Page 485 - Elements of Chemical Reaction Engineering Ebook

456 Steady-State Nonisothermal Reactor Design Chap. 8

and therefore we can solve Equation (E8-6.7) for the equilibrium conversion
KC
x, = - (E8-6.13)
l+Kc

Solution by Hand Calculation (you probably won't see this in the 4th edition)
We will now integrate Equation (E8-6.8) wing Simpson's rule after forming a table
(E8-6.1) to calculate (FAo/-rA) as a function of X. This procedure is simiiar to that
described in Chapter 2. We now carry out a sample calculation to show how Table
E8-6.1 was constructed, for example, at X = 0.2.
(a) T = 330+43.33(0.2) = 338.6 K
[ ((360N338.6) j]
(b) k = 31.1 exp 7906 338'6-360 = 31.1 exp(-1.388) = 7.76h-I

Sample calculation (c) = 3.0~~~-0 = 2.9
0412
for Table E8-6.1 Kc = 3.03 exp
2.9
(d) X, = - = G.74
1 + 2.9
mol kmol
(e) -rA = fT)(9.3)$ [I - [ 1 + &)(0.2)] = 52.8 - 52.8 -
-
-
dm3. h m3.h
FAO - (0.9 mol butane/mol tota1)(163. kmol total/h) - 2.778 m3
(0 -- -
kmol
-rA 52.8 -
m3. h
TABLE E8-6.1, HAND CALCULATION
X T(K) k(h-') Kc x, -rA(kmol/m3. h) - (m3)
- rA
0 330 4.22 3.1 0.76 39.2 3.14
0.2 338.7 1.76 2.9 0.14 52.8 2.18
0.4 341.3 13.93 2.73 0.73 58.6 2.50
0.6 356.0 24.21 2.51 0.12 37.1 3.89
0.65 358.1 21.14 2.54 0.715 24.5 5.99
0.7 360.3 31.61 2.5 0.11 4.1 35.8

(E8-6.14)

Using Equations (A-24) and (A-22), we obtain
3 0.1
V = -(0.15)[3.74 + 3(2.78) + 3(2.5) + 3.891 m3 + 7 [3.89 + 4(5.99) + 35.81 m3
8
= 1.32 m3+2.12 m3
F 3 r l