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Generalized Couette Flow
5/1/2018 1
It is the special case of flow because it is formed
from couette and poiseuille flow.
The flow in bounded in between two parallel
plates at y=0 and y=h and it is initially at rest.
The fluid is also magnetically conducted and
pass through porous media. The flow is due to
pressure gradient as well as of motion of upper
plate. The governing equation for this flow is
given as
5/1/2018 2
----- (1)
And the boundary are
5/1/2018 3
22
0
2
1 Bu u p
u u
t y x k


 
  
   
  
0
(0, ) 0
( , )
u t
u h t U


Where
*
0
*
0
*
0
,
,
,
u
u
u
x
x
x
y
y
y



* 0
*
0
0
,
,
R ,e
tu
t
d
pd
p
u
u d





5/1/2018 4
2 2
0
2
,
1
,
1
,
a B
M
k
k a
H M
k




 
M is the MHD
v/k is the porous media
5/1/2018 5
Put all these dimensional less
values in equation(1) we get
After simplification we get
5/1/2018 6
2 0* 2 * *
* *0 0 0 0 0
* 2 2 * 2 *
U U U U Uu u p
u u
d t d y d x k

  

  
   
  
* 2 * *
*
* 2 * *
u u p
Hu
t y x
  
  
  
To make it more simplify we Drop
the sign of * we get
---------------(2)
5/1/2018 7
2
2
u u p
Hu
t y x
  
  
  
Assume solution for this
phenomena is
---------(3)
5/1/2018 8
( , ) Re{ ( ) }iwt
U y t F y e
iwtp
e
x


 

Put all these values in equation(2)
we get
After applying derivative we get
5/1/2018 9
2
2
Re [Re{ ( ) }] [Re{ ( ) }] [Re{ ( ) }]iwt iwt iwt iwt
F y e F y e e M F y e
t y

 
  
 
2
2
{ ( )} [Re ]( ( ))
d
F y iw M F y
dy
   
After the arranging of above
equation we have
(4)
5/1/2018 10
2
2
{ ( )} [Re ]( ( ))
d
F y iw M F y
dy
   
And the transform boundary
conditions are:
5/1/2018 11
0
(0) 0
(1)
f
f U


The auxiliary equation of (4) is
The solution is
5/1/2018 12
2
(Re ) 0m iw M  
1 2cosh Re sinh Recy c iw My c iw My   
For particular solution we have
5/1/2018 13
'
''
0
0
p
p
p
y A
y
y



Put these values in equation(4) we get
5/1/2018 14
0 [Re ]
[Re ]
iw M A
A
iw M


   


The general solution is
Apply the boundary conditions in above
equation we get
5/1/2018 15
1 2cosh Re sinh Re
[Re ]
y c iw My c iw My
iw M

    

1
0
2
Re
cosh Re
Re Re
sinh Re
c
iw M
U iw M
iw M iw Mc
iw M

 
 

  
 

Repalce and
5/1/2018 16
2c1c
0
cosh Re
Re
cosh Re
Re Re sinh Re
sinh Re
[Re ]
y iw My
iw M
U iw M
iw M iw M iw My
iw M
iw M

 

   

  
  




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Generalized couette flow by assume solution

  • 2. It is the special case of flow because it is formed from couette and poiseuille flow. The flow in bounded in between two parallel plates at y=0 and y=h and it is initially at rest. The fluid is also magnetically conducted and pass through porous media. The flow is due to pressure gradient as well as of motion of upper plate. The governing equation for this flow is given as 5/1/2018 2
  • 3. ----- (1) And the boundary are 5/1/2018 3 22 0 2 1 Bu u p u u t y x k               0 (0, ) 0 ( , ) u t u h t U  
  • 5. 2 2 0 2 , 1 , 1 , a B M k k a H M k       M is the MHD v/k is the porous media 5/1/2018 5
  • 6. Put all these dimensional less values in equation(1) we get After simplification we get 5/1/2018 6 2 0* 2 * * * *0 0 0 0 0 * 2 2 * 2 * U U U U Uu u p u u d t d y d x k                * 2 * * * * 2 * * u u p Hu t y x         
  • 7. To make it more simplify we Drop the sign of * we get ---------------(2) 5/1/2018 7 2 2 u u p Hu t y x         
  • 8. Assume solution for this phenomena is ---------(3) 5/1/2018 8 ( , ) Re{ ( ) }iwt U y t F y e iwtp e x     
  • 9. Put all these values in equation(2) we get After applying derivative we get 5/1/2018 9 2 2 Re [Re{ ( ) }] [Re{ ( ) }] [Re{ ( ) }]iwt iwt iwt iwt F y e F y e e M F y e t y         2 2 { ( )} [Re ]( ( )) d F y iw M F y dy    
  • 10. After the arranging of above equation we have (4) 5/1/2018 10 2 2 { ( )} [Re ]( ( )) d F y iw M F y dy    
  • 11. And the transform boundary conditions are: 5/1/2018 11 0 (0) 0 (1) f f U  
  • 12. The auxiliary equation of (4) is The solution is 5/1/2018 12 2 (Re ) 0m iw M   1 2cosh Re sinh Recy c iw My c iw My   
  • 13. For particular solution we have 5/1/2018 13 ' '' 0 0 p p p y A y y   
  • 14. Put these values in equation(4) we get 5/1/2018 14 0 [Re ] [Re ] iw M A A iw M        
  • 15. The general solution is Apply the boundary conditions in above equation we get 5/1/2018 15 1 2cosh Re sinh Re [Re ] y c iw My c iw My iw M        1 0 2 Re cosh Re Re Re sinh Re c iw M U iw M iw M iw Mc iw M            
  • 16. Repalce and 5/1/2018 16 2c1c 0 cosh Re Re cosh Re Re Re sinh Re sinh Re [Re ] y iw My iw M U iw M iw M iw M iw My iw M iw M                  