Section 17.6: Diffusion and Chemical Reaction inside a porous Catalyst:

"The Effectiveness Factor"

> restart;

> A:=r->4*Pi*r^2;

> WA:=r->A(r)*NAr(r);

> eq:=WA(r)-WA(r+dr)+RA(r)*A(r)*dr;

> de:=limit(eq/(A(r)*dr),dr=0); Compare with eq. 17.6-3

> NAr:=r->-DA*D(cA)(r); eq. 17.6-4

> de; Compare to 17.6-5

> RA:=r->-kpp*a*cA(r); Volumetric rate of production of cA for 1st order rxn.

> s:=dsolve({de,cA(R)=cAs},cA(r));

> assign(s);

> cA:=unapply(cA(r),r); A little complicated, but we can deal with it.

> assume(kpp>0,a>0,DA>0,R>0);

> simplify(cA(r)); Still rather complicated!

> limit(cA(r),r=0,right); Let's see what happens as r->0

> _C2:=0: Note the infinity. Thus we must set _C1 to 0.

> simplify(cA(r)); This agrees with 17.6-9

> dif:=cA(r)/cAs-(R/r)*sinh(sqrt(kpp*a/DA)*r)/sinh(sqrt(kpp*a/DA)*R): s ubtracting off the expression given in 17.6-9

> simplify(%); There is still one more assume to do.

> WAs:=WA(R); Eq. 17.6-10

> combine(simplify(%)); Oh well, it was worth a try.

> dif:=%-4*Pi*R*DA*cAs*(1-rt*R*coth(rt*R)): Subtracting off 17.6-11

> rt:=sqrt(kpp*a/DA);

> simplify(dif); 17.6-11 is also OK.