The rate at which A must be processed is equal to
\frac{2\times 10^{6}}{0.97}lb/7000\, h=295 lb/h = 133,700g/h
Case 1: Single Stirred Tank (CSTR). From equation (8.3.38) for a single CSTR,
C_{A,out}=\frac{C_{A,in}}{1+k\tau } (A)
Now C_{A,out}=0.03C_{A,in}. Thus,
0.03=\frac{1}{1+0.8\tau }
or
\tau _{i, reactor}=\frac{1-0.03}{0.03(0.8)}=40.4 h
The volumetric feed rate may be determined from the production requirements and the density of the material:
v_{0}=133,700\frac{g}{h}\times \frac{cm^{3}}{0.9\, g}\times \frac{1\, gal}{3785\, cm^{3}}=39.3 gal/h
The reactor volume required may now be determined from the space time and the volumetric flow rate:
V_{R}=v_{0}\tau =39.3(40.4)=1586 gal
The heat transfer rate in the reactor can be determined from the energy balance equation (10.3.6):
\dot{Q}=-\frac{F_{A0}(f_{A,out}-f_{A,in})}{\nu _{A}}\Delta H_{R,at,T_{0}}+\sum \left ( F_{iF}\int_{T_{0}}^{T_{out}}\bar{C}_{pi}dT \right ) (B)
Thus,
\dot{Q}=133,700(0.97)(-83)+\int_{20}^{163}133,700(0.5)\, dT
or
\dot{Q}=-10,764,000+9,560,000=-1,204,000 calh =-4780 Btu/h
Hence, for a single continuous-flow stirred-tank reactor, the volume required will be 1586 gal, and the amount of heat that must be removed is equal to 4780 Btu/h.
Case 2: Cascade of three CSTRs. If we denote the composition leaving the nth reactor by C_{n}, it is readily shown that
C_{A1}=\frac{C_{A0}}{1+k\tau } (C)
C_{A2}=\frac{C_{A1}}{1+k\tau }=\frac{C_{A0}}{(1+k\tau )^{2}} (D)
C_{A3}=\frac{C_{A2}}{1+k\tau }=\frac{C_{A0}}{(1+k\tau )^{3}} (E)
where \tau is the ratio of the volume of a single CSTR to the input volumetric flow rate.
For 97% conversion,
\frac{C_{A3}}{C_{A0}}=0.03=\frac{1}{(1+0.8\tau )^{3}}
Rearrangement yields (1+0.8\tau )=0.3107. Thus,
\tau =\frac{1-0.3107}{0.8(0.3107)}=2.77 h
The reactor volume required per CSTR is then equal to 39.3 gal/h × 2.77 h = 109 gal.
The fraction conversion in the effluent from the first reactor can be determined from equation (C) and the definition of the fraction conversion. Because
C_{A1}=\frac{C_{A0}}{1+k\tau }=C_{A0}(1-f_{A1})
then
f_{A1}=\frac{k\tau }{1+k\tau }=\frac{0.8(2.77)}{1+0.8(2.77)}=0.689
From equation (D) and the definition of the fraction conversion,
C_{A2}=\frac{C_{A1}}{1+k\tau }=\frac{C_{A0}(1-f_{A})}{1+k\tau }=C_{A0}(1-f_{A2})
Thus,
f_{A2}=1-\frac{1-f_{A1}}{1+k\tau }=1-\frac{0.311}{1+0.8(2.77)}=0.903
The heat transfer requirements for each reactor may be determined from equations of the form of equation (B). For reactor 1,
\dot{Q}_{1}=133.700(0.689)(-83)+\int_{20}^{163}(133.700)(0.5)dT
or
\dot{Q}_{1} = -7, 646, 000 + 9, 560, 000 = 1, 914, 000 cal/h = 7595 Btu/h
For reactors 2 and 3, there will not be any sensible heat effects because the temperatures of the entering and leaving streams are identical.
\dot{Q}_{2} = 133, 700(0.970 – 0.689)(-83) = -2, 375, 000 cal/h =-9424 Btu/h
and
\dot{Q}_{3} = 133, 700(0.970 – 0.903)(-83) = -743, 500 cal/h =-2950 Btu/h
The equipment requirements that we have determined are well within the realm of technical feasibility and practicality. The heat transfer requirements are easily attained in equipment of this size. The fact that some of the heat transfer requirements are positive and others negative indicates that one should probably consider the possibility of at least partial heat exchange between the incoming cold feed stream and the effluent from the second or third reactors. The heat transfer calculations show that the sensible heat necessary to raise the cold feed to a temperature where the reaction rate is appreciable represents a substantial fraction of the energy released by reaction. These calculations also indicate that it would be advisable to investigate the possibility of adiabatic operation in a cascade of CSTRs.