Page 503 - Elements of Chemical Reaction Engineering Ebook

474 Steady-State Nonisothermal Reactor Design Chap. 8

>OO 350 400 450 500 600

T(K)
Figure ES-9.1 Determining exit conversion and temperature in the first stage.

There is no work done on the reaction gas mixture in the exchanger and the
reaction does not take place in the exchanger. Under these conditions (Fl,,,, = F,,&
the energy balance given by Equation (8-lo),
Q- Ws+C FioHi-z FiHj = 0 (8-10)

becomes
Energy balance on (E8-9.1)
the reaction gas
mixture in the heat
exchanger (E8-9.2)

(E8-9.3)

kcal
= -220- (E8-9.4)
S
We see that 220 kcal/s is removed from the reaction system mixture. The rate at
which energy must be absorbed by the coolant stream in the exchanger is
(E8-9.5)
Q = htCepc(~out - Tin)
We consider the case where the coolant is available at 270 K but cannot be heated
above 400 K and calculate the coolant flow rate necessary to remove 220 kcal/s
from the reaction mixture. Rearranging Equation (E8-9.5) and noting that the cool-
ant heat capacity is 18 cal/mol.K gives