couette flow and its generalization with MHD and Porous Media

1. 15/1/2018
CITY UNIVERSITY OF SCIENCE AND
INFORMATION TECHNOLOGY

2. GROUP MEMBERS
• Abbas Ali
• Haris Anwar
• Manzoor Ahmed
• Sehrish Amin
• Syed Wasim Shah
• Usman Khan
• Waqas Noman
2

3. • On the Unsteady unidirectional
flows generated by impulsive
motion of a boundary or sudden
application of a pressure gradient
in the presence of MHD and
porous medium .
M. Emin Erdogan
3
TOPIC OF PRESENTATION

4. INTRODUCTION
The governing equation for fluid mechanics are
the Navier-Stokes Equation. Exact solutions are
very important for many reasons. They provide a
standard for checking the accuracies of many
approximate methods. An exact solution is
defined as a solution of the Navier-Stokes
equations and the continuity equation. Most of
the exact solutions for unsteady flows are in
series form.
4

5. CONTINUED
In this paper, unsteady flows considered are
Stoke’s first problem, unsteady couette flow,
unsteady Poiseuille flow and unsteady
generalized Couette flow.
The solutions for these flows are in the form of
series.
5

6. FLOW DUE TO
IMPULSIVE
MOTION OF A
PLANE WALL
6

7. The flow over a plane wall which is initially at
rest and is suddenly moved in its own plane with
a constant velocity is termed Stoke’s first
problem. The fluid stays in the region y≥0 and
the x-axis is chosen as the plane wall, in the
presence of MHD and porous media.
7
MATHEMATICAL FORMULATION

8. The governing equation is:
• Where
• Magnetic parameter
• porousity parameter
8
22
0
2
(1)
u u
u u
t y K
 


 
   
 
2
0
K






9. • The dimensionless variable are
• , ,
9
0u
u
u 
uuu  
0
ou
y y



2
ou
t t



0
y
y
u
 
 2
0
t
t
u
 


10. • From (1)
10
2 2
0 0 0 0
0
2
2
0 0
3 3 22
0 0 0 0
02 2
3 3 22
0 0 0 0
02
( )
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
u u u u u u
u u
y Kt y
u u
u u u uu u
u u
t y K
u u u uu u
u u
t y K
 

  
 

  
 
  
  


 
 

 
 

 
 
  
 
 
    
 
 
    
 

11. • Multiplying by we get
11











u
K
Mu
y
u
t
u 1
2
2
3
0u

22
0
2 2 2
0 0
( )
uu u
u
t y u K u
   

 

 
 
   
 

12. • Let
• And
• For easy writing we use “u” instead of
12

u
2
0
2
0
2
2
0
(dim )
1
u
M ensionles
u
ku k
 






13. • Let
13
)
1
(2
2
K
Mu
y
u
t
u






H
K
M 
1
)2(2
2






Hu
y
u
t
u

14. • Dimensionless Boundary condition
• Laplace of B.C.S
14
0),(
0)0,(
1),0(
),0(
),0(
*
00
*
0





tu
yu
tu
utuu
utu
0),(
1
),0(
*
*


su
s
su

15. Taking laplace of eq (2)
15





HSDHsD
HSu
dy
ud
uHuS
dy
ud
uH
dy
ud
yusyuS
0)(
0)(
0
)0,(),(
2
__
2
__
2
____
2
__
2
__
2
__
2__

16. • Put in Eq(3)
16
00
00
),(
),(
22
2
21
__
21
__








cec
ec
ececsu
ececsyu HSyHSy

17. • Taking laplace inverse
17
HSy
HSy
e
s
syu
c
s
ecsu
ecsyu






1
),(
1
),0(
),(
__
1
0
1
__
1
__

18. And
Using convolution theorem
18

















HSy
HSy
e
s
Ltyu
e
s
LsyuL
1
),(
1
),(
1
1
__
1
1,
11




t
s
L   ss
ye
teL
s
y
HS
HSy
2
,
4
1
2




19. • Let
19
 



t s
y
HS
s
y
HS
ds
ss
ey
tyu
ss
ye
tyu
0
4
4
)4(
2
),(
2
*1
),(
2
2


ds
s
y
zdz
s
y
z 2
22
2
4
2
4



20. s
ds
z
dz
s
ds
z
zdz
s
ds
y
s
zdz



2
1
2
4
2
2
2
20
22
41
y
s
z

s
y
z
2

z
y
s
2


21. 21













t
y
z
z
Hy
t
y
z
z
Hy
t
y z
z
Hy
dzetyu
dzetyu
z
y
z
dz
e
tyu
2
4
2
4
2
4
2
2
2
2
2
2
2
2
2
2
),(
2
),(
2
)2(
2
),(




22. Let
Formula
22
24
2
2 Hy
a
Hy
a 
)()(
4 22
2
2
2
2
r
a
rerfce
r
a
rerfcedze aa
t
y
z
a
z
 

















)
2
2
()
2
2
(
2
1
),(
2
2
22
2
2
t
yt
y
erfce
t
yt
y
erfcetyu
Hy
Hy
Hy
Hy

23. Now we check the condition
23






 )
2
()
2
(
2
1
),( Ht
t
y
erfceHt
t
y
erfcetyu
yy
HH

24.  
 
 
)(1)11(
2
1
),0(
)(1)(1
2
1
),0(
)()(
2
1
),0(
)0()0(
2
1
),0( 00
satisfytu
HterfcHterfctu
HterfcHterfctu
HterfceHterfcetu




24

25. • Satisfy all the condition so our solution is ok
25
 
0),(
)()(
2
1
),(

 
tu
erfceerfcetu

26. • Porosity term in dimensionless
• And M is also Dimensionless
26
1
)( 24
24
212
24
2
0
2
 



TL
TL
LTL
TL
ku

2 2 1 2 1 2 1 3 3 2
0
2 3 2 2
0
( ) ( )L T MT A M L T A
u ML L T
 

    
 




27. 27
2
2 1 2 4 2 1 3 3 2 1 3 2 20
2
0
2
2 1 1 3 3 2 2 1 4 3 20
2
0
2
0 0 00
2
0
1
L L M T A M L T A M L L T
u
M L T
u
M L T
u
 

 

 

      
         
   
  
   

28. Unsteady
Couette Flow
5/1/2018 28

29. COUETTE FLOW
If the flow is in between two
infinite parallel plates and one
of them is moving relative to
the other plate, then this kind
of flow is called couette flow.
5/1/2018 29
CITY UNIVERSITY OF SCIENCE AND
INFORMATION TECHNOLOGY

30. MATHEMATICAL FORMULATION
Suppose that the incompressible newtonian
viscous fluid is bounded between two rigid
boundaries at y=0 and y=h. Initially the fluid is at
rest. The fluid start motion due to the
disturbance of upper plate, and the lower plate
is held stationary. Also, in the presence of MHD
and porous medium.
5/1/2018 30

31. Geometry of the Flow
5/1/2018 31

32. Mathematical modeling
And the boundary are
5/1/2018 32
(1) 
22
0
2
Bu u
u u
t y k



 
  
 
0
(0, ) 0
( , )
u t
u h t U



33. Dimensionless Variables
5/1/2018 33
*
0
* 0
,
,
u
u
u
u
y y



2
* 0
,
u
t t



34. M is the MHD
1/k is the porous media
5/1/2018 34
2
0
2
0
2
2
0
,
1
,
1
,
B
M
u
k ku
H M
k
 




 

35. Put all these dimensional less
values in equation(1)
And After simplification we get
5/1/2018 35
* 2 *
*
* 2 *
u u
Hu
t y
 
 
 

36. To make it more simplify we Drop
the sign of * we get
---------------(2)
5/1/2018 36
2
2
u u
Hu
t y
 
 
 

37. For steady flow the
Then we get
5/1/2018 37
0
u
t



1 2
2 2
1 1
2 2
c y c
u
H H
y y
 
 

38. put y(0)=0 and y(1)=1 then it
become
Put all these value in above equation
we get
5/1/2018 38
2
1
0
1
2
c
H
c

 

39. Which is the solution for the case of
steady part.
For unsteady we get an equation of
the form
5/1/2018 39
(4)
2
1
2
1
2
H
y
u
H
y
 
 
 


40. Where f(y,t) satisfies the following differential equation:
5/1/2018 40
2
1
2
( , )
1
2
H
y
u f y t
H
y
 
 
  

2
2
(1, ) 0,
(0, ) 0
f f
Hf
t y
f t
f t
 
 
 



41. As we known that
5/1/2018 41
'
'
2
''
2
( , ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
f y t Y y T t
f
Y y T t
t
f
Y y T t
y
f
Y y T t
y











42. Now the above equation become
5/1/2018 42
' ''
' ''
' ''
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
Y y T t Y y T t HY y T t
YT T Y HY
T Y HY
T Y
 
 



43. Let suppose that
The equation become
5/1/2018 43
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  

44. 5/1/2018 44
''
''
( ) 0
(0) 0
(1) 0
Y Hy y
Y H y
y
y


  
  



45. Put the initial conditions we get
5/1/2018 45
2
1 2
( ) 0
( ) cos sin
D H
D i H
Y y A H y A H y


 
  
  
   
1
2
2
0
(1) sin
sin 0
A
and
y A H
A H



 
 

46. 5/1/2018 46
2
2 2
0
sin sin
n
A
thus
H n
H n
n H
 
 
 

 
 
 

47. Thus
5/1/2018 47
'
'
sin
0
n nY a n y
T
T
T T
m





 
 
 

48. Then
5/1/2018 48
2 2
( )
( )
( )
nt
n n
n H t
n n
T t c e
T t c e



 



49. Hence
Thus
5/1/2018 49
2 2
2
( )
1
( , ) sin
n
H t
h
n n
n
f y t A e n y


  

 
2 2
( )
1
2 ( 1)
sin
n
n H t
n
u Y e n y
n




 


  

50. Unsteady
Poiseuille Flow
5/1/2018 50

51. POISEUILLE FLOW
If the flow is in between two
infinite parallel plates and the
flow is induced due to the
sudden application of pressure
gradient.
5/1/2018 51
CITY UNIVERSITY OF SCIENCE AND
INFORMATION TECHNOLOGY

52. MATHEMATICAL FORMULATION
Suppose that the incompressible newtonian
viscous fluid is bounded between two parallel
plates at y = -b and y = b, and it is initially at rest
and the fluid starts suddenly due to a constant
pressure gradient. In the presence of MHD and
porous media.
5/1/2018 52

53. Geometry of the Flow
5/1/2018 53

54. Mathematical modeling
And the boundary are
5/1/2018 54
22
0
2
1 Bu u p
u u
t y x k


 
  
   
  
(1) 
( , ) 0
( , ) 0
u b t
u b t
 
 

55. Where
5/1/2018 55
*
0
* 0
*
0
,
,
,
u
u
u
u
x x
y
y
y




* 0
*
2
0
,
,
tu
t
d
p
p
u



56. M is the MHD
1/k is the porous media
5/1/2018 56
2
0
2
0
2
2
0
,
1
,
1
,
B
M
u
k ku
H M
k
 




 

57. Put all these dimensional less
values in equation(1) we get
After simplification we get
5/1/2018 57
* 2 * *
*
* 2 * *
u u p
Hu
t y x
  
  
  

58. To make it more simplify we Drop
the sign of * we get
---------------(2)
5/1/2018 58
2
2
u u p
Hu
t y x
  
  
  

59. For steady flow the
Then we get
5/1/2018 59
0
u
t



2
1 2
2 22
1 12 1
2 22
c y cp y
u
Hy Hyx Hy
 
 
   
   
   
  

60. put y(1)=0 and y(-1)=0 then it
become
Put all these value in above equation
we get
5/1/2018 60
2
2
1
2
0
p b
c
x
c

 



61. 5/1/2018 61
(4)
Which is the
solution for the
case of study
part.
For unsteady we
get an equation
of the form
 2
2
1
1
2 1
2
p
u y
x Hy
 
 
   
   
  
  

62. Where we have
5/1/2018 62
2
2
( , )
(1, ) 0,
( 1, ) 0
f f
Hf y t
t y
f t
f t
 
 
 

 
 2
2
1
1 ( , )
2 1
2
p
u y f y t
x Hy
 
 
    
   
  
  

63. As we known that
5/1/2018 63
'
'
2
''
2
( , ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
f y t Y y T t
f
Y y T t
t
f
Y y T t
y
f
Y y T t
y











64. Now the above equation become
5/1/2018 64
' ''
' ''
''
'
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
Y y T t Y y T t HY y T t
YT T Y HY
Y HY
T T
Y
 
 



65. Let suppose that
The equation become
5/1/2018 65
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  

66. 5/1/2018 66
''
''
( ) 0
(1) 0
( 1) 0
Y Hy y
Y H y
y
y


  
  

 

67. Put the initial conditions we get
5/1/2018 67
2
1 2
( ) 0
( ) cos sin
D H
D i H
Y y A H y A H y


 
  
  
   
2
1
1
0
(1) cos
cos 0
A
and
y A H
A H



 
 

68. 5/1/2018 68
1
2 2
0
cos(2 1)
cos
2
(2 1)
2
(2 1)
4
n
A
thus
n
H
n
H
n








 

 



69. Thus
5/1/2018 69
'
'
(2 1)
cos
2
0
n n
n
Y a
T
T
T T
m






 
 
 

70. Then
5/1/2018 70
2 2
(2 1)
( )
4
( )
( )
nt
n n
n
H t
n n
T t c e
T t c e




 



71. Hence
Thus
5/1/2018 71
2
1
( , ) cosn yt
n n n
n
f y t A e y




 
2 2
( )
1
2 ( 1)
cos2 1
n
n H t
n
Y
u e n y
h n



 


  

72. Generalized Couette Flow
5/1/2018 72

73. 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 73

74. Geometry of the Flow
5/1/2018 74

75. Mathematical modeling
And the boundary are
5/1/2018 75
22
0
2
1 Bu u p
u u
t y x k


 
  
   
  
0
(0, ) 0
( , )
u t
u h t U


(1) 

76. 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 76

77. 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 77

78. Put all these dimensional less
values in equation(1) we get
After simplification we get
5/1/2018 78
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
  
  
  

79. To make it more simplify we Drop
the sign of * we get
---------------(2)
5/1/2018 79
2
2
u u p
Hu
t y x
  
  
  

80. For steady flow the
Then we get
5/1/2018 80
0
u
t



2 2
1 2
1
2
p
u y HUy c y c
y

   


81. put y(0)=0 and y(1)=1 then it
become
Put all these value in above equation
we get
5/1/2018 81
2
1
0
2
c
p
c U HU
x


  


82. After simplification we get
5/1/2018 82
(4)
2 2
[ ]
2 2
p p
u y HUy U HU y
x x 
 
    
 

83. Which is the solution for the case of
study part.
For unsteady we get an equation of
the form
5/1/2018 83
2
[ ] [ ]
2 2
p p
u HU y U HU y
x x 
 
    
 

84. Where we have
5/1/2018 84
2
[ ] [ ] ( , )
2 2
p p
u HU y U HU y f y t
x x 
 
     
 
2
2
( , )
(1, ) 0,
(0, ) 0
f f
Hf y t
t y
f t
f t

 
 
 



85. As we known that
5/1/2018 85
'
'
2
''
2
( , ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
f y t Y y T t
f
Y y T t
t
f
Y y T t
y
f
Y y T t
y











86. Now the above equation become
5/1/2018 86
' ''
' ''
''
'
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
Y y T t Y y T t HY y T t
YT T Y HY
Y HY
T T
Y
 
 



87. Let suppose that
The equation become
5/1/2018 87
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  

88. 5/1/2018 88
''
''
( ) 0
(0) 0
( ) 0
Y Hy y
Y H y
y
y h


  
  



89. Put the initial conditions we get
5/1/2018 89
2
1 2
( ) 0
( ) cos sin
D H
D i H
Y y A H y A H y


 
  
  
   
1
2
2
0
( ) sin
sin 0
A
and
y h A H
A H



 
 

90. 5/1/2018 90
2
2 2
0
sin sin
n
A
thus
H n
H n
n H
 
 
 

 
 
 

91. Thus
5/1/2018 91
'
'
sin
0
n nY a n y
T
T
T T
m





 
 
 

92. Then
5/1/2018 92
2 2
( )
( )
( )
nt
n n
n H t
n n
T t c e
T t c e



 



93. Hence
Thus
5/1/2018 93
2 2
( )
1
( , ) sinn H t
n n
n
f y t A e n y


 

 
2 2
2 ( )
1
2 ( 1)
[ ] [ ] sin
2 2
n
n H t
n
p p
u HU y U HU y e n y
x x n


  

 

  
     
 


94. ANY
QUESTION?
94

95. THANKS
95