Question

Air is separated into two product streams in a separator operating at steady state. One product stream is 98 mol% oxygen, and it contains 80% of the oxygen in the air fed to the column. The other product stream is mostly nitrogen. To a good approximation, air is 79 mol% nitrogen and 21 mol% oxygen.

Calculate the quantity of air required, in tons/day, to produce 1 ton/day of the oxygen product. In addition, calculate the mol% nitrogen in the second product stream.

Answer 1

Let F= moles/ hr of Feed air

moles of oxygen in feed = F*0.21

80% of the oxygen is in one product , amount of oxygen in one product= F*0.21*0.8 =F*0.168 moles/hr

Oxygen to be poroduced =1 ton/day= 1000kg/hr ( 1 ton=1000 kg), Moles of oxygen in product= 1000/32 ( Molecular weight of oxygen)= 31.25 moles/day

This is 98% of total product, Product rich in oxygen contains =31.25/0.98 moles.hr of Product= 31.9 kgmoles/day

moles of nitrogen in Oxygen rich strea = 31.9-31.25= 0.65 kgmoles/day

F*0.168 = 31.25 moles/hr, F= 31.25/0.168 moles/ hr=186 kg mole/day

moles of air supplied = 186 kg moles/day Molecular weight of air= 0.79*28+0.21*32=28.84kg/mole

Mass of air required= Moles* Molecular weight= 186*28.84=5364.24 kg/hr of air=5.364 tons/day

Second stream :

N2 = moles of nitrogen entering through air- moles of nitrogen in oxygen rich product=

186*0.79- 0.65 kgmoles/hr =146.35

20 mole% of oxygen has to be there with this stream oxygen in this second stream =186*0.21*0.2=7.8 kgmoles/hr

Hecnce second stream contains ; N2 =146.35 kgmoles/day and O2= 7.8 kgmoles/day

total moles in the second stream= 146.35+7.8=154.15 kgmoles/day

Composition of second stream in mole% :N2: 100*146.35/154.14= 95% and O2= 100-95= 5%