ES3C9
FLUID MECHANICS FOR MECHANICAL ENGINEERS
Example Class 1
(1) A two-dimensional incompressible flow field is defined by the following velocity
components where V and L are constants.
y
 x y
v  2V ,
u  2V   
L
L L
If they exist, find the stream function and velocity potential.
Stream function  exists if the continuity equation is verified .V  0 as the flow is
already 2D and incompressible.
du dv 2V 2V
Here .V 



 0 so  exists.
dx dy
L
L
Velocity potential  exists if the flow is irrotational   V  0
 v w
 z  y 
0

 w u 
Here   V   
 0 so  doesn’t exist.

x

z

 2V
 v u 
 x  y  L



u  y
Let determine the stream function. It should verify : 
v   

x
2
 xy y 
So   2V  
  f  x  by integrating the u-relation.
 L 2L 
Then
df

y
 y  df
 y  df
0
 2V   
so v  2V  2V   
thus
dx
x
L
 L  dx
 L  dx
 xy y 2 
And   2V  
  const.
 L 2L 
20
y
10
0
-10
-20
-20
0
x
20
(2) A disk of radius R rotates at an angular velocity  inside a disk-shaped container
filled with oil of viscosity  , as shown on Fig. 1. Assuming a linear velocity
profile and neglecting shear stress on the outer disk edges, derive a formula for the
viscous torque on the disk.
Ω
h
oil
h
R
Fig.1
R
The velocity is zero on each wall of the container.
Along the disk surface, the velocity is u  r   r
du  r 
r  0
r


dy
h
h
Let consider a small surface of disk, dA  rd dr
At a radius r, the shear stress is :   
The torque applied on both side of the disk is: dM  2 rdA  2
2 R
Then M 

 2
0 0
r 3
drd
h
r 3
R 4
drd 
h
h
(3) Consider the two-dimensional incompressible velocity potential   xy  x 2  y 2 .
(a) is it true that  2  0 ? If so, what does it mean ?
2 
 2  2

 022  0
x 2 y 2
It means that the continuity equation for an incompressible fluid is satisfied .V  0 as
V   .
(b) If it exists, find the stream function   x, y  of this flow.
y2


 2 xy  f  x 
, hence  
 y  2x 
2
x
y
x2


df
v
 x  2y  
 2 y  , hence f  x     const
2
y
x
dx
u
y 2  x2
 2 xy  const
2
(c) Find the equation of the streamline that passes through (x,y)=(2,1).
12  22
5
 2(2)(1) 
Set the constant to zero, then   2,1 
2
2
2
2
y x
5
 2 xy 
The proper streamline is  
2
2
The final stream function is thus  