Page 165 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

5.2 Absorption 139

8
x 0 ¼ 8:32 10 ðmole of H 2 S/mole of waterÞ
Then the outlet H 2 S mole fraction in the air is

y 1 50 ppmv ¼ y 0

This result indicates that the wet scrubber that operates at equilibrium governed
by Henry’s law is basically useless in absorbing the air pollutant into the liquid
phase. This is calculated based on the assumption of equilibrium at the bottom of
the system. This actually makes sense because that is what equilibrium implies—no
net mass transfer between gas and liquid phases. Practically speaking, this system
works along the minimum operating line, but it is not that effective in gas
absorption. In practice, the operating line must be different from this line.
Example 5.6: Absorption operating line
A packed bed wet scrubber is designed to remove high concentration SO 2 from the
exhaust of a sulfuric acid plant. It is expected to achieve a removal efficiency of
95 %. The incoming SO 2 concentration is 10 %. Pure water is used as an absorbent
and the solute-free liquid to gas ratio is 1.5 times the minimum ratio. Assume that
the system operates at an average temperature of 30 °C and 1 atm. The equilibrium
data for SO 2 in air and water at this temperature are as follows [18]

Partial pressure 0.6 1.7 4.7 8.1 11.8 19.7 36 52 79
(mmHg)
p SO 2
SO 2 concentration in 0.02 0.05 0.1 0.15 0.2 0.3 0.5 0.7 1
(g SO 2 /
water c SO 2
100 g water)
Plot the equilibrium line, the minimum operating line and the operating line in
the same figure.

Solution
To determine the equilibrium line Y versus X, we need to determine the mole
fraction changes in liquid and gas phases in the scrubber.
Step 1: Determine the equilibrium line by the mole fraction in gas by Dalton’s
law, Eq. (2.40) and liquid phases, respectively

P SO 2
¼
y ¼ y SO 2
P
c SO 2
64 g/mole
¼
x ¼ x SO 2 100 g
c SO 2
64 g/mole þ 18 g/mole