EXPERT ANSWER
given data
feed F=100 mol
mole fraction of ethane in feed xethane = 0.85 moles of ethane in feed = xethane F =85 mol
moles fraction of inert xi =0.15 moles of inert in feed=xiF=15mol
fractional convertion of ethane Xe = 0.501 =reacted ethane/total ethane in feed
therefore reacted ethane=0.501*85=42.585mol
unreacted ethane=85-42.585=42.415
fractional Yield of ethylene Yethylene=0.471=ethane reacted to produce ethylene/moles of ethane totally reacted
0.471=ethane reacted to produce ethylene/42.585
ethane reacted to produce ethylene=0.471*42.585=20.057535
each 1 mole of ethane reacted to produce ethylene will produce 1mol of ethylene
therefore ethylene produced=20.057535
remaining moles of reacted ethane will react to produce methane=42.585-20.057535=22.527465
1mol of ethane will produce 2 mols of methane …… from reaction
therefore methane produced= 2*22.527465 =45.05493
from reaction stoichiometry hydrogen is getting produced and reacted for methane
hydrogen in P stream = hydrogen produce – hydrogen reacted
hydrogen produce=ethylene produce from stoichiometry=20.057535
hydrogen reacted=half the amount methane produced =45.05493/2=22.527465
hydrogen in P stream=20.057535-22.527465=-2.46993
here should be problem in given data as this value comes negative
molar composition of product gas ans.
component | composition of P stream (moles) |
n1 ethane | 42.415 (unreacted) |
n2 ethylene | 20.057535 (produced) |
n3 hydrogen | -2.46993 |
n4 methane | 45.05493 (produced) |
n5 inert | 15 (from feed) |
selectivity of ethylene to methane
=moles of ethylene formed/moles of methane formed
=20.057535/45.05493
=0.445179 ans.