Page 427 - Chemical process engineering design and economics

Reactor Design 407

Bed void fraction 0.445 29
3
3
Bulk density 1300 kg/m (81.16 lb/ft ) 31
Equivalent catalyst diameter 0.005 m (0.0164 ft) 31
5
Ah R = 1.20737xl0 + 4.56 T — T in K 31
p H / K P ) — TinK 31
r A =k(p E -p s
k = 12600 exp (-11000/T) —Tin K 28
K P = 0.027 exp [0.21 (T - 773)] — T in K 28

To solve this problem, follow the procedure outlined in Table 7.15 using
the Equations listed in Table 7.14.
Substitute the molar flow rate of ethyl benzene, m Ao = 4082 kg/h (9000
Ib/h), into Equation 7.14.1, and rearrange the equation to obtain

dW c/ dx A = m Ao/ r A = (4082/106.16) / r A = 38.45 / r A

The average heat capacity, c = 244.5 kJ/kgmol (105 Btu/lbmol), is calcu-
p
lated at the reactor inlet conditions, using heat capacities taken from Reid et al.
[30] and Equation 7.14.13. The changes in temperature and composition through
the reactor will not significantly change the heat capacity.
The enthalpy of reaction (Equation 7.14.17), AhR, given by Froment and
Bischoff [31], is calculated below. We will also assume that the temperature will
not change significantly throughout the reactor. The large excess of steam will
moderate the decrease in temperature. Letting T = 873.2 K (1570 °R), the reactor
inlet temperature,

Ah R = 120737 + 4.56 T = 120737 + 4.56 (873.2)

5
4
Ah R = 1.247xl0 kJ/kgmol (5.36xl0 Btu/lbmol)
Substituting AH, m and C into the energy equation, Equation 7.14.2, we
R Ao> P
obtain
-Ah Rm Ao -124700 kJ 1 kmol-K 38.45 kgmol
dT/dx A = ————— = ————————— ————————— ______
CP mi 1 kgmol 244.5 kJ

dT/dx A =-19610/m T

According to the chemical equation there will be an increase in the molar
flow rate as the reaction proceeds. The total molar flow rate, m, is equal to mo-
T
lar flow rate into the reactor plus the increase in moles caused by the reaction.
Therefore,

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