Page 479 - Elements of Chemical Reaction Engineering Ebook

450 Steady-State Nonisothermal Reactor Design Chap. 8

2. Energy balance. Neglecting the work done by the stirrer, we combine Equa-
tions (8-42) and (8-50) to obtain

(E8-5.i)

Solving the energy balance for XEB yields

c@, ept (T - To) [ uA( T - T,) / FAO]
XEB = (E8-5.2)
-[AHL(TR) + Aep (T- TR)l
The cooling coil term in Equation (E8-5.2) is

-_ Btu [ 40 ft2 )- 92.9Btu (E8-5.3)
-
UA -
FAb h ft2. OF 43.04 lb mol/h lb mol. OF
Recall that the cooling temperature is
T, = 85°F = 545 R
The numerical values of all other terms of Equation (E8-5.2) are identical to
those given in Equation (E8-4.12):
403.3(T-535) +92.9(T-545)
= (E8-5.4)
-,
36,400 + 7(T 528)
We now have two equations [(E8-4.13) and (E8-5.4)] and two unknowns, X and T.

TABLE E8-5.1. POLYMATH: CSTR WITH HEAT EXCHANGE
Equations: Initial Values:
f (X) =X-( 403.3*( T-535) t92. Q*( T-545)) /( 36400t7* (T-528)) 0.367
f (T)=X-tau*C/( lttau*k) 564
tau=D.l22Q
A=16.96tlO**12
E=32400
R=1.987
k=Afexp( -E/( R*T))

TABLE E8-5.2. EXAMPLE 8-4 CSTR WITH BEAT EXCHANGE
Sol ut 1 on
Uar iable Value f0
We can now use the X 0.363609 -6.779~16
glass lined reactor T 563.729 -6.855~-16
tau 0.1229
A 1.696e+12
E 32400
R 1.987
k 4.64898