Difusion Maxwell Stefan

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FLUID MECHANICS AND TRANSPORT PHENOMENA

Diffusing with Stefan and MaxwellNeal R. Amundson

Dept. of Chemical Engineering and Dept. of Mathematics, University of Houston, Houston, TX 77204

Tsorng-Whay Pan and Vern I. PaulsenDept. of Mathematics, University of Houston, Houston, TX 77204

In most chemical engineering problems, diffusion is treated as an add-on to forcedadection, and the boundary conditions are the Danckwerts conditions in order tomaintain conseration. If we treat problems in which there is no applied adection orchemical reaction for a steady-state situation and with an ideal gas, it is an equi-molarprocess. Because it is considered the natural moement of molecules moing down aconcentration gradient, this requires the Stefan-Maxwell equations. In the standard 1-Dtwo-component boundaryalue problem, analogous to the Dirichlet problem, the solu-tion is direct and is probably the only one there is. In a ternary system, the solution is

already not direct and can only be obtained by numerical means, which is not seere. Ina quaternary system, it does not appear feasible to obtain a simple procedure like thatobtained for the ternary system. A different numerical scheme deeloped is robust withrapid conergence and works well with an arbitrary number of components, 60 haingbeen used in one problem. As the number of components increases, the solution profilestend to become linear and the dependence on particular diffusiities is less important.This manifests itself when using diffusiities from a random collection. The problemusing a continuous distribution of components is soled, and computationally and theo-

retically the profiles are probably linear and with a single pairwise diffusion coefficient.

Introduction

The purpose of this article is to present some manipula-

tions of the Stefan-Maxwell equations as applied to the sim- .ple one-dimensional 1-D diffusion problem in the steady

state for an ideal gas with an arbitrary number of compo-

nents. Anyone who has examined these equations becomes

frustrated by their backward presentation which manifests it-

self in some way in almost every application of their use. In

addition, they are nonlinear in the two important sets of vari-

ables although linear in each set singly. The two sets of vari-

ables are the mol fraction of each component and the corre-sponding flux of each. Manipulation of these equations re-

veals that their inversion to produce the fluxes in terms of

the gradient for a two-component system is trivial. For a

three-component system, it is doable and builds ones confi-

dence, while for a four-component system the problem is a

character builder; for a five-component system, it seems im-

Correspondence concerning this article should be addressed to N. R. Amundson.

passable and impossible in an analytical way. Some of the

earliest attempts at using the equations were carried out by . .Toor 1964 and Stewart and Prober 1964 who linearized

the equations to cast them into a usable form. This is a suc-

cessful procedure since the system is very benign, and, as will

be shown later, many of the solutions turn out to be close to

linear. Later workers developed many other schemes, and .these are discussed by Taylor and Krishna 1993 whose book

is required reading. The numerical scheme devised by

.Krishna and Standart 1976 was and still is the method ofchoice.

Our first effort will be to invert the Stefan-Maxwell equa-

tions for three and four component systems, and to attempt

to simplify these equations to make their use easier. For a

three-component system, this is relatively easy, but for a

four-component system, this is a trial and requires the intro-

duction of two sets of simple functions to present the inver-

sion in a relatively simple way. Some small surprises result

here and make its perusal worth the effort.

April 2003 Vol. 49, No. 4AIChE Journal 813


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temperature and same pressure. In one dimension the con-

centration equations are, where l is the conduit length

dNis0, 0zl , i s1, 2, 3

dz

and boundary conditions are, with x and x specified0 l

x 0 sx , x l sx . . .0 l

This simple problem will be cast in dimensionless form later,

since now the occurrence of the diffusivities is revealing. From

the conservation equation, N is a constant for all x andi

n

Ns0 iis 1

since this diffusion movement of molecules only means that

they replace each other.

In an effort to simplify Eq. 5 by removing the unity in each

row, we multiply it by the diagonal matrix

D 0 023

0 D 013

0 0 D12

and then add the rows to obtain

c dx1y D D yx D yx D yx D .13 12 1 23 2 13 3 12

x dz .

dx2

qD D yx D yx D yx D .23 12 1 23 2 13 3 12 dz

dx3qD D yx D yx D yx D .23 13 1 23 2 13 3 12

dz

sD NqD N qD N .23 1 13 2 12 3

We see that, fortuitously, the three quantities within the .parentheses are x so that from Eq. 4

dx dx dx1 2 3c D D qD D qD D13 12 23 12 23 13

dz dz dz

sD NqD N qD N .23 1 13 2 12 3

Integration between zero and l gives

x yx x yx x yx1l 10 2l 20 3l 30c q q

D D D23 13 12

N N N1 2 3s q q l /D D D D D D13 12 12 23 13 23

after dividing through by D D D . Integrating from 0 to z,12 13 23gives

x yx x yx x yx1 10 2 20 3 30c q q

D D D23 13 12

N N N1 2 3s q q z /D D D D D D13 12 12 23 13 23

so

1 dx 1 dx 1 dx1 2 3q q

D dz D dz D dz23 13 12

x yx x yx x yx 1 K1l 10 2l 20 3l 30s q q s 6 .

D D D l l23 13 12

and also

x yx x yx x yx1 10 2 20 3 30q q

D D D23 13 12

x yx x yx x yx z1l 10 2l 20 3l 30s q q /D D D l23 13 12which is not a relation one might expect for three compo-

nents.

In order to use these equations, particularly Eq. 6, let us

consider the expression for N , and write it in the form1

cNsy D D D .1 12 13 23

x .

1 dx 1 dx 1 dx 1 dx1 1 2 3 yx q q1 /D dz D dz D dz D dz23 23 13 12

which reduces to an expression with a single derivative

c 1 dx K1Nsy D D D yx .1 12 13 23 1

x D dz l . 23

If we now cast this in dimensionless form, using the Cramer .denominator Eq. 4 , we obtain

dx1 w xy qx K sZ c x qc x qc x K1 23 1 23 1 13 2 12 3 1d

with

Dz i js , c s , K sD K, K sD K, K sD K,i j 23 23 13 13 12 12

l D

N l N l N l D21 2 3Z s , Z s , Z s , K s ,1 2 3 1

D D D D Dc c c 12 13

D2 D2

K s , K s2 3D D D D12 23 13 23



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and in a similar way

dx2 w xy qx K sZ c x qc x qc x K ,2 13 2 23 1 13 2 12 3 2d

dx3 w xy qx K sZ c x qc x qc x K .3 12 3 23 1 13 2 12 3 3d

These last three equations should be simpler to apply to

the numerical problem than the original ones, but one needs

to have a good estimate of Z , Z , Z to initiate that solu-1 2 3tion. One can find such a solution by assuming that an ap-

proximate solution is a straight line connecting the appropri- .ate boundary points and using the values of x at z s0 to0

obtain

x yx1l 10 0y qx K sZ c x qc x qc x K .10 23 1 23 10 13 20 12 30 1l

x yx2l 20 0y qx K sZ c x qc x qc x K .20 13 2 23 10 13 20 12 30 2l

from which a reasonable guess should ensue. Note that.Zs 0. i

We stress that the advantage to these equations is the fact

that the lefthand sides are linear first-order operators since

K is a predetermined constant. If we write the original Ste-i jfan-Maxwell equations in the form

N xdx j jiyc qx sN i i

dz D Di j i j

the left-hand side is a linear operator in x , but its coefficienti

contains the unknown N for all j except js i. It is evidentjof course that these equations for constant values of N andi

x s 1 have an analytical solution which is not of great use isince the boundary conditions must be used to eventually de-

termine the N, which is a tortuous exercise, but doable.iIt may be interesting to consider the structure of the solu-

.tion profiles xs x , not because the solutions are richi iwith pathology, but rather in order to show their simplicity.

Diffusional processes are normally very benign as long as.they are not coupled with exothermic chemical reactions ,

since molecular motion tends to smear things out.

If we consider a ternary system, there are three equations

of which only two are linearly independent in pairs. If, for

example, dxrds0 at a point, then it follows that3

dx dx2 1q s0

d d

which is impossible since the functions defined by the corre-

sponding two SM equations are linearly independent. If we

differentiate the equation for x , we obtain1

d2x 1 dx 1 dx dx N N 1 2 3 1 2 3c sy q Nq q .12 / /D d D d d D Dd 12 13 12 13

If we assume this is zero, then, using

dxis0

di

we have

1 1 dx dx2 1y q N yN s01 2 / /D D d d12 13

which gives

dx dx2 1N sN1 2

d d

which is also impossible because of the linear independence.

Thus, the profiles of the components are strictly monotonic

between the boundaries and we shall illustrate some of these

later. The above may be generalized to higher-order equa-

tions.

.Toor 1957 in the early 1950s solved the ternary diffusionproblem by obtaining the analytical solution for fixed fluxes,

and then solved for the fluxes by trial and error. Hsu and .Bird 1960 published a massive paper for three-component

chemical reaction problems in which the catalytic reaction

occurred at one of the boundaries. These were solved in an

analytical way with computations on an IBM 650.

Quaternary Systems

In order to realize the inversion we must solve the follow-

ing set of linear simultaneous equations in N, N , N , and1 2 3N, where we use Eq. 24

yx x x x x x2 3 4 1 1 1dx1 y yc D D D D D D12 13 14 12 13 14dzx yx x x x x2 1 3 4 2 2dx2 y yc D D D D D Ds N.21 21 23 24 23 24dzx x yx x x xdx 3 3 1 2 4 33 y yc

D D D D D Ddz 31 32 31 32 34 34

0 1 1 1 1

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One could hope that the treatment of quaternary systems

might not be substantially different from ternary systems but,

of course, the fourth-order manipulations are certainly more

tedious in this Cramer solution. The challenging part of the

exercise, however, as before, is the casting of the resultant

solutions into forms which are independent of the numbering

of the species and are quasi-symmetric in the component

variables. The Cramer solution first obtained is linear in the

three derivatives dxrdz, dx rdz, and dxrdz since the last1 2 3Stefan-Maxwell equation is replaced by

4

Ns 0. ii s 1

The Cramer denominator may be written as a fourth-order

quadratic form. This is not readily apparent during the calcu-

lation, but it can be written in the form

x .

d d d d d d d d d d d d12 13 14 24 13 12 12 13 34 12 14 34

d d d d d d d d d d d d12 14 23 24 23 12 12 23 34 24 34 12Tsx x ;d d d d d d d d d d d d13 14 23 23 13 24 13 34 23 24 34 13

d d d d d d d d d d d d13 14 24 24 14 23 23 14 34 24 34 14

d sd 7 .i j ji

where d sDy1, that is, the elements in the matrix are re-i j i jciprocals of the pairwise diffusion coefficients. There are six

pairwise diffusion coefficients in a four component set, and

so there are 20 combinations of six things taken three at a

time. Clearly, four of these are missing in the above, and,

after some scrutiny, it develops that these four are those in

which one of the four digits is missing; that is, there is no

subscript form like d d d . Physically, this makes sense12 13 23since it means that every chemical component has an effect

in each term. Since the quantity obtained as the Cramer de-

terminant was not obviously a quadratic form a priori, theplacement of the terms in the matrix in a quasi-symmetric

form was somewhat arbitrary at the start and might need re-

arrangement.

The solution obtained can be put into the matrix-vector

form given below, and the placement of the functions ,i jdefined below, is restricted by the form,

4 dxiA s0

dzis 1

.where A is chosen and the functions X x and are cho-i i jsen and are defined below

c dxNsy W 8 .

x dz .

where the matrix W is

1yx Xq yx X q q yx X q q yx X q q .1 1 1 1 2 13 14 1 3 12 41 1 4 21 31yx Xq q 1yx X q yx X q q yx X q q .2 1 23 24 2 2 2 2 3 21 42 2 4 12 32yx Xq q yx X q q 1yx X q yx X q q .3 1 32 34 3 2 31 43 3 3 3 3 4 13 23yx X q q yx X q q yx X q q 1yx X q .4 1 42 43 4 2 41 34 4 3 14 24 4 4 4

Table 1. Definition of the Functions in Eqs. 8.

12 21 13 31 14 41x x x x x x x x x x x x1 2 2 1 1 3 3 1 1 4 4 1

D D D D D D D D D D D D13 24 23 14 12 34 32 14 12 43 42 13w xD sDi j ji

23 32 24 42 34 43x x x x x x x x x x x x2 3 3 2 2 4 4 2 3 4 4 3

D D D D D D D D D D D D21 34 31 24 21 43 41 23 31 42 41 32

and

x x x2 3 4Xs q q ,1

D D D D D D23 24 23 34 24 34

x x x1 3 4X s q q2

D D D D D D13 14 13 34 14 34

x x x1 2 4X s q q ,3

D D D D D D14 12 12 24 14 24

x x x1 2 3X s q q 9 .4

D D D D D D12 13 23 12 13 23

note that these have the same gross form as the ternary.Cramer denominator , and

4

s q 10 . .i i s sis s 1s i

which is the sum of the six terms in which x appears in eachiterm and is defined in Table 1.i j

It is seen that X contains no x , and each x has D D ,i i j l j k jlk ij in the denominator in Eq. 9. The Xs are quasi-isymmetric, as are the s. Note also that every has ai j i j1234 subscript property in some order, and, therefore, there

are only three different ones. If one examines the matrix so-lution carefully, one finds that it is quasi-symmetric, al-

though, in some cases, one must go back to the definitions of

the functions. It is readily apparent also that the matrix is

singular since the sum of the rows is zero.

The solution matrix can also be written in the form

c dxw xNsy wi j

x dz .

with

x q xl mw syx Xq x ; ijlm ,i j i j i

D Di j lm

x q xl mw s 1yx Xq x ; sjlm. . ii i i iD Dis lms



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If we were to assume that all the diffusivities were the same,

then

D2w syx x qx x ; ij, i j i s i ss j s i

s j

syx 1yx qx 1yx y x syx 2 , ij, . .i j i i j i

D2w s 1yx 1yx q2x 1yx s1yx 2 . . .ii i j i i i

and, hence, since the Cramer denominator equals Dy3

2 2 2 21yx yx yx yx1 1 1 12 2 2 2 dxyx 1yx yx yx2 2 2 2

NsycD ,2 2 2 2 dzyx yx 1yx yx3 3 3 32 2 2 2yx yx yx 1yx4 4 4 4

dxNsycD ,

dz

and so singularity is preserved.

Because we wish to simplify, if possible, the quaternaryproblem we will spend some time now on the structure of

what we have already established. Some surprising solutions

will surface. For example, if we examine the Cramer denomi- .nator Eq. 7 , we discover that it has the form

x .

x 2 x x x x x x1 1 2 1 3 1 4s q q q

D D D D D D D D D D D D12 13 14 12 13 24 12 13 34 12 14 34

x x x 2 x x x x2 1 2 2 3 2 4q q q q

D D D D D D D D D D D D14 21 32 12 24 23 12 23 34 12 24 34

x x x x x2

x x3 1 3 2 3 3 4q q q qD D D D D D D D D D D D13 14 23 13 24 32 13 23 34 13 24 34

x x x x x x x 24 1 4 2 4 3 4q q q q

D D D D D D D D D D D D13 14 24 14 23 24 14 23 34 14 24 34

11 .

where we use the commonality of some species in each row

and each column, and, if we look at this a little more closely,

we can, using the functions defined before, writei j

x 2 1 12 13 14 x s q q q .

D D D D D D12 13 14 12 13 14

x 2 21 2 23 24q q q q

D D D D D D21 12 23 24 23 24

x 2 31 32 3 34q q q q

D D D D D D31 32 31 32 34 34

x 241 42 43 4q q q q . 12 .

D D D D D D41 42 43 41 42 43

Using the expression for the Cramer denominator given by

Eq. 7, we can produce some other interesting results. If we .examine x , in particular, the last three rows and the last

three columns, we see that row two in the last three elements .may be written x rD X , while the third and the fourth2 12 1

.rows the last three elements become

x x3 4Xq X .1 1

D D13 14

.It then follows that x may be written

x x x q 2 3 4 12 21 x s q q Xq . 1 /D D D D12 13 14 12

q q x 213 31 14 41 1q q q . 13 .

D D D D D13 14 12 13 14

Now, to find a similar representation involving X , we look2at the first, third, and fourth rows, and first, third, and fourth

columns, and, proceeding as above, one obtains

x x x q 1 3 4 21 12 x s q q X q . 2 /D D D D21 23 24 21

q q x 223 32 24 42 2q q q . 14 .

D D D D D23 24 21 23 24

The expansion in terms of X follows from the elements of3the first, second, and fourth columns, and first, second, and

fourth rows. One obtains

x x x q 1 2 4 13 31 x s q q X q . 3 /D D D D31 32 34 31

q q x 223 32 34 43 3q q q . 15 .

D D D D D32 34 31 32 34

The expansion in terms of X follows from the elements of4the rows and columns in the upper lefthand corner and is

x x x q 1 2 3 14 41 x s q q X q . 4 /D D D D41 42 43 41

q q x 224 42 34 43 4q q q . 16 .

D D D D D42 43 41 42 43

We see also that there is a systematic scheme for the above

relations. The expansions in terms of X and X are made1 4 .directly from x . The expansion in terms of X and X2 3

requires some minor symmetric juggling, more for X than3for X .4

Simple Quaternary Representation

With the ternary system by some judicious juggling, we were

able to reduce the fundamental equations to equations which



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had a linear operator. This is desirable since the operator

determines the form of the solution and should enable one to

produce a numerical solution, which is less likely to be unsta-

ble and to converge more rapidly. With this in mind, we will

try to do the same with the quaternary system written in the

form

dxc dx dxjN

sy X

y X x

q 17 .

j x dz dz dz . j

where X is the diagonal matrix with X on the ith term ofithe diagonal and where we note that the quasi-symmetry in

in the second matrix which contains the functions belonging

to i j

q q q 1 13 14 12 41 21 31 q q q 23 2 4 2 21 42 12 32

s q q q 32 34 31 43 3 13 32 q q q 42 43 41 34 14 24 4

1

2s , 18 .

3

4

which is singular and

s

x xqx x xqx x xqx . . .1 3 4 1 2 4 1 2 31

1234 1324 1423x xqx x xqx x xqx . . .2 3 4 2 1 4 2 1 3 2

1234 1423 1324

x xqx x xqx x xqx . . .3 2 4 3 1 4 3 1 2 3

1324 1423 1234

x xqx x xqx x xqx . . .4 2 3 4 1 3 4 1 24

1423 1324 1234

19 .

where in the denominator we have only written the subscripts

of D D .i j lm

In an effort to obtain a simpler representation of the inver-sion given by Eq. 8 we will use Eq. 17 in a form by writing

first the representation for N1

dxc dx dxj1yNs X yx X q 1 1 1 j 1

x dz dz dz . j

where is the first row of . Then1

yN x dx dx .1 1 2w xs 1yx Xq y x X y y w x .1 1 1 1 2 13 1 4c dz dz

dx dx3 4w x w xy x X y y y x X y y .1 3 12 41 1 4 21 31dz dz

This equation is not as mean as it looks for all of the si jand s contain an x . A little juggling with the formulaji 1

dxjA s0

dz

will cast it into a form

yN x dx dx .1 1 2s 1yx X y x X q w x . .1 1 1 2 12

c dz dz

dx dx3 4y x X q y x X q . . .1 3 13 1 4 14

dz dz

Notice what one would call s0 and we have the follow-11ing for N, N , N , and N1 2 3 4

s q q q , s q q q ,12 12 21 31 41 21 21 12 32 42

s q q q , s q q q ,13 13 31 14 21 23 23 32 12 24

s q q q , s q q q ,14 14 41 12 13 24 24 42 23 21

s0, s0,11 22

s q q q , s q q q ,31 31 13 23 43 41 41 14 34 24

s q q q , s q q q ,32 32 23 34 13 42 42 24 43 14

s q q q , s q q q ,34 34 43 31 32 43 43 34 41 42

s0, s0.33 44

and also

1yx X yx X y yx X y yx X y .1 1 1 2 12 1 3 13 1 4 14yN x dxyx Xy 1yx X yx X y yx X y . .2 1 21 2 2 2 3 23 2 4 24

s .yx Xy yx X y 1yx X yx X y c dz .3 1 31 3 2 32 3 3 3 4 34yx Xy yx X y yx X y 1yx X .4 1 41 4 2 42 4 3 43 4 4



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The nondiagonal terms have a connection with the Cramer

determinant when multiplied by the symmetric matrix

1 1 11

D D D12 13 14

1 1 11

D D D12 23 24

1 1 11

D D D13 23 34

1 1 11

D D D14 24 34

However, taking the sum of the rows in the product does not

have the desired result which occurred with the ternary sys-

tems because the unity terms on the diagonal do not disap-

pear and so a relation analogous to Eq. 6 does not result. In

addition, the above process, which looks so promising, intro-

duces a substantial number of other terms, and, while there

is some redundancy which could be removed, the overall re-

sult is not much simpler and does not reduce to a simple

linear operator as in the ternary system.

Numerical Diffusion Problem

Solutions of the 1-D diffusion problem have been pre-

sented for the Stefan-Maxwell equations, the best known, as

mentioned earlier, being that developed by Krishna and .Standard 1976 , which involve the solution of the set of

equations treated as ordinary differential equations in the mol

fractions x for fixed values of the fluxes. To determine theifluxes, this set of equations then must be solved by some nu-

merical procedure. This has apparently produced in some

cases instability and some nonconvergence, and has been .studied by Taylor Taylor and Webb, 1981; Taylor, 1982 who

has, in addition, developed some interesting and useful re-

sults in a series of papers referenced in his book.

The purpose of this section is then to present a numerical

solution method for the standard 1-D multicomponent diffu-

sion problem with specified boundary values at z s0 and zsl for all species. The fluxes will be constant and unknown,

and we assume that there is equimolar diffusion. While the

method is applicable to n species, our illustration will be for

four. The equations to be treated are Eq. 1

dxc sBx , 0zl ,

dz

x 0 sx , x l sx . . .0 l

B is given in Eq. 1. We cast these equations in dimensionless

form using the standard notation; then, we have

Z x yZ Xdx j i i jis ; ij,

d i jj

D N li j j s ; s ; Zsi j i j ji j

D cD

where D is normally one of the D . We write these equa-i j .tions in the form where A is defined below

dxsAx , x 0 sx , 20 . .0

d

the formal solution of which is

xsexp A x 21 . .0

In the matrix A there are n unknown values of Z which areiconstant and Zs 0. Our problem is to determine the set i

i

4Z from the equationi

x sexp A x . 22 . .l 0

In order to solve this equation we will develop a different

representation for A, which follows

Z Z Z Z Z Z2 3 4 1 1 1q q y y y

12 13 14 12 13 14

Z Z Z Z Z Z2 1 3 4 2 2y q q y y

21 21 23 24 23 24As 23 .Z Z Z Z Z Z3 3 1 2 4 3

y y q q y 31 32 31 32 34 34

Z Z Z Z Z Z4 4 4 1 2 3y y y q q

41 42 43 41 42 43

which is different from the form used earlier.



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We shall decompose A into a sum of matrices which makes 4use of the linearity in the Z ; soi

1 1 10 y y y

12 13 14

10 0 0

12As Z11

0 0 013

10 0 0

14

10 0 0

12

1 1 1y 0 y y

12 23 24q Z21

0 0 023

10 0 0 24

10 0 0

13

10 0 0

23q Z31 1 1

y y 0 y 13 23 34

10 0 0

34

10 0 0

14

10 0 0

24q Z 24 .41

0 0 034

1 1 1y y y 0

14 24 34

Such a decomposition exists for any n. We write the above as

AsA Z qA Z qA Z qA Z1 1 2 2 3 3 4 4

Each of the A is singular and each has very simple eigenval-i.ues . We consider now Eq. 20, 21, and 22 and the general

procedure will involve a sequence of paired steps. Approxi-

mate values for the fluxes Z will be found by solving a set ofilinear simultaneous algebraic equations obtained by keeping

only the first term of the exponential solution. This step will

be followed by solving the differential equation Eq. 21 with

the values of Z so obtained. If we keep only the first term ofi

the exponential solution of Eq. 22, we obtain a vector

s Iq Z A x 25 . i i 0

To find the first set of Zs we set sx and s1 soi l

x y x s Z A xl 0 i i 0

This is a singular set of algebraic equations in the Z, so weireplace the last equation by Zs 0. Let us call the solution i 04so obtained Z . We then consider the set of differentiali

equations

dx1 0sA x ; x 0 sx 26 . .1 1 0d

This set of equations is also singular so that we will substitute

xs 1 for the last equation, giving the solution ii

x sexp A0 x .1 0

which we rewrite as

x sexp yA0 x .0 1

and substitute this into the solution to obtain

xsexp A exp yA0 x . . 1

which we treat as before, keeping only the first-order term in

the expansion of the exponential to give

0s Iq A ZyZ x . i i i 1

and at

s1, we obtain

x yx s A ZyZ 0 x 27 . .l 1 i i i 1

which is a new set of linear simultaneous algebraic equations 4 14for Z , which we treat as before. Call the solution Z whichi i

we use in

dx2 1sA x ; x 0 sx .2 2 0d

which has the solution

x sexp A1 x .2 0

and as before

x sexp yA1 x .0 2

Thus

x sexp A x sexp A exp yA1 x , . . .0 2

yx s A ZyZ1 x .2 i i i 2



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24and, with s1, sx , we solve for a new set of Z . Thel iprocedure then repeats itself with equations

Ak s A Zk , i idxkq1 ksA x ; x 0 sx , .kq1 kq1 0

d

x yx s A Zkq1 yZk x .l kq1 i i i kq1

with the hope that

lim x 1 sx . .k lk

The scheme described above can best be thought of as a

quasi-Newton scheme for the nonlinear equation

F Z sexp AZ x ; F: R4Rs4 . . 0

.Returning to Eq. 27, we see that our wish is to solve F Z s

x. If we apply Newtons method with an initial estimate Z sl 00, we find that

w xF 0 s A x A x A x A x . 1 0 2 0 3 0 4 0

Forgetting for the moment the problem of invertability, we

see that the next approximation to Z is

y1Z sF 0 x yx . .1 l 0

.yielding x sF Z . At the next step in Newtons method,1 1 .one needs to compute F Z . Our method approximates1

.F Z byn

w xA x A x A x A x .1 n 2 n 3 n 4 n

.How well this latter approximates F Z can be shown tondepend on how well

exp A Z qZ x . .n 0

is approximated by

exp AZ exp AZ x . .n 0

for small Z. Since matrix exponentials do not commute un-

less the respective matrices commute, we would expect that

how well this approximation works depends on the size of

commutators of AZ and AZ .nq1 nIn the implementation of the quasi-Newton method we

have applied the Runge-Kutta-Fehlberg method, described in .Burden and Faires 2001 , to solve the set of equations

dxkq1 ksA x , 01,kq1d

x 0 sx .kq1 0

for each k, k s1, 2, 3, . . . . This method adapts the numberand position of the nodes used in the approximation to en-

sure that the local truncation error is kept within a specified

bound. It consists of using a Runge-Kutta method with local

truncation error of order five to estimate the local error in a

Runge-Kutta method of order four and then adjusting the

step size to keep the local error within a specified bound; in

our examples the tolerance was 10y5. In the examples to fol-

low it took less than 20 iterations to have a convergent result.

The scheme is very robust.

Numerical ExamplesExample 1

This is taken directly from Taylor and Krishna 1993, p.. . .103 . Components are hydrogen 1 , nitrogen 2 , and carbon

.dioxide 3 .Diffusion path length ls85.9 mmTemperatures35.2C

At z s0, x s0.00000 At z sl , x s0.5012110 1lx s0.50086 x s0.4987920 2lx s0.49914 x s0.0000030 3l

Pairwise diffusion coefficients

D s83.3 m2rs 10y6 s112 12

D s48.0 m2rs 10y6 s0.816313 13

D s16.8 m2rs 10y6 s0.2016823 23

Total concentration cs39.513 molrm3

Zs 26.087Ni i

Figure 1 shows the composition profiles, and, as stated by

Taylor and Krishna, the nitrogen flux is substantially higher

Figure 1. Composition profiles with abnormal fluxes.



11/18

Table 2. Boundary Conditions in Examples 2a, 2b, and 2c.

Fig. x x x x x x10 20 30 1l 2l 3l

2a 0.55 0.05 0.40 0.01 0.45 0.542b 0.55 0.05 0.40 0.10 0.72 0.182c 0.55 0.05 0.40 0.01 0.98 0.01

than one would anticipate from the gross gradient. In fact,

the flux of nitrogen approaches that of the flux of carbon

dioxide. The profiles show some curvature and are strictlymonotonic. Here we had to solve Eq. 26 reversely by starting

at s1.

Example 2

This is also chosen from Example 4.2.4 of Taylor and

Krishna and illustrates the effect on the solution of changes . .in boundary compositions with hydrogen 1 , nitrogen 2 , and

.carbon dichloride difluoride 3 . The pairwise diffusion coef-ficients are

D s77.0 m2rs10y6 s2.329312 12

D s33.1 m2rs10y6 s1.0000013 13

D s8.1 m2rs10y6 s0.2448223 23

The different cases considered are given in Table 2.

While Figure 2a is a normal case in the sense that the fluxes

are in the directions expected, they are somewhat different

than expected. Figure 2b is an abnormal case since it appears

that the flux of component three is the reverse of the direc-

tion. Figure 2c is a really abnormal case since it shows that

component three, in spite of having a large superficial gradi-

ent, has almost no flux and substantial profile curvature.

Figure 2a. Fluxes that are normal, but with surprising

values.

Figure 2b. Abnormal flux in component three.

Example 3

This is an eight component case in which the relative pair-

wise diffusivities vary by about a factor of seven and there are

28 of them. The boundary values are given in Table 3 and the

numerical results are in Figure 3.

This is a relatively normal case since the fluxes are in the

same direction as their superficial values. It may be said,

Figure 2c. Very abnormal flux in component three.



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Table 3. Boundary Conditions in the Eight Components inExample 3 and Figure 3.

,x x x x x x x x xi 1 2 3 4 5 6 7 8

s0 0.30 0.35 0.05 0.05 0.10 0.15 0.00 0.00s1 0.05 0.00 0.00 0.10 0.00 0.25 0.10 0.50

however, that while the superficial flux of x is greater in8absolute value than that of x , the actual flux is less. The1profiles show some curvature, but most appear to be rela-

tively straight lines between boundary values, except that of

x . The chemical species used in these calculations were H8 2 . . . . . . .1 , O 2 , N 3 , CO 4 , CO 5 , CH 6 , C H 7 , and2 2 2 4 2 4

.C H 8 . Many of the pairwise coefficients were calculated2 6 .using the Fuller-Schettler-Giddings Fuller et al., 1966

scheme or were those listed in the same source.

Example 4

. . .This is a six component case with H 1 , O 2 , N 3 ,2 2 2 . . .CO 4 , CO 5 , andCH 6 , so there are 15 pairwise diffu-2 4

sion coefficients with a spread of values of over 11. There are

no real surprises here with the possible exception that the

flux of x is negative, as seen in Figure 4. Otherwise, curva-5ture in minimal.

Continuous Distribution of Components

We will now consider a slightly different approach in which

we assume that the number of components is very large and,

in fact, that there is a continuous distribution of components.

In order to do this it is easier and, perhaps, even necessary,

to change the procedure. With a large number of compo-

nents, say 60, in order to proceed as earlier, it could be nec-

essary to introduce 1,770 pairwise diffusion coefficients, all of

which probably must be estimated by the methods of Fuller .et al. 1966 . Let us now consider the problem in which there

is a continuous distribution of components, first, in which no

species is present in finite amount, and, second, in which somespecies are present in the finite amount, but the remainder

as a continuous distribution.

To start this effort, we write the Stefan-Maxwell equations

in the form

n nx Ndx j jiyc sN yx , 0zl , is1, . . .,N. i i

dz D Di j i jjs 1 js 1j i j i

28 .

If we suppose that p is a component designating variable and .x p dp is the mol fraction of that component in the mixture,

then

pgx p dps1 .H

pl

where p is the lowest p and p is the greatest p of thel gcontinuous variable, pF pFp . The total fluxes of thesel g

.variables will be N p d p where

pgN p dps0 .H

pl

Figure 3. Eight-component case with tendency toward

straight line profiles.

if we assume equimolarity of diffusion. The pairwise diffu- .sion coefficients will be D p,q where D is a function of two

.variables p and q, pFqFp . If c p d p is the concentra-l gtion of component p, then

pg

c p dpsc .Hpl

. .and c p d prcsx p d p, and c is a constant. The Stefan-Maxwell equations now become

x p x q N qp p . . .g gyc sN p dqyx p dq. . .H H

z D p ,q D p ,q . .p pl l

Figure 4. Component five with a flat profile and a nega-

tive flux.



13/18

.If we make the change of variables by normalizing D p,q asbefore, then

N p l D p ,q z . .sZ p ; p ,q s ; s , . .

D p ,p c D p ,p l . .l l l l

then Eq. 28 becomes

x p x q Z qp p . . .g gy sZ p dqyx p dq, . .H H p ,q p,q . .p pl l

29 .

and, if we assume that we have 1-D steady-state diffusion at

constant pressure, we must specify the mol fraction distribu- .tion at z s0 and zsl, or at s0 and s1. At s0 x p

0 . . l .sx p and at s1, x p sx p .Having a continuous distribution of components implies

that we must have a scheme for determining the pairwise dif-

fusion coefficients for that infinite distribution. One probably

would only be interested in such a problem when the mixture

has some organized system in its composition. One that comes

to mind immediately, of course, is that of a continuous distri-

bution of hydrocarbons. The building blocks of hydrocarbons,in general, vary from C to CH to CH with a few CH s2 3added on. Trying to compute the averageCH is a problem.For saturated hydrocarbons CH is CH . For unsaturated

2

hydrocarbons CH will be CH or CH . If we use the 2 1 .Fuller formula Taylor and Krishna, 1993, p. 68 for pairwise

diffusion constants we have

1 1q(7r4 M Mi jT

D sci j 23 3PVq V' 'i j /

where M and M are the molecular weights and V and Vi j i jare the partial molecular diffusion volumes. For carbon, the

appropriate partial diffusion volume is 15.9 and, for hydro-

gen, it is 2.31. These values come from Taylor and Krishna .1993, p. 69 . For CH, the appropriate partial diffusion vol-2ume is 20.53. For the average, we would take 20 and for the

molecular weight 13.8 p and 20 p for the approximate V.iThe Fuller formula for a hydrocarbon mixture then will be

1 1q(1.75 p pi jT 1

D p,p sc .i j 2 2r33 3P '13.8 20 .p q p' 'i j /

and, if we choose Ts500 K, Ps500,000 Pa and with csy2 1.01310 , D will be in meters squared per second multi-

y6 .ply the above by 10 ,

1 1q(p pi j

y6 2D p,p s36.8 10 mrs . .i j 23 3p q p' 'i j /

Now, with a little manipulation, one can discover a different

form for this relation with which it is easier to compute. While

we will not show a number of possibilities, let us consider one

that seems like the best and easiest. If we write

1 11 1qq (( 7r6 1r6 7r6 1r6p q p q q p1

s2 7r12 7r12 23 3 p q 1 1' 'pq q

q / 1r3 1r3

/p q

1s w

7r12 7r12p q

and, if we compute w for values of p and q from psqs5to psqs20 in various increments of p and q separately,

we will find that the average value of w will be 0.3567 with a

standard deviation of 0.0053 and the value of w for psq is' .2r4f0.3536. If in addition we normalize the D p,q by

. .D p,p then with ps 6l l l

8.09 s 30 .pq 7r12 7r12p q

In cases where we make calculations for a large number of

chemical species we will use the above formula for the com-

putation with the pairwise normalized diffusivities. The nor-

malization factor used above is specific to the numerical solu- .tions which will follow where we have used D 6,6 .

More Examples

We are now in position to compute with the Stefan-Maxwell

equations for the continuous distribution case. We will use a

lumped constant model for all of these.

Example 5

This is a problem in which the continuous distribution isdescribed at the two endpoints by

1 2py2 py16 ; 6FpF16, . .~x p ,0 s . 2,3650

0; 16FpF20,

1 py12 30yp ; 12 FpF20, . .~x p ,l s 31 . .444l

0; 6FpF12

where we think of p as a continuous variable. We will usew xonly integers in 6, 20 for p

xs x p, , is1,2,3,...,15, .i i

p s6, p s7,...,ps iq5,...,p s20.1 2 i 15

The normalization factors above are for fifteen components.

For the diffusivities, we use Eq. 30. The boundary values are

listed in Table 4 and are computed from the definitions in

the problem. The calculations in Figure 5a show what appear

to be almost straight lines. If one calculates the difference

between these solution curves and the corresponding straight

lines, one obtains Figure 5b which shows that this difference



14/18

Table 4. Boundary Conditions for the 15 Components inExample 5 and Figure 5a.

i 1 2 3 4 5 6 7 8

.x 0 .1691 .1712 .1624 .1450 .1218 .0951 .0677 .0419i .x l 0 0 0 0 0 0 0 .0383i

p 6 7 8 9 10 11 12 13i

i 9 10 11 12 13 14 15

.x 0 .0203 .0055 0 0 0 0 0i .x l .0721 .1014 .1261 .1464 .1622 .1734 .1802i

p 14 15 16 17 18 19 20i

for curves with positive slopes are concave negative while

those with negative slopes are concave positive, but not by

much. It is interesting to determine what happens if one uses

random diffusivities between 0.25 and 1.25. The profiles are

not discernibly different from those in Figure 5a and the

comparison of the fluxes given in Figure 5c with the dots rep-

resenting the formula diffusivities and the asterisks the ran-

dom diffusivities case. The random diffusivities range is well

outside the physical range of .i j

Example 6

This example is the same as Example 5. However, instead

of using 15 components as integers, we will now use 57, but

with the same distribution from ps6 to ps20 so now

iy1ps 6q , is1,2, . . . ,57.i

4

We must make it known that the normalizations must now be

different since in Example 5 we had 15 rectangles over which

to sum where now we will have 57. To accommodate this, we

will replace 2,365 in Eq. 31 with 8,866 and 444 by 1,661. Note

that these have the common ratio of 3.75. The profile struc-ture is given in Figure 6a. In Figure 6b we plot the fluxes vs.

Figure 5a. Fifteen-component case with essentially

straight line profiles.

Figure 5b. Local difference p between computational

profiles and straight line profiles.

the component number. We do this problem primarily to show

in Figure 6c that what appear to be straight lines are begin-ning to look more and more like straight lines when the

straight line comparison is made; however, comparison of

Figures 5b and 6c shows that the convergence may be slow.

In Figure 6d is the plot of flux vs. straight line slope and it

appears that k is close to 2r5 so

2 c dx 2w xN p sy D sy cD x p ,l yx p ,0 ; . . .

5 l dz 5

so that the coefficient of diffusion coefficient for this case is

2r5D where D was the normalizing diffusivity in . We willi jshow in the Theoretical section that this is a natural conse-

quence of the continuous distribution case. Figure 6d alsoindicates that the convergence to straight line profiles may be

Figure 5c. Fluxes for Example 5: formula diffusivities( ) () and randomly chosen diffusivities .



15/18

Figure 6a. Case with 57 components and three different

sets of profiles.

slow. Figure 6a is a little confusing since the profiles for 12 FpF16 do not emanate from the corners and in fact there is aprofile among 26 i40 for which it will be almost flat.

Example 7

This example is presented to illustrate the case where con-

tinuous distribution is augmented by three components which

have a finite contribution. With our lumping procedure, this

is no more difficult than the others. This time, for the lump-

ing, the continuous distribution will contain 57 species and,

with the finite three, we have

iy1ps 6q , is1,2, . . . ,57,i

4

p s3, p s 4, p s5.58 59 60

In this case rather than using integers we will again be using

jumps of one-quarter unit from ps6 to ps20, hence, ap-proximating the continuous distribution by a finer set of rect-

Figure 6b. Fluxes for 57-component case.

Figure 6c. Local difference between calculated profiles

and linear profiles.

The difference is less, but changing slowly.

angles. We will use the same distribution, but we must mod-

ify its application somewhat. At s0, we will write

2py2 py16 ; 6FpF16, . .i i iy p,0 s .i 0; 16FpF 20,ipy12 30yp ; 12 FpF 20, . .i i i

y p ,l s .i 0; 6FpF12.i57 57

l 0 . .We set Sums y p,l ; Sum s y p,0 and let i iis 1 i s 1

x p ,0 s0.8y p ,0 rSum0 , is1,2, . . . ,57, . .i i

x p,l s0.8y p ,l rSuml, is1,2, . . . ,57, . .i i

Figure 6d. Dimensionless slope vs. dimensionless flux

showing tendency toward linearity.



16/18

and with

x p ,0 s0.04; x p ,l s0.12, . .58 58

x p ,0 s0.06; x p ,l s0.02, . .59 59

x p ,0 s0.10; x p ,l s0.06. . .60 60

Also

8.09 s ; .25F F1.i j i j7r12 7r12p pi j

The results of the calculations are presented in Figure 7a,

and one immediately notices that the profiles are again

straight lines or approximately so. If one makes the same cal-

culations, but this time using diffusivities taken from a ran-

dom number generator in the interval 0.15 1.25, onei jobtains what appear to be the same profiles. The fluxes vary

obviously, depending upon the field of random numbers se-

lected. The dots are the diffusivities chosen from the Fuller

formula, while the pluses and asterisks are the two results

from different domains of random numbers. The comparison

of the three is in Figure 7b.Calculations were also made in order to determine the ef-

fect of the continuous domain on the solution profiles for the

discrete domain, if any. For a very small continuous domain,

the solution for the discrete domain is similar in shape to

Figure 7a with slightly curved profiles as before. Changing

the diffusion coefficients, such as , 58F ijF60, fromi jthese values determined by the above formula to strongly dif-

.ferent coefficients such as, the ones used in Example 2 pro-duces a strong effect on the solution profiles for the discrete

domain with little or no effect on the solution profiles for the

continuous domain.

Theoretical Justification

We have seen in the numerical solutions for the continu-

ous distribution cases that the concentration profiles are

straight lines or at least very close to straight lines. This seems

like something that should be evident from the equations.

Suppose we write Eq. 29 in the form

x p , Z qp . .gZ p s y qx p , dq . .H / p,q .pl

y1x q ,p .g

dq .H / p ,q .plThe lefthand side is not dependent upon , so the derivative

of the righthand side with respect to must be zero giving

x2 x x x x x Zp p p pg g g g

0sy dqq dqq dq dqH H H H2 p p p pl l l lx

Zp pg gyx dq dqH H

p pl l

Figure 7a. Concentration vs. mol fraction with 57 com-

ponents in the distributed section and three

finite compositions.

which can be written as

x2 x x xp p g g

dqy dqH H2Zp p pg l ldqsH xpl

x xp p g gdqyx dqH H

p pl l

Figure 7b. Dimensionless flux vs. component number

with formula based diffusivities and vs. two

sets of randomly chosen ones with total of

60 components.



17/18

The left hand side is independent of so that differentiation

with respect to , and, after some manipulation, using

x

x Ap p g gAs dq; s dq,H H

p pl l

we obtain

2 3 2 2x A A x x x A x AA yx qA y y

2 3 2 2 / 2 2 2 x A x x A

y A yx A y s02 2 2

which will certainly be equal to zero if x is a linear function

of . So, we choose as that linear function

w xx p , s x p ,1 yx p ,0 qx p,0 . . . . .

Since now the solution profiles are established, one mustcheck on the fluxes. Substitution of the linear solution in the

Stefan-Maxwell equations gives

w xy x p ,1 yx p ,0 . .

w xx q,1 yx q,0 qx q ,0 . . .PgsZ p dq .Hpl

Z qp .gw xy x p ,1 yx p ,0 qx p ,0 dq. . . . . Hp1

Again, the lefthand side is independent of , so differentia-

tion gives

w xx q ,1 yx q ,0p . .gZ p dq .H

pl

Z qp .gw xy x p ,1 yx p ,0 dqs0 . .Hp1

Thus

w xZ p syk x p ,1 yx p ,0 . . .

and because the profiles are straight lines, we must have

dx p, .Z p syk , k0. .

d

Now, since Z and the derivative are both dimensionless, k

must be dimensionless and so, reverting to dimensions again

kcD dx dxNsy sykcD ,

l d dz

where D is the normalizing diffusivity; for this to be valid, k

must be a constant since otherwise the integral sum of the

fluxes will not be zero.

We see now that the continuous distribution case for the

one-dimension diffusion will be

x p, x q , Z qp p . . .g gy sZ dqyx p, dq .H H

p ,q p ,q . .p pl l

. . .which holds for 01; and x p,0 sf p and x p,1 s0 .f p where f and f are prescribed functions of p where1 0 1

each is always non-negative and

p pg gf q dqs1, and f q dqs1 . .H H0 1

p pl l

.where f q dq stands for the mol fraction of component q.iWe know also that

pgZ q dqs0 .H

pl

which follows from the basic assumption of component bal- .ancing. We also know that p,q 0. The sort of problem

in which we are interested is in which the number of

molecules per unit of volume is always fixed and constant,

and that the reservoirs at s0 and s1 are of fixed com-position.

Because of the fact that the profiles of concentration are

straight lines, the problem may be considered to fall into two

cases. If f and f have no components in common, then all0 1of the profiles pass through either s0 or s1 and havetheir termini on s1 or s0, respectively. On the otherhand, if f and f have components in common, the corre-0 1sponding profile will run from the mol fractions on

s0 to

the mol fraction on s1. The profile will have slope zero ifthe corresponding mol fractions are also equal as it should

be.

Conclusions

Some of the following problems have been faced in this

article successfully and others somewhat less so: .1 The ternary case for the Stefan-Maxwell equations has

resulted in a successful inversion and has produced a simpler

presentation than might have been expected because of Eq.

6. .2 The quaternary case resulted in the inversion in a

quasi-symmetric form and a simpler presentation was not very

simple. .3 A numerical scheme was presented which seems to be

very robust for an arbitrary number of components with rea-

sonable convergence speed and was used on all examples

presented. .4 The case with a continuous distribution of components

was solved, and the character of the solution profiles satisfac-

torily explained. With several components, the solution pro-

files are very nearly straight lines. The lumping scheme will

always give slightly curved solutions.



18/18

Acknowledgment

The work leading to this article was not supported by any alpha-betical agencies, either public or private, except the University ofHouston whom we applaud. We thank Jiwen He for valuable consul-tations.

Literature Cited

Burden, R. L., and J. D. Faires, Numerical Analysis, BrooksrCole, .Pacific Drove, CA 2001 .

Fuller, E. N., P. D. Schettler, and J. C. Giddings, A New Methodfor Prediction of Binary Gas-Phase Diffusion Coefficients, Ind. .Eng. Chem., 58, 19 1966 .

Hsu, H. W., and R. B. Bird, Multicomponent Diffusion Problems, .AIChE J., 6 , 516 1960 .

Krishna, R., and G. L. Standard, A Multicomponent Film ModelIncorporating an Exact Matrix Method for Solution to the

.Maxwell-Stefan Equations, AIChE J., 22, 383 1976 .

Stewart, W. E., and R. Prober, Matrix Calculation of Multicompo-nent Mass Transfer in Isothermal Systems,Ind. Eng. Chem. Fun-

.dam., 3, 224 1964 .Taylor, R., and R. Krishna, Multicomponent Mass Transfer, Wiley,

.New York 1993 .Taylor, R., More on Exact Solutions of the Maxwell-Stefan Equa-

tions for the Multicomponent Film Model, Chem. Eng. Commun., .14, 361 1982

Taylor, R., and D. R. Webb, Film Models for Multicomponent MassTransfer Computational Models, Comput. Chem. Eng., 5, 61 .1981 .

Toor, H. L.,Diffusion in Three Component Gas Mixtures, AIChE .J., 3 , 198 1957 .

Toor, H. L., Solution of the Linearized Equations of Multicompo- .nent Mass Transfer, AIChE J., 10, 448 1964 .

Manuscript receied July 12, 2002 , and reision receied Oct. 10, 2002 .


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