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) C 2 H 5 OH is produced commercially by the hydration of C 2 H 4 : C 2 H 4 + H 2 O C 2 H 5 OH Some of the product is converted to (C 2 H 5 ) 2 O in an undesired side reaction: 2C 2 H 5 OH (C 2 H 5 ) 2 O + H 2 O The feed to the reactor (Stream 1) contains C 2 H 4, H 2 O, and an inert gas. A sample of the reactor effluent gas (Stream 2) is analyzed and found to contain 45.0 mol% C 2 H 4, 4.0 mol% C 2 H 5 OH, 0.15 mol% (C 2 H 5 ) 2 O, 8.2 mol% inerts, and the balance H 2 O. The molar flow rate of stream 2 is 500 mol/h. Calculate: a. (5 pts) The molar flow rate (mol/h) of C 2 H 4 in Stream 1 b. (5 pts) The molar composition of Stream 1 (mole fractions) c. (5 pts) The fractional conversion of C 2 H 4 d. (5 pts) The fractional yield of C 2 H 5 OH e. (5 pts) The selectivity of C 2 H 5 OH production relative to (C 2 H 5 ) 2 O production

MEEBAL Exam 1 October 2012 2. (25 points) C 6 H 14 at 11.96 atm and 600 K flows at a rate of 24 m 3 /min (Stream 1) into a reactor and is combined with 6.5% excess air (Stream 2). A fraction of C 6 H 14 is burned in the reactor. The product gas (Stream 3), which contains all of the unreacted C 6 H 14 and no CO, goes to a condenser in which both the water and C 6 H 14 are liquefied. The uncondensed gas (CO 2, O 2 and N 2 ) 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: P c (C 6 H 14 ) = 29.9 atm; T c (C 6 H 14 ) = 507.9 K Calculate: a. (5 pts) Molar flow rate (kmol/min) of C 6 H 14 (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 C 6 H 14 in Stream 5 d. (5 pts) Fractional conversion of C 6 H 14 e. (5 pts) Molar composition of Stream 4 (mole fractions)

MEEBAL Exam 1 October 2012 3. (25 points) CH 4 reacts with Cl 2 to produce CH 3 Cl and HCl. Once formed, CH 3 Cl reacts with Cl 2 in an undesired side reaction to form CH 2 Cl 2. Stream 2 (100 kmol/h) containing 75 mol% CH 4 and the balance Cl 2 is fed to the reactor. In the reactor, a single pass chlorine conversion of 100% is attained, and the molar ratio of CH 3 Cl to CH 2 Cl 2 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 CH 3 Cl and CH 2 Cl 2 in the reactor effluent, and a gas (Stream 4) containing all of the CH 4 and HCl in the reactor effluent. Stream 5 goes to the distillation column in which the two species (CH 3 Cl and CH 2 Cl 2 ) are separated. Stream 8 contains only CH 3 Cl, and Stream 9 contains only CH 2 Cl 2. Stream 4 is separated into two streams: one that contains only HCl (Stream 6) and the other that only contains CH 4 (Stream 7). Stream 7 is recycled to join the fresh feed (Stream 1, CH 4 and Cl 2 ). The molecular weight of CH 3 Cl is 50.49 g/mol. Calculate: a. (5 pts) Molar flow rates (kmol/h) of each component (CH 4 and Cl 2 ) 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 CH 3 Cl production rate of 10,000 kg/h

MEEBAL Exam 1 October 2012 4. (25 points) Terephthalic acid (TPA: C 8 H 6 O 4 ) is synthesized from p-xylene (PX: C 8 H 10 ) via the following reaction: C 8 H 10 + 3O 2 C 8 H 6 O 4 + 2H 2 O (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 O 2, N 2 and H 2 O 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% O 2. 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 (m 3 /h) of Stream 2 c. (5 pts) Volumetric flow rate (m 3 /h) of Stream 5 d. (5 pts) Volumetric flow rate (m 3 /h) of Stream 8 e. (5 pts) Mass flow rate (kg/h) of recycle stream (Stream 7)