Your nswerts partially omect. Try again. Methanol is yrthesned from carbon monel

Question

Your nswerts partially omect. Try again. Methanol is yrthesned from carbon monelde and hydrogen in·catalytic reactor process trntons 32.0 mo)% CO. 04.0 mol%, H2 and 4.00 mo1% N2. This stream is mixed with a recycle stream in·ratio of 4.00 mai recycle/1monest, toat to produce the The fresh feed to the fued tn the reactor, which contains 12.0 mom N streams emerge: a squid product stream containing essentisily ail of the methanol formed in the reactor, and a ges stream containing all af The resctor efmuent goes to a condenser from which two the CD, M3, and Na leaving the reator The gas stream is spit into two fractions; one is removed from the process as a purge For a methanol production r stream, and the other is the recycle stream that combines with the fresh feed to the resctor. rate of 100.0 moyh, calc ulste the fresh feed rate (moijn), the molar flow rate and compaation of the purge gas, and the overall and single-pass conversions. Fresh feed rate: 1|600 Purga rate: L27 mol fraction CO in purge: 291 T 148 mol fraction Nz in purge Overst CO conversion : 3

Explanation / Answer

Methanol production rate = 100 moles/hr

Let F= Fresh feed flow rate

It contains 4 mole% N2

Flow rate of recycle= 4* flow rate of fresh feed= 4F

Let x= mole % of nitrogen in the recycle

Making a balance across the reactor

Nitrogen from Fresh feed+ nitrogen from recycle= Nitrogen in the combined stream

F*4/100+4F*x/100 =(F+4F)*12/100

4(1+x)= 5*12

Hence x=5*12/4-1 = 14

Recycle contains combined stream of CO and H2 at =100-14= 86%

The composition of purge also is same as 86% combined CO and H2 and balance of 14% Nitrogen

Let P= purge

Nitrogen entering from Fresh feed= F*4/100

This has to leave through purge = P*14/100

F*4= P*14, P= F*4/14= F*0.285 (1)

Fresh feed is used for production of methanol according to CO+2H2----->CH3OH and to compensate for the loss of CO and H2 through purge

3 mole of CO and H2 together gives 1 mole of CH3OH

100 moles of CH3OH requires 3*100 =300 moles of CO and H2 together

F*96/100 =moles of CO and H2 entering the fresh feed

3 moles of CO and H2 together gives 1 mole of CH3OH.

Moles of fresh feed of CO and H2= moles of CO and H2 used for producing CH3OH + moles of CO and H2 in the purge

F*0.96= 300+ P*0.86

But from Eq.1 , P= 0.285*F

Hence F*0.96= 300+F*0.285*0.86

F*(0.96-0.285*0.86)= 300                       , F = 419.6 moles/hr

Recycle = 4*419.6=1678.4 moles/hr

P= 419.6*0.285 =119.586 moles/hr

Let x= mole fraction of CO in the purge

Writing overall balance of the plant

CO entering from fresh feed= 419.6*0.32 = 134.3 moles/ hr

The CO entering is used for production of   methanol and rest is for purging

Hence writing overall CO balance and noting that 1 mole of CO gives 1 mole of CH3OH

134.3= 100+119.6*x

34.3 = 119.6x

Hence x =34.3/119.6 =0.29

Hence H2 In the purge= 0.86-0.29= 0.57

Mole of CO and H2 entering the reactor = 2098 moles/hr*0.96= 2014 moles/hr

Moles of CO and H2 can be used for production of CH3OH= 2014/3=671 moles/hr

Moles of CH3OH formed =100 moles/hr

% conversion per pass = 100*(100/671)              =14.9%

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.