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Reactor Design 409

and

K P = 0.027 exp [ 0.21 (T - 773)]

Solving these last five equations simultaneously using Polymath, at a con-
version of X A = 0.45, the catalyst mass is 4164 kg (9180 Ib) and the final tem-
perature is 856.0 K (1540 °R). The decrease in temperature is only 17.2 K (31
°R), which verifies the original assumption that the temperature decrease would
be small.
Next, calculate the reactor dimensions. First, calculate the superficial ve-
locity using the Ergun Equation (Equation 7.14.5). This equation requires calcu-
lating the average viscosity and density. The mole fraction average viscosity at
5
the inlet conditions is 2.408xlO~ Pa-s (0.0241 cp). Also, the mole fraction aver-
3
3
age of the gas density at inlet conditions is 0.7996 kg/m (0.499 lb/ft ). The rec-
ommended pressure drop range across the bed to insure good flow distribution is
given by Equation 7.14.5. The smaller the reactor diameter, the greater the su-
perficial velocity, and the greater the pressure drop. If we select an average
value of (Ap) B of 0.155 psi/ft ( 3550 Pa/m), the calculated superficial velocity
from Equation 7.14.5 is 1.274 m/s (4.180 ft/s).
Now, calculate the reactor diameter. First, calculate the volumetric flow
rate using Equation 7.14.16.

491.4 kmol 1 h 0.08314 bar-m 3 873.2 K
3
V V v — ————————— — ________ __________________ __ ——————— —A T.1I7_> 111 fa ^Ito.l 1L lo)
1Q^ m /o HAS 1 ft /a\
3
1 h 3600 s 1 kmol-K 2.362 bar
Next, calculate the bed area using Equation 7.14.7.
2
A B = 4.195 /1.274 = 3.293 m (116.3 ffVs)
Finally, calculate the bed diameter using Equation 7.14.8.

1/2
D = [4 (3.293) / 3.142] = 2.047 m (6.716 ft)
According to Step 8 in Table 7.14, round off the diameter to 7.0 ft (2.134 m).

Because the bed diameter has increased, the superficial velocity will de-
crease, and therefore the bed pressure drop will decrease, according to Equation
7.14.7. The actual bed area,
2
2
2
A B = 3.142 (2.134) / 4 = 3.577 m (38.59 ft )
and from Equation 7.14.7 the actual superficial velocity,

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