Question

Methanol is synthesized from carbon monoxide and hydrogen (H2) in a catalytic reactor.

CO + 2H₂ → CH₃OH

The fresh feed to the process contains 32.0% mol of CO, 64% of H2 and 4% N2. This stream is mixed with a recirculation stream at a ratio of 5 mol recirculation / 1 mol fresh feed to the process. The mixture of these streams forms the feed stream to the reactor, which contains 13.0 mol% of N2. Low conversion is achieved in the reactor. The effluent from said reactor passes to a condenser from which two streams emerge: one of liquid product that contains all the methanol formed in the reactor, and another of gas that contains all the CO, H2, and N2 that leaves the reactor. The gas stream is divided into two fractions with different flows: one fraction is removed from the process as a purge, and the other is recirculation that is combined with fresh feed to the reactor. Assume a calculation basis of 100 mol / h of fresh feed to the process and use atomic balances to calculate methanol production rate (mol / h), molar flow and purge gas composition, and one-step conversions, and total CO.

Answer 1

Sal methand synthesis +24 - LA2 он Basi : 100 mal Ihr of fresh feed Product (R& Condenser fresh feed TE) mixed feed Reactor Recycle feed (R) Purge (P) and 4% M H2 64 mal fresh feed contains, 32% co, 64% H co 0:32 X 100 =32 mal 0.648 100 Ne = 0.04 x 100 umal mixing point F+ R=m aut Recirculation stream fresh feed 음두 R=5f 5X100 = 500 mallar mo f +5 F =6F = 600 malha feed stream to reactor Contains 13 mal % of Me M in mixed feed = 0.13 X 600 = 78 mal in reciple stream = 28-4= tumal mal fraction of my in recycle stream = 74 x 100 = 14.8% 500 in all fraction of all components receple and com will be same, in fresh feed 3 streams, purge, out of the ondenser My in Punge stream

M fraction in Punge us 0.140P stream = ollud P= 27.03 mallar In By using atonic balance method Product stream (P) contains only liquid methanol. Purge stream (P) M2 = umel co and ry = 27.03a4= 23.03 mol med of co is x then He (23.03 - ) i Carbon atom balance, Carbon in fresh feed Carbon atom in product stream at Carbon atom Punge 1 x 32 I X R, + Coxn) 32 R, tu Q - s cotom balance, ii) Hydrogen Hydrogen in frem feed atom in Hydrogen atom Product stream arom po Punge stream Hydrogen - luxR) + 2X (23.03-1) 2x64 128 EUR, + 46.06 - 2x 81.94 3 and 3 81.94 =4R, = UR, -au by solving 2 (32-R, ) 81.94 EUR, 64 +2R, 6R, R = 24.32 mal Ihr 115. a4 x = 7.68 mol med of co in in Punge of hy in mod streams Stream 7.68 mal 15:35 mol purge

and punge a methanol Produchon rare (R) = 24.32 mal ihr molar flow & tream composition. Pa med of co tmal of Hy tmol of a P = 3. 8 +15:35 - 4 - 27.03 mellhr Composition co = 7.68 -X100 28.41%. 27.03 Ha N100 56-74 = 15.35 27.03 and he is 14.8. one step Conversion 1 neill be same, in recycle stream, Composition amal. let in recycle stream Q.284 is a 500 a = 142.05 mol and red = 0.5679 4500= 283.95 mal Now in mixed feed, mal of co= 32+ 142.05= mal of ry=64t d83.95 = 344.95 mol. 174.05 mal for reaction , 24+32 mal of 4₂ on Produce, when 24.32 mol of reacred werth (2X24:32) mal of hea re reached = 2X24.32 =40.64 mal

one-step conversion 드 X 100 moles of coreacred males of co in mixed feed = 24.3.2 tag: 65 X 100 = 13.97 ol. Overall conversion X100 = moles of co reacred moles o f coin in fresh feed २५.32 32 X 100 a 76%