Page 515 - Elements of Chemical Reaction Engineering Ebook

Sec. 8.5 Nonadiabatic Reactor Operation: Oxidation of Sulfur Dioxide Example 485

4. Stoichiometric relationships and expressing -rko, as a function of X:
so,+;o, e
so,
A+;B e
c

We let A, represent SO, and vl the stoichiometric coefficient for species i:
+
(0, v,X)(RT) P (0, + v,X) P
P, = c, (RT) = CA, (E8- 10.8)
(1 +&X)(T/T,)P, = (l+EX)P,
Substituting for partial pressures in the rate law ,and combining yields

where E = -0.055, PAo = 0.22 atm, OSo2 = 1.0, OO2 = 0.91, Os0, = 0.0,
and ON2 = 7.17; FTo = 7900 lb mol/h, and FA, = 869 Ib molIh.

Per tube:
Weight 6f catalyst in one tube = W = p,nD2/4L = 28.54 lb cat./tube
869
F = - 0.188 lb mol/h.tube
=
. 4631
Substituting the values above gives us

The combined mole 1 -x 0.2-0.11xj - - [ --Iz] x
P
balance, rate law, 1 (E8-10.10)
0.055X)
1
-
Po
FAlO
and stoichiometry 1 g== 3=5.32t&{( (1 - X)Kpl
that is;-
- - (X, T, P) (E8-1Cl.11)
dx - .f
dW
The limits of integration are from zero to the weight of catalyst in one tube,
28.54 lb.
5. Energy Balance. For steady-state operation and no shaft work, Equation (8-56)
can be rewritten in terms of catalyst weight as the spatial variable, that is,

-_ - (4u/P@)(Tg- T)+ (-~A)[-~HR,{T)I (E8-10.12)
dT
dW FAO (E @I cp~ + acp)
6. Evaluating the energy balance parameters:

Heat of reaction: