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G
m2
Y
2
X
2
L
m2
G
m1
Y
1
X
1
L
m1
Slope of operating line =
L
G
m
Y
X
FIGURE 16.5
Operating line for a countercurrent flow absorber. (From USEPA,
APTI Course 415: Control
of Gaseous Emissions
, EPA 450/2-81-005, U.S. Environmental Protection Agency Air Pollution Training
Institute, Washington, DC, 1981, p. 4-17.)
Composition of the liquid into the absorber (
X
2
) = 0
Gas flow rate (
Q
) = 84.9 m
3
/min
Outlet liquid concentration (
X
1
) = ?
Liquid flow rate (
L
) = ?
H
= Henry's constant
Solution:
Sketch and label a drawing of the system (see Figure 16.6).
Y
1
= 3% by volume = 0.03, and
Y
2
= 90% reduction from
Y
1
, or only 10% of
Y
1
; therefore,
Y
2
= (0.10)(0.03) = 0.003. At the minimum
liquid rate,
Y
1
and
X
1
will be in equilibrium. The liquid will be saturated with SO
2
.
Y
1
=
H
×
X
1
From Figure 16.4,
H
= 42.7 (mole fraction SO
2
in air ÷ mole fraction SO
2
in water); thus,
0.03 = 42.7 ×
X
1
X
1
= 0.000703 mole fraction
The minimum liquid-to-gas ratio is
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