THREE-DIMENSIONAL FLUCTUATING COUETTE FLOW THROUGH THE POROUS PLATES WITH HEAT TRANSFER

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FLOW THROUGH THE POROUS PLATES WITH HEAT TRANSFER

M. GURIA AND R. N. JANA

Received 23 March 2006; Accepted 23 March 2006

Unsteady Couette flow of a viscous incompressible fluid between two horizontal porous flat plates is considered. The stationary plate is subjected to a periodic suction and the plate in uniform motion is subjected to uniform injection. Approximate solutions have been obtained for the velocity and the temperature fields, skin friction by using pertur- bation technique. The heat transfer characteristic has also been studied on taking viscous dissipation into account. It is found that the main flow velocity decreases with increase in frequency parameter. On the other hand, the magnitude of the cross-flow velocity increases with increase in frequency parameter. It is seen that the amplitude of the shear stress due to main flow decreases while that due to cross-flow increases with increase in frequency parameter. It is also seen that the tangent of phase shifts both due to the main and cross-flows decrease with increase in frequency parameter. It is observed that the temperature increases with increase in frequency parameter.

1. Introduction

Couette flow is important in numerous mechanisms involving the relative motion of two surfaces. The problem of Couette flow is considered important in transpiration cooling.

In this process several engines can be protected from the influence of hot gases. This process is used in turbojet and rocket engines, like combustion chamber walls, exhaust nozzles, and gas turbine blades. The solution is well known when both surfaces are flat and moving in their own planes. An exact solution of Navier-Stokes equations between two parallel plates without suction is well known in Schlichting [3]. Two-dimensional plane Couette flow with transpiration cooling applying uniform injection and suction at the porous plates is discussed by Eckert [1]. Gersten and Gross [2] studied the three- dimensional flow and heat transfer along a flat plate by applying periodic suction. Singh et al. [7] analyzed the three-dimensional flow and heat transfer past a vertical wall. Singh et al. [6] and Singh [4] studied the three-dimensional flow and heat transfer past a porous

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 61023, Pages1–18

DOI10.1155/IJMMS/2006/61023

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plate. Recently Singh [5] discussed the Couette flow between two parallel plates with transverse sinusoidal injection of the fluid at the stationary plate and the constant suction at the plate in motion.

However, the application of the transverse sinusoidal injection or suction velocity in the unsteady problem of transpiration cooling has not yet received much attention. In the present paper we have studied the unsteady Couette flow and heat transfer between two horizontal parallel porous flat plates with periodic suction at the stationary plate and constant injection at the plate in motion. We assume that the periodic suction velocity is time-dependent and perpendicular to the flow direction. Due to the periodic suction the flow becomes three dimensional. The main flow velocity profile, cross-flow velocity profile, and shear stress have been calculated and plotted. The heat transfer characteristic has also been studied on taking viscous dissipation into account. It is found that the main flow velocity increases with the increase in either Reynolds number or suction parameter, while it decreases with the increase in frequency parameter. On the other hand, the magnitude of the cross-flow velocity increases with the increase in either suction parameter or frequency parameter, while it increases near the stationary plate and decreases near the moving plate with the increase in Reynolds number. It is seen that the amplitude of the shear stress due to main flow increases with the increase in Reynolds number but decreases with the increase in frequency parameter. The phase shift decreases with the increase in frequency parameter. But for very small values of Reynolds number, it increases and decreases for large Reynolds number. Also it is seen that the amplitude of the shear stress due to cross-flow increases with the increase in either Reynolds number or frequency parameter. The phase shift decreases with the increase in frequency parameter while it increases with the increase in Reynolds number. It is observed that the temperature increases with the increase in frequency parameter, on the other hand, it increases near the stationary plate and decreases near the moving plate with the increase in Reynolds number.

2. Formulation of the problem

Consider the unsteady flow of a viscous incompressible fluid between two horizontal flat porous plates separated by a distanced. The upper plate moves with a uniform velocity Uin the direction of the flow. We choose a cartesian coordinate system with its origin on the lower stationary plate,x-axis is in the direction of the flow,y-axis is perpendicular to the plate, andz-axis normal to thexy-plane.

The upper plate is subjected to a constant injection−V0and the lower plate to a transverse sinusoidal suction velocity distribution of the form

v= −V0

1 +cos πz

d ⁻ct

, (2.1)

where(1) is the amplitude of the suction velocity. Denoting velocity components u,v,win the directionsx-,y-, andz-axes, respectively, the flow is governed by

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the following equations:

∂v

∂y+∂w

∂z ⁼0,

∂u

∂t +v∂u

∂y+w∂u

∂z ⁼ν ∂²u

∂y²+∂²u

∂z²

,

∂v

∂t+v∂v

∂y+w∂v

∂z^{= −} 1 ρ

∂p

∂y+ν ∂²v

∂y²+∂²v

∂z²

,

∂w

∂t +v∂w

∂y+w∂w

∂z ^{= −} 1 ρ

∂p

∂z +ν ∂²w

∂y² +∂²w

∂z²

,

(2.2)

whereνis the kinematic viscosity,ρis the density,pis the fluid pressure.

The boundary conditions of the problem are u=0, v= −V0

1 + cos

π

dz−ct

, w=0 aty=0, u=U, v= −V0, w=0 aty=d.

(2.3)

Introducing the nondimensional variables y= y

d , z=z

d, t=ct, p= p

ρU², u=u

U, v=v

U, w=w

U , (2.4) equation (2.2) becomes

∂v

∂y+∂w

∂z ⁼0, ω∂u

∂t + Re

v∂u

∂y+w∂u

∂z

=∂²u

∂y²+∂²u

∂z², ω∂v

∂t + Re

v∂v

∂y+w∂v

∂z

= −Re∂p

∂y+ ∂²v

∂y²+∂²v

∂z²

, ω∂w

∂t + Re

v∂w

∂y +w∂w

∂z

= −Re∂p

∂z+ ∂²w

∂y² +∂²w

∂z²

,

(2.5)

where Re=Ud/ν, the Reynolds number; S=V0/U, the suction parameter; and ω= cd²/ν, the frequency parameter. Using (2.4) the boundary conditions (2.3) become

u=0, v= −S[1 +cos(πz−t)], w=0 aty=0,

u=1, v= −S, w=0 aty=1. (2.6)

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3. Solution of the problem

In order to solve the diﬀerential equation (2.5), we assume the solution of the following form:

u=u0(y) +u1(y,z,t) +²u2(y,z,t) +···, v=v0(y) +v1(y,z,t) +²v2(y,z,t) +···, w=w0(y) +w1(y,z,t) +²w2(y,z,t) +···,

p=p0(y) +p1(y,z,t) +²p2(y,z,t) +···.

(3.1)

When=0 the flow becomes two-dimensional with constant suction and injection at both the plates. In this case (2.5) is reduced to

v0=0,

u0+SReu0=0, (3.2)

where the primes denote diﬀerentiation with respect toy, and the corresponding boundary conditions are

u0=0, v0= −S aty=0,

u0=1, v0= −S aty=1. (3.3)

The solutions of (3.2) using (3.3) become

v0(y)= −S, u0(y)=1−e⁻^S^Re^y

1−e⁻^SRe . (3.4)

When=0, for small values of, we take only uptoO(). Substituting (3.1) in (2.5) and comparing the coeﬃcients offrom both sides, we get

∂v1

∂y +∂w1

∂z ⁼0, (3.5)

ω∂u1

∂t + Re

−S∂u1

∂y +v1∂u0

∂y

=∂²u1

∂y² +∂²u1

∂z² , (3.6)

ω∂v1

∂t ⁻SRe∂v1

∂y ^{= −}Re∂p1

∂y +∂²v1

∂y² +∂²v1

∂z² , (3.7)

ω∂w1

∂t ⁻SRe∂w1

∂y ^{= −}Re∂p1

∂z +∂²w1

∂y² +∂²w1

∂z² . (3.8)

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The corresponding boundary conditions are reduced to

u1=0, v1= −Scos(πz−t), w1=0 aty=0,

u1=0, v1=0, w1=0 aty=1. (3.9) These are the linear partial diﬀerential equations describing the three-dimensional flow.

To solve (3.6), (3.7), and (3.8) we assumeu1,v1,w1, andp1of the following form:

u1(y,z,t)=u11(y)eⁱ⁽^πz⁻^t⁾, v1(y,z,t)=v11(y)eⁱ⁽^πz⁻^t⁾, w1(y,z,t)= i

πv11(y)eⁱ⁽^πz⁻^t⁾, p1(y,z,t)=p11(y)eⁱ⁽^πz⁻^t⁾.

(3.10)

The corresponding boundary conditions are

u11=0, v11= −S, v11 =0 aty=0,

u11=0, v11=0, v11 =0 aty=1. (3.11) Substituting (3.10) in (3.6), (3.7), and (3.8), we get the following diﬀerential equations:

u11+SReu11−

π²−iω u11=Rev11u0, v11+SRev11 −

π²−iω v11=Rep11 , v11+SRev11−

π²−iω v11=Reπ²p11.

(3.12)

Solving (3.12) under the boundary conditions (3.11), we get u1(y,z,t)=

A3e⁻^r¹^y+A4e⁻^r²^y+C1e⁻⁽^r¹⁺^S^Re)^y+C2e⁻⁽^r²⁺^S^Re)^y +C3e⁽^π⁻^S^Re)^y+C4e⁻⁽^π⁺^S^Re)^yeⁱ⁽^πz⁻^t⁾,

(3.13)

v1(y,z,t)=

Ae⁻^r¹^y+Be⁻^r²^y+Ce^{π y}+De⁻^{π y}eⁱ⁽^πz⁻^t⁾, (3.14) w1(y,z,t)= −i

π

Ar1e⁻^r¹^y+Br2e⁻^r²^y−Cπe^{π y}+Dπe⁻^{π y}eⁱ⁽^πz⁻^t⁾, (3.15)

p1(y,z,t)=

A1e^{π y}+A2e⁻^{π y}eⁱ⁽^πz⁻^t⁾, (3.16)

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where

r1=SRe +S²Re²+4π²−iω

2 , r2=SRe−

S²Re²+4π²−iω

2 ,

A=Se^ππr4+r6 +Se⁻^πr6−πr4

2r3r6−r4r5 , B=−Se^πr5+πr3 −Se⁻^πr5−πr3

2r3r6−r4r5 , C= − 1

2π

Aπ−r1 +Bπ−r2 +Sπ, D= − 1 2π

Aπ+r1 +Bπ+r2 +Sπ,

r3=e⁻^r¹− π−r1

2π e^π− π+r1

2π e⁻^π, r4=e⁻^r²− π−r2

2π e^π− π+r2

2π e⁻^π, r5=r1e⁻^r¹+1

2e^ππ−r1 −1

2e⁻^ππ+r1 , r6=r2e⁻^r²+1

2e^ππ−r2 −1

2e⁻^ππ+r2 , C1= SRe²

1−e⁻^S^Re A

r1²+SRer1−π²+iω , C2= SRe² 1−e⁻^S^Re

B

r2²+SRer2−π²+iω , C3= SRe²

1−e⁻^S^Re

C

(iω−πSRe , C4= SRe² 1−e⁻^S^Re

D (iω+πSRe), A1=SReπ+iω

Reπ C, A2=SReπ−iω Reπ D, A3= −

A4+C1+C2+C3+C4 , A4= 1

e⁻^r¹−e⁻^r² C1

e⁻^r¹⁻^S^Re−e⁻^r¹ +C2

e⁻^r²⁻^S^Re−e⁻^r¹

+C3

e^π⁻^S^Re−e⁻^r¹ +C4

e⁻^π⁻^S^Re−e⁻^r¹ .

(3.17) 4. Result and discussion

We have presented the nondimensional main flow velocityuagainstyfor diﬀerent values of Reynolds number Re, suction parameterS, and frequency parameterωforz=0.0, t=0.2,=0.2 in Figures4.1,4.2, and4.3. From the figure it is seen that the main flow velocityuincreases with increase in either Re orS, while it decreases with the increase in ω. The cross-flow velocity profile is shown in Figures4.4and4.5against yforz=0.5, t=0.2,=0.2. It is observed that the magnitude of the cross-flow velocitywincreases with the increase in eitherSorωbut it increases near the stationary plate and decreases near the moving plate with the increase in Re. This is due to the fact that suction at the stationary plate and injection at the moving plate are two exactly opposite processes. Also

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

u

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y

Re₌0.5 Re=1 Re₌1.5

Re₌2 Re=2.5

Figure 4.1. Main velocityuforω=6.0,S=1.0,t=0.2,z=0.0,=0.2.

0 0.2 0.4 0.6 0.8 1

u

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y

S=0.2 S=0.4 S=0.6 S=0.8

S=1 S= −0.5 S= −1

Figure 4.2. Main velocityuforω=6.0, Re=5.0,Z=0.0,t=0.2,=0.2.

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

u

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y

ω=2

ω=6 ω=10

ω=14

Figure 4.3. Main velocityuforS=1.0, Re=5.0,t=0.2,z=0.0,=0.2.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−10w

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y

Re=0.5 Re₌1 Re=1.5

Re=2 Re₌2.5

Figure 4.4. Cross-velocity−10wforω=6.0,S=1.0,z=0.5,t=0.2,=0.2.

the variations of cross-velocitywfor diﬀerent values ofωare shown inTable 4.1. From the table it is observed that the magnitude ofwincreases with increase inω.

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−1

−0.5 0 0.5 1 1.5

−10w

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y

S=0.2 S=0.4 S=0.6 S=0.8

S=1 S= −0.5 S= −1

Figure 4.5. Cross-velocity−10wforω=6.0, Re=5.0,z=0.5,t=0.2,=0.2.

The nondimensional shear stress components due to the main flow and cross-flow at the platey=0 are, respectively,

τx=dτ_x^∗ μU ⁼

du0

dy

y=0

+ du1

dy

y=0

= du0

dy

y=0

+ du11

dy

y=0e^i(πz⁻^t)

=τ_u0+Re1cosπz−t+φ1 , τz=dτ_z^∗

μU ⁼ dw0

dy

y=0

+ dw1

dy

y=0

=i π

dv11

dy

y=0eⁱ⁽^πz⁻^t⁾

=Re2cosπz−t+φ2 ,

(4.1)

where

Re1= X1²+Y1²

1/2

, tanφ1=Y1

X1, Re2=

X2²+Y2²

_1/2

, tanφ2=Y2

X2.

(4.2)

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Table 4.1. Cross-flow velocity forS=1.0, Re=5.0,z=0.5,t=0.2,=0.2.

y −10 w

ω=2.0 ω=6.0 ω=10.0 ω=14.0

0.00 0.000000 0.000000 0.00000 0.000000

0.20 1.109254 1.113393 1.118991 1.125880

0.40 0.907241 0.903539 0.898763 0.893008

0.60 0.568972 0.565611 0.561192 0.555843

0.80 0.282000 0.281842 0.281629 0.281395

1.00 0.000011 0.000000 −0.00500302 0.000001

Table 4.2. Shear stress due to main flow.

Re Re1 −tanφ1

ω=5.0 ω=10.0 ω=15.0 ω=5.0 ω=10.0 ω=15.0

0.4 0.3133 0.1787 0.1344 1.0171 1.2728 1.2516

0.6 0.4892 0.2891 0.2182 0.6059 0.9704 1.0421

0.8 0.6606 0.4119 0.3132 0.3236 0.7402 0.8681

1.0 0.8185 0.5452 0.4191 0.0970 0.5541 0.7193

Table 4.3. Shear stress due to cross-flow.

Re Re2 −tanφ2

ω=5.0 ω=10.0 ω=15.0 ω=5.0 ω=10.0 ω=15.0

0.4 3.9497 4.1925 4.5407 6.2946 3.3011 2.3627

0.6 4.0614 4.2989 4.6407 6.5018 3.4055 2.4331

0.8 4.1760 4.4080 4.7430 6.7209 3.5155 2.5070

1.0 4.2937 4.5198 4.8478 6.9523 3.6316 2.5847

The shear stresses due to main flow and cross-flow are shown in Tables4.2and4.3for S=1.0. From the table it is seen that the amplitude Re1of the shear stress due to main flow increases with the increase in Re but decreases with the increase inω. The magnitude of the tangent of phase shift tanφ1decreases with the increase inωwhile it increases for small values of Re and decreases for large values of Re. It is also seen fromTable 4.2that there is always a phase lag.Table 4.3shows the shear stress due to cross-flow forS=1.0.

It is observed that the amplitude Re2of shear stress due to cross-flow increases with the increase in either Re orω. It is also seen that there is always a phase lag. The magnitude of the tangent of phase angle increases with the increase in Re but decreases with the increase inω.

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5. Heat transfer

To find the temperature distribution we consider the energy equation

∂T

∂t +v∂T

∂y +w∂T

∂z ⁼α ∂²T

∂y² +∂²T

∂z²

+ μ

ρC_pΦ, (5.1) whereΦis the viscous dissipation function given by

Φ=2 ∂v

∂y 2

+ ∂w

∂z 2

+ ∂u

∂y 2

+ ∂w

∂y+∂v

∂z 2

+ ∂u

∂z 2

, (5.2)

whereCp is the specific heat at constant pressure, andμis the viscosity,αis the thermal diﬀusivity of the fluid. We assume the temperature at the lower and upper plates areT0

andT1(T1> T0), respectively.

The temperature boundary conditions are

T=T0 aty=0,

T=T1 aty=d. (5.3)

Introducing the nondimensional variables θ=T−T0

T1−T0, Pr=μ

α, Ec= U² Cp

T1−T0 (5.4)

and on using (2.4), (5.1), and (5.2) become Prω∂θ

∂t + Re Pr

v∂θ

∂y+w∂θ

∂z

=∂²θ

∂y²+∂²θ

∂z² + PrE_cΦ, (5.5) where

Φ=2 ∂v

∂y 2

+ ∂w

∂z 2

+ ∂u

∂y 2

+ ∂u

∂z 2

+ ∂w

∂y +∂v

∂z 2

. (5.6)

The corresponding boundary condition is reduced to

θ(0)=0 θ(1)=1. (5.7)

We assume the solution of the temperature equation in the form

θ=θ0(y) +θ1(y,z,t) +²θ2(y,z,t) +···. (5.8) Substitute (5.8), (3.1) in (5.13) and equate the term independent ofand the coeﬃcients of and neglect the higher order ofas1. Equating the terms independent of,

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we get

θ0+SRe Prθ0= −Pr Ecu0². (5.9) The corresponding temperature boundary conditions are

θ0(0)=0, θ0(1)=1. (5.10)

Solving (5.9) under the boundary conditions (5.10), we get θ0(y)=h1

1−e⁻^{SRe Pry} +h2

e⁻^2SRe^y−1 for Pr=2.0, θ0(y)=D1

1−e⁻²^S^Re^y +D3ye⁻²^S^Re^y for Pr=2.0, (5.11)

where

h1=1 +h2

1−e⁻²^S^Re

1−e⁻^S^{Re Pr} , h2= −Pr Ec

21−e⁻^S^Re ²(2−Pr) for Pr=2.0, D1=1−D3e⁻²^S^Re

1−e⁻^2SRe , D3= EcSRe

1−e⁻^S^Re ² for Pr=2.0.

(5.12)

Equating the coeﬃcient offrom both sides, we have

Prω∂θ1

∂t + Re Pr

v1∂θ0

∂y +v0∂θ1

∂y

=∂²θ1

∂y² +∂²θ1

∂z² + 2Pr Ecu0u1. (5.13) The corresponding conditions forθ1are

θ1(0)=0, θ1(1)=0. (5.14)

We assume the solution of the above diﬀerential equation (5.13) of the form

θ1(y,z,t)=θ11(y)e^i(πz⁻^t). (5.15) Substituting (5.15) in (5.13) and on using (3.10), we get

θ11+SRe Prθ11 −

π²−iPrω θ11=Re Prθ0v11−2Pr Ecu0u11. (5.16) The corresponding boundary conditions are reduced to

θ11(0)=0, θ11(1)=0. (5.17)

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The solution of (5.16) subject to the boundary conditions (5.17) and on using (3.4), (3.13), (3.14), and (5.11) are

θ11=Le⁻^m¹^y−e⁻^m²^y +S²Re²Pr²h1

A λ1

e⁻⁽^r¹⁺^S^{Re Pr)}^y−e⁻^m²^y+ B λ2

e⁻⁽^r²⁺^S^{Re Pr)}^y−e⁻^m²^y + C

λ3

e⁽^π⁻^S^{Re Pr)}^y−e⁻^m²^y+D λ4

e⁻⁽^π⁺^S^{Re Pr)}^y−e⁻^m²^y

−2SRe Pr

Ah2Re−EcC1

r1+SRe 1−e⁻^S^Re

1 λ5

e⁻⁽^r¹⁺²^S^Re)^y−e⁻^m²^y

+

Bh2Re−EcC2

r2+SRe 1−e⁻^S^Re

1 λ6

e⁻⁽^r²⁺²^S^Re)^y−e⁻^m²^y

+

Ch2Re +EcC 3(π−SRe) 1−e⁻^S^Re

1 λ7

e⁽^π⁻²^S^Re)^y−e⁻^m²^y

+

Dh2Re−EcC4(π+SRe) 1−e⁻^S^Re

1 λ8

e⁻⁽^π⁺²^S^Re)^y−e⁻^m²^y

+2Pr EcSRe 1−e⁻^S^Re

A3r1

λ9

e⁻⁽^r¹⁺^S^Re)^y−e⁻^m²^y

+A4r2

λ10

e⁻⁽^r²⁺^S^Re)^y−e⁻^m²^y

for Pr=2.0,

(5.18)

and for Pr=2.0,

θ11(y)=L1e⁻^m³^y+M1e⁻^m⁴^y+

2 ReAD4+C1 r1+SRe

β

1

μ1e⁻^(r¹^+2SRe)y +

2 ReBD4+C2 r2+SRe

β

1

μ2e⁻^(r²^+2SRe)y+

2 ReCD4−C3(π−SRe) β

1 μ3

×e^(π⁻^2SRe)y+

2 ReDD4+C4(π+SRe) β

1

μ4e⁻^(π+2S^Re)y +1

β A3r1

μ5 e⁻^(r¹^+S^Re)y+A4r2

μ6 e⁻^(r²^+SRe)y

−4SRe²D3

A μ1

y+2r1+SRe μ1

×e⁻⁽^r¹⁺²^S^Re)^y+ B μ2

y+2r2+SRe μ2

e⁻⁽^r²⁺²^S^Re)^y+ C μ3

×

y−2(π−SRe) μ3

e⁽^π⁻²^S^Re)^y+ D μ4

y+2(π+SRe) μ4

e⁻⁽^π⁺²^S^Re)^y

, (5.19)