978-0134663890 Chapter 1 Part 2

xi

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, P3-13

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 4.1

Hand In: Problem Set 4: Define ε, FT0, CT0, P4-2

A

.

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

xii

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 P7-10

A

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: P8-16

A

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

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 12.4

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

xiv

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 P9-19

A

In-Class Problem: 22

Study Problems: P9-12B, P9-16B, P9-21

A

28) FINAL EXAM

1-1#

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#

0

v

for#the#same#volume.#

(ii)#CA#decreases#and#CB#increases#with#an#increase#in#k#and#Ke,#and#a#decrease#in#

0

v

for#the#same#volume.#

Variable

Initial value

Minimal value

Maximal value

Final value

1

Ca

10.

2.849321

10.

2.849321

2

Cb

0

7.150679

3

k

0.23

4

Ke

3.

5

ra

–2.3

–0.1071251

6

rb

2.3

0.1071251

2.3

0.1071251

7

V

0

100.

8

v0

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 ft2

#

TSTP =491.69R

#

H=2000 ft

#

V=4 *1013 ft3

#T#=#534.7

°

R# PO#=#1atm#

R=0.7302 atm ft3

lbmol R

# yA#=#0.02###########

CS=2.04 *10−10 lbmol

ft3

#################

C#=#4*105##cars###############

FS#=#CO#in#Santa#Ana#winds##########FA#=#CO#emission#from#autos######

vA=3000 ft3

hr

##per#car#at#STP#

1-2#

P1-2)(a)))

Total#number#of#lb#moles#gas#in#the#system:)

0.73 atm.ft3

lbmol.R

#

$

&

‘

‘×534.69R

#

P1-2)(b)))

Molar#flowrate#of#CO#into#L.A.#Basin#by#cars.)

FA=yAF

T=yA⋅vACTSTP

•%no.%of%cars

#

F

T=3000 ft3

hr car ×1lbmol

359 ft3×400000 cars

## (See#appendix#B)#

FA#=#6.685#x#104#lb#mol/hr#

#

P1-2)(c)))

Wind#speed#through#corridor#is#U#=#15mph)

P1-2)(d)))

Molar#flowrate#of#CO#into#basin#from#Sant#Ana#wind.)

FS:=v0⋅CS

)

#####=#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#=#

dNCO

dt

#

FA+FS−voCco =V

dCco

dt

###########(V=constant,

Nco =CcoV

)#

#

P1-2)(f)####

#t#=#0##,##

Cco =CcoO

###)

dt

0

t

∫=V

dCco

FA+FS−voCco

CcoO

Cco

∫

#

t=V

vo

ln

FA+FS−voCcoO

FA+FS−voCco

“

#

$

%

&

‘

#

1-3#

P1-2)(g)))

Time#for#concentration#to#reach#8#ppm.)

CCO0=2.04 ×10−8lbmol

ft3

,#

CCO =2.04

4×10−8lbmol

ft3

#

From#(f),#

t=V

vo

ln FA+FS−vO.CCO0

FA+FS−vO.CCO

“

#

$

%

&

‘

=4ft3

1.673×1013 ft3

hr

ln

6.7 ×104lbmol

hr +3.4 ×103lbmol

hr −1.673×1013 ft3

hr ×2.04 ×10−8lbmol

ft3

6.7 ×104lbmol

hr +3.4 ×103lbmol

hr −1.673×1013 ft3

hr ×0.51×10−8lbmol

ft3

“

#

$

%

&

‘

#

t#=#6.92#hr#

#

P1-2)(h)###

(1)###############to####=##0#######################################tf###=##72#hrs)

##########################

co

C

##=##2.00E–10#lbmol/ft3########a###=##3.50E+04#lbmol/hr##

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

maximal value

final value

T

0

72

C

2.0E–10

2.134E–08

1.877E–08

v0

1.67E+12

A

3.5E+04

B

3.0E+04

F

341.23

V

4.0E+13

)

ODE)Report)(RKF45)#

Differential)equations)as)entered)by)the)user)

#

1-4#

P1–2)(h))Continued#

#

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

maximal value

final value

T

0

72

C

2.0E–10

2.134E–08

1.877E–08

v0

1.67E+12

A

3.5E+04

B

3.0E+04

F

341.23

V

4.0E+13

)

ODE)Report)(RKF45)#

#Differential#equations#as#entered#by#the#user#

#

1-5#

P1–2)(h))Continued#

(3))

Changing###a####!#Increasing#‘a’#reduces#the#amplitude#of#ripples#in#graph.##It#reduces#the#effect#of#the#

#

)

P1-3)(a))

Initial#number#of#rabbits,#x(0)#=#500#

Initial#number#of#foxes,#y(0)#=#200#

k4=0.04day−1

See#Polymath#program#P1-3-a.pol.#

)

POLYMATH)Results#

#Differential#equations#as#entered#by#the#user#

#[1]#d(x)/d(t)#=#(k1*x)–(k2*x*y)#

1-6#

P1-3)(a))Continued)

#

When,#tfinal#=#800#and#

30.00004 /( )k day rabbits=×

#

Plotting#rabbits#vs.#foxes#

#

)

1-7#

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))

)

1-8#

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##

#

V=

FA0−FA

−rA

#

1-9#

P1-6)(a))Continued)

#

v0

k

dCA

CA

∫=dV

V

∫

##=>#

V=

v0

k(CA0−CA)

#

P1-6)(b)##

-#rA#=#kCA#with#k#=#0.0001#s-1#

CSTR:)

We#have#already#derived#that#

#

V=

CA0v0−CAv0

−rA

##

=

v0CA0(1−0.01)

kCA

#

PFR:)

From#above#we#already#know#that#for#a#PFR#

#

dCAv0

dV =rA=−kCA

#

Integrating#

#

v0

k

dCA

CA

∫=−dV

V

∫

#

P1-6)(c)#

-#rA#=#kCA

2#with#k#=#300#dm3/mol.hr#

CSTR:)

#

V=

CA0v0−CAv0

−rA

#

=

v0CA0(1−0.01)

2

#

Substituting#all#the#values#we#get#

#

V=(10dm3/hr)(0.5mol /dm3)(0.99)

##=>#V)=)660)dm3#

PFR:)

#

dCAv0

dV =rA=−kCA

2

#

1-10#

P1-6)(c))Continued)

Integrating#

#

v0

k

dCA

2

CA

∫=−dV

V

∫

##=>

v0

k(1

CA

−1

CA0

)=V

#

P1-6)(d))

CA#=#0.001CA0#

Constant#Volume#V=V0#

t=

dCA

−rA

CA

CA0

∫

#

Zero#order:#

t=1

k

CA0−0.001CA0

“

#$

%=

.999CAo

0.05 =9.99h

#

First#order:#

t=1

k

ln

CA0

CA

!

“

#

$

%

&

&=1

0.0001

ln 1

.001

!

“

#$

%

&=69078 s=19.19h

#

Second#order:#

t=1

k

1

CA

−1

CA0

“

#

$

%

&

‘

‘=1

300

1

0.5⋅0.001 −1

0.5

“

#

$%

&

‘=6.66h

#

P1-7))Enrico#Fermi#Problem#

P1-7(a))Population#of#Chicago#=#4,000,000#

Size#of#Households#=#4#

Number#of#Households#=#1,000,000#

1-11#

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#

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#

P1-7(d))Jean#Valjean,#Les#Misérables.#

#

P1-8)

Mole#Balance:#

V”=”

F

A0 −“F

A

−r

A

#

Rate#Law#:#

−rA=kCA

2

#

Combine:#

V”=”

F

A0 −“F

A

kCA

2

#

FA0=v0CA=3dm3

s

.2molA

dm3=6molA

s

#

FA=v0CA=3dm3

s

.0.1molA

dm3=0.3molA

s

#

V”=”

(6 −0.3) mol

s

(0.03 dm3

mol.s)(0.1 mol

dm3)2

=1900dm3

#

The#incorrect#part#is#in#step#6,#where#the#initial#concentration#has#been#used#instead#of#the#exit#

concentration.#

#