0
(8.43)
where kn is the roughness of the bed. At the top of the boundary layer the velocity shall
approach the outer velocity
U (δ ) = U 0
(8.44)
which also imply that the gradient must be zero
∂U
=0
∂y
for
y =δ
(8.45)
In addition two conditions for k are required. Very close to the bed we may assume local
equilibrium between the production of energy and the dissipation. Then (8.33) can be
applied to give the boundary condition
k=
1
k
νt
∂U
∂y
for
y=
kn
30
(8.46)
Finally there is no transport of turbulence out to the outer potential flow. Since the
transport is expressed through gradient diffusion this can be written
∂k
=0
∂y
for
y =δ
(8.47)
This system of equations must be solved numerically, for example by use of a finite
difference scheme. In practice the thickness of the boundary layer is not know in
advance, but since it only enters via the boundary conditions these are just applied at the
top of the model domain.
The results of this calculation are then as follows: The velocity profiles are very
close to those we obtained in example 8.1, cf. figure 8.2. On the other hand the oneequation model gives a more realistic description of the eddy viscosity and the turbulent
kinetic energy. In figure 8.3 the dotted curves show how the one equation model predicts
the variation in the eddy viscosity during a wave period. It is seen that ν t now is always
204
positive, which describes how the eddies do not have time to die out completely before
new turbulent kinetic energy is produced in the wave motion. It is further seen that for
very rapid oscillations (small values of a/kN) ν t is not far from being constant in time
when we are just at a certain distance from the bed. Figure 8.4 shows the distribution of
the turbulent kinetic energy in the boundary layer under a wave. Since the advective
terms are not included in the analysis we can actually only calculate how k varies over
the vertical at a given location.
The picture in figure 8.4 is then made from the temporal variation by using the relation
for progressive waves
∂
1 ∂
=−
∂x
c ∂t
Fig. 8.4
(8.48)
The distribution of kinetic energy in the wave boundary layer (a/kN
= 103).
It is seen from figure 8.4 that while close to the bed k is large when the free stream
velocity is large, there is a certain delay in the maximum in k further away from the bed.
This is because there is a transport of k away from the bed, and this transport implies a
delay.
205
8.3 2-equation models (k-epsilon model)
The advantage by the one-equation model is as described in section 8.2 that k ,
which is a good measure of the turbulent velocity scale, is calculated by use of a transport
equation so that memory effects in k are included. However, it must be noted that the
turbulent
Length scale is not determined from a transport equation. Memory effects in A d can be
just as important as the effects in k, why a one-equation model often is found not to give
a better description of a flow situation than the simple mixing length theory.
In 2-equation models the kinetic energy k as well as the length scale A d are
determined from a transport equation. The eddy viscosity is then found from
νt = Ad k
(8.49)
cf. (8.31).
Often it is not A d which is calculated directly, but some quantity called z, which
is the product of k and A d raised to different powers
z = k m (A d ) n
(8.50)
The length scale A d is then determined from k and z. It is not convenient to calculate
A d directly because the diffusion of the quantity A d cannot be taken to be proportional to
∂(A d ) / ∂y .
The most frequently used combination of m and n in (8.50) is m = 1.5 and n = -1,
which gives
z=
k 3/ 2
Ad
(8.51)
so that z becomes proportional to the dissipation ε , cf. equation (6.86). An equation for
ε can be developed quite analogously to the k-equation (8.23) on the basis of the NavierStokes equation.
Generally a z-equation will read
dz ∂ ⎛ ν t ∂z ⎞ ⎡ ν t
⎟ + z ⎢c1
= ⎜
dt ∂y ⎜⎝ σ z ∂y ⎟⎠ ⎢ k
⎣
2
⎛ ∂U ⎞
k⎤
⎟⎟ − c 2 ⎥ + S z
⎜⎜
νt ⎥
⎝ ∂y ⎠
⎦
(8.52)
this equation may be compared to the k-equation (8.29), which can be written
206
⎡ν
dk
∂ ⎛ ν ∂k ⎞
⎟⎟ + k ⎢ t
= ⎜⎜ t
dt ∂y ⎝ σ k ∂y ⎠
⎢⎣ k
2
⎛ ∂U ⎞
k⎤
⎟⎟ − A ⎥
⎜⎜
νt ⎥
⎝ ∂y ⎠
⎦
(8.53)
In (8.52) the constants σ z , c1 and c2 and the term Sz must be modelled. Reference is
made to the specialised literature, e.g [5].
Example 8.4: The turbulent wave boundary layer determined by the 2-equation
model.
Finally we calculate the turbulent wave boundary layer (cf. examples 8.1 and 8.3)
by use of a k − ε model, wher the equation for k is combined with an equation for the
dissipation ε . The ε -equation is obtained from (8.52), which for
z=
k 3/ 2
Ad
(8.54)
and using
νt = Ad k
(8.55)
can be written (after multiplication by A (=0.08))
εν
∂ ⎛ ν ∂ε ⎞
dε
⎟⎟ + c1 t
= ⎜⎜ t
dt ∂y ⎝ σ ε ∂y ⎠
k
2
⎛ ∂U ⎞
ε2
⎜⎜
⎟⎟ − c1c 2
k
⎝ ∂y ⎠
(8.56)
The constants in this equation can according to [5] be chosen as c1 = 1.41, c2 = 1.92 and
σ ε = 1.3.
The k-equation can, cf. (8.40), be written
⎛ ∂U
∂ ⎛ ∂k ⎞
∂k
⎟⎟ + ν t ⎜⎜
= ⎜⎜ν t
∂t ∂y ⎝ ∂y ⎠
⎝ ∂y
2
⎞
⎟⎟ − ε
⎠
(8.57)
Finally we have the flow equation (8.39)
∂U d
∂ ⎛ ∂U d
= ⎜⎜ν t
∂y ⎝
∂y
∂t
⎞
⎟⎟
⎠
(8.58)
207
where
Ud = U - Uo
(8.59)
Equations (8.56), (8.57) and (8.58) is a complete set of equations for the three unknowns
ε , k and U. The equations have to be solved numerically with appropriate boundary
conditions. For the velocity and the turbulent kinetic energy k the boundary conditions
are the same as in example 8.3: equations (8.43), 8.44) and (8.45) for U and (8.46) and
(8.47) for k.
For ε expressions analogous to those for k can be derived. The condition for
local equilibrium near the bed will together with the logarithmic velocity distribution thus
imply that the boundary condition for ε becomes
ε = ( A) 3 / 4
k 3/ 2
κy
for
y = k N / 30
(8.60)
At the top of the boundary layer the gradient in ε is taken to be zero, that is
∂ε
=0
∂y
for
y =δ
(8.61)
A numerical result from the analysis is shown in figure 8.5, which shows the
variation in the length scale in time for a specific value of the parameter a/kN. In the
mixing length
Fig. 8.5
Time-variation in the turbulent length scale (a/kN = 53) from [6].
length theory and the one-equation model the length scale A increased with the distance
from the bed, but was constant in time. From figure 8.5 it is seen that the k- ε model
predicts a distinct time-variation of A . At present there are no measurements available to
confirm the theoretical predictions shown in figure 8.5.
208
The variation in A implies that the theoretical mean level for k becomes smaller
when the k- ε is applied in stead of the k-model, cf. figure 8.6. This is because the kequation (where k is taken to be constant) predicts less dissipation in the outer parts of the
boundary layer because the dissipation is inversely proportional to A .
On the other hand the k- ε model predicts practically the same velocity
distribution in the boundary layer as the k-model and the mixing length model do.
Fig. 8.6
The mean level in the turbulent kinetic energy, calculated from the
1- and the 2-equation model. a/kN = 1000. From [6].
209
Index
Acceleration
Advective terms
Affine boundary layer
Angular velocity
Area vector
Axi-symmetric flow
12
14
117
41
17
11
Bernoulli equation
Boundary layer
Laminar
Turbulent
Boundary layer equation
Laminar
Turbulent
45
102
165
103
166
Cascade process
Circular cylinder
Continuity equation
Convective terms
Converging flow
Correlation coefficient
Constitutive equation
Creeping flow
Current induced pressure
Current, wave-induced
Cyclic frequency
160
123, 187
34,213
14
120
163
214
127
124
96
41
Deformation
Diffusion, turbulent
Displacement thickness
Dissipation length
Dissipation
Divergence
Diverging flow
31
156
115
159
31, 215
35
120
Eddy viscosity
Energy equation, turbulent
Energy flux
Entrainment
Equilibrium layer
Error function
146
201
82
175
162, 203
111
Flow equations
Fluctuations
Free turbulence
Froude number
186, 212
138
179
55
Group velocity
74
Hydraulic gradient
Hydrostatic pressure
40
23
Ideal fluid
Intermittence
44
166
Jet, turbulent
179
k- ε -model
k-model
207
198
Laplace equation
Length scale for turbulence
47
159
Maintained turbulence
Mixing length
Momentum equation
for boundary layers
Momentum force
157
143, 193
Navier-Stokes equations
Newton's formula
Normal stress
37, 166, 214
28
17
One-equation model
198
p+
Particle path
Particle paths in waves
Perturbation
Plane flow
39
9
72
47
7
105
86
Pohlhausen's method
Poiseulle flow
Polar coordinates
Potential
Power
Pressure force
Pressure
Pressure, current induced
Production of energy
120
38
35
44
29
86
19
124
158
Reaction force
Refraction
Reynolds experiment
Reynolds number
Reynolds stresses
RMS-value
Root mean square
Rough bed
86
70
129
136
139
162
162
152
Self similar boundary layers
Separation
Set-down
Set-up
Shallow water theory
Shallow water wave
Shear stress
Sinusoidal waves
Smooth bed
Standard fluctuation
Stationary flow
Stokes length
Stream line
Strouhal number
Sudden movement of plate
Laminar flow
Turbulent flow
Suspended sand
117
124
90, 92
90, 95
60
61
16
60
147
162
8
43
65
127
Technical turbulence theory
Translation
Translatory motion
Turbulence length scale
Two-equation model
145
32
32
159
207
Uniform flow
15
109
177
155
Viscosity, dynamic
Viscosity, kinematic
Viscous sublayer
Vorticity transport equation
28
28
149
132
Wake, plane
Wave boundary layer
Laminar
Turbulent
Wave energy flux
Wave energy
Wave groups
Wave length
Wave number
Wave period
Wave table
Wave-induced current
Wind –driven flow
186
40
195
82
75
72
50
50
60
84
96
173
References
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[1]
Engelund, F.: Hydrodynamik. Den prvate Ingeniørfond, Danmarks Tekniske
Højskole, 1968.
[2]
Pedersen, Fl. Bo: Hydraulik for bygningsingeniører. Den private Ingeniørfond,
Danmarks Tekniske Højskole, 1988.
[3]
Svendsen, I. A. and Jonsson, I. G.: Hydrodynamics of Coastal Regions. Den
private Ingeniørfond, Danmarks Tekniske Højskole, 1976.
[4]
Van Dyke, M.: An Album of Fluid Motion. The Parabolic Press, Stanford,
CA, 1982.
[5]
Launder, B. E. and Spalding, D. B.: Mathematical Models of Turbulence.
Academic Press, London, 1972.
[6]
Justesen, P.: Turbulent wave boundary layers. Series Paper No. 43, ISVA,
Danmarks Tekniske Højskole, 1988.