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Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.15. Derive the von Karman boundary
layer integral equation by conserving mass and
momentum in a control volume (C.V.) of width dx
and height h that moves at the exterior flow speed
Ue(x) as shown. Here h is a constant distance that is
comfortably greater than the overall boundary layer
thickness
δ
.
Solution 10.15. Use the integral laws as applied to a
moving control volume. Conservation of mass
implies:
d
dt dx
ρ
Bdy
0
h
∫
$
%
& '
(
) +
ρ
u
[ ]
x−dx / 2 −Ue
( )
ex⋅(−exB)dy
0
h
∫
+
ρ
u
[ ]
x+dx / 2 −Ue
( )
ex⋅(exB)dy
0
h
∫+
ρ
(u−Ue)ex+vey
[ ]
y=h⋅eyBdx =0
The four terms above correspond to unsteady mass addition to the CV, mass flux on the left side
of the CV, mass flux on the right side of the CV, and mass flux on the top of the CV.
The CV is a constant size and the density is presumed to be constant, so the first term is
zero (the time derivative of a constant). Combine terms, perform the dot products, presume dx is
small, and divide by
ρ
Bdx to reach:
1
dx u
[ ]
x+dx / 2 −u
[ ]
x−dx / 2
( )
dy
0
h
∫+v
[ ]
y=h=0
, or
v
[ ]
y=h=−
∂
u
∂
xdy
0
h
∫=−d
dx udy
0
h
∫
%
&
' (
)
*
, (*)
where the final equality follows from the fact that h is constant.
Now conserve horizontal momentum using the same CV.
d
dt dx
ρ
uBdy
0
h
∫
$
%
& '
(
) +
ρ
u
[ ]
x−dx / 2 u
[ ]
x−dx / 2 −Ue
( )
ex⋅(−exB)dy
0
h
∫
+
ρ
u
[ ]
x+dx / 2 u
[ ]
x+dx / 2 −Ue
( )
ex⋅(exB)dy
0
h
∫+
ρ
u(u−Ue)ex+vey
[ ]
y=h⋅eyBdx
=−
τ
0Bdx +p
[ ]
x−dx / 2 −p
[ ]
x+dx / 2
( )
hB
where
τ
0 is the wall shear stress, and p is the pressure. Simplify and combine terms, and divide
by
ρ
Bdx to reach:
d
dt udy
0
h
∫
#
$
% &
'
( +1
dx u
[ ]
x+dx / 2
2−u
[ ]
x−dx / 2
2
{ }
dy
0
h
∫−Ue
1
dx u
[ ]
x+dx / 2 −u
[ ]
x−dx / 2
{ }
dy
0
h
∫+uv
[ ]
y=h
=−
τ
0
ρ
+1
ρ
dx p
[ ]
x−dx / 2 −p
[ ]
x+dx / 2
( )
h
Use the chain rule on the total time derivative of the unsteady term,
d
dt =dx
dt
d
dx =Ue
d
dx
, noting
that horizontal velocity of the CV (= dx/dt) is Ue; and presume dx is small:
Ue
d
dx udy
0
h
∫
#
$
% &
'
( +d
dx u2dy
0
h
∫−Ue
d
dx udy
0
h
∫+uv
[ ]
y=h=−
τ
0
ρ
−1
ρ
∂
p
∂
xh
.
u(x+dx,y)!
u(x,y)!
p(x+dx)!p(x)!
x+dx!x!
Ue!
C.V.!
Ue!
y = h!
!
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.16. For the following approximate flat-plate boundary layer profile:
u
U=sin
π
y2
δ
( )
for 0 ≤y≤
δ
1 for y>
δ
%
&
'
(
)
*
, where
δ
is the generic boundary layer thickness, determine:
a) the displacement thickness
δ
*, the momentum thicknesses
θ
, and the shape factor
H=
δ
*
θ
.
b) Use the zero-pressure gradient boundary layer integral equation to find:
δ
x
( )
Rex
1 2
,
δ
*x
( )
Rex
1 2
,
θ
x
( )
Rex
1 2
,
cfRex
1 2
, and
CDReL
1 2
for the approximate profile.
c) Compare these results to their equivalent Blasius boundary layer values.
Solution 10.16. a) The sinusoid boundary layer velocity profile,
u
U=sin
π
y
2
δ
$
%
& '
(
) =sin
π
2
ζ
$
%
& '
(
)
where
ζ
=y
δ
, is a reasonably accurate approximate laminar boundary layer profile. The momentum
thickness for this profile is:
θ
=u
U
1−u
U
$
%
& '
(
)
0
δ
∫dy =
δ
sin
π
2
ζ
$
%
& '
(
) 1−sin
π
2
ζ
$
%
& '
(
)
$
%
& '
(
)
0
1
∫d
ζ
=2
δ
π
sin
ψ
( )
1−sin
ψ
( )
( )
0
π
2
∫d
ζ
The integration is not too complicated,
sin(
ψ
)d
ψ
0
π
2
∫=1
, and
sin2(
ψ
)d
ψ
0
π
2
∫=
π
4
, so
scales and parameters produces:
b)
θ
x
Rex
1 2 =0.654
(0.664)
δ
*
x
Rex
1 2 =1.743
(1.721)
δ
x
Rex
1 2 =4.80
(5.0)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.17. An incompressible viscous fluid with kinematic viscosity
ν
flows steadily in a
long two dimensional horn with cross sectional area A(x) = Aoexp{
β
x}. At x = 0, the fluid
velocity in the horn is uniform and equal to Uo. The boundary layer momentum thickness is zero
at x = 0.
a) Assuming no separation, determine the boundary layer momentum thickness,
θ
(x), on the
lower horn boundary using Thwaites method.
b) Determine the condition on
β
that makes the no-separation assumption valid for 0 < x < L.
c) If
θ
(x = 0) was nonzero and positive, would the flow in the horn be more or less likely to
separate than the
θ
(x = 0) = 0 case with the same horn geometry?
Solution 10.17. a) As stated above, use Thwaites method to estimate the boundary layer
momentum thickness. Start with conservation of mass to determine the U(x):
U(x)A(x)=UoAo
,
so
U(x)=UoAoA(x)=Uoexp −
β
x
{ }
. Thus, the Thwaites equation becomes:
θ
2(x)=0.45
ν
U6(x)
U5(!
x)d!
x
0
x
∫=0.45
ν
Uo
6e+6
β
xUo
5e−5
β
!
xd!
x
0
x
∫
=0.45
ν
Uo
e+6
β
xe−5
β
x
−5
β
+1
5
β
$
%
&'
(
)=0.09
ν
β
Uo
e+6
β
x1−e−5
β
x
( )
b) First compute
λ
=
θ
2
ν
dU
dx =0.09
β
Uo
e+6
β
x1−e−5
β
x
( )
−
β
Uoe−
β
x
( )
=−0.09 e+5
β
x−1
( )
. Separation will
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.18. The steady two-dimensional velocity potential for a source of strength q located
a distance b above a large flat surface located at y = 0 is:
φ
(x,y)=q
2
π
ln x2+(y−b)2+ln x2+(y+b)2
( )
a) Determine U(x), the horizontal fluid velocity on y = 0.
b) Use this U(x) and Thwaites method to estimate the momentum thickness,
θ
(x), of the laminar
boundary layer that develops on the flat surface when the initial momentum thickness
θ
o is zero.
[Potentially useful information:
ξ
5d
ξ
(
ξ
2+b2)5
0
x
∫=x6(x2+4b2)
24b4(x2+b2)4
]
c) Will boundary layer separation occur in this flow? If so, at what value of x/b does Thwaites
method predict zero wall shear stress?
d) Using solid lines, sketch the streamlines for the ideal flow specified by the velocity potential
given above. For comparison, on the same sketch, indicate with dashed lines the streamlines you
expect for the flow of a real fluid in the same geometry at the same flow rate.
Solution 10.18. a)
U(x)=
∂φ
∂
x
!
"
#$
%
&
y=0
=q
2
π
x
x2+(y−b)2+x
x2+(y+b)2
!
"
#$
%
&
y=0
=q
π
x
x2+b2
!
"
#$
%
&
b) Use the Thwaites integral without the
θ
o-term.
θ
2(x)=0.45
ν
U6(x)
U5(
ξ
)d
0
x
∫
ξ
=0.45
νπ
6x2+b2
( )
6
q6x6
q5
π
5
ξ
ξ
2+b2
"
#
$%
&
'
0
x
∫
5
d
ξ
=0.45
νπ
x2+b2
( )
6
qx6
ξ
5d
ξ
(
ξ
2+b2)5
0
x
∫
Evaluate the integral using the given information:
xsep
x
y
b
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.19. A fluid-mediated particle-deposition process requires a laminar boundary layer
flow with a constant shear stress,
τ
w, on a smooth flat surface. The fluid has viscosity
µ
and
density
ρ
(both constant). The flow is steady, incompressible, and two-dimensional, and the flat
surface extends from 0 < x < L. The flow speed above the boundary layer is U(x). Ignore body
forces.
a) Assume the boundary layer thickness is zero at x = 0, and use Thwaites’ formulation for the
shear stress,
τ
w=
µ
U
θ
( )
l(
λ
)
with
λ
=
θ
2
ν
( )
dU dx
( )
, to determine
θ
(x) and U(x) in terms of
λ
,
ν
=
µρ
, x, and
τ
w
µ
= constant. [Hint: assume that
U
θ
=A
and
l(
λ
)
are both constants so
that
τ
w
µ
=Al(
λ
)
.]
b) Using the Thwaites integral (10.50) and the results of part a), determine
λ
.
c) Is boundary layer separation a concern in this flow? Explain with words or equations.
Solution 10.19. a) For Thwaites method, the shear stress is provided by the correlation
τ
w=
µ
U
θ
( )
l(
λ
)
. Thus, the simplest way to achieve constant
τ
w is for U to be proportional to
θ
,
i.e.
U=A
θ
where A is a constant, and l(
λ
) = constant, where
λ
=
θ
2
ν
( )
dU dx
( )
=
constant.
Eliminating U from this system of equations leaves:
τ
w
µ
=Al(
λ
)
, and
λ
=A
θ
2
ν
( )
d
θ
dx
( )
. (1,2)
Integrate (2) to find:
λν
x=A
θ
33
where the constant of integration is zero because
θ
= 0 at x =
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.20. The steady two-dimensional potential for incompressible flow at nominal
horizontal speed U over a stationary but mildly wavy wall is:
φ
(x,y)=Ux −U
ε
exp −ky
( )
cos kx
( )
,
where k
ε
<< 1. Here,
ε
is the amplitude of the waviness and k = 2
π
/
Λ
, where
Λ
= wavelength of
the waviness.
a) Use the potential to determine the horizontal velocity
u(x,y)
on y = 0.
b) Assume that
u(x, 0)
from part a) is the exterior velocity on the wavy wall and use Thwaites’
method to approximately determine the momentum thickness,
θ
, of the laminar boundary layer
that develops on the wavy wall when the fluid viscosity is
µ
, and
θ
= 0 at x = 0. Keep only the
linear terms in k
ε
and
ε
/x to simplify your work.
c) Is the average wall shear stress higher for
Λ
/2 ≤ x ≤ 3
Λ
/4, or for 3
Λ
/4 ≤ x ≤
Λ
.
d) Does the boundary layer ever separate when k
ε
= 0.01?
e) In 0 ≤ x ≤
Λ
, determine where the wall pressure is the highest and the lowest.
f) If the wavy surface were actually an air-water interface, would a steady wind tend to increase
or decrease water wave amplitudes? Explain.
Solution 10.20. a) The horizontal velocity u(x,y) is obtained from:
u=
∂φ ∂
x=
∂ ∂
x
( )
Ux −U
ε
exp −ky
( )
cos kx
( )
[ ]
=U+Uk
ε
exp −ky
( )
sin kx
( )
.
when evaluated on y = 0, this becomes:
u(x,y=0) =U1+k
ε
sin(kx)
( )
.
b) Evaluate the Thwaites integral with
θ
= 0 at x = 0 to find:
θ
2=0.45
ν
U6(x)
U5($
x )d$
x
0
x
∫=0.45
ν
U1+k
ε
sin(kx)
( )
61+k
ε
sin(k$
x )
( )
5d$
x
0
x
∫
≈0.45
ν
U1−6k
ε
sin(kx)
( )
1+5k
ε
sin(k&
x )
( )
d&
x
0
x
∫=0.45
ν
U1−6k
ε
sin(kx)
( )
&
x −5
ε
cos(k&
x )
[ ]
0
x
=0.45
ν
U1−6k
ε
sin(kx)
( )
x−5
ε
cos(kx)+5
ε
[ ]
≈0.45
ν
x
U1−6k
ε
sin(kx)+5(
ε
x)(1−cos(kx)
( )
Thus,
θ
≈0.45
ν
x
U1−6k
ε
sin(kx)+5(
ε
x)(1−cos(kx)
( )
'
(
)
*
+
,
1 2
.
c) First compute the correlation parameter:
λ
=
θ
2
ν
dU
dx =0.45x
U1−6k
ε
sin(kx)+5(
ε
x)(1−cos(kx)
( )
Uk 2
ε
cos(kx)
( )
.
This form can be made easier to look at if it is rearranged and the second order terms are
dropped:
λ
≈0.45(kx)(k
ε
)cos(kx) 1−6k
ε
sin(kx)+5(
ε
x)(1−cos(kx)
( )
≈0.45(kx)(k
ε
)cos(kx)
.
Then note that
τ
w=
µ
U
θ
( )
l(
λ
)
and that l(
λ
) increases monotonically with increasing
λ
.
Therefore, the shear stress will be highest where
λ
is highest. Now switch to considerations of
x!
y!
Λ
/2!
Λ
!
U!
ε
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.21. Consider the boundary layer that develops in stagnation point flow:
Ue(x)=Uox L
.
a) With
θ
= 0 at x = 0, use Thwaites method to determine
δ
*(x),
θ
(x), and cf(x).
b) This flow also has an exact similarity solution of the full Navier Stokes equations. Numerical
evaluation of the final nonlinear ordinary differential equation produces:
cfRex=2.4652
,
where
Rex=Uex
ν
=Uox2L
ν
. Assess the accuracy of the predictions for cf(x) from the
Thwaites method for this flow.
Solution 10.21. a) The Thwaites equation is:
θ
2=0.45
ν
Ue
6(x)
Ue
5(
ξ
)
0
x
∫d
ξ
+
θ
o
2Uo
Ue(x)
&
'
(
)
*
+
6
. Here,
θ
= 0
at x = 0 and
Ue(x)=Uox L
, so only the integral term contributes:
θ
2=0.45
ν
Uo
6
L6
x6Uo
5
0
x
∫
ξ
5
L5d
ξ
=0.45
ν
Uo
L
6
, or
θ
=0.2739
ν
L
Uo
.
Now compute the correlation parameter
λ
=
θ
2
ν
dUe
dx =
θ
2
ν
Uo
L=0.45
6=0.075
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.22. A laminar boundary layer develops on a large smooth flat surface under the
influence of an exterior flow velocity U(x) that varies with downstream distance, x.
a) Using Thwaites method, find a single integral-differential equation for U(x) if the boundary
layer is to remain perpetually right on the verge of separation so that the wall shear stress,
τ
w, is
zero. Assume that the boundary layer has zero thickness at x = 0.
b) Assume
U(x)=Uox L
( )
γ
and use the result of part a) to find
γ
.
c) Compute the boundary layer momentum thickness
θ
(x) for this situation.
d) Determine the extent to which the results of parts b) and c) satisfy the von Karman boundary
layer integral equation, (10.43), when
τ
w = 0 by computing the residual of this equation.
Interpret the meaning of your answer; is Von Karman’s equation well satisfied, or is the residual
of sufficient size to be problematic?
e) Can the U(x) determined for part b) be produced in a duct with cross sectional area
A(x)=Aox L
( )
−
γ
? Explain your reasoning.
Solution 10.22. a) When
τ
w is zero, then
λ
sep =
λ
≡
θ
2
ν
( )
dU dx
. Now plug this into the
Thwaites integral:
θ
2(x)=0.45
ν
U6(x)U5($
x
0
x
∫)d$
x =
νλ
sep
dU dx
, or
λ
sep =0.45
U6(x)
dU
dx U5(#
x
0
x
∫)d#
x
.
b) Plugging in
U(x)=Uox L
( )
γ
produces:
λ
sep =0.45
6x L
6
γ
γ
Uo
L
x
L
$
%
& '
(
)
γ
−1Uo
5L
5
γ
+1
x
L
$
%
& '
(
)
5
γ
+1
=0.45
γ
5
γ
+1
,
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 10.23. Consider the boundary layer that develops on a cylinder of radius a in a cross
flow
a) Using Thwaites method, determine the momentum thickness as a function of
ϕ
, the angle from
the upstream stagnation point (see drawing).
b) Make a sketch of cf versus
ϕ
.
c) At what angle does Thwaites method predict vanishing wall shear stress?
Solution 10.23. The potential for 2D ideal flow around a cylinder is:
φ
(r,
θ
)=Ur cos
θ
1+a2r2
( )
.
For the specified geometry,
θ
=
π
−
ϕ
and
cos
θ
=−cos
ϕ
so
φ
(r,
ϕ
)=−Ur cos
ϕ
1+a2r2
( )
. Thus,
the exterior velocity, Ue, outside the boundary layer that develops on the cylinder’s surface, is
given by:
Ue=1
r
∂φ
∂ϕ
%
'
(
*
=2Usin
ϕ
[ ]
x=a
ϕ
=2Usin x a
( )
, so the Thwaites integral becomes:
