Page 188 - Elements of Chemical Reaction Engineering 3rd Edition

160 Isothermal Reactor Design Chap. 4

2 X 0.0775 atdft X 60 ft (E4-5.5)
10 atm

P = 0.265Po = 2.65 atm
(E4-5.6)
AP = Po- P = 10-2.65 = 7.35 atm

Reaction with Pressure Drop
Analytical solution: Now that we have expressed pressure as a function
of catalyst weight [Equation (4-33)] we can return to the second-order isother-
a
W
mal reaction,
A-B

to relate conversion and catalyst weight. Recall our mole balance, rate law, and
I
W
stoichiometry.
Mole balance: FAO dW - -r; (2-17)
dX
-
Rate law: -r; = kCi (4-19)

Stoichiometry. Gas-phase isothermal reaction with E = 0 :
P
c, = CA0(l -X) Po - (4-35)

Using Equation (4-33) to substitute for PIP, in terms of the catalyst weight,
we obtain
Only
for c, = CA0(l -X)(1 -aW)’D
E =O
kC10
dX
Combining: - = -
(1 - X)2 [(I - ff W)”2]2
dW FAO
-
Separating variables: 7 - -(l-aW)dW
dx
kCAo (1 - x)’
Integrating with limits X = 0 when W = 0 and substituting for FAo = CAou,
yields

Solving for conversion gives