Page 200 - Elements of Chemical Reaction Engineering 3rd Edition

1 72 Isothermal Reactor Design Chap. 4

Parameter evitluation:
Recall that g, = 1 for metric units.

= G( 1 - 0.4)
l(0.032 kg/dm3)(0.02 dm)(0.4)3
(E4-8.11)
150(1 -0.4)(1.5 X 10-6kg/dm-s)+
0.02 dm
Po = t(98.87 s-')G + (25,630 dm2kg)G2] X ( 0.01 k;y$s2) 1 034-8.12)

1 I
The last term in, brackets converts (kg/dmZ. s) to (kPa/dm). Recalling other param-
eters, m = 44 kg/s, L = 27 dm, R = 30 dm, and peat = 2.6 kg/dm3.
Table E4-8.1 shows the POLYMATH input used to solve the above equations.
The MATLAB program is given as a living example problem on the CD-ROM.

TABLE E4-8.1. POLYMATH PROGRAM
Equations Initial Values

d(X) /u(t)=-ra*Ac/Fao 0
d(y) /d(z)--beta/Po/Y*(l+X) 1
Fao=440
P0=2000
CaO=O ,32
R-30
phi=0.4
kprime=O .02
L=27
rhoca t=2.6
m=44
Ca=CaO*( 1-X) *y/( 1+X)
Ac=3.1416*( R"2-( 24) -2)
V=3,1416*( z*R-2-1/3Y (2-L) *3-1/3*LA3)
G=m/Ac
ra=-kprirnerCa*rhoca tf ( 1-phi)
beta=(98.87tGt2563016^2) 10.01
W=rhocatr ( 5-phi) fV
zo = 0, 'f = 54

For the spherical reactor, the conversion and the pressure at the exit are
A comparison X = 0.81 P = 1980 kPa
between reactors
If similar calculations are performed for the tubular packed-bed reactor (PBR), one
finds that for the same catalyst weight the conversion and pressure at the exit are
X = 0.71 P = 308 kPa
Figure E4-8.2 shows how conversion, X, and dimensionIess pressure, y, vary with
catalyst weight in each reactor. Here X, and yI represent the tubular reactor and X,