Question

1. We aim to absorb 99.7% of ammonia from a gas stream containing 0.013 mole fraction of NH₃. A counter-current packed bed column is used and the inlet liquid is pure water. We wish to operate at (L/G) = 15 (L/G)_min and 20 °C.

If the slope of the equilibrium line is 2.5, and H_Oy= 0.2 m, calculate:

1. The mole fraction of ammonia at the outlet gas.

2. The mole fraction of ammonia at the outlet liquid.

3. The packing height required.

TABLE Q4 – LIST OF EQUATIONS

Overall number of transfer units based on gas phase
Mean logarithmic average of the driving forces for separation

Answer 1

Gas stream containing y1 = 0.013 mole fraction of NH3

Mole ratio of NH3, Y1 = y1/(1-y1) = 0.013/(1-0.013) = 0.0131

(L/G) =1.5 (L/G)min

Slope of equilibrium line m = 2.5

We want absorbs 99.7% of ammonia from a gas stream.

HOy = 0.2 m

inlet liquid water mole ratio of NH3, X2 = 0

a) if gas absorbs 99.7% of ammonia from a gad stream then 0.3% of ammonia exits from the tower.

Y2 = 0.003*Y1 = 0.003*0.0131 = 3.95*10^-5

Mole fraction of ammonia in outlet gas stream y2 = Y2/1(+Y2) = 3.94*10^-5

b) mole fraction of ammonia at outlet liquid,

y = 2.5x

G(y1 - y2) = Lmin(x1 - x2)

y1 is equilibrium with x1,

y1 = 2.5x1

x1 = y1/2.5 = 0.013/2.5 = 0.0052

(L/G)min = (y1 - y2) /(x1 - x2) = (0.013 - 3.94*10^-5)/(0.0052 - 0)

(L/G)min = (0.0129)/(0.0052) = 2.49

Given L/G = 1.5*(L/G)min = 1.5*2.49 = 3.73

(y1 - y2)/(x1 - x2) =  L/G

(0.013 - 3.94*10^-5)/3.73 = x1

x1 = 0.0034

Mole fraction of ammonia at the outlet liquid , x1 = 0.0034

Packing height required :

z = HOy *NOy

Overall Number of transfer unit NOy:

Mean logarithmic average of the driving forces for the separation,

NOy = (y1 - y2) /yLm

yLm = (Δye1 - Δye2) /ln(Δye1/Δye2)

Δye1 = y1 - y1e = y1 -  = 0.013 - 2.5*0.0034 = 0.0045

Δye2 = y2 - y2e = y2 - mx2 = 3.94*10^-5 - 0 = 3.94*10^-5

yLm = (0.0045 - 3.94*10^-5)/ln(0.0045/3.94*10^-5) = 0.00183

NOy = (0.013 - 3.94*10^-5)/(0.00183)

NOy = 7.07

Height of tower z = NOy *Hoy = 7.07*0.2 = 1.41 m