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382 Chapter 7
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maximum capacity, and maintain the volumetric flow rate at 5600 ft /h (21.2
m3/h). In this case, the residence time increases because of the increased reaction
volume, increasing the conversion. Choosing the third option, substitute 1069 ft3
3
(8000 gal, 30.3 m ) for the reaction volume and the same values for x 1A, V v, m 1A,
and k into the equations above for n = 1, and solve for x 2A. Using POLYMATH
[22], the conversion is now 0.4182. This design will give some flexibility. If the
demand for product increases, the feed rate can be increased, but the conversion
will decrease. The original required conversion is 0.37.
The mole balance can now be completed for one CSTR. The inlet molar flow
rate for propylene oxide is calculated above. The inlet molar flow rate of metha-
nol,
800
m 1M = ——— 0.7914 (62.43) = 1,234 Ibmol/h (560 kgmol/h)
32.04
and the inlet molar flow rate of water,
4000
= ——— 0.9941 (62.43) = 13,780 Ibmol/h (6240 kg/mol/h)
m w
18.02
The inlet flow rates (stream 1) are entered into Table 7.1.2. For x = 0.4182, the
2A
outlet flow rates are also entered into Table 7.1.2.
Next, select a heat exchanger and calculate the heat transfer area. First, cal-
culate the required heat transfer, Q, from an energy balance. Obtain the enthalpy
n
of reaction from Equation 7.4.19 and the standard enthalpies of reaction listed in
Table 7.1.1.
AH° R = - 222,600 - (- 123, 000 - 66,600)
AH° R = -33,000 Btu/lbmol (-76,760 kJ/kgmol)
The enthalpy flowing into and out of the reactor for each component is cal-
culated relative to 25 °C (77 °F), using heat capacities from Table 7.1.1. The re-
sults are contained in Table 7.1.3.
Solve for the required heat transferred, Q n, using the energy equation, Equa-
tion 7.4.2 in Table 7.4. Substituting the enthalpy of reaction, the enthalpy into the
reactor, and the enthalpy out of the reactor, obtained from Table 7.1.3, we find that
Q n = - 2,085,000 - 33,000 (308.9) + 9,465,000)
s
6
Q n = -2.814xl0 Btu/h (-2.97xlO kJ/h)
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