Hydromagnetic Flow Near A
Non-uniform Accelerating Plate In
The Presence Of Magnetic Field
Through Porous Medium
Amir Khana,b , Gul Zamana
a. Department of Mathematics, University of Malakand, Chakdara,
Dir(Lower), Khyber Pakhtunkhwa, Pakistan
b. Department of Mathematics and Statistics, University of Swat,
Khyber Pakhtunkhwa, Pakistan
Abstract
New exact solutions for unsteady magnetohydrodynamic (MHD)
flows of a generalized second-grade fluid near a non-uniform accelerating plate are derived. The generalized second-grade fluid saturates
the porous space. Fractional derivative is used in the governing equation. The analytical expressions for velocity and shear stress fields are
obtained by using Laplace transform technique for the fractional calculus. The obtained solutions are expressed in series form in terms of
Fox H-functions. Similar solutions for ordinary second-grade fluid passing through a porous space are also recovered. The characteristics of
velocity field and shear stress are analyzed graphically.
Keywords: MHD flow, Generalized second-grade fluid, Fractional
derivatives, Fox H-functions, Discrete Laplace transform.
MSC(2010): 76A05, 76A10
†
Corresponding author. Tel: +92 03339148363
E-mail addresses: [email protected],
1
1
Introduction
Interest and research activities regarding the flows of non-Newtonian fluids
have increased in the last few decades. This is due to their industrial and engineering applications as well as the interesting mathematical challenges offered
by the equations governing the flows. A large class of fluids is non-Newtonian
in which the relation between the deformation rate and shear stress is nonlinear. Since we have no model available which is considered to be a universal constitutive model and can also predict the behavior of all available nonNewtonian fluids. As a result, many constitutive models of non-Newtonian
fluids have been developed. Rivlin and Ericksen [1] introduced a subclass of
non-Newtonian fluids known as second-grade fluids for which a possibility to
obtain exact solutions exist. Exact solutions of second-grade fluids for start-up
flows have been investigated by Bandelli [2] using integral transform technique.
Tan [3] discussed flow of a suddenly moved flat plate in a generalized secondgrade fluid. Exact solutions of a generalized second grade fluid corresponding
to oscillatory flow between two cylinders have been achieved by Mahmood et
al. [4]. Tripathy [5] discussed peristaltic motion of a generalized second grade
fluid passing through a cylindrical tube. Tan [6] obtained solutions for unsteady motion between two parallel plates of a generalized second grade fluid.
In the last few decades the study of fluid motion through porous media has
received much attention due to its importance not only in the academic field
but also in industry. Many applications may be found in industrial and biological processes for example in the food industry, irrigation problems, oil
exploitation, motion of blood in the cardiovascular system [7], chemistry and
bio-engineering, soap and cellulose solutions and in biophysical sciences where
the human lungs are considered as a porous layer. Unsteady MHD flows of
viscoelastic fluids passing through porous space are of considerable interest.
In the last few years much work has been carried out on MHD flows, see [8-12]
and references contained therein.
Recently, the fractional derivative [13] approach has proved to be an important
tool for considering behaviors of such types of fluids. Many researchers investigated different problems using the fractional derivative technique for such
fluids. In their work, integer order time derivatives in the constitutive models for generalized second-grade fluids were replaced by the Riemann-Liouville
fractional derivatives. We mention only those contributions with regard to the
viscoelastic type fluids [13-20] and the references contained therein.
To the authors best knowledge no study has been carried out on MHD flow of
a generalized second-grade fluid near a non-uniform accelerating plate flowing
2
through a porous space. Our main objective in this note is to make a contribution in this field. We consider incompressible MHD flow passing through
a porous space of a generalized second-grade fluid. The Laplace transform
method is used for the fractional calculus to obtain exact solutions for the
profiles of velocity field and the corresponding shear stress. The obtained solutions satisfying all the imposed initial and boundary conditions are expressed
in terms of the Fox-H function [21]. Similar solutions for ordinary second-grade
fluids are obtained as a particular case of the general solution. Finally, the effects of different parameters on the motion are analyzed graphically.
2
Governing equations
The equation of continuity and momentum of MHD flow passing through
porous space is given by [6]
∇ · V = 0;
ρ(
dV
) = divT − σB02 V + R,
dt
(1)
where V=(u,v,w ) represents velocity vector. Electrical conductivity and density of the fluid are represented by σ and ρ respectively, B0 is the magnitude
where uniform magnetic field is applied, material time derivative is denoted by
d/dt, Cauchy stress tensor is represented by T, and R is the Darcy’s resistance
of the porous space.
For an incompressible and unsteady generalized second-grade fluid the Cauchy
stress tensor T is given by [7]:
T = S − pI;
S = µW1 + α1 W2 + α2 W21 ,
(2)
where S and pI represents the extra stress tensor and the indeterminate spherical stress, the dynamic viscosity is denoted by µ, normal stress moduli are
represented by α1 and α2 and the kinematic tensors are W1 and W2 defined
by
W1 = L + L T ,
W2 = Dtβ + W1 L + LT W1
(3)
where L is the velocity gradient and Dtβ represents the operator for fractional
differentiation whose order is β and is based on the Riemann-Liouville definition [13],
Z a
1
d
g(t)
b
Da [g(a)] =
dt,
0≤b<1
(4)
Γ(1 − b) da 0 (a − t)b
3
where the Gamma function is denoted by Γ(·). The model for an ordinary
second-grade fluid can be obtained by putting β = 1. For the compatibility of
this model with thermodynamics it is required that the material moduli should
obey the following conditions
α1 + α2 = 0,
0 ≤ α1 and µ ≥ 0.
(5)
For a second-grade fluid Darcy’s resistance satisfies the following equation:
φ
∂
R = − (1 + α1 )V
κ
∂t
(6)
where κ > 0 and φ(0 < φ < 1) are the permeability and the porosity of the
porous medium. We consider the velocity field and an extra stress of the form
V = (u(y, t), 0, 0), S = S(y, t).
(7)
where u is the velocity taken in the x -direction. Substituting Eq.(7) into Eq.(2)
and taking into account the initial condition
S(y, 0) = 0, y > 0,
(8)
the fluid being at rest up to the time t = 0, we get
Sxy = (µ + α1 Dtβ )∂y u(y, t),
(9)
where Syy = Szz = Sxz = Syz = 0, and Sxy = Syx . The balance of linear
momentum in the absence of body forces is given by:
∂y Sxy − ∂x p − σB02 u −
∂
µφ
(1 + α1 ) = ρ∂t u(y, t),
κ
∂t
∂y p = ∂z p = 0, (10)
where the pressure gradient is taken along x -axis and is denoted by ∂x p. By
substituting Sxy from Eq. (9) into Eq. (10), we obtain the governing equation
under the form
∂
φ
ρ∂t u(y, t) = (µ + α1 Dtβ )∂y2 u(y, t) − ∂x p − σB02 u(y, t) − (µ + α1 )u(y, t). (11)
κ
∂t
3
Statement of the problem
Consider unsteady incompressible flow of homogenous and electrically conducting second-grade fluid bounded by a rigid plate at y= 0. The plate is
4
taken normal to y-axis and the fluid saturates the porous medium y > 0. The
electrically conducting fluid is stressed by a uniform magnetic field B0 parallel
to the y-axis, while the induced magnetic field is neglected by choosing a small
magnetic Reynolds number. Initially, both the plate and the fluid are at rest,
and after time t = 0, it is suddenly set into motion by translating the flat plate
in its plane, with a constant velocity A. The initial and boundary conditions
of velocity field are:
u(y, 0) = 0;
y > 0,
u(0, t) = At2 ;
(12)
t > 0,
u(y, t), ∂y u(y, t) → 0 as y → ∞ and t > 0.
4
Calculation of velocity field
Employing the non-dimensional quantities
u∗ =
τ=
u
,
U
y∗ =
S
,
ρU 2
yU
,
ν
K=
t∗ =
κU 2
,
φν 2
tU 2
,
ν
M2 =
α1 U 2
, A∗
ρν 2
α∗ =
σνB02
,
ρU 2
=
A
U
(13)
∂p ν
P = − ∂x
ρU 2
Thus, the governing equations of dimensionless motion become
∂t u(y, t) = P + (1 + αDtβ )∂y2 u(y, t) −
1
∂
(1 + α )u(y, t) − M 2 u(y, t),
K
∂t
τ (y, t) = (1 + αDtβ )∂y u(y, t)
(14)
(15)
with the given conditions as
u(y, 0) = 0;
u(0, t) = At2 ;
y > 0,
t > 0,
(16)
u(y, t), ∂y u(y, t) → 0 as y → ∞, and t > 0,
where for simplicity the asterisk is omitted. We apply the Laplace transform
to Eq. (14) and use the Laplace transform formula for sequential fractional
5
derivatives [21]
u¯(y, t) =
Z
∞
u(y, t)e−st dt,
o
s ≥ 0.
(17)
Taking into account the corresponding initial and boundary conditions (16),
we obtain the following differential equation
1 + αq
P
q + M2
2
∂y u¯(y, q) −
u¯( y, q) +
+
= 0,
s ≥ 0, (18)
β
β
K(1 + αq ) 1 + αq
q(1 + αq β )
2A
; t > 0,
q3
∂y u¯(y, q) → 0 as y → ∞, and q > 0.
u¯(0, q) =
(19)
The solution of Eq.(18) satisfying the boundary conditions (19) is of the following form:
2A √
C √
C
u¯(y, q) = 3 e− By − e− By +
(20)
q
B
B
where
1
P
B=
((1 + αq) + K(q + M 2 )) and C =
(21)
β
K(1 + αq )
q(1 + αq β )
To obtain the analytical solution for the velocity field and to avoid difficult
calculations of contour integrals and residues, we apply the discrete inverse
Laplace transform method [21], but first we need to express Eq. (20) in series
form as
∞ X
∗ X
∞
X
(−1)ζ+a1 +h1 +1 αa1 +d1 +h1 +r+s M 2c1 +2g1 y e1
u¯(y, q) = P
h1 !g1 !f1 !e1 !d1 !c1 !b1 !a1 !r!s!K −b1 +e1 /2−f1 −1
a =0
h =0
1
1
Γ(b1 + a1 )Γ(c1 − b1 )Γ(d1 + b1 )Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )
×
q −a1 −b1 +c1 −f1 −h1 −d1 −βr−s+1 Γ(−1)Γ(−a1 )Γ(b1 )Γ(−b1 )Γ(e1 /2)
◦ X
∞
X
Γ(r + e1 /2)Γ(s − e1 /2)Γ(a1 + 1)
(−1)ξ+c1 αa1 +d1
×
+P
Γ(−e1 /2)Γ(e1 /2)Γ(f1 )Γ(−f1 )
a1 !b1 !c1 !d1 !
c =0
1
a1 +b1 +d1 −c1 −1
Γ(a1 + 1)Γ(b1 + 1)Γ(c1 − b1 )Γ(d1 + 1)q
K b1 +1 M 2c1
×
Γ(−1)Γ(−1)Γ(b1 )Γ(−b1 )
∞ X
∞ X
∞ X
∞ X
∞ X
∞
X
2A(−1)e1 +f1 +g1 +h1 +r+s αh1 +r+s M 2g1 y e1
+
e !f !g !h !r!s!K e1 /2−f1 q −f1 −h1 −βr−s+3
e =0 f =0 g =0 h =0 r=0 s=0 1 1 1 1
1
1
×
1
1
Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )Γ(r + e1 /2)Γ(s − e1 /2)
. (22)
Γ(e1 /2)Γ(−e1 /2)Γ(e1 /2)Γ(f1 )Γ(−f1 )
6
where
∗
X
∞ X
∞ X
∞ X
∞
∞ X
∞ X
∞ X
∞ X
X
,
=
b1 =0 c1 =0 d1 =0 e1 =0 f1 =0 g1 =0 r=0 s=0
ζ = b1 + c1 + d1 + e1 + f1 + g1 + r + s,
◦
X
=
∞
∞ X
∞ X
X
a1 =0 b1 =0 d1 =0
ξ = a1 + b 1 + d 1
Applying the discrete inverse Laplace transform to Eq. (22), we obtain
∞ X
∗ X
∞
X
(−1)ζ+a1 +h1 +1 Γ(a1 + 1)Γ(b1 + a1 )Γ(c1 − b1 )
u(y, t) = P
a1 !b1 !c1 !d1 !e1 !f1 !g1 !h1 !r!s!Γ(−1)Γ(−a1 )
a =0
h =0
1
×
+P
1
Γ(g1 − f1 )Γ(h1 + f1 )Γ(r + e1 /2)Γ(s − e1 /2)αa1 +d1 +h1 +r+s
Γ(e1 /2)Γ(f1 )Γ(−f1 )Γ(−e1 /2)Γ(e1 /2)K −b1 +e1 /2−f1 −1 M −2c1 −2g1
t−a1 −b1 +c1 −f1 −h1 −d1 −βr−s y e1 Γ(d1 + b1 )Γ(f1 − e1 /2)
×
Γ(b1 )Γ(−b1 )Γ(−a1 − b1 + c1 − f1 − h1 − d1 − βr − s + 1)
◦
∞
XX
(−1)ξ+c1 αa1 +d1 t−a1 −b1 −d1 +c1 K b1 +1 M 2c1 Γ(a1 + 1)Γ(b1 + 1)
c1 =0
a1 !b1 !c1 !d1 !Γ(−1)Γ(−1)Γ(−a1 − b1 − d1 + c1 + 1)
∞
∞
∞
∞
∞ ∞
Γ(c1 − b1 )Γ(d1 + 1) X X X X X X 2A(−1)e1 +f1 +g1 +h1 +r+s
+
×
Γ(b1 )Γ(−b1 )
e1 !f1 !g1 !h1 !r!s!
e =0 f =0 g =0 h =0 r=0 s=0
1
×
1
1
1
Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )Γ(r + e1 /2)Γ(s − e1 /2)M 2g1 y e1
Γ(e1 /2)Γ(−e1 /2)Γ(e1 /2)Γ(f1 )Γ(−f1 )K e1 /2−f1
t−f1 −h1 −βr−s+2 αh1 +r+s
×
. (23)
Γ(−f1 − h1 − βr − s + 3)
7
To write Eq. (23) in a more compact form we use the Fox H-function,
u(y, t) = P
∞ X
∗
X
(−1)ζ+a1 +1 αa1 +d1 +r+s t−a1 −b1 +c1 −f1 −d1 −βr−s y e1
a1 !b1 !c1 !d1 !e1 !f1 !g1 !r!s!K −b1 +e1 /2−f1 −1 M −2c1 −2g1
a =0
1
(−a1 , 0), (−b1 , 0), (1 − b1 − c1 , 0), (1 − d1 − b1 , 0), (1 − f1 + e1 /2, 0),
1,9 α (1 − g1 + f1 , 0), (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0).
×H
(2, 0), (2, 0), (1 − b1 , 0), (1 + b1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0),
9,11 t (0, 1), (1 + e1 /2, 0), (1 − e1 /2, 0), (a1 + b1 − c1 + f1 + d1 + βr + s, −1).
◦
X
(−1)ξ αa1 +d1 t−a1 −b1 −d1 K b1 +1
+P
a1 !b1 !d1 !
1,4 (−a1 , 0), (−b1 , 0), (1 + b1 , 1), (−d1 , 0)
2 M t
×H
(2, 0), (2, 0), (1 − b1 , 0), (1 + b1 , 0), (0, 1), (a1 + b1 + d1 , 1)
4,6
+ 2A
∞ X
∞ X
∞
∞ X
∞ X
X
(−1)e1 +f1 +g1 +r+s M 2g1 y e1 t−f1 −βr−s+2 αr+s
e1 !f1 !g1 !r!s!K e1 /2−f1
e1 =0 f1 =0 g1 =0 r=0 s=0
(1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0), (1 − f1 , 1), (1 − s + e1 /2, 0),
1,5
α (1 − r − e1 /2, 0).
×H
(1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (0, 1), (1 + e1 /2, 0),
5,7 t (1 − e1 /2, 0), (f1 + βr + s − 2, −1).
(24)
To obtain Eq. (24), the following Fox H-function property is used,
X
∞
1,s Γ(a1 + A1 r)...Γ(as + As r) r
(1 − a1 , A1 ), ..., (1 − as , As )
=
−σ H
σ .
(1, 0), (1 − b1 , B1 ), ..., (1 − bt , Bt )
r!Γ(b1 + B1 r)...Γ(bt + Bt r)
s,t+1
r=0
5
Calculation of shear stress
To obtain the shear stress we apply the Laplace transform to Eq. (15), to
obtain
τ¯(y, q) = (1 + αq β )∂y u¯(y, q),
(25)
Substituting u¯(y, q) from Eq. (20), we obtain
√ √
− By
2Ae
B P −1/2 −√By
+ B
e
τ¯(y, t) = −(1 + αq β )
.
q3
q
8
(26)
where
B=
1
((1 + αq) + K(q + M 2 )).
K(1 + αq β )
For a more compact form of τ¯( y, q), we write Eq. (26) in series form as
τ¯(y, q) = P
∞ X
∞
∗ X
∞ X
X
h1 =0 t=0 u=0
(−1)ζ+h1 +t+u
b1 !c1 !d1 !e1 !f1 !g1 !h1 !r!s!t!u!
Γ(b1 + 1/2)Γ(c1 − b1 )Γ(d1 + b1 )Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )
Γ(−1/2)Γ(b1 )Γ(−b1 )Γ(e1 /2)Γ(f1 )Γ(−f1 )Γ(−e1 /2)Γ(e1 /2)
Γ(s − e1 /2)Γ(t − 1/2)Γ(u + 1/2)Γ(r + e1 /2)K b1 −e1 /2+f1 +1 y e1 M +2c1 +2g1
×
Γ(1/2)Γ(−1/2)α−d1 −h1 −r−s−t−u q −b1 +c1 −f1 −h1 −d1 −βr−βt−s−u+1
∗∗ X
∞ X
∞
∞ X
∞ X
∞ X
∞ X
X
2A(−1)e1 +f1 +g1 +h1 +ζ1 +r+s+1
+
e !f !g !h !i !j !k !l !m1 !r!s!
r=0 s=0 1 1 1 1 1 1 1 1
e =0 f =0 g =0 h =0
×
1
1
1
Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )Γ(r + e1 /2)Γ(s − e1 /2)
Γ(e1 /2)Γ(−e1 /2)Γ(e1 /2)Γ(f1 )Γ(−f1 )K e1 /2−f1 −i1 +1/2
M 2g1 +2j1 y e1 Γ(i1 − 1/2)Γ(j1 − i1 )Γ(k1 + i1 )Γ(l1 − 1/2)Γ(m1 − 1/2)
×
(27)
Γ(1/2)Γ(1/2)Γ(1/2)Γ(i1 )Γ(−i1 )q −f1 −h1 −i1 −k1 −βl1 −m1 −βr−s+3
×
α
1
h1 +k1 +l1 +m1 +r+s
where
∗∗
X
=
∞
∞ X
∞ X
∞ X
∞ X
X
,
i1 =0 j1 =0 k1 =0 l1 =0 m1 =0
ζ1 = i1 + j1 + k1 + l1 + m1 .
9
Taking the inverse Laplace of Eq.(27), we obtain
τ (y, t) = P
∞ X
∞
∗ X
∞ X
X
(−1)ζ+h1 +t+u Γ(t − 1/2)Γ(u + 1/2)Γ(r + e1 /2)
h1 =0 t=0 u=0
×
b1 !c1 !d1 !e1 !f1 !g1 !h1 !r!s!t!u!Γ(1/2)Γ(−1/2)
Γ(b1 + 1/2)Γ(c1 − b1 )Γ(d1 + b1 )Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )
Γ(−1/2)Γ(b1 )Γ(−b1 )Γ(e1 /2)Γ(f1 )Γ(−f1 )Γ(−e1 /2)Γ(e1 /2)α−d1 −h1 −r−s−t−u
Γ(s − e1 /2)K b1 −e1 /2+f1 +1 y e1 t−b1 +c1 −f1 −h1 −d1 −βr−βt−s−u+1
×
Γ(−b1 + c1 − f1 − h1 − d1 − βr − βt − s − u + 1)M −2c1 −2g1
∗∗ X
∞ X
∞
∞ X
∞ X
∞ X
∞ X
X
2A(−1)e1 +f1 +g1 +h1 +ζ1 +r+s+1
+
e !f !g !h !i !j !k !l !m1 !r!s!
r=0 s=0 1 1 1 1 1 1 1 1
e =0 f =0 g =0 h =0
1
1
1
Γ(f1 − e1 /2)Γ(g1 − f1 )Γ(h1 + f1 )Γ(r + e1 /2)Γ(s − e1 /2)
Γ(e1 /2)Γ(−e1 /2)Γ(e1 /2)Γ(f1 )Γ(−f1 )K e1 /2−f1 −i1 +1/2
M 2g1 +2j1 y e1 Γ(i1 − 1/2)Γ(j1 − i1 )Γ(k1 + i1 )Γ(l1 − 1/2)Γ(m1 − 1/2)
×
Γ(1/2)Γ(1/2)Γ(1/2)Γ(i1 )Γ(−i1 )Γ(−f1 − h1 − i1 − k1 − βl1 − m1 − βr − s + 3)
× t−f1 −h1 −i1 −k1 −βl1 −m1 −βr−s+2 (28)
×
α
1
h1 +k1 +l1 +m1 +r+s
Finally, using the Fox H-function we obtain the stress field as,
τ (y, t) = P
∗ X
∞ X
∞
X
(−1)ζ+t+u K b1 −e1 /2+f1 +1 y e1 t−b1 +c1 −f1 −d1 −βr−βt−s−u+1
b1 !c1 !d1 !e1 !f1 !g1 !r!s!t!u!M −2c1 −2g1 α−d1 −r−s−t−u
t=0 u=0
(−b1 + 1/2, 0), (1 − c1 + b1 , 0), (1 − d1 − b1 , 0), (1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0),
1,10 α (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0), (−t + 3/2, 0), (1/2 − u, 0).
H
(3/2, 0), (1 − b1 , 0), (1 + b1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (1 + e1 /2, 0),
10,12 t (0, 1), (1 − e1 /2, 0), (1/2, 0), (3/2, 0), (b1 − c1 + f1 + d1 + βr + βt + s + u, −1).
∞ X
∗∗ X
∞ X
∞
∞ X
∞ X
X
2A(−1)e1 +f1 +g1 +ζ1 +r+s+1 y e1 t−f1 −i1 −k1 −βl1 −m1 −βr−s+2
+
e !f !g !i !j !k !l !m1 !r!s!M −2g1 −2j1 α−k1 −l1 −m1 −r−s K e1 /2−f1 −i1 +1/2
r=0 s=0 1 1 1 1 1 1 1
e1 =0 f1 =0 g1 =0
(−i1 + 3/2, 0), (1 − j1 + i1 , 0), (1 − k1 − i1 , 0), (1 − l1 + 1/2, 0), (−m1 + 3/2, 0),
1,10 α (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0), (1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0).
H
(1/2, 0), (1 − i1 , 0), (1 + i1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (1 + e1 /2, 0),
10,12 t (0, 1), (1 − e1 /2, 0), (1/2, 0), (1/2, 0), (f1 + i1 + k1 + βl1 + m1 + βr + s − 2, −1).
(29)
10
6
Limiting cases
By setting β → 1 in Eqs. (24) and (29), we obtain the velocity field and
associated shear stress of an ordinary second-grade fluid.
∞ X
∗
X
(−1)ζ+a1 +1 αa1 +d1 +r+s t−a1 −b1 +c1 −f1 −d1 −r−s y e1
u(y, t) = P
a1 !b1 !c1 !d1 !e1 !f1 !g1 !r!s!K −b1 +e1 /2−f1 −1 M −2c1 −2g1
a1 =0
(−a1 , 0), (−b1 , 0), (1 − b1 − c1 , 0), (1 − d1 − b1 , 0), (1 − f1 + e1 /2, 0),
1,9 α (1 − g1 + f1 , 0), (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0).
×H
(2, 0), (2, 0), (1 − b1 , 0), (1 + b1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0),
9,11 t (0, 1), (1 + e1 /2, 0), (1 − e1 /2, 0), (a1 + b1 − c1 + f1 + d1 + r + s, −1).
◦
X
(−1)ξ αa1 +d1 t−a1 −b1 −d1 K b1 +1
+P
a1 !b1 !d1 !
1,4 (−a1 , 0), (−b1 , 0), (1 + b1 , 1), (−d1 , 0)
2 M t
×H
(2, 0), (2, 0), (1 − b1 , 0), (1 + b1 , 0), (0, 1), (a1 + b1 + d1 , 1)
4,6
+ 2A
∞ X
∞ X
∞ X
∞
∞ X
X
(−1)e1 +f1 +g1 +r+s M 2g1 y e1 t−f1 −r−s+2 αr+s
e1 !f1 !g1 !r!s!K e1 /2−f1
e1 =0 f1 =0 g1 =0 r=0 s=0
(1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0), (1 − f1 , 1), (1 − s + e1 /2, 0),
1,5 α (1 − r − e1 /2, 0).
×H
(1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (0, 1), (1 + e1 /2, 0),
5,7 t (1 − e1 /2, 0), (f1 + r + s − 2, −1).
(30)
11
τ (y, t) = P
∗ X
∞ X
∞
X
(−1)ζ+t+u K b1 −e1 /2+f1 +1 y e1 t−b1 +c1 −f1 −d1 −r−t−s−u+1
t=0 u=0
b1 !c1 !d1 !e1 !f1 !g1 !r!s!t!u!M −2c1 −2g1 α−d1 −r−s−t−u
(−b1 + 1/2, 0), (1 − c1 + b1 , 0), (1 − d1 − b1 , 0), (1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0),
1,10 α (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0), (−t + 3/2, 0), (1/2 − u, 0).
H
(3/2, 0), (1 − b1 , 0), (1 + b1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (1 + e1 /2, 0),
10,12 t (0, 1), (1 − e1 /2, 0), (1/2, 0), (3/2, 0), (b1 − c1 + f1 + d1 + r + t + s + u, −1).
∞ X
∗∗ X
∞ X
∞
∞ X
∞ X
X
2A(−1)e1 +f1 +g1 +ζ1 +r+s+1 y e1 t−f1 −i1 −k1 −l1 −m1 −r−s+2
+
e !f !g !i !j !k !l !m1 !r!s!M −2g1 −2j1 α−k1 −l1 −m1 −r−s K e1 /2−f1 −i1 +1/2
r=0 s=0 1 1 1 1 1 1 1
e1 =0 f1 =0 g1 =0
(−i1 + 3/2, 0), (1 − j1 + i1 , 0), (1 − k1 − i1 , 0), (1 − l1 + 1/2, 0), (−m1 + 3/2, 0),
1,10
α (1 − f1 , 1), (1 − s + e1 /2, 0), (1 − r − e1 /2, 0), (1 − f1 + e1 /2, 0), (1 − g1 + f1 , 0).
H
(1/2, 0), (1 − i1 , 0), (1 + i1 , 0), (1 − e1 /2, 0), (1 − f1 , 0), (1 + f1 , 0), (1 + e1 /2, 0),
10,12 t (0, 1), (1 − e1 /2, 0), (1/2, 0), (1/2, 0), (f1 + i1 + k1 + l1 + m1 + r + s − 2, −1).
(31)
7
Numerical results and discussion
We have presented magnetohydrodynamic (MHD) flows of a generalized secondgrade fluid near a non-uniform accelerating plate. Exact analytical solutions
were established for such flow problem passing through porous medium. The
Laplace transform technique was used to obtain the solution and was expressed
in series form using Fox H-functions. Several graphs were presented for the
analysis of some important physical aspects of the obtained solutions. The numerical results showed profiles of velocity and adequate shear stress for MHD
flow. We analyzed these results by variating different parameters of interest.
In Figure 1 the effect of viscoelastic parameter α on profiles of velocity and
shear stress is shown. We show the profiles of velocity and shear stress for
three different values of α. From these figures it is observed that the profiles
of velocity and shear stress both increase with increasing α. Fig. 2 shows
the variation of the fractional parameter β. The velocity as well as the shear
stress profiles changed their monotonicity by increasing β. Fig. 3 shows the
effect of the permeability K of the porous medium. As expected, the velocity
profiles increase with increasing permeability K of the porous medium. The
consequence is that K reduces the drag force. Similarly, the profile of shear
stress also increases with increasing K. Fig. 4 shows the variation of magnetic
parameter M. Increasing the magnetic parameter M results in the velocity
12
0
0.9
a =1.1
a =1.1
a =1.5
a =1.5
a =1.9
- 0.5
a =1.9
0.8
u(y,4)
t
0.7
-1
- 1.5
0.6
0
1
2
3
0
4
1
2
3
4
y
y
Figure 1: velocity u(y,4) and shear stress τ( y, t) profiles given by Eqs. (24) and (29), K =2,
β = 0.6, t = 4s, M =0.3, P =1.2, A=1 and different values of α.
0
1.4
b =0.1
b =0.1
b =0.5
b =0.5
1.2
b =0.9
- 0.5
b =0.9
1
u(y,4)
t
0.8
-1
0.6
- 1.5
0.4
0
1
2
3
0
4
1
2
3
4
y
y
Figure 2: velocity u(y,4) and shear stress τ( y, t) profiles given by Eqs. (24) and (29), K =2,
α = 1.5, t = 4s, M =0.3, P =1.2, A=1 and different values of β.
13
0
0.9
k=2
k=2.5
k=3
- 0.5
u(y,4)
k=2
k=2.5
k=3
0.8
t
-1
0.7
- 1.5
-2
0.6
0
1
2
3
0
4
1
2
3
4
y
y
Figure 3: velocity u(y,4) and shear stress τ( y, t) profiles given by Eqs. (24) and (29), α =
1.5, β = 0.6, t = 4s, M =0.3, P =1.2, A=1 and different values of K.
0
0.9
M=0.3
M=0.9
M=1.5
- 0.5
M=0.3
M=0.9
M=1.5
0.8
u(y,4)
t
0.7
-1
- 1.5
0.6
0
1
2
3
0
4
1
2
3
4
y
y
Figure 4: velocity u(y,4) and shear stress τ( y, t) profiles given by Eqs. (24) and (29), K =2,
β = 0.6, t = 4s, P = 1.2, A= 1 and different values of M.
14
decreasing. The higher the value of M, the more prominent the reduction in
velocity. The introduction of a transverse magnetic field has a tendency to
develop a drag that resists the flow. Increasing the transverse magnetic field
results in thinning of the boundary layer thickness. Thus by increasing the
magnetic parameter M the permeability K of the porous medium shows the
opposite effect.
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17