1. Cyclohexane (C 6 H 12 ) can be made by the reaction of benzene (C 6 H 6 ) and hydrogen
gas. The products from the reactor are sent to a separator where the cyclohexane and
some of the unreacted hydrogen and benzene are removed. The rest of the unreacted
benzene and hydrogen are recycled in a stream that is 23 % benzene. If the overall
conversion of benzene is 95 % and the single pass conversion is 20 %, find the ratio of
the recycle stream to the fresh feed stream as well as the composition of the product
stream. Assume that 20 % excess hydrogen gas is used in the fresh feed.
C 6 H 6 + 3 H 2 → C 6 H 12
n1
100 C 6 H 6 n 2
360 H 2
n C6H6
n H2
n C6H12
n C6H12
n hydrogen
n benzene
n R (.23 benzene, .67 hydrogen)
equation (1 pt)
picture (2 pts)
assume basis ( 1 pt)
excess hydrogen = 360 (2 pts)
use overall conversion to find n benzene = 5 (2 pts)
use overall C balance to find n C6H12 : 600 = 30 + 6n C6H12 n C6H12 = 95 (3 pts)
use overall H balance to find n hydrogen : 1320 = 1140 + 2 n hydrogen + 30
n hydrogen = 75
(3 pts)
around reactor:
n 1 = .23 n R + 100
.20 = (.23n R + 100) – n C6H6 /(.23 n R + 100) (3 pts)
n C6H6 = 5 + .23n R (2 pts)
.20 = 95/(.23n R + 100)
n R = 1630 (1 pt)
n R /460 = 3.54 (1 pts)
Product stream:
Total = 5 + 95 + 75 = 175 (1 pt)
yhydrogen = .43 y benzene = .03 y C6H12 = .54 (1 pt each – 3 pts total)
2. The density of a certain liquid is given an equation of the following form:
ρ = (A + Bt)eCP
where ρ = density in g/cm3
t = temperature in degrees C
P = pressure in atm
a) The equation is dimensionally consistent. What are the units of A, B and C?
b) In the units above, A = 1.096, B = 0.00086 and C = 0.000953. Find A, B and C if
ρ is expressed in lb m /ft3, t in degrees R, and P in lb f /in2.
a) A: g/cm3
b)
B: g/cm3-C
A = 1.096 g/cm3
B = .00086
C = .000953
c: atm-1
(1 pt each)
A’ = 68.420 lb m /ft3
B’ = .0298 lb m /ft3 – R
C’ = 6.485 X 10-5 psi-1
2 pts
3 pts
2 pts
3. Formaldehyde (CH 2 O) is produced industrially by the catalytic oxidation of
methanol(CH 3 OH). Unfortunately, a significant portion of the formaldehyde reacts with
oxygen to produce CO and H 2 O. The two reactions are:
CH 3 OH + ½ O 2 → CH 2 O + H 2 O
CH 2 O + ½ O 2 →CO + H 2 O
Assume that methanol and 100 % excess of air are fed to the reactor. The conversion of
methanol is 90 % and the yield of formaldehyde is 75 %. Determine the composition of
the product gas leaving the reactor.
n CH3OH
n O2
n N2
n CH2O
n CO
n H2O
100 CH 3 OH
100 O 2
376 N 2
Picture (2 pts)
Basis: 100 mol methanol (1 pt)
Required: 50 moles O 2
.9 = 100 – n meth /100
100 % excess: 100 moles O 2 (2 pts)
.75 = n CH2O /100
C balance: 100 = 10 + 75 + n CO
H balance: 400 = 40 + 150 + 2n H2O
O balance: 300 = 10 + 2n O2 + 75 + 15 + 105
n CH3OH = 10
n N2 = 376
n CH2O = 75
n CO = 15
n H2O = 105
n O2 = 47.5
2 pts
2 pts
2 pts
2 pts
2 pts
2 pts
Total:
628.5
2 pts
yCH3OH = .016 y N2 = .598
(1 pt each – total 6 pts)
y CH2O = .119 y CO = .024
y H2O = .167
y O2 = .076
4. A stream of a gas containing 30 mole % CO 2 and the rest methane is fed to the bottom
of an absorption tower. The product gas leaving the top of the absorber contains 1 mole
% CO 2 and all the methane fed to the unit. The rest of the CO 2 is dissolved in methanol –
this stream leaves the bottom of the absorber and is fed to the top of a stripper. A stream
of nitrogen gas is fed to the bottom of the stripper. Ninety percent of the CO 2 in the
liquid feed to the stripper comes out of solution in the column top and exits with the
nitrogen. The liquid stream leaving the stripper is .500 mole % CO 2 and the balance
methanol, which is recycled to the absorber.
a) Taking a basis of 100 mol/h of gas fed to the absorber, draw and label a flowchart
of the process.
b) Do a degree-of-freedom analysis for the overall process as well as each unit. Do
NOT include the nitrogen gas in your calculations.
c) State your strategy for solving for the variables and then solve for them.
d) Calculate:
a. Moles absorbed of CO 2 /moles of CO 2 in gas feed
b. Molar flow rate and composition of the liquid feed to the stripping tower
e) Calculate the molar feed rate of gas to the absorber required to produce an
absorber product gas flow rate of 1000 kg/h.
Overall:
2 unknowns (n 1 , n 5 )
2 components
0 DOF
Stripper:
4 unknowns (n 2 , n 3 , n 4 ,n 6 )
2 components
1 extra information (90 % of CO 2 )
1 DOF
Strategy:
Solve for n 1 and n 5 using component balances.
Solve for n 3 using % stripped.
Solve for n 2 and n 4 around stripper.
Picture (10 pts)
overall DOF (2 pts)
absorber DOF (2 pts)
stripper DOF (2 pts)
Strategy (3 pts)
Overall:
100 = n 1 + n 5
30 = .01n 1 + n 5
n 1 = 70.7 (2 pts)
n 5 = 29.3 (2 pts)
.90n 3 = n 5
n 3 = 32.55 (2 pts)
n 3 = n 5 + .005 n 2
n 2 = 651
n 4 = .995 n 2
d) 30-.01n 1 /30 = .976
n 3 + n 4 = 680.25 (2 pts)
y meth = .952
y CO2 = .048
n 4 = 647.7 (2 pts)
(2 pts)
e)
average molecular weight of the stream: .001*44 + .99 * 16 = 16.28 g/mol
1000 kg/hr * 1000 g/kg/16.28 g/mol = 6.142 X 104 mol/h (2 pts)
6.142 X 104/70.7 = 868.74 (scaling factor) (1 pt)
100 X 868.74 = 8.69 X 104 mol/hr (1 pt)
(3 pts)
Extra credit: Look at the following equation: Z = aVbPc. Given data for Z, P and V, how
would you solve for a, b and c?