Carbon monoxide and hydrogen react catalytically to produce methanol. CO + 2HZ r

Question

Carbon monoxide and hydrogen react catalytically to produce methanol. CO + 2HZ rightarrow CH_3OH The fresh feed (100 moles/min) composition is 30 mole% carbon monoxide, a stoichiometric proportion of hydrogen and 10 mole% inert (nitrogen). A condenser (operates at 0 degree C and 1 atm) separates the unreacted gases and inert from the liquid product methanol. Part of the gas stream is recycled back into the main feed line to the reactor and the rest purged. The recycle ratio is initially set to 3 moles/min recycle stream to 1 mole/min fresh feed. Analysis on the purge stream finds it contains 25 volume% nitrogen gas. The liquid product from the condenser is essentially 100% methanol. Based on this information, and assume process in steady state, determine: a) The rate of methanol production (mol/min). The rate and composition of the purge gas. c) The overall and single pass conversation of carbon monoxide for this setup. d) The overall yield of Methanol

Explanation / Answer

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 Some of it is withdrawn as product P and in the purge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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