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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
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 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
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+12
A
3.5E+04
3.5E+04
3.5E+04
3.5E+04
B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
F
341.23
341.23
341.23
341.23
V
4.0E+13
4.0E+13
4.0E+13
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
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+12
A
3.5E+04
3.5E+04
3.5E+04
3.5E+04
B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
F
341.23
341.23
341.23
341.23
V
4.0E+13
4.0E+13
4.0E+13
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
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.#
#
