Hydrodealkylation is a process in which side chains ... - cribME!

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Given .740 − nc940.7435 − nx35nB 5 − 2⋅squig4+ squig1 nH 500 − squig1− squig2 − squig3 + squig4nT 20 − squig1+ squig2nCH4 squig1 + squig2 + squig3nx 35 − squig2 + squig3 nc9 40 − squig3 nbp squig473.724 nBntot nbp + nx + nT + nH + nCH4 + c9 + nB .00112.000 nC99.100 nXnbpntotxnxntot.820 − nT20TnTnHHntotntot348.788 nH24.000 nT151.800 nCH40.588 nBP588.020 ntotal69.900 ζ153.900 ζ228.000 ζ30.588 ζ40.015 x xylenex0.007 toluene0.593 x H20.258 x CH40.020 x C9x0.125 benzene0.001 x BPCH4nCH4ntotc9nc9ntotBnBntotbpnbpntot
2. (30 pts) Methanol is produced by reacting carbon monoxide and hydrogen. A fresh feed containingCO and H 2 joins a recycle stream and the combined stream is fed to the reactor. The reactor outletstream flows at a rate of 250 mole/min and contains 63.1 mole % H 2 , 27.4 mole % CO and 9.53 % CH 3 OH.This stream enters a cooler in which most of the methanol is condensed. The liquid methanol iswithdrawn as a product, and the gas stream leaving the condenser, containing all three species with a0.40 mole % methanol vapor is the recycle stream that combines with the fresh feed.a) Draw a picture and label all of your knowns and unknowns.b) Do a degree of freedom analysis for the overall process and the condenser.c) Find the molar flow rates of CO and H 2 in the fresh feed, the production of liquid methanol, and thesingle‐pass and overall conversions of CO.H 2CH 3 OHGiven.0953250 ⋅ n3 + .004⋅n4250 n3 + n4 n1 n32n2 ⋅4n3 ⋅ .274⋅250xn4 ⋅Find( n1, n2, n3, n4,x)→⎛⎜⎜⎜⎜⎜⎝22.91666666666666666745.83333333333333333322.916666666666666667227.08333333333333333.30165137614678899083⎞⎟⎟⎟⎠c) CO = 22.9 moles; H 2 =45.8 moles; production rate of methanol = 22.9 moles;single pass: (22.9+.3*227.8)‐.274*250 = 25 %(22.9+.3*227.8)Overall: 100 %
Page 1: Homework 51. (30 pts) Hydrodealkyla
Page 5: 4. (20 pts) A mixture of NH 3 and a

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