Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor
ID: 1009142 • Letter: M
Question
Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor.
The fresh feed to the process contains 32.0 mol% CO, 64.0 mol% H2 and 4.00 mol% N2. This stream is mixed with a recycle stream in a ratio of 10.00 mol recycle / 1 mol fresh feed to produce the feed to the reactor, which contains 10.0 mol% N2.
The reactor effluent goes to a condenser from which two streams emerge: a liquid product stream containing essentially all of the methanol formed in the reactor, and a gas stream containing all of the CO, H2, and N2 leaving the reactor.
The gas stream is split into two fractions; one is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor.
For a methanol production rate of 100.0 mol/h, calculate the fresh feed rate (mol/h), the molar flow rate and composition of the purge gas, and the overall and single-pass conversions.
Explanation / Answer
The equation is
2CO + 4H2--> 2CH3OH
32.0 mol% CO, 64.0 mol% H2 and 4.00 mol% N2
is .32 x 28 = 8.96
.64 x 2 = 1.28
.04 x 28 = 1.12
SO wt% of CO is 8.96/(8.96 +1.28+1.12) = 78.9%
wt% H2 = 11.3%
wt% N2 = 9.9%
Since the recycle stream contains 10 mol% N2 it is enriched from 4 mol%
For 2 mole of CO 4 mole of H2 is consumed and 2 moles of CH3OH is formed
When 6 mole% of CO + H2 is consumed it means 2 mol% CO and 4 mol% H2 is consumed and 2mol% CH3OH is formed.
Per pass conversion is 2mol%/32mol% x 100 = 6.25%.
To get 100 mol/h of methanol we will need 100/0.02 = 5000 passes/h
5000 mol/h of feed
purge gas composition is
30 mol% CO, (as 2 mol% consumed)
60 mol% H2 (as 4 mol% consumed)
10 mol% N2 (given)
Overall conversion 6.25% same as per pass conversion