Question - 3 An ammonia air mixture containing 2 % by volume ammonia is be scrubbed with water at 20 oc in a tower packed with 1.27 cm Raschig rings. The water and gas rates are 1170 kg/hr- m2 ,each based on empty tower cross section. Estimate the height of the tower required if 98 % of the ammonia in the entering gas is to be absorbed . the tower operates at 1 atm pressure. The equilibrium relationship is given by the following equation
Ye = 0.746 x
Where , ye = mol -fraction of ammonia air
X = mole fraction in the solution with water.
The height of transfer unit may be taken as equal to 2 m.
Solution --
Ye = Ye /1+Ye
X = x = X/1+X
Where , Ye = mole of ammonia/ mole of air
X = mole ammonia /mole water
Y1 = 2/98 = 0.0204
Y2 = 2/100 *Y1 =0.0004
Average molecular weight of gas - mixture= 0.02 *17 +0.98*29 = 28.76
Gas rate = 1170 /28.76 = 40.68 kg-mole /hr-m2
Inert gas rate = 40.68 *0.98 =39.87 kg-mole /hr-m2
From ammonia ( solute ) balance -
39.87 (0.0204 -0.0004 ) = 65*x1
X1 = x1 = 0.0123
Equilibrium relation is---
Ye1 /1+Ye1 = 0.746 *X1/1+X1
Putting the value of X1 , we get
Ye1 = 0.0092
NTU = ( 0.0204-0.0004)/(0.0204 -0.0092)-(0.0004) / ln [0.0204-0.0032/0.0004] = 6.17
Tower height = NTU *HTU = 6.17*2 = 12.34 m