Solutions Manual for Essentials of Chemical Reaction Engineering 2ed Fogler, Exercises of Chemical Principles

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Solutions Manual for

Essentials of Chemical

Reaction Engineering

Second Edition

Table of Contents

• Solutions to Chapter 1, Problems P1-1 through P1-
• Solutions to Chapter 2, Problems P2-1 through P2-
• Solutions to Chapter 3, Problems P3-1 through P3-
• Solutions to Chapter 4, Problems P4-1 through P4-
• Solutions to Chapter 5, Problems P5-1 through P5-
• Solutions to Chapter 6, Problems P6-1 through P6-
• Solutions to Chapter 7, Problems P7-1 through P7-
• Solutions to Chapter 8, Problems P8-1 through P8-
• Solutions to Chapter 9, Problems P9-1 through P9-
• Solutions to Chapter 10, Problems P10-1 through P10-
• Solutions to Chapter 11, Problems P11-1 through P11-
• Solutions to Chapter 12, Problems P12-1 through P12-
• Solutions to Chapter 13, Problems P13-1 through P13-

vii

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. A=A if 0.9 > r^2 >= 0. A=X if 0.8 > r^2 >= 0. A=F if 0.7 > r^ 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. = A7213DF

1. A => 0.9 > r^2 >= 0.
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 Perf. No. = 2 12 5 4 8 2 9 13

Final conversion = 2AB.C conversion = 242.1 = 84. Final flow rate = 2DE.F flow rate = 231.2 = 62.

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

****** CONFIDENTIAL ******

viii

ICMs with Dos ®^ interface

Module Format Interpretation Example

HETCAT

zzABzCD A=2,3,5,7: interaction done Perf. No. = 80 27 4 35 B=2,3,5,7: intro done A: Worked on interaction C=2,3,5,7: review done B: Looked at intro D denotes how much they C: Looked at review did in the interaction: D: found parameter values, didn’t find mechanism D<2 Not done 2 < D:5 4 Dependences 4 < D:5 6 Parameter values 6<D Mechanism z = random numbers Note: Performance number given only if student goes through the interaction portion of the module

HEATFX zzAzz A even: score > 85 % Perf. No. = 53 6 07 z = random numbers Score > 85 % Note: Student told they have achieved mastery if their score is greater than 85%

HEATFX

zzzAzz A even: completed interaction Perf. No. = 407 5 82 z = random numbers Interaction not completed

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

****** CONFIDENTIAL ******

x

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, sfogler@umich.edu 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 , 5 th^ 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, P1-9 (^) A, Appendix A, from the Web Chapter 2, Sections 2.1, 2.2, and 2. Hand In: Problem Set 1: P1-1 (^) A, P1-6 (^) B In-Class Problem: 1 Study Problems: P1-8 (^) A

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, P2-2 (^) A, Intro to Learncheme In-Class Problem: 2 ( Hint: Viewing the University of Alabama YouTube video “The Black Widow” noted in Problem P3-8 (^) B may help you with today's in class problem)

xi

Study Problems: P2-7 (^) A

4. Monday, September 21 Topic: Lecture 4 – Chapter 4, Stoichiometry Batch Systems Read: Chapter 4 Section 4. Hand In: Problem Set 3: Define θi , θA, θB, and δ, P2-10 (^) B, P3-5A, P3-8 (^) B, P3-11 (^) B, P3-13 (^) A In-Class Problem: 3 - Bring i>clickers (tentative) - Test Run of System in 2166 Dow Study Problems: P3-14 (^) A

5. Wednesday, September 23 Topic: Lecture 5 – Chapter 4, Stoichiometry Flow Systems Read: Chapter 4, Section 4. Hand In: Problem Set 4: Define ε, F (^) T0 , CT0 , P4-2 (^) A. In-Class Problem: 4 Study Problems: P4-1 (^) A 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: P4-1 (^) A (a) and (b) only, P4-3 (^) A, P4-4B, P4-5 (^) B. In-Class Problem: 5 Study Problems: P4-10 (^) C

7. Wednesday, September 30 Topic: Lecture 7 – Chapter 5, California Registration Exam Problem Hand In: Problem Set 6: What are you asked to find P5-18 (^) B? What is the Ergun Equation? P5-2 (^) A. In-Class Problem: 6 Study Problems: P5-1 (^) B (a) and (b)

8. Monday, October 5 Topic: Lecture 8 – Chapter 5, Pressure drop Read: Chapter 5, Sections 5.4 and 5. Hand In: Problem Set 7: P5-3 (^) A, P5-4 (^) B, P5-5 (^) A, P5-8 (^) B, P5-13 (^) B omit parts (j) and (k), P5-16 (^) B (a). In-Class Problem: 7 – Bring Laptops Study Problems: P5-9 (^) A, P5-10B (a).

9. Wednesday, October 7 Topic: Lecture 9 – Chapter 6, Membrane Reactors Read: Chapter 6 Hand In: Problem Set 8: P5-13 (^) B part (j) and (k), P5-22 (^) A. In-Class Problem: 8 – Bring Laptops Study Problems: P5-21 (^) B

10. Monday, October 12 Topic: Lecture 10 – Chapter 6, Semibatch Reactors Read: Chapter 6 Hand In: Problem Set 9: P5-1 (^) A (a), P5-11 (^) B, P6-4B delete part (c), P6-5B. In-Class Problem: 9 – Bring Laptops to carry out Polymath ODE Solver Study Problems: P6-7 (^) B

xiii

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. Hand In: Problem Set 15: P12-3 (^) B 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: P12-4 (^) A (a) and (b), P12-14B, P12-17B, P12-21 (^) B. 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: P12-26 (^) C

23. Wednesday, November 25 Topic: Lecture 22 – Multiple Steady States (MSS) Multiple Reactions with Heat Effects Read: Sections 12.6 and 12. In-Class Problem: 18 – Bring a Ruler/Straight Edge Study Problems: P13-4 (^) B

24. Monday, November 30 Topic: Lecture 23 – Safety (CSI) Read: Chapter 13 Hand In: Problem Set 18: P13-1 (^) B (b) and (f), P13-8 (^) B In-Class Problem: 19 – Bring Laptops Study Problems: P13-4 (^) B

25. Wednesday, December 2 Topic: Lecture 24 –Catalysis Reactor Safety Read: Chapter 13, Sections 13.1 through 13.3, and 13. Hand In: Problem Set 19: P10-2 (^) A part (d), P10-4 (^) B 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. Hand In: Problem Set 20: P10-3 (^) A, P10-8 (^) B, P10-10 (^) B In-Class Problem: 21 Study Problems: P10-7 (^) B, P10-9 (^) B

xiv

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

28. FINAL EXAM

P1- 2 (a)

Total number of lb moles gas in the system:

N =

P 0 V

RT

N =

1 atm ×( 4 × 1013 ft^3 )

1. 73 atm.^ ft

3

lbmol. R

× 534. 69 R

= 1.025 x 10^11 lb mol

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

FA = yAFT = yA ⋅ vA CT

STP

• %no.%of%cars

FT =

3000 ft^3 hr car

×

1 lbmol 359 ft^3

× 400000 cars (See appendix B)

FA = 6.685 x 10^4 lb mol/hr

P1- 2 (c) Wind speed through corridor is U = 15mph W = 20 miles The volumetric flowrate in the corridor is v O = U.W.H = (15x5280)(20x5280)(2000) ft^3 /hr = 1.673 x 10^13 ft^3 /hr

P1- 2 (d) Molar flowrate of CO into basin from Sant Ana wind.

FS := v 0 ⋅ CS

= 1.673 x 10^13 ft^3 /hr × 2. 04 × 10 −^10 lbmol/ft^3 = 3.412 x 10^3 lbmol/hr

P1- 2 (e)

Rate of emission of CO by cars + Rate of CO in Wind - Rate of removal of CO =

dNCO dt

FA + FS − voCco = V

dCco dt

(V=constant, Nco = CcoV )

P1- 2 (f)

t = 0 , Cco = CcoO

dt 0

t ∫ =^ V^

dCco

C^ FA^ +^ FS^ − voCco coO

Cco ∫

t =

V

vo

ln

FA + FS − voCcoO

FA + FS − voCco

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

CCO 0 = 2. 04 × 10 −^8 lbmol ft^3

, CCO = 2.^04

× 10 −^8 lbmol ft^3

From (f),

t = V vo

ln

FA + FS − vO. CCO 0 FA + FS − vO. CCO

= 4 ft

3

1. 673 × 1013 ft

3

hr

ln

6. 7 × 104

lbmol hr

+ 3. 4 × 103

lbmol hr

− 1. 673 × 1013

ft^3 hr

× 2. 04 × 10 −^8

lbmol ft^3

6. 7 × 104 lbmol hr

• 3. 4 × 103 lbmol hr

− 1. 673 × 1013 ft

3

hr

× 0. 51 × 10 −^8 lbmol ft^3

t = 6.92 hr

P1- 2 (h)

(1) to = 0 tf = 72 hrs

C co =^ 2.00E-10 lbmol/ft

(^3) a = 3.50E+04 lbmol/hr

v o = 1.67E+12 ft

(^3) /hr b = 3.00E+04 lbmol/hr

Fs = 341.23 lbmol/hr V = 4.0E+13 ft^3

a + b sin π t 6

& +^ Fs −^ voCco =^ V^

dCco dt

Now solving this equation using POLYMATH we get plot between Cco vs. t

See Polymath program P 1 - 4 - h- 1 .pol.

POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value T 0 0 72 72 C 2.0E- 10 2.0E- 10 2.134E- 08 1.877E- 08 v0 1.67E+12 1.67E+12 1.67E+12 1.67E+ A 3.5E+04 3.5E+04 3.5E+04 3.5E+ B 3.0E+04 3.0E+04 3.0E+04 3.0E+ F 341.23 341.23 341.23 341. V 4.0E+13 4.0E+13 4.0E+13 4.0E+

ODE Report (RKF45) Differential equations as entered by the user [1] d(C)/d(t) = (a+bsin(3.14t/6)+F-v0C)/V Explicit equations as entered by the user [1] v0 = 1.6710^ [2] a = 35000 [3] b = 30000 [4] F = 341. [5] V = 4*10^

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 v 0! As the value of v 0 is increased the graph changes to a “shifted sin-curve”. And as v 0 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 dt

= k 1 x − k 2 xy …………………………….(1)

dy dt

= k 3 xy − k 4 y ……………………………..(2)

Given,

k 1 = 0. 02 day −^1 k 2 = 0. 00004 / ( day × foxes ) k 3 = 0. 0004 / ( day × rabbits )

k 4 = 0. 04 day −^1

See Polymath program P 1 - 3 - a.pol.

POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value maximal value final value T 0 0 500 500 X 500 2.9626929 519.40024 4. Y 200 1.1285722 4099.517 117. k1 0.02 0.02 0.02 0. k2 4.0E- 05 4.0E- 05 4.0E- 05 4.0E- 05 k3 4.0E- 04 4.0E- 04 4.0E- 04 4.0E- 04 k4 0.04 0.04 0.04 0.

ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(t) = (k1x)-(k2xy) [2] d(y)/d(t) = (k3xy)-(k4y)

Explicit equations as entered by the user [1] k1 = 0. [2] k2 = 0. [3] k3 = 0. [4] k4 = 0.

P1- 3 (a) Continued

When, tfinal = 800 and k 3 = 0. 00004 /( day × rabbits )

Plotting rabbits vs. foxes

P1- 3 (d) Continued

P1- 4

Individualized solution

P1- 5

P1- 6 (a)

• rA = k with k = 0.05 mol/h dm^3

CSTR: The general equation is

V =

FA 0 − FA

− rA

Here CA = 0.01CA0 , v 0 = 10 dm^3 /min, FA = 5.0 mol/hr

Also we know that FA = CAv 0 and FA0 = CA0v 0 , CA0 = FA0/ v 0 = 0.5 mol/dm^3

Substituting the values in the above equation we get,

V =

C A 0 v 0 − CAv 0 k

" V = 99 dm^3

PFR: The general equation is

dFA dV

= rA = k , Now FA = CAv 0 and FA0 = CA0v 0 =>

dCAv 0 dV

= − k

Integrating the above equation we get

P1- 6 (a) Continued

v 0 k

dCA CA 0

CA ∫ =^ dV 0

V ∫ =>^ V^ =^

v 0 k

( CA 0 − CA )

Hence V = 99 dm^3

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

V =

C A 0 v 0 − CAv 0 − rA

v 0 CA 0 ( 1 − 0. 01 ) kCA

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

" V =

( 10 dm^3 / hr )( 0. 5 mol / dm^3 )( 0. 99 ) ( 0. 36 hr −^1 )( 0. 01 * 0. 5 mol / dm^3 )

=> V = 2750 dm^3

PFR:

From above we already know that for a PFR

dCAv 0 dV

= rA = − kCA

Integrating

v 0 k

dCA

C^ CA A 0

CA ∫ =^ −^ dV 0

V ∫

v 0 k

ln

CA 0

CA

= V

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 dm^3

P1- 6 (c)

• rA = kCA^2 with k = 300 dm^3 /mol.hr

CSTR:

V =

C A 0 v 0 − CAv 0 − rA

v 0 CA 0 ( 1 − 0. 01 )

kCA^2

Substituting all the values we get

V =

( 10 dm^3 / hr )( 0. 5 mol / dm^3 )( 0. 99 ) ( 300 dm^3 / mol. hr )( 0. 01 * 0. 5 mol / dm^3 )^2

=> V = 660 dm^3

PFR:

dCAv 0 dV

= rA = − kCA^2