SOLUTIONS MANUAL FOR
Fundamental
Mechanics of Fluids
Fourth Edition
by
I.G. Currie
SOLUTIONS MANUAL FOR
Fundamental
Mechanics of Fluids
Fourth Edition
by
I.G. Currie
CRC Press
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1
BASIC CONSERVATION LAWS
BASIC CONSERVATION LAWS
Problem 1.1
Inflow through x = constant:
u y z
Outflow through x x = constant:
u y z
( u y z) x
x
( u ) x y z
x
Net inflow through y = constant surfaces:
( v ) x y z
y
Net inflow through z = constant surfaces:
( w) x y z
z
But the rate at which the mass is accumulating inside the control volume is:
Net inflow through x = constant surfaces:
( x y z)
t
Then the equation of mass conservation becomes:
x y z ( u ) ( v) ( w) x y z
t
y
z
x
Taking the limits as the quantities x, y and z become vanishingly small, we get:
( v) ( w) 0
( u)
t x
y
z
Page 1-1
BASIC CONSERVATION LAWS
Problem 1.2
Inflow through R = constant:
u R R z
Outflow through R R = constant:
u R R z
( u R R z ) R
R
( R u R ) R z
R
Net inflow through = constant surfaces:
( u ) R z
Net inflow through z = constant surfaces: ( u z ) R R z
z
But the rate at which the mass is accumulating inside the control volume is:
Net inflow through R = constant surfaces:
( R R z )
t
Then the equation of mass conservation becomes:
R R z ( Ru R )
( u ) R ( u z ) R z
t
z
R
Taking the limits as the quantities R, and z become vanishingly small, we get:
1
1
( R u R )
( u ) ( u z ) 0
t R R
z
R
Page 1-2
BASIC CONSERVATION LAWS
Problem 1.3
Inflow through r = constant:
u r r 2 sin
Outflow through r r = constant:
u r r 2 sin
( r 2u r sin ) r
r
( r 2u r )sin r
r
Net inflow through = constant surfaces:
( u sin ) r r
Net inflow through = constant surfaces:
( u ) r r
But the rate at which the mass is accumulating inside the control volume is:
Net inflow through r = constant surfaces:
( r 2 sin r )
t
Then the equation of mass conservation becomes:
2
r sin r ( r 2u r ) sin r
( u sin ) r
( u ) r
t
r
Taking the limits as the quantities r, and become vanishingly small, we get:
1
1
1
2
( r 2u r )
( u sin )
( u ) 0
t r r
r sin
r sin
Page 1-3
BASIC CONSERVATION LAWS
Problem 1.4
Using the given transformation equations gives:
y
x
R
R
2R
2 x 2 R cos
cos
x
x
1 sin
1
y
and sec 2
2
sin
2
x
R cos
R
x
x
R2 x2 y2
and tan
Using these results, the derivatives with respect to x and y transform as follows:
R
cos
x x R x
R
sin
y y R y
z z
sin
R
R
cos
R
R
Using these results and the relationships between the Cartesian and cylindrical vector
components, we get the following expressions for the Cartesian terms in the continuity equation:
sin
( u ) cos
[ (u R cos u sin )]
[( (u R cos u sin )]
x
R
R
cos
( v) sin
[ (u R sin u cos )]
[( (u R sin u cos )]
y
R
R
Adding these two terms and simplifying produces the following equation:
uR
1
( u ) ( v)
( u R )
( u )
R
R
x
y
R
1
1
( R u R )
( u )
R R
R
Substituting this result into the full continuity equation yields the following result:
1
1
( R u R )
( u ) ( u z ) 0
t R R
z
R
Page 1-4
BASIC CONSERVATION LAWS
Problem 1.5
The equations that connect the two coordinate systems are as follows:
x r sin cos
r x y z
2
2
2
2
y r sin sin
y
tan
x
z r cos
z2
cos 2
(x y 2 z 2 )
2
Using the relations given above, the following identities are obtained for the various partial
derivatives:
r
sin cos
x
1
cos cos
x
r
1 sin
x r sin
r
sin sin
y
1
cos sin
y
r
1 cos
y r cos
r
cos
z
1
sin
z
r
0
z
Thus the following expressions are obtained for the various Cartesian derivatives:
r
x x r x x
1 sin
sin cos
r cos cos
r
r sin
r
y y r y y
1
1 cos
sin sin
cos sin
r r
r sin
r
z z r z z
1
cos
sin
r r
Next we need the Cartesian velocity components expressed in terms of spherical components.
This may be achieved by noting that the velocity vector may be written as follows:
u u e x v e y w e z u r e r u e u e
Then, if we express the base vectors in spherical coordinates in terms of the base vectors in
Cartesian coordinates, equating components in the previous equation will yield the required
relationships. Thus, noting that:
Page 1-5
BASIC CONSERVATION LAWS
r x e x y e y z e z r sin cos e x r sin sin e y r cos e z
Also, recalling from Appendix A that:
ei
r
x i
r
x i
it follows that
e r sin cos e x sin sin e y cos e z
e cos cos e x cos sin e y sin e z
e sin e x cos e y
Substituting these expressions into the equation obtained above for the velocity vector, and
equating coefficients of like base vectors, yields the following relationships connecting the
Cartesian and spherical velocity components:
u u r sin cos u cos cos u sin
v u r sin sin u cos sin u cos
w u r cos u sin
Using these results, and those obtained for the Cartesian derivatives, produces the following
expression for the Cartesian terms that appear in the continuity equation:
1
1
1
( u ) ( v) ( w) 2
( r 2u r )
( u sin )
( u )
x
y
z
r r
r sin
r sin
Substituting this result into the full continuity equation yields the following expression:
1
1
1
2
( r 2u r )
( u sin )
( u ) 0
t r r
r sin
r sin
Problem 1.6
From Appendix A, the following value is obtained for the convective derivative:
(a ) a
a 2
a 3 a 2
1 a1
(h2 a 2 )
(h1 a1 )
a2
a3
a1
h1 x1
x1
x1 h 2 x1
x 2
Page 1-6
a3
(h1 a1 )
(h3 a 3 ) e 1 (
h3 x 3
x1
)e2 (
)e3
BASIC CONSERVATION LAWS
Applying this result to cylindrical coordinates, we interpret the various terms as follows:
a1 u R
a 2 u
a3 u z
x1 R
x2
x3 z
h1 1
h2 R
h3 1
e1 e R
e 2 e
e3 e z
Using these results, the required term becomes:
u
u z u
u R
e R (u ) u u R
u
uz
R
R R
R
u R
u R u z
(
)
Ru
u
z
R
R
z
Simplifying the right side of this equation produces the required result:
e R (u ) u u R
u R
R
u u R
R
u
2
uz
R
u R
z
Problem 1.7
We use the same starting equation from Appendix A as in the previous problem. However, since
we are dealing with spherical coordinates here, the various terms are as follows:
a1 u r
a 2 u
a3 u
x1 r
x2
x3
h1 1
h2 r
h3 r sin
e1 e r
e 2 e
e 3 e
Using these results, the required term becomes:
u
u u
u r
u
u
e r (u ) u u r
r
r r
r
u u r
u r
(r u sin
(r u )
r sin r
r
Simplifying the right side of this equation yields the required result:
er
(u
) u u r
u r
r
u u r
r
u u
2
r
2
u
u r
r sin
Page 1-7
BASIC CONSERVATION LAWS
Problem 1.8
For a Newtonian fluid, the shear stress tensor is defined by the following equation:
ij ij
u i u j
x j x i
x k
u k
Evaluating the various terms in this expression for Cartesian coordinates ( x, y , z ) and Cartesian
velocity components ( u, v, w ) yields the following results:
u
v
w
u
u
v
w
v
u
v
w
w
xx
2
x
x y z
yy
2
y
x y z
zz
2
z
x y z
u
v
u
w
v
w
xy yx
y x
xz zx
z x
yz z y
z y
For a monotonic gas, the Stokes relation requires that 2 / 3 . Then the relations obtained
above assume the following special form:
yy
w
u
v
2
2
4
x
y
3 z
u v
yx
y x
zz
xy
u
v
w
2 2
4
y
z
3 x
u
w
v
4 2
2
x
z
3 y
xx
u
w
xz zx
z x
v w
z y
yz z y
Page 1-8
BASIC CONSERVATION LAWS
Problem 1.9
For a Newtonian fluid, the dissipation function is defined by the following equation:
u k
x k
2
u i u j
x j x i
u j
x i
Evaluating the various terms in this equation for the Cartesian coordinates ( x, y , z ) and the
Cartesian velocity components ( u, v, w ), yields the following value for :
2
u 2 v 2 w 2
u v w
2
x y z
x y z
u v 2 u w 2 v w 2
y x z x z y
For a monotonic gas, the Stokes relation requires that 2 / 3 . Then the general expression
for obtained above assumes the following special form:
2
2
2
2 u v w 2
v
u
w
2 2 2
x
z
y
3 x y z
2
2
2
u v u w v w
y x z x z y
Problem 1.10
For steady flow of an inviscid and incompressible fluid, but one for which the density is not
constant, the two-dimensional governing equations are:
( u ) ( v) 0
x
y
u
u
p
u v
x
y
x
v
v
p
u v
x
y
y
Dividing the continuity equation by 0 and using the definitions of the new velocity
components as given, we get:
Page 1-9
BASIC CONSERVATION LAWS
* *
u
v 0
y 0
x 0
u * v * * *
v
0
u
x
x 0
y 0
0 x
The last two terms in the last equation represent the steady-state form of the material derivative
of the square root of the density ratio. For an incompressible fluid, this quantity will be zero.
Then the continuity equation becomes:
u * u *
0
x
x
(1.15)
Adding the original form of the continuity equation to each of the components of the momentum
equation, and dividing throughout by the constant 0 , yields the following form of the
momentum equations:
1
2
u u v
0
x 0 y 0
2
1
v
u v
0
x 0 y 0
p
x
p
y
Using the definitions of the new velocity components, these equations become:
2
1 p
u*
u* v*
0 x
x
y
2
* *
1 p
u v
v*
0 y
x
y
Expanding the terms on the left side of this equation and using Eq. (1.15) reduces the momentum
equations to those of an incompressible fluid. The resulting equations are as follows:
u * v *
0
x
y
Page 1-10
u*
u *
u *
1 p
v*
0 x
x
y
u*
v *
v *
1 p
v*
0 y
x
y
2
FLOW KINEMATICS
FLOW KINEMATICS
Problem 2.1
The following graph was drawn using EXCEL.
__________________________________________________________________________
Problem 2.2
dy
dx
y
But x
(a)
Hence at t
v
1 t
u
(1 t ) x C
y 1 when t
0
C
0
0 the equation of the streamline is:
y
x
Page 2-1
FLOW KINEMATICS
1
dx
u
1 t
dt
x log (1 t ) C1
(b)
The condition that x
y 1 when t
dy
v 1
dt
y t C2
0 requires that C1
x 1 log (1 t )
C2
1 , so that:
y 1 t
Eliminating t between these two equations shows that the equation of the pathline is:
y
ex
1
(c) Here, the equations obtained in (b) above are required to satisfy the condition x y 1
when t
. This leads to the values C1 1 log (1 ) and C 2 1
. Hence the
parametric equations of the streakline are:
x
At time t
log (1 t ) 1 log (1
)
y t 1
)
y 1
0 these equations become:
x 1 log (1
Eliminating the parameter
for the streakline at t 0 :
between these two equations yields the following equation
y
2 e1
x
Problem 2.3
(a)
u x(1 t )
v 1
w 0
dx
x(1 t )
x C1 e(1 t ) s C1 e s at t
ds
dy
1
y s C2
ds
But x = y = 1 when s = 0 so that C1 C2 1. Hence:
x
e s and y
s 1
x
Page 2-2
ey
1
0
FLOW KINEMATICS
(b)
dx
x(1 t )
dt
dy
1
y t
dt
But x = y = 1 when t = 0 so that C1 C2
x et (1 t / 2) and
x C1 et (1
C2
1. Hence:
y t 1
x e( y
x C1 e t (1
(c) As in (b):
But x = y = 1 when t =
t / 2)
so that C1
x et (1 t / 2)
and y 1 t
x
2
1) / 2
and y
e
(1
/ 2)
(1
/ 2)
t
C2
and C2
e (1
for t
1
e
t / 2)
/ 2)
1
for t
. Hence:
0
0
(1 y )(3 y ) / 2
(d) From the continuity equation:
t
Hence:
But
0
x
( u)
y
( v)
z
D
Dt
( w)
D
u
(1 t ) since
Dt
D
i.e.
(log )
(1 t )
Dt
Therefore log
t (1 t / 2) log C
C e t (1 t / 2)
So that
when t = 0 so that C
0 . Hence:
0
e
t (1 t / 2)
u 0
u (1 t ) here.
along the streamline
Problem 2.4
The equations that define the streamlines are as follows:
dx i
ui
log x i
or x i
ds
or
s
xi
ds
(1 t )
log C i
(1 t )
C i e s /(1
dx i
t)
Page 2-3
FLOW KINEMATICS
Let x i
x i 0 when s
0 . Then:
So that:
xi
x i 0 e s /(1
x
x0
y
y0
t)
z
z0
for any time t .
The equations that define the pathlines are as follows:
dx i
dt
log x i
or x i
Let x i
x i 0 when s
ui
dx i
or
xi
(1 t )
dt
log C i
log (1 t )
C i (1 t )
0 . Then:
xi
x i 0 (1 t )
x
x0
So that:
xi
or
y
y0
(1 t )
xi 0
z
z0
as per the streamlines.
Problem 2.5
u 16 x 2
(a)
On 0
On 0
On 10
On 5
x 10, y
v 10
y
10
0:
5
y 5, x 10 :
x 0, y
y 0, x
0
0
5:
10
0
0:
5
10
u dx
0
y z2
w
0
5
v dy
0
10 dy
0
u dx
0
5
50
(16 x 2
5) dx
10 dy
50
10
v dy
16,000
3
16 x 2 dx
16,000
50
3
Then, adding the components around the counter-clockwise path gives:
50
(b) For the area specified, n e z and
A
Page 2-4
v
x
z
ω n dA
10
0
u
y
dx
5
0
1.
( 1) dy
FLOW KINEMATICS
A
ω n dA
50
This is the same result that was obtained in (a), so that Eq. (2.5) is verified for this flow.
Problem 2.6
x
u
1
1
x
2
y
1
u ( x, 1) dx
1
1
x dx
x 1
2
and
2
1
y
v
x
1
v( 1, y ) dy
y dy
11
y2
1
1
1
1
2
y2
1
u ( x, 1) dx
x dx
x 1
1
2
1
1
v( 1, y ) dy
y dy
1 y2
In the foregoing equation, we note that there are two pairs of offsetting integrals. Hence:
0
Problem 2.7
u
(a)
1
1
1
1
1
1
1
1
Hence:
u dl
ω
(b)
Hence
w
y
v
ex
z
9x2
2 yz 2
2 y v 10 x w
1
u ( x, 1) dx
(9 x 2
1
v( 1, y ) dy
1
1
1
1
1
v( 1, y ) dy
1
2 x]
2
1
1
1
u ( x, 1) dx
1
2) dx [3 x 3
10 dy [10 y ]
(9 x 2
20
1
2) dx [3 x 3
2 x]
10
1
1
( 10) dy [ 10 y ]
20
1
2 20 10 20 32
u
z
w
ey
x
ω
50 e x
v
x
8 ez
u
ez
y
2z 2 ex
8ez
on the plane z 5
Page 2-5
FLOW KINEMATICS
(c) For the plane z 5 the unit normal is n e z . Hence:
A
1
ω n dA
1
1
dx
1
8 dy
32
This agrees with the result obtained in (a) - as it should, since
u dl
A
ω n dA
Problem 2.8
y
u
1
(a)
1
4
x
2
y
1
u ( x, 1) dx
1
dx
1
2
x
1
1
dy
11
y2
1
4[tan
2
1
x
v
x
1
1
1
]
2
1
v( 1, y ) dy
1
d
1
and
2
1
y2
u ( x, 1) dx
x dx
2
x 1
1
1
1
1
v( 1, y ) dy
dy
1 y2
1
1
2
(b)
v
x
u
y
2 x2
(x 2 y 2 ) 2
1
(x 2
y 2)
(x 2
0 provided
(c)
u
x
2x y
(x
y 2)2
2
and
u 0 provided
Page 2-6
2 y2
(x 2 y 2 ) 2
1
y 2)
x and y
v
y
x and y
0
2x y
(x
y 2)2
2
0
FLOW KINEMATICS
Problem 2.9
y;
u
(a)
1
ud
1
(
v
) dx
[
4(
dx
ds
dy
dx
(c)
2
1
1
[ x]
1
1
dx
1
1
1
[
(
y]
) dy
1
1
2
x
y
y dy
or
x2
2
y2
y2
(
) dx dy
)
dy
ds
x2
1
2
y and
x2
1 and
1
[ y]
u
dx dy
y
y2
2
(d)
1
4(
n dA
A
1
dy
)
v
x
n dA
A
1
x]
2
(b)
x
1
x
x dx
c2
2
c2
c 2 where u
y and v
x.
But x = 1 when y = 0 so that c 2 = 1. Therefore;
x2
(e)
1
x2
y2
y2
1
c 2 where u
y and v
x.
But x = 0 when y = 0 so that c = 0. Therefore;
y
x
Page 2-7
FLOW KINEMATICS
Problem 2.10
The vorticity vector will be in the z direction and its magnitude will be:
(a)
uR
R
0 and u
(R u )
and
uR
R
uR
R
and
( R2)
uR
2 R
0
R 2
0
( R, ) 0 provided
Page 2-8
2 R
2 R
0 and u
(R u )
R
0
( R, )
(b)
1 u
R
1
(R u )
R R
( R, )
R
0
3
SPECIAL FORMS OF THE
GOVERNING EQUATIONS
SPECIAL FORMS OF THE GOVERNING EQUATIONS
Problem 3.1
From the definition of the enthalpy h we have e h p / . Substituting this result into the
left side of the given relation gives the expression:
D
Dh
D p
Dt
Dt
Dt
Dh Dp p D
Dt
Dt Dt
Substituting this result into the full version of the given relation produces the equation:
Dh Dp p D
p u ( k T )
Dt Dt Dt
The term immediately to the left of the equality sign and the term immediately to the right of
it cancel each other by virtue of the continuity equation. Thus the equivalent expression
becomes:
Dh Dp
( k T )
Dt Dt
Problem 3.2
From Appendix A we obtain the following vector identity:
1
(u ) u (u u) u ( u)
2
1
(u ) u (u u) u ( u)
2
1
2
(u u) ( u) ( u) u ( u)
1
(u u) u ( u) 2u
2
In the foregoing, use has been made of an identity from Appendix A for the term (u a) ,
in which a u . Also, use was made of an identity from Appendix A for the term
( u) . For an incompressible fluid u 0 so that:
1
(u ) u (u u) u ( 2u)
2
Page 3-1
SPECIAL FORMA OF THE GOVERNING EQUATIONS
Problem 3.3
From Bernoulli’s equation we have:
1
1
2
2
p (u u) p 0
U 2
In cylindrical coordinates, the velocity-squared term in this equation may be written as
follows:
2
2
u u u R u
U 1
U 2 1
U 2 1
2
2
2
a2
a2
2
2
U
cos
1
sin 2
2
2
R
R
a4
a2
2
(cos 2 sin 2 )
4
2
R
R
a4
a2
2
cos 2
4
2
R
R
Substituting this result into the Bernoulli equation gives the required expression for the
pressure at any location whose coordinates are ( R, ) :
a4
1
a2
p ( R, ) p 0 U 2 4 2 2 cos 2
2
R
R
Substituting R a in the foregoing expression produces the following value for the surface
pressure:
p ( a, ) p 0
1
2
U
2
(1 2cos 2 )
Note that this result may be expressed as a non-dimensional pressure coefficient as follows:
Cp
p ( a, ) p 0
1
2
Page 3-2
U
2
2 cos 2 1
4
TWO-DIMENSIONAL POTENTIAL FLOWS
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.1
cos F
cos(
i )
cos cos(i ) sin sin(i )
cos cosh
i sin sinh
z
x
Therefore
And
Eliminating
i y
x cos cosh
y
sin sinh
using
cos 2
x2
cosh 2
sin 2
1
y2
sinh 2
1
This is the equation of ellipses whose major and minor semi-axes are x cosh and y sinh .
1
d
cos 1 z
W ( z)
dz
1 z2
i
1
1
u 0 and v
For x = 2 and y = 0, z = 2 and we get W
i.e
3
3
3
downwards velocity.
_____________________________________________________________________________
Problem 4.2
z
x
i y
Therefore
And
Eliminating
using
cosh F
cosh(
i )
cosh cosh(i ) sinh sinh(i )
cosh cos
i sinh sin
x cosh cos
y sinh sin
cosh 2
sinh 2
2
x
cos 2
1
2
y
sin 2
1
This is the equation of a family of hyperbolas whose semi-axes are x cos and y sin .
1
d
cosh 1 z
W ( z)
dz
z2 1
1
2i
2
u 0 and v
For x = 1/2 and y = 0, z = 1/2 so that W
i.e
3/ 4
3
3
upwards velocity.
_____________________________________________________________________________
Page 4-1
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.3
m
m
log ( z i h)
log ( z i h)
2
2
m
log ( z 2 h 2 )
2
m
log ( x 2 h 2 y 2 ) i 2 x y
2
F ( z)
m
log ( x 2
2
m
tan
2
( x, y )
h2
1
x
2
y2)2
4 x2y2
i
m
tan
2
1
x
2
2x y
h2 y2
2x y
h2 y2
( x,0) 0
streamline
We can evaluate the velocity components from the complex velocity, then substitute the results
into Bernoulli's equation to get the pressure.
W ( z)
dF
dz
d m
log( z 2
dz 2
h2)
m 2z
2 z2 h2
m x( x 2 h 2 y 2 ) 2 x y 2
(x 2 h 2 y 2 )2 4 x 2 y 2
u( x, y )
m
m y( x 2 h 2 y 2 ) 2 x 2 y 2
(x 2 h 2 y 2 )2 4 x 2 y 2
x
x
2
m y( x 2 h 2 y 2 ) 2 x 2 y 2
(x 2 h 2 y 2 )2 4 x 2 y 2
m x( x 2 h 2 y 2 ) 2 x y 2
(x 2 h 2 y 2 )2 4 x 2 y 2
and v( x, y )
Hence u ( x, 0)
i
h2
and v( x, 0) 0 . Then from Bernoulli's equation:
p( x, 0)
p( x, 0)
p0
p0
1
u ( x, 0) 2
v( x, 0) 2
2
m2
2
2
x2
(x 2 h 2 ) 2
The force on the plate is obtained by integrating the pressure difference across it along the entire
surface. Noting that the pressure distribution is symmetric in x , we get:
Page 4-2
TWO-DIMENSIONAL POTENTIAL FLOWS
F
2
0
m2
2
i.e. F
p ( x, 0) dx
p0
0
m2
4 h
x 2 dx
(x 2 h 2 ) 2
( upwards )
Problem 4.4
F ( z)
m
log ( z
2
b)
m
log ( z
2
m
z
log
z
2
b
b
i
m
log
2
i
m
2
...where
m
1 z /b
log
2
1 z/b
m
log{(1 z / b)(1 z / b
2
m
log (1 2 z / b
)
2
m
log (2 z / b
)
2
m
z
b
In the expansions above it has been assumed that z / b
a way that m / b
m
2
m
2
1 z /b
1 z/b
m
log (e i )
2
b) i
i
1 ei
m
2
)}
1 . Now let b
and m
in such
U . This yields the result:
F ( z) U z
This is the complex potential for a uniform flow of magnitude U in the positive x direction.
Page 4-3
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.5
m
log ( z
2
F ( z)
b)
m
(z
log
2
(z
m
log
2
m
log ( z
2
b) ( z
b) ( z
m
log ( z
2
a 2 / b)
a 2 / b)
a 2 / b)
i
m
2
...where
m
(1 z / b) (1 a 2 / b z )
log
2
(1 z / b) (1 a 2 / b z )
m
log{(1 z / b)(1 a 2 / b z )(1 z / b
2
m
log (1 2 z / b 2 a 2 / b z
)
2
m
log (2 z / b 2 a 2 / b z
)
2
a 2 / b) i
m
2
i
1 ei
m
2
)(1 a 2 / b z
In the foregoing expansions, it has been assumed that z / b
such a way that m / b
m
log ( z
2
m
2
i
(1 z / b) (1 a 2 / b z )
(1 z / b) (1 a 2 / b z )
m
log (e i )
2
b)
}
1 . Now let b
and m
in
U . This yields the result:
a2
z
F ( z) U z
This is the complex potential for a uniform flow U past a circular cylinder of radius a .
Problem 4.6
F ( z)
m
log( z
2
b)
m
log( z
2
a 2 / b)
m
log( z
2
m
( z b)( z a 2 / b)
log
b(z
2
)( z a 2 / )
m
( z / b 1)( z a 2 / b)
log
2
(z
)( z a 2 / )
m
log
2
(z
Page 4-4
z
)( z a 2 / )
as b
a2 / )
m
log( z
2
)
m
log b
2
TWO-DIMENSIONAL POTENTIAL FLOWS
i.e. F ( z )
m
log
2
z
m
log[ z
2
1
a / )
2
(
(
a2 / )
a 2 / z)
a 2 / z )]
On the surface of the circle of radius a this result becomes:
F ( ae i )
For
a, (
a2 / )
m
a 2 / ) ae i )]
log[ae i (
2
m
m
a 2 / )]
log[2 a cos
(
log{e i [(
2
2
m
m
log{( a 2 / ) 2a cos } i
2
2
2a so that
a2 /
2a cos
and so
a2 / )
2a cos ]
m / 2 . Therefore the
circle of radius a is a streamline. The required complex potential is given by:
F ( z)
m
log
2
(z
z
)( z a 2 / )
The resulting flow field is illustrated below.
To determine the force acting on the cylinder, we calculate the complex velocity:
Page 4-5
TWO-DIMENSIONAL POTENTIAL FLOWS
m 1
2 z
W ( z)
W 2 ( z)
m
2
2
1
z2
1
(z
)
1
(z
1
(z a 2 / )
)
1
(z a 2 / ) 2
2
2
z( z )
2
z( z a 2 / )
2
)( z a 2 / )
(z
Using this result and the Blasius integral law (for a contour that includes the cylinder, but
excludes the sink at z
) produces the following expression for the force that acts on the
cylinder:
X
iY
i
i
2
2
C
W 2 ( z ) dz
2 i
(residues of W 2 ( z ) inside C )
m
2
2
2
2
a /
2
a /
2
2
2
2
(a /
)
m 2 2a 2 /
4
( a2 / )
From this result, there is no force acting on the y-direction, and the force that acts in the xdirection is:
m2
a2
2
( 2 a2)
X
Problem 4.7
F ( z)
m
log z
2
b e i(
m
log z
2
)
a2 i
e
b
m
log z
2
m
(z
log
2
(z
m
log
2
Page 4-6
(1 z / b e
(1 z / b e
i
i
)
m
log z
2
a 2 / bei )
a 2 / bei )
b e i )( z
b e i )( z
a 2 i(
e
b
i
bei
m
2
)(1 a 2 / bz e i )
)(1 a 2 / bz e i )
i
m
2
since e i
i
m
2
1
TWO-DIMENSIONAL POTENTIAL FLOWS
i.e. F ( z )
m
log (1 z / b e i )(1 a 2 / bz e i )(1 z / b e
2
m
log 1 (2 z / b e i
2a 2 / bz e i
)
2
m
2 z / be i
2a 2 / b z ei
2
In the foregoing expansions, it has been assumed that z / b
b
and m
in such a way that m / b
F ( z) U z e
i
)(1 a 2 / bz e i
1 and a 2 / bz
)
1 . Now let
U . Thus the following result is obtained:
i
a2 i
e
z
Problem 4.8
F ( z)
i
log( z
2
b)
i
log( z
2
a2
)
b
i
log( z
2
a2
)
i
log( z
2
)
i
log b
2
( z b)( z a 2 / )
i
log
2
)( z a 2 / b)
b(z
(1 z / b)( z a 2 / )
i
log
2
(z
)( z a 2 / b)
(z a 2 / )
i
log
2
(z
)z
F (a e i )
i
log
2
(
a 2 / z)
(z
)
i
log
2
(
a 2 / z)
(z
)
i
(
log
2
(
ae i )
a ei )
for b
i
log
2
i
log
2
The last result gives the value of the complex potential on the surface of the cylinder z a e i .
We now note that the argument of the first logarithm on the right side of this equation is of the
form R e i , where R 1 . Therefore, the imaginary part of the entire first term is zero, so that:
( a, )
log
constant
2
This confirms that the circle of radius a is a streamline, so that the required complex potential is:
Page 4-7
TWO-DIMENSIONAL POTENTIAL FLOWS
(z a 2 / )
i
log
2
)
z (z
F ( z)
The complex velocity for this flow field is defined by the following equation:
W ( z)
i
2
(z
1
a2 / )
1
(z
1
z
)
2
1
1
1
2
2
2
2
2
2
2
2
2
z
2
(z a / )
(z
)
( z a / )( z
) z (z a / ) z (z
)
Using this result and the Blasius integral law (for a contour that includes the cylinder, but
excludes the vortex at z
), produces the following expression for the force that acts on the
cylinder:
W 2 ( z)
X
iY
i
i
2 2
i
2
C
W 2 ( z ) dz
residues of W 2 ( z ) inside C
2 i
2
2
(a /
2
2
2
2
2
l)
2
( a / )
2
a /
2
2
(
)
a2
(
2
a2)
Hence there is no force acting in the y-direction, and the force in the x-direction is:
2
X
2
a2
(
2
a2)
Problem 4.9
F ( z) U ( z
and W ( z ) U
Page 4-8
a2
)
z
a 2U
z2
i
z
log
a
2
i 1
2 z
TWO-DIMENSIONAL POTENTIAL FLOWS
a 2U
e
R2
W ( R, ) U
i
e
2 R
i2
a 2U i 2
e
R2
and W ( R, ) U
i
ei
2 R
a 2U 2 i 2
(e
R2
W W ( R, ) U 2
i
W W ( a, )
2U 2
i
i2
e
)
a 2U
(e i
2 R3
2 U 2 cos 2
iU
(e i
2 R
e
i
)
a 4U 2
R4
2
e
i
)
4
2
R2
2
2U
sin
a
4
2
a2
Now use Bernoulli’s equation with the pressure far from the cylinder specified to be p 0 . Then:
p( R, )
p ( a, )
p0
p0
1
(U 2
WW)
2
1
U2
2
U 2 cos 2
U
sin
a
2
8
2
a2
The upward force acting on the cylinder is defined by the following expression:
FL
2
p(a, ) a sin d
0
U
sin
a
2
0
2
In the foregoing, it has been recognized that
values of m and for n
0
a sin d
sin m d
2
0
sin m cos n d
0 for all
m . Hence the value of the lift force is:
FL
U
Problem 4.10
(a)
F ( z)
m
log z i log z
2
2
m
log( Re i ) i log( Re i )
2
2
Hence
( R, )
And
( R, )
1
m log R
2
1
m
log R
2
Page 4-9
TWO-DIMENSIONAL POTENTIAL FLOWS
Therefore
m 1
1
i
2 z
2 z
m
i
e
2 R
2 R
m
2 R
W ( z)
(b)
i
(u R i u )e
uR
u
i
2 R
(c) From Bernoulli’s equation:
p
1
m2
2 4 2R 2
p
Therefore
p ( R, )
p0
1
(u R 2 u 2 )
2
p0
8
2
2
4
2
(m 2
R
p0
2
2
)
1
R2
_____________________________________________________________________________
Problem 4.11
(a)
F ( z) U 2z
a2
( z ih )
On y = 0:
F ( x) U 2 x
a2
( x ih )
Hence
( x,0)
u( x,0)
u( x,0)
(b)
Page 4-10
a2
2Ux 1
x
(x 2
( x,0)
2U 1
u
( x,0) 2U
x
a2
( x ih )
2a 2 x
(x 2 h 2 )
U 2x
Therefore
a2
( z ih )
h2)
2U 1
a2
x2
a2
x2
2Ux
h2
2a 2 x
x2
h2
2a 2 x 2
h2
2a 2 x
(x 2 h 2 )2
x2
h2
2
4a 2 x
(x 2 h 2 )2
8a 2 x 3
(x 2 h 2 )3
2
TWO-DIMENSIONAL POTENTIAL FLOWS
4a 2Ux
(x 2 h 2 )2
4x 2
(x 2 h 2 )
3
4x 2
(x 2 h 2 )
Then the maximum/minimum occurs when x = 0 or when 3
x2
0 ; that is, when
3h 2 .
Hence
u( x,0) max
And
u( x,0) min
a2
h2
2U 1
a2
8h 2
2U 1
at x = 0
3h
at x
(c) From Bernoulli’s equation:
p( x,0)
1 2
u ( x,0)
2
p( x,0)
po
So that
po
1
(2U ) 2
2
1
u 2 ( x,0)
2
2 U2
2
But from (b):
2
u ( x,0) 4U
2
a
1
x2
2
2
h2
2a x
2
2
2
x
h
2
4U
2
a 2 (h 2 x 2 )
(x 2 h 2 )2
1
2
2
2
Therefore:
p( x,0)
po
2 U2
2 U2 1
a (h
h
2
2
2
x )
x
2
2
_____________________________________________________________________________
Problem 4.12
(i)
F ( z)
m1
2
log ( z
m2
2
On the circle of radius a, z
ae
log ( z
i
1
m1
)
ae
2
i
2
)
log ( z
m2
2
ae
log ( z
i
1
ae
)
i
2
)
m3
2
log z
a e i so that:
Page 4-11
TWO-DIMENSIONAL POTENTIAL FLOWS
m1
F ( a, )
2
m2
m1
2
log a 2 e i
m2
2
m1
2
e 1)
log a (e i
2
e
ei
log a 2 e i
i
2
2
(e
m2
)
i
1
i
e
i
2
1
cos
2
)
e
i
e
m1
cos 1 )
e
i
log a ( e i
2
(e
log 2 (cos
log a ( e i
2
ei
log 2 (cos
m2
m1
i
log a (e i
2
)
e
i
2
m3
)
2
log ( a e i )
i
)
log a
m2
)
1
e
m3
i
m1
2
log a
log (a e i )
2
i
m2
2
m3
i
2
log a
m3
2
i
From the imaginary part of the last equality we deduce the result:
( a, )
(m1
2
m2
m3 )
Then the value of m 3 that makes the circle of radius a a streamline is:
m3
m1
m2
(ii) Substituting the last result into the expression for F ( z ) and differentiating with respect to z
gives:
W ( z)
m1
2
1
i
ae
z
1
z
1
ae
i
1
On the surface of the circle of radius a, z
W ( a, )
m1
2 a 1 e
1
i(
1
1)
1 e
i(
1)
1
z
m2
2
z
1
i
ae
2
z
1
ae
i 2
1
z
a e i so that:
1 e
i
m2
2 a 1 e
1
i(
1
2)
1 e
i(
2)
1 e
Using the cylindrical representation of the complex velocity gives the following expressions
for the velocity components on the surface of the cylinder:
Page 4-12
i
TWO-DIMENSIONAL POTENTIAL FLOWS
uR
i(
m1
1)
1 e
2 a 2 2 cos (
iu
i(
)
1
i(
m1
1)
1 e
2 2 cos (
2)
1 e
2 a 2 2 cos (
1
1
)
i(
2
2)
1 e
2 2 cos (
)
2
)
1
u R ( a, ) 0
sin (
m1
and u (a, )
1
)
sin (
4 a 1 cos (
1
sin (
2
m2
)
1 cos (
)
sin (
4 a 1 cos (
Then, since q
uR
2
2
u
1
2
)
)
1
)
2
)
1 cos (
2
)
, it follows that the magnitude of the velocity on the surface of
the cylinder is:
m1
q ( a, )
1
4 a
1
cot (
1
2
1
4 a
S
1
1
2
m2
1
4 a
1
)
1
2
2
2
1
m1
cosec 2 (
2
d 2q
( a,
d 2
1
m1
cos (
8 a sin
2
3 1
2
(
S
S
1
)
1
)
S
) cU R e
1
cos (
sin
2
3 1
2
(
S
S
1
1/ 2
)
1
1
cot (
2
2
1
cosec 2 (
)
1
2
1
)
) cot (
2
2
)
S
1
)
S
2
)
)
1
cosec 2 (
2
1
cosec 2 (
The condition
1
4 a
2
cosec 2 (
2
m2
m2
)
) 0 . This gives the result:
cosec 2 (
2
1
2
dq
( a,
d
(iii)We first use the condition
m1
) cot (
2
2
)
0
1
cosec 2 (
2
1
cosec 2 (
2
S
1
)
S
2
)
gives the following equation:
1
m2
cos (
8 a sin
2
3 1
2
(
S
S
2
)
2
)
1
cos (
sin
2
3 1
2
S
2
)
cU R e
(
S
2
1/ 2
)
________________________________________________________________________
Page 4-13
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.13
cn
(n 1)
z
dz
d
n 1
cn
1
n
n
c n ... for critical points
hence
That is, the critical points are located at c times the nth root of +1 in the
critical points are located at:
c e i2
m/n
where m
plane. Hence the
0,1, 2, ... , (n 1)
+ic
-c
+c
+c
n=2
-c
+c
-ic
n=4
n=3
From the equation of the mapping function it follows that:
cn
x iy
ei
e i ( n 1)
(n 1) n 1
Hence, equating real and imaginary parts on each side of this equation:
Using the relations R 2
R2
2
(cos 2
x
cos
cn
(n 1)
y
sin
cn
(n 1)
x2
Page 4-14
cos (n 1)
n 1
sin ( n 1)
y 2 and tan
sin 2 )
c 2n
( n 1) 2
n 1
2c
( n 1)
2( n 1)
y / x we get:
n
n 1
cos cos( n 1)
cos 2 ( n 1)
sin 2 ( n 1)
sin sin ( n 1)
TWO-DIMENSIONAL POTENTIAL FLOWS
cn
sin
sin (n 1)
(n 1) n 1
tan
cn
cos
cos (n 1)
(n 1) n 1
Hence the equations defining the surface in the z plane are
R
1
2
n
c
cos n
(n 1)
n
c
(n 1)
tan
n
c
(n 1)
sin (n 1)
sin
1
1/ 2
tan
cos (n 1)
cos
Expanding the results obtained above for
R
1
c
2
(n 1)
2n
1
(n 1)
1 gives:
n
cos n
n
tan
The object shape for
1
sin n
(n 1) sin cos
tan
0.7, n 3 is shown below.
y /ρ
x/ρ
Page 4-15
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.14
From the mapping function the parametric equations of the mapping are:
c2
cos
x
y
c2
sin
For the chosen system of units a circle is drawn in the
plane. From this diagram, values of the
radius
are obtained at the various values of the angle over the range 0 0 to 360 0 . The
corresponding values of x and y are then calculated from the parametric equations above. The
following diagrams were drawn using an EXCEL program to calculate the values of x and y and
to plot the results.
(a) SI Units (mm)
(b) English Units (inches)
Problem 4.15
In the following,
is the angle through which the stagnation point must be rotated in order to
satisfy the Kutta condition, a is the radius of the circle that produces the airfoil, and L is the
length of the wing element.
(a) SI Units
The chord in Prob. 4.14 was 241 mm, so that all lengths will be magnified by the ratio
3.0/0.241 = 12.448.
Page 4-16
TWO-DIMENSIONAL POTENTIAL FLOWS
tan
1
7.5
60.0 5.0
6.582 0
a 12.448 (60.0 5.0) 2
air
FL
7.5 2 mm 0.8145 m
1.225 kg / m 3
4 U a sin
293.3 m 2 / s
U L 1.225 250 293.3 1.0 N
8.982 10 4 N
FL
(b) English Units
The chord in Prob. 4.14 was 9.66 inches = 0.805 feet, so that all lengths will be magnified by
the ratio 9.0/0.805 = 11.18.
tan
1
0.3
2.4 0.2
6.582 0
a 11.18 (2.4 0.2) 2
air
FL
0.002378 slug / ft
0.3 2 ins
2.438 ft
3
4 U a sin
2, 634.3 ft 2 / s
U L (0.002378 32.2) 750 2, 634.3 3.0 lb f
4.538 10 5 lb f
FL
Problem 4.16
Consider a sector whose vertex is located at the origin in the z -plane (point A) and let the
corresponding point in the -plane be the origin (point a). Then, from the Schwarz-Christoffel
transformation:
1
1
dz
0) n
K(
d
z
nK
1
n
The constant of integration has been taken to be zero since z 0 when
0 . The scaling
constant K is undetermined in this case, and there is no loss of generality in specifying its value
to be such that n K 1 . Then:
zn
Page 4-17
TWO-DIMENSIONAL POTENTIAL FLOWS
Hence for a uniform flow of magnitude U in the
-plane:
F ( z) U z n
Problem 4.17
dz
d
K(
1)
K
2
K
z
When z 0,
constant K
(
1/2
0) 1 (
(
1
1) 1/2
1
(
2
1
1)
K cosh
1
1/2
2
(
sec
1) 1/2
1
A
1 so that the constant A = 0. Also, when z
/ . Thus the equation of the mapping is:
z
1) 1/2
1) 1/2
(
(
1) 1/2
cosh
1
sec
(1 i ) ,
1 so that the
1
The flow in the -plane is that of a source located at the origin. Since half of this flow will be in
each half-plane, the strength of the source in the -plane will be 2 U , so that the complex
potential will be:
U
F( )
log
Problem 4.18
dz
d
K(
0) 1 (
1
1
K
1)
/
Hence the differential equation of the mapping function is:
Page 4-18
(
)
/
TWO-DIMENSIONAL POTENTIAL FLOWS
dz
d
1
K
r
2n
1
The flow in the -plane is that of a source located at the origin. Since half of this flow will be in
each half-plane, the strength of the source will be 2 U H . Hence the complex potential will be:
UH
F ( z)
log
W ( z)
dF
dz
dF d
d dz
r
2n
UH
1
K
r
2n
UH
i.e. W ( z )
K
As the point D is approached, z
equation:
while
and W ( z ) U . Hence from the foregoing
UH
1
K
W ( z) U
H
K
As the point A is approached, z
W ( z)
1
0 and W ( z) U H / h . Hence:
while
UH
h
UH
K
H
h
r
2n
U
r
2n
2n
r
Page 4-19
TWO-DIMENSIONAL POTENTIAL FLOWS
Problem 4.19
dz
d
K(
0) 1 (
1
1
K
let s 2
1) 1/ 2 (
)
1/ 2
1/ 2
1
s2
then
1 s2
1
so that
1
1 s2
(
and
(1 s 2 )
s(
s2)
2s(
1)
2 2
(1 s )
2(
1)
K
2
s 2)
(1 s )(
dz
d
d
but
ds
dz dz d
ds d ds
z
K
2K
1
(1 s 2 )
1
(
2K
1
2 (1 s)
1
2 (1 s)
K log (1 s ) log (1 s )
K log
At the location C,
that s
. Also, z
1 s
1 s
1
s2)
1
2
(
s)
1
s)
1
2
log (
(
s)
s)
A
A
0 , so that A 0 . At the location B,
1
(
1
log (
s
s
log
so that s 0 . Also, z
i ( H h) so that:
K
Page 4-20
1) s 2
1 s2
1)
(H
h)
1 so
TWO-DIMENSIONAL POTENTIAL FLOWS
z
1
(
1)
(H
h) log
1 s
1 s
1
s
s
log
In order to determine the constant , we specify the velocity at the location z
to have the
value U. Then the source at z
will have strength 2U h , which will also be the strength of
the source located at
0 . Then:
F( )
W( )
(2 U h)
log
2
Uh 1
(
d
W( ) U
dz
W ( z)
1)
(H
Then using the conditions discussed above at the location
H
h
1/ 2
h
h)
1
0 produces the following result:
2
Thus the equation of the mapping becomes:
z
where s 2
H
log
1 s
1 s
h
H /h
log
H
H /h
s
s
( H / h) 2
1
Page 4-21
5
THREE-DIMENSIONAL POTENTIAL FLOWS
THREE-DIMENSIONAL POTENTIAL FLOWS
Problem 5.1
From Appendix A, the vorticity vector in spherical coordinates is defined by:
u
er
r e
r sin e
r
ur
r u
r sin u
For axisymmetric flows, the velocity component u will be zero and all derivatives with respect
to will be zero. Thus two components of the vorticity vector will be zero, and the remaining
component is defined as follows:
(r , ) r sin
u r
(r u ) r sin
r
1
1 2 2
r
sin
sin
r r 2
In the last equality, the definition of the Stokes stream function has been used from Eqs. (5.3a)
and (5.3b). Thus if the flow is irrotational, the equation to be satisfied by the Stokes stream
function is:
2
1
sin
2
r
sin
r2
0
Problem 5.2
Equation (5.18) is:
2
1
sin
0
2
r
sin
From Eq. (5.5b) the stream function for a uniform flow is:
r2
1
(r , ) U r 2 sin 2
2
U r 2 sin 2
2
r
1
2
2
sin
U r sin
sin
2
Then
And
Hence
r2
1
(r , ) U r 2 sin 2
is a solution.
2
Page 5-1
THREE-DIMENSIONAL POTENTIAL FLOWS
From Eq. (5.6b) the stream function for a source is:
Q
( r , )
(1 cos )
4
Then
And
2
0
r
r 2
1
sin
0
sin
2
( r , )
Hence
Q
(1 cos )
4
is a solution.
From Eq. (5.7b) the stream function for a doublet is:
( r , )
sin 2
4 r
r2
Then
And
Hence
sin
2
sin 2
2
r
2 r
1
sin 2
sin 2 r
( r , )
sin 2
4 r
is a solution.
Problem 5.3
Substituting the assumed form of solution into Eq. (5.18) and dividing the entire equation by the
product RT produces the following result:
r 2 d 2 R sin d 1 dT
0
R dr 2
T d sin d
Employing the usual argument of separable solutions, we now point out that since the first term
is a function of r only and the second term is a function of only, the only way that the two
terms can add up to zero for all values of r and is for each term to be constant. For
convenience, we choose the value of the constant for the term involving R to be n (n 1) . Then
the ordinary differential equation for R will be:
d 2R
r2
n (n 1) 0
dr 2
This is an equi-dimensional ordinary differential equation, so that the solution will be of the
following form:
R(r ) A r m
Then it follows that
m (m 1) n (n 1) 0
The two possible solutions to this algebraic equation are:
Page 5-2
THREE-DIMENSIONAL POTENTIAL FLOWS
m n or m (n 1)
The second possibility leads to diverging values of the velocity as the radius r becomes large.
Thus for finite values of the velocity, the solution for R will be:
R n (r ) An r n
The differential equation that is to be satisfied by T then becomes:
d 1 dT
n (n 1) T 0
d sin d
Let cos be the new independent variable. Then the differential equation becomes:
sin
d 2T
(1 ) 2 n (n 1) T 0
d
2
Next introduce a new dependent variable that is defined by the relation T (1 2 ) 1/ 2 . From
this definition, the following derivatives are obtained:
dT
d
(1 2 ) 1/ 2
(1 2 ) 1/ 2
d
d
2
d 2T
d
2 1/ 2 d
(1 )
2 (1 2 ) 1/ 2
2 (1 2 ) 3/ 2 (1 2 ) 1/ 2
2
2
d
d
d
Thus, in terms of the new dependent variable , the differential equation becomes:
d 2
d
1
(1 ) 2 2
n ( n 1)
0
d
d
1 2
2
This is a special form of the Associated Legendre equation. The general form of the Associated
Legendre equation is:
d 2
d
m2
n
n
2
(
1)
0
d 2
d
1 2
The solution of this last equation, for any values of m and n , is:
(1 2 )
mn () Bmn Pn m () C mn Qn m ()
where Pn ( ) (1 2 ) m / 2
m
and
Q n ( ) (1 )
m
2 m/2
d m Pn ( )
d m
d mQ n ( )
d m
Page 5-3
THREE-DIMENSIONAL POTENTIAL FLOWS
m
m
In the foregoing expressions, Pn and Qn are, respectively, the Associated Legendre functions
of the first and second kind of order m , while Pn and Q n are, respectively, the Legendre
functions of the first and second kind. Then the equation derived here is the Associated
Legendre equation of order one; that is, m 1 . The Legendre functions of the second kind
diverge for values of 1 so that, in order to eliminate the possibility of infinitely large
velocities, the constants C n
m
must be taken to be zero. Thus the finite solutions to the given
differential equation, for any value of n , must be of the following form:
( ) (1 2 ) 1/ 2
dPn ( )
d
Combining the elements of the solution we have, for any value of n :
(r , ) An R n (r ) (1 2 ) 1/ 2 n ( )
where cos
Substituting the foregoing results for R n (r ) and n ( ) into the expression for the stream
function yields the result:
(1 2 ) dPn ( )
n ( r , ) An
r2
d
sin d
Pn (cos )
An n
r d
This solution is valid for any value of the integer n , so that a more general solution will be
obtained by superimposing all such solutions. This produces the result:
n (r , ) An
n 1
sin d
Pn (cos )
r n d
Problem 5.4
Substituting An 0 for n 1 in the result obtained in the foregoing problem gives:
sin d
P1 (cos )
r d
But P1 (cos ) cos so that the first term in the general solution gives the following result:
(r , ) A1
(r , ) A1
sin 2
r
This is the same as Eq. (5.7b) in which A1 / 4 .
Page 5-4
THREE-DIMENSIONAL POTENTIAL FLOWS
Problem 5.5
Using the expression for the stream function for a doublet as given by Eq. (5.7b), and employing
the notation defined in Fig. 5.13, the following result is obtained:
( r , )
*
sin 2
sin 2
4
4
We now consider a point P that lies on the circle of radius r a . Then, referring to Fig. 5.13,
the following two identities follow from application of the sine rule:
a
sin ( )
a
sin ( )
sin
sin
Using these two results, the equation for the stream function on the circle of radius a becomes:
a 2 sin 3
*
( a, )
3
3
4
This result shows that the stream function will not be constant on r a unless:
3
In order to evaluate the ratio of the lengths that appear in this result, we employ the cosine rule
on the triangles shown in Fig. 5.13 to obtain the following two identities:
*
2 a 2 l 2 2 a l cos
a4
a3
2
cos
l2
l
Eliminating cos between these two equations gives the following results:
a
l
2 a2
3
a
*
l
Substituting this last result into the expression for the stream function produces the following
result for the stream function for a sphere of radius a with a doublet of strength located along
the flow axis at r l :
( r , )
a3
sin 2
sin 2
3
4
4 l
Page 5-5
THREE-DIMENSIONAL POTENTIAL FLOWS
Using Eq. (5.7a) for the velocity potential yields the following result for the two doublets:
( r , )
*
cos
cos
4 2
4 2
Using the result obtained above relating the strengths of the two doublets produces the following
expression for the velocity potential for a sphere of radius a with a doublet of strength
located along the flow axis at r l :
( r , )
a3
cos
cos
4 2
4 l 3 2
Problem 5.6
From Eq. (5.14c) the force due to a doublet of strength is given by:
u i
(5.6.1)
F
x
where u i is the velocity induced by all singularities associated with the flow field, except that of
the doublet in question. Then, from the results obtained in the previous problem, the appropriate
velocity potential, due to the doublet located at x a 2 / l , will be:
( r , )
a3
cos
4 l 3 2
But on the x -axis, for x a 2 / l , we have the following values:
0
and
(x a 2 / l)
Then along the x -axis, the appropriate velocity potential is:
( x, 0)
u i ( x, 0)
and
u i
4 l 3 ( x a 2 / l ) 2
a3
( x, 0) e x
ex
2 l 3 ( x a 2 / l ) 3
x
( x, 0)
x
u i
a3
3 a3
ex
2 l 3 ( x a 2 / l ) 4
3 a 3l
ex
2 (l 2 a 2 ) 4
x
Substituting this last result into Eq. (5.6.1) gives the following value for the force:
(l , 0)
F
Page 5-6
3 a 3l
ex
2 (l 2 a 2 ) 4
THREE-DIMENSIONAL POTENTIAL FLOWS
Problem 5.7
Integrating the partial differential equation for (r, t ) once with respect to r gives:
r2
f (t )
r
In the foregoing, f (t ) is some function of time. Applying the second of the given boundary
conditions shows that the function f (t ) has the following value:
f (t ) R 2 R
Thus the radial velocity in the fluid at any distance r from the sphere at any time t will be:
R2 R
(r , t ) 2
r
r
Integrating the foregoing equation with respect to r yields the result:
R2 R
(r , t )
g (t )
r
where g (t ) is some function of time. Since the velocity potential must be a constant, at most,
for large values of the radius r , it follows that the function g (t ) must be zero. Then the value of
the velocity potential will be:
(r , t )
R2 R
r
In order to establish an expression for the pressure, we employ Bernoulli’s equation in the form
given by Eq. (3.2c). Then, since the density is constant in this case, Bernoulli’s equation gives:
p 1
F (t )
t 2 r
Since the first and third terms on the left side of this equation are zero far from the sphere, and
given that the value of the pressure is p 0 there, the value of the quantity F (t ) is F (t ) p 0 / .
Then, using the results obtained above, the Bernoulli equation becomes:
2
p0
( R 2 R 2 R R 2 ) p (r , t ) 1 R 2 R
2
r
2 r
p( r, t ) p 0 R 2 R 2 RR 2 R 4 R 2
r
r
2r 4
2
Then on the surface of the sphere, where r R , this expression becomes:
p( r , t )
p0
3
RR R2
2
Page 5-7
THREE-DIMENSIONAL POTENTIAL FLOWS
At t 0 the foregoing expression becomes:
1 d
( R 3/2 R)
1/2
2
R dt
Integrating this equation once with respect to r produces the following result:
3
0 RR R2
R 3/2 R constant R0 3/2 R0
3/ 2
hence
and
R0 R0
dR
dt
R 3/ 2
1
dt R0 3/ 2 R0
R
3/ 2
dR
Evaluating the foregoing integrals between the limits of 0 and t for time, and R 0 and 0 for the
radius R , we get the result:
t
2 R0
5 R0
Problem 5.8
Eq. (5.9b) defines the velocity potential for a uniform flow of magnitude U approaching a
sphere of radius a in the positive x -direction. If we replace U with U in this expression, we
will get the velocity approaching the sphere in the negative x -direction. Then, if we add a
uniform flow in the positive x -direction, the result will correspond to a sphere that is moving
with velocity U in an otherwise quiescent fluid. The resulting velocity potential is:
( r , )
1U
a3
r2
2
cos
(5.8.1)
In the foregoing equation, U , r and are all considered to be functions of time. In order to
switch to coordinates x and R , where both of these coordinates do not vary with time, we use
the relations:
( x x 0 ) r cos
and
r [ R 2 ( x x 0 ) 2 ]1/ 2
Substituting these expressions into Eq. (5.8.1) we get the following result:
1
(x x0 )
2
[ R ( x x 0 ) 2 ] 3/2
( x, R ) U a 3
2
(5.8.2)
In the foregoing, U and x 0 are both considered to be functions of time, but all of the other
quantities are time-independent. To obtain an expression for the pressure, we employ
Bernoulli’s equation in the form defined by Eq. (3.2c) in which the quantity F (t ) is evaluated
Page 5-8
THREE-DIMENSIONAL POTENTIAL FLOWS
using the given pressure far from the body where temporal variations vanish, and where the
velocity components vanish. Thus the applicable form of Bernoulli’s equation is:
p
p( r, ) 1
uu
t
2
It is easier to evaluate the first term in this equation using Eq. (5.8.2), rather than using Eq.
(5.8.1). We note that only U and x 0 are functions of time, and that
x 0
U
U
U
and
t
t
Then from Eq. (5.8.2) the following expression is obtained for the temporal derivative of the
velocity potential:
3
2 3
1U a
1U a
( r , )
cos
(1 3cos 2 )
2
3
t
2 r
2 r
The velocity-squared term in Bernoulli’s equation is readily obtained from Eq. (5.8.1), producing
the following result:
2
2
1
u u
r
r
U 2a6
1
2
2
cos sin
6
4
r
Substituting these two results into the Bernoulli equation produces the following expression for
the pressure at any point in the fluid:
p( r , ) p
1
2
U a 3
r
2
cos
1
2
U 2 a 3
r
3
(1 3cos 2 )
1
2
U 2 a 6
r
6
1
(cos 2 sin 2 )
4
The force that acts on the sphere will be in the positive x -direction and its value may be
obtained by integrating the pressure over an annular area of surface, ensuring that the x component of this product is evaluated. That is, the force on the sphere will be defined by:
F 2 a 2 p(a, ) sin cos d
0
Using the result obtained above for the pressure, its value on the surface will be:
1
1
1
2
2
4
p(a, ) p U a cos U 2 (1 4cos 2 sin 2 )
Substituting this value of the surface pressure into the integral expression for the force acting on
the sphere shows that the only non-zero term will be:
F a 3 U sin cos 2 d
0
Evaluating this integral shows that the force acting on the sphere will be:
2
F a 3 U
3
Referring to Eq. (5.17), the foregoing result will be recognized as the apparent mass of the
sphere multiplied by the acceleration of the sphere. That is, this is the same result as would be
Page 5-9
THREE-DIMENSIONAL POTENTIAL FLOWS
obtained by replacing the fluid flow field with its apparent mass and considering this mass to
follow the sphere while obeying Newton’s second law of motion.
_____________________________________________________________________________
Page 5-10
6
SURFACE WAVES
SURFACE WAVES
Problem 6.1
From Eq. (6.5a)
2
c2
2 h
2 h
1
tanh
2
gh 2 h
gh
V2
Let
Then Eq. (6.1.1) becomes:
Then for
1:
c2
,
and S
gh 2
gh
2 h
S
1
V 2 1 2 tanh
S
S
V 2 1 2
(6.1.1)
(6.1.2)
Hence for a minimum value of V:
dV
S
1 2 0
d
S
2V
(6.1.3)
That is:
2 h
gh 2
Substituting the results from Eq. (6.1.3) into Eq. (6.1.2) gives:
1
V 2 S 1 1 tanh
S
That is
gh 2
c2
tanh
2
gh 2
gh
_____________________________________________________________________________
Problem 6.2
Using a frame of reference that is moving with velocity c in the positive x -direction, the
quantity z c t should be replaced by z in our earlier results. Also, from the moving frame of
reference, there will be a uniform flow of magnitude c in the negative x -direction, so that this
component must be added to the solution. Hence, using Eq. (6.7c), the required complex
potential will be:
F ( z, t ) c z
c
2
cos
( z i h)
sinh (2 h / )
We expand the ratio of the two transcendental functions as follows:
Page 6-1
SURFACE WAVES
cos
2
( z i h)
sinh (2 h / )
cos
2 z
cosh
2 h
i sin
2 z
sinh
sinh (2 h / )
2 z
2 h
2 z
coth
cos
i sin
2 z
2 z
cos
i sin
i
2 h
for 2 h /
1
2 z
e h
Hence the deep-liquid version of the foregoing result is:
F ( z, t ) c z c e
F ( z , t ) i c ( x i y ) c e
i
2
i
2 z
(x i y)
2 y
2 x
2 x
i sin
cos
Hence the stream function is defined by the following expression:
c ( x i y) c e
( x, y ) c y c e
2 y
sin
2 x
When y , 0 . Hence from the foregoing expression:
e
2 y
sin
2 x
Problem 6.3
F ( z , t ) c z c e i 2 z /
W ( z)
dF
2 i 2 z /
e
c 1 i c
dz
2 i 2 z /
e
and W ( z ) c 1 i c
2
2 i 2 z /
2 i 2 z /
2 i 2 ( z z )/
e
e
hence WW ( z ) c 2 1 i
i
2
e
2
2
2
2 2
4 y /
e
c 1 2 sin
Page 6-2
SURFACE WAVES
Substituting this result into Bernoulli’s equation produces the following result:
2
2
4 2 y /
2 x
P 1 2
2 2
4 y /
2 2
c 1
e
sin
g y c 1
e
2
2
Using this result on the free surface where y and p P , and using Eq. (6.18a), gives:
p
1
2
2
2
4
1 2
2 2
4 /
2 2
c 1
g c 1
e
2
2
Solving this equation for c 2 gives:
1
2
c2
2 g
(2 / ) (4 / ) 2 (2 / ) 2 e 4 /
2
2
Expanding the foregoing result and noting that / / gives:
2 g
c2
4
4 /
e
1
g / 2
4
1
1
g / 2
1 (2 / ) 2 2
For / 0 the foregoing result becomes:
c2
This result confirms Eq. (6.3b). Also, for /
c2
g
2
1 the same equation shows that:
g /(2 )
1 (2 / ) 2 2
The last result shows that the effect of finite amplitude is to increase the speed of the wave.
Problem 6.4
For a wave travelling at velocity c in the positive x -direction on a liquid of depth h , the
complex potential is:
Page 6-3
SURFACE WAVES
c
2
cos
( z c t i h)
sinh (2 h / )
We change the speed from c to U , and the liquid depth from h to H . At the same time, we
superimpose a uniform flow of magnitude U in the negative x -direction to account for the fact
that we are observing the flow from a frame of reference that is moving with the wave; that is,
with velocity U in the positive x -direction. Then the wave will appear to be stationary, and the
originally-quiescent liquid will appear to be approaching us in the negative x -direction. Also,
the distance z from the fixed origin will change such that z c t z . The resulting complex
potential is:
2
U
cos
(z i H )
F ( z) U z
sinh (2 H / )
F ( z)
U ( x i y)
2
U
cos
[ x i ( y H )]
sinh (2 H / )
2 x
2
2 x
2
U
cos
cosh
( y H ) i sin
sinh
( y H )
sinh (2 H / )
Hence the stream function for the flow will be:
U ( x i y)
( x, y ) U y
U
2 x
2
sin
sinh
(y H)
sinh (2 H / )
The relationship connecting the mean liquid speed, the liquid depth, and the wavelength of the
surface wave is given by Eq. (6.3a) in which c is replaced by U and the depth h is replaced by
H . This gives:
U2
2 H
tanh
g H 2 H
Problem 6.5
Put U h when y h 0 sin (2 x / ) in the expression obtained in Prob. 6.4 for the
stream function. This produces the following equation:
U
2
0 U 0
sinh
[( H h) 0 sin 2 x / ]
sinh (2 H / )
Consider 0 ( H h) and solve this equation for the amplitude ratio, to get:
sinh (2 H / )
0 sinh 2 ( H h) /
Page 6-4
1
cosh (2 h / ) coth (2 H / ) sinh (2 h / )
SURFACE WAVES
But, from Prob. 6.4, coth (2 H / ) 2 U 2 / g . Hence the expression for the amplitude ratio
becomes:
1
0 cosh (2 h / ) ( g / 2 U 2 ) sinh (2 h / )
It may be observed that the amplitude ratio will be positive (so that the surface wave will be in
phase with the bottom surface) for values of U for which:
U2
g
2 h
tanh
g h 2 h
Conversely, the amplitude ratio will be negative (so that the surface wave will be out of phase
with the bottom surface) for values of U that reverse this inequality.
Problem 6.6
Superimposing the two given waves produces the following composite wave:
1
2
2
( x, t ) sin
( x c1 t ) sin
( x c 2 t)
2
1
2
1
1
2
2
but sin A sin B 2cos ( A B) sin ( A B)
Hence the equation of the composite wave may be written in the following form:
1
1
c1 c 2
c1 c 2
1
1
x
t sin
x
t
1 2
1 2
1 2
1 2
( x, t ) cos
If c1 c 2 and 1 2 , the coefficient of time in the cosine component will be much smaller than
that of the sine component. Thus the cosine function will vary much more slowly with time than
the sine function. Hence the expression for ( x, t ) may be thought of as being of the following
form:
1
c
c
1
( x, t ) A( x, t ) sin x 1 2 t
1 2
1 2
In the foregoing equation, the amplitude A( x, t ) varies slowly with time compared with the
frequency of the sine term.
Problem 6.7
(a)
V
1
0
2
g 2 dx
Page 6-5
SURFACE WAVES
1
2
( x c t ) dx
g 2 sin 2
0
2
1
V g 2
Hence:
4
(b) From Eq. (6.7a):
2 y
2 h
2 y
( x c t ) sinh
cosh
coth
2 h
2
( x, 0) c coth
cos
(x c t)
( x, y ) c cos
2
Also
Hence:
2
2
( x, 0) c
cos
( x c t)
y
1
2
2 h
2
coth
cos 2
( x ct ) dx
T c 2 2
2
0
where c
2
2 h
g
tanh
2
1
T g 2
4
Problem 6.8
The linearized form of Bernoulli’s equation is:
p ( x, y , t )
( x, y , t )
g y F (t )
t
The function F (t ) may be considered to be incorporated into the velocity potential ( x, y, t ) . In
addition, the pressure may be considered to be measured relative to the hydrostatic value, so that
the term g y may be considered to be incorporated into the pressure term. Thus the pressure
perturbation that is induced by the wave will be represented by:
p ( x, y , t )
( x, y , t )
t
2
2
2 y
2 h
2 y
c 2
coth
sin
( x c t ) sinh
cosh
In the above, Eq. (6.7a) has been used for the velocity potential ( x, y, t ) . From this same
equation we also evaluate:
2
2
2 y
2 h
2 y
coth
( x, y , t ) c
sin
( x c t ) sinh
cosh
x
Page 6-6
SURFACE WAVES
2 y
2 h
2 y
2
2 2
( x c t ) sinh
cosh
c 3 2
coth
sin
x
2
2
cosh 2
( y h)
2
2 2
(x c t)
c 3 2
sin
2 h
sinh 2
4
2
1 cosh
( y h)
1
2
2
2
(x c t)
c 3 2
sin
2 h
2
sinh 2
Substituting this result into the expression for the work done gives:
0
WD p
dy
h
x
4
h
sinh
( y h)
2
1
4
2
2 2
c 3 2
(x c t)
sin
2
2 2 h
sinh
2
2
p
Using Eq. (6.3a) to eliminate c 2 from the foregoing equation produces the result:
2 h
2 h
2 h
sinh
cosh
1
2
(x c t)
WD g c 2 sin 2
2 h
2 h
2
sinh
cosh
WD
1
2
2 h /
g c 2 sin 2
( x c t ) 1
2
sinh (2 h / ) cosh (2 h / )
For deep liquids, the term inside the square brackets in the foregoing equation becomes unity.
Then the average work done will be:
(WD) ave
/c
0
WD dt
/c
1
2
( x c t ) dt
g c 2 sin 2
0
2
1
g 2
4
In the above, the integration has been carried out over time corresponding to one complete cycle
of the wave train. This result shows that for deep liquids:
i.e. (WD) ave
1
(WD) ave (T V )
2
Page 6-7
SURFACE WAVES
Problem 6.9
Since the fluid motion is in the x y -plane only, the vorticity vector will lie in the z -plane and
will have the following magnitude:
v u
2
x y
Hence from the given expression for the stream function:
(k 2 l 2 )
The material derivative of the total vorticity is given by the following expression:
D
u
v
v
Dt
t
x
y
t
x
In the foregoing, the nonlinear terms have been neglected as being quadratically small. Then:
D
(k 2 l 2 )
Dt
t
x
2
2
( k l )(i ) (i k )
D
i k (k 2 l 2 )
Dt
From the foregoing result, the value of that makes the material derivative of the total vorticity
equal to zero is:
k
(k l 2 )
2
Problem 6.10
The velocity potential in both regions of the flow must satisfy Laplace’s equation and the
solutions will be similar to those of section 6.12. However, the solution for 1 must be able to
satisfy the boundary condition at y , namely, that / y must vanish there. Thus the
exponential function in the solution for 1 should be replaced with the equivalent hyperbolic
function whose derivative with respect to y vanishes at y . This means that the solution
will be of the following form:
2
(x t)
i
2
y 0 :
cosh
(y )
1 ( x, y, t ) A1e
0 y:
Page 6-8
2 ( x, y , t ) A2 e
i
2
(x t)
e
2
y
SURFACE WAVES
Imposing the kinematic interface conditions given by Eqs. (6.16e) and (6.16f), with U 1 0 and
U 2 U , yields the following values for the constants:
A1
i
2
sinh
A2 i ( U )
Using these values, the expressions for the two velocity potentials become:
2
cosh
(y )
2
i
(x t)
1 ( x, y , t ) i e
2
sinh
2 ( x, y, t ) i ( U ) e
i
2
(x t)
e
2
y
Next, we impose the pressure condition at the interface between the two fluids. This condition is
defined by Eq. (6.16g), with U 1 0, U 2 U and 1 2 . This results in the following
quadratic equation for :
2 2
2
1 coth
2U U 0
Thus the coefficient of the time in the equation of the interface will be:
U
1 i coth 1/ 2
1 coth
2
2
This result shows that will have an imaginary part for all wavelengths and all flow
velocities U , so that the interface will be unstable. The wavelength that makes the imaginary
part of a maximum is 0 .
Problem 6.11
As in the last problem, we take the equation of the interface to be:
( x, t ) e
i
2
(x t)
The solution for the velocity potential in each flow region will be similar to that of section 6.12,
except that the y -dependence must be capable of satisfying the boundary conditions that
/ y 0 on both y h and y h . Hence the velocity potentials will be of the following
form:
Page 6-9
SURFACE WAVES
1 ( x, y, t ) A1 e
i
2
( x ct )
cosh
2
(h y)
2
( x ct )
i
2
2 ( x , y , t ) A2 e
cosh
(h y)
Substituting these expressions into the kinematic conditions defined by Eqs. (6.16e) and (6.16f)
in which U 1 U 2 0 , produces the following values for the constants:
A1
i
2 h
sinh
i
A2
2 h
sinh
Substituting these values into the expressions for the velocity potentials gives:
2
cosh
(h y )
2
( x ct )
i
1 ( x, y , t ) i e
2 h
sinh
2
cosh
(h y )
2
i
( x ct )
2 ( x, y , t ) i e
2 h
sinh
Next we impose the pressure condition at the interface, which is defined by Eq. (6.16g) with
U 1 U 2 0 . This gives, after dividing through by , the following expression for :
2
2 h
2
2 h
1 2
1 g 2 2
2g
coth
coth
Solving this equation for gives the following result:
1 2 g
2 h
tanh
1 2 2
This shows that the interface is stable (i.e. the amplitude will decay with time) for 1 2 , but
that it is unstable (i.e. the amplitude will grow with time) for 1 2 .
Page 6-10
7
EXACT SOLUTIONS
EXACT SOLUTIONS
Problem 7.1
The solution that will exist for large times will be that for flow between two parallel surfaces, as
defined by Eq. (7.1b). Then the solution for finite times will be of the following form:
u ( y, t ) y
u * ( y, t )
U
h
2 *
u*
u
where
t
y2
with
u * ( y, ) 0
and
u * ( y,0)
y
h
( 1) n
1
n 1
2
n y
sin
n
h
*
In the above, the velocity u ( y, ) has been made to vanish leaving the Couette flow solution
for large values of time. Also, the value of u * ( y, 0) has been made equal to the negative of the
Couette flow distribution so that the net value of the total velocity u( y, 0) is zero. The Fourier
series for the saw-tooth value of u * ( y, 0) has been used to make the initial condition compatible
with the separable solution to the partial differential equation.
The solution for u * ( y, t ) may now be obtained by separation of variables. The solution will be
trigonometric in y , to permit matching the initial condition at t 0 , and it will be exponential in
time. Thus the solution becomes:
2
2
n y
e n / h t sin
n
h
n 1
This expression satisfies the differential equation and the boundary conditions. Hence the
complete solution becomes:
u * ( y, t )
u( y , t )
U
( 1) n
y
h
1
( 1) n
1
n 1
2
e
n
n /h
2
t
sin
n y
h
Problem 7.2
The partial differential equation and boundary conditions to be satisfied are:
2
u
u
for u( y , t )
t
y2
u( y ,0) U
y
h
1
n 1
n 1
n y
2U
sin
n
h
u( y , t ) finite
Page 7-1
EXACT SOLUTIONS
In the foregoing, the Fourier series representation for the saw-tooth velocity distribution that
exists at t 0 has been deduced from Prob. 7.1.
The solution for u ( y, t ) may be obtained by separation of variables. The solution will be
trigonometric in y , to permit matching the initial condition at t 0 , and it will be exponential in
time. Also, we recall that the separation constants for both the y and t dependence must be
related to each other in such a way that they produce a solution to the partial differential
equation. Thus the solution becomes:
2U
1n
u( y , t )
n
1
n 1
n y
sin
e
h
n2 2
t
h2
Problem 7.3
Consider the solution to consist of the velocity that exists for large values of the time t, plus a second
component of the solution. For large values of the time t the flow will be independent of time, and the
velocity, call it u 1, will depend on y only. Hence the governing equation becomes:
d 2u1
0 2P
dy 2
Py 2
u1 ( y )
Then:
u1 ( 1) 0
P
A
Ay
B and u1 ( 1) 0
A 0 and B
u1 ( y )
Substituting u( y, t )
B
u2
t
t
2P
B
y2)
P(1
2
A
P
u1 ( y ) u 2 ( y, t ) into the governing equation for t
u1
P
u1
y2
2
0 gives:
u2
y2
The first term on the LHS and the first and second terms on the RHS add up to zero, so that the equation
governing the second component of the solution is:
u2
t
where u2 ( y,0)
Page 7-2
2
u2
y2
P(1
y2)
EXACT SOLUTIONS
But the Fourier series for (1
n
2
3
2
y )
4
n
n 1
2
( 1) n cos n y . Hence we look for a separation of
2
variables solution for u 2 ( y , t ) of the form:
u 2 ( y, t ) T (t ) cos n y
Substituting this expression into the PDE for u 2 ( y , t ) we get:
dT
n 2 2T
dt
2 2
T (t ) C n e n t
So that:
u 2 ( y, t )
Therefore:
C n cos n y e
n2
2
t
D
n 1
u 2 ( y,0)
But
P
2
P and C n
3
D
Therefore:
y2)
P(1
u( y , t )
y2) P
P(1
Hence:
4
where H n
n
2
2
n
2
3
P
2
3
4
n 1
n
4
n
2
2
2
2
( 1) n cos n y
( 1) n
n
H n cos n y e
n2
2
t
n 1
( 1) n
_____________________________________________________________________________
Problem 7.4
Here:
u u( y);
v w 0
For this situation, continuity is identically satisfied. Hence, noting that the gravitational force is
defined by f
g sin e x g cos e y 0 e z it follows from the Navier-Stokes equations that the
equation to be solved for u ( y) is:
d 2u
dy 2
g
sin
(a) (i) Integrating this equation twice gives u( y )
g
sin
y2
2
The conditions u(0) U and u(h) 0 give B U and A
Ay
B
gh
sin
2
U
h
Hence;
u( y ) U 1
y
h
gh 2 y
1
2 h
y
sin
h
Page 7-3
EXACT SOLUTIONS
Q
(ii)
h
0
h
u( y ) dy
0
U 1
(iii)
Q
(b) (i) Here u (0) U and
y
sin
h
dy
6U
gh 2
sin
0
gh 2 y
1
2 h
gh 3
sin
12
Uh
2
Q
y
h
du
(h) 0 (for zero stress). Hence:
dy
gh 2 y
2
2 h
u( y ) U
y
sin
h
(ii) Integrating:
Q Uh
sin
(iii) Q 0
gh 3
sin
3
3U
gh 2
which is a smaller angle than in (b)
Problem 7.5
Using Eq. (7.2c) for the velocity distribution, we get the following expression for the volumetric
flow rate, Q E , for the elliptic pipe:
QE
4
b
0
dz
a 1 z 2 /b 2
0
1 dp a 2b 2
y2
2 dx a 2 b 2 a 2
2 dp a 2b 2
ab
2
2
8
dx a
b
Denoting the flow area by A
a b and b / a by
following form:
1 dp 2
QE
A
4
dx 1
z2
b2
1 dy
, this equation may be written in the
2
(7.5.1)
1 dp 2
1
2 2
A
2
2 2
dx
4
1
(1
)
For a maximum flow rate with a given value of the area A , the derivative above must be zero.
This requires that
1 , so that:
QE
b
1
a
Page 7-4
EXACT SOLUTIONS
The velocity distribution for flow in a circular conduit is given by Eq. (7.2b). From this result
the volumetric flow rate QC in a circular conduit will be:
a
QC
0
2
1 dp 2
(a
4 dx
R 2 ) 2 R dR
dp a 4
dx 4
1 dp 2
A
8
dx
(7.5.2)
a 2 . From
In the foregoing, the flow rate has been expressed as a function of the flow area A
Eqs. (7.4.1) and (7.4.2), the ratio of the two volumetric flow rates, for a common pressure
gradient and a common flow area, is:
QE
2
2
QC 1
Hence for
4 / 3 the flow ratio becomes:
QE
QC
24
25
0.96
Problem 7.6
The partial differential equation that is to be satisfied in this case is the following:
2
2
dp
u
u
0
2
dx
y
z2
Then, substituting the assumed form of solution for the velocity u( y, z ) gives:
3 dp
6 b dx
Since this value is a constant, the assumed form of the solution is valid and the complete solution
becomes:
u ( y, z )
3 dp
z
6 b dx
b
2 3
z
3y
b
3
z
3y
b
2 3
Problem 7.7
(a)
u1 ( y )
A 1 4( y
1
2
)2
Page 7-5
EXACT SOLUTIONS
du1
1
8 A( y
dy
2
d 2u1
) and
8A
dy 2
Hence the differential equation is satisfied provided A = P so that a solution is:
u1 ( y )
u2 ( y, z )
(b)
Bn
n 1
2
n
2
2
2
Bn
n 1
u3 ( y, z )
(c)
Cn
n 1
2
(d)
u( y , z )
P 1 4( y
1
2
)2
n
u2
z2
n
2
2
z)
n2
2
u2
2
and
u3
u3
z
2
u3
sinh n (
z)
sin n y
sinh n
n 1
sinh n z
sinh n (
Bn
sin n y
Cn
sinh n
sinh n
n 1
1
Cn
0
z = α:
2
sin n y
P 1 4( y
0
P 1 4( y
1
2
1
2
z)
sin n y
1 . At z 0 and z
This expression satisfies the correct boundary conditions at y
z = 0:
n2
sinh n z
sin n y
sinh n
2
u3 ( y, z )
Hence another solution is:
and
u2
sinh n (
sinh n
u3
y
)2
sinh n z
sin n y
sinh n
u2 ( y, z )
Hence another solution is:
2
2
u2
y
1
P 1 4( y
)2
we have:
C n sin n y
n 1
)2
B n sin n y
n 1
The Fourier series for the bracketed term { } is the following:
1 4( y
1
2
)2 =
K n sin n x
where K n
n 1
16
1 ( 1) n
n3 3
Hence matching the boundary expressions requires the following values for the constants Cn and Bn:
Cn
u( y , z )
P 1 4( y
1
2
)2
PK n
P
Kn
n 1
and
Bn
PK n
sinh n z
sin n y
sinh n
P
Kn
n 1
sinh n (
sinh n
z)
sin n y
_____________________________________________________________________________
Page 7-6
EXACT SOLUTIONS
Problem 7.8
The only non-zero component of velocity in this case will be u( y, z ) and this velocity
component will be the solution to the following reduced form of the Navier-Stokes equations:
2
2
u
u
0
2
y
z2
with
u ( y, ) 0
2
n y
u( y,0) U U
[1 ( 1) n ]sin
and
b
n 1 n
In the foregoing, the boundary condition on z 0 has be written in terms of the Fourier series
for a square wave of height U . The solution to this problem may be obtained by separation of
variables. In this solution, the y -dependence will be trigonometric and the z -dependence will
be exponential. Hence, the solution will be:
u( y , t ) U
n 1
n y
2
e
[1 ( 1) n ] sin
n
b
n z
b
The volumetric flow rate Q may be evaluated by integrating the velocity to give:
Q
0
dz
b
0
U
n
2
n y
1 ( 1) n sin
e
b
1n
n z
b
dy
n z
2b
n 2
b
1
(
1)
e
dz
2 2
0
n
n 1
Thus the volumetric flow rate past any vertical plane across the flow will be:
U
2
Q U b2
n 1
n
3
3
1 ( 1) n
2
Problem 7.9
1 d
dw
R
R dR
dR
(a)
Fluid #1:
w1 ( R )
1 dp R 2
1 dz 4
Fluid #2:
w 2 ( R)
1 dp R 2
2 dz 4
The fact that w1 ( R) should be finite when R
matching the shear stresses at R
1 dp
dz
A1 log R
B1
A2 log R
B2
0 requires that the constant A1
0 . Also,
R1 produces the following result:
Page 7-7
EXACT SOLUTIONS
dw1
1
dR
dw 2
2
dp R1
dz 2
dR
dp R1
dz 2
A2
A2
R1
0
Hence the expressions for the velocity profiles become:
w1 ( R )
1 dp R 2
1 dz 4
B1
w 2 ( R)
1 dp R 2
2 dz 4
B2
The boundary conditions w1 ( R1 )
w 2 ( R1 ) and w 2 ( R 2 ) 0 then give the results:
2
1 dp R1
1 dz 4
2
1 dp R1
2 dz 4
B1
B2
2
1 dp R 2
2 dz 4
0
Therefore:
B2
2
1 dp R 2
2 dz 4
And:
B1
1 dp
( R22
2 dz
B2
1 dp 2
R1
4 1 dz
R12 )
Hence the velocity components are:
1 dp
( R22
4 2 dz
w1 ( R )
1 dp
( R22
4 2 dz
w 2 ( R)
R12 )
R2)
2
R12 )
R12
1
1 dp
R22
4 2 dz
w1 (0)
(R 2
1
1 dp
( R22
4 2 dz
w1 (0)
(b) From above:
2
R12 )
2
1 R12
1
__________________________________________________________________________
Problem 7.10
In the present case,
i
0 and
0
u ( R)
Page 7-8
0 . Then, from Eq. (7.3a):
R0
0
2
R0
2
Ri
2
R
Ri
R
2
EXACT SOLUTIONS
( R)
u
R
R R
Ri 2 R0 2
2
0
( R0
1
Ri ) R 2
2
2
In the foregoing, the shear stress has been evaluated using the formulae in Appendix C. Then,
if T0 is the torque acting on the fluid at R R 0 , we have the result:
2
2 R0 ( R0 )
T0
If T i is the torque acting on the fluid at R
Ri , then the required torques are:
2
T0
Ri R 0
4
Ti
0
( R0
2
2
2
Ri )
Problem 7.11
The velocity distribution is defined by Eq. (7.3a):
u ( R)
1
( R0 2
Ri 2 )
(
0
R0
2
2
i
Ri ) R
(
0
i
)
Ri 2 R 0 2
R2
Now let the outer radius become infinitely large in this expression; that is, let R0
u ( R)
0
R
(
0
i
Next, bring the fluid far from the origin to rest by setting
u ( R)
i
Ri
)
Ri
. Then:
2
R
0
0 . This gives the result:
2
R
The complex potential for a line vortex of the given strength is:
2
F ( z)
i i Ri log z
Thus the complex velocity will be represented by the following expression:
2
i Ri
i
i
W ( R e ) (u R i u ) e
i
e i
R
Thus the velocity resulting from the line vortex will be described by the following equation:
u ( R)
i
Ri
2
R
This is the same result as obtained for the rotating cylinder. Therefore, a line vortex may be
thought of as consisting of a very small cylinder rotating in an otherwise quiescent fluid.
Page 7-9
EXACT SOLUTIONS
Problem 7.12
The problem to be solved for the velocity u( y, t ) consists of the following partial differential
equation and boundary conditions:
2
u
u
t
y2
u(0, t ) U cos nt
u( h, t ) 0
Following the procedure employed in section 7.5, we look for a solution to this problem of the
following form:
u( y, t ) Re w( y) e i nt
In the foregoing, Re means the “real part of” as before. Substituting this assumed form of
solution into the partial differential equation produces the following ordinary differential
equation for w( y) :
d 2w
dy 2
n
i w 0
1
(1 i )
2
Hence the solution for w( y) that satisfies the homogeneous boundary condition at y
i
But
w( y )
A sinh (1 i )
n
(h
2
h is:
y)
Then the solution for the velocity u( y, t ) will assume the following form:
u( y , t )
Re A sinh (1 i )
n
(h
2
y) e i nt
Imposing the boundary condition at y 0 on this solution indicates the following value for the
constant A :
U
A
n
sinh (1 i )
h
2
Thus the solution for the velocity u( y, t ) will be:
sinh (1 i )
u( y , t ) U Re
n
(h
2
sinh (1 i )
y)
e i nt
n
h
2
Although the real part of this equation may be evaluated explicitly, it will be left in the form
given above for brevity.
Page 7-10
EXACT SOLUTIONS
Problem 7.13
Using the definition of the Reynolds number R N , Eq. (7.6) becomes:
u ( y, t )
Re i
Px
n
y
a
cosh (1 i) R N
1
e i nt
cosh (1 i) R N
(a) For any value of R N , the following identity is valid:
cosh (1 i) R N
In particular, for R N
But
cosh ( R N ) cos ( R N ) i sinh ( R N )sin ( R N )
1 , the following expansion is valid:
cosh (1 i) R N
(1
) (1
) i ( RN
) ( RN
)
1 i RN
Also
Therefore
y
cosh (1 i) R N
a
1 i RN
y
cosh (1 i ) R N
a
2
y
a
1 i RN
y
a
2
1 i RN
cosh (1 i ) R N
y
a
1 i RN 1
2
Thus the expression for the velocity u( y, t ) becomes:
u ( y, t )
Re i
Px
n
Px
n
i RN 1
y
a
1
y
a
2
e i nt
2
R N cos nt
That is, using the definition of the Reynolds number R N , we get the result:
u ( y, t )
Px a 2
2
1
y2
cos nt
a2
The solution obtained above corresponds to Couette flow between two parallel surfaces in
which the magnitude of the pressure is varying slowly with time. That is, the flow behaves
like Couette flow in which the magnitude of the pressure gradient is defined by the
instantaneous value of Px cos t .
Page 7-11
EXACT SOLUTIONS
(b) For R N
and y
a we get:
y
a
cosh (1 i) R N
0
cosh (1 i ) R N
u ( y, t )
Re i
Px
n
) e i nt
(1
That is, the velocity distribution for large values of the parameter R N will be:
Px
u ( y, t )
n
sin nt
This solution corresponds to an inviscid flow in which the velocity responds to the pressure
gradient only. That is, the temporal acceleration of the fluid is equal to the applied pressure
gradient.
Problem 7.14
The partial differential equation to be solved and the assumed form of solution are:
t
where
R R
( R, t )
2
t
R
R
f( )
R
and
t
Then the following expressions are obtained for the various derivatives indicated:
1
f
f
2 t 2t
t
t
1
2
R
R
R
R
t
2
t
f
1
R
f
t
t
t
f
In the foregoing, primes denote differentiation with respect to the variable . We now substitute
these expressions in the assumed form of solution. This produces the following ordinary
differential equation:
f
(1
1
2
)f
f
0
2
Integrating this equation term by term, and using integration by parts where appropriate, gives:
Page 7-12
EXACT SOLUTIONS
( f
f d )
1
(f
2
fd )
f
fd
A
2
In the above, A is a constant of integration. Simplifying this equation gives the result:
1
f
2
f
A
2
For small but non-zero values of t , both f and its derivative f will be zero for all areas of the
fluid away from the origin. Thus the constant A must be zero. Then the solution becomes:
2
f( )
4
Be
R2
4 t
( R, t )
Be
2 t
In order to evaluate the constant of integration B , we observe that for small values of the time t
the total circulation in the fluid will be . Thus, from Eq. (2.5):
A
0
2
t
Be
R2
4 t
or 4 B
ω n dA
2 R dR
2
e
0
1
d
R2
4 t
The value of the definite integral is a half, so that the constant B must have the value B 1/ 2 .
Then the solution for the vorticity distribution will be:
2
where
( R, t )
4
t
e
R2
4 t
From Appendix A, the only non-zero component of the vorticity vector is defined as follows:
1
( Ru )
R R
4
t
e
R2
4 t
R2
4 t
Ru
e
C
2
The constant of integration C may be evaluated by observing that u
0 as t
vortex will be fully decayed for large values of the time. This gives the value C
the velocity distribution will be defined as follows:
u ( R, t )
2 R
1 e
, since the
/2
so that
R2
4 t
Page 7-13
EXACT SOLUTIONS
The pressure distribution may be evaluated from the radial component of the momentum
equations which, for the flow field under consideration, reduces to the following form:
2
u
1 p
R
R
Hence the pressure distribution will be defined by the following integral:
2
p ( R, t )
p0
1
1 e
R3
2
4
2
p0
1
R3
2
4
2
R2
4 t
2
e
R3
dR
R2
4 t
1
e
R3
R2
2 t
dR
This result may be expressed in terms of the following exponential integral:
1 ax
E(ax)
e dx
x
ax a 2 x 2 a 3 x 3
log x
1! 2 2! 3 3!
In order to reduce the expression for p( R, t ) to the form of this standard integral we integrate
each of the integrals obtained above by parts as follows:
1
e
R3
R2
4 t
1
dR
2
2
4 t
e
1
e
2
4 t
1
1
8 t
1
e
2R2
R2
... where
d
R2
4 t
1
8 t
e
E
4 t
R2
d
R2
4 t
R2
1
1
1
R2
2 t
2 t
and
E
e
dR
e
2R2
4 t
2 t
R3
Substituting these values into the expression for the pressure obtained earlier, produces the
following result:
2
p ( R, t )
p0
8
2
R2
1
e
R2
2 t
2e
R2
4 t
R2
E
2 t
R2
2 t
E
R2
4 t
_____________________________________________________________________________
Problem 7.15
For the given velocity components, the left side of the continuity equation becomes:
uz
1
(R u R )
2a 2a 0
R R
z
Page 7-14
EXACT SOLUTIONS
That is, the continuity equation is satisfied for all values of the parameters a and K . For the
given velocity components, the R -component of the Navier-Stokes equations is:
uR
u
R
R
1 p
R
K2
R3
1 p
R
2
i.e. a R
Similarly, the
2
uR
uR
1
R
R R
R
a
R
uR
R2
a
R
p
K2
a 2R
R
R3
-component of the Navier-Stokes equations gives the result:
u
uR u
u
u
1 p
1
uR
R
R
R
R
R R
R
R2
i.e.
aK
R
aK
R
p
1 p
R
K
R3
(7.15.1)
K
R3
0
(7.15.2)
Finally, the z -component of the Navier-Stokes equations shows that:
2
uz
uz
1 p
uz
R
z
z2
1 p
i.e. 4 a 2 z
0
z
p
4 a 2z
(7.15.3)
z
Eq. (7.15.2) shows that the pressure p is independent of . Then Eqs. (7.15.1) and (7.15.3)
give, respectively:
K2
a 2R 2
p ( R, z )
g 1 ( z)
2R2
2
(2 a 2 z 2 )
p ( R, z )
g 2 ( R)
Comparing these two expressions shows that the pressure distribution is defined by the following
equation:
p ( R, z )
p0
2
a 2R 2
K2
R2
4 a 2z 2
Substituting the revised expressions for the velocity components into the Navier-Stokes
equations, as per above, leads to the following modified expressions for the pressure gradient:
Page 7-15
EXACT SOLUTIONS
p
R
K2
f
R3
p
K
R
p
z
2
a 2R
(R f
f )
aK R f
4 a 2z
The first and last of these equations pose no new restrictions on the function f . However, the
right side of the middle equation must be zero since there is no -dependence in the flow. This
leads to the following differential equation for f :
1
1
a
f
f
f 0
2
R
R
d 1
a
i.e.
f
f 0
dR R
gives:
Integrating the last equation and noting that f
1 and f
0 as R
1
a
a
f
f
R
hence f
and u
1 Ce
aR2
2
K
1 Ce
R
aR2
2
Since the velocity u must be finite as R 0 , we must choose the value of the constant in this
equation to be C
1 . This gives the following expression for the function f :
f
Page 7-16
1 e
aR2
2
8
LOW-REYNOLDS-NUMBER
SOLUTIONS
LOW-REYNOLDS-NUMBER SOLUTIONS
Problem 8.1
Eq. (8.8b) gives the expression for the pressure distribution in a fluid that is moving uniformly
past a fixed sphere. Using this result, and noting that x r cos , we get the following value for
the pressure on the surface of the sphere:
3 U
p ( a, )
cos
2 a
This pressure distribution is symmetric about the flow axis so that it is sufficient to consider the
force acting on an annular element of surface whose area is:
dS 2 a 2 sin d
The force in the positive x -direction due to the pressure distribution around the sphere will have
the following value:
F p p(a, ) cos d
0
3 U a sin cos 2 d
0
The value of the integral in the last equation is 2/3 so that the following value is obtained for the
force, or drag, on the sphere due to the pressure distribution around it:
F p 2 U a
Eq. (8.8c) gives the value of the total drag on the sphere due to both shear effects and pressure
effects. Comparing the foregoing result with Eq. (8.8c) leads us to the conclusion that two-thirds
of the total drag in Stokes flow is due to shear effects, and one-third is due to the distribution of
pressure around the surface.
Problem 8.2
For a liquid drop the boundary conditions that will exist at the interface between the outer flow
and the inner flow will be the following:
u ( a, ) u ( a, ) 0
(8.2.1a)
r
r
u ( a, ) u ( a, )
(8.2.1b)
r (a, ) r (a, )
(8.2.1c)
u u r
r r r
In the foregoing equation, the variables associated with the inner flow are indicated with a
primed superscript.
where
r ( r , ) r
The outer flow will have the same general form as for a rigid sphere, but the boundary conditions
indicated above will replace the ones used previously. Then the general solution for the outer
Page 8-1
LOW-REYNOLDS-NUMBER SOLUTIONS
flow will consist of the superposition of a uniform flow, a doublet, and a stokeslet. This gives
the following values for the velocity and the pressure:
x
x
1
1
u U e x A 3 e x 3 4 e r c e x 2 e r
r
r
r
r
1
1
3x
x
U A 3 c e x A 4 c 2 e r
r
r
r
r
x
p 2c 3
r
In order to facilitate the imposition of the boundary condition defined by Eq. (8.2.1a), we use the
relationship connecting Cartesian and spherical coordinate systems as follows:
e x cos e r sin e
Then the velocity vector for the outer flow may be written as flows:
2
2
1
1
u U A 3 c cos e r U A 3 c sin e
r
r
r
r
The first component on the right side of this equation must vanish on the surface r a by virtue
of the boundary condition (8.2.1a). This requires that the strength of the doublet is:
a3
2
U c
a
2
Then the outer flow will be described by the following equations:
2
a3
a2
u r U 1 3 cos c 1 2 cos
r
r
r
A
(8.2.2a)
1
a3
a2
(8.2.2b)
u U 1 3 sin c 1 2 sin
2r
r
r
1
(8.2.2c)
p 2 c 2 cos
r
The inner flow may be defined by solving the Stokes equations using the given form for the
pressure distribution. This is most easily done through use of the stream function formulation, as
follows. We use the following identity, valid for any Stokes flow, to give:
( u) ( u) 2u
1
2u p
Applying this result to the inner flow for which p K U x , we get:
K U e x K U cos e r K U sin e
Thus the magnitude of the vorticity must be given by the following expression:
Page 8-2
LOW-REYNOLDS-NUMBER SOLUTIONS
1
K U r sin e
2
1
This makes r 0 and K U r sin . Then:
2
1
1
( sin ) er
(r ) e
r sin
r r
K U cos e r K U sin e
...as required.
Then, from the definition of the vorticity in terms of the velocity components, this result requires
that the following equation be satisfied:
1
1 u r
1
(r u )
K U r sin
r r
r
2
We now introduce the Stokes stream function as defined by Eqs. (5.3a) and (5.3b). This
produces the following partial differential equation to be satisfied:
2 sin 1 1
2
2
2
K U r sin
2
r
r sin 2
A separable solution to this equation of the following form is sought:
(r , ) U R(r ) sin 2
Substitution of this assumed form of solution leads to the following ordinary differential
equation that is to be satisfied by R(r ) :
d 2R
R 1
2 2 K r2
2
dr
r
2
The particular solution to this differential equation is K r 4 / 20 and the complementary solution
will be of the form r n since the homogeneous equation is equi-dimensional. Substitution of
R r n reveals that n 2 or n 1 , so that the complete solution is:
1
N
R(r )
K r4 M r2
20
r
N
1
(r , ) U K r 4 M r 2 sin 2
r
20
2N
1
hence u r (r , ) U K r 2 2 M 3 cos
r
10
N
1
and u (r , ) U K r 2 2 M 3 sin
r
5
In the foregoing, K , M and N are constants of integration. Since the velocity components must
both be finite at r 0 , it follows that we must choose N 0 . Also, the radial component of
velocity must be zero on r a . This requires that M K a 2 / 20 . Thus the velocity and
pressure in the inner flow will be defined as follows:
Page 8-3
LOW-REYNOLDS-NUMBER SOLUTIONS
1
(8.2.3a )
U K (a 2 r 2 ) cos
10
1
(8.2.3b)
u ( r , )
U K (a 2 2 r 2 ) sin
10
(8.2.3c)
p(r , ) K U r cos
The outer and inner flow fields, as defined by Eqs. (8.2.2) and (8.2.3), must now be matched at
the interface, r a . From the kinematic condition (8.2.1b) and the dynamic condition (8.2.1c)
we get, respectively:
1
2
3
U a 2K c U
10
2
a
3
6
U a 2 K
c 3U
10
a
In arriving at the latter equation, it should be noted that the second term in the expression for the
shear stress is zero since u is of the form f (r ) sin where f (a) vanishes. Then, for this
particular velocity distribution, the shear stress is given by the following expression:
u
r ( a, ) r
r r r a
u r ( r , )
The solution to the two algebraic equations arising out of the boundary conditions is:
2
1 ( / )
3U a
3
c
4 1 ( / )
/
5
K 2
a 1 ( / )
The force acting on the fluid due to the stokeslet is defined by Eq. (8.6c). Thus the force acting
on the liquid drop will be:
Fdrop 8 ( c)
2
6 U a
1 ( / )
3
1 ( / )
The drag force acting on the solid sphere is given by Eq. (8.8c), so that:
2
Fdrop
Fsolid
1 ( / )
3
1 ( / )
Although not explicitly requested, the resulting velocity and pressure distributions outside and
inside the drop are defined by the following equations:
Page 8-4
LOW-REYNOLDS-NUMBER SOLUTIONS
2
1 /
a
a
a2
3U
3
u r (r , ) U 1 3 cos
1 2 cos
r
r
2 1 / r
3
2
1 /
a3
a
a2
3U
3
u (r , ) U 1 3 sin
1
sin
r2
2r
4 1 / r
2
1 /
3U
a
3
cos
p ( r , )
2
1 / r 2
/
U
(a 2 r 2 ) cos
u r ( r , )
2
2 a 1 /
/
U
(a 2 r 2 ) sin
u ( r , )
2
2 a 1 /
p(r , ) 5 U
1
r
cos
1 / a 2
Problem 8.3
One way of verifying that the given expression is a solution to the Stokes equations is to first
convert it to a fixed-base coordinate system. Thus we use the following relation:
e r sin cos e x sin sin e y cos e z
e r e x sin sin e z cos e y
Then the value of the Laplacian of the velocity vector will be:
2
1 2
1
1
2
u 2
(r e r e x ) 2
sin
(r e r e x ) 2
(r e r e x )
r
2
2
r r r
r sin
r sin
0
Hence a valid solution to the Stokes equations is:
u r er e x
p0
Consider a linear combination of the foregoing solution and the solution for a rotlet as defined by
Eq. (8.5a). Then the resulting velocity field will be defined as follows:
1
u Ar e r e x B 2 e r e x
r
The applicable boundary conditions for this case are:
Page 8-5
LOW-REYNOLDS-NUMBER SOLUTIONS
u 0 r0 e r e x
...when r r0
u i ri e r e x
...when r ri
This leads to the following algebraic equations for the constants A and B :
3
3
( 0 r0 i ri )
A
3
3
(r0 ri )
( 0 i ) r0 ri )
3
B
3
(r0 ri )
3
3
These values give the following velocity and pressure distributions for the flow between the two
spheres:
3 3
( 0 r0 3 i ri 3 )
( 0 i ) r0 ri ) 1
u
e e x
r
3
3
3
3
2 r
r
(
)
(
)
r
r
r
r
0
0
i
i
p constant
Eq. (8.5c) defines the torque that acts on the fluid, and hence the torque that acts on the sphere.
The value is:
M 8 B e x
Thus, inserting the value obtained for the constant B we get:
( 0 i ) r0 ri )
3
M 8
(r0 ri )
3
3
3
ex
If we let ri 0 in the foregoing results, the values of the velocity, the pressure and the torque
become:
u 0 r er e x
p constant
M0
Problem 8.4
For the given pressure distribution and using the assumed velocity distribution, the following
result is obtained:
u
2u
t
( 2 )
t
Hence the partial differential equation to be satisfied by the function is:
Page 8-6
LOW-REYNOLDS-NUMBER SOLUTIONS
2
1
0
t
The boundary conditions to be satisfied by the velocity on the surface of an oscillating sphere of
radius a , and also far from the sphere, are as follows:
u(a, t ) a cos t (e r )
(8.4.1a)
u(r , t ) finite as r
(8.4.1b)
In consideration of Eq. (8.4.1a) it is clear that the velocity, and the function , will both be
functions of r and t only. Furthermore, the time dependence will be defined by the time
dependence of the sphere’s motion. However, the possibility of a phase lag exists in the r dependence. Then the partial differential equation to be solved for the function will be:
1 2 1
0
r
r 2 r r t
where ( r , t ) Re{R( r ) e i t }
Substituting the assumed form of solution into the partial differential equation gives:
1 d 2 dR
r
i R0
r 2 dr dr
This differential equation may be simplified by introducing a new dependent variable r R .
Then the differential equation becomes:
d 2
i 0
dr 2
Thus the solution to the original differential equation is seen to be:
1
R( r ) Ae i / ( r a ) B e i / ( r a )
r
In order to satisfy the boundary condition (8.4.1b) we must choose B 0 . Hence the solution
for the function (r , t ) becomes:
A
( r, t ) Re e i / ( r a ) e i t
r
A
Re e /2 ( r a ) e i [ /2 ( r a )]
r
In the foregoing, we have used the fact that i (1 i ) / 2 . In terms of the function (r , t ) ,
the boundary condition (8.4.1a) is:
(a, t ) Re{a e i t }
Imposing this boundary condition on the solution establishes the value for the constant A to be
A a 2 . Then:
a 2 / ( r a ) i [ t /2 ( r a )]
e
e
r
Then the velocity distribution in the fluid will be defined by the following expressions:
( r , t ) Re
Page 8-7
LOW-REYNOLDS-NUMBER SOLUTIONS
u
a2
e
where ( r , t )
r
/ ( r a )
cos[ t / 2 ( r a )]
Problem 8.5
In spherical coordinates, the vorticity vector and the Stokes stream function for any
axisymmetric flow are defined as follows:
1
1 u r
(r u )
e
r
r r
1
r sin
1
u
r sin r
Substitution of the last two expressions into the first gives:
1 2 sin 1
2
e
r sin r 2
r sin
1
i.e.
L2 e
r sin
ur
2
2 sin 1
where L 2 2
r
r sin
The quantity L2 in the foregoing is a linear, second-order differential operator that is similar to,
but not equal to, the Laplacian operator. We next take the curl of twice and invoke standard
vector identities as follows:
1
1 2
2
(L ) e r
(L ) e
2
r sin
r sin r
2
1
r sin
sin 1 2
2 2
(L ) e
2 (L ) 2
r sin
r
1
L2 ( L2 ) e
r sin
The following vector identities, and the Stokes equations, are now used to show that the left side
of the last equation is zero:
Page 8-8
LOW-REYNOLDS-NUMBER SOLUTIONS
[ ( u) 2u]
= ( p )
0
That is, the partial differential equation that is to be satisfied by the Stokes stream function, for
any pressure distribution, is:
L2 ( L2 ) 0
where L2
2 sin 1
2
r sin
r 2
_____________________________________________________________________________
Problem 8.6
A solution to the partial differential equation obtained in Prob. 8.5 above is now sought in the
following form:
(r , ) r f ( )
Substituting this assumed form of solution into the partial differential equation shows that:
1 d4f
d2f
2 2 f 0
3
4
r dr
dr
The general form of the solution to this equation is:
f ( ) A sin B cos C sin D cos
The boundary conditions that must be satisfied by this solution are the following:
(r , 0) 0
f (0) 0
r
1
(r , 0) U
f (0) U
r
(r , / 2) 0
f ( / 2) 0
r
1
f ( / 2) 0
(r , / 2) 0
r
Imposing these boundary conditions leads to the following four algebraic equations that are to be
satisfied:
Page 8-9
LOW-REYNOLDS-NUMBER SOLUTIONS
B0
A D U
A
B C
2
2
C 0
D0
The solution to this set of equations is:
A U
B0
C U
/2
( / 2) 2 1
1
D U
( / 2) 2 1
Then the corresponding solution for the stream function will be as follows:
2
U
r
sin
sin
cos
( r , )
( / 2) 2 1 2
2
The solution obtained above exhibits the following orders of magnitude:
Ur u U
Hence the order of magnitude of the inertia terms and the viscous terms will be as follows:
inertia forces
ur
u r
U 2
r
r
U
ur
viscous forces
2
2
2
r
r
inertia forces U r
Hence Reynolds number,
RN
viscous forces
Since the Reynolds number must be small compared to unity, it follows that the radius r must
satisfy the following condition for the solution to be valid:
2
r
Page 8-10
U
9
BOUNDARY LAYERS
BOUNDARY LAYERS
Problem 9.1
For x and , the following expressions for the two first derivatives are obtained:
v
x x x
u
y y y
But the quantity h may be approximated as follows in a boundary layer:
1
1
2
2
h p u u p u 2
Then, from the foregoing results, the following expressions are obtained:
p h
u
u
x x
x
h
h
u
u
v
u
uv
p h
u
u
0
y y
y
u 1 h
y
(9.1.1)
(9.1.2)
2u u 2 h
(9.1.3)
y 2 2
Substituting Eqs. (9.1.1), (9.1.2) and (9.1.3) into the boundary layer approximation to the NavierStokes equations gives:
1 p
u
u
2u
u v
2
x
y
y
x
and
u
u 2h
u v h 1 h v h
u
u
u
u
v
u
u
v
2
Performing algebraic simplification to this equation produces the following result:
h
2h
u 2
This equation is of the same form as the one-dimensional heat conduction equation.
Problem 9.2
For flow over a flat surface the Blasius equation for f () is, from Eq. (9.4a):
d3f 1 d2f
f
0
d 3 2 d 2
(9.2.1)
Page 9-1
BOUNDARY LAYERS
(a) Consider f () F ( f ) where F d f / d . Then the following derivatives are obtained:
df
F
d
d 2 f dF df dF
dF
F
2
d
d d df
df
2
2
df d 2 F
dF
d3f
d dF df dF dF
2 d F
F
F
F
F
2
d 3 d df d df df
df 2
df
d df
Substituting these expressions into Eq. (9.2.1) and simplifying, produces the following result:
2
1 f dF
d 2 F 1 dF
0
2
df
F df
2 F df
(9.2.2)
(b) Consider F ( f ) G ( ) where G (dF / df ) / f and F / f 2 . Then the following
derivatives are obtained:
dF
G f
df
d 2 F d dG
f G
df 2 df d
dG 1 dF
F
f
2 3G
2
d f df
f
dG G
f
2 G
d f
f
dG
G 2 G
d
Substituting these expressions into Eq. (9.2.2) and simplifying produces the following result,
which is a first-order ordinary differential equation, as required:
(G 2 )
dG
G
1
G 1
0
2
d
Problem 9.3
Starting with the given ordinary differential equation, we first multiply by the integrating factor
f , then we integrate, as follows:
f 1 ( f ) 2 0
f f f ( f ) 2 f 0
Page 9-2
BOUNDARY LAYERS
1
hence
1
( f ) 2 f ( f ) 3 A
2
3
The constant of integration, A , may be evaluated by noting that along the centreline of the flow
the velocity gradient is zero due to symmetry. This leads to the following conditions:
f ( ) 1
as
f ( ) 0
as
Thus the value of the constant is A 2 / 3 and the differential equation beomes:
1
1
2
3
3
( f ) 2 f ( f ) 3
2
We now let G() f () which reduces the differential equation to the following form:
1
1
2
3
3
(G) 2 G G 3
2
hence G
2
2 3G G 3
3
so that
2
3
d
dG
2 3G G 3
dG
(1 G )(1 G )(2 G )
dG
(1 G ) (2 G )
The foregoing equation may be integrated by making the following change of variables:
let F 2 ( ) G ( ) 2
2
2 dF
d
3
(3 F 2 )
dF
dF
3 ( 3 F)
3 ( 3 F)
Integrating this equation and adding a constant of integration, B , gives the result:
2
1
1
( B) log ( 3 F )
log ( 3 F )
3
3
3
then
hence
therefore
F e
3 e
2 ( B )
1
2 ( B )
1
1
tanh
( B)
2
f ( ) 2 3 tanh
B
2
The constant B may be evaluated from the boundary condition f (0) 0 . This gives:
Page 9-3
BOUNDARY LAYERS
B
2
B
1.146
tanh
3
2
2
1.146
f ( ) 2 3 tanh
2
Solving this equation for f ( ) yields the result:
f ( ) 3tanh 2
1.146 2
2
Problem 9.4
From section 9.3, the equation to be satisfied by the stream function ( x, y) for the case of no
pressure gradient is:
2 2
3
(9.4.1)
y x dy x y 2
y 3
A solution to this partial differential equation of the following form is sought:
( x, y ) 6 x 1/3 f ( )
where ( x, y )
y
x 2/3
For this assumed form of solution, the following values are obtained for the various partial
derivatives that are required:
2 x 2/3 f 4 2 x 2/3 f
x
6 2 x 1/3 f
y
2
2 2 x 4/3 f 4 3 x 4/3 f
x y
2
6 3 x 1 f
y 2
3
6 4 x 5/3 f
3
y
Substituting these results into Eq. (9.4.1) and simplifying produces the following ordinary
differential equation for f () :
f 2 f f 2 ( f ) 2 0
(9.4.2)
The boundary conditions that are to be satisfied by the function f () are the following:
Page 9-4
BOUNDARY LAYERS
v( x, 0) 0
u
( x, 0) 0
y
u ( x, y ) 0 as y
These boundary conditions impose the following conditions on the function f () :
f (0) 0
f (0) 0
f () 0 as
(9.4.3a)
(9.4.3b)
(9.4.3c)
Integrating Eq. (9.4.2) produces the following result, where A is the constant of integration:
f 2 f f A
The boundary conditions (9.4.3a) and (9.4.3b) require that the constant A 0 . Then, integrating one
more time gives the result:
f f 2 B2
In the foregoing, the constant of integration has been taken to be B 2 . Separating variables in the last
equation produces the following expression:
df
d
(B f 2 )
2
This equation may be integrated to arrive at the following result:
f () B tanh B C
In the foregoing equation, C is the constant of integration. Applying the boundary condition (9.4.3a)
shows that C 0 , so that the solution for the function f () becomes:
f () B tanh B
Thus the solution for the stream function becomes:
( x, y ) 6 B x 1/3 tanh B
y
x 2/3
The constant B may be evaluated if the mass flow rate in the jet is known. Then, for any given value of
the flow rate, the integral of the velocity across the jet must equal the specified value, and this
requirement determines the constant B .
Problem 9.5
U k x1/ 3
dU 1 2 / 3
kx
dx 3
and U
dU k 2 1/ 3
x
dx
2
Hence the boundary layer equation becomes:
Page 9-5
BOUNDARY LAYERS
2 2 k 2 1/3
3
x
y xy x y 2
y 3
3
We assume a solution of the following form:
3
2k y
k x m f ( ) where
2
3 x1/3
Now calculate the various derivatives that are required as follows:
3
k x m1 f m f
x
2
3
k x m 1/ 3 f
y
2
1
1
k x m4 / 3 f (m ) f
3
xy
3
2
2 k m2/3
k
x
f
2
y
3
3 2 k 2 m1
x f
y 3 3
Substitute these results into the B.L. equation and remove the multiplicative brackets:
k 2 2 m5/3
1
1
x
k 2 x 2 m5/3 f f m k 2 x 2 m5/3 ( f )2
f f mk 2 x 2 m5/3 f f
3
3
3
k 2 1/3 2 2 m1
x k x f
3
3
The first and third terms cancel each other, and the quantity k 2 may be cancelled throughout the
equation. Hence the differential equation reduces to:
2 m1
1
1
x f m x 2 m5/ 3 f f m x 2 m5/ 3 ( f )2 x 1/ 3 0
3
3
3
In order for a similarity solution to exist, all x dependence must be eliminated from this equation. That is,
the following powers of x must be zero:
m 1 2m
These equations are satisfied for
m
5
1
3
3
2
3
Substituting this value of m into the differential equation gives:
1
1
f f f ( f )2 0
2
2
Page 9-6
BOUNDARY LAYERS
Problem 9.6
U 3
u
2
0 ( x) ( x, 0) 2 ( x, 0)
y
y
x
(a)
0 ( x)
2
1
U x
U 2
1/ 2
f (0)
1/ 2
f (0)
2 f (0)
Rx
2
But
0 ( x)
1
2
(b)
f ( ) 0.332 for
f ( ) 0.166 2
v( x, y )
U 2
1
0.664
Rx
1 U
U
( f f )
( f f )
x 2 x
2 Rx
v( x, y)
1
( f f )
U
2 Rx
For y ,
1 and f () 1.721
( f ) ( 1.721) 1.721
(c)
z
v( x, y ) 0.8605
U
Rx
v u
2 2
2
2 2 2
x y
x
y
y
z U
U
f ( )
x
____________________________________________________________________________________
Problem 9.7
Here
u ( x, y )
y
ab
U
The two boundary conditions that must be satisfied by this velocity distribution are:
Page 9-7
BOUNDARY LAYERS
u ( x, 0)
0
U
u ( x, )
1
U
a0
b 1
Hence the velocity distribution across the boundary layer is defined by the equation:
u( x , y )
U
...where
y
(9.7.1)
From section 9.8, the momentum integral for flow over a flat surface is:
0
d
u
(
U
u
)
dy
dx 0
0
d 1u
u
or
1
d
(9.7.2)
dx 0 U U U 2
Using Eq. (9.7.1), we obtain the following values:
u
1
u
0 U 1 U dy 0 (1 ) d 6
u
U
and 0 ( x, 0)
y
Substituting these values into Eq. (9.7.2) yields the following ordinary differential equation to be
solved for the boundary layer thickness :
1 d
6 dx U
2
x
C
12 U
We take the thickness of the boundary layer to be zero when x 0 , which requires that the
constant of integration C 0 . Thus the boundary layer thickness will be defined by:
x
12 3.464
RN
RN
The displacement thickness is defined by the following relation:
1
u
* 1 dy (1 ) d
0
0
2
U
This gives the following value for * :
*
x
3
1.732
RN
RN
The momentum thickness is defined by the following relation:
Page 9-8
BOUNDARY LAYERS
u
0 U
This gives the following value for :
u
1
U
x
1
dy 0 (1 ) d
6
1
3
1
0.577
RN
RN
Finally, the surface shear stress coefficient is defined by the following relation:
0
1
U 2
2 1
U
2
Then, using the value obtained above for we find:
0
1
U 2
1
3
1
0.577
RN
RN
2
These values compare favorably with those obtained in section 9.8, in spite of the crude velocity
profile that was used.
Problem 9.8
2
Here
u ( x, y )
y
y
y
a b c d
U
3
The four boundary conditions that must be satisfied by this velocity distribution are:
u ( x, 0) 0
u ( x, ) U
u ( x, )
0
y
2u ( x, 0)
0
y 2
These conditions yield the following values for the constants a, b, c and d .
a 0,
b 3/ 2,
c 0,
d 1/ 2
Hence the velocity distribution across the boundary layer is defined by the equation:
u( x, y ) 3
1
3
U
2
2
... where
y
(9.8.1)
From section 9.8, the momentum integral for flow over a flat surface is:
Page 9-9
BOUNDARY LAYERS
0
d
u (U u ) dy
0
dx
0
d 1u
u
or
1
d
(9.8.2)
dx 0 U U U 2
Using Eq. (9.8.1), we obtain the following values:
u
1 3
u
1 3
3
1
39
1 3 d
0 U 1 U dy 0 2 2
280
2
2
u
3 U
and 0 ( x, 0)
y
2
Substituting these values into Eq. (9.8.2) yields the following ordinary differential equation to be
solved for the boundary layer thickness :
39 d 3
280 dx 2 U
39 2 3 x
C
560
2U
We take the thickness of the boundary layer to be zero when x 0 , which requires that the
constant of integration C 0 . Thus the boundary layer thickness will be defined by:
280
13
x
1
4.641
RN
RN
The displacement thickness is defined by the following relation:
u
*
This gives the following value for :
1
3
* 1 dy (1 3 ) d
0
0
U
8
2
2
*
x
315
104
3
1
1
1.740
RN
RN
The momentum thickness is defined by the following relation:
0
1 3
u
u
1 3
3
1 3
39
1 dy 0 1 d
U U
280
2
2
2
2
This gives the following value for :
x
117
280
1
0.646
RN
RN
Finally, the surface shear stress coefficient is defined by the following relation:
Page 9-10
BOUNDARY LAYERS
0
1
U 2
3 1
U
2
Then, using the value obtained above for we find:
0
1
U 2
117
280
1
0.646
RN
RN
2
Problem 9.9
Here
y
u ( x, y )
sin
U
2
sin
...where
y
2
u ( x, 0) U
y
2
Substituting these values into the momentum integral yields the following ordinary differential
equation to be solved for the boundary layer thickness :
2 1 d
2 dx 2 U
0
Then
1 2 1 x
C
2 2 2 U
We take the thickness of the boundary layer to be zero when x 0 , which requires that the
constant of integration C 0 . Thus the boundary layer thickness will be defined by:
x
2
4
1
4.795
RN
RN
The displacement thickness is defined by the following relation:
u
2
1
2
* 1 dy (1 sin ) d 1
0
U
0
This gives the following value for :
*
*
x
( 2)
2
4
1
1.743
RN
RN
The momentum thickness is defined by the following relation:
Page 9-11
BOUNDARY LAYERS
u
u
1
U U
This gives the following value for :
2 1
2 1
dy 0 sin (1 sin ) d
2
0
4
2
x
1
0.655
RN
RN
Finally, the surface shear stress coefficient is defined by the following relation:
0
1
U 2
1
U
2
Then, using the value obtained above for we find:
0
1
U 2
4
2
1
0.655
RN
RN
2
Problem 9.10
y
u
1 e
U
u
y
U e
y
u
U
0 ( x, 0)
y
and
Then the value of the momentum integral becomes:
0
u
U
u
1
U
dy
1 e e d
1
0
1 2 e e 2
2
d
e 2
dx
x
Hence
Therefore
1
1 2 e
2
2
1 2 e e 2 U
x
Page 9-12
(taking the constant of integration to be zero)
2
1 2 e e 2
1
RN
BOUNDARY LAYERS
Problem 9.11
0
u
U
u
1
U
dy
1
1/ 7 (1 1/ 7 ) d
where y /
0
1
Kármán integral
7
7
8/ 7 9/ 7
9
8
0
7
72
1/4
0
d
U 2
where 0
dx U 2
44 U
7 d
1
72 dx 44 U
1/4
d 18
1/4
dx 77 U
1/ 4
4 5/4 18
x constant
5
77 U
The constant of integration will be zero for 0 when x 0 . Hence:
1/4
45 4/5
x
154 U
4/5
1/5
45
x 154
0
1
1
U 2 22 U
1/4
4/5
1
1/ 5
N
R
1/4
1 1 x
22 RN1/4
0.3737
1/ 5
RN
1/5
1 1 154
1/20
RN
1/4
22 RN 45
2
0
1
2
1/ 5
U 2
1 154
22 45
7
72
7 45
72 154
x
1
1/ 5
N
R
x
7
72 x
1
4/5
1/ 5
N
R
0.0581
1/ 5
RN
0.0363
1/ 5
RN
Page 9-13
BOUNDARY LAYERS
1
u
7
* 1 dy (1 1/ 7 ) d 8/ 7
0
0
8
U
0 8
1
*
1 45
x 8 154
4/5
1
1/ 5
N
R
0.0467
1/ 5
RN
Problem 9.12
From Eq. (9.12f), the momentum thickness is defined by the following equation:
0.47 x
2 ( x)
U ( ) 2 d
2 0
U ( x)
0.47 x
6 A 5 5/6 d
A x 0
0.2564
Hence
x
0.5063
A
x 5/6
A x 7/6
Substituting this result into Eq. (9.12g) gives the following equation for the pressure gradient
parameter :
2
37
2
2 dU
0.04273
dx
315 945 9072
There are three positive roots to this equation; 3.367, 21.326 and 32.829 . It follows from
Eq. (9.12b) that only the first of these values is physically reasonable. Then the following values
are obtained:
3.367
3
0.2719
10 120
37
2
0.1126
315 945 9072
Hence from Eq. (9.12d), the following value is obtained for the boundary layer thickness :
x
4.4967
A x 7/6
The value of the displacement thickness * may now be determined from Eq. (9.12c):
Page 9-14
BOUNDARY LAYERS
*
x
1.2226
A x 7/6
Finally, the value of the surface shear stress 0 may be determined from Eq. (9.12e):
U
2
6
0
2
2
1
6
U 2 U
0
2
0
1
U 2
1.1392
A x 7/6
2
Problem 9.13
The only change to the boundary layer equations for this problem is to retain the body force term
in the form of the gravitational force. Thus the equations to be solved are:
u v
0
x y
u
(9.13.1)
dU
u
u
2u
v U
2 g
dx
x
y
y
(9.13.2)
To obtain the momentum integral, we first add Eq. (9.13.1) to Eq.(9.13.2), then we integrate the
latter with respect to y between the limits y 0 and y . Following the procedure that was
used in section 9.8, this produces the following result:
2
2u
dU
2 g
(u ) (u v ) U
x
y
y
dx
0
g
0
0
0
u
dU
2
0 x (u ) dy U 0 x dy dx 0 U dy g
0
d 2
d
dU
dU
g
u
dy
U
u
dy
u
dy
U
dy
dx 0
dx 0
dx 0
dx 0
0
d
dU
g
(
)
(
)
u
U
u
dy
U
u
dy
dx 0
dx 0
2
dU
(u ) dy U v( x, )
x
dx
U dy
Page 9-15
BOUNDARY LAYERS
0
d
dU
(U 2 )
U * g
dx
dx
Using the definitions of the various thicknesses yields the following result:
That is
0
d
1 dU
g
(2 * )
2
2
dx
U dx U
U
Since the velocity profile in this problem is the same as that used in section 9.8 of the text, the
values of the various quantities of interest will be the same. Then:
u
2 2
U
u
u
2
1 dy
0 U
15
U
u
1
* 1 dy
0
3
U
u ( x, 0)
U
2
0 2
y
Substituting these values into the momentum integral obtained above, leads to the following
ordinary differential equation to be satisfied by the boundary layer thickness :
15
d
6
0
1/2
dx
x
2g x
The solution to this ordinary differential equation is of the form C x m . Substituting this form
of solution into the equation shows that:
m
1
4
and
C
12
5 (2 g ) 1/4
Then the solution for the boundary layer thickness becomes:
x
12
1.5492
3 1/4
5 (2 g x )
(2 g x 3 ) 1/4
Using the relationships connecting and * with , the following values are obtained:
Page 9-16
BOUNDARY LAYERS
*
x
x
0
1
U 2
4
0.5164
3 1/4
15 (2 g x )
(2 g x 3 ) 1/4
4
0.2066
3 1/4
(2 g x 3 ) 1/4
375 (2 g x )
20
2.582
3 1/4
3 (2 g x )
(2 g x 3 ) 1/4
2
Page 9-17
10
BUOYANCY-DRIVEN FLOWS
BUOYANCY-DRIVEN FLOWS
Problem 10.1
From section 10.6 of the text, the partial differential equations to be satisfied are as follows:
1 1 1 1 1
R
R R x R R R x R R R R R R R R
R
R x x R
R R
g
(10.1.1)
(10.1.2)
Using the given expressions, the following derivatives are obtained:
C1 x m 1 (n f m f )
x
C1C 2 x m n f
y
2
C1C 2 x m n 1[n f (m n) f ]
xy
2
2
C1C 2 x m 2 n f
2
y
3
3
C1C 2 x m 3 n f
3
y
C 3 x r 1 (n F r F )
x
C 2C 3 x r n F
y
2
2
C 2 C 3 x m 2 n F
2
y
Substituting these expressions into Eqs. (10.1.1) and (10.1.2) and simplifying yields:
( m 2 n ) C 1 2C 2 x 2 m n 1 ( f ) 2 mC 1 2C 2 x 2 m n 1 ( f f f f )
C 1 C 2 x m n ( 2 f f f ) g
C3
C2
3
x r 3 n 3 F
C1 C 2 C 3 x m n r 1r f F mC1 C 2 C 3 x m n r 1 f F C 2 C 3 x r n ( F F )
(10.1.3)
(10.1.4)
If a similarity solution exists, all of the powers of x in Eq. (10.1.3) must be the same so that x
cancels out of the equation. This leads to the equations:
m 1
and
4n r 1
Page 10-1
BUOYANCY-DRIVEN FLOWS
Similarly, the powers of x in Eq. (10.1.4) must cancel. This produces the same two equations
for m, n and r . That is, for a similarity solution to exist:
m 1
4n r 1
and
In order to determine the values of m, n and r , we must impose the condition that the total
amount of heat leaving the source has the value Q . This condition is expressed by Eq. (10.26),
which requires that m r 0 so that the total amount of heat leaving the source is independent
of x . This additional condition leads to the following values for a similarity solution to exist:
m 1
n
1
r 1
2
Problem 10.2
For Pr 1 , Eqs. (10.27a), (10.27b), (10.28a) and (10.28b) become:
f (1 f )
d 1
f F 0
d
1
F f F 0
f (0) f (0) F (0) 0
1
0
2
(a) Assume the following expressions to be valid for the functions f and F :
2
f ( ) A
a 2
1
F ( ) B
(a 2 ) 3
For these assumed values, Eq. (10.27b) becomes:
f F d
(10.27a)
(10.27b)
(10.28a)
(10.28b)
A B
6 B
0
2 4
(a )
(a 2 ) 4
It can be seen that this equation is satisfied by:
A6
(b) Using A 6 in the expression for the function f , the following quantities are obtained:
Page 10-2
BUOYANCY-DRIVEN FLOWS
f
12
12 2
(a 2 ) (a 2 ) 2
f
12
60 2
48 4
(a 2 ) (a 2 ) 2 (a 2 ) 3
f
144
432 3
288 5
(a 2 ) 2 (a 2 ) 3 (a 2 ) 4
d 1
48
48 3
f
d
(a 2 ) 2 (a 2 ) 3
Using these results, Eq. (10.28b) becomes:
d
1
2
3
0 (a )
2
4
The value of the integral in this equation is 1/(8 a ) , so that one equation relating the
constants a and B is:
a3
B
(10.2.1)
3
Substituting the various derivatives evaluated above into Eq. (10.27a) yields the following
additional equation relating a and B :
12 a B
B 96 a
(10.2.2)
From Eqs. (10.2.1) and (10.2.2) we get:
a 4 2
(4 2 ) 2
B
3
Using these values, the solutions for the functions f and F become:
f ( ) 6
2
(4 2 2 )
(4 2 ) 3
1
F ( )
3
(4 2 2 ) 3
It will be noted that these expressions satisfy the boundary condition defined by Eq.(10.28a).
Page 10-3
BUOYANCY-DRIVEN FLOWS
Problem 10.3
For Pr 2 , Eqs. (10.27a), (10.27b), (10.28a) and (10.28b) become:
d 1
f F 0
d
2
F f F 0
f (0) f (0) F (0) 0
f (1 f )
1
0
2
Assume the following expressions to be valid for the functions f and F :
2
f ( ) A
a 2
1
F ( ) B
(a 2 ) 4
For these assumed values, Eq. (10.27b) becomes:
f F d
(10.27a)
(10.27b)
(10.28a)
(10.28b)
8 B
2 A B
0
2 5
(a )
(a 2 ) 5
It can be seen that this equation is satisfied by:
A4
(10.3.1)
Using A 4 in the expression for f yields the following quantities:
8
8 3
f
(a 2 ) (a 2 ) 2
f
8
40 2
32 4
(a 2 ) (a 2 ) 2 (a 2 ) 3
f
96
288 3
192 5
(a 2 ) 2 (a 2 ) 3 (a 2 ) 4
d 1
32
32 3
f
d
(a 2 ) 2 (a 2 ) 3
Using these results, Eq. (10.28b) assumes the following form:
d
1
2 5
(a )
2
The value of the integral in the foregoing expression is 1/(10 a 5 ) , so that one of the equations
relating a and b is:
12 a B
Page 10-4
0
BUOYANCY-DRIVEN FLOWS
5 4
a
(10.3.2)
8
Substituting the various derivatives evaluated previously into Eq. (10.27a) yields the following
additional equation relating a and b :
B
B 64 a 2
From Eqs. (10.3.2) and (10.3.3) we get the values:
a 16
B
(10.3.3)
2
5
85
5
Using these values, we obtain the following solutions for the functions f and F :
f ( )
4 2
(16 2 / 5 2 )
F ( )
85
1
5 (16 2 / 5 2 ) 4
It will be noted that these expressions satisfy the boundary condition defined by Eq.(10.28a).
Problem 10.4
(a) The assumed form of solution to Eq. (10.32a) is the following:
x
U ( x) Ai e i
Substituting this assumed form of solution into Eq. (10.32a) leads to the result:
( i 2 ) 3 2 Ra 0
The solutions to this algebraic equation are as follows:
2
1/ 3
Ra
1/ 3
i 1 (1) 4
We note that the three cubic roots of (-1) are:
(1) 1/ 3 e i / 3 ;
e i ;
1/ 2
e i 5 / 3
(1 i 3) / 2;
1;
(1 i 3) / 2
Thus the values of the three exponent coefficients in the assumed form of solution consist of
one real root and one pair of complex conjugate roots. This makes the solution:
Page 10-5
BUOYANCY-DRIVEN FLOWS
R a 1/3
1 1 4
1/2
1/3
1
Ra
2 1 (1 i 3) 4
2
1/2
1/3
1
Ra
3 1 (1 i 3) 4
2
1/2
(b) Applying the six boundary conditions defined by Eq. (10.32b) produces the following six
homogeneous equations that must be satisfied:
0 C1 C 2 C 3 C 4 C 5 C 6
0 1C1 2C 2 3C 3 1C 4 2C 5 3C 6
0 ( 1 2 ) 2 C1 ( 2 2 ) 2 C 2 ( 3 2 ) 2 C 3
2
2
2
( 1 2 ) 2 C 4 ( 2 2 ) 2 C 5 ( 3 2 ) 2 C 6
2
0 C1 e
1
C 2e
0 1 C1 e
1
2
C3 e
2 C2 e
0 ( 1 2 ) 2 C1 e
2
2
1
3
2
C 4e
1
3 C3 e
C5 e
3
2
1
2
C 6e
1 C 4e
( 2 2 ) 2 C 2 e
( 1 2 ) 2 C 4 e
2
2
2
1
2 C5 e
2
( 3 2 ) 2 C 3 e
2
( 2 2 ) 2 C 5 e
2
3
2
3 C 6e
3
3
( 3 2 ) 2 C 6 e
2
3
Setting the determinant of the coefficients of the constants C i 0 requires that the following
determinant be zero:
Page 10-6
BUOYANCY-DRIVEN FLOWS
1
1
1
1
1
1
1
2
3
1
2
3
( 2 2 ) 2
( 3 2 ) 2
( 1 2 ) 2
( 2 2 ) 2
2
e
1
1e
2
e
1
( 1 2 ) 2 e
( 3 2 ) 2
2
2
2e
1
2
e
2
( 2 2 ) 2 e
( 1 2 ) 2
2
3
3e
2
e
3
( 3 2 ) 2 e
2
2
1
1e
3
2
e
1
2
2e
2
e
2
( 1 2 ) 2 e 1 ( 2 2 ) 2 e
2
2
3
3e
2
3
( 3 2 ) 2 e
2
3
Problem 10.5
For free boundaries, as with rigid boundaries, two boundary conditions that must be satisfied are
that the perturbation velocity component in the x -direction and the perturbation temperature
should both be zero on the boundaries. However, the no-slip condition should be replaced with
the condition that there should be no shear stress on the free surfaces. With reference to
Appendix C, this implies that the following components of shear stress that act on the surface x
= constant should be zero:
u v w u
0
y x x
z
Since u must be zero on all parts of the plane x = constant, it follows that derivatives of u
with respect to y and z will be zero. That is, the stress-free conditions expressed above may be
stated as follows:
v w
0
x x
Since the stability problem has been posed in terms of the velocity component in the x -direction
only, the foregoing conditions should be redefined in terms of the x -component of velocity.
This may be done by differentiating the continuity equation with respect to x , then imposing the
two conditions expressed above. This leads to the following condition:
2u
0
x 2
or D 2U 0
Page 10-7
BUOYANCY-DRIVEN FLOWS
Combining this last condition with the two unchanged conditions produces the following
boundary conditions that must be satisfied at the free surfaces:
U D 2U ( D 2 2 ) 2U 0
Page 10-8
...on
x
0 and 1
h
11
SHOCK WAVES
SHOCK WAVES
Problem 11.1
The First Law of Thermodynamics may be expressed in the following form:
de
but
p dv
p v and dq T ds
h e
hence d (h
Therefore
dq
T ds
p v) T ds
dh
p dv
v dp
s
s
dp
dT
p
T
h
h
h h( p, T )
dh
dp
dT
p
T
Substituting these expressions into the result obtained above gives:
s
s
h
h
T
dp
dT
dp
dT v dp
p
T
p
T
Equating the coefficients of dT and dp in this equation produces the following relations:
s
s ( p, T )
ds
s
T
s
p
1 h
T T
1 h
T p
v
Differentiating the first of the above equations by p and the second by T permits s to be
eliminated to produce the following result:
2
1 2h
1
h
1
h
v
v
2
T p T
T
p
T p T
T
h
v
v T
p
T
RT / p in which case the right side of this equation is
But, for an ideal gas we have v 1/
zero. That is:
h
p
0
...for p
RT
This verifies that h is independent of p for an ideal gas, and that it depends on the temperature
T only for an ideal gas.
Page 11-1
SHOCK WAVES
Problem 11.2
s
s
dv
dT
v
T
e
e
e e (v , T )
de
dv
dT
v
T
Substituting these expressions into the First Law of Thermodynamics gives:
s
s
e
e
T
dv
dT
dv
dT p dv
v
T
v
T
Equating the coefficients of dT and dv in this equation produces the following relations:
s
s (v , T )
ds
s 1 e
T T T
s 1 e
v T v
p
Differentiating the first of the above equations by v and the second by T permits s to be
eliminated to produce the following result:
2
1 2e
1
e
1
e
p
p
2
T v T
T
v
T v T
T
But, for an ideal gas we have p
e
p
T
p
v
T
RT so that the right side of this equation is zero. That is:
e
v
0
...for p
RT
This verifies that the internal energy e depends on the temperature T only for an ideal gas. A
much quicker way of arriving at this result is to use the reciprocity relation given at the end of
the Appendix on Thermodynamics.
e
p
p T
p T
R 0 if p
RT .
v
T
Problem 11.3
From Appendix E we obtain the following two relations:
dq
and
s s0
dq de
T
0
so
dq
0
Page 11-2
0
de
T
p
dv
T
0
p dv
SHOCK WAVES
s
s0
de
T
0
i.e. s
s0
Cv
0
p
dv
T
0
dT
T
Rv dv
0
Integrating this equation and using the fact that
s
Using the relation C p
s0
C v log
1/ v yields the result:
T
T0
0
R log
R in the foregoing result produces the equation:
Cv
s
T
T0
(C p R ) log
s0
C p log
T
T0
0
R log
R log
T
T0
0
Hence, using the ideal gas law, the last equation becomes:
s
s0
C p log
T
T0
R log
p
p0
Problem 11.4
u
where u
and
f [ x (u a) t ]
u ( x, t ) and a a( x, t )
u
t
f
u
x
f
(u
1
a)
u
x
u
t
a
t
t
a
t
x
u
u
u
u
a
(u a)
(u a)
(u
t
x
t
x
t
But we know that u is a function of
and that a is also a function of
consider a to be a function of u only. Hence:
a da u
t du t
a da u
x du x
Substituting these results into the differential expression above gives:
hence
a
tf
x
, so that we may
a)
Page 11-3
SHOCK WAVES
u
t
(u
u
x
a)
u
t
(u
u
x
a)
1
da
tf
du
u
u
da
(u a )
1 1
tf
0
t
x
du
Either the first curly bracket is zero or the second curly bracket is zero. In general, the latter
cannot be true since the equation is valid for all values of the time t, so that the first curly bracket
must be zero. That is, the following is the solution to the wave equation:
u( x, t )
f [x
(u
a) t ]
Problem 11.5
u
t
t
u
x
or
(u
a)
D
Dt
u
x
(u
a)
x
u
x
u
x
0
u
x
a
x
u
x
a
x
u
x
u
x
0
The partial derivative of a with respect to x may be shown to be proportional to the partial
derivative of u with respect to x as follows. In section 11.2, the following relationships were
established:
1
2
a
a0
0
d
du
a
hence
x
a
da d
u
d du x
3
1 a0
2
2
0
0
1
3
2
a0
0
0
2
u
x
u
x
Substituting this expression into the equation for the steepness of the wave front gives:
D
Dt
Page 11-4
u
x
1
2
u
x
2
SHOCK WAVES
Separating variables in the foregoing result produces the following equation:
u
D
1
x
Dt
2
2
u
x
Integrating this equation and taking the steepness to be S when t
value for the steepness at any time t :
1
1
1
t
u
S
2
x
2
or t
1
S
1
0 produces the following
1
u
x
This result gives the value of the time t to achieve any steepness
u
. In particular, to achieve
x
an infinite steepness:
2
t
1
1S
This result shows that S must be negative if the steepness is to become infinite. That is, if a
shock wave is to form, the steepness of the original wave must be negative.
Problem 11.6
s
log
Cv
2
p2
1
p1
2
We now employ Eqs. (11.8b) and (11.8c) for the density ratio and the pressure ratio,
respectively, across a normal shock wave. This leads to the following expression:
s
Cv
log
1
2
1
2
(M 1
2
1)
(
1) M 1
(
In the foregoing, we now want to express M 1 in terms of ( M 1
Hence we rewrite the second square-bracketed term as follows:
2
1) M 1
2
2
2
1) in the last square bracket.
Page 11-5
SHOCK WAVES
(
1) M 1
(
Then, denoting ( M 1
2
2
1) M 1
Cv
(
2
1) by
s
For
2
1) ( M 1
(
2
1) (
1) 1 ( M 1
2
1
2
( M 1 1)
1
2
1 ( M 1 1)
1
1)
1)
, the foregoing result assumes the following form:
2
log 1
1
1
log 1
1
log (1
)
1 , each of the foregoing logarithmic terms may be expanded as follows:
log 1
2
2
2
1
1
1
log 1
(
2
1)
(
(
1)
1)
1) 2
(
1) 2
1) 2
1(
2(
2
3
8
3(
2
3
1) 3
1) 3
1) 3
1(
3(
3
1 2 1 3
2
3
Substituting these values into the expression derived above shows that the coefficients of the
terms that are linear and quadratic in are zero. Thus the leading non-zero term is:
log (1
)
s
Cv
2
2 (
3 (
1)
1)
3
3
This result shows that weak shock waves, that is, waves for which M 1
isentropic.
1 , are almost
Problem 11.7
(a) From the given pressure/density relation:
d
hence log
dp
c
2
1
( p2
1
( p2
c
p1 ) c log
p1 )
2
u
1 1
2
2
u2
2
1
In the last equation, the pressure difference has been eliminated using the momentum
equation, Eq. (11.3b). Dividing both sides of the last equation by c and using the fact that
c / yields the following identity:
the speed of sound is given by a 2 dp / d
Page 11-6
SHOCK WAVES
2
log
M1
2
M2
2
(11.7.1)
1
Squaring Eq. (11.3a) and dividing the result by c gives:
2
2
2
2
1 u1
2 u2
c
2
2
Using the results a1 c / 1 and a 2
following result from the last equation:
c
c / 2 , where a is the speed of sound, yields the
M1
2
2
2
M2
Substituting Eq. (11.7.2) into Eq. (11.7.1) results in the following equation:
(11.7.2)
1
log
M1
2
M2
2
M1
2
M2
2
(b) From part (a) we have the result:
( p2
2
p1 ) c log
1
c log
M1
2
M2
2
( p2
so that M 2
2
2
p1 )
c
M1 e
In the foregoing, Eq. (11.7.2) has been used. Substituting this equation into the main result
of part (a) produces the following expression:
2
( p 2 p1 )
M1
2
2
c
log
M
M
e
1
1
2
M2
( p1
p2 )
c
M1
2
( p 2 p1 )
2
M1 1 e
( p1
1 e
c
p2 ) / c
( p1
p2 )/c
Problem 11.8
From section 11.2, the continuity and momentum equations for a propagating sound wave are as
follows:
Page 11-7
SHOCK WAVES
t
x
( u) 0
u
u
1 dp
u
t
x
d
x
The corresponding equations for a shallow liquid are, from section 6.5:
h
(h u ) 0
t
x
u
u
h
u
g
t
x
x
In the last two equations, h is interpreted as being the local depth of the liquid, not the mean
depth. Comparing the two sets of equations shows that the following variables are the analog of
each other:
Liquid
Gas
u
u
h
In order to make the analogy complete, the coefficients on the right side of the momentum
equation should be the same. This requires that:
1 dp
g
d
p0
p
but
c
0
dp
d
so that c
2
1
c
g
Then, in order to complete the analogy, we must have:
2
and
2
p0
2
g
0
Problem 11.9
Using the given form of the solution, the following derivatives are obtained:
Page 11-8
SHOCK WAVES
u
U p
x
p x
u
U
y
y
u
U p
U p
c
t
p t
p x
v
V p
y
y x
Substituting these results into the continuity and momentum equations, and assuming that the
partial derivative of p with respect to x is not zero, leads to the following two equations to be
satisfied by U and V :
(U
c)
U
p
U
p
U
V
y
V
y
1
0
0
For the given form of similarity solution, the following derivatives are obtained:
U 1 1/ 2 *
p U
p 2
U
y
p 1/ 2
dU *
dy
V
dV *
p 1/ 2
y
dy
Substituting these results into the equations obtained above yields the following:
dV *
dy
V*
dU *
dy
1 *
U
2
1 * *
U (U
C *)
2
1
0
0
_____________________________________________________________________________
Page 11-9
12
ONE-DIMENSIONAL FLOWS
ONE-DIMENSIONAL FLOWS
Problem 12.1
Using Eqs. (12.2a) and (12.2b) to evaluate the constant, the Riemann invariant along the line
constant shows that in region 4 the value of the velocity and the value of the pressure must
satisfy the following relation:
u 1 p 1 p1
(12.1.1)
a0 p0 p0
Similarly, using the constant characteristic we get:
u 1 p uI 1
a0 p0 a0
In the foregoing, u I represents the gas velocity at, and of, the interface. In order to evaluate this
quantity, we consider the 1 characteristic which, using Eqs. (12.2a) and (12.2b) to evaluate the
constant, gives:
u I 1 1 p1
a0 p0
u 1 p 1 p1 2
a0 p0 p0
Solving Eqs. (12.1.1) and (12.1.2) gives the following results for region 4:
u 1 p1
1
a 0 p 0
(12.1.2)
p
1
p0
Page 12-1
ONE-DIMENSIONAL FLOWS
Problem 12.2
(a) In general, the line separating regions 2 and 5 will have a different slope from the line that
separates regions 3 and 4 because the fluid temperatures are different. Noting that the time
taken for the reflected wave to travel the distance L is the same as that taken for the
transmitted wave to travel the distance L(1 ) , we have:
L L(1 )
a 01
a 02
a 02
a 01
T02
Hence
T01
1
(1 ) 2
2
(b) In regions 1 and 2, the velocity will be zero and the pressure will be p 0 . Also, from Eqs.
(12.6a) and (12.6b), the velocity and the pressure in region 3 will be defined as follows:
u3
a 01
U
a 01
p3
p0
U
1
a 01
In order to determine the velocity and pressure in regions 4 and 5, we consider a point P( x, t )
that lies on the interface between regions 4 and 5. Then, by considering the two
characteristics that pass through this point, we get:
Page 12-2
ONE-DIMENSIONAL FLOWS
1 p
1
u
U
2
a 01 p 0
a 01
u
1 p
1
a 02 p 0
In the foregoing, the Riemann invariants have been evaluated from regions 3 and 2,
respectively. Solving these two equations for the velocity and the pressure that exist in
regions 4 and 5 gives:
u 4,5
a 01
p 4,5
p0
2U / a 01
(1 a 01 / a 02 )
2 U / a 01
(1 a 01 / a 02 )
1
(c) Using the foregoing expression for the pressure, the pressure differential across the
transmitted and reflected waves may be evaluated by subtracting the pressures in regions 2
and 3, respectively. This produces the following result:
p t
p0
p r
2 U / a 01
(a 02 / a 01 1)
U (a 02 / a 01 1)
p0
a 01 (a 02 / a 01 1)
Dividing the last two equations and simplifying gives the following result for the ratio of the
pressure differentials across the reflected and transmitted waves:
p r
1 a 02
1
p t
2 a 01
From part (a), this result may be expressed in terms of only to give:
2
p r
p t
(1 2 )
2
Problem 12.3
Following the procedure employed in section 12.6 we may relate the problem to a stationary
shock wave by employing a Galilean transformation of magnitude u 2n to give:
u 1n
1
u 2 (u1n u 2 n ) u 2 n
u 2n
Page 12-3
ONE-DIMENSIONAL FLOWS
1 p2
1
p1
i.e. u 2 a 2 M 2 n
1 1 1 p 2 1
2 1 p1
In the foregoing equation, Eq. (11.4) has been used to evaluate the ratio of the gas velocities
across a normal shock wave, and the resulting expression has been simplified. Next, we use Eqs.
(11.8a) and (11.8c) to evaluate the Mach number in terms of the pressure ratio across the shock
wave.
1
2
M 1n
1 1
2
2
M 2n
1
1 M 1n 2 1
2
1 p2
2
and M 1n 1 1
1
2 1 p1
1/ 2
1/ 2
1 1 p2
p 2
1
M 2 n a 2 1
p1
2 1 p1
Substituting this last expression into the equation that was established for u 2 gives:
1 p2
1
1 1 p2
p 2
1 p1
1
u 2 a 2 1
2 1 p1
p1
1 1 1 p 2 1
2 1 p1
That is, the Mach number of the flow behind the shock wave will be described by the equation:
1/ 2
1/ 2
p 2
1 p2
1 1 p2
M2
1 1
1
2 1 p1
1 p1
p1
1/ 2
Problem 12.4
From the definition of the Mach number, we introduce its variation, dM as follows:
u2
M2
RT
dM du 1 dT
M
u 2 T
The temperature variation may be eliminated using Eq. (12.8c) to yield the following equation:
dM 1 2 du 1 q
1
M
M
u 2 C pT
Page 12-4
ONE-DIMENSIONAL FLOWS
Next, we eliminate the velocity u from this equation by using Eq. (12.9a), in which dA and f
are taken to be zero. This gives the following result:
1 M 2 q
dM
2
M
2( M 1) C pT
The heat added, q , may be eliminated from the foregoing result using the relations given
involving the pressure variation and the heat added. This produces the following result:
dM
1 M 2 dp
M
2 M 2 p
Separating variables in the foregoing result and integrating gives:
log (1 M 2 ) log p constant
(1 M 2 ) p constant
Then the relationship between p and M at any two locations will be as follows:
p2
p1
1 M1
2
1 M 2
2
Problem 12.5
From the definitions of the Mach number and the speed of sound we have the result:
2 T2
M 2 u 2 a1 1 T1
1/ 2
1/ 2
p 2 T1
p1 T2
In the foregoing, the velocity ratio has been eliminated using the continuity equation, and the
density ratio has been eliminated using the ideal gas law. Solving this equation for the
temperature ratio and using the result of Prob. 12.4 to evaluate the pressure ratio, yields the
result:
M1
u1 a 2
M2
T1 M 1
T2
2
1 M 22
1 M 12
2
To obtain an expression for the density ratio we use the ideal gas law as follows:
2 p 2 T1
1 p1 T 2
Page 12-5
ONE-DIMENSIONAL FLOWS
Using the result of Prob. 12.4, and the result obtained above, yields the following expression:
2 M1
1 M 2
2
1 M 12
1 M 22
3
Problem 12.6
Starting with the ideal gas law and employing the given expression for h , we get:
R
p RT
h
Cp
Substituting this result into the given expression for the entropy rise produces the result:
p 1
1 R h
s s1
log log
p1 C p 1
Cv
p1
In order to eliminate , we use the energy equation and the continuity equation as follows:
h
u2
h0
2
2
1 m
h
h0
2 A
1
A
[2(h0 h)]1/ 2
m
Substituting this result into our expression for the entropy change yields the result:
1
2
s s1
2
1 R 2 A
h 2 ( h 0 h)
log
Cv
p1 C p m
Separating the variable and the constant components of the last equation gives the result:
s s1
Cv
1
1
2
2
2
R
A
1
2
log h (h0 h) log
2
p1 C p m
The maximum value of the entropy s will occur when the derivative of the right side of the last
equation, with respect to h , is zero. This requires the following:
1
1
1
3
1
2
2
(
)
(
)
h
h
h
h
h
0
0
0
2
h( h 0 h ) 2
Simplifying this expression shows that it follows that:
Page 12-6
ONE-DIMENSIONAL FLOWS
h0
1
h
or
T0
1
T
2
2
1 2 1
i.e. T 1
M
T
2
2
In the last equality, the stagnation temperature has been evaluated using Eq. (12.10a). Then:
s maximum M 1
Problem 12.7
As in Prob. 12.6, we rewrite the expression for the entropy change as follows:
1 R h
s s1
log
p1 C p 1
Cv
1 R
s s1
h
i.e.
log 1 log
p1 C p
Cv
But from the momentum equation we get:
u 2 p p0
(12.7.1)
m2
R
h p0
2
A Cp
m2 C 1
C p
i.e. h p 0 2 p 2
(12.7.2)
R A R
In order to obtain Eq. (12.7.2), the velocity has been eliminated through use of the continuity
equation and the pressure has been eliminated through use of the ideal gas law. In principle Eq.
(12.7.2) can be solved for the density and the result substituted into Eq. (12.7.1) to give an
explicit expression for the entropy change in terms of the enthalpy h . However, the solution is
more compact if it is kept in the form of the two parametric equations of the form s s( ) and
h h( ) . Then, by substituting Eq. (12.7.2) into Eq. (12.7.1), we get the following result:
s s0
Cv
1
1
m 2 1
log p 0 2 log
A
p1
Cp 1
m2 1
h
p
0 2
A
R
For brevity we introduce the following notation:
Page 12-7
ONE-DIMENSIONAL FLOWS
s s1
Cp
m
G
Cv
A
R
Then the equations that define the Rayleigh line become:
1
Q 2
1
S log p 0
log
p
1
S
Q
G
Q2
p
0
The derivatives of the entropy change and the enthalpy, with respect to the density, are:
Q2
p
(
1)
0
d ( S )
d
Q2
p0
h
dh
G
Q2
2 p0 2
d
Then, eliminating d from these two equations and noting that ( p 0 Q 2 ) p from the
momentum equation, we get the following expression for the derivative of the entropy with
respect to the enthalpy:
Q2
p
d ( S )
(12.7.3)
dh
Q2
G p p
The entropy reaches a maximum when the numerator of Eq. (12.7.3) is zero; that is, when:
Q2
p
Then, since Q 2 u 2 / 2 and a 2 p / , this result reduces to the following condition:
S maximum M 1
The enthalpy will reach a maximum when the denominator of Eq. (12.7.3) is zero; then:
Q2
p
Then, since Q u / and a p / , this result reduces to the following condition:
2
2
2
2
h maximum M 1/
Page 12-8
13
MULTI-DIMENSIONAL FLOWS
MULTI-DIMENSIONAL FLOWS
Problem 13.1
The differential equation governing three-dimensional, unsteady, irrotational flow is given by
Eq. (13.1). For steady two-dimensional flow this equation reduces to the following form:
2
x
2
2
y
2
1
u (u
a2
1
u
a2
u
1
u
a2
u
)
x
v
y
x
2
ex
y
2
v
x2
ey
2
x y
ex
v
y2
2
u
x y
ey
2
2
2
1
2
2
2
u
u
v
v
a2
x2
x y
y2
Thus the differential equation governing the velocity potential for this class of flows is:
1
u2
a2
2
x2
2
uv 2
a2 x y
v2
a2
1
2
y2
0
Problem 13.2
For two-dimensional, steady flow, the continuity equation is:
( u)
( v) 0
x
y
Substituting the given expressions into the left hand side (LHS) of this equation shows that:
LHS
0
x
y
y
0
x
0 ...as required
x
( u)
y
( v) 0 ...for all ( x, y)
From the condition of irrotationality we get:
Multiplying this equation by the density
x
( v)
v
u
0
x
y
gives the result:
v
u
0
x
y
y
( u) v
x
u
y
0
Page 13-1
MULTI-DIMENSIONAL FLOWS
2
i.e.
2
v
u
0
x
y
x
y
0
0
In the last equation, use has been made of the definition of the stream function. The last two
terms in the last equation may be eliminated by use of the momentum equations in the form:
u
u
p
u
v
a2
x
y
x
v
v
p
u
v
a2
x
y
y
Using the last two equations to eliminate the derivatives of the density gives the result:
2
2
uv u
v2 u
u2 v
uv v
0
2
2
2
2
2
2
x
y
x
y
x
x
0 a
0 a
0 a
0 a
2
2
2
2
uv 2
u2 2
v2 2
0
x2
y2
a2 x y a2 x2
a2 y2
In arriving at the last result, each of the four terms in the previous equation was treated in the
following manner:
uv u 1 uv
( u) u
2
2
a
x
a
x
x
0
0
2
2
1 uv
u
0
2
x2
x
0 a
The second term inside each square bracket cancel each other, yielding the result:
u2
1
a2
2
uv 2
2 2
a x y
x2
v2
1
a2
2
0
y2
Problem 13.3
(a) Consider both the velocity potential and the stream function to be functions of x and y , and
consider infinitesimal changes in both x and y . Then the total change in the velocity
potential and the stream function will be given by the following expressions:
d
d
dx
dy u dx v dy
x
y
q cos dx q sin dy
dx
x
q sin
0
y
dy
v dx
u dy
0
q cos
dx
0
dy
0
That is, the total changes in the velocity potential and the stream function are:
Page 13-2
MULTI-DIMENSIONAL FLOWS
q (cos
d
d
dx
sin
dy )
q ( sin
dx
cos
dy )
0
From the foregoing two equations, the following identities are obtained:
cos
q cos 2
d
0
q sin cos
dx
0
sin
dy
0
q sin 2
d
q sin cos
dx
0
dy
0
Subtracting these two equations yields an expression for dx . A similar procedure leads to an
analogous expression for dy . The resulting expressions are:
dx
cos
d
q
dy
sin
d
q
0
0
sin
d
q
cos
d
q
(b) Since both the velocity potential and the stream function are considered to be functions of
q and , we have the equations:
d
d
d
dq
d
dq
d
d
d
d
dq
d
dq
d
Substituting these expressions into the equations obtained above gives the following
equations for the variation in x and y with respect to q and :
dx
cos
q
dy
sin
q
0
q
0
q
sin
q
cos
q
q
q
dq
cos
q
dq
sin
q
0
0
sin
q
d
cos
q
d
(c) We now consider x and y to be functions of q and . Then from differential calculus:
x
x
dx
dq
q
y
y
dy
dq
q
Then, equating the coefficients of dq and d in the expressions above with those obtained in
(b), leads to the following equalities:
Page 13-3
MULTI-DIMENSIONAL FLOWS
x
q
cos
q
x
cos
q
y
q
sin
q
y
sin
q
0
q
0
sin
q
sin
q
cos
q
0
q
0
q
q
cos
q
(d) Using the first two results obtained above, we form the second mixed derivative of x , then
the last two results are used to form the second mixed derivative of y . This produces the
following expressions:
2
x
sin
0 cos
q
q q
q
q
cos
q2
2
0
cos
q
y
q
0
q
sin
q2
d
dq
sin
q
0
sin
q
q
d
sin
0 cos
2
2
q
q
dq
We next add the two mixed second derivatives as follows:
2
2
x
y
1
sin
cos
q
q
q q
0
1
q2
0
1 d
q dq
cos
q
0
d 1 0
dq q
Equating the first and third equalities on the right side of the foregoing equations gives:
q
q
d 1
dq q
0
Adding the same two equations again, but in a different manner, yields the result:
2
cos
q
x
2
sin
q
y
0
1
q q
1
q2
Equating the last two equalities in the foregoing equations produces the result:
Page 13-4
MULTI-DIMENSIONAL FLOWS
0
q
q
(e) Bernoulli’s equation for the case under consideration is:
dp 1 2
contant
q
2
dp
q dq 0
dp
dq
so that
q
The following identity is now obtained:
d
dp
0
0 d
0 d
2
2
dq
dq
dp dq
We now use the definition of the speed of sound and the last two equations to give:
d
dq
0
0
From part (d) we have the following result:
d 1
q
q
dq q
q
a2
0
d
1 0
0
dq
q
Using the expression for the derivative of the density ratio in the last equation produces the
following result:
0
q
1
(1 M 2 )
q
yields:
Differentiating the last result with respect to
2
2
1
(1 M 2 ) 2
q
q
Another expression for the same quantity may be obtained from a result obtained in (d) as
follows:
0
0
q
q
2
2
0
q
q
q
0
q
0
q
q
q2
Page 13-5
MULTI-DIMENSIONAL FLOWS
2
2
0
i.e.
0
M2
0
q
q
q
q
q2
In the last identity, the result obtained above for the derivative of the density ratio with
respect to the velocity q has been used. Equating the two expressions for the second mixed
derivative of the velocity potential , then simplifying, produces our final result:
2
q
2
2
2
2
q (1 M )
q2
(1 M )
q
0
2
Problem 13.4
From section 13.2 of the text, the general form of the Janzen-Rayleigh expansion equation is as
follows:
2
2
2
a
2n
2
2n
2n
2n
n
n
n
n
M
M
M
M
M
2
x
x
a
x
x
x
x
n
n
n
n
i
i
i
j
i
j
2
a2
where
2
a
1
1
2
M
2
M
2
M
2
1
n
xi
n
2
1
1
2
0
1
xi
(
1)
0
1
xi
xi
M
2
The ratio of the two speeds of sound has to be inverted prior to substitution into the governing
equation. This may be done by considering the required ratio to be of the following form:
a2
2
(1
) 1 1
2
a
Hence the required form of the ratio of the speeds of sound is found to be:
a
a
2
2
2
1
1
2
2
0
1
xi
1
2
M
4
(
2
0
1) 1
4
xi
0
1
xi
xi
M
Also, the product of the three round-bracketed terms in the governing equation is:
0
xi
M
2
1
xi
4
M
2
0
xi
xj
2
2
0
xi x j
That is:
Page 13-6
M
M
2
2
1
xj
2
1
xi x j
M
4
2
xi x j
M
4
2
xj
4
MULTI-DIMENSIONAL FLOWS
2
0
2
0
xi
0
1
x j xi x j
2
0
0
xi x j xi x j
2
1
0
1
xi x j xi x j
0
xi x j xi x j
1
0
xi x j xi x j
1
0
0
x j xi x j
xi
1
x j xi x j
M
2
xi x j xi x j
2
2
xi
0
1
2
0
0
2
0
2
xi
2
1
2
1
2
0
0
0
x j xi x j
0
xi
2
4
M
x j xi x j
The right side of the governing equation is evaluated by multiplying the two expansions obtained
above with the square of the Mach number at infinity. This gives the following result for 3 :
2
2
2
1
3
xi xi
(
4
2
0
1) 1
4
xi
2
1
2
1
xi
1
0
xi
xi
xi
0
0
x j xi x j
2
0
0
0
xi x j xi x j
2
1
1
2
0
1
0
0
1
xi x j xi x j
1
0
xi x j xi x j
1
1
x j xi x j
1
xi x j xi x j
2
0
xi
xi
0
2
0
2
0
xi x j xi x j
2
0
2
1
2
0
0
x j xi x j
xi
2
2
0
0
x j xi x j
xi
0
2
x j xi x j
Problem 13.5
The exact version of the pressure coefficient is given by Eq. (13.4a), which is:
/(
2
M
Cp
where U 2
2
1
1
2a
uu
2U u
2
(U
2
(u ) 2
1)
u u)
(v ) 2
1
(w ) 2
/(
Cp
2
M
2
1
1
2a
2
(2U u
(u ) 2
(v ) 2
(w ) 2
1)
1
The quantity inside the square brackets is of the following form:
(1
where n
)n
n(n 1) 2
2!
n(n 1)
so that
1
2!
1
2(
1) 2
Page 13-7
MULTI-DIMENSIONAL FLOWS
Cp
2
M
2
2a
2
(u ) 2
(2 U u
(v ) 2
(w ) 2
8a
2
[4 U 2 (u ) 2
]
Simplifying this expression yields the following equation which is valid to the second order in
the primed (i.e. small) quantities:
(u ) 2 (v ) 2
U2
u
2
U
Cp
Problem 13.6
The drag force that acts on one wave-length of wall is defined by the following integral:
dy
FD
p( x, 0)
dx
0
dx
2
2 x
p( x, 0) cos
dx
0
In the foregoing equation, p( x,0) is the pressure difference across the wavy wall, and the
sinusoidal variation of the wall height has been employed. It will be assumed that the pressure
below the wall surface is the same as the reference pressure far from the wall. For subsonic flow
the pressure coefficient is given by Eq. (13.6b). Hence the required pressure distribution will be
defined as follows:
4
/
2 x 2 1 M 2y
C p ( x, y )
sin
e
2
1 M
U2/
2
p ( x, 0)
1 M
sin
2
2 x
Thus for subsonic flow the drag force per unit wavelength of wall will be:
FD
2
2
U2
1 M
i.e. FD
2
0
sin
2 x
0 for M
cos
2 x
dx
1
For supersonic flow the pressure coefficient is defined by Eq. (13.7b). Hence the required
pressure distribution will be defined as follows:
C p ( x, y )
4
M
p ( x, 0)
/
2
cos
2 x
x
M
2
1y
1
U2/
2
M
2
cos
2 x
1
Thus for supersonic flow the drag force per unit wavelength of wall will be:
Page 13-8
MULTI-DIMENSIONAL FLOWS
FD
2
2
2
U2
M
2
2
i.e. FD
2
1
2
0
cos 2
U2
M
2 x
dx
1
for M
2
1
Problem 13.7
a , the solution for the velocity potential will be of the
Considering the boundary condition at r
following separable form:
( x, r )
R(r ) cos
2 x
Substituting this expression into the differential equation gives:
2
d 2R
dR
2
2
r
r
(1 M )
r 2R 0
2
dr
dr
This is the modified Bessel equation of order zero whose solution is:
r
2 2
R(r ) A I 0 1 M
2
In the foregoing, I 0 is the modified Bessel function of the first kind of order zero. The second
solution has been suppressed because it contains a logarithmic term – which is physically
unacceptable. Then the solution for the velocity potential becomes:
r
2 x
2 2
( x, r ) A I 0 1 M
cos
a and using the result I 0 ( x)
Imposing the boundary condition at r
1
U
A
I 1 ( x) gives:
1 M
2
I1
1 M
2
2 a
Then the velocity potential for the flow field will be:
( x, r )
I0
U
1 M
2
I1
1 M
1 M
2
2
2 r
2 a
cos
2 x
Using this result, and Eq. (13.4b), the following expression is obtained for the surface pressure
coefficient on the cylinder:
Page 13-9
MULTI-DIMENSIONAL FLOWS
Cp
4 /
cylinder
2
1 M
2
1 M
I0
2
1 M
I1
2 a
2 a
sin
2 x
For a plane wall, the pressure coefficient is given by Eq. (13.6b). Thus the surface pressure
coefficient will be given by the following expression:
Cp
4 /
plane
1 M
2
2 x
sin
Thus it follows that the ratio of the two surface pressure coefficients is:
Cp
cylinder
Cp
plane
I0
1 M
2
2 a
I1
1 M
2
2 a
For large values of the argument x , the following asymptotic expansions apply:
ex
1
I 0 ( x)
1
8x
2 x
ex
3
1
8x
2 x
I 1 ( x)
I 0 ( x)
1
8x
1
I 1 ( x)
1
3
8x
1
4x
Then the ratio of the surface pressure coefficients for large values of the cylinder-radius/surfacewavelength ratio becomes:
1
Cp
cylinder
Cp
plane
1
1
1 M
2
8 a
This result shows the initial effect of surface curvature on the surface pressure coefficient.
___________________________________________________________________________
Problem 13.8
The drag coefficient, according to Ackeret's linear theory, is given by Eq. (13.9b). Then for zero
angle of attack, and for no camber, the drag coefficient will be given by the following
expression:
Page 13-10
MULTI-DIMENSIONAL FLOWS
CD
4
2
M
2
c
c
1
0
( ) 2 dx
For the wing section shown, the half-thickness function is defined as follows:
x
for 0 x c / 2
c
( x)
x
1
for c / 2 x c
c
Then the drag coefficient will have the following value:
CD
4
2
M
2
c
c/2
1
0
c
(1/ c) 2 dx
CD
4
M
c/2
( 1/ c) 2 dx
2
2
1
___________________________________________________________________________
Problem 13.9
Dividing the given equation for the upper surface of the airfoil by a 2 we get:
2
a
1
1
x
a2
c
2
2
1
c
2a
2
2
c
c
1
x
1 2
1
2
a
a
a
2
In the foregoing, the radius a has been taken to be large compared with both the half-thickness
of the wing and the chord. This linearization is required since Ackeret's theory is a linear theory,
and consistency requires that the equation of the surface be linearized. Solving the last equation
for the half-thickness produces the result:
2
c 1
c
x
2
2a
2
The radius a may now be eliminated from this result by noting that
produces the following identity, which may be used to eliminate a :
c c2
0
2 8a
c
2
Thus the half-thickness function
2
c
x
0 when x 0 . This
c
2
will be defined by the following expression:
Page 13-11
MULTI-DIMENSIONAL FLOWS
1
2
( x)
2
x
c2
4
x
c2
( x)
c
2
2
c
2
Then, according to Ackeret's theory, the drag coefficient will be defined as follows:
CD
4
2
M
2
CD
c
16
x
4
0
1 c
c
16
3 M
c
2
2
dx
2
2
1
Comparing this result with that obtained in Prob. 13.8 shows that the smooth-contour airfoil has
a larger drag than that of the sharp-edged airfoil in supersonic flow, and that the ratio of the two
drags is 4/3.
___________________________________________________________________________
Page 13-12
14
SOME USEFUL METHODS OF ANALYSIS
SOME USEFUL METHODS OF ANALYSIS
Problem 14.1
The function is periodic in the range –L ≤ x ≤ L and it is odd in x. Hence we use Eqs. (14.4a) and
(14.4b):
n x
f ( x ) bn sin
L
n 1
L
n x
2
dx
where bn f ( x) sin
0
L
L
But over the range 0 ≤ x ≤ L we have f (x) = H, so that;
2 L
n x
bn H sin
dx
0
L
L
L
2H L
n x
cos
L n
L 0
2H
1 (1) n
n
2
n x
1 ( 1) n sin
L
n 1 n
_____________________________________________________________________________
f ( x) H
Problem 14.2
The function is periodic in the range –L ≤ x ≤ L and it is even in x. Hence we use Eqs. (14.2a),
(14.2b) and (14.2c):
n x
f ( x ) a 0 a n cos
L
n 1
L
1
where a 0 f ( x) dx
L 0
n x
2 L
dx
and a n f ( x ) cos
0
L
L
But over the range 0 ≤ x ≤ L we have f (x) = H x 2/L 2, so that;
2 L x2
H
a 0 H 2 dx
3
L 0 L
n x
2 L
and
a n H cos
dx
L 0
L
2 H n
3 3 2 cos d
n 0
4H
2 2 ( 1) n
n
H
n x
4
H 2 2 ( 1) n cos
L
3
n 1 n
_____________________________________________________________________________
f ( x)
Page 14-1
SOME USEFUL METHODS OF ANALYSIS
Problem 14.3
The function is periodic in the range –L ≤ x ≤ L and it is odd in x. Hence we use Eqs. (14.4a) and
(14.4b):
n x
f ( x ) bn sin
L
n 1
L
n x
2
dx
where bn f ( x) sin
0
L
L
But over the range 0 ≤ x ≤ L we have f (x) = H x 2/L 2, so that;
n x
2 L x2
bn H 2 sin
dx
L 0 L
L
2 H n
3 3 2 sin d
n 0
2H
4H
( 1) n1 3 3 1 ( 1) n
n
n
4
2
n x
f ( x ) H ( 1) n1 3 3 1 ( 1) n sin
n
L
n 1 n
_____________________________________________________________________________
Problem 14.4
1.2
1.1
1
0.9
0.8
f(x)/H
0.7
10 Terms
0.6
20 Terms
0.5
30 Terms
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
x/L
Page 14-2
0.6
0.7
0.8
0.9
1
SOME USEFUL METHODS OF ANALYSIS
The graph shown above was drawn using EXCEL. Each curve on the graph was generated using
the number of terms specified, and for increments of 0.01 for x/L.
_____________________________________________________________________________
Problem 14.5
z2 z4
2! 4!
cos z 1 1 1 z
3
z3
z
2! z 4!
cos z 1
(a)
Therefore
Hence
(b)
Therefore
Hence
b1
1
2
pole of order 3 at z = 0
z2 z4
cosh z 1
2! 4!
cosh z 1 1 1 z
3
z3
z
2! z 4!
b1
1
2
pole of order 3 at z = 0
(c) Here the singularity is at z = α so that our expansion should be about the point z = α.
(z ) 2 (z )3
e z e {e z } e {1 ( z )
}
2!
3!
1
ez
1
1 (z )
e
Therefore
2
2
(z )
( z ) 2!
3!
(z )
Hence
b1 e
pole of order 2 at z = α.
__________________________________________________________________________
Problem 14.6
(a) There is a pole of order 3 at z = 0 here so that from Eq. (14.12b)
1 d 2 3 cos z
b1 lim
z 3
z 0 2! dz 2
z
1
lim ( cos z )
z 0 2
Hence
b1
1
2
(b) There is a pole of order 3 at z = 0 here so that from Eq. (14.12b)
Page 14-3
SOME USEFUL METHODS OF ANALYSIS
1 d 2 3 cosh z
b1 lim
z
z 0 2! dz 2
z 3
1
lim (cosh z )
z 0 2
b1
Hence
1
2
(c) There is a pole of order 2 at z = α here so that from Eq. (14.12b)
d
e z
b1 lim ( z ) 2
z dz
( z ) 2
lim e z
z
Hence
b1 e
__________________________________________________________________________
Problem 14.7
(a)
let p cos z and q sin z
This function is singular at z = 0. Then from Eq. (14.12c);
p( z 0) cos z
b1
q' ( z 0) cos z z 0
Hence
(b)
b1 1
let p z and q sinh( z )
This function is singular at z = α. Then from Eq. (14.12c);
p( z )
z
b1
q' ( z ) cosh( z )
Hence
z
b1
let p e z and q ( z )
(c) )
This function is singular at z = α. Then from Eq. (14.12c);
p( z ) e z
b1
q' ( z ) 1 z
Hence
b1 e
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Page 14-4
SOME USEFUL METHODS OF ANALYSIS
Problem 14.8
There are poles of order 2 at z = +α and at z = −α, as can be readily seen by rewriting the given
function as follows;
4 2
4 2
F ( z) 2
(z 2 ) 2 (z ) 2 (z ) 2
Then we can use Eq. (14.12c) at both singularities using m = 2.
d 4 2
8 2
Residue z lim
lim
3
z dz ( z ) 2
z ( z )
Residue z
Residue z
1
d 4 2
8 2
lim
lim
3
z dz ( z ) 2
z ( z )
Residue z
1
_____________________________________________________________________________
Problem 14.9
(a) Step 1:
Step 2:
Step 3:
(b) Step 1:
Step 2:
Step 3:
A sin
x
L
sinh ( H y )
L
H
sinh
L
sinh ( H y )
x
L
T ( x, y ) A sin
H
L
sinh
L
B sin
x
L
sinh y
L
H
sinh
L
x sinh L y
T ( x, y ) B sin
H
L
sinh
L
__________________________________________________________________________
Page 14-5
SOME USEFUL METHODS OF ANALYSIS
Problem 14.10
C sin
(a) Step 1:
H
sinh ( L x )
H
L
sinh
H
Step 2:
Step 3:
T ( x, y ) C sin
y
D sin
(b) Step 1:
sinh
H
H
( L x)
sinh
L
H
y
H
sinh x
H
L
sinh
H
Step 2:
Step 3:
y
y sinh H x
T ( x, y ) D sin
L
H
sinh
H
__________________________________________________________________________
Problem 14.11
Using the principle of superposition, we obtain the required solution by superimposing the
results of Probs. (14.9) and (14.10).
x sinh L ( H y )
x sinh L y
B sin
T ( x, y ) A sin
H
H
L
L
sinh
C sin
y
H
sinh
L
sinh
H
( L x)
sinh
L
H
D sin
L
y
sinh
H
x
H sinh L
H
If the boundary conditions on all four sides were different graphical functions, we would
represent each of them by their Fourier series and the solution would be the sum of four different
series.
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Page 14-6
SOME USEFUL METHODS OF ANALYSIS
Problem 14.12
The y component of the solution will be;
U sin
y
h
The t component of the solution will be exponential with the coefficient of t being the product of
υ with the square of the quantity π/h. That is, the t component of the solution will be;
e ( / h )
2
t
Thus the complete solution will be;
u( y, t ) U sin
y
e ( / h )
2
t
h
_____________________________________________________________________________
Problem 14.13
2
1 f dF
d 2 F 1 dF
0
2
df
F df
2 F df
(a) Let F = aF * and f = bf *. Then the differential equation above becomes;
2
1 f dF
a d 2 F a 1 dF
0
b 2 d f 2 b 2 F d f
2 F d f
If the equation is to be invariant, we must have;
2
a
F f
F
F 2
1
1
b2
F f
f 2 f 2
That is
F
f2
is an invariant coordinate.
dF
be the new dependent variable
df
f 2 F
And
be the new independent variable
We will choose n later so that f is eliminated from the differential equation.
The following derivatives are now obtained for substitution into the differential equation:
dF
f n G
df
(b) Let
G ( ) f
n
d dG
d 2F
n f ( n1)G f n
2
df
df d
d d
dF
( f 2 F ) 2 f 3 F f 2
But
df df
df
1
( n 2)
2f f
G
In the above we have used the definitions of G and ξ to eliminate F in favor of ξ and f.
Substituting this result into Eq. (*) gives;
(*)
Page 14-7
SOME USEFUL METHODS OF ANALYSIS
d 2F
n f
df 2
2 f
G
dG
d
dG
dG
n f ( n1)G 2 f ( n1)
f 2( n1)G
d
d
Substituting the above derivatives into the differential equation yields the result;
( n 1)
G f
n
f
1
( n 2)
2
dG
1
G
( n1)
( n1) dG
2( n1)
2( n1) G
G 2f
G
f
f ( n1) 0
n f
f
d
d
2
We now note that this will be an ODE for G( ) if we choose n 1. Then the ODE is;
G 2
That is
dG
dG G 2 G
G
0
d
d
2
dG
G(2G 2 1)
d
2 (G 2 )
Once this equation is solved for G( ) , we can integrate the following equation to get F ( f );
dF
fG
df
Once the equation above is solved for F ( f ) , we can integrate the following equation to get
f ( ) ;
df
F
d
That is, the problem has been reduced to a series of first order equations, for which many
analytical and numerical methods of solution exist.
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Page 14-8
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