Section 18.5 Diffusion into a Falling Liquid Film: (Gas Absorption)
>  restart;
>  Az:=W*dx; Ax:=W*dz;
>  eq:=(NAz(x,z)-NAz(x,z+dz))*Az + (NAx(x,z)-NAx(x+dx,z))*Ax; This should be zero.
>  eq2:=limit(eq/(W*dx*dz),dx=0);
>  pde:=-limit(eq2,dz=0); Compare to eq. 17.5-3
>  NAz:=(x,z)->vz(x)*cA(x,z); The convective term dominates in the z direction. It is given as xA(NAz+NBz) in eq. 18.0-1, but this can be rewritten as cA*vz as long as vz is taken to be the molar average velocity. The assumption that the molar average velocity and the mass average velocity are identical is another approximation in the problem.
>  NAx:=(x,z)->-DAB*D[1](cA)(x,z); The diffusion term is most important in the x direction
>  pde; Compare to eq. 18.5-6
>  pdsolve(pde,cA(x,z)); I keep trying this, but all it tells me is to try separation of variables.
>  cA:=(x,z)->cA0*(1-erf(x/sqrt(4*DAB*z/vmax))); Let's try the book's solution
>  vz:=x->vmax;
>  D[2](NAz)(x,z);
>  diff(NAz(x,z),z);
>  simplify(pde);
>  simplify(diff(NAz(x,z),z)+diff(NAx(x,z),x)); Thus the pde is satisfied.
>  limit(cA(x,z),z=0,right); BC 1 eq. 18.5-12
>  assume(DAB>0,vmax>0,x>0,z>0);
>  limit(cA(x,z),z=0,right);

>  cA(0,z); BC 2 eq. 18.5-13

>  limit(cA(x,z),x=infinity); BC 3 eq. 18.5-4

>  restart; This was one way to get rid of the assumptions
>  cA:=(x,z)->cA0*(1-erf(x/sqrt(4*DAB*z/vmax)));
>  NAx:=(x,z)->-DAB*D[1](cA)(x,z);
>  simplify(NAx(0,z),assume=positive); Compare to eq. 18.5-17
>  WA:=simplify(W*int(NAx(0,z),z=0...L),assume=positive);

Compare to eq. 18.5-18
>  simplify(%-W*L*cA0*sqrt(4*DAB*vmax/Pi/L),assume=positive);