fuandametals of fluid mechanics by I. G. Currie solution manual

SOLUTIONS MANUALFOR
by
Fundamental
Mechanics of Fluids
Fourth Edition
I.G. Currie

SOLUTIONS MANUAL FOR
by
Fundamental
Mechanics of Fluids
Fourth Edition
I.G. Currie

CRC Press
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BASIC CONSERVATION LAWS
Page 1-1
Problem 1.1
Inflow through x = constant: u y z  
Outflow through x x = constant: ( )u y z u y z x
x
      

 

Net inflow through x = constant surfaces: ( )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:
( )x y z
t
  


Then the equation of mass conservation becomes:
( ) ( ) ( )x y z u v w x y z
t x y z

        
    
         
Taking the limits as the quantities , andx y z   become vanishingly small, we get:
( ) ( ) ( ) 0u v w
t x y z

  
   
   
   

Page 1-2
Problem 1.2
Inflow through R = constant: Ru R z  
Outflow through R R = constant: ( )R Ru R z u R z R
R
      

 

Net inflow through R = constant surfaces: ( )RRu R z
R
   

 

Net inflow through  = constant surfaces: ( )u R z   


 

Net inflow through z = constant surfaces: ( )zu R R z
z
   

 

( )R R z
t
   


( ) ( ) ( )R zR R z Ru u R u R z
t R z


        

    
         
Taking the limits as the quantities , andR z   become vanishingly small, we get:
1 1
( ) ( ) ( ) 0R zRu u u
t R R R z


  

   
   
   

Page 1-3
Problem 1.3
Inflow through r = constant: 2
sinru r   
Outflow through r r = constant: 2
sinru r   
2
( sin )rr u r
r
    

 

Net inflow through r = constant surfaces: 2
( )sinrr u r
r
    

 

Net inflow through  = constant surfaces: ( sin )u r r    


 

Net inflow through  = constant surfaces: ( )u r r   


 

2
( sin )r r
t
    


2 2
sin ( )sin ( sin ) ( )rr r r u r u r u r
t r
 

           
 
    
         
Taking the limits as the quantities , andr   become vanishingly small, we get:
2
2
1 1 1
( ) ( sin ) ( ) 0
sin sin
rr u u u
t r r r r
 

   
   
   
   
   

Page 1-4
Problem 1.4
Using the given transformation equations gives:
2 2 2
2
2 2
and tan
2 2 2 cos cos
1 sin 1
and sec sin
cos
y
R x y
x
R R
R x R
x x
y
x x R x R

 
  
 

  
 
    
 
 
      
 
Using these results, the derivatives with respect to andx y transform as follows:
sin
cos
cos
sin
R
x x R x R R
R
y y R y R R
z z
 

 
 

 
      
   
      
      
   
      
 

 
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
( ) cos [ ( cos sin )] [( ( cos sin )]R Ru u u u u
x R R
 

       

  
   
  
cos
( ) sin [ ( sin cos )] [( ( sin cos )]R Rv u u u u
y R R
 

       

  
   
  
Adding these two terms and simplifying produces the following equation:
1
( ) ( ) ( ) ( )
1 1
( ) ( )
R
R
R
u
u v u u
x y R R R
Ru u
R R R



   

 

   
   
   
 
 
 
Substituting this result into the full continuity equation yields the following result:
1 1
( ) ( ) ( ) 0R zRu u u
t R R R z


  

   
   
   

Page 1-5
Problem 1.5
The equations that connect the two coordinate systems are as follows:
2
2 2 2 2 2
2 2 2
sin cos sin sin cos
tan cos
( )
x r y r z r
y z
r x y z
x x y z
    
 
  
    
 
Using the relations given above, the following identities are obtained for the various partial
derivatives:
sin cos sin sin cos
1 1 1
cos cos cos sin sin
1 sin 1 cos
0
sin cos
r r r
x y z
x r y r z r
x r y r z
    
  
    
    
 
  
  
  
  
     
  
  
  
  
Thus the following expressions are obtained for the various Cartesian derivatives:
1 sin
sin cos cos cos
sin
1 1 cos
sin sin cos sin
sin
1
cos sin
r
x x r x x
r
r r
r
y y r y y
r r r
r
z z r z z
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:
x y z r ru v w u u u        u e e e e e 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 1-6
sin cos sin sin cosx y z x y zx y z r r r         r e e e e e e
Also, recalling from Appendix A that:
it follows that
sin cos sin sin cos
cos cos cos sin sin
sin cos
i
i i
r x y z
x y z
x y
x x


    
    
 
 

 
  
  
  
r r
e
e e e e
e e e e
e e e
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:
sin cos cos cos sin
sin sin cos sin cos
cos sin
r
r
r
u u u u
v u u u
w u u
 
 

    
    
 
  
  
 
Using these results, and those obtained for the Cartesian derivatives, produces the following
expression for the Cartesian terms that appear in the continuity equation:
2
2
1 1 1
( ) ( ) ( ) ( ) ( sin ) ( )
sin sin
ru v w r u u u
x y z r r r r
       
   
     
    
     
Substituting this result into the full continuity equation yields the following expression:
2
2
1 1 1
( ) ( sin ) ( ) 0
sin sin
rr u u u
t r r r r
 

   
   
   
   
   
Problem 1.6
From Appendix A, the following value is obtained for the convective derivative:
1 2 3 2
1 2 3 2 2 1 1
1 1 1 1 2 1 2
3
1 1 3 3 1 2 3
3 3 1
1
( ) ( ) ( )
( ) ( ) ( ) ( )
a a a a
a a a h a h a
h x x x h x x
a
h a h a
h x x
       
                 
   
    
   
a a
e e e

Page 1-7
Applying this result to cylindrical coordinates, we interpret the various terms as follows:
1 2 3
1 2 3
1 2 3
1 2 3
1 1
R z
R z
a u a u a u
x R x x z
h h R h



  
  
  
  e e e e e e
Using these results, the required term becomes:
 ( ) ( )
R z R R z
R R z z
u u u u u u u
u u u Ru u
R R R R R z R
 
 

         
         
          
e u u
Simplifying the right side of this equation produces the required result:
 
2
( )
R R R
R R z
u u u u u
u u
R R R z
 

  
   
  
e u u
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:
1 2 3
1 2 3
1 2 3
1 2 3
1 sin
r
r
a u a u a u
x r x x
h h r h r
 
 
 

  
  
  
  e e e e e e
Using these results, the required term becomes:
 ( ) ( ) ( sin
sin
r r r
r r
u u u u u u u
u u u r u r u
r r r r r r r
   
    
  
          
          
           
e u u
Simplifying the right side of this equation yields the required result:
 
2 2
( )
sin
r r r
r r
u u u u u u u
u
r r r r
   
  
   
   
  
e u u

Page 1-8
Problem 1.8
For a Newtonian fluid, the shear stress tensor is defined by the following equation:
k i j
i j i j
k j i
u u u
x x x
  
   
       
Evaluating the various terms in this expression for Cartesian coordinates ( , ,x y z ) and Cartesian
velocity components ( , ,u v w ) yields the following results:
2
2
2
xx
y y
z z
x y y x
xz z x
y z z y
u v w u
x y z x
u v w v
x y z y
u v w w
x y z z
u v
y x
u w
z x
v w
z y
  
  
  
  
  
  
    
    
    
    
    
    
    
    
    
  
   
  
  
   
  
  
   
  
For a monotonic gas, the Stokes relation requires that 2 /3   . Then the relations obtained
above assume the following special form:
4 2 2
3
4 2 2
3
4 2 2
3
xx
y y
z z
x y y x
xz z x
y z z y
u v w
x y z
v u w
y x z
w u v
z x y
u v
y x
u w
z x
v w
z y






  
  
  
   
   
   
   
   
   
   
   
   
  
   
  
  
   
  
  
   
  

Page 1-9
Problem 1.9
For a Newtonian fluid, the dissipation function is defined by the following equation:
2
k i j j
k j i i
u u u u
x x x x
 
      
               
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 22 2
2 22
2
u v w u v w
x y z x y z
u v u w v w
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 22 2
2 22
2
2 2 2
3
u v w u v w
x y z x y z
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:
( ) ( ) 0u v
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 1-10
0 0
0 0 0
* *
* *
* *
0
0
u v
x y
u v
u v
x x x y
 
 
  
  
    
    
       
       
                    
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:
* *
0 (1.15)
u u
x x
 
 
 
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:
2
0 0 0
2
0 0 0
1
1
p
u u v
x y x
p
u v v
x y y
 
  
 
  
     
             
     
             
Using the definitions of the new velocity components, these equations become:
   
   
2
0
2
0
* * *
* * *
1
1
p
u u v
x y x
p
u v v
x y 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:
0
0
* *
* *
* *
* *
* *
0
1
1
u v
x y
u u p
u v
x y x
v v p
u v
x y y


 
 
 
  
  
  
  
  
  

FLOW KINEMATICS
Page 2-1
Problem 2.1
The following graph was drawn using EXCEL.
__________________________________________________________________________
Problem 2.2
(a) 1
(1 )
But 1 when 0 0
dy v
t
dx u
y t x C
x y t C
Hence at 0t the equation of the streamline is:
y x

FLOW KINEMATICS
Page 2-2
1 2
1
(b) 1
1
log(1 )
dx dy
u v
dt t dt
x t C y t C
The condition that 1x y when 0t requires that 1 2 1C C , so that:
1 log(1 ) 1x t y t
Eliminating t between these two equations shows that the equation of the pathline is:
1x
y e
(c) Here, the equations obtained in (b) above are required to satisfy the condition 1x y
when t . This leads to the values 1 1 log(1 )C and 2 1C . Hence the
parametric equations of the streakline are:
log(1 ) 1 log(1 ) 1x t y t
At time 0t these equations become:
1 log(1 ) 1x y
Eliminating the parameter between these two equations yields the following equation
for the streakline at 0t :
1
2 x
y e
Problem 2.3
(a) (1 ) 1 0u x t v w
(1 )
1 1(1 ) at 0t s sdx
x t x C e C e t
ds
21
dy
y s C
ds
But x = y = 1 when s = 0 so that 1 2 1C C . Hence:
and 1s
x e y s
1y
x e

FLOW KINEMATICS
Page 2-3
(b) (1 /2)
1(1 ) t tdx
x t x C e
dt
21
dy
y t C
dt
But x = y = 1 when t = 0 so that 1 2 1C C . Hence:
(1 / 2)
and 1t t
x e y t
2
( 1)/2y
x e
(c) As in (b): (1 /2)
1 2andt t
x C e y t C
But x = y = 1 when t = so that (1 /2)
1 2and 1C e C . Hence:
(1 /2) (1 /2) (1 /2)
for 0t t
x e e t
and 1 1 for 0y t t
(1 )(3 )/ 2y y
x e
(d) From the continuity equation:
( ) ( ) ( ) 0
D
u v w
t x y z Dt
u
Hence: (1 )
D
t
Dt
u since (1 )tu here.
i.e. (log ) (1 )
D
t
Dt
Therefore log (1 / 2) logt t C
So that (1 /2)t t
C e
But 0 when t = 0 so that 0C . Hence:
(1 /2)
0
t t
e along the streamline
Problem 2.4
The equations that define the streamlines are as follows:
/(1 )
(1 )
log log
(1 )
or
i i
i i
i i
s t
i i
dx dx ds
ds or
u x t
s
x C
t
x C e

FLOW KINEMATICS
Page 2-4
Let 0i ix x when 0s . Then:
/(1 )
0
s t
i ix x e
So that:
0 0 0
x y z
x y z
for any time t .
The equations that define the pathlines are as follows:
(1 )
log log(1 ) log
or (1 )
i i i
i
i i
i i
dx dx x
u or
dt dt t
x t C
x C t
Let 0i ix x when 0s . Then:
0
0
(1 ) or (1 )
i
i i
i
x
x x t t
x
So that:
0 0 0
x y z
x y z
as per the streamlines.
Problem 2.5
(a) 2 2
16 10u x y v w y z
10 10
2
0 0
5 5
0 0
0 0
2
10 10
0 0
5 5
16,000
On 0 10, 0: 16
3
On 0 5, 10: 10 50
16,000
On10 0, 5: (16 5) 50
3
On 5 0, 0: 10 50
x y udx x dx
y x vdy dy
x y udx x dx
y x vdy dy
Then, adding the components around the counter-clockwise path gives:
50
(b) For the area specified, and 1z z
v u
x y
n e .
10 5
0 0
( 1)
A
dA dx dyω n

FLOW KINEMATICS
Page 2-5
50
A
dAω n
This is the same result that was obtained in (a), so that Eq. (2.5) is verified for this flow.
Problem 2.6
2 2 2 2
and
x y
u v
x y x y
1 1 1 1
1 1 1 1
1 1 1 1
2 2 2 21 1 1 1
( , 1) ( 1, ) ( , 1) ( 1, )
1 1 1 1
u x dx v y dy u x dx v y dy
xdx y dy xdx y dy
x y x y
In the foregoing equation, we note that there are two pairs of offsetting integrals. Hence:
0
Problem 2.7
(a) 2 2
9 2 10 2u x y v x w yz
11 1
2 3
1 1 1
( , 1) (9 2) [3 2 ] 2u x dx x dx x x
11 1
1 1 1
( 1, ) 10 [10 ] 20v y dy dy y
11 1
2 3
1 1 1
( , 1) (9 2) [3 2 ] 10u x dx x dx x x
11 1
1 1 1
( 1, ) ( 10) [ 10 ] 20v y dy dy y
Hence: 2 20 10 20 32du l
(b) 2
2 8x y z x z
w v u w v u
z
y z z x x y
ω e e e e e
Hence 50 8x zω e e on the plane 5z

FLOW KINEMATICS
Page 2-6
(c) For the plane 5z the unit normal is zn e . Hence:
1 1
1 1
8 32
A
dA dx dyω n
This agrees with the result obtained in (a) - as it should, since
A
d dAu l ω n
Problem 2.8
2 2 2 2
and
y x
u v
x y x y
1 1 1 1
1 1 1 1
1 1 1 1
2 2 2 21 1 1 1
1
1 1
121
(a) ( , 1) ( 1, ) ( , 1) ( 1, )
1 1 1 1
4 4[tan ]
1
u x dx v y dy u x dx v y dy
dx dy xdx dy
x y x y
d
2
2 2
2 2 2 2 2 2 2 2 2 2
(b)
1 2 1 2
( ) ( ) ( ) ( )
v u
x y
x y
x y x y x y x y
0 provided and 0x y
2 2 2 2 2 2
2 2
(c) and
( ) ( )
u x y v x y
x x y y x y
0 provided and 0x yu

FLOW KINEMATICS
Page 2-7
Problem 2.9
;u y v x
(a)
1 1 1 1
1 1 1 1
d ( ) ( )dx dy dx dyu
1 1 1 1
1 1 1 1
2 2 2 2
[ ] [ ] [ ] [ ]y yx x
4( )
(b) ( )
A
v u
ndA dxdy dxdy
x y
4( )
A
ndA
(c) and
dx dy
y x
ds ds
or
dy x
y dy xdx
dx y
2 2 2
2 2 2
y x c
2 2 2
x y c
(d) 2 2 2
1 and 1 x y c where andu y v x .
But x = 1 when y = 0 so that c 2
= 1. Therefore;
2 2
1x y
(e) 2 2 2
1 x y c where andu y v x .
But x = 0 when y = 0 so that c = 0. Therefore;
y x

FLOW KINEMATICS
Page 2-8
Problem 2.10
The vorticity vector will be in the z direction and its magnitude will be:
1 1
( , ) ( )
u
R Ru
R R R
(a) 0 andRu u R
2
( ) ( ) 2
and 0
R
Ru R R
R R
u
( , ) 2R R
(b) 0 and
2
Ru u
R
( ) 0
2
and 0
R
Ru
R R
u
( , ) 0 provided 0R R

3
SPECIAL FORMS OF THE
GOVERNING EQUATIONS

SPECIAL FORMS OF THE GOVERNING EQUATIONS
Page 3-1
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 k T
Dt Dt Dt



      u  
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:
   
 
2
1
2
1
2
1
2
1
2
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
 
 
  
      
u u u u u u
u u u u u u
u u u u u u
u u u u u



   
     
    
    
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 0u so that:
  21
2
( ) ( ) ( )   u u u u u u
    

SPECIAL FORMA OF THE GOVERNING EQUATIONS
Page 3-2
Problem 3.3
From Bernoulli’s equation we have:
2
0
1 1
2 2
( )p p U   u u
In cylindrical coordinates, the velocity-squared term in this equation may be written as
follows:
2 2
2 22 2
2 2 2 2
2 2
4 2
2 2 2
4 2
4 2
2
4 2
1 cos 1 sin
1 2 (cos sin )
1 2 cos2
Ru u
a a
U U
R R
a a
U
R R
a a
U
R R

 
 

 
   
      
   
 
    
 
 
   
 
u u
Substituting this result into the Bernoulli equation gives the required expression for the
pressure at any location whose coordinates are ( , )R  :
4 2
2
0 4 2
1
2
( , ) 2 cos2
a a
p R p U
R R
  
 
   
 
Substituting R a in the foregoing expression produces the following value for the surface
pressure:
2
0
1
2
( , ) (1 2cos2 )Up a p    
Note that this result may be expressed as a non-dimensional pressure coefficient as follows:
0
21
2
( , )
2cos2 1p
p a p
C
U




  

4
TWO-DIMENSIONAL POTENTIAL FLOWS

Page 4-1
Problem 4.1
cosz F
cos( )
cos cos( ) sin sin( )
cos cosh sin sinh
i
i i
x i y i
Therefore cos coshx
And sin sinhy
Eliminating using 2 2
cos sin 1
2 2
2 2
1
cosh sinh
x y
This is the equation of ellipses whose major and minor semi-axes are coshx and sinhy .
1
2
1
( ) cos
1
d
W z z
dz z
For x = 2 and y = 0, z = 2 and we get
1 1
0 and
3 3 3
i
W u v i.e
downwards velocity.
_____________________________________________________________________________
Problem 4.2
coshz F
cosh( )
cosh cosh( ) sinh sinh( )
cosh cos sinh sin
i
i i
x i y i
Therefore cosh cosx
And sinh siny
Eliminating using 2 2
cosh sinh 1
2 2
2 2
1
cos sin
x y
This is the equation of a family of hyperbolas whose semi-axes are cosx and siny .
1
2
1
( ) cosh
1
d
W z z
dz z
For x = 1/2 and y = 0, z = 1/2 so that
1 2 2
0 and
3 / 4 3 3
i
W u v i.e
upwards velocity.
_____________________________________________________________________________

Page 4-2
Problem 4.3
2 2
2 2 2
2 2 2 2 2 2 1
2 2 2
1
2 2 2
( ) log( ) log( )
2 2
log( )
2
log ( ) 2
2
2
log ( ) 4 tan
2 2
2
( , ) tan
2
m m
F z z i h z i h
m
z h
m
x h y i x y
m m x y
x h y x y i
x h y
m x y
x y
x h y
( ,0) 0 streamlinex
We can evaluate the velocity components from the complex velocity, then substitute the results
into Bernoulli's equation to get the pressure.
2 2
2 2
2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2
2 2 2 2 2 2
2 2 2 2 2
2 2 2
( ) log( )
2
2
2
( ) 2 ( ) 2
( ) 4 ( ) 4
( ) 2
( , )
( ) 4
( ) 2
and ( , )
( )
dF d m
W z z h
dz dz
m z
z h
m x x h y x y m y x h y x y
i
x h y x y x h y x y
m x x h y x y
u x y
x h y x y
m y x h y x y
v x y
x h y 2 2 2
4 x y
Hence 2 2
( ,0) and ( ,0) 0
m x
u x v x
x h
. Then from Bernoulli's equation:
2 2
0
1
2
( ,0) ( ,0) ( ,0)p x p u x v x
2 2
0 2 2 2 2
( ,0)
2 ( )
m x
p x p
x h
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 4-3
0
0
2 2
2 2 2 20
2 ( ,0)
( )
F p p x dx
m x dx
x h
2
i.e. (upwards )
4
m
F
h
Problem 4.4
( ) log( ) log( )
2 2 2
log
2 2
1 /
log ...where 1
2 1 / 2
1 /
log( ) log
2 2 1 / 2
log{(1 / )(1 / )}
2
log(1 2 / )
2
log(2 / )
2
i
i
m m m
F z z b z b i
m z b m
i
z b
m z b m
i e
z b
m m z b m
e i
z b
m
z b z b
m
z b
m
z b
m
z
b
In the expansions above it has been assumed that / 1z b . Now let andb m in such
a way that /m b 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 4-4
Problem 4.5
2 2
2
2
2
2
2
( ) log( ) log( / ) log( ) log( / )
2 2 2 2 2
( )( / )
log
2 ( )( / ) 2
(1 / )(1 / )
log ...where 1
2 (1 / )(1 / ) 2
(1 / )(1
log( ) log
2 2
i
i
m m m m m
F z z b z a b z b z a b i
m z b z a b m
i
z b z a b
m z b a b z m
i e
z b a b z
m m z b a
e 2
2 2
2
2
/ )
(1 / )(1 / ) 2
log{(1 / )(1 / )(1 / )(1 / }
2
log(1 2 / 2 / )
2
log(2 / 2 / )
2
b z m
i
z b a b z
m
z b a b z z b a b z
m
z b a b z
m
z b a b z
In the foregoing expansions, it has been assumed that / 1z b . Now let andb m in
such a way that /m b U . This yields the result:
2
( )
a
F z U z
z
This is the complex potential for a uniform flow U past a circular cylinder of radius a .
Problem 4.6
2 2
2
2
2
2
2
( ) log( ) log( / ) log( / ) log( ) log
2 2 2 2 2
( )( / )
log
2 ( )( / )
( / 1)( / )
log
2 ( )( / )
log as
2 ( )( / )
m m m m m
F z z b z a b z a z b
m z b z a b
b z z a
m z b z a b
z z a
m z
b
z z a

Page 4-5
2 2
2 2
1
i.e. ( ) log
2 ( / ) / )
log[ ( / ) / )]
2
m
F z
z a a z
m
z a a z
On the surface of the circle of radius a this result becomes:
2
2 2
( ) log[ ( / ) )]
2
log[2 cos ( / )] log{ [( / ) 2 cos ]
2 2
i i i
i
m
F ae ae a ae
m m
a a e a a
2
log{( / ) 2 cos }
2 2
m m
a a i
For a , 2
( / ) 2a a so that 2
/ 2 cosa a and so / 2m . Therefore the
circle of radius a is a streamline. The required complex potential is given by:
2
( ) log
2 ( )( / )
m z
F z
z z a
The resulting flow field is illustrated below.
To determine the force acting on the cylinder, we calculate the complex velocity:

Page 4-6
2
1 1 1
( )
2 ( ) ( / )
m
W z
z z z a
2
2
2 2 2 2 2 2
1 1 1 2 2 2
( )
2 ( ) ( / ) ( ) ( / ) ( )( / )
m
W z
z z z a z z z z a z z a
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:
2
2
2
2 2 2
2 2
2
( )
2
2 (residues of ( ) inside )
2
2 2 2 2
2 / / ( / )
2 /
4 ( / )
C
X iY i W z dz
i i W z C
m
a a a
m a
a
From this result, there is no force acting on the y-direction, and the force that acts in the x-
direction is:
2 2
2 2
2 ( )
m a
X
a
Problem 4.7
2
( ) ( )
2
2
2
2
( ) log log
2 2
log log
2 2 2
( )( / )
log since 1
2 ( )( / ) 2
(1 / )(1 / )
log
2 (1 /
i i
i i
i i
i
i i
i i
m m a
F z z be z e
b
m a m m
z e z be i
b
m z be z a be m
i e
z be z a be
m z be a bz e
z be 2
)(1 / ) 2i i
m
i
a bz e

Page 4-7
2 2
2
2
i.e. ( ) log (1 / )(1 / )(1 / )(1 / )
2
log 1 (2 / 2 / )
2
2 / 2 /
2
i i i i
i i
i i
m
F z z be a bz e z be a bz e
m
z be a bz e
m
z be a b z e
In the foregoing expansions, it has been assumed that / 1z b and 2
/ 1a bz . Now let
andb m in such a way that /m b U . Thus the following result is obtained:
2
( ) i ia
F z U z e e
z
Problem 4.8
2 2
2
2
2
2
2
( ) log( ) log( ) log( ) log( ) log
2 2 2 2 2
( )( / )
log
2 ( )( / )
(1 / )( / )
log
2 ( )( / )
( / )
log for
2 ( )
(
log
2
i i a i a i i
F z z b z z z b
b
i z b z a
b z z a b
i z b z a
z z a b
i z a
b
z z
i 2
2
/ )
( )
( / )
log log
2 ( ) 2
a z
z
i a z i
z
( )
( ) log log
2 ( ) 2
i
i
i
i ae i
F ae
ae
The last result gives the value of the complex potential on the surface of the cylinder i
z ae .
We now note that the argument of the first logarithm on the right side of this equation is of the
form i
R e , where 1R . Therefore, the imaginary part of the entire first term is zero, so that:
( , ) log constant
2
a
This confirms that the circle of radius a is a streamline, so that the required complex potential is:

Page 4-8
2
( / )
( ) log
2 ( )
i z a
F z
z z
The complex velocity for this flow field is defined by the following equation:
2
2
2
2 2 2 2 2 2
1 1 1
( )
2 ( / ) ( )
1 1 1 2 2 2
( )
2 ( / ) ( ) ( / )( ) ( / ) ( )
i
W z
z a z z
W z
z a z 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:
2
2
2
2 2 2
2 2
2 2
( )
2 2
2 residues of ( ) inside
2
2 2 2 2
2 ( / ) ( / ) / ( )
2 ( )
C
i
X iY i W z dz
i i W z C
a l a a
a
a
Hence there is no force acting in the y-direction, and the force in the x-direction is:
2 2
2 2
2 ( )
a
X
a
Problem 4.9
2
( ) ( ) log
2
a i z
F z U z
z a
and
2
2
1
( )
2
a U i
W z U
z z

Page 4-9
2
2
2
2
2
2
2 2 4 2
2 2 2
2 4
2 2
3 2 2
2
2 2
2 2
( , )
2
and ( , )
2
( , ) ( ) ( )
2
( )
2 4
2
( , ) 2 2 cos2 sin
4
i i
i i
i i i i
i i
a U i
W R U e e
R R
a U i
W R U e e
R R
a U iU a U
W W R U e e e e
R R R
a U
i e e
R R
U
W W a U U
a a
Now use Bernoulli’s equation with the pressure far from the cylinder specified to be 0p . Then:
2
0
1
2
( , ) ( )p R p U W W
2
2 2
0 2 2
1
( , ) cos2 sin
2 8
U
p a p U U
a a
The upward force acting on the cylinder is defined by the following expression:
2
0
2
0
( , ) sin
sin sin
LF p a a d
U
a d
a
In the foregoing, it has been recognized that
2 2
0 0
sin sin cos 0m d m n d for all
values of and form n m . Hence the value of the lift force is:
LF U
Problem 4.10
(a) ( ) log log
2 2
m
F z z i z
log( ) log( )
2 2
i im
Re i Re
Hence
1
( , ) log
2
R m R
And
1
( , ) log
2
R m R

Page 4-10
(b)
1 1
( )
2 2
m
W z i
z z
Therefore ( )
2 2
i i
R
m
u iu e i e
R R
2
R
m
u
R
2
u
R
(c) From Bernoulli’s equation:
02 21
( )
2
R
pp
u u
Therefore
2 2
0
2 2 2 2
1
2 4 4
pp m
R R
2 2
0 2 2
1
( , ) ( )
8
p R p m
R
_____________________________________________________________________________
Problem 4.11
(a)
2 2
( ) 2
( ) ( )
a a
F z U z
z ih z ih
On y = 0:
2 2
( ) 2
( ) ( )
a a
F x U x
x ih x ih
2
2 2
2
2
( )
a x
U x
x h
Therefore
2
2 2
( ,0) 2 1
( )
a
x Ux
x h
Hence
2 2
22 2 2 2
2
( ,0) ( ,0) 2 1 2
a a x
u x x U Ux
x x h x h
2 2 2
22 2 2 2
2
( ,0) 2 1
a a x
u x U
x h x h
(b)
2 2 2 3
2 2 2 2 2 2 2 2 3
2 4 8
( ,0) 2
( ) ( ) ( )
u a x a x a x
x U
x x h x h x h

Page 4-11
2 2
2 2 2 2 2
4 4
3
( ) ( )
a Ux x
x h x h
Then the maximum/minimum occurs when x = 0 or when
2
2 2
4
3 0
( )
x
x h
; that is, when
2 2
3x h .
Hence
2
max 2
( ,0) 2 1
a
u x U
h
at x = 0
And
2
min 2
( ,0) 2 1
8
a
u x U
h
at 3x h
(c) From Bernoulli’s equation:
2 2( ,0) 1 1
( ,0) (2 )
2 2
opp x
u x U
So that 2 21
( ,0) 2 ( ,0)
2
op x p U u x
But from (b):
2
22 2 2 2 2 2
2 2 2
22 2 2 2 22 2
2 ( )
( ,0) 4 1 4 1
( )
a a x a h x
u x U U
x h x hx h
Therefore:
2
2 2 2
2 2
22 2
( )
( ,0) 2 2 1o
a h x
p x p U U
h x
_____________________________________________________________________________
Problem 4.12
1 1
2 2
1 1
2 2 3
(i) ( ) log( ) log( )
2 2
log( ) log( ) log
2 2 2
i i
i i
m m
F z z ae z ae
m m m
z ae z ae z
On the circle of radius , i
a z ae so that:

Page 4-12
1 1
2 2
1 1
2 2
1 1
2 2 3
1 2
2 32
( , ) log ( ) log ( )
2 2
log ( ) log ( ) log( )
2 2 2
log ( )
2
log ( ) log(
2 2
i ii i
i ii i i
i ii i i
i ii i i
m m
F a a e e a e e
m m m
a e e a e e ae
m
a e e e e e
m m
a e e e e e a
1 1 1
1
2 2 2 3 3
2
)
log 2(cos cos ) log
2 2
log 2(cos cos ) log log
2 2 2 2
i
e
m m m
a i
m m m m m
a i a i
From the imaginary part of the last equality we deduce the result:
1 2 3( , ) ( )
2
a m m m
Then the value of 3m that makes the circle of radius a a streamline is:
3 1 2m m m
(ii) Substituting the last result into the expression for ( )F z and differentiating with respect to z
gives:
1 1 2
1 2
2
1 1 1 1 1 1
( )
2 2i i i i
m m
W z
z z ae zz ae z ae z ae
On the surface of the circle of radius , i
a z ae so that:
1 1 2 2
1 2
( ) ( ) ( ) ( )
1 1 1 1
( , ) 1 1
2 21 1 1 1
i i
i i i i
m m
W a e e
a ae e e e
Using the cylindrical representation of the complex velocity gives the following expressions
for the velocity components on the surface of the cylinder:

Page 4-13
1 1
2 2
( ) ( )
1
1 1
( ) ( )
1
2 2
1 1
1
2 2 2cos( ) 2 2cos( )
1 1
1
2 2 2cos( ) 2 2cos( )
i i
R
i i
m e e
u iu
a
m e e
a
1 1 1
1 1
2 2 2
2 2
( , ) 0
sin ( ) sin ( )
and ( , )
4 1 cos( ) 1 cos( )
sin ( ) sin ( )
4 1 cos( ) 1 cos( )
Ru a
m
u a
a
m
a
Then, since 2 2
Rq u u , it follows that the magnitude of the velocity on the surface of
the cylinder is:
1 2
1 1 2 2
1 1 1 1
2 2 2 2
( , ) cot ( ) cot ( ) cot ( ) cot ( )
4 4
m m
q a
a a
(iii)We first use the condition ( , ) 0S
dq
a
d
. This gives the result:
1 2 2
1 1
2 2 2
2 2
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
cosec ( ) cosec ( )
4
cosec ( ) cosec ( ) 0
4
m
a
m
a
2 2
1 1
2 1
2 2
2 2
1 1
2 2
1 1
2 2
cosec ( ) cosec ( )
cosec ( ) cosec ( )
S S
S S
m m
The condition
2
1/ 2
2
( , )S e
d q
a cU R
d
gives the following equation:
1 1 2 2
1/21 2
3 3 3 3
1 1 2 2
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
cos ( ) cos ( ) cos ( ) cos ( )
8 8sin ( ) sin ( ) sin ( ) sin ( )
S S S S
e
S S S S
m m
cU R
a a
________________________________________________________________________

Page 4-14
Problem 4.13
1
( 1)
1
hence ... for critical points
n
n
n
n
n n
c
z
n
d z c
d
c
That is, the critical points are located at c times the nth root of +1 in the plane. Hence the
critical points are located at:
2 /
0,1,2, ...,( 1)i m n
ce where m n
+ic
-c +c +c -c +c
-ic
n = 2 n = 3 n = 4
From the equation of the mapping function it follows that:
( 1)
1
( 1)
n
i i n
n
c
x i y e e
n
Hence, equating real and imaginary parts on each side of this equation:
1
1
cos cos( 1)
( 1)
sin sin ( 1)
( 1)
n
n
n
n
c
x n
n
c
y n
n
Using the relations 2 2 2
and tan /R x y y x we get:
2 2 2 2
1
2
2 2
2 2( 1)
2
(cos sin ) cos cos( 1) sin sin( 1)
( 1)
cos ( 1) sin ( 1)
( 1)
n
n
n
n
c
R n n
n
c
n n
n

Page 4-15
1
1
sin sin ( 1)
( 1)
tan
cos cos( 1)
( 1)
n
n
n
n
c
n
n
c
n
n
Hence the equations defining the surface in the z plane are
1/22
2
2 1
1 cos
( 1) ( 1)
sin ( 1)
( 1)
sin
tan tan
cos( 1)
( 1)
cos
n n
n
n
c c
R n
n n
c n
n
c n
n
Expanding the results obtained above for 1 gives:
1
1 cos
( 1)
sin
tan 1 tan
( 1) sin cos
n
n
R n
n
n
n
The object shape for 0.7, 3n is shown below.
y / ρ
x / ρ

Page 4-16
Problem 4.14
From the mapping function the parametric equations of the mapping are:
2
2
cos
sin
c
x
c
y
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
0 to 360 . The
corresponding values of andx y are then calculated from the parametric equations above. The
following diagrams were drawn using an EXCEL program to calculate the values of andx 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 4-17
1 0
2 2
3
air
2
7.5
tan 6.582
60.0 5.0
12.448 (60.0 5.0) 7.5 0.8145
1.225 /
4 sin 293.3 /
1.225 250 293.3 1.0L
a mm m
kg m
U a m s
F U L N
4
8.982 10LF N
(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.
1 0
2 2
3
air
2
0.3
tan 6.582
2.4 0.2
11.18 (2.4 0.2) 0.3 2.438
0.002378 /
4 sin 2,634.3 /
(0.002378 32.2) 750 2,634.3 3.0L f
a ins ft
slug ft
U a ft s
F U L lb
5
4.538 10L fF lb
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
1
( 0) n
n
dz
K
d
z n K
The constant of integration has been taken to be zero since 0z 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 1nK . Then:
n
z

Page 4-18
Hence for a uniform flow of magnitude U in the -plane:
( ) n
F z U z
Problem 4.17
1/2
1/2 1 1/2
1/2
2 1/2
2 1/2 2 1/2
1 1
( 1)
( 1) ( 0) ( 1)
( 1)
1
( 1)
1 1
( 1) ( 1)
cosh sec
dz
K
d
K
K
z K A
When 0, 1z so that the constant A = 0. Also, when (1 ) , 1z i so that the
constant /K . Thus the equation of the mapping is:
1 1
cosh secz
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:
( ) log
U
F
Problem 4.18
1 / /
( 0) ( 1) ( )
1 1
dz
K
d
K
Hence the differential equation of the mapping function is:

Page 4-19
21 1
r
ndz
K
d
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:
( ) log
U H
F z
2
( )
1
r
n
dF dF d
W z
dz d dz
U H
K
2
i.e. ( )
1
r
nU H
W z
K
As the point D is approached, z while and ( )W z U . Hence from the foregoing
equation:
( ) 1
U H
W z U
K
H
K
As the point A is approached, z while 0 and ( ) /W z U H h. Hence:
2 2
( )
r r
n nU H U H
W z U
h K
2n
rH
h

Page 4-20
Problem 4.19
1 1/2 1/2
1/2
( 0) ( 1) ( )
1 1
dz
K
d
K
let 2
1
s
then
2
2
1
s
s
so that 2
1
1
1 s
and
2
2
( 1)
1
s
s
2
2
(1 )
( )
dz s
K
d s s
but 2 2
2 ( 1)
(1 )
d s
ds s
2 2
2 2
2( 1)
(1 )( )
1 1
2
(1 ) ( )
1 1 1 1
2
2(1 ) 2(1 ) 2 ( ) 2 ( )
dz dz d
K
ds d ds s s
K
s s
K
s s s s
1 1
log(1 ) log(1 ) log( ) log( )
1 1
log log
1
z K s s s s A
s s
K A
s s
At the location ,C so that 0s . Also, 0z , so that 0A . At the location , 1B so
that s . Also, ( )z i H h so that:
1
( )
( 1)
K H h

Page 4-21
1 1 1
( ) log log
1( 1)
s s
z H h
s s
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 2Uh , which will also be the strength of
the source located at 0. Then:
(2 )
( ) log
2
1
( )
U h
F
U h
W
1/2
( 1)
( ) ( )
( ) 1
d h
W z W U
dz H h
Then using the conditions discussed above at the location 0 produces the following result:
2
H
h
Thus the equation of the mapping becomes:
2
2
1 /
log log
1 /
( / )
where
1
H s h H h s
z
s H H h s
H h
s

5
THREE-DIMENSIONAL POTENTIAL FLOWS

Page 5-1
Problem 5.1
From Appendix A, the vorticity vector in spherical coordinates is defined by:
sin
sin
r
r
r r
r
u r u r u
 
 

 

  
 
  
e e e
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:
2
2
2
( , ) sin ( ) sin
1 1
sin
sin
ru
r r r u r
r
r
r r
   

 

  

 
 
    
    
    
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
2
2
1
sin 0
sin
r
r
 

  
   
  
   
Problem 5.2
Equation (5.18) is:
2
2
2
1
sin 0
sin
r
r
 

  
   
  
   
From Eq. (5.5b) the stream function for a uniform flow is:
2 21
2
( , ) sinr Ur  
Then
2
2 2 2
2
sinr U r
r





And 2 21
sin sin
sin
U r

 
  
  
  
  
Hence 2 21
2
( , ) sinr Ur   is a solution.

Page 5-2
From Eq. (5.6b) the stream function for a source is:
( , ) (1 cos )
4
Q
r  

  
Then
2
2
2
0r
r



And
1
sin 0
sin


  
  
 
  
Hence ( , ) (1 cos )
4
Q
r  

   is a solution.
From Eq. (5.7b) the stream function for a doublet is:
2
( , ) sin
4
r
r

  

 
Then
2
2 2
2
sin
2
r
r r
 



 

And 21
sin sin
sin 2 r
 
 
   
  
 
  
Hence 2
( , ) sin
4
r
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:
2 2
2
sin 1
0
sin
r d R d dT
R dr T d 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 andr  is for each term to be constant. For
convenience, we choose the value of the constant for the term involving R to be ( 1)n n  . Then
the ordinary differential equation for R will be:
2
2
2
( 1) 0
d R
r n n
dr
  
This is an equi-dimensional ordinary differential equation, so that the solution will be of the
following form:
( ) m
R r Ar
Then it follows that ( 1) ( 1) 0m m n n   
The two possible solutions to this algebraic equation are:

Page 5-3
or ( 1)m n m n   
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:
( ) n
n nR r A r 

The differential equation that is to be satisfied by T then becomes:
1
sin ( 1) 0
sin
d dT
n n T
d d

  
 
   
 
Let cos  be the new independent variable. Then the differential equation becomes:
2
2
2
(1 ) ( 1) 0
d T
n n T
d


   
Next introduce a new dependent variable that is defined by the relation 2 1/2
(1 )T    . From
this definition, the following derivatives are obtained:
2 1/2 2 1/2
2 2
2 1/2 2 1/2 2 2 3/2 2 1/2
2 2
(1 ) (1 )
(1 ) 2 (1 ) (1 ) (1 )
dT d
d d
d T d d
d d d

   
 
 
       
  

  
   
       
Thus, in terms of the new dependent variable  , the differential equation becomes:
2
2
2 2
1
(1 ) 2 ( 1) 0
1
d d
n n
d d
 
  
  
 
       
This is a special form of the Associated Legendre equation. The general form of the Associated
Legendre equation is:
2 2
2
2 2
(1 ) 2 ( 1) 0
1
d d m
n n
d d
 
  
  
 
       
The solution of this last equation, for any values of andm n, is:
( ) ( ) ( )
m m
mn mn n mn nB P C Q    
2 /2
2 /2
( )
where ( ) (1 )
( )
and ( ) (1 )
m
m nm
n m
m
m nm
n m
d P
P
d
d Q
Q
d

 


 

 
 

Page 5-4
In the foregoing expressions, and
m m
n nP Q are, respectively, the Associated Legendre functions
of the first and second kind of order m , while andn nP Q 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, 1m  . 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
m
nC 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:
2 1/2
( )
( ) (1 )
ndP
d

  

 
Combining the elements of the solution we have, for any value of n :
2 1/2
( , ) ( )(1 ) ( )
where cos
n n nr A R r    
 
 

Substituting the foregoing results for ( ) and ( )n nR r   into the expression for the stream
function yields the result:
2
2
( )(1 )
( , )
sin
(cos )
n
n n
n nn
dP
r A
r d
d
A P
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:
1
sin
( , ) (cos )n n nn
n
d
r A P
r d

  



   
Problem 5.4
Substituting 0nA  for 1n  in the result obtained in the foregoing problem gives:
1 1
sin
( , ) (cos )
d
r A P
r d

  

   
But 1(cos ) cosP   so that the first term in the general solution gives the following result:
2
1
sin
( , )r A
r

   
This is the same as Eq. (5.7b) in which 1 / 4A   .

Page 5-5
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:
2 2
*
( , ) sin sin
4 4
r
 
   
  
  
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:
sin ( ) sin
sin ( ) sin
a
a

  

  




Using these two results, the equation for the stream function on the circle of radius a becomes:
2 3
3 3
*sin
( , )
4
a
a
  
  
  
 
   
 
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 2 2
4 3
2 2
2
2 cos
2 cos
a l al
a a
a
l l
 
 
  
  
Eliminating cos between these two equations gives the following results:
3
*
a
l
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 :
3
2 2
3
( , ) sin sin
4 4
a
r
l
 
   
   
  

Page 5-6
Using Eq. (5.7a) for the velocity potential yields the following result for the two doublets:
2 2
*
( , ) cos cos
4 4
r
 
   
  
 
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 :
3
2 3 2
( , ) cos cos
4 4
a
r
l
 
   
   
 
Problem 5.6
From Eq. (5.14c) the force due to a doublet of strength  is given by:
(5.6.1)
i
x
 

 

u
F
where iu 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 2
/x a l , will be:
3
3 2
( , ) cos
4
a
r
l

  
 
 
But on the x -axis, for 2
/x a l , we have the following values:
2
0 ( / )and x a l   
Then along the x -axis, the appropriate velocity potential is:
3
3 2 2
3
3 2 3
3
3 2 4
3
2 2 4
( ,0)
4 ( / )
( ,0) ( ,0)
2 ( / )
3
and ( ,0)
2 ( / )
3
( ,0)
2 ( )
i x x
i
x
i
x
a
x
l x a l
a
x x
x l x a l
a
x
x l x a l
a l
l
x l a



 





 


 
 

 
 

 
 
u e e
u
e
u
e


Substituting this last result into Eq. (5.6.1) gives the following value for the force:
3
2 2 4
3
2 ( )
x
a l
l a
 



F e

Page 5-7
Problem 5.7
Integrating the partial differential equation for ( , )r t once with respect to r gives:
2
( )r 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:
2
( )f t R R
Thus the radial velocity in the fluid at any distance r from the sphere at any time t will be:
2
2
( , )
R R
r t
r r



Integrating the foregoing equation with respect to r yields the result:
2
( , ) ( )
R 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:
2
( , )
R R
r t
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:
2
1
2
( )
p
F t
t 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 0p there, the value of the quantity ( )F t is 0( ) /F t p  .
Then, using the results obtained above, the Bernoulli equation becomes:
22 2 2
0
2
1
2
( 2 ) ( , ) pR R R R p r t R R
r r 
 
    
 
2 2 4 2
0
4
( , ) 2
2
pp r t R R RR R R
r r r 
   
Then on the surface of the sphere, where r R , this expression becomes:
0 23
2
( , ) pp r t
RR R
 
  

Page 5-8
At 0t  the foregoing expression becomes:
2 3/2
1/2
3
2
1
0 ( )
d
RR R R R
R dt
  
Integrating this equation once with respect to r produces the following result:
3/2 3/2
0 0constantR R R R  
3/2
0 0
3/2
3/2
3/2
0 0
hence
1
and
R RdR
dt R
dt R dR
R R
 
  
Evaluating the foregoing integrals between the limits of 0 and t for time, and 0 and 0R for the
radius R , we get the result:
0
0
2
5
R
t
R

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:
3
2
1
2
( , ) cos (5.8.1)
U a
r
r
   
In the foregoing equation, , andU r  are all considered to be functions of time. In order to
switch to coordinates andx R, where both of these coordinates do not vary with time, we use
the relations:
2 2 1/2
0 0( ) cos and [ ( ) ]x x r r R x x    
Substituting these expressions into Eq. (5.8.1) we get the following result:
03
2 2 3/2
0
1
2
( )
( , ) (5.8.2)
[ ( ) ]
x x
x R U a
R x x


 
 
In the foregoing, 0andU x 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 5-9
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:
1
2
( , ) pp r
t
 
 

   

u u
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 0andU x are functions of time, and that
0
and
xU
U U
t t

 
 
Then from Eq. (5.8.2) the following expression is obtained for the temporal derivative of the
velocity potential:
3 2 3
2
2 3
1 1
2 2
( , ) cos (1 3cos )
U a U a
r
t r r

  

   

The velocity-squared term in Bernoulli’s equation is readily obtained from Eq. (5.8.1), producing
the following result:
2 2
2 6
2 2
6
1
4
1
cos sin
r r
U a
r
 

 
    
     
    
 
  
 
u u
Substituting these two results into the Bernoulli equation produces the following expression for
the pressure at any point in the fluid:
3 2 3 2 6
2 2 2
2 3 6
1 1 1 1
2 2 2 4
( , ) cos (1 3cos ) (cos sin )
U a U a U a
p r p
r r r
  
         
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:
2
0
2 ( , )sin cosF a p a d

     
Using the result obtained above for the pressure, its value on the surface will be:
2 2 21 1 1
2 2 4
( , ) cos (1 4cos sin )p a p Ua U         
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:
3 2
0
sin cosF a U d

     
Evaluating this integral shows that the force acting on the sphere will be:
32
3
F a U 
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 5-10
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.
_____________________________________________________________________________

SURFACE WAVES
Page 6-1
Problem 6.1
From Eq. (6.5a)
22
2
2 2
1 tanh
2
c h h
gh h gh
   
   
     
     
     
(6.1.1)
Let
2
2
2
, and
2
c
V S
gh h gh
 

 
  
Then Eq. (6.1.1) becomes: 2
2
1
1 tanh
S
V 
 
   
    
   
(6.1.2)
Then for 1 : 2
2
1
S S
V  
 
 
    
 
Hence for a minimum value of V:
2
2 1 0
dV S
V
d 
  
S  (6.1.3)
That is:
2
2 h gh
 
 

Substituting the results from Eq. (6.1.3) into Eq. (6.1.2) gives:
 2 1
1 1 tanhV S
S
 
   
 
That is
2 2
2
2 tanh
c gh
gh gh
 
 

_____________________________________________________________________________
Problem 6.2
Using a frame of reference that is moving with velocity c in the positive x -direction, the
quantity z ct 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:
2
( , ) cos ( )
sinh (2 / )
c
F z t c z z i h
h
 
  
   
We expand the ratio of the two transcendental functions as follows:

SURFACE WAVES
Page 6-2
2
2 2 2 2 2
cos ( ) cos cosh sin sinh
sinh (2 / ) sinh (2 / )
2 2 2
cos coth sin
2 2
cos sin for 2 / 1
z
i
h
z h z h
z i h i
h h
z h z
i
z z
i h
e

    
    
   
  
  
 
 
 

 

 
  

Hence the deep-liquid version of the foregoing result is:
2
( , )
z
i
F z t c z c e




  
2
( )
2
( , ) ( )
2 2
( ) cos sin
i x i y
y
F z t i c x i y c e
x x
c x i y c e i




  
 

 
 
     
 
     
 
Hence the stream function is defined by the following expression:
2
2
( , ) sin
y
x
x y c y c e



 

  
When , 0y    . Hence from the foregoing expression:
2
2
sin
y
x
e



 


Problem 6.3
2 /
( , ) i z
F z t c z c e  
 
  
2 /
2 /
2
( ) 1
2
and ( ) 1
i z
i z
dF
W z c ic e
dz
W z c ic e
 
 






 
     
 

   
 
2
2 2 / 2 / 2 2 ( )/
2
2 2 4 /
2 2 2
hence ( ) 1
2 2
1 2 sin
i z i z i z z
y
WW z c i e i e e
c e
     
 
  
  
  
 
 
 
 
 
     
   
 
    
   

SURFACE WAVES
Page 6-3
Substituting this result into Bernoulli’s equation produces the following result:
2 2
2 2 / 2 4 / 2 21 1
2 2
4 2 2 2
1 sin 1y yp x P
c e e g y c      
  
     
     
            
        
Using this result on the free surface where andy p P  , and using Eq. (6.18a), gives:
2 2
2 2 4 / 2 21 1
2 2
4 2 2
1 1c e g c   
    
  
     
          
        
Solving this equation for 2
c gives:
2
2 2 2 2 4 /
2
(2 / ) (4 / ) (2 / )
g
c
e  

        

 
Expanding the foregoing result and noting that / /    gives:
2
4 /
2 2
2
4
1
/ 2
4
1 1
/ 2
1 (2 / )
g
c
e
g
g
 

    
 
    
 
    
 
    
 
  

 
  
 


     
 

  
For / 0   the foregoing result becomes:
2
2
g
c



This result confirms Eq. (6.3b). Also, for / 1  the same equation shows that:
2
2 2
/(2 )
1 (2 / )
g
c
 
  


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:

SURFACE WAVES
Page 6-4
2
( ) cos ( )
sinh (2 / )
c
F z z ct i h
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 ct z  . The resulting complex
potential is:
2
( ) cos ( )
sinh (2 / )
2
( ) cos [ ( )]
sinh (2 / )
2 2 2 2
( ) cos cosh ( ) sin sinh ( )
sinh (2 / )
U
F z U z z i H
H
U
U x i y x i y H
H
U x x
U x i y y H i y H
H
 
  
 
  
    
     
   
     
 
        
Hence the stream function for the flow will be:
2 2
( , ) sin sinh ( )
sinh(2 / )
U x
x y U y y H
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:
2
2
tanh
2
U H
g H H
 
 

Problem 6.5
Put U h  when 0 sin(2 / )y h x     in the expression obtained in Prob. 6.4 for the
stream function. This produces the following equation:
0 0
2
0 sinh [( ) sin 2 / ]
sinh(2 / )
U
U H h x
H
 
   
  
    
Consider 0 ( )H h  and solve this equation for the amplitude ratio, to get:
 0
sinh (2 / )
sinh 2 ( )/
1
cosh (2 / ) coth (2 / ) sinh (2 / )
H
H h
h H h
  
  
     





SURFACE WAVES
Page 6-5
But, from Prob. 6.4, 2
coth (2 / ) 2 /H U g    . Hence the expression for the amplitude ratio
becomes:
2
0
1
cosh(2 / ) ( / 2 ) sinh(2 / )h g U 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:
2
2
tanh
2
U g h
g h 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
1 2
1
2
2 2
( , ) sin ( ) sin ( )x t x c t x c t
 
 
 
 
    
  
1 1
2 2
but sin sin 2cos ( ) sin ( )A B A B A B   
Hence the equation of the composite wave may be written in the following form:
1 2 1 2
1 2 1 2 1 2 1 2
1 1 1 1
( , ) cos sin
c c c c
x t x t x t   
       
          
                                    
If 1 2 1 2andc c    , 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 2
1 2 1 2
1 1
( , ) ( , )sin
c c
x t A x t x t 
   
    
               
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) 2
0
1
2
V g dx

  

SURFACE WAVES
Page 6-6
2 2
0
1
2
2
sin ( )g x ct dx
 
 

 
   

Hence: 21
4
V g  
(b) From Eq. (6.7a):
2 2 2 2
( , ) cos ( ) sinh coth cosh
y h y
x y c x ct
   
 
   
 
    
 
2 2
( ,0) coth cos ( )
h
x c x ct
 
 
 
  
Also
2 2
( ,0) cos ( )x c x ct
y
  

 

  

Hence:
2
2 2 2
0
1
2
2 2 2 2
coth cos ( ) where tanh
2
h g h
T c x ct dx c
     
 
    
  
21
4
T g  
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:
2
( , , ) ( , , )
2 2 2 2 2
sin ( ) sinh coth cosh
p x y t x y t
t
y h y
c x ct


    
 
    

 

 
   
 
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 2 2
( , , ) sin ( ) sinh coth cosh
y h y
x y t c x ct
x
     

    
  
   
  

SURFACE WAVES
Page 6-7
2 2
3 2 2
2
2
3 2 2
2
2
3 2 2
2
2 2 2 2 2
sin ( ) sinh coth cosh
2
cosh ( )
2 2
sin ( )
2
sinh
4
1 cosh ( )
1 2 2
sin ( )
22 sinh
y h y
p c x ct
x
y h
c x ct
h
y h
c x ct
h
     
 
    

   
 


   
 

    
     
    
 
  
    
   
 
 
   
    
   
 
Substituting this result into the expression for the work done gives:
0
2
3 2 2
2
4
sinh ( )
1 2 2 4
sin ( )
22 sinh
h
WD p dy
x
h y h
c x ct
h

 
   
 
 





 
   
    
   
  

Using Eq. (6.3a) to eliminate 2
c from the foregoing equation produces the result:
2 2
2 2 2
sinh cosh
1 2
sin ( )
2 22 sinh cosh
h h h
WD g c x ct
h h
  
    
 
 
 
 
   
 
 
2 21 2 2 /
sin ( ) 1
2 sinh (2 / )cosh (2 / )
h
WD g c x ct
h h
  
 
    
 
   
 
For deep liquids, the term inside the square brackets in the foregoing equation becomes unity.
Then the average work done will be:
/
ave
0
( )
c
WD WD dt

 
/
2 2
ave
0
2
1 2
i.e. ( ) sin ( )
2
1
4
c
WD g c x ct dt
g
 
 

  
 


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:
1
2
( ) ( )aveWD T V 

SURFACE WAVES
Page 6-8
Problem 6.9
Since the fluid motion is in the xy -plane only, the vorticity vector will lie in the z -plane and
will have the following magnitude:
2v u
x y
 
 
   
 
Hence from the given expression for the stream function:
2 2
( )k l  
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:
2 2
2 2
( )
( )( ) ( )
D
k l
Dt t x
k l i ik
 

  
  
  
 
  
2 2
( )
D
i k k l
Dt
  

     
From the foregoing result, the value of  that makes the material derivative of the total vorticity
equal to zero is:
2 2
( )
k
k l

 

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
( )
1 1
2 2
( )
2 2
2
0: ( , , ) cosh ( )
0 : ( , , )
i x t
i x t y
y x y t A e y
y x y t A e e



 

 

  



 
    
 

SURFACE WAVES
Page 6-9
Imposing the kinematic interface conditions given by Eqs. (6.16e) and (6.16f), with 1 0U  and
2U U , yields the following values for the constants:
1
2
2
sinh
( )
i
A
A i U
 
 

 
 
 
Using these values, the expressions for the two velocity potentials become:
2
( )
1
2 2
( )
2
2
cosh ( )
( , , )
2
sinh
( , , ) ( )
i x t
i x t y
y
x y t i e
x y t i U e e



 

 


  
 

  

 

 
 
Next, we impose the pressure condition at the interface between the two fluids. This condition is
defined by Eq. (6.16g), with 1 2 1 20, andU U U       . This results in the following
quadratic equation for  :
2 22
1 coth 2 0U U
 
 

 
    
 
Thus the coefficient of the time in the equation of the interface will be:
1/2 2
1 coth
2
1 coth
i
U
 

 




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:
2
( )
( , )
i x t
x t e



 


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
/ 0y   on both andy h y h   . Hence the velocity potentials will be of the following
form:

SURFACE WAVES
Page 6-10
2
( )
1 1
2
( )
2 2
2
( , , ) cosh ( )
2
( , , ) cosh ( )
i x ct
i x ct
x y t A e h y
x y t A e h y












 
 
Substituting these expressions into the kinematic conditions defined by Eqs. (6.16e) and (6.16f)
in which 1 2 0U U  , produces the following values for the constants:
1
2
2
sinh
2
sinh
i
A
h
i
A
h
 


 


 

Substituting these values into the expressions for the velocity potentials gives:
2
( )
1
2
( )
2
2
cosh ( )
( , , )
2
sinh
2
cosh ( )
( , , )
2
sinh
i x ct
i x ct
h y
x y t i e
h
h y
x y t i e
h





  



  





 


Next we impose the pressure condition at the interface, which is defined by Eq. (6.16g) with
1 2 0U U  . This gives, after dividing through by  , the following expression for  :
2 2
1 1 2 2
2 2 2 2
coth coth
h h
g g
   
     
   
   
Solving this equation for  gives the following result:
1 2
1 2
2
tanh
2
g h   

   
 
     
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  .

EXACT SOLUTIONS
Page 7-1
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
where
* 2 *
2
u u
t y
with *
( , ) 0u y
and * 1
1
2
( ,0) ( 1) sinn
n
y n y
u y
h 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 *
( ,0)u y has been made equal to the negative of the
Couette flow distribution so that the net value of the total velocity ( ,0)u y is zero. The Fourier
series for the saw-tooth value of *
( ,0)u y 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 0t , and it will be exponential in
time. Thus the solution becomes:
2
/* 1
1
2
( , ) ( 1) sinn h tn
n
n y
u y t e
n h
This expression satisfies the differential equation and the boundary conditions. Hence the
complete solution becomes:
2
/1
1
( , ) 2
( 1) sinn h tn
n
u y t y n y
e
U h n h
Problem 7.2
The partial differential equation and boundary conditions to be satisfied are:
2
2
1
1
for ( , )
2
( ,0) 1 sin
( , ) finite
n
n
u u
u y t
t y
y U n y
u y U
h n h
u y t

EXACT SOLUTIONS
Page 7-2
In the foregoing, the Fourier series representation for the saw-tooth velocity distribution that
exists at 0t 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 0t , and it will be exponential in
time. Also, we recall that the separation constants for both the andy 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:
2 2
21
1
2
( , ) 1 sin
n
tn h
n
U n y
u y t e
n h
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 u1, will depend on y only. Hence the governing equation becomes:
2
1
2
0 2
d u
P
dy
Then: 2
1( )u y Py Ay B
1( 1) 0u P A B and 1( 1) 0u P A B
0 andA B P
2
1( ) (1 )u y P y
Substituting 1 2( , ) ( ) ( , )u y t u y u y t into the governing equation for 0t gives:
2 2
1 2 1 2
2 2
2
u u u u
P
t t y y
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:
2
2 2
2
2
2where ( ,0) (1 )
u u
t y
u y P y

EXACT SOLUTIONS
Page 7-3
But the Fourier series for 2
2 2
1
2 4
(1 ) ( 1) cos
3
n
n
n
y n y
n
. Hence we look for a separation of
variables solution for 2 ( , )u y t of the form:
2 ( , ) ( ) cosu y t T t n y
Substituting this expression into the PDE for 2 ( , )u y t we get:
2 2dT
n T
dt
So that:
2 2
( ) n t
nT t C e
Therefore:
2 2
2
1
( , ) cos n t
n
n
u y t C n y e D
But 2
2 2 2
1
2 4
( ,0) (1 ) ( 1) cos
3
n
n
n
u y P y P n y
n
Therefore: 2 2
2 4
and ( 1)
3
n
nD P C P
n
Hence:
2 2
2
1
2 2
2
( , ) (1 ) cos
3
4
where ( 1)
n
n t
n
n
n
n
u y t P y P H n y e
H
n
_____________________________________________________________________________
Problem 7.4
Here: ( ); 0u u y v w
For this situation, continuity is identically satisfied. Hence, noting that the gravitational force is
defined by sin cos 0x y zg gf e e e it follows from the Navier-Stokes equations that the
equation to be solved for ( )u y is:
2
2
sin
d u g
dy
(a) (i) Integrating this equation twice gives
2
( ) sin
2
g y
u y Ay B
The conditions (0)u U and ( ) 0u h give and sin
2
gh U
B U A
h
Hence;
2
( ) 1 1 sin
2
y gh y y
u y U
h h h

EXACT SOLUTIONS
Page 7-4
(ii)
2
0 0
( ) 1 1 sin
2
h h y gh y y
Q u y dy U dy
h h h
3
sin
2 12
Uh gh
Q
(iii) 0Q 2
6
sin
U
gh
(b) (i) Here (0) and ( ) 0
du
u U h
dy
(for zero stress). Hence:
2
( ) 2 sin
2
gh y y
u y U
h h
(ii) Integrating:
3
sin
3
gh
Q Uh
(iii) 0Q 2
3
sin
U
gh
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, EQ , for the elliptic pipe:
2 2 2 2 2 2
1 /
2 2 2 20 0
2 2
2 2
1
4 1
2
2
8
b a z b
E
dp a b y z
Q dz dy
dx a b a b
dp a b ab
dx a b
Denoting the flow area by and / byA ab b a , this equation may be written in the
following form:
2
2
2
2
2 2 2
1
(7.5.1)
4 1
1 1 2
4 1 (1 )
E
E
dp
Q A
dx
Q dp
A
dx
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:
1
b
a

EXACT SOLUTIONS
Page 7-5
The velocity distribution for flow in a circular conduit is given by Eq. (7.2b). From this result
the volumetric flow rate CQ in a circular conduit will be:
2 2
0
4
2
1
( )2
4
2 4
1
(7.5.2)
8
a
C
dp
Q a R RdR
dx
dp a
dx
dp
A
dx
In the foregoing, the flow rate has been expressed as a function of the flow area 2
A a . From
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:
2
2
1
E
C
Q
Q
Hence for 4 / 3 the flow ratio becomes:
24
0.96
25
E
C
Q
Q
Problem 7.6
The partial differential equation that is to be satisfied in this case is the following:
2 2
2 2
0
dp u u
dx y z
Then, substituting the assumed form of solution for the velocity ( , )u y z gives:
3
6
dp
b dx
Since this value is a constant, the assumed form of the solution is valid and the complete solution
becomes:
3
( , ) 3 3
6 2 3 3 2 3
dp b b b
u y z z z y z y
b dx
Problem 7.7
(a) 2
1
1
2
( ) 1 4( )u y A y

EXACT SOLUTIONS
Page 7-6
2
1 1
2
1
2
8 ( ) and 8
du d u
A y A
dy dy
Hence the differential equation is satisfied provided A = P so that a solution is:
2
1
1
2
( ) 1 4( )u y P y
(b) 2
1
sinh
( , ) sin
sinh
n
n
n z
u y z B n y
n
2 2
2 22 2 2 2
2 22 2
and
u u
n u n u
y z
Hence another solution is: 2
1
sinh
( , ) sin
sinh
n
n
n z
u y z B n y
n
(c) 3
1
sinh ( )
( , ) sin
sinh
n
n
n z
u y z C n y
n
2 2
3 32 2 2 2
3 32 2
and
u u
n u n u
y z
Hence another solution is: 3
1
sinh ( )
( , ) sin
sinh
n
n
n z
u y z C n y
n
(d) 2
1 1
1
2
sinh sinh ( )
( , ) 1 4( ) sin sin
sinh sinh
n n
n n
n z n z
u y z P y B n y C n y
n n
This expression satisfies the correct boundary conditions at 1y . At 0 andz z we have:
z = 0: 2
1
1
2
0 1 4( ) sinn
n
P y C n y
z = α: 2
1
1
2
0 1 4( ) sinn
n
P y B n y
The Fourier series for the bracketed term { } is the following:
21
2
1 4( )y = 3 3
1
16
sin where 1 ( 1)n
n n
n
K n x K
n
Hence matching the boundary expressions requires the following values for the constants Cn and Bn:
andn n n nC PK B PK
2
1 1
1
2
sinh sinh ( )
( , ) 1 4( ) sin sin
sinh sinh
n n
n n
n z n z
u y z P y P K n y P K n y
n n
_____________________________________________________________________________

EXACT SOLUTIONS
Page 7-7
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
2 2
0
u u
y z
with ( , ) 0u y
and
1
2
( ,0) [1 ( 1) ]sinn
n
n y
u y U U
n b
In the foregoing, the boundary condition on 0z 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:
1
2
( , ) [1 ( 1) ] sin
n z
n b
n
n y
u y t U e
n b
The volumetric flow rate Q may be evaluated by integrating the velocity to give:
0 0
1
2
2 20
1
2
1 ( 1) sin
2
1 ( 1)
n z
b
n b
n
n z
n b
n
n y
Q dz U e dy
n b
b
U e dz
n
Thus the volumetric flow rate past any vertical plane across the flow will be:
22
3 3
1
2
1 ( 1)n
n
Q U b
n
Problem 7.9
(a)
1 1d dw dp
R
R dR dR dz
Fluid #1:
2
1 1 1
1
1
( ) log
4
dp R
w R A R B
dz
Fluid #2:
2
2 2 2
2
1
( ) log
4
dp R
w R A R B
dz
The fact that 1( )w R should be finite when 0R requires that the constant 1 0A . Also,
matching the shear stresses at 1R R produces the following result:

EXACT SOLUTIONS
Page 7-8
1 2 1 1 2
1 2 2
1
0
2 2
dw dw R R Adp dp
A
dR dR dz dz R
Hence the expressions for the velocity profiles become:
2
1 1
1
1
( )
4
dp R
w R B
dz
2
2 2
2
1
( )
4
dp R
w R B
dz
The boundary conditions 1 1 2 1( ) ( )w R w R and 2 2( ) 0w R then give the results:
2 2
1 1
1 2
1 2
1 1
4 4
R Rdp dp
B B
dz dz
2
2
2
2
1
0
4
Rdp
B
dz
Therefore:
2
2
2
2
1
4
Rdp
B
dz
And: 2 2 2
1 2 1 1
2 1
1 1
( )
4
dp dp
B R R R
dz dz
Hence the velocity components are:
22 2 2 2
1 2 1 1
2 1
2 2
2 2
2
1
( ) ( ) ( )
4
1
( ) ( )
4
dp
w R R R R R
dz
dp
w R R R
dz
(b) From above:
22 2 2
1 2 1 1
2 1
1
(0) ( )
4
dp
w R R R
dz
22 2
1 2 1
2 1
1
(0) 1
4
dp
w R R
dz
__________________________________________________________________________
Problem 7.10
In the present case, 00 and 0i . Then, from Eq. (7.3a):
2 2
0 0
2 2
0
( )
i
i
R R
u R R
RR R

EXACT SOLUTIONS
Page 7-9
2 2
0
0 2 2 2
0
( )
1
2
( )
i
i
u
R R
R R
R R
R R R
In the foregoing, the shear stress has been evaluated using the formulae in Appendix C. Then,
if 0T is the torque acting on the fluid at 0R R , we have the result:
2
0 0 02 ( )T R R
If iT is the torque acting on the fluid at iR R , then the required torques are:
2 2
0
0 0 2 2
0
4
( )
i
i
i
R R
T T
R R
Problem 7.11
The velocity distribution is defined by Eq. (7.3a):
2 2
02 2
0 0 02 2 2
0
1
( ) ( ) ( )
( )
i
i i i
i
R R
u R R R R
R R R
Now let the outer radius become infinitely large in this expression; that is, let 0R . Then:
2
0 0( ) ( )
i
i
R
u R R
R
Next, bring the fluid far from the origin to rest by setting 0 0. This gives the result:
2
( )
i iR
u R
R
The complex potential for a line vortex of the given strength is:
2
( ) logi iF z i R z
Thus the complex velocity will be represented by the following expression:
2
( ) ( )
i ii i i
R
R
W Re u iu e i e
R
Thus the velocity resulting from the line vortex will be described by the following equation:
2
( )
i iR
u R
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.

EXACT SOLUTIONS
Page 7-10
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
2
(0, ) cos
( , ) 0
u u
t y
u t U nt
u h t
Following the procedure employed in section 7.5, we look for a solution to this problem of the
following form:
( , ) Re ( ) int
u y t w y e
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 :
2
2
0
d w n
i w
dy
But
1
(1 )
2
i i
Hence the solution for ( )w y that satisfies the homogeneous boundary condition at y h is:
( ) sinh (1 ) ( )
2
n
w y A i h y
Then the solution for the velocity ( , )u y t will assume the following form:
( , ) Re sinh (1 ) ( )
2
i ntn
u y t A i h y e
Imposing the boundary condition at 0y on this solution indicates the following value for the
constant A :
sinh (1 )
2
U
A
n
i h
Thus the solution for the velocity ( , )u y t will be:
sinh (1 ) ( )
2
( , ) Re
sinh (1 )
2
int
n
i h y
u y t U e
n
i h
Although the real part of this equation may be evaluated explicitly, it will be left in the form
given above for brevity.

EXACT SOLUTIONS
Page 7-11
Problem 7.13
Using the definition of the Reynolds number NR , Eq. (7.6) becomes:
cosh (1 )
( , ) Re 1
cosh (1 )
N
x int
N
y
i R
P a
u y t i e
n i R
(a) For any value of NR , the following identity is valid:
cosh (1 ) cosh( )cos( ) sinh( )sin( )N N N N Ni R R R i R R
In particular, for 1NR , the following expansion is valid:
But cosh (1 ) (1 )(1 ) ( )( )
1
N N N
N
i R i R R
i R
Also
2
cosh (1 ) 1N N
y y
i R i R
a a
2
2
cosh (1 ) 1
Therefore
1cosh (1 )
1 1
N N
NN
N
y y
i R i R
a a
i Ri R
y
i R
a
Thus the expression for the velocity ( , )u y t becomes:
2
2
( , ) Re 1
1 cos
x int
N
x
N
P y
u y t i i R e
n a
P y
R nt
n a
That is, using the definition of the Reynolds number NR , we get the result:
2 2
2
( , ) 1 cos
2
xP a y
u y t nt
a
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 cosxP t .

EXACT SOLUTIONS
Page 7-12
(b) For andNR y a we get:
cosh (1 )
0
cosh (1 )
N
N
y
i R
a
i R
( , ) Re (1 )
x int
P
u y t i e
n
That is, the velocity distribution for large values of the parameter NR will be:
( , ) sin
xP
u y t nt
n
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:
where ( , ) ( )
2
and
R
t R R R
R t f
t
R
t
Then the following expressions are obtained for the various derivatives indicated:
1
2 2
1
2
1
2
f f
t t t t
f
R t t
R
R f f
R R t t t
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:
21
2
(1 ) 0f f f
Integrating this equation term by term, and using integration by parts where appropriate, gives:

EXACT SOLUTIONS
Page 7-13
21
2
( ) ( )f f d f f f d f d A
In the above, A is a constant of integration. Simplifying this equation gives the result:
21
2
f f A
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
4
( )f Be
2
4
( , )
2
R
t
R t Be
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):
2
2
4
0
0
2
2
2
2
or 4 1
where
4
A
R
t
dA
Be R dR
t
B e d
R
t
ω n
The value of the definite integral is a half, so that the constant B must have the value 1/ 2B .
Then the solution for the vorticity distribution will be:
2
4
( , )
4
R
t
R t e
t
From Appendix A, the only non-zero component of the vorticity vector is defined as follows:
2
2
4
4
1
( )
4
2
R
t
R
t
Ru e
R R t
Ru e C
The constant of integration C may be evaluated by observing that 0 asu t , since the
vortex will be fully decayed for large values of the time. This gives the value / 2C so that
the velocity distribution will be defined as follows:
2
4
( , ) 1
2
R
t
u R t e
R

EXACT SOLUTIONS
Page 7-14
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
1u p
R R
Hence the pressure distribution will be defined by the following integral:
2 2
2
4
0 2 3
1
( , ) 1
4
R
t
p R t p e dR
R
2 2
2
4 2
0 2 3 3 3
1 2 1
4
R R
t t
p e e dR
R R R
This result may be expressed in terms of the following exponential integral:
1
E( ) a x
ax e dx
x
2 2 3 3
log
1! 2 2! 3 3!
ax a x a x
x
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:
2
2
2 2
24 4
3 2
4 4
2
4
2
2
2 2
3 2
1 1
...where
2
1 1 1
2 8
1 1
E
2 8 4
1 1 1
and E
2 4 2
R
t t
t t
R
t
R R
t t
e dR e d R
R
e e d
t
R
e
R t t
R
e dR e
R R t t
Substituting these values into the expression for the pressure obtained earlier, produces the
following result:
2 2
2 2 2 2
2 4
0 2 2
( , ) 1 2 E E
8 2 2 4
R R
t t R R R
p R t p e e
R t t t
_____________________________________________________________________________
Problem 7.15
For the given velocity components, the left side of the continuity equation becomes:
1
( ) 2 2 0
z
R
u
Ru a a
R R z

EXACT SOLUTIONS
Page 7-15
That is, the continuity equation is satisfied for all values of the parameters anda K . For the
given velocity components, the R -component of the Navier-Stokes equations is:
2
2
2
2
3
2
2
3
1 1
1
i.e.
(7.15.1)
R R R
R
u u u up
u R
R R R R R R R
K p a a
a R
R R R R
p K
a R
R R
Similarly, the -component of the Navier-Stokes equations gives the result:
2
3 3
1 1
1
i.e.
0 (7.15.2)
R
R
u u u u up
u R
R R R R R R R
a K a K p K K
R R R R R
p
Finally, the z -component of the Navier-Stokes equations shows that:
2
2
2
2
1
1
i.e. 4 0
4 (7.15.3)
z z
z
u up
u
R z z
p
a z
z
p
a z
z
Eq. (7.15.2) shows that the pressure p is independent of . Then Eqs. (7.15.1) and (7.15.3)
give, respectively:
2 2 2
12
2 2
2
( , ) ( )
2 2
( , ) (2 ) ( )
K a R
p R z g z
R
p R z a z g R
Comparing these two expressions shows that the pressure distribution is defined by the following
equation:
2
2 2 2 2
0 2
( , ) 4
2
K
p R z p a R a z
R
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:

EXACT SOLUTIONS
Page 7-16
2
2 2
3
2
( )
4
p K
f a R
R R
p K
R f f a K R f
R
p
a z
z
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 :
2
1 1
0
1
i.e. 0
a
f f f
R R
d a
f f
dR R
Integrating the last equation and noting that 1 and 0 asf f R gives:
2
2
2
2
1
hence 1
and u 1
a R
a R
a a
f f
R
f C e
K
C e
R
Since the velocity u must be finite as 0R , we must choose the value of the constant in this
equation to be 1C . This gives the following expression for the function f :
2
2
1
a R
f e

8
LOW-REYNOLDS-NUMBER
SOLUTIONS

LOW-REYNOLDS-NUMBER SOLUTIONS
Page 8-1
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 cosx r  , we get the following value for
the pressure on the surface of the sphere:
3
( , ) cos
2
U
p a
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:
2
2 sindS a d  
The force in the positive x -direction due to the pressure distribution around the sphere will have
the following value:
0
2
0
( , ) cos
3 sin cos
pF p a d
Ua d


  
    
 



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:
2pF Ua 
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:
( , ) ( , ) 0 (8.2.1 )
( , ) ( , ) (8.2.1 )
( , ) ( , ) (8.2.1 )
r r
r r
u a u a a
u a u a b
a a c
 
 
 
 
   
 


where ( , )
r
r
u u
r r
r r r



  

 
  
  
In the foregoing equation, the variables associated with the inner flow are indicated with a
primed superscript.
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 8-2
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:
3 4 2
3 4 2
3
1 1
3
1 1 3
2
x x r x r
x r
x x
U A c
r r r r
x x
U A c A c
r r r r
x
p c
r

   
       
   
   
        
   

u e e e e e
e e
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:
cos sinx r   e e e
Then the velocity vector for the outer flow may be written as flows:
3 3
2 2 1 1
cos sinrU A c U A c
r r r r
 
   
        
   
u e e
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:
3
2
2
a
A U c
a
 
  
 
Then the outer flow will be described by the following equations:
3 2
3 2
3 2
3 2
2
2
1 cos 1 cos (8.2.2 )
1
1 sin 1 sin (8.2.2 )
2
1
2 cos (8.2.2 )
r
a a
u U c a
r r r
a a
u U c b
r r r
p c c
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:
2
2
( ) ( )
1
p

    
    
u u u
u
   
   
Applying this result to the inner flow for which p KU x  , we get:
cos sinx rKU KU KU       e e e 
Thus the magnitude of the vorticity must be given by the following expression:

Page 8-3
1
2
sinKUr    e
This makes
1
2
0 and sinr KUr          . Then:
1 1
( sin ) ( )
sin
cos sin ...as required.
r
r
r
r r r
KU KU
  

  
 
 
 
  
 
  
e e
e e
 
Then, from the definition of the vorticity in terms of the velocity components, this result requires
that the following equation be satisfied:
1
2
1 1
( ) sin
ru
ru KUr
r r r
 

    
 
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
2 2
2 2
1
2
sin 1
sin
sin
KU r
r r
  

  
    
  
   
A separable solution to this equation of the following form is sought:
2
( , ) ( ) sinr U R r   
Substitution of this assumed form of solution leads to the following ordinary differential
equation that is to be satisfied by ( )R r :
2
2
2 2
1
2
2
d R R
K r
dr r
 
The particular solution to this differential equation is 4
/ 20Kr and the complementary solution
will be of the form n
r since the homogeneous equation is equi-dimensional. Substitution of
n
R r reveals that 2 or 1n n   , so that the complete solution is:
4 2
4 2 2
2
3
2
3
1
( )
20
1
( , ) sin
20
1 2
hence ( , ) 2 cos
10
1
and ( , ) 2 sin
5
r
N
R r K r M r
r
N
r U K r M r
r
N
u r U K r M
r
N
u r U K r M
r

  
 
 
  
     
 
     
 
      
 

In the foregoing, , andK M N are constants of integration. Since the velocity components must
both be finite at 0r  , it follows that we must choose 0N  . Also, the radial component of
velocity must be zero on r a . This requires that 2
/ 20M K a  . Thus the velocity and
pressure in the inner flow will be defined as follows:

Page 8-4
2 2
2 2
1
( , ) ( ) cos (8.2.3 )
10
1
( , ) ( 2 ) sin (8.2.3 )
10
( , ) cos (8.2.3 )
ru r U K a r a
u r U K a r b
p r KU r c

 
 
  
   
  
 
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:
2
2
1 2 3
10 2
3 6
3
10
U a K c U
a
U a K c U
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 ( ) sinf r  where ( )f a vanishes. Then, for this
particular velocity distribution, the shear stress is given by the following expression:
( , )r
r a
u
a r
r r

  

 
  
  
The solution to the two algebraic equations arising out of the boundary conditions is:
2
2
3
1 ( / )
3
4 1 ( / )
5 /
1 ( / )
U a
c
K
a
 
 
 
 

 




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:
drop
2
3
8 ( )
1 ( / )
6
1 ( / )
F c
U a
 
 

 
 



The drag force acting on the solid sphere is given by Eq. (8.8c), so that:
drop
solid
2
3
1 ( / )
1 ( / )
F
F
 
 



Although not explicitly requested, the resulting velocity and pressure distributions outside and
inside the drop are defined by the following equations:

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