Carbon monoxide (CO) and hydrogen (H_2) react lo form methanol (CH_3OH) in a rea
ID: 482716 • Letter: C
Question
Carbon monoxide (CO) and hydrogen (H_2) react lo form methanol (CH_3OH) in a reactor, as illustrated in the diagram below. The reactor operates with an F flow rate of 1000 lbmol/hr of 1:1 CO: H_2. The reader operates with a 90% overall conversion of H_2 and an 25% single pass conversion ofH_2. The mote fraction of hydrogen in the recycle stream is 0.78. The reactor effluent is separated into a recycle stream with only unused reactants (CO and H_2) and a product stream with mostly methanol (CH_3OH), along with lower amounts of the unused reactants. Write and balance the equation. Determine the extent of reaction. Determine the product molar flowrate P in lbmol/hr and molar composition. Determine the recycle flowrate R in lbmol/hr and molar composition. Determine the recycle ratio (R/F). What is the overall conversion of CO? What could be done to increase conversion of CO?Explanation / Answer
a)
CO + 2 H2 -> CH3OH
b)
Feed F
Flow rate, F = 1000 lbmol/hr
Flow rate of CO = 500 lbmol/hr
Flow rate of H2 = 500 lbmol/hr
Product P
Overall conversion = (Flow rate of H2 in F – Flow rate of H2 in P) / Flow rate of H2 in F
0.9 = (500 – Flow rate of H2 in P) / 500
Flow rate of H2 in P = 50 lbmol/hr
Extent of reaction = (500 – 50) / 2 = 225 lbmol/hr
d)
Single pass conversion = (Flow rate of H2 in RF – Flow rate of H2 in RE) / Flow rate of H2 in RF
0.25 = [500 + R * 0.78 - Flow rate of H2 in RE] / (500 + R * 0.78) … (1)
Mass balance of H2 across separator
50 + R * 0.78 = Flow rate of H2 in RE … (2)
Solving equation (1) and (2) we get
500 + 0.78 R = 1800
R = 1666.67 lbmol/hr
Composition:
H2 = 78%
CO = 22 %
c)
Flow rate of H2 in RE = 50 + 0.78 R
= 1350 lbmol/hr
Flow rate of H2 in RF = 500 + 0.78
= 1800 lbmol/hr
Rate of moles of H2 reacted = 1800 – 1350 = 450 lbmol/hr
Flow rate of methanol in P = 450 / 2 = 225 lbmol/hr
Flow rate of CO in RF = 500 + 0.22 R = 866.67 lbmol/hr
Rate of moles of CO reacted =450 / 2 = 225 lbmol/hr
Flow rate of CO in RE =866.67 – 225 = 641.67
Flow rate of CO in P = 641.67 – 0.22 R = 275 lbmol/hr
P = 225 +50 + 275 = 550 lbmol/hr
Composition:
Methanol = 225 / 550 * 100 = 41 %
H2 = 50 / 550 * 100 = 9%
CO = 275 / 550 * 100 = 50%
e)
R / F = 1666.67 / 1000 = 1.67
f)
Overall conversion of CO = (500 – 275) / 500 * 100 = 45 %
By increasing H2 : CO in feed, conversion of CO can be increased. It is now operating at 1:1 mole ratio.