Page 322 - Introduction to chemical reaction engineering and kinetics

12.3 Design Equations for a Batch Reactor 303

In the energy balance, for isothermal operation of the reaction A + . . . -+ product(s),
dTldt = 0, so that equation 12.3-16 becomes, for the required rate of heat transfer:

0 = UA,(T, - T), = -(-AH&(-rA)V (isothermal operation) (12.3-23)

From the material balance in terms of fA, equation 2.2-4 becomes

(-r,J = h,Wdf,W (2.2-4)
Eliminating (- rA) from equation 12.3-23 with equation 2.2-4, we have

d = UA,(T, - T), = -(-AH,,)n,,(df,ldt) (12.3-24)

For simplicity, if we assume that T, is constant throughout the coil at a given instant,
then T, depends on t, from equation 12.3-24, through

T, = T _ (-A%$“” $$I --!
(12.3-25)
c

Determine Q and T, (as functions of time, t) required to maintain isothermal conditions
in the reactor in Example 12-1, if AHRA = -47,500 J mol-‘, and UA, = 25.0 W K-l.
Does b represent a rate of heat removal or heat addition?

SOLUTION

To use equations 12.3-24 and -25, we first obtain df,ldt for a first-order reaction, and use
this to eliminate fA in terms of t.
From the material balance, equation 2.2-4,

From (A), on integration,

fA = 1 - e-kAt

Substituting (A) and (B) into equation 12.3-24, we obtain (with t in min)

b = -(-AHRA)nAokAe- kAt = -(47,500)7.5(0.05/60)e-0~05’ = -297e-0,05” J s-l or W
Since 0 < 0, it represents heat removal from the system, which is undergoing an exother-
mic reaction.
Similarly, from equation 12.3-25, we obtain for the coolant temperature T, in the coil

T = 300 _ 47,500(7*5) E,-o.o% = 300 _ 11 $@.05t
c 25.0 60

with T, in K and t in min.