Page 504 - Elements of Chemical Reaction Engineering Ebook

Sec. 8.4 Equilibrium Conversion 475

Q 220,000 calls
Sizing the r&= 5 (EU-9.4)
cal
interstage heat cP, - Tm) 18 - - 270)
(400
exchanger and mol. K
coolant flow rate
= 94 molls = 1692 gls = 1.69 kgls
The necessary coolant flow rate is 1.69 kgls.
Let's next determine the countercurrent heat exchanger area. The exchanger
inlet and outlet temperatures are shown in Figure E8-9.2. The rate of heat transfer
in a countercurrent heat exchanger is given by the equation 9

Bonding with Unit (E8-9.7)
Operations

350K Reaction Mixture
Ta 400K e Tcl 270K Coolant
Figure Ea-9.2 Countercurrent heat exchanger.

Rearranging Equation (E8-9.7) assuming a value of U of 1000 cal/s.m2.K. and
then substituting the appropriate values gives

-
A=---- - -
U[(T,,Z - Td) - (Thl - T, ,)I 1000 cal [ (460 - 400) - (350 - 270)1] K
s.m2.K
-. --
-. 220 ln(0.75) m?
- 20
= 3.16 m2
Sizing the heat The heat-exchanger surface area required to accomplish this rate of heat transfer is
exchanger 3.16 m?.
Now let'? return to determine the conversion in the second reactor. The con-
ditions entering the second reactor are T = 350 and X = 0.4. The energy balance
startling from thi\ point IS shown in Figure E8-9.3. The corresponding adiabatic
equilibrium conversion is 0.63. Ninety-five percent of the equilibrium conversion is
60% and the corresponding exit temperature is T = 351) +- (0.6 - 0.4)400 = 430 K.
The heat-exchange duty to cool the reacting mixture from 430 K back to
350 K can be calculated from Equation (E8-9.4):

= -160 -
kcal
S

See page 271 of C. J. Geankoplis, Trunsporr Processes and Unir Qperurions (Upper
Saddle River, N.J.: Prentice Hall, 1993).