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 n 1 n C6H6 n H2 100 C 6 H 6 n 2 n C6H12 360 H 2 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) y hydrogen =.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)e CP where ρ = density in g/cm 3 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 /ft 3, t in degrees R, and P in lb f /in 2. a) A: g/cm 3 B: g/cm 3 -C c: atm -1 (1 pt each) b) A = 1.096 g/cm 3 A = 68.420 lb m /ft 3 2 pts B =.00086 B =.0298 lb m /ft 3 R 3 pts C =.000953 C = 6.485 X 10-5 psi -1 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 100 CH 3 OH n CH2O n CO 100 O 2 n H2O 376 N 2 Picture (2 pts) Basis: 100 mol methanol (1 pt) Required: 50 moles O 2 100 % excess: 100 moles O 2 (2 pts).9 = 100 n meth /100 n CH3OH = 10 2 pts n N2 = 376 2 pts.75 = n CH2O /100 n CH2O = 75 2 pts C balance: 100 = 10 + 75 + n CO n CO = 15 2 pts H balance: 400 = 40 + 150 + 2n H2O n H2O = 105 2 pts O balance: 300 = 10 + 2n O2 + 75 + 15 + 105 n O2 = 47.5 2 pts Total: 628.5 2 pts y CH3OH =.016 y N2 =.598 y CH2O =.119 y CO =.024 y H2O =.167 y O2 =.076 (1 pt each total 6 pts)
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 n 4 = 647.7 (2 pts) d) 30-.01n 1 /30 =.976 (2 pts) n 3 + n 4 = 680.25 (2 pts) y meth =.952 y CO2 =.048 e) average molecular weight of the stream:.001*44 +.99 * 16 = 16.28 g/mol (3 pts) 1000 kg/hr * 1000 g/kg/16.28 g/mol = 6.142 X 10 4 mol/h (2 pts) 6.142 X 10 4 /70.7 = 868.74 (scaling factor) (1 pt) 100 X 868.74 = 8.69 X 10 4 mol/hr (1 pt)
Extra credit: Look at the following equation: Z = av b P c. Given data for Z, P and V, how would you solve for a, b and c?