Question

A permeation membrane separates an inlet air stream, F, (79 mol% N₂, 21 mol% O₂), into a permeate stream, M, and a reject stream, J. The inlet stream conditions are 293 K, 0.5 MPa, and 2 mol/min; the conditions for both outlet streams are 293 K and 0.1 MPa. If the permeate stream is 50 mol% O₂, and the reject stream is 13 mol% O₂, what are the volu-metric flowrates (L/min) of the two outlet streams?

Can someone answer this as well as explain?

Answer 1

Writing overall balance gives

F= M+J

2= M+J

M = 2-J (1)

Oxygen balance gives

2*0.21= M*0.5 + J*0.13 (2)

From (1)

0.42= (2-J)*0.5+ J*0.13

0.42= 1-0.5J+0.13J

(5-0.13)J= 1-0.42= 0.58

0.37J= 0.58 J= 0.58/0.37=1.568 moles/min

M= 2-1.5680=0.432 mole/min

These streams leave at 0.1 Mpa and 293 K

From PV= nRT

Volumetric flow rate of permeate (M) and reject can be calculated from V= nRT/P

P =0.1Mpa =0.9869 atm , Moles of permeate , n = 0.432 moles/min , R =0.08206 L.atm/mole.K T=293 K

Volumetric flow rate of permeate (M)= 0.432* 0.08206*293/0.9869 L/min=10.52 L/mim

Volumetric flow rate of reject J ( where number of moles of reject =1.568 moles/min)= 1.568*0.08206*293/0.9869 =38.20076 L/min