Homework 5 1. (30 pts) Hydrodealkylation is a process in which side chains, consisting of alkyl groups, are removed from aromatics by reaction with hydrogen to form the parent aromatic compound. For instance, toluene can be converted to benzene: C6H5CH3 + H2 Î C6H6 + CH4 Xylene can be converted to toluene: C6H4(CH3)2 + H2 Î C6H5CH3 + CH4 Pseudocumene and other C9 hydrocarbons containing three CH3 groups can be converted to xylenes: C6H3(CH3)3 + H2 Î C6H4(CH3)2 + CH4 In a given application, a refinery reformate stream consisting of 5% benzene, 20% toluene, 35% xylene, and 40% C9 hydrocarbons is reacted with hydrogen. If 5 mol H2 is used per 1 mol feed, 80% conversion of toluene, 74% conversion of xylene, and 70% conversion of C9 hydrocarbons are attained. The product stream is found to contain a small amount, 0.1% of biphenyl, indicating that the side reaction 2C6H6 Î C6H5C6H5 + H2 occurs to some extent. The four reactions that are occurring can be described by the simplified form: T + H2 Î B + CH4 (1) X + H2 Î T + CH4 (2) C9 + H2 Î X + CH4 (3) 2B Î BP + H2 (4) Use the Extent of Reaction Method for multiple reactions to calculate the complete composition of the reactor outlet stream. The problem should be worked in mol/h and mole fraction. See work below: BP: .001 Benzene: .007 C9: .02 CH4: .14 H2: .71 Toluene: .12 Xylene: .015 .7
Given
nx
nB
5 − 2⋅ squig4 + squig1
nT
20 − squig1 + squig2
35 − squig2 + squig3
40 − nc9
40
35
500 − squig1 − squig2 − squig3 + squig4
nH
nCH4
nc9
35 − nx
.74
squig1 + squig2 + squig3
.8
40 − squig3
nbp
squig4
73.724 nB ntot
nbp + nx + nT + nH + nCH4 + c9 + nB
.001
12.000 nC9 9.100 nX T
nT
ntot
348.788 4.000 151.800 0.588 588.020 69.900 53.900 28.000 0.588 0.015 0.007 0.593 0.258 0.020 0.125 0.001 H
nH
ntot
CH4
nH2 nT nCH4 nBP ntotal ζ1 ζ2 ζ3 ζ4 x xylene x toluene x H2 x CH4 x C9 x benzene x BP nCH4
ntot
c9
nc9
ntot
nbp
x
ntot
B
nB
ntot
nx
ntot
bp
nbp
ntot
20 − nT
20
2. (30 pts) Methanol is produced by reacting carbon monoxide and hydrogen. A fresh feed containing CO and H2 joins a recycle stream and the combined stream is fed to the reactor. The reactor outlet stream flows at a rate of 250 mole/min and contains 63.1 mole % H2, 27.4 mole % CO and 9.53 % CH3OH. This stream enters a cooler in which most of the methanol is condensed. The liquid methanol is withdrawn as a product, and the gas stream leaving the condenser, containing all three species with a 0.40 mole % methanol vapor is the recycle stream that combines with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 H2 in the fresh feed, the production of liquid methanol, and the single‐pass and overall conversions of CO. H2 CH3OH Given
.0953⋅ 250
n3 + .004⋅ n4
250
n3 + n4
⎛ 22.916666666666666667⎞
⎜ 45.833333333333333333
⎜
⎟
Find( n1 , n2 , n3 , n4 , x) → ⎜ 22.916666666666666667
⎟ ⎜ 227.08333333333333333⎟ ⎜
⎝ .30165137614678899083⎠
n1
n3
2⋅ n2
c) CO = 22.9 moles; H2 =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 % 4⋅ n3 .274⋅ 250
x⋅ n4
3. (20 pts) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H20. Sufficient air (79 % N2 and 21 % O2) is fed to provide 35 % excess oxygen and the fractional conversion of HCl is 85 %. Calculate the mole fractions of the product stream components using atomic species balances. 4HCl + O2 → 2Cl2 + 2H2O Assume 100 moles of HCl. 1.35*100 moles HCl*(1 mole O2/4 moles HCl) = 33.75 moles O2 coming in Given
nn2
33.75⋅ 3.76
.85
100 − nhcl
2⋅ 33.75 nh2o + 2⋅ no2
n2
nn2
ntot
o2
no2
ntot
100
ntot
hcl
100
2⋅ ncl2 + nhcl
100
2⋅ nh2o + nhcl
nn2 + nhcl + ncl2 + no2 + nh2o
nhcl
ntot
cl2
ncl2
ntot
h2o
nh2o
ntot
⎛ 126.90000000000000000⎞
⎜
15.
⎜
⎟
⎜ 42.500000000000000000⎟
⎜ 42.500000000000000000⎟
⎜ 12.500000000000000000⎟
⎜
⎟
Find( nn2 , nhcl , ncl2 , nh2o , no2 , ntot , n2 , o2 , hcl , cl2, h2o ) → ⎜ 239.40000000000000000⎟
⎜ .53007518796992481203⎟
⎜
⎟
⎜ .52213868003341687552e-1
⎟
⎜ .62656641604010025063e-1
⎟
⎜ .17752715121136173768⎟
⎜
⎝ .17752715121136173768⎠
N2 = .53 O2 = .052 HCl = .063 Cl2 = .1775 H2O = .1775 4. (20 pts) A mixture of NH3 and air (21% O2, 79% N2) is fed to a reactor at rates of 2000 mol/h air and 100 mol/h NH3 where they are reacted to produce a mixture of gases comprising 3 mol% O2, 6 mol % NO, and the rest N2, NH3, H2O, and NO2. If the fractional conversion of NH3 is fNH3 = 0.8, calculate the outlet molar flow rate (mol/h) of each species. Do a degree of freedom analysis first! The following reactions are occurring: 4NH3 + 5O2 Î 4NO + 6H2O 4NH3 + 3O2 Í Î 2N2 + 6H2O 2NO + O2 Î 2NO2 Given
.8
100 − nnh3
100
.79⋅ 2⋅ 2000 + 100
nnh3 + nno + nno2 + nn2 ⋅ 2
300
3nnh3 + 2nh2o
.21⋅ 2⋅ 2000 2⋅ no2 + 2⋅ nno + nh2o + 2nno2
ntot
nnh3 + no2 + nno + nh2o + nn2 + nno2
1969.543 20.000 59.086 118.173 120.000 1469.543 182.741 total NH3 O2 NO H2O N2 NO2