me285 project mixtures

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ME 285 PROJECT: CONTINUUM THEORY OF MIXTURES

ILKER TEM IZER

December 24, 2004

Contents

1 Introduction 1

2 Balance Laws 32.1 Mass Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Linear Momentum Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Angular Momentum Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Second Law of Thermodynamics: Clasius-Duhem Inequality . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Discussion 113.1 Objectivity Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Interaction Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.4 Types of Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 Examples 134.1 Darcys Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Ficks Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Remark on Volume Additivity Constraint 15

6 Conclusion 16

1 Introduction

A mixture can be roughly dened as an aggregate which is composed of two or more materials. One can talk about, e.g., uid-uid, solid-uid, solid-solid mixtures. Flow of a uid through a porous solid, diffusion of gas thorougha solid, composites, tissue modelling are examples to topics where a mixture is to be analized. Massoudi [15] notesthat there are two types of mixtures: cases where one constituent dominates the behaviour and mass of the system,and cases where all constituents are equally important for the behavior of the mixture. The rst case is typied bydilute mixtures such as spraying of a uid through air (where air is the dominant material). On the other hand, theexamples given above to mixtures are dense mixtures . There are two continuum theories to model dense mixtures.One is the method of averaging , and the other is theory of mixtures . Averaging methods can be deterministic [16] orstatistical [18], but in all cases is based on the process of calculating effective properties of a statistically representativevolume element from the individually known behaviors of the constituents, and using the effective properties to writedown constitutive relations for the mixture. The method generally requires extensive calculation and therefore issupplemented by numerical methods such as the nite element method. This summary is on the continuum theoryof mixtures.

Consider a mixture that is enclosed in a volume R at some time t , as shown in Figure 1. If we take a small sampleof the mixture from a position x , we can observe that this sample consists of individual materials with knownbehaviour. The existence of each constituent is dense enough so that we average all properties for each contituent

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++

R

R

dV

dV dV dV

dV

m

Figure 1: Construction of the mixture theory.

separately, and assume they lled the same innitesimal volume separately. For instance, we could then refer tothe average density of a constituent at a point divided by the average density of the mixture at the same pointas the constituents concentration .Then, we assume that the inital innitesimal volume can be represented by theco-existence of these homogenized constituents in the same volume. In other words, if we have three constituents,, , we average their properties at an innitesimal volume containing the point x , and attach to this point not one particle, but one particle of each constituent. Accordingly, in the idealized conguration of the mixture R m , eachpoint x is assumed to consist of variables and properties associated with a particle of each constituent. Such anidealization obviously requires the mixture to be sufficiently dense. This approach is attributed to Truesdell [19]. Onecould alternatively think of each constituent to be represented by a different homogenized reference conguration.Such a setup is shown in Figure 2. Each constituent goes thourough a motion such that their motions overlap in aregion R m at time t.

Next, we would like to write down the equations, such as mass balance, that allow us to calculate and placerestrictions on quantities associated with the process that each constituent goes thorough. If we solve for eachconstituents equations using classical formulation with no extra terms, obviously we will not represent a mixture,but a set of materials that somehow ll the same space. For the idealization at hand to work, therefore, we needto take into account the interactions between various constituents when solving the equations for each constituent.Moreover, the classical balance laws should hold for the material enclosed in region R m , when it is taken as a blackbox. As Rajagopal [17] has stated, the modelling of the interactions between constituents is at the heart of thetheory of mixtures.

In this summary, we will state the equations that govern the motion of the mixture. As we do so, we will try togive a physical interpretation for all the terms that appear. The mixture is taken to be reacting . In other words,mass conversion among constituents is allowed. This mass conversion effect is really what complicates the equationsmore than anything else. At the end, we provide the derivation of some classical results, such as Ficks Law andDarcys Law. Among several review works and papers, the review articles by Atkin and Crains [1, 2] are the oneswe found most useful. The review works of Bowen [3] and Rajagopal [17] are also cited frequently.

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++

R 1

R 2

R 3

R

1

2

3

m

Figure 2: Representation by multiple reference congurations.

2 Balance Laws

To formulate the balance laws, we take a mixture with constituents. 1 At any point x R m , we take aninnitesimal volume element d V . Let d m be the innitesimal mass associated with constituent in this volumeelement. We dene:

=dm

dV (1)

, which represents the average density of the constituent in that region. It follows immediately that the averagedensity for the mixture is given by:

= dm

dV = dm

dV =

(2)

It should be clear that since d m = dV , and d V is the same for all constituents at a point x (in otherwords, since is the unit mass for constituent at that point), we can conduct all calculations for the mass-averagequantities associated with the mixture using the densities. Therefore we dene the barycentric velocity v of themixture by:

v =1

v (3)

, where v (as well as all other quantites to be introduced) should be understood to be the average value of thevelocity associated with tha particles of in an innitesimal volume d V about point x .

One can also dene c , the concentration with respect to mass of constituent at a point as:

c =

(4)

We dene an operator:

d

dt=

t

+

x jvj (5)

1 We use the convention of labelling constituents with Greek letters. Any denition without a superscript is meant to belong to themixture as a whole. Unless explicitly stated, we do not use the summation convention for Greek letters.

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Also, we easily see that

d

dt=

ddt

= t

+

x jvj (6)

Now, we can state the balance laws. We prefer to state the balance laws for the mixture as a whole and thenpostulate the constituent balance equations. This approach lets one see the transition and the reasoning for theterms that appear more easily. The denitions of the terms that appear as usual, unless otherwise noted. Finally, wewould like to make a note about the notation in Atkin and Craine [1]. The authors talk about the difference in theformulation of laws with respect to control volumes and material volumes. What they are really referring to is thedifference between open and closed systems. A discussion about the related concept of generalized transport theorem can be found in the book by Haupt [12]

2.1 Mass Balance

R m

V

Figure 3: A stationary control volume is used to state the equations.

First, we state the mass balance for the mixture as a whole. It is convenient to write all equations for a controlvolume V with boundary , in R m at any time t . One can alternatively write the equations for a control volume thatcoincides with R m and/or could consider a control volume with an arbitrary velocity. But it is customary to keepthe control volume stationary and more practical to take an arbitrary control volume enclosed in R m (for instance,when we want to analyze the ow of a uid through a porous solid). We state the mass balance for the material inV as follows:

t V

dv + (v n ) da = 0 (7)

, or, using equations 2 and 3, equivalently as

t V dv + (v n ) d a = 0 . (8)

To interpret the equations we write equation 7 again:

t V dv = (v n ) d a (9)

Equation 7 tells us that the rate of change of total mass enclosed in volume V is equal to the rate at which totalmass is convected into V through . From equation 8 we see that the equation for mass balance reduces to the formof mass balance for a single continuum.

Now, we will try to formulate the mass balance for a single constituent. The procedure we will apply is repeatedfor other balance laws, too. Namely, we will include some interaction terms as we separate each constituent fromequation 7. Therefore, we would expect some mass generation in V , which really is a conversion of mass from otherconstituents to, say, constituent through chemical reactions. Also, as other constituents move into V , they may bein the process of being converted to so that we need a ux term. We state the result as follows, using equation 9:

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t V dv = V dv (v n ) + (v n ) da (10)

, where by we mean the rate at which constituent is created from constituent , is the unit mass of

constituent just before becoming a part of constituent on the boundary , and v is the velocity of that constituentand . The summation is applied to all constituents except . Obviously, at a point, it could be the case that isbeing converted to and is being converted to . The terms dened are meant to include the nal conversioneffect. The eld has been introduced for purposes of discussion, since it provides a nice phyical insight to theproblem.

The fourth term of equation 10 may seem unnecessary but its existence can be justied as by the following

reasoning. The term appears because the mixture is taken to be reacting. Let, therefore,

be the unit mass of in a chemical reaction with at any time t . It will, of course, take some time for the totality of this quantity tobe converted to . By this term, we really mean the unit mass of that could possibly be involved in a chemical

reaction with in a small time interval t about t . Similarly, we dene

. Obviously, if we guessed the amount

of quantities involved in the reaction correctly, in the duration of t , the quantity

+

will remain constant.

Now, let be the portion of

that will certainly be converted to during an innitesimal time interval d t . If this quantity is on (the limit as we approach the boundary from the outside), it will surely be on + (the limitas we approach the boundary from inside V ) after d t ends and therefore be a part of V . If we do not considerthis transfer, there will be a mass increase of constituent in V that remains unaccounted for. In other words, wewill not be able to predict rate of changes of mass of any constituent correctly. This justies the presence of theterm. Also, with a similar interpretation, it would be possible to phsicallly relate the term to (for instance = d t).

In relation to the previous paragraph, we would also like to clarify the following point about equation 10. Thevelocity related to should obviously be v . But, it seems that once becomes a part of , it retains its velocityv (since in an innitesimal time d t , the change in v is innitesimal) and we get two different velocity values forconstituent at any point x + , namely v and v . But this is not so. We must not forget that our formulationis based on averaging of quantities related to constituents in innitesimal volumes around a point. As long as theprocess of mass conversion is smooth , the results of the averaging process will change smoothly. Therefore, thisaddition of with velocity v will cause an innitesimal change in the velocity of at that point. Therefore, therereally is no inconsistency in the process. This effect will be accounted for by the linear momentum balance.

We would like to clarify that this interaction by mass conversion is not to an unknown. Although one could try tomodel it, it could as just as well be a given quantity. But to maintain generality, we will provide a constitutive relationfor all interaction terms to be introduced. For instance, we can say that the rate at which constituent is beinggenerated (or, equally, destroyed) through chemical reactions with other constituents is equal to the concentrationof at that point times an arbitrary factor. Then, of course, we would have to place a restriction on the factors,which is to be explained shortly. Therefore, we simplify equation 11 as follows. We apply divergence theorem to thethird term on the right hand side and convert it to a volume integral. And, we dene:

V dv (v n ) da = V dv V div v dv=

V

( div v

)

cdv

= V c dv(11)

, so that c includes all mass conversion effects related to constituent and is more convinient to work with.(It makes more sense to provide the rate at which mass conversion takes place rather than a eld from which it willbe derived. This is reminds us of, e.g., the heat supply term in energy balance.) The c in c stands for chemical to remind us that when there are no chemical reactions, this term should disappear. We will use the same notation

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for similar terms that are to be introduced. Also, the that appears in this term is just to denote that it is arate term, and not to be understood as some sort of a material time derivative. We write the nal form of the massbalance for a constituent as

t V dv = V c dv (v n ) da . (12)

We apply divergence theorem to 12 and make use of 11 and get the point form under standard smoothness

assumptions:

t+ div( v ) = c (13)

, or

d

dt+ div(v ) = c . (14)

Also, we put 8 into point form as:

t

+ div( v ) = 0 (15)

,or

dd t

+ div(v ) = 0 (16)

Using equations 2 and 3 on 13, and comparing the result to 15, we see that an equivalent form of 16 is:

c = 0 (17)

, which is what one would expect since c is an interaction term.

2.2 Linear Momentum Balance

Like we did for balance of mass, we rst postulate the linear momentum balance for the mixture as a whole:

t V v dv = v (v n ) da + V b dv + t da . (18)

In this equation, b is the external body force acting on the particles of constituent , and t is the tractionvector due to interaction between the particles of constituent and anything external to the mixture as a whole.Therefore, b is not the total body force acting on constituent . The motivation behind this sort of a formulationis that any interaction between constituents is mutual and should not appear in the formulation. But t is thetotal traction force acting on it, and we would like to clarify this point. One could try to separate t to two partsas t = t

+ t

, where t would be an interaction force between constituent and on and t

would be

explained as an interaction between the particles of and itself or anything that is exterior to the mixture as awhole. However, although t + t = 0 is possibley true, it is not true that at a point on the boundary t

= 0 .

The reason is simple. We are summing over quantities on the boundary only so that although the -particle to which

vect is applied lies on the boundary, the -particle to which t

is applied lies remains exterior to our integrationprocess and therefore doesnot appear in the formulation. On the other hand, the use of b is justied by the fact that

this quantity appears in a volume integral and all interaction particles will be included in the integration process,and therefore volume interaction terms will disappear. But again, we do not expect, for instance, the sum of bodyinteraction terms to disappear at a point since the interaction is over the volume. This point will soon be claried.Finally, we also note that, similar to the previous arguement for traction, b does include the body force interactionbetween particles of and, say, that are exterior to volume V (hence the word external in the denition).

Therefore, we restate that equation 18 is free of interaction terms just like equation 9. Now, we follow theprocedure of the derivation of mass balance for each constituent, and write down the linear momentum balance

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equations for constituent . As before, we include interaction quantities for all the terms that appear in the balancelaw for the whole mixture. We should realize that not all terms are going to be new, and some of them are resultsof previous discussions. Therefore we state:

t V v dv = V c w dv v (v n ) d a + V b + f dv + t da . (19)

In this equation, f is volumetric interaction between at a point and all in V , such as would occur betweentwo electrically charged sets of particles (which is an interaction over the whole volume), or by drag force appliedby a viscous uid on a solid due to no-slip boundary conditions (which is a local interaction). In our case the latterinteraction would appear as a volumetric interaction although it seems it is a boundary interaction. The reason liesin our process of overlapping several continua in a region, where, in reality, they share a boundary. We assume thatthe mass generation interactions from the rst and second terms of equation 18 have been absorbed into the secondterm of equation 19. We realize that if it were possible to do the formulation in terms of a variable , we would notneed to introduce a new velocity-like quantity w . But, we remember that it is inconvinient to work in terms of .The term w is on the order of the mean velocity of constituents reacting with [5] as one could proceed to showusing eld .

Now we dene the total body force per unit mass acting on constituent in the control volume V :

b

= b +1

f

f = b +

1

f .

(20)

Using this quantity, we can rewrite equation 19 as follows:

t V v dv = V c w dv v (v n ) d a + V b dv + t da (21)

, which makes sense, since in the absence of chemical reactions, the equation is exactly in the same form as thatfor a single continuum. We will generally prefer to use the equation in the form:

t V v dv = V c w dv v (v n ) da + V ( b + f ) dv + t da (22)

, from which one can get the point forms t

( v ) = c w div( v v ) + b + f + div( T ) (23)

, where we have made use of divergence theorem, and the tetrahedron arguement to get T n = t . Making useof equation 14 we rewrite this equation as:

d v

dt= div( T ) + b + f + c (w

v ) (24)

, which is the local form of linear momentum balance for constituent .On the other hand, directly from equation 18 we get:

t

( v ) =

div( v v ) +

b +

div( T ) . (25)

We dene the total body force acting on the mixture at a point by b and the total traction t on the boundary of the mixture as:

b =1

b

t =

t (26)

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, so that the total stress tensor T is such that

t =

t =

T n = T n . (27)

In other words, the stress tensor, T , for the mixture is given by

T =

T . (28)

Now, we sum the linear momentum balance equations for each constituent. Summing equation 23 over andusing 13, together with 2, 3 and nally comparing the result with that of 25, we nd

(f + c w ) = 0 (29)

must be satised. This is an equivalent form of equation 25, and places a restriction on the interaction terms.

2.3 Angular Momentum Balance

We postulate the angular momentum balance for the mixture as follows:

t V x

v dv = (x

v ) (v n ) d a + V x

b dv + x t

da

(30)The equation tells us that the rate of change of total angular momentum in volume V is equal to the rate at

which angular momentum is convected into the volume plus the moment of the body and traction forces. Now, inpassing from the angular momentum balance of the mixture as a whole to the constituent balances, we would writedown interactions in the form of body forces and the mass transfer between components. In other words, there doesnot seem to be any new form of interaction. However, we also need to think of cases where there could be localtransfers of angular momentum balance between constituents. We would then expect the stress tensor T to begenerally non-symmetric. We write down the balance for a constituent as follows:

t V

x v dv =

V

x c w dv

(x v ) (v n ) d a +

V

x ( b + f ) dv +

x t da +

V

eijk

(31), where jk are the components of a tensor that represents the interaction we mentioned. Now, one can

remember that if the term e ijk jk were absent, by the use of previous results, we can show that T T T = 0 , i.e.skew-symmteric part of the stress tensor is zero. However, in this case we get:

(T ) (T )T = ( ) ( )T (32)

, in other words, the skew-symmetric part of T is equal to the skew-symmtric part of . One would haveto provide the form of this interaction tensor, and its sole effect is to render T skew. Therefore, without loss of generality, we can assume that is provided as skew, so that equation 32 reduces to

(T )skw = . (33)

Note that even if we did not use this convention, the sum e ijk jk would cancel off the symmteric part of anyway, so that there really is no loss of generality in our assumption. Now, from equation 30, we can easily show that

(T )skw = 0 (34)

, so that an equivalent form of this result would be (by using 33)

= 0 (35)

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, as one would expect since is related to volumetric interaction. We would like to stress that is not necessarilythe form of the interaction. It is just present to render the stress tensor non-symmteric.

2.4 Energy Balance

We state the energy balance for the mixture as a whole:

t V

+

12 v

v

dv =

+12 v

v

(v

n ) d a

+ V (r + b v ) d v + (t v h ) d a

(36)

, where we observe that there are no interaction terms. The equation simply tells us that the rate at which theenergy stored in the material in V changes is equal to the rate at which energy is transferred into the volume by theparticles entering through plus volumetric heat supply and heat input through the boundary.

Now, we use the results of our reasoning for the derivations of mass and momentum balance equations to statethe energy balance for a constituent:

t

V

+12

v v dv =

V

c (U + K ) d v

+12

v v (v n ) d a +

V

(r + b v ) d v

+ V (R + f v ) dv + (t v h ) d a V W dv(37)

, where W = ( L )sym . The term c U can be interpreted as the rate at which other constituents transfer theirinternal energy to and therefore U is on the order of at that point. Similarly, c K is the rate at which otherconsituents impart their kinetic energy to and so that K can be shown to be on the order of 12 w

w , and isindeed used as a prediction for this term [1]. R is related to the fact that constituents may exchange energy in otherforms, such as radiation or through chemical reactions. Obviously, this form is actually a sum over all constituentsindividually interacting with , but we did not write this explicitly. As before, we do not add interaction terms tothe boundary terms t and q .

We would like to note that r is an external volumetric heat supply. But, suppose all the chemical reactions wereexothermal. Obviously, this case would not be considered as an interaction and therefore we would not include the

effect in R

. Then, r would be supplemented by enegy input from the chemical reactions.Now, the presence of the term W needs to be justied, which we do in the following way. Green and Rivlin[7] have shown that under some additional invariance requirement impositions on terms appearing in the energybalance for a continuum, one can derive the mass, linear and angular momentum balances from this single equation.Green and Naghdi use this process to derive the balance equations for mixtures [9, 8]. If one did not include the termunder discussion, then we would be able to show using invariance requirements that ( T )skw = 0 , which contradictswith equation 33. Therefore, we need this term for consistency.

Finally, we use the convention of dening R :

R = R + c U (38)

, and we will prefer to use this term in our equations.Using tetrahedron arguement, we reach the conclusions:

h = q nh = q n

(39)

, and we relate the two quantities:

h =

h

q =

q .(40)

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There are other possible denitions for h (as well as t ) [1] but as Green and Naghdi [10] also note ...such denitions seem to confuse heat ux with various kinds of mechanical work and Craine, Green and Naghdi [5]note, for instance, that . . . there appear to be no grounds for dening the total stress at a surface as anything other than the sum of the partial stresses . Such alternative denitions put the point forms of mixture equations into thefamiliar form we always use. However, these denitions will cause some trouble when we want to impose boundaryconditions to the mixture, when they are already complicated enough.

We write the point form for equation 37 (as reduced to thermal energy balance through the use of linear momentum

balance), the process of which is the same as before:

d

dt= T D + r div( q ) c

+ ( w 12

v ) v K + R (41)

, where D = ( L )sym .For the whole mixture, on the other hand, we have:

d

dt=

(T D + W ) +

r

div(q )

c + ( w

12

v ) v

f v

(42)Again, comparing the two local forms, we see that the following restriction is to be place on the interaction terms:

(c K + f v W + R ) = 0 . (43)

Also, if we dene r such that:

r = r +1

R (44)

, we see that equation 41 is in the classical form of energy balance in the absence of chemical reactions.

2.5 Second Law of Thermodynamics: Clasius-Duhem Inequality

We assume that an entropy function, , exists for each constituent. Furthermore, we assume that part 2 of the 2ndlaw of thermodynamics holds for the mixture as a whole so that:

t V dv + (v n ) V r

dv +

div q

dv 0 (45)

, where is the temperature of constituent at that point. The temperature of each constituent is generallytaken to be different at a point, which may be important in cases such as high temperature ow of a gas through aporous solid.

The point form of this equation is:

d

dt

r

+ div

q

+ c

0 (46)

The proposed equation is an attempt to replicate the well known form of the inequality for a single continuum,which would be of the following form:

t V dv + (v n ) da V r dv + div q dv 0 (47)

,or in local form as:

ddt

r

+ divq

= 0 . (48)

Finally, we can dene the Helmholtz free energy function, for each constituent by:

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= . (49)

We would like to stress that the process by which the inequality is postulated is essentially what we have donethrough out the development of all balance equations. Rajagopal [17] notes that this inequality gives results thatsatisfy phyical expectations. The problem with entropy is that its existence does not make sense until it is shownto exist, such as one can do for an ideal gas or a thermoelastic solid. But in this case, we also would have entropytransfer due to mass transfer and some entropy interaction in other forms. This would prevent us from writing aninequality for each constituent separately, and it only makes sense to propose the inequality for the whole mixture,although one could easily repeat the machinery used so far to come up with constituent form of the inequality. Infact, it is exactly this possibility that Green and Naghdi [11] explored and found somewhat useful.

3 Discussion

We have analized the thermodynamics of a mixture of constituents with different constituent temperatures, andwhich are reacting with each other. As we have previously stated, the modelling of interaction terms will determinethe success of the process. But before that, we should consider the invariance requirements that should be imposedon the terms that appear.

3.1 Objectivity Requirements

First of all, it is assumed that terms such as T , q , are objective under arbitrary observer transformations. Inother words, all terms that a constituent inherits qualitatively from the balance equations for the mixture as a wholeare assumed to be objective. (It follows that is objective.) The remaining terms are due to interaction. The massproduction term c is taken to be objective, so that the mass balance for the constituent is objective. Once this isassumed, the linear momentum balance is objective, since w v is objective (it is the difference of velocities), if we also assume that the interaction force f is objective, which is reasonable. For the energy equation, we assumethat the energy interaction R is objective. We also need to assume that the term ( w 12 v

) v K is objective,so that the energy equation is also objective.

3.2 Interaction Terms

There are many ways in which the interaction terms can be modelled and the particular choices depend on the

problem at hand. For instance, the mass interaction term does not really make sense for less than 2 constituents,for a only a chemical reaction between at least two constituents that produce a third make sense (or the other wayaround). In addition, chemical reactions are restricted by stoichiometric ratios. On the other hand, just as we shalldo in the next section, one could concentrate on one of the constituents and assume that the remaining constituentsare relatively unaffected by the chemical reactions, and therefore discard the stoichiometric restrictions and massbalance equation for the mixture as a whole. Then, by assuming that c = c ( , ), a linear approximation wouldgive c = 1 + 2 , where 1 =

c , 2 =

c are constants that are to be specied.

A reasonable guess for K was noted to be 12 w w , although other forms can be provided. The term w is on

the order of average velocity of the constituents reacting with , and might be taken to be exactly this quantity, orsimply v , the barycentric velocity [17]. The interaction force, f , is possibly the most important of all interactionterms. The paper by Massoudi [15] discusses a variety of possible forms for this term. For instance, for uid (f )- solid (s) and uid-uid mixtures, motivated by the mechanics of drag force in uid mechanics (e.g. boundary

layer growth due to velocity gradients, and viscous interaction in this layer),f s

=f s

(v s v f

) can be assumed,from which a linear relationship can be derived. In general, the interaction force may be assumed to depend ondensities,temperatures, their gradients and velocity differences and their gradients.

We have not seen any discussion of energy interaction term R . But it seems reasonable to think that the formfor this term could be stated by considering , for instance, the energy release during chemical reactions between theconstituents and radiation between groups of particles of each constituent. (See Zohdi [20] for a very specic exampleon how this terms may be modelled.)

Finally, we would like to make a note on the modelling of the heat ux vector q . This term is taken to dependon the difference in the velocities of constituents at a point. This dependency is called the diffusion thermo-effect .

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A discussion of this term is given by Chapman [4] (pg.143). (Unfortunately, I could not understand the authorsreasoning. Chapmans equation 8 .41.3 on page 142 has three terms. The rst of these is the usual conduction term.The second term is really internal energy ux, which we accounted for already. The third term is the one that we areinterested in. One can argue, for the case of a common temperature, that if two constituents have different velocitiesat a point, than this difference will tend to change the temperature distribution and affect the heat ux.)

3.3 Boundary Conditions

Just as important as the interaction terms, is the process by which one splits the boundary conditions amongconstituents. Assume that the total traction, t , has been specied on some portion of the boundary, as shown ingure 4. Since t = t

, it is not clear how much of the traction goes to which constituent. The same problemexists in the splitting of the heat ux vector q . Rajagopal [17] discusses how the traction on the boundary canbe split among constituents. He uses an additional equation called the volume additivity constraint which is to bediscussed later. Therefore, we leave the discussion of boundary conditions unresolved at this point. We just notethat any method of splitting the boundary conditions would involve assumptions, and would change from case tocase.

!"

# #

$ $

t

V

Figure 4: Traction boundary condition.

3.4 Types of Mixtures

In order to solve for the equations of a mixture, one needs to provide constitutive equations for T , q , , as wellas for all the interaction terms. Obiviously, one would rst have to determine the quantities on which each of theseterms depend on. Once this is done, linear constitutive equations can be developed. The entropy inequality wouldthen be used to place restrictions that appear in the equations, together with the restrictions that were placed onthe interaction terms. Atkin and Craine [1] categorize mixtures into 4 groups, which are useful to discuss in orderto see the type of reductions that take place.

1. Chemically reacting mixture with different constituent temperatures: We need all the equations in order to solvea mixture problem.

2. Non-reacting mixture with different constituent temperatures: In this case, all the terms that appear with thecoefficient c are dropped. Accordingly, we do not need to provide forms for w , K .

3. Chemicall reacting mixture with common constituent temperature: In this case, a single temperature isassigned to each point in the mixture. We remember that in the purely mechanical theory of a single continuum,we assume that the material is under isothermal conditions, so that the energy equation is dropped, since

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an appropriate energy supply term r is needed to maintain the given condition. In the case of a commontemperature, we have a similar restriction, so that only 1 of the constituents can be allowed to go throughan arbitrary process whereas the rest of the constituents need appropriate heating to maintain a commontemperature. Accrodingly, we drop the energy equations for the 1 constituents, and the number of unknownsalso decrease by 1 temperatures. Also, instead of the remaining constituent energy equation, it is moreuseful to use the energy balance for the mixture as a whole. Finally, since in this latter equation, there areno interaction terms, we do not need to consider the restriction equation for the energy interaction terms

either, nor do we need to provide constitutive forms for them. (This is easier to see once we remember theclassical linearization = cv , where cv is the heat capacity, and q = k . We see that all the temperatureinformation is present in these forms, which can be written for each constituent. Once heating conditions arespecied, we have the energy equation to solve for the temperature, which is generally coupled to other balanceequations.)

4. Non-reacting mixture with common constituent temperature: In addition to the simplications in the previousitem, we drop the mass interaction terms and all the terms that have it as a coefficient.

4 Examples

In this section, we would like to derive two famous and widely used equations: Darcys Law and Ficks Law . Bothof these laws are widely used in engineering to model various phenomena. The more general form of these equationscan be found in the review by Atkin and Craine [2].

4.1 Darcys Law

Darcys Law states that the volumetric ow rate, q, of a uid through a porous medium (generally a solid) isproportional to the gradient of the pressure, p, in the uid, the proportionality factor, k , being related to theviscosity of the uid, and an indicator of the porosity of the solid called permeability, P o . We note that thevolumetric ow rate in a direction is nothing but the component of the velocity of the uid, v f , in that direction,when the solid is stationary, and we state the law as follows:

v f = 1k

p . (50)

This equation is used, for instance, to guess the resin ow time in production of composites. A derivation of this equation is given by Rajagopal [17] We are interested with uid behavior, so we consider its equations only. Toderive this equation, we need to make the following assumptions ( s relates to solid, and f relates to uid):

1. The solid is stationary: v s = 0 .

2. Assume isothermal conditions.

3. Ignore all inertia terms: v f 0 .

4. There are no chemical reactions: c = 0.

5. Assume that: = 0.

6. Assume that the behavior of the uid is ideal within itself : T f = pI .

7. Assume that the interaction term f is a function of ( v f v s ), and through a linear, isotropic approximationget: f f = kv f .

8. There are no body forces.

By saying that the uid behaves ideally within itself, we mean that the viscous interaction between uid particlesis ignored. But there obviously is a viscous interaction between the solid and the uid, due to no-slip boundaryconditions, otherwise we would not have an interaction term.

Directly from the linear momentum balance for the uid, we get:

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0 = div T f + f f

= p kv f (51)

, or

v f = 1k

p (52)

, which is the result we needed. We have made a very strong set of assumptions, which will certainly precludeus from applying this equation to a variety of problems. Nevertheless, the equation in its present form is widelyapplicable to many situations, by properly modelling the coefficient k . We observe that k should increase withincreasing viscosity of the uid, and should decrease as the solid becomes more porous, since then (in relation tothe nature of the mixture theory), there would be less uid-solid boundary for viscous interaction to occur. Onepossible form of k is k = P o , which is used a rough estimate to the ow time of resin into the bers during compositemanufacturing. The permeability P o can be given as a function of the volume fraction (, which corresponds to theconcentration in our case).

4.2 Ficks Laws

Ficks law governs diffusion dynamics of one constituent in another. Case hardening of a gear or diffusion of a

reacting species through a solid, or diffusion of (say) an aerosol through air are cases that can be modelled by thisequation. For purposes of discussion, we will take the diffusing species to be a uid, and the medium to be a solid.Given the concentration distribution, cf of the uid in the solid, the diffusion vector, J , of the uid is given as:

J = D cf (53)

, so that the diffusion ux, F , of the uid, which is the mass rate of diffusion of the constituent per unit area perunit mass of the mixture, is given by

F = J n (54)

, where n is the normal vector to the area at the point under consideration. This is what is usually referred to asFicks 1st Law in standard materials science textbooks. Moreover, the rate at which the concentration of the uidchanges at a point is given by:

c f

t= div( J ) = div D cf . (55)

D is called the diffusivity tensor . This equation is Ficks 2nd Law . To derive these equations, we assume all theitems of Darcys Law, and also that:

1. Assume that the density of the mixture approximately constant: f + s = constant

2. Assume that the pressure is a function of uid density only, and derive a linear approximation: p = do f

, where do = p f

3. We modify the interaction force term and assume anisotropic interaction: f f = Dv f .

Also, in many pyhsical cases, the diffusing species will have a concentration that is much less than the soliddensity, but it is not necessary to assume this now. Using a linear approximation to p, we write:

p = do f (56)

,where

do =p

f . (57)

The coefficient do is a function of (f , s ).

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A change from Ficks law is that we take the interaction term for the uid, f f , to be a function of f , but do notassume isotropy. A linear approximation gives: f f = Dv f . Now, directly from linear momentum balance, we get:

0 = p Dv f (58)

,or

v f = do D 1

f . (59)

In this equation, D represents the resistance to diffusion of the medium, and depends on the interaction betweenthe solid and the uid, which therefore would be a function of ( f , s ), as well as the viscosity of the uid . Finally,we multiply both sides of the above equation by

f

, and realize that f

vf = J , so that

J = 1

D f

= D cf (60)

, where

D = do D 1

, and cf = f

has been used together with assumption 2. Therefore we have recovered Ficks First Law. Now we

use the mass balance for the uid:

f

t+ div f v f = 0 . (61)

Upon using our previus result we get:

f

t= div D f . (62)

Using the rst assumption again, we can divide both sides by = s + f , carry this factor into the operatorsand use the fact that cf =

f

to get:

c f

t = divD

cf (63)

, thereby recovering Ficks 2nd Law.

5 Remark on Volume Additivity Constraint

Consider the mixture of two uids, F 1 and F 2 . Suppose that the density of F 1 is 1 and the density of F 2 is 2 ,in their non-homogenized state. We take an innesimal volume d V from the mixture. Let d V 1 and d V 2 be thevolume occupied by F 1 and F 2 , and d m 1 and d m 2 be the corresponding mass of each constituent. Then it followsthat d m 1 = 1 dV 1 and d m 2 = 2 dV 2 . Accordingly, d V = d V 1 + d V 2 , and d m = d m 1 + d m 2 . Using thehomogenization process, we calculate 1 = d m

1

d V , 2 = d m

2

d V and =d md V . Obviously, =

1 + 2 , as we have alreadyshown before. But also, since we can easily show that

1

1 =d V 1

d V and 2

2 =d V 2

d V , we get

1

1+

2

2= 1 . (64)

Now, assume that F 1 and F 2 are both incompressible. Then, 1 and 2 are constant. However, 1 and 2

do change because the local composition of the two uids may change with motion. Therefore, the although F 1and F 2 are incompressible, the homogenized elds are compressible . But this is not all, since equation 64 placesa restriction on the admissible processes that the mixture can go through . This is called the volume additivity constraint . We remember that the entropy inequality is also a restriction. Therefore, one can introduce equation 64

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into the entropy inequality using a Lagrange multiplier. Rajagopal [17], Atkin and Craine [2] give examples to theuse of this restriction.

We note that the volumetric composition of the mixture may be specied, for instance, by the introduction of an additional eld [0, 1], so that 1 = 1 and 2 = (1 )2 can be calculated initially. Accordingly, the eld = d V

1

d V changes with the motion at every point, and the volume additivity constraint is always satised.Although the volume additivity constraint seems to apply to all mixtures, there are clearly cases where it is not

necessary, such as in cases where the application of Ficks law is valid, or cases where the mixture theory is applied

to analize the behavior of one of the many constituents.The volume additivity constraint can be used to resolve one issue partially: that of splitting the boundaryconditions. We build upon the present example using the presentation of Rajagopal [17]. If we take an intinitesimalportion a of the boundary of V , a portion a 1 of a will be occupied by F 1 , and a portion a 2 of a will beoccupied by F 2 . Let us assume that at that point, a

1

a d V 1

d V = 1

1 anda 2

a d V 2

d V = 2

2 , or by summing the twoand using the volume additivity we get a

1

a +a 2

a = 1. Now, we mutliply both sides by the traction t at that pointto get t

1

1 + t 2

2 = t . Now, we assume that :

t 1 = t1

1, t 2 = t

2

2(65)

, which lets one easily split the boundary tractions, and resolves the issue of application of boundary condi-tions. Obviously, the assumptions of equation 65 is restricted in applicability, and requires experimental verication.Nonetheless, this example shows that there are ways in which the boundary conditions can be dealt with.

6 Conclusion

The given summary of the continuum theory of mixtures was focused on the theory. The two examples given arerather trivial, although they are very useful. The process of solving more complicated problems is quite tediousand we refer the reader to Atkin and Craine [2] and Rajagopal [17] for many examples, although no examples of achemically reacting mixture is involved. For an application where such a process is modelled, see Zohdi [20].

The major review articles are very useful in understanding the theory. But going through older papers helps onesee the development of the theory more clearly, and it is possible to nd more detailed discussions on some of theinteraction terms in the older papers.

As noted many times, the most interesting portion of the theory lies in the application, and therefore modelling

of the interaction terms. Although the theory seems to involve many terms to be determined, it simplies manyproblems a lot. Consider, for example, the ow of a uid through a porous solid. One could try to obtain the exactstructure of the porous solid and then make the model of the uid owing through this solid, just as one makes themodel of a uid owing through a pipe. Obviously, for a very ne structure, this model problem would be very hardto solve. On the other hand, once the use of the mixture theory is proved to be valid, the equations take a muchsimpler form. Basically, one than has two continua that are coupled to one another through interaction terms, whichis much easier to solve than the rst model.

Finally, we would like to make a note of some papers that are related to ME 285 material. In the paper byKrishnaswamy and Batra [14] , the authors go through the process that we followed in class, and develop the entropyfunctions for each constituent of a solid-uid mixture. In addition, there is a paper by Klisch [13], where he considersinternally constrained mixtures of elastic continua, making use of the notion of equivalence classes, as we did inclass. Gray and Svendsen [6] discuss the cases of surfaces of discontinuity in a mixture. (The paper itself is mainlyinterested in the modeling of phase transitions.) Rajagopal [17] also considers this topic.

NOTE: Some names that one sees frequently in this eld are: Green, Naghdi, Atkin, Craine, M uller, Bowen,Rajagopal (with Massoudi), Rivlin, Truesdell and Gurtin.

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References

[1] Atkin R.J., Craine R.E., Continuum Theories of Mixtures: Basic Theory and Historical Development , QuarterlyJournal of Mechanics and Applied Mathematics (1976a) 29 , 209-244

[2] Atkin R.J., Craine R.E., Continuum Theories of Mixtures: Applications , Journal of the Institute of Mathematicsand its Applications (1976b) 17 , 153-207

[3] Bowen R.M., Theory of Mixtures , Continuum Physics (Ed. A.C. Eringen), Vol.3, Academic Press, New York,1971 [QC173.7 .C661 v.3 PHYSICS]

[4] Chapman S., Cowling T.G., The Mathematical Theory of Non-Uniform Gases , 3rd ed., Cambridge UniversityPress, Cambridge, 1970 [QC 175.C5 1970 MATH]

[5] Craine R.E., Green A.E., Nagdi P.M., A Mixture of Viscous Elastic Materials with Different Constituent Tem-peratures , Quarterly Journal of Mechanics and Applied Mathematics (1970) 23 (2), 171-184

[6] Gray J.M.N.T., Svendsen B., Interaction Models for Mixtures with Application to Phase Transitions , Interna-tional Journal of Engineering Science, vol.35, No.1, pp.55-74, 1997

[7] Green A.E., Rivlin R.S., On Cauchys Equations of Motion , ZAMP (Zeitschrift fur angewandte Mathematikund Physik), Vol.15(3), Pag.217-328, 1964

[8] Green A.E., Naghdi P.M., A Theory of Mixtures , Archive for Rational Mechanic and Analysis, Volume 24,Number 4, 1967, P. 243-263

[9] Green A.E., Naghdi P.M., A Dynamical Theory of Interacting Continua , International Journal of EngineeringScience, Vol.3, pp. 231-241, 1965

[10] Green A.E., Naghdi P.M., On Basic Equations for Mixtures , Quarterly Journal of Mechanics and AppliedMathematics 22 (4) 427-438, 1969

[11] Green A.E., Naghdi P.M., Entropy Inequalities for Mixtures , Quarterly Journal of Mechanics and AppliedMathematics 24 (4) 473-485, 1970

[12] Haupt P., Continuum Mechanics and Theory of Materials , Springer-Verlag, Berlin, 2000

[13] Klisch S., Internally Constrained Mixtures of Elastic Continua , Mathematics and Mechanics of Solids 4: , 481-498,1999

[14] Krishnaswamy S., Batra R.C., A Thermomechanical Theory of Solid-Fluid Mixtures , Mathematics and Mechan-ics of Solids 2: 143-151, 1997

[15] Massoudi M., Constitutive Relations for the Interaction Force in Multicomponent Particulate Flows , Interna-tional Journal of Non-Linear Mechanics 38 (2001) 313-336

[16] Nemat-Nasser S., Hori M., Micromechanics: Overall Properties of Heterogeneous Materials , 2nd ed., Elsevier,Amsterdam, 1999

[17] Rajagopal K.R., Tao L., Mechanics of Mixtures , World Scientic, Singapore, 1995 [QA911.R27 1995 MATH]

[18] Torquato S.,Random Heterogeneous Materials: Microstructure and Macroscopis Properties

, Springer-Verlag,New York, 2002

[19] Truesdell C., Toupin R., The Classical Field Theories , Handbuch der Physik (Ed. S. Fl ugge) Vol. III/1 p.226(Springer-Verlag, Berlin, 1960)

[20] Zohdi T.I., Modeling and simulation of variably coupled time-transient thermo-chemo-mechanical processes in multiphase solids , (to appear in ZAMP)

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me285 project mixtures

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