# Unsteady Solutions in a Third-Grade Fluid Filling the Porous Space

Authors: T. Hayat; H. Mambili-Mamboundou; F. M. Mahomed
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

An analysis is made of the unsteady flow of a third-grade fluid in a porous medium. A modified Darcy's law is considered in the flow modelling. Reduction and solutions are obtained by employing similarity and numerical methods. The effects of pertinent parameters on the flow velocity are studied through graphs.


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## Body

## 1. Introduction

Recently, it has been recognised in industrial and
technological applications that non-Newtonian fluids are more appropriate than
viscous fluids. However, there is no model which  can alone predict the
behaviour of all non-Newtonian fluids. The governing equations of non-Newtonian
fluids are  of higher order than the Navier-Stokes equations. Therefore, the
adhering boundary conditions are not sufficient, and one needs additional
boundary conditions for a unique solution. Excellent critical reviews in this
direction have been given by Rajagopal [1, 2],
Rajagopal et al. [3],
and Rajagopal and Kaloni [4]. Amongst the several non-Newtonian fluid models, much
attention has been paid to the simplest subclass of viscoelastic fluids known
as the second grade. However, this model is not capable of describing the shear
thinning and thickening phenomena for steady flow over a rigid boundary. The
third-grade fluid model represents a further, although inconclusive, attempt
towards a more comprehensive description of the behaviour of viscoelastic
fluids. Also, the flows of such fluids in porous medium are quite prevalent in
many engineering fields such as enhanced oil recovery, paper and textile
coating, and composite manufacturing processes. Some important studies dealing
with the flows of non-Newtonian fluids are made by Rajagopal and Na [5, 6], Rajagopal and Gupta
[7], Hayat et al. [8–11], Hayat and Ali [12], Ariel et al. [13], Hayat and Kara [14], Abdel-Malek et al. [15], Wafo Soh [16] and Chen et al. [17], and Fetecau and Fetecau [18–20].Recently, Tan and Masuoka [21] analysed the Stokes' first
problem for a second-grade fluid in a porous medium. In another paper, Tan and
Masuoka [22] studied
the Stokes' first problem for an Oldroyd-B fluid in a porous medium. In these
investigations, the authors have used the modified Darcy's law.The main goal of this paper is to determine analytical
solutions for an unsteady flow of a third-grade fluid over a moved plate. The
relevant problem is formulated using modified Darcy's law of a third-grade
fluid. Two types of analytical solutions are presented and discussed. A
numerical solution is also presented.


## 2. Problem Formulation

Let us
introduce a Cartesian coordinate system OXYZ withy-axis in the upward direction. The third-grade
fluid fills the porous space y>0 and is in contact with an infinite moved plate
at y=0.
For unidirectional flow, the velocity field is(2.1)V=(u(y,t),0,0), where the above
definition of velocity automatically satisfies the incompressibility condition.
The equation of motion in a porous medium without body forces is(2.2)ρdVdt=divT+r, where ρ is the fluid density, d/dt is the material time differentiation, T is the Cauchy stress tensor, and r is the Darcy's resistance in a porous
space. The Cauchy stress tensor of an incompressible third-grade fluid has the
form [23](2.3)T=−pI+μA1+α1A2+α2A12+β1A3+β2(A1A2+A2A1)+β3(trA12)A1, in which p is the pressure, I is the identity tensor, αi (i=1,2) and βi (i=1−3) are the material constants, and Ai (i=1−3) are the first Rivlin-Ericksen tensors [24] which may be defined
through the following equations:(2.4)A1=(gradV)+(gradV)T,An=dAn−1dt+An−1(gradV)+(gradV)TAn−1;n>1.
In studying
fluid dynamics, it is assumed that the flow meets the Clausius-Duhem inequality
and that the specific Helmholtz free energy of the fluid is a minimum at
equilibrium when [25](2.5)μ≥0,α1≥0,β1=β2=0,β3≥0,|α1+α2|≤24μβ3.
On the basis of
constitutive equation in an Oldroyd-B fluid, the following expression in a
porous medium has been proposed [26]:(2.6)(1+λ∂∂t)∇p=−μϕk(1+λr∂∂t)V, where λ and λr are the relaxation and retardation times, and ϕ and k are the porosity and permeability of the
porous medium, respectively. It should be pointed out that for λr=0,
(2.6) reduces to the expression which holds for a Maxwell fluid [26] and when λ=0,
it reduces to that of second-grade fluid [22].Keeping the analogy of (2.6) with the constitutive equation
of an extra stress tensor in an Oldroyd-B fluid, the following expression in
the present problem has been suggested:(2.7)∂p∂x=−ϕk[μ+α1∂∂t+2β3(∂u∂y)2]u. Since  the pressure gradient in (2.7) can also be
interpreted as a measure of the flow resistance in the bulk of the porous
medium, and rx is the measure of the flow resistance offered
by the solid matrix in x-direction,
then
(2.8)rx=−ϕk[μ+α1∂∂t+2β3(∂u∂y)2]u. From
(2.1) to (2.5)
and (2.8), we have(2.9)ρ∂u∂t=μ∂2u∂y2+α1∂3u∂y2∂t+6β3(∂u∂y)2∂2u∂y2−[μ+α1∂∂t+2β3(∂u∂y)2]ϕuk. The relevant
boundary and initial conditions are(2.10)u(0,t)=u0V(t),t>0,(2.11)u(∞,t)=0,t>0,(2.12)u(y,0)=g(y),y>0, in which u0 is the reference velocity.


## 3. Solutions of the Problem

We rewrite (2.9)
as(3.1)∂u∂t=μ*∂2u∂y2+α∂3u∂y2∂t+γ1(∂u∂y)2∂2u∂y2−γ2u(∂u∂y)2−ϕ1u, where(3.2)μ*=μρ+α1(ϕ/k),α=α1ρ+α1(ϕ/k),γ1=6β3ρ+α1(ϕ/k),γ2=2β3(ϕ/k)ρ+α1(ϕ/k),ϕ1=μ(ϕ/k)ρ+α1(ϕ/k).


## 3.1. Lie Symmetry Analysis

The Lie
symmetry analysis reveals that (3.1) admits two sets of symmetry generators
depending on the value of ϕ1.
The Appendix provides details of the symmetry analysis of (3.1).Case 1 (ϕ1≠μ*/α).
We obtain a two-dimensional Lie algebra generated by(3.3)X1=∂∂y,X2=∂∂t.Case 2 (ϕ1=μ*/α).
We find a three-dimensional Lie algebra generated by(3.4)X1=∂∂y,X2=∂∂t,X3=e(2μ*/α)t∂∂t−μ*αe(2μ*/α)tu∂∂u.


## 3.2. Travelling Wave Solutions

We now look for invariant solutions under the operatorX1−cX2,
which represents wave-front-type travelling wave solutions with constant wave
speed c.
The invariant is given by(3.5)u(y,t)=U(x1),wherex1=y+ct. Substituting (3.5)
into (3.1) yields a third-order ordinary differential equation for U(x1),(3.6)ϕ1U(x1)=−cdUdx1−γ2U(x1)(dUdx1)2+μd2Udx12+γ1(dUdx1)2d2Udx12+cαd3Udx13. It can be seen
that this equation admits the solution(3.7)U(x1)=u0exp(γ2x1−γ1) provided
that
(3.8)γ2γ1(μ−cαγ2γ1)+(cαγ2γ1−ϕ1)=0, and hence
(3.1) subject to (2.10)–(2.12) admits the
solution(3.9)u(y,t)=u0exp(γ2(y+ct)−γ1). This solution
is plotted in Figures 1 and 4 for various values of the emerging parameters.Figure 1
Travelling wave solutions
varyingt.On the other hand, we find
group-invariant solutions corresponding to operators
which give meaningful physical solutions of the initial and boundary value
problems (2.9) to (2.12). This means X2 and X3.


## 3.3. Group-Invariant Solutions Corresponding toX2

The invariant solution admitted byX2 is the steady-state solution(3.10)u(y,t)=F(y). The
substitution of (3.10) into (3.1) yields the second-order ordinary differential
equation for F(y):(3.11)γ1(F′(y))2F′′(y)+μ*F′′(y)−γ2F(y)(F′(y))2−ϕ1F(y)=0, subject to
boundary conditions(3.12)F(0)=v0,(3.13)F(l)=0,l>0, where l is  sufficiently large, and u0V=v0 is a constant (V is taken to be a constant).
Let(3.14)K(F)=dFdy,
then
(3.11) transforms to(3.15)γ1K(F)3K′(F)+μ*K(F)K′(F)−γ2FK(F)2−ϕ1F=0. The integration
of (3.15) gives(3.16)γ1γ2K(F)+μ*γ2−γ1ϕ1γ2γ2ϕ1,Arctan[γ2γ1K(F)]=12y2+C, where C is a constant. Equation (3.16) is equivalent to
the following first-order ODE in F:(3.17)γ1γ2F′(y)+μ*γ2−γ1ϕ1γ2γ2ϕ1,Arctan[γ2γ1F′(x)]=12y2+C. One can solve
this numerically subject to the boundary conditions (3.12) and
(3.13). This
solution is plotted in Figure 5.


## 3.4. Group-Invariant Solutions Corresponding toX3

The invariant solution admitted byX3 is (3.18)u=u0exp(−μ*αt)B(y), where B(y) as yet is an
undetermined function of y.
Substituting (3.18) into (3.1) yields the linear second-order ordinary
differential equation(3.19)B′′−γ2γ1B=0. From (2.10) to
(2.11), the appropriate boundary conditions for
(3.19) are(3.20)B(0)=1,B(l)=0,l→∞, where(3.21)V(t)=u0exp(−μ*αt).We solve (3.19) subject to the boundary conditions (3.20)
for positive γ2/γ1 and obtain(3.22)B=exp(−γ2γ1y).The solutions (3.18) are plotted for positive γ2/γ1 in Figure 6. This solution is similar to (3.9)
except that we do not have a condition like (3.8)
here.


## 3.5. Numerical Solution

We present the
numerical solution of (3.1) subject to the initial and boundary conditions:(3.23)u(0,t)=u0V(t),u(∞,t)=0,t>0,u(y,0)=g(y),y>0, where g(y) is an arbitrary function of y.This solution is plotted using Mathematica's solverNDSolve.


## 4. Results and Discussion

In order to see
the variation of various physical parameters on the velocity, Figures1–7 have
been plotted.Figure 2
Travelling wave solutions varyingc.Figure 3
Travelling wave solutions
varyingγ1.Figure 4
Travelling wave solutions
varyingγ2.Figure 5
Numerical solution of (3.11) or
(3.17) subject to the boundary conditions (3.12) and (3.13).Figure 6
Analytical solutions
forμ*=1, u0=1.Figure 7
Numerical solution of (3.1), with V(t)=e−t, g(y)=e−y2, μ*=2.5, u0=1, α=2, γ1=1.5, γ2=2.6, ϕ1=0.8.The effect of unsteadiness on the velocity profile is
shown in Figure1. This figure depicts that velocity decreases for large values
of time. Clearly, the variation of velocity is observed for 0≤t<3.8.
For t≥3.8,
the velocity profile remains the same. In other words, one can say that
steady-state behaviour is achieved for t≥3.8.The influence of the wave speedc on the velocity profile has been presented in
Figure 2. It is revealed that velocity decreases by increasing c.
Moreover, the effects of the fluid parameters γ1 and γ2 are given in Figures 3 and 4, respectively.
These figures depict that γ1 and γ2 have opposite roles on the velocity. These
figures show that velocity increases for large values of γ1 whereas it decreases for increasing γ2.
In Figure 5, the steady-state solution is plotted, and the velocity profile is
the same as observed in the case of travelling wave solution.Further, the analytical solutions (3.19) for μ*/α>0 is plotted in Figure 6. Here as indicated in
Figure 6, the velocity profile decreases for large values of t.
Ultimately when t≥3.8,
there is almost no variation in velocity.Finally in Figure7, we have plotted numerically the
velocity profile for small variations of time, and it is observed that the
velocity decreases as time increases, which is the the same observation made
previously for the analytical solutions.



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*Source: 139560-2008-06-16.xml*