CHE 205 Exam 1 Solutions
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1 CHE 05 Exam 1 Solutions Problem 1 a) Figure +5 Reactor, Still, Condenser, Combustion Heater, Mixing point +1 dots on flowrates +1 for and x for molar composition + for correctl labeled streams b) Degree of Freedom Analsis: + Unknowns: n 8; n 9; O,9 - Equations: O, C, H balance +1 for correct number of unknowns + for correct number of equations +1 for relation +1 correct D.O.F -1 Relation: ethanol,9 + CO,9 + CO,9 + H O,9 + O,9 1 D.O.F -1 c) Molar composition of the feed and percent conversion of ethlene Carbon balance on reactor: + for correct carbon balance +1 for n eehaaaa, relationship +1 for n eeh, relationship +1 for putting the correct values into carbon balance +1 for correct n eeh,1 calculation + correct conversion relationship +1 for correct numerical values 1
2 CHE 05 Exam 1 Solutions inert gas, ethlene, HO, d) Molar flow rate of fresh ethlene and steam to reactor + for ethlene balance +1 for n eeh answer +1 for oxgen balance +1 for n HO answer for e) Combustion reactions C H 5 OH + O CO + H O C H 5 OH + O CO + H O +4 each reaction f) Percent excess molecular oxgen Carbon balance on combustion heater: + for carbon balance +1 for n 9 answer +4 oxgen balance +1 for the relationship,9 sum to 1 + theoretical oxgen equation +1 correct answer
3 CHE 05 Exam 1 Solutions n ethanol,6 mol C 1 mol C H 5 OH n mol C 9 ethanol,9 1 mol C H 5 OH + 1 mol C 1 mol C CO,9 + 1 mol CO CO,9 1 mol CO n mol/h Oxgen balance on combustion heater: n ethanol,6 1 mol O 1 mol C H 5 OH + n mol O O,8 1 mol O 1 mol O n 9 ethanol,9 1 mol C H 5 OH + mol O 1 mol O CO,9 + 1 mol CO CO,9 1 mol CO + mol O 1 mol O O,9 + 1 mol O H O,9 1 mol H O In which: O,9 1 ethanol,9 CO,9 CO,9 H O,9 7.41% n O, mol/h mol O Theor O n ethanol,6 90 mol/h 1 mol C H 5 OH % excess O n O,8 Theor O Theor O 98% AN ALTERNATIVE SOLUTION to Problem 1: +5 Reactor, Still, Condenser, Combustion Heater, Mixing point +1 dots on flowrates +1 for and x for molar composition + for correctl labeled streams b) Degree of Freedom Analsis:
4 CHE 05 Exam 1 Solutions + Unknowns: n 8; n 9; O,9-5 Species balances: C H 5 OH; CO ; CO; H O; O + Chemical reactions: C H 5 OH + O CO + H O +1 for correct number of unknowns + for correct number of balance equations +1 for relation C H 5 OH + O CO + H O D.O.F -1 c) Reaction in reactor: C H 4 + H O C H 5 OH ξ ξ n ( CH50H9 ) 60 mol/hr n inert,1 90 mol/hr n inert, n HO, n HO,1 ξ n HO, 4740 mol/hr n ethlene, n CH4,1 ξ n ethlene, 4590 mol/hr inert gas, for reaction/extent of reaction +1 extent of reaction equation +1 for n HO, relationship +1 for n eeh, relationship +1 for correct n eeh,1 calculation + correct conversion relationship +1 for correct numerical values ethlene, HO, d) Molar flow rate of fresh ethlene and steam to reactor 4
5 CHE 05 Exam 1 Solutions + for ethlene balance +1 for n eeh answer +1 for oxgen balance +1 for n HO answer for e) Combustion reactions C H 5 OH + O CO + H O C H 5 OH + O CO + H O (I) (II) +4 each reaction f) Percent excess molecular oxgen O,9 1 ethanol,9 CO,9 CO,9 H O,9 7.41% Suppose the extents of reaction (I) and (II) are ξ 1 and ξ, respectivel, then we will have Species n i,in n i,out Change C H 5 OH n ethanol,6 0 mol/h n 9 ethanol,9 ξ 1 ξ CO 0 n 9 CO,9 ξ 1 CO 0 n 9 CO,9 ξ H O 0 n 9 H O,9 ξ 1 + ξ O n O,8 n 9 O,9 ξ 1 ξ C H 5 OH balance: n 9 ethanol,9 n ethanol,6 ξ 1 ξ (1) CO balance: n 9 CO,9 ξ 1 () CO balance: 5
6 CHE 05 Exam 1 Solutions n 9 CO,9 ξ () (1)* + () + (): n 9 ethanol,9 n ethanol,6 + n 9 CO,9 + n 9 CO,9 0 Substitute n 9 into equations () and (): O balance: n mol/h ξ 1.1 mol/h ξ 9. mol/h + for in, out and change chart +4 for spices balance equations +1 for n 9 answer +1 for the relationship,9 sum to 1 + theoretical oxgen equation +1 correct answer n 9 O,9 n O,8 ξ 1 ξ n O, mol/h mol O Theor O n ethanol,6 90 mol/h 1 mol C H 5 OH % excess O n O,8 Theor O Theor O 98% 6
7 CHE 05 Exam 1 Solutions Problem ) Two gas-phase reactions occur in the reactor: Reaction 1: Reaction : CO + H CH OH; K e, atm CO + H O CO + H ; K e, Let the rate of the extent of reaction for reaction 1 be ζ 1 and that for reaction beζ. +5 if Correct D.O.F. analsis (including correct labeling of all components) Answer for Part (a) Degree of freedom analsis: +.5 if Partiall Correct D.O.F. with a majorit of numbers and terms correctl labeled. + 6 unknown labeled variables ( H, CO, HO, 4 + independent chemical reactions K ) N, P, eq, - 5 chemical species in the reactions (H, CO, H O, CO, CH OH) - specification equations (equilibrium equations Keq,1 Keq,; sum of mole fractions in Stream 4) The sum of all values provides the number of degrees of freedom. There are 0 degrees of freedom and the problem is solvable. 7
8 CHE 05 Exam 1 Solutions CO 1 H 1 1 HO CH OH 1 CO T 1 1 CH OH 1 ζ + ζ + + ζ ζ ζ ζ ζ ζ CH OH T ζ ζ1 ζ ζ1 ζ1 0.( given) 10.1 ζ1 ζ points for correct mole table, including correct notation and a correct total mole balance equation 8
9 CHE 05 Exam 1 Solutions CO CO T ζ 1+ + ζ1 ζ ζ1 ζ 0.01( given) 10.1 ζ1 ζ CO ζ1 ζ CO + + ζ H T (.01) 0.48 H ζ1+ ζ + + ζ 5 (.01) (.01) T ( ) H O CO H CH OH CO
10 CHE 05 Exam 1 Solutions Given that K CHOH e,1 10.5atm CO * H P P CH OH CO H *0.171 *10.5 * *10.5 Answers for Part (b) P 1.49atm Stream 4: CO 0.48 H CHOH 0. CO 0.01 HO ζ 10.1 (.01) 6.08 kmol / h T 1 1 Now, Part B: +5 for correct calculation of P +10 total for correct calculation of ξ 1, ξ, and unknown mole fractions (+ for each correct term) K e, * CO H * CO H O 0.01* * Part C: +1.5 for correct smbolic expression of K e,. +.5 for correct numerical value of K e, Partial Credit is awarded for the smbolic expression where appropriate Answer for Part (c) K e,
11 CHE 05 Exam 1 Solutions AN ALTERNATIVE SOLUTION to Problem : Two gas-phase reactions occur in the reactor: Reaction 1: Reaction : CO + H CH OH; K e, atm CO + H O CO + H ; K e, Answer for Part (a) Degree of freedom analsis: + 6 unknown labeled variables ( - atomic species balances (H, C, O) H, CO, HO, 4 K ) N, P, eq, - specification equations (equilibrium equations Keq,1 Keq,; sum of mole fractions in Stream 4) The sum of all values provides the number of degrees of freedom. There are 0 degrees of freedom and the problem is solvable. +5 if Correct D.O.F. analsis (including correct labeling of all components) +.5 if Partiall Correct D.O.F. with a majorit of numbers and terms correctl labeled. 11
12 CHE 05 Exam 1 Solutions CH OH 0. (given) CO 0.01 (given) Atomic Species Balances: C Balance: 1kmolC kmolco COkmolCO 1kmolC CO kmolco 1kmolC * ( * + * kmolco min kmoltotal kmolco kmoltotal kmoleco CHOH kmolchoh 1kmolC 4kmoltotal + * ) kmoltotal kmolch OH min 5 ( + 0.4) CO 4 (a) H Balance: kmolh kmolh kmolh kmolh O * 1 + * kmolh min kmolh O min kmolh HkmolH kmolh HO kmolho 4kmolH CHOH kmolchoh 4kmoltotal ( + + ) kmolh kmoltotal kmolh O kmoltotal kmolch OH kmoltotal min 10. ( ) H HO 4 (b) O balance: 1kmolO kmolco 1kmolO * + * kmolco min kmolh O kmolh O min COkmolCO 1kmolO CO kmolco kmolo CHOH kmolchoh 1kmolO ( * + * + * kmoltotal kmolco kmoltotal kmolco kmoltotal kmolchoh HO kmolho 1kmolO 4kmoltotal + * ) kmoltotal kmolh O min 5.1 ( ) CO H O 4 (c) 1
13 CHE 05 Exam 1 Solutions Sum of mole fractions H H O CO CO CH OH d) H H O CO 0.66 We have 4 equations (a), (b), (c) and (d) and 4 unknowns, namel, H, CO, equations simultaneousl, we get HO, N 4. Solving the H CO HO kmol / h Now, to compute the pressure, we can use the specification equation of K eq,1 K CHOH e,1 10.5atm CO * H P P CH OH CO H *0.171 *10.5 * *10.5 Answers for Part (b) P 1.49atm Unknowns in Stream 4: CO H HO T kmol / h Part B: +5 for correct calculation of pressure +7 for each correct calculation of unknown mole fraction (+4 for each correct atomic species balance if mole fractions not completed) +4 for correct calculation of total moles in outlet stream 1
14 CHE 05 Exam 1 Solutions And finall, CO P* HP Ke, P* P CO HO * CO H * CO HO 0.01* * Part C: +1.5 for correct smbolic expression of K e, +.5 for correct numerical calculation of K e, Partial Credit is awarded for the smbolic expression where appropriate Answer for Part (c) K e,
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