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Elements
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Solutions Manual for
Elements of Chemical Reaction Engineering Fourth Edition
Brian Vicente Max Nori H. Scott Fogler
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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 1010 lb/year
1-14
Annual Production rate of ethylene oxide for year 2002 is 7.6 x109 lb/year
Pl-12 (e) Because the basic raw material 'coal and petroleum' for organic chemicals is very limited and their production is not increasing as production of raw material for inorganic chemicals.
1-15
Pl-13 Type
Characteristics
Phases
Usage
Advantage
Disadvantage
Batc h
All the reactants fed into the reactor. During reaction nothing is added or removed . Easy heating or cooling. Continuous flow of reactants and products . Uniform composition tluoughout
1. Liquid phase 2" Gas phase 3" Liquid Solid
1, Small scale pdn. 2. Used for lab experimentation. 3. Pharmaceuticals 4. Felmentation
1. High Operating cost 2. Variable product quality.
L Liquid phase 2. Gasliquid 3. Solid -, liquid
L Used when agitation required. 2. Series Configuration possible for different configuration streams
PFR
One long reactor or number of CSTR's in series . No radial variations . Conc. changes along the length,
1. Primarily gas Phase
1. Large Scale pdn . 2 . Fast reactions 3. Homogenous reactions 4, Heterogeneous reactions 5" Continuous pdn,
PBR
Tubular reactor that is packed with solid catalyst particles"
L Gas Phase (Solid Catalyst) 2.Gas -solid reactions.
1. Used plimarily in the heterogeneous gas phase reaction with solid catalyst e . g Fischer tropsch synthesis.
L High Conversion per unit volume. 2. Flexibility of using for multiple reactions" 3. Easy to clean 1. Continuous Operation" 2.Good Temperature Control 3 . Two phase reactions possible. 4.Good Control 5. Simplicity of construction. 6. Low operating cost 7. Easy to clean 1. High conversion per unit volume 2. Easy to maintain (No moving parts) 3 . low operating cost 4. continuous operation 1. High conversion per unit mass of catalyst 2" low operating cost 3., Continuous operation
CST R
""
_.
1·16
1. Lowest conversion per unit volume. 2. By passing possible with poor agitation. 3 High power Input reqd.
1. Undesired thelmal gradient 2. Poor temperature control 3" Shutdown and cleaning expensive" -
L Undesired thelmal gradient 2" Poor temperature control 3. Channeling 4" Cleaning .~~pensiv£:. __
Pl-14 Given
A
= 2 *1010 ft 2
TSTP
= 491.69R
T = 534 . 7°R
Po= latm
cS = 2.04 *10-10 lbm301
=0.7302 atm ft3
R
H = 2000ft
ft
lbmolR C = 4*105 cars
FS = CO in Santa Ana wind
VA
FA = CO emission from autos
ft3 = 3000-·- per car at STP hr
Pl-14 (a) Total number of Ib moles gas in the system: POV N :=-R-T
latmx(4xlO 13 ft3) N= (
3
J
= L025
X
1011 lb mol
0.73 atm·ft_ x534.69R lbmol.R
Pl-14 (b) Molar flowrate of CO into L.A. Basin by cars .
F F A -YA T -YA
VA CPo ·Rr STP
= ~~00ft3
F T
FA
hr car
= 6 . 685
X
X llbmol
x400000 cars
359 ft3
(See appendix B)
104 lb mollhr
Pl-14 (C) Wind speed through corridor is v = l.5mph W = 20 miles The volumetric flowrate in the conidor is Vo = v.W.H = (l5x5280)(20x5280)(2000) fe/ill l3 = L673 x lO fe/ill 1-17
Pl-14 (d) Molar flowrate of CO into basin from Sant Ana wind FS:= vOCS = 1.673 x 1013 ft 3fhr Fs = 3412 x 103lbmolflu
x2.04xlO- 1O Ibmolfft3
Pl-14 (e) Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO
=
V dC co dt
(V=constant, Nco
Pl-14 (0 t= 0 ,
Ceo
= Ccoo
Pl-14 (g) Time for concentration to reach 8 ppm,
C = 2.04 X 10-8 ~bmol C, = 3..:94 X 10-8 ~~mol coo ft3 ' co 4 ft3 From (f),
_ V In (FA +Fs--vo.CcooJ t--VO FA + Fs --VoCca
=
4ft 3
--In
1.673xlO13
t=
ft' hr
6.92 hr
Pl-14 (h) (1)
Ceo = 2.00E-1O IbmoUft3
tl = 72lus a = 3.50E+041bmoUln
v0
b
to
0 L67E+ 12 fe flu
= 3,OOE+04 IbmoUln
1-18
dN
=-
co
dt
= Cco V )
Fs
a + b
=
v = 4.0E+13
341.231bmol/hr
Sin( ~ JZ"
)
ft 3
dCco = v
+ Fs
dt
Now solving this equation using POLYMATH we get plot between Ceo vs
See Polymath program Pl-14···h·-l.pol. POLYMATH Results Calculated values of the DEQ variahles Var'iable t C
vO a b F
V
initial value 0 2.0E-10 1.67E+12 3.5E+04 3 . 0E+04 341..23 4.0E+13
minimal value
maximal value
0 2 . 0E-10 1 . 67E+12 3.5E+04 3 . 0E+04 341 . 23 4 . 0E+13
final value
72
72
2.134E-08 1 . 67E+12 3.5E+04 3 . 0E+04 341..23 4.0E+13
1.877E-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-vO*C)/V Explicit equations as entered by the user [1] vO=1 . 67*10A12 [2] a = 35000 [3] b = 30000 [4] F = 341 . 23 [5] V = 4*10A13 3.0e-8,---------------
24e··8
80
(2)
tf= 48lus
F,=O
a + bSin(
JZ"~)
Now solving this equation using POLYMATH we get plot between Ceo vs t
1·19
v dCco dt
See Polymath program P 1-14-h--2.pol. POL YMA TH Results Calculated values of the DEQ variables Variable t C
vO a b V
initial value 0 2 . 0E-I0 1 . 67E+12 3 . 5E+04 3 . 0E+04 4.0E+13
minimal value 0 2 . 0E-I0 1 . 67E+12 3 . 5E+04 3 . 0E+04 4 . 0E+13
maximal value 48 1 . 904E-08 1 . 67E+12 3 . 5E+04 3 . 0E+04 4 . 0E+13
final value 48 1 . 693E-08 1.67E+12 3 . 5E+04 3.0E+04 4.0E+13
ODE Report (RKF45) Differential equations as entered by the user t 1] d(C)/d(t) = (a+b*sin(3 . 14*t/6)-vO*C)/V Explicit equations as entered by the user [1] vO = 1.67*10A12 [2] a = 35000 [3] b = 30000 [4] V = 4*10A13 2 Oe-8
r-------- .----------,
50
(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.
-+ The amplitude of ripples is directly proportional to 'b' .
Changing b
As b decreases amplitude decreases and graph becomes smooth. Changing Vo
-+ As the value of Vo is increased the graph changes to a "shifted sin-curve" And as Vo is decreased graph changes to a smooth increasing curve .
------------------ - - - - - - - - - - -
PI-IS (a) - rA =
k with k = OD5 mol/h dm3
CSTR: The general equation is 1-20
v = FAO -FA -rA 3
Here C A= D"DIC Ao , Vo = 10 dm /min, FA = 5 . 0. mollhr 3 Also we know that FA = CAVO and F Ao = CAOVo, CAO = FAoI Vo = 0..5 mol/dm Substituting the values in the above equation we get,
V =
CAOVO -CAvo
= (0.5)10-0.01(0.5)10
k
0.05
-7 V=99dm3 PFR: The general equation is
dFA dCAv O --=r =-·k A =k,NowFA=CAvoandFAo=CAovo=> dV dV Integrating the above equation we get V
CA
k
V
V
-~ JdC A = JdV
=>
0.
C AO
V =--.!l(CAG ,,- C A ) k
Hence V = 99 dm3 Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of concentration.
PI-IS (b) - rA = kC A with k = 0.,,0.0.0.1 s"! CSTR: We have already derived that
V
= CAOVO -CAvo - rA
_ vo C Ao (1-0.01) kC A
k = D. DDDls"! = 0. . 0.0.0.1 x 360.0. hr"!= 0..36 hI'''!
-7
3 V = (10"dm / hr )(0.5mal / dny3 )(O.~?) (0.36hr -1 )(0.01 * 0.5mal / dm 3 )
PFR: From above we already know that for a PFR
dCAv O -rA -- kCA dV Integrating
dC f _A =- JdV V
C _VG
k Vo
k
CAO
CA
0.
In C AG =V CA
Again k = D.DDDls"! = 0..0.0.0.1 x 360.0. hI != 0..36 ru-! o
1-21
=> V
=2750 dm3
Substituing the values in above equation we get V = 127.9 dm3
PI-IS (c) - fA
= kC A2 with k = 3 dm3/moLhr
CSTR: -rA Substituting all the values we get
V = (lOdm (3dm
3
3
/
/
3
hr)(O.5mol / dm )(O.99)
hr)(O.Ol * O.5mol / dm
3
=> V = 66000 dm3
)2
PFR:
dCAv O _- r -kC 2 A A dV Integrating CA
V
--~
k
f
dC _A
C/o
CAD
=> V
vII
V
=- fdV =>~-(----)=V
=
k CA
3
10dm I hr ( 1 3 3dm lmol.hr O.OlCAO
CAO
_~_) CAO
PI-IS (d)
t-
t
. dN -- r V A
AO
WA
Constant Volume V=Vo
t=
fAO
dC A
1A -
rA
Zero order:
t
.999C Ao = -1 [CAO -O.OOlCAO ] = - - = 9.99h k
0.05
FiIst Older: 1-22
= 660 dm3
=~ln(CADJ
t
CA
k
=_I_ln(_I_) = 0.001 .001
6908s
Second order: t
=k1[1 C
A -
1]
CAD
1 ="31[ 0.0005 -
1] 0.5 = 666 h
PI-16 Individualized Solution PI·17 (a)
Initial number of rabbits, x(O) =500 Initial number of foxes, y(O) = 200 Number of days = 500
dx
dt = k1x-k2 ·xy .................................... (1)
dy 'dt
= k3 XY _. k4 y
.............. .
................ (2)
Given,
kl = 0.02day--1
= 0.00004/(dayx foxes) k3 = 0.0004/(dayx rabbits) k4 = 0.04day-1 k2
See Polymath program P 1 . 17 -a.pol. POLYMATH Results Calculated values of the DEQ variables Variable t x Y
kl k2 k3 k4
initial value 0 SOO 200 0 . 02 4 . 0E-OS 4.0E-04 0 . 04
minimal value 0 2.9626929 1.128S722 0 . 02 4.0E-OS 4.0E-04 0 . 04
maximal value SOO S19.40024 4099.S17 0 . 02 4.0E--OS 4.0E-04 0 . 04
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(x)/d(t) (k1 *x)-(k2*x*y) [ 2 J d(y)/d(t) = (k3*x*y)-(k4 *y)
=
1-23
final value SOO 4.2199691 117.62928 0.02 4.0E-OS 4.0E-04 0.04
Explicit equations as entered by the user [1] k1 = 0 . 02 [2] k2 = 0 . 00004 [.3] k3 = 0.0004 [4] k4=0 . 04 5000.------------------
4000
3000
2000
o
o
200
When, tfinal= 800 and
160
kj
t
300
400
500
= O.00004/(dayxrabbits)
320 t
480
640
800
Plotting rabbits Vs . foxes
1-24
160lli-----------------
o 378
720
1061
rabddPl
1744
2086
PI·I7 (b) POLYMATII Results
See Polymath program Pl-17-b.pol. POLYMA TH Results
NLES Solution Var~able
Value
Ini Guess
f(x)
X------- 2.3850387 --'253E-iI" 3.7970279
y
2 2
1.72E-12
NLES Report (safenewt) Nonlinear equations ( 1] f(x) = xA3*y-4*yA2+3*x-1 = 0 [ 2 1 f(y)
=6*yJ\2-9*x*y-5 =0
PI-IS (a) No preheatmg of the benzene feed wtli dlmmlsh the rate of reactIon and thus lesser converSIOns achIeved .
WIll
PI-IS (b) An lDterpolation can be done on the logarithnnc scale to find the desued values from the gIven data. Now we can lDterpolate to the get the cost at 6000 gallons and 15000 gallons Cost of 6000 gal reactor = 1 905 x lOS $ Cost of 15000 gal reactor = 5 623 X 105 $ Coat va Volume of reactor (log - log plot)
PI-IS (c) We are given C A IS 0 . 1% of minal concentration ~ CA =OOOlCAO
..
. 1
--~-
--
., 1-25
"--t---
-
--~-"
1011 volume (1I8110na)
..
••
be
Also from Example 1.3,
Substituting vo= 10 dm3/min and k = 0.23 min-! we get
v = 300dm 3 which is three times the volume of reactor used in Example 1-3.
PI-18 (d) Safety of Plant PI-19 Enrico Fermi Problem - no definite solution PI-20 Enrico Fermi Problem -- no definite solution PI-21 Individualized solution . CDPI-A (a) How many moles of A are in the reactor initially? What is the initial concentration of A? If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite simple. We insert our variables into the ideal gas equation:
n=
!.V =
(20~tm)(200dm3)
(lOL3~~a) = 97.5moles
3
(8.3145 kPa.dm )(500) molK
RT
latm
Knowing the mole fraction of A (y Ao) is 75%, we multiply the total number of moles (N1o) by the YA:
molesA = N Ao
= 0.75x97.5 = 73.1
The initial concentration of A (C Ao) is just the moles of A divided by the volume:
CAo
=
moles volume
= N Ao =73 .1mol~~ = 0.37 moles / dm 3 V
200dm
CDPI-A(b) Time (t) for a 1st order reaction to consume 99% of A.
dC dt
r =_-A A
OUf
first order rate law is:
-rA =kCA
1-26
mole balance: dCA =--kC dt A
-kt =
In(
-k t fdt =>
= CfA dCA CAO
o
C
A
CA ) , knowing CA=O.OI C Ao and our rate constant (k=O.1 min-I), we can solve CAO
for the time of the reaction: t
= -..!.In(O.OI) = k
4.61 1 O.lmin-
= 46. 1min
CDPI-A (c) Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.
rate law: 1 1 =>-kt=--+CA CAo
dC mole balance: :. _2. = --kC~ dt
We can solve for the time in terms of our rate constant (k = 0.7) and our initial concentration (C Ao):
-kt = _~ +_1_ CAo CAO
t=_4_= 4 ----=15.4min.Todetermine kCAo (0.7dm 3 / mol min)( 0.37moll dm 3 ) the pressure of the reactor following this reaction, we will again use the ideal gas law. First, we determine the number of moles in the reactor:
NB =Nc =0.8NAo NT = (0.25)N yo + (0.2+0.8+0.8)N Ao
= 0.25(97.6) + (1.8)(73.2) = 156.1moles
3
(156. Imole) (0.082 dm atm] (500K) N 1RT molK p =_ _ = _ .___ ---''---__ = 32atm V 200dm 3
CDPI-B Given:
Liquid phase reaction in a foam reactor, A ~B
Consider a differential element, ~ V of the reactor: By material balance
FA -(FA +~FA)=-rA(l-e)~V 1-27
Where, (1- e)~ V = fraction of reactor element which is liquid. or:
-FA
=-rA(1-e)~V
dF
=r
_A
dV
(I-e) A
Must relate (- rA) to FA ' where, FA is the total (gas +liquid) molar flow rate of A.
-rA=rate of reaction (g mol A per cubic cm. of liquid per sec . ); e flow rate of A (g mol/sec . ); V = volume of reactor
1-28
= volume fraction of gas; FA = molar
Solutions for Chapter 2 - Conversion and Reactor Sizing Synopsis General: The overall goal of these problems is to help the student realize that if they have -rA=j{X) they can "design" or size a large number of reaction systems. It sets the stage for the algorithm developed in Chapter 4"
P2-l.
This problem will keep students thinking about writing down what they learned every chapter.
P2-2.
This "forces" the students to determine their learning style so they can better use the resources in the text and on the CDROM and the web.
P2-3.
ICMs have been found to motivate the students learning.
P2-4.
Introduces one of the new concepts of the 4th edition whereby the students "play" with the example problems before going on to other solutions.
P2-S.
This is a reasonably challenging problem that reinforces Levenspiels plots.
P2-6.
Novel application of Levenspiel plots from an article by Professor Alice Gast at Massachusetts Institute of Technology in CEE.
P2-7.
Straight forward problem alternative to problems 8, 9, and 12.
P2-S.
To be used in those courses emphasizing bio reaction engineering.
P2-9.
The answer gives ridiculously large reactor volume. The point is to encourage the student to question their numerical answers.
P2-l0.
Helps the students get a feel of real reactor sizes.
P2-ll.
Great motivating problem. Students remember this problem long after the course is over.
P2-l2.
Alternative problem to P2-7 and P2-9.
CDP2-A
Similar to 2-9 2-1
CDP2-B
Good problem to get groups started working together (e.g. cooperative learning.
CDP2-C
Similar to problems 2-8, 2-9, 2-12.
CDP2-D
Similar to problems 2-8,2-9,2-12. Summary
• • • •
P2-1 P2-2 P2-3 P2-4 P2-.5 P2-6 P2-7 P2-8 P2·-9 P2-10 P2-11 P2-12 CDP2-A CDP2-B CDP2-C CDP2-D
Assigned 0 A A 0 0 S AA S AA S AA AA 0 0 0 0
Alternates
8,9,12 7,9,12
7,8,9 9,B,C,D 9,B,C,D 9,B,C,D 9,B,C,D
Difficulty
M M FSF FSF SF SF SF SF FSF FSF FSF FSF
Time (min) 15 30 30 7.5 75
60 4.5 45 45 15 1
60 5 30 30 45
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 problems, either the problem with a dot or anyone of the alternates are always assigned.
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) 2-2
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-2 Straight forward
1,2,3,4,10
Fairly straight forward
7,9,12
More difficult
5,6,8
Open-ended
6
Comprehensive
4,5,6,7,8,9,12
Critical thinking
P2-9
-
P2-l Individualized solution . P2-2 Individualized solution. P2-3 Solution is in the decoding algorithm given with the modules . P2-4 (a) Example 2-1 through 2-3 If flow rate FAD is cut in half. VI = v/2, F I = FAd2 and CAD will remain same . Therefore, volume of CSTR in example 2-3,
F;.X 1 FAOX 1 VI =----=---=-6.4=3.2 - rA 2 -r A 2 If the flow rate is doubled, F2 = 2FAD and CAD will remain same, Volume ofCSTR in example 2-3, V 2 = F2X1-rA = 12 . 8 m3
P2-4 (b) Example 2-5 2-3
--
Levenspiel Plot
Faol2
45-.------------------------------------4 ~;==;==;==;===~ 35~~~
J--O+X overall
'"';- 25
o '" LL
2 15
05 o+-~~r=~~~~~T=~-=~--~
Faol2
o
02
04
06
08
Conversion
Now, F AO = 0.,4/2 = 0...2 molls,
F
New Table: Divide each term ~ in Table 2-3 by 2.
cx:= ---.--/
-- 'A
0.-·10.·-1- . - - -0..2--'--, , 0-.4- . - - ,0.-.6 0..445 TI545 0..665 1.0.25 1.77
~-rAJ(m3)
Reactor 2 3 V 2 = 1.2 m
Reactor 1
VI = 0.. 82m3 V = (FAol-IA)X
0.82 =
(_~~o J (Xl) A
XI
By trial and error we get: and X 2 = 0. . 8 Overall conversion XOver.1I = (l/2)XI + (l/2)X2 = (0..546+0.._8)/2 = 0..673 XI
= 0..546
P2-4 (C) Example 2-6 Now, F AO = 0..4/2 = 0..2 molls, Fao/2
X1
....--+X overall
Faol2
2-4
_Jj)4J.~J
_ ,/ 0.---,,-.7
~.53
k.-J
F
New Table: Divide each term ~ in Table 2-3 by 2 .
-rA
0.1
0.6 1.77
0.4 1.025
0.2 0.665
3
VI = 0.551m
V; = FAO
x dX
I--
o -rA
Plot FAof-rA versus conversion. Estimate outlet conversions by computing the integral of the plotted function.
Levenspiel Plot 45 4 3.5
...ca•
"0 ca
LL.
3 25 2 1.5 1 0.5 0
0. 2
0
06
0.4
0.8
Conversion [ XI = 0603 f~r VI = 0.551m "---X;-=-O. 89 for Vz = L614m3 Overall conversion Xo= (l/2)XI + (1I2)Xz = (0 . 603+0.89)/2 = 0.746 3
-
._----
-.~
_J
Levenspiel Plot
P2-4 (d) Example 2-7
25
(1)
2
ForPFR, ~ 15
c5
'"
IL
1 05
= 0.222m
3 0 0
0.1
02
For first CSTR,
03
04
Convelsion
2-5
05
06
07
F
X 2 = 0..6, _AO.
-rA
_
3
-1.32m ,
For second CSTR,
F 3 X3 = 0.65, -AO - = 2.0m ,
-rA
V 3-
FAO (X 3 -X 2 )
.
-rA
= 01. m3
(2) First CSTR remains unchanged ForPFR:
V=
05(F } X' f.
_AO
02
rA
Using the Levenspiel Plot
Levenspiel Plot 25 2+-----------------~---
*
15+----------------------
VPPR = 0..22
"0
.f ForCSTR,
1
o5
i=====~=iO;;;;;:~------~ +-------~----
o +---,---'-'-'--o
01
02
0.3
04
0.5
06
07
Conversion
(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR.
'[J_O~23J)-IIT53 ~
-r)- (
Conversion Original Reactor Volumes ~ Worst Ana"itgement_,,___ Xl = 0..20 VI = 0.188 (CSTR) VI = 0..23 (PFR) X2 = 0..60 V2 = 0.,38 (PFR) V2 = 0.53 (CSTR) LX:.::.::..3_=-=0..:.:.6:.::.5_ _ _ _ _ _ _ _V.3 = 0.10 (CSTR) _____ y3 = 0.10 (CSTR)
ForPFR, Xl = 0..2
V. =
f( ~~J1X
Using trapezoidal rule, Xo = 0,1, Xl = 0.1
2-6
~ = (X~:D)[f(XD)+f(Xl)J = 0.2 [1.28+0.98]m 3
2 =0.23m 3 ForCSTR,
FAD
FAD V2 = -( X2 - ) Xl = 1.32(0.6- 0.2) = 0.53 ill3
For X 2 = 0 . 6, - - = 1.32m , 3
-rA
-rA
nd
For 2 CSTR,
FAD -rA
3
For X3 = 0.65, - - = 2m ,
P2-4 (e) Example 2-8 T Vo
= 5hrs
= 1dm3/min = 60dm3/hr CA = 2.5 mol/dm3X = 0.8
For CSTR,
T
V
= .-
Vo V= 300dm3 (1)
--r A= CAOX = 2.5xO.8 moll dm 3 hr T 5 3 = OAmol / dm hr
(2) V = 300dm3 (3)C A = CAO(1-X) = 0.5 mol/dm3
P2-5
m3 )
0
0.1
0..2
0.8 9
1.0 8
13 3
-
OA
0.6
0..7
O.
2. 0 5
3.5
5.0 6
8 8.. 0 ._.
4
Levenspiel plot
_
..............- .... _.. ... _._......_..- .... _"
P2-5 (a) Two CSTRs in series For first CSTR, V = (FAd-rAXl) XI =>Xl =OA4 For second CSTR, V = (FAd-rAX2) (X2 ·- XI) => X2 = 0.67
~5~----------·-------------------~L----.~
g
~4~--------------------
01
02
03
05
04
x
06
07
08
09
P2-5 (b) Two PFRs in series 4
v~ 1( ~~:}x +)( ~~:}X By extrapolating and solving, we get Xl = 0..50. X 2 = 0.74
P2-5 (C)
3
----f--------~
+------------
F"o -rA 2 (m') 1+-----------~~----------------~ .....-
Two CSTRs in parallel with the feed, F AO, divided equally between two reactors, FANEW/IAXI = o.5FAd-IAXI V = (o..5FAd-rAXI) Xl Solving we get, X out =0.,,60.
----_ ..
--_.._....
..
o +------,-----.------.------.----~ 0,,8 o 0,2 0,4 0.6 1 X
P2-5 (d) Two PFRs in parallel with the feed equally divided between the two reactors, FANEW/-rAXI = o.5FAOI-rAXI By extrapolating and solving as part (b), we get
X out = 0..74
P2-5 (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.,60. + 0.,74) 12 = 0.,67
P2-5 (0 A PFR followed by a CSTR, X PFR = 0...50. (using palt(b» V = (FAd-rA-XCSTR) (XcSTR - X PFR ) Solving we get, X CSTR = 0.,70.
P2-5 (g) A CSTR followed by a PFR, X CS1R = 0.,44 (using part(a»
2-8
V
=
X pFR
f
F -.MLdX
X CSTR -rA
XPFR = 0. . 72
By extrapolating and solving, we get
P2-5 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs,
ForPFR, XPFR = 0..50. (using part(b» 3 CSTR 1: V = (FAd-rA-XCSTR) (X CSTR - XPFR) = 0..5 m XCSTR = 0..63 3 CSTR2: V = (FAd-rA-XCSTR2) (XCSTR2 - XCSTRI) = 0..5 m X CSTR2 =0..72
P2-6 (a) Individualized Solution 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) = Y2 Fao (mother). The Levenspiel Plot is drawn for the baby hippo below .
Autocatalytic Reaction 5
-
.•....................................... __ ..._•....•.•. .....••.•............................,
4~------------------~--4 35~r----------------~----4
i
e
-.l.
o
~
3 -l-\---------------+------1-'-------:-:--::-~ 2 5 t-~~--------------j~----I
2t\-~~---------~~-.I_~r~--~-~ 1.5 t=~==~S;;;;;;;;o.."","'--~_r-----;
05
02
04
06
08
Conversion
2-9
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.
Age =
4 ..5 years 2
= 2 .25. years
2) IfVmax and mao are both one halfofthe mother's then
Catalytic Reaction
and since 2
then
1 -vrnaxCA
..=2:....-.._ _
-rAM2
KM
bllby
+C
A
mAo (
-rAM2
J baby
1
=- - r .
2
o
AM2mothef
1 '-m 2 Ao ----1 - 2 rAM2
Conversion
mAo
=
JS= 0.4
(
-
rAM
~
J mother
mother
m -rAM2
~ 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 040
P2-6 (C) VSlomach = 0.2 m
3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
m
- - AO
= 127X4 _ 172.J6X3 + 100.l8X2 - 28354X + 4.499
-rAMI
2·10
· r.n AO And smce Vstomach = - - - X, 5
solve V= 127X Xstomach = . 067.
-
-rAMI 4
2
172.36X + 1OG.18X3 - 28..354X + 4 . 499X = 0.2 m
3
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 . Autocatalytic Reaction
Catalyti c Reaction
5 45
+--------
4+\---35~--------------+_
i
----+---·l
3-··
n:t
!!: 25 fil E 2
-f-..'~-----------j'-------i
2
1.5 1-·······----
I
t - - - - - - - . - - - -.. - - -
05
o
-----,-----r-------.,..------/
o
X~0067
02
04
06
o
08
Conversion
X=0178 Con v er·si on
P2-6 (d) PFR~ CSTR PFR: Outlet conversion of PFR = 0 . 111 Autocatalytic Reaction
Catalytic Readion 5 45 4 3.5
,...
..
:E ca
2
-.!. 2.5 0
ca
E
2 1.5
05
o
X= 0.111 Conversion
0 0
0.2
0.4
Conversion
CSTR: 2-11
0.6
08
We must solve
v = 0..46 = (X-QJ 1l)(127X4 -
172.36X3 + 100. J 8X2 - 28,354X + 4.499)
X=O,42 Since the hippo gets a conversion over 30.% it will survive"
P2-7 Exothermic reaction: A ~ B + C 3
r(mol/dm ,min
X
1/-r(dm 3.min/mol) 1 0.,,6 0..2 0.,,2 0..2 0..2 0. . 8 1.1
) 0. 0. . 20. 0..40. 0..45 0..50. 0..60. 0..80. 0..90.
1 1.67 5 5 5 5 1 . 25 0..91
P2-7 (a) To solve this problem, first plot 1/-rA vs" X hom the chart above. Second, use mole balance as given below,
CSTR:
Mole balance:
= FAo
V CSTR
!
= (300mollmin)(0.4) => 3
(Smol / dm .min)
- 'A
=>V CSTR = 24 dm 3 PFR:
x dX
VpFR = F AO Mole balance:
f--r
0
A
0.4
= 3QQ(area under the curve)
0.6
X
V PFR = 72 dm3
P2-7 (b) 2-12
For a feed stream that enters the reaction with a previous conversion of DAD 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.
0040
CSTR
0.60
P2-' (C) 3
V CSTR = 105 dm
Mole balance:
FAOX VCSlR = - -rA
x ._. fA
105dm 3 - - - . = O.35dm 3 mini mol 300mollmin
Use trial and error to find maximum conversion.. At X= 0..70.,
1/-fA = 0..5, and X/-rA = 0..35 dm3 . minlmol
Maximum conversion = 0..70.
.
P2-' (d) From part (a) we know that Xl = DAn.
".
1
Xl PFR
II
1
Use trial and error to find X 2 .
X2
CSTR
2-13
--
.....
-
Reananging, we get
X 2 - 0040 - rAlx2
At X 2 = 0.64,
= ~ = 0.008 FAD
X 2 -OAO - rA
I
= 0.008
X 2
Conversion = 0.64
P2-7 (e)
From part (a), we know that Xl = OAO. Use trial and error to find X2
Mole balance: VPFR
= 72 = FAO
dX f --.r = 300 f-r
X2
dX
040 -
A
X2
040 -
At X 2 = 0 . 908, V = 300 x (area under the curve) => V = 300(024) = 72dm3 Conversion = 0..908 .
P2-7 (£) See Polymath program P2·7fpoL
2-14
A
-lIrA
0.65
6.0 5. 0
052
Q
0.39·
GJ
·to 3.0
0.26 2.0 0.13
1.0
0.00 0
20
40
V
60
0. 0
100
80
0
17
33
50
V
67
83
100
P2-8 (a) FsoX
V=---rs Fso = 1000 glhr
1
At a conversion of 40% - -
-rs
Therefore
V
3
dm hr = 0.15··--g
= (0.15)(1000)(0.40) = 60dm 3
P2-8 (b) 1 _. rs
At a conversion of 80%, - -
3
dm hr = 0.8----g
Fso = 1000 glhr Therefore
V
= (0.8)(1000)(0.80) = 640dm 3
P2-8 (C) x dX
VPFR = Fso
f--
o -rs
From the plot of 1/-rs Calculate the area under the curve such that the area is equal to VIPso = 80/ 1000 = 0..08 X= 12%
For the 80 dm3 CSTR,
F X V = 80 dm 3 =~ -f
s
XI-rs = 0.08.. From guess and check we get X = 55%
2-15
P2-8 (d) To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of 1I-rs Next is a PFR with the necessary volume to achieve the 80% conversion following the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion For two CSTR's in series, the optimum arrangement would still include a CSTR with the volume to achieve a conversion of about 45%, or the conversion that cOllesponds to the minimum value of 1I-rs, first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR
_ kC s (O.l[C so - C s ]+ 0.001)
-r
l
KM
+ Cs
Let us first consider when Cs is small Cso is a constant and if we group together the constants and simplify
since Cs < KM
1
-
KM
klC~ + k 2 Cs
r,
which is consistent with the shape of the graph when X is large (if Cs is small X is
large and as Cs grows X decreases), Now consider when Cs is large (X is small) As Cs gets larger Cc appwaches 0:
kCsCc 1 then - KM + Cs r, As Cs grows large!, Cs » KM
=
If - r
l
KM + C s = ---''-'---
kCsC c
2-16
And since Cc is becoming very small and approaching 0 at X = 0, 1/-1s should be increasing with Cs (or decreasing X)., This is what is observed at small values of X_ At intermediate levels of Cs and X, these driving forces are competing and why the curve of 1/-rs has a minimum.
P2-9 Irreversible gas phase reaction 2A+B ~2C
See Polymath program P2"-9.poJ.
P2-9 (a) PFR volume necessary to achieve 50% conversion Mole Balance
V=FAO
X2
dX
Xl
(-rA )
f---
Volume = Geometric area under the curve of (FAoI-rA) vsX)
V V
= GX400000XO.5) + (100000 X0.5)
= 150000 m3 x
P2-9 (b) CSTR Volume to achieve 50% conversion Mole Balance
V=!AO~_ (-rA ) V = 0.5 X 100000 V = 50000m3
-I- -+"o.s 0
J."~X.:;;;; 20.281.9 s o ""fA
80% conversion in PFR: fA
is not known for X>O.60 - can not do.
CDP2-D (e)
50 % in CSTR: 't' :::;
X
C Ao ~- :::; 30,422.9 s - fA.
V:::; v 0"' ;;;; (30,422.9 s { \
I
~~:~ Y.2 m>Imin)-;;;:: 1014.1 m 3
DOS
J
CDP2-D (1) 50 to 60% conversion in CSTR:
C.\o(X) -Xl) :::::: 8112.8 .
t -;;;:: --.:...--.....:::."".....- fAl
V = v.' = (8112.8S{To':-}2 m'/min)= 270.4 m' CDP2-D (g)
2-31
Rate of Reaction vs. VolUl:oe IOE,05 ,"-""'.'--"---
5 OE
'--1
I
061 .._, ---, --
2 SE 06
O.OEtOO
06
I
I ... --- "--, -,--
7 5E 06
?
.,
Conversion vs. Volurne
0.4 0.2
~
-, - -
L__,_._"~ ____ ,_~_,,,_J o
400
200
600
~--~-
0,
800
o
200
3 Volnme(m )
400
Volume (m
600 3 )
CDP2-D (h) Critique Answers are Valid: 1. Constant Temperature and Pressure No heat effects No pressure drop 2. Single interpolation to X .. :;;; 0.15, 030,0,45, and 0.50 allowable 3. tiuge volume (the size of the LA Basin)! Raise T? Raise P?
CDP2-E For the CSTR : VI
F X
=: , . ",::'l£, . .L ::;;;; .,['....
F...o ( Area) ]
Alea == VI == 1200 drri A ----
From the graph we can see thaI Xl :: 0.,60
4000
For the PFR:
Fio
XI) == FAo (Area und.er curve) V7. :::: .F"o(Xl ---,"._,
orA
''''''rA
We
2000 1000
J
Area under curve == V, == 600 dm From the graph
3000
can see that X z ::;;;; 0. . 80
2-32
poiu:::t
800
CDP2-F (a) Find the conversion for the CSTR and PFR connected in series ·,rA X 0 0.2 0.1 0.0167 0.00488 0.4 0.7 0.00286 0.9 0.00204
400 L CSTR and 100 L PFR
lI(-rA) 5 59.9 204.9 349.65 490.19
Feed is 41 % A, 41 % B, and 18% L
T :::;:; 2270 C = 500 K
P ;:: 10 attn
10 atm = ..-P ;;;:;.--..' .,...----.-...... ,-----...--",.,.".,--- = 0_244 mollL.
C
RT
' To
(0.082 I.. . atmfmol "K)( 500 K)
CA. ;;:: OAlC,o;:: 0.41(0.244 maUL)::::: 0.1 mol/I..
FAo
:;;; t)OC AO :::::
1 Us(O.l moUL)::::: 0.1 molls::: 6 mol/min
There are tv.'o possible arrangements of the system: 1. CSTR followed by the PFR 2. PFR followed by the CSTR Case 1: CSTR -.,. PFR
CSTR:
VI = F. . )Area) V
400
Area =: """,l". = '.... , ,:;;:;: 66.67 FA., 6 From the graph PFR :
V1
:;;:;:
~
XI = 0.36
FA., (Area under curve) V,
100
Area under curve:;: ,-- ;;: -""".,.,:;:; 16.667 F:. .., 6 From the graph .. X~ = 0.445
-
....... '.' ..~' ...... '.~-'- . ~-..-.,...-"."..,.,..-~--........
Case 1:
CSTR ••> PFR
Case 2;
500r-----~-------.,,~_,
•
..
-~ ~-
PFR --> CSTR
500~-------------r~
400 ~
.
400 ,.: :3 00 ,+.••,..."...",...
.
300
;;::; 200
::::; lQO +." .."-",,.,
100
100
o 0,0 0.2
o 0.4
0.6 X
C = 12.4 KJ/mol and a = 0..32 Using these values, and .1HR = - 6 kcaUmol, we get E = 10.48 KJ/mol
CDP3-C (a) A-7B Rate law at low temperature: -
rA = kC A
The rate law t higher temperature must: 1) Satis(y thermodynamics relationships at equilibrium, and 2) Reduce to irleversible rate law when the concentration of one or more of the reaction products is zero. Also, We know,
KC
__CBe -
CAe
Reauanging, we get
C
Ae
- C Be =0 K
c 3-25
So, lets assume rate law as
- rA
=kA(C ~:J A -
Also when CB = 0, it satisfies the given rate law, Hence the proposed rate law is conect.
CDP3-C (b) A+2B72D
- rA =kC A
Rate law at low temperature:
1/2
CB
Here,
But it does not satisfy the irreversible late law at low temperatures" Hence it is not COlrect So, taking square root of Kc
fj{
-V 1\' C = -
C De
C
112' CAe C Be
112 C
Ae
(c
- rA = k A
_ CDe = 0 ~Kc
Be
De Ae 1I2CBe - ~CK C
J
Which satisfies the irreversible rate law, Hence it is the required rate law.,
CDP3-C (C) A+B-7C+D
.
, k P A PB = ...::::--1 + KAPA + KBPB PCPD PCPD Kp = or PBPA -· =0 Kp PBPA
IrreversIble rate law: -, r A
We know,
Hence assume rate law as:
PCPDJ k PP -'--. K ( A B p - rA = . ."----1 + KAPA + KBPB + KcPc + KDPD Which satisfies both the above mentioned conditions,
3-26
CDP3-D y a)
- 0.15
A$$umi~1
C
b)
NlIJ • O
-
eo
CNH
ideal gas law
~Q
...
Va
3 ,.0
~
8.2 Itra ' .. 0.2 ,mol/l · OKCSOOK) 0.082 latln . ,:no 1
.. -
H:ro
.. y~ r_~ ~ (0.15)(0.2 ,mol/1) .. 0.03 gmol/1 ........3.0 '"""I ..
c)
Compound.
S:r=bol
~!p. ...
J.
;)
Il1it.i .. l
Chan,,,
0.15
-O.lS:.t 5
Find
0.15(1-X)
°l
B
0.18
- 4 1:kCAO =
ForPFR,
X 2 = O.~ =1.111 (I-X) 0.6
1:=-1-1 dX 2 kCAO 0 (1- X)
=_1_~ kCAO 1- X
=> X =1- _ _ 1_.=1-_1_=.526
1+1:kCAO
2.111
So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second order reaction.
P4-8 (d) 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 B, which have same conversion as particle size at A. But when we decrease the particle size by 20%, we reach at point C, so a decrease in conversion is noticed" Also when we increase the particle size from position A, we reach at point D, again there is a decrease in the conversion,
1 x
A
n
dp
P4-9 AB Kco (300K)= 3.0 V = lOOOgal = 3785.4 dm3 Mole balance:
V =!.AO X - rA
4-21
Rate law: StOIchIometry:
Now using:
v =(-Z) Z
where
X
[(I-X)'
=>
_of]
K C
R To
(Z) __-.!___ Z
[
X2] (1-X)2 -'K c
z=.!. = exp(E[J_ . _!]) ko
f (X) = 0 =V _
and
T.
= K co exp(MI~[_!._!..]) R T T o
Solving usmg polymath to get a table of values of X Vs T..
See Polymath program P4-9 po] POLYMA TH Results
NLE Solution variable X To T z V E R Y Kco Hrx Kc
Value
0.422~453
f(x) 3.638E··12
Ini Guess . O.S
300 305.5 2902.2 3785.4 1.SE+04 2 1.5684405 3 ·-2 SE+04 1 4169064
NLE Report (safenewt)
043
Nonlinear equations ( 1 J f(X) = (z/y)*x/«1-X)"2 - X"21Kc) .. V = 0
042+---
- ..... - -..' •.' -..
---------.-.---~
,
--------,
i
---+--.-------~
041
x
04 - - 0.39 038
4-22
- - - - - - -..- - - - -.. -----.-----
0.37
o36
-··---~--~I------_.,_-·-·__r_-~--
Explicit equations [1] To = 300 [2] T =305.5 [ 3] z =2902.2 [4] V =3785..4 [5] E = 15000 [6] R=2 Y = exp(E/R*(1rro-1rr)) [8] Kco=3 [ 9 ] Hrx = -25000 [10] Kc = Kco*exp(HrxlR*(1rro-1rr)) [7]
----
T(in K)
X
300
0.40
301
0.4075
303 304 f--
305
--
0.4182
---
-._--
--
0.4213 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.
P4-10 (a) For substrate:
4-23
Fso - Fs + rs V = 0 (Cso - Cs)vo = rg VY S1C =
VYSIC[ ,urn.xCS Cc ] KM +CS
P4-10 (b)
= YC/s [C so -
Cc
Cs ]
(C so - Cs )Vo - VYS1C [,uMAXCS Cc ] KM+Cs
=
°
(C so ---Cs )Vo - VYCIS [,uMAXCS ]( Cso - Cs ) = KM+Cs
=>
°
(30--C s )5_25XO.Sx[O.5XCs ](30-Cs ) = 0 5+Cs
Solving we get Cs = 5.0 g1dm or 30 g/dm if C s = C so no reaction has occurred so the only valid answer is C s = 0.5 g1dm3• 3
3
P4-10 (C) Cc = Y ClS(C SO - C s) = 0..8(30 - 5.0)g1dm3 = 20 g1dm3
.
P4-10 (d) 3 V new = vJ2 = 25 dm /h
3
3
Using equation trom above, we get C s = 167 g/dm and Cc = 22 . 67 g1dm
P4-10 (e) 3
V new = V J3 = 25/3 dm Using equation from above, we get C s = 1.0 g1dm3 and Cc = 216 g/dm 3
P4-10 (f) For batch reactor: C so = 30 g/dm3 Ceo = OJ g/dm 3 Cc = Cco + Y ClS(C SO - C s) V = lOdm3
See Polymath program P4-1O-f.pol. POLYMA T!lRe§ults Calculated values of the DEQ variables Variable t Cs Cso Yes KIn
Umax Ceo Ce
initial value
o
minimal value
o
maximal value 15 30 30
final value 15 0.0382152 30 0.8
30 30
0 . 0382152 30
0. 8 5 o. 5
0. 8 5
0.8 5
0. 5
0.5
0. 1 0. 1
0. 1
0.1
o. 1
0.1
24.069428
24.069428
4-24
5
0.5
rg rs negative_
0.0428571 -0.0535714 0.0535714
0.0912841 -0.1141052 0.1141052
5.4444349 -0.0535714 6.8055436
0.0428571 -6.8055436 0.0535714
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = rs
Explicit equations as entered by the user [1] Cso = 30 [2] Ycs=0.8 [3] Km=5 [4] Umax = 0.5 [5] Cco =0.1 [6 J Cc = Cco+Ycs*(Cso-Cs) [7] rg = (Umax*Cs/(Km+Cs»*Cc [8J rs=-(1IYcs)*rg [ 9 ] negative_rs = ·'rs 30.0r---==:=::;:;;::::::--·-------1
7.0 . . . . - - - - - - - - - - - - - - - - - - ,
24.0 .
5.6
-] c:: "
18.0 .
4.2
12.0
2.8
6,,0 .
1.4
0,,0
0.0 0.00
3.00
6.00 t
9.00
12,00
15,,00
tU
"~!O
'-="""'"'~==
0,00
3.00
__ ____ ___.!J ~
~
6.00 t
9,00
1200
15,00
P4-10 (g) Graphs should look the same as part (f) since reactor volume is not in the design equations for a constant volume batch reactor.
P4-11 Gaseous reactant in a tubular reactor: A
~
B
-rA =kCA k = 0.0015 min-l at 80 P 0
E
= 25 , OOO~gmol
X
=0.90
M B
=1000~ hr
lb MWA =MWB =58--lbmol
Dt =1 inch (I.D.)
L = 10ft
P = 132 psig = 146.7 psia
T = 260° F = 720° R
nt = number of tubes
4-25
1721lbmol F = FB = ' hr = 19 .1 1b mol AO X 0.9 hr
1000!! FB = hr =17.21Ibmol 58 lb hr lbmol For a plug flow reactor:
V
2
= nt 1lDt L = F °f·9 dX AO
4
° -fA
6=1-1=0
c
= PA AO
(E(
1-1 k2 =k1 exp - R ~ T2
RT
=~ RT
)J =0.0015exp (25000( . - - - -1- - -1)) - =53.6mm 540 720 1.104
k2 = 53.6min-- 1 = 3219hr-1 19.11bmol)(10.73 it3 psia )(720 R) V = FAORT In 10 = . hr lb mo~~_R_____ In 10 1 kP (3219hr- )(146.7psia) 0
(
V
= 0.72ft'
V
= !!L,!Dt L
2
4 4V 4(0.72/t n =--=-_. t
1£D2L t
1£
3 )
(112ftJ2 (10ft)
13.2
Therefore 14 pipes are necessary..
P4-12 A
--4
BI2
4-26
-1
Stoichiometry 1
(5
2
!/AO
1 2
1
FAU
FAO +Pno+ ,--
IE
11 AO()
1 2
-
1/4
_v .. v
.1 X
V p I;' R
-
FAO
o
dX ---7'11
Rate Law (elementary reaction) .M._. ____ .................
....
_~
.....
~._
.• .,.¥ .. _" ...
~._
..••.. "._._..
¥ ............... " . _ . _
_.,.~,
_ _ . ••
(for the int.egratitJll, refer Appendix A) from the Idea.l (jus assumption,
= YAoCm 0.8 (),nd E ~1/4 to Eq. (4) CAO
Substituting
Eq.(5)~
X
yields~
VPFRky~oC~O = 2(--1 / 4) (1-1/ 4) In(1- 0.8) + (-1/ 4) 2 0.8 + (1-1/ 4)20.8 = 2.9, ... ".. .. ,(6) FAO 1-0.8
--'-'-'~"""---':..:;:.,-
1\1oh\1' flow 1a to of A cut iu haH,
F.~1.0
-
.F~~to
1! ~'1A_O
c' lihHll
IF' 2,., .-'10
F~~. :~""FBU"':f~ Fc;;'; ~
-
Y~1UJ =1/6
Eq. (4),
FAO i{
2E/(1 -+ t')ln(l
4-27
1
:3
VPFR ky ,2AD C'2TO_ F'2AD
1" I/G)2 X'
2( --'-1/6)(1 - 1/6)ln(l -- X') + ( l/fi) 2 X'
Polymath ='fol1-Liw:cll' Eqnation Solver, X f
I
XI
:.::c
(}.758
P4-13 Given: The metal catalyzed isomerization
CB-rA = kl ( CA - Keq
A q B , liquid phase reaction
J.
WIth Keq = 5.8
For a plug flow reactor with YA = LO, Xl = 0.55 Casel: an identical plug flow reactor connected in series with the original reactor.
Since YA = 10,
e = 0 . For a liquid phase reaction e A = e AO (1- X) and e = e AOX B
B
-r, = kCAo((l- X)-
i::J
For the first reactor,
~ =F
AO
Xl
dX
o
-fA
dX
Xl
f-. "= F f kC ((1- X)-~.----"-J" AO
0
AO
or
K
eq
4-28
\
In[I-(1 +_I_JX 1 ] = -0.853In(.355) =0.883
1+Keq
Keq
Take advantage of the fact that two PFR' s in series is the same as one PFR with the volume of the two combined.
~CAOVF._ = X F AO
dX
Jl-(l+_l_Jx
0
o
Keq
kCAOVF = 2 kCAOl-i FAO FAO
=
1+~1+++:JX2] Keq
2
.
kCAO l-i =2(0.883)=1.766 FAO
1.766 =.
.)X
11 In[I-(I+_1 1+--. 5.8 5.8
2]
X2 = 0..74 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. 4-29
kC AO
F
l+_l_ln 1- 1+ Keq1 JXI ]
v; -
1
AO
[(
Keq By performing a material balance on the separator, F AO ,2 = FAO(l-X I )
Since pure A enters both the first and second reactor CAO ,2 = C AO , CBO,2 = 0, 9B =
C A=CAO (1 - X) X2
V2
= FAo,2
dX
0
CB =C AOX for the second reactor.
f- =
FAO (1- X)
Xl
kCAO
0
o -rA
dX
f (l-X)-X Keq
and since VI = V2
V
kC AO 2 _ _kCAOV;
FAO
FAO
or
---.!-1 In[l--(l +-}-Jx 1+eq
l]
K
~
=-
1- ~l In[l-(l +_1 JX2] 1+Keq K ~
1-(1+;~ Jx, +-(1+ ;Jx,t' X
1- 1- ( [ _ 2 -
1 J ]l-~l l+K. Xl q
1+_1_
1
= 1- (0.356)045 = 0.766 1.174
Keq Overall conversion for this scheme:
.- _ FAO - FAO ,2 (1 - X 2) _ FAO -- FAO (1 _. Xl) (1 - X 2) _ ( )( ) X - - - - F- - - -1- I-Xl I-X2 AO
FAO
X =0.895
P4-14 4-30
Given: Ortho- to meta- and para- isomerization of xylene.
M M
>P >0
kj k2
o
>P(neglect)
Pressure = 300 psig T = 750°F V = 1000 cat.
fe
Assume that the reactions are irreversible and rust order. Then:
--rM = kjCM + k 2CM = kCM k = kl +k2
£=0 Check to see what type of reactor is being used . Case 1: Vo
= 2500 gal hr
X = 0.37
Case 2: V
o
=1667 gal hr
X =0.50
Assume plug flow reactor conditions:
FMOdX =--rMdV or dX J-o
x
V=FMO V=
-rM
Xc
f
"0
d x MOVO X =vo f-'~.-'= VO• In (I-X) -rM 0 k(I-X) k
C MO , k, and V should be the same for Case 1 and Case 2. Therefore,
(kV)casel = -( Vo )casel In (1- X casej ) = -2500 ~:IIn [1-0.37] = 1155 ~~ (kV)case2 = -( vO)caSel in (1- Xcase2) = -1667 ~~ In [1-0.50] = 1155 ~~~ The reactor appears to be plug flow since (kVkase 1 = (kV)Case 2 As a check, assume the reactor is a CSTR
FMOX
= CMOVOX =-rMV 4-31
vX
kV=_oI-X
or
-rM
Again kV should be the same for both Case I and Case 2.
( kV)
( kV)
=(
V )
X
0 Casel
Casel
=
I-X Casel
Casel
=
(V ) 0 Case2
X Case2
1-· X Case 2
Case2
gal ( 0.37 ) 25001 hr = 1468 ga 1-037 hr •
gal ( 0.50 ) 16671 = hr = 1667 ga 1---050 hr •
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 .
kV =1155 gal hr 1155 gal k= hr _=1.55 gal 1000 ft3 cat. hr It 3 cat For the new plant, with vo =5500 gal / hr, XF = 046, the required catalyst volume is:
-5500 gal_ h
-v
V = _ 0 In (1- X F) = ~ In (1--0.46) = 2931ft3 cat k 1.155--.-K~ hr ft cat This assumes that the same hydrodynamic conditions are present in the new reactor as in the old .
P4-15 A-4 B in a tubular reactor
Tube dimensions: L = 40 ft, D = 0 ..75 in. n t =50 2
V
= nt 1[D 4
2
(50)n- 0.75
L=
(
4
12
)
40 = 6. 14ft3 4-32
500~
F =~= AO MWA 73
hr =6.861bmol lb hr lbmol
x dX V=FAofo -rA
-r = kCAo (I-X).=kC (I-X) A 1+8X AO V = FAO SdX.= FAO S dX. o-rA okCAo (I-·X) . P YAOP wIth CAO = - = - RT RT V = FAORT kyAOP
In (_1_)
or
k
=
FAO In(_I_) kCAO I-X
I n (_1_) VYAOP
= FAOR~
1- X
1- X
Assume An'henius equation applies to the rate constant. -E o
At T J = 600 R, kJ = 0.00152
= Ae
R1 ;
-E
At T2 = 760 oR, k2 = 0 . 0740
= Ae RT2
~ =ex+~If(:' -t)] In ~ = -:(k- n= !T~;; E = TrT2 ._ In k2 =(660)(760) In 0.740 =19,500oR R Tr - T2 kl 100 0.00152 A = kl exp
'0
[~-] RTr
k=k\ex
p[ -
!u -n]
From above we have
k
= .FAoRT Vy. AO P
so
In (_1_) l--X
FAoR~P In (_1)= kl exp [- E (~-~J] VYAO
I-X
R T
Tr
Dividing both sides by T gives: 4-33
(
6.S61bmal)(10.73 psiaft3 ) hr lb maze R
-In 5 =
sec)( 6.14ft3 )(114.7 psia) ( .00152-)(3600 sec hr Evaluating and simplifying gives:
exp [-19500 (! - _1_)] T 660 R T 0
0.030So R-1 =
Solving for T gives:
P4-16 Reversible isomerization reaction m-Xylene -> p-Xylene Xe is the equilibrium conversion., Rate law:
-rm
= k(Cm _ Ck p ) e
At equilibrium,
C
p C=m ke
-rm = 0 =>
K=~e
l--X e
1+ _1_ = (1 + 1- Xe ) = Xe + 1- Xe Ke
-rm
Xe
Xe
= _I_ Xe
= kCAD (1-~-) X e
P4-16 (a) 4-34
ex [-19500(!- 1 )] P T 660 R 0
T
For batch reactor,
dX
Mole balance:
= -rmV = kCAO
dt T=X e1n ( k
N mO
Xe Xe --X
CAO
(1- XeXJ
J
ForPFR,
V =FAO = va
k
fdX
f
dX
1-(1+ :JX 1
TpFR
-rm
=-;;
f
dX
( 1J 1- 1+- X Ke
= Xe
T
k
PFR
In--~ X-X e
P4-16 (b) ForCSTR,
V= Fma X -rm X
T CSlR
=
~k[;::-1--(-:-1' +-K-1e-J-X-=]
Putting the value of Ke,
TCSIR
X( XeXe-X J =k
P4-16 (C) Xe
In(.
Xe . )
In(-~-) X --X
vOlumeefficieney~qV=T::~ ~(; )=(~)=('x }{X,~£) T,
k
k
X -X
X e --X
4-35
X e -X
X -X
X
1] v
=
1-(~)
1 X
Xe In
X
1--· Xe
Xe
Following is the plot of volume efficiency as a function of the ratio (x/Xe),
See Polymath program P4-16-c.pol. 1.0 , ; : : : - - - - - - - - - - - - - - ,
Efficiency
0.8 0.6 04 0.2 0.0
0.010
- - - - - "- OA06v.\.1.604 0.802 1.000
0. 208
P4-16 (d) Efficiency = VPFR I VCSTR = 1 from problem statement, which is not possible because conversion will not be the same for the CSTR' s in series as for the PFR
-------_.
P4-17 (a) A--7Y2B E
= -1/2, X = 0.3, W = 1 kg, Yexit = 0..25
ForPBR,
C =Co (l-X)y A (1+ eX)
and
dX
rA
dW
FAO
dX
(1-x)2l
dW
(1+eX)2
--=z
kC o (l-
XY y2
let
vJl+eXY and
kC Ao
Z =---Vo
dy =-!:'.(l+eX) dW 2y
Solving for z by trial and error in Polymath to match x and y at exit, Yo = 1 and Yf = 5/20 = 0..25 X = 0.3 we get: a = 1.043 kg- l and z = 0.7 kg'l
See Polymath program P4-I7-al.pol. POLYMA TH Results 4-36
Calculated values of the DEQ variables variable
initial value 0 0
W
x
1
Y
esp alfa
-0.5 1.043 0.7
Z
minimal value 0 0 0 . 2521521 -0 . 5 1.043 0.7
maximal value 1
0.302004 1 -0.5 1.043 0.7
Differential equations as entered by the user [1] d(x)/d(W) = Z*«1-x)*y/(1+esp*x»)A2 [2) d(y)/d(W) =·alfa*(1+esp*x)/(2*y) Explicit equations as entered by the user [1] esp = -0.5 [2] alfa = 1.043 [3] Z=.7
Now for CSTR:
_ FoX _ X(l+cXY W -----_._--T
z(l-XY
A
Solving we get for W = lkg and z = 0..7 kg·' X=OAO
See Polymath program P4·17-a2.pol. POLYMA TH . . Results
NLE Solution Variable
x W esp Z
Value 0.396566 1 -0 . 5 0.7
fix)
-1.142E-13
Ini Guess 0.5
NLE Report (fastnewt) Nonlinear equations [1 J f(x) = W*Z*«1-x)/(1 +esp*x)}A2-x = 0 Explicit equations [1] W = 1 [ 2 J esp = -0 . 5 [3J Z = 0.7
P4-17 (b) For turbulent flow:
G2
a=(constant)Dp
4-37
final value 1 0.302004 0 . 2521521 -0.5 1.043 0. 7
= _a = 0.0326 j
a
=>:. Z2
and
32
2
= 4xO.7 = 2.8
Now solving using polymath:
See Polymath program P4-17-b.poI. POLYMATH Results Calculated values of the DEQ variables Variable w x
minimal value 0 0 0,9887079 -0.5 0 . 0326 2. 8
initial value 0 0 1
Y
esp alfa
-0 . 5 0.0326 2 "8
z
maximal value 1 0.8619056 1 -0.5 0.0326 2.8
ODE Report (STIFF) Differential equations as entered by the user [1 J d(x)/d(w) = Z*((1-x)*y/(1 +esp*x))"2 [2 J d(y)/d(w) = -alfa*(1 +esp*x)/(2*y) Explicit equations as entered by the user [ 1] esp = -0,,5 [2] alfa = 0,,0326 [3]
Z = 2.8
So, convelsion in PBR, X = 0,,862
P4-17 (C) Individualized solution P4-17 (d) Individualized solution P4-18 Given a Fluidized Bed ('SIR
w=
F[~&X
'fA
0= 0'
E =0,
thcll PA
=
p J'\o(t- X) p n
No pressure drop ill the CSTR
PA == l\o(I . . . X)
'vV =f"AO)l: kPAO(IX) ,
4-38
final value 1 0.8619056 0 . 9887079 -0.5 0 . 0326 2.8
k & F A(} unknmvl1 .. group into a constant, use values from 1st case,
~o C~x,) PA:\V ~j~20(150) =
~o ~t;1~~~)~~t ==
=
Put PFR downstream" less \Nasted volume
a)
'IS
P4-18 (b) b) PBR;
.,~~ dW
=
tAo:i~" == --r,~ = kPJ\o(l·- X)~)::;: kPAO(1-· X)(l ,,--nW)1l2(since e =0) P~Qk (I _ X)(l- aW)li2 FAO
PAOk rW ( . )112 l·-uW dW F;\0 ()
X z dX JxII····· X
·----"=--'JI
When X :;:: XI , \V.:::: 0
In } .- ~L .1 ",., X 2
=_~_I~.Ao. (~_?_ )[ I .m (I ' ', u W) 3/ 2 ]
1"",0. 5 1
F\o
3a
In [ l"::x-~. ""'atmkg 20dtm lO3
'."
2
3("0.oi8)""
[ '.' 1~ 1-,,(I-.O.018kg )Okg)
k(yb X2
=0.756 4-39
3{2]
P4-18 (c) C)
P == PO(!··' (lVV')
I·) ito
(
;=
1
.')t(Z
20 aIm J" 0018kg- (50kg) , = 63 alm
P4-18 (d) For turbulent flow
(~J2 = 0.0142kg-l l Dpz J2 (Ac.lAc2 J2 =(0.018kg- l )(,,~)2 1 1.5
a 2 = a (D Pl
P exit = P o(1- UW)05 = (20atm)(1-0.0142kg·!(50kg»1!2 = 10 ..7 atm 3
In[1-0.5]= 10-. 1-X2
atmkg
)][1-(1-0.0142kg - l (50kg l 3 0.0142kg-
20atm( (
2
)t2]
X 2 = 0.,77
P4-19 Production of phosgene in a microreactor . CO + Cl 2 -> COCl2 (Gas phase reaction) A+B ->C
See Polymath program P4·19.pol. .POLX,MA I!lResults Calculated values of the DEQ variables Variable W X
Y
e FAO FBO Fa Fb vO v Fe Ca Cb a k
rA Ce
initial value 0 0 1 -0.5 2"OE-05 2,OE-05 2.0E-05 2"OE-05 2,,83E-07 2,,83E-07 0 70,,671378 70,,671378 3.55E+05 0,,004 -19.977775 0
minimal value 0 0 0.3649802 -0,,5 2"OE-05 2"OE-05 4.32E-06 4,,32E-06 2,83E-07 2" 444E-·07 0 9,1638708 9.1638708 3,,55E+05 0,,004 -19,,977775 0
maximal value 3,,5E-06 0,,7839904 1 ··0" 5 2.0E-05 2.0E-05 2,OE-05 2"OE-05 2,,83E-07 4,,714E-07 1,,568E-05 70.671378 70,,671378 3.55E+05 0.004 -0.3359061 53,532416
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -rA/FAQ r2 J d(y)/d(W) = -a*(1 +e*X)/(2*y) Explicit equations as entered by the user
4-40
final value 3.5E-06 0.7839904 0,,3649802 -0.,5 2"OE-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
[1]
e
=-.5 =
FAO = 2e-5 Fa = FAO*(1-X) [6] vO = 2 . 83e-7 [8] Fe FAO*X [ 10] Cb Fblv [2] [4]
[3] FBO FAO [5] Fb = FBO-FAO*X [7] V vO*(1 +e*X)/y [9] Ca Fa/v [11] a = 3.55e5 [13] rA = -k*Ca*Cb
= =
= =
[12] [14]
k=.004 Ce
= Fe/v
P4-19 (a) 0 . 0 ..- - - - - - - - . - - - - - - - - - ,
1.0
0.0
0.8
0.0
0.6
[j Fa. - Fb - Fe
0.0
0.4
0.2
0.0
0.0
...--------------_=____,
IL--_ _ _ _ _
OOe+O
7.0('·· '7
~
__
~
_ _ _ ____l
lAe-6 W 2.1e-6
2.8e-6
0.0
.3.5l'-6
L -_ _ _ _ __
O.Oe+O
7.0l'-7
1.4e-6\V2.1e-6
2.8e-6
P4-19 (b) The outlet conversion of the reactor is 0 . 784 The yield is then MW*FA *X = 99 g/mol * 2 e-5 molls * 0 . 784 = .001.55 g/s = 48 . 9.5 g/ year. Therefore 10,000 kg/year / 48 . 95 kg/ year = 204 reactors are needed.
P4-19 (C) Assuming laminar flow, a - D p ·2, therefore
a z = a1 .D~1 = (3 ..55x105kg -1) 4 = 14.2x105kg- 1 DP2
4-41
3.5e-6
-5.---------------,
2.0 ...
16...-5
10 . - - - - - - - - - - - - - - - - ,
0.8
lj
1.2\'-5
80\'-6
"F:l •. lih
0.6 .
- Fe
OA
~." W
':'()e-':'
OOHO
l..4e-6 W2.1e-6
2.8e-6
0. 01'+0
_
_
'
_
~
_
-
0.0
3.5('-(;
L
O.OHOIL-----~-----·----l
I
02 _ _ _ _ _ _ _ _ _ __ _ l
':'.Oe-':'
1.41'-6 ",,2 ..1e-6
2.8e-6
3 . 5e-6
P4-19 (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"
P4-19 (e) Individualized solution P4-19 (f) Individualized solution P4-19 (g) Individualized solution
P4-20 (a) Mole Balance
_4~ =z lA' IF ... dW . . .o Rate Law
k k[!:r(
k' :.:;;; 11 ;;;;
t!>
com t!> ""'1)J
':' c Dp
4-42
k' '''''-'''''~'--'''''-''''-~'---- i1 - 1 k' = Tl k, k = -11 1.0·H;:::::::::::;1..-.._
Largen..
cD p
~
Increasing Particle Size
Then
3 . "" 11 == 3 :;;:: -_ ~-,
cDp
when Dp::: 2 mm, k' ::: 0.06, 11:::;:
{,= :~6,:::;; 0.02
3
0.02=: c(2)
,:=75
[~_~J~ For turbulent flow:
2{3 " PoAop(l- fjJ)
a =-
constant =>a=-----Dp
a
aoDpo Dpl
=--"-----.:..:..
P4-20 (b) See Polymath program P420"b.pol.
4-43
k'
= -1 = k" = 3
3.0
2.4
Q
1.8
1.2 0.6
0.0
0.0
04
0.8
16
Dp 12
2.0
P4-20 (C) Gas, £=0, C A =C"Q(l-X)y
~~. iy' whele a~a{~;) Ct,
~ pJi~)p,.~ =2oru;;;6---Jl%:~"s'i:~!ki:S2d;;;'r 708 x 10' kg' and
?Y
;:;;;
dW
a 2)'
See Polymath program P4 . 20·e.pol. POLYMA TH Results Calculated values of the DEQ variables Variable w X
Y
Dp Q Fao
initial value 0 0 1 0 . 0075 0.5625 5
minimal value 0 0 O. 2366432 0 . 0075 0 . 5625 5
maximal value 100 0 . 5707526
4-44
1
0 . 0075 0 . 5625 5
final value 100 0 . 5707526 0 . 2366432 0 . 0075 0 . 5625 5
alpha Cao kprime
0 . 00944 0.207 2.9385672 -0.1259147
ra
0 . 00944 0.207 2 . 9385672 -0.1259147
0 . 00944 0.207 2 . 9385672 -0.0012992
0.00944 0 . 207 2.9385672 -0 . 0012992
ODE Report (RKF45) 0.90
Differential equations as entered by the user [1] d(X)/d(w) = -ra/Fao [2] d(y)/d(w) = -alpha/21y
0.72
Explicit equations as entered by the user [lJ Dp = . 0075 [2J Q 75*Dp [3J Fao 5 [ 4 J alpha = .0000708/Dp [5J Cao .207 [ 6 J kprime = 3*(3/QA2)*(Q*coth(Q)-1) [7 J ra = -kprime*(Cao*(1·X)*y)A2
=
0.54
= =
036
0.J8
0.00 0.0
0.1
0.2
Dp 03
0.4
0.5
P4-20 (d) Individualized solution P4-20 (e) Individualized solution P4-20 (1) Individualized solution P4-20 (g) Individualized solution P4-20 (h) Individualized solution P4-21 (a) Assume constant volume batch reactor Mole balance:
dX CAD -
dt
= -rA
Rate law and stoichiometry: - rA
= kCA = kCAD (1- X)
Specific reaction rate: k ( 25° C) = 0.0022 weeks- l Combine:
dX --1 f----·= --In (1- X) kC (1- X) k
x
t=
CAD
D
AD
--}
52.2 weeks = 1 In (1- X) 0.0022 weeksX =0.108
CA = CAD (1-·· X) but since volume and molecular weight are constant the equation can be written as: m A = mAD (1- X) 4-45
6500IU mAO
= mAO (1-0.108)
= 7287IU
%OU = CAO -CA *100 = 7287 -6500 *100 = 12.1 % CA 6500
P4-21 (b)
10,000,000 lbs/yr = 458 * 109 glyr of cereal Serving size = 30g Number of servings per year = 458 * 109 /30 =1.51 * 109 servings/)'1 Each serving uses an excess of787 IU = 4 . 62 * 10-4 = 1.02 * 1O-6 1b Total excess per year = (L51 * 108 servings/yr) * (102 * 1O-6 Ibs/serving) = 154111b/yr Total overuse cost = $100/lb * 154.1 Ilb/yr = $15411 /)'1. (trivial cost)
P4-21 (C) If the nutrients are too expensive, it could be more economical to store the cereal at lower temperatures where nutrients degrade more slowly, therefore lowering the amount of overuse . The cost of this storage could prove to be the more expensive alternative. A cost analysis needs to be done to determine which situation would be optimaL
P4-21 (d) k ( 40° C) = 0.0048 weeks -)
6 months = 26 weeks
-1 fo kCAOdX(1- X )= -In (1- X ) k
x
t = CAO
-1 26weeks =-)In(1- X) 0.0048 weeksX =0.12 CA = CAO (1 - X) but since volume and molecular weight are constant the equation can be written as: m A =mAo(1-X) 6500IU =mAo(1-0.12) mAO
= 7386IU
%OU = CAO -CA *100= 7386-6500_*100=l3.6% CA 6500
P4-22 No solution necessary P4-23 CH20HCH2CI + NaHC0 3
-7
(CH2 0Hh + NaCI + CO 2
4-46
A FAa
+
c
B
= 0.1 mol/min = 6 mol/hI'
+ D
+ E
P4-23 (a) Mole balance:
dC A dt
= r + Vo (C V
!:!CB = r
dt
- C ) AO
+ V0
V r A = -kCACB
Rate law:
A
(_
C )
dCc dt
B
= -r + Vo (-- C V
V =Vo +Vot
See Polymath program P4-23-a.pol. POLYMA ~H Results Calculated values of the DEQ variables Variable t Ca Cb Cc Faa Caa k
Va va V r Xb Nc
initial value 0 0 0.75 0 6 1.5 5.1 1500 4 1500 0 0 0
minimal value 0 0 1. 395E-13 0 6 1.5 5.1 1500 4 1500 -0.0039398 0 0
maximal value 250 0.15 0.75 0.829586 6 1.5 5.1 1500 4 2500 0 1 2073.9649
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = r+(voN)*(Cao-Ca) [2] d(Cb)/d(t) = r+(voN)*(-Cb) [3 J d(Cc)/d(t) = -r+(voN)*Cc Explicit equations as entered by the user [1] Fao =6 [2J Cao = 1..5 [3 J k = 5.1 [4] Vo = 1500 [5] vo = Fao/Cao [6] V=Vo+vo*t [7] r = -k*Ca*Cb [8] Xb = 1-Cb*V/(O.75*1500) [9] Nc = Cc*V
4-47
final value 250 0.15 1.395E-13 0.829586 6 1.5 5.1 1500 4 2500 -·1.067E-13 1 2073.9649
) C
LO
O,Oe+O"-j-----------~>
....-----------~
D
0. 8
-8.0e-4
0.6
-1.6e-3
04
-2 Ae-·31
/I
II
II
l I !
".
/
/
/
"P,""";'#'/
-3.2e-3 \
1
/
~/~
1" ¢~,#,,r 1 _-I.Oe_3l.!s:::..,"-~--~--~--- __- - : ' ,
0.0
0
50
100 t
150
200
,/.i'
o
250
50
100 t
150
200
250
3000.---------------,
0.90....------· - Ca
2400
0.72 ' 054'
036 600
1118 o,ool.::------------~~;...
o
50
100 t
150
200
250
o I)
50
100 t
150
200
P4-23 (b) Taking average time for activities like charging, heating, cleaning = 4..5 hr So, if there is one batch per day, the time for reaction = 24-4.5 hr = 19.5 hr Since at the temperature at which we are operating the reactor no side reactions occur, the quickest way to run the reaction will be at the highest flow rate of A (2 mol/min or 120 mol/hour) But at 19.5 hours and a flow rate of 120 mol/hr, 1560 dm3 would be added to the reactor.. Because the volume of the reactor is 2500 dm3 , and there is already 1500 dm3 of B in the reactor, the reaction cannot not run for 19.5 hours at a flow rate of 120 mol/hr (2 mol/min) Also, because there is 1500 dm3 ofB at a concentration of 0.75 M, there are only 1125 moles ofB to react. This means about 750 dm3 of 15 M A is all that can be reacted. More A may be added to the reactor to keep the reaction rate high as the concentration of B drops, but adding twice the necessary volume would be a waste of time and material (on top of being physically impossible !).. To add 1125 moles of A at a rate of 120 mol/min takes 9375 hours. Using the Polymath code from part (a) and changing Fao to 120 and the time to 9375 gives 1107 moles of C. Allowing the reaction to go for 9.5 hours results in 1115 moles of C and a reaction time of 10 hours gives 1124 moles of C. Now consider multiple batches per day.. If two batches are run, then there will be 9 hours of downtime, meaning that the time for the reaction will be 15 hours - split between the two batches, with a maximum
4-48
batch time of 10 hours., The following table shows the tluee possible times for reactions and the moles of C that are formed from the two batches. Maximum Batch Time (hr) 10 9
Total Moles of C Formed 1723 1790 1795 ..-
8
So the best setup will be to run 2 batches per day. One batch will run for 8 hours and the second batch for 7 hours. Both will be run with a flow rate of 2 mol/min of A. (If tlu'ee batches are run there will be 135 hows of downtime and only 105 hours for the reaction. A maximum of 1257 moles of C can be formed if the time is split evenly for each batch (3.5 hows).)
See the Polymath code from part (a) and vary time and flow rate.
P4-23 (C) F AO = 0.15 mol/min = 9 mol flu vo = FaofCa = 9 molJlu f 1.5 mol/dm3 = 6 dm3flu 3
1000 dm3 is needed to fill the reactor. At 6 dm flu it will take 166.67 hows
Now solving using the code from part (a) with the changed equations:
See Polymath program P4-23-e.pol. 1.0 , - - - - - - - . - - - - ,
0.8 0.6 0.4
66
t
99
132
165
P4-23 (d) Individualized solution P4-24 NaOH + CH3 COOC2 H s ------t CH 3 COO- Na+ + C2 H sOH A+B~C+D
4·49
Mole balance:
dec =-r+ va dt V
(-c) C
Rate law:
V=Va+Va t k =koex p (
!Uo -nJ
To produce 200 moles of D, 200 moles of A and 200 moles of B are needed. Because the concentration of A must be kept low, it makes sense to add A slowly to a large amount of R Therefore, we will start with pure B in the reactor. To get 200 moles of B, we need to fill the reactor with at least 800 dm3 of pure R Assume it will take 6 hours to fill, heat, etc . the reactor. That leaves 18 hours to cany out the reaction. We will need to add 1000 dm3 of A to get 200 moles in the reactor. We need to check to make sure the reactor can handle this volume if only 1 batch per day is to be used . Since we add 1800 dm3 or 18 m3 and the reactor has a volume of 4.42 m3 we can safely carry out a single batch per day and achieve the necessary output of ethanol. Now vary the initial amount of B in the reactor, the flow rate of A, and the temperature to find a solution that satisfies all the constraints . The program below shows one possible solution .
See Polymath program P4--24.pol. POLY1VIA TH Results Calculated values of the DEQ variables Variable t Ca
Cb Cc Cd ko
Fao Cao Vo vo
initial value
o o
0 . 25
minimal value
o o
0 . 0068364
o o
o o
5.2E-05 0.04
5 . 2E-05 0 . 04 0.2 1200
0. 2
1200 0.2 308 1.224E-04
ra
o
308 1..224E-04
V Nc
1200
1200
o
o
T k
maximal value 6 . 5E+04 0.1688083 0 . 25 0 . 0151725 0 . 0151725 5.2E-05 0 . 04 0.2
1200 0. 2 308 1 . 224E-04 1.397E--06 1 . 42E+04 202 . 92284
0.2
o
ODE Report (RKF45) Differential equations as entered by the user [11 d(Ca)/d(t) = -ra+(vo/V)*(Cao-Ca) [2] d(Cb)/d(t) = -ra-(vo/V)*Cb [3] d(Cc)/d(t) = ra-(vo/V)*Cc [ 4] d(Cd)/d(t) = ra-(voN)*Cd Explicit equations as entered by the user
4-50
final value 6.5E+04 0 . 1688083 0 . 0068364 0.0142903 0 . 0142903 5.2E-05 0 . 04 0. 2 1200 0.2 308 1 . 224E-04 1.. 412E-07 1 . 42E+04 202 . 92284
[ 1] ko == 5 . 2e-5 [2] Fao == .04 [3] Cao ==.2 [4] Vo == 1200 [5] vo == Fao/Cao [ 6] T == 35+273 [7] k == ko*exp«42810/8.3144)*(1/293-1fT)) [8] ra == k*Ca*Cb [9] V == Vo+vo*t [10] Nc == Cc*V
P4-25 (a) A ~ B + 2C 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
dFA =r -·R dV A A dFB =r -R dV B B dFc --=r. dV c Next we express the rate law: First-order reversible reaction
Transport out the sides of the reactor:
kACroF
A RA =kAC A = -.!.!......!..!!.-!.!.. FI
kBCroF Fr
B RB = kB CB = --"'--'-""--':::.. Stoichiometery:
Combine and solve in Polymath code:
See Polymath program P4-25-·a.pol. POLYMA TH Results Calculated values of the DEQ variables 4-51
Variable v Fa Fb Fe Ke Ft Co K Kb ra Ka Ra Rb Fao
initial value 0 100 0 0 0.01 100 1 10 40 -10 1 1 0 100 0
X
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
maximal value 20 100 9.0599877 61.916043 0.01 122.2435 1 10 40 -0.542836 1 1 2.9904791 100 0.4278998
minimal value 0 57 . 210025 0 0 0.01 100 1 10 40 -10 1 0 . 472568 0 100 0
ODE Rel!ort {RKF4S1 Differential equations as entered by the user [1 ] d(Fa)/d(v) ra .. Ra [2] d(Fb)/d(v) ··ra .. Rb [3 ] d(Fe)/d(v) -2*ra
= = =
Explicit equations as entered by the user [1] Ke = 0..Q1 [2 J Ft = Fa+ Fb+ Fe [3 ] Co = 1 [4] [5] [6] [7 ] [8 ] [9]
3
K= 10 Kb = 40 ra = - (K*Co/Ft)*(Fa- CoI\2*Fb*FeI\2/(Ke*FtI\2)) Ka = 1 Ra = Ka*Co*Fa/Ft Rb = Kb*Co*Fb/Ft
r-...,....,---------.-----
100
2
80
2
60
1
40
~---
1
0
0
..j
8
~.~ 16 20
v
20 ..., ..........--.-~-.......... (I
12
o
4
8
y
12
16
P4-25 (b) The setup is the same as in part (a) except there is no transport out the sides of the reactor .
See Polymath program P4-25-b.pol. POL YMAIH Results 4-52
20
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
final value 20 84.652698 15.347302 30.694604 0.01 130 . 6946 1 10 -3.598E--09 100 0.153473
maximal value 20 100 15.347302 30 . 694604 0.01 130.6946 1 10 -3.598E-09 100 0.153473
minimal value 0 84.652698 0 0 0.01 100 1 10 -10 100 0
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- CoA2*Fb*Fc"'2/(Kc*FtA2»
-------------------
0.50 0 . 45 0.40 0.35 0.30
[ --] - X PFR
0.25
... X l\fembmn.
0.20
-.---..----.-
............
0 . 15
~------
............-.--.-...
--.. .....--_........_---_... ""
0.10 0 . 05 0 . 00
t -_ _ _ _ _ _ _ _ .. ________
0
2
4
6
8
~
10
___
V 12
14
16
P4-25 (C) Conversion would be greater ifC were diffusing out P4-25 (d) Individualized solution 4-53
18
20
P4-26
co
+ H20 ...... CO2 + H2 A+B ...... C +D Assuming catalyst distributed uniformly over the whole volume
dF
dF
Mole balance:
A =r ._-
Rate law:
r = rA = rB=C -.y'
dW
RH2
=
KH2
dF
B =r --
dF dW
e = -r --
dW
- -D= - r - R
dW
= -1D = -k
[c
H2
C _ CeC AB K D] eq
CD
Stoichiometry:
FD CD=C ro FT FT
= FA + FB
+ Fe + FD
Solving in polymath:
See Polymath program P4-26.pol. POLYMA TH Results Calculated values of the DEQ variables variable ---W
Fa Fb Fe Fd Keq Ft Cto Ca Cb Kh Ce Cd Rh k r
initial value 0 2 2 0 0 1,,44 4 0,,4 0,,2 0.2 0,,1 0
°°
1.. 37 -0,,0548
minimal value
maximal 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
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
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d{Fa)/d{W) = r [2 J d{Fb)/d{W) = r [ 3 J d{Fc)/d{W) = -r [ 4 J 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
4-54
final value 100 0.7750721 0,7750721 1.2249279 0,5536716 1. 44 3,,3287437 0,,4 0" 0931369 0,,0931369
°"
1 0,,147194 0.0665322 0.0066532 1..37 -0,,002567
[4] Ca = Cto*FalFt [ 5] Cb =Cto*Fb/Ft [6] Kh=O,,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
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
= CeCD = C~OX2
K eq
CACB
=_
C~o(1 __ X)2
X_2_ = 1.44
(l_X)2
solve this for X, X= 5454 This is the maximum conversion that can be achieved in a normal PFR.
If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of 459 2 . 0 r;;----;::====;_
1.6
_..,,,.,---_.....
1.2
--
...
0.8
04
20
·40
W 60
80
100
P4-27 Individualized solution P4-28 (a) Assume isothermal and £=0
therefore,
P=Po(l-aWl5
1=10 (1-.01 glW).5
W=99g
P4-28 (b) 4-55
dX -rA dW= FAO .!..=(l-Wa)-' Po
-rA =kCA CA ==CAo(l-X) PlPo
CA =CAo(1 -X) (1 - 0.01 W).5
Integrate from X=O to X=.9
ax = kCAO (1-X) (l-.01W).5
dW
FAo
W=59.88g
Fust 5% conversion integrate from X=O to X=.OS W=1.31 g
Last 5% conversion integrate from X==.85 to X=.90 W:alO.85 g
P4-29 Individualized solution P4-30 (a) First order gas phase reaction, C6HjCH(CH3h -7 C6~ + C3~ YAO = 1, g = 1 ~ [; = 1
P
kC
Y = - , a = -AO -
Po
FAO
For aPBR:
dX dW
a(1-X)
= (1 +
iT
dy
(I+X)
dW
2y
-=
y ........................ (1)
a" ....................... (2)
X = 0.064 and y = 0 . 123, Solving (1) and (2) by trial and enOl on polymath we get, a = 0.000948 (kg of catalystr1 a = 0.000101 dm- 3
4-56
Now solve for a fluidized bed with 8000 kg of catalyst. FAOX
Mole balance: W = - -rA Rate law: -fA = kCA
I-X I+cX
Stoichiometry: CA = CAO - - Combine: W
= X (I + eX) = 8000 = ___. X (I + X) a(1-X) O.OOOlOI(I-X)
X =0.37
P4-30 (b) ForaPBR:
dX _ a(1-X) dW - (I+XfY dy _ (I+X) ---- - ' - " - - a
dW
2y
where
a = kCAO FAO
From chapter 12 we see that k will increase as Dp decreases_ We also know that for turbulent flow
I a-· - .. This means that there are competing fOIces on conversion when Dp is changed . Dp We also know that alpha is dependant on the cross-sectional area of the pipe:
I
a - ._--
Ac
But alpha is also a function of superficial mass velocity (G). If the entering mass flow rate is held constant, then increasing pipe diameter (or cross-sectional area) will result in lower superficial mass velocity . The relationship is the following for turbulent flow:
I 2 G - - and a-G
Ac
I
therefOIe, a _. -'2 .
~
If we combine both effects on alpha we get the following:
I
I
a--Ac~ I
a-~ So increasing pipe diameter will lower alpha and increase conversion and lower pressure drop .
4·57
For Laminar flow:
1
a-·D2 p
so decreasing the particle diameter has a larger effect on alpha and will increase pressure drop resulting in a lower conversion
1
For Laminar flow a·- G and so a .-. -2 .
A;
This means increasing pipe diameter will have the same trends for pressure drop and conversion but will result in smaller changes.
P4-30 (c) Individualized Solution P4-31 (a) K = 0. . 0.5
= 0..33(1-3) =-0..666 P AO = 0..333*10 F Ao = 13.33 Mole balance: dXldW = -ralFao rA = -KPB Rate law: P A= P AO(l - X)/(l - o..666X) Pc = P AoXl(l - o. . 666X) For a = 0., Y= l(no pressure dlOp) I>
10..---------------#_-
0. 00
8
-0.08
6
-0.16
.-----------.--$-.__---:=
1
4
-0.24
2
-032/
o I)
-040
20
·to
W 60
80
/
L -_ _
o
100
~
_____
20
40
~_.
\V 60
See Polymath program P4-31-a . pol. POL):''l~'IA!!1
Results Calculated values of the DEQ variables Variable
initial value -------
X
o
K
0.05 3 . 33 3 . 33 6 . 66
0 . 05 3 . 33 0.0406732 0 . 0813464
Pao Pa
Pb Pc Fao ra
o
minimal value
o o
W
o
13 . 33 -0 ...333
o
13 . 33 -0.333
maximal value 100 0.995887 0 . 05 3.33 3 . 33 6 . 66 9.8482838 13 . 33 -0 . 0040673
4·58
final value 100 0.995887 0 . 05 3 . 33 0 . 0406732 0.0813464 9.8482838 13.33 -0 . 0040673
_ _ _....J
80
100
esp
-0.666
-0 . 666
-0 . 666
-0 . 666
ODE Report (RKF45) Differential equations as entered by the user [1 j d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [1] K = 0.05 [2] Pao = 0 . 333*10 [3] Pa = Pao*(1 - X)/(1 - 0 . 666*X) [4] Pb = 2*Pa [5] Pc = Pao*XI(1 - 0 . 666*X) [6] Fao = 13 . 33 [7] ra = -K*Pb [8] esp = -0.666
For first 5% conversion weight required = WI = 2 kg For last 5% conversion weight required = W 2 = 38 kg Ratio = Wi WI = 19 Polymath solution (4-34 a)
P4-31 (b) For a = 0 . 027 kg· l , Polymath code with pressure drop equation:
See Polymath program P4·31-b.pol. POLYMA TH Results Calculated values of the DEQ variables Variable w
X Y K
Pao Pa Pb Pc Pao ra esp alfa
initial value 0 0 1 0.05 3 . 33 3.33 6 . 66 0 13.33 -0.333 -0.666 0 . 027
minimal value 0 0 0.1896048 0.05 3 . 33 0.4867002 0.9734003 0 13 . 33 -0 . 333 -0.666 0.027
maximal value 30 0.4711039 1 0.05 3.33 3.33 6 . 66 1.0583913 13.33 -0.04867 -0.666 0.027
final value
~----
0.4711039 0.1896048 0.05 3.33 0.4867002 0.9734003 0.4335164 13 . 33 -0 . 04867 -0 . 666 0.027
ODE Report (STIFF) Differential equations as entered by the user [1] d(X)/d(w) = ·ra/Fao [2] d(y)/d(w) = -alfa*(1 - esp*X)/(2*y) Explicit equations as entered by the user [1] K=0 . 05 [2] Pao = 0.333*10 [3] Pa = Pao*(1 - X)*y/(1 - 0 . 666*X) [4] Pb = 2*Pa [5 J Pc = Pao*X*y/(1 - 0.666*X) [6] Fao=13.33
7.0
5.6
42 2.8 14
4-59 0.0 OC""""-~---'---~--~---'30 6
12
W 18
24
ra = -K*Pb esp -0.666 alfa = 0,,027
[7
=
[8 [9
0.00,---------------,
1.0 . . . . . , . . . - - - - - - . - - - - - - - - ,
0.8 0.6
0.4 0.2 -040~-~--~--~-----~
o
6
12
W 18
2-t
0.0
L -_ _ _ _ _
o
30
6
~
_ _ _ _ _ _ _- - '
12
W 18
2·4
30
P4-31 (C) 1) For laminar flow: Diameter of pipe = D and diameter of particle = Dp Now DIlDo = 3/2 so G I = 4/9Go a = (constant)(GID/)(lIAc) So a = ao(G/Go)(Dpof Dpl)2(DofDI)2 = ao(4/9)(2/3)2(2/3)2 = 0 . 00237 kg· 1 Less pressure drop and more conversion for same weight of catalyst as in part (b). 2) For turbulent flow: ~
G2IDp a = (constant)(G2IDp)(lIAc) So, a = ao(G I/Go)2(Dpof D pl )(DofDl)2 = ao(4/9)2(2/3)(2/3)2 = 0.0016 kg·! Again less pressure drop and more conversion for the same catalyst weight eX)
It is better to have a larger diameter pipe and a shorter reactor, assuming the flow remains the same as through the smaller pipe.
P4-32 (a) At equilibrium, r = 0 =>
cAeB
4-60
1.O,-:::::==========; 0.8
0.6
04
0.2
OO~~~------~----------~
0081:0
=>
CAO (l-X(CBO ~t-C V AO o
AO => t= VOC [X2
VOC BO ,Kc(l'-X)
+
1 0..1:5 t 1 56E5
5.181:4
2.07E5
2591:5
xJ= (CAOXY K
C
x]
Now solving In polymath.
See Polymath program P4-32-,a.pol.
P4-32 (b) See Polymath program P4-32-b.pol. POLYMATH Results Calculated values of the DEQ variables Var~able
t
Ca Cb Cc Cd Kc k
ra va Va V X
initial value 0 7.72 10.93 0 0 1.08 9.0E-Os -0.0075942 0.05 200 200 0
minimal value 0 0 2074331 7.6422086 0 0 1. 08 9 OE-05 --0.0075942 0.05 200 200 0
---
maximal value 1.SE+04 772 10.93 3.2877914 3.2877914 1 08 9.0E-OS ··1.006E-Os 0.05 200 950 0.9731304
final value '-L5E+O-40.2074331 9.51217 1.41783 1. 41783 1 08 9.0E-Os -·1.006E-OS 0.05 200 950
0.9731304
1.0e+o ,
8.0e--l
ODE Report (RKF4S) Differential equations as entered by the user
6.0e-l
..,.' . /',;',."
..
.,.~ ~,,,"~ ~~
.........
-.--
~-"
..
~ ... ~- . . .
¥ ...
~~
[oJ --_ ..
I
f
".
!
-,
..
I
I
40e--l 2.0e--l
4-61
O.Oe+o 0
2976
5952t
8928
11904 14880
[1 [2
[3 [4
d(Ca)/d(t) = ra - Ca*voN d(Cb )/d(t) = ra - voN*(Cb- 10 . 93) d(Cc)/d(t) = ora - vo*CcN d(Cd)/d(t) = ora - vo*CdN
Explicit equations as entered by the user [1 J 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 2(le+l . . . . - - - - - - - - - - - - . , -1.6e-3
.' c;:J< • ell
1.6e+1
D
-3.2e-3
.~-n
L2e+l
I\'-_--------l
-4.8e-3
8.Ue+O
·6 ...:/e-3 '
'.O'ffi~
-8.0e-3
Lo 2976
5952
t
8928
t),Ue+O
1190-1 1,4880
o
2976
5952
t
8928
11904 1-1880
Polymath solution
P4-32 (c) Change the value of Vo and CAO in the Polymath program to see the changes
1 II , - - - - - - - .
08
P4-32 (d) As ethanol evaporates as fast as it forms: Now using part (b) remaining equations, Polymath code:
CD=O 11,6
04
See Polymath program P4·32··d.po!. POLYMATH Results Calculated values of the DEQ variables
11 . 2
Variable initial value minimal value maximal value final value t 0 o 7.72 0.0519348 Ca 10.,93 6.9932872 Cb
0.0
k
ra va Va
9.,OE-05 --0" 0075942 0,,05 200
9,OE-05 -0,0075942 0,,05 200
'----~---------~----'
I)
6000 7 . 72 10,,93 9"OE-05 -3,,69E-05 0.05 200
4-62
1189
2378
t
3567
6000 0,,0519348 7.8939348 9"OE-05 -3 .. 69E-05 0.05 200
4756
5945
V X
200
200 0
500 0_9932727
500 0_9932727
o
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ra - Ca*voN [2] d(Cb)/d(t) ra - voN*(Cb- 10.93)
=
Explicit equations as entered by the user
=
[1 J k 0.00009 [2] ra = -k*Ca*Cb [3] vo = 0.05
[4J Vo = 200 [5] V [6] X
=Vo + vo*t = 1 - Cal7. 72 ------.-------
20
O.Oe+O~-----
16
12
8
""'--.".,,~.....
--
-... ..-.... "
.........................--.'.........,._..... .,..
4
-8.0e-.3 0
{I
-~---~::-::--~-------
1189
2378
t .3567
4756
5945
o
P4-32 (e) Individualized solution P4-32 (0 Individualized solution P4-33 (a) Mole balance on reactor 1:
- dNAl CAlv-rAlV - - dt C dN -.AQ. v -C v - r V=--..A!... 2 0 Ai Al dt CAOV AO -
.
with
1 2
_ V AO - - Vo
Liquid phase reaction so V and v are constant CAO
CAl
2r
r
--- - ---- - r Al
_ dCAl
- -----
dt
Mole balance on reactor 2:
4-63
--~-:----....::::::::::::::;:====~
1189
2378 t
3567
____--!
4756
5945
CAl C A2 , _ dCA2 ------r --
T
T
dt
A2
Mole balance for reactor 3 is similar to reactor 2:
, _ dN C A2V O - CA3VO -rA3V - - -A3dt C
C
_ dC
A3 - A2 - - - -A3- r - - -
T
T
A3
dt
Rate law:
--rAi
=kCAiCBi
Stoichiometry For parts a, b, and c C Ai = CBi so that -rAi
= kC~i
See Polymath program P4-33"pol. POL YMA TH Results Calculated values of the DEQ variables Variable t Cal Ca2 Ca3 k
Cao tau X
initial value 0 0 0 0 0" 025 2 10 1
minimal value 0 0 0 0 0,,025 2 10 0,3890413
maximal value 100 0,8284264 0,,7043757 0.6109587 0.025 2 10 1
ODE Report (RKF45) Differential equations as entered by the user
4-64
final value 100 0.8284264 0.7043757 0.6109587 0.025 2 10 0" 3890413
d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1A2 [2] d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2A2 [3] d(Ca3)/d(t) (Ca2 - Ca3)/tau -k*Ca3A2 [1]
=
Explicit equations as entered by the user [1] k = 0,,025 [2] Cao=2 [3] tau 10 [4] X = 1 - 2*Ca3/Cao
=
From Polymath, the steady state conversion of A is approximately 0.39
P4-33 (b) 99% of the steady state concentration of A (the concentration of A leaving the third reactor) is: (0.99)(0.611) = 0.605 This occurs at t =
P4-33 (C) The plot was generated from the Polymath program given above. 0.90 r - - - - - - - - - - - - - - - - - - ,
0.72
0.54
- Cal 0.18
- Ca2 - (:a3
o00 ~:....--........---~-----'------'.------I .
0
20
40
t
60
80
100
P4-33 (d) We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole balance on species A will change slightly. Because species B is added to two different reactors we will also need a mole balance for species B . Mole balance on reactor 1 species A: CAOVAO -CAlv-rAIV
dNAI
.
= - - - wIth
dt dN AI -_·v -C v-r V=---
VAO
2
200
3
15
=--Vo and Vo = - -
2CAO
3
0
Al
M
~
Liquid phase reaction so V and v are constant
4-65
CAO - - -CAl - - r =dCAI -27: 7: Al dt
Mole balance on reactor 1 species B: , _ dN BI CBOV BO -CBIV-yBIV - - - and
dt
Stoichiometry has not changed so that -rAi = --lBi and it is a liquid phase reaction with V and v constant CBO C , _ dC BI - - - - -BI- Y - - 37: 7: Al dt
Mole balance on reactor 2 species A: We are adding more of the feed of species B into this reactor such that V2 = Vo + VBO = 20
CA2V 2 -
CAlVO -
dNA2
rA2 V
=--
dt
Mole balance for reactor 3 species A:
-rA3V
C A2 V 2 - CA3 V 2
C
C
7: 2
3
, _ dCA3
A2- - -A3- Y --
7: 2
dNA =dt
A3
---
dt
Mole balance for reactor 3 species B: C
, _ dN B3 V --C V --,Y V B2 2 B3 2 A3
dt
C B2 C , _ dCB3 -· - - -B3- Y -A3
7:2
7:2
dt
Rate law:
4-66
See Polymath program P4-J3-d.po1. POLYMA TH Results Calculated vaiues of the DEQ variables Variable t Cal Ca2 Ca3 Cb1 Cb2 Cb3 k Cao tau X
tau2 V
vbo
initial value
o o
o o o
minimal value
o
o o o o
o o
o o
0 . 025 2 13.333333 1 10 200 5
0.025 2 13.333333 0.3721856 10 200 5
maximal value 100 1.1484734 0.7281523 0 . 6278144 0.4843801 0.7349863 0.6390576 0.025 2 13.333333 1
10 200 5
ODE Report (RKF45) Differential equations as entered by the user [1 J d(Ca1 )/d(t) = (2*Cao/3 -Ca1 )/tau -k*Ca1 *Cb1 [2 J d(Ca2)/d(t) = Ca1/tau - Ca2/tau2 -k*Ca2*Cb2 [3 J d(Ca3)/d(t) = (Ca2 - Ca3)/tau2 -k*Ca3*Cb3 [4] d(Cb1 )/d(t) = (1 *Cao/3-Cb1 )/tau-k*Ca1 *Cb1 [5] d(Cb2)/d(t) = Cb1/tau+Cao*vboN-Cb2/tau2-k*Ca2*Cb2 [6 J d(Cb3)/d(t) = (Cb2-Cb3)/tau2-k*Ca3*Cb3 Explicit equations as entered by the user [1] k=0.025 [2J Cao =2 [ 3 1 tau = 200/15 [ 4 J X = 1 - 2*Ca3/Cao [5] tau2 = 10 [6] V = 200 [7] vbo = 5
Equilibrium conversion is 0.372 . This conversion is reached at t = 85.3 minutes .
4-67
final value 100 1.1484734 0 . 7281523 0.6278144 0_4821755 0.7291677 0.6309679 0.025 2 13.333333 0.3721856 10 200 5
2.0 . - - - - - - - - - - - - - - - - - - , - Cd
1.6
- Cal
1.2
0.8
---,_."._" ......"" . ".,,---1
0.4
0.0
"""""=--~---'-----~--~----'
o
P4-33 (e)
20
40
60
80
100
Individualized solution
CDP4-A CH3I + AgCl04 0.7 molll CE31 0.5 moli1 AgCI0
V '" 30 dm o
;=
4
~
CH3Cl04 + AgI
CBO ~
CAD
3
3/2
C tCH 1 ~ -k C;CH IAgCI0 3 4 S k ~ 0.00042 (dm 3!mol)3!2(sec)-1 T=298 K I "" 0.93 Integration for! "" 0.98, solved lll!lllerically 1 . t =
~---.- ••.• ~
(24.18)
0.00042 (O.5)3fb t "
v = Vo
CDP4-B a)
4-68
1.628:t10
5
sec '" 4S.:2 hr
d=". -"-'=1:"' dt - -
dr
dX
= -FAO -
4-76
dr
X) = CAO (1- X)
= rA (2mh)
J(21lkC
=
2 AO
)rdr
FAO
Ro
x
By Integration we get: - - -
I-X
b) Now, with the pressure drop, C A= CB = CAO(l-X)y Hence,
-rA =kCA0 20-X)2y2
dX _ 2mhkCAO dr Where
2
0_ -
y = (1- aW)1I2
2
X) Y
"",,----_._-----------, ..,......
=>
1~"
2
G"phTItl.
tJ2&3
and
W = PbJZ"(r 2 - RO 2 )h Using polymath, following graph is obtained: Differential equations d(X)/d(r) =2*3 . 1416*h*k*Ca0"2/Fao*(1-X)1\2*yI\2*r y = (1-alfa*W)"O . 5 W = density*3_1416*h*(rI\2- 1'0"2) ro 0,,1 density = 2 Cao = 0.1 Fao 10 k=0,,6 h=O.4 alfa = 0 . 07 variable name: r initial value: 0.1
oro.
'--""'---_~---'-_~
~
~
~
~
~
_ _ _ _. ~
,
~
~
m
=
=
c) Increasing the value of k increases conversion while decreasing it decreases the conversion" Increasing FAO will decrease the conversion and decreasing it will increase the conversion. Increasing C AO causes a dramatic increase of conversion.. Similarly, decreasing CAoresults in a large decrease in conversion. Increasing the height will only slightly increase the conversion and decreasing the height causes a real small decrease in the conversion. Increasing Ro decreases the volume ofthe reactor and hence decreases the conversion. Increasing RL will increase the conversion as volume increases .
4-77
CDP4-H Liquid phase rexrion A+B--rC
vellmixed no inflov or outtloll BATCH
L Mole balance on batch reactor
2. Ratelaw -1'A "'" Ie eACa
3. SroichiCll:!l!'!try liquid phase
V "" V" (batCh) (if flow u
=u",]
4. C.ommne
4-78
Table of reaction imegrals can be found in Appendix A-lO. a)
I. Niok baIa...,ce on CSTR
2. RateLaw
3. Stoichiometry liquid phase
u ::::: U o
4, Combine:
5. Parameter evaluation 10 moUmin (0.9) V=: - - - - - - - . ..._--
(o,m d:a:J!l/mol min) (2 lllOvd,m3f(1-0.9)2
b)
V"", 22500 4m3
1. Mole bala.'1ce onPFR FAO JiX.::::: 'dV
-fA
if!lQ pressure OJ.' phase change (liquid phase) therefore
r
V=F~o ~~ 2. Ratelaw
4-79
5, P = = evaluation
C:A
~
:;=
fM ~l·:~~., = (~AD Cl--A)
U
___ :-QQ.rooVrrrin}· ___ (01 cio:r'Jmol min)(2 rnolJdm'J
F,,,,,ieR_hX\
:: __:...~.:__.,.....1t., ....~_;;:; (:AD (I···X)
C::s ::::
e)
"rA=kfcACS··~ t
KcJ
Arequilibrium
Kc - _Cce ....
(A
V""rAO
I
;j.~: .
li·X)'-
10
=
- CAe CJl"
""
x
~=4
_X,.= X. KCN)
X2-2.25X-t 1=0
PFR .dX.",.::!A.""kC rC (1-X'I1.X.j""kC2 0 [(lXf·,:;. X ..J dV FAG AOl AO I Kc A \ KcCAol
f6 .-. . ."...")-''''X",.,",....X'_. £.,i o 1·· • • JA,
+, •
V", 250 ~.[~O).,. 4~, 15) -+ l~ ,3) + 4~,45) .. i\ 6}]::::
150
Q'j 5[11 4{ 141)t 2(241).· 4{5.26) +100J
V", 250 ~[r{O) + 4~.,15) + 2~,.3)+4ft.45)+ ~6)J
.:;: 250 llir ;~11.411 +-li2Al) + 4{5 26) +100J 3 c1., "'\ • '" 1656 c:n'
CDP4-I (a)
4-80
,·,,,,=0
,.,.CAolL.= __ ..,..x._.._ CAd)-Xf
C'idl-xf
KCAo=l~. 1 .,. 2X +- X2
V", 250
.
d) Solution si::::ni1:at· to part (a) ro (e)
-UO
'J
·····,i·f1g·
v = .....
DO
4
Benzene is A -rA =k A (C A2
for equilibrium - rA
=0
C CBC K ) c
_
:.
4Kc:{1-Xc'f =Xc
2
2
4Kc - SKcXc: + 4KcXc = Xc
2
0.2X2 - 2.4Xc: + 1.2 = 0 Xc: =0.52
CDP4-I (b) PFR
Design Equation:
V = FAD
I -rA dX
-rA =k A (C 2A
RateLaw:
_
C CBC K ) C
Stoichiometry :
C Ao
=~ = RT
C
,,= CAD (1 - X)
CB
=C 2 X Ao
C
c
5 atm = 0.OO371bmol (0.73 ·atm K1859.67-R)· ft3 lbmol·- R )
fe
4-81
=C ADX 2
Combine:
v=
FAo
kC!c,
r[O-Xy _ X dX
2
]
4K,
V_
rX
(IOlbmoUminXlminl60s)
dX
- (ISOOfellbmol.sXO.OO37Y Jo [(I-XY -O.S33X 1 ]
v - 6 76ft 3rX -.
dX
Jo &-2X+O.167X z]
1 In[O.523(X-11.45)~) - . \. (O.617X11.45-0.523) 11.45 X-O.523 U
v- 671
v =13.5fe CDP4-I (c) CSTR
Design Equation:
RateLaw: Stoichiometry: Combine:
V=
V = FAOX -rA
-r.
=k.(
C! _
CA =CAo(l-X) V=
C~~c ) Ca
FAo
=C~X
X
k.C~[(l-X)'-(
(lOXO.51X1l60) .
(1800XO.oo37)'[(l-0.51)' -(
Cc
=C~X
!:J
4~~))]
= 147.73ft 3
CDP4-1 (d) Amount processed in CSTR ; (10 IbmoVminX60 minlhr X24 hrlday)= 14,400 IbmoVday
4-82
Batch
t-C -
Ao
Jd.X - kCC J -::;-A
Ao 2
[
Ao
d.X
2
X ] (l-xf-4Kc
1 In[0.s23(X-l1.45)~) - kC Ao (O.167Xl1.45-0.523) 11.45 x -0.523 U
t __ I (
=
t 0.30s Taking into account the time it takes to clean the reactor and other down time assume that the total time per run is 4 hours. Assuming that the reactor can be used twenty-four hours a day there can be 6 runs per day.
14,400 Ibmollday =2400 Ibmollrun 6 runs/day
= N Ao V
f -rA dX
V - N Ao
1 (
t
- t kC!o
:.
J
V = N Ao dX t -rA
In)
1 In[0.523 (X -11.45 (0.167Xl1.45-0.523) 11.45 X-O.S23 U
2400 ( 1 1 [O.523(X-ll.45)~) - (4X1800XO.0037Y (O.l67XU.45-0.523) n 11.45 X-O.S23 U
v-
V = 48,690fe
CDP4-1 (e) E
=30,202 btullbmol
for X =0,
-rA =kC!o
- rA(800) k800C!o k800 - rA(l400) = k'400 C!o = k l400
E( 1
=exp R .
1)
T2 - Tl
-
- rA (800) = ex 30,202 btu/lbmol ( 1 _ 1 ) -tA(I400) P 1.987btullbmol·-R 1259..6rR 185?67·R
-r
A(800)
= 49
- r A (1400)
4-83
CDP4-J PBR
?vfB:
Combine:
ME:
w ::::;~~Q.>!: -fA
Combine:
X W r; .•..f'...o ---. .... _ ...... _ . kC llo (1 . . . . X) 1000::;:: ---
1
X (I- X)
X =.18 b) We know, s = 0, so
P = Po (1- aW)l!2
=>
1 = 20(1-- aW)l!2
4
=> a = 9.98 x 10. c) For conversion to be maximum, a should be minimum 2
G
Also, we know that a is proportional to - - - - = AcDp
G
And it is proportional to - - - 2 - = AcDp
1 6
2 for turbulent flow
D pipe Dp
1 4
2 for laminar flow
D pipe Dp
Hence for minimum a, Dp and D pipe should be increased" d) Yes, we can increase Dp and decrease D pipe (or vice versa) according to the above equation to get the same value of a.e) For assumed turbulent flow,a is proportional to .-
1
6
D pipe Dp
4-84
2
Now,
6
al
2
_ Dpipe2 Dp2
-
6
a2
-1
2-
D pipe! D pI
Dpipel ---''---=--4
and Dp1 = 0.5 cm
6
Dpipe2
=>D p2 = 0.044 em
.
1
t) We're gIven, k ~ Dp
k2
Hence -
kl
Dpl
=> k2 = kl - Dp2
0.5
= - - - = 11.4 0.044 rA
g) Rate law: -
Stoichiometry:
= kCA
CA = CAO (1 -, X) Y dX ·-r
Mole balance: -
...
dW dX
Combining:--
=_-A FAO
=
k C 2(1_ X)(l-a W)1I2 2
k2 _
-
- 11.4 and
2
AO
dW
,- (forE = 0)
FAO CAOkl
= 2 X 10
--4
..
..
Integratmg WIth hrruts: (X -70 to X; W -7 0 to 1000) => X
FAO
kl
CDP4-K 1.
Mole Balance: dz
2,
Rate Law :
lA
FM
;;:;-·k'C:\pc (1--
CA = CB Hence, rate law becomes -rA = kC~C; = kC~2+n) Observation table 2: for
C AO
=0.01 and CBO = 0..1
C A (molJdm3) 1.0 '(J.278 . 0.95. r-.---0.816 1.389 0.707 2.78 -0.50 8.33 -0.37 16.66 t (h) 0
t
1 C(l-C2+n)) = ___ AO k
C(l-C2+n)) _ _ 1 (O.l)C-l-n) A
(1-(2+n))
k
C C- 1- n ) A
(-I-n))
5-8
See Polymath program P5-7-a--2.pol. POL YMA TH Results Nonlinear regression (L-M) Model: t
= (1"(-·1-·n)··Ca"(--1-·n))/(k*(--1-n)) Ini guess 3 2
Variable n
k
Value 0.8319298 0.1695108
95% confidence 0.0913065 0.0092096
Nonlinear regression settings Max # iterations 64
=
Precision R"2 R"2adj = Rmsd Variance =
= 0.9999078 0.9998848
= 0,,0233151
0,,0048923
Therefore, n = 0,,8 Hence rate law is:
__ r = 0 • 17 CA2 CBO.8 mol 3 A dmh
P5-7 (b) Individualized solution P5-8 (a) At t = 0, theIe is only (CH3hO. At t = 00, there is no(CH3hO. Since for every mole of (CH3hO consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure"
P(oo) = 3Po 931 = 3Po Po :::o31OmmHg
P5-8 (b) Constant volume reactor at T
=.504°C =777 K hase: "] 1195
562
(CH 3 h O ~ CH 4 + H2 + CO YAO
=1
£5=3-1=2
e=£5y. AO =2 V
=Vo( ; ) (1- eX) =Va
because the volume is constant.
P=Po(1+eX) at t = 00, X = X AF = 1
5-9
1 dNA N AO dX ---=-----=r V dt Vo dt A st Assume -rA = kCA (i . e . 1 order)
CA = CAO (1- X) (V is constant) dX =kCAO (1- X) dt po_oR
Then: CAO and X
=
0
cPo dX 1 dP Therefore: =- - dt cPo dt _1 :!P = k[l- P-Po]
cPo dt or dP
dt
cPo
=~([l+e]Pa -p) cPo·
= k ([1 + e] Po - p)
dP f-----= fkdt [l+e]Po--P
P
t
R
0
o
Integrating gives:
In [ ( cPo ) - ] 1+ e Po --- P
Therefore, if a plot of In
= In [2Po] = In [624 ] = kt 3Pa - P
936 - P _
624 vel sus 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:
---"y_=_0_.0_0_0_48_X_-_O_.O_29_0_7_~~/ j ~
1.6 1.2
+ -__
...
0.8
+---------7"'''---------------1
0.4
0+£-------------------1 -~4
I
o P5-8 (c)
1000
2000
3000
Individualized solution
5-10
4000
P5-8 (d) The rate constant would increase with an increase in temperatme, 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 fro colder temperatures,
P5-9 Photochemical decay of bromine in bright sunlight: t (min) CA (ppm)
20 1.7 4
10 2..4 5
30 1.2 3
40 0.8 8
50 0.6 2
60 0,,4 4
P5-9 (a) Mole balance: constant V
dCA =r =-kC a dt A A
Differentiation T (min) ~t (min) CA (ppm) ~CA (ppm)
10
20 10
2,45
.L\CA ( pp.m ) L\t mIll
30
40
10 1.74
50
10
10 0.88
1.23
0,,62
·0.51
-0..35
-0.26
-0.18
-0,071
-0,,051
-0,,035
-0.026
-0.018
.:,:,::~
0.,1)4
0 .. 01
.,
10
.,..,---,_._. )0
40
5·11
50
.._'••"__-.1.-.;),
50
10 0..44
-071
0,,06
.t
60 10
-}.a
-4.0
3.£
After piI ' an oUmg -dCA/dt In( -dCA/dt) InCA
1
erentIatmg by equa area 0.061 0.082 -2.501 -2.797 0.896 0.554 .-
0.042 -3.170 0.207
0.030 ··3.507 -0.128
1.0
0.014 -4.269 -0.821
0.0215 -3.840 ··0.478
-
Using linear regression: a = 1.0 in k = -3.3864 k = 0 ..0344 min- l
P5-9 (b) !!NA=Vr =F lit A B rA = FB =
·-0.0344 p~m -0.0344 lmm m~ . mm (25000 gal) (0.0344 lmm m~ J(60 min_)(_lL'J(3.?~51lJ( IlbS_J hr 1000mg 19a1 453.6g =
at CA= 1 ppm
P5-9 (C) Individualized solution P5-10 (a) Gas phase decomposition A7B+2C Determine the reaction order and specific reaction rate for the reaction
dC A dt
-kCA
Assume the rate law as: - - =
Integrating:
t = __ 1- [
k(n-l) C
1n A
-1
-
n
C
1n
AO
--1
J 5-12
=
0.426
lbs hr
n 1
-lJ
=>In(tl/2) = In 2 ( (n -l)k + (1- n)In(CAO)
Run #
CAO (gmolllt) 0.025 0.0133 0.01 0.05 0.075
1 _. 2 3 4 5
t (min.) - ~.1 7.7 9.8 1.96 1.3
See Polymath program P5··10··a.pol. POLYMATH Results Linear Regression Report Model: Int = aO + a1*lnCaO
a:o--
Var'iable
Value -2:"'3528748
a1
-1.0128699
95% confidence
0.1831062 0.0492329
General Regression including free parameter Number of observations = 5 Statistics RJ\2 = RJ\2adj = Rmsd= Variance =
0.9993004 0 . 9990673 0 . 0091483 6 . 974E-04
3.0 - - . - - - - - - - - - - - - - .
24 1.8 1.2
0.6 0. 0
L - ._ _ _ _ _
-4 . 61
-4.20
~
__
~_.
-3 . 80 In(dojO
-2 . 99
-2.59
From linearization, n = 1- slope = 2 ..103;:::;; 2
(2 a - 1 _l)e-interLePI it k=-· . =10 ..52 - - - a··-l gmol.min
5-13
-
In t 1.410987 2.0412203 2.2823824 0.67294447 0.26236426
InC Ao -3.6888795 -4.3199912 -4.6051702 -2.9957323 -2.5902672
_ dCA =1O.5C/gmolllt.rnin dt
.c'"
i:.
\1 ~
C aO
See Polymath program P5-· lO··a,pol. POLYMA TH Results Nonlinear regression (L-M) Model: t = ((2J\(a·1 ))··1 )/(k*(a··1 ))*(1/CaOJ\(a·· 1)) variable a
Ini guess 2 10.52
k
Value 1. 9772287 8.9909041
95% confidence 0.093057 3 . 9974498
Precision R"2 = 0 . 9986943 R A 2adj 0.9982591 Rrnsd 0 . 0531391 Variance = 0.0235313
= =
dCA = -----
Rate law:
dt
9 .OC 2 gmol 11' t.mm A
PS-IO (b) We know,
( 2 a--l
-
1) (
1
k = t1l2 (a·- i) C AO a-I
J
Solving for k at 110° C
k'= (2(2-1) -1)( __1_)=20 it 2(2-1) 0.025(2-1) gmol.rnin
Rln k2 From these values, E = kl
U-;J
(8.314 =
J mol.K
)In. 20 10.5 = 76 ..5 kJ/mol
(37~K - 38~K) 5-14
~,
ij ~
P5-11 0 3 + wall ~ loss of 0 3 0 3 + alkene ~ products Rate law:
-~
0
3
dCo3 =kJ = __
.
kJ k2
C Coz
+k2~
dt
Using polymath nonlinear regression we can find the values of kl and k2 ozone rate (mol/s.dm3) Ozra
Run #
Ozone concentration (mol/s.dm3) C03
le-12 le-ll
0.01 0.02 0.015 0 . 005 0.001 O..oIS
1.5e-7 3.2e-7 3.5e-7 5.0e-7 S.Se-7 4.7e-7
1 2 3 4 5 6
Butene concentration (mol/s.dm3) Cbu
le-lO le-09 Ie-OS le-09
See Polymath program PS· 1 l.pol. POLYMA TH Results Nonlinear regression (L-M) Model: Ozra
= k1 +k2*Cbu/C03 lni guess 2.0E-07 0.1
Variable
k"l-k2
Value 3 . 546E-07 0.0528758
95% confidence 4.872E-ll 1.193E-·05
Nonlinear regression settings Max # iterations 300
=
Precision R"2 R"2adj Rrnsd Variance ==
= 0.7572693 = 0.6965866 = 4.531E-08 1 .848E-14
Rate law:
-ro,
= (3.5xlO-7 )+(O.05) ~BU mol/dm 3 .s 03
P5-12 Given: Plot of percent decomposition of N0 2 vs VIFAO
X= % Decomposition of N0 2 100 Assume that -rA
= kC~ 5-15
ForaCSTR
or
F X
V =~ -rA
x x v --=--=-FAD
kC~
-rA
V
withn=O, X =k-FAD
V
X has a linear relationship with - - as FAD
shown in the figure . Therefore the reaction is zero order .
P5-13 Si02 + 6HF ~ H 2 SiF6 + 2H 2 0 N s = moles of Si0 2 = AcPsD MWs
Ac = cross···sectional area Ps = silicon dioxide density MWs = molecular weight of silicon dioxide = 60.0 o= depth of Si
N F = molesofHF=
wpV lOOMWF
w = weight percentage of HF in solution P = density of solution V = volume of solution MWF = molecular weight ofHF = 20 . 0 Assume the rate law is
dN
Mole balance: _ _S
dt
-rs = kC;
= rs V
_ l\;Ps do _ k( wV MWs dt lOOV MWF __ do dt
]a V
= _ kMWs __ (~]a Vw lOOa !\Ps
MWF
- do = fJwa where fJ = ~MWs dt
a _(--.£ __ ]a V MW
lOOa !\Ps
F
In(- ~~) = InfJ + aln w
5-16
In(- ~~)
-16,629
In w
-15.425 2,,996
2.079
where (-
dJ) dt
is in
-1432 3.497
-13.816
-13.479 3.871
3689
~ rom
FlOm linear regression between
In (- ~~) and In w we have:
slope = a = 1.775 intercept = In ~ = -20,462 or ~ = 1.2986 * 10-9
~13 .,..------:-""'-----------------,
-:g -Il,}
"a "C
2.5
..14
3.5
3
y ;: 1.7746x - 2Q.461
-15
t
c: ..16 ·17 In w kMWs
(p
fJ = 1001775 AcpsMW:;-
Jl775 V
Ac = (10 *10-6 m) (lOm)( 2 sides )( 1000 wafers) = 0.2 m 2 g
MWs =60-gmol
Ps
= 2.32 L = 2.32 *106 -.1] ml
m
(Handbook of Chemistry and Physics, 57 th ed, p"B-155)
pzl-~=106L3 ml m g
MWF =20--
gmol
= 0.5 dm 3 = 0.0005 m 3 fJ = 1.2986 *10-9
V
5-17
1.775 106~
1.2986 *10-
9
= _--.!!L g 20--
gmol 3
k =3.224*10-'7 ~ ( gmol Final concentration of HF
J0775
=
min-I
5-2.316 (0.2) = 0.107 5
weight fraction = 10.7% weight fraction = 20%
Initial concentration of HW = 0.2 (given)
Mole balance for
_
=
pV .dw 100MWF dt
f
_lO 7_
dw
WI
775
20
dN
dN
dt
dt
F HF: - = 6--s
6k( 100wpMW Ja V F
=6k('_~J0775 tfdt 100MW ° F
_1_( __1_) 10.7 0.775
W0
775
where u = 1.775
= 6(3.224*10-7 ) (
20
__~_,( 1 _1_) = 2 0.775 10.7°775 t=331 min
20°775
o775 6 10 J . t 20*100
389*1O-4t .
P 5-14 (a) A + 3B -7 C + 2D + E Observation table for differential reactor Cone., Of Temperature(K) A(molJdm3)
rm-
SQ,--
c--.
Cone, Of B(molJdm3) 0.10 0.10 0.10
0.10 333 343 -,- 0.05 0.10 353 ,-- ~~--,0.01 0.20 363 0.01 0.01 363 Space time for differential reactor = 2 min
V= F;, = VoCp --rp
-rp
5-18
-
Cone., Of C(molJdm3) 0.002 0.006 0.008 0.02 0.02 0.01
--
Rate (mol/dm3 .min) 0.001 0.003 0.004 ,-0.01 0.01 -,0.005
--
r, p
= Cp = CCzH4
r
2
Rate law:
rc =AJ-BIT)C:C~ Cc =Ae(-BIT)CC 2 A B Where, A is Arrhenius constant B = activation energy/R x is the order of reaction WIt A Y is order of reaction WIt B CA is the concentration of C2H 4Br CB is the concentration of KI Now using data for temperature 323K, 333K, and 363K, for finding the approximate value of B because, at these temperature, the concentration of A and B are the same. Using polymath, the rough value of B = 55o.o.K While using polymath for solving the rate law apart from guessing the initial values of n, m, and A , we change the value of B in the model to get the optimum solution . So after trial and enOf we got B = 65o.o.K
See Polymath program P5-14-a.pol. POLYMAI!!. Results Nonlinear regression (L-M) Model: r = A*exp(··6500/T)*GaA x*CbA y variable Ini guess Value A---3 . 6E+05 3 . 649E+06 x 0.25 0.2508555 Y 0.2 0 . 2963283 Precision R"2 = 0,,9323139 R"2adj = 0.8871898 Rmsd = 3 . 615E-04 Variance 1.568E-06
95% confidence 2.928E+04 0.0032606 0.0020764
=
Hence, by nonlinear regression using polymath A = 3649E+o.6(mole/dm3 2 6(1/s) E = 65o.o.R = 54,,0.15 KJ/mol x = 0..25 y= 0..30. hence,
r
rc = 3.64E + 06e(-54015 / RT)C~25 C~30 mole/dm3.min
P 5-14 (b) Individualized solution
P5-15 (a) 5-19
Modell: Monod equation
dCc = r = JlmaxCsCc --
----'=='---"--~
dt
Ks + C s
g
See Polymath program P5-15··a.pol. POL YMA TH Results Nonlinear regression (L-M)
=
Model: rg (umax)*Cs*Cc/(Ks+Cs) variable lni guess Value umax 1 0 . 3284383 Ks 1 1.694347 Precision R"2 = 0.9999439 R"2adj = 0.9999327 Rmsd = 0.0038534 Variance 1A55E-04
95% confidence 0 . 00686 2.2930643
=
P5-15 (b) Model 2: Tessier Equation Tg
=tL={I-ex
p( - ;
J]Cc
See Polymath program P5···15-b.pol. POLYMATH Results Nonlinear regression (L-M) Model: rg = umax*(1··exp(-Cs/k))*Cc
Variable umax
lni guess 0.5 100
k
Precision R"2 RA2adj Rmsd Variance
Tg
= = = =
Value 0.3258202 20.407487
95% confidence 0.0034969 5.7120407
0 . 9999454 0 . 9999345 0 . 0038004 1.415E-04
=0.33[I-ex{ ;~~ J]CC
Wdm'h
P5-15 (C) Model 3: Moser Equation
=
r g
f.1rruJ.x Cc 1+ kC --Y S
See Polymath program P5··1Sc.pol. POLYMATH Results Nonlinear regression (L-M)
5-20
Model: rg = umax*Cc/(1+k*CsJ\(-y» Variable umax
k
y Precision RA2 RA2adj
Rmsd Variance
Ini guess 0.3 1.6 1
Value 0 . 3265614 162,,599 2,,0892232
95% confidence 6.984E-04 34.273983 0.0461489
= 0 . 9999447 = 0 . 999917 = 0.0038269
= 1.794E-04
O.33C 1 + 162.6Cs (-2.1)
c ---...:::.....,--gldm3 .h
P5-16 Thermal decomposition of isopropyl isocynate in a differential reactor .
Run 1 2 3 4 5 6
--r------
Rate (mol/s.dm3) 4.9 x 10-4 1.1 X 10-4 2.4 x 10-32.2 X 10-2 1.18 x 10- 1 1.82 x 10-2
Rate law:
--rA -- A e(-EIRT)CAn Where, A is Arrhenius constant E is the activation energy n is the order of reaction C A is the concentration of isopropyl isocynate
See Polymath program P5···16 . pol.. POLYMATH Results Nonlinear regression (L-M) Model: rA = A*exp(-E/(8 . 314*T»*(CA)An var'iable A E
Ini guess 100--1000
n
1
Value 1.01E+04 5.805E+04 1.7305416
95% confidence 327.35758 237.32096 0 . 0134196
Nonlinear regression settings Max # iterations = 64 Precision RA2 RA2adj
= 0 . 6690419 = 0.4484032
Rmsd
:::: 0 . 0097848
--
Concentration Temperatwe (mol/dm3) -(K) 0.2 700 0.02 750 1-------0.05 800 0.08 850 0.1 900 -0.06 950
5-21
Variance = 0.,0011489
Hence, by nonlinear regression using polymath A = 10100 (mole/dm3 2 6(1/S) E = 58000 J/mol n = 1.7 therefore,
r
-YA =lOlOOexp ( -6:76)C~7
mole/dm3 s
CDP5-A Given the reaction P + NH;PH -? NHzOHP where P is Penicillin and NH10HP is hydroxylamine acid (denoted by subscript HA) Let A """ Absorbency, then CHA, ::; KA where K is some constant. Cp
"""
C HA
Cp., (1,,,· X) (s = 0 for liquid phase reaction.)
=Cp"X "'" KA
Att=-
A
•
KA .', X= ~- when X C v"
=1 and
A = A..
C
-
=,.......l:2. K
- rp' := kC;
Assume reaction is irreversible:
:=
1 dN
kC;" (1- X)*'
-~
For a batch. constant volume reactor:
Vdt
dC? =-,= To dt
l."
or
,~~e.=Cl''''~''''''
or
'~'=k(~~rl(A_'A)"=K(A_.'Ar, WhereK-k(~:r-l
dt
dt
dA"",.-kC;o(l-X)"=,-kC;"(l-A/A_)"
A_ dt
Try integral analysis first. Assume that reaction is zero Older:
Then
,~~;;;;;: K dt
or
rdA = A = Kt
Jo
a plot of A vs. t should be linear if
reaction is zero order. FlOm the plot below, it is evident that the reaction is not zero order; 06
05 4) (,,)
c
4-
0.4 '
«I
...
..Q (:)
03"
.-
II)
..Q
<
::L_o
10
20
30
40
Time
5-22
50
Next, assume that the reaction is first order;
dA - :;:: K( A_ - A) or dt
J(A. ..ciA_. A) :;: JKdt ::: In (A. -A-A) == - Kt ... A
t
0
0
. A plot of (A_ - A)
VS. t
on semi -log paper should be linear. -.. ---...-..-, •.•....-..-.-. ..'-•...... .............. .~:- .....w... .......-2C--.. - ..... 30-............. --.40-............. *--_.. _. . . . . . _................_..................._. . . . . . . . . . . . . . . . . . . . . . . . --.. ••••• _ •• _ ...... _ _ .. _ _
-
~._
.. -.. ....."- ..
Time
A
0
0
10 t-20"-
A.,.-A
-0.348
0.433 0.495 0.539 0.561 0.685
0.252 0.190 0.146 0.124 0
30 40 50
"'"
.
0.685
"'om
~
-
.
•• _ .................. _ . . . . . . . _ _ _ _ _ _ •• _
•• _ .
.......• -,....
~
or _... •.!. ..-.-
(A_-A)
o
--1'0· ·'-20
o
1.460 "-'2.967 . . 68--
0.33'1" 0..
_,
A_
A.. .. A
1 .. A plot of-----vs. t should be linear
(A_-A)
s ........__. -.. -.. _........................................... 8
•
7
30
•
'"40 0.539 50
0.56 f
_ _
,-
0.1 ................- .. - .....- ,-. ,..................................................- ..,..... .."..........................._"' ......-
: : : ...!_. +- Kt
A
~
-.-.--~.-.-
..
It is evident from the plot that the reaction is not first order. Try second order:
dt
•• _ ••••••••• _ . . . . . . . . .
. Time
?A "" K(A_ .. Al
"W'
-8])65 2
o o
10
20
Time 30
4U
50
From the plot, it is evedent that linear relationship exists between (VAx, - A) and time; Therefore the reaction is second order .
CDP5-B Determine the reaction order and specific rate constant for the isomerization reaction: A7B Rate law:
a dC A -rA=kCA =-_.dt
5-23
..--.:r;-...-..- . -..----.--..-.-......--..-.....- .....- ...-.--.--.---....... . -f,.CA/f,.t .-dCA/dt........____ .._
Iime_(min) .._ _ _. C A (mol/dm) o 4
0 . 39 0.37
2.89
3
035 0.32
225
5
0..30
0 . 267 1.45
8
0.25 0.225
to
0 . 21 0\75
12
065
0.15 0.133
0.25
15
0.1
0.072
17.5
0.07
0 . 06
Plot of log --dCA/dt vs log C A shows a. "" 0.5
de.,
kA
=···~iitM.;:; O~~S019·d;:11~1:
CDP5-C Ethane hydrolysis over a commercial nickel catalyst in a stirred contained solid reactor.
H2 + C2H6 -7 2CH4
PA = CART =CAoRTCl- X) =PAoCl- X) PB =CBRT=CAORTCBB -X)=PAoCBB -X) X =!SFTO =~= Yp Cl+BB) 2FAO 2YAO 2 I
I
FAOX
Fp
-r A =-r B=- W--'--2W = .- r'A
Y pFAO 2W
= kPA a pB P
InC-r'A) = Ink + aInPA + fJInPB y = Ao
+ Al Xl + A2 X 2
FTo(gmollh)
PAO(atm)
PBO(atm)
1.7
0 ..5
05
1.2
05
05
0.6
0.5
0.5
0.3 0.75
04 0. 6
06 0.6
2.75 0.6 0.4 POLYMATH Results Nonlinear regression {L.M}
YCH4
0.0 5 0.0 7 0.1 6 01 6 0.1 0.0 6
X
0. 0 5 0. 0 7 01 6 0.2 0.1 0.0 5
Q
PA(atm)
PB(atm)
-rA(gmollkg . h)
0.475
0.475
1.0625
0.465
0465
L05
1
042
042
L2
L5 1 0.6 7
032 0.54
0..52 0 ..54
0.6 0 . 9375
057
0.37
20625
1
Model: ra = k*(PaAalfa)*(PbAbeta)
Variable
lni guess
- -Value --_.
95% confidence
5-24
0.1
k
alfa beta Precision
0 . 1124446 0.152574 0 . 1668241
0.5068635 0.9828027 -1. 9669749
1 1
:::: 0.999213 :::: 0.9986883
R"2 R"2adj
RInsd :::: 0.0051228 Variance :::: 3.149E-04
k = 0.5 atm gmollhr kg; hence, the rate law is:
a == 1;
P == -2
-r~ = 0.5 ~ gmol / kg .hr B
CDP5-D Since oxygen is found in excess, we assume that.-r·1iO is dependent only on ~. This gives us a rate law of the form:
- rHO
=kC:.o
.
we
From the units of the specific reaction [ate, assume that ex == 3. Now, using equation (5-18) from chapter S, we can solve for the desired balf-.Jives.
(a)
For ~ == 3000ppm:
t",
(b)
~ 2\iA"iO}""".'I~(3000~)' )= 119.05_
For C"'No
t",
=1 ppm:
=2\1.4,,10. 3pp;;;:>J;;;i~{(1~y L071x 10' mID
)=
CDP5-E Given the data.. postulate a rate law.
=kc:.
·rA Then write the design equation in tenns of the data given, in this case volume and time. CA=NAV
NA =N"o(l-X) V=
Yo(l + EX)
V-· V. X=·· . ···. . _J1.
Vue
V-v. =__ . Yoe
1-·_··· ........2..
C "
V
.5-2.5
Plug that into the design equation:
de
--2..=kC: dt A
5-26
The following grapb is made.
Once that is dorie it is ready to be graphed. The following graph is the natUral log the derivitive ofthc volunle function against the natUral 10 . of the volume function.
From the graph we see that ex. is 2. We can also find k: 9 k == :9 =.018 J
5
N .. 0
303.39*.2 020 1 =YAO NTO =. 85* 8.314*313 =. rna es
k' .018 k =. -N-a - I = M!, •.9 .v...
.40
The following rate law is found: 2
-r:A - .9*cA
Now, to decennine the volume of the CSTR. we must use the design equation and stoichiometry:
,
V;: F,..oX. -rA
CA
:;:
CAo(l- X)
C"o =!MJ
RTa
~o ;: Y"ofQ =.6 * 1013.25kPa == 607.95 Combining these with the rate law just determined. the followinl! volume is found.
5-27
CDP5-F Asr't, '" U
"riP. dZ
Z
P
0
129
1.5 2,5
70 50
45.3 28,,63 19.4
:l.O
30
9.5
6.5
18
L3
9.0
16
(1.7)
_de. AsHi "'" 3.0
AsHi = 3.0
Z
P
dZ
0
129
1..5
45
90 34.0
2.5 4.0
22
15.4
10
2.0
_df.
Z
P
dZ
0
129
1.5
95
17.4 15 . 0
PFR A ... B ::C
-8
I
1.5 2.0
I
)
6
9
,
I 4
PM PAl.
Task I, Re\\l'I'ite the design equation (i.e.. mole balance) in. te:rms of me measurement variables. Recall V ::: Ac Z. then
ax. = :r,,& dZ
FAo
5-28
For iSOthe."!I1ai operation and no pressure drop. X) rr:.. eX)' (l ..
CA =
CAO
CA
'=
P,JRT
(1. X)
= PA 0(1 +"e;C)
PA
(I + EX)
PA = PAO (1 - X)
Now we have the differential. mole balance in terms of the measured variables PA and
Z.
Posrulate -tA '=
k[PA ~ -~~j
Task 2. Look for simplifications. A) Sc:e if volume change can be neglected.
a therefore neglect volume change.
E ::
forr..:ns 1 and 3 where PAsH} "" 1.5 and 3.0 torr. respectively. [har most of tht E!;In is consumed. indicating the equilibrium is reasonably fJr to the ri giu. Conse:quen ely. the reverse reaction is negligible in the fU"St pan of the
B) \Ve see
L1:U
reactor. i.e.
C)
\'v:: ::lise
see i..l-tat for nlnS 1 and 3 that B is in excess and that far excess B II
5-29
-r,
• ,,,,\
:::::
k
,
R pi-' DO.. 80 .. A
=
lK' pet ~ r\
Algorlcnm Task -' Co1=ul>te (0
~;) Jnd plot vs Z to find (0 ~)
. dP
Plod ~ dZ-'\) vs (PA) on log-log paper to find alph3..
Task 4.
T~e
Task 5.
Ev:uua~e k" wd K?
r:ldo of inidal races at PAsH]
=:
3.0 torr and PA1l:1J
=:
L5 corr
From AjHl of 3,,0 corr we see equilibrium is reached at P Ae :=:.0 { PC For PFR, Mole balance: - - = - dX -YA
~V
= lOxO.162x
dX
0.99
f
2
1/2
074 (k 1C A
+k2 C A +k3 C A )
= 92.8dm
3
P6-6 (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 k[, k2, and k3 in terms of temperature. From the given data we know:
ki =
A exp(I:~~~ )
Since we have the constants given at T = 300 K, we can solve for Ai'
.004
" = - - - - - - - = 1.4ge12
--20000
exp ( 1.98 (300)
~ = ___ . 3 ··-10000
exp ( 1.98 (300)
J
=5.7ge6
J
6-13
(.25 )~1.798e21 -30000
A, ~
exp 1.98(300) Now we use a mole balance on species A
V= FAO -FA -rA
V
v( CAO -CA)
=---'-----=-::..::.-.-:.:.:...
A mole balance on the other species gives us:
F; =vC =ljV j
Ci =7lj 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 ofB 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 P6-6-f.pol. POLYMA TIl Results NLE Solution Variable Ca T R
k1 k2 Cao Cb k3
tau Cx Cy Sbxy
Value 0.01.70239 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
NLE Report (safenewt) Nonlinear equations [1] f(Ca} = (Cao-Ca}/(k1 *CaA ,,5+k2*Ca+k3*CaA 2}-1 0 = 0 Explicit equations T=306 [2] R= 1.987 [3] k1 = 1.4ge12*exp(-20000/Rff) [ 4] k2 = 5790000*exp( -1 OOOO/Rff} [5] Cao =.1 [1]
6-14
6] Cb = 10*k2*Ca 71 k3 = 1.798e21*exp(-30000IRfT) 8] tau = 10 9] Cx = tau*k1 *Ca/\5 lO} Cy = tau*k3*Ca"2 11] Sbxy = Cb/(Cx+Cy)
P6-6 (g) Concenttation is proportional to pressure in a gas-phase system Therefore:
S B / XY
~
~
p
Vp+p2
which would suggest that a low pressure would be ideal. But as before the ttadeoff is
lower production of B. A moderate pressure would pIObably be best
P6-7 US legal limit: 0 8 gil Sweden legal limit: 0.5 gil A~B k2 >C Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stteam
dCA =-kC dt ) A dCB =kC -k dt ) A 2 k)
= lOhr- 1 g
2.0
k2 =0.J92-L hr Two tall martinis = 80 g of ethanol Body fluid = 40 L
C AD
16
1.2
= 80g =2 g
40L
----------.-------
L
Now we can put the equations into Polymath
0.8
See Polymath program P67 .poL 04 PO'LY~fAT!LReslilts
Calculated values of the DEO variables 0 . 0 0'---2----,--4~·
Vaziable initial value minimal value maximal value final value ----
t O O 2 7 .131E-44
Ca Cb
o
o
kl
10
k2
0 . 192
10 0.192
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = ··k1 *Ca
6-15
6
10
10
2 1 . 8901533 10 0.192
0.08 10 0 . 192
7 131E-44
8
10
[2]
d(Cb)/d(t)
=-k2+k1 *Ca
Explicit equations as entered by the user [1] k1 = 10 [2] k2 =0 . 192
P6-7 (a) In the US the legal limit it 0.8 gIL. This occurs at t = 6.3 hours ..
P6-7 (b) In Sweden C B
= 0.5 gil , t = 7.8 hrs.
P6-7 (C) In Russia CB = 0 . 0 gil, t = 10.5 hrs P6-7 (d) For this situation we will use the original Polymath code and change the initial concentration of A to 1 gIL . 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 gil will be added to the final concentration of A after a half an hour. At a half an hour CA = 0.00674 gIL and CB = 0.897 gIL . The Polymath code for after the second drink is shown below.
See Polymath program P6·7··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
final value 10 5.394E-42 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
for the US t = 62 hours Sweden: t = 7 . 8 hours Russia: t =10.3 hours .
P6-7 (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 glhr·. Since volume is not changing the rate of change in concentration due to the incoming ethanol is 2 gIL/hr. For the first hour the differential equation for CA becomes:
6-16
dC A = -kl CA + 2t after that it reverts back to the original equations . dt See Polymath program P6-7 -e . pol. POLYMATH Results Calculated values of the DEQ variables initial value
Variable t Ca
°
°10°0.192
Cb k1 k2
minimal value
°
°
-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 Q..OOICD then,
- 45399.9CAC B -2.7CD
S DIU
-
49.7CACB
45399 -913
_ - -
49.7
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.
P6-9 (I) A + B --7 D
S
and
-'iA = 10exp(-8000K IT)CACB
A ---7 U
and
-'2A = 26 exp(-IO, 800K IT)CA
U ---7 A
and
-r3U =1000exp(-1.5,000K IT)Cu
- rD
DIU -
-_____ lOexp(-8000K IT)CACB ru - 26exp(-'}0,800K IT)CA-1000exp( -15,000K IT)Cu
6-22
We want high concentrations ofB and U in the reactor. Also low temperatures will help keep the selectivity high. If we use pseudo equilibrium and set -rA = 0 .
r-
-rA =lOexp ( -SOOO) CACB + 26exp (-lOS00) T CA-1000exp (-15000) T Cu =0
C, _ lOex p ( -8~OO
)C + 26exp ( -1~OO) 8
lOooex p ( -1~OO)
Cu -
CA =_1_ exp (7000)C Cu 100 T B
+~exp(4200) 1000
P6-9 (g) A+B--7D
and
T
-·liA =SOOexp(-SOOOK IT)C~5CB
A+B---tU1
and
--r2B = 10exp(-300K IT)CACB
D+B--7U 2
and
-13D
= 106 exp(-SOOOK IT)CDCB
(1)
SOOexp( -SOOOIT) C~5CB . SO exp ( -SOOOIT) SDlU = lOexp(-300IT)CACB - = exp( -300IT)C~5 j
AtT = 300
S
DIU
2.09S *10- 10
j
=----0.36SC~5
At T = 1000
= ____ 29.43
S DlU
0.740SC~5
j
To keep this selectivity high, low concentrations of A, and high temperatures should be used.
S
DlU2
=
SOO exp ( -SOOO IT) C~5 CB SOOC~5 =106 exp(-SOOOIT) CDCB 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 dependant on temperature . To keep optimize the reaction, run it at a low temperature to maximized allows only D to diffuse out. (2)
SOOexp(-SOOOIT)C~5CB
_
SDIUIU2 -
(
) 6 (._--) lOexp -3001T CACB + 10 exp -SOOOIT CDCB
SOO exp ( -SOOO IT) C~5
SDlU1U2
= lOexp-300lT ( )C
6 ---'--'--(...:..:........_-)-
A +10 exp -BOOOIT CD
At T = 300
6-23
SDiUl
in a membrane reactor that
S
DIV1U2
2.09 *10-9 COS A::::O 3.67 CA+ 2.62 *10-6 CD
=
At T = 1000 and very low concentrations of D
=
S DIU1U2
0.268C~s 7.408CA+335CD
.03617 C~5
If temperatme is the only parameter that can be varied, then the highest temperatme possible will result in the highest selectivity. Also removing D will help keep selectivity high.
P6-9 (h) No solution will be given P6-9 (i) exp(-7000K IT)CA1I2 -r 1OC/ 2 rD
v
dF - -A= r dV A dFv --=r. dV v
dF dV
B
dF dV
D
+R
B _.-=r,
B
- -D= r
R = FAO B
v:
T
A+B~D
liA = --1Oexp(-8000K IT)CACB
A+B~U
r2A = -1OOexp(-1OOOK IT)CA1I2C/12
rA = rB = liA + r2A
FA
C A =Cro -FI
mol CIO =0.4--3 dm FAo = Crovo
s
These equations are entered into Polymath and the plots below are for the membrane reactor . The code can be modified to compare with the PFR results.
See Polymath program P6··9i.poJ. POL YMA TH Results Calculated values. of the DEQ variables Variable V
Fa Fb Fd Fu Cto
initial value
o
4
minimal value
o
0.5141833
o o
o
o o o
0.4 600 4
Cb
o
0.4 600 4
Ca
0.4
0.0241566
T
Ft
maximal value 10 4
4.5141833 3.034E-06 .3.4858137 0 .. 4 600 8.5141833 0.2120783 0.4
o
6-24
final value 10 0.5141833 4.51418.33 3 . 034E-06 3.4858137 0.4 600 8.5141833 0.2120783 0 . 0241566
0 4,S7SE-07 0 0 0 0 ,4461944 S 4 0.8 S '. 423E+OS
-4.S7SE-07 0 -0.4461944 -0.4461948 -0.4461948 0 S 4 0.8 2.894E-07
0 0 0 0 0 0 S 4
rIa rd r2a ra rb ru Vt Fao Rb Sdu
o '. 8 S . 423E+OS
-8 . 297E-08 8.297E-08 -0.2867066 -0.2867066 -0.2867066 0,.2867066 S 4 0.8 2.894E-07
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb )/d(V) = rb+Rb [3] d(Fd)/d(V) = rd [4) d(Fu)/d(V) = ru Explicit equations as entered by the user [1] Cto =..4 [2) T=600 [3) Ft = Fa+Fb+Fd+Fu [4) Cb = Cto*Fb/Ft [5) Ca = Cto*FalFt [ 6] r1 a = -1 O*exp( -8000IT)*Ca*Cb [7) rd= -r1a [8) r2a = -1 OO*exp( -1 OOOIT)*CaA 5*CbI\1 ,,5 [ 9) ra = r1 a+r2a [10] rb = ra [ 11] ru = -r2a [12] Vt=5 [1.3] Fao = 4 [14] Rb = FaolVt [IS] Sdu = exp(-7000IT)*CaA ,5/(10*Cbl\,,5+,,00000000001)
5
6.0e+5
4
48e+5
3
3,6e+5
2
2 -Ie+5
1
1.2e+5
2
.:I
V
6
8
10
O,Oe+O
P6-9 (j) No solution will be given P6-9 (k) No solution will be given P6-9 (I) No solution will be given
6-25
0
.-.......--.. 2
---.--.-~-
.:I
V 6
. ---.-. s 10
P6-10 (a) Spectl00 and k2
: :; ; k3 Cg
X3
-ra ""'-rsl 'r fC + I'D::::: - k t CA + leI CB +- k::3 Cs . ~~ ::: -ki CA.O
e-lc,t ....
(k2 + k3) CB
~oodCB . ::::: -{0.01)(O.2) e·~·Olt + (0. 003 d!
::d~B. == -0.002 (1
t·
0.0(2) CB
- c·.o Oll} + 0.005 Cs
Using a Runge.,Kut:u Gill numerical so1.unon. we find that. for t::::: 2 min;: 120 s. CB:::
O. I 36 gmolJdm3.. (c)
!!;:.
= 'c = x, C. ;
£~. = k, C.
; C"
="
L
c. dl
From the solunon in patt (b). we have values of C:s at intervals as small as 5 sec, so we can use Simpson's rule to obcain t ""
(:'c:
1 min =: 60 sec.
Cc "" 0.003 Pf)[O + 4(0.0449) + 2(0.0782) + 4{O.102j + 0.1183] Cc =0.0129 gmolJdm3
6-73
t
= 2 min == 120 sec
Cc:= 0.003 (3f)(0 + 4(0.0782)'1- 2{0.1l83) + 4(0.134) + 0.1355]
Cc == 0.0366 gmol/dm3 T 0 20
(d)
I
40 60 KO
C",
CB
0.2 0.164 0.134
0
a.lOK
0.01:199 0.0736 0.0602
100 120 140
0.0572 0.0950 0.118 0.131
0.136 0.135 0.131
24(
0.0493 0.0404 0.0331 0.0271 0.0222 0.0181
260 280
0.0149 0·9122
JUU
--0:00.)(:\ 0.00996 0.0082 0.0566 0.0067 0.0499
160
lsu
2()(
I
22!
:320 340
I
0.125 O.lib
0.107 0.0982 0.0890 ---o.()~Ol 0.0717
Cc 0
Co
(J.(J()1~
0~12
0.()()65 O.OLlY
O,t.J(J43 0.0081)
0.0205 0.0285 0.0366 0.0447 0.0523
O.Oij{) 0.O19()
u.u~95
U.0392 0.0442 0.041:12 0.0520
0
0.0244 0.029g u.~~
O~Q§g
0.0724 0,r}}7,! 0.Q~~9
U~5~ O·tl,:s~
0.U1:I74 0.U914
O.()bU':/
0.0949 0.0632 0.0979 .-1~_. 0.0635'-·
020 •
o.ts
012
FAo -, FA +fAV =0 1,)
,..
CAO - 1.) CA - kl CA V ::::: 0
CAO -
CA ~
CAO -
CA (l -+- kl
CA CA.O
:::::
kl
CA 1: ::; 0
---1__, __ (1 +
ki
t)
1:) ;;:: 0
FBO -
FB
+ fa V ::::: 0
·u CB + (kl CA " kl CH - k3 C:a) V::::: 0 ,CHi-' kl C A t .. k2 Cg
1: -
k3 CB 1: == 0
Cg (1 + kz t + k3 "t) ;;;; kl t C A
CA =~a
Cso = Cs =
=>
DMAJ(
--·k
Cc = 0
In(I--~)
=,urnax (1-
,urnax
e-Csolk)=
Ihr- I (I- e-20gldm3 ISgldm#
Cs=-kln(l-
DCc=DYm(Cso-C s ) DCc = DYos ( C so
+k+- ,u:-)J 7-38
)= O.918hr-
.u:-)
1
7.0
Now
For
d(DC c )
Dmaxprod '
Dmax
prod =
dD
:::
0
0,628 hr-!
P7-17 (a) Individualized solution P7 -17 (b) Individualized solution P7-18 (a) rg = JlC c
Jl::: Jlmax CS KM +C s
ForCSTR,
DCc:::rg
D(Cso-Cs):::-rs
-rs:::YsICrg
s
C :::Cso (1-X):::lOg/dm 3(1-0.9)=lg/dm 3 Cc ::: YCIS (c so - C s )::: 0.5(10 -l)g / dm 3
:::
D=
v;{
4.5g / dm 3
DC ::: r ::: JlmaxCsCc c g K M +C s Vo
=>
= 0.8hr- 1 xlg / dm 3 x4.5g / dm 3
4.5g / dm3
(4+1)g/dm 3
V V ::: 6250dm 1
P7-18 (b) Flow of cells out = Flow of cells in
Fc ::: VoCc Cell Balance:
dC Fc + r~ V - Fc ::: V· __cdt
dCc=r __ dt g dC
Substrate Balance: _2_::: voCso
dt
r g
- voCs - Ys/ r~ /lC
= JlmaxCsCC
~=----"---"'-
KM +Cs
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"
P7-18 (C) TwoCSTR's For 1sl CSTR, V = 5000dm3 ,
DCc
= rg
D(C-C so s )=-rs
7-39
POLYMA TH Results NLES Solution Variable Ce Cs urnax KIn
Csoo Cso Yse rg r's V vo D X
Ceo
fIx) 9 . 878E-12 1.976E-ll
Value 4.3333333 1. 3333333 0.8 4 10 10 2 0.8666667 -1.7333333 5000 1000 0.2 0 . 8666667 4 . 33
Ini Guess 4
5
NLES Report (safenewt) Nonlinear equations f(Cc) = D*(Cc)-rg = 0 f(Cs) = D*(Cso-Cs)+rs = 0
[1) [2)
Explicit equations [1) umax = 0.8 [2) Km=4 [3) Csoo = 10
Cso = 10 Ysc = 2 (6) rg = umax*Cs*Cc/(Km+Cs) [7) rs=-Ysc*rg [8) V = 5000 (9) vo = 1000 [10) D=voN [ 11) X = 1-Cs/Csoo [12] Ceo = 4 . 33 (4) (5)
x = 0 . 867
Ccl = 4.33 g/dm3 3 CSt = 1.33 g/dm
CPt
= YP/CCCt =0 . 866 g/dm3
nd
D(CC2 -CCI)= rg See Polymath program P7-18·. c-2cstr..pol.
For2 CSTR,
POL YMATH Results NLES Solution Variable Ce Cs urnax KIn
Csoo Cs1
Value 4.9334151 0 . 1261699 O. B 4 10
fIx) 3 . 004E-I0 6.00BE-10
Ini Guess 4 5
1.333
7-40
2 0.120683 -0.241366 5000 1000 0. 2 0.987383 4.33
Yse rg
rs
v
vo D X
eel
NLES Report (safenewt) Nonlinear equations [1] [2]
f(Cc) = O*(Cc-Cc1 )-rg = 0 f(Cs) = O*(Cs1-Cs)+rs = 0
Explicit equations [ 1] [2] [3] [4]
[51 [6] [7] [8] [9]
umax = 0.8 Km=4 Csoo = 10 Cs1=1.333 Ysc=2 rg = umax*Cs*Cc/(Km+Cs) rs = -Ysc*rg V = 5000 vo = 1000
[10] 0 [ 11]
=voN
X = 1-Cs/Csoo
[121 Cc1 = 4 . 33
CC2 = 4.933 g/dm 3 C S2 = 126 g/dm
3
X= 0.987 C P1 = YP/CCCl =0 . 9866 g/dm3
P7-18 (d) For washout dilution rate, Cc = 0
=
D
max
DMAXPROD
= Jimax
[
1-
JimaxCso KM
+ C so
~--]
~
-KM +C so
Production rate = Ccvo(24hr)
3 1 J!.:_8hr- x 109 / dm 4g/dm 3 +lOg/dm 3
= O. 8hr
=4 . 85
= O.57hr-1
-I[1-- ~·-·4g/dm3
._-] 4g/dm +lOg/dm 3
3
---·---3
3
g / dm xlOOOdm Jlnx24hr = 116472.56g/day
P7-18 (e) For batch reactor,
dC c_=r __ dt
g
= 0.37hr
dCs
---=r dt s
Ceo = 0.5 g/dm3 C so = 109/dm}
ro = .J!:max CS C o
See Polymath program P7-18e.pol. POL Y1\1ATH Results
7-41
K
M
+C S
C
-I
Calculated values of the DED variables Variable t Cc Cs Km Ysc umax rg
rs
initial value
o
0.5 10 4 2 0.8 0.2857143 -·0.5714286
minimal value
o 0 .. 5 0.1417155 4 2 0.8 0.1486135 -2.8064061
maximal value 6 5.4291422 10 4 2 0.8 1. 403203 -0.2972271
final value 6 5.4291422 0.14171.55 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 1 umax = 0.8 [4] rg = umax*Cs*Cc/(Km+Cs) [5] rs = -Ysc*rg
For t =6hrs, C c =5.43g1dm3• 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 gldm3 x 3 x 2 x 500dm3 = 16290glday.
P7 -18 (g) Individualized solution P7 -18 (f) Individualized solution
7-42
Fogler 7 -19 Solution Problem Statement: Lactic acid is produced by a Lactobacillus species cultured in a CSTR. To increase the cell concentration and production rate, most of the cells in the reactor outlet are recycled to the CSTR, such that the cell concentration in the product stream is 10 % of cell concentration in the reactor. Find the optimum dilution rate that will maximize the rate of lactic acid production in the reactor. How does this optimum dilution rate change if the exit cell concentration fraction is changed? [rp = (a /-l + ~)Cc J /-lmax = 0.5 h-1, ~ = O.lg/g..h,
Ks =2.0 giL, Y x/s = 0.2 gig,
a
=0.2 gig,
Yp/ s =
Cs,o= 50 gIL
0.3 gig
Solution:
CS,o = 50 IL 0.1 Cc
CS,Cc
A steady state material balance on the Bioreactor (including the recycle device) gives: Accumulation =
outlet +
inlet
Cells:
o
=
Substrate:
o
= D *Cs o
Product:
o
=
o
- D*O.l * Cc + -D
* Cs
o
/-l*Cc
(7 ..19.1)
-/-l*Cc/Yx/s -- r p /Y p/s(7.19.2)
+ 7- 43
generation
(7..19.3)
From equation (7.19.1) :
C
s
= 0.1 * D* Ks /I rmax
.
Fromeq uatlOn(7.19.2):Cc ,
D*(Cso-Cs )
=(- - +
J=(
J1
aJ1 + fl
YX1S
YPIS
I
(7.19.4)
-0.1 *D
D*(Cso-Cs ) 0.1 * D a * 0.1 * D + flJ' + --------"'YXIS Y P1S
,
(7.19.5) Rate of production
rp = (a Jl + B)Cc =
(a*O.l*D+jJ)*D*(C s,o- O.l*D*:s) J1max -0.1 D
(7.19.6)
(9.1Y * D + a * 0.1Yp/s* D + flJ XIS
Differentiating equation (7.19.6) w.r.t. the dilution rate D, one can determine the optimum dilution rate that will maximize the rate of production. For the given parameter values in the problem statement, the substrate and cell concentrations and the rate of lactic acid production can be calculated from the above equations and plotted versus the dilution rate. The optimum dilution rate = 3.76.5 In,-I. -_._-------------------------
Recycle CSTR - Fogler 7-19 ----------_._--_._--_._----------
80 70
-----------
60
~'7_"""--'----
50
40 30
20 10
-
=EUbstrate cone Cell Cone ------
J
-"'- Rate of Production
--------1-
------
_~_._._
o ~---~=--='-:-::..-:-~========~~======~====::= - - , - · - - - - - , - 1 ..--.---'L, o 1 2 3 4 5 Dilution Rate, 1/h ----------
7-43a
P7-20 (a) XI + S -.-> More XI + PI X 2 + XI -> More X 2 + P2
ForCSTR,
dCs - -_ D (C50 dt
-
. C S ) - Ys I x rgX I
I
dC x? =D-'C ( ) +r ._-dt Xl gX 2 r gX,
= Jil C XI
See Polymath program P7-20-a.pol. POLYMATH Results Calculated values of the DEQ variables Variable
initial value
minimal value
maximal value
t O O
1
7-44
final value 1
Cs Cx1 Cx2 Km1 Km2 u1max u2max rgx1 rgx2 Yx1s Ysx1 Yx2x1 Yx1x2 Cso D
10 25 7 10 10 0.5 0 . 11 6.25 0.55 0.14 7.1428571 0.5 2 250 0.04
1. 2366496 25 7 10 10 0.5 0.11 1 . 4008218 0.55 0.14 7.1428571 0. 5 2 250 0.04
10 25.791753 7.2791882 10 10 0.5 0.11 6.25 0.5748833 0.14 7.1428571 0. 5 2 250 0.04
1.2366496 25.456756 7.2791882 10 1.0 0. 5 0.11 1.4008218 0.5748833 0.14 7.1428571 0.5 2 250 0.04
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = D*(Cso-Cs)-Ysx1 *rgx1 [2] d(Cx1 )/d(t) = D*(-·Cx1 )+rgx1-Yx1 x2*rgx2 [3] d(Cx2)/d(t) = D*(-Cx2)+rgx2 Explicit equations as entered by the user [1] Km1=10 [2] Km2 = 10 [3] u1max = 0.5 [ 4] u2max = 0.11 [5] rgx1 = u1 max*Cs*Cx1/(Km1+Cs) [ 6] rgx2 = u2max*Cx1 *Cx2/(Km2+Cx1) [7] Yx1s = 0.14 [8] Ysx1=1lYx1s [9] Yx2x1 = 0 . 5 [101 Yx1x2 = 11Yx2x1 [11] Cso = 250 [12] D = 0 . 04
8.2 6.4 4. 6 2. 8
10
0.0
0.2
04
t
0.6
0.8
10
7.30,---
7.24
~J
718 712 7.06 25.00'---~-.
0.0
0.2
_ _ _ _ _ _ _ ._ _ _...J 04 t 0.6 0.8 10
7.00
IL--_~. _
0.0
7-45
Q2
_~_
Q4 t
Q6
0.8
10
0..580 r - - - - - - - - - - - - - - - - - - , 7.0 0.574 0.568 . 0562 0.556 0.550 t-~
j for X"
W J YS.
080
0.50 -,
I;
005~ 0,00
t,._"~~,...;-:..=..::"~~:::;=_ _ _""..""'' .. ' ' ' ' ' _.
o Pj
20 ;::::
40
j
80
60
100
M,,(l . p)Zpjl =M o (1' p)2p9
Use the above equation to generate a graph of P IO V$, p:
°
D2
0.1.
P
0,6
0,8
{c)
Use these equations to generate graphs of Yj and
Wj VS,
W j
VS.
j. j for p :::;; 0.80, 0.90, 0.95
0,,07 -r--·~"···"···"·-·-·--·'·'·---·'-·'·"'---"--· 0,,06 OSo. ;------1'=o,90
..-'-'" ."""'' ----,
+=---p:::; l i\"
0.05 i
~
i
004
I i
O.oJ
"
\"
,."
p =0.95 .. , ----,
',",."'-'"
\
t! /"'\,:" "
!
:: t~s::~~: ~-'~: ;~J OO'J-i.
o
7-62
\"
20
40
60
80
100
We must find the value of p.
ill
~ =(~~:~,"nex{=~~) -I]; -0.067
wf = 2.803 I2 :::;:
12o cxp( -kat):::;: (O.OOl)exp[( -L4 X 10,3)(14,400)1 == 2.794x 10 . 9 k~M
P == fJ =: - - - - - - - -....- ..- ...- - - - - . . . . kpM + kmM + kee + ksS + ~2ktkof(IJ
. = 0.99991
This can then be used to calculate the desired values:
_.104 __ . 6 NI ::;:: .-_._--._ . . _. . . . . . . . :: l.b) x 10 Jl
1.- 0.99991
(e) Mole fraction of polystyrene of chain length 10 (y 10)'
Use the above equations t.o plot Yo VS.t::
7-63
y 10
VS.
t
o° ~: r------'--------------------'.--,.'----·------------------l
O
0.03
I
O.O!)
-
I
;'002
\
0.,015
~
I
O.Ol 0005
.
..:::::::~
2
________ .. _,J1
t (hr)
5
3
CDP7-K Reaction
R.} + I . .", . &i", PJ (a) and (b)
...;;-' r.:;:; L}
r r +- R' (k l"l+ k S S + k rt.[') , m
.. [
_
l
l
X +~ +~. tk;h+ k,,~r ~ (1~rr0 + N
(c)
k,
+
From the above derivation we know that
rp
Neglecting the solvent term and rearranging yields: 1
:::::
i .\,·t
-1M
Substituting in for
-'1 M
and Ii and simplifying:
7-64
{:}1+ kk:fi
knI z)!?vf(-r =1. . . _- .2k,J(I . . . . _ . . _-_ . . . . _. . . .M_,) + .krn- + _.,.-
XN'
k!(2kof(IzJlkl)
kp
kpM
---
. k m ..,.,.., --,_ krjI..... 1 _ .....kJ-r;.t) -:;'."" "'--...,.... XN k;(lVl) kp kpM To determine rate law parameters experimentally from a CSTR both the [mal XI'! value and the final concentrations of wI and I must be recorded. l11ese data can be ~-
(d)
..;,.,-
..
used in the above equation to frnd values for the parameters.
An increase in temperature would cause an increase in all three primary steps of
(e)
free-radical polymerization (initiation. propagation, and termination). By looking at the overall rate law:
l2k:(I,W
r . ;;;:;; k M (._-_._.::_. M
?~.
k~
it can be seem that the greatest effect of temperature would be on propagation. Overall, there would be an increase in monomer disappearance and an increase in pol ymerization.
CDP7-L PFR:
a)
_dlvl .._-:::::: r dr »',1 )"-~.--.-
12koL,f
riM : : : -kp~M ~~;;. Plug those into POLYMATH to get this graph. §~ati2E:!':' dIm) Id(tau);rm
!£!i t. i~!
d(i) /d(cau}=:·:.
n .01
kp=lG ko""Je- :)
t.au
~~S!.!1..V~!.~ 0
~~~. :;:!!s~
=; . t:.au
n
kp"m·sqrt(2*ko~i"f/ktl
:; {)t
~b~~ ... :!~:·u~
~'in4-l
VAll.].f~
7'99$
,999~
'.1'H~
l
r:i;;-.. ko*i.
Y~±.:±~
:3
2 'S'!J.2"J.
0.0:
.J 01
S. %! ::. .. ·",8
S .06~:';,,,, .. Ja
:~
t·t-
10
0 OUl
14 0.001
0.5
{)
s,,·n7
$e+0 1
Se·,;].";
··t.e··05
•. :5
~:.@,
,,1
7-65
j4J)~~~ ..
i)$
~
-- -
C S
5 Ofjl.'1l~···41.
.. < 99lSa,,·23
-.
'4 iTO
!400.
1400.
1400.[400.
is ,!k
iO.133
10.0651696
,0.133
6 IvO 17
20. '0.2
1
[kr
!
:0.0651696
,~
1"
:5.
18 IUarho
i20. 10.0622149 ,5.
,;20.10.0622149 ,
20.
t
iO.2
1
,~5.
!323.
15. ]"323.
11.013E+06
I~.,013E+06
1304.6819
" 1304.6819
,1304.6819
1304.6819
/304.6819
'292.8948
!304.6819
1292.8948
10
0
117.40323
!17.40323
14IC6
10
0
;17.40323
117.40323
15frA '.
'-40.52269 !
-40.52269
1-0.2446624
1-0.2446624
19
IT~
b23. ··11.013E+06
10jPO
llicAO 12ICA 13 Icc
T
!
,
;
~
~
:323.
~
Differential equations i d(X)jd(W) = -rA j vO j CAO
8-75
J1.013E+06
~j d(T)/d(W)
= (Uarho * (Ta - T) + rA * 20000) / vO / CAO /
40
Explicit equations
= 400 k = 0.133 * exp(31400 / 8.314 * (1 / TO - 1/ T)) vO = 20 kr = 0.2 * exp(51400 / 8.314 * (1/ TO - 1/ T)) Uarho = 5 ;§;:Ta = 323 TO
PO = 1013250
= PO /8.314/ TO CA = CAO * (1 - X) / (1 + X) * TO / T $~l CC = CAO * X / (1 + X) * TO / T CB = CAO * X / (1 + X) * TO / T rA = -(k * CA - kr * CB * CC) CAO
400
0.060
394
Q
388 382 376
0012
---~-- ...
0. 000
....J.-~---.I-_
0
10
20
W 30
:;0
40
370
0
10
20
W 30
---.--,
40
50
P8-23 First note that ~Cp = 0 for both reactions. This means that MIRx(T) = ~HRx 0 for both reactions. Now strut with the differential energy balance for a PFR; dT _ Ua(Ta - T)+ dV -
,L lij (-MiRxij) _ Ua(Ta -T)+ 'i.c (-MiRxIC) + r2D (_.Mi Rx2D)
,LFjCpj
-
,LFjCpj If we evaluate this differential equation at its maximum we get dT --- = 0 and therefore, Ua(~ - T) -- 'i.c Mi RxIC -- r2D Mi Rx2D = 0 dV We can then solve for rIC from this information.
8-76
MI RxIC 'ic
=
10(325 - 500) - 0.4 (0.2) (0.5) (5000) -50000
=0.039
'ic = 0.039 = klCCA CB = klC (0.1)( 0.2) klC =1.95
~C(500) = ~C( ~= R E
lr o.22 ( 0.91-9· 025 }_ 0,0025 1 1-0.05:>(0.05) (1-0.05fK~J = k{t) [0.848 -
Q.Q~~O~l
(4)
8-98
Energy balance:
FAO (l:9i c;,i + X .6.C p) Tlr - FAo (Lei Cpi + .6.Cp X) Tlr+Ar + rA ~W(AHR) - (2) (AA) U(r) (T-T A) = 0
or: FAO (re j Cp; + x ACp) Tlr - FAO (rej CPi + ACp
x) Tlr+Ar
. + rA 2raIhp (.1HR) 1M - (2) (2xr tu) U(r} (T-T.J =0 Taking the limit as tu -+ 0:
FAO (rei ~ + x ~Cp) ~ = -U(r) 4m (T-TA) + (-.rA) (-MIR) 2xrhp Rearranging:
dI.
=-U{r) 4m- (T-TA) + (-rA) (-AHR) 27trhp
dr
(5)
FAO(:r9i CPi +X~Cp)
-U(r) 1h.(T- TA ) + (-rAl (-~HR)
or: .dI.. dW
=------p~------'!""--
(6)
FAO{:rei CPi + X acp)
Assume that: U(r) = U(ro) (.It.)1f2 ; ~
•
Therefore: U(r) = U(ro}
l) = l)1-ro (1 +£X)J.. ~o
(!f-)l'\f.yn {I - O.055XA)1f2
From example 8-10. we have:
K.p =exp (42R~ 1 _ 11.24] (Kp in ann- 1f2 • T in OR) k = exp [
(1)
T - 110.1 In T + 912.84]
-176008
(8)
MiRm = -42,471 - (1.563)(T - 1260) + (1.36 x
1O-3)(T2 - i26(2)
- (2 459 x lO-7){T3 - 126(3) where AliR in Btu . 3 l~le
8-99
(9)
l:8i Cy. = 57.23 + 0.014T - 1.788 x 1()-6 T2
(10)
Since equations (1) to (10) must be solved together as two pairs of coupled differential equations. they must be solved on a computer. employing numerical methods such as Runge-Kuaa. The results follow:
i
T
I
_ - - -~ 1.,0".
,;'-
....---.·.-/·. .- - - - - -.. .E: 8
12
16
zo
24
n
'8
y"",.
T
---~~~--------------i~
--'
.-'
.'
.4
~
________________ •
I:
"
.0.24
12
%I
.,
,- /----------
0." ------------~O
16
~
...
U
48
"
60
(I~'I
i
I AGO
T
t
X
1"0_
t
12110
lOGO .,;' --~
eoo
0
(J..:
I
a
:go
---
.
r.i.
.'
0 ..
'
I
~t.
I
:
"
4
a
12
16
:0
44
:t
n
16
.a
oM
41
~
." l....j
l71
P8-30 (d) 8-100
U
60
;
1I
.~~ = ::2m-~£t~.~4). }~o
dr
~rA ={qB~f~ J c = c =C A
B
.'10
(1'"- X'l_!Q. 'T
TO
Cc =2CAO XT
dT _ U(r)47rr(TA -T)+(--rA)(-LlliRJ dr FAO ( CpA + CPB )
u=u(ro)(;
)"(;,r
Now put these equations into Polymath to generate the plots. See Polymath program P8 . 30·d.pol. POL YMA TH Results Calculated values of the DEQ variables Variable -r X
T Ta dHrx Fao Cpa Cpb h
ro To E R
Cao U
k
Ca ra
initial value 0. 5 0 350 373 -2.0E+04 1200 25 15 0.5 0.5 350 7.0E+04 8 . 314 0. 1 33.3 30 . 955933 0. 1 -3 . 0955933
minimal value 0.5 0 350 373 -2.0E+04 1200 25 15 0. 5 0. 5 350 7 . 0E+04 8 . 314 0.1 0 . 769425 30.955933 -2 . 608E-10 -3.0955933
maximal - - -value -1000 1 493.0235 373 -2 . 0E+04 1200 25 15 0.5 0. 5 350 7 . 0E+04 8 . 314 1 33.3 3 . 398E+04 0. 1 2.09E-06
°.
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(X)/d(r) = -2*3 . 1416*h*ra/Fao [2] d(T)/d(r) = (U*4*3 . 1415*r*(Ta-T)+(-ra)*(-dHrx))/(Fao*(Cpa+Cpb)) Explicit equations as entered by the user [1] Ta = 373 [2] dHrx = ··20000 [ 3 ] Fao = 1200 [4] Cpa = 25 [5] Cpb = 15
8-101
final value 1000 1
373 373 -·2 . OE+04 1200 25 15 0. 5 0. 5 350 7 . 0E+04 8 . 314 0.1 0.769425 136.44189 -3 . 125E-17 4 . 264E-15
(6) h =.5 (7) ro =.5
To=350 (9) E = 70000 (10) R=8.314 (11) Cao =.1 (12) U = 33.3*(ro/r)I\.5*(TfTo)I\.5 [13] k 035*exp«E/R)*(1/273-1fT)) [14] Ca = Cao*(1-X)/(1+X)*(TofT) [15] ra = -k*Ca (8)
=.
P8-31 (a)
Mole balances:
v = F.4.0 - I:.. -rA V=Fs=Fc rB re
rate laws:
= ","CA rB =k.CA -~CB rc = k"CB -rA
Stoichiometry:
c.=F;I.. I
Va
To
Fa =10Fe .5 = .05-~
0.5
F;. = .025 11Fe
=.025
Fe =.023
FB =.023
From this we can use two of the mole balances to solve for T
8"·102
~o-~
1G~ 1 A1e- E1 KI
F;
=-
~Fs 0.1
-
~e-EI KI
T=269°F P8-31 (b)
Knowing the temperature we can then solve for the Volume:
P8-31 (c) We then need the energy balance:
UA(~ - T)- ~oCJHl(T - Yo)+ v[(MfR1)('iA) + MfR2 ('iB)] = 0 Solve for A and we get: A = 399 ft2
P8-31 (d) In order to get multiple steady-states, the kappa, tau and feed temperature had to be changed. l( == 0.1, 't == 0.0005 and To would be changed around. Tlus first graph is G(T), R(T) vs T at To == 2000 of 20000 " ,--,-,,----"""'-,,---.--..,------. 15000 10000 5000 "
-5000
1000
15 0
-,10000 .....- - - , . - - , - " -- - - - - - -.......
As can be seen there are three steady-states.
8-103
Ts vs To 1200 . 1000 800
~600
•• • • •• • • • •• • •
400 4 200 .... 0
10000
5000
0
To
P8-32 (a) Energy balance
.!L= Ual PII(r. -T)+(-rAX-AHR(~)] ~o(L8iCpi+XACp)
dW
tIT _
Ual P.(7;, - T)+(-rAX-MIR(TR))
dW
I:..oC,..
dX
Mole balance: dW
-r =-4. ~o
Pressure drop: dy =~1i(l+X) dW 2y T Rate law:
Stoichiometry: CA =
CAo(lXXI..} l+X To
Evaluating the parameters:
(-1-_1.)1 =ex]3776.7J'2... _1.)1 T J ,. \. 450 T J
k = ex] E ~R 450
Plugging these equations into POLYMAlH we get the following plots. See Polymath program P8·32··a.pol. OL YMATH Results Calculated values of the DEQ variables 8-104
initial value 0 1 450 0 5 0.007 450 300 -2 . 0E+04 5 40 1 0.25 0 . 25 -0 . 25
},:ariable W Y
T X U
a To Ta dHrx Fao Cpa k
Cao Ca ra
minimal value 0 0 . 6823861 310 . 69106 0 5 0.007 450 300 -2 . 0E+04 5 40 0 . 0232094 0.25 0 . 0825702 -0 . 25
maximal value 50 1
450 0 . 175758 5 0.007 450 300 -2.0E+04 5 40 1 0 . 25 0 . 25 -0.0019164
final value 50 0 . 6823861 310 . 69106 0 . 175758 5 0.007 450 300 -2.0E+04 5 40 0.0232094 0 . 25 0.0825702 -0.0019164
ODE Report (RKF45) Differential equations as entered by the user [1] d(y)/d(W) = -a/(2*y)*(TofT)*(1 +X) [2 J d(T)/d(W) = (U*(Ta-T)-ra*dHrx)/Fao/Gpa [3] d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [IJ U = 5 [2] a = . 007 [3J To=450 [4] Ta = 300 [ 5 J dHrx = -20000 [6J Fao = 5 [7J Cpa = 40 [S J k = exp(3776 . 76*(1/450-1fT)) [9J Cao = . 25 [101 Ca = Cao*((1-X)/(1-+X))*(T/450)*y [111 ra = -k*Ca 1.0
...-.=-------- -----..,
50U r - - - - - - - - - - - -
460
0. 8
IT] - v
0.6
420
- X
380
0.4
0.2
.
---"-~
............. " ..."'" ... " ......................... ~ .... '~ .................. ~ ... ~ ..... """
340
~--.-..,.--10
20
W 30
40
50
300
(I
10
20
W 30
40
50
P8-32 (b) From the Polymath summaty table, it is appatent that the maximum value for -rA occurs at the beginning of the reactor.
P8-32 (c) 8-105
The maximum value for the temperature also occurs at the beginning of the reactor.
P8-32 (d) Doubling the heat-transfer coefficient causes a decrease in the temperature, the conversion, and the pressure drop. Halving the heat transfer coefficient casuses all three to increase.
P8-33 Mole balances V = vo(C AO -CAl -rA
Rate laws: -rA
=k1CA +k2 CS
-rs =k1CA +k2C S Energy balance:
ru =k2CS
-FAO[c;,A(T -TAO) +CpB(T - TBO)] + V(rAIX.MiR1(TR}+ 6Cp1 (T - TR») + V(rA2XllliR2(TR)+6Cp2(T - TR})
Evaluating the parameters: T=400K kl =1000exp(- 2~)=6.73
t..C p1 =50-20-30=0
k2 =2000exp( -
3~OO)=1.11
t..C p2 =40-30-20=-10
FAo 60 dm 3 vo =-=-=6000Coo =C AO C AO .01 min Simplifying: C A = 't "rA +CAO C s = 't "rs +C so Co = 't" ro
C u = 't It ru
171000 V=-------20190· CA."!" 6660· C~
We can plug those into POLYMATH and find the exit concentrations of U and D and find the volume of the CSTR. See Polymath program P833.pol. POLYMA TH Results
NLES Solution
8-106
variable Ca Cb Cd Cu V
Cao Cbo vo
k1 k2 tau rd ru ra rb
Value 0.0016782 0.0016782 0 . 0071436 0.0011782 3794.94 0 . 01 0.01 6000 6.73 1.11 0.63249 0.0112944 0.0018628 -0 . 0131572 -0 . 0131572
f{x)
-1. 472E-16 -1.472E-16 -1.154E-16 -2 . 017E-17 1. 018E-09
Ini Guess 0 . 0017 0 . 0017 0.0072 0 . 0012 3794
NLES Report (safenewt) Nonlinear equations [ 1] f(Ca) = tau*ra+Cao-Ca = 0
f(Cb) = tau*rb+Cbo-Cb = 0 f(Cd) = Cd-tau*rd = 0 f(Cu) = Cu··tau*ru = 0 [ 5] f(V) = 171 000/(20190*Ca+6660*Cb)- V = 0 [2]
[3] [4)
Explicit equations [1] Cao = . 01 [2] Cbo = . 01 [3] vo = 6000 [4J k1 = 6..73 [5] k2=1..11 [6) tau = Vivo [7J rd=k1*Ca [ 8] ru = k2*Cb [9] ra = -k1 *Ca-k2*Cb [10] rb=ra
P8-33 (a) Cu = .0012 Co = .0072
P8-33 (b)
V=3794dm3 P8-33 (c) Individualized solution
8-107
Solutions for Chapter 9 - Unsteady State Nonisothermal Reactor Design P9-1 Individualized solution P9-2 (a) Example 9·1 The new TO of 20 of (497 OR) gives a new AHRn and T. With T=497+89 . 8X the polymath program of example 9-1 gives t= 8920 s for 90 % conversion.
P9-2 (b)
Example 9·2
To show that no explosion occurred without cooling failure . Isothermal operation throughout (T 17S0C)
=
~:laximum
cooling rate:
Qr == UAI448-· 298] == 142 * 150
::: 21300 BTU/min Maximum Qg at t == 0 (maximum concentration and reaction rate)
Qg := k _liclQ1!!.!.Q. V2 * V(··MI R~ ) == O.OOO1l67[.?.~'~~~J2..34 *' 10 6 . 5J 19 . ::: 15914.2 BTU/min For all t:
Qg < Or
No explosion
9-1
To show Ll}at no explosion occurs with cooling shllt down for 10 mi after 12 P..IS Isothermal operation for 12 hrs (at T:::: 175°C)
t ==
[~~-Ie~~::2}n~(i:::~~)
') x__ == 1 276 __e,,· iL-=:: 8 s (l·-x) 3 64--2x == 1276*3.64(1····· x) x:::: 0..38
Qg at
t :::;
12 hrs .
Qg "" k.~~JL '-" 0..000
. X.llVzo{~l!.:_~.:l(._AH ) V &
11671··~:O~~{1.:~:~~X~..!:::_~*:~?112..34 '* 106 "
= 77844
5..119
.
BTU/min..
Adiabatic operation for 10 min. UA:::: 0. After 10. minutes
x == 0.385, T == 184"C Qg :::: 10000 BTU/min. When we restart the cooling flow rate Q,I"",. == 21,30.0 BTlJlrnin. Temperature will dIop to 175"C NQ~~lQ:iiQ11
P9-2 (c) Example 9-3 Decreasing the electric heating rate (Tedot in polymath program from example 9-3) by a factor of 10 gives a conversion of 9.72 % at the onset temperature . For a decrease by a factor 10 the conversion is 2,49 %. The higher conversion of a lower heating rate is logical since the time it takes to reach the onset temperature is longer and the reactants have a longer time to react. P9-2 (d) Example 9-4 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 aT max 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.
P9-2 (e) Example 9-5 Using the same code as seen in Example 9",4, we were able to change the various parameters. The two graphs we have show To;::: 70 and 120°C, T i :;;;; 160 and
40°C, and Caj
=:: .. 1 and ,,2. Each set of parameters had a Temperature time trajectory and a teIuperature- 0,4'
c::: g
0,.3
--
0.2
..
0.1
0""------"'"---,---··.,.-----·,.,...·,,--,·---,.-··..-,,-·,,,····,,,. · · · · · . . ·. . ·"....""""'t----·. . . T···"·" .. ·~··
66
68
70
72
'74
To
9-4
76
78
80
82
_
..... ...........
_-_ _-_._-_ _ ..
............
................
_._..........__....__--_...................................................._._-------..
x vs coolant flow
1 ...-...._ _................-._...............- .....--.
------.----~
0.9 0.8 .
c 0:7 .E 0.6 ....coQ,) 0.5 c> 0.4 o
o
--_._-___ __ _--
0.3 .
•.
..
0,2 .
0.1
. . . . .-r------r,..-.. . . . . -----1
o -!---.----,------......,...,--, o
2000
..
..
6000
4000
8000
10000
Coolant flow, mollh
180
...-_ ..•..-
T vs. coolant flow ---"' ...................._............................... __.,-------------
160 .
u.. 140
....c) 120 .
.aa:s
100··
~
80·
--_._-_ __ _---_._-....
M
....
E 60 Q,) • i-
40 20·
o ..-.. . . . . . . . . . . .--.,.....--_. _. . . . . · · ·. . . . · · . r······_·_··. ······. •·....·---r-·_-'--.,.....--- . . . ._ . . o
2000
------_
...
__._---"
6000
4000
8000
10000
Coolant flow mol/h ........................
P9-2 (g) Example 9-7 The temperature trajectory changes significantly., Instead of a maximum in temperature, there is now a minimum. The concentration profiles also change. CB no longer goes through a maximum and Cc does not rise above 0 . It also appears that there is no maximum for C A .. It appears that because the reactor is kept much cooler with the increase heat transfer and lower coolant temperature the second reaction does not occur .
P9-2 (h) Example RE9-1 Using the code from Example RE9·1 • we can determine the value of k. for which the reacwi""willtall to the lower steady··state and when it becomes unstable. The following two graphs show those points when ~;; ..2 and 24 respectively. The third graph shows what happens when To;; 65. It becomes unstable ata much lower temperature. 9-5
t 1"" •
KEY:
--T
,
.:J-co-'-
1 i
1 ~c",~cO;-·
t
T 0=65,
IntE'gr al
1
