Under steady state conditions, all the purge will be in the purge (stream-7)
Writing nitrogen balance Nitrogen entering the system= nitrogen leaving in the purge
100*10/100 = P1* 25/100, P1= 100*10/25= 40 moles/min , P1 is flow rate of purge
Flow rate of recycle = 3*100 =300 moles/min
Feed contains 30 mole CO and 60 mole H2
Moles of CO and H2 entering = 90 moles/min
Since purge contains 25% purge, rest has to be CO,H2 and CH3OH
Let R= recycle, x= mole fraction of CO, 2x= mole fraction of H2, then CH3OH= 0.75-x
Moles of CH3OH to be formed = 90/3= 30 moles/min. Some of it is withdrawn as product P and in the purge
30 = P+ 40*(0.75-3x) (1)
Writing overall CO balance
30 = P +40*x (2)
Hence subtracting 1 and 2 P+40x= P+30-120x 160x= 30, x= 30/160 = 0.1875
Purge composition ( as well as recycle composition) : CO=0.1875, H2= 2*0.1875= 0.375, CH3OH= 0.75-0.375-0.1875=0.1875
Hence CH3OH in recycle = 40*0.1875= 7.5 moles/min
Hence CH3OH with drawn as product= 30-7.5= 22.5 moles/min
Let R= recycle =3*100 = 300 moles/min
CO and H2 entering from recycle= 300*(0.1875+0.375)= 168.75
CO and H2 entering =90 moles/min
Total of CO and H2 entering = 168.75+90= 258.75 moles/min
Moles of CH3OH that can be formed =258.75/3 =86.25 moles/min
Formed CH3OH =30 moles/min
Conversion per pass =100*30/86.25=34.78%
From the reaction ,1 mole of CO gives one mole of CH3OH
% percent conversion of CO = 100*30/78.35 = 38.3
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