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05_chap_wang.qxd 05/05/2004 3:46 pm Page 231
Wet and Dry Scrubbing 231

ABS =

B. Again, using Fig. 2, determine the point of flooding ordinate (ORD).
ORD =

C. Based on the chosen packing, use published data (11) to find the packing constants.
a = , e =
Using ref. 1, the viscosity, µ , is
L
µ = centipoise or cp
L
D. Using Eq (8), the G (mass flow of tower at flooding) is now determined:
area,f
3
G = {[(ORD) D D g ] / [(a/e )(µ 0.2) ]} 0.5 (8)
area, f G L c L
G = lb/s-ft 2
area, f
E. As previously discussed, a fraction of flooding for design purposes is typically 0.6 <
f < 0.75.
f = (chosen value)
So Eq. (9) can be used to determine G .
area
G = f G (9)
area area, f
G = lb/h-ft 2
area
F. The cross-sectional area of the column (tower) is now determined from Eq (10):
A = G / (3600G ) (10)
column area
A = ft 2
column
G. Equation (11) will now yield the column (tower) diameter.
D = [(4/π(A )] 0.5 = 1.13(A ) 0.5 (11)
column column column
D = ft
column
Note, as previously discussed, a design should always be made to a readily available stan-
dard size of tower, normally a whole foot diameter. So this exercise is normally, in real
practice, repeated for several iterations until such a tower diameter is calculated.

Example 5
Again, using Emission Stream 5 from Table 11, a packed tower (wet scrubber; Fig 1b) is
determined to be the most effective control solution for the given pollutant/gas system.
Determine the column (tower) diameter using the calculated results from Example 3 and
the procedures in Example 4.
Solution
Using Eqs. (6a), (6), and (8)–(11),
L = (MW )(L ) = (18)(967) = 17,410 lb/h
solvent mol
G = (G ) MW = (465)(28.4) = 13,200 lb/h
mol e
D = PM/RT (6a)
G
D = (1.0)(28.4) / [(0.7302)(460 + 85)] = 0.071 lb/ft 3
G
From ref. 1 at 85°F,
3
D = 62.18 lb/ft at 85°F
L