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Worked Out Examples
(Continuity Equation)
Example 1. (Evaluation of (x,y)):
Determine the family of  functions that will yield the velocity field



V  ( x 2  y 2 )  i  2 xy  j .
Solution
1. Statement of the Problem
a) Given



 Velocity field: V  ( x 2  y 2 )  i  2 xy  j  u
b) Find
 Family of  functions from the given velocity field.
 x2  y2
&
v  2 xy
2. System Diagram
It is not necessary for this problem.
3. Assumptions
 Steady state condition
 Incompressible fluid flow
 2 - D problem
4. Governing Equations
Stream function (incompressible fluid flow version) definition:
u

 &
v
x
y
5. Detailed Solution
With the definition of stream function and the given velocity components:
u

 x2  y2 … 
y
v

 2 xy … 
x
    ( x 2  y 2 )  y 
    ( x
2
 y 2 )  y    x 2 y 
1 3
y  f ( x) ,
3
where f(x) is any function of x including constants.


df ( x)
 2 xy  0 
that will be
…
x
dx
x
df ( x)
 0 , that represents f(x) = constant.
Comparing  with ,
dx
1
Finally,   x 2 y  y 3  const.
3
Using this  obtained, take
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6. Critical Assessment
The stream function, , exists; therefore, the velocity field satisfies the continuity equation,
u v

 0 , and also it can be said that it is a valid velocity field.
x y
Example 2. (Use of  and its properties):
In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly
from zero at y = 0 to 100 ft/s at y = 4 ft. Determine an expression for the stream function, .
Also determine the y coordinate above which the volume flow rate is half the total between y
= 0 and y = 4 ft.
1. Statement of the Problem
a) Given
 1 - D flow parallel to the positive x direction.
 Velocity varies linearly, u(y = 0 ft) = 0 ft/s & u(y = 4 ft) = 100 ft/s.
b) Find
 An expression for the stream function, .
 y coordinate above which the volume flow rate is half the total between y = 0 and y =
4 ft.
2. System Diagram
y
4 ft
B
u = u(y)
x
0
A
3. Assumptions
 Steady state condition
 Incompressible fluid flow
 2 - D problem
4. Governing Equations
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Stream function (incompressible fluid flow version) definition: u   &
y
v

x
5. Detailed Solution
Find the velocity from the given information, u(y = 0 ft) = 0 ft/s & u(y = 4 ft) = 100 ft/s.
Since the velocity varies linearly, u  u ( y )   (100 ft / s )  (0 ft / s )   y  25 y .



(4 ft )  (0 ft )

There is no flow going on in the y direction.  v = 0 ft/s.
Using the definition of stream function,

 25 y … 
y

v
0 …
x
u
 becomes:   25 y  y 
    25 y  y   
25 2
y  f ( x)
2
where f(x) is any function of x including constants.


df ( x)
 0
that will be
.…
x
dx
x
df ( x)
 0 .  f(x) = constant.
Comparing  with ,
dx
25 2
y  const .
Finally,  
2
Using this , take
The volume flow rate, Q, across AB in the diagram can be evaluated as follows:
For a unit depth (dimension perpendicular to the xy plane), the flow rate across AB is
yB
yB
yA
yA
Q   udy  

dy
y
Along AB, x = constant, and d 

dy . Therefore,
y
B

dy   d   B   A
y A y
A
Q
yB
The total volume flow rate is
 25
  25

QTotal   B ( y  4)   A ( y  0)   4 2   const .   0 2   const .  200 ft 2 / s / ft
2
 2

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The half of the total volume flow rate is then,
200
 25
  25

  y   A   y 2  const .   (0 2 )  const .  12.5 y 2
2
2
 2

 y
200
 2.828 ft
2  12.5
6. Critical Assessment
The stream function, , exists; therefore, the velocity field satisfies the continuity
equation,
u v

 0 . Also it can be said that it is a valid velocity field for 2x y
Dimensional incompressible flows. This problem also demonstrates how volumetric flow



rate may be computed alternatively (without using Q  V  dA ) by the use of a 
property)