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Solutions Manual for

Essentials of Chemical Reaction Engineering Second Edition

H. Scott Fogler Ame and Catherine Vennema Professor of Chemical Engineering and The Arthur F. Thurnau Professor at the University of Michigan, Ann Arbor, Michigan

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The author and publisher have taken care in the preparation of this work, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein. Visit us on the Web: informit.com Copyright © 2018 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. ISBN-13: 978-0-13-466396-8 ISBN-10: 0-13-466396-9 October 2017

Acknowledgments

The following students participated in the solutions to the end-of-chapter problems in the second edition of Essentials of Chemical Reaction Engineering: Max Nori Brian Vicente Sombudda Ghosh Nihat Gürman Yongzhong Lui Duc Ahn Nguyen Vishal Chaudhary Ravi Kapoor

Manosij Basu Arpit Gupta Sneh Shriyansh Utkarsh Prasad Darshan Shah Anamika Singh Sravya Jangareddy Kaushik Nagaraj

Fan Zhang Keyvan Edrisi Richa Motwani Prafful Bhansali Maithri Venkat Yimeng Lyu Sheng Mark Zheng

Table of Contents Screen Shot of Web Home Page .......................................................................................... i Screen Shot of Interactive Computer Games (ICGs) .......................................................... ii Algorithm to Decode ICGs ................................................................................................. v Screen Shot of Polymath Living Example Problems (LEPs)........................................... viii Sample Course Syllabus...................................................................................................... x Solutions to Chapter 1, Problems P1-1 through P1-8 Solutions to Chapter 2, Problems P2-1 through P2-10 Solutions to Chapter 3, Problems P3-1 through P3-15 Solutions to Chapter 4, Problems P4-1 through P4-10 Solutions to Chapter 5, Problems P5-1 through P5-26 Solutions to Chapter 6, Problems P6-1 through P6-13 Solutions to Chapter 7, Problems P7-1 through P7-11 Solutions to Chapter 8, Problems P8-1 through P8-18 Solutions to Chapter 9, Problems P9-1 through P9-21 Solutions to Chapter 10, Problems P10-1 through P10-23 Solutions to Chapter 11, Problems P11-1 through P11-10 Solutions to Chapter 12, Problems P12-1 through P12-27 Solutions to Chapter 13, Problems P13-1 through P13-9

WEB HOME PAGE

i

INTERACTIVE COMPUTER GAMES (ICGs)

ii

iii

iv

ALGORITHM TO DECODE ICGs **** CONFIDENTIAL **** UNIVERSITY OF MICHIGAN INTERACTIVE COMPUTER MODULES FOR CHEMICAL ENGINEERING CHEMICAL REACTION ENGINEERING MODULES H. Scott Fogler, Project Director M. Nihat Gürmen, Project Manager (2002-2004) Susan Montgomery, Project Manager (1991-1993) Department of Chemical Engineering University of Michigan Ann Arbor, MI 48109-2136 ©2005 Regents of the University of Michigan - All Rights Reserved -

INTERPRETATION OF PERFORMANCE NUMBERS Students should record their Performance Number for each program, along with the name of the program, and turn it in to the instructor. The Performance Number for each program is decoded as described in the following pages. The official site for the distribution of the modules is http://www.engin.umich.edu/~cre/icm

Please report problems to [email protected].

v

**** CONFIDENTIAL **** ICMs with Windows® interface Module

Format

Interpretation

Example

KINETIC CHALLENGE I Score = 1.5 * AB.C z = random numbers

CzBzzAzz

Perf. No. = 75241692 Score = 1.5*(62.7) = 94

% Note: 75% constitutes mastery. KINETIC CHALLENGE II Score = 2.0 * AB.C z = random numbers

CzBzzAzz

Perf. No. = 03776467 Score = 2.0*(47.0) = 94

% Note: 75% constitutes mastery. MURDER MYSTERY A even: Killer and victim correctly identified A odd: Killer and victim not identified z = random numbers

zzAzz

Perf. No. = 50732 Score: No credit

Note: An even number for the middle digit constitutes mastery. TIC TAC TOE Score = 4.0 * AB.C z = random numbers

zDzCzBzA

Perf. No. = 77803581 Score = 4*(15.0) = 60 configuration 7

completed Configurations

8

X X X

X X X

X X X

1

2

3

4 5 6 7

Note: Student receives 20 points for every square answered correctly. A score of 60 is needed for mastery of this module. GREAT RACE zzzCzABz

Score = 6.0 * AB.C z = random numbers

Perf. No. = 77738078 Score = 6*(07.3) = 44

Note: A score of 40 is needed for mastery of this module.

vi

**** CONFIDENTIAL **** ECOLOGY AzBCzaaD

z = random numbers a = random characters

A gives info on r^2 value of the student’s linearized plot A=Y if r^2 >= 0.9 A=A if 0.9 > r^2 >= 0.8 A=X if 0.8 > r^2 >= 0.7 A=F if 0.7 > r^2 A=Q if Wetland Analysis/Simulator portion has not been completed B gives info on alpha B=1 to 4 => student's alpha < (simulator's alpha  0.5) B=5 to 9 => student's alpha > (simulator's alpha  0.5) B=X if Wetland Analysis/Simulator portion has not been completed C indicates number of data points deactivated during analysis C=number of deactivated data points if at least 1 point has been deactivated C=a randomly generated letter from A to Y if 0 points deactivated C=Z if Wetland Analysis/Simulator portion has not been completed D gives info on solution method used by student D=1 if polynomial regression was used D=2 if differential formulas were used D=3 if graphical differentiation was used D=4 to 9 if Wetland Analysis/Simulator portion has not been completed Perf No. = A7213DF2 1) A => 0.9 > r^2 >= 0.8 2) 2 => student’s alpha < (simulator’s alpha  0.5) 3) 1 => one data point was deactivated 4) 2 => differential formulas were used STAGING zCBzAFzED

z = random numbers

Final conversion = 2*AB.C Final flow rate = 2*DE.F

Perf. No. = 2125482913 conversion = 2*42.1 = 84.2 flow rate = 2*31.2 = 62.4

Please make a pass/fail criterion based on these values.

vii

**** CONFIDENTIAL **** ICMs with Dos® interface Module

Format

Interpretation

Example

A=2,3,5,7: interaction done B=2,3,5,7: intro done

Perf. No. = 8027435 A: Worked on

C=2,3,5,7: review done D denotes how much they did in the interaction:

B: Looked at intro C: Looked at review D: found parameter

HETCAT zzABzCD interaction

values, didn’t find mechanism D 85 %

Note: Student told they have achieved mastery if their score is greater than 85% HEATFX2 zzzAzz

A even: completed interaction z = random numbers

Perf. No. = 407582 Interaction not

completed Note: Performance number given only if student goes through the interaction portion of the module.

viii

LIVING EXAMPLE PROBLEMS (LEPs)

ix

SAMPLE COURSE SYLLABUS ChE 344: CHEMICAL REACTION ENGINEERING Fundamentals of chemical reaction engineering. Rate laws, kinetics, and mechanisms of homogeneous and heterogeneous reactions. Analysis of rate data, multiple reactions, heat effects, bioreactors. Design of industrial reactors. Prerequisite: ChE 330, ChE 342 Fall 2015 Lectures: M,W 8:40 (Sharp) to 10:30 (not so sharp) – Room: 1013 Dow Instructor: Professor H. Scott Fogler 3168 DOW, 763-1361, [email protected] Office Hours: M,W 10:30a to 11:30a Course assistants include: Instructional aids, tutor, proctors, and graders Text Required Elements of Chemical Reaction Engineering, 5th edition, H. Scott Fogler Web site: www.umich.edu/~elements/5e Recommended Reading List • Problem Solving in Chemical and Biochemical Engineering with POLYMATH, Excel, and MATLAB, 2nd Edition 2008, Cutlip & Shacham • The Elements of Style, Strunk and White • Strategies for Creative Problem Solving, 3rd Edition 2014, Fogler, LeBlanc & Rizzo (for OEP’s) Schedule *Note - all ICGs (Interactive Computer Games) are Individual* 1) Wednesday, September 9 Topic: Lecture 1 – Chapter 1, Introduction, POLYMATH, Mole balances Read: Preface, Prerequisites, Appendix B In-Class Problem: No In-Class Problem 2) Monday, September 14 Topic: Lecture 2 – Chapter 2, Design equations, Levenspiel plots, Reactor staging Read: Chapter 1, P1-9A, Appendix A, from the Web Chapter 2, Sections 2.1, 2.2, and 2.3 Hand In: Problem Set 1: P1-1A, P1-6B In-Class Problem: 1 Study Problems: P1-8A 3) Wednesday, September 16 Topic: Lecture 3 – Chapter 3, Rate laws Read: Chapter 2, Chapter 3 Hand In: Problem Set 2: Define terms in the Arrhenius Equation, P2-2A, Intro to Learncheme In-Class Problem: 2 (Hint: Viewing the University of Alabama YouTube video “The Black Widow” noted in Problem P3-8B may help you with today's in class problem)

x

Study Problems:

P2-7A

4) Monday, September 21 Topic: Lecture 4 – Chapter 4, Stoichiometry Batch Systems Read: Chapter 4 Section 4.1 Hand In: Problem Set 3: Define θi, θA, θB, and δ, P2-10B, P3-5A, P3-8B, P3-11B, P3-13A In-Class Problem: 3 - Bring i>clickers (tentative) - Test Run of System in 2166 Dow Study Problems: P3-14A 5) Wednesday, September 23 Topic: Lecture 5 – Chapter 4, Stoichiometry Flow Systems Read: Chapter 4, Section 4.1 Hand In: Problem Set 4: Define ε, FT0, CT0, P4-2A. In-Class Problem: 4 Study Problems: P4-1A parts (c) and (d) 6) Monday, September 28 Topic: Lecture 6 – Chapter 5, Isothermal reactor design Read: Chapter 5, Chapter 5 Summary Notes on the Web site Hand In: Problem Set 5: P4-1A (a) and (b) only, P4-3A, P4-4B, P4-5B. In-Class Problem: 5 Study Problems: P4-10C 7) Wednesday, September 30 Topic: Lecture 7 – Chapter 5, California Registration Exam Problem Hand In: Problem Set 6: What are you asked to find P5-18B? What is the Ergun Equation? P5-2A. In-Class Problem: 6 Study Problems: P5-1B (a) and (b) 8) Monday, October 5 Topic: Lecture 8 – Chapter 5, Pressure drop Read: Chapter 5, Sections 5.4 and 5.5 Hand In: Problem Set 7: P5-3A, P5-4B, P5-5A, P5-8B, P5-13B omit parts (j) and (k), P5-16B (a). In-Class Problem: 7 – Bring Laptops Study Problems: P5-9A, P5-10B (a). 9) Wednesday, October 7 Topic: Lecture 9 – Chapter 6, Membrane Reactors Read: Chapter 6 Hand In: Problem Set 8: P5-13B part (j) and (k), P5-22A. In-Class Problem: 8 – Bring Laptops Study Problems: P5-21B 10) Monday, October 12 Topic: Lecture 10 – Chapter 6, Semibatch Reactors Read: Chapter 6 Hand In: Problem Set 9: P5-1A (a), P5-11B, P6-4B delete part (c), P6-5B. In-Class Problem: 9 – Bring Laptops to carry out Polymath ODE Solver Study Problems: P6-7B

xi

11) Wednesday, October 14 Topic: Lecture 11 – Chapter 7, Analysis of Rate Data/Chapter 9, Pseudo Steady State Read: Chapter 7, Chapter 9, Section 9.1 and the cobra web module Hand In: Problem Set 10: LEP for Example 6-1, P6-2B, P6-11B omit part (c) In-Class Problem: 10 – Bring Laptops to carry out Polymath Regression Study Problems: P7-6B. 12) Monday, October 19 Topic: No Classes – Fall Study Break 13) Wednesday, October 21 Topic: Lecture 12 – Chapter 8, Multiple Reactions Read: Chapter 8, Sections 8.1, 8.2, 8.3 and 8.4; Hand In: Problem Set 11: P7-7A, P7-8A. In-Class Problem: 11 Study Problems P7-10A 14) Monday, October 26 Topic: Lecture 13 – EXAM I – Covers Chapters 1 through 7 Closed book, web, notes, in-class problems and home problems. 15) Wednesday, October 28 Topic: Lecture 14 – Chapter 8: Multiple Reactions Read: Chapter 8, Sections 8.5, 8.6, 8.7 and 8.8 In-Class Problem: 12 – Bring Laptops Hand In: Problem Set 12: P8-1A (a) part (1) only, P8-1A (b), P8-1A (c) part (1) only, P8-2B, P8-6B, P8-7C (a), (b) and (c) Study Problems P8-10B 16) Monday, November 2 Topic: Lecture 15 – Derivation of Energy Balance Read: Chapter 11, Sections 11.1, 11.2 and 11.3 Hand In: Problem Set 13: P8-12B. Comprehensive Problem In-Class Problem: 13 – Bring Laptops Study Problems: P8-16A 17) Wednesday, November 4 Topic: Lecture 16 – Chapter 11: Adiabatic Equilibrium Conversion and Reactor Staging Read: Finish Reading Chapter 11, Equilibrium conversion appendix In-Class Problem: 14 Study Problems P11-6B 18) Monday, November 9 Topic: Lecture 17 – Heat Exchange, Adiabatic Reactors ICPs Read: Chapter 12 Sections 12.1 through 12.2 Hand In: Problem Set 14: P11-1A (b), P11-3B, P11-4A. In-Class Problem: 15 Study Problem: P12-6A

xii

19) Wednesday, November 11 Topic: Lecture 18 – Trends in Conversion and Temperature Profiles Applications of the Energy Balance to PFRs Read: Chapter 12, Section 12.3 and 12.4 Hand In: Problem Set 15: P12-3B LEP In-Class Problem: 16 – Bring Laptops 20) Monday, November 16 Topic: Lecture 19 – Multiple Reactions with Heat Effects This topic is a major goal of this course, to carry out calculations for non isothermal multiple reactions. Applications of the Energy Balance to PFRs Hand In: Problem Set 16: P12-4A (a) and (b), P12-14B, P12-17B, P12-21B. In-Class Problem: 17 – Bring Laptops Study Problem: P12-19B, i>clicker questions handed out in class 21) Wednesday, November 18 Topic: Lecture 20 – CSTR and Review for Exam II 22) Monday, November 23 Topic: Lecture 21 – EXAM II – Chapters 8, 11 and 12. Book and notecard are the only materials allowed Hand In: Problem Set 17: P12-26C 23) Wednesday, November 25 Topic: Lecture 22 – Multiple Steady States (MSS) Multiple Reactions with Heat Effects Read: Sections 12.6 and 12.7 In-Class Problem: 18 – Bring a Ruler/Straight Edge Study Problems: P13-4B 24) Monday, November 30 Topic: Lecture 23 – Safety (CSI) Read: Chapter 13 Hand In: Problem Set 18: P13-1B (b) and (f), P13-8B In-Class Problem: 19 – Bring Laptops Study Problems: P13-4B 25) Wednesday, December 2 Topic: Lecture 24 –Catalysis Reactor Safety Read: Chapter 13, Sections 13.1 through 13.3, and 13.5 Hand In: Problem Set 19: P10-2A part (d), P10-4B In-Class Problem: 20 Study Problems: P12-16B 26) Monday, December 7 Topic: Lecture 25 – Catalysis Read: Chapter 10, Sections 10.1 through 10.2.2 Hand In: Problem Set 20: P10-3A, P10-8B, P10-10B In-Class Problem: 21 Study Problems: P10-7B, P10-9B

xiii

27) Wednesday, December 9 Topic: Lecture 23 – PSSH and Enzyme Read Chapter 9 Hand In: Problem Set 21: P9-4A, P9-5B, P9-9B, P9-14B P9-19A In-Class Problem: 22 Study Problems: P9-12B, P9-16B, P9-21A 28) FINAL EXAM

xiv

Solutions for Chapter 1 – Mole Balances P1-1 (a) Example 1-3 (i) CA decreases and CB increases with an increase in k, and a decrease in v0 for the same volume. (ii) CA decreases and CB increases with an increase in k and Ke, and a decrease in v0 for the same volume. (iii) Individualized solution (iv) Refer to the polymath report below POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

10.

2.849321

10.

2.849321

2 Cb

0

0

7.150679

7.150679

3 k

0.23

0.23

0.23

0.23

4 Ke

3.

3.

3.

3.

5 ra

-2.3

-2.3

-0.1071251

-0.1071251

6 rb

2.3

0.1071251

2.3

0.1071251

7 V

0

0

100.

100.

8 v0

10.

10.

10.

10.

Differential equations 1 d(Ca)/d(V) = ra / v0 2 d(Cb)/d(V) = rb / v0 Explicit equations 1 k = 0.23 2 Ke = 3 3 ra = -k * (Ca-Cb/Ke) 4 rb = -ra 5 v0 = 10

P1-2 Given

A = 2*1010 ft 2

TSTP = 491.69R

H = 2000 ft

V = 4*1013 ft 3

T = 534.7 ° R

PO = 1atm

R = 0.7302

atm ft 3 lbmol R

yA = 0.02

C S = 2.04*10−10

lbmol ft

3

C = 4*105 cars

FS = CO in Santa Ana winds FA = CO emission from autos v A = 3000 1-1

ft 3 per car at STP hr

P1-2 (a) Total number of lb moles gas in the system: PV N= 0 RT N =

1atm×(4 ×1013 ft 3 ) = 1.025 x 1011 lb mol " % 3 $ 0.73 atm. ft ' ×534.69R $ lbmol.R '& #

P1-2 (b) Molar flowrate of CO into L.A. Basin by cars.

FA = y AFT = y A ⋅v A CT FT =

•%no.%of%cars STP



3000 ft 3 1lbmol × ×400000 cars hr car 359 ft 3

(See appendix B)

FA = 6.685 x 104 lb mol/hr P1-2 (c) Wind speed through corridor is U = 15mph W = 20 miles The volumetric flowrate in the corridor is vO = U.W.H = (15x5280)(20x5280)(2000) ft3/hr = 1.673 x 1013 ft3/hr P1-2 (d) Molar flowrate of CO into basin from Sant Ana wind. FS := v0 ⋅C S = 1.673 x 1013 ft3/hr ×2.04 ×10−10 lbmol/ft3 = 3.412 x 103lbmol/hr P1-2 (e) Rate of emission of CO by cars + Rate of CO in Wind - Rate of removal of CO =

dC FA + FS − voCco = V co (V=constant, Nco = CcoV ) dt



P1-2 (f) t = 0 , Cco = CcoO t

∫ dt 0

= V

Cco

∫

CcoO

dCco

FA + FS −voCco



V "$ FA + FS −voCcoO %' t = ln vo $# FA + FS −voCco '&

1-2

dNCO dt



P1-2 (g) Time for concentration to reach 8 ppm.

CCO0 = 2.04 ×10−8

lbmol ft 3

, CCO =

2.04 lbmol ×10−8 4 ft 3

From (f), % V " F + F −v .C t = ln$$ A S O CO0 '' vo # FA + FS −vO .CCO & " $ 6.7×104 lbmol + 3.4 ×103 lbmol −1.673×1013 3 $ hr hr 4 ft = ln$ ft 3 $ lbmol lbmol 1.673×1013 6.7×104 + 3.4 ×103 −1.673×1013 $ hr # hr hr

% ft 3 lbmol ' ×2.04 ×10−8 hr ft 3 ' ' ft 3 −8 lbmol ' × 0.51×10 ' hr ft 3 &

t = 6.92 hr P1-2 (h) (1)

to = 0 tf = 72 hrs

Cco = 2.00E-10 lbmol/ft3 a = 3.50E+04 lbmol/hr v o = 1.67E+12 ft3 /hr b = 3.00E+04 lbmol/hr Fs = 341.23 lbmol/hr V = 4.0E+13 ft3

! t$ dC a + bsin# π & + Fs − voCco = V co dt " 6% Now solving this equation using POLYMATH we get plot between Cco vs. t See Polymath program P1-4-h-1.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value T 0 0 C 2.0E-10 2.0E-10 v0 1.67E+12 1.67E+12 A 3.5E+04 3.5E+04 B 3.0E+04 3.0E+04 F 341.23 341.23 V 4.0E+13 4.0E+13

maximal value 72 2.134E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+13



ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)+F-v0*C)/V Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] F = 341.23 [5] V = 4*10^13

1-3

final value 72 1.877E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+13

P1-2 (h) Continued



! t$ dC (2) tf = 48 hrs Fs = 0 a + bsin# π & − voCco = V co dt " 6% Now solving this equation using POLYMATH we get plot between Cco vs t See Polymath program P1-4-h-2.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value T 0 0 C 2.0E-10 2.0E-10 v0 1.67E+12 1.67E+12 A 3.5E+04 3.5E+04 B 3.0E+04 3.0E+04 F 341.23 341.23 V 4.0E+13 4.0E+13

maximal value 72 2.134E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+13



ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V

Explicit equations as entered by the user [1] v0 = 1.67*10^12 [2] a = 35000 [3] b = 30000 [4] V = 4*10^13

1-4

final value 72 1.877E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+13

P1-2 (h) Continued (3) Changing a ! Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of the sine function by adding to the baseline. Changing b ! The amplitude of ripples is directly proportional to ‘b’. As b decreases amplitude decreases and graph becomes smooth. Changing v0 ! As the value of v0 is increased the graph changes to a “shifted sin-curve”. And as v0 is decreased graph changes to a smooth increasing curve. P1-3 (a) Initial number of rabbits, x(0) = 500 Initial number of foxes, y(0) = 200 Number of days = 500

dx = k x − k2 xy …………………………….(1) dt 1 dy = k3 xy − k4 y ……………………………..(2) dt Given,

k1 = 0.02day −1 k2 = 0.00004 / (day × foxes) k3 = 0.0004 / (day × rabbits)



k4 = 0.04day −1 See Polymath program P1-3-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value T 0 0 500 X 500 2.9626929 519.40024 Y 200 1.1285722 4099.517 k1 0.02 0.02 0.02 k2 4.0E-05 4.0E-05 4.0E-05 k3 4.0E-04 4.0E-04 4.0E-04 k4 0.04 0.04 0.04 ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(t) = (k1*x)-(k2*x*y) [2] d(y)/d(t) = (k3*x*y)-(k4*y) Explicit equations as entered by the user [1] k1 = 0.02 [2] k2 = 0.00004 [3] k3 = 0.0004 [4] k4 = 0.04

1-5

final value 500 4.2199691 117.62928 0.02 4.0E-05 4.0E-04 0.04

P1-3 (a) Continued

When, tfinal = 800 and k3 = 0.00004 /(day × rabbits)

Plotting rabbits vs. foxes



1-6

P1-3 (b)

P1-3 (c) We would have to change k2 and k4 for the plot to become a circle from an oval. P1-3 (d)



1-7

P1-3 (d) Continued



P1-4 Individualized solution P1-5



P1-6 (a) – rA = k with k = 0.05 mol/h dm3 CSTR: The general equation is

F −F V = A0 A −rA

Here CA = 0.01CA0 , v0 = 10 dm3/min, FA = 5.0 mol/hr Also we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3 Substituting the values in the above equation we get,

C v −C v (0.5)10 − 0.01(0.5)10 V = A0 0 A 0 = k 0.05 " V = 99 dm3 PFR: The general equation is

dFA

= rA = k , Now FA = CAv0 and FA0 = CA0v0 =>

dV Integrating the above equation we get

1-8

dC Av0 dV

= −k

P1-6 (a) Continued

CA

v0 k

∫

C A0

V

dC A =

v

∫ dV => V = k0 (C A0 −C A ) 0

Hence V = 99 dm3 Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration. P1-6 (b) - rA = kCA with k = 0.0001 s-1 CSTR: We have already derived that

C v −C v v C (1− 0.01) V = A0 0 A 0 = 0 A0 −rA kC A



k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1



" V =

(10dm3 / hr)(0.5mol / dm3 )(0.99) 3

−1

(0.36hr )(0.01*0.5mol / dm )

=> V = 2750 dm3

PFR: From above we already know that for a PFR

dC Av0



v0

dV Integrating



k

CA

∫

C A0

= rA = −kC A

dC A CA

V

= − dV

∫ 0

v0

C ln A0 =V k CA

Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1 Substituting the values in above equation we get V = 127.9 dm3 P1-6 (c) - rA = kCA2 with k = 300 dm3/mol.hr CSTR:

v C (1− 0.01) C v −C v V = A0 0 A 0 = 0 A0 −rA kC 2 A

Substituting all the values we get

V=

(10dm3 / hr)(0.5mol / dm3 )(0.99) 3

32

(300dm / mol.hr)(0.01*0.5mol / dm )

=> V = 660 dm3

PFR:

dC Av0 dV

= rA = −kC 2A 1-9

P1-6 (c) Continued Integrating

v0



k

CA

∫

C A0

=> V =



V v 1 1 ) =V = − dV => 0 ( − 2 k C A C A0 CA 0

dC A



∫

10dm3 / hr

1 1 − ) = 6.6 dm3 3 0.01C C 300dm / mol.hr A0 A0 (

P1-6 (d) CA = 0.001CA0

t=

NA0

∫N

A

dN −rAV

Constant Volume V=V0

t=

C A0 dC A

∫C

A

−rA



Zero order:

.999C Ao 1 t = "#C A0 − 0.001C A0 $% = = 9.99h k 0.05 First order: ! 1 $ 1 !C $ 1 t = ln## A0 && = ln# & = 69078s = 19.19h k " C A % 0.0001 " .001 % Second order: 1" 1 1 % 1 " 1 1 % '= t= $ − − $ ' = 6.66h k $# C A C A0 '& 300 # 0.5⋅0.001 0.5 & P1-7 Enrico Fermi Problem P1-7(a) Population of Chicago = 4,000,000 Size of Households = 4 Number of Households = 1,000,000 Fraction of Households that own a piano = 1/5 Number of Pianos = 200,000 Number of Tunes/year per Piano = 1 Number of Tunes Needed Per Year = 200,000 Tunes per day = 2 Tunes per year per tuner =

250!days 2 × = 500/yr/tuner yr day

200,000!tunes 1 × = 400 Tuners yr 500!tunes / yr / tuner 1-10

P1-7(b) Assume that each student eats 2 slices of pizza per week. Also, assume that it is a 14” pizza, with 8 pieces. Hence, the area of 1 slice of pizza = 19.242 inch2 = 0.012414 m2 Thus, a population of 20000, over a span of 4 months, eats 20000 * 2 slices * 4 months * 4 weeks/month = 640000 slices of pizza, with a total area of 640000 * 0.012414 m2 = 7945 m2 of pizza in the fall semester. P1-7(c) Assume you drink 1L/day Assume you live 75 years*365days/year = 27375 days 1L/day*27375 days = 27375 L drank in life Bathtub dimensions: 1m*0.7m*0.5m = 0.35m3 = 350L/tub Bathtubs drunk = 27375L*1tub/350L = 78 tubs P1-7(d) Jean Valjean, Les Misérables. P1-8 Mole Balance: F − "F V"=" A0 A −rA Rate Law :

−rA = kC 2A Combine: F − "F V"=" A0 A kC 2A

FA0 = v0C A = 3 FA = v0C A = 3

dm3 2molA 6molA . = s s dm3

dm3 0.1molA 0.3molA . = s s dm3

mol s V"=" = 1900dm3 3 dm mol 2 (0.03 )(0.1 ) mol.s dm3 The incorrect part is in step 6, where the initial concentration has been used instead of the exit concentration. (6 − 0.3)

1-11

Solutions for Chapter 2 – Conversion and Reactor Sizing P2-1 (a) Example 2-1 through 2-3 If flow rate FAO is cut in half. v1 = v/2 , F1= FAO/2 and CAO will remain same. Therefore, volume of CSTR in example 2-1, If the flow rate is doubled, F2 = 2FAO and CAO will remain same, Volume of CSTR in example 2-1, V2 = F2X/-rA = 12.8 m3 P2-1 (b) No solution will be given P2-1 (c) No solution will be given P2-1 (d) Example 2-4



Now, FAO = 0.4/2 = 0.2 mol/s, Table: Divide each term

in Table 2-3 by 2.

X [FAO/-rA](m3) Reactor 1 V1 = 0.82m3 V = (FAO/-rA)X

0 0.445



!F $ 0.82 = ## A0 && X1 " −rA %X

( )

1

0.1 0.545

0.2 0.665

0.4 1.025



Reactor 2 V2 = 3.2 m3



"F % 3.2 = $$ A0 '' X2 # −rA &X

0.6 1.77

( )

2

By trial and error we get: X1 = 0.546 and X2 = 0.8 Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673 2-1

0.7 2.53

0.8 4

P2-1 (e) Example 2-5 (1) For first CSTR, At X=0.2,

FA0 = 0.94 m3 ; V1 = 0.94&m3 0.2 = 0.188&m3 − rA

(

)(

)

From previous example; V1 (volume of first CSTR) = 0.188 m3 Also the next reactor is PFR, Its volume is calculated as follows

For next CSTR, X3 = 0.65,



, V3 =



(2) Now the sequence of the reactors remain unchanged. But all reactors have same volume. First CSTR remains unchanged Vcstr = .1 = (FA0/-rA )*X1 ! X1 = .088



Now For PFR:

By estimation using the Levenspiel plot X2 = .183 For CSTR, VCSTR2 = => X3 = .316





(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR.



2-2

P2-1 (e) Example 2-5 Continued Conversion X1 = 0.20 X2 = 0.60 X3 = 0.65



Original Reactor Volumes V1 = 0.188 (CSTR) V2 = 0.38 (PFR) V3 = 0.10 (CSTR)

Worst Arrangement V1 = 0.23 (PFR) V2 = 0.53 (CSTR) V3 = 0.10 (CSTR)

For PFR, X1 = 0.2

Using trapezoidal rule, XO = 0, X1 = 0.2

= 0.2(1.28+0.94)/2 = 0.222 m3 For CSTR,

F At X2 = 0.6 A0 = 1.32 m3 , nd

−rA

V2 =

For 2 CSTR,

F At X3 = 0.65, A0 = 2 m3 , −rA



V3 =

FA0 −rA FA0 −rA

(X2 − X1 ) = 1.32(0.6 − 0.2) = 0.53-m3 (X3 − X2 ) = 2)m3 (0.65− 0.6) = 0.1)m3

P2-1 (f) No solution will be given. P2-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. P2-3





X 0 0.1 0.2 0.4 0.6 0.7 0.8 3 FAO/-rA (m ) 0.89 1.08 1.33 2.05 3.54 5.06 8.0 V = 1 m3 2-3

P2-3 (a) Two CSTRs in series For first CSTR, V = (FAo/-rAX1) X => X1 = 0.435 For second CSTR, V = (FAo/-rAX2) (X2 – X1) => X2 = 0.66 P2-3 (b) Two PFRs in series

By extrapolating and solving, we get X1 = 0.565 X2 = 0.775 P2-3 (c) Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 V = (0.5FAO/-rAX1) X1 Solving we get, Xout = 0.585 P2-3 (d) Two PFRs in parallel with the feed equally divided between the two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1 By extrapolating and solving as part (b), we get Xout = 0.775 P2-3 (e) A CSTR and a PFR are in parallel with flow equally divided Since the flow is divided equally between the two reactors, the overall conversion is the average of the CSTR conversion (part C) and the PFR conversion (part D) Xo = (0.585 + 0.775) / 2 = 0.68 P2-3 (f) A PFR followed by a CSTR, XPFR = 0.565 (using part (b)) V = (FAo/-rA-XCSTR) (XCSTR – XPFR) Solving we get, XCSTR = 0.732 P2-3 (g) A CSTR followed by a PFR, XCSTR = 0.435 (using part(a))

By extrapolating and solving, we get XPFR = 0.71 2-4

P2-3 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs, For PFR, XPFR = 0.565 (using part (b)) CSTR1: V = (FAo/-rA-XCSTR) (XCSTR – XPFR) = 0.5 m3 XCSTR = 0.673 CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 – XCSTR1) = 0.5 m3 XCSTR2 = 0.75 P2-4 Exothermic reaction: A → B + C X 0 0.20 0.40 0.45 0.50 0.60 0.80 0.90

r(mol/dm3.min) 1 1.67 5 5 5 5 1.25 0.91

1/-r(dm3.min/mol) 1 0.6 0.2 0.2 0.2 0.2 0.8 1.1

P2-4 (a) To solve this problem, first plot 1/–rA vs. X from the chart above. Second, use mole balance as given below. CSTR: Mole balance:

=>

=>VCSTR = 24 dm3 PFR:

Mole balance: = 300(area under the curve) VPFR = 72 dm3 P2-4 (b) For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is constant over this conversion range.



2-5

P2-4 (b) Continued



P2-4 (c) VCSTR = 105 dm3 Mole balance:



Use trial and error to find maximum conversion. At X = 0.70, 1/–rA = 0.5, and X/–rA = 0.35 dm3.min/mol Maximum conversion = 0.70 P2-4 (d) From part (a) we know that X1 = 0.40. Use trial and error to find X2. Mole balance:



Rearranging, we get



At X2 = 0.64



, Conversion = 0.64 P2-4 (e) From part (a), we know that X1 = 0.40. Use trial and error to find X2. Mole balance:



At X2 = 0.908, V = 300 x (area under the curve) => V = 300(0.24) = 72dm3 Conversion = 0.908. 2-6

P2-4 (f) See Polymath program P2-4-f.pol.

P2-5 We must first find a CSTR up to X = 0.2 to minimize the volume, and hence the cost of reactor. We can either use a CSTR or a PFR for X = 0.2 to X = 0.6 since the rate is independent of X in that range. A PFR for X > 0.6 is preferable since it will reduce the volume required and the cost. Let’s calculate the volumes for the different cases. i) X = 0 to X = 0.2 Volume for CSTR = 20*0.2 = 4 dm3 Volume for PFR = 0.5*(50+20)*0.2 = 7 dm3 ii) X = 0.2 to X = 0.6 Volume for CSTR = Volume for PFR = 0.4*20 = 8 dm3 iii) X > 0.6 Volume for PFR < Volume for CSTR; for the same conversion. Hence a PFR would minimize the cost. First, we use the 4 dm3 CSTR worth $2000. Conversion at this stage = 0.2 After this, we have $8000, and we see that the PFR minimizes the cost (So, we can have more conversion with the same money) Hence, we can use a 12 dm3 PFR followed by a 4 dm3 PFR, totally worth $8000, which will give the highest conversion. The first 8 dm3 of PFR would give conversion up to 0.6, and the rest (16 – 8) dm3 = 8 dm3 would give the conversion of:

Therefore, X = 0.8 Final conversion = 0.8 P2-6 (a) Individualized Solution



2-7

P2-6 (b) 1) In order to find the age of the baby hippo, we need to know the volume of the stomach. The metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of what the mother eats then Fao (baby) = ½ Fao (mother). The Levenspiel Plot is shown:

Autocatalytic Reaction

Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the baby’s stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.

P2-6 (b) 2) If Vmax and mao are both one half of the mother’s then



and since

then



will be identical for both the baby and mother. Assuming that like the stomach the intestine volume is proportional to age then the volume of the intestine would be 0.75 m3 and the final conversion would be 0.40 2-8

P2-6 (c) Vstomach = 0.2 m3 From the web module we see that if a polynomial is fit to the autocatalytic reaction we get: = 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499 And since Vstomach =



X, solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3 Xstomach = .067. For the intestine, the Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178. Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.



P2-6 (d) PFR→ CSTR PFR: Outlet conversion of PFR = 0.111

CSTR: We must solve V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499) X=0.42 Since the hippo gets a conversion over 30% it will survive. 2-9

P2-7 Irreversible gas phase reaction See Polymath program P2-7.pol. 2A + B → 2C P2-7 (a) PFR volume necessary to achieve 50% conversion Mole Balance

Volume = Geometric area under the curve of (FA0/-rA) vs X)

V = 150000 m3 P2-7 (b) CSTR Volume to achieve 50% conversion Mole Balance







3

V = 50000m P2-7 (c) Volume of second CSTR added in series to achieve 80% conversion



V2 = 150000m3 P2-7 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion

VPFR = 90000m

3







2-10

P2-7 (e) For CSTR, V = 60000 m3 (CSTR) Mole Balance





X = 0.463 For PFR, V = 60000 m3 (PFR) Mole balance





X = 0.134



P2-7 (f) Real rates would not give that shape. The reactor volumes are absurdly large. P2-8 Problem 2-8 involves estimating the volume of three reactors from a picture. The door on the side of the building was used as a reference. It was assumed to be 8 ft high. The following estimates were made: CSTR h = 56ft d = 9 ft 2 2 V = πr h = π(4.5 ft) (56 ft) = 3562 ft3 = 100,865 L PFR Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(11) = 3036 ft D = 1 ft V = πr2h = π(0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L Answers will vary slightly for each individual. P2-9 No solution necessary.

2-11



P2-10 (a) The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in series and containing equal amounts of catalyst can be calculated from the figure below.

The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR. This figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR. See Polymath program P2-10.pol. P2-10 (b) Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining the area of the shaded region in the figure below.

The area of the rectangle is approximately 23.2 kg of catalyst.

2-12



P2-10 (c) The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of the shaded rectangle shown in the figure below.

The area of the rectangle is approximately 7.6 kg of catalyst. P2-10 (d) The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the shaded region in the figure below.

The necessary catalyst weight is approximately 22 kg.

2-13



P2-10 (e) The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating the area of the shaded region in the graph below.



The necessary catalyst weight is approximately 13 kg. P2-10 (f)





2-14

Solutions for Chapter 3 – Rate Laws P3-1 (a) (i) Individualized solution (ii) 2550 K (iii) Individualized solution P3-1 (b) (i) The equilibrium concentration changes, but the equilibrium conversion remains the same in all the three cases (50%). The time taken to attain equilibrium remains the same in all the three cases.

(ii) The trajectories remain similar, but the time taken to attain equilibrium changes, as the rate constants are lower.

(iii) The forward/reverse reaction goes to completion.

(iv) This is the nature of a stochastic simulation. The number of molecules in the simulation are very less, as compared to the large number in the deterministic model (number of molecules are of the order of the Avogadro number). The fluctuations reduce as you increase the number of molecules, and the stochastic model is identical to a deterministic model when there are infinite molecules.

(v) The fluctuations in the trajectories reduce.

(vi) No, equilibrium only means that the forward reaction rate is equal to the reverse reaction rate. It does not mean that they stop occurring. This can be seen from the fact that there is always a small fluctuation about the equilibrium concentration. P3-1 (c)

3-1

P3-1 (c) Continued ln 𝑘 = ln 𝐴 −

𝐸 1 𝑅 𝑇

From the graph of ln k vs 1/T above, we get: 𝐸 = −10889 𝐾 𝑅

ln 𝐴 = 27.577, Thus, 𝐴 = 9.474 ∗ 10!! 𝑠 !! ,

𝐸 = 90.531 𝑘𝐽/𝑚𝑜𝑙 10889 !! 𝑘 = 9.474 ∗ 10!! exp − 𝑠 𝑇

P3-1 (d) Example 3-1 For, E = 60kJ/mol







For, E = 240kJ/mol



T (K)

k (1/sec)

310

1/T





T (K)

k (1/sec)

1023100 0.003226 13.83918



310

4.78E-25 0.003226 -56.0003

315

1480488 0.003175

14.2087



315

2.1E-24

320

2117757 0.003125 14.56667



320

8.77E-24 0.003125 -53.0903

325

2996152 0.003077 14.91363



325

3.51E-23 0.003077 -51.7025

330

4194548

15.25008



330

1.35E-22

335

5813595 0.002985 15.57648



335

4.98E-22 0.002985 -49.0511

0.00303

ln(k)



1/T

ln(k)

0.003175 -54.5222

0.00303

-50.3567





3-2

P3-1 (e) No solution will be given P3-1 (f)

1 1 A+ B→ C 2 2 Rate law:

and



𝟏 𝒅𝒎𝟑 𝒌𝑪 = 𝒌𝑩 =12.5 𝒔 𝒎𝒐𝒍

𝟐



P3-2 (a) Refer to Fig 3-4 The fraction of molecular collisions having energies less than or equal to 35 Kcal is given by the area under the curve, f(E,T)dE from EA = 0 to 35 Kcal. P3-2(b) The fraction of molecular collisions having energies between 10 and 20 Kcal is given by the area under the curve f(E,T) from EA = 10 to 20 Kcal. P3-2 (c) The fraction of molecular collisions having energies greater than the activation energy EA= 25 Kcal is given by the area under the curve f(E,T) from EA =25 to 50 Kcal. P3-3 (a)

(a)



(b)

P3-4 No solution will be given 3-3



P3-5 (a) The fraction of collisions having energy between E = 3 and E = 5 is the area under the graph between those two boundaries = 2*0.5*(0.1875 + 0.25) = 0.4375 P3-5 (b)

P3-5 (c)

P3-5 (d) Since f(E,T) = 0 for E > 8 kcal, the fraction with energies greater than 8 kcal = 0 P3-6 (a) Note: This problem can have many solutions as data fitting can be done in many ways. Using Arrhenius Equation For Fire flies: T(in K) 1/T Flashes/min ln(flashes/min) 294 0.003401 9 2.197 298 0.003356 12.16 2.498 303 0.003300 16.2 2.785

Plotting ln (flashes/min) vs. 1/T, We get a straight line.

For Crickets: T(in K) 1/T x103 287.2 293.3 300

3.482 3.409 3.333

chirps/min ln(chirps/min) 80 126 200

4.382 4.836 5.298



Plotting ln (chirps/min) Vs 1/T, We get a straight line. ➔Both, Fireflies and Crickets data follow the Arrhenius Model. ln y = A + B/T , and have the similar activation energy.

3-4

P3-6 (b) For Honeybee: T(in K) 1/T x103 298 3.356 303 3.300 308 3.247

V(cm/s) 0.7 1.8 3

ln(V) -0.357 0.588 1.098



Plotting ln (V) vs. 1/T, almost straight line. ln (V) = 44.6 – 1.33E4/T

At T = 40oC (313K) V = 6.4cm/s At T = -5oC (268K) V = 0.005cm/s (But bee would not be alive at this temperature) P3-6 (c) For Ants: T(in K) 1/T x103 V(cm/s) ln(V) 283 3.53 0.5 -0.69 293 303

3.41 3.30

2 3.4

0.69 1.22

311

3.21

6.5

1.87



Plotting ln (V) vs. 1/T, We get almost a straight line.

So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same.

Insect Cricket Firefly Ant Honeybee

Activation Energy 52150 54800 95570 141800

P3-6 (d) There is a limit to temperature for which data for any one of the insect can be extrapolate. Data which would be helpful is the maximum and the minimum temperature that these insects can endure before death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless. P3-6 (e) 1) The rate at which the beetle can push a ball of dung is directly proportional to its rate constant, therefore -rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant 𝑘 = 𝐴𝑒𝑥𝑝

3-5

!!! !"



P3-6 (e) Continued From the data given -rA T(K) 6.5 300 13 310 18 313



1/T 0.003333 0.003226 0.003195

ln k 1.871802 2.564949 2.890372

3.5 3 2.5

ln(k)

2

1.5

y = -7120.4x + 25.593 R² = 0.9893

1 0.5 0 0.00318 0.00321 0.00324 0.00327 0.0033 0.00333 0.00336

1/T (K-1) Refer to P3-8 (similar procedure) Therefore, A = 1.299X1011 E = 59195.68 J/mol k = 1.299X1011 exp(-7120/T) Now at T = 41.5 C = 314.5 K k = 19.12 cm/s Therefore, beetle can push dung at 19.12 cm/s at 41.5 C P3-6 (e) 2) Individualized solution P3-7 There are two competing effects that bring about the maximum in the corrosion rate: Temperature and HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the column and consequently the rate of corrosion should increase. However, the HCN concentrations (and the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion rate somewhere around the middle of the column. 3-6

P3-8 Antidote did not dissolve from glass at low temperatures. P3-9 (a) If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1. Then with k = Ae-E/RT in general,



and



or

, or







Therefore:

which can be approximated by

. Consequently, for this doubling rate fule of thumb to

be valid, the temperature at which the doubling will take place must be related to the activation energy by this relationship. P3-9 (b) Individualized solution P3-10 From the given data -rA(dm3/mol.s) T(K) k= -rA/(4*1.5) ln (k) 1/T (𝐾 !! ) 0.002 300 0.00033333 0.003333 -8.00637 0.046

320

0.00766667

0.003125

-4.87087

0.72

340

0.12

0.002941

-2.12026

8.33

360

1.38833333

0.002778

0.328104

Plotting ln(k) vs (1/T), we have a straight line:

Since,𝑙𝑛𝑘 = 𝑙𝑛𝐴 −

!" !"

therefore ln A = 41.99 and E/R = 14999 3-7

P3-10 (a) Activation energy (E), ⇒ E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol P3-10 (b) Frequency Factor (A), ln A = 41.99 ⇒ 𝐴 = 1.72 ∗ 10!" P3-10 (c)

!"! !

.

!"# ! !

𝑘 = 1.72 ∗ 10!" exp (

!!"### !

) (1)

Given T0 = 300K Therefore, putting T = 300K in (1), we get 𝑘 𝑇! = 3.33 ∗ 10!! Hence,

𝐸 1 1 𝑘 = 𝑘 𝑇! exp − 𝑅 300 𝑇

!"! !

.

!"# ! !

𝑑𝑚! 1 . 𝑚𝑜𝑙 ! 𝑠

P3-11 (a) –ra=kCACB P3-11 (b) –ra=k P3-11 (c) –ra=kPAPB P3-11 (d) –ra=kCA P3-12 (a) C2H6 → C2H4 + H2



Rate law: -rA =

P3-12 (b) C2H4 + 1/2O2 → C2H4O



Rate law: -rA =



, k = [ s-1 ]

, k = [ dm1.5/s.mol0.5 ]

P3-12 (c) (CH3)3COOC(CH3)3 ↔ C2H6 + 2CH3COCH3 A ↔ B + 2C Rate law: -rA = k[CA – CBCC2/KC] , k = [ s-1 ] P3-12 (d) n-C4H10 ↔ I- C4H10



Rate law: -rA = k[



–

/Kc], k = [ s-1 ]

P3-12 (e) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH A + B ↔ C + D Rate law: -rA = k[CACB – CCCD/KC], k = [ dm3/mol.s ] P3-12 (f) 2CH3NH2 ↔ (CH3)2NH + NH3

Rate law: -r’A = kP2CH3NH2 , k = [ mol/kg cat.s.Pa2 ]

3-8

P3-12 (g) (CH3CO)2O + H2O → 2CH3COOH A +

B → 2C





Rate law: -rA = kCACB , k = [ dm3/mol.s ]

P3-13 (a) (1) –rA = kACACB2 (2) –rA = kACB (3) –rA = kA (4) –rA = kACA/CB P3-13 (b) (1) H2 + Br2 → 2HBr

(2) H2 + I2 → 2HI

Rate law: -rHBr =

Rate law: -rA =

P3-14 (a)











We need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (B->A) is = then:





From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so: Solving for KC gives:





P3-14 (b) If we assume the rate law for the reverse reaction

3-9

is



P3-14 (b) Continued then:







From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so: Solving for KC gives:

c)







If we assume the rate law for the reverse reaction

is

then:





From Appendix C we know that for a reaction at equilibrium: KC At equilibrium, rnet 0, so: Solving for KC gives:



3-10



P3-15 Assuming the reactions to be elementary: 2 Anthracene -> Dimer # & C ⇒ −rA = β k+ %%C 2Anthracene − Dimer (("where,"Kc = k+ k− Kc ' $ Similarly for the second reaction: Norbornadiene → Quadricyclane # CQuadricyclane & % (!where,!K = k k ⇒ −rNorbornadiene = β k+ CNorbornadiene − c + − % ( K c $ ' where β is greater than 1.0 and is ralted to the intensity of the light.

3-11

Solutions for Chapter 4 – Stoichiometry P4-1 (a) Example 4-4 (i) The critical value is around 0.5 atm (ii) Kp has the least effect, and KSO3 has the greatest effect (iii) 0.25 (iv) Individualized Solution (v) See Polymath program P4-1-av.pol. POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

epsilon

-0.14

-0.14

-0.14

-0.14

2

Fao

3.

3.

3.

3.

3

k

9.7

9.7

9.7

9.7

4

KO2

38.5

38.5

38.5

38.5

5

KP

930.

930.

930.

930.

6

KSO3

42.5

42.5

42.5

42.5

7

PO2

2.214

1.476695

2.214

1.476695

8

PSO2

4.1

2.605932

4.1

2.605932

9

PSO20

4.1

4.1

4.1

4.1

10 PSO3

0

0

1.737288

1.737288

11 ra

-0.5651727

-0.5651727

-0.0045241

-0.0045241

12 t

0

0

0.4

0.4

13 thetaB

0.54

0.54

0.54

0.54

14 X

0

0

0.4

0.4

15 yaxis

5.308112

5.308112

663.1111

663.1111

Differential equations 1 d(X)/d(t) = 1 Explicit equations 1

Fao = 3

2

PSO20 = 4.1

3

epsilon = -0.14

4

PSO2 = PSO20* (1-X)/(1+epsilon*X)

5

PSO3 = PSO20* X/(1+epsilon* X)

6

thetaB = 0.54

7

PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)

4-1

8

k = 9.7

9

KO2 = 38.5

10 KSO3 = 42.5 11 KP = 930 12 ra = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 13 yaxis = Fao/(-ra) General Total number of equations

14

Number of differential equations 1 Number of explicit equations

13

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

At X = 0.4 FA0/-rA’ = 663.11 W = (FA0/-rA’)X at X = 0.4 Hence, W = 265.24 g (vi) See Polymath program P4-1-avi-1.pol. This program is to find Xe POLYMATH Report Nonlinear Equation

05-Jun-2017

Calculated values of NLE variables Variable Value 1 X

f(x)

Initial Guess

0.997577 -1.251E-09 0.5 ( 0 < X < 1. )

Variable Value 1

epsilon

-0.14

2

k

9.7

3

KO2

38.5

4

KP

930.

5

KSO3

42.5

6

PO2

0.1963961

7

PSO2

0.0115472

8

PSO20

4.1

9

PSO3

4.754015

10 thetaB

0.54

Nonlinear equations 1 f(X) = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 = 0 Explicit equations

4-2

1

PSO20 = 4.1

2

epsilon = -0.14

3

PSO2 = PSO20* (1-X)/(1+epsilon*X)

4

thetaB = 0.54

5

PSO3 = PSO20* X/(1+epsilon* X)

6

k = 9.7

7

KO2 = 38.5

8

KSO3 = 42.5

9

KP = 930

10 PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X) General Settings Total number of equations

11

Number of implicit equations 1 Number of explicit equations 10 Elapsed time

0.0000 sec

Solution method

SAFENEWT

Max iterations

150

Tolerance F

0.0000001

Tolerance X

0.0000001

Tolerance min

0.0000001

Hence, Xe = 0.9976 See Polymath program P4-1-avi-2.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

epsilon

-0.14

-0.14

-0.14

-0.14

2

Fao

1000.

1000.

1000.

1000.

3

k

9.7

9.7

9.7

9.7

4

KO2

38.5

38.5

38.5

38.5

5

KP

930.

930.

930.

930.

6

KSO3

42.5

42.5

42.5

42.5

7

PO21

2.214

1.669102

2.214

1.669102

8

PO22

2.214

0.2197559

2.214

0.2197559

9

PSO20

4.1

4.1

4.1

4.1

10 PSO21

4.1

2.995825

4.1

2.995825

11 PSO22

4.1

0.0588832

4.1

0.0588832

12 PSO31

0

0

1.283925

1.283925

13 PSO32

0

0

4.698973

4.698973

14 ra1

-0.5651727

-0.5651727

-0.0092831

-0.0092831

15 ra2

-0.5651727

-0.5651727

-5.276E-06

-5.276E-06

16 t

0

0

0.3

0.3

17 thetaB

0.54

0.54

0.54

0.54

18 X1

0

0

0.3

0.3

4-3

19 X2

0

0

0.987624

0.987624

20 Xe

0.9976

0.9976

0.9976

0.9976

21 yaxis1

1769.371

1769.371

1.077E+05

1.077E+05

22 yaxis2

1769.371

1769.371

1.895E+08

1.895E+08

Differential equations 1 d(X1)/d(t) = 1 2 d(X2)/d(t) = 0.99*Xe/0.3 Explicit equations 1

Fao = 1000

2

PSO20 = 4.1

3

epsilon = -0.14

4

PSO21 = PSO20* (1-X1)/(1+epsilon*X1)

5

PSO31 = PSO20* X1/(1+epsilon* X1)

6

thetaB = 0.54

7

PO21 = PSO20* (thetaB-X1/2)/(1+epsilon* X1)

8

k = 9.7

9

KO2 = 38.5

10 KSO3 = 42.5 11 KP = 930 12 ra1 = -k *(PSO21* Sqrt(PO21)-PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2 13 yaxis1 = Fao/(-ra1) 14 Xe = 0.9976 15 PSO22 = PSO20* (1-X2)/(1+epsilon*X2) 16 PSO32 = PSO20* X2/(1+epsilon* X2) 17 PO22 = PSO20* (thetaB-X2/2)/(1+epsilon* X2) 18 ra2 = -k *(PSO22* Sqrt(PO22)-PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2 19 yaxis2 = Fao/(-ra2)

At X = 0.3 FA0/-rA’ = 107,700 W = (FA0/-rA’)X at X = 0.3 Hence, W = 32,310 g = 32.31 kg At X = 0.99Xe FA0/-rA’ = 1.895 x 108 W = (FA0/-rA’)X at X = 0.99Xe Hence, W = 1.872 x 108 g = 1.872 x 105 kg P4-1 (b) Example 4-5 (i) 4-4

X 1.0 0.8 0.6 0.4 0.2 0.0 0.0

0.2

0.4

0.6

0.8

1.0

yA0

(ii) Kc = 0.4 and CT0 = 0.09 (iii) When Kc is minimum and CT0 is maximum (iv)



Xeb Xef 1.0 0.9 0.8 0.7 0.6 0.5 0.0

0.2

0.4

0.6

0.8

1.0

yA0

P4-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. P4-3 2A ! B CA0 = 4 mol/dm3 Xe = 0.6 P4-3 (a) Kc = CB/ CA2 ε = -1/2 CA = CA0(1-X)/(1+ εX) = 4*0.4/0.7 = 2.286 mol/dm3 CB = 4*0.3/0.7 = 1.714 mol/dm3 Therefore Kc = 0.328 dm3/mol 4-5

P4-3 (b) For a liquid phase reaction, CA = 1.6 mol/dm3 CB = 1.2 mol/dm3 Thus, Kc = 0.469 dm3/mol P4-3 (c) CA = CA0(1-X)/(1+ εX) CB = CA0(0.5*X)/(1+ εX) -rA = k(CA^2- CB/Kc) -rA = k((CA0(1-X)/(1+εX)) ^2 – (CA0(0.5*X)/(1+ εX) )/Kc) -rA = 2((2.0(1-X)/(1- 0.5X)) ^2 – (2.0(0.5X)/(1-0.5X) )/0.5) P4-3 (d) The type of reactor does not change the expression for -rA. P4-4

1 1 B→ C 2 2 1 ⎛ 1 ⎞⎡ 1 1 ⎤ ε = y A0δ = ⎜ ⎟⎢ − − 1⎥ = − 2 ⎝ 2 ⎠⎣ 2 2 ⎦ A+





P4-4 (a)

C B0 = C A0 = 0.1

mol dm 3

P4-4 (b)

€

C A = C A0

(1− X) = 0.1 (1− 0.25) = (0.1)(0.75) = 0.086 mol ( )⎛ 1 ⎞ 0.875 dm 3 (1+ εX) 1− X ⎜ ⎟ ⎝

€

2 ⎠

⎛X⎞ ⎛ 0.25 ⎞ ⎜ ⎟ ⎜ ⎟ ( 0.1)( 0.125) = 0.0143 mol 2 2 ⎠ CC = C A0 ⎝ ⎠ = ( 0.1) ⎝ = 0.875 dm3 ⎛ 1 ⎞ (1 + ε X ) 1 − X ⎜ ⎟ ⎝ 2 ⎠ P4-4 (c)

C B = C A0 P4-4 (d)

⎛ 1 ⎞ ⎛ 1 ⎞ ⎜1 − X ⎟ C A 0 ⎜ 1 − X ⎟ 2 ⎠ ⎝ 2 ⎠ = C = 0.1 mol =⎝ = A0 (1 + εX ) dm 3 ⎛ 1 ⎞ ⎜1 − X ⎟ ⎝ 2 ⎠

4-6



C B = 0.1

mol dm 3

P4-4 (e) €

rA rC = , rA = −2rC = −4 mol dm 3 min −4 2

P4-4 (f)

€

2A + B ! C ε = δyA0 = -0.5 CA0 = 2 mol/dm3 Irreversible rate of reaction –rA = kCA2CB Reversible rate of reaction

–rA = k (CA2CB – CC/KC)



Hence



P4-4 (g) –rA = -0.296 mol/dm3s P4-5 (a) Liquid phase reaction, O CH2--OH CH2 - CH2 + H2O → CH2--OH A + B → C CAO = 16.13mol/dm3 P4-5 (a) Continued Stoichiometric Table: Species Ethylene oxide Water Glycol

CBO = 55.5 mol/dm3

Symbol A

Initial CAO=16.13 mol/dm3

Change - CAOX

B

CBO= 55.5 mol/dm3, θ B =3.441

-CAOX

Remaining CA= CAO(1-X) = (1-X) mol/dm3 CB = CAO( θ B -X)

0

CAOX

=(3.441-X) mol/dm3 CC = CAOX mol/dm3

C

Rate law:

-rA = kCACB

Therefore,

2 -rA = k C AO (1-X) ( θ B -X) = k (16.13)2(1-X) (3.441-X)

At 300K

E = 12500 cal/mol, k = 0.1dm3/mol.s

X = 0.9, 4-7

τ CSTR =

(16.13)( 0.9 ) C AO X = = 2.196sec 2 − rA ( 0.1)(16.13) (1 − 0.9 )(3.441 − 0.9 )

and, V = τ *vo = 2.196 sec X 200 liters/sec = 439.2 liters At 350K, k2 = k exp((E/R)(1/T-1/T2))= 0.1exp((12500/1.987)(1/300-1/350)) = 1.99 dm3/mol.s Therefore, τ CSTR =

(16.13)( 0.9 ) C AO X = = 0.110sec , 2 − rA (19.99 )(16.13) (1 − 0.9 )(3.441 − 0.9 )

and, V = τ* vo = 0.110 X 200 liters = 22 liters P4-5 (b) Isothermal, isobaric gas-phase pyrolysis, C2H6 C2H4 + H2 A " B + C Stoichiometric table: Species C 2H 6 C 2H 4 H 2

symbol A B C



Entering FAO 0 0 FTO=FAO

Change -FAOX +FAOX +FAOX



ε = yao δ = 1(1+1-1) = 1 v = vo(1+ ε X) => v = vo(1+X) P4-5 (b) Continued

P RT (1)(6atm )

CAO = yAO CTO = yAO =

3

= 0.067 kmol/m3 = 0.067 mol/dm3

⎛ m atm ⎞ ⎜ 0.082 ⎟ (1100 K ) K .kmol ⎠ ⎝ (1 − X ) mol/dm3 F (1 − X ) F CA = A = AO = C AO v vO (1 + X ) (1 + X ) CB =

F (X ) FB X mol/dm3 = AO = CAO v vO (1 + X ) 1 + X ( )

CC =

FC F (X ) X mol/dm3 = AO = CAO v vO (1 + X ) (1 + X )

Rate law: 4-8

Leaving FA=FAO(1-X) FB=FAOX FC=FAOX FT=FAO(1+X)

-rA = kCA= kCAO

(1 − X ) =0.067 k (1 − X ) (1 + X ) (1 + X )

If the reaction is carried out in a constant volume batch reactor, =>( ε = 0) CA = CAO(1-X) mol/dm3 CB = CAO X mol/dm3 CC = CAO X mol/dm3 P4-5 (c) Isothermal, isobaric, catalytic gas phase oxidation,

1 O2 " C2H4O 2 1 A + B " C 2 C2H4 +

-rA’ = kApApB0.5 pA = yA0P = yA0CRT = CART pB = yB0P = yB0CRT = CBRT Therefore, -rA’ = kACACB0.5(RT)1.5 (1) For a fluidized batch reactor, V is constant CA = CA0(1-X) CB = CA0(θB – b/a X) θB = 0.5 (stoichiometric feed) b/a = 0.5 Therefore, CB = CA0(1-X)/2 Therefore, -rA’ = kACA01.5(1-X)1.5(RT)1.5/20.5 P4-5 (c) Continued (2) Stoichiometric table: Species Symbol Entering Change Leaving C 2H 4 A FAO -FAOX FA=FAO(1-X) O 2 B FBO 1 FB=FAO( θ B -X/2) - FAOX

2

C2H4O

C

0

+FAOX

FC=FAOX



1 F FBO 2 AO 1 θ B = = = FAO FAO 2

y AO =

FAO FAO 2 = = FTO FAO + FBO 3



(6atm ) P 2 mol = = 0.092 3 3 RT 3 ⎛ dm atm.dm ⎞ ⎜ 0.082 ⎟ (533K ) mol.K ⎠ ⎝ F (1 − X ) C AO (1 − X ) 0.092 (1 − X ) F C A = A = AO = = v vO (1 + ε X ) (1 − 0.33 X ) (1 − 0.33 X )

C AO = y AO CTO = y AO

4-9

X⎞ ⎛ FAO ⎜ θ B − ⎟ F 2 ⎠ 0.046 (1 − X ) ⎝ CB = B = = v vO (1 + ε X ) (1 − 0.33 X ) CC =

0.092 ( X ) FC FAO X = = v vO (1 + ε X ) (1 − 0.33 X )

If the reaction follow elementary rate law 0.5

Rate law: -rA’ = kACACB

0.5

⎧⎪ 0.092 (1 − X )⎫⎪ ⎧⎪ 0.046 (1 − X )⎫⎪ 1.5 (RT) , T = 533 K ⇒ − rA = k ⎨ ⎬⎨ ⎬ (RT) ⎪⎩ (1 − 0.33 X ) ⎪⎭ ⎪⎩ (1 − 0.33 X ) ⎪⎭ 1.5

P4-5 (d) Isothermal, isobaric, catalytic gas-phase reaction in a PBR C6H6 + 2H2 → C6H10 A + 2B → C Given:

ν 0 = 50 dm 3 / min

Stoichiometric Table: Species Symbol C 6H 6 A H 2 B C6H10 C P4-5 (d) Continued

θB =

CB = CC =

Change -FA0X -2FA0X FA0X

Leaving FA=FA0(1-X) FB=FA0(θB-2X) Fc=FA0X

FB 0 2 FA0 F FA0 1 1 2 = = 2 y A0 = A0 = = ε = y A0δ = (1 − 2 − 1) = − 3 3 FA0 FA0 FT 0 FA0 + FB 0 3

C A0 = CT 0 y A0 =

CA =

Entering FA0 FB0=2FA0 0

FA

ν FB

ν FC

ν

P ⎛1⎞ 6atm mol ⎛1⎞ ⎜ ⎟= ⎜ ⎟ = 0.055 3 dm3 ⋅atm RT ⎝ 3 ⎠ 0.0821 mol⋅K * (443K ) ⎝ 3 ⎠ dm

=

FA0 (1 − X ) (1 − X ) = C A0 ν 0 (1 + εX ) (1 − 23 X )

=

FA0 (θ B − 2 X ) (2 − 2 X ) (1 − X ) = C A0 = 2 * C A 0 ν 0 (1 + εX ) (1 − 23 X ) (1 − 23 X )

=

FA0 X X = C A0 ν 0 (1 + εX ) (1 − 23 X )

Rate Law: NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or concentration units are used. See below for an example. 2 − rA ' = kPA PB

mol mol [=] * atm * atm 2 3 kgcat ⋅ min kgcat ⋅ min⋅ atm 4-10

Notice that if you use concentrations in this rate law, the units will not work out.

PA = y A0 P = ( y A0 C ) * RT = C A RT 2

2

− rA = kPA PB = kC AC B ( RT ) 3 = 4kC A0

3

(1 − X ) 3 ( RT ) 3 3 2 (1 − 3 X )

Design Equation for a fluidized CSTR:

W= W= W=

FA 0 X − rA ' FA0 X (1 − 23 X ) 3 3

4kC A0 (1 − X ) 3 ( RT ) 3

ν 0 X (1 − 23 X ) 3 2

4kC A0 (1 − X ) 3 ( RT ) 3



Evaluating the constants:

k = 53

mol at 300K kgcat ⋅ min⋅ atm3

P4-5 (d) Continued At 170°C (443K),

k 443

J ⎡ ⎤ ⎢ 80000 mol ⎛ 1 ⎡ EA ⎛ 1 1 ⎞⎤ 1 ⎞⎥ mol ⎟⎟⎥ = 53 exp⎢ = k 300 exp⎢ ⎜⎜ − − ⎜ ⎟⎥ = 1663000 J R T T 300 K 443 K kgcat ⋅ min ⋅ atm ⎝ ⎠ 300 443 ⎝ ⎠ ⎣ ⎦ ⎢ 8.314 ⎥ mol ⋅ K ⎣ ⎦

Plugging in all the constants into the design equation: X = 0.8 3

3 2 50 dm min ⋅ 0.8 ⋅ (1 − 3 0.8) W= = 5.25 ⋅ 10 −7 kgcat 3 3 mol mol 2 dm3 ⋅atm 4 ⋅ 1663000 kgat⋅min ⋅atm3 (0.055 dm3 ) (1 − 0.8) (0.0821 mol⋅K ⋅ 443K )

At 270°C (543K),

k 543

J ⎡ ⎤ 80000 ⎢ ⎡ EA ⎛ 1 ⎤ 1 ⎞ mol mol ⎛⎜ 1 − 1 ⎞⎟⎥ = 90790000 ⎟⎟⎥ = 53 exp⎢ = k 300 exp⎢ ⎜⎜ − ⎥ kgcat ⋅ min⋅ atm ⎣ R ⎝ T300 T543 ⎠⎦ ⎢ 8.314 J ⎝ 300K 543K ⎠⎥ mol ⋅ K ⎣ ⎦

Plugging in all the constants into the design equation: X = 0.8 3

3 2 50 dm min ⋅ 0.8 ⋅ (1 − 3 0.8) W= = 5.22 ⋅10 −9 kgcat 3 3 mol mol 2 dm3 ⋅atm 4 ⋅ 9079000 kgat⋅min ⋅atm3 (0.055 dm3 ) (1 − 0.8) (0.0821 mol⋅K ⋅ 543K )

P4-6 (a) Let A = ONCB

C = Nibroanaline 4-11



B = NH3

D = Ammonium Chloride

A + 2B ⎯⎯ → C+D -rA = kC ACB



P4-6 (b) Species A B

Entering FA0 FB0 = ΘBFA0 =6.6/1.8 FA0 0 0

C D P4-6 (c) For batch system, CA=NA/V P4-6 (d)

Change - FA0X -2 FA0X FA0X FA0X

Leaving FA0(1-X) FB= FA0(ΘB – 2X) FC=FA0X FD=FA0X

-rA = kNANB/V2



-rA = kCACB FA =

N A N A N A0 F F = = (1 − X ) = C A0 (1 − X ), C A = A = A = C A0 (1 − X ) V V0 V0 v v0

FB =

N B N B N A0 F = = (θ B − 2 X ) = C A0 (θ B − 2 X ), CB = B = C A0 (θ B − 2 X ) V V0 V0 v0

−rA = kCA20 (1 − X )(θ B − 2 X )

θB =

CB 0 6.6 = = 3.67 C A0 1.8

C A0 = 1.8

kmol m3 2

−rA = k (1.8) (1 − X )(3.67 − 2 X ) P4-6 (e) 1) At X = 0 and T = 188°C = 461 K 2

− rA0 = kC A2 0 Θ B = 0.0017

− rA0 = 0.0202

m3 kmol ⎛ kmol ⎞ ⎜1.8 3 ⎟ 3.67 = 0.0202 3 kmol min ⎝ m ⎠ m min

kmol m3 min

2) At X = 0 and T = 250C = 298K

⎛ E ⎛ 1 1 ⎞⎞ ⎜ ⎟⎟ ⎜ R ⎜ T − T ⎟⎟ ⎠⎠ ⎝ ⎝ O

k = k O exp⎜

4-12

cal ⎛ 11273 ⎜ m mol k = 0.0017 exp⎜ cal kmol. min ⎜ ⎜ 1.987 mol .K ⎝ m3 = 2.03 × 10 −6 kmol. min 3

−rA = 2.03 ×10−6

3

m ⎛ kmol ⎞ ⎜1.8 3 ⎟ kmol min ⎝ m ⎠

2

⎞ 1 ⎞ ⎟⎟ ⎛ 1 − ⎜ ⎟ ⎝ 461 298 ⎠ ⎟ ⎟ ⎠

kmol ⎞ ⎛ ⎜ 3.67 3 ⎟ m ⎠ ⎝

-rAO = kCAOaCBO = 2.41 X 10-5 kmol/m3min 3)

⎡ E ⎛ 1 1 ⎞⎤ k = k0 exp ⎢ ⎜ − ⎟⎥ ⎣⎢ R ⎝ T0 T ⎠⎦⎥ P4-6 (e) Continued

⎡ ⎤ cal ⎢ 11273 ⎥ ⎛ ⎞ m3 1 1 mol k = 0.0017 exp ⎢ − ⎜ ⎟⎥ kmol min ⎢1.987 cal ⎝ 461 K 561 K ⎠ ⎥ ⎢⎣ ⎥⎦ mol K

k = 0.0152

m3 kmol min

− rA0 = kC A0 2CB 0 m3 ⎛ kmol ⎞ −rA = 0.0152 ⎜1.8 3 ⎟ kmol min ⎝ m ⎠ −rA = 0.1807

2

kmol ⎞ ⎛ ⎜ 3.67 3 ⎟ m ⎠ ⎝

kmol m3 min

P4-6 (f) rA = kCAO2(1-X)(θB-2X) At X = 0.90 and T = 188C = 461K 1) at T = 188 C = 461 K 2

⎛ ⎞⎛ kmol ⎞ m3 ⎟⎜1.8 3 ⎟ (1 − 0.9)(3.67 − 2(0.9 )) − rA = ⎜⎜ 0.0017 kmol. min ⎟⎠⎝ m ⎠ ⎝ kmol = 0.00103 3 m min 2) at X = 0.90 and T = 25C = 298K 2



⎛ ⎞⎛ kmol ⎞ m3 ⎟⎜1.8 3 ⎟ (1 − 0.9)(3.67 − 2(0.9)) − rA = ⎜⎜ 2.03 × 10 −6 kmol. min ⎟⎠⎝ m ⎠ ⎝ kmol = 1.23 × 10 −6 3 m min 4-13

3) at X = 0.90 and T = 288C = 561K 2

⎛ ⎞⎛ kmol ⎞ m3 ⎟⎟⎜1.8 3 ⎟ (1 − 0.9 )(3.67 − 2(0.9)) − rA = ⎜⎜ 0.0152 kmol . min m ⎠ ⎝ ⎠⎝ kmol = 0.0092 3 m min P4-6 (g) FAO = 2 mol/min 1) For CSTR at 25oC -rA = 1.23 × 10 −6

kmol m 3 min

P4-6 (g) Continued

V =

=

FAO (1 − X ) − rAMX =0.9

2mol / min× 0.1 = 162.60 m 3 mol 1.23 × 10 −3 3 m min

2) At 288oC, -rA = 0.0092

V = =

kmol m 3 min

FAO (1 − X ) − rAMX =0.9 2mol / min× 0.1 = 21.739m 3 mol 0.0092 3 m min

P4-7 No solution will be given P4-8 (a) Isothermal gas phase reaction.

1 3 N 2 + H 2 → NH 3 2 2 Making H2 as the basis of calculation:

1 2 H 2 + N 2 → NH 3 3 3 1 2 A + B → C 3 3 Stoichiometric table: Species Symbol Initial change H 2 A FAO -FAOX N 2 B FBO= θ B FAO -FAOX/3 NH3

C

0

Leaving FA=FAO(1-X) FB=FAO( θ B -X/3)

+2FAOX/3 FC=(2/3)FAOX 4-14

P4-8 (b)

⎛2 1

2

⎞

δ = ⎜ − − 1⎟ = − 3 ⎝3 3 ⎠



1 ⎛ 2⎞ ε = y AOδ = 0.5 × ⎜ − ⎟ = − 3 ⎝ 3⎠ (16.4atm ) = 0.2 mol/dm3 C AO = 0.5 3 ⎛ atm.dm ⎞ ⎜ 0.082 ⎟ (500 K ) mol.K ⎠ ⎝ P4-8 (b) Continued

C AO (1 − X ) 0.2 (1 − 0.6 ) = = 0.1mol / dm3 ⎛ 0.6 ⎞ (1 + ε X ) ⎜1 − ⎟ 3 ⎠ ⎝ 2 C ( X ) 2 0.2 ( 0.6 ) CNH3 = CC = × AO = × = 0.1mol / dm3 0.6 3 (1 + ε X ) 3 ⎛ ⎞ ⎜1 − ⎟ 3 ⎝ ⎠ P4-8 (c) CH 2 = C A =

kN2 = 40 dm3/mol.s (1) For Flow system: 1

−rN2 = k N2 ⎡⎣CN2 ⎤⎦ 2 ⎡⎣CH 2 ⎤⎦

⎡⎛ ⎢ ⎜1 − 2 ⎝ = 40 (C AO ) ⎢ ⎢ ⎛1 − ⎢⎣ ⎜⎝

X 3 X 3

⎞⎤ ⎟⎥ ⎠⎥ ⎞⎥ ⎟⎥ ⎠⎦

1

2

3

2 3

⎡ ⎤ 2 ⎢ (1 − X ) ⎥ ⎢ ⎥ ⎢ ⎛1 − X ⎞ ⎥ ⎟ ⎢⎣ ⎜⎝ 3 ⎠ ⎥⎦

3

⎡ ⎤2 ⎢ (1− X ) ⎥ ⎥ −rN 2 = 1.6⎢ ⎢⎛⎜1− X ⎞⎟ ⎥ ⎢⎣⎝ 3 ⎠ ⎥⎦

€

4-15

(2) For batch system, constant volume. 1

3

2

2

− rN2 = k N2 ⎡⎣CN2 ⎤⎦ ⎡⎣CH 2 ⎤⎦ N (1 − X ) N N C A = A = A = A0 V V0 V0 C A = C A0 (1 − X ) N CB = B = V

X⎞ ⎛ N A0 ⎜ Θ B2 − ⎟ 3⎠ ⎝ V0

⎛ X⎞ = C H 2O ⎜ 1 − ⎟ 3⎠ ⎝ 1

3 X ⎞⎤ 2 ⎡⎛ 2 − rN2 = 40 ( C A0 ) ⎢⎜1 − ⎟ ⎥ ⎡⎣(1 − X )⎤⎦ 3 ⎠⎦ ⎣⎝

⎡ X⎤ = 1.6 ⎢1 − ⎥ ⎣ 3⎦

1

2

[1 − X ]

3

2

2

P4-9 (a) P0 = PA0 = CA0RT PA = PA0(1-Xe)/(1 + ɛXe) PB = PA0Xe/2(1 + ɛXe) CA = PA / RT CB = PB / RT Kc = CB/CA2 CA0 = 4 mol/dm3 Xe = 0.6 ɛ = -1 Substituting the values of CA0, Xe and ɛ, we get CA = 4 mol/dm3 CB = 3 mol/dm3 Kc = CB/CA2 = 0.1875 dm3/mol (b) Kc = CB/CA2 CA = CA0(1-Xe) CB = CA0Xe CA0 = 4 mol/dm3 Xe = 0.6 Substituting the values of CA0 and Xe, we get CA = 1.6 mol/dm3 CB = 1.2 mol/dm3 Kc = CB/CA2 = 0.46875 dm3/mol P4-10 No solution will be given.



4-16

P4-11 No solution will be given. P4-12

Given: Gas phase reaction A + B " 8C in a batch reactor fitted with a piston such that V = 0.1P0 3

k = 1.0

2

( ft )

lb mol 2 sec



− rA = kC A2 C B NA0 = NB0 at t = 0 V0 = 0.15 ft3 T = 140°C = 600°R = Constant

P4-12 (a)

y A0 =

N A0 = 0.5 N A0 + N B 0

δ = 8 −1−1 = 6 ε = y A0δ = 3 Now V =

V0 P0 (1 + ε X ) and T = 1, P0 = 10V0 , and P = 10V ⎛T ⎞ T0 P⎜ ⎟ ⎝ T0 ⎠

Therefore V =

10V02 (1 + ε X ) or V 2 = V02 (1 + ε X ) 10V

N A = N A0 [1 − X ]



N B = N A0 [θ B − X ] 3



θB =

N B0 = 1 N A0

kN 3 [1 − X ] ⎛y P ⎞ kN 2 N N A0 = ⎜ A0 0 ⎟ V0 −rA = kC CB = A3 B = A0 3 V ⎝ RT ⎠ V03 (1 + ε X )2 2 A

Therefore 3

⎛ y P ⎞ [1 − X ] −rA = k ⎜ A0 0 ⎟ 3 ⎝ RT ⎠ (1 + ε X )2 3

−rA = 5.03*10

−9

[1 − X ] 3 (1 + 3 X )2



lb mol ft 3 sec



P4-12 (b)

V 2 = V02 (1 + ε X ) 0.22 = 0.152 (1 + ε X ) 4-17

X = 0.259

−rA = 8.63*10−10

lb mol ft 3 sec



4-18

Solutions for Chapter 5 – Isothermal Reactor Design- Conversion P5-1 (a) Example 5-3 For 50% conversion, X = 0.5 and k = 3.07 sec-1 at 1100 K (from Example 5-3) FB = 200X106 / (365 X 24 X 3600 X 28) lbmol/sec = 0.226 lbmol/sec FAO =



Now, we have from the example

Also,









3

= 35.47 X 0.886 ft = 31.44 ft3 Now, n = 31.44 ft3/0.0205 ft2 X40 ft = 38.33 So, we see that for lower conversion and required flow rate the volume of the reactor is reduced. P5-1 (b) Example 5-4 Individualized Solution P5-1 (c) Example 5-5 New Dp = 3D0/4 Because the flow is turbulent









p



Now,

, so too much pressure drop P = 0 and the flow stops. P5-1 (d) Example 5-6 For turbulent flow

5-1

P5-1 (d) Example 5-6 Continued

Therefore there is no change. P5-1 (e) Example 5-7 (i) X increases and f decreases with an increase in k’ for the same W, while p remains unchanged, and vice versa. This is an expected observation, because as k’ increases, the rate of the reaction increases, and naturally, X increases. Since the conversion down the reactor increases, more product is formed, lesser amount of reactants are present. Since ɛ < 0, the volumetric flow rate decreases with an increase in conversion. (ii) As α and FA0 increase, X and p decrease for the same W, while f increases. (iii) Individualized solution P5-1 (f) Example 5-8 Individualized solution P5-1 (g) Example 5-3, Using ASPEN, we get (Refer to Aspen Program P5-2g from polymath CD) (1) At 1000K, for the same PFR volume we get only 6.2% conversion. While at 1200K, we get a conversion of nearly 100%. This is because the value of reaction constant ‘k’ varies rapidly with reaction temperature. (2) Earlier for an activation energy of 82 kcal/mol we got approx. 81% conversion. For activation energy of 74 kcal/mol keeping the PFR volume the same we get a conversion of 71.1%. While for an activation energy of 90 kcal/mol we get a conversion of 89.93%. (3) On doubling both flow rate and pressure we find that the conversion remains the same. P5-1 (h) Individualized solution. P5-1 (i) Individualized solution P5-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. P5-3 (a) (3) 50% (Liquid phase reactions do not depend on pressure) P5-3 (b) (2) 0.234 2

−rA = kC 2A0 1− X p

(

)

−rA =∝ kp2 p2 < 1





ktrue p2 = k = 0.234 0.234 ktrue = > 0.234 p2 P5-4 (a)

1 D Increase D, decrease, increase X.

α~



P5-4 (b)

Ans. (1) X > 0.5

Ans. (3) Xe = 0.75





P5-4 (c)

Ans. (3) remain the same



Kc =



P5-4 (d) P5-4 (e)

CA

C Ae

=

(

)

C A0 1− Xe p 1− X = C A0 Xe p Xe

Xe is not a function of Ans. (3) remain the same Xe is not a function of Ans. (4) insufficient information to tell

P5-5 r = - k CA2





5-3

P5-5 Continued Thus, k = 5 For a CSTR,

P5-6 To = 300K

KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3

Mole balance:



Rate law:





Stoichiometry:



and

















( " + k E $ 1 1 %'* Z = 2902.2 dm , Let z = = exp $ − ' *) R # T0 T &-, ko 3

Now using:



where

and

Solving using polymath to get a table of values of X Vs T. See Polymath program P5-6.pol.

5-4



P5-6 Continued POLYMATH Results NLE Solution Variable Value f(x) X 0.4229453 3.638E-12 To 300 T 305.5 z 2902.2 V 3785.4 E 1.5E+04 R 2 y 1.5684405 Kco 3 Hrx -2.5E+04 Kc 1.4169064 NLE Report (safenewt) Nonlinear equations [1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0 Explicit equations [1] To = 300 [2] T = 305.5 [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 = 3 [9] Hrx = -25000 [10] Kc = Kco*exp(Hrx/R*(1/To-1/T))

Ini Guess 0.5

T(in K)

X

300

0.40

301

0.4075

303

0.4182

304

0.4213

305

0.4228

305.5

0.4229

305.9

0.4227

307

0.421

310

0.4072

315

0.3635

We get maximum X = 0.4229 at T = 305.5 K.

5-5

P5-7 PFR













CSTR

















5-6

P5-8

Base Case PFR above. Find k Mole Balance Rate Law



Stoichiometry

Combine



CSTR added upstream of PFR

X1 = 0.5



Now find X2





5-7

P5-8 Continued



P5-9 (a)

Using the Arrhenius equation at the CSTR temperature of 300 K yields the new specific reaction rate.

The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal. So, we get 0.57 as conversion instead of 0.5.









X=0.85 So, considering the above results, we will choose a CSTR. 5-8

P5-9 (b)



P5-9 (c)



P5-9 (d) 1) CSTR and PFR are connected in series: Solving the quadratic equation, XCSTR = 0.44 For PFR,













2) when CSTR and PFR are connected in parallel, XCSTR = 0.56



For PFR,





5-9

P5-9 (d) Continued XPFR = 0.92 Hence, final conversion X =



= 0.74

P5-9 (e) To process the same amount of species A, the batch reactor must handle

If the reactants are in the same concentrations as in the flow reactors, then

So the batch reactor must be able to process 14400 dm3 every 24 hours. Now we find the time required to reach 90% conversion. Assume the reaction temperature is 300K.

, and since



Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tf = 3 hours ttotal = tR + tf ttotal = 2.14hours + 3 hours = 5.14 hours. Therefore, we can run 4 batches in a day and the necessary reactor volume is Referring to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be approximately $85,000 for the reactor. P5-9 (f) The points of the problem are: 1) To note the significant differences in processing times at different temperature (i.e. compare part (b) and (c)). 2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to react and 180 to fill and empty. 3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black. 5-10

P5-9 (g) Individualized solution P5-10 (a) The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal. So, we get 0.57 as conversion instead of 0.5.



















5-11



P5-10 (b) CAO = 2 mol/dm3 A → B Assuming 1st order reaction, For CSTR,







–rA = kCA = kCAO(1-X)





=>





For PFR,



, =>



= 1 – exp(-0.67) = 0.486

Now assuming 2nd order reaction, For CSTR, Now, assuming 2nd order reaction, For CSTR,







=



=>



For PFR,

=>



So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second order reaction. P5-10 (c) A graph between conversion and particle size is as follows: Originally we are at point A in graph, when particle size is decreased by 15%, we move to point C, which have same conversion as particle size at A. But when we decrease the particle size by 20%, we reach at point D, so a decrease in conversion is noticed. Also when we increase the particle size from position A, we reach at point B, again there is a decrease in the conversion.

5-12

P5-11 Reaction: A + B → C + D Mole Balance:

FA0

dX = −rA" dW

Rate Law:

−rA" = KCACB Stoichiometry:

(

)

12 F FA0 1− X P CA = A = CA0 1− X p((;(( 1− αW υ υ P0

(

)(

CA = CA0 1− X 1− αW

(

)

(

)

12

)

(

)(

Equimolar flow rate: CB = CA = CA0 1− X 1− αW

12

)



Combine:

−rA" = kC2A0 1− X

(

2

) (1− αW) 2 2 dX kCA0 (1− X) (1− αW) =

dW

FA0

Evaluate:

" % " αW % X $ FA0 ' = W $1− ' 1− X $ kC2 ' 2 & # # A0 &



Solution: First we must solve for KW:

5-13







5-14

P5-12



P5-12 (a)



P5-12 (b)



5-15

P5-12 (c)

P5-12 (d) The conversion will remain unchanged since the reactor is well-mixed ideal CSTR. P5-13 (a) V = 5 dm3 Mass of one capsule = 18 g M.wt. of Iocane = 56.25 g/mol Moles of Iocane in one capsule (NA0) = 3

!" !".!"

= 0.32 mol

CA0 = NA0 / V = 0.064 mol/dm P5-13 (b) For a batch reactor,

Solving the differential equation, we get:

𝑑𝐶! = 𝑟! 𝑑𝑡 𝑑𝐶! = −𝑘! 𝐶! 𝑑𝑡 ln

𝐶! = 𝑘! 𝑡 𝐶!!

For concentration of Optoid to be 0.01 mol/dm3, CA = 0.064 – 0.01 = 0.054 mol/dm3 Substituting the values of kA, CA and CA0, we get t = 18 hr So, Ambercromby should arrest Ms. Patel. P5-13 (c) 𝑘! 𝐸 1 1 = exp − − 𝑘!! 𝑅 𝑇 𝑇! 5-16

P5-13 (c) Continued Substituting values of kA0 = 0.00944 hr-1, E = 20500 cal/mol, R = 1.986 cal/mol K, T = 311.7 K and T0 = 310 K, we get: kA = 0.01131 hr-1 𝐶! ln = 𝑘! 𝑡 𝐶!! For concentration of Optoid to be 0.01 mol/dm3, CA = 0.064 – 0.01 = 0.054 mol/dm3 Substituting the values of kA, CA and CA0, we get t = 15 hr If Shoemaker had a fever, then Ambercromby arrested the wrong suspect. He should arrest Mr. Jenkins. P5-14 (a) POLYMATH Report Ordinary Differential Equations Calculated values of DEQ variables Variable

Initial value

Minimal value

Maximal value

Final value

1 w

0

0

2000.

2000.

2 x

0

0

0.9978192

0.9978192

3 FBuO

50.

50.

50.

50.

4 PBuO

10.

10.

10.

10.

5 PBu

10.

0.0109159

10.

0.0109159

6 k

0.054

0.054

0.054

0.054

7 K

0.32

0.32

0.32

0.32

-0.0421872

-0.0005854

-0.0005854

8 rBu -0.0306122 Differential equations 1 d(x)/d(w) = -rBu / FBuO Kgcat-1 Explicit equations 1 FBuO = 50 Kmol.hr-1 2 PBuO = 10 atm

3 PBu = PBuO * (1 - x) / (1 + x) atm 4 k = 0.054 kmol.kgcat-1.hr-1.atm-1 5 K = 0.32 atm-1 6 rBu = -k * PBu / (1 + K * PBu) ^ 2 Kmol.kgcat-1.hr-1

5-17

P5-14 (a) continued



Hence, weight of catalyst required for 80% conversion is 1054.1 kg. P5-14 (b) dX " = −rBu dW

Differential Mole Balance:

FA0 =

PBu = PBuO

Here,

Rate equation:







–rBuʹ = K*PBu



5-18



P5-14 (b) continued For Fluidized CSTR:





Therefore,







w = FBuo*X



Given, X = 0.8 FBuO = 50 Kmol.hr-1 KBu = 0.32 atm-1 PBuO = 10 atm



Therefore, putting the values in the above equation w = 50*0.8

= 1225 Kg

Therefore, 1225 Kg of fluidized CSTR catalyst weight is required to achieve 80% conversion. P5-14 (c) With pressure drop parameter alpha = 0.0006 kg-1 POLYMATH Report Ordinary Differential Equations Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value

1 w

0

0

1100.

1100.

3 y

1.

0.2689385

1.

0.2689385

2 x

4 FBuO 5 f

6 PBuO 7 PBu 8 k

9 K

10 rBu

11 alpha

0

50. 1.

10. 10.

0.054 0.32

0

50. 1.

10.

0.3408456 0.054 0.32

-0.0306122 -0.0421871 0.0006

0.0006

0.7750365 50.

6.600157 10. 10.

0.054 0.32

-0.0149635 0.0006

0.7750365 50.

6.600157 10.

0.3408456 0.054 0.32

-0.0149635 0.0006

Differential equations 1 d(x)/d(w) = -rBu / FBuO

Kgcat-1

2 d(y)/d(w) = -alpha * (1 + x) / 2 / y

Kgcat-1

5-19

P5-14 (c) continued Explicit equations 1 FBuO = 50

Kmol.hr-1

2 f = (1 + x) / y 3 PBuO = 10

atm

4 PBu = PBuO * (1 - x) / (1 + x) * y

atm

5 k = 0.054

kmol.kgcat-1.hr-1.atm-1

6 K = 0.32

atm-1

7 rBu = -k * PBu / (1 + K * PBu) ^ 2

Kmol.kgcat-1.hr-1

8 alpha = 0.0006

Kg-1





5-20

P5-14 (c) continued

So, we see the maximum rate in case with pressure drop is at catalyst weight equal to around 600 Kg.

To achieve 70% conversion, catalyst weight required is 932.3 kg .

In case of (a), 915.5 kg of catalyst is required to achieve 70% conversion.

5-21

P5-14 (d) Individualized solution P5-14 (e) Individualized solution P5-15 Gaseous reactant in a tubular reactor: A → B

















For a plug flow reactor:











At T2 = 260°F = 720°R, with k1 = 0.0015 min-1 at T1 = 80°F = 540°R,





Therefore 14 pipes are necessary.



5-22

P5-16 (a) A → B/2





Combining

(for the integration, refer to Appendix A) from the Ideal Gas assumption. Substituting Eqn. (5), X = 0.8 and ε = –1/4 to Eqn. (4) yields,

……….(6)











5-23

P5-16 (b) Individualized solution. P5-17 (a) Given: The metal catalyzed isomerization



liquid phase reaction

with Keq = 5.8 For a plug flow reactor with yA = 1.0, X1 = 0.55 Case 1: an identical plug flow reactor connected in series with the original reactor.

Since yA = 1.0, ∪B = 0. For a liquid phase reaction



and





For the first reactor,

or



Take advantage of the fact that two PFR’s in series is the same as one PFR with the volume of the two combined. VF = V1 + V2 = 2V1 and at VF, X = X2



5-24

P5-17 (a) Continued



X2 = 0.74 P5-17 (b) Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,

The analysis for the first reactor is the same as for case 1.

By performing a material balance on the separator, FA0,2 = FA0(1-X1) Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0, ∪B = 0

for the second reactor.



and since V1 = V2

or 5-25

P5-17 (b) continued





Overall conversion for this scheme:

P5-17 (c) Individualized Solution P5-18 Given: Meta- to ortho- and para- isomerization of xylene. Pressure = 300 psig T = 750°F V = 1000 ft3 cat. Assume that the reactions are irreversible and first order. Then: Check to see what type of reactor is being used. Case 1: 5-26

P5-18 Continued Case 2: Assume plug flow reactor conditions: or



CM0, k, and V should be the same for Case 1 and Case 2. Therefore,

The reactor appears to be plug flow since (kV)Case 1 = (kV)Case 2 As a check, assume the reactor is a CSTR.



or Again kV should be the same for both Case 1 and Case 2.



kV is not the same for Case 1 and Case 2 using the CSTR assumption, therefore the reactor must be modeled as a plug flow reactor.



5-27

P5-18 Continued For the new plant, with v0 = 5500 gal / hr, XF = 0.46, the required catalyst volume is:

This assumes that the same hydrodynamic conditions are present in the new reactor as in the old. P5-19 A→ B in a tubular reactor



Tube dimensions: L = 40 ft, D = 0.75 in. nt = 50





with



or



Assume Arrhenius equation applies to the rate constant. At T1 = 600°R, k1 = 0.00152 At T2 = 760°R, k2 = 0.0740

5-28

P5-19 continued





so



From above we have so



Dividing both sides by T gives:



Evaluating and simplifying gives:

Solving for T gives: T = 738°R = 278°F P5-20

5-29

P5-20 Continued





5-30

P5-20 (b)



P5-20 (c)





P5-21 Production of phosgene in a micro reactor CO + Cl2

COCl2 (Gas phase reaction)

A + B

C

The equations that need to be solved are as follows: d(X)/d(W) = -rA/FA0 d(y)/d(W) = -α*(1+Ɛ*X)/(2*y) ( from 4.30) P5-21 Continued FB0 = FA0; Fb = FB0-FA0*X; Fc = FA0*X See Polymath program P5-22.pol.

5-31

P5-21 Continued POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value W 0 0 X 0 0 y 1 0.3649802 e -0.5 -0.5 FA0 2.0E-05 2.0E-05 FB0 2.0E-05 2.0E-05 Fa 2.0E-05 4.32E-06 Fb 2.0E-05 4.32E-06 v0 2.83E-07 2.83E-07 v 2.83E-07 2.444E-07 Fc 0 0 Ca 70.671378 9.1638708 Cb 70.671378 9.1638708 a 3.55E+05 3.55E+05 k 0.004 0.004 rA -19.977775 -19.977775 Cc 0 0 ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -rA/FA0 [2] d(y)/d(W) = -a*(1+e*X)/(2*y) Explicit equations as entered by the user [1] e = -.5 [3] FB0 = FA0 [5] Fb = FB0-FA0*X [7] v = v0*(1+e*X)/y [9] Ca = Fa/v [11] a = 3.55e5 [13] rA = -k*Ca*Cb

maximal value 3.5E-06 0.7839904 1 -0.5 2.0E-05 2.0E-05 2.0E-05 2.0E-05 2.83E-07 4.714E-07 1.568E-05 70.671378 70.671378 3.55E+05 0.004 -0.3359061 53.532416

final value 3.5E-06 0.7839904 0.3649802 -0.5 2.0E-05 2.0E-05 4.32E-06 4.32E-06 2.83E-07 4.714E-07 1.568E-05 9.1638708 9.1638708 3.55E+05 0.004 -0.3359061 33.259571

[2] FA0 = 2e-5 [4] Fa = FA0*(1-X) [6] v0 = 2.83e-7 [8] Fc = FA0*X [10] Cb = Fb/v [12] k = .004 [14] Cc = Fc/v

P5-21 (a)

P5-21 (b) The outlet conversion of the reactor is 0.784 The yield is then MW*FA*X = 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 g/s = 48.95 g/ year. Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed. 5-32



P5-21 (c) Assuming laminar flow, α ~ Dp-2, therefore



P5-21 (d) A lower conversion is reached due to equilibrium. Also, the reverse reaction begins to overtake the forward reaction near the exit of the reactor. P5-21 (e) Individualized solution P5-21 (f) Individualized solution P5-21 (g) Individualized solution P5-22 No solution necessary P5-23 (a) P5-23 (b) Elementary with respect to partial pressure, so the rate law is:



P5-23 (c) Feed is pure A, so

.



5-33

P5-23 (c) Continued Therefore

There is no change in P or T, and = 0, so Stoichiometric Table: Species Methylamine Dimethylamine Ammonia

Symbol A B C



Initial FA0 0 0

Change - FA0X 0.5 FA0X 0.5 FA0X

P5-23 (d) By ideal gas law & Stoichiometric table:







P5-23 (e) Substituting for the partial pressures:



Finally,

P5-23 (f) At equilibrium,















, so



Therefore:

P5-23 (g) Mole Balance for PBR:









Rate Law: 5-34

Remaining FA0(1-X) 0.5 FA0X 0.5 FA0X

P5-23 (g) Continued Stoichiometry: Feed is pure A, so



& A is the limiting reactant;

Therefore

There is no change in P or T, so Therefore:



;

Combine:

Separate and integrate: Let’s look at the left hand side first:



Notice that the equation is now in quadratic form, where





The solution to the integral is, from Appendix A (equation A-12): Where p and q are the roots of the quadratic equation: p, q =



So







Now we move the constants from the right hand side to the left:

5-35

P5-23 (g) Continued So







(*)

P5-23 (h) We want the catalyst weight at X = 0.9 XeXe = 0.684 Using equation (*), we have:

P5-23 (i) For a fluidized bed reactor (CSTR), when X = 0. 9 XeXe = 0.684:





We can put smaller CSTRs in series to reduce the total catalyst weight. P5-23 (j) Since







It can be easily seen that





Pressure fall below 1 atm:



5-36



P5-23 (k)

Pressure fall below 1 atm:

P5-24















1. Mole Balance – Use Differential Form dX −rA = dW FA0



2. Rate Law – Psuedo Zero Order in B



−rA = kCACB0 = kCA

" T %" P % 3. Stoichiometry – Gas: v = v0 1+εX $$ ''$$ 0 '' Isothermal, therefore T = T0 # T0 &# P &

(

(

)

) )

( (

) )

FA0 1− X P 1− X P F 0 =C CA = A = p**,**p = 0 A0 v v 1+εX P P 1+εX 0

(



$1' ε = yA0δ = & ) 1+1−1−1 = 0 %2(



dy α = where W = 0 then p = 1 and X = 0 dW 2p

( CA = CA0 (1− X)p

)



5-37

P5-24 Continued

Integrating



p = 1− αW

(

12

)



Part (a) Now since P/P0 >0.1 . So p= .1 for minimum value of W Now (1 – αW)1/2 = .1 => W = 99 kg 4. Combine



( ) ( 12 dX kCA0 = 1− X) (1− αW) ( dW F

)(

−rA = kCA = kCA0 1− X p = kCA0 1− X 1− αW

12

)



A0

Separate Variables



when W = 0, X = 0











5-38

P5-24 Continued W

Y

1/y

X

P

0

1

1

0

10 atm

1

1

1

0.47



2.5

0.99

1.01

0.11



5

0.975

1.02

0.21



15

0.92

1.08

0.5



25

0.87

1.16

0.67



50

0.71

1.41

0.87



75

0.5

2.0

0.94



90

0.32

3.16

0.955



99

0.1

10

0.96

1 atm

X = 0.9





W = 57g for 5% W = 1.05g for last 5% (85 to 90%) W = 57 – 45 = 12 g

Ratio =

= 11.4

Polymath







dX = −rprime FA0 dW dy = −alpha 2 p dW





(

)

CA = CA0 ∗ 1− X p



k = 1.2dm3 g⋅min alpha = 0.01g−1

See Polymath program P5-24.pol.

5-39

P5-24 Continued

p

p

p





5-40

P5-24 Continued



P5-25

δ = 0 ","ε = 0

(

∴p = 1− αW

1 2

)

Mole Balance/Design Equation

5-41

P5-25 Continued Rate Law Stoichiometry

( )

CA = CA0 1− X p Combining

FA0

2 dX = kC2A0 1− X p2 dW

(

)



P5-26 Individualized solution

5-42

Solutions for Chapter 6 – Isothermal Reactor Design- Molar Flow rates P6-1 (a) Example 6-1 (i) The conversion increases with an increase in both E and T. This is because the rate constant is higher for higher E and T. (ii) For pressure doubled and temperature decrease by 20 C, CTO = 2*Po/RT and T = 678K See Polymath program P6-1-aii.pol.

POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.5822357

0.0292158

0.5822357

0.0292158

2

Cto

0.5822357

0.5822357

0.5822357

0.5822357

3

E

2.4E+04

2.4E+04

2.4E+04

2.4E+04

4

Fa

2.26E-05

1.659E-06

2.26E-05

1.659E-06

5

Fao

2.26E-05

2.26E-05

2.26E-05

2.26E-05

6

Fb

0

0

2.094E-05

2.094E-05

7

Fc

0

0

1.047E-05

1.047E-05

8

Ft

2.26E-05

2.26E-05

3.307E-05

3.307E-05

9

k

164.7179

164.7179

164.7179

164.7179

10 ra

-55.83911

-55.83911

-0.1405976

-0.1405976

11 rateA

55.83911

0.1405976

55.83911

0.1405976

12 rb

55.83911

0.1405976

55.83911

0.1405976

13 rc

27.91956

0.0702988

27.91956

0.0702988

14 T

678.

678.

678.

678.

15 Tau

0

0

0.2576264

0.2576264

16 V

0

0

1.0E-05

1.0E-05

17 vo

3.882E-05

3.882E-05

3.882E-05

3.882E-05

18 X

0

0

0.9265741

0.9265741

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc

6-1

Explicit equations 1

T = 678

2

Cto = 2*1641/8.314/T

3

E = 24000

4

Ft = Fa+Fb+Fc

5

Ca = Cto*Fa/Ft

6

k = 0.29*exp(E/1.987*(1/500-1/T))

7

Fao = 0.0000226

8

vo = Fao/Cto

9

Tau = V/vo

10 ra = -k*Ca^2 11 X = 1-Fa/Fao 12 rb = -ra 13 rc = -ra/2 14 rateA = -ra General Total number of equations

17

Number of differential equations 3 Number of explicit equations

14

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

(iii) See Polymath program P6-1-aiii.pol. POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.2827764

0.1426944

0.2827764

0.1426944

2

Cb

0

0

0.093388

0.093388

3

Cc

0

0

0.046694

0.046694

4

Cto

0.2827764

0.2827764

0.2827764

0.2827764

5

E

2.4E+04

2.4E+04

2.4E+04

2.4E+04

6

Fa

2.26E-05

1.366E-05

2.26E-05

1.366E-05

7

Fao

2.26E-05

2.26E-05

2.26E-05

2.26E-05

8

Fb

0

0

8.94E-06

8.94E-06

9

Fc

0

0

4.47E-06

4.47E-06

10 Ft

2.26E-05

2.26E-05

2.707E-05

2.707E-05

11 k

274.4284

274.4284

274.4284

274.4284

12 Kc

0.02

0.02

0.02

0.02

13 ra

-21.94397

-21.94397

-1.497E-10

-1.497E-10

14 rateA

21.94397

1.497E-10

21.94397

1.497E-10

15 rb

21.94397

1.497E-10

21.94397

1.497E-10

6-2

16 rc

10.97199

7.487E-11

10.97199

7.487E-11

17 T

698.

698.

698.

698.

18 Tau

0

0

0.1251223

0.1251223

19 V

0

0

1.0E-05

1.0E-05

20 vo

7.992E-05

7.992E-05

7.992E-05

7.992E-05

21 X

0

0

0.3955739

0.3955739

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc Explicit equations 1

T = 698

2

Cto = 1641/8.314/T

3

E = 24000

4

Kc = 0.02

5

Ft = Fa+Fb+Fc

6

Ca = Cto*Fa/Ft

7

Fao = 0.0000226

8

vo = Fao/Cto

9

Tau = V/vo

10 Cb = Cto*Fb/Ft 11 X = 1-Fa/Fao 12 k = 0.29*exp(E/1.987*(1/500-1/T)) 13 Cc = Cto*Fc/Ft 14 ra = -k*(Ca^2 - Cb^2*Cc/Kc) 15 rb = -ra 16 rc = -ra/2 17 rateA = -ra General Total number of equations

20

Number of differential equations 3 Number of explicit equations

17

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

6-3



(iv)

!

𝑝 = 1 − ∝! 𝑉 ! 𝐶! =

𝐶!! 𝐹! 𝑝𝑇! 𝐹! 𝑇

For isothermal operation, 1 − 99000𝑉 𝐶! = 𝐹!

! !

𝐶!! 𝐹!



See Polymath program P6-1-aiv.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.2827764

0.005108

0.2827764

0.005108

2

Cto

0.2827764

0.2827764

0.2827764

0.2827764

3

E

2.4E+04

2.4E+04

2.4E+04

2.4E+04

4

Fa

2.26E-05

5.616E-06

2.26E-05

5.616E-06

5

Fao

2.26E-05

2.26E-05

2.26E-05

2.26E-05

6

Fb

0

0

1.698E-05

1.698E-05

7

Fc

0

0

8.492E-06

8.492E-06

8

Ft

2.26E-05

2.26E-05

3.109E-05

3.109E-05

9

k

274.4284

274.4284

274.4284

274.4284

10 p

1.

0.1

1.

0.1

11 ra

-21.94397

-21.94397

-0.0071603

-0.0071603

12 rateA

21.94397

0.0071603

21.94397

0.0071603

13 rb

21.94397

0.0071603

21.94397

0.0071603

14 rc

10.97199

0.0035801

10.97199

0.0035801

15 T

698.

698.

698.

698.

6-4

16 Tau

0

0

0.1251223

0.1251223

17 V

0

0

1.0E-05

1.0E-05

18 vo

7.992E-05

7.992E-05

7.992E-05

7.992E-05

19 X

0

0

0.7514895

0.7514895

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc Explicit equations 1

T = 698

2

Cto = 1641/8.314/T

3

E = 24000

4

p = (1 - 99000*V)^0.5

5

k = 0.29*exp(E/1.987*(1/500-1/T))

6

Ft = Fa+Fb+Fc

7

Fao = 0.0000226

8

vo = Fao/Cto

9

Tau = V/vo

10 Ca = p*Cto*Fa/Ft 11 X = 1-Fa/Fao 12 ra = -k*Ca^2 13 rb = -ra 14 rc = -ra/2 15 rateA = -ra General Total number of equations

18

Number of differential equations 3 Number of explicit equations

15

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001



6-5

P6-1 (b) Example 6-2 (i) The conversion increases with increase in KC, k, CT0, and kc. This is because in each of the cases, the rate of consumption of A increases. (ii) Same as (i) (iii) Individualized solution (iv) RB is maximum when KC, k, CT0, and kc are at their maximum values (v) Individualized solution (vi) See Polymath program P6-1-b.pol. POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Cto

0.2

0.2

0.2

0.2

2

Fa

10.

3.338718

10.

3.338718

3

Fao

10.

10.

10.

10.

4

Fb

0

0

3.687651

3.421463

5

Fc

0

0

6.661282

6.661282

6

Ft

10.

10.

13.68765

13.42146

7

k

0.7

0.7

0.7

0.7

8

kc

0.2

0.2

0.2

0.2

9

Kc

0.05

0.05

0.05

0.05

10 p

1.

0

1.

0

11 ra

-0.14

-0.14

0

0

12 Rate

0.14

0

0.14

0

13 Rb

0

0

0.0107766

0.010197

14 V

0

0

500.

500.

15 X

0

0

0.6661282

0.6661282

6-6



Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = -ra-kc*Cto*(Fb/Ft)*p 3 d(Fc)/d(V) = -ra Explicit equations 1

p = (1 - 0.002*V)^0.5

2

Ft = Fa+Fb+Fc

3

k = 0.7

4

Cto = 0.2

5

Kc = 0.05

6

kc = 0.2

7

ra = -k*Cto*((Fa/Ft)*p - Cto/Kc*(Fb/Ft)*(Fc/Ft)*p*p)

8

Rate = -ra

9

Rb = kc*Cto*(Fb/Ft)

10 Fao = 10 11 X = (Fao-Fa)/Fao General Total number of equations

14

Number of differential equations 3 Number of explicit equations

11

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

The conversion goes up from 60% to 66.6% (vii) Individualized solution (viii) Individualized solution

6-7

P6-1 (c) Example 6-2 See Polymath program P6-1-c.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Cto

0.2

0.2

0.2

0.2

2

Fa

10.

0.4432696

10.

0.4432696

3

Fao

10.

10.

10.

10.

4

Fb

0

0

6.058466

5.126694

5

Fc

0

0

9.55673

9.55673

6

Ft

10.

10.

16.05847

15.12669

7

k

0.7

0.7

0.7

0.7

8

kc

0.2

0.2

0.2

0.2

9

Kc

0.05

0.05

0.05

0.05

10 p

1.

0

1.

0

11 ra

-0.14

-0.14

0

0

12 Rate

0.14

0

0.14

0

13 Rb

0

0

0.015091

0.0135567

14 V

0

0

500.

500.

15 X

0

0

0.955673

0.955673

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = -ra-kc*Cto*(Fb/Ft)*p 3 d(Fc)/d(V) = -ra Explicit equations 1

p = (1 - 0.002*V)^0.5

2

Ft = Fa+Fb+Fc

3

k = 0.7

4

Cto = 0.2

5

Kc = 0.05

6

kc = 0.2

7

ra = -k*Cto*(Fa/Ft)*p

8

Rate = -ra

9

Rb = kc*Cto*(Fb/Ft)

10 Fao = 10 11 X = (Fao-Fa)/Fao

6-8

General Total number of equations

14

Number of differential equations 3 Number of explicit equations

11

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

P6-1 (d) Example 6-3 See Polymath program P6-1-d.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.05

0.0052864

0.05

0.0052864

2

Cao

0.05

0.05

0.05

0.05

3

Cb

0

0

0.0177864

0.0177864

4

Cbo

0.025

0.025

0.025

0.025

5

Cc

0

0

0.0041087

0.0030469

6

Cd

0

0

0.0041087

0.0030469

7

k

2.2

2.2

2.2

2.2

8

Kc

0.1

0.1

0.1

0.1

9

ra

0

-0.0001295

0

-2.62E-06

10 rate

0

0

0.0001295

2.62E-06

11 t

0

0

500.

500.

12 V

5.

5.

30.

30.

13 Vo

5.

5.

5.

5.

14 vo

0.05

0.05

0.05

0.05

15 X

0

0

0.3656277

0.3656277

Differential equations 1 d(Ca)/d(t) = ra- vo*Ca/V 2 d(Cb)/d(t) = ra+ (Cbo-Cb)*vo/V 3 d(Cc)/d(t) = -ra-vo*Cc/V 4 d(Cd)/d(t) = -ra-vo*Cd/V Explicit equations 1

vo = 0.05

2

Vo = 5

3

V = Vo+vo*t

4

k = 2.2

5

Cbo = 0.025

6-9

6

Kc = 0.1

7

ra = -k*(Ca*Cb - Cc*Cd/Kc)

8

Cao = 0.05

9

rate = -ra

10 X = (Cao*Vo-Ca*V)/(Cao*Vo)

General Total number of equations

14

Number of differential equations 4 Number of explicit equations

10

Elapsed time

0.000 sec

Solution method

RKF_45

Step size guess. h

0.000001

Truncation error tolerance. eps 0.000001

On comparison with the solution obtained from the irreversible case, we find that the conversion after t = 500 s drops down from 99.9% to about 36%. This is an expected result as the reverse reaction also takes place. P6-1 (e) Example 6-4 CA0 = 0.042 for exit pressure drop to go to atmospheric pressure P6-1 (f) Example 6-5 Individualized solution P6-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. P6-3 Polymath d(Cb) / d(t) = rb Cb(0) = 2 d(Ca) / d(t) = ra Ca(0) = 2 d(Cc) / d(t) = rc Cc(0) = 0 t(0) = 0 t(f) = 100 k = 0.0445 Ca0 = 2 ra = -k*Ca*Cb rb = ra rc = -ra X = (Ca0-Ca)/Ca0

The conversion after 10, 50, 100 minutes are: At 10 mins: X = 0.47

6-10

P6-3 Continued At 50 mins: X = 0.82 At 100 mins: X = 0.9 For reactor 1: Where

, then

, so

For reactor 2: Where

, so

, then

Therefore the design equation and concentration profiles are the same as in part a) for the two reactors. As they have the same concentration profile at any time, the conversions are the same for the two reactors as in part a). At 10 mins: X = 0.47, Ca = Cb = 1.06 mol/dm3, Cc = 0.94 mol/dm3. At 50 mins: X = 0.82, Ca = Cb = 0.37 mol/dm3, Cc = 1.63 mol/dm3. At 80 mins: X = 0.9, Ca = Cb = 0.25 mol/dm3, Cc = 1.75 mol/dm3. At 10 mins: X = 0.47, Ca = Cb = 1.06 mol/dm3, Cc = 0.94 mol/dm3. At 50 mins: X = 0.82, Ca = Cb = 0.37 mol/dm3, Cc = 1.63 mol/dm3. The conversion in the two reactors are the same. At 50 mins: X = 0.82 6-11

P6-4 (a)

1

P6-4 (b)

P6-5









6-12

P6-5 Continued

Using Polymath to solve the differential equation gives a volume of 290 dm3 See Polymath program P6-5-c.pol. PFR with pressure drop: Alter the Polymath equations from part (c). See Polymath program P6-5-c-pressure.pol.







6-13

P6-5 continued Calculated values of the DEQ variables Variable v x y Kc alfa Cao k esp fo r

initial value 0 0 1 0.025 0.001 0.3 0.044 2 2.5 -0.0132

minimal value 0 0 0.3181585 0.025 0.001 0.3 0.044 2 2.5 -0.0132

maximal value 500 0.5077714 1 0.025 0.001 0.3 0.044 2 2.5 -1.846E-04

final value 500 0.5077714 0.3181585 0.025 0.001 0.3 0.044 2 2.5 -1.846E-04



ODE Report (STIFF) Differential equations as entered by the user [1] d(x)/d(v) = -r/fo [2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y)

Explicit equations as entered by the user [1] Kc = .025 [2] alfa = 0.001 [3] Cao = 0.3 [4] k = 0.044 [5] esp = 2 [6] fo = 2.5 [7] r = -(k*Cao/(1+esp*x))*(4*Cao^2*x^3/((1+esp*x)^2*Kc))



At V = 500 dm3 X = 0.507 and y = 0.381 Note: In the Polymath Program y = P = y = p = P6-6 (a)

P P0



To plot the flow rates down the reactor we need the differential mole balance for the three species, noting that BOTH A and B diffuse through the membrane

Next we express the rate law: First-order reversible reaction

6-14

P6-6 (a) Continued Rename Transport out the sides of the reactor: RA = kACA =





RB = kBCB =



Stoichiometery: –rA = rB =1/2 rC Combine and solve in Polymath code: See Polymath program P6-6-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable v Fa Fb Fc Kc Ft Co K Kb ra Ka Ra Rb Fao X

initial value 0 100 0 0 0.01 100 1 10 40 -10 1 1 0 100 0

minimal value 0 57.210025 0 0 0.01 100 1 10 40 -10 1 0.472568 0 100 0

maximal value 20 100 9.0599877 61.916043 0.01 122.2435 1 10 40 -0.542836 1 1 2.9904791 100 0.4278998

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(v) = ra - Ra [2] d(Fb)/d(v) = -ra - Rb [3] d(Fc)/d(v) = -2*ra

Explicit equations as entered by the user [1] Kc = 0.01 [2] Ft = Fa+ Fb+ Fc [3] Co = 1 [4] K = 10 [5] Kb = 40 [6] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2)) [7] Ka = 1 [8] Ra = Ka*Co*Fa/Ft [9] Rb = Kb*Co*Fb/Ft

6-15

final value 20 57.210025 1.935926 61.916043 0.01 121.06199 1 10 40 -0.542836 1 0.472568 0.6396478 100 0.4278998

P6-6 (a) Continued

P6-6 (b) The setup is the same as in part (a) except there is no transport out the sides of the reactor. See Polymath program P6-6-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable v Fa Fb Fc Kc Ft Co K ra Fao X

initial value 0 100 0 0 0.01 100 1 10 -10 100 0

minimal value 0 84.652698 0 0 0.01 100 1 10 -10 100 0

maximal value 20 100 15.347302 30.694604 0.01 130.6946 1 10 -3.598E-09 100 0.153473

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(v) = ra [2] d(Fb)/d(v) = -ra [3] d(Fc)/d(v) = -2*ra

Explicit equations as entered by the user [1] Kc = 0.01 [2] Ft = Fa+ Fb+ Fc [3] Co = 1 [4] K = 10 [5] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))



6-16

final value 20 84.652698 15.347302 30.694604 0.01 130.6946 1 10 -3.598E-09 100 0.153473



P6-6 (b) Continued



P6-6 (c) Conversion would be greater if C were diffusing out. P6-6 (d) Individualized solution P6- 7 (a)

Assuming catalyst distributed uniformly over the whole volume Mole balance:







P6- 7 (a) Rate law:









Stoichiometry: Solving in polymath: See Polymath program P6-7.pol.



6-17





P6-7 (a) Continued POLYMATH Results Calculated values of the DEQ variables Variable W Fa Fb Fc Fd Keq Ft Cto Ca Cb Kh Cc Cd Rh k r

initial value 0 2 2 0 0 1.44 4 0.4 0.2 0.2 0.1 0 0 0 1.37 -0.0548

minimal value 0 0.7750721 0.7750721 0 0 1.44 3.3287437 0.4 0.0931369 0.0931369 0.1 0 0 0 1.37 -0.0548

maximal value 100 2 2 1.2249279 0.7429617 1.44 4 0.4 0.2 0.2 0.1 0.147194 0.0796999 0.00797 1.37 -0.002567

final value 100 0.7750721 0.7750721 1.2249279 0.5536716 1.44 3.3287437 0.4 0.0931369 0.0931369 0.1 0.147194 0.0665322 0.0066532 1.37 -0.002567



ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(W) = r [2] d(Fb)/d(W) = r [3] d(Fc)/d(W) = -r [4] d(Fd)/d(W) = - r -Rh

Explicit equations as entered by the user [1] Keq = 1.44 [2] Ft = Fa+Fb+Fc+Fd [3] Cto = 0.4 [4] Ca = Cto*Fa/Ft [5] Cb = Cto*Fb/Ft [6] Kh = 0.1 [7] Cc = Cto*Fc/Ft [8] Cd = Cto*Fd/Ft [9] Rh = Kh*Cd [10] k = 1.37 [11] r = -k*(Ca*Cb-Cc*Cd/Keq)

For 85% conversion, W = weight of catalyst = 430 kg P6-7 (b) In a PFR no hydrogen escapes and the equilibrium conversion is reached.

solve this for X, X = .5454 This is the maximum conversion that can be achieved in a normal PFR. 6-18

P6-7 (c) If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of .459

P6-8 Suppose the volumetric flow rate could be increased to as much as 6,000 dm3/h (9,000 mol/h) and the total time to fill, heat, empty and clean is 4.5 hours. What is the maximum number of moles of ethylene glycol (CH2OH)2 you can make in one 24 hour period? The feed rate of ethylene chlorohydrin will be adjusted so that the volume of fluid at the end of the reaction time will be 2500 dm3. Now suppose CO2 leaves the reactor as fast as it is formed.

A+ B → C + D +CO2 Mole Balance

dNA



dt dNB dt dNC

= rAV = FB0 + rBV



= rCV dt ND = NC Overall Mass Balance Accumulation = In – Out

dm  CO = ρυ0 − m 2 dt m = ρV !Assume!constant!density  CO m dV 2 = υ0 − dt ρ

The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).  m = F •MWCO CO2



CO2

2

FCO •MWCO2 dV = υ0 − 2 = υ0 − υCO 2 dt ρ FCO •MWCO2 υCO = 2 2 ρ 6-19

P6-8 Continued Rate Law and Relative Rates −rA = kC ACB

rB = rA



rC = −rA Stoichiometry



N CA = A V NB CB = V







6-20



P6-8 continued



P6-8(c)



FA0 – 0.15 mol/min = 9 mol/hr vo = Fao/cao = (9 mol/hr) / (1.5 mol/dm3) = 6 dm3/hr 1000 dm3 is needed to fill the reactor. At 6 dm3/hr it will take 166.67 hours Now solving using the code form part (a) with the changed equations: See Polymath program P6-9

(

2

)(

)

π D2L π 1.5m 2.5m VT = = 4 4 3 VT = 4.42m = 4,420)dm3

We don’t know V0 or v0. First try equal number of moles of A and B added to react.

6-21



P6-9 continued

NA = NB C A0V0 = CB0 ΔV = CB0 #$VT −V0 %& V0 =



VT

C A0 CB0

= +1

4420 = 2456 0.8 +1.0

ΔV = 1964

υ0 =

ΔV tR

k0 = 0.000052*dm3 mol s = 0.187*dm3 mol h For 1 batch tR = 24 – 3 = 21



υ0 =

VT −V0 21



(1) For one batch we see that only 198 moles of C are made so one batch will not work. (2) For two batches, we have a down time of 2 × 3 = 6 h and therefore each batch has a reaction time of 18h/2 = 9 h. We see that 106 moles of C are made in one batch therefore 2 × 106 = 212 moles/day are made in two batches.









6-22



P6-9 continued P6-9 Two Batches

P6-9 Two Batches



P6-10 Individualized solution

6-23

P6-11 (a)

C C At equilibrium, r = 0 => C ACB = C D KC

V = V0 + vot CA =

(

NAO 1− X V

) = C AO (1− X ) = C AO (1− X ) V VO

" v % $1+ o t ' $ V ' # O &

v CBO O t −C AO X VO CB = " v % o $1+ t ' $ V ' # O &

C AO X CC = C D = ! v $ o #1+ t & # V & " O %

(

" % C X v C AO 1− X $$CBO O t −C AO X '' = AO VO KC # &

(

)

2

)



" % VOC AO $ X 2 t= + X' ' vOCBO $ K 1− X # C &

(

)

See Polymath program P6-11-a.pol. Solving in Polymath POLYMATH Report Nonlinear Equation Calculated values of NLE variables

Variable Value f(x) Initial Guess 0 0 0.495 ( 0 < x < 0.99 )

1 x

Nonlinear equations

1 f(x) = 2825.25*(x^2/1.08/(1-x)+x) = 0

6-24

P6-11 (a) continued



If we solve in Excel 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 0.99

0 148.1466374 311.591358 493.0338235 695.8486111 924.3101852 1183.914286 1481.84765 1827.692593 2234.515909 2720.611111 3312.40216 4049.525 4994.264683 6250.42963 8004.875 10631.31111 15001.7287 23732.1 49902.28611 259188.435



6-25

P6-11 (a) continued X t (sec)



P6-11 (b) See Polymath program P6-11-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb Cc Cd Kc k ra vo Vo V X

initial value 0 7.72 10.93 0 0 1.08 9.0E-05 -0.0075942 0.05 200 200 0

minimal value 0 0.2074331 7.6422086 0 0 1.08 9.0E-05 -0.0075942 0.05 200 200 0

maximal value 1.5E+04 7.72 10.93 3.2877914 3.2877914 1.08 9.0E-05 -1.006E-05 0.05 200 950 0.9731304

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93) [3] d(Cc)/d(t) = -ra - vo*Cc/V [4] d(Cd)/d(t) = -ra - vo*Cd/V

Explicit equations as entered by the user [1] Kc = 1.08 [2] k = 0.00009 [3] ra = -k*(Ca*Cb - Cc*Cd/Kc) [4] vo = 0.05 [5] Vo = 200 [6] V = Vo + vo*t [7] X = 1 - Ca/7.72

6-26

final value 1.5E+04 0.2074331 9.51217 1.41783 1.41783 1.08 9.0E-05 -1.006E-05 0.05 200 950 0.9731304

P6-11 (b) continued

Polymath solution P6-11 (c) As ethanol evaporates as fast as it forms: Now using part (b) remaining equations, Polymath code: See Polymath program P6-11-c.pol.

CD = 0

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k ra vo Vo V X

initial value 0 7.72 10.93 9.0E-05 -0.0075942 0.05 200 200 0

minimal value 0 0.0519348 6.9932872 9.0E-05 -0.0075942 0.05 200 200 0

maximal value 6000 7.72 10.93 9.0E-05 -3.69E-05 0.05 200 500 0.9932727



ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*vo/V [2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)

Explicit equations as entered by the user [1] k = 0.00009 [2] ra = -k*Ca*Cb [3] vo = 0.05 [4] Vo = 200 [5] V = Vo + vo*t [6] X = 1 - Ca/7.72



6-27

final value 6000 0.0519348 7.8939348 9.0E-05 -3.69E-05 0.05 200 500 0.9932727

P6-11 (c) continued



P6-11 (d) Change the value of vo, CAO, etc. in the Polymath program to see the changes. P6-11 (e) Individualized solution P6-11 (f) Individualized solution P6-12 (a) No solution will be given P6-12 (b) No solution will be given P6-13 Reaction:

𝐴 ⇄ 𝐵 with

!

𝑘! = 𝑘 = 0.4

𝐾! = 4

!

Reactor - mass balance (1): 𝑚!,!" = 𝐶!! ∙ 𝑣! + 𝐶!! ⋅ 𝑣! = 100 ⋅ (12 + 𝑣! ) Reactor - component A balance (2): input = output + reaction; 𝑚!,!" = 𝑚!,!"# + (−𝑟! 𝑉) Rate law in the reaction term (3): −𝑟! 𝑉 = 𝑘 𝐶! −

!! !!

Stoichiometry (4): 𝐶! = 𝐶!! − 𝐶! 6-28

𝑉

P6-13 Continued Combine (3) and (4), substitute for numerical values of CA0, k and KC to get (5): −𝑟! 𝑉 = 30𝐶! − 600 Combine (5) and (1) with (2) and evaluate numerically to get: 𝑐! = 𝑚!,!"# (𝑣! ) = 𝐶! ⋅ 𝑣! + 𝑣! =

!"##!!""!! !"!!!

; 𝑐! = 100 − 𝑐! =

2400(12 + 𝑣! ) 42 + 𝑣!

𝑃𝑟𝑜𝑓𝑖𝑡 𝑣! = 𝐼𝑛𝑐𝑜𝑚𝑒 𝑣! − 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐶𝑜𝑠𝑡𝑠(𝑣! ) 𝑃𝑟𝑜𝑓𝑖𝑡 𝑣! = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑜𝑓 𝐵 × 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐵 −

𝑐𝑜𝑠𝑡 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 × 𝑣𝑜𝑙𝑢𝑚. 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑚!

𝑃 𝑣! = 𝑚!,!"# (𝑣! )×2 − 50×(𝑣! + 𝑣! ) 𝑃 𝑣! =

2400(12 + 𝑣! ) ×2 − 50×(12 + 𝑣! ) 42 + 𝑣!

Optimization: extremum (maximum) in profit when the derivative 𝑑𝑃 𝑣! /𝑑𝑣! is zero: 𝑑𝑃 𝑣! 4800 ⋅ 30 = − 50 ⇒ 𝑣! = 11.6656 𝑚 ! /ℎ 𝑑𝑣! 42 + 𝑣! ! Into the definition of the recycle ratio, f: 𝑓=

𝑣! = 0.89 𝑣! + 𝑣! − 𝑣!

Where 𝑣! is calculated from the overall mass balance of A : 𝑣! + 𝑣! 𝑐! = 𝑣! + 𝑣! − 𝑣! ⋅ 𝑐!! ⇒ 𝑣! = 10.58 𝑚 ! /ℎ

6-29

!"## !"!!!



Solutions for Chapter 7 – Collection and Analysis of Rate Data P7-1 (a) Example 7-4 α rate law: rCH = kPCO PHβ2 4

Regressing the data r’(gmolCH4/gcat.min) 5.2e-3 13.2e-3 30e-3 4.95e-3 7.42e-3 5.25e-3

PCO (atm) 1 1.8 4.08 1 1 1

PH2 (atm) 1 1 1 0.1 0.5 4

See Polymath program P7-1-a.pol. POLYMATH Results Nonlinear regression (L-M) Model: r = k*(PCO^alfa)*(PH2^beta) Variable Ini guess Value k 0.1 0.0060979 alfa 1 1.1381455 beta 1 0.0103839 Precision R^2 = 0.9869709 R^2adj = 0.9782849 Rmsd = 4.176E-04 Variance = 2.093E-06

95% confidence 6.449E-04 0.0850634 0.1951217

Therefore order of reaction = 1.14 Again regressing the above data putting β = 1 POLYMATH Results Nonlinear regression (L-M) Model: r = k*(PCO^0.14)*(PH2) Variable Ini guess Value k 0.1 0.0040792 Precision R^2 = -0.8194508 R^2adj = -0.8194508 Rmsd = 0.0049354 Variance = 1.754E-04

95% confidence 0.0076284



Therefore, k = 0.004 (gmolCH4/(gcat.min.atm1.14)) P7-2 Individualized solution P7-3 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. 7-1

P7-4 (a) For the reaction, HbO2 → Hb + O2 Limiting Reactant A, or HbO2 Reactor Type tubular Dependent Variable X Design Equation dFA/dV = -γA Rate Equation Assume power law, -rA = kCAn Stoichiometry FA = voCA, dV = ACdz Combine dCA/dz = -(kAC /vo)CAn Assume the reaction 1st order, n = 1.0. Integrate with initial conditions X(z=0) = 0, we get Ln(1/(1-X)) = (kAC /vo)z Electrode Position Position (cm) Conversion(XA) ln(1/(1-XA))

1

2

3

4

5

6

7

0 0 0

5 0.0193 0.0195

10 0.0382 0.0389

15 0.0568 0.0585

20 0.0748 0.0777

25 0.0925 0.0971

30 0.110 0.1165

Plot ln(1/(1-XA) vs. z, we find a perfect linear relationship 0.14 0.12

ln(1/

0.1

0.08 y = 0.0039x + 9E-05 R² = 1

0.06 0.04 0.02 0 0

10

20

30

40

z (cm)

Slope = kAC /vo = 0.039 AC = πd2 = 0.0196 cm2 k = 0.039 vo/AC = (0.039 cm-1)X 19.6 cm3/s/(0.0196cm2) k = 3.9 s-1 Hence, the rate law is, –rA = 3.9CA mol/L/s P7-4 (b) First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z) See Polymath program P7-4-b.pol.

7-2

P7-4 (b) Continued POLYMATH Results Polynomial Regression Report Model: X = a0 + a1*z + a2*z^2 + a3*z^3 + a4*z^4 + a5*z^5 + a6*z^6 Variable a0 a1 a2 a3 a4 a5 a6

Value 2.918E-14 0.0040267 -6.14E-05 7.767E-06 -5.0E-07 1.467E-08 -1.6E-10

95% confidence 0 0 0 0 0 0 0

Next we differentiate our expression of X(z) to find dX/dz and knowing that

⎛ dX ⎞ ln ⎜ ⎟ = ln a + n ln (1 − X ) ⎝ dz ⎠ n kC Ao Ac where, a = FAo

⎛ dX ⎝ dz

Linear regression of ln ⎜

⎞ ⎟ as a function of ln (1 − X ) gives us similar vaules of slope and intercept ⎠

as in the finite differences. POLYMATH Results Linear Regression Report Model: ln(dxdz) = a0 + a1*ln(1-X) Variable a0 a1

Value -5.531947 1.2824279

95% confidence 0.0241574 0.3446187

n = 1.28 ln a = -5.53, a = 0.00396

k=

FAO a n C AO AC

=

45.7 ×10−6 moles / s 3.96 ×10−3 cm = 4.0s−1 2.3×10−6 moles / cm3 × 0.0196cm2

(

Hence rate law is,

−rA = 4.0C 1.28 A

mol dm3 is



7-3

)

P7-5 Constant volume batch reactor: A! B Mole balance:

dC A = kC Aα dt

−

Integrating with initial condition when t = 0 and CA = CAO for α ≠ 1.0 (1−α ) − CA(1−α ) ……………substituting for initial concentration C 1 CAO 1 (2)(1−α ) − CA(1−α ) 3 AO = 4 mol/dm . t= = k (1 − α ) k (1 − α )

CA (mol/dm3) 4 2.25 1.45 1 0.65 0.25 0.06 0.008

t (min.) 0 5 8 10 12 15 17.5 20

See Polymath program P7-5.pol. POLYMATH Report Nonlinear Regression (L-M)

Model: t = ((4^(1-alpha))-Ca^(1-alpha))/(k*(1-alpha)) Variable Initial guess Value

95% confidence

alpha

2.

0.532479

0.034828

k

1.

0.1963504 0.0031774

Nonlinear regression settings Max # iterations = 64 Precision R^2

0.9989802

R^2adj 0.9988102 Rmsd

0.0699673

Variance 0.0522179 General Sample size 8 Model vars 2 Indep vars 1 Iterations

11

7-4

P7-5 Continued Source data points and calculated data points Ca

t

t calc

Delta t

1 4

0

0

0

2 2.25

5

4.912288 0.0877122

3 1.45

8

7.867621 0.1323793

4 1

10

9.934279 0.0657208

5 0.65

12

11.92141 0.0785877

6 0.25

15

15.13018 -0.1301777

7 0.06

17.5 17.90411 -0.4041085

8 0.008 20

19.688

0.3120008

Hence, k= 0.2 (mol/dm3)0.5min-1 and α = 0.5 P7-6 (a) Liquid phase irreversible reaction: A ! B + C ; CAO = 2 mole/dm3

C AO − C A

τ

= kC αA

⎛ C − CA ⎞ ln ⎜ AO ⎟ = ln k + α ln C A τ ⎝ ⎠ Space time ( τ )min. 15 38 100 300 1200

CA(mol/dm3) 1.5 1.25 1.0 0.75 0.5

ln(CA) 0.40546511 0.22314355 0 -0.28768207 -0.69314718

ln((CAO-CA)/ τ ) -3.4011974 -3.9252682 -4.6051702 -5.4806389 -6.6846117

By using linear regression in polymath: See Polymath program P7-5-a.pol.

7-5

P7-6 (a) Continued POLYMATH Results Linear Regression Report Model: y = a0 + a1*lnCa

⎛ C − CA ⎞ ln ⎜ AO ⎟ = ln k + α ln C A τ ⎝ ⎠ Variable Value 95% confidence a0 -4.6080579 0.0162119 a1 2.9998151 0.0411145 Statistics R^2 = 0.9999443 R^2adj = 0.9999258 Rmsd = 0.003883 Variance = 1.256E-04

Hence,

α = slope ≡ 3 ln(k) = intercept = -4.6 therefore, k = 0.01 mole-2min-1. Rate law: −

dC A = 0.01C A3 mol / dm3 min dt

P7-6 (b) Individualized solution P7-6 (c) Individualized solution P7-7 (a) Constant voume batch reactor: A! B +C Mole balance:

−

dC A = kC Aα dt

Integrating with initial condition when t = 0 and CA = CAO for α ≠ 1.0

t=

(1−α ) − CA(1−α ) 1 (2)(1−α ) − CA(1−α ) 1 CAO ……………substituting for initial concentration CAO = 2 mol/dm3. = k (1 − α ) k (1 − α )

CA (mol/dm3) 2 1.6 1.35 1.1 0.87 0.70 0.53 0.35

t (min.) 0 5 9 15 22 30 40 60

See Polymath program P7-7-a.pol. 7-6

P7-7 (a) Continued POLYMATH Results Nonlinear regression (L-M) Model: t = (1/k)*((2^(1-alfa))-(Ca^(1-alfa)))/(1-alfa) Variable Ini guess Value 95% confidence k 0.1 0.0329798 3.628E-04 alfa 2 1.5151242 0.0433727 Precision R^2 = 0.9997773 R^2adj = 0.9997327 Rmsd = 0.1007934 Variance = 0.0995612

K= 0.03 (mol/dm3)-0.5min-1 and α = 1.5 P7-7 (b) Individualized solution P7-7 (c) Individualized solution P7-7 (d) Individualized solution P7-8 (a) At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure. P(∞) = 3P0 931 = 3P0 P0 ≈ 310 mm Hg P7-8 (b) Constant volume reactor at T = 504°C = 777 K Data for the decomposition of dimethylether in a gas phase: Time PT(mm Hg)

0 312

390 408

777 488

1195 562



(CH 3 ) 2 O → CH 4 + H 2 + CO

y A0 = 1 δ = 3 −1 = 2 ε = δ y A0 = 2 ⎛P ⎞ V = V0 ⎜ 0 ⎟ (1 + ε X ) = V0 because the volume is constant. ⎝P⎠ P = P0 (1 + ε X ) at t = ∞, X = XAF = 1

N dX 1 dN A = − A0 = rA V dt V0 dt 7-7

3155 799

∞ 931

P7-8 (b) Continued Assume −rA

= kCA (i.e. 1st order)

CA = CA0 (1 − X ) (V is constant) dX = kC A0 (1 − X ) dt P − P0 and X = ε P0 Then: C A0

Therefore:

dX 1 dP = dt ε P0 dt

⎡ P − P0 ⎤ 1 dP k = k ⎢1 − ([1 + ε ]P0 − P ) ⎥= ε P0 dt ε P0 ⎦ ε P0 ⎣ dP or = k ([1 + ε ]P0 − P ) dt P

t

dP ∫P [1 + ε ]P0 − P = ∫0 kdt 0

⎤ ⎡ 2 P0 ⎤ ε P0 ⎡ 624 ⎤ ⎥ = ln ⎢ ⎥ = ln ⎢ ⎥ = kt ⎣ 936 − P ⎦ ⎣ 3P0 − P ⎦ ⎣⎢ (1 + ε )P0 − P ⎦⎥ ⎡

Integrating gives: ln ⎢

Therefore, if a plot of ln

624 versus time is linear, the reaction is first order. From the figure below, 936 − P

we can see that the plot is linear with a slope of 0.00048. Therefore the rate law is:

−rA = 0.00048C A 1.6 y = 0.00048x - 0.02907

1.2 0.8 0.4 0 -0.4 0

1000

2000

3000

4000



7-8

P7-8 (c) Individualized solution P7-8 (d) The rate constant would increase with an increase in temperature. This would result in the pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true for colder temperatures. P7-9 (a) Photochemical decay of bromine in bright sunlight: t (min) CA (ppm)

10 2.45

20 1.74

30 1.23

40 0.88

50 0.62

60 0.44

Mole balance: constant V

dC A dt

= rA = −kC αA

" dC % ln$$ − A '' = ln k + αln C A # dt &

()

( )

Differentiation T (min) Δt (min) CA (ppm) ΔCA (ppm)



ΔC A ⎛ ppm ⎞ ⎜ ⎟ Δt ⎝ min ⎠

10

20 10

2.45

30 10

1.74

40 10

1.23

50 10

0.88

60 10

0.62

10 0.44

-0.71

-0.51

-0.35

-0.26

-0.18

-0.071

-0.051

-0.035

-0.026

-0.018

7-9

P7-9 (a) continued

After plotting and differentiating by equal area



-dCA/dt ln(-dCA/dt) ln CA

0.082 -2.501 0.896

0.061 -2.797 0.554

0.042 -3.170 0.207

0.030 -3.507 -0.128

0.0215 -3.840 -0.478

0.014 -4.269 -0.821

Using linear regression: α = 1.0 ln k = -3.3864 k = 0.0344 min-1 P7-9 (b)

dNA dt

= −VrA −FB = 0**,**FB = –rA V

rA = −0.0344

ppm mg at CA = 1 ppm = −0.0344 min l min

⎛ mg ⎞ ⎛ min ⎞ ⎛ 1 g ⎞⎛ 3.7851l ⎞⎛ 1lbs ⎞ lbs FB = (25000 gal )⎜ 0.0344 ⎟ ⎜ 60 ⎟⎜ ⎟⎜ ⎟ = 0.426 ⎟⎜ l min ⎠ ⎝ hr ⎠ ⎝ 1000 mg ⎠⎝ 1 gal ⎠⎝ 453.6 g ⎠ hr ⎝ P7-9 (c) Individualized solution P7-10 For the reaction,

Oz + wall ! loss of O3 k1



Oz + alkene ! products k2

Rate law:



−rO z =

dCO z dt

m n = k 1CO z + k 2CO C z Bu

Using polymath nonlinear regression we can find the values of m, n, k1 and k2 7-10

€

P7-10 Continued See Polymath program P7-10.pol.





7-11

P7-11 Given: Plot of percent decomposition of NO2 vs V/FA0

X=

%DecompositionofNO2

100 Assume that − rA = kC An For a CSTR V = or



FA0 X −rA

V X X = = n FA0 −rA kC A

with n = 0, X = k

V FA0

X has a linear relationship with

V as FA 0

shown in the figure. Therefore the reaction is zero order. P7-12



See Polymath program P7-12.pol





7-12

P7-12 Continued



7-13

Solutions for Chapter 8 – Multiple Reactions P8-1 (a) Example 8-1 (i) k1 affects the selectivity and conversion the most. (ii) No solution will be given. (iii) For PFR (gas phase with no pressure drop or liquid phase),

dC A = − k1 − k 2 C A − k 3 C A2 dτ dC B = k 2C A dτ









dC X = k1 dτ dC Y = k 3 C A2 dτ

In PFR with V = 1566 dm3 we get τ = V/v0 = 783 X = 0.958 , SB/XY (instantaneous) = 0.244, and SB/XY (overall) = 0.624

also at τ = 350, SB/XY (instantaneous) is at its maximum value of 0.84 See polymath problem P8-1-a.pol



Calculated values of the DEQ variables Variable initial value tau 0 Ca 0.4 Cx 1.0E-07 Cb 0 Cy 1.0E-06 Cao 0.4 X 0 k1 1.0E-04 k2 0.0015 k3 0.008 Sbxy_over 0 Sbxy_inst 0.4347826

minimal value 0 0.0166165 1.0E-07 0 1.0E-06 0.4 0 1.0E-04 0.0015 0.008 0 0.2438609

maximal value 783 0.4 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6437001 0.8385161

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2 [2] d(Cx)/d(tau) = k1 [3] d(Cb)/d(tau) = k2*Ca [4] d(Cy)/d(tau) = k3*Ca^2

8-1

final value 783 0.0166165 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6238731 0.2438609

P8-1 (a) Example 8-1 Continued Explicit equations as entered by the user [1] Cao = 0.4 [2] X = 1-Ca/Cao [3] k1 = 0.0001 [4] k2 = 0.0015 [5] k3 = 0.008 [6] Sbxy_overall = Cb/(Cx+Cy) [7] Sbxy_instant = k2*Ca/(k1+k3*Ca^2)

(iv) PFR - Pressure increased by a factor of 100. (a) Liquid phase: No change, as pressure does not change the liquid volume appreciably. (b) Gas Phase: Now CA0 = P/RT = 100 (P0_initial)/RT = 0.4 × 100 = 40 mol/dm3 Running with CA0 = 40 mol/dm3, we get: X = 0.998 , SB/XY (instantaneous) = 0.681, and SB/XY (overall) = 0.024

As a practice, students could also try similar problem with the CSTR.



P8-1 (b) Example 8-2 (a) CSTR: intense agitation is needed, good temperature control. (b) PFR: High conversion attainable, temperature control is hard – non-exothermic reactions, selectivity not an issue (c) Batch: High conversion required, expensive products (d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low, to control the conversion of a reactant. (f) and (g) Tubular with side streams: selectivity i.e. to keep a reactant concentration high, to achieve higher conversion of a reactant. (h) Series of CSTR’s: To keep a reactant concentration high, easier temperature control than single CSTR. (i) PFR with recycle: Low conversion to reuse reactants, gas reactants, for highly exothermic reactions (j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid reactants , for highly exothermic reactions (k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product , used for thermodynamically limited reactions where the equilibrium shifts on left side. (l) Reactive Distillation: when one product is volatile and the other is not ; for thermodynamically limited gas phase reactions. 8-2

P8-1 (c) Example 8-3 (i)-(iii) Individualized Solution (iv) We know

𝐶! = 𝐶!! 𝑒 !!! ! and

𝑑𝐶! + 𝑘! 𝐶! = 𝑘! 𝐶!! 𝑒 !!! ! 𝑑𝑡

Using the integrating factor gives:

𝑑(𝐶! 𝑒 !! ! ) = 𝑘! 𝐶!! 𝑒 (!! !!! )! 𝑑𝑡

For 𝑘! = 𝑘! ,

𝑑(𝐶! 𝑒 !! ! ) = 𝑘! 𝐶!! 𝑑𝑡

At time t = 0, 𝐶! = 0. Solving the above equation gives

𝐶! = 𝑘! 𝐶!! 𝑡𝑒 !!! ! To find the time of optimal yield we need to differentiate 𝐶! with respect to time and set it to zero: 𝑑𝐶! = 0 = 𝑘! 𝐶!! (𝑒 !!! ! −𝑘! 𝑡𝑒 !!! ! ) 𝑑𝑡 ! Therefore: 𝑡!"# = = 4 ℎ ! !

P8-1 (d) Example 8-4 (i)-(ii) Individualized Solution (iii) According to the plot of CB versus T the maximum temperature is T = 310.52 0C.



τ ʹk1C A0 CB = (τ ʹk 2 + 1)(τ ʹk1 + 1)

8-3



P8-1 (d) Example 8-4 Continued POLYMATH Results See polymath problem P8-1-d.pol Calculated values of the DEQ variables Variable t T Cao tau k1o k2o E1 E2 R k1 k2 Cb

initial value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.6944444

minimal value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.0052852

maximal value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.8040226

final value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.0052852

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 1 Explicit equations as entered by the user [1] Cao = 5 [2] tau = .5 [3] k1o = .4 [4] k2o = .01 [5] E1 = 10 [6] E2 = 20 [7] R = .001987 [8] k1 = k1o*exp((-E1/R)*(1/T-1/300)) [9] k2 = k1o*exp((-E2/R)*(1/T-1/300)) [10] Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)

P8-1 (e) Individualized solution P8-1 (f) Individualized solution P8-1 (g) Individualized solution P8-1 (h) Example 8-8 (i) Individualized Solution (ii) Membrane Reactor PFR

SD/U Original Problem 2.58 0.666

SD/U P8-2 h 1.01 0.208



Doubling the incoming flow rate of species B lowers the selectivity.

The selectivity becomes 6.52 when the first reaction is changed to A+2B ! D

8-4

P8-1 (i) CDROM Example

Original Case – CDROM example







The reaction does not go as far to completion when the changes are made. The exiting concentration of D, E, and F are lower, and A, B, and C are higher. See Polymath program P8-1-i.pol. P8-1 (j) For equal molar feed in hydrogen and mesitylene. CHO = yHOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3 CMO = 0.016 lbmol/ft3 Using equations from example, solving in Polymath, we get

τ opt = 0.38 hr. At τ = 0.5 hr all of the H2 is reacted and only the decomposition of X takes place. XH CH CM CX

τ

~

S X /T

CD-ROM example 0.50 0.0105 0.0027 0.00507 0.2hr 0.596

This question 0.99 0.00016 0.0042 0.0077 0.38hr 1.865

See Polymath program P8-1-j.pol. POLYMATH Results Calculated values of the DEQ variables Variable tau CH CM CX k1 k2 r1M r2T r1H r2H r1X r2X

initial value 0 0.016 0.016 0 55.2 30.2 -0.1117169 0 -0.1117169 0 0.1117169 0

minimal value 0 1.64E-06 0.0041405 0 55.2 30.2 -0.1117169 0 -0.1117169 -0.0159818 2.927E-04 -0.0159818

maximal value 0.43 0.016 0.016 0.0077216 55.2 30.2 -2.927E-04 0.0159818 -2.927E-04 0 0.1117169 0

8-5

final value 0.43 1.64E-06 0.0041405 0.0077207 55.2 30.2 -2.927E-04 2.986E-04 -2.927E-04 -2.986E-04 2.927E-04 -2.986E-04

P8-1 (j) Continued ODE Report (RKF45) Differential equations as entered by the user [1] d(CH)/d(tau) = r1H+r2H [2] d(CM)/d(tau) = r1M [3] d(CX)/d(tau) = r1X+r2X Explicit equations as entered by the user [1] k1 = 55.2 [2] k2 = 30.2 [3] r1M = -k1*CM*(CH^.5) [4] r2T = k2*CX*(CH^.5) [5] r1H = r1M [6] r2H = -r2T [7] r1X = -r1M [8] r2X = -r2T ~

Increasing θH decreases τopt and S X/T. P8-1 (k) From CD-ROM example Polymath code: See Polymath program P8-1-k.pol. POLYMATH Results NLES Solution Variable CH CM CX tau K1 K2 CHo CMo

Value 4.783E-05 0.0134353 0.0023222 0.5 55.2 30.2 0.016 0.016

f(x) -4.889E-11 -1.047E-11 -9.771E-12

Ini Guess 1.0E-04 0.013 0.002

NLES Report (safenewt) Nonlinear equations [1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0 [2] f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0 [3] f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau-CX = 0 Explicit equations [1] tau = 0.5 [2] K1 = 55.2 [3] K2 = 30.2 [4] CHo = 0.016 [5] CMo = 0.016

A plot using different values of τ is given. For τ =0.5, the exit concentration are CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3 CX =0.00232 lbmol/ft3

8-6

P8-1 (k) Continued The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for τ =0.5:

YMX =

FX CX 0.00232 0.89mole ⋅ xylene ⋅ produced = = = FMO − FM C MO − C M 0.016 − 0.0134 mole ⋅ mesitylene ⋅ reacted

The overall selectivity of xylene relative to toluene is:

F 8.3mole ⋅ xylene ⋅ produced ~ S X /T = X = FT mole ⋅ toluene ⋅ produced CH CM CX

τ YMX SX/T

CD-ROM example 0.0089 0.0029 0.0033 0.5 0.41 0.7

This Question 4.8 x 10-5 0.0134 0.00232 0.5 0.89 8.3

P8-1 (l) Individualized solution P8-1 (m)

At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. At the same time the reactions that consume the two components begin to accelerate and the concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again. Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions. P8-1 (n) Individualized solution P8-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. 8-7

P8-3 (a) Plot of CA, CD and CU as a function of time (t): See Polymath program P8-3-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cd Cu k1 k2 K1a K2a Cao X

initial value 0 1 0 0 1 100 10 1.5 1 0

minimal value 0 0.0801802 0 0 1 100 10 1.5 1 0

maximal value 15 1 0.7995475 0.5302179 1 100 10 1.5 1 0.9198198

final value 15 0.0801802 0.7995475 0.1202723 1 100 10 1.5 1 0.9198198

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)-k2*(Ca-Cu/K2a)) [2] d(Cd)/d(t) = k1*(Ca-Cd/K1a) [3] d(Cu)/d(t) = k2*(Ca-Cu/K2a) Explicit equations as entered by the user [1] k1 = 1.0 [2] k2 = 100 [3] K1a = 10 [4] K2a = 1.5 [5] Cao = 1 [6] X = 1-Ca/Cao

To maximize CD stop the reaction after a long time. The concentration of D only increases with time P8-3 (b) Conc. Of U is maximum at t = 0.31 min.(CA = 0.53) P8-3 (c) Equilibrium concentrations: CAe = 0.08 mol/dm3 CDe = 0.8 mol/dm3 CUe = 0.12 mol/dm3 8-8

P8-3 (d) See Polymath program P8-3-d.pol. POLYMATH Results NLES Solution Variable Ca Cd Cu Ca0 k1 k2 K1a K2a t

Value 0.0862762 0.7843289 0.1293949 1 1 100 10 1.5 100

f(x) -3.844E-14 -2.631E-14 6.478E-14

Ini Guess 1 0 0

NLES Report (safenewt) Nonlinear equations [1] f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0 [2] f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0 [3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0 τ

1 min

10 min

100min

CAexit

0.295

0.133

0.0862

CDexit

0.2684

0.666

0.784

CUexit

0.436

0.199

0.129

X

0.705

0.867

0.914

Explicit equations [1] Ca0 = 1 [2] k1 = 1 [3] k2 = 100 [4] K1a = 10 [5] K2a = 1.5 [6] t = 100 [7] X = 1-Ca/Ca0

P8-4 (a) 1/2



! mol $ k1 = 0.004 # 3 & min " dm %

rB = k2C A



k2 = 0.3min −1

rY = k3C A2



k3 = 0.25

A→ X



rX = k1C

A→ B



A→Y



1/2 A

dm3 mol.min

Sketch SBX, SBY and SB/XY as a function of CA See Polymath program P8-4-a.pol. It shows the table of SBX, SBY and SB/XY in correspondence with values of CA. (1)

r k C SB/X = B = 2 A rX

k1C A1 / 2

=

k2 C A1 / 2 k1

8-9

P8-4 (a) Continued



(2)

r k C k2 SB/Y = B = 2 A = 2 rY k 3C A k 3C A

(3)

rB k 2C A SB/XY = = 1 rX + rY k1C A / 2 + k 3C A 2



P8-4 (b) Volume of first reactor can be found as follows We have to maximize SB/XY From the graph above, maximum value of SBXY = 10 occurs at CA* = 0.040 mol/dm3 So, a CSTR should be used with exit concentration CA* Also, CA0 = PA/RT = 0.162 mol/dm3 8-10

P8-4 (b) Continued 1/ 2

And − rA = rX + rB + rY = (k1C A *

v (C − C A ) => V = 0 A0 = − rA

P8-4 (c) Effluent concentrations:

+ k 2 C A + k 3C A 2 )

v0 (C A0 − C A* ) * 1/ 2

(k1 (C A )

*

* 2

+ k 2C A + k3 (C A ) )

= 92.4dm 3

CB C mol = B ⇒ C B * = 0.11 3 rB k 2C A dm mol mol = 0.007 3 and CY * = 0.0037 3 dm dm

We know, τ = 9.24 min => τ = Similarly: C X

*

P8-4 (d) Conversion of A in the first reactor:

C A0 − C A = C A0 X ⇒ X = 0.74

P8-4 (e) A CSTR followed by a PFR should be used. Required conversion = 0.99 => For PFR, Mole balance:

dV FA0 = dX − rA

0.99

⇒ V = 10 × 0.162 × ∫

1/ 2 0.74 ( k1C A

dX 2

+ k 2 C A + k 3C A )

= 92.8dm 3

P8-4 (f) If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B, decrease as temperature drops. So we have to compromise between high selectivity and production. To do this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know:

⎛ − Ei ki = Ai exp ⎜ ⎝ 1.98T

⎞ ⎟ ⎠

Since we have the constants given at T = 300 K, we can solve for Ai.

A1 =

.004

= 1.49e12 ⎛ −20000 ⎞ exp ⎜⎜ ⎟⎟ ⎝ 1.98 (300 ) ⎠ .3 A2 = = 5.79e6 ⎛ −10000 ⎞ exp ⎜⎜ ⎟⎟ ⎝ 1.98 (300 ) ⎠ .25 A3 = = 1.798e21 ⎛ −30000 ⎞ exp ⎜⎜ ⎟⎟ ⎝ 1.98 (300 ) ⎠ 8-11

P8-4 (f) Continued Now we use a mole balance on species A

FA0 − FA − rA v (C A0 − C A ) V= −rA C − CA C A0 − C A τ = A0 = 0.5 −rA k1C A + k2C A + k3C A2 V=

A mole balance on the other species gives us: Fi = vCi = rV i

Ci = τ ri



Cb versus temp

290 292 294 296 298 300 302 304

Using these equations we can make a Polymath program and by varying the temperature, we can find a maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state for certain that the optimal temperature will be equal to or less than 306 K.

See Polymath program P8-4-f.pol. POLYMATH Results NLE Solution Variable Ca T R k1 k2 Cao Cb k3 tau Cx Cy Sbxy

Value 0.0170239 306 1.987 0.0077215 0.4168076 0.1 0.070957 0.6707505 10 0.0100747 0.0019439 5.9039386

f(x) 3.663E-10

Ini Guess 0.05

8-12

P8-4 (f) Continued NLE Report (safenewt) Nonlinear equations [1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0 Explicit equations [1] T = 306 [2] R = 1.987 [3] k1 = 1.49e12*exp(-20000/R/T) [4] k2 = 5790000*exp(-10000/R/T) [5] Cao = .1 [6] Cb = 10*k2*Ca [7] k3 = 1.798e21*exp(-30000/R/T) [8] tau = 10 [9] Cx = tau*k1*Ca^.5 [10] Cy = tau*k3*Ca^2 [11] Sbxy = Cb/(Cx+Cy)

P8-4 (g) Concentration is proportional to pressure in a gas-phase system. Therefore:

S B/ XY ~

PA PA + PA2

which would suggest that a low pressure would be ideal. But as before the tradeoff is

lower production of B. A moderate pressure would probably be best. We know that: SB/XY =

rB k 2C A = 1 rX + rY k1C A / 2 + k 3C A 2

Substituting PA = CA R T ; in the above co – relation we get

SB/XY

rB k2 ( PA / RT ) = 1/2 2 = rX + rY k1 ( PA / RT ) + k3 ( PA / RT )

See Polymath program P8-4-g.pol.

8-13

P8-4 (g) Continued

Thus we find that the maximum is obtained at P= 1 atm . P8-5 US legal limit: 0.8 g/l Sweden legal limit: 0.5 g/l k1 k2 A ⎯⎯ → B ⎯⎯ →C

Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream

dC A = −k1C A dt dCB = k1C A − k2 dt k1 = 10 hr −1

k2 = 0.192

g L hr

Two tall martinis = 80 g of ethanol Body fluid = 40 L

C A0 =

80 g g =2 40 L L

Now we can put the equations into Polymath. See Polymath program P8-5.pol.

8-14



P8-5 Continued POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 2 0 10 0.192

minimal value 0 7.131E-44 0 10 0.192

maximal value 10 2 1.8901533 10 0.192

final value 10 7.131E-44 0.08 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca

Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

P8-5 (a) In the US the legal limit it 0.8 g/L. This occurs at t = 6.3 hours. P8-5 (b) In Sweden CB = 0.5 g/l , t = 7.8 hrs. P8-5 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs P8-5 (d) For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L. Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half an hour. At a half an hour CA = 0.00674 g/L and CB = 0.897 g/L. The Polymath code for after the second drink is shown below. See Polymath program P8-5-d.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0.5 1.0067379 0.8972621 10 0.192

minimal value 0.5 5.394E-42 0.08 10 0.192

maximal value 10 1.0067379 1.8069769 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

8-15

final value 10 5.394E-42 0.08 10 0.192

P8-5 (d) Continued for the US t = 6.2 hours Sweden: t = 7.8 hours Russia: t =10.3 hours. P8-5 (e) The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g of ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the rate of change in concentration due to the incoming ethanol is 2 g/L/hr. For the first hour the differential equation for CA becomes:

dC A = −k1C A + 2 after that it reverts back to the original equations. dt See Polymath program P8-5-e.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 0 0 10 0.192

minimal value 0 0 -1.1120027 10 0.192

maximal value 11 0.1785514 0.7458176 10 0.192

final value 11 6.217E-45 -1.1120027 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = if(t 1000CD

S D / U 1U 2

1.18 × 10 −14 C A − 1.94 × 10 −18 C D 1.18 ≈ = 1.96 = 1.94 × 10 −18 C D + 6.03 × 10 −15 C A .603

At T = 1000K k1 = 0.026 &

k2 = 3.1 X 10-3 & k3 = 5.3 x 10-4

If we keep CA > 1000CD

S D / U 1U 2

0.026C A − 3.1 × 10 −3 C D .026 ≈ = 49 = −3 −4 3.1 × 10 C D + 5.3 × 10 C A .00053

Here, in order to lower U1 use low temperature and high concentration of A But low temperature and high concentration of A favors U2 So, it’s a optimization problem with the temperature and concentration of A as the variables . Membrane reactor in which D is diffusing out can be used. P8-7 (e) A +

B → D

S D /U

- r1 A

= 10 9 exp(−10000 K / T )C AC B

A + B

and - r2 D

= 20 exp(−2000K / T )C D

B →U

and - r3 A

= 10 3 exp(−3000 K / T )C AC B

D → A +

and

rD 10 9 exp(−10000 K / T )C A C B − 20 exp(−2000 K / T )C D = = rU 10 3 exp(−3000 K / T )C A C B

At T = 300K k1 = 3.34 x 10-6 &

k2 = 0.025

& k3 = 0.045

8-21

P8-7 (e) Continued The desired reaction lies very far to the left and CD is probably present at very low concentrations so that:

S D /U =

3.34 × 10 −6 C A C B − 0.025C D 0.000334 ≈ = 0.000074 4.5 0.045C A C B

SD /U ≈ 0 At T = 1000K k1 = 45399.9 &

k2 = 2.7 & k3 = 49.7

If we assume that CACB > 0.001CD then,

S D /U =

45399.9C AC B − 2.7C D 45399 ≈ = 913 49.7 49.7C AC B

Here we need a high temperature for a lower reverse reaction of D and lower formation of U Also we need to remove D as soon as it is formed so as to avoid the decomposition. P8-7 (f) A +

B → D

−r1 A = 800 exp(−8000 K / T )C A0.5CB



and

A + B → U1



and −r2 B

D + B → U 2



and − r3 D = 10 exp( −8000 K / T )CD CB

= 10exp(−300K / T )CACB 6

(1)

S D /U1 =

800exp (−8000 / T )C A0.5CB 10exp (−300 / T )C ACB

=

80exp (−8000 / T ) exp (−300 / T )CA0.5



At T = 300

S D /U1 =

2.098*10−10 0.368C A0.5

At T = 1000

S D / U1 =

29.43 0.7408C A0.5

To keep this selectivity high, low concentrations of A, and high temperatures should be used.

SD /U 2 =

800exp (−8000 / T )C A0.5CB 106 exp (−8000 / T )CDCB

=

800C A0.5 106 CD

To keep this selectivity high, high concentrations of A and low concentrations of D should be used. Try to remove D with a membrane reactor or reactive distillation. The selectivity is not dependent on temperature. To keep optimize the reaction, run it at a low temperature to maximized SD/U1 in a membrane reactor that allows only D to diffuse out. 8-22

P8-7 (f) Continued (2)

S D / U 1U 2 = S D / U 1U 2 =

800 exp (−8000 / T )C A0.5CB 10 exp (−300 / T )C ACB + 106 exp (−8000 / T )CD CB 800 exp (−8000 / T )C A0.5



10 exp (−300 / T )C A + 106 exp (−8000 / T )CD

At T = 300

S D / U 1U 2 =

2.09*10−9 C A0.5 ≈ 0 3.67C A + 2.62*10−6 CD

At T = 1000 and very low concentrations of D

S D / U 1U 2 =

0.268C A0.5 .03617 = 7.408C A + 335CD C A0.5

If temperature is the only parameter that can be varied, then the highest temperature possible will result in the highest selectivity. Also removing D will help keep selectivity high. P8-8 No solution will be given P8-9 (a) Species A :

dC A = rA ; dt

Species B:

-rA = k1 * Ca

dC B = rb dt

Species C:

rB = k1 * Ca – k2 * Cb

dCC = rc dt

rC = k2*Cb

8-23

P8-9 (a) continued See polymath program P8-9-a.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca

1.6

6.797E-18

1.6

6.797E-18

2 Cb

0

0

1.455486

0.6036996

3 Cc

0

0

0.9963004

0.9963004

4 k1

0.4

0.4

0.4

0.4

5 k2

0.01

0.01

0.01

0.01

6 ra

-0.64

-0.64

-2.719E-18

-2.719E-18

7 rb

0.64

-0.0132415

0.64

-0.006037

8 rc

0

0

0.0145549

0.006037

100.

100.

9 t 0 0 Differential equations 1 d(Cc)/d(t) = rc 2 d(Cb)/d(t) = rb 3 d(Ca)/d(t) = ra Explicit equations 1 k2 = 0.01 2 rc = k2*Cb 3 k1 = 0.4 4 ra = - k1*Ca 5 rb = k1*Ca - k2*Cb



P8-9 (b) Now the rate laws will change - ra = k1-1Cb – k1*Ca rb =k1*Ca – k1-1Cb – k2*Cb rc = k2*Cb 8-24

P8-9 (b) continued See polymath program P8-9-b.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.6

0.3949584

1.6

0.3949584

2

Cb

0

0

0.8736951

0.5191343

3

Cc

0

0

0.6859073

0.6859073

4

k1

0.4

0.4

0.4

0.4

5

k1b

0.3

0.3

0.3

0.3

6

k2

0.01

0.01

0.01

0.01

7

ra

-0.64

-0.64

-0.0022431

-0.0022431

8

rb

0.64

-0.0047669

0.64

-0.0029483

9

rc

0

0

0.008737

0.0051913

0

0

100.

100.

10 t

Differential equations 1 d(Ca)/d(t) = ra 2 d(Cb)/d(t) = rb 3 d(Cc)/d(t) = rc Explicit equations 1 k1 = 0.4 2 k2 = 0.01 3 k1b = 0.3 4 ra = k1b*Cb - k1*Ca 5 rb = k1*Ca - k1b*Cb - k2*Cb 6 rc = k2*Cb



8-25

P8-9 (c) rC = k2*CB – k-2*Cc See polymath program P8-9-c.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.6

0.4500092

1.6

0.4500092

2

Cb

0

0

0.8742432

0.5953978

3

Cc

0

0

0.554593

0.554593

4

k1

0.4

0.4

0.4

0.4

5

k1b

0.3

0.3

0.3

0.3

6

k2

0.01

0.01

0.01

0.01

7

k2c

0.005

0.005

0.005

0.005

8

ra

-0.64

-0.64

-0.0013843

-0.0013843

9

rb

0.64

-0.0044688

0.64

-0.0017967

10 rc

0

0

0.0085186

0.003181

11 t

0

0

100.

100.

Differential equations 1 d(Ca)/d(t) = ra 2 d(Cb)/d(t) = rb 3 d(Cc)/d(t) = rc Explicit equations 1 k1 = 0.4 2 k2 = 0.01 3 k1b = 0.3 4 k2c = 0.005 5 ra = k1b*Cb - k1*Ca 6 rb = k1*Ca - k1b*Cb - k2*Cb + k2c*Cc 7 rc = k2*Cb - k2c*Cc

8-26

P8-9 (d) Individualized Solution P8-9 (e) When k1 > 100 and k2 < 0.1 the concentration of B immediately shoots up to 1.5 and then slowly comes back down , while CA drops off immediately and falls to zero . This is because the first reaction is very fast and the second reaction is slower with no reverse reactions. When k2 = 1 then the concentration of B spikes again and remains high, while very little of C is formed. This is because after B is formed it will not further get converted to C because the reverse reaction is fast. When k-2 = 0.25 , B shoots up , but does not stay as high because for the second reaction in the reverse direction is a slightly slower than seen before , but still faster than the reaction in the forward direction. P8-9 (f) Individualized Solution P8-10 (a) Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCl2 catalyst reacted with K-terephthalate in an autocatalytic reaction step:

k1 k2 A ⎯⎯ → R ⎯⎯ → S Series k3 R + S ⎯⎯ → 2S Autocatalytic P 110kPa = = = 0.02mol / dm3 3 RT ⎛ kPa.dm ⎞ ⎜ 8.314 ⎟ (683K ) mol.K ⎠ ⎝





C AO

Maximum in R occurs at t = 880 sec. See Polymath program P8-10-a.pol. POLYMATH Results Variable t A R S k1 k2 k3 0.00159

initial value 0 0.02 0 0 0.00108 0.00119 0.00159 0.00159

minimal value 0 0.003958 0 0 0.00108 0.00119 0.00159

maximal value 1500 0.02 0.0069892 0.0100382 0.00108 0.00119

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 1.08e-3 [2] k2 = 1.19e-3 [3] k3 = 1.59e-3



8-27

final value 1500 0.003958 0.005868 0.0100382 0.00108 0.00119

P8-10 (b) (1) T = 703 K CAO = 0.019 mol/dm3

⎛ E1 ⎛ 1 1 ⎞ ⎞ ⎜ − ' ⎟⎟ ⎝ R ⎝ T T ⎠⎠

k1' = k1 exp ⎜

⎛ (42600cal / mol ) ⎛ 1 1 ⎞⎞ −3 −1 k1' = (1.08 ×10−3 s −1 ) exp ⎜ − ⎜ ⎟ ⎟ = 2.64 ×10 s ⎝ (1.987cal / mol.K ) ⎝ 683K 703K ⎠ ⎠ Similarly,

k2' = 3.3 × 10−3 s −1 '

−3

3

And, k3 = 3.1× 10 dm / mol.s Maximum in R occurs at around t =320 sec. See Polymath program P8-10-b1.pol. POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 0.00264 0.0033 0.0031

minimal value 0 3.622E-04 0 0 0.00264 0.0033 0.0031

maximal value 1500 0.019 0.0062169 0.0174625 0.00264 0.0033 0.0031

final value 1500 3.622E-04 8.856E-04 0.0174625 0.00264 0.0033 0.0031

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 2.64e-3 [2] k2 = 3.3e-3 [3] k3 = 3.1e-3

(2) T = 663 K CAO = 0.19 mol/dm3

⎛ (42600cal / mol ) ⎛ 1 1 ⎞⎞ −3 −1 k1' = (1.08 ×10−3 s −1 ) exp ⎜ − ⎜ ⎟ ⎟ = 0.42 ×10 s ⎝ (1.987cal / mol.K ) ⎝ 683K 663K ⎠ ⎠ ' −3 −1 k2 = 0.4 × 10 s

k3' = 0.78 ×10−3 dm3 / mol.s See Polymath program P8-10-b2.pol.

8-28

P8-10 (b) continued POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 4.2E-04 4.0E-04 7.8E-04

minimal value 0 2.849E-04 0 0 4.2E-04 4.0E-04 7.8E-04

maximal value 10000 0.019 0.0071414 0.016889 4.2E-04 4.0E-04 7.8E-04

final value 10000 2.849E-04 0.0012573 0.016889 4.2E-04 4.0E-04 7.8E-04

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S)

Explicit equations as entered by the user [1] k1 = 0.42e-3 [2] k2 = 0.4e-3 [3] k3 = 0.78e-3

Independent variable variable name : t initial value : 0 final value : 10000

Maxima in R occurs around t = 2500 sec. P8-10 (c) Use the Polymath program from part (a) and change the limits of integration to 0 to 1200. We get: CAexit = 0.0055 mol/dm3 CRexit = 0.0066 mol/dm3 CSexit = 0.0078 mol/dm3 P8-11 (a)

P8-11 (c)















P8-11 (b)



P8-11 (d)

8-29







P8-11 (e)

P8-11 (f)



P8-11 (g)







P8-11 (h) Mole balance:

C AO − C A = (− rA )τ







CC = (rC )τ







CD = (rD )τ

Rate law:



1 ⎡ ⎤ rA = − ⎢k1 A C A + k 2 D C A CC2 ⎥ 3 ⎣ ⎦







2 ⎡1 ⎤ rC = ⎢ k1 A C A − k 2 D C A CC2 − k 3 E CC C D ⎥ 3 ⎣3 ⎦







4 ⎡ ⎤ rD = ⎢k 2 D C A CC2 − k 3 E CC C D ⎥ 3 ⎣ ⎦

Solving in polymath:

C A = 0.0069M CB = 0.96M CC = 0.51M CD = 0.004M



SB/D = rB/rD = 247







See Polymath program P8-11-h.pol.

8-30

SB/C = 1.88





P8-11 (h) continued POLYMATH Results NLES Solution Variable Value Ca 0.0068715 Cb 0.9620058 Cc 0.5097027 Cd 0.0038925 Ce 0.2380808 kd 3 ka 7 rb 0.0160334 ra -0.0498855 ke 2 rc 0.008495 rd 6.488E-05 re 0.003968 tau 60 Cao 3

f(x) -2.904E-10 -1.332E-15 -1.67E-08 -2.391E-08 1.728E-08

Ini Guess 3 0 0 0 0

NLES Report (safenewt) Nonlinear equations [1] f(Ca) = Cao-Ca+ra*tau = 0 [2] f(Cb) = Cb - rb*tau = 0 [3] f(Cc) = Cc-rc*tau = 0 [4] f(Cd) = Cd-rd*tau = 0 [5] f(Ce) = Ce - re*tau = 0 Explicit equations [1] kd = 3 [2] ka = 7 [3] rb = ka*Ca/3 [4] ra = -(ka*Ca+kd/3*Ca*Cc^2) [5] ke = 2 [6] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc [7] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc [8] re = ke*Cd*Cc [9] tau = 60 [10] Cao = 3

P8-11 (i) For PFR and gas phase: Mole balance: Rate law:











dFA dFB = rA = rB dV dV ⎡ ⎤ 1 rA = −⎢k1ACA + k2DCACC2 ⎥ ⎣ ⎦ 3

dFC = rC dV

⎡1 ⎤ rB = ⎢ k1AC A ⎥ ⎣3 ⎦ ⎡1 ⎤ 2 rC = ⎢ k1AC A − k2 DC ACC2 − K 3ECC C D ⎥ 3 ⎣3 ⎦ ⎡ ⎤ 4 rD = ⎢k2 DC ACC2 − k3ECC C D ⎥ 3 ⎣ ⎦ rE = ⎡⎣k3ECC C D ⎤⎦ 8-31

dFD = rD dV

dFE = rE dV

P8-11 (i) continued Stoichiometry: C A = CTO

F FA F p CC = CTO C p C D = CTO D p FT FT FT





FT = FA + FB + FC + FD + FE





dp −α FT = dV 2 p FTO

Plot of CB and CC are overlapping. See Polymath program P8-11-i.pol. POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc Fd Fe y Ft Cto Cc ka kd ke Ca rb ra Cd Fto rc rd re alfa X

initial value 0 20 0 0 0 0 1 20 0.2 0 7 3 2 0.2 0.4666667 -1.4 0 20 0.4666667 0 0 1.0E-04 0

minimal value 0 9.147E-04 0 0 0 0 0.9964621 13.330407 0.2 0 7 3 2 1.367E-05 3.191E-05 -1.4 0 20 -1.923E-05 -7.012E-05 0 1.0E-04 0

maximal value 100 20 6.6638171 6.6442656 0.0201258 0.0043322 1 20 0.2 0.0993605 7 3 2 0.2 0.4666667 -9.586E-05 3.0E-04 20 0.4666667 8.653E-04 5.908E-05 1.0E-04 0.9999543

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc [4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(y)/d(V) = -alfa*Ft/(2*y*Fto)

8-32

final value 100 9.147E-04 6.6638171 6.6442167 0.0171261 0.0043322 0.9964621 13.330407 0.2 0.0993325 7 3 2 1.367E-05 3.191E-05 -9.586E-05 2.56E-04 20 -1.923E-05 -6.742E-05 5.087E-05 1.0E-04 0.9999543

P8-11 (i) continued Explicit equations as entered by the user [1] Ft = Fa+Fb+Fc+Fd+Fe [2] Cto = 0.2 [3] Cc = Cto*Fc/Ft*y [4] ka = 7 [5] kd = 3 [6] ke = 2 [7] Ca = Cto*Fa/Ft*y [8] rb = ka*Ca/3 [9] ra = -(ka*Ca+kd/3*Ca*Cc^2) [10] Cd = Cto*Fd/Ft*y [11] Fto = 0.2*100 [12] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc [13] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc [14] re = ke*Cd*Cc [15] alfa = 0.0001 [16] X = 1-Fa/20

P8-11 (j) Changes in equation from part (i):

dFC = rC − RC dV





RC = k diffuse CC k diffuse = 2 min −1

See Polymath program P8-11-j.pol.







8-33

P8-12 (a)





See Polymath program P8-12-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.5

0.3100061

1.5

0.3100061

2

Cb

2.

0.3384816

2.

0.3384816

3

Cc

0

0

0.1970387

0.105508

4

Cd

0

0

0.6751281

0.6516263

5

Ce

0

0

0.1801018

0.1801018

6

Cf

0

0

0.3621412

0.3621412

7

kd1

0.25

0.25

0.25

0.25

8

ke2

0.1

0.1

0.1

0.1

9

kf3

5.

5.

5.

5.

10 ra

-1.5

-1.5

-0.0694818

-0.0694818

11 rb

-3.

-3.

-0.0365984

-0.0365984

12 rc

1.5

-0.0490138

1.5

-0.0085994

13 rd

1.5

-0.0128609

1.5

-0.0126825

14 rd1

1.5

0.0088793

1.5

0.0088793

15 re

0

0

0.0523329

0.0202008

16 re2

0

0

0.0523329

0.0202008

17 rf

0

0

0.2571468

0.0188398

18 rf3

0

0

0.2571468

0.0188398

19 V

0

0

50.

50.

20 vo

10.

10.

10.

10.

8-34

P8-12 (a) Differential equations 1 d(Ca)/d(V) = ra/vo 2 d(Cb)/d(V) = rb/vo 3 d(Cc)/d(V) = rc/vo 4 d(Cd)/d(V) = rd/vo 5 d(Ce)/d(V) = re/vo 6 d(Cf)/d(V) = rf/vo Explicit equations 1

vo = 10

2

kf3 = 5

3

ke2 = .1

4

kd1 = 0.25

5

rf3 = kf3*Cb*Cc^2

6

rd1 = kd1*Ca*Cb^2

7

re2 = ke2*Ca*Cd

8

rf = rf3

9

re = re2

10 rd = rd1-2*re2+rf3 11 ra = -rd1-3*re2 12 rb = -2*rd1-rf3 13 rc = rd1+re2-2*rf3



8-35

P8-12 (a) Continued 0.9 0.8 0.7 0.6 X

0.5 0.4

Conversion

0.3 0.2 0.1 0 0

10

20

30

40

50

60

Time

P8-12 (b) Determine the effluent concentration and conversion from a 50 dm3 CSTR. Mole Balance:











8-36

P8-12 (b) Continued

P8-12 (c) V0 = 40 dm3 Semi-Batch reactor. (1) A is fed to B, (2) B is fed to A (Case 1) A is fed to B,





8-37

P8-12 (c)





8-38

P8-12 (c)







P8-12 (d) As θB increases the outlet concentration of species D and F increase, while the outlet concentrations of species A, C, and E decrease. When θB is large, reactions 1 and 3 are favored and when it is small the rate of reaction 2 will increase.

8-39

P8-12 (e) When the appropriate changes to the Polymath code from part (a) are made we get the following. See Polymath program P8-12-e.pol. POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc Fd Fe Ff vo Ft Cto kd1 ke2 kf3 Cc Cd Cb Ca rd1 re2 rf3 re rf rd ra rb rc Scd Sef

initial value 0 20 20 0 0 0 0 100 40 0.4 0.25 0.1 5 0 0 0.2 0.2 0.002 0 0 0 0 0.002 -0.002 -0.004 0.002 1 0

minimal value 0 18.946536 18.145647 0 0 0 0 100 38.931546 0.4 0.25 0.1 5 0 0 0.1864364 0.1946651 0.0016916 0 0 0 0 0.0014393 -0.0021989 -0.004 0.0016889 1 0

maximal value 500 20 20 0.9342961 0.8454829 0.0445942 0.0149897 100 40 0.4 0.25 0.1 5 0.0095994 0.0086869 0.2 0.2 0.002 1.691E-04 8.59E-05 1.691E-04 8.59E-05 0.002 -0.002 -0.003469 0.002 1.1734311 83.266916

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc [4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(Ff)/d(V) = rf Explicit equations as entered by the user [1] vo = 100 [2] Ft = Fa+Fb+Fc+Fd+Fe+Ff [3] Cto = .4 [4] kd1 = 0.25 [5] ke2 = .1 [6] kf3 = 5 [7] Cc = Cto*Fc/Ft [8] Cd = Cto*Fd/Ft [9] Cb = Cto*Fb/Ft [10] Ca = Cto*Fa/Ft [11] rd1 = kd1*Ca*Cb^2 [12] re2 = ke2*Ca*Cd [13] rf3 = kf3*Cb*Cc^2 [14] re = re2 [15] rf = rf3 [16] rd = rd1-2*re2+rf3 [17] ra = -rd1-3*re2

8-40

final value 500 18.946536 18.145647 0.9342961 0.8454829 0.0445942 0.0149897 100 38.931546 0.4 0.25 0.1 5 0.0095994 0.0086869 0.1864364 0.1946651 0.0016916 1.691E-04 8.59E-05 1.691E-04 8.59E-05 0.0014393 -0.0021989 -0.003469 0.0016889 1.1734311 1.9686327

P8-12 (e) continued [18] [19] [20] [21]

rb = -2*rd1-rf3 rc = rd1+re2-2*rf3 Scd = rc/(rd+.0000000001) Sef = re/(rf+.00000000001)

P8-12 (f) The only change from part (e) is:

dFD = rD − kcDCD dV

See Polymath program P8-12-f.pol.



P8-12 (g) The only change from part (e) is:

F dFB = rB − B 0 where VT = 500 dm3 and FB0 = 20 mol/min dV VT

See Polymath program P8-12-g.pol.



8-41

P8-13 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst.

⎡ CC ⎤ ⎯⎯ →B +C r ' = k1C ⎢C A − B C ⎥ A ←⎯ ⎯ 1C K1C ⎦ ⎣ ' A ! D r2 D = k 2 D C A '

2

2C + D ! 2E r3 E = k3 E CC CD

C AO =

P 24.6atm = = 0.6mol / dm3 3 RT (0.082dm atm / mol.K )(500 K )

Fa(0) = 10 mol/min Fb(0) = Fc(0) = Fd(0) = Fe(0) = 0 See Polymath program P8-13-a.pol Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 alfa

0.008

0.008

0.008

0.008

2 Ca

0.6

0.0029024

0.6

0.0029024

3 Cb

0

0

0.1525032

0.0075281

4 Cc

0

0

0.1425644

0.0541764

5 Cd

0

0

0.0254033

0.0036956

6 Ce

0

0

0.274725

0.235079

7 Cto

0.6

0.6

0.6

0.6

8 Fa

10.

0.0702103

10.

0.0702103

9 Fb

0

0

3.391481

0.1821054

10 Fc

0

0

3.16936

1.310531

11 Fd

0

0

0.5564071

0.0893965

12 Fe

0

0

5.686575

5.686575

13 Ft

10.

7.338818

12.8179

7.338818

14 Fto

10.

10.

10.

10.

15 k1c

2.

2.

2.

2.

16 K1c

0.2

0.2

0.2

0.2

17 k2d

0.4

0.4

0.4

0.4

18 k3e

400.

400.

400.

400.

19 kb

1.

1.

1.

1.

20 r1c

1.2

0.0017264

1.2

0.0017264

21 r2d

0.24

0.001161

0.24

0.001161

22 r3e

0

0

0.200243

0.0043387

23 ra

-1.44

-1.44

-0.0028874

-0.0028874

24 rb

1.2

0.0017264

1.2

0.0017264

25 rc

1.2

-0.0492057

1.2

-0.0026123

26 rd

0.24

-0.0164816

0.24

-0.0010084

27 re

0

0

0.200243

0.0043387

28 W

0

0

100.

100.

29 y

1.

0.5056359

1.

0.5056359

8-42

P8-13 (a) continued Differential equations 1 d(Fa)/d(W) = ra 2 d(Fb)/d(W) = rb-(kb*Cb) 3 d(Fc)/d(W) = rc 4 d(Fd)/d(W) = rd 5 d(Fe)/d(W) = re 6 d(y)/d(W) = -alfa*Ft/(2*Fto*y) Explicit equations 1 k2d = 0.4 2 K1c = 0.2 3 Ft = Fa+Fb+Fc+Fd+Fe 4 Cto = 0.6 5 Cb = Cto*(Fb/Ft)*y 6 Ca = Cto*(Fa/Ft)*y 7 Cd = Cto*(Fd/Ft)*y 8 Cc = Cto*(Fc/Ft)*y 9 kb = 1 10 k1c = 2 11 r2d = k2d*Ca 12 k3e = 400 13 r1c = k1c*(Ca-(Cb*Cc/K1c)) 14 ra = -r1c-r2d 15 r3e = k3e*(Cc^2)*Cd 16 rd = r2d-(r3e/2) 17 rb = r1c 18 rc = r1c-r3e 19 re = r3e 20 Ce = Cto*(Fe/Ft)*y 21 alfa = 0.008 22 Fto = 10

P8-13 (b) Species B, C, D, and E all go though a maximum. The concentration of species are affected by two factors: reaction and pressure drop. We can look at species B for example: Species B: Production by reaction 1 => tends to increase concentration of B Pressure drop => tends to decrease concentration of B At W0.0001)then (Fc/Fd) else (0) 23 Sae = if (V>0.0001)then (Fa/Fe) else (0) 24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)





8-48

P8-14 (a) continued







8-49

P8-14 (a) continued







8-50

P8-14 (a) continued









8-51

P8-14 (b) Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.098

0.0951293

0.098

0.0951293

2

Cb

0.049

0.0475647

0.049

0.0475647

3

Cc

0

0

0.0032152

0.0032152

4

Cd

0

0

0.0001412

0.0001411

5

Ce

0

0

0.0002229

0.0002229

6

Cg

0

0

0.0005038

0.0005038

7

Cto

0.147

0.147

0.147

0.147

8

Cw

0

0

0.0002229

0.0002229

9

Fa

10.

9.63739

10.

9.63739

10 Fb

5.

4.818695

5.

4.818695

11 Fc

0

0

0.3257277

0.3257277

12 Fd

0

0

0.0143026

0.014299

13 Fe

0

0

0.0225834

0.0225834

14 Fg

0

0

0.0510394

0.0510394

15 Ft

15.

14.89232

15.

14.89232

16 Fw

0

0

0.0225834

0.0225834

17 k1

0.014

0.014

0.014

0.014

18 k2

0.007

0.007

0.007

0.007

19 k3

0.014

0.014

0.014

0.014

20 k4

0.45

0.45

0.45

0.45

21 ra

-0.0003709

-0.0003709

-0.0003538

-0.0003538

22 rb

-0.0001855

-0.0001855

-0.0001769

-0.0001769

23 rc

0.0003037

0.0003037

0.000336

0.000309

24 rd

6.723E-05

-1.674E-07

6.723E-05

-1.674E-07

25 re

0

0

4.501E-05

4.501E-05

26 rg

0

0

6.352E-05

6.351E-05

27 rw

0

0

4.501E-05

4.501E-05

28 Sae

0

0

7.875E+05

426.7466

29 Scd

0

0

22.77975

22.77975

30 Sce

0

0

594.3536

14.42332

31 Sdg

0

0

18.10358

0.2801563

32 V

0

0

1000.

1000.

33 Yc

0

0

0.0337983

0.0337983

Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg

8-52

P8-14 (b) continued Explicit equations 1

Cto = 0.147

2

k1 = 0.014

3

Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg

4

k2 = 0.007

5

Cb = Cto*(Fb/Ft)

6

k3 = 0.014

7

k4 = 0.45

8

Ca = Cto*(Fa/Ft)

9

Cc = Cto*(Fc/Ft)

10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 11 rb = ra/2 12 re = k3*Cc 13 Cd = Cto*(Fd/Ft) 14 Ce = Cto*(Fe/Ft) 15 Cg = Cto*(Fg/Ft) 16 Cw = Cto*(Fw/Ft) 17 rg = k4*Cd 18 rw = k3*Cc 19 rd = k2*(Ca)^2-k4*Cd 20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 21 Yc = Fc/Fa 22 Scd = if (V>0.0001)then (Fc/Fd) else (0) 23 Sae = if (V>0.0001)then (Fa/Fe) else (0) 24 Sdg = if (V>0.0001)then (Fd/Fg) else (0) 25 Sce = if (V>0.0001)then (Fc/Fe) else (0)

Then taking different values of FA0 and FB0, we find the following data: FA0 FB0 Θ= FA0/FB0 YC SC/E SD/G SC/D 0.01 1 5 10 100 1000

5 5 5 5 5 5

0.002 0.2 1 2 20 200

0.139038

4.424789

0.070678

3.48E+04

0.111343

5.379092

0.088876

321.5455

0.057955

9.342066

0.168248

52.61217

0.033798

14.42332

0.280156

22.77975

0.001968

107.1563

2.87001

1.931864

5.99E-05

1001.268

30.05269

0.413009



8-53

P8-14 (b) continued

Yc

ΘO2



SC/E

ΘO2



SD/G

ΘO2



SD/G

ΘO2



8-54

P8-14 (c) Now we have a pressure drop parameter α = 0.002, so we modify our polymath program as follows: Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

alpha

0.002

0.002

0.002

0.002

2

Ca

0.098

0.0041647

0.098

0.0041647

3

Cb

0.049

0.0020823

0.049

0.0020823

4

Cc

0

0

0.0004581

3.917E-05

5

Cd

0

0

5.56E-05

3.278E-06

6

Ce

0

0

8.655E-06

9.533E-07

7

Cg

0

0

3.75E-05

3.755E-06

8

Cto

0.147

0.147

0.147

0.147

9

Cw

0

0

8.655E-06

9.533E-07

10 Fa

10.

9.896873

10.

9.896873

11 Fb

5.

4.948437

5.

4.948437

12 Fc

0

0

0.0930717

0.0930717

13 Fd

0

0

0.0082669

0.0077896

14 Fe

0

0

0.0022655

0.0022655

15 Fg

0

0

0.0089238

0.0089238

16 Ft

15.

14.95962

15.

14.95963

17 Ft0

15.

15.

15.

15.

18 Fw

0

0

0.0022655

0.0022655

19 k1

0.014

0.014

0.014

0.014

20 k2

0.007

0.007

0.007

0.007

21 k3

0.014

0.014

0.014

0.014

22 k4

0.45

0.45

0.45

0.45

23 ra

-0.0003709

-0.0003709

-2.782E-06

-2.782E-06

24 rb

-0.0001855

-0.0001855

-1.391E-06

-1.391E-06

25 rc

0.0003037

3.587E-06

0.0003037

3.587E-06

26 rd

6.723E-05

-4.638E-06

6.723E-05

-1.354E-06

27 re

0

0

6.413E-06

5.483E-07

28 rg

0

0

2.502E-05

1.475E-06

29 rw

0

0

6.413E-06

5.483E-07

30 Sae

0

0

3.344E+06

4368.521

31 Scd

0

0

11.94813

11.94813

32 Sce

0

0

1218.255

41.08224

33 Sdg

0

0

37.46977

0.8729093

34 V

0

0

500.

500.

35 y

1.

0.0428242

1.

0.0428242

36 Yc

0

0

0.0094042

0.0094042

8-55

P8-14 (c) continued Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg 8 d(y)/d(V) = -alpha/2/y*(Ft/Ft0) Explicit equations 1

alpha = 0.002

2

Ft0 = 15

3

Cto = 0.147

4

k1 = 0.014

5

Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg

6

k2 = 0.007

7

Cb = Cto*(Fb/Ft)*y

8

k3 = 0.014

9

k4 = 0.45

10 Ca = Cto*(Fa/Ft)*y 11 Cc = Cto*(Fc/Ft)*y 12 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 13 rb = ra/2 14 re = k3*Cc 15 Cd = Cto*(Fd/Ft)*y 16 Ce = Cto*(Fe/Ft)*y 17 Cg = Cto*(Fg/Ft)*y 18 Cw = Cto*(Fw/Ft)*y 19 rg = k4*Cd 20 rw = k3*Cc 21 rd = k2*(Ca)^2-k4*Cd 22 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 23 Yc = Fc/Fa 24 Scd = if (V>0.0001)then (Fc/Fd) else (0) 25 Sae = if (V>0.0001)then (Fa/Fe) else (0) 26 Sdg = if (V>0.0001)then (Fd/Fg) else (0) 27 Sce = if (V>0.0001)then (Fc/Fe) else (0)

8-56

P8-14 (c) continued







8-57

P8-14 (c) continued







8-58

P8-14 (c) continued







8-59

P8-14 (c) continued



P8-14 (d) Since C(Formic acid) is our desired product, so temperature corresponding to the maximum yield of C should be recommended.

8-60

P8-14 (d) continued Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.098

3.689E-08

0.098

3.689E-08

2

Cb

0.049

1.844E-08

0.049

1.844E-08

3

Cc

0

0

0.0150519

1.497E-12

4

Cd

0

0

0.0089776

1.018E-10

5

Ce

0

0

0.0496431

0.0496431

6

Cg

0

0

0.0477138

0.0477138

7

Cto

0.147

0.147

0.147

0.147

8

Cw

0

0

0.0496431

0.0496431

9

Fa

10.

7.43E-06

10.

7.43E-06

10 Fb

5.

3.715E-06

5.

3.715E-06

11 Fc

0

0

1.854

3.016E-10

12 Fd

0

0

1.030757

2.05E-08

13 Fe

0

0

9.999993

9.999993

14 Fg

0

0

9.611354

9.611354

15 Ft

15.

14.91758

29.61135

29.61135

16 Fw

0

0

9.999993

9.999993

17 k1

0.014

0.014

30.64368

30.64368

18 k10

0.014

0.014

0.014

0.014

19 k2

0.007

0.007

7.341E+07

7.341E+07

20 k20

0.007

0.007

0.007

0.007

21 k3

0.014

0.014

6.707E+04

6.707E+04

22 k30

0.014

0.014

0.014

0.014

23 k4

0.45

0.45

984.9755

984.9755

24 k40

0.45

0.45

0.45

0.45

25 ra

-0.0003709

-0.0625305

-1.0E-07

-1.0E-07

26 rb

-0.0001855

-0.0312653

-5.002E-08

-5.002E-08

27 rc

0.0003037

-0.0238556

0.0236439

-5.93E-12

28 rd

6.723E-05

-0.0093009

0.0120515

-3.804E-10

29 re

0

0

0.0773459

1.004E-07

30 rg

0

0

0.0627128

1.003E-07

31 rw

0

0

0.0773459

1.004E-07

32 Sae

0

0

5.205E+05

7.43E-07

33 Scd

0

0

3.394658

0.0147083

34 Sdg

0

0

18.42613

2.133E-09

35 T

300.

300.

550.

550.

36 V

0

0

1000.

1000.

37 Yc

0

0

0.4483122

4.059E-05

8-61

P8-14 (d) continued Differential equations 1 d(Fa)/d(V) = ra 2 d(Fb)/d(V) = rb 3 d(Fc)/d(V) = rc 4 d(Fd)/d(V) = rd 5 d(Fe)/d(V) = re 6 d(Fw)/d(V) = rw 7 d(Fg)/d(V) = rg 8 d(T)/d(V) = 0.25 Explicit equations 1

k20 = 0.007

2

k10 = 0.014

3

k1 = k10*exp((10000/1.97)*(1/300-1/T))

4

k2 = k20*exp((30000/1.97)*(1/300-1/T))

5

k30 = 0.014

6

k3 = k30*exp((20000/1.97)*(1/300-1/T))

7

k40 = 0.45

8

Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg

9

Cto = 0.147

10 Ca = Cto*(Fa/Ft) 11 Cb = Cto*(Fb/Ft) 12 Cc = Cto*(Fc/Ft) 13 Cd = Cto*(Fd/Ft) 14 Ce = Cto*(Fe/Ft) 15 Cg = Cto*(Fg/Ft) 16 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2) 17 rb = ra/2 18 re = k3*Cc 19 k4 = k40*exp((10000/1.97)*(1/300-1/T)) 20 Cw = Cto*(Fw/Ft) 21 rg = k4*Cd 22 rw = k3*Cc 23 rd = k2*(Ca)^2-k4*Cd 24 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd 25 Yc = Fc/Fa 26 Scd = if (V>0.0001)then (Fc/Fd) else (0) 27 Sae = if (V>0.0001)then (Fa/Fe) else (0) 28 Sdg = if (V>0.0001)then (Fd/Fg) else (0)

8-62

P8-14 (d) continued



Temperature corresponding to maximum yield = 367.8 K P8-15 (1) C2H4 + 1/2O2 → C2H4O (2) C2H4 + 3O2 → 2CO2 + 2H2O E + 1/2O → D E + 3O → 2U1 + 2U2

FIO = 0.82 FTO = 0.007626 P8-15 (a) Selectivity of D over CO2

S=

FD FU 1

See Polymath program P8-15-a.pol.

8-63

P8-15 (a) POLYMATH Results Variable W Fe Fo Fd Fu1 Fu2 Finert Ft K1 K2 Pto Pe Po k1 k2 X S r1e r2e

initial value 0 5.58E-04 0.001116 0 1.0E-07 0 0.007626 0.0093001 6.5 4.33 2 0.1199987 0.2399974 0.15 0.088 0 0 -0.0024829 -0.0029803

minimal value 0 1.752E-10 4.066E-05 0 1.0E-07 0 0.007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0 0 -0.0024829 -0.0029803

maximal value 2 5.58E-04 0.001116 2.395E-04 6.372E-04 6.371E-04 0.007626 0.0093001 6.5 4.33 2 0.1199987 0.2399974 0.15 0.088 0.9999997 0.4101512 -3.692E-10 -8.136E-10

final value 2 1.752E-10 4.066E-05 2.395E-04 6.372E-04 6.371E-04 0.007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0.9999997 0.3758225 -3.692E-10 -8.136E-10

Differential equations as entered by the user [1] d(Fe)/d(W) = r1e+r2e [2] d(Fo)/d(W) = 1/2*r1e + 3*r2e [3] d(Fd)/d(W) = -r1e [4] d(Fu1)/d(W) = -2*r2e [5] d(Fu2)/d(W) = -2*r2e Explicit equations as entered by the user [1] Finert = 0.007626 [2] Ft = Fe+Fo+Fd+Fu1+Fu2+Finert [3] K1 = 6.5 [4] K2 = 4.33 [5] Pto = 2 [6] Pe = Pto*Fe/Ft [7] Po = (Pto*Fo/Ft) [8] k1 = 0.15 [9] k2 = 0.088 [10] X = 1 - Fe/0.000558 [11] S = Fd/Fu1 [12] r1e = -k1*Pe*Po^0.58/(1+K1*Pe)^2 [13] r2e = -k2*Pe*Po^0.3/(1+K2*Pe)^2

X = 0.999 and S = 0.376(mol of ethylene oxide)/(mole of carbon dioxide)

8-64

P8-15 (b) Changes in equation from part (a):











dFO 1 and FO (o ) = 0 = r1E + 3r 2 E + RO dW 2 0.12 × 0.0093 0.001116 mol RO = = W 2 kg .s 𝐹! 0 = 0.0093 1 − 0.12 ∗ 0.06 = 4.91𝐸 − 04 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒

See Polymath program P8-15-b.pol. From Polymath program: X = 0.83 S = 0.05 (mol of ethylene oxide)/(mole of carbon dioxide) P8-15 (c) Changes in equation from part (a):

dFE = r1E + r 2 E + R E and FE (o) = 0 dW 0.06 × 0.0093 0.000558 mol RE = = W 2 kg .s











𝐹! 0 = 0.0093 1 − 0.06 ∗ 0.12 = 1.05𝐸 − 03 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒

See Polymath program P8-15-c.pol. From Polymath program: P8-16 The reactions are

X = 0.95 S = 0.46(mol of ethylene oxide)/(mole of carbon dioxide)

1. C + W ! 6 H2 + 6CO 2. L + 7W ! 13 H2 + 10 CO Rate laws are : - r1C = k1C CC CW -r2L = k2L CL Cw2 CT0 = P0/RT = ( 1 atm )/(0.082)(1473) = 0.00828 mol/dm3 FC0 = 0.00411 mol/s FL0 = 0.0185 mol/s FW0 = 0.02 mol/s

8-65

P8-16 (a) See Polymath program P8-16.pol Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

ConvC

0

0

0.8337198

0.8337198

2

ConvL

0

0

0.7379696

0.7379696

3

Ct0

0.0082791

0.0082791

0.0082791

0.0082791

4

Fc

0.0041152

0.0006843

0.0041152

0.0006843

5

Fc0

0.0041152

0.0041152

0.0041152

0.0041152

6

Fco

0

0

0.0342518

0.0342518

7

Fh

0

0

0.0383516

0.0383516

8

Fl

0.0018519

0.0004852

0.0018519

0.0004852

9

Fl0

0.0018519

0.0018519

0.0018519

0.0018519

10 Ft

0.0259671

0.0259671

0.0807757

0.0807757

11 Fw

0.02

0.0070028

0.02

0.0070028

12 k1c

3.0E+04

3.0E+04

3.0E+04

3.0E+04

13 k2l

1.4E+07

1.4E+07

1.4E+07

1.4E+07

14 Mc

0.6666673

0.1108536

0.6666673

0.1108536

15 Mco

0

0

0.9590499

0.9590499

16 Mh

0

0

0.0767032

0.0767032

17 Ml

0.333333

0.0873434

0.333333

0.0873434

18 Mw

0.36

0.1260502

0.36

0.1260502

19 P

1.

1.

1.

1.

20 R

0.082

0.082

0.082

0.082

21 r1c

-0.2509955

-0.2509955

-0.0015102

-0.0015102

22 r2l

-0.3361048

-0.3361048

-0.0003587

-0.0003587

23 T

1473.

1473.

1473.

1473.

24 V

0

0

0.417

0.417



Differential equations 1 d(Fc)/d(V) = r1c 2 d(Fl)/d(V) = r2l 3 d(Fw)/d(V) = r1c+7*r2l 4 d(Fh)/d(V) = -6*r1c-13*r2l 5 d(Fco)/d(V) = -6*r1c-10*r2l Explicit equations 1 T = 1473 2 R = 0.082 3 P=1 4 k2l = 14000000 5 k1c = 30000 6 Ft = Fc+Fl+Fw+Fh+Fco 7 Ct0 = P/(R*T) 8 Fc0 = 0.00411523 9 Fl0 = 0.00185185

8-66

P8-16 (a) continued 10 Mc = Fc*162 11 Ml = Fl*180 12 Mh = Fh*2 13 Mco = Fco*28 14 Mw = Fw*18 15 r1c = -k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft)) 16 r2l = -k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2) 17 ConvC = (Fc0-Fc)/Fc0 18 ConvL = (Fl0-Fl)/Fl0



Plot of Fc vs V



Plot of Fl vs V 8-67

P8-16 (a) continued



Plot of Fw vs V



Plot of Fh vs V

8-68

P8-16 (a) continued

Plot of Fco vs V P8-16 (b) W is the key reactant YC = (-dCC/-dCW) = -r1C/(-r1C -7*r2L) The modification is made in the polymath code for Problem 8.16(a) The plot of YC versus V is as follows;





For plotting YW versus volume; since W is our key reactant then, Yw = -dW/(-dW) = 1 8-69

P8-16 (b) continued Now YL = -dL/(-dW) = -r2L/(-r1C – 7*r2L) The plot is as follows :

Overall selectivity SCO / H 2 = N CO = exit molar flow rate of CO/exit molar flow rate of H2 = N H2



CCO |exit /CH2 | exit at any point in the reactor.



P8-16 (c) Individualized solution

8-70

P8-17 (a) The reactions are (i) L + 3 W ! 3 H2 + 3 CO + char (ii) Ch + 4 W ! 10 H2 + 7CO rate laws at 1200 0C are : -r1L = k1LCLCw2 ; k1L = 3721 (dm3/mol)2/sec -r2CH = k2ChCchCw2 ; k2CH = 1000 ( dm3/mol)2/sec Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

ConL

0

0

0.9899505

0.9899505

2

Ct0

0.2

0.2

0.2

0.2

3

Fch

0

0

0.0075887

0.0047166

4

Fco

0

0

0.0887476

0.0887476

5

Fh

0

0

0.1111269

0.1111269

6

Fl

0.0123

0.0001236

0.0123

0.0001236

7

Fl0

0.0123

0.0123

0.0123

0.0123

8

Ft

0.1233

0.1233

0.2493464

0.2493464

9

Fw

0.111

0.0446317

0.111

0.0446317

10 k1l

3721.

3721.

3721.

3721.

11 k2ch

1000.

1000.

1000.

1000.

12 r1l

-2.406642

-2.406642

-0.0004728

-0.0004728

13 r2ch

0

-0.0997755

0

-0.0048484

14 V

0

0

0.417

0.417

Differential equations 1 d(Fl)/d(V) = r1l 2 d(Fch)/d(V) = r2ch-r1l 3 d(Fw)/d(V) = 3*r1l+4*r2ch 4 d(Fh)/d(V) = -3*r1l-10*r2ch 5 d(Fco)/d(V) = -3*r1l-7*r2ch Explicit equations 1 k2ch = 1000 2 k1l = 3721 3 Ft = Fch+Fl+Fw+Fh+Fco 4 Ct0 = 0.2 5 r1l = -k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2) 6 r2ch = -k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2) 7 Fl0 = 0.0123

8 ConL = (Fl0-Fl)/(Fl0)

8-71

P8-17 (a) continued

P8-17 (b)



8-72

P8-17 (c) SCO/Ch Let S1 = S~CO/Ch = Fco/Fch The plot is as follows

Considering W as the principle reacting species; Yw = 1 and Yl = (rl)/(rw) = (r1l)/(3*r1l + 4*r2ch) the plot is as follows:

P8-17 (d) The molar flow rate for char is maximum at about V= 0.02085 dm3. P8-18 (a) Individualized Solution P8-18 (b) Individualized Solution 8-73





Solutions for Chapter 9 – Reaction Mechanisms, Pathways, Bioreactions and Bioreactors P9-1 (a) Example 9-1 (i) The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the slope of line will decrease. (ii) Intensity increases with increase in concentration of M. (iii) k3 = 3 P9-1 (b) Example 9-2 (i) It is a constant at very low KM (ii) Vmax = 2 kmol/m3s, and KM = 0.04 kmol/m3 (iii) The inhibitor shows competitive inhibition. See Polymath program P9-1-b.pol.



P9-1 (c) Example 9-3 (i) Infinity

(ii) Now Curea = 1 mol/dm3 and t = 15 min = 900 sec.





900 =

!.!"## !.!∗!"!!

ln

! !!!

+

! !.!∗!"!!



Solving, we get X = 0.2136 (iii) For t = 30 min = 1800 sec 1800 =

𝐾! 1 0.1𝑋 ln + 2.4 ∗ 10!! 1−𝑋 2.4 ∗ 10!!

Putting X = 0.5 KM = 0.551 mol/dm3 Thus, KM must be greater than 0.551 mol/dm3 for conversion to be below 50% after half an hour.

9-1

P9-1 (d) Example 9-4





Yes there is disparity as substrate is also used in maintenance. P9-1 (e) Example 9-5 (i) YP/C (ii) 107 g/dm3 (iii) 0.25 h-1 (iv) 15.5 g/g See Polymath program P9-1-e.pol. Calculated values of the DEQ variables Variable t Cc Cs Cp rd Ysc Ypc Ks m umax rsm kobs rg vo Cso Vo V mp

initial value 0 2.0E-04 5.0E-04 0 2.0E-06 12.5 5.6 1.7 0.03 0.33 6.0E-06 0.33 1.941E-08 0.5 5 1 1 0

minimal value 0 1.276E-04 5.0E-04 0 1.277E-06 12.5 5.6 1.7 0.03 0.33 3.83E-06 0.3299706 1.941E-08 0.5 5 1 1 0

maximal value 24 0.002742 4.5794959 0.0159354 2.742E-05 12.5 5.6 1.7 0.03 0.33 8.226E-05 0.33 6.598E-04 0.5 5 1 13 0.20716

ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg-rd-vo*Cc/V [2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V [3] d(Cp)/d(t) = rg*Ypc-vo*Cp/V



9-2

final value 24 0.002742 4.5794959 0.0159354 2.742E-05 12.5 5.6 1.7 0.03 0.33 8.226E-05 0.3299706 6.598E-04 0.5 5 1 13 0.20716

P9-1 (e) Example 9-5 Continued Explicit equations as entered by the user [1] rd = Cc*.01 [2] Ysc = 1/.08 [3] Ypc = 5.6 [4] Ks = 1.7 [5] m = .03 [6] umax = .33 [7] rsm = m*Cc [8] kobs = (umax*(1-Cp/93)^.52) [9] rg = kobs*Cc*Cs/(Ks+Cs) [10] vo = 0.5 [11] Cso = 5 [12] Vo = 1 [13] V = Vo+vo*t [14] mp = Cp*V











(vi) Change the cell growth rate law:

All other equations are the same as in part (v).

9-3

P9-1 (e) Example 9-5 Continued





Notice that uncompetitive inhibition by the substrate causes the concentration of cells to decrease with time, whereas without uncompetitive inhibition, the concentration of cells increased with time. Consequently, the concentration of product is very low compared to the case without uncompetitive inhibition. (vii) Change the observed reaction rate constant: All other equations are the same as in part (v). Since CP is so small, the factor

whether CP* = 93 g/dm3 or CP* = 10,000 g/dm3. Thus the

plots for this part are approximately the same as the plots in part (v). P9-1 (f) Example on the Professional Reference Shelf R9.1 For t = 0 to t = 0.35 sec, PSSH is not valid as steady state not reached. And at low temperature PSSH results show greatest disparity. See Polymath program P9-1-f.pol.

9-4



P9-1 (f) Continued POLYMATH Results Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value t 0 0 12 C1 0.1 2.109E-04 0.1 C2 0 0 1.311E-09 C6 0 0 3.602E-09 C4 0 0 2.665E-07 C7 0 0 0.0979179 C3 0 0 0.0012475 C5 0 0 0.0979179 C8 0 0 6.237E-04 CP5 0 0 0.0979123 CP1 0.1 2.166E-04 0.1 k5 3.98E+09 3.98E+09 3.98E+09 T 1000 1000 1000 k1 0.0014964 0.0014964 0.0014964 k2 2.283E+06 2.283E+06 2.283E+06 k4 9.53E+08 9.53E+08 9.53E+08 k3 5.71E+04 5.71E+04 5.71E+04

12 2.109E-04 1.311E-09 3.602E-09 1.276E-08 0.0979179 0.0012475 0.0979179 6.237E-04 0.0979123 2.166E-04 3.98E+09 1000 0.0014964 2.283E+06 9.53E+08 5.71E+04

ODE Report (STIFF) Differential equations as entered by the user [1] d(C1)/d(t) = -k1*C1-k2*C1*C2-k4*C1*C6 [2] d(C2)/d(t) = 2*k1*C1-k2*C1*C2 [3] d(C6)/d(t) = k3*C4-k4*C6*C1 [4] d(C4)/d(t) = k2*C1*C2-k3*C4+k4*C6*C1-k5*C4^2 [5] d(C7)/d(t) = k4*C1*C6 [6] d(C3)/d(t) = k2*C1*C2 [7] d(C5)/d(t) = k3*C4 [8] d(C8)/d(t) = 0.5*k5*C4^2 [9] d(CP5)/d(t) = k3*(2*k1/k5)^0.5*CP1^0.5 [10] d(CP1)/d(t) = -k1*CP1-2*k1*CP1-(k3*(2*k1/k5)^0.5)*(CP1^0.5)

P9-1 (g) Individualized solution P9-2 (a) The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual. P9-2 (b) Individualized Solution P9-2 (c) Applying the pseudo-steady state hypothesis, ! 𝑟!"#∗ = 𝑘! 𝐶!"# + 𝑘! 𝐶!"# 𝐶! − 𝑘! 𝐶!"# 𝐶!"#∗ − 𝑘! 𝐶!"#∗ − 𝑘! 𝐶!"#∗ 𝐶! = 0

𝐶!"#∗ =

! 𝑘! 𝐶!"# + 𝑘! 𝐶! 𝐶!"# 𝑘! 𝐶!"# + 𝑘! + 𝑘! 𝐶!

𝑟!! !! = 𝑘! 𝐶!"#∗ = 𝑘! 9-5

! 𝑘! 𝐶!"# + 𝑘! 𝐶! 𝐶!"# 𝑘! 𝐶!"# + 𝑘! + 𝑘! 𝐶!

P9-3 Burning:









9-6

P9-3 Continued











9-7

P9-3 Continued



P9-4 (a)



















Active intermediates:









9-8

P9-4 (a) Continued









Plugging in expression for





:







Now, substitute expressions for



into equation for







P9-4 (b)



When



P9-4 (c)

CH3CHO





k 1 k 4 C2 H 6

CH3•





CHO•



k 2

k 3



CH4

CO

H2

9-9

:

P9-5 (a)











P9-5 (b)

9-10

P9-5 (b) Continued



add rCOCl to rCl



P9-5 (c)



The proposed mechanism for this reaction is:

Through this mechanism, it may be deduced that the net rate of formation of HBr (product) is given by: where the concentrations of the intermediates may be determined by invoking the steady state approximation, i.e.:

9-11

P9-5 (c) Continued and:

i.e. from (ii) + (i) we obtain: i.e.:

which substituted back in (i) gives:

i.e.:

i.e.:

or: dividing 'top' and 'bottom' by k`b :

Thus, the net rate of formation of HBr may be written solely in terms of reactants, i.e.,:

9-12

P9-5 (c) Continued

which simplifies to:

as predicted by the empirical rate law:

where:



9-13

P9-6 (a)















9-14

P9-6 (a) Continued



P9-6 (b) Low temperatures with anti-oxidant













9-15

P9-6 (b) Continued



P9-6(c) If the radicals are formed at a constant rate, then the differential equation for the concentration of the radicals becomes: and

The substitution in the differential equation for R· also changes. Now the equation is: and solving and substituting gives:



Now we have to look at the balance for RO2·. and if we substitute in our expression for [R·] we get





which we can solve for [RO2·].

Now we are ready to look at the equation for the motor oil. and making the necessary substitutions, the rate law for the degradation of the motor oil is:





9-16

P9-6 (d) Without antioxidants

With antioxidants

P9-6 (e) Individualized solution P9-7 (a)





9-17

P9-7 (b)

P9-7 (c)









P9-7 (d) See Polymath program P9-7-d.pol.

9-18

P9-7 (d) Continued

Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is much slower than the rate at which a healthy person becomes ill. Eventually everyone is ill and people start dying. P9-7 (e) Individualized solution P9-8 (a) By applying PSSH for the complex [E.S], we have













9-19

P9-8 (a) Continued Since E is not consumed, we have

where



is a constant

So,



where





P9-8 (b)













since E is not consumed:

or



Insert this into the equation for rE·S and solve for the concentration of the intermediate:







9-20

P9-8 (c) (1)



(2)



If we add these two rates we get: (3)



From equation (2) we get







Plug this into equation 3 and we get:











9-21



P9-8 (d)



P9-8 (e) No solution will be given

9-22

P9-9 (a) The enzyme catalyzed reaction of the decomposition of hydrogen peroxide. For a batch reactor:











9-23

P9-9 (a) continued





P9-9 (b)





9-24

P9-9 (c) Individualized solution P9-9 (d) Individualized solution P9-10 (a)











9-25

P9-10 (b)





9-26

P9-10 (b) continued





P9-10 (c)





If ET is reduced by 33%, -rS will also decrease by 33%. From the original plot, we see that if the curve –rS is decreased by 33%, the straight line from the CSTR calculation will cross the curve only once at approximately CS = 40 mmol/L

P9-10 (d) Individualized solution P9-10 (e) Individualized solution P9-11 (a)



9-27

P9-11 (a) continued









P9-11 (b)



9-28

P9-11 (b) continued





P9-11 (c) Individualized solution P9-11 (d) Individualized solution P9-12 For No Inhibition, using regression,

Equation model:







a0 = 0.008

a1 = 0.0266

For Maltose,

Equation model:







a0 = 0.0098

a1 = 0.33



For α-dextran,

Equation model:







a0 = 0.008

a1 = 0.0377



⇒ Maltose show non-competitive inhibition as slope and intercept, both changing compared to no inhibition case.

⇒ α-dextran show competitive inhibition as intercept same but slope increases compared to no inhibition case. 9-29

P9-13 (a)





Now plug the value of (EH) into rP +

At very low concentrations of H (high pH) rS approaches 0 and at very high concentrations of H+ (low pH) rS also approaches 0. Only at moderate concentrations of H+ (and therefore pH) is the rate much greater than zero. This explains the shape of the figure. P9-13 (b) Individualized solution P9-13 (c) Individualized solution P9-14 No solution will be given P9-15 No solution will be given

9-30

P9-16 –rs = (μmax*Cs*Cc)/(Km+Cs) μmax= 1hr-1 Km = 0.25 gm/dm3 Yc/s = 0.5g/g P9-16 (a) Cc0 = 0.1g/dm3 Cs0 = 20 g/dm3 Cc=Cc0 + Yc/s(Cs0-Cs) See polymath problem P9-16-a.pol Calculated values of DEQ variables Variable Initial value Minimal value

Maximal value

Final value

1 Cc

0.1

0.1

10.1

10.1

2 Cc0

0.1

0.1

0.1

0.1

3 Cs

20.

6.301E-11

20.

6.301E-11

4 Cs0

20.

20.

20.

20.

5 Km

0.25

0.25

0.25

0.25

6 rc

0.0493827

1.273E-09

4.043487

1.273E-09

7 rs

-0.0987654

-8.086974

-2.546E-09

-2.546E-09

8 t

0

0

10.

10.

9 umax

1.

1.

1.

1.

0.5

0.5

0.5

10 Ycs 0.5 Differential equations 1 d(Cs)/d(t) = rs Explicit equations 1 Cc0 = 0.1 2 Ycs = 0.5 3 Cs0 = 20 4 umax = 1 5 Cc = Cc0+Ycs*(Cs0-Cs) 6 Km = 0.25 7 rs = -umax*Cs*Cc/(Km+Cs) 8 rc = -rs*Ycs

9-31

P9-16 (a) continued

Plot of Cc and Cs versus time

Plot of rs and rc with time P9-16 (b) Change the polymath code to include rg = μmax*(1-Cc/C∞)*Cc C∞= 1 g/dm3

9-32

P9-16 (b) continued Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Cc

0.1

0.1

0.5751209

0.5751209

2 Cc0

0.1

0.1

0.1

0.1

3 Cinf

1.

1.

1.

1.

4 Cs

20.

19.04976

20.

19.04976

5 Cs0

20.

20.

20.

20.

6 Km

0.25

0.25

0.25

0.25

7 rc

0.0225

0.0225

0.0624999

0.0610892

8 rg

0.09

0.09

0.2499995

0.2443569

9 rs

-0.045

-0.1249997

-0.045

-0.1221784

10 t

0

0

10.

10.

11 umax

1.

1.

1.

1.

0.5

0.5

0.5

12 Ycs 0.5 Differential equations 1 d(Cs)/d(t) = rs Explicit equations 1 Cc0 = 0.1 2 Ycs = 0.5 3 Cs0 = 20 4 umax = 1 5 Cinf = 1 6 Cc = Cc0+Ycs*(Cs0-Cs) 7 Km = 0.25

8 rg = umax * (1-Cc/Cinf) * Cc 9 rs = -Ycs*rg 10 rc = -rs*Ycs

9-33

P9-16 (b) continued



Plot of Cc with time P9-16 (c) Cs0 = 20g/dm3 Cc0 = 0 The dilution rate at which wash-out occurs will be by setting Cc=0 in equation; Cc = Ycs*(Cs0 – (DKs)/(umax – D)) Dmax =



⇨ Dmax =





= 0.987 hr-1

Thus dilution rate at which washout occurs is 0.987hr-1. P9-16 (d) ,

,

,



Divide by CCV,



Now





⇒ DCc = D YC/S (CS0 –

) 9-34



P9-16 (d) continued Now, for



,



⇒ Dmax,prod = 0.88 hr-1

Using this value of D we can find the value of Cs Cs =



= 1.83 g/dm3



= 9.085g/dm3 rs = D(Cs0 – Cs) = 15.98 g/dm3/hr P9-16 (e) Cell death cannot be neglected. Kd =0.02 hr-1 DCc = rg –rd And D(CS0 –Cs)= rS For steady state operation to obtain mass flow rate of cells out of the system, Fc FC =CCv0 = (rg-rd)V= (µ-kd) CCV After dividing by CcV; D=µ-kd Now since maintenance is neglected. Substituting for µ in terms of substrate concentration; Cs =



The stoichiometry equation can be written as : -rs = rg YS/C CC = Yc/s



Now the dilution rate at can be found by substituting Cc =0; = 0.96 hr-1

Thus Dmax =

Similarly the expression for dilution rate for maximum production is given by solving the equation ; (D+kd) Cc = (D+kd ) YC/S (CS0 - Now for





,

-1

Thus we obtain Dmax,prod = 0.86 hr 9-35

)

P9-16 (f) In this case the maintenance cannot be neglected. m = 0.2 g/hr/dm3 The correlation for steady substrate concentration will remain the same. Cs =

But the cell maintenance cannot be neglected. Thus the stoichiometry equation will be changed. The equation will be - -rs = Ys/Crg + mCc => -rs = rg/YS/C + mCc => (Cs0 – Cs) = Cc/YC/S +



Also we know that by mass balance - Using this relation, the stoichiometric equation for substrate consumption changes to - ⇨ Cc =



Now for finding the dilution rate at which wash out occurs, Cc =0; So Dwashout = 0.98 hr-1 Similarly to calculate



,

Thus we obtain Dmax,prod = 0.74 hr-1



P9-16 (g) Individualized solution P9-16 (h) Individualized solution P9-17 Tessier Equation,







Redoing P9-16 part (a) For batch reaction,









,

,

See Polymath program P9-16-a.pol. 9-36



P9-17 continued POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 7 Cs 20 0.0852675 20 Cco 0.1 0.1 0.1 Ycs 0.5 0.5 0.5 Cso 20 20 20 Cc 0.1 0.1 10.057366 k 8 8 8 umax 1 1 1 Ysc 2 2 2 rg 0.0917915 0.0917915 3.8563479 rs -0.183583 -7.7126957 -0.183583 Rates 0.183583 0.183583 7.7126957

7 0.0852675 0.1 0.5 20 10.057366 8 1 2 0.1066265 -0.213253 0.213253



ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = rs

Explicit equations as entered by the user [1] Cco = 0.1 [2] Ycs = 0.5 [3] Cso = 20 [4] Cc = Cco+Ycs*(Cso-Cs) [5] k = 8 [6] umax = 1 [7] Ysc = 2 [8] rg = umax*(1-exp(-Cs/k))*Cc [9] rs = -Ysc*rg [10] RateS = -rs





Redoing P9-16 part (b)





m = CC v0 = rg V = µCCV

( )

Divide by CCV,









9-37



P9-17 continued Now





For dilution rate at which wash out occur, CC = 0



CSO = CS









Redoing P9-16 part (c)











Now



For







,



= 0.628 hr-1



P9-17 (a) Individualized solution P9-17 (b) Individualized solution P9-18 (a) rg = µCC



For CSTR,













7





















9-38

P9-18 (b) Flow of cells out = Flow of cells in Cell Balance:





Substrate Balance:





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 finite sized reactor. Either a cell death rate must be included or the cells cannot be recycled. P9-18 (c) Two CSTR’s For 1st CSTR, V = 5000 dm3,











rg =



See Polymath program P9-18-c-1cstr.pol. POLYMATH Results NLES Solution Variable Value f(x) Ini Guess Cc 4.3333333 9.878E-12 Cs 1.3333333 1.976E-11 umax 0.8 Km 4 Csoo 10 Cso 10 Ysc 2 rg 0.8666667 rs -1.7333333 V 5000 vo 1000 D 0.2 X 0.8666667 Cco 4.33 NLES Report (safenewt) Nonlinear equations [1] f(Cc) = D*(Cc)-rg = 0 [2] f(Cs) = D*(Cso-Cs)+rs = 0

4 5

9-39







P9-18 (c) continued Explicit equations [1] umax = 0.8 [2] Km = 4 [3] Csoo = 10 [4] Cso = 10 [5] Ysc = 2 [6] rg = umax*Cs*Cc/(Km+Cs) [7] rs = -Ysc*rg [8] V = 5000 [9] vo = 1000 [10] D = vo/V [11] X = 1-Cs/Csoo [12] Cco = 4.33

CC1 = 4.33 g/dm3 X = 0.867 CS1 = 1.33 g/dm3 CP1 = YP/CCC1 =0.866 g/dm3 nd For 2 CSTR, See Polymath program P9-18-c-2cstr.pol.

POLYMATH Results NLES Solution Variable Value f(x) Ini Guess Cc 4.9334151 3.004E-10 Cs 0.1261699 6.008E-10 umax 0.8 Km 4 Csoo 10 Cs1 1.333 Ysc 2 rg 0.120683 rs -0.241366 V 5000 vo 1000 D 0.2 X 0.987383 Cc1 4.33

4 5



NLES Report (safenewt) Nonlinear equations [1] f(Cc) = D*(Cc-Cc1)-rg = 0 [2] f(Cs) = D*(Cs1-Cs)+rs = 0 Explicit equations [1] umax = 0.8 [2] Km = 4 [3] Csoo = 10 [4] Cs1 = 1.333 [5] Ysc = 2 [6] rg = umax*Cs*Cc/(Km+Cs) [7] rs = -Ysc*rg [8] V = 5000 [9] vo = 1000 [10] D = vo/V [11] X = 1-Cs/Csoo [12] Cc1 = 4.33

9-40

P9-18 (c) continued CC2 = 4.933 g/dm3 3 CS2 = 1.26 g/dm P9-18 (d) For washout dilution rate,



X = 0.987 CP1 = YP/CCC1 =0.9866 g/dm3

CC = 0









Production rate = CCvO (24hr) = 4.85 g/dm3 x1000dm3/hrx24hr = 116472.56g/day P9-18 (e) For batch reactor,

V = 500dm3,























CCO = 0.5 g/dm3 CSO = 10g/dm3



rg =

See Polymath program P9-18-e.pol.



POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 6 Cc 0.5 0.5 5.4291422 Cs 10 0.1417155 10 Km 4 4 4 Ysc 2 2 2 umax 0.8 0.8 0.8 rg 0.2857143 0.1486135 1.403203 rs -0.5714286 -2.8064061 -0.2972271

6 5.4291422 0.1417155 4 2 0.8 0.1486135 -0.2972271



ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg [2] d(Cs)/d(t) = rs

Explicit equations as entered by the user [1] Km = 4 [2] Ysc = 2 [3] umax = 0.8 [4] rg = umax*Cs*Cc/(Km+Cs) [5] rs = -Ysc*rg

For t = 6hrs, CC = 5.43g/dm3. So we will have 3 cycle of (6+2) hrs each in 2 batch reactors of V = 500dm3. Product rate = CC x no. of cycle x no. of reactors x V = 5.43 g/dm3 x 3 x 2 x 500dm3 = 16290g/day. 9-41

P9-18 (f) Individualized solution P9-18 (g) Individualized solution P9-19 (a)





2

CS gm /dm D(day-1) CS/D

1 1 1

3 1.5 2

4 1.6 2.5

10 1.8 5.6

Slope = 0.5 day intercept = 0.5 gm/dm3 µmax = 2 KS = 0.5*2 = 1 P9-19 (b)

Inserting values from dataset 4













9-42

P9-19 (c) Individualized solution P9-19 (d) Individualized solution P9-20 No solution will be given P9-21 (a) See Polymath program P9-20-a.pol. Calculated values of DEQ variables Variable Initial value

Minimal value

Maximal value

Final value

1 Cc

1.

1.

1.774022

1.774022

2 f

0

0

0.99985

0

3 t

0

0

48.

48.

Differential equations 1 d(Cc)/d(t) = f* Cc * 0.9 / 24 Explicit equations 1 f = (If (((t > 0) And (t < 12)) Or ((t > 24) And (t < 36))) Then (sin(3.14 * t / 12)) Else (0))



P9-21 (b) Given the initial concentration as Co = 0.5 mg/liter we have, CC dCC t # πt & = sin% (µ+dt Co C 0 $ 12 ' C

∫

∫

# # πt && C # 12 & ln C = % (µ+dt %%1− cos % ((( Co $ π ' $ 12 '' $



*,# 12 & # # πt &&., CC = Co exp +% (µ %%1− cos % (((/ $ 12 ''0, -,$ π ' $ 9-43

P9-21 (b) Continued µ = 0.9 day -1 = 0.0375 hr–1 From 6am to 6pm, t= 12 hrs

+-# 12 & # π∗12 &/⇒ CC = Co exp ,% (µ %1− cos (0 12 '-1 -.$ π ' $ +-# 12 & /⇒ CC = Co exp ,% (µ 2 0 -1 .-$ π '

()



+-# 12 & /⇒ CC = Co exp ,% (∗0.0375∗ 2 0 -.$ π ' -1

()

⇒ CC = 1.332/Co Thus the concentration progresses as = 1.332(conc. of previous day)

3

Therefore the time taken to reach the concentration of 200 mg/dm using MS Excel is 22 days. P9-21 (c) Now 𝜇 is a function of Cc.

" C % rg = µCC = µ0 $$1− C ''CC # 200 & rg decreases with increasing cell concentration. It will take an infinitely long time to reach 200 mg/dm3 because the cell density blocks the sunlight, inhibiting growth. If CC ever actually reached 200 mg/dm3, the growth rate would be zero. As we approach 200 mg/dm3, the growth rate slows so much that it is effectively zero, so we would never reach 200 mg/dm3 in a reasonable timescale. P9-21 (d) From part (b) Concentration at the end of 1st day = 1.33Co Concentration at the start of 2nd day =1.33Co/2 + 100 (mg/lit) Concentration at the end of 2rd day= 1.33(1.33Co/2 + 100) And so forth. Thus the concentration progresses as = 1.332(conc. of previous day)+100 Solving for the progression on MS Excel we find the concentration to converge to 299.4011mg/lit Thus, at the end of 365 days the conc. becomes = 299.4011mg/lit 9-44

P9-21 (e) Assuming that dilution and removal of algae is done at the end of the day after the growth period is over. Then, Rate of decrease of conc. =

dCC

= –k CC where, k = 1 (day–1)

dt At t=0, Co = 200mg/lit (i.e. maximum concentration allowed for maximum productivity at the end of the day) So, removal per day:- Cc= Co exp (-kt) = 200 exp (-1 day-1 x 1 day) = 200 exp (-1) ⇒ Cc= 73.5 mg/dm3/day ⇒ Volume of the pond = 5000 gallons or 18940.5 dm3 ⇒ Total mass flow rate of algae = 1392.1 g/day P9-21 (f) Now, since CO2 will affect the rate of growth of algae too, then let the reaction be: → more algae CO2 + Algae !! Hence, A + C !! →2C

Where,&&A :&&CO2



C :&&algae Now, rate of growth of algae should be rc = k sin (πt/12) CaCc Earlier, Ca = 1.69gm/kg of water = 1.69 gm/dm3 of water (density of water = 1kg/lit) We have rate law as rc = µ sin (πt/12) Cc (CO2 concentration was assumed to be a constant = 1.69g /dm3 of water) Hence, µ = k × 1.69 gm / dm3 k = (.9 per day) ÷ (1.69 gm / dm3) k = 0.5325 dm3/g day or 0.022 dm3/g hr

A + C !! →2C At t=0 Ca0 Cc0 0 Let conversion with respect to C is X Then, at t=t Ca0 – XCc0 Cc0 (1-X) 2Cc0X thus, total number of moles of C Cc= Cc0 (1 + X) Ca= Ca0 – XCc0 = Cc0 (M – X) where, M= Ca0/Cc0 Hence,

⇒

dCC dt

= k sin (πt/12) CaCc

( ( )) = k sin (πt/12) Cc0 (1+X) Cc0 (M-X)

d CC 0 1+ X dt

dX = Cc0 k sin (πt/12) (1+X) (M-X) dt X t $ πt ' dX = k Cc0 sin& )dt ⇒ 0 1+ X M− X 0 % 12 ( ⇒

( )∫ ( )( ) )+ M(1+ X) # πt &-+# 12 & = (M+1)kCc0 *1− cos % (.% ( ⇒ ln +, $ 12 '+/$ π ' (M− X) ∫

9-45

P9-21 (f) Continued # & 0 0 )+ )+ # πt &-+# 12 &3 # πt &-+# 12 &3 ⇒ X %1+ exp2k M+1 Cc0 *1− cos % (.% (5 M( = exp2k M+1 Cc0 *1− cos % (.% (5 −1 % ( 21 +, 21 +, $ 12 '+/$ π '54 $ 12 '+/$ π '54 $ '

(

)

(

)

0 # πt &+-# 12 &3 +) exp2k M+1 Cc0 *1− cos % (.% (5 −1 21 $ 12 '/+$ π '54 ,+ ⇒ X = 0 )+ # πt &-+# 12 &3 1+ exp2k M+1 Cc0 *1− cos % (.% (5 M +, 21 $ 12 '+/$ π '54

(

)

(

Now, X= 1 –

)

Cc Cc0

0 # πt &+-# 12 &3 +) exp2k M+1 Cc0 *1− cos % (.% (5 −1 21 $ 12 '/+$ π '54 ,+ Cc Hence, = 1− 0 )+ # πt &-+# 12 &3 Cc0 1+ exp2k M+1 Cc0 *1− cos % (.% (5 M 21 $ 12 '/+$ π '54 ,+ 3 We have, Ca0 = Ks = 2g/dm & Cc0 = 1mg/ dm3 (assuming initial seeding value of algae from (c)) M= Ca0/Cc0 = 2000

(

)

(

)

Substituting the values we have the concentration profile i.e.,

Cc vs. t Cc0

P9-21 (g) Let A: the species of unwanted algae C: the species of desired algae Then, Rate of growth of A = 2 × rate of growth of C

dCa dCc = 2 × dt dt ⇒ µ for A = 2 × µ for C Now overall density of the medium at any time t = Ca + Cc Hence, from given conditions, Ca=0.5(Ca + Cc) So, at that time Ca=Cc Now, ⇒

dCc = µ Cc dt ⇒ Cc = Cc0 exp (µt) ⇒

dCa = 2 µ Ca dt ⇒ Ca = Ca0 exp (2 µt) Assuming, Cco = initial seed concentration of desired algae= 1 mg/liter & Cao = 0.1 mg/liter (given). Now, since the concentration is very less assuming there is no constraint of sunlight ⇒

9-46

P9-21 (g) Continued ⇒ Ca = Cc ⇒ Cc0 exp (µt) = Ca0 exp (2 µt) ⇒ ln (Cco/Cao) = exp (µt) where, µ = 0.9 per day Putting the values, we get t = 2.56 days or 61.4 hrs. Since the number of days is coming less than 4.347 days, which was calculated in part (c), so the effect of daylight can be neglected. Hence, the initial assumption is verified. P9-22 No solution will be given

9-47

Solutions for Chapter 10 – Catalysis and Catalytic Reactors P10-1 (a) Example 10-1 (i)

CT .S Cv KT PT KT PT 0 (1 − X ) KT (1 − X ) = = = CB.s Cv K B PB K B PT 0 X KB X



K B = 1.39 CT .S

Therefore, C

B.s

KT = 1.038

= f(X) can be plotted.



(ii) KB = 3

(iii) KB = 1.2 P10-1 (b) Example 10-2 (i) Increasing the pressure will increase the conversion for the same catalyst weight. At 40 atm, we had 68.2% conversion for a catalyst weight of 10000kg. While, at 80 atm we have 85.34% conversion for the same weight of the catalyst. At 1 atm, we get 0.95% conversion for 10000 kg of catalyst. However, it’s not practically possible to operate at inlet pressure of 1 atm, because there will be no flow of feed into the PFR, due to absence of pressure difference.

(ii)

If the flow rate is decreased the conversion will increase for two reasons: (a) Smaller pressure drop (b) Reactants spend more time in the reactor.



(iii)-(iv) Individualized solution

(v) From figure E10-3.1 we see that when X = 0.6, W = 5800 kg.

(vi) k has to be varied P10-1 (c) Example 10-3 (1) With the new data, model (a) best fits the data (a) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe) Variable Ini guess Value 95% confidence k 3 3.5798145 0.0026691 Kea 0.1 0.1176376 0.0014744 Ke 2 2.3630934 0.0024526 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9969101 0.9960273 0.0259656 0.0096316

(b) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe) Variable Ini guess Value 95% confidence k 3 2.9497646 0.0058793 Ke 2 1.9118572 0.0054165

10-1

P10-1 (c) Example 10-3 (1) (b) Continued Precision R^2 R^2adj Rmsd Variance

= = = =

0.9735965 0.9702961 0.0759032 0.0720163

(c) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1+Ke*Pe)^2) Variable Ini guess Value 95% confidence k 3 1.9496445 0.319098 Ke 2 0.3508639 0.0756992 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9620735 0.9573327 0.0909706 0.1034455

(d) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe^a*Ph^b Variable Ini guess Value 95% confidence k 3 0.7574196 0.2495415 a 1 0.2874239 0.0955031 b 1 1.1747643 0.2404971 Precision R^2 R^2adj Rmsd Variance

= = = =

0.965477 0.9556133 0.0867928 0.107614

P10-1 (c) Example 10-3 (2) (e) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1+Ka*Pea+Ke*Pe)^2) Variable Ini guess Value 95% confidence k 3 2.113121 0.2375775 Ka 1 0.0245 0.030918 Ke 1 0.3713644 0.0489399 Precision R^2 R^2adj Rmsd Variance

= = = =

0.9787138 0.9726321 0.0681519 0.0663527

(f) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ka*Pea) Variable Ini guess Value 95% confidence k 3 44.117481 7.1763989 Ka 1 101.99791 16.763192

10-2

P10-1 (c) Example 10-3 (2) (f) Continued Precision R^2 R^2adj Rmsd Variance

= -0.343853 = -0.5118346 = 0.5415086 = 3.6653942

Model (e) at first appears to work well but not as well as model (a). However, the 95% confidence interval is larger than the actual value, which leads to a possible negative value for Ka. This is not possible and the model should be discarded. Model (f) is the worst model of all. In fact it should be thrown out as a possible model due to the negative R^2 values. P10-1 (d) Example 10-4 (i) About 90% Conversion 1.0 0.8 0.6 0.4 0.2 0.0 0

100

200

300

time 500

400

(ii) Individualized solution (iii) Individualized solution

(iv) As the activation energy increases, it becomes more difficult for the reaction or decay process to occur (from Arrhenius law, rate constant reduces). Hence, with increase in Ed, the conversion increases and with an increase in E, conversion decreases. (v) 1st order reaction, 2nd order decay

Xd =1−

1 (1 + kd t ) k / k d

10-3

P10-1 (d) Example 10-4 (v) Continued

(vi) 2nd order reaction, 1st order decay Rate Law:

N A0

dX d = −rA'W dt

For 2nd order reaction: −r ' = k 'a(t )C 2 A

A

For 1st order decay

da = −kd a , at t = 0, a = 1. dt −k t

So a = e d Stoichiometry:

C A = C A0 (1 − X d ) =

N A0 (1 − X d ) V

Combine:

dX d N W = k '(1 − X d )2 a A02 dt V Let k

= k ' N A0W / V 2 . Substituting for we have

dX d = k (1 − X d ) 2 e − k t dt X t 1 −k t dX = k ∫0 (1 − X )2 d ∫0 e dt d d

d

X k = (1 − e − k t ) 1 − X kd d

X=

1 − e− k t kd / k + 1 − e − k t d

d

10-4

P10-1 (d) Example 10-4 (vi) Continued

(vii) 1st order reaction, 1st order decay Rate Law:

N A0

dX d = −rA'W dt

For 1st order reaction:

−r ' = k 'a(t )CA A

For 1st order decay

da = −kd a , at t = 0, a = 1. dt −k t

So a = e d Stoichiometry:

C A = C A0 (1 − X d ) =

N A0 (1 − X d ) V

Combine:

dX d W = k '(1 − X d )a dt V Let k

= k 'W / V . Substituting for we have

dX d −k t = k(1− X d )e d dt X

t

∫

1 −k t dX = k e d dt 1− X d d

ln

1 k −k t = (1− e d ) 1− X kd

0

∫ 0

X = 1− e



k −kd t (e −1) kd

10-5

P10-1 (d) Example 10-4 (vii) Continued

(viii) 2nd order reaction, 2nd order decay Rate Law:

N A0

dX d = −rA'W dt

For 2nd order reaction:

−r ' = k 'a(t )C 2 A

A

nd

For 2 order decay

da = −kd a 2 , at t = 0, a = 1. dt 1 So a = 1 + kd t Stoichiometry:

C A = C A0 (1 − X d ) =

N A0 (1 − X d ) V

Combine:

dX d N W = k '(1 − X d )2 a A02 dt V Let k

= k ' N A0W / V 2 . Substituting for we have

dXd (1− Xd )2 =k dt 1+ kd t X

t

1 dXd = k dt 2 1+ kd t (1− Xd ) 0 0

∫

1



∫

10-6

P10-1 (d) Example 10-4 (viii) Continued X k = ln(1+ kd t) 1− X kd ln(1+ kd t) X= kd / k + ln(1+ kd t)

P10-1 (e) (i) See polymath program P10-1-e.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

a

1.

1.

1.

1.

2

Ca

0.075

0.0215827

0.075

0.0215827

3

Cao

0.075

0.075

0.075

0.075

4

Fao

30.

30.

30.

30.

5

k

600.

600.

600.

600.

6

kd

0

0

0

0

7

raprime

-3.375

-3.375

-0.2794886

-0.2794886

8

Us

10.

10.

10.

10.

9

W

0

0

22.

22.

0

0

0.7122302

0.7122302

10 X

10-7

P10-1 (e) Continued Differential equations 1 d(a)/d(W) = -kd*a/Us 2 d(X)/d(W) = a*(-raprime)/Fao Explicit equations 1 Us = 10 2 kd = 0 3 Fao = 30 4 Cao = 0.075 5 Ca = Cao*(1-X) 6 k = 600 7 raprime = -k*Ca^2

Hence, X = 0.712 (ii) If the solids and reactants are fed from opposite ends,

da kd a at W = We, a = 1 = dW U S ln a =

kd W + C1 US



C1 =

kdWe US

⎡k ⎤ a = exp ⎢ d (W − We )⎥ ⎣U S ⎦

FA0

dX 2 = kC A2 0 (1 − X ) a dW We

⎡ −k W ⎤ U ⎡k W ⎤ kC 2 X = A0 exp ⎢ d e ⎥ S exp ⎢ d ⎥ 1− X FA0 ⎣ U S ⎦ kd ⎣ US ⎦ 0 kC 2 U X = A0 S 1− X kd FA0

⎛ ⎡ −kdWe ⎤ ⎞ ⎜⎜1 − exp ⎢ ⎥ ⎟⎟ ⎣ US ⎦ ⎠ ⎝

This gives the same expression for conversion as in the example. (iii) Second order decay

a=

1 kdW 1+ US

10-8

P10-1 (e) Continued

kC A2 0U S ⎛ kdW ⎞ X = ln ⎜1 + ⎟ 1− X kd FA0 US ⎠ ⎝ 2

( 0.6)(0.075) U S ⎛ ( 0.72) 22000 ⎞ 1.24 = ln ⎜1 + ⎟ US ( 0.72)(30) ⎝ ⎠ Solve for US by trial and error or a non-linear equation solver. US = 0.902 (iv) If ε = 2 2

(1 − X ) a dX FA0 = kC A2 0 2 dW (1 + ε X ) 2ε (1 + ε ) ln (1 − X ) + ε

12ln (1 − X ) + 4 X +

2

(1 + ε ) X+

2

1− X

X

kC A2 0U S = kd FA0

⎛ ⎡ −kdWe ⎤ ⎞ ⎜⎜1 − exp ⎢ ⎥ ⎟⎟ U ⎣ S ⎦⎠ ⎝

9X = 1.24 1− X

X = 0.372 P10-1 (f)



10-9

P10-1 (f) Continued







P10-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games (ICGs) is given at the front of this Solutions Manual.

10-10

P10-3



P10-3 (a)



P10-3 (b) Adsorption of isobutene limited





10-11

P10-3 (b) Continued



P10-3 (c) Site balance: CT = CV ( 1 + KI CI + KTBACTBA)





P10-3 (d)





P10-3 (e) Individualized solution

10-12

P10-4 (a) Species A & C. P10-4 (b) Figure B tells us that the reaction is irreversible because when PC0 = 1 atm, increasing the product B does not change the rate. If the reaction were reversible with B and C present increasing B would drive the reaction in the reverse direction. P10-4 (c) Assume surface reaction is the rate limiting step

→ A •S A + S← A •S + A •S → B(g) + C •S →C + S C •S← $ & & r €A0 ~ 0 & kA & & C A•S = K APAC V & & rDI kPA2 & −rA = ≈0 & k DI [1+ K APA + KC PC ] 2 % & CC•S = KC PC C V k = k SK 2AC 2t & & −rA = rS = k SK 2APA2C 2V & & C t = C V + K APAC V + KC PC C V & & Ct & CV = &' [1+ K APA + K BPC ] −rA = rS = k SC 2A•S

P10-4 (d)

PB 0 = PC 0 = 0 −rA0 =

€

kPA2 (1 + K APA ) 2



1 1 + K APA = −rA0 k PA 1 1 1 KA = + −rA0 k PA k So one can plot

1 1 versus PA −rA0

to linearize the intial rate data in Figure A.

10-13

P10-4 (e)

C A•S = CV K A PA = CV K A PA 0

(1 − X ) (1 + ε X )

CC •S = CV K C PC = CV K C PA 0

X (1 + ε X )

C A•S = CC •S K A (1 − X ) = K C X 1 − X K C 0.25 = = = 0.5 X K A 0.5 1 = 0.5 X + X = X (1 + 0.5 ) X=

1 = 0.66 1.5

P10-5 (a)

H = H2 E = Ethylene A = Ethane



H2 + C2H4 !cat! → C2H6 H+E !→ !A Because neither H2 or C2H6 are in the denominator of the rate law they are either not adsorbed or weakly adsorbed. Assume H2 in the gas phase reacts with C2H6 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

" C % rE = k AD $PEC V − E•S ' KE '& $#

E•S +H !→ ! A+S

rS = k SCE•SPH 2

Assume surface reaction is rate limiting:

CE•S = KEPEC V rS = k S !"CE•SPH #$ C T = C V + CE•S



! PP # −rA& = k SKEC T ' E H ( '" 1+KEPE ($ P10-5 (b) Individualized solution 10-14

P10-6

⎯⎯ → 2O • S O2 + 2S ←⎯ ⎯





⎯⎯ →2A• S A2 + 2S ←⎯ ⎯

C3 H 6 + O • S → C3 H5OH • S



B + A• S → C • S

⎯⎯ → C3 H 5OH + S C3 H 6OH • S ←⎯ ⎯



C•S →C +S

−rB = rS = k3 PBC A•S ⎡ C2 ⎤ rAD = k A ⎢ PA2 CV2 − A•S ⎥ KA ⎦ ⎣ Assume surface reaction is the rate-limiting step

rAD =0 kA

C AgS = CV K A PA2 −rB = rS = k3 PBCV K A PA2 ⎡ PC ⎤ rC •S = kD ⎢CC •S − C V ⎥ = kD [CC •S − KC PC CV ] KD ⎦ ⎣

rC • S = 0 kD

CC •S = KC PC CV CT = CV + C A•S + CC •S = CV ⎡⎣1 + K A PA 2 + K C PC ⎤⎦

−rB = rS =

k3CT PB K A PA2 1 + K A PA2 + KC PC



P10-7 (a)

10-15

P10-7 (a) Continued







P10-7 (b) From the figure,





10-16

P10-7 (b) Continued





P10-7 (c)



P10-7 (d) Individualized solution P10-7 (e) Individualized solution P10-7 (f) Individualized solution

10-17

P10-8 2ME → DME + W P10-8 (a)



From the plot we can conclude:

1. The partial pressure of DME is greater initially, while the generation rate of water and DME are the same according to stoichiometry.

→ Water must be adsorbed on the surface. → DME is not adsorbed on the surface, otherwise it would take up sites and would not exit for a while. 2. When we reach steady state after a period of time, the exit concentration of DME and water are the same, which is consistent with the stoichiometry. 3. The partial pressure of DME decreases after ~ 80 minutes, this is because water occupies a number of the sites so that there are fewer for ME to adsorb upon and react. P10-8 (b) H2O & ME took longer than others to exit the reactor. For ME, this is because it’s adsorbed on the surface and consumed in the reaction. For H2O, this is because it’s adsorbed on the surface. P10-8 (c) H2O & ME are adsorbed on surface. P10-8 (d) Yes, DME is not adsorbed on the surface. P10-8 (e) With the information above, a probable mechanism could be (assuming surface reaction is irreversible): ME + S ↔ ME•S 2ME•S → DME + W•S + S W•S ↔ W + S Assume surface reaction is the rate-limiting step: ! 𝑟! = 𝑘! 𝐶!"•!

𝐶!"•! = 𝐾!" 𝑃!" 𝐶! 𝐶!•! = 𝐾! 𝑃! 𝐶! 10-18

P10-8 (e) Continued

𝐶! = 𝐶! + 𝐶!"•! + 𝐶!•! ! Therefore, 𝑟!" = 𝑟! =

P10-8 (f) Ans: (4) P10-9

! !!!"

(!!!! !! !!!" !!" )!

! , where 𝑘 = 𝑘! 𝐾!" 𝐶!!



P10-9 (a)













10-19

P10-9 (b)







P10-10





P10-10 (a)

10-20



P10-10 (a) Continued













10-21

P10-10 (b)









Substituting the expressions for CV and CA·S into the equation for –r’A





P10-10 (c) Individualized solution 10-22

P10-10 (d) First we need to calculate the rate constants involved in the equation for –r’A in part (a). We can rearrange the equation to give the following







Thus from the slope and intercept data











See Polymath program P10-10-d.pol. 10-23

P10-10 (d) continued POLYMATH Results Calculated values of the DEQ variables Variable W X e Pao Pa k1 k2 Fao ra rate

initial value 0 0 1 10 10 560 2.04 600 -12.228142 12.228142

minimal value 0 0 1 10 0.0042521 560 2.04 600 -68.5622 2.3403948

maximal value 23 0.9991499 1 10 10 560 2.04 600 -2.3403948 68.5622

final value 23 0.9991499 1 10 0.0042521 560 2.04 600 -2.3403948 2.3403948

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 [3] Pa = 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 = -ra





P10-10 (e) Individualized solution

10-24

P10-10 (f)

Use these new equations in the Polymath program from part (d). See Polymath program P10-10-f.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value W 0 0 X 0 0 y 1 0.0746953 e 1 1 Pao 10 10 Pa 10 7.771E-05 k1 560 560 k2 2.04 2.04 Fao 600 600 ra -12.228142 -68.584462 rate 12.228142 0.0435044 alpha 0.03 0.03

maximal value 23 0.9997919 1 1 10 10 560 2.04 600 -0.0435044 68.584462 0.03

final value 23 0.9997919 0.0746953 1 10 7.771E-05 560 2.04 600 -0.0435044 0.0435044 0.03

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao [2] d(y)/d(W) = -alpha*(1+X)/2/y Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 [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 = -ra [9] alpha = .03





10-25

P10-11 (a)

















P10-11 (b)



10-26

P10-11 (b) Continued

P10-11 (c) The estimates of the rate law parameters were given to simplify the search techniques to make sure that it converged on a false minimum. In real life, one should make a number of guesses of the rate law parameters and they should include a large range of possibilities P10-11 (d)

A ⎯⎯ →B+ H ra =

−kPA

(1 + K A PA + K B PB + K H PH )

PA =

2



PA0 (1 − X ) P X P X , PB = A0 , PH = A0 here, ε = y A0δ = 1(2 − 1) = 1 (1 + X ) (1 + X ) (1 + X )

Where,

k = 0.00137 K A = 4.77 K B = 0.259 K H = 0.424



PA0 = 15atm FA0 = 10mol / sec Now, the design equation for the PFR,

dX −rA = dW FA0 Solving the above equations using Polymath (Refer to P10-11-d.pol)

10-27

P10-11 (d) continued Calculated values of DEQ variables Variable

Initial value

Minimal value

Maximal value

Final value

1

Fao

10.

10.

10.

10.

2

k

0.00137

0.00137

0.00137

0.00137

3

Ka

4.77

4.77

4.77

4.77

4

Kb

0.262

0.262

0.262

0.262

5

Kh

0.423

0.423

0.423

0.423

6

Pa

15.

3.064E-09

15.

3.064E-09

7

Pao

15.

15.

15.

15.

8

Pb

0

0

7.5

7.5

9

Ph

0

0

7.5

7.5

10 ra

-3.904E-06

-1.26E-05

-1.114E-13

-1.114E-13

11 W

0

0

2.0E+06

2.0E+06

12 X

0

0

1.

1.



Differential equations 1 d(X)/d(W) = -ra/Fao Explicit equations 1

Pao = 15

2

k = 0.00137

3

Ka = 4.77

4

Kb = 0.262

5

Kh = 0.423

6

Fao = 10

7

Ph = Pao*X/(1+X)

8

Pa = Pao*(1-X)/(1+X)

9

Pb = Pao*X/(1+X)

10 ra = -k*Pa/((1+Ka*Pa+Kb*Pb+Kh*Ph)^2)



From the above graph, we get the weight of catalyst required for 85% conversion is 1.2083x106 kg. The required catalyst weight is so high because of very low value of reaction constant ’k’. 10-28



P10-12 (a)

Changing Variable Run Pa 1 1 3 10 6 20 7 0.1

Pa Pb 1 1 1 1

Pc 2 2 2 2

reaction rate 0.114 0.18 0.186 0.0243





Changing Variable Run Pa 1 1 2 1 4 1

Pb Pb 1 10 20

Pc 2 2 2

reaction rate 0.114 1.14 2.273



-rA' (mol/gcat s)

2.5

Pa = 1 atm Pc = 2 atm

2

1.5

1

0.5

0 0

5

Changing Variable Run Pa 4 1 5 1

10

15

Pb (atm)

Pc Pb 20 20

Pc 2 10

20

25



reaction rate 2.273 0.926

10-29



-rA' (mol/gcat s)

2.5

2

1.5

Pa = 1 atm Pb = 20 atm

1

0.5

0 0

2

4

6

Pc (atm)

8

10

12



Suggested rate law according to the plots:

−rA ' =

kPA PB 1 + K APA + K C PC



P10-12 (b) Species A and C are on the surface P10-12 (c)

A + S = AgS AgS + B = C gS C gS = C +S −rA ' = rS = kS C AgS PB C AgS = K A PACV CC gS = K C PC CV



Ct = CV + C AgS + CC gS = CV (1 + K A PA + K C PC ) CV =

Ct 1 + K A PA + K C PC

Finally,

−rA ' = rS = kS C AgS PB −rA ' =

kS Ct K A PA PB kPA PB = 1 + K A PA + KC PC 1 + K A PA + KC PC

Where k

= kS Ct K A

10-30



P10-12 (d)

C A•S = CV K A PA = CV K A PA 0 CC •S = CV K C PC = CV K C PA 0

(1 − X ) (1 + ε X ) X (1 + ε X )

CA•S K A (1 − X ) (4)(1 − 0.8) 1 = = = CC •S KC X (13)(0.8) 13

P10-12 (e)

C A•S = CV K A PA = CV K A PA 0

(1 − X ) (1 + ε X )

CC •S = CV K C PC = CV K C PA0

X (1 + ε X )

C A•S = CC •S



K A (1 − X ) = K C X 1 − X K C 13 = = X KA 4 X = 0.235

P10-12 (f)

X dX 1 + K A PA + K C PC = FA0 ∫ dX kPA PB 0 − rA ' 0

X

W = FA0 ∫

1− X X PA = PB = PA 0 ( ), PC = PA 0 ( ) 1+ ε X 1+ ε X ε = 1 − 1 − 1 = −1 1− X X 1 + K A PA0 ( ) + K C PA0 ( ) 1 + ε X 1 + ε X W = FA0 ∫ dX 1− X 2 0 kPA 0 ( ) 1+ ε X X 0.9 1 + (4)(2) + (13)(2) 1 − X dX = 2∫ (2.5)(2) 0 0.9 17 26 1 = 2∫ − + ( )dX 5 5 1− X 0 X

17 26 1 X + ln( )]0.9 0 5 5 1− X = 8.9 gcat = 2[−

10-31

P10-13 (a) Proposed Single site Mechanism:

⎯⎯ → S '• H 2O S '+ H 2O ←⎯ ⎯ 1 S '• H 2O ⎯⎯ → S '• H 2 + O2 2 ⎯⎯ → S '+ H 2 S '• H 2 ←⎯ ⎯ 1 S '+ O2 ⎯⎯ →S 2 Rate of adsorption: rAD = k AD (Cv ' PH O − 2

CS '• H 2O K H 2O

)

Rate of surface reaction: rs = k s CS '• H 2O ---------- (1) Rate of desorption: rD = k D (CS '• H 2 − K H 2 PH 2 Cv ' ) Assuming, surface reaction to be rate limiting, we get

⇒

rAD rD ≅ ≅ 0 k AD k D

∴ CS '• H 2O = K H 2O Cv ' PH 2O and CS '• H 2 = K H 2 PH 2 Cv ' From total site balance, we get

Cv ' =

Ct 1 + K H 2O PH 2O + K H 2 PH 2



Putting all the values in equation (1), we get

rs =

ks K H 2OCt PH 2O 1 + K H2O PH 2O + K H 2 PH 2

⇒ rs =



kPH 2O 1 + K H2O PH 2O + K H 2 PH 2

Where, k = k s K H 2O Ct



P10-13 (b) Proposed Mechanism:

⎯⎯ → S '• H 2O S '+ H 2O ←⎯ ⎯ ⎯⎯ → S • H2 S '• H 2O ←⎯ ⎯



⎯⎯ → S + H2 S • H 2 ←⎯ ⎯

Rate of adsorption: rAD = k AD (Cv ' PH O − 2

CS '• H 2O K H 2O

)

10-32

P10-13 (b) Continued Rate of surface reaction: rs = k s (CS '• H 2O − CS • H 2 / K S ) ---------- (1) Rate of desorption: rD = k D (CS • H 2 − K H 2 PH 2 Cv ) Assuming, surface reaction to be rate limiting, we get

⇒

rAD rD ≅ ≅ 0 k AD k D

∴ CS '• H 2O = K H 2O Cv ' PH 2O and CS • H 2 = K H 2 PH 2 Cv From total site balance, we get

Cv ' =

Ct ' 1 + K H 2O PH 2O

Cv =

Ct 1 + K H 2 PH 2

Putting all the values in equation (1), we get

K H O PH OCt ' K H 2 PH 2 Ct rs = ks ( 2 2 − ) 1 + K H2O PH2O K S (1 + K H 2 PH 2 )

P10-13 (c) Proposed Mechanism: Step (1)

⎯⎯ → S '• O2 S + hυ ←⎯ ⎯ ⎯⎯ → S '+ O2 S '• O2 ←⎯ ⎯



⎯⎯ → S '• H 2O S '+ H 2O ←⎯ ⎯ Step (2)

⎯⎯ → S • H2 S '• H 2O ←⎯ ⎯



⎯⎯ → S + H2 S • H 2 ←⎯ ⎯ For Step (1) Rate of adsorption of O2: rAD

O2

= k D O 2 (Cv −

CS '•O2 KS

)

Rate of Reaction: rSO 2 = k SO 2 (CS '•O2 − K O2 PO2 Cv ' )

Q Rate of surface reaction of water is rate controlling, we get ⇒

rAD O 2 kD O 2

≅

rS O 2 kS O 2

≅ 0

10-33

P10-13 (c) Continued

∴ CS '•O2 = Cv K S ----------- (2) CS '•O2 = K O2 PO2 Cv ' ------- (3) Equating (2) and (3), we get

Cv =

K O2 PO2 Cv ' KS

-------- (4)

Now, here the site balance gets modified because oxygen is also getting adsorbed.

Ct = Cv + Cv ' + CS '•O2 + CS '• H 2O + CS • H 2 Putting, the values and replacing Cv using equation (4), we get

Cv ' =

Ct KO2 PO2 KO2 K H 2 PO2 PH 2 (1 + KO2 PO2 + K H 2O PH 2O + + ) KS KS

And, Cv

=

K O2 PO2 Cv ' KS

=

KO2 PO2 Ct KO2 PO2 KO2 K H 2 PO2 PH 2 K S (1 + KO2 PO2 + K H 2O PH 2O + + ) KS KS

Putting all the values in equation (1), we get

rs = ks (

K H 2O PH 2OCt K H 2 PH 2 KO2 PO2 Ct − ) KO2 PO2 KO2 K H 2 PO2 PH 2 KO2 PO2 KO2 K H 2 PO2 PH 2 (1 + KO2 PO2 + K H 2O PH 2O + + ) K S (1 + KO2 PO2 + K H 2O PH 2O + + ) KS KS KS KS

P10-14 Assume the rate law is of the form rDep = At high temperatures K ↓

2 kPVTIPO 2 1 + KPVTIPO

2 > 1 2

rDep =

2 kPVTIPO k = 2 KPVTIPO K

This fits the high pressure data At PVTIPO = 1.5, r = 0.095 and at PVTIPO = 2, r = 0.1 Now find the activation energy At low pressure and high temperature k = 11.2 At low pressure and low temperature k = 0.4

⎛ k ⎞ E ⎛ 1 1 ⎞ E ⎛ T −T ⎞ ln ⎜ 2 ⎟ = ⎜ − ⎟ = ⎜ 2 1 ⎟ 1 2 ⎠ ⎝ k1 ⎠ R ⎝ T1 T2 ⎠ R ⎝ TT ⎛ 11.2 ⎞ E ⎛ 473 − 393 ⎞ ln ⎜ ⎟ ⎟= ⎜ ⎝ 0.4 ⎠ R ⎜⎝ (473)(393) ⎟⎠

E = 7743 R E = 15330.65

cal mol



10-35

P10-15



P10-16 (a)

Using Polymath non-linear regression few can find the parameters for all models: See Polymath program P10-16.pol. (1) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM^a*PH2^b Variable k a b Precision R^2 R^2adj Rmsd Variance

Ini guess 1 0.1 0.1

= = = =

Value 1.1481487 0.1843053 -0.0308691

95% confidence 0.1078106 0.0873668 0.1311507

0.7852809 0.7375655 0.0372861 0.0222441

α = 0.184 β = -0.031 k = 1.148

10-36

P10-16 (a) Continued (2) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM/(1+KM*PM) Variable Ini guess k 1 KM 2

Value 12.256274 9.0251862

95% confidence 2.1574162 1.8060287

Model: rT = k*PM*PH2/((1+KM*PM)^2) Variable Ini guess Value k 1 8.4090333 KM 2 2.8306038

95% confidence 18.516752 4.2577098

Precision R^2 R^2adj Rmsd Variance

= = = =

0.9800096 0.9780106 0.0113769 0.0018638

k = 12.26 KM = 9.025 (3) POLYMATH Results Nonlinear regression (L-M)

Precision R^2 R^2adj Rmsd Variance

= -4.3638352 = -4.9002187 = 0.1863588 = 0.5001061

k = 8.409 KM = 2.83 (4) POLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2 Variable Ini guess Value k 1 101.99929 KM 2 83.608282 KH2 2 67.213622

95% confidence 4.614109 7.1561591 5.9343217

Nonlinear regression settings Max # iterations = 300 Precision R^2 R^2adj Rmsd Variance

= -3.2021716 = -4.1359875 = 0.1649487 = 0.4353294

k = 102 KM = 83.6 KH2 = 67.21

10-37

P10-16 (b) We can see from the precision results from the Polymath regressions that rate law (2) best describes the data. P10-16 (c) Individualized solution P10-16 (d) We have chosen rate law (2) Proposed Mechanism

⎯⎯ →M • S M + S ←⎯ ⎯



M • S → H 2( g ) + T( g )



Rate of adsorption: rAD = k AD (Cv PM − Rate of surface reaction: rs

CM • S ) KM

= ks CM •S ---------- (1)

Assuming, surface reaction to be rate limiting:

⇒

rAD ≅ 0 k AD

⇒ CM •S = Cv PM K M Putting, the value of CM •S in Equation (1)

rs = ksCv PM K M ----------------- (2) Now, applying site balance

Ct = Cv + CM •S ⇒ Ct = Cv + Cv PM K M ⇒ Cv =

Ct 1 + PM K M

Putting in Equation (2), we get

⇒ rs =

ks K M Ct PM 1 + K M PM

⇒ rs =

kPM Where, k = ks K M Ct 1 + K M PM





10-38

P10-17 (a)

––A better solution is to put this equation directly into polymath.









10-39

P10-17 (b)



assume CA changes very slowly w.r.t. “a” changing







10-40

P10-17 (c) Find the new equation for a: again assuming CB changes slowly w.r.t. “a”. A better solutions is to put

da = –kaC B into the Polymath program dτ





10-41



P10-17 (d)







10-42

P10-17 (e)



da kd C Aa Not a good solution. Just put = into Polymath program. dW Us





10-43

P10-17 (e) Continued







P10-18 (a)

da = −kD dt

W = USt

dt da −k D = dW dt U S If 0 = 1 −



a = 1−

dW = U S dt

k DW US

U k DW .5 then W = S = = 2.5 kg US k D .2 2

akC A2 0 (1 − X ) dX −rA a = = ( −rA ( 0 )) = dW FA0 FA0 FA0 2

dX ⎛ kDW ⎞ kC A0 (1 − X ) = ⎜1 − ⎟ dW ⎝ US ⎠ v0 dX

∫ (1 − X )

2

=

kC A0 k W 1 − D dW ∫ v0 US

10-44

P10-18 (a) Continued Activity is zero for W > 2.5 kg, so the catalyst weight only goes to the effective weight. 2 0.2 ( 2.5 ) ⎤ kC A0 ⎡ k DWe2 ⎤ 1( 0.2 ) ⎡ X = ⎢ 2.5 − ⎥ = 0.25 ⎢We − ⎥= 1− X v0 ⎣ 2U s ⎦ 1 ⎢ 2*0.5 ⎥ ⎣ ⎦

X = 0.2 P10-18 (b)

P10-18 (c) For infinite catalyst loading a = 1. 2

dX kC A0 (1 − X ) = dW v0

kC X = A0 W = 1 1− X v0 X = 0.5

P10-18 (d)

kC X = A0 1− X v0

⎡ k DW 2 ⎤ W − ⎢ ⎥ 2 U s ⎦ ⎣

⎡ 0.2*25 ⎤ 0.4 = 0.2 ⎢5 − ⎥ 1 − 0.4 2 U s ⎣ ⎦

U S = 1.5

kg s

P10-18 (e)

a = 1−

k DW US

0 = 1−

k DW US

U S = kDW = 0.2*5 = 1

kg s

P10-18 (f) a = 0 means there is no reaction is taking place. Activity can never be less than 0. 10-45

P10-18 (g)

U = −U S da k D = dW U S



a=

k DW +C US

1=

k DWt +C US

a=

k DWe k W +1− D t US US

when W

= Wt , a = 1

Now find We.

0=

k DWe k W +1− D t US US

We =

U S ⎡ kDWt ⎤ .5 ⎡ .2*5 ⎤ − 1⎥ = ⎢ − 1⎥ ⎢ kD ⎣ U S .2 .5 ⎣ ⎦ ⎦

We = 2.5 2

⎞ kC (1 − X ) dX ⎛ kDWt = ⎜1 − + W ⎟ A0 dW ⎝ US v0 ⎠ ⎛ k W ⎞ kC X = A0 ∫ ⎜1 − D t + W ⎟dW 1− X v0 ⎝ US ⎠ ⎡ ⎛ kDWt ⎞ Wt 2 − We2 ⎤ W − W ⎢( t ⎥ ⎟+ e ) ⎜1 − U 2 ⎢⎣ ⎥⎦ S ⎝ ⎠ ⎡ X ⎛ 0.2*5 ⎞ 25 − 6.25 ⎤ = 0.2 ⎢( 5 − 2.5 ) ⎜1 − ⎟+ ⎥ 1− X 0.5 ⎠ 2 ⎝ ⎣ ⎦

kC X = A0 1− X v0

X = 1.875 1− X X = 0.65 P10-18 (h)

$ = 160 FA0 X − 10U S a = 1−

k DW US

⎛ k W dX = ka = k ⎜1 − D dW US ⎝

⎞ ⎟ ⎠

kk DW 2 X = kW − 2U S 10-46

P10-18 (h) Continued To maximize profit, a maximum in profit is reached and so we set the differential of profit equal to 0.

d$ dX = 0 = 160 FA0 − 10 dU S dU S kk W 2 dX = D 2 dU S 2U S kkDW 2 160 FA0 = 10 2U S2

8FA0 kk DW 2 = U S2 U S = 8FA0 kk DW 2 = 8 ( 2 )(.2 )(.2 )( 25 )

US = 4

kg min

P10-19 (a)



10-47

P10-19 (b)



P10-19 (c)





10-48

P10-19 (d)





P10-19 (e)



10-49

P10-19 (e) Continued



P10-20 (a)

P10-20 (b) P10-20 (c)



10-50

P10-20 (c) Continued



P10-20 (d)



10-51

P10-20 (d) Continued





10-52

P10-20 (e)

P10-21 (a)

10-53

P10-21 (a) Continued



P10-21 (b)



10-54

P10-21 (b) Continued

P10-22 (a)









10-55

P10-22 (a) continued







10-56

P10-22 (b)





P10-23





10-57

P10-23 continued





10-58

Solutions for Chapter 11 – Non-isothermal Reactor Design-The Steady-State Energy Balance and Adiabatic PFR Applications P11-1 (a) Example Table 11-2 (i) T0 and CPA have the greatest effect, FA0 and k0 have the least effect on the temperature profiles. (ii) -3500 Cal/mol (iii) The maximum of the rate increases with an increase in T0 and the distance down the reactor where this maximum is achieved decreases. (iv) Xe reduces and X increases for the same distance down the reactor with an increase in T0. Since the reaction is exothermic, according to Le Chatelier’s principle, the equilibrium conversion reduces. However, with an increase in inlet temperature, the rate of the reaction increases. Hence, the conversion is expected to increase. (v) Individualized answer P11-1 (b) Example 11-3 (i) No solution will be provided (ii) A minimum feed temperature of 320 K must be maintained. (iii) The critical value is 0.6. (iv) At low values of heat of reaction, the heat that is released is negligible. As a result, the reaction mixture does not get heated so much. Hence, the temperature along the reactor remains virtually a constant. (v) T0 affects the rate the most. The maximum of the rate increases with an increase in T0 and the distance down the reactor where it is achieved reduces. (vi) Set Vfinal = 0.8 m3 See Polymath program P11-1-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable V X Ca0 Fa0 T Kc k Xe ra rate

initial value 0 0 9.3 146.7 330 3.099466 4.2238337 0.7560658 -39.281653 39.281653

minimal value 0 0 9.3 146.7 330 2.852278 4.2238337 0.7404133 -56.196156 39.281653

maximal value 0.8 0.2603491 9.3 146.7 341.27312 3.099466 9.3196276 0.7560658 -39.281653 56.196156

ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fa0 Explicit equations as entered by the user [1] Ca0 = 9.3 [2] Fa0 = .9*163 [3] T = 330+43.3*X

11-1

final value 0.8 0.2603491 9.3 146.7 341.27312 2.852278 9.3196276 0.7404133 -56.196156 56.196156

[4] [5] [6] [7] [8]

Kc = 3.03*exp(-830.3*((T-333)/(T*333))) k = 31.1*exp(7906*(T-360)/(T*360)) Xe = Kc/(1+Kc) ra = -k*Ca0*(1-(1+1/Kc)*X) rate = -ra

(vii) Change the entering temperature T0 in the polymath code above (highlighted in yellow), and record the exiting conversion: T X

330 0.26

340 0.54

350 0.68

370 0.66

390 0.65

420 0.62

450 0.59

500 0.55

600 0.48

X

(viii) We see that there is a maximum in the exit conversion corresponding to T0 = 350K. Therefore, the inlet temperature we would recommend is 350K. (ix)



P11-1 (c) Aspen Problem No solution will be provided.

11-2

P 11-1 (d) (i)

(ii) At high To, the graph becomes asymptotic to the X-axis, that is the conversion approaches 0. At low To, the conversion approaches 1. (iii) On addition of equal molar inerts, the energy balance reduces to:

(



)

CP T −T0 +CP (T −T0 ) A I #  & %$ΔHRx (' CP = CP A I

XEB =

⇒ XEB =

(

2CP T −T0 A

)

#  & ΔH $% Rx '(

(iv) It is observed that at the same inlet temperature, the equilibrium conversion on the addition of inerts is greater. This also follows from the Le Chatelier’s principle.

11-3

P11-1 (e) Example 11-5

 P ΔT Q = mC Q = 220kcal / s = ⇒m



220 ×103 18(400 −270) = 854.7mol / s

For the2nd case: Hot Stream: 460K ! 350K Cold Stream: 270K ! 400K LMTD = 69.52 oC ⇒ A =

220 ×103 100 ×69.52 = 31.64m2

P11-1 (f) Example 11-6 (i) The optimum feed temperature is 480 K. (ii) The optimum entering temperature does not change. However, the conversion increases when compared to the base case. (iii) When θI is set to 3, the new optimum feed temperature becomes 510 K. (iv) Addition of inert increases both the conversion and equilibrium conversion. To achieve 50% conversion, θI must be set to 1.5 (v) Temperature in the reactor increases with the heat of reaction initially, and then decreases after a point. The exit temperature is maximum for a heat of reaction of -13,000 (vi) Individualized solution P11-2

C rA = k(C A − B ) = 0 KC C X rA = k(C A0 (1− X)− A0 ) KC rA = 0, then



X (1− X e ) = e KC KC =

Xe

1− X e

11-4

P11-2 Continued

Since KC (T2 ) = KC (T1 )exp[ We have

X e2

1− X e2

=

X e1

! ΔHRx 1 1 ( − )] R T1 T2

1− X e1

exp[

! ΔHRx 1 1 ( − )] R T1 T2



!

ΔH 0.8 0.5 1 1 = exp[ Rx ( − )] 1− 0.8 1− 0.5 R (227+273) (127+273) ! ΔHRx = −23.05 kJ / mol

P11-3 (a)

True. Endothermic & adiabatic so T decreases, for endothermic reaction, Xe decreases as T decreases. Once X reaches Xe, T doesn’t change and therefore Xe reaches a plateau.

P11-3 (b) False. If the reaction is endothermic and heated along the length, T will decrease and then increase along the reactor. Therefore, Xe will decrease, reach a minimum, and then increase along the reactor for an endothermic reaction. P11-3 (c)

False. If the reaction is exothermic and cooled along the length, T will increase and then decrease along the reactor. Therefore, Xe will decrease, reach a minimum, and then increase along the reactor for an exothermic reaction.

P11-3 (d) True. Exothermic & adiabatic so T increases, for exothermic reaction, Xe decreases as T increases. Once X reaches Xe, T doesn’t change and therefore Xe reaches a plateau. P11-4

A+ B → C Since the feed is equimolar, CA0 = CB0 = .1 mols/dm3 CA = CA0(1-X) CB = CB0(1-X) Adiabatic:

T = T0 +

X[−ΔHR (T0 )] θ C + XΔC

∑i

pi

P



ΔCP = C pC −C pB −C pA = 30 −15−15 = 0

(

)

ΔHR (T) = HC − HB − HA = − 41000 −(−15000 − −20000 = −6000 cal / mol A cal ∑θiCpi = CpA +θBCpB = 15+15 = 30 mol K T = 300 +

6000 X = 300 +200X 30 11-5



P11-4 Continued

−rA = kC 2A0 (1− X)2 = .01k(1− X)2 VPFR = FA0

dX

∫ −r



A

P11-4 (a)

For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3 See Polymath program P11-4-a.pol. Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca0

0.1

0.1

0.1

0.1

2 Fa0

0.2

0.2

0.2

0.2

3 k

0.01

0.01

7.85772

7.85772

4 ra

-0.0001

-0.0018941

-4.3E-20

-4.3E-20

5 T

300.

300.

500.

500.

6 T0

300.

300.

300.

300.

7 V

0

0

3.441E+09

3.441E+09

8 X

0

0

1.

1.

Differential equations 1 d(V)/d(X) = -Fa0 / ra Explicit equations 1 Ca0 = .1 2 Fa0 = .2 3 T0 = 300 4 T = T0 + 200 * X 5 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T)) 6 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)

The conversion and temperature both increase down the reactor volume because the reaction is exothermic. They increase faster because the reaction rate increases as the temperature increases. 11-6

P11-4 (b) T = T0 + 200X T0 = T - 200X = 550 – 200(1) = 350 K P11-4 (c) T and therefore k is constant under isothermal condition

V = FA0

dX

∫ −r

A 2 −rA = kC A0 (1− X)2

V= V= X=

FA0

kC 2A0 FA0

kC 2A0

∫ (

dX (1− X)2

1 −1) 1− X



1 FA0

kVC 2A0

+1

Qr = FA0 X(−ΔHR (T0 )) =

−FA0 ΔHR (T0 ) FA0

kVC 2A0

+1

=

−(0.2)(−6000) 1200 = (0.2) 200,000 +1 +1 2 V (0.01)V(0.01)



11-7

P11-4 (d)

C rA = k(C A CB − c ) = 0 KC C X rA = k(C2A0(1− X)2 − A0 ) KC rA = 0, then 2+ Xe =

1 1 2 −[(2+ ) − 4]0.5 KC C A0 KC C A0

2 See Polymath program P11-4-d.pol. Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value 1

Ca0

0.1

0.1

0.1

0.1

2

dHrx

-6000.

-6000.

-6000.

-6000.

3

Fa0

0.2

0.2

0.2

0.2

4

k

0.01

0.01

0.010583

0.010583

5

Kc

22.29685

22.11524

22.29685

22.11524

6

ra

-0.0001

-0.0001047

-0.0001

-0.0001047

7

T

300.

300.

301.0234

301.0234

8

T0

300.

300.

300.

300.

9

V

0

0

10.

10.

10 X

0

0

0.005117

0.005117

11 Xe

0.518003

0.5166571

0.518003

0.5166571

Differential equations 1 d(X)/d(V) = -ra / Fa0 Explicit equations 1 Ca0 = .1 2 Fa0 = .2 3 T0 = 300 4 T = T0 + 200 * X 5 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T)) 6 dHrx = -6000 7 Kc = 10*exp((dHrx/8.314)*(1/450-1/T)) 8 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2-Ca0*X/Kc) 9 Xe = 0.5*(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)

11-8

P11-4 (d) Continued

The conversion and temperature are both lower than in part (a), because the reaction rate is lower due to reversible reaction.



P11-5 (a)

11-9

P11-5 (a) Continued







P11-5 (b) Individualized solution P11-5 (c) dX V = FA0 −rA

∫

−rA = kC A0 (1− X) / (1+ X) V=

FA0

kC A0

1+ X

∫ 1− X dX

v 1 V = 0 (2ln − X) k 1− X



11-10

P11-5 (c) Continued

Qr = FA0 X(−ΔHR (T0 )), so X= v V = 0 (2ln k

1−

1 Qr

Qr

FA0 (−ΔHR (T0 )) −

FA0 (−ΔHR (T0 ))

Pv ),where FA0 = 0 0 FA0 (−ΔHR (T0 )) RT0 Qr

Qr vs V can be plotted according to the above equation. Qr vs. V 30 25

Qr

20

15 10 5 0 0

10

20

30

V

P11-5 (d) & (e)

40

50







11-11

P11-5 (d) & (e) continued









11-12

P11-5 (d) & (e) continued









11-13

P11-5 (d) & (e) continued











P11-5 (f) Individualized Solution

P11-6 (a)

A → B +C C A = CT

θI =

FA FT

FI FA

CT = C A + C I FT = FA + FI



(

C A01 = C A0 + C I 0 C A01 =

C A0 + C I 0 θ I +1

FA0 + FI 0 A0

)F

11-14

P11-6 (b) Mole balance:

dX −rA = dV FA0 Rate law: −rA = kC A Stoichiometry: C A = C A01

1− X T0 1+ ε X T

ε = y A0δ

δ = 1+1−1= 1 y A0 =

ε= T=

FA0 FA0 1 = = FT 0 FA0 + Fi0 1+ θ i

1 1+ θ i

(

)

− X ΔH RX + C PA + θ iC Pi T0 C PA + θ iC Pi



Enter these equations into Polymath See Polymath program P11-6-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value V 0 0 X 0 0 Cao 0.0221729 0.0221729 Cio 0.0221729 0.0221729 theta 100 100 Fao 10 10 Cao1 4.391E-04 4.391E-04 e 0.009901 0.009901 To 1100 1100 dHrx 8.0E+04 8.0E+04 Cpa 170 170 Cpi 200 200 T 1100 1098.3458 k 25.256686 24.100568 ra -0.0110894 -0.0110894

maximal value 500 0.417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1100 25.256686 -0.0061524



ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fao



11-15

final value 500 0.417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.100568 -0.0061524

P11-6 (b) Continued Explicit equations as entered by the user [1] Cao = 2/(.082*1100) [2] Cio = Cao [3] theta = 100 [4] Fao = 10 [5] Cao1 = (Cao+Cio)/(theta+1) [6] e = 1/(1+theta) [7] To = 1100 [8] dHrx = 80000 [9] Cpa = 170 [10] Cpi = 200 [11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi) [12] k = exp(34.34-34222/T) [13] ra = -k*Cao1*(1-X)*To/(1+e*X)/T





P11-6 (c) There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the temperature low to favor the endothermic reaction. As θ increases beyond 8, there is so much more inert than reactants that the rate law becomes the limiting factor. P11-6 (d) The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The new code is not shown, but the plots are below. 11-16

P11-6 (d) Continued See Polymath program P11-6-d.pol.



The maximum conversion occurs at low values of theta (θ < 8) because the reaction is now exothermic. This means heat is generated during the reaction and there is no advantage to adding inerts as there was in the endothermic case. P11-6 (e) We need to alter the equations from part (c) such that − rA = kC A2 and CA0 = 1 A plot of conversion versus theta shows a maximum at about θ = 5. See Polymath program P11-6-e.pol.



11-17

P11-6 (f)

" CC % We need to alter the equations from part (c) such that −rA = k $C A − B C ' KC '& $#

1− X T0 . Now we need expressions for CB and CC. From stoichiometry 1+ ε X T X T0 we can see that CB = CC. In terms of CA0 we find that: CB = CC = C A0 1+ ε X T ) ΔH # 1 1 &, ) 80000 # 1 1 &, − (. We also need an equation for KC: KC = KC1 exp+ RX %% − ((. = 2exp+ % +* 8.314 $ 1100 T '.+* R $ T1 T '.When we enter these into Polymath we find that the maximum conversion is achieved at approximately θ = 8. See Polymath program P11-6-f.pol. We already know that C A = C A0



P11-6 (g) See Polymath program P11-6-g.pol.



11-18

P11-7 (a) Adiabatic Mole Balance

(1)

dX = −rAʹ FA0 dW







W = ρbV







rʹ ρ r dX =− A B =− A dV FA0 FA0

Rate Law

(2)

⎡ C ⎤ rA = −k ⎢C A − B ⎥ KC ⎥⎦ ⎢⎣





(3)

⎡ ⎛ ⎤ E 1 1⎞ k = k1 exp⎢ ⎜⎜ − ⎟⎟⎥ ⎢⎣ R ⎝ T1 T ⎠⎥⎦





(4)

⎡ ΔH ⎛ ⎤ 1 1⎞ KC = KC2 exp⎢ Rx ⎜⎜ − ⎟⎟⎥ ⎢⎣ R ⎝ T2 T ⎠⎥⎦

Stoichiometry

(5)

C A = C A0 1− X y T0 T



(6)



(

)( ) CB = C A0 Xy (T0 T )

dy α FT ⎛⎜ T ⎞⎟ α ⎛⎜ T ⎞⎟ = = − dW y FT ⎜⎝ T0 ⎟⎠ 2y ⎜⎝ T0 ⎟⎠ 0





W = ρV



αρ ⎛ T ⎞ dy = − b ⎜⎜ ⎟⎟ dV 2y ⎝ T0 ⎠ Parameters

(7) – (15)

FA0 , k1 , E, R, T1 , KC2 , ΔHRx , T2 , C A0 , T0 , a, rb

Energy Balance Adiabatic and ΔCP = 0



(16A)

T = T0 +

(−ΔHRx ) X ∑Θi CP

i

T0 , ∑Θi CP = CP + ΘICP i A I



Additional Parameters (17A) & (17B)



Heat Exchange











⎡ ⎤ ∑Fi CP = FA0 ⎢∑Θi CP + ΔCP X ⎥ , if ΔCP = 0 then ⎣ ⎦ i i







(16B)

−r −ΔHRx −Ua T −Ta dT = A dV FA0 ∑Θi CP i

( )(

)

(

)

−r −ΔHRx −Ua T −Ta dT = A dV ∑Fi CP i

( )(

11-19

)

(

)

P11-7 (a) continued

Adiabatic

Adiabatic



P11-7 (b) Individualized Solution

11-20

P11-7 (c)

FA0

dX = −rA ' dW

−rA ' = k(1− α W)0.5 C A0 [(1− X)−

∫ (1− α W)0.5 dW = FA0 ∫

X ] KC

1 1 1−(1+ )X KC

dX

F 2 1 [1−(1− α W)1.5 ] = − A0 ln[1−(1+ )X] 3α 1 KC (1+ ) KC 2(1+

1 ) KC

1 )X] = − [1−(1− α W)1.5 ] KC 3αFA0 1 2(1+ ) KC 1 1−(1+ )X = exp(− [1−(1− α W)1.5 ]) KC 3αFA0 ln[1−(1+

2(1+ 1− exp(− X=

1 ) KC

3αFA0

1+

[1−(1− α W)1.5 ])

1 KC



2(1+ Qr = FA0 X(−ΔHR (T0 )) =

FA0 (−ΔHR (T0 ))(1− exp(−

3αFA0

1+

Qr = FA0 X(−ΔHR (T0 )) =

[1−(1− α W)1.5 ]))

1 KC

2(1+ FA0 (−ΔHR (T0 ))(1− exp(−

1 ) KC

1 ) KC

3αFA0

1+



[1−(1− αρb V)1.5 ]))

1 KC

Qr vs V can be plotted according to the above equation.

11-21

P11-7 (c) Continued

Qr vs. V 120

Qr (kcal/min)

100 80 60

40 20 0 0

10

20

30

40

V (dm3)



P11-8 (a) (1)

dX −rA" = dW FA0

(2)

" CC2 % −rA = k $C ACB − ' KC '& $#

(3)

( E " 1 1 %+ k = k1 exp* $$ − ''- *) R # T1 T &-,

(4)

k1 = 0.0002

(5)

T1 = 310

(6)

E = 25,000 ) ΔH # 1 1 &, KC = KC2 exp+ Rx %% − ((. +* R $ T2 T '.-

(7) (8)

ΔHRx = −20,000

(9)

T2 = 305

(10)

KC 2 = 1,000

(11)





δ = 2−1−1 = 0

(12)





ε = y A0δ = 0.5 2−1−1 = 0

(13)



C A = C A0 1− X p

(14)





(15)



CB = C A0 2− X

( −20,000 " 1 1 %+ KC = 1000exp* − '- $ *) 1.987 # 305 T &-,

(

(

)

T0 T



ΘB = 2

(

T0

)T p 11-22

)

P11-8 (a) Continued

T0

(16)



CC = 2C A0 X

(17)



dp −α T = dW 2p T0

(18) P11-8 (b)



α = 0.00015

T

( * * −rA = k *C A0 1− X * * *)

(

p

2

" T % 0

)$$# p T ''&C ( A0

" T % $$2C A0 Xp 0 '' " T % # T& 2− X $$ p 0 '' − KC # T&

)

+ - -,

2

" 4X 2 %( T0 + '* p −rA = kC $ 1− X 2− X − KC &'*) T -, $# 2 A0

(

)(

)

P11-8 (c) Equilibrium Conversion

KC =

2 4C A0 X e2 2 C A0 1− X e 2− X e

(

)(

KC 4





2

KC

)

=



X e2

(1− X )(2− X ) e

−

3KC

Xe +

e

KC X

2 e

= X e2



4 4 4 "K % 3K K $$ C −1'' X e2 − C X e +2 C = 0 4 4 #4 & 2

Equilibrium Conversion (19)

" 3K % "K % 3KC − $$ C '' −2KC $$ C −1'' 4 # 4 & #4 & Xe = "K % 2$$ C −1'' #4 &

Xe

Xe

T

X Section 1.01

W

W

Case 1 Adiabatic

11-23



P11-8 (d) Energy Balance

(16)

T = T0 +



(−ΔH ) X Rx

∑Θi CP

i

Energy Balance

(17)



T0 = 325







ΔHRx = ΔHR + ΔCP T −TR







ΔCP = 2CP −CP −CP = 2 20 −20 −20 = 0







ΔCP = 0







∑Θi CP = CP + ΘBCP + ΘC CP + ΘICP i A B C I







= 20 +2 20 + 0 20 + 1 40

Energy Balance

(18)



∑Θi CP = 100#cal mol ⋅K





(

C

A

)

( )

B

Everything that enters

( ) ( )( ) ( )( ) i



P11-8 (e)

T = T0 +

(−ΔH ) X = 400 + 20,000X = 400 +200X Rx

∑Θi CP

100

i



P11-8 (f) Adiabatic

0

11-24



Adiabatic

P11-8 (f) Continued Gas Phase Adiabatic

Gas Phase Adiabatic

P11-8 (g) For isothermal solution, set dT/dW to be zero in the polymath code provided in P11-8(f) and add an equation for heat that must be removed along the reactor as Q = -DH*X*Fa0 to get the plot. P11-9 KC =

CC CD X e2 = C ACB (1 − X e )2

⇒ Xe =

T = T0 −

KC 1 + KC

(−30000) X = 300 + 600 X ΔH R X = 300 − CPA + CPB (25 + 25)

See Polymath program P11-9.pol. 11-25



P11-9 Continued Calculated values of NLE variables Variable Value 1 Xe

f(x)

Initial Guess

0.9997644 3.518E-11 0.5 ( 0 < Xe < 1. )

Variable Value 1 T

300.

2 Kc

1.8E+07

Nonlinear equations 1 f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 = 0 Explicit equations 1 T = 300 2 Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 - 1 / T))

T

X

300

1

320

0.999

340

0.995

360

0.984

380

0.935

400

0.887

420

0.76

440

0.585

460

0.4

480

0.26

500

0.1529

520

0.091

540

0.057

560

0.035 Xe

1

Xe (Eqm Conversion)

0.9 0.8 0.7 0.6 Xe

0.5 0.4 0.3 0.2 0.1 0 300

350

400

450

500

Temp (K)

11-26

550

P11-10 (a) For first reactor,

KC =

X e1 KC or X e1 = 1 − X e1 1 + KC

For second reactor

KC =

θ B 2+ X e 2 1 − X e2

orX e 2 =

KC − θ B 2 1 + KC

orX e3 =

KC − θ B3 1 + KC

For 3rd reactor

KC =

θ B 3+ X e 3 1 − X e3

1st reactor: in first reaction Xe = 0.3 So, FB = FA01(.3) 2nd reactor: Moles of A entering the 2nd reactor: FA02 = 2FA01 - FA01(.3) = 1.7FA01

θB2 =

.2 FA01 = .12 1.7 FA0

− FA02 ΣCPi θi (T − T0 ) + FA02 X (−ΔH R ) = 0 X=

(C

pA

+ θ B C pB )(T − T 0 ) −ΔH R

Slope is now negative 3rd reactor:

X e2 = 0.3 Say

θ B = (.2 FA01 ) + .3FA02 = FA01 (.2 + .3 ×1.8) FA03 = FA01 + FA02 (1 − X e 2 ) = FA01 + 1.8FA01 (1 − X e 2 ) = 1 + 1.8(1 − .3) FA01 FA03 = 2.26 FA01

Feed to the reactor 3:

(2 FA01 ) + .3FA02 = FA01 (.2 + .3 ×1.8) = 0.7FA01

θ B3 =

.74 2.26



Feed Temperature to the reactor 2 is (520+450)/2 = 485 K Feed Temperature to reactor 3 is 480 K Xfinal = .4 Moles of B = .2FA01 + .3FAo2 + .4FA03 = FA01(.2 + .54 + (.4)(2.26)) = 1.64 FA01 X = FB/3FA01 = .54

11-27



P11-10 (a) continued

P11-10 (b) The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is now 450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the temperature is (520+450)/2 = 485 K Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3 would be (510+510+450)/3 = 490 K For any reactor j,

− FA0 j ΣCPi θi (T − T0 ) + FA0 j X (−ΔH R ) = 0 X=

(C

pA

+ θ B C pB )(T − T 0 )



−ΔH R

and θB for reactor 1 = 0. For reactor 2, θB > 0. This means that the slope of the conversion line from the energy balance is larger for reactor 2 than reactor 1. And similarly θB for reactor 3 > θB for reactor 2. So the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass balance equations are the same as in part (b) and so the plot of equilibrium conversion will decrease from reactor 1 to reactor 2, and, likewise, from reactor 2 to reactor 3.





11-28

Solutions for Chapter 12 – Steady-State Nonisothermal Reactor Design P12-1 (a) (i)-(vii) Individualized solution (viii) They separate at θI = 1.1. At θI = 1.2, we find that X and Xe profiles meet at low α but separate at higher α. (ix) θI = 1.8 (x-xi) θI (xii-xiv) Individualized solution (xv) For countercurrent flow, the only equation which changes is: d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) Note that the right hand side of the equation has been multiplied by -1. Now, we have to guess a value of Ta such that it matches Ta0 at W = 4500. We get Ta = 323.1302 K, for which Ta = Ta0 = 320 K at W = 4500.

See Polymath program P12-1a-1.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

alpha

0.0002

0.0002

0.0002

0.0002

2

Ca

0.1

0.0111851

0.1

0.0111851

3

Cao

0.1

0.1

0.1

0.1

4

Cb

0.1

0.0111851

0.1

0.0111851

5

Cc

0

0

0.0673836

0.0258241

6

CpA

20.

20.

20.

20.

7

CpB

20.

20.

20.

20.

8

Cpcool

18.

18.

18.

18.

9

CpI

40.

40.

40.

40.

10 Cto

0.3

0.3

0.3

0.3

11 Ea

2.5E+04

2.5E+04

2.5E+04

2.5E+04

12 Fao

5.

5.

5.

5.

13 Hr

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

14 k

0.046809

0.0207345

11.53755

0.0207345

15 Kc

66.01082

0.805727

126.6264

126.6264

16 mc

1000.

1000.

1000.

1000.

17 p

1.

0.2359323

1.

0.2359323

18 ra

-0.0004681

-0.012037

-2.485E-06

-2.485E-06

19 sumcp

80.

80.

80.

80.

20 T

330.

323.0995

385.7156

323.0995

12-1

21 Ta

323.1302

320.

323.1302

320.

22 thetaB

1.

1.

1.

1.

23 thetaI

1.

1.

1.

1.

24 To

330.

330.

330.

330.

25 Uarho

0.5

0.5

0.5

0.5

26 W

0

0

4500.

4500.

27 X

0

0

0.5358329

0.5358329

28 Xe

0.8024634

0.3097791

0.849089

0.849089

29 yao

0.3333333

0.3333333

0.3333333

0.3333333

Differential equations 1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) 2 d(p)/d(W) = -alpha/2*(T/To)/p 3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp) 4 d(X)/d(W) = -ra/Fao Explicit equations 1

Hr = -20000

2

alpha = .0002

3

To = 330

4

Uarho = 0.5

5

mc = 1000

6

Cpcool = 18

7

Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))

8

Fao = 5

9

thetaI = 1

10 CpI = 40 11 CpA = 20 12 thetaB = 1 13 CpB = 20 14 Cto = 0.3 15 Ea = 25000 16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4)) 17 k = .004*exp(Ea/1.987*(1/310-1/T)) 18 yao = 1/(1+thetaB+thetaI) 19 Cao = yao*Cto 20 sumcp = (thetaI*CpI+CpA+thetaB*CpB) 21 Ca = Cao*(1-X)*p*To/T 22 Cb = Cao*(thetaB-X)*p*To/T 23 Cc = Cao*2*X*p*To/T 24 ra = -k*(Ca*Cb-Cc^2/Kc)

12-2

P12-1 (a) Continued Plot of X, Xe and p versus W



Plot of T and Ta versus W

(xvi) For constant Ta, we have to use the same Polymath program in (xv) but multiply the right hand side by 0 in the program; i.e., d(Ta)/d(W) = Uarho*(T-Ta)/(mc*Cpcool)*0 See Polymath program P12-1a-2.pol

12-3

P12-1 (a) Continued POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

alpha

0.0002

0.0002

0.0002

0.0002

2

Ca

0.1

0.011675

0.1

0.011675

3

Cao

0.1

0.1

0.1

0.1

4

Cb

0.1

0.011675

0.1

0.011675

5

Cc

0

0

0.0659781

0.026488

6

CpA

20.

20.

20.

20.

7

CpB

20.

20.

20.

20.

8

Cpcool

18.

18.

18.

18.

9

CpI

40.

40.

40.

40.

10 Cto

0.3

0.3

0.3

0.3

11 Ea

2.5E+04

2.5E+04

2.5E+04

2.5E+04

12 Fao

5.

5.

5.

5.

13 Hr

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

14 k

0.046809

0.021188

8.254819

0.021188

15 Kc

66.01082

1.053204

124.4536

124.4536

16 mc

1000.

1000.

1000.

1000.

17 p

1.

0.2441145

1.

0.2441145

18 ra

-0.0004681

-0.0080133

-2.769E-06

-2.769E-06

19 sumcp

80.

80.

80.

80.

20 T

330.

323.2791

381.7968

323.2791

21 Ta

320.

320.

320.

320.

22 thetaB

1.

1.

1.

1.

23 thetaI

1.

1.

1.

1.

24 To

330.

330.

330.

330.

25 Uarho

0.5

0.5

0.5

0.5

26 W

0

0

4500.

4500.

27 X

0

0

0.5314832

0.5314832

28 Xe

0.8024634

0.3391177

0.8479767

0.8479767

29 yao

0.3333333

0.3333333

0.3333333

0.3333333

Differential equations 1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) *0 2 d(p)/d(W) = -alpha/2*(T/To)/p 3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp) 4 d(X)/d(W) = -ra/Fao

12-4

P12-1 (a) Continued Explicit equations 1

Hr = -20000

2

alpha = .0002

3

To = 330

4

Uarho = 0.5

5

mc = 1000

6

Cpcool = 18

7

Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))

8

Fao = 5

9

thetaI = 1

10 CpI = 40 11 CpA = 20 12 thetaB = 1 13 CpB = 20 14 Cto = 0.3 15 Ea = 25000 16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4)) 17 k = .004*exp(Ea/1.987*(1/310-1/T)) 18 yao = 1/(1+thetaB+thetaI) 19 Cao = yao*Cto 20 sumcp = (thetaI*CpI+CpA+thetaB*CpB) 21 Ca = Cao*(1-X)*p*To/T 22 Cb = Cao*(thetaB-X)*p*To/T 23 Cc = Cao*2*X*p*To/T 24 ra = -k*(Ca*Cb-Cc^2/Kc)

Plot of X, Xe and p versus W

12-5

P12-1 (a) Continued Plot of T and Ta versus W

Ta will remain constant with W For adiabatic operation, Using the Polymath program of part (xv) and making the parameter Ua =0; we have; See Polymath program P12-1a-3.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

alpha

0.0002

0.0002

0.0002

0.0002

2

Ca

0.1

0.0153303

0.1

0.0153303

3

Cao

0.1

0.1

0.1

0.1

4

Cb

0.1

0.0153303

0.1

0.0153303

5

Cc

0

0

0.0403759

0.0105141

6

CpA

20.

20.

20.

20.

7

CpB

20.

20.

20.

20.

8

Cpcool

18.

18.

18.

18.

9

CpI

40.

40.

40.

40.

10 Cto

0.3

0.3

0.3

0.3

11 Ea

2.5E+04

2.5E+04

2.5E+04

2.5E+04

12 Fao

5.

5.

5.

5.

13 Hr

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

14 k

0.046809

0.046809

22.60961

22.60961

15 Kc

66.01082

0.470375

66.01082

0.4703751

16 mc

1000.

1000.

1000.

1000.

17 p

1.

0.2456997

1.

0.2456997

18 ra

-0.0004681

-0.0120578

1.617E-08

-2.451E-18

19 sumcp

80.

80.

80.

80.

12-6



20 T

330.

330.

393.8384

393.8384

21 Ta

320.

320.

320.

320.

22 thetaB

1.

1.

1.

1.

23 thetaI

1.

1.

1.

1.

24 To

330.

330.

330.

330.

25 Uarho

0

0

0

0

26 W

0

0

4000.

4000.

27 X

0

0

0.2553537

0.2553537

28 Xe

0.8024634

0.2553537

0.8024634

0.2553537

29 yao

0.3333333

0.3333333

0.3333333

0.3333333

Differential equations 1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) *0 2 d(p)/d(W) = -alpha/2*(T/To)/p 3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp) 4 d(X)/d(W) = -ra/Fao

Explicit equations 1

Hr = -20000

2

alpha = .0002

3

To = 330

4

Uarho = 0

5

mc = 1000

6

Cpcool = 18

7

Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))

8

Fao = 5

9

thetaI = 1

10 CpI = 40 11 CpA = 20 12 thetaB = 1 13 CpB = 20 14 Cto = 0.3 15 Ea = 25000 16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4)) 17 k = .004*exp(Ea/1.987*(1/310-1/T)) 18 yao = 1/(1+thetaB+thetaI) 19 Cao = yao*Cto 20 sumcp = (thetaI*CpI+CpA+thetaB*CpB) 21 Ca = Cao*(1-X)*p*To/T 22 Cb = Cao*(thetaB-X)*p*To/T 23 Cc = Cao*2*X*p*To/T 24 ra = -k*(Ca*Cb-Cc^2/Kc)

12-7

P12-1 (a) Continued Plot of X, Xe and p versus W;

Plot of T and Ta versus W

Ta remains a constant P12-1 (b)





(i) Ua brings the temperature profiles close together. (ii) FA0 separates X and Xe the most. (iii) Varying CA0 keeps X and Xe profiles closest together. (iv) The specific heat of the coolant has to be increased to 80 kJ/kg K. Conversion increases very slightly. (v) Conversion increases with increase in heat of reaction initially, and then decreases. Conversion at the exit of the reactor is maximum for a heat of reaction of about -25000 kJ/mol. (vi) Individualized solution (vii) The terms Qg = ra*δHRx and Qr = Ua*(T-Ta) are added in the Polymath code. 12-8

P12-1 (b) Continued

(viii) Individualized solution (ix) Ta = 306 K (x) Individualized solution (xi) See Polymath program P12-1-b-1.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca0

1.86

1.86

1.86

1.86

2

Cpc

28.

28.

28.

28.

3

Cpo

159.

159.

159.

159.

4

deltaH

-2.0E+04

-2.0E+04

-2.0E+04

-2.0E+04

5

Fa0

14.67

14.67

14.67

14.67

6

k

0.5927441

0.5927441

17.40568

4.824093

7

Kc

1817.59

649.9456

1817.59

960.3748

8

m

500.

500.

500.

500.

9

ra

-1.102504

-9.223172

-0.1358393

-0.1358393

10 rate

1.102504

0.1358393

9.223172

0.1358393

11 T

305.

305.

350.7305

331.8405

12 Ta

315.

314.3716

331.1465

331.1465

13 Ua

5000.

5000.

5000.

5000.

14 V

0

0

5.

5.

15 X

0

0

0.9838366

0.9838366

16 Xe

0.9994501

0.9984638

0.9994501

0.9989598

12-9

P12-1 (b) Continued Differential equations 1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc 2 d(X)/d(V) = -ra/Fa0 3 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0 Explicit equations 1

Cpc = 28

2

m = 500

3

Ua = 5000

4

Ca0 = 1.86

5

Fa0 = 0.9*163*.1

6

deltaH = -20000

7

k = 31.1*exp((7906)*(T-360)/(T*360))

8

Kc = 1000*exp((deltaH/8.314)*((T-330)/(T*330)))

9

Xe = Kc/(1+Kc)

10 ra = -k*Ca0*(1-(1+1/Kc)*X) 11 Cpo = 159 12 rate = -ra

Plot of X and Xe versus V



12-10

P12-1 (b) Continued Plot of T and Ta versus V

(xii) For isothermal operation, the same code in part (vii) is used, with one change: d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0*0

Note that the right hand side of the equation is multiplied by 0. See Polymath program P12-1-b-2.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca0

1.86

1.86

1.86

1.86

2

Cpc

28.

28.

28.

28.

3

Cpo

159.

159.

159.

159.

4

deltaH

-3.45E+04

-3.45E+04

-3.45E+04

-3.45E+04

5

Fa0

14.67

14.67

14.67

14.67

6

k

0.5927441

0.5927441

0.5927441

0.5927441

7

Kc

9.512006

9.512006

9.512006

9.512006

8

m

500.

500.

500.

500.

9

Qr

-5.0E+04

-5.0E+04

-8383.862

-8383.862

10 ra

-1.102504

-1.102504

-0.7278292

-0.7278292

11 rate

1.102504

0.7278292

1.102504

0.7278292

12 T

305.

305.

305.

305.

13 Ta

315.

306.6768

315.

306.6768

14 Ua

5000.

5000.

5000.

5000.

15 V

0

0

5.

5.

16 X

0

0

0.3075111

0.3075111

17 Xe

0.9048707

0.9048707

0.9048707

0.9048707

12-11

P12-1 (b) Continued Differential equations 1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc 2 d(X)/d(V) = -ra/Fa0 3 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0*0

Explicit equations 1

Cpc = 28

2

m = 500

3

Ua = 5000

4

Ca0 = 1.86

5

Fa0 = 0.9*163*.1

6

deltaH = -34500

7

k = 31.1*exp((7906)*(T-360)/(T*360))

8

Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))

9

Xe = Kc/(1+Kc)

10 ra = -k*Ca0*(1-(1+1/Kc)*X) 11 Cpo = 159 12 rate = -ra 13 Qr = Ua*(T-Ta)

Plot of Ta versus V to maintain isothermal operation



12-12

P12-1 (b) Continued Plot of Qr versus V to maintain isothermal operation

(xiii) As the magnitude of heat of reaction increases, it is found that the point where X and Xe curves come close together decreases. This can be explained from the Van’t Hoff equation. As the enthalpy of the reaction becomes more exothermic, the equilibrium conversion is expected to drop. (xiv) No solution will be provided (xv) Ua can be varied so as to have the desired conversion and keep the temperature under 370 K. See Polymath program P12-1-b-3.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca0

1.86

1.86

1.86

1.86

2

Cpc

28.

28.

28.

28.

3

Cpo

159.

159.

159.

159.

4

deltaH

-3.45E+04

-3.45E+04

-3.45E+04

-3.45E+04

5

Fa0

14.67

14.67

14.67

14.67

6

k

0.5927441

0.5927441

54.76627

1.362998

7

Kc

9.512006

0.884224

9.512006

6.144184

8

m

500.

500.

500.

500.

9

ra

-1.102504

-43.8873

-0.319219

-0.319219

10 rate

1.102504

0.319219

43.8873

0.319219

11 T

305.

305.

369.5214

315.1228

12 Ta

340.3

314.8105

344.917

314.8105

13 Ua

4.3E+04

4.3E+04

4.3E+04

4.3E+04

14 V

0

0

5.

5.

15 X

0

0

0.751735

0.751735

16 Xe

0.9048707

0.4692775

0.9048707

0.860026

12-13



P12-1 (b) Continued Differential equations 1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc 2 d(X)/d(V) = -ra/Fa0 3 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0 Explicit equations 1

Cpc = 28

2

m = 500

3

Ua = 43000

4

Ca0 = 1.86

5

Fa0 = 0.9*163*.1

6

deltaH = -34500

7

k = 31.1*exp((7906)*(T-360)/(T*360))

8

Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))

9

Xe = Kc/(1+Kc)

10 ra = -k*Ca0*(1-(1+1/Kc)*X) 11 Cpo = 159 12 rate = -ra

In the above report, it can be seen that Ua is the only changed parameter to get a conversion that is greater than 75% and the temperature is maintained under 370 K. P12-1 (c) (i) The reaction dies out near the reactor entrance when varying the heat capacity of A. (ii) The heat capacity of A is the variable which most drastically changes the profiles. (iii) It increases with an increase in heat capacity of A. (iv) Heat transfer coefficient could have changed. (v) Individualized solution (vi) The terms Qg = ra*δHRx and Qr = Ua*(T-Ta) are added in the Polymath code. See Polymath program P12-1-c-1.pol

12-14

P12-1 (c) Continued Case1: adiabatic operation Qr=0 in this case



Case 2: For constant heat exchange conditions



12-15

Case 3: Co-current heat exchange



Case 4 : counter current heat exchange : We need to guess a value of Ta such that at exit Ta = 1250K If we take Ta(0) = 995.15K then this can be done.

(vii) Now, V = 0.5 m3 T0 = 1050 K Ta0 = 1250K Case 1: adiabatic operation; Substitute; Ua = 0 in the Polymath code (viii) Substitute, the volume V= 5 m3 in the Polymath code and get the results. (ix) Plot of Qg for all 4 cases against volume

12-16



P12-1 (c) Continued

Plot of Qr versus volume for all cases



Plot of -ra (rate) versus V

It can be seen that the rate for constant heat exchange fluid temperature Ta, is higher than the rest of cases because the difference between heat generated and heat removed in this case is highest. 12-17

P12-1 (c) Continued The rate of reaction for all cases is decreasing because the temperature of the system is decreasing with volume. The rate of reaction for counter-current heat exchanger system is a U shaped curve plotted against volume. At the front of the reactor, the reaction takes place very rapidly, drawing energy from the sensible heat of the gas causing the gas temperature to drop because the heat exchanger cannot supply energy at an equal or greater rate. This drop in temperature, coupled with the consumption of reactants, slows the reaction rate as we move down the reactor. (x) Introduction of inert will introduce a change in energy balance equation and the value of Ѳ1 as well. (xi) (1) ѲI = 0 All the plots will remain as for part (vi) (2) ѲI = 1.5 Instead of the energy equation which was used previously the equation will change. Now we have to change the value of Now the value will be ∑Cp =



*Cp + C = 1.5×50 + 1×163 = 238 I I pA

Value of Ѳ will change as well Ѳ =

=







(3) Ѳ I = 3 ∑Cp = I*CpI + CpA = 3×50 + 1×163 = 313J/molA/K Ѳ=

=



Incorporating these changes in the code and plotting X versus V for different cases. The analysis is as follows: Case I: adiabatic operation:

12-18

P12-1 (c) Continued Case2: Constant heat exchange fluid temperature Ta



Case 3: Co-current heat exchange

The Polymath program for reference is for the case of co-current heat exchange with ѲI =3. By changing values of ∑Cp and ε variables as shown we can change the program for various cases and sub cases. Case 4: Countercurrent heat exchange

12-19

P12-1 (c) Continued Remarks: Thus we can see that for all cases when inert gas concentration is more, then the reaction proceeds faster but then the overall yield is less as well. In the case of adiabatic operation this phenomenon is very significant. In case of constant heat exchange fluid temperature the effect of inert gas is negligible. (xii) Here we will change the Polymath program as entered in P12-2(c) part (vi). The Ta value will be changed and the program will be tested for following values of Ta. 1000K, 1175K, 1350K Case 1: adiabatic conditions

Case 2: Constant heat exchange fluid temperature Ta



Case 3: Co-current conditions



12-20

P12-1 (c) Continued Case 4: Countercurrent conditions We need to enter Ta (V =0) values such that at V=Vf, Taf = 1000K, 1175K and 1350K respectively Ta (V=0) (K) Ta(V=Vf) (K) 983.75 992 999

1000 1175 1350





P12-1 (d) Aspen problem No solution will be provided P12-1 (e) (i) There are at least two solutions for 16415 𝑅
2665 kCal/K (iii) No explosion would occur for UA > 37.2 kCal/min C (iv) To show that no explosion occurred without cooling failure.



13-3

P13-1 (b) Continued (v)





P13-1 (c) Example 13-3

(i) 1650 lbmol/hr (ii) No, it is not possible. (iii) The temperature trajectory exceeds practical stability limit for T0 = 120 F, CAi = 0 (iv) See Polymath program P13-1civ.pol

13-4

P13-1 (c) Continued POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

0.1

0.0074672

0.1

0.0309593

2

Ca0

0.1812152

0.1812152

0.1812152

0.1812152

3

Cb

3.45

2.114934

3.45

2.114934

4

Cb0

2.26519

2.26519

2.26519

2.26519

5

Cc

0

0

0.1502559

0.1502559

6

Cm

0

0

0.226519

0.226519

7

Cm0

0.226519

0.226519

0.226519

0.226519

8

dh

-3.6E+04

-3.6E+04

-3.6E+04

-3.6E+04

9

Fa0

80.

80.

80.

80.

10 Fb0

1000.

1000.

1000.

1000.

11 Fm0

100.

100.

100.

100.

12 k

41.71035

31.59807

144.4831

32.06998

13 mc

1000.

1000.

1000.

1000.

14 Na

6.680919

0.4988795

6.680919

2.068369

15 Nb

230.4917

141.2971

230.4917

141.2971

16 Nc

0

0

10.03847

10.03847

17 NCp

4382.683

3372.614

4382.683

3372.614

18 Nm

0

0

15.13355

15.13355

19 Qg

1.003E+07

2.339E+06

1.003E+07

2.388E+06

20 Qr

2.586E+06

2.377E+06

3.578E+06

2.388E+06

21 Qr1

1.706E+06

1.564E+06

2.383E+06

1.571E+06

22 Qr2

8.798E+05

8.133E+05

1.195E+06

8.168E+05

23 ra

-4.171035

-4.171035

-0.9725584

-0.9928658

24 rb

-4.171035

-4.171035

-0.9725584

-0.9928658

25 rc

4.171035

0.9725584

4.171035

0.9928658

26 T

150.

143.7291

179.7336

144.0607

27 t

0

0

4.

4.

28 T0

75.

75.

75.

75.

29 Ta1

67.

67.

67.

67.

30 Ta2

115.8777

112.1848

133.3874

112.3801

31 tau

0.1513355

0.1513355

0.1513355

0.1513355

32 ThetaCp 284.375

284.375

284.375

284.375

33 UA

1.6E+04

1.6E+04

1.6E+04

1.6E+04

34 V

66.80919

66.80919

66.80919

66.80919

35 v0

441.464

441.464

441.464

441.464

13-5

P13-1 (c) Continued Differential equations 1 d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra 2 d(Cb)/d(t) = 1/tau*(Cb0-Cb)+rb 3 d(Cc)/d(t) = 1/tau*(0-Cc)+rc 4 d(Cm)/d(t) = 1/tau*(Cm0-Cm) 5 d(T)/d(t) = (Qg-Qr)/NCp Explicit equations 1

Fa0 = 80

2

T0 = 75

3

V = (1/7.484)*500

4

UA = 16000

5

dh = -36000

6

Ta1 = 67

7

k = 16.96e12*exp(-32400/1.987/(T+460))

8

Fb0 = 1000

9

Fm0 = 100

10 mc = 1000 11 ra = -k*Ca 12 rb = -k*Ca 13 rc = k*Ca 14 Nm = Cm*V 15 Na = Ca*V 16 Nb = Cb*V 17 Nc = Cc*V 18 ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5 19 v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54 20 Ta2 = T-(T-Ta1)*exp(-UA/(18*mc)) 21 Ca0 = Fa0/v0 22 Cb0 = Fb0/v0 23 Cm0 = Fm0/v0 24 Qr2 = mc*18*(Ta2-Ta1) 25 tau = V/v0 26 NCp = Na*35+Nb*18+Nc*46+Nm*19.5 27 Qr1 = Fa0*ThetaCp*(T-T0) 28 Qr = Qr1+Qr2 29 Qg = ra*V*dh

From the above program, we see that by varying Ta1, the maximum possible value such that practical stability limit is not exceeded is 67 F. 13-6

P13-1 (c) Continued (v) See polymath program p13-1cv.pol





(vi)



13-7

P13-1 (c) Continued







(vii) No solution will be provided (viii) Individualized solution

13-8



P13-1 (c) Continued (ix) Time to reach steady state increases with increase in tank volume. When the tank volume is at its minimum, practical stability limit is not exceeded for any of the three sets of initial conditions. However, as we increase the volume, it is observed that for Ti = 340 K and CAi = 1.4 M, the practical stability is exceeded. (x) The maximum feed temperature such that practical stability limit is not exceeded is 284 K. (xi) Individualized solution (xii) The number of oscillations increase with increasing EA initially, and then decrease later on. (xiii) Individualized solution P13-1 (d) Example 13-4 (i) The maximum in CC increases with an increase in CB0. (ii) mc and CPc have virtually no effect on the temperature trajectory. UA, CP, CPc and mc have virtually no effect on the concentration trajectory. (iii) Ta1 (iv) 275 K (v) On running the polymath code provided in the CRE Web Site, we find that NC attains its maximum at t = 270 s and remains constant thereafter. CC is maximum at t = 250 s. (vi) Decreasing the coolant rate to 10 kg/s gives a weak-cooling effect and the maximum temperature in the reactor becomes 315 K. An increase of the coolant rate to 1000 kg/s gives a Tmax of 312 K. A big change to the coolant rate has, in this case, only a small effect on the temperature, and because the temperature does not change significantly the conversion will be kept about the same. P13-1 (e) Example 13-5 (i) One of the possible combinations is CA0 = 8 mol/dm3 and 𝑣! = 240 dm3/h. (ii) It is intuitive that the concentrations increase with an increase in CA0 and 𝑣! . (iii) The parameter is 𝑣! . It occurs at a value of 50 dm3/h. (iv) The conversion of A increases with an increase in the volumetric flow rate. We also find that the final concentration of C increases with an increase in the flow rate. The temperature also increases. (v) See polymath program P13-1ev.pol

13-9

P13-1 (e) Continued

We can see from the graph that NA and NB become constant at large times. (vi) See polymath program P13-1evi.pol POLYMATH Report Ordinary Differential Equations

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

Ca

1.

0.0786053

2.036072

0.0786053

2

Cao

4.

4.

4.

4.

3

Cb

0

0

0.8803026

0.247476

4

Cc

0

0

5.018042

5.018042

5

k1a

0.2664781

0.2664781

27.26262

3.288066

6

k2b

0.08

0.08

2.421438

0.5095541

7

Na

100.

29.21648

413.9672

290.8397

8

Nb

0

0

915.6613

915.6613

9

ra

-0.2664781 -21.60838

-0.2584595

-0.2584595

10 rb

0.133239

-0.7139195

9.023013

0.0031273

11 rc

0

0

6.394794

0.3783073

12 t

0

0

15.

15.

13 T

290.

290.

403.1775

342.148

14 V

100.

100.

3700.

3700.

15 vo

240.

240.

240.

240.

Differential equations 1 d(Ca)/d(t) = ra+(Cao-Ca)*vo/V 2 d(Cb)/d(t) = rb-Cb*vo/V 3 d(Cc)/d(t) = rc-Cc*vo/V 4

d(T)/d(t) = (35000*(298-T)-Cao*vo*30*(T-305)+((-6500)*(-k1a*Ca)+(8000)*(k2b*Cb))*V)/((Ca*30+Cb*60+Cc*20)*V+100*35)

13-10

P13-1 (e) Continued Explicit equations 1

Cao = 4

2

vo = 240

3

k1a = 1.25*exp((9500/1.987)*(1/320-1/T))

4

k2b = 0.08*exp((7000/1.987)*(1/290-1/T))

5

ra = -k1a*Ca

6

V = 100+vo*t

7

rc = 3*k2b*Cb

8

rb = k1a*Ca/2-k2b*Cb

9

Na = Ca*V

10 Nb = Cb*V

We find that after running the code for a final time of 10 hours, we get NA = 290.8397 mol and NB = 915.6613 mol. 𝐶!! 𝑣! 240 =4∗ = 291.962 𝑚𝑜𝑙 𝑘!! 3.2881 𝐶!! 𝑣! 240 =4∗ = 942.008 𝑚𝑜𝑙 2𝑘!! 2 ∗ 0.50955 (vii) From the polymath code given in the previous problem, it can be seen that at t = 10 h, the volume is 3700 L. Hence, if the reactor had no lid, then there would be an overflow. ! ! (viii) Since NB = !! ! at long times, we can maximize NB by increasing the volumetric flow rate.

!!!!

P13-1 (f) Example 13-6 (i) The pressure shoots up when the volumetric flow rate is 3976 dm3. (ii) E1A has to be varied. (iii) Yes, it is possible to reach temperature limitation before pressure limitation. An instance of this is when E1A = 134,000 J/mol K and all other parameters are set to their maximum values. (iv) -270,000 J/mol (v) Individualized solution (vi) Individualized solution (vii) The transition to runaway occurs over a very small range of Ua. If U a value is changed such that Ua is taken from 2.1 ×104 to 1.9 ×104 J/hr/K the runaway occurs. Thus it occurred over a very narrow range of Ua values. (viii) The modeling is done and the changes are incorporated in the polymath code for Example 13.6 A new switch type variable is introduced.

Sw2 such that Sw2 = 1 if 300 455K = 0 for T< 455K

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1

A1A

4.0E+14

4.0E+14

4.0E+14

4.0E+14

2

A2S

1.0E+84

1.0E+84

1.0E+84

1.0E+84

3

CA

4.3

0.0457129

4.3

0.0457129

4

CB

5.1

0.8457129

5.1

0.8457129

5

CS

3.

2.999997

3.

2.999997

6

Cv1

3360.

3360.

3360.

3360.

7

Cv2

5.36E+04

5.36E+04

5.36E+04

5.36E+04

8

DHRx1A -4.54E+04

-4.54E+04

-4.54E+04

-4.54E+04

9

DHRx2S

-3.2E+05

-3.2E+05

-3.2E+05

-3.2E+05

10 E1A

1.28E+05

1.28E+05

1.28E+05

1.28E+05

11 E2S

8.0E+05

8.0E+05

8.0E+05

8.0E+05

12 FD

2467.445

61.27745

8874.034

61.27745

13 Fvent

2467.445

61.27745

8874.034

61.27745

14 k1A

0.0562573

0.0562573

1.82787

0.7925113

15 k2S

8.428E-16

8.428E-16

2.367E-06

1.276E-08

16 P

4.4

4.4

4.4

4.4

17 r1A

-1.233723

-4.437017

-0.0306385

-0.0306385

18 r2S

-2.529E-15

-7.102E-06

-2.529E-15

-3.829E-08

19 SumNCp 1.26E+07

1.26E+07

1.26E+07

1.26E+07

20 SW1

1.

1.

1.

1.

21 Sw2

0

0

0

0

22 T

422.

422.

466.4882

454.9731

23 t

0

0

4.

4.

24 UA

0

0

2.77E+06

0

25 V0

4000.

4000.

4000.

4000.

26 VH

5000.

5000.

5000.

5000.

Differential equations 1 d(CA)/d(t) = SW1*r1A mol/dm3/hr 2 d(CB)/d(t) = SW1*r1A change in concentration of cyclomethylpentadiene 3 d(CS)/d(t) = SW1*r2S change in concentration of diglyme 4 d(P)/d(t) = SW1*((FD-Fvent)*0.082*T/VH) 5 d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp) + SW1*Sw2*4/60

13-12

P13-1 (f) Continued Explicit equations 1 V0 = 4000 dm3 2 VH = 5000 dm3 3 DHRx1A = -45400 J/mol Na 4 DHRx2S = -3.2E5 J/mol of Diglyme 5 SumNCp = 1.26E7 J/K 6 A1A = 4E14 per hour 7 E1A = 128000 J/kmol/K 8 k1A = A1A*exp(-E1A/(8.31*T)) rate constant reaction 1 9 A2S = 1E84 per hour 10 E2S = 800000 J/kmol/K 11 k2S = A2S*exp(-E2S/(8.31*T)) rate constant reaction 2 12 SW1 = if (T>600 or P>45) then (0) else (1) 13 r1A = -k1A*CA*CB mol/dm3/hour (first order in sodium and cyclomethylpentadiene) 14 r2S = -k2S*CS mol/dm3/hour (first order in diglyme) 15 FD = (-0.5*r1A-3*r2S)*V0 16 Cv2 = 53600 17 Cv1 = 3360 18 Fvent = if (FD300) and (T