Title | Elements of Chemical Reactor Engineering_4th (Soution Manual)-Fogler.pdf |
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~CD-ROM ~INClUDED Brian Vicente / Max Nori / H. Scott Fogler Soludons Manual for Elements of Chemical , , Reaction Engineering Fourth Edition ~ Prentice Hallitaternational Series in the Phys!ptl and Chemical E.ngineering Sciences ....- ••• ••• II!I!!III •• A Complimentary Copy from •• PRENTI CE HAL...
~CD-ROM
~INClUDED
Brian Vicente / Max Nori / H. Scott Fogler
Soludons Manual for
Elements
of Chemical Reaction Engineering Fourth Edition ~
Prentice Hallitaternational Series in the Phys!ptl and Chemical E.ngineering Sciences
....••• •••
A Complimentary Copy from
II!I!!III
•• ••
PRENTI CE
HALL
FT Prentice Hall
www"pearsoned-asia"com
Enquiries: 3181 0368
,
,
Solutions Manual for
Elements of Chemical Reaction Engineering Fourth Edition
Brian Vicente Max Nori H. Scott Fogler
~
PRENTICE
HALL
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ISBN 0-13-·186383-5 Text printed in the United States at OPM in Laflin, Pennsylvania. First printing, November 2005 ______________...________ .________ :
**** CONFIDENTIAL **** UNIVERSITY OF MICHIGAN INTERACTIVE COMPUTER MODULES FOR CHEMICAL ENGINEERING CHEMICAL REACTION ENGINEERING MODULES H. Scott Fogler, Project Director M . Nihat GUIllen, Project Manager (2002-2004) Susan Montgomery, Project Manager (1991-1993) Department of Chemical Engineering University of Michigan Ann Arbor, MI48109-2136 ©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 tum 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].
PerfOimance NumbeI InteIpIetation: eRE modules
iii
**** CONFIDENTIAL **** ICMs with Windows® interface Module
Format
Example
Interpretation
KINETIC CHALLENGE I CzBzzAzz
Score = 1.5 * AB.C z =random numbers
Perf. No. =15.641~92 Score = 1.5 *( 62.7) =94 %
Note: 75% constitutes mastery.
KINETIC CHALLENGE II CzBzzAzz
Score = 2.0 * ARC z = random numbers Note: 75% constitutes mastery.
Perf.. No. =Q3176.167 Score = 2.0*(47.0) = 94 %
A even: Killer and victim correctly identified A odd: Killer and victim not identified z = random numbers
Perf. No. = 50132 Score: No credit
MURDER MYSTERY zzAzz
Note: An even number for the middle digit constitutes mastery.
TIC TAC TOE zDzCzBzA
Score =4.0 * AB.C z = random numbers
Perf. No. = 718Q3281 Score =4*(15 . 0) =60 configuration 7 completed
Configurations
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. = 777J.8Q18 Score = 6*(07.3) = 44
Note: A score of 40 is needed for mastery of this module. PeIformance NumbeI Interpletation: eRE modules
iv
**** CONFIDENTIAL **** ECOLOGY AzBCzaaD
z = random numbers
a =random characters
A gives info on rl\2 value of the student's linearized plot A=Y if rl\2 >= 0.9 A=A if 0.9 > rl\2 >= 0.8 A=X if 0..8> rl\2 >= 0.7 A=F if 0..7 > rl\2 A=Q if Wetland Analysis/Simulator portion has not been completed B gives info on alpha B=l 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=l 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 > rl\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*ARC Final flow rate = 2*DE.F
Perf. No.. =
2125.:!8~913
conversion = 2*42 ..1 = 84.2 flow rate = 2*31.2 = 62.4
Please make a passlfiil criterion based on these values .
Performance Number Interpretation: eRE modules
v
**** CONFIDENTIAL **** ICMs with Dos® interface Module
Format
Interpretation
Example
A=2,3,5,7: interaction done B=2,3,5,7: intra done C=2,3,5,7: review done D denotes how much they did in the interaction:
Perf. No. = 8027435 A: Worked on interaction B: Looked at intra C: Looked at review D: found parameter values, didn't find mechanism
HETCAT zzABzCD
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. = 407.2.82 Interaction not completed
Note: Performance number given only if student goes through the interaction portion of the module.
PerfclImance NumbeI InteIpretation: eRE modules
vi
Solutions for Chapter 1 - Mole Balances
Synopsis General: The goal of these problems are to reinforce the definitions and provide an understanding of the mole balances of the different types of reactors . It lays the foundation for step 1 of the algorithm in Chapter 4.,
PI-I.
This problem helps the student understand the course goals and objectives.
PI-2.
Part (d) gives hints on how to solve problems when they get stuck. Encourages students to get in the habit of writing down what they learned from each chapter. It also gives tips on problem solving.
PI-3.
Helps the student understand critical thinking and creative thinking, which are two major goals of the course.
PI-4.
Requires the student to at least look at the wide and wonderful resources available on the CD-ROM and the Web.
PI-S.
The ICMs have been found to be a great motivation for this material.
PI-6.
Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives the student an idea of things to come in terms of sizing reactors in chapter 4. An alternative to PI-IS.
PI-7.
Straight forward modification of Example 1-1.
PI-S.
Helps the student review and member assumption for each design equation.
PI-9 and PI-IO. The results of these problems will appear in later chapters. Straight forward application of chapter 1 principles. PI-II.
Straight forward modification of the mole balance. Assigned for those who emphasize bioreaction problems.,
PI-12.
Can be assigned to just be read and not necessarily to be worked, It will give students a flavor of the top selling chemicals and top chemical companies.
1·6
PI-13.
Will be useful when the table is completed and the students can refer back to it in later chapters. Answers to this problem can be found on Professor Susan Montgomery's equipment module on the CD-ROM. See PI-17.
PI-14.
Many students like this straight forward problem because they see how CRE principles can be applied to an everyday example. It is often assigned as an inclass problem where parts (a) through (0 are printed out from the web. Part (g) is usually omitted.
PI-IS.
Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with PI-6.
PI-16.
Open-ended problem.
PI-17.
I always assign this problem so that the students will learn how to use POLYMATHIMaLab before needing it for chemical reaction engineering problems.
PI-IS.
Parts (a) and (b) are open-ended problem.
PI-19 and PI-20. Help develop critical thinking and analysis. CDPI-A
Similar to problems 3, 4, 11, and 12.
CDPI-B
Points out difference in rate per unit liquid volume and rate per reactor volume. Summary
• • •
PI--l PI-2 PI-3 PI-4 PI-5 PI-6 PI-7 PI-8 PI-9 Pl-IO Pl-11
Assigned AA I 0 0 AA AA I S S S 0
Alternates
Difficulty
1-15
SF SF SF SF SF SF SF SF SF SF FSF
1-7
Time (min) 60 30 30 30 30
15 15 15 15 15 15
•
PI-12 PI-13 PI-14 PI-IS PI-16 PI-17 PI-18 PI-19 PI-20 CDPI-A CDPI-B
I I 0 0 S AA S 0 0 AA I
SF SF FSF SF SF SF SF
- Read Only
FSF FSF FSF
30 1 30 60 15 60 30 30 15 30 30
Assigned
• =Always assigned, AA =Always assign one from the group of alternates,
o = Often, I = Infrequently, S = Seldom, G = Graduate level
Alternates In problems that have a dot in conjunction with AA means that one of the problem, either the problem with a dot or anyone of the alternates are always assigned. Time Approximate time in minutes it would take a B/B+ student to solve the problem. Difficulty SF =Straight forward reinforcement of principles (plug and chug) FSF = Fairly straight forward (requires some manipulation of equations or an intermediate calculation). IC =Intermediate calculation required M =More difficult OE = Some parts open-ended.
*Note the letter problems are found on the CD-ROM. For example A == CDPI-A.
Summary Table Ch-l ---------------,
Review of Definitions and Assumptions
1,5,6,7,8,9
Introduction to the CD-ROM
1,2,3,4
1-8
Make a calculation
6
Open-ended
8,16
PI-I Individualized solution. PI-2 Individualized solution. PI-3 Individualized solution. PI-4 Individualized solution. PI-5 Solution is in the decoding algorithm given with the modules . PI-6 The general equation for a CSTR is:
V
= F AO
--FA
- rA Here rA is the rate of a first order reaction given by: fA=·- kC A Given: C A= O.lCAO , k = 0.23 min-I, Vo = 10dm3 min-I, FA = 5.0 moVhr And we know that FA = CAVo andFAo = CAOVo 3 => C AO = F Aol Vo = 0.5 moVdm Substituting in the above equation we get:
v = CAOVO -CAVO = (0.5mol/dm 3 )(lOdm 3 lmin)-0.1(0.5mq!ldm 3 )(lOdm 3 I min) (0.23 min--I)(O. 1(0.5mol I dm 3 »
kC A V =391.3 dm3
PI-7 t
= INA
-~-kN
dNA
A
NAO k = 0.23 min-l
dNA
From mole balance:
-=IA' V dt 1-9
Rate law:
Combine:
at T:::: 0, N Ao = 100 mol and T:::: T, NA = (O ..01)NAo
1 (N AO J -+ t= "kIn NA 1
:::: -In(lOO) min 0.23
t = 20 min
Pl-8 a)
The assumptions made in deriving the design equation of a batch reactor are: Closed system: no streams carrying mass enter or leave the system. Well mixed, no spatial variation in system properties Constant Volume or constant pressure.
b)
The assumptions made in deriving the design equation of CSTR, are: Steady state. No spatial variation in concentration, temperature, or reaction rate throughout the vesseL
c)
The assumptions made in deriving the design equation ofPFR are: Steady state . No radial variation in properties of the system.
d)
The assumptions made in deriving the design equation of PBR are: Steady state . No radial variation in properties of the system.
e) For a reaction, A-7 B -rA is the number of moles of A reacting (disappearing) per unit time per unit volume (dm3 s) 1-10
[=J moles/
··rA' is the rate of disappearance of species A per unit mass (or area) of catalyst [=] moles/ (time. mass of catalyst).. rA' is the rate of formation (generation) of species A per unit mass (or area) of catalyst [=] moles/ (time. mass catalyst).. -rA is an intensive property, that is, it is a function of concentration, temperature, pressure, and the type of catalyst (if any), and is defined at any point (location) within the system. It is independent of amount On the other hand, an extensive property is obtained by summing up the properties of individual subsystems within the total system; in this sense, -rA is independent of the 'extent' of the system.
P 1-9 Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per second . It is an Intensive property and the concentration, temperature and hence the rate varies with spatial coordinates.
rA on the other hand is defined as g mol of A reacted per gm. of the catalyst per second.. Here mass of catalyst is the basis as this is what is important in catalyst reactions and not the reactor volume. Applying general mole balance we get:
dN. dt
--.L=F· o -F. + fr.dV J
]
J
No accumulation and no spatial variation implies
o= F·Jo -
F.] + fr J.dV Also Ij = Pb rj' and W = VPb where Pb is the bulk density of the bed. =>
f
0 = (FjO - F) + r;CPb dV )
Hence the above equation becomes
Fo-F W=_l __,_l --rj We can also just apply the general mole balance as
-dN j = (FjO dt
Fj ) + f'r/dW)
Assuming no accumulation and no spatial variation in rate, we get the same form above:
FO-F W =._1_._,_1
-rj
1-11
as
Fj
,/ i' F JO
PI-IO Mole balance on species j is:
v
Po -P + fr.dV I
I
o
I
dN.
=__ 1 dt
Let M j = molecular wt. of species j
= W jO
Then FjoM j
N jM j
= mass flow rate of j into reactor:
= m j = mass cif species j in the reactor
Multiplying the mole balance on species j by M
v
FjOM j
FjM I +M I frjdV = M j
-
o
Now M
I
dN
1
._ _
dt
is constant:
F M. -PM. + vfM rdV 10
I
I
I
I
o
I I
= d0! j N j ) = d(m1l dt
dt
v
w jO -- Wj
+
fM jrjdV =
o
PI-II Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 7), so there is no cell growth and the nutrients are used in making product Let's do pari c first
1-12
[In flowrate (moles/time)] penicillin + [generation rate (molesltime)]penicillin - [Out flowrate (molesltime)] penicillin =
[rate of accumulation (moles/time)]penicillin Fp,in + Gp - Fp,out =
dNp
dt ,., .,(because no penicillin inflow)
Fp,in = 0 """ ''''', ... v Gp =
fr~.dV
Therefore,
frp .dV -
dNp
v
Fp,out =
dt
Assuming steady state for the rate of production of penicillin in the cells stationary state,
dNp
--=0
dt And no variations
Or,
Similarly, fOI Corn Steep Liquor with Fe = 0
Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.
Pl-12 (a) Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below:
.--'
Rank 2002 1 2
3
Rank 1995 I 2 4
4 5
9
6
-
3
Chemical H 2SO4 N2 C2H 4 O2 C3H 6 H2 1-13
6 NH3 10 8 Ch 9 P20S 10 C2H 2Clz -+ Cherrucals like H2 , P 20 S , C2H2Cl 2 has come III top 10 cherrucals and C3H6 has jumped to rank 5 now then rank 9 in 1995 . 7
Pl-12 (b) Ranking of top 10 chemical companies in sales in yeru 2003 and 2002: 2003 I----
1 2 3 4 5
+-. 8 9 10
Chemical Sales
Company
2002
($ million 2003)
1 2 3 4 8 5 6 7 11
9
-
Dow Chemical DuPont ExxonMobil General Electric Chevron Phillips Huntsman Corp. PPG Industries Equistar Chemicals Air Products Eastman Chemicals
32632 30249 20190 8371 7018 6990 6606 6545 6029 5800
----
--
SOllIce: Chemical and Engineering News may 17,2004
-+ We have Chevron Phillips whichjumped to 5 rank in 2003 from 8th rank in 2002 and Air Products coming to 9th rank in 2003 flom 11 th in 2001. -+Chemical sales of each company has increased compared to year 2002 from 9%(Eastman Chemical) t028.2%(Chevron Phillips) but Huntsman Corp . has a decrease by 2 . 9%.
Pl-12 (C) Sulfuric acid is prime importance in manufacturing. It is used in some phase of the manufacture of neruly all industrial products It is used in production of every other strong acid. Because of its lru·ge number of uses, it's the most produced chemicaL Sulfuric acid uses are: -+ It is consumed in production of fertilizers such as ammonium sulphate (NH4hS04 and superphosphate (Ca(H2P0 4)2) , which is formed when rock phosphate is treated with sulfUric acid . -+Used as dehydrating agent. -+Used in manufacturing of explosives, dyestuffs, other acids, pruchment paper, glue, purification of petroleum and picking of metals. -+ Used to remove oxides from iron and steel before galvanizing or electroplating. -+ Used in non-ferrous metallurgy, in production of rayon and film. -+as laboratory reagent and etchant and in storage batteries -+ It is also general purpose food additive.
Pl-12 (d) Annual Production rate of ethylene for year 2002 is 5.21x 1010 lb/yeru Annual Production rate of benzene for yeru 2002 is 1.58 x 10...