ChE 344 Winter 011 Final Exam + Solution Monday, April 5, 011 Open Book, Notes, and Web Name Honor Code (Please sign in the space provided below) I have neither given nor received unauthorized aid on this examination, nor have I concealed any violations of the Honor Code. (Signature) The Basics 1) / 5 pts ) / 5 pts 3) / 5 pts 4) / 5 pts 5) / 5 pts 6) / 5 pts 7) / 5 pts Applications 8) / 5 pts 9) /10 pts 10) /10 pts 11) /10 pts Professional 1) /10 pts 13) /0 pts Total /100 pts
(5 pts) 1) Mole Balances, Chapter 1 The reaction A + B C takes place in an unsteady CSTR. The feed is only A and B in equimolar proportions. Which of the following set of equations gives the correct mole balances on A, B and C. Species A and B are disappearing and Species C is being formed. Circle the correct answer where all the mole balances are correct (a) F B0 F A V r A dv = dn A F B0 F B V r A dv = dn B Wrong sign for r A, should be + V r A dv (b) F C + F A0 F A + V V r A dv = dn C r A dv = dn A F A0 F B + V r A dv = dn B All are correct. (c) F C V V F A0 F A + 0 r A dv = dn C r A dv = dn A F A0 F B + V r A dv = dn B Wrong sign for F C, should be F C (d) F C + F B0 F A V V r C dv = dn C r A dv = dn A F B0 F B V r A dv = dn B Wrong sign for r A, should be + V r A dv F C + V r C dv = dn C Solution Answer is (b). 1
(5 pts) ) Circle the correct answer, true (T), False (F) or (CT) Can t tell from the information given. T F CT (a) Multiple steady states can exist for an irreversible endothermic first order reactions. T F CT (b) A steady-state CSTR operates at 150 C. The reactor effluent is at the same temperature as the reactor contents. Is the reactor operating isothermally? T F CT (c) Multiple steady states can only exist for adiabatic reactions. T F CT (d) Reactor staging is only used for irreversible reactions. T F CT (e) The detrimental effect of pressure drop in gas phase reactions is more pronounced for adiabatic- exothermic reactions than for adiabatic endothermic reactions. Solution (a) False. R T G (b) Insufficient information to answer definitively. Need to know the feed temperature. T 0 = 300 K T = 300 K T = 300 K (c) False. See Equation 1-8 (d) False. See Figure 11-4 (e) True. dy dw = α y Exothermic T y Endothermic T y ( 1+ εx) T T 0
(5 pts) 3) Circle the correct answer. Consider the following Levenspiel plot (1 pt) (a) The equilibrium conversion is (1) X e < 0.6 () X e = 0.8 (3) X e > 0.8 (4) Can t tell from the information given (1 pt) (b) The flow rate to an 8 dm 3 CSTR corresponding to Figure E4-1 where 80% conversion is achieved is (1) F A0 = 0.8 mol/s () F A0 = 10 mol/s (3) F A0 = 1 mol/s (4) Can t tell from the information given (1 pt) (c) If the conversion achieved in a single 8 dm 3 CSTR is 80%, what would the conversion be if the flow is equally divided into to two CSTRs in parallel with first reactor having a volume of 4 dm 3 each (same total volume). " 0 " 0 " 0 " 0 vs. 8 dm 3 X=0.8 4 dm 3 4 dm 3 X=_?_ X=_?_ The total reactor volume is constant at 8 dm 3. The conversion for the two reactors in parallel is (1) X > 0.8 () X < 0.8 (3) X = 0.8 (4) Can t tell from the information given ( pts) (d) If the conversion achieved in a single 8 dm 3 CSTR is 80%, what would the conversion be if two CSTRs are connected in series with first reactor having a volume of approximately 3.0 dm 3 and the second reactor having a volume of 0.6 dm 3. 3
υ 0 vs. 8 dm 3 3.0 dm 3 X=0.8 0.6 dm 3 X=_?_ The conversion for the two reactors in series is (1) X > 0.8 () X < 0.8 (3) X = 0.8 (4) Can t tell from the information given Solution (a) Ans. (3) X e > 0.8 (b) Ans. (d) Can t tell from information given (c) Ans. (3) X = 0.8. See p161. (d) Ans. () X < 0.8. Try X = 0.6 3 dm 3 = 5 dm 3 x 0.6 = 3 dm 3 checks Try ΔV = 0.1 between X = 0.6 and 0.7 V = 0.1 x 6 dm 3 = 0.6 dm 3 checks F A0 r A (dm 3 ) 3dm 3 10 9 8 7 6 5 4 3 1 0. 0.4 0.6 0.8 X 4
(5 pts) 4) Consider the following reaction for parts (a), (b) and (c) A + B C Write the rate law in terms of the specific reaction rate and species concentration when (a) The reaction is irreversible and second order in A, and independent of the concentration of C, and overall first order. r A = (b) The reaction is elementary and reversible r A = (c) Now consider the case when the reaction is first order in A and first order in B at high concentrations of A and B and is first order in A and second order in B at low concentrations of B. The rate law is r A = (d) What is the rate law for the reaction CH 3 CHO CH 4 + CO? r CH 3 CHO = Solution A + B C (a) r A = k A C A C B (b) r A = k A C A C B C C (c) r A = k 1 C A C B 1+ k C B K C (d) r A = kc A 3 (see page 80) 5
(5 pts) 5) The following figure shows the energy distribution function at 300 K for the reaction A + B C f(e,t) (kcal) 1 0.5 0. 0.15 0.1 0.05 0 1 3 4 5 6 7 8 E (kcal) (a) What fraction of the collisions have energies between 3 and 5 kcal? (b) What fraction of collisions have energies greater than 5 kcal? Solution (a) Between 0 and 4 k cal f( E,T) = 0.5 E 4 Between 4 and 8 kcal 0.5 0.5 f( E,T) = E 4 0.5 f(e,t) (kcal) 1 0. 0.15 0.1 0.05 at E = 3 f( E,T) = 3.5 = 0.188 4 at E = 5 f( E,T) = 3.5 = 0.188 4 (b) 3.5 4 ( ) 0.5 0 1 3 4 5 6 7 8 E (kcal) Area =1 ( 0.5) + ( 1) 3 4.5 = 0.44 = 44% f(e,t) (kcal) 1 0. 0.15 0.1 0.05 Area = 3.5 4 = 0.8 = 8% ( ) 3 1 = 0 1 3 4 5 6 7 8 E (kcal) 6
(5 pts) 6) Consider the following elementary gas phase reaction A + B C Write r A solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, reversible, gas phase, isothermal reaction with no pressure drop with an equal molar feed with C A0 =.0, k A = and K C = 0.5 all in proper units. r A = Solution Elementary r A = k C A C B C C K C A is the limiting reactant A + B C y A0 = 1 (Equimolar) ε = y A0 δ = 1 1 1 1 = 1 C A = C A0 1 X C A0 Θ B X X C A0, C B =, C C = 1+ εx 1+ εx r A = k ( ) ( ) 3 C A0 ( ) ( 1 X ) Θ B X ( 1+ εx) 3 C A0 X K C 1+ εx ( ) 1+ εx ( ) 1 r A = k C A0( 1 X) C A0 1 1 1 X X C 1 1 A0 X X K C 1 1 X r A = C 3 A0 1 X 1 1 X ( ) C A0 X K C 1 1 X 7
dm 3 k =, C A0 = mol mol s dm 3, Θ B =1, K C = 0.5 dm3 mol r A = 8( 1 X) 1 1 X X 1 1 X 8
(5 pts) 7) The elementary gas phase isomerization exothermic reaction A B cat is carried out isothermally at 400K in a PBR where the pressure drop occurs with α = 0.001 kg 1. The flow is laminar. Currently 50% conversion is achieved. The equilibrium constant at this temperature is 3.0. (a) For a fixed mass flow rate m, if the reactor diameter is increased by a factor of 4, the conversion X is (1) X > 0.5 () X < 0.5 (3) X = 0.5 (4) insufficient information to tell. (b) For a fixed mass flow rate m, the equilibrium conversion X e is (1) X e = 0.5 () X e = 0.667 (3) X e = 0.75 (4) insufficient information to tell. (c) For a fixed mass flow rate m, if the reactor diameter is increased by a factor of, the equilibrium conversion X e will (1) increase () decrease (3) remain the same (4) insufficient information to tell (d) For a fixed mass flow rate m, if the particle size is increased the equilibrium conversion X e will (1) increase () decrease (3) remain the same (4) insufficient information to tell (e) Consider the case where an adiabatic endothermic reaction becomes frozen in a PFR [cf. p.541]. The actual conversion, X, can be greater than the equilibrium conversion, X e, near the point where the reaction becomes frozen. (1) True () False (3) Depends on feed condition Solution (a) Ans. (1) X > 0.5 α ~ 1 D Increase D, decrease α, increase X. (b) Ans. (3) X e = 0.75 X e = K C 1+ K C = 3 1+ 3 = 0.75 9
(c) Ans. (3) remain the same K C = C A = C A0( 1 X e )y C Ae C A0 X e y X e is not a function of pressure drop parameter α. = 1 X e X e (d) Ans. (3) or (4) Exothermic D P then K then X e X e is not a function of particle size or any other geometry. But will a change in pressure drop affect the achievable conversion, and therefore temperature? X e is a function of temperature. So probably accept unchanged OR Can t tell (e) Ans. () X e can never be less than X, (i.e., X can never be greater than X e). 10
(5 pts) 8) Suppose inerts are added to the system in Example 13-. The dashed line represents the relationship after the inerts were added. Which figure represents how the relationship between the line the reactor failed after start up, ts, and the down time, t d, would change? The solid line represents the case without inerts? Solution Answer: B Explanation: dt = Qg Qr N i C Pi = Qg Qr N A C PA + N B C PB + N C C PC + N I C PI We see the rate of temperature increase,, decreases as the amount of inert, N I, increases. Therefore, the temperature will not rise as rapidly during the adiabatic period (down time, td) and we can have longer down times without having an explosion. 11
(10 pts) 9) P10-5A Study Problem. The rate law for the hydrogenation (H) of ethylene (E) to form ethane (A) H + C H 4 C cat H 6 over a cobalt-molybdenum catalyst [Collection Czech. Chem. Commun., 51, 760 (1988)] is r E ʹ = kp EP H 1+ K E P E Suggest a mechanism and rate-limiting step consistent with the rate law and then derive the rate law. Solution H = H E = Ethylene A = Ethane H + C H 4 H + E A C cat H 6 Because neither H or C H 6 are not in the denominator of the rate law they are either not adsorbed or weakly adsorbed. Assume H in the gas phase reacts with C H 6 adsorbed on the surface and ethane goes directly into the gas phase. Then check to see if this mechanism agrees with the rate law Eley Rideal E + S E S r AD = k AD P E C V C E S E S + H A + S r S = k S C E S P H E S A + S Assume surface reaction C E S = K E P E C V r S = k S [ C E S P H ] C T = C V + C E S P r A ʹ = k S K E C E P H T 1+ K E P E K E 1
(10 pts) 10) P7-10 B Study Problem. In order to study the photochemical decay of aqueous bromine in bright sunlight, a small quantity of liquid bromine was dissolved in water contained in a glass battery jar and placed in direct sunlight. The following data were obtained at 5 C: Time (min) Ppm Br, C A 0 10 0 30 40 50 60 3.45.45 1.74 1.3 0.88 0.6 0.44 (a) Determine whether the reaction rate is zero, first, or second order in bromine, and calculate the reaction rate constant in units of your choice. (Hint: Preliminary calculations suggest the reaction may be first order.) (b) Assuming identical exposure conditions, calculate the required hourly rate of injection of bromine (grams per hour) into a very large sunlit body of water, 5,000 gal (94,600 dm ) in volume, in order to maintain a sterilizing level of bromine of 1.0 ppm. Note: ppm = parts of bromine per million parts of brominated water by weight. In dilute aqueous solutions, 1 ppm = 1 milligram per liter.) (From California Professional Engineers Exam.) 3 13
Solution P7-10 (a) Photochemical decay of bromine in bright sunlight: t (min) CA (ppm) 0 3.45 10.45 0 1.74 30 1.3 40 0.88 50 0.6 Mole balance: constant V dc A = ra = kcαa dc A α = 1, therefore = kc A dc A C = kc A, ln A0 = kt CA After plotting and differentiating by equal area Time (min) 0 10 0 30 3.45.45 1.74 1.3 Ppm Br, CA dca/ 0.08 0.061 0.04 -.501 -.797-3.170 ln( dca/) 0.896 0.554 0.07 ln CA 0 0.34 0.68 1.03 ln CA0/CA 40 0.88 0.030-3.507-0.18 1.34 First order ln C A0 CA t 14 60 0.44 50 0.6 0.015-3.840-0.478 1.7 60 0.44 0.014-4.69-0.81.05
Slope of Plot of ln (C A0/C A) versus t.0 Slope = 58 min k = 0.0345 min 1 P7-10 (b) dn A dn A = Vr A = F A0 0 + r A V Steady state dn A F A0 = r A V at CA = 1 ppm r A = 0.344 ppm min = 0 F A0 94,600 dm 3 = 0.344 mg l min ( ) 0.034 mg 1g dm 3 min 1,000mg 60m =195 g h h 15
This problem is continued from Problem (6) on MidTerm Exam II (Complete parts (e) and (f)) (10 pts) 11) The temperature and conversion in a virtually infinitely long PFR are shown below as a function of the reactor volume. The reactor is surrounded by a jacket for heat transfer. The value of Ua is 100 cal/(sec m 3 K) with T a being constant. The gasphase, reversible reaction is A B + C and pure A is fed to the reactor at a concentration C A0 = 1.0 mol/m 3 and a molar flow rate of 10 mol/s. The absolute value of the heat of reaction is 5,000 cal/mol of A at 500K, and the heat capacities of A, B, and C are each 10 cal/mol/k. T(K) 800 700 600 500 400 300 0 10 0 30 100 V(m 3 ) (a) Using what is the equilibrium constant at 500 K? K e (500) = 0.11 (From MidTerm Exam II solution) (b) What is the rate of disappearance of A, r A, at 10 m 3? r A = mol/m 3 s (From MidTerm Exam II solution) (6 pt) (c) What is the equilibrium constant at 400 K? K e (400) = (8 pt) (d) What is the specific reaction rate at V = 10 m 3? k = (6 pt) (e) What is the total amount of heat added/removed to the entire reactor per mol of A feed? Include proper sign in your numerical answer if possible. Q F A0 = 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0.1 X 16
Solution T(K) 800 700 600 500 400 300 0 10 0 30 V(m 3 ) 100 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0.1 X (a) K e (500) = _? at V = 10 m 3 Q g = Q r A 1 B+ 1 C δ = 0 ε = 0 C A = C A0 ( 1 X) C B = C C = C A0 X at 500 K X = X e = 0.4 K e = C BC C C A K e = 0.11 = ( ) = ( 0.4) 4( 1 0.4) = 0.16 4( 0.36) = 0.11 C A0 X e C A0 4 1 X e (b) r A = _? dx dw = r A F A0 Q g Q r!#" # $!#" # $ dt dv = ( r A) ( ΔH Rx ) Ua( T T a ) F i C Pi At minimum dt dv = 0 Q g = Q r At large V no further change in temperature so assume T = T a = 500 K and that T at minimum is 400 K. At minimum 17
Q r = Ua (T T a) = 100 (400 500) = 10,000 cal/s m 3 Q g = ( r A )( ΔH Rx ) ( r A )( ΔH Rx ) = ( r A )( 5,000) = Q r = 0,000 cal m 3 s r A = 10,000 cal s m 3 5,000 cal mol (c) K e (400) = _? K e ( 400) = K C 500 K e 400 = mol m 3 s ( )exp ΔH Rx 1 1 R T 1 T ( ) = 0.11exp 5,000 = 0.11exp 1 1.987 500 1 400 100 400 ( )( 500) 5,000 1.987 K e = 0.11exp( 1.5) = ( 0.11) ( 0.8) = 0.031 (d) k = _? r A = k C A0 C BC C K e A B + C C A = C A0 ( 1 X) C B = C C = C A0 X at V = 10 m 3 T = 400, X = 0., K e = 0.031 18
r A = k C A0 1 X = kc A0 1 X = k 1 1 0. ( ) C A0 ( ) X X 4K C 4K e ( ) ( 0.) ( 4) ( 0.031) = k 0.64 0.04 0.14 r A = mol mol m 3 = 0.3 s m 3 k = k[ 0.64 0.3] (e) k = m3 = 6.5 0.3 mol s Q =? F A0 Q F A0 Θ i C Pi T T 0 Per Mole A Q F A0 = C PA T T 0 A B + C! ( ) ΔH Rx [ + ΔC P ( T T a )]F A0 X = 0 Eqn. (11-8) [ ( )X] [ ] + ΔH Rx + ΔC P T T R ΔC P = C P C + C P B C P A = 10 + 10 10 = 0 = 0 Q F A0 =10 500 600 ( ) + +5,000 [ ] 0.4 [ ] = 1,000 +,000 =1,000 cal mol 19
(10 pts) 1) Let s revisit homework Problem P1-3 where the reaction B C A + B is carried out in a packed bed reactor. Match the following temperature and conversion profiles for the 4 different heat exchange cases adiabatic, constant T, co-current exchange and counter current exchange. a Figure 1 Figure Figure 3 Figure 4 0
Figure A Figure B Figure C Figure D (a) Figure 1 matches Figure (b) Figure matches Figure (c) Figure 3 matches Figure (d) Figure 4 matches Figure Solution (a) Figure 1 matches Figure _C_ (b) Figure matches Figure _A_ (c) Figure 3 matches Figure _D_ (d) Figure 4 matches Figure _B_ 1
In Class Problem 14 Solution (a) Adiabatic
Constant T In Class Problem 14 Solution (b) Constant T Heat Exchange a 3 a
In Class Problem 14 Solution (c) Variable T Co-Current Heat Exchange a 4
In Class Problem 14 Solution (d) Variable T Counter Current Heat Exchange a Matches 5
(0 pts) 13) The following reactions are taking place in a,000 dm 3 liquid phase batch reactor under a pressure of 400 psig k 1A A + B C ΔH Rx1B = 5,000 cal mol r 1A = k 1A C A C B k A 3C + A D ΔH RxC = +10,000 cal mol r A = k A C A C C B+ 3C k 3C E ΔH Rx3B = 50,000 cal mol r 3C = k 3C C B C C The initial temperture is 450 K and the initial concentrations of A, B and C are 1.0, 0.5 and 0. mol/dm 3 respectively. The coolant flow rate was at it s maximum value so that T a1 = T a = T a = 400 K so that the product the exchange area and overall heat transfer coefficient, UA, is UA = 100 cal/s K. (a) What is Q r at t = 0? (b) What is Q g at t = 0? (c) If Q r > Q g at time t = 0, and there is no failure of the heat exchange system, is there any possibility that reactor will run away? Explain (d) What is the initial rate of increase in temperature, (dt/) at t = 0? dt = (e) Suppose that the ambient temperature T a is lowered from 400 K to 350 K, what is the initial rate of reactor temperature change? dt = (f) A suggestion was made to add 50 mole of inerts at a temperature of 450 K. Will the addition of the inerts make runaway more likely or less likely? How? Show quantitatively. Additional information As a first approximation, assume all heats of reaction are constant ( ΔC P 0) Specific reaction rates at 450 K are ( ) s k 1A =1 10 3 dm 3 mol k A = 1 ( 3 10 3 dm 3 mol) s k 3C = 0.6 10 3 ( dm 3 mol) s 6
C PA =10cal mol K C PB =10cal mol K C PD = 80cal mol K C PE = 50cal mol K Solution Part (a) Part (b) C PC = 50cal mol K Q r = UA( T T a ) =100 cal cal [ 450 400]K = 5,000 5 K s Q g = V[ r 1B ΔH Rx1B + r C ΔH RxC + r 3B ΔH Rx3B ] Initially T = 350 K Reaction 1: Reaction : r 1A 1 = r 1B = r 1C 1 r A = r C 3 = r D 1 r 1B = r 1A r C = 3 r A Part (c) Reaction 3: r 3B 1 = r 3C 3 = r 3E 1 Q g = V[ k 1A C A C B ] ΔH Rx1B r 3B = 1 3 r 3C [ ] + V 3 k AC A C C ΔH RxC [ ] + V 1 3 k 3CC B C C = (,000) [( ) ( 10 3 ) ( 1 ) ( 0.5 ) 5,000]!### "#### $ +,000 3 1 3 10 3 ( 1) ( 0.) [ 10,000] +!####"#### $ 5,000 ( ) 0.5 +,000 1 0.6 10 3 ( )( 0.) 3 [ 50,000 ] = 5,000!##### "###### $ Q g = 5,000cal s dt = dt = +,000 Q r Q g N A0 C PA + N B0 C PB + N C0 C PC = Q g Q r N A C PA + N B C PB + N C C PC,000 5,000 5,000 N A0 C PA + N B0 C PB + N C0 C PC = 0 ΔH Rx3C [ ] If Q r > Q g then the temperature can only decrease causing the specific reaction rates k i to decrease, hence runaway is unlikely. 7
Part (d) Part (e) N A0 C PA = C A0 VC PA = ( 1) (,000) ( 10) = 0,000 N B0 C PB = C B0 VC PB = ( 0.5) (,000) ( 10) =10,000 N C0 C PC = C C0 VC PC = ( 0.) (,000) ( 50) = 0,000 dt = Q g Q r 50,000 Drop T a by 50 Q r = UA( T T a ) =100( 450 350) =10,000 dt Part (f) = 5,000 10,000 50,000 = 0.1 [ ] UA( T T a ) dt = Q g Q r = r 1B VΔH Rx1B + r C VΔH RxC + r 3A VΔH Rx3A N i C Pi N A C PA + N B C PB + N C C PC + N D C PD + N E C PE + N Inerts C PInerts Inerts (N Inerts) will not change Q g or Q r, they will only show the rate of temperature increase or decrease. 8
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