Question

Ethylene oxide (C2H4O) is made by the catalytic reaction between ethylene (C2H4) and oxygen.Fresh ethylene, pure oxygen, and recycled ethylene are mixed to give a reactor feed gas containing 55%C2H4 which enters the reactor at 1 atm absolute and 300 oC. From test experiments, the single-pass conversion of ethylene in the reactor is 54% and, of this, 72% goes to ethylene oxide and the balance decomposes to CO2 and H2O. Most of the unreacted ethylene is separated from the reactor exit stream and is recycled, a pure ethylene oxide product stream is produced, and the remaining materials leave in a waste stream which analyses 9.0% ethylene. The production rate of ethylene oxide is 26,430 kg/day.Determine: a) the molar flow rate of fresh ethylene feed; b) the flow rate of the recycle stream; c) the fraction of feed oxygen wasted; and d) the percent excess oxygen

Answer 1

Let F= Fresh feed, R= Recycle (C2H4) and M= mixed feed, molar mass of C2H4O= 46, moles of C2H4O= 26430/46= 574.5 kg moles/day

THe reactions are C2H4 +0.5O2----> C2H4O and C2H4+3O2---->2CO2+3H2O

C2H4 in mixed feed = M*0.55 Single pass conversion = M*0.55*0.54

C2H4O produced = M*0.55*0.54*0.72= 574.5, M= 2687 kg moles/day = same as moles of C2H4 consumed

Oxygen requires = 574.5/2 = 287.25 kg moles/day

Oxygen entering is coming from fresh feed = 2687*0.45= 1209 kg moles/day

Oxygen consumed for producing C2H4O = 2687/2 = 1343.5 Kg moles/day

C2H4 converted to CO2 and H2O = 2687*0.55*0.54*0.28 = 223.45 kg moles/day

Oxygen required = 3 times   C2H4 = 3*223.45 =679.35 kg moles/day

Total oxygen consumed = 679.35+287.25=966.6 Kg moles/day

Oxygen remaining = 1209-966.6 = 251.4 kg moles/day

This along with CO2 and H2O ( 91% is goinf as waster )

CO2, H2O and O2 in waste= 4*223.45 ( for every moles of C2H4 reacts ,4 moles of CO2 and H2O form) +251.4 =1145.2 kg moles/day

This is 91% of Waster ,. W*0.91= 1145.2

W= 1258.4 kg moles/day

unreacted C2H4= C2H4 in waster + Recyle C2H4

M*0.54*0.55*0.46= 1258.4*9/100 +R

R= 220.5 kg moles/day

C2H4 balance across the reactor

M*0.55= 220.5+C2H4 in fresh feed

C2H4 in fresh feed = 1257.35 kg moles/day

Total oxygen to be supplied for complete conversion of C2H4 to C2H4O= 2687*0.55/2 = 739 kg moles/day

% excess oxygen = 100*(1209-739)/739 =63.59%