Page 181 - Elements of Chemical Reaction Engineering Ebook

152 Isothermal Reactor Design Chap. 4

0.425 lb molls
V=
(3.07/s)( 0.004 15 lb mollft3)
(E4-4.8)
= 33.36ft3 [ 2 In (LIXJX] -

For X = 0.8,

-
V = 33.36 ft3 [z In [A) 0.81
= 80.7 f? = (2280 dm3 = 2.28 m3)

. It was decided to use a bank of 2-in. schedule 80 pipes in parallel that are
40ft in length. For pipe schedule 80, the cross-sectional area is 0.0205 ft2. The
number of pipes necessary is

The number of PFRs n= 80.7 ft3 = 98.4
in parallel (0.0205 ft2)(40 ft)

Equation (E4-4.7) was used along with Ac = 0.0205 ft2 and Equation$ (M-4.3) and
(M-4.4) to obtain Figure €3-4.1. Using a bank of 100 pipes will give us the reactor

- 0.8
- 0.7

- 0.6
- 0.5 C
0
n .-
5 E
- 0.4
>
c
- 0.3 0
V
- 0.2
- 0.1
0 I I, 1 I I 1 1 I. I IO
0 5 IO 15 20 25 30 35 40 45 50
Distance down the reoctor,L (ft)

Figure. E4-4.1 Conversion and concenwation profiles.

volume necessary to make 300 million pounds per year of ethylene from ethane.
The concentration and conversion profiles down any one of the pipes are shown in
Figure E4-4.1.