dr. sohail presentation.(2010)

8/6/2019 Dr. Sohail Presentation.(2010)

1/63

Peristaltic flow with an endoscope

By


2/63

Geometry of the problem


3/63

For an incompressible fluid the balance of mass andmomentum are given by

,div

,0divf

V

V

V V !

!

T pdt d

1

2

The constitutive equation for A six-constant Jeffrey's fluidmodel is given by

T 2 1 dT d t ! W .T T .W d T . D D.T bT : DI cDtr T 29 D 2 2 d Dt ! W . D D.W 2d D. D bD : DI

3


4/63

D symmetric part of velocity gradient V

VT

,

W antisymmetric part of velocity gradient V ! VT

.

W here

b, d and c are material constants of six constant

Jeffrey fluid model and is the relaxation time and

is the delay time


5/63

S pecial cases

The transition from the eight-constant Oldroyd model to the six-constant generalized Jeffrey's model occurs under the followingconditions

9 1

!a8 1,

9 !c8 1, : 1 b8 1,

9 2

!a8 2, : 2 b8 2.

Further Maxwell, De W itt and W hite-Metzner models, as wellas three and four-constants Oldroyd models can be treated asparticular cases of the generalized six-constant Jeffrey's model


6/63

The geometry of the wall surface is

R1 a1,

2 a 2 b1 s i 2@ Z ! ct ,

4

5

0,

The governing equations in the fixed frame for anincompressible flow are

6

A t

! p 1 RR R R Z ! 5 5 , 7

C ontinuity equation

R- C omponent of momentum equation


7/63

Z -C omponent of momentum equation

A W W ! p 1 RT R Z T Z Z .

8

z c 1t , r R ,

z Z 1

t , r R ,

Transformations between two reference frames are defined

w W c 1, ,

9

10

The corresponding boundary conditions are the no-slip atboth the walls

0 w c1 , at r r 1 ,

!c dr 2 z , w !c1 , at r r 2 a b sin2@ z .

11

12


8/63

R Ra 2 , r r

a 2 , Z Z 8

, z z 8

, W W c 1 , wwc 1 , T

a 2 T c 1 9 ,

U 8 a 2 c 1 , u8

a 2 c 1 , P a 2

2 P

c 1 8 9 , t c 1 t

8 , 0

a 2

8 , Re Ac 1 a 24 0 ,

r 2r 2a 2 1 F sin 2 @ z , 8 1

2 1 c 1a 2 , 8 2

2 2 c 1a 2 , 2

1

a 2 1 .

Dimensionless variables are defined

13U sing the non-dimensional quantities and transformation theresulting equations can be written as

u

r

u

r

w

z 0 ,

C ontinuity equation

r- C omponent of momentum equation

Re0 3 u r w z u ! pr

0 r r

rT rr 0 2

z T rz ! 0 T 55 r ,

14

15


9/63

z -C omponent of momentum equation

Re 0 ur

w z

w ! p z

1

r r rT rz 0

zT zz , 16

T r z

wr

1 8 18 2 1! d d b ! c2 2d 3b wr 2

1 8 12 1! d d b ! c2 2d 3b wr 2 ,

T rr 8

2 wr

21 d b

! 8 1 w

r 1 d b T r z ,

T zz 8 2 wr

2

!1 d b ! 8 1 wr

!1 d b T r z ,

T 55 8 2 wr

2b! 8 1 w

r bT r z ,

W here

17


10/63

U nder the assumption of long wave length approximation andlow Reynolds number above equations take the form

p 0 ,

p z

1r r

r wr

1r r

,, 1 wr

3

, , wr

5,

, 1 ! d d b ! c 2 d 3 b , , 1 8 1 8 2 ! 8 1 , , 2 !8 13 8 2 .where

18

19

C orresponding boundary conditions in dimensionless form are

w 1 , at r r 1 ,

w 1, t r r 2 1 sin 2 z .

20

21


11/63

S olution of the problem

Perturbation solution

For perturbation solution, we expand w, F, p by takingas perturbation parameter

E

w w 0 , w 1 , 2 w 2 O, 3 ,

p p0

, p1

, 2

p2 O

, 3 ,

0 , 1 , 2

2 , 3

.

,ln

ln

lnln41

6766863

66

46160

502

594

586

578

561 0

552

111 0252

44

321

2

xx

!

ar ar

ar

ar

ar

ar ar ar ar ar ar a

ar ar ar ar a z par a

r w

E

E

The perturbation results for small parameter E22

S atisfying boundary condition can be written as

23


12/63

dp z 2 F r 2

2

!r 12

! , a69

! , 2 a7 0

a6 8 ,

Axial pressure gradient

24

W here all constant are presented as follows

a1r 12 ! r 22

4 ln r 2 ! ln r 1 , a2 !r 12 ln r 2 ! r 22 ln r 14 ln r 2 ! ln r 1 , a3 !,

1

32dp 0d z

3

,

a4 !3, 1a18dp 0d z

3, a5 , 1a1

3

8dp 0d z

3, a6 !3, 1a1

2

2dp 0d z

3,

a7 a3r 14 a4 r 12 a5r 12

a6 ln r 1 , a8 a3r 24 a4 r 22 a5r 2

2 a6 ln r 2 ,

a9 a7 ! a8ln r 2 ! ln r 1 , a10 a9 a6 , a11 !a7 ! ln r 1a9 , a12 16 a3

2 ,

a13 16 a3a4 4a3 dp 1d z

, a14 14

dp 1d z

2

2a4 dp 1d z

8a3a1 dp 1d z

8a3a10 4a42 dp 1 z

,


13/63

a 1 a 1dp 1d z

2

a 10dp 1d z

4 a 4 a 1dp 1d z

4 a 4 a 10 ! 16a 3 a 5 ,

a 16 !4 a 4 a 5 a 102

a 12 dp 1

d z

2

2 a 5dp 1d z 2 a 10 a 1

dp 1d z ! 4 a 5 a 4 ,

a 17 !4 a 1 a 5 dp 1d z ! 4 a 5 a 10 , a 1 4 a 52 , a 19 1

2

dp 1d z

2 a 4 , a 2 0 4 a 3 ,

a 2 1 a 1dp 1d z

a 10 , a 22 !2 a 5 , a 23 a 2 0 a 12 , a 2 4 a 1 2 a 19 a 13 a 2 0 ,a 25 a 12 a 2 1 a 2 0 a 1 4 a 13 a 19 , a 2 6 a 1 2 a 22 a 2 1 a 13 a 14 a 19 a 15 a 2 0 ,

a 2 7 a 13 a 22 a 2 1 a 1 4 a 15 a 19 a 16 a 2 0 , a 28 a 14 a 22 a 2 1 a 15 a 16 a 19 a 17 a 2 0 ,a 2 9 a 15 a 22 a 2 1 a 16 a 17 a 19 a 18 a 2 0 , a 3 0 a 16 a 22 a 18 a 19 a 17 a 2 1 ,

a 31 a18a21 a17 a22 , a32 a18 a22 , a33 132dp 0d z

5

, a34 a18

3a116

dp 0d z

5

,

a3510a12

8dp 0d z

5,

a3610a13

4dp 0d z

5

, a375a14

2dp 0d z

5

a38 a15 dp 0

d z

5

,

a39 3, 1a204

dp 0d z

2, a40 3, 1a19

43, 1a1a20 dp 0

d z

2

,

a41 3, 1a214

3, 1a12a20 3, a1a19 dp 0d z

2

, a42 3, 1a224

3, 1a12 a19 3, a1a21 dp 0d z

2

,

a43 3, 1a22 a1 3, 1a12 a21 dp 0 z

2

, a44 3, 1a12 a22 dp 0 z

2


14/63

a 4 5 , 1 a 23 , a 4 6 , 1 a 2 4 , a 4 7 , 1 a 25 , 2 a 33 a 39 , a 4 8 , 1 a 2 6 , 2 a 3 4 a 4 ,a 4 9 , 1 a 2 7 , 2 a 35 a 4 1 , a 5 , 1 a 28 , 2 a 3 6 a 4 2 , a 5 1 , 1 a 29 , 2 a 3 7 a 4 3 ,a 52 , 1 a 3 , 2 a 38 a 44 , a 53 , 1 a 3 1 , a 5 4 , 1 a 32 , a 55 !a 4 5

10, a 5 6 !a 4 6

8,

a 5 7 !a 4 76 , a 58 !a 4 8

4, a 59 !a 4 9

2, a 60 a 5 1

2, a 61 a 52

4, a 62 a 53

6,

a 63 a 5 48

, a 64 a 55 r 110 a 5 6 r 18 a 5 7 r 1

6 a 58 r 14 a 59 r 12 a 5 ln r 1 a60

r 12a 61r 14

a 62

r 16

a 63

r 18 ,

a65 a55 r 210 a56 r 28 a57 r 26 a58 r 24 a59 r 22 a50 ln r 2 a6 0r 2

2a6 1r 2

4a6 2r 2

6

a6 3r 2

8 ,

a66 a6 4 ! a6 5ln r 2

!ln r 1

, a6 7 !a66 ln r 1

a6 8 r 24 ! r 14

8a1 r 22 ln r 2 ! r 12 ln r 1 ! a12 r 2

2 ! r 12 a2 r 22 ! r 12 ,

a69 2 a3 r 26 ! r 16

8a44

r 24 ! r 14 a5 ln r 2 ! ln r 1 a10 r 2

2 ln r 2 ! r 12 ln r 12 !

r 22 ! r 12

4,

a11 r 22 ! r 122

.


15/63

The expression for pressure rise and frictional forces are givenby

# p0

1dpd z

d z ,

F 0

:0

1

r 12

!dpd z

d z ,

F i

:0

1

r 22

!dpd z d z ,

25

26

27

W here d p is defined in equation ( 21)


16/63

HAM solution

In this section, we have found the HAM solution of Eqs.(15 ) to ( 18 ). For that we choose

w0 ! 1 r 2 a1 lnr a2 p z .

28

as the initial guess. Further, the auxiliary linear operator for

the problem is taken as

w r w 1r r r

w 0

r 29

which satisfy

wr w 0 0 . 3 0


17/63

W e can define following zeroth-order deformation problemsthe

1 qwr

wr

,q w0 r

qw N

wr w

r

,q , 3 1

w r ,q ! 1, at r r 1,w r ,q , at r r 2 ,

C orresponding boundary conditions are

3 2

In Eq. ( 3 0) denote the non-zero auxiliary parameter ,

is the embedding parameter and,

.15

311)],([5

222

242

2

2

22

1

3

12

2

d z d p

r

wr r

wr

wr

wr

wr

wr r

wr

wr

qr w N wr

xx

xx

xx

xx

xx

xx

xx

xx

EEEE

EEEE

33


18/63

Obviously

r , 0 w 0, r , 1 w r , 34Expanding

,in Taylor's series with respect to

embedding parameter q.

r ,q

w0 r n 1

-

wm r qm

, 35

W here

w m 1m !

m

w r ,qq m q 0 3 6

Differentiating the zeroth order deformation m-times withrespect to q and dividing by m! and finally setting q= 0 , we

get the following mth order deformation problem


19/63

w wm r G mwm 1 r w Rwr r , 3 7

.1

1

5

3

11

0001

1

0

2

2

0001

1

0

22

01

1

01

01

1

0111

mii j

j

i jl

l

jl k

k

l k m

m

k

ii j

j

i jl

l

jl k

k

l k m

m

k

l l k

k

l k m

m

k

l l k

k

l k m

m

k mmwr

d z dp

wwwwwr

wwwww

www

wwwr

wr

w R

G EE

EE

EE

EE!

dd

!

d

!

d

!

d

!

ddd

!

d

!

d

!

d

!

ddd

!

d

!

dd

!

d

!

ddd

where

3 8

Gm0 , m 1,

1, m 1.

where

3 9


20/63

The solution of the above equation with the help of Mathematica can be calculated and presented as follows

w r lim M -m 0

M

a m ,00

n 1

2 M 1

m n! 1

2 M

k 1

2 m 1! n

a m ,nk r n ln r

lim M -

n 1

2 M 1

m n! 1

2 M

k 0

2 m 1! n

a m ,nk r 4n 2 ,

a m,00W here and a m ,k are constants

4 0


21/63

Fig .a. Represents h- curve for velocity profile.

H-curve


22/63

Expressions for the five considered wave forms

1) S inusoidal wave

h z 1 sin 2 z

2 ) Triangular wave

h z 1 F @

3n 1

- !1 n 12 n ! 1 si 2 @2 n ! 1 z

3 ) S quare wave

h z 1 F 4@ n 1

-! 1

n 1

2n ! 1cos 2@ 2n ! 1 z

4 1

4 2

43


23/63

4 ) Trape z oidal wave

h z 1 F 32@

2n 1

- sin@8 2n ! 12n 1 2

sin2@ 2n ! 1 z

5 ) M ulti sinusoidal wave

h z 1 sin 2 m z

44

45


24/63

N umerical solution

To see the validity of perturbation and homotopy

analysis method we have also calculated thesolution of governing equations numerically byemploying shooting method. The numerical resultsare also compared with the perturbation and HAMresults. The comparison of different types of solutions is shown through table and figure, wehave found a good agreement between all theresults.


25/63

Table. 1

Table 1. C omparison of solutions by various methods .


26/63

C omparison of velocity profile

0.2 0.4 0.6 0.8 1-1.09

-1.08

-1.07

-1.06

-1.05

-1.04

-1.03

-1.02

-1.01

-1

-0.99

r

w ( r

, z )

Num er i ca l so lu t ion

Pe r tu rba t ion so lu t ion

H A M s o l u t i o n

Fig .a. C omparison of velocity profile for different

solutions


27/63

Velocity profile

0.2 0.4 0.6 0.8 1

-1.09

-1.08

-1.07

-1.06

-1.05

-1.04

-1.03

-1.02

-1.01

-1

-0.99

r

( r , z

)

P1 = 0.4

P1 = 1.2

P1 = 1.6

0.2 0.4 0.6 0.8 1

1.09

1.08

1.07

1.06

1.05

1.04

1.03

1.02

1.01

1

0.99

z

P2 = 1

P2 = 5

P2 = 1 0

8

8 2

Fig. 3 . Velocity field for different values of retardation time

Fig. 3 . Velocity field for different values of relaxation time


28/63

0.2 0.4 0.6 0.8 1-1.09

-1.08

-1.07

-1.06

-1.05

-1.04

-1.03

-1.02

-1.01

-1

-0.99

E = 0.0

E = 0.4

E = 0.8

- 1.5 - 1 - 0.5 0 0.5 1 1.5

-1 5

-1 0

-5

0

5

10

15

Q

( P

E = 0.0

E = 0.1

E = 0.3

E = 0.5

Fig. 3 . Velocity field for different values of perturbation parameter

Fig. 4 . Pressure rise versusflow rate for differentvalues of perturbationparameter ,

,


29/63

-1 5 -1 -0 5 0 0 5 1 1 5 -20

-15

-10

-5

0

5

10

15

20

Q

( P

P1 0 15

P1 0 20

P1 0 25

P1 0 30

-1

5 -1 -0

5 0 0

5 1 1

5

-8

-6

-4

-2

0

2

4

6

8

Q

( P

P 2 0 15

P 2 0 20

P 2 0 25

P 2 0 30

Fig.6. Pressure rise versusflow rate for differentvalues of relaxationtime

Fig. 5 . Pressure rise versusflow rate for differentvalues of retardationtime

2


30/63

-

.

-

-

.

.

.

-

-

-

Q

F ( i )

E = . E = . E = .

E = .

-

.

-

-

.

.

.

-8

-6

-4

-2

2

4

6

8

Q

F ( i )

P

=

.

P

=

.2

P

=

.2

P

=

.

Fig.7. Frictional forces versusflow rate for outer tubefor different values of perturbation parameter

Fig.8. Frictional forces versusflow rate for outer tubefor different values of retardation time

,


31/63

-!

."

-!

-#

."

# #

."

! !

."

-8

-6

-4

-$

#

$

4

6

8

Q

F ( i )

P%

=#

.!

"

P%

=#

.$ #

P%

=#

.$

"

P%

=#

.&

#

-!

."

-!

-#

." # #

."

! !

."

-#

.#

&

-#

.# $

-#

.#

!

#

#

.#

!

#

.# $

#

.#

&

Q

F (

'

)

( =#

.#

( =#

.!

( =#

.&

( =#

."

Fig.9. Frictional forces versusflow rate for outer tubefor different values of relaxation time 2

Fig. 10 . Frictional forces versusflow rate for inner tubefor different values of

perturbation parameter ,


32/63

Fig. 11 . Frictional forces versusflow rate for inner tubefor different values of retardation time

-1.5 -1 -0.5 0 0.5 1 1.5

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Q

F ( 0 )

P1 ) 0.15

P1 ) 0.20

P1 ) 0.25

P1 ) 0.30

-1.5 -1 -0.5 0 0.5 1 1.5

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Q

0 ( 0

)

P2 1 0.15

P2 1 0.20

P2 1 0.25

P2 1 0.30

Fig. 12 . Frictional forces versusflow rate for inner tubefor different values of relaxation time 8 2


33/63

0 0.5 1 1.5 -8

-7.5

-7

-6.5

-6

-5.5

-5

-4.5

-4

-3.5

z

d P / d z

J 2 0.05

J 2 0.07

J 2 0.09

J 2 0.11

0 0.5 1 1.5 -

3

0

-10

0

10

3

0

30

40

z

d P / d z

J 4 0.05

J 4 0.07

J 4 0.09

J 4 0.11

Fig. 13 . Pressure gradient versusz for sinusoidal wavefor different values of amplitude ratio

Fig. 14 . Pressure gradient versusz for square wave for different values of amplitude ratio


34/63

0 0 .5

6 6

.5

-7

0

-6

0

0

10

20

30

40

z

d P / d

z

J = 0 .05

J = 0 .07

J = 0 .09

J = 0 .11

0 0 .5 1 1 .5 2

4

6

8

10

12

14

16

18

z

d P / d z

J = 0 .05

J = 0 .07

J = 0 .09

J = 0 .11

Fig. 15 . Pressure gradient versusz for trape z oidal wavefor different values of

Fig. 16. Pressure gradient versusz for triangular wavefor different values of amplitude ratio


35/63

0 0 .5 1 1 .5 -20

-10

0

10

20

30

40

z

d P / d z

J = 0 .05

J = 0 .07

J = 0 .09

J = 0 .11

Fig. 17. Pressure gradient versusz for multi sinusoidal wavefor different values of amplitude ratio

Fig. 18. S treamlines for sinusoidal wave


36/63

Fig. 2 0. S treamlines for trape z oidal wave

Fig. 19. S treamlines for square wave


37/63

Fig. 2 1. S treamlines for

triangular wave

Fig. 22 . S treamlines for multi sinusoidalwave


38/63


39/63

Peristaltic flow of a non-Newtonian fluidwith variable viscosity in an endoscope


40/63


41/63

T r z

wr 9 r 1 8 18 2 1! d d b ! c2 2d 3b wr

2

1 8 12

1! d d b !c2 2d 3b

wr

2 ,

T rr 8 2 9 r wr

21 d b ! 8 1 w

r 1 d b T r z ,

T zz 8 2 9 r wr

2

!1 d b ! 8 1 wr

!1 d b T r z ,

T 55 8 2 9 r wr

2b! 8 1 w

r bT r z .

where

4U nder the assumption of long wave length approximation andlow Reynolds number above equations take the form

,0!xxr p

p z

1r r

9 r r wr

1r r

,, 1 9 r wr

3

, 2 , 2 9 r w

r

5

5

6


42/63

, 1 ! d d b ! c 2 d 3 b , , 1 8 1 8 2 ! 8 1 , , 2 !8 13 8 2 .where

C orresponding boundary conditions in dimensionless form are

w 1 , at r r 1 ,

w 1, at r r 2 1 F sin2 z.

7

8

S olution of the problem

Perturbation solution

For perturbation solution, we expand w, F, p by takingas perturbation parameter

E

w w 0 , w 1 , 2 w 2 O, 3 ,

w P 0 , P 1 , 2 P 2 O, 3 ,

0 , 1 , 2 2 , 3 ,

9


43/63

The perturbation results for small perturbation parameter satisfying

boundary condition can be written as

w !1 3 r 2 2-r 3 12 a 1 l r -r 12 a 2 112 P

z , a 2 0 r 7

a 2 1 r 6 a 22 r 5 a 23 r 4 a 2 4 r 3 a 25 r 2 a 2 6 r a 2 7 l r a 28r a 29r 2

a 3 1 l r

-r a 33

, 2 a 64 r 19 a 65 r 1 a 66 r 17 a 67 r 16

a 6 r 15 a 69 r 1 4 a 70 r 13 a 71 r 1 2 a 72 r 11 a 73 r 10 a 74 r 9 a 7 5 r 8

a 76 r 7 a 77 r 6 a 7 r 5 a 79 r 4 a 8 0 r 3 a 8 1 r 2 a 8 2 r a 8 3 l r a 8 4

r a 8 5

2

a 8 63

a 8 74

a 885

a 8 96

a 9 07

a 9 18 a 63 .

Velocity profile

10A xial pressure gradient

d P

z

2 F r 22

! r 12

! , a 9 3 ! , 2

a 94

a 9 2, 11


44/63

The expression for pressure rise and frictional forces are givenby

# p0

1 dpd z

d z ,

F 0 :0

1

r 12 ! dpd z d z ,

F i

:0

1

r 22 ! dpd z

d z,

12

13

14

is defined in equation ( 21)dpW here


45/63

N umerical solution

Equations ( 5 ) and (6 ) are also solved numerically by employingshooting method. C omparison of both the perturbation andnumerical solution have been presented by the following table

r Numer ica lsol Perturb a tionsol Err or 0.1 -1 .00000 -1 .00000 0 .000000.2 -1 .04474 -1 .04455 0.000 180.3 -1 .06692 -1 .06655 0.000 340.4 -1 .08 045 -1 .0778 4 0.002420.5 -1 .08 498 -1 .08 21 4 0.002620.6 -1 .08 432 -1 .081 09 0.002980.7 -1 .078 45 -1 .07558 0.00266

0.8 -1 .06907 -1 .0661 3 0.002750.9 -1 .0561 0 -1 .05309 0.0028 51 .0 -1 .03772 -1 .03668 0.00 1 001 .1 -1 .01 773 -1 .01 707 0.0000 61 .18 -1 .00000 -1 .00000 0 .00000

Table 1. C omparison of solutions by various methods .


46/63

0.2 0.4 0.6 0.8 1-1.09

-1.08

-1.07

-1.06

-1.05

-1.04

-1.03

-1.02

-1.01

-1

-0.99

r

w

( r , z

)

Numerical solution

P er turbat ion so lu t ion

C omparison of velocity profile

Fig .a. C omparison of velocity profile for differentsolutions


47/63

-5 - 4 - 3 - 2 -1 0 1 2 3- 15

- 10

- 5

0

5

10

15

20

25

30

35

Q

( P

E 8 0.0

E 8 0.3

E 8 0.6

E 8 0.9

-5 -4 -3 -2 -1 0 1 2 3-15

-10

-5

0

5

10

15

20

25

Q

( P

F 9 0.00

F 9 0.15

F 9 0.30

F 9 0.45

Fig. 2 . Pressure rise versusflow rate for differentvalues of viscosityparameter

Fig. 1. Pressure rise versusflow rate for different

values of perturbationparameter


48/63


49/63

-5 -4 -3 -2 -1 0 1 2 3-15

-10

-5

0

5

10

15

20

25

Q

( P

B

= 0.0 B

= 0.2 B

= 0.3B

= 0.4

-5 -4 -3 -2 -1 0 1 2 3-20

-10

0

10

20

30

40

Q

( P

I = 0.1

I = 0.3

I = 0.4

I = 0.5

Fig.6. Pressure rise versusflow rate for differentvalues radius ratio

Fig. 5 . Pressure rise versusflow rate for different

values of amplituderatio F


50/63

-5 -4 -3 -2 -1 0 1 2 3-6

-5

-4

-3

-2

-1

0

1

2

3

Q

F ( 0 )

E = 0 .0

E = 0 .3

E = 0 .6

E = 0 .9

- 5 - 4 - 3 -2 - 1 0 1 2 3- 6

- 5

- 4

- 3

- 2

- 1

0

1

2

3

Q

F ( 0 )

F = 0 .00

F = 0 .15

F = 0 .30

F = 0 .4 5

Fig.8. Frictional forces for inner tube versusflow rate for differentvalues of viscosityparameter

Fig.7. Frictional forces for inner tube versus

flow rate for differentvalues of perturbationparameter

-

,


51/63

-5 -4 -3 -2 -1 0 1 2 3-8

-6

-4

-2

0

2

4

Q

F ( 0 )

P2 C 0.2

P2 C 0.4

P2 C 0.6

P2 C 0.8

- 5 -4 -3 -2 - 1 0 1 2 3-20

-15

-10

-5

0

5

Q

F ( 0 )

P1 D 1

P1 D 2

P1 D 3

P1 D 4

Fig. 10 . Frictional forces for inner tube versusflow rate for differentvalues of retardationtime

Fig.9. Frictional forces for inner tube versus

flow rate for differentvalues of relaxationtime 8 2


52/63

-5 -4 -3 -2 -1 0 1 2 3-4

-3

-2

-1

0

1

2

Q

F ( 0 )

J E 0.0

J E 0.2

J E 0.3

J E 0.4

-5 -4 -3 -2 -1 0 1 2 3-4

-3

-2

-1

0

1

2

Q

F ( 0 )

F E 0.1F E 0.3F E 0.4F E 0.5

Fig. 12 . Frictional forces for inner tube versusflow rate for differentvalues radius ratio

Fig. 11 . Frictional forces for inner tube versus

flow rate for differentvalues of amplituderatio


53/63


54/63

-5 -4 -3 -2 -1 0 1 2 3-25

-20

-15

-10

-5

0

5

10

15

Q

I

P

Q

R

P2 S 0.2

P2 S 0.4

P2 S 0.6

P2 S 0.8

-5 - 4 -3 - 2 -1 0 1 2 3- 70

- 60

- 50

- 40

- 30

- 20

- 10

0

10

20

Q

T

P

Q

R

P1

U 1

P1 U 2

P1 U 3

P1 U 4

Fig. 16. Frictional forces for outer tube versusflow rate for differentvalues of retardationtime

Fig. 15 . Frictional forces for outer tube versus

flow rate for differentvalues of relaxationtime

8

8 2


55/63

-5 -V

-3 -2 -1 0 1 2 3-20

-15

-10

-5

0

5

10

W

FX

Y

`

J a 0.0

J = 0.2

J = 0.3

J = 0 .V

-5 -b

-3 -2 -1 0 1 2 3-25

-20

-15

-10

-5

0

5

10

c

Fd

e

f

I = 0.1

I = 0.3

I = 0.b

I = 0.5

Fig. 18. Frictional forces for outer tube versusflow rate for differentvalues radius ratio

Fig. 17. Frictional forces for outer tube versus

flow rate for differentvalues of amplituderatio


56/63

0 0.5 1 1.5 4

6

8

10

12

14

16

18

20

22

z

d P / d z

J = 0.20

J = 0.25

J = 0.30

J = 0.35

0 0.5 1 1.5 -140

-120

-100

-80

-60

-40

-20

z

d P / d z

I = 0.20

I = 0.25

I = 0.30

I = 0.35

Fig. 2 0. Pressure gradientversus z for differentvalues radius ratio

Fig. 19. Pressure gradient versusz for different values of

amplitude ratio


57/63

0 0 .5 1 1 .5 3

4

5

6

7

8

9

10

11

12

z

d P

g d z

Q h 0 .50 Q h 0 .55 Q h 0 .60 Q h 0 .65

0 0 .5 1 1 .5 0

20

40

60

80

100

120

140

Q

d p

i

d z

P1 h 0 .0

P1 h 0 .1

P1

h 0 .2

P1 h 0 .3

Fig. 2 1. Pressure gradient versusz for different values of

flow rate Q

Fig. 22 . Pressure gradient versusz for different values of retardation time


58/63

0 0 .5 1 1 .5 0

5

10

15

20

25

Q

d p

p

d z

P 2 q 0 . 0

P 2 q 0 . 1

P 2 q 0 . 2

P 2 q 0 . 3

0 0 .5 1 1 .5 1.4

1.6

1.8

2

2 .2

2 .4

2 .6

2 .8

3

3.2

z

d

r

s

d

z

E t 0 .1

E t 0 .2

E t 0 .3

E t 0 .4

Fig. 24 . Pressure gradientversus z for differentvalues of perturbation

parameter

Fig. 23 . Pressure gradient versusz for different values of

relaxation time 8

,


59/63

-0 .2 -0 .1 0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7

0 .2

0 .4

0 .6

0 .8

1

1 .2 (e )

-0 .2 -0 .1 0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7

0 .2

0 .4

0 .6

0 .8

1

1 .2 (f )

Fig. 25 . S treamlines for differentvalues of perturbationparameter


60/63

-0 .u -0 . v 0 0 . v 0 . u 0 . w 0 . x 0 . y 0 . 0 .

0 .u

0 .x

0 .

0 .

v

v .u i

-0 .u -0 . v 0 0 . v 0 . u 0 . w 0 . x 0 . y 0 . 0 .

0 .u

0 .x

0 .

0 .

v

v .u j

Fig. 2 6. S treamlines for different

values of viscosityparameter -


61/63

-0 . -0 . 0 0 . 0 . 0 . 0 . 0 . 0 . 0 .

0 .

0 .

0 .

0 .

. k

-0 . -0 . 0 0 . 0 . 0 . 0 . 0 . 0 . 0 .

0 .

0 .

0 .

0 .

. l

Fig. 2 7. S treamlines for different

values of relaxationtime 2


62/63

-0 . -0 . 0 0 . 0 . 0 . 0 . 0 . 0 . 0 .

0 .

0 .

0 .

0 .

. g

j

-0 .k -0 . l 0 0 . l 0 . k 0 . m 0 . n 0 . o 0 . 0 .

0 .k

0 .n

0 .

0 .

l

l .k h

j

Fig. 2 8. S treamlines for differentvalues of retardation

time


63/63

dr. sohail presentation.(2010)

Documents