Ghosh - 550
Page 1
2/13/2016
Worked Out Examples
(Ideal Flows)
Example 1. (Superposition of Source + U.F.):
Consider the flow field formed by combining a uniform flow in the positive x direction and
a source located at the origin. Obtain expressions for the stream function, velocity potential,
and velocity field for the combined flow. If U = 25 m/s, determine the source strength if the
stagnation point is located at x = -0.5 m. Plot the stagnation streamline. Evaluate the
locations of the branches of the stagnation streamline far downstream.
1. Statement of the Problem
a) Given
 Flow field formed by combining a uniform flow in the positive x direction and a
source located at the origin.
 U = 25 m/s
 Stagnation point is located at x = -0.5 m.
b) Find
 Expressions for the stream function (), velocity potential (), and velocity field for
the combined flow.
 Source strength if U = 25 m/s and the stagnation point is located at x = -0.5 m.
 Plot the stagnation streamline.
 Evaluate the locations of the branches of the stagnation streamline far downstream.
2. System Diagram
Flow Field is a superposition of Uniform Flow + Source Flow = Flow past a half-body
+
=
Ghosh - 550
Page 2
2/13/2016
3. Assumptions
 Steady state condition
 Incompressible fluid flow
 Inviscid fluid flow
 2 - D problem
4. Governing Equations
 Stream Function Definition
1 
r 

V  
r
Vr 

Velocity Potential Definition (2 - D case)

r
1 
V  
r 
Vr  

Elementary Plane Flows
 Uniform Flow (positive x direction)
  Ur sin(  )
  Ur cos( )

Source Flow (from origin)
q

2
q
   ln( r )
2

5. Detailed Solution
(Note: Stream function, Velocity Potential, Velocity field are all required)
Since this is an elementary plane flow problem, the stream function and the velocity potential
of the flow field can be obtained by superposition technique:
Flow field = Uniform flow + Source flow

Stream Function
  Ur sin(  ) 

q

2
Velocity Potential
Ghosh - 550
Page 3
2/13/2016
q
 q



ln( r )   Ur cos( ) 
ln( r )
2
 2



   Ur cos( )  
Velocity field can be obtained using either stream function definition or velocity potential
definition. Here, use velocity potential definition:
Vr  

  
q
q 1

   Ur cos 
ln r   U cos 
r
r  
2
2 r

V  
1 
1 

r 
r 


 
q
 1
 Ur cos  2 ln r   r  Ur sin   0  U sin 

 


q 1 

 er   U sin    e
 V  Vr  er  V  e  U cos  
2 r 

Source Strength
Stagnation point is located at x = -0.5 m  Vr = V = 0 at r = 0.5 m &  = .
q 1
q 1
q
 U cos  
 U   0
2 r
2 0.5

V  U sin   U sin    0
Vr  U cos 
 U 
q

 0  q  U  q  25 m2/s
Stagnation streamline
Now, the stream function can be written as
  Ur sin(  ) 
q
U


  Ur sin  
  U r sin   
2
2
2

and the stream function value ss at the stagnation point is





 ss  U r sin     U 0.5sin      U
2
2
2




Stagnation streamline is   U r sin  






  ss  U
2 
2



 
 U r sin     U
 r sin   
 r sin     r 
2 2
2 2
2 sin 
2
2

Ghosh - 550
Page 4
2/13/2016
Using MatLab, the stagnation streamline plot looks like:
Stagnation Streamline
2
1.5
1
y
0.5
Stagnation streamline
Origin & Source location
Stagnation point
0
-0.5
-1
-1.5
-2
-2
0
2
4
6
x
8
10
12
14
Locations of the branches of the stagnation streamline far downstream
Stagnation streamline equation was (See above)
r sin  
 
2
r sin  (in polar coordinate system) is equivalent to y (in rectangular coordinate system).
Thus,
y
 
2
Branches of the stagnation streamline far downstream occur when   0 & 2 .
y  lim
 0
 


2
2
 

y  lim

  2
2
2
Therefore, the locations of the branches of the stagnation streamline far downstream is at
Ghosh - 550
Page 5
y
2/13/2016

2
.
6. Critical Assessment
Since this is a superposition of an elementary plane flow problem, it doesn't include
viscosity. It gives an exact solution to a real case problem as long as the viscous effects
are small. In other words, if we are only interested in the ideal pressure flow field, this
should be the approach. But viscosity must be included to obtain a more realistic
solution, which will be discussed in later chapters.
Example 2. (Force Calculation in Ideal Flows):
The flow over a Quonset hut may be approximated by the velocity distribution,

  R 2 
  R 2 


V  U 1     cos   er  U 1     sin    e with 0     . During a storm the
  r  
  r  
wind speed reaches 100 km/hr; the outside temperature is 5 C. A barometer inside the hut reads
720 mm of mercury; pressure p is also 720 mmHg. The hut has a diameter of 6 m and a length of
18 m. Determine the net force tending to lift the hut off its foundation.
U
Quonset hut
p
Tout
R
pin

1. Statement of the Problem
a) Given
 Velocity distribution:

  R 2 
  R 2 


V  U 1     cos   er  U 1     sin    e for 0    
  r  
  r  
 U = 100 km/hr = 27.778 m/s
 Tout = 5 C  air = 1.27 kg/m3
 p = pin = 720 mmHg = 95.992 kPa
 D=6mR=3m
 L = 18 m
b) Find
 Net force tending to lift the hut off its foundation, FL.
Ghosh - 550
Page 6
2/13/2016
2. System Diagram
p
U

p
pin
R

Tout
pin
3. Assumptions
 Steady state condition
 Incompressible fluid flow
 Inviscid fluid flow  Irrotational flow
Actually, this can be checked by the rotation  z 

1  1 rV  1 Vr 

0
2  r r
r  
4. Governing Equations

Bernoulli's Equation:
p


V2
 gz  const.
2
Restrictions:
(1) Steady flow
(2) Incompressible flow
(3) Frictionless flow
(4) Flow along a streamline
5. Detailed Solution
Velocity components are:
  R 2 
Vr  U 1     cos 
  r  
  R 2 
V  U 1     sin  
  r  

  R 2 
  R 2 


from the velocity distribution V  U 1     cos   er  U 1     sin    e .
  r  
  r  
Note that Vr  0 along r = R, so this represents flow around the hut. Flow is irrotational, so
the Bernoulli equation may be applied between any two points. Applying the equation
between a point far upstream and a point on the surface of the hut (neglecting elevation
differences), we obtain
p

Thus

U2
p V2
 gz1  
 gz 2
2
 2
Ghosh - 550
Page 7
p  p 
2/13/2016

1
 U 2 V 2
2

Along the surface of the hut, r = R, and
  R 2 
Vr  U 1     cos   0
  R  
  R 2 
V  U 1     sin    2U sin  
  R  
V 2  Vr2  V2  0 2   2U sin    4U 2 sin 2  
2
Substituting this expression into the Bernoulli equation,
p  p 




1
1
 U 2  4U 2 sin 2    U 2 1  4 sin 2  … 
2
2
Lift is the force component normal to the free stream flow direction. (By convention,
positive lift is an upward force.) The infinitesimal lift is given by (See the system diagram)
dFL   pin  p   sin   dA … 
where dA  R  d  L .
Substituting  into ,

1

dFL   pin   U 2 1  4 sin 2   p   sin   dA
2








1
dFL   U 2 1  4 sin 2   sin   dA since p = pin.
2
1
dFL   U 2 1  4 sin 2   sin   R  d  L
2
Finally, integrate this expression over 0     to obtain the lift:
Ghosh - 550
Page 8

2/13/2016



1
FL   dFL   U 2 RL  1  4 sin 2  sin   d
0
0
2


1
  U 2 RL   sin   d  4  sin 3   d 
0
 0

2


 cos 3 
 
1

2

  U RL  cos 0  4 
 cos  

2
3

 0 

After substituting values, FL = 88.196 kN.
6. Critical Assessment
This problem illustrates the use of integration to compute lift in a cylindrical system.
Make sure you understand the choice of area element and the lift formulation. It shows that
lift (as well as drag) may be non-zero in an ideal flow if the flow symmetry is lost.
Continue