Page 255 - Air Pollution Control Engineering

05_chap_wang.qxd 05/05/2004 3:46 pm Page 234
234 Lawrence K. Wang et al.

Solution

A. Calculation of N (number of gas transfer units) using Eq. (13) and with an assumed
og
value for AF of 1.6:
HAP = 20,000 ppmv
e
RE = 98%
HAP = HAP [1 − (RE/100)] (14)
o e
HAP = 20,000 [1 − (98/100)] = 400 ppmv
o
N = ln{[(HAP / HAP ) (1 − (1/AF)) + (1/AF)]} / (1 − 1/AF) (13)
og e o
N = ln{[ (20,000 / 400) (1 − (1/1.6)) + (1/1.6)]} / (1 − 1/1.6)
og
N = 7.97
og
B. Calculation of H using Eqs. (15)–(18).
og
L'' = L/A (18)
column
Use the following values:
L = 17,410 lb/h
2
A = 10.8 ft from Example 5
column
Then, substitute these values into Eq. (18).
L'' = 17,410 / 10.8 = 1,612 lb/h-ft 2
Use the following value
2
G = 0.34 lb/s-ft from Example 5
area
3600G = (3600)(0.34) = 1224 lb/h-ft 2
area
With these values for L'' and 3,600 G , the packing factors are obtained from Tables
area
1 and 2. The following constants (used for determining height of a gas film transfer
units) are obtained from Table 1 for 2 in. Raschig rings:
b = 3.82, c = 0.41, d = 0.45
2
Because 1224 lb/hr-ft is outside the range for 3600 G of Table 1, one must pro-
area
ceed on the assumption that the above packing factors do apply and that any error
introduced into the calculation will be minimal. For 2 in. Raschig rings, the following
constants (used for determining height of a liquid film transfer units) are obtained
from Table 2:
Y = 0.0125 and s = 0.22
From Tables 3 and 4, the Schmidt numbers are obtained:
Sc = 0.66 and Sc = 570
G L
Also,
µ '' = 0.85 centipoises (cP) at 85°F (see Example 5)
L
µ '' = (0.85)(2.42) = 2.06 lb/h-ft (the conversion factor from cP to lb/h-ft is 2.42)
L
Therefore, H (height of gas transfer unit) and H (height of liquid transfer unit) are
G L
determined from Eqs. (16) and (17), respectively.
c
d
H = [b(3600G ) / (L'') ](Sc ) 0.5 (16)
G area G
H = [3.82(1224) 0.41 /(1612) 0.45 ](0.66) 0.5
G
H = 2.06 ft
G