Page 192 - Elements of Chemical Reaction Engineering 3rd Edition

164 Isothermal Reactor Design Chap. 4

lb mol lb
28
hAo = 1.08 - - 30.24 Ibh
X
=
h lb mol
bmol x 32 -
hBo = 0.54 - lb - 17.28 Ibh
-
h lb mol
Ib mol
lb
mIo = 2.03 - - 56.84 Ibh
28
X
=
h lb mol
The total mass flow rate is
lb
hTo = 104.4 -
h
This is essentially the same superficial mass velocity, temperature, and pres-
sure as in Example 4-5. Consequently, we can use the value of Po calculated
in Example 4-5.
atm
Po = 0.0775 -
ft
280 - (2)(0.0775) atdft
-
a=
Ac(l - +)pcPo (0.01414 ft2)(0.55)(120 Ib cat/ft3)(10 atmj
- 0.0166
_-
Ib cat
Substituting into Equation (E4-6.12) yields

lb mol 1
l-ll-' o.0266 G h
L
W=
0.0166Ab cat
= 45.4 lb of catalyst per tube
or 45,400 lb of catalyst total
This catalyst weight corresponds to a pressure drop of approximately 5 atm.
If we had neglected pressure drop, the result would have been

1
w=--; (1+&)h-----EX
k
FAo[ 1-x
1
0.6
(1-0.15) In --(-0.15)(0.6)
1
-
0.0266
Neglecting L
pressure drop = 35.3 Ib of catalyst per tube (neglecting pressure drop) ,
results in poor
design (here 53% and we would have had insufficient catalyst to achieve the desired conpersion.
vs. 60% Substituting this catalyst weight (Le., 35,300 Ib total) into Equation (E4-6.10)
gives a conversion of only 53%.