MEEBAL Exam 1
October 2012
Show all work in your blue book. Points will be deducted if steps leading to answers are not
shown. No work outside blue books (such as writing on the flow sheets) will be considered.
No outgoing text messages are allowed during the exam. You must use given stream
numbering in the problems (no points given if different numbering system is used). Report
all answers with three significant digits. You are allowed to use one page of notes and a
calculator (no textbooks, computers or tablets such as iPads are allowed). You must pass in
your test sheet with your blue book for your exam to be graded (put your name on the
exam sheet).
1. (25 points)
C2H5OH is produced commercially by the hydration of C2H4:
C2H4 + H2O → C2H5OH
Some of the product is converted to (C2H5)2O in an undesired side reaction:
2C2H5OH → (C2H5)2O + H2O
The feed to the reactor (Stream 1) contains C2H4, H2O, and an inert gas. A sample of the reactor
effluent gas (Stream 2) is analyzed and found to contain 45.0 mol% C2H4, 4.0 mol% C2H5OH,
0.15 mol% (C2H5)2O, 8.2 mol% inerts, and the balance H2O. The molar flow rate of stream 2 is
500 mol/h.
Calculate:
a. (5 pts) The molar flow rate (mol/h) of C2H4 in Stream 1
b. (5 pts) The molar composition of Stream 1 (mole fractions)
c. (5 pts) The fractional conversion of C2H4
d. (5 pts) The fractional yield of C2H5OH
e. (5 pts) The selectivity of C2H5OH production relative to (C2H5)2O production
MEEBAL Exam 1
October 2012
2. (25 points)
C6H14 at 11.96 atm and 600 K flows at a rate of 24 m3/min (Stream 1) into a reactor and is
combined with 6.5% excess air (Stream 2). A fraction of C6H14 is burned in the reactor. The
product gas (Stream 3), which contains all of the unreacted C6H14 and no CO, goes to a
condenser in which both the water and C6H14 are liquefied. The uncondensed gas (CO2, O2 and
N2) leaves the condenser as Stream 4. The liquid condensate (Stream 5) contains 2.1 kmol/min of
water. Assume that the gases are non-ideal and obey the compressibility factor equation of state.
Critical constant data: Pc (C6H14) = 29.9 atm; Tc (C6H14) = 507.9 K
Calculate:
a. (5 pts) Molar flow rate (kmol/min) of C6H14 (Stream 1)
b. (5 pts) Molar flow rate (kmol/min) of air entering the reactor (Stream 2)
c. (5 pts) Molar flow rate (kmol/min) of C6H14 in Stream 5
d. (5 pts) Fractional conversion of C6H14
e. (5 pts) Molar composition of Stream 4 (mole fractions)
MEEBAL Exam 1
October 2012
3. (25 points)
CH4 reacts with Cl2 to produce CH3Cl and HCl. Once formed, CH3Cl reacts with Cl2 in an
undesired side reaction to form CH2Cl2. Stream 2 (100 kmol/h) containing 75 mol% CH4 and the
balance Cl2 is fed to the reactor. In the reactor, a single pass chlorine conversion of 100% is
attained, and the molar ratio of CH3Cl to CH2Cl2 in the product (Stream 3) is 5:1. The product
(Stream 3) flows to the condenser. Two streams emerge from the condenser: the liquid
condensate (Stream 5), which contains all of the CH3Cl and CH2Cl2 in the reactor effluent, and a
gas (Stream 4) containing all of the CH4 and HCl in the reactor effluent. Stream 5 goes to the
distillation column in which the two species (CH3Cl and CH2Cl2) are separated. Stream 8
contains only CH3Cl, and Stream 9 contains only CH2Cl2. Stream 4 is separated into two
streams: one that contains only HCl (Stream 6) and the other that only contains CH4 (Stream 7).
Stream 7 is recycled to join the fresh feed (Stream 1, CH4 and Cl2). The molecular weight of
CH3Cl is 50.49 g/mol.
Calculate:
a. (5 pts) Molar flow rates (kmol/h) of each component (CH4 and Cl2) in Stream 1
b. (5 pts) Molar flow rate (kmol/h) of Stream 6
c. (5 pts) Molar flow rate (kmol/h) of Stream 7
d. (5 pts) Molar flow rate (kmol/h) of Stream 8
e. (5 pts) Molar flow rate (kmol/h) of Stream 1 required to achieve a CH3Cl production rate
of 10,000 kg/h
MEEBAL Exam 1
October 2012
4. (25 points)
Terephthalic acid (TPA: C8H6O4) is synthesized from p-xylene (PX: C8H10) via the following
reaction:
C8H10 + 3O2 → C8H6O4 + 2H2O
(PX)
(TPA)
A fresh feed of pure liquid PX (Stream 1) combines with a recycle stream (Stream 7) containing
PX and a catalyst solution (S). The combined stream (Stream 3), which contains S and PX in a
3:1 mass ratio, is fed to the reactor in which 90% of the PX is converted to TPA. A stream of air
(Stream 2) at 25 °C and 6.0 atm absolute is also fed to the reactor. A liquid stream (Stream 4)
containing all of the PX, TPA and solution (S) that entered the reactor goes to a separator in
which all of the TPA is removed in Stream 6. Stream 7 contains all of the S and PX that was in
Stream 4. A gas stream (Stream 5) containing O2, N2 and H2O leaves the reactor at 105 °C and
5.5 atm absolute, and goes to a condenser in which all the water is condensed and removed in
Stream 8. The uncondensed gas (Stream 9) contains 4.0 mol% O2. 100 kmol/h of TPA is
produced (Stream 6). Assume that the gases are ideal. The molecular weights of PX and water
are 106 and 18 g/mol, respectively. The density of water is 1 kg/L.
Calculate:
a. (5 pts) Molar flow rate (kmol/h) of Stream 1
b. (5 pts) Volumetric flow rate (m3/h) of Stream 2
c. (5 pts) Volumetric flow rate (m3/h) of Stream 5
d. (5 pts) Volumetric flow rate (m3/h) of Stream 8
e. (5 pts) Mass flow rate (kg/h) of recycle stream (Stream 7)