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Nuclear Engineering and Design 243 (2012) 20– 32
Contents lists available at SciVerse ScienceDirect
Nuclear Engineering and Design
jo ur n al homep age : www.elsev ier .com/ locate /nucengdes
n exact solutions of Stokes second problem for MHD Oldroyd-B fluid
asood Khan, Misbah Arshad, Asia Anjum ∗
epartment of Mathematics, Quaid-i-Azam University, Islamabad 44000, Pakistan
r t i c l e i n f o
rticle history:eceived 9 October 2011ccepted 21 November 2011
a b s t r a c t
The present study is an attempt to look at the exact solutions of Stokes second problem. Constitutiveequations for an Oldroyd-B fluid have been taken into consideration. The fluid is electrically conductingunder the influence of a uniform transverse magnetic field. The hydromagnetic flow is generated by theoscillations of an infinite flat plate. Employing the Laplace transform, the expressions for the velocity fieldand the associated tangential stress are obtained. These solutions are presented as a sum of steady stateand transient solutions. The results are graphically displayed and the influence of various parameters isdiscussed. Further, the results under the limiting conditions are found to be in good agreement with theexisting ones.
© 2011 Elsevier B.V. All rights reserved.
. Introduction
Many fluids of industrial importance (e.g., polymer melts, suspensions, liquid crystals or biological fluids) exhibit viscoelastic charac-eristics such as shear-thinning, shear-thickening, stress relaxation, normal stress differences, yield stress and so forth. These propertiesead to nonlinear viscoelastic behaviour that cannot be simply described by the classical Navier–Stokes theory. Such fluids are known ason-Newtonian fluids. Amongst many models which have been used to describe the non-Newtonian behaviours, the fluids of differentialype have received special attention as well as much controversy (Dunn and Rajagopal, 1995). These fluids cannot predict the stress relax-tion phenomena exhibited by many polymeric liquids. Amongst non-Newtonian models which are capable of describing such phenomenare rate type models. Some interesting studies regarding these fluids are presented in (Rajagopal, 1992; Rajagopal and Bhatnagar, 1995;etecau and Fetecau, 2003; Fetecau et al., 2007, 2009; Chen et al., 2004; Hayat et al., 2004; Qi and Xu, 2007; Khan et al., 2009) and theeferences therein. One of the popular rate type models to account for stress relaxation phenomenon is the Oldroyd model (Oldroyd, 1950).n this paper, we are interested to discuss Stokes second problem for an Oldroyd-B fluid which encounters both the elasticity and memoryffects exhibited by many polymeric liquids.
The above studies have not addressed the important features of hydromagnetic flow. Considerable practical applications of such flowsnclude plasma aerodynamics, MHD energy systems, nuclear engineering control, mechanical engineering manufacturing process and soorth. Some of the recent studies on the topic include those of Hayat et al. (2008, 2008), Khan et al. (2009, 2010) and Wang et al. (2008).
In the present article we shall discuss Stokes second problem for magnetohydrodynamic (MHD) Oldroyd-B fluid. The fluid is electricallyonducting under the influence of a uniform magnetic field and occupies the half space over an infinite flat plate. The flow induced by theosine and sine oscillations of an infinite plate is considered. The governing equations are solved by using Laplace transform method andhe expressions for the velocity field and shear stress are obtained. The corresponding solutions for hydrodynamic flow appear as limitingases of the present analysis. The similar solutions for a Newtonian fluid, performing the same motion, are also deduced as limiting cases.
. Governing equations and mathematical model
The equations which govern the unsteady incompressible flow of a MHD fluid are (Khan et al., 2007)
�dVdt
= div T−�B20V, (1)
div V = 0. (2)
∗ Corresponding author. Tel.: +92 51 90643059.E-mail address: asia [email protected] (A. Anjum).
029-5493/$ – see front matter © 2011 Elsevier B.V. All rights reserved.oi:10.1016/j.nucengdes.2011.11.024
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M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 21
n the above equations V is the velocity vector, � the density of the fluid, T the Cauchy stress tensor, � the electrical conductivity, B0 theagnitude of applied magnetic field B0, and d/dt the material time derivative.The constitutive equations for an Oldroyd-B fluid are (Khan et al., 2007)
T = −pI + S, S+�(
dSdt
− LS − SLT)
= �[
A+�r
(dAdt
− LA − ALT)]
, (3)
here p is the pressure, I the identity tensor, S the extra stress tensor, A = L + LT the first Rivlin–Ericksen tensor, � the dynamic viscosity, Lhe velocity gradient, � and �r(< �) are the relaxation and retardation times, respectively.
In the following, we shall assume velocity field and extra stress of the form
V = V(y, t) = u(y, t)i, S = S(y, t), (4)
here i is the unit vector in the x-direction of the Cartesian coordinate system. For this velocity field the constraint of incompressibility2) is automatically satisfied.
Substituting Eq. (4) into Eqs. (1) and (3), assuming that there is no pressure gradient in the flow direction, and taking into account thenitial condition
S(y, 0) = 0; y > 0, (5)
e obtain the relevant partial differential equations
�∂u
∂t= ∂T(y, t)
∂y− �B2
0u; y, t > 0, (6)
(1 + �
∂∂t
)T(y, t) = �
(1 + �r
∂∂t
)∂u
∂y, (7)
here T(y, t) = Sxy(y, t) is the tangential stress which is different from zero.Let us consider an incompressible and electrically conducting Oldroyd-B fluid occupying the half space y > 0. The fluid is bounded by a
igid plate at y = 0 and permeated by an applied magnetic field normal to the plate. For t > 0, the plate starts to oscillate in its own planeith velocities V cos(ωt) or V sin(ωt). Due to the shear the fluid above the plate is gradually moved. Its velocity is of the form (4) and the
overning equations are (6) and (7). The appropriate initial and boundary conditions are
u(y, 0) = T(y, 0) = 0; y > 0, (8)
u(0, t) = V cos(ωt) or u(0, t) = V sin(ωt); t > 0, (9)
u(y, t), T(y, t) → 0 as y → ∞ and t > 0, (10)
here V is the amplitude and ω the imposed frequency of the velocity of the plate.Eqs. (6)–(10) can be written in dimensionless form by introducing
U = u
V, � = t
�, ω = ω�, = y
c�, = �r
�(≤ 1), S = T
�cV, M2 = �B2
0�
�and c =
√�/(��). (11)
Substituting and rearranging in a straight forward way lead to dimensionless problem(1 + ∂
∂�
)S(, �) =
(1 + ˛
∂∂�
)∂U(, �)
∂; , � > 0, (12)
∂U(, �)∂�
= ∂S(, �)∂
− M2U(, �); , � > 0, (13)
U(, 0) = S(, 0) = 0; > 0, (14)
U(0, �) = cos(ω�) or U(0, �) = sin(ω�); � > 0, (15)
U(, �), S(, �) → 0 as → ∞. (16)
. Solution of the problem
In order to solve the above problem, we shall use the Laplace transform method. Consequently, applying the Laplace transform to Eqs.12) and (13) and taking into account the corresponding initial and boundary conditions (14)–(16), we obtain
(1 + q)S(, q) = (1 + ˛q)∂U(, q)
∂, (17)
∂S(, q)∂
= (q + M2)U(, q), (18)
U(0, q) = qor U(0, q) = ω
, (19)
q2 + ω2 q2 + ω2U(, q), S(, q) → 0 as → ∞, (20)
here U(, q) and S(, q) are the Laplace transforms of U(, �) and S(, �), respectively, with q as the transform parameter.
2
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2 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
Differentiating Eq. (17) with respect to we get
(1 + q)∂S(, q)
∂= (1 + ˛q)
∂2U(, q)∂2
. (21)
Elimination of S(, q) between Eqs. (18) and (21) gives
∂2U(, q)∂2
− (q + M2)(1 + q)(1 + ˛q)
U(, q) = 0. (22)
he solutions of the above differential equation subject to the boundary conditions (19) and (20) are given by
U(, q) = q
q2 + ω2exp
(−
√(q + M2)(1 + q)
(1 + ˛q)
), (23)
espectively
U(, q) = ω
q2 + ω2exp
(−
√(q + M2)(1 + q)
(1 + ˛q)
). (24)
nvoking Eqs. (23) and (24) into (18) immediately give the following expressions
S(, q) = −q(q + M2)q2 + ω2
√(1 + ˛q)
(q + M2)(1 + q)exp
(−
√(q + M2)(1 + q)
(1 + ˛q)
), (25)
espectively
S(, q) = −ω(q + M2)q2 + ω2
√(1 + ˛q)
(q + M2)(1 + q)exp
(−
√(q + M2)(1 + q)
(1 + ˛q)
). (26)
.1. Calculation of the velocity field
In order to find the inverse Laplace transform of Eqs. (23) and (24), following the methodology of references (Fetecau et al., 2009; Khant al., 2010), we can decompose U(, q) as
U(, q) = U1(q)U2(, q), (27)
here
U1(q) = q
q2 + ω2W(q), U2(, q) = 1
W(q)exp
(− √
W(q)
), (28)
espectively
U1(q) = ω
q2 + ω2W(q), U2(, q) = 1
W(q)exp
(− √
W(q)
), (29)
ith
W(q) = (1 + ˛q)(q + M2)(1 + q)
. (30)
If U1(�) = L−1{U1(q)} and U2(, �) = L−1{U2(, q)} then we can write the velocity field as a convolution product
U(, �) = (U1 ∗ U2)(�) =∫ �
0
U1(� − s)U2(, s) ds =∫ �
0
U1(s)U2(, � − s) ds. (31)
The Laplace inversion of Eqs. (28) and (29) results in the following expressions
U1(�) = ˇ1e−� + ˇ2M2e−M2� + ˇ3 cos(ω�) + ˇ4ω sin(ω�), (32)
espectively
U1(�) = −ˇ1ωe−� − ˇ2ωe−M2� + ˇ3 sin(ω�) − ˇ4ω cos(ω�), (33)
here
ˇ1 = − 1(M2 − 1)(1 + ω2)
, ˇ2 = 1 − ˛M2
(M2 − 1)(M4 + ω2), ˇ3 = [M2 − ω2(1 − ˛(1 − M2))]
(1 + ω2)(M4 + ω2)and ˇ4 = [1 − M2( − 1) + ˛ω2]
(1 + ω2)(M4 + ω2). (34)
F
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M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 23
The function U2(, �) can be determined using an inversion formula of compound functions (Robert and Kaufman, 1968). Choosing(, q) = (1/q) exp(−(/
√q)), we get
f (, �) = L−1{F(, q)} = 1
2�√
�
∫ ∞
0
z3/2 exp
(−z2
4�
)J1
(2√
z)
dz, (35)
nd hence
U2(, �) =∫ ∞
0
f (, u)g(u, �) du = 1
2√
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
g(u, �) dz du, (36)
here g(u, �) = L−1{exp[−uW(q)]} and J1(·) is the Bessel function of the first kind of order one.For g(u, �), we decompose W(q) in the form
W(q) = (1 + ˛q)(q + M2)(1 + q)
= − − 1(M2 − 1)(1 + q)
+ ˛M2 − 1(M2 − 1)(q + M2)
, (37)
nd exp [− uW(q)] can be written as follows
exp[−uW(q)] = exp(
( − 1)u(M2 − 1)(1 + q)
)exp
(− (˛M2 − 1)u
(M2 − 1)(q + M2)
),
= 1 − H1(q) + H2(q) − H1(q)H2(q),(38)
here
H1(q) = 1 − exp
(− (˛M2 − 1)u
(M2 − 1)(q + M2)
)and H2(q) = exp
(( − 1)u
(M2 − 1)(1 + q)
)− 1. (39)
Denoting by
h1(�) =√
(˛M2 − 1)u(M2 − 1)�
e−M2� J1
(2
√(˛M2 − 1)u�
M2 − 1
), (40)
nd
h2(�) =√
( − 1)u(M2 − 1)�
e−� I1
(2
√( − 1)u�
M2 − 1
), (41)
he inverse Laplace transforms of H1(q) and H2(q), respectively, we find that
g(u, �) = ı(�) − h1(�) + h2(�) − (h1 ∗ h2)(�), (42)
here ı(·) is the Dirac delta function and I1(·) is the modified Bessel function of the first kind of order one. From Eqs. (40) and (41), we canrite
g(u, �) = ı(�) + g1(u, �), (43)
here
g1(u, �) = −√
(˛M2 − 1)u(M2 − 1)�
e−M2� J1
(2
√(˛M2 − 1)u�
M2 − 1
)
+√
( − 1)u(M2 − 1)�
e−� I1
(2
√( − 1)u�
M2 − 1
)
−√
( − 1)(˛M2 − 1)
1 − M2
∫ �
0
u√s(� − s)
e−M2s−(�−s)
×J1
(2
√(˛M2 − 1)us
M2 − 1
)I1
(2
√( − 1)u(� − s)
M2 − 1
)ds.
(44)
Substituting Eq. (43) and (44) into Eq. (36), in combination with Eq. (31), and using the result
(f ∗ ı)(�) = (ı ∗ f )(�) = f (�), (45)
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4 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
e finally get the velocity field
U(, �) = 1
2√
U1(�)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
dz du
− 1
2√
(M2 − 1)
∫ �
0
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u
)J1
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×U1(� − s) dz du ds −√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ �
0
∫ s
0
∫ ∞
0
∫ ∞
0
z√
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J1
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)U1(� − s) dz du d� ds,
(46)
here U1(�) is given through Eqs. (32) and (33), respectively.For M = 0, the above equation can be reduced to the similar solution for hydrodynamic flow as obtained by Khan et al. (2010) for an
ldroyd-B fluid.The above equation can be written in the equivalent form as
U(, �) = 1
2√
U1(�)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
dz du
− 1
2√
(M2 − 1)
∫ ∞
0
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u
)J1
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×U1(� − s) dz du ds + 1
2√
(M2 − 1)
∫ ∞
�
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u
)J1
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×U1(� − s) dz du ds −√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
0
∫ s
0
∫ ∞
0
∫ ∞
0
z√
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J1
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)U1(� − s) dz du d� ds
+√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
�
∫ s
0
∫ ∞
0
∫ ∞
0
z√
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J1
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
(47)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)U1(� − s) dz du d� ds.
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M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 25
The starting solution (47), presented as sum of the steady state and transient solutions, describes the motion of the fluid for small andarge times. For large values of time, the transients gradually disappear and the motion of the fluid is described by the steady-state solutioniven by
Us(, �) = 1
2√
U1(�)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
dz du
− 1
2√
(M2 − 1)
∫ ∞
0
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u
)J1
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×U1(� − s) dz du ds −√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
0
∫ s
0
∫ ∞
0
∫ ∞
0
z√
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J1
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)U1(� − s) dz du d� ds,
(48)
here
U1(�) = ˇ3 cos(ω�) + ˇ4ω sin(ω�), (49)
espectively
U1(�) = ˇ3 sin(ω�) − ˇ4ω cos(ω�), (50)
hich is periodic in time and independent of initial conditions.The corresponding transient solution is
Ut(, �) = 1
2√
(M2 − 1)
∫ ∞
�
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u
)J1
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×U1(� − s) dz du ds +√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
�
∫ s
0
∫ ∞
0
∫ ∞
0
z√
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J1
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)U1(� − s) dz du d� ds,
(51)
hich vanishes when �−→ ∞.
.2. Calculation of the tangential stress
Following the same way as before, we write S(, �) as
S(, q) = −S1(q)S2(, q), (52)
here
S1(q) = q(q + M2)q2 + ω2
W(q), S2(, q) = 1√W(q)
exp
(− √
W(q)
), (53)
espectively
S1(q) = ω(q + M2)q2 + ω2
W(q), S2(, q) = 1√W(q)
exp
(− √
W(q)
). (54)
2
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6 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
If S1(�) = L−1{S1(q)} and S2(, �) = L−1{S2(, q)} then
S(, �) = −(S1 ∗ S2)(�) = −∫ �
0
S1(� − s)S2(, s) ds = −∫ �
0
S1(s)S2(, � − s) ds. (55)
Adopting the methodology of references (Fetecau et al., 2009; Khan et al., 2010), the Laplace inversion of the above equations result in
S(, �) = − 12√
S1(�)
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u
)J0
(2√
z)
dz du
+ 1
2√
(M2 − 1)
∫ �
0
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u
)J0
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
×S1(� − s) dz du ds +√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ �
0
∫ s
0
∫ ∞
0
∫ ∞
0
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J0
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)S1(� − s) dz du d� ds,
(56)
here
S1(�) = ˇ5e−� + ˇ6 cos(ω�) + ˇ7 sin(ω�), (57)
espectively
S1(�) = −ˇ5ωe−� + ˇ6 sin(ω�) − ˇ7 cos(ω�), (58)
ˇ5 = − 1(1 + ω2)
, ˇ6 = 1 + ˛ω2
(1 + ω2)and ˇ7 = ω(1 − ˛)
(1 + ω2). (59)
imilar to the velocity field, the above equations are the sum of the steady state and transient solutions. Hence
Ss(, �) = − 12√
S1(�)
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u
)J0
(2√
z)
dz du
+ 1
2√
(M2 − 1)
∫ ∞
0
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u
)J0
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
× S1(� − s) dz du ds +√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
0
∫ s
0
∫ ∞
0
∫ ∞
0
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J0
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)(60)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)S1(� − s) dz du d� ds,
w
r
4
4
w
r
FN
a
r
M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 27
St(, �) = − 1
2√
(M2 − 1)
∫ ∞
�
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u
)J0
(2√
z)
×[√
˛M2 − 1e−M2sJ1
(2
√(˛M2 − 1)us
M2 − 1
)− √
− 1e−sI1
(2
√( − 1)us
M2 − 1
)]
× S1(� − s) dz du ds −√
( − 1)(˛M2 − 1)
2√
(1 − M2)
∫ ∞
�
∫ s
0
∫ ∞
0
∫ ∞
0
z√u�(s − �)
× exp
(−z2
4u− s + �(1 − M2)
)J0
(2√
z)
J1
(2
√(˛M2 − 1)u�
M2 − 1
)
×I1
(2
√( − 1)u(s − �)
M2 − 1
)S1(� − s) dz du d� ds,
(61)
here
S1(�) = ˇ6 cos(ω�) + ˇ7 sin(ω�), (62)
espectively
S1(�) = ˇ6 sin(ω�) − ˇ7 cos(ω�). (63)
. Limiting cases
.1. The case = 1 (Newtonian fluid)
Letting = 1 into Eqs. (32), (33) and (46), the velocity fields for a Newtonian fluid take the form
U(, �) = 1
2√
U1(�)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
dz du − 1
2√
∫ �
0
∫ ∞
0
∫ ∞
0
z√
z
u√
s
× exp
(−z2
4u− M2s
)U1(� − s)J1
(2√
z)
J1(
2√
us)
dz du ds,
(64)
here
U1(�) =[− M2
M4 + ω2e−M2� + M2(1 − ω2)
(1 + ω2)(M4 + ω2)cos(ω�) + ω
M4 + ω2sin(ω�)
], (65)
espectively
U1(�) =[
ω
M4 + ω2e−M2� + M2(1 − ω2)
(1 + ω2)(M4 + ω2)sin(ω�) − ω
M4 + ω2cos(ω�)
]. (66)
or M = 0, the above equations can be reduced to the similar solutions for hydrodynamic flow as obtained by Khan et al. (2010) for aewtonian fluid.
Following the methodology of the reference (Khan et al., 2010), we again decompose the above velocity field as a sum of the steady-statend the transient solutions given by
Us(, �) = ω
2√
(M4 + ω2)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u− M2u
M4 + ω2
)sin(
ω� + uω
M4 + ω2
)
×J1
(2√
z)
dz du + M2(1 − ω2)
2√
(1 + ω2)(M4 + ω2)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u− M2u
M4 + ω2
)
×cos(
ω� + uω
M4 + ω2
)J1
(2√
z)
dz du,
(67)
espectively
Us(, �) = − ω
2√
(M4 + ω2)
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u− M2u
M4 + ω2
)cos(ω� + uω
M4 + ω2)
( √ )M2(1 − ω2)
∫ ∞ ∫ ∞z√
z(
−z2 M2u)
×J1 2 z dz du +2√
(1 + ω2)(M4 + ω2) 0 0 u√
uexp
4u−
M4 + ω2
× sin(ω� + uω
M4 + ω2)J1(
2√
z)
dz du,
(68)
2
a
r
T
r
H
r
a
r
4
a
8 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
nd
Ut(, �) = 1
2√
∫ ∞
�
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u− M2s
)
×U(� − s)J1(
2√
z)
J1(
2√
us)
dz du ds,
(69)
espectively
Ut(, �) = 1
2√
∫ ∞
�
∫ ∞
0
∫ ∞
0
z√
z
u√
sexp
(−z2
4u− M2s
)
×U(� − s)J1(
2√
z)
J1(
2√
us)
dz du ds.
(70)
he corresponding expressions for tangential stress are
S(, �) = − 12√
cos(ω�)
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u
)J0
(2√
z)
dz du
+ 12√
∫ �
0
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u− M2s
)cos(ω(� − s))
×J0
(2√
z)
J1(
2√
us)
dz du ds,
(71)
espectively
S(, �) = − 12√
sin(ω�)
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u
)J0
(2√
z)
dz du
+ 12√
∫ �
0
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u− M2s
)sin(ω(� − s))
×J0
(2√
z)
J1(
2√
us)
dz du ds.
(72)
ence
Ss(, �) = − 12√
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u− M2u
M4 + ω2
)cos(
ω� + uω
M4 + ω2
)J0
(2√
z)
dz du, (73)
espectively
Ss(, �) = − 12√
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u− M2u
M4 + ω2
)sin(
ω� + uω
M4 + ω2
)J0
(2√
z)
dz du, (74)
nd
St(, �) = − 12√
∫ ∞
�
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u− M2s
)cos(ω(� − s))J0
(2√
z)
J1(
2√
us)
dz du ds, (75)
espectively
St(, �) = − 12√
∫ ∞
�
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u− M2s
)sin(ω(� − s))J0
(2√
z)
J1(
2√
us)
dz du ds. (76)
.2. The case ω → 0 (first problem of Stokes)
Finally making ω → 0 into Eqs. (64) and (71), we obtain the solutions of the first problem of Stokes for MHD flow of a Newtonian fluids
U(, �) = (1 − e−M2�)
2M2√
∫ ∞
0
∫ ∞
0
z√
z
u√
uexp
(−z2
4u
)J1
(2√
z)
dz du
− 1√ ∫ � ∫ ∞ ∫ ∞z√
z√ (1 − e−M2(�−s)) (77)
2M2 0 0 0 u s× exp
(−z2
4u− M2s
)J1
(2√
z)
J1(
2√
us)
dz du ds,
M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 29
0 1 2 3 4 5-0. 6
-0. 4
-0. 2
0
0.2
0.4
ξ
U
α=0.1α=0.4α=0.7α=1.0
(a) τ = 2
0 2 4 6 8-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
ξ
U
α=0.1α=0.4α=0.7α=1.0
(b) τ = 18
Fig. 1. Profiles of the velocity U(, �) for different values of elasticity parameter for the cosine oscillations when M = 0.5 and ω = 1.
0 1 2 3 4 5-0.2
0
0.2
0.4
0.6
0.8
1
ξ
U
α=0.1α=0.4α=0.7α=1.0
(a) τ = 2
0 1 2 3 4 5 6 7-0.8
-0.6
-0.4
-0.2
0
0.2
ξ
U
α=0.1α=0.4α=0.7α=1.0
(b) τ = 18
a
r
R
5
aa
bFmro
toc
o
Fig. 2. Profiles of the velocity U(, �) for different values of elasticity parameter for the sine oscillations when M = 0.5 and ω = 1.
nd
S(, �) = − 12√
∫ ∞
0
∫ ∞
0
z
u√
uexp
(−z2
4u
)J0
(2√
z)
dz du
+ 12√
∫ �
0
∫ ∞
0
∫ ∞
0
z
u√
sexp
(−z2
4u− M2s
)
×J0
(2√
z)
J1(
2√
us)
dz du ds,
(78)
espectively.For M = 0, the above equations can be reduced to the similar solutions for hydrodynamic flow as obtained in Khan et al. (2010) and
obert and Kaufman (1968).
. Numerical results and discussion
In order to investigate the effects of various parameter of the flow problem, we have plotted the profiles of the velocity field and thessociated tangential stress in Figs. 1–10. The graphs are plotted for both cosine and sine oscillations. We compare profiles for Newtoniannd Oldroyd-B fluids. A comparison between hydromagnetic and hydrodynamic flows is also made.
Figs. 1 and 2 display the velocity profiles for various values of the elasticity parameter for both cosine and sine oscillations of theoundary at two different times. The results corresponding to the value = 1 describe the Newtonian flow which contains no elastic effect.rom these figures, it is observed that for both type of oscillations, when the parameter increases from = 0.1 to = 1, a decrease in theagnitude of the velocity profiles is seen. This decrease in the magnitude of the velocity profiles is much in the case of cosine oscillation
ather than the sine oscillations. Compared with the Newtonian fluid, the difference of velocity profiles of Oldroyd-B fluid is much morebvious. The velocity profile of Newtonian fluid is smaller when compared with Oldroyd-B fluid.
Figs. 3 and 4 are prepared for the variation of magnetic parameter M on the velocity distribution for the cosine and sine oscillations ofhe boundary. These figures also display a comparison between hydromagnetic and hydrodynamic flows. It is noted that the magnitudef the velocity profiles decreases with a rise in magnetic parameter M. Such an effect may also be expected, because under the conditions
onsidered the magnetic force is a resistance to the flow and hence the magnitude of the flow near the plate decreases.Figs. 5 and 6 record velocity profiles at different instants for both the cosine and sine oscillations. It is shown that the velocity profilesscillate periodically. From these figures, the maximum amplitude of oscillation is seen to be at the plate surface and it gradually decays
30 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
0 1 2 3 4-0.6
-0.4
-0.2
0
0.2
0.4
ξ
U
M=0.0
M=0.5
M=0.9
M=1.5
(a) τ=2
0 2 4 6 8-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
ξ
U
M=0.0
M=0.5
M=0.9M=1.5
(b) τ=18
Fig. 3. Profiles of the velocity U(, �) for different values of magnetic parameter M for the cosine oscillations when = 0.3 and ω = 1.
0 1 2 3 40
0.2
0.4
0.6
0.8
1
ξ
U
M=0.0
M=0.5
M=0.9
M=1.5
(a) τ=2
0 1 2 3 4 5 6 7-0.8
-0.6
-0.4
-0.2
0
0.2
ξ
U
M=0.0
M=0.5
M=0.9M=1.5
(b) τ=18
Fig. 4. Profiles of the velocity U(, �) for different values of magnetic parameter M for the sine oscillations when = 0.3 and ω = 1.
0 1 2 3 4 5 6-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
ξ
U
τ=2τ=3τ=4τ=5
(a) M=0
0 1 2 3 4 5 6-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
ξ
U
τ=2τ=3τ=4τ=5
(b) M=0.5
Fig. 5. Profiles of the velocity U(, �) for different values of times � for the cosine oscillations when = 0.3 and ω = 1.
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
ξ
U
τ=2τ=3τ=4τ=5
(a) M=0
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
ξ
U
τ=2τ=3τ=4τ=5
(b) M=0.5
Fig. 6. Profiles of the velocity U(, �) for different values of times � for the sine oscillations when ˛ = 0.3 and ω = 1.
M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32 31
0 1 2 3 40
0.2
0.4
0.6
0.8
1
ξ
S
M=0.0
M=0.5
M=0.9
M=1.5
(a) τ=2
0 2 4 6 8-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
ξ
S
M=0.0
M=0.5
M=0.9M=1.5
(b) τ=18
Fig. 7. Profiles of the tangential stress S(, �) for different values of magnetic parameter M for the cosine oscillations when = 0.3 and ω = 0.5.
0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
1.2
1.4
ξ
S
M=0.0
M=0.5
M=0.9M=1.5
(a) τ=2
0 2 4 6 8 10-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
ξ
S
M=0.0
M=0.5
M=0.9
M=1.5
(b) τ=18
Fig. 8. Profiles of the tangential stress S(, �) for different values of magnetic parameter M for the sine oscillations when = 0.3 and ω = 0.5.
0 1 2 3 4 5 6 7-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
ξ
S
τ=2τ=3τ=4τ=5
(a) M=0
0 1 2 3 4 5 6 7
-0.6
-0.4
-0.2
0
0.2
0.4
ξ
S
τ=2τ=3τ=4τ=5
(b) M=0.5
Fig. 9. Profiles of the tangential stress S(, �) for different values of times � for the cosine oscillations when = 0.3 and ω = 0.5.
0 1 2 3 4 5 6 7-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
ξ
S
τ=2τ=3τ=4τ=5
(a) M=0
0 1 2 3 4 5 6 7-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
ξ
S
τ=2τ=3τ=4τ=5
(b) M=0.5
Fig. 10. Profiles of the tangential stress S(, �) for different values of times � for the sine oscillations when = 0.3 and ω = 0.5.
3
at
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2 M. Khan et al. / Nuclear Engineering and Design 243 (2012) 20– 32
way from the plate towards zero. A comparison reveals that the decay of the amplitude of oscillation in case of cosine oscillation is rapidhan the sine oscillation.
It is well-known that the tangential stress at the boundary plays an important role that provides useful information concerning theature of dissipation at the boundary. The tangential stress profiles for different values of magnetic parameter M and time � are plotted inigs. 7–10. It is observed that the tangential stress oscillates periodically. Further, as it was expected that, the strongest tangential stressccurs near the boundary in both cases and decreases far away from the boundary.
. Concluding remarks
In this paper exact solutions for velocity field and tangential stress corresponding to the oscillatory motion of a magnetohydrodynamicldroyd-B fluid are investigated. These solutions, presented as a sum of steady-state and transient solutions, are obtained using Laplace
ransform method. The variations of the velocity and tangential stress are also analysed graphically. The conclusions can be listed as follows:
. The velocity profile decreases with the increase of elasticity parameter ˛.
. The decrease in the magnitude of the velocity profiles is much in case of the cosine oscillation rather than the sine oscillation.
. The velocity profile of Newtonian fluid is smaller when compared with Oldroyd-B fluid.
. A monotonic decrease in the magnitude of the velocity profiles is seen with a rise in magnetic parameter M.
. It is observed that the strongest tangential stress occurs near the oscillating boundary.
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i, H.T., Xu, M.Y., 2007. Stokes first problem for a viscoelastic fluid with the generalized Oldroyd-B model. Acta Mech. Sin. 23, 463–469.ajagopal, K.R., Bhatnagar, R.K., 1995. Exact solutions for some simple flows of an Oldroyd-B fluid. Acta Mech. 113, 223–239.ajagopal, K.R., 1992. Flows of viscoelastic fluids between rotating discs. Theor. Comput. Fluid Dyn. 3, 185–206.obert, G.E., Kaufman, H., 1968. Table of Laplace Transforms. W.B. Saunders Company, Philadelphia, London.ang, Y., Hayat, T., Ali, N., Oberlack, M., 2008. Magnetohydrodynamic peristaltic motion of a Sisko fluid in a symmetric or asymmetric channel. Physica A 387, 347–362.
