HEAT TRANSFER ANALYSIS ON ROTATING FLOW
OF A SECOND-GRADE FLUID PAST A POROUS PLATE
WITH VARIABLE SUCTION
T. HAYAT, ZAHEER ABBAS, AND S. ASGHAR
Received 2 November 2004
We deal with the study of momentum and heat transfer characteristics in a second-grade
rotating flow past a porous plate. The analysis is performed when the suction velocity
normal to the plate, as well as the external flow velocity, varies periodically with time. The
plate is assumed at a higher temperature than the fluid. Analytic solutions for velocity,
skin friction, and temperature are derived. The effects of various parameters of physical
interest on the velocity, skin friction, and temperature are shown and discussed in detail.
1. Introduction
The study of non-Newtonian fluids has attracted much attention, because of their practical applications in engineering and industry particularly in extraction of crude oil from
petroleum products, food processing, and construction engineering. Due to complexity
of fluids, various models have been proposed. The equations of motion of nonNewtonian fluids are highly nonlinear and one order higher than the Navier-Stokes equations. Finding accurate analytic solutions to such equations is not easy. There is a particular class of non-Newtonian fluids namely the second-grade fluids for which one can
reasonably hope to obtain an analytic solution. Important studies of second-grade fluids
in various contexts have been given in the references [1, 3, 6, 7, 9, 10, 11, 12, 13, 17, 19,
20, 21, 22, 24].
Since the pioneering work of Lighthill [16] there has been a considerable amount of
research undertaken on the time-dependent flow problems dealing with the response of
the boundary layer to external unsteady fluctuations about a mean value. Important contributions to the topic with constant and variable suction include the work of Stuart [25],
Messiha [18], Kelley [15], Soundalgekar and Puri [23], and Hayat et al. [8].
Despite the above studies, no attention has been given to the study of the simultaneous effects of the rotation and heat transfer characteristics on the non-Newtonian flow
with variable suction. Such work seems to be important and useful for gaining our basic understanding of such flow and partly for possible applications to geophysical and
astrophysical problems. Also, heat transfer plays an important role during the handling
and processing of non-Newtonian fluids. The understanding of heat transfer in boundary
Copyright © 2005 Hindawi Publishing Corporation
Mathematical Problems in Engineering 2005:5 (2005) 555–582
DOI: 10.1155/MPE.2005.555
556
Rotating second-grade flow with suction
layer flows of non-Newtonian fluids is of importance in many engineering applications
such as the design of thrust bearings and radial diffusers, transpiration cooling, drag reduction, thermal recovery of oils, and so forth. The primary purpose of the present paper
is to make an investigation of the combined effects of rotation, and heat transfer characteristics on the flow of a second-grade fluid past a porous plate with variable suction.
This work is concerned with a boundary value problem in a rotating flow. The analytical solution of the velocity field, skin-friction, and temperature distribution is obtained.
Special attention is given to finding the analytical solutions and to describe the physical
nature. Finally, in order to see the variations of different emerging parameters, the graphs
are sketched and discussed.
2. Mathematical formulation
Let us consider an incompressible second-grade fluid past a porous plate. The plate and
the fluid rotate in unison with an angular velocity Ω about the z -axis normal to the plate.
The plate is located at z = 0 having temperature T0 . The flow far away from the plate is
uniform and temperature of the fluid is T∞ .
For the problem under question, we consider the velocity and temperature fields as
V = u (z ,t ),v (z ,t ),w (z ,t ) ,
T = T(z ,t ),
(2.1)
(2.2)
in which u , v , and w are the velocity components in x , y , and z directions, respectively,
and T indicates the temperature.
The governing equations in absence of body forces and radiant heating are
ρ
div V = 0,
dV
+ 2Ω × V + Ω × (Ω × r) = div T,
dt de
ρ = T · L − div q.
dt
(2.3)
(2.4)
(2.5)
In above equations d/dt , ρ , e, L, and q are, respectively, the material derivative, density, the specific internal energy, the gradient of velocity, the heat flux vector, and the radial distance r 2 = x2 + y 2 . The Cauchy stress T in an incompressible homogeneous fluid
of second grade is of the form
T = − pI + µA1 + α1 A2 + α2 A21 ,
A2 =
(2.6)
A1 = (grad V) + (gradV) ,
(2.7)
dA1
+ A1 (grad V) + (grad V) A1 ,
dt
(2.8)
T. Hayat et al. 557
where µ, − pI, α j ( j = 1,2), A1 , and A2 are, respectively, the dynamic viscosity, spherical stress, normal stress moduli, and first two Rivlin-Ericksen tensors. The thermodynamic analysis of model (2.6) has been discussed in detail by Dunn and Fosdick [4]. The
Clausius-Duhem inequality and the assumption that the Helmholtz free energy is a minimum in equilbrium provide the following restrictions [5]:
µ ≥ 0,
α1 + α2 = 0.
α1 ≥ 0,
(2.9)
It is evident from (2.1) and (2.3) that
∂w
= 0.
∂z
(2.10)
The above equation shows that w is a function of time. Following Messiha [18] and
Soundalgekar and Puri [23] we take
w = −W0 1 + Aeiω t .
(2.11)
In above equation W0 is nonzero constant mean suction velocity, A is real positive
constant, is small such that A ≤ 1, and negative sign indicates that suction velocity
normal to the plate is directed towards the plate. From (2.1), (2.4), (2.6), (2.8), and (2.11)
we get
∂2 u
1 ∂ p
∂3 u
∂u
∂u
− W0 1 + Aeiω t
− 2Ωv = − + υ 2 + α∗ 2 ∂t
∂z
ρ ∂x
∂z
∂z ∂t
3 ∂ u
− α∗ W0 1 + Aeiω t
,
∂z 3
∂2 v 1 ∂ p
∂3 v ∂v
∂v
− W0 1 + Aeiω t
2Ωu = − + υ 2 + α∗ 2 ∂t
∂z
ρ ∂y
∂z
∂z ∂t
3 ∂ v
− α∗ W0 1 + Aeiω t
,
∂z 3
1 ∂ p
∂w
=− ,
∂t ρ ∂z
(2.12)
(2.13)
(2.14)
subject to the boundary conditions
u = v = 0 at z = 0,
u −→ U (t ),
v −→ 0
(2.15)
as z −→ ∞,
(2.16)
where U (t ) is the free stream velocity and modified pressure
∂u
1
p = p − ρ Ω2 r 2 − 2α1 + α2
2
∂z
µ
α
υ= ,
α∗ = 1 .
ρ
ρ
2
+
∂v
∂z
2 ,
(2.17)
558
Rotating second-grade flow with suction
In view of (2.11) and (2.14), ∂ p/∂z is small in the boundary and hence can be ignored
[8, 18, 23]. The modified pressure p is assumed constant along any normal and is given
by its value outside the boundary layer. Equations (2.12) and (2.13) for the free stream
yields
1 ∂ p dU =
,
ρ ∂x
dt 1 ∂ p
− = 2ΩU .
ρ ∂y
−
(2.18)
(2.19)
Making use of (2.18) and (2.19) into (2.12) and (2.13), we have
3 ∂2 u
dU ∂u
iω t ∂u
∗ ∂ u
−
W
1
+
Ae
−
2Ωv
=
+
υ
+
α
0
∂t ∂z
dt ∂z2
∂z2 ∂t 3 ∂ u
− α∗ W0 1 + Aeiω t
,
∂z 3
∂v
∂2 v ∂3 v ∂v
− W0 1 + Aeiω t
2Ωu = 2ΩU + υ 2 + α∗ 2 ∂t
∂z
∂z
∂z ∂t
3 ∂ v
− α∗ W0 1 + Aeiω t
,
∂z 3
(2.20)
(2.21)
where U is periodic free stream velocity given by
U (t ) = U0 1 + eiω t ,
(2.22)
where U0 is the reference velocity.
With the help of (2.22), (2.20), (2.21), and boundary conditions (2.15) become
∂F ∂F
− W0 1 + Aeiω t
+ 2iΩF ∂t
∂z
∂2 F ∂3 F = U0 iω eiιω t + υ 2 + 2iΩU0 1 + eiω t + α∗ 2 ∂z
∂z ∂t
3
∂ F
− α∗ W0 1 + Aeiω t
,
∂z 3
F = 0 at z = 0,
F = Uo 1 + eiω t
as z −→ ∞,
(2.23)
(2.24)
where
F = u + iv .
(2.25)
Introducing the nondimensional variables
η=
z Wo
,
υ
t=
u=
u
,
Uo
Wo2 t ,
4υ
v=
ω=
v
,
Uo
4υω
U
U = ,
2 ,
Wo
Uo
F
F = ,
Uo
(2.26)
T. Hayat et al. 559
the boundary value problem consisting of (2.23) and conditions (2.24) yields
∂F
∂2 F
1 ∂F 1
− 1 + Aeiωt
+ 2iNF = iωeιωt + 2iN 1 + eiωt + 2
4 ∂t
∂η
4
∂η
3
1 ∂3 F
iωt ∂ F
+α
−
1
+
Ae
,
4 ∂η2 ∂t
∂η3
(2.27)
F = 0 at η = 0,
F −→ 1 + eiωt
as η −→ ∞,
(2.28)
where
α=
α∗ Wo2
,
υ2
N=
Ων
.
Wo 2
(2.29)
3. Analytical solution
The solution of (2.27) subject to conditions (2.28) is written as
F(η,t) = f1 (η) + eiωt f2 (η).
(3.1)
Using above equation into (2.27) and separating the harmonic and nonharmonic
terms we obtain
α
d3 f1 d2 f1 df1
−
−
+ iN f1 = iN,
dη3
dη2
dη
d 3 f2
df1
d 3 f1
iαω d2 f2 df2
α 3 − 1+
−
f
=
iN
+
A
−
αA
,
+
iN
1
2
1
dη
4 dη2
dη
dη
dη3
(3.2)
where
N1 = N +
ω
.
4
(3.3)
The corresponding boundary conditions are
f1 = 0 at η = 0,
f1 −→ 1 as η −→ ∞,
f2 = 0 at η = 0,
f2 −→ 1
as η −→ ∞.
(3.4)
560
Rotating second-grade flow with suction
It is worth emphasizing that (3.2) for second-grade fluid are third order (one order
higher than the Navier-Stokes equation). Thus, one needs three conditions for the unique
solution whereas two conditions are prescribed. One possible way to overcome this difficulty is to employ a perturbation analysis (as in Beard and Walters [2], Soundalgekar and
Puri [23], Kaloni [14], and Hayat et al. [8]) and take the solution as follows
f1 = f01 + α f11 + o α2 ,
(3.5)
f2 = f02 + α f12 + o α2 .
Substituting (3.5) into (3.2), (3.4), equating the coefficients of α, and then solving the
corresponding problems we have for f1 and f2
f1 = 1 − (1 + αηL)e−hη ,
f2 = 1 − Se−gη − (1 − S)e−hη + α c5 e−gη − ηMe−gη − ηc3+ c5 e−hη ,
(3.6)
and so from (3.1),


1 − Se−gη − (1 − S)e−hη
F = 1 − (1 + αηL)e−hη + eiωt  −gη
 ,
+α c5 e − ηMe−gη − ηc3+ c5 e−hη
(3.7)
which upon separating real and imaginary parts gives
u = u0 + eiωt u1 = 1 − e−hr η 1 + αηLr coshi η + αηLi sinhi η

−g r η
4Ahr
4Ahi
1−
cosgi η +
singi η
ω
ω


 1−e






4Ah
4Ah


i
r
−
h
η
r


+e
η
−
η
cosh
sinh
i
i


ω
ω


iωt 
,
+ e 

−g r η
+αe
c5r cosgi η + c5i singi η








−
g
η
r


−αηe
M
cosg
η
+
M
sing
η
r
i
i
i




−h r η
ηc3r + c5r coshi η + ηc3i + c5i sinhi η
−αe
(3.8)
T. Hayat et al. 561
v = v0 + eiωt v1 = e−hr η 1 + αηLr sinhi η + αηLi coshi η

e −g r η
1−
4Ahi
4Ahr
singi η −
cosgi η
ω
ω









4Ahr
4Ahi


−
h
η
r


−
e
η
+
η
sinh
cosh
i
i


ω
ω


iωt 
,
+ e 

−g r η
+αe
c5i cosgi η − c5r singi η










−αηe−gr η Mi cosgi η − Mr singi η




−h r η
−αe
ηc3i + c5i coshi η − ηc3r + c5r sinhi η
(3.9)
where
√
1 + 1 + 4iN
,
2
h = hr + ihi =
hr =
a=
1
2
1 + 1 + 16N 2
2
2
r = a +b =
2
1
2
1 + 1 + 16N12
2
b1 =
,
r1 = a21 + b12 = 1 + 16N12 ,
Sr = 1 +
4Ahi
,
ω
Si =
 2
2
,
− 1 + 1 + 16N 2
1 + 1 + 4iN1
g = gr + igi =
,
2
2
1 1
gi =
,
− 1 + 1 + 16N12
2 2
1 + 16N 2 ,
2
1 1 1 1 + 1 + 16N12 ,
+
2 2 2
a1 =
1
b=
,
2
1 1
,
− 1 + 1 + 16N 2
2 2
hi =
gr =
2
1 1 1 1 + 1 + 16N 2 ,
+
2 2 2
1
2
2
,
− 1 + 1 + 16N12
4iAh
,
ω
h3
L = Lr + iLi = √
,
1 + 4iN
S = Sr + iSi = 1 −
4Ahr
,
ω

1
1 r + 1 1/2
r +1
1+
+ − 2N 2 

4
2
2
2


1
,
Lr = 


r

7 r − 1 r + 1 1/2
−
4
2
2
 1 r −1
2

4
1
Li = 
r

1/2 1−
3 r +1
+
2 2
r −1
2
r −1
2

1/2
+
5N

2 
,


g 2 g + iω/4 1 − 4iAh/ω
M = Mr + iMi =
,
1 + 4iN1
562
Rotating second-grade flow with suction


r + 1 1/2 r1
1 r1 + 1 1/2 1 + r1 + 1 1
+

2
8
2
2 

 4


1/2
1/2


r1 + 1
r
+
1
3
1
2


r
−
1
−
N
+
1
1


8
2
2




1/2 1/2  


r
+
1
r
+
1
r
+
1
1
1
1

+

 

2
2
2

 


 

1/2
 
r1 + 1
 Ahi 
3
 +2r1 + r1 − 1
 
 +

 
1
ω
2
2

 

,
Mr = 

 
1/2
r1 

 

r
+
1
1
2


−4N1


2 

1/2 


3
r1 − 1




2N
r
+
1
+
r
+
1


1
1
1




2
2




 Ahr  r − 1 1/2 r − 1 r − 1 1/2 
1
1
1

+


−
+


 ω 


2
2
2



1/2



2 r1 − 1
+4N1
2


1/2
r + 1 1/2
3
1 r1 − 1
+ r1 + 1 1

 2N1 r1 −


4
2
8
2


1/2


r1 − 1 1/2
3
r
−
1


1
2


+
r
−
1
+
N
1
1


8
2
2




1/2




r
−
1
1


8N
+
4N
r
+


1
1
1




2








1/2
1/2

 Ahi  r1 − 1

r
−
1
r
−
1
1
1
2




+
+ 4N1
1  ω +


2
2
2

,
Mi = 


r1 


1/2
r1 − 1


5


+ r1 + 1




2
2






−3N1 r1 − 1 − 2r1












 1/2 


r
+
1
r
+
1
1
1
Ahr −

1+


+
2
2


ω 

1/2


r1 + 1
+4N12
c5 = c5r + ic5i = c1r + c2r + c4r
2
+ i c1i + c2i + c4i ,
4h3 A − (1 − S) ,
ιω
3
4Ahi
Ahr 3
2
c1r = 4 hi − 3hr hi A +
hr − 3h2i hr ,
− 16
ω
ω
3
4Ah
Ah
i
r 3
c1i = −4 hr − 3h2i hr A +
hr − 3h2r hi ,
− 16
ω
ω
4A h3 + L
,
c2 = c2r + ic2i =
iω
4A 3
Li hr − 3h2i hr − Lr h3i − 3h2r hi ,
c2r =
ω
c1 = c1r + ic1i =








T. Hayat et al. 563
4A 3
Lr hr − 3h2i hr + Li h3i − 3h2r hi ,
ω
4AhL
c3 = c3r + ic3i =
,
iω
4A 4A hi L r + hr L i ,
c3i = −
hr L r − hi L i ,
c3r =
ω
ω
16AhL(1 − 2h)
c4 = c4r + ic4i =
,
ω2
16A c4r = 2 1 − 2hr hr Lr − hi Li + 2hi hr Li + hi Lr ,
ω
16A c4i = 2 1 − 2hr hr Li + hi Lr − 2hi hr Lr − hi Li .
ω
c2i = −
(3.10)
The drag Pxz and lateral stress P yz at the plate in nondimensional form can be written,
respectively, as
∂2 u
Px z
∂u α ∂2 u
−
− 4 1 + Aeiωt
,
=
U0 W0 ρ
∂η 4 ∂η∂t
∂η2
P y z
2
∂v α ∂2 v
iωt ∂ v
P yz = =
−
− 4 1 + Ae
.
U0 W0 ρ
∂η 4 ∂η∂t
∂η2
Pxz =
(3.11)
The above equations after using (3.8) and (3.9) give

8Ahi hr
4A gr −
hi g r + hr g i −



ω
ω



2
2


−
α
g
c
−
g
c
+
αA
h
−
h
r 5r
i 5i


i
r




αM
+
α
h
c
−
h
c
αc
−
−
r
r
5r
i
5i
3r




2
2Ahi hr 
g
A
2
iωt 
r
,
Pxz = α hi − hr − hr − αLr + e 
−
iαω
−
h
g
+
h
g
−
i r
r i


4 ω
ω




 


2
8Ah
g
g
4Ah
r
i
r
i
  g − g2 1 −
+

r

  i
ω
ω


+α 




 
12Ahi h2r 4Ah3i
−
+
ω
ω


4A 2
4A gi +
hr g r − hi g i +
hr − h2i


ω
ω






−α gr c5i + gi c5r − 2αAhi hr






−αMi + α hr c5i + hi c5r − αc3i



4A 2

4A P yz = hi − αLi − 2αhi hr + eiωt 
hr g r − hi g i +
hr − h2i 
.
−iαω gi +


ω
ω





g
g
8Ah
4Ah
i i r
r

−2gi gr +
−
gr2 − gi2  




ω
ω


 +α 





12Ahr h2i 4Ah3r
+
−

ω
ω
(3.12)
(3.13)
564
Rotating second-grade flow with suction
The above equations can also be written as
Pxz = α h2i − h2r − hr − αLr + |B | cos(ωt + β),
(3.14)
P yz = hi − αLi − 2αhi hr + B1 cos(ωt + γ),
where

8Ahi hr
4A hi g r + hr g i −
ω
ω
−α gr c5r − gi c5i + αA h2i − h2r
gr −















−αMr + α hr c5r − hi c5i − αc3r






,
B = Br + iBi = 
g
A
2Ah
h
r


i r
−
hi g r + hr g i −
 −iαω



4 ω
ω
 



8Ahr gi gr 
4Ahi


2
2
+
  gi − gr 1 −

 

ω
ω

+α 

 

12Ahi h2r 4Ah3i
−
+
ω

ω

8Ahi hr
4A hi g r + hr g i −
ω
ω
−α gr c5r − gi c5i + αA h2i − h2r
gr −














−αMr + α hr c5r − hi c5i − αc3r
,
Br = 
 



8Ahr gi gr 
4Ahi


2
2
+
  gi − gr 1 −

 

ω
ω

+α 

 

12Ahi h2r 4Ah3i
−
+
ω
Bi = αω

ω
2Ahi hr
gr A −
hi g r + hr g i −
,
4 ω
ω
4A 2
4A gi +
hr g r − hi g i +
h − h2i
ω
ω r
−α gr c5i + gi c5r − 2αAhi hr















−αMi + α hr c5i + hi c5r − αc3i





4A 2

4A ,
2
B1 = B1r + iB1i = 
−iαω gi +

h
g
−
h
g
+
h
−
h
r
r
i
i
i
r


ω
ω





g
g
8Ah
4Ahr 2


i i r
2

−
2g
g
+
−
g
−
g
i
r
i 
r



ω
ω


 +α 






12Ahr h2i 4Ah3r
+
−
β = tan−1
Bi
,
Br
ω
γ = tan−1
ω
B1i
,
B1r
T. Hayat et al. 565


4A 2
4A gi +
hr g r − hi g i +
hr − h2i


ω
ω






α
g
c
+
g
c
2αAh
h
−
−
r
5i
i
5r
i
r






−
αM
+
α
h
c
+
h
c
αc
−
i
r
5i
i
5r
3i

B1r = 
 ,
 


 −2gi gr + 8Ahi gi gr − 4Ahr g 2 − g 2 
i
r
 
ω
ω

+α 



2
3

 
12Ahr hi 4Ahr
+
−
ω
ω
4A 2
4A hr g r − hi g i +
hr − h2i .
B1i = αω gi +
ω
ω
(3.15)
We now proceed to derive the energy equation appropriate for the problem under
consideration. We start with the energy equation (2.5). It follows from (2.5), (2.6), (2.7),
(2.8), and (2.9) and L = grad V that

T·L = µ
2
∂u
∂z
+
2 ∂v
∂z

2 ∂u ∂2 u
∂ u
 ∂z ∂t ∂z + w ∂z2 



+ α
2 v  .
 ∂v ∂2 v ∂
+ +
w
∂z ∂t ∂z
∂z2
(3.16)
Following the thermodynamical considerations given in Dunn and Fosdick [4] for fluids of second grade and representing q by Fourier’s law with a constant thermal
conductivity, k, (2.5) reduces to


2 ∂u ∂2 u
∂ u
+
w
2 2 
2
∂T
∂T
∂u
∂v
∂z2 
 ∂z ∂t ∂z

∂T

,
+
w
k
=
µ
+
+
α
−
ρ c
1
2

 ∂v ∂2 v ∂t ∂z
∂z2
∂z
∂z
∂ v
+ +
w
∂z ∂t ∂z
∂z2
(3.17)
where c is the specific heat. The boundary conditions for the temperature are
T = T0
T −→ T∞
at z = 0,
as z −→ ∞.
(3.18)
Using
θ=
T − T0
,
T∞ − T0
(3.19)
566
Rotating second-grade flow with suction
(3.17) and boundary conditions (3.18) become


∂u ∂2 u ∂v ∂2 v
 ∂η ∂η∂t + ∂η ∂η∂t 



2 2 2u 


∂
∂u
∂2 θ
∂θ
∂u
∂v
P
∂θ
,
− 1 + Aeιωt
− 2 − Pr 1 + Aeiωt
= Ec
+
+P
+ r


2
∂η
∂η
∂η
∂η 4 ∂t
∂η
∂η




2

∂v ∂ v 
ιωt
− 1 + Ae
∂η ∂η2
(3.20)
θ = 0 at η = 0,
(3.21)
θ −→ 1 at η −→ ∞,
in which
Pr =
k∗ U02
,
Ec = T∞ − T0
µc
,
k
(3.22)
αU 2 µ
.
P= 0
k T∞ − T0
(3.23)
θ = θ0 + eiωt θ1 .
(3.24)
We further assume that
Substituting (3.24) into (3.20) and boundary conditions (3.21), and equating the coefficients of the harmonic and nonharmonic term after neglecting the coefficients of 2 ,
we get
dθ0
du1 2
dv1 2
du1 d2 u1 dv1 d2 v1
d2 θ0
+
P
=
−
E
+
+P
+
,
r
c
2
dη
dη
dη
dη
dη dη2
dη dη2
d2 θ1
dθ0
dθ1 Pr
du1 du2 dv1 dv2
+
P
−
=
−
P
A
−
2E
iωθ
+
r
1
r
c
dη2
dη
4
dη
dη dη
dη dη


du1 du2 dv1 dv2
 iω dη dη + dη dη



 
2u
2v


dv
du
d
d
1
2
1
2
−
 ,
+


2
2
dη dη
dη dη



−P
2
2


dv
du
u
v
d
d
1
1
1
1
−A

+

2
2 
dη
dη
dη
dη



 
du2 d2 u1 dv2 d2 v1 
−
+
dη dη2
dη dη2
θ0 = 0 at η = 0,
θ0 −→ 1
at η −→ ∞,
θ1 = 0 at η = 0,
θ1 −→ 0
at η −→ ∞.
(3.25)
(3.26)
(3.27)
T. Hayat et al. 567
Solving (3.25) and (3.26) along with the boundary conditions (3.27), we obtain
θ0 = 1 − 1 + d7 e−Pr η + d7 + d8 η e−2hr η ,
(3.28)
θ1 = −m16 e− f η + m7 + m9 + m8 η e−2hr η
+ m10 + m14 + m12 η e−(hr +gr )η cos hi − gi η
(3.29)
+ m11 + m15 + m13 η e−(hr +gr )η sin hi − gi η,
where
d1 = −Ec h2r + h2i − 2αLi hi ,
d2 = −2Ec αLr h2r + h2i ,
d3 = P − h3r + 3αLr h2r − hr h2i + 2αLi hi hr + αLr h2i ,
d4 = −2PαLr hr h2r + h2i ,
d7 =
d5
2
4hr − 2Pr hr
d5 = d1 + d3 ,
d6 4hr − Pr
2 ,
4h2r − 2Pr hr
−
d6 = d2 + d4 ,
d
,
d8 = 2 6
4hr − 2Pr hr


4Ahr 4Ahi 2
−
h
g
+
h
+
1
−
h
g
−
h
g
r
r
i
r
r
i
i


ω
ω




 +α h g c + g c − h g c − g c

r r 5i
i 5r
i r 5r
i 5i





d9 = 
4αAhr ,

 +α hr Mi − hi Mr +

g
L
−
g
L
r
r
i
i


ω






4Ahi +α 1 −
gi Lr − gr Li

ω

4Ahi Lr gr + Li gi − αhr 1 −
Lr gi − Li gr 
−
ω
ω






−α hr gr Mi + gi Mr − hi gr Mr − gi Mi
,
d10 = 




 4αAhr hi 
4Ahi gi Lr − gr Li + αhi 1 −
gr Lr − gi Li
−
4αAh2r ω
ω
ω

4Ahr hi 4Ahi −
h
−
g
+
α
1
−
L
g
−
L
g
r
r
r
r
i
i


ω
ω






−
α
h
g
c
−
g
c
h
g
c
+
g
c
−
r r 5r
i 5i
i r 5i
i 5r


,
d11 = 


 4αAhr 
4Ah
i +
gi Lr − gr Li + 1 −
g r hr + g i hi 


ω
ω


−α Mr hr + Mi hi


4αAh2r 4Ahi Lr gi − Li gr + αhr 1 −
Lr g r + Li g i 
− ω
ω





,
d12 = 
−α hr gr Mr + gi Mi − hi gr Mi + gi Mr



 4αAhr hi 
4Ahi
gr Lr − gi Li + αhi 1 −
gi Lr − gr Li
−

ω
568
Rotating second-grade flow with suction

−8Ahr hi hr − αLr + αc5r h2r + h2i



ω


,
d13 = 
 4Ahi 
4αAL
i
2
2
2
2
+
hr − hi −
hr − hi − α hr c3r + hi c3i
ω
ω


4αA 2
−8Ahr hi 2
h
L
−
h
L
+
h
−
h
h
L
−
h
L
r
r
i
i
r
i
i
r
i
r


ω
ω
d14 = 
,
2
2
+αc3r hr + hi


4Ahr 2
4Ahi 2
hr − αLr 
 ω gr − gi hr − αLr + 2gr gi 1 − ω






−
2α
h
g
M
−
g
M
h
g
M
−
g
M
−
r
r
i
i
r
i
r
r
i
i




d15 = 






−α gr2 − gi2 hr c5i − hi c5r
4Ahi 2
gr − gi2 hi − αLi
− 1−
ω
8Ahr gr gi hi − αLi
−2αgr gi hr c5r + hi c5i +


,






ω

4αAhr 2
2
g
−
g
h
L
−
h
L
+
2αg
g
h
M
+
g
M
r r
i i
r i r r
i i 

i
r
 ω



 
g
g
8αAh
r
r
i
+α g 2 − g 2 h M − h M +
hr L i + hi L r 
r i
i r


i
r
ω


,
d16 = 


4Ah
i
2 − g2 h L + h L


−
α
1
−
g
r
i
i
r
i
r


ω






4Ahi +2αgr gi 1 −
hr L r − hi L i
ω


4Ahr 2
4Ahi gr − gi2 hi − αLi + 2gr gi 1 −
hi − αLi 
−
ω
ω






+2α
h
g
M
−
g
M
+
h
g
M
−
g
M
r
r
r
i
i
i
r
i
i
r




,
d17 = 
2
8Ahr gr gi 2


+α
g
−
g
h
c
+
h
c
+
h
+
αL
r 5r
i 5i
r
r
i
r


ω


 


4Ahi 2
2
− 1−
gr − gi hr − αLr − 2αgr gi hr c5i − hi c5r
ω


8αAhr gr gi 4αAhr 2
gr − gi2 hr Li + hi Lr +
hr L r − hi L i 
−
ω
ω






4Ah
i


−
2αg
g
1
−
h
L
+
h
L
r i
r i
i r


ω
,
d18 = 


2
 +2αgr gi hr Mi − hi Mr − α g − g 2 hr Mr + hi Mi

i
r






4Ahi 2
2
g r − g i hr L r − hi L i
−α 1 −
ω


3
8Ahr hi 2 2 3 2h c
h
+
h
α
h
c
+
h
c
+
αh
−
5r
5i
i
5i
i
i
r
r
r
 ω





4αA 
12αAh
h
r
i
d19 = 
3
3 ,
−
h
L
−
h
L
+
L
h
−
L
h
r r
i i
r i
i r 

ω
ω


+2αc3r h2r + h2i + αhr hi hr c5i − hi c5r

T. Hayat et al. 569


8αAhr hi Lr 2 2 4αALi 4
hr + hi −
hr − h4i 

ω
d20 =  ω
,
−α h3r c3r + h3i c3i + αhr hi hr c3i − hi c3r
 
4Ahi 2
2
2
hr − hi − αhr gr c5i + gi c5r + 2αhr hi Mr 
gi 1 −
ω






4Ah
i
 +2αhr hi gr c5r − gi c5i − 2hr gr 1 −

h
−
αL
i
i


ω




2
 4Ahr gr 
8αAh
g
L
r
r
r
2
2
2
,
d21 = 
hr − hi + αhi gr c5i + gi c5r −
 + ω

ω




2g 

8Ah
4Ahi
r i
 +

h
−
αL
2αL
1
−
h
g
−
h
g
−
i
i
r
r
i
i
r


ω
ω






8αAhr hi 4Ahi
−
Lr gi − Li gr + 2αLi hi gi 1 −
ω
ω


α gr Mi + gi Mr h2r − h2i


8αAhr 2


2


−
h
−
h
L
g
+
L
g
r
r
i
i
i
r


ω




8αAhr hi


+2αhr hi gr Mr − gi Mi +
L
g
− Li gr 
r
i
,
d22 = 
ω




2
4Ahi 

2
+α
1
−
L
g
−
L
g
h
g
−
h


r i
i r
i
i
r


ω




4Ahi −2αhr hi 1 −
Lr g r + Li g i
ω


4Ahi 2
hr − h2i + αgr c5r h2r − h2i 
−gr 1 − ω



2
8Ah2r gr 


2
 +αgi c5i hr + hi −

hi − αLi


ω




4Ahr gi 2


2


+
h
−
h
+
2αh
h
M
r i i
i
r


ω



,
d23 = 
+2αhr hi gr c5i + gi c5r





4Ah
i


−2hr gi 1 −
hi + αLi


ω




2g L


8αAh
h
8αAh
i
r
r
i
r
 −

L
g
+
L
h
+
r r
i i


ω
ω




4Ahi +2αhi 1 −
Lr gi − Li gr
ω


4Ahi 2
−α 1 −
hr − h2i Lr gr + Li gi


ω


 2
8αAhr hi 


2
2
Lr g r + Li g i 
−α hr − hi Mr gr − Mi gi −


ω


,
2
4αAh
r
d24 = 
2


+
L
g
−
L
g
h
−
h
r i
i r
i
r


ω 



4Ahi


−2αhr hi 1 −
Lr gi − Li gr




ω
−2αhr hi Mi gr + Mr gi
570
Rotating second-grade flow with suction


3
2
8αAhr Li 2
3 2
2
−
α
h
c
+
h
c
+
αc
h
−
h
+
h
−
h
5r
5i
3r

i
i 
i
r
r
r
ω


,
d25 = 
3


8αAhi Lr
+2αhr hi c3i − αhr hi hr c5i + hi c5r −
ω


4αALi 4 4 8αAh2i Li h2r
−
h
+
h
−


i
r
ω
ω
d26 = 
,
3
3 −α hr c3r + hi c3i − αhr hi hr c3i + hi c3r
d27 = − h3r + 3αh2r Lr − hr h2i + 2αhr hi Li + αLr h2i ,
d28 = −2αLr hr h2r + h2i ,
−Pr A Pr 1 + d7 − 2hr d7 + d8 + PAd27
,
m1 = m1r + im1i =
− 2Ec + iωP d13 + P d19 + d25
2Pr Ahr d8 − 2E
+ iωP d14
c
,
m2 = m2r + im2i =
+PAd28 + P d20 + d26
m3 = m3r + im3i = − 2Ec + iωP d9 + P d15 + d21 ,
m4 = m4r + im4i = − 2Ec + iωP d10 + P d16 + d22 ,
m5 = m5r + im5i = − 2Ec + iωP d11 + P d17 + d23 ,
m6 = m6r + im6i = − 2Ec + iωP d12 + P d18 + d24 ,
m7r =
m7 = m7r + im7i ,
m1r 4h2r − 2hr Pr − m1i ωPr /4
m1i 4h2 − 2hr Pr + m1r ωPr /4
m7i = r
2 2 ,
2 2 ,
4h2r − 2hr Pr + ωPr /4
4h2r − 2hr Pr + ωPr /4
m8 = m8r + im8i ,
m2r 4h2 − 2hr Pr − m2i ωPr /4
m8r = r
2 2 ,
4h2r − 2hr Pr + ωPr /4
m2i 4h2 − 2hr Pr + m2r ωPr /4
m8i = r
2 2 ,
4h2r − 2hr Pr + ωPr /4


2
2
ωPr 2
4h
−
2h
P
+


r r
r


4
,
n1 = n1r + in1i = 

2
ωPr 
−2i 4hr − 2hr Pr
n1r =
m9r =
2
4h2r − 2hr Pr +
4
2
ωPr
ωPr
,
n1i = −2 4h2r − 2hr Pr
,
4
4
m9 = m9r + im9i ,
4hr − Pr n1r m2r − m2i n1i
,
n21r + n21i

m9i =

2
− hi − g i − P r hr + g r


n2 = n2r + in2i =  ωP  ,
+i − 2 hr + gr hi − gi + Pr hi − gi − r
n2r =
hr + g r
2
hr + g r
n2i = −2 hr + gr
2
− hi − g i
2
− P r hr + g r
,
ωPr
hi − g i + P r hi − g i −
,
4
4hr − Pr n1i m2r + m2i n1r
,
n21r + n21i
4
T. Hayat et al. 571
m10 = m10r + im10i ,
n2r m5r − m3r n2i
,
n22r + n22i
m10r =
n2r m5i − m3i n2i
,
m11 = m11r + im11i ,
n22r + n22i
n2i m5r + m3r n2r
n2i m5i + m3i n2r
,
m
=
,
m11r =
11i
n22r + n22i
n22r + n22i
n2r m6r − m4r n2i
m12r =
,
m12 = m12r + im12i ,
n22r + n22i
n2r m6i − m4i n2i
,
m13 = m13r + im13i ,
m12i =
n22r + n22i
n2i m6r + m6r n2r
n2i m6i + m6i n2r
,
m13i =
,
m13r =
n22r + n22i
n22r + n22i
n3 = n3r + in3i = 2 hr + gr − Pr + i − 2 hi − gi ,
m10i =
n3r = 2 hr + gr − Pr ,

n3i = −2 hi − gi ,
n4 = n4r + in4i ,

4 4
hr + g r + hi − g i +
hi − g i


2






2 2 ω2 Pr2


−
6
h
+
g
h
−
g
+
r
r
i
i

,
n4r = 
16



+P 2 h + g 2 − h − g 2 − ωP h + g h − g 
 r
r
r
i
i
r r
r
i
i 


3
2 +2Pr − hr + gr + 3 hr + gr hi − gi


3 3
−4 hr + gr hi − gi + 4 hr + gr hi − gi



2 3 ωPr2 

+2Pr 3 hr + gr hi − gi − hi − gi −
hr + g r 
,
n4i = 


2




ωP
2
2
r
2
−2Pr hr + gr hi − gi −
hr + g r − hi − g i
ωPr2 2
m14 = m14r + im14i ,
m6r n4r n3r + n4i n3i + m4r n4r n3i − n4i n3r
,
n24r + n24i
m6i n4r n3r + n4i n3i + m4i n4r n3i − n4i n3r
,
m14i =
n24r + n24i
m15 = m15r + im15i ,
m14r =
m15r =
−m6r n4r n3i − n4i n3r + m4r n4r n3r + n4i n3i
,
n24r + n24i
−m6i n4r n3i − n4i n3r + m4i n4r n3r + n4i n3i
,
m14i =
n24r + n24i
m16 = m16r + im16i = m7 + m9 + m10 + m14 ,
m16r = m7r + m9r + m10r + m146r ,
f = fr + i fi =
m16i = m7i + m9i + m10i + m146i ,
Pr + Pr2 + iωPr
2
,
572
Rotating second-grade flow with suction
2
4
2 2
P r a2 P r 1 P r + P r + ω P r
fr = + = +
2
2
2 2
2
a2 =
2
4
2 2
b
1 −Pr + Pr + ω Pr
fi = 2 =
2
2
2
Pr2 + Pr4 + ω2 Pr2
1/2
,
2
b2 =
1/2
,
1/2
,
1/2
−Pr2 + Pr4 + ω2 Pr2
2
,
r2 = a22 + b22 = Pr4 + ω2 Pr2 .
(3.30)
From (3.24), (3.28), and (3.29), we can write
θ = θ0 + θ1r cosωt − θ1i sinωt ,
(3.31)
in which
θ1r = −e− fr η m16r cos fi η − m16i sin fi η + m7r + m9r + m8r η e−2hr η
+
m + m14r + m12r η cos hi − gi η −(hr +gr )η
10r
e
,
+ m11r + m15r + m13r η sin hi − gi η
−f η
θ1i = −e
r
(3.32)
m16r sin fi η + m16i cos fi η + m7i + m9i + m8i η e−2hr η
m
+
m
+
m
η
cos
h
− gi η
10i
14i
12i
i
e−(hr +gr )η .
+ + m11i + m15i + m13i η sin hi − gi η
(3.33)
4. Discussion of results
In this paper, we consider the problem of heat transfer in rotating flow of an incompressible fluid of second grade. A perturbation procedure has been used to obtain the analytic
solution. The effects of various parameters such as Ω, Pr , and Ec on the real and imaginary parts of velocity (u,v) and temperature (θr ,θi ) distributions are studied and the
results have been presented by several graphs.
To study the effect of Ω on the velocity components, we have plotted u and v against
η in Figures 4.1, and 4.2 for Newtonian and second-grade fluids. From Figure 4.1(a), it
is observed that near the plate u increases with the increase of Ω. Figure 4.1(b) indicates
that u increases very near to the plate and then fluctuates through an increase in Ω. The
comparison of these two figures reveals that u in case of second-grade fluid is greater than
that of Newtonian fluid. Also, the velocity boundary layer thickness for second-grade
fluid is larger than the Newtonian fluid. It is also seen from Figures 4.2(a) and 4.2(b)
that v increases near the plate and then decreases for large value of Ω. The fluctuations in
second-grade fluid are more visible than that of Newtonian fluid. Also, the value of v for
second-grade fluid is smaller than in the case of Newtonian fluid.
Figures 4.3 and 4.4 show the effect of Ω on the real (θr ) and imaginary (θi ) parts of
temperature distributions. Figure 4.3(a) shows that with the increase of Ω, θr decreases
near the wall. As shown in Figure 4.3(b), we can see that as Ω increases, θr increases near
T. Hayat et al. 573
1
0.8
u
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
2
2.5
η
Ω = 0.5
Ω=1
Ω = 1.5
Ω=2
Ω = 2.5
(a)
1.5
1.25
1
u
0.75
0.5
0.25
0
0
0.5
1
1.5
η
Ω=2
Ω = 2.5
Ω = 0.5
Ω=1
Ω = 1.5
(b)
Figure 4.1. Effect of Ω on real part of velocity profile u versus η. In (a) for Newtonian fluid at α = 0,
ωt = π/2, A = 0.2, = ω = 0.5, W0 = −0.5, ν = 0.1. In (b) for second-grade fluid at α = 0.4.
574
Rotating second-grade flow with suction
0.3
0.25
v
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
2
2.5
2
2.5
η
Ω = 0.5
Ω=1
Ω = 1.5
Ω=2
Ω = 2.5
(a)
0.2
0.15
v
0.1
0.05
0
0
0.5
1
1.5
η
Ω = 0.5
Ω=1
Ω = 1.5
Ω=2
Ω = 2.5
(b)
Figure 4.2. Effect of Ω on imaginary part of velocity profile v versus η. In (a) for Newtonian fluid at
α = 0, ωt = π/2, A = 0.2, = ω = 0.5, W0 = −0.5, ν = 0.1. In (b) for second-grade fluid at α = 0.1.
T. Hayat et al. 575
5
4
θr
3
2
1
0
−1
0
0.5
1
1.5
η
Ω=1
Ω=2
Ω=3
2
2.5
3
Ω=4
Ω=5
(a)
12
10
8
θr
6
4
2
0
0
1
2
η
Ω=1
Ω=2
Ω=3
3
4
Ω=4
Ω=5
(b)
Figure 4.3. Effect of Ω on real part of temperature profile θr versus η. In (a) for Newtonian fluid at
α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Pr = 1.5, Ec = 5.0, k = 0.2, P = 0.3. In (b) for
second-grade fluid at α = 0.04.
576
Rotating second-grade flow with suction
4
3
2
θi
1
0
−1
−2
0
1
2
η
Ω=1
Ω=3
Ω=5
3
4
3
4
Ω=7
Ω = 10
(a)
12
10
8
θi
6
4
2
0
0
1
2
η
Ω = 0.5
Ω=1
Ω = 1.5
Ω=2
Ω = 2.5
(b)
Figure 4.4. Effect of Ω on imaginary part of temperature profile θi versus η. In (a) for Newtonian
fluid at α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Pr = 1.5, Ec = 5.0, k = 0.2, P = 0.3. In
(b) for second-grade fluid at α = 0.05.
T. Hayat et al. 577
6
4
θr
2
0
−2
0
0.5
1
1.5
η
Pr = 1
Pr = 2
Pr = 3
2
2.5
3
2.5
3
Pr = 4
Pr = 5
(a)
8
6
θr
4
2
0
−2
0
0.5
1
1.5
η
Pr = 1
Pr = 2
Pr = 3
2
Pr = 4
Pr = 5
(b)
Figure 4.5. Effect of Pr on real part of temperature profile θr versus η. In (a) for Newtonian fluid at
α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Ω = 3.0, Ec = 5.0, k = 0.2, P = 0.3. In (b) for
second-grade fluid at α = 0.05.
578
Rotating second-grade flow with suction
10
7.5
5
θi
2.5
0
−2.5
−5
−7.5
0
0.5
1
1.5
2
2.5
3
η
Pr = 1
Pr = 2
Pr = 4
Pr = 6
Pr = 8
(a)
10
7.5
θi
5
2.5
0
−2.5
−5
−7.5
0
1
2
η
Pr = 1
Pr = 2
Pr = 4
3
4
Pr = 6
Pr = 8
(b)
Figure 4.6. Effect of Pr on imaginary part of temperature profile θi versus η. In (a) for Newtonian
fluid at α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Ω = 2.5, Ec = 5.0, k = 0.2, P = 0.3. In
(b) for second-grade fluid at α = 0.04.
T. Hayat et al. 579
8
6
θr
4
2
0
−2
−4
0
0.5
1
1.5
η
Ec = 2
Ec = 4
Ec = 8
2
2.5
3
2.5
3
Ec = 10
Ec = 15
(a)
1
0
−1
θr
−2
−3
−4
−5
0
0.5
1
1.5
η
Ec = 1
Ec = 2
Ec = 3
2
Ec = 4
Ec = 5
(b)
Figure 4.7. Effect of Ec on real part of temperature profile θr versus η. In (a) for Newtonian fluid at
α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Ω = 4.0, Pr = 5.0, k = 0.2, P = 0.3. In (b) for
second-grade fluid at α = 0.04.
580
Rotating second-grade flow with suction
10
8
θi
6
4
2
0
0
0.5
1
1.5
2
2.5
2
2.5
η
Ec = 2
Ec = 4
Ec = 6
Ec = 8
Ec = 10
(a)
2
θi
0
−2
−4
0
0.5
1
1.5
η
Ec = 2
Ec = 4
Ec = 6
Ec = 8
Ec = 10
(b)
Figure 4.8. Effect of Ec on imaginary part of temperature profile θi versus η. In (a) for Newtonian
fluid at α = 0, ωt = π/2, A = = ω = 0.5, W0 = −0.5, ν = 0.1, Ω = 2.5, Pr = 5.0, k = 0.2, P = 0.3. In
(b) for second-grade fluid at α = 0.05.
T. Hayat et al. 581
the plate, and then at a distance of η = 1, the θr begins to decrease. That is, the behavior of
θr is quite opposite for Newtonian and second-grade fluids near the plate. Figure 4.4(a)
shows the variation of Ω on θi . It can be seen that as Ω increases, the value of θi decreases
at a distance of approximately η = 0.8 and then increases. Figure 4.4(b) indicates that θi
increases near the wall for Ω > 1.
In order to illustrate the variation of Pr on θr and θi , we have prepared Figures 4.5
and 4.6. Figures 4.5(a) and 4.6(a) explain the effect of Pr on θr and θi , respectively, for
Newtonian fluid case. From these figures, it is revealed that near the plate, θr decreases
and θi increases for Pr > 2. The thermal boundary layer thickness in θr increases where as
for θi decreases. For second-grade fluid, we note that from Figures 4.5(b) and 4.6(b) that
for Pr > 2, θr decreases near the wall and increases far away. Also θi decreases for Pr > 2.
Figures 4.7 and 4.8 show the effect of Ec on θr and θi . From Figures 4.7(a) and 4.7(b),
we observe that θr decreases near the wall with the increase in Ec and increases far away.
The thermal boundary layer thickness increases for large Ec . Moreover, it can be seen
from Figure 4.8(a) that θi increases for large values of Ec . From Figure 4.8(b), it can be
seen that with the increase in the values of Ec , the temperature θi decreases near the plate
and increases far away. The thermal boundary layer thicknesses in both fluids increases.
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T. Hayat: Department of Mathematics, Quaid-i-Azam University, 45320 Islamabad, Pakistan
E-mail address: t pensy@hotmail.com
Zaheer Abbas: Department of Mathematics, Quaid-i-Azam University, 45320 Islamabad, Pakistan
E-mail address: za qau@yahoo.com
S. Asghar: Department of Mathematics, Quaid-i-Azam University, 45320 Islamabad, Pakistan
Current address: Department of Mathematical Sciences, COMSATS Institute of Information Technology, Plot no. 30, Sector H-8, Islamabad, Pakistan
E-mail address: s asgharpk@yahoo.com