THEORY OF MIXTURES

COURSE LECTURE NOTES

JOSEF MÁLEK, ONDREJ SOUCEK

Charles UniversityPrague2019

WE THANK KAREL TUMA, MICHAL HABERA, VOJTECH PATOCKA AND MARK DOSTALÍK FOR THEIR HELP

AND REMARKS.

Contents

1 Theory of interacting continua - introduction 2

2 Reminder of the concepts of classical equilibrium thermodynamics 32.1 Physical postulates and mathematical model for a macroscopic system in thermodynamic equi-

librium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Physical interpretation of the mathematical model . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Other thermodynamic potentials - Legendre transform . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Reminder of some basic concepts of classical continuum thermodynamics and mechanicsof single continuum 113.1 Basic cornerstones of continuum thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Local equilibrium thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 Entropic representation (for fluid mixtures) . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2.2 Energetic representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2.3 Other thermodynamic potentials - Enthalpy, Helmholtz and Gibbs free energy . . . . . . 18

3.3 Structure of local equilibrium thermodynamics directly from the local form of fundamentalrelation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Molar-based quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5 Constitutive theory of continuum thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Kinematical description of mixtures 26

5 Balance equations 295.1 Auxiliary definitions - mass, volume other measures . . . . . . . . . . . . . . . . . . . . . . . . . . 295.2 General form of a balance law in the bulk (in Eulerian description) . . . . . . . . . . . . . . . . . 305.3 Balance of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.4 Balance of linear momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.5 Balance of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.6 Balance of energy (for non-polar mixture of non-polar constituents) . . . . . . . . . . . . . . . . . 375.7 Balance of entropy (Second law of thermodynamics) . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.8 Classification of the mixture theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6 Class I mixtures 416.1 Fick-Navier-Stokes-Fourier model (ψ= ψ(ϑ, 1

ρ, c)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.2 Constraints: Incompressibility and Quasi-incompressibility . . . . . . . . . . . . . . . . . . . . . . 476.2.1 Incompressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2.2 Quasi-incompressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.3 Cahn-Hilliard-NSF and Allen-Cahn-NSF model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.4 Allen-Cahn and Cahn-Hilliard models as gradient flows . . . . . . . . . . . . . . . . . . . . . . . . 596.5 Chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6.5.1 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.5.2 Mixture of ideal gasses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666.5.3 Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.5.4 Chemical kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

7 Class II mixtures 767.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 767.2 Balance laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.3 Interaction forces - structure inferred from balance laws . . . . . . . . . . . . . . . . . . . . . . . . 777.4 Interaction forces - macroscopic mechanical analogies . . . . . . . . . . . . . . . . . . . . . . . . . . 79

7.4.1 Flow around a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 797.5 Darcy’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

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7.5.1 Reduction of two-component momentum balance . . . . . . . . . . . . . . . . . . . . . . . . 837.5.2 Derivation from macroscopic analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.5.3 Derivation through homogenization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7.6 Generalizations of Darcy’s law: models of Brinkman and Forchheimer . . . . . . . . . . . . . . . . 877.7 A thermodynamic framework for a mixture of two liquids . . . . . . . . . . . . . . . . . . . . . . . 89

7.7.1 Compressible case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.7.2 Incompressible case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

8 Multiphase continuum thermodynamics 958.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958.2 General balance laws under presence of discontinuities - single continuum . . . . . . . . . . . . . 968.3 Jump conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.4 A thermodynamic framework for boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . 1018.5 General balance laws under presence of discontinuities - multicomponent continuum . . . . . . 1038.6 Framework of the multiphase continuum theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048.7 Averaging theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

1 Theory of interacting continua - introduction

What is a mixture? Oxford English dictionary: mixture: “A product of mixing, a complex unity or aggre-gate (material or immaterial) composed of various ingredient or constituent parts mixed together”.and goes on as follows:“... mixed state or condition; co-existence of different ingredients or different groups or classes of thingsmutually diffused through each other.”

Goal of the course

• Basic understanding of (derivation of) models that might be capable of describing the following pro-cesses (examples of mixtures around us):

– Geophysics - Thermohaline circulation - (convection due to both temperature and concentrationvariations), Porous media flow - flow of water and transport of solubles through soils, rocks,...,Avalanches, Pollution spreading and transport through environment, Melting of ice and freezingof meltwater in glaciers (phase transitions), Mineral phase transitions in rocks, Generation andextraction of magma, Liquefaction (change of strength of porous water saturated solids underseismic forcing),..

– Astrophysics - Plasmas and gaseous mixtures in the stars

– Biology - Flow of tracers/fluids through biological tissues, Processes at cell membranes, Biochem-ical reactions, Remodulation of bones, Blood coagulation, Cloth formation and dissolution, Flow ofblood plasma,...

– Chemistry - Chemical reactions, Chemical equilibrium, Chemical kinetics

– Material sciences - Multi-phase flow, Flow of aerosols, dyes, Deformation of composite materials

– ...

• Understanding of the assumptions required for the derivation of the models and developing frameworkfor their possible generalizations.

• To arrive at three-dimensional thermodynamically consistent material models of mixtures which in-clude temperature effects.

• To understand how the models of Fick, Darcy, Forchheimer, Biot, Brinkman, Allen-Cahn, Cahn-Hilliardand Stephan fit into the general framework of the mixture theory.

• To provide a sufficiently colorful “palette” of material models to cover the real-world processes.

2

Some names & literature

• Fick (1855), On liquid diffusion.

• Darcy (1856), Les fontaines publiques de la ville de Dijon.

• Muskat (1937), The Flow of Homogeneous Fluids through Porous Media.

• Truesdell and Toupin (1960), The Classical Field Theories.

• Truesdell and Noll (1965), The non-linear field theories of mechanics.

• Atkin and Craine (1976), Continuum Theories of Mixtures: Applications

• Bowen (1976), Theory of mixtures.

• Truesdell (1984), Rational thermodynamics.

• Müller (1985), Thermodynamics

• Samohýl (1982), Racionální termodynamika chemicky reagujících smesí.

• de Groot and Mazur (1984), Non-equilibrium Thermodynamics.

• Rajagopal and Tao (1995), Mechanics of Mixtures.

• Drew and Passman (1998), Theory of Multicomponent Fluids.

• Pekar and Samohýl (2014), The thermodynamics of linear fluids and fluid mixtures.

• Bothe and Dreyer (2015), Continuum thermodynamics of chemically reacting fluid mixtures.

Two main approaches within the theory of interacting continua

• Approach 1 - a concept based on the principle of equi-presence (all components of the mixture areassumed to co-exist simultaneously in each point of the continuum; the foundations of this approachhave been laid by Truesdell et al. (e.g. Truesdell and Toupin, 1960) and is sometimes referred to asContinuum mixture theory

• Approach 2 - a concept relying on postulating classical single-component balance laws together withinterface conditions at intermediate (sufficiently fine spatial) scale, where components of the mixtureare distinguishable (e.g. at the scale of the pore-space in case of porous media flow). This is followedby a suitable averaging procedure, which delivers a continuum-like mixture framework; one of theclassical references to this approach is Drew and Passman (1998). This approach is often referred to asMulti-phase theory.

In this lecture, we shall mostly rely on the first approach, with the exception of Chapter 8, where founda-tions of the second approach will be presented.

2 Reminder of the concepts of classical equilibrium thermodynamics

In this section, we first address as a reminder the formal structure of classical equilibrium thermodynamics,following closely Evans’ Lecture notes entropy and PDEs. (Evans, 2017). The development of classical phys-ical notions is here performed in a mathematically formal way, a more "physical" introduction to the topiccan be found, for example in Callen (1960). In the next chapters, these concepts are used under so-calledassumption of local equilibrium used to transfer the thermodynamic relation to the realm of continuumphysics.

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2.1 Physical postulates and mathematical model for a macroscopic system in thermo-dynamic equilibrium

Let us nevertheless start with the physical postulates (c.f. Callen, 1960, p. 283-284):

A physical model for a system in thermal equilibrium

• There exist particular states (called equilibrium states) that, macroscopically are characterized com-pletely by the specification of internal energy E and set of extensive parameters X1, . . . Xm

• There exists a function (called entropy) of the extensive parameters, defined for all equilibrium states,with the following property: the value assumed by the extensive parameters in the absence of con-straints, are those that maximize the entropy over the manifold of equilibrium states

• The entropy of a composite system is addition over the constituent subsystems (hence the entropyof each system is a homogeneous first order function of the extensive parameters. The entropy iscontinuous and differentiable and is monotonically increasing function of energy

• The entropy of any system vanishes in the state for which T := ∂E∂S = 0

Let us translate these physical postulates into a set of mathematical postulates

A mathematical model for a system in thermal equilibriumLet us suppose we are given

• an open convex subset Σ of Rm+1

• a C1 function S :Σ→R such that

(i) S is concave

(ii) ∂S∂E > 0

(iii) S is positively homogeneous of degree 1:

S(λE,λX1, . . . , Xm)=λS(E, X1, . . . , Xm) λ> 0 (2.1)

We call Σ the state space and S the entropy of the system

S def= S(E, X1, . . . , Xm) Fundamental thermodynamic equation (2.2)

depending on the internal energy E and other extensive quantities X1, . . . , Xm (examples of such quantitiesare for example the volume V , number (or molar number) of particles of individual atoms or molecules Nα ortheir masses Mα, electric polarization P, magnetization M, etc.)

Owing to (ii), we can invert relation (2.2) and obtain a C1 function E as a function

E = E(S, X1, . . . , Xm). (2.3)

We define:

• Thermodynamic temperature T

T = T = ∂E∂S

(2.4)

• Generalized force (or pressure)

Pk = Pk =− ∂E∂Xk

(2.5)

Lemma 2.1. (i) The function E is positively homogeneous of degree 1:

E(λS,λX1, . . . ,λXm)=λE(S, X1, . . . , Xm) λ> 0. (2.6)

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(ii) The functions T, Pk are positively homogeneous of degree 0:

T(λS,λX1, . . . ,λXm)= T(S, X1, . . . , Xm) (2.7)

Pk(λS,λX1, . . . ,λXm)= Pk(S, X1, . . . , Xm) λ> 0 (2.8)

These properties will leads to distinguishing so-called extensive parameters: S,E (and also Xk), which-so-to-speak depend on how “big” the system is and intensive parameters T, Pk, which do not depend onthe “size” of the system.

Proof. 1. Clearly, it must hold for all E,X1, . . . , Xm:

E = E(S(E, X1, . . . Xm), X1, . . . , Xm). (2.9)

Thus for all λ> 0:

λE = E(S(λE,λX1, . . .λXm),λX1, . . . ,λXm)

= E(λS(E, X1, . . . , Xm),λX1, . . . ,λXm) (2.10)

due to (2.1)

2. Since S is C1, so is E. Differentiate (2.6) with respect to S:

λ∂E∂S

(λS,λX1, . . . ,λXm)︸ ︷︷ ︸T(λS,λX1,...,λXm)

=λ ∂E∂S

(S, X1, . . . , Xm)︸ ︷︷ ︸T(S,X1,...,Xm)

(2.11)

Lemma 2.2. It holds

∂S∂E

= 1T

,∂S∂Xk

= Pk

Tk = 1, . . . ,m (2.12)

Proof. We defined T as T = ∂E∂S and the function S(E, X1, . . . , Xm) as an inverse of E = E(S, X1, . . . , Xm). Thus

it holds

E = E(S(E, X1, . . . , Xm), X1, . . . , Xm) . (2.13)

Therefore

1= dEdE

= ∂E∂S

∂S∂E

=⇒ T = ∂E∂S

=(∂S∂E

)−1

. (2.14)

Similarly, by taking a derivative with respect to Xk, we get

0= ∂E∂S︸︷︷︸T

∂S∂Xk

+ ∂E∂Xk︸ ︷︷ ︸−Pk

k = 1, . . . ,m . (2.15)

The definitions (2.4) and (2.5) can be summarized by the so-called Gibbs’s formula

dE = TdS−m∑

k=1PkdXk (2.16)

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2.2 Examples

The extensive parameters X1, . . . , Xm can represent various physical quantities. Let us discuss some proto-typical examples

1. Simple fluid. Possibly the simplest example of an equilibrium thermodynamic system is a homoge-neous simple fluid, for which the extensive parameters are:

• E - internal energy

• X1 =V - volume

• X2 =N - number of particles

consequently, the fundamental thermodynamic relation for such system reads S = S(E,V ,N ), and theassociated intensive parameters are

• T = ∂E∂S - thermodynamic temperature

• P1 = P =− ∂E∂V - pressure

• P2 =−µ=− ∂E∂N - (molar) chemical potential

The Gibbs formula for homogeneous simple fluid thus reads

dE = TdS−PdV +µdN .

2. Multicomponent simple fluid. If the fluid is composed of N constituents, we are in the followingsetting. The extensive parameters are now

• E - internal energy

• X1 =V - volume

• X2, . . . , XN+1 =N1, . . . ,NN - numbers of particles of the individual constituents

consequently, the fundamental thermodynamic relation for such system reads S = S(E,V ,N1, . . . ,NN ),and the associated intensive parameters are

• T = ∂E∂S - thermodynamic temperature

• P1 = P =− ∂E∂V - pressure

• P2, . . .PN+1 : Pα+1 =−µα =− ∂E∂Nα

- chemical potentials α= 1, . . . , N

The Gibbs formula for such a multicomponent simple fluid thus reads

dE = TdS−PdV +N∑α=1

µαdNα .

3. Homogeneous single-component dielectric fluid in a homogeneous electrostatic field withintensity E. From the principle of virtual work, one can deduce that the change of total energy of thevolume filled with the dielectric is

dE = TdS−PdV +µdN +∫Ω

E ·dD .

where D is the electric induction vector. Assuming a homogenous field, this relation can be written as

dE = TdS−PdV +µdN +VE ·dD .

For an isotropic linear dielectric body reads D= ε0E+P, where P is the so-called polarization vector andε0 is the permitivity of vacuum. Often one introduces a reduced energy E := E −V |E|2

8π which results

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is obtained by subtracting the energy of the electrostatic field (it cannot be however interpreted asintrinsic energy of the dielectric since the value of E is affected by the presence of the dielectric).

The Gibbs relation in terms of E reads

dE = TdS−PdV +µdN +VE ·dP

and, by introducing the extensive quantity total polarization P :=PV , we can rewrite it as

dE = TdS− PdV +µdN +E ·dP ,

with P = P + ε02 |E|2 +E ·P , and consider fundamental relation E = E(S,V ,N ,P ).

2.3 Physical interpretation of the mathematical model

Let us spend some time with an attempt of deeper explanation of the mathematical assumptions laid in ourmodel and their physical interpretations and consequences.

• The assumption ∂S∂E > 0 implies T > 0, i.e. positivity of the thermodynamic temperature

• The 1-homogeneity assumption of S is equivalent to additivity of entropy over subsystems ofan equilibrium system. Indeed, consider a body B in equilibrium, with internal energy E, entropyS and other extensive variables Xk, k = 1, . . . ,m. Let us consider a subregion B(1), which represents aλ fraction of B, and consequently, its extensive parameters take values: E(1) =λE, X (1)

1 =λX1 . . . X (1)m =

λXm, and also S(1) =λS, due to 1-homogeneity of S, since

S(1) = S(E(1), X (1)1 , . . . , X (1)

m )= S(λE,λX1, . . . ,λXm)=λS(E, X1, . . . , Xm)=λS .

The complementary subregion B(2) (such that B =B(1)∪B(2)), has extensive parameters E(2) = (1−λ)E,X (2)

1 = (1−λ)X1 . . . X (2)m = (1−λ)Xm and thus S(2) = (1−λ)S by 1-homogeneity of S. Consequently, we get

S = S(1) +S(2). The argument can be reversed and by assuming additivity of entropy over subsystemsof an equilibrium system, we would deduce 1-homogeneity of S: Take division of B into N subsystemseach of the same size, i.e. 1

N fraction of B. By the additivity assumption, it must hold

S =N∑

i=1S(i) = NS

(1N

E,1N

X1, . . . ,1N

Xm

),

which must hold for all N ∈N. Using the same argument for subsystems of the size MN , M ∈ 1, . . . , N, we

obtain that qS(E, X1, . . . , Xm)= S(qE, qX1, . . . , qXm) for all q ∈Q, q > 0. By the assumption of continuityof S and by the density argument of Q in R, we obtain the desired result.

• On the other hand, we have proven the intensive parameters T,P1, . . . ,Pm, to be 0-homogeneous. Thisimplies that these parameters take the same value irrespective of the size of the subregion, i.e.

T = T(1) = T(2) = . . . ., Pk = P(1)k = P(2)

k = . . . . k = 1, . . . ,m .

• Concavity of S. Let us now consider two isolated bodies made out of the same substance B(1) andB(2) which do not form subregions of one equilibrium system, but each of them is in equilibrium. Let

S(1) = S(E(1), X11 , . . . , X1

m) S(2) = S(E(2), X21 , . . . , X2

m)

The total entropy of the two systems in this configuration is S(1)+(2) = S(1) +S(2). If we now combinethe two bodies into one (e.g. by allowing transfer of all extensive parameters such as energy, mass,...)but without doing any work, neither allowing heat transfer from outside, we end up with a system B(3)

with extensive parameters E(3) = E(1) +E(2), X (3)k = X (1)

k + X (2)k , k = 1, . . . ,m. Since irreversible processes

may have taken place inside the system B(3) before it equilibrated, we cannot assume more than

S(3) ≥ S(1) +S(2)

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i.e. we assume that the final equilibrium entropy of the joint system is greater or equal than the sumof the two entropies of the originally isolated subsystems. So this means that

S(3) = S(E(1) +E(2), X (1)1 + X (2)

1 , . . . , X (1)m + X (2)

m )≥ S(1) +S(2) = S(E(1), X (1)1 , . . . , X (1)

m )+ S(E(2), X (2)1 , . . . , X (2)

m ) .

This however implies that S is concave, since taking any 0<λ< 1:

S(λE(1) + (1−λ)E(2),λX (1)

1 + (1−λ)X (2)1 , . . . ,λX (1)

m + (1−λ)X (2)m

)≥ S

(λE(1),λX (1)

1 , . . . ,λX (1)m

)+ S

((1−λ)E(2), (1−λ)X (2)

1 , . . . , (1−λ)X (2)m

)=λS

(E(1), X (1)

1 , . . . , X (1)m

)+ (1−λ)S

(E(2), X (2)

1 , . . . , X (2)m

)where in the last equality, we employed the 1-homogeneity of S.

• Convexity of E as a consequence of concavity of S. Let us take any S(1), X (1)1 , . . . X (1)

m and S(2), X (2)1 , . . . X (2)

mand any 0<λ< 1. Define

E(1) = E(S(1), X (1)1 , . . . X (1)

m ) E(2) = E(S(2), X (2)1 , . . . X (2)

m )

and thus

S(1) = S(E(1), X (1)1 , . . . X (1)

m ) S(2) = S(E(2), X (2)1 , . . . X (2)

m ) .

Since S is concave, it must hold

S(λE(1) + (1−λ)E(2),λX (1)

1 + (1−λ)X (2)1 , . . . ,λX (1)

m + (1−λ)X (2)m

)≥λS(E(1), X (1)

1 , . . . , X (1)m )+ (1−λ)S(E(2), X (2)

1 , . . . , X (2)m ) .

Since it holdsE = E(S(E, X1, . . . , Xm), X1, . . . , Xm)

and since ∂E∂S = T ≥ 0, we obtain

λE(1) + (1−λ)E(2) = E(S(λE(1) + (1−λ)E(2), . . . ,λX (1)

k + (1−λ)X (2)k , . . . ), . . . ,λX (1)

k + (1−λ)X (2)k , . . .

)≥ E

(λS(E(1), . . . , X (1)

k , . . . )+ (1−λ)S(E(2), . . . , X (2)k , . . . ), . . . ,λX (1)

k + (1−λ)X (2)k , . . .

)= E(λS(1) + (1−λ)S(2), . . . ,λX (1)

k + (1−λ)X (2)k , . . . ).

So we obtained

λE(S(1), X (1)1 , . . . , X (1)

m )+ (1−λ)E(S(2), X (2)1 , . . . , X (2)

m )≥ E(λS(1) + (1−λ)S(2), . . . ,λX (1)k + (1−λ)X (2)

k , . . . )

which implies that E is convex.

• Entropy maximization and energy minimization for thermally isolated systems.

– Entropy maximization principle - The equilibrium value of any unconstrained internal pa-rameter is such as to maximize the entropy of the system for the given value of total internalenergy.

– Energy minimization principle - The equilibrium value of any unconstrained internal param-eter is such as to minimize the energy for the given value of the total entropy.

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Figure 1: Entropy function S(E, . . . , Xk, . . . ) and visualization of the entropy maximization (left) end energyminimization (right) principle. The unconstrained variable Xk attains equilibrium value X∗

k , such that eitherentropy is maximal provided that the energy is fixed at value E∗ (left), or such that the energy is minimalprovided that the entropy is fixed at value S∗.

2.4 Other thermodynamic potentials - Legendre transform

Assume that H: Rn → (−∞,+∞] is a convex, lower semicontinuous function, which is proper (i.e. not identi-cally equal to infinity).

Definition 1. The Legendre transform of H(p) denoted H∗(q) is defined as

H∗(q)= supp∈Rn

(p ·q−H(p)) (2.17)

One can show, that also H∗(q) is convex, lower semicontinuous and proper. Furthermore,

(H∗)∗ = H , (2.18)

i.e. H is a Legendre transform of H∗, that’s why we call the pair H, H∗ as dual convex functions.Provided that on top of the assumptions before, H is moreover also a C2 and strictly convex, then for each

q ∈Rn there exists a unique point p where (p ·q−H(p)) is maximal, namely the unique point p, where

q= DpH(p) , providing relation q= q(p) (2.19)

which can be inverted to givep= p(q). (2.20)

The Legendre transform then readsH∗(q)= p(q) ·q−H(p(q)) (2.21)

As a consequence, one obtains also

DqH∗(q)=p+ (q−DpH(p)

)Dqp=p , (2.22)

so to summarize

q= DpH(p) p= DqH∗(q) (2.23)

So far, we have seen S = S(E, X1, . . . , Xm) andE = E(S, X1, . . . , Xm). Now Legendre transform is used to defineother thermodynamic potentials

• The Helmholtz free energy F is

F(T,V , . . . ) def= infS

(E(S,V , . . . )−TS) (2.24)

9

• The enthalpy H isH(S,P, . . . ) def= inf

V(E(S,V , . . . )+PV ) (2.25)

• The Gibbs free energy G is

G(T,P, . . . ) def= infS,V

(E(S,V , . . . )−TS+PV ) (2.26)

Assuming E is strictly convex and C2 and that the infima are attained at a unique point in the domain of E,we can rewrite the potentials as

• The Helmholtz free energy

F(T,V , . . . )= E−TS, where S = S(T,V , . . . ) solves T = ∂E(S,V , . . . )∂S

(2.27)

• The enthalpy

H(S,P, . . . )= E+PV , where V = V (S,P, . . . ) solves P =−∂E(S,V , . . . )∂V

(2.28)

• The Gibbs free energy

G(T,P, . . . )= E−TS+PV , where S = S(T,P, . . . ) solves T = ∂E(S,V , . . . )∂S

, P =−∂E(S,V , . . . )∂V

(2.29)

Based on the definitions and the assumption E is strictly convex and C2, we can prove the following

Lemma 2.3. 1. E(S,V , . . . ) is locally strictly convex in (S,V )

2. F(T,V , . . . ) is locally strictly concave in T, locally strictly convex in V.

3. H(S,P, . . . ) is locally strictly concave in P, locally strictly convex in S.

4. G(T,P, . . . ) is locally strictly concave in (T,P).

Proof. First, the statement (1) is already assumed hence it is true. To prove (2), let us recall the definition ofF

F(T,V , . . . )= E(S(T,V , . . . ),V , . . . )−TS(T,V , . . . ) ,

with

T = ∂E(S(T,V , . . . ),V , . . . )∂S

(2.30)

This implies

∂F∂T

= ∂E∂S

∂S∂T

− S−T∂S∂T

=−S(T,V , . . . )

∂F∂V

= ∂E∂S

∂S∂V

+ ∂E∂V

−T∂S∂V

= ∂E∂V

(S(T,V , . . . ),V , . . . )=−P(T,V , . . . )

which implies

∂2F∂T2 =−∂S(T,V , . . . )

∂T(2.31)

∂2F∂V 2 =−∂P(T,V , . . . )

∂V= ∂2E∂S∂V

∂S(T,V , . . . )∂V

+ ∂2E∂V 2 (2.32)

10

Differentiating (2.30) with respect to (T,V ), we obtain

1= ∂2E∂S2

∂S(T,V , . . . )∂T

0= ∂2E∂S2

∂S(T,V , . . . )∂V

+ ∂2E∂V∂S

Thus (2.31) and (2.32) imply

∂2F∂T2 =−

(∂2E∂S2

)−1

∂2F∂V 2 = ∂2E

∂V 2 −(∂2E∂S∂V

)2 (∂2E∂S2

)−1

and since we assumed that E is strictly convex in (S,V , . . . ), it holds

∂2E∂S2 > 0,

∂2E∂V 2 > 0

(∂2E∂S2

)(∂2E∂V 2

)−

(∂2E∂S∂V

)2

> 0 ,

and thus∂2F∂T2 < 0 ,

∂2F∂V 2 > 0 . (2.33)

The remaining two statements are proved analogously.

Exercise 1. Prove the statements (3) and (4).

3 Reminder of some basic concepts of classical continuum thermody-namics and mechanics of single continuum

3.1 Basic cornerstones of continuum thermodynamics

• A priori homogenization - continuum mechanics introduces the notion of a material point as a point-wise representative of some sufficiently small/sufficiently large control volume of real material; mate-rial properties assigned to such material point are averages (volume/time/stochastic) of the propertiesof the real material contained in the control volume.

X

referential conf.

x= χ(X, t)

current conf.

B

κ0(B)κt(B)

Figure 2: Essential kinematical concept of continuum mechanics - notion of motion χ of an abstract body B,viewed as a mapping from some reference configuration κ0(B) ⊂ R3 to the current configuration κt(B) ⊂ R3

at given time t.

11

• KinematicsMotion - mapping

χ(·, t) : κ0(B)→ κt(B) (3.1)

R3 ×R→R3 : x=χ(X, t) xk =χk(XK , t) , (3.2)

which is assumed to be sufficiently smooth and invertible.

– Spatial gradient

GradΦ(X, t) = ∂Φ(X, t)∂X

(3.3)

gradφ(x, t) = ∂φ

∂x(3.4)

– Velocity

V(X, t) = ∂χ

∂t

∣∣∣∣X

in Lagrangean description (3.5)

v(x, t) = V(χ−1(x, t), t) in Eulerian description (3.6)

– Material time derivativeDDtΦ(X, t) = ∂Φ(X, t)

∂t

∣∣∣∣X

(3.7)

DDt

φ(x, t) = ∂φ(χ(X, t), t)∂t

∣∣∣∣X= ∂φ(x, t)

∂t

∣∣∣∣x+v(x, t) ·gradxφ(x, t) (3.8)

– Deformation gradient

F(X, t)= ∂χ(X, t)∂X

(F)kK (X, t)= ∂χk(X, t)

∂XK (3.9)

– Green deformation tensor

C(X, t) = FTF (3.10)

(C)IJ = (F)iI (F) j

Jδi j (3.11)

(3.12)

– Cauchy deformation tensor

c(x, t) = F−TF−1 (3.13)

(c)i j = (F−1)Ii(F

−1)Jj δIJ (3.14)

– Finger deformation tensor

B(X, t) = FFT (3.15)

(B)i j = (F)iI (F) j

JδIJ (3.16)

– Piola deformation tensor

b(x, t) = F−1F−T (3.17)

(b)IJ = (F−1)Ii(F)J

j δi j (3.18)

– Velocity gradient

L(x, t) = gradv (3.19)

(L)ij = ∂vi

∂x j (3.20)

12

Exercise 2. Show that F= LF .Exercise 3. Show that B= LB+BLT .

• Volume and mass measures, constraintsConsider a mixture body Ω and let P ⊂Ω open subset, then we define

∀P t “nice”M (P ) . . . mass contained in P

V (P ) . . . volume of P

Constraint: ∀P t: 0 = ddtV (P t) (using the substitution theorem and identity ˙det(F) = divvdetF), we

obtain the incompressibility constraint 0= ∫P t

divv= 0, localization gives divv= 0

• Balance equations (in Eulerian description)

– Balance of mass∂ρ

∂t+div(ρv)= 0 , (3.21)

where ρ is the density, v is the velocity.

– Balance of linear momentum

∂(ρv)∂t

+div(ρv⊗v)= divT+ρb , (3.22)

where T is the Cauchy stress, b is the intensity of the body forces.

– Balance of angular momentum (for non-polar continuum)

T=TT . (3.23)

– Balance of total energy

∂

∂t

(ρ

(e+ 1

2|v|2

))+div

(ρ

(e+ 1

2|v|2

)v)= div(Tv−q)+ρb ·v+ρr , (3.24)

where e is the specific internal energy, and r is the energy supply (e.g. radiation).

In the framework of continuum thermodynamics, one has to consider also

– Balance of entropy

∂(ρη)∂t

+div(ρηv

)+divqη+ρrη = ξ , where ξ≥ 0 . (3.25)

Here η is the specific entropy, qη is the entropy flux, rη is the specific entropy supply, and ξ is theentropy production, which, by the second law of thermodynamics must be non-negative,

Exercise 4. Show that (3.21)-(3.24) allow to rewrite (3.24) in a more compact form (balance of internalenergy)

ρ e =T :D−divq+ρr . (3.26)

Assuming the body forces b and energy supply r are given, the balance equations (3.21), (3.22), (3.24)represent system of of 5 evolutionary equations for unknowns ρ, e,v = (v1,v2,v3), T = (T11,T12, T13,T22, T23, T33), q= (q1,q2,q3) (14 unknowns), and it is known that it is in general impossible to predictthe evolution of the T and q from this system knowing the initial state and boundary conditions. Itis necessary to provide closure relations - relations describing how T and q depend on the mechanicaland thermal state of the system.

13

3.2 Local equilibrium thermodynamics

In this course we shall rarely leave the realm of so called local equilibrium thermodynamics - a thermody-namic theory in which each material point (representing sufficiently small subsystem of the whole system),is assumed to be in local thermodynamic equilibrium. By saying local, we mean, that this equilibrium onlyconcerns the subsystem, but naturally, this material point behaves as an open subsystem, exchanging mass,energy, entropy etc., with the neighbouring points, with which equilibrium may not have been reached.

Based on this idea, we can exploit the notions and relations that have been derived for macroscopicsystems in thermodynamic equilibrium and simply apply them on some small representative volume of oursystem. We will then deduce some local relations, and postulate them to hold point-wise in our continuummodel. In the following application we will be exclusively interested in the description of fluid mixtures inthe absence of electric and magnetic fields. We shall often tacitly assume this without further emphasizingthis point explicitly.

3.2.1 Entropic representation (for fluid mixtures)

So consider this volume element dΩ, assumed to be in thermodynamic equilibrium, for which we apply theformal equilibrium structure, briefly reminded in Section 2. So again, we postulate existence of a functionS called “entropy”, which is a function of energy of the system, and it extensive variables - in most cases wewill restrict ourselves to volume, and masses of individual components of the system:

S = S(E,V ,Mα) , (3.27)

or more preciselyS(dΩ)= S(E(dΩ),V (dΩ),Mα(dΩ)), (3.28)

where E(dΩ), V (dΩ) and Mα(dΩ) are the energy, volume and masses of component in the subdomain dΩ.The function S was postulated (see (2.1)) to be

• positive 1−homogeneous with respect to the extensive variables, meaning

S(λE,λV ,λMα)=λS(E,V ,Mα) ∀λ> 0 . (3.29)

• increasing function of energy∂S∂E

> 0 (3.30)

• S is concave

From (3.27), we get by differentiating

dS = ∂S∂E

dE+ ∂S∂V

dV +N∑α=1

∂S∂Mα

dMα . (3.31)

In chapter 2, we defined the thermodynamic temperature ϑ, thermodynamic pressure p and chemical poten-tials µα by relations

1ϑ= ∂S∂E

,pϑ= ∂S∂V

, −µαϑ

= ∂S∂Mα

, (3.32)

so we can rewrite (3.31) in the classical form

ϑdS = dE+ pdV −N∑α=1

µαdMα . (3.33)

Differentiating (3.29) w.r.t. λ at λ= 1, we obtained the corresponding Euler relation

S(E,V ,Mα)= ∂S∂E

E+ ∂S∂V

V +N∑α=1

∂S∂Mα

Mα , (3.34)

14

or, equivalently, using the introduced definitions

ϑS = E+ pV −N∑α=1

µαMα . (3.35)

Differentiating (3.35) and using (3.31), we obtained the Gibbs-Duhem relation:

Sdϑ=V dp−N∑α=1

Mαdµα . (3.36)

Let us now explore the consequences of 1-homogeneity of entropy:

• Let us first take λ := 1M

, with M :=∑Nα=1 Mα:

S(

EM

,VM

,Mα

M

)= 1

MS(E,V ,Mα) . (3.37)

Recognizing in the arguments of the function on the l.h.s., the specific energy e, inverse of density ofthe mixture 1

ρand concentrations (mass fractions) cα, defined as

e def= EM

,1ρ

def= VM

, cαdef= Mα

Mα= 1, . . . , N , (3.38)

and observing that the function on the l.h.s. is a specific entropy η, we can write this relation as

1M

S(E,V ,Mα)= η def= η(e,1ρ

, c1, . . . , cN ) , (3.39)

differentiating the last equality yields

dη= ∂η

∂ede+ ∂η

∂ 1ρ

d(

1ρ

)+

N∑α=1

∂η

∂cαdcα (3.40)

Dividing the Gibbs-Duhem relation (3.36) by M , we get

ηdϑ= 1ρ

dp−N∑α=1

cαdµα . (3.41)

Dividing the Euler relation (3.35) by M , we obtain its local form

ϑη= e+ pρ−

N∑α=1

µαcα . (3.42)

Finally, differentiating (3.42) and subtracting (3.41), we recover the local version of (3.33):

ϑdη= de+ pd(

1ρ

)−

N∑α=1

µαdcα . (3.43)

Note that (3.43) implies the following local definitions of temperature, pressure and chemical potential:

1ϑ= ∂η

∂e

∣∣∣∣ 1ρ

,c1,...cN

,pϑ= ∂η

∂ 1ρ

∣∣∣∣∣e,c1,...,cN

, −µαϑ

= ∂η

∂cα

∣∣∣∣e, 1

ρ,cβ 6=α

(3.44)

15

Remark 1. In the definition of η (3.39) we did not employ the constraint∑Nα=1 cα = 1, which follows from

the definition of mass fractions cα in (3.38). Sometimes it is more convenient to employ this constraintwhen defining the specific entropy η and eliminate one of the concentrations (withou loss of generalitycN ) and define reduced specific entropy η

η(e,1ρ

, c1, . . . , cN−1) def= η

(e,

1ρ

, c1, . . . , cN−1,1−N−1∑β=1

cβ

). (3.45)

Immediatelly, we then obtain the following relations

1ϑ= ∂η

∂e

∣∣∣∣ 1ρ

,c1,...cN−1

,pϑ= ∂η

∂ 1ρ

∣∣∣∣∣e,c1,...,cN−1

, −µα−µN

ϑ= ∂η

∂cα

∣∣∣∣e, 1

ρ,cβ β=1,...,N−1&β6=α

. (3.46)

Note that the only difference when employing this reduction arises in the last set of relations, i.e. defin-ing the partial derivatives of η with respect to reduced (independent) set of concentrations c1, . . . , cN−1,which define the difference of chemical potentials with respect to the eliminated component.

Clearly one can also employ the constraint in the Gibbs relation (3.43) and get

ϑdη= de+ pd(

1ρ

)−

N−1∑α=1

(µα−µN )dcα . (3.47)

If one defines the relative chemical potentials

µαdef= µα−µN , (3.48)

one can simply use the reduced description (i.e. consider η= η(e, 1

ρ, c1, . . . , cN−1

)the Gibbs relation as

ϑdη= de+ pd(

1ρ

)−

N−1∑α=1

µαdcα . (3.49)

• Now, let’s consider λ := 1V

. Then 1-homogeneity of S yields

S(

EV

,1,Mα

V

)= 1

VS(E,V ,Mα) . (3.50)

Recognizing in the arguments of the function on the l.h.s., the volume density of energy ρe, and partialdensity ρα, and observing that the function on the l.h.s. is a volume density of entropy ρη, we can writethis relation as

1V

S(E,V ,Mα)= ρη= ρη(ρe,ρ1, . . . ,ρN ) , (3.51)

differentiating the last equality yields

d(ρη)= ∂ρη

∂(ρe)d(ρe)+

N∑α=1

∂ρη

∂ραdρα . (3.52)

Dividing the Gibbs-Duhem relation (3.36) by V , we get

ρηdϑ= dp−N∑α=1

ραdµα . (3.53)

Dividing the Euler relation (3.35) by V , we get

ϑρη= ρe+ p−N∑α=1

µαρα . (3.54)

16

Finally, differentiating (3.54) and subtracting (3.53), we recover the local version of (3.33):

ϑd(ρη)= d(ρe)−N∑α=1

µαdρα . (3.55)

Note that (3.55) implies the following alternative local definitions of temperature, and chemical poten-tial:

1ϑ= ∂ρη

∂(ρe)

∣∣∣∣ρα

, −µαϑ

= ∂ρη

∂ρα

∣∣∣∣ρe,ρβ6=α

(3.56)

Note: One might be puzzled now, where has the pressure vanished from our description in (3.55) and(3.56). In this setting we can introduce the pressure through the Euler relation (3.54). A naturalquestion is whether thermodynamic pressure defined in such a manner coincides with (3.44).

Exercise 5. Confirm by calculation that two discussed definitions of thermodynamic pressure are com-patible. HINT: Assume that you start from fundamental thermodynamic relation ρη= ρη(ρe,ρ1, . . . ,ρN )and realize that the introduced definitions imply

η

(e,

1ρ

, c1, . . . , cN

)= 1

MS(E,V ,M1, . . . ,MN )= V

M

1V

S(E,V ,M1, . . . ,MN )= VM

S(

EV

,1,M1

V, . . . ,

MN

V

)= 1ρρη(ρe,ρ1, . . . ,ρN ) .

Use this relation, define pressure through the Euler relation (3.54) and try to compute what is

ϑ∂η

∂ 1ρ

∣∣∣∣e,c1,...,cN

=ϑ ∂

∂ 1ρ

∣∣∣∣e,c1,...,cN

(ρηρ

)(ρe,ρ1, . . . ,ρN ) .

3.2.2 Energetic representation

Alternatively, using the fact, that ∂S∂E = 1

ϑ> 0, assuming sufficient smoothness of S we could invert

(3.27) and write insteadE = E(S,V ,Mα) . (3.57)

This function was proved in 2.1 to be also 1-homogeneous, i.e.

E(

SM

,VM

,Mα

M

)= 1

ME(S,V ,Mα) . (3.58)

Differentiating this relation and comparing with (3.33), we obtained alternative definitions of absolutetemperature, pressure and chemical potential:

ϑ= ∂E∂S

, p =−∂E∂V

, µα = ∂E∂Mα

. (3.59)

Both the Euler relation (3.35) and the Gibbs-Duhem relation (3.36) are recovered in the same form,and one can proceed to the local forms as before, by applying the 1-homogeneity w.r.t total mass andtotal volume. The same form of local Euler and Gibbs-Duhem relations are recovered, plus we obtainthe following two sets of alternative definitions:

ϑ= ∂e∂η

∣∣∣∣1ρ

,c1,...cN

, −p = ∂e∂ 1ρ

∣∣∣∣∣η,c1,...,cN

, µα = ∂e∂cα

∣∣∣∣η, 1

ρ,cβ 6=α

(3.60)

fore = e(η,

1ρ

, c1, . . . , cN ) , (3.61)

and

ϑ= ∂ρe∂(ρη)

∣∣∣∣ρα

, µα = ∂ρe∂ρα

∣∣∣∣ρe,ρβ6=α

(3.62)

forρe = ρe(ρη,ρ1, . . . ,ρN ) . (3.63)

17

3.2.3 Other thermodynamic potentials - Enthalpy, Helmholtz and Gibbs free energy

The other thermodynamic potentials can now be obtained as in the classical theory by means of Legendretransform. In particular, we define

• Thermodynamic potentials associated with e(η, 1ρ

, c1, . . . , cN )

– The specific Helmholtz free energy ψ is

ψ

(ϑ,

1ρ

, c1 . . . , cN

)def= inf

η

e(η,

1ρ

, c1, . . . , cN

)−ϑη

(3.64)

which for e being a C2 strictly convex function can be rewritten as

ψ

(ϑ,

1ρ

, c1 . . . , cN

)= e−ϑη , where η= η(ϑ,

1ρ

, c1 . . . , cN ) (3.65)

solves ϑ=∂e(η, 1

ρ, c1 . . . , cN )

∂η(3.66)

From the above definition, we obtain

∂ψ

∂ϑ= ∂

∂ϑ

(e(η(ϑ,

1ρ

, cα),1ρ

, cβ)−ϑη

(ϑ,

1ρ

, cα))

= ∂e∂η︸︷︷︸=ϑ

∂η

∂ϑ− η−ϑ∂η

∂ϑ=−η ,

∂ψ

∂ 1ρ

= ∂

∂ 1ρ

(e(η(ϑ,

1ρ

, cα),1ρ

, cβ)−ϑη

(ϑ,

1ρ

, cα))

= ∂e∂η︸︷︷︸=ϑ

∂η

∂ 1ρ

+ ∂e∂ 1ρ︸︷︷︸

=−p

−ϑ ∂η∂ 1ρ

=−p ,

∂ψ

∂cγ= ∂

∂cγ

(e(η(ϑ,

1ρ

, cα),1ρ

, cβ)−ϑη

(ϑ,

1ρ

, cα))

= ∂e∂η︸︷︷︸=ϑ

∂η

∂cγ+ ∂e∂cγ︸︷︷︸=µγ

−ϑ ∂η

∂cγ=µγ .

In summary, we get relations

−η= ∂ψ

∂ϑ

∣∣∣∣ 1ρ

,c1,...cN

, −p = ∂ψ

∂ 1ρ

∣∣∣∣∣ϑ,c1,...,cN

, µα = ∂ψ

∂cα

∣∣∣∣ϑ, 1

ρ,cβ 6=α

(3.67)

– The specific enthalpy h is

h(η, p, c1 . . . , cN ) def= inf1ρ

e(η,

1ρ

, c1, . . . , cN

)+ pρ

(3.68)

which for e a C2 strictly convex function gives

h(η, p, c1 . . . , cN

)= e+ pρ

where1ρ=

(1ρ

)(η, p, c1 . . . , cN ) (3.69)

solves − p =∂e(η, 1

ρ, c1 . . . , cN )

∂(

1ρ

) (3.70)

By analogous computations as for the Helmholtz free energy, we obtain the following expresionsfor the partial derivatives of enthalpy

ϑ= ∂h∂η

∣∣∣∣∣p,c1,...cN

,1ρ= ∂h∂p

∣∣∣∣∣η,c1,...,cN

, µα = ∂h∂cα

∣∣∣∣∣η,p,cβ 6=α

(3.71)

18

Exercise 6. Proof the above relations (3.71).

– The specific Gibbs free energy g is

g (ϑ, p, c1 . . . , cN ) def= infη, 1

ρ

e(η,

1ρ

, c1, . . . , cN

)−ϑη+ p

ρ

(3.72)

which for e being a C2 strictly convex function can be rewritten as

g (ϑ, p, c1 . . . , cN )= e−ϑη+ pρ

(3.73)

where η= η(ϑ,1ρ

, c1 . . . , cN ) solves ϑ=∂e(η, 1

ρ, c1 . . . , cN )

∂η, (3.74)

and1ρ=

(1ρ

)(η, p, c1 . . . , cN ) solves − p =

∂e(η, 1ρ

, c1 . . . , cN )

∂(

1ρ

) . (3.75)

Consequently, we arrive at the following set of expressions for the partial derivatives of Gibbs’free energy

−η= ∂ g∂ϑ

∣∣∣∣p,c1,...cN

,1ρ= ∂ g∂p

∣∣∣∣ϑ,c1,...,cN

, µα = ∂ g∂cα

∣∣∣∣ϑ,p,cβ 6=α

(3.76)

• Thermodynamic potentials associated with ρe(ρη,ρ1, . . . ,ρN )

– The Helmholtz free energy (per unit volume) ρψ is

ρψ(ϑ,ρ1 . . . ,ρN

) def= infρη

ρe

(ρη,ρ1, . . . ,ρN

)−ϑρη (3.77)

which for ρe being a C2 strictly convex function can be rewritten as

ρψ(ϑ,ρ1 . . . ,ρN

)= ρe−ϑρη , where ρη= ρη(ϑ,ρ1 . . . ,ρN ) (3.78)

solves ϑ= ∂ρe(ρη,ρ1 . . . ,ρN )∂(ρη)

(3.79)

– The enthalpy (per unit volume) ρh is cannot be obtained as a Legendre transform of ρe in thiscase, but can be defined as

ρh(ρη,ρ1 . . . ,ρN )= ρe+ p (3.80)

where p is given by the Euler relation, i.e. understood as (3.42) , which implies

ρh(ρη,ρ1 . . . ,ρN )=ϑρη+N∑α=1

µαρα , (3.81)

where we understand ϑ= ∂ρe∂ρη

(ρη,ρ1, . . . ,ρN ) and µα = ∂ρe∂ρα

(ρη,ρ1, . . . ,ρN ).

– Similarly, the Gibbs free energy (per unit volume) ρg cannot be obtained as a Legendre trans-form of ρe but can be defined by

ρg(ϑ,ρ1 . . . ,ρN

) def= ρψ+ p (3.82)

where p is given by the Euler relation (3.42), meaning

ρg(ϑ,ρ1 . . . ,ρN

)= N∑α=1

µαρα , where µα = ∂ρψ

∂ρα. (3.83)

19

3.3 Structure of local equilibrium thermodynamics directly from the local form of fun-damental relation

In the previous section, we have employed the classical equilibrium thermodynamics applied on some smallcontrol volume dΩ in the continuum, for which we assumed that all the relaxation processes within thecontrol volume have taken place, i.e. should this control volume be suddenly isolated from the rest of thevolume, it would not evolve anymore - it would already be in equilibrium.

In continuum mechanics, it would be convenient not to make the digression to the world of macroscopicthermodynamics and not to invoke the associated concepts anymore. In the entropic representation, we havearrived at two possible forms of the local fundamental relation, namely

ρη= ρη(ρe,ρ1, . . . ,ρN ) , (3.84)

andη= η(e,

1ρ

, c1, . . . , cN ) (3.85)

We have seen that the Gibbs relations can be directly obtained from these relations by taking a differential(or time derivative). Indeed, taking the time derivative and using the definitions (3.60) and (3.62), we obtain

d(ρη)= 1ϑ

d(ρe)−N∑α=1

µα

ϑdρα (3.86)

and

dη= 1ϑ

de+ pϑ

d(

1ρ

)−

N∑α=1

µα

ϑdcα (3.87)

which are relations (3.55) and (3.43), respectively.But to arrive at the Euler relations and Gibbs-Duhem relations, we had to rely on the macroscopic equi-

librium thermodynamics for some representative volume and employ the 1-homogeneity. Let us show that infact, the 1-homogeneity is already present it the fundamental relation written as above, and can be directlyused to retrieve both the Euler and the Gibbs-Duhem relations. Assume we have the fundamental relationwritten first as

ρη= ρη(ρe,ρα) (3.88)

and defineη(e,

1ρ

, c1, . . . , cN ) def= ρη(ρe,ρc1, . . .ρcN )ρ

(3.89)

Naturally, it holdsρη= ρη. (3.90)

Notice

1ϑ

def= ∂ρη

∂(ρe)= ∂η

∂e(3.91)

−µαϑ

def= ∂ρη

∂ρα= ∂η

∂cα

∣∣∣∣e, 1

ρ,cβ6=α

(3.92)

If we take a material time derivative of (3.90), we obtain

ρη+ρ∂η∂e

e+ρ ∂η

∂(

1ρ

) ˙(1ρ

)+

N∑α=1

ρ∂η

∂cαcα = ∂ρη

∂(ρe)˙(ρe)+

N∑α=1

∂ρη

∂ραρα

= ∂η

∂e(ρe+ρ e

)+ N∑α=1

∂η

∂cα(ρcα+ρ cα), (3.93)

which implies

ρ

η− 1ρ

∂η

∂(

1ρ

) − ∂η

∂ee−

N∑α=1

∂η

∂cαcα

= 0. (3.94)

20

Since this holds for arbitrary ρ and the terms in parenthesis are independent of ρ, the parenthesis must beequal to zero. Employing the above definitions of thermodynamic temperature and of the chemical potentials,and defining the thermodynamic pressure as

pϑ= ∂η

∂(

1ρ

) , (3.95)

and multiplying by ϑ, we obtain the Euler relation

ϑη= e+ pρ−

N∑α=1

µαcα. (3.96)

Differentiating this Euler relation and subtracting the differential of the fundamental relations η= η(e, 1ρ

, cα)and/or ρη= ρη(ρe,ρα), we obtain the Gibbs-Duhem relation

ρηdϑ= dp−N∑α=1

ραdµα (3.97)

3.4 Molar-based quantities

Let us also mention an important and practical variant of thermodynamic potentials and their independentvariables, based on counting the number of molecules (or moles which are more practical unit in this case).Starting from the fundamental relation (3.27) but expressing now the masses Mα as

Mα = nαMα , understood in the sense Mα(dΩ)= nα(dΩ)Mα , (3.98)

where nα(dΩ) is the number of moles in dΩ, and Mα is the molar mass of αth constituent, one can define

S = S(E,V ,nα) def= S(E,V , Mαnα) (3.99)

The function S inherits the positive 1-homogeneity of S, i.e.

S(λE,λV ,λnα)=λS(E,V ,nα) ∀λ> 0 (3.100)

and one may now proceed analogously as in the mass-based case by exploiting this 1-homogeneity for twoparticular weights.

• λ := 1n , with the total molar number n defined by n def=

N∑α=1

nα :

S(

En

,Vn

,nαn

)= 1

nS(E,V ,nα) . (3.101)

The arguments define molar-based quantities

– Molar energy eM

eM def= E(dΩ)n(dΩ)

, (3.102)

– Molar concentration cM

cM def= n(dΩ)V (dΩ)

, (3.103)

– Molar fractions yMα

yMα

def= nα(dΩ)n(dΩ)

, α= 1, . . . , N . (3.104)

– Molar entropy ηM by

ηM def= S(dΩ)n(dΩ)

, (3.105)

21

Then (3.101) reads

ηM = ηM(eM,1cM

, yM1 , . . . , yM

N ) (3.106)

Going back to definition (3.99), we infer that

η(e,1ρ

, c1, . . . , cN )= 1M

S(E,V ,Mα)= 1M

S(E,V ,nα)

= nM

S(

En

,Vn

,nαn

)= n

MηM(eM,

1cM

, yMα)= cM

ρηM(eM,

1cM

, yMα) , (3.107)

meaning thatρη= cMηM . (3.108)

In the same spirit, from (3.102), we conclude

ρe = cMeM . (3.109)

Since

ρα =Mαnα , we obtain ρ =N∑α=1

ρα =N∑α=1

MαnαV

=N∑α=1

MαcMα , (3.110)

where we introduced molar concentrations cMα

cMα

def= cM yMα

(= nα(dΩ)

V (dΩ)

), (3.111)

we can define ηM(eM, 1cM , yM

1 , . . . , yMN ) with the already defined η(e, 1

ρ, c1, . . . , cN ) as follows

ηM(eM,1cM

, yMα) def=

(N∑α=1

MαyMα

)︸ ︷︷ ︸

= ρ

cM

η

eM∑N

α=1 MαyMα︸ ︷︷ ︸

=e

,1cM

1∑Nα=1 MαyM

α︸ ︷︷ ︸= 1ρ

,MαyM

α∑Nβ=1 MβyM

β︸ ︷︷ ︸=cα

, (3.112)

which implies the following relations between the corresponding derivatives:

∂ηM

∂eM= ∂η

∂e= 1ϑ

,∂ηM

∂ 1cM

= ∂η

∂ 1ρ

= pϑ

(3.113)

and finally

∂ηM

∂yMγ

=Mγη+ (N∑β=1

MβyMβ )

(∂η

∂e−eMMγ

(∑Nβ=1 MβyM

β)2

+ ∂η

∂ 1ρ

−Mγ

cM(∑Nβ=1 Mβ)2

+N∑α=1

∂η

∂cα

Mβδγα∑Nβ=1 MβyM

β

− MαyMαMγ

(∑Nβ=1 MβyM

β)2

)

=Mγη−Mγ∑N

β=1 MβyMβ

(∂η

∂eeM + ∂η

∂ 1ρ

1cM

+N∑α=1

∂η

∂cαMαyM

α

)+ ∂η

∂cγMγ

=Mγη−Mγ∑N

β=1 MβyMβ

(eM

ϑ+ pϑcM

−N∑β=1

µα

ϑMαyM

α

)− µγ

ϑMγ

=Mγη−Mγ1ϑ

(e+ p

ρ−

N∑α=1

µαcα

)− µγ

ϑMγ

=−µγϑ

Mγ (3.114)

where in the last equality, we employed the Euler relation (3.42). So defining the molar chemicalpotential µM

α by

µMα

def= µαMα, α= 1, . . . , N , (3.115)

22

we can summarize the obtained result as

∂ηM

∂yMα

= ∂η

∂cαMα =−µα

ϑMα =−µ

Mα

ϑα= 1, . . . , N . (3.116)

With these identities and definitions, we can rewrite the Gibbs, Euler and Gibbs-Duhem relations interms of molar quantities. In particular, from (3.43), one arrives at

ϑdηM = deM + pd(

1cM

)−

N∑α=1

µMαd yM

α Molar Gibbs’ relation. (3.117)

From (3.42), using (3.108), (3.109), (3.110) and (3.115), one obtains the molar version of the Eulerrelation

ϑηM = eM + pcM

−N∑α=1

µMα yM

α Molar Euler relation . (3.118)

Taking a differential of this relation and substracting the molar Gibbs’ relation (3.117), yields the moalrversion of the Gibbs-Duhem relation

ηMdϑ= 1cM

dp−N∑α=1

yMαdµM

α . (3.119)

For the definition of molar counterparts of other thermodynamic potentials and for derivation of rela-tions among them, one proceeds analogously.

Remark 2. Similarly as the mass fractions cα, also the molar fraction yMα are by deifnition bound by the

costraintN∑α=1

yMα = 1 . (3.120)

In perfect analogy with the transition from η to η made in remark 1, we can reduce ηM by eliminatingone of the molar fractions using the constraint (3.120) and obtain reduced ηM as

ηM def= ηM

(eM,

1cM

, yM1 . . . , yM

N−1,1−N−1∑β=1

yMβ

). (3.121)

The consequences, in perfect analogy with the case discussed in remark 1 yields the following relations

∂ηM

∂eM= 1ϑ

,∂ηM

∂(

1cM

) = pϑ

,∂ηM

∂yMα

=−µMα−µM

N

ϑ=− µ

Mα

ϑ, α= 1, . . . , N −1 , (3.122)

where we define

µMα

def= µMα−µM

N , α= 1, . . . , N −1 . (3.123)

The molar Gibbs relation can then be written as

ϑdηM = deM + pd(

1cM

)−

N−1∑α=1

µMαd yM

α . (3.124)

• λ= 1V FIXME

23

3.5 Constitutive theory of continuum thermodynamics

1. Assumption of local thermodynamic equilibrium - we assume that there exists a function η ofstate variables, one of which is internal energy e, which is continuously differentiable and is increasingw.r.t energy e. This function is called specific entropy. I.e. we assume

η= η(e, y1, . . . ) , (3.125a)∂η

∂e> 0 . (3.125b)

This allows to invert (3.125a) and write

e = e(η, y1, . . . ) . (3.126)

We define thermodynamic temperature ϑ by

ϑ= ∂e∂η

. (3.127)

2. We assume that yi, i ∈ (1,2, . . . ) is known (we have them from balance equations or from kinematics).Let’s apply material time derivative to (3.126) and multiply by ρ:

ρ e = ρϑη+∑iρ∂e∂yi

yi , (3.128)

leading to

ρη+divqη = 1ϑ

∑b

Jb ·Ab , (3.129)

where the entropy flux qη is in many cases related to energy flux q through qη = qϑ

. Second law ofthermodynamics (in analogy with the classical equilibrium relation dS ≥ dQ

T ) implies

ρη+divqη = ξ≥ 0 , (3.130)

that is, the rate of entropy production ξ is non-negative. We can see that we can identify ξ with theright-hand side of (3.129). If we succeed in identifying the products Ja ·Aa in such a way that theyrepresent independent dissipative (=entropy producing mechanisms), we can, for example satisfy thesecond law by the following choice

3. SettingJa = γaAa , where γa ≥ 0 ∀a , (3.131)

we obtain ∑a

Ja ·Aa =∑aγa|Aa|2 =

∑a

1γa

|Ja|2 ≥ 0 , (3.132)

and the validity of the second law of thermodynamics is ensured.

Note: The above procedure has many weaknesses, it is rather formal, it may be difficult to identify theproducts Ja ·Aa, or ensure their independence, ... It serves merely as a possible guide/view at the consti-tutive theory. It must be supplemented with observations, measurements and further verifications in eachparticular situation.

Examples of the above constitutive procedure

1. Compressible Navier-Stokes-Fourier (NSF) fluids

η= η(e,B)→ η(e, tr(B), tr(B2),detB)→ η(e,ρ)↔ e = e(η,ρ) (3.133)

We define

ϑ= ∂e∂η

thermodynamic temperature, p = ρ2 ∂e∂ρ

thermodynamic pressure. (3.134)

24

By the procedure outlined above, we obtain

ρ e = ρϑη+ 1ρ

pρ , (3.135)

and employing the mass balance (3.21) and energy balance (3.26), we arrive at

ρϑη=T :D−divq+ρr+ pdivv (3.136)

We denote the deviatoric (traceless) part of any tensor by ()d, i.e. Ad =A− 13 tr(A)I, and note that

T :D= (Td + 13

trTI) : (Dd + 13

trD︸︷︷︸divv

I)=Td :Dd + 13

trTdivv (3.137)

The quantity m= 13 trT is so-called mean normal stress. We have thus arrived at

ρϑη=Td :Dd + (p+m)divv−divq+ρr . (3.138)

Dividing by temperature ϑ and rearranging, we obtain

ρη+div(qϑ

)− ρrϑ

= 1ϑ

[Td :Dd + (m+ p)divv−q · ∇ϑ

ϑ

](3.139)

Let us set the energy sources equal to zero, i.e. set r ≡ 0. We can see that we have the right-hand sidein the desired form of products related to independent processes, which suggests to set

Td = 2νDd , ν≥ 0 , (3.140a)

(m+ p)= 2ν+3λ3

divv , 2ν+3λ≥ 0 , (3.140b)

q=−k∇ϑϑ

, k ≥ 0 . (3.140c)

Putting all together (and setting κ def= kϑ≥ 0), we obtain

T= mI+Td =−pI+λdivvI+2νD Compressible NS fluid, (3.141a)

q=−κ∇ϑ Fourier law. (3.141b)

2. Incompressible NSF fluidsWe add a constraint divv= 0, i.e. we consider a material, which in all processes moves isochorically. Thederivation is analogous to the compressible case, except the dependence of thermodynamic potential ondensity, which is now suppressed due to the incompressibility assumption, since from the mass balance(3.21), we have ρ = 0, and thus, provided at initial time, ρ is constant, it remains so during the wholeprocess. Analogously as before, we obtain

ρη+div(qϑ

)= 1ϑ

[Td :Dd −q · ∇ϑ

ϑ

]+ ρrϑ

(3.142)

Note that the mean normal stress m has vanished from the entropy identity, i.e. now we cannot con-strain it thermodynamically as before. Constraining ourselves again to simple piece-wise relationshipsbetween the entropy producing fluxes and affinities, we obtain

T = mI+2νD Incompressible NS fluid, (3.143)

q = −k∇ϑ Fourier law. (3.144)

3. Elastic solidLet us consider

η= η(e,B)↔ e = e(η,B) ↔ e = e(η, trB, trB2,detB) , (3.145)

25

as a consequence of the principle of material frame indifference. Let us take, for simplicity, only

e = e(η,ρ, trB) , (3.146)

and let’s compute

ρ e = ρϑη+ 1ρ

pρ+ρ ∂e∂trB︸ ︷︷ ︸=:µ

˙trB (3.147)

Exercise 7. trB= 2B :DExercise 8. We obtain (verify):

ρη+div(qϑ

)= 1ϑ

[(Td −2µρBd) :Dd + (m+ p− 2µ

3ρtrB)divv−q · ∇ϑ

ϑ

]. (3.148)

Asserting for a perfectly elastic material that the entropy production is identically zero, we get

Td = 2µρBd , (3.149)

m = −p+ 2µ3ρtrB , (3.150)

q = 0 , (3.151)

i.e.T=−pI+2µρB Neo-Hookean solid . (3.152)

4. Kelvin-Voigt solid FIXME (add pixture)

Dd = 12ν

(Td −2µBd) (3.153)

divv = 32ν+3λ

(m+ p+ 2µ

3ρtrB

)(3.154)

which implies

T = −pI+λdivvI+2µρB+2νD (3.155)

Exercise 9. Go through derivation of 3 and 4 for an incompressible material.

5. Burger’s model FIXME

Note: So far, we were dealing with linear constitutive relations, this assumption can be further relaxed, wewill see some examples later.

4 Kinematical description of mixtures

Assumption of coexistence (co-occupancy) In each material point of a mixture with N components, allthe components are present at the same time, see Fig. 8.

This assumption is a generalization of the continuum hypothesis from the traditional single-componentcontinuum mechanics. It can be justified by the same means - assuming we can distinguish between thecomponents and describe them independently, the averaging procedure over some sufficiently small repre-sentative volume allows to assign to each point in a physical space some averaged values of all the fieldquantities related to the particular component of the mixture.

Realize that instead of a single reference configuration, as in the traditional continuum mechanics, nowwe have in principle N independent reference configurations between which we need to distinguish1. Let usassume for simplicity, that all N reference frames coincide and all the N spatial frames coincide. We willassume that the two (reference and actual) may differ, and we will distinguish between them (in the compo-nent description) by assigning capital letters to the referential and by small letters to the actual description.The particular components of the the mixture will be distinguished by Greek indices.

1This has severe implications on the role between the Lagrangian - referential and Eulearian - actual description. The differenceis now much bigger than for single component materials in favor of the Eulerian approach.

26

κt(B)

x

κ01(B)

X1χ1

κ0α(B)

Xα

χα

κ0N (B)

XN

χN

Figure 3: The assumption of coexistence and generalization of the classical continuum concept of motion.For mixtures with N components, N mappings χα, α= 1, . . . , N are assumed to exist, each being a mappingfrom a reference configuration of the particular component to the current configuration. One could call themapping χα the motion of the αth component. Each point in the current configuration is thus assumed to bean image of N preimages (material points of the individual mixture components).

Remark 3. Let us note that the mixture concept of co-occupancy can be also extended to a somewhat extremecase, when some of the substances are considered to be locally infinitely diluted. If we considered, for examplea liquid and its vapor as the two components of a binary mixture, then we can use the framework of themixture theory for the description of the motion, development and interaction of macroscopic bubbles of thevapor in the liquid, thinking of the two phases as these end-member cases of the mixture - liquid being themixture in which the vapor phase is infinitely diluted, and vice-versa, the vapor as the mixture phase, wherethe liquid is infinitely diluted. This extension of the view makes the mixture concept a very versatile tool forattacking many physical problems.

Motion of αth component is traditionally defined as a mapping

χα :R3 ×R→R3, i.e. xα =χα(Xα, t) , xkα =χk

α(XKα , t) , (4.1)

which is assumed to be sufficiently smooth and invertible. In the actual configuration it is natural to look atone particular point in the space-time (x, t), where x= xα , α= 1. . . N. To this point there are N correspondingreferential points determined by the inverse mappings

Xα(x, t)=χ−1α (x, t) α= 1, . . . , N . (4.2)

The assumption of co-occupancy can be then expressed as follows. For any point x ∈Ω(t) = (current state ofmixture), and time t > 0, there are material points Xα, α= 1, . . . , N, such that x=χα(Xα, t).

With the notion of motion of the αth component, we can now introduce the following referential (La-grangian) and spatial (Eulerian) kinematic quantities:

• Gradient

GradαΦ(Xα, t) = ∂Φ(Xα, t)∂Xα

referential , (4.3)

gradφ(x, t) = ∂φ

∂xspatial . (4.4)

• Velocity of the αth component

Vα(Xα, t) = ∂χα

∂t

∣∣∣∣Xα

in referential description , (4.5)

vα(x, t) = Vα(χ−1α (x, t), t) in spatial description . (4.6)

27

• Material time derivative with respect to the αth component

Dα

DtΦ(Xα, t) = ∂Φ(Xα, t)

∂t

∣∣∣∣Xα

, (4.7)

Dα

Dtφ(x, t) = ∂φ(χα(Xα, t), t)

∂t

∣∣∣∣Xα

= ∂φ(x, t)∂t

∣∣∣∣x+vα(x, t) ·gradxφ(x, t) . (4.8)

• Deformation gradient of the αth component

Fα(Xα, t)= ∂χα(Xα, t)∂Xα

(Fα)kK (Xα, t)= ∂χk

α(Xα, t)∂XK

α

(4.9)

• right Cauchy-Green stretch tensor of the αth component

Cα(Xα, t)= FTαFα , (Cα)IJ = (Fα)i

I (Fα) jJδi j (4.10)

• Cauchy deformation tensor of the αth component

cα(x, t)= F−Tα F−1

α , (cα)i j = (F−1)Ii(F

−1)Jj δIJ (4.11)

• left Cauchy-Green stretch tensor of the αth component

Bα(Xα, t)= FαFTα , (Bα)i j = (Fα)i

I (Fα) jJδ

IJ (4.12)

• Piola deformation tensor of the αth component

bα(x, t)= F−1α F−T

α , (bα)IJ = (F−1α )I

i(Fα)Jj δ

i j (4.13)

• Velocity gradient of the αth component

Lα(x, t)= gradvα , (Lα)ij =

∂viα

∂x j (4.14)

Due to the fact that for each material point in the actual configuration there correspond N referenatialpoints as preimages of this point in the N reference configurations of individual components, the calcula-tions of even simple kinematic quantities in referential description can become very cumbersome, see alsoExercise(10). From this point of view, we find the Eulerian description preferable, and we will formulateall the theory in the actual (Eulerian) configuration. Also, in the following we will consider both the coor-dinate frame to be Cartesian and we will thus stop distinguishing betweenthe covariant and contravariantcomponents of tensors for simplicity.

Remark 4. The fact that we distinguish N different reference configurations has an important impact on thenotion of material symmetry. Recall this notion from the basic course on continuum mechanics.

Exercise 10. Difficulties with Lagrange description. Since the concept of an N-component mixture requires inprinciple the notion of N different motion mappings and N associated reference configurations, one may ex-pect that Lagrange (referential description) may be more cumbersome than in the case of a single-componentmaterial. This is indeed so, let’s demonstrate it on a very simple example. It is natural, for component αof the mixture, to consider some associated field quantity, say Ψα, which in the simplest case, depends onlyon Xα and t, i.e. we have Ψα =Ψα(Xα, t). While it is natural to consider DαΨα

Dt , i.e. material time derivativeof Ψα following the motion of α material point, one may easily be forced to evaluate also terms of the typeDγΨα

Dt for γ 6=α. Switching for a moment to the Eulerian coordinates by considering ψα(x, t)=Ψα(Xα, t) (withx=χα(Xα, t), it is straightforward to show (try it) that

DγΨα

Dt= Dγψα

Dt= ∂ψα

∂t

∣∣∣∣x+vγ ·gradψα = ∂ψα

∂t

∣∣∣∣x+vα ·gradψα+ (vγ−vα) ·gradψα = Dαψα

Dt+ (vγ−vα) ·gradψα .

(4.15)Now try to prove the same statement, i.e. DγΨα

Dt = DαΨα

Dt + (vγ−vα) · gradψα directly from the Lagrangiandefinition of the material time derivative.HINT: Start by considering Ψα(Xα, t)=Ψα(χ−1

α (x, t), t)=Ψα(χ−1α (χγ(Xγ, t), t), t).

28

5 Balance equations

5.1 Auxiliary definitions - mass, volume other measures

We start by defining several useful measures. Consider a mixture body Ω and let P ⊂Ω open subset, thenwe defineM (P ) . . . mass of the mixture as a whole contained in P ,Mα(P ) . . . mass of the α component of the mixture contained in P ,V (P ) . . . volume of the mixture as a whole contained in P ,Vα(P ) . . . volume of the α component of the mixture contained in P

Remark 5. The notion of the partial volume Vα may be a bit puzzling in view of the co-occupancy notion.And indeed, it will become more clear in Chapter 8, when talking about the multiphase theory. It is usefulnevertheless to define all these measures in one place. The concept of Vα should be understood in the follow-ing sense. If we exclude solutions, where the mixing among the components happens really at microscopic(molecular) level, and consider a material with internal structure, i.e. material, where after sufficient zoom-ing, we can distinguish the particular components, then the partial volume denotes the volume occupied byindividual component.

We postulate the following (and natural) relations of absolute continuity (¿) between the defined mea-sures

1. M (P )¿ V (P ). The Radon-Nikodym theorem implies the existence of density ρ, such that

M (P )=∫PρdV . (5.1)

2. Mα(P )¿ V (P ). By the same argument we arrive at partial density ρα, such that

Mα(P )=∫PραdV . (5.2)

3. Vα(P )¿ V (P ), implying the existence of a volume fraction φα such that

Vα(P )=∫PφαdV . (5.3)

4. Mα(P )¿M (P ), which allows to define the concentration (mass fraction) cα such, that

Mα(P )=∫P

cαdM(5.1)=

∫PρcαdV . (5.4)

5. Mα(P )¿ Vα(P ), and thus there exists material (true) partial density ρmα such that

Mα(P )=∫Pρmα dVα

(5.3)=∫Pρmα φαdV . (5.5)

By comparing (5.2) with (5.4), we obtain

cα = ρα

ρ, (5.6)

while (5.2) and (5.5) implyρα =φαρm

α . (5.7)

Employing the following natural assumption called mass additivity constraint, which simply statesthat all components of the mixture have been taken into account, i.e.,

N∑α=1

Mα(P )=M (P ) , (5.8)

29

we obtainN∑α=1

ρα = ρ . (5.9)

This together with (5.6), also implies thatN∑α=1

cα = 1 . (5.10)

We will sometimes employ also the volume additivity constraint

N∑α=1

Vα(P )= V (P ) , (5.11)

from which we obtain (taking into account also (5.3)) that∑α

φα = 1 . (5.12)

Remark 6. While the mass additivity constraint is in fact a natural universal requirement, expressing ba-sically the conservation of mass, the volume additivity constraint represents a much stronger assumptionwhich does not hold universally. We know from observation - e.g. by putting salt into water, that volumeof the salted water changes only very slightly - in the solution the ions have just occupied voids betweenthe molecules of water and the total volume of the solution has not changed. Examples of cases, where thevolume additivity constraint applies are typically materials without interstitial voids.

5.2 General form of a balance law in the bulk (in Eulerian description)

General form of a balance law for a field quantity ψα(x,t) (some scalar or vectorial of the α component),assuming that no singular surfaces are present in the bulk, will be postulated in the following integral form

ddt

∫Ωα(t)

ψαdΩ=∫∂Ωα(t)

ψαΦ ·n dS+∫Ωα(t)

ψαξdΩ+∫Ωα(t)

ψαΣdΩ +∫Ωα(t)

ψαΠdΩ , (5.13)

where Ωα(t) = χα(Ω0α, t), and where ψαΦ is the density of flux of ψα across the boundary ∂Ωα (with outer

unit normal n), ψαξ denotes the density of production of ψα inside Ωα, ψαΣ is the corresponding density ofsupply (from outside the system) of ψα and ψαΠ is the interaction term - supply of ψα by interaction with theremaining components of the mixture inside Ωα. This integral relation is assumed to hold for any materialvolume Ωα(t). Note that compared to the analogous statement in single-continuum mechanics, here we haveto distinguish the control material volumes corresponding to the individual components since they evolveindependently. In the above statement the appropriate material volume is then the one following the motionof the α component.

The above integral balance law is localized in the standard way, using the Reynolds’ transport theoremand Gauss’ theorem. Assuming for now that the material does not contain any discontinuities, the corre-sponding local form reads

∂ψα

∂t+div(ψα⊗vα)−divψαΦ−ψαξ−ψαΣ−ψαΠ= 0 , (5.14)

or, equivalently,

Dαψα

Dt+ψαdivvα−divψαΦ−ψαξ−ψαΣ−ψαΠ= 0. (5.15)

A more general case, when material or immaterial discontinuities are present will be discussed later. Thisway we have formulated a general balance law for a quantity ψα. By postulating that the nature of the termsψαΠ is truly interactional, we assert that in total, when summed over all components of the mixture theseterms add up to zero:

N∑α=1

ψαΠ= 0 , (5.16)

30

which leads, by summing (5.14) to the following local form of the balance of quantity ψ over the wholemixture:

∂∑Nα=1ψα

∂t+div

(N∑α=1

ψα⊗vα

)−div

(N∑α=1

ψαΦ

)−

N∑α=1

ψαξ−N∑α=1

ψαΣ= 0 , (5.17)

or, equivalently

N∑α=1

Dαψα

Dt+

N∑α=1

ψαdivvα−div

(N∑α=1

ψαΦ

)−

N∑α=1

ψαξ−N∑α=1

ψαΣ= 0. (5.18)

With this general form of the balance laws for the particular components of the mixture and for themixture as a whole, we can now proceed to the particular physical balance laws, by merely identifying (orpostulating) the corresponding field quantities, their fluxes, productions, supplies and interaction terms.

5.3 Balance of mass

Setting ψα = ρα, since the volume in the integral balance law is assumed to be ‘(α)‘-material, there is no flux,i.e. ραΦ= 0, neither is there supply, i.e. ραΣ= 0, or production ραξ= 0, but due to chemical reactions, or phasetransitions, there may be a non-zero transfer of mass of α through conversion process from other componentsof the mixture. We denote this conversion term as mα = ραΠ, and the local mass balance for α component ofthe mixture thus reads

∂ρα

∂t+div(ραvα)= mα α= 1, . . . , N , (5.19)

or equivalently

Dαρα

Dt+ραdivvα = mα α= 1, . . . , N . (5.20)

The interaction terms - mass exchanges due to chemical reaction (or phase changes) must sum up to zero,when summed over all components of the mixture, so that mass is neither produced, nor destroyed in thesereactions, i.e. it must hold

N∑α=1

mα = 0 . (5.21)

Consequently, the mass balance for the mixture as a whole, obtained by summing (5.19) over all components,reads

∂(∑N

α=1ρα)

∂t+div

(N∑α=1

ραvα

)= 0, (5.22)

or, equivalently

N∑α=1

Dαρα

Dt+

N∑α=1

ραdivvα = 0. (5.23)

For all balances for the mixture as a whole, one may attempt to reformulate them in the form of a correspond-ing single-component body under suitable definition of the whole-mixture properties from the componentproperties. It is particularly easy in the case of the balance of mass. By (5.9), we already have the naturaldefinition of the total mixture density ρ. If we define the mixture velocity as a barycentric velocity through

v= 1ρ

N∑α=1

ραvα , (5.24)

from (5.22), we arrive at the familiar single-component form of the mass balance for the mixture as a whole

∂ρ

∂t+div(ρv)= 0 , (5.25)

31

or, if we define the barycentric time derivative as

D•Dt

= ∂•∂t

+v ·grad•, (5.26)

we can write equivalently

DρDt

+ρdivv= 0 . (5.27)

Remark 7. Note that according to the previous observation, if we take the single-component form of the bal-ance law as a certain selective rule, the barycentric velocity becomes the privileged definition of the mixturevelocity. Also note that alternative formulation might be possible, and is indeed useful in applications, whenthe weighting functions are not mass fractions, but rather volume fractions, for example.

Exercise 11. Consider an N-component mixture of incompressible fluids, meaning that ρmα = const ∀α. Define

the mixture velocity alternatively as vφ = ∑Nα=1φαvα, where φα = ρα

ρmα

is the volume fraction of α componentof the mixture. Derive from the local balance law (5.19) the global balance law for the mixture as a whole interms of this new velocity and show under which conditions is the resultant motion isochoric (=divergence-free).

5.4 Balance of linear momentum

Setting ψα = ραvα, since the volume in the integral balance law is assumed to be ‘(α)‘-material, there is onlymomentum flux by action of surface forces represented by the (partial) Cauchy stress tensor Tα, i.e. ραvαΦ=Tα. The linear momentum supply is realized through long-range body forces (gravity, elmag.) ραvαΣ= ραbα.We postulate that there is no production of linear momentum ραvαξ = 0. There is however, a possibilityof linear momentum transfer between the components of the mixture by interaction, with a contributionarising from the mass transfer (when there is a mass transfer between two components of the mixture,which are moving, it is necessarily accompanied by a corresponding linear momentum transfer) and anothercontribution arising from mechanical contact between the constituents represented by interaction forces Iα.We thus postulate

ραvαΠ= vαmα+Iα α= 1, . . . , N . (5.28)

The local (Eulerian) form of the linear momentum balance for α component of the mixture thus reads

∂(ραvα)∂t

+div(ραvα⊗vα)= divTα+ραbα+mαvα+Iα . (5.29)

We can rewrite the left-hand side as(∂ρα

∂t+div(ραvα)

)vα+ρα Dαvα

Dt= mαvα+ρα Dαvα

Dt,

where the mass balance (5.19) was applied. Thus, we can rewrite the linear momentum balance as follows

ραDαvα

Dt= divTα+ραbα+Iα α= 1, . . . , N . (5.30)

The corresponding balance of linear momentum is obtained again by summing (5.29) over all componentsand using the conservation assumption on the interaction terms (5.16), which in our case reads

N∑α=1

(mαvα+Iα)= 0 , (5.31)

and, correspondingly, the linear momentum balance for the mixture as a whole reads

N∑α=1

(∂(ραvα)

∂t+div

(ραvα⊗vα

))= div

(N∑α=1

Tα

)+

N∑α=1

ραbα , α= 1, . . . , N . (5.32)

32

As in the case of the mass balance, we may be interested in whether a corresponding single-componentequivalent of the linear momentum balance for the mixture as a whole can be identified. For this purpose,let us express the velocity as

vα = v+ (vα−v)= v+uα , α= 1, . . . , N . (5.33)

The term

uαdef= vα−v , (5.34)

is called diffusive velocity of the α component (with respect to the mixture barycentric velocity v). Usingthis splitting, the definitions of the barycentric velocity (5.24) and mixture density (5.9), and the fact that∑Nα=1ραuα = 0, we can rewrite the left-hand-side of (5.32) as follows

N∑α=1

(∂(ραvα)

∂t+div(ραvα⊗vα)

)= ∂(ρv)

∂t+div(ρv⊗v)+div

(N∑α=1

ραuα⊗uα

).

The first two terms are in the desired form already, we put the the third term on the right-hand side andrewrite (5.32) as

∂(ρv)∂t

+div(ρv⊗v)= div

(N∑α=1

(Tα−ραuα⊗uα)

)+

N∑α=1

ραbα . (5.35)

We thus see, that a natural definition of the mixture stress tensor, is not just the sum of partial stress tensors,but contains also a contribution due to diffusion. This is a typical feature of mixture theory when velocitiesof the constituents are distinguished. Defining the mixture stress tensor as

T=N∑α=1

(Tα−ραuα⊗uα) , (5.36)

and the mixture body force as

ρb=N∑α=1

ραbα , (5.37)

equation (5.35) takes the form of a traditional single-component linear momentum balance

∂(ρv)∂t

+div(ρv⊗v

)= divT+ρb , (5.38)

or, with the use of the mixture mass balance (5.27)

ρDvDt

= divT+ρb . (5.39)

Exercise 12. Derive the balances of mass and linear momentum for α component of the mixture in termsof φα (using the relation: ρα = φαρ

mα ). Define now the mixture velocity as vφ := ∑N

α=1φαvα and derive thebalance of linear momentum for the mixture as a whole in terms of this (volume-averaged) velocity.

Remark 8. How should one understand the concept of a partial stress? Can it be obtained intuitively atleast in some simple situation? Let us consider a two-component material composed of a porous matrix withconnected pores filled with a fluid as in Fig. 4.

Let us consider an equilibrium rest state, in which case the Cauchy stress in the fluid is the isotropicpressure Tm

f =−pmf I. The superscript “m” refers to the “material” or true value, we used this notation talking

about material density, meaning that is the stress state (pressure) that we would measure with a sufficientlysmall barometer inside an individual pore. What is the corresponding “upscaled” partial fluid stress tensor?Consider a surface element dS with a normal n cross-cutting the porous structure. The meaning of the par-tial fluid Cauchy stress is that T f ndS should give the force exerted on the surface dS by the fluid component.Clearly this is

T f ndS =Tmf ndS f =−pm

f dS f ,

33

Figure 4: A two component material composed of a porous matrix with pores filled with a fluid at pressurepm

f . Control surface dS, the part dS f of the the surface dS is exposed to the fluid.

where dS f is the fraction of the surface dS exposed to the fluid. If we knew everything about the geo-metric structure of the porous matrix, we could identify a function ϕ depending on the position and on theorientation of the surface that would provide the relation

dS f (x,n)=ϕ f (x,n)dS .

In practice, this information is never available and one has at best information about partial volume, (orporosity) of the particular phase, which we denoted φα. For special periodic structures, one can relate thevolume fraction and the surface fraction φα and ϕα, in particular, if we consider parallel tubular pores in oursituation, it would hold

ϕ f =φ f ,

i.e. the two are equal and the surface fraction does not depend on the orientation of the surface. Consequently,one gets

T f =−pmf φ f I ,

i.e. the corresponding “partial pressure” is the true pressure in the fluid times the volume fraction of thefluid component. Relationship of this type involving volume fraction (but sometimes also mass fraction)as the weighting factor between the actual material stress state and the mixture partial stress state areoften used in more general situations and for more general geometries, with a tacit understanding that suchrelationships are approximate.

5.5 Balance of angular momentum

We will consider a slightly extended setting with respect to the typical assumptions used in continuum me-chanics. There, one typically deals with the so called non-polar materials, in which case the balance ofangular momentum merely reduces to the assumption of symmetry of the Cauchy stress tensor. For materi-als with internal structure, which is often the case for mixtures, one may be forced to adopt a more generalview. For example, for a mixture composed of microscopic dispersed spheres of solid in some liquid, apartfrom translation of the spheres, one must also consider their rotations. This, in the averaged continuumview, gives rise to the notion of intrinsic angular momentum, the so called spin (which is a purely mechanicalproperty in the sense explained above, not to be confused with the quantum mechanical property of micropar-ticles). Materials which possess intrinsic angular momentum (spin) are called polar materials. The theoryof continuum mechanics of polar materials goes back to Cosserat brothers (1909) (the theory is sometimescalled Cosserat mechanics) and later extended and formalized by Eringen (1967), Toupin (1964) and manyothers. Here we give a very brief and formal review of this theory, for multi-component materials (mixtures).

Cosserat mechanics introduces several novel concepts into the classical continuum mechanics.

• Spin sα is the internal specific angular moment - ραsαdΩ represents the internal angular momentumof the volume element dΩ - that is, angular momentum hidden in the micro-rotations inside dΩ.

• Couple stress Mα - analogy of the Cauchy stress with respect to stress couples. It is introduced analogi-cally as the Cauchy stress, by a tetrahedron argument, looking for the stress couple acting on a surfaceelement and deriving linearity of such mapping with respect to the orientation of the element (given byits unit normal). The stress couple is then given by Mα ·n

34

Moreover, we shall also consider some outer specific spin supply term lα and interaction couple pα.The balance of angular momentum in the local form in Eulerian description is then obtained in our formal

balance framework by settingψα = x×ραvα+ραsα, identifying the corresponding non-convective flux as beingcomposed of moment of the partial surface force (expressed through the partial Cauchy stress) plus the couplestress, i.e. by setting x×ραvα+ραsαΦ = x×Tα+Mα, where (x×Tα)i j := εiabxa(Tα)b j. The angular momentumsupply is realized through moment of long-range body forces (gravity, elmag.) plus external spin supply,i.e. we set x×ραvα+ραsαΣ = x×ραbα+ραlα. We postulate that there is no production of angular momentumx×ραvα+ραsαξ = 0. Angular momentum transfer between the components is possible by interaction, with acontribution arising from (i) moment of interaction forces, (ii) as a result of mass transfer (when there is amass transfer between two components of the mixture, which have some non-zero angular momentum, itis necessarily accompanied by a corresponding angular momentum transfer) and (iii) another contributionarising from mechanical contact between the constituents realized through interaction couple pα. We thuspostulate

x×ραvα+ραsαΠ= x× (mαvα+Iα)+pα+mαsα , α= 1, . . . , N. (5.40)

This leads to the angular balance for polar component α in the following form:

∂

∂t

x×ραvα︸ ︷︷ ︸moment of linear momentum

+ραsα︸ ︷︷ ︸spin

+div((

x×ραvα+ραsα)⊗vα

)= div

x×Tα︸ ︷︷ ︸moment of surface forces

+ Mα︸︷︷︸couple stress

+ x×ραbα︸ ︷︷ ︸

moment of body forces

+ ραlα︸ ︷︷ ︸external spin supply

+ x× (mαvα+Iα︸ ︷︷ ︸moment of interaction forces

+ pα︸︷︷︸internal spin exchange

+ mαsα︸ ︷︷ ︸spin exch. due to mass exch.

, α= 1, . . . , N . (5.41)

In index notation this reads

∂

∂t(εi jkx jραvαk +ραsα i

)+ ∂

∂xl

((εi jkx jραvαk +ραsα i

)vαl

)= ∂

∂xl

(εi jkx jTαkl +Mα il

)+εi jkx jραbαk +ραlα i +εi jkx j(mαvαk +Iαk) +pα i +mαsα i , α= 1, . . . , N . (5.42)

which can be recast to the following form

εi jkx j

(∂(ραvα)

∂t+div(ραvα⊗vα)−divTα−ραbα−mαvα−Iα

)k+ ∂(ραsα i)

∂t+ραεi jkvαkvα j︸ ︷︷ ︸

=0

+ ∂

∂xl

(ραsα ivαl

)= εi jkTαk j +∂Mα il

∂xl+ραlα i +pα i +mαsα i , α= 1, . . . , N . (5.43)

where εi jk is the Levi-Civita permutation symbol. Introducing the vector Aα is defined as

Aα i = εi jk(Tα)k j , α= 1, . . . , N . (5.44)

and employing the balance of linear momentum (5.30), we obtain the following local form of the spin balance

∂(ραsα)∂t

+div(ραsα⊗vα)= divMα+ραlα+Aα+pα+mαsα , α= 1, . . . , N . (5.45)

or, using also the mass balance (5.19):

ραDαsα

Dt= divMα+ραlα+Aα+pα , α= 1, . . . , N . (5.46)

The vector Aα is closely related with the antisymmetric part of the partial Cauchy stress tensor as can beseen by multiplying (5.44) by εilm, (summing over i) and using the identity

εi jkεilm = δ jlδkm −δ jmδkl , (5.47)

35

which gives(Tα)ml − (Tα)lm = εilm(Aα)i, α= 1, . . . , N . (5.48)

In particular, for a mixture of non-polar constituents, i.e. setting sα = 0, Mα = 0, lα = 0, pα = 0, (5.46) implies

Aα = 0⇐⇒Tα =TTα , α= 1, . . . , N , (5.49)

that the partial Cauchy stresses are symmetric.Now we can formulate the angular balance for the mixture as a whole. Summing (5.45) and postulating

in accord with (5.16)N∑α=1

(x× (Iα+mαvα)+pα+mαsα))= 0 , (5.50)

which due to (5.31) implies alsoN∑α=1

(pα+mαsα)= 0 , (5.51)

we obtain∂(∑N

α=1ραsα)

∂t+div

(N∑α=1

(ραsα⊗vα)

)= div

(N∑α=1

Mα

)+

N∑α=1

ραlα+N∑α=1

Aα . (5.52)

Defining the total internal spin of the mixture as a whole as

ρs=N∑α=1

ραsα , (5.53)

the mixture spin outer supply by

ρl=N∑α=1

ραlα , (5.54)

vector A for the mixture as a whole

A=N∑α=1

Aα , (5.55)

and the total couple stress of the mixture as a whole by

M=N∑α=1

(Mα−ραsα⊗uα) , (5.56)

we arrive at the following single-component form of the spin balance for the mixture as a whole

∂(ρs)∂t

+div(ρs⊗v)= divM+ρl+A . (5.57)

A possible and sometimes considered case is that of a material, which behaves as a non-polar as a whole, i.e.the mixture Cauchy stress is symmetric, but interaction couples are possible among the constituents, leadingto non-symmetry of the partial Cauchy stresses. This can be achieved by neglecting partial spins (sα = 0),partial couples stresses (Mα = 0), partial spin supplies (lα = 0) in (5.46), which then reduces to

Aα =−pα, α= 1, . . . , N . (5.58)

or, equivalently in view of (5.48)Tα−TT

α =Pα , α= 1, . . . , N . (5.59)

where Pα is the antisymmetric tensor associated with the vector pα, i.e.

(Pα)i j = εi jk(pα)k , α= 1, . . . , N . (5.60)

In the limit of vanishing interaction couples, we recover the symmetry of the partial Cauchy stresses.Remark 9. The case described by (5.59) might seem a bit puzzling - we do not consider any spin in thematerial, yet we consider the interaction couples. Think, for example, of arrangement of molecules, whichhave some electric dipole moment. As a result of interaction among these electric dipoles, not only forces,but also momenta of forces act among the molecules. If now all the molecules are, for example, fixed insome crystallographic structure, and thus cannot rotate to accommodate to the exterted momenta, we areexactly in the discussed situation - the momenta have to be compensated by the momenta of contact forces.Consequently, this gives rise to the asymmetry of the Cauchy stress tensor.

36

5.6 Balance of energy (for non-polar mixture of non-polar constituents)

We start with the balance of total energy Eα for a mixture component. We split the total energy density into akinetic part and internal energy (with possible further non-trivial structure - e.g. containing potential energyin outer force fields, potential energy due to long-range interaction between the particles of the mixture and‘thermal’ energy due to thermal thermal motion of the particles of the particular mixture component), thatis, we set

ψα = ραEα = ρα(eα+ 1

2|vα|2

), α= 1, . . . , N . (5.61)

The total energy flux across a material boundary is composed of the power of surface forces acting thereand heat flux, i.e. we set ραEαΦ = Tαvα−qα, we postulate non-presence of energy production, i.e. we setραEαξ = 0, and the energy supply is realized through supply of mechanical power and other outer energysupply mechanisms, which we denote rα, i.e. we set ραEαΣ = ραbα ·vα+ραrα. Finally, the energy exchangebetween the components of the mixture is realized through: (i) mechanical power due to interaction forces,(ii) exchange of energy due to exchange of mass and (iii) the remaining interaction mechanisms denoted εα,i.e we set ραEαΠ= Iα ·vα+mα

(eα+ |vα|2

2

)+εα. With such choices, for the mixture constituents α= 1, . . . , N the

total energy balance reads:

∂

∂t

(ρα

(eα+

|vα|22

))+div

(ρα

(eα+

|vα|22

)vα

)= div(Tαvα−qα)

+ραbα ·vα+Iα ·vα+ραrα+εα+mα

(eα+

|vα|22

), α= 1, . . . , N. (5.62)

where we summarize: eα is the internal specific energy of α component, qα is the (partial) heat flux of the αcomponent, rα represent the specific energy sources (e.g. radiation) corresponding to α component, εα is theinteraction energy term representing the rate of energy exchange between α component and the rest of themixture due to internal processes in the material. We can see three novel terms (w.r.t. the classical single-continuum theory - cf. (3.24)) on the right-hand side of (5.62), namely Iα ·vα, εα, mα

(eα+ |vα|2

2

). All these

terms are interaction terms representing energy exchange between α component and the rest of the mixturedue to internal processes in the mixture. From this point of view, all these three terms could be absorbedinto one interaction term, say εα. Nevertheless, we choose to explicitly impose the given structure of theseterms, since we can a-priori pin out two particular of all the energy exchange mechanisms. The first one -the term Iα ·vα corresponds to the work performed by the interaction (e.g. frictional) forces Iα, the secondterm mα

(eα+ |vα|2

2

)represents the energy transfer directly related to mass transfer mα. To explicitly expand

the interaction terms as we did is motivated merely by the aim to arrive at simple relation for the internalenergy balance which will follow.

The interaction energy terms must again sum up to zero, i.e. we assert

N∑α=1

(εα+Iα ·vα+mα

(eα+ |vα|2

2

))= 0 . (5.63)

We will now derive a more compact form of the energy balance starting from (5.62) by getting rid of thekinetic energy contribution. Taking a scalar product of the partial linear momentum balance (5.30) with vα,we obtain

ρα

(∂

∂t|vα|2

2+vα ·∇ |vα|2

2

)= div(Tαvα)−Tα :∇vα+ραbα ·vα+Iα ·vα . (5.64)

Now the left-hand side of (5.62) can be written as follows (using also (5.19) and (5.64)):

(LHS) of (5.62) =(∂ρα

∂t+div(ραvα)

)(eα+ |vα|2

2

)+ρα

(∂eα∂t

+vα ·∇eα)+ρα

((∂

∂t|vα|2

2+vα ·∇ |vα|2

2

))= mα

(eα+ |vα|2

2

)+ρα

(∂eα∂t

+vα ·∇eα)+div(Tαvα)−Tα :∇vα+ραbα ·vα+Iα ·vα ,

(5.65)

37

Comparing with (5.62), we obtain

ρα

(∂eα∂t

+vα ·∇eα)=−divqα+Tα :∇vα+ραrα+εα , α= 1, . . . , N . (5.66)

or, using again the mass balance (5.19):

∂(ραeα)∂t

+div(ραeαvα)=−divqα+Tα :∇vα+ραrα+εα +mαeα , α= 1, . . . , N . (5.67)

The balance of total energy for the mixture as a whole is obtained by summing the partial balances (5.62)over all components of the mixture, employing (5.63). We arrive at

∂

∂t

(N∑α=1

ρα

(eα+|vα|2

2

))+div

(N∑α=1

ρα

(eα+|vα|2

2

)vα

)=div

(N∑α=1

(Tαvα−qα)

)+

N∑α=1

ραbα·vα+N∑α=1

ραrα (5.68)

Since energy is additive, the natural definition of the total specific energy of the mixture E is

ρE =N∑α=1

ραEα =N∑α=1

ρα

(eα+ |vα|2

2

). (5.69)

We expand the kinetic energy terms

N∑α=1

ρα|vα|2

2=

N∑α=1

12ρα(v+uα) · (v+uα)= ρ |v|

2

2+v ·

N∑α=1

ραuα︸ ︷︷ ︸=0

+N∑α=1

ρα|uα|2

2, (5.70)

and thus

ρE =N∑α=1

ραeα+ 12

N∑α=1

ρα|uα|2 + 12ρ|v|2 . (5.71)

Let us rewrite (5.68) using the definition of E and splitting the velocity vα = v+uα and the definition of themixture body force (5.37). We obtain

∂(ρE)∂t

+div

(ρEv+

N∑α=1

ραEαuα

)= div

(N∑α=1

Tαv+N∑α=1

Tαuα−N∑α=1

qα

)+

N∑α=1

ραbα ·vα+N∑α=1

ραrα , (5.72)

Let us recall how we defined the total mixture stress T in (5.36) and rewrite the term on the right-hand sideof (5.72)

div

(N∑α=1

Tαv

)= div(Tv)+div

((

N∑α=1

ραuα⊗uα)v

), (5.73)

and on the left-hand side we rewrite the term

div

(N∑α=1

ραEαuα

)= div

(N∑α=1

ραeαuα

)+div

N∑α=1

ρα

2(v+uα) · (v+uα)︸ ︷︷ ︸|v|2+|uα|2+2v·uα

uα

= div

(N∑α=1

ραeαuα

)+div

|v|22

N∑α=1

ραuα︸ ︷︷ ︸=0

+div(

12ρα|uα|2uα

)+div

((

N∑α=1

ραuα⊗uα)v

), (5.74)

Plugging (5.73) and (5.74) into (5.72) and noticing that the last terms in both equations cancel out, afterslight rearranging of the terms, we arrive at

∂(ρE)∂t

+div(ρEv)= div(Tv)−divN∑α=1

(qα−Tαuα+ρα(eα+ 1

2|uα|2)uα

)+ρb ·v+

N∑α=1

(ραrα+ραbα ·uα) . (5.75)

38

We can see that provided we define the mixture heat flux q is

q=N∑α=1

(qα−Tαuα+ρα(eα+ 12|uα|2)uα) , (5.76)

and the mixture energy sources r as

ρr =N∑α=1

(ραrα+ραbα ·uα) , (5.77)

we obtain the single-component form of the total mixture energy balance

∂(ρE)∂t

+div(ρEv)= div(Tv)−divq+ρb ·v+ρr . (5.78)

It remains to identify the condensed form (the internal energy balance) by subtracting the mixture kineticenergy balance. Multiplying the mixture linear momentum balance (5.35) by v, we get

v ·

∂ρ∂t+div(ρv)︸ ︷︷ ︸

=0

+ρ(∂

∂t|v|22

+v ·∇ |v|22

)= div(Tv)−T :∇v︸ ︷︷ ︸

=(divT)·v+ρb ·v , (5.79)

or, using the mass balance again on the left-hand side

∂

∂tρ|v|2

2+div

(ρv

|v|22

)= div(Tv)−T :∇v︸ ︷︷ ︸

=(divT)·v+ρb ·v , (5.80)

We subtract this identity from (5.78) and obtain

∂

∂t

(N∑α=1

ρα(eα+ |uα|22

)

)+div

(v

(N∑α=1

ρα(eα+ |uα|22

)

))=T :∇v−divq+ρr , (5.81)

which suggests to introduce the internal energy e of the mixture as

ρe =N∑α=1

ρα(eα+ |uα|22

) , (5.82)

in which case the reduced energy balance (internal energy balance) for the mixture as a whole reduces to itstraditional single-component form

∂(ρe)∂t

+div(ρev)=T :∇v−divq+ρr , (5.83)

or, using the mixture mass balance (5.27)

ρDeDt

=T :∇v−divq+ρr . (5.84)

Sometimes it is also useful to consider the balance obtained by summation of (5.67):

∂(∑Nα=1ραeα)∂t

+div(N∑α=1

ραeαvα)=−divN∑α=1

qα+N∑α=1

Tα :∇vα+N∑α=1

ραrα+N∑α=1

εα +N∑α=1

mαeα , (5.85)

Using the constraint on the interaction terms (5.63), we can rewrite this relation as

∂(∑Nα=1ραeα)∂t

+div(N∑α=1

ραeαvα)=−divN∑α=1

qα+N∑α=1

Tα :∇vα+N∑α=1

ραrα−N∑α=1

Iα ·vα−N∑α=1

mα|vα|2

2, (5.86)

39

This can still be simplified as follows:

N∑α=1

Iα ·vα+N∑α=1

mα|vα|2

2= |v|2

2

N∑α=1

mα︸ ︷︷ ︸=0

+v ·N∑α=1

(Iα+mαvα)︸ ︷︷ ︸=0

+N∑α=1

Iα ·uα+N∑α=1

mα|uα|2

2

=N∑α=1

Iα ·uα+N∑α=1

mα|uα|2

2. (5.87)

Using this relation in (5.88), we can rewrite it as

∂(∑Nα=1ραeα)∂t

+div(N∑α=1

ραeαvα)=−divN∑α=1

qα+N∑α=1

Tα :∇vα+N∑α=1

ραrα−N∑α=1

Iα ·uα−N∑α=1

mα|uα|2

2. (5.88)

5.7 Balance of entropy (Second law of thermodynamics)

Setting ψα = ραηα, where ηα is the partial specific entropy of α component, denoting the partial entropyflux as ραηαΦ = qηα, considering entropy exchange due to mass exchange and other mechanism denoted byξ, i.e. setting ραηαΠ= mαηα+ ξα and denoting the entropy production ραηαξ= ξα, and the entropy supply byραηαΣ= ραrηα, the balance of entropy for a particular α component of the mixture reads

∂(ραηα)∂t

+div(ραηαvα

)=−divqηα+ξα+ ξα +mαηα +ραrηα , α= 1, . . . , N . (5.89)

As in the case of energy balance, we have explicitly split the entropy exchange into two terms: ξα and mαηα,the second one corresponding to entropy exchange due to mass exchange between the components of themixture. The interaction exchange terms must by definition sum up to zero, i.e. we assert

N∑α=1

(ξα+mαηα)= 0 . (5.90)

The statement of the second law of thermodynamics for a particular component would be postulate of non-negativity of the partial entropy productions:

ξα ≥ 0 , α=1, . . . , N . (5.91)

Balance of entropy for the mixture as a whole is obtained again by summing up the partial entropybalances (5.89) over all component of the mixture, with the use of (5.90), we obtain:

∂

∂t

(N∑α=1

ραηα

)+div

(N∑α=1

(ραηα)v

)=−div

(N∑α=1

qηα+ραηαuα

)+

N∑α=1

ξα +N∑α=1

ραrηα . (5.92)

This expression suggests to identify

ρη =N∑α=1

ραηα, (5.93)

qη =N∑α=1

(qηα+ραηαuα) , (5.94)

ξ =N∑α=1

ξα , (5.95)

ρrη =N∑α=1

ραrηα . (5.96)

in which case, we recover the traditional single-component entropy balance:

∂(ρη)∂t

+div(ρηv)=−divqη+ξ+ρrη , (5.97)

40

or, using (5.27)

ρDηDt

=−divqη+ξ+ρrη , (5.98)

and the statement of the second law of thermodynamics in view of (5.95) and (5.91) reads

ξ≥ 0 . (5.99)

5.8 Classification of the mixture theories

Certain classification of the mixture theories is possible based on the chosen complexity at the level of balancelaws. The motivation is that we do not always need to distinguish between the constituents at all levels. Asan example, we can be interested in describing the evolution of concentrations of dispersed tracers in asolvent, that is, to solve the balances of mass for all constituents, but concerning the mechanical and thermalbehavior, we are only concerned with the behavior of the mixture as a whole, that is not considering individualbalances. Similarly, we can proceed to further levels by increasing the complexity, giving rise to the followingclassification:

Following the classification introduced by Hutter and Jöhnk (2004), we divide the mixtures based on thedescriptive level into four classes, depending on whether partial component-wise properties of the mixtureare formulated, or just properties of the mixture as a whole are considered:

Class IV Class III Class II Class Iρα ρα ρα ραvα vα vα veα eα e eηα η η η

Class I. N balances of mass for constituents, one balance of linear momentum for the mixture as a wholeand one balance of energy and entropy for the mixture as a whole. Typical examples of class I mixtures:Fickian systems, dissolved species in solvent, mixtures of dilute gases, pollution in air, water, ...

Class II. N balances of mass for individual components, N balances of linear momentum for components,but only one balance of energy and entropy for the mixture as a whole. Typical examples: Flow of fluidsin porous media (Darcy, Forchheimer, Brinkman), partial melting and melt extraction in geologicalmaterials, chemical reactors (bubbly flow),. . .

Class III. N balances of mass, N balances of lin. momentum and N balances of energy, one balance ofentropy for the mixture as a whole. Typical examples: Plasmas, Systems of melt + matrix far fromequilibrium: need to distinguish different temperatures of the species,. . .

In principle, one can consider balances of entropy at the level of individual constituents, i.e the Class IV.mixtures. Such formulation is, however, not often met in practice, where in most cases, the second law canbe reasonably formulated only for the mixture as a whole.

6 Class I mixtures

As defined in the previous section, within the Class I mixture theory, we employ the partial mass balancesfor the mixture constituents, but consider only linear momentum, angular momentum, energy (and entropy)balance for the mixture as a whole. The system of governing equations (without entropy balance which will

41

be employed later in the constitutive theory for class I mixtures) reads

∂ρα

∂t+div(ραvα)= mα , α= 1, . . . , N , (6.1a)∑

α

mα = 0 , (6.1b)

∂(ρv)∂t

+div(ρv⊗v)= divT+ρb , (6.1c)

T=TT , (6.1d)

∂

∂t

(ρ

(e+ |v|2

2

))+div

(ρ

(e+ |v|2

2

)v)= div(Tv−q)+ρb ·v+ρr , (6.1e)

where

ρ =∑α

ρα , ρv=N∑α=1

ραvα .

This is still not the final form, since these equations contain the partial velocities vα, for which we do nothave any evolution equations within the Class I, (we only have one eq. for the barycentric velocity v). Let usthus rewrite the system (6.1) by introducing the diffusive fluxes (with respect to barycenter)

jαdef= ρα(vα−v) , α= 1, . . . , N . (6.2)

and rewrite the system (6.1) as follows

∂ρα

∂t+div(ραv)= mα−divjα , α= 1, . . . , N , (6.3a)∑

α

mα = 0 , (6.3b)

∂(ρv)∂t

+div(ρv⊗v)= divT+ρb , (6.3c)

T=TT , (6.3d)

∂

∂t

(ρ

(e+ |v|2

2

))+div

(ρ

(e+ |v|2

2

)v)= div(Tv−q)+ρb ·v+ρr . (6.3e)

This system of equations is now perceived as a system for unknowns ρα, v, e, which needs to be complementedby constitutive equations determining T, mα, q and jα, i.e. we ignore the structure (6.2) and determine jαconstitutively. We only need to bear in mind that any constitutive relations for jα must satisfy the constraint

N∑α=1

jα = 0 , (6.4)

which on the level (6.2) follows directly from the definition (of the barycentric velocity).Instead of using the set of independent variables ρα,v, e, we can equivalently work in terms of variables

cα,ρ,v, e (we could also start with another equivalent set containing e.g. the volume fractions φα), where, letus remind

cα = ρα

ρ.

Let us evaluate the material time derivative of cα with respect to barycentric velocity and multiply by ρ

ρ cα = ρ∂cα∂t

+ρv ·∇cα(5.25)= ∂

∂t(ρcα

)+div(ρcαv

)== ∂ρα

∂t+div

(ραv

) (5.19)= mα+div(ρα(v−vα)

).

we arrive at the following evolution equation for concentration cα:

ρ cα+divjα = mα , α= 1, . . . , N . (6.5)

Note that since∑Nα=1 cα ≡ 1, and

∑Nα=1 jα = 0, only N −1 equations from (6.5) are independent. To obtain a

system equivalent with (5.25), we must accompany (6.5), with the mixture mass balance (5.19).

42

The system of governing equations for class I mixtures, written in terms of concentrations reads

ρ cα = mα−divjα , α= 1, . . . , N , (6.6a)∑α

mα = 0 , (6.6b)∑α

jα = 0 , (6.6c)

∂ρ

∂t+div(ρv)= 0 , (6.6d)

∂(ρv)∂t

+div(ρv⊗v)= divT+ρb , (6.6e)

T=TT , (6.6f)

∂

∂t

(ρ

(e+ |v|2

2

))+div

(ρ

(e+ |v|2

2

)v)= div(Tv−q)+ρb ·v+ρr . (6.6g)

(6.6h)

In the following, we will proceed in the similar manner as for the single continuum, i.e. starting froma fundamental thermodynamic relation for the Helmholtz free energy ψ = ψ(ϑ, 1

ρ, c1, . . . , cN ), or its reduced

form ψ= ψ(ϑ, 1ρ

, c1, . . . , cN−1), and we will attempt to derive the constitutive relations. We will moreover alsodiscuss how to impose various additional constraint (incompressibility,...) into these models and discuss theirconsequences.

We will in the following, for simplicity, for a while consider only a special case of a binary mixture, i.e. thecase N = 2. Since we know that

N∑α=1

cα = 1⇒ c1 + c2 = 1, and thus c2 = 1− c1 ,

denoting c def= c1, m def= m1 = 1−m2, j def= j1 =−j2, we obtain the following system

ρ c+divj= m , (6.7a)

ρ =−ρdivv , (6.7b)

ρv= divT+ρb , (6.7c)

T=TT , (6.7d)

ρE = div(Tv−q)+ρb ·v+ρr . (6.7e)

or using the reduced form of the energy balance (internal energy balance):

ρ c+divj= m , (6.8a)

ρ =−ρdivv , (6.8b)

ρv= divT+ρb , (6.8c)

T=TT , (6.8d)

ρ e =T :∇v−divq+ρr . (6.8e)

We will now investigate the following special cases of binary Class I mixtures based on various choicesand constraints on the constitutive equations, which will allow to derive as special cases the constitutivemodels of Fick, Cahn-Hilliard, Allen-Cahn and their generalizations. Throughout most of this chapter, wewill be work with fundamental relation for the Helmholtz free energy, see (3.64) in order to work with anatural set of variables involving thermodynamic temperature ϑ, density ρ concentrations cα. Particularchoices of the fundamental relation for ψ will lead to several classical models

1. ψ= ψ(ϑ, 1ρ

, c) ⇒ Fick-Navier-Stokes-Fourier model

2. ψ= ψ(ϑ, 1ρ

, c,∇c), j6=0, m=0 ⇒ Cahn-Hilliard-Navier-Stokes-Fourier model

3. ψ= ψ(ϑ, 1ρ

, c,∇c), j=0, m 6=0 ⇒ Allen-Cahn-Navier-Stokes-Fourier model

43

6.1 Fick-Navier-Stokes-Fourier model (ψ= ψ(ϑ, 1ρ

, c))

We start from the fundamental relation for the Helmholtz free energy

ψ= e−ϑη= ψ(ϑ,1ρ

, c) ,

where, for simplicity, we work with the reduced form of the fundamental relation (denoted by the bar symbol)which reflects the fact that one of the concentrations was eliminated. Taking a time derivative, and recalling(3.67)2

−η= ∂ψ

∂ϑ, −p = ∂ψ

∂ 1ρ

µ= ∂ψ

∂c, (6.9)

we obtain using the balances of mass and energy (setting for simplicity the energy sources equal to zeror ≡ 0):

ρϑη=T :D−divq+ pdivv−µm+µdivj , (6.10)

or. equivalentlyρϑη=Td :Dd + (m+ p)divv−µm−div(q−µj)− j ·∇µ . (6.11)

Dividing (6.11) by temperature and absorbing the term 1ϑ

in the divergence term, we obtain the balance ofentropy in the following form

ρη+div(q−µjϑ

)= 1ϑ

[Td ·Dd + (m+ p)divv− (q−µj) · ∇ϑ

ϑ− j ·∇µ−µm

]︸ ︷︷ ︸

ξ

. (6.12)

If we want to identify independent entropy-producing mechanism and provide piece-wise linear closures,we now have more options, how to treat the terms involving j ·∇ϑ. Let us consider two end-member variants.

Variant A

Assuming the terms in (6.12) represent these mutually independent mechanism, if we restrict ourselves asbefore to only linear relations, we obtain constitutive closures in the form

Td = 2νDd , (6.13a)

m+ p = 2ν+3λ3

divv , (6.13b)

q−µj=−κ∇ϑϑ

, (6.13c)

j=−α∇µ , (6.13d)

m =−βµ , (6.13e)

where for ν ≥ 0, 2ν+ 3λ ≥ 0, κ ≥ 0, α ≥ 0, β ≥ 0, we obtain that automatically the entropy production isnon-negative, i.e. ξ

ϑ≥ 0 and we obtain the final form of the constitutive relations

T=−pI+λdivv+2νD , (6.14a)

j=−α∇µ , (6.14b)

q=−κ∇ϑϑ

+µj , (6.14c)

m =−βµ . (6.14d)

2Note that based on our definitions from Sec. 3.2.2, for a binary mixture we would have ψ= ψ(ϑ, 1ρ , c1, c2), with the definition of

the chemical potential as µα = ∂ψ∂cα . Since in this section, we defined our Helmholtz energy as ψ(ϑ,ρ, c) = ψ(ϑ, 1

ρ , c,1−c), we can see

that in fact µ = ∂ψ∂c = ∂ψ

∂c1− ∂ψ∂c2

= µ1−µ2 , that is the “chemical potential” we are working with is in fact the difference between thechemical potentials of the two component of the mixture.

44

Note that if ψ= ψ0(ϑ, 1ρ

)+ ψ1(c), we obtain µ= ψ′1

j=−αψ′′1(c)∇c , (6.15)

and the evolution equation for concentration reads

ρ c−αdiv(ψ′′1∇c)=−βψ′

1 . (6.16)

The relation (6.15), written in the more general form

j=−k(c)∇c , (6.17)

represents the Fick’s law - analogy of Fourier’s law of heat conduction relating the heat flux and the tem-perature gradient.

Remark 10. Note that should ψ1 be quadratic (and convex) in c, the evolution equation for concentrationreads

ρ c−k1∆c+k2c = 0 , (6.18)

where k1 ≥ 0 and k2 are constants. This is an example of the so-called advection-diffusion-reaction equation(the three terms corresponding to these three mechanisms).

Remark 11. A perhaps more realistic, despite still rather special example could be considered as following.Let’s have a binary mixture, composed of substance A which is highly diluted in substance B which inturn represents vast majority of the mixture. Let’s assume that mass conversion (reaction) between thetwo substances is possible, i.e. reaction A ↔ B takes place. Let, for simplicity, the molar masses of bothsubstances be the same (equal to M). Then, as we shall see later, we can write the chemical potentialµ = µA −µB = (µ0

A −µ0B)(ϑ,ρ)+ Rϑ

M (ln cA − ln(1−cA)) .= (µ01−µ0

2)(ϑ,ρ)+ RϑM ln cA, the last equality due to the

assumption of diluteness. With the same piece-wise linear closure relations as before, and assuming somegiven isothermal conditions and neglecting volume changes of the mixture as a whole (i.e. ϑ=const., ρ=const),we obtain the following evolution equation for concentration c def= cA:

ρ c− RM

div(α

c∇c)=−β

((µ0

A−µ0B)(ϑ,ρ)+ Rϑ

Mln c

),

we could consider α0 > 0 and α=α0c ≥ 0, which yields

ρ c− α0︸︷︷︸def=α0

RM

∆c =−β((µ0

A−µ0B)(ϑ,ρ)+ Rϑ

Mln c

).

On the left-hand side, we obtained the classical advection-diffusion operator. The term on the right-hand siderepresents the rate of the reaction and, while β≥ 0, the bracket does not have any a-priori given sign. Its sign,which determines whether the reaction goes forward or backward, depends on whether the concentrationis larger or smaller than the actual equilibrium composition (such c for which the bracket is zero). Thisequilibrium value, as we can infer, can be a very intricate function of temperature and pressure (and ittypically is the case in real-world chemical reactions).

Variant B

Alternatively, we can rewrite (6.12) as follows

ρη+div(q−µjϑ

)= 1ϑ

[Td :Dd + (m+ p)divv−q · ∇ϑ

ϑ−µm

]− j ·∇

(µϑ

)︸ ︷︷ ︸

ξϑ

. (6.19)

45

Assuming now the grouped terms represent the individual entropy-producing mechanisms, ignoring the pos-sible mutual couplings (cross-effects), for a class of piece-wise linear closure relations, we obtain:

Td = 2νDd , (6.20a)

m+ p = 2ν+3λ3

divv , (6.20b)

q=−κ∇ϑϑ

, (6.20c)

j=−α∇(µϑ

), (6.20d)

m =−βµ , (6.20e)

where for ν ≥ 0, 2ν+ 3λ ≥ 0, κ ≥ 0, α ≥ 0, β ≥ 0, we obtain that automatically the entropy production isnon-negative, i.e. ξ≥ 0 and we obtain the final form of the constitutive relations

T=−pI+λdivv I+2νD , (6.21a)

j=−α∇(µϑ

), (6.21b)

q=−κ∇ϑϑ

, (6.21c)

m =−βµ . (6.21d)

Variant C

Alternatively, if we want to remain indecisive as for the splitting, we can introduce the following convexcombination for a free parameter δ ∈ (0,1), use the following identity

−j ·∇µ+µj · ∇ϑϑ

=−j ·(δ∇µ+ (1−δ)ϑ∇

(µϑ

))+δµj · ∇ϑ

ϑ(6.22)

with the help of which, we can rewrite (6.12) as follows

ρη+div(q−µjϑ

)= 1ϑ

[Td :Dd + (m+ p)divv− (q−δµj) · ∇ϑ

ϑ− j ·

((1−δ)ϑ∇

(µϑ

)+δ∇µ

)−µm

]︸ ︷︷ ︸

ξϑ

. (6.23)

Assuming now the grouped terms represent the individual entropy-producing mechanisms, ignoring the pos-sible mutual couplings (cross-effects), for a class of piece-wise linear closure relations, we obtain:

Td = 2νDd , (6.24a)

m+ p = 2ν+3λ3

divv , (6.24b)

q−δµj=−κ∇ϑϑ

, (6.24c)

j=−α((1−δ)ϑ∇

(µϑ

)+δ∇µ

), (6.24d)

m =−βµ , (6.24e)

where for ν ≥ 0, 2ν+ 3λ ≥ 0, κ ≥ 0, α ≥ 0, β ≥ 0, we obtain that automatically the entropy production isnon-negative, i.e. ξ≥ 0 and we obtain the final form of the constitutive relations

T=−pI+λdivv I+2νD , (6.25a)

j=−α((1−δ)ϑ∇

(µϑ

)+δ∇µ

), (6.25b)

q=−κ∇ϑϑ

+δµj , (6.25c)

m =−βµ . (6.25d)

46

Remark 12. In all cases the chemical reaction rates were written simply as m =−βµ. This is, however, veryunrealistic in real applications. Chemical reactions typically require non-linear closures, we will investigatechemical reactions in more detail in Sec. 6.5.4.

6.2 Constraints: Incompressibility and Quasi-incompressibility

Our starting system of governing equations is still

ρ c =−divj+m , (6.26a)

ρ =−ρdivv , (6.26b)

ρv= divT+ρb , (6.26c)

T=TT , (6.26d)

ρ e =T :D−divq . (6.26e)

We will now consider two special types of constrains - incompressibility constraint, which will be formulatedas an assumption of isochoricity of the motion of the mixture as a whole, and so called quasi-incompressibility,which will be based on an explicit relation between the density of the mixture and the concentration.

6.2.1 Incompressibility

By incompressibility, we shall understand the assumption that the motion of the mixture as a whole isisochoric, meaning

divv= 0 . (6.27)

Starting from the same (i.e. compressible) constitutive assumptions as before, and by just plugging (6.27)into (6.10), we arrive at

ρϑη=Td :Dd −divq−µm+µdivj . (6.28)

From here we can proceed in the same lines as in the compressible case and obtain the three variantsA,B,C. In particular, for variant C (A and B follow setting δ= 1 and 0, respectively), we get:

Td = 2νDd , (6.29a)

q−δµj=−κ∇ϑϑ

, (6.29b)

j=−α((1−δ)ϑ∇

(µϑ

)+δ∇µ

), (6.29c)

m =−βµ . (6.29d)

As we can see the entropy inequality no longer provides any information about the isotropic part of theCauchy stress tensor (mean normal stress m def= 1

3 trT). Therefore, concerning the final closures, we can onlywrite,

T=mI+2νDd , (6.30a)

j=−α((1−δ)ϑ∇

(µϑ

)+δ∇µ

), (6.30b)

q=−κ∇ϑϑ

+δµj , (6.30c)

m =−βµ , (6.30d)

where now the mean-normal stress m is a constitutively undetermined variable, which has to be sought bysolving the corresponding initial-boundary value problem.

47

6.2.2 Quasi-incompressibility

The quasi-incompressibility will be a certain assumption of explicit relation between the mixture densityand concentration, i.e. relation

ρ = ρ(c) . (6.31)

Taking time derivative of (6.31) and using (6.26a) and (6.26b), we obtain relation

divv=(

1ρ(c)

)′(m−divj). (6.32)

Let us remind that we have in mixture the following measures ρα (partial density), cα (concentration), φα(volume fraction), ρm

α (true material density) and let us now assume that the following assumptions hold inthe material:

(P1) mass additivity constraint M (P )=∑Nα=1 Mα(P ), which implies ρ =∑N

α=1ρα ,

(P2) volume additivity constraint V (P )=∑Nα=1 Vα(P ), which implies

∑αφα = 1 ,

(P3) ρmα are constant for all α= 1, . . . , N .

Under the assumptions (P1)–(P3) we obtain for an N-component mixture the following expression for 1ρ

interms of concentrations cα and (constant) material densities ρm

α

1ρ

(c1, . . . , cN )=N∑α=1

cαρmα

. (6.33)

which follows immediately if we write down the volume additivity constraint as

1=N∑α=1

φα =N∑α=1

ρα

ρmα

= ρN∑α=1

cαρmα

. (6.34)

In particular, for a binary mixture, we get

1ρ(c)

= cρm

1+ 1−cρm

2. (6.35)

We have demonstrated that the assumption ρmα constant + volume additivity ⇒ Quasi-incompressibility. In

the following we will be using this stronger assumption, so that not only there is some relation between ρ

and cα, but that the particular formula (6.33) holds.Coming back to the binary mixture, having ρm

1 ,ρm2 ∈R+, ρm

2 > ρm1 , we can explicitly compute(

1ρ(c)

)′=

(ρm

2 −ρm1

ρm1 ρ

m2

)def= r∗ > 0 , (6.36)

and our quasi-incompressible constraint thus reads

divv= r∗(m−divj) . (6.37)

Exercise 13. How does analogue of (6.37) look like for a N-component quasi-incompressible mixture from?Solution:

divv=N∑α=1

mα−divjαρmα

Let us discuss the constraint (6.37) a bit more. The terms involved are connected with three dissipativeprocesses - roughly speaking divv corresponds to compaction of the material, divj to diffusion and m to massconversion by reactions (chemical, phase transitions). Let’s get back to our constitutive procedure. Startingfrom the internal energy in the “Fickian” form for simplicity ψ= ψ(ϑ,ρ, c). We can again obtain the followingbalance

ρϑη=Td :Dd + (m+ p)divv−µm+µdivj−divq . (6.38)

Now we want to plug in the constraint (6.37), the question is how? There are several possibilities

48

(A) “Compaction is a dependent mechanism”: we substitute in (6.38) for divv from (6.37)

(B) “Compaction is an independent mechanism”: we substitute in (6.38) for m−divj from (6.37)

(C) Both mechanisms are present and (to-some extent) independent

Variant A

Substituting from (6.37) into (6.38) for divv gives us

ρϑη=Td :Dd − (µ−r∗(m+p))m+ (µ−r∗(m+p))divj−divq . (6.39)

Denoting auxiliary “chemical potential”µ

def= µ−r∗(m+p) , (6.40)

and comparing with the incompressible Fick-NSF case (6.28), we see, we are in fact in the same setting, onlywith µ replaced by µ. We can immediately “copy-and-paste” the closure relations:

Td = 2νDd , (6.41a)

q−δµj=−κ∇ϑϑ

, (6.41b)

j=−α((1−δ)ϑ∇

(µ

ϑ

)+δ∇µ

), (6.41c)

m =−βµ , (6.41d)

for some δ ∈ ⟨0,1⟩. Writing explicitly the final closures, we get for some ν≥ 0, α≥ 0, κ≥ 0, β≥ 0:

T=mI+2νD , (6.42a)

j=−α((1−δ)ϑ∇

(µ−r∗(m+p)

ϑ

)+δ∇(µ−r∗(m+p))

), (6.42b)

q=−κ∇ϑϑ

+δ(µ−r∗(m+p))j , (6.42c)

m =−β(µ−r∗(m+p)) . (6.42d)

It is instructive to look at the structure of the balance equations after plugging in these closures. Let us, forsimplicity, consider an isothermal setting and write down the balances of mass and momentum. We obtain

divv= r∗(−β+α∆)(µ−r∗(m+p)) , (6.43a)

ρv=∇m+div(2νDd) , (6.43b)

ρ c−div(α∇(µ−r∗(m+p))

)=−β(µ−r∗(m+p)) . (6.43c)

These are 5 equations for 5 unknowns v, c, m, and the system should be understood in the sense, that inrelations for both µ and p, density was substituted from the constraint (6.31), i.e. µ= µ(c), p = p(c), it is moreelucidating to rewrite the system as follows

divv= r2∗(β−α∆)m+ r∗(−β+α∆)( àµ−r∗p)(c) , (6.44a)

ρv=∇m+div(2νDd) , (6.44b)

ρ c−div(α∇( àµ−r∗p)(c)

)+β( àµ−r∗p)(c)= r∗(β−α∆)m . (6.44c)

Remark 13. (Mathematical remark on quasi-incompressibility) The incompressible Navier-Stokes systemreads (with ρ = 1 since this is a mathematical remark):

divv= 0 (6.45)∂v∂t

+div(v⊗v)=∇m+ν∆v (6.46)

49

R. Temam in his “1st book on NSEs in the West” (J. Málek) replaced this system by the following quasi-compressible approximation

divvε =−ε∆m (6.47)∂vε

∂t+div(vε⊗vε)=∇mε+ν∆vε . (6.48)

Let us inspect the advantages of such a system in terms of a-priori estimates. Multiplying the momentumeq. by v, we get:

∂ |vε|22

∂t+div

( |vε|22

vε)+ν|∇vε|2 +div

(νvε ·∇vε

)−div(mvε)+mdivvε = 0 (6.49)

substituting in the last term from (6.45), integrating over Ω and assuming all the boundary integrals vanish,we obtain

∂t‖vε‖2L2(Ω) +‖ν∇vε‖2

L2(Ω) +ε‖∇mε‖2L2(Ω) = 0 , (6.50)

integrating over time from 0 to some T, we see, that in this case, one obtains apart from the standardestimates on velocity (vε ∈ L2((0,T),H1(Ω))∩L∞((0,T),L2(Ω)), for a fixed ε, we get also ∇mε ∈ L2((0,T),L2(Ω)),suggesting the usefulness of the concept of mathematical quasi-compressibility.

Now inspecting (6.44a), we can see that the structure of the r.h.s. with respect to m is analogous (withε ∼ αr2∗). This indicates that the mathematical concept can interpreted in terms of the physical notion ofquasi-incompressibility, which we have defined.

Variant B

Substituting from (6.37) into (6.61) for m−divj gives us

ρϑη=Td :Dd + (m+(p−r−1∗ µ))divv−divq . (6.51)

Denotingp def= p−r−1

∗ µ , (6.52)

and comparing with (3.136), we can see the same structure as for compressible NSF system. We can again“copy-and-paste” the corresponding closure relations (3.140), only with p replaced by p:

Td = 2νDd , ν≥ 0 , (6.53a)

(m+ p)= 2ν+3λ3

divv , 2ν+3λ≥ 0 , (6.53b)

q=−k∇ϑϑ

, k ≥ 0 . (6.53c)

Putting all together (and setting κ def= kϑ≥ 0), we obtain

T=−(p−r−1∗ µ)I+λdivvI+2νD , (6.54a)

q=−κ∇ϑ . (6.54b)

The chemical potential is assumed to be expressed now as a function of η and ρ, obtained by substitutingfor c from inversion of (6.31). Again, in the isothermal setting, we can write down the balances of mass andmomentum

ρ+ρdivv = 0 , (6.55)

ρv = ∇( áp−r−1∗ µ)(ϑ,ρ)+∇(λdivv)+div(2νD) , (6.56)

so it is indeed compressible NSF system, only with a modified definition of the thermodynamic pressure.Concentration is obtained from ρ (solution of the corresponding IBVP) simply by inverting relation the in-compressibility constraint (6.31).

50

Remark 14. Let us shortly comment on some interesting mathematical properties of the system (6.55) Assum-ing ideal mixing contribution of concentration to the chemical potential, we can write (up to some constants,but this is mathematical detour):

µ(c)= (ln(c)− ln(1−c)) (6.57)

Assuming without loss of generality ρm2 ≥ ρm

1 and setting ρm2 = 1, ρm def= ρm

1 , we get r∗ = 1−ρm

ρm =⇒ 1ρm = 1+ r∗

and c(ρ)= ρr∗+(ρ−1)ρr∗

. Hence

µ(c(ρ))= ln((1+ r∗)ρ−1

)− ln(1−ρ)=⇒ ρ ∈(

11+ r∗

,1)

, (6.58)

so in the quasi-incompressible setting, we obtain structure formally similar to compressible Navier-Stokesequations, but due to arising singular pressure, we obtain boundedness of density ρ, which otherwise is notguaranteed in the compressible case.

Variant C

If we want to keep both mechanisms present, we will split

v= vA +vB , (6.59a)

j= jA + jB , (6.59b)

m = mA +mB , (6.59c)

and we will assume that both sets A and B satisfy the constraint, i.e.

divvA = r∗(mA −divjA) , (6.60a)

divvB = r∗(mB −divjB) . (6.60b)

We write the initial balance as

ρϑη=Td :Dd + (m+p)(divvA+divvB)− (mA+mB)µ+ (divjA+divjB)µ−divq , (6.61)

and we substitute for divvA from (6.60a) and for mB−divjB from (6.60b). We obtain

ρϑη=Td :Dd −divq− (µ−r∗(m+p))(mA−divjA)+ (m+p−r−1∗ µ)divvB . (6.62)

Denoting as before

µdef= µ−r∗(m+p) , (6.63a)

p def= p−r−1∗ µ , (6.63b)

we obtain

ρϑη=Td :Dd −divq+ (m+p)divvB − µmA + µdivjA , (6.64)

and comparing with the compressible Fick-NSF case (6.10), we see, we are in fact in the same setting, onlywith v, j, m, p, and µ replaced by vB, jA, mA, p and µ, respectively. We can immediately “copy-and-paste”the closure relations:

Td = 2νDd , (6.65a)

m+ p = 2ν+3λ3

divvB , (6.65b)

q−δµjA =−κ∇ϑϑ

, (6.65c)

jA =−α((1−δ)ϑ∇

(µ

ϑ

)+δ∇µ

), (6.65d)

mA =−βµ , (6.65e)

51

where ν,2ν+3λ,κ,α,β ≥ 0, we obtain that automatically the entropy production is non-negative, i.e. ξ ≥ 0.We can not identify the full closures, (e.q. for j or m), but the given ones allow to close the system of balances.

Consider again, an isothermal setting, for simplicity. Now can compute

divv = divvA +divvB = r∗(mA−divjA)+divvB

= r∗(−βµ+div(α∇µ))+ 32ν+3λ

r∗(m+ p) , (6.66)

m−divj = (mA −divjA)+ r−1∗ divvB

= −βµ+div(α∇µ)+ r−1∗

32ν+3λ

(m+p) . (6.67)

6.3 Cahn-Hilliard-NSF and Allen-Cahn-NSF model

Motivation: The Cahn-Hilliard and Allen-Cahn models are one of the simplest models which allows to de-scribe a phenomena of phase-separation, i.e. spontaneous separation of two component of a binary mixtureinto distinct macroscopic phases. They are typical representatives of the class of the so-called diffuse interfacemodels. These models allow to deal with multi-phase continua without the necessity to track and explicitlycapture the behaviour of the interfaces between the phases. In the diffuse interface framework, the inter-faces are described intrinsically as regions of steep gradient of some order parameter (e.g. concentration).This viewpoint is also more physical, true interfaces between phases or between different materials exhibit,in fact, some transition layers (with typical thickness of few molecular diameters).History - capillarity effects - Young (1805); van der Waals (1893); Korteweg (1901); Cahn and Hilliard (1958,1959); Lowengrub and Truskinowsky (1998). The crucial conceptual idea behind the models of Cahn-Hilliardand Allen-Cahn is (in the simplest two-component setting) related to the variational (minimization) problemfor the free energy functional F of the form:

F =∫Ω

f (c)+ σ

2|∇c|2dx . (6.68)

The first term - called phobic term, typically looks as in Fig.5 the second term ∼ |∇c|2 is called phyllic term.The coefficient σ is proportional to the surface tension and is non-negative.

c

f (c)

c = 0 c = 1

Figure 5: Typical “double-well” structure of the phobic part of the free energy in the Allen-Cahn and Cahn-Hilliard models.

As we will see, both the Cahn-Hilliard and the Allen-Cahn model can be viewed as certain evolutionequations, that ensure minimization of the free-energy functional. Since the phobic part has two local min-ima, the system subject to such evolution will try to separate into the two phases corresponding to thesetwo minima (c = 0 and c = 1 for the case depicted in Fig5), thereby the name phobic. On its own, this termwould lead to a separation into regions with c = 1 and c = 0, with arbitrarily fine structure, depending onthe initial conditions. Consequently, there would be lot of “interfaces” i.e. regions at the contact of these

52

two types of regions, with a very steep (in the limit infinite) gradient of concentration. This situations isprevented by the presence of the second term - phyllic term - which penalizes the occurrence of such regionswith high gradients and during the evolution is thus responsible for the so-called coarsening - i.e. tendencyof the system to evolve in such a manner that the amount of interfaces between the phases is minimal. Howsuch an evolution can look like can be seen, for example on the snapshots in Fig.6.

Let us first present the “historical” form of the various equations related to the problems of phase sepa-ration and phase change.

Figure 6: Example of numerical simulation of a model of Cahn-Hilliard type. The evolution starts from arandom distribution (top left corner) and continues to the right by separation into two almost pure phasesand gradual coarsening and coalescence (second line from left to right).

• Cahn and Hilliard (1958): “Free energy of a nonuniform system. I. Interfacial free energy,” J. Chem.Phys. 28, 258 (1958).

∂c∂t

− ∆( f ′(c)−σ∆c)︸ ︷︷ ︸4th order operator!!!

= 0 , (6.69a)

can be extended to take into account mechanical coupling with flow (Cahn-Hilliard-Navier-Stoke model),e.g. Lowengrub and Truskinowsky (1998):

∂(ρv)∂t

+div(ρv⊗v)−div(2νD(v))+∇(p(ρ)−λdivv)+div( σ∇c⊗∇c︸ ︷︷ ︸Korteweg-like stress

)= 0 , (6.70a)

∂(ρc)∂t

+div(ρcv)−∆( f ′(c)−σ∆c)= 0 . (6.70b)

• Allen and Cahn (1979): “A microscopic theory for multiphase boundary motion and its application tomultiphase domain coarsening”, Acta Mechanica, 27, 1085-1095.

- “order parameter” c leads to∂c∂t

=−α f ′(c)+ασ∆c , (6.71)

(first term - responsible for conversion of one phase to another, second term - diffusion term, penaliza-tion of the interface). Original Allen-Cahn (Fourier) system (for unknown c and ϑ):

∂c∂t

−β∆c+α f ′(c)= 0 , (6.72a)

∂

∂t(γϑ+λc

)−div(κ∇ϑ)= 0 . (6.72b)

53

• Stefan (1891) : Über die Theorie der Eisbildung, insbesondere über die Eisbildung im Polarmeere,Annalen der Physik, 278, 2, 269–286- describing eg. advance interface between phases, i.e. surface of ice during melting occurring at thesurface.

∂tγ(ϑ)−div(κ(ϑ)∇ϑ))= 0 . (6.73)

Natural questions:

1. What is the “order parameter”

2. Balance equations of single continuum - where does the equation for c come from?

3. Are these models O.K. from thermodynamic point of view?

4. How to include mechanical effects?

5. How to include thermal effects?

6. What are natural BC’s for such class of models?

7. Is ασ∆c in A-C model indeed connected with diffusive flux?

We will be able to provide answers to these questions and derive such class of models in a thermodynami-cally consistent manner in a quite simple setting of binary mixtures in the Class I framework. Our approachwill also in natural way allow to include thermal phenomena.

Again, the system of balance equations (neglecting for simplicity the energy sources, i.e. r ≡ 0) reads:

ρ c+divj= m , (6.74a)

ρ+ρdivv= 0 , (6.74b)

ρv= divT+ρb , (6.74c)

T=TT , (6.74d)

ρ e =T :D−divq , (6.74e)

where ρ = ρ1+ρ2 and c = ρ1/ρ. Let us now assume the fundamental relation for the Helmholtz free energy inthe following form

ψ= ψ(ϑ,ρ, c,∇c),

and let us compute the material time derivative

ψ= ∂ψ

∂ϑϑ+ ∂ψ

∂ρρ+ ∂ψ

∂cc+ ∂ψ

∂∇c· ∇c . (6.75)

Using the already defined quantities: thermodynamic pressure p = ρ2 ∂ψ∂ρ

, chemical potential µ = ∂ψ∂c and

relation −η= ∂ψ∂ϑ

, and employing also the thermodynamic relation ψ= e−ϑη, we get:

ρϑη= ρ e− pρ

ρ−µρ c−µc · ∇c , (6.76)

where we defined a new quantity

µcdef= ρ

∂ψ

∂∇c. (6.77)

In the following, we shall consider the following more explicit form of ψ:

ψ=ψ0(ϑ,ρ)+ ψ1(ϑ, c)ρ

+ s(ϑ)2ρ

|∇c|2 . (6.78)

54

Then it holds

p = ρ2 ∂ψ0

∂ρ−ψ1 − s(ϑ)

2|∇c|2 , (6.79a)

µ= 1ρ

∂ψ1

∂c, (6.79b)

µc = s(ϑ)∇c . (6.79c)

A particular example of function ψ is as follows:

ψ=ψ0(ϑ,ρ)+ 12σ(ϑ)c2(1−c)2

ερ+ 3

4ρσ(ϑ)ε|∇c|2 , (6.80)

i.e. we have ψ1 = 12σc2(1−c)2

ε, s = 3

2σε. With such choice of parameters, σ represents the surface tension be-tween the two materials (interpreted as corresponding surface free energy density of the interaction betweenthe two materials) and ε is a parameter characterizing the (equilibrium) thickness of the transition zone.

Remark 15. Motivation for this particular choice comes from a formula used in the literature for the freeenergy functional (e.g. Boyer and Lapuerta, 2006)

Fσ,ε =∫Ω

12σ

εc2(1− c)2︸ ︷︷ ︸ρψ1

+34σε|∇c|2dx .

The first term ρψ1 = 12σε

c2(1−c)2 has a double-well structure with two local minima at c = 0 and c = 1. In theprocess of minimization of Fσ,ε, where the minimum corresponds to an equilibrium of the system, this termis thus responsible for separation of the material into the two “phases”. The second terms is non-negativepenalty term measuring the “volume of interfaces”, in the minimization process this term is responsible forthe so-called coarsening, i.e. growth of spatial scale of the two-phase structures. This particular formula hasseveral beautiful properties, namely it allows to give the parameter σ a precise physical meaning - identifyit with the surface density of energy stored in the interfacial zone = surface tension.

To see this, let us consider only one-dimensional case in the whole R, and let us minimize the free energy

min∫ ∞

−∞12σ

εc2(1− c)2 + 3

4σε|c′|2dx , (6.81)

with the additional constraint limx→−∞ c(x)= 0, limx→+∞ c(x)= 1 (i.e. “one phase on the left, the other on theright”). The corresponding Euler-Lagrange equation for the minimizer c0(x) reads (try as an exercise):

−32σεc′′(x)+24

σ

εc0(1− c0)(1−2c0) = 0 ,

limx→−∞ c0(x) = 0 ,

limx→∞ c0(x) = 1 .

An analytical solution of this equation can be found and it reads

c0(x)= 12

(1+ tanh

(2xε

)), (6.82)

from where we can see that the parameter ε has the interpretation of a intrinsic length scale for the thicknessof the interface. Finally, computing the total energy corresponding to the found solution, we get∫ ∞

−∞12σ

εc2

0(1− c0)2 + 34σε|c′0|2dx =σ . (6.83)

Let us assume that in a 3-dimensional case we can zoom in the vicinity of the interface obtaining a locallyplanar geometry. Taking now the 1-d solution as an approximation of the equilibrium concentration with xnow representing the distance along the normal to the interface, the obtained result indicates that σ wouldbe energy of the interface integrated over the whole interface, which is nothing but the surface energy densityassigned to the interface. This quantity is in physics denoted as the surface tension.

55

Exercise 14. Verify that c0 given by (6.82) is indeed the sought minimizer of (6.81) and that (6.83) holds.

Remark 16. The polynomial form of the phobic part of the free energy used above is in fact not very physical.A more physical relation is obtained by considering an ideal solution theory, in which case the correspondingphobic part of the potential reads

ψ1 =ω1c(1− c)+ω2kϑ(c ln c+ (1−c) ln(1−c)) , (6.84)

where ω1, ω2 are some parameters, k is the Boltztmann constant. The polynomial approximation is howeveroften sufficient, especially if one is not interested in accurate description of the concentration distributionin the interfacial region and merely wants to have a model that ensures phase separation and apropriatellycaptures the average interfacial energy density (surface) tension.

Exercise 15. From the balance (6.74a), obtain evolution equation for ∇c:

∇c =−(∇v)T∇c+∇(

m−divjρ

). (6.85)

Proceeding further, and plugging into (6.76) balances of mass (6.74a), (6.74b), and relation (6.85), we obtain

ρϑη=Td :Dd + (m+ p)divv−divq−µ(m−divj)+ (µc ⊗∇c) : (∇v)T −µc ·∇(

m−divjρ

), (6.86)

Exercise 16. The dependence of the Helmholtz free energy ψ(ϑ,ρ, c,∇c) on ∇c must, by the principle of mate-rial frame indifference, reduce to dependence only on the modulus of ∇c, i.e. ψ= ψ(ϑ,ρ, c, |∇c|). Show that inthat case the dyadic product µc ⊗∇c is a symmetric tensor.

We employ the identity

µc ·∇(

m−divjρ

)= div

(m−divj

ρµc

)− (m−divj)

ρdivµc (6.87)

We substitute from (6.87) into (6.86) and introduce the Korteweg tensor K

K=µc ⊗∇c , (6.88)

and obtain

ρϑη= (Td+Kd):Dd +(m+p+ trK

3

)divv−div

(q+m−divj

ρµc

)−

(µ−divµc

ρ

)m

+(µ− divµc

ρ

)divj︸ ︷︷ ︸

=div((µ− divµc

ρ

)j)−j·∇

(µ− divµc

ρ

), (6.89)

Denoting

µcdef=

(µ−divµc

ρ

), (6.90)

dividing (6.89) by ϑ and rearranging, we obtain the balance of entropy in the form

ρη+div

q+m−divjρ

µc−µcj

ϑ

= 1ϑ

[(Td+Kd) :Dd +

(m+p+ trK

3

)divv−µcm−

(q+m−divj

ρµc−µcj

)· ∇ϑϑ

− j ·∇µc

]With the terms involving j we now have the same choice as in the Fickian case. Using the same identity asbefore:

−j ·∇µc +µcj · ∇ϑϑ

=−j ·(δ∇µc + (1−δ)ϑ∇

(µc

ϑ

))+δµcj · ∇ϑ

ϑ, (6.91)

56

for some δ ∈ ⟨0,1⟩, we arrive at the final form of the entropy balance:

ρη+div

q+m−divjρ

µc−µcj

ϑ

= 1ϑ

[(Td+Kd) :Dd+

(m+p+ trK

3

)divv−µcm−

(q+m−divj

ρµc−δµcj

)· ∇ϑϑ

−j ·(δ∇µc+(1−δ)ϑ∇

(µc

ϑ

))],

(6.92)

Now we will consider two particular situations

Allen-Cahn-NSF model

Consider first the case j= 0, m 6= 0. Consider piece-wise linear closure relations, we obtain from (6.92):

Td+Kd = 2νDd , (6.93a)

m+p+ trK3

= 2ν+3λ3

divv , (6.93b)

m =−βµc , (6.93c)

q+ mρµc =−κ∇ϑ

ϑ, (6.93d)

for ν ≥ 0, 3λ+2ν≥0,β≥0, κ≥0, the second law of thermodynamics is automatically satisfied, i.e. the r.h.s. of(6.92) representing the entropy production is non-negative. These closures can be rewritten as

T=−pI+λdivvI+2νD−K , (6.94a)

m =−β(µ− divµc

ρ

), (6.94b)

q=−κ∇ϑϑ

− mρµc , (6.94c)

or, for our particular choice of the Helmholtz free energy (6.78), using (6.79), we obtain

T=−(ρ2 ∂ψ0

∂ρ−ψ1− s(ϑ)

2|∇c|2

)I+λdivvI+2νD− s(ϑ)∇c⊗∇c , (6.95a)

m =−βρ

(∂ψ1

∂c−div(s(ϑ)∇c)

), (6.95b)

q=−κ∇ϑϑ

+ βs(ϑ)ρ2

(∂ψ1

∂c−div(s(ϑ)∇c)

)∇c . (6.95c)

Reduction: Allen-Cahn - Fourier model

In order to compare with the original Allen-Cahn system, let us consider a special situation v=0. Conse-quently from (6.74b), ρ = ρ0, which we set to be constant for simplicity. Let us also ignore the mechanicalcoupling, i.e. the balance of momentum, and let us write down the balance equation for concentration (6.74a)and for internal energy (6.74e). Plugging in the constitutive relations (6.95), we obtain

ρ∂c∂t

− β

ρdiv(s∇c)+ β

ρ

∂ψ1

∂c= 0 , (6.96a)

ρ∂e∂t

−div(κ∇ϑϑ

− βsρ2

(∂ψ1

∂c−div(s∇c)

)∇c

)= 0 . (6.96b)

It remains to express e in terms of ψ, but that is easy. We know from the definition of the Helmholtz potentialthat

e(ϑ,ρ, c,∇c)=ψ+ϑη = ψ(ϑ,ρ, c,∇c)−ϑ∂ψ(ϑ,ρ, c,∇c)∂ϑ

,

57

where we used (3.64). For our choice of Helmholtz free energy (6.78), this yields

e(ϑ,ρ, c,∇c)=(ψ0−ϑ∂ψ0

∂ϑ

)+ 1ρ

(ψ1−ϑ∂ψ1

∂ϑ

)+ 1

2ρ

(s−ϑ ∂s

∂ϑ

)|∇c|2 , (6.97)

so unless ψ1 and s are linear functions of temperature, dependence of c and ∇c persists also in e. Differenti-ating e(ϑ,ρ, c,∇c) in (6.96b), we obtain (using ρ = const and v=0):

ρ

(∂e∂ϑ

∂ϑ

∂t+ ∂e∂c

∂c∂t

+ ∂e∂∇c

· ∂∇c∂t

)−div

κ∇ϑϑ

−βsρ2

(∂ψ1

∂c−div(s∇c)

)︸ ︷︷ ︸

s ∂c∂t

∇c

= 0 , (6.98)

which can be with the aid of (6.96a) written in the following form

ρ

(∂e∂ϑ

∂ϑ

∂t+

[∂e∂c

−div(s∇c)ρ

+(∂e∂∇c

− s∇cρ

)·∇

]∂c∂t

)−div

(κ∇ϑϑ

)= 0 . (6.99)

If we denote γ = ρ ∂e∂ϑ

and the differential operator (!) λ = ρ[∂e∂c−div(s∇c)

ρ+

(∂e∂∇c− s∇c

ρ

)·∇

], β = β

ρ2 , κ = κϑρ

, as-suming further that s is constant (!) and denoting α = βs, we can write the derived system in the form

∂c∂t

− α∆c+ β∂ψ1

∂c= 0 , (6.100a)

∂

∂t(γϑ+ λc

)−div(κ∇ϑ)= 0 , (6.100b)

which is the original Allen-Cahn (Fourier) system, cf. (6.72). We can see that the main difference is, that ourλ is not a parameter or function, but it is a differential operator, because of the term ∂e

∂∇c− s∇cρ

. Expressingthis term explicitly from (6.97), we obtain

∂e∂∇c

− s∇cρ

=−1ρϑ∂s∂ϑ

∇c = 0 , (6.101)

under our assumption that s is constant, λ thus reduces to a function (still depending on ∆c, however).

Cahn-Hilliard - NSF model

Assuming now in contrast to the Allen-Cahn model situation, j 6= 0, but setting m=0, we obtain from (6.92)assuming piece-wise linear closure relations

(T+K)d = 2νDd , (6.102a)

m+ trK3

+ p = 2ν+3λ3

divv , (6.102b)

q− divjρ

µc−δ(µ−divµc

ρ

)j=−κ∇ϑ

ϑ, (6.102c)

j=−β(δ∇µc+(1−δ)ϑ∇

(µc

ϑ

)). (6.102d)

which choice, under the assumption ν≥ 0, 2ν+3λ≥ 0, κ≥ 0, β≥ 0 implies that the second law is automaticallysatisfied (the entropy production is non-negative). The closure relations can be rewritten as follows

T=−pI+λdivvI+2νD−K , (6.103a)

q=−κ∇ϑϑ

+ divjρ

µc +δ(µ− divµc

ρ

)j , (6.103b)

j=−β[∇

(µ−divµc

ρ

)− (1−δ)

(µ−divµc

ρ

) ∇ϑϑ

]. (6.103c)

Plugging these equations into the balance laws (6.74), we obtain the explicit form of the Cahn-Hiliard-NSFmodel.

58

Reduction: Cahn-Hilliard - NS model

Such system is already quite complex and thus, for the sake of simplicity, let us consider simpler isothermalsetting and let us investigate the balances of mass and momenta. Considering β constant, we obtain

∂(ρv)∂t

+div(ρv⊗v)−div2νD+∇(p−λdivv)+div(s∇c⊗∇c)= 0 , (6.104a)

ρ+ρdivv= 0 , (6.104b)∂(ρc)∂t

+div(ρcv)−β∆(µ−div(s∇c)

ρ

)= 0 . (6.104c)

Reduction: Cahn-Hilliard model

In 1954 the original system of Cahn-Hilliard equations was derived with the additional assumption v = 0(Note that in that case assuming ρ(·, t = 0) = ρ0 constant implies that ρ(·, t) = ρ0, ∀t ≥ 0). In this case,inspecting the equation for concentration, we obtain

∂c∂t

− β∆(∂ψ1

∂c− s∆c

)= 0 , (6.105)

with β= β

ρ2 , which is the original form of the Cahn-Hilliard equation.Remark 17. Lowengrub and Truskinowsky (1998) obtained a model with an incompressiblity assumptiondivv = 0 and so-called quasi-incompressibilty (which has been discussed in Sec. 6.2), assuming dependenceρ = ρ(c).Remark 18. Instead of the initial assumption on the fundamental thermodynamic relation of the form ψ =ψ(ϑ,ρ, c,∇c) we could have started from e.g. ψ= ψ(ϑ,ρ,ρ1,∇ρ1), for details see Heida et al. (2012).

6.4 Allen-Cahn and Cahn-Hilliard models as gradient flows

Both the Allen-Cahn and Cahn-Hilliard systems can be derived as so-called gradient flows (e.g. Cowan, 2004).The idea behind is to postulate evolution of some field quantity (concentration c in our example), providedwe are given corresponding energy functional F on some Hilbert space H as the following abstract ODE onthis Hilbert space:

∂c∂t

=−kgradX F , k ≥ 0 , (6.106)

where the symbol gradX F denotes X -constrained gradient, is understood in the sense

(gradX F ,v)H =⟨δF

δc,v

⟩H ∗,H

= dds

∣∣∣∣s=0

F (c+ sv) , ∀v ∈ X ⊂H , (6.107)

i.e. as a Riesz representant of the constrained differential (we assume that F is Fréchet differentiable,the derivative thus being a member of H ∗). An important feature of such setting is that such gradient flowproduce evolution of c in such a manner that the associated energy F (c) decreases in time. Indeed, assumingsufficient regularity of all objects to justify the calculation, and ignoring for a moment the constraint spaceX , we get formally

ddt

F (c(t))=⟨δF

δc,∂c∂t

⟩H ∗,H

=(gradF ,

∂c∂t

)H

=−k‖gradF (c(t))‖2H ≤ 0 . (6.108)

Such evolution thus tends to evolve the system to a local (global) energetic minimum.Let us consider the energy functional in the form corresponding to the concentration-dependent part of

energy (6.78):

Fdef=

∫Ωψ1 + s

2|∇c|2 dx , (6.109)

then we obtain (formally, provided all the formulae make sense):⟨δF

δc,v

⟩H ∗,H

=∫Ω

∂ψ1

∂cv+ s∇c ·∇v dx (6.110)

The two cases - Allen-Cahn and Cahn-Hilliard models will now differ only by the choice of the correspondingHilbert space. Since we are interested in the original A-C and C-H setting, we consider s constant.

59

• Allen-Cahn: H = L2(Ω), X =C ∞c (Ω)

Seeking for a representative in L2(Ω) is easy, just using the Green theorem, we get⟨δF

δc,v

⟩L2(Ω),L2(Ω)

=(∂ψ1

∂c− s∆c,v

)L2(Ω)

, (6.111)

i.e. the sought gradient is

gradXL2F (c)= ∂ψ1

∂c− s∆c . (6.112)

Taking k = β, the corresponding gradient flow is thus

∂c∂t

+ β∂ψ1

∂c− βs∆c = 0 , (6.113)

which is the original Allen-Cahn equation for concentration, cf (6.100a).

• Cahn-Hilliard: H = H−1(Ω)= (H10)∗, X =C ∞

c (Ω)The inner product in H−1 is defined as follows

(u,v)H−1def= (∇u∗,∇v∗)L2(Ω) , (6.114)

where the starred functions are associates of the unstarred in the sense of the unique solution of thefollowing problem:

∆u∗ = u in Ω ,

u∗ = 0 at ∂Ω ,

i.e. formally, u∗ =∆−1u . So the sought gradXH−1F should satisfy

(gradXH−1F ,v)H−1 =

(∇(gradX

H−1F )∗,∇v∗)

L2=

⟨δF

δc,v

⟩H ∗,H

(6.110)=∫Ω

∂ψ1

∂cv+ s∇c ·∇v dx

=∫Ω

∂ψ1

∂c∆v∗+ s∇c ·∇∆v∗ dx =

∫Ω−∇∂ψ1

∂c·∇v∗+ s∇∆c∗ ·∇∆v∗ dx

=∫Ω∇

(−∆

(∂ψ1

∂c

)∗)+ (s∆2c∗)

·∇v∗ dx =

(−∆∂ψ1

∂c+ s∆2c,v

)H−1

,

and thus the corresponding gradient flow reads, taking again k = β:

∂c∂t

− β∆(∂ψ1

∂c− s∆c

)= 0 , (6.115)

which is the original Cahn-Hilliard equation for concentration, cf. (6.105).

6.5 Chemical reactions

Motivation We remain within the Class I with the balance equations, i.e. we consider the following systemof balance equations:

∂ρ

∂t+div(ρv)= 0 , (6.116a)

ρ cα+divjα = mα , α= 1, . . . N , (6.116b)N∑α

cα = 1 ,N∑α=1

jα = 0 ,N∑α=1

mα = 0 , (6.116c)

∂(ρv)∂t

+div(ρv⊗v)= divT+ρb , (6.116d)

T=TT , (6.116e)∂(ρE)∂t

+div(ρEv)= div(Tv−q)+ρb ·v , (6.116f)

60

and in this whole chapter we will solely focus on the reaction term mα. In the previous text we have alreadyprovided closure relations for mα, but we merely restricted ourselves to linear closures of the type mα =−βαµα. Let us briefly recall what we were doing (for the simplest “Fickian” setting). We considered anN-component mixture with specific internal or free Helmholtz/Gibbs’ energy in the form

e = e(η,ρ, cα) , ψ= ψ(ϑ,ρ, cα) , g = g(ϑ, p, cα) ,

and by the standard procedure (taking time derivative and plugging the balance laws), we obtained theentropy balance in the form

ρη + div

(q−∑N

α=1 jαµαϑ

)= 1

ϑ

[Td :Dd + (m+ p)divv−

(q−∑

α

jαµα)· ∇ϑϑ

−∑α

jα ·∇µα−∑α

mαµα

]. (6.117)

We will focus on the last term in the dissipation, which we will call chemical dissipation and denote ξCH.Ignoring for the moment again any possible cross-effects and assuming thus independence of the chemicalreactions on the remaining dissipative processes, we will be interested in ensuring the non-negativity ofchemical contribution to entropy production, i.e. in ensuring validity of

0≤ ξCH =−∑α

mαµα , (6.118)

where we recall the definition of the chemical potential

µα = ∂e∂cα

∣∣∣∣η,ρ,cβ6=α

= ∂ψ

∂cα

∣∣∣∣ϑ,ρ,cβ6=α

= ∂ g∂cα

∣∣∣∣ϑ,p,cβ6=α

, (6.119)

depending on the set of independent thermodynamic variables we want to work with. The linear relations,which one might be tempted to impose, i.e. mα =−βαµα, (no summation), s.t. β≥0 are quite inapplicable inpractice. In chemistry in particular, one needs to step out of the realm of linear force-flux relations to capturethe kinetics realistically.

Another deficiency of such type of constitutive closure is that in the process of its derivation, we haven’t atany stage employed the information about the atomic/molecular structure of the matter, in particular the factthat in chemical reactions, only bonds between the atoms and molecules become disrupted or created, but theatoms themselves do not transform (unlike in nuclear reactions). This additional assumption of conservationof atoms or some of their combinations, will impose additional restrictions on the sought constitutive theory,stronger than the assumption of the conservation of mass in the reactions

∑Nα=1 mα = 0. Our aim will be to

enforce this statement/observation formally into our theory and we will see the important consequences itwill have on the closure relations and also on the notion of chemical equilibrium.

We will focus on the following topics

• Stoichiometry (= mathematical description of chemical reactions, using the apparatus of linear alge-bra).

• Mixture of ideal gases as the simplest possible mixture model will serve us to identify explicitly theform of the chemical potential and to infer its general structure in more general cases.

• Chemical equilibrium will be studied as a special process in which there is no dissipation due toreactions, and we will derive the equilibrium mass action law

• Outside the equilibrium we will study chemical kinetics, i.e. look for constitutive relations for thereaction rates.

6.5.1 Stoichiometry

Stoichiometry is a discipline providing the mathematical description of chemical reactions. Let us assumethat we are dealing with a N-component mixture with components indexed by α = 1, . . . , N. Let us assumethat all the components are composed of a set of Z different elements, for which we shall use index σ= 1, . . . , Z.

61

Definition 2 (Molar mass). Let us remind the basic unit of atomic substances 1 mole .= 6,022×1023 of unitsand let us define the molar mass of α component of the mixture as

Mα =Z∑

σ=1TσαA σ , α= 1, . . . , N . (6.120)

where A σ denotes the atomic (molar) mass of the “element” σ and T is a matrix the columns of whichcaptures the composition of each component in terms of the present elements.

Definition 3 (Molar production of α component). Let us define the molar production Jα as follows

mα =MαJα (no summation) , α= 1, . . . , N . (6.121)

Clearly Jα expresses the amount of number of moles of α component that are produced (or consumed) duringall ongoing chemical reactions per unit time and per unit volume of the mixture.

Theorem 6.1 (Conservation of number of atoms in chemical reactions). In chemical reactions, the atomicstructure of the mass by definition does not change, we can express this with the use of the above definitions asfollows ∑

α

TσαJα = 0 , σ= 1, . . . , Z . (6.122)

Naturally, conservation of number of atoms is much stronger assumption than conservation of total massduring the reactions, not surprisingly we can prove

Lemma 6.2. Conservation of number of atoms implies conservation of mass in chemical reactions.

Proof.N∑α=1

mα =N∑α=1

MαJα =N∑α=1

Z∑σ=1

TσαA σJα =Z∑

σ=1A σ

N∑α=1

TσαJα︸ ︷︷ ︸=0

= 0 ,

which is the desired statement - sum of mass productions over all component of the mixture is zero.

Remark 19. Note by inspecting (6.122) that if the matrix Tσα has rank equal to N, then it must hold Jα =0,∀α, which means that no chemical reactions take place. For chemically reacting system, we must thereforehave rank(Tσα)<N .

Let us assume, in view of the previous remark that H def= rank(T)<N. By excluding linearly dependentrows, we can construct from matrix Tσα ∈RZ×N sub-matrix Sσα ∈RH×N and it holds

N∑α=1

TσαJα = 0 , σ= 1, . . . , Z ⇐⇒N∑α=1

SσαJα = 0 , σ= 1, . . . ,H . (6.123)

and for the molar masses

Mα =Z∑

σ=1TσαA σ =

H∑σ=1

SσαE σ, (6.124)

where E σ is so-called atomic substance (certain linear combination of atoms A σ). In view of this, a minimalversion of the conservation law for atoms concerns only the independent atomic substances.

Let us have N components of the mixture, and let us introduce an N-dimensional linear vector space U

with an orthonormal basis eαNα=1, eαN

α=1,eα = eα, ∀α. Let us define the vector of molar masses M

M=N∑α=1

Mαeα , (6.125)

and the vector of molar reaction speed J:

J=N∑α=1

Jαeα . (6.126)

62

Let us further define H linearly independent vectors

fσdef=

N∑α=1

Sσαeα , σ= 1, . . . ,H , (6.127)

where H is the number of linearly independent rows of matrix S and let us define subspace Wdef= spanfσH

σ=1.Its orthogonal complement, denoted V represents the so-called reaction subspace of U :

U =W ⊕V , V ⊥W . (6.128)

It holds M ∈W and J ∈ V . Indeed

M=N∑α=1

Mαeα =N∑α=1

H∑σ=1

SσαE σeα =H∑σ=1

E σN∑α=1

Sσαeα︸ ︷︷ ︸=fσ

∈W .

Furthermore, as a direct consequence of the law of conservation of atomic substances, we get

J · fσ =(

N∑α=1

Jαeα

)·(

N∑α=1

Sσαeα)=

N∑α=1

JαSσα = 0 , σ= 1, . . .H , (6.129)

i.e. J ∈W ⊥ = V .

Remark 20. The statement just proved implies immediately conservation of mass in chemical reactions.Indeed, we have

0=J ·M=(

N∑α=1

Jαeα

)·(

N∑α=1

Mαeα)=

N∑α=1

MαJα =N∑α=1

mα = 0 . (6.130)

Let now gpN−HP=1 be the basis of the reaction space V

gp =N∑α=1

P pαeα =⇒ P pα = gp ·eα . (6.131)

and let gqN−Hq=1 be the associated dual base in the sense:

gp ·gq = δpq ∀p, q = 1, . . . , N−H . (6.132)

with P pα the corresponding expansion coefficients in the eαNα=1 basis. The orthogonality relation between

V and W then can be rewritten as:

0= fσ ·gq = (N∑α=1

Sσαeα) · (N∑β=1

Pqβeβ)=N∑α=1

SσαPqα ∀q = 1, . . . , N−H,∀σ= 1, . . . ,H . (6.133)

Since M ∈W , it must hold

0=M ·gq =(

N∑α=1

Mαeα)·(

N∑α=1

Pqαeα

)=

N∑α=1

MαPqα , ∀q = 1, . . . , N−H . (6.134)

The above relation, when the molar masses are replaced with the symbol of the reacting molecules (or atoms)represents transcription of the N−H independent chemical reactions, with the i-th rows of the matrix Pcorresponding to the stoichiometric coefficients of i-th (independent) chemical reaction.

Remark 21. It is customary to interpret the molecules with positive stoichiometric entry in P as reactionproducts and write them on the right-hand side of a chemical reaction, and molecules with negative corre-sponding entry in P as reactants and write them on the left-hand side of the chemical reaction.

63

Example: Let us consider a binary mixture of molecules of NO2 a N2O4:

N = 2 :α= 1. . .NO2 ,

α= 2. . .N2O4 ,

Z = 2 :σ= 1. . .N ,

σ= 2. . .O .

Then our newly defined quantities read as follows:

T =(1 22 4

), S = (1 2).

The orthogonal subspace is generated e.g. by P = (2,−1), from where we can directly read the stoichiometriccoefficients of the corresponding reaction:

2NO2 −N2O4 = 0 ⇐⇒N2O4 → 2NO2 .

Example: Let us consider a ternary mixture of atomic oxygen O, molecular oxygen O2 and ozone moleculesO3:

N = 3 :α= 1. . .O ,

α= 2. . .O2 ,

α= 3. . .O3 ,

Z = 1 :σ= 1. . .O .

Then our newly defined quantities read as follows:

T = (1 2 3

), S = (1 2 3) ,

i.e. H = 1, N−H = 2. To find the orthogonal complement in general can be achieved by the following simpletrick (in chemistry sometimes called as Hooyman method):

Remark 22. Let the rank of S be H. Let us look for P in the following special form

P =

1 0 . . . 0 P1 N−H+1 . . . P1N

0 1 . . . 0 P2 N−H+1 . . . P2N

0 0. . . 0

... . . ....

0 0 . . . 1 PN−R N−H+1 . . . PN−HN

i.e. unit matrix of dimension N−H complemented with H× (N−H) unknowns, and we are solving the follow-ing orthogonality relation:

N∑α=1

SσαPqα = 0 σ= 1, . . . ,H, q = 1, . . . , N−H ,

which represent H× (N−H) independent equations.

In our particular example, we look for P in the form

P =(1 0 P13

0 1 P23

)and the orthogonality relations are

S11P11 +S12P12 +S13P13 = 0S11P21 +S12P22 +S13P23 = 0

=⇒ 1+3P13 = 02+3P23 = 0

=⇒ P =(1 0 −1

30 1 −2

3

),

64

or, equivalently

P =(3 0 −10 3 −2

),

which corresponds to the following two independent chemical reactions:

O3 → 3O ,

2O3 → 3O2 .

These are the independent reactions of the system in the sense that any other reaction can be written as alinear combination of these ones. (This also includes inverse reactions, naturally). The choice of independentreactions is by no means unique, of course.

Exercise 17. Consider a 6-component mixture with reactants CH4, O2, CO2, H2O, CO and H2. Identify thestoichiometric matrices Tσα, and Sσα and using the ansatz for the reaction matrix as in remark (22) above,identify the set of independent chemical reactions in this system.

Since J ∈ V , we have

J=N−H∑p=1

Jpgp =N∑α=1

N−H∑p=1

JpP pαeα , (6.135)

i.e.

Jα =N−H∑q=1

JqPqα . (6.136)

The above expression determines the molar reaction rate for α component in terms of independent chem-ical reactions. Plugging this relation back into (6.121), yields a standard formula for the reaction rates inchemistry literature:

mα =N−H∑q=1

MαPqαJq (6.137)

Remark 23. Let us reinspect the above formula from a different perspective. Let us assume that we are apriori given the information about the chemical reactions that take place in the system. Let the system ofN −H independent chemical reactions be

ζ1,1f X1 +ζ1,2

f X2 +·· ·+ζ1,Nf XN ζ

1,1b X1 +ζ1,2

b X2 +·· ·+ζ1,Nb XN ,

ζ2,1f X1 +ζ2,2

f X2 +·· ·+ζ2,Nf XN ζ

2,1b X1 +ζ2,2

b X2 +·· ·+ζ2,Nb XN ,

...

ζN−H,1f X1 +ζN−H,2

f X2 +·· ·+ζN−H,Nf XN ζ

N−H,1b X1 +ζN−H,2

b X2 +·· ·+ζN−H,Nb XN ,

where X1, . . . , XN are the components and ζq,βf , ζq,β

b are the forward and backward stoichiometric coeffi-

cients of q-th reaction. Denoting Pqβ = ζqβb −ζqβ

f , and considering a fixed reacting (unit) volume, with theN (X1), . . .N (XN ), the numbers of moles of the individual species, we can see that in each (q-th) reaction itmust hold

1Pq1

dN (X1)dt

= 1Pq2

dN (X2)dt

= ·· · = 1Pq N

dN (XN )dt

def= Jq , q = 1, . . . , N −H . (6.138)

Consequently, the amount of species Xα produced or consumed per unit of time and unit of volume in q-threaction is MαPqαJq, with Jq denoting the so called rate of advancement of the q-th reaction. Summingfor each component these contributions over all N −H independent chemical reactions, we obtain formula(6.137), and we can indeed see that the interpretation of Pqα as stoichiometric coefficients of independentchemical reactions occurring advancing with the rates Jq is meaningful.

65

Let us return to the chemical dissipation rate and let’s employ the stoichiometry. We have

ξCH =−∑α

mαµα =−∑α

JαMαµα︸ ︷︷ ︸µMα

=−∑α

JαµMα =−J ·µM (6.139)

where we introduced the molar chemical potential and the associated vector

µMα

def= Mαµα , µM def=N∑α=1

µMαeα . (6.140)

We denote the (unique) projections of µM into the reaction space V and its orthogonal complement W as Aand B:

µM = A︸︷︷︸∈V

+ B︸︷︷︸∈W

, (6.141)

and we call A vector of chemical affinity. We find the components of A explicitly from the definition, i.e. asthe projection:

A=µM ·N−H∑q=1

gqgq︸ ︷︷ ︸PV

=N∑α=1

µMαeα ·

N−H∑q=1

gqgq =N−H∑q=1

(N∑α=1

µMαPqα)gq =

N−H∑q=1

Aqgq , (6.142)

hence for the contravariant components of A (w.r.t to the basis gq) reads

Aq =N∑α=1

µMαPqα , q = 1, . . . , N −H. (6.143)

Since we already know that J ∈ V and W ⊥ V , it holds

ξCH =−J ·µM =−J ·A=−N−H∑q=1

JqAq . (6.144)

This is one of the crucial outcomes of the stoichiometry, i.e. of considering the atomic composition of matter- it is not the whole chemical potential, but only its projection to the reaction subspace - the chemical affinity,that drives the chemical reactions and determines the chemical reaction rates.

So in order to comply with the second law of thermodynamics, we will require J·A≤ 0 and we will look forconstitutive relations for J. In order to do so, we shall first inspect how affinity and molar chemical potentiallook like for a special, though in some sense very representative case of the so-called ideal gas mixture.

6.5.2 Mixture of ideal gasses

In a mixture of ideal gasses with common temperature, each component behaves as an ideal gas, in particularwe have the following state equations

• Partial pressures:

pα = pα(ϑ,ρα)= RϑMα

ρα =⇒ ρα = ρα(ϑ, pα)= Mα

Rϑpα , α= 1, . . . , N , (6.145)

where R is the universal gas constant (R = 8.3144598 J K−1 mol−1).

• Specific internal energy

eα = eα(ϑ)= zαRϑMα

+βα , α= 1, . . . , N , (6.146)

where βα are constants and zα is the “equi-partitioning” term (e.g. 32 for a monoatomic gas).

66

• Entropy

ηα = ηα(ϑ,ρα)= zαR

Mαlnϑ− R

Mαlnρα+γα , α= 1, . . . , N , (6.147)

where γα are constants.

Note that we don’t have any of the thermodynamic potentials in its natural variables. But we can constructHelmholtz free energy of the mixture as a whole ψ(ϑ,ρα) = ψ(ρ,ϑ, cα) from these. Assuming ρe =∑N

α=1ραeα(i.e. ignoring the diffusive kinetic energy terms in (5.82) and setting the mixture entropy in accord with (5.93)as ρη=∑N

α=1ραηα, we can compute

ρψ=∑α

ραψα =∑α

ρα(eα−ϑηα)=∑α

ρα

(zα

RϑMα

+βα)−ραϑ

(zα

RMα

lnϑ− RMα

lnρα+γα)

. (6.148)

Now we can evaluate the chemical potential µα:

µα = ∂ρψ(ϑ,ρ1, . . . ,ρN )∂ρα

∣∣∣∣ϑ,ρβ6=α

=(zα

RϑMα

+βα)

︸ ︷︷ ︸eα(ϑ)

−ϑ(zα

RϑMα

lnϑ− RMα

lnρα+γα)

︸ ︷︷ ︸ηα(ϑ,ρα)

+ RϑMα︸︷︷︸

pα(ϑ,ρα)ρα

=eα(ϑ)−ϑηα(ϑ,ρα)+ pα(ϑ,ρα)ρα

.

Inverting the relation for pα = pα(ϑ,ρα) from (6.145), and writing ρα = ρα(ϑ, pα), and defining function gα(specific Gibbs’ free energy also called free enthalpy of (pure) α component) as follows:

gα(ϑ, pα)= eα(ϑ, pα)−ϑηα(ϑ, pα)+ pαρα(ϑ, pα)

, (6.149)

we can see that we obtained relationµα = gα(ϑ, pα) , (6.150)

with the notion e(ϑ, pα) def= e(ϑ, ρα(ϑ, pα)), and η(ϑ, pα) def= η(ϑ, ρα(ϑ, pα)).We would like to replace the partial pressures with the total pressure of the mixture, so that we have

the chemical potential expressed in terms of ϑ, p, which are typically experimentally well-accessible andwell-controllable quantities.

Exercise 18. Confirm for our Helmholtz free energy (6.148), that the Dalton’s law holds, i.e. defining thethermodynamic pressure as always according to p = ρ2 ∂ψ

∂ρ, yields

p =N∑α=1

pα . (6.151)

Let us now recall the definition of molar concentration (see (3.111))

cMα

def= ρα

Mα= lim

|V |→0+Mα(V )|V |Mα

= lim|V |→0+

nα(V )|V | , (6.152)

where nα(V ) is the number of moles of substance α in the volume V of the mixture. Hence the name of thequantity. So now we can rewrite

µα = gα(ϑ, pα)= gα(ϑ, p

pαp

)= gα

(ϑ, p

cMα∑N

β=1 cMβ

), α= 1, . . . , N , (6.153)

where we employed (6.145) in order to subtitute for pαp .

Exercise 19. Verify that

gα

(ϑ, p

cMα∑N

β=1 cMβ

)= gα(ϑ, p)+ Rϑ

Mαln

(cMα∑N

β=1 cMβ

), α= 1, . . . , N .

Hint: p in fact only appears in the expression for entropy ηα(ϑ,•) = η(ϑ, ρα(ϑ,•)) and in the fraction •ρα(ϑ,•) .

And it holds

67

1.xp

ρα(ϑ, xp)= Rϑ

Mα∀x ,

2.

ln ρα

(ϑ, p

cMα∑N

β=1 cMβ

)= ln

(Mα

Rϑp

cMα∑N

β=1 cMβ

)= ln

(Mα

Rϑp)+ ln

(cMα∑N

β=1 cMβ

).

So we summarize that we obtained

µα = µα(ϑ, p, cMβ)= gα(ϑ, p)+ Rϑ

Mαln

(cMα∑N

β=1 cMβ

), α= 1, . . . , N . (6.154)

We can rewrite it a bit more as follows. Using

N∑β=1

cMβ =

N∑β=1

ρβ

Mβ= 1

Rϑ

N∑β=1

pβ = pRϑ

, (6.155)

so that we can absorb

µα = gα(ϑ, p)− RϑMα

ln( pRϑ

)︸ ︷︷ ︸

=:µ0α(p,ϑ)

+RϑMα

ln cMα , α= 1, . . . , N , (6.156)

i.e.

µα(p,ϑ, cMα)=µ0

α(p,ϑ)+ RϑMα

ln cMα , α= 1, . . . , N , (6.157)

or in terms of the molar potential

µMα(p,ϑ, cM

α)=µMα

0(p,ϑ)+Rϑ ln cMα , α= 1, . . . , N . (6.158)

Remark 24. This expression is widely used, despite being slightly confusing. While cMα∑N

β=1 cMβ

is a dimensionless

quantity as an argument of the logarithm, cMα is not. One should always keep in mind original formula

(6.157), when working with (6.158).

The expression (6.158) is a characteristic formula for the chemical potential of so-called ideal solutions.Despite we derived it for a very special model of ideal mixture of gasses, the applicability of this formula isin fact very large and holds in many practical situations. In cases, when ideality of the solution is no-longera viable approximation, a generalization of such formula is used, introducing the so-called activity aα(cM

α):

µMα(p,ϑ,aα)=µM

α0(p,ϑ)+Rϑ lnaα(cM

α) , α= 1, . . . , N , (6.159)

In the ideal case, activity aα = cMα. Another popular notion is the so-called activity coefficient γα defined as

aα = γα(cMα)cM

α, being one in ideal solutions.

6.5.3 Chemical equilibrium

In this chapter we will use the explicit form of the chemical potential which we obtained in the previoussection in order to investigate a special class of chemical process - the chemical equilibrium. We have seenthat the chemical dissipation reads

ξCH =−J ·A , (6.160)

In equilibrium, which we will denote by symbol +, no dissipation takes place, i.e. ξCH+ = 0. This implies thateither

J+ = 0 , or A+ = 0 . (6.161)

68

Here, we postulate the chemical equilibrium to be the process in which

A+ = 0 and J+(A+)= 0 . (6.162)

Let us investigate the first condition A+ = 0 that is

0=N−H∑q=1

(Aq)+gq =N−H∑q=1

N∑α=1

µMα+Pqαgq =

N−H∑q=1

(N∑α=1

(µMα

0++Rϑ lna+α

)Pqα

)gq , (6.163)

so it must holdN∑α=1

(µMα

0++Rϑ lna+α

)Pqα = 0 q=1, . . . , N−H . (6.164)

Let us define the so called equilibrium constant Kq(ϑ, p)

−Rϑ lnKq(p,ϑ) def=N∑α=1

µMα

0+Pqα, q=1, . . . , N−H , (6.165)

and let us plug it in (6.163), we obtain an important relation for the chemical equilibrium, so called equilib-rium mass action law

Kq(ϑ, p)=N∏α=1

(a+α)Pqα

, q = 1, . . . , N−H . (6.166)

Exercise 20. Let us consider the following reaction, called Haber-Bosch synthesis of amoniac:

N2 +3H2 2NH3 ,

which takes place in a closed container. Let the initial composition of the mixture be as follows:

nN2 = 1 mole ,

nH2 = 3 moles ,

nNH3 = 0 moles .

What is the resultant equilibrium composition of the mixture provided we know the equilibrium constant ofthe reaction at the given pressure and temperature?

Solution: So we have N = 3 (α=1 ∼ N2, α=2 ∼ H2, α=3 ∼ NH3) and Z = 2 (σ=1 ∼ N, σ=2 ∼ O). The matrixT thus reads

T =(2 0 10 2 3

)H = rank(T)= 2 ,

i.e. S=T , and thus P matrix has rank 1 and it can be picked e.g. as

P = (−1 −3 2)

,

which indeed corresponds to the considered reaction N2 +3H2 → 2NH3, the only independent reaction in thesystem. Now the balances mass read

∂ρα

∂t+divραvα = mα , α= 1,2,3.

Let us integrate these relations over the container Ω, the volume of which is V , and employ the assumptionof closedness of the container =⇒ vα ·n= 0, ∀α at ∂Ω. We get, assuming mα to be uniform in Ω:

∂Mα(Ω, t)∂t

=V mα =V MαP1αJ1(t) , α= 1,2,3 ,

with Mα(Ω, t) the total mass of α component in Ω at time t. Dividing by molar masses Mα and integratingw.r.t time from the initial time (t=0) to arbitrary t≥0:

nα(Ω, t)P1α − nα(Ω,0)

P1α =V∫ t

0J1 dt︸ ︷︷ ︸

advancement of reaction

α= 1,2,3 ,

69

where nα(Ω, t) in the number of moles of α component in Ω at time t. When interested in the equilibriumcomposition, we do not need to know J1 i.e. the chemical kinetics. We can eliminate the common (unknown)advancement of reaction and write

n1(Ω, t)P11 − n1(Ω,0)

P11 = n2(Ω, t)P12 − n2(Ω,0)

P12 = n3(Ω, t)P13 − n3(Ω,0)

P13 ,

which are two (!) independent reactions for 3 unknowns n1(Ω, t), n2(Ω, t), n3(Ω, t). In particular, for theequilibrium values, using the initial conditions, the two independent relations (switching from indexing bynumbers to stoichiometric labels for better clarity) read

n+N2

−1=−n+

NH3

2n+

H2

3−1=−

n+NH3

2.

The missing relation, available only in equilibrium (!) is the equilibrium mass action law, which in this casereads: (

cMNH3

+)2

(cM

N2

+)(cM

H2

+)3 = K1(ϑ, p) .

Expressingn+α =V (cM

α)+ , α= 1,2,3 ,

we get (n+

NH3

)2

n+N2

(n+

H2

)3 = 1V 2 K1(ϑ, p) .

Solution of this nonlinear algebraic system yields the equilibrium composition.

6.5.4 Chemical kinetics

So far we have only investigated equilibrium process (chemical equilibrium), in which J+ = 0, A+ = 0. Now wewill finally investigate the actual chemical reaction, i.e. the case when J 6= 0. We will discuss two approaches.

Taylor expansion constrained by equilibrium relations In the first approach we will proceed in sev-eral consequent steps:

1. We perform ansatz of the molar reaction rates in the form of a Taylor expansion (around zero), inactivities or in molar concentrations.

2. We employ equilibrium mass action law and condition of equilibrium in order to reduce the number ofcoefficients of the expansion.

3. We constrain the reduced form by investigating the validity of the second law.

Let us attempt to clarify the whole procedure first on a simple example and then come back to the generalcase. Consider again two molecules: NO2 (α=1) and N2O4 (α=2). We already know, that only one independentreaction may take place, e.g. N2O4 →2NO2, in which case the reaction matrix reads P = (2,−1). Let us nowlook for the corresponding rate of this reaction J1 in the form

J1 = J1(ϑ, p,a1,a2) .

In equilibrium, the concentrations must satisfy the mass action law:

K1(ϑ, p)= (a+1 )2(a+

2 )−1 , (6.167)

70

with K1 the corresponding equilibrium constant. Let us look for J1 in the form of a polynomial of maximaldegree D = 2 in the two activity coefficients a1, a2 i.e. let the ansatz for the reaction kinetics be

J1 = k00 +k10a1 +k01a2 +k20(a1)2 +k11a1a2 +k02(a2)2 , (6.168)

where the coefficients ki j may, possibly still depend on pressure and temperature, but do no longer dependon activities. Let us now employ the equilibrium assumptions J1(a+)= 0:

0= k00 +k10a+1 +k01a+

2 +k20(a+1 )2 +k11a+

1 a+2 +k02(a+

2 )2

= k00 +k10a+1 +k01K−1

1 (a+1 )2 +k20(a+

1 )2 +k11a+1 K−1

1 (a+1 )2 +k02K−2

1 (a+1 )4 ,

where in the second equality, we substituted from the mass action law for a+2 . By collecting the terms of same

power in a+1 , and realizing that the equilibrium concentrations of a+

1 are arbitrary (can be arbitrarily variedby the choice of initial concentrations, for example), all the coefficients at all powers of a+

1 must be zero. Weget the following constraints on the coefficients of the polynomial

k00 = 0 , k10 = 0 , k01K−11 +k20 = 0 , k11 = 0 , k02 = 0 .

Plugging now back to the original ansatz formula for J1 (6.168) we obtain

J1 = k01a2 −k01K−11 (a1)2 .

Finally, let us check whether, or under which conditions this satisfies the second law. I.e. let us inspect

ξCH =−J1A1 .

Let us evaluate the affinity A1:

A1 =N∑α=1

µMαP1α = (2µM

10 −µM

20)︸ ︷︷ ︸

=−Rϑ lnK1

+Rϑ ln

(a2

1

a2

)= Rϑ ln

(K−1

1a2

1

a2

), (6.169)

and thus

ξCH =−k01(a2 −K−1

1 a21)Rϑ ln

(K−1

1a2

1

a2

)= k01Rϑ

(a2−K−1

1 a21)(

ln(a2)− ln(K−11 a2

1))

, (6.170)

and thus we see, thanks to the fact that logarithm is a monotone increasing function, that

ξCH ≥ 0⇐⇒ k01 ≥ 0 .

So, under the assumption k01 ≥ 0, we obtained a thermodynamically consistent description of reaction kinet-ics of the reaction N2O4 → 2 NO2 in the form

JN2O4→2NO2 = k(aN2O4 −K−11 a2

NO2) .

Note that the sign of the parenthesis depends on which “side” from the equilibrium given by the eq. massaction law the system is. If we have more products than corresponding to equilibrium, the parenthesisbecomes negative, and the reaction goes backwards. So kinetics describes in fact both cases - the forward andbackward reaction simultaneously, which is sometimes expressed by writing the reaction as N2O4 2 NO2and we shall thus write

JN2O42NO2 = k(aN2O4 −K−11 a2

NO2) .

In reality both equations the forward and the backward one happen simultaneously and it is only the over-all balance of these two processes which determines whether in total the number of “reactants” decreasesand number of “products” increases (i.e. reaction goes forwards) or the other way around (reaction goesbackwards). The definition of reactants and products is thus only formal and depends on our choice of theindependent chemical reaction (i.e. the choice of matrix P). Structure of the formula for reaction rate whichwe obtained reflects this idea of two reactions happening simultaneously: the forward reaction (dissociation

71

of N2O4) with the rate J f = k f aN2O4 and the backward reaction (synthesis of NO2) with the rate Jb = kba2NO2

,

where k f and kb are kinetic parameters of the forward and backward reactions, which satisfy k fkb

= K , K be-ing the equilibrium constant of the reaction. This type of chemical kinetics is quite generic and is sometimesreferred to as “non-equilibrium mass action law”.

Let us consider now still one reaction, but with more reactants and products. For a reaction of the form

ζ1,1f X1 +ζ1,2

f X2 +·· ·+ζ1,Nf XN ζ

1,1b X1 +ζ1,2

b X2 +·· ·+ζ1,Nb XN

where X1, . . . , XN are the reacting species and ζ1βf , ζ1β

b are the forward and backward stoichiometric coeffi-

cients, we recall the sign convention P1β = ζ1βb −ζ1β

f , i.e. “products” having positive stoichiometric coefficients,“reactants” having negative ones. Let us look for the reaction kinetics in the form of a polynomial of degreeD in the chemical activities, i.e.

J1 =Q∑β=0

kνβ

N∏α=1

aνβαα , s.t. multi-indices νβ satisfyN∑α=1

νβα ≤ D ∀β=1, . . . ,Q , (6.171)

where

Q =D∑

k=0

(N+k−1)!(N−1)!k!

, (6.172)

by this formula we mean the following:

J1 = k0...0

+k10...0a1 +k010...0a2 +·· ·+k00...01aN

+k110...0a1a2 +k101...0a1a3 . . .

+ . . . .

In equilibrium, it must hold

0=Q∑β=0

kνβ

N∏α=1

(a+α)νβα , (6.173)

and for the given reaction, we also have the equilibrium mass action law in the form

K1 =N∏α=1

(a+α)P1α

. (6.174)

Let us from here express

a+N = K

1P1N

1

N−1∏α=1

(a+α)−

P1α

P1N , (6.175)

and let us substitute for a+N from here to (6.173):

0=Q∑β=0

kνβKνβNP1N

1

N−1∏α=1

(a+α)νβα−

(P1α

P1N

)νβN =

Q∑β=0

kνβKνβNP1N

1

N∏α=1

(a+α)νβα−

(P1α

P1N

)νβN . (6.176)

Now there are two possibilities, each power νβα−(

P1α

P1N

)νβN either occurs exactly once, or there are multiple

of them in the expansion. In the first case, the corresponding coefficient kνβmust be identically zero. In the

second case, let us have

νβα−(

P1α

P1N

)νβN =νκ1α−

(P1α

P1N

)νκ1N

=νκ2α−(

P1α

P1N

)νκ2N

...

=νκkα−(

P1α

P1N

)νκk N (6.177)

72

for some κ1 . . . ,κk. Now all such νκi N must differ from each other and from νβN , for otherwise, the aboverelation would imply equality of the whole multi-indices whenever they would have the same N-th compo-nent, and since each multi-index is unique in the original expansion, this is not possible. We can rewrite therelations (6.177) as

νβ−νκ1 =P1νβN −νκ1N

P1N ,

νβ−νκ2 =P1νβN −νκ2N

P1N ,

...

νβ−νκk =P1νβN −νκk N

P1N , (6.178)

i.e. all such multi-indices differ by a non-zero multiple of the reaction vector P1. So now the equilibriumrelation (6.176) reads

0= ∑groups

(kνβ

KνβNP1N

1 +kνκ1Kνκ1NP1N

1 + . . .kνκkKνκk N

P1N

1

)N∏α=1

(a+α)νβα− P1α

P1N νβN ,

where the first sum is over groups of terms with the same multi-indices. Consequently, we obtain the follow-ing relations for the coefficients

kνβ=−kνκ1

Kνκ1N−νβN

P1N

1 · · ·−kνκkKνκk N−νβN

P1N

1

Plugging these relations back to (6.171), we can group terms as before and write

J1 =∑

groups

kνβ

N∏α=1

aνβ

α +kνκ1

N∏α=1

aνκ1α +·· ·+kνκk

N∏α=1

aνκkα

= ∑groups

N∏α=1

aνβ

α

kνβ

+kνκ1

N∏α=1

aP1ανκ1N−νβN

P1Nα · · ·+kνκk

N∏α=1

aP1α

νκk N−νβN

P1Nα

= ∑groups

N∏α=1

aνβ

α

kνκ1

(N∏α=1

aP1ανκ1N−νβN

P1Nα −K

νκ1N−νβNP1N

1

)· · ·+kνκk

(N∏α=1

aP1α

νκk N−νβN

P1Nα −K

νκk N−νβN

P1N

1

),

where for the second equality, we employed relations (6.178). This kinetic formula complies with our equi-librium assumptions, since each factor in the round brackets vanishes in equilibrium. The resultant for-mulae represent the generic form of the non-equilibrium mass action law for one reaction expanded aboutzero to degree D. The exponents of the type

νκk N−νβN

P1N are not a-priori neither positive nor whole num-bers. Provided P1N =±1 (which can always be achieved by suitable normalization), one can group the termswith positive and negative coefficients (which now are whole numbers), and using the formula xk − yk =(x− y)(xk−1 + xk−2 y+·· ·+ xyk−2 + yk−1), we obtain the form

J1 =P1(a1, . . . ,aN )

(N∏α=1

aP1α

α −K1

)+P2(a1, . . . ,aN )

(N∏α=1

a−P1α

α −K−11

), (6.179)

where P1 and P2 are functions of a1, . . . ,aN . Employing again the forward and backward stoichiometriccoefficients, i.e. writing P1α = ζ1,α

b −ζ1,αf , we can further write

J1 =

P2

N∏α=1

a−ζ1,α

bα −K1P1

N∏α=1

a−ζ1,α

fα

(N∏α=1

aζ

1,αfα −K−1

1

N∏α=1

aζ1α

bα

), (6.180)

or, in short

J1 = k(a1, . . . ,aN )

(N∏α=1

aζ

1,αfα −K−1

1

N∏α=1

aζ1α

bα

), (6.181)

73

where on the right-hand side we again recognize the products corresponding to forward and backward reac-tions, respectively. Denoting the forward kinetic coefficient k1

f = k and the backward coefficient k1b = kK−1

1 ,we obtain as in the example the kinetics of the reaction in the form

J1 = k1f (a1, . . . ,aN )

N∏α=1

aζ

1,αfα −k1

b(a1, . . . ,aN )N∏α=1

aζ

1,αbα . (6.182)

It remains to inspect the validity of the second law. The affinity

A1 =N∑α=1

µMαP1α =

N∑α=1

µM0αP1α

︸ ︷︷ ︸−RϑK1

+Rϑ ln

(N∏α=1

aP1α

α

)= Rϑ ln

(K−1

1

N∏α=1

aP1α

α

),

and thus, using (6.179),

ξCH =−J1A1 =−RϑP1(a1, . . . ,aN )

(N∏α=1

aP1α

α −K1

)(ln

N∏α=1

aP1α

α − lnK1

)

+RϑP2(a1, . . . ,aN )

(N∏α=1

a−P1α

α −K−11

)(ln

N∏α=1

a−P1α

α − lnK−11

), (6.183)

by monotone increasing property of logarithm, we thus see that a sufficient condition for the requirementξCH ≥ 0 is P1 ≤ 0 & P2 ≥ 0, which is the sought constraint on the functions P1, P2. Let us come back to thegeneral case a let us look for J in the form of a polynomial of maximum degree D in chemical activities.

J=Q∑β=0

kνβ

N∏α=1

aνβαα , s.t. multi-indices νβ satisfyN∑α=1

νβα ≤ D ∀β=1, . . . ,Q , (6.184)

where

Q =D∑

k=0

(N+k−1)!(N−1)!k!

. (6.185)

Let us note that the coefficients kνβ can still depend on temperature, pressure, etc, but no-longer on activitiesaα (or molar concentrations cM

α in the ideal case).Analysis such as the one in the example above can be performed for the general polynomial ansatz.

Exercise 21. Try the same polynomial ansatz as in the above example for reaction N2O4→2NO2, but this timefor linear (D=1) and cubic (D=3) case. Can you still recast the result to the form given by the non-equilibriummass action law? (Result: for D=1 we get J1 = 0, for M = 3 one has J1 = (k01 +k11a1 +k02a2)

(a2 −K−1

1 a21).)

For a system with N reacting components in which N−H independent reactions take place:

ζ1,1f X1 +α1,2

f X2 +·· ·+ζ1,Nf XN ζ

1,1b X1 +ζ1,2

b X2 +·· ·+ζ1,Nb XN ,

ζ2,1f X1 +ζ2,2

f X2 +·· ·+ζ2,Nf XN ζ

2,1b X1 +ζ2,2

b X2 +·· ·+ζ2,Nb XN ,

...

ζN−H,1f X1 +ζN−H,2

f X2 +·· ·+ζN−H,Nf XN ζ

N−H,1b X1 +ζN−H,2

b X2 +·· ·+ζN−H,Nb XN ,

where X1, . . . , XN are the reacting species and ζp,βf , ζp,β

b are the forward and backward stoichiometric coef-

ficients of p-th reaction, respectively with the sign convention P pβ = ζpβb −ζpβ

f . One can now proceed in thesame spirit as in the case of one reaction. I.e., apply the N−H independent equilibrium mass action laws

Kq =N∏α=1

(a+α)Pqα

, q = 1, . . . , N −H , (6.186)

to express certain N−H equilibrium activities in terms of the remaining H ones. By plugging these relationsinto the polynomial ansatz for equilibrium, one arrives at relations among the coefficients, some of which

74

must be identically zero, some, being linearly dependent. Finally, these relations are used outside equilibriumfor the polynomial ansatz, and, one arrives at formulae of the form

J= ·· ·+kντN∏α=1

aνκαα

(N∏γ=1

a(ντγ−νκγ)γ −K

)+ . . . ,

which is a typical term in the final expansion. By the same procedure as before, one can show that thecoefficient ντγ−νκγ is some linear combination of the stoichiometric coefficients of the chosen independentreactions, i.e. are coefficients of some possible chemical reaction, and K is the corresponding equilibriumconstant of this reaction. It is not straightforward to discuss the constraints implied by the second law onthe coefficients of these formulae in the general case, one has to inspect case-by case (in the same spirit asfor one reaction).

Non-linear closure based on the structure of the chemical contribution to entropy productionautomatically satisfying the second law. An alternative to the procedure just described above is morein the spirit we are used to - we start from inspecting the structure of the chemical dissipation contribution tothe entropy production and infer directly certain closure relation which will automatically satisfy the secondlaw of thermodynamics.

Let us recall the chemical dissipation term in the entropy production

ξCH =−N∑α=1

mαµα =−N∑α=1

N−H∑q=1

MαµαPqαJq =−N∑α=1

N−H∑q=1

µMαPqα︸ ︷︷ ︸Aq

Jq

=−N−H∑q=1

N∑α=1

µM0αPqα

︸ ︷︷ ︸=−Rϑ lnKq

+Rϑ ln

(N∏α=1

aPqα

α

)Jq =−RϑN−H∑q=1

ln

(K−1

q

N∏α

aPqα

α

)Jq . (6.187)

Expressing the stoichiometric coefficients as differences of the corresponding backward and forward contri-butions, i.e.

Pqα = ζqαb −ζqα

f , q = 1, . . . , N −H , (6.188)

we obtain

ξCH =−RϑN−H∑p=1

ln

K−1q

∏Nα=1 a

ζqαbα∏N

α=1 aζ

pαfα

Jq

= RϑN−H∑q=1

(ln

(k f

q

N∏α=1

aζ

qαfα

)− ln

(kb

q

N∏α=1

aζ

qαbα

))Jq , (6.189)

where k fq, kb

q are such thatkb

q

k fq= K−1

q , q=1, . . . , N−H . (6.190)

The revealed structure of ξCH shows that a constitutive closure

Jq = k fq

N∏α=1

aζ

qαfα −kb

q

N∏α=1

aζ

qαbα , q=1, . . . , N−H , (6.191)

automatically satisfies ξCH ≥ 0 due to the fact that logarithm is monotonously increasing function. The twoterms in (6.191) can be interpreted as the molar reaction rates of the forward and backward reaction. Notethat in equilibrium, these two really cancel out due to the equilibrium mass action law, and the equilibriumreaction rate is thus zero.

Add remark about non-constant reaction coefficientsAdd remark about elementary and non-elementary chemical reactions

75

7 Class II mixtures

7.1 Motivation

Mixture models of class II in the sense of definition in Chapter 5.8, include such multicomponent materials,for which mechanical interaction of the constituents becomes important. Some typical problem that can beaddressed within the class II mixture framework are:

• Porous media flow (engineering hydrology, oil industry, fracking, ... )

• Bubble flows (chemical reactors)

• Swelling (of biological tissues, or geological materials,....)

Let us remind, that in the class II mixture framework, we formulate N balances of mass for N componentsof the mixture and also N balances of linear momentum for N components of mixture. As in class II setting,we only formulate one balance of energy and one balance of entropy for the mixture as a whole.

Remark 25. Class II mixture involve a concept of partial stresses. This brings in a serious problem withspecification of boundary conditions, as it is often practically impossible to assess the partial stresses (partialpressures) experimentally. It appears that for class II mixtures, this problem is a fundamental one.

The framework of class II mixtures will turn out to be very useful for derivation and understanding ofsimplified models of the type Darcy, Brinkman, Forchheimer or Biot

We will proceed in the following steps.

• We recall the class II balance laws and identify the quantities that need to be specified by constitutiverelations.

• We infer some a-priori knowledge about the structure of the interaction force for a binary mixture byrewriting suitably the momentum balance.

• Next, we discuss possible forms of the interaction force based on macroscopic analogy - examiningforces exerted on a sphere immersed in a flow

• We derive Darcy’s law from in three ways - (i) as a reduced form of the class II momentum balance, (ii)from macroscopic analogy considering Poiseuille flow in channels, (iii) from formal two-scale homoge-nization

• We discuss the generalizations of Darcy’s model, namely the models of Brinkman, and the Darcy-Forchheimer model

• We present a class II thermodynamic framework based on the formulation of constitutive relationsfor the Helmohltz free energy ψ and rate of entropy production ξ in a close analogy with relations forsingle-component continuum (i.e. working with quantities related to the mixture as a whole).

76

7.2 Balance laws

Balance laws in class II comprise the balances of mass for individual component, balances of momenta forindividual components, and one balance of energy for the mixture as a whole.

∂ρα

∂t+div(ραvα)= mα , α=1, . . . , N , (7.1a)

N∑α=1

mα = 0 , (7.1b)

∂(ραvα)∂t

+div(ραvα⊗vα)= divTα+ραbα+Iα+mαvα , α= 1, . . . , N , (7.1c)

N∑α=1

Iα+mαvα = 0 , (7.1d)

TTα =Tα , α=1, . . . , N , (7.1e)

∂

∂t

(ρ

(e+ |v|2

2

))+div

(ρ

(e+ |v|2

2

)v)= div(Tv−q)+ρb ·v+ρr . (7.1f)

The terms in these balances, which are novel with respect to the system of balance equations for Class I,and for which now constitutive closures must be provided, are the partial Cauchy stresses Tα and the newinteraction forces Iα.

7.3 Interaction forces - structure inferred from balance laws

A lot about the possible structure of the interaction force can be gained by investigating macroscopic me-chanical interaction in two typical situations - (i) flow of fluid around obstacles and (ii) flow of fluid throughchannels. Before doing so, let us gain some information about the interaction force just by rewriting themomentum balances in a slightly different manner. Let us consider a binary mixture, for simplicity. ForClass I mixtures, we have derived in Sec. 6 two equivalent sets of mass balances, one in terms of partialdensities ρα, the other in terms of the mixture density ρ, which had the classical single-component form andwas supplemented with evolution equations for the N−1 independent concentrations. In Section 5.4, we haveshown that in Class II, we can arrive at single-component balance of momentum for the mixture as a whole,which, under suitable definition of the mixture Cauchy stress and mixture body forces, has the same formas in the case of a single-component continuum. One might be interested whether we can supplement thisequation with a set of equations for some N−1 quantities, such that the final system would be equivalentwith the system of for the partial linear momenta. This is indeed so. Let us consider a binary, non-reactingmixture (i.e. setting mα ≡ 0) for simplicity. The two momentum balances then read

∂(ρ1v1)∂t

+div(ρ1v1 ⊗v1

)= divT1 +ρ1b1 +I , (7.2a)

∂(ρ2v2)∂t

+div(ρ2v2 ⊗v2

)= divT2 +ρ2b2 −I , (7.2b)

where we denoted I def= I1 = −I2. Equivalently, assuming moreover (b1 = b2 - which hold, for example if thebody force is gravity), the two momentum balances read

ρ1D1v1

Dt= divT1 +ρ1b+I , (7.3a)

ρ2D2v2

Dt= divT2 +ρ2b−I , (7.3b)

reminding the definition of the convective time derivative with respect to the α component Dα()Dt = ∂

∂t +[∇()]vα.As we already know, by adding (7.2a) and (7.2b), with the definitions of mixture density ρ = ρ1+ρ2 mixturevelocity v= 1

ρ(ρ1v1+ρ2v2), mixture body force b= 1

ρ(ρ1b1+ρ2b2) we get the momentum balance in the standard

single-component form

∂(ρv)∂t

+div(ρv⊗v

)= divT+ρb , or using the mass balance ρDvDt

= divT+ρb , (7.4)

77

provided that we define the mixture stress tensor as T=T1+T2−ρ1u1 ⊗u1−ρ2u2 ⊗u2, where uα = vα −v,α = 1,2 are the diffusive velocities. Now we want a complementary evolution equation, which would giveus together with the total momentum balance (7.4) an equivalent system to the system (7.2). The quantitycomplementary to the mixture velocity v can be for example the relative velocity between the componentsv1−v2. Observing

v1 −v2 =u1 −u2 = j1

ρc− j2

ρ(1−c)= j1

ρc(1−c), (7.5)

(where we used j1+j2=0), we can equivalently choose j def= j1 as complementary variable to v instead of v1−v2.Let us now investigate the evolution equation for j.

DjDt

= DDt

(ρc(v1−v)

)= ρc(v1−v︸ ︷︷ ︸jρc

)+ρ c(v1−v︸ ︷︷ ︸jρc

)+ρc(

Dv1

Dt− Dv

Dt

), (7.6)

Using the mass balances in the form

ρ =−ρdivv , (7.7)

ρ c =−divj , (7.8)

realizing that

cρDv1

Dt= cρ

(∂v1

∂t+ [∇v1]v

)= cρ

(∂v1

∂t+ [∇v1]v1

)− cρ[∇v1]u1 = cρ

D1v1

Dt− [∇v1]j (7.9)

(7.3a)= (divT1 +ρcb+I

)− [∇v1]j = (divT1 +ρcb+I

)− [∇v]j −[∇

(jρc

)]j (7.10)

and employing the momentum balance for the mixture as a whole (7.4) we can recast (7.6) to the form

DjDt

=−jdivv− jρc

divj+divT1 +I− [∇v]j−[∇

(jρc

)]j− cdivT . (7.11)

Let us further express

T=T1 +T2 − j1 ⊗ j1

ρc− j2 ⊗ j2

ρ(1− c)=T1 +T2 − j⊗ j

ρ

(1c+ 1

1− c

)=T1 +T2 − j⊗ j

ρc(1− c), (7.12)

which, when plugged into (7.11) yields

DjDt

=− (divvI+ [∇v])j+I−div(j⊗ jρc

)+ (1−c)divT1 − cdivT2 + cdiv

(j⊗ j

ρc(1−c)

)=− (divvI+ [∇v])j+I−div

(j⊗ jρ

(1c− 1

1−c

))+ (1−c)divT1 − cdivT2 −

(j⊗ j

ρc(1−c)

)·∇c

=− (divvI+ [∇v])j+I−div(j⊗ jρ

(1c− 1

1−c

))+div((1−c)T1 − cT2)+

T1 +T2 −(

j⊗ jρc(1−c)

)︸ ︷︷ ︸

=T

·∇c . (7.13)

So to summarize we have shown that the complementary balance to the mixture momentum balance (7.4)for the binary non-reactive mixture can be written in the form of an evolution equation for diffusive flux j inthe form

j+ ((divv)I+∇v)j+div(

1ρ

(1c− 1

1−c

)j⊗ j

)= div((1−c)T1 − cT2)+T∇c+I . (7.14)

Remark 26. One can show (Soucek et al., 2014), that under some conditions this evolution equation reducesto Fick’s law j = −k∇c, or in fact to its “Maxwell-Cattaneo” form. Maxwell-Cattaneo equation is so calledextended thermodynamics counterpart of the classical Fourier law for heat transfer: τ∂q

∂t +q = −κ∇ϑ. Thisequation yields for very small τ a very fast relaxation of q to the stationary form given by the classical Fourierlaw q=−κ∇ϑ, but formally, it completely changes the mathematical character of the associated energy-balance– heat equation, making it hyperbolic instead of parabolic. Such form of the heat law thus solves the paradoxof infinite speed of propagation of heat, known for the parabolic Fourier law.

78

Let us now neglect in equation (7.14) the non-linear terms j⊗j, (divv)j, (∇v)j, and consider on top of it itsstationary limit. We arrive at

0' div((1−c)T1 − cT2)+T∇c+I , (7.15)

in equilibrium, the partial Cauchy stresses reduce only to isotropic parts - partial pressures, for a binaryfluid mixture, these can be shown to be of the form

T1 =−p1I=−cpI , T2 =−p2I=−(1−c)pI , (7.16)

where p is the “mixture pressure”. Under such assumptions, we obtain equilibrium relation

I= p∇c . (7.17)

This is quite an important result: the interaction force, contains term p∇c and essentially reduces to it inequilibrium. This will be confirmed also in a different (thermodynamic) framework in Section 7.7.

7.4 Interaction forces - macroscopic mechanical analogies

What other terms except the equilibrium term p∇c can be part of the interaction force? We will employmacroscopic considerations, studying first the problem of flow of a fluid around a sphere and also the problemof flow of a fluid through channels.

7.4.1 Flow around a sphere

Let us consider flow around a sphere (located in the origin) with radius a, in a velocity field only perturbedby the presence of the sphere, which at infinity reads v∞ = (v∞r ,0,0), vr being the relative velocity betweenthe fluid (at infinity) and the sphere. In the dimensionless form, the balance of momentum becomes theNavier-Stokes equation

∂v∂t

+div(v⊗v)=−∇p+ 1Re∆v , (7.18)

where Re is the Reynolds’ number given here as Re = inertial forcesviscous forces = |div(ρ f v⊗v)|

|div(µ fD(v))| =ρ f v2

rL

µ fvrL2

∼ 2aρ f vrµ f

, µ f being the

dynamic viscosity of the fluid. We are now interested in the total force exerted on the sphere by the fluid

f =∫∂ΩTnda . (7.19)

Several approximations are now considered

1. Stokes problem - is an approximation valid provided we can neglect the inertial terms on the l.h.s.(i.e. for slow laminar flows with small Re), it reads

0=−∇p+ 1Re∆v , (7.20)

an analytical solution exists and the drag force is given by the famous Stokes formula

f StokesD = 6πaµ f v∞

r , (7.21)

valid approximately for Re ≤ 0.2.

Remark 27. To gain some insight, for a 1mm sphere immersed in water with viscosity and density µ=10−3Pa s, ρ f = 103 kg m−3, Reynolds’ number 0.2 corresponds to relative flow velocities vr = Re·µ f

2aρ f= 10−4

ms−1!

2. Oseen correction - valid for slightly higher Re (up to Re ∼ 2), is

v∞r ·∇v=−∇p+ 1

Re∆v , (7.22)

79

can also be solved analytically and yields the drag force

f OseenD = 1

2ρπa2CD |vr|vr , (7.23)

with the drag coefficient

CD = 24Re

(1+ 3

16Re

), (7.24)

where CD = 24/Re corresponds to for the Stokes’ formula.

3. Slip shear lift - if the imposed flow field to which the sphere is immersed is not uniform, but if thereis a horizontal gradient ∂vr

∂z 6= 0, and if the sphere is not allowed to rotate additionally to the drag forcea lift force (Saffman, 1965,1968) arises in the direction perpendicular to the flow v∞

r . The force is

f Shear liftL = fLez = 6.46

√µ f ρ f a2

√∣∣∣∣∂vr

∂z

∣∣∣∣|vr|sgn(∂vr

∂z

)ez . (7.25)

4. Spin lift (Magnus force) - if the sphere is rotating with a given angular rotational vector ω additionalforce is exerted

f MagnusL =−C(vr)ω×v∞

r , (7.26)

where C(vr) is a lift parameter.

Remark 28. This force is responsible of many famous goals in soccer, when mastered, it can be a dia-bolically devious weapon in table tennis, it is used in airsoft guns to extend their reach, but was alsoused for “serious” purposes to drive ships equipped with rotating cylinders instead of sails.

5. Virtual mass effect - so far we considered only stationary cases, if the sphere would be acceleratingin the fluid, or the other way round, if the surrounding flow was not stationary, the sphere would apartfrom the classical inertial force be subjected to additional drag force due to the necessity to acceleratethe flow of the surrounding fluid (a phenomenon known to anybody who has ever been paddling in acanoe). A typical form of the total force associated with acceleration is then

f virtual massD = (ms +mvm)a12 , (7.27)

where ms is the mass of the sphere (this is the classical inertial term), and the virtual mass mvm for asphere reads mvm = 1

2ρ f43πa3 and a12 is the relative acceleration between the two phases.

Remark 29. What is the class of relative accelerations, which are objective (frame indifferent)? Ananalysis in the context of two-phase flow has been done by Drew et al. (1979), showing that a generalframe-indifferent class of relative accelerations between two phases is of the following form:

a= ∂(v f −vs)∂t

+∇(v f −vs)v f +[(λ−2)∇v f + (1−λ)∇vs

](v f −vs),λ ∈R (7.28)

The derivation proceeds as follows. Consider the change of frame described by a general time-dependenttranslation and rotation, i.e. by the mapping x∗(t) = Q(t)(x(t)−x0)+y(t), t∗ = t+ t′, where t′ ∈ R andQ(t)QT (t)=Q(t)TQ(t)=I. Let us consider two continua and in each of them pick a material particle, sayX1 and X2, their motion being described by the mappings x1(X1, t) and x2(X2, t) giving velocities

vα = Dαxα(Xα, t)Dt

, α= 1,2 .

In the starred frame, these velocities read

v∗α(Xα, t∗)= ∂x∗

α(Xα, t∗)∂t∗

=Q(t)vα(Xα, t)+ Q(t)xα(Xα, t)+ y(t) , α= 1,2. (7.29)

Taking time derivatives of the velocities yields accelerations

aα = Dαvα(Xα, t)Dt

, α= 1,2

80

and their transformations read

a∗α(Xα, t∗)= ∂v∗

α(Xα, t∗)∂t∗

=Q(t)aα(Xα, t)+2Q(t)vα(Xα, t)+ Qxα(Xα, t)+ y(t) , α= 1,2. (7.30)

Switching from reference description to spatial, i.e. defining two spatial velocity fields

vα(x, t) def= vα(X−1α (x, t), t) α= 1,2,

and accelerationsaα(x, t) def= aα(X−1

α (x, t), t) α= 1,2,

eq. (7.29) and (7.30), then read

v∗α(x∗, t∗)=Q(t)vα(x, t)+ Q(t)x+ y(t) , α= 1,2, (7.31)

a∗α(x∗, t∗)=Q(t)aα(x, t)+2Q(t)vα(x, t)+ Qx+ y(t) , α= 1,2. (7.32)

These relations imply that relative velocity is an objective vector, since it transforms as

(v∗1 −v∗

2 )(x∗, t∗)=Q(t)(v1−v2)(x, t) , (7.33)

but also that relative acceleration is not an objective vector, since

(a∗1 −a∗

2 )(x∗, t∗)=Q(t)(a1 −a2)(x, t)+2Q(t)(v1 −v2)(x, t) . (7.34)

The troublemaker is the Coriolis term 2Q(v1 −v2). In order to come up with an objective measure ofmutual acceleration, let us take the spatial gradient of (7.31) using grad()= grad∗()Q, expressing

Q= grad∗v∗α(x∗, t∗)Q(t)−Q(t)gradvα(x, t) ,α= 1,2.

Since for the same quantity Q, we have two expressions (for α= 1,2), let us take their affine combinationwith some parameter λ ∈R and express

2Q(t)(v1 −v2)= 2(1−λ)grad∗v∗

1 (x∗, t∗)Q(t)−Q(t)gradv1(x, t)(v1 −v2))

+2λgrad∗v∗

2 (x∗, t∗)Q(t)−Q(t)gradv2(x, t)(v1 −v2) . (7.35)

Substituting (7.35) into (7.34) yields after some manipulation

D∗1v∗

1

Dt∗− D∗

2v∗2

Dt∗− [∇∗v∗

1 ](v∗1 −v∗

2 )− [∇∗v∗2 ](v∗

1 −v∗2 )+ (1−λ)[∇∗(v∗

2 −v∗1 )](v∗

1 −v∗2 )

=Q(t)

D1v1

Dt− D2v2

Dt− [∇v1](v1 −v2)− [∇v2](v1 −v2)+ (1−λ)[∇(v2 −v1)](v1 −v2)

, (7.36)

i.e. the quantity on the left-hand side transforms as an objective vector. The expression can be recast tothe following form of the objective relative acceleration

a12 =(

D1v1

Dt−[∇v1](v1−v2)

)−

(D2v2

Dt+[∇v2](v1−v2)

)+ (1−λ)[∇(v2 −v1)](v1 −v2) , (7.37)

which is equivalent to (7.28). Another equivalent forms is

a12 = D2v1

Dt− D1v2

Dt+ (1−λ)[∇(v2 −v1)](v1−v2) . (7.38)

Postulating the form of the virtual-mass force as between the fluid and dispersed phase (droplets, bub-bles, ...) as

fvirtual massD = ρlCvma12 ,

the analysis of two cases with extreme values of the volume fractions of the dispersed phase yields thecorresponding λ values: very low volume fraction of the dispersed phase, i.e. limφ→0 corresponds toλ=2, while the other extreme limφ→1 implies λ= 0. For an intermediate value λ= 1 one gets a relativelysimple and symmetric form

a12 = D2v1

Dt− D1v2

Dt.

81

6. Basset force - results from the viscous flow field created by the particle when accelerating, or equiva-lently, in an unsteady flow (Basset, 1888). For a laminar unsteady flow around a sphere the force reads

f BassetD =−6ρa2√

πµ f

∫ t

t0

1pt− t′

dvr

dt′dt′ . (7.39)

7. Buoyancy can be computed simply in the hydrostatic approximation

(f buoyancyL )i =−

∫∂Ω

pnida =−p(0)∫∂Ω

nida− ∂p∂x j

(0)∫∂Ω

x jnida+·· · =− ∂p∂xi

(0)∫Ω

dx+ . . . , (7.40)

where (0) is some mid-value which gives the buoyancy lift force

f buoyancyL '−∇pVΩ . (7.41)

For a fluid in hydrostatic equilibrium ∇p =−ρ f gez, and combined with the gravity force exerted on thesphere itself

fgravity =−ρsVΩgez , (7.42)

the combined force gravity+buoyancy reads

fgravity+buoyancyL =−(ρs−ρ f )VΩgez . (7.43)

These interaction mechanisms can now be used to construct continuum counterparts of the interaction forcebetween a solid phase composed of small particles (moving with velocity vs) dispersed in a liquid phasemoving with velocity v f . The procedure is essentially to prescribe the force as φs

fparticle

V particle (where φs is thevolume fraction of the solid phase), simultaneously replacing all kinematic quantities by their suitably chosencontinuum counterparts. In constructing the individual terms one must also respect the basic principles ofcontinuum constitutive theory such as principle of material frame indifference. Consequently, the followingformulas can be identified

Interaction mechanism Macroscopic analogue Continuum counterpartdrag 1

2ρ f πa2CD |vr|vr α1(φ,ν f ,ρ f ,a, |vs−v f |)(vs−v f )

shear lift 6.46pν f ρ f a2√∣∣∣∂vr

∂z

∣∣∣|vr|sgn(∂vr∂z

)ez α2(φ,ν f ,ρ f ,a,‖D f ‖)D f (vs−v f )

spin lift −C(vr)ω×v∞r α3(φ,ν f ,ρ f ,a)(Ws−W f )(vs−v f )

virtual mass (ms +mvm)ar α4(φ,ρ f ,ρs,a)(

D f vsDt − Dsv f

Dt

)buoyancy −∇pVΩ −φ∇p

Based on the extensions and generalizations of the above drag and lift contributions to the interactionforce, the following ansatz can be found in the literature for interaction force between solid phase movingwith velocity vs dispersed in liquid moving with velocity v f :

I∼α1(vs −v f )+α2D f (vs −vs)+α3(Ws −W f )(vs −v f )+α4

(D f vs

Dt− Dsv f

Dt

)+α5∇φ+ p∇φ , (7.44)

where D f = 12(∇v f +∇Tv f

), W f = 1

2(∇v f −∇Tv f

), Ds = 1

2(∇vs +∇Tvs

), Ws = 1

2(∇vs −∇Tvs

). Note that the

expression does not contain the buoyancy force, which enters the momentum balance through the partialCauchy stress, and it contains non-classical diffusion term α5∇φ and also the last term −p∇φ, which corre-sponds to the equilibrium remainder derived in the last section (only with concentration replaced by volumefraction).

7.5 Darcy’s law

Darcy’s law is a law describing the flow of fluids through porous media. Formulated by Henry Darcy in 1856,based on experiments of flow of water through beds of sand, it is commonly viewed as a relationship betweenthe fluid mass flux and pressure gradient in the medium. We will show three ways how to infer the form ofthis law, which will also give us some insight into possible generalizations of this law.

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7.5.1 Reduction of two-component momentum balance

Consider a two-component medium composed of solid porous matrix and a fluid. Let the two components notinteract chemically (i.e. no mass transfer between the components). In the context of class-II mixture theory,we have the following generic form of the partial momentum balances for the components:

∂(ρ f v f )∂t

+div(ρ f v f ⊗v f

)= divT f +ρ f g+I f , (7.45a)

∂(ρsvs)∂t

+div(ρsvs ⊗vs

)= divTs +ρsg−I f , (7.45b)

where I f =−Is. Let us now completely ignore the second momentum balance, which is possible, if the solidmatrix is for example some rock, in such case the deformation of the rock resulting from the flow of the fluidthrough the pores is negligible and we can even set vs ≡ 0. Let us, moreover, neglect the inertial forces on theleft-hand side of momentum balance for the fluid (this is possible for small enough Reynolds numbers) and letus, in addition, keep only the isotropic part of the partial Cauchy stress T f . If the (real measurable) pressurein the fluid in the pores is p, then the effective partial pressure exerted by the fluid on some surface of theporous medium would be p times fraction of the surface exposed to the fluid. Often a good approximation isto consider average volume fraction of the fluid φ f at the given point (though the relation between surfacefraction and volume fraction is definitely much more complicated and depends in general on the detailedstructure of the pores, their shape, etc,.). So under these assumptions, the partial Cauchy stress of the fluidreads

T f.=−φ f pI . (7.46)

If we use ρ f = ρmf φ f , the momentum balance for the fluid has reduced to

0=−∇(φ f p)+ρmf φ f g+I f , (7.47)

if we now keep in the interaction force only the drag term and the equilibrium term, i.e. if we consider

I .=−α1v f + p∇φ f , (7.48)

we obtain Darcy’s law in the formα1v f =−φ f (∇p−ρm

f g) . (7.49)

7.5.2 Derivation from macroscopic analogy

Let’s consider a porous material and a fluid (water flowing through a bed of sand). Let us simplify the poresas channels with circular cross-section. Let us consider stationary laminar Poiseuille flow through a straightsegment of a pore. Under an imposed pressure gradient in the pore, the flow - the volume of water expelledfrom the segment per second - would be

h = πa4

8µ f|∇p| , (7.50)

where a is the radius of the pore and µ f is the dynamic viscosity of the fluid. A vectorial generalization,which takes into account that the pore is not straight reads

h=− πa4

8µ f X∇p , (7.51)

where X is so-called tortuosity, which is essentially a geometric factor only. Let us now move back fromthis idealized pore to the sand bed and consider some particular channel (pore). Let us denote d the char-acteristic dimension of the sand grains, and φ f ∼ a2

d2 the fluid volume fraction, the flux through the channelbecomes

hd2 =−

φ2f d2

µ f X∇p . (7.52)

Let us denote h the flux of liquid per unit surface, then

h=−k(φ)µ f

∇p, (7.53)

83

where the permeability k(φ)= k0φ2, (typically in applications k(φ)= k0φ

n with n = 2. . .3), and where k0 := d2

X .From the definition of the volume flux (assuming the porous matrix static)

h=φ f v f ,

we obtain

v f =−k(φ f )φ f µ f

∇p =− k0

µ fφ f ∇p . (7.54)

To include gravity, we only replace the pressure p with the so-called hydraulic head:

p → p+ρmf gz︸ ︷︷ ︸

hydraulic head

,

then Darcy’s law becomes

v f =− k0

µ fφ f (∇p−ρm

f g) , (7.55)

compare with (7.49) and observe that we have identified the drag coefficient

α1 =µ f

k0. (7.56)

The assumption of laminarity of the flow can be translated into an assumption on the associated Reynolds’number:

Re = 2ρ f vraµ f

∼ 2ρ f hdµ fφ f

ad︸︷︷︸

∼pφ f

= 2ρ f k|∇p|dµ2

f√φ f

=φ3/2

f

X

(ρ f

√gdd

µ f

)2

,

where in the last equality, we took the scale of the pressure gradient |∇p| ∼ ρ f g, i.e. hydrostatic. For water,(µ f ∼ 10−3 Pa s, ρ f ∼ 103 kg m−3), taking we obtain Re ∼ 10

X (100φ f )3/2 × (d [in mm])3. For porosities of theorder of units of percent, and for critical Reynolds number Recrit ∼ 103 we obtain the critical pore size dcrit ofthe order of milimeters.

7.5.3 Derivation through homogenization

Let us consider a Stokes flow of a viscous incompressible fluid through channels in a porous medium with acharacteristic dimension l, onto which is imposed a pressure gradient of the order δp/l, and which has porousmicro-structure at scale d, as in Figure7. Let us assume a periodic structure as in Fig.7, i.e. Ω being viewed

Figure 7: A porous medium (left) with a spatial scale l, which is assumed to possess an internal structureoccurring at a scale d.

as a union of cells Ωcelli , i.e. Ω=∪iΩ

celli , and each cell Ωcell

i is composed of a fluid part Ω fi and solid part Ωs

i ,such that Ωcell

i =Ω fi ∪Ωs

i , ∀i. Let us define the fluid and solid parts of Ω by

Ω f def= ∪iΩfi , Ωs def= ∪iΩ

fi . (7.57)

84

Then we can formulate the Stokes flow problem in the fluid part of the domain Ω f as follows

divv= 0 in Ω f , (7.58a)

0=−∇p+µ f∆v , in Ω f , (7.58b)

supplemented with boundary conditions, e.g.

v= 0 , at ∂Ω f . (7.58c)

Let us first rewrite the above system in a dimensionless form. Let us introduce the following scale for theflow velocity:

[v]= d2[δp]lµ f

, (7.59)

the motivation here being naturally the Darcy’s law, which we want to derive. Note that this is a typicalscenario with homogenization - by the “right” choice of scales, you obtain the “right” results. Let us writedown the Stokes problem in a dimensionless form, assuming the scaling of the spatial gradient with themacroscopic scale l (i.e. taking ∇= 1

l ∇):

divv= 0 in Ω f (7.60a)

0=−δpl∇p+ d2δp

l3 ∆v in Ω f , (7.60b)

v= 0 , at ∂Ω f , (7.60c)

where the tildes denote the dimensionless counterparts of dimensional quantities (i.e. for each quantity ψ

we consider its scale [ψ] and a dimensionless ψ such that ψ= [ψ]ψ). Introducing a small parameter ε := d/l,and omitting the tilde symbol for brevity, we obtain

divvε = 0 in Ω f (7.61a)

0=−∇pε+ε2∆vε in Ω f , (7.61b)

vε = 0 , at ∂Ω f , (7.61c)

where we explicitly added the subscript ε to the unknowns vε, pε in order to underline the dependence ofthese fields on the considered micro-structure.

Now we want to employ the technique of asymptotic expansions. Consider for a given ε any field variableassociated with the problem, denotedψε(x, t). Now the basic idea of formal asymptotic expansion is to expressthe quantity ψε as a series in powers of ε:

ψε =∞∑

i=0εiψ(i)

ε , (7.62)

where ψ(i)ε are yet to be specified. Since the problem is characterized by two structures: macro-structure Ω

and micro-structure Ωcell, it seems reasonable to introduce

ψ(i) :Ω×Ωcell →Rk , (7.63)

withψ(i)ε (x, t) def= ψ(i)(x,

xε

, t) (7.64)

such that ψ(i) is periodic in the second variable ξ def= xε.

Remark 30. How should one understand this? Think of ψ being the density, for instance. It is clear, thatthe density can exhibit very strong oscillations (density of the fluid inside the channel can be very differentfrom the density of the surrounding rock), but at the same time, when "averaged" over the characteristicsmall-scale d, it may be only a mildly spatially varying function, or even a constant. So it makes sense totry to separate the behavior at the two spatial scales from each other by defining ρε(x, t) def= ρ(x,ξ, t). For agiven x and t, the function ρ(x, ·, t) describes the variation of density within the cell (micro-structure), this

85

corresponding to density variations due to the fact that the cell is composed of two materials - say water androck. The small-scale variable ξ therefore cares for the description of density variations on the fine scale.

What about the dependence of the two-scale function ψ(x,ξ, t) on x? If we take the original (and possiblyvery fast varying density field ρε(x, t)) and define an average value over the cell centered at x by

⟨ρ⟩(x, t) def= 1|Ωcell|

∫Ωcell(x)

ρε(x′, t) dx′ , (7.65)

we get

⟨ρ⟩(x, t)= 1|Ωcell|

∫Ωcell(x)

ρ(x,ξ, t)dξ , (7.66)

which is the sought interpretation - as expected the dependence of ρ(x,ξ, t) on x captures the “slow” (aver-aged) spatial dependence of the density.

So adopting the two-scale separation (7.63), the asymptotic expansion (7.62) takes the form

ψε(x, t)=∞∑

i=0εiψ(i)(x,ξ, t) . (7.67)

In view of (7.63), we thus obtain the following identity:

∇xψε(x, t)=(∇x + 1

ε∇ξ

)ψ(x,ξ, t) , (7.68)

the first part of the operator measuring the changes in the macroscopic coordinate x, the second measuringvariations at the microscopic coordinate ξ.

Coming back to our Stokes flow problem, the two-scale asymptotic expansions of velocity vε and pressurepε read

vε(x, t)= v(0)(x,ξ, t)+εv(1)(x,ξ, t)+ε2v(2)(x,ξ, t)+ . . . (7.69)

pε(x, t)= p(0)(x,ξ, t)+εp(1)(x,ξ, t)+ε2 p(2)(x,ξ, t)+ . . . (7.70)

Plugging these expansions into the mass balance (7.61a) yields

divx(v(0) +εv(1) +ε2v(2) + . . .

)+ 1ε

divξ(v(0) +εv(1) +ε2v(2) + . . .

)= 0 , (7.71a)

and, similarly, the balance of linear momentum (7.61b) becomes

0=−(∇xx+ 1

ε∇ξ

)(p(0) +εp(1) +ε2 p(2) + . . . )+ε2

(∇x + 1

ε∇ξ

)·(∇x + 1

ε∇ξ

)(v(0) +εv(1) +ε2v(2) + . . .

). (7.71b)

Both the above equations were formulated in Ω f , so now in terms of the two-scale expansion, they hold inΩ×Ω f

cell. The boundary condition (7.61c) becomes

0= v(0)(x,ξ, t)+εv(1)(x,ξ, t)+ε2v(2)(x,ξ, t)+ . . . in Ω×∂Ω fcell . (7.71c)

The idea of formal asymptotic expansion is now to gather in each balance equation the terms multiplied bythe same power of ε and write down a hierarchy of corresponding sub-problems. The justification of such ahierarchy is within homogenization technique ensured by a (here strictly formal) limit procedure limε→0+.

Looking at the leading-order terms standing at ε−1 in the mass balance (7.71a), we obtain:

divξv(0) = 0 . (7.72)

In the momentum balance (7.71b) the leading order term (ε−1) yields

∇ξp(0) = 0 , (7.73)

which only tells us that p(0)(x,ξ, t)= p(0)(x, t) which is valuable but not sufficient information. Let us inspectthe terms at ε0:

−∇x p(0) −∇ξp(1) +∆ξv(0) = 0 , in Ω×Ω fcell . (7.74)

86

We observe that we can rewrite the problem at this level for each fixed x as a Stokes system (in variables ξ, t)on the given cell Ωcell(x) for (p(1),v(0)) under known p(0)(x, t).

divξv(0) = 0 in Ω fcell (7.75a)

−∇ξp(1) +∆ξv(0) +∇x p(0) = 0 in Ω fcell (7.75b)

and the boundary condition (7.61c) becomes

v(0) = 0 at ∂Ω fcell . (7.75c)

Note that the term ∇x p0 is a constant vector in the system thanks to independence on ξ, see (7.73). Thesolution of the so-called cell problem (7.75) can thus be found by solving three auxiliary cell sub-problems

divξv(0)α = 0 in Ω f

cell (7.76a)

−∇ξp(1)α +∆ξv(0)

α +eα = 0 in Ω fcell (7.76b)

v(0) = 0 at ∂Ω fcell . (7.76c)

for α= 1,2,3, since then by linearity, we get the sought solution of (7.75) in the form

v(0)(x,ξ, t)=−3∑

α=1

∂p(0)(x, t)∂xα

v(0)α (x,ξ, t) . (7.77)

Denoting the volume average over the cell centered at x by ⟨v(0)⟩(x, t):

⟨v(0)⟩(x, t) def= 1|Ωcell

∫Ωcell(x)

v(0)(x,ξ, t)dξ , (7.78)

we get

⟨v(0)⟩(x, t)=−3∑

α=1⟨v(0)

α ⟩(x, t)∂p(0)

∂xα(x, t) . (7.79)

We obtained Darcy’s law as a relation between the leading-order terms of the two-scale asymptotic expansionsof velocity and pressure:

⟨v(0)⟩(x, t)=−k(φ,x, t)µ fφ

∇x p(0)(x, t) . (7.80)

where the (in general) tensorial permeability k is defined as

k(φ,x, t) def= µ fφ3∑

α=1⟨v(0)

α ⟩(x, t)⊗eα . (7.81)

Note that we obtained characterization of the permeability in terms of (averaged) solution of the cell-problem,which is computable (e.g by numerical simulation), provided that we explicitly describe the geometry of thecell problem. This is the strongest point of the homogenization technique - the possibility to evaluate the“upscaled” material parameters (such as permeability) by solving some associated cell problem and analysingits solution.

Remark 31. The physically weak point of the homogenization procedure is the limit procedure l imε→0+which we used to “justify” the validity of the hierarchy of subproblems at different orders of ε. In reality, themicroscale ε may be small, but it is certainly fixed and does not converge to zero.

7.6 Generalizations of Darcy’s law: models of Brinkman and Forchheimer

A large number of various generalizations of Darcy’s model exist in the literature. Let us briefly mention twoimportant examples:

87

• Brinkmann’s model deals with calculation of a viscous force exerted between a flowing fluid anda dense swarm of spherical particles immersed in the fluid (Brinkman, 1949). It is considered as amodification of the Stokes problem in which the spherical particle with radius a is immersed in aporous medium. The local force balance then reads

∇p =−µ f

kvr +µ′∆vr , (7.82)

where vr is the relative “mean rate of flow”, relative mean velocity (i.e. volume-averaged fluid velocityin the frame connected with the solid particle) and µ′ generally differs from the fluid viscosity µ f .Brinkman derived analytical solution of the problem (7.82) and the corresponding drag force in theform

fBrinkmanD = 6πµ′v∞

r a(1+λa+ λ2a2

3

), (7.83)

i.e. as a correction of the Stokes drag formula, where λ=√

µ fkµ′ .

In our framework the generalization (7.82) is easily accessible as well. Let us take, for instance, thederivation of Darcy’s law from section 7.5.1, i.e. one based on the reduction of the momentum balancefor fluid in a class II binary mixture, and let relax the assumption adopted that on the form of the fluidpartial Cauchy stress. I.e. let us instead of (7.46) assume

T f =φ(−pI+2µ fD(v f )

), (7.84)

i.e. φ-weighted Cauchy stress of an incompressible Newtonian fluid. Then following the same procedureas in section 7.5.1, we arrive at the following extension of Darcy’s law (after adding also the contributionof gravity force):

α1v f =−φ(∇p−ρmf g)+φµ f∆v f . (7.85)

Note that we have identified in section 7.5.2 the drag coefficient as α1 = µ fk0

, see eq (7.56) so, comparingwith (7.82), we have indeed obtained Brinkmann’s form of Darcy’s law, where Brinkmann’s “mean ve-locity of flow” is now φv f , provided that we consider constant porosity φ. So to summarize, Brinkmann’smodel has been recovered within the same framework as Darcy’s law, just by relaxing slightly the as-sumptions used while reducing the momentum balance in section 7.5.1, namely by keeping also theviscous part of the partial Cauchy stress tensor.

• Darcy-Forchheimer extension of Darcy’s law (Forchheimer, 1901) contains a correction term thatrepresents the effect of inertial forces, and can be written in the following form

φv f +kk1φ2ρ f

µ f|v f |v f =− k

µ f(∇p−ρm

f g) , (7.86)

where k1 is so-called inertial permeability.

This model can also be obtained within the framework of class II mixtures. We only need to replace thesimple linear drag formula by a nonlinear one, valid for higher Reynolds numbers, i.e. taking in (7.44)again just the first drag term, but with Oseen’s quadratic correction to the linear Stokes drag term,(see (7.22) and (7.23)), i.e. with:

I=−α1vr −α′1|vr|vr , (7.87)

where vr = v f −vs. For a fixed solid matrix, vr = v f and indeed, we obtain terms formally coincidingwith the Forchheimer formula (7.86):

α1vr +α′1|vr|vr =−φ(∇p−ρm

f g) (7.88)

This makes the relationship between the pressure gradient in the medium and the flux non-linear.

88

7.7 A thermodynamic framework for a mixture of two liquids

In this chapter, we make a little excursion to the realm of class III framework, in order to derive a thermo-dynamic framework for a mixture of two liquids. In fact, the final model will be a model belonging to classII, but its derivation will require to start from considering energy balance for the individual component andthen summing them up, so, effectively, we stay within class II. The derivation will rely on the procedure ofmaximization of rate of entropy production as formulated in Rajagopal and Srinivasa (2004).

Let us start from complete class III mixture description, i.e. let us consider the following system ofbalance equations

∂ρα

∂t+div(ραvα)= mα , α= 1, . . . , N , (7.89a)

∂(ραvα)∂t

+div(ραvα⊗vα)= divTα+ραbα+Iα+mαvα , α= 1, . . . , N , (7.89b)

∂(ραEα)∂t

+div(ραEαvα)= div(Tαvα−qα)+ραbα ·vα+ραrα+εα+Iα ·vα+mαEα , α= 1, . . . , N ,

(7.89c)

where Eα = eα+ 12 |vα|2, and the above relations are supplemented with the constraints on the interaction

terms

N∑α=1

mα = 0 , (7.90a)

N∑α=1

(Iα+mαvα)= 0 , (7.90b)

N∑α=1

(εα+Iα ·vα+mαEα)= 0 . (7.90c)

Multiplying (7.89b) by vα, we obtain after some manipulation (cf Sec. 5.6) the balance of kinetic energy:

∂

∂t

(ρα|vα|2

2

)+div

(ρα|vα|2

2vα

)= div(Tαvα)−Tα :∇vα+ραbα ·vα+Iα ·vα+mα

|vα|22

, (7.91)

Subtracting this balance from (7.89c), we get reduced balance of energy in the form of internal energy balance

∂(ραeα)∂t

+div(ραeαvα)=−divqα+ραrα+Tα :∇vα+εα+mαeα , (7.92)

summing over α and using constraints (7.90), we arrive at

∂

∂t

(N∑α=1

ραeα

)+div

(N∑α=1

ραeαvα

)=−div

(N∑α=1

qα

)+

N∑α=1

ραrα+N∑α=1

Tα :∇vα+N∑α=1

(εα+mαeα) , (7.93)

which can be recast to the form (cf. eq. 5.88)

∂

∂t

(N∑α=1

ραeα

)+div

(N∑α=1

ραeαv

)+div

(N∑α=1

ραeαuα

)=−div

(N∑α=1

qα

)+

N∑α=1

ραrα+N∑α=1

Tα :∇vα

−N∑α=1

Iα ·uα−N∑α=1

mα|uα|2

2, (7.94)

Let us now consider a binary mixture, i.e. take N = 2. Since by definition of the diffusive velocities uα itholds

ρ1u1 +ρ2u2 = 0 , (7.95)

we havev1 −v2 =u1 −u2 =−

(ρ2

ρ1+1

)u2 =

(1+ ρ1

ρ2

)u1 , (7.96)

89

which implies

u1 = ρ2

ρ(v1 −v2)= (1− c)(v1 −v2) , (7.97)

u2 =−ρ1

ρ(v1 −v2)=−c(v1 −v2) , (7.98)

where c := c1. Consequently, it holds

N∑α=1

ραeαuα = ρc(1−c)(e1−e2)︸ ︷︷ ︸=:E12

(v1−v2) . (7.99)

Let us define

qI :=N∑α=1

qα , ρr I :=N∑α=1

ραrα , ρe I :=N∑α=1

ραeα , (7.100)

we can rewrite the internal energy balance as follows

∂

∂t(ρe I )+div

(ρe Iv

)=−divqI +ρr I +m1 divv1 +m2 divv2 +T1 :Dd1 +T2 :Dd

2 −∇E12 · (v1−v2)

−E12(divv1−divv2)− ((1−c)I1 − cI2) · (v1−v2)−(m1

2(1−c)2 + m2

2c2

)(v1−v2)2︸ ︷︷ ︸

= m12 (1−2c)(v1−v2)2

, (7.101)

and so we can write

ρ e I =−divqI +ρr I + (m1−E12)divv1 + (m2+E12)divv2 +T1 :Dd1 +T2 :Dd

2

−((1−c)I1 − cI2 +∇E12 + m

2(1−2c)(v1−v2)

)· (v1−v2) , (7.102)

where m := m1. Let us assume that both components have the same temperature ϑ = ϑ1 = ϑ2. Let usdistinguish the compressible and incompressible cases.

7.7.1 Compressible case

Let us assume that the (internal) Helmholtz free energy of the mixture is simply

ΨI = ΨI (ϑ,ρ, c)= cΨ1(ϑ,ρ, c)+ (1−c)Ψ2(ϑ,ρ, c) , (7.103)

where Ψ1, Ψ2 are the partial Helmholtz free energies (Legendre transforms of e1, e2, respectively). Let usdefine

ηI :=−∂ΨI

∂ϑ, (7.104)

and let us assume η= ηI . Then we get

ρΨI = ρ ˙(e I −ϑη

)= ρ e I −ρηϑ−ρϑη , (7.105)

and also, by definition,

ρΨI = ρ(∂ΨI

∂ϑϑ+ ∂ΨI

∂ρρ+ ∂ΨI

∂cc)

, (7.106)

comparing these two relations, and using (7.104) together with mass balances in the ρ-c form (cf. FIXME)

ρϑη= ρ e I + pdivv−µ(m−divj) , (7.107)

where we defined as usually the thermodynamic pressure p and the chemical potential µ as

p := ρ2 ∂ΨI

∂ρ, µ := ∂ΨI

∂c, (7.108)

90

and where the diffusive flux j can be written as

j= ρcu1 = ρc(1−c)(v1−v2) . (7.109)

Let us plug-in the internal energy balance (7.101), and the identity

divv= div(cv1+(1−c)v2)= cdivv1 + (1−c)divv2 +∇c · (v1−v2) , (7.110)

we obtain

ρϑη=−divqI +ρr I + (m1−E12+cp)divv1 + (m2+E12+(1−c)p)divv2 +T1 :Dd1 +T2 :Dd

2 +div(µj)−µm

−((1−c)I1 − cI2 +∇E12 + m

2(1−2c)(v1−v2)− p∇c+ρc(1−c)∇µ

)· (v1−v2) . (7.111)

Dividing by ϑ, we recast the above equation into the form of entropy balance:

ρη+div(qI −µjϑ

)+ ρr I

ϑ

= 1ϑ

(m1−E12+cp)divv1 + (m2+E12+(1−c)p)divv2 +T1 :Dd

1 +T2 :Dd2 −µm− q−µj

ϑ·∇ϑ

−((1−c)I1 − cI2 +∇E12 + m

2(1−2c)(v1−v2)− p∇c+ρc(1−c)∇µ

)· (v1−v2)

, (7.112)

from where we interpret the terms contributing to entropy change as entropy flux, entropy supply and theterm in ...

ϑas entropy production ξ

ϑ. Let us now assume slight extension of the class-I formula for entropy

production in a binary mixture of heat conducting and chemically-reacting linear fluids, the extension beinginclusion of one more term corresponding to class-II feature – dissipation due to drag force between the twofluids, i.e. let us postulate

ξ= ξ= 3λ+2ν3

(divv)2 +2ν‖Dd‖2 +κ|∇ϑ|2 +βµ2 +α|v1−v2|2 , (7.113)

where3λ+2ν≥ 0,ν≥ 0,κ≥ 0,β≥ 0,α≥ 0 , (7.114)

which ensure non-negativity of the entropy production in accord with the second law of thermodynamics.Note that we consider the very simplest setting, piece-wise quadratic form (even for the chemical reaction),which as we know corresponds to piece-wise linear closure relations between the associated thermodynamicfluxes and affinities in the classical case. Let us now identify the closure relations by employing the principleof maximization of entropy production (PMEP), i.e. let us define the Lagrange function and let us nowmaximize with respect to the “class-II” affinities, i.e. let us solve

maxDd

1 ,Dd2 ,divv1,divv2,v1−v2,∇ϑ,µ

L := ξ+ l1(ξ− . . . ) . (7.115)

Let us first perform some auxiliary computations Since

v= cv1 + (1−c)v2 , (7.116)

we have

∇v= c∇v1 + (1−c)∇v2 + (v1−v2)⊗∇c , (7.117a)

divv= cdivv1 + (1−c)divv2 + (v1−v2) ·∇c , (7.117b)

Dd = cDd1 + (1−c)Dd

2 + 12

((v1−v2)⊗∇c+∇c⊗ (v1−v2))− 13

(v1−v2) ·∇c I . (7.117c)

91

We get

∂ξ

∂Dd1

= 4νDd :∂Dd

∂Dd1

= 4νcDd , (7.118a)

∂ξ

∂Dd1

= 4νDd :∂Dd

∂Dd2

= 4ν(1−c)Dd , (7.118b)

∂ξ

∂divv1= 2

(3λ+2ν

3

)∂divv∂divv1

= 2c(

3λ+2ν3

)divv , (7.118c)

∂ξ

∂divv2= 2

(3λ+2ν

3

)∂divv∂divv2

= 2(1−c)(

3λ+2ν3

)divv , (7.118d)

∂ξ

∂(v1−v2)= 4νDd :

∂Dd

∂(v1−v2)+2

(3λ+2ν

3

)divv

∂divv∂(v1−v2)

+2α(v1−v2)

= 4νDd∇c+2(

3λ+2ν3

)divv∇c+2α(v1−v2) , (7.118e)

∂ξ

∂∇ϑ = 2κ∇ϑ , (7.118f)

∂ξ

∂µ= 2βµ . (7.118g)

Now the necessary conditions for maximum of the Lagrange function are vanishing first derivatives, whichimplies

Td1 = 1+l1

l14νcDd , (7.119a)

Td2 = 1+l1

l14ν(1−c)Dd , (7.119b)

m1−E12+cp = 1+l1

l12c

(3λ+2ν

3

)divv , (7.119c)

m2+E12+(1−c)p = 1+l1

l12(1−c)

(3λ+2ν

3

)divv , (7.119d)(

(1−c)I1 − cI2 +∇E12 + m2

(1−2c)(v1−v2)− p∇c+ρc(1−c)∇µ)=−1+l1

l14νDd∇c+2

(3λ+2ν

3

)divv∇c

+2α(v1−v2) , (7.119e)q−µjϑ

=−1+l1

l12κ∇ϑ , (7.119f)

m =−1+l1

l12βµ . (7.119g)

The value of the Lagrange multiplier l1 is found by multiplying the above relations by Dd1 , Dd

2 , divv1, divv2,v1−v2, ∇ϑ, and µ, respectively, one arrives at

1+l1

l12ξ= ξ =⇒ 1+l1

l1= 1

2=⇒ l1 =−2 . (7.120)

Using the relationI1 +I2 +m1v1 +m2v2 = 0, (7.121)

we can rewrite

(1−c)I1 − cI2 + m2

(1−2c)(v1−v2)= I+ m2

(v1−v2) , (7.122)

92

where we denoted I := I1. So finally, we arrive at the following closure relations

Td1 = 2νcDd , (7.123a)

Td2 = 2ν(1−c)Dd , (7.123b)

m1 =−cp+E12 + 3λ+2ν3

cdivv , (7.123c)

m2 =−(1−c)p−E12 + 3λ+2ν3

(1−c)divv , (7.123d)

I= p∇c−∇E12 −ρc(1−c)∇µ−2νDd∇c−(

3λ+2ν3

)divv∇c−α(v1−v2)− m

2(v1−v2) , (7.123e)

qI =−κ∇ϑ+µρc(1− c)(v1−v2) , (7.123f)

m =−βµ . (7.123g)

The first four equations imply the following form of the partial Cauchy stresses

T1 = (E12−cp) I+λcdivvI+2νcD , (7.124a)

T2 = (−E12−(1−c)p) I+λ(1−c)divvI+2ν(1−c)D . (7.124b)

DefiningTI :=T1 +T2 , (7.125)

we obtainTI =−pI+λdivv I+2νD , (7.126)

and we know, that TI differs from the mixture Cauchy stress only by a quadratic term∑Nα=1ραuα⊗uα, which

can be often ignored. Under such assumptions, we can see that the retrieved mixture Cauchy stress reducedto the classical compressible fluid. Note that in view of (7.126), the formula for the interaction force (7.127)can be rewritten as follows

I=−∇E12 −ρc(1−c)∇µ−(α+ m

2

)(v1−v2)−TI∇c . (7.127)

7.7.2 Incompressible case

For an incompressible mixture, we impose additional constraint

divv= 0 , (7.128)

into our thermodynamic framework. Let us impose analogous ansatz for the (internal) Helmholtz free energyΨI , only neglecting the dependence on mixture density (which is now constant provided it was constant atsome initial time, which we assume here), i.e. let us assume

ΨI = ΨI (ϑ, c)= cΨ1(ϑ, c)+ (1−c)Ψ2(ϑ, c) . (7.129)

Let us define as in the compressible case

ηI :=−∂ΨI

∂ϑ, (7.130)

and let us assume η= ηI . Then we get

ρΨI = ρ ˙(e I −ϑη

)= ρ e I −ρϑ−ρϑη , (7.131)

and also, by definition,

ρΨI = ρ(∂ΨI

∂ϑϑ+ ∂ΨI

∂cc)

, (7.132)

comparing these two relations, and using (7.104) together with mass balances in the ρ-c form (cf. FIXME)

ρϑη= ρ e I −µ(m−divj) , (7.133)

93

where we defined the chemical potential µ as before

µ := ∂ΨI

∂c, (7.134)

and where the diffusive flux j can be written as

j= ρcu1 = ρc(1−c)(v1−v2) . (7.135)

Let us plug-in the internal energy balance (7.101), we obtain

ρϑη=−divqI +ρr I + (m1−E12)divv1 + (m2+E12)divv2 +T1 :Dd1 +T2 :Dd

2 +div(µj)−µm

−((1−c)I1 − cI2 +∇E12 + m

2(1−2c)(v1−v2)+ρc(1−c)∇µ

)· (v1−v2) . (7.136)

Dividing by ϑ, we recast the above equation into the form of entropy balance:

ρη+div(qI −µjϑ

)+ ρr I

ϑ

= 1ϑ

(m1−E12)divv1 + (m2+E12)divv2 +T1 :Dd

1 +T2 :Dd2 +div(µj)−µm− qI −µj

ϑ

−((1−c)I1 − cI2 +∇E12 + m

2(1−2c)(v1−v2)+ρc(1−c)∇µ

)· (v1−v2)

, (7.137)

from where we interpret the terms contributing to entropy change as entropy flux, entropy supply and theterm in ...

ϑas entropy production ξ

ϑ. Let us now assume slight extension of the class-I formula for entropy

production in a binary mixture of heat conducting and chemically-reacting linear fluids, which is moreoverincompressible, again extended by an inclusion of dissipation due to drag force between the two fluids, i.e.let us postulate

ξ= ξ= 2ν‖Dd‖2 +κ|∇ϑ|2 +βµ2 +α|v1−v2|2 , (7.138)

whereν≥ 0,κ≥ 0,β≥ 0,α≥ 0 , (7.139)

which ensures the non-negativity of entropy production, in accord with the second law of thermodynamics.Let us now identify the closure relations by employing the principle of maximization of entropy production(PMEP), but now with additional constrain due to the incompressibility assumption. Let us perform againmaximization with respect to the “class-II” affinities, i.e. let us solve

maxDd

1 ,Dd2 ,divv1,divv2,v1−v2,∇ϑ,µ

L := ξ+ l1(ξ− . . . )+ l2 divv . (7.140)

Employing the identity

divv= div(cv1+(1−c)v2)= cdivv1 + (1−c)divv2 +∇c · (v1−v2) , (7.141)

we can proceed as in the compressible case and write down the set of necessary optimality conditions:

Td1 = 1+l1

l14νcDd , (7.142a)

Td2 = 1+l1

l14ν(1−c)Dd , (7.142b)

m1−E12 =− l2

l1c , (7.142c)

m2+E12 =− l2

l1(1−c) , (7.142d)(

(1−c)I1 − cI2 +∇E12 + m2

(1−2c)(v1−v2)+ρc(1−c)∇µ)=−1+l1

l14νDd∇c+2α(v1−v2) + l2

l1∇c (7.142e)

qI−µjϑ

=−1+l1

l12κ∇ϑ , (7.142f)

m =−1+l1

l12βµ . (7.142g)

94

The value of the Lagrange multiplier l1 is found by multiplying the above relations by Dd1 , Dd

2 , divv1, divv2,v1−v2, ∇ϑ, and µ, respectively, and summing these relations together, one arrives at

1+l1

l12ξ= ξ+ l2

l1(cdivv1+(1−c)divv2+(v1−v2) ·∇c)︸ ︷︷ ︸

=divv=0

=⇒ 1+l1

l1= 1

2=⇒ l1 =−2 . (7.143)

Summing (7.142c ) and (7.142d) yields

m :=m1 +m2 =− l2

l1. (7.144)

Using the relationI1 +I2 +m1v1 +m2v2 = 0 , (7.145)

we can rewrite

(1−c)I1 − cI2 + m2

(1−2c)(v1−v2)= I+ m2

(v1−v2) , (7.146)

where we denoted I := I1. So finally, we arrive at the following closure relations

Td1 = 2νcDd , (7.147a)

Td2 = 2ν(1−c)Dd , (7.147b)

m1 = cm+E12, (7.147c)

m2 =−(1−c)m−E12 , (7.147d)

I=−∇E12 −ρc(1−c)∇µ−2νDd∇c−(

3λ+2ν3

)divv∇c−α(v1−v2)− m

2(v1−v2) , (7.147e)

qI =−κ∇ϑ+µρc(1− c)(v1−v2) , (7.147f)

m =−βµ . (7.147g)

The first four equations imply the following form of the partial Cauchy stresses

T1 = (E12+cm) I+2νcD , (7.148a)

T2 = (−E12+(1−c)m) I+2ν(1−c)D . (7.148b)

Defining as beforeTI :=T1 +T2 , (7.149)

we obtainTI =mI+2νD , (7.150)

and we know, that TI differs from the mixture Cauchy stress only by a quadratic term∑Nα=1ραuα⊗uα, which

can be often ignored. Under such assumptions, we can see that the retrieved mixture Cauchy stress reducedto the classical incompressible fluid. Note that in view of (7.150), the formula for the interaction force (7.151)can be rewritten as follows

I=−∇E12 −ρc(1−c)∇µ−(α+ m

2

)(v1−v2)−TI∇c , (7.151)

which in this form coincides with (7.127)

8 Multiphase continuum thermodynamics

8.1 Introduction

In this chapter we will introduce yet another viewpoint leading to continuum mixture theory, taking a slightlydifferent perspective. Let us consider a situation of a multicomponent material as in Fig.8. Looking at thematerial from distance, it appears as homogeneous, but inspecting it closer by zooming to a sufficiently finescale, we find out that it is composed of distinct subregions, each having its own composition and physical

95

α= 1α= 2

Figure 8: Basic idea behind the concept of multiphase continuum mechanics. We assume that at sufficientlyfine scale, we can distinguish individual phases of the material, which themselves on this scale can be de-scribed by a single-or-multicomponent mixture theory.

properties. These distinguishable states of the material will be denoted as phases in this chapter and suchcomposite material is called a multiphase material. Now each of the phases may be either a single-componentmaterial, or it can be a mixture of several components in the sense in which we understood the word mixtureuntil now. So the main difference between a mixture and a multiphase material is in the presence of some(small but still not microscopic) spatial scale, at which, we distinguish different states of the material.

Facing a task of describing the properties of such a composite material, we could either zoom in andformulate on each of the subdomains mixture equations of appropriate type and look for suitable transitionconditions between the phases and this is in fact what we will do. But staying at that detailed level of de-scription is often not necessary and mostly even counterproductive and most importantly basically impossiblesince the information about the internal structure of the material and thus of the geometrical organisationof the various phases of the composite is out of our reach. A natural way to deal with this lack of informationis to define some averaging procedure, which will smear out the exact geometrical structure, and will leaveus with balance equations for the smeared out quantitites. This way we end up, as we shall see, again withclassical mixture balance equations. Since the starting point for the derivation is a finer description of thematerial (at the mesoscale with distinguishable phases), we expect to obtain some extra information by theend of the day. This is indeed so, in particular, we will gain better insight into the interaction terms betweenthe phases - they will be (averaged) interface interaction mechanisms.

8.2 General balance laws under presence of discontinuities - single continuum

In this chapter, we will first generalize the important analytical tools we have been using in deriving thelocal form of our balance laws from their integral representations: the Reynolds’ transport theorem andGauss theorem. So far, we have always tacitly assumed that the balanced quantity is distributed in thebulk continuously or even with higher smoothness so that all the relations written in the form of (strong)PDEs are meaningful. In many situations, it is necessary to relax such assumption and to include materialand immaterial interfaces representing discontinuities into the description. Moreover, it may also happen,that the properties of the thin contact zone between two bulk materials are in some sense singular, leadingto substantial concentration of some bulk quantities into this zone and consequently, to distinct transportphenomena inside this interfacial zone. In such situations, it is often convenient to represent the zone as an−1 dimensional manifold at which we introduce surface counterparts of all considered bulk fields. In such acase, one has to reinspect carefully the formulation of general balance laws, including these new terms, andarrive, as we shall see, to a combination of bulk balance laws in "regular" points outside the discontinuitiesand surface/interfacial balance laws at the discontinuities/boundaries. Let us immediately note that such aviewpoint gives a nice and natural way of dealing with boundary conditions, since they can be interpretedand treated as (discontinuous) interfaces between the inner bulk of the domain and the "rest of the world".

Let us consider a single component body in the framework of classical continuum mechanics. Let Ψ(Ω)denote a generic additive quantity (such as mass, momentum, energy, etc. contained in a given controlmaterial volume Ω). Let us consider the general integral balance law for such quantity, i.e. evaluate the rateof change of Ψ, which is considered to result from three generic independent processes (i) the flux throughboundaries of the domain ψF , (ii) through internal production ψP and (iii) as a result of outer supply ψS ,

96

Figure 9: Control volume Ω and discontinuity Γ.

i.e the general balance is postulated in the form

ddtΨ=−ψF +ψP +ψS . (8.1)

Quantity Ψ is assumed to be composed of a bulk contribution and a surface contribution localized at aninternal discontinuity (interface) Γ, and both the bulk and the surface contributions are assumed to be rep-resentable by corresponding densities. We thus assume to have the following representation:

• Additive quantity Ψ:

Ψ=∫Ω+∪Ω−

ψΩ dx+∫ΓψΓ dS (8.2)

• Flux ψF :

ψF =∫

S+∪S−ψΦΩ ·n dS+

∫∂Γ

ψΦΓ ·νdl (8.3)

where n is the outer unit boundary to ∂Ω and ν is a outer unit normal to the line ∂Γ (lying in Γ).

• Production ψP :

ψP =∫Ω+∪Ω−

ψξΩ dx+∫Γ

ψξΓ dS (8.4)

• Supply ψS

ψS =∫Ω+∪Ω−

ψΣΩ dx+∫Γ

ψΣΓ dS (8.5)

We will distinguish between the bulk contribution ψΩ and its surface counterpart ψΓ, in the sense thatin general ψΩ|Γ 6= ψΓ, that is, the trace of the bulk quantity need not coincide with the corresponding sur-face quantity. This assumption corresponds to the fact that interfaces are in general zones where materialproperties may undergo abrupt changes - the transition layers being typically very thin, it is reasonable totreat them as n−1 dimensional manifolds, n being the dimension of the “bulk” space, and the correspond-ing averaged (over the thickness of the layer) bulk quantities are then taken as the independent surfacecounterparts.

Remark 32. Interpretation of surface quantities - example - surface density ρΓ. Let us provide at least oneexplicit interpretation of the concept of surface quantity - density. Consider a body composed of two parts Ω+

and Ω−, separated by an interfacial region ΩΓ as in Fig. 10 (left). Let the density in Ω+ be a constant ρ+ and

97

Figure 10

in Ω− constant ρ− and let us assume that in the interfacial region ΩΓ, density varies smoothly between thetwo values. Let us introduce the so called excess density ρE

Γ defined the interfacial ΩΓ region ρEΓ

ρEΓ

def=ρ−ρ+ in Ω+

Γρ−ρ− in Ω−

Γ

(8.6)

The surface density ρΓ is now defined by an assumption∫Γ∩V

ρΓdS =∫ΩΓ∩V

ρEΓ dx (8.7)

for any control volume V of the type as in Fig. 10 (left). It turns out (see e.g. Slattery et al., 2007) that ρΓ canbe then expressed as follows

ρΓ(xΓ) .=∫ +ε

−ερEΓ (xΓ,ξ)(1−κ1ξ)(1−κ2ξ)dξ . (8.8)

Here xΓ denotes the position on Γ, ξ is the position along the local normal to Γ originating from xΓ, and κ1and κ2 are principal curvatures of Γ at xΓ. Often the interfacial region is thin enough that it holds |κ1ξ|¿ 1and |κ2ξ|¿ 1 and we obtain

ρΓ(xΓ) .=∫ +ε

−ερEΓ (xΓ,ξ)dξ . (8.9)

Surface density ρΓ is then just the integrated excess density ρEΓ along the interfacial region (integration

taking place along local normal direction). The original density distribution (Fig.10 left) is thus replaced bya density distribution as in Fig.10 on the right. The two bulk regions Ω+ and Ω− now extend all the way tothe surface Γ and the density values in the Ω+

Γ and Ω−Γ regions are defined as interpolants of the values from

Ω+ and Ω−, i.e. in our case ρ+ and ρ−, respectively, and all the density variation in the original interfacialregionΩΓ is now localized on the surface Γ and expressed through the surface density ρΓ. For further details,see Slattery et al. (2007).

In order to obtain the local form of the general integral balance, we will employ the following theorems.For their proofs, see, for instance Slattery et al. (2007).

Theorem 8.1 (Generalized Reynold’s transport theorem). Let Ω(t) be a material domain composed oftwo subdomains Ω+(t) and Ω−(t) separated by a non-material surface (e.g. a phase-change front) Γ(t), whichmoves in the normal direction with velocity v⊥Γ as in Fig.9. Then it holds

ddt

∫Ω(t)+∪Ω−(t)

ψΩ dx =∫Ω(t)+∪Ω−(t)

∂ψΩ

∂t+div

(ψΩ⊗v

)dx+

∫Γ(t)

[ψΩ⊗ (v−v⊥ΓnΓ)

] ·nΓ dS (8.10)

Theorem 8.2. Generalized Gauss theorem∫Ω(t)+∪Ω(t)−

divψΦΩdx =∫

S+(t)∪S−(t)

ψΦΩ ·nΓdS−∫Γ(t)

[ψΦΩ

] ·nΓ dS. (8.11)

98

Theorem 8.3 (Surface transport theorem).

ddt

∫Γ(t)

ψΓ dS =∫Γ(t)

DΓψΓ

Dt+ψΓdivΓvΓ dS (8.12)

=∫Γ(t)

DΓψΓ

Dt+ψΓdivΓv‖

Γ−2KMψΓv⊥Γ dS (8.13)

Theorem 8.4 (Surface Gauss theorem). Let tΓ be a tangent surface vector field. Then it holds∫Γ(t)

divΓ tΓdS =∫∂Γ(t)

tΓ ·n∂Γ dl , (8.14)

where n∂Γ is the outer normal to ∂Γ(t) - considered as a tangent vector in Γ(t).

In particular, this implies that ∫Γ(t)

divΓψΓv‖ΓdS =

∫∂ΓψΓv‖

Γ ·n∂Γ dl . (8.15)

By using the generalized Reynolds’ transport theorem, and the surface transport theorem, together with thebulk and surface Gauss’ theorem, under an additional assumption of sufficient smoothness of all the involvedquantities, we can derive the following local form of the integral balances (Slattery et al., 2007)

• In regular points in the bulk, i.e. in Ω\Γ:

∂ψΩ

∂t+div

(ψΦΩ+ψΩv

)−ψξΩ−ψΣΩ = 0 , (8.16)

where v is the material velocity.

• At singular points at discontinuity Γ

DΓψΓ

Dt+ψΓdivΓv‖

Γ−2KMψΓv⊥Γ +divΓψΦΓ−ψξΓ−ψΣΓ =−[ψΦΩ+ψΩ(v−v⊥ΓnΓ)

] ·nΓ , (8.17)

where vΓ is the surface velocity, v‖Γ is its projection to the surface Γ, and v⊥Γ is its component normal to Γ, KM

is the mean curvature of the surface, and [] denotes the jump of the bracketed quantity across the surface Γdefined by [·] def= ·+−·− where the sign convention is given by the normal vector nΓ oriented from − to + side ofthe surface. The operator DΓ

Dt denotes the surface material time derivative and divΓ is the surface divergence(see, e.g. Slattery et al., 2007).

Let us explicitly state all the surface balance laws corresponding to (8.17), where for ψΓ, we take thefields of density, linear and angular momentum (for non-polar continuum), energy and entropy:

• Surface balance of mass

DΓρΓ

Dt+ρΓ

(divΓv‖

Γ−2KMv‖Γ

)=−[ρ(v−v⊥ΓnΓ)] ·nΓ , (8.18)

where ρΓ is the surface mass density.

Surface balance of linear momentum

DΓ(ρΓvΓ)Dt

+ρΓvΓ(divΓv‖

Γ−2KMv‖Γ

)−divΓTΓ−ρΓbΓ =−[ρv⊗ (v−v⊥ΓnΓ)−T] ·nΓ , (8.19)

where TΓ is the surface Cauchy stress tensor and bΓ is the surface specific force.

Surface balance of angular momentum

TΓ =TTΓ , (8.20)

99

Surface energy balanceThe surface total energy balance reads:

DΓ

(ρΓEΓ

)Dt

+ρΓEΓ(divΓv‖

Γ−2KMv‖Γ

)+divΓqΓ+divΓ(TΓvΓ)−ρΓrΓ−ρΓbΓ ·vΓ

=−[ρE(v−v⊥ΓnΓ)−v ·T+q] ·nΓ , (8.21)

whereEΓ

def= eΓ+ 12|vΓ|2 (8.22)

is the surface total energy composed of the internal surface energy eΓ and the surface kinetic energy.

As in the bulk, we can obtain a reduced form of this balance by subtracting the surface kinetic energybalance (obtained from the surface momentum balance by scalar multiplication by the surface velocityvΓ). The reduced form of surface energy balance - surface internal energy balance then reads:

DΓ

(ρΓeΓ

)Dt

+ρΓeΓ(divΓv‖

Γ−2KMv‖Γ

)+divΓqΓ−TΓ :∇ΓvΓ−ρΓrΓ−ρΓbΓ ·vΓ

=−[ρ

(e+ 1

2|v−vΓ|2

)(v−v⊥ΓnΓ)− (v−vΓ) ·T+q

]·nΓ , (8.23)

• Surface entropy balance and the second law of thermodynamics

DΓ(ρΓηΓ)Dt

+ρΓηΓ(divΓv‖

Γ−2KMv‖Γ

)+divΓ(qη)Γ−ρΓ(rη)Γ+ [ρη(v−v⊥ΓnΓ)+qη] ·nΓ = ξΓ , (8.24)

where ηΓ is the surface (specific) entropy, (qη)Γ is the surface entropy flux, (rη)Γ is the surface entropysupply and, finally ξΓ is the surface entropy production , which must be non-negative according to thesecond law of thermodynamics:

ξΓ ≥ 0 . (8.25)

8.3 Jump conditions

A particular situation arises, if the singular surface does not have any structure, and no accumulation ofmass, momentum, energy or entropy occurs there. In such a case the single-component and multi-componentgeneral balance laws at singular surfaces (8.17) and (8.49) take the form of so-called jump conditions ex-pressing continuity of the normal component of bulk convective and non-convective fluxes. Since these jumpconditions are of particular importance per se, let us write them down explicitly. Recalling the definitions ofψΩ and corresponding non-convective fluxes ψΦΩ for the individual balances, we obtain the following set ofjump conditions associated to the particular balance laws:[

ψΦΩ+ψΩ(v−v⊥ΓnΓ)] ·nΓ = 0 (8.26)

• Mass jump condition [ρ(v−v⊥ΓnΓ)

] ·nΓ = 0 . (8.27)

This condition expresses the simple fact of continuity of the normal mass flux (i.e. conservation of mass)through the discontinuity of density (e.g. a phase change front). Indeed the quantity (v−v⊥ΓnΓ) ·nΓ isthe relative motion of the material with respect to the moving discontinuity, multiplied by density, weobtain the mass flux. What mass flows “in”, must flow “out” is what the condition tells us.

• Momentum jump condition [T−ρv⊗ (v−v⊥ΓnΓ)

]nΓ = 0 . (8.28)

If the interface is material, i.e. if v± ·nΓ = v⊥Γ , one obtains the continuity of traction condition

[T]nΓ = 0, (8.29)

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but not that for non-material surfaces (e.g. phase-change fronts), traction force need not be continuousany more due to the momentum transfer accross the interface. Let us inspect one more case, whichcan often be encountered in practice. This concern a situation, when in the general interfacial balancelaw (8.49), for linear momentum, we keep the surface stress tensor on the left-hand-side, i.e. we keepρvΦΓ =−TΓ. Moreover, assuming that the interface acts as an isotropic membrane, we assume that thesurface Cauchy stress is simply

TΓ =σIΓ ,

where σ is the surface tension and IΓ is the surface projection tensor (surface indentity), which isdefined as

IΓdef= I−nΓ⊗nΓ . (8.30)

Then a generalization of (8.28) is obtained:[T−ρv⊗ (v−v⊥ΓnΓ)

]nΓ = divΓ(σIΓ) . (8.31)

The right-hand side can be further expanded into two terms

divΓ(σIγ)=∇Γσ+σdivΓ IΓ =∇Γσ−2KMnΓ , (8.32)

where KM =−12 divΓnΓ is the mean curvature.

• Total energy jump condition[q−Tv+ρ

(e+ 1

2|v|2

)(v−v⊥ΓnΓ)

]·nΓ = 0 . (8.33)

• Entropy jump condition [qη+ρη(v−v⊥ΓnΓ)

] ·nΓ = 0 . (8.34)

8.4 A thermodynamic framework for boundary conditions

Let us demonstrate on a simple example how the thermodynamic framework that combines balance laws andsecond law of thermodynamics both in the bulk and at internal discontinuities can be used for thermodynamicderivation of boundary conditions.

The key idea is to adopt a viewpoint that boundary ∂Ω of a domain Ω is actually a discontinuity separat-ing Ω from its exterior. Thus in view of the previous chapters, the boundary ∂Ω becomes the discontinuityΓ, at which surface counterparts of bulk quantities can be defined, and for which balance laws can be formu-lated.

Example: Navier-Stokes-Fourier fluid Consider the simplest setting, in which the domain Ω is occupiedby a Newtonian heat-conducting fluid, and the boundary ∂Ω=Γ is really “just a boundary” in the sense thatnone of the bulk fields have any excess value at Γ i.e. all the surface fields are identically zero:

ρΓ ≡ 0,vΓ ≡ 0, eΓ ≡ 0,ηΓ ≡ 0, (8.35)

the same holds for the surface fluxes:

TΓ ≡ 0,qΓ ≡ 0, (qη)Γ ≡ 0 , (8.36)

and, finally, also the external surface supply terms are zero

bΓ = 0, rΓ = 0,(rη)Γ = 0 . (8.37)

The only term which we allow to be non-zero (and of course, non-negative), is the surface entropy productionξΓ ≥ 0. In the bulk, let us consider the standard procedure. i.e. take some fundamental thermodynamicpotential, for instance the bulk Helmoltz free energy

e−ϑη=ψ= ψ(ρ,ϑ) ,

101

take the material time derivative, and use the fact that ∂ψ∂ϑ

=−η, and ρ2 ∂ψ∂ρ

=−p. After plugging in the bulkinternal energy balance and the bulk mass balance, the standard calculation yields the bulk entropy balancein the form

ρη=−div(qϑ

)+ 1ϑ

((m+ p)divv+Td :Dd − q ·∇ϑ

ϑ

),

where the first term on the right-hand side is interpreted as the divergence of the bulk entropy flux and thesecond term as the bulk entropy production, i.e. we assume that

qη = qϑ

,

ξ= 1ϑ

((m+ p)divv+Td :Dd − q ·∇ϑ

ϑ

).

Piece-wise linear closures for the “fluxes” m+p, Td and q in terms of the “affinities” divv, Dd and ∇ϑϑ

, togetherwith the assumption of non-negativity of the rate of entropy production in order to comply with the secondlaw of thermodynamics, yield the classical Navier-Stokes-Fouries model

T=−3λ+2µ3

divv+2µD ,

q=−k∇ϑ ,

where 3λ+2µ≥ 0, µ≥ 0 and k ≥ 0.Let us now inspect the surface balances, assuming that the boundary is material and fixed, which implies

v ·nΓ = v⊥Γ = 0, where nΓ is taken as the outer unit normal (i.e. oriented outside from Ω). The bulk quantitiesinside Ω should thus be written with superscript -, which will omit, however.

The balances of momentum, energy and entropy read

[T] ·nΓ = 0 (8.38)

[q−v ·T] ·nΓ = 0 (8.39)

[qη] ·nΓ = ξΓ (8.40)

The surface balance of mass is satisfied trivially. The surface balance of entropy at Γ with the above assump-tions reads

0≤ ξΓ = [qη] ·nΓ =[qϑ

]·nΓ.

Let us employ the following relation for a jump of a product of two quantities

[ab]= ⟨a⟩[b]+ [a]⟨b⟩ where ⟨x⟩ def= 12

(x++ x−). (8.41)

We obtain0≤

⟨1ϑ

⟩[q] ·nΓ+

[1ϑ

]⟨q⟩ ·nΓ . (8.42)

In the first term, we substitute from the energy jump condition (8.39), expressing:

[q] ·nΓ = [v ·T]nΓ = [(v‖+v⊥nΓ︸ ︷︷ ︸=0

) ·T]nΓ = [v‖] ·TnΓ, (8.43)

where we employed the momentum jump condition, i.e. continuity of traction (8.38) (0 = [T] ·nΓ⇒T+ ·nΓ =T− ·nΓ) and we can thus take T ·nΓ out of the jump bracket. We also split the jump of velocity into normalcomponent (identically zero) and tangential part. Now the product [v‖] ·TΓ is frictional power - indeed, it isa product of the traction force at the interface TnΓ and slip velocity along the surface vslip

def= [v‖]. So finally,the surface entropy balance can be recast to the following form

0≤ ξΓ =⟨

1ϑ

⟩vslip ·TnΓ+⟨q⟩ ·nΓ

[1ϑ

]. (8.44)

102

The two dissipative mechanisms have clear physical meaning - the first term corresponds to mechanicaldissipation at Γ due to friction, the second one is transfer of heat across the interface. Postulating piece-wiselinear closures, yields the following model for two dissipative surface processes:

TnΓ =αΓvslip , (8.45)

⟨q⟩ ·nΓ = kΓ[

1ϑ

], (8.46)

where αΓ ≥ 0 and kΓ ≥ 0 in order to comply with the second law of thermodynamics. The first constitutiveclosure is the Navier-slip boundary condition, and the second term, which expresses the average normalheat flux across the interface in terms of temperature difference is sometimes called Kapitza condition.

8.5 General balance laws under presence of discontinuities - multicomponent contin-uum

The above results can be easily extended for multicomponent materials - mixtures. Assuming an N-componentmixture, we introduce index α= 1, . . . , N denoting the mixture components, and we can immediately formu-late the mixture counterparts to (8.16) and (8.17):

• In regular points in the bulk

∂ψαΩ

∂t+div

(ψαΦΩ+ψαΩvα

)−ψαξΩ−ψαΣΩ−ψαΠΩ = 0 α= 1, . . . , N , (8.47)

balancing the quantity ψαΩ with a corresponding flux ψαΦΩ, a production ψαξΩ, an outer supply ψαΣΩand as expected - a novel term ψαΠΩ corresponding to interaction with other components of the mixture.These interaction terms must satisfy the constraint

N∑α=1

ψαΠΩ = 0 , (8.48)

expressing the (local) conservation of ψαΩ during the exchange processes.

• At singular points at discontinuities

dψαΓ

dt+ψαΓdivΓv‖

αΓ−2KMψαΓv⊥Γ +divΓψαΦΓ−ψαξΓ−ψαΣΓ−ψαΠΓ =−[ψαΦΩ+ψαΩ(vα−v⊥ΓnΓ)

] ·nΓ ,

(8.49)balancing the surface quantity ψαΓ with a corresponding surface flux ψαΦΓ, a surface production ψαξΓ,an outer surface supply ψαΣΓ and surface interaction terms ψαΠΓ, which satisfy the constraint

N∑α=1

ψαΠΓ = 0 , (8.50)

expressing the (local) conservation of ψαΓ during the exchange processes.

Remark 33. Note that while we distinguish individual tangential velocities of the components v‖αΓ

α = 1, . . . , N, we only use one v⊥Γ being the (normal) velocity of the discontinuity Γ. The discontinuityis thus understood as being common to all mixture components. This assumption is without loss ofgenerality in fact, if we adopt the following viewpoint. Whenever any material property of any ofthe components undergoes a jump or concentration at some surface, we define such a surface as adiscontinuity for the whole mixture. The motion of such a surface is then given by the quantity v⊥Γ andwe keep in mind, that some quantities (e.g. properties corresponding to several mixture components)can be continuous across the surface.

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8.6 Framework of the multiphase continuum theory

8.7 Averaging theorems

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