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Reactor Design 383

Thus, heat must be transferred out of the reactor to maintain the reaction
temperature at 100 °F (37.8 °C).
3
Next, calculate the heat transfer for a jacket, Qj, for the 8000 gal (30.3 m )
standard reactor from Equations 7.4.7 to 7.4.9. The average jacket temperature,

Tj = (5+15)/2 = 10°C(50°F)

Selecting an approximate overall heat-transfer coefficient is a problem be-
cause of insufficient data. Although there are correlations available for calculating
the individual heat-transfer coefficients and hence the overall heat-transfer coeffi-
cients, at the preliminary stage of the process design, we try to avoid detailed cal-
culations. The best we can do is to select a coefficient that best matches the condi-
tions in the CSTR. Because the jacket liquid is water, and the reactor liquid is a
dilute aqueous solution, we find that from Table 7.6, Uj varies from 60 to 110
2
2
Btu/h-ft -°F (341 to 625 W/m -°F) The average value is 85 Btu/h ft 2 °F (483
3
2
W/m -K). From Equation 7.4.9, we find that the standard 8000 gal (30.3 m ) reac-
2
2
tor has a jacket area of 466 ft (43.3 m ). From Equation 7.4.7, the heat that can be
transferred to the jacket,
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Qj= 85 (466) (100 - 50) = 1.981xl0 Btu/h (2.09xl0 kJ/h)
which is insufficient according to Equation 7.4.10 because we are required to re-
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move 2.814xl0 Btu/h (2.97xl0 kJ/h), but the jacket is only capable of removing
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1.981xl0 Btu/h (2.08xl0 kJ/h).
Next, determine if the heat-transfer rate for a coil, Qc, will be sufficient.
From Table 7.6, the closest match we can find for an overall heat-transfer coeffi-
cient is for an aqueous solution in a coil and water in the reactor. From Table 7.6,
2
2
U c varies from 80 to 120 Btu/h-ft -^ (454 to 681 W/m -K), the average being 100
2
2
Btu/h-ft -°F (568 W/m -K). The heat-transfer area for a coil is given by Equation
7.4.13.
2
2
3
273
= 4.6 [3.785xlO~ (8000)] = 44.69 m (480.9 ft )
A c
From Equation 7.4.11, the heat transfer rate for a coil,
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Qc = 100 (480.9) (100 - 50) = 2.405xl0 Btu/h (2.537xl0 kJ/h)
Clearly, a coil alone is also insufficient. Now, if we add the jacket and coil
heat transfer rates,
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6
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Qc + QJ = 1.981X10 + 2.405X10 = 4.386xl0 Btu/h (4.63xl0 kJ/h)
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