Page |7 Q3 (30 points) The process diagram below reflects a Methanol (CH,0H) Pro

Question

Page |7 Q3 (30 points) The process diagram below reflects a Methanol (CH,0H) Production process as per the following chemical reaction: CO + 2 H2 CH3OH A fresh feed stream containing CO and H2 joins recycle stream and the cobined stream is fed to the reactor (see figure). The reactor outlet streams flows at a rate of 350 mol/min and contains: 63.0 mol% H2. 27.0% co, and 9.5% CH30H. The stream enters the cooler unit next, where Methanol is removed as a product. And a second gas stream which contains CO , H2 and 0.40 mol% methanol vapor is recyled and mixed with the feed n, (mol/min) XR-0.004 x,n-(1-0.004-Xco) 350 moU/min 0,10 mole Cll,Ollmol 0.27mol CO/mol n(mole CH,OH/mi Separator-_ Reactor D, (mol CO/min) n, (mol CO/min) 0.63 mole H, mol (mole H/min) n, (mole H/min) Ds(mole CH,OH/min) Based on the provided information, calculate the following: 1. The molar flowrate of CO and H2 in the fresh feed (ni and n2) 2. The production rate of methanol (n7) 3. Single pass and Overall converesion for CO o.lx 350

Explanation / Answer

Moles of CH3OH in the reactor outlet = 0.10 x 350 = 35 mol/min

Moles of H2 in the reactor outlet = 0.63 x 350 = 220.5 mol/min

Moles of CO in the reactor outlet = 0.27 x 350 = 94.5 mol/min

Only CH3OH is at the outlet of separator

Moles of CO in the recycled stream = 94.5 mol/min

Moles of H2 in the recycled stream = 220.5 mol/min

Overall balance around the separator

350 = n7 + n6

CH3OH balance around the separator

35 = n7 + 0.004 n6

Solve the above equations simultaneously

315 = 0.996 n6

n6 = 316.265 mol/min

n7 = 350 - 316.265 = 33.735 mol/min

From the atomic balance of the overall system

Carbon balance

n1 = n7 = 33.735 mol/min

Hydrogen balance

2 n2 = 4 n7

n2 = 2 n7 = 2* 33.735 = 67.47 mol/min

Part 1

n1 = 33.735 mol/min

n2 = 67.47 mol/min

Part 2

n7 = 33.735 mol/min

Part 3

Mol fraction of CO in the recycled stream = moles of CO/n6

= 94.5/316.265 = 0.2988 = 0.3

Single pass conversion for CO

= (n3 - moles of CO at reactor outlet) / n3

= [(33.735 + 0.3*316.265) - (0.27*350)] / [33.735 + 0.3*316.265]

= (128.6145 - 94.5 )/ (128.6145)

= 0.2652 = 26.52%

Overall conversion for CO =(n3 - 0*n7) *100% / n3 = 100%

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