Page 179 - Elements of Chemical Reaction Engineering 3rd Edition

Sec. 4.3 Tubular Reactors 151

Gas phase, constant T and P:

u=u0-= FT uo(l+EX):
FTO
1-x
(EI-4.3)

cx
c -AO (Et-4.4)
- (1 +EX)
4. We now combine Equations (E4-4.1) through (E4-4.3) to obtain

Combining the = X dX =
design equation, kCA0 (1 - x)/ (1 + EX) 0 kCAO(1-X)
rate law, and (E41-4.5)
stoichiometry (1 +EX)dX

Since the reaction is carried out isothermally, we can take k outside the inte-
gral sign and use Appendix A.l to cany out our integration.

Analytical solution

5. Parameter evaiuation:

= 0.00415 -
lb mol
ft3

E = yAos = (1)(1+ 1 - 1) = I

We need to calculate k at 1100 K.

(E4-4.7)

0.072 82, OOO caVg mol( 1100 - 1OOO) K 1
1.987 cal/(g mol * K)( lo00 K)( 1 100 K)
_I
= 3.07 s-I
Substituting into Equation (E4-4.6) yields