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Peristaltic flow with an endoscope
By
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Geometry of the problem
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
,
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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
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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
.
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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)
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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
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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
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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
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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
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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
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Fig .a. Represents h- curve for velocity profile.
H-curve
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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
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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
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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.
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Table. 1
Table 1. C omparison of solutions by various methods .
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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
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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
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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 ,
,
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-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
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-
.
-
-
.
.
.
-
-
-
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
,
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-!
."
-!
-#
."
# #
."
! !
."
-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 ,
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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
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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
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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
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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
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Fig. 2 0. S treamlines for trape z oidal wave
Fig. 19. S treamlines for square wave
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Fig. 2 1. S treamlines for
triangular wave
Fig. 22 . S treamlines for multi sinusoidalwave
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Peristaltic flow of a non-Newtonian fluidwith variable viscosity in an endoscope
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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
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, 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
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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
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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
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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 .
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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
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-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
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-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
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-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
-
,
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-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
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-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
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-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
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-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
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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
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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
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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
,
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-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
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-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 -
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-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
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-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
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