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1-1 Solutions for Chapter 1 – Mole Balances Synopsis General: The goal of these problems are to reinforce the definitions and provide an understanding of the mole balances of the different types of reactors. It lays the foundation for step […]

10-81 CDP10-I 10-82 CDP10-J 10-83 CDP10-K 10-84 CDP10-L 10-85 CDP10-M 10-86 10-87 CDP10-N 10-88 CDP10-O 10-89 CDP10-P 10-90 10-91 10-92

10-93 CDP10-Q 10-94 CDP10-R 10-95 10-96 10-97 CDP10-S 10-98 CDP10-T 10-99 CDP10-U 10-100 CDP10-V 10-101 CDP10-W 10-102 CDP10-X 10-103 10-104

10-41 P10-18 (c) 10-42 10-43 P10-18 (d) 10-44 P10-18 (e) 10-45 P10-19(a) D da k dt =! S W U t= S dW U dt= D S k dt da dW dt U ! = 1D S k W a […]

10-21 Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 [3] Pa = y*Pao*(1-X)/(1+e*X) P10-10 (a) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2) [8] rate […]

10-1 Solutions for Chapter 10 – Catalysis and Catalytic Reactors P10-1 Individualized solution P10-2 (a) Example 10-1 (1) Pentane isomerization Pt nP iP!!“ Assume that Pt is the catalyst used. Maximum f = 5 molecules/site/sec Maximum rate: Minimum f = […]

11-21 M=XAMA+XBMB CB=XBC VA=NA/CA VB=NB/CB=0 V*=NA+NB C=7.17 *10“3 V=nA+nB =nA/ “ “ JA=CA(VA“V*)=NA“XANA=XBNA JB=CB(VB“V*)=“XBNA=“JA (*105) (*103) z XB XA CB CA ρΒ ρA ρ ωΒ ωA ω 0 0.329 0.671 1.3 2.65 0.376 2.02 2.4 0.158 0.842 60.7 5 0.434 […]

12-17 P12-10 (c) Let’s try again with “0 = 10 “=1+10 ( ) 20.12#1 [ ] =1#1020.99 ( ) “=1#99 =#99 not possible ψ will be negative for any value of “0 greater than one. P12-10 (d) We now need […]

12-1 Solutions for Chapter 12 – Diffusion and Reaction in Porous Catalysts P12-1 Individualized solution P12-2 (a) (a) “=5 De~ DAB ~T P Dex ~ De1 ~P1 P T2 T1 “ # $ % & ‘ 1.75 (lines and angles […]

13-21 ( ) ( ) ( ) ( ) ( ) ( ) ! ! ! ! ! ! ! ! ! ! d F eE e C C d F E k d F E k Ao A“# $ […]

13-41 X =41%. Maximum Mixedness ( ) ( )X F E C r d dX Ao A ! ! ! “ += 1 Rate Law : 2 BAA CkCr =! ( ) XCC AoA !=1 ( ) XCC BoB !=1 […]

13-58 3 /998.3 dmmoleC A!= 004.0= CD S 3 /0065.0 dmmolCB= 368.3= DE S 3 /000277.0 dmmolCC= 286.4= EF S 3 /068.0 dmmolCD= 3 /0202.0 dmmolCE= 3 /0047.0 dmmolCF= Segregation Maximum Mixedness CSTR PFR 00196.0= CD S 0009.0= CD S […]

13-1 Solutions for Chapter 13 – Distributions of Residence Times for Chemical Reactors P13-1 No solution will be given. P13-2 (a) The area of a triangle (h=0.044, b=5) can approximate the area of the tail :0.11 P13-2 (b) ~0.11 P13-2 […]

14-1 Solutions for Chapter 14 – Models for Non-ideal Reactors P14-1 Individualized solution P14-2 (a) Approximated formula for Segregation Model (1st reaction) Error tk kte kt ++!= ! 2 1 22 . The error is then o((kt)3). Approximating the !3 […]

P14-17 The presence of a minimum (run2) imply the presence of a radial temperature profile that effects the reaction rate and determines the deviation from the concentration profile given by Eq. (14.51). The enthalpy balance allows the identification of the […]

14-21 P14-7 Tubular Reactor 1st order, irreversible, pulse tracer test → 2 ! = 65 2 s and m t = 10 s For a 1st order reaction, PFR: X =1– e-kτ = 0.98 We need τ and k. There […]

14-54 I. Variation of Da number Conversion Damköhler number/ Da Closed-vessel Open-vessel Parameter 0.69 0.432 0.378 8*U0 1.38 0.713 0.641 4*U0 2.76 0.959 0.911 2*U0 5.52 1 1 U0 11.04 / / U0/2 22.08 / / U0/4 44.16 / / […]

2-18 P2-9 (f) Real rates would not give that shape. The reactor volumes are absurdly large. P2-10 Problem 2-10 involves estimating the volume of three reactors from a picture. The door on the side of the building was used as […]

2-1 Solutions for Chapter 2 – Conversion and Reactor Sizing Synopsis General: The overall goal of these problems is to help the student realize that if they have –rA=f(X) they can “design” or size a large number of reaction systems. […]

3-21 ( )( ) 3 0.305 1 0.39 0.19 A mol C dm =!= ( )( ) 3 0.305 0.39 0.36 C mol C dm = = P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law […]

3-1 Solutions for Chapter 3 – Rate Law and Stoichiometry P3-1 Individualized solution. P3-2 (a) Example 3-1 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 310 315 320 325 330 335 T (K) k (1/s) For E = 60kJ/mol […]

4-74 4-75 CDP4-I (a) CDP4-I (b) 4-76 CDP4-I (c) 4-77 CDP4-I (e) CDP4-I (d) 4-78 CDP4-J (a) b) We know, ε = 0, so 2/1 0)1( WPP ! “= => 2/1 )1(201 W ! “= => α = 9.98 x […]

4-21 Now using: ( ) ( ) ( ) 2 2 2 2 0 1 1 C C Z X Z X V f X V z z X X X X K K ! “ ! “ =#= = […]

4-61 ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2 [2] d(Ca2)/d(t) = (Ca1 – Ca2)/tau -k*Ca2^2 Explicit equations as entered by the user [1] k = 0.025 [2] Cao = 2 [3] […]

4-41 POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value w 0 0 100 100 X 0 0 0.5707526 0.5707526 Q 0.5625 0.5625 0.5625 0.5625 Fao 5 5 5 5 alpha 0.00944 […]

5-18 60 S g MW gmol = 6 3 2.32 2.32 *10 S g g ml m ! = = (Handbook of Chemistry and Physics, 57th ed., p.B-155) 6 3 1 10 g g ml m ! “= 20 F […]

5-1 Solutions for Chapter 5 – Collection and Analysis of Rate Data P5-1 (a) Individualized solution P5-1 (b) Individualized solution P5-1 (c) Individualized solution P5-1 (d) Individualized solution P5-1 (e) Individualized solution P5-1 (f) Individualized solution P5-1 (g) Individualized solution […]

6-61 P6-21 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst. A  B + C ‘ 1 1 1 B C C C A C C C r k C K ! “ =# $ % […]

Solutions for Chapter 6 – Multiple Reactions P6-1 Individualized solution P6-2 (a) Example 6-2 For PFR, 6-1 In PFR with τ = V/vo = 783 sec we get X = 0.958 and SB/XY = 0.624 also at τ = 560.24 […]

6-76 The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3 6-77 CDP6-C 6-78 6-79 CDP6-D 6-80 CDP6-E 6-81 6-82 6-83 CDP6-F (a) Mole balances: dFA/dV = -rD-rU dFD/dV = rD dFU/dV = […]

P7-10 (a) 7-21 =Et [ ] k1k3S [ ] “k2k4P [ ] k2+k3+k1S [ ] +k4P [ ] $ % ‘ ( =k3Et [ ] S [ ] “k2k4 k1k3 P [ ] k2+k3 ( ) k1 +S [ ] […]

7-41 max S C g M S C C rK C µ =+ This would result in the Cell concentration growing exponentially. This is not realistic as at some point there will be too many cells to fit into a […]

7-1 Solutions for Chapter 7 – Reaction Mechanisms, Pathways, Bioreactions and Bioreactors P7-1 (a) Example 7-1 The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the slope P7-1 (b) Example 7-2 For t […]

8-41 9 P0 1.013E+06 1.013E+06 1.013E+06 1.013E+06 10 CA0 270.8283 270.8283 270.8283 270.8283 11 CA 270.8283 258.1071 270.8283 258.1071 12 kr 0.2 0.076962 0.2 0.076962 13 CC 0 0 15.77362 15.77362 14 CB 0 0 15.77362 15.77362 15 rA -36.02017 […]

8-21 P8-5 8-22 2A B C+! A B C Fio lb mole hr ! ” # $ % & ‘ 10 10 0.0 Tio(F) 80 80 – ~ Pio Btu C lb mole F ! “ # $ ° % […]

8-1 Solutions for Chapter 8 – Steady-State Nonisothermal Reactor Design P8-1 Individualized solution P8-2 (a) Example 8-1 For CSTR V=FA0X “r A =X # 0k1“X ( ) X= $ k 1+ $ k= $ Ae“E RT 1+Ae“E RT One equation, […]

9-1 Solutions for Chapter 9 – Unsteady State Nonisothermal Reactor Design P9-1 Individualized solution P9-2 (a) Example 9-1 The new T0 of 20 ˚F (497 ˚R) gives a new ΔHRn and T. With T=497+89.8X the polymath program of example 9-1 […]

9-41 To 530 530 530 530 E 3.0E+04 3.0E+04 3.0E+04 3.0E+04 k 0.3478362 0.3478362 164.49672 164.49672 Ca 0.479966 0.0241085 0.479966 0.0241085 Cp 0.75 0.75 0.75 0.75 Gt 2.404E+05 2.404E+05 5.711E+06 5.711E+06 Rt -5.25E+05 -5.25E+05 6.825E+06 6.825E+06 ODE Report (RKF45) Differential […]

P9-10 (c) The unsteady energy balance for a batch reactor is showed in equation 9-11. 9-21 ( ) ( ) ! “#“$#++ = ii ARxs CpN VrHWQ dt dT & & Adiabatic operation ( Q & =0), neglecting s W […]

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