I\'m not sure why I\'m having problems answering this simple problem. Answers: a
ID: 947177 • Letter: I
Question
I'm not sure why I'm having problems answering this simple problem.
Answers:
a) (Mass fraction for purge stream)
xCO=0.2
xH2=0.768
xCH4=0.032
b) 32.5-1.25=31.25 mol/hr of methanol production
CO + 2H2 CH3OH Fresh feed is combined with a recycle stream and fed to a reactor to form CH3OH. The reactor effluent is fed to a separator that removes effectively all of the methanol as a pure stream. The remaining products are recycled, but part of this stream is purged. A diagram is shown below: Fresh Feed Product Reactor Separator Mole % 100% CHOH 67.3 H2 32.5 CO 0.2 CH4 Recycle Purge CO CH4 The synthesis gas has a small amount of methane present as a contaminant, which does not participate in the reaction. The single pass conversion of CO in the reactor is 18%. Finally, the purge stream is used to eliminate both methane and excess hydrogen, and is set to have a methane fraction of 3.2 mol%. For simplicity, assume a 100 mole/hour total Fresh Feed rate. (Hint: Answer the following questions in whatever order seems simplest to you. Also, this problem gets a lot simpler if you think of the excess hydrogen in the feed as an inert.) (a) What is the composition of the purge stream? (b) What is the molar production rate of methanol product?Explanation / Answer
Basis : 100 moles/ hr of Fresh Feed.
Moles of methane is the Fresh Feed= 100*0.2/100= 0.2 moles
This has to be removed through purge. Let Purge be P. P contains 3.2 mole% CH4
P*3.2/100 = 0.2 P= 100*0.2/3.2=6.25 moles
CH4 in purge stream =0,2 moles
CO entering = CO leaving through reaction +CO in purge
CH3OH formed = x
CO entering is leaving through CH3OH as well as purge. H2 entering is leaving through purge as well as through CH3OH and mole fraction of H2 and CO in purge =1-3.2/100 =0.968 .Let y= mole fraction of CO in purge
32.5= x+6.25*y (1) (CO balance)
67.3= 2x+6.25*(0.968-y) (2) (H2 balance)
Multiplying Eq.1 with 2
65 = 2x+12.5y (1A)
67.3= 2x+6.25*(0.968-y) (2)
Eq.2- Eq.1A 2.3= 6.25*(0.968-y)-12.5y
(12.5+6.25)y= 6.25*0.968-2.3=3.75
Therefore CO in purge =6.25*0.2= 1.25 moles/ hr and hydrogen = 6.25*(0.968-0.2) =4.8 moles/hr
Composition of purge stream : CO= 0.2 , H2= 0.968-0.2 = 0.768 and CH4= 0.032
CO lost in purge = 1.25 moles/ hr CO remaiinng =32.5-1.25= 31.25 moles/ hr
CH3OH formed = 31.25moles/hr