Gas Emission Control - Mathematics and Statistics for the Environment - page 456

Y 2 = 0.003

L = ?
X 2 = 0

Q = 84.9 m 3 /min
Y 1 = 0.03

X 1 = ?

FIGURE 16.6 Material balance for 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. 4-20.)

−= (

)

( )

YY LG XX −

1 2 mm 1 2

( )

YY
XX

−

( ) =

1 2

LG

mm

( )

−

1 2

.03 0 003
0 000703

0 − .

=

. − 0

= 38 4 . g-molwater/g-mol air)

Compute the minimum required liquid flow rate. First, convert m 3 of air to g-mol:

At 0°C (273 K) and 101.3 kPa, there are 0.0224 m 3 /g-mol of an ideal gas.
At 20°C (293 K),

3

m
g-mol

293 K
273 K

=

3 /g-mol

0 0224 . 0 024 . m

3

m
min

g-molair
0.024 m

=

G m = 84 9 . 3538 g-m ol air/min

3

L m / G m = 38.4 (g-mol water/g-mol air) at minimum conditions

L m = 38.4 × 3538 = 136.0 kg-mol water minimum

In mass units,

L = 136 kg-mol/min × 18 kg/kg-mol = 2448 kg/min