# Some Effects of Side Walls on Unsteady Flow of a Viscous Fluid over a Plane Wall

Authors: M. Emin Erdoğan; C. Erdem İmrak
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

Four types of unsteady flows of a viscous fluid over a plane wall bounded by two side walls are considered. They are flow caused by impulsive motion of a plate, flow due to oscillation of a plate, flow induced by constantly accelerating plate, and flow imposed by a plate that applies a constant tangential stress to the fluid. In order to solve these problems, the sine and cosine transformations are used, and exact solutions for the velocity distribution are found in terms of definite integrals. The cases for which the time goes to infinity and the distance between two side walls goes to infinity are compared with the cases for flows over a plane wall in the absence of the side walls. These provide to know the required time to attain the steady-state and what is the distance between the side walls for which the measured value of the velocity or the stress would be unaffected by the presence of the side walls.


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## Body

## 1. Introduction

The flow of a viscous fluid over a plane wall with different boundary and initial conditions has been investigated by many authors. The fluid over a plane wall is initially at rest and is set in motion under the application of a body force considered by Erdoğan [1]. He has shown that there are similarities between the flows considered and Stokes’ first and second problems. The flow over a plate, which is initially at rest and is suddenly moved in its own plane with different boundary and initial conditions, has been considered in [2–5]. The flows over a plane wall can be realized due to the impulsive motion of a plate, generated by the constant acceleration of a plate, caused by the oscillation of a plate and due to the application of the tangential stress to the fluid. The extension of these flow systems to the flows in a rotating system [6–9], to the flows of non-Newtonian fluid [10–18] was made by many authors. In [1–18], the authors assumed that the plate extends to infinity. It is well known that the plate is bounded by two side walls and the effect of the side walls is very important. If one wishes to know what is the distance between the side walls for which the measured value of the velocity or stress or temperature would be unaffected by the presence of the side walls, it is necessary to investigate the flows over the plate between two side walls. The effects of the side walls for some flows have been considered in [19–25].The aim of this paper is to investigate the effects of the side walls on the unsteady flow over a plane wall. Four illustrative examples are considered. They are flow due to the impulsive motion of a plate, flow caused by oscillation of a plate, flow due to the constant acceleration of the plate and the flow generated by tangential stress on the fluid. It is well known that the solution of the governing equation for the flow of second-grade fluid over a plane wall obtained by the Laplace transform method is failed [16, 26, 27]. For the flow over a plane wall bounded by two side walls, the shear stress at the bottom wall cannot be calculated by the expression obtained by the Laplace transform method [28–31]. For this reason, in this paper, the Fourier transform method is used.The first problem considered is the flow due to impulsive motion of a plate. The velocity and the flux across a plane normal to the flow are given in terms of definite integrals. In the limiting cases when time goes to infinity, the solution reduces to the steady-state, and when the distance between two side walls goes to infinity, the solution reduces to the flow over a plate. The second problem considered is the flow due to the oscillation of a plane wall. Two solutions are given in terms of definite integrals. It is important fact that the starting solution can be represented as the sum of the steady-state and transient solutions. For large values of time, the transient solutions disappear. The limiting cases when time goes to infinity, the distance between two side walls goes to infinity and the frequency of the oscillation goes to zero are discussed. The third problem is the flow induced by a constantly accelerating plate. In the limiting case when the distance between the side walls goes to infinity, the solution reduces to the flow over a plate, which can be expressed in terms of a tabulated function. The fourth problem is the flow induced by a plate that applies a constant stress to the fluid. The limiting cases when the time goes to infinity and when the distance between the side walls goes to infinity are discussed. It is a very important fact that these four examples show that the required time to attain the steady-state is affected by the side walls.


## 2. Flow due to the Impulsive Motion of a Plane Wall

The fluid is over a plane wall and between two side walls perpendicular to the plate, Thex-axis is taken along the plate, the y-axis and z-axis are perpendicular to the x-axis. Flow geometry and coordinate system are illustrated in Figure 1. The distance between the side walls is 2b and z-axis extends to infinity. The governing equation isFigure 1
Flow geometry and coordinate system.(2.1)∂u∂t=ν(∂2u∂y2+∂2u∂z2),
where u(y,z,t) is the velocity; ν is kinematic viscosity of the fluid and t is time. The boundary and initial conditions are(2.2)u(±b,z,t)=0∀t,u(y,0,t)=Ufort>0,-b<y<b,u(y,∞,t)=0∀t,-b≤y≤b,u(y,z,0)=0for-b≤y≤b,z>0.
The shear stress at the bottom wall cannot be calculated by the expression for velocity obtained by the Laplace transform method [28–31]. For this reason, an expression for velocity by using the sine transform method is obtained.The first boundary condition suggests thatu(y,z,t) can be written in the following form:(2.3)uU=∑n=0fn(z,t)coskny,
where kn=(2n+1)π/2b. The sine transform of fn(z,t) [32] is(2.4)f̅n=∫0∞fnsinλzdz,
and f̅n(0)=0. The problem reduces to the solution of the following ordinary differential equation:(2.5)f̅n′+ν(kn2+λ2)f̅n=4(-1)nνλ(2n+1)π,
where a prime denotes differentiation with respect to t. The solution is(2.6)f̅n=4(-1)n(2n+1)πλλ2+kn2[1-e-(kn2+λ2)νt].
The inverse of f̅n is given by the relation [32](2.7)fn=2π∫0∞f̅nsinλzdλ,
and inserting the expression for f̅n into this equation, one finds(2.8)fn=8(-1)n(2n+1)π2∫0∞λsinλzλ2+kn2dλ-8(-1)n(2n+1)π2∫0∞λsinλzλ2+kn2e-(kn2+λ2)νtdλ.
The first integral is(2.9)∫0∞λsinλzλ2+kn2dλ=π2e-knz,
and since [28](2.10)4π∑n=0(-1)n2n+1e-knzcoskny=1-2π∫0∞coshλyλcoshλbsinλzdλ,
the velocity becomes(2.11)uU=1-2π∫0∞coshλyλcoshλbsinλzdλ-4π∑n=0(-1)ncosknyknb∫0∞λsinλzλ2+kn2e-(kn2+λ2)νtdλ,
or(2.12)uU=2πtan-1cos(πy/2b)sinh(πz/2b)-2π∑n=0(-1)ncoskny2n+1×{2e-knz-[e-knzerfc(z2νt-knνt)+eknzerfc(z2νt+knνt)]}.
In the limiting case, when t goes to infinity, (2.11) reduces to(2.13)uU=1-2π∫0∞coshλyλcoshλbsinλzdλ.
This is the expression for velocity in the case of the steady-state.Whenb goes to infinity in (2.11), since(2.14)∫0∞sinλzλe-λ2νtdλ=π2erfz2νt,
one finds(2.15)uU=erfcz2νt.
The variation of u/U with z/b for various values of νt/b2, at y=0, with and without side walls is illustrated in Figure 2. It is clearly seen from the figure that the required time to attain the steady-state in the presence of the side walls is shorter than that of the case in the absence of the side walls.Figure 2
The variation ofu/U with z/b for various values of νt/b2, at y=0, with and without side walls; R=Ub/ν.The volume flux across a plane normal to the flow is given by(2.16)Q=∫-bb∫0∞udydz.
Inserting (2.11) into this equation and using the integral [33](2.17)∫0∞sinλzdλ=1z
and the integral(2.18)∫0∞e-(λ2+kn2)νtλ2+kn2dλ=π2knerfc(knνt),
one finds(2.19)QUb2=4πF(∞)-4∑n=01b3kn3erfc(knνt),
where F(∞) is given by a definite integral(2.20)F(∞)=∫0∞ξ-tanhξξ3dξ,
and the numerical value of F(∞) is 0.8525867976 correct to ten decimal places [30]. The comparison of the values of the volume flux in the case of the steady-state gives(2.21)F(∞)=8π2∑n=0(2n+1)-3.The shear stress at the bottom wall can be calculated by (2.11). Taking the derivative of (2.11) with respect to zand then putting z=0, one finds(2.22)(σxz)z=0ρU2=-[νUbcosπ(y/(2b))+2νU∑n=0(-1)ncosknybknνtierfc(knνt)],
where(2.23)ierfcx=∫x∞erfcξdξ
is the integral of the complementary error function, which can be obtained from the tables [34]. When t goes to infinity since ierfc(∞)=0, (2.22) reduces to the expression for the steady-state(2.24)(σxz)z=0ρU2=-νUbcosπ(y/(2b)).
In the absence of the side walls, namely, when b goes to infinity, since ierfc(0)=1/π, (2.22) reduces to(2.25)(σxz)z=0ρU2=-1Uνπt.


## 3. Flow due to Oscillation of a Plane Wall

The flow over a plane wall which is initially at rest and the plate begins to oscillate in its own plane is termed Stokes’ second problem [35]. The x-axis is taken along the plate; the y-axis and z-axis are perpendicular to the x-axis. The distance between the side walls is 2b, and the z-axis extends to infinity. The governing equation is (2.1). The boundary and initial conditions are(3.1)u(±b,z,t)=0∀t,u(y,0,t)=UcosωtorUsinωtfort>0,-b<y<b,u(y,∞,t)=0∀t,-b≤y≤b,u(y,z,0)=0for-b≤y≤b,-z>0.
The first boundary condition suggests that u(y,z,t) can be written in the following form:(3.2)uU=∑n=0fn(z,t)coskny,
where kn=(2n+1)π/2b. The sine transform of fn(z,t) is(3.3)f̅n=∫0∞fnsinλzdz
and f̅n(0)=0. The problem reduces to the solution of the following ordinary differential equation:(3.4)f̅n′+ν(kn2+λ2)f̅n=4(-1)nνλ(2n+1)πcosωt,
where a prime denotes differentiation with respect to t. The solution is(3.5)f̅n=4(-1)n(2n+1)π{ν2λ(λ2+kn2)ω2+ν2(λ2+kn2)2cosωt+ωνλω2+ν2(λ2+kn2)2sinωt-ν2λ(λ2+kn2)ω2+ν2(λ2+kn2)2e-(λ2+kn2)νt},
respectively,(3.6)f̅n=4(-1)n(2n+1)π{-ωνλω2+ν2(λ2+kn2)2cosωt+ν2λ(λ2+kn2)ω2+ν2(λ2+kn2)2sinωt+ωνλω2+ν2(λ2+kn2)2e-(λ2+kn2)νt},
The inverse of f̅n is given by the relation(3.7)fn=2π∫0∞f̅nsinλzdλ,
and inserting the expression for f̅n into this equation, one finds(3.8)fn=8(-1)n(2n+1)π2[(cosωt)∫0∞ν2λ(kn2+λ2)ω2+ν2(kn2+λ2)2sinλzdλ+(sinωt)∫0∞ωνλω2+ν2(kn2+λ2)2sinλzdλ-∫0∞ν2λ(kn2+λ2)ω2+ν2(kn2+λ2)2e-(kn2+λ2)νtsinλzdλ],
respectively,(3.9)fn=8(-1)n(2n+1)π2[-(cosω)t∫0∞ωνλω2+ν2(kn2+λ2)2sinλzdλ+(sinωt)∫0∞ν2λ(kn2+λ2)ω2+ν2(kn2+λ2)2sinλzdλ+∫0∞ωνλω2+ν2(kn2+λ2)2e-(kn2+λ2)2νtsinλzdλ].
Since [36](3.10)∫0∞λsinλzω2+ν2(kn2+λ2)2dλ=π2ωνe-AzsinBz,∫0∞λ(kn2+λ2)sinλzω2+ν2(kn2+λ2)2dλ=π2ν2e-AzcosBz,
where(3.11)2A2=kn4+ω2ν2+kn2,2B2=kn4+ω2ν2-kn2,
inserting the integrals into (3.8) and (3.9), one finds(3.12)fn=4(-1)n(2n+1)π[e-Azcos(ωt-Bz)-2π∫0∞ν2λ(kn2+λ2)sinλzω2+ν2(kn2+λ2)2e-(kn2+λ2)νtdλ],
respectively,(3.13)fn=4(-1)n(2n+1)π[e-Azsin(ωt-Bz)+2π∫0∞ωνλsinλzω2+ν2(kn2+λ2)2e-(kn2+λ2)νtdλ].
In the limiting case, where t goes to infinity, (3.8) and (3.9) reduce to(3.14)fn=4(-1)n(2n+1)π[e-Azcos(ωt-Bz)],
respectively,(3.15)fn=4(-1)n(2n+1)π[e-Azsin(ωt-Bz)].
When b goes to infinity, (3.8) and (3.9) reduce to [5](3.16)uU=e-(ω/2ν)zcos(ωt-ω2νz),
respectively,(3.17)uU=e-(ω/2ν)zsin(ωt-ω2νz).
The variation of u/U with (ω/ν)1/2z for various values of ωt at y=0, with and without side walls is illustrated in Figure 3. It is obvious that the required time to attain the steady-state for the cosine oscillation of the plate is shorter than for the sine oscillation of it. There is an appreciable effect of the side walls on the disappearance of the transients. More details were given in [2, 31]. The first terms in (3.12) and (3.13) show the steady-state and the second terms denote transients. When t  goes to infinity the transients disappear. The integrals in (3.12) and (3.13) can be written asFigure 3
The variation ofu/U with (ω/ν)1/2z for various values of  ωt, at y=0, with and without side walls. R=ωb2/ν.(3.18)∫0∞λ(kn2+λ2)sinλzω2+ν2(kn2+λ2)2e-(kn2+λ2)νtdλ=e-kn2νt2v2(F+G),∫0∞λsinλzω2+ν2(kn2+λ2)2e-(kn2+λ2)νtdλ=ie-kn2νt2ωv(F-G),
where(3.19)F=∫0∞λsinλzλ2+β2e-λ2νtdλ,G=∫0∞λsinλzλ2+α2e-λ2νtdλα2=kn2-iων,β2=kn2+iων,∫0∞xsinbxc2+x2e-ax2dx=π4eac2[2e-bc-ebcerfc(b2a+ca)-e-bcerfc(b2a-ca)].


## 4. Flow Induced by a Constantly Accelerating Plate

The plate is initially at rest and has, after time zero, a constant accelerationA. The governing equation is given by (2.1). The boundary and initial conditions are(4.1)u(±b,z,t)=0∀t,u(y,0,t)=Atfort>0,-b<y<b,u(y,∞,t)=0∀t,-b≤y≤b,u(y,z,0)=0for-b≤y≤b,z>0.
The first boundary condition suggests that  u(y,z,t) can be written in the following form:(4.2)u=A∑n=0fn(z,t)coskny,
where kn=(2n+1)π/2b. The sine transform of fn(z,t) is f̅n(t) and the initial condition becomes f̅n(0)=0. The problem reduces to the solution of the following ordinary differential equation:(4.3)f̅n′+ν(kn2+λ2)f̅n=4(-1)n(2n+1)πλνt,
where a prime denotes differentiation with respect to t. The solution is
(4.4)f̅n=4(-1)n(2n+1)π{λtkn2+λ2-λν(kn2+λ2)2[1-e-(kn2+λ2)νt]}.
The inverse of f̅n is given by the relation(4.5)fn=2π∫0∞f̅nsinλzdλ,
and inserting the expression for f̅n into this equation, one finds(4.6)fn=8(-1)n(2n+1)π2[π2t-(kn2t)∫0∞sinλzλ(kn2+λ2)dλ-1ν∫0∞λsinλz(kn2+λ2)2[1-e-(kn2+λ2)νt]dλ].
In the limiting case when b goes to infinity, one obtains(4.7)limb→∞uA=t-2π∫0∞sinλzνλ3(1-e-λ2νt)dλ
or in nondimensional form(4.8)u(Aν)1/3=4τi2erfcζ2τ,
where τ=(A2/ν)1/3t, ζ=(A/ν2)1/3z and (Aν)1/3 is the characteristic velocity, (ν2/A)1/3 is the characteristic length and (ν/A2)1/3 is the characteristic time. i2erfcx is the integral of the complementary error function, which can be obtained from the tables [34]. The variation of u/(Aν)1/3 with (A/ν2)1/3z for various values of (ν/A2)1/3t, at y=0 is illustrated in Figure 4. It is clearly seen from the figure that the required time to attain to the flow without the side walls is short.Figure 4
The variation ofu/(Aν)1/3 with (A/ν2)1/3z for various values of (ν/A2)1/3t, at y=0, with and without side walls.  R=b/(ν2/A)1/3.


## 5. Flow Induced by a Plate that Applies a Constant Stress

Suppose that the plane wall suddenly applies a constant tangential stressσ to the fluid. The governing equation is (2.1), and the boundary and initial conditions are(5.1)u(±b,z,t)=0∀t,u(y,∞,t)=0∀t,-b≤y≤b,u(y,z,0)=0for-b≤y≤b,z>0,μ(∂u∂z)z=0=-σ.
The minus sign appears, and σ represents the tangential stress exerted on the fluid by the plate. The first boundary condition suggests that u(y,z,t) can be written in the following form:(5.2)u=∑n=0fn(z,t)coskny,
where kn=(2n+1)π/2b. The cosine transform of fn(z,t) is(5.3)f̅n=∫0∞fncosλzdz,
and the initial condition is f̅n(0)=0. The problem reduces to the solution of the following ordinary differential equation:(5.4)f̅n′+ν(kn2+λ2)f̅n=4(-1)n(2n+1)πσρ,
where a prime denotes differentiation with respect to t. The solution is(5.5)f̅n=4(-1)n(2n+1)πσρν1kn2+λ2[1-e-(kn2+λ2)νt].
The inverse of f̅n is given by the relation [32](5.6)fn=2π∫0∞f̅ncosλzdλ,
and inserting the expression for f̅n into this equation, one finds(5.7)fn=8(-1)nσ(2n+1)π2ρν∫0∞cosλzkn2+λ2[1-e-(kn2+λ2)νt]dλ.
The velocity u becomes(5.8)u=8σρνπ2∑n=0(-1)ncoskny2n+1∫0∞cosλzkn2+λ2[1-e-(kn2+λ2)νt]dλ
or(5.9)u=8σρνπ2∑n=0(-1)ncoskny2n+1[π2kn(e-knz-1)+∫0∞1-e-(kn2+λ2)νtcosλzkn2+λ2dλ].The variation ofu/(σ/ρ)1/2 with z/(ρν2/σ)1/2 for various values of (σ/ρν)t, at y=0, is illustrated in Figure 5. It is clearly seen from the figure that for large times, the flow becomes steady in the case of the side walls, but the flow remains time-dependent in the absence of the side walls. When b goes to infinity, (5.9) becomesFigure 5
The variation ofu/(σ/ρ)1/2 with z/(ρν2/σ)1/2 for various values of (σ/ρν)t, at y=0, with and without side walls. R=b/(ρν2/σ)1/2.(5.10)u=σρν(-z+2π∫0∞1-e-λ2νtcosλzλ2dλ).
By using the equality(5.11)∫0∞1-e-λ2νtcosλzλ2dλ=π2z+πνtierfcz2νt,
one finds(5.12)u=2σρννtierfcz2νt,
where ierfc x is he integral of the complementary error function, which can be obtained from the tables [34].


## 6. Conclusions

Four types of unsteady flows of incompressible viscous fluids over a plane wall bounded by two side walls are considered. In order to understand the effects of the side walls, the cases for which time goes to infinity and the distance between two side walls goes to infinity are compared with the cases for flows over a plane wall in the absence of the side walls. These provide to know the required time to attain the steady-state and what is the distance between the side walls for which the measured value of the velocity or the stress or the temperature would be unaffected by the presence of the side walls. The flow problems which are the flow generated by impulsive motion of a plate, flow due to oscillation of the plate, flow induced by constantly accelerating plate, and flow due to imposed by a plate that applies a constant tangential stress to the fluid are solved by the application of the Fourier transform method to the governing equation of motion. The solutions can be expressed in terms of the definite integrals. It is shown that the starting solution for the flow of the oscillating plate can be presented as the sum of the steady-state and the transient solutions. The limiting cases when the time goes to infinity and the distance between two side walls goes to infinity are obtained and discussed. For flow induced by a plate that applies a constant stress, for large times the flow becomes steady in the case of the side walls, but the flow remains time-dependent in the absence of the side walls.



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