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P1-6. Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the student an idea of things to come in terms of sizing reactors in chapter 4. An alternative to P1-15. P1-7. Straight forward modification of […]

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. Minimum: ( ) 3 7 1 1 3*10 0.5 7.69 […]

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

P10-25 10-61 P10-26 (a) 10-62 10-63 P10-26 (b) Using the same program we can see that the maximum conversion is 0.887 P10-26 (c) CDP10-A 10-64 10-65 CDP10-B 10-66 10-67 CDP10-C 10-68 10-69 The thickness on these wafers can be obtained […]

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

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

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

Solutions for Chapter 11 – External Diffusion Effects on Heterogeneous Reactions P11-1 Individualized solution P11-2 (a) WB=“2WA WA=cDAB 1+yA ( ) dyA dz =“cDABdln 1+yA ( ) dz Integrating with yA=0 at z=“ WA=cDAB ” # z ( ) ln […]

11-21 M=XAMA+XBMB CB=XBC (*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 0.566 1.71 2.24 0.495 1.71 2.21 0.226 0.774 56 10 0.573 […]

12-1 Solutions for Chapter 12 – Diffusion and Reaction in Porous Catalysts P12-1 Individualized solution P12-2 (a) (a) “=5 P12-2 (b) (1) First Order Reaction Kinetics d2# d$2“Da#=0 Da =kb2 De #=A cosh Da$+B sinh Da$ d# d$=ADa sinh Da$+BDa […]

12-17 P12-10 (c) Let’s try again with “0 = 10 P12-10 (d) We now need to resolve the problem with the fact that there is a critical value of λ, λc, for which both ψ = 0 and d“ d#=0 […]

13-41 Maximum Mixedness ( ) ( )X F E C r d dX Ao A ! ! ! “ += 1 Rate Law : 2 BAA CkCr =! ( ) XCC AoA !=1 ( ) XCC BoB !=1 ( )32 […]

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 For […]

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

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) ! k Error 0.1 -0.0001667 1 -0.1667 10 -166.7 P14-2 (b) Parameters Dispersion Model Closed-Closed dispersion model ( […]

14-41 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 […]

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 II. Variation of Pe number Conversion Peclet number/Pe Closed-vessel Open-vessel Parameter 1.04e3 1 1 100* DAB 1.04e4 1 1 […]

14-21 P14-7 Tubular Reactor 1st order, irreversible, pulse tracer test → 2 ! = 65 2 s and m t = 10 s Calculate, k k = ! X“1 1 ln = 3.91/25 s = 0.156 s-1 Therefore assume closed […]

P2-8. To be used in those courses emphasizing bio reaction engineering. P2-9. The answer gives ridiculously large reactor volume. The point is to encourage the student to question their numerical answers. P2-10. Helps the students get a feel of real […]

h = 56ft d = 9 ft V = πr2h = π(4.5 ft)2(56 ft) = 3562 ft3 = 100,865 L Length of entire reactor = (23 ft)(12)(11) = 3036 ft D = 1 ft V = πr2h = π(0.5 ft)2(3036 […]

3-21 dm P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction): 3 C A A C C r k C K ! “ #=# $ % & ‘ ( )( ) 01 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 0.005 310 315 320 325 330 335 T (K) k (1/s) 0.006 0.007 0.008 For E = 60kJ/mol For […]

[1] Dp = .0075 [2] Q = 75*Dp [3] Fao = 5 [4] alpha = .0000708/Dp [5] Cao = .207 [6] kprime = 3*(3/Q^2)*(Q*coth(Q)-1) [7] ra = -kprime*(Cao*(1-X)*y)^2 P4-20 (f) Individualized solution P4-20 (g) Individualized solution P4-20 (h) Individualized solution […]

P4-2 (b) Example 4-1 There would be no error! The initial liquid phase concentration remains the same. FAO = 6.137 12.27 / min 0.5 C Flbmol X= = Also, 3 12.27 min AO AO AO Fft vC = = vo […]

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

4-61 ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1^2 P4-33 (b) 99% of the steady state concentration of A (the concentration of A leaving the third reactor) is: (0.99)(0.611) = 0.605 This […]

4-21 z 2902.2 V 3785.4 E 1.5E+04 R 2 y 1.5684405 Kco 3 Hrx -2.5E+04 Kc 1.4169064 [3] z = 2902.2 [4] V = 3785.4 [5] E = 15000 [6] R = 2 [7] y = exp(E/R*(1/To-1/T)) [8] Kco = […]

The Finite differences method uses mathematical estimates to calculate the rate. It is only possible to use this when the time interval of each data point is uniform. The graphical method uses polynomial regression to approximate CA as a function […]

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) Final concentration of HF ( ) 5 2.316 0.2 0.107 weight fraction […]

6-21 P6-8 (d) If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also drag the drug to the intestines and may limit its effectiveness. This effect […]

6-76 The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3 CDP6-C 6-77 6-78 CDP6-D 6-79 6-80 CDP6-E 6-81 6-82 6-83 CDP6-F (a) Mole balances: dFA/dV = -rD-rU fa 0.06705 0.0018652 0.06705 0.0018652 […]

6-41 ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 0.1 Explicit equations as entered by the user [1] E2 = 10000 0 0.005 0.01 0.015 0.02 400 600 800 1000 1200 1400 1600 T [K] […]

6-1 Solutions for Chapter 6 – Multiple Reactions P6-1 Individualized solution P6-2 (a) Example 6-2 For PFR, 0 0.05 0.1 0.15 0.2 0.25 0 200 400 600 800 ! CB CX CY 0 0.1 0.2 0.3 0.4 0 200 400 […]

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

7-21 P7-10 (a) E+Sk1 “ # “ E•S E•Sk2 “ # “ E+S E•Sk3 “ # “ P+E P+Ek4 “ # “ E•S E # & [ ] [ ] [ ] [ ] k2+k3 # $ & ‘ E […]

7-41 C g dC r dt = V = 5000dm3 , gC rDC = ( ) SSSO rCCD !=! gCSS rYr / =! rg = C SM SC CK C + max µ V v DO = See Polymath program […]

C4 0 0 2.665E-07 1.276E-08 C7 0 0 0.0979179 0.0979179 C3 0 0 0.0012475 0.0012475 C5 0 0 0.0979179 0.0979179 C8 0 0 6.237E-04 6.237E-04 CP5 0 0 0.0979123 0.0979123 CP1 0.1 2.166E-04 0.1 2.166E-04 k5 3.98E+09 3.98E+09 3.98E+09 3.98E+09 […]

7-59 CDP7-C (e) From part (c), we have: 7-60 7-61 CDP7-D No solution will be given. 7-62 CDP7-E CDP7-F No solution will be given. CDP7-G 7-63 7-64 CDP7-H 7-65 CDP7-I No solution will be given. CDP7-J (a) No solution will […]

8-61 P8-16 (a) G(T)=“HRX X= “ k 1+ “ k,k=6.6 #10$3exp E R 1 350 $1 T % & ‘ ( ) * + , – . / 0 R(T)=Cp01+ “ ( ) T#Tc ( ) Cp0=“iCpi =50 # “ […]

8-96 T 550 225.22683 550 225.22683 Cao 0.064 0.064 0.064 0.064 To 550 550 550 550 k 1.949E+05 5.047E-05 1.949E+05 5.047E-05 Kc 1.01E+06 1841.4832 1.01E+06 1841.4832 f 1 1 2.4419826 2.4419826 Ca 0.064 0.0547713 0.064 0.0547713 Cb 0.064 0.0547713 0.064 […]

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 Differential equations 1 d(X)/d(W) = –rA / v0 / CA0 2 d(T)/d(W) = (Uarho * (Ta – T) + rA * 20000) / v0 / CA0 / […]

8-21 8-22 P8-5 2A B C+! A B C Fio lb mole hr ! ” # $ % & ‘ 10 10 0.0 Tio(F) 80 80 – 20, 000 R Btu H lb mol A != , Energy balance with […]

P8-25 Mole balance: C A B A B C dF dF dF r r r dW dW dW = = = Rate Laws: 1 2 3 1 1 2 3 3 B A B C A A A B B […]

Solutions for Chapter 8 – Steady-State Nonisothermal Reactor Design P8-1 Individualized solution P8-2 (a) Example 8-1 For CSTR X= $ k 1+ $ k= $ Ae“E RT 1+Ae“E RT One equation, two unknowns Adiabatic energy balance T=T0“#HRx X CPA In […]

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

9-1 Solutions for Chapter 9 – Unsteady State Nonisothermal Reactor Design P9-1 Individualized solution P9-2 (a) Example 9-1 P9-2 (b) Example 9-2 To show that no explosion occurred without cooling failure. The new T0 of 20 ˚F (497 ˚R) gives […]

[4] vo = 400 [5] UA = 250*150 [6] Ta = 530 [7] rho = 50 [8] To = 530 [9] E = 30000 [10] k = 2*7.08e11*exp(-E/1.987/T) [11] tau = V/vo [12] Ca = Cao/(1+tau*k) [13] Cp = .75 […]