> restart;

Binary System => xA + xB = 1

> xB(z):=1-xA(z);

We can make a steady state assumption that:

which simplifies equation 18.4-15 to equation 18.5-17. Combining this result with the analogous equation for the second species (B), we can write the following ordinary differential equation:

> deq:=(g*Omega/(R*T))*(MB*VA-MA*VA)=-VA*diff(ln(xB(z)),z)+VB*diff(ln(xA(z)),z);

Solving the differential equation for the extreme separation of A in the centrifuge:

> s:=dsolve({deq,xA(0)=xA0},xA(z)):

> assign(s): xA:=unapply(xA(z),z);

Second boundary condition expresses extreme separation of B:

> eq1:=xA(z)=xA:

Set this equal to one so that it can be compared with the BSL answer:

> e1:=eq1/xA:

This is the BSL's answer to the problem (Equation 18.5-18):

> BSLeq:=(xA/xA0)^(VB)*(xB/xB0)^VA=exp((VA*MB-VB*MA)*(g*Omega*z/(R*T)));

Set this equal to one so that it can be compared with our answer:

> BSLe:=BSLeq/((xA/xA0)^(VB)*(xB/xB0)^VA);

Now test the solution obtained here with BSL's.

> simplify(rhs(e1) - lhs(BSLe));

Thus, we have obtained Equation 18.5-18, giving the distribution of two components at steady-state in terms of their partial molal volumes, pressure gradient, and linear position in the cell.