EXACT SOLUTIONS FOR UNSTEADY UNIDIRECTIONAL FLOWS
OF SECOND GRADE FLUID
FARAH NADZIRAH BINTI JAMRUS
A dissertation submitted in partial fulfillment of the
requirements for the award of the degree of
Master of Science (Mathematics)
Faculty of Science
Universiti Teknologi Malaysia
APRIL 2010
iii
ACKNOWLEDGEMENT
Alhamdulillah, thanks to Allah for giving me opportunity to finish this thesis
although always facing with some difficulties. I would like to express my deep and
sincere gratitude to my supervisor; Prof. Dr. Norsarahaida S. Amin. Her wide
knowledge and her logical way of thinking have been of great value for me. Her
understanding, encouraging and personal guidance have provided a good basis for the
present thesis. Thank you for your encouragement guidance and advice.
I owe my loving thanks to my family. Without their encouragement and
understanding, it would have been impossible for me to finish this work.
Lastly, to my friends. Thanks for your help and support. I hope that we will
always success in our life.
iv
ABSTRACT
Second grade fluid or second grade fluid is one of the most popular used model
to describe non-Newtonian fluids. In study, the problems involving unsteady
unidirectional flows of second grade fluid, flows of second grade fluid through a porous
medium and flows of heated second grade fluid are discussed. The unidirectional flows
of second grade fluid considered in this study are Couette flow, Stokes’ first flow
problem and plane Poiseuille flow. The exact solutions for Couette flow and Poiseuille
flow are obtained using the Laplace transform while the Fourier transform method is
used to obtain the exact solution of the Stokes’ first flow problem. The momentum
equation for the flows of second grade fluid through a porous medium is derived based
on the modified Darcy’s law. The momentum equation for flows of second grade fluid
through a porous medium has an extra term, which is the viscous damping force caused
by the micro-pore structures of the porous medium. It is found that the velocity of the
flows decrease when the porosity, ϕ and permeability, K parameter are increased. For
flows of heated second grade fluids, the velocity and temperature distribution are
determined using the simple and double Fourier sine transforms.
v
ABSTRAK
Bendalir peringkat kedua atau bendalir gred kedua merupakan salah satu model
yang sering digunakan untuk mewakili bendalir bukan Newtonan. Perbincangan
meliputi masalah yang berkaitan aliran bendalir peringkat kedua, aliran bendalir
peringkat kedua melalui media poros dan aliran bendalir peringkat kedua yang
dipanaskan. Aliran bendalir yang dikaji ialah aliran Couette, aliran Poiseuille dan
masalah aliran pertama Stoke. Penyelesaian tepat untuk aliran Couette dan Poiseuille
diperolehi menggunakan teknik jelmaan Laplace manakala teknik jelmaan Fourier
digunakan untuk mendapat penyelesaian tepat bagi masalah aliran pertama Stoke.
Berdasarkan hukum Darcy, persamaan momentum untuk bendalir peringkat kedua
melalui media poros diterbitkan. Persamaan momentum untuk bendalir peringkat kedua
melalui media poros mempunyai satu fungsi yang di mana fungsi itu mengandungi daya
kelikatan yang disebabkan struktur media yang poros. Dalam kajian yang dilakukan,
didapati bahawa taburan halaju bagi aliran bendalir peringkat kedua melalui media poros
menurun apabila nilai parameter resapan dan keporosan meningkat. Seterusnya,
penyelesaian tepat untuk taburan halaju dan suhu bagi aliran bendalir panas peringkat
kedua diperolehi dengan mengaplikasikan teknik jelmaan sin Fourier mudah dan
berganda.
vi
TABLE OF CONTENTS
CHAPTER
ITEMS
TITLE
DECLARATION
ACKNOWLEDGEMENT
ABSTRACT
ABSTRAK
CONTENTS
1
Introduction
Background of Research
Statement of Problem
Objectives of Research
Scope of Research
Significance of Research
Organization of Research
1
2
6
6
7
7
9
THE GOVERNING EQUATIONS
2.1
2.2
2.3
2.4
3
i
ii
iii
iv
v
vi
INTRODUCTION
1.1
1.2
1.3
1.4
1.5
1.6
1.7
2
PAGE
Introduction
Continuity Equation
Momentum Equation
Energy Equation
10
11
12
17
UNSTEADY UNIDIRECTIONAL FLOWS OF
SECOND GRADE FLUID
3.1 Introduction
3.2 The Governing Equation
3.3 Unsteady Couette Flow
3.3.1 Flow driven by the motion of the upper
plate and the lower plate is fixed.
3.3.2 Flow driven by the motion of lower
plate and the upper plate is free.
3.4 Poiseuille Flow
3.5 Stokes’ First Flow Problem
3.6 Unsteady Generalized Couette Flow
23
24
28
33
35
36
39
vii
4
5
FLOWS OF SECOND GRADE
THROUGH A POROUS MEDIUM
FLUID
4.1 Introduction
4.2 The Governing Equation
4.3 Unsteady Couette Flow with Moving Bottom
Plate
4.4 Unsteady Couette Flow with Oscillating Plate
41
42
47
4.5 Stokes’ First Flow Problem
4.6 Analysis of Results
52
55
FLOWS OF HEATED SECOND GRADE
FLUID
4.1 Introduction
4.2 The Governing Equation
4.3 The Stokes’ First Flow Problem on A Heated
Flat Plate
4.4 The Stokes’ First Flow Problem within A
Heated Edge
6
50
59
60
63
66
CONCLUSION
6.1 Introduction
6.2 Summary of Research
6.3 Suggestion for Further Research
68
68
73
REFERENCES
74
APPENDIX
CHAPTER I
INTRODUCTION
1.1
Introduction
Fluids can be divided into Newtonian fluid and non-Newtonian fluid. Newtonian
fluids are fluids for which a linear relation exists between stress and spatial variation of
velocity, while non-Newtonian fluids are the otherwise, where the stress at a given
temperature and pressure is not a linear function of the spatial variation of velocity.
Therefore, a constant coefficient of viscosity cannot be defined.
Non-Newtonian fluids form an extremely wide class of different materials,
whose only common features are fluidity and a failure to obey Newton’s law of friction.
Recently, the interest in flows of non-Newtonian fluids has grown considerably because
of their vast applications in daily life. It is well-known in fact that non-Newtonian fluids
are more appropriate than Newtonian fluids in industrial and technological applications.
A large variety of consumer goods containing high concentration of glass or carbon
fibres, paints and lubricants containing polymer additives, many foodstuffs and
biological fluids are non-Newtonian.
Non-Newtonian biologically important fluids response is typically observed in
concentrated suspensions and in high molecular-weight materials. The rheology of blood
2
has received much study. Blood is rheologically complex on two counts: it is a
suspension because erythrocytes with characteristic dimensions of several micrometers
are present, and the suspending fluid itself exhibits non-Newtonian behavior because of
the presence of high molecular weight protein. Flows of elastic solutions and of those
containing long-chain polymers, including coatings, as well as flows in extruders, molds,
and other processing equipment, dominate rheology today.
Many industrial problems involve rheological concerns. These include the need
to understand the transport of foams and yield-stress fluids in oil drilling and enhanced
oil recovery, and the importance of understanding the behavior of biological
macromolecules in micro fluidic devices for lab-on-a-chip applications. Geoscientists
apply fluid mechanics in studies of volcanism and the convection through Earth’s mantle
and outer core since they are dealing with non-Newtonian fluid; lava.
Many materials such as clay coatings, drilling muds, certain oils and greases, and
many emulsions have been treated as non-Newtonian fluids. It is difficult to suggest a
single model, which exhibits all properties of non-Newtonian fluids as, is done for the
Newtonian fluids. They cannot be described in a simple model by the Navier-Stokes
equation as for the Newtonian fluids and there has been much confusion over the
classification of non-Newtonian fluids.
The class of Newtonian flows for which there exists analytic solution of the
Navier-Stokes equations is very restricted. In fact, the resulting equations of nonNewtonian fluids are higher order, more non-linear and complicated than the NavierStokes equations. Thus, this makes the task of obtaining the accurate solutions is
difficult. Being scientifically challenging, flows of non-Newtonian fluids have been
studied extensively.
3
1.2
Background of Research
Non-Newtonian fluids may be classified as (i) fluids for which the shear stress
depends only on the rate of shear; (ii) fluids for which the relation between shear stress
and shear rate depends on time; (iii) the visco-elastic fluids, which possess both elastic
and viscous properties [11-13]. Given this, the term non-Newtonian fluid is open to
interpretation [23].
Because of the great diversity in the physical structure of non-Newtonian fluids,
it is not possible to recommend single constitutive equations. By the term constitutive
equation, it means an equation relating the stress on some material to the motion of the
material [23]. For this reason, many non-Newtonian models or constitutive equations
have been proposed and most of them are empirical or semi empirical. Although many
constitutive equations have been suggested, many problems are still unsolved. Some of
the models do not give satisfactory results in accordance with the available experimental
data. For this reason, in many practical application, empirical or semi empirical
equations have been applied.
For previous solved problems which the flow is slow enough in the visco-elastic
sense, the results given by Oldroyd’s constitutive equations will be substantially equal to
those of the second- or third order Rivlin-Ericksen constitutive equations [11]. Thus, it
would seem reasonable to use the second- or third-order constitutive equations in
carrying out the calculations. This is particularly so in view of the fact that the
calculation will generally simpler. For this reason, the second-order fluid model is used.
The constitutive relation in the following form is for to the second grade fluid
[7]:
σ = − pI + µ A1 + α1 A2 + α 2 A12
%
%
%
%
%
where σ is the stress tensor and p is the pressure. This model has three coefficients
%
where µ is the coefficient of viscosity, and α 1 and α 2 are the material moduli which are
4
usually referred to as the visco-elasticity and cross-viscosity parameter respectively.
While A1 and A2 are the kinematical tensors. There are some restrictions on these
%
%
coefficients due to the Clasius-Duhem inequality and also due to the assumption that
Helmholtz free energy is minimum in equilibrium (see Appendix). A comprehensive
discussion on the restrictions for these three coefficients has been given by Dunn and
Fosdick [8].
In the studies of non-Newtonian fluids, the non-linearities and in certain cases
the order of the differential equations are increased. The equation of motion of
incompressible second grade fluids is of higher order than the Navier-Stokes equation.
The Navier-Stokes equation is a second order partial differential equation, but the
equation of motion of a second grade fluid is a third order partial differential equation.
Thus, in order to obtain a determinate solution in a second grade fluid one requires either
a boundary (initial condition) or some similar assumption in addition to the boundary
conditions required for the Navier-Stokes equations.
Unsteady flows of a second grade fluid in a bounded region have been studied by
Ting (1963). He is the first author who studied the unsteady flows of a fluid of second
grade [10]. Then Hayat et al. (2000) has carried out a study on unsteady unidirectional
flows of a non-Newtonian fluid which is the second grade fluid. In their study, they
obtained the exact analytic solutions for a class of unsteady unidirectional flows and the
frictional forces of an incompressible second grade fluid. The solutions are obtained by
the method of separation of variables.
Then Hayat et al. (2004) extended the study by Hayat et al. (2000) to obtain the
solutions for both large and small times. As for solutions for small times, the Laplace
transform technique is particularly well studied. By knowing the unidirectional
flows’velocity, the volume flux and frictional forces can be calculated. They found that
5
large and small times solutions are dependent on the coefficient of visco-elasticity, α 1 .
As the visco-elastic parameter increases, the value of velocity at the distance y
decreases.
Erdogan and Imrak (2007) conducted a study about some unsteady flows of
second grade fluid where the unsteady unidirectional flows are considered. The correct
condition for a free surface is given in this study. In their study, they have obtained
velocity and stress solutions for each flows. Then they compared the stress at t=0 on the
stationary boundary for flows generated by the impulsive motion of a boundary for
second grade fluid with stress at t=0 for Newtonian fluid. They found that this stress is
finite for second grade fluid and infinite for a Newtonian fluid.
Erdogan and Imrak (2005) continued their study on unsteady unidirectional
flows of a second grade fluid. They have discovered that for small values of time some
interesting results occur. The stress at the boundary show a different character for flow
generated by impulsive motion of a boundary compared to that for a Newtonian fluid.
Another property of a second grade fluid is that the required time to attain the
asymptotic value of a second grade is longer than that for a Newtonian fluid.
The solution of the governing equation for unsteady unidirectional flows of
second grade fluids in unbounded regions obtained by the Laplace transform method
shows some undesirable results. For this reason, in order to solve the governing equation
the use of the other transform methods are discussed by Erdogan and Imrak (2007). It is
shown in their study that the solution obtained by the Laplace transform method does
not satisfy the initial condition. For this reason, the Fourier transform method for the
unsteady flows of second grade fluids in unbounded regions has been used instead of the
Laplace transform method.
Stokes first solved Stokes’ first problem in 1856, which involved a viscous
Newtonian fluid [15]. Jordan and Puri (2003) conducted a study on Stokes’ first problem
involving a second grade fluid in a porous half-space. The open half space of the
6
original Stokes’ first problem has been replaced with a porous half space of permeability
and porosity. Based on the Darcy’s law, the Stokes’ first flow problem was solved.
Tan and Masuoka (2005) developed a modified Darcy model (see Appendix) for
visco-elastic fluids in a porous media, and then investigated the Stokes’ first problem for
visco-elastic fluids in a porous half-space. They obtained exact solutions of the velocity
and temperature using Fourier sine transforms.
In the same year, research about the unsteady Couette flow of a second grade
fluid in a layer of porous medium were conducted by Hayat et al. (2005). In this study,
the two Couette flows of a second grade fluid were discussed when (i) bottom plate
moves suddenly, (ii) bottom plate oscillates. The flow problems were solved using the
modified Darcy’s law and Laplace transform method.
Bandelli (1995) has carried out a research on the heated boundaries of the
unsteady unidirectional flows of second order fluids. He derived the energy equation for
second grade fluids and obtained exact solution for some flows.
Fatecau (2002) has conducted the Rayleigh-Stokes problem for heated second
grade fluids. The Stokes’ first flow for a heated flat plate and Rayleigh-Stokes flow for a
heated edge were considered in this paper. The temperature distribution of a second
grade fluid subject to a linear flow on a heated flat plate is determined using the Fourier
sine transform.
7
1.3
Statement of Problem
Recently the interest in non-Newtonian fluids has grown considerably because of
their practical and fundamental importance associated with many technological
applications. Therefore, this research is conducted to study one of the most popular
models for non-Newtonian fluids is the model that is called the second grade fluid. This
study will review what are the mathematical studies that have been done on second
grade fluids and the methods used to obtain the solution of the problems involving the
second grade fluid.
1.4
Objective of Research
The objectives of this project are:
1. To derive the governing equations of motions for flows of second grade
fluids.
2. To derive the exact solution for the problems involving unsteady
unidirectional flows of second grade fluids.
3. To solve the problems involving unsteady unidirectional flows of second
grade fluids through a porous medium.
4. To solve the problems of unsteady unidirectional flows for heated second
grade fluids.
1.5
Scope of Research
This work is a series of exact solutions of unidirectional flows of second grade
fluid. The unsteady Couette flow and Stokes’ first flow problem of a second grade fluid
8
in a layer of porous medium and the Rayleigh-Stokes flow and the Poiseuille flow for
heated second grade fluid will be reviewed.
1.6
Significance of Research
The study of non-Newtonian fluid dynamics is important in relation with
engineering, geophysics, physiology and pharmaceutics. In engineering, it affects the
production and use of polymeric materials, but plasticity theory has been similarly
important for the design of metal forming processes. Many industrially important
substances such as concrete, paint and chocolate have complex flow characteristics.
Geophysics includes the flow of lava, but in addition measures the flow of solid Earth
materials over long time scales: those that display viscous behavior, e.g. granite, are
known as rheids. While in physiology, many bodily fluids have complex compositions
and thus flow characteristics. In particular, there is a specialist study of blood flow
called hemorheology. The term biorheology is used for the wider field of study of the
flow properties of biological fluids. Food rheology is important in the manufacture and
processing of food products.
1.7
Organization of Dissertation
Chapter 1 contains general introduction to non-Newtonian fluid. The
background, objective, statement and the significance of this research are stated in this
chapter too.
The derivation of the governing equation of second grade fluid will be shown in
Chapter 2. The constitutive equation of an incompressible second grade fluid and the
discussion on the material moduli are considered.
9
As going through to Chapter 3, the problems involving of the unsteady
unidirectional flows of second grade fluids will be studied. The governing equation for
the unsteady unidirectional flows of second grade fluid will be derived. The method how
to obtain the velocity distribution, frictional force and volume flux for the respective
flows will be reviewed in this chapter.
While in Chapter 4, the heated unsteady unidirectional flows of a second grade
fluid will be discussed. The governing equations for second grade fluid, which are the
momentum and energy equations will be derived. The method to get the velocity and
temperature distributions for the Stokes’ first flow problem will be reviewed
Chapter 5 contains the review of the unsteady unidirectional flows problems
through a porous medium. The governing equation derived in Chapter 3 cannot be used
to govern the unidirectional flows of second grade fluid through a porous medium.
Therefore, the governing equation for the unsteady unidirectional flows of second grade
fluid through a porous medium will be derived in this chapter. Then, the method to
obtain the velocity distribution for the unsteady unidirectional flows will be reviewed.
Lastly is the conclusion chapter. In this chapter, it will conclude and summarize
all the results obtained from the beginning until the end of the research.
CHAPTER II
THE GOVERNING EQUATIONS
2.1
Introduction
The derivations of equations that govern the flow of second grade fluid are
presented in this chapter. In Section 2.2, the continuity equation will be derived based on
the physical principle which is the mass is conserved and in Section 2.3, from the
Newton’s second law, the derivation of the momentum equation is shown. Lastly, the
energy equation will be derived in Section 2.4.
11
2.2
Continuity Equation
In obtaining the continuity equation, we must used the appropriate physical
principle which is the mass is conserved. The fundamental physical principle that mass
is conserved means
Net mass flow out of
the control volume
through surface
=
Time rate of decrease of
mass inside the control
volume
The mass conservation principle can be expressed in the form
Mass =

D 
 ∫ ρ dV  = 0
Dt V

(2.1)
By applying the Reynolds transport theorem in Equation (2.1), Equation (2.1) becomes;
 ∂ρ

∫  ∂t + ∇ ⋅ ( ρ v% ) dV = 0
(2.2)
V
Note that this integral condition on ρ and v can be satisfied for an arbitrary volume
%
element only if the integral is identically zero, that is;
∂ρ
+ ∇ ⋅ ( ρv ) = 0
%
∂t
(2.3)
which is the continuity equation. By using the chain rule, Equation (2.3) can be written
as;
∂ρ
+ ( v ⋅ ∇ ρ ) + ( ρ∇ ⋅ v ) = 0
%
%
∂t
(2.4)
12
The first two terms on the left side of Equation (2.4) are simply the substantial derivative
of density. Hence, Equation (2.4) becomes
Dρ
+ ρ∇ ⋅ v = 0
%
Dt
(2.5)
But in this study, the motion of second order incompressible fluid is considered.
The approximation of incompressibility in fluid mechanics is usually limited to the
statement that the statement that the density ρ is independent of the pressure. Thus, if an
incompressible fluid is isothermal, its density must be constant. In this case, the
derivative
Dρ
is identically equal to zero, and the continuity equation in Equation (2.5)
Dt
takes the simpler form as follows [1];
∇ ⋅ v = div v = 0
%
%
2.3
Momentum Equation
From Newton’s second law of motion, the total external force acting on the
control volume given as;
Φ
(2.6)
where Φ is the momentum of the mass of the fluid is contained in control volume and ϕ
is the local velocity u [14]. While represents the sum of the body force acting
%
throughout the control volume and the surface forces acting on the boundary of the
control volume. In this case, the body force is not considered for simplicity.
13
To calculate the surface forces, let consider a part of the control surface and the
stresses acting on it as shown in Figure 2.1.
CS
Figure 2.1: Stress Components on a Control Surface
The x and y component of the force acting on the element ds are
and
The unit normal vector acting on the element ds is
Hence, the element traction vector on the control surface is;
where is the stress tensor.
. The total surface force is;
. (2.7)
14
By applying the Gauss divergence theorem in Equation (2.7), the surface integral can be
converted into the volume integral;
· (2.8)
Substitute Equation (2.8) into Equation (2.6) and obtain the external force acting on the
control volume as
Φ
·
(2.9)
Then, the application of the Reynold’s Transport theorem at the right hand side of
equation (2.9) gives;
∂

) dV = ∫ ∇ ⋅σ dV
∫  ∂t ( ρu% ) + ∇ ⋅ ( ρuu
% % 
V
(2.10)
V
Rearrange Equation (2.10) and obtained;
∂

) − ∇ ⋅σ  dV = 0
∫  ∂t ( ρu% ) + ∇ ⋅ ( ρuu
%%
%
(2.11)
V
Since the initial choice for V is arbitrary, it follows that Equation (2.11) can be satisfied
if and only if the integrand is equal to zero for each point in the fluid, that is;
∂
( ρ u ) + ∇ ( ρ uu ) = ∇ ⋅ σ
%%
%
∂t %
(2.12)
15
By using the chain rule, Equation (2.12) can be written as;
∂u
∂t
 ∂ρ

+ ∇ ⋅ ( ρ u )  =∇ ⋅ σ
%  ∂t
% 
%
ρ % + ρ u ⋅ ∇u + u 
%
%
(2.13)
The term in the bracket denotes the continuity equation where it is equal to zero. Thus,
Equation (2.13) becomes;
 ∂u
 ∂t

%
ρ  % + u ⋅ ∇ u  =∇ ⋅ σ
%
%
(2.14)
Using the definition of the substantial derivative, Equation (2.14) reduces to
ρ
Du
% =∇ ⋅ σ
%
Dt
(2.15)
Equation (2.15) is known as Cauchy’s first law of motion. Cauchy’s Law of
motion holds for every continuum, so it holds for every fluid. It is the starting point for
the calculation of fluid mechanics problems. Using the constitutive equation, that is the
relationship between stress tensor and the motion, Cauchy’s equation of motion is
changed to a specific equation of motion for the fluid under observation.
16
The constitutive equation for fluid of second grade is [7];
σ = − pI + µ A1 + α1 A2 + α 2 A12
%
%
%
%
%
(2.16)
where − pI denotes the indeterminate spherical stress, µ is the coefficient of viscosity,
%
α 1 and α 1 are referred to as the visco-elasticity and cross-viscosity parameter
respectively and A1 and A2 are the kinematic tensors. defined through;
%
%
(2.17a)
T
A1 = ( ∇u ) + ( ∇u )
%
%
%
D
T
A2 =
A1 + A1 ( ∇u ) + ( ∇u ) A1
%
%
Dt %
%
% %
(2.17b)
In our analysis, we shall consider the model represented by Equation (2.16) as an
exact model. If this model is required to be compatible with thermodynamics [8], then
the material moduli must meet the following restrictions;
µ ≥ 0, α 1 ≥ 0, and α 1 + α 2 = 0
(2.18)
The momentum equation for the second grade fluid can be obtained by substituting
Equation (2.16) into Equation (2.15). The resulting equation is the momentum equation
for a second grade fluid expressed as follows;
ρ
Du
2
% = −∇p + µ∇A1 + α1∇A2 + α 2∇A1
%
%
%
Dt
(2.19)
17
3.4
Energy Equation
We begin by considering an arbitrary material control volume as it moves along
with the fluid, and in this case consider the change in its total energy with respect to
time. From the continuum point of view, the total energy of an arbitrary material control
volume is written as [9];
∫
Vm ( t )
 ρ (u ⋅ u )

 % % + ρ e  dV
 2

where u ⋅ u = u 2 is the local ‘speed’ of the continuum motion (kinetic energy) and e is
% %
the internal energy per unit mass.
The rate at which the total energy changes with time is determined by the
principle of energy conservation for the material volume element, according to which;
Rate of work done
 ρu 2

D
the
material +
+ ρ e  dV = on

∫
V
(
t
)
m
Dt
control volume by
 2

external forces
Rate of at which
heat is transferred
to the material
control volume
(2.20)
Let denotes;
A= rate of work done on the material control volume by external forces
B= rate of at which heat is transferred to the material control volume.
First, work can be done on the material control volume only as a consequence of
forces acting on it. The applied forces that appear are the surface force and body force.
The power of the surface forces σ dS is u ⋅ σ dS , while the body force ρ g dV is
%
% %
%
u ⋅ ρ g dV .
% %
18
Therefore, the rate of work done on the material control volume is given by;
A=
∫∫∫ u% ⋅ ρ g% dV
+
V (t )
∫∫ u% ⋅σ%
dS
(2.21)
S (t )
Heat flux is due to (1) volumetric heating such as absorption and (2) heat transfer
across the surface due to temperature gradients, i.e., thermal conduction. Define q& as
the rate of volumetric heat addition per unit mass and the heat transferred by thermal
conduction through the material control volume of the surface is defined by − q ⋅ n . The
% %
minus sign is chosen so that inflowing energy is counted as positive. Hence, it can be
written
B=
∫∫∫ ρ q& dV
V (t )
−
∫∫ q% ⋅ n% dS
(2.22)
S (t )
By substituting Equation (2.21) and (2.22) into (2.20), we obtain the mathematical form
of Equation (2.20);
 ρu 2

D
+ ρ e  dV =

∫
Dt Vm (t )  2

∫∫∫ u% ⋅ ρ g% dV + ∫∫ u% ⋅ (σ% ⋅ n% ) dS + ∫∫∫ ρ q& dV
V (t )
S (t )
V (t )
−
∫∫ q% ⋅ n% dS
S (t )
To obtain the differential form, we apply the Reynolds transport theorem to the
left hand side and the divergence theorem to the right hand side. With all terms then
expressed as volume integrals over the arbitrary material control volume Vm ( t ) ,since the
integrand is assumed to be continuous, and the domain of integration is arbitrary, the
integrand must vanish, and we obtain the differential form of the energy conservation
principle;

D  u2
ρ  + e  = ρ g ⋅ u + ∇ ⋅ ( u ⋅ σ ) + ρ q& − ∇ ⋅ q ⋅ n
% %
Dt  2
% %
% %

( )
( )
(2.23)
19
It appears from Equation (2.23) that contributions from any of the terms on the
right-hand side will lead to a change in the sum of kinetic energy and internal energy.
This equation can be further expressed in terms of the internal energy. To do it, we
substitute the inner product of momentum equation with u into Equation (2.23), we
%
obtain;
ρ
De
= σ ⋅ ( ∇u ) − ∇ ⋅ q + ρ q&
%
Dt %
%
(2.34)
Equation (2.34) is a form of energy equation where the substantial derivative on
the left hand side is strictly in terms of the internal energy only. The kinetic energy and
the body force terms have dropped out. It is important to emphasize that the energy
equation when written in terms of e alone does not explicitly contain the body force.
The heat flux due to the thermal conduction, from the Fourier’s law of heat
conduction, is proportional to the local temperature gradient;
q = −k∇T
%
%
In more general form, we can state the energy equation in term of the internal energy as;
ρ
De
= ρ q& + k ∇ 2T + σ . ( ∇u )
%
%
Dt
(2.35)
In our case, we will use the general form of the energy equation to derive the energy
equation for the second grade fluid. Substituting Equation (2.16) into Equation (2.35),
we obtained
ρ
De
= ρ q& + k ∇ 2T + ( −∇p + µ∇A1 + α1∇A2 + α 2∇A12 ) . ( ∇u )
%
%
%
%
Dt
(2.36)
20
By using Equation (2.17a), (2.17b) and (2.18), the third term in the right hand side of
Equation (2.36) become as follows;
( −∇p+µ∇A% +α ∇A% +α ∇A% ) .( ∇u%) = −pI% +µA% +α A% +α A% ⋅ ∂∂xu
1
1
2
2
2
1
1
2
2 1
1 2
i
j
 ∂ ∂u ∂uj ∂u 
∂ ∂u ∂uj ∂u 
  i +  i  +uj  i +  i 
∂u ∂u ∂u  ∂t ∂xj ∂xi 
∂xj  ∂xj ∂xj ∂xi 
∂xj 
= µ  i + j  i  +α1 

∂xj   ∂ui ∂uj ∂ui  ∂ui ∂ui ∂uj 

∂xj ∂xi 
+∂x + ∂x ∂x  + ∂x ∂x + ∂x 

 j   j 
 j

i 
i 
  j

2
∂u ∂uj 
+α2  i + 
∂xj ∂xi 
∂ui 
 
∂xj 
 ∂u
= µ i
 ∂x
 j
 ∂u
= µ i
 ∂x
 j
2

 ∂u
 + α1  i

 ∂x j
2




∂  ∂u
 + α1  i
∂t  ∂x j


∂  ∂u
 + α1  i
∂t  ∂x j

µ 2 α D
2
= A1 + 1
A1
2 %
4 Dt %
2
3

 ∂u
 + α 2  i

 ∂x j



3
2
Then the energy equation, which is Equation (2.36), becomes;
ρ
De µ
α D
2
2
= A1 + 1
A1 + k ∇ 2θ + ρ q&
Dt 2 %
4 Dt %
(2.37)
In order for the second grade fluid model to be compatible with thermodynamics [8], the
specific Helmholtz free energy that characterizes the fluid has to take the form [8];
ψ = ψ (θ , A1 , A2 ) = ψ (θ , 0 ) +
α1
2
A1
4ρ %
(2.38)
and the specific entropy must be defined through;
η = −ψ θ
(2.39)
21
where the subscript denotes the partial differentiation with respect to that variable. Since
the internal energy is related to the Helmholtz free energy through
e = ψ + θη
(2.40)
It follows from (2.38)-(2.40), given that
2
α  ∂u 
e = ϕ (θ ) + 1  i  − θϕθ
ρ  ∂x j 
where
ϕ (θ ) = ψ (θ , 0 )
Thus,
ρ
D
De
α D
2
A1 
= ρ  (ϕ (θ ) − θϕθ ) + 1
Dt
4 ρ Dt % 
 Dt
Next, note that (2.39) and (2.40) imply that
eθ =
D
(ϕ (θ ) − θϕθ ) = −θϕθ = c
Dθ
where c is called the specific heat. Thus;
ρ
 Dθ α1 D
De
2
= ρ c
+
A1 
Dt
 Dt 4 ρ Dt % 
Then, substitute Equation (2.41) into Equation (2.37) and yields;
ρc
Dθ α1 D
µ 2 α D
2
2
+
A1 = A1 + 1
A1 + k ∇ 2T + ρ q&
Dt 4 Dt %
2 %
4 Dt %
(2.41)
22
Rearrange back the above equation;
ρc
Dθ
2
= µ A1 + k ∇ 2T + ρ q&
%
Dt
which is formally the energy equation for the second grade fluid.
(2.42)
CHAPTER III
UNSTEADY UNIDIRECTIONAL FLOWS OF SECOND GRADE FLUID
3.1
Introduction
In this chapter, we will discuss the solutions of the unidirectional flows of second
grade fluid in bounded and unbounded region. The momentum equation for
unidirectional flow of second grade fluid will be derived in the Section 3.2.
In the section 3.3-3.6, the velocity distributions for each considered flows will be
reviewed. The methods that will be used in order to obtain the velocity distributions are
the Laplace transform and the Fourier transform method depend on the type of the
flows’ boundaries. After obtaining the velocity distribution, frictional forces and volume
fluxes of each considered flow will be yielded by direct computational using the velocity
distribution.
24
3.2
The Governing Equation
The purpose of derivation of the momentum equation is to solve in order to get
the velocity distribution for the discussed unidirectional flows in this chapter. Recall the
momentum equation for second grade fluid as given by Equation (2.19) as follows;
ρ
For
unsteady
Du
2
% = −∇p + µ∇A1 + α1∇A2 + α 2∇A1
%
%
%
Dt
unidirectional
flow,
the
velocity
field
(2.19)
is
in
the
form
u = u ( y, t ) , v = 0, and w = 0. which u is the x-component of the velocity.
x-component of the momentum equation:
 ∂u  ∂u
∂u
∂u  
∂p
+  u + v + w   = − + µ∇A1 + α1∇A2 + α 2∇A12
%
%
%
∂y
∂z  
∂x
 ∂t  ∂x
ρ
(3.1a)
Substitute Equation (2.17a) and (2.17b) into Equation (3.1a) and finds that;
 ∂ui ∂u j 
+


%
 ∂x j ∂xi 
∂ 2u
= µ 2i
∂x j
µ∇A1 = µ
∂
∂x j
 ∂ 2 u ∂ 2u ∂ 2 u 
=µ 2 + 2 + 2
∂y
∂z 
 ∂x
∂ 2u
=µ 2
∂y
(3.1b)
25
α1∇A2 = α1
%
∂
∂x j
2
 ∂ 2ui
∂ 2u j
∂ 2ui ∂ 2ui ∂ u j 
+
+
u
+
+


j
∂x 2j ∂x 2j
∂xi2 
 ∂t ∂x j ∂t ∂xi
2
 ∂  ∂ u
= α1   2i

 ∂t  ∂x j

∂ 3u ∂ 3u 
 + u j 3i + 3i 
∂x j ∂x j 

2
2
2
∂
∂   ∂ 2u ∂ 2u ∂ 2u   ∂ 3u ∂ 3u ∂ 3u  
 ∂  ∂ u ∂ u ∂ u   ∂
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3  
∂y
∂z   ∂x
∂y
∂z   ∂x
∂y
∂z   ∂x ∂y
∂z  
 ∂t  ∂x
∂ 3u
∂ 3u
= α1
+ α1 3
(3.1c)
2
∂t ∂y
∂y
 ∂ 2 ui ∂ 2 u j 
 2 + 2 
%
∂xi 
 ∂x j
 ∂ 3ui ∂ 3u j 
= α2  3 + 3 
∂xi 
 ∂x j
α 2∇A12 = α 2
∂
∂x j
 ∂ 3u ∂ 3u ∂ 3u 
= α2  3 + 3 + 3 
∂z 
 ∂x ∂y
∂ 3u
= α2 3
∂y
(3.1d)
Then substitute Equation (3.1b)-(3.1d) into Equation (3.1a) and reduce to;
ρ
∂u
∂p
∂ 2u
∂ 3u
∂ 3u
∂ 3u
= − + µ 2 + α1
+
α
+
α
1
2
∂t
∂x
∂y
∂t ∂y 2
∂y 3
∂y 3
(3.1e)
Based on the restriction that have been given in Chapter 2 given by Equation (2.18),
Equation (3.1e) become as follows;
ρ
∂u
∂p
∂ 2u
∂ 3u
= − + µ 2 + α1
∂t
∂x
∂y
∂t ∂y 2
(3.1)
y-component of the momentum equation:
 ∂v  ∂v
∂v
∂v  
∂p
+  u + v + w   = − + µ∇A1 + α1∇A2 + α 2∇A12
%
%
%
∂y
∂z  
∂y
 ∂t  ∂x
ρ
(3.2a)
26
Substitute Equation (2.17a) and (2.17b) into Equation (3.2a) and finds that;
 ∂ui ∂u j 
+


%
 ∂x j ∂xi 
∂ 2u
= µ 2i
∂x j
µ∇A1 = µ
∂
∂x j
 ∂ 2v ∂ 2v ∂ 2v 
=µ 2 + 2 + 2
 ∂x ∂y ∂z 
=0
∂
α1∇A2 = α1
%
∂x j
(3.2b)
2
 ∂2ui ∂2u j
∂2ui ∂2ui ∂ u j 
+
+ uj 2 + 2 + 2 

∂x j ∂x j ∂xi 
 ∂t∂x j ∂t∂xi
 ∂  ∂2u 
∂3ui ∂3ui 
i
= α1   2  + u j 3 + 3 


∂x j ∂x j 
 ∂t  ∂x j 
 ∂  ∂2v ∂2v ∂2v   ∂
∂
∂   ∂2v ∂2v ∂2v   ∂3v ∂3v ∂3v 
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3 
∂z   ∂x ∂y ∂z   ∂x ∂y ∂z 
 ∂t  ∂x ∂y ∂z   ∂x ∂y
(3.2c)
=0
α 2∇A12 = α 2
%
∂
∂x j
 ∂ 2ui ∂ 2u j 
 2 + 2 
∂xi 
 ∂x j
 ∂ 3ui ∂ 3u j 
= α2  3 + 3 
∂xi 
 ∂x j
 ∂ 3 v ∂ 3v ∂ 3v 
= α2  3 + 3 + 3 
 ∂x ∂y ∂z 
=0
(3.2d)
Then, substitute Equation (3.2b)-(3.2d) into Equation (3.2a) and obtain;
−
∂p
=0
∂y
(3.2)
27
z-component of the momentum equation:
 ∂w  ∂w
∂w
∂w  
∂p
+ u
+v
+ w   = − + µ∇A1 + α1∇A2 + α 2∇A12
%
%
%
∂y
∂z  
∂z
 ∂t  ∂x
ρ
(3.3a)
Substitute Equation (2.17a) and (2.17b) into Equation (3.3a) and finds that;
µ∇A1 = µ
%
∂
∂x j
=µ
 ∂ui ∂u j 
+


 ∂x j ∂xi 
∂ 2 ui
∂x 2j
 ∂2w ∂2w ∂2w 
=µ 2 + 2 + 2 
∂y
∂z 
 ∂x
=0
∂
α1∇A2 = α1
%
∂x j
(3.3b)
2
 ∂2ui ∂2u j
∂2ui ∂2ui ∂ u j 
+
+ uj 2 + 2 + 2 

∂x j ∂x j ∂xi 
 ∂t∂x j ∂t∂xi
 ∂  ∂2u 
∂3ui ∂3ui 
i
= α1   2  + u j 3 + 3 


∂x j ∂x j 
 ∂t  ∂x j 
∂
∂   ∂2 w ∂2w ∂2w   ∂3v ∂3v ∂3v 
 ∂  ∂2 w ∂2 w ∂2 w   ∂
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3 
∂z   ∂x ∂y ∂z   ∂x ∂y ∂z 
 ∂t  ∂x ∂y ∂z   ∂x ∂y
(3.3c)
=0
∂
α 2 ∇A = α 2
%
∂x j
2
1
 ∂ 2ui ∂ 2u j 
 2 + 2 
∂xi 
 ∂x j
 ∂ 3u ∂ 3u 
= α 2  3i + 3j 
∂xi 
 ∂x j
 ∂3 w ∂3 w ∂3w 
= α2  3 + 3 + 3 
∂y
∂z 
 ∂x
=0
(3.3d)
28
Then, substitute Equation (3.3b)-(3.3d) into Equation (3.3a) and obtain;
−
∂p
=0
∂z
Equation (3.2) and Equation (3.3) imply that p = p ( x ) since −
(3.3)
∂p
∂p
=−
= 0.
∂y
∂z
From Equation (3.1), (3.2) and (3.3), the governing equation for the unidirectional flows
of second grade fluid is as follows;
∂u
1 ∂p
∂ 2u
∂ 3u
=−
+ν 2 + β
∂t
∂y
∂t ∂y 2
ρ ∂x
where ν =
(3.4)
µ
α
is the kinetic viscosity and β = 1 is the elastic coefficient.
ρ
ρ
3.3 Unsteady Couette flow
3.3.1 Flow driven by the motion of the upper plate and the lower plate is fixed.
Consider that the second grade fluid between two parallel plates whose sides are at
y = h and y = 0 . The upper plate is moving with velocity U and the lower plate is at
rest. The governing equation is Equation 3.4 with the pressure is neglected, the boundary
and the initial conditions are
u ( h, t ) = U for t > 0,
u ( 0, t ) = 0 for all t ,
u ( y , 0 ) = 0 for 0 ≤ y < h.
(3.5)
29
The solution of the governing equation subject to the boundary and the initial condition
given by Equation (3.5) will be obtained by the used of the Laplace transform method.
The Laplace transform of u ( y, t ) is defined by the equation
∞
u = ∫ u e − st dt
0
Then the governing equation and the boundary and initial conditions become the
ordinary differential equation
 s 
u ′′ − 
 u = 0,
ν + β s 
u ( 0, s ) = 0, u ( h, s ) =
(3.6a)
U
.
s
(3.6b)
where in Equation (3.6a), primes denote the differentiation with respect to y.
Solving the ordinary differential equation and obtained;
 s 
m2 − 
=0
ν + β s 
 s 
m2 = 

ν + β s 
1
 s  2
m =
 .
ν + β s 
Therefore the general solution of Equation (3.6a) is
u ( y, s ) = A cosh qy + B sinh qy
with
1
 s 2
q=
 .
ν + β s 
(3.7)
30
In order to obtain the particular solution of Equation (3.7), we apply the boundary
condition u ( 0, s ) = 0 to Equation (3.7) and becomes;
u ( y, s ) = B sinh qy
Then, we apply the boundary condition u ( h, s ) =
(3.8)
U
to Equation (3.8) and obtain;
s
u ( y, s ) sinh qy
=
U
s sinh qh
(3.9)
In order to obtain u ( y, t ) , we must find the inverse of Equation (3.9). The
inverse of Equation (3.9) is not tabulated and it is obtained by the usual residue
principle. The inversion theorem for Laplace transform (see Appendix) gives
u ( y, t )
U
γ + i∞
1
sinh qy est
=
ds
2π i γ −∫i∞ s sinh qh
(3.10)
In equation (3.10) s = 0 is a simple pole. Therefore residue at s = 0 is

sinh qy 
Re s ( y; s = 0 ) = lim ( s − 0 ) e st ×
s →0
s sinh qh 

 y cosh qy 
= lim  ×
s →0 h
cosh qh 

y
= .
h
(3.11)
The other singular points are the zeros of
sinh qh = 0
qh = nπ i
nπ i
qn =
, n = 1, 2, 3,...., ∞.
h
(3.12)
31
Then,
−µ n2π 2
ρβ
sn =
, n = 1, 2, 3,...., ∞ where α1 = 2 .
2
2 2
h
( ρ h + α1n π )
are the poles. Since all q n 's are symmetrically placed about origin on the real axis, all
poles ( s n ) lie on the negative real axis. These are the simple poles and at all these poles
can be obtained as

sinh qy 
Re s ( y; s = sn ) = lim ( s − sn ) e st ×
s → sn
s sinh qh 

=
2(−1) n +1 nπ (ν + β sn )
2
(ν h ) s
n
2
sin
nπ
y.
h
(3.13)
Adding Res ( y; s = 0) and Res ( y; s = sn ) , a complete solution which is the velocity
distribution is obtained as
2
s t
n
u
2π ∞ (−1) n (ν + β sn ) e n
nπ
= y−
× sin
y
∑
U
ν h n=1
sn
h
h
where sn =
(3.14)
−µ n2π 2
, n = 1, 2, 3,...., ∞ .
( ρ h2 + α1n2π 2 )
From the study that have been conducted by Erdogan (2007), it is shown that in
order to obtain a solution in bounded region, instead of using the Laplace transform, the
use of the form of the boundary condition and the Fourier transform also can be used.
Since the expressions of the velocity from each method are same, Erdogan has proved
that it is not possible to obtain a solution for small values of times by the Laplace
transform method for a fluid of second grade fluid as is done for a Newtonian fluid.
32
Now, we will find the volume flux, which the volume flux across a plane normal
to the flow is;
h
Q = ∫ u ( y, t ) dy
(3.15)
0
Then, insert Equation (3.14) into Equation (3.15), we find
h
U
2π U
Q=∫ y−
νh h
0 h
∞
∑
n
( −1) n (ν + β sn )
2
e snt
sn
n =1
× sin
nπ
y
h
h

U y 2 2π U
=
−
 h 2 νh h

nπ  

− cos
y
∞
(−1) n n(ν + β sn )2 e sn t 
h 
×
 nπ

∑
sn
n =1

h 

  0

U h 2 2π U
=
−
 h 2 νh h

( −1) n n(ν + β sn ) 2 e snt
∑
sn
n =1
∞

2π U
− 0 +
 νh h
Uh 4U
=
+
2 νh
∞
∑
 − cos nπ
×
 nπ
h

( −1) n n(ν + β sn ) 2 e snt
∑
sn
n =1
∞
(ν + β s2n+1 )




 h 
×

 nπ  
2
s2 n+1
n =0
Finally, the volume flux can be written as;
8
Q
= 1+ 2
Uh
νh
2
∞
(ν + β s2 n +1 )
n =0
s2 n +1
∑
(3.16)
The frictional force per unit area on the moving plate at y=h is
 ∂ 2u 
 ∂u 
τ h = µ   + α1 

 ∂y  y = h
 ∂y∂t  y = h
(3.17)
33
Substitute Equation (3.14) into Equation (3.17) and obtain;
U 2π U ∞ ( −1) n n (ν + β sn )2 e snt nπ

n
×
×
−
1
τh = µ  −

(
)
∑
sn
h
 h ν h h n =1

nπ
 2π U ∞
2
n
n
+ α1  −
( −1) n (ν + β sn ) e snt × × ( −1) 
∑
h
 ν h h n =1

Or the frictional force can be written as;
τh
µU
= 1−
h
1 α 
2π 2 ∞ 2
2
n (ν + β sn ) e snt  + 1 
2 ∑
ν h n=1
 sn µ 
(3.18)
3.3.2 Flow driven by the motion of the lower plate and the upper plate is being free.
Suppose that the second grade fluid is in a region bounded by one rigid boundary
at y = 0 and one free boundary at y = h . The fluid is initially at rest and starts suddenly
due to the motion of the plate y = 0 . The governing equation is Equation 3.4 where the
pressure is neglected, the boundary and the initial conditions are
u ( 0, t ) = U for t > 0,
σ xy ( h, t ) = 0 for all t ,
u ( y, 0 ) = 0 for 0 < y ≤ h.
(3.19)
The condition for the stress at the free surface for second grade fluid is different
from that for Newtonian fluid. The velocity distribution obtained by Hayat et al. (2000)
is not correct since the surface condition was used is the surface condition for
Newtonian fluid. Therefore, Erdogan (2007) has introduced the correct condition for the
34
stress at the free surface for flows of second grade fluid and obtained the correct
expression of the velocity.
Using the same method of solution as in Section 3.3.1, we obtained the expressions for
velocity distribution, volume flux and the frictional force respectively.
u
4
1
= 1− ∑
e
U
π n = 0 ( 2 n + 1)
∞
where ε1 =
β
h2
−
( 2 n +1)2 π 2 (ν t
h2
)
4
2
1+ ε1 ( 2 n +1) π 2 4
× sin ( 2 n + 1)
πy
2h
(3.20)
.
The volume flux across a plane normal to the flow will be obtained by substituting
Equation (3.20) into Equation (3.15). Therefore, after doing the integration, the volume
flux for the flow is;
Q
8
= 1− 2
Uh
π
∞
∑
n =0
−
1
( 2 n + 1)
2
e
( 2 n +1)2 π 2 (ν t
h2
2
2
1+ ε1 ( 2 n +1) π
)
4
4
(3.21)
While the frictional force exerted by the suddenly moving plate y=0 is given by;
τ0
µU h
−
∞
1
n=0
1 + ε 1 ( 2 n + 1) π 2 4
= −2∑
2
e
( 2 n +1)2 π 2 (ν t
2
h2
)
1+ ε 1 ( 2 n + 1) π 2 4
4
(3.22)
35
3.4
Poiseuille Flow
Flow of a second grade fluid between two stationary parallel plates due to sudden
application of a constant pressure gradient is termed the Poiseuille flow of second grade
fluid or shortly the Poiseuille flow.
Suppose that the fluid is bounded by two parallel plates at y=b and y=-b, and
initially at rest. The fluid starts flowing suddenly due to the application of a constant
pressure gradient. The governing equation and the initial condition are
∂u
1 ∂p
∂ 2u
∂ 3u
=−
+ν 2 + β
∂t
∂y
∂t ∂y 2
ρ ∂x
u ( ±b, t ) = 0 for all t ,
(3.23)
u ( y, 0 ) = 0 for − b ≤ y ≤ b,
where u ( y, t ) is the velocity, ν and β are the material moduli, and
∂p
is the pressure
∂x
gradient. The solution of the governing equation subject to the boundary and the initial
condition given by Equation (3.23) will be obtained by using the same method that has
been used in the previous section which is the Laplace transform method.
This expression of the velocity has been given by Erdogan (2007) [10] using the
Laplace transform method, one finally obtains the velocity distribution for Poiseuille
flow;
n
( 2 n +1)2 π 2ν t
4 b2
−
2
u
y 2 32 ∞ ( −1)
πy
1+ ε ( 2 n +1) π 2 4
= 1− 2 − 3 ∑
e
× cos ( 2n + 1)
3
1 ∂p 2
b π n = 0 ( 2n + 1)
2b
−
b
2 µ ∂x
(3.24)
In order to get the volume flux for the Poiseuille flow, we insert Equation (3.24) into the
below equation;
36
b
Q = ∫ u ( y, t ) dy
−b
and obtained the volume flux across a plane normal to the flow and per unit width of the
plane;
 ( 2 n +1)2 π 2 ν t b 2
−
 1+ ε ( 2 n +1)2 π 2 4

(
Q
96
−4
= 1 − 4 ∑ ( 2n + 1) e
π n =0
− ( 2b 3µ ) ∂p ∂x
∞
)
4



(3.25)
3
Then the frictional force per unit area that exerted by the fluid at the initial time
on one of the plates, for example at y=b, is
τb
∂p ∂x
= 1−
8
π
2
1
∞
∑
n =0
( 2n + 1)
2
×
1
2
1 + ε ( 2n + 1) π 2 4
(3.26)
The above expression shows that the stress at t=0 on the plate is non zero for a fluid of
second grade compare to the Newtonian fluid that has a zero value.
3.5
Stokes’ First Flow Problem
The flow over a plane wall, which is initially at rest and is suddenly moved in its
own plane with a constant velocity, is termed as Stokes’ first flow problem or Rayleigh
flow. The second grade fluid stays in the region y ≥ 0 and x-axis is chosen as the plane
wall. The governing equation is
∂u
∂ 2u
∂ 3u
=ν 2 + β 2
∂t
∂y
∂ y∂t
where u ( y, t ) is the velocity, ν and β are the material constants and t is the time.
The boundary and initial conditions are:
37
u ( 0, t ) = U for t > 0,
u ( y, 0 ) = 0 for y > 0,
u → 0 for y → ∞,
(3.27)
where U is the velocity of the plate at y=0.
The solution of the governing equation subject to the boundary and the initial condition
given by Equation (3.27) will be obtained by the used of the sine Fourier transform
method.
The Fourier sine transform of u ( y, t ) is [24];
∞
u = ∫ u sin λ y dy
0
and the initial condition becomes
u ( 0) = 0.
The problem reduces to the solution of the following ordinary differential equation:
2
2
(1 + βλ ) u ′ +νλ u = νλU
where prime denotes differentiation with respect to t. The solution of the differential
equation is
νλ
−
(1+ βλ 2 ) 
u 1
= 1 − e t
.
U λ 

2
(3.28)
The inverse of u is given by the relation;
u=
2
π∫
∞
0
u sin λ y dy,
(3.29)
38
and inserting Equation (3.28) into Equation (3.29), one finds
u
2
= 1−
U
π
U2 
U
where τ = 
 t, η = 
ν
ν 
∫
∞
0
e
− ξ 2 τ 1+ εξ 2
(
) sin ηξ
dξ
ξ

ν 
 y, ξ =   λ , and

U 
ε =β
U2
ν2
(3.30)
. The solution given by
Equation (3.30) satisfies the boundary and the initial conditions given by Equation
(3.27) and the governing equation given by Equation (3.4).
Then, substitute the expression of velocity given by Equation (3.30) into
Equation (3.17) and the frictional force is obtained in the following form:
τ
ρU 2
=−
2
π
∫
∞
0
1
1 + εξ 2
e
−ξ 2τ
2
(1+ εξ )
cos ηξ d ξ
(3.31)
Then the stress at t=0 on the plane wall is
τ ( 0, 0 ) = −
ρU 2
ε
In research that has been conducted by Erdogan in 2007, this shows that the
stress for the Rayleigh flow at t=0 on the plane wall is finite for a fluid of second grade
fluid, but infinite for a Newtonian fluid [11].
3.6
Unsteady Generalized Couette Flow
39
The generalized Couette flow is the superposition of the simple Couette flow
over the two dimensional Poiseuille flow [11]. Suppose that the fluid is bounded by two
parallel plates at y=0 and y=h and it is initially at rest. The second grade fluid starts
suddenly due to a constant pressure gradient and by the motion of the upper plate.
The governing equation is Equation (3.4), the boundary and the initial conditions
are
u ( h, t ) = U for t > 0,
(3.32)
u ( 0, t ) = 0 for all t ,
u ( y, 0 ) = 0 for 0 ≤ y < h,
where h is the distance between two parallel plates.
Using the Laplace transform method, the solution subject to conditions in (3.32)
can be written as
2 2
2
n
 n π (ν t h )
 y y 2  2 ∞  ( −1)
2λ 
u y
πy
n  − 1+ ε n 2π 2
= +λ − 2 + ∑
− 3 2 1 − ( −1)   e
sin n


U h
nπ
h
 h h  π n =1  n

(3.33)
 dp  2
− h
β
dx 
, ε = 2 and λ is the dimensionless pressure gradient.
where λ = 
h
2 µU
Inserting the expression of velocity given by Equation (3.33) into Equation (3.17), the
frictional force is obtained in the following form;
∞
−
τ xy
y
2λ
n
n 


= 1 + λ 1 − 2  + 2∑ ( −1) − 2 2 1 − ( −1)   × e

µU h
h
nπ 

n =1 
n 2π 2 ν t h 2
(
2 2
1+ε n π
) cos n π y
h (3.34)
1 + ε n 2π 2
40
The stress on the stationary plate is
n2π 2 ν t h2
( )
∞
1+ ε n π
τ xy ( 0, t )
λ
2
e
n
n


= 1 + λ + 2∑ ( −1) − 2 2 1 − ( −1)   ×
2 2


µU h
nπ
 1+ ε n π
n =1 
−
2 2
(3.35)
While the stress on the plate y=h is
n 2π 2 ν t h 2
−
(
)
1+ε n 2π 2
τ xy ( h, t )
2λ
n
n  e

= 1 − λ + 2∑ ( −1) − 2 2 1 − ( −1)   ×
  1 + ε n2π 2
µU h
nπ 
n =1 
∞
(3.36)
The volume flux across a plane normal to the unsteady generalized Couette flow is

 −
Q
λ 8
1
4λ
= 1+ − 2 ∑
1
−

×e
Uh 2
3 π n =0 ( 2n + 1) 2  ( 2n + 1) 2 π 2 
∞
( 2 n +1)2 π 2 (ν t
2
h2
1+ε ( 2 n +1) π 2
)
(3.37)
It is shown that the Laplace transform method was used to obtain the exact
solution for unidirectional flows in bounded region. The Laplace transform is not
suitable to find the exact solution for flows in unbounded region. This is because of the
obtained solution does not satisfies the initial condition [10]. Therefore, the Fourier
transform method is used to obtain the exact solution for the flows in unbounded region.
CHAPTER IV
FLOWS OF SECOND GRADE FLUID THROUGH A POROUS MEDIUM
4.1
Introduction
Problems of fluids through porous media play an indispensable role in countless
practical applications, but very little efforts have so far been made to discuss these
problems. Examples of these applications are filtration process, biomechanics, packed
bed reactors, ceramic processing, enhanced oil recovery, chromatography and many
others [13].
In section 4.2, the governing equation for the unidirectional flows in a porous
medium will be derived. After getting the governing equation, the unsteady
unidirectional Couette flow and the Stokes’ first problem in a porous medium will be
discussed and the velocity distribution will be obtained in the next sections. A procedure
of Laplace transform will be used to obtain the exact solutions of the problems. The
effects of the porous medium on the velocity field also will be discussed in section 4.6.
The space between the two plates is porous with constant permeability and porosity. To
incorporate the effects of the pores on the velocity field, the modified Darcy’s law will
be used [17].
42
4.2
The Governing Equation
The momentum equation for flows of second grade fluid through a porous
medium has one extra term compared to the momentum equation of second grade fluid
flows as given by Equation (2.19).
The relation between the pressure drop and velocity for a second grade fluid in
porous media is
∇p = −
ϕ
∂
 µ + α1  u
K
∂t  %
(4.1)
where K ( < 0 ) and ϕ ( 0 < ϕ < 1) are the permeability and porosity parameter respectively.
But Equation (4.1) cannot be directly used to analyze the flow problems in a porous
space. Thus, the modified Darcy’s law based on the local volume averaging technique
will be considered in a porous layer.
Under consideration of the balance of forces acting on a volume element of fluid,
the local volume average balance of linear momentum is given by
ρ
Du
2
% = −∇p + µ∇A1 + α1∇A2 + α 2∇A1 + r
%
%
%
%
Dt
(4.2)
where r is the Darcy’s resistance for a second grade fluid in a porous space. r is also a
%
%
measure of the flow resistance offered by the solid matrix; then r satisfies the following
%
equation [17];
ϕ
∂
r = −  µ + α1  u
%
K
∂t  %
(4.3)
43
Substitute (5.3) into (5.2), we have
ρ
Du
ϕ
∂
2
% = −∇p + µ∇A1 + α1∇A2 + α 2∇A1 −  µ + α1  u
%
%
%
Dt
K
∂t  %
(4.4)
The velocity field of the unidirectional flow is u = u ( y, t ) , v = 0, and w = 0 which u is
the x component of the velocity.
x-component of the momentum equation:
 ∂u  ∂u
∂u
∂u  
∂p
ϕ
∂
+ u
+ v + w   = − + µ∇A1 + α1∇A2 + α 2∇A12 −  µ + α1  u
%
%
%
∂y
∂z  
∂x
K
∂t 
 ∂t  ∂x
ρ
(4.5a)
Substitute Equation (2.17a) and (2.17b) into Equation (4.5a) and finds that;
 ∂ui ∂u j 
+


%
 ∂x j ∂xi 
∂ 2u
= µ 2i
∂x j
µ∇A1 = µ
∂
∂x j
 ∂ 2 u ∂ 2u ∂ 2 u 
=µ 2 + 2 + 2
∂z 
 ∂x ∂y
∂ 2u
=µ 2
∂y
α1∇A2 = α1
%
∂
∂x j
(4.5b)
2
 ∂ 2ui
∂ 2u j
∂ 2u ∂ 2u ∂ u 
+
+ u j 2i + 2i + 2j 

∂x j ∂x j
∂xi 
 ∂t ∂x j ∂t ∂xi
2
 ∂  ∂ u
= α1   2i

 ∂t  ∂x j

∂ 3u ∂ 3u 
 + u j 3i + 3i 
∂x j ∂x j 

 ∂  ∂ 2u ∂ 2u ∂ 2u   ∂
∂
∂   ∂ 2u ∂ 2u ∂ 2u   ∂ 3u ∂ 3u ∂ 3u  
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3  
∂y
∂z   ∂x
∂y
∂z   ∂x
∂y
∂z   ∂x ∂y
∂z  
 ∂t  ∂x
= α1
∂ 3u
∂ 3u
+
α
1
∂t ∂y 2
∂y 3
(4.5c)
44
 ∂ 2 ui ∂ 2 u j 
 2 + 2 
∂xi 
 ∂x j
 ∂ 3ui ∂ 3u j 
= α2  3 + 3 
∂xi 
 ∂x j
∂
α 2 ∇A = α 2
%
∂x j
2
1
 ∂ 3u ∂ 3u ∂ 3u 
= α2  3 + 3 + 3 
∂z 
 ∂x ∂y
∂ 3u
= α2 3
∂y
(4.5d)
Then substitute Equation (4.5b)-(4.5d) into Equation (4.5a) and reduce to
∂u
∂p
∂ 2u
∂ 3u
∂ 3u
∂ 3u ϕ 
∂
= − + µ 2 + α1
+ α1 3 + α 2 3 −  µ + α1  u
ρ
2
∂t
∂x
∂y
∂t ∂y
∂y
∂y K 
∂t 
(4.5e)
Based on the restriction that have been given in Chapter II given by Equation (2.18),
Equation (4.5e) become as follows;
ρ
∂u
∂p
∂ 2u
∂ 3u
ϕ
∂
= − + µ 2 + α1
−  µ + α1  u
2
∂t
∂x
∂y
∂t ∂y
K
∂t 
(4.5)
y-component of the momentum equation:
 ∂v  ∂v
∂v
∂v  
∂p
ϕ
+  u + v + w   a = − + µ∇A1 + α1∇A2 + α 2∇A12 −
%
%
%
∂y
∂z  
∂y
K
 ∂t  ∂x
ρ
∂

 µ + α1  v
∂t 

Substitute Equation (2.17a) and (2.17b) into Equation (4.6a) and finds that;
(4.6a)
45
 ∂ui ∂u j 
+


%
 ∂x j ∂xi 
∂ 2ui
=µ 2
∂x j
µ∇A1 = µ
∂
∂x j
 ∂ 2v ∂ 2v ∂ 2v 
=µ 2 + 2 + 2
 ∂x ∂y ∂z 
=0
(4.6b)
2
 ∂ 2 ui ∂ 2 u j
∂ 2 ui ∂ 2 ui ∂ u j 
+
+
u
+
+


j
%
∂x 2j ∂x 2j
∂xi2 
 ∂t∂x j ∂t ∂xi
 ∂  ∂ 2u 
∂ 3u ∂ 3u 
= α1   2i  + u j 3i + 3i 


∂x j ∂x j 
 ∂t  ∂x j 
∂
∂   ∂ 2v ∂ 2v ∂ 2 v   ∂ 3v ∂ 3v ∂ 3v  
 ∂  ∂ 2 v ∂ 2 v ∂ 2v   ∂
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3  
∂z   ∂x
∂y
∂z   ∂x ∂y ∂z   ∂x ∂y ∂z  
 ∂t  ∂x ∂y
=0
(4.6c)
α1∇A2 = α1
∂
∂x j
∂
α 2 ∇A = α 2
%
∂x j
2
1
 ∂ 2 ui ∂ 2 u j 
 2 + 2 
∂xi 
 ∂x j
 ∂ 3u ∂ 3u j 
= α 2  3i + 3 
∂xi 
 ∂x j
 ∂ 3 v ∂ 3v ∂ 3v 
= α2  3 + 3 + 3 
 ∂x ∂y ∂z 
=0
(4.6d)
Then, substitute Equation (4.6b)-(4.6d) into Equation (4.6a) and obtain;
−
∂p
=0
∂y
(4.6)
46
z-component of the momentum equation:
 ∂w  ∂w
∂w
∂w  
∂p
ϕ
∂
2
µ + α1  w
+ u
+v
+w
1 + α1∇A2 + α 2∇A1 −
  = − + µ∇A

%
%
%
∂y
∂z  
∂z
K
∂t 
 ∂t  ∂x
ρ
(4.7a)
Substitute Equation (2.17a) and (2.17b) into Equation (4.7a) and finds that;
µ∇A1 = µ
%
∂
∂x j
=µ
 ∂ui ∂u j 
+


 ∂x j ∂xi 
∂ 2 ui
∂x 2j
 ∂2w ∂2w ∂2w 
=µ 2 + 2 + 2 
∂y
∂z 
 ∂x
=0
α1∇A2 = α1
%
∂
∂x j
(4.7b)
2
 ∂ 2ui
∂ 2u j
∂ 2ui ∂ 2ui ∂ u j 
+
+
u
+
+


j
∂x 2j
∂x 2j
∂xi2 
 ∂t ∂x j ∂t ∂xi
2
 ∂  ∂ u
= α1   2i

 ∂t  ∂x j

∂ 3u ∂ 3u 
 + u j 3i + 3i 
∂x j ∂x j 

 ∂  ∂ 2 w ∂ 2 w ∂ 2 w   ∂
∂
∂   ∂ 2 w ∂ 2 w ∂ 2 w   ∂ 3v ∂ 3v ∂ 3v  
= α1   2 + 2 + 2  +  u + v + w   2 + 2 + 2  +  3 + 3 + 3  
∂y
∂z   ∂x
∂y
∂z   ∂x
∂y
∂z   ∂x ∂y ∂z  
 ∂t  ∂x
=0
(4.7c)
∂
α 2 ∇A = α 2
%
∂x j
2
1
 ∂ 2 ui ∂ 2 u j 
 2 + 2 
∂xi 
 ∂x j
 ∂ 3u ∂ 3u j 
= α 2  3i + 3 
∂xi 
 ∂x j
 ∂3 w ∂3 w ∂3 w 
= α2  3 + 3 + 3 
∂y
∂z 
 ∂x
(4.7d)
=0
Then, substitute Equation (4.7b)-(4.7d) into Equation (4.7a) and obtain;
−
∂p
=0
∂z
(4.7)
47
Equation (4.6) and (4.7) imply that p = p ( x ) since −
∂p
∂p
=−
= 0 . Based on Equation
∂y
∂z
(4.5)-(4.7), the governing equation for unidirectional flows of second grade fluid
through a porous medium is as follows;
∂u
1 ∂p
∂ 2u
∂ 3u
ϕ
∂
=−
+ν 2 + d 2
− ν + d 2  u
2
ρ ∂x
∂t
∂y
∂t ∂y
K
∂t 
where ν =
(4.8)
µ
is the kinematic viscosity, ρ d 2 = α1 , the elastic coefficient and pressure
ρ
gradient is neglected which is reasonable when there is no applied pressure gradient.
4.3
Couette Flow with Moving Bottom Plate
This section deals with the solution of a second grade fluid in a porous layer in
absence of the pressure gradient. The flow is induced due to motion of the lower plate
i.e., for t>0, the plate at y=0 starts to slide in its own plane with a constant speed U0, the
velocity of the plate is given by (U0, 0, 0). The upper plate, at y=h is kept fixed. Under
this situation, the boundary and initial conditions are given by;
u (0, t ) = U 0θ ( t ) ,
u ( y, 0) = 0
u (h, t ) = 0,
(y > 0)
where θ ( t ) denotes the Heaviside unit step function and the governing equation is
Equation (4.8).
48
Defining the dimensionless quantities
u′ =
νt
ϕ
u
y
h 2ω
d
, y′ = , t ′ = 2 , ω ′ =
, l = and β = h
ν
U0
h
h
h
K
The governing problem becomes
2
(1 + β l ) ∂∂ut − ∂∂yu − l
2 2
2
2
u ( 0, t ) = θ ( t ) ,
u ( y, 0 ) = 0,
∂ 3u
+ β 2u = 0,
∂y 2 ∂t
(4.9a)
u (1, t ) = 0,
(4.9b)
y > 0.
where the primes have been suppressed for simplicity.
For the solution of (4.9a) subject to (4.9b), we define
∞
u ( y, s ) = L u ( y, t )  = ∫ e− st u ( y, t ) dt
0
as the Laplace transform of u ( y, t ) (where s is a Laplace transform parameter).
Taking Laplace transform of (4.9a) and (4.9b), we obtain
2
2 2
∂ 2u  β + s (1 + β l ) 
 u = 0,
−

∂y 2 
1 + sl 2


1
u ( 0, s ) = ,
u (1, s ) = 0.
s
Solving the above problem, we have
u ( y, s ) =
where
sinh q (1 − y )
s sinh q
(4.10)
49
 β 2 + s (1 + β 2l ) 

q=
1 + sl 2


Taking inverse Laplace transform of (4.10), we obtain
u ( y, t ) = θ ( t )
1 r +i∞ sinh q (1 − y )
ds
2π i ∫r −i∞ s sinh q
(4.11)
In order to obtain the solution, we have to solve the integral in Equation (4.11).
For that, we use the complex variable theory. It is seen that s=0 is a simple pole.
Therefore the residue at s=0 can be find using below equation
Res ( s = sn ) = lim ( s − sn ) est ⋅ u ( y, s )
s → sn
(4.12)
Then, we obtain
Res ( 0 ) =
sinh β (1 − y )
sinh β
(4.13)
The other singular points are the zeros of
sinh q = 0
Setting q = iλ , we find
sin λ = 0
If λn = nπ , n = 1, 2,3,....., ∞ are the zeros of (4.14), then


β 2 + n 2π 2
 , n = 1, 2,3,....., ∞
sn = − 
2
2 2
2
1 + ( β + n π ) l 
are the poles.
(4.14)
50
The residue at all these poles is obtained as
n
2
2 2



2 ( −1) nπ e snt
β
π
+
n
 = 2
Res  sn = − 
sin nπ (1 − y )
2
2 2
2

 ( β + n 2π 2 ) 1 + ( β 2 + n 2π 2 ) l 2
β
π
1
+
+
n
l


(
)



(
)
(4.15)
Adding Equation (4.13) and (4.15), a complete solution is given by
n
 sinh β (1 − y )

∞
−1) ne sn t
(
u ( y, t ) = θ ( t ) 
+ 2π ∑ 2
sin
n
1
−
y
π
( ) .
2 2
2
2 2
2
 sinh β

n =1 ( β + n π ) 1 + ( β + n π ) l


(
)
(4.16)
If we let t → ∞ , the steady state will be obtained in this following form:
u=
sinh β (1 − y )
sinh β
which holds for large t.
4.4
Couette Flow with Oscillating Plate
In this section, we will discuss the flow of second grade fluid in a porous layer
0<y<h when no pressure gradient is applied. The plate y=0 is oscillating in its own
plane with frequency ω. The oscillating plate at y=0 is responsible for generating the
motion of the flow. The plate at y=h is at rest. The corresponding boundary and initial
conditions are:
51
u (0, t ) = U 0 eiωt ,
u ( y, 0) = 0
u (h, t ) = 0,
(y > 0).
(4.17)
and the governing equation is given by Equation (4.8).
Using the dimensionless quantities that have been defined in Section 4.3 and
after suppressing the primes, the governing problem becomes
(1 + β 2l 2 )
∂u ∂ 2u 2 ∂ 3u
− 2 −l
+ β 2u = 0,
2
∂t ∂y
∂y ∂t
u ( 0, t ) = eiωtθ ( t ) ,
u (1, t ) = 0,
u ( y, 0 ) = 0
( y > 0) .
(4.18)
Note that the mathematical problem is the same as in Section 4.3 except that the
boundary condition. Therefore, employing the same procedure as in Section 4.3, we
obtain
n
∞
 sinh m (1 − y ) eiωt

−1) ne sn t
(
u ( y, t ) = θ ( t ) 
+ 2π ∑
sin nπ (1 − y )  , (4.19)
sinh m
D
n =1


where
1
 β 2 (1 + ω 2l 4 ) + ω 2l 2 + iω  2
 ,
m=
1 + ω 2l 2


D = 1 + ( β 2 + n 2π 2 ) l 2  ( β 2 + n 2π 2 )(1 + iωl 2 ) + iω  .
52
The steady state solution is of the following form
 sinh m (1 − y ) eiωt 
u=

sinh m


4.5
(4.20)
Stokes’ First Flow Problem
Stokes’ first problem is also known as Rayleigh’s flow. Jordan and Puri have
been conducted a study about Stokes’ first problem in a porous half space and obtained
an exact solution using Laplace transform [15]. However, the solution does not satisfy
the initial condition of the flow model.
Therefore, Tan and Masouka (2005) have obtained the exact solution of Stokes’
first problem that satisfies the initial condition by using Fourier sine transforms. The
drawback of solution not satisfying the initial condition [15] has been removed.
For Stokes’ first problem, the pressure gradient in the x-direction can be
neglected. The governing equation that given by Equation (4.8) becomes:
∂u
∂ 2u
∂ 3u
ϕ
∂
=ν 2 + d 2
− ν + d 2  u
2
∂t
∂y
∂t ∂y
K
∂t 
(4.21)
53
The corresponding initial and boundary conditions are
u ( y, 0 ) = 0,
u ( y, t ) → 0,
u ( 0, t ) = U 0 ,
∂u ( y, t )
→ 0 as y → ∞.
∂y
(4.22)
In order to solve the problem, we employed the non-dimensional quantities
u′ =
yU 0
tU 02
u
′
′
,y =
,t =
U0
ν
ν
and obtain the dimensionless motion equation as follows (for simplicity, the primes are
omitted):
(1 + α )
∂u ∂ 2u 2 ∂ 3u
= 2 +l 2
− β 2u ,
∂t ∂y
∂ y∂t
u ( y, 0 ) = 0,
(4.22a)
u ( 0, t ) = 1,
∂u ( y, t )
u ( y, t ) → 0,
→ 0 as y → ∞.
∂t
(4.22b)
where
U K 
ϕ
 0
dU 0
d2
α = ϕ ,l =
,β = 

ν
ν
K




−1
Then we multiply both sides of Equation (4.22a) by sin sy, and then integrate with
respect to y from 0 to ∞ and use the conditions (4.22b), we obtain
∂u
s2 + β 2
s
+
u−
= 0,
2 2
∂t 1 + α + l s
1 + α + l 2s2
u ( s, 0 ) = 0,
where u ( s, t ) denotes the Fourier sine transform of u ( y, t ) with respect to y.
(4.23)
54
The solution of the ordinary differential equation (4.23) subject to the initial condition is
s +β

−
t
1+α + l 2 s 2
1 − e



2
s
u ( s, t ) = 2
s +β2
2
(4.24)
Inverting Equation (4.24) by means of Fourier sine transform, we have
s +β
−
t 
s 
1+α + l 2 s 2

1
−
e
 sin sy ds

s 2 + β 2 

2
u ( y, t ) =
2
π
∫
∞
0
2
(4.25)
From the above equation, obviously (4.25) which is the velocity distribution of Stokes’
first flow problem satisfies the initial condition.
By letting t → ∞ , Equation (4.25) can be reduced to the steady state solution as follows
u ( y ) = e− β y
4.6
(4.26)
Analysis of Results
Hayat et al. (2005) has studied the effects of the porous medium which are the
porosity and permeability parameter, β on the velocity field of unsteady Couette flows.
The other parameters such as time t, elastic coefficient l, and frequency ω for the flows
are kept fixed where t=0.1, l=0.1 and ω =0.1.
55
1
0.8
0.6
u
β=0.5
β=1.0
0.4
β=5.0
0.2
β=10.0
0
0
0.2
0.4
0.6
0.8
1
y
Figure 4.1: Effect of the porosity and permeability parameter, β on the
velocity distribution of the unsteady Couette flow with sudden motion of
bottom plate
1
0.8
0.6
u
β=0.5
β=1.0
0.4
β=5.0
0.2
β=10.0
0
0
0.2
0.4
0.6
0.8
1
y
Figure 4.2: Effect of the porosity and permeability parameter, β on the
velocity distribution of the unsteady Couette flow with oscillation of bottom
plate
56
Figure 4.1 and 4.2 show the effects of porosity and permeability, β on the
velocity distributions for . From these figures, the velocity decreases with an increase in
β . Although the equations 4.16 and 4.19 are different (with or without oscillation term,
ω ), figures 4.1 and 4.2 show the negligible effect of the velocity profile graphs.
Hayat et al. (2005) also studied the effect of time on the velocity distribution.
They have plotted u against y for fixed values of l, β and ω .
1
0.8
0.6
u
t=0.01
0.4
t=0.05
0.2
t=0.1
0.5
0
0
0.2
0.4
0.6
0.8
1
y
Figure 4.3: Effect of t on the unsteady Couette flow with sudden motion of
bottom plate
1
t=0.5
t=2.0
t=4.0
t=6.0
u
0.5
0
0
0.2
0.4
0.6
0.8
1
-0.5
-1
y
Figure 4.4: Effect of t on the unsteady Couette flow with oscillation of
bottom plate
57
From Figure 4.3, it is seen that the increase of t, the velocity approaches a steady
state. Here in the first flow problem, the steady state of the velocity distribution is
obtained at t=0.5. For the second problem, the transient part velocity will decay after a
certain time [17]. It is found that velocity given by Equation (4.20) becomes the steady
state periodic velocity at t=6. Thus as a conclusion, the steady state in the first flow
problem is achieved much earlier when compared with the second flow problem.
Tan et al. (2005) has studied the Stokes’ first problem through half space porous
medium. Then, they have plotted the velocity distribution of a second grade fluid in the
porous half space for fixed β and l. It can be seen in Figure 4.4, the flow does not need
too long time for the flow to get to the steady state, where the velocity profile stops
changing with time.
Figure 4.5: u vs y for β = 2 and l = 2.
Figure 4.6: u vs y for β = 2 and l = 2.
58
In special case, when K → ∞, it implies α → 0 and β → 0 , the solution of
Stokes’ first flow problem for second grade fluid in non-porous case will be obtained.
u ( y, t ) =
2
π
∫
∞
0
− 2 2t 
1
1 − e 1+l s  sin sy ds

s 

s2
(4.27)
In this case, the graph for Stokes’ first flow problem has been plotted for fixed l
where l=2 as shown in Figure 4.6.
Figure 4.6: u vs y for β = 0, α = 0 and l = 2.
From Figure 4.6, it can be seen that the velocity gradient near the flat plate is
very high. Comparing Figure 4.5 and 4.6, it can be seen that the velocity of the second
grade fluid in non-porous case is absolutely an increasing function of t, which is
different from that of second grade fluid in porous case.
CHAPTER V
FLOWS OF HEATED SECOND GRADE FLUID
5.1
Introduction
The determination of the temperature distribution in a fluid layer when the
internal friction is not negligible is of utmost importance in many industrial fields, such
as oil exploitation, chemical and food processing and bio-engineering. The purpose of
this chapter is to review the work done by Fatecau (2002) where the temperature
distribution for second grade fluid will be obtained.
To derive the governing equations for this problem, in section 5.2 we consider
the three equations of motions in Chapter III, which are the continuity, momentum and
energy equations. In this chapter, the unsteady unidirectional flows that will be
considered are the Stokes’ first flow problem on a heated flat plate and on a heated edge.
After getting the governing equation, the velocity and temperature distributions
for the Stokes’ first flow problem on a heated flat plate and in a heated edge is
determined by the use of the simple and double sine Fourier transform.
60
5.2
The Governing Equation
5.2.2
The Momentum Equation
The momentum equation that will be used in this chapter has been derived in
Chapter III. Recall the momentum equation for unidirectional flows of second grade
fluid as given by Equation (3.14) as follows;
∂u
1 ∂p
∂ 2u
∂ 3u
=−
+ν 2 + β
ρ ∂x
∂t
∂y
∂t ∂y 2
5.2.3
(3.14)
The Energy Equation
The energy equation for second grade fluid has been given by Equation (2.42)
which is
ρc
Dθ
2
= µ A1 + k ∇ 2T + ρ q&
%
Dt
(2.42)
From Equation (2.42), the energy equation for unidirectional flows of second grade fluid
will be derived.
For unsteady unidirectional flow, the velocity field is in the form
u = u ( y, t ) , v = 0, and w = 0. which u is the x-component of the velocity.
61
x-component of the energy equation:
 ∂u
 ∂θ
∂θ
∂θ
∂θ 
+v
+w  = µ i
ρc  + u

∂x
∂y
∂z 
 ∂t
 ∂x j
 ∂u
= µ
 ∂x
 ∂u
= µ
 ∂y
2

2
 + k ∇ θ + ρ q&

2
 ∂ 2θ ∂ 2θ ∂ 2θ
∂u ∂u 
k
+
+
+

 2+ 2+ 2
∂y
∂z 
∂y
∂z
 ∂x

 ∂ 2θ 
 + k  2  + ρ q&
 ∂y 


 + ρ q&

(5.1a)
Hence, Equation (5.1a) can be simplified as
ρc
 ∂u 
 ∂ 2θ 
∂θ
= µ
 + k  2  + ρ q&
∂t
 ∂y 
 ∂y 
(5.1)
y-component of the energy equation:
 ∂u
 ∂θ
∂θ
∂θ
∂θ 
ρc  + u
+v
+w  = µ i
 ∂x
∂x
∂y
∂z 
 ∂t
 j
2

2
 + k ∇ θ + ρ q&

2
 ∂v ∂v ∂v 
 ∂ 2θ ∂ 2θ ∂ 2θ
= µ
+
+
+
k

 2+ 2+ 2
x
y
z
∂
∂
∂
∂y
∂z
 ∂x


=0

 + ρ q&

(5.2a)
Equation (5.2a) can be written as
ρc
∂θ
=0
∂t
(5.2)
62
z-component of the energy equation:
 ∂u
 ∂θ
∂θ
∂θ
∂θ 
ρc  + u
+v
+w  = µ i
 ∂x
∂x
∂y
∂z 
 ∂t
 j
2

2
 + k ∇ θ + ρ q&

2
 ∂w ∂w ∂w 
 ∂ 2θ ∂ 2θ ∂ 2θ
k
= µ
+
+
+

 2+ 2+ 2
∂y
∂z 
∂y
∂z
 ∂x
 ∂x
=0

 + ρ q&

(5.3a)
Equation (5.3a) can be written as
ρc
∂θ
=0
∂t
(5.3)
Based on Equation (5.1)-(5.3), we can conclude that the energy equation for
unidirectional flows of second grade fluid is Equation (5.1). Therefore, after simplifying,
we obtained the energy equation for unidirectional flows of second grade fluid as
follows
2
 ∂ 2θ  ν  ∂u  q& ∂θ
β  2 + 
 + =
 ∂y  c  ∂y  c ∂t
where ν =
µ
k
and β =
.
ρ
ρc
Or we can write the energy equation in other form as follows;
β
where g ( y, t ) =
ν  ∂u 
2
q&
+


c  ∂y  c
∂ 2θ
∂θ
+ g ( y, t ) =
2
∂y
∂t
(5.4)
63
5.3
The Stokes’ First Problem on A Heated Flat Plate
Suppose that a second grade fluid, at rest, occupies the space above an infinitely
extended plate in the x-plane. At time t=0+, the plate starts suddenly to slide, in its plane,
with constant speed, V. Let T(t) and f(y) denote the temperature of the plate t ≥0 and the
temperature of the fluid at the moment t=0. By the influence of shear and of heat
conduction the fluid, above the plate, is gradually moved and heated. The velocity field
will be of the form;
V = u ( y, t ) i
%
%
where i denotes a unit vector along the x-coordinate direction and the temperature field
%
in the form θ = θ ( y, t ) .
In order to find the temperature distribution of the Stokes’ first problem for a
heated flat plate, firstly we need to find the flow’s velocity distribution. The
corresponding governing differential equations, which are the momentum and energy
equations respectively;
∂u
∂ 2u
∂ 3u
=ν 2 + d 2
∂t
∂y
∂t ∂y 2
(5.5a)
∂ 2θ
∂θ
+ g ( y, t ) =
2
∂y
∂t
(5.5b)
β
with initial and boundary conditions that need to be satisfied;
u ( y , 0 ) = 0, y > 0; u ( 0, t ) = V , t > 0,
u ( y, t ) ,
∂u
→ 0 as y → ∞
∂y
(5.6a)
64
θ ( y, 0 ) = f ( y ), y > 0; θ ( 0, t ) = T (t ), t > 0,
θ ( y, t ) ,
(5.6b)
∂θ
→ 0 as y → ∞
∂y
Firstly, we have to find the velocity distribution of the flow in order to obtain the
flow’s temperature distribution. The solution of the linear partial differential equation
(4.5a) with the initial and boundary conditions (5.6a) has the simple form using the
Fourier transform method that we used before in Chapter III.
 2 ∞ sin ( yξ ) − νξ 2 
u ( y, t ) = U 1 − ∫
e 1+αξ dξ 
0
ξ
 π

2
(5.7)
In order to determine the temperature distribution in fluid, both sides of Equation (5.5b)
multiplied by
2
π
sin ( yξ ) and obtain;
 ∂ 2θ

2
∂θ
sin ( yξ )  β 2 + g ( y , t )  =
sin ( yξ )
π
π
∂t
 ∂y

2
(5.8)
Then, we integrate (5.8) with respect to y from 0 to ∞
∞
2
0
π
β∫
sin ( yξ )
∞
∞
∂ 2θ
2
2
∂θ
dy + ∫
sin ( yξ ) g ( y , t ) dy = ∫
sin ( yξ )
dy (5.9)
2
0
0
π
π
∂y
∂t
Having in mind the boundary and initial conditions (5.6b), we find that (5.9) become
∂θ s (ξ , t )
2
+ βξ 2θ s (ξ , t ) =
βξ T ( t ) + g s (ξ , t )
∂t
π
65
with the initial condition
θ s (ξ , 0 ) = f s ( ξ )
where θ s (ξ , t ) , g s (ξ , t ) and f s (ξ ) are the Fourier sine transform of θ ( y, t ) , g ( y, t ) and
respectively.
Then we solve the above ordinary differential equation, subject to the initial
condition and obtain

t
 2
θ s (ξ , t ) = e − βξ t  f s ( ξ ) + ∫ 
0
2

 π

2

βξ T (τ ) + g s (ξ ,τ )  e βξ t dτ 


(5.10)
The result (5.10) is inverted by means of Fourier’s sine formula, integrating by
parts the term containing T (τ ) ; we obtain that the temperature distribution for Stokes’
first flow problem for a heated plate as follows:

 t
1 − erf
′
+
T
τ
(
)

 ∫0

2
β
t






θ ( y, t ) = T 1 − erf 

+
2
π∫
∞
0
y


y

  dτ


 2 β ( t − τ ) 
t
2
2
e − βξ t sin ( yξ )  f s (ξ ) + ∫ g s (ξ ,τ )e βξ t dτ  d ξ

0

(5.11)
66
5.4
The Stokes’ First Flow Problem Within A Heated Edge
Now, let us consider a second grade fluid at rest occupying the space of first dial
of a rectangular edge ( x ≥ 0, −∞ < y < ∞, z ≥ 0) . At the moment t=0+ the extended edge is
impulsively brought to the constant speed V. the two walls of the edge will be again
maintained to the temperature T(t). The velocity field will be in the form;
V = u ( y, z , t ) i
%
%
and the temperature field θ = θ ( y, z , t ) .
The momentum and energy equations respectively reduce to;
2
∂u 
∂ 2u 
2 ∂  ∂ u
= ν + d
+


∂t 
∂t   ∂y 2 ∂z 2 
 ∂ 2θ ∂ 2θ 
∂θ
+ 2  + g ( y, z , t ) =
2
∂z 
∂t
 ∂y
β
(5.12a)
(5.12b)
with the initial and boundary conditions,
u ( y , z , 0 ) = 0, y > 0, z > 0; u ( 0, z , t ) = u ( y , 0, t ) = V , t > 0,
(5.13a)
θ ( y, z , 0 ) = f ( y, z ), y > 0, z > 0; θ ( 0, z , t ) = θ ( y , 0, t ) = T (t ), t ≥ 0,
(5.13b)
where the function f ( y, z ) represents the temperature distribution of the second grade
fluid at the moment t = 0.
67
The Rayleigh flow for a heated edge is a flow problem involving three
independent variables where the double Fourier sine transform need to be applied in
order to reduce this problem into the ordinary differential equation. The Laplace
transform method cannot be used to solve this problem due to the solution obtained does
not satisfies the initial condition of the flow [10].
The solutions of the linear partial differential equations (5.12a) and (5.12b) which are
the velocity and temperature distributions and with the initial and boundary conditions
(5.13a) and (5.13b) are respectively:

4

u ( y, z , t ) = U 1 − 2
π




y 
 erf
 2 βt 
θ ( y, z, t ) = T 1 − erf 

+
2
∞
∞ − β ξ 2 +η 2 t
0
0
π∫ ∫
e
(
)
ν ξ 2 +η 2
∫
∞
0
sin yξ
ξ
∫
∞
0
−
sin zη
×e
η
(
)
1+α ξ 2 +η 2
(

t
) dη d ξ 

(5.14)





 z  ∞
y
 erf

 + ∫0 T ′ (τ ) 1 − erf 



 2 β t 
 2 β ( t −τ ) 

sin ( yξ ) sin ( zη ) ×  f s (ξ ,η ) + ∫ gs (ξ ,η ,τ ) e

0
t
− β ξ 2 +η 2 t
(
)


z

 dτ


 2 β ( t −τ ) 
dτ  dξ dη

(5.15)
where f s (ξ ,η ) and g s (ξ ,η ,τ ) are the double Fourier sine transforms of the functions
f ( y, z ) and g ( y, z , t ) with respect to the variables y and z.
The computations show that the temperature distributions, which are Equations
(5.11) and (5.15), as well as the velocity distributions which are Equations (5.7) and
(5.14), satisfy the matching partial differential equations and all boundary and initial
conditions.
CHAPTER VI
CONCLUSION
6.1
Introduction
This chapter will summarize and conclude the results that have been obtained
from Chapter I to Chapter V. Lastly, the suggestion to further this research also
presented in this chapter.
6.2
Summary of Research
Flows of second grade fluid have been reviewed and discussed in this study. The
basic definitions, examples and the advantages of second grade fluid have been
discussed in Chapter I. The governing equation that describes the flows of Newtonian
fluid is the Navier-Stokes equation. But second grade fluid cannot be described by the
Navier-Stokes equation. In Chapter II, the governing equations for second grade fluid
have been derived using the constitutive equation of a second grade fluid.
69
In Chapter III, the work done on the unidirectional flows of second grade fluid
have been reviewed. The unsteady unidirectional second grade fluid flows in bounded
region and unbounded region have been considered. It is shown that the Laplace
transform method is the method that used to solve the flows in bounded region while the
Fourier transform method is the method to solve the flows in unbounded region (see
Table 6.1(a) and Table 6.1(b) ). The solution for flows in unbounded region have been
obtained by using the Laplace transform does not satisfy the initial conditions and
therefore the Fourier transform method is used.
In Chapter IV, the unsteady unidirectional flows of second grade fluid through a
porous medium have been presented. The solutions obtained in paper done by Hayat
(2005) and Tan (2005) have been reviewed. It is shown in Figure 4.1 and 4.2, the
velocity profiles decrease with an increase in the porosity, ϕ and permeability
parameter, K. In special case, when K → ∞, it implies α → 0 and β → 0 , the solution
of Stokes’ first flow problem for second grade fluid in non-porous case will be obtained.
The velocity of the second grade fluid in non-porous medium is an absolutely increasing
function of t compared to the velocity of second grade fluid in porous case.
Paper done by Fatecau (2002) is about the methods of how to obtain the velocity
and temperature distributions of the Stokes’ first flow problem of heated second grade
fluid that have been reviewed in Chapter V. In this chapter, the energy equation for
unsteady unidirectional flows of second grade fluid has been derived (see Table 6.2).
The energy equation is used to obtain the temperature distribution of the unsteady
unidirectional flows of second grade fluid. The Stokes’ first flow problem on a heated
flat plate and in a heated edge is determined by means of the simple and double Fourier
sine transforms respectively. This sine Fourier transform method is convenient for flows
in unbounded regions of second grade fluid because the obtained solution satisfy the
initial condition.
CHAPTER VI
CONCLUSION
6.1
Introduction
This chapter will summarize and conclude the results that have been obtained
from Chapter I to Chapter V. Lastly, the suggestion to further this research also
presented in this chapter.
6.2
Summary of Research
Flows of second grade fluid have been reviewed and discussed in this study. The
basic definitions, examples and the advantages of second grade fluid have been
discussed in Chapter I. The governing equation that describes the flows of Newtonian
fluid is the Navier-Stokes equation. But second grade fluid cannot be described by the
Navier-Stokes equation. In Chapter II, the governing equations for second grade fluid
have been derived using the constitutive equation of a second grade fluid.
69
In Chapter III, the work done on the unidirectional flows of second grade fluid
have been reviewed. The unsteady unidirectional second grade fluid flows in bounded
region and unbounded region have been considered. It is shown that the Laplace
transform method is the method that used to solve the flows in bounded region while the
Fourier transform method is the method to solve the flows in unbounded region (see
Table 6.1(a) and Table 6.1(b) ). The solution for flows in unbounded region have been
obtained by using the Laplace transform does not satisfy the initial conditions and
therefore the Fourier transform method is used.
In Chapter IV, the unsteady unidirectional flows of second grade fluid through a
porous medium have been presented. The solutions obtained in paper done by Hayat
(2005) and Tan (2005) have been reviewed. It is shown in Figure 4.1 and 4.2, the
velocity profiles decrease with an increase in the porosity, ϕ and permeability
parameter, K. In special case, when K → ∞, it implies α → 0 and β → 0 , the solution
of Stokes’ first flow problem for second grade fluid in non-porous case will be obtained.
The velocity of the second grade fluid in non-porous medium is an absolutely increasing
function of t compared to the velocity of second grade fluid in porous case.
Paper done by Fatecau (2002) is about the methods of how to obtain the velocity
and temperature distributions of the Stokes’ first flow problem of heated second grade
fluid that have been reviewed in Chapter V. In this chapter, the energy equation for
unsteady unidirectional flows of second grade fluid has been derived (see Table 6.2).
The energy equation is used to obtain the temperature distribution of the unsteady
unidirectional flows of second grade fluid. The Stokes’ first flow problem on a heated
flat plate and in a heated edge is determined by means of the simple and double Fourier
sine transforms respectively. This sine Fourier transform method is convenient for flows
in unbounded regions of second grade fluid because the obtained solution satisfy the
initial condition.
6.2 Summary of Research
Table 6.1 (a): Velocity Distributions for Unidirectional Flows of Second Grade Fluid
Type of
flows
Unsteady
Couette
flow with
moving
bottom
plate and
lower plate
is fixed
Unsteady
Couette
flow with
lower plate
is moving
and upper
surface is
free
Poiseuille
flow
Rayleigh
flow
Governing Equation
Boundary and Initial
Conditions
Method of
solving
Velocity Distribution
2
u ( h , t ) = U for t > 0,
2
3
∂u
∂u
∂u
=ν 2 + β
∂t
∂y
∂t∂y 2
The Laplace
transform
u ( 0, t ) = 0 for all t ,
method is used since
u ( y , 0 ) = 0 for 0 ≤ y < h.
the flow
is in bounded region.
st
n
u
2π ∞ (−1) n (ν + β sn ) e n
nπ
= y−
× sin
y
∑
U
ν h n=1
sn
h
h
− µ n 2π 2
, n = 1, 2,3,...., ∞ .
where sn =
( ρ h2 + α1n2π 2 )
u
4
1
= 1− ∑
e
U
π n= 0 ( 2n + 1)
∞
∂u
∂ 2u
∂ 3u
=ν 2 + β
∂t
∂y
∂t ∂y 2
∂u
1 ∂p
∂ 2u
∂3u
=−
+ν 2 + β
ρ ∂x
∂t
∂y
∂t∂y2
∂u
∂ 2u
∂ 3u
=ν 2 + β
∂t
∂y
∂t∂y 2
u ( 0, t ) = U for t > 0,
σ xy ( h, t ) = 0 for all t,
u ( y,0) = 0 for 0 < y ≤ h.
u ( ±b, t ) = 0 for all t,
u ( y,0) = 0 for − b ≤ y ≤ b,
u ( 0 , t ) = U fo r t > 0,
u ( y, 0 ) = 0
fo r y > 0,
u → 0 fo r y → ∞ ,
The Laplace
transform
method is used since
the flow
is in bounded region.
The Laplace
transform
method is used since
the flow
is in bounded region.
where
ε1 =
β
h2
−
( 2 n +1)2 π 2 (ν t
h2
2
2
1+ ε1 ( 2 n +1) π
)
4
4
× sin ( 2n + 1)
πy
2h
.
n
( 2 n +1)2 π 2ν t
4 b2
−
2
u
y 2 32 ∞ ( −1)
1+ ε 2 n +1 π 2 4
e ( )
= 1− 2 − 3 ∑
× cos ( 2n + 1) π y 2b
3
1 ∂p 2
π
b
n = 0 ( 2 n + 1)
−
b
2 µ ∂x
−ξ τ (1+εξ ) sin ηξ
u
2
The Fourier
= 1−
e
dξ
0
transform
U
π
ξ
method is used since
U 2 
U 
ν
the flow
where τ = 
 t , η =   y, ξ = 
is in unbounded
ν 
U
ν 
region.
.
∫
∞
2
2
2
U

 λ , and ε = β ν 2

70
2 2
Unsteady
Generalized
Couette
flow
The Laplace
transform
method since the
flow
is in bounded region.
u ( h, t ) = U for t > 0,
∂u
1 ∂p
∂2u
∂3u
=−
+ν 2 + β
2
ρ ∂x
∂t
∂y
∂t∂y u ( 0, t ) = 0 for all t ,
u ( y, 0 ) = 0 for 0 ≤ y < h,
2
n
 n π (νt h )
u y  y y2  2 ∞ ( −1) 2λ 
πy
n  − 1+εn2π2
= +λ − + ∑
−
1−( −1) e
sinn

U h  h h2  π n=1  n n3π2 
h

where
 dp  2
− h
dx 
2 µU
λ=
and ε = β .
h2
Table 6.1 (b): Frictional Forces and Volume Fluxes for Unidirectional Flows of Second Grade Fluid
Type of flows
Frictional Force
Unsteady Couette
flow with moving
bottom plate and
lower plate is
fixed
Frictional force on y=h
Unsteady Couette
flow with lower
plate is moving
and upper plate is
being free
Frictional force on y=0
τh
µU
2
= 1−
h
2π
ν h2
∞
∑ n (ν + β s )
2
n
n =1
2
Volume Flux
1 α 
e snt  + 1 
 sn µ 
2
τ0
µU h
n =0
−
1
∞
= −2 ∑
2
Q
8 ∞ (ν + β s2 n +1 )
= 1+ 2 ∑
Uh
s2 n+1
ν h n=0
2
1 + ε1 ( 2n + 1) π 2 4
e
2
( 2 n +1) π (ν t
h
2
)
2
1+ ε1 ( 2 n +1) π 2 4
4
Q
8
= 1− 2
Uh
π
∞
∑
n =0
−
1
( 2n + 1)
2
( 2 n +1)2 π 2 (ν t
e
h2
)
4
2
1+ ε1 ( 2 n +1) π 2 4
Frictional force on y=b
τb
= 1−
8
2
∞
∑
1
2
1
×
2
Poiseuille flow
∂p ∂x
Rayleigh flow
−ξ τ (1+εξ )
τ
2 ∞ 1
=− ∫
e
cosηξ d ξ
2
2
0
ρU
π 1 + εξ
π
n=0
 ( 2 n +1)2 π 2 ν t b 2
−
 1+ε ( 2 n +1) 2 π 2 4

(
( 2n + 1)
2
Q
96
−4
= 1 − 4 ∑ ( 2n + 1) e
3
π n =0
− ( 2b 3µ ) ∂p ∂x
∞
2
1 + ε ( 2n + 1) π 4
) 4 


2
71
Unsteady
Generalized
Couette flow
∞
−
τ xy
y
2λ
n
n 


= 1 + λ  1 − 2  + 2∑ ( −1) − 2 2 1 − ( −1)   × e


h
nπ
µU h


n =1 
n 2π 2 ν t h 2
(
1+ ε n 2π 2
) cos n π y

 −
Q
λ 8
1
4λ
= 1+ − 2 ∑
1−
×e
2 
2
Uh 2
3 π n= 0 ( 2n + 1)  ( 2n + 1) π 2 
∞
h
1 + ε n 2π 2
( 2 n +1)2 π 2 (ν t
h2
)
2
1+ ε ( 2 n +1) π 2
Table 6.2: Velocity Distributions for Flows of Heated Second Grade Fluid
Type of
flows
Momentum
Equation
Velocity Distribution
Energy Equation
β
Rayleigh
flow on a
heated flat
plate
∂u
∂ 2u
∂ 3u
=ν 2 + d 2
∂t
∂y
∂t ∂y 2
 2 ∞ sin ( yξ )
u ( y, t ) = U 1− ∫
e
ξ
 π 0
νξ 2
−
1+αξ 2

dξ 

Temperature Distribution
∂ 2θ
∂θ
+ g ( y, t ) =
2
∂y
∂t



 y  t
y
 dτ
 + ∫0 T′ (τ ) 1− erf 
 2 β ( t −τ ) 

 2 βt 




θ ( y, t ) = T 1− erf 

where
g ( y, t ) =
ν  ∂u 
2
q&
+


c  ∂y 
c
2
∞
π ∫0
+
t
2
2
e−βξ t sin ( yξ )  fs (ξ ) + ∫ gs (ξ ,τ )eβξ t dτ  dξ

0

 ∂2θ ∂2θ 
∂θ
+ 2  + g( y, z,t) =
2
∂t
 ∂y ∂z 
β
Rayleigh
flow
within a
heated
edge
∂u 
∂   ∂2u ∂2u 
= ν + d 2   2 + 2 
∂t 
∂t   ∂y ∂z 
ν(ξ2+η2)


−
t
 4 ∞sin yξ ∞ sinzη 1+α(ξ2+η2)

u( y, z,t) =U1− 2 ∫
×
e
d
η
d
ξ

π 0 ξ ∫0 η






 

 y   z  ∞
y
z
 erf 
 dτ
 erf 
 + ∫0 T′ (τ ) 1− erf 






2
β
t
2
β
t
−
−
2
t
2
t
β
τ
β
τ
(
)
(
)


 


 



θ ( y, z, t ) = T 1− erf 

where
2
ν ∂u ∂u q&
g( y, z,t) =  +  +
c ∂y ∂z  c
+
2
) sin yξ sin zη ×  f ξ ,η + t g ξ ,η,τ e−β(ξ +η )t dτ  dξ dη
( ) ( )  s( ) ∫ s(
)

∞ ∞ −β ξ 2 +η2 t
π ∫0
∫
0
e
(
2

0
2

72
Table 6.3: Velocity Distributions for Flows of Second Grade Fluid through a Porous Medium
Type of
flows
Governing Equation
Boundary and Initial Conditions
u (0, t ) = U 0θ ( t ) ,
Unsteady
Couette flow
with moving
bottom plate
u ( y, 0) = 0
2
Unsteady
Couette flow
with oscillating
bottom plate
∂u
1 ∂p
∂u
=−
+ν 2
ρ ∂x
∂t
∂y
+ d2
∂ 3u
∂
ϕ
− ν + d 2  u
2
∂t ∂y
K
∂t 
where
θ ( t ) is
(y > 0)
the Heaviside unit step
 sinh β (1− y)

+


β
sinh


u ( y, t ) = θ ( t )  ∞
n
snt
.
ne
−
1
( )
2π ∑

π
sin
n
1
−
y
(
)
 n=1 ( β 2 + n2π 2 ) 1+ ( β 2 + n2π 2 ) l 2



(
)
function
u (0, t ) = U 0eiωt ,
u ( y, 0) = 0
where
u (h, t ) = 0,
(y > 0)
n
∞
sinh m(1− y) eiωt
( −1) nesnt sin nπ 1− y  ,
+ 2π ∑
u ( y, t ) =θ ( t ) 
( )
sinh m
D
n=1


ω is the frequency.
u ( y, 0 ) = 0,
Rayleigh flow
u ( h, t ) = 0,
Velocity Distribution
u ( 0, t ) = U 0 ,
∂u ( y, t )
u ( y, t ) → 0,
→ 0 as y → ∞.
∂y
s +β

−
t 
2 2
 1 − e 1+α + l s  sin sy ds




2
u ( y, t ) =
2
π
∫
∞
0
s
s2 + β 2
2
73
73
6.3
Suggestion for Further Research
The study presented in this dissertation gives some suggestions for future areas
of research especially for the problem related with second grade fluid. This study will be
a reference to those interested in the problem of second grade fluid.
In this study, work done by Hayat et al. (2005) and Erdogan et al. (2005) has
been reviewed. Hayat et al. (2005) conducted a research involving Couette flow through
a porous medium while Erdogan et al. (2005) conducted a study on unsteady
unidirectional flows of second grade fluid. Therefore, from the above stated researches,
a study on a heated unsteady generalized Couette flow through a porous medium can be
suggested for future research.
74
REFERENCES
[1]
Anderson, J. D. (1995). Computational Fluid Dynamics. McGraw-Hill, Inc.
[2]
Fatecau, C., Unsteady Couette Flows of Second Grade Fluids In Heated
Cylindrical Domains. Acta Mechanica 150 (2001) 127-134.
[3]
Fetecau, C., The Rayleigh-Stokes Problem for Heated Second Grade Fluids.
International Journal of Non-Linear Mechanics. 37 (2002) 1011-1015.
[4]
Xue, C. and Nie, J., Exact solutions of The Rayleigh–Stokes Problem for A
Heated Generalized Second Grade Fluid in A Porous Half-space. Applied
Mathematical Modelling. 33 (2009) 524-531.
[5]
Nield, D. A. and Bejan, A., (1999), Convection in Porous Media, Second
Edition, Springer, Berlin.
[6]
Tan, F. S., Zhao, W., , Y., Masuoka, T., The Rayleigh–Stokes Problem for A
Heated Generalized Second Grade Fluid With Fractional Derivative Model.
Nonlinear Analysis: Real World Applications. 7 (2006) 1072 – 1080.
[7]
Astarita, G. and Marucci, G. (1974). Principles of Non-Newtonian Fluid
Mechanics, McGraw-Hill, Inc.
[8]
Dunn, J. E. and Fosdick, R. E., Thermodynamics, Stability and Boundedness of
Fluids of Complexity 2 and Fluids of Second Grade. Archieve for Rational
Mechanics and Analysis. 56 (1974) 191-252.
[9]
Leal, L. G., (2007). Advanced Transport Phenomena, Cambridge University
Press.
[10]
Erdogan, M. E. and Imrak, C. E., On The Comparison of The Methods Used for
The Solutions of The Governing Equation for Unsteady Unidirectional Flows of
Second Grade Fluids. International Journal of Enginering Science. 45 (2007)
786-796.
[11]
Erdogan, M. E. and Imrak, C. E., On Some Unsteady Flows of A Non-Newtonian
Fluid. Applied Mathematical Modelling. 31 (2007) 170-180.
[12]
Erdogan, M. E. and Imrak, C. E., On Unsteady Unidirectional Flows of A Second
Grade Fluid. International Journal of Non-Linear Mechanics. 40 (2005) 12381251.
[13]
Hussain, M., Hayat, T., Asghar, S. and Fetecau, C., Oscillatory Flows of Second
Grade Fluid in A Porous Space. Nonlinear Analysis: Real World
Applications.(2009).
75
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Muralidhar, K. and Sundararajan, T. (2003). Computational Fluid Flow and Heat
Transfer. 2nd. ed. India: Alpha Science.
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Jordan, P. M. and Puri, P., Stokes’ Problem for A Rivlin-Ericksen Fluid of
Second Grade in A Porous Half-space. International Journal of Non-Linear
Mechanics. 38 (2003) 1019-1025
[16]
Paterson, A. R. (1983), A First Course in Fluid Dynamics. Cambridge.
[17]
Hayat, T., Masood Khan, Siddiqui, A. M. and Ayub, M., The Unsteady Couette
Flow of A Second Grade Fluid in A Layer of Porous Medium. Archive of
Mechanics. 57 (2005) 405-416.
[18]
Hayat, T., Masood Khan, Siddiqui, A. M and Ashgar, S., Transient Flows of A
Second Grade Fluid. International Journal of Non-Linear Mechanics. 39 (2004)
1621-1633.
[19]
Hayat, T., Siddiqui, A. M. and Ashgar, S., Some Unsteady Unidirectional Flows
of A Non–Newtonian Fluid. International Journal of Engineering Science. 38
(2000) 337-346.
[20]
Fosdick, R. L. and Rajagopal, K. R., Anomalous Features in The Model of
Second Order Fluids. Archive for Rational Mechanics and Analysis. 70 (1979)
145-152.
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Bandelli, R., Unsteady Unidirectional Flows of Second Grade Fluids in Domains
With Heated Boundaries. International Journal of Non-Linear Mechanics. 30
(1995) 263-269.
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Tan, W. and Masuoka, T.,Stokes’ First Problem For A Second Grade Fluid in A
Porous Half-space With Heated Boundary. International Journal of Non-Linear
Mechanics. 40 (2005) 515-522.
[23]
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1974.
APPENDIX
In order the second grade fluid model to satisfy the Clausius-Duhem inequality and the
assumption that the specific Helmholtz free energy be a minimum in equilibrium, the following
conditions must be hold;
µ≥0 , α1≥0 , α1+α2=0
Therefore, the definition of Helmholtz free energy and the Clausius-Duhem inequality will be
explained.
Helmholtz free energy
In thermodynamics, the Helmholtz free energy is a thermodynamic potential which measures the
“useful” work obtainable from a closed thermodynamic system at a constant temperature and
volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the
maximum amount of work extractable from a thermodynamic process in which temperature and
volume are held constant. Under these conditions, it is minimized at equilibrium. The Helmholtz
free energy was developed by Hermann von Helmholtz.
The Helmholtz free energy is defined by
(1)
where E is the energy, T is the temperature, and S is the entropy. When a system changes its
thermodynamic state, the change in Helmholtz free energy is therefore given by
(2)
If T and V are constant, the (2) reduces to
(3)
But the combined law of thermodynamics states that
(4)
and, since we have stipulated dV = 0, this becomes
(5)
Therefore
(6)
The Helmholtz free energy is intimately related to the equilibrium constant at constant volume
via
(7)
where k is Boltzmann's constant.
The Clausius-Duhem Inequality
In classical continuum mechanics, by combining the momentum balance law, the two laws of
thermodynamics and using the free energy, one gets an inequality
r
r
σ d : D U − Q ⋅ gradT ≥ 0,
( )
r
r
where σ d is the dissipative stress, D U is the strain rate, T is the temperature and TQ is the heat
( )
r
flux vector, Q being the entropy flux vector. This inequality is called the Clausius-Duhem
inequality. Scientists do not agree on the status to be given to this inequality. Instead of focusing
on the inequality which is the subject of some discussions, let us remark that its expression
involves two quantities,
•
the mechanical dissipation
r
σd :D U
( )
which is the opposite of the power of the dissipative interior forces. This quantity results
directly from the choice of the power of the interior forces;
•
the thermal dissipation
−Q ⋅ ∇ T
%
which results from the first law of thermodynamics
The Modified Darcy’s Law
It is well known that in flows of viscous Newtonian fluid at low speed through porous
medium the pressure drop caused by the frictional drag is proportional to the velocity. This is the
familiar Darcy’s law, which relates the pressure drop and velocity in an unbounded porous
medium.
However, there are few studies in viscoelastic flow in porous medium. By analogy with
Oldroyd-B model, the following phenomenological model, which relates the pressure drop and
velocity for a viscoelastic fluid in unbounded porous medium, has been introduced;
(1)
where K is permeability, λp the relaxation time, λv the retardation time and α1 = µλv . VD is the
Darcian velocity, which is related to the usual velocity vector. V by VD=φV. φis porosity of the
porous media.
From (1), we can see that the constitutive equation for the second grade fluid can be
obtained from that of Oldroyd-B by letting λp =0 for unidirectional flows. As we are only
interested in unidirectional flow, the relations between the pressure drop and velocity for a
second grade fluid in porous media can be inferred from (1) as follows;
(2)
Equation (2) is known as the modified Darcy’s law. It is also can be regarded as an
approximate form of an empirical momentum equation for the second grade fluid in an
unbounded porous medium.
The Inversion Theorem ( for the Laplace transform )
Refer to Tranter (1974), the required inversion formula for the Laplace transform is
f ( x) =
1 γ +i∞ xp
e f ( p ) dp
2π i ∫γ −i∞
The Double Fourier Sine Transform
Fourier transforms can be used in a similar way to reduce the number of independent
variables in partial differential equations. The form of the boundary conditions at the lower limit
of the variable selected for exclusion decides the choice of sine or cosine transform. The sine
transform is employed to find solution of partial differential equations in a semi-infinite domain.
If a problem involving three independent variables, two transforms have to applied on the partial
differential equation to an ordinary differential equation. Therefore, the double Fourier sine
transform means that we have to apply the sine Fourier transform to the partial differential
equation twice to reduce into ordinary differential equation.