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15/1/2018
CITY UNIVERSITY OF SCIENCE AND
INFORMATION TECHNOLOGY
GROUP MEMBERS
• Abbas Ali
• Haris Anwar
• Manzoor Ahmed
• Sehrish Amin
• Syed Wasim Shah
• Usman Khan
• Waqas Noman
2
• 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
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
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
FLOW DUE TO
IMPULSIVE
MOTION OF A
PLANE WALL
6
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
The governing equation is:
• Where
• Magnetic parameter
• porousity parameter
8
22
0
2
(1)
u u
u u
t y K
 


 
   
 
2
0
K





• 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
 

• 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
 

  
 

  
 
  
  


 
 

 
 

 
 
  
 
 
    
 
 
    
 
• 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
   

 

 
 
   
 
• 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
 





• Let
13
)
1
(2
2
K
Mu
y
u
t
u






H
K
M 
1
)2(2
2






Hu
y
u
t
u
• 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
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__
• Put in Eq(3)
16
00
00
),(
),(
22
2
21
__
21
__








cec
ec
ececsu
ececsyu HSyHSy
• Taking laplace inverse
17
HSy
HSy
e
s
syu
c
s
ecsu
ecsyu






1
),(
1
),0(
),(
__
1
0
1
__
1
__
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



• 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


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













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
),(



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
Now we check the condition
23






 )
2
()
2
(
2
1
),( Ht
t
y
erfceHt
t
y
erfcetyu
yy
HH
 
 
 
)(1)11(
2
1
),0(
)(1)(1
2
1
),0(
)()(
2
1
),0(
)0()0(
2
1
),0( 00
satisfytu
HterfcHterfctu
HterfcHterfctu
HterfceHterfcetu




24
• Satisfy all the condition so our solution is ok
25
 
0),(
)()(
2
1
),(

 
tu
erfceerfcetu
• 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
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
 

 

 

      
         
   
  
   
Unsteady
Couette Flow
5/1/2018 28
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
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
Geometry of the Flow
5/1/2018 31
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


Dimensionless Variables
5/1/2018 33
*
0
* 0
,
,
u
u
u
u
y y



2
* 0
,
u
t t


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
 




 
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
 
 
 
To make it more simplify we Drop
the sign of * we get
---------------(2)
5/1/2018 36
2
2
u u
Hu
t y
 
 
 
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
 
 
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

 
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
 
 
 

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
 
 
 


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










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
 
 


Let suppose that
The equation become
5/1/2018 43
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  
5/1/2018 44
''
''
( ) 0
(0) 0
(1) 0
Y Hy y
Y H y
y
y


  
  


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



 
 
5/1/2018 46
2
2 2
0
sin sin
n
A
thus
H n
H n
n H
 
 
 

 
 
 
Thus
5/1/2018 47
'
'
sin
0
n nY a n y
T
T
T T
m





 
 
 
Then
5/1/2018 48
2 2
( )
( )
( )
nt
n n
n H t
n n
T t c e
T t c e



 


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




 


  
Unsteady
Poiseuille Flow
5/1/2018 50
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
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
Geometry of the Flow
5/1/2018 53
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
 
 
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


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
 




 
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
  
  
  
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
  
  
  
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
 
 
   
   
   
  
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

 


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
 
 
   
   
  
  
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
 
 
    
   
  
  
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










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
 
 


Let suppose that
The equation become
5/1/2018 65
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  
5/1/2018 66
''
''
( ) 0
(1) 0
( 1) 0
Y Hy y
Y H y
y
y


  
  

 
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



 
 
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








 

 


Thus
5/1/2018 69
'
'
(2 1)
cos
2
0
n n
n
Y a
T
T
T T
m






 
 
 
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




 


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



 


  
Generalized Couette Flow
5/1/2018 72
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
Geometry of the Flow
5/1/2018 74
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) 
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
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
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
  
  
  
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
  
  
  
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

   

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


  

After simplification we get
5/1/2018 82
(4)
2 2
[ ]
2 2
p p
u y HUy U HU y
x x 
 
    
 
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 
 
    
 
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

 
 
 


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










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
 
 


Let suppose that
The equation become
5/1/2018 87
' ''
T Y
H
T Y
   
'
''
T
T
Y
H
Y


 
  
5/1/2018 88
''
''
( ) 0
(0) 0
( ) 0
Y Hy y
Y H y
y
y h


  
  


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



 
 
5/1/2018 90
2
2 2
0
sin sin
n
A
thus
H n
H n
n H
 
 
 

 
 
 
Thus
5/1/2018 91
'
'
sin
0
n nY a n y
T
T
T T
m





 
 
 
Then
5/1/2018 92
2 2
( )
( )
( )
nt
n n
n H t
n n
T t c e
T t c e



 


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


  

 

  
     
 

ANY
QUESTION?
94
THANKS
95

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Couette flow

  • 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       
  • 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 __
  • 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  
  • 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                                     
  • 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           
  • 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          
  • 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) 
  • 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                     