# A Note on the Unsteady Incompressible MHD Fluid Flow with Slip Conditions and Porous Walls

Authors: H. Zaman; Z. Ahmad; M. Ayub
Journal: ISRN Mathematical Physics
 (2013)

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## Abstract

This work is concerned with the influence of uniform suction or injection on flow and heat transfer analysis of unsteady incompressible magnetohydrodynamic (MHD) fluid with slip conditions. The resulting unsteady problem for velocity and heat transfer is solved by means of Laplace transform. The characteristics of the transient velocity, overall transient velocity, steady state velocity and heat transfer at the walls are analyzed and discussed. Graphical results reveal that the magnetic field, slip parameter, and suction (injection) have significant influences on the velocity, and temperature distributions, which also changes the heat transfer behaviors at the two plates. The results of Fang (2004) are also recovered by keeping magnetic field and slip parameter absent.


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## Body

## 1. Introduction

Navier-Stokes equations are the basic equations of fluid mechanics. Exact solutions of Navier-Stokes equations are rare due to their inherent nonlinearity. Exact solutions are important because they serve as accuracy checks for numerical solutions. Complete integration of these equations is done by computer techniques, but the accuracy of the results can be established only by comparison with exact solutions. In the literature, there are a large number of Newtonian fluid flows for which exact solutions are possible [1–6]. The effects of transverse magnetic field on the flow of an electrically conducting viscous fluid have been studied extensively in view of numerous applications to astrophysical, geophysical, and engineering problems [7–15]. If the working fluid contains concentrated suspensions, then the wall slip can occur [16]. Khaled and Vafai [3] studied the effect of the slip on Stokes and Couette flows due to an oscillating wall. However, the literature lacks studies that take into account the possibility of fluid slippage at the walls. Applications of these problems appear in microchannels or nanochannels and in applications where a thin film of light oil is attached to the moving plates or when the surface is coated with special coating such as a thick monolayer of hydrophobic octadecyltrichlorosilane [17]. Yu and Ameel [18] imposed nonlinear slip boundary conditions on flow in rectangular microchannels. Erdogan [6] studied deeply the solution to the Stokes problem under nonslip conditions at the wall. Ayub and Zaman [19] studied the effects of suction and blowing for orthogonal flow impinging on a wall. Khan et al. [20] discussed the flow of Sisko fluid through a porous medium. Ariel et al. [21] considered the flow of elasticoviscous fluid with partial slip. Raptis et al. [22–25] studied steady and unsteady free convection and mass transfer flow through a porous medium. Penton [26] presented the transient solution for the flow due to the oscillating plate.In this note, the flow of an incompressible, unsteady, viscous, MHD fluid with slip conditions is considered. Unsteady means time dependent flow, and we are looking for the effects of different parameters on flow with the variation of time. Unsteady and steady velocity profiles with mass transfer will be presented and solved exactly. There is mass injection from one plate and the same amount of suction on the other plate. The steady state temperature is also solved and discussed. When the fluid motion is set up from rest, the velocity field contains transients, determined by the initial conditions which gradually disappear in time. The effect of magnetic field and time on the transient velocity and on overall transient velocity has been seen graphically for both injection and suction. The effect of slip parameter on steady state velocity for injection/suction is shown graphically. Steady state temperature profiles and heat transfer rate at the walls (Nusselt number at the walls) are also discussed for injection/suction for different Prandtl and Reynolds numbers. The results of Fang [1] are recovered by taking magnetic field parameter and slip parameter to be zero.


## 2. Theoretical Derivation

### 2.1. Transient Velocity Profiles

Consider an incompressible, viscous, unsteady flow problem, in which there is slip between the bottom wall and fluid and also between top wall and fluid. There are mass injection velocityvwat the bottom wall and mass suction velocityvwat the top wall; vw>0corresponds to injection andvw<0corresponds to suction. The governing equation for this problem can be obtained as [2]
(1)∂u(y,t)∂t+vw∂u(y,t)∂y=ν∂2u(y,t)∂y2-Nu(y,t),
whereu(y,t)is the velocity of the fluid in they-direction which is along the wall direction,yis the distance alon y-axis g,tis the time,ν=μ/ρ,N=σB02/ρ,μis the dynamic viscosity,νis the kinematic viscosity,ρis the density of the fluid,σis the electric conductivity of the fluid,B0is the applied magnetic field, andNis the MHD factor or parameter. For the boundary conditions we consider the existence of slip between the velocity of the fluid at the walls and speed of the walls:
(2)u(0,t)-λ∂u(y,t)∂y|y=0=0,u(h,t)+λ∂u(y,t)∂y|y=h=U0.
Initial condition is
(3)u(y,0)=0,
whereλis the slip parameter (λ=0gives the usual no slip condition at the wall) andU0is the velocity at the upper wall. The problem exactly reduces to the problem of Fang [1] if we takeN=0,λ=0.Equations (1), (2), and (3) can be made dimensionless by defining
(4)U=uU0,Y=yh,T=tτc=tνh2.
Then (1), (2), and (3) become
(5)∂U∂T+Re∂U∂Y=∂2U∂Y2-N¯U,(6)U(0,T)-λ¯∂U(Y,T)∂Y|Y=0=0,(7)U(1,T)+λ¯∂U(Y,T)∂Y|Y=1=1,(8)U(Y,0)=0,
whereN¯=N(h2/ν),λ¯=λ/h, andRe=vw(h/ν)(Reynolds number). DecomposingU(Y,T)into two parts, say, transient part and a steady state part,
(9)U(Y,T)=Ut(Y,T)+Us(Y,T).
Then we have two separate problems and the steady state part will be
(10)RedUsdY=d2UsdY2-N¯Us,(11)Us(0,T)-λ¯∂Us(Y,T)∂Y|Y=0=0,(12)Us(1,T)+λ¯∂Us(Y,T)∂Y|Y=1=1.
The solution of (10) with BCs (11) and (12) can be obtained as
(13)Us(Y)=eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)]ea[(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)],
wherea=Re/2andb1=Re2/4+N¯. Whenλ-=0 (no slip) and N¯=0 (no magnetic field), then (13) becomes
(14)Us(Y)=e(Re/2)Ysinh(Re/2)YeRe/2sinh(Re/2)=1-eReY1-eRe.
Equation (14) the result number (7) of Fang [1]. If there is no mass transfer at the walls, thenRe=0, soa=0andb1=N¯, and (13) becomes
(15)Us(Y)=sinh(N¯Y)+λ¯N¯cosh(N¯Y)(1+λ¯2N¯)sinh(N¯)+2λ¯N¯cosh(N¯).
If there is no mass transfer at the walls and magnetic field is absent, thenRe=0,N¯=0, and slip parameterλ¯=0then (10), (11), and (12) will collapse into
(16)d2Us(Y)dY2=0,Us(0)=0,Us(1)=1.
Its solution isUs(Y)=Y, which is conventional Couette flow. WhenRe→-∞,N¯=0, and λ¯=0, then (13) givesUs(Y)=1, except at the bottom wall (at the bottom wallY=0andUs(0)=0).WhenRe→∞,N¯=0, and λ¯=0, then (13) givesUs(Y)=0, except at the upper wall (at the upper wall Y=1andUs(1)=1). The transient part problem becomes
(17)∂Ut∂T+Re∂Ut∂Y=∂2Ut∂Y2-N¯Ut,(18)Ut(0,T)-λ¯∂Ut(Y,T)∂Y|Y=0=0,(19)Ut(1,T)+λ¯∂Ut(Y,T)∂Y|Y=1=1,(20)Ut(Y,0)=-Us(Y).
The solution can be derived by using Laplace transformation techniques [27]. The Laplace transform pair is defined by
(21)U¯(Y,s)=L[U(Y,T)]=∫0∞‍U(Y,T)e-sTdT,U(Y,T)=L-1[U¯(Y,s)]=12πι∫λ1-ι∞λ1+ι∞‍U¯(Y,s)esTds,
over that range of values ofsfor which the integrals exist. Here,sis a parameter, real or complex,Lis the operator that transformsU(Y,T)intoU¯(Y,s), called Laplace transform operator, andL-1is the inverse Laplace transform operator. The solution can be shown as
(22)Ut(Y,T)=L-1[×sinh(b)+2λ¯bcosh(b)λ¯2))-1(eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY)))×(eas((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)λ¯2))-1]-Us,
wherea=Re/2andb=Re2/4+s+N¯. In inverse Laplace transform of the above equation we have simple pole ats=0and infinite number of poles (located on the negative real axis) atsn=-(ln2+a2+N¯),(n=1,2,3,…), whereln=ibn,(bn=Re2/4+sn+N¯)and are given by(23)tan(ln)=-2λ¯ln1-λ¯2(a2+ln2).
The transient part velocity will be
(24)Ut(Y,T)=∑n=1∞Res[U¯t(Y,s)esT]s=sn+Res[U¯t(Y,s)esT]s=0-Us,
whereResstands for residue andU¯t(Y,s)is given by
(25)U¯t(Y,s)=eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY))eas((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)).
We have
(26)∑n=1∞Res[U¯t(Y,s)esT]s=sn=∑n=1∞G1(ln,Y)G2(ln)esnT,
where
(27)G1(ln,Y)=eaY((1-λ¯a)sin(lnY)hhhh+λ¯lncos(lnY)),G2(ln)=easn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))cos(ln)2ln).
The residue ats=0gives steady velocity
(28)Res[U¯t(Y,s)esT]s=0=lims→0[sU¯t(Y,s)esT]=lims→0(eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY))esT)×(ea((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)(1-λ¯2(a2-b2))))-1,lims→0b=lims→0Re24+s+N¯=Re24+N¯=b1.
Therefore,
(29)Res[U¯t(Y,s)esT]s=0=(eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)])×(ea[(1-λ¯2(a2-b12))×sinh(b1)+2λ¯b1cosh(b1)(1-λ¯2(a2-b12))])-1.
By using (13), we get
(30)Res[U¯t(Y,s)esT]s=0=Us.The transient part velocity from (24) becomes
(31)Ut(Y,T)=∑n=1∞G1(ln,Y)G2(ln)esnT,Ut(Y,T)=∑n=1∞[cos(ln)2ln(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))×(cos(ln)2lncos(ln)2lneasn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))×cos(ln)2ln))-1esnT].
Therefore, the overall transient solution from (9) becomes
(32)U(Y,T)=∑n=1∞{(cos(ln)2lneaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY))×((eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))easn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))×cos(ln)2ln))-1)esnT(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))}+(eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)])×(ea[(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)(1-λ¯2(a2-b12))])-1.
To recover Fang [1] result number (13) we substituteλ¯=0,N¯=0,a=Re/2, andb1=Re/2into (32), and we get
(33)U(Y,T)=∑n=1∞{sin(lnY)esnTe(Re/2)Y-eRe/2sncos(ln)/(2ln)}+sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2).
Sinceλ¯=0, so (23) gives
(34)tan(ln)=0⟹sin(ln)=0⟹ln=arcsin(0)=nπ,n=1,2,3,….
Thereforecos(ln)=cos(nπ)=(-1)n; also we haveln=ibn,(bn=Re2/4+sn+N¯), butN¯=0, so
(35)ln=iRe24+sn⟹nπ=iRe24+sn⟹n2π2=-(Re24+sn)⟹sn=-(n2π2+Re24).
Usingln=nπ,cos(ln)=(-1)n, andsn=-(n2π2+Re2/4)into (33) we obtain
(36)U(Y,T)=2πe-(Re2/4)Te(Re/2)YeRe/2×∑n=1∞{n(-1)nsin(nπY)e-n2π2T(n2π2+Re2/4)}+sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2).
Consider
(37)sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2)=e(Re/2)Y(e(Re/2)Y-e-(Re/2)Y)eRe/2(eRe/2-e-Re/2)=(eReY-1)(eRe-1)=(1-eReY)(1-eRe).
So (36) becomes
(38)U(Y,T)=2πe-(Re2/4)Te(Re/2)YeRe/2×∑n=1∞{n(-1)nsin(nπY)e-n2π2T(n2π2+Re2/4)}+(1-eReY)(1-eRe).
Result number (13) of Fang [1] is exactly followed in (38).


### 2.2. Steady State Temperature

It is too difficult to exactly solve the transient energy equation with viscous dissipation for this problem by usingU(Y,T). We solve the steady state energy equation, which in dimensionless form is given by [1]
(39)PrRedσdY=d2σdY2+EcPr(dUsdY)2,σ(0)=0,σ(1)=1,
wherePr=ν/αis the Prandtl number,Ec=U02/cp(θ1-θ2)is the Eckert number,θ1 is temperature of the bottom fixed plate,θ2 is temperature of the top moving plate,σis dimensionless temperature(σ=(θ-θ1)/(θ2-θ1)), and Usis the steady state velocity given by (13). The solution of (39) with BCs (35) forPr≠1, is
(40)σ(Y)=(ePrReY-1)(ePrRe-1)+EcPrM[(ePrReY-1)(ePrRe-1)×{N1(e2(a+b1)-1)+N2(e2(a-b1)-1)-N3(e2a-1)}+{N1(1-e2(a+b1)Y)+N2(1-e2(a-b1)Y)-N3(1-e2aY)}(ePrReY-1)(ePrRe-1)],
where
(41)N1=M1L1,N2=M2L2,N3=M3L3,L1=4(a+b1)2-2PrRe(a+b1),L2=4(a-b1)2-2PrRe(a-b1),L3=Re2(1-Pr),M=1[ea{(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)}]2,M1=(B1+b12)2,M2=(B1-b12)2,M3=(B12-b122),B1=(a-λ¯(a2-b12)).
When viscous dissipation term is negligible, then energy equation (39) is
(42)d2σdY2-PrRedσdY=0,σ(0)=0,σ(1)=1.
Its solution is
(43)σ(Y)=(ePrReY-1)(ePrRe-1).
So the first term in (40) is the solution of energy equation when viscous dissipation term is negligible and the second term in (40) is the temperature profile from viscous dissipation and MHD. It is found from (40) that the temperature profile is linearly dependent upon the Eckert number. The Nusselt number at the walls will beNu=|dσ/dY|w. The Nusselt number for the bottom wall, Y=0, is
(44)Nu1=|dσdY|Y=0,Nu1=|PrRe(ePrRe-1)[1+EcPrM{N1(e2(a+b1)-1)hhhhhhhhhhhhhhhhhhhhhhhh+N2(e2(a-b1)-1)hhhhhhhhhhhhhhhhhhhhhhhh-N3(e2a-1)}{N1(e2(a+b1)-1)]-2EcPrM{N1(a+b1)hhhhhhhh+N2(a-b1)-N3a}PrRe(ePrRe-1)PrRe(ePrRe-1)|.
The Nusselt number for the upper wall,Y=1, is
(45)Nu2=|dσdY|Y=1,Nu2=|PrReePrRe(ePrRe-1)×[1+EcPrM{N1(e2(a+b1)-1)hhhhhhhhhhhhhhh+N2(e2(a-b1)-1)hhhhhhhhhhhhhhh-N3(e2a-1)}]-2EcPrM{N1(a+b1)e2(a+b1)hhhhhhhhhhhh+N2(a-b1)e2(a-b1)-N3ae2a}PrRe(ePrRe-1)PrRe(ePrRe-1)|.
WhenPr=1, then the energy equation is
(46)d2σdY2-RedσdY=-EcM{M1e2(a+b1)Y+M2e2(a-b1)Y-M3e2aY},σ(0)=0,σ(1)=1.
The solution of (46) is
(47)σ(Y)=(eReY-1)(eRe-1)[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhhhhh-M3Ree2a}]+2EcM{M3YReN4(1-e2(a+b1)Y)hhhhhhh+N5(1-e2(a-b1)Y)+M3YRee2aY},
where
(48)N4=M1L4,N5=M2L5,L4=4(a+b1)2-2Re(a+b1),L5=4(a-b1)2-2Re(a-b1).
The Nusselt number in this case is
(49)Nu1=|Re(eRe-1)×[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhh-M3Ree2a}]-2EcM{M3ReN4(a+b1)hhhhhhhhh+N5(a-b1)-M3Re}Re(eRe-1)|,Nu2=|ReeRe(eRe-1)×[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhh-M3Ree2a}]-2EcM{M3ReN4(a+b1)e2(a+b1)hhhhhhhhh+N5(a-b1)e2(a-b1)-M3Reae2a}ReeRe(eRe-1)|.
It is worth recalling in the vicinity of (40) that whenλ¯=0,N¯=0, then result number (18) of Fang [1] can be derived.


## 2.1. Transient Velocity Profiles

Consider an incompressible, viscous, unsteady flow problem, in which there is slip between the bottom wall and fluid and also between top wall and fluid. There are mass injection velocityvwat the bottom wall and mass suction velocityvwat the top wall; vw>0corresponds to injection andvw<0corresponds to suction. The governing equation for this problem can be obtained as [2]
(1)∂u(y,t)∂t+vw∂u(y,t)∂y=ν∂2u(y,t)∂y2-Nu(y,t),
whereu(y,t)is the velocity of the fluid in they-direction which is along the wall direction,yis the distance alon y-axis g,tis the time,ν=μ/ρ,N=σB02/ρ,μis the dynamic viscosity,νis the kinematic viscosity,ρis the density of the fluid,σis the electric conductivity of the fluid,B0is the applied magnetic field, andNis the MHD factor or parameter. For the boundary conditions we consider the existence of slip between the velocity of the fluid at the walls and speed of the walls:
(2)u(0,t)-λ∂u(y,t)∂y|y=0=0,u(h,t)+λ∂u(y,t)∂y|y=h=U0.
Initial condition is
(3)u(y,0)=0,
whereλis the slip parameter (λ=0gives the usual no slip condition at the wall) andU0is the velocity at the upper wall. The problem exactly reduces to the problem of Fang [1] if we takeN=0,λ=0.Equations (1), (2), and (3) can be made dimensionless by defining
(4)U=uU0,Y=yh,T=tτc=tνh2.
Then (1), (2), and (3) become
(5)∂U∂T+Re∂U∂Y=∂2U∂Y2-N¯U,(6)U(0,T)-λ¯∂U(Y,T)∂Y|Y=0=0,(7)U(1,T)+λ¯∂U(Y,T)∂Y|Y=1=1,(8)U(Y,0)=0,
whereN¯=N(h2/ν),λ¯=λ/h, andRe=vw(h/ν)(Reynolds number). DecomposingU(Y,T)into two parts, say, transient part and a steady state part,
(9)U(Y,T)=Ut(Y,T)+Us(Y,T).
Then we have two separate problems and the steady state part will be
(10)RedUsdY=d2UsdY2-N¯Us,(11)Us(0,T)-λ¯∂Us(Y,T)∂Y|Y=0=0,(12)Us(1,T)+λ¯∂Us(Y,T)∂Y|Y=1=1.
The solution of (10) with BCs (11) and (12) can be obtained as
(13)Us(Y)=eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)]ea[(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)],
wherea=Re/2andb1=Re2/4+N¯. Whenλ-=0 (no slip) and N¯=0 (no magnetic field), then (13) becomes
(14)Us(Y)=e(Re/2)Ysinh(Re/2)YeRe/2sinh(Re/2)=1-eReY1-eRe.
Equation (14) the result number (7) of Fang [1]. If there is no mass transfer at the walls, thenRe=0, soa=0andb1=N¯, and (13) becomes
(15)Us(Y)=sinh(N¯Y)+λ¯N¯cosh(N¯Y)(1+λ¯2N¯)sinh(N¯)+2λ¯N¯cosh(N¯).
If there is no mass transfer at the walls and magnetic field is absent, thenRe=0,N¯=0, and slip parameterλ¯=0then (10), (11), and (12) will collapse into
(16)d2Us(Y)dY2=0,Us(0)=0,Us(1)=1.
Its solution isUs(Y)=Y, which is conventional Couette flow. WhenRe→-∞,N¯=0, and λ¯=0, then (13) givesUs(Y)=1, except at the bottom wall (at the bottom wallY=0andUs(0)=0).WhenRe→∞,N¯=0, and λ¯=0, then (13) givesUs(Y)=0, except at the upper wall (at the upper wall Y=1andUs(1)=1). The transient part problem becomes
(17)∂Ut∂T+Re∂Ut∂Y=∂2Ut∂Y2-N¯Ut,(18)Ut(0,T)-λ¯∂Ut(Y,T)∂Y|Y=0=0,(19)Ut(1,T)+λ¯∂Ut(Y,T)∂Y|Y=1=1,(20)Ut(Y,0)=-Us(Y).
The solution can be derived by using Laplace transformation techniques [27]. The Laplace transform pair is defined by
(21)U¯(Y,s)=L[U(Y,T)]=∫0∞‍U(Y,T)e-sTdT,U(Y,T)=L-1[U¯(Y,s)]=12πι∫λ1-ι∞λ1+ι∞‍U¯(Y,s)esTds,
over that range of values ofsfor which the integrals exist. Here,sis a parameter, real or complex,Lis the operator that transformsU(Y,T)intoU¯(Y,s), called Laplace transform operator, andL-1is the inverse Laplace transform operator. The solution can be shown as
(22)Ut(Y,T)=L-1[×sinh(b)+2λ¯bcosh(b)λ¯2))-1(eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY)))×(eas((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)λ¯2))-1]-Us,
wherea=Re/2andb=Re2/4+s+N¯. In inverse Laplace transform of the above equation we have simple pole ats=0and infinite number of poles (located on the negative real axis) atsn=-(ln2+a2+N¯),(n=1,2,3,…), whereln=ibn,(bn=Re2/4+sn+N¯)and are given by(23)tan(ln)=-2λ¯ln1-λ¯2(a2+ln2).
The transient part velocity will be
(24)Ut(Y,T)=∑n=1∞Res[U¯t(Y,s)esT]s=sn+Res[U¯t(Y,s)esT]s=0-Us,
whereResstands for residue andU¯t(Y,s)is given by
(25)U¯t(Y,s)=eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY))eas((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)).
We have
(26)∑n=1∞Res[U¯t(Y,s)esT]s=sn=∑n=1∞G1(ln,Y)G2(ln)esnT,
where
(27)G1(ln,Y)=eaY((1-λ¯a)sin(lnY)hhhh+λ¯lncos(lnY)),G2(ln)=easn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))cos(ln)2ln).
The residue ats=0gives steady velocity
(28)Res[U¯t(Y,s)esT]s=0=lims→0[sU¯t(Y,s)esT]=lims→0(eaY((1-λ¯a)sinh(bY)+λ¯bcosh(bY))esT)×(ea((1-λ¯2(a2-b2))sinh(b)+2λ¯bcosh(b)(1-λ¯2(a2-b2))))-1,lims→0b=lims→0Re24+s+N¯=Re24+N¯=b1.
Therefore,
(29)Res[U¯t(Y,s)esT]s=0=(eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)])×(ea[(1-λ¯2(a2-b12))×sinh(b1)+2λ¯b1cosh(b1)(1-λ¯2(a2-b12))])-1.
By using (13), we get
(30)Res[U¯t(Y,s)esT]s=0=Us.The transient part velocity from (24) becomes
(31)Ut(Y,T)=∑n=1∞G1(ln,Y)G2(ln)esnT,Ut(Y,T)=∑n=1∞[cos(ln)2ln(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))×(cos(ln)2lncos(ln)2lneasn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))×cos(ln)2ln))-1esnT].
Therefore, the overall transient solution from (9) becomes
(32)U(Y,T)=∑n=1∞{(cos(ln)2lneaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY))×((eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))easn(cos(ln)2lnλ¯(λ¯+1)sin(ln)-(1+2λ¯-λ¯2(a2+ln2))×cos(ln)2ln))-1)esnT(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))(eaY((1-λ¯a)sin(lnY)+λ¯lncos(lnY)))}+(eaY[(1-λ¯a)sinh(b1Y)+λ¯b1cosh(b1Y)])×(ea[(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)(1-λ¯2(a2-b12))])-1.
To recover Fang [1] result number (13) we substituteλ¯=0,N¯=0,a=Re/2, andb1=Re/2into (32), and we get
(33)U(Y,T)=∑n=1∞{sin(lnY)esnTe(Re/2)Y-eRe/2sncos(ln)/(2ln)}+sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2).
Sinceλ¯=0, so (23) gives
(34)tan(ln)=0⟹sin(ln)=0⟹ln=arcsin(0)=nπ,n=1,2,3,….
Thereforecos(ln)=cos(nπ)=(-1)n; also we haveln=ibn,(bn=Re2/4+sn+N¯), butN¯=0, so
(35)ln=iRe24+sn⟹nπ=iRe24+sn⟹n2π2=-(Re24+sn)⟹sn=-(n2π2+Re24).
Usingln=nπ,cos(ln)=(-1)n, andsn=-(n2π2+Re2/4)into (33) we obtain
(36)U(Y,T)=2πe-(Re2/4)Te(Re/2)YeRe/2×∑n=1∞{n(-1)nsin(nπY)e-n2π2T(n2π2+Re2/4)}+sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2).
Consider
(37)sinh((Re/2)Y)e(Re/2)YeRe/2sinh(Re/2)=e(Re/2)Y(e(Re/2)Y-e-(Re/2)Y)eRe/2(eRe/2-e-Re/2)=(eReY-1)(eRe-1)=(1-eReY)(1-eRe).
So (36) becomes
(38)U(Y,T)=2πe-(Re2/4)Te(Re/2)YeRe/2×∑n=1∞{n(-1)nsin(nπY)e-n2π2T(n2π2+Re2/4)}+(1-eReY)(1-eRe).
Result number (13) of Fang [1] is exactly followed in (38).


## 2.2. Steady State Temperature

It is too difficult to exactly solve the transient energy equation with viscous dissipation for this problem by usingU(Y,T). We solve the steady state energy equation, which in dimensionless form is given by [1]
(39)PrRedσdY=d2σdY2+EcPr(dUsdY)2,σ(0)=0,σ(1)=1,
wherePr=ν/αis the Prandtl number,Ec=U02/cp(θ1-θ2)is the Eckert number,θ1 is temperature of the bottom fixed plate,θ2 is temperature of the top moving plate,σis dimensionless temperature(σ=(θ-θ1)/(θ2-θ1)), and Usis the steady state velocity given by (13). The solution of (39) with BCs (35) forPr≠1, is
(40)σ(Y)=(ePrReY-1)(ePrRe-1)+EcPrM[(ePrReY-1)(ePrRe-1)×{N1(e2(a+b1)-1)+N2(e2(a-b1)-1)-N3(e2a-1)}+{N1(1-e2(a+b1)Y)+N2(1-e2(a-b1)Y)-N3(1-e2aY)}(ePrReY-1)(ePrRe-1)],
where
(41)N1=M1L1,N2=M2L2,N3=M3L3,L1=4(a+b1)2-2PrRe(a+b1),L2=4(a-b1)2-2PrRe(a-b1),L3=Re2(1-Pr),M=1[ea{(1-λ¯2(a2-b12))sinh(b1)+2λ¯b1cosh(b1)}]2,M1=(B1+b12)2,M2=(B1-b12)2,M3=(B12-b122),B1=(a-λ¯(a2-b12)).
When viscous dissipation term is negligible, then energy equation (39) is
(42)d2σdY2-PrRedσdY=0,σ(0)=0,σ(1)=1.
Its solution is
(43)σ(Y)=(ePrReY-1)(ePrRe-1).
So the first term in (40) is the solution of energy equation when viscous dissipation term is negligible and the second term in (40) is the temperature profile from viscous dissipation and MHD. It is found from (40) that the temperature profile is linearly dependent upon the Eckert number. The Nusselt number at the walls will beNu=|dσ/dY|w. The Nusselt number for the bottom wall, Y=0, is
(44)Nu1=|dσdY|Y=0,Nu1=|PrRe(ePrRe-1)[1+EcPrM{N1(e2(a+b1)-1)hhhhhhhhhhhhhhhhhhhhhhhh+N2(e2(a-b1)-1)hhhhhhhhhhhhhhhhhhhhhhhh-N3(e2a-1)}{N1(e2(a+b1)-1)]-2EcPrM{N1(a+b1)hhhhhhhh+N2(a-b1)-N3a}PrRe(ePrRe-1)PrRe(ePrRe-1)|.
The Nusselt number for the upper wall,Y=1, is
(45)Nu2=|dσdY|Y=1,Nu2=|PrReePrRe(ePrRe-1)×[1+EcPrM{N1(e2(a+b1)-1)hhhhhhhhhhhhhhh+N2(e2(a-b1)-1)hhhhhhhhhhhhhhh-N3(e2a-1)}]-2EcPrM{N1(a+b1)e2(a+b1)hhhhhhhhhhhh+N2(a-b1)e2(a-b1)-N3ae2a}PrRe(ePrRe-1)PrRe(ePrRe-1)|.
WhenPr=1, then the energy equation is
(46)d2σdY2-RedσdY=-EcM{M1e2(a+b1)Y+M2e2(a-b1)Y-M3e2aY},σ(0)=0,σ(1)=1.
The solution of (46) is
(47)σ(Y)=(eReY-1)(eRe-1)[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhhhhh-M3Ree2a}]+2EcM{M3YReN4(1-e2(a+b1)Y)hhhhhhh+N5(1-e2(a-b1)Y)+M3YRee2aY},
where
(48)N4=M1L4,N5=M2L5,L4=4(a+b1)2-2Re(a+b1),L5=4(a-b1)2-2Re(a-b1).
The Nusselt number in this case is
(49)Nu1=|Re(eRe-1)×[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhh-M3Ree2a}]-2EcM{M3ReN4(a+b1)hhhhhhhhh+N5(a-b1)-M3Re}Re(eRe-1)|,Nu2=|ReeRe(eRe-1)×[1+EcM{M3ReN4(e2(a+b1)-1)hhhhhhhhhhhhh+N5(e2(a-b1)-1)hhhhhhhhhhhhh-M3Ree2a}]-2EcM{M3ReN4(a+b1)e2(a+b1)hhhhhhhhh+N5(a-b1)e2(a-b1)-M3Reae2a}ReeRe(eRe-1)|.
It is worth recalling in the vicinity of (40) that whenλ¯=0,N¯=0, then result number (18) of Fang [1] can be derived.


## 3. Graphs and Discussion

In this part we discuss the variation of the transient part velocityUt, overall transient velocityU, and steady state velocityUswith distance from the wallYfor different values of Reynolds numberRe, magnetic field parameterN¯, slip parameter λ¯, and timeT.Figures1 and 2 show the variation of transient part velocityUtwith distance from the wallYfor several values of magnetic parameterN¯, by keepingλ¯andTfixed. Figure 1 shows that when there is mass suctionRe<0at the top wall, with increase in magnetic field, transient part velocity decreases in magnitude. Figure 2 shows that when there is mass injectionRe>0at the bottom wall, with increase in magnetic field, transient part velocityUtdecreases and will become weaker as compared to the case of suction. From Figure 3 it is observed that for suction at top wall and for fixed values ofλ¯,N¯transient part velocityUtdecreases in magnitude with increase in time. Figure 4 shows that for injection at bottom wall and for fixed values ofλ¯,N¯transient part velocityUtdecreases in magnitude with increase in time. From Figures 3 and 4 it is seen that the transient part velocity will decay with time, which is consistent with what we expected. From Figures 3 and 4 it is clear that, after a certain time, the transient part velocity will die away and velocity will become developed. Figures 5 and 6 indicate variation of overall transient velocityUwithYfor fixed values ofλ¯andT. Figure 5 shows that, forRe=-5<0, with increase in magnetic parameterN¯, overall transient velocityUdecreases. Figure 6 shows that, forRe=5>0, with increase in magnetic parameterN¯, overall transient velocityUdecreases but weaker. The overall transient velocities forRe=5at different times are depicted in Figure 7. Figure 7 shows that for mass injection at the bottom wall, overall transient velocityUincreases with time.Figure 1
Variation ofUtwithYfor several values ofN-withRe=-5,  λ-=0.1, and T=0.05.Figure 2
Variation ofUtwithYfor several values ofN-withRe=5,  λ-=0.1, and T=0.05.Figure 3
Variation ofUtwithYfor several values of  TwithRe=-5,  λ-=0.1, and N-=0.5.Figure 4
Variation ofUtwithYfor several values of  TwithRe=5,λ-=0.1, and  N-=0.5.Figure 5
Variation ofUwithYfor several values of  N-withRe=-5,λ-=0.1, and  T=0.05.Figure 6
Variation ofUwithYfor several values of  N-withRe=5,λ-=0.1, and  T=0.05.Figure 7
Variation ofUwithYfor several values ofTwithRe=5,λ-=0.1, and  N-=0.5.Figures8 and 9 illustrate the variation of steady state velocityUswithYfor several values of slip parameterλ¯and for fixed value ofN¯. Figure8shows that, for suction at the top wall with increase in slip parameterλ¯, steady state velocityUsincreases. Figure 9 shows that, for injection at the bottom wall with increase in slip parameterλ¯, steady state velocityUsdecreases.Figure 8
Variation ofUswithYfor several values ofλ-withRe=-5,N-=0.5.Figure 9
Variation ofUswithYfor several values ofλ-withRe=5,N-=0.5.Now we discuss the variation of the steady state temperature distributionσwith distance from the surfaceYfor different values of five dimensionless parameters: Reynolds numberRe, Prandtl numberPr, Eckert numberEc, slip parameterλ¯, and magnetic parameterN¯. Variation of the Nusselt number at walls for different Prandtl numbers is also discussed. Figures 10 and 11 elucidate the variation of steady state temperature distributionσwithYfor several values ofPrand for fixed values ofEc,λ¯, andN¯. Figure 10 shows that forRe=5>0, (injection) temperatureσat a point decreases with increase inPr. Figure 11 shows that, forRe=-5<0, (suction) temperature σ with increase in Pr increases at a point. Comparison of Figures 10 and 11 shows that, for mass suction at the bottom wall, the maximum temperature will exist in the fluid because of viscous dissipation.Figure 10
Variation ofσwithYfor several values ofPrwithRe=5,Ec=2,N-=0.5, and λ-=0.1.Figure 11
Variation ofσwithYfor several values ofPrwithRe=-5,Ec=2,N-=0.5, and λ-=0.1.Figures12 and 13 illustrate the variation of Nusselt numbers at the bottom wall withRe, for fixed values ofEc,λ¯, andN¯and for several values ofPr. Figure 12 shows that when there is mass suction at the bottom wall, heat transfer rateNu1increases with increase inPr. Figure 13 shows that when there is mass injection at the bottom wall, heat transfer rateNu1decreases with increase inPr. Figures 14 and 15 depict the variation of Nusselt numbers at the top wall withRe, for fixed values ofEc,λ¯, andN¯and for several values ofPr. Figure 14 shows that when there is mass suction at the top wall, heat transfer rateNu2increases in magnitude with increase inPr. Figure 15 shows that when there is mass injection at the top wall, heat transfer rateNu2increases with increase inPr.Figure 12
Variation ofNu1withRe:-5.5→0.9for several values ofPrwithEc=2,N-=0.5, and λ-=0.1.Figure 13
Variation ofNu1withRe:1→5.5for several values ofPrwithEc=2,N-=0.5, and  λ-=0.1.Figure 14
Variation ofNu2withRe:-5.5→2.2for several values ofPrwithEc=2,N-=0.5, and λ-=0.1.Figure 15
Variation ofNu2withRe:2→6for several values ofPrwithEc=2,N-=0.5, and λ-=0.1.


## 4. Final Remarks

In this study exact solutions for the velocity field and temperature field in the presence of magnetic field, porosity, and slip parameter are constructed. A uniform magnetic field is applied transversely to the flow. The expressions of the velocity field and temperature field for flow subjected to the slip conditions between the two parallel plates and fluid are obtained by means of Laplace transform. The so-obtained solutions, depending on the initial and the boundary conditions, are presented as sum of the steady state transient solutions. The results of Fang [1] are also recovered by taking λ¯=0,N¯=0. Graphical results for mass transfer reveal that it has significant influence on the velocity distribution, temperature distribution, and heat transfer rate at the walls. The current analysis will be useful in dealing with real engineering problems.



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*Source: 705296-2013-12-04.xml*