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5.2 Absorption 135

20 10 6
x 0 6
X 0 ¼ ¼ 6 20 10
1 x 0 1 20 10
x 1
X 1 ¼ ¼ 0
1 x 1

Substitute the mole ratios and the carrier gas mole flow rate of G = 80 (mole/s)
into the mass balance Eq. (5.26), and we have:

ð Y 0 Y 1 Þ ð 50 10Þ
L
L

G ¼ ð X 0 X 1 Þ ! 80 ¼ ð 20 0Þ ¼ 2
So the solute-free water flow rate is 160 mol/s. Assuming water molar weight of
18 g/mole, we can calculate the water mass flow rate and it is 2.88 kg/s.
This example also shows quantitatively that for the cases with very low mole
fractions of y and x; 1 y 1 and 1 x 1, then X ¼ x and Y ¼ y. Then the mass
balance Eq. (5.22) becomes
y 0 y 1
L

G ¼ x 0 x 1 ðfor x 1; y 1Þ ð5:27Þ

5.2.2 Absorption Equilibrium Line and Operating Line

Equation (5.24) can be rewritten as

L
Y ¼ ð X X 1 Þ þ Y 1 ð5:28Þ
G
This equation implies that the gas phase mole ratio Y is a linear function of
L
liquid phase mole ratio X if , X 1 ; and Y 1 are constant. Therefore, it describes the
G
relationship between the gas phase mole ratio and liquid phase mole ratio at any
elevation in an operating tower. If we plot a Y versus X in a figure, it is thereby
called the absorption operating line.

5.2.2.1 Absorption Equilibrium Line

A special operating line is equilibrium line. If the absorption tower operates in such
a manner that the gas-liquid system at any elevation reaches equilibrium, then for
any gas mole ratio Y, there is a corresponding equilibrium liquid mole ratio X*.
And Y can be related to X* according to the Henry’s law. The corresponding linear
function defines the absorption equilibrium line.