Question

A gas mixture having 75% by mass Ammonia (NH3) and the remainder Argon (Ar) is passed through a Membrane Filter to recover the Ar. There are two streams, (i) recovered stream with 60% by mass Ar and (ii) waste gas. The recovered stream comprises twenty-five percent (25%) of the inlet gas mass. The inlet flow rate is 100 kmol /hr of gas (total flow in). Calculate the composition and mass flow rate of the waste gas.

Answer 1

Moles od Inlet gas =100 kmol/hr

Basis : 100 kg/hr of Ammonia and Argon

Mass of NH3= 75kg and mass of Argon= 25 kgs

Moles of NH3= 75/17= 4.412 ( 17 is the molecular weight of NH3) and moles of Ar= 25/40= 0.625

total moles of mixture = 4.412+0.625=5.037 kmoles/ hr

5.037 kmoles/hr of mixture correspond to 100 kg/hr

100 kmoles/hr correspond to 100*100/5.037 =1985.039 kg/hr

Let F= Feed =1985.039 Kg/hr

This is sent to membrane fileter to get Argon rich steam (R) and Waste stream (W)

F= R+W

R = 1985.039*25/100=496.32 Kg/hr, W= F-R= 1985.039- 496.032 = 1488.982 Kg/hr, Inlet Argon = 1985.039*25/100= 496.32 Kg/hr

Argon in the Recovered stream = 496.32*60/100=297.792 Kg/hr

Argon in the waste stream =Argon in the feed- argon in the recovered stream= 496.32-297.792 =199.128 kg/hr

NH3 in waste stream= total waste- Argon in waste= 1488.982-199.128= 1289.854 Kg/hr

Composition ;NH3= 100*(NH3 in waste/ total waste)= 100*1289.854/1498.982=86% and Argon =100-86= 14%