Example 18.5-5 3 Component Diffusion with Heterogeneous Chemical Reaction

> restart;

> x:=array(1...3); N:=array(1...3); Dif:=array(1...3,1...3);eq:=array(1...3);

Using arrays to hold the mol fractions, fluxes, diffusion coefficients and equations.

> for i from 1 to 3 do

> eq[i]:=diff(x[i](z),z)=sum((x[i](z)*N[j]-x[j](z)*N[i])/(c*Dif[i,j]),j=1...3); od; The Stefan-Maxwell Equations for a three component system: 18.4-19

> N[2]:=0; N[3]:=-N[1]/n;x[3](z):=1-x[1](z)-x[2](z);

Component 2 is stationary and component 3 comes from nA1->A3 at the catalytic surface.

> eq[1];

> eq[2];

> s:=dsolve({eq[2],x[2](0)=x20},x[2](z)); Unfortunately Maple does not handle DEs among arrays. Let's finish confirming eqs 18.5-29 and 30.

> N[1]:=nu32*c*Dif[3,2]/delta; nu32 is what BS&L call nuAnB

> Dif[3,2]:=Dif[2,3]; Dif[2,1]:=Dif[1,2]; The diffusivity array is symmetric since those are the binary diffusion coefficients.

> Dif[2,3]:=Dif[1,2]*rB; Dif[1,3]:=Dif[3,2]/rAn; Introducing rB and rAn used in eqs 18.5-29 and 30.

> eqA:=simplify(delta*eq[1]/nu32); This should be 18.5-29. Note that z/delta is what they call zeta.

> eqB:=simplify(delta*eq[2]/nu32); This obviously agrees with eq. 18.5-30

> coeff(rhs(eqA),x[1](z)); This is the coefficient of x[1](z) in 18.5-29 and the coefficient of x[2](z) obviously agrees with what we see in eq. 18.5-29.

So we have derived eqs 18.5-29 & 30, but we can not solve them in the form we derived them.