Continuum Thermomechanics

Progress in Mathematical Physics Volume 37

Editors-in-Chief Anne Boutet de Monvel, Université Paris VII Denis Diderot Gerald Kaiser, The Virginia Center for Signals and Waves Editorial Board D. Bao, University of Houston C. Berenstein, University of Maryland, College Park P. Blanchard, Universität Bielefeld A.S. Fokas, Imperial College of Science, Technology and Medicine C. Tracy, University of California, Davis H. van den Berg, Wageningen University

Alfredo Bermúdez de Castro

Continuum Thermomechanics

Birkhäuser Verlag Basel · Boston · Berlin

Author: Alfredo Bermúdez de Castro Facultad de Matemáticas Universidade de Santiago de Compostela Campus Universitario Sur 15782 Santiago de Compostela Spain e-mail : [email protected]

2000 Mathematics Subject Classification 74A, 74J, 76A, 76N, 80A

A CIP catalogue record for this book is available from the Library of Congress, Washington D.C., USA Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at .

ISBN 3-7643-7265-6 Birkhäuser Verlag, Basel – Boston – Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use whatsoever, permission from the copyright owner must be obtained. © 2005 Birkhäuser Verlag, P.O. Box 133, CH-4010 Basel, Switzerland Part of Springer Science+Business Media Printed on acid-free paper produced of chlorine-free pulp. TCF ∞ Printed in Germany ISBN-10: 3-7643-7265-6 ISBN-13: 978-3-7643-7265-1 987654321

www.birkhauser.ch

To my wife, Ana

Contents Preface 1 General Definitions, Conservation Laws 1.1 Motion of a Body . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . 1.3 Balance of Linear and Angular Momentum . . . . . . . . . 1.4 Balance of Energy. First Principle of Thermodynamics . . . 1.5 Second Principle of Thermodynamics. The Clausius-Duhem Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xi

. . . .

1 1 4 5 7

. . . .

11

2 Lagrangian Coordinates 2.1 The Piola-Kirchhoff Stress Tensors . . . . . . . . . . . . . . . . . . 2.2 The Conservation Equations in Lagrangian Coordinates . . . . . .

13 13 14

3 Constitutive Laws 3.1 Thermodynamic Process. Material Body . . . . . . . . . . . . . . . 3.2 Coleman-Noll Materials . . . . . . . . . . . . . . . . . . . . . . . .

17 17 18

4 The Principle of Material Frame-Indifference 4.1 Change in the Observer. The Indifference Principle . . . . . . . . . 4.2 Consequences for Coleman-Noll Materials . . . . . . . . . . . . . .

27 27 28

5 Replacing Entropy with Temperature 5.1 The Conservation Equations in Terms of Temperature . . . . . . .

33 33

6 Isotropy 6.1 The Extended Symmetry Group . . . . . . . . . . . . . . . . . . . 6.2 Isotropic Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 39

7 Equations in Lagrangian Coordinates

43

. . . .

. . . .

. . . .

viii

Contents

8 Linearized Models 8.1 Linear Approximation of the Motion Equation . . . . . . . . . . . 8.2 Linear Approximation of the Energy Equation . . . . . . . . . . . 8.3 Isotropic Linear Thermoviscoelasticity . . . . . . . . . . . . . . . .

47 47 52 54

9 Quasi-static Thermoelasticity 9.1 Statement of the Equations . . . . . . . . . . . . . . . . . . . . . . 9.2 Time Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 A Particular Case . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57 57 57 59

10 Fluids 10.1 The Concept of Fluid, First Properties . . . . . . . . 10.2 Motion Equation. Thermodynamic Pressure . . . . . 10.3 Energy Equation, Enthalpy . . . . . . . . . . . . . . 10.4 Thermodynamic Coefficients and Equalities . . . . . 10.5 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . 10.6 Statics of Fluids . . . . . . . . . . . . . . . . . . . . 10.7 The Boussinesq Approximation, Natural Convection

. . . . . . .

61 61 63 64 66 77 77 78

11 Linearized Models for Fluids, Acoustics 11.1 General Equations, Dissipative Acoustics . . . . . . . . . . . . . . 11.2 The Isentropic Case, Non-Dissipative Acoustics . . . . . . . . . . . 11.3 Linearized Models under Gravity . . . . . . . . . . . . . . . . . . .

81 81 85 87

12 Perfect Gases 12.1 Definition, General Properties . . . . . . . . 12.2 Entropy and Free Energy . . . . . . . . . . 12.3 The Compressible Navier-Stokes Equations 12.4 The Compressible Euler Equations . . . . .

. . . .

93 93 94 97 98

13 Incompressible Fluids 13.1 Isochoric Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

101 101 101 103

. . . .

. . . .

. . . .

. . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

. . . . . . .

. . . .

14 Turbulent Flow of Incompressible Newtonian Fluids 105 14.1 Turbulence Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 14.2 The k −
Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 15 Mixtures of Coleman-Noll Fluids 109 15.1 General Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 15.2 Mixture of Perfect Gases . . . . . . . . . . . . . . . . . . . . . . . . 113

Contents

ix

16 Chemical Reactions in a Stirred Tank 119 16.1 Chemical Kinetics. The Mass Action Law . . . . . . . . . . . . . . 119 16.2 Conservation of Chemical Elements . . . . . . . . . . . . . . . . . . 122 16.3 Reacting Mixture of Perfect Gases . . . . . . . . . . . . . . . . . . 123 17 Chemical Equilibrium of a Reacting Mixture of Perfect Gases in a Stirred Tank 17.1 The Least Action Principle for the Gibbs Free Energy . . . . . . . 17.2 Equilibrium for a Set of Reversible Reactions, Equilibrium Constants . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 The Stoichiometric Method . . . . . . . . . . . . . . . . . . . . . .

125 125 126 131

18 Flow of a Mixture of Reacting Perfect Gases 18.1 Mass Conservation Equations . . . . . . 18.2 Motion Equation . . . . . . . . . . . . . 18.3 Energy Conservation Equation . . . . . 18.4 Conservation of Elements . . . . . . . . 18.5 Equilibrium Chemistry . . . . . . . . . . 18.6 The Case of Low Mach Number . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

135 135 137 137 140 141 142

19 The Method of Mixture Fractions 19.1 General Equations . . . . . . . . . . 19.2 Examples . . . . . . . . . . . . . . . 19.3 The Adiabatic Case . . . . . . . . . 19.4 The Case of Equilibrium Chemistry

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

145 145 148 149 149

. . . .

. . . .

20 Turbulent Flow of Reacting Mixtures of Perfect Gases, The PDF Method 153 20.1 Elements of Probability . . . . . . . . . . . . . . . . . . . . . . . . 153 20.2 The Mixture Fraction/PDF Method . . . . . . . . . . . . . . . . . 155 A Vector and Tensor Algebra A.1 Vector Space. Basis . . A.2 Inner Product . . . . . A.3 Tensors . . . . . . . . A.4 The Affine Space . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

161 161 163 164 170

B Vector and Tensor Analysis B.1 Differential Operators . . . . . . . . . . . . . . B.2 Curves and Curvilinear Integrals . . . . . . . . B.3 Gauss’ and Green’s Formulas. Stokes’ Theorem B.4 Change of Variable in Integrals . . . . . . . . . B.5 Transport Theorems . . . . . . . . . . . . . . . B.6 Localization Theorem . . . . . . . . . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

173 173 175 177 178 178 179

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

x

Contents B.7 Differential Operators in Coordinates B.7.1 Cartesian Coordinates . . . . B.7.2 Cylindrical Coordinates . . . B.7.3 Spherical Coordinates . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

179 179 182 184

C Some Equations of Continuum Mechanics in Curvilinear Coordinates 189 C.1 Mass Conservation Equation . . . . . . . . . . . . . . . . . . . . . 189 C.2 Motion Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 C.3 Constitutive Law for Newtonian Viscous Fluids in Cooordinates . 191 D Arbitrary Lagrangian-Eulerian (ALE) Formulations of the Conservation Equations 195 D.1 ALE Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 D.2 Conservative ALE Form of Conservation Equations . . . . . . . . . 197 D.2.1 Mixed Conservative ALE Form of the Conservation Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 D.3 Mixed Nonconservative Form of ALE Conservation Equations . . . 200 Bibliography

203

Index

205

Preface This book is intended to be an extension of Gurtin’s book on continuum mechanics [5] by including the laws of thermodynamics and thus making it possible to study the mechanical behaviour of material bodies, the response of which involves variables such as entropy or temperature. In order to do that our departure point is Coleman and Noll’s article [3] on the thermodynamics of elastic materials with heat conduction and viscosity which has been extended for the purpose at hand to the case of nonhomogeneous materials. The present book has been used for many years as a textbook for graduate and undergraduate mathematics students at the University of Santiago de Compostela. The first Chapter revisits the conservation principles of continuum thermomechanics, that is, the conservation of mass, linear and angular momentum balance and the first two principles of thermodynamics: namely, energy conservation and entropy inequality. All principles are introduced in integral form and in Eulerian coordinates. Local forms consisting of partial differential equations are then obtained. Writing these local equations in Lagrangian coordinates is the subject of Chapter 2. Chapter 3 deals with the constitutive laws of continuum thermomechanics. First the notion of a material body characterised by its constitutive class is given. Then we introduce a general material body defined by Coleman and Noll in the above referenced article. By imposing the second principle of thermodynamics, we prove some relations to be satisfied by the response functions of such a material. Then, in Chapter 4, the principle of material frame-indifference is introduced and its consequences for the response functions of the Coleman-Noll materials are established. In Chapter 5, the partial differential equations governing a thermodynamic process are written replacing entropy with temperature. Chapter 6 is devoted to isotropy. By using the representation theorems for isotropic tensor and vector-valued functions, we obtain simple forms for the response functions of Coleman-Noll materials. In Chapter 7, the equations satisfied by each thermodynamic process of these materials are written in Lagrangian coordinates. We also show that inviscid Coleman-Noll materials are hyperelastic. The linear approximations of these equations about a static reference state are deduced in Chapter 8, assuming that the gradient of the displacement and the difference of temperature with respect to a reference state are both small. This is rigorously done through careful computation of the derivatives of the response functions. Isotropic materials are specifically considered. Thus, we obtain the partial differential system for linear thermoviscoelasticity; its numerical solution by incremental methods, in the inviscid quasi-static case, is addressed in Chapter 9. Fluids are the subject of Chapter 10 where they are introduced as particular Coleman-Noll materials when the extended symmetry group is the unimodular group. We define the classical thermodynamic variables like specific heat, sound speed, volumetric thermal expansion, and write the conservation equations

xii

Preface

in terms of them. We also include a section dealing with fluid statics under a gravity field, and finally the approximate Boussinesq model for natural convection is introduced. The linearized models obtained for general Coleman-Noll materials in Chapter 8 are specialized for fluids in Chapter 11. Thus, the standard models for dissipative and non-dissipative acoustics are properly deduced. Assuming that the body force is gravity force, we also deduce the equations for internal gravity waves. Particular fluids called perfect gases are studied in Chapter 12 where the compressible Navier-Stokes and Euler systems of partial differential equations are deduced. Incompressible fluids are briefly examined in Chapter 13, but since these materials are not Coleman-Noll materials, they need to be studied separately. In particular, the incompressible Navier-Stokes and Euler equations are introduced. A quick overview of turbulence models is the goal of Chapter 14. Thermodynamics of mixtures of Coleman-Noll fluids and, more specifically, of mixtures of perfect gases is the subject of Chapter 15. In Chapters 16 and 17 some concepts of chemical kinetics are given, with special emphasis placed on modelling the chemical equilibrium by using Gibbs free energy. By using this methodology we study the flow of a mixture of reacting perfect gases in Chapter 18. In particular, the standard equations for modelling combustion are properly written both for finite-rate and equilibrium chemistry. Chapter 19 deals with a computational method widely used in combustion: the mixture fraction method. The important case of equilibrium chemistry is specifically considered. Finally, in Chapter 20, we give a short introduction to the interesting subject of turbulent reacting flows by recalling the fundamentals of the probability density function (PDF) method. The appendices include some mathematical background in tensor algebra and analysis and then the formulation of conservation equations in so-called arbitrary Lagangian-Eulerian (ALE) coordinates is given. From a computational point of view, this formulation is very useful when dealing with free boundary flows or fluid-structure interaction problems.

Acknowledgements I wish to thank R. Ohayon and P. G. Ciarlet for their encouragement to publish the book, and P. Quintela and O. L. Pouso for their careful reading of the manuscript and their many useful suggestions.

Chapter 1

General Definitions, Conservation Laws In this chapter we recall some notations and results from continuum mechanics which will be extensively used along these notes. Further details can be found, for instance, in [5] (see also Appendices A and B).

1.1 Motion of a Body Let E be an affine space on a three-dimensional Euclidean vector space V. Let us denote by Lin the vector space of endomorphisms of V (which is isomorphic to the space of second order tensors) and by Sym the subspace of those which are symmetric. Definition 1.1.1. A body B is a regular region of the Euclidean space E. Sometimes we will refer to B as the reference configuration. Elements in B are called material points. Definition 1.1.2. A deformation of B is a smooth one-to-one mapping f which maps B onto a closed region in E and satisfies det(∇f ) > 0.

(1.1)

Definition 1.1.3. A motion of B is a class C 3 mapping X:B×R→E with X(·, t), a deformation of B for each fixed t.

(1.2)

2

Chapter 1. General Definitions, Conservation Laws

We refer to x = X(p, t) as the place occupied by the material point p at time t while (1.3) Bt = X(B, t) is the region of space occupied by the body at time t (see Figure 1.1).

X( ,t)

B

P( ,t)

Bt

Figure 1.1: Motion and reference map.

Definition 1.1.4. We denote by trajectory of the motion the set T = {(x, t) : x ∈ Bt , t ∈ R}. The vector u(p, t) = X(p, t) − p represents the displacement of p at time t while the gradient of the motion is the tensor field defined by F (p, t) := ∇X(p, t),

(1.4)

so that we have F (p, t) = I + ∇u(p, t). The gradient gives information on the local deformation of the body. In continuum mechanics other tensor fields are also used to represent the state of local deformation. Let us mention C := F t F and B := F F t , which are called the right and left Cauchy-Green strain tensors, respectively, and the Green-Lagrange (sometimes called Green-Saint Venant) strain tensor defined by 1 1 G = (C − I) = (F t F − I). (1.5) 2 2 It is straightforward to see that G=

1 (∇u + ∇ut + ∇ut ∇u). 2

(1.6)

The linear part of this expression, namely E=

∇u + ∇ut , 2

(1.7)

is called infinitesimal strain tensor. Moreover, we recall that, from the Polar Decomposition Theorem, we can factorize F as F = RU = V R, (1.8)

1.1. Motion of a Body

B×R

-

Ψ

@ φm @ @ R @

3

T

T

Ψ−1 -

@ ϕs @ @ R @

φ W

B×R ϕ

W

Figure 1.2: Material and spatial descriptions.

where R is a rotation tensor field (i.e., orthogonal with determinant equal to 1) and U and V are symmetric positive definite tensor fields given by U = B 1/2 ,

V = C 1/2 .

(1.9)

Let P(·, t) : Bt → B be the inverse mapping of X(·, t). Then the mapping P:T →B

(1.10)

gives the material point which occupies the place x at time t. It is called the refer˙ ¨ ence map of the motion. We call X(p, t) the velocity and X(p, t) the acceleration of the material point p at time t. Fields defined in T are called spatial (or Eulerian) fields while those defined in B × R are called material (or Lagrangian) fields. Let Ψ : B × R → T be the mapping given by Ψ(p, t) := (X(p, t), t),

(1.11)

the inverse of which is Ψ−1 (x, t) = (P(x, t), t). By using these mappings any spatial field can be transformed into a material field and vice versa. Let φ be a spatial field. Its material description is defined by φm := φ ◦ Ψ.

(1.12)

Similarly, let ϕ be a material field. Its material description is defined by ϕs := ϕ ◦ Ψ−1 .

(1.13)

In particular, the spatial description of the velocity is ˙ v(x, t) := X(P(x, t), t).

(1.14)

4

Chapter 1. General Definitions, Conservation Laws Throughout this book, the following notations for differential operators are

used: • Material fields. ∂φ , φ˙ = ∂t Divφ = divergencep φ,

∇φ = gradientp φ,

(1.15)

Curlφ = Curlp φ.

(1.16)

• Spatial fields. ∂ϕ , ∂t div ϕ = divergencex ϕ, ϕ
=

grad φ = gradientx ϕ,

(1.17)

curlϕ = curlx ϕ,

(1.18)

ϕ˙ = material derivative of ϕ.

(1.19)

1.2 Conservation of Mass We assume a mass distribution for B defined by a reference density, ρ0 : B → R+ , in the reference configuration. The mass conservation law implies that the density in the motion X, ρ(x, t), must satisfy ρ(x, t) det(F (p, t)) = ρ0 (p), x = X(p, t).

(1.20)

From (1.20), the following local equations of the mass conservation law can be deduced: ρ
+ div(ρv) = 0,

(1.21)

ρ˙ + ρ div v = 0,

(1.22)

which are called the conservative and the non-conservative forms, respectively.

Weak Formulation It is very easy to obtain a so-called weak formulation of the mass conservation law. Let t be any time and z : Bt → R be any square integrable scalar field. Multiplying (1.22) by z and integrating in Bt , we get

ρz ˙ dVx + ρ div vz dVx = 0. (1.23) Bt

Bt

The above weak formulation can be used for finite element approximation. However, for finite volume methods the following equation is used:

d ρ dVx + ρv · n dAx = 0. (1.24) dt R ∂R

1.3. Balance of Linear and Angular Momentum

5

It can be easily obtained by integrating the conservative form (1.21) in a control volume R and then applying Gauss’ Theorem. The following useful result is a straightforward consequence of the Reynolds’ transport Theorem and the mass conservation principle. Lemma 1.2.1. Let Φ be a smooth spatial field defined in the trajectory of a motion X. Let us assume that Φ is either scalar or vector-valued. Then for any part P and time t

d ˙ dVx . ρΦ dVx = ρΦ (1.25) dt Pt Pt

1.3 Balance of Linear and Angular Momentum Let us denote by N the set of unit vectors of the linear space V. Definition 1.3.1. A system of forces for a body B during the motion X is a pair (s, b) of vector fields s : N × T → V, b : T → V, (1.26) with 1. s(n, x, t), for each n ∈ N and t, a smooth function of x in Bt , 2. b(x, t), for each t, a continuous function of x in Bt . Vector field s represents the density of surface force exerted across an oriented surface S with positive unit normal vector n at point x by the material on the positive side of S upon the material on the negative side (Cauchy’s hypothesis). Vector field b is called the body force and gives the force, per unit volume, exerted by the environment of the body on x. The balance of linear and angular momentum laws state that, for every part P and time t, the following equalities hold:


d vρ dVx = s(n) dAx + b dVx , (1.27) dt Pt ∂Pt Pt


d r×vρ dVx = r×s(n) dAx + r×b dVx . (1.28) dt Pt ∂Pt Pt According to the Cauchy Theorem, there exists a symmetric second order tensor field T : T → Sym (called the Cauchy stress tensor) such that s(n, x, t) = T (x, t)n. Furthermore, the following equations hold: (ρv)
+ div(ρv ⊗ v) = div T + b, ρv˙ = div T + b,

(1.29) (1.30)

which are the conservative and the non-conservative motion equation, respectively.

6

Chapter 1. General Definitions, Conservation Laws

In (1.30) v˙ denotes the material time derivative which is defined, for a vector field w, by ∂ ∂w w(X(p, t), t)|p=P(x,t) = (x, t) + gradw(x, t)v(x, t). ∂t ∂t The following result is known as the Theorem of Power Expended: ˙ w(x, t) :=

Theorem 1.3.2. For every part P and time t we have



|v|2 d dVx + ρ T · D dVx = s(n) · v dAx + b · v dVx , dt Pt 2 Pt ∂Pt Pt

(1.31)

(1.32)

where D=

L + Lt 2

(1.33)

and L := gradv. Proof. Firstly, by using a Green’s formula we get


div T · v dVx = T n · v dAx − T · gradv dVx Pt ∂Pt Pt

= s(n) · v dAx − T · D dVx ,

(1.34)

Pt

∂Pt

because T is symmetric. Moreover, from Lemma 1.2.1,

d |v|2 dVx . ρv˙ · v dVx = ρ dt Pt 2 Pt

(1.35)

Let us make the scalar product of the motion equation (1.30) by v and integrate in Pt . We obtain


˙ ρv · v dVx = div T · v dVx + b · v dVx . (1.36) Pt

Pt

Pt

Now this equality easily yields (1.32) by using (1.34) and (1.35).  Term Pt T · D dVx in equation (1.32) is called the power stress.




Weak Formulation. Boundary and Initial Conditions As for the mass conservation, we can obtain a weak formulation for the balance of momentum. Let t be any time and w : Bt → V be any smooth vector field. By making the scalar product of (1.30) with w and using a Green’s formula we get



ρv˙ · w dVx = − T · gradw dVx + s(n) · w dAx + b · w dVx . Bt

Bt

∂Bt

Bt

(1.37)

1.4. Balance of Energy. First Principle of Thermodynamics

7

We have to give an initial condition, the velocity field at initial time, and boundary conditions. The latter may have different forms depending on the particular material. At this stage we consider two general ones. Let us divide the boundary of Bt into two parts: ∂Btv and ∂Bts . Let us suppose the velocity is given on ∂Btv while the surface force density is prescribed on ∂Bts. More precisely, ¯ on ∂Btv , v=v s(n) = ¯s on ∂Bts ,

(1.38) (1.39)

¯ and s(n) = ¯s are given boundary fields. In order to incorporate these where v boundary conditions into the weak formulation (1.37), firstly, we only take test functions, w, vanishing on ∂Btv . Then the surface integral in (1.37) becomes


¯s · w dAx . s(n) · w dAx = s(n) · w dAx = (1.40) ∂Bts

∂Bt

∂Bts

Finally, the weak formulation consists of the following equalities:



¯s · w dAx + ρv˙ · w dVx + T · gradwdVx = b · w dVx , Bt

∂Bts

Bt

Bt

v

=

∀w null on ∂Btv , ¯ on ∂Btv , v

v

=

v0 in B0 at t = 0.

(1.41) (1.42) (1.43)

Moreover, we can also obtain a formulation of the motion equation to be used for finite volume approximation. For this purpose we integrate the conservative form (1.29) in a control volume R and use Gauss’ Theorem. We get



d ρv dVx + ρvv · n dAx = T n dAx + b dVx . (1.44) dt R ∂R ∂R R

1.4 Balance of Energy. First Principle of Thermodynamics We begin by stating Cauchy’s hypothesis concerning the surface heat which is similar to that for surface forces. It consists in assuming the existence of a surface heat density g(n, x, t) defined for each unit vector n and every (x, t) in the trajectory of the motion, having the following property: let S be an oriented surface in Bt with positive unit normal n at x. Then g(n, x, t) is the heat, per unit area and time, flowing from the negative side of S to the positive side of S. Besides this surface heat, we suppose there is another one supplied at interior points of Bt . Such a heat is determined by a scalar field f defined on T : f (x, t) gives the heat, per unit volume and time, supplied by the environment at point x at time t.

8

Chapter 1. General Definitions, Conservation Laws The previous discussion motivates the following:

Definition 1.4.1. A system of heat for a body B during a motion (with trajectory T ) is a pair (g, f ) of functions g : N × T → R, f : T → R where 1. g(n, x, t), for each n ∈ N and t, is a smooth function of x on Bt , 2. f (x, t), for each t, is a continuous function of x on Bt . We call g the surface heat and f the body heat. Definition 1.4.2. The heat rate supplied into the part P at time t is given by

Q(P, t) = − g(n) dAx + f dVx . (1.45) ∂Pt

Pt

Let us consider a system of forces (s, b) and a system of heat (g, f ) during a motion X of a body B. The energy conservation law (which is also called First Principle of Thermodynamics) asserts that there exists a scalar field E, the specific total energy, such that for every part P and time t,


d ρE dVx = s(n) · v dAx + b · v dVx dt Pt ∂Pt Pt

− g(n) dAx + f dVx . (1.46) Pt

∂Pt

We have the following fundamental result. Theorem 1.4.3. (Cauchy). Suppose that the momentum balance laws hold. Then a necessary and sufficient condition that the energy conservation law be satisfied is the existence of a spatial vector field q (called heat flux vector) such that 1. For each unit vector n,

2.

g(n, x, t) = q(x, t) · n,

(1.47)

ρE˙ = div(T v) + b · v − div q + f.

(1.48)

Proof. We only prove that assuming (1.47) the local form (1.48) holds. To see that we first replace s by T n and g by q · n in (1.46). Next we use Lemma 1.2.1 to transform the left-hand side of (1.46). We get




ρE˙ dVx = T n · v dAx + b · v dVx − q · n dAx + f dVx . Pt

∂Pt

Pt

∂Pt

Pt

(1.49) Then by using Gauss’ Theorem we deduce


ρE˙ dVx = div(T v) dVx + b · v dVx Pt Pt Pt

− div q dVx + f dVx . Pt

Pt

(1.50)

1.4. Balance of Energy. First Principle of Thermodynamics

9

Now (1.48) follows by applying the Localization Theorem.




Corollary 1.4.4. Let us assume the energy conservation law holds. Then, for each t ∈ R and any smooth scalar field z : Bt → R, we have



˙ ρEz dVx + T v · gradz dVx = s(n) · vz dAx + b · vz dVx Pt Pt Pt
∂Pt
− g(n)z dAx + f z dVx , (1.51) Pt

∂Pt

∀P part of B. Definition 1.4.5. We denote by specific internal energy the scalar field e=E−

|v|2 . 2

(1.52)

The next Proposition provides both integral and local forms of the energy conservation law in terms of the specific internal energy. Proposition 1.4.6. We have



d 1. ρe dVx = T · D dVx − g(n) dAx + f dVx , dt Pt Pt ∂Pt Pt

(1.53)

for all parts P and times t, where D is defined by (1.33). ρe˙ = T · D − div q + f.

2.

(1.54)

Proof. By subtracting equality (1.32) from (1.46) we immediately get (1.53). Now, in order to prove the second part of the Proposition, we replace g(n) by q · n in (1.53) and then use Gauss’ Theorem. We obtain d dt





Pt

ρe dVx =

Pt


T · D dVx −

Pt


div q dVx +

Pt

f dVx .

Now (1.54) follows from the Localization Theorem.

(1.55)


Remark 1.4.7. Equations (1.48) and (1.54) are local non-conservative versions of the energy conservation law. We can also write conservative forms which are obtained by adding the mass conservation equation (1.21) multiplied by E and e to (1.48) and (1.54), respectively. We get (ρE)
+ div(ρEv) = div(T v) + b · v − div q + f,

(1.56)

(ρe)
+ div(ρev) = T · D − div q + f.

(1.57)

10

Chapter 1. General Definitions, Conservation Laws

Weak Formulation. Boundary and Initial Conditions Let us obtain a weak formulation for the energy balance in terms of the internal energy. Let t be any time and y : Bt → R be any smooth scalar field. By making the product of (1.54) with y, integrating in Bt , and using a Green’s formula in the boundary integral including the heat flux, we get
Bt


ρey ˙ dVx =

Bt


T · Dy dVx +




Bt

q · grady dVx −

g(n)y dAx ∂Bt




+ Bt

f y dVx .

(1.58)

In order to state a well-posed mathematical problem, we have to prescribe an initial condition, the internal energy at initial time, and boundary conditions. We consider two types of boundary conditions. Let us divide the boundary of Bt into two parts: ∂Bte and ∂Btg . Let us suppose the internal energy is given on ∂Bte while the heat flux is prescribed on ∂Btg . That is, we impose e = e¯ on ∂Bte , g(n) = g¯ on ∂Btg .

(1.59) (1.60)

In order to include these boundary conditions in the weak formulation (1.58), we only take test functions y null on ∂Bte . Then the boundary integral transforms as follows:



g(n)y dAx =


g(n)y dAx =

∂Btg

∂Bt

∂Btg

g¯y dAx .

(1.61)

Thus the weak formulation of the energy equation is
Bt


ρey ˙ dVx =

Bt


T · Dy dVx

+ Bt


q · grady dVx −

∂Btg


g¯y dAx +

∀y null on ∂Bet , e e

= =

∂Bet ,

e¯ on e0 in B0 at t = 0.

Bt

f y dVx , (1.62) (1.63) (1.64)

For finite volume discretization we have to use the conservation forms (1.56) or (1.57). They are first integrated in a control volume R and then transformed by

1.5. Second Principle of Thermodynamics. The Clausius-Duhem Inequality using Gauss’ Theorem. We get



d ρEdVx + ρEv · n dAx = T n · v dAx + b · v dVx dt ∂R ∂R ∂R R

− g(n) dAx + f dVx , ∂R


R d ρe dVx + ρev · n dAx = T · D dVx dt R ∂R R

− g(n) dAx + f dVx , ∂R

11

(1.65)

R

(1.66) respectively.

1.5 Second Principle of Thermodynamics. The Clausius-Duhem Inequality In this section we introduce the Second Principle of Thermodynamics. It asserts that there exist a scalar field s, the specific entropy, and a strictly positive scalar field θ, the absolute temperature, such that


d q·n f dAx + dVx ρs dVx ≥ − (1.67) dt Pt θ ∂Pt Pt θ for all parts P and times t. The quantity


S(P, t) =

Pt

ρs dVx

is called entropy of the part P at time t, while the sum

q·n f dAx + dVx − θ ∂Pt Pt θ

(1.68)

(1.69)

is the entropy rate supplied to the part P at time t. Thus the inequality (1.67) states that entropy can not be conserved because the entropy growth rate of a part can be greater than the rate of supplied entropy from its environment. The following local form, called the Clausius-Duhem inequality can be easily obtained from (1.67) by using the Reynolds, the Gauss and the Localization Theorems: q f − ≥ 0. (1.70) ρs˙ + div θ θ

12

Chapter 1. General Definitions, Conservation Laws

Moreover, by multiplying (1.70) by θ and then subtracting (1.54) we obtain 1 ρθs˙ − ρe˙ + T · D − q · gradθ ≥ 0. θ

(1.71)

Finally, we write the second principle of thermodynamics in terms of the free energy. Definition 1.5.1. We denote by specific Helmholtz free energy the scalar field ψ defined by ψ = e − sθ. (1.72) Since we have ˙ ψ˙ = e˙ − sθ ˙ − sθ,

(1.73)

by using this expression in (1.71) we get another local form of the second principle of thermodynamics, namely, 1 ρsθ˙ + ρψ˙ − T · D + q · gradθ ≤ 0. θ

(1.74)

Chapter 2

Lagrangian Coordinates Sometimes it is convenient to write the conservation principles in material (also called Lagrangian) coordinates rather than in spatial (also called Eulerian) ones. This reveals some new tensor fields related with the Cauchy stress tensor.

2.1 The Piola-Kirchhoff Stress Tensors Let us introduce the First Piola-Kirchhoff stress tensor (also called Boussinesq stress tensor) which is defined, in the reference configuration, by S(p, t) := det(F (p, t))T (x, t)F −t (p, t), x = X(p, t).

(2.1)

Similarly, the Second Piola-Kirchhoff stress tensor (also called Piola-Lagrange stress tensor) is defined by P (p, t) := det(F (p, t))F −1 (p, t)T (x, t)F −t (p, t), x = X(p, t).

(2.2)

We notice that P is symmetric but S is not. We can write the stress power representing the rate of mechanical internal energy (see (1.53)) in two Lagrangian alternative forms. Proposition 2.1.1. We have


T · D dVx = S · ∇u˙ dVp = P · G˙ dVp . Pt

P

(2.3)

P

Proof. We need the following Lemmas which can be found in [5]. Lemma 2.1.2. Let A, B and C be three tensors. Then A · BC = AC t · B = B t A · C.

(2.4)

14

Chapter 2. Lagrangian Coordinates

Lemma 2.1.3. We have F˙ = Lm F.

(2.5)

Now, let us prove the Proposition. Since T is symmetric we deduce

T · D dVx = T · L dVx Pt

(2.6)

Pt

and, by making the change of variable x = X(p, t),


T · L dVx = Tm · Lm det(F ) dVp = Tm · F˙ F −1 det(F ) dVp Pt P P

−t ˙ = Tm F · F det(F ) dVp = S · ∇u˙ dVp . (2.7) P

P

Moreover, we can compute the time derivative of the Green-Lagrange strain tensor,  1 1 1 G˙ = (F t F ). = (F˙ t F + F t F˙ ) = (Lm F )t F + F t Lm F 2 2 2  1 1 t t F Lm F + F t Lm F = F t (Lm + Ltm )F = F t Dm F. = 2 2 Then we have
T · D dVx


=

Pt




˙ −1 det(F ) dVp Tm · F −t GF

−1 −t ˙ F Tm F · G det(F ) dVp = P · G˙ dVp . (2.9) P P
P

=

(2.8)

Tm · Dm det(F ) dVp =

P

2.2 The Conservation Equations in Lagrangian Coordinates By using the change of variable x = X(p, t), we obtain the equalities (see Lemma 1.2.1 and Appendix B)


d ρv dVx = ρv˙ dVx = ρ(X(p, t), t)¨ u(p, t)det(F (p, t)) dVp , (2.10) dt Pt Pt P

b dVx = b(X(p, t), t)det(F (p, t)) dVp , (2.11) Pt P

T (x, t)n(x) dA(x) = det(F (p, t))T (X(p, t), t)F −t (p, t)m(p) dAp , (2.12) ∂Pt

∂P

where n(x) (respectively m(p)) denotes the outward unit normal vector to ∂Pt (respectively to ∂P) at point x (respectively p). Then we replace these equalities in (1.27) and apply the Gauss and the Localization Theorems. We get ¨ = Div S + b∗ , ρ0 u

(2.13)

2.2. The Conservation Equations in Lagrangian Coordinates

15

where b∗ is given by b∗ (p, t) = b(X(p, t), t) det(F (p, t)).

(2.14)

Analogous computations allow us to obtain, from (1.49), the following form of the energy equation in Lagrangian coordinates: ˙ + b∗ · u˙ − Divq∗ + f∗ , ρ0 E˙ m = Div(S u)

(2.15)

q∗ (p, t) = det(F (p, t))F −1 (p, t)q(X(p, t), t), f∗ (p, t) = f (X(p, t), t) det(F (p, t)).

(2.16) (2.17)

where

Similarly, in terms of the internal energy, the energy equation can be written in Lagrangian coordinates as ρ0 e˙ m = S · ∇u˙ − Divq∗ + f∗ , by using (2.3).

(2.18)

Chapter 3

Constitutive Laws In what follows we use simultaneously material and spatial fields. As we did in the previous chapter, sometimes subscript m (respectively, s) is used to denote the material (respectively, spatial) description of a spatial (respectively, material) field. But, for the sake of simplicity in notation, they will be omitted in what follows if no ambiguity arises.

3.1 Thermodynamic Process. Material Body Firstly, we extend the concept of material body given in reference [5]. We begin with the following: Definition 3.1.1. A thermodynamic process for a body with a mass distribution ρ0 is a set of eight mappings: X : B × R → E,

(3.1)

T : T → Sym, b : T → V,

(3.2) (3.3)

e : T → R, θ : T → R+ ,

(3.4) (3.5)

q : T → V, f : T → R,

(3.6) (3.7)

s : T → R,

(3.8)

such that X is a motion, T its trajectory, T ∈ C (T ; Sym), b ∈ C (T ; V), e ∈ C 1 (T ; R), θ ∈ C 1 (T ; R), q ∈ C 1 (T ; V), f ∈ C 0 (T ; R), s ∈ C 1 (T ; R) and, furthermore, the following equations hold: 1

0

ρv˙ = div T + b,

(3.9)

ρe˙ = T · D − div q + f,

(3.10)

18

Chapter 3. Constitutive Laws

where ρ(x, t) det(F (p, t)) = ρ0 (p), with x = X(p, t).

(3.11)

Remark 3.1.2. We notice that, in order to define a thermodynamic process, it is enough to prescribe the six mappings X, T, e, θ, q and s, because the two remaining ones b and f can be deduced from (3.9) and (3.10), respectively. With the definition above we can introduce an extended concept of material body. Definition 3.1.3. A material body is a triple (B, ρ0 , C) consisting of a body B, a mass distribution ρ0 and a family C of thermodynamic processes called the constitutive class of the body.

3.2 Coleman-Noll Materials Let us denote by Lin+ the subset of second order tensors with positive determinant, i.e., (3.12) Lin+ = {G ∈ Lin : det(G) > 0}. Now we introduce a particular material body. Definition 3.2.1. A hyperelastic material with heat conduction and viscosity is a material body the constitutive class of which consists of all thermodynamic processes satisfying T (x, t) = Tˆ(F (p, t), s(x, t), p) + ˆl(F (p, t), s(x, t), p)(L(x, t)), e(x, t) = eˆ(F (p, t), s(x, t), p), ˆ (p, t), s(x, t), p), θ(x, t) = θ(F ˆ (F (p, t), s(x, t), gradθ(x, t), p), q(x, t) = q

(3.13) (3.14) (3.15) (3.16)

with x = X(p, t), for some “smooth enough” mappings Tˆ : Lin+ × R × B → Sym,

(3.17)

ˆ l : Lin+ × R × B → L(Lin, Sym),

(3.18)

eˆ : Lin × R × B → R, ˆ θ : Lin+ × R × B → R+ ,

(3.19)

ˆ : Lin × R × V × B → V, q

(3.21)

+

+

(3.20)

which are called response mappings of the body. In what follows a hyperelastic body with heat conduction and viscosity will be also called a Coleman-Noll material. In order to prove the next theorem we need the following result:

3.2. Coleman-Noll Materials

19

Lemma 3.2.2. Let ϕ be defined by ϕ(A) = det(A).

(3.22)

Then ϕ is smooth and, if A is invertible, then ∂ϕ (A) (U ) = det(A) tr(U A−1 ), ∂A

(3.23)

for each U ∈ Lin.


Proof. See [5].

Corollary 3.2.3. Let S be a mapping defined in an open set D ⊂ R and valued in the set of invertible tensors. Then ˙ −1 . (det(S))· = (det(S)) trSS

(3.24)


Proof. See [5]. We have the following fundamental result.

Theorem 3.2.4. Let us consider a Coleman-Noll material with constitutive class C. We make the following assumption: (H1) There exists a “smooth enough” function, sˆ : Lin+ × R+ × B → R, such that if s ∈ R, F ∈ Lin+ , θ ∈ R+ and p ∈ B, then s = sˆ(F, θ, p)

ˆ s, p). if and only if θ = θ(F,

(3.25)

Then all elements in C satisfy the second law of thermodynamics if and only if e ˆ s, p) = ∂ˆ (F, s, p), θ(F, ∂s ρ0 (p) ∂ˆ e Tˆ (F, s, p) = (F, s, p)F t , det(F ) ∂F ˆl(F, s, p)(L) · L ≥ 0 (dissipation inequality), ˆ (F, s, w, p) · w ≤ 0, q

(3.26) (3.27) (3.28) (3.29)

for all F ∈ Lin+ , s ∈ R, p ∈ B, L ∈ Lin and w ∈ V. Remark 3.2.5. Notice that assumption (H1) implies that Lin+ , θ ∈ R+ and p ∈ B. Indeed, from equality ˆ s, p), p) s = sˆ(F, θ(F,

∂ˆ s ∂θ (F, θ, p)

= 0 ∀ F ∈ (3.30)

and the chain rule, we deduce 1=

∂ θˆ ∂ˆ s (F, θ, p) (F, s, p), ∂θ ∂s

ˆ s, p), which implies the result. with θ = θ(F,

(3.31)

20

Chapter 3. Constitutive Laws

The following Lemmas will be used in the proof of Theorem 3.2.4. Lemma 3.2.6. Let F ∈ Lin+ and L ∈ Lin. Then there exists F˜ : R → Lin+ such that F˜ (0) = F and F˜˙ (t)F˜ (t)−1 = L. Moreover det(F˜ (t)) = det(F ) e tr(L)t > 0.

(3.32)

Proof. Let us consider the Cauchy problem: F˜˙ (t) = LF˜ (t), F˜ (0) = F.

(3.33) (3.34)

Existence and uniqueness of solution to this problem, defined for all t ∈ R, is an immediate consequence of the Picard-Lipschitz Theorem. Since F˜ is continuous and det(F˜ (0)) = det(F ) > 0, the set A = {λ ∈ R : det(F˜ (t)) > 0, 0 ≤ t ≤ λ}

(3.35)

is nonempty. Let τ = sup {A}. We are going to prove that τ = ∞. Indeed, otherwise det(F˜ (τ )) = 0 because of continuity. Moreover, det (F˜ (t)) > 0, ∀t ∈ [0, τ ). Hence there exists F˜ (t)−1 and, from (3.33), L = F˜˙ (t)F˜ (t)−1 . Furthermore, (det(F˜ ))· (t) = det(F˜ (t)) tr(F˜˙ (t)F˜ (t)−1 ) = det(F˜ (t)) tr(L) ∀t ∈ [0, τ ),

(3.36)

from which (3.32) follows for t ∈ [0, τ ). By using continuity arguments, (3.32) yields det(F˜ (τ )) = det(F ) e tr(L)τ > 0, which is a contradiction. Then τ = ∞ and so F˜ (t) ∈ Lin+ and (3.32) holds ∀t ∈ R.
Following Gurtin[5], we denote by Skw the set of skew second order tensors and by Orth+ the set of rotations. ˜ be the solution of the Cauchy Lemma 3.2.7. Let W ∈ Skw and Q ∈ Orth+ . Let Q problem ˜˙ ˜ Q(t) = W Q(t), ˜ Q(0) = Q.

(3.37) (3.38)

˜ ∈ Orth+ ∀t ∈ R. Then Q(t) ˜ Q(t) ˜ t . We have Proof. Let Z(t) = Q(t) ˙ ˜˙ Q(t) ˜ t + Q(t) ˜ Q(t) ˜˙ t = W Q(t) ˜ Q(t) ˜ t − Q(t) ˜ Q(t) ˜ tW Z(t) = Q(t) = W Z(t) − Z(t)W.

(3.39)

˜ Q(0) ˜ t = QQt = I, Z is a solution of the Cauchy problem Since Z(0) = Q(0) Z˙ = W Z − ZW,

(3.40)

Z(0) = I,

(3.41)

3.2. Coleman-Noll Materials

21

which has unique solution. But it is obvious that the identity I is also a solution. ˜ ∈ Orth. On the other hand, from Hence we deduce Z(t) = I ∀t ∈ R and then Q(t) the previous Lemma we deduce ˜ det(Q(t)) = det(Q) e tr(W )t = 1,

(3.42)


which completes the proof.

The next Lemma allows us to build up particular thermodynamic processes in the constitutive class of a Coleman-Noll material. Lemma 3.2.8. Let us consider a Coleman-Noll material with constitutive class C. Suppose (H1) holds. Then, for given F ∗ ∈ Lin+ and L∗ ∈ Lin, there exists a motion X such that F (p, 0) = F ∗ and gradv = L∗ . Let us denote by T its trajectory. Moreover, let s∗ ∈ R, p∗ ∈ B and θ ∈ C 1 (T ; R+ ) satisfying ˆ ∗ , s∗ , p∗ ), x∗ = X(p∗ , 0). θ(x∗ , 0) = θ∗ := θ(F

(3.43)

Then there exists a thermodynamic process (X, T, b, e, θ, q, f, s) ∈ C such that F (p, 0) = F ∗ , ∀p ∈ B, ∗

L(x, t) = L , ∀(x, t) ∈ T , s(x∗ , 0) = s∗ .

(3.44) (3.45) (3.46)

Proof. Let F˜ (t) be given as in Lemma 3.2.6, for F = F ∗ and L = L∗ . Let us define X by X(p, t) = o + F˜ (t)(p − o), (3.47) where o is any point in E. We have F (p, t) = F˜ (t)

(3.48)

L(x, t) = F˜˙ (t)F˜ (t)−1 = L∗ .

(3.49)

and, furthermore, Now let us consider the given θ and define s by s(x, t) = sˆ(F˜ (t), θ(x, t), p),

(3.50)

ˆ ∗ , s∗ , p∗ ), from assumption (H1) we have with x = X(p, t). Then, since θ∗ = θ(F s(x∗ , 0) = sˆ(F ∗ , θ(x∗ , 0), p∗ ) = sˆ(F ∗ , θ∗ , p∗ ) = s∗ .

(3.51)

Finally, let T (x, t) = Tˆ (F˜ (t), s(x, t), p) + ˆl(F˜ (t), s(x, t), p)(L∗ ), e(x, t) = eˆ(F˜ (t), s(x, t), p), ˜ ˆ (F (t), s(x, t), gradθ(x, t), p), q(x, t) = q

(3.52) (3.53) (3.54)

22

Chapter 3. Constitutive Laws

and define b and f by b = ρv˙ − div T, f = ρe˙ − T · D + div q,

(3.55) (3.56)

ρ0 (p) with ρ(x, t) = det( and x = X(p, t). F˜ (t)) Then the thermodynamic process (X, T, b, e, θ, q, f, s) belongs to the constitutive class of the material and satisfies the requirements (3.44)–(3.46).


Now we are ready to prove Theorem 3.2.4. Proof. We start with the following form of the second principle (see (1.71)): 1 ρθs˙ − ρe˙ + T · D − q · gradθ ≥ 0. θ

(3.57)

By using (3.14) we deduce e(x, ˙ t) =

∂ˆ e ∂ˆ e (F (p, t), s(x, t), p)s(x, ˙ t) + (F (p, t), s(x, t), p) · F˙ (p, t), ∂s ∂F

with x = X(p, t), which can be used in (3.57) to get     ∂ˆ e 1 ∂ˆ e ˙ ρ θ− · F − q · gradθ ≥ 0. s˙ + T · D − ρ ∂s ∂F θ

(3.58)

(3.59)

Since T is symmetric we can replace T ·D by T ·L. Moreover, F˙ = LF (see Lemma 2.1.3). Then (3.59) becomes     ∂ˆ e 1 ∂ˆ e t ρ θ− F · L − q · gradθ ≥ 0, (3.60) s˙ + T − ρ ∂s ∂F θ where we have used the Lemma 2.1.2. Now we choose a particular thermodynamic process. For this we apply Lemma 3.2.8 for L∗ = 0, any given F ∗ , s∗ , p∗ , and θ satisfying (3.43), gradθ ≡ 0 as well as θ
(x∗ , 0) = a, a being any real number. Then (3.60) becomes   ∂ˆ e ρ θ− s˙ ≥ 0. (3.61) ∂s But s˙ =

∂ˆ s ˙ ∂θ θ

because F˙ ≡ 0 and hence   s˙ ∂ˆ e ∂ˆ θ ≥ 0. ρ θ− ∂s ∂θ

Then we take p = p∗ and t = 0 in (3.62). We get   s ∗ ∗ ∗ ∂ˆ e ∗ ∗ ∗ ∂ˆ ∗ ∗ ∗ ˆ ρ θ(F , s , p ) − (F , s , p ) (F , θ , p )a ≥ 0, ∂s ∂θ

(3.62)

(3.63)

3.2. Coleman-Noll Materials

23

from which (3.26) follows just by taking, successively, a = 1 and a = −1, and ∂ˆ s (F ∗ , θ∗ , p∗ ) = 0 (see Remark 3.2.5). using the fact that ∂θ Now we apply Lemma 3.2.8 with the following choices: L∗ = αL, α being any real number and L being any second order tensor, any given F ∗ , s∗ , p∗ and θ satisfying (3.43) with gradθ = 0. Then (3.60) becomes   ∂ˆ e ˜ Tˆ (F˜ (t), s(x, t), p) − ρ (F (t), s(x, t), p)F˜ t (t) · (αL) ∂F +ˆl(F˜ (t), s(x, t), p)(αL) · (αL) ≥ 0 (3.64) and, by taking t = 0 and p = p∗ , we have s(x∗ , 0) = s∗ and then   e ∗ ∗ ∗ ρ0 (p∗ ) ∂ˆ ∗ ∗ ∗ ∗ t ˆ α T (F , s , p ) − (F , s , p )(F ) · L det(F ∗ ) ∂F +α2 ˆl(F ∗ , s∗ , p∗ )(L) · L ≥ 0.

(3.65)

In order for the previous inequality to hold for all α in R, it is necessary and sufficient that (3.27) and (3.28) be satisfied. Finally, we consider the thermodynamic process provided by Lemma 3.2.8 for L∗ = 0, any given F ∗ , s∗ , p∗ and θ ≥ 0 satisfying (3.43) and gradθ(x∗ , 0) = w, with w ∈ V. Then, from (3.60), we deduce −ˆ q(F˜ (t), s(x, t), gradθ(x, t), p) · gradθ(x, t) ≥ 0,

(3.66)

because θ > 0. By taking t = 0 and p = p∗ we obtain (3.29). To prove sufficiency, we notice that (3.57) is equivalent to (3.60) for ColemanNoll materials. But it is obvious that if (3.26)–(3.29) are satisfied then the latter holds.
Remark 3.2.9. In what follows we assume that (H1) and the Second Law of Thermodynamics hold. Corollary 3.2.10. For a Coleman-Noll material we have the following form of the energy equation: ρθs˙ = l(L) · D − div q + f.

(3.67)

Proof. Firstly, e(x, ˙ t)

= = =

∂ˆ e ∂ˆ e (F (p, t), s(x, t), p)s(x, ˙ t) + (F (p, t), s(x, t), p) · F˙ (p, t) ∂s ∂F 1 ˆ θ(x, t)s(x, ˙ t) + T (F (p, t), s(x, t), p)F −t (p, t) · F˙ (p, t) ρ(x, t) 1 ˆ T (F (p, t), s(x, t), p) · L(x, t). (3.68) θ(x, t)s(x, ˙ t) + ρ(x, t)

By replacing this expression for e˙ in (3.10) we get (3.67).




24

Chapter 3. Constitutive Laws

Remark 3.2.11. We have q  q q = gradθ · + θ div . div q = div θ θ θ θ

(3.69)

Hence (3.67) can be written as ρs˙ =

q f 1 1 l(L) · D − 2 gradθ · q − div + , θ θ θ θ

(3.70)

which yields ρs˙ + div

q θ

−

f 1 = θ θ

1 l(L) · D − gradθ · q . θ

(3.71)

Thus, the scalar field Φ := l(L) · D −

1 gradθ · q, θ

(3.72)

which is called dissipation rate, is responsible for production of non-reversible entropy. From the Clausius-Duhem inequality we easily deduce that Φ ≥ 0 in T . Remark 3.2.12. Property (3.29) means that heat goes from hot to cold regions in the following sense: let (x, t) ∈ T and n be a unit normal vector. Let S be any surface containing x and having n as a normal vector at this point. Then we recall that the heat per unit surface and time flowing across S from the negative side of gradθ(x,t) . Then we S to its positive side is given by q(x, t) · n. Let us take n = | gradθ(x,t)| have ˆ (F (p, t), s(x, t), gradθ(x, t), p) · q(x, t) · n = q

gradθ(x, t) ≤ 0, | gradθ(x, t)|

(3.73)

by using (3.29) for w = gradθ(x, t). Now we recall that gradθ(x, t) is the direction of highest growth of θ(·, t). Definition 3.2.13. A thermodynamic process is called isentropic if s˙ ≡ 0. Remark 3.2.14. This means that material points conserve entropy during motion, i.e., that there exists a function s0 : B → R such that s(X(p, t), t) = s0 (p).

(3.74)

In particular, if s0 is constant then s(x, t) = s0 ∀(x, t) ∈ T . Some authors say that an isentropic process is a thermodynamic process such that s(x, t) is constant in T. The next result is a straightforward consequence of the above definition and (3.67).

3.2. Coleman-Noll Materials

25

Proposition 3.2.15. A thermodynamic process of a Coleman-Noll material is isentropic if and only if the following equation holds: ˆl(F (p, t), s0 (p), p)(L) · D − div q + f = 0.

(3.75)

Moreover, the motion equation becomes  ρv˙ − div Tˆ (F (p, t), s0 (p), p) − div ˆl(F (p, t), s0 (p), p)(L) = b.

(3.76)

Remark 3.2.16. We notice that for an isentropic process the mass and the motion equations can be solved independently of the energy equation. Isentropic thermodynamic processes are of great importance in continuum mechanics. An important field where they arise is in standard (i.e., non-dissipative) linear acoustics. This subject will be studied in Chapter 11. Definition 3.2.17. A thermodynamic process is called adiabatic if q ≡ 0 and f ≡ 0. Definition 3.2.18. A thermodynamic process is called Eulerian if l ≡ 0. Proposition 3.2.19. Any Eulerian adiabatic thermodynamic process is isentropic. Proof. It is an immediate consequence of (3.67) and the above Definitions.




Chapter 4

The Principle of Material Frame-Indifference In this chapter we state the principle of material frame-indifference and prove that it implies that the response functions of Coleman-Noll material bodies have particular forms.

4.1 Change in the Observer. The Indifference Principle We start by recalling some definitions. Definition 4.1.1. Two thermodynamic processes (X, T , b, e, θ, q, f , s) and (X∗ , T ∗ , b∗ , e∗ , θ∗ , q∗ , f ∗ , s∗ ) are related by a change in the observer if there exist two C 3 functions q : R → E and Q : R → Orth+ such that X∗ (p, t) = q(t) + Q(t)(X(p, t) − o), ∗

∗

t

(4.1)

T (x , t) = Q(t)T (x, t)Q(t) , e∗ (x∗ , t) = e(x, t),

(4.2) (4.3)

θ∗ (x∗ , t) = θ(x, t), q∗ (x∗ , t) = Q(t)q(x, t),

(4.4) (4.5)

s∗ (x∗ , t) = s(x, t),

(4.6)

for some o ∈ E, with x = X(p, t) and x∗ = X∗ (p, t). Definition 4.1.2. We say that the response of a material body is independent of the observer if its constitutive class, C, has the following property: If (X, T, b, e, θ, q, f, s) ∈ C and (X∗ , T ∗ , b∗ , e∗ , θ∗ , q∗ , f ∗ , s∗ ) is another thermodynamic process which is related to the previous one by a change in the observer, then (X∗ , T ∗ , b∗ , e∗ , θ∗ , q∗ , f ∗ , s∗ ) ∈ C.

28

Chapter 4. The Principle of Material Frame-Indifference

Definition 4.1.3. We say that a material body satisfies the material frame-indifference principle if its response is independent of the observer, in the sense of the previous definition.

4.2 Consequences for Coleman-Noll Materials Firstly we can prove the following: Proposition 4.2.1. Let us consider a Coleman-Noll material. If it satisfies the material frame-indifference principle, then the following conditions hold: eˆ(F, s, p) = eˆ(QF, s, p), ˆ s, p) = θ(QF, ˆ θ(F, s, p),

(4.7) (4.8) (4.9)

QTˆ(F, s, p)Q = Tˆ (QF, s, p), ˆ Q ˆl(F, s, p)(L)Qt = l(QF, s, p)(QLQt + W ), t

(4.10) (4.11)

ˆ (QF, s, Qw, p), Qˆ q(F, s, w, p) = q

for all F ∈ Lin+ , L ∈ Lin, s ∈ R, p ∈ B, Q ∈ Orth+ and W ∈ Skw. The converse is also true. Proof. Let (X, T, b, e, θ, q, f, s) ∈ C. For arbitrarily given C 3 functions q : R → E and Q : R → Orth+ , we introduce another thermodynamic process as follows, X∗ (p, t) = q(t) + Q(t)(X(p, t) − o), ∗

∗

(4.12)

t

T (x , t) = Q(t)T (x, t)Q(t) , e∗ (x∗ , t) = e(x, t),

(4.13) (4.14)

θ∗ (x∗ , t) = θ(x, t), q∗ (x∗ , t) = Q(t)q(x, t),

(4.15) (4.16)

s∗ (x∗ , t) = s(x, t),

(4.17)

with b∗ and f ∗ defined so as to satisfy the motion and the energy equations. According to Definition 4.1.1, this process is related to (X, T, b, e, θ, q, f, s) ∈ C by a change in the observer. Hence it must belong to C, which means T ∗ (x∗ , t) = Tˆ (F ∗ (p, t), s∗ (x∗ , t), p) + ˆl(F ∗ (p, t), s∗ (x∗ , t), p)(L∗ (x∗ , t)),

(4.18)

e∗ (x∗ , t) = eˆ(F ∗ (p, t), s∗ (x∗ , t), p), ˆ ∗ (p, t), s∗ (x∗ , t), p), θ∗ (x∗ , t) = θ(F

(4.19)

∗

∗

∗

∗

∗

∗

∗

∗

(4.20)

ˆ (F (p, t), s (x , t), grad θ (x , t), p), (4.21) q (x , t) = q

4.2. Consequences for Coleman-Noll Materials

29

and therefore Tˆ (Q(t)F (p, t), s(x, t), p)

  t ˙ +ˆl(Q(t)F (p, t), s(x, t), p) Q(t)L(x, t)Q(t)t + Q(t)Q(t) = Q(t)Tˆ (F (p, t), s(x, t), p)Q(t)t + Q(t)ˆl(F (p, t), s(x, t), p)(L(x, t))Q(t)t , (4.22) eˆ(Q(t)F (p, t), s(x, t), p) = eˆ(F (p, t), s(x, t), p), (4.23) ˆ ˆ (p, t), s(x, t), p), θ(Q(t)F (p, t), s(x, t), p) = θ(F (4.24) ˆ (Q(t)F (p, t), s(x, t), Q(t) grad θ(x, t), p) q = Q(t)ˆ q(F (p, t), s(x, t), grad θ(x, t), p),

(4.25)

where we have used that (see [5]) F ∗ (p∗ , t) = Q(t)F (p, t), t ˙ , L∗ (x∗ , t) = Q(t)L(x, t)Q(t)t + Q(t)Q(t)

(4.26) (4.27)

and the fact that grad∗ θ∗ = Q gradθ. Now we use Lemma 3.2.8 to build a particular thermodynamic process in C. Let us take L∗ = 0 and any F ∗ , s∗ , p∗ , and θ satisfying (3.43) with gradθ(x∗ , 0) = w ∈ V. Moreover, let Q(t) = Q with Q ∈ Orth+ . Then, from (4.22)–(4.25) we have Tˆ (QF ∗ , s(x, t), p) = QTˆ (F ∗ , s(x, t), p)Qt , eˆ(QF ∗ , s(x, t), p) = eˆ(F ∗ , s(x, t), p), ∗ ˆ ˆ ∗ , s(x, t), p), θ(QF , s(x, t), p) = θ(F ∗

∗

ˆ (QF , s(x, t), Q gradθ(x, t), p) = Qˆ q q(F , s(x, t), gradθ(x, t), p),

(4.28) (4.29) (4.30) (4.31)

and taking t = 0 and p = p∗ , Tˆ (QF ∗ , s∗ , p∗ ) = QTˆ (F ∗ , s∗ , p∗ )Qt , ∗

∗

∗

∗

∗

∗

(4.32)

eˆ(QF , s , p ) = eˆ(F , s , p ), ∗ ∗ ∗ ˆ ∗ , s∗ , p∗ ), ˆ , s , p ) = θ(F θ(QF

(4.33)

ˆ (QF ∗ , s∗ , Qw, p∗ ) = Qˆ q q(F ∗ , s∗ , w, p∗ ).

(4.35)

(4.34)

Now, by using (4.32) in (4.22) we obtain   t ˆl(Q(t)F (p, t), s(x, t), p) Q(t)L(x, t)Q(t)t + Q(t)Q(t) ˙ = Q(t)ˆl(F (p, t), s(x, t), p)(L(x, t))Q(t)t .

(4.36)

Let us consider the thermodynamic process given by Lemma 3.2.8 for the following choices: any F ∗ ∈ Lin+ , L∗ ∈ Lin, s∗ ∈ R, p∗ ∈ B and θ satisfying (3.43). This

30

Chapter 4. The Principle of Material Frame-Indifference

process satisfies F (p, 0) = F ∗ , L(x, t) = L∗ ,

(4.37) (4.38)

s(x∗ , 0) = s∗ , θ(x∗ , 0) = θ∗ ,

(4.39) (4.40)

with x∗ = X(p∗ , 0). ˜ Moreover, let Q(t) be defined by Lemma 3.2.7 for any given Q ∈ Orth+ and W ∈ Skw. Then (4.36) yields   ∗ ˜ ˆl(Q(t)F ˜ ˜ ˜˙ Q(t) ˜ t Q(t)t + Q(t) (p, t), s(x, t), p) Q(t)L ˜ t. ˜ ˆl(F (p, t), s(x, t), p)(L∗ )Q(t) = Q(t) Finally, by taking t = 0 and p = p∗ we get (4.10).

(4.41)


As a consequence of this important Proposition we obtain very simple expressions for l. Corollary 4.2.2. We have ˆl(F, s, p)(L) = ˆl(F, s, p)(D), where D denotes the symmetric part of L, i.e., D = Proof. It is enough to take Q = I and W = − (L−L 2

t

)

(4.42)

(L+Lt ) . 2

in (4.10).




The next Corollary is also a straightforward consequence of (4.10). Corollary 4.2.3. The following equality holds for any Coleman-Noll material satisfying the material-frame indifference principle: ˆ l(QF, s, p)(QDQt ) = Qˆl(F, s, p)(D)Qt ∀Q ∈ Orth+ .

(4.43)

Proof. It is enough to use the previous Corollary and to take into account in (4.10)
that the symmetric part of QLQt + W is QDQt . Proposition 4.2.4. For any Coleman-Noll material satisfying the material-frame indifference principle, if furthermore ˆl(QF, s, p) = ˆl(F, s, p) ∀Q ∈ Orth+ ,

(4.44)

then there exist two functions ηˆ : Lin+ × R × B → R, ξˆ : Lin+ × R × B → R,

(4.45) (4.46)

4.2. Consequences for Coleman-Noll Materials

31

such that ˆl(F, s, p)(D) = 2ˆ ˆ s, p) tr(D)I. η(F, s, p)D + ξ(F,

(4.47)

ˆ s, p) = ξ(QF, ˆ Moreover, ηˆ(F, s, p) = ηˆ(QF, s, p) and ξ(F, s, p), for all Q ∈ Orth+ . Proof. It follows immediately from Corollary 4.2.3 and the representation theorem for isotropic linear tensor functions (see Appendix A). This result would be applied, for any fixed F, s, p, to the mapping G = ˆl(F, s, p).
Proposition 4.2.5. Suppose that ˆl(F, s, p)(D) = 2ˆ ˆ s, p) tr(D)I. η(F, s, p)D + ξ(F,

(4.48)

Then the dissipation inequality holds if and only if ηˆ ≥ 0,

2ˆ η + 3ξˆ ≥ 0.

(4.49)

Proof. We have tr(D) = I · D and then | tr(D)|2 ≤ |I|2 |D|2 = 3|D|2 .

(4.50)

ˆ tr(D))2 . l(L) · L = l(D) · D = 2ˆ η |D|2 + ξ(

(4.51)

Moreover

Let us assume that (4.49) holds. Then l(L) · L = l(D) · D ≥

η + 3ξˆ 2ˆ η ˆ tr(D))2 = 2ˆ ( tr(D))2 + ξ( ( tr(D))2 ≥ 0. 3 3 (4.52)

Conversely, let D = 0 with tr(D) = 0. We have 0 ≤ l(D) · D = 2ˆ η |D|2

(4.53)

and then ηˆ ≥ 0. Now we take D = I. We obtain ˆ tr(I))2 = (2ˆ ˆ 0 ≤ l(I) · I = 2ˆ η |I|2 + ξ( η + 3ξ)3, from which (4.49) follows.

(4.54)


Chapter 5

Replacing Entropy with Temperature The equations we have deduced from the conservation and the constitutive laws involve unknowns ρ, X, s and θ. While θ can be eliminated in terms of X and ˆ s, p), in most cases it is more interesting to s through the response function θ(F, retain this field instead of entropy. In this chapter we use the Helmholtz free energy defined in (1.72) and assumption (H1) to eliminate entropy in the motion and the energy equations.

5.1 The Conservation Equations in Terms of Temperature Firstly we have the following: Proposition 5.1.1. Let us write the free energy ψ as a function of F and θ, namely, ˆ θ, p). Then ψ = ψ(F, ∂ ψˆ (F, θ, p) = −s, ∂θ ∂ ψˆ ∂ˆ e det(F ) ˆ (F, θ, p) = (F, s, p) = T (F, s, p)F −t , ∂F ∂F ρ0 (p)

(5.1) (5.2)

with s = sˆ(F, θ, p). Proof. We apply the chain rule to the mapping ˆ θ(F, ˆ s, p), p) + θ(F, ˆ s, p) s eˆ(F, s, p) = ψ(F,

(5.3)

34

Chapter 5. Replacing Entropy with Temperature

and use (3.26) to deduce e ∂ ψˆ ∂ θˆ ∂ θˆ ˆ s, p) = ∂ˆ θ(F, (F, s, p) = (F, θ, p) (F, s, p) + (F, s, p)s ∂s ∂θ ∂s ∂s ˆ s, p), +θ(F, from which it follows that 

∂ θˆ ∂ ψˆ (F, θ, p) + s (F, s, p) = 0 ∂θ ∂s

(5.4)

(5.5)

and then (5.1). Moreover, by taking the derivative with respect to F in (5.3) and using (5.1) we obtain ∂ ψˆ ∂ ψˆ ∂ θˆ ∂ θˆ ∂ˆ e (F, s, p) = (F, θ, p) + (F, θ, p) (F, s, p) + (F, s, p) s ∂F ∂F ∂θ ∂F ∂F ∂ ψˆ (F, θ, p). = ∂F

(5.6)


From (5.2) we obtain the following: Corollary 5.1.2. We have ρ0 (p) ∂ ψˆ Tˆ (F, θ, p) = (F, θ, p)F t , det(F ) ∂F

(5.7)

where x = X(p, t). If we know the free energy response function ψˆ or, more precisely, its partial derivatives with respect to F and θ, then we can state the equations to describe any thermodynamic process of a Coleman-Noll material. Indeed, according to (5.7) and using equation (3.67), we can write the mass, momentum and energy conservation equations as

ρv˙ = div ρ  −ρθ



ρ˙ + ρ div v = 0,

∂ ψˆ (F, θ, p)F t + div(l(D)) + b, ∂F ·

∂ ψˆ (F, θ, p) ∂θ

= l(D) · D − div q + f.

(5.8) (5.9)

(5.10)

Definition 5.1.3. The specific heat at constant deformation is the scalar field defined by cF (x, t) = cˆF (F (p, t), θ(x, t), p),

(5.11)

5.1. The Conservation Equations in Terms of Temperature

35

with cˆF (F, θ, p) :=

∂ˆ e (F, θ, p). ∂θ

(5.12)

∂ˆ s (F, θ, p). ∂θ

(5.13)

Proposition 5.1.4. We have cˆF (F, θ, p) = θ Proof. Indeed, ∂ˆ e ∂ˆ s ∂ˆ s ∂ˆ e (F, θ, p) = (F, sˆ(F, θ, p), p) (F, θ, p) = θ (F, θ, p). ∂θ ∂s ∂θ ∂θ

(5.14)


Now we transform equation (5.10) by using the specific heat. Assuming enough regularity we have ·  2ˆ ∂ 2 ψˆ ∂ ψˆ ˙ + ∂ ψ (F, θ, p) · F˙ . (F, θ, p) = (5.15) (F, θ, p) θ ∂θ ∂θ2 ∂F ∂θ Moreover, from (5.1) we obtain ∂ 2 ψˆ ∂ˆ s (F, θ, p) = − (F, θ, p), 2 ∂θ ∂θ

(5.16)

∂ 2 ψˆ ∂ 2 ψˆ det(F ) ∂ Tˆ (F, θ, p) = (F, θ, p) = (F, θ, p)F −t , ∂F ∂θ ∂θ∂F ρ0 (p) ∂θ

(5.17)

and, on the other hand,

where we have used (5.2). Then, by using (5.13) we obtain  · ∂ ψˆ 1 det(F ) ∂ Tˆ (F, θ, p) (F, θ, p)F −t (p, t) · F˙ = − cF θ˙ + ∂θ θ ρ0 (p) ∂θ 1 det(F ) ∂ Tˆ = − cF θ˙ + (F, θ, p) · L. θ ρ0 (p) ∂θ

(5.18)

By replacing this equality in the energy equation (5.10), we deduce ∂ Tˆ (F, θ, p) · L + l(D) · D − div q + f, ∂θ

(5.19)

∂ Tˆ ρcF θ˙ = θ (F, θ, p) · D + l(D) · D − div q + f, ∂θ

(5.20)

ρcF θ˙ = θ or, equivalently,

because Tˆ is Sym-valued and then

∂ Tˆ ∂θ (F, θ, p)

∈ Sym.

Chapter 6

Isotropy Isotropy is a property which leads to some interesting simplifications in the constitutive laws. It is satisfied by many Coleman-Noll materials, in particular, by the fluids to be considered in Chapter 10.

6.1 The Extended Symmetry Group Definition 6.1.1. Let us consider a Coleman-Noll material. A symmetry transformation at p ∈ B is a tensor H ∈ Lin+ such that Tˆ (F, s, p) = Tˆ(F H, s, p),

(6.1)

eˆ(F, s, p) = eˆ(F H, s, p), ˆ s, p) = θ(F ˆ H, s, p), θ(F,

(6.2) (6.3)

ˆ l(F, s, p) = ˆl(F H, s, p), ˆ (F, s, w, p) = q ˆ (F H, s, w, p), q

(6.4) (6.5)

∀F ∈ Lin+ , s ∈ R, w ∈ V. Roughly speaking, a symmetry transformation is an orientation-preserving change in the reference configuration which leaves the response of the material unaltered. We denote by Hp the set of all symmetry transformations at p. It is very easy to see that Hp is a subgroup of Lin+ called the extended symmetry group at p. Remark 6.1.2. We notice that, if the material satisfies the Second Principle, then (6.2) implies (6.3). It also implies (6.1) if det(H) = 1. We are also interested in replacing s by θ as an independent variable. For this purpose we need the next Lemmas.

38

Chapter 6. Isotropy

Lemma 6.1.3. Let us assume the material indifference principle holds. Then QTˆ (F, θ, p)Qt = Tˆ(QF, θ, p), ˆ (QF, θ, Qw, p), Qˆ q(F, θ, w, p) = q

(6.6) (6.7)

Tˆ(QF, θ, p) = Tˆ(QF, sˆ(QF, θ, p), p) = QTˆ(F, sˆ(QF, θ, p), p)Qt = QTˆ (F, sˆ(F, θ, p), p)Qt = QTˆ (F, θ, p)Qt ,

(6.8)

for all Q ∈ Orth+ . Proof. We have

where we have used (4.9) and (4.8) and assumption (H1). Similarly, from (4.11), (4.8) and assumption (H1), we have ˆ (QF, θ, Qw, p) q

= =

ˆ (QF, sˆ(QF, θ, p), Qw, p) = Qˆ q q(F, sˆ(QF, θ, p), w, p) Qˆ q(F, sˆ(F, θ, p), w, p) = Qˆ q(F, θ, w, p), (6.9)


which completes the proof. Lemma 6.1.4. The following equalities hold for all H ∈ Hp : Tˆ (F, θ, p) = Tˆ(F H, θ, p), ˆ (F, θ, w, p) = q ˆ (F H, θ, w, p). q

(6.10) (6.11)

Tˆ (F H, θ, p) = Tˆ(F H, sˆ(F H, θ, p), p) = Tˆ (F, sˆ(F H, θ, p), p) = Tˆ (F, sˆ(F, θ, p), p) = Tˆ (F, θ, p),

(6.12)

Proof. We have

where we have used (6.1) and (6.3), and assumption (H1). Moreover, ˆ (F, sˆ(F H, θ, p), w, p) ˆ (F H, θ, w, p) = q ˆ (F H, sˆ(F H, θ, p), w, p) = q q ˆ (F, sˆ(F, θ, p), w, p) = q ˆ (F, θ, w, p). =q

(6.13)


As a consequence of these two last Lemmas the results given below concerning the response functions Tˆ and qˆ are true for y replaced by either specific entropy or temperature. Firstly we are able to prove the following: ˆ (F, y, w, p) are invariant under Proposition 6.1.5. The Functions Tˆ (F, y, p) and q Hp ∩ Orth+ , i.e., Tˆ (QF Qt , y, p) = QTˆ (F, y, p)Qt , t

ˆ (QF Q , y, Qw, p) = Qˆ q q(F, y, w, p), ∀Q ∈ Hp ∩ Orth+ .

(6.14) (6.15)

6.2. Isotropic Bodies

39

Proof. By using successively Lemma 6.1.3 and Lemma 6.1.4 we deduce QTˆ(F, y, p)Qt = Tˆ(QF, y, p) = Tˆ (QF Qt , y, p),

(6.16)

because if Q ∈ Hp ∩ Orth+ , then Qt = Q−1 ∈ Hp ∩ Orth+ . Similarly, ˆ (QF, y, Qw, p) = q ˆ (QF Qt , y, Qw, p). Qˆ q(F, y, w, p) = q

(6.17)


6.2 Isotropic Bodies Definition 6.2.1. A material body is said to be isotropic at p if its symmetry group at this point contains all rotations, i.e., Orth+ ⊂ Hp .

(6.18)

Corollary 6.2.2. Let us assume that the material is isotropic at p. Then, for each (y, p), the mappings Tˆ (·, y, p) and q(·, y, ·, p) are isotropic, that is, Tˆ (QF Qt , y, p) = QTˆ (F, y, p)Qt ,

(6.19)

ˆ (QF Qt , y, Qw, p) = Qˆ q q(F, y, w, p),

(6.20)

∀Q ∈ Orth+ . Now we recall the Theorem of Representation of isotropic tensor functions (see for instance [5]). Theorem 6.2.3. A mapping G : A ⊂ Sym → Sym

(6.21)

is isotropic if and only if there exist three scalar functions, ϕi : I(A) → R

i = 0, 1, 2,

(6.22)

such that G(A) = ϕ0 (IA )I + ϕ1 (IA )A + ϕ2 (IA )A2

(6.23)

for every A ∈ A, where IA denotes the set of the three principal invariants of tensor A. We notice that this Theorem cannot be directly applied to Tˆ(F, y, p) because, in general, F is not symmetric. However, with this result and Corollary 6.2.2 we easily prove the following:

40

Chapter 6. Isotropy

Proposition 6.2.4. Let us assume the material at p is isotropic. Then the response function Tˆ (F, y, p) can be written in the form Tˆ (F, y, p) = β0 (IB , y, p)I + β1 (IB , y, p)B + β2 (IB , y, p)B 2 ,

(6.24)

where we recall that B := F F t is the left Cauchy-Green strain tensor. Proof. Let us consider the polar decomposition of F , namely, F = RU = V R, R ∈ Orth+ , U, V ∈ P sym.

(6.25)

We recall that V = B 1/2 = RU Rt . Since the material at p is isotropic we have Tˆ(F, y, p) = Tˆ (V R, y, p) = Tˆ (V, y, p) = Tˆ(B 1/2 , y, p).

(6.26)

Let us introduce the mapping T˜(B, y, p) := Tˆ(B 1/2 , y, p).

(6.27)

We are going to prove that T˜ (B, y, p) is isotropic. Indeed, for every Q ∈ Orth+ , T˜ (QBQt , y, p) = T˜ (QB 1/2 Qt QB 1/2 Qt , y, p) = Tˆ (QB 1/2 Qt , y, p) = QTˆ (B 1/2 , y, p)Qt = QT˜ (B, y, p)Qt .

(6.28)

Then, for y ∈ R we can apply Theorem 6.2.3 to the tensor mapping G(B) := T˜ (B, y, p),

(6.29)


which yields the result.

Since the eigenvalues of B and C are the same, their corresponding principal invariants are also the same and then we can easily prove the following: Corollary 6.2.5. Let us assume the material is isotropic at p. Then the response function Pˆ (F, y, p) for the Second Piola-Kirchhoff stress tensor can be written in the form Pˆ (F, y, p) = P˜ (C, y, p) = γ0 (IC , y, p)I + γ1 (IC , y, p)C + γ2 (IC , y, p)C 2 .

(6.30)

Proof. Pˆ (F, y, p) = det(F )F −1 Tˆ(F, y, p)F −t = det(F )F

−1

(6.31) 2

[β0 (IB , y, p)I + β1 (IB , y, p)B + β2 (IB , y, p)B ]F

−t

(6.32) 1/2

= i3 (C)

[β0 (IC , y, p)C

−1

+ β1 (IC , y, p)I + β2 (IC , y, p)C].

(6.33)

The result follows by using the Cayley-Hamilton Theorem, allowing us to write
C −1 in terms of I, C and C 2 .

6.2. Isotropic Bodies

41

Definition 6.2.6. A Coleman-Noll material is called a Saint Venant-Kirchhoff material if its response function for the second Piola-Kirchhoff stress is of the form Pˆ (F, y, p) = λ tr(G)I + 2µG

(6.34)

where G = 12 (F t F − I) is the Green-Saint Venant strain tensor and λ and µ are two constants. For further results in nonlinear elasticity we refer to [2]. ˆ in the case of isotropy we recall another In order to give a simpler form of q Theorem of Representation for vector valued functions (see [8]) Theorem 6.2.7. A function g : A × C ⊂ Sym × V → V

(6.35)

is isotropic if and only if there exist three scalar functions, φi : A × C ⊂ Sym × V → R

i = 0, 1, 2

(6.36)

of the form φi (A, w) = ϕi (IA , w · w, w · Aw, w · A2 w), i = 0, 1, 2,

(6.37)

g(A, w) = φ0 (A, w)w + φ1 (A, w)Aw + φ2 (A, w)A2 w

(6.38)

such that

for every A ∈ A, w ∈ C. ˆ (F, y, w, p) but we have Again, this theorem cannot be directly applied to q a similar result to Proposition 6.2.4: Proposition 6.2.8. Let us assume that the material at p is isotropic. Then the ˆ (F, y, w, p) can be written as response function q ˆ (F, y, w, p) = α0 (B, w, y, p)w + α1 (B, w, y, p)Bw + α2 (B, w, y, p)B 2 w, q (6.39) where αi , i = 0, 1, 2 are functions of the following form: αi (B, w, y, p) = γi (IB , w · w, w · Bw, w · B 2 w, y, p).

(6.40)

Proof. Since the material at p is isotropic, ˆ (F, y, w, p) = q ˆ (V R, y, w, p) = q ˆ (V, y, w, p) q ˆ (B 1/2 , y, w, p). =q

(6.41)

42

Chapter 6. Isotropy

Let us introduce the mapping ˜ (B, y, w, p) := q ˆ (B 1/2 , y, w, p). q

(6.42)

˜ (·, y, w, p) is isotropic. Indeed, for every Q ∈ Orth+ , Now we prove that q ˜ (QBQt , y, Qw, p) = q ˆ (QB 1/2 Qt , y, Qw, p) = =q

˜ (QB 1/2 Qt QB 1/2 Qt , y, Qw, p) q Qˆ q(B 1/2 , y, w, p) = Q˜ q(B, y, w, p).

(6.43)

Then, for every y ∈ R we can apply Theorem 6.2.7 to the vector mapping ˜ (B, y, w, p), g(B, w) := q

(6.44)


which yields the result.

ˆ does Corollary 6.2.9. Let us assume that the material is isotropic at p and that q not depend on F . Then there exists a mapping kˆ : R × R+ × B → R such that ˆ |w|2 , p)w. ˆ (F, y, w, p) = −k(y, q

(6.45)

Proof. It is enough to take F = I in (6.39). Function kˆ is given by   ˆ |w|2 , p) = − γ0 (|w|2 , y, p) + γ1 (|w|2 , y, p) + γ2 (|w|2 , y, p) . k(y,

(6.46)


Remark 6.2.10. We notice that, under the assumption of the above Corollary, the heat flux q is given by ˆ | gradθ|2 , p) gradθ, q = −k(y,

(6.47)

which is a generalized version of the Fourier law. Moreover, from (3.29) we deduce ˆ | gradθ|2 , p)| gradθ|2 ≤ 0 q · gradθ = −k(y,

(6.48)

ˆ | gradθ|2 , p) ≥ 0, k(y,

(6.49)

and then if gradθ = 0. ˆ The scalar field k(x, t) = k(y(x, t), | gradθ(x, t)|2 , p) is called thermal conductivity.

Chapter 7

Equations in Lagrangian Coordinates In previous chapters we have introduced some partial differential equations which should be solved in order to determine the thermodynamic process followed by a material body. The difficulty of this system is that it takes place in the deformed configuration along time, i.e., in the trajectory T which, in many cases, is not known in advance. This is generally true for solids. For fluids, since they usually move in domains bounded by solids which can be considered as rigid, the set T is a data of the problem. Of course, we must exclude free boundary flows from this situation (see Appendix D).

From Eulerian to Lagrangian Coordinates In this section we show how one can change from spatial (Eulerian) to material (Lagrangian) coordinates, in order to obtain a new system of partial differential equations which holds in the (known) reference configuration B. We can introduce a response function for the first Piola-Kirchhoff stress tensor (see (2.1)) from the response function for the Cauchy stress tensor, namely ˆ S(F, H, θ, p) := Sˆelas (F, θ, p) + Sˆvisc (F, H, θ, p),

(7.1)

Sˆelas (F, θ, p) := det(F )Tˆ (F, θ, p)F −t

(7.2)

Sˆvisc (F, H, θ, p) := det(F ) ˆl(F, θ, p)(HF −1 )F −t .

(7.3)

with and We notice that, in order to compute the viscous part of S, we have to replace H with F˙ in the above formula. Then HF −1 = F˙ F −1 = L (see Lemma 2.1.3).

44

Chapter 7. Equations in Lagrangian Coordinates

Then, by performing the change of variable x = X(p, t) in (1.30) and (5.20), the following equations, stated in the reference configuration, can be obtained: Theorem 7.1. Any thermodynamic process of a Coleman-Noll material satisfies ˆ F˙ , θ, p) + b∗ , ¨ = Div S(F, ρ0 u ∂ Sˆelas (F, θ, p) · ∇u˙ + Sˆvisc (F, F˙ , θ, p) · ∇u˙ ρ0 cF θ˙ = θ ∂θ ˆ ∗ (F, θ, ∇θ, p) + f∗ , − Div q

(7.4)

(7.5)

ˆ ∗ and f∗ are given by where b∗ , q b∗ (p, t) = b(X(p, t), t) det(F (p, t)), ˆ (F, θ, F −t w, p), ˆ ∗ (F, θ, w, p) = det(F )F −1 q q f∗ (p, t) = f (X(p, t), t) det(F (p, t)).

(7.6) (7.7) (7.8)

The proof of this theorem relies upon the following well-known results on change of variable in integrals (see Appendix B): Lemma 7.2. Let w : T → V be a continuous vector field and R : T → Lin a continuous tensor field. For any part P of B we have

w(x, t) dVx = w(X(p, t), t) det(F (p, t)) dVp , (7.9) Pt

P







det(F (p, t))F −1 (p, t)w(X(p, t), t) · m(p) dAp ,

w(x, t) · n(x) dA(x) = ∂Pt

∂P

(7.10)





det(F (p, t))R(X(p, t), t)F −t (p, t)m(p) dAp , (7.11)

R(x, t)n(x) dA(x) = ∂Pt

∂P

where n(x) (respectively m(p)) denotes the outward unit normal vector to ∂Pt (respectively to ∂P) at point x (respectively p). Proof of Theorem 7.1. Firstly we recall that equation (7.4) has been already obtained in (2.13) by making the change of variable x = X(p, t) in (1.27). In order to prove (7.5) we begin with the energy equation written in the form (5.20). By integrating in Pt and then using the Gauss Theorem we get
Pt


ρcF θ˙ dVx =

θ Pt


∂ Tˆ (F, θ, p) · D dVx + l(L) · D dVx ∂θ Pt

− q · n dA(x) + f dVx . ∂Pt

Pt

(7.12)

45 Now we perform the change of variable x = X(p, t). We get

ρcF θ˙ dVx = ρ0 cF m θ˙m dVp , Pt

(7.13)

P


∂ Tˆ ∂ Tˆ (F, θ, p) · D dVx = (F, θ, p) · L dVx θ θ Pt ∂θ Pt ∂θ    

ˆ ∂ Tˆ ∂ T = θ · (F˙ F −1 ) det(F ) dVp = det(F )θ F −t · F˙ dVp ∂θ ∂θ P P m m
∂ Sˆelas · ∇u˙ dVp , = θ (7.14) ∂θ
P

l(D) · D dVx = l(L) · L dVx = l(F˙ F −1 ) · (F˙ F −1 ) det(F ) dVp Pt Pt P
˙ −1 )F −t · ∇u˙ dVp , = det(F )l(∇uF (7.15) P


q · n dA(x) = det(F )F −1 qm · m dAp = Div(det(F )F −1 qm ) dVp ,


∂Pt

∂P


Pt

P

(7.16)


f dVx =

P

fm det(F ) dVp .

(7.17)

From these equalities and the definitions (7.1) and (7.6)–(7.8) we easily get (7.5), by using the Gauss and the Localization Theorems.
The following important result is a consequence of (5.7). Corollary 7.3. We have ∂ ψˆ Sˆelas (F, θ, p) = ρ0 (p) (F, θ, p). ∂F

(7.18)

Remark 7.4. From (7.18) we see that, if l = 0, then the response function Sˆ is the partial derivative with respect to F of the Helmholtz free energy density function ˆ θ, p). In this sense, any inviscid Coleman-Noll material is σ ˆ (F, θ, p) =: ρ0 (p)ψ(F, hyperelastic (see for instance [5, Section 28]).

Chapter 8

Linearized Models Firstly we emphasize that this chapter applies to any Coleman-Noll material, no matter whether it is a solid or a fluid. A particularization to fluids of the results we obtain below will be done in Chapter 11. Equations (7.4), (7.5) constitute a nonlinear evolutionary system of partial differential equations, the unknowns of which are the fields X and θ. Hence its analysis and numerical solution are difficult tasks. However, in many practical situations, we may assume that only small deformations and changes of temperature arise and then this system can be approximated by a linear one. Throughout this chapter we suppose that the gradient of displacement from an initial equilibrium state, ∇u, is small and that variation of temperature from this state is also small.

8.1 Linear Approximation of the Motion Equation We choose as reference configuration the initial equilibrium position of the body, i.e., B = B0 . Let b0 , f0 , ρ0 , T0 , θ0 and q0 be the initial values of body sources, density, Cauchy stress tensor, temperature and heat flux. Assuming initial equilibrium we have DivT0 + b0 = 0, − Divq0 + f0 = 0,

(8.1) (8.2)

where both equations hold in B. Now we consider a “small” thermodynamic process from this initial state. This means, in particular, that X(p, 0) = p so F (p, 0) = I and furthermore Sˆelas (I, θ0 (p), p) = Tˆ (I, θ0 (p), p) = T0 (p), ˆ ∗ (I, θ0 (p), ∇θ0 (p), p) = q ˆ (I, θ0 (p), ∇θ0 (p), p) = q0 (p). q We need the following technical results:

(8.3) (8.4)

48

Chapter 8. Linearized Models

ˆ be the mapping defined in the set of invertible tensors by Lemma 8.1.1. Let K −1 ˆ ˆ is differentiable and, furthermore, K(A) = A . Then K ˆ ∂K (A) (H) = −A−1 HA−1 . ∂A

(8.5)

Proof. We assume the differentiability and prove (8.5). Let us consider the auxilˆ defined in Lin by M ˆ (A) = A. Then M ˆ (A)K(A) ˆ iary mapping M = I. By using the product rule of differential calculus we get ˆ ˆ ∂M ˆ ˆ (A) ∂ K (A) (H) = 0. (A) (H) K(A) +M ∂A ∂A ˆ (A) is linear, Since M

ˆ ∂M ∂A (A) (H)

ˆ (H) from which the result follows. =M




Lemma 8.1.2. The response function Tˆ (F, y, p), where y represents either temperature or specific entropy, satisfies ∂ Tˆ (F, y, p) (W F ) = W Tˆ (F, y, p) + Tˆ(F, y, p)W t , ∂F

(8.6)

for all F ∈ Lin+ , W ∈ Skw and y ∈ R. Proof. Let us consider the ordinary differential equation, Q˙ = W Q, with initial condition Q(0) = I, where W ∈ Skw. From Lemma 3.2.7 we know that Q(t) is a ˙ rotation for all t and Q(0) = W. Moreover, by taking the derivative of equation (6.6) with respect to t we get   ∂ Tˆ ˙ ˙ Tˆ (F, y, p)Qt (t) + Q(t)Tˆ (F, y, p)Q˙ t (t). = Q(t) (QF, y, p) QF ∂F Finally, by choosing t = 0 in the above equation we obtain (8.6).

(8.7)


If F = I, the previous Lemma yields the following: Corollary 8.1.3. ∂ Tˆ (I, y, p) (W ) = W Tˆ (I, y, p) + Tˆ(I, y, p)W t . ∂F

(8.8)

From now on let us denote by E and W the symmetric and skew parts of tensor ∇u defined by ∇u + ∇ut , 2 t ∇u − ∇u W = . 2 E=

We recall that E ∈ Sym is the infinitesimal strain tensor (see [5]).

(8.9) (8.10)

8.1. Linear Approximation of the Motion Equation Proposition 8.1.4. We have ∂ Tˆ ∂ Sˆelas 1. (I, y, p)(∇u) = (I, y, p)(E) + DivuTˆ (I, y, p) ∂F ∂F −Tˆ (I, y, p)E + W Tˆ (I, y, p).

49

(8.11)

∂ Sˆvisc ∂l (I, H, y, p)(∇u) = (I, y, p)(∇u)(H) + Divu ˆl(I, y, p)(H) ∂F ∂F (8.12) −ˆl(I, y, p)(H)∇ut − ˆl(I, y, p)(H∇u).

2.

3.

∂ Tˆ ∂ Sˆelas (I, y, p) = (I, y, p). ∂y ∂y

(8.13)

4.

∂ Sˆvisc ∂l (I, H, y, p) = (I, y, p)(H). ∂y ∂y

(8.14)

Proof.

1. We have Selas (F, y, p)F t = det(F ) Tˆ (F, y, p)

(8.15)

and, by using the product rule of the differential calculus, we get ∂ Sˆelas (F, y, p) (U ) F t + Sˆelas (F, y, p)U t ∂F ∂ Tˆ (F, y, p) (U ) . (8.16) = det(F ) tr(U F −1 )Tˆ(F, y, p) + det(F ) ∂F If F = I and U = ∇u, by taking into account that Sˆelas (I, y, p) = Tˆ (I, y, p) we obtain ∂ Sˆelas (I, y, p) (∇u) = Divu Tˆ(I, y, p) ∂F ∂ Tˆ + (I, y, p) (∇u) − Tˆ (I, y, p)∇ut . (8.17) ∂F By writing ∇u = E + W and ∇ut = E − W , we deduce ∂ Sˆelas (I, y, p) (∇u) = Divu Tˆ(I, y, p) ∂F ∂ Tˆ (I, y, p) (E + W ) − Tˆ (I, y, p)(E − W ). (8.18) ∂F Since the differential is a linear mapping, and using Corollary 8.1.3, the previous equality yields +

∂ Sˆelas ∂ Tˆ (I, y, p) (∇u) = Divu Tˆ(I, y, p) + (I, y, p) (E) ∂F ∂F +W Tˆ (I, y, p) + Tˆ (I, y, p)W t − Tˆ (I, y, p)E + Tˆ (I, y, p)W. t

(8.19)

As W + W = 0 because W is skew, then equation (8.19) leads to (8.11).

50

Chapter 8. Linearized Models

2. Firstly we have Sˆvisc (F, H, y, p)F t = det(F ) ˆl(F, y, p)(HF −1 ).

(8.20)

By applying the product rule and Lemma 8.1.1 we obtain ∂ Sˆvisc (F, H, y, p) (U ) F t + Sˆvisc (F, H, y, p)U t ∂F ∂l (F, y, p)(U )(HF −1 ) = det(F ) tr(U F −1 )ˆl(F, y, p)(HF −1 ) + det(F ) ∂F + det(F )ˆl(F, y, p)(−HF −1 U F −1 ). (8.21) Let us take F = I and U = ∇u. Since Sˆvisc (I, H, y, p) = ˆl(I, y, p)(H), we finally deduce (8.12). 3. It follows from the fact that Sˆelas (I, y, p) = Tˆ (I, y, p). 4. It follows from the fact that Sˆvisc (I, H, y, p) = ˆl(I, y, p)(H).




The following Corollary is an immediate consequence of the differentiability of both Sˆelas and Sˆvisc with respect to F and y at F = I and y = y0 . Corollary 8.1.5. We have ∂ Tˆ ˆ (I, y0 , p)(E) + DivuT0 − T0 E + W T0 S(F, H, y, p) = T0 + ∂F ∂ Tˆ ∂l (I, y0 , p)(y − y0 ) + ˆl(I, y0 , p)(H) + (I, y0 , p)(∇u)(H) + ∂y ∂F ˆ y0 , p)(H∇u) ˆ y0 , p)(H) − ˆl(I, y0 , p)(H)∇ut − l(I, + Divu l(I, +

∂l (I, y0 , p)(H)(y − y0 ) + o(∇u) + o(y − y0 ). ∂y

(8.22)

Definition 8.1.6. The linear operator C(θ0 , p) ∈ L(Lin, Lin) defined by C(θ0 , p) :=

∂ Tˆ (I, θ0 , p) ∂F

(8.23)

is called a (fourth order) elasticity tensor at point p and temperature θ0 . Moreover, we use the notation Y (θ0 , p) :=

∂ Tˆ (I, θ0 , p). ∂θ

(8.24)

The next properties follow from the fact that Tˆ is symmetric tensor valued. Proposition 8.1.7. We have C(θ0 , p)(H) ∈ Sym ∀H ∈ Lin,

(8.25)

Y (θ0 , p) ∈ Sym.

(8.26)

8.1. Linear Approximation of the Motion Equation

51

Proposition 8.1.8. Let us suppose the initial stress, T0 , is null. Then the elasticity tensor C(θ0 , p) is symmetric, i.e., H · C(θ0 , p)(G) = C(θ0 , p)(H) · G ∀H, G ∈ Lin.

(8.27)

Proof. Firstly, if T0 = 0 then C(θ0 , p) =

∂ Sˆelas (I, θ0 , p) ∂F

(8.28)

and hence

∂ 2 ψˆ (I, θ0 , p). ∂F 2 The result follows from the symmetry of the second derivative. C(θ0 , p) = ρ0 (p)

(8.29)


∂ Tˆ ∂F (I, θ0 , p)

Let us assume that is invertible when considered as a linear mapping from Sym into Sym. Then, by the Implicit Function Theorem, the equation T = Tˆ(F, θ, p) defines F as a function of T and θ in a neighborhood of F = I, θ = θ0 and T = T0 . Furthermore, the derivative of this function, which is called tensor of thermal expansion at constant (elastic) stress, is given by ∂ Fˆ (T, θ, p) = − ∂θ



−1 ∂ Tˆ ∂ Tˆ (F, θ, p) (F, θ, p) ∂F ∂θ

(8.30)

with T = Tˆ (F, θ, p). By evaluating this expression at F = I, θ = θ0 , and hence at T = T0 , we have A(θ0 , p) :=

∂ Fˆ (T0 , θ0 , p) = −C(θ0 , p)−1 (Y (θ0 , p)) . ∂θ

(8.31)

Now we assume that the thermodynamic process is a small perturbation of the reference state, more precisely, ∇u = O(ε),

θ − θ0 = O(ε),

F˙ = ∇u˙ = O(ε),

(8.32)

ε being a small parameter. Then, from Corollary 8.1.5 we can approximate the Piola-Kirchhoff stress tensor: S = T0 + C(θ0 , p)(E) + DivuT0 −T0 E + W T0 + Y (θ0 , p)(θ − θ0 ) + ˆl(I, θ0 , p)(F˙ ) + o(ε). (8.33) If θ = θ0 , ˆ l ≡ 0 and the initial state is stress free, i.e., T0 = 0, then this approximation becomes S = C(θ0 , p)(E),

(8.34)

52

Chapter 8. Linearized Models

which is called the generalized Hooke law. By replacing (8.33) in (7.4) we get the linearized motion equation ¨ = Div [T0 + DivuT0 − T0 E + W T0 ] ρ0 u + Div [C(θ0 , p)(E − A(θ0 , p)(θ − θ0 ))]  ˙ + b∗ . + Div ˆl(I, θ0 , p)(∇u)

(8.35)

If T0 = 0 then the above equation simplifies to ¨ = Div [C(θ0 , p)(E − A(θ0 , p)(θ − θ0 ))] ρ0 u  ˙ + b∗ . + Div ˆl(I, θ0 , p)(∇u)

(8.36)

We recall that, under assumption (4.44), l is given by (4.47) and then ˆ ˆ θ0 , p) DivuI. ˙ = 2ˆ ˙ l(I, θ0 , p)(∇u) η(I, θ0 , p)E˙ + ξ(I,

(8.37)

Moreover, if the reference state is spatially homogeneous, ¨ = Div [C(θ0 , p)(E − A(θ0 , p)(θ − θ0 ))] ρ0 u +η0 ∆u˙ + (η0 + ξ0 )∇ Divu˙ + b∗ .

(8.38)

Remark 8.1.9. Usually, body force b is gravity force. In this situation, by using the mass conservation equation (1.20), we get b∗ (p, t) = det F (p, t)b(X(p, t), t) = det(F (p, t))ρ(X(p, t), t)g(X(p, t)) = ρ0 (p)g(X(p, t)) = ρ0 (p)g(p + u(p, t)).

(8.39)

Thus b∗ can be approximated by neglecting the o(u(p)) term in the following equality: b∗ (p, t) = ρ0 (p) [g(p) + ∇g(p)u(p, t) + o(u(p, t))] .

8.2 Linear Approximation of the Energy Equation Let us go to the energy equation. Firstly we linearize the heat flux.

(8.40)

8.2. Linear Approximation of the Energy Equation

53

Proposition 8.2.1. We have ˆ ˆ∗ ∂q ∂q ˆ (I, y, w, p) (I, y, w, p)(∇u) = (I, y, w, p)(∇u) + Divu q ∂F ∂F ˆ ∂q ˆ (I, y, w, p) − −∇u q (I, y, w, p)∇ut w. ∂w

1.

(8.41)

2.

ˆ ˆ∗ ∂q ∂q (I, y, w, p) = (I, y, w, p). ∂y ∂y

(8.42)

3.

ˆ∗ ˆ ∂q ∂q (I, y, w, p) = (I, y, w, p). ∂w ∂w

(8.43)

Proof. From (7.7) we have ˆ ∗ (F, y, w, p) = det(F )ˆ Fq q(F, y, F −t w, p)

(8.44)

and then ˆ∗ ∂q (F, y, w, p)(U ) = det(F ) tr(U F −1 )ˆ q(F, y, F −t w, p) ∂F ˆ ˆ ∂q ∂q + det(F ) (F, y, F −t w, p)(U ) + det(F ) (F, y, F −t w, p)(−F −t U t F −t w). ∂F ∂w (8.45)

ˆ ∗ (F, y, w, p) + F Uq

Then (8.41) follows by taking F = I and U = ∇u. Equalities (8.42) and (8.43) follow from the fact that ˆ ∗ (I, y, w, p) = q ˆ (I, y, w, p). q

(8.46)


Corollary 8.2.2. We have ˆ ∗ (F, y, w, p) = q

qˆ(I, y0 , w0 , p) +

ˆ ∂q (I, y0 , w0 , p)(∇u) + Divu q0 ∂F

ˆ ∂q (I, y0 , w0 , p)∇ut w0 ∂w ˆ ˆ ∂q ∂q (I, y0 , w0 , p)(w − w0 ) + (I, y0 , w0 , p)(y − y0 ) + ∂y ∂w +o(∇u) + o(y − y0 ) + o(w − w0 ). (8.47) −∇u q0 −

Assuming that ∇u = O(ε), F˙ = ∇u˙ = O(ε), y − y0 = O(ε) and w − w0 = ˆ ∗ (F, y, w, p) by neglecting the three last terms in the O(ε), we can approximate q right hand side of the above equality. Moreover, in order to linearize the energy equation (7.5), we take y = θ, ˆelas ˆelas (F, θ, p), cF (F, θ, p), θ ∂ S∂θ (F, θ, p) and w = ∇θ and approximate terms ∂ S∂y

54

Chapter 8. Linearized Models

Sˆvisc (F, F˙ , θ, p) by their values at the reference configuration, i.e., F = I, F˙ = 0 and θ = θ0 . Thus we get ρ0 cˆF (I, θ0 , p)θ˙

=

θ0 Y (θ0 , p) · ∇u˙  ˆ ∂q −Div q0 + (I, θ0 , ∇θ0 , p)(∇u) + Divu q0 ∂F ˆ ∂q (I, θ0 , ∇θ0 , p)∇ut ∇θ0 −∇u q0 − ∂w ˆ ∂q + (I, θ0 , ∇θ0 , p)(θ − θ0 ) ∂θ  ˆ ∂q (I, θ0 , ∇θ0 , p)(∇θ − ∇θ0 ) + f∗ . + (8.48) ∂w

We notice that the viscous dissipation term has been neglected in this equation ˙ · ∇u, ˙ is second order in ∇u. ˙ Moreover, the because its main part, ˆl(I, θ0 , p)(∇u) heat source f∗ can be approximated by f∗ = (1 + Divu)f according to the Lemma below. Lemma 8.2.3. We have det(F ) = 1 + Divu + o(∇u).

(8.49)

Proof. From Lemma 3.2.2, since det(F ) is differentiable at F = I, we get det(F ) = det(I) + det(I) tr(F − I) + o(F − I),

(8.50)

from which the result follows because F − I = ∇u and tr(∇u) = Divu.




ˆ is given by the generalized Fourier law (see (6.47)): Let us assume that q ˆ |w|2 , p)w ˆ (F, θ, w, p) = −k(θ, q

(8.51)

and that the reference temperature θ0 is spatially homogeneous. Then q0 = 0 and it is straightforward to see that equation (8.48) becomes  ˆ 0 , 0, p)∇θ + f∗ , ρ0 cˆF (I, θ0 , p)θ˙ = θ0 Y (θ0 , p) · ∇u˙ + Div k(θ

(8.52)

which is the standard linearized energy equation for thermoelasticity.

8.3 Isotropic Linear Thermoviscoelasticity If the body is isotropic, operators C and A become much simpler as a consequence of the following.

8.3. Isotropic Linear Thermoviscoelasticity

55

Proposition 8.3.1. Let Hp be the extended symmetry group at p. Then 1. C is invariant under Hp ∩ Orth+ , that is, QC(θ0 , p)(U )Qt = C(θ0 , p)(QU Qt ).

(8.53)

2. Y (θ0 , p) commutes with each Q ∈ Hp ∩ Orth+ . Proof. The first part holds because Tˆ is invariant under Hp ∩ Orth+ (see (6.14)) and the derivative of an invariant mapping is also invariant (see Gurtin[5]). For the second part we take the derivative of (6.14) with respect to θ. We get ∂ Tˆ ∂ Tˆ Q (F, θ, p)Qt = (QF Qt , θ, p). (8.54) ∂θ ∂θ By taking F = I and θ = θ0 we obtain QY (θ0 , p)Qt = Y (θ0 , p).

(8.55)


As a consequence we have the important Proposition 8.3.2. Let us assume the body at p is isotropic. Then there exist funcˆ 0 , p), µ tions λ(θ ˆ(θ0 , p), yˆ(θ0 , p) and χ(θ ˆ 0 , p) defined in R+ × B such that ˆ 0 , p) tr(E)I, C(θ0 , p)(E) = 2ˆ µ(θ0 , p)E + λ(θ

(8.56)

Y (θ0 , p) = yˆ(θ0 , p)I, ˆ 0 , p)I. A(θ0 , p) = χ(θ

(8.57) (8.58)

Moreover, χ(θ ˆ 0 , p) = −

yˆ(θ0 , p) . ˆ 0 , p) 2ˆ µ(θ0 , p) + 3λ(θ

(8.59)

Proof. The first assertion follows from the Representation Theorem for Linear Isotropic Tensor Functions (see Appendix A). The second one is a consequence of the fact that symmetric tensors which commute with any rotation must be isotropic, i.e., proportional to the identity tensor (see also Appendix A). Finally, the expression for A(θ0 , p) and equality (8.59) can be obtained from (8.31).
ˆ 0 , p) and µ Definition 8.3.3. Functions λ(θ ˆ(θ0 , p) are called Lam´e’s coefficients of the body at point p at temperature θ0 , while χ(θ ˆ 0 , p) is called the coefficient of linear thermal expansion at constant stress of the body at point p at temperature θ0 .

56

Chapter 8. Linearized Models

Definition 8.3.4. We say that a fourth order tensor D ∈ L(Lin, Sym) is positive definite if E · D(E) > 0 (8.60) for all symmetric tensors E = 0. Proposition 8.3.5. Let us assume that the material at p is isotropic. Then C(θ0 , p) is positive definite if and only if the Lam´e coefficients satisfy the inequalities ˆ 0 , p) > 0. µ ˆ (θ0 , p) > 0, 2ˆ µ(θ0 , p) + 3λ(θ Proof. See [5].

(8.61)


We end this section by writing the equations for linearized thermoviscoelasticity under the following assumptions: • The body is an isotropic Coleman-Noll material. • The heat flux is given by the generalized Fourier law. • The initial state is spatially homogeneous, i.e., initial fields are constant functions.

¨ = µ0 ∆u + (µ0 + λ0 )∇ Divu − (2µ0 + 3λ0 )χ0 ∇θ ρ0 u +η0 ∆u˙ + (η0 + ξ0 )∇ Divu˙ + b∗ , ˙ ρ0 cF 0 θ = −θ0 (2µ0 + 3λ0 )χ0 Divu˙ + k0 ∆θ + f∗ .

(8.62) (8.63)

Chapter 9

Quasi-static Thermoelasticity In this chapter we state the equations for quasi-static thermoelasticity and give an approximated method to solve them.

9.1 Statement of the Equations ¨ , Sˆvisc and In the case of quasi-static thermoelasticity terms ρ0 u θ

∂ Sˆ (F, θ, p) · ∇u˙ ∂θ

(9.1)

may be neglected and then (7.4) and (7.5) yield Div Sˆelas (F, θ, p) + b∗ = 0, ρ0 cF θ˙ + Div qˆ(F, θ, ∇θ, p) = f∗ .

(9.2) (9.3)

9.2 Time Discretization In order to write a discretization in time we use the idea underlying incremental methods in nonlinear elasticity. For the sake of simplicity, we restrict ourselves to the case of small displacements. However, we do not make any assumption on temperature. Let us take the derivative of equation (9.2) with respect to time. We get   ˆelas ∂ S ∂ Sˆelas (F, θ, p) (F˙ ) + (F, θ, p) θ˙ + b˙ ∗ = 0. (9.4) Div ∂F ∂θ We assume small deformations, namely, F − I = ∇u = O(ε).

(9.5)

58

Chapter 9. Quasi-static Thermoelasticity

Then we have ∂ Sˆelas ∂ Sˆelas (F, θ, p) = (I, θ, p) + O(ε) ∂F ∂F

(9.6)

and similarly ∂ Sˆelas ∂ Sˆelas (F, θ, p) = (I, θ, p) + O(ε). ∂θ ∂θ Thus (9.4) can be approximated by   ˆelas ∂ Sˆelas ∂ S Div (I, θ, p) (F˙ ) + (I, θ, p) θ˙ + b˙ ∗ = 0. ∂F ∂θ

(9.7)

(9.8)

Let us assume that the heat flux obeys the generalized Fourier law. The previous equation can be discretized by using an implicit Euler method: we compute sequences u1 , . . . , uN and θ1 , . . . , θN by solving

n+1 n+1

 n n ˆelas F θ ∂ Sˆelas − F − θ ∂ S Div (I, θn+1 , p) (I, θn+1 , p) + ∂F δt ∂θ δt bn+1 − bn∗ ∗ = 0, (9.9) δt   n+1 n −θ θ ˆ n , |∇θn |, p)∇θn+1 = f n+1 , − Div k(θ ρ0 cˆF (F n , θn , p) ∗ δt (9.10)  T −t0  n n n n where δt is a time step, N = δt , F = I + ∇u , b∗ = b∗ (tn ), f∗ = f∗ (tn ) and tn = t0 + nδt. By using (8.11), (8.23) and (8.24), and introducing the notations +

δun := un+1 − un ,

(9.11)

δθn := θn+1 − θn , − bn∗ , δbn∗ := bn+1 ∗

(9.12) (9.13)

discretized equation (9.9) becomes  Div C(θn+1 , p) (E(δun )) + Divδun Tˆ (I, θn+1 , p) − Tˆ(I, θn+1 , p)E(δun ) +W (δun )Tˆ(I, θn+1 , p) + Y (θn+1 , p)δθn + δbn∗ = 0. (9.14) In terms of the tensor of thermal expansion at constant stress this equation can be written as follows:    Div C(θn+1 , p) E(δun ) − A(θn+1 , p)δθn + Divδun Tˆ (I, θn+1 , p) −Tˆ (I, θn+1 , p)E(δun ) + W (δun )Tˆ(I, θn+1 , p) + δbn∗ = 0. (9.15)

9.3. A Particular Case

59

We notice that if the body is isotropic then (8.56) and (8.57) hold and they can be replaced in (9.15). Moreover, after solving (9.15) and (9.10) for n = 0, . . . , N − 1, the displacements can be computed by n+1

u

0

=u +

n 

δuk ,

(9.16)

k=0

while stresses are given by S n+1 = Tˆ (I, θ0 , p) +

n     C(θk+1 , p) E(δuk ) − A(θk+1 , p)δθk k=0

+Divδu Tˆ(I, θ k

k+1

, p) − Tˆ (I, θk+1 , p)E(δuk ) + W (δuk )Tˆ(I, θk+1 , p) . (9.17)

9.3 A Particular Case Sometimes, instead of the tensor of thermal expansion A, we know the tensor of the accumulated thermal expansion between two temperatures θ0 and θ, to be denoted by Υ(θ0 , θ, p). Parameters A and Υ are related by A(θ, p) =

∂Υ 0 (θ , θ, p). ∂θ

(9.18)

Thus A is the inverse of a temperature while Υ is non-dimensional. A discrete version of (9.18) is A(θn+1 , p) ≈

Υ(θ0 , θn+1 , p) − Υ(θ0 , θn , p) θn+1 − θn

(9.19)

and therefore n−1  k=0

A(θ

k+1

, p)(θ

k+1

−θ )≈ k

n−1 

  Υ(θ0 , θk+1 , p) − Υ(θ0 , θk , p)

k=0

= Υ(θ0 , θn , p) − Υ(θ0 , θ0 , p) = Υ(θ0 , θn , p).

(9.20)

Now let us assume that tensor C is independent of temperature and that the terms due to the initial stresses can be neglected. Then, by summing up equations (9.15) from 0 to n − 1 we get    (9.21) Div C(p) ∇un − Υ(θ0 , θn )I + bn∗ = 0. This means that, under the above assumption, we can compute the state at time tn by making only one step from the initial solution.

Chapter 10

Fluids In this chapter we study a subclass of Coleman-Noll materials called fluids.

10.1 The Concept of Fluid, First Properties Definition 10.1.1. A fluid is a Coleman-Noll material having the unimodular group U nim = {H ∈ Lin+ : det(H) = 1}

(10.1)

as extended symmetry group. Remark 10.1.2. Since Orth+ ⊂ U nim, any fluid is isotropic in the sense of Definition 6.1.1. The following Lemma characterizes the functions which are invariant by unimodular transformations. ˆ : Lin+ → W Lemma 10.1.3. Let W be a finite dimensional normed space and G a mapping such that ˆ H) = G(F ˆ ) ∀H ∈ U nim ∀F ∈ Lin+ . G(F

(10.2)

Then, for any given constant ρ0 , there exists a mapping gˆ : R → W such that

where ν :=

det(F ) ρ0

ˆ ) = gˆ(ν), G(F

(10.3)

ˆ dG (F )(U ) = νF −t · U gˆ
(ν). dF

(10.4)

and

Furthermore, if W = R then ˆ dG (F )(U ) = νˆ g
(ν)F −t · U. dF

(10.5)

62

Chapter 10. Fluids 1/3

Proof. For any given F ∈ Lin+ , let us choose H = (det(F )) U nim and

F −1 . Then H ∈

1/3 ˆ ) = G(F ˆ H) = G(F ˆ (det(F ))1/3 F −1 ) = G((det(F ˆ G(F )) I).

Let gˆ : R → W be defined by

  ˆ (ρ0 r)1/3 I . gˆ(r) = G

(10.6)

(10.7)

Then (10.3) holds. Indeed, gˆ(ν) = gˆ(

det(F ) 1/3 ˆ ˆ ). ) = G((det(F )) I) = G(F ρ0

(10.8)

Now the second part of the Lemma is a consequence of the chain rule:



ˆ det(F ) 1 dG (F )(U ) = gˆ
det(F ) tr(U F −1 ) dF ρ0 ρ0 = νF −t · U gˆ
(ν).

(10.9)


In what follows we deal with a Coleman-Noll material which is a fluid in the sense of Definition 10.1.1. According to Definition 3.1.3 and the previous Lemma there must exist functions still denoted with “hat”: Tˆ : R+ × R × B → Sym,

(10.10)

ˆl : R × R × B → L(Lin, Sym), eˆ : R+ × R × B → R, θˆ : R+ × R × B → R+ ,

(10.11) (10.12)

+

ˆ : R × R × V × B → V, q +

(10.13) (10.14)

“smooth enough” and satisfying T (x, t) = Tˆ (ν(x, t), s(x, t), p) + ˆl(ν(x, t), s(x, t), p)(L),

(10.15)

e(x, t) = eˆ(ν(x, t), s(x, t), p), ˆ θ(x, t) = θ(ν(x, t), s(x, t), p),

(10.16) (10.17)

ˆ (ν(x, t), s(x, t), gradθ(x, t), p), q(x, t) = q

(10.18)

with x = X(p, t). Now we recall assumption (H1) introduced in Theorem 3.2.4. In the case of a fluid it becomes: There exists a function sˆ : R+ × R+ × B → R such that ˆ s, p) ∀ν ∈ R+ ∀θ ∈ R+ ∀p ∈ B. s = sˆ(ν, θ, p) ⇐⇒ θ = θ(ν,

(10.19)

As a consequence of Theorem 3.2.4 and equality (10.4) we have the following:

10.2. Motion Equation. Thermodynamic Pressure

63

Theorem 10.1.4. Assume (H1) holds. Then all thermodynamic processes in the constitutive class C of a fluid satisfy the second law of thermodynamics if and only if e ˆ s, p) = ∂ˆ (ν, s, p), θ(ν, ∂s ∂ˆ e Tˆ (ν, s, p) = (ν, s, p)I, ∂ν ˆ l(ν, s, p)(L) · L ≥ 0 (dissipation inequality), ˆ (ν, s, w, p) · w ≤ 0, q

(10.20) (10.21) (10.22) (10.23)

for all ν ∈ R+ , s ∈ R, p ∈ B, L ∈ Lin and w ∈ V.

10.2 Motion Equation. Thermodynamic Pressure Definition 10.2.1. We denote by thermodynamic pressure the scalar field π(x, t) = π ˆ (ν(x, t), s(x, t), p), with π ˆ given by π ˆ (ν, s, p) = −

∂ˆ e (ν, s, p). ∂ν

(10.24)

Remark 10.2.2. Since fluids are isotropic Coleman-Noll materials, we deduce from the above theorem that functions in (6.24) satisfy π(ν, s, p), β0 (IB , s, p) = β0 (det(B), s, p) = −ˆ β1 (IB , s, p) = 0,

β2 (IB , s, p) = 0.

(10.25) (10.26)

In what follows we assume that the Principle of Material Frame-Indifference holds. As an immediate consequence of Proposition 4.2.4 we have the following Proposition 10.2.3. For any fluid there exist two functions ηˆ : R+ × R × B → R, ξˆ : R+ × R × B → R,

(10.27)

ˆl(ν, s, p)(L) = ˆl(ν, s, p)(D) = 2ˆ ˆ s, p) tr(D)I. η (ν, s, p)D + ξ(ν,

(10.29)

(10.28)

such that

Proof. It follows from Proposition 4.2.4 and the fact that property (4.44) holds for any fluid. Indeed, let us suppose F ∈ Lin+ and Q ∈ Orth+ . Then ν(QF ) = det(QF ) ) = det(F = ν(F ).
ρ0 ρ0

64

Chapter 10. Fluids

Definition 10.2.4. Scalar field η(x, t) = ηˆ(ν(x, t), s(x, t), p) is called dynamic visˆ cosity while the scalar field ξ(x, t) = ξ(ν(x, t), s(x, t), p) is called second viscosity coefficient. Moreover, the scalar field 2 ζ = ξ + η, 3

(10.30)

is called bulk viscosity. From the dissipation inequality and Proposition 4.2.5 we deduce η ≥ 0, ζ ≥ 0. (10.31) Corollary 10.2.5. For a fluid, the motion equation becomes ρv˙ + gradπ = div(l(D)) + b,

(10.32)

with l(D) given by (10.29).

10.3 Energy Equation, Enthalpy ˆ satisfies Proposition 10.3.1. For a fluid the response function q ˆ y, |w|2 , p)w ˆ (ν, y, w, p) = −k(ν, q

(10.33)

where y can be replaced by s or θ, ˆ only depends on det(F ), Proof. It follows from Proposition 6.2.8. Indeed, since q equality (6.39) yields ˆ (ν, y, w, p) = α0 (ν, w, y, p)w = γ0 (ν, w · w, y, p)w, q

(10.34)

ˆ y, |w|2 , p) = −γ0 (ν, w · w, y, p). which is (10.33) for k(ν, Definition 10.3.2. We denote by specific enthalpy the scalar spatial field defined by π(x, t) h(x, t) := e(x, t) + . (10.35) ρ(x, t) The next Proposition provides a local form of the energy conservation law in terms of the specific enthalpy. Proposition 10.3.3. The following equality holds: ρ h˙ − π˙ = l(D) · D − div q + f.

(10.36)

e˙ = h˙ − (πν)· = h˙ − πν ˙ − π ν˙ = h˙ − πν ˙ − πν div v,

(10.37)

Proof. We have

where we have used the mass conservation equation (1.22) to get the last equality. Then ρe˙ = ρh˙ − π˙ − π div v and, by replacing this equality in (3.10), we easily obtain (10.36).


10.3. Energy Equation, Enthalpy

65

The following Proposition gives the so-called Crocco form of the motion equation. Proposition 10.3.4. Under the assumptions • eˆ is independent of p, • the thermodynamic process is Eulerian, • the body force is conservative, i.e.,

b ρ

= − gradβ, for some scalar field β,

the following equation holds: |v|2 + β) − θ grads = 0. 2

(10.38)

∂ˆ e π ∂ˆ e gradν + grads = 2 gradρ + θ grads. ∂ν ∂s ρ

(10.39)

v
+ curlv × v + grad(h + Proof. We have grade = On the other hand

π 1 π gradπ − 2 gradρ = 2 gradρ + θ grads ρ ρ ρ π 1 1 + gradπ − 2 gradρ = θ grads + gradπ, ρ ρ ρ

gradh = grade +

(10.40)

which implies 1 gradπ = gradh − θ grads. ρ

(10.41)

By replacing (10.41) in the motion equation v˙ +

1 b gradπ = , ρ ρ

(10.42)

and using the equality v˙ = v
+ grad

|v|2 + curlv × v, 2

(10.43)


we deduce (10.38). − ρ1

Remark 10.3.5. From the motion equation, v˙ = gradπ − gradβ, and (10.41) we deduce that, if the specific entropy is spatially homogeneous, then the acceleration is the gradient of a potential, namely, −(h + β). In this case, the important Theorems of Lagrange-Cauchy, Kelvin and Transport of Vorticity apply (see [5], Section 11). Definition 10.3.6. The scalar field H = h + thalpy.

|v|2 2

is called specific stagnation en-

66

Chapter 10. Fluids

Definition 10.3.7. Suppose the body force is conservative. A thermodynamic process is called homentropic if (H + β)˙ = 0. Definition 10.3.8. A thermodynamic process (X, T, b, e, θ, q, f, s) ∈ C is said to be steady if X is a steady motion (see [5]) and all fields T, b, e, θ, q, f, s are independent of time. We have the following: Proposition 10.3.9. Let us consider a steady thermodynamic process satisfying the assumptions of Proposition 10.3.4. Then it is isentropic if and only if it is homentropic. Proof. We make the scalar product of (10.38) with v. Since v
= 0 we obtain v · grad(H + β) − θ grads · v = 0

(10.44)

(H + β)˙− θs˙ = 0.

(10.45)

and then

The conclusion follows because θ > 0.




Theorem 10.3.10. (Crocco). Let us consider an Eulerian homentropic steady thermodynamic process under a conservative force, with H + β constant in T . Then, if s is not constant in T , the process cannot be irrotational. Proof. Since H + β is constant, equation (10.38) becomes curlv × v = θ grads. Then curlv = 0 implies grads = 0.

(10.46)


Remark 10.3.11. We notice that the converse is not always true. Indeed, there may exist steady isentropic thermodynamic processes which are not irrotational. For this it is enough that curlv and v be parallel.

10.4 Thermodynamic Coefficients and Equalities In what follows we use standard notation in books on thermodynamics: let us suppose that a thermodynamic variable z is a function of any two other called x and y, that is, z = zˆ(x, y). Then  the  partial derivatives of zˆ with respect to x and  ∂z  ∂z and . Accordingly, we will also write z instead y will be denoted by ∂x ∂y y x of zˆ.

10.4. Thermodynamic Coefficients and Equalities

67

Proposition 10.4.1. There exists a mapping sˆ : R+ × R × B → R such that s = sˆ(ν, e, p). Moreover,

∂s 1 (10.47) = , ∂e ν θ

∂s π (10.48) = . ∂ν e θ   Proof. Since ∂e ˆ(ν, s, p) is strictly increasing ∂s ν = θ > 0, the mapping s ∈ R → e so it is injective for any given ν ∈ R+ and p ∈ B. Then it has an inverse defined in the image set. The inverse mapping theorem immediately gives (10.47) from (10.20). To prove (10.48) it is enough to apply the chain rule to the identity e = eˆ(ν, sˆ(ν, e, p), p). Indeed, by taking the derivative with respect to ν, we obtain



∂e ∂e ∂s 0= + , ∂ν s ∂s ν ∂ν e

(10.49)

(10.50)


which yields (10.48).

Now we notice that, under assumption (H1), we can write the specific internal energy as a function of the specific volume and the temperature. Next, we can replace ν by ρ1 to get e = eˆ(ρ, θ, p) which gives sense to the following: Definition 10.4.2. We denote by specific heat at constant volume the scalar field cv defined by cv (x, t) = cˆv (ρ(x, t), θ(x, t), p) with cˆv given by cˆv (ρ, θ, p) :=

∂ˆ e (ρ, θ, p). ∂θ

(10.51)

We notice that, for fluids, the specific heat at constant volume coincides with the specific heat at constant deformation defined in (5.11). Now we make the following assumption: (H2) For any given p, the mapping Φp : R+ × R → R+ × R defined by

∂ˆ e ∂ˆ e (ν, s, p), − (ν, s, p) (10.52) (θ, π) = Φp (ν, s) := ∂s ∂ν has an inverse allowing us to write ν and s as functions of θ and π. Definition 10.4.3. We denote by coefficient of volumetric thermal expansion at constant pressure the scalar field α defined by α(x, t) = α ˆ (θ(x, t), π(x, t), p) with α ˆ given by 1 ∂ νˆ α ˆ (θ, π, p) := (θ, π, p). (10.53) νˆ(θ, π, p) ∂θ

68

Chapter 10. Fluids

This coefficient is related with the tensor of thermal expansion at constant elastic stress given in (8.31). Indeed, since νˆ(θ, π, p) =

det Fˆ (θ, π, p) , ρ0

(10.54)

by using the chain rule and Lemma 3.2.2 we get   det Fˆ (θ, π, p) ∂ νˆ ∂ Fˆ −1 (θ, π, p) = (θ, π, p)(Fˆ (θ, π, p)) tr ∂θ ρ0 ∂θ 

and then α ˆ (θ, π, p) = tr

∂ Fˆ (θ, π, p)(Fˆ (θ, π, p))−1 ∂θ

(10.55)

 .

(10.56)

By replacing θ with θ0 and π with π0 , and using that Fˆ (θ0 , π0 , p) = I (see Chapter 8) and that fluids are isotropic materials we deduce ∂ νˆ (θ0 , π0 , p) = ν0 3χ(θ ˆ 0 , p) ∂θ

(10.57)

ˆ 0 , p). α ˆ (θ0 , π0 , p) = 3χ(θ

(10.58)

and hence Definition 10.4.4. We denote by coefficient of isothermal compressibility the scalar field κ defined by κ(x, t) = κ ˆ(θ(x, t), π(x, t), p) with κ ˆ given by κ ˆ (θ, π, p) := −

1 ∂ νˆ (θ, π, p). νˆ(θ, π, p) ∂π

Remark 10.4.5. We notice that, since ν = ρ1 , we have

1 ∂ρ , α=− ρ ∂θ π

1 ∂ρ . κ= ρ ∂π θ

(10.59)

(10.60) (10.61)

By using α and κ we can write the mass conservation equation as follows: Corollary 10.4.6. We have −αθ˙ + κπ˙ + div v = 0. Proof. From (10.60) and (10.61) we deduce



∂ρ ∂ρ ˙ ρ˙ = π˙ = −ραθ˙ + ρκπ. ˙ θ+ ∂θ π ∂π θ

(10.62)

(10.63)

Then the result can be obtained by replacing this expression for ρ˙ in the mass conservation equation (1.22).


10.4. Thermodynamic Coefficients and Equalities

69

Under assumption (H2), the specific enthalpy can be written as a function of θ and π. Indeed π ν (θ, π, p), sˆ(θ, π, p), p) + νˆ(θ, π, p)π. (10.64) h = e + = eˆ (ˆ ρ Hence the following definition makes sense. Definition 10.4.7. We denote by specific heat at constant pressure the scalar field cπ defined by cπ (x, t) = cˆπ (θ(x, t), π(x, t), p) with cˆπ given by ∂ ˆh (θ, π, p). (10.65) ∂θ The next Proposition gives the partial derivatives of s when it is considered as a function of θ and π. cˆπ (θ, π, p) :=

Proposition 10.4.8. We have



∂s cπ = , ∂θ π θ



∂s 1 ∂ρ α = 2 =− , ∂π θ ρ ∂θ π ρ



∂h θ ∂ρ 1 = + 2 . ∂π θ ρ ρ ∂θ π

(10.66) (10.67) (10.68)

Proof. Firstly we make use of Proposition 10.4.1 and assumption (H2) to write s as a function of θ and π: sˆ(θ, π, p) = sˆ(ˆ ν (θ, π, p), eˆ(θ, π, p), p)   ˆ π, p) − πˆ = sˆ νˆ(θ, π, p), h(θ, ν (θ, π, p), p . Then,



∂s ∂θ





∂s ∂ν ∂h ∂s ∂ν = −π + ∂e ∂θ ∂θ ∂ν ∂θ π ν π  π e π ∂s ∂s ∂s ∂h ∂ν = + −π + ∂e ν ∂θ π ∂e ν ∂ν e ∂θ π  π π ∂ν

1 1 = cπ + − + = cπ . θ θ θ ∂θ π θ



(10.69)



(10.70)

Similarly,





∂s ∂ν ∂s ∂h ∂s ∂ν = −ν −π + ∂π θ ∂e ν ∂π θ ∂π θ ∂ν e ∂π θ 

 



∂s ∂s ∂s ∂h ∂ν = − νˆ(θ, π, p) + − π+ ∂e ν ∂π θ ∂e ν ∂ν e ∂π θ 

 ∂h 1 1 = − . θ ∂π θ ρ

(10.71)

70

Chapter 10. Fluids

Now we compute the partial derivatives of (10.70) and (10.71) with respect to π and θ, respectively. We obtain



1 ∂ ∂s ∂h ∂ = , (10.72) ∂π ∂θ π θ θ ∂π ∂θ π θ





∂ ∂s ∂h ∂ 1 1 = − ∂θ ∂π θ π ∂θ θ ∂π θ ρ π





∂ρ 1 ∂h 1 1 ∂ ∂h 1 =− 2 + 2+ + 2 . (10.73) θ ∂π θ ρθ θ ∂θ ∂π θ π θρ ∂θ π By Schwarz’s Theorem these two second derivatives are equal. Hence







∂ρ ∂h 1 ∂ 1 ∂h 1 1 ∂ ∂h 1 =− 2 + 2+ + 2 θ ∂π ∂θ π θ θ ∂π θ ρθ θ ∂θ ∂π θ π θρ ∂θ π (10.74) and using again Schwarz’s Theorem



∂ ∂ ∂h ∂h = ∂π ∂θ π θ ∂θ ∂π θ π

(10.75)

we can deduce (10.68). Finally, by replacing (10.68) in (10.71) we obtain (10.67), which completes the proof.
Corollary 10.4.9. Let us assume that cˆπ (θ, π, p) = 0. Then we have



∂ρ ∂θ θ θα =− 2 = . ∂π s ρ cπ ∂θ π ρcπ

(10.76)

Proof. Firstly, from assumption (H2) we can write the specific entropy as a function of temperature and pressure: s = sˆ(θ, π, p).

(10.77)

As a consequence of the Implicit Function Theorem, for any given s and p ∈ B we ∂ˆ s can obtain θ as a function of π, because ∂θ (θ, π, p) = 0 according to (10.66) and the assumption of the Corollary. Finally, by using (10.66) and (10.67), we deduce    ∂s  ∂ρ 1



2 ρ ∂θ ∂θ ∂ρ θ θ π  = −  ∂π = − = − . (10.78) cπ ∂s 2c ∂π s ρ ∂θ π π θ ∂θ π
Corollary 10.4.10. The following form of the energy equation holds: ρcπ θ˙ − αθπ˙ = l(D) · D − div q + f.

(10.79)

10.4. Thermodynamic Coefficients and Equalities Proof. From (10.66) and (10.67) we deduce



∂s ∂s α cπ ˙ ˙ π˙ = θ˙ − π. θ+ s˙ = ∂θ π ∂π θ θ ρ By replacing this equality in (3.67) we get the result.

71

(10.80)


The next Proposition gives expressions for the partial derivatives of s as a function of ρ and θ. Proposition 10.4.11. We have 1. 2. 3.



∂s ∂θ



1 cv . θ ρ



∂s 1 ∂π =− 2 . ∂ρ θ ρ ∂θ ρ



∂π ∂e 1 = 2 π−θ . ∂ρ θ ρ ∂θ ρ =

(10.81) (10.82)

(10.83)

Proof. We first write s as a function of ρ and θ: sˆ(ρ, θ, p) = sˆ(ˆ ν (ρ), eˆ(ρ, θ, p), p). Then we use the chain rule to compute the partial derivative:



∂s ∂s ∂e 1 = = cv , ∂θ ρ ∂e ν ∂θ ρ θ which proves (10.81). Similarly,





∂s ∂s ∂e ∂ν ∂s = + ∂ρ θ ∂e ν ∂ρ θ ∂ν e ∂ρ θ



1 ∂e π 1 = + − 2 . θ ∂ρ θ θ ρ

(10.84)

(10.85)

(10.86)

By calculating the derivatives of (10.85) and (10.86) with respect to ρ and θ we get   

 1 ∂cv 1 ∂ ∂e ∂ ∂s = = , (10.87) ∂ρ ∂θ ρ θ ∂ρ θ θ ∂ρ ∂θ ρ θ θ

 ∂π  





θ − π ∂ ∂s 1 ∂e 1 ∂ ∂e 1 ∂θ ρ =− 2 + − . (10.88) ∂θ ∂ρ θ ρ θ ∂ρ θ θ ∂θ ∂ρ θ ρ ρ2 θ2

72

Chapter 10. Fluids

By Schwarz’s Theorem, these two derivatives must be equal. Hence, 



1 ∂ ∂e 1 ∂e 1 ∂ ∂e =− 2 + θ ∂ρ ∂θ ρ θ ∂ρ θ θ ∂θ ∂ρ θ ρ θ

 ∂π   1 ∂θ ρ θ − π − 2 , ρ θ2 and using the fact that



∂ ∂ρ



∂e ∂θ



=

ρ

θ

∂ ∂θ



∂e ∂ρ

(10.89)



, θ

(10.90)

ρ

we immediately deduce (10.83) from (10.89). Finally, (10.82) can be easily obtained by replacing (10.83) in (10.86). This completes the proof.
Corollary 10.4.12. Let us assume that cˆv (ρ, θ, p) = 0. Then we have



∂π ∂θ θ = 2 . ∂ρ s ρ cv ∂θ ρ

(10.91)

Proof. Firstly, we write the specific entropy as a function of density and temperature: s = sˆ(ρ, θ, p). (10.92) Now, for any given s and p ∈ B, we can apply the Implicit Function Theorem to ∂ˆ s obtain θ as a function of ρ, because ∂θ (ρ, θ, p) = 0 according to (10.81) and the assumption of the Corollary. Finally, by using (10.81) and (10.82) we obtain     ∂s



− ρ12 ∂π ∂ρ θ ∂θ ∂π θ ∂θ ρ = −  ∂s  = − = 2 . (10.93) cv ∂ρ s ρ c ∂θ ρ v θ ∂θ ρ
From now on let us assume that cv and cπ are always strictly positive. The next definition makes sense if ∂∂ρπˆ (ρ, s, p) ≥ 0. Definition 10.4.13. We denote by sound speed the scalar field c defined by c(x, t) := cˆ(ρ(x, t), s(x, t), p) with cˆ defined by  ∂π ˆ cˆ(ρ, s, p) = (ρ, s, p). (10.94) ∂ρ Proposition 10.4.14. The following equality holds:

cπ ∂π 2 c = . cv ∂ρ θ

(10.95)

10.4. Thermodynamic Coefficients and Equalities

73

Proof. As in the proof of Corollary 10.4.9, we can write θ as a function of π and s. More precisely, ˆ s, p) = θ(ˆ ˆ π (ρ, s, p), s, p)). θ = θ(ρ, (10.96) Hence, by the chain rule,

∂θ ∂ρ



=

s

∂θ ∂π

s

∂π ∂ρ

,

(10.97)

s

from which it follows that c2 =

∂π ∂ρ





∂θ ∂ρ

 θ ρ2 cv

=  ∂θ  s = s

−θ ρ2 cπ

∂π s

 ∂π  ∂θ ρ



∂ρ ∂θ

 ,

(10.98)

π

by taking (10.76) and (10.91) into account. Now, we use the fact that ρ = ρˆ(π, θ, p) to write π=π ˆ (ˆ ρ(π, θ, p), θ, p).

(10.99)

By taking the derivative of this equality with respect to θ we get



∂π ∂π ∂ρ 0= + ∂ρ θ ∂θ π ∂θ ρ and then



∂π ∂ρ

 ∂π 



= − θ

∂θ ρ

 .

∂ρ ∂θ

(10.101)

π

By replacing this equality in (10.98) we immediately get (10.95). Proposition 10.4.15. The following equality holds:



∂π θ ∂π = . ∂s ρ cv ∂θ ρ Proof. Firstly, we prove that

∂π ∂s

(10.100)



=− ρ

∂θ ∂ν




(10.102)

.

(10.103)

s

Indeed, let us recall that π=−

∂e ∂ν



, s

θ=

∂e ∂s

. ν

(10.104)

74

Chapter 10. Fluids

Moreover, by Schwarz’s Theorem



∂ ∂e ∂ ∂e = . ∂s ∂ν s ν ∂ν ∂s ν s Then we immediately deduce (10.103) from (10.104). Now, by using (10.91) we obtain





∂θ ∂θ ∂π θ dρ =− 2 = ρ2 ∂ν s ∂ρ s dν ρ cv ∂θ ρ

θ ∂π =− . cv ∂θ ρ

(10.105)

(10.106)


From the previous Proposition we can easily prove the following: Proposition 10.4.16. We have

∂π ∂s



c2 ραθ . cπ

= ρ

(10.107)

Proof. From (10.95) and (10.53) we deduce





c2 ραθ ∂ν 1 ∂π 1 ∂ν 1 ∂π = ρ θ=− θ. cπ cv ∂ρ θ ν ∂θ π cv ∂ν θ ∂θ π

(10.108)

Now, by taking the derivative with respect to θ in the equality π=π ˆ (ˆ ν (θ, π, p), θ, p)

(10.109)

we get 0=

∂π ∂ν

θ

∂ν ∂θ



+ π

∂π ∂θ

,

(10.110)

ν

and using this in (10.108) c2 ραθ θ = cπ cv





∂π ∂θ

ν

θ = cv



∂π ∂θ

.

(10.111)

ρ




Now the results follows from (10.107). Corollary 10.4.17. We have

∂ρ ∂s

=− π

ραθ . cπ

(10.112)

10.4. Thermodynamic Coefficients and Equalities

75

Proof. We apply the Implicit Function Theorem to get ρ as a function of π from equality π = π ˆ (ρ, s, p). We have

∂ρ ∂s

 ∂π 



= − π

∂s ρ ∂π ∂ρ

 .

(10.113)

s

Now the result follows by using (10.107) and (10.94).




Corollary 10.4.18. We have

where γ :=

∂π ∂θ

= ρ

∂π ∂ρ



c2 ρα , γ

(10.114)

c2 , γ

(10.115)

= θ

cπ cv .

Proof. Equality (10.114) follows immediately from (10.111) while (10.115) is a consequence of (10.95).
Now, we can obtain new forms of the motion and energy equations which are useful, for instance, in dissipative acoustics. Proposition 10.4.19. Let us assume that function eˆ(ρ, θ, p) does not depend on p. Then we have ρv˙ +

c2 c2 ρα gradρ + gradθ = div l(D) + b. γ γ

(10.116)

Proof. It is enough to write π as a function of ρ and θ. From the previous Corollary we deduce



∂π ∂π gradπ = gradρ + gradθ ∂ρ θ ∂θ ρ =

c2 c2 ρα gradρ + gradθ. γ γ

(10.117)

By replacing this equality in the motion equation (10.32) we deduce (10.116).
Proposition 10.4.20. We have αθc2 ρcv θ˙ − ρ˙ = l(D) · D − div q + f. γ

(10.118)

76

Chapter 10. Fluids

Proof. Let us consider s as a function of ρ and θ. From (10.81), (10.82) and (10.114) we deduce



∂s ∂s cv ˙ α c2 ρ˙ + θ. (10.119) ρ˙ + θ˙ = − s˙ = ∂ρ θ ∂θ ρ ργ θ Now the result follows by replacing this equality in the energy equation (3.67).
Remark 10.4.21. As we said above, for fluids, the cF defined by (5.11) coincides with cv . Actually, equation (10.118) is nothing but (5.20) for fluids. The next Proposition gives two useful relations between some of the thermodynamic fields we have introduced in this section. Proposition 10.4.22. The following equalities hold: γ = ρκc2 .

1.

cπ − cv =

2.

(10.120)

2 2

c α θ . γ

(10.121)

Proof. Equality (10.120) is a straightforward consequence of (10.61), (10.115) and the fact that

−1 ∂π ∂ρ = . (10.122) ∂ρ θ ∂π θ To prove (10.121) we first write the specific internal energy as a function of ρ and θ. From (10.35) we deduce ˆ π (ρ, θ, p), θ, p) − e = eˆ(ρ, θ, p) = h(ˆ

π ˆ (ρ, θ, p) . ρ

Now we take the partial derivative with respect to θ,







∂h ∂h ∂π ∂e 1 ∂π = + − ∂θ ρ ∂π θ ∂θ ρ ∂θ π ρ ∂θ ρ



 ∂h ∂π ∂h 1 = + − . ∂θ π ∂θ ρ ∂π θ ρ From the definitions of the specific heats and (10.68) we deduce



∂π θ ∂ρ cv = cπ + . ∂θ ρ ρ2 ∂θ π

(10.123)

(10.124)

(10.125)

Then, from (10.114) and (10.60) we get cv = cπ + which finishes the proof.

c2 ρα θ c2 α2 θ , (−ρα) = cπ − 2 γ ρ γ

(10.126)


10.5. Gibbs Free Energy

77

10.5 Gibbs Free Energy Definition 10.5.1. We denote by specific Gibbs free energy the scalar spatial field defined by G(x, t) := h(x, t) − θ(x, t)s(x, t). (10.127) From (1.72) we deduce the expression G(x, t) = ψ(x, t) +

π(x, t) ρ(x, t)

(10.128)

relating the Gibbs free energy to the Helmholtz free energy. Proposition 10.5.2. We have 

 ∂G = ν, ∂π θ   ∂G = −s. ∂θ

(10.129)

(10.130)

π

Proof. By using Proposition 10.4.8 and formula (10.127) we get           ∂G ∂h ∂s 1 θ ∂ρ θ ∂ρ 1 = −θ = + 2 − 2 = = ν. (10.131) ∂π ∂π ∂π ρ ρ ∂θ ρ ∂θ ρ θ

θ

θ

π

π

Similarly, by taking into account the definition of cπ (see Definition 10.4.7) and equality (10.66) we have       ∂h ∂s cπ ∂G = −s. (10.132) = −s+θ = cπ − s + θ ∂θ ∂θ ∂θ θ π

π

π




10.6 Statics of Fluids In this section we study the fluids at rest in a gravitational field.

Mechanical Equilibrium Firstly we have the following: Definition 10.6.1. A steady thermodynamic process is a mechanical equilibrium if v ≡ 0.

78

Chapter 10. Fluids For a fluid in mechanical equilibrium, the motion equation becomes gradπ = b.

(10.133)

Thus, a necessary condition for a thermodynamics process to be a mechanical equilibrium is that the body force be a gradient. Let us consider a mechanical equilibrium in a gravitational field, i.e., b = ρg with g = −g(x3 )e3 . Then the equilibrium equation (10.133) implies ∂π ∂π = =0 ∂x1 ∂x2

(10.134)

and, furthermore, ρ = ρ(x3 ) = −

1 ∂π . g(x3 ) ∂x3

(10.135)

Therefore, if a fluid is in mechanical equilibrium in a gravitational field, then pressure and density are functions of the altitude only. Furthermore, since temperature can be obtained from pressure and density through a constitutive law, it will depend only on x3 . Conversely, if temperature is different at points with the same altitude x3 , then the thermodynamic process cannot be a mechanical equilibrium, in other words, the fluid cannot be at rest. Definition 10.6.2. A steady thermodynamic process is a thermomechanical equilibrium if v ≡ 0 and the temperature is a constant field. It is obvious that any thermomechanical equilibrium is also a mechanical equilibrium. Moreover, since the heat flux q must be identically null (see (10.33)), we deduce from energy equation (10.118) that for any thermomechanical equilibrium the body heat, f , must be identically null.

10.7 The Boussinesq Approximation, Natural Convection In this section we obtain an approximate model for thermodynamic processes close to a mechanical equilibrium under gravity. For the sake of simplicity, we omit the explicit dependency on p of the response functions.

Linearization Let us denote the fields corresponding to the mechanical equilibrium with a 0 subscript. We have gradπ0 = ρ0 g, div q0 = f0 ,

(10.136) (10.137)

ρ0 = ρˆ(π0 , θ0 ).

(10.138)

10.7. The Boussinesq Approximation, Natural Convection

79

Let us consider a thermodynamic process close to the previous one in the following sense: div v = O(ε), α0 (θ − θ0 ) = O(ε), κ0 (π − π0 ) = O(ε),

(10.139)

where ε is a small parameter and α0 and κ0 denote, respectively, the coefficients of thermal expansion and isothermal compressibility at the initial equilibrium state. From (10.60) and (10.61) we obtain ρ = ρ0 − α0 ρ0 (θ − θ0 ) + κ0 ρ0 (π − π0 ) + o(θ − θ0 ) + o(π − π0 ).

(10.140)

Let us denote by πE the Eulerian fluctuation of pressure defined by πE (x, t) = π(x, t) − π0 (x).

(10.141)

gradπ = gradπ0 + gradπE = ρ0 g + gradπE .

(10.142)

Then

By using this equality and (10.140) the motion equation can be approximated by ρ0 v˙ + gradπE − div (2η0 D) = ρ0 [−α0 (θ − θ0 ) + κ0 (π − π0 )] g = ρ0 [α0 (θ − θ0 ) − κ0 (π − π0 )] g(x3 )e3 ,

(10.143)

where η0 = ηˆ(π0 , θ0 ). Moreover, according to assumptions (10.139), we can approximate the mass conservation equation by the incompressibility condition, div v = 0.

(10.144)

Equations (10.143) and (10.144) are similar to incompressible Navier-Stokes equations (13.18) and (13.2). Since (10.143) involves temperature not only on the righthand side but also in the viscosity coefficient, we also need to solve the energy equation. We use the form (10.79) which can be approximated in our case by ρ0 cπ0 θ˙ − α0 θ0 π˙ = 2η0 |D|2 − div(k0 gradθ) + f.

(10.145)

Usually, the terms involving the pressure and the viscous dissipation in this equation are small compared to the other terms, so they are neglected. Equations (10.143), (10.144) and (10.145) constitute the Boussinesq model for natural convection.

Chapter 11

Linearized Models for Fluids, Acoustics In this chapter we specialize the linearized thermoviscoelastic equations obtained in Chapter 8 to the case of fluids. These equations will be used, in particular, to model acoustic waves in fluids so they are the fundamental equations of sound propagation theory. Moreover, they also constitute a suitable model in cases where a linear approximation of the general models can be used in the neighbourhood of an initial state. Thus, their applicability is not merely to sound propagation. For instance they can deal with internal gravity waves in fluids. As in Chapter 8, we assume the fluid is at rest at the initial state. A similar linearization procedure could be done to obtain the approximate equations for small perturbed motion from a fluid flow (see [1]).

11.1 General Equations, Dissipative Acoustics In this section we obtain linearized models for small perturbations from an initial state in which the fluid is at rest (i.e., in mechanical equilibrium). We recall that, F for fluids, Tˆ (F, θ, p) = −ˆ π (ν, θ, p)I with ν = 1/ρ = det ρ0 . Proposition 11.1.1. The following equalities hold for a fluid: ∂ Tˆ ρ0 c20 1. (I, θ0 , p)(U ) = C(θ0 , p)(U ) = tr(U )I, ∂F γ0

(11.1)

2.

∂ Tˆ ρ0 α0 c20 (I, θ0 , p) = Y (θ0 , p) = − I, ∂θ γ0

(11.2)

3.

∂ Tˆ (I, s0 , p)(U ) = ρ0 c20 tr(U )I, ∂F

(11.3)

82

Chapter 11. Linearized Models for Fluids, Acoustics ρ0 α0 θ0 c20 ∂ Tˆ (I, s0 , p) = − I, ∂s cπ0

4.

(11.4)

ˆ∗ ˆ ∂q ∂q ˆ (ρ0 , y, w, p) (I, y, w, p)(∇u) = −ρ0 (ρ0 , y, w, p) Divu + Divu q ∂F ∂ρ ˆ ∂q ˆ (ρ0 , y, w, p) − −∇u q (ρ0 , y, w, p)∇ut w, (11.5) ∂w

5.

where y denotes either temperature or specific entropy. Proof.

1. By using the Lemma 10.1.3 we have ∂ Tˆ ∂π ˆ (F, θ, p)(U ) = −ν (ν, θ, p)(F −t · U )I ∂F ∂ν ˆ 1 ∂π (ρ, θ, p)(F −t · U )I. = ν ∂ρ

(11.6)

By taking F = I and θ = θ0 , in which case ν = ν0 = 1/ρ0 , and using (10.115) we deduce that the elasticity tensor C is given by C(θ0 , p)(U ) =

∂π ˆ ∂ Tˆ (I, θ0 , p)(U ) = ρ0 (ρ0 , θ0 , p) tr(U )I ∂F ∂ρ ρ0 c20 tr(U )I. = γ0

(11.7)

2. From (10.114) we deduce ∂π ˆ ραc2 ∂ Tˆ (F, θ, p) = − (ν, θ, p)I = − I. ∂θ ∂θ γ

(11.8)

The result follows by taking F = I and θ = θ0 . 3. From Lemma 10.1.3 we have ∂π ˆ ∂ Tˆ (F, s, p)(U ) = −ν (ν, s, p)(F −t · U )I ∂F ∂ν ˆ 1 ∂π (ρ, s, p)(F −t · U )I. = ν ∂ρ

(11.9)

By taking F = I, s = s0 and using Definition 10.4.13 we deduce ∂ Tˆ ∂π ˆ (I, s0 , p)(U ) = ρ0 (ρ0 , s0 , p) tr(U )I ∂F ∂ρ = ρ0 c20 tr(U )I.

(11.10)

11.1. General Equations, Dissipative Acoustics

83

4. From (10.107) we get ∂ Tˆ ∂π ˆ ραθc2 (F, s, p) = − (ν, s, p)I = − I. ∂s ∂s cπ

(11.11)

The result follows by taking F = I and s = s0 . 5. By using again Lemma 10.1.3 we have ˆ ˆ ˆ ∂q ∂q ∂q (F, y, w, p)(U ) = ν (ν, y, w, p)F −t · U = −ρ (ρ, y, w, p)F −t · U. ∂F ∂ν ∂ρ (11.12) Then equality (11.5) follows from Proposition 8.2.1 by using the previous equality for F = I and U = ∇u .
Remark 11.1.2. We recall that fluids are isotropic Coleman-Noll materials so (8.56)–(8.59) hold. In particular, (11.1) shows that the Lam´e coefficients for a fluid are µ(θ0 , p) = 0,

λ(θ0 , p) =

ρ0 (p)c20 (p) . γ0 (p)

(11.13)

Moreover, by using the previous equalities, (8.59) and (11.2) yield χ(θ0 , p) =

α0 (p) . 3

(11.14)

Now we can obtain approximations for the Piola-Kirchhoff stress tensor and the heat flux vector in a fluid in terms of either absolute temperature or specific entropy. Let us assume that the thermodynamic process is a small perturbation of the reference state, more precisely, ∇u = O(ε),

F˙ = ∇u˙ = O(ε),

θ − θ0 = O(ε),

∇θ − ∇θ0 = O(ε),

(11.15)

ε being a small parameter. As an inmediate consequence of Corollaries 8.1.5 and 8.2.2 and of the above Proposition, we can get approximations of the Piola-Kirchhoff stress tensor of the first kind: Corollary 11.1.3. We have 2 ˆ F˙ , θ, p) = −π0 I − Divuπ0 I + π0 ∇ut + ρ0 c0 DivuI S(F, 1. γ0 2 ρ0 α0 c0 ˙ + o(ε). (θ − θ0 )I + ˆl(ρ0 , θ0 , p)(∇u) − γ0 (11.16)

84 2.

Chapter 11. Linearized Models for Fluids, Acoustics ˆ F˙ , s, p) = −π0 I − Divuπ0 I + π0 ∇ut + ρ0 c20 DivuI S(F, ρ0 α0 θ0 c20 ˙ + o(ε). (s − s0 )I + ˆl(ρ0 , s0 , p)(∇u) − cπ0 (11.17)


Corollary 11.1.4. For a fluid we have ˆ ∂q ˆ (ρ0 , y0 , w0 , p) (ρ0 , y0 , w0 , p) Divu+ Divu q ∂ρ ˆ ˆ ∂q ∂q ˆ (ρ0 , y0 , w0 , p) − (ρ0 , y0 , w0 , p)∇ut w0 + (ρ0 , y0 , w0 , p)(y − y0 ) −∇u q ∂w ∂y ˆ ∂q + (ρ0 , y0 , w0 , p)(w − w0 ) + o(∇u) + o(y − y0 ) + o(w − w0 ). ∂w (11.18)

ˆ ∗ (F, y, w, p) = q ˆ (ρ0 , y0 , w0 , p)−ρ0 q

By replacing (11.16) or (11.17) in (7.4) we get linearized motion equations for a fluid in terms of temperature or specific entropy, respectively:   ρ0 c20 ρ0 α0 c20 t ¨ = Div −π0 I − Divuπ0 I + π0 ∇u + DivuI − (θ − θ0 )I ρ0 u γ0 γ0  ˙ + b∗ , + Div ˆl(ρ0 , θ0 , p)(∇u) (11.19)   ρ0 α0 θ0 c20 t 2 ¨ = Div −π0 I − Divuπ0 I + π0 ∇u + ρ0 c0 DivuI − (s − s0 )I ρ0 u cπ0  ˙ + b∗ . + Div ˆl(ρ0 , s0 , p)(∇u) (11.20) By using the equalities of vector calculus ∇(ϕψ) = ϕ∇ψ + ψ∇ϕ, Div(ϕA) = A∇ϕ + ϕ DivA,

(11.21) (11.22)

Div(a ⊗ c) = (∇a)c + a Divc,

(11.23)

∇(a · c) = (∇a) c + (∇c) a,

(11.24)

t

t

we easily deduce Div(− Divuπ0 I + π0 ∇ut ) = − Divu∇π0 + ∇ut ∇π0 = Div(−∇π0 ⊗ u + ∇π0 · uI).

(11.25)

These equalities allow us to write the linearized motion equations (11.19) and (11.20) in another form. For instance, in terms of temperature we have

ρ0 c20 t ¨ = −∇π0 − Divu∇π0 + ∇u ∇π0 + ∇ Divu ρ0 u γ0

 ρ0 α0 c20 ˙ + b∗ , −∇ (θ − θ0 ) + Div ˆl(ρ0 , θ0 , p)(∇u) (11.26) γ0

11.2. The Isentropic Case, Non-Dissipative Acoustics

85

as well as the conservative form

ρ0 c20 ¨ = −∇π0 + Div(−∇π0 ⊗ u+∇π0 · uI)+∇ ρ0 u Divu γ0

 2 ρ0 α0 c0 ˙ +b∗ . −∇ (θ − θ0 ) + Div ˆl(ρ0 , θ0 , p)(∇u) γ0

(11.27)

If the reference state is spatially homogeneous, then the above equation becomes ¨= ρ0 u

ρ0 c20 ρ0 α0 c20 ∇ Divu − ∇θ + η0 ∆u˙ + (η0 + ξ0 )∇ Divu˙ + b∗ , γ0 γ0 (11.28)

where we have used a similar result to (10.29) by replacing specific entropy with temperature. Now we deal with the energy equation. By using (11.2) and Corollary 11.1.4, equation (8.48) becomes  ρ α c2 ρ0 cv0 θ˙ = −θ0 0 γ00 0 Divu˙ − Div q0 − ρ0 ∂∂ρqˆ (ρ0 , θ0 , ∇θ0 , p) Divu ˆ ∂q (ρ0 , θ0 , ∇θ0 , p)∇ut ∇θ0 + ∂w  ˆ ˆ ∂q ∂q (ρ0 , θ0 , ∇θ0 , p)(θ − θ0 ) + (ρ0 , θ0 , ∇θ0 , p)(∇θ − ∇θ0 ) ∂θ ∂w +f∗ , (11.29) where cv0 =: cˆv (ρ0 , θ0 , p). If the reference state is spatially homogeneous, this equation simplifies to + Divu q0 − ∇u q0 −

ρ0 cv0 θ˙ = −θ0

ρ0 α0 c20 Divu˙ + k0 ∆θ + f∗ , γ0

(11.30)

ˆ 0 , θ0 , 0, p). where we have used (10.33) and k0 := k(ρ Equations (11.28) and (11.30) constitute a suitable model for dissipative acoustics.

11.2 The Isentropic Case, Non-Dissipative Acoustics Sometimes it turns out that the thermodynamic process can be considered as isentropic. This is, for instance, the usual assumption in the fundamental case of standard (non-dissipative) acoustics. It is obvious that, in such a situation, it is more convenient to write the motion and energy equations in terms of specific enthalpy rather than temperature.

86

Chapter 11. Linearized Models for Fluids, Acoustics By using (11.25), equation (11.20) can be written as   ¨ = −∇π0 − Divu∇π0 + ∇ut ∇π0 + ∇ ρ0 c20 Divu ρ0 u

  ρ0 α0 θ0 c20 ˙ + b∗ , (s − s0 ) + Div ˆl(ρ0 , s0 , p)(∇u) −∇ cπ0

(11.31)

as well as in conservative form:   ¨ = −∇π0 + Div (−∇π0 ⊗ u + ∇π0 · uI) + ∇ ρ0 c20 Divu ρ0 u

  ρ0 α0 θ0 c20 ˙ + b∗ . −∇ (s − s0 ) + Div ˆl(ρ0 , s0 , p)(∇u) cπ0

(11.32)

If the thermodynamic process is isentropic, then s(p, t) = s0 (p) (see (3.74)) and (11.31) becomes   ¨ = −∇π0 − Divu∇π0 + ∇ut ∇π0 + ∇ ρ0 c20 Divu ρ0 u   ˙ + b∗ . + Div ˆl(ρ0 , s0 , p)(∇u) (11.33) Remark 11.2.1. From Proposition 3.2.19 we easily deduce that any Eulerian and adiabatic thermodynamic process is isentropic. However, these conditions are not strictly necessary for a thermodynamic process to be isentropic as far as equality (3.75) holds. When b∗ can be approximated by b0 = ∇π0 (see (8.1)) and the viscous effects can be neglected, (11.33) becomes   ¨ = − Divu∇π0 + ∇ut ∇π0 + ∇ ρ0 c20 Divu . ρ0 u (11.34) This is the Galbrun equation. In standard acoustics, the first two terms on the right-hand side are also neglected and then (11.34) becomes   ¨ = ∇ ρ0 c20 Divu . (11.35) ρ0 u Let us define the Lagrangian fluctuation of pressure by πL (p, t) = π(X(p, t), t) − π0 (p).

(11.36)

Then πL (p, t) can be approximated by −ρ0 c20 Divu(p, t). Indeed, ∂π ˆ (I, s0 (p), p)(∇u(p, t)) ∂F +o(∇u(p, t)) = π0 (p) − ρ0 (p)c20 (p) Divu(p, t) + o(∇u(p, t)). (11.37)

π(X(p, t), t) = π ˆ (F (p, t), s0 (p), p) = π0 (p) +

(see the Proof of point 3 in Proposition 11.1.1). By using this equality and (11.36), equation (11.35) can be written as ¨ = −∇πL . ρ0 u

(11.38)

11.3. Linearized Models under Gravity

87

Now we apply the divergence operator after dividing by ρ0 . We get 1 ∇πL ) ρ0

(11.39)

1 ∇πL ) = 0. ρ0

(11.40)

Div¨ u = − Div( and using (11.37) π ¨L − ρ0 c20 Div(

The above equation is known as the Pekeris equation. Finally, if the initial state is spatially homogeneous, ρ0 and c20 are constant and this equation becomes the standard wave equation π ¨L − c20 ∆πL = 0.

(11.41)

11.3 Linearized Models under Gravity Let us go back to equation (11.33) and consider the case where the body force is the gravity force, that is b(x, t) = ρ(x, t)g(x), g being the intensity of the gravity field. Then b0 (p) = ρ0 (p)g(p) and, according to Remark 8.1.9, b∗ can be written as b∗ (p, t) = b0 (p) + ρ0 (p)∇g(p)u(p, t) + o(u(p, t)).

(11.42)

Moreover, the initial state satisfies the equations ∇π0 = ρ0 g, − Divˆ q(ρ0 , θ0 , ∇θ0 , p) + f0 = 0,

(11.43) (11.44)

π0 = π ˆ (ρ0 , θ0 , p),

(11.45)

which can be solved to get ρ0 , θ0 and π0 . In this case, equation (11.27) becomes

ρ0 c20 ¨ = −ρ0 g − Div(ρ0 g ⊗ u) + ∇(ρ0 g · u) + ∇ ρ0 u Divu γ0

  ρ0 α0 c20 ˙ + b∗ . −∇ (θ − θ0 ) + Div ˆl(ρ0 , θ0 , p)(∇u) (11.46) γ0 By using the vector calculus equalities (11.23) and ∇(ϕa) = ϕ∇a + a ⊗ ∇ϕ,

(11.47)

we deduce Div(ρ0 g ⊗ u)

= ∇(ρ0 g)u + ρ0 g Divu = ρ0 ∇gu + (g ⊗ ∇ρ0 )u + ρ0 g Divu = ρ0 ∇gu + u · ∇ρ0 g + ρ0 Divug.

(11.48)

88

Chapter 11. Linearized Models for Fluids, Acoustics

Let us assume u(p, t) = O(ε). By using (11.42) and (11.48) in (11.46), we get

ρ0 c20 ¨ = −(u · ∇ρ0 + ρ0 Divu)g + ∇ ρ0 g · u + Divu ρ0 u γ0

  2 ρ0 α0 c0 ˙ . −∇ (θ − θ0 ) + Div ˆl(ρ0 , θ0 , p)(∇u) (11.49) γ0 Moreover ρ0 ρ0 (u · ∇ρ0 + ρ0 Divu)g = (u · ∇ρ0 − 2 u · g)g + (ρ0 Divu + 2 g · u)g c c0

0 ρ0 ρ0 (11.50) = g ⊗ (∇ρ0 − 2 g) u + (ρ0 Divu + 2 g · u)g. c0 c0 The tensor B=

1 ρ0

ρ0 g ⊗ (∇ρ0 − 2 g) , c0

(11.51)

has dimension inverse of a square of time. It is called the V¨ais¨al¨a-Brunt tensor. By replacing (11.50) in (11.49) we obtain

ρ0 ρ0 c20 ¨ = −ρ0 Bu − (ρ0 Divu + 2 g · u)g + ∇ ρ0 g · u + Divu ρ0 u c0 γ0

  ρ0 α0 c20 ˙ . −∇ (θ − θ0 ) + Div ˆl(ρ0 , θ0 , p)(∇u) (11.52) γ0 Let us now look at some considerations about the V¨ ais¨ al¨ a-Brunt tensor. From the state equation, π = π ˆ (ρ, s), we have ∇π0 =

∂π ˆ ∂π ˆ (ρ0 , s0 )∇ρ0 + (ρ0 , s0 )∇s0 , ∂ρ ∂s

(11.53)

and then ρ0 g = c20 ∇ρ0 +

c20 ρ0 α0 θ0 ∇s0 , cπ0

(11.54)

where we have used (10.94) and (10.107). From (11.54) we deduce ∇ρ0 −

ρ0 ρ0 α0 θ0 g=− ∇s0 , 2 c0 cπ0

(11.55)

so the V¨ ais¨ al¨ a-Brunt tensor is also given by B=−

α0 θ0 (g ⊗ ∇s0 ) . cπ0

(11.56)

11.3. Linearized Models under Gravity

89

This shows that, if the specific entropy at the reference state is constant, then the V¨ ais¨ al¨ a-Brunt tensor is null. Let us suppose g(p) = −g(p3 )e3 . From (11.43) we deduce that π0 only depends on the third coordinate p3 . Moreover, this equality yields ρ0 (p) = −

1 ∂π0 (p) g(p3 ) ∂p3

(11.57)

and hence ρ0 only depends on p3 . Consequently ∇ρ0 =

∂ρ0 e3 . ∂p3

(11.58)

From this result and (11.54) we get a similar property for the specific entropy, namely, ∇s0 =

∂s0 e3 . ∂p3

(11.59)

Then (11.51) and (11.56) become

1 ∂ρ0 ρ0 α0 θ0 g ∂s0 B=− g + 2 g e3 ⊗ e3 = e3 ⊗ e3 . ρ0 ∂p3 c0 cπ0 ∂p3

(11.60)

The number N defined by 1 N =− g ρ0 2



∂ρ0 ρ0 + 2g ∂p3 c0

=

α0 θ0 g ∂s0 cπ0 ∂p3

(11.61)

is called V¨ais¨al¨a-Brunt frequency. We notice that it is a function of p3 only. Moreover, N will be real or imaginary depending on whether the specific entropy, s0 , is increasing or decreasing with altitude, respectively. Now the first term on the right-hand side of (11.52), ρ0 Bu, can also be written as ρ0 Bu = ρ0 N 2 u · e3 e3 = ρ0 N 2 u3 e3 .

(11.62)

The Isentropic Case Let us consider the particular case where the process is isentropic. Firstly, similar computations to those made above allow us to transform equation (11.32) into the following one:   ρ0 ¨ = −ρ0 Bu − (ρ0 Divu + 2 g · u)g + ∇ ρ0 g · u + ρ0 c20 Divu ρ0 u c0

  ρ0 α0 θ0 c20 ˙ . −∇ (s − s0 ) + Div ˆl(ρ0 , s0 , p)(∇u) cπ0

(11.63)

90

Chapter 11. Linearized Models for Fluids, Acoustics

  ρ α θ c2 If the thermodynamic process is isentropic, then term ∇ 0 c0π00 0 (s − s0 ) disappears. Let us recall that the Eulerian fluctuation of pressure is πE (p, t) = π(p, t) − π0 (p).

(11.64)

πE (p, t) = π(p, t) − π(X(p, t), t) + π(X(p, t), t) − π0 (p) = −∇π(p, t) · u(p, t) + o(u(p, t) + πL (p, t)

(11.65)

We have

and then a useful approximation is given by (see (11.37)) πE = −∇π0 · u − ρ0 c20 Divu = −ρ0 g · u − ρ0 c20 Divu.

(11.66)

Using this equality in (11.63) we obtain the following mixed formulation (in terms of displacement and pressure): ¨ = −ρ0 Bu + ρ0 u

  1 ˙ , πE g − ∇πE + Div ˆl(ρ0 , s0 , p)(∇u) 2 c0

(11.67)

which, if g = −ge3 , can also be written as ¨ = −ρ0 N 2 u · e3 e3 + ρ0 u

  1 ˙ . πE g − ∇πE + Div ˆl(ρ0 , s0 , p)(∇u) 2 c0

(11.68)

We end this section by obtaining an energy conservation principle. For this purpose we make the scalar product of (11.68) with u˙ and integrate in P, a part of the body. We get 1 d 2 dt



1 1 d 2 2 ˙ dVp = − ˙ ρ0 |u| ρ0 N |u · e3 | dVp + 2 πE g · u dVp 2 dt c P P 0

P   ˙ · u˙ dVp . − ∇πE · u˙ dVp + Div ˆl(ρ0 , s0 , p)(∇u) (11.69)




2

P

P

By replacing the expression (11.66) for πE we get
1 d ˙ dVp = − ρ0 |u| ρ0 N 2 |u · e3 |2 dVp 2 dt P P

1 2 ˙ − (ρ g · u + ρ c Divu)g · u dV + ∇(ρ0 g · u + ρ0 c20 Divu) · u˙ dVp 0 0 0 p 2 c P 0 P
  ˙ · u˙ dVp + Div ˆl(ρ0 , s0 , p)(∇u) (11.70) 1 d 2 dt




2

P

11.3. Linearized Models under Gravity

91

and using Green’s formulas in the two rightmost terms we obtain


1 d ρ0 1 d 1 d ˙ 2 dVp = − ρ0 |u| ρ0 N 2 |u · e3 |2 dVp − |g · u|2 dVp 2 dt P 2 dt P 2 dt P c20


− ρ0 Divug · u˙ dVp + ρ0 g · u u˙ · n dAp − ρ0 g · u Divu˙ dVp P ∂P

P ˙ dAp − + ρ0 c20 Divuu.n ρ0 c20 Divu Divu˙ dVp ∂P P

ˆl(ρ0 , s0 , p)(∇u) ˆl(ρ0 , s0 , p)(∇u) ˙ n · u˙ dAp − ˙ · ∇u˙ dVp . + (11.71) P

∂P

By using again (11.66) we easily deduce 
 1 d 1 2 ˙ 2 + ρ0 N 2 |u · e3 |2 + |π | ρ0 |u| dVp E 2 dt P ρ0 c20

ˆl(ρ0 , s0 , p)(∇u) ˙ · ∇u˙ dVp = − + πE u˙ · n dAp P ∂P
ˆl(ρ0 , s0 , p)(∇u) ˙ n · u˙ dAp . +

(11.72)

∂P

The sum W =

  1 1 2 ˙ 2 + ρ0 N 2 |u · e3 |2 + |π | ρ0 |u| E 2 ρ0 c20

(11.73)

represents the density of mechanical energy, while the vector field I = πE u˙ is called acoustic intensity. The physical meaning of I is the following: the inner product I·n is the rate at which the acoustic energy is transported across a surface element normal to n, per unit area. If viscous effects are neglected (l ≡ 0), by using the Localization Theorem in (11.72) we can deduce the following energy conservation equation: ∂W = − DivI. (11.74) ∂t

Chapter 12

Perfect Gases In this chapter we study particular Coleman-Noll fluids: the so-called perfect gases.

12.1 Definition, General Properties Definition 12.1.1. A perfect gas is a fluid for which the following state equation holds: π = ρRθ. (12.1) More precisely, π ˆ (ρ, θ, p) = ρR(p)θ, where R has a specific value for each perfect R gas. Actually, R(p) = M(p) , where R is a universal constant (R = 8314 J/(kmol K)) and M (p) is the molecular mass of the gas at material point p. Proposition 12.1.2. The specific internal energy of a perfect gas only depends on temperature. Therefore, the same is true for the specific heat at constant volume. Proof. By using (12.1) in (10.83) we get

∂e 1 = 2 [ρRθ − θρR] = 0. ∂ρ θ ρ The second part follows immediately from the definition of cv .

(12.2)


A similar result is valid for the specific enthalpy: Proposition 12.1.3. The specific enthalpy of a perfect gas only depends on temperature. Therefore, the same is true for the specific heat at constant pressure. Proof. It is a consequence of the definition of h. We have h=e+

π = eˆ(θ, p) + R(p)θ. ρ

(12.3)


94

Chapter 12. Perfect Gases

Corollary 12.1.4. For a perfect gas, the following relation between the specific heats, called Mayer equation, holds: cˆπ (θ, p) = cˆv (θ, p) + R(p).

(12.4)

Proof. It is enough to take the partial derivative of (12.3) with respect to θ.




Proposition 12.1.5. The sound speed of a perfect gas only depends on temperature. More precisely we have:  (12.5) c = γRθ, where γ is the ratio of specific heats, γ =

cπ cv .

Proof. From Proposition 10.4.14 and equality (12.1) we deduce

cπ ∂π cπ 2 c = = Rθ = γRθ. cv ∂ρ θ cv

(12.6)


The next Proposition easily follows from (12.1). Proposition 12.1.6. For a perfect gas, the coefficients of thermal expansion α and isothermal compressibility κ are given by 1 , θ 1 κ= . π

(12.7)

α=

(12.8)

Corollary 12.1.7. For a perfect gas we have the following expressions for internal energy and enthalpy:
θ 1. eˆ(θ, p) = eˆ(θ0 , p) + cˆv (r, p) dr, (12.9) θ0

2.




θ

ˆ p) = h(θ ˆ 0 , p) + h(θ,

cˆπ (r, p) dr.

(12.10)

θ0

ˆ 0 , p) = eˆ(θ0 , p) + R(p)θ0 . Moreover, h(θ

12.2 Entropy and Free Energy Now we obtain an expression for the entropy of a perfect gas. For this we notice that (10.66) and (10.67) yield

∂s cπ (12.11) = , ∂θ π θ

  ∂s R 1 1 1 =− . (12.12) = − =− ∂π θ θ ρ ρθ π

12.2. Entropy and Free Energy

95

Therefore


θ

sˆ(θ, π, p) = sˆ(θ0 , π0 , p) + θ0

cˆπ (r, p) dr − r




π

R(p) dr. r

(12.13)

cv , θ

(12.14)

1 R ρR = − . ρ2 ρ

(12.15)

πo

Similarly, from (10.81) and (10.82) we have

∂s ∂ρ

θ

1 =− 2 ρ



∂π ∂θ

=− ρ

∂s ∂θ

= ρ

Then we have


θ

sˆ(ρ, θ, p) = sˆ(ρ0 , θ0 , p) + θ0

cˆv (r, p) dr − r




ρ

ρo

R(p) dr. r

(12.16)

Definition 12.2.1. A perfect gas is called calorically perfect if cˆv , and hence cˆπ , do not depend on θ. Monoatomic gases are examples of calorically perfect gases. In this case cˆπ (p) = 52 R(p) and cˆv (p) = 32 R(p). Then, for monoatomic gases γ = 53 . Diatomic gases can be considered as calorically perfect at low temperature. Actually, the specific heat at constant pressure of a diatomic gas is given by cˆπ (θ, p) =

 2 θν (p)/(2θ) 7 R(p) + R(p) , 2 sinh(θν (p)/(2θ))

(12.17)

where θν is the so-called characteristic temperature of vibration. For O2 , θν = 2230 K while, for N2 , θν = 3340 K. For low temperature the second term in (12.17) can be neglected. Hence cπ can be approximated by 72 R and, according to (12.4), cv = 52 R and γ = 1.4. In the case of calorically perfect gases, expression (12.16) becomes sˆ(ρ, θ, p)

θ ρ = sˆ(ρ0 , θ0 , p) + cˆv (p) ln( ) − R(p) ln( ) θ0 ρ0

(θ/θ0 )cˆv (p) = sˆ(ρ0 , θ0 , p) + ln (ρ/ρ0 )R(p)

((θρ)/(θ0 ρ0 ))cˆv (p) = sˆ(ρ0 , θ0 , p) + ln (ρ/ρ0 )R(p)+ˆcv (p)   (θρ)/(θ0 ρ0 ) = sˆ(ρ0 , θ0 , p) + cˆv (p) ln R(p)+ˆ cv (p) (ρ/ρ0 ) cˆv (p)

(θρ)/(θ0 ρ0 ) = sˆ(ρ0 , θ0 , p) + cˆv (p) ln , (ρ/ρ0 )γ(p)

(12.18)

96

Chapter 12. Perfect Gases

because cˆπ (p) − cˆv (p) + cˆv (p) cˆπ (p) R(p) + cˆv (p) = = = γ(p). cˆv (p) cˆv (p) cˆv (p)

(12.19)

From equation (12.18) we can obtain the pressure as a function of ρ and s; indeed, we have



(θρ)/(θ0 ρ0 ) sˆ(ρ, θ, p) − sˆ(ρ0 , θ0 , p) (π/π0 ) = ln = ln . (12.20) cˆv (p) (ρ/ρ0 )γ(p) (ρ/ρ0 )γ(p) Then s ˆ(ρ,θ,p)−ˆ s(ρ0 ,θ0 ,p) (π/π0 ) c ˆv (p) =e γ(p) (ρ/ρ0 )

(12.21)

and finally π0

ργ(p) e γ(p)

π=

s ˆ(ρ,θ,p)−ˆ s (ρ0 ,θ0 ,p) c ˆv (p)

.

(12.22)

ρ0

Remark 12.2.2. Any inviscid (i.e., l ≡ 0) calorically perfect gas undergoing an isentropic thermodynamic process behaves like an elastic fluid in the sense of [5]. Indeed, since each material point conserves entropy during the process, we deduce from (12.22) π0 π = γ(p) ργ(p) = D(p)ργ(p) (12.23) ρ0 with D(p) =

π0 γ(p) . ρ0

Actually, in [5], elastic fluids having a constitutive law as

(12.23) are called ideal gases (see exercise 19.1 of this reference). Finally we obtain an expression of the specific Gibbs free energy for a perfect gas. From Proposition 10.5.2 we deduce





θ

ˆ π) = G(θ ˆ 0 , π0 ) − G(θ,

π

sˆ(r, π0 ) dr + θ0

νˆ(θ, r) dr.

(12.24)

π0

By using (12.13) we obtain ˆ π) = G(θ ˆ 0 , π0 ) − G(θ,

θ θ0

 sˆ(θ0 , π0 ) +

ˆ 0 , π0 ) − sˆ(θ0 , π0 )(θ − θ0 ) − = G(θ

θ θ0



z θ0

z θ0

 cˆπ (r) r

dr dz +

 cˆπ (r) r

dr dz +

π

π

Rθ π0 r

π0

νˆ(θ, r) dr

dr.

(12.25)

The previous expression can also be written in the form ˆ π) = G ˆ 0 (θ) + Rθ ln ( π ), G(θ, π0

(12.26)

12.3. The Compressible Navier-Stokes Equations

97

ˆ π0 ) is the specific Gibbs free energy of formation at tempeˆ 0 (θ) = G(θ, where G rature θ and at reference pressure π0 . We have 
θ 
z c ˆ (r) π 0 ˆ (θ) = G(θ ˆ 0 , π0 ) − sˆ(θ0 , π0 )(θ − θ0 ) − G dr dz. (12.27) r θ0 θ0 ˆ 0 (θ) is called standard specific Usually π0 is taken to be 1 atm in which case G Gibbs free energy at temperature θ.

12.3 The Compressible Navier-Stokes Equations Now we write the equations to be satisfied for any thermodynamic process of a perfect gas. Firstly, in the energy equation, we write the internal energy and the enthalpy in terms of the temperature. To do that, we recall that for a perfect gas ˙ t), e(x, ˙ t) = cˆv (θ(x, t), p)θ(x, ˙ ˙ t). h(x, t) = cˆπ (θ(x, t), p)θ(x,

(12.28) (12.29)

We use these equalities in (1.54) and in (10.36). They yield, respectively, ρˆ cv (θ, p)θ˙ = T · D − div q + f, ρcˆπ (θ, p)θ˙ − π˙ = l(D) · D − div q + f.

(12.30) (12.31)

By using (10.19), we can express viscosity coefficients η and ξ in terms of θ rather ˆ given by (10.33). Thus we get the than s. Moreover, we use the expression for q following equations: ρ˙ + ρ div v = 0,   ˆ θ, p) div vI = b, ρv˙ + gradπ − div 2ˆ η (ν, θ, p)D + ξ(ν,

(12.32)

π = ρRθ,

(12.34)

ˆ θ, p)(div v)2 η(ν, θ, p)|D|2 + ξ(ν, ρˆ cv (θ, p)θ˙ = −π div v + 2ˆ ˆ θ, | gradθ|2 , p) gradθ) + f + div(k(ν,

(12.35)

ˆ θ, p)(div v)2 η (ν, θ, p)|D|2 + ξ(ν, ρcˆπ (θ, p)θ˙ − π˙ = 2ˆ ˆ θ, | gradθ|2 , p) gradθ) + f. + div(k(ν,

(12.36)

(12.33)

and as energy equation, either

or

This system is called the compressible Navier-Stokes equations. Moreover, it is easy to obtain the following expression for the dissipation rate η Φ = | gradv + gradvt |2 + ξ(div v)2 + kθ−1 | gradθ|2 . (12.37) 2

98

Chapter 12. Perfect Gases

12.4 The Compressible Euler Equations This system arises when we neglect the viscous and heat conduction effects in the compressible Navier-Stokes equations. We get ρ˙ + ρ div v = 0, ρv˙ + gradπ = b,

(12.38) (12.39)

π = ρRθ,

(12.40)

and either ρˆ cv (θ, p)θ˙ = −π div v + f

(12.41)

ρcˆπ (θ, p)θ˙ − π˙ = f.

(12.42)

or

Sometimes, for numerical purposes it is more convenient to write this system in a different equivalent way. Let us go back to the energy equation written in terms of the specific total energy E. Under the above assumptions, equation (1.48) becomes ρE˙ + div(πv) = 0, (12.43) where we have supposed b = 0 and f = 0 for simplicity. Then we replace (12.41) or (12.42) by (12.43) and the algebraic equation, E=

1 2 |v| + eˆ(θ, p). 2

(12.44)

We recall that, since we are dealing with perfect gases, function eˆ(., p) is nothing but a primitive of cˆv (·, p). Equation (12.44) allows us to obtain θ once E has been computed. In order to use finite volume methods, which are the most used discretization techniques for solving the compressible Euler equations, we must rewrite them in a conservative form. By using (1.21), (1.29) and (1.56) we get ρ
+ div(ρv) = 0, (ρv)
+ div(ρv ⊗ v + πI) = 0,

(12.45) (12.46)

(ρE)
+ div ((ρE + π)v) = 0, π = ρRθ, 1 E = |v|2 + eˆ(θ, p). 2

(12.47) (12.48) (12.49)

Now ρE + π in (12.47) can be written in terms of the specific stagnation enthalpy. Indeed, by using Definitions 10.3.2 and 10.3.6 we get ρE + π = ρ

|v|2 |v|2 + ρe + π = ρ + ρh = ρH. 2 2

(12.50)

12.4. The Compressible Euler Equations

99

If the thermodynamic process is stationary the above equations yield

div(ρv) = 0, div(ρv ⊗ v + πI) = 0,

(12.51) (12.52)

div (ρHv) = 0,

(12.53)

π = ρRθ, 1 2 π H = |v| + eˆ(θ, p) + . 2 ρ

(12.54)

div (ρHv) = H div (ρv) + ρv · gradH = ρv · gradH,

(12.56)

(12.55)

Moreover

by using (12.51). Thus equation (12.53) yields H˙ = H
+ v · gradH = 0 and hence the thermodynamic process is homentropic. This means that H is constant along streamlines. In particular, if all streamlines arise from the same uniform flow, then H will be given and constant everywhere, in which case (12.53) can be eliminated. Moreover if the gas is calorically perfect, then e = cv θ and (12.54) yields π = ρR

e . cv

(12.57)

By using the Mayer equation (12.4), R = cπ − cv is also constant and the state law becomes cπ − cv e = ρ(γ − 1)e. (12.58) π=ρ cv Replacing (12.48) or (12.54) by (12.58) and eˆ(θ, ρ) by e in (12.49) or (12.55) allows us to eliminate temperature, which can be computed afterwards from e, by solving the algebraic equation e = eˆ(θ, p).

Chapter 13

Incompressible Fluids In this chapter we study material bodies which are not Coleman-Noll materials in the sense of the Definition 3.2.1. They are called incompressible fluids.

13.1 Isochoric Processes They are volume preserving thermodynamic processes: Definition 13.1.1. A thermodynamic process is said to be isochoric if, for each t and part P,
d dVx = 0. (13.1) dt Pt It is easy to prove that a thermodynamic process is isochoric if and only if div v = 0

(13.2)

(see, for instance, [5]).

13.2 Newtonian Fluids Viscous fluids are called Newtonian fluids when the viscous stress depends linearly on the velocity gradient. Let us denote by Lin0 (respectively by Sym0 ) the space of tensors (respectively, symmetric tensors) with null trace. Definition 13.2.1. An incompressible Newtonian fluid is a material body, the constitutive class of which consists of all isochoric thermodynamic processes such that

102

Chapter 13. Incompressible Fluids

there exist mappings ˆl : R × B → L(Lin0 , Sym0 ),

(13.3)

eˆ : R × B → R,

(13.4)

θˆ : R × B → R ,

(13.5)

ˆ :R×V ×B →V q

(13.6)

+

“smooth enough” and satisfying T d (x, t) = ˆl(s(x, t), p)(L),

(13.7)

e(x, t) = eˆ(s(x, t), p), ˆ θ(x, t) = θ(s(x, t), p),

(13.8) (13.9)

ˆ (s(x, t), gradθ(x, t), p), q(x, t) = q

(13.10)

with x = X(p, t), where T d is the deviatoric part of T , i.e., Td = T −

1 tr(T )I. 3

(13.11)

Analogous to Theorem 3.2.4 we have the following Theorem 13.2.2. Let us consider an incompressible fluid with constitutive class C. We make the following assumption: (H1∗ ) There exists a function sˆ : R+ × B → R, “smooth enough”, such that ˆ p) ∀θ ∈ R+ ∀s ∈ R ∀p ∈ B. s = sˆ(θ, p) ⇐⇒ θ = θ(s,

(13.12)

Then all elements in C satisfy the second law of thermodynamics if and only if e ˆ p) = ∂ˆ (s, p), θ(s, ∂s ˆl(s, p)(L) · L ≥ 0 (dissipation inequality), ˆ (s, w, p) · w ≤ 0, q

(13.13) (13.14) (13.15)

for all s ∈ R, p ∈ B, L ∈ Lin0 and w ∈ V. Proof. It is a straightforward adaptation of that of Theorem 3.2.4.




Definition 13.2.3. The scalar field π defined by 1 π(x, t) = − tr(T (x, t)) 3

(13.16)

is called the pressure of the thermodynamic process. Assuming the material-frame indifference principle, we can prove an analogous result to Proposition 10.2.3:

13.3. Ideal Fluids

103

Proposition 13.2.4. There exists a function ηˆ : R×B → R, called dynamic viscosity of the fluid, such that ˆl(s, p)(D) = 2ˆ η(s, p)D. (13.17)


Proof. It follows from the fact that tr(D) = div v.

Now we write the equations to be satisfied by the thermodynamic processes of incompressible Newtonian fluids. First of all, we use (13.12) to replace s by θ in the constitutive laws. Then, we notice that the mass conservation equation (1.20) yields ρ(x, t) = ρ0 (p), with x = X(p, t), because det F = 1. Furthermore we have (13.2). Hence, the motion equation becomes ρ0 (p)v˙ + gradπ − div {2ˆ η(θ, p)D} = b.

(13.18)

Concerning the energy equation, we use the fact that e = eˆ(θ, p) to obtain ˙ t), e(x, ˙ t) = cˆv (θ, p)θ(x,

(13.19)

with x = X(p, t). ˆ | gradθ|2 , p) gradθ Finally we assume q obeys the Fourier’s law, i.e., q = −k(θ, and write the energy equation as follows: ˆ | gradθ|2 , p) gradθ) + f, ρ0 (p)ˆ cv (θ, p)θ˙ = 2ˆ η (θ, p)|D|2 + div(k(θ,

(13.20)

because div v = 0. Remark 13.2.5. Unlike general Coleman-Noll materials, equations (13.2) and (13.18) (which are called the non-homogeneous incompressible Navier-Stokes equations) can be solved independently of the energy equation, as far as ηˆ does not depend on temperature.

13.3 Ideal Fluids Now we introduce another material body, the constitutive class of which is a subclass of the constitutive class of the incompressible Newtonian fluid given in Definition 13.2.1 above. Definition 13.3.1. An ideal fluid is an incompressible fluid characterized by the following additional requirements: 1. The mass distribution ρ0 is a constant function. 2. All thermodynamic processes in its constitutive class are Eulerian and adiabatic.

104

Chapter 13. Incompressible Fluids

Consequently, the partial differential equations satisfied by any thermodynamic process of an ideal fluid are the following: ρ0 v˙ + gradπ = b, div v = 0, ρ0 cv θ˙ = 0.

(13.21) (13.22) (13.23)

Notice that a similar Remark to 13.2.5 applies for an ideal fluid: the equations (13.21), (13.22) which are called the incompressible Euler equations, can be solved independently of the energy equation. Moreover, the energy equation is equivalent to θ˙ = 0. Hence material points conserve temperature during motion. Furthermore, if the process is steady, then temperature is constant on streamlines.

Chapter 14

Turbulent Flow of Incompressible Newtonian Fluids When the Reynolds number (see, for instance, [5]) goes beyond a threshold, flows become turbulent. Turbulent flows are characterized by exhibiting fluctuating velocity fields which also produce fluctuations in the other fields: density, pressure, temperature and composition. Moreover, the scale of these fluctuations is so small that it makes impossible its numerical simulation, due to the big size of the required computational mesh. Thus we are led to use turbulence models which consist of equations (algebraic or partial differential equations) allowing us to determine the so-called turbulent viscosity. In this way we are able to simulate the macroscopic flow. One of the most used turbulence models is the k −
model which will be introduced below (see, for instance, [7]).

14.1 Turbulence Models We suppose there are two scales: the large one corresponds to the macroscopic flow we are interested in simulating and the small one corresponds to the fluctuations. In practice, we are not interested in knowing these fluctuations in detail except for the fact that they strongly influence the large scale. It is this influence that should be determined by using turbulence models. Let us decompose the velocity field as v = v + v
= V + v
,

(14.1)

where   is a filter (i.e., some particular average) giving the main part of the velocity field and v
is the fluctuation with respect to it.

106

Chapter 14. Turbulent Flow of Incompressible Newtonian Fluids

For the sake of simplicity, let us suppose ρ0 and η are constant. Then the incompressible Navier-Stokes equations (13.18) can be written in the form ρ0

∂v + ρ0 div(v ⊗ v) + gradπ − η∆v = 0. ∂t

(14.2)

By replacing (14.1) in (14.2) and then taking the average of the obtained equation it is not difficult to get ρ0

∂V + ρ0 div(V ⊗ V) + gradΠ − η∆V + ρ0 divv
⊗ v
 = 0, ∂t

(14.3)

where π = Π + π
, Π being the average of π. The tensor τ = −ρ0 v
⊗ v


(14.4)

is called the Reynolds stress tensor. On the other hand, the incompressibility condition (13.2) yields div V = 0.

(14.5)

In order to have a closed problem it is necessary to be able to express tensor τ as a function of the averaged fields V and/or Π. The Boussinesq hypothesis consists in assuming that this tensor has the form τ=

1 tr(τ )I + 2ηT D(V), 3

(14.6)

where D(V) =

gradV + gradVt 2

(14.7)

and ηT is called eddy dynamic viscosity; it has to be determined by so-called closure models. Since we are dealing with incompressible flows, the isotropic term tr(τ )I does not need to be computed because it can simply be absorbed into the mean pressure Π which is then replaced by Π∗ = π − 13 tr(τ ). Thus we are interested in determining ηT . For this purpose several models exist. Let us mention, among others, algebraic models like the Smagorinsky model: √ ηT = ρ0 l02 2|D(V)|,

(14.8)

where l0 is the mixing length for sub-grid scales, or two equation models like the celebrated k −
model.

14.2.

14.2

The k −
Model

107

The k −
Model

In this model the eddy dynamic viscosity is assumed to be of the form ηT = ρ0 cµ

k2 ,


(14.9)

where k := − 2ρ10 tr(τ ) = 12 |v
|2  is the turbulent kinetic energy,
is the turbulent dissipation rate η
:= | gradv
|2 , (14.10) ρ0 and cµ is a constant parameter to be chosen below. Scalar fields k and
are obtained from the following system of partial differential equations: ρ0

∂k + ρ0 V · gradk ∂t

− −

ρ0

∂
+ ρ0 V · grad
− ∂t −

ηT | gradV + ( gradV)t |2 2

ηT div (η + ) gradk + ρ0
= 0, δk c1 ρ0 k| gradV + ( gradV)t |2 2


2 ηT div (η + ) grad
+ c2 ρ0 = 0. δ

k

(14.11)

(14.12)

The constants in this system are obtained from experiments. In particular they must be consistent with the time decay of turbulence and the logarithmic wall law at boundary layer. The following values are often proposed: cµ = 0.09, c1 = 0.126, c2 = 1.92, δk = 1 and δ = 1.3. Finally we summarize the whole system of equations modelling turbulent flows of Newtonian incompressible fluids and write them in a more compact way: ˙ + ∇Π∗ − 2 div ([η + ηT ]D(V)) = 0, ρ0 V

(14.13)

div V = 0,

η T ρ0 k˙ − div [η + ] gradk − 2ηT |D(V)|2 + ρ0
= 0, δk


2 ηT ρ0
˙ − div [η + ] grad
− 2c1 ρ0 k|D(v)|2 + c2 ρ0 = 0, δ

k

(14.14) (14.15) (14.16)

where the dot stands for the material time derivative with respect to the mean velocity field V.

Chapter 15

Mixtures of Coleman-Noll Fluids In this chapter we consider the flow of a mixture of (possibly chemically reacting) species. More precisely, we suppose that the mixture consists of Coleman-Noll fluids as those studied in Chapter 10. As we will see, a mixture of ColemanNoll fluids is not a Coleman-Noll fluid so we must adapt to mixtures the theory developed in the previous chapters.

15.1 General Definitions Let us consider the flow of a mixture of N reacting species Ei , 1≤i≤N . We assume that each species is a Coleman-Noll fluid and use the following notations for species Ei : 1. ρi : density, 2. Mi : molecular mass, 3. vi : velocity, 4. πi : (partial) pressure, 5. ei : specific internal energy, 6. si : specific entropy, 7. hi : specific enthalpy, 8. ψi : specific Helmholtz free energy, 9. Gi : specific Gibbs free energy, 10. cv,i : specific heat at constant volume,

110

Chapter 15. Mixtures of Coleman-Noll Fluids

11. cπ,i : specific heat at constant pressure. We also assume local thermodynamic equilibrium which implies that, at each (x, t), the temperature of all species is the same. Definition 15.1.1. The density of the mixture is defined by ρ=

N 

ρi .

(15.1)

i=1

Definition 15.1.2. The mass fraction of species Ei is defined by ρi . ρ

Yi =

(15.2)

Lemma 15.1.3. The following equality holds N 

Yi = 1.

(15.3)

i=1

Proof.

N  i=1

Yi =

N  ρi i=1

ρ

N =

i=1

ρ

ρi

=

ρ = 1. ρ




Definition 15.1.4. The velocity of the mixture is defined by v=

N 

Yi vi ,

(15.4)

i=1

while the diffusion velocity of species Ei is defined by Vi = vi − v.

(15.5)

Definition 15.1.5. We introduce the following fields for the mixture. • specific internal energy: e=

N 

Yi ei ,

(15.6)

Yi si ,

(15.7)

Yi hi ,

(15.8)

i=1

• specific entropy: s=

N  i=1

• specific enthalpy: h=

N  i=1

15.1. General Definitions

111

• specific Helmholtz free energy: ψ=

N 

Yi ψi ,

(15.9)

Yi Gi ,

(15.10)

Yi cv,i ,

(15.11)

i=1

• specific Gibbs free energy: G=

N  i=1

• specific heat at constant volume: cv =

N  i=1

• pressure (Dalton’s law) π=

N 

πi .

(15.12)

i=1

From the definition above, we can see that the specific internal energy cannot be written as a function of ν, s and p only. Thus the mixture itself is not a ColemanNoll fluid. However, according to (15.6) we have e = eˆ(ρ, θ, Y1 , . . . , YN ) =

N 

Yi eˆi (ρYi , θ)

(15.13)

i=1

and then

∂e ∂θ

= ρ,Y1 ,...,YN

N  i=1

Yi

∂ei ∂θ

= ρi

N 

Yi cv,i = cv .

(15.14)

i=1

Thus, the defined specific heat at constant volume for the mixture satisfies the same relation as for one single Coleman-Noll fluid, i.e., it is also the partial derivative of the specific internal energy with respect to θ, at constant density and mass fractions. Definition 15.1.6. We define concentration of species Ei to be the field ci defined by ρi . (15.15) ci = Mi Remark 15.1.7. If the molecular mass Mi is given in g/mol, then ci is given in k − mol/m3 .

112

Chapter 15. Mixtures of Coleman-Noll Fluids

Definition 15.1.8. We define the concentration of the mixture by c=

N 

ci .

(15.16)

i=1

Definition 15.1.9. We define molecular mass of the mixture to be the scalar field M such that N  1 Yi = . (15.17) M M i i=1 The following Lemma gives a justification to this Definition. Lemma 15.1.10. We have c=

ρ . M

(15.18)

Proof. c=

N N N    ρi ρYi Yi ρ . = =ρ = Mi Mi Mi M i=1 i=1 i=1


Definition 15.1.11. We define molar fraction of species Ei to be the non-dimensional scalar field Xi given by ci (15.19) Xi = . c The next Lemma relates the mass fraction to the molar fraction. Lemma 15.1.12. The following equality holds: Yi Xi . = Mi M Proof.

(15.20)

Yi ci Xi ρi /ρ ci = = . = = Mi Mi ρ cM M




Definition 15.1.13. The scalar field ni given by ni =

Yi Mi

(15.21)

represents the k-moles of species Ei per kg of mixture. We notice that ni is related to ci by ci = ρni and to Xi by Xi = Mni .  Lemma 15.1.14. Let n := N i=1 ni . Then n= Proof.

N  i=1

ni =

N  Yi 1 . = Mi M i=1

1 . M

(15.22)


15.2. Mixture of Perfect Gases

113

15.2 Mixture of Perfect Gases We consider the particular case where the mixture consists of perfect gases. We recall that this means each species Ei obeys the state law πi = ρi Ri θ, where Ri =

(15.23)

R Mi .

Proposition 15.2.1. The state equation (15.23) can be written in the following equivalent forms: 1.

πi = ci Rθ,

(15.24)

2.

πi = ρni Rθ.

(15.25)

Proof. πi = ρi Ri θ = ρi

R Yi θ = ci Rθ = ρ Rθ = ρni Rθ. Mi Mi




Proposition 15.2.2. The following state law holds for the mixture: π = cRθ.

(15.26)

Proof. By using Dalton’s law and (15.24) we get π=

N  i=1

πi =

N 

ci Rθ = cRθ.

(15.27)

i=1


Remark 15.2.3. We observe that the state equation for the mixture is similar to the one for a perfect gas. However the mixture itself is not a perfect gas because R is not constant; actually, it depends on the composition of the mixture R= M through the mass fractions (see (15.17)). Proposition 15.2.4. The following equalities hold: 1.

πi ci = , π c

(15.28)

2.

πi = Xi , π

(15.29)

3.

ni πi = . π n

(15.30)

114

Chapter 15. Mixtures of Coleman-Noll Fluids

Proof. By using (15.24), (15.26) and (15.19) we obtain c cRθ 1 π = = = . πi ci Rθ ci Xi Then, from (15.21) and (15.22), we get c π = = πi ci

ρ M ρYi Mi

=

1 M Yi Mi

=

n . ni




Now we write the specific Gibbs free energy as a function of θ, π, n1 , . . . , nN and compute its partial derivatives with respect to the two first variables. For this purpose, we notice that (15.30) yields ni πi = π N

nj

j=1

,

(15.31)

which allows us to express partial pressures in terms of the above independent variables. Thus the specific Gibbs free energy of the mixture can be written as a function of θ, π, n1 , . . . , nN , namely, ˆ π, n1 , . . . , nN ) = G(θ,

N 

ˆ i (θ, πi ) = M i ni G

N 

i=1

N  ˆ i (θ, π ni /( M i ni G nj )).

i=1

j=1

(15.32) Proposition 15.2.5. The partial derivatives of   by ∂G 1. ∂θ π,nj  ∂G 2. ∂π

ˆ with respect to θ and π are given G = −s,

(15.33)

= ν.

(15.34)

θ,nj

Proof. By using (10.129) and (10.130) we have     N N   1. ∂G ∂Gi = M i ni = Mi ni (−si ) = −s, ∂θ ∂θ i=1 i=1 

2.

∂G ∂π



π,nj

=

N 



M i ni

i=1

θ,nj

=

N  i=1

∂Gi ∂π



πi

= θ

N  i=1



M i ni

∂Gi ∂πi

 θ

1 ni  1 ni = = = ν. Yi n ρY n ρ i i=1

∂πi ∂π

(15.35)

θ,nj

N

Mi ni νi

(15.36)


15.2. Mixture of Perfect Gases

115

We recall that, for perfect gases, the specific Gibbs free energy is given by (see (12.26)) ˆ i (θ, πi ) = G ˆ 0 (θ) + Ri θ ln ( πi ), G (15.37) i π0 ˆ 0 (θ) being the specific Gibbs free energy of formation at temperature θ and G i pressure π0 . Then the specific Gibbs free energy for the mixture is   N N   πi 0 ˆ ˆ ˆ G(θ, π, n1 , . . . , nN ) = ni Mi Gi (θ, πi ) = ni Mi Gi (θ) + Ri θ ln ( ) π0 i=1 i=1   N N   ni ˆ 0 (θ) + Rθ ln ( π ) + Rθ (15.38) = ni M i G ni ln ( ), i π0 n i=1 i=1 where we have used (15.30). Remark 15.2.6. In most books the standard specific Gibbs free energy of formation is given in J/k-mol, that is, they include values of ˆ 0 (θ), µ ˆ0i (θ) = Mi G i

(15.39)

ˆ 0 (θ), for π0 = 1 atm. rather than G i Finally we have the following expression for the Gibbs free energy of a mixture of perfect gases:  ni ˆ π, n1 , . . . , nN ) = G ˆ 0 (θ, n1 , . . . , nN )+Rθ ln ( π )+Rθ G(θ, ni ln ( ), (15.40) π0 n i=1 N

where we have used the fact that N 

ni R =

i=1

R = R, M

(15.41)

and introduced the specific Gibbs free energy of formation of the mixture at temperature θ and pressure π0 by ˆ 0 (θ, n1 , . . . , nN ) := G

N 

ni µ ˆ0i (θ).

(15.42)

i=1

Definition 15.2.7. We denote by chemical potential of species Ei the scalar field µi defined by µi (x, t) = µ ˆi (θ(x, t), π(x, t), n1 (x, t), . . . , nN (x, t)), where µ ˆi (θ, π, n1 , . . . , nN ) =

ˆ ∂G (θ, π, n1 , . . . , nN ). ∂ni

(15.43)

116

Chapter 15. Mixtures of Coleman-Noll Fluids

Proposition 15.2.8. We have the following expressions for the chemical potential of species Ei : π ni 1. (15.44) µi = µ0i + Rθ ln ( ) + Rθ ln ( ), π0 n µi = µ0i + Rθ ln (

2.

πi ). π0

(15.45)

Proof. ˆ ∂G (θ, π, n1 , . . . , nN ) = ∂ni  N   δij n − nj n +Rθ nj = n2 nj j=1

µ ˆ0i (θ) + Rθ ln (

π ni ) + Rθ ln ( ) π0 n

µ ˆ0i (θ) + Rθ ln (

π ni ) + Rθ ln ( ), π0 n

because N  δij n − nj j=1

n

=

n−

N j=1

n

nj

= 0,

(15.46)

(15.47)

from which (15.44) follows. To get (15.45) it is enough to replace in (15.44) ni /n by πi /π (see (15.30)).
Corollary 15.2.9. The following expression for the specific Gibbs free energy of the mixture holds: ˆ π, n1 , . . . , nN ) = G(θ,

N 

ni µ ˆi (θ, π, n1 , . . . , nN ).

(15.48)

i=1

Proof. It follows from the fact that

πi 0 ˆ i (θ, πi ) = M−1 µ G (θ) + Rθ ln( ) = M−1 ˆi (θ, π, n1 , . . . , nN ), ˆ i i i µ π0 which is an immediate consequence of (15.37) and (15.45).

(15.49)


Now we consider other thermodynamic variables and compute their partial derivatives. Let us start with the specific internal energy. We have the following Proposition 15.2.10.



∂e ∂ni

 = µi . ν,s,nj

(15.50)

15.2. Mixture of Perfect Gases

117

Proof. From the definition of G we have e = G + θs − π/ρ

(15.51)

and then ˆ s, n1 , . . . , nN ), π ˆ θ(ν, eˆ(ν, s, n1 , . . . , nN ) = G( ˆ (ν, s, n1 , . . . , nN ), n1 , . . . , nN ) ˆ s, n1 , . . . , nN ) − ν π ˆ (ν, s, n1 , . . . , nN ). + s θ(ν, (15.52) By taking the partial derivative with respect to ni we get           ∂G ∂G ∂e ∂θ ∂π = + ∂ni ∂θ ∂ni ∂π ∂ni ν,s π,nj ν,s,nj θ,nj ν,s,nj       ∂G ∂θ ∂π + +s −ν . ∂ni ∂ni ∂ni θ,ν,nj

ν,s,nj

(15.53)

ν,s,nj

By using (15.33) and (15.34) in the above equality, the result is easily obtained.
In a similar way, we can compute the partial derivatives below. Proposition 15.2.11. We have     1. ∂e ∂e = θ, ∂s ∂ν ν,nj



2.

3.

4.





∂h ∂s

∂ψ ∂θ

∂s ∂e



 = θ, π,nj



 = −s, ν,nj

 ν,nj

1 = , θ



∂h ∂π

∂ψ ∂ν

∂s ∂ν

 = −π, s,nj



 = ν, s,nj



∂e ∂ni

∂h ∂ni 

= −π, θ,nj

 e,nj

π =− , θ



 (15.54)

= µi ,

(15.55)

ν,s,nj

 π,s,nj

∂ψ ∂ni

∂s ∂ni

= µi ,

 = µi ,

(15.56)

µi . θ

(15.57)

ν,θ,nj

 =− ν,e,nj

Chapter 16

Chemical Reactions in a Stirred Tank In this chapter we consider a mixture of reacting Coleman-Noll fluids in a stirred tank B0 . The latter means that fields do not depend on the space variable but only on time. We also assume that the tank is closed so that its total mass is conserved and hence density is constant.

16.1 Chemical Kinetics. The Mass Action Law Let us assume the mixture undergoes a chemical process involving L reactions: l EN −→ λl1 E1 + · · · + λlN EN , ν1l E1 + · · · + νN

1 ≤ l ≤ L.

Coefficients νil and λli are called stoichiometric coefficients. The evolution of concentrations of different species is governed by a system of ordinary differential equations, namely, dci  l = (λi − νil )δl , dt L

(16.1)

l=1

where δl is the velocity of the l-th chemical reaction. According to the mass action law, it is given by N  νl cj j , (16.2) δl = kl j=1

kl being a function of temperature through the Arrhenius law: El

kl (θ) = Bl θαl e− Rθ .

(16.3)

120

Chapter 16. Chemical Reactions in a Stirred Tank

In the previous expression, Bl is the frequency (or pre-exponential) factor, αl the pre-exponential exponent and El the activation energy. In many cases constant αl is null. El is non-dimensional, θl = ERl has dimension Remark 16.1.1. Since the exponent Rθ of temperature. Actually, it is called activation temperature. In the particular case of combustion reactions, El as well as Bl are very large. Then, for values of temperature below θl , kl and, correspondingly, the velocity of reaction, δl , are very small. On the contrary, for temperatures of the order of θl or higher, this parameter becomes very large which means that reaction progresses very rapidly. By replacing expression (16.2) into (16.1) we get the nonlinear system of ordinary differential equations L N El  ν l dci  l l αl − Rθ (λi − νi )Bl θ e cj j . = dt j=1

(16.4)

l=1

In order to solve it we need to specify the initial concentrations of species, ci (0) = ci,init ,

(16.5)

as well as another equation allowing us to determine the evolution of the mixture temperature. For this purpose several different assumptions can be made concerning the heat exchange between the tank and its environment. Here we suppose there is a system of heat supplied to the tank (see Definition 1.4.1). This leads to the following form of the energy conservation principle:
d ρe dV = Q, (16.6) dt B0 where the function of time Q(t) is the heat rate supplied into the tank B0 at time t (see Definition 1.4.2). Since ρ is constant and e does not depend on the space variable x, we deduce from (16.6) 1 de = Q, dt m

(16.7)

m being the (constant) mass of the mixture. We notice that m = ρ vol(B0 ). Moreover, from (15.6) we deduce the equation: e=

N 

Yi eˆi (ρi , θ),

(16.8)

i=1

where e(t) = einit +

1 m




t

Q(s)ds, 0

(16.9)

16.1. Chemical Kinetics. The Mass Action Law

121

einit being the given initial specific internal energy of the mixture. The “algebraic” equation (16.8) allows us to compute temperature along the time. Alternatively, one can obtain an ordinary differential equation for θ as follows. By replacing (16.8) in (16.7) we get N  dYi i=1

dt

eˆi (ρi , θ) + ρ

N  i=1

1 ∂ˆ ei dYi  dθ + = Q, (ρi , θ) Yi cˆv,i (ρi , θ) ∂ρi dt dt m i=1 N

Yi

(16.10) ei because ρ is constant. We recall that cˆv,i (ρi , θ) = ∂ˆ ∂θ (ρi , θ) is the specific heat at constant volume of species Ei . According to (15.11), the specific heat at constant volume of the mixture is

cv = cˆv (ρ, θ, Y1 , . . . , YN ) =

N 

Yi cˆv,i (ρYi , θ).

(16.11)

i=1

Then equation (16.10) can be written as  N  dYi i=1

dt

ρYi

 1 ∂ˆ ei dθ = Q. (ρi , θ) + eˆi (ρYi , θ) + cˆv (ρ, θ, Y1 , . . . , YN ) ∂ρi dt m (16.12)

Now we rewrite (16.4) in terms of the mass fractions rather than the concentrations. We have

ν l L N El  dYi ρYj j Mi  l l αl − Rθ = (λi − νi )Bl θ e . (16.13) dt ρ Mj j=1 l=1

In order to obtain a system of ordinary differential equations in normal form we replace (16.13) in (16.12). It yields   N  dθ 1 ∂ˆ ei =− Mi ρYi (ρi , θ) + eˆi (ρYi , θ) dt ρˆ cv (ρ, θ, Y1 , . . . , YN ) i=1 ∂ρi ⎡ ⎤⎫

ν l L N  El  1 ρYj j ⎦⎬ l l αl − Rθ ⎣ × Q. + (λi − νi )Bl θ e ⎭ M mˆ c (ρ, θ, Y j v 1 , . . . , YN ) j=1 l=1

(16.14) The system (16.13), (16.14) can now be integrated for the initial conditions (16.5) and θ(0) = θinit , (16.15) θinit being the initial temperature of the mixture.

122

Chapter 16. Chemical Reactions in a Stirred Tank

16.2 Conservation of Chemical Elements Since atoms are conserved in chemical reactions, we can write linear algebraic equations establishing this fact. In this way, we get a reduction in the number of equations of the ordinary differential system (16.13). Let us suppose there are K different elements involved in species, named Ak , 1 ≤ k ≤ K. Let the formula of species Ei be Ei = (A1 )a1i · · · (AK )aKi , 1 ≤ i ≤ N .

(16.16)

Thus aki denotes the number of atoms of element Ak in a molecule of species Ei . In terms of mass, there are aki k-atoms of element Ak per k-mol of species Ei . At time t we have a composition of the mixture defined, for instance, in terms of n1 , . . . , nN . Then aki ni gives the k-atoms of element Ak in species Ei per kg of mixture. Let Zk,init be the number of k-atoms of element Ak per kg of mixture at initial time. Since this quantity is conserved, we must have that N 

aki ni =

i=1

N  i=1

aki

Yi = Zk,init , 1 ≤ k ≤ K, Mi

(16.17)

at any time. In fact we can prove that any solution of the ordinary differential system (16.1) satisfies (16.17) at any time. Indeed, firstly equations (16.17) can be written in terms of species concentrations, in the form N 

aki ci = ρZk,init , k = 1, . . . , K.

(16.18)

i=1

Let us multiply (16.1) by ak,i and then add all equations for i = 1, . . . , N . We have   L   N N N L      d l l l l (aki ci ) = aki (λi − νi )δl = δl aki (λi − νi ) , dt i=1 i=1 i=1 l=1

l=1

k = 1, . . . , K.

(16.19)

Moreover, since atoms of each element are conserved in the l-th reaction, we must have N N   aki λli = aki νil , k = 1, . . . , K (16.20) i=1

i=1

and then (16.19) yields d dt



N  i=1

 aki ci

= 0, k = 1, . . . , K.

(16.21)

16.3. Reacting Mixture of Perfect Gases

123

N Hence i=1 aki ci , k = 1, . . . , K are constant in time so equal to their respective initial values. Constraints (16.18) can be considered as a linear system of equations for ci , i = 1, . . . , N, which allows us to write K of them in terms of the remaining N − K. Thus we can eliminate the K corresponding equations in (16.1).

16.3 Reacting Mixture of Perfect Gases Let us suppose all components of the mixture are perfect gases. In this case the specific internal energy of each component is a function of temperature only (see Proposition 12.1.2) and then (16.8) becomes N 

Yi eˆi (θ) = e.

(16.22)

i=1

The specific internal energy of species Ei is now given by
θ eˆi (θ) = e0i + cˆv,i (s)ds,

(16.23)

θ0

where e0i = eˆi (θ0 ) is the specific internal energy of formation of species Ei at temperature θ0 . If this temperature is 25o C, then e0i is called the standard specific internal energy of formation of species Ei . By replacing (16.23) in (16.22) we obtain
θ N  0 Yi [ei + cˆv,i (s)ds] = e, (16.24) i=1

θ0

with e still given by (16.9). We can include this equation into the ordinary differential system (16.13) to get a closed problem. It allows us to compute the temperature and the mass fractions of species along time. Instead, we can also use the ordinary differential equation (16.14) which, for a mixture of perfect gases, becomes ⎤ ⎡

νjl N L N   El  ρY 1 dθ j ⎦ =− Mi eˆi (θ) ⎣ (λli − νil )Bl θαl e− Rθ dt ρˆ cv (θ, Y1 , . . . , YN ) i=1 M j j=1 l=1

+

1 Q. mˆ cv (θ, Y1 , . . . , YN ) (16.25)

Usually, equation (16.22) is written in terms of enthalpy rather than internal energy. We recall that πi hi = e i + = ei + Ri θ, (16.26) ρi

124

Chapter 16. Chemical Reactions in a Stirred Tank

with hi given by




θ

ˆ i (θ) = h0 + hi = h i

cˆπ,i (s)ds,

(16.27)

θ0

hi (θ0 ) being the specific enthalpy of formation of species Ei at temand h0i = ˆ perature θ0 . If this temperature is 25o C, then h0i is called the standard specific enthalpy of formation. The standard specific enthalpy of formation of chemical elements is taken to be zero. Thus the standard specific enthalpy of formation represents the change in enthalpy that accompanies the formation of one kg of the species from its elements at the standard specified temperature, when each reactant and each product is in its standard state. By using (16.26) and (16.27), equation (16.22) can be rewritten in the form
θ N  0 Yi [hi + cˆπ,i (s)ds − Ri θ] = e. (16.28) i=1

θ0

Alternatively, by taking the derivative of this equation with respect to time, we can obtain an ordinary differential equation for θ. Indeed, firstly we have
θ N N   1 dYi 0 dθ (hi + = Q, cˆπ,i (s)ds − Ri θ) + Yi [ˆ cπ,i (θ) − Ri ] dt dt m θ0 i=1 i=1 (16.29) from which it follows that
θ N  dYi 0 ˆ 1 , . . . , YN )] dθ = 1 Q, (hi + cˆπ,i (s)ds−Ri θ)+[ˆ cπ (θ, Y1 , . . . , YN )− R(Y dt dt m θ 0 i=1 (16.30) where N  cˆπ (θ, Y1 , . . . , YN ) := Yi cˆπ,i (θ) (16.31) i=1

ˆ 1 , . . . , YN ) is its gas is the specific heat at constant pressure of the mixture and R(Y constant given by R ˆ 1 , . . . , YN ) = R(Y , (16.32) M(Y1 , . . . , YN ) M(Y1 , . . . , YN ) being the molecular mass of the mixture (see Definition 15.1.9). Finally we replace (16.13) in (16.30) to get – j» Z θ N X dθ 1 Mi cˆπ,i (s)ds − Ri θ =− h0i + ˆ 1 , . . . , YN )] dt ρ[ˆ cπ (θ, Y1 , . . . , YN ) − R(Y θ0 i=1 " L „ «νjl #) N X l Y El Q ρYj × (λi − νil )Bl θαl e− Rθ + . ˆ M j m[ˆ c (θ, Y , . . . , Y π 1 N ) − R(Y1 , . . . , YN )] j=1 l=1

Chapter 17

Chemical Equilibrium of a Reacting Mixture of Perfect Gases in a Stirred Tank In this Chapter we study the chemical equilibrium of a reacting mixture of perfect gases in a stirred tank.

17.1 The Least Action Principle for the Gibbs Free Energy In this Section we assume the tank is closed both to mass and energy transfer. At initial time we introduce a certain amount of each species Ei . Then they react and attain chemical equilibrium (see, for instance, [9]). The problem consists in determining the composition of the mixture as well as its pressure and temperature. The data are the initial composition defined by ni,init , i = 1, . . . , N , (from which we can easily compute the corresponding numbers Zk,init , k = 1, . . . , K, giving the k-atoms of element Ak per kg of mixture), and the initial specific internal energy of the mixture. We notice that, since density does not change with time, it is also known at equilibrium and hence we could write the pressure as a function of temperature and composition through the state law. The main result in chemical equilibrium theory is the least action principle for the free energy: The system is in equilibrium if and only if the specific Gibbs free energy, ˆ π, n1 , . . . , nN ), attains a minimum subject to the following constraints: G(θ,

126 • mass

Chapter 17. Chemical Equilibrium of a Reacting Mixture

N 

N 

aki ni = Zk,init :=

i=1

aki ni,init , 1 ≤ k ≤ K,

(17.1)

i=1

• energy N  i=1




θ

Mi ni [h0i +

cˆπ,i (s)ds − Ri θ] = einit ,

(17.2)

θ0

• state law π = ρR

N 

ni θ,

(17.3)

i=1

• positivity

ni ≥ 0, i = 1, . . . , N.

(17.4)

Remark 17.1.1. We recall that, in this constrained optimization problem, ρ, einit and Zk,init , k = 1, . . . , K are known. Alternatively, if θ is given instead of einit , then we remove constraint (17.2). Remark 17.1.2. In order to compute the equilibrium state, the above constrained optimization problem must be solved by mathematical programming techniques. It is important to notice that we do not need to know the chemical reactions involved, but only the species that are present in the mixture at equilibrium. In what follows, we suppose that these chemical reactions are also given and show that, in this case, the equilibrium problem can be more easily solved.

17.2 Equilibrium for a Set of Reversible Reactions, Equilibrium Constants For the sake of clarity we first consider a particular case consisting of only one reversible reaction. Moreover, we assume that θ and π are given so we can remove constraints (17.2) and (17.3) from the above optimization problem characterizing the equilibrium state. Accordingly, the density is not imposed which means that the volume of the tank may change. In fact, it can be obtained from the state law (17.3) after the composition of the mixture has been determined by solving the minimization problem. In the framework of Section 16, one reversible reaction corresponds to two reactions (L = 2). Let us introduce λi := νi2 = λ1i ,

(17.5)

λ2i

(17.6)

νi :=

=

νi1 .

17.2. Equilibrium for a Set of Reversible Reactions, Equilibrium Constants 127 By using this notation, the reversible reaction can be written as −→

ν1 E1 + · · · + νN EN ← λ1 E1 + · · · + λN EN , so that we have δ1 = k1

N 

ν

cj j

and

δ2 = k2

j=1

N 

λ

cj j

(17.7)

(17.8)

j=1

for the velocities of the forward and backward reactions, respectively. The ordinary differential system (16.1) has to be replaced by ρ

N N   dni ν λ = (λi − νi )(k1 cj j − k2 cj j ). dt j=1 j=1

(17.9)

By introducing δ = (k1

N 

ν

cj j − k2

j=1

N 

λ

cj j ),

(17.10)

j=1

equation (17.9) yields dni δ 1 = . (λi − νi ) dt ρ

(17.11)

By integrating this system we deduce nN (t) − nN,init n1 (t) − n1,init = ··· = = ξ(t), (λ1 − ν1 ) (λN − νN )

(17.12)

where ni,init , i = 1, . . . , N correspond to the initial composition of the mixture and ξ, the reaction extent, is given by
ξ(t) = 0

t

δ(s) ds. ρ(s)

(17.13)

The interest of (17.12) is that we can write the ni , i = 1, . . . , N, in terms of one single function ξ. Thus the specific Gibbs free energy is given by ˆ π, n1,init + (λ1 − ν1 )ξ, . . . , nN,init + (λN − νN )ξ). G = gˆ(ξ) := G(θ,

(17.14)

In this way, the mass conservation equations (17.1) are automatically satisfied. Then the only constraint remaining is (17.4) and so the equilibrium composition can be obtained by solving the following one-dimensional constrained optimization problem: min

(λi −νi )ξ+ni,int ≥0

gˆ(ξ).

(17.15)

128

Chapter 17. Chemical Equilibrium of a Reacting Mixture

If gˆ attains a minimum at ξ, two possibilities exist: either (λi − νi )ξ + ni,int = 0 for some i or gˆ
(ξ) = 0. The latter can be written as follows: gˆ
(ξ) =

N N   ˆ dni ∂G = (θ, π, n1 , . . . , nN ) (λi − νi )ˆ µi (θ, π, n1 , . . . , nN ) = 0, ∂ni dξ i=1 i=1

(17.16) with ni = ni,init + (λi − νi )ξ, i = 1, . . . , N. Definition 17.2.1. The function a ˆ(θ, π, n1 , . . . , nN ) = −

N 

(λi − νi )ˆ µi (θ, π, n1 , . . . , nN )

(17.17)

i=1

is called chemical affinity of reaction (17.7). Since species Ei are perfect gases we can replace the expression for µ ˆi given by (15.45) in (17.16). It becomes

πi 0 (λi − νi ) µ ˆi (θ) + Rθ ln ( ) = 0. π0 i=1

N 

(17.18)

Let us introduce Kπ0 , the equilibrium constant based on partial pressures for chemical reaction (17.7). It is given as a function of temperature by 

Kπ0 (θ)

 N 1  0 = exp − (λi − νi )ˆ µi (θ) . Rθ i=1

(17.19)

Then, from (17.18), Kπ0 (θ)

=

λ −ν N  πi i i i=1

π0

.

(17.20)

The quantity ∆G0 (θ) =

N 

(λi − νi )ˆ µ0i (θ)

(17.21)

i=1

is called change of specific free energy in the reaction at temperature θ and pressure π0 . Then (17.19) yields Kπ0 (θ)



1 0 ∆G (θ) . = exp − Rθ

(17.22)

17.2. Equilibrium for a Set of Reversible Reactions, Equilibrium Constants 129 We can define another equilibrium constant based on concentrations. For this purpose, we rewrite (17.20) making use of (15.24). We obtain Kπ0 (θ)

=

λ −ν N  ci Rθ i i π0

i=1

=

Rθ π0

PN N i=1 (λi −νi ) 

cλi i −νi .

(17.23)

i=1

By introducing the equilibrium constant based on concentrations, Kc0 (θ) = Kπ0 (θ)

Rθ π0

PN i=1 (νi −λi ) ,

(17.24)

we deduce from (17.23) Kc0 (θ) =

N 

cλi i −νi .

(17.25)

i=1

Finally we recall that we can choose π0 to be the standard pressure, i.e., 1 atm. In this case it can be removed from the computations above as far as partial pressures are also given in atm. We come back to the solution of the equilibrium problem. Since the change of specific free energy in the reaction can be computed from the specific free energy of species by (17.21), we can obtain the equilibrium constant Kπ0 from (17.22). Then, using (15.30) and (17.12) in (17.20) we get Kπ0 (θ)



λi −νi π[ni,init + (λi − νi )ξ] = = N π0 n π0 i=1 [ni,init + (λi − νi )ξ] i=1 i=1 λi −νi  PN N i=1 (λi −νi )  π ni,init + (λi − νi )ξ = . (17.26) N π0 i=1 i=1 [ni,init + (λi − νi )ξ]

λ −ν N  πni i i

N 

This equation can be solved for ξ. Let ξ ∗ be a solution. If ξ ∗ satisfies all of the constraints then it allows us to compute the composition at equilibrium by means of (17.12). Now we can consider the more general case of J reversible reactions between species E1 , . . . , EN : −→

j ν1j E1 + · · · + νN EN ← λj1 E1 + · · · + λjN EN , j = 1, . . . , J.

(17.27)

We recall that each of them leads to two reactions in the framework introduced in Section 16. We denote by ξj the extent of the j-th reversible reaction given by (17.27). Then, in a similar way to (17.12), we can obtain ni = ni,init +

J  j=1

(λji − νij )ξj , i = 1, . . . , N.

(17.28)

130

Chapter 17. Chemical Equilibrium of a Reacting Mixture

Hence the specific Gibbs free energy can be written in terms of the extents ξj , j = 1, . . . , J instead of ni , i = 1, . . . , N . Let gˆ be defined by ⎞ ⎛ J J   ˆ ⎝θ, π, n1,init + (λj −ν j )ξj , . . . , nN,init + (λj −ν j )ξj ⎠ . gˆ(ξ1 , . . . , ξJ ) := G 1 1 N N j=1

j=1

(17.29) Since now the ni , i = 1, . . . , N are restricted to those of the form (17.28), the mass conservation constraints (17.1) automatically hold. Indeed, N 

aki ni

=

i=1

N 

aki [ni,init +

i=1

= Zk,init

+

J  N 

J 

(λji − νij )ξj ]

j=1

aki (λji − νij )ξj = Zk,init , 1 ≤ k ≤ K,

(17.30)

j=1 i=1

because of (16.20). Then the equilibrium problem is equivalent to the following constrained optimization problem: ni,init +

gˆ(ξ1 , . . . , ξJ ). PJmin j j j=1 (λi −νi )ξj ≥0

(17.31)

Let us suppose that the solution to this problem does not saturate any positivity constraint, i.e., the corresponding ni , i = 1, . . . , N are all strictly positive. Then the partial derivatives of gˆ must vanish, that is, N  ˆ ∂ˆ g ∂G dni (ξ1 , . . . , ξJ ) = (θ, π, n1 , . . . , nN ) ∂ξj ∂n dξj i i=1

=

N 

(λji − νij )ˆ µi (θ, π, n1 , . . . , nN ) = 0, j = 1, . . . , J,

(17.32)

i=1

 with ni = ni,init + Jj=1 (λji − νij )ξj , i = 1, . . . , N . This is a system of J equations for the J unknowns ξj , j = 1, . . . , J. Since species Ei are perfect gases, we can replace in equations (17.32) the expression for µ ˆi given by (15.45). They become

N  πi j j 0 (λi − νi ) µ ˆi (θ) + Rθ ln ( ) = 0, j = 1, . . . , J. (17.33) π0 i=1 0 Let us introduce Kπ,j , the equilibrium constant based on partial pressures for the j-th chemical reaction in (17.27). It is given as a function of temperature by   N 1  j j 0 0 (λ − νi )ˆ µi (θ) . (17.34) Kπ,j (θ) = exp − Rθ i=1 i

17.2. Equilibrium for a Set of Reversible Reactions, Equilibrium Constants 131 Then, from (17.33),

λ −ν N  πi i i j

0 (θ) Kπ,j

=

j

, j = 1, . . . , J.

π0

i=1

(17.35)

The quantity ∆G0,j (θ) =

N 

(λji − νij )ˆ µ0i (θ)

(17.36)

i=1

is called change of specific free energy in the j-th reaction at temperature θ and pressure π0 . Using (17.21) in (17.34) we get

1 0 0,j Kπ,j (θ) = exp − ∆G (θ) . (17.37) Rθ As in the case of one reversible reaction, we can define other equilibrium constants based on concentrations. For this purpose, we rewrite (17.35) making use of (15.24). We obtain

λ −ν N  ci Rθ i i j

0 (θ) Kπ,j

=

i=1



j

π0

=

Rθ π0

j j

PN N i=1 (λi −νi ) 

λj −νij

ci i

.

(17.38)

i=1

By introducing the equilibrium constant based on concentrations for the j-th reaction, j j

PN i=1 (νi −λi ) Rθ 0 0 Kc,j (θ) = Kπ,j (θ) , (17.39) π0 we deduce from (17.38) 0 Kc,j (θ) =

N 

λj −νij

ci i

.

(17.40)

i=1

We come back to the solution of the equilibrium problem. Since the change of specific free energy of the reactions can be computed from the specific free energy of 0 by using (17.37). species by (17.36), we can obtain the equilibrium constants Kπ,j Then we use (15.30) and (17.28) in (17.35) to write 

λji −νij J j j + (λ − ν )ξ ] π[n i,init j i j=1 i 0 Kπ,j (θ) = = J N j π n 0 π [n + (λ − νij )ξj ] 0 i=1 i=1 i=1 i,init j=1 i λji −νij  J j j PN N i=1 (λi −νi )  ni,init + j=1 (λji − νij )ξj π = . N J j j π0 i=1 [ni,init + j=1 (λi − νi )ξj ] i=1 (17.41) This system of equations can be solved for ξ1 , . . . , ξJ and then the composition at equilibrium can be computed by means of (17.28).

λ −ν N  πni i i j

j

N 

132

Chapter 17. Chemical Equilibrium of a Reacting Mixture

17.3 The Stoichiometric Method An important conclusion of the previous section is that, when the chemical reactions are known, we can eliminate the mass conservation constraints and reduce the number of unknowns in the equilibrium problem. Our present goal is to show that, even if we do not know these chemical reactions, it is possible to introduce some virtual ones which also allow us to reduce the number of unknowns and to eliminate the mass constraints in the equilibrium minimization problem. Let A be the matrix A = (aki )k=1,...,K, i=1,...,N . Then (16.17) can be written as (17.42) An = Zinit . where n = (n1 , . . . , nN )t and Zinit = (Z1,init , . . . , ZK,init )t . Let us denote by R the rank of matrix A, that is, the maximum number of independent rows or columns. Then the dimension of the kernel of A is given by J = dim Ker(A) = N − R.

(17.43)

Zinit = Aninit ,

(17.44)

Since then the linear system (17.42) is compatible which implies that rank(A|Zinit ) = R.

(17.45)

Furthermore its general solution is n = ninit +

R 

ξj γ j ,

(17.46)

j=1

where {ξj , j = 1, . . . , J} are arbitrary real numbers and the set of vectors in RN , {γ j , j = 1, . . . , J}, is a basis of the kernel of A which is nothing but the subspace of RN of the solutions of the homogeneous linear system An = 0. In order to obtain a basis for Ker(A) we can use the next algorithm. Since rank(A) = R there must exist an invertible square matrix P of order K, and such that B = PA has the block form

IR C B= , (17.47) 0 0 IR being the identity matrix of order R and C a non-null rectangular matrix of order R × J. Matrix B can be obtained by Gaussian elimination. Then an admissible choice for {γ j , j = 1, . . . , J} are the columns of the N × J matrix G defined by

−C G= . (17.48) IJ

17.3. The Stoichiometric Method

133

Indeed, we have

−C IR A G = P −1 B = P −1 IJ 0

C 0



−C IJ



= P −1



0 0

= (0). (17.49)

Now we come back to the general situation where {γ j , j = 1, . . . , J} is any basis of Ker(A). Let us consider the matrix or rank J obtaining by putting these vectors as columns, i.e.,   (17.50) Γ = γ 1 | . . . |γ J . We notice that Γ = G in the previous discussion. This matrix is called full stoichiometric matrix and defines a so-called full set of chemical reactions. These reactions are j 0 = γ1j E1 + · · · + γN EN , j = 1, . . . , J. (17.51) One can consider as reactants those species Ei for which γij are negative and as products those for which these coefficients are positive. Comparing with (17.7) we have (17.52) γij = λji − νij , i = 1, . . . , N, j = 1, . . . , J. Thus, in order to solve the equilibrium problem, firstly we determine vectors γ j , j = 1, . . . , J. Secondly, we replace the expressions for ni , i = 1, . . . , N given ˆ π, n1 , . . . , nN ) and in the constraints by (17.46) in the Gibbs free energy, G(θ, (17.2)–(17.4). Thirdly, we minimize the obtained function ˆ π, n1,init + gˆ(θ, π, ξ1 , . . . , ξJ ) =: G(θ,

J 

γ1j ξj , . . . , nN,init

j=1

+

J 

j γN ξj )

(17.53)

j=1

under the constraints (17.2)–(17.4), with ni replaced by ni,init +

R 

ξj γ ji , i = 1, . . . , N.

(17.54)

j=1

For this purpose it is convenient to compute the partial derivatives of gˆ. A straightforward application of the chain rule leads to the following results: Proposition 17.3.1. We have 1.

2.





∂g ∂θ

∂g ∂π



 = π,ξj



 = θ,ξj

∂G ∂θ

∂G ∂π

 .

(17.55)

.

(17.56)

π,ni

 θ,ni

134 3.

Chapter 17. Chemical Equilibrium of a Reacting Mixture 

∂g ∂ξl

 = θ,π,ξj

N 

γil µ ˆi (θ, π, n1,init +

i=1

J 

γ1j ξj , . . . , nN,init +

j=1

J 

j γN ξj ).

j=1

(17.57) The quantity a ˆl (θ, π, ξ1 , . . . , ξJ ) = −

N  i=1

γil µ ˆi (θ, π, ninit +

J 

ξj γ j )

j=1

is the chemical affinity of the l-th virtual reaction (see (17.17)).

(17.58)

Chapter 18

Flow of a Mixture of Reacting Perfect Gases In this chapter we establish the equations for general flows of a mixture of reacting perfect gases. Thus we remove the assumption of stirred tank and from now on all functions characterizing the system depend on position and time, i.e., they are spatial fields. This is the case, in particular, for those defining the composition of the mixture. We consider N superposed motions Xi , i = 1, . . . , N, of N bodies B i , i = 1, . . . , N (the species Ei ) such that all of them occupy the same position at each time t, i.e., (18.1) Bti = Bt ∀i = 1, . . . , N. Hence, each species has its own spatial velocity vi and we can define the velocity of the mixture, v, by (15.4) and the diffusion velocity of species Ei , Vi , by (15.5).

18.1 Mass Conservation Equations The mass conservation equation of species Ei is ∂ρi + div(ρi vi ) = ωi + mi , ∂t

(18.2)

where ωi and mi are the source of mass for species Ei due to chemical reactions and to an evaporating or combusting condensed phase, respectively. An example of the latter is a liquid or solid fuel in a combustion chamber. According to (16.1), ωi is given by L  (λli − νil )δl , (18.3) ωi = Mi l=1

136

Chapter 18. Flow of a Mixture of Reacting Perfect Gases

where δl is the velocity of the l-th reaction (see (16.2)). In its turn, term mi would be determined by a model for the condensed phase. In order to get a conservation equation for the total mass of the mixture we add (18.2) from i = 1 to N : N N N N N     ∂ρi ∂  + div(ρi vi )] = ( [ ρi ) + div( ρi vi ) = ωi + mi . ∂t ∂t i=1 i=1 i=1 i=1 i=1

(18.4)

We prove that the first term on the right-hand side is null. Indeed, N  i=1

ωi =

N  i=1

Mi

L 

(λli − νil )δl =

l=1

L  l=1

δl

N 

Mi (λli − νil ) = 0,

(18.5)

i=1

because the total mass is conserved in each chemical reaction, i.e., N  i=1

Mi λli =

N 

Mi νil , l = 1, . . . , L.

(18.6)

i=1

By using this, and definitions (15.1) and (15.4), (18.4) yields  ∂ρ + div(ρv) = m := mi . ∂t i=1 N

(18.7)

Now we come back to the mass conservation equation for species (18.2). We do not intend to compute the velocity of each species so, in order to get a closed system, we have to write the diffusion velocity in terms of any other fields remaining in the model. This can be done by using Fick’s law: Vi = −Di grad(lnYi ) = −

Di gradYi , Yi

(18.8)

where Di is the mass diffusion coefficient of species Ei into the mixture. By replacing (18.8) in (18.2) after using (15.5) we get ∂ (ρYi ) + div(ρYi v) − div(ρDi gradYi ) = ωi + mi , ∂t

(18.9)

which is the mass conservation equation for species Ei in conservative form. By multiplying (18.7) by Yi and then subtracting from (18.9) we can also obtain the non-conservative form: ρY˙ i − div(ρDi gradYi ) = ωi + mi − mYi .

(18.10)

18.2. Motion Equation

137

18.2 Motion Equation The balance of linear and angular momentum for the mixture can be written as (1.27) and (1.28) where now the body force b includes all external body forces as, for instance, gravity forces or the rate of linear momentum supplied to the mixture by a condensed phase. Moreover, we assume a constitutive law for T similar to that for a ColemanNoll fluid, namely T = −πI + l(D), (18.11) where π is related to the partial pressures of species by Dalton’s law (see (15.12)) and l(D), the viscous part of the Cauchy stress tensor, is given by (10.29). Then we deduce the following motion equation in conservative form: ∂(ρv) + div(ρv ⊗ v) + gradπ = div(l(D)) + b. ∂t

(18.12)

By multiplying (18.7) by v and then subtracting from (18.12), we get the non-conservative form: ρv˙ + gradπ = div(l(D)) + b − mv.

(18.13)

18.3 Energy Conservation Equation We assume the energy conservation law (1.46) where we recall that f accounts for the external contribution of heat, in particular from the condensed phase. Then, in terms of the specific internal energy, the conservative form of the energy equation is ∂(ρe) + div(ρev) = T · D − div q + f. (18.14) ∂t From the definitions of internal energy and enthalpy of the mixture and the Dalton law it is easy to deduce an equation for the mixture analogous to (10.35). It allows writing the energy equation in terms of enthalpy, namely ∂(ρh) + div(ρhv) − π˙ = l(D) · D − div q + f. ∂t

(18.15)

Moreover, the constitutive law for the conductive heat flux q is taken to be q = −k gradθ +

N 

ρi Vi hi ,

(18.16)

i=1

where the second term on the right-hand side represents the enthalpy flux due to the diffusion velocity of species. We make use of Fick’s law to replace the diffusion

138

Chapter 18. Flow of a Mixture of Reacting Perfect Gases

velocities in the above expression. This leads to N  ∂π ∂(ρh) +div(ρhv)− −div(πv) = l(D)·D−div(−k gradθ−ρ Di hi gradYi )+f. ∂t ∂t i=1 (18.17) Equation (18.17) can be written in a more compact form in which temperature does not appear in an explicit way. For this, we have to make the following assumptions:

1. The mass diffusion coefficient is the same for all species, i.e., Di = D, i = 1, . . . , N. 2. The Lewis number of the mixture is equal to 1. More precisely, we can replace the thermal conductivity, k, by ρDcπ in (18.16). Then we have



ρD gradh = ρD grad

N 

 Yi hi

= ρD

i=1

= ρD

N  i=1

gradYi hi + ρD

N 

N 

gradYi hi + ρD

i=1

Yi cπ,i gradθ = ρD

i=1

N 

Yi gradhi

i=1 N 

gradYi hi + ρDcπ gradθ,

i=1

(18.18) and hence q = −ρD gradh.

(18.19)

Now we replace this expression for q in equation (18.15). It yields ∂(ρh) + div(ρhv) − π˙ = l(D) · D + div(ρD gradh) + f. ∂t

(18.20)

Sometimes it is more convenient to handle this energy conservation equation in non-conservative form. For this we multiply (18.7) by h and subtract from (18.20). Since we have ∂ρ ∂h ∂ρ ∂(ρh) + h div(ρv) = ρ + ρv · gradh + h + h div(ρv) = + div(ρvh), ∂t ∂t ∂t ∂t (18.21) it yields ρh˙ − π˙ = l(L) · D + div(ρD gradh) + f − mh. (18.22)

ρ h˙ + h

Sometimes equation (18.20) is written in terms of the specific thermal enthalpy rather than the specific (total) enthalpy. For this we recall that, because of (15.8) and (16.27), the specific enthalpy of the mixture is given by h=

N  i=1

Yi h0i + hT ,

(18.23)

18.3. Energy Conservation Equation

139

ˆ i (θ0 ) is the enthalpy of formation of species Ei at where we recall that h0i = h temperature θ0 and hT is the thermal enthalpy of the mixture with respect to this reference temperature. The latter is given, in terms of temperature and mass fractions, by
θ N  ˆ hT = hT (θ, Y1 , . . . , Yn ) = Yi cπ,i (s) ds. (18.24) θ0

i=1

By using (18.23) we obtain ∂ (ρh) ∂t div(ρvh)

=

N    ∂ ∂ h0i (ρYi ) , (ρhT ) + ∂t ∂t i=1

= div(ρv

N 

Yi h0i ) + div(ρvhT ),

i=1



− div(ρD gradh) = − div ρD gradhT

 −

N 

  h0i div ρD gradYi .

(18.25)

(18.26)

(18.27)

i=1

Let us multiply the mass conservation equation for species Ei , (18.9), by h0i and then add for i = 1, . . . , N . We get N 

N N N    ∂ (ρYi ) + h0i div(ρvYi ) − h0i div(ρD gradYi ) = h0i (ωi + mi ). ∂t i=1 i=1 i=1 i=1 (18.28) Then we add equations (18.25), (18.26) and (18.27), and use (18.28). We get

h0i

∂ ∂ (ρh) + div(ρvh) − div(ρD gradh) = (ρhT ) + div(ρvhT ) ∂t ∂t N  − div(ρD gradhT ) + h0i (ωi + mi ).

(18.29)

i=1

Now, we use this equality in (18.20) to obtain a partial differential equation for thermal enthalpy in conservative form: N  ∂ (ρhT )+div(ρvhT )−π−div(ρD ˙ gradhT ) = l(D)·D+f − h0i (ωi +mi ). (18.30) ∂t i=1

In order to obtain the corresponding non-conservative equation it is enough to subtract the mass conservation equation (18.7) multiplied by hT from (18.30). It yields ρh˙T − π˙ − div(ρD gradhT ) = l(D) · D + f −

N  i=1

h0i (ωi + mi ) − hT m.

(18.31)

140

Chapter 18. Flow of a Mixture of Reacting Perfect Gases

Finally, we summarize the equations to be satisfied by any thermodynamic process of a reacting mixture of perfect gases. Under the assumptions above concerning the mass diffusion coefficients and Lewis number, they are, in conservative form, ∂ρ + div(ρv) = m, ∂t

(18.32)

∂(ρv) + div(ρv ⊗ v) + gradπ = div(l(D)) + b, ∂t

(18.33)

∂(ρh) + div(ρhv) − π˙ − div(ρD gradh) = l(D) · D + f, ∂t

(18.34)

∂(ρYi ) + div(ρYi v) − div(ρD gradYi ) = ωi + mi , i = 1, . . . , N, ∂t

(18.35)

together with π

=

R = l(D) = h

=

ρRθ,

(18.36) N 

R Yi =R , M Mi i=1 2ηD + ξ div vI,
N  ˆ i (θ0 ) + Yi [h

=

Mi

(18.38) θ

cˆπ,i (s) ds],

(18.39)

θ0

i=1

ωi

(18.37)

L 

(λli − νil )δl , i = 1, . . . , N,

(18.40)

l=1

ν N  ρYi j l

δl

=

kl

j=1

Mi

, l = 1, . . . , L.

(18.41)

18.4 Conservation of Elements In Section 16.2 we have defined some linear combinations of the mass fractions of species which are conserved in a stirred tank, namely Zk =

N  i=1

aki ni =

N  i=1

aki

Yi , 1 ≤ k ≤ K. Mi

(18.42)

They represent the k-atoms of element Ak per kg of mixture. The situation is not the same in the case of a general flow: these fields are not conserved anymore at each particular position x because the mixture is moving

18.5. Equilibrium Chemistry

141

and element mass fractions may be different from one point to another. Nevertheless, they satisfy partial differential equations similar to (18.35) but without the reaction terms ωi on the right-hand side. In other words, they are transported like passive scalars. In order to obtain these equations, let us multiply (18.35) by aki /Mi and then add from i = 1, . . . , N . We get  aki ∂ (ρZk ) + div(ρZk v) − div(ρD gradZk ) = mi , k = 1, . . . , K, ∂t Mi i=1 N

(18.43)

where we have used the fact that (see (16.19)) N  aki ωi = 0, k = 1, . . . , K. M i i=1

(18.44)

Thus, in a first step, we can solve these partial differential equations for Zk , k = 1, . . . , K. Next we can construct, for each (x, t), the linear system (cf. (18.42)) N  i=1

aki

Yi (x, t) = Zk (x, t), 1 ≤ k ≤ K. Mi

(18.45)

In general, this system is under-determined because K < N so it cannot be solved to obtain the mass fractions of species. Nevertheless, it allows us to replace K nonlinear partial differential equations in the system (18.35) by K linear algebraic equations, which are much easier to solve.

18.5 Equilibrium Chemistry Sometimes chemical reactions are very fast compared to the residence time so we can assume local equilibrium chemistry at each point x and time t. As we will show below, in this case numerical computations are highly reduced because none of equations (18.35) needs to be solved. Instead, it is enough to solve the (linear) uncoupled equations (18.43) to determine fields Zk . Indeed, according to the previous analysis in Section 17, the local equilibrium problem to be solved at each (x, t) is the following: Find θ(x, t), π(x, t), n1 (x, t), . . . , nN (x, t) minimizing the specific Gibbs free energy ˆ π, n1 , . . . , nN ) G(θ,

(18.46)

under the following constraints: • mass

N  i=1

aki ni = Zk (x, t),

1 ≤ k ≤ K,

(18.47)

142

Chapter 18. Flow of a Mixture of Reacting Perfect Gases

• enthalpy N 




θ

Mi ni [h0i +

cˆπ,i (s)ds] = h(x, t),

(18.48)

θ0

i=1

• state law π = ρ(x, t)R

N 

ni θ,

(18.49)

i=1

• positivity

ni ≥ 0, i = 1, . . . , N,

(18.50)

where fields ρ(x, t), h(x, t) and Zk (x, t), k = 1, . . . , K are the solution of the partial differential equations system (18.32), (18.33), (18.34) and (18.43).

18.6 The Case of Low Mach Number For low Mach number, one can show that the spatial fluctuations of pressure are small compared to the pressure itself. In this case pressure may be considered as spatially constant in the state equation while, in the motion equation, the term gradπ has to be retained. Moreover, sometimes the mean pressure is given as a function of time and its value, to be denoted by π ¯=π ¯ (t), can be used in the state equation as well as in the specific Gibbs free energy. Then the problem to be solved is the following: For each (x, t), to find θ(x, t), n1 (x, t), · · · , nN (x, t) minimizing the specific Gibbs free energy ˆ π G(θ, ¯ (t), n1 , . . . , nN )

(18.51)

under the constraints: • mass

N 

1 ≤ k ≤ K,

(18.52)

cˆπ,i (s)ds] = h(x, t),

(18.53)

aki ni = Zk (x, t),

i=1

• enthalpy N  i=1


Mi ni [h0i

θ

+ θ0

18.6. The Case of Low Mach Number • positivity

ni ≥ 0, i = 1, . . . , N,

143

(18.54)

where fields h(x, t) and Zk (x, t), k = 1, . . . , K, are solution of partial differential equations (18.34) and (18.43). In the former, π˙ must be replaced by π ¯˙ , which is given. Remark 18.6.1. The rest of the equations, namely, (18.32), (18.33) and the state equation π ¯ = ρRθ, (18.55) are solved independently in order to determine density, velocity and pressure. More precisely, after computation of θ and Yi , i = 1, . . . , N , from the above equilibrium problem, we can use (18.55) to obtain density. Then we solve (18.32), (18.33) to get velocity and pressure. Actually, equation (18.32) can be viewed as a constraint on the velocity allowing us to determine pressure in equation (18.33) as a Lagrange multiplier. In this sense, problem (18.32), (18.33) is quite similar to the incompressible Navier-Stokes equations (13.18) and (13.2).

Chapter 19

The Method of Mixture Fractions We introduce a method based on the conservation of chemical elements which is very useful to reduce the number of equations to be solved.

19.1 General Equations We restrict ourselves to the steady state case. This means that fields do not depend on time and therefore equations (18.32)–(18.35) have to be replaced by div(ρv) = m,

(19.1)

div(ρv ⊗ v) + gradπ = div(l(D)) + b,

(19.2)

div(ρhv) − v · gradπ − div(ρD gradh) = l(D) · D + f,

(19.3)

div(ρYi v) − div(ρD gradYi ) = ωi + mi , i = 1, . . . , N.

(19.4)

Similarly, instead of equations (18.43) for Zk , we have div(ρZk v) − div(ρD gradZk ) =

N  i=1

aki

mi , Mi

1 ≤ k ≤ K.

(19.5)

We are going to see that, in some particular cases, namely, if boundary conditions have the particular form described below, then the solution of partial differential equations (19.5) can be obtained in an advantageous way.

146

Chapter 19. The Method of Mixture Fractions

Let us suppose the motion takes place in a fixed domain B0 , the boundary of which, Γ, consists of several parts: the inlets Γ1 , . . . , ΓP , the outlet Γo and the rest Γw = Γ \ (∪P (19.6) p=1 Γp ∪ Γo ), which will be called the wall. We make the following assumptions: 1. There are S sources defined by scalar fields ϕs , s = 1, . . . , S, such that mi (x) =

S 

bis ϕs (x), i = 1, . . . , N.

(19.7)

s=1

Function ϕs represents the rate of total mass per unit volume provided by the s-th source to the gas phase. Thus we assume that N 

bis = 1, s = 1, . . . , S,

(19.8)

i=1

and hence S 

ϕs =

S  N 

s=1

s=1 i=1

bis ϕs =

N  S 

bis ϕs =

i=1 s=1

N 

mi = m.

(19.9)

i=1

2. Boundary conditions for equations (19.4) are of the following form: Yi ∂Yi ∂n

=

Yi,p on Γp , p = 1, . . . , P, i = 1, . . . , N,

(19.10)

=

0 on Γo ∪ Γw , i = 1, . . . , N,

(19.11)

for some given constants Yi,p . As a consequence of this, boundary conditions for equations (19.5) are Zk ∂Zk ∂n

=

Zk,p on Γp , p = 1, . . . , P, k = 1, . . . , K,

(19.12)

=

0 on Γo ∪ Γw , k = 1, . . . , K,

(19.13)

where constants Zk,p denote the k-atoms of element Ak per kg of mixture entering the domain through inlet Γp . They can be computed from the mass fractions of species at Γp , Yi,p , by Zk,p =

N  i=1

aki

Yi,p , p = 1, . . . , P, k = 1, . . . , K. Mi

(19.14)

19.1. General Equations

147

In order to solve (19.5), let us define the scalar steady fields fp (x), p = 1, . . . , P , to be the solution of the partial differential equations div(ρfp v) − div(ρD gradfp ) = fp (x) ∂fp ∂n

0 in B0 ,

(19.15)

=

δnp on Γn , n = 1, . . . , P,

(19.16)

=

0 on Γo ∪ Γw ,

(19.17)

and the scalar fields gs , s = 1, . . . , S, as the solution of the boundary value problems div(ρgs v) − div(ρD gradgs )

= ϕs in B0 ,

(19.18)

= 0 on Γn , n = 1, . . . , P,

(19.19)

= 0 on Γo ∪ Γw .

(19.20)

S N mi (x)   bis = ( aki )ϕs (x), Mi Mi s=1 i=1

(19.21)

gs (x) ∂gs ∂n Since N 

aki

i=1

it is straightforward to see that the solution of equation (19.5) is given by Zk (x) =

P 

Zk,p fp (x) +

p=1

S 

αks gs (x) ∀x ∈ B0 , k = 1, . . . , K,

(19.22)

s=1

where αks is defined by αks :=

N  i=1

aki

bis . Mi

(19.23)

Moreover, we have P 

fp (x) +

p=1

S 

gs (x) = 1 ∀x ∈ B0 .

(19.24)

s=1

Indeed, by adding (19.15) from p = 1, . . . , P , together with (19.18) for s = 1, . . . , S, we obtain     P S P S     div ρv( fp + gs ) − div ρD grad( fp + gs ) p=1

s=1

p=1

=

s=1 S  s=1

ϕs = m.

(19.25)

148

Chapter 19. The Method of Mixture Fractions

Furthermore, by adding the boundary conditions, we get P 

fp +

p=1

S 

gs = 1 on Γp , p = 1, . . . , P,

(19.26)

s=1

P S ∂( p=1 fp + s=1 gs ) ∂n

= 0 on Γo ∪ Γw .

(19.27)

Finally, by using (19.1), it is straightforward to see that w(x) = 1 is the unique solution of the boundary-value problem div(ρwv) − div(ρD gradw) = m, w = 1 on Γp , p = 1, . . . , P, ∂w = 0 on Γo ∪ Γw , ∂n

(19.28) (19.29) (19.30)

from which (19.24) follows. Scalar fields fp and gs are called mixture fractions. Inlets and sources are called streams. We notice that, if the number of streams, P + S, is less than the number of chemical elements, K, then it is advantageous to solve the partial differential equations for fp and gs , instead of those for Zk , and then to compute the latter by (19.22). Moreover, it is enough to solve P + S − 1 of them and then to get the remaining stream by using (19.24). Finally we emphasize the fact that we do not require that each part of the boundary Γp be connected. Actually, each Γp may consist of several different physical inlets. We only need that input gases have the same uniform composition at these inlets (cf. (19.10)).

19.2 Examples In this section we consider some simple particular cases: 1. P=1, S=0 (one stream). f1 (x) ≡ 1,

Zk (x) = Zk,1 .

(19.31)

2. P=2, S=0 (two streams). f2 (x) = 1 − f1 (x),

Zk (x) = Zk,1 f1 (x) + Zk,2 (1 − f1 (x)).

(19.32)

Zk (x) = Zk,1 f1 (x) + αk1 (1 − f1 (x)).

(19.33)

3. P=1, S=1 (two streams). g1 (x) = 1 − f1 (x),

19.3. The Adiabatic Case

149

4. P=2, S=1 (three streams). g1 (x) = 1 − f1 (x) − f2 (x), Zk (x) = Zk,1 f1 (x) + Zk,2 f2 (x) + αk1 (1 − f1 (x) − f2 (x)).

(19.34) (19.35)

5. P=1, S=2 (three streams). g2 (x) = 1 − f1 (x) − g1 (x),

(19.36)

Zk (x) = Zk,1 f1 (x) + αk1 g1 (x) + αk2 (1 − f1 (x) − g1 (x)).

(19.37)

19.3 The Adiabatic Case Moreover, under the assumptions below, the specific enthalpy can also be obtained from the mixture fractions instead of solving (19.3) directly. Let us suppose that H1. pressure can be considered as spatially constant (isobaric process), H2. viscous dissipation l(D) · D may be neglected, H3. body heating, f , is null, H4. there is no heat flux by conduction through Γ0 ∪ Γw . In this case we say that the flow is adiabatic. We recall that the first assumption holds for low Mach number flows (see (18.6)). Moreover, we notice that assumption H3 is not compatible with the existence of a condensed phase. Under H1 to H4 the energy equation (19.3) becomes div(ρhv) − div(ρD gradh) = 0.

(19.38)

Furthermore, if boundary conditions on inlets are h

=

hp on Γp , p = 1, . . . , P,

(19.39)

for some constants hp , p = 1, . . . , P , then the specific enthalpy of the mixture is given by h(x) =

P 

hp fp (x) ∀x ∈ B0 .

(19.40)

p=1

19.4 The Case of Equilibrium Chemistry In this section we still suppose steady state. Furthermore, we assume local equilibrium chemistry and isobaric process in the sense that, except in the motion equation, pressure can be taken equal to a given constant π ¯.

150

Chapter 19. The Method of Mixture Fractions

We will see that we can make some previous computations allowing us to save computational effort in solving the whole model for the reacting flow of a mixture. Indeed, we notice that giving the mixture fractions f1 (x), . . . , fP (x), g1 (x), . . . , gS (x) and the specific enthalpy h(x), at any point x, we can determine temperature, density and composition of the mixture at equilibrium, at this point. For this purpose, we first determine the element mass fractions Zk , k = 1, . . . , K, by (19.22) and then temperature and composition are obtained by solving the equilibrium chemistry. The latter is done by minimizing ˆ π G(θ, ¯ , n1 , . . . , nN ),

(19.41)

under the constraints • mass N 

aki ni = Zk (x) =

P 

Zk,p fp (x) +

p=1

i=1

S 

αks gs (x), 1 ≤ k ≤ K,

(19.42)

s=1

• enthalpy N  i=1


Mi ni [h0i

θ

+

cˆπ,i (s)ds] = h(x),

(19.43)

θ0

• positivity ni ≥ 0, i = 1, . . . , N.

(19.44)

Next the density of the mixture at point x can be obtained from the state law by π ¯ ρ(x) = (19.45) R(x)θ(x) with R(x) = R

N  Yi (x) i=1

Mi

(19.46)

and Yi (x) = Mi ni (x). Thus, for given f1 , . . . , fP , g1 , . . . , gS and h we can compute θ, ρ and Yi , i = 1, . . . , N without solving any partial differential equation, just by solving the above constrained optimization problem.

19.4. The Case of Equilibrium Chemistry

151

Let us suppose that specific enthalpy is between two given values hmin and hmax . Then, in a first step, we may tabulate the “response” functions θˆ : A −→ R

(19.47)

ˆ 1 , . . . , fP , g1 , . . . , gS , h), (f1 , . . . , fP −1 , g1 , . . . , gS , h) −→ θ = θ(f ρˆ : A −→ R (19.48) (f1 , . . . , fP −1 , g1 , . . . , gS , h) −→ ρ = ρˆ(f1 , . . . , fP , g1 , . . . , gS , h), Yˆi : A −→ R

(19.49)

(f1 , . . . , fP −1 , g1 , . . . , gS , h) −→ Yi = Yˆi (f1 , . . . , fP , g1 , . . . , gS , h), where A = [0, 1]P +S−1 × [hmin , hmax ].

(19.50)

Then these tables are used as follows: we compute f1 (x), . . . , fP −1 (x), g1 (x), . . . , gS (x) (or f1 (x), . . . , fP (x), g1 (x), . . . , gS−1 (x)) and h(x) by solving partial differential equations (19.15)–(19.20) and (19.3), and, after that, we determine θ(x), Yi (x), i = 1, . . . , N and ρ(x) from the tables. Then this density is replaced in (19.1) and (19.2) which are solved to get velocity and pressure. Since ρ and v appear in these partial differential equations, some iterative procedure needs to be introduced. In practice (19.15), (19.18) and (19.3), with their respective boundary conditions, are solved by using numerical methods and, usually, the steady state is obtained as the limit of an evolutionary process as time goes to infinity, which is solved step by step by discretization schemes. We notice that the number of partial differential equations to be solved is P + S which is usually much smaller than N + 1 and even than K + 1. Finally, we notice that, under the additional assumptions H1 to H4, h(x) can be obtained by using (19.40) instead of solving the partial differential equation (19.3). Thus, h(x) is also given by means of a response function of the mixture fractions, namely h(x) = ˆ h(f1 , . . . , fP −1 ) :=

P 

hp fp (x).

(19.51)

p=1

ˆ ρˆ and Yˆi can be defined in terms of This also implies that response functions θ, f1 , . . . , fP −1 and g1 , . . . , gS only.

Chapter 20

Turbulent Flow of Reacting Mixtures of Perfect Gases, The PDF Method Throughout this chapter we make the assumptions introduced at the beginning of Chapter 19, namely, (19.7) and the particular boundary conditions (19.10) and (19.11). We recall that, when the flow is turbulent, the velocity induces fluctuations in the remaining physical variables: density, pressure, temperature and composition. However, as in Chapter 14, we are interested in determining the mean values of these fields. An approach to compute them is the following: instead of writing partial differential equations to directly determine these mean values, as we did for velocity and pressure in Chapter 14, we consider the mixture fractions as random variables with some prescribed probability density functions (PDF). Then, since density, temperature and mass fractions are functions of these mixture fractions through the response functions (19.47)–(19.49), we can get their mean values at each particular point x ∈ B0 by using classical results in probability theory.

20.1 Elements of Probability Let us first recall some basic notions on random variables and vectors. We consider a probability space, that is, a triple (Ω, F , P) consisting of a set Ω, a σ-field M of subsets of Ω and a probability measure P on (Ω, F ). Definition 20.1.1. A random variable Y is a function Y : Ω −→ R with the property that the set [Y ≤ y] := {ω ∈ Ω : Y (ω) ≤ y} belongs to F ∀y ∈ R.

154

Chapter 20. Turbulent Flow of Reacting Mixtures of Perfect Gases

Definition 20.1.2. The distribution function of a random variable Y is the function Φ : R −→ [0, 1] given by Φ(y) = P([Y ≤ y]). (20.1) Definition 20.1.3. The random variable Y is said to be continuous if its distribution function can be expressed as
y φ(s) ds, ∀y ∈ R (20.2) Φ(y) = −∞

for some integrable function φ : R −→ [0, ∞) called the probability density function of Y . Definition 20.1.4. The mean value or expectation of a continuous random variable Y with density function φ is the number
∞ Y¯ = E[Y ] := yφ(y) dy, (20.3) −∞

whenever this integral exists. Definition 20.1.5. The variance of a continuous random variable Y with probability density function φ is the number σ = E[(Y − E[Y ])2 ]

(20.4)

whenever it exists. Definition 20.1.6. A random vector Y = (Y1 , . . . , YN ) is a mapping Y : Ω −→ RP such that [Y ≤ y] := {ω ∈ Ω : Yi (ω) ≤ yi , i = 1, . . . , P } belongs to F ∀y = (y1 , . . . , yP ) ∈ RP . Definition 20.1.7. The (joint) distribution function of a random vector Y on the probability space (Ω, F , P) is the function Φ : RP −→ [0, 1], Φ(y) = P([Y ≤ y]).

(20.5)

Definition 20.1.8. A random vector is (jointly) continuous if their joint distribution function can be expressed as
y1
yP ··· φ(s1 , · · · , sP ) ds1 . . . dsP ∀(y1 , . . . , yP ) ∈ RP . Φ(y1 , . . . , yP ) = −∞

−∞

(20.6)

Definition 20.1.9. The random variables Y1 , . . . , YP are called independent if the events [Y1 ≤ y1 ], . . . , [YP ≤ yP ] are independent ∀(y1 , . . . , yP ) ∈ R, i.e., P(

P (

i=1

[Yi ≤ yi ]) =

P  i=1

P([Yi ≤ yi ]).

(20.7)

20.2. The Mixture Fraction/PDF Method

155

Proposition 20.1.10. Let us assume that Y1 , . . . , YP are independent continuous random variables with density functions φ1 , . . . , φP , respectively. Then the random vector Y = (Y1 , . . . , YP ) is continuous and its joint density function, φ is given by φ(y1 , . . . , yP ) = φ1 (y1 ) · · · φP (yP ). Proposition 20.1.11. Let (Y1 , . . . , YP ) be a random vector and F : RP −→ R a continuous function. Then the mapping ω ∈ Ω −→ F (Y1 (ω), . . . , YP (ω)) ∈ R

(20.8)

is a random variable. Furthermore its expectation is given by
∞
∞ E[F (Y1 , . . . , YP )] = ··· F (y1 , . . . , yP )φ(y1 , · · · , yP ) dy1 . . . dyP . −∞

−∞

(20.9) Corollary 20.1.12. Under the assumptions of the previous Proposition if, furthermore, Y1 , . . . , YP are independent continuous random variables with density functions φ1 , . . . , φN , respectively, then the expectation of F (Y1 , . . . , YP ) is given by
∞
∞ ··· F (y1 , . . . , yP )φ1 (y1 ) · · · φP (yP ) dy1 . . . dyP . E[F (Y1 , . . . , YP )] = −∞

−∞

(20.10)

20.2 The Mixture Fraction/PDF Method Let us come back to the turbulent reacting flow. We postulate that, at each point x ∈ B0 , the mixture fractions f1 (x), . . . , fP −1 (x) and g1 (x), . . . , gS (x) (or f1 (x), . . . , fP (x), g1 (x), . . . , gS−1 (x)) are independent continuous random variables with density functions φ1 (x, ·), . . . , φP −1 (x, ·) and ψ1 (x, ·), . . . , ψS (x, ·), respectively, which are completely determined by their respective expectations f¯1 (x), . . . , f¯P −1 (x) and g¯1 (x), . . . , g¯S (x), and variances σ1 (x), . . . , σP −1 (x) and ε1 (x), . . . , εS (x). Moreover, we assume that these expectations and variances are the solutions of the following so-called Favre averaged partial differential equations (see also Chapter 14): ηT ] gradf¯p ) = 0, (20.11) div(ρf¯p V) − div([ρD + δf f¯p = δpn on Γn , n = 1, . . . , P, (20.12) ∂ f¯p = 0 on Γo ∪ Γw , ∂n div(ρσp V) − div([ρD +

ηT ηT
] gradσp ) − 2 | gradf¯p |2 + Cd ρ σp = 0, δf δf k σp = 0 on Γn , n = 1, . . . , P, ∂σp = 0 on Γo ∪ Γw , ∂n

(20.13) (20.14) (20.15) (20.16)

156

Chapter 20. Turbulent Flow of Reacting Mixtures of Perfect Gases ηT ] grad¯ gs ) = ϕs , δg g¯s = 0 on Γn , n = 1, . . . , P, ∂¯ gs = 0 on Γo ∪ Γw , ∂n

div(ρ¯ gs V) − div([ρD +

div(ρεs V) − div([ρD +

ηT ηT
] gradεs ) − 2 | grad¯ gs |2 + Cd ρ εs = 0, δg δg k εs = 0 on Γn , n = 1, . . . , P, ∂εs = 0 on Γo ∪ Γw , ∂n

(20.17) (20.18) (20.19) (20.20) (20.21) (20.22)

where σf and σg are the turbulent Prandtl/Schmidt numbers and the turbulence constant Cd is taken to be 2. Concerning density functions φp (x, ·) and ψs (x, ·), they are usually chosen of the same form. Several possibilities are considered in the bibliography, two of which are given below (see [6]): 1. The double delta function  ⎧ if y = f¯i (x) − σi (x), ⎨ 0.5 φi (x, y) = 0.5 if y = f¯i (x) + σi (x), ⎩ 0 otherwise.

(20.23)

2. The β-function φi (x, y) =  1 0

y αi (x)−1 (1 − y)βi (x)−1 sαi (x)−1 (1 − s)βi (x)−1 ds

where αi (x) = f¯i (x) and

,

 ¯ fi (x)(1 − f¯i (x)) −1 σi (x)

 ¯ fi (x)(1 − f¯i (x)) ¯ βi (x) = (1 − fi (x)) −1 . σi (x)

(20.24)

(20.25)

(20.26)

Similar expressions can be taken for ψs (x, ·), s = 1, . . . , S. Now, in order to compute the local mean values of density, temperature and mass fractions of species, we use (20.10) with F replaced by the response functions

20.2. The Mixture Fraction/PDF Method

157

ρˆ, θˆ and Yˆi , i = 1, . . . , P, as defined in (19.47)–(19.49). We have
ρ¯(x)




1

1







1

···

=

1

···

0

0

0

ρˆ(y1 , . . . , yP , z1 , . . . , zS , h(x)) 0

φ1 (x, y1 ) · · · φP (x, yP )ψ1 (x, z1 ) · · · ψS (x, zS ) dy1 · · · dyP dz1 · · · dzS ,
¯ θ(x)

=

1




···

= 0

Y¯i (x)




1




1

ˆ 1 , . . . , yP , z1 , . . . , zS , h(x)) θ(y

··· 0

0

(20.27) 1

0

φ1 (x, y1 ) · · · φP (x, yP )ψ1 (x, z1 ) · · · ψS (x, zS ) dy1 · · · dyP dz1 · · · dzS , (20.28)
1
1
1
1 Yˆi (y1 , . . . , yP , z1 , . . . , zS , h(x)) ··· ··· 0

0

0

0

φ1 (x, y1 ) · · · φP (x, yP )ψ1 (x, z1 ) · · · ψS (x, zS ) dy1 · · · dyP dz1 . . . dzS , (20.29) ∀i = 1, . . . , N, where h is the solution of the turbulent version of the energy equation (19.3), namely div(ρhV) − div([ρD +

ηT ] gradh) = l(D) · D + f. δh

(20.30)

We recall that the response functions can be previously tabulated . Now we give a schematic idea on how to solve the whole problem. We distinguish several steps: 1. Tabulate the mean values of density, temperature and mass fractions as functions of the mean values and variances of the mixture fractions and the specific total enthalpy (the latter only in the case where (19.40) cannot be used). The data for this step are the following: • The mass fractions of species at each inlet Γp : Yi,p , i = 1, . . . , N, p = 1, . . . , P.

(20.31)

• Either the values of hp or the values of θp at inlets, if assumptions H1 to H4 hold, so that (19.40) can be applied to determine the specific enthalpy. • The mean pressure π ¯. This task consists of the following ones: (a) Tabulate the response functions (19.47)–(19.49):

158

Chapter 20. Turbulent Flow of Reacting Mixtures of Perfect Gases i. Compute the mass fractions of elements at inlets by Zk,p =

N 

aki

i=1

Yi,p , k = 1, . . . , K, p = 1, . . . , P, Mi

(20.32)

and then the right-hand side of (19.42), namely, P  p=1

Zk,p fp +

S 

αks gs , k = 1, . . . , K,

(20.33)

s=1

for the chosen values of f1 , . . . , fP −1 and g1 , . . . , gS (or f1 , . . . , fP , g1 , . . . , gS−1 ). ii. If (19.40) can be used and the values of temperature at inlets are given, then compute hp by  
θp P  Yi,p hi (θ0 ) + cˆπ,i (s) ds , p = 1, . . . , P. (20.34) hp = i=1

θ0

iii. If (19.40) can be used, compute the right-hand side of (19.43), namely, P  hp f p . (20.35) h= p=1

iv. Compute θ and n1 , . . . , nN as the solution of the equilibrium problem (19.41)–(19.44). v. Compute the molecular mass of the mixture by 1 M = N i=1

ni

(20.36)

and then ρ by

π ¯M . (20.37) Rθ (b) Use the tables for the response functions calculated above to obtain tables of mean values of density, temperature and mass fractions, against the mean values and variances of mixture fractions, and the specific enthalpy if (19.40) cannot be used. More precisely, for chosen values of numbers f¯1 , . . . , f¯P −1 , σ1 , . . . , σP −1 and g¯1 , . . . , g¯S , ε1 , . . . , εS , build the PDF functions φ1 , . . . , φP , ψ1 , . . . , ψS and then use (20.27), (20.28), (20.29), with (19.47), (19.48), (19.49). ρ=

2. Solve the turbulent flow equations (which are compressible versions of (14.13)–(14.16)), to determine the mean velocity V and pressure Π.

20.2. The Mixture Fraction/PDF Method

159

3. Compute the mean values of mixture fractions f¯1 , . . . , f¯P −1 and g¯1 , . . . , g¯S , and their variances σ1 , . . . , σP −1 and ε1 , . . . , εS by solving equations (20.11)– (20.16) and (20.17)–(20.22), and h by (20.30), for the case where we cannot use (19.40). 4. For each point x ∈ B0 , determine the mean values of ρ, θ and Yi , i = 1, . . . , N , at point x by using (20.27), (20.28) and (20.29), and the tables obtained in step 1. Remark 20.2.1. We notice that all of the equations to be solved in steps 2 and 3 need the density field. However, this field is calculated in step 4 making use of the mean values and variances of the mixture fractions. This means that some iterative process should be introduced in order to solve the whole coupled problem.

Appendix A

Vector and Tensor Algebra A.1 Vector Space. Basis A real vector space is an algebraic structure consisting of a set V endowed with two operations. One of them is an internal operation denoted by + and satisfying the following properties: i) Associativity. a + (b + c) = (a + b) + c ∀ a, b, c. ii) Existence of a neutral element 0 such that a + 0 = 0 + a = a ∀ a. iii) Existence of a symmetric element : for each a there exists b such that a + b = b + a = 0. iv) Commutativity. a + b = b + a ∀ a, b. Thus, V with the + operation is a commutative group. The second operation is an external one, the product by real numbers, satisfying the following properties: v) λ(a + b) = λa + λb. vi) λ(µa) = (λµ)a. vii) (λ + µ)a = λa + µa.

162

Appendix A. Vector and Tensor Algebra

viii) 1a = a. Elements of V are called vectors. A set of vectors {ei }N i=1 is said to be • linearly independent if N 

λi ei = 0 ⇒ λi = 0 ∀i ∈ {1, . . . , N },

(A.1)

i=1

• a system of generators if ∀v ∈ V ∃λi , i = 1, . . . , N, such that v =

N 

λi ei ,

(A.2)

i=1

• a basis if it is both linearly independent and a system of generators. If B = {e1 , . . . , eN } is a basis of V, then each v ∈ V can be written as a linear combination,

v=

N 

vi ei ,

i=1

in a unique way. The vi ∈ R, i = 1, . . . , N, are called coordinates of v with respect to basis B. N N Let us consider two bases in V, {ei }i=1 and {Ei }i=1 . In particular, vectors Ei can be written as linear combinations of vectors {ei }N i=1 , namely, El =

N 

Cli ei .

(A.3)

i=1

Now, let v be any vector in V. Let us denote by vie and viE , i = 1, . . . , N , the N N coordinates of v with respect to basis {ei }i=1 and {Ei }i=1 , respectively. If we e E denote by v the column vector of the former and by v the column vector of the latter, it is easy to see that v E = [C]v e , where [C] is the matrix [C]ij = Cij with Cij given in (A.3).

(A.4)

A.2. Inner Product

163

A.2 Inner Product A mapping ϕ : V × V −→ R is called an inner product if the following properties hold: 1) ϕ is bilinear : 1.1) ϕ(λ1 v1 + λ2 v2 , w) = λ1 ϕ(v1 , w) + λ2 ϕ(v2 , w), 1.2) ϕ(v, λ1 w1 + λ2 w2 , w) = λ1 ϕ(v, w1 ) + λ2 ϕ(v, w2 ). 2) ϕ is symmetric: ϕ(v, w) = ϕ(w, v). 3) ϕ is positive definite: ϕ(v, v) > 0 ∀ v = 0.

Finite dimensional vector spaces with an inner product are called Euclidean vector spaces. For them we can define a norm by 1

|v| = ϕ(v, v) 2 . In what follows we write v · w instead of ϕ(v, w). Two vectors v and w are said to be orthogonal if v · w = 0. A basis B = {e1 , . . . , eN } is said to be orthogonal if ei · ej = 0 ∀ i = j and is said to be orthonormal if  ei · ej = δij =

1 0

if i = j, if i =

j.

If B is orthonormal, then the i-th coordinate of a vector v is given by vi = v · ei . Moreover, the matrix to change coordinates with respect to two orthonormal basis is orthogonal, namely, (A.5) [C][C]t = [I], where [I] is the identity matrix.

164

Appendix A. Vector and Tensor Algebra

A.3 Tensors A tensor of order p is any p-linear mapping ← p →

T :V × ... × V→ R. The set of tensors of order p is denoted by Lp (V). When endowed with the two algebraic operations: • Sum (f + g)(v) = f (v) + g(v)

∀v ∈ V,

• Product by scalars (λf )(v) = λf (v),

∀λ ∈ R

∀v ∈ V,

it is a vector space. Given a basis B = {e1 , ..., eN } of V, we can associate to any f ∈ Lp (V) the p-dimensional array fi1 ,...,ip := f (ei1 , ..., eip ),

1 ≤ i1 , ..., ip ≤ N.

(A.6)

These numbers are called coordinates of f with respect to basis B. A tensor is called isotropic if its coordinates do not depend on the basis. For p = 1, the space L1 (V) of first order tensors is nothing but the dual space of V which is isomorphic to V. Each f ∈ L1 (V) is identified to the unique v ∈ V such that f (w) = v · w ∀w ∈ V. For any given orthonormal basis, B, one can show that the coordinates of tensor f , defined by A.6 with p = 1, coincides with the coordinates of vector v. Moreover, the only isotropic first order tensor is f ≡ 0. For p = 2, the space L2 (V) is isomorphic to Lin, the space of endomorphisms of V. A tensor f ∈ L2 (V) is identified to the unique endomorphism S ∈ Lin such that f (v, w) = Sw · v ∀v, w ∈ V. It is easy to see that, given a basis, the entries of the matrix associated with the endomorphism S are the coordinates of f given by (A.6) with p = 2. Moreover, one can show that f is isotropic if and only if S = QSQt ∀Q ∈ Orth+ . For p = 4, the space L4 (V) is isomorphic to L(Lin, Lin), the space of endomorphisms of Lin. Indeed, each f ∈ L4 (V) can be identified to the unique endomorphism of Lin, l, such that f (v1 , v2 , v3 , v4 ) = l(v3 ⊗ v4 ) · (v1 ⊗ v2 ) ∀vi ∈ V, i = 1, . . . , 4. In what follows, due to the above identifications, the elements of Lin will be called (second order) tensors and the endomorphisms of Lin will be called fourth order tensors.

A.3. Tensors

165

Second Order Tensors Let a and b be two vectors in V. The tensor (a ⊗ b)v := v · b a is called the tensor product of a and b. If e is a unit vector, then (e ⊗ e) v = (v · e) e is the orthogonal projection of v on the straight line generated by e. Moreover, (I − e ⊗ e)v = v − (v · e) e,

(A.7)

is the projection of v on the orthogonal plane to e. If B = {e1 , . . . , eN }, the set of tensors {ei ⊗ ej : 1 ≤ i, j ≤ N } is a basis of Lin. More precisely, we have S=

N 

Sij ei ⊗ ej ,

i,j=1

where Sij = S ej · ei are the coordinates of S with respect to basis B. We notice that the coordinates of a tensor S can be arranged as a matrix, ⎞ ⎛ S11 . . . S1N ⎜ .. ⎟ . .. [S] = ⎝ ... . . ⎠ S N 1 . . . SN N For example, the matrix of coordinates of tensor a ⊗ b is the matrix product ⎛ ⎞ a1 ⎜ a2 ⎟ ⎜ ⎟ (A.8) ⎜ .. ⎟ (b1 b2 · · · bN ), ⎝ . ⎠ aN where a =

N  i=1

ai ei and b =

N 

b i ei .

i=1

Let us see how tensor coordinates change when we change the basis in vector e E space V. Let S ∈ Lin be a tensor. Let us denote by Sij and Sij , i, j = 1, . . . , N ,

166

Appendix A. Vector and Tensor Algebra N

N

its coordinates with respect to basis {ei }i=1 and {Ei }i=1 , respectively. Then it is not difficult to see that the following relation holds: [S E ] = [C][S e ][C]t .

(A.9)

In vector space Lin we can introduce another internal operation which is the mapping composition: given two tensors S and T , ST is the tensor defined by (ST )v = S(T v). It is easy to see that the matrix of coordinates of ST is the product of those corresponding to S and T , i.e., [ST ] = [S][T ].

(A.10)

The transpose tensor of S is the unique tensor S t satisfying Sa · b = a · S t b ∀ a, b ∈ V. A tensor is called symmetric if S = S t and skew if S = −S t . The subspace of symmetric tensors will be denoted by Sym and that of skew tensors by Skw. We can also define the inner product of tensors by S · T = tr(S t T ), where tr denotes the trace operator which is the unique linear operator from Lin into R satisfying tr(a ⊗ b) = a · b. We have ⎛ tr(S) = tr ⎝

N 

⎞ Sij ei ⊗ ej ⎠ =

i,j=1

N 

Sij ei · ej =

i,j=1

N  i,j=1

Sij δij =

N  i=1

and hence S·T =

N 

Sij Tij .

i,j=1

If T ∈ Sym and W ∈ Skw then it easy to show that T · W = 0.

Sii ,

A.3. Tensors

167

Any tensor S is the sum of one symmetric tensor E and one skew tensor W . More precisely, E and W are defined by  1 S + St , 2  1 S − St . W = 2 E=

(A.11) (A.12)

Tensor E is called the symmetric part of S while W is called the skew part of S. Furthermore, this decomposition is unique. More precisely, Lin = Sym

⊥ .

Skw.

The following equalities can be easily proved: (a ⊗ b)(c ⊗ d) = (b · c)a ⊗ d. R · (ST ) = (S t R) · T = (RT t ) · S

∀R, S, T ∈ Lin.

(A.13) (A.14)

1 · S = tr(S).

(A.15)

S · (a ⊗ b) = a · Sb.

(A.16)

(a ⊗ b)(c ⊗ d) = (a · c)(b · d).

(A.17)

Let N = 3 and {e1 , e2 , e3 } be a positively oriented orthonormal basis in V. We define the cross product of vectors a and b by a×b=

3 

Eijk aj bk ei .

i,j,k=1

where Eijk

⎧ ⎨ 1 −1 = ⎩ 0

if ijk is an even permutation of (1 2 3), if ijk is an odd permutation of (1 2 3), otherwise .

One can prove the following equality: a × (b × c) = (a · c)b − (a · b)c

(A.18)

showing that the cross product is not associative. There is an isomorphism between the vector space V and the vector space of skew tensors. It is defined by V w

−→ Skw −→ W

168

Appendix A. Vector and Tensor Algebra

with W v := w × v ∀v ∈ V. Vector w is called the axial vector associated to W . A tensor S is invertible if there exists another tensor S −1 , called the inverse of S, such that SS −1 = S −1 S = I. (A.19) A tensor Q is orthogonal if it preserves the inner product, that is, Qu · Qv = u · v ∀u, v ∈ V.

(A.20)

A necessary and sufficient condition for Q to be orthogonal is that QQt = Qt Q = I,

(A.21)

Qt = Q−1 .

(A.22)

which is equivalent to

In particular, det Q must be 1 or −1. A rotation is an orthogonal tensor with positive determinant. The set of rotations is a subspace of Lin which will be denoted by Orth+ . One can show that a tensor S is isotropic if and only if it commutes with any rotation, that is, S = QSQt ∀Q ∈ Orth+ . Furthermore (see, for instance, [5]), this condition is satisfied if and only if S is spherical, i.e., S = ωI for some scalar ω. A tensor S is called deviatoric if it is symmetric and traceless. We denote by Dev the subspace of deviatoric tensors. Any symmetric tensor S ∈ Sym can be uniquely decomposed as the sum of a spherical tensor and a deviatoric tensor. Indeed, the tensor 1 tr(S)I SD = S − N is a deviatoric tensor, and we can rewrite this equality in the form S=

1 tr(S)I + SD . N

Moreover the subspace of spherical tensors is orthogonal to the subspace of deviatoric tensors because I · S = tr(S) = 0 for any S ∈ Dev. Hence, we have the orthogonal direct sum decomposition Sym = Sph

⊥ .

Dev,

where Sph denotes the subspace of spherical tensors. A tensor S is positive definite if v · Sv > 0

∀v = 0.

(A.23)

A.3. Tensors

169

The principal invariants of a tensor S are the three numbers ı1 (S) = tr(S), 1 ı2 (S) = [( tr(S))2 − tr(S 2 )], 2 1 ı3 (S) = det(S) = [( tr(S))3 + 2( tr(S))2 − 3 tr(S 2 ) tr(S)]. 6

(A.24) (A.25) (A.26)

The set of principal invariants of S will be denoted by IS .

Fourth Order Tensors Let T, S ∈ Lin. The fourth order tensor T ⊗ S ∈ L(Lin, Lin) defined by (T ⊗ S)R = S · RT

∀R ∈ Lin,

is called the tensor product of T and S. For the basis B of V, the set {(ei ⊗ ej ) ⊗ (ek ⊗ el ),

1 ≤ i, j, k, l ≤ N }

is a basis of the space of fourth order tensors. In fact, for any fourth order tensor l we have N  l= lijkl (ei ⊗ ej ) ⊗ (ek ⊗ el ), i,j,k,l=1

where lijkl = l(ek ⊗ el ) · (ei ⊗ ej )

(A.27)

are the coordinates of l with respect to basis B. One can show that l is isotropic (i.e., its coordinates are independent of the basis) if and only if l(QSQt ) = Ql(S)Qt

∀Q ∈ Orth+

∀S ∈ Lin.

(A.28)

The following result has important applications in continuum mechanics: Proposition 1.3.1. Let l ∈ L(Lin, Lin) be a fourth order tensor. Then l is isotropic if and only if there exist three scalars α, β and γ such that l(S) = α tr(S)I + βS + γS t .

(A.29)

Proof. If l is of the form (A.29) then it is isotropic because (A.28) holds. Indeed, let Q ∈ Orth+ . We have l(QSQt ) = α tr(QSQt )I + βQSQt + γ(QSQt )t   = Q α tr(QSQt )I + βS + γS Qt = Ql(S)Qt ,

(A.30) (A.31)

170

Appendix A. Vector and Tensor Algebra

because tr(QSQt ) = tr(S) ∀Q ∈ Orth. The proof of this follows immediately by using (A.14): tr(QSQt ) = I · QSQt = Qt · SQt = QQt · S = I · S = tr(S). The proof of the converse result is much more difficult and will be omitted here (see [4]).
Corollary 1.3.2. Let l ∈ L(Lin, Lin) be a fourth order tensor. Then, if l is isotropic, there exist two scalars λ and µ such that l(E) = λ tr(E)I + 2µE ∀E ∈ Sym. Proof. It is enough to take λ = α and µ =

β+γ 2

in (A.29).




Corollary 1.3.3. The isotropic fourth order tensor l given by (A.29) satisfies l(S) ∈ Sym ∀S ∈ Lin if and only if β = γ in which case

S + St l(S) = l ∀S ∈ Lin. 2 Proof. It follows from the fact that tr(S) = tr(S t ) ∀S ∈ Lin.




It is immediate to see that the coordinates of l given by (A.29) are, in any orthonormal basis, lijkm = αδij δkm + βδik δmj + γδim δjk . Indeed, according to (A.27), lijkl = l(ek ⊗ el ) · (ei ⊗ ej ) = (α tr(ek ⊗ em )I + βek ⊗ em + γem ⊗ ek ) · ei ⊗ ej = αek · em ei · ej + βek · ei em · ej + γem · ei ek · ej ,

(A.32)

and the result follows.

A.4 The Affine Space An affine space is a triple (E, V, +) consisting of a set E, a vector space V and a mapping + E × V −→ E (p, v) −→ p + v satisfying the following property: For any p, q ∈ E there exists a unique v ∈ V such that p + v = q. → This vector v is usually denoted by − pq or q − p.

A.4. The Affine Space

171

If V is a Euclidean vector space, then the corresponding affine space is called Euclidean affine space. In a Euclidean affine space we can define a distance by E ×E (p, q)

d −→ R → −→ d(p, q) = |− pq|.

Thus (E, d) is a metric space and hence a topological space. In a Euclidean affine space, a cartesian coordinate frame is a pair (o, {ei , i = 1, . . . , N }) where o is any fixed point in E called the origin and {ei , i = 1, . . . , N } is an orthonormal basis of the Euclidean vector space V. For any point p ∈ E its coordinates with respect to the above cartesian frame are the numbers pi , i = 1, . . . , N, such that − → op =

N  i=1

pi ei .

Appendix B

Vector and Tensor Analysis A field is any mapping defined in a subset of E and valued in R, V or Lin, in which case it is called a scalar, vector or tensor field, respectively.

B.1 Differential Operators Let W be a normed vector space (in practice W = R, V, Lin). A mapping f : R ⊂ E → W defined in an open set R of E is said to be differentiable at point x ∈ R if there exists a linear mapping X : V −→ W such that f(x + h) − f(x) − X (h) = o(h) as h → 0, which means lim

h→0

f(x + h) − f(x) − X (h) = 0. h

If f is differentiable at x, then linear mapping X is called the differential of f at x. It is denoted by Df (x). Particular cases: 1) Let us assume W = R, i.e., f = ϕ is a scalar field differentiable at x. Then Dϕ(x) : V −→ R is a linear mapping and the unique vector ∇ϕ(x) such that Dϕ(x)a = ∇ϕ(x) · a ∀a ∈ V is called the gradient of ϕ at x. If ϕ is differentiable at all points in R we can define the vector field

174

Appendix B. Vector and Tensor Analysis

R⊂E x

−→ V −→ ∇ϕ(x),

which is called the gradient of ϕ. 2) Let us consider the case where W = V, i.e., f = u is a vector field. Then Du(x) : V −→ V, that is, Du(x) ∈ Lin. If u is differentiable at each point in R, we can define the tensor field R⊂E x

−→ Lin −→ Du(x),

which is called the gradient of u. In order to denote this differential operator we will use ∇u or gradu, instead of Du, Now we introduce other differential operators. Given a differentiable vector field u, the scalar field div u : R ⊂ E x

−→ R −→ div u(x) =: tr(∇u(x))

is called the divergence of u. Let S : R ⊂ E −→ Lin be a smooth tensor field. The divergence of S at point x ∈ R is defined as the unique vector, div S(x), such that div S(x) · a = div(S t a)(x) ∀ a ∈ V. Let u : R ⊂ E −→ V be a smooth vector field. The curl of u at x is the unique vector, curl u(x), such that, curl u(x) × a = (∇u(x) − ∇ut (x))a ∀ a ∈ V. Let S : R ⊂ E −→ Lin be a smooth tensor field. We define the curl of S at point x as the unique tensor curl S(x) such that curl S(x)a = curl(S t a)(x) ∀ a ∈ V. For a scalar field ϕ : R ⊂ E −→ R, its Laplacian is the scalar field ∆ϕ(x) = div(∇ϕ)(x).

B.2. Curves and Curvilinear Integrals For a vector field u : R ⊂ E −→ V, its Laplacian is the vector field ∆u(x) = div(∇u)(x). The following equalities hold: ∇(ϕv)

=

ϕ∇v + v ⊗ ∇ϕ,

div(ϕv)

=

ϕ div v + v · ∇ϕ,

∇(v · w) =

(∇w)t v + (∇v)t w,

div(v ⊗ w) =

v div w + (∇v)w,

t

div(S v)

=

div(ϕS) = div(v × w) =

S · ∇v + v · div S, ϕ div S + S∇ϕ, w · curl v − v · curl w,

div ∇vt

=

∇(div v),

curl(ϕv)

=

∇ϕ × v + ϕ curl v,

curl(v × w) = div curl v

∇vw − w div v + v div w − ∇wv,

=

0,

curl ∇ϕ =

0,

∆v

=

∇(div v) − curl(curl v),

∆(∇ϕ)

=

∇(∆ϕ),

div(∆v)

=

∆(div v),

curl(∆v)

=

∆(curl v),

∆(ϕψ)

=

ϕ∆ψ + ψ∆ϕ + 2∇ϕ · ∇ψ.

B.2 Curves and Curvilinear Integrals A (regular) curve c in R is a smooth map c : [0, 1] −→ R such that c˙ (σ) = 0 ∀σ ∈ [0, 1]. The curve is closed if c(0) = c(1). The length of c is the number

175

176

Appendix B. Vector and Tensor Analysis


1

|˙c(σ)| dσ.

length(c) = 0

Let v : R ⊂ E −→ V be a continuous vector field. The curvilinear integral of v along c is the scalar defined by





1

v(x) · dx =

v(c(σ)) · c˙ (σ) dσ. 0

c

Similarly, let S : R ⊂ E −→ Lin be a continuous tensor field. The curvilinear integral of S along c is the vector defined by

1 S(x)dx = S(c(σ))˙c(σ) dσ. 0

c

If v is the gradient of a scalar field ϕ, then








v(x) · dx = c


= 0

In particular

 c

1

∇ϕ(x) · dx = c 1

∇ϕ(c(σ)) · c˙ (σ)dσ 0

d (ϕ ◦ c)(σ) dσ = ϕ(c(1)) − ϕ(c(0)). dσ

(B.1)

∇ϕ(x) · dx = 0 whenever c is closed.

A subset R ⊂ E is simply connected if any closed curve in R can be continuously deformed to a point without leaving R. An open region is any connected subset of E. The closure of an open region will be called a closed region. We designate by regular region any closed region with smooth boundary ∂R. Let R be a closed region and Φ a field defined in R. We say that Φ is C1 (R) if it is continuously differentiable in the interior of R, and Φ and ∇Φ have continuous extensions to all of R. We end this section on vector calculus by recalling some fundamental results. Theorem 2.2.1. Potential Theorem. Let v be a smooth vector field on an open or closed simply connected region R and assume that, curl v = 0. Then there is a C2 scalar field ϕ : R −→ R such that, v = ∇φ.

B.3. Gauss’ and Green’s Formulas. Stokes’ Theorem

177

B.3 Gauss’ and Green’s Formulas. Stokes’ Theorem Theorem 2.3.1. Divergence Theorem. Let R be a bounded regular region and let ϕ : R −→ R, v : R −→ V and S : R −→ Lin be smooth fields. Then,

ϕn dAx = ∇ϕ dVx , 1.


R

∂R




v ⊗ n dAx =

2.


∂R


v · n dAx =

3.


∂R

4.


Sn dAx =

∂R

div v dVx ,

R

R

∇v dVx ,

R

div S dVx .

Theorem 2.3.2. Green’s formulas. Let R be a bounded regular region and let ϕ, ψ : R −→ R, v, w : R −→ V and S : R −→ Lin be smooth fields. Then,


∂ϕ ψ dAx = 1. ∇ϕ · ∇ψ dVx + ∆ϕ ψ dVx , ∂R ∂n R R


2. Sn · v dAx = S · ∇v dVx + v · div S dVx ,


R

∂R

(Sn) ⊗ v dAx =

3.


∂R




∂R




∂R

R

R




div v ψ dVx +

R

R

R





R

∂R

div S ⊗ v dVx ,

v · ∇ψ dVx ,

(div w) v dVx +

(v × w) · n dAx =

6.

S∇v dVx +




(w · n)v dAx =

5.




t


v · nψ dAx =

4.

R




R

(∇v)w dVx ,


w · curl v dVx −

R

v · curl w dVx .

Theorem 2.3.3. Stokes’ Theorem. Let v (respectively, S) be a smooth vector (respectively, tensor) field on an open set R ⊂ E. Let S be a smooth surface in R bounded by c, supposed to be a closed curve. At each point of S we choose a unit normal vector n consistent with the direction of traversing c in that it causes a right-handed screw to advance along n. Then,



S

curl v · n dAx =

v · dlx , c







t

S

(curl S) n dAx =

S dlx . c

178

Appendix B. Vector and Tensor Analysis

B.4 Change of Variable in Integrals Theorem 2.4.1. Let f be a deformation of a body B and ϕ : f (B) → R a smooth scalar field. Let c (resp. S) be a smooth curve (resp. surface) in B and P a part of B. Then we have

1. ϕt dlx = ϕ ◦ f F k dlp ,


f (c)

c




2.


f (S)

3.


ϕn dAx =

f (c)







ϕ ◦ f F k dlp ,

ϕ dlx =

4. f (S)

5. f (P)

ϕ ◦ f detF F −t m dAp ,

S

c




ϕ dAx =

S

ϕ ◦ f detF F −t m dAp ,


ϕ dVx =

P

ϕ ◦ f detF dVp ,

where k (resp. t) denotes a unit tangent vector to the curve c (resp. f (c)) and m (resp. n) denotes a unit normal vector to surface S (resp. f (S)).

B.5 Transport Theorems Theorem 2.5.1. Let X be a motion of a body B. Let c, S and P be a curve, a surface and a part of B. Let ϕ : T → R and w : T → V be a scalar field and a vector field, respectively. Then we have

d ϕt dlx = (ϕt ˙ + ϕLt) dlx , 1. dt ct ct

  d 2. ϕn ˙ + ϕ divv n − ϕLt n dAx , ϕn dAx = dt St St


d 3. ϕ dVx = (ϕ˙ + ϕ divv) dVx = (ϕ
+ div(ϕv)) dVx dt Pt Pt P
t

= ϕ dVx + ϕv · n dAx , 4. 5.

d dt d dt

d 6. dt

Pt





w ⊗ t dlx = ct

˙ ⊗ t + w ⊗ Lt) dlx , (w ct







  ˙ · t + Lt w · t dlx , w

w · t dlx = ct




St

∂Pt

ct

w ⊗ n dAx =


St

  ˙ ⊗ n + divv w ⊗ n − w ⊗ Lt n dAx , w

B.6. Localization Theorem 7.

d dt

d 8. dt





St

w · n dAx =

˙ · n + divv w · n − ∇vw · n) dAx (w
St = (w
· n + divw v · n + curl(w × v) · n) dAx ,


Pt

179


wdVx =

St




˙ + divv w) dVx = (w Pt
=

(w
+ div(w ⊗ v)) dVx
w
dVx + wv · n dAx ,

Pt

Pt

∂Pt

where t denotes a unit tangent vector to ct and n is a unit normal vector to St .

B.6 Localization Theorem Theorem 2.6.1. Let Φ be a continuous scalar or vector valued mapping defined in an open set R ⊂ E. If
Φ dVx ≥ 0 B

for every closed ball B ⊂ R, then Φ ≥ 0 in R. Hence, if
Φ dVx = 0 B

for every closed ball B ⊂ R, then Φ ≡ 0 in R.

B.7 Differential Operators in Coordinates B.7.1 Cartesian Coordinates Definition f : Ω = R3 → R3 f (u1 , u2 , u3 ) = (u1 , u2 , u3 ).

(B.2) (B.3)

g1 = e1 ,

g2 = e2 ,

g3 = e3 .

(B.4)

g 1 = e1 ,

g 2 = e2 ,

g 3 = e3 .

(B.5)

Basis • Contravariant • Covariant • Physical e1 ,

e2 ,

e3 .

(B.6)

180

Appendix B. Vector and Tensor Analysis

Christoffel symbols of the second kind ⎛ Γ1jk

Γ2jk

Γ3jk

0 0 =⎝ 0 0 0 0 ⎛ 0 0 =⎝ 0 0 0 0 ⎛ 0 0 =⎝ 0 0 0 0

⎞ 0 0 ⎠, 0 ⎞ 0 0 ⎠, 0 ⎞ 0 0 ⎠. 0

(B.7)

(B.8)

(B.9)

Differential Operators in the Physical Basis • Gradient of scalar field ∇φ =

∂φ ∂φ ∂φ e1 + e2 + e3 . ∂x1 ∂x2 ∂x3

(B.10)

• Gradient of a vector field ∂w1 ∂w1 ∂w1 e1 ⊗ e1 + e1 ⊗ e2 + e1 ⊗ e3 ∂x1 ∂x2 ∂x3 ∂w2 ∂w2 ∂w2 + e2 ⊗ e1 + e2 ⊗ e2 + e2 ⊗ e3 ∂x1 ∂x2 ∂x3 ∂w3 ∂w3 ∂w3 + e3 ⊗ e1 + e3 ⊗ e2 + e3 ⊗ e3 . ∂x1 ∂x2 ∂x3

∇w =

(B.11)

• Convective term

  ∂w1 ∂w1 ∂w1 (∇w)w = w1 + w2 + w3 e1 ∂x1 ∂x2 ∂x3   ∂w2 ∂w2 ∂w2 + w1 + w2 + w3 e2 ∂x1 ∂x2 ∂x3   ∂w3 ∂w3 ∂w3 + w1 + w2 + w3 e3 . ∂x1 ∂x2 ∂x3

• Curl of a vector field

(B.12)



∂w3 ∂w2 curl w = − e1 ∂x2 ∂x3

∂w1 ∂w3 + − e2 ∂x3 ∂x1

∂w2 ∂w1 + − e3 . ∂x1 ∂x2

(B.13)

B.7. Differential Operators in Coordinates

181

• Divergence of a vector field div w =

∂w2 ∂w3 ∂w1 + + . ∂x1 ∂x2 ∂x3

(B.14)

• Divergence of a tensor field 

 ∂S11 ∂S12 ∂S13 + + div S = e1 ∂x1 ∂x2 ∂x3   ∂S21 ∂S22 ∂S23 + + + e2 ∂x1 ∂x2 ∂x3   ∂S31 ∂S32 ∂S33 + + + e3 . ∂x1 ∂x2 ∂x3

(B.15)

• Laplacian of a scalar field ∆φ =

∂2φ ∂2φ ∂2φ + + . ∂x21 ∂x22 ∂x23

(B.16)

• Laplacian of a vector field ∆w = ∆w1 e1 + ∆w2 e2 + ∆w3 e3 .

(B.17)

Measure elements for integration • Line dx1 , dx2 , dx3 .

(B.18)

dx2 δx3 , dx1 dx3 , dx1 dx2 .

(B.19)

dx1 dx2 dx3 .

(B.20)

• Surface

• Volume

182

Appendix B. Vector and Tensor Analysis

B.7.2 Cylindrical Coordinates Definition f : Ω = (0, ∞) × (0, 2π) × (−∞, ∞) → R3 f (r, θ, z) = (r cos θ, r sin θ, z).

(B.21) (B.22)

⎧ ⎨ g1 (r, θ, z) = cos θe1 + sin θe2 , g2 (r, θ, z) = −r sin θe1 + r cos θe2 , ⎩ g3 (r, θ, z) = e3 .

(B.23)

Basis • Contravariant

• Covariant

• Physical

⎧ 1 ⎨ g (r, θ, z) = cos θe1 + sin θe2 , g2 (r, θ, z) = − r1 sin θe1 + 1r cos θe2 , ⎩ 3 g (r, θ, z) = e3 . ⎧ ⎨ er (r, θ, z) = cos θe1 + sin θe2 , eθ (r, θ, z) = − sin θe1 + cos θe2 , ⎩ ez (r, θ, z) = e3 .

(B.24)

(B.25)

Christoffel symbols of the second kind ⎛

0 Γ1jk = ⎝ 0 0 ⎛ 0 Γ2jk = ⎝ 1r 0 ⎛ 0 Γ3jk = ⎝ 0 0

⎞ 0 0 ⎠, 0 ⎞ 1 0 r 0 0 ⎠, 0 0 ⎞ 0 0 0 0 ⎠. 0 0

0 −r 0

(B.26)

(B.27)

(B.28)

Differential operators in the physical basis • Gradient of scalar field ∇φ =

∂φ 1 ∂φ ∂φ er + eθ + ez . ∂r r ∂θ ∂z

(B.29)

B.7. Differential Operators in Coordinates

183

• Gradient of a vector field   1 ∂wr 1 ∂wr ∂wr er ⊗ er + − wθ er ⊗ eθ + er ⊗ ez ∇w = ∂r r ∂θ r ∂z   1 ∂wθ 1 ∂wθ ∂wθ eθ ⊗ er + + wr eθ ⊗ eθ + eθ ⊗ ez + ∂r r ∂θ r ∂z ∂wz 1 ∂wz ∂wz ez ⊗ er + ez ⊗ eθ + ez ⊗ ez . + ∂r r ∂θ ∂z

(B.30)

• Convective term   1 ∂wr (wθ )2 ∂wr ∂wr wr + wθ + wz − er ∂r r ∂θ ∂z r   1 ∂wθ wr wθ ∂wθ ∂wθ + wθ + wz + + wr eθ ∂r r ∂θ ∂z r   1 ∂wz ∂wz ∂wz + wθ + wz + wr ez . ∂r r ∂θ ∂z

∇ww =

(B.31)

• Curl of a vector field

1 ∂wz ∂wθ − er r ∂θ ∂z

∂wr ∂wz + + − eθ ∂r ∂z

1 ∂ 1 ∂wr (rwθ ) − + ez . r ∂r r ∂θ

curl w =

(B.32)

• Divergence of a vector field

div w =

∂wz 1 ∂wθ 1 ∂ (rwr ) + + . r ∂r r ∂θ ∂z

• Divergence of a tensor field   1 ∂ ∂Srz 1 ∂Srθ div S = (rSrr ) + +r − Sθθ er r ∂r ∂θ ∂z r   ∂Sθr 1 ∂Sθθ ∂Sθz 1 1 + + + Srθ + Sθr eθ + ∂r r ∂θ ∂z r r   ∂Szz 1 ∂ ∂Szθ (rSzr ) + +r + ez . r ∂r ∂θ ∂z

(B.33)

(B.34)

184

Appendix B. Vector and Tensor Analysis

• Laplacian of a scalar field 





 1 ∂ ∂φ ∂ 1 ∂φ ∂ ∂φ ∆φ = r + + r r ∂r ∂r ∂θ r ∂θ ∂z ∂z

2 2 1 ∂ ∂φ 1 ∂ φ ∂ φ = r + 2 2 + 2. r ∂r ∂r r ∂θ ∂z

(B.35)

• Laplacian of a vector field     1 2 ∂wθ 1 2 ∂wr ∆w = ∆wr − 2 wr − 2 er + ∆wθ − 2 wθ + 2 eθ + ∆wz ez . r r ∂θ r r ∂θ (B.36) Measure elements for integration • Line dr, r dθ, dz.

(B.37)

r dθdz, drdz, r drdθ.

(B.38)

r drdθdz.

(B.39)

f : Ω = (0, ∞) × (0, π) × (0, 2π) → R3 , f (r, θ, ϕ) = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ).

(B.40)

• Surface

• Volume

B.7.3 Spherical Coordinates Definition

Basis • Contravariant ⎧ ⎨ g1 (r, ϕ, θ) = sin θ cos ϕe1 + sin θ sin ϕe2 + cos θe3 , g2 (r, ϕ, θ) = r cos θ cos ϕe1 + r cos θ sin ϕe2 − r sin θe3 , ⎩ g3 (r, ϕ, θ) = −r sin θ sin ϕe1 + r sin θ cos ϕe2 . • Covariant ⎧ 1 ⎨ g (r, ϕ, θ) = sin θ cos ϕe1 + sin θ sin ϕe2 + cos θe3 , g2 (r, ϕ, θ) = 1r cos θ cos ϕe1 + 1r cos θ sin ϕe2 − 1r sin θe3 , ⎩ 3 1 1 g (r, ϕ, θ) = − r sin θ sin ϕe1 + r sin θ cos ϕe2 .

(B.41)

(B.42)

B.7. Differential Operators in Coordinates

185

• Physical ⎧ ⎨ er (r, ϕ, θ) = sin θ cos ϕe1 + sin θ sin ϕe2 + cos θe3 , eθ (r, ϕ, θ) = cos θ cos ϕe1 + cos θ sin ϕe2 − sin θe3 , ⎩ eϕ (r, ϕ, θ) = − sin ϕe1 + cos ϕe2 .

(B.43)

Christoffel symbols of the second kind ⎛ ⎞ 0 0 0 ⎠, 0 Γ1jk = ⎝ 0 −r 2 0 0 −r sin θ ⎛ ⎞ 0 0 1r ⎠, Γ2jk = ⎝ 1r 0 0 0 0 − 21 sin 2θ ⎛ ⎞ 1 0 0 r 0 cot θ ⎠ . Γ3jk = ⎝ 0 1 cot θ 0 r

(B.44)

(B.45)

(B.46)

Differential operators • Gradient of a scalar field ∇φ =

∂φ 1 ∂φ 1 ∂φ ∂φ 1 ∂φ 2 ∂φ 3 g + g + g = er + eθ + eϕ . ∂r ∂θ ∂ϕ ∂r r ∂θ r sin θ ∂ϕ

(B.47)

• Gradient of a vector field     ∂wr 1 ∂wr 1 1 ∂wr 1 er ⊗ er + − wθ er ⊗ eθ + − wϕ er ⊗ eϕ ∇w = ∂r r ∂θ r r sin θ ∂ϕ r     1 ∂wθ 1 1 ∂wθ 1 ∂wθ eθ ⊗ er + + wr eθ ⊗ eθ + − cot θ wϕ eθ ⊗ eϕ + ∂r r ∂θ r r sin θ ∂ϕ r   1 ∂wϕ 1 ∂wϕ 1 ∂wϕ 1 eϕ ⊗ er + eϕ ⊗ eθ + + wr + cot θwθ eϕ ⊗ eϕ . + ∂r r ∂θ r sin θ ∂ϕ r r (B.48) • Convective term / (∇w)w = / +

2 w2 + wϕ 1 ∂wr wϕ ∂wr ∂wr wr + wθ + − θ ∂r r ∂θ r sin θ ∂ϕ r

0 er

2 wϕ cot θ 1 ∂wθ wϕ ∂wθ wr wθ ∂wθ + wθ + + − wr ∂r r ∂θ r sin θ ∂ϕ r r

0 eθ

  1 ∂wϕ wϕ ∂wϕ wr wϕ wθ wϕ cot θ ∂wϕ + wθ + + + + wr eϕ . ∂r r ∂θ r sin θ ∂ϕ r r (B.49)

186

Appendix B. Vector and Tensor Analysis

• Curl of a vector field



1 ∂ ∂wθ (wϕ sin θ) − er r sin θ ∂θ ∂ϕ

1 ∂ 1 ∂wr − (rwϕ ) eθ + r sin θ ∂ϕ r ∂r

1 ∂ ∂wr (rwθ ) − + eϕ . r ∂r ∂θ

curl w =

(B.50)

• Divergence of a vector field div w =

1 ∂ 2 ∂ 1 1 ∂wϕ (r wr ) + (sin θwθ ) + . 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

(B.51)

• Divergence of a tensor field divS  ∂Srr 1 ∂Srθ 1 ∂Sϕr 1 = + + + (2Srr − Sθθ − Sϕϕ + Srθ cot θ) er ∂r r ∂θ r sin θ ∂ϕ r   ∂Srθ 1 ∂Sθθ 1 ∂Sθϕ 1 + + + + ((Sθθ − Sϕϕ ) cot θ + 3Srθ ) eθ ∂r r ∂θ r sin θ ∂ϕ r   ∂Sϕr 1 ∂Sθϕ 1 ∂Sϕϕ + + + (B.52) eϕ . ∂r r ∂θ r sin θ ∂ϕ • Laplacian of a scalar field 





 1 ∂ 1 ∂φ ∂φ ∂ ∂φ ∂ 2 ∆ϕ = 2 r sin θ + sin θ + r sin θ ∂r ∂r ∂θ ∂θ ∂ϕ sin θ ∂ϕ ∂2φ 1 ∂ ∂φ 1 ∂φ 1 ∂ (sin θ ) + 2 2 . (B.53) = 2 (r2 ) + 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ2 • Laplacian of a vector field   2 2 2 ∂wθ 2 ∂wϕ − 2 wθ cot θ − 2 ∆w = ∆wr − 2 wr − 2 er r r ∂θ r r sin θ ∂ϕ   1 2 ∂wr 2 cos θ ∂wϕ − 2 2 wθ − 2 + ∆wθ + 2 eθ r ∂θ r sin2 θ ∂ϕ r sin θ   1 2 ∂wr 2 cos θ ∂wθ + ∆wϕ − 2 2 wϕ + 2 + 2 eϕ r sin θ ∂ϕ r sin2 θ ∂ϕ r sin θ

(B.54)

where the Laplacian operators in the right-hand side are given by (B.53).

B.7. Differential Operators in Coordinates

187

Measure elements for integration • Line dr, r dθ, r sin θ dϕ.

(B.55)

r2 sin θ dθdϕ, r sin θ drdϕ, r drdθ.

(B.56)

r2 sin θ drdθdϕ.

(B.57)

• Surface

• Volume

Appendix C

Some Equations of Continuum Mechanics in Curvilinear Coordinates In this section we write some partial differential equations of continuum mechanics in cylindrical and spherical coordinates.

C.1 Mass Conservation Equation It is given by ∂ρ + div(ρv) = 0, ∂t where ρ is density and v is velocity.

(C.1)

• Cartesian coordinates ∂ρ ∂(ρv1 ) ∂(ρv2 ) ∂(ρv3 ) + + + = 0. ∂t ∂x1 ∂x2 ∂x3

(C.2)

• Cylindrical coordinates ∂ρ 1 ∂ 1 ∂(ρvθ ) ∂(ρvz ) + (rρvr ) + + = 0. ∂t r ∂r r ∂θ ∂z

(C.3)

• Spherical coordinates ∂ρ 1 ∂ ∂ 1 1 ∂(ρvϕ ) + 2 (r2 ρvr ) + (sin θρvθ ) + = 0. ∂t r ∂r r sin θ ∂θ r sin θ ∂ϕ

(C.4)

190

Appendix C. Some Equations in Curvilinear Coordinates

C.2 Motion Equation The general motion equation has the form ρ(

∂v + (grad v)v) = div T + b, ∂t

(C.5)

where T is the Cauchy stress tensor and b is the body force. • Cartesian coordinates  ρ

∂v1 ∂t

+ =

 ρ

∂v2 ∂t

+ =

 ρ

∂v3 ∂t

+ =

 ∂v1 ∂v1 ∂v1 v1 + v2 + v3 ∂x1 ∂x2 ∂x3   ∂T11 ∂T12 ∂T13 + + + b1 , ∂x1 ∂x2 ∂x3  ∂v2 ∂v2 ∂v2 v1 + v2 + v3 ∂x1 ∂x2 ∂x3   ∂T21 ∂T22 ∂T23 + + + b2 , ∂x1 ∂x2 ∂x3  ∂v3 ∂v3 ∂v3 v1 + v2 + v3 ∂x1 ∂x2 ∂x3   ∂T31 ∂T32 ∂T33 + + + b3 . ∂x1 ∂x2 ∂x3

(C.6)

(C.7)

(C.8)

• Cylindrical coordinates  ρ

∂vr ∂t

+ =

 ρ

∂vθ ∂t

+ =

 ρ

∂vz ∂t

+ =

 vθ ∂vr vθ2 ∂vr ∂vr vr + + vz − ∂r r ∂θ ∂z r   ∂Trz 1 ∂Trθ 1 ∂ (rTrr ) + +r − Tθθ + br , (C.9) r ∂r ∂θ ∂z r  vθ ∂vθ vr vθ ∂vθ ∂vθ vr + + vz + ∂r r ∂θ ∂z r   ∂Tθr 1 ∂Tθθ ∂Tθz 1 1 + + + Trθ + Tθr + bθ , (C.10) ∂r r ∂θ ∂z r r  vθ ∂vz ∂vz ∂vz vr + + vz ∂r r ∂vθ ∂z   ∂Tzz 1 ∂ ∂Tzθ (rTzr ) + +r (C.11) + bz . r ∂r ∂θ ∂z

C.3. Constitutive Law for Newtonian Viscous Fluids in Cooordinates

191

• Spherical coordinates  ρ 

∂vr ∂t

∂Trr = ∂r +br ,

0

+

v 2 + vϕ2 1 ∂vr vϕ ∂vr ∂vr vr + vθ + − θ ∂r r ∂θ r sin θ ∂ϕ r

+

1 ∂Trθ 1 ∂Tϕr 1 + + (2Trr − Tθθ − Tϕϕ + Trθ cot θ) r ∂θ r sin θ ∂ϕ r



(C.12)  ρ  =

vϕ2 cot θ ∂vθ 1 ∂vθ vϕ ∂vθ vr vθ vr + vθ + + − ∂r r ∂θ r sin θ ∂ϕ r r

0

∂vθ ∂t

+

∂Trθ ∂r

+

 1 ∂Tθθ 1 ∂Tθϕ 1 + + ((Tθθ − Tϕϕ ) cot θ + 3Trθ ) + bθ , r ∂θ r sin θ ∂ϕ r (C.13)

+

vr

 ∂vϕ ρ ∂t  ∂Tϕr = ∂r

+

1 ∂vϕ vϕ ∂vϕ vr vϕ vθ vϕ cot θ ∂vϕ + vθ + + + ∂r r ∂θ r sin θ ∂ϕ r r  1 ∂Tθϕ 1 ∂Tϕϕ + + bϕ . r ∂θ r sin θ ∂ϕ



(C.14)

C.3 Constitutive Law for Newtonian Viscous Fluids in Cooordinates In terms of the stretching D, it is given by T = 2ηD + ξ div vI,

(C.15)

with D=

1 (grad v + grad vt ). 2

(C.16)

192

Appendix C. Some Equations in Curvilinear Coordinates

• Cartesian coordinates ∂v1 , ∂x1 ∂v2 , T22 = ξ div v + 2η ∂x2 ∂v3 T33 = ξ div v + 2η , ∂x3

∂v1 ∂v2 T12 = T21 = η + , ∂x2 ∂x1

∂v1 ∂v3 + , T13 = T31 = η ∂x3 ∂x1

∂v2 ∂v3 T23 = T32 = η + , ∂x3 ∂x2 T11 = ξ div v + 2η

(C.17) (C.18) (C.19) (C.20) (C.21) (C.22)

with div v =

∂v2 ∂v3 ∂v1 + + . ∂x1 ∂x2 ∂x3

(C.23)

• Cylindrical coordinates ∂vr Trr = ξ div v + 2η , ∂r

1 ∂vθ vr Tθθ = ξ div v + 2η + , r ∂θ r ∂vz , Tzz = ξ div v + 2η ∂z    1 ∂vr ∂ vθ Trθ = Tθr = η +r , r ∂θ ∂r r

∂vz ∂vr + , Trz = Tzr = η ∂z ∂r

∂vθ 1 ∂vz + Tθz = Tzθ = η , ∂z r ∂θ

(C.24) (C.25) (C.26) (C.27) (C.28) (C.29)

with div v =

∂vz 1 ∂vθ 1 ∂ (rvr ) + + . r ∂r r ∂θ ∂z

(C.30)

C.3. Constitutive Law for Newtonian Viscous Fluids in Cooordinates

193

• Spherical coordinates ∂vr Trr = ξ div v + 2η , ∂r

1 ∂vθ vr + Tθθ = ξ div v + 2η , r ∂θ r

1 ∂vϕ vr vθ cot θ + + , Tϕϕ = ξ div v + 2η r sin θ ∂ϕ r r   1 ∂vr ∂  vθ  +r Trθ = Tθr = η , r ∂θ ∂r r   1 ∂vr ∂  vϕ  + , Trϕ = Tϕr = η r ∂r r r sin θ ∂ϕ   1 ∂vθ sin θ ∂  vϕ  + Tθϕ = Tϕθ = η , r sin θ ∂ϕ r ∂θ sin θ

(C.31) (C.32) (C.33) (C.34) (C.35) (C.36)

with div v =

∂ 1 1 ∂vϕ 1 ∂ 2 (r vr ) + (sin θvθ ) + . r2 ∂r r sin θ ∂θ r sin θ ∂ϕ

(C.37)

Appendix D

Arbitrary Lagrangian-Eulerian (ALE) Formulations of the Conservation Equations In some fluid-structure interaction problems, the domain where the motion of the fluid is taking place changes with time because the structure interacting with the fluid also moves. This is the case, for instance, of aeroelasticity problems. Moreover, there are free boundary flows for which the sets Bt , i.e., the positions occupied by the body, change with time and are a priori unknown. In these two situations, numerical methods based on so-called Arbitrary Lagrangian-Eulerian (ALE) formulations can be very useful. The main goal of this chapter is to write several ALE formulations of the conservation laws.

D.1 ALE Configuration The situation is summarized in Figure D.1. Recall that B is the reference or Lagrangian configuration, Bt is the Eulerian configuration and Ω denotes the ALE configuration. Y(·, t) is a motion transforming Ω into Bt at each time t and Z(·, t) = [Y(·, t)]

−1

◦ X(·, t).

Hence, Y(Z(p, t), t) = X(p, t).

(D.1)

Fields can be written in terms of Eulerian (spatial), Lagrangian (material) or ALE configuration. More precisely, if Φ is an Eulerian field, we can define its Lagrangian

196

Appendix D. ALE Formulations of the Conservation Equations

X( ,t) B

Bt

Z( ,t)

Y( ,t)

W Figure D.1: Lagrangian, Eulerian and ALE configurations

and ALE descriptions by Φm (p, t) = Φ(X(p, t), t), Φa (z, t) = Φ(Y(z, t), t), ˙ respectively. We recall that, in particular, vm = X. Similarly, for a Lagrangian field Ψ we can define its Eulerian and ALE counterparts by Ψs (x, t) = Ψ(P(x, t), t), Ψa (z, t) = Ψ(PZ (z, t), t), where P is the reference map for motion X, i.e., P(·, t) = [X(·, t)]−1 , and PZ (z, t) = [Z(·, t)]−1 . Hence, PZ (z, t) = P(Y(z, t), t). ˙ s. In particular v = X ∂ =
for the corresponding difWe will use the notations grada , diva and ∂t ferential operators applied to fields in ALE configuration. Let us denote w the spatial field defined by wa = ∂Y ∂t . Field wa is called the ALE velocity. We also   introduce d := ∂Z . ∂t a By taking the time derivative of (D.1) we obtain ∂X ∂Y ∂Z (p, t) = (Z(p, t), t) + grada Y(Z(p, t), t) (p, t). ∂t ∂t ∂t

(D.2)

Let p = PZ (z, t). Then z = Z(p, t) and (D.2) yields va (z, t) = wa (z, t) + G(z, t)d(z, t),

(D.3)

D.2. Conservative ALE Form of Conservation Equations

197

where G denotes the tensor field grada Y. For a part P of B, let us denote by Pta the subset of Ω given by Pta = Z(P, t). Then, from the Reynolds Transport Theorem we get Theorem 4.1.1. Let ϕ and Φ be an ALE scalar field and an ALE vector field, respectively. Then


d ϕ dVz = ϕ
dVz + diva (ϕd) dVz , (D.4) dt Pta Pta Pta


d Φ dVz = Φ
dVz + diva (Φ ⊗ d) dVz . (D.5) dt Pta Pta Pta Corollary 4.1.2. We have


  d 1. ϕ dVz = ϕ
dVz + diva ϕG−1 (va − wa ) dVz , dt Pta Pta Pta 2.

d dt


Pta


Φ dVz =

Pta

Φ
dVz +


Pta

(D.6)

1 2 diva [Φ ⊗ (va − wa )] G−t dVz . (D.7)

Proof. It is an immediate consequence of (D.3). Indeed, we get d = G−1 (va − wa ) and then   diva (Φ ⊗ d) = diva Φ ⊗ G−1 (va − wa ) 1 2 = diva [Φ ⊗ (va − wa )] G−t . (D.8)


D.2 Conservative ALE Form of Conservation Equations Now we are in a position to obtain the ALE formulations of the conservation equations in conservative form. For this purpose we recall the conservation principles in integral form and Eulerian coordinates, namely, • Mass

d dt

• Linear Momentum d dt


ρ dVx = 0.





Pt

Pt

ρv dVx =


T n dAx +

∂Pt

(D.9)

Pt

b dVx .

(D.10)

198

Appendix D. ALE Formulations of the Conservation Equations

• Energy




d ρE dVx = T n · v dAx + b · v dVx − q · n dAx + f dVx . dt Pt ∂Pt Pt ∂Pt Pt (D.11) The corresponding ALE formulations are obtained by making, first, the change of variable x = Y(z, t) in the integrals. Let J(z, t) = det G(z, t) and Pt = Y(Pta , t). We have • Mass

d dt


Pta

ρa J dVz = 0.

(D.12)

By using (D.11) for ϕ = ρa J we obtain
1  2 (ρa J)
+ diva ρa JG−1 (va − wa ) dVz = 0,

(D.13)

and hence, from the Localization Theorem,   (ρa J)
+ diva ρa JG−1 (va − wa ) = 0.

(D.14)

Pta

• Linear momentum


d ρa J dVz = JTa G−t m dAz + Jba dVz . dt Pta ∂Pta Pta

(D.15)

We transform this equality by using (D.7) for Φ = ρa Jva and Gauss’ Theorem. We obtain
 1 2 (ρa va J)
+ diva ρa J[va ⊗ (va − wa )]G−t dVz (D.16) Pta




  diva JTa G−t dVz +

= Pta


Pta

Jba dVz ,

and then, the Localization Theorem yields 1 2 (ρa va J)
+ diva J[ρa va ⊗ (va − wa ) − Ta ]G−t = Jba . • Energy


+ Pta

d dt


Pta


ρa Ea J dVz =

Jba · va dVz −


∂Pta

∂Pta

JTa va · G−t m dAz

Jqa · G−t m dAz +

(D.17)

(D.18)

(D.19)


Pta

Jfa dVz .

(D.20)

D.2. Conservative ALE Form of Conservation Equations

199

By using (D.11) for ϕ = ρa Ea J and Gauss’ Theorem, this equality becomes
1  2 (ρa Ea J)
+ diva ρa Ea JG−1 (va − wa ) dVz Pta




= Pta

  diva JG−1 Ta va dVz +
−

Pta


Pta

  diva JG−1 qa dVz +

Jba · va dVz
Pta

Jfa dVz ,

and, from the Localization Theorem, 2 1 (ρa Ea J)
+ diva JG−1 [ρa Ea (va − wa ) − Ta va + qa ] = Jba · va + Jfa .

(D.21)

(D.22)

Similar computations allow us to deduce the ALE formulation of the energy equation in terms of the internal energy, e = E − 12 |v|2 , namely, 1 2 (ρa ea J)
+ diva JG−1 [ρa ea (va − wa ) + qa ] = JTa G−t · grada va + Jfa .

(D.23)

We notice that ALE partial differential equations (D.14), (D.18), (D.22) and (D.23) hold in the fixed domain Ω.

D.2.1 Mixed Conservative ALE Form of the Conservation Equations Now, we are going to deduce another conservative form to be called mixed form because it includes fields in both ALE and spatial configuration while only partial differential operators in the spatial configuration are involved. For this purpose we will make extensive use of the following results: Lemma 4.2.1. Let g be a vector spatial field and S a tensor spatial field. We have   (D.24) diva JG−1 ga = G−t · grada ga = J (div g)a ,   −t = J (div S)a . (D.25) diva JSa G Proof. We use the following equalities: diva (Rt y) = R · grada y + y · diva R,

(D.26)

which holds for any smooth ALE tensor field R and vector field y, and diva (JG−t ) = 0

(Piola’s identity).

(D.27)

200

Appendix D. ALE Formulations of the Conservation Equations

Firstly, from (D.26) and (D.27) we get     diva JG−1 ga = diva JG−t · ga + JG−t · grada ga = JG−t · grada ga . (D.28) Moreover, by the chain rule, grada ga = ( gradg)a G

(D.29)

( gradg)a = grada ga G−1 .

(D.30)

and hence

From this equality we deduce   (div g)a = tr ( gradg)a = tr grada ga G−1 = G−t · grada ga .

(D.31)

Then, equality (D.24) follows by using this equality in (D.28). In order to prove (D.25) we recall that, from the definition of the divergence operator of a tensor field, we have,     diva (JSa G−t ) · e = diva JG−1 Sat e = diva JG−t · Sat e   +JG−t · grada (Sat e) = J div(S t e) a = J (div S)a · e,

(D.32)

for all vectors e, where we have used (D.26), (D.27), and (D.24) for y = S t e and R = JG−1 .
From this Lemma, and (D.14), (D.18) and (D.22) we easily get (ρa J)
+ J (div [ρ(v − w)])a = 0, (ρa va J)
+ J (div [ρv ⊗ (v − w) − T ])a = Jba ,

(ρa Ea J)
+ J (div [ρE(v − w) − T v + q])a = Jba · va + Jfa ,

(D.33) (D.34) (D.35)

respectively.

D.3 Mixed Nonconservative Form of ALE Conservation Equations Finally, we want to obtain mixed nonconservative ALE formulations of the conservation equations. • Mass. Firstly we have, (ρa J)
= ρ
a J + ρa J
= ρ
a J + ρa J (div w)a

(D.36)

D.3. Mixed Nonconservative Form of ALE Conservation Equations

201

and J (div [ρ(v − w)])a = J [( gradρ)a · (va − wa ) + ρa (div(v − w))a ] = JGt ( gradρ)a · G−1 (va − wa ) + Jρa (div(v − w))a = J grada ρa · d + Jρa (div(v − w))a .

(D.37)

Replacing these expressions in (D.33) we get J (ρ
a + grada ρa · d) + Jρa (div v)a = 0

(D.38)

and finally ρ˙a + ρa (div v)a = 0,

(D.39)

where ρ˙a denotes the material time derivative of ALE field ρa with respect to motion Z. • Momentum. Firstly, we have (ρa Jva )
= ρa Jva
+ (ρa J)
va

(D.40)

J (div [ρv ⊗ (v − w)])a = Jva (div [ρ(v − w)])a +Jρa ( gradv)a (va − wa ) .

(D.41)

and

We add these equalities and then subtract the mass conservation equation (D.33) multiplied by va . We get (ρa Jva )
+ J (div [ρv ⊗ (v − w)])a = Jρa va
+ Jρa ( gradv)a (va − wa ) = Jρa va
+ Jρa ( gradv)a GG−1 (va − wa ) = Jρa va
+ Jρa grada va d = Jρa v˙ a . (D.42) By replacing this equality in (D.34) we finally obtain ρa v˙ a − (div T )a = ba .

(D.43)

(ρa JEa )
= ρa JEa
+ (ρa J)
Ea

(D.44)

• Energy. Firstly we have

and J (div [ρE(v − w)])a = JEa (div [ρ(v − w)])a + Jρa ( gradE)a · (va − wa ) .

(D.45)

202

Appendix D. ALE Formulations of the Conservation Equations We add these equalities and then subtract the mass conservation equation (D.33) multiplied by Ea . We get (ρa JEa )
+ J (div [ρE(v − w)])a = Jρa Ea
+ Jρa ( gradE)a · (va − wa ) = Jρa Ea
+ Jρa ( gradE)a · GG−1 (va − wa ) = Jρa Ea
+ Jρa grada Ea · d = Jρa E˙ a . (D.46) By replacing this equality in (D.35) we finally obtain ρa E˙ a − (div [T v])a + (div q)a = ba · va + fa .

(D.47)

Bibliography [1] A. Berm´ udez, Obtaining the linear equations for the small perturbations of a flow. Mat. Notae XLI (2001/02), 123–138. [2] P. G. Ciarlet, Mathematical Elasticity. North Holland, 1988. [3] B. D. Coleman, W. Noll, Thermodynamics of viscosity, heat conduction and elasticity. Arch. for Rational Mech. and Anal., 13 (1963), 167–178. Also in The Foundations of Mechanics and Thermodynamics. Springer. Berlin, 1974. [4] Z. H. Guo, The representation theorem for isotropic, linear asymmetric stressstrain relations. J. Elasticity 13 (1983), 121–124. [5] M. E. Gurtin, An Introduction to Continuum Mechanics. Academic Press, New York, 1981. [6] K. K. Kuo, Principles of Combustion. John Wiley and Sons, Hoboken, N.J., 2005. [7] B. Mohammadi, O. Pironneau, Analysis of the K-Epsilon Turbulence Model. John Wiley and Sons, 1994. [8] W. Noll, Representations of certain isotropic tensor functions, Arch. Math. 21 (1970), 87–90. [9] W. R. Smith, R. W. Missen, Chemical Reaction Equilibrium Analysis: Theory and Algorithms. John Wiley and Sons, 1982.

Index β-function, 156 absolute temperature, 11 acceleration, 3, 65 accumulated thermal expansion, 59 acoustic intensity, 91 activation energy, 120 activation temperature, 120 affine space, 170 ALE configuration energy equation, 198 linear momentum equation, 198 mass equation, 198 velocity, 196 Arbitrary Lagrangian-Eulerian (ALE) configuration, 195 Arrhenius law, 119 associativity, 161 atoms, 122 axial vector, 168 balance of linear and angular momentum, 5 basis, 162 bilinear, 163 body, 1 body force, 5, 137 body heat, 8 Boussinesq hypothesis, 106 Boussinesq model, 79 bulk viscosity, 64 calorically perfect, 95 cartesian coordinate frame, 171

cartesian coordinates, 179 constitutive law for Newtonian viscous fluids, 192 mass conservation equation, 189 motion equation, 190 Cauchy’s hypothesis, 7 Cauchy’s theorem, 8 Cauchy-Green strain tensors, 2 change of specific free energy, 128, 129, 131 characteristic temperature of vibration, 95 chemical affinity, 128, 134 chemical equilibrium, 125 chemical potential, 115 Clausius-Duhem inequality, 11 closure models, 106 coefficient of isothermal compressibility, 68 of linear thermal expansion at constant stress, 55 of volumetric thermal expansion at constant pressure, 67 Coleman-Noll material, 18 commutative group, 161 commutativity, 161 compatible, 132 compressible Euler equations, 98 compressible Navier-Stokes equations, 98 concentration of species Ei , 111 of the mixture, 112 conductive heat flux, 137 coordinates, 162, 165, 169

206 Crocco’s theorem, 66 cross product, 167 curl, 174 curve, 175 closed, 175 length, 175 curvilinear integral, 176 cylindrical coordinates, 182 constitutive law for Newtonian viscous fluids, 192 mass conservation equation, 189 motion equation, 190 Dalton’s law, 111 deformation, 1 density, 150 in the motion, 4 of the mixture, 110 reference, 4 deviatoric, 168 differentiable, 173 differential, 173 diffusion velocity, 110, 135 displacement, 2 dissipation inequality, 19 dissipation rate, 24 dissipative acoustics, 81 distance, 171 distribution function, 154 divergence, 174 divergence theorem, 177 double delta function, 156 dynamic viscosity, 64, 103 eddy dynamic viscosity, 106, 107 elastic fluid, 96 elasticity tensor, 50 element mass fractions, 150 endomorphisms, 1 energy, 126 energy equation conservative, 137 linearized for thermoelasticity, 54 non-conservative, 138

Index enthalpy, 64, 138, 142, 150 of formation, 139 of the mixture, 110 standard, 124 entropy, 11 of the mixture, 110 equilibrium constant, 129–131 based on concentrations, 129 based on partial pressures, 128 Euclidean affine space, 171 Euclidean vector space, 163 Eulerian field, 3 Eulerian fluctuation, 90 Eulerian fluctuation of pressure, 79 expectation, 154, 155 extended symmetry group, 37 extent, 127, 130 Favre averaged, 155 Fick’s law, 136, 137 field, 173 filter, 105 first principle of thermodynamics, 8 flow adiabatic, 149 fluid, 61 energy conservation law, 64 energy equation, 70 motion equation, 64 Fourier’s law, 42 free energy, 12, 129 frequency factor, 120 full stoichiometric matrix, 133 generalized Hooke’s law, 52 Gibbs free energy, 77, 114, 127, 130, 141, 142 of formation, 97, 115 of the mixture, 111, 114 standard, 97 gradient, 173 of the motion, 2 gravity force, 87 Green’s formulas, 177

Index Green-Lagrange strain tensor, 2 Green-Saint Venant strain tensor, 2, 41 heat flux, 8 heat rate, 8 Helmholtz free energy, 12 of the mixture, 111 hyperelastic material with heat conduction and viscosity, 18 ideal fluid, 103 ideal gases, 96 incompressible Euler equations, 104 incompressible Navier-Stokes equations, 106 incompressible Newtonian fluid, 101 incremental methods, 57 infinitesimal strain tensor, 2, 48 inlets, 146 inner product, 163, 166 internal energy, 9 of formation, 123 of the mixture, 110, 121 standard, 123 invertible, 168 isobaric process, 149 isotropic, 164 k-atoms, 122, 140, 146 k-moles, 112 kernel, 132 Lagrangian coordinates, 13 Lagrangian fields, 3 Lam´e’s coefficients, 55 Laplacian, 174 least action principle, 125 Lewis number, 138 linearly independent, 162 local equilibrium chemistry, 149 local equilibrium problem, 141 localization theorem, 179

207 Mach number, 142, 149 mass, 126, 141, 142, 150 of the mixture, 120 mass action law, 119 mass conservation equation, 135 conservative, 136 mass diffusion, 136 mass diffusion coefficient, 138 mass distribution, 4 mass fraction, 110, 140, 146 material body, 18 isotropic, 39 material fields, 3 material frame-indifference principle, 28 material points, 1 material time derivative, 6 Mayer equation, 94 mean value, 154 mechanical energy, 91 mechanical equilibrium, 77 mixture fractions, 148, 150, 155 molar fraction, 112 molecular mass, 93 of the mixture, 112 motion, 1 motion equation conservative, 5, 137 linear approximation of the , 52 non-conservative, 5, 137 natural convection, 78 neutral element, 161 non-dissipative acoustics, 85 norm, 163 observer change, 27 origin, 171 orthogonal, 163, 168 orthonormal, 163 outlet, 146 passive scalars, 141 Pekeris equation, 87

208 perfect gas, 93 coefficient of isothermal compressibility, 94 coefficient of thermal expansion, 94 enthalpy, 93 entropy, 94 internal energy, 93 sound speed, 94 specific heat at constant pressure, 93 positive definite, 56, 163 positivity, 126, 142, 143, 150 potential theorem, 176 power stress, 6 pre-exponential exponent, 120 pressure, 102 Lagrangian fluctuation, 86 of the mixture, 111 standard , 129 principal invariants, 169 probability density functions (PDF), 153, 154 probability space, 153 random variable, 153 continuous, 154 independent, 154 random vector, 154 (jointly) continuous, 154 distribution function, 154 rank, 132 rate of total mass per unit volume, 146 ratio of specific heats, 94 real vector space, 161 reference configuration, 1 reference map, 3 region closed, 176 open, 176 regular, 176 response mappings, 18 rotation, 168

Index Saint Venant-Kirchhoff material, 41 Schmidt numbers, 156 second principle of thermodynamics, 11 second viscosity coefficient, 64 simply connected, 176 skew, 166 skew part, 167 Smagorinsky’s model, 106 small deformations, 57 sound speed, 72 source of mass, 135 sources, 146 spatial fields, 3, 135 species, 109 density, 109 enthalpy, 109 entropy, 109 Gibbs free energy, 109 Helmholtz free energy, 109 internal energy, 109 molecular mass, 109 pressure, 109 specific heat at constant pressure, 110 specific heat at constant volume, 109 velocity, 109 specific heat at constant deformation, 34 at constant pressure, 69 of the mixture, 124 at constant volume, 67, 121 of the mixture, 111, 121 spherical, 168 spherical coordinates, 184 constitutive law for Newtonian viscous fluids, 193 mass conservation equation, 189 motion equation, 191 stagnation enthalpy, 65, 98 state law, 125, 126, 142, 150 stirred tank, 119, 125, 140 stoichiometric coefficients, 119

Index stoichiometric method, 132 Stokes’ theorem, 177 streams, 148 stress tensor Boussinesq, 13 Cauchy, 5 First Piola-Kirchhoff, 13 Piola-Lagrange, 13 Reynolds, 106 Second Piola-Kirchhoff, 13 surface force, 5 surface heat, 8 symmetric, 163, 166 symmetric element, 161 symmetric part, 167 symmetry transformation, 37 system of forces, 5 system of generators, 162 system of heat, 8, 120 tensor, 164 positive definite, 168 tensor of thermal expansion at constant stress, 51 tensor product, 165, 169 the response of a material body is independent of the observer, 27 theorem of power expended, 6 thermal conductivity, 42, 138 thermal enthalpy, 139 thermodynamic pressure, 63 thermodynamic process, 17 adiabatic, 25 Eulerian, 25 homentropic, 66 isentropic, 24 isochoric, 101 steady, 66 thermomechanical equilibrium, 78 total energy, 8, 98 trace, 166 trajectory, 2 transpose, 166

209 turbulence models, 105 turbulent dissipation rate, 107 turbulent kinetic energy, 107 turbulent Prandtl, 156 V¨ ais¨ al¨ a-Brunt frequency, 89 tensor, 88 variance, 154, 155 vectors, 162 velocity, 3, 119 of the mixture, 110, 135 wall, 146 wave equation, 87