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 velocity vw at the bottom wall and mass suction velocity vw at the top wall; vw>0 corresponds to injection and vw<0 corresponds 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),
where u(y,t) is the velocity of the fluid in the y-direction which is along the wall direction, y is the distance alon y-axis g, t is the time, ν=μ/ρ, N=σB02/ρ, μ is the dynamic viscosity, ν is the kinematic viscosity, ρ is the density of the fluid, σ is the electric conductivity of the fluid, B0 is the applied magnetic field, and N is 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,
where λ is the slip parameter (λ=0 gives the usual no slip condition at the wall) and U0 is the velocity at the upper wall. The problem exactly reduces to the problem of Fang [1] if we take N=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,
where N¯=N (h2/ν), λ¯=λ/h, and Re=vw(h/ν) (Reynolds number). Decomposing U(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)],
where a=Re/2 and b1=Re2/4+N¯. When λ-=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, then Re=0, so a=0 and b1=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, then Re=0, N¯=0, and slip parameter λ¯=0 then (10), (11), and (12) will collapse into
(16)d2Us(Y)dY2=0, Us(0)=0, Us(1)=1.
Its solution is Us(Y)=Y, which is conventional Couette flow. When Re→-∞, N¯=0, and λ¯=0, then (13) gives Us(Y)=1, except at the bottom wall (at the bottom wall Y=0 and Us(0)=0). When Re→∞, N¯=0, and λ¯=0, then (13) gives Us(Y)=0, except at the upper wall (at the upper wall Y=1 and Us(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 of s for which the integrals exist. Here, s is a parameter, real or complex, L is the operator that transforms U(Y,T) into U¯(Y,s), called Laplace transform operator, and L-1 is 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,
where a=Re/2 and b=Re2/4+s+N¯. In inverse Laplace transform of the above equation we have simple pole at s=0 and infinite number of poles (located on the negative real axis) at sn=-(ln2+a2+N¯), (n=1,2,3,…), where ln=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,
where Res stands for residue and U¯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 at s=0 gives 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 substitute λ¯=0, N¯=0, a=Re/2, and b1=Re/2 into (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).
Since λ¯=0, so (23) gives
(34)tan(ln)=0⟹sin(ln)=0⟹ln=arcsin(0)=nπ, n=1,2,3,….
Therefore cos(ln)=cos(nπ)=(-1)n; also we have ln=ibn,(bn=Re2/4+sn+N¯), but N¯=0, so
(35)ln=iRe24+sn⟹nπ=iRe24+sn⟹n2π2=-(Re24+sn)⟹sn=-(n2π2+Re24).
Using ln=nπ, cos(ln)=(-1)n, and sn=-(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 using U(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,
where Pr=ν/α is 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, σ is dimensionless temperature (σ=(θ-θ1)/(θ2-θ1)), and Us is the steady state velocity given by (13). The solution of (39) with BCs (35) for Pr≠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 be Nu=|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)|.
When Pr=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 when λ¯=0, N¯=0, then result number (18) of Fang [1] can be derived.