Page 195 - Elements of Chemical Reaction Engineering Ebook

166 1,sothermal Reactor Design Chap. 4

TABLE EX-7.1. POLYMATH SCREEN SHOWING EQUATIONS TYPED
IN AND READY TO BE SOLVED.
Equations Initial Values

d(y)/d(w)=-alpha*(l+eps*x) /2/Y 1
d[x)/d(w) =rate/faO 0
faO=1.00
alpha=0.0166
eps=-O. 15
kprime=O. 0266
f=(l+eps*x) /y
rate=kprime*( (l-x)/(l+ePS*X))*Y
w0 = a, w f = 60

3‘000 T
Scale:

0.000 12.00a 24. ooo 36.000 48.000 60, ooc
U
Figure E4-7.1 Reaction rate profile down the PBR.

However, larger errors will result if large values of EX are neglected! By taking into
account the change in the volumetric flow rate (Le., E = -0.15) in the pressure
drop term, we see that 44.0 lb of catalyst is required per tube as opposed to 45.4 lb
when E was neglected in the analytical solution, Equation (E4-7.4). Why was less
catalyst required when E was not neglected in Equation (E4-7.4)? The reasonis that
the numerical solution accounts for the fact that the pressure drop will be less
because E is negative.