Page 256 - Air Pollution Control Engineering

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Wet and Dry Scrubbing 235

s
H = Y(L''/µ '') (Sc ) 0.5 (17)
L L L
H = 0.0125 (1612 / 2.06) 0.22 (570) 0.5
L
H = 1.29 ft
L
Use AF = 1.6. Then Eq. (15) can be used to determine H (height of the overall gas
og
transfer unit):
H = H + (1/AF) H (15)
og G L
H = 2.06 + (1/1.6) 1.29 = 2.87 , or 2.9
og
C. Calculation of Ht using Eq. (12).
Ht = N H (12)
column og og
Ht = (7.97)(2.9) = 23.1, or 23 ft
column
D. Now use Eq. (19) to determine Ht
total
Ht = Ht + 2 + (0.25D ) (19)
total column column
Ht = 23 + 2 + [(0.25) (4)] = 26 ft
total
E. The volume of packing needed to fill the tower is determined from Eq. (20).
2
V = (π/4)(D ) (Ht ) (20)
packing column column
2
V = (0.785)(4) (23) = 290 ft 3
packing
Example 8
In this example, a step-by-step procedure for determining the pressure drop of the packed
bed of the wet scrubber (Fig. 1b) is presented.
Solution

A. Use Eq. (21) to the calculate pressure drop, P , select a packing, and determine the
a
constants (g and r) using Table 5.
g = r =
P = ( g × 10 −8 )[ 10 ( rL D L ) ](3600 G ) 2 D G (21)
′′
area
a
P = pressure drop (lb/ft -ft)
2
a
B. Use L'', D , G , and D from previous exercises.
L area G
C. Use Eq. (22) to calculate P . The total pressure drop through a packed tower (Fig.
total
1b) or wet scrubber is
P = P Ht (22)
total a column
P = lb/ft 2
total
P /5.2 = in. H O
total 2
Example 9
Emission Stream 5 (Table 11) is again a candidate for pollution control using a packed
tower (wet scrubber, Fig. 1b). The total pressure drop through the entire scrubber tower is
determined from the results of Examples 1, 3, 5, and 7 and the procedure just explained in
Example 8.
Solution
From Table 5, the packing constants g = 11.13 and r = 0.00295 and