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) =U1− 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 [14] Muralidhar, K. and Sundararajan, T. (2003). Computational Fluid Flow and Heat Transfer. 2nd. ed. India: Alpha Science. [15] 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. [21] Bandelli, R., Unsteady Unidirectional Flows of Second Grade Fluids in Domains With Heated Boundaries. International Journal of Non-Linear Mechanics. 30 (1995) 263-269. [22] 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] William R. S., Mechanics of Non-Newtonian Fluids, Pergamon Press. 1978. [24] Tranter, C.J., Integral Transforms in Mathematical Physics, Methuen & Co. Ltd. 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.