Page 191 - Elements of Chemical Reaction Engineering 3rd Edition

Sec. 4.4 Pressure Drop in Reactors 163

Separating variables to form the integrals yields

Integrating gives us

Solving for W, we obtain

1 - [I - (3aFAO/2kr)((l $-E) In[ 1/(1 -x)] - Ex)]u3
W= (E4-6.11)
a
6. Parameter evaluation per tube (i.e., divide feed rates by 1000):

Ethylene: FAo = 3 X lb mob's = 1.08 lb movh
Oxygen: FBo = 1.5 X lb moYs = 0.54 lb mom
0.79 mol N,
I = inerts = N,: F, = 1.5 X 10-4 lb molVs X 0.21 mol 0,

Fi = 5.64 X lop4 Ib moUs = 2.03 ib moh
Summing: F, = FAo + FBo + Fx = 3.65 ib moyh

3
E = yAoG = (0.3)(1 - - 1) = -0.15
PA0 = yaoPo = 3.0 atm
lb lb mol
k' = kPA0(i)2/3 = 0.0141 X 3 atm X 0.63 = 0.0266 -
atm.lb catch h Ib cat
1 - [1 -(3aFAo/2k'){(l -0.15) ln[l/(l -0.6)]-(-0.!5)(0.6)}12/3
w=-
a
For 60% conversion, Equation (E4-6.11) becomes

1 - (1 - 1.303aFA0/k')2'3
W= (E4-6.12)
ci
In order to calculate a,

Evaluating the 2PO
pmsure drop
parameters = A,(1 - +)P,PO
we need the superficial mass velocity, G. The mass flow rates of each entering
species are: