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Separator Design 329

m v = 1.5 (65.59) = 98.39 Ibmol/h (44.6 kgmol/h)
Now, calculate the number of equilibrium stages, N, from Equation 6.21.5S,
e
and then divide the result by the column efficiency to obtain the actual number of
trays, N A.

From Equation 6.21.4, the absorption factor,
4
A A = 1.919xl0 / 289.5 (98.39) = 0.6734
From Equation 6.21.5S, we obtain.

5
2.119xl(T -0
Ne
(l/A A) = ———————— (1 - 0.6734) + 0.6734 = 33.33
7
2.119xlO~ -0
log 33.33
N e = ———————— = 8.868
log (I/0.6734)

The column height, given by Equation 6.21.7T for tray columns or by Equa-
tion 6.21.7P for packed columns, depends on the column diameter, which we will
now calculate. From the ideal gas law, the volumetric air flow rate,

98.39 Ibmol 0.7302 atm-ft 3 560 °R
3
3
3
4
V v = —————— ————————— ———— =4.023xl0 ft /h(1.14xl0 m /h)
1 h 1 lbmol-°F 1 arm
Assume a superficial velocity of 2 ft/s (0.61 m/s). Calculate a preliminary
column cross-sectional area.
4.023x10 4 1
2
2
A = ————— — = 5.588 ft (0.5191 m )
3600 2
and from Equation 6.23.3 the column diameter,

(4 (5.588) V 5
D= ————— I =2.667 ft (0.7981m)
I 71 )

Because D > 2.5 ft (0.7620 m), select a tray column.
Next, determine an actual superficial velocity, using Figure 6.18. The flow
paramater is out of the range of Figure 6.18. Select 2.0, which means the air flow

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