Total Molal Volume of Gas Mixture at 60 O F and 1 Atm= 359*( 460+32)/(460 +32 )= 379 Ft3/Lb

Srubber problem-

QUESTION - 21 A mixture of 25% ammonia gas and 75 % air ( dry basis) is passed upward through a vertical scrubbing tower , to the top of which water is pumped. Scrubbed gas containing 0.5 % ammonia leaves the top of the tower, and an aqueous solution containing 10 % ammonia by weight leaves the bottom. Both entering and leaving gas streams are saturated with water vapour. The gas enters the tower at 100 o F and leaves at 70 o F. The pressure of the of both streams and throughout the tower is 15 lbf/in2 gauge. The air – ammonia mixture enters the tower at a rate of 1000 ft3/min, measured as dry gas at 60 Of and 1 atm. What % of ammonia entering the tower is not absorbed by water ? How many gallons of water per minute are pumped to the top of the tower ?

Solution - it is the problem based on the mixture of gases . so ideal gas law formula may be used . temperature and pressure vary, and use of lb/moles will simplify the problem for the material balance.

Total molal volume of gas mixture at 60 o F and 1 atm= 359*( 460+32)/(460 +32 )= 379 ft3/lb mole

As we know - 1 lb /mole gas contains 359 ft3 space at NTP ( AT 32 Of and 1 atm pressure )

And total ammonia and air entering the tower = 1000/379 = 2.64 lb/moles -min

The ammonia entering the tower = 25/100*2.64 =0.660lbmoles/min

Dry air = 75/100 *2.64 =1.98 moles /min

The molal ratio of ammonia/air in the effluent gas = 0.5/100/99.5/100 = 0.005/0.995

And the unabsorbed ammonia = 1.98 *0.005/0.995 = 0.0099 moles/min

The fraction of entering ammonia = 0.0099 /0.66*100 = 1.5 %

The vapour pressure of water at 100 o F and at 70 o F IS 0.949 and 0.363 lb/in2 respectively (

from the chart ) .

the total pressure of the gas= 14.7 +15 .0 = 29.7 lb/in2

the partial pressure of water in the entering gas = 0.949 lb/in2

dry gas partial pressure = 29.7- 0.949 = 28.75 lb/in2

as the partial pressure ratio = molal ratio

water in inlet gas = (0.949 /28.75 )*2.64 = 0.0872 moles/ min

since the air and ammonia in the effluent gas are = 1.98 +0.0099 = 1.99 moles/min

water in this stream = 1.99*0.363 /( 29.7 - 0.363 ) = 0.0246 mole/min

the water condensed from the gas stream = ( 0.0872 - 0.0246 )*18*60 = 67 lb/hr

the ammonia liquor leaving the tower = 10% by weight . and carries = 0.66 – 0.0099 =0.650 moles of ammonia /min

the total water in the liquor = 0.650 *17 *0.9/0.1 = 99.4 lb/min

of this , 67 lb/hr , or 1.1 lb/min , is absorbed from the entering air and = 99.4 -1.1 = 98.3 lb/min is pumped to the tower.

The mass of 1 gallon water = 8.33 lb

The volume of water to the tower = 98.3/8.33 =11.8 gallons/min

And the water condensed from the gas = 1% of the total water in the system may be neglected.