> restart;
Start with equation 16.3_1 on page 505:
> eq:=p*DAB/((pcA*pcB)^(1/3)*(TcA*TcB)^(5/12)*(1/MA+1/MB)^(1/2))=a*(T/sqrt(TcA*TcB))^b;
Solve for DAB, and use unapply to turn it into a function of T and p.
> DAB:=solve(eq,DAB);
> DAB:=unapply(DAB,T,p);
Constants a and b for non-polar gas pairs are given on p 505. Problem statement gives us experimental value of (pDAB)0 at 293 K.
> a:=2.745e-4: b:=1.823: pDAB293_0:=.163*atm*cm^2/s:
Use equation 16.3-1 to find new (pDAB)0 at 313 K.
> ratio:=p*DAB(313,p)/(p*DAB(293,p));
> pDAB313_0:=ratio*pDAB293_0;
Input pressure and temperature of mixture. Input critical properties of components (from Table B-1) and composition of mixture.
> p:=136*atm: T:=313*K:
> TcCH4:= 190.7*K: pcCH4:= 45.8*atm:
> TcC2H6:= 305.4*K: pcC2H6:= 48.2*atm:
> xCH4:=0.8: xC2H6:=0.2:
Use critical properties of components to find pseudo-critical properties of mixture.
> pc:=pcCH4*xCH4+pcC2H6*xC2H6;
> Tc:=TcCH4*xCH4+TcC2H6*xC2H6;
Now find pseudo-reduced properties of mixture.
> pr:=p/pc; Tr:=T/Tc;
Using matlab program based on Figure 16.3-1 (p 506), the ratios of pDAB to (pDAB)0 is found to be 0.7315. Use this ratio to find pDAB.
> pDAB:=0.7315*pDAB313_0;
Divide by pressure to find DAB.
> DAB:=pDAB/p;
>